a radically modern approach to introductory physics

A Radically Modern Approach to

Introductory Physics — Volume 1

David J. RaymondPhysics DepartmentNew Mexico Tech

Socorro, NM 87801

July 2, 2008

ii

Copyright c!1998, 2000, 2001, 2003, 2004,2006 David J. Raymond

Permission is granted to copy, distribute and/or modify this document underthe terms of the GNU Free Documentation License, Version 1.1 or any laterversion published by the Free Software Foundation; with no Invariant Sec-tions, no Front-Cover Texts and no Back-Cover Texts. A copy of the licenseis included in the section entitled ”GNU Free Documentation License”.

Contents

Preface to April 2006 Edition ix

1 Waves in One Dimension 1

1.1 Transverse and Longitudinal Waves . . . . . . . . . . . . . . . 11.2 Sine Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Types of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3.1 Ocean Surface Waves . . . . . . . . . . . . . . . . . . . 41.3.2 Sound Waves . . . . . . . . . . . . . . . . . . . . . . . 61.3.3 Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Superposition Principle . . . . . . . . . . . . . . . . . . . . . . 71.5 Beats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6 Interferometers . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6.1 The Michelson Interferometer . . . . . . . . . . . . . . 141.7 Thin Films . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.8 Math Tutorial — Derivatives . . . . . . . . . . . . . . . . . . . 171.9 Group Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.9.1 Derivation of Group Velocity Formula . . . . . . . . . . 191.9.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2 Waves in Two and Three Dimensions 29

2.1 Math Tutorial — Vectors . . . . . . . . . . . . . . . . . . . . . 292.2 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.3 Superposition of Plane Waves . . . . . . . . . . . . . . . . . . 36

2.3.1 Two Waves of Identical Wavelength . . . . . . . . . . . 382.3.2 Two Waves of Di!ering Wavelength . . . . . . . . . . . 392.3.3 Many Waves with the Same Wavelength . . . . . . . . 44

2.4 Di!raction Through a Single Slit . . . . . . . . . . . . . . . . 46

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iv CONTENTS

2.5 Two Slits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.6 Di!raction Gratings . . . . . . . . . . . . . . . . . . . . . . . . 492.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3 Geometrical Optics 57

3.1 Reflection and Refraction . . . . . . . . . . . . . . . . . . . . . 573.2 Total Internal Reflection . . . . . . . . . . . . . . . . . . . . . 603.3 Anisotropic Media . . . . . . . . . . . . . . . . . . . . . . . . 603.4 Thin Lens Formula and Optical Instruments . . . . . . . . . . 613.5 Fermat’s Principle . . . . . . . . . . . . . . . . . . . . . . . . 663.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

4 Kinematics of Special Relativity 73

4.1 Galilean Spacetime Thinking . . . . . . . . . . . . . . . . . . . 744.2 Spacetime Thinking in Special Relativity . . . . . . . . . . . . 764.3 Postulates of Special Relativity . . . . . . . . . . . . . . . . . 78

4.3.1 Simultaneity . . . . . . . . . . . . . . . . . . . . . . . . 794.3.2 Spacetime Pythagorean Theorem . . . . . . . . . . . . 82

4.4 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.5 Lorentz Contraction . . . . . . . . . . . . . . . . . . . . . . . 854.6 Twin Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

5 Applications of Special Relativity 91

5.1 Waves in Spacetime . . . . . . . . . . . . . . . . . . . . . . . . 915.2 Math Tutorial – Four-Vectors . . . . . . . . . . . . . . . . . . 925.3 Principle of Relativity Applied . . . . . . . . . . . . . . . . . . 945.4 Characteristics of Relativistic Waves . . . . . . . . . . . . . . 965.5 The Doppler E!ect . . . . . . . . . . . . . . . . . . . . . . . . 975.6 Addition of Velocities . . . . . . . . . . . . . . . . . . . . . . . 985.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

6 Acceleration and General Relativity 103

6.1 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1036.2 Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 1056.3 Acceleration in Special Relativity . . . . . . . . . . . . . . . . 1066.4 Acceleration, Force, and Mass . . . . . . . . . . . . . . . . . . 1086.5 Accelerated Reference Frames . . . . . . . . . . . . . . . . . . 110

CONTENTS v

6.6 Gravitational Red Shift . . . . . . . . . . . . . . . . . . . . . . 1136.7 Event Horizons . . . . . . . . . . . . . . . . . . . . . . . . . . 1156.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

7 Matter Waves 119

7.1 Bragg’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1197.2 X-Ray Di!raction Techniques . . . . . . . . . . . . . . . . . . 121

7.2.1 Single Crystal . . . . . . . . . . . . . . . . . . . . . . . 1217.2.2 Powder Target . . . . . . . . . . . . . . . . . . . . . . 121

7.3 Meaning of Quantum Wave Function . . . . . . . . . . . . . . 1237.4 Sense and Nonsense in Quantum Mechanics . . . . . . . . . . 1247.5 Mass, Momentum, and Energy . . . . . . . . . . . . . . . . . . 126

7.5.1 Planck, Einstein, and de Broglie . . . . . . . . . . . . . 1267.5.2 Wave and Particle Quantities . . . . . . . . . . . . . . 1287.5.3 Non-Relativistic Limits . . . . . . . . . . . . . . . . . . 1297.5.4 An Experimental Test . . . . . . . . . . . . . . . . . . 130

7.6 Heisenberg Uncertainty Principle . . . . . . . . . . . . . . . . 1307.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

8 Geometrical Optics and Newton’s Laws 135

8.1 Fundamental Principles of Dynamics . . . . . . . . . . . . . . 1358.1.1 Pre-Newtonian Dynamics . . . . . . . . . . . . . . . . 1368.1.2 Newtonian Dynamics . . . . . . . . . . . . . . . . . . . 1368.1.3 Quantum Dynamics . . . . . . . . . . . . . . . . . . . 137

8.2 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . 1378.2.1 Gravity as a Conservative Force . . . . . . . . . . . . . 139

8.3 Work and Power . . . . . . . . . . . . . . . . . . . . . . . . . 1398.4 Mechanics and Geometrical Optics . . . . . . . . . . . . . . . 1418.5 Math Tutorial – Partial Derivatives . . . . . . . . . . . . . . . 1438.6 Motion in Two and Three Dimensions . . . . . . . . . . . . . 1448.7 Kinetic and Potential Momentum . . . . . . . . . . . . . . . . 1488.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

9 Symmetry and Bound States 153

9.1 Math Tutorial — Complex Waves . . . . . . . . . . . . . . . . 1549.2 Symmetry and Quantum Mechanics . . . . . . . . . . . . . . . 157

9.2.1 Free Particle . . . . . . . . . . . . . . . . . . . . . . . . 1579.2.2 Symmetry and Definiteness . . . . . . . . . . . . . . . 157

vi CONTENTS

9.2.3 Compatible Variables . . . . . . . . . . . . . . . . . . . 1609.2.4 Compatibility and Conservation . . . . . . . . . . . . . 1609.2.5 New Symmetries and Variables . . . . . . . . . . . . . 161

9.3 Confined Matter Waves . . . . . . . . . . . . . . . . . . . . . . 1619.3.1 Particle in a Box . . . . . . . . . . . . . . . . . . . . . 1619.3.2 Barrier Penetration . . . . . . . . . . . . . . . . . . . . 1649.3.3 Orbital Angular Momentum . . . . . . . . . . . . . . . 1669.3.4 Spin Angular Momentum . . . . . . . . . . . . . . . . 168

9.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

10 Dynamics of Multiple Particles 173

10.1 Momentum and Newton’s Second Law . . . . . . . . . . . . . 17310.2 Newton’s Third Law . . . . . . . . . . . . . . . . . . . . . . . 17410.3 Conservation of Momentum . . . . . . . . . . . . . . . . . . . 17510.4 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

10.4.1 Elastic Collisions . . . . . . . . . . . . . . . . . . . . . 17610.4.2 Inelastic Collisions . . . . . . . . . . . . . . . . . . . . 178

10.5 Rockets and Conveyor Belts . . . . . . . . . . . . . . . . . . . 18010.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

11 Rotational Dynamics 189

11.1 Math Tutorial — Cross Product . . . . . . . . . . . . . . . . . 18911.2 Torque and Angular Momentum . . . . . . . . . . . . . . . . . 19111.3 Two Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . 19411.4 The Uneven Dumbbell . . . . . . . . . . . . . . . . . . . . . . 19611.5 Many Particles . . . . . . . . . . . . . . . . . . . . . . . . . . 19711.6 Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19811.7 Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19811.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

12 Harmonic Oscillator 205

12.1 Energy Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 20512.2 Analysis Using Newton’s Laws . . . . . . . . . . . . . . . . . . 20712.3 Forced Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 20812.4 Quantum Mechanical Harmonic Oscillator . . . . . . . . . . . 21012.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

CONTENTS vii

A Constants 213

A.1 Constants of Nature . . . . . . . . . . . . . . . . . . . . . . . 213A.2 Properties of Stable Particles . . . . . . . . . . . . . . . . . . 213A.3 Properties of Solar System Objects . . . . . . . . . . . . . . . 214A.4 Miscellaneous Conversions . . . . . . . . . . . . . . . . . . . . 214

B GNU Free Documentation License 215

B.1 Applicability and Definitions . . . . . . . . . . . . . . . . . . . 216B.2 Verbatim Copying . . . . . . . . . . . . . . . . . . . . . . . . . 217B.3 Copying in Quantity . . . . . . . . . . . . . . . . . . . . . . . 217B.4 Modifications . . . . . . . . . . . . . . . . . . . . . . . . . . . 218B.5 Combining Documents . . . . . . . . . . . . . . . . . . . . . . 221B.6 Collections of Documents . . . . . . . . . . . . . . . . . . . . . 221B.7 Aggregation With Independent Works . . . . . . . . . . . . . 221B.8 Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222B.9 Termination . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222B.10 Future Revisions of This License . . . . . . . . . . . . . . . . . 223

C History 225

viii CONTENTS

Preface to April 2006 Edition

This text has developed out of an alternate beginning physics course at NewMexico Tech designed for those students with a strong interest in physics.The course includes students intending to major in physics, but is not limitedto them. The idea for a “radically modern” course arose out of frustrationwith the standard two-semester treatment. It is basically impossible to in-corporate a significant amount of “modern physics” (meaning post-19th cen-tury!) in that format. Furthermore, the standard course would seem to bespecifically designed to discourage any but the most intrepid students fromcontinuing their studies in this area — students don’t go into physics to learnabout balls rolling down inclined planes — they are (rightly) interested inquarks and black holes and quantum computing, and at this stage they arelargely unable to make the connection between such mundane topics and theexciting things that they have read about in popular books and magazines.

It would, of course, be easy to pander to students — teach them superfi-cially about the things they find interesting, while skipping the “hard stu!”.However, I am convinced that they would ultimately find such an approachas unsatisfying as would the educated physicist.

The idea for this course came from reading Louis de Broglie’s NobelPrize address.1 De Broglie’s work is a masterpiece based on the principles ofoptics and special relativity, which qualitatively foresees the path taken bySchrodinger and others in the development of quantum mechanics. It thusdawned on me that perhaps optics and waves together with relativity couldform a better foundation for all of physics than does classical mechanics.

Whether this is so or not is still a matter of debate, but it is indisputablethat such a path is much more fascinating to most college freshmen interestedin pursing studies in physics — especially those who have been through the

1Reprinted in: Boorse, H. A., and L. Motz, 1966: The world of the atom. Basic Books,New York, 1873 pp.

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x PREFACE TO APRIL 2006 EDITION

usual high school treatment of classical mechanics. I am also convinced thatthe development of physics in these terms, though not historical, is at leastas rigorous and coherent as the classical approach.

The course is tightly structured, and it contains little or nothing that canbe omitted. However, it is designed to fit into the usual one year slot typicallyallocated to introductory physics. In broad outline form, the structure is asfollows:

• Optics and waves occur first on the menu. The idea of group velocity iscentral to the entire course, and is introduced in the first chapter. Thisis a di"cult topic, but repeated reviews through the year cause it toeventually sink in. Interference and di!raction are done in a reasonablyconventional manner. Geometrical optics is introduced, not only forits practical importance, but also because classical mechanics is laterintroduced as the geometrical optics limit of quantum mechanics.

• Relativity is treated totally in terms of space-time diagrams — theLorentz transformations seem to me to be quite confusing to studentsat this level (“Does gamma go upstairs or downstairs?”), and all desiredresults can be obtained by using the “space-time Pythagorean theorem”instead, with much better e!ect.

• Relativity plus waves leads to a dispersion relation for free matterwaves. Optics in a region of variable refractive index provides a power-ful analogy for the quantum mechanics of a particle subject to potentialenergy. The group velocity of waves is equated to the particle velocity,leading to the classical limit and Newton’s equations. The basic topicsof classical mechanics are then done in a more or less conventional,though abbreviated fashion.

• Gravity is treated conventionally, except that Gauss’s law is introducedfor the gravitational field. This is useful in and of itself, but also pro-vides a preview of its deployment in electromagnetism. The repetitionis useful pedagogically.

• Electromagnetism is treated in a highly unconventional way, thoughthe endpoint is Maxwell’s equations in their usual integral form. Theseemingly simple question of how potential energy can be extended tothe relativistic context gives rise to the idea of potential momentum.

xi

The potential energy and potential momentum together form a four-vector which is closely related to the scalar and vector potential ofelectromagnetism. The Aharonov-Bohm e!ect is easily explained usingthe idea of potential momentum in one dimension, while extension tothree dimensions results in an extension of Snell’s law valid for matterwaves, from which the Lorentz force law is derived.

• The generation of electromagnetic fields comes from Coulomb’s law plusrelativity, with the scalar and vector potential being used to produce amuch more straightforward treatment than is possible with electric andmagnetic fields. Electromagnetic radiation is a lot simpler in terms ofthe potential fields as well.

• Resistors, capacitors, and inductors are treated for their practical value,but also because their consideration leads to an understanding of energyin electromagnetic fields.

• At this point the book shifts to a more qualitative treatment of atoms,atomic nuclei, the standard model of elementary particles, and tech-niques for observing the very small. Ideas from optics, waves, andrelativity reappear here.

• The final section of the course deals with heat and statistical mechanics.Only at this point do non-conservative forces appear in the contextof classical mechanics. Counting as a way to compute the entropyis introduced, and is applied to the Einstein model of a collection ofharmonic oscillators (conceptualized as a “brick”), and in a limited wayto an ideal gas. The second law of thermodynamics follows. The bookends with a fairly conventional treatment of heat engines.

A few words about how I have taught the course at New Mexico Tech arein order. As with our standard course, each week contains three lecture hoursand a two-hour recitation. The book contains little in the way of examplesof the type normally provided by a conventional physics text, and the styleof writing is quite terse. Furthermore, the problems are few in number andgenerally quite challenging — there aren’t many “plug-in” problems. Therecitation is the key to making the course accessible to the students. I gener-ally have small groups of students working on assigned homework problemsduring recitation while I wander around giving hints. After all groups have

xii PREFACE TO APRIL 2006 EDITION

completed their work, a representative from each group explains their prob-lem to the class. The students are then required to write up the problems ontheir own and hand them in at a later date. In addition, reading summariesare required, with questions about material in the text which gave di"cul-ties. Many lectures are taken up answering these questions. Students tendto do the summaries, as their lowest test grade is dropped if they completea reasonable fraction of them. The summaries and the associated questionshave been quite helpful to me in indicating parts of the text which needclarification.

I freely acknowledge stealing ideas from Edwin Taylor, Archibald Wheeler,Thomas Moore, Robert Mills, Bruce Sherwood, and many other creativephysicists, and I owe a great debt to them. My colleagues Alan Blyth andDavid Westpfahl were brave enough to teach this course at various stages ofits development, and I welcome the feedback I have received from them. Fi-nally, my humble thanks go out to the students who have enthusiastically (oron occasion unenthusiastically) responded to this course. It is much, muchbetter as a result of their input.

There is still a fair bit to do in improving the text at this point, such asrewriting various sections and adding an index . . . Input is welcome, errorswill be corrected, and suggestions for changes will be considered to the extentthat time and energy allow.

Finally, a word about the copyright, which is actually the GNU “copy-left”. The intention is to make the text freely available for downloading,modification (while maintaining proper attribution), and printing in as manycopies as is needed, for commercial or non-commercial use. I solicit com-ments, corrections, and additions, though I will be the ultimate judge as towhether to add them to my version of the text. You may of course do whatyou please to your version, provided you stay within the limitations of thecopyright!David J. RaymondNew Mexico TechSocorro, NM, [email protected]

Chapter 1

Waves in One Dimension

The wave is a universal phenomenon which occurs in a multitude of physicalcontexts. The purpose of this section is to describe the kinematics of waves, i.e., to provide tools for describing the form and motion of all waves irrespectiveof their underlying physical mechanisms.

Many examples of waves are well known to you. You undoubtably knowabout ocean waves and have probably played with a stretched slinky toy,producing undulations which move rapidly along the slinky. Other examplesof waves are sound, vibrations in solids, and light.

In this chapter we learn first about the basic properties of waves andintroduce a special type of wave called the sine wave. Examples of wavesseen in the real world are presented. We then learn about the superpositionprinciple, which allows us to construct complex wave patterns by superim-posing sine waves. Using these ideas, we discuss the related ideas of beatsand interferometry. Finally, The ideas of wave packets and group velocityare introduced.

1.1 Transverse and Longitudinal Waves

With the exception of light, waves are undulations in some material medium.For instance, ocean waves are (nearly) vertical undulations in the position ofwater parcels. The oscillations in neighboring parcels are phased such that apattern moves across the ocean surface. Waves on a slinky are either trans-verse, in that the motion of the material of the slinky is perpendicular to theorientation of the slinky, or they are longitudinal, with material motion in the

1

2 CHAPTER 1. WAVES IN ONE DIMENSION

Transverse wave

Longitudinal wave

Figure 1.1: Example of displacements in transverse and longitudinal waves.The wave motion is to the right as indicated by the large arrows. The smallarrows indicate the displacements at a particular instant.

direction of the stretched slinky. (See figure 1.1.) Some media support onlylongitudinal waves, others support only transverse waves, while yet otherssupport both types. Deep ocean waves are purely transverse, while soundwaves are purely longitudinal.

1.2 Sine Waves

A particularly simple kind of wave, the sine wave, is illustrated in figure 1.2.This has the mathematical form

h(x) = h0 sin(2!x/"), (1.1)

where h is the displacement (which can be either longitudinal or transverse),h0 is the maximum displacement, sometimes called the amplitude of the wave,and " is the wavelength. The oscillatory behavior of the wave is assumed tocarry on to infinity in both positive and negative x directions. Notice thatthe wavelength is the distance through which the sine function completesone full cycle. The crest and the trough of a wave are the locations of themaximum and minimum displacements, as seen in figure 1.2.

So far we have only considered a sine wave as it appears at a particulartime. All interesting waves move with time. The movement of a sine wave

1.2. SINE WAVES 3

x

h 0

hλ

crest

trough

Figure 1.2: Definition sketch for a sine wave, showing the wavelength " andthe amplitude h0 and the phase # at various points.

to the right a distance d may be accounted for by replacing x in the aboveformula by x! d. If this movement occurs in time t, then the wave moves atvelocity c = d/t. Solving this for d and substituting yields a formula for thedisplacement of a sine wave as a function of both distance x and time t:

h(x, t) = h0 sin[2!(x ! ct)/"]. (1.2)

The time for a wave to move one wavelength is called the period of thewave: T = "/c. Thus, we can also write

h(x, t) = h0 sin[2!(x/"! t/T )]. (1.3)

Physicists actually like to write the equation for a sine wave in a slightlysimpler form. Defining the wavenumber as k = 2!/" and the angular fre-quency as $ = 2!/T , we write

h(x, t) = h0 sin(kx ! $t). (1.4)

We normally think of the frequency of oscillatory motion as the number ofcycles completed per second. This is called the rotational frequency, and isgiven by f = 1/T . It is related to the angular frequency by $ = 2!f . Therotational frequency is usually easier to measure than the angular frequency,but the angular frequency tends to be used more in theoretical discussions.As shown above, converting between the two is not di"cult. Rotational fre-quency is measured in units of hertz, abbreviated Hz; 1 Hz = 1 s!1. Angular


frequency also has the dimensions of inverse time, e. g., inverse seconds, butthe term “hertz” is generally reserved only for rotational frequency.

The argument of the sine function is by definition an angle. We refer tothis angle as the phase of the wave, # = kx!$t. The di!erence in the phaseof a wave at fixed time over a distance of one wavelength is 2!, as is thedi!erence in phase at fixed position over a time interval of one wave period.

Since angles are dimensionless, we normally don’t include this in the unitsfor frequency. However, it sometimes clarifies things to refer to the dimen-sions of rotational frequency as “rotations per second” or angular frequencyas “radians per second”.

As previously noted, we call h0, the maximum displacement of the wave,the amplitude. Often we are interested in the intensity of a wave, which isdefined as the square of the amplitude, I = h2

0.The wave speed we have defined above, c = "/T , is actually called the

phase speed. Since " = 2!/k and T = 2!/$, we can write the phase speedin terms of the angular frequency and the wavenumber:

c =$

k(phase speed). (1.5)

1.3 Types of Waves

In order to make the above material more concrete, we now examine thecharacteristics of various types of waves which may be observed in the realworld.

1.3.1 Ocean Surface Waves

These waves are manifested as undulations of the ocean surface as seen infigure 1.3. The speed of ocean waves is given by the formula

c =

!

g tanh(kH)

k

"1/2

, (1.6)

where g = 9.8 m s!2 is a constant related to the strength of the Earth’sgravity, H is the depth of the ocean, and the hyperbolic tangent is defined

1.3. TYPES OF WAVES 5

H

Figure 1.3: Wave on an ocean of depth H . The wave is moving to the rightand the particles of water at the surface move up and down as shown by thesmall vertical arrows.

as1

tanh(x) =exp(x) ! exp(!x)

exp(x) + exp(!x). (1.7)

As figure 1.4 shows, for |x| " 1, we can approximate the hyperbolictangent by tanh(x) # x, while for |x| $ 1 it is +1 for x > 0 and !1 forx < 0. This leads to two limits: Since x = kH , the shallow water limit,which occurs when kH " 1, yields a wave speed of

c # (gH)1/2, (shallow water waves), (1.8)

while the deep water limit, which occurs when kH $ 1, yields

c # (g/k)1/2, (deep water waves). (1.9)

Notice that the speed of shallow water waves depends only on the depthof the water and on g. In other words, all shallow water waves move atthe same speed. On the other hand, deep water waves of longer wavelength(and hence smaller wavenumber) move more rapidly than those with shorterwavelength. Waves for which the wave speed varies with wavelength arecalled dispersive. Thus, deep water waves are dispersive, while shallow waterwaves are non-dispersive.

For water waves with wavelengths of a few centimeters or less, surfacetension becomes important to the dynamics of the waves. In the deep water

1The notation exp(x) is just another way of writing the exponential function ex. Weprefer this way because it is prettier when the function argument is complicated.


-1.00

-0.50

0.00

0.50

-4.00 -2.00 0.00 2.00tanh(x) vs x

Figure 1.4: Plot of the function tanh(x). The dashed line shows our approx-imation tanh(x) # x for |x| " 1.

case the wave speed at short wavelengths is actually given by the formula

c = (g/k + Ak)1/2 (1.10)

where the constant A is related to an e!ect called surface tension. For anair-water interface near room temperature, A # 74 cm3 s!2.

1.3.2 Sound Waves

Sound is a longitudinal compression-expansion wave in a gas. The wavespeed is

c = (%RTabs)1/2 (1.11)

where % and R are constants and Tabs is the absolute temperature. Theabsolute temperature is measured in Kelvins and is numerically given by

Tabs = TC + 273" (1.12)

where TC is the temperature in Celsius degrees. The angular frequency ofsound waves is thus given by

$ = ck = (%RTabs)1/2k. (1.13)

The speed of sound in air at normal temperatures is about 340 m s!1.

1.4. SUPERPOSITION PRINCIPLE 7

1.3.3 Light

Light moves in a vacuum at a speed of cvac = 3 % 108 m s!1. In transparentmaterials it moves at a speed less than cvac by a factor n which is called therefractive index of the material:

c = cvac/n. (1.14)

Often the refractive index takes the form

n2 # 1 +A

1 ! (k/kR)2, (1.15)

where k is the wavenumber and kR and A are positive constants characteristicof the material. The angular frequency of light in a transparent medium isthus

$ = kc = kcvac/n. (1.16)

1.4 Superposition Principle

It is found empirically that as long as the amplitudes of waves in most mediaare small, two waves in the same physical location don’t interact with eachother. Thus, for example, two waves moving in the opposite direction simplypass through each other without their shapes or amplitudes being changed.When collocated, the total wave displacement is just the sum of the displace-ments of the individual waves. This is called the superposition principle. Atsu"ciently large amplitude the superposition principle often breaks down —interacting waves may scatter o! of each other, lose amplitude, or changetheir form.

Interference is a consequence of the superposition principle. When two ormore waves are superimposed, the net wave displacement is just the algebraicsum of the displacements of the individual waves. Since these displacementscan be positive or negative, the net displacement can either be greater or lessthan the individual wave displacements. The former case is called construc-tive interference, while the latter is called destructive interference.

Let us see what happens when we superimpose two sine waves with dif-ferent wavenumbers. Figure 1.5 shows the superposition of two waves withwavenumbers k1 = 4 and k2 = 5. Notice that the result is a wave with aboutthe same wavelength as the two initial waves, but which varies in amplitude


-1.00

-0.50

0.00

0.50

-10.0 -5.0 0.0 5.0sine functions vs x

-2.00

-1.00

0.00

1.00

-10.0 -5.0 0.0 5.0sum of functions vs x

k1, k2 = 4.000000 5.000000

Figure 1.5: Superposition (lower panel) of two sine waves (shown individuallyin the upper panel) with equal amplitudes and wavenumbers k1 = 4 andk2 = 5.

depending on whether the two sine waves are in or out of phase. When thewaves are in phase, constructive interference is occurring, while destructiveinterference occurs where the waves are out of phase.

What happens when the wavenumbers of the two sine waves are changed?Figure 1.6 shows the result when k1 = 10 and k2 = 11. Notice that thoughthe wavelength of the resultant wave is decreased, the locations where theamplitude is maximum have the same separation in x as in figure 1.5.

If we superimpose waves with k1 = 10 and k2 = 12, as is shown in figure1.7, we see that the x spacing of the regions of maximum amplitude hasdecreased by a factor of two. Thus, while the wavenumber of the resultantwave seems to be related to something like the average of the wavenumbersof the component waves, the spacing between regions of maximum waveamplitude appears to go inversely with the di!erence of the wavenumbers ofthe component waves. In other words, if k1 and k2 are close together, theamplitude maxima are far apart and vice versa.

We can symbolically represent the sine waves that make up figures 1.5,1.6, and 1.7 by a plot such as that shown in figure 1.8. The amplitudes andwavenumbers of each of the sine waves are indicated by vertical lines in this


-1.00

-0.50

0.00

0.50


-2.00

-1.00

0.00

1.00


k1, k2 = 10.000000 11.000000

Figure 1.6: Superposition of two sine waves with equal amplitudes andwavenumbers k1 = 10 and k2 = 11.

-1.00

-0.50

0.00

0.50


-2.00

-1.00

0.00

1.00


k1, k2 = 10.000000 12.000000

Figure 1.7: Superposition of two sine waves with equal amplitudes andwavenumbers k1 = 10 and k2 = 12.


Δk Δk

k 1 k 2

k 0

ampl

itude

wavenumber

2 waves

Figure 1.8: Representation of the wavenumbers and amplitudes of two su-perimposed sine waves.

figure.

The regions of large wave amplitude are called wave packets. Wave pack-ets will play a central role in what is to follow, so it is important that weacquire a good understanding of them. The wave packets produced by onlytwo sine waves are not well separated along the x-axis. However, if we super-impose many waves, we can produce an isolated wave packet. For example,figure 1.9 shows the results of superimposing 20 sine waves with wavenumbersk = 0.4m, m = 1, 2, . . . , 20, where the amplitudes of the waves are largestfor wavenumbers near k = 4. In particular, we assume that the amplitudeof each sine wave is proportional to exp[!(k ! k0)2/#k2], where k0 = 4 and#k = 1. The amplitudes of each of the sine waves making up the wave packetin figure 1.9 are shown schematically in figure 1.10.

The quantity #k controls the distribution of the sine waves being super-imposed — only those waves with a wavenumber k within approximately #kof the central wavenumber k0 of the wave packet, i. e., for 3 & k & 5 inthis case, contribute significantly to the sum. If #k is changed to 2, so thatwavenumbers in the range 2 & k & 6 contribute significantly, the wavepacketbecomes narrower, as is shown in figures 1.11 and 1.12. #k is called thewavenumber spread of the wave packet, and it evidently plays a role similarto the di!erence in wavenumbers in the superposition of two sine waves —the larger the wavenumber spread, the smaller the physical size of the wavepacket. Furthermore, the wavenumber of the oscillations within the wavepacket is given approximately by the central wavenumber.


-5.00

-2.50

0.00

2.50

-10.0 -5.0 0.0 5.0displacement vs x

k0 = 4.000000; dk = 1.000000; nwaves = 20

Figure 1.9: Superposition of twenty sine waves with k0 = 4 and #k = 1.

k 0

Δk

ampl

itude

0 4wavenumber8 12 16 20

Figure 1.10: Representation of the wavenumbers and amplitudes of 20 su-perimposed sine waves with k0 = 4 and #k = 1.


-10.0

-5.0

0.0

5.0

-10.0 -5.0 0.0 5.0displacement vs x

k0 = 4.000000; dk = 2.000000; nwaves = 20

Figure 1.11: Superposition of twenty sine waves with k0 = 4 and #k = 2.

k0

Δk

ampl

itude

0 4wavenumber8 12 16 20

Figure 1.12: Representation of the wavenumbers and amplitudes of 20 su-perimposed sine waves with k0 = 4 and #k = 2.

1.5. BEATS 13

We can better understand how wave packets work by mathematicallyanalyzing the simple case of the superposition of two sine waves. Let us definek0 = (k1 + k2)/2 where k1 and k2 are the wavenumbers of the componentwaves. Furthermore let us set #k = (k2 ! k1)/2. The quantities k0 and #kare graphically illustrated in figure 1.8. We can write k1 = k0 ! #k andk2 = k0 + #k and use the trigonometric identity sin(a + b) = sin(a) cos(b) +cos(a) sin(b) to find

sin(k1x) + sin(k2x) = sin[(k0 ! #k)x] + sin[(k0 + #k)x]

= sin(k0x) cos(#kx) ! cos(k0x) sin(#kx) +

sin(k0x) cos(#kx) + cos(k0x) sin(#kx)

= 2 sin(k0x) cos(#kx). (1.17)

The sine factor on the bottom line of the above equation produces the oscil-lations within the wave packet, and as speculated earlier, this oscillation hasa wavenumber k0 equal to the average of the wavenumbers of the componentwaves. The cosine factor modulates this wave with a spacing between regionsof maximum amplitude of

#x = !/#k. (1.18)

Thus, as we observed in the earlier examples, the length of the wave packet#x is inversely related to the spread of the wavenumbers #k (which in thiscase is just the di!erence between the two wavenumbers) of the componentwaves. This relationship is central to the uncertainty principle of quantummechanics.

1.5 Beats

Suppose two sound waves of di!erent frequency impinge on your ear at thesame time. The displacement perceived by your ear is the superposition ofthese two waves, with time dependence

A(t) = sin($1t) + sin($2t) = 2 sin($0t) cos(#$t), (1.19)

where we now have $0 = ($1 + $2)/2 and #$ = ($2 ! $1)/2. What youactually hear is a tone with angular frequency $0 which fades in and outwith period

Tbeat = !/#$ = 2!/($2 ! $1) = 1/(f2 ! f1). (1.20)


The beat frequency is simply

fbeat = 1/Tbeat = f2 ! f1. (1.21)

Note how beats are the time analog of wave packets — the mathematicsare the same except that frequency replaces wavenumber and time replacesspace.

1.6 Interferometers

An interferometer is a device which splits a beam of light into two sub-beams, shifts the phase of one sub-beam with respect to the other, andthen superimposes the sub-beams so that they interfere constructively ordestructively, depending on the magnitude of the phase shift between them.In this section we study the Michelson interferometer and interferometrice!ects in thin films.

1.6.1 The Michelson Interferometer

The American physicist Albert Michelson invented the optical interferome-ter illustrated in figure 1.13. The incoming beam is split into two beams bythe half-silvered mirror. Each sub-beam reflects o! of another mirror whichreturns it to the half-silvered mirror, where the two sub-beams recombine asshown. One of the reflecting mirrors is movable by a sensitive micrometerdevice, allowing the path length of the corresponding sub-beam, and hencethe phase relationship between the two sub-beams, to be altered. As figure1.13 shows, the di!erence in path length between the two sub-beams is 2xbecause the horizontal sub-beam traverses the path twice. Thus, construc-tive interference occurs when this path di!erence is an integral number ofwavelengths, i. e.,

2x = m", m = 0,±1,±2, . . . (Michelson interferometer) (1.22)

where " is the wavelength of the light and m is an integer. Note that m isthe number of wavelengths that fits evenly into the distance 2x.

1.6. INTERFEROMETERS 15

d

d + x

Michelson interferometer

half-silvered mirror

movable mirror

fixed mirror

Figure 1.13: Sketch of a Michelson interferometer.

d

n > 1

A

n = 1

B

C

incident wave fronts

Figure 1.14: Plane light wave normally incident on a transparent thin filmof thickness d and index of refraction n > 1. Partial reflection occurs atthe front surface of the film, resulting in beam A, and at the rear surface,resulting in beam B. Much of the wave passes completely through the film,as with C.


1.7 Thin Films

One of the most revealing examples of interference occurs when light interactswith a thin film of transparent material such as a soap bubble. Figure 1.14shows how a plane wave normally incident on the film is partially reflectedby the front and rear surfaces. The waves reflected o! the front and rearsurfaces of the film interfere with each other. The interference can be eitherconstructive or destructive depending on the phase di!erence between thetwo reflected waves.

If the wavelength of the incoming wave is ", one would naively expectconstructive interference to occur between the A and B beams if 2d were anintegral multiple of ".

Two factors complicate this picture. First, the wavelength inside the filmis not ", but "/n, where n is the index of refraction of the film. Constructiveinterference would then occur if 2d = m"/n. Second, it turns out thatan additional phase shift of half a wavelength occurs upon reflection whenthe wave is incident on material with a higher index of refraction than themedium in which the incident beam is immersed. This phase shift doesn’toccur when light is reflected from a region with lower index of refraction thanfelt by the incident beam. Thus beam B doesn’t acquire any additional phaseshift upon reflection. As a consequence, constructive interference actuallyoccurs when

2d = (m + 1/2)"/n, m = 0, 1, 2, . . . (constructive interference) (1.23)

while destructive interference results when

2d = m"/n, m = 0, 1, 2, . . . (destructive interference). (1.24)

When we look at a soap bubble, we see bands of colors reflected back froma light source. What is the origin of these bands? Light from ordinary sourcesis generally a mixture of wavelengths ranging from roughly " = 4.5%10!7 m(violet light) to " = 6.5 % 10!7 m (red light). In between violet and redwe also have blue, green, and yellow light, in that order. Because of thedi!erent wavelengths associated with di!erent colors, it is clear that for amixed light source we will have some colors interfering constructively whileothers interfere destructively. Those undergoing constructive interferencewill be visible in reflection, while those undergoing destructive interferencewill not.

1.8. MATH TUTORIAL — DERIVATIVES 17

tangent tocurve at point A

Δy

Δy

xΔ

xΔ

x

y

slope = /

y = y(x)

A

B

Figure 1.15: Estimation of the derivative, which is the slope of the tangentline. When point B approaches point A, the slope of the line AB approachesthe slope of the tangent to the curve at point A.

Another factor enters as well. If the light is not normally incident onthe film, the di!erence in the distances traveled between beams reflected o!of the front and rear faces of the film will not be just twice the thicknessof the film. To understand this case quantitatively, we need the conceptof refraction, which will be developed later in the context of geometricaloptics. However, it should be clear that di!erent wavelengths will undergoconstructive interference for di!erent angles of incidence of the incominglight. Di!erent portions of the thin film will in general be viewed at di!erentangles, and will therefore exhibit di!erent colors under reflection, resultingin the colorful patterns normally seen in soap bubbles.

1.8 Math Tutorial — Derivatives

This section provides a quick introduction to the idea of the derivative. Oftenwe are interested in the slope of a line tangent to a function y(x) at somevalue of x. This slope is called the derivative and is denoted dy/dx. Sincea tangent line to the function can be defined at any point x, the derivative


itself is a function of x:

g(x) =dy(x)

dx. (1.25)

As figure 1.15 illustrates, the slope of the tangent line at some point onthe function may be approximated by the slope of a line connecting twopoints, A and B, set a finite distance apart on the curve:

dy

dx#

#y

#x. (1.26)

As B is moved closer to A, the approximation becomes better. In the limitwhen B moves infinitely close to A, it is exact.

Derivatives of some common functions are now given. In each case a is aconstant.

dxa

dx= axa!1 (1.27)

d

dxexp(ax) = a exp(ax) (1.28)

d

dxlog(ax) =

1

x(1.29)

d

dxsin(ax) = a cos(ax) (1.30)

d

dxcos(ax) = !a sin(ax) (1.31)

daf(x)

dx= a

df(x)

dx(1.32)

d

dx[f(x) + g(x)] =

df(x)

dx+

dg(x)

dx(1.33)

d

dxf(x)g(x) =

df(x)

dxg(x) + f(x)

dg(x)

dx(product rule) (1.34)

d

dxf(y) =

df

dy

dy

dx(chain rule) (1.35)

The product and chain rules are used to compute the derivatives of com-plex functions. For instance,

d

dx(sin(x) cos(x)) =

d sin(x)

dxcos(x) + sin(x)

d cos(x)

dx= cos2(x) ! sin2(x)

andd

dxlog(sin(x)) =

1

sin(x)

d sin(x)

dx=

cos(x)

sin(x).

1.9. GROUP VELOCITY 19

1.9 Group Velocity

We now ask the following question: How fast do wave packets move? Sur-prisingly, we often find that wave packets move at a speed very di!erent fromthe phase speed, c = $/k, of the wave composing the wave packet.

We shall find that the speed of motion of wave packets, referred to as thegroup velocity, is given by

u =d$

dk

#

#

#

#

#

k=k0

(group velocity). (1.36)

The derivative of $(k) with respect to k is first computed and then evaluatedat k = k0, the central wavenumber of the wave packet of interest.

The relationship between the angular frequency and the wavenumber fora wave, $ = $(k), depends on the type of wave being considered. Whateverthis relationship turns out to be in a particular case, it is called the dispersionrelation for the type of wave in question.

As an example of a group velocity calculation, suppose we want to find thevelocity of deep ocean wave packets for a central wavelength of "0 = 60 m.This corresponds to a central wavenumber of k0 = 2!/"0 # 0.1 m!1. Thephase speed of deep ocean waves is c = (g/k)1/2. However, since c ' $/k,we find the frequency of deep ocean waves to be $ = (gk)1/2. The groupvelocity is therefore u ' d$/dk = (g/k)1/2/2 = c/2. For the specified centralwavenumber, we find that u # (9.8 m s!2/0.1 m!1)1/2/2 # 5 m s!1. Bycontrast, the phase speed of deep ocean waves with this wavelength is c #10 m s!1.

Dispersive waves are waves in which the phase speed varies with wavenum-ber. It is easy to show that dispersive waves have unequal phase and groupvelocities, while these velocities are equal for non-dispersive waves.

1.9.1 Derivation of Group Velocity Formula

We now derive equation (1.36). It is easiest to do this for the simplestwave packets, namely those constructed out of the superposition of just twosine waves. We will proceed by adding two waves with full space and timedependence:

A = sin(k1x ! $1t) + sin(k2x ! $2t) (1.37)


After algebraic and trigonometric manipulations familiar from earlier sec-tions, we find

A = 2 sin(k0x ! $0t) cos(#kx ! #$t), (1.38)

where as before we have k0 = (k1+k2)/2, $0 = ($1+$2)/2, #k = (k2!k1)/2,and #$ = ($2!$1)/2. Again think of this as a sine wave of frequency $0 andwavenumber k0 modulated by a cosine function. In this case the modulationpattern moves with a speed so as to keep the argument of the cosine functionconstant:

#kx !#$t = const. (1.39)

Di!erentiating this with respect to t while holding #k and #$ constantyields

u 'dx

dt=

#$

#k. (1.40)

In the limit in which the deltas become very small, this reduces to the deriva-tive

u =d$

dk, (1.41)

which is the desired result.

1.9.2 Examples

We now illustrate some examples of phase speed and group velocity by show-ing the displacement resulting from the superposition of two sine waves, asgiven by equation (1.38), in the x-t plane. This is an example of a spacetimediagram, of which we will see many examples latter on.

Figure 1.16 shows a non-dispersive case in which the phase speed equalsthe group velocity. The regions with vertical and horizontal hatching (shortvertical or horizontal lines) indicate where the wave displacement is large andpositive or large and negative. Large displacements indicate the location ofwave packets. The positions of waves and wave packets at any given time maytherefore be determined by drawing a horizontal line across the graph at thedesired time and examining the variations in wave displacement along thisline. The crests of the waves are indicated by regions of short vertical lines.Notice that as time increases, the crests move to the right. This correspondsto the motion of the waves within the wave packets. Note also that the wavepackets, i. e., the broad regions of large positive and negative amplitudes,move to the right with increasing time as well.


0.00

2.00

4.00

6.00

8.00

-10.0 -5.0 0.0 5.0time vs x

k_mean = 4.500000; delta_k = 0.500000

omega_mean = 4.500000; delta_omega = 0.500000

phase speed = 1.000000; group velocity = 1.000000

Figure 1.16: Net displacement of the sum of two traveling sine waves plottedin the x! t plane. The short vertical lines indicate where the displacement islarge and positive, while the short horizontal lines indicate where it is largeand negative. One wave has k = 4 and $ = 4, while the other has k = 5and $ = 5. Thus, #k = 5 ! 4 = 1 and #$ = 5 ! 4 = 1 and we haveu = #$/#k = 1. Notice that the phase speed for the first sine wave isc1 = 4/4 = 1 and for the second wave is c2 = 5/5 = 1. Thus, c1 = c2 = u inthis case.


0.00

2.00

4.00

6.00

8.00

-10.0 -5.0 0.0 5.0time vs x

k_mean = 5.000000; delta_k = 0.500000

omega_mean = 5.000000; delta_omega = 1.000000

phase speed = 1.000000; group velocity = 2.000000

Figure 1.17: Net displacement of the sum of two traveling sine waves plottedin the x-t plane. One wave has k = 4.5 and $ = 4, while the other hask = 5.5 and $ = 6. In this case #k = 5.5!4.5 = 1 while #$ = 6!4 = 2, sothe group velocity is u = #$/#k = 2/1 = 2. However, the phase speeds forthe two waves are c1 = 4/4.5 = 0.889 and c2 = 6/5.5 = 1.091. The averageof the two phase speeds is about 0.989, so the group velocity is about twicethe average phase speed in this case.

Since velocity is distance moved #x divided by elapsed time #t, theslope of a line in figure 1.16, #t/#x, is one over the velocity of whateverthat line represents. The slopes of lines representing crests (the slanted lines,not the short horizontal and vertical lines) are the same as the slopes oflines representing wave packets in this case, which indicates that the twomove at the same velocity. Since the speed of movement of wave crests isthe phase speed and the speed of movement of wave packets is the groupvelocity, the two velocities are equal and the non-dispersive nature of thiscase is confirmed.

Figure 1.17 shows a dispersive wave in which the group velocity is twicethe phase speed, while figure 1.18 shows a case in which the group velocity isactually opposite in sign to the phase speed. See if you can confirm that thephase and group velocities seen in each figure correspond to the values forthese quantities calculated from the specified frequencies and wavenumbers.


0.00

2.00

4.00

6.00

8.00

-10.0 -5.0 0.0 5.0time vs x

k_mean = 4.500000; delta_k = 0.500000

omega_mean = 4.500000; delta_omega = -0.500000

phase speed = 1.000000; group velocity = -1.000000

Figure 1.18: Net displacement of the sum of two traveling sine waves plottedin the x-t plane. One wave has k = 4 and $ = 5, while the other has k = 5and $ = 4. Can you figure out the group velocity and the average phasespeed in this case? Do these velocities match the apparent phase and groupspeeds in the figure?


1.10 Problems

1. Measure your pulse rate. Compute the ordinary frequency of your heartbeat in cycles per second. Compute the angular frequency in radiansper second. Compute the period.

2. An important wavelength for radio waves in radio astronomy is 21 cm.(This comes from neutral hydrogen.) Compute the wavenumber of thiswave. Compute the ordinary and angular frequencies. (The speed oflight is 3 % 108 m s!1.)

3. Sketch the resultant wave obtained from superimposing the waves A =sin(2x) and B = sin(3x). By using the trigonometric identity given inequation (1.17), obtain a formula for A + B in terms of sin(5x/2) andcos(x/2). Does the wave obtained from sketching this formula agreewith your earlier sketch?

4. Two sine waves with wavelengths "1 and "2 are superimposed, makingwave packets of length L. If we wish to make L larger, should we make"1 and "2 closer together or farther apart? Explain your reasoning.

5. By examining figure 1.9 versus figure 1.10 and then figure 1.11 versusfigure 1.12, determine whether equation (1.18) works at least in anapproximate sense for isolated wave packets.

6. The frequencies of the chromatic scale in music are given by

fi = f02i/12, i = 0, 1, 2, . . . , 11, (1.42)

where f0 is a constant equal to the frequency of the lowest note in thescale.

(a) Compute f1 through f11 if f0 = 440 Hz (the “A” note).

(b) Using the above results, what is the beat frequency between the“A” (i = 0) and “B” (i = 2) notes? (The frequencies are givenhere in cycles per second rather than radians per second.)

(c) Which pair of the above frequencies f0 ! f11 yields the smallestbeat frequency? Explain your reasoning.

1.10. PROBLEMS 25

microwavesource

microwavedetector

Figure 1.19: Sketch of a police radar.

7. Large ships in general cannot move faster than the phase speed ofsurface waves with a wavelength equal to twice the ship’s length. Thisis because most of the propulsive force goes into making big wavesunder these conditions rather than accelerating the ship.

(a) How fast can a 300 m long ship move in very deep water?

(b) As the ship moves into shallow water, does its maximum speedincrease or decrease? Explain.

8. Given the formula for refractive index of light quoted in this section, forwhat range of k does the phase speed of light in a transparent materialtake on real values which exceed the speed of light in a vacuum?

9. A police radar works by splitting a beam of microwaves, part of whichis reflected back to the radar from your car where it is made to interferewith the other part which travels a fixed path, as shown in figure 1.19.

(a) If the wavelength of the microwaves is ", how far do you have totravel in your car for the interference between the two beams togo from constructive to destructive to constructive?

(b) If you are traveling toward the radar at speed v = 30 m s!1, usethe above result to determine the number of times per secondconstructive interference peaks will occur. Assume that " = 3 cm.

10. Suppose you know the wavelength of light passing through a Michelsoninterferometer with high accuracy. Describe how you could use theinterferometer to measure the length of a small piece of material.


partially reflecting surfacesd

Figure 1.20: Sketch of a Fabry-Perot interferometer.

11. A Fabry-Perot interferometer (see figure 1.20) consists of two parallelhalf-silvered mirrors placed a distance d from each other as shown. Thebeam passing straight through interferes with the beam which reflectsonce o! of both of the mirrored surfaces as shown. For wavelength ",what values of d result in constructive interference?

12. A Fabry-Perot interferometer has spacing d = 2 cm between the glassplates, causing the direct and doubly reflected beams to interfere (seefigure 1.20). As air is pumped out of the gap between the plates,the beams go through 23 cycles of constructive-destructive-constructiveinterference. If the wavelength of the light in the interfering beams is5% 10!7 m, determine the index of refraction of the air initially in theinterferometer.

13. Measurements on a certain kind of wave reveal that the angular fre-quency of the wave varies with wavenumber as shown in the followingtable:

$ (s!1) k (m!1)5 120 245 380 4125 5

(a) Compute the phase speed of the wave for k = 3 m!1 and fork = 4 m!1.

(b) Estimate the group velocity for k = 3.5 m!1 using a finite di!er-ence approximation to the derivative.

1.10. PROBLEMS 27

AB

D

k

ω

C E

Figure 1.21: Sketch of a weird dispersion relation.

14. Suppose some type of wave has the (admittedly weird) dispersion rela-tion shown in figure 1.21.

(a) For what values of k is the phase speed of the wave positive?

(b) For what values of k is the group velocity positive?

15. Compute the group velocity for shallow water waves. Compare it withthe phase speed of shallow water waves. (Hint: You first need to derivea formula for $(k) from c(k).)

16. Repeat the above problem for deep water waves.

17. Repeat for sound waves. What does this case have in common withshallow water waves?

Chapter 2

Waves in Two and Three

Dimensions

In this chapter we extend the ideas of the previous chapter to the case ofwaves in more than one dimension. The extension of the sine wave to higherdimensions is the plane wave. Wave packets in two and three dimensionsarise when plane waves moving in di!erent directions are superimposed.

Di!raction results from the disruption of a wave which is impingent uponan object. Those parts of the wave front hitting the object are scattered,modified, or destroyed. The resulting di!raction pattern comes from thesubsequent interference of the various pieces of the modified wave. A knowl-edge of di!raction is necessary to understand the behavior and limitations ofoptical instruments such as telescopes.

Di!raction and interference in two and three dimensions can be manipu-lated to produce useful devices such as the di!raction grating.

2.1 Math Tutorial — Vectors

Before we can proceed further we need to explore the idea of a vector. Avector is a quantity which expresses both magnitude and direction. Graph-ically we represent a vector as an arrow. In typeset notation a vector isrepresented by a boldface character, while in handwriting an arrow is drawnover the character representing the vector.

Figure 2.1 shows some examples of displacement vectors, i. e., vectorswhich represent the displacement of one object from another, and introduces

29

30 CHAPTER 2. WAVES IN TWO AND THREE DIMENSIONS

A y

C y

B y

A x B xC x

xMary

A

BC

George

Pauly

Figure 2.1: Displacement vectors in a plane. Vector A represents the dis-placement of George from Mary, while vector B represents the displacementof Paul from George. Vector C represents the displacement of Paul fromMary and C = A + B. The quantities Ax, Ay, etc., represent the Cartesiancomponents of the vectors.

2.1. MATH TUTORIAL — VECTORS 31

x

y

θ

A

A = A cos

A = A sin y

x

θ

θ

Figure 2.2: Definition sketch for the angle & representing the orientation ofa two dimensional vector.

the idea of vector addition. The tail of vector B is collocated with the headof vector A, and the vector which stretches from the tail of A to the head ofB is the sum of A and B, called C in figure 2.1.

The quantities Ax, Ay, etc., represent the Cartesian components of thevectors in figure 2.1. A vector can be represented either by its Cartesiancomponents, which are just the projections of the vector onto the Cartesiancoordinate axes, or by its direction and magnitude. The direction of a vectorin two dimensions is generally represented by the counterclockwise angle ofthe vector relative to the x axis, as shown in figure 2.2. Conversion from oneform to the other is given by the equations

A = (A2x + A2

y)1/2 & = tan!1(Ay/Ax), (2.1)

Ax = A cos(&) Ay = A sin(&), (2.2)

where A is the magnitude of the vector. A vector magnitude is sometimesrepresented by absolute value notation: A ' |A|.

Notice that the inverse tangent gives a result which is ambiguous relativeto adding or subtracting integer multiples of !. Thus the quadrant in whichthe angle lies must be resolved by independently examining the signs of Ax

and Ay and choosing the appropriate value of &.To add two vectors, A and B, it is easiest to convert them to Cartesian

component form. The components of the sum C = A + B are then just thesums of the components:

Cx = Ax + Bx Cy = Ay + By. (2.3)


AB

θ

Ax

y

x

Figure 2.3: Definition sketch for dot product.

Subtraction of vectors is done similarly, e. g., if A = C !B, then

Ax = Cx ! Bx Ay = Cy ! By. (2.4)

A unit vector is a vector of unit length. One can always construct aunit vector from an ordinary vector by dividing the vector by its length:n = A/|A|. This division operation is carried out by dividing each of thevector components by the number in the denominator. Alternatively, if thevector is expressed in terms of length and direction, the magnitude of thevector is divided by the denominator and the direction is unchanged.

Unit vectors can be used to define a Cartesian coordinate system. Conven-tionally, i, j, and k indicate the x, y, and z axes of such a system. Note that i,j, and k are mutually perpendicular. Any vector can be represented in termsof unit vectors and its Cartesian components: A = Axi+Ayj+Azk. An alter-nate way to represent a vector is as a list of components: A = (Ax, Ay, Az).We tend to use the latter representation since it is somewhat more economicalnotation.

There are two ways to multiply two vectors, yielding respectively whatare known as the dot product and the cross product. The cross product yieldsanother vector while the dot product yields a number. Here we will discussonly the dot product.

Given vectors A and B, the dot product of the two is defined as

A · B ' |A||B| cos &, (2.5)

where & is the angle between the two vectors. An alternate expression forthe dot product exists in terms of the Cartesian components of the vectors:

A ·B = AxBx + AyBy. (2.6)

2.1. MATH TUTORIAL — VECTORS 33

θ

x

x’

yy’

RY’

Y

X’

Xa

b

Figure 2.4: Definition figure for rotated coordinate system. The vector R hascomponents X and Y in the unprimed coordinate system and componentsX # and Y # in the primed coordinate system.

It is easy to show that this is equivalent to the cosine form of the dot productwhen the x axis lies along one of the vectors, as in figure 2.3. Notice inparticular that Ax = |A| cos &, while Bx = |B| and By = 0. Thus, A · B =|A| cos &|B| in this case, which is identical to the form given in equation (2.5).

All that remains to be proven for equation (2.6) to hold in general is toshow that it yields the same answer regardless of how the Cartesian coor-dinate system is oriented relative to the vectors. To do this, we must showthat AxBx + AyBy = A#

xB#

x + A#

yB#

y, where the primes indicate componentsin a coordinate system rotated from the original coordinate system.

Figure 2.4 shows the vector R resolved in two coordinate systems rotatedwith respect to each other. From this figure it is clear that X # = a + b.Focusing on the shaded triangles, we see that a = X cos & and b = Y sin &.Thus, we find X # = X cos & + Y sin &. Similar reasoning shows that Y # =!X sin & + Y cos &. Substituting these and using the trigonometric identitycos2 & + sin2 & = 1 results in

A#

xB#

x + A#

yB#

y = (Ax cos & + Ay sin &)(Bx cos & + By sin &)

+ (!Ax sin & + Ay cos &)(!Bx sin & + By cos &)

= AxBx + AyBy (2.7)


wave fronts

λ|k|

wave vector

Figure 2.5: Definition sketch for a plane sine wave in two dimensions. Thewave fronts are constant phase surfaces separated by one wavelength. Thewave vector is normal to the wave fronts and its length is the wavenumber.

thus proving the complete equivalence of the two forms of the dot productas given by equations (2.5) and (2.6). (Multiply out the above expression toverify this.)

A numerical quantity which doesn’t depend on which coordinate systemis being used is called a scalar. The dot product of two vectors is a scalar.However, the components of a vector, taken individually, are not scalars,since the components change as the coordinate system changes. Since thelaws of physics cannot depend on the choice of coordinate system being used,we insist that physical laws be expressed in terms of scalars and vectors, butnot in terms of the components of vectors.

In three dimensions the cosine form of the dot product remains the same,while the component form is

A · B = AxBx + AyBy + AzBz. (2.8)

2.2. PLANE WAVES 35

2.2 Plane Waves

A plane wave in two or three dimensions is like a sine wave in one dimensionexcept that crests and troughs aren’t points, but form lines (2-D) or planes(3-D) perpendicular to the direction of wave propagation. Figure 2.5 showsa plane sine wave in two dimensions. The large arrow is a vector calledthe wave vector, which defines (1) the direction of wave propagation by itsorientation perpendicular to the wave fronts, and (2) the wavenumber by itslength. We can think of a wave front as a line along the crest of the wave.The equation for the displacement associated with a plane sine wave in threedimensions at some instant in time is

A(x, y, z) = sin(k · x) = sin(kxx + kyy + kzz). (2.9)

Since wave fronts are lines or surfaces of constant phase, the equation defininga wave front is simply k · x = const.

In the two dimensional case we simply set kz = 0. Therefore, a wavefront,or line of constant phase # in two dimensions is defined by the equation

k · x = kxx + kyy = # (two dimensions). (2.10)

This can be easily solved for y to obtain the slope and intercept of thewavefront in two dimensions.

As for one dimensional waves, the time evolution of the wave is obtainedby adding a term !$t to the phase of the wave. In three dimensions thewave displacement as a function of both space and time is given by

A(x, y, z, t) = sin(kxx + kyy + kzz ! $t). (2.11)

The frequency depends in general on all three components of the wave vector.The form of this function, $ = $(kx, ky, kz), which as in the one dimensionalcase is called the dispersion relation, contains information about the physicalbehavior of the wave.

Some examples of dispersion relations for waves in two dimensions are asfollows:

• Light waves in a vacuum in two dimensions obey

$ = c(k2x + k2

y)1/2 (light), (2.12)

where c is the speed of light in a vacuum.


• Deep water ocean waves in two dimensions obey

$ = g1/2(k2x + k2

y)1/4 (ocean waves), (2.13)

where g is the strength of the Earth’s gravitational field as before.

• Certain kinds of atmospheric waves confined to a vertical x ! z planecalled gravity waves (not to be confused with the gravitational wavesof general relativity)1 obey

$ =Nkx

kz(gravity waves), (2.14)

where N is a constant with the dimensions of inverse time called theBrunt-Vaisala frequency.

Contour plots of these dispersion relations are plotted in the upper pan-els of figure 2.6. These plots are to be interpreted like topographic maps,where the lines represent contours of constant elevation. In the case of fig-ure 2.6, constant values of frequency are represented instead. For simplicity,the actual values of frequency are not labeled on the contour plots, but arerepresented in the graphs in the lower panels. This is possible because fre-quency depends only on wave vector magnitude (k2

x + k2y)

1/2 for the first twoexamples, and only on wave vector direction & for the third.

2.3 Superposition of Plane Waves

We now study wave packets in two dimensions by asking what the super-position of two plane sine waves looks like. If the two waves have di!erentwavenumbers, but their wave vectors point in the same direction, the re-sults are identical to those presented in the previous chapter, except thatthe wave packets are indefinitely elongated without change in form in thedirection perpendicular to the wave vector. The wave packets produced inthis case march along in the direction of the wave vectors and thus appear toa stationary observer like a series of passing pulses with broad lateral extent.

1Gravity waves in the atmosphere are vertical or slantwise oscillations of air parcelsproduced by buoyancy forces which push parcels back toward their original elevation aftera vertical displacement.

2.3. SUPERPOSITION OF PLANE WAVES 37

(k + k )x y2 2 1/2 (k + k )x y2 2 1/2

k x k x

k z

k x

k y k y

θ

ωωω

θ

−π/2 π/2

Light waves (2−D) Deep ocean waves Gravity waves

Figure 2.6: Contour plots of the dispersion relations for three kinds of wavesin two dimensions. In the upper panels the curves show lines or contoursalong which the frequency $ takes on constant values. Contours ar drawnfor equally spaced values of $. For light and ocean waves the frequencydepends only on the magnitude of the wave vector, whereas for gravity wavesit depends only on the wave vector’s direction, as defined by the angle & inthe upper right panel. These dependences for each wave type are illustratedin the lower panels.


Superimposing two plane waves which have the same frequency resultsin a stationary wave packet through which the individual wave fronts pass.This wave packet is also elongated indefinitely in some direction, but thedirection of elongation depends on the dispersion relation for the waves beingconsidered. One can think of such wave packets as steady beams, whichguide the individual phase waves in some direction, but don’t themselveschange with time. By superimposing multiple plane waves, all with the samefrequency, one can actually produce a single stationary beam, just as onecan produce an isolated pulse by superimposing multiple waves with wavevectors pointing in the same direction.

2.3.1 Two Waves of Identical Wavelength

If the frequency of a wave depends on the magnitude of the wave vector,but not on its direction, the wave’s dispersion relation is called isotropic. Inthe isotropic case two waves have the same frequency only if the lengths oftheir wave vectors, and hence their wavelengths, are the same. The first twoexamples in figure 2.6 satisfy this condition. In this section we investigatethe beams produced by superimposing isotropic waves of the same frequency.

We superimpose two plane waves with wave vectors k1 = (!#kx, ky)and k2 = (#kx, ky). The lengths of the wave vectors in both cases arek1 = k2 = (#k2

x + k2y)

1/2:

A = sin(!#kxx + kyy ! $t) + sin(#kxx + kyy ! $t). (2.15)

If #kx " ky, then both waves are moving approximately in the y direction.An example of such waves would be two light waves with the same frequenciesmoving in slightly di!erent directions.

Applying the trigonometric identity for the sine of the sum of two angles(as we have done previously), equation (2.15) can be reduced to

A = 2 sin(kyy ! $t) cos(#kxx). (2.16)

This is in the form of a sine wave moving in the y direction with phase speedcphase = $/ky and wavenumber ky, modulated in the x direction by a cosinefunction. The distance w between regions of destructive interference in the xdirection tells us the width of the resulting beams, and is given by #kxw = !,so that

w = !/#kx. (2.17)


ky

kxΔ

λ

2y

x

α

w

Figure 2.7: Wave fronts and wave vectors of two plane waves with the samewavelength but oriented in di!erent directions. The vertical bands showregions of constructive interference where wave fronts coincide. The verticalregions in between the bars have destructive interference, and hence definethe lateral boundaries of the beams produced by the superposition. Thecomponents #kx and ky of one of the wave vectors are shown.

Thus, the smaller #kx, the greater is the beam diameter. This behavior isillustrated in figure 2.7.

Figure 2.8 shows an example of the beams produced by superposition oftwo plane waves of equal wavelength oriented as in figure 2.7. It is easy toshow that the transverse width of the resulting wave packet satisfies equation(2.17).

2.3.2 Two Waves of Di!ering Wavelength

In the third example of figure 2.6, the frequency of the wave depends onlyon the direction of the wave vector, independent of its magnitude, which isjust the reverse of the case for an isotropic dispersion relation. In this casedi!erent plane waves with the same frequency have wave vectors which pointin the same direction, but have di!erent lengths.

More generally, one might have waves for which the frequency dependson both the direction and magnitude of the wave vector. In this case, twodi!erent plane waves with the same frequency would typically have wavevectors which di!ered both in direction and magnitude. Such an example is


0.0

10.0

20.0

30.0

-30.0 -20.0 -10.0 0.0 10.0 20.0y vs x

(k_x1, k_x2, k_y1, k_y2) = (-0.10, 0.10, 1.00, 1.00)

Figure 2.8: Example of beams produced by two plane waves with the samewavelength moving in di!erent directions. The wave vectors of the two wavesare k = (±0.1, 1.0). Regions of positive displacement are illustrated byvertical hatching, while negative displacement has horizontal hatching.


1 2

k0

k1k2

k−ΔkΔ

y

xwλ λ

Figure 2.9: Wave fronts and wave vectors (k1 and k2) of two plane waves withdi!erent wavelengths oriented in di!erent directions. The slanted bands showregions of constructive interference where wave fronts coincide. The slantedregions in between the bars have destructive interference, and as previously,define the lateral limits of the beams produced by the superposition. Thequantities k0 and #k are also shown.

illustrated in figures 2.9 and 2.10. We now investigate the superposition ofnon-isotropic waves with the same frequency.

Mathematically, we can represent the superposition of these two waves asa generalization of equation (2.15):

A = sin[!#kxx + (ky + #ky)y ! $t] + sin[#kxx + (ky !#ky)y !$t]. (2.18)

In this equation we have given the first wave vector a y component ky +#ky

while the second wave vector has ky ! #ky. As a result, the first wavehas overall wavenumber k1 = [#k2

x + (ky + #ky)2]1/2 while the second hask2 = [#k2

x + (ky ! #ky)2]1/2, so that k1 (= k2. Using the usual trigonometricidentity, we write equation (2.18) as

A = 2 sin(kyy ! $t) cos(!#kxx + #kyy). (2.19)

To see what this equation implies, notice that constructive interference be-tween the two waves occurs when !#kxx + #kyy = m!, where m is aninteger. Solving this equation for y yields y = (#kx/#ky)x + m!/#ky,which corresponds to lines with slope #kx/#ky. These lines turn out to


0.0

10.0

20.0

30.0

-30.0 -20.0 -10.0 0.0 10.0 20.0y vs x

(k_x1, k_x2, k_y1, k_y2) = (-0.10, 0.10, 1.00, 0.90)

Figure 2.10: Example of beams produced by two plane waves with wavevectors di!ering in both direction and magnitude. The wave vectors of thetwo waves are k1 = (!0.1, 1.0) and k2 = (0.1, 0.9). Regions of positivedisplacement are illustrated by vertical hatching, while negative displacementhas horizontal hatching.

be perpendicular to the vector di!erence between the two wave vectors,k2 ! k1 = (2#kx,!2#ky). The easiest way to show this is to note thatthis di!erence vector is oriented so that it has a slope !#ky/#kx. Compari-son with the #kx/#ky slope of the lines of constructive interference indicatesthat this is so.

An example of the production of beams by the superposition of two waveswith di!erent directions and wavelengths is shown in figure 2.10. Notice thatthe wavefronts are still horizontal, as in figure 2.8, but that the beams arenot vertical, but slant to the right.

Figure 2.11 summarizes what we have learned about adding plane waveswith the same frequency. In general, the beam orientation (and the lines ofconstructive interference) are not perpendicular to the wave fronts. This onlyoccurs when the wave frequency is independent of wave vector direction.


orientation ofbeam

orientation ofwave fronts

Δk−Δk

ky

kx

k1k2

k0

const.frequencycurve

Figure 2.11: Illustration of factors entering the addition of two plane waveswith the same frequency. The wave fronts are perpendicular to the vector av-erage of the two wave vectors, k0 = (k1+k2)/2, while the lines of constructiveinterference, which define the beam orientation, are oriented perpendicularto the di!erence between these two vectors, k2 ! k1.


y

kx

kα

Figure 2.12: Illustration of wave vectors of plane waves which might be addedtogether.

2.3.3 Many Waves with the Same Wavelength

As with wave packets in one dimension, we can add together more than twowaves to produce an isolated wave packet. We will confine our attention hereto the case of an isotropic dispersion relation in which all the wave vectorsfor a given frequency are of the same length.

Figure 2.12 shows an example of this in which wave vectors of the samewavelength but di!erent directions are added together. Defining 'i as theangle of the ith wave vector clockwise from the vertical, as illustrated infigure 2.12, we could write the superposition of these waves at time t = 0 as

A =$

i

Ai sin(kxix + kyiy)

=$

i

Ai sin[kx sin('i) + ky cos('i)] (2.20)

where we have assumed that kxi = k sin('i) and kyi = k cos('i). The param-eter k = |k| is the magnitude of the wave vector and is the same for all thewaves. Let us also assume in this example that the amplitude of each wavecomponent decreases with increasing |'i|:

Ai = exp[!('i/'max)2]. (2.21)

The exponential function decreases rapidly as its argument becomes morenegative, and for practical purposes, only wave vectors with |'i| & 'max

contribute significantly to the sum. We call 'max the spreading angle.Figure 2.13 shows what A(x, y) looks like when 'max = 0.8 radians and

k = 1. Notice that for y = 0 the wave amplitude is only large for a small


0.0

10.0

20.0

30.0

-30.0 -20.0 -10.0 0.0 10.0 20.0y vs x

Half-angle of diffraction = 0.800000

Wavenumber = 1

Figure 2.13: Plot of the displacement field A(x, y) from equation (2.20) for'max = 0.8 and k = 1.

0.0

10.0

20.0

30.0

-30.0 -20.0 -10.0 0.0 10.0 20.0y vs x

Half-angle of diffraction = 0.200000

Wavenumber = 1

Figure 2.14: Plot of the displacement field A(x, y) from equation (2.20) for'max = 0.2 and k = 1.


region in the range !4 < x < 4. However, for y > 0 the wave spreads into abroad semicircular pattern.

Figure 2.14 shows the computed pattern of A(x, y) when the spreadingangle 'max = 0.2 radians. The wave amplitude is large for a much broaderrange of x at y = 0 in this case, roughly !12 < x < 12. On the other hand,the subsequent spread of the wave is much smaller than in the case of figure2.13.

We conclude that a superposition of plane waves with wave vectors spreadnarrowly about a central wave vector which points in the y direction (as infigure 2.14) produces a beam which is initially broad in x but for which thebreadth increases only slightly with increasing y. However, a superpositionof plane waves with wave vectors spread more broadly (as in figure 2.13)produces a beam which is initially narrow in x but which rapidly increasesin width as y increases.

The relationship between the spreading angle 'max and the initial breadthof the beam is made more understandable by comparison with the results forthe two-wave superposition discussed at the beginning of this section. Asindicated by equation (2.17), large values of kx, and hence ', are associatedwith small wave packet dimensions in the x direction and vice versa. Thesuperposition of two waves doesn’t capture the subsequent spread of thebeam which occurs when many waves are superimposed, but it does lead toa rough quantitative relationship between 'max (which is just tan!1(kx/ky) inthe two wave case) and the initial breadth of the beam. If we invoke the smallangle approximation for ' = 'max so that 'max = tan!1(kx/ky) # kx/ky #kx/k, then kx # k'max and equation (2.17) can be written w = !/kx #!/(k'max) = "/(2'max). Thus, we can find the approximate spreading anglefrom the wavelength of the wave " and the initial breadth of the beam w:

'max # "/(2w) (single slit spreading angle). (2.22)

2.4 Di!raction Through a Single Slit

How does all of this apply to the passage of waves through a slit? Imagine aplane wave of wavelength " impingent on a barrier with a slit. The barriertransforms the plane wave with infinite extent in the lateral direction into abeam with initial transverse dimensions equal to the width of the slit. Thesubsequent development of the beam is illustrated in figures 2.13 and 2.14,

2.5. TWO SLITS 47

narrow slitbroad spreading

broad slitnarrow spreading

Figure 2.15: Schematic behavior when a plane wave impinges on a narrowslit and a broad slit.

and schematically in figure 2.15. In particular, if the slit width is comparableto the wavelength, the beam spreads broadly as in figure 2.13. If the slit widthis large compared to the wavelength, the beam doesn’t spread as much, asfigure 2.14 illustrates. Equation (2.22) gives us an approximate quantitativeresult for the spreading angle if w is interpreted as the width of the slit.

One use of the above equation is in determining the maximum angu-lar resolution of optical instruments such as telescopes. The primary lensor mirror can be thought of as a rather large “slit”. Light from a distantpoint source is essentially in the form of a plane wave when it arrives at thetelescope. However, the light passed by the telescope is no longer a planewave, but is a beam with a tendency to spread. The spreading angle 'max

is given by equation (2.22), and the telescope cannot resolve objects withan angular separation less than 'max. Replacing w with the diameter of thelens or mirror in equation (2.22) thus yields the telescope’s angular resolu-tion. For instance, a moderate sized telescope with aperture 1 m observingred light with " # 6 % 10!7 m has a maximum angular resolution of about3 % 10!7 radians.


slit A

slit B

L 1

L 2

L 0

d sin θ

d

L θ

z

incidentwave front

m = −1

m = +1

m = 0

m = +1.5

m = +0.5

m = −0.5

m = −1.5

Figure 2.16: Definition sketch for the double slit. Light passing through slitB travels an extra distance to the screen equal to d sin & compared to lightpassing through slit A.

2.5 Two Slits

Let us now imagine a plane sine wave normally impingent on a screen withtwo narrow slits spaced by a distance d, as shown in figure 2.16. Since theslits are narrow relative to the wavelength of the wave impingent on them,the spreading angle of the beams is large and the di!raction pattern fromeach slit individually is a cylindrical wave spreading out in all directions, asillustrated in figure 2.13. The cylindrical waves from the two slits interfere,resulting in oscillations in wave intensity at the screen on the right side offigure 2.16.

Constructive interference occurs when the di!erence in the paths traveledby the two waves from their originating slits to the screen, L2 ! L1, is aninteger multiple m of the wavelength ": L2 ! L1 = m". If L0 $ d, the linesL1 and L2 are nearly parallel, which means that the narrow end of the darktriangle in figure 2.16 has an opening angle of &. Thus, the path di!erencebetween the beams from the two slits is L2!L1 = d sin &. Substitution of thisinto the above equation shows that constructive interference occurs when

d sin & = m", m = 0,±1,±2, . . . (two slit interference). (2.23)

Destructive interference occurs when m is an integer plus 1/2. The integer

2.6. DIFFRACTION GRATINGS 49

-1.00

-0.50

0.00

0.50

-20.0 -10.0 0.0 10.0amplitude vs d*k*sin(theta)

0.00

0.25

0.50

0.75

-20.0 -10.0 0.0 10.0intensity vs d*k*sin(theta)

Number of grating lines = 2Wave packet envelope

Intensity

Figure 2.17: Amplitude and intensity of interference pattern from a di!rac-tion grating with 2 slits.

m is called the interference order and is the number of wavelengths by whichthe two paths di!er.

2.6 Di!raction Gratings

Since the angular spacing #& of interference peaks in the two slit case dependson the wavelength of the incident wave, the two slit system can be used as acrude device to distinguish between the wavelengths of di!erent componentsof a non-sinusoidal wave impingent on the slits. However, if more slits areadded, maintaining a uniform spacing d between slits, we obtain a moresophisticated device for distinguishing beam components. This is called adi!raction grating.

Figures 2.17-2.19 show the amplitude and intensity of the di!raction pat-tern for gratings with 2, 4, and 16 slits respectively. Notice how the interfer-ence peaks remain in the same place but increase in sharpness as the numberof slits increases.

The width of the peaks is actually related to the overall width of thegrating, w = nd, where n is the number of slits. Thinking of this width asthe dimension of large single slit, the single slit equation, 'max = "/(2w),


-2.00

-1.00

0.00

1.00


0.00

1.00

2.00

3.00



Intensity


-8.00

-4.00

0.00

4.00


0.0

16.0

32.0

48.0



Intensity


2.7. PROBLEMS 51

tells us the angular width of the peaks.Whereas the angular width of the interference peaks is governed by the

single slit equation, their angular positions are governed by the two slit equa-tion. Let us assume for simplicity that |&| " 1 so that we can make the smallangle approximation to the two slit equation, m" = d sin & # d&, and askthe following question: How di!erent do two wavelengths have to be in orderthat the interference peaks from the two waves not overlap? In order forthe peaks to be distinguishable, they should be separated in & by an angle#& = m#"/d, which is greater than the angular width of each peak, 'max:

#& > 'max. (2.24)

Substituting in the above expressions for #& and 'max and solving for #", weget #" > "/(2mn), where n = w/d is the number of slits in the di!ractiongrating. Thus, the fractional di!erence between wavelengths which can bedistinguished by a di!raction grating depends solely on the interference orderm and the number of slits n in the grating:

#"

">

1

2mn. (2.25)

2.7 Problems

1. Point A is at the origin. Point B is 3 m distant from A at 30" coun-terclockwise from the x axis. Point C is 2 m from point A at 100"

counterclockwise from the x axis.

(a) Obtain the Cartesian components of the vector D1 which goesfrom A to B and the vector D2 which goes from A to C.

(b) Find the Cartesian components of the vector D3 which goes fromB to C.

(c) Find the direction and magnitude of D3.

2. For the vectors in the previous problem, find D1 · D2 using both thecosine form of the dot product and the Cartesian form. Check to seeif the two answers are the same.

3. Show graphically or otherwise that |A + B| (= |A| + |B| except whenthe vectors A and B are parallel.


/4π

λ

y

x

k

Figure 2.20: Sketch of wave moving at 45" to the x-axis.

4. A wave in the x-y plane is defined by h = h0 sin(k · x) where k =(1, 2) cm!1.

(a) On a piece of graph paper draw x and y axes and then plot a linepassing through the origin which is parallel to the vector k.

(b) On the same graph plot the line defined by k · x = kxx + kyy =0, k · x = !, and k · x = 2!. Check to see if these lines areperpendicular to k.

5. A plane wave in two dimensions in the x!y plane moves in the direction45" counterclockwise from the x-axis as shown in figure 2.20. Determinehow fast the intersection between a wavefront and the x-axis movesto the right in terms of the phase speed c and wavelength " of thewave. Hint: What is the distance between wavefronts along the x-axiscompared to the wavelength?

6. Two deep plane ocean waves with the same frequency $ are movingapproximately to the east. However, one wave is oriented a small angle( north of east and the other is oriented ( south of east.

(a) Determine the orientation of lines of constructive interference be-tween these two waves.

(b) Determine the spacing between lines of constructive interference.

2.7. PROBLEMS 53

ky

k x

kckb

ka

ω = 4

ω = 8

ω = 12

1 2-1

1

-1

Figure 2.21: Graphical representation of the dispersion relation for shallowwater waves in a river flowing in the x direction. Units of frequency are hertz,units of wavenumber are inverse meters.

7. An example of waves with a dispersion relation in which the frequencyis a function of both wave vector magnitude and direction is showngraphically in figure 2.21.

(a) What is the phase speed of the waves for each of the three wavevectors? Hint: You may wish to obtain the length of each wavevector graphically.

(b) For each of the wave vectors, what is the orientation of the wavefronts?

(c) For each of the illustrated wave vectors, sketch two other wave vec-tors whose average value is approximately the illustrated vector,and whose tips lie on the same frequency contour line. Deter-mine the orientation of lines of constructive interference producedby the superimposing pairs of plane waves for which each of thevector pairs are the wave vectors.

8. Two gravity waves have the same frequency, but slightly di!erent wave-


lengths.

(a) If one wave has an orientation angle & = !/4 radians, what is theorientation angle of the other? (See figure 2.6.)

(b) Determine the orientation of lines of constructive interference be-tween these two waves.

9. A plane wave impinges on a single slit, spreading out a half-angle 'after the slit. If the whole apparatus is submerged in a liquid withindex of refraction n = 1.5, how does the spreading angle of the lightchange? (Hint: Recall that the index of refraction in a transparentmedium is the ratio of the speed of light in a vacuum to the speedin the medium. Furthermore, when light goes from a vacuum to atransparent medium, the light frequency doesn’t change. Therefore,how does the wavelength of the light change?)

10. Determine the diameter of the telescope needed to resolve a planet2% 108 km from a star which is 6 light years from the earth. (Assumeblue light which has a wavelength " # 4 % 10!7 m = 400 nm. Also,don’t worry about the great di!erence in brightness between the twofor the purposes of this problem.)

11. A laser beam from a laser on the earth is bounced back to the earthby a corner reflector on the moon.

(a) Engineers find that the returned signal is stronger if the laser beamis initially spread out by the beam expander shown in figure 2.22.Explain why this is so.

(b) The beam has a diameter of 1 m leaving the earth. How broad isit when it reaches the moon, which is 4 % 105 km away? Assumethe wavelength of the light to be 5 % 10!7 m.

(c) How broad would the laser beam be at the moon if it weren’tinitially passed through the beam expander? Assume its initialdiameter to be 1 cm.

12. Suppose that a plane wave hits two slits in a barrier at an angle, suchthat the phase of the wave at one slit lags the phase at the other slit byhalf a wavelength. How does the resulting interference pattern changefrom the case in which there is no lag?

2.7. PROBLEMS 55

1 m

laserbeam spreader

Figure 2.22: Sketch of a beam expander for a laser.

13. Suppose that a thin piece of glass of index of refraction n = 1.33 isplaced in front of one slit of a two slit di!raction setup.

(a) How thick does the glass have to be to slow down the incomingwave so that it lags the wave going through the other slit by aphase di!erence of !? Take the wavelength of the light to be" = 6 % 10!7 m.

(b) For the above situation, describe qualitatively how the di!ractionpattern changes from the case in which there is no glass in frontof one of the slits. Explain your results.

14. A light source produces two wavelengths, "1 = 400 nm (blue) and"2 = 600 nm (red).

(a) Qualitatively sketch the two slit di!raction pattern from this source.Sketch the pattern for each wavelength separately.

(b) Qualitatively sketch the 16 slit di!raction pattern from this source,where the slit spacing is the same as in the two slit case.

15. A light source produces two wavelengths, "1 = 631 nm and "2 =635 nm. What is the minimum number of slits needed in a gratingspectrometer to resolve the two wavelengths? (Assume that you arelooking at the first order di!raction peak.) Sketch the di!raction peakfrom each wavelength and indicate how narrow the peaks must be toresolve them.

Chapter 3

Geometrical Optics

As was shown previously, when a plane wave is impingent on an aperturewhich has dimensions much greater than the wavelength of the wave, di!rac-tion e!ects are minimal and a segment of the plane wave passes through theaperture essentially unaltered. This plane wave segment can be thought ofas a wave packet, called a beam or ray, consisting of a superposition of wavevectors very close in direction and magnitude to the central wave vector ofthe wave packet. In most cases the ray simply moves in the direction de-fined by the central wave vector, i. e., normal to the orientation of the wavefronts. However, this is not true when the medium through which the lightpropagates is optically anisotropic, i. e., light traveling in di!erent directionsmoves at di!erent phase speeds. An example of such a medium is a calcitecrystal. In the anisotropic case, the orientation of the ray can be determinedonce the dispersion relation for the waves in question is known, by using thetechniques developed in the previous chapter.

If light moves through some apparatus in which all apertures are muchgreater in dimension than the wavelength of light, then we can use the aboverule to follow rays of light through the apparatus. This is called the geomet-rical optics approximation.

3.1 Reflection and Refraction

Most of what we need to know about geometrical optics can be summarizedin two rules, the laws of reflection and refraction. These rules may both beinferred by considering what happens when a plane wave segment impinges on

57

58 CHAPTER 3. GEOMETRICAL OPTICS

mirror

θR

θ I

Figure 3.1: Sketch showing the reflection of a wave from a plane mirror. Thelaw of reflection states that &I = &R.

a flat surface. If the surface is polished metal, the wave is reflected, whereasif the surface is an interface between two transparent media with di!eringindices of refraction, the wave is partially reflected and partially refracted.Reflection means that the wave is turned back into the half-space from whichit came, while refraction means that it passes through the interface, acquiringa di!erent direction of motion from that which it had before reaching theinterface.

Figure 3.1 shows the wave vector and wave front of a wave being reflectedfrom a plane mirror. The angles of incidence, &I , and reflection, &R, aredefined to be the angles between the incoming and outgoing wave vectorsrespectively and the line normal to the mirror. The law of reflection statesthat &R = &I . This is a consequence of the need for the incoming and outgoingwave fronts to be in phase with each other all along the mirror surface. Thisplus the equality of the incoming and outgoing wavelengths is su"cient toinsure the above result.

Refraction, as illustrated in figure 3.2, is slightly more complicated. SincenR > nI , the speed of light in the right-hand medium is less than in the left-hand medium. (Recall that the speed of light in a medium with refractiveindex n is cmedium = cvac/n.) The frequency of the wave packet doesn’tchange as it passes through the interface, so the wavelength of the light on

3.1. REFLECTION AND REFRACTION 59

A

B D

C

θ I

θ R

n = n n = nI R

Figure 3.2: Sketch showing the refraction of a wave from an interface betweentwo dielectric media with n2 > n1.

the right side is less than the wavelength on the left side.Let us examine the triangle ABC in figure 3.2. The side AC is equal to

the side BC times sin(&I). However, AC is also equal to 2"I , or twice thewavelength of the wave to the left of the interface. Similar reasoning showsthat 2"R, twice the wavelength to the right of the interface, equals BC timessin(&R). Since the interval BC is common to both triangles, we easily seethat

"I

"R=

sin(&I)

sin(&R). (3.1)

Since "I = cIT = cvacT/nI and "R = cRT = cvacT/nR where cI and cR arethe wave speeds to the left and right of the interface, cvac is the speed of lightin a vacuum, and T is the (common) period, we can easily recast the aboveequation in the form

nI sin(&I) = nR sin(&R). (3.2)

This is called Snell’s law, and it governs how a ray of light bends as it passesthrough a discontinuity in the index of refraction. The angle &I is called theincident angle and &R is called the refracted angle. Notice that these anglesare measured from the normal to the surface, not the tangent.


3.2 Total Internal Reflection

When light passes from a medium of lesser index of refraction to one withgreater index of refraction, Snell’s law indicates that the ray bends towardthe normal to the interface. The reverse occurs when the passage is in theother direction. In this latter circumstance a special situation arises whenSnell’s law predicts a value for the sine of the refracted angle greater thanone. This is physically untenable. What actually happens is that the incidentwave is reflected from the interface. This phenomenon is called total internalreflection. The minimum incident angle for which total internal reflectionoccurs is obtained by substituting &R = !/2 into equation (3.2), resulting in

sin(&I) = nR/nI (total internal reflection). (3.3)

3.3 Anisotropic Media

Notice that Snell’s law makes the implicit assumption that rays of light movein the direction of the light’s wave vector, i. e., normal to the wave fronts.As the analysis in the previous chapter makes clear, this is valid only whenthe optical medium is isotropic, i. e., the wave frequency depends only onthe magnitude of the wave vector, not on its direction.

Certain kinds of crystals, such as those made of calcite, are not isotropic— the speed of light in such crystals, and hence the wave frequency, dependson the orientation of the wave vector. As an example, the angular frequencyin an isotropic medium might take the form

$ =

%

c21(kx + ky)2

2+

c22(kx ! ky)2

2

&1/2

, (3.4)

where c1 is the speed of light for waves in which ky = kx, and c2 is its speedwhen ky = !kx.

Figure 3.3 shows an example in which a ray hits a calcite crystal orientedso that constant frequency contours are as specified in equation (3.4). Thewave vector is oriented normal to the surface of the crystal, so that wavefronts are parallel to this surface. Upon entering the crystal, the wave frontorientation must stay the same to preserve phase continuity at the surface.However, due to the anisotropy of the dispersion relation for light in the

3.4. THIN LENS FORMULA AND OPTICAL INSTRUMENTS 61

ω = const.

direction ofray

k

Dispersion relation for crystal

crystal

k x

ky

Figure 3.3: The right panel shows the fate of a light ray normally incident onthe face of a properly cut calcite crystal. The anisotropic dispersion relationwhich gives rise to this behavior is shown in the left panel.

crystal, the ray direction changes as shown in the right panel. This behavioris clearly inconsistent with the usual version of Snell’s law!

It is possible to extend Snell’s law to the anisotropic case. However, wewill not present this here. The following discussions of optical instrumentswill always assume that isotropic optical media are used.

3.4 Thin Lens Formula and Optical Instru-

ments

Given the laws of reflection and refraction, one can see in principle how thepassage of light through an optical instrument could be traced. For each ofa number of initial rays, the change in the direction of the ray at each mirrorsurface or refractive index interface can be calculated. Between these points,the ray traces out a straight line.

Though simple in conception, this procedure can be quite complex inpractice. However, the procedure simplifies if a number of approximations,collectively called the thin lens approximation, are valid. We begin with thecalculation of the bending of a ray of light as it passes through a prism, asillustrated in figure 3.4.

The pieces of information needed to find &, the angle through which the


α

θθ

θ

θ θ

3

4

21

Figure 3.4: Bending of a ray of light as it passes through a prism.

ray is deflected are as follows: The geometry of the triangle defined by theentry and exit points of the ray and the upper vertex of the prism leads to

'+ (!/2 ! &2) + (!/2 ! &3) = !, (3.5)

which simplifies to' = &2 + &3. (3.6)

Snell’s law at the entrance and exit points of the ray tell us that

n =sin(&1)

sin(&2)n =

sin(&4)

sin(&3), (3.7)

where n is the index of refraction of the prism. (The index of refraction ofthe surroundings is assumed to be unity.) One can also infer that

& = &1 + &4 ! '. (3.8)

This comes from the fact that the the sum of the internal angles of the shadedquadrangle in figure 3.4 is (!/2 ! &1) + ' + (!/2 ! &4) + (! + &) = 2!.

Combining equations (3.6), (3.7), and (3.8) allows the ray deflection &to be determined in terms of &1 and ', but the resulting expression is verymessy. However, great simplification occurs if the following conditions aremet:

• The angle ' " 1.

• The angles &1 " 1 and &4 " 1.


θ

α

d do i

o ir

Figure 3.5: Light ray undergoing deflection through an angle & by a lens.The angle ' is the angle between the tangents to the entry and exit pointsof the ray on the lens.

With these approximations it is easy to show that

& = '(n ! 1) (small angles). (3.9)

Generally speaking, lenses and mirrors in optical instruments have curvedrather than flat surfaces. However, we can still use the laws for reflection andrefraction by plane surfaces as long as the segment of the surface on whichthe wave packet impinges is not curved very much on the scale of the wavepacket dimensions. This condition is easy to satisfy with light impinging onordinary optical instruments. In this case, the deflection of a ray of light isgiven by equation (3.9) if ' is defined as the intersection of the tangent linesto the entry and exit points of the ray, as illustrated in figure 3.5.

A positive lens is thicker in the center than at the edges. The angle 'between the tangent lines to the two surfaces of the lens at a distance rfrom the central axis takes the form ' = Cr, where C is a constant. Thedeflection angle of a beam hitting the lens a distance r from the center istherefore & = Cr(n ! 1), as indicated in figure 3.5. The angles o and i sumto the deflection angle: o + i = & = Cr(n ! 1). However, to the extentthat the small angle approximation holds, o = r/do and i = r/di where do isthe distance to the object and di is the distance to the image of the object.Putting these equations together and cancelling the r results in the thin lens


d d

S

S

o i

o

i

Figure 3.6: A positive lens producing an image on the right of the arrow onthe left.

formula:1

do+

1

di= C(n ! 1) '

1

f. (3.10)

The quantity f is called the focal length of the lens. Notice that f = di if theobject is very far from the lens, i. e., if do is extremely large.

Figure 3.6 shows how a positive lens makes an image. The image isproduced by all of the light from each point on the object falling on a corre-sponding point in the image. If the arrow on the left is an illuminated object,an image of the arrow will appear at right if the light coming from the lensis allowed to fall on a piece of paper or a ground glass screen. The size ofthe object So and the size of the image Si are related by simple geometry tothe distances of the object and the image from the lens:

Si

So=

di

do. (3.11)

Notice that a positive lens inverts the image.An image will be produced to the right of the lens only if do > f . If

do < f , the lens is unable to converge the rays from the image to a point,as is seen in figure 3.7. However, in this case the backward extension ofthe rays converge at a point called a virtual image, which in the case of apositive lens is always farther away from the lens than the object. The thinlens formula still applies if the distance from the lens to the image is takento be negative. The image is called virtual because it does not appear on aground glass screen placed at this point. Unlike the real image seen in figure3.6, the virtual image is not inverted.


ddi

o

objectvirtual image

Figure 3.7: Production of a virtual image by a positive lens.

object virtual image

dd

oi

Figure 3.8: Production of a virtual image by a negative lens.


dod i

object image

Figure 3.9: Production of a real image by a concave mirror.

A negative lens is thinner in the center than at the edges and producesonly virtual images. As seen in figure 3.8, the virtual image produced by anegative lens is closer to the lens than is the object. Again, the thin lensformula is still valid, but both the distance from the image to the lens andthe focal length must be taken as negative. Only the distance to the objectremains positive.

Curved mirrors also produce images in a manner similar to a lens, asshown in figure 3.9. A concave mirror, as seen in this figure, works in analogyto a positive lens, producing a real or a virtual image depending on whetherthe object is farther from or closer to the mirror than the mirror’s focallength. A convex mirror acts like a negative lens, always producing a virtualimage. The thin lens formula works in both cases as long as the angles aresmall.

3.5 Fermat’s Principle

An alternate approach to geometrical optics can be developed from Fermat’sprinciple. This principle states (in it’s simplest form) that light waves of agiven frequency traverse the path between two points which takes the leasttime. The most obvious example of this is the passage of light througha homogeneous medium in which the speed of light doesn’t change withposition. In this case shortest time is equivalent to the shortest distancebetween the points, which, as we all know, is a straight line. Thus, Fermat’sprinciple is consistent with light traveling in a straight line in a homogeneous

3.5. FERMAT’S PRINCIPLE 67

h1

w

y

w-y

h 2

B

A

θ I

θR

Figure 3.10: Definition sketch for deriving the law of reflection from Fermat’sprinciple. &I is the angle of incidence and &R the angle of reflection as in figure3.1.

medium.Fermat’s principle can also be used to derive the laws of reflection and

refraction. For instance, figure 3.10 shows a candidate ray for reflection inwhich the angles of incidence and reflection are not equal. The time requiredfor the light to go from point A to point B is

t = [{h21 + y2}1/2 + {h2

2 + (w ! y)2}1/2]/c (3.12)

where c is the speed of light. We find the minimum time by di!erentiating twith respect to y and setting the result to zero, with the result that

y

{h21 + y2}1/2

=w ! y

{h22 + (w ! y)2}1/2

. (3.13)

However, we note that the left side of this equation is simply sin &I , while theright side is sin &R, so that the minimum time condition reduces to sin &I =sin &R or &I = &R, which is the law of reflection.

A similar analysis may be done to derive Snell’s law of refraction. Thespeed of light in a medium with refractive index n is c/n, where c is its speedin a vacuum. Thus, the time required for light to go some distance in such a


h 1

y

w-y

A

θ I

w

B

θR

h 2

n > 1

Figure 3.11: Definition sketch for deriving Snell’s law of refraction from Fer-mat’s principle. The shaded area has index of refraction n > 1.

medium is n times the time light takes to go the same distance in a vacuum.Referring to figure 3.11, the time required for light to go from A to B becomes

t = [{h21 + y2}1/2 + n{h2

2 + (w ! y)2}1/2]/c. (3.14)

This results in the condition

sin &I = n sin &R (3.15)

where &R is now the refracted angle. We recognize this result as Snell’s law.Notice that the reflection case illustrates a point about Fermat’s principle:

The minimum time may actually be a local rather than a global minimum —after all, in figure 3.10, the global minimum distance from A to B is still justa straight line between the two points! In fact, light starting from point Awill reach point B by both routes — the direct route and the reflected route.

It turns out that trajectories allowed by Fermat’s principle don’t strictlyhave to be minimum time trajectories. They can also be maximum timetrajectories, as illustrated in figure 3.12. In this case light emitted at pointO can be reflected back to point O from four points on the mirror, A, B, C,and D. The trajectories O-A-O and O-C-O are minimum time trajectorieswhile O-B-O and O-D-O are maximum time trajectories.

3.5. FERMAT’S PRINCIPLE 69

O

A

C

D B

Figure 3.12: Ellipsoidal mirror showing minimum and maximum time raysfrom the center of the ellipsoid to the mirror surface and back again.

O I

Figure 3.13: Ray trajectories from a point O being focused to another pointI by a lens.


n n n1 2 3

θ1

θ2

θ2

θ3

Figure 3.14: Refraction through multiple parallel layers with di!erent refrac-tive indices.

Figure 3.13 illustrates a rather peculiar situation. Notice that all the raysfrom point O which intercept the lens end up at point I. This would seemto contradict Fermat’s principle, in that only the minimum (or maximum)time trajectories should occur. However, a calculation shows that all theillustrated trajectories in this particular case take the same time. Thus, thelight cannot choose one trajectory over another using Fermat’s principle andall of the trajectories are equally favored. Note that this inference appliesnot to just any set of trajectories, but only those going from one focal pointto another.

3.6 Problems

1. The index of refraction varies as shown in figure 3.14:

(a) Given &1, use Snell’s law to find &2.

(b) Given &2, use Snell’s law to find &3.

(c) From the above results, find &3, given &1. Do n2 or &2 matter?

2. A 45"-45"-90" prism is used to totally reflect light through 90" as shownin figure 3.15. What is the minimum index of refraction of the prismneeded for this to work?

3. Show graphically which way the wave vector must point inside thecalcite crystal of figure 3.3 for a light ray to be horizontally oriented.Sketch the orientation of the wave fronts in this case.

3.6. PROBLEMS 71

Figure 3.15: Refraction through a 45"-45"-90" prism.

x

y

Figure 3.16: Focusing of parallel rays by a parabolic mirror.

n > 1n = 1 n = 1

Figure 3.17: Refraction through a wedge-shaped prism.


4. The human eye is a lens which focuses images on a screen called theretina. Suppose that the normal focal length of this lens is 4 cm andthat this focuses images from far away objects on the retina. Let usassume that the eye is able to focus on nearby objects by changing theshape of the lens, and thus its focal length. If an object is 20 cm fromthe eye, what must the altered focal length of the eye be in order forthe image of this object to be in focus on the retina?

5. Show that a concave mirror that focuses incoming rays parallel to theoptical axis of the mirror to a point on the optical axis, as illustrated infigure 3.16, is parabolic in shape. Hint: Since rays following di!erentpaths all move from the distant source to the focal point of the mirror,Fermat’s principle implies that all of these rays take the same time todo so (why is this?), and therefore all traverse the same distance.

6. Use Fermat’s principle to explain qualitatively why a ray of light followsthe solid rather than the dashed line through the wedge of glass shownin figure 3.17.

7. Test your knowledge of Fermat’s principle by using equation (3.14) toderive Snell’s law.

Chapter 4

Kinematics of Special

Relativity

Albert Einstein invented the special and general theories of relativity earlyin the 20th century, though many other people contributed to the intellec-tual climate which made these discoveries possible. The special theory ofrelativity arose out of a conflict between the ideas of mechanics as developedby Galileo and Newton, and the ideas of electromagnetism. For this reasonrelativity is often discussed after electromagnetism is developed. However,special relativity is actually a generally valid extension to the Galilean worldview which is needed when objects move at very high speeds, and it is onlycoincidentally related to electromagnetism. For this reason we discuss rela-tivity before electromagnetism.

The only fact from electromagnetism that we need is introduced now:There is a maximum speed at which objects can travel. This is coincidentallyequal to the speed of light in a vacuum, c = 3 % 108 m s!1. Furthermore, ameasurement of the speed of a particular light beam yields the same answerregardless of the speed of the light source or the speed at which the measuringinstrument is moving.

This rather bizarre experimental result is in contrast to what occurs inGalilean relativity. If two cars pass a pedestrian standing on a curb, oneat 20 m s!1 and the other at 50 m s!1, the faster car appears to be movingat 30 m s!1 relative to the slower car. However, if a light beam moving at3% 108 m s!1 passes an interstellar spaceship moving at 2% 108 m s!1, thenthe light beam appears to be moving at 3 % 108 m s!1 to occupants of thespaceship, not 1 % 108 m s!1. Furthermore, if the spaceship beams a light

73

74 CHAPTER 4. KINEMATICS OF SPECIAL RELATIVITY

World Line

Time

Past

PositionEvent

SimultaneityLine of

Future

Figure 4.1: Spacetime diagram showing an event, a world line, and a line ofsimultaneity.

signal forward to its (stationary) destination planet, then the resulting beamappears to be moving at 3%108 m s!1 to instruments at the destination, not5 % 108 m s!1.

The fact that we are talking about light beams is only for convenience.Any other means of sending a signal at the maximum allowed speed wouldresult in the same behavior. We therefore cannot seek the answer to thisapparent paradox in the special properties of light. Instead we have to lookto the basic nature of space and time.

4.1 Galilean Spacetime Thinking

In order to gain an understanding of both Galilean and Einsteinian relativ-ity it is important to begin thinking of space and time as being di!erentdimensions of a four-dimensional space called spacetime. Actually, since wecan’t visualize four dimensions very well, it is easiest to start with only onespace dimension and the time dimension. Figure 4.1 shows a graph withtime plotted on the vertical axis and the one space dimension plotted on thehorizontal axis. An event is something that occurs at a particular time anda particular point in space. (“Julius X. wrecks his car in Lemitar, NM on21 June at 6:17 PM.”) A world line is a plot of the position of some objectas a function of time (more properly, the time of the object as a functionof position) on a spacetime diagram. Thus, a world line is really a line in

4.1. GALILEAN SPACETIME THINKING 75

spacetime, while an event is a point in spacetime. A horizontal line parallelto the position axis is a line of simultaneity — i. e., all events on this lineoccur simultaneously.

In a spacetime diagram the slope of a world line has a special meaning.Notice that a vertical world line means that the object it represents does notmove — the velocity is zero. If the object moves to the right, then the worldline tilts to the right, and the faster it moves, the more the world line tilts.Quantitatively, we say that

velocity =1

slope of world line. (4.1)

Notice that this works for negative slopes and velocities as well as positiveones. If the object changes its velocity with time, then the world line iscurved, and the instantaneous velocity at any time is the inverse of the slopeof the tangent to the world line at that time.

The hardest thing to realize about spacetime diagrams is that they rep-resent the past, present, and future all in one diagram. Thus, spacetimediagrams don’t change with time — the evolution of physical systems isrepresented by looking at successive horizontal slices in the diagram at suc-cessive times. Spacetime diagrams represent evolution, but they don’t evolvethemselves.

The principle of relativity states that the laws of physics are the same inall inertial reference frames. An inertial reference frame is one that is notaccelerated. Reference frames attached to a car at rest and to a car movingsteadily down the freeway at 30 m s!1 are both inertial. A reference frameattached to a car accelerating away from a stop light is not inertial.

The principle of relativity is not cast in stone, but is an educated guessor hypothesis based on extensive experience. If the principle of relativityweren’t true, we would have to do all our calculations in some preferred ref-erence frame. This would be very annoying. However, the more fundamentalproblem is that we have no idea what the velocity of this preferred framemight be. Does it move with the earth? That would be very earth-centric.How about the velocity of the center of our galaxy or the mean velocity ofall the galaxies? Rather than face the issue of a preferred reference frame,physicists have chosen to stick with the principle of relativity.

If an object is moving to the left at velocity v relative to a particularreference frame, it appears to be moving at a velocity v# = v ! U relativeto another reference frame which itself is moving at velocity U . This is the


x, x’

t t’

world line of objectmoving at speed v

primed coordsys movingatspeed U

x, x’

t’t

world line of objectmoving at speedv’ = v - U

unprimed coordsys moving atspeed -U

Figure 4.2: The left panel shows the world line in the unprimed referenceframe, while the right panel shows it in the primed frame, which moves tothe right at speed U relative to the unprimed frame. (The “prime” is just alabel that allows us to distinguish the axes corresponding to the two referenceframes.)

Galilean velocity transformation law, and it is based in everyday experience— i. e., if you are traveling 30 m s!1 down the freeway and another carpasses you doing 40 m s!1, then the other car moves past you at 10 m s!1

relative to your car. Figure 4.2 shows how the world line of an object isrepresented di!erently in the unprimed (x, t) and primed (x#, t#) referenceframes.

4.2 Spacetime Thinking in Special Relativity

In special relativity we find that space and time “mix” in a way that theydon’t in Galilean relativity. This suggests that space and time are di!erentaspects of the same “thing”, which we call spacetime.

If time and position are simply di!erent dimensions of the same abstractspace, then they should have the same units. The easiest way to arrangethis is to multiply time by the maximum speed, c, resulting in the kind ofspacetime diagram shown in figure 4.3. Notice that world lines of light haveslope ±1 when the time axis is scaled this way. Furthermore, the relationship

4.2. SPACETIME THINKING IN SPECIAL RELATIVITY 77

ct

x

world linesof light

C

A

B

D

O

FUTURE

PAST

ELSEWHERE

ELSEWHERE

Figure 4.3: Scaled spacetime diagram showing world lines of light passingleft and right through the origin.

between speed and the slope of a world line must be revised to read

v =c

slope(world line). (4.2)

Notice that it is physically possible for an object to have a world line whichconnects event O at the origin and the events A and D in figure 4.3, sincethe slope of the resulting world line would exceed unity, and thus representa velocity less than the speed of light. Events which can be connected by aworld line are called timelike relative to each other. On the other hand, eventO cannot be connected to events B and C by a world line, since this wouldimply a velocity greater than the speed of light. Events which cannot beconnected by a world line are called spacelike relative to each other. Noticethe terminology in figure 4.3: Event A is in the past of event O, while eventD is in the future. Events B and C are elsewhere relative to event O.


ct

x

A

C

X

IcT

B

Figure 4.4: Triangle for Pythagorean theorem in spacetime.

4.3 Postulates of Special Relativity

As we learned previously, the principle of relativity states that the laws ofphysics are the same in all inertial reference frames. The principle of relativityapplies to Einsteinian relativity just as it applies to Galilean relativity.

Notice that the constancy of the speed of light in all reference framesis consistent with the principle of relativity. However, as noted above, it isinconsistent with our notions as to how velocities add, or alternatively, howwe think the world should look from reference frames moving at di!erentspeeds. We have called the classical way of understanding the view fromdi!erent reference frames Galilean relativity. The new way that reconcilesthe behavior of objects moving at very high speeds is called Einsteinianrelativity. Einstein’s great contribution was to discover the laws that tell ushow the world looks from reference frames moving at high speeds relative toeach other. These laws constitute a geometry of spacetime, and from themall of special relativity can be derived.

All of the observed facts about spacetime can be derived from two pos-tulates:

• Whether two events are simultaneous depends on the reference framefrom which they are viewed.

• Spacetime obeys a modified Pythagorean theorem, which gives the dis-

4.3. POSTULATES OF SPECIAL RELATIVITY 79

x

ct ct’

x’

B

C

X

cT

E D

A

Figure 4.5: Sketch of coordinate axes for a moving reference frame, x#, ct#.The meanings of the events A-E are discussed in the text. The lines tiltedat ±45" are the world lines of light passing through the origin.

tance, I, in spacetime as

I2 = X2 ! c2T 2, (4.3)

where X, T , and I are defined in figure 4.4.

Let us discuss these postulates in turn.

4.3.1 Simultaneity

The classical way of thinking about simultaneity is so ingrained in our every-day habits that we have a great deal of di"culty adjusting to what specialrelativity has to say about this subject. Indeed, understanding how relativitychanges this concept is the single most di"cult part of the theory — onceyou understand this, you are well on your way to mastering relativity!

Before tackling simultaneity, let us first think about collocation. Twoevents (such as A and E in figure 4.5) are collocated if they have the same


x value. However, collocation is a concept that depends on reference frame.For instance, George is driving from Boston to Washington. Just as he passesNew York he sneezes (event A in figure 4.5). As he drives by Baltimore, hesneezes again (event D). In the reference frame of the earth, these two sneezesare not collocated, since they are separated by many kilometers. However, inthe reference frame of George’s car, they occur in the same place — assumingthat George hasn’t left the driver’s seat!

Notice that any two events separated by a timelike interval are collocatedin some reference frame. The speed of the reference frame is given by equation(4.2), where the slope is simply the slope of the world line connecting thetwo events.

Classically, if two events are simultaneous, we consider them to be simul-taneous in all reference frames. For instance, if two clocks, one in New Yorkand one in Los Angeles, strike the hour at the same time in the earth refer-ence frame, then classically these events also appear to be simultaneous toinstruments in the space shuttle as it flies over the United States. However,if the space shuttle is moving from west to east, i. e., from Los Angeles to-ward New York, careful measurements will show that the clock in New Yorkstrikes the hour before the clock in Los Angeles!

Just as collocation depends on one’s reference frame, this result showsthat simultaneity also depends on reference frame. Figure 4.5 shows howthis works. In figure 4.5 events A and B are simultaneous in the rest orunprimed reference frame. However, in the primed reference frame, events Aand C are simultaneous, and event B occurs at an earlier time. If A and Bcorrespond to the clocks striking in Los Angeles and New York respectively,then it is clear that B must occur at an earlier time in the primed frame ifindeed A and C are simultaneous in that frame.

The tilted line passing through events A and C in figure 4.5 is called theline of simultaneity for the primed reference frame. Its slope is related to thespeed, U , of the reference frame by

slope = U/c (line of simultaneity). (4.4)

Notice that this is the inverse of the slope of the world line attached to theprimed reference frame. There is thus a symmetry between the world lineand the line of simultaneity of a moving reference frame — as the referenceframe moves faster to the right, these two lines close like the blades of a pairof scissors on the 45" line.

4.3. POSTULATES OF SPECIAL RELATIVITY 81

x

ctct

x

A B

A

B

O1 O2 O1 O2S S

Stationary source and observers Moving source and observers

X X

Figure 4.6: World lines of two observers (O1, O2) and a pulsed light source(S) equidistant between them. In the left frame the observers and the sourceare all stationary. In the right frame they are all moving to the right at halfthe speed of light. The dashed lines show pulses of light emitted simultane-ously to the left and the right.

In Galilean relativity it is fairly obvious what we mean by two events beingsimultaneous — it all boils down to coordinating portable clocks which aresitting next to each other, and then moving them to the desired locations.Two events separated in space are simultaneous if they occur at the sametime on clocks located near each event, assuming that the clocks have beencoordinated in the above manner.

In Einsteinian relativity this doesn’t work, because the very act of movingthe clocks changes the rate at which the clocks run. Thus, it is more di"cultto determine whether two distant events are simultaneous.

An alternate way of experimentally determining simultaneity is shown infigure 4.6. Since we know from observation that light travels at the samespeed in all reference frames, the pulses of light emitted by the light sourcesin figure 4.6 will reach the two equidistant observers simultaneously in bothcases. The line passing through these two events, A and B, defines a line ofsimultaneity for both stationary and moving observers. For the stationaryobservers this line is horizontal, as in Galilean relativity. For the movingobservers the light has to travel farther in the rest frame to reach the observerreceding from the light source, and it therefore takes longer in this frame.Thus, event B in the right panel of figure 4.6 occurs later than event A in


the stationary reference frame and the line of simultaneity is tilted. We seethat the postulate that light moves at the same speed in all reference framesleads inevitably to the dependence of simultaneity on reference frame.

4.3.2 Spacetime Pythagorean Theorem

The Pythagorean theorem of spacetime di!ers from the usual Pythagoreantheorem in two ways. First, the vertical side of the triangle is multiplied by c.This is a trivial scale factor that gives time the same units as space. Second,the right side of equation (4.3) has a minus sign rather than a plus sign. Thishighlights a fundamental di!erence between spacetime and the ordinary xyzspace in which we live. Spacetime is said to have a non-Euclidean geometry— in other words, the normal rules of geometry that we learn in high schooldon’t always work for spacetime!

The main consequence of the minus sign in equation (4.3) is that I2 canbe negative and therefore I can be imaginary. Furthermore, in the specialcase where X = ±cT , we actually have I = 0 even though X, T (= 0 — i.e., the “distance” between two well-separated events can be zero. Clearly,spacetime has some weird properties!

The quantity I is usually called an interval in spacetime. Generally speak-ing, if I2 is positive, the interval is called spacelike, while for a negative I2,the interval is called timelike.

A concept related to the spacetime interval is the proper time ) . Theproper time between the two events A and C in figure 4.4 is defined by theequation

) 2 = T 2 ! X2/c2. (4.5)

Notice that I and ) are related by

) 2 = !I2/c2, (4.6)

so the spacetime interval and the proper time are not independent concepts.However, I has the dimensions of length and is real when the events definingthe interval are spacelike relative to each other, whereas ) has the dimen-sions of time and is real when the events are timelike relative to each other.Both equation (4.3) and equation (4.5) express the spacetime Pythagoreantheorem.

If two events defining the end points of an interval have the same t value,then the interval is the ordinary space distance between the two events. On

4.4. TIME DILATION 83

ct

x’

x

ct’ ct ct’

x’

x

A A

View fromstationaryframe

View fromco-movingframe

BC B

CX

cT’cT’ cT

X

cT

Figure 4.7: Two views of the relationship between three events, A, B, and C.The left panel shows the view from the unprimed reference frame, in which Aand C are collocated, while the right panel shows the view from the primedframe, in which A and B are collocated.

the other hand, if they have the same x value, then the proper time is justthe time interval between the events. If the interval between two events isspacelike, but the events are not simultaneous in the initial reference frame,they can always be made simultaneous by choosing a reference frame inwhich the events lie on the same line of simultaneity. Thus, the meaning ofthe interval in that case is just the distance between the events in the newreference frame. Similarly, for events separated by a timelike interval, theproper time is just the time between two events in a reference frame in whichthe two events are collocated.

4.4 Time Dilation

Stationary and moving clocks run at di!erent rates in relativity. This isillustrated in figure 4.7. The triangle ABC in the left panel of figure 4.7can be used to illustrate this point. Suppose that the line passing throughthe events A and C in this figure is the world line of a stationary observer.At zero time another observer moving with velocity V passes the stationaryobserver. The moving observer’s world line passes through events A and B.


We assume that events B and C are simultaneous in the rest frame, soABC is a right triangle. Application of the spacetime Pythagorean theoremthus yields

c2T #2 = c2T 2 ! X2. (4.7)

Since the second observer is moving at velocity V , the slope of his world lineis

c

V=

cT

X, (4.8)

where the right side of the above equation is the slope calculated as the riseof the world line cT over the run X between events A and B. Eliminating Xbetween the above two equations results in a relationship between T and T #:

T # = T (1 ! V 2/c2)1/2 ' T/%, (4.9)

where

% =1

(1 ! V 2/c2)1/2. (4.10)

The quantity % occurs so often in relativistic calculations that we give it thisspecial name. Note by its definition that % ) 1.

Equation (4.9) tells us that the time elapsed for the moving observer isless than that for the stationary observer, which means that the clock of themoving observer runs more slowly. This is called the time dilation e!ect.

Let us view this situation from the reference frame of the moving observer.In this frame the moving observer becomes stationary and the stationaryobserver moves in the opposite direction, as illustrated in the right panel offigure 4.7. By symmetric arguments, one infers that the clock of the initiallystationary observer who is now moving to the left runs more slowly in thisreference frame than the clock of the initially moving observer. One mightconclude that this contradicts the previous results. However, examinationof the right panel of figure 4.7 shows that this is not so. The interval cT isstill greater than the interval cT #, because such intervals are relativisticallyinvariant quantities. However, events B and C are no longer simultaneous,so one cannot use these results to infer anything about the rate at whichthe two clocks run in this frame. Thus, the relative nature of the concept ofsimultaneity saves us from an incipient paradox, and we see that the relativerates at which clocks run depends on the reference frame in which these ratesare observed.

4.5. LORENTZ CONTRACTION 85

ct’

x’

x

A

ct ct ct’

A

cT’ X

cT’

B

Co-movingframe

Stationaryframe

C

x’

xB

C

X’

X’

X

Figure 4.8: Definition sketch for understanding the Lorentz contraction. Theparallel lines represent the world lines of the front and the rear of a movingobject. The left panel shows a reference frame moving with the object, whilethe right panel shows a stationary reference frame.

4.5 Lorentz Contraction

A similar argument can be made to show how the postulates of relativityresult in the Lorentz contraction. Figure 4.8 compares the length X # of amoving object measured in its own reference frame (left panel) with its lengthX as measured in a stationary reference frame (right panel). The lengthof a moving object is measured by simultaneously measuring the positionsof the front and the rear of the object and subtracting these two numbers.Events A and C correspond to these position measurements for the stationaryreference frame since they are respectively on the rear and front world linesof the object. Thus, the interval AC, which is equal to X, is the length ofthe object as measured in the stationary frame.

In the left panel, X is the hypotenuse of a right triangle. Therefore, bythe Pythagorean theorem of spacetime, we have

X = I = (X #2 ! c2T #2)1/2. (4.11)

Now, the line passing through A and C in the left panel is the line of si-multaneity of the stationary reference frame. The slope of this line is !V/c,where V is the speed of the object relative to the stationary reference frame.


x

ct

world lineof stationarytwin

world line oftravelingtwin

O

A

B

C

Figure 4.9: Definition sketch for the twin paradox. The vertical line is theworld line of the twin that stays at home, while the traveling twin has thecurved world line to the right. The slanted lines between the world linesare lines of simultaneity at various times for the traveling twin. The heavylines of simultaneity bound the period during which the traveling twin isdecelerating to a stop and accelerating toward home.

Geometrically in figure 4.8, the slope of this line is !cT #/X #, so we find byequating these two expressions for the slope that

T # = V X #/c2. (4.12)

Finally, eliminating T # between (4.11) and (4.12) results in

X = X #(1 ! V 2/c2)1/2 = X #/%. (4.13)

This says that the length of a moving object as measured in a stationaryreference frame (X) is less than the actual length of the object as measuredin its own reference frame (X #). This apparent reduction in length is calledthe Lorentz contraction.

4.6 Twin Paradox

An interesting application of time dilation is the so-called twin paradox, whichturns out not to be a paradox at all. Two twins are initially the same age.

4.7. PROBLEMS 87

One twin travels to a distant star on an interstellar spaceship which movesat speed V , which is close to the speed of light. Upon reaching the star, thetraveling twin immediately turns around and heads home. When reachinghome, the traveling twin has aged less than the twin that stayed home. Thisis easily explained by the time dilation e!ect, which shows that the propertime elapsed along the world line of the traveling twin is (1!V 2/c2)1/2 timesthe proper time elapsed along the world line of the other twin.

The “paradox” part of the twin paradox arises from making the symmetricargument in which one assumes the reference frame of the traveling twin to bestationary. The frame of the earth-bound twin must then travel in the senseopposite that of the erstwhile traveling twin, which means that the earth-bound twin must age less rather than more. This substitution is justified onthe basis of the principle of relativity, which states that the laws of physicsmust be the same in all inertial reference frames.

However, the above argument is fallacious, because the reference frame ofthe traveling twin is not inertial throughout the entire trip, since at variouspoints the spaceship has to accelerate and decelerate. Thus, the principle ofrelativity cannot be used to assert the equivalence of the traveler’s referenceframe to the stationary frame.

Of particular importance is the period of deceleration and accelerationnear the distant star. During this interval, the line of simulaneity of thetraveling twin rotates, as illustrated in figure 4.9, such that the twin stayingat home rapidly ages in the reference frame of the traveler. Thus, eventhough the acceleration of the traveling twin may occupy only a negligiblesegment of the twin’s world line, the overall e!ect is not negligible. In fact,the shorter and more intense the period of acceleration, the more rapidly theearth-bound twin ages in the traveling frame during this interval!

4.7 Problems

1. Sketch your personal world line on a spacetime diagram for the last24 hours, labeling by time and location special events such as meals,physics classes, etc. Relate the slope of the world line at various timesto how fast you were walking, riding in a car, etc.

2. Spacetime conversions:

(a) What is the distance from New York to Los Angeles in seconds?


From here to the moon? From here to the sun?

(b) What is one nanosecond in meters? One second? One day? Oneyear?

3. Three events have the following spacetime coordinates: A is at (x, ct) =(2 m, 1 m); B is at (x, ct) = (!2 m, 0 m); C is at (x, ct) = (0 m, 3 m).

(a) A world line for an object passes through events B and C. Howfast and in which direction is the object moving?

(b) A line of simultaneity for a coordinate system passes throughevents A and B. How fast and in which direction is the coordi-nate system moving?

(c) What is the invariant interval between events A and B? B and C?A and C?

(d) Can a signal from event B reach event A? Can it reach event C?Explain.

Hint: Draw a spacetime diagram with all the events plotted beforetrying to answer the above questions.

4. In the following problem be sure to indicate the slope of all pertinentlines drawn.

(a) In a spacetime diagram, sketch a line of simultaneity for a referenceframe moving to the left at V = c/2, where c is the speed of light.

(b) Sketch the world line of an object which is initially stationary, butwhich accelerates to a velocity of v = c/3.

(c) If the slopes of the world lines of the observers in the right panel offigure 4.6 are both 1/(, find the slope of their line of simultaneity,AB.

5. Suppose that an interstellar spaceship goes a distance X = 100 light yearsrelative to the rest frame in T # = 10 years of its own time.

(a) Draw a spacetime diagram in the rest frame showing X, T #, andthe time T needed for this journey relative to the rest frame.

(b) Compute T , using your spacetime diagram as an aid.

4.7. PROBLEMS 89

(c) Compute the speed of the spaceship.

6. If an airline pilot flies 80 hr per month (in the rest frame) at 300 m s!1

for 30 years, how much younger will she be than her twin brother (whohandles baggage) when she retires? Hint: Use (1 ! *)x # 1 ! x* forsmall *.

7. A mu particle normally lives about 2%10!6 sec before it decays. How-ever, muons created by cosmic rays 20 km up in the atmosphere reachthe Earth’s surface. How fast must they be going?

8. The Stanford Linear Accelerator accelerates electrons to a speed suchthat the 3 km long accelerator appears to be 8 cm long to the electron,due to the Lorentz contraction. How much less than the speed oflight is the electron traveling? Hint: It is best to first develop anapproximation for the relationship between % = (1!v2/c2)!1/2 and thedi!erence between c and v for a particle moving close to the speed oflight.

9. How fast do you have to go to reach the center of our galaxy in yourexpected lifetime? At this speed, what does this distance appear tobe? (We are about 30000 ly from the galactic center.)

10. Two identical spaceships pass each other going in the opposite directionat the same speed.

(a) Sketch a spacetime diagram showing the world lines of the frontand rear of each spaceship.

(b) Indicate an interval on the diagram corresponding to the rightward-moving spaceship’s length in its own reference frame.

(c) Indicate an interval corresponding to the leftward-moving space-ship’s length in the reference frame of the rightward-moving space-ship.

(d) Indicate an interval equal to the length of either spaceship in therest frame.

11. George and Sally are identical twins initially separated by a distance dand at rest. In the rest frame they are initially the same age. At timet = 0 both George and Sally get in their spaceships and head to the


x

ct

A B

C

d

SallyGeorge

Figure 4.10: Sketch for moving identical twins. Line AC is the line of simul-taneity for a reference frame moving with Sally and George.

right at velocity U . Both move a distance d to the right and decelerateto a halt. (See figure 4.10.)

(a) When both are moving, how far away is Sally according to George?

(b) How much older or younger is Sally relative to George while bothare moving?

(c) How much older or younger is Sally relative to George after bothstop?

Hint: Draw the triangle ABC in a reference frame moving with Georgeand Sally.

Chapter 5

Applications of Special

Relativity

In this chapter we continue the study of special relativity. Three importantapplications of the ideas developed in the previous chapter are made here.First, we show how to describe waves in the context of spacetime. We thensee how waves which have no preferred reference frame (such as that of amedium supporting them) are constrained by special relativity to have adispersion relation of a particular form. This dispersion relation turns out tobe that of the relativistic matter waves of quantum mechanics. Second, weinvestigate the Doppler shift phenomenon, in which the frequency of a wavetakes on di!erent values in di!erent coordinate systems. Third, we show howto add velocities in a relativistically consistent manner.

A new mathematical idea is presented in the context of relativistic waves,namely the spacetime vector or four-vector. Writing the laws of physicstotally in terms of relativistic scalars and four-vectors insures that they willbe valid in all inertial reference frames.

5.1 Waves in Spacetime

We now look at the characteristics of waves in spacetime. Recall that a wavein one space dimension can be represented by

A(x, t) = A0 sin(kx ! $t), (5.1)

where A0 is the (constant) amplitude of the wave, k is the wavenumber, and$ is the angular frequency, and that the quantity # = kx ! $t is called

91

92 CHAPTER 5. APPLICATIONS OF SPECIAL RELATIVITY

Wave4-vector

/cω

ct

x

Wave fronts

k

Figure 5.1: Sketch of wave fronts for a wave in spacetime. The large arrowis the associated wave four-vector, which has slope $/ck. The slope of thewave fronts is the inverse, ck/$. The phase speed of the wave is greater thanc in this example. (Can you tell why?)

the phase of the wave. For a wave in three space dimensions, the wave isrepresented in a similar way,

A(x, t) = A0 sin(k · x ! $t), (5.2)

where x is now the position vector and k is the wave vector. The magnitude ofthe wave vector, |k| = k is just the wavenumber of the wave and the directionof this vector indicates the direction the wave is moving. The phase of thewave in this case is # = k · x ! $t.

In the one-dimensional case # = kx!$t. A wave front has constant phase#, so solving this equation for t and multiplying by c, the speed of light in avacuum, gives us an equation for the world line of a wave front:

ct =ckx

$!

c#

$=

cx

up!

c#

$(wave front). (5.3)

The slope of the world line in a spacetime diagram is the coe"cient of x, orc/up, where up = $/k is the phase speed.

5.2 Math Tutorial – Four-Vectors

Also shown in figure 5.1 is a spacetime vector or four-vector which representsthe frequency and wavenumber of the wave, which we refer to as the wave

5.2. MATH TUTORIAL – FOUR-VECTORS 93

four-vector. It is called a four-vector because it has 3 spacelike componentsand one timelike component when there are 3 space dimensions. In the caseshown, there is only a single space dimension. The spacelike component ofthe wave four-vector is just k (or k when there are 3 space dimensions),while the timelike component is $/c. The c is in the denominator to give thetimelike component the same dimensions as the spacelike component. Fromfigure 5.1 it is clear that the slope of the line representing the four-vector is$/ck, which is just the inverse of the slope of the wave fronts.

Let us define some terminology. We indicate a four-vector by underliningand write the components in the following way: k = (k,$/c), where k isthe wave four-vector, k is its spacelike component, and $/c is its timelikecomponent. For three space dimensions, where we have a wave vector ratherthan just a wavenumber, we write k = (k,$/c).

Another example of a four-vector is simply the position vector in space-time, x = (x, ct), or x = (x, ct) in three space dimensions. The c multipliesthe timelike component in this case, because that is what is needed to giveit the same dimensions as the spacelike component.

In three dimensions we define a vector as a quantity with magnitudeand direction. Extending this to spacetime, a four-vector is a quantity withmagnitude and direction in spacetime. Implicit in this definition is the notionthat the vector’s magnitude is a quantity independent of coordinate systemor reference frame. We have seen that the invariant interval in spacetime isI = (x2 ! c2t2)1/2, so it makes sense to identify this as the magnitude of theposition vector. This leads to a way of defining a dot product of four-vectors.Given two four-vectors A = (A, At) and B = (B, Bt), the dot product is

A · B = A ·B ! AtBt (dot product in spacetime). (5.4)

This is consistent with the definition of invariant interval if we set A = B = x,since then x · x = x2 ! c2t2 = I2.

In the odd geometry of spacetime it is not obvious what “perpendicular”means. We therefore define two four-vectors A and B to be perpendicular iftheir dot product is zero: A · B = 0.

The dot product of two four-vectors is a scalar result, i. e., its valueis independent of coordinate system. This can be used to advantage onoccasion. For instance, consider the dot product of a four-vector A whichresolves into (Ax, At) in the unprimed frame. Let us further suppose that thespacelike component is zero in some primed frame, so that the components


in this frame are (0, A#

t). The fact that the dot product is independent ofcoordinate system means that

A · A = A2x ! A2

t = !A#2t . (5.5)

This constitutes an extension of the spacetime Pythagorean theorem to four-vectors other than the position four-vector. Thus, for instance, the wavenum-ber for some wave may be zero in the primed frame, which means that thewavenumber and frequency in the unprimed frame are related to the fre-quency in the primed frame by

k2 ! $2/c2 = !$#2/c2. (5.6)

5.3 Principle of Relativity Applied

Returning to the phase of a wave, we immediately see that

# = k · x ! $t = k · x ! ($/c)(ct) = k · x. (5.7)

Thus, a compact way to rewrite equation (5.2) is

A(x) = sin(k · x). (5.8)

Since x is known to be a four-vector and since the phase of a wave isknown to be a scalar independent of reference frame, it follows that k is alsoa four-vector rather than just a set of numbers. Thus, the square of thelength of the wave four-vector must also be a scalar independent of referenceframe:

k · k = k · k ! $2/c2 = const. (5.9)

Let us review precisely what this means. As figure 5.2 shows, we can re-solve a position four-vector into components in two di!erent reference frames,x = (X, cT ) = (X #, cT #). However, even though X (= X # and T (= T #, thevector lengths computed from these two sets of components are necessarilythe same: x · x = X2 ! c2T 2 = X #2 ! c2T #2.

Applying this to the wave four-vector, we infer that

k2 ! $2/c2 = k#2 ! $#2/c2 = const., (5.10)

where the unprimed and primed values of k and $ refer to the componentsof the wave four-vector in two di!erent reference frames.

5.3. PRINCIPLE OF RELATIVITY APPLIED 95

x

ct’ct

X

x’

cT

X’

cT’

Figure 5.2: Resolution of a four-vector into components in two di!erentreference frames.

Up to now, this argument applies to any wave. However, waves canbe divided into two categories, those for which a “special” reference frameexists, and those for which there is no such special frame. As an example ofthe former, sound waves look simplest in the reference frame in which thegas carrying the sound is stationary. The same is true of light propagatingthrough a material medium with an index of refraction not equal to unity.In both cases the speed of the wave is the same in all directions only in theframe in which the material medium is stationary.

The following argument can be made. Suppose we have a machine thatproduces a wave with wavenumber k and frequency $ in its own rest frame.If we observe the wave from a moving reference frame, the wavenumber andfrequency will be di!erent, say, k# and $#. However, these quantities will berelated by equation (5.10).

Up to this point the argument applies to any wave whether a special refer-ence frame exists or not; the observed changes in wavenumber and frequencyhave nothing to do with the wave itself, but are just consequences of how wehave chosen to observe it. However, if there is no special reference frame forthe type of wave under consideration, then the same result can be obtainedby keeping the observer stationary and moving the wave-producing machinein the opposite direction. By moving it at various speeds, any desired valueof k# can be obtained in the initial reference frame (as opposed to some otherframe), and the resulting value of $# can be computed using equation (5.10).

This is actually an amazing result. We have shown on the basis of the


principle of relativity that any wave type for which no special reference frameexists can be made to take on a full range of frequencies and wavenumbersin any given reference frame, and furthermore that these frequencies andwavenumbers obey

$2 = k2c2 + µ2. (5.11)

Equation (5.11) comes from solving equation (5.10) for $2 and the constantµ2 equals the constant in equation (5.10) times !c2. Equation (5.11) relatesfrequency to wavenumber and therefore is the dispersion relation for suchwaves. We call waves which have no special reference frame and thereforenecessarily obey equation (5.11) relativistic waves. The only di!erence in thedispersion relations between di!erent types of relativistic waves is the valueof the constant µ. The meaning of this constant will become clear later.

5.4 Characteristics of Relativistic Waves

Light in a vacuum is an example of a wave for which no special referenceframe exists. For light, µ = 0, and we have (taking the positive root) $ = ck.This simply states what we know already, namely that the phase speed oflight in a vacuum is c.

If µ (= 0, waves of this type are dispersive. The phase speed is

up =$

k= (c2 + µ2/k2)1/2. (5.12)

This phase speed always exceeds c, which at first seems like an unphysicalconclusion. However, the group velocity of the wave is

ug =d$

dk=

c2k

(k2c2 + µ2)1/2=

kc2

$=

c2

up, (5.13)

which is always less than c. Since wave packets and hence signals propagateat the group velocity, waves of this type are physically reasonable even thoughthe phase speed exceeds the speed of light.

Another interesting property of such waves is that the wave four-vector isparallel to the world line of a wave packet in spacetime. This is easily shownby the following argument. As figure 5.1 shows, the spacelike componentof a wave four-vector is k, while the timelike component is $/c. The slopeof the four-vector on a spacetime diagram is therefore $/kc. However, the

5.5. THE DOPPLER EFFECT 97

wave fronts

world line of movingobserver

cT’ cτ

ct

x

X

cT

X

world line ofstationarysource

Figure 5.3: Definition sketch for computing the Doppler e!ect for light.

slope of the world line of a wave packet moving with group velocity ug isc/ug = $/(kc).

Note that when k = 0 we have $ = µ. In this case the group velocity ofthe wave is zero. For this reason we call µ the rest frequency of the wave.

5.5 The Doppler E!ect

You have probably heard how the pitch of a train horn changes as it passesyou. When the train is approaching, the pitch or frequency is higher thanwhen it is moving away from you. This is called the Doppler e!ect. Asimilar, but distinct e!ect occurs if you are moving past a source of sound.If a stationary whistle is blowing, the pitch perceived from a moving car ishigher while moving toward the source than when moving away. The firstcase thus has a moving source, while the second case has a moving observer.

In this section we compute the Doppler e!ect as it applies to light movingthrough a vacuum. Figure 5.3 shows the geometry for computing the timebetween wave fronts of light for a stationary and a moving reference frame.The time in the stationary frame is just T . Since the world lines of the wavefronts have a slope of unity, the sides of the shaded triangle have the samevalue, X. If the observer is moving at speed U , the slope of the observer’s


world line is c/U , which means that c/U = (cT + X)/X. Solving this for Xyields X = UT/(1!U/c), which can then be used to compute T # = T+X/c =T/(1!U/c). This formula as it stands leads to the classical Doppler shift fora moving observer. However, with relativistic velocities, one additional factorneeds to be taken in into account: The observer experiences time dilationsince he or she is moving. The actual time measured by the observer betweenwave fronts is actually

) = (T #2 ! X2/c2)1/2 = T(1 ! U2/c2)1/2

1 ! U/c= T

!

1 + U/c

1 ! U/c

"1/2

, (5.14)

where the last step uses 1!U2/c2 = (1!U/c)(1 + U/c). From this we inferthe relativistic Doppler shift formula for light in a vacuum:

$# = $

!

1 ! U/c

1 + U/c

"1/2

, (5.15)

where the frequency measured by the moving observer is $# = 2!/) and thefrequency observed in the stationary frame is $ = 2!/T .

We could go on to determine the Doppler shift resulting from a movingsource. However, by the principle of relativity, the laws of physics shouldbe the same in the reference frame in which the observer is stationary andthe source is moving. Furthermore, the speed of light is still c in that frame.Therefore, the problem of a stationary observer and a moving source is con-ceptually the same as the problem of a moving observer and a stationarysource when the wave is moving at speed c. This is unlike the case for, say,sound waves, where the stationary observer and the stationary source yielddi!erent formulas for the Doppler shift.

5.6 Addition of Velocities

Figure 5.4 shows the world line of a moving object from the point of view oftwo di!erent reference frames, with the primed frame (left panel) moving tothe right at speed U relative to the unprimed frame (right panel). The goalis to calculate the velocity of the object relative to the unprimed frame, v,assuming its velocity in the primed frame, v# is known. The classical resultis simply

v = U + v# (classical result). (5.16)

5.6. ADDITION OF VELOCITIES 99

ct’

Δc T

XΔ

x’ xO O

A B

B

x’cT’

X’

cT’

AX’

Boost by U = X/T

X

cT

E

D

World line(velocity v’)

World line(velocity v)

ct’ ct

Figure 5.4: Definition sketch for relativistic velocity addition. The two panelsshow the world line of a moving object relative to two di!erent referenceframes moving at velocity U with respect to each other. The velocity of theworld line in the left panel is v# while its velocity in the right panel is v.

However, this is inconsistent with the speed of light being constant in allreference frames, since if we substitute c for v#, this formula predicts that thespeed of light in the unprimed frame is U + c.

We can use the geometry of figure 5.4 to come up with the correct rela-tivistic formula. From the right panel of this figure we infer that

v

c=

X + #X

cT + c#T=

X/(cT ) + #X/(cT )

1 + #T/T. (5.17)

This follows from the fact that the slope of the world line of the object inthis frame is c/v. The slope is calculated as the ratio of the rise, c(T + #T ),to the run, X + #X.

From the left panel of figure 5.4 we similarly see that

v#

c=

X #

cT #. (5.18)

However, we can apply our calculations of Lorentz contraction and timedilation from the previous chapter to triangles ABD and OAE in the rightpanel. The slope of AB is U/c because AB is horizontal in the left panel, soX # = #X(1!U2/c2)1/2. Similarly, the slope of OA is c/U since OA is vertical


in the left panel, and T # = T (1!U2/c2)1/2. Substituting these formulas intothe equation for v#/c yields

v#

c=

#X

cT. (5.19)

Again using what we know about the triangles ABD and OAE, we see that

U

c=

c#T

#X=

X

cT. (5.20)

Finally, we calculate #T/T by noticing that

#T

T=

#T

T

c#X

c#X=

'

c#T

#X

( '

#X

cT

(

=U

c

v#

c. (5.21)

Substituting equations (5.19), (5.20), and (5.21) into equation (5.17) andsimplifying yields the relativistic velocity addition formula:

v =U + v#

1 + Uv#/c2(special relativity). (5.22)

Notice how this equation behaves in various limits. If |Uv#| " c2, thedenominator of equation (5.22) is nearly unity, and the special relativisticformula reduces to the classical case. On the other hand, if v# = c, thenequation (5.22) reduces to v = c. In other words, if the object in questionis moving at the speed of light in one reference frame, it is moving at thespeed of light in all reference frames, i. e., for all possible values of U . Thus,we have found a velocity addition formula that 1) reduces to the classicalformula for low velocities and 2) gives the observed results for very highvelocities as well.

5.7 Problems

1. Sketch the wave fronts and the k four-vector in a spacetime diagram forthe case where $/k = 2c. Label your axes and space the wave frontscorrectly for the case k = 4! m!1.

2. If the four-vector k = (0, 1 nm!1) in the rest frame, find the space andtime components of k in a frame moving to the left at speed c/2.

3. Let’s examine the four-vector u = (ug, c)/(1 ! (2)1/2 where ( = ug/c,ug being the velocity of some object.

5.7. PROBLEMS 101

wave fronts

c τ

x

ct

cT’cT

X

X

world line of stationaryobserver

world line ofmoving source

Figure 5.5: Doppler e!ect for a moving light source.

(a) Show that u is parallel to the world line of the object.

(b) Show that u · u = !c2.

(c) If ug is the group velocity of a relativistic wave packet, show thatk = (µ/c2)u, where k is the central wave four-vector of the wavepacket.

The four-vector u is called the four-velocity.

4. Find the Doppler shift for a moving source of light from figure 5.5,roughly following the procedure used in the text to find the shift for amoving observer. (Assume that the source moves to the left at speedU .) Is the result the same as for the moving observer, as demanded bythe principle of relativity?

5. Suppose you shine a laser with frequency $ and wavenumber k on amirror moving toward you at speed v, as seen in figure 5.6. Whatare the frequency $# and wavenumber k# of the reflected beam? Hint:Find the frequency of the incident beam in the reference frame of themirror. The frequency of the reflected beam will be the same as that ofthe incident beam in this frame. Then transform back to the referenceframe of the laser.


’ωk’,

ωk,

v

v

incident beam

reflected beam

laser

mirror

Figure 5.6: Laser beam reflecting o! of a moving mirror.

6. Suppose the moving twin in the twin paradox has a powerful telescopeso that she can watch her twin brother back on earth during the entiretrip. Describe how the earthbound twin appears to age to the travellingtwin compared to her own rate of aging. Use a spacetime diagram toillustrate your argument and consider separately the outbound andreturn legs. Remember that light travels at the speed of light! Hint:Does the concept of Doppler shift help here?

7. Find the velocity of an object with respect to the rest frame if it ismoving at a velocity of 0.1c with respect to another frame which itselfis moving at 0.1c relative to the rest frame using

(a) the Galilean formula and

(b) the formula of special relativity.

Determine the fractional error made in using the Galilean formula.

8. Each stage of a high performance 3 stage rocket can accelerate to aspeed of 0.9c from rest. If the rocket starts from rest, how fast doesthe final stage eventually go?

9. An interstellar spaceship is going from Earth to Sirius with speed U =0.8c relative to the rest frame. It passes a spaceship which is going fromSirius to Earth at a speed of 0.95c in the reference frame of the firstspaceship. What is the velocity (direction and speed) of the secondspaceship in the rest frame?

Chapter 6

Acceleration and General

Relativity

General relativity is Einstein’s extension of special relativity to include grav-ity. An important aspect of general relativity is that spacetime is no longernecessarily flat, but in fact may be curved under the influence of mass. Un-derstanding curved spacetime is an advanced topic which is not easily acces-sible at the level of this course. However, it turns out that some insight intogeneral relativistic phenomena may be obtained by investigating the e!ectsof acceleration in the flat (but non-Euclidean) space of special relativity.

The central assumption of general relativity is the equivalence principle,which states that gravity is a force which arises from being in an acceleratedreference frame. To understand this we must first investigate the concept ofacceleration. We then see how this leads to phenomena such as the gravi-tational red shift, event horizons, and black holes. We also introduce in apreliminary way the notions of force and mass.

6.1 Acceleration

Imagine that you are in a powerful luxury car stopped at a stoplight. Asyou sit there, gravity pushes you into the comfortable leather seat. The lightturns green and you “floor it”. The car accelerates and and an additionalforce pushes you into the seat back. You round a curve, and yet anotherforce pushes you toward the outside of the curve. (But the well designed seatand seat belt keep you from feeling discomfort!)

103

104 CHAPTER 6. ACCELERATION AND GENERAL RELATIVITY

a

Δtv

Δθ

Linear Motion

A

B

O

C D

E

t

x

v

aΔtv

vΔθ

v

r r

Circular Motion

Figure 6.1: Definition sketches for linear and circular motion.

Let us examine the idea of acceleration more closely. Considering firstacceleration in one dimension, the left panel of figure 6.1 shows the positionof an object as a function of time, x(t). The velocity is simply the time rateof change of the position:

v(t) =dx(t)

dt. (6.1)

The acceleration is the time rate of change of velocity:

a(t) =dv(t)

dt=

d2x(t)

dt2. (6.2)

In the left panel of figure 6.1, only the segment OA has zero velocity. Velocityis increasing in AB, and the acceleration is positive there. Velocity is constantin BC, which means that the acceleration is zero. Velocity is decreasing inCD, and the acceleration is negative. Finally, in DE, the velocity is negativeand the acceleration is zero.

In two or three dimensions, position r, velocity v, and acceleration a areall vectors, so that the velocity is

v(t) =dr(t)

dt(6.3)

while the acceleration is

a(t) =dv(t)

dt. (6.4)

Thus, over some short time interval #t, the changes in r and v can be written

#r = v#t #v = a#t. (6.5)

6.2. CIRCULAR MOTION 105

v

Inertial reference frame Frame moving with object

centripetalacceleration centrifugal

force

r

tension in string tension in string

Figure 6.2: Two di!erent views of circular motion of an object. The leftpanel shows the view from the inertial reference frame at rest with the centerof the circle. The tension in the string is the only force and it causes anacceleration toward the center of the circle. The right panel shows the viewfrom an accelerated frame in which the object is at rest. In this frame thetension in the string balances the centrifugal force, which is the inertial forcearising from being in an accelerated reference frame, leaving zero net force.

These are vector equations, so the subtractions implied by the “delta” op-erations must be done vectorially. An example where the vector nature ofthese quantities is important is motion in a circle at constant speed, whichis discussed in the next section.

6.2 Circular Motion

Imagine an object being swung around in a circle on a string, as shown infigure 6.2. The right panel in figure 6.1 shows the position of the objectat two times spaced by the time interval #t. The position vector of theobject relative to the center of the circle rotates through an angle #& duringthis interval, so the angular rate of revolution of the object about the centeris $ = #&/#t. The magnitude of the velocity of the object is v, so theobject moves a distance v#t during the time interval. To the extent that this


distance is small compared to the radius r of the circle, the angle #& = v#t/r.Solving for v and using $ = #&/#t, we see that

v = $r (circular motion). (6.6)

The direction of the velocity vector changes over this interval, even thoughthe magnitude v stays the same. The right sub-panel in figure 6.1 shows thatthis change in direction implies an acceleration a which is directed towardthe center of the circle. The magnitude of the vectoral change in velocityin the time interval #t is a#t. Since the angle between the initial and finalvelocities is the same as the angle #& between the initial and final radiusvectors, we see from the geometry of the triangle in the right sub-panel offigure 6.1 that a#t/v = #&. Solving for a results in

a = $v (circular motion). (6.7)

Combining equations (6.6) and (6.7) yields the equation for centripetalacceleration, i. e., the acceleration toward the center of a circle:

a = $2r = v2/r (centripetal acceleration). (6.8)

The second form is obtained by eliminating $ from the first form using equa-tion (6.6).

6.3 Acceleration in Special Relativity

As noted above, acceleration is just the time rate of change of velocity. Weuse the above results to determine how acceleration transforms from onereference frame to another. Figure 6.3 shows the world line of an acceleratedreference frame, with a time-varying velocity U(t) relative to the unprimedinertial frame. Defining #U = U(T ) ! U(0) as the change in the velocityof the accelerated frame (relative to the unprimed frame) between events Aand C, we can relate this to the change of velocity, #U #, of the acceleratedframe relative to an inertial frame moving with the initial velocity, U(0), ofthe accelerated frame. Applying the equation for the relativistic addition ofvelocities, we find

U(T ) = U(0) + #U =U(0) + #U #

1 + U(0)#U #/c2. (6.9)

6.3. ACCELERATION IN SPECIAL RELATIVITY 107

CB

cT

X

cT’

A

World line ofacceleratedreference frame

ct

x

Figure 6.3: World line of the origin of an accelerated reference frame.

We now note that the mean acceleration of the reference frame betweenevents A and C in the unprimed reference frame is just a = #U/T , whereasthe mean acceleration in the primed frame between the same two events isa# = #U #/T #. From equation (6.9) we find that

#U =#U #[1 ! U(0)2/c2]

1 + U(0)#U #/c2, (6.10)

and the acceleration of the primed reference frame as it appears in the un-primed frame is

a =#U

T=

#U #[1 ! U(0)2/c2]

T [1 + U(0)#U #/c2]. (6.11)

Since we are interested in the instantaneous rather than the average ac-celeration, we let T become small. This has three consequences. First, #Uand #U # become small, which means that the term U(0)#U #/c2 in the de-nominator of equation (6.11) can be ignored compared to 1. This meansthat

a ##U #[1 ! U(0)2/c2]

T, (6.12)

with the approximation becoming perfect as T * 0. Second, the “triangle”with the curved side in figure 6.3 becomes a true triangle, with the resultthat T # = T [1 ! U(0)2/c2]1/2. The acceleration of the primed frame with


respect to an inertial frame moving at speed U(0) can therefore be written

a# =#U #

T #=

#U #

T [1 ! U(0)2/c2]1/2. (6.13)

Third, we can replace U(0) with U , since the velocity of the accelerated framedoesn’t change very much over a short time interval.

Dividing equation (6.12) by equation (6.13) results in a relationship be-tween the two accelerations:

a = a#(1 ! U2/c2)3/2, (6.14)

which shows that the apparent acceleration of a rapidly moving object, a, isless than its acceleration relative to a reference frame in which the object isnearly stationary, a#, by the factor (1 ! U2/c2)3/2. We call this latter accel-eration the intrinsic acceleration. This is purely the result of the geometryof spacetime, but it has interesting consequences.

Identifying a with dU/dt, we can integrate the acceleration equation as-suming that the intrinsic acceleration a# is constant and that the velocityU = 0 at time t = 0. We get the following result (verify this by di!erentiat-ing with respect to time):

a#t =U

(1 ! U2/c2)1/2, (6.15)

which may be solved for U/c:

U

c=

a#t/c

[1 + (a#t/c)2]1/2. (6.16)

This is plotted in figure 6.4. Classically, the velocity would reach the speedof light when a#t/c = 1. However, as figure 6.4 shows, the rate at which thevelocity increases with time slows as the object moves faster, such that Uapproaches c asymptotically, but never reaches it.

6.4 Acceleration, Force, and Mass

We have a good intuitive feel for the concepts of force and mass becausethey are very much a part of our everyday experience. We think of force

6.4. ACCELERATION, FORCE, AND MASS 109

0.00

0.20

0.40

0.60

0.80

0.00 0.50 1.00 1.50 2.00 2.50U/c vs a’t/c

Figure 6.4: Velocity over the speed of light as a function of the product ofthe time and the (constant) acceleration divided by the speed of light.


as how hard we push on something. Mass is the resistance of an object toacceleration if it is otherwise free to move. Thus, pushing on a bicycle ona smooth, level road causes it to accelerate more readily than pushing on acar. We say that the car has greater mass. We can formalize this relationshipwith the statement

F = ma (6.17)

where F is the total force on an object, m is its mass, and a is the accelerationresulting from the force.

Three provisos apply to equation (6.17). First, it only makes sense inunmodified form when the velocity of the object is much less than the speedof light. For relativistic velocities it is best to write this equation in a slightlydi!erent form which we introduce later. Second, the force really must be thetotal force, including all frictional and other incidental forces which mightotherwise be neglected by an uncritical observer. Third, it only works in areference frame which itself is unaccelerated. We deal below with acceleratedreference frames.

6.5 Accelerated Reference Frames

Referring back to the forces being felt by the occupant of a car, it is clear thatthe forces associated with accelerations are directed opposite the accelera-tions and proportional to their magnitudes. For instance, when acceleratingaway from a stoplight, the acceleration is forward and the perceived forceis backward. When turning a corner, the acceleration is toward the cornerwhile the perceived force is away from the corner. Such forces are calledinertial forces.

The origin of these forces can be understood by determining how accel-eration changes when one observes it from a reference frame which is itselfaccelerated. Suppose that the primed reference frame is accelerating to theright with acceleration A relative to the unprimed frame. The position x# inthe primed frame can be related to the position x in the unprimed frame by

x = x# + X, (6.18)

where X is the position of the origin of the primed frame in the unprimedframe. Taking the second time derivative, we see that

a = a# + A, (6.19)

6.5. ACCELERATED REFERENCE FRAMES 111

where a = d2x/dt2 is the acceleration in the unprimed frame and a# =d2x#/dt2 is the acceleration according to an observer in the primed frame.

We now substitute this into equation (6.17) and move the term involvingA to the left side:

F ! mA = ma#. (6.20)

This shows that equation (6.17) is not valid in an accelerated reference frame,because the total force F and the acceleration a# in this frame don’t balanceas they do in the unaccelerated frame — the additional term !mA messesup this balance.

We can fix this problem by considering !mA to be a type of force, inwhich case we can include it as a part of the total force F . This is the inertialforce which we mentioned above. Thus, to summarize, we can make equation(6.17) work when objects are observed from accelerated reference frames ifwe include as part of the total force an inertial force which is equal to !mA,A being the acceleration of the reference frame of the observer and m themass of the object being observed.

The right panel of figure 6.2 shows the inertial force observed in thereference frame of an object moving in circular motion at constant speed. Inthe case of circular motion the inertial force is called the centrifugal force.It points away from the center of the circle and just balances the tension inthe string. This makes the total force on the object zero in its own referenceframe, which is necessary since the object cannot move (or accelerate) in thisframe.

General relativity says that gravity is nothing more than an inertial force.This was called the equivalence principle by Einstein. Since the gravitationalforce on the Earth points downward, it follows that we must be constantlyaccelerating upward as we stand on the surface of the Earth! The obviousproblem with this interpretation of gravity is that we don’t appear to bemoving away from the center of the Earth, which would seem to be a naturalconsequence of such an acceleration. However, relativity has a surprise instore for us here.

It follows from the above considerations that something can be learnedabout general relativity by examining the properties of accelerated referenceframes. Equation (6.16) shows that the velocity of an object undergoingconstant intrinsic acceleration a (note that we have dropped the “prime”


xA

B

ct

O

x’

ct’

AcceleratedWorld Line

Zone""Twilight

Line of Simultaneity

World LineTangent

Eventhorizon

Figure 6.5: Spacetime diagram showing the world line of the origin of areference frame undergoing constant acceleration.

from a) is

v =dx

dt=

at

[1 + (at/c)2]1/2, (6.21)

where t is the time and c is the speed of light. A function x(t) which satisfiesequation (6.21) is

x(t) = (c2/a)[1 + (at/c)2]1/2. (6.22)

(Verify this by di!erentiating it.) The interval OB in figure 6.5 is of lengthx(0) = c2/A.

The slanted line OA is a line of simultaneity associated with the unaccel-erated world line tangent to the accelerated world line at point A. This lineof simultaneity actually does go through the origin, as is shown in figure 6.5.To demonstrate this, multiply equations (6.21) and (6.22) together and solvefor v/c:

v/c = ct/x. (6.23)

From figure 6.5 we see that ct/x is the slope of the line OA, where (x, ct)are the coordinates of event A. Equation (6.23) shows that this line is indeedthe desired line of simultaneity, since its slope is the inverse of the slope ofthe world line, c/v. Since there is nothing special about the event A, we inferthat all lines of simultaneity associated with the accelerated world line passthrough the origin.

6.6. GRAVITATIONAL RED SHIFT 113

ct

x

X’

B

A

X’X

O

Light ray

World line of observer

Line of simultaneityof observer

C

L

Figure 6.6: Spacetime diagram for explaining the gravitational red shift.Why is the interval AC equal to the interval BC? L is the length of theinvariant interval OB.

We now inquire about the length of the invariant interval OA in figure6.5. Recalling that I2 = x2 ! c2t2 and using equation (6.22), we find that

I = (x2 ! c2t2)1/2 = (c4/a2)1/2 = c2/a, (6.24)

which is the same as the length of the interval OB. By extension, all eventson the accelerated world line are the same invariant interval from the origin.Recalling that the interval along a line of simultaneity is just the distancein the associated reference frame, we reach the astonishing conclusion thateven though the object associated with the curved world line in figure 6.5 isaccelerating away from the origin, it always remains the same distance (in itsown frame) from the origin. In other words, even though we are acceleratingaway from the center of the Earth, the distance to the center of the Earthremains constant!

6.6 Gravitational Red Shift

Light emitted at a lower level in a gravitational field has its frequency reducedas it travels to a higher level. This phenomenon is called the gravitational red


shift. Figure 6.6 shows why this happens. Since experiencing a gravitationalforce is equivalent to being in an accelerated reference frame, we can use thetools of special relativity to view the process of light emission and absorptionfrom the point of view of the unaccelerated or inertial frame. In this referenceframe the observer of the light is accelerating to the right, as indicated by thecurved world line in figure 6.6, which is equivalent to a gravitational force tothe left. The light is emitted at point A with frequency $ by a source whichis stationary at this instant. At this instant the observer is also stationary inthis frame. However, by the time the light gets to the observer, he or she hasa velocity to the right which means that the observer measures a Dopplershifted freqency $# for the light. Since the observer is moving away from thesource, $# < $, as indicated above.

The relativistic Doppler shift is given by

$#

$=

!

1 ! U/c

1 + U/c

"1/2

, (6.25)

so we need to compute U/c. The line of simultaneity for the observer at pointB goes through the origin, and is thus given by line segment OB in figure 6.6.The slope of this line is U/c, where U is the velocity of the observer at pointB. From the figure we see that this slope is also given by the ratio X #/X.Equating these, eliminating X in favor of L = (X2 ! X #2)1/2, which is theactual invariant distance of the observer from the origin, and substitutinginto equation (6.25) results in our gravitational red shift formula:

$#

$=

!

X ! X #

X + X #

"1/2

=

!

(L2 + X #2)1/2 ! X #

(L2 + X #2)1/2 + X #

"1/2

. (6.26)

If X # = 0, then there is no redshift, because the source is collocated with theobserver. On the other hand, if the source is located at the origin, so X # = X,the Doppler shifted frequency is zero. In addition, the light never gets to theobserver, since the world line is asymptotic to the light world line passingthrough the origin. If the source is at a higher level in the gravitational fieldthan the observer, so that X # < 0, then the frequency is shifted to a highervalue, i. e., it becomes a “blue shift”.

6.7. EVENT HORIZONS 115

x

t

A B

E

F

CD

Figure 6.7: Position of an object as a function of time.

6.7 Event Horizons

The 45" diagonal line passing through the origin in figure 6.5 is called theevent horizon for the accelerated observer in this figure. Notice that lightfrom the “twilight zone” above and to the left of the event horizon cannotreach the accelerated observer. However, the reverse is not true — a lightsignal emitted to the left by the observer can cross the event horizon into the“twilight zone”. The event horizon thus has a peculiar one-way character —it passes signals from right to left, but not from left to right.

6.8 Problems

1. An object moves as described in figure 6.7, which shows its position xas a function of time t.

(a) Is the velocity positive, negative, or zero at each of the points A,B, C, D, E, and F?

(b) Is the acceleration positive, negative, or zero at each of the pointsA, B, C, D, E, and F?

2. An object is moving counterclockwise at constant speed around thecircle shown in figure 6.8 due to the fact that it is attached by a stringto the center of the circle at point O.

(a) Sketch the object’s velocity vectors at points A, B, and C.


O

C

B

A

Figure 6.8: Object in circular motion.

(b) Sketch the object’s acceleration vectors at points A, B, and C.

(c) If the string breaks at point A, sketch the subsequent trajectoryfollowed by the object.

3. How fast are you going after accelerating from rest at a = 10 m s!2 for

(a) 10 y?

(b) 100000 y?

Express your answer as the speed of light minus your actual speed.Hint: You may have a numerical problem on the second part, whichyou should try to resolve using the approximation (1 + *)x # 1 + x*,which is valid for |*| " 1.

4. An object’s world line is defined by x(t) = (d2 + c2t2)1/2 where d is aconstant and c is the speed of light.

(a) Find the object’s velocity as a function of time.

(b) Using the above result, find the slope of the tangent to the worldline as a function of time.

(c) Find where the line of simultaneity corresponding to each tangentworld line crosses the x axis.

5. A car accelerates in the positive x direction at 3 m s!2.

(a) What is the net force on a 100 kg man in the car as viewed froman inertial reference frame?

6.8. PROBLEMS 117

(b) What is the inertial force experienced by this man in the referenceframe of the car?

(c) What is the net force experienced by the man in the car’s (accel-erated) reference frame?

6. A person is sitting in a comfortable chair in her home in Bogota,Columbia, which is essentially on the equator.

(a) What would the rotational period of the earth have to be to makethis person weightless?

(b) What is her acceleration according to the equivalence principle inthis situation?

7. At time t = 0 a Zork (a creature from the planet Zorkheim) acceler-ating to the right at a = 103 m s!2 in a spaceship accidently drops itsstopwatch o! of the spaceship.

(a) Describe qualitatively how the hands of the watch appear to moveto the Zork as it observes the watch through a powerful telescope.

(b) After a very long time what does the watch read?

Hint: Draw a spacetime diagram with the world lines of the spaceshipand the watch. Then send light rays from the watch to the spaceship.

8. Using a spacetime diagram, show why signals from events on the hiddenside of the event horizon from an accelerating spaceship cannot reachthe spaceship.

Chapter 7

Matter Waves

We begin our study of quantum mechanics by discussing the di!raction un-dergone by X-rays and electrons when they interact with a crystal. X-raysare a form of electromagnetic radiation with wavelengths comparable to thedistances between atoms. Scattering from atoms in a regular crystallinestructure results in an interference pattern which is in many ways similarto the pattern from a di!raction grating. We first develop Bragg’s law fordi!raction of X-rays from a crystal. Two practical techniques for doing X-raydi!raction are then described.

It turns out that electrons have wave-like properties and also undergoBragg di!raction by crystals. Bragg di!raction thus provides a crucial bridgebetween the worlds of waves and particles. With this bridge we introducethe classical ideas of momentum and energy by relating them to the wavevector and frequency of a wave. The properties of waves also give rise to theHeisenberg uncertainty principle.

Table 7.1 shows a table of the Nobel prizes associated with the ideaspresented in this chapter. This gives us a feel for the chronology of thesediscoveries and indicates how important they were to the development ofphysics in the early 20th century.

7.1 Bragg’s Law

Figure 7.1 schematically illustrates interference between waves scatteringfrom two adjacent rows of atoms in a crystal. The net e!ect of scatteringfrom a single row is equivalent to partial reflection from a mirror imagined to

119

120 CHAPTER 7. MATTER WAVES

Year Recipient Contribution

1901 W. K. Rontgen Discovery of X-rays1906 J. J. Thomson Discovery of electron1914 M. von Laue X-ray di!raction in crystals1915 W. and L. Bragg X-ray analysis of crystal structure1918 M. Planck Energy quantization1921 A. Einstein Photoelectric e!ect1922 N. Bohr Structure of atoms1929 L.-V. de Broglie Wave nature of electrons1932 W. Heisenberg Quantum mechanics1933 Schrodinger and Dirac Atomic theory1937 Davisson and Thomson Electron di!raction in crystals

Table 7.1: Selected Nobel prize winners, year of award, and contribution.

d

θ

θ

h

Figure 7.1: Schematic diagram for determining Bragg’s law.

7.2. X-RAY DIFFRACTION TECHNIQUES 121

be aligned with the row. Thus, the angle of “reflection” equals the angle ofincidence for each row. Interference then occurs between the beams reflectingo! di!erent rows of atoms in the crystal.

For the two adjacent rows shown in figure 7.1, the path di!erence be-tween beams is 2h = 2d sin &. For constructive interference this must be aninteger number of wavelengths, m", where the integer m is called the orderof interference. The result is Bragg’s law of di!raction:

m" = 2d sin &, m = 1, 2, 3 . . . (Bragg’s law). (7.1)

If only two rows are involved, the transition from constructive to destruc-tive interference as & changes is gradual. However, if interference from manyrows occurs, then the constructive interference peaks become very sharp withmostly destructive interference in between. This sharpening of the peaks asthe number of rows increases is very similar to the sharpening of the di!rac-tion peaks from a di!raction grating as the number of slits increases.

7.2 X-Ray Di!raction Techniques

Two types of targets are used in Bragg di!raction experiments, single crystalsand powder targets.

7.2.1 Single Crystal

In a single crystal setup, an X-ray detector is mounted as shown in figure7.2. A mechanical device keeps the detector oriented so that the angle ofincidence equals the angle of reflection for the desired crystal plane. Peaksin the X-ray detection rate are sought as the angle & is varied.

The advantage of this type of apparatus is that di!raction peaks fromonly the selected crystal plane are observed.

7.2.2 Powder Target

The powder in a powder target is really a conglomeration of many tiny crys-tals randomly oriented. Thus, for each possible Bragg di!raction angle thereare crystals oriented correctly for Bragg di!raction to take place. The de-tector is usually a photographic plate or an equivalent electronic device as


θ2θ

single crystalsourceX-ray

X-ray detectorcrystal plane

Figure 7.2: Setup for single crystal Bragg di!raction.

end view

sourceX-ray

powder target

side view

Figure 7.3: Setup for powder target Bragg di!raction.

7.3. MEANING OF QUANTUM WAVE FUNCTION 123

illustrated in figure 7.3. For each Bragg di!raction angle one sees a ring onthe plate concentric with the axis of the incident X-ray beam.

The advantage of this type of system is that no a priori knowledge isneeded of the crystal plane orientations. Furthermore, a single large crystalis not required. However, all possible Bragg scattering angles are seen atonce, which can lead to confusion in the interpretation of the results.

7.3 Meaning of Quantum Wave Function

Bragg di!raction illustrates the most di"cult thing to understand aboutquantum mechanics, namely that particles can have wave-like properties andwaves can have particle-like properties.

The variation of X-ray intensity with angle seen in a Bragg di!raction ap-paratus is very di"cult to explain in any terms other than wave interference.Yet, X-rays are typically detected by a device such as a Geiger counter whichproduces a pulse of electricity for each X-ray particle, or photon, which hitsit. Thus, X-rays sometimes act like particles and sometimes like waves.

Light isn’t alone in having both particle and wave properties. Davissonand Germer and later G. P. Thomson (son of J. J. Thomson, the discovererof the electron) showed that electrons also can act like waves. They did thisby demonstrating that electrons undergo Bragg di!raction in crystals, muchas X-rays do.

Most physicists (including Albert Einstein) have found quantum mechan-ics to be extremely bizarre, so if you feel the same way, you are in good com-pany! However, there is a useful interpretation of quantum mechanics whichat least allows us to get on with using it to solve problems, even though itmay not satisfy our intuitive reservations about the theory.

The displacement of a quantum mechanical wave associated with a par-ticle is usually called the wave function, +. Like light, it is not at all clearwhat + is a displacement of, but its use is straightforward. The absolutesquare of the wave function, |+(x, t)|2, is proportional to the probability offinding the associated particle at position x and time t. The absolute squareis taken because under many circumstances the wave function is actuallycomplex, i. e., it has both real and imaginary parts. The reasons for this willbe discussed later.

Due to the interpretation of the wave function, quantum mechanics isa probabilistic theory. It does not tell us with certainty what happens to


a particular particle. Instead, it tells us, for instance, the probabilities forthe particle to be detected in certain locations. If many experiments aredone, with one particle per experiment, the numbers of experiments withparticles being detected in the various possible locations are in proportion tothe quantum mechanical probabilities.

7.4 Sense and Nonsense in Quantum Mechan-

ics

The essential mystery of quantum mechanics becomes clearer when discussedin the conceptually simpler context of two slit interference. If light andelectrons can have both particle and wave properties, then one might askthrough which of the two slits the particle passed. However, in physics aquestion simply doesn’t make sense if it cannot be answered by experiment.Let us see if the above question makes sense.

One can indeed perform an experiment to determine which slit the X-rayphoton or electron passes through in the two slit experiment. However, bythe very act of making this measurement, the form of the associated waveis altered. In particular, since the absolute square of the wave displacementrepresents the probability of finding the particle, once the particle has def-initely been found passing through one or the other of the slits, the wavefunction collapses into a very small wave packet located at the observed po-sition of the particle. Thus, the wave displacement becomes zero at the slitit didn’t pass through. However, the interference pattern results from thesuperposition of waves emanating from two slits. If no wave comes from oneof the slits (because the wave displacement is zero there), then there can beno interference pattern!

We can now make the inverse argument. If there is an interference pat-tern, then we know that the wave displacement is non-zero at both slits.From the probability interpretation of the wave displacement, we concludethat we cannot say, even in principle, through which slit the particle passed.It is not just that we don’t know the answer to this question; there is simplyno experiment which can give us an answer without destroying the interfer-ence pattern. In other words, the question “Which slit did the particle passthrough?” is a nonsensical question in the case where an interference patternis actually produced.

7.4. SENSE AND NONSENSE IN QUANTUM MECHANICS 125

The American physicist Richard Feynman noticed that the above behav-ior can be interpreted as violating the normal laws of probability. These lawssay that the probability of an event is the sum of the probabilities of alternateindependent ways for that event to occur. For instance, the probability for aparticle to reach point A on the detection screen of a two slit setup is just theprobability P1 for the particle to reach point A after going through slit 1, plusthe probability P2 for the particle to reach point A after going through slit 2.Thus, if P1 = P2 = 0.1, then the probability for the particle to reach pointA irrespective of which slit it went through should be Ptotal = P1 + P2 = 0.2.However, if point A happens to be a point of destructive interference, thenwe know that Ptotal = 0.

Feynman proposed that the above rule stating that alternate independentprobabilities add, is simply incorrect. In its place Feynman asserted thatprobability amplitudes add instead, where the probability amplitude in thiscase is just the wave function associated with the particle. The probability isobtained by adding the alternate probability amplitudes together and takingthe absolute square of the sum.

In the two slit case, let us call +1 the amplitude for the particle to reachpoint A via slit 1, while +2 is the amplitude for it to reach this point via slit 2.The total amplitude is therefore +total = +1 + +2 and the overall probabilityfor the particle to reach point A is Ptotal = |+1 + +2|2. The wave functions+1 and +2 at point A are proportional respectively to sin(kL1) and sin(kL2),where k is the wavenumber and L1 and L2 are the respective distances fromslit 1 and slit 2 to point A. It is easy to see that destructive interferenceoccurs if the di!erence between L1 and L2 is half a wavelength, thus yieldingPtotal = 0.

The above treatment of two slits in terms of probability amplitudes yieldsresults which are identical to the conventional treatment of the two slit prob-lem for waves. However, it is conceptually clearer in the case of quantummechanics because we no longer need to ask whether “things” are really par-ticles or waves. In particular, the particle/wave is replaced by just a particleplus the assertion that probability amplitudes rather than probabilities add.The laws of quantum mechanics then boil down to rules for how probabil-ity amplitudes behave. In the case of particles, the amplitudes behave likewaves, which explains why the study of wave kinematics and dynamics is soimportant to our understanding of quantum mechanics.


7.5 Mass, Momentum, and Energy

In this section we relate the classical ideas of mass, momentum and energyto what we have done so far. Historically, these connections were first madeby Max Planck and Louis de Broglie with help from Albert Einstein. Braggdi!raction by electrons is invoked as an experimental test of the Planck andde Broglie relations.

Technically, we don’t need the ideas of mass, momentum, and energy todo physics – the notions of wavenumber, frequency, and group velocity aresu"cient to describe and explain all observed phenomena. However, mass,momentum, and energy are so firmly embedded in physics that one couldn’ttalk to other physicists without an understanding of these quantities!

7.5.1 Planck, Einstein, and de Broglie

Max Planck was the first to make a connection between the energy of aparticle and the frequency of the associated wave. Planck developed hisformula in his successful attempt to explain the emission and absorptionof electromagnetic radiation from warm material surfaces. It su"ces hereto convey Planck’s formula, which was originally intended to apply only toparcels of electromagnetic radiation, or photons:

E = hf = h$ (Planck relation), (7.2)

where E is the energy of the photon, f is the rotational frequency of theassociated light wave, and $ is its angular frequency. The constant h =6.63 % 10!34 kg m2 s!1 is called Planck’s constant. The related constanth = h/2! = 1.06 % 10!34 kg m2 s!1 is also referred to as Planck’s constant,but to avoid confusion with the original constant, we will generally refer toit as “h bar”.

Notice that a new physical dimension has appeared, namely mass, withthe unit kilogram, abbreviated “kg”. The physical meaning of mass is muchlike our intuitive understanding of the concept, i. e., as a measure of theresistance of an object to its velocity being changed. The precise scientificmeaning will emerge shortly.

Einstein showed that Planck’s idea could be used to explain the emissionof electrons which occurs when light impinges on the surface of a metal.This emission, which is called the photoelectric e!ect, can only occur whenelectrons are supplied with a certain minimum energy EB required to break

7.5. MASS, MOMENTUM, AND ENERGY 127

them loose from the metal. Experiment shows that this emission occurs onlywhen the frequency of the light exceeds a certain minimum value. This valueturns out to equal $min = EB/h, which suggests that electrons gain energyby absorbing a single photon. If the photon energy, h$, exceeds EB, thenelectrons are emitted, otherwise they are not. It is much more di"cult toexplain the photoelectric e!ect from the classical theory of light.

Louis de Broglie proposed that Planck’s energy-frequency relationshipbe extended to all kinds of particles. In addition he hypothesized that themomentum " of the particle and the wave vector k of the correspondingwave were similarly related:

" = hk (de Broglie relation). (7.3)

Note that this can also be written in scalar form in terms of the wavelengthas $ = h/".

De Broglie’s hypothesis was inspired by the fact that wave frequency andwavenumber are components of the same four-vector according to the theoryof relativity, and are therefore closely related to each other. Thus, if the en-ergy of a particle is related to the frequency of the corresponding wave, thenthere ought to be some similar quantity which is correspondingly related tothe wavenumber. It turns out that the momentum is the appropriate quan-tity. The physical meaning of momentum will become clear as we proceed.

We will also find that the rest frequency, µ, of a particle is related to itsmass, m:

Erest ' mc2 = hµ. (7.4)

The quantity Erest is called the rest energy of the particle.From our perspective, energy, momentum, and rest energy are just scaled

versions of frequency, wave vector, and rest frequency, with a scaling factorh. We can therefore define a four-momentum as a scaled version of the wavefour-vector:

$ = hk. (7.5)

The spacelike component of $ is just ", while the timelike part is E/c.Planck, Einstein, and de Broglie had extensive backgrounds in classical

mechanics, in which the concepts of energy, momentum, and mass have pre-cise meaning. In this text we do not presuppose such a background. Perhapsthe best strategy at this point is to think of these quantities as scaled ver-sions of frequency, wavenumber, and rest frequency, where the scale factor is


h. The significance of these quantities to classical mechanics will emerge bitby bit.

7.5.2 Wave and Particle Quantities

Let us now recapitulate what we know about relativistic waves, and how thisknowledge translates into knowledge about the mass, energy, and momentumof particles. In the following equations, the left form is expressed in waveterms, i. e., in terms of frequency, wavenumber, and rest frequency. Theright form is the identical equation expressed in terms of energy, momentum,and mass. Since the latter variables are just scaled forms of the former, thetwo forms of each equation are equivalent.

We begin with the dispersion relation for relativistic waves:

$2 = k2c2 + µ2 E2 = $2c2 + m2c4. (7.6)

Calculation of the group velocity, ug = d$/dk, from the dispersion relationyields

ug =c2k

$ug =

c2$

E. (7.7)

These two sets of equations represent what we know about relativistic waves,and what this knowledge tells us about the relationships between the mass,energy, and momentum of relativistic particles. When in doubt, refer backto these equations, as they work in all cases, including for particles with zeromass!

It is useful to turn equations (7.6) and (7.7) around so as to express thefrequency as a function of rest frequency and group velocity,

$ =µ

(1 ! u2g/c

2)1/2E =

mc2

(1 ! u2g/c

2)1/2, (7.8)

and the wavenumber as a similar function of these quantities:

k =µug/c2

(1 ! u2g/c

2)1/2$ =

mug

(1 ! u2g/c

2)1/2. (7.9)

Note that equations (7.8) and (7.9) work only for particles with non-zeromass! For zero mass particles you need to use equations (7.6) and (7.7) withm = 0 and µ = 0.

7.5. MASS, MOMENTUM, AND ENERGY 129

The quantity $ ! µ indicates how much the frequency exceeds the restfrequency. Notice that if $ = µ, then from equation (7.6) k = 0. Thus,positive values of $k ' $!µ indicate |k| > 0, which means that the particleis moving according to equation (7.7). Let us call $k the kinetic frequency :

$k =

%

1

(1 ! u2g/c

2)1/2! 1

&

µ K =

%

1

(1 ! u2g/c

2)1/2! 1

&

mc2. (7.10)

We call K the kinetic energy for similar reasons. Again, equation (7.10) onlyworks for particles with non-zero mass. For zero mass particles the kineticenergy equals the total energy.

7.5.3 Non-Relativistic Limits

When the mass is non-zero and the group velocity is much less than the speedof light, it is useful to compute approximate forms of the above equationsvalid in this limit. Using the approximation (1 + *)x # 1 + x*, we find thatthe dispersion relation becomes

$ = µ +k2c2

2µE = mc2 +

$2

2m, (7.11)

and the group velocity equation takes the approximate form

ug =c2k

µug =

$

m. (7.12)

The non-relativistic limits for equations (7.8) and (7.9) become

$ = µ +µu2

g

2c2E = mc2 +

mu2g

2(7.13)

andk = µug/c

2 $ = mug, (7.14)

while the approximate kinetic energy equation is

$k =µu2

g

2c2K =

mu2g

2. (7.15)

Just a reminder — the equations in this section are not valid for masslessparticles!


7.5.4 An Experimental Test

How can we test the above predictions against experiment? The key pointis to be able to relate the wave aspects to the particle aspects of a quantummechanical wave-particle. Equation (7.9), or equation (7.14) in the non-relativistic case, relates a particle’s wavenumber k to it’s velocity ug. Bothof these quantities can be measured in a Bragg’s law experiment with elec-trons. In this experiment electrons are fired at a crystal with known atomicdimensions at a known speed, which we identify with the group velocityug. The Bragg angle which yields constructive interference can be used tocalculate the wavelength of the corresponding electron wave, and hence thewavenumber and momentum. If the momentum is plotted against group ve-locity in the non-relativistic case, a straight line should be found, the slopeof which is the particle’s mass. In the fully relativistic case one needs toplot momentum versus ug/(1 ! u2

g/c2)1/2. Again, a straight line indicates

agreement with the theory and the slope of the line is the particle’s mass.

7.6 Heisenberg Uncertainty Principle

Classically, we consider the location of a particle to be a knowable piece ofinformation. In quantum mechanics the position of a particle is well knownif the wave packet representing it is small in size. However, quantum me-chanics imposes a price on accurately knowing the position of a particle interms of the future predictability of its position. This is because a smallwave packet, which corresponds to accurate knowledge of the correspondingparticle’s position, implies the superposition of plane waves corresponding toa broad distribution of wavenumbers. This translates into a large uncertaintyin the wavenumber, and hence the momentum of the particle. In contrast, abroad wave packet corresponds to a narrower distribution of wavenumbers,and correspondingly less uncertainty in the momentum.

Referring back to chapters 1 and 2, recall that both the longitudinal(along the direction of motion) and transverse (normal to the direction ofmotion) dimensions of a wave packet, #xL and #xT , can be related to thespread of longitudinal and transverse wavenumbers, #kL and #kT :

#kL#xL # 1, (7.16)

#kT#xT # 1. (7.17)

7.6. HEISENBERG UNCERTAINTY PRINCIPLE 131

We have omitted numerical constants which are order unity in these approx-imate relations so as to show their essential similarity.

The above equations can be interpreted in the following way. Since theabsolute square of the wave function represents the probability of finding aparticle, #xL and #xT represent the uncertainty in the particle’s position.Similarly, #kL and #kT represent the uncertainty in the particle’s longitudi-nal and transverse wave vector components. This latter uncertainty leads touncertainty in the particle’s future evolution — larger or smaller longitudinalk results respectively in larger or smaller particle speed, while uncertainty inthe transverse wavenumber results in uncertainty in the particle’s directionof motion. Thus uncertainties in any component of k result in uncertaintiesin the corresponding component of the particle’s velocity, and hence in itsfuture position.

The equations (7.16) and (7.17) show that uncertainty in the present andfuture positions of a particle are complimentary. If the present position isaccurately known due to the small size of the associated wave packet, thenthe future position is not very predictable, because the wave packet dispersesrapidly. On the other hand, a broad-scale initial wave packet means that thepresent position is poorly known, but the uncertainty in position, poor asit is, doesn’t rapidly increase with time, since the wave packet has a smalluncertainty in wave vector and thus disperses slowly. This is a statement ofthe Heisenberg uncertainty principle.

The uncertainty principle also applies between frequency and time:

#$#t # 1. (7.18)

This shows up in the beat frequency equation 1/Tbeat = #f = #$/2!. Thebeat period Tbeat may be thought of as the size of a “wave packet in time”.The beat frequency equation may be rewritten as #$Tbeat = 2!, which is thesame as equation (7.18) if the factor of 2! is ignored and Tbeat is identifiedwith #t.

The above forms of the uncertainty principle are not relativistically in-variant. A useful invariant form may be obtained by transforming to thecoordinate system in which a particle is stationary. In this reference framethe time t becomes the proper time ) associated with the particle. Fur-thermore, the frequency $ becomes the rest frequency µ. The uncertaintyprinciple thus becomes

#µ#) # 1 (7.19)


in this reference frame. However, since #µ and #) are relativistic invariants,this expression of the uncertainty principle is valid in any reference frame.

It is more common to express the uncertainty principle in terms of themass, momentum, and energy by multiplying equations (7.16) - (7.19) by h.Lumping the momentum equations, we find

#$#x # h, (7.20)

#E#t # h, (7.21)

and#(mc2)#) # h. (7.22)

Classical mechanics is the realm of quantum mechanics in which the di-mensions of the system of interest are much larger than the wavelengths ofthe waves corresponding to the particles constituting the system. In thiscase the uncertainties induced by the uncertainty principle are unimportant.This limit is analogous to the geometrical optics limit for light. Thus, wecan say that classical mechanics is the geometrical optics limit of quantummechanics.

7.7 Problems

1. An electron with wavelength " = 1.2%10!10 m undergoes Bragg di!rac-tion from a single crystal with atomic plane spacing of d = 2%10!10 m.

(a) Calculate the Bragg angles (all of them!) for which constructiveinterference occurs.

(b) Calculate the speed of the electron.

2. Suppose that electrons impinge on two slits in a plate, resulting in atwo slit di!raction pattern on a screen on the other side of the plate.The amplitude for an electron to pass through either one of the slitsand reach point A on the screen is +. Thus, the probability for anelectron to reach this point is |+|2 if there is only a single slit open.

(a) If there are two slits open and A is a point of constructive inter-ference, what is the probability of an electron reaching A?

(b) If there are two slits open and A is a point of destructive interfer-ence, what is the probability of an electron reaching A?

7.7. PROBLEMS 133

(c) If there are two slits open, what is the probability for an electron toreach point A according to the conventional rule that probabilitiesadd? (This is the result one would expect if, for instance, theparticles were machine gun bullets and the slits were, say, 5 cmapart.)

(d) If the slit separation is very much greater than the electron wave-length, how does this a!ect the spacing of regions of constructiveand destructive interference? Explain how the results of parts (a)and (b) become approximately consistent with those of part (c)in this case.

3. Compute the (angular) rest frequency of an electron and a neutron.(Look up their masses.)

4. How does the dispersion relation for relativistic waves simplify if therest frequency (and hence the particle mass) is zero? What is the groupvelocity in this case?

5. X-rays are photons with frequencies about 2000 times the frequen-cies of ordinary light photons. From this information and what youknow about light, infer the approximate velocity of electrons whichhave Bragg di!raction properties similar to X-rays. Are the electronsrelativistic or non-relativistic?

6. Electrons with velocity v = 0.6c are di!racted with a 0.2 radian half-angle of di!raction when they hit an object. What is the approximatesize of the object? Hint: Di!raction of a wave by an object of a certainsize is quite similar to di!raction by a hole in a screen of the same size.

7. Work out an approximate formula for the kinetic energy of a particle asa function of mass m and velocity ug which is valid when u2

g " c2. Hint:Use the approximation (1 + *)x # 1 + x*, which is valid for |*| " 1. Asug/c becomes larger, how does this approximate formula deviate fromthe exact formula?

8. Work out an approximate formula for the momentum of a particle asa function of m and ug in the case where u2

g " c2. You may wish touse the approximation mentioned in the previous problem.


9. If a photon is localized to within a distance #x, what is the uncertaintyin the photon energy?

10. If an electron is localized to within a distance #x, what is the un-certainty in the electron kinetic energy? Hint: As long as #$ " $,#$2 # 2$#$. To see why, compute d$2/d$.

11. A grocer dumps some pinto beans onto a scale, estimates their massas 2 kg, and then dumps them o! after 5 s. What is the quantummechanical uncertainty in this measurement? For this problem only,assume that the speed of light is 10 m s!1 (speed of a fast buggy) andthat h = 1 kg m2 s!1.

12. Mary’s physics text (mass 0.3 kg) has to be kept on a leash (length0.5 m) to prevent it from wandering away from her in Quantum World(h = 1 kg m2 s!1).

(a) If the leash suddenly breaks, what is the maximum speed at whichthe book is likely to move away from its initial location?

(b) In order to reduce this speed, should Mary make the new leashshorter or longer than the old one? Explain.

13. A proton (mass M = 1.7 % 10!27 kg) is confined to an atomic nucleusof diameter D = 2 % 10!15 m.

(a) What is the uncertainty in the proton’s momentum?

(b) Roughly what kinetic energy might you expect the proton to have?

Planck’s constant is h = 1.06 % 10!34 kg m2 s!1. You may use thenon-relativistic equation for the energy.

Chapter 8

Geometrical Optics and

Newton’s Laws

The question that motivates us to study physics is “What makes things go?”The answers we conceive to this question constitute the subject of dynamics.This is in contrast to the question we have primarily addressed so far, namely“How do things go?” As noted earlier, the latter question is about kinematics.Extensive preparation in the kinematics of waves and particles in relativisticspacetime is needed to intelligently address dynamics. This preparation isnow complete.

In this chapter we outline three di!erent dynamical principles based re-spectively on pre-Newtonian, Newtonian, and quantum mechanical thinking.We first discuss the Newtonian mechanics of conservative forces in one di-mension. Certain ancillary concepts in mechanics such as work and powerare introduced at this stage. We then show that Newtonian and quantummechanics are consistent with each other in the realm in which they overlap,i. e., in the geometrical optics limit of quantum mechanics. For simplicity,this relationship is first developed in one dimension in the non-relativisticlimit. Higher dimensions require the introduction of partial derivatives, andthe relativistic case will be considered later.

8.1 Fundamental Principles of Dynamics

Roughly speaking, there have been three eras of physics, characterized bythree di!erent answers to the question of what makes things go.

135

136 CHAPTER 8. GEOMETRICAL OPTICS AND NEWTON’S LAWS

8.1.1 Pre-Newtonian Dynamics

Aristotle expounded a view of dynamics which agrees closely with our every-day experience of the world. Objects only move when a force is exerted uponthem. As soon as the force goes away, the object stops moving. The act ofpushing a box across the floor illustrates this principle — the box certainlydoesn’t move by itself!

8.1.2 Newtonian Dynamics

In contrast to earthly behavior, the motions of celestial objects seem e!ort-less. No obvious forces act to keep the planets in motion around the sun. Infact, it appears that celestial objects simply coast along at constant velocityunless something acts on them. The Newtonian view of dynamics — objectschange their velocity rather than their position when a force is exerted onthem — is expressed by Newton’s second law :

F = ma (Newton’s second law), (8.1)

where F is the force exerted on a body, m is its mass, and a is its acceleration.Newton’s first law, which states that an object remains at rest or in uniformmotion unless a force acts on it, is actually a special case of Newton’s secondlaw which applies when F = 0.

It is no wonder that the first successes of Newtonian mechanics were inthe celestial realm, namely in the predictions of planetary orbits. It tookNewton’s genius to realize that the same principles which guided the planetsalso applied to the earthly realm as well. In the Newtonian view, the tendencyof objects to stop when we stop pushing on them is simply a consequence offrictional forces opposing the motion. Friction, which is so important on theearth, is negligible for planetary motions, which is why Newtonian dynamicsis more obviously valid for celestial bodies.

Note that the principle of relativity is closely related to Newtonian physicsand is incompatible with pre-Newtonian views. After all, two referenceframes moving relative to each other cannot be equivalent in the pre-Newtonianview, because objects with nothing pushing on them can only come to restin one of the two reference frames!

Einstein’s relativity is often viewed as a repudiation of Newton, but thisis far from the truth — Newtonian physics makes the theory of relativitypossible through its invention of the principle of relativity. Compared with

8.2. POTENTIAL ENERGY 137

the di!erences between pre-Newtonian and Newtonian dynamics, the changesneeded to go from Newtonian to Einsteinian physics constitute minor tinker-ing.

8.1.3 Quantum Dynamics

In quantum mechanics, particles are represented by matter waves, with theabsolute square of the wave displacement yielding the probability of find-ing the particle. The behavior of particles thus follows from the reflection,refraction, di!raction, and interference of the associated waves. The connec-tion with Newtonian dynamics comes from tracing the trajectories of matterwave packets. Changes in the speed and direction of motion of these packetscorrespond to the accelerations of classical mechanics. When wavelengthsare small compared to the natural length scale of the problem at hand, thewave packets can be made small, thus pinpointing the position of the as-sociated particle, without generating excessive uncertainty in the particle’smomentum. This is the geometrical optics limit of quantum mechanics.

8.2 Potential Energy

We now address Newtonian mechanics in the case where the force on a par-ticle is conservative. A conservative force is one that can be derived from aso-called potential energy U . We assume that the potential energy of the par-ticle depends only on its position. The force is obtained from the potentialenergy by the equation

F = !dU

dx. (8.2)

Using this equation we write Newton’s second law as

!dU

dx= ma. (8.3)

We then notice that the acceleration can be written in terms of the x deriva-tive of v2/2:

a =dv

dt=

dv

dx

dx

dt=

dv

dxv =

1

2

dv2

dx. (8.4)

The last step in the above derivation can be verified by applying the prod-uct rule: dv2/dt = d(vv)/dt = v(dv/dt) + (dv/dt)v = 2v(dv/dt). Putting


x

ener

gy

EK

turning points

U(x)

Figure 8.1: Example of spatially variable potential energy U(x) with fixedtotal energy E. The kinetic energy K = E ! U is zero where the E and Ulines cross. These points are called turning points.

equations (8.3) and (8.4) together, we find that d(mv2/2 +U)/dt = 0, whichimplies that mv2/2+U is constant. We call this constant the total energy Eand the quantity K = mv2/2 the kinetic energy. We thus have the principleof conservation of energy for conservative forces:

E = K + U = constant. (8.5)

Recall that in quantum mechanics the momentum is related to the groupvelocity ug by

$ = mug (momentum) (8.6)

in the nonrelativistic case. Equating the group velocity to v and eliminatingit in the kinetic energy results in an alternate expression for this quantity:

K '1

2mu2

g =$2

2m(kinetic energy). (8.7)

Since the total energy E is constant or conserved, increases in the poten-tial energy coincide with decreases in the kinetic energy and vice versa, as isillustrated in figure 8.1. In Newtonian dynamics the kinetic energy cannotbe negative, since it is the product of half the mass and the square of thevelocity, both of which are positive. Thus, a particle with total energy E andpotential energy U is forbidden to venture into regions in which the kineticenergy K = E ! U is less than zero.

The points at which the kinetic energy is zero are called turning points.This is because a particle decreases in speed as it approaches a turning point,

8.3. WORK AND POWER 139

stops there for an instant, and reverses direction. Note also that a particlewith a given total energy always has the same speed at some point x regard-less of whether it approaches this point from the left or the right:

speed = |ug| = | ± [2(E ! U)/m]1/2|. (8.8)

8.2.1 Gravity as a Conservative Force

An example of a conservative force is gravity. An object of mass m near thesurface of the earth has the gravitational potential energy

U = mgz (gravity near earth’s surface) (8.9)

where z is the height of the object and g = 9.8 m s!2 is the local value ofthe gravitational field near the earth’s surface. Notice that the gravitationalpotential energy increases upward. The speed of the object in this case is|ug| = [2(E !mgz)/m]1/2. If |ug| is known to equal the constant value u0 atelevation z = 0, then equations (8.8) and (8.9) tell us that u0 = (2E/m)1/2

and |ug| = (u20 ! 2gz)1/2.

There are certain types of questions which energy conservation cannotdirectly answer. For instance if an object is released at elevation h withzero velocity at t = 0, at what time will it reach z = 0 under the influenceof gravity? In such cases it is often easiest to return to Newton’s secondlaw. Since the force on the object is F = !dU/dz = !mg in this case,we find that the acceleration is a = F/m = !mg/m = !g. However,a = du/dt = d2z/dt2, so

u = !gt + C1 z = !gt2/2 + C1t + C2 (constant gravity), (8.10)

where C1 and C2 are constants to be determined by the initial conditions.These results can be verified by di!erentiating to see if the original acceler-ation is recovered. Since u = 0 and z = h at t = 0, we have C1 = 0 andC2 = h. With these results it is easy to show that the object reaches z = 0when t = (2h/g)1/2.

8.3 Work and Power

When a force is exerted on an object, energy is transferred to the object. Theamount of energy transferred is called the work done on the object. However,


energy is only transferred if the object moves. The work W done is

W = F#x (8.11)

where the distance moved by the object is #x and the force exerted on it isF . Notice that work can either be positive or negative. The work is positiveif the object being acted upon moves in the same direction as the force, withnegative work occurring if the object moves opposite to the force.

Equation (8.11) assumes that the force remains constant over the full dis-placement #x. If it is not, then it is necessary to break up the displacementinto a number of smaller displacements, over each of which the force canbe assumed to be constant. The total work is then the sum of the worksassociated with each small displacement.

If more than one force acts on an object, the works due to the di!erentforces each add or subtract energy, depending on whether they are positiveor negative. The total work is the sum of these individual works.

There are two special cases in which the work done on an object is relatedto other quantities. If F is the total force acting on the object, then W =F#x = ma#x by Newton’s second law. However, a = dv/dt where v is thevelocity of the object, and #x = (#x/#t)#t # v#t, where #t is the timerequired by the object to move through distance #x. The approximationbecomes exact when #x and #t become very small. Putting all of thistogether results in

Wtotal = mdv

dtv#t =

d

dt

!

mv2

2

"

#t = #K (total work), (8.12)

where K is the kinetic energy of the object. Thus, when F is the only force,W = Wtotal is the total work on the object, and this equals the change inkinetic energy of the object. This is called the work-energy theorem, and itdemonstrates that work really is a transfer of energy to an object.

The other special case occurs when the force is conservative, but is notnecessarily the total force acting on the object. In this case

Wcons = !dU

dx#x = !#U (conservative force), (8.13)

where #U is the change in the potential energy of the object associated withthe force of interest.

8.4. MECHANICS AND GEOMETRICAL OPTICS 141

The power associated with a force is simply the amount of work done bythe force divided by the time interval #t over which it is done. It is thereforethe energy per unit time transferred to the object by the force of interest.From equation (8.11) we see that the power is

P =F#x

#t= Fv (power), (8.14)

where v is the velocity at which the object is moving. The total power is justthe sum of the powers associated with each force. It equals the time rate ofchange of kinetic energy of the object:

Ptotal =Wtotal

#t=

dK

dt(total power). (8.15)

8.4 Mechanics and Geometrical Optics

Louis de Broglie1 made an analogy between matter waves and light waves,pointing out that wave packets of light change their velocity as the resultof spatial variations in the index of refraction of the medium in which theyare travelling. This behavior comes about because the dispersion relationfor light traveling through a medium with index of refraction n is $ = kc/n,so that the group velocity, ug = d$/dk = c/n. Thus, when n increases, ug

decreases, and vice versa.2

In this section we pursue de Broglie’s analogy to see if we can comeup with a theory of matter waves which gives the same results as classicalmechanics in the geometrical optics limit of these waves. We pretend that wedon’t know anything about classical mechanics and make a simple guess as tohow to modify the free particle dispersion relation for matter waves, limitingconsideration initially to the one-dimensional, non-relativistic case. As inmany situations in physics, the simple guess turns out to be correct! Thisleads to a connection between wave dynamics and the Newtonian mechanicsof conservative forces.

1See Louis de Broglie’s 1929 Nobel Prize address, reproduced in Boorse, H. A., and L.Motz, 1966: The World of the Atom, Basic Books.

2This group velocity calculation ignores the possible dependence of index of refractionon wavenumber. If n is a function of k, the calculation is more complicated, but theprinciple is the same.


The dispersion relation for free matter waves is $ = (k2c2 +µ2)1/2. In thenon-relativistic limit k2c2 " µ2 and this equation can be approximated as

$ = µ(1 + k2c2/µ2)1/2 # µ + k2c2/(2µ). (8.16)

Let us modify this equation by adding a term S # which depends on x,and which plays a role analogous to the index of refraction for light:

$ = S #(x) + µ + k2c2/(2µ) = S(x) + k2c2/(2µ). (8.17)

The rest frequency has been made to disappear on the right side of the aboveequation by defining S = S # + µ. This is done to simplify the notation. Atthis point we don’t know what the physical meaning of S is; we are simplyexploring the consequences of a hunch in the hope that something sensiblecomes out of it.

Let us now imagine that all parts of the wave governed by this dispersionrelation oscillate in phase. The only way this can happen is if $ is constant,i. e., it takes on the same value in all parts of the wave.

If $ is constant, the only way S can vary with x in equation (8.17) is ifthe wavenumber varies in a compensating way. Thus, constant frequency andspatially varying S together imply that k = k(x). Solving equation (8.17)for k yields

k(x) = ±

%

2µ[$ ! S(x)]

c2

&1/2

. (8.18)

Since $ is constant, the wavenumber becomes smaller and the wavelengthlarger as the wave moves into a region of increased S.

In the geometrical optics limit, we assume that S doesn’t change muchover one wavelength so that the wave remains reasonably sinusoidal in shapewith approximately constant wavenumber over a few wavelengths. However,over distances of many wavelengths the wavenumber and amplitude of thewave are allowed to vary considerably.

The group velocity calculated from the dispersion relation given by equa-tion (8.17) is

ug =d$

dk=

kc2

µ=

!

2c2($ ! S)

µ

"1/2

(8.19)

where k is eliminated in the last step with the help of equation (8.18). Theresulting equation tells us how the group velocity varies as a matter wave

8.5. MATH TUTORIAL – PARTIAL DERIVATIVES 143

traverses a region of slowly varying S. Thus, as S increases, ug decreases andvice versa.

We can now calculate the acceleration of a wave packet resulting fromthe spatial variation in S. We assume that x(t) represents the positionof the wave packet, so that ug = dx/dt. Using the chain rule dug/dt =(dug/dx)(dx/dt) = (dug/dx)ug, we find

a =dug

dt=

dug

dxug =

du2g/2

dx= !

c2

µ

dS

dx= !

h

m

dS

dx. (8.20)

The group velocity is eliminated in favor of S by squaring equation (8.19)and substituting the result into equation (8.20).

Comparison of this equation with equation (8.3) solves the mystery as tothe identity of the variable S; it is simply the potential energy divided byPlanck’s constant:

U = hS. (8.21)

Thus, the geometrical optics approach to particle motion is completely equiv-alent to the classical mechanics of a particle moving under the influence of aconservative force, at least in one dimension. We therefore have two ways ofsolving for the motion of a particle subject to a potential energy U(x). Wecan apply the principles of classical mechanics to get the force and the accel-eration of the particle, from which we can derive the motion. Alternatively,we can apply the principles of geometrical optics to compute the spatiallyvariable velocity of the wave packet using equation (8.19). The results arecompletely equivalent, though the methods are conceptually very di!erent.

8.5 Math Tutorial – Partial Derivatives

In order to understand the generalization of Newtonian mechanics to two andthree dimensions, we first need to understand a new type of derivative calledthe partial derivative. The partial derivative is used in functions of more thanone variable. It is just like an ordinary derivative, except that when takingthe derivative of the function with respect to one of the variables, the othervariables are held constant. As an example, let us consider the function

f(x, y) = Ax4 + Bx2y2 + Cy4 (8.22)


where A, B, and C are constants. The partial derivative of f with respectto x is

,f

,x= 4Ax3 + 2Bxy2 (8.23)

and the partial derivative with respect to y is

,f

,y= 2Bx2y + 4Cy3. (8.24)

That’s it! Note that a special symbol “,” is used in place of the normal“d” for the partial derivative. This is sometimes called a “curly d”.

8.6 Motion in Two and Three Dimensions

When a matter wave moves through a region of variable potential energy inone dimension, only the wavenumber changes. In two or three dimensions thewave vector can change in both direction and magnitude. This complicatesthe calculation of particle movement. However, we already have an exampleof how to handle this situation, namely, the refraction of light. In that caseSnell’s law tells us how the direction of the wave vector changes, while thedispersion relation combined with the constancy of the frequency gives usinformation about the change in the magnitude of the wave vector. Formatter waves a similar procedure works, though the details are di!erent,because we seek the consequences of a change in potential energy ratherthan a change in the index of refraction.

Figure 8.2 illustrates the refraction of matter waves at a discontinuity inthe potential energy. Let us suppose that the discontinuity occurs at x = 0.If the matter wave to the left of the discontinuity is +1 = sin(k1xx+k1yy!$1t)and to the right is +2 = sin(k2xx + k2yy ! $2t), then the wavefronts of thewaves will match across the discontinuity for all time only if $1 = $2 ' $and k1y = k2y ' ky. We are already familiar with the first condition fromthe one-dimensional problem, so the only new ingredient is the constancy ofthe y component of the wave vector.

In two dimensions the momentum is a vector: " = mu when |u| " c,where u is the particle velocity. Furthermore, the kinetic energy is K =m|u|2/2 = |"|2/(2m) = ($2

x + $2y)/(2m). The relationship between kinetic,

potential, and total energy is unchanged from the one-dimensional case, sowe have

E = U + ($2x + $2

y)/(2m) = constant. (8.25)

8.6. MOTION IN TWO AND THREE DIMENSIONS 145

k1

k2k 1x

k 1y

k 2y

k 2x

U1 2Uy

x

Figure 8.2: Refraction of a matter wave by a discontinuity in potential energy.The component of the wave vector parallel to the discontinuity, ky, doesn’tchange, so k1y = k2y.

The de Broglie relationship tells us that " = hk, so the constancy of ky

across the discontinuity in U tells us that

$y = constant (8.26)

there.Let us now approximate a continuously variable U(x) by a series of steps

of constant U oriented normal to the x axis. The above analysis can beapplied at the jumps or discontinuities in U between steps, as illustrated infigure 8.3, with the result that equations (8.25) and (8.26) are valid acrossall discontinuities. If we now let the step width go to zero, these equationsthen become valid for U continuously variable in x.

An example from classical mechanics of a problem of this type is a ballrolling down an inclined ramp with an initial velocity component across theramp, as illustrated in figure 8.4. The potential energy decreases in the downramp direction, resulting in a force down the ramp. This accelerates the ballin that direction, but leaves the component of momentum across the rampunchanged.


x

y

θ

> > > > >U1 U2 U3 U4 U5 U6

uu x

yu

Figure 8.3: Trajectory of a wave packet through a variable potential energy,U(x), which decreases to the right.

Figure 8.4: Classical mechanics example of the problem illustrated in figure8.3.

8.6. MOTION IN TWO AND THREE DIMENSIONS 147

Using the procedure which we invoked before, we find the force compo-nents associated with U(x) in the x and y directions to be Fx = !dU/dxand Fy = 0. This generalizes to

F = !

!

,U

,x,,U

,y,,U

,z

"

(3-D conservative force) (8.27)

in the three-dimensional case where the orientation of constant U surfacesis arbitrary. It is also valid when U(x, y, z) is not limited to a simple rampform, but takes on a completely arbitrary structure.

The definitions of work and power are slightly di!erent in two and threedimensions. In particular, work is defined

W = F ·#x (8.28)

where #x is now a vector displacement. The vector character of this expres-sion yields an additional possibility over the one dimensional case, where thework is either positive or negative depending on the direction of #x relativeto F. If the force and the displacement of the object on which the force isacting are perpendicular to each other, the work done by the force is actu-ally zero, even though the force and the displacement both have non-zeromagnitudes. The power exhibits a similar change:

P = F · u. (8.29)

Thus the power is zero if an object’s velocity is normal to the force beingexerted on it.

As in the one-dimensional case, the total work done on a particle equalsthe change in the particle’s kinetic energy. In addition, the work done by aconservative force equals minus the change in the associated potential energy.

Energy conservation by itself is somewhat less useful for solving problemsin two and three dimensions than it is in one dimension. This is becauseknowing the kinetic energy at some point tells us only the magnitude ofthe velocity, not its direction. If conservation of energy fails to give us theinformation we need, then we must revert to Newton’s second law, as we didin the one-dimensional case. For instance, if an object of mass m has initialvelocity u0 = (u0, 0) at location (x, z) = (0, h) and has the gravitationalpotential energy U = mgz, then the force on the object is F = (0,!mg).


The acceleration is therefore a = F/m = (0,!g). Since a = du/dt = d2x/dt2

where x = (x, z) is the object’s position, we find that

u = (C1,!gt + C2) x = (C1t + C3,!gt2/2 + C2t + C4), (8.30)

where C1, C2, C3, and C4 are constants to be evaluated so that the solutionreduces to the initial conditions at t = 0. The specified initial conditions tellus that C1 = u0, C2 = 0, C3 = 0, and C4 = h in this case. From these resultswe can infer the position and velocity of the object at any time.

8.7 Kinetic and Potential Momentum

If you have previously taken a physics course then you have probably noticedthat a rather odd symbol is used for momentum, namely ", rather than themore commonly employed p. The reason for this peculiar usage is thatthere are actually two kinds of momentum, kinetic momentum and totalmomentum, just as there are two kinds of energy, kinetic and total.3 Thesymbol " represents total momentum while p represents kinetic momentum.In non-relativistic mechanics we don’t need to distinguish between the twoquantities, as they are generally equal to each other. However, we will findlater in the course that it is crucial to make this distinction in the relativisticcase. As a general rule, the total momentum is related to a particle’s wavevector via the de Broglie relation, " = hk, while the kinetic momentum isrelated to a particle’s velocity, p = mu/(1 ! u2/c2)1/2.

8.8 Problems

1. Suppose the dispersion relation for a matter wave under certain con-ditions is $ = µ + (k ! a)2c2/(2µ) where k is the wavenumber of thewave, µ = mc2/h, m is the associated particle’s mass, a is a constant,c is the speed of light, and h is Planck’s constant divided by 2!.

(a) Use this disperson relation and the Planck and de Broglie relationsto determine the relationship between energy E, momentum $,and mass m.

3In advanced mechanics the total momentum is called the canonical momentum andthe kinetic momentum is the ordinary momentum.

8.8. PROBLEMS 149

Re( )ψx

Figure 8.5: Real part of a wave function in which the wavelength varies.

(b) Compute the group velocity of the wave and use this to determinehow the group velocity depends on mass and momentum in thiscase.

2. A matter wave function associated with a particle of definite (constant)total energy E takes the form shown in figure 8.5. Make a sketchshowing how the kinetic, potential, and total energies of the particlevary with x.

3. Compute the partial derivatives of the following functions with respectto x and y. Other symbols are constants.

(a) f(x, y) = ax2 + by3

(b) f(x, y) = ax2y2

(c) f(x, y) = (x + a)/(y + b)

4. Given a potential energy for a particle of mass M of the form U(x) =Ax3 ! Bx where A and B are positive constants:

(a) Find the force on the particle.

(b) Find the values of x where the force is zero.

(c) Sketch U(x) versus x and graphically compare minus the slopeof U(x) to the force computed above. Do the two qualitativelymatch?

(d) If the total energy of the particle is zero, where are its turningpoints?

(e) What is the particle’s speed as a function of position?

5. Given a potential energy function U(x, y) = A(x2 + y2) where A is apositive constant:


(a) Sketch lines of constant U in the x-y plane.

(b) Compute the components of force as a function of x and y anddraw sample force vectors in the x-y plane on the same plot usedabove. Do the force vectors point “uphill” or “downhill”?

6. Do the same as in the previous question for the potential energy func-tion U(x, y) = Axy.

7. Suppose that the components of the force vector in the x-y plane areF = (2Axy3, 3Ax2y2) where A is a constant. See if you can find apotential energy function U(x, y) which gives rise to this force.

8. You are standing on top of a cli! of height H with a rock of mass M .

(a) If you throw the rock horizontally outward at speed u0, what willits speed be when it hits the ground below?

(b) If you throw the rock upward at 45" to the horizontal at speed u0,what will its speed be when it hits the ground?

Hint: Can you use conservation of energy to solve this problem? Ignoreair friction.

9. A car of mass 1200 kg initially moving 30 m s!1 brakes to a stop.

(a) What is the net work done on the car due to all the forces actingon it during the indicated period?

(b) Describe the motion of the car relative to an inertial referenceframe initially moving with the car.

(c) In the above reference frame, what is the net work done on thecar during the indicated period?

Is work a relativistically invariant quantity?

10. A soccer player kicks a soccer ball, which is caught by the goal keeperas shown in figure 8.6. At various points forces exerted by gravity,air friction, the foot of the o!ensive player, and the hands of the goalkeeper act on the ball.

(a) List the forces acting on the soccer ball at each of the points A,B, C, D, and E.

8.8. PROBLEMS 151

A

BC D

E

Figure 8.6: The trajectory of a soccer ball.

(b) State whether the instantaneous power being applied to the soccerball due to each of the forces listed above is positive, negative, orzero at each of the labeled points.

11. A cannon located at (x, z) = (0, 0) shoots a cannon ball upward atan angle of & from the horizontal at initial speed u0. Hint: In orderto solve this problem you must first obtain the x and z componentsof acceleration from Newton’s second law. Second, you must find thevelocity components as a function of time from the components of ac-celeration. Third, you must find x and z as a function of time from thethe components of velocity. Only then should you attempt to answerthe questions below.

(a) How long does it take the cannon ball to reach its peak altitude?

(b) How high does the cannon ball go?

(c) At what value of x does the cannon ball hit the ground (z = 0)?

(d) Determine what value of & yields the maximum range.

Chapter 9

Symmetry and Bound States

Until now we have represented quantum mechanical plane waves by sine andcosine functions, just as with other types of waves. However, plane matterwaves cannot be truly represented by sines and cosines. We need insteadmathematical functions in which the wave displacement is complex ratherthan real. This requires the introduction of a bit of new mathematics, whichwe tackle first. Using our new mathematical tool, we are then able to exploretwo crucially important ideas in quantum mechanics; (1) the relationshipbetween symmetry and conservation laws, and (2) the dynamics of spatiallyconfined waves.

When quantum mechanics was first invented, the dynamical principlesused were the same as those underlying classical mechanics. The initial de-velopment of the field thus proceeded largely by imposing quantum laws onclassical variables such as position, momentum, and energy. However, asquantum mechanics advanced, it became clear that there were many situ-ations in which no classical analogs existed for new types of quantum me-chanical systems, especially those which arose in the study of elementaryparticles. To understand these systems it was necessary to seek guidancefrom novel sources. One of the most important of these sources was the ideaof symmetry, and in particular the relationship between symmetry and con-served variables. This type of relationship was first developed in the early20th century by the German mathematician Emmy Nother in the context ofclassical mechanics. However, her idea is easier to express and use in quan-tum mechanics than it is in classical mechanics. Emmy Nother showed thatthere is a relationship between the symmetries of a system and conserveddynamical variables. This idea is naturally called Nother’s theorem.

153

154 CHAPTER 9. SYMMETRY AND BOUND STATES

Wave confinement can happen in a number of ways. We first look atthe so-called “particle in a box” in one spatial dimension. We find thatconfined particles can take on only discrete energy values. When confinementisn’t perfect we see how a quantum mechanical particle can leak through apotential energy barrier which is classically impenetrable. Movement of aparticle on a circular ring leads us to another form of confinement and theintroduction of angular momentum. This brings us finally to a discussion ofthe intrinsic or spin angular momentum of elementary particles.

9.1 Math Tutorial — Complex Waves

A complex number z is the sum of a real number and an imaginary number.An imaginary number is just a real number multiplied by i ' (!1)1/2. Thus,we can write z = a + ib for any complex z, where a and b are real.

Quantum mechanics requires wave functions to be complex, i. e., topossess real and imaginary parts. Plane waves in quantum mechanics actuallytake the form + = exp[i(kx!$t)] rather than, say, cos(kx!$t). The reasonfor this is the need to distinguish between waves with positive and negativefreqencies. If we replace k and $ by !k and !$ in the cosine form, we getcos(!kx+$t) = cos[!(kx!$t)] = cos(kx!$t). In other words, changing thesign of k and $ results in no change in a wave expressed as a cosine function.The two quantum mechanical states, one with wavenumber and frequency kand $ and the other with !k and !$, yield indistinguishable wave functionsand therefore would represent physically indistinguishable states. The cosineform is thus insu"ciently flexible to represent quantum mechanical waves.On the other hand, if we replace k and $ by their negatives in the complexexponential form of a plane wave we get + = exp[!i(kx ! $t)], which isdi!erent from exp[i(kx! $t)]. These two wave functions are distinguishableand thus correspond to distinct physical states.

It is not immediately obvious that a complex exponential function pro-vides the oscillatory behavior needed to represent a plane wave. However,the complex exponential can be expressed in terms of sines and cosines usingEuler’s equation:

exp(i#) = cos(#) + i sin(#) (Euler’s equation). (9.1)

As noted above, any complex number z may be expressed in terms of tworeal numbers as z = a+ib. The quantities a and b are the real and imaginary

9.1. MATH TUTORIAL — COMPLEX WAVES 155

b

aφ

rRe(z)

Im(z)

Figure 9.1: Graphical representation of a complex number z as a point inthe complex plane. The horizontal and vertical Cartesian components givethe real and imaginary parts of z respectively.

parts of z, sometimes written Re(z) and Im(z). If we define r = (a2 + b2)1/2

and # = tan!1(b/a), then an alternate way of expressing a complex numberis z = r exp(i#), which by Euler’s equation equals r cos(#) + ir sin(#). Com-parison shows that a = r cos(#) and b = r sin(#). Thus, a complex numbercan be thought of as a point in the a-b plane with Cartesian coordinates aand b and polar coordinates r and #. The a-b plane is called the complexplane.

We now see how the complex wave function represents an oscillation. If+ = exp[i(kx ! $t)], the complex function +(x, t) moves round and roundthe unit circle in the complex plane as x and t change, as illustrated in figure9.1. This contrasts with the back and forth oscillation along the horizontalaxis of the complex plane represented by cos(kx ! $t).

We will not present a formal proof of Euler’s equation — you will even-tually see it in your calculus course. However, it may be helpful to note thatthe # derivatives of exp(i#) and cos(#) + i sin(#) have the same behavior:

d

d#exp(i#) = i exp(i#); (9.2)

d

d#[cos(#) + i sin(#)] = ! sin(#) + i cos(#)

= i[cos(#) + i sin(#)]. (9.3)


(In the second of these equations we have replaced the!1 multiplying the sinefunction by i2 and then extracted a common factor of i.) The # derivative ofboth of these functions thus yields the function back again times i. This is astrong hint that exp(i#) and cos(#)+i sin(#) are di!erent ways of representingthe same function.

We indicate the complex conjugate of a complex number z by a super-scripted asterisk, i. e., z$. It is obtained by replacing i by !i. Thus,(a + ib)$ = a ! ib. The absolute square of a complex number is the numbertimes its complex conjugate:

|z|2 = |a + ib|2 ' (a + ib)(a ! ib) = a2 + b2. (9.4)

Notice that the absolute square of a complex exponential function is one:

| exp(i#)|2 = exp(i#) exp(!i#) = exp(i#! i#) = exp(0) = 1. (9.5)

In quantum mechanics the absolute square of the wave function at any pointexpresses the relative probability of finding the associated particle at thatpoint. Thus, the probability of finding a particle represented by a plane waveis uniform in space. Contrast this with the relative probability associatedwith a sine wave: | sin(kx ! $t)|2 = sin2(kx ! $t). This varies from zeroto one, depending on the phase of the wave. The “waviness” in a complexexponential plane wave resides in the phase rather than in the magnitude ofthe wave function.

One more piece of mathematics is needed. The complex conjugate ofEuler’s equation is

exp(!i#) = cos(#) ! i sin(#). (9.6)

Taking the sum and the di!erence of this with the original Euler’s equationresults in the expression of the sine and cosine in terms of complex exponen-tials:

cos(#) =exp(i#) + exp(!i#)

2sin(#) =

exp(i#) ! exp(!i#)

2i. (9.7)

We aren’t used to having complex numbers show up in physical theoriesand it is hard to imagine how we would measure such a number. However,everything observable comes from taking the absolute square of a wave func-tion, so we deal only with real numbers in experiments.

9.2. SYMMETRY AND QUANTUM MECHANICS 157

9.2 Symmetry and Quantum Mechanics

The idea of symmetry plays a huge role in physics. We have already usedsymmetry arguments in the theory of relativity — applying the principle ofrelativity to obtain the dispersion relation for relativistic matter waves is justsuch an argument. In this section we begin to explore how symmetry can beused to increase our understanding of quantum mechanics.

9.2.1 Free Particle

The wave function for a free particle of definite momentum $ and energy Eis given by

+ = exp[i(kx ! $t)] = exp[i($x ! Et)/h] (free particle). (9.8)

For this wave function |+|2 = 1 everywhere, so the probability of finding theparticle anywhere in space and time is uniform. This contrasts with the prob-ability distribution which arises if we assume a free particle to have the wavefunction + = cos[($x!Et)/h]. In this case |+|2 = cos2[(!x!Et)/h], whichvaries with position and time, and is inconsistent with a uniform probabilitydistribution.

9.2.2 Symmetry and Definiteness

Quantum mechanics is a probabilistic theory, in the sense that the predictionsit makes tell us, for instance, the probability of finding a particle somewherein space. If we know nothing about a particle’s previous history, and if thereare no physical constraints that would make it more likely for a particle tobe at one point along the x axis than any another, then the probabilitydistribution must be P (x) = constant.

This is an example of a symmetry argument. Expressed more formally,it states that if the above conditions apply, then the probability distributionought to be subject to the condition P (x+D) = P (x) for any constant valueof D. The only possible P (x) in this case is P = constant. In the languageof physics, if there is nothing that gives the particle a higher probability ofbeing at one point rather than another, then the probability is independentof position and the system is invariant under displacement in the x direction.


The above argument doesn’t su"ce for quantum mechanics, since as wehave learned, the fundamental quantity describing a particle is not the prob-ability distribution, but the wave function +(x). Thus, the wave functionrather than the probability distribution ought to be the quantity which isinvariant under displacement, i. e., +(x + D) = +(x).

This condition turns out to be too restrictive, because it implies that+(x) = constant, whereas we know that a one-dimensional plane wave, whichdescribes a particle with a uniform probability of being found anywhere alongthe x axis, has the form +(x) = exp(ikx). (For simplicity we temporarilyignore the time dependence.) If we make the substitution x * x + D in aplane wave, we get exp[ik(x + D)] = exp(ikx) exp(ikD). The wave functionis thus technically not invariant under displacement, in that the displacedwave function is multiplied by the factor exp(ikD). However, the probabilitydistribution of the displaced wave function still equals one everywhere, sothere is no change in what we observe. Thus, in determining invarianceunder displacement, we are allowed to ignore changes in the wave functionwhich consist only of multiplying it by a complex constant with an absolutevalue of one. Such a multiplicative constant is called a phase factor.

It is easy to convince oneself by trial and error or by more sophisticatedmeans that the only form of wave function +(x) which satisfies the condition+(x+D) = +(x)%(phase factor) is +(x) = A exp(ikx) where A is a (possiblycomplex) constant. This is just in the form of a complex exponential planewave with wavenumber k. Thus, not only is the complex exponential wavefunction invariant under displacements in the manner defined above, it is theonly wave function which is invariant to displacements. Furthermore, thephase factor which appears for a displacement D of such a plane wave takesthe form exp(iC) = exp(ikD), where k is the wavenumber of the plane wave.

As an experiment, let us see if a wave packet is invariant under displace-ment. Let’s define a wave packet consisting of two plane waves:

+(x) = exp(ik1x) + exp(ik2x). (9.9)

Making the substitution x * x + D in this case results in

+(x + D) = exp[ik1(x + D)] + exp[ik2(x + D)]

= exp(ik1x) exp(ik1D) + exp(ik2x) exp(ik2D)

(= [exp(ik1x) + exp(ik2x)] % (phase factor). (9.10)

The impossibility of writing +(x + D) = +(x) % (phase factor) lends plausi-bility to the assertion that a single complex exponential is the only possible

9.2. SYMMETRY AND QUANTUM MECHANICS 159

form of the wave function that is invariant under displacement.Notice that the wave packet does not have definite wavenumber, and

hence, momentum. In particular, non-zero amplitudes exist for the associatedparticle to have either momentum $1 = hk1 or $2 = hk2. This makes sensefrom the point of view of the uncertainty principle — for a single plane wavethe uncertainty in position is complete and the uncertainty in momentumis zero. For a wave packet the uncertainty in position is reduced and theuncertainty in the momentum is non-zero. However, we see that this ideacan be carried further: A definite value of momentum must be associatedwith a completely indefinite probability distribution in position, i. e., withP = constant. This corresponds to a wave function which has the form ofa complex exponential plane wave. However, such a plane wave is invariantunder displacement D, except for the multiplicative phase factor exp(ikD),which has no physical consequences since it disappears when the probabilitydistribution is obtained. Thus, we see that invariance under displacement ofthe wave function and a definite value of the momentum are linked, in thateach implies the other:

invariance under displacement +, definite momentum (9.11)

The idea of potential energy was introduced in the previous chapter. Inparticular, we found that if the total energy is constant, then the momentumcannot be constant in the presence of spatially varying potential energy. Thismeans that the wavenumber, and hence the wavelength of the oscillations inthe wave function also vary with position. The spatial inhomogeneity of thepotential energy gives rise to spatial inhomogeneity in the wave function, andhence an indefinite momentum.

The above argument can be extended to other variables besides momen-tum. In particular since the time dependence of a complex exponential planewave is exp(!i$t) = exp(!iEt/h), where E is the total energy, we have byanalogy with the above argument that

invariance under time shift +, definite energy. (9.12)

Thus, invariance of the wave function under a displacement in time impliesa definite value of the energy of the associated particle.

In the previous chapter we assumed that the frequency (and hence theenergy) was definite and constant for a particle passing through a region ofvariable potential energy. We now see that this assumption is justified only


if the potential energy doesn’t change with time. This is because a time-varying potential energy eliminates the possibility of invariance under timeshift.

9.2.3 Compatible Variables

We already know that definite values of certain pairs of variables cannot beobtained simultaneously in quantum mechanics. For instance, the indefinite-ness of position and momentum are related by the uncertainty principle —a definite value of position implies an indefinite value of the momentum andvice versa. If definite values of two variables can be simultaneously obtained,then we call these variables compatible. If not, the variables are incompatible.

If the wave function of a particle is invariant under the displacementsassociated with both variables, then the variables are compatible. For in-stance, the complex exponential plane wave associated with a free particleis invariant under displacements in both space and time. Since momentumis associated with space displacements and energy with time displacements,the momentum and energy are compatible variables for a free particle.

9.2.4 Compatibility and Conservation

Variables which are compatible with the energy have a special status. Thewave function which corresponds to a definite value of such a variable is in-variant to displacements in time. In other words, the wave function doesn’tchange under this displacement except for a trivial phase factor. Thus, if thewave function is also invariant to some other transformation at a particulartime, it is invariant to that transformation for all time. The variable associ-ated with that transformation therefore retains its definite value for all time— i. e., it is conserved.

For example, the plane wave implies a definite value of energy, and is thusinvariant under time displacements. At time t = 0, it is also invariant underx displacements, which corresponds to the fact that it represents a particlewith a known value of momentum. However, since momentum and energyare compatible for a free particle, the wave function will represent the samevalue of momentum at all other times. In other words, if the momentum isdefinite at t = 0, it will be definite at all later times, and furthermore willhave the same value. This is how the conservation of momentum (and by

9.3. CONFINED MATTER WAVES 161

extension, the conservation of any other variable compatible with energy) isexpressed in quantum mechanics.

9.2.5 New Symmetries and Variables

In modern quantum physics, the discovery of new symmetries leads to newdynamical variables. In the problems we show how that comes about forthe symmetries of parity (x * !x), time reversal t * !t), and chargeconjugation (the interchange of particles with antiparticles). One of thekey examples of this was the development of the quark theory of matter,which came from the observation that the interchange of certain groups ofelementary particles left the universe approximately unchanged, meaningthat the universe was (approximately) symmetric under these interchanges.

9.3 Confined Matter Waves

Confinement of a wave to a limited spatial region results in rather peculiarbehavior — the wave can only fit comfortably into the confined region if thewave frequency, and hence the associated particle energy, takes on a limitedset of possible values. This is the origin of the famous quantization of energy,from which the “quantum” in quantum mechanics comes. We will exploretwo types of confinement, position confinement due to a potential energywell, and rotational confinement due to the fact that rotation of an objectthrough 2! radians returns the object to its original orientation.

9.3.1 Particle in a Box

We now imagine how a particle confined to a region 0 & x & a on the xaxis must behave. As with the displacement of a guitar string, the wavefunction must be zero at x = 0 and a, i. e., at the ends of the guitar string.A single complex exponential plane wave cannot satisfy this condition, since| exp[i(kx!$t)]|2 = 1 everywhere. However, a superposition of leftward andrightward traveling waves creates a standing wave, as illustrated in figure 9.2:

+ = exp[i(kx ! $t)] ! exp[i(!kx ! $t)] = 2i exp(!i$t) sin(kx). (9.13)

Notice that the time dependence is still a complex exponential, which meansthat |+|2 is independent of time. This insures that the probability of finding


x = 0 x = a

n = 1

n = 3

n = 2ψ

Figure 9.2: First three modes for wave function of a particle in a box.

the particle somewhere in the box remains constant with time. It also meansthat the wave packet corresponds to a definite energy E = h$.

Because we took a di!erence rather than a sum of plane waves, the condi-tion + = 0 is already satisfied at x = 0. To satisfy it at x = a, we must haveka = n!, where n = 1, 2, 3, . . .. Thus, the absolute value of the wavenumbermust take on the discrete values

kn =n!

a, n = 1, 2, 3, . . . . (9.14)

(The wavenumbers of the two plane waves take on plus and minus this abso-lute value.) This implies that the absolute value of the particle momentumis $n = hkn = n!h/a, which in turn means that the energy of the particlemust be

En = ($2nc2 + m2c4)1/2 = (n2!2h2c2/a2 + m2c4)1/2, (9.15)

where m is the particle mass. In the non-relativistic limit this becomes

En =$2

n

2m=

n2!2h2

2ma2(non-relativistic) (9.16)

where we have dropped the rest energy mc2 since it is just a constant o!set.In the ultra-relativistic case where we can ignore the particle mass, we find

En = |$n|c =n!hc

a(zero mass). (9.17)

In both limits the energy takes on only a certain set of possible values.This is called energy quantization and the integer n is called the energy


0

5

10

15

20

E/E0

n = 1

n = 2

n = 3

n = 4

non-relativistic limitparticle in a box

Figure 9.3: Allowed energy levels for the non-relativistic particle in a box.The constant E0 = !2h2/(2ma2). See text for the meanings of symbols.

quantum number. In the non-relativistic limit the energy is proportionalto n2, while in the ultra-relativistic case the energy is proportional to n.

We can graphically represent the allowed energy levels for the particle ina box by an energy level diagram. Such a diagram is shown in figure 9.3 forthe non-relativistic case.

One aspect of this problem deserves a closer look. Equation (9.13) showsthat the wave function for this problem is a superposition of two plane wavescorresponding to momenta $1 = +hk and $2 = !hk and is therefore a kindof wave packet. Thus, the wave function is not invariant under displacementand does not correspond to a definite value of the momentum — the mo-mentum’s absolute value is definite, but its sign is not. Following Feynman’sprescription, equation (9.13) tells us that the amplitude for the particle inthe box to have momentum +hk is exp[i(kx ! $t)], while the amplitude forit to have momentum !hk is ! exp[i(!kx ! $t)]. The absolute square ofthe sum of these amplitudes gives us the relative probability of finding theparticle at position x:

P (x) = |2i exp(!i$t) sin(kx)|2 = 4 sin2(kx). (9.18)

Which of the two possible values of the momentum the particle takes on isunknowable, just as it is impossible in principle to know which slit a particlepasses through in two slit interference. If an experiment is done to measure


d

d

)Re(ψ )Re(ψ

xx

E EU(x) U(x)barrier barrier

Figure 9.4: Real part of wave function Re[+(x)] for barrier penetration. Theleft panel shows weak penetration occurring for a large potential energy bar-rier, while the right panel shows stronger penetration which occurs when thebarrier is small.

the momentum, then the wave function is irreversibly changed, just as theinterference pattern in the two slit problem is destroyed if the slit throughwhich the particle passes is unambiguously determined.

9.3.2 Barrier Penetration

Unlike the situation in classical mechanics, quantum mechanics allows thekinetic energy K to be negative. This makes the momentum $ (equal to(2mK)1/2 in the nonrelativistic case) imaginary, which in turn gives rise toan imaginary wavenumber.

Let us investigate the nature of a wave with an imaginary wavenumber.Let us assume that k = i- in a complex exponential plane wave, where - isreal:

+ = exp[i(kx ! $t)] = exp(!-x ! i$t) = exp(!-x) exp(!i$t). (9.19)

The wave function doesn’t oscillate in space when K = E!U < 0, but growsor decays exponentially with x, depending on the sign of -.

For a particle moving to the right, with positive k in the allowed region,- turns out to be positive, and the solution decays to the right. Thus, aparticle impingent on a potential energy barrier from the left (i. e., whilemoving to the right) will have its wave amplitude decay in the classically


forbidden region, as illustrated in figure 9.4. If this decay is very rapid, thenthe result is almost indistinguishable from the classical result — the particlecannot penetrate into the forbidden region to any great extent. However, ifthe decay is slow, then there is a reasonable chance of finding the particle inthe forbidden region. If the forbidden region is finite in extent, then the waveamplitude will be small, but non-zero at its right boundary, implying thatthe particle has a finite chance of completely passing through the classicalforbidden region. This process is called barrier penetration.

The probability for a particle to penetrate a barrier is the absolute squareof the amplitude after the barrier divided by the square of the amplitudebefore the barrier. Thus, in the case of the wave function illustrated inequation (9.19), the probability of penetration is

P = |+(d)|2/|+(0)|2 = exp(!2-d) (9.20)

where d is the thickness of the barrier.The rate of exponential decay with x in the forbidden region is related to

how negative K is in this region. Since

!K = U ! E = !$2

2m= !

h2k2

2m=

h2-2

2m, (9.21)

we find that

- ='

2mB

h2

(1/2

(9.22)

where the potential energy barrier is B ' !K = U ! E. The smaller B is,the smaller is -, resulting in less rapid decay of the wave function with x.This corresponds to stronger barrier penetration. (Note that the way B isdefined, it is positive in forbidden regions.)

If the energy barrier is very high, then the exponential decay of the wavefunction is very rapid. In this case the wave function goes nearly to zero atthe boundary between the allowed and forbidden regions. This is why wespecify the wave function to be zero at the walls for the particle in the box.These walls act in e!ect as infinitely high potential barriers.

Barrier penetration is important in a number of natural phenomena. Cer-tain types of radioactive decay and the fissioning of heavy nuclei are governedby this process. In addition, the field e!ect transistors used in most com-puter microchips control the flow of electrons by electronically altering thestrength of a potential energy barrier.


Rθ

M

Π

Figure 9.5: Illustration of a bead of mass M sliding (without friction) on acircular loop of wire of radius R with momentum $.

9.3.3 Orbital Angular Momentum

Another type of bound state motion occurs when a particle is constrainedto move in a circle. (Imagine a bead sliding on a circular loop of wire, asillustrated in figure 9.5.) We can define x in this case as the path lengtharound the wire and relate it to the angle &: x = R&. For a plane wave wehave

+ = exp[i(kx ! $t)] = exp[i(kR& ! $t)]. (9.23)

This plane wave di!ers from the normal plane wave for motion along a Carte-sian axis in that we must have +(&) = +(& + 2!). This can only happen ifthe circumference of the loop, 2!R, is an integral number of wavelengths,i. e., if 2!R/" = m where m is an integer. However, since 2!/" = k, thiscondition becomes kR = m.

Since $ = hk, the above condition can be written $mR = mh. Thequantity

Lm ' $mR (9.24)

is called the angular momentum, leading to our final result,

Lm = mh, m = 0,±1,±2, . . . . (9.25)

We see that the angular momentum can only take on values which are integermultiples of h. This represents the quantization of angular momentum, and m


in this case is called the angular momentum quantum number. Note that thisquantum number di!ers from the energy quantum number for the particle inthe box in that zero and negative values are allowed.

The energy of our bead on a loop of wire can be expressed in terms ofthe angular momentum:

Em =$2

m

2M=

L2m

2MR2. (9.26)

This means that angular momentum and energy are compatible variablesin this case, which further means that angular momentum is a conservedvariable. Just as definite values of linear momentum are related to invari-ance under translations, definite values of angular momentum are related toinvariance under rotations. Thus, we have

invariance under rotation +, definite angular momentum (9.27)

for angular momentum.We need to briefly address the issue of angular momentum in three di-

mensions. Angular momentum is actually a vector oriented perpendicular tothe wire loop in the example we are discussing. The direction of the vectoris defined using a variation on the right-hand rule: Curl your fingers in thedirection of motion of the bead around the loop (using your right hand!).The orientation of the angular momentum vector is defined by the directionin which your thumb points. This tells you, for instance, that the angularmomentum in figure 9.5 points out of the page.

In quantum mechanics it turns out that it is only possible to measuresimultaneously the square of the length of the angular momentum vectorand one component of this vector. Two di!erent components of angularmomentum cannot be simultaneously measured because of the uncertaintyprinciple. However, the length of the angular momentum vector may be mea-sured simultaneously with one component. Thus, in quantum mechanics, theangular momentum is completely specified if the length and one componentof the angular momentum vector are known.

Figure 9.6 illustrates the angular momentum vector associated with abead moving on a wire loop which is tilted from the horizontal. One compo-nent (taken to be the z component) is shown as well. For reasons we cannotexplore here, the square of the length of the angular momentum vector L2 isquantized with the following values:

L2l = h2l(l + 1), l = 0, 1, 2, . . . . (9.28)


L z

z

L

Figure 9.6: Illustration of the angular momentum vector L for a tilted loopand its z component Lz.

One component (say, the z component) of angular momentum is quantizedjust like angular momentum in the two-dimensional case, except that l actsas an upper bound on the possible values of |m|. In other words, if thesquare of the length of the angular momentum vector is h2l(l + 1), then thez component can take on the values

Lzm = hm, m = !l,!l + 1, . . . , l ! 1, l. (9.29)

The quantity l is called the angular momentum quantum number, while m iscalled the orientation or magnetic quantum number, the latter for historicalreasons.

9.3.4 Spin Angular Momentum

The type of angular momentum discussed above is associated with the move-ment of particles in orbits. However, it turns out that even stationary parti-cles can possess angular momentum. This is called spin angular momentum.The spin quantum number s plays a role analogous to l for spin angular mo-mentum, i. e., the square of the spin angular momentum vector of a particleis

L2s = h2s(s + 1). (9.30)

9.4. PROBLEMS 169

The spin orientation quantum number ms is similarly related to s:

Lzs = hms, ms = !s,!s + 1, . . . , s ! 1, s. (9.31)

The spin angular momentum for an elementary particle is absolutely con-served, i. e., it can never change. Thus, the value of s is an intrinsic prop-erty of a particle. The major di!erence between spin and orbital angularmomentum is that the spin quantum number can take on more values, i. e.,s = 0, 1/2, 1, 3/2, 2, 5/2, . . ..

Particles with integer spin values s = 0, 1, 2, . . . are called bosons afterthe Indian physicist Satyendra Nath Bose. Particles with half-integer spinvalues s = 1/2, 3/2, 5/2, . . . are called fermions after the Italian physicistEnrico Fermi. As we shall see later in the course, bosons and fermions playvery di!erent roles in the universe.

9.4 Problems

1. Suppose that a particle is represented by the wave function + = sin(kx!$t) + sin(!kx ! $t).

(a) Use trigonometry to simplify this wave function.

(b) Compute the probability of finding the particle by squaring thewave function.

(c) Explain what this result says about the time dependence of theprobability of finding the particle. Does this make sense?

2. Repeat the above problem for a particle represented by the wave func-tion + = exp[i(kx ! $t)] + exp[i(!kx ! $t)].

3. Determine if the wavefunction +(x) = exp(iCx2) is invariant underdisplacement in the sense that the displaced wave function di!ers fromthe original wave function by a multiplicative constant with an absolutevalue of one.

4. Just as invariance under the substitution x * x+D is associated withmomentum, invariance under the substitution x * !x is associatedwith a quantum mechanical variable called parity, denoted P . However,unlike momentum, which can take on any numerical value, parity can


take on only two possible values, ±1. The parity of a wave function+(x) is +1 if +(!x) = +(x), while the parity is !1 if +(!x) = !+(x).If +(x) satisfies neither of these conditions, then it has no definite valueof parity.

(a) What is the parity of + = sin(kx)? Of + = cos(kx)? The quantityk is a constant.

(b) Is +(x) = cos(kx) invariant under the substitution x = x + D forall possible values of D? Does this wave function have a definitevalue of the momentum?

(c) Show that a wave function with a definite value of the momentumdoes not have a definite value of parity. Are momentum and paritycompatible variables?

5. Realizing that cos(kx ! $t) can be written in terms of complex expo-nential functions, give a physical interpretation of the meaning of theabove cosine wave function. In particular, what are the possible valuesof the associated particle’s momentum and energy?

6. The time reversal operation T makes the substitution t * !t. Similarto parity, time reversal can only take on values ±1. Is symmetry of awave function under time reversal, i. e., +(!t) = +(t), consistent witha definite value of the energy? Hint: Any wave function correspondingto a definite value of energy E must have the form + = A exp(!iEt/h)where A is not a function of time t. (Why?)

7. The operation C takes the complex conjugate of the wave function, i.e., it makes the substitution i * !i. In modern quantum mechanicsthis corresponds to interchanging particles and antiparticles, and iscalled charge conjugation. What does the combined operation CPT doto a plane wave?

8. Make an energy level diagram for the case of a massless particle in abox.

9. Compare |$| for the ground state of a non-relativistic particle in abox of size a with #$ obtained from the uncertainty principle in thissituation. Hint: What should you take for #x?

9.4. PROBLEMS 171

λλ’

(x)]ψRe[

U(x)

E

K

Figure 9.7: Real part of the wave function +, corresponding to a fixed totalenergy E, occurring in a region of spatially variable potential energy U(x).Notice how the wavelength " changes as the kinetic energy K = E ! Uchanges.

10. Imagine that a billiard table has an infinitely high rim around it. Forthis problem assume that h = 1 kg m2 s!1.

(a) If the table is 1.5 m long and if the mass of a billiard ball isM = 0.5 kg, what is the billiard ball’s lowest or ground stateenergy? Hint: Even though the billiard table is two dimensional,treat this as a one-dimensional problem. Also, treat the problemnonrelativistically and ignore the contribution of the rest energyto the total energy.

(b) The energy required to lift the ball over a rim of height H againstgravity is U = MgH where g = 9.8 m s!2. What rim heightmakes the gravitational potential energy equal to the ground stateenergy of the billiard ball calculated above?

(c) If the rim is actually twice as high as calculated above but is only0.1 m thick, determine by what factor the wave function decreasesgoing through the rim.

11. The real part of the wave function of a particle with positive energy Epassing through a region of negative potential energy is shown in figure9.7.


(a) If the total energy is definitely E, what is the dependence of thiswave function on time?

(b) Is the wave function invariant under displacement in space in thiscase? Why or why not?

(c) Does this wave function correspond to a definite value of momen-tum? Why or why not?

(d) Is the momentum compatible with the energy in this case? Whyor why not?

12. Assuming again that h = 1 kg m2 s!1, what are the possible speedsof a toy train of mass 3 kg running around a circular track of radius0.8 m?

13. If a particle of zero mass sliding around a circular loop of radius Rcan take on angular momenta Lm = mh where m is an integer, whatare the possible kinetic energies of the particle? Hint: Remember thatL = $R.

Chapter 10

Dynamics of Multiple Particles

So far we have considered only the dynamics of a single particle subject to anexternally imposed potential energy. The particle has no way of influencingthis external agent. In the real world particles interact with each other. Inthis chapter we learn how this happens.

We first rewrite Newton’s second law in terms of momentum. This is use-ful in the subsequent consideration of Newton’s third law, which leads to theprinciple of the conservation of momentum. Collisions between particles andthe behavior of rockets and conveyor belts are then studied as applicationsof the conservation laws to more than one particle.

10.1 Momentum and Newton’s Second Law

Up to this point we have stated Newton’s second law in its conventional form,F = ma. However, in the non-relativistic case ma = mdu/dt = d(mu)/dt,so we can also write Newton’s second law as

F =dp

dt(Newton’s second law) (10.1)

where p = mu is the non-relativistic kinetic momentum. This form of New-ton’s second law is actually closer to Newton’s original statement of the law.It also has the advantage that it is correct even in the relativistic case whenthe relativistic definition of kinetic momentum, p = mu/(1 ! u2/c2)1/2, issubstituted.

173

174 CHAPTER 10. DYNAMICS OF MULTIPLE PARTICLES

A

B

C

Figure 10.1: Interactions between three particles, A, B, and C. A and B areconsidered to be part of the system defined by the dashed line.

10.2 Newton’s Third Law

Newton’s third law states that if particle A exerts a force F on particle B,then particle B exerts a force !F on particle A. Newton’s third law makesit possible to apply Newton’s second law to systems of particles withoutconsidering the detailed interactions between particles within the system.For instance, if we (arbitrarily) define the system in figure 10.1 to be theparticles A and B inside the box, then we can divide the forces acting onthese particles into internal and external parts,

FA = FA!internal + FA!external =dpA

dt, (10.2)

FB = FB!internal + FB!external =dpB

dt. (10.3)

Adding these equations together results in

FA!internal + FA!external + FB!internal + FB!external =d

dt(pA + pB). (10.4)

However, the internal interactions in this case are A acting on B and B actingon A. These forces are equal in magnitude but opposite in direction, so theycancel out, leaving us with the net force equal to the sum of the externalparts, Fnet = FA!external + FB!external. The external forces in figure 10.1

10.3. CONSERVATION OF MOMENTUM 175

are the force of C on A and the force of C on B. Defining the total kineticmomentum of the system as the sum of the A and B momenta, ptot = pA+pB,the above equation becomes

Fnet =dptot

dt, (10.5)

which looks just like Newton’s second law for a single particle, except thatit now applies to the system of particles (A and B in the present case) asa whole. This argument easily generalizes to any number of particles insideand outside the system. Thus, for instance, even though a soccer ball consistsof billions of atoms, we are sure that the forces between atoms within thesoccer ball cancel out, and the trajectory of the ball as a whole is determinedsolely by external forces such as gravity, wind drag, friction with the ground,and the kicks of soccer players.

Remember that for two forces to be a third law pair, they have to beacting on di!erent particles. Furthermore, if one member of the pair is theforce of particle A acting on particle B, then the other must be the force ofparticle B acting on particle A.

10.3 Conservation of Momentum

If all external forces on a system are zero, then equation (10.5) reduces to

ptot = const (isolated system). (10.6)

A system of particles with no external forces acting on it is called isolated.Newton’s third law thus tells us that the kinetic momentum of an isolatedsystem doesn’t change with time. This law is called the conservation ofmomentum.

10.4 Collisions

Let us now consider the situation in which two particles collide with eachother. There can be several outcomes to this collision, of which we will studytwo:

• The two particles collide elastically, in essence bouncing o! of eachother.


p1

p’1 p’2

p 2 p1

p’1 p’2

p2x x

ct ctm 1 = m2 m 1 > m2

Figure 10.2: One-dimensional elastic collisions of two particles in the centerof momentum frame as seen in spacetime diagrams.

• The two particles stick together, resulting in the production of a singleparticle, or a single particle breaks apart into two particles. These areinelastic processes.

In both of the above cases energy and momentum are conserved. Weassume that the forces acting between the particles are short range, so thatexcept in the instant of collision, we need not worry about potential energyor potential momentum — all energy is in the form of rest plus kinetic energyexcept in this short interval, and all momenta are kinetic momenta.

Because of the principle of relativity, we are free to consider collisions inany convenient reference frame. We can then transform the results to anyreference frame we please. Generally speaking, the most convenient referenceframe to consider is the one in which the total momentum of the two particlesis zero. In the non-relativistic limit this is called the center of mass referenceframe. However, the idea of center of mass is not useful in the relativisticcase, so we refer instead to this frame as the center of momentum frame. Forthe sake of simplicity we only consider collisions in one dimension.

10.4.1 Elastic Collisions

Suppose a particle with mass m1 and initial velocity u1 in the center ofmomentum frame collides elastically with another particle of mass m2 with

10.4. COLLISIONS 177

initial velocity u2. The momenta of the two particles are

p1 =m1u1

(1 ! u21/c2)1/2

p2 =m2u2

(1 ! u22/c2)1/2

. (10.7)

In the center of momentum frame we must have

p1 = !p2. (10.8)

Figure 10.2 shows what happens when these two particles collide. Thefirst particle acquires momentum p#1 while the second acquires momentump#2. The conservation of momentum tells us that the total momentum afterthe collision is the same as before the collision, namely zero, so

p#1 = !p#2. (10.9)

In the center of momentum frame we know that |p1| = |p2| and we knowthat the two momentum vectors point in opposite directions. Similarly, |p#1| =|p#2|. However, we as yet don’t know how p#1 is related to p1. Conservation ofenergy,

E1 + E2 = E #

1 + E #

2, (10.10)

gives us this information. Notice that if p#1 = !p1, then E #21 = p#21 c2 +m2

1c4 =

p21c

2 + m21c

4 = E21 . Assuming positive energies, we therefore have E #

1 = E1.If p#2 = !p2, then we can similiarly infer that E #

2 = E2. If these conditionsare satisfied, then so is equation (10.10). Therefore, a complete solution tothe problem is

p1 = !p#1 = !p2 = p#2 ' p (10.11)

and

E1 = E #

1 = (p2c2 + m21c

4)1/2 E2 = E #

2 = (p2c2 + m22c

4)1/2. (10.12)

In other words, the particles just exchange momenta.The left panel of figure 10.2 shows what happens in a collision when the

masses of the two colliding particles are equal. If m1 = m2, then the incomingand outgoing velocities of the two particles are the same, as indicated by theinverse slopes of the world lines. On the other hand, if m1 > m2, then thevelocity of particle 2 is greater than the velocity of particle 1, as is illustratedin the right panel of figure 10.2.

Suppose we wish to view the results of an elastic collision in a referenceframe in which particle 2 is initially stationary. All we have to do is to


m 1= m2

1 2 2x x

ctct

1

general case

Figure 10.3: Elastic collisions viewed from a reference frame in which oneparticle is initially stationary.

transform the velocities into a reference frame moving with the initial velocityof particle 2, as illustrated in figure 10.3. We do this by relativistically addingU = !u2 to each velocity. (Note that the velocity U of the moving frameis positive since u2 is negative.) Using the relativistic velocity translationformula, we find that

v1 =u1 + U

1 + u1U/c2v#

1 =u#

1 + U

1 + u#

1U/c2v#

2 =u#

2 + U

1 + u#

2U/c2(10.13)

where u1, u#

1, u2, and u#

2 indicate velocities in the original, center of momen-tum reference frame and v1, v#

1, etc., indicate velocities in the transformedframe.

In the special case where the masses of the two particles are equal to eachother, we have v1 = 2U/(1 + U2/c2), v#

1 = 0, and v#

2 = 2U/(1 + U2/c2) = v1.Thus, when the masses are equal, the particles simply exchange velocities.

If the velocities are nonrelativistic, then the simpler Galilean transforma-tion law v = u + U can be used in place of the relativistic equations invokedabove.

10.4.2 Inelastic Collisions

An inelastic collision is one in which the particles coming out of the collisionare not the same as the particles going into it. Inelastic collisions conserveboth total momentum and energy just as elastic collisions do. However,

10.4. COLLISIONS 179

xx

ctct3

21 3

2 1

Figure 10.4: Building blocks of inelastic collisions. In the left panel twoparticles collide to form a third particle. In the right panel a particle breaksup, forming two particles.

unlike elastic collisions, inelastic collisions generally do not conserve the totalkinetic energy of the particles, as some rest energy is generally created ordestroyed.

Figure 10.4 shows the fundamental building blocks of inelastic collisions.We can consider even the most complex inelastic collisions to be made up ofcomposites of only two processes, the creation of one particle from two, andthe disintegration of one particle into two.

Let us consider each of these in the center of momentum frame. In bothcases the single particle must be stationary in this frame since it carriesthe total momentum of the system, which has to be zero. By conservation ofmomentum, if particle 1 in the left panel of figure 10.4 has momentum p, thenthe momentum of particle 2 is !p. If the two particles have masses m1 andm2, then their energies are E1 = (p2c2 + m2

1c4)1/2 and E2 = (p2c2 + m2

2c4)1/2.

The energy of particle 3 is therefore E3 = E1 + E2, and since it is at rest, allof its energy is in the form of “mc2” or rest energy, and so the mass of thisparticle is

m3 = (E1 + E2)/c2 (center of momentum frame)

= (p2/c2 + m21)

1/2 + (p2/c2 + m22)

1/2

= m1[1 + p2/(m21c

2)]1/2 + m2[1 + p2/(m22c

2)]1/2. (10.14)

The last line in the above equation shows that m3 > m1 + m2 because itis in the form m1A + m2B where both A and B are greater than one. Thus,


rest energy is created in the amount #Erest = (m3 ! m1 ! m2)c2.Actually, it is easy to calculate the mass of particle 3 in the above case

from any reference frame as long as the momenta and energies of particles 1and 2 are known in this frame. By conservation of energy and momentum,E3 = E1 + E2 and p3 = p1 + p2. Furthermore, E2

3 = p23c

2 + m23c

4, so we cansolve for m3:

m3 = [(E1 + E2)2/c4 ! (p1 + p2) · (p1 + p2)/c

2]1/2 (any frame). (10.15)

The right panel of figure 10.4 shows the process of particle decay. Thisis just the inverse of the particle creation process, and all of the analysis wehave done for creation is valid for particle decay except that rest energy isconverted to kinetic energy rather than vice versa.

10.5 Rockets and Conveyor Belts

Normally when we define a system to which Newton’s second law is to beapplied, the system is closed in the sense that mass cannot enter or exit thesystem. However, sometimes it is convenient to work with open systems forwhich this is not true. The classic example is the rocket, where exhaust gasesleave the system, thus decreasing the mass of the rocket with time.

Open systems can be analyzed if momentum is considered to be a quantitywhich is accounted for much as money is accounted for in a bank account.The bank account can change in three ways; money can be deposited in theaccount, it can be withdrawn from the account, and the amount can growor shrink as a consequence of interest payments or fees. Analogously, theamount of momentum in a system can change as the result of mass enteringthe system, mass leaving the system, and forces acting on the system. Thetime rate of change of momentum in a system is therefore

dp

dt= F +

!

dp

dt

"

in

!

!

dp

dt

"

out

, (10.16)

where F is the net force on the system, (dp/dt)in is the momentum perunit time added by mass entering the system, and (dp/dt)out is the amountlost per unit time by mass exiting the system. In the non-relativistic case,(dp/dt)in = uin(dm/dt)in and (dp/dt)out = uout(dm/dt)out, where (dm/dt)in

is the mass entering the system per unit time with velocity uin and (dm/dt)out

is the mass per unit time exiting the system with velocity uout.

10.5. ROCKETS AND CONVEYOR BELTS 181

mR = -dm/dt

xV - u V

Figure 10.5: Rocket moving with velocity V while expelling gas at a rate Rwith velocity V ! ux.

For non-relativistic velocities, the momentum of the system can be writtenas p = mu so that

dp

dt=

dm

dtu + m

du

dt. (10.17)

To complete the analysis, we need an accounting of the mass entering andleaving the system:

dm

dt=

!

dm

dt

"

in

!

!

dm

dt

"

out

. (10.18)

Let us see how to apply this to a rocket for which all velocities are non-relativistic. As figure 10.5 indicates, a rocket spews out a stream of exhaustgas. The system is defined by the dashed box and includes the rocket and thepart of the exhaust gas inside the box. The reaction to the momentum carriedo! in this stream of gas is what causes the rocket to accelerate. We note that(dm/dt)in = 0 since no mass is entering the system, and (dm/dt)out = R, i.e., R is the rate at which mass is ejected by the rocket in the form of exhaustgas. The rocket is assumed to be moving to the right at speed V and the gasis ejected at a speed ux relative to the rocket, which means that its actualvelocity after ejection is V !ux. We call ux the exhaust velocity. Notice thatV ! ux may be either positive or negative, depending on how big V is.

Equating the mass of the rocket to the system mass, we find that R =!dm/dt. The momentum balance equation (10.16) becomes dp/dt = !(V !ux)R. The force on the rocket is actually zero, so the force term does notenter the momentum balance equation. This is non-intuitive, because we areused to acceleration being the result of a force. However, nothing, includingthe ejected gas, is actually pushing on the system, so we must indeed conclude


F

R = dm/dt

V

Figure 10.6: Sand is dumped on a conveyor belt and in turn is dumped o!the end of the belt.

that there is no force — all of the change in the system’s momentum arisesfrom the ejection of gas with the opposite momentum.1

Finally, we see that dp/dt = (dm/dt)V + m(dV/dt) = !RV + m(dV/dt).Equating this to the results of the momentum balance calculation gives us!RV + m(dV/dt) = !(V ! ux)R. Solving for the acceleration dV/dt resultsin

dV

dt=

uxR

m(rocket acceleration). (10.19)

Thus, the acceleration of the rocket depends on the exhaust velocity of theejected gas, the rate at which the gas is being ejected, and the mass of therocket.

Figure 10.6 illustrates another type of open system problem. A hopperdumps sand on a conveyor belt at a rate of R kilograms per second. Theconveyor belt is moving to the right at (non-relativistic) speed V and thesand is dumped o! at the end. What force F is needed to keep the conveyorbelt moving at a constant speed, assuming that the conveyor belt mechanismitself is frictionless? In this case (dm/dt)in = (dm/dt)out = R. Furthermore,since the system outlined by the dashed line is in a steady state, dp/dt = 0.

1The back pressure of the gas outside the system on the gas inside the system isnegligible once the gas exits the nozzle of the rocket engine. If we took the inside of thecombustion chamber to be part of the system boundary, the results would be di!erent,as the gas pressure there is non-negligible. At this point the gas is indeed exerting asignificant force on the rocket. However, though this viewpoint is conceptually simpler, itis computationally more di"cult, which is why we define the system as we do.

10.6. PROBLEMS 183

Mgblock

plate

b

p

N

N

Figure 10.7: Block of mass M subject to gravitational force Mg while restingon a plate. The force of the block on the plate is Np while the force of theplate on the block is Nb.

The key to understanding this problem is that the sand enters the systemwith zero horizontal velocity, but exits the system with the horizontal velocityof the conveyor belt, V . The momentum balance equation is thus

0 = F ! V R (10.20)

and the force isF = V R (force on conveyor belt). (10.21)

This force serves to accelerate the sand up to the velocity of the conveyorbelt.

10.6 Problems

1. Imagine a block of mass M resting on a plate under the influence ofgravity, as shown in figure 10.7.

(a) Determine the force of the plate on the block, Nb, and the forceof the block on the plate, Np.

(b) State which of the three forces, Mg, Nb, and Np, form a Newton’sthird law pair.

2. Repeat the previous problem assuming that the block and the plate arein an elevator accelerating upward with acceleration a.


FBFB

pusher boatbarge #2barge #1 v

Figure 10.8: Barges being pushed by a pusher boat on the Mississippi. Eachbarge experiences a drag force Fb.

3. Straighten out the misunderstanding of Newton’s third law implicit inthe question “If the force of the horse on the cart equals the force of thecart on the horse, why does anything ever go anywhere”? Examine inparticular the conditions under which the horse-cart system accelerates.

4. A pusher boat (mass M) on the Mississippi is pushing two barges (eachmass m) at a steady speed as shown in figure 10.8. Each barge issubject to a drag force by the water of FB. Consider only horizontalforce components in the following.

(a) What is the total horizontal force of the water on the barge-boatsystem? Explain.

(b) What is the direction and magnitude of the force of the pusherboat on barge 1? Explain.

5. A train with an engine of mass M and 2 freight cars, each of mass m,is accelerating to the right with acceleration a on a horizontal track.Assume that the two freight cars roll with negligible friction. Consideronly horizontal force components below.

(a) Find the direction and magnitude of the force of the rails on theengine and specify the system to which F = Ma is applied.

(b) Find the direction and magnitude of the force of the engine on thefirst car and specify the system to which F = Ma is applied.

(c) Find the direction and magnitude of the force of the first car onthe second car and specify the system to which F = Ma is applied.

(d) Find the direction and magnitude of the force of the second caron the first car and specify the law used to obtain this force.

10.6. PROBLEMS 185

#2 m #1 m engine Ma

Figure 10.9: An engine and two freight cars accelerating to the right.

mM

θ

Figure 10.10: A car and a trailer going down a hill.

6. A car and trailer are descending a hill as shown in figure 10.10. As-sume that the trailer rolls without friction and that air friction can beignored.

(a) Compute the force of the road on the car if the car-trailer systemshown in figure 10.10 is moving down the hill at constant speed.

(b) Compute the force of the trailer on the car in the above conditions.

(c) If the driver takes his foot o! the brake and lets the car coastfrictionlessly, recompute the force of the trailer on the car.

7. Consider a one-dimensional elastic collision between particles of massesm1 and m2. If particle 2 is initially stationary, what must the ratiom1/m2 be for the initial particle to rebound backwards along its initialtrack after the collision? (Do this problem non-relativistically.)

8. A stationary pion (mass M) decays into a muon (mass m < M) and aneutrino (massless).

(a) What is the (fully relativistic) momentum of the muon after thedecay?

(b) What is the energy of the neutrino?


u1

u 2

approachingplanet

receding fromplanet

V V

Figure 10.11: A space probe approaches a planet, curves around it, and headso! in the opposite direction.

9. In an elastic collision viewed in the center of momentum frame, theenergy of each particle is conserved individually. Is this true for thesame process viewed from a reference frame in which one of the particlesis initially stationary?

10. A space probe approaches a planet in the !x direction, curves aroundit under the influence of the planet’s powerful gravity (a conservativeforce) and recedes from the planet in the +x direction, as seen in figure10.11. The planet is moving in the +x direction at speed V , whilethe space probe is initially moving in the !x direction at speed u1.What is its speed u2 in the +x direction after this close approach to theplanet? Treat this problem nonrelativistically. Hint: First transform tothe center of mass frame in which the planet is essentially stationary.Work out the interaction between the probe and the planet in thisframe. Then transform back to the original reference frame.

11. Two asteroids, each with mass 1010 kg and initial speed 105 m s!1,collide head on. The whole mess congeals into one large mass. Howmuch rest mass is created?

12. Two equal objects, both with mass m, collide and stick together. Beforethe collision one mass is stationary and the other is moving at speedv. In the following, assume that velocities are fully relativistic.

(a) Compute the total momentum and energy (including rest energy)of the two masses before the collision.

(b) Compute the mass M of the combined system after the collision,taking the conversion of energy into mass into account.

10.6. PROBLEMS 187

scale

V, R

Figure 10.12: A bottle being filled with soft drink at a rate R. The liquidenters the bottle with velocity V .

13. Explain qualitatively why a fireman needs to push forward on a firehoseto keep it stationary. Hint: The water is flowing faster after it comesout of the nozzle of the hose than before.

14. Solve equation (10.19) for V as a function of m, assuming that V = 0and m = m0 at t = 0. Hint: Since R = !dm/dt, we have R/m =!d ln(m)/dt.

15. Bottles are filled with soft drink at a bottling plant as shown in figure10.12. The bottles sit on a scale which is used to determine when toshut o! the flow of soft drink. If the desired mass of the bottle plussoft drink after filling is M , what weight should the scale read whenthe bottle is full? The rate at which mass is being added to the bottleis R and its velocity entering the bottle is V .

16. An interstellar space probe has frontal area A, initial mass M0, andinitial velocity V0, which is non-relativistic. The tenuous gas betweenthe stars has mass density .. These gas molecules stick to the probewhen they hit it. Find the probe’s acceleration. Hint: In a frame ofreference in which the gas is stationary, does the momentum of thespace probe change with time? Does its mass?

17. A light beam with power J hits a plate which is oriented normally tothe beam. Compute the force required to hold the plate in place if


(a) the plate completely absorbs the light, and

(b) the plate completely reflects the light.

Hint: Photons are massless, so the momentum of a photon with energyE is E/c. Thus, the momentum per unit time hitting the plate is J/c.

18. Solve the rocket problem when the exhaust “gas” is actually a laserbeam of power J . Assume that the rocket moves at non-relativisticvelocities and that the decrease in mass due to the loss of energy in thelaser beam is negligible.

Chapter 11

Rotational Dynamics

We have already seen the quantum mechanical treatment of angular momen-tum and rotational dynamics. In this section we study these subjects in aclassical, non-relativistic context. We first define the concepts of torque andangular momentum in order to understand the orbital motion of a singleparticle. Next we examine two particles in arbitrary motion and learn howkinetic energy and angular momentum are partitioned between orbital andinternal components. Two particles fixed to the ends of a light rod consti-tute a dumbbell, which serves as a prototype for the rotation of rigid bodies.We then see how what we learned for two particles extends to an arbitrarynumber of particles. Finally, we explore the physics of structures in staticequilibrium.

Before we begin, we need to extend our knowledge of vectors to the crossproduct.

11.1 Math Tutorial — Cross Product

There are two ways to multiply two vectors together, the dot product andthe cross product. We have already studied the dot product of two vectors,which results in a scalar or single number.

The cross product of two vectors results in a third vector, and is writtensymbolically as follows:

C = A% B. (11.1)

As illustrated in figure 11.1, the cross product of two vectors is perpendic-ular to the plane defined by these vectors. However, this doesn’t tell us

189

190 CHAPTER 11. ROTATIONAL DYNAMICS

A

C

B

θ

Figure 11.1: Illustration of the cross product of two vectors A and B. Theresulting vector C is perpendicular to the plane defined by A and B.

whether the resulting vector in figure 11.1 points upward out of the plane ordownward. This ambiguity is resolved using the right-hand rule:

1. Point the uncurled fingers of your right hand along the direction of thefirst vector A.

2. Rotate your arm until you can curl your fingers in the direction of thesecond vector B.

3. Your stretched out thumb now points in the direction of the cross prod-uct vector C.

The magnitude of the cross product is given by

|C| = |A||B| sin(&), (11.2)

where |A| and |B| are the magnitudes of A and B, and & is the angle betweenthese two vectors. Note that the magnitude of the cross product is zerowhen the vectors are parallel or anti-parallel, and maximum when they areperpendicular. This contrasts with the dot product, which is maximum forparallel vectors and zero for perpendicular vectors.

Notice that the cross product does not commute, i. e., the order of thevectors is important. In particular, it is easy to show using the right-handrule that

A% B = !B % A. (11.3)

An alternate way to compute the cross product is most useful when thetwo vectors are expressed in terms of components, i. e., A = (Ax, Ay.Az)

11.2. TORQUE AND ANGULAR MOMENTUM 191

M

r p

F

O

Figure 11.2: A mass M located at r relative to the origin O has momentum p

and has a force F applied to it. By the right-hand rule the torque ! = r%F

points out of the page, while the angular momentum L = r % p points intothe page.

and B = (Bx, By, Bz):

Cx = AyBz ! AzBy

Cy = AzBx ! AxBz

Cz = AxBy ! AyBx. (11.4)

Notice that once you have the first of these equations, the other two can beobtained by cyclically permuting the indices, i. e., x * y, y * z, and z * x.This is useful as a memory aid.

11.2 Torque and Angular Momentum

Torque is the action of a force F on a mass M which induces it to revolveabout some point, called the origin. It is defined

! = r % F, (11.5)

where r is the position of the mass relative to the origin, as illustrated infigure 11.2.

Notice that the torque is zero in a number of circumstances. If the forcepoints directly toward or away from the origin, the cross product is zero,resulting in zero torque, even though the force is non-zero. Likewise, if


r = 0, the torque is zero. Thus, a force acting at the origin produces notorque. Both of these limits make sense intuitively, since neither induces themass to revolve around the origin.

The angular momentum of a mass M relative to a point O is

L = r % p, (11.6)

where p is the ordinary kinetic momentum of the mass.1 The angular mo-mentum is zero if the motion of the object is directly towards or away fromthe origin, or if it is located at the origin.

If we take the cross product of the position vector and Newton’s secondlaw, we obtain an equation that relates torque and angular momentum:

r % F = r %dp

dt=

d

dt(r% p) !

dr

dt% p. (11.7)

The second term on the right side of the above equation is zero because dr/dtequals the velocity of the mass, which is parallel to its momentum and thecross product of two parallel vectors is zero. This equation can therefore bewritten

! =dL

dt(Newton’s second law for rotation). (11.8)

It is the rotational version of Newton’s second law.For both torque and angular momentum the location of the origin is

arbitrary, and is generally chosen for maximum convenience. However, itis necessary to choose the same origin for both the torque and the angularmomentum.

For the case of a central force, i. e., one which acts along the line ofcenters between two objects (such as gravity), there often exists a particularlyconvenient choice of origin. Imagine a planet revolving around the sun, asillustrated in figure 11.3. If the origin is placed at the center of the sun (whichis assumed not to move under the influence of the planet’s gravity), then thetorque exerted on the planet by the sun’s gravity is zero, which means thatthe angular momentum of the planet about the center of the sun is constantin time. No other choice of origin would yield this convenient result.

1In the presence of a potential momentum we would have to distinguish between totaland kinetic momentum. This in turn would lead to a distinction between total and kineticangular momentum. We will assume that no potential momentum exists here, so that thisdistinction need not be made.

11.2. TORQUE AND ANGULAR MOMENTUM 193

O F

p

Figure 11.3: A convenient choice of origin for a planet (right-hand sphere)revolving around the sun is simply the center of the sun. In this case thetorque of the sun’s gravitational force on the planet is zero.

F12

F21M1

M2O

Figure 11.4: Scenario for the non-conservation of angular momentum.

We already know about two fundamental conservation laws — those ofenergy and linear momentum. We believe that angular momentum is simi-larly conserved in isolated systems. In other words, particles can exchangeangular momentum between themselves, but the vector sum of the angularmomentum of all the particles in a system isolated from outside influencesmust remain constant.

Conservation of angular momentum is not an automatic consequence ofthe conservation of linear momentum, even though the governing equation(11.8) for angular momentum is derived from Newton’s second law. As anexample, figure 11.4 shows a hypothetical situation in which the force of M1

on M2 is equal in magnitude but opposite in sign to the force of M2 on M1, i.


r ’1

r ’2d1

d2

M1

M2

center of mass O

r

r

Rcm

1

2

Figure 11.5: Two particles of mass M1 and M2 with M2 > M1.

e., Newton’s third law holds, and the sum of the momenta of the two massesis conserved. However, because the forces are non-central, the masses revolvemore and more rapidly with time about the origin, and angular momentum isnot conserved. This scenario is impossible if angular momentum is conserved.

11.3 Two Particles

Suppose we wish to apply Newton’s second law to two particles consideredtogether as a single system. As we showed previously, only external forcesact on the total momentum, ptotal = p1 + p2, of the two particles:

Fexternal =dptotal

dt. (11.9)

Let’s write the total non-relativistic momentum of the two particles in aspecial way:

ptotal = M1v1 + M2v2 = Mtotal

'

M1v1 + M2v2

Mtotal

(

' MtotalVcm, (11.10)

where Mtotal = M1 + M2. The quantity Vcm is the velocity of the center ofmass and can be expressed as the time derivative of the position of the centerof mass, Rcm,

Vcm =dRcm

dt, (11.11)

where

Rcm =M1r1 + M2r2

Mtotal. (11.12)

11.3. TWO PARTICLES 195

We now see how the kinetic energy and the angular momentum of thetwo particles may be split into two parts, one having to do with the motionof the center of mass of the two particles, the other having to do with themotion of the two particles relative their center of mass. Figure 11.5 showsgraphically how the vectors r#1 = r1 ! Rcm and r#2 = r2 ! Rcm are defined.These vectors represent the positions of the two particles relative to the centerof mass. Substitution into equation (11.12) shows that M1r

#

1 + M2r#

2 = 0.This leads to the conclusion that M1d1 = M2d2 in figure 11.5. We also definethe velocity of each mass relative to the center of mass as v#

1 = dr#1/dt andv#

2 = dr#2/dt, and we therefore have M1v#

1 + M2v#

2 = 0.The total kinetic energy is just the sum of the kinetic energies of the two

particles, K = M1v21/2 + M2v2

2/2, where v1 and v2 are the magnitudes of thecorresponding velocity vectors. Substitution of v1 = Vcm + v#

1 etc., into thekinetic energy formula and rearranging yields

Ktotal = Ktrans + Kintern = [MtotalV2cm/2] + [M1v

#21 /2 + M2v

#22 /2]. (11.13)

Terms like Vcm · v#

1 cancel out because M1v#

1 + M2v#

2 = 0.The first term on the right side of equation (11.13) in square brackets

is the kinetic energy the two particles would have if all of the mass wereconcentrated at the center of mass. The second term is the kinetic energy itwould have if it were observed from a reference frame in which the center ofmass is stationary. The first is called the translational kinetic energy of thesystem while the second is called the internal kinetic energy.

The angular momentum of the two particles is just the sum of the angularmomenta of the two masses: Ltotal = M1r1 % v1 + M2r2 % v2. By reasoningsimilar to the case of kinetic energy, we can rewrite this as

Ltotal = Lorb +Lspin = [MtotalRcm%Vcm]+ [M1r#

1%v#

1 +M2r#

2%v#

2]. (11.14)

The first term in square brackets on the right is called the orbital angularmomentum while the second term is called the spin angular momentum. Theformer is the angular momentum the system would have if all the mass wereconcentrated at the center of mass, while the latter is the angular momentumof motion about the center of mass.

Interestingly, the idea of center of mass and the corresponding split ofkinetic energy and angular momentum into orbital and spin parts has nouseful relativistic generalization. This is due to the factor of % ' (1 !


v1

v2

d1

M1M2

d2

ω

Figure 11.6: Definition sketch for the rotating dumbbell attached to an axlelabeled $. The axle attaches to the crossbar at the center of mass.

v2/c2)!1/2 in the relativistic definition of momentum, which means that

dmx

dt(= p (relativistic case). (11.15)

11.4 The Uneven Dumbbell

So far we have put no restrictions on the movements of the two particles.An interesting special case occurs when the particles are connected by alightweight, rigid rod, giving us a dumbbell. In order to further simplifythings, we assume that the rod is connected rigidly to a fixed axle at thecenter of mass of the two particles, as shown in figure 11.6. The massesconstituting the ends of the dumbbell are therefore free to revolve in circlesabout the axle, but they are prevented from executing any other motion.The key e!ect of this constraint is that both masses rotate about the axlewith the same angular frequency $.

If the particles are respectively distances d1 and d2 from the axle, thentheir speeds are v1 = d1$ and v2 = d2$. Thus the kinetic energy of therotating dumbbell is

Kintern =1

2M1v

21 +

1

2M2v

22 =

1

2I$2 (fixed axle), (11.16)

where I = M1d21 + M2d2

2 is called the moment of inertia. Similarly, themagnitude of the spin angular momentum, which is a vector parallel to the

11.5. MANY PARTICLES 197

axle, is

Lspin = M1d1v1 + M2d2v2 = I$ (fixed axle). (11.17)

Finally, Newton’s second law for rotation becomes

) =dLspin

dt=

dI$

dt= I

d$

dt(fixed axle), (11.18)

where ) is the component of torque along the rotation axis.Note that the rightmost expression in equation (11.18) assumes that I is

constant, which only is true if d1 and d2 are constant – i. e., the dumbbellmust truly be rigid.

11.5 Many Particles

The generalization from two particles to many particles is quite easy in prin-ciple. If a subscripted i indicates the value of a quantity for the ith particle,then the center of mass is given by

Rcm =1

Mtotal

$

i

Miri (11.19)

where

Mtotal =$

i

Mi. (11.20)

Furthermore, if we define r#i = ri ! Rcm, etc., then the kinetic energy is just

Ktotal = MtotalV2cm/2 +

$

i

Miv#2i /2 (11.21)

and the angular momentum is

Ltotal = MtotalRcm % Vcm +$

i

Mir#

i % v#

i. (11.22)

In other words, both the kinetic energy and the angular momentum can beseparated into a part due to the overall motion of the system plus a part dueto motions of system components relative to the center of mass, just as forthe case of the dumbbell.


11.6 Rigid Bodies

For a rigid body rotating about a fixed axle, the moment of inertia is

I =$

i

Mid2i , (11.23)

where di is the perpendicular distance of the ith particle from the axle. Equa-tions (11.16)-(11.18) are valid for a rigid body consisting of many particles.Furthermore, the moment of inertia is constant in this case, so it can betaken out of the time derivative:

) =dI$

dt= I

d$

dt= I' (fixed axle, constant I). (11.24)

The quantity ' = d$/dt is called the angular acceleration.The sum in the equation for the moment of inertia can be converted to

an integral for a continuous distribution of mass. We shall not pursue thishere, but simply quote the results for a number of solid objects of uniformdensity:

• For rotation of a sphere of mass M and radius R about an axis piercingits center: I = 2MR2/5.

• For rotation of a cylinder of mass M and radius R about its axis ofsymmetry: I = MR2/2.

• For rotation of a thin rod of mass M and length L about an axisperpendicular to the rod passing through its center: I = ML2/12.

• For rotation of an annulus of mass M , inner radius Ra, and outer radiusRb about its axis of symmetry: I = M(R2

a + R2b)/2.

11.7 Statics

If a rigid body is initially at rest, it will remain at rest if and only if thesum of all the forces and the sum of all the torques acting on the body arezero. As an example, a mass balance with arms of di!ering length is shownin figure 11.7. The balance beam is subject to three forces pointing upwardor downward, the tension T in the string from which the beam is suspendedand the weights M1g and M2g exerted on the beam by the two suspended

11.7. STATICS 199

M1g M2g

MM1 2

d1 d2

T

asymmetric balancebeampivot point for

torque calculation

Figure 11.7: Asymmetric mass balance. We assume that the balance beamis massless.

masses. The parameter g is the local gravitational field and the balancebeam itself is assumed to have negligible mass. Taking upward as positive,the force condition for static equilibrium is

T ! M1g ! M2g = 0 (zero net force). (11.25)

Defining a counterclockwise torque to be positive, the torque balance com-puted about the pivot point in figure 11.7 is

) = M1gd1 ! M2gd2 = 0 (zero torque), (11.26)

where d1 and d2 are the lengths of the beam arms.The first of the above equations shows that the tension in the string must

beT = (M1 + M2)g, (11.27)

while the second shows thatM1

M2

=d2

d1

. (11.28)

Thus, the tension in the string is just equal to the weight of the massesattached to the balance beam, while the ratio of the two masses equals theinverse ratio of the associated beam arm lengths.


RR’ vv’

Figure 11.8: Trajectory of a mass on a frictionless table attached to a stringwhich passes through a hole in the table. The string is drawing the mass in.

11.8 Problems

1. Show using the component form of the cross product given by equation(11.4) that A %B = !B % A.

2. A mass M is sliding on a frictionless table, but is attached to a stringwhich passes through a hole in the center of the table as shown in figure11.8. The string is gradually drawn in so the mass traces out a spiralpattern as shown in figure 11.8. The initial distance of the mass fromthe hole in the table is R and its initial tangential velocity is v. Afterthe string is drawn in, the mass is a distance R# from the hole and itstangential velocity is v#.

(a) Given R, v, and R#, find v#.

(b) Compute the change in the kinetic energy of the mass in goingfrom radius R to radius R#.

(c) If the above change is non-zero, determine where the extra energycame from.

3. A car of mass 1000 kg is heading north on a road at 30 m s!1 whichpasses 2 km east of the center of town.

(a) Compute the angular momentum of the car about the center oftown when the car is directly east of the town.

(b) Compute the angular momentum of the car about the center oftown when it is 3 km north of the above point.

11.8. PROBLEMS 201

F

g M

a

b

fixed axle

Figure 11.9: A crank on a fixed axle turns a drum, thus winding the ropearound the drum and raising the mass.

4. The apparatus illustrated in figure 11.9 is used to raise a bucket of massM out of a well.

(a) What force F must be exerted to keep the bucket from fallingback into the well?

(b) If the bucket is slowly raised a distance d, what work is done onthe bucket by the rope attached to it?

(c) What work is done by the force F on the handle in the abovecase?

5. Derive equations (11.13) and (11.14).

6. A mass M is held up by the structure shown in figure 11.10. Thesupport beam has negligible mass. Find the tension T in the diagonalwire. Hint: Compute the net torque on the support beam about pointA due to the tension T and the weight of the mass M .

7. A system consists of two stars, one of mass M moving with velocityv1 = (0, v, 0) at position r1 = (d, 0, 0), the other of mass 2M with zerovelocity at the origin.

(a) Find the center of mass position and velocity of the system of twostars.

(b) Find the spin angular momentum of the system.


M

θ

dg

support beamTA

Figure 11.10: A mass is supported by the tension in the diagonal wire. Thesupport beam is free to pivot at point A.

M

T T21

2d d

d

Figure 11.11: A mass is supported by two strings.

11.8. PROBLEMS 203

F

B

A

M

LD

θ

Figure 11.12: A ladder leaning against a wall is held in place by friction withthe floor.

(c) Find the internal kinetic energy of the system.

8. A solid disk is rolling down a ramp tilted an angle & from the horizontal.Compute the acceleration of the disk down the ramp and compare itwith the acceleration of a block sliding down the ramp without friction.

9. A mass M is suspended from the ceiling by two strings as shown infigure 11.11. Find the tensions in the strings.

10. A man of mass M is a distance D up a ladder of length L which makesan angle & with respect to the vertical wall as shown in figure 11.12.Take the mass of the ladder to be negligible. Find the force F neededto keep the ladder from sliding if the wall and floor are frictionless andtherefore can only exert normal forces A and B on the ladder.

Chapter 12

Harmonic Oscillator

Figure 12.1 illustrates the prototypical harmonic oscillator, the mass-springsystem. A mass M is attached to one end of a spring. The other end ofthe spring is attached to something rigid such as a wall. The spring exerts arestoring force F = !kx on the mass when it is stretched by an amount x.This is called Hooke’s law and k is called the spring constant.

12.1 Energy Analysis

The potential energy of the mass-spring system is

U(x) = kx2/2 (12.1)

since !d(kx2/2) = !kx is the Hooke’s law force. This is shown in figure 12.2.Since a potential energy exists, the total energy E = K + U is conserved, i.

k

M

xF = - kx

Figure 12.1: Illustration of a mass-spring system.

205

206 CHAPTER 12. HARMONIC OSCILLATOR

U(x)

E

U

K

x

turning points

Figure 12.2: Potential, kinetic, and total energy of a harmonic oscillatorplotted as a function of spring displacement x.

e., is constant in time. If the total energy is known, this provides a usefultool for determining how the kinetic energy varies with the position x of themass M : K(x) = E ! U(x). Since the kinetic energy is expressed (non-relativistically) in terms of the velocity u as K = Mu2/2, the velocity at anypoint on the graph in figure 12.2 is

u = ±

!

2(E ! U)

M

"1/2

. (12.2)

Given all this, it is fairly evident how the mass moves. From Hooke’slaw, the mass is always accelerating toward the equilibrium position, x = 0.However, at any point the velocity can be either to the left or the right. Atthe points where U(x) = E, the kinetic energy is zero. This occurs at theturning points

xTP = ±'

2E

k

(1/2

. (12.3)

If the mass is moving to the left, it slows down as it approaches the leftturning point. It stops when it reaches this point and begins to move to theright. It accelerates until it passes the equilibrium position and then begins todecelerate, stopping at the right turning point, accelerating toward the left,etc. The mass thus oscillates between the left and right turning points. (Notethat equations (12.2) and (12.3) are only true for the harmonic oscillator.)

How does the period of the oscillation depend on the total energy of thesystem? Notice that from equation (12.2) the maximum speed of the mass

12.2. ANALYSIS USING NEWTON’S LAWS 207

(i. e., the speed at x = 0) is equal to umax = (2E/M)1/2. The average speedmust be some fraction of this maximum value. Let us guess here that it ishalf the maximum speed:

uaverage #umax

2=

'

E

2M

(1/2

(approximate). (12.4)

However, the distance d the mass has to travel for one full oscillation istwice the distance between turning points, or d = 4(2E/k)1/2. Therefore,the period of oscillation must be approximately

T #d

uaverage= 4

'

2E

k

(1/2 '

2M

E

(1/2

= 8'

M

k

(1/2

(approximate). (12.5)

12.2 Analysis Using Newton’s Laws

The acceleration of the mass at any time is given by Newton’s second law:

a =d2x

dt2=

F

M= !

kx

M. (12.6)

An equation of this type is known as a di!erential equation since it involvesa derivative of the dependent variable x. Equations of this type are generallymore di"cult to solve than algebraic equations, as there are no universaltechniques for solving all forms of such equations. In fact, it is fair to saythat the solutions of most di!erential equations were originally obtained byguessing!

We already have the basis on which to make an intelligent guess for thesolution to equation (12.6) since we know that the mass oscillates back andforth with a period that is independent of the amplitude of the oscillation. Afunction which might fill the bill is the sine function. Let us try substitutingx = sin($t), where $ is a constant, into this equation. The second derivativeof x with respect to t is !$2 sin($t), so performing this substitution resultsin

!$2 sin($t) = !k

Msin($t). (12.7)

Notice that the sine function cancels out, leaving us with !$2 = !k/M . Theguess thus works if we set

$ =

!

k

M

"1/2

. (12.8)


d = d sin ω0 F t

k

M

xF = - k (x - d)

Figure 12.3: Illustration of a forced mass-spring oscillator. The left end ofthe spring is wiggled back and forth with an angular frequency $F and amaximum amplitude d0.

The constant $ is the angular oscillation frequency for the oscillator,from which we infer the period of oscillation to be T = 2!(M/k)1/2. Thisagrees with the earlier approximate result of equation (12.5), except that theapproximation has a numerical factor of 8 rather than 2! # 6. Thus, theearlier guess is only o! by about 30%!

It is easy to show that x = A sin($t) is also a solution of equation (12.6),where A is any constant and $ = (k/M)1/2. This confirms that the oscillationfrequency and period are independent of amplitude. Furthermore, the cosinefunction is equally valid as a solution: x = B cos($t), where B is anotherconstant. In fact, the most general possible solution is just a combination ofthese two, i. e.,

x = A sin($t) + B cos($t). (12.9)

The values of A and B depend on the position and velocity of the mass attime t = 0.

12.3 Forced Oscillator

If we wiggle the left end of the spring by the amount d = d0 sin($F t), as infigure 12.3, rather than rigidly fixing it as in figure 12.1, we have a forcedharmonic oscillator. The constant d0 is the amplitude of the imposed wigglingmotion. The forcing frequency $F is not necessarily equal to the natural orresonant frequency $ = (k/M)1/2 of the mass-spring system. Very di!erentbehavior occurs depending on whether $F is less than, equal to, or greaterthan $.

12.3. FORCED OSCILLATOR 209

ω F/ω

x0 /d 0

-1

-2

1.5 2.0

2

1

00.5

Figure 12.4: Plot of the ratio of response to forcing vs. the ratio of forced tofree oscillator frequency for the mass-spring system.

Given the above wiggling, the force of the spring on the mass becomesF = !k(x ! d) = !k[x ! d0 sin($F t)] since the length of the spring is thedi!erence between the positions of the left and right ends. Proceeding as forthe unforced mass-spring system, we arrive at the di!erential equation

d2x

dt2+

kx

M=

kd0

Msin($F t). (12.10)

The solution to this equation turns out to be the sum of a forced part inwhich x - sin($F t) and a free part which is the same as the solution to theunforced equation (12.9). We are primarily interested in the forced part ofthe solution, so let us set x = x0 sin($F t) and substitute this into equation(12.10):

!$2F x0 sin($F t) +

kx0

Msin($F t) =

kd0

Msin($F t). (12.11)

Again the sine factor cancels and we are left with an algebraic equation forx0, the amplitude of the oscillatory motion of the mass.

Solving for the ratio of the oscillation amplitude of the mass to the am-plitude of the wiggling motion, x0/d0, we find

x0

d0

=1

1 ! $2F/$2

, (12.12)


where we have recognized that k/M = $2, the square of the frequency of thefree oscillation. This function is plotted in figure 12.4.

Notice that if $F < $, the motion of the mass is in phase with the wig-gling motion and the amplitude of the mass oscillation is greater than theamplitude of the wiggling. As the forcing frequency approaches the natu-ral frequency of the oscillator, the response of the mass grows in amplitude.When the forcing is at the resonant frequency, the response is technicallyinfinite, though practical limits on the amplitude of the oscillation will in-tervene in this case — for instance, the spring cannot stretch or shrink aninfinite amount. In many cases friction will act to limit the response of themass to forcing near the resonant frequency. When the forcing frequency isgreater than the natural frequency, the mass actually moves in the oppositedirection of the wiggling motion — i. e., the response is out of phase withthe forcing. The amplitude of the response decreases as the forcing frequencyincreases above the resonant frequency.

Forced and free harmonic oscillators form an important part of manyphysical systems. For instance, any elastic material body such as a bridgeor an airplane wing has harmonic oscillatory modes. A common engineeringproblem is to ensure that such modes are damped by friction or some otherphysical mechanism when there is a possibility of exitation of these modesby naturally occurring processes. A number of disasters can be traced to afailure to properly account for oscillatory forcing in engineered structures.

12.4 Quantum Mechanical Harmonic Oscilla-

tor

The quantum mechanical harmonic oscillator shares the characteristic ofother quantum mechanical bound state problems in that the total energycan take on only discrete values. Calculation of these values is too di"-cult for this course, but the problem is su"ciently important to warrantreporting the results here. The energies accessible to a quantum mechanicalmass-spring system are given by the formula

En = (n + 1/2)h(k/M)1/2, n = 0, 1, 2, . . . . (12.13)

In other words, the energy di!erence between successive quantum mechan-ical energy levels in this case is constant and equals the classical resonantfrequency for the oscillator, $ = (k/M)1/2.

12.5. PROBLEMS 211

L

M

θ

x

Figure 12.5: The pendulum as a harmonic oscillator.

12.5 Problems

1. Consider the pendulum in figure 12.5. The mass M moves along an arcwith x denoting the distance along the arc from the equilibrium point.

(a) Find the component of the gravitational force tangent to the arc(and thus in the direction of motion of the mass) as a functionof the angle &. Use the small angle approximation on sin(&) tosimplify this answer.

(b) Get the force in terms of x rather than &. (Recall that & = x/L.)

(c) Use Newton’s second law for motion in the x direction (i. e., alongthe arc followed by the mass) to get the equation of motion forthe mass.

(d) Solve the equation of motion using the solution to the mass-springproblem as a guide.

2. An oscillator (non-harmonic) has the potential energy function U(x) =Cx4, where C is a constant. How does the oscillation frequency dependon energy? Explain your reasoning.

3. A massless particle is confined to a box of length a. (Think of a photonbetween two mirrors.) Treating the particle classically, compute the


period of one round trip from one end of the box to the other and backagain. From this compute an angular frequency for the oscillation ofthis particle in the box. Does this frequency depend on the particle’senergy?

4. Compute the ground state energy Eground of a massless particle in abox of length a using quantum mechanics. Compare Eground/h withthe angular frequency computed in the previous problem.

Appendix A

Constants

This appendix lists various useful constants.

A.1 Constants of Nature

Symbol Value Meaning

h 6.63 % 10!34 J s Planck’s constanth 1.06 % 10!34 J s h/(2!)c 3 % 108 m s!1 speed of lightG 6.67 % 10!11 m3 s!2 kg!1 universal gravitational constantkB 1.38 % 10!23 J K!1 Boltzmann’s constant/ 5.67 % 10!8 W m!2 K!4 Stefan-Boltzmann constantK 3.67 % 1011 s!1 K!1 thermal frequency constant*0 8.85 % 10!12 C2 N!1 m!2 permittivity of free spaceµ0 4! % 10!7 N s2 C!2 permeability of free space (= 1/(*0c2)).

A.2 Properties of Stable Particles


e 1.60 % 10!19 C fundamental unit of chargeme 9.11 % 10!31 kg = 0.511 MeV mass of electronmp 1.672648% 10!27 kg = 938.280 MeV mass of protonmn 1.674954% 10!27 kg = 939.573 MeV mass of neutron

213

214 APPENDIX A. CONSTANTS

A.3 Properties of Solar System Objects


Me 5.98 % 1024 kg mass of earthMm 7.36 % 1022 kg mass of moonMs 1.99 % 1030 kg mass of sunRe 6.37 % 106 m radius of earthRm 1.74 % 106 m radius of moonRs 6.96 % 108 m radius of sunDm 3.82 % 108 m earth-moon distanceDs 1.50 % 1011 m earth-sun distanceg 9.81 m s!2 earth’s surface gravity

A.4 Miscellaneous Conversions

1 lb = 4.448 N1 ft = 0.3048 m1 mph = 0.4470 m s!1

1 eV = 1.60 % 10!19 J

Appendix B

GNU Free Documentation

License

Version 1.1, March 2000

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216 APPENDIX B. GNU FREE DOCUMENTATION LICENSE

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Appendix C

History

• Prehistory: The text was developed to this stage over a period of about5 years as course notes to Physics 131/132 at New Mexico Tech byDavid J. Raymond, with input from Alan M. Blyth. The course wastaught by Raymond and Blyth and by David J. Westpfahl at NewMexico Tech.

• 14 May 2001: First “copylefted” version of this text.

• 14 May 2003: Numerous small changes and corrections were made.

• 25 June 2004: More small corrections to volume one.

• 14 December 2004: More small corrections to volume one.

• 9 May 2005: More small corrections to volume two.

• 7 April 2006: More small corrections to both volumes.

• 7 July 2006: Finish volume 2 corrections for the coming year.

• 2 July 2008: Finish volume 1 corrections for fall 2008.

•

225

A Radically Modern Approach toIntroductory Physics — Volume 2

David J. RaymondPhysics DepartmentNew Mexico Tech

Socorro, NM 87801

January 4, 2008

ii

Copyright c!1998, 2000, 2001, 2003, 2004,2006 David J. Raymond

Permission is granted to copy, distribute and/or modify this document underthe terms of the GNU Free Documentation License, Version 1.1 or any laterversion published by the Free Software Foundation; with no Invariant Sec-tions, no Front-Cover Texts and no Back-Cover Texts. A copy of the licenseis included in the section entitled ”GNU Free Documentation License”.

Contents

Preface to April 2006 Edition ix

13 Newton’s Law of Gravitation 223

13.1 The Law of Gravitation . . . . . . . . . . . . . . . . . . . . . 22313.2 Gravitational Field . . . . . . . . . . . . . . . . . . . . . . . . 22413.3 Gravitational Flux . . . . . . . . . . . . . . . . . . . . . . . . 22513.4 Flux from Multiple Masses . . . . . . . . . . . . . . . . . . . . 22813.5 E!ects of Relativity . . . . . . . . . . . . . . . . . . . . . . . . 22913.6 Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 23013.7 Use of Conservation Laws . . . . . . . . . . . . . . . . . . . . 23213.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

14 Forces in Relativity 239

14.1 Potential Momentum . . . . . . . . . . . . . . . . . . . . . . . 23914.2 Aharonov-Bohm E!ect . . . . . . . . . . . . . . . . . . . . . . 24114.3 Forces from Potential Momentum . . . . . . . . . . . . . . . . 243

14.3.1 Refraction E!ect . . . . . . . . . . . . . . . . . . . . . 24314.3.2 Time-Varying Potential Momentum . . . . . . . . . . . 245

14.4 Lorentz Condition . . . . . . . . . . . . . . . . . . . . . . . . . 24614.5 Gauge Theories and Other Theories . . . . . . . . . . . . . . . 24614.6 Conservation of Four-Momentum Again . . . . . . . . . . . . . 24714.7 Virtual Particles . . . . . . . . . . . . . . . . . . . . . . . . . 24814.8 Virtual Particles and Gauge Theory . . . . . . . . . . . . . . . 25014.9 Negative Energies and Antiparticles . . . . . . . . . . . . . . . 25114.10Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

15 Electromagnetic Forces 259

15.1 Electromagnetic Four-Potential . . . . . . . . . . . . . . . . . 259

iii

iv CONTENTS

15.2 Electric and Magnetic Fields and Forces . . . . . . . . . . . . 26015.3 A Note on Units . . . . . . . . . . . . . . . . . . . . . . . . . 26015.4 Charged Particle Motion . . . . . . . . . . . . . . . . . . . . . 261

15.4.1 Particle in Constant Electric Field . . . . . . . . . . . . 26115.4.2 Particle in Conservative Electric Field . . . . . . . . . 26115.4.3 Torque on an Electric Dipole . . . . . . . . . . . . . . . 26215.4.4 Particle in Constant Magnetic Field . . . . . . . . . . . 26415.4.5 Crossed Electric and Magnetic Fields . . . . . . . . . . 265

15.5 Forces on Currents in Conductors . . . . . . . . . . . . . . . . 26615.6 Torque on a Magnetic Dipole and Electric Motors . . . . . . . 26815.7 Electric Generators and Faraday’s Law . . . . . . . . . . . . . 26915.8 EMF and Scalar Potential . . . . . . . . . . . . . . . . . . . . 27215.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273

16 Generation of Electromagnetic Fields 279

16.1 Coulomb’s Law and the Electric Field . . . . . . . . . . . . . . 27916.2 Gauss’s Law for Electricity . . . . . . . . . . . . . . . . . . . . 280

16.2.1 Sheet of Charge . . . . . . . . . . . . . . . . . . . . . . 28116.2.2 Line of Charge . . . . . . . . . . . . . . . . . . . . . . 282

16.3 Gauss’s Law for Magnetism . . . . . . . . . . . . . . . . . . . 28316.4 Coulomb’s Law and Relativity . . . . . . . . . . . . . . . . . . 28416.5 Moving Charge and Magnetic Fields . . . . . . . . . . . . . . 284

16.5.1 Moving Line of Charge . . . . . . . . . . . . . . . . . . 28516.5.2 Moving Sheet of Charge . . . . . . . . . . . . . . . . . 288

16.6 Electromagnetic Radiation . . . . . . . . . . . . . . . . . . . . 28916.7 The Lorentz Condition . . . . . . . . . . . . . . . . . . . . . . 29216.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

17 Capacitors, Inductors, and Resistors 297

17.1 The Capacitor and Ampere’s Law . . . . . . . . . . . . . . . . 29717.1.1 The Capacitor . . . . . . . . . . . . . . . . . . . . . . . 29717.1.2 Circulation of a Vector Field . . . . . . . . . . . . . . . 30017.1.3 Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . 301

17.2 Magnetic Induction and Inductors . . . . . . . . . . . . . . . . 30317.3 Resistance and Resistors . . . . . . . . . . . . . . . . . . . . . 30617.4 Energy of Electric and Magnetic Fields . . . . . . . . . . . . . 30717.5 Kirchho!’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . 30917.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

CONTENTS v

18 Measuring the Very Small 315

18.1 Continuous Matter or Atoms? . . . . . . . . . . . . . . . . . . 31518.2 The Ring Around the Moon . . . . . . . . . . . . . . . . . . . 31818.3 The Geiger-Marsden Experiment . . . . . . . . . . . . . . . . 32018.4 Cosmic Rays and Accelerator Experiments . . . . . . . . . . . 322

18.4.1 Early Cosmic Ray Results . . . . . . . . . . . . . . . . 32218.4.2 Particle Accelerators . . . . . . . . . . . . . . . . . . . 32318.4.3 Size and Structure of the Nucleus . . . . . . . . . . . . 32318.4.4 Deep Inelastic Scattering of Electrons from Protons . . 32618.4.5 Storage Rings and Colliders . . . . . . . . . . . . . . . 32618.4.6 Proton-Antiproton Collisions . . . . . . . . . . . . . . . 32818.4.7 Electron-Positron Collisions . . . . . . . . . . . . . . . 329

18.5 Commentary . . . . . . . . . . . . . . . . . . . . . . . . . . . 32918.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330

19 Atoms 333

19.1 Fermions and Bosons . . . . . . . . . . . . . . . . . . . . . . . 33319.1.1 Review of Angular Momentum in Quantum Mechanics 33319.1.2 Two Particle Wave Functions . . . . . . . . . . . . . . 334

19.2 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . 33619.3 The Periodic Table of the Elements . . . . . . . . . . . . . . . 33819.4 Atomic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . 34019.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342

20 The Standard Model 345

20.1 Quarks and Leptons . . . . . . . . . . . . . . . . . . . . . . . 34520.2 Quantum Chromodynamics . . . . . . . . . . . . . . . . . . . 34720.3 The Electroweak Theory . . . . . . . . . . . . . . . . . . . . . 35220.4 Grand Unification? . . . . . . . . . . . . . . . . . . . . . . . . 35420.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356

21 Atomic Nuclei 359

21.1 Molecules — an Analogy . . . . . . . . . . . . . . . . . . . . . 35921.2 Nuclear Binding Energies . . . . . . . . . . . . . . . . . . . . . 36021.3 Radioactivity . . . . . . . . . . . . . . . . . . . . . . . . . . . 36521.4 Nuclear Fusion and Fission . . . . . . . . . . . . . . . . . . . . 36821.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

vi CONTENTS

22 Heat, Temperature, and Friction 375

22.1 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37522.2 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378

22.2.1 Specific Heat . . . . . . . . . . . . . . . . . . . . . . . 37922.2.2 First Law of Thermodynamics . . . . . . . . . . . . . . 37922.2.3 Heat Conduction . . . . . . . . . . . . . . . . . . . . . 38022.2.4 Thermal Radiation . . . . . . . . . . . . . . . . . . . . 381

22.3 Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38322.3.1 Frictional Force Between Solids . . . . . . . . . . . . . 38322.3.2 Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . 384

22.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386

23 Entropy 389

23.1 States of a Brick . . . . . . . . . . . . . . . . . . . . . . . . . 39123.2 Second Law of Thermodynamics . . . . . . . . . . . . . . . . . 39723.3 Two Bricks in Thermal Contact . . . . . . . . . . . . . . . . . 39723.4 Thermodynamic Temperature . . . . . . . . . . . . . . . . . . 40023.5 Specific Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . 40123.6 Entropy and Heat Conduction . . . . . . . . . . . . . . . . . . 40123.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402

24 The Ideal Gas and Heat Engines 405

24.1 Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40524.1.1 Particle in a Three-Dimensional Box . . . . . . . . . . 40724.1.2 Counting States . . . . . . . . . . . . . . . . . . . . . . 40924.1.3 Multiple Particles . . . . . . . . . . . . . . . . . . . . . 41024.1.4 Entropy and Temperature . . . . . . . . . . . . . . . . 41124.1.5 Work, Pressure, and Gas Law . . . . . . . . . . . . . . 41224.1.6 Specific Heat of an Ideal Gas . . . . . . . . . . . . . . 414

24.2 Slow and Fast Expansions . . . . . . . . . . . . . . . . . . . . 41524.3 Heat Engines . . . . . . . . . . . . . . . . . . . . . . . . . . . 41724.4 Perpetual Motion Machines . . . . . . . . . . . . . . . . . . . 42024.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422

A Constants 425

A.1 Constants of Nature . . . . . . . . . . . . . . . . . . . . . . . 425A.2 Properties of Stable Particles . . . . . . . . . . . . . . . . . . 425A.3 Properties of Solar System Objects . . . . . . . . . . . . . . . 426

CONTENTS vii

A.4 Miscellaneous Conversions . . . . . . . . . . . . . . . . . . . . 426

B GNU Free Documentation License 427

B.1 Applicability and Definitions . . . . . . . . . . . . . . . . . . . 428B.2 Verbatim Copying . . . . . . . . . . . . . . . . . . . . . . . . . 429B.3 Copying in Quantity . . . . . . . . . . . . . . . . . . . . . . . 429B.4 Modifications . . . . . . . . . . . . . . . . . . . . . . . . . . . 430B.5 Combining Documents . . . . . . . . . . . . . . . . . . . . . . 433B.6 Collections of Documents . . . . . . . . . . . . . . . . . . . . . 433B.7 Aggregation With Independent Works . . . . . . . . . . . . . 433B.8 Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434B.9 Termination . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434B.10 Future Revisions of This License . . . . . . . . . . . . . . . . . 435

C History 437

viii CONTENTS

Preface to April 2006 Edition

This text has developed out of an alternate beginning physics course at NewMexico Tech designed for those students with a strong interest in physics.The course includes students intending to major in physics, but is not limitedto them. The idea for a “radically modern” course arose out of frustrationwith the standard two-semester treatment. It is basically impossible to in-corporate a significant amount of “modern physics” (meaning post-19th cen-tury!) in that format. Furthermore, the standard course would seem to bespecifically designed to discourage any but the most intrepid students fromcontinuing their studies in this area — students don’t go into physics to learnabout balls rolling down inclined planes — they are (rightly) interested inquarks and black holes and quantum computing, and at this stage they arelargely unable to make the connection between such mundane topics and theexciting things that they have read about in popular books and magazines.

It would, of course, be easy to pander to students — teach them superfi-cially about the things they find interesting, while skipping the “hard stu!”.However, I am convinced that they would ultimately find such an approachas unsatisfying as would the educated physicist.

The idea for this course came from reading Louis de Broglie’s NobelPrize address.1 De Broglie’s work is a masterpiece based on the principles ofoptics and special relativity, which qualitatively foresees the path taken bySchrodinger and others in the development of quantum mechanics. It thusdawned on me that perhaps optics and waves together with relativity couldform a better foundation for all of physics than does classical mechanics.

Whether this is so or not is still a matter of debate, but it is indisputablethat such a path is much more fascinating to most college freshmen interestedin pursing studies in physics — especially those who have been through the

1Reprinted in: Boorse, H. A., and L. Motz, 1966: The world of the atom. Basic Books,New York, 1873 pp.

ix

x PREFACE TO APRIL 2006 EDITION

usual high school treatment of classical mechanics. I am also convinced thatthe development of physics in these terms, though not historical, is at leastas rigorous and coherent as the classical approach.

The course is tightly structured, and it contains little or nothing that canbe omitted. However, it is designed to fit into the usual one year slot typicallyallocated to introductory physics. In broad outline form, the structure is asfollows:

• Optics and waves occur first on the menu. The idea of group velocity iscentral to the entire course, and is introduced in the first chapter. Thisis a di"cult topic, but repeated reviews through the year cause it toeventually sink in. Interference and di!raction are done in a reasonablyconventional manner. Geometrical optics is introduced, not only forits practical importance, but also because classical mechanics is laterintroduced as the geometrical optics limit of quantum mechanics.

• Relativity is treated totally in terms of space-time diagrams — theLorentz transformations seem to me to be quite confusing to studentsat this level (“Does gamma go upstairs or downstairs?”), and all desiredresults can be obtained by using the “space-time Pythagorean theorem”instead, with much better e!ect.

• Relativity plus waves leads to a dispersion relation for free matterwaves. Optics in a region of variable refractive index provides a power-ful analogy for the quantum mechanics of a particle subject to potentialenergy. The group velocity of waves is equated to the particle velocity,leading to the classical limit and Newton’s equations. The basic topicsof classical mechanics are then done in a more or less conventional,though abbreviated fashion.

• Gravity is treated conventionally, except that Gauss’s law is introducedfor the gravitational field. This is useful in and of itself, but also pro-vides a preview of its deployment in electromagnetism. The repetitionis useful pedagogically.

• Electromagnetism is treated in a highly unconventional way, thoughthe endpoint is Maxwell’s equations in their usual integral form. Theseemingly simple question of how potential energy can be extended tothe relativistic context gives rise to the idea of potential momentum.

xi

The potential energy and potential momentum together form a four-vector which is closely related to the scalar and vector potential ofelectromagnetism. The Aharonov-Bohm e!ect is easily explained usingthe idea of potential momentum in one dimension, while extension tothree dimensions results in an extension of Snell’s law valid for matterwaves, from which the Lorentz force law is derived.

• The generation of electromagnetic fields comes from Coulomb’s law plusrelativity, with the scalar and vector potential being used to produce amuch more straightforward treatment than is possible with electric andmagnetic fields. Electromagnetic radiation is a lot simpler in terms ofthe potential fields as well.

• Resistors, capacitors, and inductors are treated for their practical value,but also because their consideration leads to an understanding of energyin electromagnetic fields.

• At this point the book shifts to a more qualitative treatment of atoms,atomic nuclei, the standard model of elementary particles, and tech-niques for observing the very small. Ideas from optics, waves, andrelativity reappear here.

• The final section of the course deals with heat and statistical mechanics.Only at this point do non-conservative forces appear in the contextof classical mechanics. Counting as a way to compute the entropyis introduced, and is applied to the Einstein model of a collection ofharmonic oscillators (conceptualized as a “brick”), and in a limited wayto an ideal gas. The second law of thermodynamics follows. The bookends with a fairly conventional treatment of heat engines.

A few words about how I have taught the course at New Mexico Tech arein order. As with our standard course, each week contains three lecture hoursand a two-hour recitation. The book contains little in the way of examplesof the type normally provided by a conventional physics text, and the styleof writing is quite terse. Furthermore, the problems are few in number andgenerally quite challenging — there aren’t many “plug-in” problems. Therecitation is the key to making the course accessible to the students. I gener-ally have small groups of students working on assigned homework problemsduring recitation while I wander around giving hints. After all groups have

xii PREFACE TO APRIL 2006 EDITION

completed their work, a representative from each group explains their prob-lem to the class. The students are then required to write up the problems ontheir own and hand them in at a later date. In addition, reading summariesare required, with questions about material in the text which gave di"cul-ties. Many lectures are taken up answering these questions. Students tendto do the summaries, as their lowest test grade is dropped if they completea reasonable fraction of them. The summaries and the associated questionshave been quite helpful to me in indicating parts of the text which needclarification.

I freely acknowledge stealing ideas from Edwin Taylor, Archibald Wheeler,Thomas Moore, Robert Mills, Bruce Sherwood, and many other creativephysicists, and I owe a great debt to them. My colleagues Alan Blyth andDavid Westpfahl were brave enough to teach this course at various stages ofits development, and I welcome the feedback I have received from them. Fi-nally, my humble thanks go out to the students who have enthusiastically (oron occasion unenthusiastically) responded to this course. It is much, muchbetter as a result of their input.

There is still a fair bit to do in improving the text at this point, such asrewriting various sections and adding an index . . . Input is welcome, errorswill be corrected, and suggestions for changes will be considered to the extentthat time and energy allow.

Finally, a word about the copyright, which is actually the GNU “copy-left”. The intention is to make the text freely available for downloading,modification (while maintaining proper attribution), and printing in as manycopies as is needed, for commercial or non-commercial use. I solicit com-ments, corrections, and additions, though I will be the ultimate judge as towhether to add them to my version of the text. You may of course do whatyou please to your version, provided you stay within the limitations of thecopyright!David J. RaymondNew Mexico TechSocorro, NM, [email protected]

222 PREFACE TO APRIL 2006 EDITION

Chapter 13

Newton’s Law of Gravitation

In this chapter we study the law which governs gravitational forces betweenmassive bodies. We first introduce the law and then explore its consequences.The notion of a test mass and the gravitational field is developed, followed bythe idea of gravitational flux. We then learn how to compute the gravitationalfield from more than one mass, and in particular from extended bodies withspherical symmetry. We finally examine Kepler’s laws and learn how theselaws plus the conservation laws for energy and angular momentum may beused to solve problems in orbital dynamics.

13.1 The Law of Gravitation

Of Newton’s accomplishments, the discovery of the universal law of gravita-tion ranks as one of the greatest. Imagine two masses, M1 and M2, separatedby a distance r. The force has the magnitude

F =M1M2G

r2, (13.1)

where G = 6.67 ! 10!11 m3 kg!1 s!2 is the universal gravitational constant.The gravitational force is always attractive and it acts along the line of centersbetween the two masses.

223

224 CHAPTER 13. NEWTON’S LAW OF GRAVITATION

g1

g2

g

r1

r2

test point

M1

M2

Figure 13.1: Sketch showing the addition of gravitational fields at a test pointresulting from two masses.

13.2 Gravitational Field

The gravitational field at any point is the gravitational force on some testmass placed at that point, divided by the mass of the test mass. The di-mensions of the gravitational field are length over time squared, which isthe same as acceleration. For a single mass M (other than the test mass),Newton’s law of gravitation tells us that

g = "GMr

r3(point mass), (13.2)

where r is the position of the test point relative to the mass M . Note thatwe have written this equation in vector form, reflecting the fact that thegravitational field is a vector. Thus, r = xtest " xmass, where xtest and xmass

are the position vectors of the test point and the mass M . The vector rpoints from the mass to the test point. The quantity r = |r| is the distancefrom the mass to the test point.

If there is more than one mass, then the total gravitational field at atest point is obtained by computing the individual fields produced by eachmass at the test point, and vectorially adding these fields. This process isschematically illustrated in figure 13.1.

13.3. GRAVITATIONAL FLUX 225

θ

S g

Figure 13.2: Definition sketch for the gravitational flux through the directedarea S.

Sg

1

2

S

Figure 13.3: Two areas with the same projected area normal to g. Is theflux through area 2 greater than, less than, or equal to the flux through area1? (The two areas are being viewed edge-on and are assumed to have somedimension d in the direction normal to the page.)

13.3 Gravitational Flux

The next concept we need to discuss is the gravitational flux. Figure 13.2shows a rectangular area S with a vector S perpendicular to the rectangle.The vector S is defined to have length S, so it is a compact way of representingthe size and orientation of a rectangle in three dimensional space. The vectorS could point either upward or downward, and the choice of directions turnsout to be important. This is why we say that S represents a directed area.

Figure 13.2 also shows a vector g, representing the gravitational field onthe surface of the rectangle. It’s value is assumed here not to vary withposition on the rectangle. The angle ! is the angle between the vectors Sand g.


The gravitational flux through the rectangle is defined as

#g = S · g = Sg cos ! = Sgn, (13.3)

where gn = g cos ! is the component of g normal to the rectangle. The flux isthus larger for larger areas and for larger gravitational fields. However, onlythe component of the gravitational field normal to the rectangle (i. e., parallelto S) counts in this calculation. A consequence is that the gravitational fluxthrough area 1, S1 · g, in figure 13.3 is the same as the flux through area 2,S2 · g.

The significance of the directedness of the area is now clear. If the vectorS pointed in the opposite direction, the flux would have the opposite sign.When defining the flux through a rectangle, it is necessary to define whichway the flux is going. This is what the direction of S does — a positive fluxis defined as going from the side opposite S to the side of S.

An analogy with flowing water may be helpful. Imagine a rectangularchannel of cross-sectional area S through which water is flowing at velocityv. The flux of water through the channel, which is defined as the volumeof water per unit time passing through the cross-sectional area, is #w = vS.The water velocity takes the place of the gravitational field in this case, andits direction is here assumed to be normal to the rectangular cross-section ofthe channel. The field thus expresses an intensity (e. g., the velocity of thewater or the strength of the gravitational field), while the flux expresses anamount (the volume of water per unit time in the fluid dynamical case). Thegravitational flux is thus the amount of some gravitational influence, whilethe gravitational field is its strength. We now try to more clearly understandto what this amount really refers.

We need to briefly consider the case in which the gravitational field variesfrom one point to another on the rectangular surface. In this case a propercalculation of the flux through the surface cannot be made using equation(13.3) directly. Instead, we must break the surface into a grid of sub-surfaces.If the grid is su"ciently fine, the gravitational field will be nearly constantover each sub-surface and equation (13.3) can be applied separately to eachof these. The total flux is then the sum of all the individual fluxes.

There is actually no need for the area in figure 13.2 to be rectangular. Wecan calculate the gravitational flux through the surface of a sphere of radiusR with a mass M at the center. As illustrated in figure 13.4, the gravitationalfield points inward toward the mass. It has magnitude g = GM/R2, so if we

13.3. GRAVITATIONAL FLUX 227

R

g

M

Figure 13.4: Calculation of the gravitational flux through the surface of asphere with a mass at the center.

desire to calculate the gravitational flux out of the sphere, we must introducea minus sign. Finally, the area of a sphere of radius R is S = 4"R2, so theflux is

#g = "gS = "(GM/R2)(4"R2) = "4"GM. (13.4)

Notice that this flux doesn’t depend on how big the sphere is — the factorof R2 in the area cancels with the factor of 1/R2 in the gravitational field.This is a hint that something profound is going on. The size of the surfaceenclosing the mass is unimportant, and neither is its shape — the answer isalways the same — the gravitational flux outward through any closed surfacesurrounding a mass M is just #g = "4"GM ! This is an example of Gauss’slaw applied to gravity.

It is possible to formally prove this result using arguments like those posedin figure 13.3, but perhaps the easiest way to understand this result is viathe analogy with the flow of water. If we think of the mass as somethingwhich destroys water at a certain rate, then there must be an inward flowof water through the surfaces in the left and center examples in figure 13.5.Furthermore, the volume of water per unit time flowing inward through thesesurfaces is the same in the two examples, because the rate at which wateris being destroyed is the same. In the right case the mass is not contained


GMGMΦ = −4π Φ = −4π Φ = 0

Figure 13.5: Three cases of a mass M and a closed surface. In the left andcenter examples the mass is inside the closed surface and the outward fluxthrough the surface is #g = "4"GM . In the right example the mass isoutside the surface and the outward flux through the surface is zero.

inside the surface and though water flows into the volume bounded by thesurface, it also flows out the other side, resulting in a net outward (or inward)volume flux through the surface of zero.

13.4 Flux from Multiple Masses

Gauss’s law extends trivially to more than one mass. As figure 13.6 shows,the outward flux through a closed surface is just

#g = "4"G!

inside

Mi (Gauss’s law). (13.5)

In other words, all masses inside the closed surface contribute to the flux,while no masses outside the surface contribute. This is the most generalstatement of Gauss’s law as it applies to gravity.

An important application of Gauss’s law is to show that the gravitationalfield outside of a spherically symmetric extended mass M is exactly the sameas if all the mass were concentrated at a point at the center of the sphere.The proof goes as follows: Imagine a sphere concentric with the center of theextended mass, but with larger radius. The gravitational flux from the massis just #g = "4"GM as before. However, because of the assumed sphericalsymmetry, we know that the gravitational field points normally inward atevery point on the spherical surface and is equal in magnitude everywhere

13.5. EFFECTS OF RELATIVITY 229

M

M

M

M

M 1

3

4

5

2

1 + M2 + M3)G(MgΦ = −4π

Figure 13.6: Gauss’s law applied to more than one mass. The masses M1,M2, and M3 contribute to the outward gravitational flux through the surfaceshown. The masses M4 and M5 don’t contribute.

on the sphere. Thus we can infer that #g = "4"R2g, where R is the radiusof the sphere and g is the magnitude of the gravitational field at radius R.From these two equations we immediately infer that the field magnitude is

g =GM

R2. (13.6)

Expressing this in vector form for arbitrary radius r, and remembering thatthe gravitational field points inward, we find that

g = "GMr

r3, (13.7)

which is precisely the equation for g resulting from a point mass M . Recallthat r points from the mass to the test point.

13.5 E!ects of Relativity

So far our discussion of gravity has been completely non-relativistic. Wewill not explore in detail how the theory of gravity changes in a completelyrelativistic treatment. As we noted earlier in the course, Einstein’s generaltheory of relativity covers this, and the mathematics are formidable. Weconfine ourselves to two comments:


• As noted previously, gravity is locally equivalent to being in an accel-erated reference frame. However, unlike the simple example which westudied earlier, there is in general no universal frame of reference towhich we can transform which is everywhere inertial.

• Space is even more non-Euclidean in general relativity than in specialrelativity. In particular, there is no such thing as a straight line inthe geometry of general relativistic spacetime. This is true becausespacetime itself is curved. An example of a curved space is the surfaceof a sphere. Clearly, a straight line cannot be embedded in this space.The closest equivalent to a straight line in this geometry is a greatcircle. This is an example of a geodesic curve. In general relativityobjects subject only to the force of gravity move along geodesic curves.

One potentially observable prediction of relativity is the existence of grav-itational waves. Imagine two stars revolving around each other. The grav-itational field from these stars will change periodically due to this motion.However, this change propagates outward only at the speed of light. As aresult, ripples in the field, or gravitational waves, spread outward from the re-volving stars. E!orts are currently under way to develop apparatus to detectgravitational waves produced by violent cosmic events such as the explosionof a supernova.

13.6 Kepler’s Laws

Johannes Kepler, using data compiled by Tycho Brahe, inferred three lawsgoverning the motions of planets in the solar system:

1. Planets move in elliptical orbits with the sun at one focus.

2. Equal areas are swept out in equal times by the line connecting the sunand the planet.

3. The square of the period of revolution of the planet around the sun isproportional to the cube of the semi-major axis of the ellipse.

These laws were instrumental in the development of modern mechanics andthe universal law of gravitation by Isaac Newton.

Showing that the first law is consistent with Newtonian mechanics ismathematically more di"cult than we can undertake in this course. However,

13.6. KEPLER’S LAWS 231

perihelion aphelion

21

R dA

dsdx

sun a

b

v

Figure 13.7: Illustration of elliptical orbit of a planet with the sun at theleft focus. The semi-major and semi-minor axes are denoted by a and b.The shaded triangular area element is needed for the discussion of Kepler’ssecond law. Perihelion and aphelion are respectively the points on the orbitnearest and farthest from the sun. Note that at perihelion and aphelion thevelocity is purely tangential, i. e., the velocity component along the radiusvector is zero.

the second law turns out to be a simple consequence of the conservation ofangular momentum. Figure 13.7 shows an elliptical orbit with the area sweptout as a planet moves from position 1 to position 2. We estimate this areaas dA = Rdx/2, where we have ignored the small unshaded part of the areato the right of the shaded triangle. The distance traveled by the planet intime dt is ds, so the magnitude of the velocity is v = ds/dt. However, incomputing the angular momentum, we need the tangential component of thevelocity, i. e., the component normal to the radius vector R. This is simplyvt = dx/dt. The angular momentum is L = mRvt = mRdx/dt, where m isthe mass of the planet. Combining this with the formula for dA results in

dA

dt=

L

2m. (13.8)

Since gravitation is a central force, angular momentum is conserved, whichmeans that dA/dt is constant. Thus, we have shown that conservation ofangular momentum is equivalent to Kepler’s second law.

Kepler’s third law turns out to be a consequence of the universal law ofgravitation. We can prove this for circular orbits. We know that a planet


moving in a circular orbit around the sun is accelerating toward the sun withthe centripetal acceleration a = v2/R, where v is the speed of the planet’smotion in its orbit and R is the orbit’s radius. This acceleration is caused bythe gravitational force, so we can equate the force divided by the planetarymass to a, resulting in

v2

R=

GM

R2, (13.9)

where M is the mass of the sun. This may be solved for v:

v ="

GM

R

#1/2

. (13.10)

Eliminating v in favor of the period of revolution T = 2"R/v results in

T 2 =4"2R3

GM. (13.11)

This agrees with Kepler’s third law since the semi-major axis of a circle issimply the radius R.

13.7 Use of Conservation Laws

The gravitational force is conservative, so two point masses M and m sepa-rated by a distance r have a potential energy:

U = "GMm

r. (13.12)

It is easily verified that di!erentiation recovers the gravitational force.The conservation of energy and angular momentum in planetary motions

can be used to solve many practical problems involving motion under theinfluence of gravity. For instance, suppose a bullet is shot straight upwardfrom the surface of the moon. One might ask what initial velocity is neededto insure that the bullet will escape from the gravity of the moon. Since totalenergy E is conserved, the sum of the initial kinetic and potential energiesmust equal the sum of the final kinetic and potential energies:

E = Kinitial + Uinitial = Kfinal + Ufinal. (13.13)

For the bullet to escape the moon, its kinetic energy must remain positive nomatter how far it gets from the moon. Since the potential energy is always

13.7. USE OF CONSERVATION LAWS 233

negative, asymptoting to zero at infinite distance (i. e., Ufinal = 0), theminimum total energy consistent with this condition is zero. For zero totalenergy we have

mv2initial

2= Kinitial = "Uinitial = +

GMm

R, (13.14)

where m is the mass of the bullet, M is the mass of the moon, R is theradius of the moon, and vinitial is the minimum initial velocity required forthe bullet to escape. Solution for vinitial yields

vinitial ="

2GM

R

#1/2

. (13.15)

This is called the escape velocity. Notice that the escape velocity from agiven radius is a factor of 21/2 larger than the velocity needed for a circularorbit at that radius (see equation (13.10)).

An object is energetically bound to the sun if its kinetic plus potentialenergy is less than zero. In this case the object follows an elliptical orbitaround the sun as shown by Kepler. However, if the kinetic plus potentialenergy is zero, the object follows a parabolic orbit, and if it is greater thanzero, a hyperbolic orbit results. In the latter two cases the sun also resides ata focus of the parabola or hyperbola. Figure 13.8 shows a typical hyperbolicorbit. The impact parameter, defined in this figure, is the closest the objectwould have come to the center of the sun if it hadn’t been deflected by gravity.

Sometimes energy and angular momentum conservation can be used to-gether to solve problems. For instance, suppose we know the energy and an-gular momentum of an asteroid of mass m and we wish to infer the maximumand minimum distances of the asteroid from the sun, the so called aphelionand perihelion distances. Since the asteroid is gravitationally bound to thesun, it is convenient to characterize the total energy by Eb = "E, the so-called binding energy. If v is the orbital speed of the asteroid and r is itsdistance from the sun, then the binding energy can be written in terms ofthe kinetic and potential energies:

"Eb =mv2

2"

GMm

r. (13.16)

The magnitude of the angular momentum of the asteroid is L = mvtr,where vt is the tangential component of the asteroid’s velocity. At aphelion


b

sun

hyperbolicorbit

asymptotesof hyperbola

Figure 13.8: Example of a hyperbolic orbit of a positive energy object passingby the sun. The sun sits at the focus of the hyperbola. The quantity b iscalled the impact parameter.

13.8. PROBLEMS 235

and perihelion the radial part of the velocity of the asteroid is zero andthe speed equals the tangential component of the velocity, v = vt. Thus, ataphelion and perihelion we can eliminate v in favor of the angular momentum:

"Eb =L2

2mr2"

GMm

r(aphelion and perihelion). (13.17)

This can be rearranged into a quadratic equation

r2 "GMm

Ebr +

L2

2mEb= 0, (13.18)

which can be solved to yield

r =1

2

$

%

GMm

Eb±

&

G2M2m2

E2b

"2L2

mEb

'1/2(

) . (13.19)

The larger of the two solutions yields the aphelion value of the radius whilethe smaller yields perihelion.

Equation (13.19) tells us something else interesting. The quantity insidethe square root cannot be negative, which means that we must have

L2 #G2M2m3

2Eb. (13.20)

In other words, for a given value of the binding energy Eb there is a maximumvalue for the angular momentum. This maximum value makes the square rootzero, which means that the aphelion and the perihelion are the same — i. e.,the orbit is circular. Thus, among all orbits with a given binding energy, thecircular orbit has the maximum angular momentum.

13.8 Problems

1. Assume a mass M is located at ("2 m, 0 m) and a mass 2M is lo-cated at (0 m, 3 m). Find the (vector) gravitational field at the point(1 m, 1 m).

2. If two equal masses M are located at x1 = ("3 m,"4 m) and x2 =("3 m, +4 m), determine where a third mass M must be placed toresult in zero gravitational force at the origin.


M

M

2M

8M

M

2M

Figure 13.9: Various masses inside and outside a Gaussian surface.

3. Given the value of g at the Earth’s surface, the radius of the Earth(look it up), and the universal gravitational constant G, determine themass of the Earth.

4. Given the situation in figure 13.9:

(a) What is the gravitational flux through the illustrated surface?

(b) Explain why you cannot use this information to compute the grav-itational field on the surface in this case.

5. Suppose mass is distributed uniformly with density µ kg m!1 in a thinline along the z axis. Try to figure out a way of using Gauss’s law plussymmetry arguments to predict the gravitational field resulting fromthis mass distribution.

6. If the Earth is of uniform density, #, use Gauss’s law to determine thegravitational field inside the Earth as a function of distance from thecenter.

7. Using the results of the above problem, determine the motion of anobject moving through an evacuated hole drilled through the center ofthe Earth.

13.8. PROBLEMS 237

8. Two infinite thin sheets of mass, each with $ mass per unit area, arealigned perpendicular to each other. Determine the gravitational fieldfrom this combination. Hint: Compute g from each sheet separatelyand add vectorially.

9. Suppose that the universal law of gravitation says that the (attractive)gravitational force takes the form F = "M1M2Gr, where r is theseparation between the two masses M1 and M2 and G is a constant.Find the relationship between the orbital radius and the period for acircular orbit of a planet around the sun in this case.

10. An alien spaceship enters the solar system at distance D from the sunwith speed v0. (D may be considered to be very far from the sun.) Itcoasts through the solar system, approaching within a distance d $ Dof the sun.

(a) Find its speed at the point of closest approach.

(b) Find the angular momentum of the spaceship with respect to thecenter of the sun.

(c) What was the tangential component of the spaceship’s velocity (i.e., the component normal to the radius vector) when it enteredthe solar system at distance D?

11. As a result of tidal torques, the spin angular momentum of the Earth isgradually being converted into orbital angular momentum of the moon,which causes the radius of its (circular) orbit to increase. Hint: Recallthat for a solid sphere the moment of inertia is I = 2mr2/5.

(a) Obtain a relationship between the moon’s orbital velocity and itsdistance from the Earth, assuming that the orbit is circular.

(b) If the Earth’s rotation rate is cut in half due to this e!ect, whatwill the new radius of the moon’s orbit be?

Chapter 14

Forces in Relativity

In this chapter we ask an apparently simple question: How can the idea ofpotential energy be extended to the relativistic case? The answer to this ques-tion is unexpectedly complex, but it leads us to immensely fruitful results.In particular, it prompts us to investigate the idea of potential momentum,which results ultimately in gauge theory, of which electricity and magnetismis an example.

Along the way we show that conservation of four-momentum has an un-expected consequence — the idea of force at a distance is inconsistent withthe theory of relativity. This means that momentum and energy must becarried between interacting particles by another type of particle which wecall an intermediary particle. These particles are virtual in the sense thatthey don’t have their real-world mass when acting in this role.

In relativistic quantum mechanics, we find that particles can take onnegative energies. Feynman’s interpretation of this fact is discussed, whichleads us to a model for antiparticles.

14.1 Potential Momentum

For a free, non-relativistic particle of mass m, the total energy E equals thekinetic energy K and is related to the momentum " of the particle by

E = K =|"|2

2m(free, non-relativistic). (14.1)

(Note that we have ignored the contribution of the rest energy to the totalenergy here.) In the non-relativistic case, the momentum is " = mv, where

239

240 CHAPTER 14. FORCES IN RELATIVITY

v is the particle velocity.If the particle is not free, but is subject to forces associated with a po-

tential energy U(x, y, z), then equation (14.1) must be modified to accountfor the contribution of U to the total energy:

E " U = K =|"|2

2m(non-free, non-relativistic). (14.2)

The force on the particle is related to the potential energy by

F = "

&

%U

%x,%U

%y,%U

%z

'

. (14.3)

For a free, relativistic particle, we have

E = (|"|2c2 + m2c4)1/2 (free, relativistic). (14.4)

The obvious way to add forces to the relativistic case is by rewriting equation(14.4) with a potential energy, in analogy with equation (14.2):

E " U = (|"|2c2 + m2c4)1/2 (incomplete!). (14.5)

Unfortunately, equation (14.5) is incomplete, because we have subtractedsomething (U) from the energy E without subtracting something from themomentum" as well. However, $ = (", E/c) is a four-vector, so an equationwith something subtracted from just one of the components of this four-vectoris not relativistically invariant. In other words, equation (14.5) doesn’t obeythe principle of relativity, and therefore cannot be correct!

How can we fix this problem? One way is to define a new four-vectorwith U/c being its timelike part and some new vector Q being its spacelikepart:

Q % (Q, U/c) (potential four-momentum). (14.6)

We then subtract Q from the momentum ". When we do this, equation(14.5) becomes

E " U = (|"" Q|2c2 + m2c4)1/2 (non-free, relativistic). (14.7)

The quantity Q is called the potential momentum and Q is the potentialfour-momentum.

14.2. AHARONOV-BOHM EFFECT 241

Some additional terminology is useful. We define

p % ""Q (kinetic momentum) (14.8)

as the kinetic momentum for reasons discussed below. In order to avoidconfusion, we rename " the total momentum.1 Thus, the total momentumequals the kinetic plus the potential momentum, in analogy with energy.

So far, we have shown that the introduction of a potential momentumcomplements the potential energy so as to make the energy-momentum re-lationship for a particle relativistically invariant. However, we as yet haveno idea what causes potential momentum and what it does to the a!ectedparticle. We shall put o! answering the former question and address onlythe latter at this point. A hint comes from the corresponding behavior ofenergy. The total energy of a particle is related to the quantum mechanicalfrequency & of the particle, and the total momentum is related to its wavevector k:

E = h& " = hk. (14.9)

However, the kinetic energy2 and the kinetic momentum are related to theparticle’s velocity v:

E " U =mc2

(1 " v2/c2)1/2p = "" Q =

mv

(1 " v2/c2)1/2, (14.10)

where v = |v|.The relationship between kinetic momentum and velocity can be proven

by dividing equation (14.7) by h to obtain a dispersion relation and then com-puting the group velocity, which we equate to the particle velocity. However,we will not do this here.

14.2 Aharonov-Bohm E!ect

Let us now study a phenomenon which depends on the existence of potentialmomentum. If the potential energy of a particle is zero and both the kinetic

1In advanced mechanics, ! is called the canonical momentum.2In relativity, the quantity E " U is actually equal to the kinetic plus the rest energy.

This quantity ought to have a separate name but it does not.


in interferenceparticle

potential momentum

Figure 14.1: Setup for the Aharonov-Bohm e!ect. The particle movesthrough a channel which has a divided segment with non-zero potential mo-menta pointing in opposite directions in the two sub-channels. The verticalline segments show the wave fronts for the particle.

and potential momenta point in the ±x direction, the total energy equation(14.7) for the particle becomes

E = [($ " Q)2c2 + m2c4]1/2 = (p2c2 + m2c4)1/2. (14.11)

Since its total energy E is conserved, the magnitude of the kinetic momentump of the particle doesn’t change according to the above equation. Thus, if aregion of non-zero potential momentum is encountered, the total momentumof the particle must change so as to keep the kinetic momentum constant.This results in a change in the wavelength of the matter wave associatedwith the particle. In particular, if the potential momentum points in thesame direction as the kinetic momentum, the total momentum is increasedand the wavelength decreases, while a potential momentum pointing in thedirection opposite the kinetic momentum results in an increase in wavelength.

Figure 14.1 illustrates what might happen to a particle moving througha channel which splits into two sub-channels for an interval. If we arrange tohave non-zero potential momenta pointing in opposite directions in the sub-channels, the wavelength of the particle will be di!erent in the two regions.At the end of the interval, the waves recombine, interfering constructively ordestructively, depending on the magnitude of the phase di!erence betweenthem. If destructive interference occurs, then the particle cannot pass. Thepotential momentum thus acts as a valve controlling the flow of particles

14.3. FORCES FROM POTENTIAL MOMENTUM 243

through the channel. This is an example of the Aharonov-Bohm e!ect.

14.3 Forces from Potential Momentum

In the Aharonov-Bohm e!ect the potential momentum didn’t result in anyforce on the particle — its only manifestation was to change the particle’swavelength. In such situations the potential momentum’s presence is onlyrevealed by quantum mechanical e!ects.

The potential momentum has more of an influence on the non-quantumworld when the problem is two or three-dimensional or when the potentialmomentum is changing with time. The total force on a particle due to allpossible e!ects involving the potential energy and the potential momentumis given by

F = "

&

%U

%x,%U

%y,%U

%z

'

"%Q

%t+ u !P, (14.12)

where u is the particle velocity and P is a vector obtained from the potentialmomentum vector as follows:

P %

&

%Qz

%y"%Qy

%z,%Qx

%z"%Qz

%x,%Qy

%x"%Qx

%y

'

. (14.13)

This is unexpectedly complicated. However, equation (14.12) consists ofthree parts. The first part involves derivatives of the potential energy and isexactly the same as in the non-relativistic case. The new e!ects are confinedto the second and third parts, "%Q/%t and u!P. A full derivation of theseequations involves rather complex mathematics. However, it is possible tounderstand the origin of these additional contributions to the force by lookingat a couple of simple examples.

14.3.1 Refraction E!ect

A matter wave impingent on a discontinuity in potential momentum is re-fracted, just as it is refracted by a discontinuity in potential energy. Re-fraction of a matter wave packet means that the velocity of the associatedparticle changes as it moves across the interface. This means that the particleundergoes an acceleration, implying that it is subject to a force.

As in the case of Snell’s law for optics, the frequency of a matter wavedoesn’t change as it crosses such a discontinuity in potential momentum.


x

y

12

34

56

QQ

QQ

QQ

p

Figure 14.2: Trajectory of a wave packet through a region of variable poten-tial momentum Q. Q points in the y direction and increases in magnitudeby steps with increasing x. The kinetic momentum p indicates the directionof motion of the associated particle at each point along the trajectory.

Furthermore, neither does the component of the wave vector parallel to thediscontinuity. These two conditions together insure phase continuity at theinterface.

Figure 14.2 shows an example of what happens when a wave encountersa series of parallel slabs with increasing values of Q. The y component of thewave vector doesn’t change as the wave crosses each of the interfaces betweenslabs, for reasons discussed above. Hence, $y = hky doesn’t change either,which means that d$y/dx = 0. The y component of kinetic momentum,py = $y " Qy, must therefore decrease as Qy increases, as illustrated infigure 14.2.

Newton’s second law tells us that the y component of the force on the par-ticle associated with the wave is just the time derivative of the y componentof the kinetic momentum:

Fy =dpy

dt=

dpy

dx

dx

dt=

dpy

dxux = "

dQy

dxux. (14.14)

14.3. FORCES FROM POTENTIAL MOMENTUM 245

y

x

Moving particle,stationary field pattern

Stationary particle,moving field pattern

v

-v

Figure 14.3: A moving particle and a stationary pattern of potential momen-tum Q must be equivalent to a stationary particle and a moving pattern ofpotential momentum according to the principle of relativity.

In the last step we used the fact that d$y/dx = 0.The x component of the force can be obtained by similar reasoning, using

the additional information that the speed, and hence the magnitude of thekinetic momentum, p2 = p2

x + p2y, doesn’t change under the influence of the

potential momentum:

Fx =dpx

dt=

dpx

dx

dx

dt= "

py

(p2 " p2y)

1/2

dpy

dxux =

py

px

dQy

dxux =

dQy

dxuy. (14.15)

Aside from assuming that p2 = constant, we have used the relationshipspx = (p2"p2

y)1/2 and py/px = uy/ux. Equations (14.14) and (14.15) constitute

a special case of equations (14.12) and (14.13) which is valid when Q pointsin the y direction and is a function only of x.

14.3.2 Time-Varying Potential Momentum

The necessity for the term "%Q/%t in equation (14.12) is easily understoodfrom the following argument, which is illustrated in figure 14.3. The examplein the previous section showed that a particle moving in the +x directionwith velocity ux through a field of increasing Qy (left panel of figure 14.3)experiences a force in the "y direction equal to Fy = "(dQy/dx)ux. However,viewing this same process from a reference frame in which the particle isstationary (right panel of this figure), we see that the potential momentumat the position of the particle increases with time at the rate dQy/dt. The


particle is not moving in this reference frame, so the term u!P = 0. However,the stationary particle must still experience the above force in this referenceframe in order to satisfy the principle of relativity.

Noting that dQy/dt = (dQy/dx)ux, we see that equation (14.12) providesthis force via the term "%Q/%t in the reference frame moving with the parti-cle. Thus, the time derivative term in equation (14.12) is needed to maintainthe principle of relativity — the same force occurs in the two di!erent ref-erence frames but originates from the term u ! P in the original referenceframe and the term "%Q/%t in the frame moving with the particle.

14.4 Lorentz Condition

It turns out that the four components of the potential four-momentum arenot independent, but are subject to the condition

%Qx

%x+%Qy

%y+%Qz

%z+

1

c2

%U

%t= 0. (14.16)

This is called the Lorentz condition. The physical meaning of this conditionwill become clear when we study electromagnetism.

14.5 Gauge Theories and Other Theories

The theory of potential momentum is only one of three ways in which the ideaof potential energy can be extended to the relativistic case. This theory iscalled gauge theory for obscure historical reasons. Gauge theory is importantbecause electromagnetism as well as the theories of weak and strong sub-nuclear interactions are all of this type.

Gravity is the only fundamental force which does not take the form ofa gauge theory. Instead, gravity takes the form of one of two other possi-ble relativistic extensions of potential energy. This theory is called generalrelativity. The gravitational force in general relativity can be interpreted ge-ometrically as a consequence of the curvature of spacetime. Mathematically,it is far too di"cult to pursue here.

The third relativistic extension of potential energy considers potentialenergy to be a field which alters the rest energy of particles. High energyphysicists believe that the elementary particles gain their mass by this mech-anism. The field is called the “Higgs field” after the English physicist who

14.6. CONSERVATION OF FOUR-MOMENTUM AGAIN 247

first proposed this theory, Peter Higgs. However, this theory has yet to beexperimentally tested.

14.6 Conservation of Four-Momentum Again

We earlier introduced the ideas of energy and momentum conservation. Inother words, if we have a number of particles isolated from the rest of the uni-verse, each with momentum pi and energy Ei, then particles may be createdand destroyed and they may collide with each other.3 In these interactionsthe energy and momentum of each particle may change, but the sum totalof all the energy and the sum total of all the momentum remains constantwith time:

E =!

i

Ei = const p =!

i

pi = const. (14.17)

The expression is simpler in terms of four-momentum:

p =!

i

pi. (14.18)

At this point a statement such as the one above should ring alarm bells.Just what does it mean to say that the total energy and momentum remainconstant with time in the context of relativity? Which time? The time inwhich reference frame?

Figure 14.4 illustrates the problem. Suppose two particles exchange four-momentum remotely at the time indicated by the fat horizontal bar in theleft panel of figure 14.4. Conservation of four-momentum implies that

pA

+ pB

= p"A

+ p"B, (14.19)

where the subscripted letters correspond to the particle labels in figure 14.4.Primed values refer to the momentum after the exchange while no primesindicates values before the exchange.

Now view the exchange from the reference frame in the right panel offigure 14.4. A problem with four-momentum conservation exists in the re-gion between the thin horizontal lines. In this region particle B has already

3We use the symbol p for kinetic momentum here. However, in collisions we assumethat the potential momentum and energy are only non-zero when the particles are veryclose together. Thus, when the particles are reasonably well separated, the distinctionbetween kinetic and total momentum is unimportant.


B

B’

A

A’

A

A’

B

B’

x

ct

x’

ct’

x

ctct’

x’

Figure 14.4: The trouble with action at a distance. View of the remoteexchange of four-momentum from the point of view of two di!erent coordi-nate systems. The fat line in both pictures is the line of simultaneity in theunprimed frame which is coincident with the exchange of four-momentumbetween the two particles.

transferred its four-momentum, but it has yet to be received by particle A.In other words, four-momentum is not conserved in this reference frame!

This problem is so serious that we must eliminate the concept of actionat a distance from the repertoire of physics. The only way to have particlesinteract remotely and still conserve four-momentum in all reference framesis to assume that all remote interactions are mediated by another particle,as indicated in figure 14.5. In other words, momentum and energy are trans-ferred from particle A to particle B in a two step process. First, particle Aemits particle C in a manner which conserves the four-momentum. Second,particle C is absorbed by particle B in a similarly conservative interaction.Four-momentum is conserved at all times in all reference frames in this pic-ture.

14.7 Virtual Particles

Another problem is evident from figure 14.5. As drawn, the velocity of themediating particle exceeds the speed of light. This is reflected in the factthat di!erent reference frames yield contradictory results as to whether the

14.7. VIRTUAL PARTICLES 249

CC

A’

A

A’B’

BA

B

B’

x

ct

x’

ct’

x

ctct’

x’

Figure 14.5: Mediation of action at a distance by a third particle. Noticethat the world line of the mediating particle has a slope less than unity, whichmeans that it is nominally moving faster than the speed of light.

mediating particle moves from A to B or B to A. These di"culties turn outto be much less severe than those arising from non-locality. Let us addressthem in sequence.

For sake of definiteness, let us view the emission of particle C by particleA in a reference frame in which the velocity of particle A is just reversed inthe emission process. In this case the four-momentum before the emissionis p

A= (p, E/c), where E = (p2c2 + m2c4)1/2. After the emission we have

p"A

= ("p, E/c). Conservation of four-momentum in the emission processrequires that

pA

= p"A

+ q (14.20)

where q is the four-momentum of particle C. From the above assumptions itis clear that

q = pA" p"

A= (p + p, E/c " E/c) = (2p, 0). (14.21)

Suppose that the real, measured mass of particle C is mC . This conflictswith the apparent or virtual mass of this particle in its flight from A to B,which is

m" = ("q · q)1/2/c % iq/c, (14.22)

where q % |q| is the momentum transfer. Note that the apparent mass isimaginary because the four-momentum is spacelike.


Classically, this discrepancy in the apparent and actual masses of theparticle C would simply indicate that the process wasn’t possible. However,recall that the uncertainty principle allows there to be an uncertainty in themass if it doesn’t persist for too long in terms of the proper time intervalalong the particle’s world line. The statement of this law is %µ%' & 1.Expressed in terms of mass, this becomes

%m%' & h/c2. (14.23)

Let us convert the proper time to an interval since the world line of particleC is horizontal in the reference frame in which we are viewing it. Ignoringthe factor of i, %' = %I/c. We finally compute the absolute value of themass discrepancy as follows: |mC " iq/c| = [(mC " iq/c)(mC + iq/c)]1/2 =(m2

C + q2/c2)1/2. Solving for I yields the approximate maximum invariantinterval particle C can move from its source point while keeping its erroneousmass hidden by the uncertainty principle:

%I &h

(m2Cc2 + q2)1/2

. (14.24)

A particle forced into having an apparent mass di!erent from its actualmass is called a virtual particle. The interaction shown in figure 14.5 canonly take place if particles A and B come closer to each other than thedistance %I. This argument thus produces an estimate for the “range” of aninteraction with momentum transfer 2p and mediating particle mass mC .

Two distinct possibilities exist. If the mediating particle is massless (aphoton, for instance), then the range of the interaction is inversely related tothe momentum transfer: %I & h/q. Thus, small momentum transfers canoccur at large distances. An interaction of this type is called “long range”.On the other hand, if the mediating particle has mass, the range is simply%I & h/mCc when q $ mCc. The range is thus constant and inversely pro-portional to the mass of the mediating particle for low momentum transfers.For large momentum transfer, i. e., when q ' mCc, the range decreases fromthis value with increasing momentum transfer, as in the case of a masslessmediating particle.

14.8 Virtual Particles and Gauge Theory

According to quantum mechanics, particles are represented by waves. Theabsolute square of the wave amplitude represents the probability of finding

14.9. NEGATIVE ENERGIES AND ANTIPARTICLES 251

the particle. In gauge theory the potential four-momentum performs this rolefor the virtual particles mediating interactions. Thus a larger potential four-momentum at some point means a higher probability of finding the relatedvirtual particles at that point.

14.9 Negative Energies and Antiparticles

Figure 14.5 illustrates another oddity in the role of mediating particles incollisions. In the unprimed frame, particle C appears to be emitted by parti-cle A and absorbed by particle B. In the primed frame the reverse is true; itappears to be emitted by B and absorbed by A. These judgements are basedon the fact that the A vertex occurs earlier than the B vertex in the unprimedframe, while the B vertex occurs earlier in the primed frame. However, sincethese distinctions are based on time ordering in di!erent reference frames ofevents separated by a spacelike interval, they are inherently not relativisti-cally invariant. Since the principle of relativity states that physical laws arethe same in all inertial reference frames, we have a conceptual problem toovercome.

A related problem has to do with the computation of energy from massand momentum. Solution of equation E2 = p2c2 + m2c4 for the energy has asign ambiguity which we have so far ignored:

E = ±(p2c2 + m2c4)1/2. (14.25)

A natural tendency would be to omit the minus sign and just consider positiveenergies. However, this would be a mistake — experience with quantummechanics indicates that both solutions must be considered.

Richard Feynman won the Nobel Prize in physics largely for developinga consistent interpretation of the above negative energy solutions, whichwe now relate. Notice that the four-momentum points backward in timein a spacetime diagram if the energy is negative. Feynman suggested that aparticle with four-momentum p is equivalent to the corresponding antiparticlewith four-momentum "p. Thus, we interpret a particle with momentum pand energy E < 0 as an antiparticle with momentum "p and energy "E > 0.

Antiparticles are known to exist for all particles. If a particle and itsantiparticle meet, they can annihilate, creating one or more other parti-cles. Correspondingly, if energy is provided in the right form, a particle-antiparticle pair can be created.


A A AA A A A A

B B BB

B

A

A

Figure 14.6: Equivalence of di!erent processes according to Feynman’s pic-ture.

Suppose a particular kind of particle, call it an A particle, produces a Bparticle when it annihilates with its antiparticle A. This is illustrated in theleft panel in figure 14.6. In Feynman’s view, this process is equivalent to thescattering of an A particle backward in time by a B particle, the scatteringof an A backward in time by a B particle, the creation of an AA pair movingbackward in time by a B particle (an antiB), and the emission of a B particleby an A particle moving forward in time.

The statement “moving backward in time” has stimulated generationsof physics students to contemplate the possibility that Feynman’s picturemakes time travel possible. As far as we know, this is not so. The key phraseis equivalent to. In other words, causality still works forward in time as wehave come to expect.

The real utility of the “backward in time” picture is that it makes cal-culations easier, since processes which are normally thought of as being verydi!erent turn out to have the same mathematical form.

Returning to the ambiguity shown in figure 14.5, it turns out that it doesnot matter whether the picture in the left or right panel is chosen. Accordingto the Feynman view the two processes are equivalent if one small correctionis made — if the mediating particle going from left to right is a C particle,then the mediating particle going from right to left in the other picture isa C particle, or an antiC. It is immaterial whether the arrow representingeither the C or the C points forward or backward in time. The key pointis that if an arrow points into a vertex, the four-momentum of that particlecontributes to the input side of the momentum-energy budget for that vertex.If an arrow points away from a vertex, then the four-momentum contributesto the output side.

14.10. PROBLEMS 253

14.10 Problems

1. An alternate way to modify the energy-momentum relation while main-taining relativistic invariance is with a “potential mass”, H(x):

E2 = p2c2 + (m + H)2c4.

If |H| $ m and p2 $ m2c2, show how this equation may be approxi-mated as

E = something + p2/(2m)

and determine the form of “something” in terms of H . Is this theorydistinguishable from the theory involving potential energy at nonrela-tivistic velocities?

2. For a given channel length L and particle speed in figure 14.1, determinethe possible values of potential momentum ±Q in the two channelswhich result in destructive interference between the two parts of theparticle wave.

3. Show that equations (14.14) and (14.15) are indeed recovered fromequations (14.12) and (14.13) when Q points in the y direction and isa function only of x.

4. Show that the force F = u!P is perpendicular to the velocity u. Doesthis force do any work on the particle? Is this consistent with the factthat the force doesn’t change the particle’s kinetic energy?

5. Show that the potential momentum illustrated in figure 14.2 satisfiesthe Lorentz condition, assuming that U = 0. Would the Lorentz con-dition be satisfied in this case if Q pointed in the x direction?

6. A mass m moves at non-relativistic speed around a circular track ofradius R as shown in figure 14.7. The mass is subject to a potentialmomentum vector of magnitude Q pointing counterclockwise aroundthe track.

(a) If the particle moves at speed v, does it have a longer wavelengthwhen it is moving clockwise or counterclockwise? Explain.


R

Q

v

m

Figure 14.7: The particle is constrained to move along the illustrated trackunder the influence of a potential momentum Q.

(b) Quantization of angular momentum is obtained by assuming thatan integer number of wavelengths n fits into the circumference ofthe track. For given |n|, determine the speed of the mass (i) if it ismoving clockwise (n < 0), and (ii) if it is moving counterclockwise(n > 0).

(c) Determine the kinetic energy of the mass as a function of n.

7. Suppose momentum were conserved for action at a distance in a partic-ular reference frame between particles 1 cm apart as in the left panel offigure 14.4 in the text. If you are moving at velocity 2!108 m s!1 rela-tive to this reference frame, for how long a time interval is momentumapparently not conserved? Hint: The 1 cm interval is the invariantdistance between the kinks in the world lines.

8. An electron moving to the right at speed v collides with a positron (anantielectron) moving to the left at the same speed as shown in figure14.8. The two particles annihilate, forming a virtual photon, whichthen decays into a proton-antiproton pair. The mass of the electron ism and the mass of the proton is M = 1830m.

(a) What is the mass of the virtual photon? Hint: It is not 2m. Why?

14.10. PROBLEMS 255

antiprotonproton

photon

electron positron

Figure 14.8: Electron-positron annihilation leading to proton-antiproton pro-duction.

(b) What is the maximum possible lifetime of the virtual photon bythe uncertainty principle?

(c) What is the minimum v the electron and positron need to have tomake this reaction energetically possible? Hint: How much energymust exist in the proton-antiproton pair?

9. A muon (mass m) interacts with a proton as shown in figure 14.9, sothat the velocity of the muon before the interaction is v, while after theinteraction it is "v/2. The interaction is mediated by a single virtualphoton. Assume that v $ c for simplicity.

(a) What is the momentum of the photon?

(b) What is the energy of the photon?

10. A photon with energy E and momentum E/c collides with an electronwith momentum p = "E/c and mass m. The photon is absorbed, cre-ating a virtual electron. Later the electron emits a photon with energyE and momentum "E/c. (This process is called Compton scatteringand is illustrated in figure 14.10.)

(a) Compute the energy of the electron before it absorbs the photon.

(b) Compute the mass of the virtual electron, and hence the maximumproper time it can exist before emitting a photon.


protonmuon

photon

vel = v

vel = -v/2

Figure 14.9: Collision of a muon with a proton, mediated by the exchange ofa virtual photon.

(c) Compute the velocity of the electron before it absorbs the photon.

(d) Using the above result, compute the energies of the incoming andoutgoing photons in a frame of reference in which the electron isinitially at rest. Hint: Using Ephoton = h& and the above velocity,use the Doppler shift formulas to get the photon frequencies, andhence energies in the new reference frame.

11. The dispersion relation for a negative energy relativistic particle is

& = "(k2c2 + µ2)1/2.

Compute the group velocity of such a particle. Convert the result intoan expression in terms of momentum rather than wavenumber. Com-pare this to the corresponding expression for a positive energy particleand relate it to Feynman’s explanation of negative energy states.

12. The potential energy of a charged particle in a scalar electromageticpotential ( is the charge times the scalar potential. The total energyof such a particle at rest is therefore

E = ±mc2 + q(

where q is the charge on the particle and ±mc2 is the rest energy, the± corresponding to positive and negative energy states. Assume that|q(| $ mc2.

14.10. PROBLEMS 257

electron

photon

photon

Figure 14.10: Compton scattering.

(a) Given that a particle with energy E < 0 is equivalent to thecorresponding antiparticle with energy equal to "E > 0, what isthe potential energy of the antiparticle?

(b) From this, what can you conclude about the charge on the an-tiparticle?

Hint: Recall that the total energy is always rest energy plus kineticenergy (zero in this case) plus potential energy.

Chapter 15

Electromagnetic Forces

In this chapter we begin the study of electromagnetism. The forces oncharged particles due to electromagnetic fields are introduced and relatedto the general case of force on a particle by a gauge field. The principlesof electric motors and generators are then addressed as an example of suchforces in action.

15.1 Electromagnetic Four-Potential

Electromagnetism is a gauge theory. Particles which have a property calledelectric charge are subject to forces exerted by the gauge fields of electro-magnetism. The potential four-momentum Q = (Q, U/c) of a particle withcharge q in the presence of the electromagnetic four-potential a is just

Q = qa. (15.1)

In the simplest case the four-potential represents the amplitude for findingthe intermediary particle associated with the electromagnetic gauge field.This particle has zero mass and is called the photon. If more than onephoton is present, the interpretation of a becomes more complicated. Thisissue will be considered later.

The four-potential has space and time components A and (/c such thata = (A,(/c). The quantity A is called the vector potential and ( is called thescalar potential. The scalar and vector potential are related to the potentialenergy U and potential momentum Q of a particle of charge q by

U = q( Q = qA. (15.2)

259

260 CHAPTER 15. ELECTROMAGNETIC FORCES

The Lorentz condition written in terms of A and ( is

%Ax

%x+%Ay

%y+%Az

%z+

1

c2

%(

%t= 0. (15.3)

15.2 Electric and Magnetic Fields and Forces

The forces caused by electric and magnetic fields are mostly what we canactually measure in electromagnetism. These vector quantities are related tothe scalar and vector potentials as follows:

E = "

&

%(

%x,%(

%y,%(

%z

'

"%A

%t(electric field) (15.4)

B %

&

%Az

%y"%Ay

%z,%Ax

%z"%Az

%x,%Ay

%x"%Ax

%y

'

(magnetic field). (15.5)

By comparison of equations (15.4) and (15.5) with the general expressionfor force in gauge theory, we find that the electromagnetic force on a particlewith charge q is

Fem = "

&

%U

%x,%U

%y,%U

%z

'

"%Q

%t+ v !P

= "q

&

%(

%x,%(

%y,%(

%z

'

" q%A

%t+ qv !B

= qE + qv !B (Lorentz force) (15.6)

where v is the velocity of the particle and where we have used equations(15.2) and (15.4). For historical reasons this is called the Lorentz force.

15.3 A Note on Units

We use the International System of units, often called SI units after theFrench translation of “International System”. In SI units the meter (m),kilogram (kg), and second (s) are fundamental units of length, mass, andtime. As previously noted, the unit of force is the Newton (1 N = 1 kg m s!2)and the unit of energy is the Joule (1 J = 1 N m = 1 kg m2 s!2).

Electromagnetism introduces a new fundamental unit, the Coulomb (C),which is the unit for electric charge. The Lorentz force law tells us that the

15.4. CHARGED PARTICLE MOTION 261

electric field has the units N C!1. The magnetic field has its own derivedunit, i. e., one which can be expressed in terms of fundamental units, namelythe Tesla (T): 1 T = 1 N s C!1 m!1. The vector potential has units T m,while the scalar potential again has its own derived unit, the volt (V): 1 V =1 N m C!1 = 1 J C!1. The electric field can also be expressed in units ofV m!1. A commonly used unit for magnetic field is the Gauss (G). This

non-SI unit is related to the Tesla as follows: 1 G = 10!4 T.The charge on the electron is "e = "1.60!10!19 C. (The sign is arranged

so that e is positive.) A commonly used non-SI unit for energy is the electronvolt (eV). This is the energy gained by an electron passing through a potentialdi!erence of 1 V. Thus, 1 eV = 1.60 ! 10!19 J. Commonly used multiplesof the electron volt are 1 KeV = 103 eV, 1 MeV = 106 eV, 1 GeV = 109 eV,and 1 TeV = 1012 eV.

The electric current is the amount of charge passing some point per unittime. The unit of current is the Ampere (A): 1 A = 1 C s!1.

15.4 Charged Particle Motion

We now explore some examples of the motion of charged particles under theinfluence of electric and magnetic fields.

15.4.1 Particle in Constant Electric Field

Suppose a particle with charge q is exposed to a constant electric field Ex

in the x direction. The x component of the force on the particle is thusFx = qEx. From Newton’s second law the acceleration in the x directionis therefore ax = Fx/m = qEx/m where m is the mass of the particle.The behavior of the particle is the same as if it were exposed to a constantgravitational field equal to qEx/m.

15.4.2 Particle in Conservative Electric Field

If %A/%t = 0, then the electric force on a charged particle is

Felectric = "q

&

%(

%x,%(

%y,%(

%z

'

,%A

%t= 0. (15.7)


z +

z -

E F = qEF = -q Ed/2-

O

θ

z

d/2

-q

q

Figure 15.1: Definition sketch for an electric dipole. Two charges, q and "qare connected by an uncharged bar of length d. The vectors d/2 and "d/2give the positions of the two charges relative to the central point betweenthem. The two forces are due to the electric field E.

This force is conservative, with potential energy U = q(. Recalling that thetotal energy, E = K + U , of a particle under the influence of a conservativeforce remains constant with time, we can infer that the change in the kineticenergy with position of the particle is just minus the change in the potentialenergy, %K = "%U . Notice in particular that if the particle returns to itsinitial position, the change in the potential energy is zero and the kineticenergy recovers its initial value.

If %A/%t (= 0, then there is the possibility that the electric force is notconservative. Recall that the magnetic field is derived from A. Interestingly,a necessary and su"cient criterion for a non-conservative electric force is thatthe magnetic field be changing with time. This result was first inferred ex-perimentally by the English physicist Michael Faraday in 1831 and at nearlythe same time by the American physicist Joseph Henry. It will be furtherexplored later in this chapter.

15.4.3 Torque on an Electric Dipole

Let us now imagine a “dumbbell” consisting of positive and negative chargesof equal magnitude q separated by a distance d, as shown in figure 15.1. Ifthere is a uniform electric field E, the positive charge experiences a forceqE, while the negative charge experiences a force "qE. The net force on thedumbbell is thus zero.

The torque acting on the dumbbell is not zero. The total torque acting


about the origin in figure 15.1 is the sum of the torques acting on the twocharges:

! = ("q)("d/2) ! E + (q)(d/2) ! (E) = qd !E. (15.8)

The vector d can be thought of as having a length equal to the distancebetween the two charges and a direction going from the negative to thepositive charge.

The quantity p = qd is called the electric dipole moment. (Don’t confuseit with the momentum!) The torque is just

! = p! E. (15.9)

This shows that the torque depends on the dipole moment, or the productof the charge and the separation, but not either quantity individually. Thus,halving the separation and doubling the charge results in the same dipolemoment.

The tendency of the torque is to rotate the dipole so that the dipolemoment p is parallel to the electric field E. The magnitude of the torque isgiven by

' = pE sin(!), (15.10)

where the angle ! is defined in figure 15.1 and p = |p| is the magnitude ofthe electric dipole moment.

The potential energy of the dipole is computed as follows: The electro-static potential associated with the electric field is ( = "Ez where E isthe magnitude of the field, assumed to point in the +z direction. Thus, thepotential energy of a single particle with charge q is U = q( = "qEz. Thetotal potential energy of the dipole is the sum of the potential energies of theindividual charges:

U = (+q)("Ez+) + ("q)("Ez!) = "qE(z+ " z!)

= "qEd cos(!) = "pE cos(!) = "p ·E, (15.11)

where z+ and z! are the z positions of the positive and negative charges. Theequating of z+ " z! to d cos(!) may be verified by examining the geometryof figure 15.1.

The tendency of the electric field to align the dipole moment with itselfis confirmed by the potential energy formula. The potential energy is lowestwhen the dipole moment is aligned with the field and highest when the twoare anti-aligned.


charged particle trajectory

magnetic field

end view

Figure 15.2: Spiraling motion of a charged particle in the direction of themagnetic field. This is composed of a circular motion about the field vectorplus a translation along the field.

15.4.4 Particle in Constant Magnetic Field

The magnetic force on a particle with charge q moving with velocity v isFmagnetic = qv ! B, where B is the magnetic field. The magnetic force isdirected perpendicular to both the magnetic field and the particle’s velocity.Because of the latter point, no work is done on the particle by the magneticfield. Thus, by itself the magnetic force cannot change the magnitude of theparticle’s velocity, though it can change its direction.

If the magnetic field is constant, the magnitude of the magnetic forceon the particle is also constant and has the value Fmagnetic = qvB sin(!)where v = |v|, B = |B|, and ! is the angle between v and B. If theinitial velocity is perpendicular to the magnetic field, then sin(!) = 1 andthe force is just Fmagnetic = qvB. The particle simply moves in a circlewith the magnetic force directed toward the center of the circle. This forcedivided by the mass m must equal the particle’s centripetal acceleration:v2/R = a = Fmagnetic/m = qvB/m in the non-relativistic case, where R isthe radius of the circle. Solving for R yields

R = mv/(qB). (15.12)

The angular frequency of revolution is

& = v/R = qB/m (cyclotron frequency). (15.13)

Notice that this frequency is a constant independent of the radius of theparticle’s orbit or its velocity. This is called the cyclotron frequency.

If the initial velocity is not perpendicular to the magnetic field, then theparticle still has a circular component of motion in the plane normal to thefield, but also drifts at constant speed in the direction of the field. The net


qelectricFv

Fmagnetic

E

B

Figure 15.3: With crossed electric E and magnetic B fields (i. e., fieldsperpendicular to each other), a charged particle can move at a constantvelocity v with magnitude equal to v = |E|/|B| and direction perpendicularto both E and B. This is because the electric and magnetic forces, Felectric =qE and Fmagnetic = qv ! B, balance each other in this case.

result is a spiral motion in the direction of the magnetic field, as illustratedin figure 15.2. The radius of the circle is R = mvp/(qB) in this case, wherevp is the component of v perpendicular to the magnetic field.

15.4.5 Crossed Electric and Magnetic Fields

If we have perpendicular electric and magnetic fields as shown in figure 15.3,then it is possible for a charged particle to move such that the electric andmagnetic forces simply cancel each other out. From the Lorentz force equa-tion (15.6), the condition for this happening is E + v ! B = 0. If E and B

are perpendicular, then this equation requires v to point in the direction ofE!B (i. e., normal to both vectors) with the magnitude v = |E|/|B|. This,of course, is not the only possible motion under these circumstances, just thesimplest.

It is interesting to consider this situation from the point of view of areference frame which is moving with the charged particle. In this referenceframe the particle is stationary and therefore not subject to the magneticforce. Since the particle is not accelerating, the net force, which in thisframe consists only of the electric force, is zero. Hence, the electric fieldmust be zero in the moving reference frame.

This argument shows that the electric field perceived in one reference


_a_a

φ ’/c

ct

x

ct’ct’

x’

a’x

x

ct

x’

Figure 15.4: The four-potential a and its components in two di!erent refer-ence frames. In the unprimed frame the four-potential is purely space-like.The primed frame is moving in the x direction at speed U relative to theunprimed frame. The four-potential points along the x axis.

frame is not necessarily the same as the electric field perceived in anotherframe. Figure 15.4 shows why this is so. The left panel shows the situation inthe reference frame moving to the right, which is the unprimed frame in thispicture. The charged particle is stationary in this reference frame. The four-potential is purely spacelike, having no time component (/c. Assuming thata is constant in time, there is no electric field, and hence no electric force.Since the particle is stationary in this frame, there is also no magnetic force.However, in the primed reference frame, which is moving to the left relativeto the unprimed frame and therefore is equivalent to the original referenceframe in which the particle is moving to the right, the four-potential has atime component, which means that a scalar potential and hence an electricfield is present.

15.5 Forces on Currents in Conductors

So far we have talked mainly about point charges moving in free space. How-ever, many practical applications of electromagnetism have charges movingthrough a conductor such as copper. A conductor is a material in whichelectrically charged particles can freely move. An insulator is a material inwhich charged particles are fixed in place. Practical conductors are often sur-

15.5. FORCES ON CURRENTS IN CONDUCTORS 267

Δx

-

+

+

+

+

-

- -

- +

Figure 15.5: Fixed positive charge and negative charge moving to the rightwith speed v in blown up segment of wire.

rounded by insulators in order to confine the motion of charge to particularpaths.

The current through a wire is defined as the amount of charge passingthrough the wire per unit time. When defining current, one needs to decidewhich direction constitutes a positive current for the problem at hand, i.e., the direction in which positive charge is moving. If the current consistsof particles carrying negative charge, then the direction of the current isopposite the direction of the motion of the particles.

Metals tend to be good conductors, while glass, plastic, and other non-metallic materials are usually insulators. All materials contain both positiveand negative charges. In metals, negatively charged electrons can escapefrom atoms and are free to move about the material. When atoms lose oneor more electrons, they become positively charged. Atoms tend to be fixedin place. Since the electron charge is negative, the current in a wire actuallyhas a direction opposite the direction of motion of the electrons, as notedabove.

If a conductor is in the form of a wire, we can compute the magnetic forceon the wire if we know the number of mobile particles per unit length of wireN , the charge on each particle q, and the speed v with which they are movingdown the wire. The total force on a length of wire L is F = qNLvn ! B,where n is a unit vector pointing in the direction of motion of the particlesthrough the wire. The quantity i % qNv is called the current in the wire. Itequals the amount of charge per unit time flowing down the wire. Writtenin terms of the current, the force on a length L of the wire is

F = iLn !B. (15.14)


w

d

i

F

F

m

B θ

1

2

4

3 θ

BF

F

Perspective view Side view

m

axle

torque

Figure 15.6: Perspective and side views of a rectangular loop of wire mountedon an axle in a magnetic field. Forces on the currents in loop segments 1 and3 generate a torque about the axle.

15.6 Torque on a Magnetic Dipole and Elec-

tric Motors

Figure 15.6 shows a rectangular loop of wire mounted on an axle in a magneticfield. A current i exists in the loop as shown. The currents in loop segments2 and 4 experience a force parallel to the axle. These forces generate no nettorque. However, the magnetic forces on loop segments 1 and 3 are eachF = idB in magnitude, where B = |B| is the magnitude of the magneticfield. Together these forces generate a counterclockwise torque about theaxle equal to ' = 2F (w/2) sin(!) = iwdB sin(!). This can be represented invector form as

! = m! B, (15.15)

where m is a vector with magnitude iwd and direction normal to the loopas shown in figure 15.6. The vector m is called the magnetic dipole moment.

The loop can actually be any shape, not just rectangular. In the generalcase the magnitude of the magnetic moment equals the current i times thearea S of the loop:

|m| = iS (magnetic dipole moment). (15.16)

In the above example the area is S = wd. The direction of m is determined

15.7. ELECTRIC GENERATORS AND FARADAY’S LAW 269

by the right hand rule; curl the fingers on your right hand around the loopin the direction of the current and your thumb points in the direction of m.

In analogy with the electric dipole in an electric field, the potential energyof a magnetic dipole in a magnetic field is

U = "m · B. (15.17)

Figure 15.6 illustrates the principle of an electric motor. A motor consistsof multiple loops of wire on an axle carrying a current in a magnetic field.The torque on the axle turns the loops so that the magnetic moment isparallel to the field. The angular momentum of the loops carries the rotationof the axle through the zero torque region, which occurs when the magneticmoment is either perfectly parallel or perfectly anti-parallel (i. e., pointingin the opposite direction) to the field. At this point either the magnetic fieldis reversed by some mechanism or the magnetic dipole is reversed by makingthe current circulate around the loops in the opposite direction. The torquedue to the magnetic force then turns the axle through another half-turn,whereupon the field or the magnetic moment is again reversed, and so on.

15.7 Electric Generators and Faraday’s Law

As was shown earlier, the electric field is derived from two di!erent sources,spatial derivatives of the scalar potential and time derivatives of the vectorpotential:

Ex = "%(

%x"%Ax

%tEy = "

%(

%y"%Ay

%tEz = "

%(

%z"%Az

%t. (15.18)

In time independent situations the vector potential part drops out and weare left with a dependence only on the scalar potential. In this case a particlewith charge q has an electrostatic potential energy U = q(, which means thatthe electric force is conservative. However, in the time dependent situationthere is no guarantee that the part of the electric field derived from the vectorpotential will be conservative.

An example of a non-conservative electric field occurs when we have

A = (Cyt,"Cxt, 0) ( = 0 (15.19)

where C is a constant. In this case the electric and magnetic fields are

E = ("Cy, Cx, 0) B = (0, 0,"2Ct). (15.20)


q

R

Figure 15.7: Illustration of the electric field pattern E = ("Cy, Cx, 0) (smallarrows). A charged particle moving in a circle as shown continually gainsenergy.

The magnetic field points in the "z direction and increases in magnitudewith time. The electric field vectors are shown in figure 15.7. Notice thata positively charged particle moving in a counterclockwise circle as shownis continually being accelerated in the direction of motion, and is thereforecontinually gaining energy. This is impossible with a conservative force.

How much energy is gained by a particle with charge q moving in acomplete circle of radius R under the above circumstances? The magnitudeof the electric field at this radius is E = CR, so the force on the particle isF = qCR. The circumference of the circle is 2"R, so the total work done bythe electric field in one revolution is just %W = 2"RF = 2"qCR2 = 2qCS,where S = "R2 is the area of the circle. Let us define %V = %W/q = 2CS.For historical reasons this is called the electromotive force or EMF. This isdeceptive terminology, because in fact %V doesn’t have the dimensions offorce — it is really just the work per unit charge done on a particle makinga single loop around the circle in figure 15.7.

Recall that the z component of the magnetic field in this case is Bz ="2Ct. Note that the time derivative of the magnetic field is just %Bz/%t =

15.7. ELECTRIC GENERATORS AND FARADAY’S LAW 271

motion of charge

dB/dt

Figure 15.8: Sketch for Faraday’s law. The arrows passing through the loopindicate the direction of the time rate of change of the magnetic field. Thearrow going around the loop indicates the direction a positive charge wouldbe pushed by the electric field.

"2C. Comparison with the equation for electromotive force shows us that

%V = "%Bz

%tS = "

%BzS

%t, (15.21)

where the area is brought inside the time derivative since it is constant intime. This is a special case of a general law in electromagnetism calledFaraday’s law.

Notice that the argument of the time derivative in the above equation isthe component of B perpendicular to the plane of the loop. The loop areatimes the normal component of B is the magnetic flux through the loop:#B % BnormalS. Faraday’s law is expressed most compactly as

%V = "d#B

dt(Faraday’s law), (15.22)

and it turns out to be valid for arbitrary loops and arbitrary magnetic fieldconfigurations, not just for the simple loop we have been investigating. Themost general statement of the law is that the EMF around a closed loopequals minus the time rate of change of magnetic flux through the loop.

The minus sign in equation (15.22) means the following: If the fingers onyour right hand curl around the loop in the direction opposite to the directionwhich causes a positive charge to gain energy, then your thumb points in thedirection of the time rate of change of the magnetic flux passing through theloop. This is illustrated in figure 15.8.


w

d

i

B θ

1

2

4

3axle

normal

Figure 15.9: Rotating wire loop in a magnetic field. At the instant illustratedthe magnetic flux is increasing with time, which means that an EMF tendsto drive a current as illustrated.

The electric generator is perhaps the best known application of Faraday’slaw. Figure 15.9 shows a rectangular loop of wire fixed to an axle whichrotates at an angular rate &. The magnetic flux through the loop thus varieswith time according to #B = wdB cos(!) = wdB cos(&t). The EMF aroundthe loop is thus

%V = "d#B

dt= &wdB sin(&t). (15.23)

In a real generator there are many loops forming a coil of wire and the endsof the coil are brought out through the axle so that the resulting current canbe tapped for practical use.

15.8 EMF and Scalar Potential

The EMF %V has the same units as the scalar potential (. What is thedi!erence between the two quantities? Both represent work done per unitcharge by the electric field on a particle moving through the field. However,recall that the electric field is composed of two parts:

E = "

&

%(

%x,%(

%y,%(

%z

'

"%A

%t. (15.24)

15.9. PROBLEMS 273

%( = (2"(1 is minus the work done on the particle in going from point 1 topoint 2 by the part of the electric field associated with the scalar potential.%V is (plus) the work done by the part of the electric field associated withthe time derivative of the vector potential.

Aside from the di!erent sign conventions, there is one other fundamentaldi!erence between the two quantities: %( is always zero for closed paths, i.e., paths in which the particle returns to its initial point. This is becausepoint 1 is then the same as point 2, so (1 = (2. This condition doesn’tnecessarily apply to the EMF. %V often is non-zero for closed particle paths.The electric generator which we have just discussed is an important case inpoint. The total work done per unit charge by the electric field on a chargedparticle moving along some path is thus %V " %(. The %( term drops outif the path is closed.

15.9 Problems

1. Given a four-potential a = (Cyt,"Cxt, 0, 0) where C is a constant:

(a) Determine whether this four-potential satisfies the Lorentz condi-tion.

(b) Compute the electric and magnetic fields from this four-potential.

2. Given a1 = (C1zt, 0, 0, C2x), find the electric and magnetic field compo-nents. Compare with the fields you get from a2 = (C1zt, C3y,"C3z, C2x).C1, C2, and C3 are constants. Can one have more than one four-potential field giving rise to the same electric and magnetic fields?

3. Suppose that in the rest frame we have a four-potential of the forma = (0, 0, 0, Ky) where K is a constant.

(a) Find the electric and magnetic fields in this frame.

(b) Find the components of a in a reference frame moving in the "xdirection at speed U . Hint: Draw a spacetime diagram showingthe a vector and resolve into components in the moving frameusing the spacetime Pythagorean theorem.

(c) Find the electric and magnetic fields in the moving frame.


4. Assume a four-potential of the form a = (A,(/c), where A = (Ky, 0, 0)and ( = 0 in the rest frame, K being a constant.

(a) Compute the electric and magnetic fields in the rest frame.

(b) Find the components of the four-potential in a reference framemoving in the "x direction at speed U .

(c) Compute the electric and magnetic fields in the moving frameusing the above results.

5. Using the right-hand rule, show that the electric torque acting on anelectric dipole tries to align the dipole so that it is in its state of lowestpotential energy.

6. The net electric force on an electric dipole is zero in a uniform electricfield. However, if the field varies with position, this is not necessarilytrue. Consider an electric field which has the form E = E0(1 + )z)kalong the z axis, where E0 and ) are positive constants.

(a) An electric dipole consisting of charges ±q spaced by a distance dis centered at the origin. If the dipole is aligned with the electricfield, determine the direction and magnitude of the net force onthe dipole.

(b) Determine the force on the dipole if it is anti-aligned with (i. e.,pointing in the opposite direction from) the electric field.

7. Suppose that a charged particle is moving under the influence of electricand magnetic fields such that it periodically returns to some point P. Ifthe four-potential is independent of time, will the kinetic energy of theparticle be the same or di!erent every time it returns to P? Explain.

8. Given constant electric and magnetic fields E = Ej and B = Bk:

(a) Find the velocity (magnitude and direction) of a charged particlefor which the Lorentz force is zero.

(b) Using this result, describe how you would build a setup to selectout only those particles in a beam moving at a certain velocity.

9. Determine qualitatively how a charged particle moves in crossed elec-tric and magnetic fields in the general case in which it is not moving at

15.9. PROBLEMS 275

i

B45o

View lookingdown

wire

Figure 15.10: Horizontal wire with current i in a magnetic field.

z

i

BB

Figure 15.11: Magnetic dipole (current loop) in an inhomogeneous magneticfield.

constant velocity. For the sake of definiteness, assume that the mag-netic field points in the +z direction and the electric field in the +xdirection. Hint: Is there a reference frame in which the electric fieldvanishes? If there is, describe the motion in this reference frame andthen determine how this motion looks in the original reference frame.

10. A horizontal wire of mass per unit length 0.1 kg m!1 passes through ahorizontal magnetic field of strength B = 0.1 T with an orientation of45# to the field as shown in figure 15.10. What current must the wirecarry for the magnetic force on the wire to just balance gravity?

11. Figure 15.11 shows a current loop in a magnetic field. The magneticfield diverges with increasing z, so that its magnitude decreases with


v

w

d = vt

B

Figure 15.12: A moving crossbar on a U-shaped wire in a magnetic field.

height.

(a) Which way does the magnetic dipole vector due to the currentloop point?

(b) Is this dipole oriented so as to have maximum or minimum poten-tial energy, or is it somewhere in between?

(c) Is there a net force on the dipole? If so, what direction does itpoint? Hint: Determine the direction of the v ! B force at eachpoint on the current loop. What direction does the sum of allthese forces point?

12. A charged particle moving in a circle in a magnetic field constitutes acircular current which forms a magnetic dipole.

(a) Determine whether the dipole moment produced by this currentis aligned or anti-aligned with the initial magnetic field.

(b) Do charged particles moving in a non-uniform magnetic field asshown in figure 15.11 tend to accelerate toward regions of strongeror weaker field?

13. Why do electric motors have many turns of wire around the loop whichcuts the magnetic field instead of just one? Hint: Magnetic fields

15.9. PROBLEMS 277

aq

Figure 15.13: The charged bead continuously accelerates around the loopdue to electromagnetic fields.

in normal motors are of order 0.1 T and currents are typically a fewamps. Estimate the torque on a reasonably sized current loop for theseconditions. Compare this to the torque you could expect to exert withyour hand acting on a 1 m moment arm.

14. Imagine a stationary U-shaped conductor with a moving conducting barin contact with the U as shown in figure 15.12. A uniform magneticfield exists normal to the plane of the U and has magnitude B. Thebar is moving outward along the U at speed v as shown.

(a) Using the fact that the charged particles in the moving bar aresubject to a Lorentz force due to the motion of the bar througha magnetic field, compute the EMF around the closed loop con-sisting of the bar and the U. Hint: Recall that the EMF is thework done per unit charge on a charged particle moving aroundthe loop.

(b) Compute the EMF around the above loop using Faraday’s law. Isthe answer the same as obtained above?

15. A bead on a loop has charge q and accelerates continuously aroundthe loop in the counterclockwise direction, as shown in figure 15.13.Explain qualitatively what this information tells you about

(a) the vector potential in the vicinity of the loop, and

(b) the magnetic flux through the loop.

Chapter 16

Generation of ElectromagneticFields

In this section we investigate how charge produces electric and magneticfields. We first introduce Coulomb’s law, which is the basis for everythingelse in the section. We then discuss Gauss’s law for the electric and magneticfield, drawing on what we learned while using it on the gravitational field.Coulomb’s law plus the theory of relativity together show that magneticfields are generated by moving charge. We then use this fact to computethe magnetic fields from some simple charge distributions. We finish with adiscussion of electromagnetic waves.

16.1 Coulomb’s Law and the Electric Field

A stationary point electric charge q is known to produce a scalar potential

( =q

4"*0r(16.1)

a distance r from the charge. The constant *0 = 8.85 ! 10!12 C2 N!1 m!2

is called the permittivity of free space. The vector potential produced by astationary charge is zero.

The potential energy between two stationary charges is equal to the scalarpotential produced by one charge times the value of the other charge:

U =q1q2

4"*0r. (16.2)

279

280 CHAPTER 16. GENERATION OF ELECTROMAGNETIC FIELDS

Notice that it doesn’t make any di!erence whether one multiplies the scalarpotential from charge 1 by charge 2 or vice versa – the result is the same.

Since r = (x2 + y2 + z2)1/2, the electric field produced by a charge is

E = "

&

%(

%x,%(

%y,%(

%z

'

=qr

4"*0r3(16.3)

where r = (x, y, z) is the vector from the charge to the point where theelectric field is being measured. The magnetic field is zero since the vectorpotential is zero.

The force between two stationary charges separated by a distance r isobtained by multiplying the electric field produced by one charge by theother charge. Thus the magnitude of the force is

F =q1q2

4"*0r2(Coulomb’s law), (16.4)

with the force being repulsive if the charges are of the same sign, and attrac-tive if the signs are opposite. This is called Coulomb’s law.

Equation (16.4) is the electric equivalent of Newton’s universal law ofgravitation. Replacing mass by charge and G by "1/(4"*0) in the equa-tion for the gravitational force between two point masses gives us equation(16.4). The most important aspect of this result is that both the gravita-tional and electrostatic forces decrease as the square of the distance betweenthe particles.

16.2 Gauss’s Law for Electricity

The electric flux is defined in analogy to the gravitational flux as

#E = S · E (electric flux) (16.5)

where S is the directed area through which the flux passes. Since the electricfield obeys an inverse square law, Gauss’s law applies to the electric flux #E

just as it applies to the gravitational flux. In particular, since the magnitudeof the outward electric field a distance r from a charge q is E = q/(4"*0r2),the electric flux through a sphere of radius r (and area 4"r2) concentric withthe charge is ES = [q/(4"*0r2)] ! (4"r2) = q/*0. This generalizes to anarbitrary distribution of charge as in the gravitational case:

#E = qinside/*0 (Gauss’s law for electricity), (16.6)

16.2. GAUSS’S LAW FOR ELECTRICITY 281

xh

sheet of charge

electric field

Figure 16.1: Definition sketch for use of Gauss’s law to obtain the electricfield due to an infinite sheet of surface charge. The dashed line shows theGaussian box, which is of height h and depth d into the page.

where #E in this equation is the outward electric flux through a closed surfaceand qinside is the net charge inside this surface. This is an expression ofGauss’s law for the electric field. Since Gauss’s law for electricity and forgravitation are so similar, we can use all our insights from studying gravityon the electric field case.

16.2.1 Sheet of Charge

Figure 16.1 shows how to set up the Gaussian surface to obtain the electricfield emanating from an infinite sheet of charge. We assume a charge densityof $ Coulombs per square meter, which means that the amount of chargeinside the box is qinside = $hd, where the box has height h and depth d intothe page. The total electric flux out of the left and right faces of the box is#E = 2Ehd, where E is the magnitude of the electric field on these surfaces.The field is assumed to point away from the charge, and hence out of thebox on both faces. Due to the assumed direction of the electric field, thereis no electric flux out of any of the other faces of the box.


d

R

perspective viewend view of Gaussian cylinder

line of charge

R

line of charge

Figure 16.2: Definition sketch for use of Gauss’s law to obtain the electricfield due to an infinite line charge oriented normal to the page. The dashedline shows the Gaussian cylinder, which is of radius R and length d into thepage. The outward-pointing arrows show the electric field.

Applying Gauss’s law, we infer that 2Ehd = $hd/*0, which means thatthe electric field emanating from a sheet of charge with charge density perunit area $ is

E =$

2*0. (16.7)

The scalar potential associated with this electric field is easily obtainedby realizing that equation (16.7) gives the x component of this field — theother components are zero. Using E = "%(/%x, we infer that

( = "$|x|

2*0. (16.8)

The absolute value signs around x take account of the fact that the directionof the electric field for negative x is opposite that for positive x.

16.2.2 Line of Charge

Similar reasoning is used to obtain the electric field due to a line of charge. Asketch of the expected electric field vectors and a Gaussian cylinder coaxial

16.3. GAUSS’S LAW FOR MAGNETISM 283

B

closed surface open surface

Figure 16.3: Illustration for Gauss’s law for magnetism. The net flux out ofthe closed surface is zero, but the flux through the open surface is not.

with the line of charge is shown in figure 16.2. If the charge per unit lengthis +, the amount of charge inside the cylinder is qinside = +d, where d is thelength of the cylinder. The outward electric flux at radius r is #E = 2"rdE.Gauss’s law therefore tells us that the electric field at radius r is just

E =+

2"*0r. (16.9)

In this case E = "%(/%r, so that the scalar potential is

( = "+

2"*0ln(r). (16.10)

16.3 Gauss’s Law for Magnetism

By analogy with Gauss’s law for the electric field, we could write a Gauss’slaw for the magnetic field as follows:

#B = Cqmagnetic inside, (16.11)

where #B is the outward magnetic flux through a closed surface, C is a con-stant, and qmagnetic inside is the “magnetic charge” inside the closed surface.Extensive searches have been made for magnetic charge, generally called a


magnetic monopole. However, none has ever been found. Thus, Gauss’s lawfor magnetism can be written

#B = 0 (Gauss’s law for magnetism). (16.12)

This of course doesn’t preclude non-zero values of the magnetic flux throughopen surfaces, as illustrated in figure 16.3.

16.4 Coulomb’s Law and Relativity

The equation (16.1) for the scalar potential of a point charge is valid only inthe reference frame in which the charge q is stationary. By symmetry, thevector potential must be zero. Since ( is actually the time-like component ofthe four-potential, we infer that the four-potential due to a charge is tangentto the world line of the charged particle.

A consequence of the above argument is that a moving charge produces amagnetic field, since the four-potential must have space-like components inthis case.

16.5 Moving Charge and Magnetic Fields

We have shown that electric charge generates both electric and magneticfields, but the latter result only from moving charge. If we have the scalarpotential due to a static configuration of charge, we can use this result tofind the magnetic field if this charge is set in motion. Since the four-potentialis tangent to the particle’s world line, and hence is parallel to the time axisin the reference frame in which the charged particle is stationary, we knowhow to resolve the space and time components of the four-potential in thereference frame in which the charge is moving.

Figure 16.4 illustrates this process. For a particle moving in the +xdirection at speed v, the slope of the time axis in the primed frame is justc/v. The four-potential vector has this same slope, which means that thespace and time components of the four-potential must now appear as shownin figure 16.4. If the scalar potential in the primed frame is (", then in theunprimed frame it is (, and the x component of the vector potential is Ax.Using the spacetime Pythagorean theorem, ("2/c2 = (2/c2"A2

x, and relating

16.5. MOVING CHARGE AND MAGNETIC FIELDS 285

x

φ/c φ’/c

x

x’

ct ct’A

Figure 16.4: Finding the space and time components of the four-potentialproduced by a particle moving at the velocity of the primed reference frame.The ct" axis is the world line of the charged particle which generates thefour-potential.

slope of the ct" axis to the components of the four-potential, c/v = ((/c)/Ax,it is possible to show that

( = ,(" Ax = v,("/c2 (16.13)

where

, =1

(1 " v2/c2)1/2. (16.14)

Thus, the principles of special relativity allow us to obtain the full four-potential for a moving configuration of charge if the scalar potential is knownfor the charge when it is stationary. From this we can derive the electric andmagnetic fields for the moving charge.

16.5.1 Moving Line of Charge

As an example of this procedure, let us see if we can determine the magneticfield from a line of charge with linear charge density in its own rest frameof +", aligned along the z axis. The line of charge is moving in a directionparallel to itself. From equation (16.10) we see that the scalar potential adistance r from the z axis is

(" = "+"

2"*0ln(r) (16.15)


z

r

v

Amoving line ofcharge

Figure 16.5: Vector potential from a moving line of charge. The distributionof vector potential around the line is cylindrically symmetric.

in a reference frame moving with the charge. The z component of the vectorpotential in the stationary frame is therefore

Az = "+"v,

2"*0c2ln(r) (16.16)

by equation (16.13), with all other components being zero. This is illustratedin figure 16.5.

We infer that

Bx =%Az

%y= "

+"v,y

2"*0c2r2By = "

%Az

%x=

+"v,x

2"*0c2r2Bz = 0, (16.17)

where we have used r2 = x2 + y2. The resulting field is illustrated in fig-ure 16.6. The field lines circle around the line of moving charge and themagnitude of the magnetic field is

B = (B2x + B2

y)1/2 =

+"v,

2"*0c2r. (16.18)

There is an interesting relativistic e!ect on the charge density +", whichis defined in the co-moving or primed reference frame. In the unprimedframe the charges are moving at speed v and therefore undergo a Lorentzcontraction in the z direction. This decreases the charge spacing by a factorof , and therefore increases the charge density as perceived in the unprimedframe to a value + = ,+".

16.5. MOVING CHARGE AND MAGNETIC FIELDS 287

x

y

Figure 16.6: Magnetic field from a moving line of charge. The charge ismoving along the z axis out of the page.

We also define a new constant µ0 % 1/(*0c2). This is called the per-meability of free space. This constant has the assigned value µ0 = 4" !10!7 N s2 C!2. The value of *0 = 1/(µ0c2) is actually derived from this as-signed value and the measured value of the speed of light. The reasons forthis particular way of dealing with the constants of electromagetism are ob-scure, but have to do with making it easy to relate the values of constantsto the experiments used in determining them.

With the above substitutions, the magnetic field equation becomes

B =µ0+v

2"r. (16.19)

The combination +v is called the current and is symbolized by i. The currentis the charge per unit time passing a point and is a fundamental quantity inelectric circuits. The magnetic field written in terms of the current flowingalong the z axis is

B =µ0i

2"r(straight wire). (16.20)


16.5.2 Moving Sheet of Charge

As another example we consider a uniform infinite sheet of charge in thex"y plane with charge density $". The charge is moving in the +x directionwith speed v. As we showed in the section on Gauss’s law for electricity, theelectric field for this sheet of charge in the co-moving reference frame is inthe z direction and has the value

E "

z =$"

2*0sgn(z) (16.21)

where we define

sgn(z) %

*

+

,

+

-

"1 z < 00 z = 01 z > 0

. (16.22)

The sgn(z) function is used to indicate that the electric field points upwardabove the sheet of charge and downward below it (see figure 16.7).

The scalar potential in this frame is

(" = "$"|z|

2*0. (16.23)

In the stationary reference frame in which the sheet of charge is movingin the x direction, the scalar potential and the x component of the vectorpotential are

( = ",$"|z|

2*0= "

$|z|

2*0Ax = "

v,$"|z|

2*0c2= "

v$|z|

2*0c2, (16.24)

according to equation (16.13), where $ = ,$" is the charge density in thestationary frame. The other components of the vector potential are zero.We calculate the magnetic field as

Bx = 0 By =dAx

dz= "

v$

2*0c2sgn(z) Bz = 0 (16.25)

where sgn(z) is defined as before. The vector potential and the magneticfield are shown in figure 16.7. Note that the magnetic field points normal tothe direction of motion of the charge but parallel to the sheet. It points inopposite directions on opposite sides of the sheet of charge.

16.6. ELECTROMAGNETIC RADIATION 289

x

y

v

z

B

A

A

B

E

E

σ

Figure 16.7: Vector potential A, electric field E, and magnetic field B froma moving sheet of charge. The charge is moving in the x direction.

16.6 Electromagnetic Radiation

We have found so far that stationary charge produces an electric field whilemoving charge produces a magnetic field. It turns out that accelerated chargeproduces electromagnetic radiation. Electromagnetic radiation is nothingmore than one or more photons which have zero mass, and are therefore real,not virtual.

Acceleration of a charged particle is needed to produce radiation becauseof the conservation of energy and momentum. The left panel of figure 16.8shows why. Since a photon carries o! energy and momentum, conservationmeans that the energy and momentum of the emitting particle change dueto the emission of a photon. This corresponds in classical mechanics to anacceleration.

The process in the left panel of figure 16.8 actually cannot occur if par-ticles A and B have the same mass. If the mass of the outgoing particle Bis less than the mass of the incoming particle A, then this reaction can anddoes occur. An example is the decay of an atom from a higher energy stateto a lower energy state (and hence lower mass), accompanied by the emissionof a photon.

Another type of reaction which can generate radiation occurs when two


A

B

real photon

virtualelectron

virtualphoton

Figure 16.8: Feynman diagrams for two processes which potentially mightproduce real photons and hence electromagnetic radiation. The process inthe left panel turns out to be impossible if the masses of particles A and Bare the same, for reasons discussed in the text. The process in the right paneloccurs commonly. Solid lines represent electrons while dashed lines representphotons. Particles are taken to be real unless otherwise labeled.

charged particles (say, electrons) collide, as illustrated in the right panel offigure 16.8. In an elastic collision both electrons are real both before andafter the photon transfer. However, it is possible for one of the electrons tohave a virtual mass which is greater than the normal electron mass after thecollision, which means that it is free to decay to a real electron plus a realphoton.

We now try to understand the characteristics of free electromagnetic radi-ation. In our studies of waves we found it easiest to examine plane waves. Wewill follow this path here, writing the four-potential for an electromagneticplane wave moving in the x direction as

a = (A,(/c) = (A0,(0/c) cos(kxx " &t), (16.26)

where a0 = (A0,(0/c) is a constant four-vector representing the direction andmaximum amplitude of the four-potential and kx and & are the wavenumberand the angular frequency of the wave. Since the real photon is massless, wehave & = kxc in this case. Virtual photons are not subject to this constraint.

By substituting A and ( from equation (16.26) into the Lorentz condition,we find that

kxAx " &(/c2 = 0 (Lorentz condition for plane wave). (16.27)

Thus, the Lorentz condition requires that the scalar potential ( be relatedto the x or longitudinal component of the vector potential, Ax, i. e., the

16.6. ELECTROMAGNETIC RADIATION 291

View from y axis of B field vectorsPerspective view of E and B field orientation

x

x

y

z

B

E

Figure 16.9: Electric and magnetic fields in a horizontally polarized planewave, i. e., with Az = 0, moving in the direction of the large arrow. The leftpanel shows how the electric and magnetic fields point, while the right panelshows the distribution of the magnetic field in space.

component pointing in the direction of wave propagation. The transversecomponents, Ay and Az, are unconstrained by the Lorentz condition, sincethey don’t depend on y and z.

Using equations for the electric and magnetic field as well as equations(16.26) and (16.27), we can now find E and B in an electromagnetic planewave:

B = (0, kxA0z ,"kxA0y) sin(kxx " &t) (16.28)

E = (kx(0 " &A0x,"&A0y,"&A0z) sin(kxx " &t). (16.29)

The electric field has a longitudinal or x component proportional to kx(0 "&A0x = "&((0/c " A0x). However, comparison with equation (16.27) showsthat Ex = 0 as long as &/kx = c, i. e., as long as the photons travel at thespeed of light, c. Thus, virtual photons, i. e., those which have a non-zeromass and therefore travel at a speed other than that of light, can have a non-zero longitudinal component of the electric field, but real photons cannot.

The dot product of the electric and magnetic fields in a plane wave isE ·B = 0, as can be verified from equations (16.28) and (16.29). This meansthat E and B are perpendicular to each other. Furthermore, both E and Bare perpendicular to the direction of wave motion for real photons.


Figure 16.9 shows the electric and magnetic fields for real photons in thespecial case where Az = 0. The electric field points in the same direction asthe transverse part of the vector potential, while the magnetic field pointsin the other transverse direction. The ratio of the magnitudes of the electricand magnetic fields is easily inferred from equations (16.28) and (16.29):

|E||B|

=&(A2

y + A2z)

1/2 sin(kxx " &t)

kx(A2z + A2

y)1/2 sin(kxx " &t)

=&

kx= c. (16.30)

Notice that the electric and magnetic fields for a wave do not dependon the longitudinal component of the vector potential, Ax. This is becausethe Lorentz condition forces Ax to cancel with the term containing ( in theexpression for Ex.

16.7 The Lorentz Condition

We are now in a position to see what the Lorentz condition means. For anisolated stationary charge, the scalar potential is given by equation (16.1)and the vector potential A is zero. The Lorentz condition reduces to

1

c2

%(

%t=

1

4"*0rc2

dq

dt= 0. (16.31)

From this we see that the Lorentz condition applied to the four-potentialfor a point charge is equivalent to the statement that the charge on a pointparticle is conserved, i. e., it doesn’t change with time. This is extended toany stationary distribution of charge by the superposition principle.

We thus see that the Lorentz condition is a consequence of charge conser-vation for the four-potential of any charge distribution in the reference framein which the charge is stationary. If we can further show that the Lorentzcondition is an equation which is equally valid in all reference frames, thenwe will have demonstrated that it is true for the four-potential produced bymoving charged particles as well.

If the Lorentz condition is valid in one reference frame, it is valid in allframes for the special case of a plane electromagnetic wave. This follows fromsubstituting the four-potential for a plane wave into the Lorentz condition, aswas done in equation (16.27) in the previous section. In this case the Lorentzcondition reduces to k · a = 0. Since the dot product of two four-vectors is a

16.8. PROBLEMS 293

A B C

σ −σ

Figure 16.10: Two parallel sheets of charge, one with surface charge density$, the other with "$.

relativistic scalar, the Lorentz condition is equally valid in all frames. Sincewe believe that charge is indeed conserved in all circumstances, the Lorentzcondition must always be satisfied.

16.8 Problems

1. Imagine that an electron actually consists of two point charges, eachwith charge e/2, separated by a distance D, where e is the charge on theelectron. Compute D such that the potential energy of the two chargesequals the rest energy of the electron. Look up the constants andcompute a numerical value for D. Finally, compute the force betweenthe two charges and compare to the gravitational force between twomasses each equal to half the electron mass separated by this distance.

2. Verify that the equations for the scalar potentials associated with asheet and a line of charge, (16.8) and (16.10), yield the correspondingelectric fields.

3. Two sheets of charge, one with charge density $, the other with "$,are aligned as shown in figure 16.10. Compute the electric field in eachof the regions A, B, and C.

4. Positive charge is distributed uniformly on the upper surface of aninfinite conducting plate with charge per unit area $ as shown in figure16.11. Use Gauss’s law to compute the electric field above the plate.Hint: Is there any electric field inside the plate?


+ + + + + + + +conductor

Figure 16.11: A charged metal plate.

B

Figure 16.12: Hypothesized magnetic field. Does it satisfy Gauss’s law formagnetism?

5. Suppose a student proposes that a magnetic field can take the formshown in figure 16.12. Is the proposed form of the magnetic field con-sistent with Gauss’s law for magnetism? Explain.

6. The magnetic flux through the sides of the cone illustrated in figure16.13 is zero. The magnetic field may be assumed to be approximatelynormal to the ends of the cone and the magnetic flux into the left endis #B. The areas of the left and right ends of the cone are Sa and Sb.

(a) What is the magnetic flux out of the right end of the cone?

(b) What is the value of the magnetic field B on the left end of thecone?

(c) What is the value of B on the right end?

7. In the lab frame a wire has negative charge with linear charge density"+ moving at speed "U corresponding to a current i = +U as shownin figure 16.14. Positive charge is stationary, and has charge density +,so the net charge is zero.

16.8. PROBLEMS 295

aSbS

side

side

sideB B

Figure 16.13: Converging magnetic field passing through a closed surface.

lab frame−U

moving frame U

Figure 16.14: A horizontal wire with current i viewed in two di!erent refer-ence frames.

(a) What are the electric and magnetic fields produced by the chargein the wire in the stationary frame?

(b) In a reference frame moving at velocity "U in the x direction, suchthat the negative charge is stationary, what is the apparent linearcharge density of (1) the negative charge, and (2) the positivecharge? Hint: The Lorentz contraction must be taken into accounthere.

(c) What is the electric field produced by the charge in the wire in themoving frame? Hint: Do the charge densities from the positiveand negative charge cancel in this frame?

(d) What is the current in the wire in the moving frame, and hence,what is the magnetic field around the wire in this frame? Hint: Isthe positive or negative charge causing the current in this frame?

(e) Explain why the net force on a separate charged particle somedistance from the wire and stationary in the lab frame is zero inboth reference frames.


8. The left panel of figure 16.8 shows a real charged particle A emittinga real photon, turning into a possibly di!erent real particle B afterthe emission. If particle A and particle B have the same mass, showthat this process is energetically impossible. Hint: Work in a referenceframe in which particle A is stationary.

9. Given the four-potential for an electromagnetic plane wave, show whythe longitudinal component of the magnetic field is zero.

10. Referring to figure 16.9, show that the vector E ! B points in thedirection of propagation of a plane electromagnetic wave.

11. Referring to figure 16.9, what direction and speed must a charged par-ticle move in the presence of a free electromagnetic wave such that thenet electromagnetic force on it is zero?

Chapter 17

Capacitors, Inductors, andResistors

Various electronic devices are considered in this section. This is useful notonly for understanding these devices but also for revealing new aspects ofelectromagnetism. The capacitor is first discussed and Ampere’s law is in-troduced. The theory of magnetic inductance is then developed. Ohm’s lawand the resistor are treated. The energy associated with electric and mag-netic fields is calculated and Kirchho!’s laws for electric circuits are brieflydiscussed.

17.1 The Capacitor and Ampere’s Law

In this subsection we first discuss a device which is commonly used in elec-tronics called the capacitor. We then introduce a new mathematical ideacalled the circulation of a vector field around a loop. Finally, we use thisidea to investigate Ampere’s law.

17.1.1 The Capacitor

The capacitor is an electronic device for storing charge. The simplest typeis the parallel plate capacitor, illustrated in figure 17.1. This consists of twoconducting plates of area S separated by distance d. Positive charge q resideson one plate, while negative charge "q resides on the other.

The electric field between the plates is E = $/*0, where the charge per

297

298 CHAPTER 17. CAPACITORS, INDUCTORS, AND RESISTORS

xd

E

Side viewPerspective view

+

+

+

+ -

-

-

-

Gaussian box

S

Figure 17.1: Two views of a parallel plate capacitor.

unit area on the inside of the left plate in figure 17.1 is $ = q/S. The densityon the right plate is just "$. All charge is assumed to reside on the insidesurfaces and thus contribute to the electric field crossing the gap between theplates.

The above formula for the electric field comes from applying Gauss’s lawto the sheet of charge on the positive plate. The factor of 1/2 present inthe equation for an isolated sheet of charge is absent here because all of theelectric flux exits the Gaussian surface on the right side — the left side ofthe Gaussian box is inside the conductor where the electric field is zero, atleast in a static situation.

There is no vector potential in this case, so the electric field is relatedsolely to the scalar potential (. Integrating Ex = "%(/%x across the gapbetween the conducting plates, we find that the potential di!erence betweenthe plates is %( = Exd = qd/(*0S), since Ex is known to be constant in thiscase. This equation indicates that the potential di!erence %( is proportionalto the charge q on the left plate of the capacitor in figure 17.1. The constantof proportionality is d/(*0S), and the inverse of this constant is called thecapacitance:

C =*0S

d(parallel plate capacitor). (17.1)

The relationship between potential di!erence, charge, and capacitance is thus

%( = q/C or C = q/%(. (17.2)

17.1. THE CAPACITOR AND AMPERE’S LAW 299

d

z

x y

zq -q

S

i i

AE

R

B

Figure 17.2: Parallel plate capacitor with circular plates in a circuit withcurrent i flowing into the left plate and current i flowing out of the rightplate. The magnetic field which occurs when the charge on the capacitoris increasing with time is shown at right as vectors tangent to circles. Theradially outward vectors represent the vector potential giving rise to thismagnetic field in the region where x > 0. The vector potential points radiallyinward for x < 0. The y axis is into the page in the left panel while the xaxis is out of the page in the right panel.

The equation for the capacitance of the illustrated parallel plates containsjust a fundamental constant (*0) and geometrical factors (area of plates,spacing between them), and represents the amount of charge the parallelplate capacitor can store per unit potential di!erence between the plates. Aword about signs: The higher potential is always on the plate of the capacitorwhich which has the positive charge.

Note that equation (17.1) is valid only for a parallel plate capacitor.Capacitors come in many di!erent geometries and the formula for the capac-itance of a capacitor with a di!erent geometry will di!er from this equation.However, equation (17.2) is valid for any capacitor.

We now show that a capacitor which is charging or discharging has amagnetic field between the plates. Figure 17.2 shows a parallel plate capacitorwith a current i flowing into the left plate and out of the right plate. Thisis necessarily accompanied by an electric field which is changing with time:Ex = q/(*0S) = it/(*0S). Such an electric field can be derived from a scalarpotential which is a function of time: ( = "itx/(*0S). However, the Lorentz


condition%Ax

%x+%Ay

%y+%Az

%z+

1

c2

%(

%t= 0 (17.3)

demands that some component of the vector potential A be non-zero underthese circumstances, since %(/%t is non-zero.

How much can we infer about the vector potential from the geometry ofthe capacitor and equation (17.3)? Substituting ( = "itx/(*0S) into thisequation results in

%Ax

%x+%Ay

%y+%Az

%z=

ix

*0c2S, (17.4)

which suggests a number of di!erent possibilities for A. For instance, A =(0, ixy/(*0c2S), 0) and A = [0, 0, ixz/(*0c2S)] both satisfy equation (17.4).However, neither of these trial choices is satisfactory by itself, as they arenot consistent with the cylindrical symmetry of the capacitor about the xaxis.

A choice of vector potential which is consistent with the shape of thecapacitor and which satisfies the Lorentz condition is obtained by combiningthese two trial solutions:

A = [0, ixy/(2*0c2S), ixz/(2*0c

2S)]. (17.5)

This vector potential leads to the magnetic field

B = [0,"iz/(2*0c2S), iy/(2*0c

2S)]. (17.6)

These fields are illustrated in the right-hand panel of figure 17.2.

17.1.2 Circulation of a Vector Field

We have already seen one example of the circulation1 of a vector field, thoughwe didn’t label it as such. In the previous section we computed the workdone on a charge by the electric field as it moves around a closed loop in thecontext of the electric generator and Faraday’s law. The work done per unitcharge, or the EMF, is an example of the circulation of a field, in this casethe electric field, 'E . Faraday’s law can be restated as

'E = "d#B

dt(Faraday’s law). (17.7)

1The terminology comes from fluid dynamics where the concept is used with the fluidvelocity field. The idea of circulation is so useful in fluid dynamics that it seems worthwhileto generalize it to vector fields in other areas of physics.

17.1. THE CAPACITOR AND AMPERE’S LAW 301

r

R

x

By

x = a x = b

y = c

y = dA

Figure 17.3: Two examples of circulation paths in a vector field.

In the simple case of a circular loop with the field directed along the loop,the circulation is just the magnitude of the field times the circumference ofthe loop, as illustrated in the left panel of figure 17.3. In more complicatedcases in which the field points in a direction other than the direction of theloop, just the component in the direction of traversal around the loop entersthe circulation. If this component varies as one progresses around the loop,the calculation must be broken into pieces. The total circulation is thenobtained by adding up the contributions from segments of the loop in whichthe value of the field component parallel to the motion around the loop isconstant. An example of this type is the calculation of the EMF around asquare loop of wire in an electric generator. Another is illustrated in theright panel of figure 17.3.

17.1.3 Ampere’s Law

The magnetic circulation 'B around the periphery of the capacitor in theright panel of figure 17.2 is easily computed by taking the magnitude of B

in equation (17.6). The magnitude of the magnetic field on the inside of thecapacitor is just B = ir/(2*0c2S), since r = (y2 + z2)1/2 in figure 17.2. Thus,at the periphery of the capacitor, r = R, and B = iR/(2*0c2S) there. Thearea of the capacitor plates is S = "R2 and *0c2 = 1/µ0, as we discussedpreviously. Thus, the magnetic field is B = µ0i/(2"R) at the periphery. Ifthe periphery is traversed in the counter-clockwise direction, the magnetic


circulation around the capacitor is 'B = 2"RB = µ0i.Let us now compute the magnetic circulation around a wire carrying a

current. The magnetic field a distance r from a straight wire carrying acurrent i is B = µ0i/(2"r). The magnetic field points in the direction of acircle concentric with the wire. The magnetic circulation around the wire isthus 'B = 2"rB = µ0i.

Notice that the same magnetic circulation is found to be the same aroundthe wire and around the periphery of the capacitor. Furthermore, this circu-lation depends only on the current in the wire and the constant µ0.

One further item needs to be calculated, namely the electric flux acrossthe gap between the capacitor plates. This is just the electric field E = $/*0times the area S, or #E = S$/*0 = q/*0. The current into the capacitor isthe time rate of change on the capacitor, so i = dq/dt = *0d#E/dt.

We are now in a position to understand Ampere’s law:

'B = µ0

&

i + *0d#E

dt

'

(Ampere’s law). (17.8)

This states that the magnetic circulation around a loop equals the sum oftwo contributions, (1) µ0 times the electric current through the loop and(2) µ0*0 times the time rate of change of the electric flux through the loop.In the above example the first term dominates when the loop is around thewire, while the second term acts when the loop is around the gap betweenthe capacitor plates.

Ampere actually formulated an incomplete version of the law named afterhim — he included only the first term containing the current. The Scottishphysicist James Clerk Maxwell added the second term, based primarily ontheoretical reasoning. Maxwell’s additional term solved a serious internal in-consistency in electromagnetic theory — in our terms, the Lorentz conditionrequires a magnetic field to exist if the scalar potential ( is time-dependent.This magnetic field is only predicted by Ampere’s law if Maxwell’s term isincluded. The quantity *0d#E/dt was called the displacement current byMaxwell since it has the dimensions of current and is numerically equal tothe current entering the capacitor. However, it isn’t really a current — it isjust the time-changing electric flux!

Gauss’s law for electricity and magnetism, Faraday’s law, and Ampere’slaw are collectively called Maxwell’s equations. Together they form the basisfor electromagnetism as it developed historically. However, our formulation

17.2. MAGNETIC INDUCTION AND INDUCTORS 303

B

y

x

z

d

w

l

d

Perspective view

Side view

i

iA

Figure 17.4: Magnetic field and vector potential for two parallel plates car-rying equal currents in opposite directions.

of electromagnetism in terms of the four-potential, the dispersion relationfor free electromagnetic waves, the Lorentz condition, and Coulomb’s law, isprecisely equivalent to Maxwell’s equations, and is much closer to the modernapproach to electromagnetism.

17.2 Magnetic Induction and Inductors

Inductance is the tendency of a current in a conductor to maintain itself inthe face of changes in the potential di!erence driving the current. Figure17.4 shows a parallel plate inductor in which a current i passes through thetwo plates in opposite directions. The vector potential between the plates is

A ="

"µ0iz

w, 0, 0

#

, (17.9)

where w is the width of the plates, as illustrated in figure 17.4.Let us try to understand how this vector potential is constructed from

what we already know. The vector potential for a single current sheet in the


i

iz

x

from top plate

from bottom plate combined vector potential

d

Figure 17.5: Illustration of the addition of the vector potentials from twocurrent sheets with the left-moving current located above the x axis and theright-moving current below. The sum is obtained by the vector addition ofthe two components. Notice how the vector potential varies with z betweenthe current sheets, but is constant outside of them.

x-y plane at z = 0 moving in the x direction was computed in the previouschapter as Ax = "v$|z|/(2*0c2), with Ay = Az = 0. The quantity $ is thecharge per unit area on the sheet and v is the velocity of the charge sheet inthe x direction. We use the relationship 1/(*0c2) = µ0 and also realize thatif each plate has a width w, then the current in each plate is i = v$w, whichmeans that we can rewrite Ax = "µ0i|z|/(2w) for a single plate.

To proceed further, we first need to understand that |z| in the aboveequation is only valid if the charge sheet is at z = 0. If the sheet is located adistance a from the origin, then we must replace |z| by |z" a|. We also needto call on the definition of absolute value to realize that |z " a| = z " a ifz > a, and |z " a| = "z + a if z < a. Figure 17.5 shows how the profiles ofAx from each of the charge sheets add together to form a combined profilefor the two sheets together.

The resulting magnetic field between the plates can be computed fromthe vector potential:

B ="

0,"µ0i

w, 0

#

. (17.10)

Above and below the plates the magnetic field is zero because the vectorpotential is constant.

17.2. MAGNETIC INDUCTION AND INDUCTORS 305

Let us now ask what happens when the current through the device in-creases or decreases with time. Assuming initially that no scalar potentialexists, the x component of the electric field in the device is

Ex = "%Ax

%t=

µ0z

w

di

dt, (17.11)

while Ey = Ez = 0. Substituting the z values for each plate, we see that

Ex!upper = +µ0d

2w

di

dt(upper plate)

Ex!lower = "µ0d

2w

di

dt(lower plate). (17.12)

The work done by this electric field on a unit charge moving from the rightend of the upper plate, around the wire loops at the left end, and back tothe right end of the lower plate is %V = Ex!upper("l) + Ex!lower(+l) ="(µ0dl/w)(di/dt), where l is the length of the plate, as illustrated in figure17.4.

The minus sign means that the electric field acts so as to oppose a changein the current. However, in order for the current i to flow through theinductor, an external potential di!erence %( must be imposed between theinput and output wires of the inductor which just balances the e!ects of theinternally generated electric field:

%( =µ0dl

w

di

dt(parallel plate inductor). (17.13)

If this potential di!erence is positive, i. e., if the input wire of the inductoris at a higher potential then the output wire, then the current through theinductor will increase with time. If it is lower, the current will decrease.

As with capacitors, inductors come in many shapes and forms. The aboveequation is valid only for a parallel plate inductor, but the relationship

%( = Ldi

dt(17.14)

is valid for any inductor, assuming that the inductance L is known. Compar-ison of the above two equations reveals that the inductance for the parallelplate inductor shown in figure 17.4 is just

L =µ0dl

w(parallel plate inductor). (17.15)


hi i

lw

Figure 17.6: Rectangular resistor with a current i flowing through it.

17.3 Resistance and Resistors

Normal conducting materials require an electric field to keep an electric cur-rent flowing through them. The electric field causes a force on the electronsin the material, which is balanced by the energy loss that occurs when theelectrons collide with the atoms forming the material. Most objects exhibita linear relationship between the current i through them and the potentialdi!erence %( applied to them. This relationship is called Ohm’s law,

%( = iR (R constant), (17.16)

where the constant of proportionality R is called the resistance. The quantity%( is sometimes called the voltage drop across the resistor.

For certain materials such as semiconductors, the resistance depends onthe current. For such materials, the above equation defines resistance, butsince the resistance doesn’t remain constant when the current changes, thesematerials don’t obey Ohm’s law.

Figure 17.6 illustrates a rectangular resistor. The resistance of such aresistor can be written

R =l

wh# (17.17)

where the resistivity # is characteristic only of the material and not its shapeor size.

Unlike capacitors and inductors, resistors are dissipative devices. Thework done on a charge q passing through a resistor is just q%(. This energyis converted to heat. The work done per unit time, which equals the powerdissipated by a resistor is therefore

P = i%( = i2R = %(2/R. (17.18)

17.4. ENERGY OF ELECTRIC AND MAGNETIC FIELDS 307

ΔφφΔ

i i

L

V VG Gcurrentsource

voltagesourceC

V = 0 V = 0

Figure 17.7: Capacitor (left) and inductor (right) being charged respectivelyby constant sources of current and voltage.

17.4 Energy of Electric and Magnetic Fields

In this section we calculate the energy stored by a capacitor and an inductor.It is most profitable to think of the energy in these cases as being stored in theelectric and magnetic fields produced respectively in the capacitor and theinductor. From these calculations we compute the energy per unit volumein electric and magnetic fields. These results turn out to be valid for anyelectric and magnetic fields — not just those inside parallel plate capacitorsand inductors!

Let us first consider a capacitor starting in a discharged state at timet = 0. A constant current i is caused to flow through the capacitor by somedevice such as a battery or a generator, as shown in the left panel of figure17.7. As the capacitor charges up, the potential di!erence across it increaseswith time:

%( =q

C=

it

C. (17.19)

The EMF supplied by the generator has to increase to match this value.The generator does work on the positive charges moving around the cir-

cuit in the direction indicated by the arrow. We assume that %( equals theEMF or work per unit charge done by the generator VG, so the work done intime dt by the generator is dW = VGdq = VGidt. Using the equation for thepotential di!erence across a capacitor, we see that the power input is

P =dW

dt= %(i =

i2t

C. (17.20)

Integrating this in time yields the total energy UE supplied to the capacitorby the generator:

UE =i2t2

2C=

q2

2C(capacitor). (17.21)


Assuming that we have a parallel plate capacitor, let’s insert the formulafor the capacitance of such a device, C = *0S/d. Let us further recall thatthe electric field in a parallel plate capacitor is E = $/*0 = q/(*0S), so thatq = *0ES and

UE =(E*0S)2

2(*0S/d)=*0E2Sd

2. (17.22)

The combination Sd is just the volume between the capacitor plates. Theenergy density in the capacitor is therefore

uE =UE

Sd=*0E2

2(electric energy density). (17.23)

This formula for the energy density in the electric field is specific to a parallelplate capacitor. However, it turns out to be valid for any electric field.

A similar analysis of a current increasing from zero in an inductor yieldsthe energy density in a magnetic field. Imagine that the generator in theright panel of figure 17.7 produces a constant EMF, VG, starting at timet = 0 when the current is zero. The work done by the generator in time dtis dW = VGdq = VGidt so that the power is

P =dW

dt= VGi = Li

di

dt=

d

dt

&

Li2

2

'

. (17.24)

We have assumed that the EMF supplied by the generator, VG, balances thevoltage drop across the inductor: VG = %( = L(di/dt).

If we integrate the above equation in time, we get the energy added tothe inductor as a result of increasing the current through it. Substitutingthe formula for the inductance of a parallel plate inductor, L = µ0dl/w, wearrive at the equation for the energy stored by the inductor:

UB =Li2

2=

µ0dli2

2w(parallel plate inductor). (17.25)

Finally, using the relationship between the current and the magnetic fieldin a parallel plate inductor, B = µ0i/w, we can eliminate the current i andwrite

UB =dlwB2

2µ0

. (17.26)

The volume between the inductor plates is just dlw, so again we can writean energy density, this time for the magnetic field:

uB =UB

dlw=

B2

2µ0

(magnetic energy density). (17.27)

17.5. KIRCHHOFF’S LAWS 309

Though we only proved this equation for the magnetic field inside a parallelplate inductor, it turns out to be true for any magnetic field.

The total energy density is just the sum of the electric and magneticenergy densities:

uT = uE + uB =*0E2

2+

B2

2µ0

. (17.28)

17.5 Kirchho! ’s Laws

In the above discussion of energy we made two assumptions about electriccircuits:

• The current entering one end of a wire connecting two devices is equalto the current leaving the other end. Thus, the current out of thegenerator in the left panel of figure 17.7 is assumed to equal the currentinto the capacitor.

• The net work done on a charged particle passing completely arounda circuit loop is zero. Thus, the positive work done on charge pass-ing through the generator in the right panel of figure 17.7 is exactlybalanced by the potential di!erence across the inductor.

These are called Kirchho! ’s laws. They are used extensively in electroniccircuit design.

It is important to realize that Kirchho!’s laws are only approximationswhich hold when the currents and potentials in a circuit change slowly withtime. For steady currents and constant potentials they are precisely true,since imbalances in charge entering and leaving a junction between deviceswould result in the indefinite buildup of charge in the junction with timeand therefore an increasing electrostatic potential, which would violate thesteady state assumption. Furthermore, a non-zero EMF around a closed loopwould result in net acceleration of charge around the loop and a constantlyincreasing current.

If currents and potentials are changing with time, Kirchho!’s laws areapproximately valid only if the capacitance, inductance, and resistance ofthe wires connecting circuit elements are much smaller than the capacitance,inductance, and resistance of the circuit elements themselves. For very highfrequency operation, the e!ects of these “parasitic” properties are not smalland must be included in the design of the circuit.


17.6 Problems

1. Compute the capacitance of an isolated conducting sphere of radius R.Hint: Consider the other electrode to be a spherical shell surroundingthe conducting sphere at very large radius.

2. Given a parallel plate capacitor with plate area S, charge ±q on theplates, and the possibly variable plate separation x:

(a) Is the force between the plates attractive or repulsive?

(b) Compute the magnitude of the force of each plate on the other.Hint: You know both the electric field and the charge.

(c) Make an alternate computation of the force as follows: Computethe energy U in the electric field between the plates. The force isF = "dU/dx.

(d) You probably found that the above two calculations of the forcedidn’t agree. Which is correct? Explain. Hint: In doing part (b),what part of the electric field acting on (say) the negative chargeis due to itself, and what part is due to the positive charge? Onlythe latter part can exert a net force on the negative charge!

3. Compute the circulation of the vector field around the illustrated circlein the left panel of figure 17.3. Assume that the magnitude of the vectorfield equals Kr where K is a constant.

4. Compute the circulation of the vector field around the illustrated rect-angle in the right panel of figure 17.3. Assume that the x componentof the vector field equals Ky where K is a constant.

5. The solar wind consists of a plasma (a gas consisting of charged particleswith equal amounts of positive and negative charge) streaming out fromthe sun. In certain sectors of the solar wind the magnetic field pointsaway from the sun while in other sectors it points toward the sun.What is the magnitude and direction of the current flowing through theloop defined by the dashed rectangle which spans a sector boundary asshown in figure 17.8?

6. A superconducting parallel plate inductor with plate dimensions 0.1 mby 0.1 m and spacing 0.01 m is held together by connectors with max-

17.6. PROBLEMS 311

L

B

B

W

Figure 17.8: Magnetic field at a solar wind sector boundary.

L

switch

battery

+V

V = 0

Figure 17.9: Battery in parallel with an inductor.

imum breaking strength 500 N and has the input and the output con-nected by a superconducting wire. A current i is circulating throughthe inductor.

(a) Is the force between the plates attractive or repulsive?

(b) What is the maximum magnetic field that the inductor can havebetween the plates without blowing apart? Hint: Find the energyin the magnetic field as a function of plate separation and computethe force between the plates as for the capacitor. The magneticflux through the inductor remains constant as the plates move inthis case, which means that the current can change.

(c) What is the current corresponding to the above maximum field?

7. Use Kirchho!’s laws to compute the net resistance of

(a) resistors in parallel, and

(b) resistors in series,


R1

R1 R2

R2

seriesparallel

Figure 17.10: Resistors in parallel and in series.

H

L

R

Figure 17.11: Circuit consisting of a shorted resistor.

as shown in figure 17.10. Hint: In the first case the voltage drop acrossthe resistors is the same, in the second, the current through the resistorsis the same. Recall that Ohm’s law relates the current through a deviceto the voltage drop across it. (If you already know the answers, derivethem, don’t just write them down.)

8. Try to explain in physical terms why doubling the length of a resistordoubles its resistance, while doubling its cross-sectional area halves itsresistance. Use this argument to justify equation (17.17).

9. Describe qualitatively what happens when

(a) the switch is closed in the circuit in figure 17.9, and

(b) when it is abruptly opened.

The battery produces a voltage di!erence V , but also may be thoughtof as having a small internal resistance R.

10. Given the circuit shown in figure 17.11:

(a) What do Kirchho!’s laws tell you about %( across the resistor?

17.6. PROBLEMS 313

(b) Suppose a time-varying magnetic field B = B0 sin(&t) is appliednormal to the circuit loop, where B0 is a constant. What is the(time-dependent) voltage drop %( across the resistor in this situ-ation?

(c) Given the above %(, what is the current through the resistor asa function of time?

You may ignore the e!ect of the current in creating an additional mag-netic field.

Chapter 18

Measuring the Very Small

To begin our study of matter we discuss experiments in the late 19th andearly 20th centuries which led to proof of the existence of atoms and theirconstituents. We then introduce a fundamental idea about scattering ofwaves using the di!raction of light by small particles as a prototype. Thefamous Geiger-Marsden experiment which lead to the idea of the atomicnucleus is discussed. Finally, we examine some of the crucial experimentsdone with modern particle accelerators and the physical principles behindthem.1

18.1 Continuous Matter or Atoms?

From the time of the ancient Greeks there have been debates about theultimate nature of matter. One of these debates is whether matter is infinitelydivisible or whether it consists of fundamental building blocks which arethemselves indivisible. However, it wasn’t until the late 19th century thatreal progress began to be made on this question.

Advancements in our understanding of matter have largely been coupledto the development of machines to accelerate atomic and sub-atomic parti-cles. The original accelerator was developed in the 19th century and is calledthe Crookes tube.

J. J. Thomson measured the charge to mass ratio for both electrons andpositive ions in the Crookes tube in the following way: If a potential dif-

1Many of the ideas in this chapter were taken from Aitchison, I. J. R., and A. J. G.Hey, 1989: Gauge theories in particle physics. IOP Publishing, 571 pp.

315

316 CHAPTER 18. MEASURING THE VERY SMALL

to vacuum pump

electrons

positive ions

greenglow

glass vessel

- +

Figure 18.1: Crookes tube, the original particle accelerator. When potentialsare applied to the plates as shown, electrons are emitted by the left electrodeand accelerated to the right, some of which pass through holes in the rightelectrode. Positive ions, which are atoms missing one or more electrons, arecreated by collisions between electrons and residual gas atoms. These areaccelerated to the left.

ference %( is applied between the electrodes, then by energy conservationa particle of charge q starting from rest will acquire a kinetic energy mov-ing from electrode to electrode of K = mv2/2 = q%(. Solving for v, wefind v = (2q%(/m)1/2. If a magnetic field B is then imposed normal tothe electron beam after it has passed the positive electrode, the beam bendswith a radius of curvature of R = mv/(qB). Since R and B are known, thecharge to mass ratio can be computed by eliminating v and solving for q/m:q/m = 2%(/(BR)2. Thomson found that positive ions typically had chargeto mass ratios several thousand times smaller than the electrons. Further-more, the ions were positively charged, while the electrons were negativelycharged. If the ions and the electrons have electrical charges equal in mag-nitude (plausible, since the ions are neutral atoms with at least one electronremoved), the ions have to be much more massive than the electrons.

Robert Millikan made the first direct measurement of electric charge.He did this by suspending electrically charged oil drops in a known electricfield against gravity. The size of an oil drop is directly measured using amicroscope, leading to a calculation of its mass, and hence the gravitationalforce, mg. This is then balanced against the electric force, qE, leading toq = mg/E. Occasionally an oil drop loses an electron due to photoelectricemission caused by photons from an ultraviolet lamp. This disrupts the force

18.1. CONTINUOUS MATTER OR ATOMS? 317

balance, and causes the oil drop to move up or down. If the electric field isquickly adjusted, this motion can be arrested. The change in the charge canbe related to the change in the electric field: %q = mg%(1/E). If only asingle electron is emitted, then %q is equal to the electronic charge.

Between the work of Thomson and Millikan, the masses and the charges ofsub-atomic particles were accurately measured for the first time. Ironically,this work also showed that the “atom”, which means “indivisible” in Greek,in fact isn’t. Atoms consist of positive charges with large mass, or protons,in conjunction with low mass electrons of negative charge. Electrons andprotons have opposite charges, so they attract each other to form atoms inthis picture.

Geiger and Marsden did an experiment which strongly suggested thatatoms consist of very small, positively charged atomic nuclei, surrounded bya cloud of circling, negatively charged electrons. This is called the Rutherfordmodel of the atom after Ernest Rutherford.

Chadwick completed our picture of the atom with the discovery of aneutral particle of mass comparable to the proton, called the neutron. Theneutron is a constituent of the atomic nucleus along with the proton. Thenumber of protons in a nucleus is denoted Z while the number of neutronsis N . We define A = Z + N to be the total number of nucleons (protonsplus neutrons). The parameter Z is often called the atomic number while Ais called the atomic mass number.

Marie and Pierre Curie and Henri Becquerel were the first to discover amore fundamental divisibility of atoms in the form of the radioactive decay,though the implications of their results did not become clear until muchlater. Radioactive decay of atomic nuclei comes in three common forms,alpha, beta, and gamma decay. Alpha decay is the spontaneous emissionof a helium-4 nucleus, called an alpha particle by a heavy nucleus such asuranium or radium. The alpha particle consists of two protons and twoneutrons, so the emission decreases both Z and N by 2. Beta decay is theemission of an electron or its antiparticle, the positron, by a nucleus, withan accompanying change in the electric charge of the nucleus. For electronemission Z increases by 1 while N decreases by 1. The opposite occurs forpositron emission. Gamma decay is the emission of a high energy photon bya nucleus. The values of Z and N remain unchanged. The energy releasedby these decays is typically of order a few million electron volts.

Of the three forms of decay, beta decay is the most interesting, since itinvolves the transformation of one sub-atomic particle into another. In the


case of neutron decay, a neutron is converted into a proton, an electron, andan antineutrino. For proton decay, a proton becomes a neutron, a positron,and a neutrino. (Only the neutron form occurs for an isolated particle. How-ever, the energetics inside atomic nuclei can result in either form, dependingon the nucleus in question.) The neutrino is one of the great theoreticalpredictions of modern physics. Careful studies of beta decay, which at thetime was thought to result only in the emission of a proton and an electronfor the neutron form of the reaction, showed apparent non-conservation ofenergy and angular momentum. Rather than accept this rather unpalatableconclusion, Wolfgang Pauli proposed that a third particle named a neutrino,or little neutral particle, is emitted in the decay, thus accounting for themissing energy and angular momentum. The presumed electrical neutralityof the particle explained the di"culty of detecting it. Over 25 years passedbefore Frederick Reines and Clyde Cowan from Los Alamos observed thiselusive particle.

The three forms of radioactive decay are associated with three of thefour known fundamental forces of nature. Gamma decay is electromagneticin nature, while alpha decay involves the breaking of bonds produced bythe nuclear or strong force. Beta decay is a manifestation of the so-calledweak force. (The fourth force is gravity, which plays a negligible role on thesub-atomic scale, as far as we know.)

Beta decay gives us a strong hint that even particles such as protonsand neutrons, which make up atomic nuclei, are not “atomic” in the senseof the original Greek, since neutrons can change into protons in beta decayand vice versa. We now have excellent evidence that protons, neutrons, andmany other sub-nuclear particles, are made up of particles called quarks.Quarks and electrons are currently thought to be fundamental in that theyare supposedly indivisible, and are hence the true “atoms” of the universe.However, who knows, perhaps someday we will discover that they too arecomposed of even more fundamental constituents!

18.2 The Ring Around the Moon

Sometimes at night one sees a di!use disk of light around the moon if ithappens to be shining through a thin layer of cloud. This disk consists oflight di!racted by the water or ice particles in the cloud. The diameter ofthe disk contains information about the size of the cloud particles doing the

18.2. THE RING AROUND THE MOON 319

d

α

λ

Figure 18.2: Scattering of an incident plane wave by a water droplet. Theopening half-angle of the scattered wave is ) & +/(2d).

di!raction. In particular, if the particles have diameter d and the light haswavelength +, then the di!raction half-angle shown in figure 18.2 is approx-imately

) & +/(2d). (18.1)

This equation comes from the problem of passage of light through a holeor slit of diameter or width d. This problem was treated in the section onwaves, and the above formula was concluded to hold in that case. One canthink of the di!raction of light by a particle to be the linear superposition ofa plane wave minus the di!raction of light by a hole in a mask, as illustratedin figure 18.2. The angular spread of the di!racted light is the same in bothcases.

The interesting point about equation (18.1) is that the opening angle ofthe di!raction cone is inversely proportional to the diameter of the di!ractingparticles. Thus, for a given wavelength, smaller particles cause di!ractionthrough a wider angle.

Note that when the wavelength exceeds the diameter of the particle bya significant amount, equation (18.1) fails, since scattering through an anglegreater than " doesn’t make physical sense. In this case the di!racted pho-tons tend to be isotropic, i. e., they are scattered with equal probability intoany direction.

If one wishes to measure the size of an object by observing the di!ractionof a wave around the object, the lesson is clear; the wavelength of the wave


θalpha particlesource collimator

gold foil

screenscintillation

Figure 18.3: Schematic of Geiger-Marsden experiment. The radioactivesource produces alpha particles which are collimated into a beam and di-rected at a gold foil. The alpha particles scatter o! the foil and are detectedby a flash of light when they hit the scintillation screen.

must be less than or equal to the dimensions of the object — otherwise thescattering of the wave by the object is largely isotropic and equation (18.1)yields no information. Since wavelength is inversely related to momentum bythe de Broglie relationship, this condition implies that the momentum mustsatisfy

p = h/+ > h/d (18.2)

in order that the size of an object of diameter d be resolved.

18.3 The Geiger-Marsden Experiment

In 1908 Hans Geiger and Ernest Marsden, working with Ernest Rutherfordof the Physical Laboratories at the University of Manchester, measured theangular distribution of alpha particles scattered from a thin gold foil in anexperiment illustrated in figure 18.3. In order to understand this experiment,we need to compute the de Broglie wavelength of alpha particles resultingfrom radioactive decay. Typical alpha particle kinetic energies are of order5 MeV = 8!10!13 J. Since the alpha particle consists of two protons and twoneutrons, its mass is about M! = 6.7 ! 10!27 kg. This implies a velocity ofabout v = 1.1!107 m s!1, a momentum of about p = mv = 7.4!10!20 N s,and a de Broglie wavelength of about + = h/p = 9.0 ! 10!15 m.

18.3. THE GEIGER-MARSDEN EXPERIMENT 321

pi

pfpi

pf qθ

alpha particle nucleus

Figure 18.4: Illustration of alpha particle trajectory in Rutherford’s model ofthe atom. The momentum transfer q from the nucleus to the alpha particleis equal to the change in the alpha particle’s momentum.

Other evidence indicates that atoms have dimensions of order 10!10 m, sothe de Broglie wavelength of an alpha particle is about a factor of 104 smallerthan a typical atomic dimension. Thus, the typical di!raction scatteringangle of alpha particles o! of atoms ought to be very small, of order ) =+/(2d) & 10!4 radian & 0.01#.

Imagine the surprise of Geiger and Marsden when they found that whilemost alpha particles su!ered only small deflections when passing throughthe gold foil, a small fraction of the incident particles scattered throughlarge angles, some in excess of 90#!

Ernest Rutherford calculated the probability for an alpha particle, con-sidered to be a positive point charge, to be scattered through various anglesby a stationary atomic nucleus, assumed also to be a positive point charge.The calculation was done classically, though interestingly enough a quantummechanical calculation gives the same answer. The relative probability forscattering with a momentum transfer to the alpha particle of q is propor-tional to |q|!4 % q!4 according to Rutherford’s calculation. (Do not confusethis q with charge!) As figure 18.4 indicates, a larger momentum transfercorresponds to a larger scattering angle. The maximum momentum transferfor an incident alpha particle with momentum p is 2|p|, or just twice the ini-tial momentum. This corresponds to a head-on collision between the alphaparticle and the nucleus followed by a recoil of the alpha particle directly


backwards. Since this collision is elastic, the kinetic energy of the alpha par-ticle after the collision is approximately the same as before, as long as thenucleus is much more massive than the alpha particle.

Rutherford’s calculation agreed quite closely with the experimental re-sults of Geiger and Marsden. Though the probability for scattering througha large angle is small even in the Rutherford theory, it is still much largerthan would be expected if there were no small scale atomic nucleus.

18.4 Cosmic Rays and Accelerator Experi-

ments

18.4.1 Early Cosmic Ray Results

Earlier we indicated that particles interacted with each other via the exchangeof a virtual intermediary particle which interchanges energy, momentum,and other physical properties between the interacting particles. This ideaoriginated with the Japanese physicist Hideki Yukawa in 1935 in an e!ort tounderstand the forces between nucleons. Yukawa hypothesized that the forcewhich holds nucleons together is associated with the exchange of a boson, i.e., a particle with integer spin, with rest energy mc2 & 100 MeV. The rangeof this force at low momentum transfers is I & h/(mc) & 2 ! 10!15 m, orcomparable to the observed size of an atomic nucleus.

In 1947 two new particles were discovered in cosmic rays, the negativelycharged muon with a rest energy of 106 MeV, and the pion, which comes inthree varieties, the "+, the "!, and the "0, which respectively have positive,negative, and zero charge. The rest energies of the "+ and "! are 140 MeVwhile that of the "0 is 135 MeV. All of these particles are unstable in thatthey decay into other, more stable particles in a tiny fraction of a second.In particular, the negative pion decays into a muon plus an antineutrino,while the neutral pion decays into two gamma rays, or high energy photons.The antineutrino which results from pion decay is actually distinct from theantineutrino emitted in nuclear beta decay; it is called the mu antineutrinosince it is associated with the muon in the same way that the antineutrinoin beta decay is associated with the electron. To further distinguish betweenthe two, the latter is called the electron antineutrino. The muon itself decaysinto an electron, a mu neutrino, and an electron antineutrino.

The muon and its associated neutrino are rather peculiar. In all respects

18.4. COSMIC RAYS AND ACCELERATOR EXPERIMENTS 323

except mass the muon appears to be identical to the electron. The physicistI. I. Rabi is reputed to have responded “Who ordered that?” upon learning ofthe properties of the muon. Furthermore, the electron neutrino only interactswith the electron and the muon neutrino only interacts with the muon. Thisis the first hint that elementary particles occur in families which appear tobe replicated at higher energies.

Since the muon is a fermion with spin 1/2, it can’t be Yukawa’s interme-diary particle since all intermediary particles are bosons with integral spin.Furthermore, as with the electron, it is not subject to the nuclear force. Thepions are more promising candidates for being intermediary particles of thenuclear force, since they are bosons with spin 0. However, as we shall see, thesituation is more complex than Yukawa imagined, and the force between nu-cleons cannot be so simply treated. However, Yukawa’s idea of intermediaryparticle exchange lives on in today’s theories of sub-nuclear particles.

18.4.2 Particle Accelerators

Soon after the discovery of muons and pions in cosmic rays, a whole plethoraof unstable particles was uncovered. Central to these discoveries was theparticle accelerator. In these devices, charged particles, typically electronsor protons, are accelerated to high energy and then smashed into a target.Detectors of various sorts are used to examine the particles created by thecollisions of the accelerated particles and the atomic nuclei with which theycollide. Sometimes an elastic collision occurs, in which the accelerated par-ticle simply “bounces o!” of the target particle, transferring a good bit ofits momentum to this particle. However, under many circumstances the col-lision results in the production of new particles which didn’t exist before thecollision. This is referred to as an inelastic collision.

The simplest type of target is liquid hydrogen since the nucleus consistsof a single proton. The orbital electrons of the target atoms are so lightthat they are generally just “brushed aside” without greatly a!ecting thetrajectories of the accelerated particles. However, a variety of targets areused under di!erent circumstances.

18.4.3 Size and Structure of the Nucleus

In the late 1950s and early 1960s Robert Hofstadter of Stanford Universityextended the Geiger-Marsden experiment to much shorter de Broglie wave-


log(q)

log(P)

P = q -4

log[F(q)]

probabilityactual

Figure 18.5: Schematic illustration of Robert Hofstadter’s results for scat-tering of electrons o! of atomic nuclei. The solid line shows the relativeprobability (in log-log coordinates) of elastic scattering as a function of themomentum transfer. The dashed curve illustrates the observed probabilitydistribution. The di!erence between the curves is the logarithm of the formfactor, F (q).


lengths using high energy electrons from an accelerator rather than alphaparticles as the probe. The type of results obtained by Hofstadter are shownin figure 18.5. After accounting for some e!ects having to do with the elec-tron spin, these experiments should agree with the Rutherford formula if thenucleus is truly a point particle. However, the actual results show proba-bilities which drop o! more rapidly with increasing momentum transfer qthan is predicted by the Rutherford model. The ratio of the actual to theRutherford probability distributions is called the form factor, F (q), for thisprocess:

Pobs(q) = F (q)PRuth(q) ) F (q)q!4. (18.3)

Taking the logarithm of this equation results in

log[Pobs] = log[F (q)] " 4 log(q) + const. (18.4)

These results are related to the fact that the nucleus is actually of fi-nite size. The di!raction e!ects discussed in the section on the scatteringof moonlight come into play here, in that little scattering takes place forscattering angles larger than roughly +/(2d), where + is the de Broglie wave-length of the probing particle and d is the diameter of the target. For smallscattering angle (which we now call !), it is clear from figure 18.4 that

! & q/p, (18.5)

where p is the momentum of the incident electron and q is the momentumtransfer. If qmax is the maximum momentum transfer for which there issignificant scattering, then we can write

qmax/p = !max & +/d, (18.6)

where the factor of 2 in the denominator on the right side has been droppedsince this is an approximate analysis. However, since + = h/p, we find that

qmax &h

d. (18.7)

Thus, the momentum transfer for which the measured form factor becomessmall compared to one gives us an immediate estimate of the diameter of anatomic nucleus: d & h/qmax. The results obtained by Hofstadter show thatnuclear diameters are typically a few times 10!15 m.

More than just size information can be extracted from the form factor.Hofstadter’s experiments also led to a great deal of information about theinternal structure of atomic nuclei.


18.4.4 Deep Inelastic Scattering of Electrons from Pro-

tons

The construction of the Stanford Linear Accelerator Center (SLAC), whichaccelerates electrons up to 40 GeV, allowed experiments like Hofstadter’s tobe carried out at much higher energies. At these energies many of the colli-sions between electrons and protons and neutrons are inelastic — generallya great mess of short-lived particles is spewed out, and are very di"cult tointerpret. However, the so-called deep inelastic collisions, where the electronscatters through a large angle and therefore transfers a large momentum, q,to the proton, yield very interesting results. In particular, these collisionsoccur essentially with a probability proportional to q!4 — just as in theGeiger-Marsden experiment!

The electron is a point particle as far as we know. However, previousexperiments showed the proton to have a finite size, of order 10!15 m. There-fore, the scattering probability should drop o! more rapidly with increasingmomentum transfer q than q!4, as in the earlier Hofstadter experiments.

James Bjorken and Richard Feynman showed a way out of this dilemma.They proposed that the proton actually consists of a small number of pointparticles bound together by weakly attractive forces. A su"ciently energeticphoton is able to knock a single one of these particles out of the proton, asillustrated in the right panel of figure 18.6. This leads to a subsequent setof reactions which produce the profusion of particles seen in the left panel ofthis figure. Feynman called the particles which make up the proton partons.However, we now know that they are actually quarks, spin 1/2 particles withfractional electronic charge which are thought to be the fundamental buildingblocks of matter, and gluons, the massless spin 1 intermediary particles whichcarry the strong force.

18.4.5 Storage Rings and Colliders

An alternate way to create interesting collisions is to crash particles and an-tiparticles of the same energy into each other. This is done via a storagering, as shown in figure 18.7. A set of magnets forces particles and antiparti-cles (which have opposite charges) to move in opposing circles within a highvacuum. The circles are slightly o!set so that the beams cross at only twopoints. Collisions occur at these points and are observed by various types ofexperimental equipment.


ignorance

x

ct ct

xelectron proton

photon

lots ofstuff

electron

photon

proton

partonsqq

Initial view Bjorken-Feynman theory

Figure 18.6: Deep inelastic scattering of a high energy electron by a protonoccurs when the momentum transfer q is large and many particles are pro-duced. According to the Bjorken-Feynman theory of this process, the protonconsists of a number of partons flying in “loose formation”. A su"cientlyenergetic photon, i. e., with large momentum transfer q, kicks out just oneof these partons, leaving the others undisturbed.

factory factory

experimentalareas

particle antiparticle

Figure 18.7: Schematic model of a particle-antiparticle collider. The particlesand antiparticles are injected into the storage rings shown and are made to goin a circle by magnetic fields. The beams cross at two points and apparatusis set up around these points to observe the products of collisions.


proton

antiproton

θ

quark 1quark2quark 3

antiquark 1antiquark 2antiquark 3gluon

Figure 18.8: Illustration of what happens in a high energy collision between aproton and an antiproton according to the Bjorken-Feynman parton model.

An alternate type of collider has two storage rings which intersect at onlyone point. This type of system can be used to collide particles of the sametype together, e. g., protons colliding with protons.

18.4.6 Proton-Antiproton Collisions

If collisions occur by the exchange of a single intermediary particle of zeromass between point particles, the q!4 dependence of the collision probabilityon momentum transfer will occur in proton-antiproton collisions as in theGeiger-Marsden experiment. However, if the colliding particles are not pointparticles, a form factor which decreases for increasing momentum transferwill occur as with the Hofstadter experiments.

When collisions between protons and antiprotons of a few hundred GeVare arranged, certain types of events called two jet events are recorded. Inthese events, two jets, each containing many particles, are emitted in oppositedirections at wide angles (i. e., with large momentum transfer) from thecolliding beams. Furthermore, these jets show a probability distribution asa function of momentum transfer very close to q!4. This indicates thatthe colliding particles are point-like, at least down to the minimum spatialresolutions available to today’s accelerators.

According to the Bjorken-Feynman parton model of the proton, the col-

18.5. COMMENTARY 329

θ

electronpositron

x

ctantiquarkquark

positronelectron

photon

Figure 18.9: Two jet events resulting from the annihilation of high energyelectrons and positrons. The virtual photon decays into a quark-antiquarkpair which in turn generate the oppositely pointing jets of particles.

lision between highly energetic protons and antiprotons should operate asshown in figure 18.8. The actual collision is between individual partons.Figure 18.8 illustrates the collision between a quark in the proton and anantiquark in the antiproton. The result of this interaction is the scatteringof these particles out of the incident particles, resulting ultimately in a twojet event as described above.

18.4.7 Electron-Positron Collisions

Two jet events can also be created by the collision of high energy electronsand positrons. Figure 18.9 shows how this process is thought to work. Theannihilation of the electron and positron results in a virtual photon, whichin turn decays into a quark-antiquark pair. The quarks then produce thejets. These results suggest that quarks can indeed occur outside of protons,at least if they occur in quark-antiquark pairs.

18.5 Commentary

We have examined a selected set of experiments performed over the last100 years. Though complicated in detail, we have seen that they can beunderstood in their essence using one idea, namely the uncertainty principle.


This principle underlies the di!raction angle formula and also turns out (inan argument which we have not made) to be central to the q!4 dependence ofscattering probability for point particles. For momentum transfers of order1000 GeV/c, we are able to probe spatial scales of order 10!17 m, or a factorof 100-500 less than the scale of the atomic nucleus. Even on this scale itappears that both the electron and the quark act like point particles. Theythus appear to be the ultimate “atoms” of matter in the original sense of theword. However, it is possible that experiments at even higher momentumtransfers would show the electron or the quark to have some kind of internalstructure. Perhaps this heirarchy of structure, of which we have noted theatom, the atomic nucleus, nucleons, and quarks, goes on forever.

18.6 Problems

1. If possible, observe the moon through a thin cloud layer and estimatethe angular size of the disk of scattered light around the moon. Fromthis, estimate the size of the particles doing the scattering.

2. Which particle can be used to investigate smaller scales, a proton oran electron, at

(a) the same velocity, and

(b) at the same kinetic energy? (Work non-relativistically in bothcases.)

(c) Now consider ultra-relativistic protons and electrons with the sametotal energy. Is there a significant di!erence between their abilityto investigate very small scales?

3. Electron microscope:

(a) What kinetic energy (in electron volts) must electrons in an elec-tron microscope have to match the resolution of an optical mi-croscope? (The resolutions match when the wavelengths of theelectrons and the light are the same.)

(b) If the electrons have kinetic energy 50 KeV, how much betterresolution does the electron microscope have than the best opticalmicroscope?

18.6. PROBLEMS 331

Hint: Use the non-relativistic kinetic energy and check whether thisassumption is valid in retrospect.

4. Integrated circuits are made by a system in which the circuit patternis engraved on a silicon wafer using a photochemical process workingwith an optical imaging device which projects the circuit image on thewafer.

(a) Assuming visible light is used, estimate the size of the smallestfeature which could be produced on the silicon by this system.

(b) Do the same for 1 KeV X-rays.

Hint: Recall that the smallest feature resolvable by a wave is of orderthe wavelength.

5. The rest energy of two colliding particles is just c2 times the mass ofthe single particle created by the colliding particles sticking together.

(a) Compute the rest energy (in GeV) of a particle resulting from a100 GeV energy proton colliding with a stationary proton.

(b) Compute the rest energy of the particle resulting from two 50 GeVprotons colliding head-on.

Hint: These calculations are relativistic, since the rest energy of theproton is about 0.9 GeV.

6. Relativistic charged particle in magnetic field: Assume that a relativis-tic particle of mass m and charge e is moving in a circle under theinfluence of the magnetic field B = (0, 0,"B). The position of theparticle as a function of time is given by x = [R cos(&t), R sin(&t), 0].

(a) Compute the (vector) velocity of the particle and show that itsspeed is v = &R.

(b) Compute the (relativistic) momentum (again in vector form) ofthe particle using the above results.

(c) Compute the magnetic force F on the particle.

(d) Using the relativistic version of Newton’s second law, F = dp/dt,determine how the rotational frequency & depends on the speedof the particle, the magnetic field B, and the particle’s charge andmass. Examine particularly the limits where v $ c and v & c.


(e) Eliminate & between the above result and the speed formula toget an equation for the radius R of the circle. Show that this takesthe particularly simple form R = p/(eB) when written in termsof the magnitude of the momentum p = mv,.

7. A 30 GeV electron is scattered by a virtual photon through an angleof 60# without changing its energy.

(a) Compute its momentum vector before and after the scattering.

(b) Compute the momentum transfer to the electron by the photonin the scattering event.

(c) Compute the wavelength of the virtual photon.

(d) What is the virtual photon’s energy?

(e) What is the virtual photon’s mass?

8. Find ), -, and , such that h!c"G# has the units of length. (G is theuniversal gravitational constant.) Compute the numerical value of thislength, which is called the Planck length. Compare this value to theresolution available today in the highest energy accelerators.

Chapter 19

Atoms

In this section we investigate the structure of atoms. However, before we canunderstand these, we first need to review some facts about angular momen-tum in quantum mechanics.

19.1 Fermions and Bosons

19.1.1 Review of Angular Momentum in Quantum Me-

chanics

As we learned earlier, angular momentum is quantized in quantum mechan-ics. We can simultaneously measure only the magnitude of the angular mo-mentum vector and one component, usually taken to be the z component.Measurement of the other two components simultaneously with the z com-ponent is forbidden by the uncertainty principle.

The magnitude of the orbital angular momentum of an object can takeon the values |L| = [l(l + 1)]1/2h where l = 0, 1, 2, . . .. The z component canlikewise equal Lz = mh where m = "l,"l + 1, . . . , l.

Particles can have an intrinsic spin angular momentum as well as anorbital angular momentum. The possible values for the magnitude of thespin angular momentum are |S| = [s(s + 1)]1/2h and the z component of thespin angular momentum Sz = msh where ms = "s,"s+1, . . . , s. Spin di!ersfrom orbital angular momentum in that the spin can take on half-integeras well as integer values: s = 0, 1/2, 1, 3/2, . . . are possible spin quantumnumbers.

333

334 CHAPTER 19. ATOMS

Spin is an intrinsic, unchangeable quantity for an elementary particle.Particles with half-integer spins, s = 1/2, 3/2, 5/2, . . ., are called fermions,while particles with integer spins, s = 0, 1, 2, . . . are called bosons. Fermionscan only be created or destroyed in particle-antiparticle pairs, whereas bosonscan be created or destroyed singly.

19.1.2 Two Particle Wave Functions

We learned in quantum mechanics that a particle is represented by a wave,.(x, y, z, t), the absolute square of which gives the relative probability offinding the particle at some point in spacetime. If we have two particles, thenwe must ask a more complicated question: What is the relative probability offinding particle 1 at point x1 and particle 2 at point x2? This probability canbe represented as the absolute square of a joint wave function .(x1, x2), i. e.,a single wave function that represents both particles. If the particles are notidentical (say, one is a proton and the other is a neutron) and if they are notinteracting with each other via some force, then the above wave function canbe broken into the product of the wave functions for the individual particles:

.(x1, x2) = .1(x1).2(x2) (non-interacting dissimilar particles). (19.1)

In this case the probability of finding particle 1 at x1 and particle 2 atx2 is just the absolute square of the joint wave amplitude: P (x1, x2) =P1(x1)P2(x2). This is consistent with classical probability theory.

The situation in quantum mechanics when the two particles are identicalis quite di!erent. If P (x1, x2) is, say, the probability of finding one electron atx1 and another electron at x2, then since we can’t tell the di!erence betweenone electron and another, the probability distribution cannot change if weswitch the electrons. In other words, we must have P (x1, x2) = P (x2, x1).There are two obvious ways to make this happen: Either .(x1, x2) = .(x2, x1)or .(x1, x2) = ".(x2, x1).

It turns out that the wave function for two identical fermions is anti-symmetric to the exchange of particles whereas for two identical bosons itis symmetric. In the special case of two non-interacting particles, we canconstruct the joint wave function with the correct symmetry from the wavefunctions for the individual particles as follows:

.(x1, x2) = .1(x1).2(x2)".1(x2).2(x1) (non-interacting fermions) (19.2)

19.1. FERMIONS AND BOSONS 335

0.00

0.50

0.00 0.50x2 vs x1

0.00

0.50

0.00 0.50x2 vs x1

0.00

0.50

0.00 0.50x2 vs x1

Figure 19.1: Joint probability distributions for two particles, one in theground state and one in the first excited state of a one-dimensional box.Left panel: non-identical particles. Middle panel: identical fermions. Rightpanel: identical bosons. The curved lines are contours of constant probabil-ity. The vertical hatching shows where the probability is large.

for fermions and

.(x1, x2) = .1(x1).2(x2) + .1(x2).2(x1) (non-interacting bosons) (19.3)

for bosons.Figure 19.1 shows the joint probability distribution for two particles in

di!erent energy states in an infinite square well: P (x1, x2) = |.(x1, x2)|2.Three di!erent cases are shown, non-identical particles, identical fermions,and identical bosons. Notice that the probability of finding two fermions atthe same point in space, i. e., along the diagonal dotted line in the centerpanel of figure 19.1, is zero. This follows immediately from equation (19.2),which shows that .(x1, x2) = 0 for fermions if x1 = x2. Notice also thatif two fermions are in the same energy level (say, the ground state of theone-dimensional box) so that .1(x) = .2(x), then .(x1, x2) = 0 everywhere.This demonstrates that the two fermions cannot occupy the same state. Thisresult is called the Pauli exclusion principle.

On the other hand, bosons tend to cluster together. Figure 19.1 showsthat the highest probability in the joint distribution occurs along the linex1 = x2, i. e., when the particles are colocated. This tendency is accentuatedwhen more particles are added to the system. When there are a large numberof bosons, this tendency creates what is called a Bose-Einstein condensate inwhich most or all of the particles are in the ground state. Bose-Einstein con-


densation is responsible for such phenomena as superconductivity in metalsand superfluidity in liquid helium at low temperatures.

19.2 The Hydrogen Atom

The hydrogen atom consists of an electron and a proton bound together bythe attractive electrostatic force between the negative and positive chargesof these particles. Our experience with the one-dimensional particle in a boxshows that a spatially restricted particle takes on only discrete values of thetotal energy. This conclusion carries over to arbitrary attractive potentialsand three dimensions.

The energy of the ground state can be qualitatively understood in termsof the uncertainty principle. A particle restricted to a region of size a byan attractive force will have a momentum equal at least to the uncertaintyin the momentum predicted by the uncertainty principle: p & h/a. Thiscorresponds to a kinetic energy K = mv2/2 = p2/(2m) & h2/(2ma2). Forthe particle in a box there is no potential energy, so the kinetic energy equalsthe total energy. Comparison of this estimate with the computed groundstate energy of a particle in a box of length a, E1 = h2"2/(2ma2), shows thatthe estimate di!ers from the exact value by only a numerical factor "2.

We can make an estimate of the ground state energy of the hydrogenatom using the same technique if we can somehow take into account thepotential energy of this atom. Classically, an electron with charge "e movingin a circular orbit of radius a around a proton with charge e at speed vmust have the centripetal acceleration times the mass equal to the attractiveelectrostatic force, mv2/a = e2/(4"*0a2), where m is the electron mass. (Theproton is so much more massive than the electron that we can assume it tobe stationary.) Multiplication of this equation by a/2 results in

K =mv2

2=

e2

8"*0a= "

U

2, (19.4)

where U is the (negative) potential energy of the electron and K is its kineticenergy. Solving for U , we find that U = "2K. The total energy E is thereforerelated to the kinetic energy by

E = K + U = K " 2K = "K (hydrogen atom) (19.5)

19.2. THE HYDROGEN ATOM 337

Since the total energy is negative in this case, and since U = 0 whenthe electron is infinitely far from the proton, we can define a binding energywhich is equal to minus the total energy:

EB % "E = K = "U/2 (virial theorem). (19.6)

The binding energy is the minimum additional energy which needs to beadded to the electron to make the total energy zero, and thus to remove itto infinity. Equation (19.6) is called the virial theorem, and it is even truefor non-circular orbits if the energies are properly averaged over the entiretrajectory.

Proceeding as before, we assume that the momentum of the electron isp & h/a and substitute this into equation (19.4). Solving this for a % a0

yields an estimate of the radius of the hydrogen atom:

a0 =4"*0h

2

e2m=

&

4"*0hc

e2

' &

h

mc

'

(19.7)

This result was first obtained by the Danish physicist Niels Bohr, using an-other method, in an early attempt to understand the quantum nature ofmatter.

The grouping of terms by the large parentheses in equation (19.7) issignificant. The dimensionless quantity

) =e2

4"*0hc&

1

137(fine structure constant) (19.8)

is called the fine structure constant for historical reasons. However, it isactually a fundamental measure of the strength of the electromagnetic in-teraction. The Bohr radius can be written in terms of the fine structureconstant as

a0 =h

)mc= 5.29 ! 10!11 m (Bohr radius). (19.9)

The binding energy predicted by equations (19.4) and (19.6) is

EB = "U

2=

e2

8"*0a0

= )hc

2a0

=)2mc2

2= 13.6 eV. (19.10)

The binding energy between the electron and the proton is thus proportionalto the electron rest energy times the square of the fine structure constant.


n = 1

n = 2

n = 3

l = 1 l = 2

2*1

2*1

2*1

2*3

2*3 2*5E = 0 l = 0

8

2

18

Figure 19.2: Energy levels of the hydrogen atom. Energy increases upwardand angular momentum increases to the right. The numbers above each levelindicate the spin orientation times the orbital orientation degeneracy for eachlevel. The numbers at the right show the total degeneracy for each value ofn. Only the first three values of n are shown.

The above estimated binding energy turns out to be precisely the groundstate binding energy of the hydrogen atom. The energy levels of the hydrogenatom turn out to be

En = "EB

n2= "

)2mc2

2n2, n = 1, 2, 3, . . . (hydrogen energy levels), (19.11)

where n is called the principal quantum number of the hydrogen atom.

19.3 The Periodic Table of the Elements

The energy levels of the hydrogen atom whose energies are given by equation(19.11) are actually degenerate, in that each energy has more than one stateassociated with it. Three extra degrees of freedom are associated with angularmomentum, expressed by the quantum numbers l, m, and ms. For energylevel n, the orbital angular momentum quantum number can take on the

19.3. THE PERIODIC TABLE OF THE ELEMENTS 339

values l = 0, 1, 2, . . . , n " 1. Thus, for the ground state, n = 1, the onlypossible value of l is zero. For a given value of l, there are 2l + 1 possiblevalues of the orbital z component quantum number, m = "l,"l + 1, . . . , l.Finally, there are two possible values of the spin orientation quantum number,ms. Thus, for the nth energy level there are

Nn = 2n!1!

l=0

(2l + 1) (19.12)

states. In particular, for n = 1, 2, 3, . . ., we have Nn = 2, 8, 18, . . .. This issummarized in figure 19.2.

These results have implications for the character of atoms with more thanone proton in the nucleus. Let us imagine how such atoms might be built.The binding energy of a single electron in the ground state of a nucleus withZ protons is Z2 times the binding energy of the electron in the ground stateof a hydrogen atom. If the force between electrons can be ignored comparedto the force between an electron and the nucleus (a very poor but initiallyuseful assumption which we will discuss below), then we could constructan atom by dropping Z electrons one by one into the potential well of thenucleus. The Pauli exclusion principle prevents all of these electrons fromfalling into the ground state. Instead, the available states will fill in orderof lowest energy first until all Z electrons are added and the atom becomeselectrically neutral. From figure 19.2 we see that Z = 2 fills the n = 1 levels,with two electrons, one spin up and one spin down, both with zero orbitalangular momentum. For Z = 10 the n = 2 levels fill such that two electronshave l = 0 and six have l = 1.

As electrons are added to an atom, previous electrons tend to shieldsubsequent electrons from the nucleus, since their negative charge partiallycompensates for the nuclear positive charge. Thus, binding energies are con-siderably less than would be expected on the basis of the non-interactingelectron model. Furthermore, the binding energies for states with higherorbital angular momentum are smaller than those with lower values, sinceelectrons in these states tend to be more e!ectively shielded from the nucleusby other electrons. This e!ect becomes su"ciently important at higher Z todisrupt the sequence in which states are filled by electrons — sometimes leveln+1 states with low l start to fill before all the level n states with large l arefull. Accurate calculations of atomic properties in which electron-electroninteractions are taken into account are possible, but are computationallyexpensive.


19.4 Atomic Spectra

The best evidence for atomic energy levels comes from the emission of lightby atoms in a gas at low pressure. If the atoms are put in an excited stateby some mechanism, say, collisions with energetic electrons accelerated bya potential di!erence between electrodes, then light is emitted at particularfrequencies called spectral lines. These frequencies can be separated by adevice called a spectroscope. Spectroscopes use either a prism or a di!ractiongrating plus ancillary optics to make the separation visible to the eye.

The frequency of a spectral line is equal to the energy di!erence betweentwo states divided by Planck’s constant. This is a consequence of the conser-vation of energy — the energy released when an atom undergoes a transitionfrom a state with energy E2 to a state with energy E1 is just the di!erencebetween these energies. The frequency of the emitted photon is then derivedfrom the Planck formula. In terms of the angular frequency,

&21 =E2 " E1

h. (19.13)

Figure 19.3 shows the possible transitions between the lowest four energylevels of hydrogen plus the ionized state in which the electron is initiallya large distance from the hydrogen nucleus. Transitions from any state tothe ground state form a series called the Lyman series, while transitions tothe first excited state are called the Balmer series, transitions to the secondexcited state are called the Paschen series, and so on. Within each series,increasing frequencies are labeled using the Greek alphabet, so the transitionfrom n = 2 to n = 1 is called the Lyman-) spectral line, etc.

Atoms can absorb as well as emit radiation. For instance, if hydrogenatoms in the ground state are bombarded with photons of energy equal tothe energy di!erence between the ground state and some excited state, someof the atoms will absorb these photons and undergo transitions to the excitedstate. If white light (i. e., many photons with a continuous distribution offrequencies) irradiates such atoms, just those photons with the right energieswill be absorbed. Examination of the light with a spectroscope after it passesthrough a gas of atoms will show absorption lines where the photons with thecritical energies have been removed. This is one of the main ways in whichastrophysicists learn about the elemental constitution of stars and interstellargases.

Atoms in excited states emit photons spontaneously. However, a process

19.4. ATOMIC SPECTRA 341

n = infinity

n = 3

n = 2

n = 1

n = 4

α β γ

α β

inf

inf

α infPaschen series

Balmer series

Lyman series

energy

E2

E3E4

E1

0

Figure 19.3: Spectral lines from transitions between electron energy statesin hydrogen.


called stimulated emission is also possible. This occurs when a photon withenergy equal to the di!erence between two atomic energy levels interacts withan atom in the higher energy state. The amplitude for this process is equalto the spontaneous emission amplitude times n+1, where n is the number ofincident photons with energy equal to the energy of the photon which wouldbe spontaneously emitted. If a beam of photons with the right energy shineson atoms in an excited state, the beam will gain energy at a rate which isproportional to the initial intensity of the beam. For intense beams, thisstimulated emission process overwhelms spontaneous emission and a largeamount of energy can be rapidly extracted from the excited atoms. This ishow a laser works.

19.5 Problems

1. The wave function for three non-identical particles in a box of unitlength with one particle in the ground state, the second in the firstexcited state, and the third in the second excited state is

.(x1, x2, x3) = sin("x1) sin(2"x2) sin(3"x3).

(a) From this write down the wave function for three identical bosonsin the above mentioned states.

(b) Do the same for three identical fermions.

Hint: In each case there are six terms corresponding to the six per-mutations of x1, x2, and x3. Exchanging any two particles leaves .unchanged for bosons but changes the sign for fermions.

2. Two identical particles with equal energies collide nearly head-on, sothat they are both deflected through an angle !, as shown in the sketchbelow. A physicist calculates the amplitude . as a function of ! forthis deflection to take place (using very advanced theory!), resultingin the solid curve shown in figure 19.4. However, measurements showthat the actual amplitude as a function of ! (not probability!) is givenby the dashed curve.

(a) What did the physicist forget to take into account? Explain.

(b) Are the particles fermions or bosons? Explain.

19.5. PROBLEMS 343

#1 #2

ψ(θ)

π/2 π

#1

#2

θ

θ

prediction

observation

θ

Figure 19.4: Incorrectly calculated and observed scattering amplitude for acollision between two identical particles.

Hint: If the outgoing particles (but not the incoming particles) areinterchanged, how does the apparent deflection angle change?

3. Following the analysis made for the hydrogen atom, compute the “Bohrradius” and the ground state binding energy for an “atom” consistingof Z protons in the nucleus and one electron.

4. Upper and lower bounds on the binding energy of the last (outermost)electron in the sodium atom may be obtained by assuming (a) that theother electrons have no e!ect, or (b) that the other electrons neutralizeall but one proton in the nucleus. Compute the binding energy of thelast electron in sodium in these two limits. (The actual binding energyof the last electron in sodium is 5.139 eV.)

5. A uranium atom (Z = 92) has all its electrons stripped o! except thefirst one.

(a) What is the first electron’s binding energy in electron volts?

(b) What is the ground state radius of the electron orbit in this case?

6. The energy levels of a particle in a box are given by En = E0n2 =E0, 4E0, 9E0, . . . where E0 is the ground state energy for the particle.Find the lowest possible total energy of a group of particles, expressedas a multiple of E0, for the following particles in the box:

(a) 5 identical spin 0 particles.


(b) 5 identical spin 1/2 particles.

(c) 5 identical spin 1 particles.

(d) 5 identical spin 3/2 particles.

7. A charged particle in a 1-D box has energy levels at En = E0n2 =E0, 4E0, 9E0, 16E0, 25E0, . . ., where E0 is the ground state energy ofthe particle. If the particle can absorb a photon with any of the energies5E0, 12E0, 21E0, . . ., what can you infer about the initial energy of theparticle? Explain.

8. The X-rays in your dentist’s o"ce are produced when an energeticbeam of free electrons knocks the most tightly bound electrons (n = 1)completely out of the target atoms. Electrons from the next level up(n = 2) then drop into the n = 1 level.

(a) Estimate the energy in electron volts of the resulting photons fora copper target (Z = 29). Hint: For the inner electrons, you mayignore the e!ects of the other electrons to reasonable accuracy.

(b) What minimum energy must the electron beam have in this case?

9. What is the shortest ultraviolet wavelength usable in astronomy? Hint:UV photons more energetic than the binding energy of the electron inhydrogen are strongly absorbed by this gas.

10. In the naive periodic table model, the first three closed shells occur forZ = 2, 10, 28. However, the first three noble gases have Z = 2, 10, 18.Explain why this is so.

Chapter 20

The Standard Model

In this section we learn about the most fundamental known particles of theuniverse, and how they act as building blocks for everything that we know.The theory describing this scheme is called the standard model. Speculationsexist about possible more fundamental structures in the universe, such asthe constructs of string theory. However, with the standard model we havereached the frontier of what is known with any degree of certainty.

20.1 Quarks and Leptons

The standard model of quarks and leptons is a united set of quantum me-chanical theories encompassing electromagnetism, the weak force, which isresponsible for beta decay, and the strong force, which holds atomic nucleitogether. Before investigating the standard model, we need to describe thestate of a!airs previous to its development. The creation of high energy par-ticle accelerators led to the discovery of a plethora of particles in addition tothose already known. These particles fall into the following categories:

• Leptons are spin 1/2 particles which do not interact via the strongforce. The electron, muon, and the electron and muon neutrinos areexamples.

• Hadrons are particles which interact via the strong force. They aredivided into two sub-categories depending on their spin:

– Baryons are hadrons with half-integral spin, mainly 1/2 and 3/2.The proton and neutron are well known examples. The neutral

345

346 CHAPTER 20. THE STANDARD MODEL

lambda particle is another.

– Mesons are hadrons with integral spin, mainly 0 and 1. Examplesare the pions and kaons.

• Strange particles are baryons and mesons which are unstable, but havemuch longer half-lives than other particles of similar mass and spin.This is interpreted to mean that such particles possess a property calledstrangeness which is conserved by strong processes, thus making strangeparticles stable against strong decay into non-strange particles. How-ever, strangeness is not conserved by weak processes, allowing strangeparticles to decay via the weak interaction. This explains their anoma-lously long half-lives. Strange particles are always created in pairsby strong processes in such a way that the total strangeness remainszero. For instance, if one particle has strangeness +1 then the othermust have strangeness "1. An example of strange particle productionis when a negative pion collides with proton, giving rise to a neutrallambda particle and a neutral kaon.

• Intermediary particles are those which transfer energy, momentum,charge, and other properties from one particle to another in associ-ation with one of the four fundamental forces.

– Photons transmit the electromagnetic force and have zero massand spin 1.

– Gravitons are thought to transmit the gravitational force, thoughthey have not been directly observed. The graviton is postulatedto have zero mass and spin 2.

We will discover additional intermediary particles in our discussion ofthe standard model.

• Antiparticles exist for all particles. These have the same mass and spinbut opposite values of the electric charge and various other quantumnumbers such as lepton number or baryon number. The lepton numberis the number of leptons minus the number of antileptons, with a similardefinition for baryon number. Thus, a lepton has lepton number 1 anda baryon has baryon number 1. Their antiparticles have lepton number"1 and baryon number "1. As far as we know, baryon number andlepton number are absolutely conserved, which means that baryons and

20.2. QUANTUM CHROMODYNAMICS 347

Type Charge Rest energy s c b tdown (d) "1/3 0.333 0 0 0 0up (u) +2/3 0.330 0 0 0 0

strange (s) "1/3 0.486 "1 0 0 0charm (c) +2/3 1.65 0 +1 0 0bottom (b) "1/3 4.5 0 0 "1 0

top (t) +2/3 176 0 0 0 +1

Table 20.1: Table of quark types, charge (as a fraction of the proton charge),rest energy (in GeV), and the four “exotic” flavor quantum numbers.

leptons can only be created or destroyed in particle-antiparticle pairs.1

Antiparticles are represented by the symbol of the particle with anoverbar.

20.2 Quantum Chromodynamics

The standard model postulates that all known particles are either fundamen-tal point particles or are composed of fundamental point particles accordingto a remarkably small set of rules. Just as atoms are bound states of atomicnuclei and electrons, atomic nuclei are bound states of protons and neutrons.Atomic nuclei are discussed in the next section. In this section we delve onestep deeper in the heirarchy of the universe. We now believe that all hadronsare actually bound states of fundamental spin 1/2 particles called quarks.Whereas all other known particles have an electric charge equal to ±e wheree is the proton charge, quarks have electric charges equal to either "e/3 or+2e/3. Leptons themselves are considered to be fundamental, so the leptonsand the quarks form the basic building blocks of all matter in the universe.

Quarks are subject to electromagnetic forces via their charge, but interactmost strongly via the so-called strong force. The strong force is carried bymassless, uncharged, spin 1 bosons called gluons.

1Actually, lepton conservation is even more restrictive, with conversion between elec-trons, muons, and tau particles being apparently forbidden. However, recent work showsthat electron, muon, and tau neutrinos convert into each other on slow time scales. Wealso know from this work that neutrinos have small, but non-zero mass. The implicationsof these results are still being explored by the physics community.


Type Charge Rest energy Spin Composition Mean lifeProton +1 938.280 1/2 uud stableNeutron 0 939.573 1/2 udd 898Lambda 0 1116 1/2 uds 3.8 ! 10!9

positive pion +1 140 0 ud 2.6 ! 10!8

negative pion "1 140 0 ud 2.6 ! 10!8

neutral pion 0 135 0 uu - dd 8.7 ! 10!17

positive rho +1 770 1 ud 4 ! 10!24

positive kaon +1 494 0 us 1.24 ! 10!8

neutral kaon 0 498 0 ds 8.6 ! 10!11

J/psi 0 3097 1 cc 1.5 ! 10!20

Table 20.2: Sample hadrons. The charge is specified as a fraction of theproton charge, the rest energy is in MeV, and the mean life (1.44 times thehalf life) is in seconds. The composition is presented in terms of the flavorsof quarks and antiquarks which make up the particle.

When Murray Gell-Mann and George Zweig first proposed the quarkmodel in 1963, they needed to postulate only three types or flavors of quarks,up, down, and strange. These were su"cient to explain the constitution ofall hadrons known at the time. We currently know of six di!erent flavorsof quarks. Their properties are listed in table 20.1. The properties charm,topness, and bottomness are analogous to strangeness — these properties areconserved in strong interactions. Weak interactions, discussed in the nextsection, can turn quarks of one flavor into another flavor. However, thestrong and electromagnetic forces cannot do this.

Just as the proton and the neutron have antiparticles, so do quarks. An-tiquarks of a particular type have strong and electromagnetic charges of thesign opposite to the corresponding quarks. Quarks have baryon number equalto 1/3, while antiquarks have "1/3. Thus combining three quarks results ina baryon number equal to 1, while together a quark plus an antiquark havebaryon number zero. All baryons are thus combinations of three quarks,while all mesons are combinations of a quark and an antiquark. Table 20.2lists a sampling of hadrons and some of their properties. Notice that thesame combination of quarks can make up more than one particle, e. g., thepositive pion and the positive rho. The positive rho may be considered as anexcited state of the ud system, while the positive pion is the ground state of

20.2. QUANTUM CHROMODYNAMICS 349

white

green

antiblue

red

antired

blue

antigreen

Figure 20.1: The color table of quantum chromodynamics. Quarks havecolors red, green, and blue, while antiquarks have colors antired (cyan), anti-green (magenta), and antiblue (yellow). The combination of a quark and itscorresponding antiquark is colorless or white, as is the combination of threequarks (or antiquarks) of three di!erent colors.

this system.

Yet to be mentioned is the quantum number color, which has nothing todo with real colors, but has analogous properties. Each flavor of quark cantake on three possible color values, conventionally called red, green, and blue.This is illustrated in figure 20.1. Antiquarks can be thought of as having thecolors antired, antigreen, and antiblue, also known as cyan, magenta, andyellow. Because of this, the theory of quarks and gluons is called quantumchromodynamics. Counting all color and flavor combinations, there are 6 !3 = 18 known varieties of quarks.

As in electromagnetism, the strong force has associated with it a “strongcharge”, gs. However, this charge is somewhat more complicated than elec-


tromagnetic charge, in that there are three kinds of strong charge, one foreach of the strong force colors. Each color of charge can take on positiveand negative values equal to ±gs. As with electromagnetism, positive andnegative charges (of the same color) cancel each other. However, in quantumchromodynamics there is an additional way in which charges can cancel. Acombination of equal amounts of red, green, and blue charges results in zeronet strong charge as well.

Gluons, the intermediary particles of the strong interaction come in eightdi!erent varieties, associated with di!ering color-anticolor combinations. Sincegluons don’t interact via the weak force, there is no flavor quantum numberfor gluons — quarks of all flavors interact equally with all gluons.

The quark model of matter has led to extensive searches for free quarkparticles. However, these searches for free quarks have proven unsuccessful.The current interpretation of this result is that quarks cannot exist in afree state, basically because the attractive potential energy between quarksincreases linearly with separation. This appears to be related to the fact thatgluons, the intermediary particles for the strong force, can interact with eachother as well as with quarks. This leads to a series of increasingly complexprocesses as quarks move farther and farther apart. The result is called quarkconfinement — apparently, individual quarks can never be observed outsideof the confines of the observable particles which contain them.

Confinement works not only on single quarks, but on any “colored” com-binations of quarks and gluons, e. g., a red up quark combined with a greendown quark. It appears that long range inter-quark forces only vanish for in-teractions between “white” or “color-neutral” combinations of quarks. Thisis why only color-neutral combinations of quarks — three quarks of threedi!erent colors or a quark-antiquark pair of the same color — are actuallyseen as observable particles.

The strong equivalent of the fine structure constant is the coupling con-stant for the strong force:

)s =g2

s

4"hc. (20.1)

Note that )s is dimensionless. The binding energy between quarks is compa-rable to the rest energies of the quarks themselves. In other words, )s & 1.Furthermore, as we have noted, the potential energy between two quarks ap-pears to increase indefinitely with separation. Though forces exist betweencolor-neutral particles, they are weak and of short range compared to theforces between quarks or colored combinations of quarks. However, they are

20.3. THE ELECTROWEAK THEORY 351

d u u u d

s u dd

−

−s

lambdaneutral kaon

negative pion

u d

u u u−

−

d−

positive pionneutral pion

proton positive rho

gg

Figure 20.2: Some sample strong interactions illustrated in terms of gluonemission and absorption. The process on the left shows the reaction "!+p *K0 + (0, while the one on the right shows the decay #+ * "+ + "0. Quarksare labeled with solid lines while gluons are shown by dashed lines.

still relatively strong compared to, say, electromagnetic forces. As we shallsee later, these residual strong forces are responsible for nuclear processes.

Interactions between hadrons can be thought of as resulting from inter-actions between the individual quarks making up the hadrons. Two samplestrong interactions are shown in figure 20.2. Virtual gluons can be emittedand absorbed by quarks much as virtual photons can be emitted and ab-sorbed by electrically charged particles. Particles unstable to strong decayprocesses (such as the positive rho particle) typically live only about 10!23 s,whereas particles stable to strong decay but unstable to weak decay live oforder 10!10 s or longer, depending strongly on how much energy is liberatedin the decay. Particles subject to electromagnetic decay processes, such asthe neutral pion, take on mean lifetimes intermediate between strong andweak values, typically of order 10!18 s.


Type Charge Rest energy Mean lifeelectron (e!) -1 0.000511 stable

electron neutrino (/e) 0 & 0 stablemuon (µ!) -1 0.106 2.2 ! 10!6

mu neutrino (/µ) 0 & 0 stabletau (') -1 & 1.7 3.0 ! 10!13

tau neutrino (/$ ) 0 & 0 stable

Table 20.3: Table of lepton types, charge (as a fraction of the proton charge),rest energy (in GeV), and mean life (in seconds).

20.3 The Electroweak Theory

The strong force acts only on quarks and the strong force carrier, the gluon.It does not act on leptons, e. g., electrons, muons, or neutrinos. Table 20.3shows all of the known leptons. The so-called weak force acts on leptons aswell as on quarks.

In 1979 Sheldon Glashow, Abdus Salam, and Steven Weinberg won theNobel Prize for their electroweak theory, which unites the electromagneticand weak interactions. Unlike the strong and electromagnetic forces, theintermediary particles of the weak interaction, the W+, the W!, and theZ0, have rather large masses. In particular, the rest energy of the W± is81 GeV while that of the Z0 is 92 GeV. Electroweak theory considers elec-tromagnetism and the weak interactions to be di!erent aspects of the sameforce. A key aspect of the theory is the explanation of why three out offour of the intermediary particles of the electroweak force are massive. (Thephoton is the massless one.) Unfortunately, the details of why this is so arehighly technical, so we cannot delve into this subject here. We only note thatthe explanation requires the existence of a highly massive (several thousandGeV) spin zero boson called the Higgs particle. Due to its large mass, wehave not yet determined whether the Higgs particle exists.

The weak force has certain bizarre properties not shared by the otherforces of nature:

• The weak interaction can change quark flavors. For instance, the betadecay of a neutron converts a down quark into an up quark. On theother hand, the weak interaction is “colorblind”, i. e., it is insensitive

20.3. THE ELECTROWEAK THEORY 353

W-

n

pe

p

n

W+

beta decay antineutrino detection

_e

_ν

νe

e

d d u

u

W-

neutron --> proton

p

n

Figure 20.3: Illustration of two weak reactions. The left panel shows betadecay while the middle panel shows how electron antineutrinos can be de-tected by conversion to a positron. The right panel shows how W! emissionworks according to the quark model, resulting in the conversion of a downquark to an up quark and the resulting transformation of a neutron into aproton.

to quark colors.

• The weak interaction is not left-right symmetric. In other words, thephysical laws governing the weak interaction look di!erent when seenin a mirror.

• The weak interaction is slightly asymmetric to the interchange of par-ticles and antiparticles in certain situations.

The prototypical weak interaction is the decay of the neutron into a pro-ton, an electron, and an antineutrino. This decay is energetically possiblebecause the neutron is slightly more massive than the proton, and is illus-trated in the left panel of figure 20.3. Note that this figure is drawn as if aneutrino moving backward in time absorbs a W! particle, with a resultingelectron exiting the reaction forward in time. However, we know that thisis equivalent to an electron and an antineutrino both exiting the reactionforward in time according to the Feynman interpretation of negative energystates.


Generation Leptons Quarks1 electron down

electron neutrino up2 muon strange

mu neutrino charm3 tau bottom

tau neutrino top

Table 20.4: Generations of leptons and quarks. Members of each generationtend to fit together.

The weak interaction is called “weak” because it appears to be so in com-monly observed processes. For instance, the range of a relativistic electronin ordinary matter is of order centimeters to meters. This is because theelectromagnetic force between the charge of the electron and the charges onatomic nuclei are strong enough to rapidly cause the energy of the electronto be dissipated. However, the range in matter of a neutrino produced bybeta decay is many orders of magnitude greater than that of an electron.This is not because the weak force is intrinsically weak — the value of the“fine structure constant” for the weak force is

)w & 10!2 (20.2)

according to the standard model, and is actually larger than ) for electro-magnetism.

The real reason for the apparent weakness of the weak force is the largemass of the intermediary particles. As we have seen, large mass translatesinto short range for a virtual particle at low momentum transfers. This shortrange is what causes the weak force to appear weak for momentum transfersmuch less than the masses of the W and Z particles, i. e., for q $ 100 GeV.For leptons and quarks with energies E ' 100 GeV, the weak force acts withmuch the same strength as the electromagnetic force.

20.4 Grand Unification?

The standard model is a great achievement, but it leaves a number of ques-tions unanswered. As table 20.4 shows, nature seems to have produced more

20.4. GRAND UNIFICATION? 355

1010 1015 10201051

.01

.1

1

.001

α

electromagnetic

weak

strong

momentum transfer (GeV)

Figure 20.4: Speculated behavior of the dependence of the coupling constant) on momentum transfer for each of the forces. Extrapolation based on cur-rent measurements suggests that these constants come together to a commonvalue at very high momentum transfer.

particles than are needed to construct the universe. Virtually everything weknow of is composed of electrons, electron neutrinos, up quarks, and downquarks. These four particles seem to fall naturally together in a family orgeneration. Why then are there apparently unneeded additional generations?What role do muons, taus, and the exotic quark forms play in the universe?

Another question concerns the dichotomy between leptons and quarks.Electrons and electron neutrinos can be converted into each other by weakinteractions, as can up and down quarks. Why then can’t quarks be convertedinto leptons and vice versa?

In the standard model, electromagnetic and weak forces are truly unitedas aspects of a single phenomenon. However, quantum chromodynamicsstands more on its own. One could imagine further advances which wouldshow that the electroweak and strong forces were in fact di!erent aspects ofthe same phenomenon. This could be characterized as a grand unification ofthe forces of nature.

As previously noted, the strong force coupling constant, )s, gets smallerwith increasing momentum transfer. It turns out that the weak couplingconstant, )w, exhibits similar behavior, while the electromagnetic couplingconstant, the fine structure constant ), becomes stronger at higher energies.


This behavior is illustrated in figure 20.4, though it is based on data only upto about 103 GeV/c. Figure 20.4 is thus largely speculative. However, if theobserved trends do continue to very high momentum transfers, this would beevidence in favor of grand unification.

A number of speculative grand unification theories have been proposed.Most such theories view leptons and quarks as being di!erent states of thesame particle and also predict that leptons can turn into quarks and viceversa, albeit at very low rates. One of the consequences of such theories isthat the proton would be an unstable particle, but with a very long lifetime,of order 1030 yr. Experiments have been done to detect the decay of theproton, but so far without success. These experiments are su"cient to ruleout some but not all of the proposed grand unification theories.

One task which would not be accomplished by grand unification is theincorporation of gravity into a common framework with the strong, weak andelectromagnetic forces. Creation of a satisfactory quantum theory of gravityhas been a very di"cult problem and is unsolved to this day, though manypeople are working on it.

20.5 Problems

1. Verify that the quark model predicts the correct electric charge forthe proton, the neutron, and all the pions. Also check to see if thespin angular momentum of each of these particles is consistent with itsquark composition.

2. Draw a picture of how the negative pion decays into a muon and amu antineutrino in terms of the quark model of the pion and our ideasabout the weak interaction.

3. Draw a picture of how the muon decays into a mu neutrino, an electron,and an electron antineutrino in terms of our ideas about the weakinteraction.

4. A mu antineutrino hits a proton, turning it into a neutron.

(a) What other particle must be emitted from this reaction?

(b) Could you use this result to distinguish between electron and muantineutrinos?

20.5. PROBLEMS 357

pionpositive

electron proton

photon

neutron

Figure 20.5: An example of inelastic electron-proton scattering.

(c) What minimum total energy in the center of momentum framewould you expect of the mu antineutrino for this reaction to bepossible?

5. Suppose that the electron had a rest energy of M = 500 MeV ratherthan & 0.5 MeV. Describe as best you can the many ways in whichthis would change the world and universe in which we live.

6. In the reaction shown in figure 20.5, specify what actually happens atthe vertex in the shaded region in terms of the quark model of hadrons.

7. A solar neutrino detector in South Dakota consists of a huge tank ofcleaning fluid, which has a large concentration of chlorine-37 (Z =17, A = 37).

(a) Will an electron neutrino more likely interact with a proton or aneutron in the chlorine-37 nucleus?

(b) If this interaction occurs, what will the final products be?

Note: Z = 16 is sulfur and Z = 18 is argon.

8. An electron collides with an antimuon, resulting in the apparent dis-appearance of both particles. This seems to indicate that energy is notconserved.


(a) What do you, the Sherlock Holmes of particle physics, suggestactually happened?

(b) Is this likely to be a very common event? Why or why not?

9. A ( particle consists of 3 quarks with flavors u, d, s. A possible decaymode is ( * p + "!.

(a) Is the ( a fermion or a boson? Explain.

(b) Draw a Feynman diagram showing how the above decay can hap-pen at the quark level.

(c) Is the above decay a strong or a weak process?

Reminder: p = u, u, d; "! = u, d.

Chapter 21

Atomic Nuclei

Atomic nuclei are composite particles made up of protons and neutrons.These two particles are collectively known as nucleons. In order to betterunderstand atomic nuclei, we first make an analogy with molecules. We theninvestigate the binding energies of atomic nuclei. This information is centralto the subjects of radioactive decay as well as nuclear fission and fusion.

21.1 Molecules — an Analogy

Molecules are bound states of two or more atoms. In chemistry we identifyseveral modes of molecular binding, e. g., covalent and ionic bonds, the hy-drogen bond, and binding at low temperatures due to the van der Waal’sforce. All of these bonds involve electromagnetic forces, but all (except ar-guably the ionic bond) are relatively subtle residual forces between atomswhich are electrically neutral. The ways in which atoms form molecules aretherefore complex and resistent to accurate calculation.

Atomic nuclei are the nuclear equivalent of molecules, in that they arebound states of nucleons, which are themselves “uncharged” composite parti-cles. The charge we refer to here is not the electric charge (nuclei do of coursepossess this!), but the strong or color charge. As we discovered in the pre-vious section, nucleons are color-neutral combinations of quarks. Thus, the“strong” forces between nucleons are subtle residuals of inter-quark forces.This is reflected in the binding energies; quark-quark binding energies areon the order of the rest energies of the quarks themselves. However, nuclearbinding energies are typically of order 10 MeV per nucleon, or about 1% of

359

360 CHAPTER 21. ATOMIC NUCLEI

the rest energy of a nucleon.The residual nature of nuclear forces makes them complex and di"cult to

calculate from our basic knowledge of quantum chromodynamics for the samereasons that intermolecular forces are di"cult to calculate. An empiricalapproach is thus needed in order to understand their e!ects.

In contrast to molecules and atomic nuclei, atoms are relatively easy tounderstand. This is true for two reasons: (1) Electrons appear to be trulyfundamental point particles. (2) Though the atomic nucleus itself is a verycomplex system, little of this complexity spills over into atomic calculations,because on the atomic scale the nucleus is very nearly a point particle. Thus,both main ingredients in atoms are “simple” from the point of view of atomiccalculations.

The above result is true because by some accident of nature, the mass ofthe electron is so much less than the masses of quarks. It would be interestingto speculate what atomic theory would be like if this weren’t true — therewould be no scale separation between the atomic and nuclear scales, and theworld would be a very di!erent place!

21.2 Nuclear Binding Energies

It is impossible to specify an accurate inter-nucleon force valid under allcircumstances, but figure 21.1 gives an approximate representation of thepotential energy associated with the strong force as the function of nucleonseparation. The binding energy is of order 2 MeV, with an attractive forcefor separations greater than about 2!10!15 m and an intense repulsive forcefor smaller separations. At large distances the potential energy decays expo-nentially with distance rather than according to the r!1 law of the Coulombpotential.

The short range of the inter-nuclear force means that atomic nuclei canbe thought of as conglomerations of “sticky billiard balls”. The nuclear forceis essentially a contact force and each nucleon simply binds to all its nearestneighbors. When nucleons are close-packed, the binding energy per nucleondue to the strong force is simply the number of nearest neighbors for eachnucleon, times the binding energy per nucleon pair, divided by 2. The factorof 1/2 accounts for the fact that each nuclear bond is shared by two nucleons.

Several other e!ects need to be accounted for in the nucleus. The nucleonson the surface of the nucleus do not have as many bonds as nucleons in the

21.2. NUCLEAR BINDING ENERGIES 361

U(r)r

2 fm

B = 2 MeV

Figure 21.1: Approximate sketch of the strong force potential energy betweentwo nucleons. 1 fm = 10!15 m. The binding energy B is the energy requiredto separate the two nucleons. If the nucleons are bound together, the restenergy of the resulting combination, Mcomboc2 is less than the sum of the restenergies of the two nucleons, M1c2, M2c2, by the amount B: Mcomboc2 =M1c2 + M2c2 " B.


protons neutrons protons neutrons

3

1

2

energy

Figure 21.2: E!ect of Pauli exclusion principle on two nuclei, each with 8nucleons. The total energy of the nucleus on the left, which has an equalnumber of protons and neutrons is 2! (1+2)+2! (1+2) = 12. The nucleuson the right has total energy 2 ! (1) + 2 ! (1 + 2 + 3) = 14.

interior. Thus, to compute the nuclear binding energy of a nucleus with afinite number of nucleons, a correction must be made for this e!ect. Thiscontributes negatively to the nuclear binding energy in proportion to thesurface area of the nucleus, which scales as the number of nucleons to thetwo-thirds power.

In addition to the nuclear force, the repulsive electrostatic force betweenprotons needs to be accounted for. Since the electrostatic force is a longrange force, the (negative) contribution to the binding energy of the nucleusgoes as the square of the number of protons divided by the radius of thenucleus. The latter goes as the cube root of the number of nucleons.

The Pauli exclusion principle operates in nuclei so as to favor equal num-bers of protons and neutrons. This e!ect is illustrated in figure 21.2. If aproton is converted into a neutron in a nucleus in which equal numbers ofthe two particles occur, then the exclusion principle forces these nucleonsto move to a higher energy level than they previously occupied. The bind-ing energy of the nucleus is correspondingly decreased. This e!ect opposesthe weaker repulsive Coulomb potential which occurs when there are moreneutrons and fewer protons.

The net result of all these e!ects is a nuclear binding energy equationwith four terms representing the four above-mentioned e!ects:

B(Z, A) = avA " asA2/3 " acZ

2/A1/3 " aa(2Z " A)2/A (21.1)

where Z is the atomic number or the number of protons, N is the number of

21.2. NUCLEAR BINDING ENERGIES 363

0.0

20.0

40.0

60.0

80.0

0 30 60 90 120Z vs N

Binding Energy Per NucleonContour Interval: 1 MeVNegative Values: horizontal hatchingValues exceeding 8 MeV: vertical hatching

Figure 21.3: Nuclear binding energy per nucleon B(Z, A)/A, calculated fromequation (21.1). The curved line starting near the origin gives the line ofstability for atomic nuclei.

neutrons, and A = Z +N is the atomic mass number, or number of nucleons.Equation (21.1) represents the binding energy of the entire nucleus. Thebinding energy per nucleon is just B/A.

Fitting equation 21.1 to observed binding energies in nuclei yields thefollowing values for the coe"cients of the above equation: av & 16 MeV,as & 17 MeV, ac & 0.70 MeV, and aa & 23 MeV. A contour plot of bindingenergy per nucleon, B/A, is shown in figure 21.3. We note that this equationdoesn’t work well for nuclei with only a few nucleons. For instance, thehelium nucleus with A = 4 is more stable than the lithium nucleus withA = 6, and there is no stable nucleus at all with A = 5.

Part of the reason for the problem at small A is that even numbers of pro-tons and neutrons tend to bind more strongly together than nuclei containingodd numbers of either. This is because spin up-spin down pairs of protons orneutrons fully occupy nuclear states while an odd nucleon occupies a stateby itself with energy greater than that of all the other occupied states. Thisbehavior can be approximately accounted for by adding the term ap/A1/2 to


5.00

6.00

7.00

8.00

9.00

0 50 100 150 200B/A (MeV per nucleon) vs A

Binding Energy per Nucleon on Line of Stability

Figure 21.4: Binding energy per nucleon along line of stability according toequations (21.3) and (21.2).

equation (21.1), where ap = 12 MeV if N and Z are both even, ap = 0 ifeither N or Z is odd, and ap = "12 MeV if both are odd. We leave this termo! even though it is sometimes quite important, in order to make equation(21.1) a smooth function of Z and A and thus representative of the generaltrend of binding energy.

For a given value of A, it is easy to demonstrate that the maximumnuclear binding energy in equation (21.1) occurs when

Z =A

2(1 + acA2/3/4aa). (21.2)

This formula confirms the trend seen in figure 21.3 that the most stablenuclear configuration contains an increasing fraction of neutrons as A in-creases. The function Z(N) given by equation (21.2) and illustrated by thecurve starting near the origin in figure 21.3 defines the line of stability foratomic nuclei.

Figure 21.4 shows the binding energy per nucleon as a function of nucleonnumber A along the line of stability. The rapid increase in binding energyfor small A reflects the decreasing surface e!ect as the number of nucleons

21.3. RADIOACTIVITY 365

increases. The subsequent decrease is a result of the combined e!ects ofCoulomb repulsion of protons and the Pauli exclusion principle. Notice thatthe maximum binding energy per nucleon occurs near A = 60.

The chemical properties of the atom associated with an atomic nucleus aredetermined by the number of protons, Z, in the nucleus. In many cases thereexists more than one stable nucleus with a given value of Z. These nucleidi!er in their neutron number, N . Nuclei with the same Z and di!ering Nare called isotopes of the element defined by the specified value of Z. Forinstance, there are three isotopes of the element hydrogen, normal hydrogen,deuterium, and tritium, with zero, one, and two neutrons respectively.

21.3 Radioactivity

Radioactive decay is the emission of some particle from an atomic nucleus,accompanied by a change of state or type of the nucleus, depending on thetype of radioactivity.

Gamma rays or photons are emitted when a nucleus decays from an ex-cited state to its ground state. No transformation of the nuclear type occurs.Photons are often emitted when some other form of radioactive decay leavesthe resulting nucleus in an excited state.

Beta minus decay is the conversion of a neutron into a proton, an electron,and an electron antineutrino. This and the inverse reaction, beta plus decay,or conversion of a proton into a neutron, a positron, and an electron neutrino,were described in the last chapter. These processes occur in the nucleonscontained in nuclei when they are energetically possible.

Alpha particle emission occurs in heavy elements where it is energeticallypossible. Since an alpha particle is just a helium 4 nucleus containing twoprotons and two neutrons, the values of Z and N of the emitting nucleus areeach reduced by two.

The rest energy of a nucleus (ignoring atomic e!ects) is just the sum ofthe rest energies of all the nucleons minus the total binding energy for thenucleus:

M(Z, A)c2 = ZMpc2 + NMnc2 " B(Z, A), (21.3)

where Mpc2 = 938.280 MeV is the rest energy of the proton and Mnc2 =939.573 MeV is the rest energy of the neutron.


-15.0

-10.0

-5.0

0.0

0 50 100 150 200Q for alpha decay (MeV) vs A

-15.0

-10.0

-5.0

0.0

0 50 100 150 200Q for alpha decay (MeV) vs A

Energy Release in Alpha Decay

Figure 21.5: Approximate curve for the energy released in alpha decay of anucleus on the line of stability. Decay is only possible if Q > 0.

Energy conservation requires that

M(Z, A)c2 = M(Z " 2, A " 4)c2 + M(2, 4)c2 + Q (21.4)

for the alpha decay of a nucleus. If Q > 0, then the decay is energeticallypossible. The excess energy, Q, goes into kinetic energy of the new nucleusand the alpha particle, mainly the latter. Substitution of equation (21.3)into equation (21.4) yields

Q = B(Z " 2, A " 4) + B(2, 4) " B(Z, A) (alpha decay). (21.5)

The binding energy of the alpha particle is not accurately represented byequation (21.1), but is known to be about B(2, 4) = 28.3 MeV. On the otherhand, the heavy elements are generally well represented by equation (21.1).The curve of Q versus A is plotted in figure 21.5, and it shows that alphadecay for nuclei along the line of stability is energetically impossible (i. e.,Q < 0) for nuclei with A less than about 175.

Figure 21.6 shows schematically how alpha and beta decay transformatomic nuclei in the N -Z plane. As previously indicated, alpha decay de-creases both Z and N by two. Ordinary beta decay (i. e., n * p+ + e! + /e)

21.3. RADIOACTIVITY 367

N

Z

Z = -2Z = -1

N = 1Z = 1

N = -2

Δ

Δ ΔΔ

Δ

-β decay

α decay

ΔN = -1

β+ decay

Figure 21.6: Schematic illustration of the paths of nuclear transformationsin the N -Z plane due to alpha and beta decay. The thick line represents theline of stability. -+ is the decay of a proton into a neutron, positron, andelectron neutrino, while -! is the decay of a neutron into a proton, electron,and electron antineutrino.

decreases N by one and increases Z by one. This is sometimes called -!

decay since it produces an electron with negative charge. Though the protonin isolation is stable, the energetics of atomic nuclei are such that a nucleuswith a higher proton-neutron ratio than specified by the line of stability cansometimes release energy by the reaction p+ * n + e+ + /e. This is called-+ decay since it produces a positively charged positron.

Certain isotopes of very heavy elements are at the head of a chain ofradioactive decays. This chain consists of a combination of alpha decaysinterspersed with -! decays. The latter are needed because the alpha decayscreate nuclei with too low a ratio of protons to neutrons relative to the lineof stability, as illustrated in figure 21.6. The beta decays push the chainback toward this line. An example of a chain is one which starts with theelement thorium (Z = 90, A = 232) and ends with lead (Z = 82, A =208). Radioactive decay thus accomplishes what medieval alchemists tried,but failed to do, namely transmute elements from one type into another.Unfortunately, no radioactive chain ends at the element gold!

Radioactive decay is governed by a simple law, namely that the rate atwhich nuclei decay is proportional to the number of remaining nuclei. In


mathematical terms, this is expressed as follows:

dN

dt= "+N, (21.6)

where N(t) is the number of remaining nuclei at time t and + is called thedecay rate. This di!erential equation has the solution

N(t) = N(0) exp("+t) (radioactive decay), (21.7)

which shows that the number of nuclei decreases exponentially with time.The half-life, t1/2 of a certain nuclear type is the time required for half

the nuclei to decay. Setting N(t1/2) = N(0)/2, we find that

t1/2 =ln(2)

+(half-life). (21.8)

The nature of exponential decay means that half the particles are left afterone half-life, a quarter after two half-lives, an eighth after three half-lives,etc. The actual value of +, and hence t1/2, depends on the character of thenucleus in question, with half-lives ranging from a small fraction of a secondto many billions of years.

21.4 Nuclear Fusion and Fission

From figure 21.4 it is clear that atomic nuclei with A < 60 can combine toform more tightly bound nuclei and in so doing release energy. This is callednuclear fusion and it is the process which powers stars.

It is not easy to fuse two nuclei. As figure 21.7 shows, the nuclear force,which is attractive but short in range, and the Coulomb force, which isrepulsive, combine to create a potential barrier which must be surmountedin order to release energy from fusion. Nuclei must therefore somehow attainlarge kinetic energy for fusion to take place. We shall discover later thattemperature is a measure of the translational kinetic energy of atoms andnuclei. Therefore, one way to create fusion is to heat the appropriate materialto a very high temperature. The interiors of ordinary stars are hot enough tofuse hydrogen into helium. Somewhat hotter stars can create slightly heavierelements. However, we believe that only the interior of a type of explodingstar called a supernova is hot enough to create the heavy elements we find

21.4. NUCLEAR FUSION AND FISSION 369

ground state

potential barrier

U(r)

energy released by fusion

nuclear potential

r

combined potential

Coulomb potential

Figure 21.7: Combined nuclear and Coulomb potentials between two lightnuclei. The resulting potential barrier repels the two nuclei unless theirkinetic energy is very large. However, if the nuclei are able to overcome thisbarrier, substantial energy is released.

Nucleus Z A B (MeV)deuterium 1 2 2.22tritium 1 3 8.48helium-3 2 3 7.72helium-4 2 4 28.30lithium-6 3 6 32.00lithium-7 3 7 39.25

Table 21.1: Binding energies of light nuclei.


energy of alpha particle in nucleus

5 MeV

potential energy ofalpha particle

r

U(r)

Figure 21.8: Spontaneous fission of a heavy nucleus into a slightly lighternucleus and an alpha particle occurs when the alpha particle penetrates thepotential barrier illustrated by the shading and leaves the nucleus. Othertypes of spontaneous fission occur in a similar manner. Compared to the caseof two light nuclei in figure 21.7, the Coulomb potential is more importanthere, which makes the resultant force more repulsive.

in the universe. Thus, the iron in your automobile engine and the copper inyour electrical wiring were created in some of the most spectacular explosionsin the universe!

In computing energy balances for light nuclei, it is important to use exactvalues of binding energies, not the approximate values obtained from thebinding energy formula given by equation (21.1), as the values given by thisequation for small A can be o! by a large amount. Sample values for suchnuclei are given in table 21.1.

It is possible for a heavy nucleus such as uranium, with atomic numberand atomic mass number (Z, A) to spontaneously fission or split into twolighter nuclei with (Z ", A") and (Z " Z ", A " A") if there is a net energyrelease from this process:

Q % B(Z"Z ", A"A")+B(Z ", A")"B(Z, A) > 0 (fission possible). (21.9)

An energy of order 160 MeV per nucleus can be released by causing uranium(Z = 92) or plutonium (Z = 94) to fission.

Even if Q > 0, spontaneous fission generally occurs at a very slow rate.This is because a potential energy barrier of order 5 MeV typically must beovercome for this split to occur. Barrier penetration allows fission to occur

21.5. PROBLEMS 371

spontaneously in the absence of the energy needed to overcome this barrier,as illustrated in figure 21.8, but is generally a slow process. Alpha decay isan example of spontaneous fission of a heavy nucleus by barrier penetrationin which Z " = 2 and A" = 4.

If a heavy nucleus collides with an energetic particle such as a neutron,photon, or alpha particle, it can be induced to fission if the energy transferredto the nucleus exceeds the approximate 5 MeV needed to breach the potentialbarrier.

If the heavy nucleus has an odd number of neutrons, another way forfission to occur is for the nucleus to capture a slow neutron, i. e., one withenergy much less than the 5 MeV needed to directly overcome the potentialbarrier. In this case neutron capture actually converts the nucleus fromatomic number and mass (Z, A) to atomic number and mass (Z, A + 1).

The binding energy per nucleon of a nucleus with an even number ofneutrons is greater than the binding energy per nucleon of one with an oddnumber, since in the former case all neutron spins are paired. Thus, if theinitial nucleus has an odd number of neutrons, the capture of a slow neutronmakes it more tightly bound than if the initial nucleus has an even number ofneutrons. If the di!erence in binding energy between the initial nucleus andthe nucleus modified by neutron capture exceeds the 5 MeV needed to over-come the potential barrier for spontaneous fission, then energy conservationleaves the new nucleus in a su"ciently high excited state that it instantlyfissions. Examples of nuclei subject to fission by slow neutron absorptionare uranium 235 and plutonium 239. Note that both have odd numbers ofneutrons. In contrast, uranium 238 has an even number of neutrons and slowneutron bombardment does not cause fission.

21.5 Problems

1. How would nuclear physics be di!erent if the weak interactions didn’texist?

2. Suppose one started with 1020 radioactive atoms with a half life of 2 hr.How many half lives would one have to wait to be reasonably sure thatnone of the atoms were left?

3. One possible laboratory fusion reaction is d+d * )+Q where d repre-sents a deuteron (Z = 1, A = 2), ) an alpha particle, and Q the released


energy. Given the binding energies for the deuteron (2.22 MeV) and forthe alpha particle (28.30 MeV), find the energy released by this reac-tion. For the purposes of this problem you may ignore the rest energyof the electrons and their binding energy.

4. Fusion in the sun is a complicated process, but the net e!ect is theconversion of four protons into an alpha particle, or a helium-4 nucleus.This is what powers the sun.

(a) How much energy is released for every helium-4 nucleus created?

(b) How many and what kind of neutrinos or antineutrinos are re-leased for every helium-4 nucleus created?

(c) At the earth’s orbit we get about 1400 J m!2 s!1 from the sun.How many neutrinos or antineutrinos do we expect to get fromthe sun per square meter per second from solar fusion?

5. A neutrino has to pass within a distance D & h/(Mc) of a quark tohave a chance of )2

w interact with it, where M is the mass of a Wparticle and )w is the weak “fine structure constant”.

(a) What is the area of the circular “target” centered on the quarkthrough which the neutrino has to pass in order to interact withthe quark?

(b) If the quarks are located in the nuclei of water molecules, howmany quarks are there per molecule with which the neutrino caninteract? Hint: The neutrino can only interact with d quarks inneutrons. Why?

(c) Imagine a cylindrical water tank of end cross-sectional area A andlength L, with neutrinos passing through the tank in a directionparallel to the axis of the cylinder. How many quarks of the rightkind are needed in the tank to give a neutrino passing throughthe tank a 50% probability of interacting with a quark?

(d) How big must L be in this case? Water has a density of about1000 kg m!3.

6. Suppose that fission of a uranium-235 nucleus induced by absorbing aslow neutron ultimately results in two equal nuclei plus two neutrons.

21.5. PROBLEMS 373

(a) How much energy is released for each fissioned uranium-235 nu-cleus? Hint: The fission products must beta decay until they reachthe line of stability on the N-Z plot. Thus, the final state consistsof the two free neutrons, two nuclei with the same value of A asthe fission products, but with some of the neutrons converted toprotons, and the resulting electrons and neutrinos.

(b) How many neutrinos or antineutrinos are released per second bya 100 MW nuclear power plant?

Chapter 22

Heat, Temperature, andFriction

Human beings have long had an intuitive understanding of heat and tem-perature from personal experience. We sense that di!erent things often havedi!erent temperatures and we know that objects tend to acquire the sametemperature after being placed in physical contact for some time. We viewthis equilibration process as a flow of “heat” (whatever that is) from thewarmer body to the cooler body.

A need for a more precise understanding of the behavior of heat andtemperature was felt with the development of the steam engine. The scienceof thermodynamics arose out of this need. Thermodynamics was developedbefore we understood the atomic nature of matter. More recently the ideasof thermodynamics were related to mechanical processes happening on theatomic scale. Today we understand the phenomena of heat and temperatureto be aspects of the collective mechanical behavior of large numbers of atomsand molecules.

22.1 Temperature

We measure temperature by a variety of means. The most primitive measure-ment is direct sensing by the human body. We immediately discern whethersomething we touch is hot or cold relative to our own body. Furthermore, wecan detect a hot stove from a distance by the feeling of warmth on our skin.In the case of direct contact, heat is transferred to our hand by conduction,

375

376 CHAPTER 22. HEAT, TEMPERATURE, AND FRICTION

W + WΔ

L + LΔ

T + TΔ

W

L

T

Figure 22.1: Most solid bodies expand by the same fractional amount in alldirections when their temperature increases, so that %L/L = %W/W . Thus,the ratio ) = %L/(LdT ) is the same for all objects constructed of the samematerial, generally over a considerable range of temperature.

whereas in the latter case the transfer takes place by thermal radiation. Ourbody considers something to be hot if heat is transferred from the object toour body, whereas it is perceived as being cold if the transfer of heat is fromour body to the object.

A more objective measure of temperature is obtained by using the factthat ordinary material objects expand when they become warmer and con-tract when they cool. Empirically it is found that the fractional change inthe length of a solid body, %L/L, is related to the change in temperature%T , as illustrated in figure 22.1:

%L

L= )%T, (22.1)

where ) is called the linear coe"cient of thermal expansion.For liquids the fractional change in volume, %V/V , is easier to relate to

the change in temperature than the fractional change in linear dimension:

%V

V= -%T, (22.2)

where - is the volume coe"cient of thermal expansion. The quantities )and - depend on the material properties and on the temperature scale beingused. The ordinary thermometer is based on the thermal expansion of aliquid such as mercury.

The most commonly used temperature scales in science are the Celsiusand Kelvin scales. Roughly speaking, water freezes at 0# C and it boils (atsea level) at 100# C. More precise definition of the Celsius scale depends

22.1. TEMPERATURE 377

Material ) (K!1) - (K!1)steel 12 ! 10!6 —

copper 16 ! 10!6 —aluminum 23 ! 10!6 —

invar 0.7 ! 10!6 —glass 9 ! 10!6 —lead 29 ! 10!6 —

methyl alcohol — 1.22 ! 10!3

glycerine — 0.53 ! 10!3

mercury — 0.182 ! 10!3

water (15# C) — 0.15 ! 10!3

water (35# C) — 0.35 ! 10!3

water (90# C) — 0.70 ! 10!3

Table 22.1: Values of the linear coe"cient of thermal expansion for commonsolids and the volume coe"cient of expansion for common liquids. Invar isan alloy which is specificially formulated to have a low coe"cient of thermalexpansion.

on a detailed understanding of the phase changes of water which we won’tdevelop here.

There is a limit to how cold something can be. The Kelvin scale is de-signed to go to zero at this minimum temperature. The relationship betweenthe Kelvin temperature T and the Celsius temperature TC is

T = TC + 273.15. (22.3)

Thus, water freezes at about 273 K and boils at about 373 K. (Notice thatthe little circle or degree sign is used for Celsius temperatures but not Kelvintemperatures.) Unless otherwise noted, we will use the Kelvin scale. Table22.1 gives values of ) and - for some common materials.

Accurate temperature measurements depend in practice on a knowledgeof the properties of materials under temperature changes. However, we shallfind later that the concept of temperature can be defined in a way that iscompletely independent of material properties.


xΔ

FT

Figure 22.2: Conversion of internal energy of gas in the cylinder to macro-scopic energy. The work done by the force of the gas on the piston as itmoves outward results in a decrease in temperature of the gas.

22.2 Heat

Two types of experiments suggest that heating is a form of energy transfer.First of all, on the macroscopic or everyday scale of things, there are forceswhich are apparently nonconservative. This is in marked contrast to the mi-croscopic world, where forces are either conservative (gravity, electrostatics),or don’t change a particle’s energy (magnetic force), or convert energy fromone known form to another (non-static electric forces). With these funda-mental forces all energy is accounted for — it is neither created or destroyed.

In contrast, macroscopic energy routinely disappears in the everydayworld. Cars once set in motion don’t continue in motion forever on a levelroad once the engine is stopped; a soccer ball once kicked eventually comesto rest; electrical energy powering a light bulb appears to be lost. Carefulmeasurements show that whenever this type of energy loss is found, heat-ing occurs. Since we believe that macroscopic forces are really just largescale manifestations of fundamental microscopic forces, we do not believethat energy really disappears as a result of these forces — it must simplybe converted from a form visible to us into an invisible form. We now knowthat such forces convert macroscopic energy to internal energy, a form of en-ergy which is just the kinetic and potential energy of atomic and molecularmotions. Thus, the apparent disappearance of macroscopic energy is just aconsequence of the conversion of this energy into microscopic form.

The second type of experiment which suggests that heating convertsmacroscopic energy to internal energy is one in which this energy is con-verted back to macroscopic form. An example of this process is illustrated in

22.2. HEAT 379

Material C (J kg!1 K!1)brass 385glass 669ice 2092

steel 448methyl alcohol 2510

glycerine 2427water 4184

Table 22.2: Specific heats of common materials.

figure 22.2. As the piston moves out of the cylinder under the force exertedon it by the gas, work is done which can be stored or used by, say, compress-ing a spring or running an electric generator. As the piston moves out, thegas in the cylinder decreases in temperature, which indicates that the gas islosing microscopic energy.

22.2.1 Specific Heat

Conversion of macroscopic energy to microscopic kinetic energy thus tendsto raise the temperature, while the reverse conversion lowers it. It is easyto show experimentally that the amount of heating needed to change thetemperature of a body by some amount is proportional to the amount ofmatter in the body. Thus, it is natural to write

%Q = MC%T (22.4)

where M is the mass of material, %Q is the amount of energy transferredto the material, and %T is the change of the material’s temperature. Thequantity C is called the specific heat of the material in question and is theamount of heating needed to raise the temperature of a unit mass of materialby one degree. C varies with the type of material. Values for commonmaterials are given in table 22.2.

22.2.2 First Law of Thermodynamics

We now address some questions of terminology. The use of the terms “heat”and “quantity of heat” to indicate the amount of microscopic kinetic energy


AL

T1 T2

T2 < T1

heat flow

Figure 22.3: Geometry of heat flow problem. Heat flows from higher to lowertemperature.

inhabiting a body has long been out of favor due to their association withthe discredited “caloric” theory of heat. Instead, we use the term internalenergy to describe the amount of microscopic energy in a body. The wordheat is most correctly used only as a verb, e. g., “to heat the house”. Heatthus represents the transfer of internal energy from one body to anotheror conversion of some other form of energy to internal energy. Taking intoaccount these definitions, we can express the idea of energy conservation insome material body by the equation

%E = %Q "%W (first law of thermodynamics) (22.5)

where %E is the change in internal energy resulting from the addition of heat%Q to the body and the work %W done by the body on the outside world.This equation expresses the first law of thermodynamics. Note that the signconventions are inconsistent as to the direction of energy flow. However, theseconventions result from thinking about heat engines, i. e., machines whichtake in heat and put out macroscopic work. Examples of heat engines aresteam engines, coal and nuclear power plants, the engine in your automobile,and the engines on jet aircraft.

22.2.3 Heat Conduction

As noted earlier, internal energy may be transferred through a material fromhigher to lower temperature by a process known as heat conduction. Therate at which internal energy is transferred through a material body is known

22.2. HEAT 381

Material 0 (W m!1 K!1)brass 109brick 0.50

concrete 1.05ice 2.2

paper 0.050steel 46

Table 22.3: Values of thermal conductivity for common materials.

empirically to be proportional to the temperature di!erence across the body.For a rectangular body it is also known to scale in proportion to the crosssectional area of the body perpendicular to the temperature gradient andto scale inversely with the distance over which the temperature di!erenceexists. This is known as the law of heat conduction and is expressed in thefollowing mathematical form:

dQ

dt=0A%T

L(22.6)

where A is the cross sectional area of the body normal to the internal energyflow direction, L is the length of the body in the direction of heat flow, %Tis the temperature di!erence along its length, and 0 is a constant charac-teristic of the material known as the thermal conductivity. The geometry isillustrated in figure 22.3 and the thermal conductivities of common materialsare shown in table 22.3.

22.2.4 Thermal Radiation

Energy can also be transmitted though empty space by thermal radiation.This is nothing more than photons with a mixture of frequencies near afrequency &thermal which is a function only of the temperature T of the bodywhich is emitting them:

&thermal = KT, (22.7)

where the constant K = 3.67!1011 s!1 K!1. The amount of thermal energyper unit area per unit time emitted by a material surface is called the fluxof radiation and is given by Stefan’s law

JE = 1$T 4 (Stefan’s law) (22.8)


incident

reflected

emitted

incident

emitted

reflectedJtot

Jtot(1 - ε )

T4σε

JtotJtot(1 - ε )

T4σε

T T

Figure 22.4: Two surfaces facing each other, each with emissivity 1 andtemperature T .

where $ = 5.67 ! 10!8 W m!2 K!4 is the Stefan-Boltzmann constant and1 is the emissivity of the material surface. The emissivity lies in the range0 # 1 # 1 and depends on the type of material and the temperature of thesurface.

Surfaces which emit thermal radiation at a particular frequency can alsoreflect radiation at that frequency. If JI is the flux of radiation incident onthe surface, then the reflected radiation is just

JR = (1 " 1)JI (reflected radiation) (22.9)

and the balance of the radiation is absorbed by the surface:

JA = 1JI (absorbed radiation). (22.10)

Thus, high thermal emissivity goes along with high absorbed fraction and viceversa. A little thought indicates why this has to be so. If the emissivity werehigh and the absorption were low, then the object would spontaneously coolrelative to its environment. If the reverse were true, it would spontaneouslywarm up. Thus, the universally observed behavior that internal energy flowsfrom higher to lower temperatures would be violated.

Imagine two surfaces of equal temperature T facing each other. Theradiation emitted by one surface is partially absorbed and partially reflected

22.3. FRICTION 383

from the other surface, as illustrated in figure 22.4. The total radiative flux,Jtot, coming from each surface is the sum of the reflected radiation originatingfrom the other surface, (1" 1)Jtot, and the emitted thermal radiation, 1$T 4.Thus,

Jtot = (1 " 1)Jtot + 1$T 4. (22.11)

Solving for Jtot, we find that

Jtot % JBB = $T 4. (22.12)

Note that the total radiation originating from each surface, Jtot, is indepen-dent of the emissivity of the surfaces and depends only on the temperature.This radiative flux is called the black body flux. We give it the special nameJBB. Because it no longer depends on 1, it is independent of the character ofthe material making up the emitting surfaces. Di!erent materials result indi!erent fractions of thermal and reflected radiation, but the sum is alwaysequal to the black body flux if both surfaces are at the same temperature.Planck’s arguments which led to the energy-frequency relationship of quan-tum mechanics, E = h&, came from his attempt to explain black bodyradiation. The laws of black body radiation presented here can be derivedfrom quantum mechanics.

22.3 Friction

In this section we consider the quantitative forms of non-conservative forceson the macroscopic level. We first examine the frictional force between twosolid bodies and then consider viscosity in liquids.

22.3.1 Frictional Force Between Solids

The frictional force Fk between two solid objects in contact obeys an empiricallaw.1 If the two objects are sliding over each other, the frictional force on eachobject acts so as to oppose the relative motion of the two objects. (See figure22.5.) The frictional force is proportional to the normal force N pressing theobjects together:

Fk = µkN (kinetic friction). (22.13)

1An empirical law is one which we cannot justify in terms of the fundamental principlesof physics, but which is observed to be true in a wide variety of situations.


vFk

FextN

Figure 22.5: The kinetic frictional force Fk is exerted on the upper body bythe stationary lower body. The upper body is moving with velocity v and ispressed together with the lower body by a normal force N . It may also beacted upon by an additional non-normal external force Fext.

The dimensionless quantity µk is called the coe"cient of kinetic friction.This quantity is di!erent for di!erent pairs of materials rubbing together.It is typically of order one, but may be much less for particularly slipperymaterials.

Equation (22.13) is only valid if the two objects are moving relative toeach other. If they are not in relative motion, but if some other force is beingexerted on one of them, a static frictional force Fs will precisely counteractthis force so as to result in zero net force on the object. However, the staticfrictional force will keep the bodies from slipping only up to some limit definedby

|Fs| # µsN (static friction), (22.14)

where µs is the coe"cient of static friction. Generally we find that µs > µk, sogradually increasing the external force on an object in static frictional contactwith another object will cause it to suddenly break loose and accelerate whenthe maximum sustainable static frictional force is exceeded. Once the objectis in motion, a lesser external force is needed to keep it moving at a constantvelocity.

22.3.2 Viscosity

If two objects are not in physical contact but are separated by a thin layerof fluid (i. e., a liquid or a gas), there is still a frictional or viscous drag forcebetween the two objects but its behavior is di!erent. Figure 22.6 tells the

22.3. FRICTION 385

Fdrag = - µSA

viscous

y

dfluid

x

v(y) = Sy

Figure 22.6: Two solid plates separated by a distance d, the gap being filledby a viscous fluid. The lower plate is stationary and the upper plane ismoving to the right at speed vp = v(d) = Sd. The fluid is sheared, with thefluid moving according to v(y) = Sy. The fluid velocity matches that of theplates where the fluid touches the plates. The upper plate experiences a dragforce Fdrag = "µSA where µ is the viscosity of the fluid and A is the area ofthe plate.

story: The viscous drag force in this case is

Fdrag = "µSA (22.15)

where S = vp/d is the shear in the fluid, A is the area of the plates, and µ isthe viscosity of the fluid. (Don’t confuse this parameter with the static anddynamic coe"cients of friction!) The parameter vp is the velocity of the topplate with respect to the bottom plate and d is the separation between theplates.

Viscosity has the dimensions mass per length per time. The most commonunit of viscosity is the Poise: 1 Poise = 1 g cm!1 s!1. The viscosity of watervaries from 0.0179 Poise at 0# C to 0.0100 Poise at 20# C to 0.0028 Poise at100# C. The viscosity of water thus decreases with increasing temperature,which is typical of liquids. In contrast, the viscosity of a gas is independentof the density of the gas and is proportional to the square root of its absolutetemperature. The viscosity of a gas thus increases with temperature, incontrast to the viscosity of a liquid. For air at 20# C, the viscosity is 1.81 !10!4 Poise.

Thin layers of oil between moving parts are commonly used in machineryto reduce friction, since the resulting viscous drag is generally much less


than the corresponding kinetic friction which would occur if the parts werein direct contact. The ways in which the layer of oil is maintained betweenmoving parts are fascinating, but beyond the scope of this course.

22.4 Problems

1. The George Washington bridge, which spans the Hudson River betweenNew York and New Jersey, is 4760 feet long and is made out of steel.How much does it expand in length between winter and summer? (Pickreasonable winter and summer temperatures.)

2. A volume coe"cient of expansion - can be defined for solids as well asliquids. Show that - = 3) in this case, where ) is the linear coe"cientof expansion. Hint: Imagine a cube which increases the length of a sideby a fractional amount )%T $ 1 when the temperature increases by%T . Compute the fractional change in the volume of the cube.

3. Equal masses of brass and glass are put in the same insulating con-tainer, the brass initially at 300 K, the glass at 350 K. Assumingthat the interior of the container has negligible heat capacity, whattemperature does the material in the container reach after coming toequilibrium?

4. The gravitational potential energy of water going over Niagara Falls(60 m high) is converted to kinetic energy in the fall and then dissipatedat the bottom. How much warmer does the water get as a result?

5. A normal-sized house has concrete walls and roof 0.1 m thick. Abouthow much does it cost per month to heat the house electrically if elec-tricity costs $0.10 per kilowatt-hour? Estimate the wall and roof areasof a typical house and typical inside-outside temperature di!erences inwinter.

6. Compute the thermal frequency &thermal and the power per unit areaemitted by a surface with emissivity * = 1 for

(a) T = 3 K (cosmic background temperature),

(b) T = 300 K (earth’s temperature),

(c) T = 6000 K (sun’s surface),

22.4. PROBLEMS 387

MMg sin

Mg cos

N

θ

θ

θ

F

x

Figure 22.7: Mass M subject to gravity, friction (F ), and a normal force (N)on a ramp tilted at an angle ! with respect to the horizontal.

(d) and T = 2 ! 107 K (sun’s interior).

7. Derive an equation for the light pressure (force per unit area) actingon the walls of a box whose interior is at temperature T . Assume forsimplicity that all photons being emitted and absorbed by a wall movein a direction normal to the wall. Compute this pressure for the interiorof the sun. Hint: Recall that a photon with energy E has momentumE/c, and that both emitted and absorbed photons transfer momentumto the wall.

8. Imagine two plates each at temperature T as in figure 22.4, except thatthe left plate has emissivity *L and the right plate has emissivity *R.Show that the radiative energy flux incident on each plate from theother is still $T 4.

9. Two parallel plates facing each other, one at temperature T1, the otherat temperature T2, each have emissivity * = 1. Assuming that T1 =200 K and T2 = 300 K, compute the net radiative transfer of energyper unit area per unit time from plate 2 to plate 1.

10. Imagine a mass sliding down a ramp subject to frictional and normalforces as shown in figure 22.7. If the coe"cient of kinetic friction is µk,determine the acceleration down the ramp.

11. Suppose the mass in the previous problem has been given a push so


that it is sliding up the ramp. Determine its acceleration down theramp.

12. If the coe"cient of static friction is µs, compute the maximum anglefor which the mass in figure 22.7 will remain stationary.

Chapter 23

Entropy

So far we have taken a purely empirical view of the properties of systemscomposed of many atoms. However, as previously noted, it is possible tounderstand such systems using the underlying principles of mechanics. Theresulting branch of physics is called statistical mechanics. J. Willard Gibbs,a late 19th century American physicist from Yale University, almost single-handedly laid the groundwork for the modern form of this subject. Interest-ingly, the quantum mechanical version of statistical mechanics is much easierto understand than the version based on classical mechanics which Gibbsdeveloped. It also gives correct answers where the Gibbs version fails.

A system of many atoms has many quantum mechanical states in whichit can exist. Think of, say, a brick. The atoms in a brick are not stationary— they are in a continual flurry of vibration at ordinary temperatures. Thekinetic and potential energies associated with these vibrations constitute theinternal energy of the brick.

Though the details of each state are unimportant, the number of statesturns out to be a crucial piece of information. To understand why this isso, let us imagine two bricks, brick A with internal energy between E andE + %E, and brick B with energy between 0 and %E. Think of %E asthe uncertainty in the energy of the bricks — we can only observe a brickfor a finite amount of time %t, so the uncertainty principle asserts that theuncertainty in the energy is %E & h/%t.

The brick is a complex system consisting of many atoms, so in generalthere are many possible quantum mechanical states available to brick A inthe energy range E to E + %E. It turns out, for reasons which we will seelater, that significantly fewer states are available to brick B in the energy

389

390 CHAPTER 23. ENTROPY

range 0 to %E than are available to brick A.Roughly speaking, the larger the internal energy of an object, the higher

is its temperature. Thus, we infer that brick A has a much higher tempera-ture than brick B. What happens when we bring the two bricks into thermalcontact? Our experience tells us that heat (i. e., internal energy) immedi-ately starts to flow from one brick to the other, ultimately resulting in anequilibrium state in which the temperature is the same in the two bricks.

We explain this process as follows. Statistical mechanics hypothesizesthat any system of atoms (such as a brick) is free to roam through all quan-tum mechanical states which are energetically available to it. In fact, thisroaming is assumed to be continually taking place. Given this picture andthe assumption that the roaming between states is completely random, onewould expect equal probabilities for finding the system in any particularstate.

Of course, this probability argument assumes that we don’t know any-thing about the initial state of the system. If the system is known to be insome particular state at time t = 0, then it will take some time for the sys-tem to evolve in such a way that it has “forgotten” the initial state. Duringthis interval our knowledge of the initial state and the quantum mechanicaldynamics of the system can be used (in principle) to follow the evolution ofthe system. Eventually the uncertainty in our initial knowledge of the systemcatches up with us and we cannot predict the future evolution of the systembeyond this point. The brick develops “amnesia” and its probability of beingin any of the energetically allowed states is then uniform.

Something like this happens to the two bricks if they are brought intothermal contact. Initially brick A has virtually all of the energy and brickB has only a tiny amount. When the bricks are brought into contact, theyeventually can be treated as a single brick of twice the size. However, it takestime for the new, larger brick to evolve to the point where it has forgotten thefact that it started out as two separate bricks at di!erent temperatures. Inthis interval the temperature of brick A is decreasing while the temperatureof brick B is increasing as a result of internal energy flowing from one to theother. This evolution continues until equilibrium is reached.

Even though the combined brick has forgotten its initial state, there is asmall chance that it will return to this state, since the probability of findingthe brick in any state, including the original one, is non-zero. Thus, accordingto the postulates of statistical mechanics, one might suddenly find the brickagain in a state in which virtually all of the internal energy is concentrated

23.1. STATES OF A BRICK 391

Figure 23.1: “Inner-spring mattress” model of the atoms in a solid body.Interatomic forces act like miniature springs connecting the atoms. As aresult the whole system oscillates like a bunch of harmonic oscillators.

in former brick A. Actually, the issue is slightly more complicated than this.Brick A actually had many states available to it before being brought togetherwith brick B. Thus, a more interesting problem is to find the probability ofthe system suddenly finding itself in any of the states in which (virtually)all of the energy is concentrated in former brick A. Given the randomnessassumption of statistical mechanics, this probability is simply the numberof states which correspond to all of the energy being in brick A, divided bythe total number of states available to the combined brick. Computing thisnumber is the task we set for ourselves.

23.1 States of a Brick

In this section we demonstrate the above assertions by making a crude modelof the quantum mechanical states of a brick. We approximate the atoms ofthe brick as a collection of harmonic oscillators, three oscillators per atom,since each atom can oscillate in three dimensions under the influence of in-teratomic forces (see figure 23.1). For simplicity we assume that all of theoscillators have the same classical oscillation frequency, &0, so that the energyof each oscillator is given by

En = (n + 1/2)h&0 % (n + 1/2)E0, n = 0, 1, 2, . . . . (23.1)

This assumption is a rather poor approximation to the behavior of a solidbody when the total amount of internal energy is so small that many of the


E2 / E0

E1 / E0

E2 / E0

E1 / E0

E3 0/

Two oscillatorsThree oscillators

E2 4

0

1

2

3

4

10 3

Figure 23.2: Diagrams for counting states of systems of two (left panel) andthree (right panel) harmonic oscillators with the same classical oscillationfrequency.

harmonic oscillators are in their ground state. However, it is adequate forsituations in which the energy per oscillator is several times the ground stateoscillator energy.

We further assume that each oscillator is weakly coupled to its neighbor.This allows a slow transfer of energy between oscillators without appreciablya!ecting the energy levels of each oscillator.

The next step is to calculate the number of states of a system of harmonicoscillators for which the total energy is less than some maximum value E.This calculation is easy for a system consisting of a single oscillator. Fromequation (23.1) we infer that the number of states, N , of one oscillator withenergy less than E is

N = E/E0 (one oscillator) (23.2)

since the states are evenly spaced in energy with spacing E0.The calculation for a system of two oscillators is slightly more compli-

cated. The dots in the left panel of figure 23.2 show the states available toa two oscillator system. Each dot corresponds to a unique pair of values ofthe quantum numbers n1 and n2 for the two oscillators. The total energy ofthe two oscillators together is Etotal = E1 + E2 = (n1 + n2 + 1)E0.

The line defined by the equation E/E0 = E1/E0 + E2/E0 is illustratedby the hypotenuse of the shaded triangle in the left panel of figure 23.2. The


number of states with total energy less than E is obtained by simply countingthe dots inside this triangle. An easy way to do this “counting” is to notethat there is one dot per unit area in the plot, so that the number of dotsapproximately equals the area of the triangle:

N =1

2

"

E

E0

#2

(two oscillators). (23.3)

For a system of three oscillators the possible states of the system forma cubical grid in a three-dimensional space with axes E1/E0, E2/E0, andE3/E0, as shown in the right panel of figure 23.2. The dots representingthe states are omitted for clarity, but one state per unit volume exists inthis space. The dark-shaded oblique triangle is the surface of constant totalenergy E defined by the equation E1/E0 + E2/E0 + E3/E0 = E/E0, so thevolume of the tetrahedron formed by this surface and the coordinate axisplanes equals the number of states with energy less than E. This volume iscomputed as the area of the base of the tetrahedron, (E/E0)2/2, times itsheight, E/E0, times 1/3. We get

N =1

2 · 3

"

E

E0

#3

(three oscillators). (23.4)

There is a pattern here. We infer that there are

N (E) =1

1 · 2 · 3 . . .N

"

E

E0

#N

=1

N !

"

E

E0

#N

(N oscillators) (23.5)

states available to N oscillators with total energy less than E. The notationN ! is shorthand for 1 · 2 · 3 . . . N and is pronounced N factorial.

Let us summarize what we have accomplished. N (E) is the number ofstates of a system of harmonic oscillators, taken together, with total energyless than E. What we need is an estimate of the number of states betweentwo energy limits, say E and E + %E. This is easily obtained from N (E)as follows: N (E) is the number of states with energy less than E, whileN (E + %E) is the number of states with energy less than E + %E. Wecan obtain the number of states with energies between E and E + %E bysubtracting these two quantities:

%N = N (E +%E)"N (E) =N (E + %E) "N (E)

%E%E &

%N

%E%E. (23.6)


N %N (r = 5) %N (r = 10)1 1 12 5 103 50 2004 563 45005 6667 1066676 81381 26041677 1012500 648000008 12765734 16340138899 162539683 4161015873010 2085209002 106762700892811 26911444555 2755731922398612 349006782021 714765889577822

Table 23.1: Number of states %N available to N identical harmonic oscilla-tors between energies E and E + %E, where E = rNE0 and where we havechosen %E = E0. Results are shown for two di!erent values of r.

For N harmonic oscillators we find that

%N =N

N !

EN!1

EN0

%E =1

(N " 1)!

"

E

E0

#N!1 %E

E0

. (23.7)

Table 23.1 shows the number of states of a system of a small numberof harmonic oscillators with energy between E and E + %E where we havechosen %E = E0. Results are shown for systems up to N = 12 (i. e.,“microbricks” with up to 4 atoms, each with 3 modes of oscillation). Thequantity r is defined to be the average value of the quantum number n ofall the harmonic oscillators in the system; r = E/(NE0). Thus, rE0 is theaverage energy per oscillator. Recall that our calculation is only valid if r isappreciably greater than one. The number of available states is computedfor r = 5 and 10.

We see that a few atoms considered jointly have an astonishingly largenumber of possible states. For instance, a system of 4 atoms (i. e., 12oscillators) with r = 5 has about 3.5 ! 1011 states. Suppose we now confinethis energy to only 2 of the atoms or 6 oscillators. In this case r doubles toa value of 10 since the same amount of internal energy is now spread amonghalf the number of oscillators. Table 23.1 shows that this reduced system


has only about 2.6 ! 106 states. The probability of having all of the energyof the 4 atom system in these 2 atoms is the ratio of the number of states inthe 2 atom case to the total number of possible states of the 4 atom system,or 2.6 ! 106/3.5 ! 1011 = 7.4 ! 10!6. This is a rather small number, whichmeans that it is rare to find the system with all internal energy concentratedin two atoms.

We now determine how the number of states available to a system ofharmonic oscillators behaves for a very large number of oscillators such asmight be found in a real brick. Values of %N become so large in this casethat it is useful to work in terms of the natural logarithm of %N . For largeN we can safely approximate N"1 by N . Using the properties of logarithms,we get

ln(%N ) = ln

&

(E/E0)N!1

(N " 1)!

%E

E0

'

& ln

&

(E/E0)N

N !

%E

E0

'

= N ln(E/E0) " ln(N !) + ln(%E/E0). (23.8)

A useful mathematical result for large N is the Stirling approximation1:

ln(N !) & N ln(N) " N (Stirling approximation). (23.9)

Substituting this into equation (23.8), using the fact that N ln(E/E0) "N ln N = N ln[E/(NE0)], and rearranging results in

ln(%N ) = N.

ln"

E

NE0

#

+ 1/

+ ln"

%E

E0

#

(N oscillators). (23.10)

We now return to the original question, which we state in this form:What fraction of the states of a brick corresponds to the special situationwith all of the internal energy in half of the brick? A real brick has of order3 ! 1025 atoms or about N = 1026 oscillators. Half of the brick thus hasN " = 5 ! 1025 oscillators. If as before we assume that r = 5 when theinternal energy is distributed throughout the brick, then we have r" = 10when all the energy is in half of the brick. Therefore the logarithm of the

1To derive the Stirling approximation note that ln(N !) = ln(1)+ln(2)+. . .+ln(N). This

sum can be approximated by the integral0 N

1ln(x)dx = N ln(N)"N + 1 & N ln(N)"N .


total number of available states is ln(%N ) = N [ln(r) + 1] + ln(%E/E0),while the logarithm of the number of states available when all the energy isin half of the brick is ln(%N ") = N "[ln(r") + 1] + ln(%E/E0). Putting in thenumbers, we find that the probability of finding all the energy in half of thebrick is

%N "/%N = exp[ln(%N ") " ln(%N )]

= exp[N " ln(r") + N " + ln(%E/E0)

" N ln(r) " N " ln(%E/E0)]

= exp("0.96N) = exp("9.6 ! 1025) = 10!4.2$1025

.(23.11)

This probability is extremely small, and is zero for all practical purposes.Notice that %E, which we haven’t specified, cancels out. This typically

happens in the theory when measurable quantities are calculated, and itshows that the actual value of %E isn’t important. Furthermore, for verylarge values of N typical of normal bricks, the term in equation (23.10)containing %E is always negligible for any reasonable values of %E. Wetherefore drop it in future calculations.

The variable ln(%N ) is proportional to a quantity which we call theentropy, S. The actual relationship is

S = kB ln(%N ) (definition of entropy) (23.12)

where kB = 1.38!10!23 J K!1 is called Boltzmann’s constant. Ludwig Boltz-mann was a 19th century Austrian physicist who played a pivotal role in thedevelopment of the concept of entropy. The entropy of a brick containing Noscillators is therefore

S = NkB

.

ln"

E

NE0

#

+ 1/

(entropy of N oscillators). (23.13)

As with the speed of light and Planck’s constant, Boltzmann’s constantis not really needed for a complete development of statistical mechanics. It’sonly role is to convert entropy and related quantities to everyday units. Theconventional dimensions of entropy are thus the same as those of Boltzmann’sconstant, or energy divided by temperature. However, more fundamentally,we consider entropy (without Boltzmann’s constant) to be a dimensionlessquantity since it is just the logarithm of the number of available states.

23.2. SECOND LAW OF THERMODYNAMICS 397

23.2 Second Law of Thermodynamics

What use is entropy? In our example we found that the number of statesfor the situation in which all of the internal energy of a brick is restrictedto half of the brick is much less than the number of states available whenno restrictions are put upon the distribution of the same amount of internalenergy through the entire brick. Thus, the entropy, which is just proportionalto the logarithm of the number of available states, is less in the restrictedcase than it is in the unrestricted case.

This turns out to be generally true. Any measurable restriction we placeon the distribution of internal energy in the brick turns out to result in a muchsmaller number of available quantum mechanical states and hence a smallervalue for the entropy. Once such a restriction is lifted, all possible statesbecome available, and according to the postulates of statistical mechanics thebrick eventually evolves to the point where it is roaming randomly throughthese states. The probability of the brick revisiting the original restrictedset of states is so small as to be completely ignorable once it forgets itsinitial state, because these states form only a miniscule fraction of the statesavailable to the brick. Thus, with a very high degree of certainty, one cansay that the entropy of the brick increases when the restriction is lifted.

Strictly speaking, our definition of entropy is only valid after the brickhas reached equilibrium, i. e., when the initial state has been forgotten.The entropy during the equilibration period according to our definition istechnically undefined.

Our inferences about a brick can be extended to any isolated system, i. e.,any system which doesn’t exchange mass or energy with the outside world:The entropy of any isolated system consisting of a large number of atoms willnot spontaneously decrease with time. This principle is called the second lawof thermodynamics.

23.3 Two Bricks in Thermal Contact

Where does the idea of temperature fit into the picture? This concept hascome up informally, but we need to give it a precise definition. If two objectsat di!erent temperatures are placed in contact with each other, we observethat internal energy flows from the warmer object to the cooler object, asillustrated in figure 23.3.


>

heat flowTA

TB

TA TB

Figure 23.3: Two bricks in thermal contact, one at temperature TA, the otherat temperature TB. If TA > TB, internal energy flows from brick A to brickB.

We wish to see if the role of temperature di!erences in the flow of internalenergy can be related to the ideas developed in the previous section. Let usconsider two bricks as before, but possibly of di!erent size, and thereforecontaining di!erent numbers of harmonic oscillators. Suppose brick A hasNA oscillators and energy EA while brick B has NB oscillators and energyEB. The two bricks have entropies

SA = kBNA

.

ln"

EA

NAE0

#

+ 1/

(23.14)

and

SB = kBNB

.

ln"

EB

NBE0

#

+ 1/

. (23.15)

If the two bricks are thermally isolated from each other but are never-theless considered together as one system, then the total number of statesavailable to this combined system is just the product of the numbers of statesavailable to each brick separately:

%N = %NA%NB. (23.16)

To make an analogy, the total number of ways of arranging two coins, each ofwhich may either be heads up or tails up, is 4 = 2!2, or heads-heads, heads-tails, tails-heads, and tails-tails. We compute the states of the combinedsystem just as we compute the total number of ways of arranging the coins,i. e., by taking the product of the numbers of states of the individual systems.

Taking the logarithm of N and multiplying by Boltzmann’s constantresults in an equation for the combined entropy S of the two bricks:

S = SA + SB. (23.17)

23.3. TWO BRICKS IN THERMAL CONTACT 399

0

equilibrium

EEA

TA = T

TA > TBTB > T

A) = SA(EA) + SB(E - EA)Entropy S(E

B

A

Figure 23.4: Total entropy of two systems for fixed total energy E = EA+EB

as a function of EA, the energy of system A.

In other words, the combined entropy of two (or more) isolated systems isjust the sum of their individual entropies.

We can determine how the total entropy of the two bricks depends on thedistribution of energy between them by using equations (23.14) and (23.15).Plotting the sum of the entropies of the two bricks SA(EA) + SB(EB) ver-sus the energy EA of brick A under the constraint that the total energyE = EA + EB is constant yields a curve which typically looks somethinglike figure 23.4. Notice that the total entropy reaches a maximum for somecritical value of EA. Since the slope of S(EA) is zero at this point, we candetermine the corresponding value of EA by setting the derivative to zero ofthe total entropy with respect to EA, subject to the condition that the totalenergy is constant. Under the constraint of constant total energy E, we havedEB/dEA = d(E " EA)/dEA = "1, so

%S

%EA=%SA

%EA+%SB

%EA=

%SA

%EA+%SB

%EB

dEB

dEA=%SA

%EA"%SB

%EB= 0. (23.18)

(The partial derivatives indicate that parameters besides the energy are heldconstant while taking the derivative of entropy.) Thus,

%SA

%EA=

%SB

%EB(equilibrium condition) (23.19)

at the point of maximum entropy.


Once the equilibrium values of EA and EB are found, we can calculatethe total entropy S = SA + SB of two thermally isolated bricks. We nowassert that this entropy doesn’t change when two bricks in equilibrium arebrought into thermal contact. Why is this so?

The derivative of the entropy of a system with respect to energy turnsout to be one over the temperature of the system. Thus, the temperaturesof the bricks can be found from

1

T%

%S

%E(definition of temperature). (23.20)

The condition for equilibrium (23.19) therefore reduces to 1/TA = 1/TB,or TA = TB. This is consistent with observations of the behavior of realsystems. Thus, at the equilibrium point the temperatures of the two bricksare the same and bringing them together causes no heat flow to occur. Theprocess of bringing two bricks at the same temperature into thermal contactis thus completely reversible, since separating them leaves each with the sameamount of energy it started with.

The temperature of a brick is easily calculated using equation (23.20):

T =E

kBN(temperature of N harmonic oscillators). (23.21)

We see that the temperature of a brick is just the average energy per harmonicoscillator in the brick divided by Boltzmann’s constant.

23.4 Thermodynamic Temperature

Equation (23.20) provides us with a physical definition of temperature whichis independent of specific material properties such as the thermal expansioncoe"cient of some particular metal. Though di!erent materials have dif-ferent dependences of entropy on internal energy, the derivative of entropywith respect to energy will be the same for any two materials in thermalequilibrium with each other.

Note that the unit of temperature is the Kelvin degree according to thistheory. If we had left o! Boltzmann’s constant in the definition of entropy, thedimensions of temperature would be that of energy. Boltzmann’s constantis thus simply a scaling factor which changes temperature to energy just asmultiplication by the speed of light converts time to distance.

23.5. SPECIFIC HEAT 401

23.5 Specific Heat

How can we compute the specific heat of a collection of harmonic oscillators?Starting from the temperature of a brick, as given by equation (23.21), wesolve for the brick’s internal energy:

E = NkBT (internal energy of N oscillators). (23.22)

Recall that the specific heat is the heat required per unit mass to increasethe temperature of the brick by one degree. For a solid body, essentially allthe heat added to the body goes into increasing its internal energy. Thus, ifthe mass of the brick is M = Nm where m is the mass per oscillator, thenthe predicted specific heat of the brick is

C %1

M

dQ

dT&

1

M

dE

dT=

kB

m(specific heat of harmonic oscillators). (23.23)

This formula is in reasonable agreement with measurements when the tem-perature is high enough so that all the harmonic oscillators are in excitedstates. (We equate dQ = dE using the first law of thermodynamics, since nowork is being done by the brick.)

23.6 Entropy and Heat Conduction

Though entropy is formally not defined in a system which is not in ther-modynamic equilibrium, one can imagine situations in which elements of asystem interact only weakly with other elements. Each element is thereforevery close to internal equilibrium, so that the entropy of each element canbe defined. However, the elements are not in equilibrium with each other.

Figure 23.5 shows an example of such a situation. Since 1/T = %S/%E,one can write

%S1 = "%Q/T1 (23.24)

since heat flowing out of region 1 results in a decrease in internal energy%E1 = "%Q. Likewise, we find that

%S2 = %Q/T2, (23.25)

since the internal energy of region 2 increases by %E2 = %Q. The totalchange of entropy of the system is therefore

%S = %S1 + %S2 = %Q"

1

T2

"1

T1

#

. (23.26)


T1 T2heat flow = Q

1 = − Q/T1Δ

Δ

SΔ = Q/TΔ 22 SΔ

Figure 23.5: The two regions at temperatures T1 and T2 < T1 are connectedby a thin bar which conducts heat slowly from the first to the second region.For heat %Q transferred, the entropy of region 1 decreases according to%S1 = "%Q/T1, while the entropy of region 2 increases by %S2 = %Q/T2.

From our experience, we know that heat will only flow from region 1 toregion 2 if T1 > T2. However, equation (23.26) shows that the net entropychange is positive when this is true. Conversely, if T1 < T2, then the netentropy change would be negative and heat would be flowing spontaneouslyfrom lower to higher temperatures. Thus, the statement that heat cannotspontaneously flow from lower to higher temperatures is equivalent to thestatement that the entropy of an isolated system must not decrease. Analternative statement of the second law of thermodynamics is therefore heatcannot spontaneously flow from lower to higher temperatures.

If entropy increases in some process, we call it irreversible. Spontaneousheat flow is always irreversible. However, in the limit in which the temper-ature di!erence is very small, the entropy increase due to heat flow is alsosmall. Of course, the rate of flow of heat is also quite slow in this case. Nev-ertheless, this situation forms a useful idealization. In the idealized limit ofvery small, but nonzero temperature di!erence, the flow of heat is said to bereversible because the generation of entropy is negligible.

23.7 Problems

1. Compute an approximate value for NN/N ! using the Stirling approx-imation. (This gives the essence of %N for N harmonic oscillators.)From this show that ln(%N ) ) N .

2. States of a pair of distinguishable dice (i. e., one is red, the other isgreen):

23.7. PROBLEMS 403

(a) List all of the possible states of a pair of dice, i. e., all the possiblecombinations of face-up numbers.

(b) Given that each of the dice has six faces, does the total numberof states equal that given by equation (23.16)?

3. There are N !/[M !(N"M)!] ways of arranging N pennies with M headsup. Verify this for 2, 3, and 4 pennies. (Note that by definition 0! = 1.)

4. Suppose we have N pennies on a shaking table which bounces the pen-nies around, flipping them over at random. The pennies are weightedso that the gravitational potential energy of a penny is zero with tailsup and U with heads up.

(a) If M heads are up, what is the total energy E?

(b) How many “states”, %N , are there with M heads up? Hint:Compute this directly from the statement of the previous problem,not by computing dN /dE as we did for N harmonic oscillators.What does the energy interval %E correspond to in this case?

(c) Compute the entropy of the system as a function of E and N .Hint: You will need to use the Stirling approximation to do thispart.

(d) Compute the temperature as a function of E and N .

(e) Invert the temperature equation derived in the previous step toobtain E as a function of T and N . To understand this result,approximate it in the low and high limits, i. e., kBT/U $ 1and kBT/U ' 1. Try to think of an explanation of the behaviorof the pennies in these limits which would make sense to (say)an 8th grade student. In particular, how is the intensity of theshaking of the table related to the “temperature”? Hint: In thelow temperature limit note that exp(U/kBT ) ' 1, while in thehigh temperature limit exp(U/kBT ) & 1.

5. Suppose that two systems, A and B, have available states %NA = EXA

and %NB = EYB , where E = EA + EB = 2. Compute and plot %N =

%NA%NB as a function of EA over the range 0 < EA < 2 for:

(a) X = Y = 1;

(b) X = Y = 5;


(c) X = Y = 25;

(d) X = 2; Y = 8 — explain the position of the peak in terms of thevalues of X and Y .

How does the width of the peak change as X and Y get larger? Explainthe consequences of this result for the reliability of the second law ofthermodynamics as a function of the number of particles in each system.

6. Suppose we have a system of mass M in which kBT = AE1/2, where T isthe temperature, E is the internal energy, kB is Boltzmann’s constant,and A is a constant.

(a) Derive a formula for the entropy of the system as a function ofinternal energy. Hint: Remember the thermodynamic definitionof temperature.

(b) Compute the specific heat of this system.

Chapter 24

The Ideal Gas and HeatEngines

All heat engines have the common property of turning internal energy intouseful macroscopic energy. They extract internal energy from a high tem-perature reservoir, convert part of this energy to useful work, and transferthe rest to a low temperature reservoir. The second law of thermodynamicsimposes a firm limit on the fraction of the initial internal energy which canbe converted to macroscopic energy.

Almost all heat engines work by means of expansions and contractions ofa gas. A simple theoretical model called the ideal gas model quite accuratelypredicts the behavior of the gases in most heat engines of this type.

Our first task is to build the ideal gas model using the techniques learnedin the previous section. We then use this model to understand the operationof heat engines. We are particularly interested in determining the maximumtheoretical e"ciency at which these devices can convert heat to useful work.

24.1 Ideal Gas

An ideal gas is an assembly of atoms or molecules which interact with eachother only via occasional collisions. The distance between molecules is muchgreater than the range of inter-molecular forces, so gas molecules behave asfree particles most of the time. We assume here that the density of moleculesis also low enough to make the probability of finding more than one moleculein a given quantum mechanical state very small. For this reason it doesn’t

405

406 CHAPTER 24. THE IDEAL GAS AND HEAT ENGINES

S

S

S’ > 2S

S

Sdecreasesincreases

entropy entropy

Figure 24.1: Consequence of the incorrect classical calculation of entropy ofan ideal gas by Gibbs. Two parts of a container separated by a divider eachcontain the same type of gas at the same temperature and pressure. Thetotal entropy is 2S where S is the classically calculated entropy of each half.If the divider is removed, the classical calculation yields an entropy for theentire body of gas S " > 2S. Reinserting the divider returns the container tothe initial state in which the total entropy is 2S.

matter whether the molecules are bosons or fermions for our calculations.

J. Willard Gibbs tried computing the entropy of an ideal gas using hisversion of statistical mechanics, which was based on classical mechanics. Theresult was wrong in a very fundamental way — the calculated amount ofentropy was not proportional to the amount of gas. In fact, the amount ofentropy of an ideal gas at fixed temperature and pressure is calculated tohave a non-linear dependence on the number of gas molecules. In particular,doubling the amount of gas more than doubles the entropy according to theGibbs formula.

The significance of this error is illustrated in figure 24.1. Imagine a con-tainer of gas of a certain type, temperature, and pressure which is dividedinto two equal parts by a sheet of material. The total entropy of this stateis 2S, where S is the entropy calculated separately for each half of the bodyof gas. This follows because the two halves are completely separate systems.

If the divider is now removed, a calculation of the entropy of the full bodyof gas yields S " > 2S according to the Gibbs formula, since the calculatedentropy doesn’t scale with the amount of gas. Furthermore, replacing thedivider restores the system to the initial state in which the total entropy is 2S.Thus, simply inserting or removing the divider, an operation which transfers

24.1. IDEAL GAS 407

no heat and does no work on the system, is able to increase or decrease theentropy of the gas at will according to Gibbs. This is at variance with thesecond law of thermodynamics and is known not to occur. Its prediction bythe formula of Gibbs is called the Gibbs paradox. Gibbs was well aware of theserious nature of this problem but was unable to come up with a satisfyingsolution.

The resolution of the paradox is simply that the Gibbs formula for theentropy of an ideal gas is incorrect. The correct formula is only obtainedwhen the quantum mechanical version of statistical mechanics is used. Thefailure of Gibbs to obtain the proper entropy was an early indication thatclassical mechanics had problems on the atomic scale.

We will now calculate the entropy of a body of ideal gas using quantumstatistical mechanics. In order to reduce the di"culty of the calculation, wewill take a shortcut and assume that the amount of entropy is proportionalto the amount of gas. However, the more rigorous calculation confirms thatthis actually is true.

24.1.1 Particle in a Three-Dimensional Box

The quantum mechanical calculation of the states of a particle in a three-dimensional box forms the basis for our treatment of an ideal gas. Recall thata non-relativistic particle of mass M in a one-dimensional box of width a canonly support wavenumbers kl = ±"l/a where l = 1, 2, 3, . . . is the quantumnumber for the particle. Thus, the possible momenta are pl = ±h"l/a andthe possible energies are

El = p2l /(2M) = h2"2l2/(2Ma2) (one-dimensional box). (24.1)

If the box has three dimensions, is cubical with edges of length a, andhas one corner at (x, y, z) = (0, 0, 0), the quantum mechanical wave functionfor a single particle which satisfies . = 0 on all the walls of the box is athree-dimensional standing wave,

.(x, y, z) = sin(lx"/a) sin(my"/a) sin(nz"/a), (24.2)

where the quantum numbers l, m, n are positive integers. You can verify thisby examining . for x = 0, x = a, etc.

Equation (24.2) is a solution in which the x, y, and z wavenumbers arerespectively kx = ±l"/a, ky = ±m"/a, and kz = ±n"/a. The corresponding


E/E0 E/E0

0

30

20

10

0

30

20

10

1 (1)

2 (1)

3 (1)

5 (1)

4 (1)

111 (1)

122 (3)

222 (1)123 (6)

333 (1), 115 (3)

223 (3)

112 (3)

233 (3)

113 (3)

114 (3)

124 (6)

134 (6)

234 (6)

133 (3)

224 (3)

3-D box1-D box 125 (6)

Figure 24.2: Energy levels for a non-relativistic particle in a one-dimensionaland a three-dimensional box, each of side length a. The value E0 is theground state energy of the one-dimensional particle in a box of length a.The numbers to the right of the levels respectively give the values of l forthe 1-D oscillator and the values of l, m, and n for the 3-D oscillator. Thenumbers in parentheses give the degeneracy of each energy level (see text).

components of the kinetic momentum are therefore px = hkx, etc. Thepossible energy values of the particle are

Elmn =p2

2M=

p2x + p2

y + p2z

2M=

h2"2(l2 + m2 + n2)

2Ma2

=h2"2L2

2MV 2/3% E0L

2 (three-dimensional box). (24.3)

In the last line of the above equation we have eliminated the linear dimensiona of the box in favor of its volume V = a3 and have adopted the shorthandnotation L2 = l2 + m2 + n2 and E0 = h2"2/(2Ma2) = h2"2/(2MV 2/3). Thequantity E0 is the ground state energy for a particle in a one-dimensionalbox of size a.

Figure 24.2 shows the energy levels of a particle in a one-dimensional anda three-dimensional box. Di!erent values of l, m, and n can result in thesame energy in the three-dimensional case. For instance, (l, m, n) = (1, 1, 2),(1, 2, 1), (2, 1, 1) all yield L2 = 6 and hence energy 6E0. This energy level is

24.1. IDEAL GAS 409

0

1

3

0 1 2 3 4 5

2

4

5

m

l

L

Figure 24.3: The states of a particle in a two-dimensional box. The dotsindicate particular states associated with allowed values of the x and y di-rection quantum numbers, l and m. The pie-shaped segment bounded bythe arc of radius L and the l and m axes encompasses all of the states withl2 + m2 # L2.

thus said to have a degeneracy of 3. Similarly, the states (1, 2, 3), (2, 3, 1),(3, 1, 2), (3, 2, 1), (2, 1, 3), (1, 3, 2) all have the same value of L2, so this levelhas a degeneracy of 6. However, the state (1, 1, 1) is unique and thus has adegeneracy of 1. From this we see that the degeneracy of an energy level is thenumber of di!erent physically distinguishable states which have that energy.Counting the e!ects of degeneracy, the particle in a three-dimensional boxhas 60 distinct states for E # 30E0, while the one-dimensional box has 5.As the limiting value of E/E0 increases, this ratio becomes even larger.

24.1.2 Counting States

In order to compute the entropy of a system, we need to count the numberof states available to the system in a particular band of energies. Figure 24.3shows how to count the states with energy less than some limiting value fora particle in a two-dimensional box. The pie-shaped segment bounded bythe arc of radius L and the l and m axes has an area equal to one fourth thearea of a circle of radius L, or "L2/4. The dots represent allowed values of


the l and m quantum numbers. One dot, and hence one state exists per unitarea in this graph, so the above expression tells us how many states N existwith l2 + m2 # L2.

In two dimensions the particle energy is E = (l2 + m2)E0. Thus, thenumber of states with energy less than or equal to some maximum energy Eis

N ="L2

4="

4

"

E

E0

#

(two-dimensional box). (24.4)

Similar arguments can be made to calculate the number of states of aparticle in a three-dimensional box. The equivalent of figure 24.3 would bea plot with three axes, l, m, and n representing the x, y, and z quantumnumbers. The volume of a sphere with radius L is then 4"L3/3 and theregion of the sphere with l, m, n > 0, i. e., an eighth of the sphere, containsreal physical states. The result is that

N =4"L3

3 · 8="

6

"

E

E0

#3/2

(three-dimensional box) (24.5)

states exist with energy less than E.

24.1.3 Multiple Particles

An ideal gas of only one molecule isn’t very interesting. Calculating the num-ber of states available to many particles in a box is a bit complex. However,by analogy with the case of multiple harmonic oscillators, we guess that thenumber of states of an N -particle gas is the number of states available to asingle particle to the Nth power times some as yet unknown function of N ,F (N):

N = F (N)"

E

E0

#3N/2

(N particles in 3-D box). (24.6)

(Note that the ("/6)N from equation (24.5) has been absorbed into F (N).)Substituting E0 = h2"2/(2MV 2/3) results in

N = F (N)

&

2MEV 2/3

h2"2

'3N/2

. (24.7)

Now, "2h2/(2M) has the units of energy times volume to the two-thirdspower, so we write this combination of constants in terms of constant refer-ence values of E and V :

"2h2/(2M) = ErefV2/3

ref . (24.8)

24.1. IDEAL GAS 411

Given the above assumption, we can rewrite the number of states with energyless than E as

N = F (N)

&

E

Eref

'3N/2 &

V

Vref

'N

. (24.9)

We now argue that the combination F (N) must take the form KN!5N/2

where K is a dimensionless constant independent of N . Substituting thisassumption into equation (24.9) results in

N = K

&

E

NEref

'3N/2 &

V

NVref

'N

. (24.10)

It turns out that we will not need the actual values of any of the threeconstants K, Eref , or Vref .

The e!ect of the above hypothesis is that the energy and volume occuronly in the combinations E/(NEref ) and V/(NVref). First of all, these com-binations are dimensionless, which is important because they will becomethe arguments of logarithms. Second, because of the N in the denominatorin both cases, they are in the form of energy per particle and volume perparticle. If the energy per particle and the volume per particle stay fixed,then the only dependence of N on N is via the exponents 3N/2 and N inthe above equation. Why is this important? Read on!

24.1.4 Entropy and Temperature

Recall now that we need to compute the number of states in some smallenergy interval %E in order to get the entropy. Proceeding as for the caseof a collection of harmonic oscillators, we find that

%N =%N%E

%E =3KN%E

2E

&

E

NEref

'3N/2 &

V

NVref

'N

. (24.11)

The entropy is therefore

S = kB ln(%N ) = NkB

1

3

2ln

&

E

NEref

'

+ ln

&

V

NVref

'2

(ideal gas)

(24.12)where we have dropped the term kB ln[3KN%E/(2E)]. Since this term isnot multiplied by the number of particles N , it is unimportant for systemsmade up of lots of particles.


Δx

F = pA

Figure 24.4: Gas in a cylinder with a movable piston. The force F exertedby the gas on the piston is the area A of the face of the piston times thepressure p.

Notice that this equation has a very important property, namely, thatthe entropy is proportional to the number of particles for fixed E/N andV/N . It thus satisfies the criterion which Gibbs was unable to satisfy inhis computation of the entropy of an ideal gas. However, we cannot claimthat our calculation is superior to his, because we cheated ! The reason weassumed that F (N) = KN!5N/2 is precisely so that we would obtain thisresult.

The temperature is the inverse of the E-derivative of entropy:

1

T=

%S

%E=

3NkB

2E=+ T =

2E

3NkBor E =

3NkBT

2. (24.13)

24.1.5 Work, Pressure, and Gas Law

The pressure p of a gas is the normal force per unit area exerted by the gason the walls of the chamber containing the gas. If a chamber wall is movable,the pressure force can do positive or negative work on the wall if it moves outor in. This is the mechanism by which internal energy is converted to usefulwork. We can determine the pressure of a gas from the entropy formula.

Consider the behavior of a gas contained in a cylinder with a movablepiston as shown in figure 24.4. The net force F exerted by gas moleculesbouncing o! of the piston results in work %W = F%x being done by the gasif the piston moves out a distance %x. The pressure is related to F and thearea A of the piston by p = F/A. Furthermore, the change in volume of thecylinder is %V = A%x.

24.1. IDEAL GAS 413

If the gas does work %W on the piston, its internal energy changes by

%E = "%W = "F%x = "F

AA%x = "p%V, (24.14)

assuming that %Q = 0, i. e., no heat is added or removed during the changein volume. Solving this for p results in

p % "%E

%V. (24.15)

The assumption that %Q = 0 normally implies that the entropy S doesnot change as well. Thus, in the evaluation of %E/%V , the entropy is heldconstant. This turns out to be a non-trivial assumption and the conditionsunder which it is true are discussed in the next section. For now we shallassume that this assumption is valid.

We can determine the pressure for an ideal gas by solving equation (24.12)for E and taking the V derivative. This equation may be written in compactform as

E =N5/3ErefV

2/3

ref exp[2S/(3NkB)]

V 2/3=

B

V 2/3(ideal gas) (24.16)

where B contains all references to entropy, number of particles, etc. Forisentropic (i. e., constant entropy) processes, we therefore have

EV 2/3 = const (constant entropy processes). (24.17)

The pressure can be rewritten as

p = "%E

%V=

2

3

B

V 5/3=

2E

3V(ideal gas) (24.18)

where B is eliminated in the last step using equation (24.16). Employingequation (24.13) to eliminate the energy in favor of the temperature, thiscan be written

pV = NkBT (ideal gas law), (24.19)

which relates the pressure, volume, temperature, and particle number of anideal gas.

This equation is called the ideal gas law and jointly represents the ob-served relationships between pressure and volume at constant temperature


(Boyle’s law) and pressure and temperature at constant volume (law ofCharles and Gay-Lussac). The fact that we can derive it from statisticalmechanics is evidence in favor of our quantum mechanical model of a gas.

The formula for the entropy of an ideal gas (24.12), its temperature(24.13), and the ideal gas law (24.19) summarize our knowledge about idealgases. Actually, the entropy and temperature laws only apply to a particulartype of ideal gas in which the molecules consist of single atoms. This is calleda monatomic ideal gas, examples of which are helium, argon, and neon. Themolecules of most gases consist of two or more atoms. These molecules havevibrational and rotational degrees of freedom which can store energy. Thecalculation of the entropy of such gases needs to take these factors into ac-count. The most common case is one in which the molecules are diatomic,i. e., they consist of two atoms each. In this case simply replacing factors of3/2 by 5/2 in equations (24.12) and (24.13) results in equations which applyto diatomic gases.

24.1.6 Specific Heat of an Ideal Gas

As previously noted, the specific heat of any substance is the amount ofheating required per unit mass to raise the temperature of the substance byone degree. For a gas one must clarify whether the volume or the pressureis held constant as the temperature increases — the specific heat di!ersbetween these two cases because in the latter situation the added energyfrom the heating is split between the production of internal energy and theproduction of work as the gas expands.

At constant volume all heating goes into increasing the internal energy,so %Q = %E from the first law of thermodynamics. From equation (24.13)we find that %E = (3/2)NkB%T . If the molecules making up the gas havemass M , then the mass of the gas is NM . Thus, the specific heat at constantvolume of an ideal gas is

CV =1

NM

3NkB

2=

3kB

2M(specific heat at const vol) (24.20)

As noted above, when heat is added to a gas in such a way that thepressure is kept constant as a result of allowing the gas to expand, the addedheat %Q is split between the increase in internal energy %E and the workdone by the gas in the expansion %W = p%V such that %Q = %E + p%V .In a constant pressure process the ideal gas law (24.19) predicts that p%V =

24.2. SLOW AND FAST EXPANSIONS 415

tuning peg

12th fret

Figure 24.5: Harmonics on a guitar. Plucking a string while a finger restslightly on the string at the 12th fret results in excitation of the first harmonicon the guitar string. Only one string is shown for clarity.

NkB%T . Using this and the previous equation for %E results in the specificheat of an ideal gas at constant pressure:

CP =1

NM

&

3NkB

2+ NkB

'

=5kB

2M(specific heat at const pres). (24.21)

24.2 Slow and Fast Expansions

How does the entropy of a particle in a box change if the volume of thebox is changed? The answer to this question depends on how rapidly thevolume change takes place. If an expansion or compression takes place slowlyenough, the quantum numbers of the particle don’t change.

This fact may be demonstrated by the tuning of a guitar. A guitar stringis tuned in frequency by adjusting the tension on the string with the tuningpeg. If the first harmonic mode (corresponding to quantum number n = 2 forparticle in a one-dimensional box) is excited on a guitar string as illustratedin figure 24.5, changing the tension changes the frequency of the vibration butit does not change the mode of vibration of the string — for instance, if thefirst harmonic is initially excited, it remains the primary mode of oscillation.

Slowly changing the volume of a gas consisting of many particles, eachwith its own set of quantum numbers, results in the same behavior — chang-ing the dimensions of the box results in no switching of quantum numbers


E

V

compression

rapid expansion

rapid compression

slow expansion orinitial state

Figure 24.6: The curved line indicates the reversible adiabatic curve E )V !2/3 for an ideal gas in a box. The two straight line segments indicate whathappens in a rapid expansion or compression.

beyond that which would normally take place as a result of particle collisions.As a consequence, the number of states available to the system, %N , andhence the entropy doesn’t change either.

A process which changes the macroscopic condition of a system but whichdoesn’t change the entropy is called isentropic or reversible adiabatic. Theword “isentropic” means at constant entropy, while “adiabatic” means thatno heat flows in or out of the system.

If the entropy doesn’t change as a result of a change in volume, thenEV 2/3 = const. Thus, the energy of the gas increases when the volume isdecreased and vice versa. This behavior is illustrated in figure 24.6. Thechange in energy in both cases is a consequence of work done by the gason the walls of the container as it changes volume — positive in expansion,meaning that the gas loses energy, and negative in contraction, meaning thatthe gas gains energy. This type of energy transfer is the means by whichinternal energy is converted to useful work.

A rapid expansion of the box has a completely di!erent e!ect. If theexpansion is so rapid that the quantum mechanical waves trapped in the boxundergo negligible evolution during the expansion, then the internal energyof the particles in the box does not change. As a consequence, the particle

24.3. HEAT ENGINES 417

quantum numbers must change to compensate for the change in volume.Equation (24.12) tells us that if the volume increases and the internal energydoesn’t change, the entropy must increase.

A rapid compression has the opposite e!ect — it does extra work on thematerial in the box, thus adding internal energy to the gas at a rate in excessof the reversible adiabatic rate. The entropy increases in this type of processas well. Both of these e!ects are illustrated in figure 24.6.

24.3 Heat Engines

Heat engines typically operate by transferring internal energy (i. e., heat) toand from a volume of gas and by compressing or expanding the gas. If theseoperations are done in a particular order, internal energy can be converted touseful work. We therefore seek to understand how an ideal gas reacts to theaddition and subtraction of internal energy and to the change in the volumeof the gas.

The equation for the entropy of an ideal gas and the ideal gas law containthe information we need. The entropy of an ideal gas is a function of its in-ternal energy E and its volume V . (We assume that the number of moleculesin the gas remains fixed.) Thus, a small change %S in the entropy can berelated to small changes in the energy and volume as follows:

%S =%S

%E%E +

%S

%V%V. (24.22)

We know that %S/%E = 1/T . Using equation (24.12) we can similarlycalculate %S/%V = NkB/V = p/T , where the ideal gas law is used in thelast step to eliminate NkB in favor of p. Substituting these into the aboveequation, multiplying by T , and solving for p%V results in

p%V = %W = T%S " %E (work by ideal gas) (24.23)

where we have recognized p%V = %W to be the work done by the gas onthe piston.

We are now in a position to investigate the conversion of internal energyto useful work. If the gas is allowed to push the piston out in a reversibleadiabatic manner, then %S = 0 and energy is converted with 100% e"ciencyfrom internal form to work. This work could in principle be used to run anelectric generator, stretch springs, power an automobile, etc.


E

S

S S

E

E1

2

1 2

B

DA

C

Figure 24.7: Plot of Carnot cycle for an ideal gas in a cylinder. Entropy-energy coordinates are used.

Unfortunately, a piston in a cylinder which can only extract energy duringsingle expansion wouldn’t be very useful — it would be like an automobileengine which only worked for half a turn of the crankshaft and then hadto be replaced! If the piston is simply pushed back into the cylinder, thenthe macroscopic energy gained from the initial expansion would be convertedback into internal energy of the gas, resulting in zero net creation of usefulwork.

The trick to obtaining non-zero useful work from the expansion and con-traction of a gas is to add heat to the gas before the expansion and extractheat from it before the recompression. This makes the gas cooler in thecompression than in the expansion. The pressure is therefore less in thecompression and the work needed to compress the gas is less than that pro-duced in the expansion.

Figure 24.7 shows a particular way of executing a complete cycle of expan-sion and compression of the gas which results in a net conversion of internalenergy to useful work. Assuming that the gas has initial entropy and inter-nal energy S1 and E1 at point A in figure 24.7, the gas is first compressed inreversible adiabatic fashion to point B. The entropy doesn’t change in thiscompression but the internal energy increases from E1 to E2. The work doneby the gas is negative and equals WAB = E1 " E2.

The gas then is allowed to slowly expand (so that the expansion is re-

24.3. HEAT ENGINES 419

versible), moving from point B to point C in figure 24.7 in such a waythat its internal energy doesn’t change. From equation (24.23) we see thatWBC = T2(S2 " S1) for this segment of the expansion. However, heat mustbe added to the gas equal in amount to the work done in order to keep theinternal energy fixed: Q2 = T2(S2 " S1).

From point C to point D the gas expands further but in this segment theexpansion is reversible adiabatic so that the entropy change is again zero.Thus, WCD = E2 " E1.

Finally, the gas is slowly compressed from point D to point A in a constantinternal energy process. Keeping the internal energy fixed means that the(negative) work done by the gas in this segment is WDA = T1(S1 " S2).Furthermore, heat equal to the work done on the gas by the piston must beremoved from the gas in order to keep the internal energy constant: Q1 ="WDA = T1(S2 " S1). The net work done by the gas over the full cycle isobtained by adding up the contributions from each segment:

W = WAB + WBC + WCD + WDA

= (E1 " E2) + T2(S2 " S1) + (E2 " E1) + T1(S1 " S2)

= (T2 " T1)(S2 " S1) (Carnot cycle). (24.24)

The energy source for this work is internal energy at temperature T2. Asdemanded by energy conservation, W = Q2"Q1. The fraction of the internalenergy Q2 which is converted to useful work in this cycle is

* =W

Q2

=(T2 " T1)(S2 " S1)

T2(S2 " S1)= 1 "

T1

T2

(thermodynamic e"ciency).

(24.25)This quantity * is called the thermodynamic e"ciency of the heat engine.Notice that it depends only on the ratio of the lower and upper temperatures,expressed in absolute or Kelvin form. The smaller this ratio, the larger thethermodynamic e"ciency.

Heat engines normally work via repeated cycling around some loop such asdescribed above. The particular cycle we have discussed is called the Carnotcycle after the 19th century French scientist Sadi Carnot. Heat is acceptedfrom a high temperature heat source, created, for example, by burning coalin a power plant. Excess heat is disposed of in the atmosphere or in somesource of running water such as a river. Notice that the ability to get ridof excess heat at low temperature is as important to a heat engine as thesupply of heat at a high temperature.


Many cycles for converting heat to work are possible — these are repre-sented by di!erent closed trajectories in the S-E plane. However, the Carnotcycle is special for two reasons: First, all heat absorbed by the system is ab-sorbed at a single temperature, T2, and all heat rejected from the system isrejected at a single temperature T1. This allows the expression of the e"-ciency simply in terms of the two temperatures. Second, the Carnot cycle isreversible, which means that no net entropy is generated.

A Carnot engine running backwards acts as a refrigerator. Heat %Q1

is extracted at temperature T1 from the box being cooled with the aid ofexternally supplied work W . An amount of heat Q2 = W + Q1 is thentransferred to the environment at temperature T2 > T1. Equation (24.25)gives the ratio of W to Q2 in this case as well as when the heat engine is runin the forward direction. This may be verified by tracing the cycle in figure24.7 in reverse.

24.4 Perpetual Motion Machines

Perpetual motion machines are devices which are purported to create usefulwork for “nothing” by violating some physical principle. Generally they aredivided into two types, perpetual motion machines of the first and secondkinds. Perpetual motion machines of the first kind violate the conservationof energy, while perpetual motion machines of the second kind violate thesecond law of thermodynamics. It is the latter type that we address here.

We commonly hear talk of an “energy crisis”. However, it is clear toall physicists that such a crisis, if it exists, is actually an “entropy crisis”.Energy beyond the most extravagant projected needs of mankind exists inthe form of internal energy of the earth. Furthermore, one cannot possibly“waste” energy, because energy can neither be created nor destroyed.

The real problem is that internal energy can only be tapped if two reser-voirs of internal energy exist, one at high temperature and one at low tem-perature. Heat engines depend on this temperature di!erence to operate andif all internal energy exists at the same temperature, no conversion of internalenergy to useful work is possible, at least using the Carnot cycle.

One naturally inquires as to whether some cycle exists which is moree"cient than the Carnot cycle. In other words, does there exist a heat engineoperating between temperatures T2 and T1 which extracts more work fromthe high temperature heat input Q2 than *Q2? Recall that * = 1 " T1/T2 is

24.4. PERPETUAL MOTION MACHINES 421

T2

T1

Carnot Super-XWQ2

Q1

Q4

Q3 < T2

Figure 24.8: Perpetual motion machine of the second kind. The Super-Xmachine is advertised as having a thermodynamic e"ciency greater than aCarnot engine. The output of the Super-X machine runs the Carnot enginebackwards as a refrigerator, resulting in net transfer of heat from the lowertemperature to the higher temperature reservoir.

the thermodynamic e"ciency of the Carnot engine.

Let’s suppose that an inventor has presented us with the “Super-X ma-chine”, which is purported to have a thermodynamic e"ciency greater than aCarnot engine. Figure 24.8 shows how we could set up an experiment in ourlaboratory to test the inventor’s claim. The Carnot engine runs in reverse asa refrigerator, emitting heat Q2 to the upper reservoir, absorbing Q1 from thelower reservoir, and using the work W = Q2 " Q1 = *Q2 from the Super-Xmachine. The Super-X machine is operated in heat engine mode, emittingQ3 to the lower reservoir and absorbing Q4 = Q3 + W < W/* from the up-per reservoir. The inequality indicates that the ratio of work produced andheat extracted from the upper reservoir, W/Q4, is greater for the Super-Xmachine than for an equivalent Carnot engine.

Let us examine the net heat flow out of the upper reservoir, Qupper =Q4 " Q2. Since Q4 < W/* = Q2, we find that Qupper < 0. In other words,the Super-X machine is extracting less heat from the upper reservoir thanthe Carnot engine is returning to this reservoir using the work produced bythe Super-X machine. The source of this energy is the lower reservoir, fromwhich an equivalent amount of heat is being extracted. The net e!ect ofthese two machines working together is a spontaneous transfer of heat froma lower to a higher temperature, since no outside energy source or entropysink is needed to make it operate. This is a violation of the second law of


thermodynamics. Therefore, the Super-X machine, if it truly works, is aperpetual motion machine of the second kind.

Though there have been many claims, no perpetual motion machine hasbeen convincingly demonstrated. Thus, heat engines are apparently inca-pable of converting all of the internal energy supplied to them to useful work,as this would require either an infinite input temperature or a zero outputtemperature. As we have demonstrated, this source of ine"ciency is intrin-sic to all heat engines and is in addition to the usual sources of ine"ciencysuch as friction and heat loss from imperfect insulation. No heat engine, nomatter how perfectly designed, can overcome this intrinsic ine"ciency.

As a result of the second law of thermodynamics, we see that real heatengines, which are always less e"cient than Carnot engines, produce usefulwork, W ,

W < *Q2 (real heat engines), (24.26)

where Q2 is the amount of heat energy extracted from the upper reservoir.On the other hand, refrigerators transfer heat Q2 to the upper reservoir inthe amount

Q2 < W/* (real refrigerators), (24.27)

where W is the work done on the refrigerator.

24.5 Problems

1. Following the procedure for a three-dimensional gas, do the followingfor a two-dimensional gas in a box of area A = a2, where a is the sidelength of the box.

(a) Find N for N particles. Eliminate a in favor of the box area A.

(b) Compute the entropy for this gas.

(c) Compute the temperature T , as a function of N and the internalenergy E. Invert to obtain the internal energy as a function of Nand T .

(d) Solve the entropy equation for energy and compute the “two-dimensional pressure”, ' = "%E/%A. What units does ' have?

(e) Find the two-dimensional analog to the ideal gas equation.

24.5. PROBLEMS 423

These calculations are relevant to atoms which can move freely aroundon a surface, but cannot escape it for energetic reasons.

2. Suppose your house has interior volume V . There are a few small airleaks, so that the inside air pressure p always equals the outside airpressure, which is assumed not to change.

(a) Compute the internal energy of the air in your house.

(b) Your roommate, trying to impress you with his knowledge ofphysics, says that he is going to turn up the thermostat to in-crease the internal energy of the air in the house. Will this work?Explain.

3. It has been proposed to extract useful work from the ocean by exploit-ing the temperature di!erence between deep ocean water at & 0# Cand tropical surface water at & 30# C to run a heat engine. Whatthermodynamic e"ciency would this process have?

4. Suppose your house is heated by a Carnot engine working as a re-frigerator between an outdoor temperature of 273 K and an indoortemperature of 303 K. (This means you are cooling the outdoors toheat the indoors! Such devices are called heat pumps.) If your houseloses heat at a rate of 5 kW, how much electrical energy must be usedto power the (perfectly e"cient) electrical motor running the Carnotengine? Compare the monthly cost of running this Carnot engine tothe cost of direct electric heating, i. e., via a big resistor.

5. Suppose an airplane engine is a heat engine which works between tem-peratures Tair and Tair + %T , where Tair is the air temperature, andwhere %T is fixed. Other things being equal, is this engine more ther-modynamically e"cient in the summer or winter? Explain.

6. Suppose the spring constant k of a spring varies with temperature, sothat k = CT , where C is a constant and T is the (Kelvin) temperature.Describe how this spring could be used to construct a heat engine.

7. Suppose a monatomic ideal gas at initial temperature T is allowed toexpand very rapidly so that it’s new volume is twice its original volume.It is then compressed isentropically (i. e., at constant entropy, which


means it is done slowly) back to the original volume. What is its newtemperature?

8. An inventor claims to have a refrigerator that extracts 100 W of heatfrom its interior which is kept at 150 K, rejecting the heat at roomtemperature (300 K). He claims that the refrigerator only consumes10 W of externally supplied power. If this device works, does it violatethe second law of thermodynamics?

Appendix A

Constants

This appendix lists various useful constants.

A.1 Constants of Nature


h 6.63 ! 10!34 J s Planck’s constanth 1.06 ! 10!34 J s h/(2")c 3 ! 108 m s!1 speed of lightG 6.67 ! 10!11 m3 s!2 kg!1 universal gravitational constantkB 1.38 ! 10!23 J K!1 Boltzmann’s constant$ 5.67 ! 10!8 W m!2 K!4 Stefan-Boltzmann constantK 3.67 ! 1011 s!1 K!1 thermal frequency constant*0 8.85 ! 10!12 C2 N!1 m!2 permittivity of free spaceµ0 4" ! 10!7 N s2 C!2 permeability of free space (= 1/(*0c2)).

A.2 Properties of Stable Particles


e 1.60 ! 10!19 C fundamental unit of chargeme 9.11 ! 10!31 kg = 0.511 MeV mass of electronmp 1.672648! 10!27 kg = 938.280 MeV mass of protonmn 1.674954! 10!27 kg = 939.573 MeV mass of neutron

425

426 APPENDIX A. CONSTANTS

A.3 Properties of Solar System Objects


Me 5.98 ! 1024 kg mass of earthMm 7.36 ! 1022 kg mass of moonMs 1.99 ! 1030 kg mass of sunRe 6.37 ! 106 m radius of earthRm 1.74 ! 106 m radius of moonRs 6.96 ! 108 m radius of sunDm 3.82 ! 108 m earth-moon distanceDs 1.50 ! 1011 m earth-sun distanceg 9.81 m s!2 earth’s surface gravity

A.4 Miscellaneous Conversions

1 lb = 4.448 N1 ft = 0.3048 m1 mph = 0.4470 m s!1

1 eV = 1.60 ! 10!19 J

Appendix B

GNU Free DocumentationLicense

Version 1.1, March 2000

Copyright c, 2000 Free Software Foundation, Inc.59 Temple Place, Suite 330, Boston, MA 02111-1307 USAEveryone is permitted to copy and distribute verbatim copies of this licensedocument, but changing it is not allowed.

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This License is a kind of “copyleft”, which means that derivative worksof the document must themselves be free in the same sense. It complementsthe GNU General Public License, which is a copyleft license designed for freesoftware.

We have designed this License in order to use it for manuals for freesoftware, because free software needs free documentation: a free programshould come with manuals providing the same freedoms that the software

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Appendix C

History

• Prehistory: The text was developed to this stage over a period of about5 years as course notes to Physics 131/132 at New Mexico Tech byDavid J. Raymond, with input from Alan M. Blyth. The course wastaught by Raymond and Blyth and by David J. Westpfahl at NewMexico Tech.

• 14 May 2001: First “copylefted” version of this text.

• 14 May 2003: Numerous small changes and corrections were made.

• 25 June 2004: More small corrections to volume one.

• 14 December 2004: More small corrections to volume one.

• 9 May 2005: More small corrections to volume two.

• 7 April 2006: More small corrections to both volumes.

• 7 July 2006: Finish volume 2 corrections for the coming year.

• 4 January 2008: Finish volume 1 corrections for fall 2008.

437