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TRANSCRIPT
Introduction to Special Relativity
Dr. Vern Lindberg, RIT
August 21, 2013
ii
Contents
1 Introduction to Special Relativity, Measuring Time and Space in theSame Units, Intelligent Observers, Event and Space-time Diagrams 1
1.1 What is Relativity, and Why is it Special? . . . . . . . . . . . . . . . . . . . 1
1.2 The Parable of the Surveyors . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Completing the analogy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Time in meters, Space in seconds . . . . . . . . . . . . . . . . . . . . . . . . 5
1.5 Intelligent Observers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.6 Events and Event Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.7 Synchronization of Clocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.8 More on Events and Intelligent Observers . . . . . . . . . . . . . . . . . . . 12
1.9 Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.10 From Event Diagrams to Space-Time Diagrams . . . . . . . . . . . . . . . . 16
1.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2 Reference Frames, Basic Postulates of Relativity, Time Dilation, ProperObservers, Proper Lengths, Simultaneity 19
2.1 Inertial or “Free-float” frames . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.2 What is the Frame of a Reference Frame . . . . . . . . . . . . . . . . . . . . 21
2.3 Einstein’s Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.4 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.5 Some Nice Advanced EXCEL Features . . . . . . . . . . . . . . . . . . . . . 25
2.6 Proper Observers, Space-time Interval . . . . . . . . . . . . . . . . . . . . . 26
2.7 But how do we know transverse width is the same for Louise and Ralph? . 27
2.8 Lengths of Objects Along Direction of Motion . . . . . . . . . . . . . . . . . 28
2.9 Simultaneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3 Lorentz Transformations, Velocity Transformations, Doppler Shifts, Ap-proximations 37
3.1 Mapping with Intervals Only . . . . . . . . . . . . . . . . . . . . . . . . . . 37
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iv CONTENTS
3.2 The Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.3 Using the Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . . 42
3.4 The Velocity Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.4.1 Non-Relativistic Galilean Transformation of Velocities . . . . . . . . 45
3.4.2 Vectors and Components . . . . . . . . . . . . . . . . . . . . . . . . 45
3.4.3 Relativistic Velocity Transformation . . . . . . . . . . . . . . . . . . 46
3.5 Using Velocity Transformations . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.6 Doppler Shifts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.6.1 Dopper Equation for Sound . . . . . . . . . . . . . . . . . . . . . . . 49
3.6.2 One-Dimensional Relativistic Doppler Effect . . . . . . . . . . . . . 49
3.6.3 Redshifts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.7 Approximations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
3.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4 Paradoxes and Their Resolution, Moving Along a Curved Worldline 57
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.2 The equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.3 Seeing versus Measuring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.4 Seeing the Length of a Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.5 Scissors Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
4.6 What If I Could Move Faster Than Light? . . . . . . . . . . . . . . . . . . . 60
4.7 Change of Measured Orientation of Objects . . . . . . . . . . . . . . . . . . 60
4.8 The Headlight Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
4.9 The Rising Stick. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
4.10 Paradox of the Identically Accelerated Twins . . . . . . . . . . . . . . . . . 62
4.11 The Twin Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.12 Position and Time Not Absolute . . . . . . . . . . . . . . . . . . . . . . . . 65
4.13 General Worldline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4.14 Time Kept By Traveller on a World Line . . . . . . . . . . . . . . . . . . . 68
4.15 Maximal or Minimal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.16 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
5 Time-like, Space-like, Light-like Intervals, Invariant Hyperbolae,Minkowski Diagrams 73
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
5.2 Time-like, Space-like, and Light-like Intervals . . . . . . . . . . . . . . . . . 73
5.3 Invariant Hyperbolae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
5.4 Special Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
5.5 Minkowski Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
5.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
CONTENTS v
6 Energy and Momentum 836.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.2 Four Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.3 Classical Energy and Momentum . . . . . . . . . . . . . . . . . . . . . . . . 846.4 Does This Definition of Momentum Work Relativistically? . . . . . . . . . . 856.5 Dealing With Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 856.6 New Definitions for Momentum and Energy . . . . . . . . . . . . . . . . . . 866.7 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876.8 Evidence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 886.9 Massless Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 906.10 The Compton Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 916.11 Low Speed Approximations . . . . . . . . . . . . . . . . . . . . . . . . . . . 926.12 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926.13 Electric and Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 936.14 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
vi CONTENTS
List of Figures
1.1 Alan, Beth, and location of sparks (X and Y). Alan is equidistant from thetwo sparks and receives light from the sparks at the same time. . . . . . . . 9
1.2 Alan and a Beeper. Alan wants to determine the time at which the beeperbeeped. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Alan hears a horn and a beeper at exactly the same time. . . . . . . . . . . 10
1.4 Alan and Beth Prepare to Synchronize Watches. . . . . . . . . . . . . . . . 12
1.5 Spaceship flyby showing Event 1 Viewed by Alan . . . . . . . . . . . . . . . 13
1.6 Alan, Andy, Beth and Becky making measurements for a train. . . . . . . . 14
2.1 A large railway car dropped near the earth (a) horizontally (b) vertically.A ball is located at each end, and is free of the railway car. If the frameis NOT inertial (free-float, Lorentz), the separation of the balls will changeduring the duration of the experiment. This depends on the care with whichwe can take measurements. . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2 Event Diagrams for two views of a gedanken experiment, (a) Louise (L) and(b) Ralph(R). Events E1: Pulse leaves Ralph E2: Pulse reaches mirror E3:pulse returns to Ralph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.3 Multiple exposure of the Event diagram. The heavy lines are the pathstaken by the pulse as seen by Louise and Ralph. . . . . . . . . . . . . . . . 24
2.4 Railroad car on tracks (a) Both at rest (b) Tracks at rest, car moving intopaper (c) Car at rest, tracks moving out of paper. Diagrams are drawnas if the transverse dimension changes with motion. The text shows thatthis assumption is absurd, and that the transverse dimension must remainunchanged as measured by different observers. . . . . . . . . . . . . . . . . . 28
2.5 Event diagrams for measuring the length of a rod (a) at rest (b) moving.Various events are marked. . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.6 Basic Set-up for simultaneity example . . . . . . . . . . . . . . . . . . . . . 32
2.7 Wavefronts have just arrived at Alan . . . . . . . . . . . . . . . . . . . . . . 33
2.8 Sparks have just occurred—not to scale! . . . . . . . . . . . . . . . . . . . . 33
vii
viii LIST OF FIGURES
3.1 Making a map knowing only the distances between cities. The arcs are la-beled with the two cities involved, e.g. RP = Rochester-Philadelphia distance. 39
3.2 Coordinate systems of Ursula the Unprimed and Pete the Primed as seenby (a)Ursula and (b) Pete. They are drawn as event diagrams with timeincreasing as you move upwards. The lower diagrams are for t = t′ = 0, andthe systems are shown with a slight offset so that you can see both. . . . . . 40
3.3 Galilean (“non-relativistic”) Transformation . . . . . . . . . . . . . . . . . . 41
3.4 A two dimensional vector can be described either by components, x, y, orby length and angle, r, θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.5 We use this for our conventional picture for the Doppler effect. The source isto the left and the observer to the right. Velocities to the right are positive,to the left are negative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.7 Red and blue shifts at the center of a rotating galactic core. These are shiftsrelative to the overall red- or blue-shift due to the overall motion of the galaxy. 52
3.6 Red shift of the Hydrogen Balmer series for quasar 3C 273. Vertical barsare the locations of lines for hydrogen in the lab. Shifts in the observedspectrum are shown by the arrows. This quasar has a red shift of 0.158, anda speed of 0.146. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.1 Seeing versus Measuring: Event diagram for a rocket heading eastward ona flat earth. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
4.2 Scissors Paradox. The horizontal rod is at rest, top rod is tilted and falling.The point of intersection of the rods can move faster than the speed of light. 59
4.3 The Rising Stick: Observers move along x-axis, Stick is rising along y-axisfor Ursula. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
4.4 On left, surveyor’s parable. Arrows connect points at the same distance fromthe center. On the right, an hyperbola representing something happening inthe future. The arrows connect events at the origin with different events onthe hyperbola. All arrows have the same value for the space-time interval,STI. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4.5 Space-time diagram showing a worldline for a particle. In general the world-line must be within 45◦ of the t-axis. At the top end of the line the lightcone is drawn showing possible past path and allowed future path of theparticle. The worldline on the right is impossible—you should be able toexplain why. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.6 Worldlines for Alan and Beth with three events marked. How do the elapsedtimes between E1 and E2 compare for the two observers? . . . . . . . . . . 70
LIST OF FIGURES ix
5.1 (a) Hyperbolae for time-like events, one at the origin and one on the hy-perbolae. The two events are strictly ordered in time, but can occur to theleft, at the same location, or to the right of each other. (b) Hyperbolae forspace-like events, one at the origin and one on the hyperbolae. The twoevents are strictly ordered in space, but can occur before, simultaneous to,or after each other. (c) Hyperbolae (lines) for light-like events. x = ±t . . . 76
5.2 Worldline of some particle, showing the light cone and several events aroundthe current time-space coordinate. . . . . . . . . . . . . . . . . . . . . . . . 78
5.3 Two special lines. (a) x′ = 0, this is the t′−axis. (b) t′ = 0, this is the x′−axis. 795.4 Minkowski diagrams for Beth (t′, x′) moving to the (a) right at 0.20, α =
11.3◦, and (b) left at −0.60, α = −31◦. Dashed lines show how to readvalues of (t′, x′). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
5.5 (a) Using Minkowski Diagram for Length of a rod (b) Coordinates of an event 81
6.1 Head on 1D elastic collision of two particles . . . . . . . . . . . . . . . . . . 846.2 A moving electron meets positron at rest. They annihilate leaving two photons. 916.3 Compton Effect: Scattering of a photon by an electron through an angle θ . 92
x LIST OF FIGURES
List of Tables
1.1 Surveying Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Connecting Parable and Special Relativity . . . . . . . . . . . . . . . . . . . 5
1.3 Using the relativistic interval . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.1 Values of γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.2 Observations from example of previous section. Values in µs. . . . . . . . . 26
3.1 Distances in miles between cities. Notice that no directions are given. . . . 38
3.2 Text and Lindberg Transformations Compared . . . . . . . . . . . . . . . . 42
3.3 Comparing Velocity Transformation Equations: vz transformations are likethe vy transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4.1 Andrew and Brad take a trip. Event coordinates measured by Mom andDad on earth in units of days. . . . . . . . . . . . . . . . . . . . . . . . . . . 62
4.2 Coordinates of Beth, measured by Alan . . . . . . . . . . . . . . . . . . . . 69
5.1 Measuring the length of a rod, Time and position in meters. . . . . . . . . . 74
5.2 Various Events for t′ = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
6.1 Classical Calculation of Billiard Ball Velocities Seen in Lab and by an Ob-server moving at 0.50. Velocities for second observer are found using velocitytransformations. While Energy and momentum are conserved for the Labobserver, they are not for the second observer as is shown in the text. . . . 86
6.2 One Dimensional Collision of billiard balls treated relativistically. Velocitiesand dilation factors for Lab and Second Observer. . . . . . . . . . . . . . . 88
6.3 Momenta and energy for relativistic collisions of two billiard balls. . . . . . 89
6.4 Although energy and momentum values vary for different observers, themass, Equation 6.17, is an invariant for all observers . . . . . . . . . . . . . 89
xi
xii LIST OF TABLES
6.5 Energy and Momentum in Positron Annihilation. In the unprimed system anelectron moving at 0.50 toward a stationary positron annihilates to producetwo photons. The primed values are tose seen by an observer moving to theright at 0.267946. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
Chapter 1
Introduction to Special Relativity,Measuring Time and Space in theSame Units, Intelligent Observers,Event and Space-time Diagrams
1.1 What is Relativity, andWhy is it Special?
Suppose we are trying to describe the worldas we see it. We would need to tell the lo-cation of objects in our world, the velocitiesof the objects, and how these change withtime.
Consider two observers, let us call them Al-fred (A) and Betsy (B). Alfred and Betsycould be at rest relative to each other, inwhich case they would see a constant dis-tance and direction of one relative to theother—the vector displacement between thetwo would be constant. This is a pretty bor-ing scenario.
Instead let us suppose that Betsy is mov-ing relative to Alfred. According to Alfred,Betsy:
(a) could be moving in a straight line at a
constant speed, with a changing separa-tion but in the same direction
(b) could be moving in a straight line at achanging speed, i.e. accelerating, butmoving in the same direction
(c) could be moving in a circle around Al-fred, with a constant separation, but ev-erchanging direction
(d) any of an infinite number of other pos-sibilities
Alfred could record the location and veloc-ity of Betsy in intervals of 1 second. Alfredwould know that he is at rest with himself.Betsy would know that she is at rest withherself. She could then record the locationand velocity of Alfred in intervals of 1 sec-ond.
We say that Alfred and Betsy are in relativemotion, and relativity is a mechanism of de-termining Betsy’s numbers knowing Alfred’s
1
2 1.2 The Parable of the Surveyors
numbers, or vice versa.
There is nothing new about this basic idea ofrelativity. The cavemen Akoba and Loana1
would have known that if Akoba saw Loanain the East, then Loana would see Akoba inthe west.
Fast forward to the time of Galileo who pos-tulated a Relativity Hypothesis2:
Any two observers moving at con-stant speed and direction with re-spect to each other will obtain thesame results for all mechanical ex-periments.
By the latter half of the 19th centuryMaxwell had shown that electricity and mag-netism were interrelated by 4 equations, andthat these predicted electromagnetic wavesthat travel at the speed of light. Thisposed problems for Galilean relativity (wewill discuss these in a bit) and scientistslike FitzGerald, Lorentz, and Poincare foundvarious patches that made things work bet-ter3.
Einstein published his paper “On the Elec-trodynamics of Moving Bodies” in 1905,showing how the difficulties can be resolvedby simply extending Galileo’s Relativity Hy-pothesis from mechanical experiments to allexperiments. Subsequent papers expandedon this new theory now called Special Rela-tivity.
Initially we will discuss our friends Alfredand Betsy and have one move at a constantvelocity (speed and direction) relative to the
1From the movie One Million Years B.C.2Dialogue Concerning the Two Chief World Sys-
tems, 16323http://en.wikipedia.org/wiki/History of special relativity
other. Special Relativity works equally wellif one of our friends is accelerating rela-tive to the other, as we will discuss muchlater in the course. Special relativity doesnot, however, discuss gravity, and assumes“flat space-time”—we’ll discuss space-timeshortly, and touch on curved space-time atthe end of the course.
General Relativity is Einstein’s theory thatdiscusses gravity, and describes masses as“curving the very fabric of space.” GeneralRelativity requires a great deal of mathemat-ical and physical sophistication, and will beleft for other courses.
This course will discuss Special Relativity,including Reference Frames, Space-Time di-agrams, invariant space-time intervals, Iner-tial and Free-floating reference frames, thePrincipal of Special Relativity, consequencesof the theory on the measurement of distanceintervals and time intervals, transformationsbetween coordinates measured in differentsystems, velocity and frequency transforma-tions, Time and space intervals for acceler-ated motion, space-like time-like and light-like intervals, and ending with the ideas ofenergy and momentum.
1.2 The Parable of the Sur-veyors
Spacetime Physics4 begins with an excellentanalogy involving surveyors on a flat earth.Paraphrased this is:
In the Institute of RIT the direction called
4Spacetime Physics, Second Edition, Taylor andWheeler, W.H. Freeman and Company 1992
1.2 The Parable of the Surveyors 3
North was sacred, and distances measurednorthward (and southward) were measuredin the sacred unit, the mile. Distances east-ward were measured in the not-so-sacredunit, the meter.
A surveyor named Anahita (Persian God-dess of day) does her work only in daytime.She uses a magnetic compass to find the di-rection North. She measures the coordinatesof every important building in the Insti-tute, relative to the center the Sentinel, andrecords the Northward (in miles) and east-ward (in meters) coordinates of each. (SeeFigure 1-1 in Spacetime Physics.) Anahitais making more careful measurements of theobjects measured by her grandfather Horus(yeah, I know he is Egyptian not Persian.)She is able to get readings with more signif-icant figures than Horus.
Unbeknownst to Anahita, Nyx (goddess ofnight, Greek) has been doing her survey-ing at night, using Polaris to tell her wherenorth is. She measures the same buildingsas Anahita, starting from the same loca-tion, the center of the Sentinel. And sheis checking the numbers of her grandfatherTezcatlipoca (Aztec).
One day Anahita is studying for a Physicsexam and decides to pull an all nighter.When she stops for some coffee she meetsNyx, and the two begin to compare num-bers. Alas, alack their numbers disagree,not by much, but definitely different. Thedifferences would not have been obvious inthe lower precision numbers of their grand-fathers.
After several more pots of coffee arguing overthe numbers, a mutual friend Koit (Esto-nian God of Dawn) stops by with a radical
suggestion: make a unit conversion to makethe northward distance end up in units ofmeters. To get North in meters, multiplyNorth in miles by a constant k = 1609.344meters/mile.5
The numbers in Table 1.1 show some of therecords of Anahita and Nyx. Together withKoit they made a startling observation. InSpacetime Physics this is written
(k North(mi))2 + (East(m))2 = (distance)2
(1.1)The distance, and the (distance)2 are exactlythe same number whether the daytime mea-surements of Anahita or the nighttime mea-surements of Nyx are used.
This is even easier if we use the distancesmeasured in the same unit, meters.
(North(m))2 + (East(m))2 = (distance)2
(1.2)
For example, for Stake A
distance2 = (4010.1)2 + (2949.9)2 =24783000 m2 = (3950.0)2 + (3029.9)2
where we have taken care to write the an-swer to 5 significant figures, the same as thenumbers used in the computation.
The example works for the coordinates ofeach point, but we can also define the dis-tance between two points by making the fol-lowing definitions.
∆N12 = North2 (meters)−North1 (meters)
∆E12 = East2 (meters)− East1 (meters)
5Alternately convert all measurements into milesby multiplying the easterly distance in meters by1/k = 6.213712× 10−5
4 1.3 Completing the analogy
Table 1.1: Surveying ResultsDaytime Anahinda Nighttime Nyx
East North North East North (miles) North (m)(m) (miles) (m) (m) (miles) (m)
Sentinel 0 0 0 0 0 0A 4010.1 1.8330 2949.9 3950.0 1.8827 3029.9B 5010.0 1.8268 2939.9 4950.0 1.8890 3040.1C 4000.0 1.2117 1950.0 3960.0 1.2614 2030.0D 5000.0 1.2054 1939.9 4960.0 1.2676 2040.0
We’ll call the two quantities the North In-terval and the East Interval. Then
(Displacement Interval)2 = (∆N)2 + (∆E)2
(1.3)
For example the north interval from A to Dis
∆NAD = 1939.9 − 2949.9 =−1010.0 m
and the east interval is
∆EAD = 5000.0 − 4010.1 = 989.9 m usingdaytime numbers,
so the displacement interval is
(Displacement Interval)2 = (−1010.0)2 +(989.9)2 = 2000000 m2.
You can try the same calculation usingnighttime numbers, with different values for∆NAD and ∆EAD, but the same value forthe displacement interval.
The advantage to using intervals is that weno longer are required to use one point as areference (The Sentinel), but that each per-son can use a separate reference point fromwhich to measure.
Together the three goddesses published theirresults and described them as the ANK In-
variance Principle: the displacement intervaldefined in Equation (1.3) is invariant, thatis it is the same for all observers.
1.3 Completing the anal-ogy
Table 1.2 makes a comparison of the Para-ble’s important results with the results forspecial relativity.
The last two rows have almost identicalequations. Do you understand the differ-ence?
Prior to the full acceptance of special relativ-ity, people believed in absolute time but rel-ative space. Let’s make this statement moreclear.
Imagine boarding an airplane in Rochesterand deplaning in Seattle. For you on theplane, the distance from the place where yousat down in Rochester to the exit door inSeattle was perhaps 100 feet. For me stay-ing in Rochester the distance between thetwo points is 2698 miles. The space intervalis dependent on the observer, i.e. relativeto the observer. This causes us no confu-
1.4 Time in meters, Space in seconds 5
Table 1.2: Connecting Parable and Special RelativityParable Special Relativity
North measured in miles Time measured in SecondsEast measured in meters Distance measured in meters
Conversion Factor k = 1609.344 m/mi Conversion Factor c = 2.99792458× 108 m/sN(m) = kN(mi) t(m) = ct(s)E(mi) = E(m)/k x(s) = x(m)/c
North Interval ∆N = N2 −N1 Space Interval (1D) ∆x = x2 − x1
East Interval ∆E = E2 − E1 Time Interval ∆t = t2 − t1(Displacement Interval)2 = (k ∆N)2 + (∆E)2 (Space-Time Interval)2 = (c∆t)2 − (∆x)2
(Displacement Interval)2 = (∆N)2 + (∆E)2 (Space-Time Interval)2 = (∆t)2 − (∆x)2
sion.
For you on the airplane the flight may havetaken 5 hours. For me on the ground, theflight would also have taken 5 hours. Nor-mally we think of time intervals as not de-pendent on the observer, but absolute. Thiswas assumed to be true until 1905.
A direct check of whether time intervals arerelative or absolute requires very accurateclocks, clocks that were not available un-til 1949. Measurements with less preciseclocks could not distinguish between abso-lute and relative time intervals. we mightbelieve time to be absolute, but if we hadvery accurate clocks, we would find that thetime interval measured by you on the planeand me on the ground would be different.This is one of the wonderfully weird conse-quences of relativity that we will explore indetail. And yes, that exact experiment, fly-ing atomic clocks on airplanes, was done in1975 and confirmed the predictions of rela-tivity6.
6Both Special and General Relativity were neededto analyze the results, SR because of the motion ofthe plane and GR since the airplane is at a differentaltitude that the ground where the gravitational field
The final line in the table is interesting.For the parable the displacement intervalis just the Pythagorean Theorem. In spe-cial relativity we use a space-time inter-val (STI) with STI2 calculated by subtract-ing the square of the space interval fromthe square of the time interval. The nega-tive sign will turn out to have profound ef-fects.
1.4 Time in meters, Space inseconds
In the parable we had different units fornorthward and eastward positions. In spe-cial relativity we have different units for po-sition (meters) and for time (seconds). Inthe parable, we found it more natural to ex-press the northward and eastward positionsin the same units. Likewise in Special Rela-tivity we find it useful to have the same unitsfor space and time.
You all know about light-years. A light yearis a unit of distance equal to the distance
is different.
6 1.4 Time in meters, Space in seconds
travelled by light in one year. Essentially weare measuring distance in a unit of time. Toconvert from meters to seconds we divide bythe speed of light—and for convenience wewill use the value 3.00× 108 m/s.
Example Convert the following distancesinto time units. Use approprite prefixesrather than scientific notation. (a) 30cm (about 1 foot) (b) 6378 km (radiusof earth) (c) 1 a.u. = 93 million miles(distance to sun) (d) 30 a.u. (distanceto Neptune)
(a) 0.30 m3.0×108 m/s
= 1× 10−9 s = 1 ns
(b) 6378 km3.0×108 m/s
= 6378000 m3.0×108 m/s
= 21.3 ms
(c) First convert 93 million miles intometers. Here is a lazy way to do it: en-ter “93e6 miles in m” into Google andget the answer 1.50 × 1011 m. Then1.50×1011 m3.0×108 m/s
= 500 s
(d) Multiply the answer to (c) by 30 toget 15000 s = 15 ks (or 4 hours, 10 min.)
Likewise we can convert time from units ofseconds into units of meters.
Example Convert the following into unitsof meters, using the appropriate prefixand not scientific notation. (a) 0.025 s(this would be a 30 cm lead at the endof a 100 m dash) (b) 50 minutes (c) 1day
(a) (0.025 s)(3 × 108 m/s) = 7.5 ×106 m = 7.5 Mm
(b) 50 minutes =3000 s, so (3000 s)(3 × 108 m/s) =9× 1011 m = 900 Gm
(c) 1 day = 86400 s, so (86400 s)(3 ×
108 m/s) = 2.59× 1013 m = 25.9 Tm
There are two ways to write equations inrelativity. One uses conventional units suchas m for space, s for time with an equationlike
Space− time Interval =√
(c∆t)2 − (∆x)2
(1.4)and the other assumes that a single set ofunits is used so that
Space− Time Interval =√
(∆t)2 − (∆x)2
(1.5)
You should be able to distinguish betweenthe two based on the context of the prob-lems being worked on. We will tend to usethe same units for space and time, unlesswe have to match up to something in thelab.
Example We have not given any justifica-tion for the Special Relativity formulasyet, but here is how we could use them.Suppose a rocket carries a laser that itpulses periodically. For Ralph on therocket, the space interval is zero, butfor Grace on the ground the laser willhave moved between flashes. Supposewe have the following data. Fill in thequantities marked ??
Table 1.3: Using the relativistic interval∆tr Rckt ∆tg Gnd ∆xg Gnd ∆xg GndTime Time Space SpaceInterval Interval Interval Interval(ns) (ns) (ns) (m)
(a) ?? 15 ?? 3.6(b) 15 ?? 20 ??(c) 40 58 ?? ??
1.5 Intelligent Observers 7
(a) Find the ground space interval inns, ∆xg = 3.6 m/3 × 108 m/s =12 ns. Then use (Space-TimeInterval)2 = (∆tr)
2 − (∆xr)2 =
(∆tg)2 − (∆xg)
2.
(∆tr)2 − 02 = 152 − 122 so that
∆tr = 9 ns
(b) (∆tr)2 − 02 = (∆tg)
2 − (∆xg)2 so
(∆tg)2 = (∆tr)
2 + (∆xg)2 giving
∆tg = 25 ns.
For ∆xg in meters, ∆xg = (3 ×108 m/s)(20× 10−9 s = 6.0 m
(c) (∆tr)2 − 02 = (∆tg)
2 − (∆xg)2 so
(∆xg)2 = (∆tg)
2 − (∆tr)2 giving
∆xg = 42 ns.
For ∆xg in meters, ∆xg = (3 ×108 m/s)(42× 10−9 s = 12.6 m
1.5 Intelligent Ob-servers
Much of the confusion that arises in relativ-ity comes from the basic meaning of observa-tion, and we must spend some time on thisseemingly trivial and boring subject7. Let’sfirst carefully define some terms: object, lo-cation, instant, and event.
• Object: This includes an observer(Alan), another object (rocket), butalso things like the crest of a wave. Theobject can be point-like and have a sin-gle location, or may be large and spreadout.
7The following exercises are adapted from RachelScherr PhD thesis.
• Location: This is a set of specific spacecoordinates (x, y, z) but the time is notdefined.
• Instant: This has a definite time coordi-nate, t, but can apply to many differentlocations.
• Event: This refers to a specific locationand time, that is a definite set of valuesfor (x, y, z, t).
Consider the following items: define themas object, location, instant, event, or noneof the above.
(a) The Declaration of Independence.
(b) The dot over the “i” in BenjaminFranklin’s signature on the Declaration.
(c) Babe Ruth’s final home run.
(d) The opening of the Olympic games inChina.
(e) Your friend honks her horn.
(f) The sound travels from your friend toyou.
(g) You hear the beep.
(h) Two successive beeps on your friend’shorn.
What ambiguities do you need toworry about in using these terms?How can you resolve these ambigui-ties?
Next consider the values that two observerswill see for the positions and instants ofsome events. Here use just ordinary non-relativistic physics, the stuff that operatesin almost all of what you see.
8 1.5 Intelligent Observers
For now we will talk about the space sep-aration and the time separation of twoevents.
1. A straight runway is 100 m long. Asmall explosion occurs at the east end ofthe runway (Event 1); 10 seconds later,an explosion occurs at the west endof the runway (Event 2). An airplanemoves from west to east with speed 25m/s relative to the runway.
How far apart in space are the locations(i.e. what is the space separation) of theexplosions:
• in the frame of the runway? Ex-plain.
• in the frame of the airplane? Ex-plain.
2. Two physics students, Alan and Beth,are shown in Figure 1.1. Alan and Bethhave measured their exact relative dis-tances from points X and Y.
Sparks jump at the points marked Xand Y. When each spark jumps, it emitsa flash of light that expands outward ina spherically symmetric pattern. Alan,who is equidistant from points X and Y,receives the wavefront from each sparkat the same instant.
Answer each of the following questionsfor the observers listed.
(a) Does Alan conclude that for hisreference frame, the spark thatjumped at point X jumps before,after, or at exactly the same timeas the spark that jumped at pointY? Explain your reasoning.
(b) Does Beth receive the wavefrontfrom the spark that jumped atpoint X before, after, or at ex-actly the same time as the wave-front from the spark that jumpedat point Y? Explain your reason-ing.
(c) Does Beth conclude that for herreference frame, the spark thatjumped at point X jumps before,after, or at exactly the same timeas the spark that jumped at pointY? Explain your reasoning.
(d) Does Beth’s answer in (b) agreewith Alan’s observation that he re-ceives the sparks at he same time?If they do not, explain why theyget different answers.
(e) Do the answers of Alan and Bethabout when the sparks occur, parts(a) and (c), agree? If they do not,explain why they get different an-swers.
3. A physics student named Alan and abeeper are arranged as shown in Fig-ure 1.2. The beeper has just emitted abeep (the arcs show the sound progress-ing through space), and Alan wants todetermine the exact time of the beep.However, he is a long way from thebeeper and unable to travel to it.
(a) Alan is equipped a large number ofaccurate meter sticks and clocks,and has a lot of friends (Alex, Al-fred, Andy, . . . ) who will assisthim if necessary. Neither the me-ter sticks nor the clocks are af-fected by being moved.
1.5 Intelligent Observers 9
Figure 1.1: Alan, Beth, and location of sparks (X and Y). Alan is equidistant from the twosparks and receives light from the sparks at the same time.
Figure 1.2: Alan and a Beeper. Alan wantsto determine the time at which the beeperbeeped.
i. Describe a procedure by whichAlan can determine the dis-tance to the beeper. Remem-ber he cannot move.
ii. Describe a set of measure-ments by which Alan can de-termine the time at whichthe beep is emitted using hisknowledge of the speed ofsound in air.
iii. Describe a method by whichAlan can determine the timeat which the beep is emittedwithout knowing or measur-ing the speed of sound first.( Hint: Alan’s assistants are
free to stand at any location.)
(b) A fugitive from justice is at largein Rochester. His identity and ex-act whereabouts are unknown. Areporter has reason to believe thatthe fugitive will soon confess to hiscrime, and wishes to record as ex-actly as possible the time and placeof the confession. Her funding forthis project is excellent
i. Describe an arrangement ofobservers and equipment withwhich the reporter may recordthe position and time of theconfession.
ii. An observer’s reference frameis an arrangement of assistantsand equipment with which theobserver may record the posi-tion and time of anything thatoccurs.
Justify the claim that the re-porter’s arrangement of ob-servers and equipment is thereporter’s reference frame.
(c) A horn is located between Alan
10 1.6 Events and Event Diagrams
and the beeper. The beeper beepsonce and the horn honks once.Alan hears the two sounds at thesame instant in time.
Figure 1.3: Alan hears a horn and a beeperat exactly the same time.
i. In Alan’s reference frame, isthe beep emitted before, after,or at the same instant as thehonk is emitted? Explain.
ii. Describe a method by whichAlan can measure the timeseparation between the emis-sion of the beep and the emis-sion of the honk in his refer-ence frame without knowing ormeasuring the speed of soundfirst.
An intelligent observer is equippedwith measuring devices (such as me-ter sticks, clocks, and assistants) andis able to use them to make correctand accurate observations of whereand when something occurs. Intelli-gent observers correct for any trans-mission time taken by signals. All ob-servers in the study of relativity areintelligent observers.
1.6 Events and Event Dia-grams
Event diagrams are a first, very pictorial,method of showing how events occur. Theyare useful for 1D and 2D situations. Con-sider a 1D situation. Objects are spacedalong a single axis that we boringly call x,and by convention x increases as we moveto the right. A single drawing can show allthe objects and their locations for one spe-cific instant in time. It is a snapshot of theworld.
If nothing moves, one such diagram is all weneed. When objects move we need to haveseveral snapshots at different times—i.e. weneed frames of a movie.
By convention we stack the frames one abovethe other, so that on a piece of paper thefirst snapshot is at the bottom of the paper,and successive snapshots are placed abovethe first. Thus time increases as we move upthe diagram. The group of frames is like aflip book.
An event can be shown by marking andlabeling a particular location on a partic-ular frame (snapshot), and the entire dia-gram just discussed is called an Event Dia-gram.
Example Recall the events described in thelast section in connection with Figure1.3. Draw an event diagram for this sit-uation.
At the bottom of the page, sketch apicture showing Alan, the beeper, thehorn, and any other objects of interestat the instant the beeper beeps. In-
1.7 Synchronization of Clocks 11
dicate the location of the event “thebeeper beeps” on the picture. Label itE1.
Above that picture, sketch another pic-ture showing the objects of interest atthe instant the horn honks. Indicate theposition of the event “the horn honks”on this picture. Label it E2.
Above that picture, sketch picture(s)showing the objects of interest at the in-stant(s) of the remaining events. Sketcha separate picture for each different in-stant of time. Indicate the location ofeach event on the appropriate picture.
Be sure you understand the meaning of thediagram you just drew by answering the fol-lowing questions.
Does the first frame, the picture you drew atthe bottom of the page, represent an object,a location, an instant, an event, several ornone of these?
Can more than one object appear in a singleframe? Can more than one event occur in asingle frame?
Can an object occur in several differentframes? Can an event occur in several dif-ferent frames?
1.7 Synchronization ofClocks
We require intelligent observers to havemethods to ensure that they are measuringquantities in the same units, and with thesame origin. In 1999 a Mars orbiter was lost
because NASA used SI units, while Lock-heed Martin used British Engineering units,and they failed to do a proper conversion.We must have a method to avoid such prob-lems.
Setting up the coordinate system for dis-tances is relatively easy. We lay out a grid ofmetersticks. We can produce them at a cen-tral plant (in order to ensure that they areidentical) and disperse them throughout theuniverse. Figure 2-6 in Spacetime Physicsshows a latticework grid.
We are confident that moving a meter stickand then bringing it back to rest will notchange the length of the stick, and we willrequire that the sticks touch each other, soa grid is established. Suppose we are at theorigin of the grid. We can then be confidentthat when our friend records a location ona meter stick near her, that her reading willtell us how far that location is from us.
Clocks are another matter. We can establisha calibration method for setting the tick rateof a clock so that we are confident that ourfriend’s clock has the same tick rate as ours.But how do we synchronize the two clocks,that is how do we ensure that when our clockreads noon, her clock reads noon? Here is amethod to try.
Alan and Beth are exactly 10 meters apartrelative to the floor as shown in Figure 1.4.Each of them wears a watch. Both watchesare extremely accurate, run at the same rate,and measure time in meters. (Remember,one meter of time is the amount of time ittakes light to travel one meter.) However,the reading on Beth’s watch is not the sameas the reading on Alan’s watch.
12 1.8 More on Events and Intelligent Observers
Figure 1.4: Alan and Beth Prepare to Syn-chronize Watches.
1. Determine the amount of time, in me-ters, that it will take a light signal totravel from Alan to Beth.
2. Beth and Alan decide in advance thatat the instant Alan’s watch reads 50meters, Alan’s laser pointer will emit apulse of light in Beth’s direction.
What time will Alan’s watch read at theinstant Beth first receives the light fromthe laser pointer?
3. Describe a method by which Beth couldsynchronize her watch with Alan’s (i.e.make her watch have the same readingas Alan’s at every instant.)
4. Another physics student, Caroline, is atrest with respect to Alan and Beth butis very far from them. Caroline looks atthe reading on Alan’s watch with a pow-erful telescope, and finds that at everyinstant the reading she sees on Alan’swatch through the telescope is identicalto the reading on her watch.
Is Caroline’s watch synchronized withAlan’s? Explain why or why not.
1.8 More on Events and In-telligent Observers
The following exercises should help you bet-ter understand events, event diagrams, andmeasurement by intelligent observers.
1. Two spaceships, A and B, pass veryclose to each other. Alan and Andy ridespaceship A, Alan at the front of shipand Andy at the rear. Beth and Beckyride ship B, with Beth at the front of hership and Becky at the rear. Accordingto Alan, ship B moves to the left withspeed v = 3 m/s and ships A and B eachhave length 12 m.
Define three events as follows:
• Event 1: Alan and Beth are adja-cent
• Event 2: Andy and Beth are adja-cent
• Event 3: Alan and Becky are ad-jacent
The first frame of an event diagram isshown in Figure 1.5.
Define ∆xAlan12 = xAlan2 − xAlan1 , wherethe superscript tells who is making theobservation, and the subscripts identifythe event or events of interest. Thisquantity can be either positive or nega-tive.
Determine numerical values for the fol-lowing ratios
• ∆xBeth12
∆xAlan12
• ∆xAlan13
∆xBeth12
Be sure to explain your rea-
1.8 More on Events and Intelligent Observers 13
Figure 1.5: Spaceship flyby showing Event 1 Viewed by Alan
soning in words.
2. A train moves with constant nonrel-ativistic speed along a straight track.The train is 12 meters long.
Alan and Andy stand 12 meters apartat rest on the track (see figure). Bethand Becky stand at rest at the front andrear of the train, respectively.
Define these three events.
• Event 1: Alan and Beth pass eachother.
• Event 2: Andy and Beth pass eachother.
• Event 3: Alan and Becky pass eachother.
(a) On a large sheet of paper, sketchan event diagram showing Alan,Andy, Beth, and Becky at the in-stants of events 1, 2, and 3 inAlan’s frame. (That is, sketch aseparate picture for each differentinstant; sketch pictures for succes-sive instants one above the other;and indicate the location of eachevent on the appropriate picture.)
i. What feature(s) of your eventdiagram can be used to indi-cate that it is a diagram forAlan’s frame?
ii. How would an event diagramfor Andy’s reference framecompare to the one you drewabove? Explain.
iii. What procedure could Alan
14 1.8 More on Events and Intelligent Observers
Figure 1.6: Alan, Andy, Beth and Becky making measurements for a train.
(or Andy) follow to measurethe distance between the loca-tions of events 1 and 2?
iv. How far apart in space are thelocations of the following pairsof events in Alan’s frame?
• Events l and 2
• Events 2 and 3
• Events 1 and 3
(b) Sketch an event diagram showingevents 1, 2, and 3 in Beth’s frame.Be sure your diagram correctlyrepresents the motion of the trainin this frame.
How far apart in space are the lo-cations of the following pairs ofevents in Beth’s reference frame?
• Events l and 2
• Events 2 and 3
• Events 1 and 3
(c) How does Beth’s procedure for
measuring the distance betweenthe positions of two events com-pare to Alan’s procedure?
(d) On the basis of your answersabove, develop a general rule thatuses an event diagram to deter-mine how far apart the locationsof two events are in a given refer-ence frame.
3. Give interpretations for the magnitudeof each of the following quantities; thatis, tell the meaning of the number inthis physical context. One has been pro-vided as an example. Some quantitiesmay have more than one interpretation.
(a) ∆xAlan12 This is the displacement ofBeth (train) as measured by Alan(Andy).
(b) ∆xAlan13
(c) ∆xAlan23
(d) ∆xBeth12
(e) ∆xBeth13
1.9 Velocity 15
(f) ∆xBeth23
4. A train of unknown length moves withconstant nonrelativistic speed on thesame track. Alan and his assistantsstand shoulder-to-shoulder on the track.
(a) Describe a method by which Alanall by himself can determine thelength of the train in his frame ifhe knows the speed of the train inhis frame. Specify two events asso-ciated with this measurement pro-cedure.
• Event a:
• Event b:
(b) Describe a method by which Alancan determine the length of thetrain in his frame without knowingor measuring its speed first. Spec-ify two events associated with thismeasurement procedure.
• Event c
• Event d:
5. Suppose event 4 occurs at the front ofa long ship, and event 5 occurs at therear of the same ship. Describe the cir-cumstances in which the absolute valueof ∆xwho?45 is equal to the length of theship—be sure to identify who is the ob-server in each case.
• in the frame S in which the ship isat rest
• in frame F, in which the ship ismoving
Draw event diagrams to support youranswers.
1.9 Velocity
For most of the course we will focus on thesimplest of motion, motion in a straight line(1D motion) at a constant speed. We includethe speed (magnitude) and direction (wecould use left/right, East/West, but mostcommonly ± to describe the two directions)in the velocity of an object, v.8
If a single observer measures a space inter-val ∆x and a time interval ∆t, then we canwrite
∆x = v∆t (1.6)
Sometimes it is more useful to think of thisas
v =∆x
∆t=dx
dt(1.7)
where the last form is the calculus ver-sion.
It is imperitive that you think of the equa-tions as written, with intervals, and NOTas x = v t which assumes that the startingpoint is the origin at t = 0.
Conventionally we have units like m/s whenwe use Equation (1.7). What happens ifwe measure ∆x and ∆t in the same units?Equation (1.7) then says that the velocityhas no units. It is easier to think of this interms of the speed as a fraction of the speedof light. In dimensionless units c = 1.
8Velocity is a vector and in 2D and 3D we willneed to write it as −→v . A two dimensional vectorvelocity will have components, vx = v cos θ and vy =v sin θ where v is the speed and θ is the angle withthe x-axis.
16 1.10 From Event Diagrams to Space-Time Diagrams
1.10 From Event Diagramsto Space-Time Dia-grams
Event diagrams are messy when we want todraw many objects on a diagram and in-dicate events. A 1D space-time diagramcontains the same information in a morecompact form, as well as in a quantitativeform.
Consider a graph where the horizontal axisis space, x, and the vertical axis is time, t.A horizontal line represents many locationsat one instant of time. A vertical line repre-sents one location at many instants of time.Note that this is the reverse of whatyou usually draw in physics.
We use the same units for the vertical andhorizontal axes. For convenience we will usethe same scale so that when 5 squares =20 ns on the horizontal axis (space), then5 squares = 20 ns on the vertical axis (time)also.
An event is a point somewhere on thisgraph—it has a specific position and timecoordinate.
A line on the graph represents the object atdifferent times. Suppose the object is notmoving. Then its location is constant, andon the graph we draw a vertical line at theleft and right ends. A vertical line repre-sents zero velocity.
What would we draw for an object movingto the right at a velocity v = +0.2c? Firstwe should recognize that the proper way togive the velocity for this graph is with di-mensionless units, so β = +0.2 where we
use β to represent the dimensionless veloc-ity.
Recall Equation (1.6), ∆x = β∆t, where wemeasure time and space in the same unitsand speed is dimensionless. Remember thatslope on a graph is rise/run, so the slope onour space-time graph is
slope =∆t
∆x=
1
β(1.8)
Thus on a space-time diagram, steep linesare slow speeds. You should have pur-chased a pad of engineering note paper thatis green with a coarse grid. Each squarecould represent 1 ns. Make the origin some-where in the middle of the paper. On it drawand label a horizontal axis for space, and avertical axis for time. Then draw lines rep-resenting:
(a) Oscar, an object at x = 1 ns that is notmoving.
(b) Ralph, an object that is located at x =−10 ns at t = 0 ns and is moving to theright at βR = +0.40.
(c) Lorraine, an object located at x = +15ns at t = −5 ns and is moving to the leftat βL = −0.80.
(d) A pulse of light sent to the right by Os-car at t = 0 ns
The lines you have drawn are called theworld lines for the objects. Be sure you labelthe world lines Oscar, Ralph, Lorraine, andLight.
On the space-time diagram you just made,mark the following events. You can use thegraph to determine and record the positionand time for each event. Alternately you
1.11 Summary 17
can use algebra to find the answer. The firstproblem is done for you.
(a) Ralph blows a bubble at t = −5 ns. Onthe graph this event occurs at x = −12ns.
We can use algebra to get the space andtime coordinates more precisely.
We can expand Equation(1.6) as
x− x0 = βR(t− t0) = βR t− v t0 (1.9)
or rewrite it as
∆t =1
βR∆x (1.10)
and expand to get
t− t0 =1
βR(x− x0) =
1
βRx− 1
βRx0
(1.11)
So for Ralph, x−(−10) = 0.4 (−5) = −2and solving we get x = −12 ns.
(b) Lorraine snaps her fingers at t = 0 ns.
(c) Ralph and Lorraine meet.
Verify that your coordinates for the otherevents are correct.
1.11 Summary
a. It is convenient to measure time andspace using the same units, either unitsof length like meters, or units of time likeseconds (sometimes called light seconds.)The conversion factor is c = 3.00 × 108
m/s. YET TO BE ANSWERED: Whyuse this speed, and not another speed likethe speed of sound?
b. We begin by concentrating on space inter-vals, ∆x and time intervals ∆t betweenevents.
c. In special relativity the space-time inter-val, is invariant, that is the same for allobservers, while both the space and timeintervals are relative. YET TO BE AN-SWERED: Why is this true? What doesthis imply?
d. We talk about objects, events, locations,and instants of time in relativity.
e. We can draw either a pictorial event di-agram or a more precise space-time di-agram (graph) to represent objects andevents.
f. A line on a space-time diagram is calledthe world-line. The velocity of an objectcan be found as the inverse of the slope ofa line on the space-time diagram. Steeplines mean slow speeds.
g. All observers are intelligent observers.They have ways to ensure that clocks aresynchronized, and can use friends and areference frame to find the location andtime of any event. The transmission timeof a signal from the event to the observercan be corrected for, and is not the reasonfor the weirdness of relativity.
h. We can get a rough idea of what is hap-pening by drawing world lines on a space-time diagram, or by doing simple algebrausing Equations (1.9) or (1.11)
18 1.11 Summary
Chapter 2
Reference Frames, BasicPostulates of Relativity, TimeDilation, Proper Observers,Proper Lengths, Simultaneity
2.1 Inertial or “Free-float”frames
In Chapter 1, I mentioned that Special Rela-tivity (SR) does not deal with gravity, whileGeneral Relativity (GR) does. Chapter 2 ofSpacetime Physics deals with how to tell ifwe must use GR or not.
Imagine that we are doing an experimentwhile on an airplane that is flying at a con-stant altitude above the earth. We releasea pencil and it falls, evidence that gravity isacting.
If the airplane suddenly enters a steep dive,things change. The Vomit Comet 1 de-liberately enters a parabolic path so thatpassengers feel the effects of “weightless-
1Or Weightless Wonder as NASA prefers tocall it, http://en.wikipedia.org/wiki/Vomit cometAbout 1/3 of the passengers get violently ill, 1/3feel mild distress, and 1/3 are fine.
ness”.
The airplane and its passengers are in a stateof “free fall”, or as Taylor and Wheeler pre-fer to call it, “free float”. For the durationof the parabola the passengers cannot distin-guish their environment from a truly gravity-free environment. The space shuttle is like-wise a free float frame of reference, contin-uously falling around the earth in a circularorbit.
The free float frames that we have discussedare local frames, and if we look carefullyenough, we can tell whether a gravitationalforce acts. Consider an initially horizon-tal railway car dropped so that it appearsto be a free float frame as shown in Figure2.1(a).
Suppose you are inside the car and watchtwo balls, one at each end of the car. Forsome period of time you feel no effects of theearth—that is until the car hits the ground.
19
20 2.1 Inertial or “Free-float” frames
Figure 2.1: A large railway car dropped near the earth (a) horizontally (b) vertically. Aball is located at each end, and is free of the railway car. If the frame is NOT inertial(free-float, Lorentz), the separation of the balls will change during the duration of theexperiment. This depends on the care with which we can take measurements.
Ouch!
Careful measurements of the positions of theballs reveals that they come closer togetherduring the duration of the fall. Likewise inthe case of the vertical fall, Figure 2.1(b),careful measurements reveal that the ballsmove farther apart. There appears to besome mysterious force that brings the twoballs closer together when the car is hori-zontal, and farther apart when it is verti-cal.
This is the result of a non-uniform gravi-tational field around the earth. In Figure2.1(a) the balls fall toward the center of theearth, while in Figure 2.1(b) the ball initiallyfarther away from the earth moves less thanthe ball closer to the earth due to the vari-
ation of gravitational force with distance.(Here I have used the language of classicalphysics and not the language of GR.)
With this backdrop we can describe a localfree float frame of reference. The referenceframe is said to be “free float” or “in-ertial” or “Lorentz” providing that ina certain region of space and for a cer-tain amount of time, throughout theregion, and within the measurementerror, Newton’s First Law is valid: i.e.a free particle at rest remains at rest,and a free particle having a velocitymaintains that velocity, both size anddirection.
What does this mean in practice. If the rail-way car in Figure 2.1(a) is 20 m long and is
2.3 Einstein’s Postulates 21
dropped from a height of 315 m with no airresistance, then it will hit the surface in 8 s= 2.4 × 109 m of time. During the fall theballs will move toward each other a distanceof 1 mm. So the car can be considered aninertial frame providing we focus on only 8 sof its world-line and make measurements ofdistances only to the nearest cm.
If it is dropped as in Figure Figure 2.1(b)with the lower edge of the box 315 abovethe surface, then in the 8 seconds before ithits the earth the balls will separate by 2mm. Again the car can be considered aninertial frame providing we focus only of 8 sof its world-line and make measurements ofdistances only to the nearest cm.
2.2 What is the Frame of aReference Frame
In Chapter 1 we described a method of set-ting up a grid of metersticks throughoutthe universe, and a method of synchroniz-ing clocks for observers in a single referenceframe.
Suppose that we have a local inertial refer-ence frame “A”, Albert, and inside this localframe is a second observer “B”, Betsy, mov-ing with a constant velocity with respect toA. Then B will also be a local inertial refer-ence frame subject to the same constraintson time and accuracy of measurement as thefirst. (In the examples that were given in thelast section this was for 8 sec and positionaccurate to the nearest cm.
Albert and Betsy will each have a lattice ofmeter sticks and clocks at rest with himself
or herself. They will make measurements us-ing their own meter sticks and clocks. Spe-cial Relativity will allow us to compare thereadings that the two observers make.
2.3 Einstein’s Postu-lates
Einstein’s Postulates of Special Relativityare generally given as
1. All Laws of Physics are the same in allfree float inertial reference frames. E.g.∑ ~F = m ~a for all inertial frames.
2. The speed of light in vacuum is the samefor all observers.
We will initially talk about two observers indifferent reference frames, one moving withdimensionless speed β with respect to theother (β = v/c), and we will ask about whatis the same for both observers and what isdifferent.
What is different for the two observers? Spe-cific numbers for things like space inter-vals, time intervals, velocities, accelerations,forces, electric, and magnetic fields will bedifferent. For example, Al and Becky aretwo observers who measure the intervals be-tween two events. For these two events, Almeasures a space interval of +51 m and atime interval of +60 m. For the same events,Becky measures a space interval of +5 m anda time interval of +32 m.
What is the same? The Laws of Physicssuch as Newton’s Laws and the electromag-netic Maxwell Equations are the same. Fun-damental constants—things like Avogadro’s
22 2.4 Time Dilation
number, the charge of the electron, and theconstants ε0 and µ0 from electricity andmagnetism have exactly the same numericvalue in the two inertial reference frames.The events are the same—if Al sees anevent of a spark jumping a gap, so doesBecky.
Let’s take a brief detour into electricity andmagnetism. In the 1780’s, Charles Au-gustin de Coulomb developed the law ex-plaining electric forces. In SI units thereis a fundamental constant ε0 (epsilon-sub-zero), the permittivity of free space, ε0 =8.854×10−12 C2/N/m2. In about 1820 JeanBabtiste Biot and Felix Savart found a lawto explain the magnetic force that includesa constant µ0 (mu-zero), the permeability offree space, µ0 = 4π × 10−7N/A2.
Using Einstein’s First postulate we knowthat the values of ε0 and µ0 are the same fordifferent inertial reference frames. In 1861,James Clerk Maxwell took the equations ofelectricity and magnetism and showed thatby combining them he could explain thatlight was a combination of an electric anda magnetic field, and that it should have aspeed of
c =
√1
ε0µ0(2.1)
Using the values for the constants given
above. c =√
1(8.854×10−12)(4π×10−7)
=
2.998× 108
with units given by√(Nm2/C2)(A2/N) =
√m2(C/s)2/C2 = m/s
The First Postulate therefore says that sinceε0 and µ0 are fundamental constants, theyhave the same value in all reference frames,and therefore the speed of light in vacuum
is the same for observer’s in all referenceframes. Thus Einstein’s Second Postulate isa specific case of the First Postulate.
2.4 Time Dilation
Einstein figured out his theories not by do-ing real experiments, but rather by doingthought experiments (Gedankenexperimentin German) and we shall begin our develop-ment of the relativity equations with a clas-sic gedankenexperiment.
Louise (L) sits in a lab and watches Ralph(R) move to the right in a rocket with speedβ. Ralph has a laser pointer that he aimsat the far side of the rocket where there isa mirror. Event 1 is Ralph briefly sendinga pulse of light. Event 2 is the light pulsereaching the mirror, and Event 3 is the lightpulse returning to Ralph. Event diagramsas seen by Louise and Ralph are in Figure2.2
For the sake of subsequent calculation, wecan draw the three “frames” superimposedas is done in Figure 2.3.
Call the width of the rocket, between Ralphand the mirror, W/2. According to Ralphthe pulse has travelled a total distance Wbefore he sees it again, and the time inter-val between events E1 and E3 is ∆tRalph13 =W/c = W , where we have used the fact thatin dimensionless units, c = 1. Recall thatthe superscript tells who is measuring thetime interval between the events, in this caseRalph.
Now consider the view of Louise. For her,Ralph and the rocket are moving to the right
2.4 Time Dilation 23
Figure 2.2: Event Diagrams for two views of a gedanken experiment, (a) Louise (L) and(b) Ralph(R). Events E1: Pulse leaves Ralph E2: Pulse reaches mirror E3: pulse returnsto Ralph
with a speed β. The time interval betweenevents 1 and 2 is half the total interval,∆tLouise13 /2, and during this time the rocketincluding Ralph and the mirror have movedto the right a distance ∆x/2 = β∆tLouise13 /2.The light according to Louise has travelledalong the diagonal taking a time
∆tLouise13
2=diagonal
c= diagonal. (2.2)
Now we use the Pythagorean equation tosay
diagonal2 =
(W
2
)2
+
(∆x
2
)2
=
(W
2
)2
+
(β∆tLouise13
2
)2
(2.3)
Combining Equations 2.2 and 2.3 and doing
some algebra we end up with
∆tLouise13 =W√
1− β2=
∆tRalph13√1− β2
(2.4)
The factor in Equation 2.4 occurs so fre-quently that we give it the symbol gamma.If β < 1, then γ > 1.
γ =1√
1− β2(2.5)
and therefore we can write
∆tLouise13 = γ∆tRalph13 (2.6)
We will use γ throughout the course. It isinstructive, therefore, to calculate γ for sev-eral velocities. Fill in values in Table 2.1.
Example Suppose Ralph measures a timeinterval of 24 µs and according to
24 2.4 Time Dilation
Figure 2.3: Multiple exposure of the Event diagram. The heavy lines are the paths takenby the pulse as seen by Louise and Ralph.
Speed β Dilation Factor γ
0.10000.30000.50000.60000.80000.90000.99000.99900.9999
Table 2.1: Values of γ
Louise the rocket is moving to the rightat β = 0.6. What is the time intervalmeasured by Louise?
γ = 1/√
1− .36 = 1.25 so ∆tLouise =1.25(24 µs) = 30 µs
Example Suppose Louise sees the rocketmoving at β = 0.06. What will she mea-sure for the time interval?
γ = 1/√
1− 0.0036 = 1.0018 so∆tLouise = 1.0018(24 µs) = 24.04 µs
Example Suppose Louise sees the rocketmoving at β = 0.99. What will she mea-sure for the time interval?
γ = 1/√
1− 0.9801 = 7.09 so∆tLouise = 7.09(24 µs) = 170 µs
Notice that reducing the speed by a factor of10 has a huge effect on the time interval seenby Louise. For a speed of 0.06 the differencebetween Louise and Ralph is small, but fora speed of 0.60 it is large and unmistakable.As the speed approaches the speed of light,β → 1, γ →∞, the effect is huge!
The time interval measured by Louise in thiscase is larger than that measured by Ralph.The common term used for the increase inthe time interval is time dilation2. Louisesees Ralph’s clock moving, and often timedilation is described as “Moving clocks runslow.”
Special relativity contains a number of seem-ing paradoxes, that is situations that viewedfrom two different viewpoints appear to givecontradictory results. The paradoxes allowus to sharpen our thinking, because whenwe think about them carefully the paradoxdisappears.
Consider the paradox that relates to the sit-uation just analyzed. Louise says that Ralphmoves. We have Equation 2.6 that says thetime interval measured by Louise is largerthan the time interval measured by Ralph.(Ralph’s clock runs slow.)
2Or if you want to sound British, dilatation.
2.5 Some Nice Advanced EXCEL Features 25
But what does Ralph think? Ralph seesLouise moving, and therefore would saythat his time interval should be larger thanLouise’s time interval. (Louise’s clock runsslow.)
Only one of these can be correct. How canthe paradox be resolved? Can you see wherethe asymmetry comes from? Hint: Thinkabout the idea of intelligent observers thatwe discussed earlier.
2.5 Some Nice AdvancedEXCEL Features
Having to do the calculations over and overwith a calculator is tiring, and prone to er-ror. Even if you can program a specialcalculation pattern into you calculator, youcannot easily print the results. Excel al-lows more reliable calculations—at least ifwe program it right.
Named cells
Traditionally a cell is addressed by Columnand Row such as C6. This is a relative ad-dress, meaning that if you enter a formulainto cell B3 like =C6^2, and then copy theformula into cell C3, the formula changes to= D6^2. This can be very useful if we wantto apply the same formula to several differ-ent values.
Sometimes however we might want to referto the exact same cell. If cell A1 containeda value for acceleration, and we wanted tocompute c = a× t2 for several different val-ues of t in cells in column B, we could not
copy the formula unless we went back andchanged the reference to the acceleration. Soif we typed =A1*B2^2 into cell C2 and copiedit to cell C3, it would become =A2*B3^2, andwe would manually need to change A2 to A1,as was done in the table below.
A B C
1 5.5 t c
2 2 =A1*B2^2
3 4 =A1*B3^2
4 6 =A1*B4^2
Instead we could use an absolute reference,$A$1 so that the equation is = $A$1*B2^2.When this is copied, only the relative cellreference, B2, changes.
Even nicer would be to name the cell witha word like “accel”. One way to do this isto find the cell address up in the tool bar.With the cursor in cell A1, the cell addressshould be A1.
Click in the cell address in the tool bar andtype “accel” and enter. Now the cell hasthat name, and we can write an equation=accel*B2^2.
Writing your own functions
Excel has a wealth of functions likesqrt, abs, sin, exp, max, average—the whole list can be seen usingInsert Function from the Insert menu orthe fx shortcut. You can add User Definedfunctions also.
Suppose that you want a function that willcalculate γ. Here are the steps.
(a) Go to
26 2.6 Proper Observers, Space-time Interval
Tools/Macro/Visual Basic Editor
Two windows called Projects andProperties should appear.
(b) Go to Insert/Module and a Workbook1-Module1 (Code) window should open.
(c) Click in the Workbook area. The firstcommand defines the function.
Function gamma(beta) which startsthe definition of a function called gammathat will take a single argument, beta.
When you hit return, an End Functionstatement will appear.
(d) Now type the function. It can haveseveral lines if needed, and have othervariables besides beta and gamma, butsomewhere you must have gamma = ...
For this function a single line suffices,gamma = 1 / Sqr(1 - beta ^ 2).Click back in the spreadsheet and see ifit works.
Look at the function list and you should beable to find your function in the User Definedcategory.
Your test on writing functions. If youare given a value of γ, what is the value ofβ? First work this out algebraically, thenwrite a function for it called beta(gamma).This can appear in the same module asgamma(beta).
Can I Save User-Defined Func-tions
If you save the Excel workbook, and user-defined functions are saved with it. To ac-
cess these functions you must reopen theworkbook.
2.6 Proper Observers,Space-time Interval
Equation 2.6 applies to very special situa-tions where a time interval is measured be-tween two events. One observer (Ralph)must be present at both events. The ob-server present at both events is called theproper observer and sees ∆x = 0 and mea-sures the smallest value for ∆t.
This is consistent with the space-time in-terval STI =
√∆t2 −∆x2. Extending the
first example in Section 2.4 to include Syd-ney, whom Louise sees moving to the leftat β = −0.6. The events are the sendingand receiving of the flash in Ralph’s rocket.Ralph is the proper observer of time.
Observer ∆t ∆x STI
Ralph β = 0.6 24 0 24Louise 30 18 24Sydney β = −0.6 51 45 24
Table 2.2: Observations from example ofprevious section. Values in µs.
The data in the table show that different ob-servers will see different values for the timeinterval and the space interval, but the sameinvariant value for the space-time interval(STI).
According to Louise, Sydney has γS = 1.25.Can we use Equation 2.6 to relate the timeintervals of Louise and Sydney? If you trythe numbers you see that
2.7 But how do we know transverse width is the same for Louise and Ralph?27
∆tS = 51 µs 6= γS × ∆tL = 1.25(24) =30 µs.
Evidently Equation 2.6 must ONLY be usedwhen one of the observers is the proper ob-server.
Example Louise sees Ralph move at β =0.9 and measures a time interval be-tween two events of ∆tL = 50 ms.Ralph is the proper observer of the twoevents. What does Ralph measure forthe time interval?
First method Calculate γ = 2.294.Then 50 = 2.294(∆tR) and wesolve to get ∆tR = 21.8 ms.
Second Method Louise sees Ralphmove ∆xL = βt = 0.9(50) = 45ms. Equating the expressions forintervals, 502 − 452 = (∆tR)2 − 0so ∆tR = 21.8 ms.
Both methods give the same results, as theymust.
Example Suppose we look at eventsE1=Ralph briefly sending a pulse oflight and E2 = light pulse reaching themirror in our Louise/Ralph scenario.Who, if anyone, is a proper observer ofthe time interval?
2.7 But how do we knowtransverse width is thesame for Louise andRalph?
In the derivation of the time dilation for-mula, Equation 2.6 we assumed that both
Louise and Ralph measured the same widthfor the rocket, the dimension transverse(perpendicular) to the relative velocity. Howcan we make this assumption?
The proof of this relies on the technique ofreductio ad absurdum. We will make an as-sumption, follow it to its logical conclusion,and if that conclusion is absurd we will rejectthe original assumption.
Assume that the lateral dimension of amoving object is less than its lateral dimen-sion when at rest.
Imagine that we stand next to railway trackswatching a train riding on tracks headingstraight away from us (into the page.) Whenthe train is at rest, its wheel base justmatches the track spacing, Figure 2.4(a).Now suppose the train is moving into the pa-per, and therefore shrinks in size as shown inFigure 2.4(b). This is weird, but we expectweird from relativity.
Finally imagine that we are on the train. Wewould see the train at rest and the tracksmoving out of the paper. By the originalassumption we would then see the spacingof the track reduced as shown in Figure2.4(c).
In Figure 2.4(b) the car would leave marksinside the tracks, in Figure 2.4(c) the carwould leave marks outside the tracks. Wecannot have a single set of marks both in-side and outside the tracks—this is absurd.So we must reject the original assumption.We could try a new assumption that thelater dimension of a moving object increases,but it would also lead to an absurd conclu-sion.
We are left to conclude that there is no
28 2.8 Lengths of Objects Along Direction of Motion
Figure 2.4: Railroad car on tracks (a) Both at rest (b) Tracks at rest, car moving intopaper (c) Car at rest, tracks moving out of paper. Diagrams are drawn as if the transversedimension changes with motion. The text shows that this assumption is absurd, and thatthe transverse dimension must remain unchanged as measured by different observers.
change in lateral dimension of a moving ob-ject.
2.8 Lengths of ObjectsAlong Direction of Mo-tion
Consider a rigid object. It has a definitelength that does not change with time. Howdo we go about measuring its length?
If the object is at rest with respect to us wecan walk to the left end and make note ofits position on our reference frame of metersticks. Then we can walk to the right endand note its position, then we subtract thetwo readings to get a length. It does notmatter that the measurements are made atdifferent times. Alternatively we could geta friend and ask them to stand at the rightend while we stand at the left end, and eachrecord the positions, at the same time or dif-ferent times.
If the rigid object is at rest with respect
to us we say that we measure its properlength.
What will be the length of the object as mea-sured by intelligent observers in Ralph’s ref-erence frame?
Suppose that Louise sees a rigid object ofproper length LLouise and sees Ralph moveto the right at speed β. The events of inter-est are E1: Ralph at left end of object andE2: Ralph at right end of object.
What is the time interval between theevents? Louise uses numbers obtained in herreference frame and calculates the time ofthe trip to be
∆tLouise =LLouise
β(2.7)
Ralph says that Louise is moving to the leftat −β, and that he measures the properinterval between the two events (Ralph ispresent at both) of ∆tRalph. In this timeLouise has moved a magnitude of
LRalph = | − β∆tRalph| (2.8)
2.8 Lengths of Objects Along Direction of Motion 29
But we know from the time dilation formulathat ∆tLouise = γ∆tRalph. Combining Equa-tions 2.7 and 2.8 we get the length contrac-tion formula,
LRalph =1
γLLouise (2.9)
This formula says that moving rods aremeasured to be shorter than their restlength.
Example An electron enters a tube travel-ing at β = 0.90. In the lab the tube is100 m long.
(a) What is the length of the tube asmeasured by an observer movingwith the electron?
The tube is at rest in thelab, so the lab measures properlength. We first calculate γ =1/√
1− 0.92 = 2.294.
Then LImproper = 100 m/2.294 =43.6 m.
(b) For the lab, how long does it takefor the electron to pass through thetube?
∆tLab = LLab/β = (100 m)/0.9 =111 m
(c) For the electron, how long doesit take for the tube to pass com-pletely over it?
∆telectron = Lelectron/β =(43.6 m)/0.9 = 48.4 m
I can check using the relation be-tween proper and improper time:
∆tLab = γ∆telectron =(2.294)(48.4) = 111 m It agreeswith (b)
Example Louise in the lab has a tube atrest with respect to her that she mea-sures to be 60.0 m long. She sees Al-bert moving. Albert measures the timefor the tube to pass over him to be 0.267µs. What is the relative speed of Albertand Louise?
Albert measures ∆tAlbert = 0.267µs. Isthis proper or improper time?
Louise measures LLouise = 60 m. Is thisproper or improper length?
A naıve approach would be to just di-vide these two numbers,
vWrong = 60.0 m/0.267 µ s = 2.25×108
m/s, i.e. βWrong = 0.75.
This is wrong because we are usingnumbers from different observers. Wemust use numbers from a single ob-server.
Since β is dimensionless, we convertthe time into meters, ∆tAlbert = (3 ×108 m/s)(0.267× 10−6 s) = 80.0 m.
We can either compute β asβ = LAlbert/∆tAlbert or β =LLouise/∆tLouise. I’ll do it withLouise. To get the time interval mea-sured by Louise and her assistants, weuse ∆tLouise = γ∆tAlbert.
30 2.9 Simultaneity
Then
β =LLouise
∆tLouise
=60 m
γ 80.0 m
=
√(1− β2)(60.0m)
80.0 m)(2.10)
Square both sides and rearrange to get
β2 =(60/80)2
1 + (60/80)2= 0.360 (2.11)
so that β = 0.600
How do we actually measure the length ofa moving object? It is important to under-stand this clearly. Consider the object atrest as shown in Figure 2.5(a). Events E1and E3 are for recording the position of theleft end of the rod, while events E2 and E4are for recording the position of the rightend of the rod. While the times will be dif-ferent for E1 and E3, the position will be thesame. We can use either of the even eventscombined with either of the odd events todetermine the length. In symbols
Lrest = (x4−x1) = (x2−x3) = (x4−x3) = (x2−x1)(2.12)
Now consider a moving rod, Figure 2.5(b).From the diagram it should be clear that thelength of the moving rod is
Lmoving = (x2 − x3) = (x4 − x5) (2.13)
To measure the length of a moving ob-ject we must measure the two endsat the same time, i.e. simultane-ously.
Space intervals like (x3 − x1) and (x4 − x1)are certainly valid intervals, and may be ofinterest in specific applications, but they doNOT represent the length of the rod.
Also be sure you know the difference be-tween what we measure, using our friendsthroughout the universe, and what we seebased on light entering our eye. We will dis-cuss this again in terms of space-time dia-grams.
2.9 Simultaneity
Now for the most confusing part of relativity,the relativity of simultaneity. What you willdeduce is that two events that occur at thesame time but at different locations accord-ing to one observer will NOT be simultane-ous for a second observer who moves relativeto the first.
Example Al and his friends Aaron andAndy are standing equally spaced alongthe ground, with Aaron to the left, thenAl, then Andy on the right, as shownin Figure 2.6. They are at rest witheach other, and 50 meters apart. Im-mediately in front of Al is a bulb that isflashed at t = 0. This is event E1. Sometime later Aaron sees the flash, eventE2, and Andy sees the flash, event E3.
(a) On the top half of engineering pa-per draw a space-time diagram asseen by Al. Show labeled world-lines for each of the three men, andfor the light that moves left andright from the bulb. Use a dashedline for the worldline of light. La-bel the events E1, E2, and E3.
2.9 Simultaneity 31
Figure 2.5: Event diagrams for measuring the length of a rod (a) at rest (b) moving.Various events are marked.
(b) According to Al, who sees the lightfrom the bulb first, Aaron Andy, orthey see it at the same time? Ex-plain with reference to your space-time diagram..
(c) At the time that Al flashed thebulb, Beth was at his location andmoving to the right at β = 0.6.Draw and label the worldline forBeth on the spacetime diagram.
(d) Now draw an event diagram asseen by Beth on the bottom halfof the paper. Here are the steps.
Calculate γ in this case, and thenfind the separation of Aaron, Al,and Andy as seen by Beth. Usethe length contraction formula.
(e) Draw and label the worldlines forBeth, Al, Aaron, and Andy on thediagram.
(f) Finally, add worldlines for the lightthat moves left and right. Labelevents E1, E2, and E3 on the dia-gram.
(g) According to Beth, who sees thelight from the bulb first, AaronAndy, or they see it at the sametime? Explain.
What is up with this? Can it really betrue? Could we actually imagine measuringthis?
Example Alan, Andy, Alex, and the usualbunch of guys are riding on SpaceshipAlpha. Beth, Beatrix, Betsy, and therest of the ladies ride on Spaceship Beta.(Will they ever meet?) Alan and Bethstand at the center of their respectiveships. The spaceships may or may notbe the same length—you will find outshortly. The ships are transparent solight can pass through to the observers.
Event E1: When the front end ofAlan’s ship is at the same loca-tion as the rear of Beth’s ship,a spark jumps between them andlight waves spread out. Charmarks from the spark are left onboth spaceships.
Event E2: Likewise when the rear of
32 2.9 Simultaneity
Figure 2.6: Basic Set-up for simultaneity example
Alan’s ship and the front of Beth’sship pass, a spark jumps betweenthem emitting a light pulse, andleaving char marks.
Event E3: Alan is equidistant fromthe char marks left on his space-ship, and receives the pulses oflight from both sparks at thesame time (simultaneously). Fig-ure 2.7 shows this for the instantof time when the wavefronts arriveat Alan.]
A) 1. In Alan’s frame, does spark 1(from event E1) jump before,after, or at the same time asspark 2 (from event E2)? Ex-plain your reasoning.
2. Alan’s friend Andy standsnear Char mark 2 (in theirship). According to Andy,does spark 1 jump before, af-ter, or at the same time as
spark 2 (on the right of pic-ture)? Explain your reason-ing.
3. What does the distance be-tween the char marks in Alan’sship tell you about Beth’sship? Explain your reasoning.
4. The diagram below representsthe situation in Alan’s framea short time after the sparks.Show the wavefronts at thistime.
B) 1. On gridded paper, draw thespace-time diagram as seen byAlan. Time and distance willbe measured in the same units,“squares.” Assume that Alanmeasures the length of Space-ship Alpha to be 20 squares.Alan sees Beth move to theright at β = 0.60. On your di-agram identify events E1, E2,
2.9 Simultaneity 33
Figure 2.7: Wavefronts have just arrived at Alan
Figure 2.8: Sparks have just occurred—not to scale!
E3, and E4, Alan and Beth areexactly opposite each other.
2. Draw and label the world linesfor the char marks and for thewavefronts.
3. What is the time separa-tion, as measured by Alan, insquares, between
Event 1 and Event 2
Event 1 and Event 4
Event 1 and Event 3
You can get the answer eitherfrom algebra or from the dia-gram.
C) 1. Who measures the properlength of Spaceship Alpha—Alan, Beth, both, or neither.Explain
2. Who measures the properlength of Spaceship Beta—Alan, Beth, both, or neither.Explain
3. Alan measures the length of
34 2.9 Simultaneity
Spaceship Alpha to be 20squares. What does Alanmeasure for the length ofSpaceship Beta? Explain.
4. What does Beth measure forthe lengths of Spaceship Al-pha and Spaceship Beta? Ex-plain.
D) Now use another piece of paperand draw the space-time diagramas seen by Beth.
Start about 10 squares up from thebottom of the paper, and draw thediagram corresponding to Event 4.Show the locations of the front,back, and center of both ships,and draw world lines for the front,back, and center of both ships.Also show the world lines of thelight emitted from the sparks.
Extend the lines as necessary anddetermine the times and locationsof events E1, E2, and E3. Markthe events on the diagram.
Using the diagram, and using al-gebra, determine the time separa-tion, in squares, between
Event 1 and Event 2, Event 1 andEvent 4, Event 1 and Event 3
E) Carefully draw an event diagramas seen by Beth—that is draw pic-toral diagrams like those on thefirst page. Time is positive up-wards. Show all 4 events, as well asthe spaceships and the wavefronts.
One Final Example Does the Lack of Si-multaneity make sense? It probably
won’t make sense in terms of being anatural way of thinking, at least notnow, but is it at least consistent? Wehave the same situation as before withAlan (and friend’s) and Beth.
A. A CD burner sits at Beth’s feet. InAlan’s frame, when the wavefrontfrom spark E2 reaches the burner,it starts to burn a legal copy of anEric Clapton CD. When the wave-front from E1 reaches the burner itshuts off. If the wavefronts reachthe burner at the same time, orthe wavefront from E1 reaches theburner first, no music is recorded.
In Alan’s frame, is there any musicburned onto the CD?
In Beth’s frame, is there any mu-sic burned onto the CD? Explainhow your spacetime diagrams areconsistent with this.
B. Is your answer consistent withyour answers to the following ques-tion?
Later in the day the CD is removedfrom the burner, and Beth travelswith it via courier rocket to Alan.They play the CD. Will they hearClapton or not?
C. Alan also has a CD burner that op-erates just like Beth’s. In Alan’sframe, is there any music burnedonto his CD?
In Beth’s frame, is there any musicburned onto Alan’s CD? Explainhow your spacetime diagrams areconsistent with this.
2.10 Summary 35
D. Is your answer consistent withyour answers to the following ques-tion?
Later in the day when Beth hastraveled via courier rocket to Alan,they play Alan’s CD. Will theyhear Clapton or not?
2.10 Summary
Our observers are in two inertial frames withdimensionless relative speed β. and dilationfactor γ = 1/
√1− β2
Proper time intervals are measured by ob-servers present at both events. ∆tImproper =γ∆tProper
Proper lengths of a rigid object are measuredby an observer for whom the object is at rest.LImproper = LProper/γ
Two events that are simultaneous to one ob-server are not simultaneous for a differentobserver.
36 2.10 Summary
Chapter 3
Lorentz Transformations, VelocityTransformations, Doppler Shifts,Approximations
3.1 Mapping with IntervalsOnly
What information do we need to have in or-der to make a space-time diagram? Consider1D space, i.e. all events occur along a singleline. Well, we need to pick a starting point—this can be anywhere we choose, suppose itis Event A with coordinates for Louise ofxLouise = 0 = tLouise.
Now we have another event, Event B. If weknow ∆tLouise and ∆xLouise, then we canindicate where event B is on Louise’s space-time map.
If we do not have ∆tLouise but rather∆tRalph, then the calculation is more diffi-cult. If one of the observers is a proper ob-server of the time interval we can use timedilation to find the improper time interval,then use the invariant space-time intervalto find the space interval. We would needto know the relative speeds of the two ob-servers, β.
There are two other alternatives. We couldset up coordinate systems for the differentobservers and give a way to transform po-sitions and times of events (NOT intervals)measured by one observer to positions andtimes measured by a second observer. Wewill discuss Lorentz Transformations in thenext section.
Another approach is to record the invari-ant space-time intervals between events—not position and time separately— and usethose invariant quantities to map the events.We will defer detailed discussion of this, butwill mention the equivalent situation for thecase of the Parable Surveyors of Chapter1.
Suppose we know the distance between allpoints in our world, but NOT the directions.Can we make an accurate map?
The invariant quantity for our two surveyorsAnahinda and Nyx was the distance betweenpoints. Suppose we have the following data.
37
38 3.2 The Lorentz Transformation
To → Roch- Phila- Boston Mon-From↓ ester delphia treal
Roch- 0 254 335 329esterPhila- 0 273 461delphiaBoston 0 283Mon- 0treal
Table 3.1: Distances in miles between cities.Notice that no directions are given.
If the distance between two points is d, andwe use one point as the origin, then the otherpoint lies somewhere on a circle of radius d =√x2 + y2. Circles are easy to draw.
Let’s arbitrarily pick Rochester as the ori-gin. Philadelphia is 254 miles away. sowe draw an arc of a circle of radius 2.54inches (Scale factor 1 inch = 100 miles)and mark Philadelphia somewhere on thatarc (it doesn’t really matter where on thearc.)
Now to locate Montreal. Draw an arc of ra-dius 3.29 inches (329 miles) with Rochesteras the center, and an arc of radius 4.61 inches(461 miles) with Philadelphia as the center.The two arcs intersect in two locations, andwe arbitrarily pick one of them and mark thelocation of Montreal.
Finally to locate Boston, draw arcs withRochester (radius = 3.29 inches), Philadel-phia (radius= 2.73 inches), and Montreal(radius = 2.83 inches) The three circles meetat a single point, Boston.
The procedure is shown in Figure 3.1.
What would have happened if we had used
the other intersection of arcs to locate Mon-treal? The resulting map would be topo-logically equivalent—if we located Montrealabove Rochester, and looked at the map inthe mirror it would look the same as Figure3.1.
In the case of Special Relativity the in-variant is the space-time interval, STI =√
∆t2 −∆x2, and this is the equation of anhyperbola. If we had a table of several eventsand the STI between all pairs of events wecould draw a space-time diagram by usinghyperbolas rather than circles . These aremuch harder to draw, but conceptually theidea is the same as for the Parable of theSurveyors.
3.2 The Lorentz Transfor-mation
Often it is much simpler to define our refer-ence system, including directions, right atthe start, and then record positions andtimes of the events. From the positions andtimes it is easy to get the space and timeintervals as well as the invariant space-timeinterval between any two events.
We consider two observers Ursula the Un-primed and Pete the Primed who are bothare in inertial reference frames. All mea-surements made by Ursula will be unprimed(x, y, z, t, β) and all measurements made byPete will be primed (x′, y′, z′, t′, β′).
Since Pete and Ursula are both in iner-tial reference frames, they must have a con-stant relative velocity—size and direction—and we choose to call this line of relativemotion the x-axis, with the x′-axis parallel
3.2 The Lorentz Transformation 39
Figure 3.1: Making a map knowing only the distances between cities. The arcs are labeledwith the two cities involved, e.g. RP = Rochester-Philadelphia distance.
to this. Perpendicular to x are y and z axes(and parallel y′ and z′ axes).
According to Ursula, Pete is moving with avelocity β along the x-axis. If β is positive,Pete is moving to the right, if β is negativePete is moving to the left.
If Ursula sees Pete move to the right, Petesees Ursula move to the left. We can say thatPete sees Ursula move with velocity alongthe x-axis of
β′ = −β. (3.1)
Ursula has an origin somewhere, as doesPete. We ask Pete to choose the origin ofhis coordinate system to lie along Ursula’sx-axis. (It is moving of course, accordingto Ursula. According to Pete it is Ursulaand her coordinate system that are moving.)Let us define the event E0 = Origins coin-cide.
Let’s also ask Pete and Ursula to set theirrespective clocks to a time of zero at this
event. Since we are talking about a singleevent, we can do all of these things withoutworrying about relativity. Event diagramsas seen by Ursula and Pete are shown in Fig-ure 3.2.
The discussion so far means that an eventoccurring at the origin at time zero has thesame coordinates in both reference frames.That is
x0 = y0 = z0 = t0 = 0 = x′0 = y′0 = z′0 = t′0(3.2)
Suppose Ursula and Pete look at a ballthat can be moving and accelerating. Asingle event has coordinates, (x, y, z, t) and(x′, y′, z′, t′) as measured by Ursula andPete. We want a method to take Ur-sula’s results and figure out Pete’s results,a way to transform from one observer to theother.
Consider the directions y and z that aretransverse to the motion. In Chapter 2 we
40 3.2 The Lorentz Transformation
Figure 3.2: Coordinate systems of Ursula the Unprimed and Pete the Primed as seen by(a)Ursula and (b) Pete. They are drawn as event diagrams with time increasing as youmove upwards. The lower diagrams are for t = t′ = 0, and the systems are shown with aslight offset so that you can see both.
argued that relative motion will not changelengths in the transverse direction, and thismeans
y′ = y and z′ = z (3.3)
To get the transformations for x and t willtake more work. First lets look at the non-relativistic transformation between twoobservers—i.e. Galilean relativity. The twoobservers will have coordinate systems thatcoincide at t = 0 and they both look ata ball. At time t this is shown in Figure3.3.
From Figure 3.3 we can easily see thatthe NON-RELATIVISTIC transformationis
t′1 = t1
x′1 = x1 − βt1y′1 = y1 Non− relativistic (3.4)
All of these equations are linear relations,that is x, y, and t occur raised to the firstpower only, and we never see cross termscontaining the products like x t. We willrequire this of our relativistic transformsalso.
t = B x′ +D t′ (3.5)
x = G x′ +H t′ (3.6)
Situation 1: Consider these two events, E1:Pete sends a spark at x′ = 0, t′ = 0. Ursulawill record x = 0, t = 0 for this event. E2:At a later time Pete sends a second spark,x′ = 0, t′ = t′.
Pete sees the proper time interval betweenthese events, ∆t′ = t′ − 0. Ursula sees animproper time interval ∆t = t − 0 = γ ∆t′.Hence for this situation t = γ t′ and we cansay that D = γ in Equation 3.5.
Also we can write the location of E2 for Ur-
3.2 The Lorentz Transformation 41
Figure 3.3: Galilean (“non-relativistic”) Transformation
sula, x = β t and substituting t = γ t′,we have x = β γ t′. This expression isa mixed expression—on the right we havetime measured by Pete, but relative veloc-ity measured by Ursula. Use Equation 3.1to make all terms on the right quantitiesmeasured by Pete to get for this situationx = −β′ γ t′, meaning that in Equation 3.6,H = −β′ γ.
Situation 2: To get the other two constantswe consider a third event E3 that has coor-dinates (x, t) for Ursula and (x′, t′) for Pete.Consider the interval for E1 to E3. Then∆x = x − 0, and ∆t = t − 0. We use ourestablished fact that the space-time intervalis invariant to write
∆t2 −∆x2 = ∆t′2 −∆x′2 (3.7)
and using Equations 3.5 and 3.6,
(B x′ + γ t′)2 − (G x′ − β′ γ t′)2 = t′2 − x′2(3.8)
Expand the squares and collect like terms toget (
γ2(1− β′2))t′2 + 2 γ(B +G β′)x′t′
+ (B2 −G2)x′2
= t′2 − x′2 (3.9)
In the first term we have(γ2(1− β′2)
)= 1
by the definition of γ, and we can cancel thet′2 from both sides of the equation. HenceEquation 3.9 becomes
2 γ(B +G β′)x′t′ + (B2 −G2)x′2 = −x′2(3.10)
This is not an ordinary equation that is truefor one particular choice of (x, t) but must betrue for all values1 of (x, t) chosen indepen-dently of each other. This means that
0 = 2 γ(B +G β′)
B = −G β′ (3.11)
and that(B2 −G2) = −1 (3.12)
Combining Equations 3.11 and 3.12 we endup with G = γ,B = −γ β′ making our finaltransformation equations
t = γ(t′ − β′ x′) (3.13)
andx = γ(x′ − β′ t′) (3.14)
Why are my Lorentz transforma-tions different than the ones in thetext?
1“For all” is denoted by ∀ in mathematics.
42 3.3 Using the Lorentz Transformation
The text chooses to only use β, Ursula’smeasure of the relative velocity, while I useboth β and β′. The text must then definetwo transformations, the “Lorentz Transfor-mation” and the “Inverse Lorentz Transfor-mation”. By using the two relative velocitiesβ and β′, I need only one transformation.
Table 3.2: Text and Lindberg Transforma-tions Compared
Text Lindberg
L10a t = γ β x′ + γ t′ 3.13 t = γ t′ − γ β′ x′x = γ x′ + γ β t′ 3.14 x = γ x′ − γ β′ t′
L11a t′ = −γ β x+ γ t 3.13 t′ = γ t− γ β xx′ = γ x− γ β t 3.14 x′ = γ x− γ β t
In my version of the transformations, there isno change of sign between the forward andinverse transformations, all I need to do iskeep all unprimed variables together, and allprimed variables together. Also space andtime appear symmetrically so that it is easyto get the second equation from the first justby switching the roles of x and t.
At some time you may want to de-fine an Excel function lorentz(beta,a,b)
that does the transformation2. Onlyone function is need in my formula-tion, since t′ = lorentz(beta, t, x), x′ =lorentz(beta, x, t), t = lorentz(beta′, t′, x′),x = lorentz(beta′, x′, t′),
2The same form of transformation transforms en-ergy and momentum from one frame to another.
3.3 Using the Lorentz Trans-formation
Example Ursula measures the coordinatesof E1 to be x1 = 100, t1 = 500 m. Petemeasures the coordinates of E2 to bex′2 = 600, t′2 = 900 m. Ursula sees Petemove to the left at 0.60.
(a) Find the coordinates of E1 as seenby Pete.
Using the conventional choice ofright as being positive, we haveβ = −0.60 and γ = 1.25.
Then x′1 = (1.25)(100) −(1.25)(−.60)(500) = 500 mand
t′1 = (1.25)(500) −(1.25)(−.60)(100) = 700 m
(b) Find the coordinates of E2 as mea-sured by Ursula.
β′ = −β = +0.60
So x2 = (1.25)(600) −(1.25)(0.60)(900) = 75 m
and t2 = (1.25)(900) −(1.25)(0.60)(600) = 675 m
(c) Find the space interval, the timeinterval, and the space-time inter-val between E1 and E2 as mea-sured by Ursula.
∆x = (75)− (100) = −25 m,∆t = (675)− (500) = 175 m, and
STI =√
1752 − (−25)2 = 173.2m
3.3 Using the Lorentz Transformation 43
(d) Find the space interval, the timeinterval, and the space-time inter-val between E1 and E2 as mea-sured by Pete.
∆x′ = (600) − (500) = 100 m,∆t = (900)− (700) = 200 m, and
STI =√
2002 − 1002 = 173.2 m
Notice that the space and time in-tervals for the two observers aredifferent, but the STI is the same.
(e) Are either Pete or Ursula theproper observer for the time inter-val between the events? If neitherof these are observers of the propertime interval, who is?
To observe the proper time in-terval, ∆x = 0, and this is notthe case for either Ursula or Pete.Consider Daisy as the proper ob-server. For Daisy, ∆x′′ = 0. Callthe velocity of Daisy relative to Ur-sula, β′′ and use the Lorentz trans-formations for Events E1 and E2.
x′′1 = γD(x1 − βD t1) = γD(100 −βD 500)
x′′2 = γD(x2 − βD t2) = γD(75 −βD 675)
Equate the two, since the eventsoccur at the same location for theobserver of proper time.The γD’scancel, leaving an equation that iseasy to solve for βD,
(100− βD 500) = (75− βD 675)
βD = −25/175 = −0.143.
Daisy must move to the left at a
speed of 0.143 in order that E1and E2 occur at the same loca-tion for her and she will be theproper observer for the time in-terval between these two events.The proper time interval is (∆t′′ =√
∆t2 −∆x2 = 173.2 m.
(f) What is the velocity of Daisy asmeasured by Pete?
We can use the same approach asthe previous part, but using Pete’sdata.
(500− β′D 700) = (600− β′D 900)
β′D = 100/200 = 0.500.
Example Imaginary Space-Time Interval?
Ursula sees event E3 with x3 =500, t3 = 700 m and E4 with x4 =688, t4 = 813 m. She sees Pete moveto the right at 0.60.
(a) What does Pete see for coordinatesof E3?
x3 = 1.25(500 − 0.60(700)) = 100m, t3 = 1.25(700 − 0.60(500)) =500 m
(b) What does Pete see for coordi-nates of E4? x4 = 1.25(688 −0.60(813)) = 250 m, t4 =1.25(813− 0.60(688)) = 500 m
(c) Calculate the space-time intervalbetween E3 and E4.
∆t = 813 − 700 = 113 m, ∆x =688− 500 = 188 m
So STI =√
1132 − 1882 =√−22575 = (150 ı) m, an imag-
inary answer. What does this
44 3.4 The Velocity Transformation
mean? Certainly the events canoccur wherever they want to, butthen some space-time intervals willbe real, some imaginary, and somezero.
We will discuss this further inthe next chapter, and describetime-like, space-like, and light-likespace-time intervals
Example Ursula stands at the rear of a shipthat she measures to be L = 100 m longwith her friend Ulla at the front. Theywatch Pete fly by moving to the left at0.60; Pete is at the front of his ship andPaul is at the rear. At t = 0 m Ursulameasures two events, with the help ofher friend Ulla. E1: At x = 0 m a sparkis sent from the front of Pete’s ship tothe rear of Ursula’s ship. At x = L =100 m a spark is sent from the rear ofPete’s ship to the front of Ursula’s ship.
(a) According to Ursula, how long isher ship? Is this a proper length?
(b) Draw a space-time diagram as seenby Ursula, and on it mark the twoevents and appropriate world lines.
(c) Use the Lorentz transformation tofind the x′ and t′ coordinates of thetwo events as measured by Pete.
(d) According to Ursula, how long isPete’s ship? Is this a properlength?
(e) According to Pete, how long is Ur-sula’s ship?
(f) According to Pete, how long is hisship?
(g) Draw a space-time diagram as seenby Pete. On it mark the twoevents.
(h) Who sees the proper time betweenthe two events, Pete or Ursula?
(i) Draw an event diagram (picture)of the events as seen by Pete.
(j) Ursula would say “The spark be-tween Pete and me occurs simul-taneously with the spark betweenPaul and Ulla.” What would Petesay about these two events?
(k) According to Pete, simultaneouswith the spark between Ulla andPaul, Ursula sends a spark. Thisis event E3. Find the coordinatesof this event according to Pete.
(l) On the two space-time diagrams,mark E3.
(m) Using each space-time diagrams,determine at what fraction of thelength of Pete’s ship the char markfrom the spark in E3 occurs. Dothe two diagrams give the same re-sult?
3.4 The Velocity Transfor-mation
Ursula and Pete are watching a third rocketin which Ted is riding. Ursula measures thevelocity of Pete, Pete measures the velocityof Ted. From these two velocities we wouldlike to know what result Ursula would mea-sure for the velocity of Ted.
3.4 The Velocity Transformation 45
Notation: We will use v for velocity, butmust keep track of who is doing the mea-suring. A couple of notations are useful.We could add two subscripts, one for theobject that is moving, and one for the ob-server. Or we could use a superscript, nonefor Ursula’s measurement, and a prime, v′
for Pete’s measurement. To indicate thevelocity of Ted as measured by Ursula wecould write vTU = vT . For the velocityof Ted as measured by Pete we could usevTP = v′T .
If Ted is the only object being measured wecould omit the T subscript entirely. Eventu-ally we will look in 2D and 3D and will needanother subscript for the component of thevelocity.
The relation between velocities measured byUrsula and Pete is known by a variety ofnames: Velocity Addition Formula, Law ofCombination of Velocities, or my choice, Ve-locity Transformation.
I’ll begin with the non-relativistic GalileanTransformation.
3.4.1 Non-Relativistic GalileanTransformation of Veloci-ties
Start with Equation 3.4, x′ = x−βt. Here isa non-calculus derivation. An object movesin 1D from position x1 to x2 in some timeinterval ∆t.
Write the coordinates of two different eventsas measured by Pete,
x′2 = x2 − β t2 (3.15)
x′1 = x1 − β t1 (3.16)
and subtract
∆x′ = ∆x− β ∆t (3.17)
Now divide by the time interval,
∆x′
∆t=
∆x
∆t− β (3.18)
or
v′ = v − β Non− Relativistic (3.19)
Or we can start with Equation 3.4 and dothe derivative from calculus
dx′
dt=dx
dt− β (3.20)
This gives the same result, Equation3.19.
So if Ursula measures Pete fly past in anairplane at 90 m/s east and Pete measures aball thrown at 5 m/s east, then Ursula knowsthat she would measure the ball moving at95 m/s east. This works great as long as thespeeds are small compared with the speed oflight.
3.4.2 Vectors and Compo-nents
We have been discussing the position of anobject in 2D and have given its coordinates,relative to an origin, as x and y. It is alsopossible to give the position in terms of adistance and an angle, r and θ. The rela-tion between the two ways of presenting thedata are shown in Figure 6.1, and in symbolsare
r =√x2 + y2 (3.21)
46 3.4 The Velocity Transformation
Figure 3.4: A two dimensional vector can bedescribed either by components, x, y, or bylength and angle, r, θ
θ = tan−1 y
x(3.22)
If we know the distance and the anglewe can use trigonometry to get the compo-nents.
x = r cos θ (3.23)
y = r sin θ (3.24)
The position is an example of a vector. Sim-ilar expressions are true for the vector veloc-ity, we can give its components, vx, vy or itsspeed and angle v, θ where
v =√v2x + v2
y (3.25)
θ = tan−1 vyvx
(3.26)
and
vx = v cos θ (3.27)
vy = v sin θ (3.28)
3.4.3 Relativistic Velocity Trans-formation
To determine the x component of velocityof Ted we can use vx,ave = ∆x/∆t and
take small time intervals so the average isthe instantaneous velocity. First use theLorentz transformations to get expressionsfor the space and time intervals between twoevents.
x2 = γ(x′2 − β′ t′2)
x1 = γ(x′1 − β′ t′1)
∆x = γ(∆x′ − β′ ∆t′) (3.29)
t2 = γ(t′2 − β′ x′2)
t1 = γ(t′1 − β′ x′1)
∆t = γ(∆t′ − β′ ∆x′) (3.30)
Now the velocity in Ursula’s reference framecan be written
vx =∆x
∆t=γ(∆x′ − β′ ∆t′)γ(∆t′ − β′ ∆x′)
=∆x′/∆t′ − β′ ∆t′/∆t′)∆t′/∆t′ − β′ ∆x′/∆t′)
=v′x − β′
1− β′v′x(3.31)
We can similarly get the transformation forthe transverse velocity components, using∆y = ∆y′.
vy =∆y
∆t=
∆y′
γ(∆t′ − β′ ∆x′)
=∆y′/∆t′
γ(∆t′/∆t′ − β′ ∆x′/∆t′)
=v′y
γ(1− β′v′x)(3.32)
Similarly
vz =v′z
γ(1− β′v′x)(3.33)
3.5 Using Velocity Transformations 47
The velocity transformation for x-components involves only the relativemotion (assumed to be in x direction) andthe x-components of velocity. Howeverthe transverse velocities involve velocitycomponents in both the transverse direction(y or z) and the velocities along the xdirection, β and vx. This will lead tocurious results.
Differences between my VelocityTransformation and that of SpacetimePhysics.
Equation L-13 in Spacetime Physics looksdifferent than my Equation 3.31. This is be-cause my formula uses only Pete’s primedmeasurements on the right rather than mix-ing up Pete’s primed and Ursula’s unprimedmeasurements.
Notice that by using my notation there is noneed for an inverse transformation, you justreplace primed quantities with un-primedquantities and vice versa. This is summa-rized in Table 3.3.
I would suggest writing Excel for-mulas for the velocity transfor-mations, vxprime(beta, vx) andvyprime(beta, vx, vy).
3.5 Using Velocity Transfor-mations
The key to using Equations 3.31 and 3.32 isto carefully determine which are the primedand which are the unprimed velocities, aswell as to remember that β′ = −β. The firstset of examples will be for 1D motion.
Example Ursula measures Pete moving tothe left at a speed of 0.60 and Ted mov-ing to the right at a speed of 0.50. Whatis the velocity of Ted according to Pete?
β = −0.60 vx = +0.50
Then v′x = vx−β1−βvx = 0.50−(−0.60)
1−(0.50)(−0.60) =1.11.3 = 0.85
Pete measures Ted moving to the rightat 0.85.
Example Ursula measures Pete moving tothe left at 0.60. Pete measures Tedmoving to the right at 0.50. What doesUrsula measure for the velocity of Ted?
β = −0.60 β′ = +0.60 v′x = +0.50
vx = v′x−β′1−β′v′x
= 0.50−0.601−(0.50)(0.60) = −0.14
Ursula sees Ted moving to the left at0.14.
Example Ursula measures Pete moving (a)to the right at 0.95 (b) to the left at0.95. Pete turns on a flashlight andmeasures the light moving to the rightat 1.00. What does Ursula measure forthe velocity of the light?
(a) β = +0.95 β′ = −0.95 v′x =+1.00
vx = v′x−β′1−β′v′x
= 1.00−(−0.95)1−(−0.95)(1.00) =
+1.00
Ursula also sees the light movingto the right at 1.00.
(b) β = −0.95 β′ = +0.95 v′x =+1.00
vx = v′x−β′1−β′v′x
= 1.00−0.951−(0.95)(1.00) =
+1.00
48 3.5 Using Velocity Transformations
Table 3.3: Comparing Velocity Transformation Equations: vz transformations are like thevy transformations
Equation Text Equation Lindberg
x-component L-13 vx = v′x+β1+v′xβ
3.31 vx = v′x−β′1−β′v′x
Inverse x v′x = vx−β1−vxβ 3.31 v′x = vx−β
1−βvx
vy Not given 3.32 vy =v′y
γ(1−β′v′x)
v′y Not given 3.32 v′y =vy
γ(1−βvx)
Ursula still sees the light movingto the right at 1.00.
These examples give us a hint that thespeed of light in vacuum seems to be thefastest possible speed that any object canhave.
Now consider 2D motion, with the trans-verse component being the y-axis. Ursulaand Pete are still moving relative to eachother along the x-axis, it is Ted that is mov-ing in 2D.
Example Ursula measures Pete moving tothe left at 0.60. Pete shines a flash-light and measures the beam moving at1.00 at an angle of 37◦ to the x-axis.What is the velocity—size and angle(direction)—that Ursula measures forthe light?
β = −0.60 β′ = +0.60 γ = 1.25
v′x = +1.00 cos 37◦ = +0.799
v′y = +1.00 sin 37◦ = +0.602
vx = v′x−β′1−β′v′x
= 0.799−0.6001−(0.799)(0.600) = 0.382
vy =v′y
γ(1−β′v′x) = 0.6021.25(1−(0.60)(0.799) =
0.925
This means that v =√v2x + v2
y =√
0.3822 + 0.9252 = 1.00
Light still moves at the speed of light inagreement with Einstein’s Second Pos-tulate. The angle with respect to thex-axis is
θ = tan−1 vyvx
= tan−1 0.9250.382 =
tan−1 2.42 = 67.6◦
Example Ursula measures Pete moving tothe right at 0.80 and Ted moving at 0.50at 30◦ above the horizontal. What doesPete measure for Ted’s velocity?
β = +0.80 β′ = −0.80 γ = 1.667
vx = 0.50 cos 30◦ = +0.433
vy = +0.50 sin 30◦ = +0.250
v′x = vx−β1−βvx = 0.433−0.800
1−(0.443)(0.800) = −0.561
v′y =vy
γ(1−βvx) = 0.2501.667(1−(0.80)(0.433) =
0.229
This means that v′ =√v′2x + v′2y =√
(−0.5612 + 0.2292 = 0.606
3.6 Doppler Shifts 49
at an angle θ = tan−1 vyvx
=
tan−1 0.229−0.561 = 157.7◦
3.6 Doppler Shifts
Imagine Pete carrying with him a periodicsource of pulses—this could be a tuning forkin the case of sound, or a laser in the caseof light. Pete measures the period and fre-quency of the source, T = 1/f . Ursula seesPete moving, and we will consider motion tobe straight toward or straight away from Ur-sula. When Ursula receives the pulses, whatis the received frequency?
We will define the following conventional pic-ture. The source is on the left and the ob-server is on the right. Velocities to the rightare considered to be positive, velocities tothe left are considered to be negative.
Figure 3.5: We use this for our conventionalpicture for the Doppler effect. The sourceis to the left and the observer to the right.Velocities to the right are positive, to the leftare negative.
3.6.1 Dopper Equation forSound
You may have heard of the Doppler effectfor sound. When a sound source approachesyou, you hear a higher frequency, when thesound source recedes, you hear a lower fre-quency. The equation is
fobserved =1− vobserver/vsound1− vsource/vsound
for sound
where vsound is the speed of the sound.
This formula works very well since soundmoves in a medium, typically air. For along time scientists believed in the ether(also spelled æther), a medium in whichlight was assumed to propagate. Duringthe 19th century this ether came to needstranger and stranger properties—it had tobe rigid enough to let light propagate, butfluid enough to let the earth orbit the sun.The ether needed to be stationary in one spe-cial reference frame, that where Maxwell’sequations were perfectly correct.
Michelson and Morley did an experimentin 1887 in Cleveland Ohio to determinethe speed of the earth with respect to theether. They found no result, one of the firstpieces of evidence that led to the demise ofthe ether, and the rise of Special Relativ-ity.
3.6.2 One-Dimensional Relativis-tic Doppler Effect
You can follow the following outline to derivethe One-Dimensional Relativistic DopplerEffect. We will assume that Ursula is at the
50 3.6 Doppler Shifts
origin of her coordinate system and begin-ning at t = 0 sends pulses of light to theright separated by a period T = 1/fs = 2m. For Ursula, the time T is a proper timebetween her sending pulses, and thereforeshe measures the proper frequency, fs, of thesource.
At t = 0 Ursula measures Pete to be atxinit = 5 m and moving with speed β = 0.60to the right.
(a) Start by drawing a space-time diagramusing 1 small square as 1 m. Show theworld lines for Ursula and Pete. Thendraw the world lines for first three lightpulses that Ursula sends.
Label the events that correspond to Petereceiving the pulses E0, E1, and E2.The coordinates for these events are(t0, x0), (t1, x1), (t2, x2).
(b) Consider the light pulses. For the firstpulse traveling at the speed of light,1.00, x0 = t0. Write similar equationsfor x1 and x2, then generalize to the N th
pulse. Your answers should include Tand N .
(c) Now consider Pete. He started at xinitand received the first pulse at t0, so wecan write x0 = xinit+β t0. Write similarexpressions for the positions x1 and x2.
(d) Equating the two expressions for x0 wecan solve for t0.
t0 = xinit + β t0 (3.34)
t0 =xinit1− β
and so
x0 =xinit1− β
(3.35)
Repeat this process to get expressionsfor t1, t2, x1, and x2, then extrapolate fortN and xN . The expressions can containβ, xinit, N , and T .
(e) Now lets switch to Pete’s viewpoint. UseLorentz transformations to get the timest′0 and t′N , and subtract to get the inter-val between receiving pulse 0 and pulseN as measured by Pete. This differenceis N T ′.
(f) Use the fact that γ = 1√1−β2
=
1√(1−β)(1+β)
to simplify the expression,
and switch from period to frequency toget
fobserved =
(√1− β1 + β
)fs (3.36)
This equation relates the source frequency fs(where the light is generated), measured fora source at rest, to the frequency fobservedmeasured by an observer moving with re-spect to the source. The source frequency,fs, is the proper frequency measured by thesource, and the observed frequency, fobserved,is the proper frequency measured by the ob-server. The relative velocity β in the equa-tion is the speed of the source as measuredby the observer, and is positive when theobserver moves away from the source (or thesource moves away from the receiver.).
When you use Equation ??, it is easy tocheck that you used the correct sign for β.If the source and observer are approaching,fobserved > fs while if they are receding,fobserved < fs.
If the source is moving, the same equationis valid and its use will be shown in the last
3.6 Doppler Shifts 51
example below.
Example Ursula carries a green laserthat has a wavelength of 532 nmand a frequency fs = c/λ =(3× 108 m/s
)/(532× 10−9 m
)=
5.64 × 1014 Hz. Pete is moving awayfrom Ursula at a speed of 0.10. Whatare the frequency and wavelengthmeasured by Pete?
fobserved =
(√1− .10
1 + .10
)5.64× 1014
= 5.10× 1014 Hz
λobserved = c/fobserved = 588 nm
Pete sees the light as being red. Thisis an example of red-shift. If Pete andUrsula were approaching the frequencywould increase, the wavelength woulddecrease, and we would call it blue-shifted light.
Example If we observe a frequency 1/3 ofthe emitted frequency, (a) is the sourceapproaching or receding and (b) what isthe speed?
(a) Since the frequency is lower we mustbe receding.
(b) Use Equation 3.36 to get β = +0.80.The positive sign confirms the directionof motion.
Example If we observe a frequency threetimes (3×) the emitted frequency, (a) isthe source approaching or receding and(b) what is the speed?
(a) Since frequency is increasing, we areapproaching the source.
(b) Use Equation 3.36 to get β = −0.80.The negative sign confirms the directionof motion.
Example Jack sees Paris moving to theright at 0.60 and George moving to theright at 0.20, with George to the rightof Paris. Paris is carrying a laser withwavelength of 700 nm. What is thewavelength measured by George?
Both are moving, and are moving closer.Therefore we know that the frequencyobserved by George is higher than thefrequency emitted. We also know thatc = f λ and with wavelengths Equation3.36 becomes
λobserved =
(√1 + β
1− β
)λs (3.37)
and if the observed frequency is higher,the observed wavelength is shorter.
We must use the velocity transforma-tion equations to find β, the velocity ofParis seen by George.
β =0.60− 0.20
1− 0.60(0.20)= 0.455 (3.38)
Then we get the shorter wavelength by usinga negative sign with β.
λobserved =
(√1 + β
1− β
)
λs =
(√1 + (−0.455)
1− (−0.455)
)700
= 428 nm. (3.39)
52 3.6 Doppler Shifts
3.6.3 Redshifts
If we know the frequency of the source inits reference frame, we can use the DopplerEffect to find speeds of approach or reces-sion. Let’s apply this to stars. It is verydifficult to bring a known laser to even thenearest star, so how can we know the sourcefrequency?
Fortunately nature gives us this informationin terms of the emission spectra (and ab-sorption spectra) of elements. For example,prominent lines emitted by Hydrogen (theBalmer series) are red, Hα, 656 nm, bluegreen, Hβ, 486 nm, and two violet lines at434 nm, Hγ, and 410 nm, Hδ. The rela-tive spacing of the wavelengths identifies thesource as hydrogen, even if the wavelengthsare Doppler shifted far away from the visi-ble.
Astronomers define the red shift as
zredshift =λobserved − λemitted
λemitted
=femitted − fobserved
fobserved(3.40)
Figure 3.6 shows the visible Balmer lines asvertical bars, and the observed spectrum ofquasar 3C 273.3 The peaks in the observedspectrum are shifted toward longer wave-lengths as are shown by the arrows. Thecomputed red shift for this quasar is 0.158and that translates to a speed of 0.146. Ifwe look closely at an active galaxy such asM87, the nucleus will be rotating at a high
3This is one of the nearest quasars, discovered in1959, and is now known to be the active galactic nu-clei surrounding a massive black hole. For more in-formation see http://en.wikipedia.org/wiki/3C_
273
Figure 3.7: Red and blue shifts at the centerof a rotating galactic core. These are shiftsrelative to the overall red- or blue-shift dueto the overall motion of the galaxy.
speed. The overall galaxy may be red shiftedby some average amount, but the part of thegalaxy approaching us will be blue shiftedrelative to the average, while the part of thegalaxy receding from us will be red shiftedrelative to the average. This is shown in Fig-ure 3.7.
Red shifts are also caused by the expan-sion of the universe and by strong gravita-tional fields (General Relativity), and thesebecome important in the interpretation ofsome red-shift data. We will only considerred shifts due to Doppler effects.
Most stars and galaxies in the universe arered-shifted, but some nearby galaxies likeAndromeda are blue-shifted. Andromedahas a velocity of -0.001 and a z-shift of -
3.7 Approximations 53
0.001.
Example Find the red shift for a star (a)approaching at 0.40 (b) receding at0.90.
(a) β = −0.40.
You can show that
z =
√1 + β −
√1− β√
1− β= −0.34
(b) β = +0.90 and using the equationwe get z = 3.36.
The largest red-shifts corresponding toquasars have z = 7.
3.7 Approximations
Relativistic effects are very small if we aremoving at “low” speed, β << 1. However“low” relativistically is still very fast for usterrestrial observers. When we move slowlywe can make approximations.
Probably the most common approximationis the Binomial Approximation for (1 + x)n
when x << 1.
(1+x)n ≈ 1+n x+n(n− 1)
2x2 + · · · (3.41)
In calculus you can prove this using the Tay-lor series, but for now just consider one ex-ample with x = 0.01 and n = 3:
(1 + x)3 =1 + 3x+ 3x2 + x3
= 1 + 3(0.01) + 3(0.01)2 + 0.013
= 1 + 0.03 + 0.0003 + 0.000001
≈ 1.030
Clearly the higher order terms in x con-tribute very little to the total and can beignored. We are making a first-order approx-imation to the binomial.
For small values of β the binomial expansionof γ is
γ =1√
1− β2
= (1− β2)−1/2
≈ 1− (−1
2)β2
= 1 + β2/2 (3.42)
Try β = 0.3. The exact value is γ = 1.0483and the approximation is 1.0450, quite close.For slower speeds the result is even better.Try β = 0.1.
Example Ursula sees a proper time intervalof 100.00000 s. She sees Pete movingat 1380 m/s (Speed of space shuttle.)What is the time interval according toPete?
First lets get the dimensionless speedand γ.
β = 1380/3 × 108 = 4.6 × 10−6 γ ≈1 + (4.6× 10−6)2/2 = 1 + 1.058× 10−11
If I try to sum these on my (cheap) cal-culator I just get 1, so I keep the termsseparate. The improper time intervalmeasured by Pete is 1.058 ns longerthan the proper interval measured byUrsula.
54 3.8 Summary
3.8 Summary
It is often convenient to describe positionand time of events, rather than interval be-tween two events. Lorentz Transformationsallow the conversion of values from one iner-tial observer to another.
t = γ(t′ − β′ x′)x = γ(x′ − β′ t′)y = y′ z = z′ (3.43)
From the Lorentz Transformations we canget the Velocity Transformations. Thevelocity transformations show that noth-ing can travel faster than the speed oflight.
vx =v′x − β′
1− β′v′x
vy,z =v′y,z
γ(1− β′v′x)(3.44)
Also we can find the Relativistic DopplerEffect Formula that explains frequencychanges due to motion along the line join-ing source and observer.
f ′ =
(√1− β1 + β
)f (3.45)
In astrophysics, overall motion of a galaxyrelative to us, the observers, results in aDoppler Shift, toward blue for approachinggalaxies (rare) or toward red for recedinggalaxies. The red shift z is frequently usedto describe the motion. Other sources ofred shift also exist, but are not discussedhere.
3.8 Summary 55
Figure 3.6: Red shift of the Hydrogen Balmer series for quasar 3C 273. Vertical bars arethe locations of lines for hydrogen in the lab. Shifts in the observed spectrum are shownby the arrows. This quasar has a red shift of 0.158, and a speed of 0.146.
56 3.8 Summary
Chapter 4
Paradoxes and Their Resolution,Moving Along a Curved Worldline
4.1 Introduction
In this chapter we will examine a wholebunch of problems and paradoxes that willhelp us understand special relativity kine-matics. At the end we will discuss time alonga general world-line, one that allows an ob-server to accelerate rather than move at con-stant velocity.
4.2 The equations
Here are the equations that we have used inthe first three chapters. I have used my ver-sions of the Lorentz Transformations, Equa-tions 4.6, and the Velocity Transformations,Equations 4.7, that have only primed quan-tities on the right hand side, and unprimedon the left.
The two observers may be considered Ursula(unprimed) and Pete (primed.) Pete is mov-ing with a constant velocity β with respectto Ursula. The primed and unprimed axesare parallel, motion is along the x− x′ axesand the origins coincide at time 0. Both are
inertial reference frames to the limit of ac-curacy of our measurements.
β′ = −β (4.1)
γ =1√
1− β2(4.2)
∆timproper = γ ∆tproper (4.3)
Limproper = L/γ (4.4)
STI2 = ∆t2 −∆x2 (4.5)
t = γ(t′ − β′ x′
)x = γ
(x′ − β′ t′
)(4.6)
y = y′ z = z′
vx =v′x − β′
1− β′v′x
vy =v′y
γ(1− β′v′x)(4.7)
vz =v′z
γ(1− β′v′x)
f ′ =
(√1− β1 + β
)f (4.8)
57
58 4.4 Seeing the Length of a Rod
4.3 Seeing versus Measur-ing
Spacetime Physics problem 3-15.
You are located in Rochester on a flat earth.A rocket is heading toward you, with ve-locity β. As it passes the Golden Gatebridge in San Francisco it sends out a pulseof light. It sends a second pulse of lightwhen it passes under the Gateway Arch inSt. Louis. Based on unintelligent observa-tion, what would you see for the velocity ofthe rocket?
An event diagram is shown below. Assumethat for you on the earth, the second pulseoccurs a time ∆t after the first.
(a) What is the distance between San Fran-cisco and Saint Louis in terms of ∆t?
(b) At the time of the second pulse, whereis the first pulse located, in terms of ∆t?
(c) What is the space separation of the twopulses? What is the time separation be-tween the two pulses?
(d) A naıve calculation of the rocket speedwould be to take the known distance be-tween the two cities and divide by thetime interval between when you see thetwo pulses. What doe this give for vseen?
(e) Suppose the rocket were traveling at β =0.5. What would be vseen?
(f) Suppose that vseen = 4. What is thetrue speed of the rocket?
(g) Suppose that your friend were in SanFrancisco. What would she see for thevelocity of the rocket in each of the last
two cases? Hint: you can either rean-alyze the situation, or use the equationalready derived and a little cleverness toget the new equation.
(h) Resolve the paradox. Why do we see therocket moving faster than light speed?
4.4 Seeing the Length of aRod
Pete rides on the primed rocket carrying astick that is parallel to the x′-axis and oflength L′ = 100 m. His friend Paris is tohis right and proposes the following methodby which Ursula can determine the length ofthe stick. Ursula is at the origin of her coor-dinate system, and she sees Pete and Parismove to the right at β = 0.60.
“At time t′ = 10 m Pete and I will stand ateither end of the stick (Pete will be at x′ =50 m) and send light pulses towards Ursula.Ursula will measure the time between thepulses and knowing that the dimensionlessspeed of light is 1.00 can say that the lengthof the rod equals the time difference betweenreceiving pulses.”
(a) Without any calculation, can you see theflaw in the logic? What is it?
(b) Write the coordinates of the two events(Pete sends flash, Paris sends flash) asmeasured by Pete.
(c) Do a Lorentz transformation to find thecoordinates of the events as measured byUrsula.
(d) At what times will the two pulses arriveat Ursula who is standing at the origin
4.5 Scissors Paradox 59
Figure 4.1: Seeing versus Measuring: Event diagram for a rocket heading eastward on aflat earth.
of her coordinate system?
(e) Using Paris’s method, what would bethe apparent length of the stick as seenby Ursula?
(f) What is the actual length of the stickas measured by the intelligent observerUrsula?
4.5 Scissors Paradox
Spacetime Physics problem 3-14 a. We sim-ulate a closing scissors by having one hori-zontal fixed rod, and a second rod tilted atangle θ to the horizontal and moving verti-cally downwards with speed β. Consider thepoint where the two rods cross. This pointmoves to the right at some speed vA. De-termine vA in terms of β and θ. Find theconditions under which this speed is larger
than 1 (i.e. faster than light speed.)
Figure 4.2: Scissors Paradox. The horizon-tal rod is at rest, top rod is tilted and falling.The point of intersection of the rods canmove faster than the speed of light.
On the figure, draw the rod after it has fallenfor a short amount of time ∆t. Write expres-sions for distances and then use trigonome-try.
In Section 4.3 we had a situation where theobserved speed of an object exceeded 1, how-ever this was the result of NOT using intel-ligent observers. In the Scissors Paradox wedo not have that problem—the intersection
60 4.7 Change of Measured Orientation of Objects
point can travel at any speed up to infin-ity. Does this violate Einstein’s postulates.If not, why not?
4.6 What If I Could MoveFaster Than Light?
Spacetime Physics Sample problem L-2.Here is another case of proof by reductio adabsurdum.
Assume that an object can have velocitiesboth less than and greater than the speedof light in vacuum. What does this lead to?We will see that it leads to the absurd con-clusion that you could be killed before youare born. Therefore we must reject the ini-tial assumption and say rather that objectscan only have speeds 0 < v < c.
Event E1: At a time −4 years on theplanet Klingon, the Peace Treaty of Shal-imar was signed between Klingon and theFederation. Immediately upon signing thetreaty, a Federation ship set out on impulsepower at a speed of 0.60 to bring the treatyto Earth.
The murderous Klingons immediatelystarted work on a super drive that wouldallow them to move faster than the speedof light. They perfected their drive and. . .
Event E2: . . . at time 0 the Klingonslaunched a ship at speed 3.00 to overtake theFederation ship and destroy it. Event 3 isthe destruction of the Federation ship.
(a) Determine the position and time coordi-nates of each of the events as measuredby an observer on Klingon.
(b) Draw a space-time diagram from thevantage point of the Klingon homeplanet.
(c) Use Lorentz transformations to find thecoordinates of each of the events as mea-sured by the Federation crew.
(d) Draw a space-time diagram from thevantage point of the Federation ship.
(e) For observers on the planet Klingon, theFederation ship was destroyed after itwas launched. For observers on the Fed-eration ship, what is the order of events?
By working through this exercise youshould come to the conclusion of SpacetimePhysics,
“What have we here? A confusionof cause and effect, a confusion thatcannot be straightened out as longas we assume that the Super—orany other material object—travelsfaster than the speed of light in avacuum.
“Why does no signal and no objecttravel faster than light in a vac-uum? Because if either signal orobject did so, the entire networkof cause and effect would be de-stroyed, and science as we know itwould not be possible.”
4.7 Change of MeasuredOrientation of Ob-jects
Spacetime Physics Problem L-6a.
4.9 The Rising Stick. 61
Pete rides a rocket carrying a stick of lengthL′ that is tilted at an angle θ′ from the x′-axis. Ursula in the lab sees Pete move to theright at β.
(a) What are the x′ and y′ components ofthe stick’s ends?
(b) What are the x and y components ofthe stick’s ends as measured by Ursula—assume an intelligent observer?
(c) What is the length of the stick as mea-sured by Ursula?
(d) What is the angle the stick makes withthe horizontal as measured by Ursula?
(e) Evaluate your expressions for (c) and (d)for the case of L′ = 100 m, θ′ = 30◦, andβ = 0.60.
4.8 The Headlight Ef-fect
Spacetime Physics Problem L-9.
As usual, Pete is riding a rocket moving tothe right at β as measured by Ursula. Petesends a pulse of light oriented at an angle θ′
to the x′-axis.
Show that for Ursula the light makes an an-gle θ with
cos θ =cos θ′ − β′
1− β′ cos θ′(4.9)
This should be a really easy answer!
Now consider the light emerging from a lightbulb that Pete carries. According to Petethe light is evenly spread in all directions,with half of the light emerging with angles
between −π/2 and π/2. We say the half-angle is π/2.
Find the expression for the half-angle in Ur-sula’s reference frame.
Evaluate in the case of β′ = −0.95.
The headlight effect says that a moving lightemits light that is concentrated in a smallcone in the forward direction.
4.9 The Rising Stick.
Spacetime Physics Problem L-10.
A stick of proper length L = 20 m is parallelto Ursulas’s x-axis and is rising at a dimen-sionless speed vy = 0.5. Pete is seen to moveto the right at β = 0.6. These numbers aremeasured by Ursula.
Figure 4.3: The Rising Stick: Observersmove along x-axis, Stick is rising along y-axis for Ursula.
(a) Explain without doing calculations whythe stick should be tilted according toPete.
(b) According to Ursula, the left end of thestick is at x = 0, y = 0 m at t = 0 m.This is event E1: left end crossing the x-axis. What are the coordinates for event
62 4.10 Paradox of the Identically Accelerated Twins
E2 (x, y and t), the right end crossingthe axis according to Ursula?
(c) What are the x′- and y′-components ofthe stick’s velocity according to Pete?
(d) What are the coordinates (x′, y′ and t′)of the two events according to Pete?
(e) Now consider a third event, the locationof the left end of the stick at the sametime as event E2. What are the x′ andy′ values for this event?
(f) From the results get the horizontallength of the stick as measured by Pete.Does this agree with the length contrac-tion formula?
(g) What is the angle from the horizontal,θ′, that Pete measures for the stick?
(h) If you have done this using numbers, goback and do it with symbols (all parts),ending up with tan θ′ = γβvy.
Application: Consider problem L-11 inSpacetime Physics. Here is Ursula’s view.A stick that is exactly 1 m (proper length)long moves horizontally at β. A plate witha hole of diameter exactly 1 m rises verti-cally at vy. The stick is Lorentz contractedand therefore should fall through the hole.But consider the viewpoint of Pete ridingon the stick. For him it is the hole that isLorentz contracted and the stick should notfall through the hole. How can this paradoxbe resolved?
4.10 Paradox of the Iden-tically AcceleratedTwins
Problem L-13 in Spacetime Physics TwinsAndrew and Brad own identical spaceships.Andrew’s ship is on the left and Brad’s ison the right separated by some distance. Attime 0 in the earth frame they are 20 yearsold and accelerate to the right until they runout of fuel. According to Mom and Dadon earth, the twins run out of fuel at ex-actly the same time, and then have the samespeed vfinal, and they are separated by ex-actly the same distance as they were at thestart.
Andrew and Brad compare notes and agreethat they had the same history of accelera-tions, but find that Brad is older than An-drew. How can this happen?
We will look at a simpler situation where theacceleration occurs instantaneously by hav-ing the twins jump from one ship to another.Here is the trip as measured by Mom andDad, with several events given.
Table 4.1: Andrew and Brad take a trip.Event coordinates measured by Mom andDad on earth in units of days.
Event x t
E0 Andrew on ship A 0 0E4 Brad on ship D 10 0E1 Andrew jumps A→ B 1 5E5 Brad jumps D → E 11 5E2 Andrew jumps B → C 4 10E6 Brad jumps E → F 14 10E3 Andrew on C, fuel gone 8.5 15E7 Brad on F, fuel gone 18.5 15
4.11 The Twin Paradox 63
This means that at t = 0 days Andrew andBrad stepped on spaceships A and D trav-eling at some speed to the right. After 5days they each jump onto faster spaceshipsB and E for another 5 days. Finally theyjump on spaceships C and F that are go-ing even faster, until those ships run out offuel.
(a) Find the velocities of each of the space-ships A through F.
(b) How much does each twin age during thetrip? This is NOT 15 days!
(c) How far apart are the twins at the startand at the end of the trip?
(d) Draw a space-time diagram for theevents as seen by Mom and Dad. La-bel each event on the diagram. Drawthe world line for each twin by joiningthe points.
(e) Now consider the events as seen by thepilots of ships C and F. Find the coor-dinates of the 8 events by these pilots.
(f) Draw a space-time diagram as seen bythese pilots. Draw the world line foreach twin.
(g) In order to measure the separation of thetwins, you need to measure two eventsthat occur at the same time. Extend theworld lines so that you can find the sep-aration at the earliest and latest timesthat you found by using the Lorentztransformations. This is as measured bythe pilots of the last ships. What are theseparations at these two times.
(h) If the twins were of identical age atevents E0 and E4, by how much havethey aged when you compare their ages
at the end of the trips, at the sametimes.
The seemingly absurd results that occur inproblems like this can be verified experimen-tally. In this problem, as in so many otherparadoxes, the resolution to the paradox liesin remembering that events that are simulta-neous in one reference frame are no simulta-neous in a different reference frame movingwith respect to the first. As strange as itseems, we must embrace this experimentalreality in order to solve problems in specialrelativity.
4.11 The Twin Paradox
Spacetime Physics, Chapter 4 discusses thisin detail. Read it carefully to complementwhat I will discuss.
The most famous of Special Relativity Para-doxes is the Twin Paradox, in part because itwas at the center of the 1963 novel La planetedes singes by Pierre Boulle, more familiar tous as Planet of the Apes, source for innumer-able movies beginning in 1968.
As Wikipedia summarizes, “Ulysse beginsby explaining that he was friends withProfessor Antelle, a genius scientist onEarth, who invented a sophisticated space-ship which could travel at nearly the speed oflight. Ulysse, the professor, and a physiciannamed Levain fly off in this ship to exploreouter space. They travel to the nearest starsystem that the professor theorized mightbe capable of life, the red sun Betelgeuse,which would take them about 350 years toreach. Due to time dilation, however, the
64 4.11 The Twin Paradox
trip only seems two years long to the travel-ers.”1
We will consider a famous pair of twins,Danny DiVito and Arnold Schwarzenegger.2
Danny stays on earth while Arnold blastsoff making a trip to Proxima Centauri, 4.2years away, traveling at β = 0.9 for whichγ = 2.29. Danny says that the trip takestD = 4.2/0.9 = 4.67 years, while Arnold,the proper observer, says the trip takes4.67/2.29 = 2.03 years.
Upon reaching Proxima Centauri, Arnoldimmediately returns at the same speed.Danny says that the round trip takes 9.33years while Arnold claims that it takes 4.06years. By this analysis Arnold is youngerthan Danny at the end of the trip.
Well that is strange enough, but Arnoldmakes the following claim. “In my refer-ence frame I am at rest, and it is Dannyand the rest of the universe that moves firstat β′ = −0.9, then at β′ = 0.9. I see theimproper distance travelled during the firsthalf of the trip by Danny and the stars tobe LA = 4.2/2.29 = 1.83 years and thistakes 1.83/0.9 = 2.03 years. Danny mea-sures the proper time of 2.03/2.29 = 0.89years. I claim the round trip will take 4.06years but Danny will measure 1.78 years,and Danny is younger than me when wemeet again.”
There is a problem—only one of the two canactually be older. But who is correct? Can
1http://en.wikipedia.org/wiki/Planet of the ApesBoulle underestimated the distance to Betelgeusewhich is approximately 427 years. In either case,the rocket must travel at β > 0.99998.
2The movie Twins, 1988, “Only their mother cantell them apart.”
you see the resolution to this problem, i.ecan you tell whether Danny or Arnold isolder?
The resolution lies in deciding who sees anacceleration. From riding in cars, you knowthat you can sense an acceleration. Dannynever feels an acceleration, while Arnolddoes feel an acceleration when he switchesdirection at Proxima Centauri. ThereforeDanny’s analysis is correct, and Arnoldis younger than Danny when they meetagain.
Now imagine a slightly more complicatedsituation, Danny and Arnold both boardrockets and make round trips traveling atdifferent speeds so as to return to earthat the same time. Both have experiencedaccelerations. Who is older and who isyounger?
We will examine the Twin paradox closelyin order to answer this question. Notice thatSpecial Relativity is perfectly well equippedto deal with situations of accelerations. Westart with a simple situation.
Example Consider the twin paradox in thisway, using space-time diagrams. Thetwins will be called Ursula and Pete.Ursula stays on earth the entire time,Pete makes a trip to Proxima Centauri,located 4 light years away from earth,and back. Heading out he rides onSpaceship Alpha that travels at β =0.80, and on the return he rides onSpaceship Omega that travels at β =−0.80. All the numbers mentioned sofar are measured by Ursula.
Pete departs on July 4 2006. He andUrsula agree that they will each send a
4.12 Position and Time Not Absolute 65
light pulse to the other every July 4.
(a) According to Ursula, how long willit take for Pete to reach ProximaCentauri?
(b) Use graph paper and draw thespace-time diagram as seen by Ur-sula. On it you should have worldlines for Ursula and Pete (i.e. bothspaceships), and for the pulses sentby Ursula. Mark the events whenPete receives the pulses from Ur-sula. How many pulses has Petereceived, including the final onewhen he returns to earth? This isthe age of Ursula when they meetagain.
(c) Now compute quantities as seen bythe Captain Ahab of Alpha. Whatis the velocity of Spaceship Omegaas seen by Ahab? How long is Peteon Spaceship Alpha?
(d) Draw the space-time diagram forAhab, including world lines forboth Spaceships and Ursula andthe pulses sent by Pete while he ison Alpha. Mark the events whenUrsula receives a pulse from Pete.How many pulses has Ursula re-ceived on the outbound leg?
(e) By symmetry, on the inbound legUrsula will receive the same num-ber of pulses. What is the totalnumber of pulses received by Ur-sula for the round trip? This is theage of Pete when they meet again.
4.12 Position and Time NotAbsolute
Spacetime Physics, Chapter 5 will next bediscussed.
Consider an event, you at the start of thisclass. The event has a definite time anda definite location. However the numer-ical values of the time and location arenot unique, but vary depending on the ob-server.
Consider two cities, Rochester and Cleve-land, and two observers, Alan and Beth.Alan and Beth start in Cleveland and agreeto call the position there ‘0’. They wantto know the distance from Cleveland toRochester. Alan drives along I90 passingby Buffalo and when he gets to Rochesterhis odometer reads 261 miles. Beth vis-its friends at UPitt, Penn State, and Cor-nell on her way and her odometer reads 449miles.
Both Alan and Beth make valid read-ings. Both agree that Cleveland andRochester have definite locations, howeverthey get different distances between the twocities.
In the same way we have established therules that show that the time interval be-tween two events is not unique, but dependson the observer.
For relativity we have an invariant space-time interval, STI2 = ∆t2 − ∆x2 − ∆y2 −∆z2.
On a space-time drawing, two events, E1and E2, that occur at the same locationoccur on a vertical line, two events, E1
66 4.12 Position and Time Not Absolute
and E3, that occur at the same time—i.e. simultaneously— occur on a horizontalline.
If we switch to the frame of a different ob-server events E1 and E2 will no longer occurat the same location. Events E1 and E3 willno longer occur at the same time, i.e. nolonger be simultaneous. The STI betweenany two events will, however, be the samefor all observers.
Let’s return to our parable of the surveyorsfrom Chapter 1. Different observers founddifferent coordinates for parts of the city,however the distance between buildings wasinvariant, i.e. the same regardless of ob-server. In Section 1 of Chapter 3 we de-scribed a method to produce a map of thecity given only distances between buildings.How does this translate to the case of specialrelativity?
The invariant quantity for the surveyor’s wasthe distance, ∆r =
√∆x2 + ∆y2 This is the
equation of a circle. The arrows shown onthe left of Figure 4.4 all are the same dis-tance from the center. In the case of specialrelativity the invariant is the STI, in one di-mension
STI2 = ∆t2 −∆x2 (4.10)
This is the equation of an hyperbola, andassuming that ∆t > ∆x its principal axis istime, shown on the right of Figure 4.4.
Recall that in the surveyor’s parable, the twosurveyors had different direction for North.However both would agree that the samecircle represents points a distance 5 apart.In the same way in special relativity, dif-ferent observers will measure different time
and space intervals, but all agree that thehyperbola represents events separated by aSTI of 5. Unlike the case of the surveyors,the arrows showing STI appear of differentlength, but the computed STI is exactly thesame.
Here is another way to view the hyperbola.Imagine that a rocket, Alpha, moves to theright. Alpha carries a light flasher and a mir-ror, as shown in Spacetime Physics Figure5-2. Consider two events: E1 = flash pro-duced, it bounces off mirror and E2 = flashreceived back at location of flasher. Alan,in the rocket, can draw an hyperbola show-ing an STI between the events. For Alan∆x = 0, so the vertical arrow connects thetwo events as measured by him.
Consider Beth in the lab. She can measurethe space and time intervals, compute theSTI, and will get the same result as Alan.She will draw an identical hyperbola as didAlan to show points at the same STI. For her∆x > 0, so the arrow pointing up and to theright connects the two events as measuredby her.
Finally consider Sam, riding in a super-rocket moving faster than Alpha to the right.Sam will measure time and space intervalsbetween the events and get the same STI andtherefore draw the same hyperbola as Alanand Beth. Sam will measure a space intervalthat is negative, and the arrow pointing upand to the left connects the events as mea-sured by Sam.
So different observers will draw exactly thesame hyperbola, the invariant hyperbola, fora given STI.
We can check to see that the hyperbola and
4.13 General Worldline 67
Figure 4.4: On left, surveyor’s parable. Arrows connect points at the same distance fromthe center. On the right, an hyperbola representing something happening in the future.The arrows connect events at the origin with different events on the hyperbola. All arrowshave the same value for the space-time interval, STI.
the Lorentz transformations yield the sameresults. Consider Alan who sees (∆t =5,∆x = 0). Alan draws the hyperbola inFigure 4.4, with the vertical arrow connect-ing the two events. The coordinates of thesecond event as seen by Beth are (let’s pick)(∆t = 7.81,∆x = 6.00) m.
What is Beth’s velocity, according to Alan?Beth must be moving to the left, and withvelocity β = −∆x/∆t = −6.00/7.81 =−0.768.
Now we can check the Lorentz transforma-tions to see if they give the same results.E1 has coordinates (0,0) for both Alan andBeth. E2 has coordinates (5,0) m as mea-sured by Alan. We compute γ = 1.562,
then
tBeth = γ(tAlan − βxAlan)
= 1.562(5− (−0.768)0)
= 7.81m
xBeth = γ(xAlan − βtAlan)
= 1.562(0− (−0.768)5)
= 6.00m (4.11)
So our results are consistent.
4.13 General Worldline
Previously we drew worldlines for objectsmoving with constant velocities. But life isnot really like that, velocities will increaseand decrease as acceleration occurs. What isthe most general form of a worldline?
Recall that on a space-time diagram, withtime vertical, position horizontal, the slope
68 4.14 Time Kept By Traveller on a World Line
of a straight line equals the inverse of thevelocity:
Slope = ∆t/∆x = 1/β (4.12)
and that the speed of an object must notexceed the speed of light, β ≤ 1. This meansthat the slope, 1/β ≥ 1.
Frequently we use the same scale for timeand position, in which case the above state-ments tell us that the slope of a worldlinemust have an angle with the time axis
−π/4 = −45◦ ≤ θ ≤ π/4 = 45◦ (4.13)
This is shown in Figure 4.5, where the world-line on the left is possible. At the end of theworldline I have drawn two lines at 45◦ thatdelineate possible futures of the worldline.Regions shaded are not allowed for the par-ticle. The two dotted lines define the lightcone, and the worldline (past or future) canonly exist inside the lightcone. You shouldbe able to explain why the diagram on theright is impossible.
4.14 Time Kept By Trav-eller on a WorldLine
Let’s say that Figure 4.6 shows a curvedworldline for Beth as measured by Alan, andtwo events, E1 and E2. Each of them car-ries a clock (either a real clock or a biolog-ical clock). The space interval is the samefor both Alan and Beth, ∆x = 0, and byour ideas of proper time interval, both reada proper time interval between the events.We shall see that these two proper times aredifferent, however.
The proper time interval that Alan reads isthe time interval read on a clock held byhim, the proper time that Beth reads is theproper time read on a different clock heldby her. How do the two time intervals com-pare?
To answer this, break the curved path upinto short pieces where the velocity is almostconstant. E1 and E3 are two events thatfit this description. Only Beth will see theproper time for the interval, and
∆tAlan13 = γ∆tBeth13 (4.14)
Alternately we could look at the space-timeinterval, recalling that ∆xBeth13 = 0,
(∆tBeth13 )2 = (∆tAlan13 )2 − (∆xAlan13 )2 (4.15)
If we add all the proper time intervals forBeth between E1 and E2 we will get ashorter time interval for Beth’s clock thanAlan would measure for his clock. Space-time Physics refers to this as the Principleof Maximal Aging: Between two events anobserver that travels on a straight line on aspace-time diagram will measure the largesttime interval.
Let’s look at a specific example.
Example From Alan’s viewpoint, he is atrest, Beth undertakes the path de-scribed by the following table of data,and Carl moves to the right at a con-stant speed of 0.30.
(a) Draw a space-time diagram fromAlan’s viewpoint. Label the worldlines of Alan, Beth, and Carl (as-sume he starts at x = 0, t = 0).
4.14 Time Kept By Traveller on a World Line 69
Figure 4.5: Space-time diagram showing a worldline for a particle. In general the worldlinemust be within 45◦ of the t-axis. At the top end of the line the light cone is drawn showingpossible past path and allowed future path of the particle. The worldline on the right isimpossible—you should be able to explain why.
Table 4.2: Coordinates of Beth, measuredby Alan
t (m) x (m)
0 01 02 0.23 0.84 1.45 1.66 1.47 0.88 0.29 010 0
(b) For each small interval, compute∆t,∆x as measured by Alan, andcompute the proper time intervalmeasured by Beth.
(c) At the end Alan claims that 10 mof time have passed. What doesBeth claim? Does this agree with
the Principle of Maximal Aging?
(d) Now consider Carl’s measurementsfor the world lines of Alan andBeth. Do Lorentz transformationsof Beth’s coordinates and of Alan’scoordinates.
(e) Draw a space-time diagram fromCarl’s viewpoint, and label theworldlines of Alan, Beth, and Carl.
(f) Using Carl’s measurements forAlan, repeat the steps of part (b)to get the proper time intervals.Then repeat for Carl’s measure-ments of Beth.
(g) What does Carl compute for theproper time interval between E1and E2 for Alan? for Beth? Howdo these compare with what Alancomputes? Do the results agreewith the Principle of Maximal Ag-ing?
70 4.15 Maximal or Minimal
Figure 4.6: Worldlines for Alan and Beth with three events marked. How do the elapsedtimes between E1 and E2 compare for the two observers?
So apparently everything is self-consistent,if a bit weird. We can also make the con-nection to a free particle, a free particle, byNewton’s first law, will have a constant ve-locity, and thus move along a straight world-line. So we can state the principle of max-imal aging as “A free particle will travelthe worldline in which it ages by the largestamount, larger than the aging if it follows apath where it is accelerated.”
The Principle of Maximal Aging carries overto General Relativity where it can be used toexplain the curvature of light near a massivestar.
4.15 Maximal or Mini-mal
Spacetime Physics, Chapter 5 Section 8points out an important area where confu-sion can arise.
Suppose Beth carries a laser that she flashestwice, one after the other separated by a
time ∆tBeth. Alan watches Beth move tothe right at β.
Case 1: What Alan measures for thetime interval between flashes (and remem-ber, as an intelligent observer he corrects forany transit time effects) is a time ∆tAlan.According to our ideas of time dilation,∆tAlan > ∆tBeth since Beth is the properobserver of the time interval, being presentat both events. This is true whatever thespeed of Beth (as measured by Alan), andtherefore we can say that Beth measures theproper time interval and it is the minimumtime interval.
Case 2: But we have just made a big dealabout the time interval measured by an ob-server traveling along a straight line beingthe maximum. Which is it, minimum ormaximum?
Case 1 compares time intervals measured bytwo different observers, only one of whommeasures the proper time interval. Theproper time interval is always shorter thanthe improper time interval, hence the propertime interval is a minimum. The ratio of
4.16 Summary 71
the two measured time intervals is alwaysγ.
Case 2 also compares time intervals mea-sured by two observers. In this case bothobservers see ∆x = 0 m, and thereforeboth observers measure a proper time ontheir watches. However their watches willdisagree, and the worldline joining the twoevents that appears shortest on a space-timediagram will measure the longer time, withthe maximum time measured by the worldline that is a straight line. There is no sim-ple expression for the ratio of time inter-vals.
4.16 Summary
Paradoxes are a good way to check your un-derstaqnding of special relativity. To removethe paradox, you must make very careful useof the rules of special relativity.
On a 1D space-time diagram, hyperbolaswith foci on the time axis connect eventsthat all have the same space-time intervalfrom the origin.
World lines are the trajectory of an objectthrough space-time, and must always havean angle of magnitude less than or equal to45◦ from the time-axis.
Principle of Maximal Aging: Betweentwo events an observer that travels on astraight line on a space-time diagram willmeasure the largest time interval.
72 4.16 Summary
Chapter 5
Time-like, Space-like, Light-likeIntervals, Invariant Hyperbolae,Minkowski Diagrams
5.1 Introduction
In this chapter we will look at cause-and-effect in relativity, time-like, space-like, andlight-like intervals, the light cone (again),and another way to consider the space-timediagram, Minkowski diagrams.
5.2 Time-like, Space-like,and Light-like Inter-vals
We have discussed the invarient space-timeinterval, STI, which in one space dimensionis
STI2 = ∆t2 −∆x2 (5.1)
Thus if class is held on a bus and the groundbased observer, Gwen, measures the startat (t1 = 40, x1 = 20) min and the end at(t2 = 80, x1 = 45) min, the STI is STI =√
(80− 40)2 − (45− 20)2 = 31.2 min. And
we expect that there is some proper ob-server, Pete, for whom the two events occurat the same location.
What is the speed of the Pete, the properobserver, according to Gwen?
β = ∆x/∆t = 25/40 = 0.625 (5.2)
You can verify that this works by doing theLorentz transformations, and find that thetwo events both occur at x = −6.4 min andat times 35.2 and 66.5 min. The proper timeinterval for the class is 31.3 min.
Now consider a second pair of events for astick moving past you. Event 3 is the leftend of the stick at your eyes, (t3 = 0, x3 = 0)m, and Event 4 is the right end of the sticknear your toes, (t4 = 5, x4 = 15) m. Whenwe compute the STI we get
STI2 = (5)2 − (15)2 = −200 (5.3)
What does it mean when the STI2 is nega-tive?
73
74 5.2 Time-like, Space-like, and Light-like Intervals
Here is a simpler way to get a negativeSTI2: Consider two events that occur si-multaneously but at different locations, say(t = 0, x = 0) m and (t = 0, x = L) m,Then
STI2 = 02 − L2 = −L2 (5.4)
and this is invariant for all observers.
The two events just described could refer tomeasuring a proper length. For an object ofsome non-zero length, there is no observerwho can see the both ends at the same loca-tion, that is ∆x > 0 for all observers. Thiscan be made more clear by the following ex-ample.
Example Alan holds a rod with its left endat (0, 0) and it right end at (0, 10), thatis he can simultaneously measure thetwo ends and find a length of 10 m forthe rod. Alan sees Beth move past atβ = 0.80. Since the rod is at rest withrespect to Alan, he could measure theends at any other combinations as isshown in Table 5.1.
Table 5.1: Measuring the length of a rod,Time and position in meters.
Event Alan BethNumber t x t′ x′
E1 0.00 0.00 0.00 0.00E2 0.00 10.00 -13.33 16.67E3 10.00 0.00E4 5.00 0.00E5 5.00 10.00E6 10.00E7 20.00 0.00E8 20.00 10.00E9 10.00
Notice that events E1, E4, and E7 refer
to measuring the left end of the rod atdifferent times, while events E2, E3, E5,E6, E8, and E9 refer to measuring theright end.
Since the rod is at rest with respectto Alan, he can ignore the times com-pletely and compute the length of therod as LAlan = 10− 0 = 10 m.
For Beth’s measurements of the coordi-nates of the events we use the Lorentztransformations. The result is shownfor E2, and you can fill in the table forE4, E5, E7, and E8.
What are events E3, E6, and E9? IfBeth wants to measure the length ofa moving rod, she must simultaneously(for her) measure the positions of theleft and right ends. And while E1 andE2 are simultaneous to Alan, they areNOT simultaneous for Beth. Insteadwe must determine where the right endof the rod is simultaneous to measuringthe left end of the rod, simultaneous forBeth. Look at E1. Beth measures atime of 0 m so we need to know wherethe right end is at this same tBeth = 0.
We can use the Lorentz transformation,
tBeth =γ(tAlan − βxAlan)
0 =1.6667(tAlan − 0.80(10))
tAlan = 8.0m (5.5)
and hence
xBeth = 1.6667(10− 0.80(8)) = 6.0m(5.6)
So Beth measures the rod to have lengthof 6.0 m.
5.3 Invariant Hyperbolae 75
Compute the rest of the table, and seethat regardless of when Beth makes hermeasurement, she determines the rod tobe 6.0 m long. This of course agreeswith the length contraction formula.
Intervals where ∆t > ∆x are calledtime-like, with STI2 > 0. For eventsseparated by a time-like interval, it isalways possible to find an observer whowill see the events occur at the samelocation. Also the temporal order ofthe events in time is the same for allobservers, E2 will always be after E1,however the relative locations are flexi-ble, ∆x = x2−x1 can be positive, zero,or negative depending on the observer.The observer who sees a space intervalof 0 measures a proper time interval,and this is the smallest interval that canbe measured.
Intervals where ∆t < ∆x are calledspace-like, with STI2 < 0. For eventsseparated by a space-like interval, it isalways possible to find an observer whowill see the events occur at the sametime. Also the spatial order of theevents in time is the same for all ob-servers, E2 will always be to the rightof E1. However the timing of the evntsis flexible, ∆t = t2 − t1 can be positive,zero, or negative depending on the ob-server. The observer who sees a timeinterval of 0, i.e. measures the eventssimultaneously, will measures a properlength interval, and this is the largestlength interval that can be measured.
Intervals where ∆t = ∆x are calledlight-like. For events separated by light-like intervals, something moving at the
speed of light will measure ∆t = ∆x =0.The only things that travel at thespeed of light at our current under-standing of physics are photons, thequantum particles of light. Gravitons,hypothesized but never yet detected,would also travel at the speed of light.Neutrinos were long thought to move atthe speed of light, but since 1998 fairlystrong evidence that they move slightlyslower than c emerged.
5.3 Invariant Hyperbo-lae
In the last chapter we discussed the invarianthyperbola. If event E1 is at the origin, thenthe hyperbola shows all events E2 that havethe same STI. This arises from
STI2 = ∆t2 −∆x2 = Constant (5.7)
If E1 is at the origin, this becomes
t2 − x2 = constant (5.8)
For time-like intervals the constant is posi-tive and the hyperbolae are shown in Figure5.1(a). There are two curves, one in the pastand one in the future that have the sameSTI. Notice that for the future hyperbola,E2 occurs after E1 for all observers, howeversome observers say E1 occurs to the left ofE2, some say to the right, and some (properobservers) say that the two events occur atthe same location.
For space-like intervals the constant is neg-ative and the hyperbolae are shown in Fig-ure 5.1(b). Consider event E2 on the right
76 5.3 Invariant Hyperbolae
Figure 5.1: (a) Hyperbolae for time-like events, one at the origin and one on the hyperbolae.The two events are strictly ordered in time, but can occur to the left, at the same location,or to the right of each other. (b) Hyperbolae for space-like events, one at the origin andone on the hyperbolae. The two events are strictly ordered in space, but can occur before,simultaneous to, or after each other. (c) Hyperbolae (lines) for light-like events. x = ±t
hyperbola. All observers will see E2 to theright of E1 (the origin), however the time or-der is not restricted, t2 > t1 or t2 = t1 or t2 <t1.
Figure 5.1(c) shows the hyperbolae—actually two lines—for light-like eventswhere the constant is 0 so that ∆x =±∆t.
Think about what this implies. Two eventsthat are separated by a space-like intervalhave no definite time order.
For example Ethyl measures that Alan wasborn at (t = 0, x = 0) m and Beth was bornat (t = 8, x = 10) m, then STI2 = 82 −102 = −36m2. Can we answer the question,“Who was born first?” Certainly we can,but we will get different answers accordingto different observers. For Ethyl, Alan wasborn before Beth.
But consider Tammy who moves at β = 0.80.Applying the Lorentz transformations we get
for Tammy’s measurements (t = 0, x = 0) mand (t = 0, x = 6) m, so Tammy says thatAlan and Beth were born simultaneously. Orconsider Ivan who moves at β = 0.95. Ap-plying the Lorentz transformations we getfor Ivan’s measurements (t = 0, x = 0) mand (t = −4.8, x = 7.7) m, so Ivan says thatBeth was born before Alan. So there is nostrict time-order to the births, however allobservers will agree that Beth was born tothe right of Alan.
If we change the scenario to Alan firing a gunand Beth dying of a gunshot we can see thatthere cannot be a cause-and-effect relationbetween the two events.
Notice that in our example, Tammy sees thesmallest spatial separation between the twoevents, and for Tammy, ∆t = 0 m. We candetermine the speed required of Tammy asfollows. We have E1 at (0, 0) m and E2 at
5.4 Special Lines 77
(t, x). For Tammy, tTammy = 0 so
tTammy =γ(tEthyl − βxEthyl)0 =γ(tEthyl − βxEthyl)
tEthyl =βxEthyl
β =tEthyl/xEthyl = 8/10 = 0.8
or if the first event is not at the origin,
βspace−like =∆t/∆x (5.9)
Wait a minute you might be saying, velocityis defined as distance over time, not timeover distance. Certainly that is true, but weare not using this definition, we are using therequirement of the Lorentz transformationto get our answer.
Consider time-like intervals where ∆t2 −∆x2 > 0. We can ask, “What is the ve-locity of a “proper” observer who would seethe two events occur at the same location?”Again we first consider E1 to be at the ori-gin. For the proper observer,
x′ =γ(x− βt)0 =γ(x− βt)β =x/t
or if the first event is not at the origin,
βtime−like =∆x/∆t (5.10)
Two events separated by a time-like intervalcould be a cause-and-effect pair. For exam-ple the pair of events E1 = me throwing a piethrown at coordinates (t = 0, x = 0) and E2= a pie hitting you in the face at (50,30) mare separated by a time-like interval. HenceI could have been the cause of you being hitby a pie. Of course there is no guarantee thatjust because two events are separated by atime-like interval, that they are cause and
effect—I may have thrown an apple pie, andyou may have been hit by a coconut creampie thrown by Dr. Axon.
On a space-time diagram it is easy to distin-guish space-like and time-like intervals. InChapter 4 we showed a worldline for a par-ticle. This is repeated in Figure 5.2. Thisshows a particle at some definite location intime and space, plus the past history (world-line) of the particle. A light cone is centeredon the current coordinates of the particle,and several other events are shown.
All events that are in the shaded regions (E2,E3, E4) are separated from the particle byspace-like intervals. Event E1 is separatedfrom the particle by a time-like interval andis an event in the particle’s past. E1 couldhave caused an effect on the particle. EventE5 is separated from the particle by a time-like interval and is an event in the particle’sfuture. The particle could cause an effect atE5.
As the worldline progresses through time,the (imaginary) light cone travels with it,separating space-like and time-like eventsfrom the particle.
5.4 Special Lines
In addition to the invariant hyperbolae thatwe have already discussed, it is helpful todiscuss two other lines on a space-time dia-gram.
We consider the space-time diagram of Alanwho sees Beth moving to the right at β =0.20. As always, we choose the origins ofAlan’s and Berth’s coordinate systems to
78 5.4 Special Lines
Figure 5.2: Worldline of some particle, showing the light cone and several events aroundthe current time-space coordinate.
coincide—x′ = x = 0 when t′ = t = 0.
The first special line is the worldline of theorigin of Beth’s coordinate system, that iswhere Beth’s origin will be at points in thefuture. This suggests a time-like interval,and we can use Equation 5.10 to get x =βt.
I’ll try to say it in words—As Beth movesforward in time she always says that her ori-gin is at x′ = 0, but Alan says that her originat time t is found at x = βt. This is shownin Figure 5.3(a). The line makes an angle αwhere
tanα = x/t = β (5.11)
For β = 0.20, α = 11.3◦.
Note that for this line x′ = 0, meaningthat it is in essence the t′-axis as seen byAlan.
Now for the second special line. In words,“At t′ = 0 Beth can measure various posi-tions along her x′ axis. What will these looklike in Alan’s reference frame?”
This suggests space-like intervals, and wehad Equation 5.9, t = βx. This is shown
in Figure 5.3(b). It also makes an angle αbut with Alan’s x-axis.
If this seems confusing, try this—I’ll use theexample numbers for β = 0.20. Here areBeth’s coordinates for several points alongher x′ axis at t′ = 0. I use the Lorentz trans-formation with β′ = −0.20 to get Alan’smeasurements, and note that t = βx doesindeed work.
Table 5.2: Various Events for t′ = 0Beth Alanx′ t′ x t
0 0 0 01 0 1.02 0.2042 0 2.04 0.4083 0 3.06 0.612
Note that for the second line t′ = 0, mean-ing that it is in essence the x′-axis as seenby Alan. These two special lines will formthe basis of a powerful graphical way tolook at special relativity, Minkowski dia-grams.
5.5 Minkowski Diagrams 79
Figure 5.3: Two special lines. (a) x′ = 0, this is the t′−axis. (b) t′ = 0, this is the x′−axis.
5.5 Minkowski Dia-grams
Minkowski diagrams are a variation of space-time diagrams that include all referenceframes on a single diagram, rather thanhaving a different diagram for each ob-server.
As we saw in Section 5.4, there are two spe-cial lines that are Beth’s x′- and t′-axes asviewed by Alan, the unprimed observer. Thetwo axes make the same angle α with the re-spective x- and t-axes. Equation ?? tells ushow to compute the angle, tanα = β.
The angle is in the range −90◦ < α < 90◦.As is shown in Figure 5.4, positive anglesoccur for Beth moving to the right, and neg-ative for Beth moving to the left.
The remaining complication is to determinewhere the tick marks occur on the rotatedaxes.
Consider equally spaced marks on the x′
axis. For all these points t′ = 0. We
can then use the Lorentz Transformationsto get the coordinates in Alan’s unprimedframe. Of course (t′ = x′ = 0) transforms to(t = 0, x = 0). For (t′ = 0, x′ = 1),
t = γ(0− β′1) = −γβ′ = γβ
x = γ(1− β′0) = γ (5.12)
The distance from the origin to this point is,by the Pythagorean Theorem,√
(γβ)2 + γ2 = γ√
1 + β2 (5.13)
Example Suppose β = 0.6, γ = 1.25Then the spacing on the (t′, x′) axes is1.25√
1 + .62 = 1.46. On the Minkowskidiagram we can then place tick marksfor the primed axes. This is shown onFigure 5.5.
What can the Minkowski diagram tell us?The relativity of simultaneity is obvious.Look at Figure 5.5(a). The solid and hollowdots occur at different times in Alan’s frame,but at the same time in Beth’s frame.
Figure 5.5(a) also shows length contraction.The solid rod is at rest in Alan’s frame, with
80 5.6 Summary
Figure 5.4: Minkowski diagrams for Beth (t′, x′) moving to the (a) right at 0.20, α = 11.3◦,and (b) left at −0.60, α = −31◦. Dashed lines show how to read values of (t′, x′).
a length of 1 unit. Since the rod is at restwith respect to him, he can measure the twoends at different times and still get a correctlength. Beth sees the rod as moving, so forher to get a correct length she must measurethe lengths at the same time t′—these arethe hollow and solid dots at t′ = 0. Readingthe length on the x′ axis gives a length ofabout 0.8, and that agrees with length con-traction.
Figure 5.5(b) shows how the coordinatesread from the diagram agree with theLorentz transformation (no surprise, sincethe transformation was used to produce thediagram.) For Alan, the solid dot has co-ordinates (t = 1, x = 2.5) m, while forBeth the coordinates are indicated with thehollow dots and are (t = −0.64, x = 2.3)m by my crude measurement on my dia-gram. The Lorentz transformations give(t = −0.63, x = 2.38), in agreement withthe graphical measurement.
5.6 Summary
The relation between two events can betime-like, light-like, or space-like.
For events E1 and E2 with a time-like in-terval, the square of the space-time interval,STI2 = ∆t2 − ∆x2 is positive and eventE2 occurs after event E1 for all observers,∆t > 0. Different observers will measure dif-ferent values for ∆x: some will see event E2occur to the left of E1, some will see eventE2 occur to the right of E1, and one ob-server will see the events occur at the samelocation. Earlier we described this as theproper observer of the time interval, and forthis proper observer the time interval is aminimum.
For events with light like intervals, the space-time interval is 0. There is no physical ob-server who can see the events occur at thesame location or the same time. Only raysof light have this property.
5.6 Summary 81
Figure 5.5: (a) Using Minkowski Diagram for Length of a rod (b) Coordinates of an event
For events with space-like intervals, thesquare of the space-time interval is negative,and if E2 occurs to the right of E1 for oneobserver, it occurs to the right for all ob-servers, ∆x > 0. Different observers willmeasure different values for ∆t: some willsee event E2 occur before E1, some will seeevent E2 occur after E1, and one observerwill see the events occur at the same time(simultaneous).
On a space-time diagram with one event atthe origin, hyperbolas with foci on the time-axis connect time-like events with the samespace-time interval. Hyperbolas with foci ofthe space-axis connect space-like events withthe same space-time interval. Diagonal linesat 45◦ connect light-ike events.
For an event on a world line, we can draw a“light cone” that makes it easy to categorizeother events in relation to the point on theworld-line as space-like, light-like, or time-like. Cause-and-effect events must be time-
like (however being time-like does not implycausality.)
In regular space-time diagrams we use onediagram for each observer. Minkowski di-agrams allow a single diagram to be usedfor different observers. If a second observerhas a velocity β relative to you, the axes forthat observer are given by t = x/β for thet′-axis and by t = βx for the x′-axis. theseare rotated by and angle α = tan−1 β fromthe t- and x- axes. Coordinates of an eventin your frame are determined by projectinglines parallel to your t- and x- axes. Coordi-nates in the other frame are determined byprojecting lines parallel to the t′- and t′-axes,but the scale is different on the two differentaxes.
82 5.6 Summary
Chapter 6
Energy and Momentum
6.1 Overview
In this final chapter we will discuss the rel-ativistic expressions for energy, E, and mo-mentum ~p and explain the meaning of per-haps the most famous physics equation ofthem all. E = mc2. First we will discuss aformalism that is useful when dealing withspace-time as well as energy-momentum,four vectors.
6.2 Four Vectors
First consider a classical vector like posi-tion, ~r. We typically describe a vector assomething having a size (magnitude) and adirection (one or more angles). Thus on a2D surface, a position could be described as“5.0 m at a direction of 25◦ north of eastfrom me.”
Usually we define x- and y-axes correspond-ing to East and North, and then find vectorcomponents along these directions If the vec-tor has size r and angle θ from the x-axis,then
x = r cos θ
y = r sin θ (6.1)
Often in math, and sometimes in physics, avector is written by giving the componentsinside parentheses like (3, 4,−5).
If we have the components we can get themagnitude by
r =√x2 + y2 2D
r =√x2 + y2 + z2 3D (6.2)
In our parable of the Surveyors, the dis-tance, calculated from Equation 6.2 wasthe invariant quantity, the same for all ob-servers.
For relativity, the space-time interval, STI,was the invariant.
STI =√
∆t2 −∆x2 −∆y2 −∆z2 (6.3)
By parallel construction to a classicalvector, a 4-vector can be written as(∆t,∆x,∆y,∆z) with Equation 6.3 usedto calculate the magnitude of the 4-vector1.
1A second way to get the signs correct is
83
84 6.3 Classical Energy and Momentum
6.3 Classical Energy andMomentum
In high school physics you should have dis-cussed the ideas of energy and momentum,and discussed the Laws of Conservation ofEnergy and of Momentum.
Energy is a scalar (simple number) thatcomes in many forms, kinetic energy, KE,potential energy, PE, chemical energy,Echem . . . .
For a point object, kinetic energy is definedclassically in terms of the mass and speed ofthe object as
KE =1
2mv2 Classical Kinetic Energy
(6.4)
The total energy of a system is defined asthe sum of all types of energies,
E =∑
KE +∑
PE +∑
Echem + · · ·(6.5)
The Law of Conservation of Energy thensays
For an isolated system of particles,
the total energy is constant for all time.(6.6)
Momentum (technically linear momentum)is a vector that is defined classically interms of a particles mass, m and velocity~v as
~p = m~v (6.7)
to use ı =√−1 and write the 4-vector as
(ı∆t, ı∆x, ı∆y, ı∆z) with the magnitude beingSTI =
√(∆t2 + (ı∆x)2 + (ı∆y)2 + (ı∆z)2)
Once we have a coordinate system we can de-termine the components of the momentum,(px, py, pz).
For a system of particles, the total momen-tum is
~ptot =∑
~pi (6.8)
and the Law of Conservation of Momentumis
For an isolated system of particles,
the total momentum is constant
for all time. (6.9)
Consider a one-dimensional, elastic collisionbetween two billiard balls on a horizontalpool table shown in Figure 6.1. The firstball has mass m and initially moves to theright at v0 = 0.60 towards the second ball, ofmass 2m, which is at rest. The balls collidehead on and separate in a situation whereonly kinetic energy needs to be considered.
Figure 6.1: Head on 1D elastic collision oftwo particles
Find the speeds of the two balls after thecollision.
6.5 Dealing With Units 85
Writing conservation of momentum,
m(0.60) + 0 = mv1 + (2m)v2
0.60 = v1 + 2v2
v2 =0.60− v1
2(6.10)
and conservation of energy,
1
2m(0.60)2 + 0 =
1
2mv2
1 +1
2(2m)v2
2
0.36 = v21 + 2v2
2 (6.11)
Now substitute Equation 6.10 into Equation6.11 and do some algebra (and a quadraticequation) to get
v1 = 0.60 v2 = 0 or
v1 = −0.20 v2 = 0.40 (6.12)
The first solution is our initial condition,and the second is the answer after the colli-sion.
6.4 Does This Definition ofMomentum Work Rela-tivistically?
Another reductio ad absurdum proof. As-sume that the solution to the collision is Sec-tion 6.3 is correct for the observer in thelab. Then by the postulates of relativityit should be correct for someone moving atconstant velocity measured by the lab, sayβ = 0.50.
Using the velocities previously found, weuse velocity transformation equations to getthe velocities as measured by the movingobserver. These are summarized in Table6.1.
In the Lab frame momentum is con-served,
Before m(0.600) + 2m(0) = m(0.600)
After m(−.200) + 2m(0.400) = m(0.600)(6.13)
The lab observer sees momentum conservedusing the classical expression. However theSecond Observer has a problem,
Before m(0.143) + 2m(−.50) = m(−0.857)
After m(−0.636) + 2m(−0.125) = m(−0.886)(6.14)
The lab observer sees momentum conserved,while the moving observer says momentum isnot conserved. Similar results hold true forenergy conservation, total kinetic energy isconserved in teh Lab, but not for the SecondObserver.
The Conservation Laws are very powerful inclassical physics, and we wold like to havesimilar laws in special relativity. To do so wewill need to redefine momentum and kineticenergy.
6.5 Dealing With Units
In Chapter 1 we found a way to have timeand position have the same units. Similarlyin relativity we would like to have energyand momentum have the same units. Thesame conversion factor c will suffice.
Classical units of energy are (kg m2
s2) or
joule, J.
86 6.6 New Definitions for Momentum and Energy
Before Lab Measures Second Observer Measures
Ball of mass m 0.60 0.143Ball of mass 2m 0.00 −0.50
After
Ball of mass m −0.20 −0.636Ball of mass 2m 0.40 −0.125
Table 6.1: Classical Calculation of Billiard Ball Velocities Seen in Lab and by an Observermoving at 0.50. Velocities for second observer are found using velocity transformations.While Energy and momentum are conserved for the Lab observer, they are not for thesecond observer as is shown in the text.
Classical units of momentum are(kg m
s ).
By multiplying the momentum by c = 3.00×108 m/s it will have the same units as energy.Or we could divide the energy by c2 to getunits of kg and divide the momentum by cto also get kg.
Example An rock has a momentum of 6.0×10−6 kg m/s. What is its momentum injoules?
p = 6.0× 10−6(3.00× 108) = 180 J
Example A stick has energy of 1800 J andmomentum of 9.0×10−5 kg m/s. Whatare these quantities in kg?
E = 1800 J/(3.00 × 108 m/s)2 = 2.0 ×10−14 kg
p = (9.0 × 10−5 kgm/s)/(3.00 ×108m/s) = 3.0× 10−13 kg
The most common unit for energy and mo-mentum used in special relativity is theelectron-volt, eV. You will discus this whenyou take Modern Physics.
6.6 New Definitions for Mo-mentum and Energy
Spacetime Physics has a very elegantintroduction to relativistic energy andmomentum—so elegant that I am not sure itis easy to learn until you already are familiarwith the answer. So I will just present therelativistic formulas and discuss them.
Here are the relativistic results.
(a) Mass, m, is invariant. For example themass of an electron is 9.11 × 10−31 kgwhen it is at rest or when it is movingrelative to you.
(b) For a particle of mass m moving withvelocity ~v Momentum is defined as
~p = γm~v
px = γmvx
py = γmvy
pz = γmvz (6.15)
(c) The total energy of the particle is
E = γmc2 (6.16)
6.7 An Example 87
(d) There is an invariant quantity relatingmass, energy and momentum,
(mc2)2 = E2 − (pxc)2 − (pyc)
2 − (pzc)2
Or if all quantities have the same units,
m2 = E2 − p2x − p2
y − p2z (6.17)
This is applied to a single particle at atime, not a system.
(e) Relativistic kinetic energy is
KE = E −mc2 = mc2(γ − 1) (6.18)
With these definitions we maintain the Lawsof Conservation of Energy and Momen-tum.
6.7 An Example
Let’s solve the same problem that we didclassically in Section 6.3, shown in Figure6.1.
Consider a collision between two billiardballs on a horizontal pool table. The firstball has mass m = 1 and initially moves tothe right at v0 = 0.60 towards the secondballof mass 2m = 2, which is at rest. Theballs collide head on and separate in a situ-ation where only kinetic energy needs to beconsidered.
Find the speeds of the two balls after thecollision.
Now use the equations of Section 6.6, withsubscripts 0, t for before and 1, 2 for afterthe collision.
Conserve momentum.
γ0(1)(v0) + γt(2)(vt) = γ1(1)(v1) + γ2(2)(v2)
1.25(1)(0.6) + 1(2)(0) = γ1(1)(v1) + γ2(2)(v2)
0.75 = γ1(1)(v1) + 2γ2(v2)(6.19)
Conserve energy.
γ0(1) + γt(2) = γ1(1) + γ2(2)
1.25(1) + 1(2) = γ1(1) + γ2(2)
3.25 = γ1(1) + 2γ2 (6.20)
Solving the simultaneous Equations 6.19 and6.20 is difficult since the unknown velocitiesare in the γs as well as appearing by them-selves. Here is how I did it. First I rear-ranged Equation 6.20 for γ1
γ1 = 3.25− 2γ2 (6.21)
Put this into Equation 6.19
0.75 = (3.25− 2γ2)v1 + 2γ2v2
v1 =0.75− 2γ2v2
3.25− 2γ2(6.22)
Use this to write an expression for γ21
γ21 =
1
1− v21
=1
1−(
0.75− 2γ2v2
3.25− 2γ2
)2
=(3.25− 2γ2)2
(3.25− 2γ2)2 − (0.75− 2γ2v2)2
=(3.25− 2γ2)2
10.5625− 13γ2 + 4γ22 − 0.5625 + 3γ2v2 − 4γ2
2v22
=(3.25− 2γ2)2
10− γ2(13− 3v2) + 4γ22(1− v2
2)
=(3.25− 2γ2)2
10− γ2(13− 3v2) + 4
=(3.25− 2γ2)2
14− γ2(13− 3v2)(6.23)
88 6.8 Evidence
We can equate this to the square of Equation6.21
(3.25− 2γ2)2 =(3.25− 2γ2)2
14− γ2(13− 3v2)
1 =1
14− γ2(13− 3v2)
14− γ2(13− 3v2) = 1
γ2 =13
13− 3v2(6.24)
Finally, we square both sides in express γ2
in terms of v2
1
1− v22
=169
169− 78v2 + 9v22
169− 78v2 + 9v22 = 169(1− v2
2) = 169− 169v22
178v22 − 78v2 = 0
v2 = 0 or v2 =78
178= 0.438
(6.25)
Using Equation 6.22 we find that v1 =0.60 or v1 = −0.220. The first answersare the initial velocities and the second an-swers are the final velocities.
Now we can check to see whether these ex-pressions maintain the Conservation Lawsfor an observer moving at β = 0.50.
Putting these values into Equations 6.19 and6.20 we get the results in Table 6.3 (all quan-tities in the same unit).
So each observer (lab and moving) are happybecause momentum is conserved and energyis conserved. They get different values forthe total energy and the total momentum,but these values are not supposed to beinvariant. Instead the quantity
√E2 − p2
should be invariant, see Table 6.4.
Initial Values Lab Moving Observer
v0 0.600 0.143γ0 1.250 1.010vt 0.000 -0.500γt 1.000 1.155
Final Values
v1 -0.220 -0.648γ1 1.025 1.313v2 0.438 -0.079γ2 1.112 1.003
Table 6.2: One Dimensional Collision of bil-liard balls treated relativistically. Velocitiesand dilation factors for Lab and Second Ob-server.
Indeed√E2 − p2 = m is the invariant mass,
and invariant means the same before or aftera collision and for all observers.
6.8 Evidence
The most direct evidence of the validityof these equations comes from radioactivedecay. Among the early pioneers of ra-dioactivity (a term she coined) was MariaSklodowska who became Marie Curie whenshe married. She chemically separated someradioactive nuclides2, starting with severaltons of uranium ore (pitchblende) and ob-taining less than a milligram of a nuclideshe named after her home country of Poland,Polonium.
2Nuclide refers to a unique combination of pro-tons and neutrons in a nucleus. Isotopes are severalnuclides that have the same number of protons butdifferent numbers of neutrons. Thus He4, O16, P b206
are three nuclides, while He4, He5, and He6 arethree isotopes of helium (and also three nuclides.
6.8 Evidence 89
Initial Lab Moving Observer Lab Moving Observer
p0 0.750 0.144 E0 1.250 1.010pt 0.000 -1.154 Et 2.000 2.309ptotal 0.750 -1.010 Etot 3.250 3.319
Final
p1 −0.225 −0.851 E1 1.025 1.313p2 0.975 −0.159 E2 2.225 2.006ptotal 0.750 −1.010 Etot 3.250 3.319
Table 6.3: Momenta and energy for relativistic collisions of two billiard balls.
Lab Moving
E p m =√E2 − p2 E p m =
√E2 − p2
Mass 1 before 1.250 0.750 1.000 1.010 0.144 1.000Mass 2 before 2.000 0.000 2.000 2.309 −1.155 2.000Mass 1 after 1.025 −0.225 1.000 1.313 −0.852 1.000Mass 2 after 2.225 0.975 2.000 2.006 −0.159 2.000
Table 6.4: Although energy and momentum values vary for different observers, the mass,Equation 6.17, is an invariant for all observers
Scientists can now measure the mass of thenuclides very accurately, and with these re-sults we can see that mass is not conservedbut that the relativistic energy-momentumequations do explain the process. Considerthe alpha decay of the most abundant iso-tope of Polonium3,
Po210 → Pb206 +He4 (6.26)
Careful measurements of the masses of thethree nuclides, in “u”, in Equation 6.26yield
3This isotope is often found in antistatic brushesused in photography. It also is the isotope that killedAlexander Litvinenko in 2006 (a lethal dose is lessthan 1 µg!) This isotope is also used in remote powersources (e.g. the moon.)
Nuclide Mass (u) Uncertainty (u)
Po210 209.9828737 0.0000013Pb206 205.9744653 0.0000013He4 4.00260325415 0.00000000006
If we check to see if mass is conserved we seethat
209.9828737 6= 205.9744653 + 4.0026033
= 209.9770686 (6.27)
where we have rounded to the same numberof digits after the decimal point.
Evidently the original nuclide has more massthan the total mass of the products of thedecay. What has happened to the extra massthat we started out with? The extra massends up as kinetic energy of the two daughternuclides, so we expect that the total kineticenergy of this decay will be (209.9828737 -209.9770686) u = 0.00580515 u.
90 6.9 Massless Particles
This is using a mass unit for energy. We caneasily find the conversion from u to kg, 1u =1.66053886×10−27 kilograms and convert toget a kinetic energy of 1.66053886×10−30 kg,and then use c = 299792458 m/s to get theresult in joules, 8.6637× 10−13 J. The mostcommon unit of energy when dealing withatoms and nuclei is the electron Volt, 1 eV= 1.60217646×10−19 J resulting in a kineticenergy of 5.407×106 eV = 5.407 MeV.
The measured value for this decay is 5.407MeV, in exact agreement.
6.9 Massless Particles
When you take Modern Physics you will dis-cuss light not only as a wave (with wave-length and frequency) but as a particle (withmomentum and energy.) When light istreated as a particle it is called a photon.The invariant mass of a photon is zero.
How do we treat the energy and momentumof a photon? From the relativistic expres-sions in Equations 6.15, 6.16, and 6.17 wesee that in order for momentum to be non-zero when mass is zero, the value of γ mustbe infinite, and hence the speed of the par-ticle must be exactly 1. Also for the photonEquation 6.17 tells us that
Ephoton = |pphoton| (6.28)
If you watch Star Trek you have heard theword positron (Data had a positronic brain).A positron has the same mass as an electronbut is positively charged. When a positronmeets an electron the two annihilate to pro-duce photons, usually 2, rarely 3 or more.The symbols for the electron and positron
are e−, e+ and for the photon γ—yes this isconfusing since we have used γ for the dila-tion factor.
Consider the annihilation shown in Figure6.2. An electron moving at 0.50 annihilateswith a positron at rest. After the collisionthere are two photons moving in oppositedirections. We want to find the energy andmomentum of the two photons.
All quantities will be measured in the sameunit, the MeV. The mass of the electron andpositron is 0.511 MeV. For a speed of 0.50,γ = 1.155.
Energy Conservation yields
1.155(0.511) + 1.000(0.511) = E1 + E2
1.101 = E1 + E2
(6.29)
Momentum conservation gives
1.155(0.511)(0.50) + 0 = p1 + p2
0.295 = −|p1|+ |p2|0.295 = −E1 + E2 (6.30)
Solving the last two equations we get
E1 = |p1| = 0.698 MeV
E2 = |p2| = 0.403 MeV (6.31)
Table 6.5 summarizes the energy and mo-mentum of the objects before and after theannihilation and shows that both energyand momentum are conserved. For the mo-ment focus on the unprimed momenta andenergy—the primed quantities will be dis-cussed in Section 6.12.
6.10 The Compton Effect 91
Figure 6.2: A moving electron meets positron at rest. They annihilate leaving two photons.
6.10 The Compton Ef-fect
One final example of using energy and mo-mentum conservation is the derivation of theCompton Formula.
In 1918 Ohio-born Arthur Compton beganstudying the energy of scattered X-rays. Hefound that the X-rays scattered off “freeelectrons” and was able to use the ideas ofquantum mechanics and special relativity todeduce the theory. For this achievement hewon the Nobel prize in physics in 1927.
Assume a photon of energy and momentumE1 = p1 moves along the x-axis toward anelectron , mass m, at rest. After the colli-sion the photon has energy and momentumE2 = p2 and moves at the angle θ while theelectron has energy and momentum E and pand moves at the angle φ, as shown in Figure6.3.
In the lab it is relatively easy to measurethe photon energies before and after the col-lision, and to measure the angle throughwhich the photon scatters. Our goal is tofind an expression for the final photon energyas a function of initial photon energy, E1, theelectron mass, m, and the angle θ.
First write down Conservation of En-
ergy.
E1 +m = E2 + E
E = E1 +m− E2 (6.32)
For the horizontal component of momen-tum, using the fact that for a photon energyequals momentum,
E1 + 0 = E2 cos θ + p cosφ
p cosφ = E1 − E2 cos θ (6.33)
and in the y-direction
0 + 0 = E2 sin θ − p sinφ
p sinφ = E2 sin θ (6.34)
Now square Equations 6.33 and 6.34 andadd, using the fact that sin2 φ + cos2 φ = 1to get
p2 = E22 + E2
1 − 2E1E2 cos θ (6.35)
Square Equation 6.32 to get
E2 = E21 +E2
2 +m2−2E1E2 +2mE1−2mE2
(6.36)
But E2 − p2 = m2, and putting Equations6.36 and 6.35 into this and doing some alge-bra yields
−E1E2 +mE1 −mE2 + E1E2 cos θ = 0(6.37)
92 6.12 Lorentz Transformations
Figure 6.3: Compton Effect: Scattering of a photon by an electron through an angle θ.
and with a bit more algebra this can be writ-ten in the form
m
E2− m
E1= 1− cos θ (6.38)
Compton measured wavelength rather thanenergy, using the quantum mechanical re-lation (also from Einstein) E = hc/λ andwrote the equation (in SI units, with the c’s)as
λ2 − λ1 =h
mc(1− cos θ) (6.39)
The change in wavelength is very small, sinceh/mc = 2.42630994 × 10−12 m and X-rayshave wavelengths in the range of 1 × 10−8
to 1 × 10−11 m. The effect is unmistak-able when we use gamma-ray photons withshorter wavelengths, such as a gamma rayfrom the decay of a common nuclide, cobalt-60 with wavelength 1.2× 10−12 m.
6.11 Low Speed Approxima-tions
For 300 years the expressions of classical me-chanics worked perfectly fine to explain en-ergy and momentum. Let’s chsck to see ifthe relativistic equations become the sameas classical at low speeds.
Momentum
At low speed γ = 1 and the expression formomentum reduces as
p = γmv → mv (6.40)
which is the classical expression.
Kinetic energy in relativity is defined asE − m. Does this approach our classicallimit?
KE = E −m= γm−m
= m(
(1− v2)−1/2 − 1)
≈ m(
1 +
(−1
2
)(−v2
)− 1
)≈ 1
2mv2 (6.41)
Success! At low speeds the relativisticexpression becomes the classical expres-sion.
6.12 Lorentz Transforma-tions
Earlier we had the Lorentz Transformationsfor time and position. The transformations
6.13 Electric and Magnetic Fields 93
for energy and momentum are similar, andrelate the measurements of energy and mo-mentum recorded by one observer to the en-ergy and momentum recorded by a secondobserver moving along the x-axis at velocityβ.
E′ = γ(E − βpx) E = γ(E′ − β′p′x)
p′x = γ(px − βE) px = γ(p′x − β′E′)p′y = py p′z = pz (6.42)
Example Consider the positron-electronannihilation of Section 6.9 as viewedby an observer moving to the right atβ = 0.267946, γ = 1.03795389. Whatdoes this observer measure for the en-ergy and momentum of the two pho-tons?
Table 6.5 shows the results of the origi-nal calculations and the Lorentz trans-formations. Notice that although thetwo observers see different values for thetotal energy and the total momentum,they both see that momentum and en-ergy conservation work.
6.13 Electric and MagneticFields
The culmination of special relativity is thedescription of electric and magnetic fields.Here I will just describe the invariant quan-tities and give the Lorentz Transformations.This may be covered in a junior level E&Mcourse.
Michael Fowler lays out the basic is-sue at http://galileo.phys.virginia.
edu/classes/252/rel\_el\_mag.html. Aneutral wire carries a current of movingelectrons, all having a velocity v to theright. Outside the wire is a positive charge,also moving right and with the same speedv.
In this frame there is no electric force, butthere is a magnetic force on the charge, ~F =q~v × ~B.
Imagine the view of an observer moving atthe same velocity as the external charge. Forher, the external charge is at rest, the elec-trons are at rest, and the positive charge inthe wire moves to the left at speed v. Thereis no magnetic force, and a classical view ofthe wire would suggest no electric force sincethe wire is still expected to be neutral.
This suggests a paradox that can be resolvedby considering length contraction. In thecorrect analysis the wire is no longer neutralin the second reference frame and there is anelectric force on the external charge.
Just as with space/time, and en-ergy/momentum, the equations are easier ifwe use the same units for both quantities.In the SI system electric field is in units of(N/C) while magnetic field is in units oftesla, T. To get the two fields in the sameunits, (N/C), multiply the magnetic fieldby our conversion factor c. In the equationsbelow I assume the same units for E andB.
Here are the Lorentz Transformations forelectric and magnetic fields—notice thatthey are similar but not identical to ourother Lorentz Transformations.
94 6.14 Summary
Before the Annihilation (MeV)Electron Positron Total
px 0.295026 0.000000 0.295026p′x 0.142121 -0.142117 0.000004E 0.590052 0.511000 1.101052E′ 0.530395 0.530394 1.060790
After the Annihilation (MeV)Photon γ1 Photon γ2 Total
px -0.403013 0.698039 0.295026p′x -0.07106 0.07106 0.00000E 0.403013 0.698039 1.101052E′ 0.530395 0.530395 1.060790
Table 6.5: Energy and Momentum in Positron Annihilation. In the unprimed system anelectron moving at 0.50 toward a stationary positron annihilates to produce two photons.The primed values are tose seen by an observer moving to the right at 0.267946.
E′x = Ex B′x = Bx (6.43)
E′y = γ(Ey − βBz) B′y = γ(By + βEz)
(6.44)
E′z = γ(Ez + βBy) B′z = γ(Bz − βEy)(6.45)
For space/time and momentum/energy welook for invariant quantities. For elec-tric and magnetic phenomena the invariantsare
• Charge
• E2 −B2
• ~E · ~B = ExBx + EyBy + EzBz
All applications of these will be left to futurecourses.
6.14 Summary
In order to keep the Law of Conservation ofMomentum and the Law of Conservation ofEnergy valid, we must redefine some quanti-ties.
It is useful to have the same unit for energy,mass, and momentum. If we use our tradi-tional unit of energy, the conversion factorsfor momentum and mass are
• Mass in units of energy = mc2 where mis the mass in units like kg.
• Momentum in units of energy = pcwhere p is in units like kg m/s.
Relativistic momentum is defined as −→p =γm−→v where we can treat p as having unitsof kg m/s providing v is in m/s and m isin kg, or we can treat p and m as havingrelativistic units of energy, wth v being thedimensionless velocity.
Relativistic total energy is defined as E =
6.14 Summary 95
γm if m is in relativistic energy units, orE = γmc2 for m in kg.
With these definitions the Laws of Con-servation of Momentum and Energy stillwork.
The quantity m2 = E2 − p2 (relativisticunits) or (mc2)2 = E2 − (pc)2 (SI units) isan invariant quantity, the same for all ob-servers.
Kinetic energy is defined in relativistic unitsas KE = E −m = (γ − 1)m
In the limit of low speeds the expressions formomentum and kinetic energy revert to theclassical expressions.
Relativity allows for the existence of parti-cles with no mass, with the only experimen-tally observed such creature being the pho-ton. Massless particles travel at the speed oflight, and for them E = p.
Two observers will measure different quanti-ties for energy and momentum. The quanti-ties are related by Lorentz transformations.As always we assume that the observersmove along the x-axis.
E′ = γ(E − βpx) E = γ(E′ − β′p′x)
p′x = γ(px − βE) px = γ(p′x − β′E′)p′y = py p′z = pz (6.46)
More complicated transformation equationsare available for electric and magneticfields.