chapter 8 electromagnetism - harvey mudd collegesaeta/courses/p111/uploads/y2013/chap08.pdfchapter 8...

Chapter 8

Electromagnetism

While gravity was the first of the fundamental forces to be quantified and atleast partially understood – all the way back in the 17th century – it took anadditional 200 years for physicists to unravel the secrets of a second funda-mental force, the electromagnetic force. Ironically, it is the electromagneticforce that is by far the stronger of the two, and at least as prevalent in ourdaily lives. The fact that atoms and molecules stick together to form the mat-ter you are made of, the contact forces you feel when you touch any objectaround you, and virtually all modern technological advances of the twentiethcentury, all these rely on the electromagnetic force. In this chapter, we in-troduce the subject within the Lagrangian formalism and demonstrate somefamiliar as well as unfamiliar aspects of this fascinating fundamental forcelaw of Nature.

8.1 Gravitation, revisited

So far, we have studied gravity as a force between two (or more) point par-ticles. Perhaps we are instead interested in finding the gravitational forceexperienced by a probe immersed well into a star’s gaseous interior, deepbelow the surface. If we were outside the star, we would intuitively representthe star as a point particle located at its center and compute the gravitationalforce between the probe of mass M and the star of mass m as

F = −GM m

r2r (8.1)

303

8.1. GRAVITATION, REVISITED

where r is the distance between the center of the star and the probe. Whenwe move the probe inside the star, we may guess that the best strategy goesas follows: we conceptually divide up the star into small bits, we compute theforce between each point-like bit and the point probe using equation (8.1),and sum up all these forces to account for the contribution from every smallbit of the star. And this would be a correct approach, since Newtoniangravity is a force law that adds linearly – it obeys the force superpositionprinciple.

There is however a more elegant approach that uses the notion of a vectorfield: a mathematical object associating a vector with every point of space.We introduce a gravitostatic vector field g cast around some source massdistribution as a book-keeping tool. We quantify the source mass distributionthrough a volume mass density function ρm that is not necessarily constant.We then write the relations

∇ · g = −4πGρm , ∇× g = 0 . (8.2)

and define the newtonian gravitational force on a probe mass M as

Fg = M g . (8.3)

We can then check that for ρm = mδ3(r), a source point mass m, one recoversthe Newtonian force between two masses m and M (8.1). Here δ3(r) is athree-dimensional delta function, which is infinite at r = 0, zero elsewhere,and whose volume integral is unity.

EXAMPLE 8-1: Gravitostatic field of a point mass

Given (8.2) and (8.3), we want to show that the gravitational force between two pointmasses is indeed given by (8.1). We start with the source mass represented by ρm = mδ3(r).We can check that the total mass of this mass distribution is∫

ρmdV = m

∫δ3(r)dV = m (8.4)

as expected. We then integrate the divergence equation from (8.2) over all space∫∇ · g dV = −4πG

∫ρmdV (8.5)

304

CHAPTER 8. ELECTROMAGNETISM

FIGURE 8.1 : A Gaussian surface probing the gravity around a point mass.

and use Gauss’ law while employing the spherical symmetry in the problem, as depicted in Fig-ure 8.1. This gives∫

∇ · g dV =

∮g · dA =

∮(gr) · (r2 sin θdθdϕ r) = 4πr2g = −4πGm (8.6)

where we used a spherical Gaussian surface centered at the source mass. This then gives

g = −Gmr2r ⇒ Fg = M g = −GmM

r2r (8.7)

as promised. Since equations (8.2) and (8.3) are linear in g and ρm, the gravitostatic field fromsay two point masses can be correctly built up from this point mass result by summing thefields from each individual point mass. And from this, we can go to three or four, or infinitelymany point masses. Hence, equations (8.2) and (8.3) constitute an equivalent description ofNewtonian gravity to that of (8.1) and the force superposition principle.

Equations (8.2) also satisfy the existence and uniqueness theorem of dif-ferential equations: given a mass density ρm and appropriate boundary con-ditions for g, there is a unique solution for g. We then conclude that, armedwith (8.1), we can in principle tackle any gravitostatic problem involving anycomplicated mass distribution. Let us see how our motivational problem ofa probe inside a star could be handled using this general methodology.

305

8.1. GRAVITATION, REVISITED

FIGURE 8.2 : A Gaussian surface probing the gravity inside a star of uniform volume massdensity.

EXAMPLE 8-2: Gravity inside the body of a star

As an application of our new gravitostatic field method, consider a spherical star of radiusR, mass m, and constant volume density ρ0 = m/(4πR3/3). We would like to find the forcefelt by a probe of mass M that penetrates the star. Starting from (8.2), we integrate over aspherical Gaussian surface of radius r < R, as shown in Figure 8.2∫

∇ · g dV = −4πG

∫ρ0dV = −4πGρ0

(4

3πr3

)= −4πGm

r3

R3. (8.8)

That is, only the fraction of the mass inside the sphere of radius r contributes. The left handside is handled through Gauss’ law∫

∇ · g dV =

∮g · dA = 4πr2g . (8.9)

We then have

g = −Gm r2

R3r . (8.10)

This gives a radial force on the probe of mass M

Fg = −GmMr2

R3r (8.11)

which increase with r until we reach the surface at r = R.

306


Outside the star, the computation proceeds in a similar manner except that the Gaussiansurface now encompasses the entire star mass. We then get the point particle result∫

∇ · g dV =

∮g · dA = 4πr2g = −4πGm⇒ g = −Gm

r2r . (8.12)

As far as gravity is concerned, the star looks like a point mass from the outside. Note that atthe star surface r = R, the two gravitostatic fields (8.10) and (8.12) match up.

8.2 The Lorentz force law

In the mid-nineteenth century, the Scottish physicist James C. Maxwell(1831-1879) combined the results of many experimental observations, havingto do with currents, magnets, and even fuzzy cats, and formulated a set ofequations describing a new force law, the electromagnetic force. Maxwell’sequations in vacuum can be written in Gaussian units as

∇ ·E = 4πρ ∇×E = −1

c

∂B

∂t

∇ ·B = 0 ∇×B =4π

cJ +

1

c

∂E

∂t. (8.13)

Here E and B are the space-time dependent electric and magnetic vectorfields. Through perfect vacuum they can relay a force between objects thatcarry an attribute called electric charge. The quantities ρ and J are thecharge density and current density of the charged stuff that is causing thecorresponding E andB fields. For example, a single, isolated, and stationarypoint particle of electric charge q located at a position r0 has charge andcurrent densities

ρ = q δ3(r − r0) , J = 0 , (8.14)

where δ3(r − r0) is the three-dimensional delta function centered at r = r0.The corresponding electric and magnetic fields are obtained from Maxwell’sequations (8.13) as (with R ≡ r − r0)

E =q

|R|2R , B = 0 (8.15)

known as the Coulomb field. This computation is very similar to thecase of the gravitostatic field of a point mass encountered in the previous

307

8.2. THE LORENTZ FORCE LAW

FIGURE 8.3 : The electrostatic Coulomb force between two charged particles.

section. Another classical example is that of a charge q moving with constantvelocity v. At a position R away from the charge, one finds the Biot-Savartmagnetic field

B =qv ×RcR3

. (8.16)

Given a probe particle of mass M and electric charge Q in the presenceof electric and magnetic fields – generated by other nearby charges describedby ρ and J – the force that the probe particle feels is given by

Fem = QE +Q

cv ×B , (8.17)

known as the Lorentz force. For example, taking the environment of theprobe as consisting of the point charge q of (8.14) and (8.15), the probe feelsa force given by

F =Qq

|r − r0|2r (8.18)

with the probe charge Q located at r and the source charge q located at r0

(see Figure 8.3). This force law should look very familiar.Remember that the gravitational force experienced by a probe mass M

located at r in the vicinity of a source mass m at r0 is given by

F = −G M m

|r − r0|2r . (8.19)

308


Instead of the product −GM m in (8.19), the strength of the electromagneticforce is proportional to the product of the charges q Q as in (8.18). The rest,the inverse square distance law, is the same. This is not a coincidence.Both forces have a geometrical origin and are tightly constrained by similarsymmetries of Nature.

Equations (8.13) consist of eight differential equations for the six fieldstucked within E and B – sourced by some charge distribution described byρ and J . An existence and uniqueness theorem of the theory of differentialequations guarantees that, given ρ and J , equations (8.13) always determineE and B uniquely. In turn, each of the charges making up ρ and J experi-ences the Lorentz force (8.17) and evolves accordingly; which in turn changesthe electric and magnetic fields via (8.13). Hence, we have a coupled set ofdifferential equations for E, B, and the position of the charges – equationsthat are to be solved in principle simultaneously.

In practice, this is a very hard problem. Fortunately, in many practicalcircumstances, there is a clear separation of roles between the charges in-volved in the electromagnetic interactions. Some of the charges – called thesource charges – have fixed and given dynamics and can be used to computethe electric and magnetic fields in a region of interest: that is, given ρ andJ , one uses (8.13) to find E and B. The remaining charges of interest arecalled probe charges. The electromagnetic fields they generate are negligi-ble compared to the ones from the source charges, and their dynamics isdescribed by the Lorentz force law (8.17) with given E and B backgroundfields. This approximation scheme decouples the set of differential equationsinto two separate, more tractable, sets. In this text, we focus on the secondproblem: the mechanics of probe charges in the background of given electricand magnetic fields.

The task at hand is then the following. Given some E and B fields, wewant to study the dynamics of point charges using the Lagrangian formalism.Hence, we want to incorporate the Lorentz force law (8.17) into a Lagrangian– and hence a variational principle.

We start by rewriting the background electric and magnetic fields in termsof new fields that make the underlying symmetries of Maxwell’s equationsmore apparent. From (8.13), we know that the magnetic field is divergence-less. This implies that we can write

B = ∇×A, (8.20)

309

8.2. THE LORENTZ FORCE LAW

due to the identity ∇· (∇×A) = 0 for any vector field A – thus trading theB field for a new vector field A we call the vector potential. Using also(8.13), we can therefore write

∇×(E +

1

c

∂A

∂t

)= 0 (8.21)

which implies that

E +1

c

∂A

∂t= −∇φ⇒ E = −∇φ− 1

c

∂A

∂t(8.22)

due to the general identity ∇×∇φ = 0 for any function φ – thus introducinga new scalar field φ which we call the scalar potential. Hence, we havetraded the six fields in E and B for four fields A and φ. The fact that wecan do so is a reflection of a deep and foundational symmetry underlying theelectromagnetic force law. Given E and B, even A and φ are not unique.We can apply the following transformations to A and φ without changing Eand B – and hence the force law and corresponding physics –

A→ A+∇f , φ→ φ− 1

c

∂f

∂t(8.23)

with any arbitrary function f(t, r). Thus, there is even less physical infor-mation in A and φ. In fact, one can show that, when the dust settles, oneis left with only two physical fields quantifying electrodynamics: from theoriginal six fields in E and B, to the four in A and φ, now to only two.This remarkable symmetry of the theory is called gauge symmetry. In-deed, it is possible to derive electromagnetism based solely on the principlesof relativity and gauge symmetry.

EXAMPLE 8-3: Potentials of a point charge

The electromagnetic fields from a point charge were given in (8.15). We can now find thescalar and vector potentials for a point charge – noting that the answer is not unique due tothe gauge symmetry (8.23). Since B = 0, we can choose

A = 0 (8.24)

satisfying the equation (8.20). Note however we could have chosen for A a vector field of theform A = ∇f for an arbitrary function f – since we have the identity ∇ × ∇f = 0. The

310


A = 0 choice corresponds to the simplest case for which f = constant. To find the scalarpotential φ, we need to do more work. We start from equation (8.22)

E = −∇φ− 1

c

∂A

∂t= −∇φ (8.25)

and we integrate both sides along any path between two points a and b∫ b

a

E · dr = −∫ b

a

∇φ · dr = −∫ b

a

dφ = −φ(rb) + φ(ra) . (8.26)

This then gives

φ(R) =q

R(8.27)

up to an arbitrary constant. The easiest way to check this is to use a simple radial path ofintegration.

EXAMPLE 8-4: Vector potential for a uniform magnetic field

Consider a region of space filled with a uniform magnetic field B = B0z, where B0 is aconstant, and arbitrary static electric field E. We write E = −∇φ from equation (8.22) andlook for a time independent vector potential A satisfying equation (8.20). Again, the choiceis not unique. A particularly useful choice is

A =1

2B × r . (8.28)

We can verify that the curl of this vector potential is indeed B, as long as B is uniform. Todo this, simply use the vector identity

∇× (V1 × V2) = V1(∇ · V2)− V2(∇ · V1) + (V2 · ∇)V1 − (V1 · ∇)V2 . (8.29)

For our case of a magnetic field pointing in the z direction, we then get

A = −1

2Byx+

1

2Bxy . (8.30)

This is a vector field circling about the z axis counter-clockwise, lying in the x-y plane.

311

8.3. THE LAGRANGIAN FOR ELECTROMAGNETISM

8.3 The Lagrangian for electromagnetism

Maxwell’s equations – inspired by a series of experiments – are supposed todescribe a law of Physics in an inertial reference frame. Given that theyimply that light propagates in vacuum with a universal speed c, the speedof light must then be the same in all inertial reference frames. This impliesthat Galilean transformations must be modified and hence Lorentz transfor-mations are established to link the perspectives of inertial observers. This ishow Relativity was historically developed.

The question we now want to address is the following: if an inertialobserver O measures an electric field E and a magnetic field B, what arethe electric field E′ and magnetic field B′ as measured by an observer O′ –moving as usual with constant velocity V along the common x or x′ axis?We assume that the transformations relating these fields are linear in thefields, much like the Lorentz transformation of four-position or four-velocity;we also expect that they are linear in the relative velocity V . We start withexpressions of the form

E′ = a1 ·E + a2 ·B , B′ = a3 ·E + a4 ·B (8.31)

where the ai are four 3× 3 matrices whose components can depend on V atmost linearly. And we require that the Lorentz force law (8.17) fits as thelast three components of a four-force, as seen from Chapter (2). With allthese conditions in place, one finds a unique solution for the ai’s. One canshow that, given E and B as measured by an inertial observer O, anotherinertial observer O′ moving with respect to O with velocity V measuresdifferent electric and magnetic fields E′ and B′ given by

E′ = γ

(E +

V

c×B

)+ (1− γ)

E · VV 2

V (8.32)

B′ = γ

(B − V

c×E

)+ (1− γ)

B · VV 2

V (8.33)

These rather complicated relations become more transparent when writtenin terms of the potentials φ, A and φ′, A′. Introduce a four vector

Aµ = (φ,A) . (8.34)

Then one can show that we have simply

Aµ′ = Λµ′µAµ (8.35)

312


where Λµ′µ is the usual Lorentz transformation matrix of Chapter (2). Hence,the information about the electromagnetic fields is now packaged in a four-vector field Aµ that transforms in a simple manner under Lorentz transfor-mations.

We are now ready to develop a variational principle for the electromag-netic force law using Lorentz symmetry. We start with the familiar relativisticaction

S = −mc2

∫dτ = −mc2

∫dt

√1− v2

c2(8.36)

for a free point mass m. Now we want to add a term to this action suchthat the particle experiences the Lorentz force law as if it had a charge Q insome background φ and A fields; and this combined action must be Lorentzinvariant1. Looking back at the Lorentz force law (8.17), noting in particularthat it is linear in the particle velocity and the background fields, we can writeonly a single Lorentz-invariant integral consistent with these statements andLorentz symmetry:∫

Aµηµνdxν

dτdτ . (8.37)

Adding an appropriate multiplicative constant, we get the full action

S = −mc2

∫dτ +

Q

c

∫Aµηµνdx

ν . (8.38)

We then expand the second term in more detail to get

Q

c

∫Aµηµνdx

ν =Q

c

∫Aµηµν

dxν

dtdt =

Q

c

∫ (−φ c dt+A · dr

dtdt

)= Q

∫dt(−φ+A · v

c

). (8.39)

The full action then becomes

S = −mc2

∫dt

√1− v2

c2+Q

∫dt(−φ+A · v

c

). (8.40)

1In fact, as mentioned earlier in the book, the principle of invariance of this actionunder Lorentz transformations was the guiding principle for developing relativity and theassociated Lorentz transformations. It may appear as a chicken and egg problem; in reality,one should think of the Lorentz symmetry (and gauge symmetry) as the fundamentalrequirements, and the physical consequences as Relativity and Electromagnetism.

313

8.4. THE TWO-BODY PROBLEM, ONCE AGAIN

We leave it as an exercise for the reader to check that the resulting equationsof motion from (8.38) reproduce the Lorentz force law

dp

dt=

d

dt(γvmv) = Q

(E +

v

c×B

)= Fem (8.41)

We thus have a Lagrangian formulation of the electromagnetic force law!In the non-relativistic limit, the action (8.38) becomes

L =1

2mx2

i − qφ+q

cAixi (8.42)

with equations of motion

ma = qE +q

cv ×B , (8.43)

by dropping terms quadratic in v/c while keeping linear terms. This set ofnon-relativistic equations will be our focus in the next several examples.

8.4 The two-body problem, once again

We begin with the familiar two-body problem, now with electromagneticrather than gravitational interactions. Two point particles of masses m1 andm2, located at r1 and r2 respectively, carry electric charges q1 and q2. Let usfirst roughly estimate the relative importance of the various forces involved.Electromagnetic fields propagate with the speed of light in vacuum. Thisimplies that if the particles are moving slowly compared to the speed of light,we may think of the electromagnetic fields cast about them as propagatingessentially instantaneously – always reflecting their instantaneous positions.The electric field E from the point charge q2 a distance r = |r1 − r2| awaygenerates a force on q1 of the order Fel = q1E ∼ q1q2/r

2. If q2 is moving witha speed v2 c while q1 is moving with v1 c, q1 experiences a magneticforce Fm ∼ q1v1B/c ∼ q1q2v1v2/c

2 r2, where we used the Biot-Savart lawfrom (8.16). Accelerating charges radiate electromagnetic energy which canadd a level of complication. However, once again, this effect is much smallerin our non-relativistic regime. Finally, the two masses interact gravitationallywith a force of the order of Fg ∼ Gm1m2/r

2. Putting things together, wesummarize

Fel : Fm : Fg ∼q1q2

r2:q1q2v1v2

c2r2:Gm1m2

r2. (8.44)

314


Since v1v2 c2, we see that the magnetic force between the charges is lessthan the electric force by a factor v2/c2. To compare the electric force tothe gravitational one, consider the case of two electrons with mass m '9× 10−28 g and charge q ' 5× 10−10 esu. We get an estimate for Fe : Fg ∼1 : 10−43... Phrasing things gently, we need not care about gravitationalforces! Gravity normally becomes relevant only when we are dealing withmacroscopic quantities of electrically neutral matter.

Thus, care only about the electrostatic force acting between the two pointcharges. We write the Lagrangian, using (8.43),

L =1

2m1r

21 +

1

2m2r

22 − U(r) (8.45)

where U(r) = q1φ(r), with φ(r) being the scalar potential due to sourcecharge q2 at the location of the probe charge q1

φ =q2

|r1 − r2|=q2

r. (8.46)

This gives the electric potential energy

U(r) =q1q2

r. (8.47)

We now have a familiar two-body problem with a central potential! We canthen import the entire machinary developed in chapter (7). We then willnot repeat the analysis, but instead step through the main benchmarks. Wefirst factor away the trivial center of mass motion and write a Lagrangian forthe interesting relative motion tracked by r = r1 − r2

L =1

2µr2 − q1q2

r(8.48)

where µ = m1m2/(m1 + m2) is the reduced mass. This quickly leads to aone-dimensional problem in the radial direction with fixed energy

E =1

2µr2 + Ueff(r) (8.49)

and effective potential

Ueff(r) =l2

2µr2+q1q2

r(8.50)

where l is the conserved angular momentum. All is very similar to the prob-lem of two point masses interacting gravitationally, except for the followingimportant observations:

315

8.4. THE TWO-BODY PROBLEM, ONCE AGAIN

• If q1q2 < 0, i.e., the charges are of opposite signs, the electric forcebetween the point particles is attractive. Our entire analysis of orbitsand trajectories from the gravitational analogue goes through with thesimple substitution

−Gm1m2 → q1q2 (8.51)

in all equations. We will then find closed orbits consisting of circlesand ellipses, and open orbits consisting of parabolas and hyperbolas.For example, the radius of a stable circular orbit would be given by(see (7.58))

rp =|Gm1m2|

2|E|→ |q1q2|

2|E|. (8.52)

For a hydrogen atom with energy |E| ' 13.6 eV, we would get rp ∼10−10 m, a good estimate for the size of the ground state atom.

• If q1q2 > 0, the electric force cannot be attractive. Unlike the gravita-tional force, the electromagnetic force allows for repulsive effects. Letus then look at this case a little closer. When q1q2 > 0, the formal-ism developed in Chapter (7) still goes through, except we need tobe careful with certain signs. Figure 8.4 depicts the effective potentialof the problem, noting that the dip in the potential has disappearingsignaling the absence of bound orbits. The orbit trajectory becomes

r =(l2/Gm1m2) (1/m1)

1 + e cos θ→ (l2/q1q2m1)

1 + e cos θ(8.53)

with the eccentricity

ε =

√1 +

2El2

(G2m21m

22)m1

→

√1 +

2El2

(q21q

22)m1

. (8.54)

However, since q1q2 > 0, we see fromFigure 8.4 that we necessarily haveE > 0, and hence

ε > 1 . (8.55)

316


FIGURE 8.4 : The effective Coulombic potential between two charges of the same sign forvarious angular momenta.

This implies that only hyperbolas arises as trajectories. Obviously,with a repulsive force, we may not have bound orbits. The interestingphysics problem is that of particle scattering.

8.5 Coulomb scattering

Consider a point charge q1 projected with some initial energy from infinityonto point charge q2 as shown in Figure 8.5. We say that the repulsiveelectrostatic force from q2 scatters the probe of charge q1 at an angle thatcan be read off from (8.53) as

r →∞⇒ cos θ1,2 → −1

ε(8.56)

as shown in the figure. The scattering angle Θ is defined as

θ1 − θ2 = 2θ0 = π −Θ . (8.57)

We then get

cos Θ =2El2/m1 − q2

1q22

2El2/m1 + q21q

22

(8.58)

317

8.5. COULOMB SCATTERING

FIGURE 8.5 : Hyperbolic trajectory of a probe scattering off a charged target.

using (8.54). It is convenient to write the angular momentum l in terms ofthe so-called impact parameter b shown in the figure. Note that b wouldbe the distance of closest approach between the probe and target if therewere no force between them. Looking at the initial configuration at r →∞,we have the angular momentum l given by

l = m1v0b = b√

2m1E (8.59)

where v0 is the initial speed of the probe, related to the constant energy

E =1

2mv2

0 (8.60)

evaluated here at initial infinite separation. We then get

cos Θ =4 b2E2 − q2

1q22

4 b2E2 + q21q

22

. (8.61)

Using the trigonometric identity

cot2 Θ

2=

(1 + cos Θ)2

(1− cos Θ)2 , (8.62)

we can simplify this expression further to

cotΘ

2=

2 bE

q1q2

. (8.63)

318


FIGURE 8.6 : Definition of the scattering cross section in terms of change in impact area2πbdb and scattering area 2π sin ΘdΘ on the unit sphere centered at the target.

This relation tells us the scattering deflection angle a point charge q1 wouldexperience if projected from infinity with energy E and impact parameterb onto another point charge q2. A scattering process such as this one is apowerful experimental probe into atomic structure and was instrumental indiscovering the constituents of atoms. Put simply, one uses the electromag-netic force to poke into the electrically charge universe of the atom – bythrowing charged particles at it.

While the initial energy E of the probe can be controlled, the impactparameter b is in practice impossible to measure on a per scattering atombasis. It is then useful to describe scattering processes through a quantitycalled the scattering cross-section: we look at the change in the areawithin which a probe scatters in relation to a change in the impact parameter,as shown in Figure 8.6. The scattering cross section σ(Θ) is defined as thechange in initial impact area per change in scattering area on a unit spherecentered at the target

σ(Θ) ≡∣∣∣∣ 2πbdb

2π sin ΘdΘ

∣∣∣∣ =b

sin Θ

∣∣∣∣ dbdΘ

∣∣∣∣ . (8.64)

If you think of a stream of incident probe particles falling onto the targetat various unknown impact parameters b with a uniform distribution in b,σ(Θ) is then proportional to the probability of finding a scattered probe at

319

8.5. COULOMB SCATTERING

FIGURE 8.7 : The Rutherford scattering cross section. The graph shows log σ(Θ) as afunction of log Θ superimposed on actual data in scattering of protons off gold atoms.

an angle Θ on the unit sphere centered at the target. In the celebrated caseof Coulomb scattering described in this example, one gets, using (8.63), theso-called Rutherford scattering cross section

σ(Θ) =1

4

(q2

1q22

2E

)2

csc4 Θ

2(8.65)

As shown in Figure 8.7, this probability is sharply peaked in the forwardΘ = 0 direction. It also has a strong dependence on the charges througha fourth power. σ(Θ) can be readily measured, and given the initial probeenergy E, used to determine the charge of the target. A scattering processis effectively a way to looking into atoms using charges – as we look into sayneutral biological tissue using scattered light and a microscope.

EXAMPLE 8-5: Snell scattering

As an illustration of the concept of scattering cross section, consider light scattered by aperfectly polished bead whose surface acts like a mirror. The bead’s radius is R, as shown inFigure 8.8. We want to find the scattering cross section of this bead as parallel light falls onit.

320


FIGURE 8.8 : Scattering of light off a reflecting bead.

We start from equation (8.64). We then need to find b(Θ) using Snell’s law. Looking atthe figure, note that

b(Θ) = R sin θ = R sin

(π

2− Θ

2

)= R cos

(Θ

2

). (8.66)

Using this in (8.64), we get

σ(Θ) =1

4bR csc

Θ

2=

1

4

bR

sin θ=

1

4R2 =

1

4π

(πR2

). (8.67)

The total scattering cross section is then

σT =

∫ 2π

0

∫ π

0

σ(Θ) sin ΘdΘdΦ = 4π1

4π

(πR2

)= πR2 , (8.68)

which is simply the cross sectional area of the bead! This is because the scattering occurs on

contact. As the interaction of the in-falling probe with the target becomes more long-ranged,

the total scattering cross section increases: it is as if the cross section size seen by probes

expands because of the longer range of the interactions.

8.6 Uniform magnetic field

Consider a non-relativistic point particle of massmmoving in the backgroundof some given static magnetic field with no electric fields. The Lagrangian isthen given by equation (8.43)

L =1

2m(x2 + y2 + z2

)+q

cAixi . (8.69)

321

8.6. UNIFORM MAGNETIC FIELD

If the background magnetic field B is uniform as in

B = Bz , (8.70)

we can write a vector potential corresponding to this magnetic field as

A = −1

2Byx+

1

2Bxy , (8.71)

as seen previously in (8.30), with φ = 0. The Lagrangian then becomes

L =1

2m(x2 + y2 + z2

)− q B

2 cxy +

q B

2 cyx , (8.72)

with corresponding equations of motion

mx =q B

cy , my = −q B

cx , mz = 0 . (8.73)

The dynamics in the z direction decouples and is rather trivial. Setting initialconditions

z(0) = 0 , z(0) = Vz (8.74)

we get

z(t) = Vzt , (8.75)

so the particle moves with uniform speed along the z axis as a free particle,along the magnetic field direction. In the x and y directions, however, wehave a more interesting scenario. We can integrate the x and y equationsonce immediately, and get

x = ω0 (y − y0) , y = −ω0 (x− x0) (8.76)

where

ω0 ≡q B

mc; (8.77)

while x0 and y0 are constants of integration whose role will become apparentshortly. The new equations (8.76) suggest the change of variable

X ≡ x− x0 , Y ≡ y − y0 , (8.78)

322


(a) (b)

FIGURE 8.9 : (a) Top view of a charged particle in a uniform magnetic field; (b) The helicaltrajectory of the charged particle.

to yield a somewhat simpler set of coupled equations

X = ω0Y , Y = −ω0X . (8.79)

These can be decoupled quickly by differentiating with respect to time

X = −ω20X , Y = −ω2

0Y (8.80)

leading us to familiar harmonic oscillators. We now see that the point chargecircles in the x, y plane, about the point (x0, y0), as shown in Figure 8.9(a).If X > 0, we have Y < 0 with ω0 > 0. This implies that if q B > 0, thecircling is in the clockwise direction in the x− y plane, as seen looking downthe z axis.

Let us choose a set of convenient initial conditions: first,

Y (0) = 0⇒ X(0) = 0 (8.81)

since X = ω0Y . Next,

X(0) = R⇒ Y (0) = Vy = −ω0R (8.82)

since Y = −ω0X, where we have denoted the radius of the circular trajectoryas R. We then get the trajectory

X(t) = R cos (ω0t) , Y (t) = −R sin (ω0t) . (8.83)

323


In the original coordinates, it is given by

x(t)− x0 = R cos (ω0t) , y(t)− y0 = −R sin (ω0t) ; (8.84)

As promised, this is a circle centered about (x0, y0), with radius R(x− x0

R

)2

+

(y − y0

R

)2

= 1 . (8.85)

The radius R and initial speed Vy are not independent and we have therelation from (8.82)

R =mcVyq B

. (8.86)

Hence, the radius of the circle is given by the initial speed in the x−y plane;the larger the B field, the tighter the radius. This means we can use thissetup to measure attributes of fundamental charged particles, such as theircharge or speed, by measuring the radius of their circular trajectory in knownuniform magnetic fields. The bubble chambers of mid-twentieth century werebased on this simple principle.

Superimposing the x, y motion onto the dynamics in the z direction, weget the celebrated spiral trajectory depicted in Figure 8.9(b) of a chargedparticle in a uniform magnetic field. The phenomenon of charges circling inuniform magnetic field was one of the first tools used by particle physicists toidentify and measure properties of sub-atomic particles. Figure 8.10 shows aphotograph from a Bubble Chamber. Unknown charged particles enter abox immersed in a known external magnetic field. The box is also filled witha dilute fluid that interacts with the particles relatively weakly – creating atrail of bubbles as they pass through. The spirals shown in the figure areactual trajectory of electrons and other more exotic sub-atomic particles!The spiral direction tells us the sign of the charge of the unknown particle.The radius of the spiral can be related to the particle’s speed

q v B = mv2

r⇒ q

mB r = v (8.87)

As the particles moves through the fluid in the box, it loses energy (and hencespeed); thus the decreasing radius of the spiral. The device can be used tomeasure the charge to mass ratio q/m of many sub-atomic particles. This

324


FIGURE 8.10

rather simple device spearheaded the golden age of experimental particlephysics. Unfortunately, it is rather useless for neutral particles like the π0

meson...

EXAMPLE 8-6: Crossed electric and magnetic fields

Consider a non-relativistic point charge q of mass m moving in the background of uniformelectric and magnetic fields given by

E = E0y , B = B0z . (8.88)

We want to describe the trajectory of the particle. We start by finding the potentials. Usingthe example of a uniform magnetic field and an arbitrary electric field from earlier, we canwrite

φ = −E0y , A =1

2B0xy −

1

2B0yx (8.89)

where we used equation (8.28). The Lagrangian follows from (8.43) and is given by

L =1

2m(x2 + y2 + z2

)+ q E0y +

q

2 cB0xy −

q

2 cB0yx . (8.90)

We first identify the canonical momenta

px =∂L

∂x= mx− q B0

2 cy , py =

∂L

∂y= my +

q B0

2 cx , pz = mz . (8.91)

325


Noting immediately that the system is invariant under translations in the z direction, momen-tum pz is conserved and we can describe the evolution along the z axis as motion with constantspeed

mz = 0 . (8.92)

Hence, we can focus on the two-dimensional evolution in the x-y plane by switching to anappropriate inertial frame moving in the z direction along with the particle.

Looking for other constants of motion, we note that the Lagrangian is time independent,so the Hamiltonian

H =1

2m(x2 + y2

)− qE0y (8.93)

is conserved, where we have eliminated the z motion, as described above. Notice that themagnetic field does not appear in the Hamiltonian, due to the fact that the magnetic forcedoes no work. We may have realized this from the form of the magnetic force law

dE

dt= Fm · v = (qv ×B) · v = 0, (8.94)

since the cross product of two vectors is perpendicular to both. The Hamiltonian is a constantof motion and can be used to help find the trajectory of the particle.

The equations of motion in the x and y directions take a simple form

mvx −q B0

cvy = 0 , mvy +

q B0

cvx = q E0 (8.95)

and do very much reflect the presence of the magnetic field. We have written these equationsin terms of the velocity instead of the position since they are easier to decouple in this form:we take the time derivative of the first equation

mvx −q B0

cvy = 0 (8.96)

and now use the second equation to eliminate vy

vx +q2B2

0

m2c2vx −

q2E0B0

m2c= 0 . (8.97)

This equation is now fully decoupled and can be easily solved with the ansatz

vx(t) = A1 cos (ω0t+A2) + C . (8.98)

Here A1 and A2 are integration constants, but ω0 and C are determined by the differentialequations. Substituting this into the second-order differential equation, we get

ω0 =q B0

mc, C =

E0

B0c . (8.99)

326


Using equation (8.95), we can now find vy,

vy(t) = −A1 sin (ω0t+A2) . (8.100)

We have thus determined the velocity of the particle in terms of two constants of integration.These constants can be used to fix the initial velocity at say t = 0. Let us choose vx(0) =vy(0) = 0, leading to

A1 = −E0

B0c , A2 = 0 . (8.101)

Integrating the velocities, we can get to x(t) and y(t). However, it is easier to used theconserved Hamiltonian for this task. Substituting the velocities into the Hamiltonian, wequickly get

y(t) = y0 +E0mc2

q B20

cos (ω0t) (8.102)

where y0 is an integration constant in which we have also absorbed the constant value of theHamiltonian H. Choosing y(0) = 0, we can write more neatly

y(t) = −E0mc2

q B20

(1− cos (ω0t)) . (8.103)

Finally, we can integrate one of the equations of motion to get x(t) from y(t)

x(t) =E0

B0

(c t− c

ω0sin (ω0t)

). (8.104)

Note in particular the term linear in time! To unravel the shape of the trajectory, we firstnote that the velocity vanishes periodically, with a period of T = 2π/ω0. Furthermore, they motion is bounded with amplitude E0c/(B0ω0), while the x motion is unbounded, growinglinearly in time as x → E0c t/B0. However, both x and y oscillate with period T , with xadvancing in discrete steps of E0v T/B0. Figure 8.11 shows the trajectory, with the variousattributes of the motion we just summarized. While the circling pattern about the magneticfield looks familiar, the drift in the x direction is somewhat counter-intuitive since the electricfield points in the y direction. We can understand why this is happening by breaking up theevolution in hypothetical steps: first, the electric field accelerates the positive charge in thedirection of the field; but then the magnetic field forces it into a circle; yet again, the electricfields speeds up the charge; but this leads to a tighter circle imposed by the magnetic field.And as the charge picks up speed, it is forced to curl more and more until it flips around atthe cusps of the trajectory and starts a new cycle. This becomes more apparent if we were tochoose non-zero velocity at the initial time. The figure depicts the results of such scenarios aswell.

327


FIGURE 8.11 : The trajectory of a particle in uniform crossed electric and magnetic fields.

EXAMPLE 8-7: Mirror mirror on the wall

The previous two examples have taught us that probe charges like to spiral in a planetransverse to magnetic vector fields. But both of these examples involved uniform fields. Aninteresting new twist (no pun intended) kicks in when one has non-uniformity in the backgroundfield. Figure 8.12 shows a non-uniform magnetic field profile of interest. The setup has axialsymmetry, hence we adopt cylindrical coordinates ρ, ϕ, and z as depicted in the figure. Dueto this symmetry, the magnetic field cannot depend on ϕ. Furthermore, the figure indicatesthat we want Bϕ = 0. We then have two components to worry about: Bρ(ρ, z) and Bz(ρ, z).But we know from Maxwell’s equations that (the no-magnetic monopole statement)

∇ ·B =1

ρ

∂

∂ρ(ρBρ) +

1

ρ

∂Bϕ∂ϕ

+∂Bz∂z

= 0 (8.105)

where we write the divergence in cylindrical coordinates. In our case, since Bϕ = 0, we have

∂

∂ρ(ρBρ) = −ρ∂Bz

∂z. (8.106)

Integrating this equation, one gets

ρBρ = −∫ ρ

ρ′∂Bz∂z

dρ′ + f(z) (8.107)

with some arbitrary function f(z) arising from the ρ integration. To proceed further, we wantto make some simplifications. The key new ingredient in the setup is the desired non-uniformityalong the z direction – we want ∂Bz/∂z 6= 0. Let us then assume that Bz is arranged so thatit is independent of ρ, Bz(z); and let us arrange so that f(z) = 0. These conditions can be

328


FIGURE 8.12 : A non-uniform magnetic field profile with axial symmetry.

achieved, at least approximately, by a proper alignment and geometry of magnets surroundingthe region of interest. We then have

ρBρ = −ρ2

2

∂Bz∂z⇒ Bρ(ρ, z) = −ρ

2

∂Bz∂z

. (8.108)

Hence, if we want ∂Bz/∂z > 0, we must have a component of the magnetic field vector pointradially inward – with a larger radial component the further away we are from the z axis, asdepicted in the figure. This configuration is known as a magnetic mirror.

We are now ready to tackle the problem of a point charge moving in this backgroundmagnetic field profile. We already expect that spiraling about the z axis will be part of thedynamics; but there is also a new effect to be expected from the region where ∂Bz/∂z 6= 0.To write the Lagrangian, we first need to find the potentials. With no electric field and a staticmagnetic field, we can write φ = 0 using equation (8.22). We are then left with equation (8.20)written in cylindrical coordinates(

1

ρ

∂Az∂ϕ− ∂Aϕ

∂z

)ρ+

(∂Aρ∂z− ∂Az

∂ρ

)ϕ+

(1

ρ

∂

∂ρ(ρAϕ)− ∂Aρ

∂ϕ

)z

= Bρ(ρ, z)ρ+Bz(z)z . (8.109)

Since the choice for A is not unique, we need to specify an ansatz based on an educatedguess. From the axial symmetry, we want no dependence on ϕ. Furthermore, from the case ofuniform magnetic field encountered in earlier examples, we expect that Aϕ may play a centralrole as A whirls around the magnetic field. Let us then try the ansatz

Aρ(ρ, z) 6= 0 , Aϕ(ρ, z) 6= 0 , Az = 0 . (8.110)

Substituting this ansatz into (8.109), we get

∂Aρ∂z

= 0 ,∂Aϕ∂z

= −Bρ(ρ, z) ,1

ρ

∂

∂ρ(ρAϕ) = Bz(z) (8.111)

329


We can then immediately guess at a solution

Aϕ =ρ

2Bz(z) , Aρ = 0 (8.112)

which then implies

∂Aϕ∂z

=ρ

2

∂Bz(z)

∂z= −Bρ(ρ, z) . (8.113)

Checking back with equation (8.108), we see that this is the no-magnetic-monopole conditionencountered earlier. Hence, we have a good choice for a vector potential describing the desiredmagnetic field profile

A =ρ

2Bz(z)ϕ . (8.114)

We now can write the Lagrangian of a probe charge of mass m and charge q using equa-tion (8.43)

L =1

2m(ρ2 + ρ2ϕ2 + z2

)+q

c

ρ2

2ϕBz(z) (8.115)

expressed in cylindrical coordinates. The absences of ϕ and of time from this Lagrangian implythe conservation of canonical angular momentum and the Hamiltonian respectively. We get

pϕ = mρ2ϕ+q

2 cρ2Bz(z) (8.116)

for the canonical angular momentum pϕ = constant; and

H =1

2m(ρ2 + ρ2ϕ2 + z2

)(8.117)

for the Hamiltonian H = constant. Note that the Hamiltonian is the kinetic energy. Hence,the kinetic energy of the probe charge is conserved in the problem! As discussed before, themagnetic force does no work: it can deflect and scramble the probe in complicated ways, butit cannot change its energy – in this case the probe speed’s remains constant.

Next, we look at the equations of motion in the ρ and z direction. We get

mρ = mρϕ2 +q

cρϕBz(z) (8.118)

and

mz =q

2 cρ2ϕ

dBzdz

. (8.119)

We can use the canonical angular momentum conservation equation (8.116) to eliminate ϕfrom these two equations. We have

ϕ =pϕ −mρ2ω0

mρ2(8.120)

330


where we have defined

ω0 ≡q Bz2mc

, (8.121)

the natural circling angular frequency of the probe about the magnetic field, seen already inearlier examples. Note that this ω0 in z dependent. The ρ and z equations become

ρ =p2ϕ −m2ω2

0ρ4

m2ρ3=

(pϕ −mρ2ω0

) (pϕ +mρ2ω0

)m2ρ3

=ϕ

mρ

(pϕ +mρ2ω0

)(8.122)

and

z =ω0

m

1

Bz

dBzdz

(pϕ −mρ2ω0

)= ρ2ω0ϕ

1

Bz

dBzdz

. (8.123)

Note that, for uniform magnetic field dBz/dz = 0, choosing initial conditions such thatpϕ = −mρ2ω0 leads to ϕ = −qBz/mc, ρ = 0, and z = 0 – i.e. the expected spiralmotion about the axis of symmetry. With non-uniform magnetic field, we can then alreadysee why the setup may be called a magnetic mirror. If we start off the particle so thatϕ < 0 (for example by choosing pϕ ∼ −mρ2ω0), as the particle travels from the regionnear z = 0 towards the magnetic funnel of Figure 8.13, the non-uniformity in the magneticfield kicks in through equation (8.123): this is a force in the negative z direction – assumingdBz/dz > 0 and ϕ < 0 – indicating that the probe is pushed back towards negative z.Meanwhile, equation (8.122) suggests that the probe is pushed towards ρ = 0, focusing alongthe z axis. The problem is however somewhat more intricate than this qualitative analysismay suggest: ω0 is in fact a function of z in the region where the magnetic field is changing,and indeed it increases as the particle penetrates this region. Our set of equations of motion,equations (8.120), (8.122), and (8.123), are tightly entangled. For example, equation (8.120)and equation (8.121) suggest that the probe will circle the magnetic field lines faster and fasteras it slows down its advance in the z direction.

To analyze the dynamics more quantitatively, it helps to make an approximation: weassume that the probe experiences the changes in the magnetic field very slowly compared tothe timescale associated with its circling of the field lines. The latter faster time is determinedby ω0. Hence, at short timescales, the particle behaves much like a probe in a uniform magneticfield – spiraling around the field lines – with this magnetic field sampled from the vicinity ofthe probe. As the particle moves along the z direction, it samples larger and larger values ofmagnetic field, which increases ω0 and hence the spinning frequency through equation (8.121).This is known as an adiabatic regime: a slowly changing background parameter graduallyshifting the evolution of a fast motion. We can arrange to be close to this regime by choosing

pϕ ' −mρ2ω0 ⇒ ϕ ' −q Bzmc

(8.124)

which starts off the particle in a circular trajectory around the symmetry axis z at the spin ratewe already know from the case of a probe in a uniform magnetic field. We can then write theconserved Hamiltonian as

H ' 1

2mρ2 +

1

2mz2 +

1

2mρ2ϕ

(−q Bzmc

)=

1

2mρ2 +

1

2mz2 + µBz (8.125)

331


FIGURE 8.13 : A probe particle bouncing off a region of non-uniform magnetic field. Thisphenomenon is known as a magnetic mirror.

where in the last step, we defined a new quantity called the magnetic moment of the circlingprobe

µ ≡ −1

2

q

cρ2ϕ . (8.126)

The purpose of this juggling of terms is to show that µ is approximately conserved in theadiabatic regime. To see this, we take the time derivative of the Hamiltonian, which we knowis conserved,

dH

dt= mρρ+mzz + µBz + µBz = 0 . (8.127)

Looking back at equation (8.122), we see that ρ ' 0 in this regime since pϕ ' −mρ2ω0. Wethen have

dH

dt= mzρ2ω0ϕ

1

Bz

dBzdz

+ µBz + µdBzdz

z = 0 (8.128)

where we used equation (8.123). In the last term, we used the chain rule to write

Bz =dBzdz

z . (8.129)

Substituting for the magnetic moment in the first term of equation (8.128), we then get

dµ

dt= 0 Adiabatic regime . (8.130)

Since µ ∼ ρ2ϕ, this implies that the traditional angular momentum about the z axis isconserved. As the particle samples stronger magnetic fields, it must spin faster by increasing

332


ϕ: as a result of this adiabatic conservation statement, we must then have the particle focustowards the z axis by decreasing ρ. We can determine the trajectory of the probe in the ρ-zplane easily. We take the time derive of µ explicitly

dµ

dt= 0⇒ d(ρ2ϕ)

dt= 0⇒ 2ρρϕ+ ρ2ϕ = 0 . (8.131)

Using equation (8.120), we get

2ρρϕ− 2ρ2ω0 ' 0 (8.132)

where we dropped the ρ as compared to ρϕ and used pϕ ' −mρ2ω0. Looking back at equa-tion (8.121), we find

ω0 =q Bz2mc

=q z

2mc

dBzdz

. (8.133)

Finally, substituting this and equation (8.124) in equation (8.132), we get

ρ

z= −1

2ρ

1

Bz

dBzdz

=BρBz

(8.134)

using equation (8.108). This implies that the probe tracks the magnetic field lines as theycurved towards the z axis! All the while, spinning faster and faster.

Through equation (8.123), we already argued that the probe is pushed away from theregion of non-uniform magnetic field. In the Problems section, you will find the maximumextent the probe penetrates the region of dense magnetic fields before bouncing back. Thismirror effect can be very useful for trapping charged particles. If we arrange two such magneticfield profiles as in Figure 8.14, we have a magnetic bottle in which charged particles canbounce back and forth. Indeed, the magnetic field lines of the earth create such a naturalbottle as shown in the Figure. A plasma of space particles get trapped between the North andSouth pole as they bounce back and forth. At the poles, when the particles decelerate, theyradiate electromagnetic waves in the visible spectrum – putting on a marvelous show of colorsin the skies known as the aurora.

EXAMPLE 8-8: Ion trapping

Imagine trapping a few ions – or even a single ion – in a small enclosure, poking it around,and observing the intricate physics within it as it reacts to external perturbations. This issomething that physicists do regularly using the electromagnetic forces that rule the realmof atomic physics. While many such situations are most interesting because of the quantum

333


Trapping region

FIGURE 8.14 : Various artificial and natural examples of magnetic bottles.

physics they allow one to probe, the basic trapping mechanism can be understood using classicalphysics.

The task is to trap an ion of charge q using external electric and magnetic fields that wecan tune arbitrarily. The simplest setup perhaps would involve pure electrostatic fields that wecan generate by some arrangement of charges far away from the ion. This is however not thecase, as clarified by Earnshaw’s theorem: it is not possible to construct a stable stationarypoint for a probe charge using only electrostatic or only magnetostatic fields in vacuum. Tosee this for the case of electrostatic fields, consider a region of space where the ion probe is tosit and where we have some external electrostatic fields. There are no source charges in thisregion since these are far away from the trapping region. We then have ∇ ·E = 0; and usingE = −∇φ, we get Laplace’s equation for the electric potential φ

∇2φ =∂2φ

∂x2+∂2φ

∂y2+∂2φ

∂z2= 0 . (8.135)

The potential energy of the ion would then be U = qφ, which then implies that ∇2U = 0.For trapping the ion, we then need a minimum in this potential. This implies we would need

∂2U

∂x2> 0 ,

∂2U

∂y2> 0 ,

∂2U

∂z2> 0 while

∂2U

∂x2+∂2U

∂y2+∂2U

∂z2= 0 (8.136)

We immediately see that this is not possible! A saddle surface – where say ∂2U/∂x2 < 0 while∂2U/∂y2 > 0 and ∂2U/∂z2 > 0 – is the best one can do: the ion would quickly find a waydown such a potential, running away to infinity along the x-axis.

There are several ways to circumvent this unfortunate situation. One is to consider timevarying electric fields. Imagine a saddle surface that is, say, spinning fast enough that everytime the ion ventures a little down the potential, it is quickly pushed back into the middle.Along this idea, one builds what is know as a Paul trap. We leave this case to the Problems

334


section of this chapter. In this example, we discuss instead the so-call Penning trap, whichinvolves both electrostatic and magnetostatic fields.

The idea of the Penning trap is to start with a uniform magnetic field

B = Bz (8.137)

which, as we now know, leads to a spiral trajectory of an ion, circling in the x− y plane withangular speed

ω0 =qB

mc. (8.138)

To confine the ion in the z direction as well, we add an electrostatic field described by theelectric potential

φ =φ0

D2

(z2 − x2 + y2

2

)(8.139)

where D is some length associated with the geometry of the setup. Note that this electricpotential satisfies, as it must, the Laplace equation

∇2φ = 0 . (8.140)

The non-relativistic Lagrangian for the ion of charge q then becomes, using (8.43),

L =1

2m(x2 + y2 + z2

)− q B

2 cxy +

q B

2 cyx− q φ0

D2

(z2 − x2 + y2

2

). (8.141)

It is advantageous to switch to cylindrical coordinates ρ, θ, and z given the symmetries of thepotential. We then get

L =1

2m(ρ2 + ρ2θ2 + z2

)+q B

2 cρ2θ − q φ0

D2

(z2 − ρ2

2

). (8.142)

The dynamics in the z direction is then that of a simple harmonic oscillator

z = −ω2zz (8.143)

with

ω2z =

2 qφ0

mD2. (8.144)

Note that we would need

q φ0 > 0 (8.145)

to make sure the ion is trapped in the z direction. We also can write an energy conservationstatement

1

2mz2 +

1

2mω2

zz2 = Ez (8.146)

335


FIGURE 8.15 : The effective Penning potential. At the minimum, we have a stable circulartrajectory. In general however, the radial extent will oscillate with frequency ω0.

for some constant Ez. To see this, multiply (8.143) by z and integrate. For the θ equation ofmotion, one gets the angular momentum conservation law

mρ2θ +qB

2 cρ2 = l (8.147)

where l denotes the angular momentum constant. Instead of looking at the ρ equation ofmotion, we realize we have conservation of the Hamiltonian since ∂L/∂t = 0; we then get

H =1

2m(ρ2 + ρ2θ2 + z2

)+ q

φ0

D2

(z2 − ρ2

2

)(8.148)

for a constant H. This allows us to write an effective potential for a one dimensional problemin the radial direction ρ – akin to the central force problem we have already seen

1

2mρ2 + Ueff(ρ) = H (8.149)

Ueff(ρ) = Ez −1

2l ω0 c+

l2

2mρ2+

1

8mρ2

(ω2

0 − 2ω2z

)(8.150)

where we also eliminated θ in favor of l using (8.147). This potential is shown in Figure 8.15.We identify a minimum at ρ = ρ0

∂Ueff

∂ρ

∣∣∣∣ρ0

= 0⇒ ρ20 =

2 l

m

(ω2

0 − 2ω2z

)−1/2, (8.151)

336


with the curvature near the minimum given by

∂2Ueff

∂ρ2

∣∣∣∣ρ0

= m(ω2

0 − 2ω2z

)> 0 (8.152)

which is positive since typically the oscillation frequency ωz is much shorter than ω0

ωz ω0 . (8.153)

We may then write

ρ0 '√

2 l

mω0. (8.154)

At this critical radius, we can find the angular speed θ using (8.147)

θ∣∣∣ρ0

= −ω0

2+

1

2

√ω2

0 − 2ω2z ' −

ω2z

2ω0≡ −ωm (8.155)

where in the last step, we used ωz ω0. Hence, ωm ωz ω0. The ion circles withradius ρ0 in the x-y plane very slowly at frequency ωm, while oscillating a little bit faster inthe z direction at frequency ωz. To see the role of the third frequency ω0, we note that thegeneral trajectory implied by the effective potential shown in Figure 8.15 involves also radialoscillation. The frequency of this oscillation is given by (8.152)

Ueff(ρ) ' Ueff(ρ0) +1

2m(ω2

0 − 2ω2z

)(ρ− ρ0)

2. (8.156)

This means that ρ oscillates with a frequency√ω2

0 − 2ω2z ' ω0. This is the third shortest

frequency in the problem, tuned by the strength of the external magnetic field. The combinedmotion is shown in Figure 8.16.

It involves a slow circular trajectory in the x, y plane of large radius, on top of which we

superimpose a slightly faster vertical oscillation in the z direction; and on top of these, we

superimpose fast epicycles with tight radii. This setup can in practice achieve ion trapping

that last days. But eventually, the configuration is unstable, and other considerations, such as

energy leak through electromagnetic radiation, invalidate the analysis.

8.7 Contact forces

Consider a square block of mass M resting on an horizontal floor. Newtonianmechanics tells us that the block experiences two forces: a downward pullgiven by M g of gravitational origin. And an upward push by the floor called

337

8.7. CONTACT FORCES

FIGURE 8.16 : The full trajectory of an ion in a Penning trap. A vertical oscillation alongthe z axis with frequency ωz is superimposed onto an fast oscillation of frequency ω0, whilethe particle traces a large circle with characteristic frequency ωm.

the normal force N . A static scenario implies that N = M g, so the forcessum to zero and there is no acceleration. Hence, the normal force adjusts itsstrength as needed to counteract the gravitational pull M g until it matchesit. If the balance succeeds, the block stays put on the floor. If however thenormal push of the floor is not enough because the block is too heavy, thefloor would disintegrate and the block would fall through... The normal forceis an example of a contact force. It arises by virtue of two objects beingin physical contact with each other; in this case, the block and the floor.Contact forces are always electromagnetic in origin. As two entities toucheach other, the atoms at the contact interface push against each other throughelectromagnetic forces. Each of these tiny pushes may be negligible, but with1023 atoms reinforcing the effect, we get a net, effective, macroscopic forcewe call a contact force. Hence, contact forces are not fundamental forces:they are approximate forces that quantify the effect of many complicatedmicroscopic interactions within a simple, effective, phenomenological forcelaw. The normal force, the tension force, and the friction force are examplesof such force laws. Underlying all of them we find intricate electromagneticinteractions between large numbers of constituent atoms.

Our square block rests on the floor. The net effect of the contact forcethat is the normal force is to constraint its vertical motion: the block canonly move sideways and not up and down. The dynamics of the center of

338


mass of the box is now reduced to two degrees of freedom, from the originalthree. Similarly, the tension force in the rope of a blob pendulum implementsa constraint that assures that the ball at the end of the rope does not fly waybeyond a distance equal to the length of the rope. Many more such contactforces translate into statements of constraints on the degrees of freedom.In this section, we develop the technology of solving mechanics problem –with various types of constraints.

Consider a mechanical system parameterized by N coordinates qk, k =1 . . . N . However, due to some constraint forces, there are P algebraic rela-tions amongst these coordinates, given by

Cl(q1, q2, · · · , qN , t) = 0 (8.157)

where l = 1 · · ·P . This means we have effectively N − P generalized coordi-nates or degrees of freedom, instead of the usual N . For example, if we takethe simplest example of a block of mass m resting on a horizontal floor, westart with N = 3 coordinates x, y, z; then we specify P = 1 constraint, theequation z = 0, where z is the vertical direction and the center of mass ofthe block rests at z = 0.

At this point, we have two choices. We could try to use the P rela-tions (8.157) to eliminate P qk’s, and write the Lagrangian in terms of N−Pgeneralized coordinates. This is in a sense what we have been doing so far.For the example at hand, we would write

L =1

2m(x2 + y2 + z2

)−mg z =

1

2m(x2 + y2

). (8.158)

Alternatively, we may want to delay the constraint implementation. Thereare two good reasons for this. First, it may be difficult or inconvenient toeliminate P qk’s using the constraints (8.157). Second, we may be interestedin finding the constraint forces underlying these constraint relations. Forexample, we may want to find out the normal force on a block sliding alonga curved rail; when this force vanishes, the block is losing contact with therail, and we would be able to determine this critical point in the evolution.

Hence, we now focus on a method that delays implementing the con-straints in a mechanical problem, instead dealing with all N qk’s in theproblem. The complication arises because the variational formalism assumesindependent qk’s: the qk’s must not have relations amongst them. Otherwise,in the process of extremizing the action, we would get to the step

δI =

∫ (d

dt

(− ∂L∂qk

)+∂L

∂qk

)δqkdt = 0 , (8.159)

339

8.7. CONTACT FORCES

and, since the δqk’s are not independent due to (8.157), we cannot concludethat the parenthesized expression – that is the Lagrange equations of motion– are necessarily satisfied individually for every k.

Instead, let us consider the new Lagrangian defined as

L′ = L+P∑

l=1

λlCl (8.160)

where we have introduced P additional degrees of freedom labeled λl withl = 1 · · ·P , each multiplying a related constraint equation from (8.157).

For example, with the block on a horizontal floor example, we would write

L =1

2m(x2 + y2 + z2

)−mg z + λ1z . (8.161)

We now assume that the constraint equations (8.157) are not satisfied a priori.We then have N+P degrees of freedom: N qk’s, and P λl’s. Correspondingly,we have N + P equations of motion. For the example at hand, we haveN + P = 3 + 1 = 4 variables left: x, y, z, and λ1.

The equations of motion for (8.160) are

• P equations for the λl’s

d

dt

(∂L′

∂λl

)− ∂L′

∂λl= 0⇒ ∂L′

∂λl= 0⇒ Cl = 0 . (8.162)

These are simply the original constraints (8.157)! But they now arisedynamically through the equations of motion and need not be imple-mented from the outset. The P parameters labeled λl are called La-grange multipliers. For our simple example, we have one Lagrangemultiplier λ1 with equation of motion z = 0 as needed.

• N equations for the qk’s

d

dt

(∂L′

∂qk

)− ∂L′

∂qk= 0 . (8.163)

In terms of the original L, these look like

d

dt

(∂L

∂qk

)− ∂L

∂qk− λl

∂Cl∂qk

= 0 . (8.164)

340


Note the additional terms involving the λl’s (l is repeated and hencesummed over). We can rewrite these as

d

dt

(∂L

∂qk

)− ∂L

∂qk= Fk ≡ λl

∂Cl∂qk

(8.165)

where we defined the generalized constraint forces Fk that essen-tially enforce the constraints onto the qk dynamics. For the block onthe horizontal floor problem, these become

mx = 0 , m y = 0 , m z = λ1 . (8.166)

How do we relate the generalized constraint forces – and hence the λ’s– to the actual forces? For every object i in the problem located atposition ri, denote the total constraint force acting on it by Fi. We alsoknow the relations ri(qk, t) that connect the position of every objectto the generalized coordinates qk. Using all this, we can relate theLagrange multipliers to the constraint forces by noting that the righthand side of the new Lagrange equations must be given by

Fk = Fi ·∂ri∂qk

= λl∂Cl∂qk

. (8.167)

The easiest way to see this is to start with Fi = −∇iU for every particlei, and apply the chain rule to write it in terms of ∂U/∂qk = Fk. Oncewe determine the Lagrange multipliers λl, we can use this relation toread off constraint forces Fi. For the example at hand, we have

λ1 = F · ∂r∂z

= Fz = N . (8.168)

Hence λ1 is the expected normal force.

Before we apply this general and abstract treatment to particular exam-ples, we note that this method of constraints can easily be generalized a bitfurther. Looking back at (8.159), we see that we only need that the constraintis in the form of a variation. That is, we only need to add the constraint inthe infinitesimal change of the action

δI =

∫ (d

dt

(− ∂L∂qk

)+∂L

∂qk+ λlalk

)δqkdt = 0 , (8.169)

341

8.7. CONTACT FORCES

for some functions alk of qk and t. That is, if the constraints on the generalizedcoordinates qk can be written in the form

alkδqk + altδt = 0 (8.170)

where alk and alt are arbitrary functions of qk and t, we can write the followingset of N + P equations of motion

d

dt

(∂L

∂qk

)− ∂L

∂qk= λlalk (8.171)

alkqk + alt = 0 . (8.172)

This is a useful generalization of the original formulation of the problem givenby (8.162) and (8.165) because we do not necessarily have

∂alk

∂t6= ∂alt

∂qk. (8.173)

To see the further reach of this approach, note that if the constraints couldbe written as before in terms of P algebraic relations Cl(t, q) = 0, we couldwrite

0 = dCl =∂Cl∂qk

dqk +∂Cl∂t

dt = alkdqk + altdt (8.174)

reading off the needed alk and alt as functions of qk and t. However, in thisspecial class of constraints, we would have

∂alk

∂t=∂alt

∂qk(8.175)

because of the commutativity of derivatives ∂2Cl/∂qk∂t = ∂2Cl/∂t∂qk. Alge-braic constraints of the form Cl(q, t) = 0 are called holonomic constraints.The type of constraints that cannot be expressed in this form but insteadcan be written in variational form (8.172) are called non-holonomic con-straints. In either case, equations (8.171) and (8.171) gives us a powerfulrecipe to tackle the constrained problem.

Hence, to summarize, when dealing with a mechanics problem involvingconstraints of the form (8.170), we may choose to delay the implementationof the constraints in an effort to extract constraint forces acting on the sys-tem. To do so, we would need to solve a set of N + P differential equations

342


FIGURE 8.17 : A hoop rolling down an inclined plane without slipping.

given by (8.171) and (8.172). In addition to finding qk(t), this procedurealso leads to P Lagrange multipliers that can be related to constraint forcesthrough (8.167). The best way to learn this technology is once again throughexamples.

EXAMPLE 8-9: Rolling down the plane

Consider a hoop of radius R and mass M rolling down an inclined plane as shown inFigure 8.17. To describe the hoop, we may prescribe three variables: a position in twodimensions r and rotational angle θ

r = xx+ yy , θ (8.176)

where the coordinate system is set up tilted as shown in the figure for convenience. However,we know of two potential constraints. First, the hoop is prevented from falling through theincline because of the normal force. This contact force enforces the constraint

y = 0⇒ dy = 0 . (8.177)

Furthermore, if there is friction involved, the hoop may roll down without slipping. Thefrictional contact force enforces the constraint

Rdθ = dx (8.178)

343

8.7. CONTACT FORCES

At the end of the day, these two constraint suggest that the problem involves only one degreeof freedom, not three. The kinetic energy is

T =1

2M(x2 + y2

)+

1

2MR2θ2 (8.179)

while the potential energy is

V = −Mgx sinϕ . (8.180)

We may then proceed as usual by implementing the constraint from the outset. This leads tothe Lagrangian with one degree of freedom, which we may choose to be the x coordinate

L = T − V = Mx2 +Mgx sinϕ . (8.181)

The equation of motion then tells us the acceleration down the incline

x =g

2sinϕ . (8.182)

What if we are interested in knowing the size of the friction force? We still do notcare about the normal force. This means we will implement the normal force constraint givenby (8.177)) from the outset, eliminating the y coordinate; however, we will delay implementingthe frictional constraint given by (8.178). This leaves us with two of the original coordinates,x and θ; and, a new degree of freedom, λ1, a Lagrange multiplier that will measure the sizeof the friction force. We then will need three differential equations. Looking back at (8.178)and mapping it onto the general form (8.172), we read off

a1θ = R , a1x = −1 , a1t = 0 . (8.183)

We then write the Lagrangian in terms of x and θ only

L =1

2mx2 +

1

2MR2θ2 +Mgx sinϕ ; (8.184)

and use the modified equations of motion given by (8.171). This gives for the x direction

Mx−Mg sinϕ = −λ1 = Fx . (8.185)

And for the θ direction, one gets

MR2θ = λ1R = Fθ (8.186)

We now have a set of three differential equations given by (8.185)), (8.186), and

Rθ = x (8.187)

which follows from the constraint (8.178). Solving this system of equations, we get

x =g

2sinϕ , λ1 =

Mg

2sinϕ . (8.188)

344


FIGURE 8.18 : Two barrels stacked on top of each other. The lower barrel is stationary,while the upper one rolls down without slipping.

The novelty is of course the determination of λ1; which we can now relate to the friction forcethrough

Fθ = λ1R =MgR

2sinϕ = F · ∂r

∂θ= F · ∂x

∂θx = FxR = τ . (8.189)

That is, the friction force M g sinϕ/2 applies a torque equal to M gR sinϕ/2. Hence, the

method of Lagrangian multipliers allowed us to selectively extract forces of constraint in a me-

chanical problem without abandoning the powerful and elegant machinery of the Lagrangian

formalism.

EXAMPLE 8-10: Stacking barrels

Consider the problem of two cylindrical barrels on top of each other, as shown in Fig-ure 8.18. The bottom barrel is fixed in position and orientation, but the top one, of mass M ,is free to move. It starts slipping off from its initial position at the top, rolling down withoutslipping due to friction between the barrels. The problem is to find the point along the lowerbarrel where the top barrel loses contact with it. That is, we need to find out the momentwhen the normal force acting on the top barrel vanishes.

We are tracking the motion of the top barrel. Hence, we have a priori three variables tokeep track of, two positions r and θ, and one rotational angle ϕ

r = rr , ϕ (8.190)

345

8.7. CONTACT FORCES

where we use polar coordinates centered on the bottom barrel to track the position of the topbarrel. A normal force acting on the top barrel enforces the constraint

R+ a = r . (8.191)

If the top barrel rolls without slipping, the friction force enforces the constraint

adϕ = Rdθ (8.192)

The full kinetic energy in terms of r, θ, ϕ is then

T =1

2m(r2 + r2θ2

)+

1

2ma2ϕ2 (8.193)

with potential energy

V = mg r cos θ . (8.194)

Note that we included the rotational kinetic energy of the hoop given by (1/2)ma2ϕ2. Sincewe are only interested in the normal force, we implement the frictional constraint (8.192) fromthe outset to eliminate ϕ in favor of θ. We then get

L = T − V =1

2m(r2 + r2θ2

)+

1

2mR2θ2 −mg r cos θ (8.195)

We do not implement the normal force constraint (8.191), which we now write in canonicalform

dr = 0 . (8.196)

This means we will have three degrees of freedom: r, θ, and a Lagrange multiplier λ1 associatedwith (8.191)). We can now read off the relevant coefficients of (8.172)

a1r = 1 a1θ = a1t = 0 . (8.197)

The equations of motion follow from (8.171). For the r direction, we get

mr −mrθ2 +mg cos θ = λ1 = Fr = F · ∂r∂r

= Fr = N (8.198)

while for the θ direction, we get

mr2θ +mR2θ −mgr sin θ = 0 (8.199)

The third and final equation follows from the constraint (8.191), which we now write as

r = 0 . (8.200)

We may now proceed solving this set of three differential equations. It is however slightly moreelegant to realize that we do have energy conservation in this problem. Hence, we can write

E = T + V =1

2m(

(R+ a)2θ2)

+1

2mR2θ2 +mg(R+ a) cos θ (8.201)

346


with r = R + a from the constraint and E a constant. Arranging for the initial conditionsθ(0) = 0 and θ(0) = 0, we have E = mg (R+ a). This implies

θ2 =2g (R+ a)

2R2 + a2 + 2Ra(1− cos θ) . (8.202)

From (8.198), we have

−m (R+ a) θ2 +mg cos θ = λ1 = N . (8.203)

And setting N = 0 at the moment when the top barrel loses contact with the bottom one, weget

θ2c =

g cos θcR+ a

(8.204)

where θc denotes the critical angle at which this condition is satisfied. Using (8.202), we find

cos θc =2

3 + R2

(R+a)2

. (8.205)

There are two interesting limiting cases within this expression. If the top barrel is tiny, we havea/R 1, which leads to

cos θc =1

2. (8.206)

On the other hand, taking the opposite regime a/R 1, one gets

cos θc =2

3. (8.207)

EXAMPLE 8-11: On the rope

A classic problem is that of a blob pendulum consisting of a point mass m at the end ofa rope of length l swinging in a plane (see Figure 8.19). We would like to determine thetension in the rope as a function of the angle θ. We start with two variables

r = rr (8.208)

using a polar coordinate system centered at the pivot as shown in the figure. Hence, ourvariables are r and θ. However, we have a constraint enforced by the tension in the rope

r = l⇒ dr = 0 ⇒ a1r = 1 , a1t = 0 (8.209)

347

8.7. CONTACT FORCES

FIGURE 8.19 : A blob pendulum with one constraint given by the finite length of the rope.

which we immediately used to read off the relevant coefficients for (8.172). The kinetic energyis

T =1

2m(r2 + r2θ2

)(8.210)

while the potential energy is simply

V = −mgr cos θ . (8.211)

The Lagrangian becomes

L = T − V =1

2m(r2 + r2θ2

)+mgr cos θ (8.212)

where we keep track of both r and θ as independent degrees of freedom at the cost ofintroducing one Lagrange multiplier λ1 associated with the constraint. The equation of motionfor r comes from (8.171) and looks like

mr +mrθ2 −mg cos θ = λ1 = F · ∂r∂r

= Fr = T . (8.213)

While that for θ looks like

d

dt

(mr2θ

)+mgr sin θ = 0 . (8.214)

Our three degrees of freedom are associated with three differential equations; and the thirdcomes of course from the constraint (8.209), which we now write as

r = 0 . (8.215)

348


This is enough to determine all three variables of interest. Once again however, it is easier touse energy conservation

E = T + V =1

2ml2θ2 −mg l cos θ . (8.216)

We start with initial conditions given by

θ(0) = θ0 , θ(0) = 0 ⇒ E = −mg l cos θ0 . (8.217)

We then have

θ2 =2g

l(cos θ − cos θ0) . (8.218)

This allows us to find the Lagrange multiplier λ1 in terms of θ

λ1 = −mg cos θ + 2mg (cos θ − cos θ0) = −mg (cos θ − 2 cos θ0) = T , (8.219)

the tension in the rope.

EXAMPLE 8-12: A rubber wheel on a roadA wheel rolls on a two dimensional surface without slipping. If the wheel cannot skid in anydirection, we have an interesting constrained problem. Assuming that the wheel does not tipover, we can describe its state by specifying the coordinates x, y on the plane, and the angleθ measured from the x axis: that is, the wheel’s position on the plane and its orientation.The rolling angle can be related to x and y and hence can be eliminated from the outset.However, the wheel cannot move sideways! yet, it can make gradual turns... Thus, there isindeed another constraint in the problem. How can we quantify it? If we write the vectorperpendicular to the wheel as k = (sin θ,− cos θ), we want to say that the velocity of thewheel is always perpendicular to the k

k · (x, y) = 0⇒ x sin θ − y cos θ = 0 . (8.220)

The orientation angle θ is a freedom that we may take control of by steering the wheel bysome external mechanical method. If we lock the orientation so that θ = 0, equation (8.220)can be integrated and we have an algebraic relation between x and y. With such a holonomicconstraint, the problem now involves only one degree of freedom: θ is fixed to a constant,while x and y are related to each other algebraically; the wheel rolls in a straight line along asingle direction.

What is however we steer the wheel in pursuit of a moving target! Say we aim the wheeltowards a running person, with the wheel’s orientation thus determined by the line joining itsposition and the position of the runner. This implies that theta is not constant, but instead isgenerally given by a function of x, y, and t: θ(x, y, t). In general, equation (8.220) becomes

349

8.7. CONTACT FORCES

non-integralable – we have a non-holonomic constraint. As seen in our general treatmentearlier however, we can still handle this problem. We write the Lagrangian of the wheel as thesum of its rotational and translational kinetic energies; and we then use equation (8.171) andequation (8.172). We leave the treatment as a problem for the reader.

350


Problems

PROBLEM 8-1: Consider an infinite wire carrying a constant linear charge density λ0. Writethe Lagrangian of a probe charge Q in the vicinity, and find its trajectory.

PROBLEM 8-2: Consider the oscillating Paul trap potential

U(z, r) =U0 + U1 cos Ωt

r20 + 2 z2

0

(2 z2 +

(r20 − r2

))(8.221)

written in cylindrical coordinates. (a) Show that this potential satisfies Laplace’s equation.(b) Consider a point particle of charge Q in this potential. Analyze the dynamics using aLagrangian and show that the particle is trapped.

PROBLEM 8-3: Show that the Coulomb gauge ∇ ·A = 0 is a consistent gauge condition.

PROBLEM 8-4: Find the residual gauge freedom in the Coulomb gauge.

PROBLEM 8-5: Show that the Lorentz gauge ∂µAνηµν = 0 is a consistent gauge condition.

PROBLEM 8-6: Find the residual gauge freedom in the Lorentz gauge.

PROBLEM 8-7: A particle of mass m slides inside a smooth hemispherical bowl of radiusR. Use spherical coordinates r, θ and ϕ to describe the dynamics. (a) Write the Lagrangiangeneralized coordinates and solve the dynamics. (b) Repeat the exercise using a Lagrangemultiplier. What does the multiplier measure in this case?

PROBLEM 8-8: A pendulum consisting of a ball at the end of a rope swings back and forthin a two dimensional vertical plane, with the angle θ between the rope and the vertical evolvingin time. However, the rope is pulled upward at a constant rate so that the length l of thependulum’s arm is decreasing as in dl/dt = −α ≡ constant. (a) Find the Lagrangian for thesystem with respect to the angle θ. (b) Write the corresponding equations of motion. (c)Repeat parts (a) and (b) using Lagrange multipliers.

PROBLEM 8-9: A particle of mass m slides inside a smooth paraboloid of revolution whosesurface is defined by z = αρ2, where z and ρ are cylindrical coordinates. (a) Write theLagrangian for the three dimensional system using the method of Lagrange multipliers. (b)Find the equations of motion.

PROBLEM 8-10: In certain situations, it is possible to incorporate frictional effects withoutintroducing the dissipation function. As an example, consider the Lagrangian

L = eγt(

1

2mq2 − 1

2kq2

). (8.222)

351

8.7. CONTACT FORCES

(a) Find the equation of motion for the system. (b) Do a coordinate change s = eγt/2q.Rewrite the dynamics in terms of s. (c) How would you describe the system?

PROBLEM 8-11: A massive particle moves under the acceleration of gravity and withoutfriction on the surface of an inverted cone of revolution with half angle α. (a) Find theLagrangian in polar coordinates. (b) Provide a complete analysis of the trajectory problem,mimicking what we did for the case of Newtonian potential in class. Do not integrate the finalorbit equation, but explore circular orbits in detail. (c) Repeat using Lagrange multipliers.

PROBLEM 8-12: A toy model for our expanding universe during the inflationary epoch con-sists of a circle of radius r(t) = r0e

ωt with our miserable lives confined on the one dimensionalworld that is the circle. To probe the physics, imagine two point masses of identical mass mfree to move on this circle without friction, connected by a spring of spring constant k andrelaxed length zero, as depicted in the figure. (a) Write the Lagrangian for the two particlesystem in terms of the common radial coordinate r, and the two polar coordinates θ1 and θ2.Do NOT implement the radial constraint r(t) = r0e

ωt yet. (b) Using a Lagrange multiplierfor the radial constraint, write four equations describing the dynamics. In process, you bettershow that

a1r = 1 , a1t = −ωr . (8.223)

(c) Consider the coordinate relabeling

α ≡ θ1 + θ2 , β ≡ θ1 − θ2 . (8.224)

Show that the equations of motion of part (b) for the two angle variables θ1 and θ2 can berewritten in a decoupled form as

α = C1α ; (8.225)

β = C2β + C3β ; (8.226)

where C1, C2 and C3 are constants that you will need to find.

PROBLEM 8-13: For the previous problem, (a) identify a symmetry transformation δt, δα, δβfor this system. Find the associated conserved quantity. What would you call this conservedquantity? (b) Then find α(t) and β(t) using equations (8.225) and (8.226). Use the boundaryconditions:

α(0) = α0 , α(0) = 0 , β(0) = 0 , β(0) = C , (8.227)

What is the effect of the expansion on the dynamics? NOTE: This conclusion is the same as inthe more realistic three dimensional cosmological scenario! (c) Find the force on the particlesexerted by the expansion of the universe. Write this as a function of α(t), β(t), and r(t); andthen show that its limiting form for later times in the evolution is given by

2mω2 r(t) . (8.228)

352


k

k'

!

m

r

R

θ

Θ

PROBLEM 8-14: The figure above shows a mass m connected to a spring of spring constantk along a wooden track. The mass is restricted to move along this track without friction. Thewhole setup is mounted on a toy wagon of zero mass resting on a track along a secondfrictionless beam. The wagon is connected by a spring of spring constant k′ to an axleabout which the whole darn thing is spinning with constant angular speed ω. The figure isa top down view, with gravity pointing into the page, and the rest length of each spring iszero. Take a deep breath while reciting the Latin alphabet backwards. (a) First, write theLagrangian of the system in terms of the four variables r, θ, R, and Θ shown on the Figure,without implementing any constraints. (b) Identify two constraint equations. Implement theone keeping the two tracks perpendicular to each other into the result of part (a) by eliminatingR. Do NOT implement the one that spins things at constant angular speed ω. (c) Introducinga Lagrange multiplier for the constraint having to do with the spin, write four differentialequations describing the system. (d) Identify the force on the mass m due to the spin of thesetup, and find all conditions for which this force vanishes!

PROBLEM 8-15: Consider the system shown in the figure below. The particle of mass m2

moves on a vertical axis without friction and the whole contraption rotates about this axiswith a constant angular speed Ω. The frictionless joint near the top assures that the threemasses always lie in the same plane; and the rods of length a are all rigid. (HINT: Use theorigin of your coordinate system at the upper frictionless joint.) (a) Find the equation ofmotion in terms of one degree of freedom θ. (b) Using the method of Lagrange multipliers,find the torque on the masses m1 due to the rotational motion. (c) Find a static solutionin θ and identify the corresponding angle in terms of m1, m2, g, a, and Ω. Consider some

353

8.7. CONTACT FORCES

a

a

a

a

m1m1

m2

frictionless joint!

!

limits/inequalities in your result and comment on whether they make sense. (d) Is the solutionin part (c) stable? if so, what is the frequency of small oscillations about the configuration.(HINT: Use ξ = cos θ and work on the Lagrangian instead of the equation of motion.)

PROBLEM 8-16: Show that the action (8.42) leads to the equations of motion given by (8.43)

PROBLEM 8-17: Show that the action (8.42) leads to the equations of motion given by (8.43)

PROBLEM 8-18: Find the equations of motion for the example in the text of a wheel chasinga moving target using the non-holonomic constraint.

PROBLEM 8-19: Consider the example of the wheel from the example in the text; but now,we have no control on the wheel’s steering except of course at time zero. We start the wheelat some position on the plane, we give it an initial roll ω0, and an initial spin θ0. Describe thetrajectory, assuming that the wheel does not tip away from the vertical.

PROBLEM 8-20: Show that

z2 = v20 − (ρ2 + ρ2ϕ2)

BρBz

(8.229)

for the example of a probe in the background of a non-uniform magnetic field, where v0 is thespeed of the particle; and use this result to demonstrate that the probe will bounce back whilecircling in the same direction.

PROBLEM 8-21: Determine the maximum extent a probe penetrates a region of non-uniformmagnetic field as a function of the initial conditions of the probe. Work in the adiabaticapproximation as in the example of the text.

354

BEYOND EQUATIONS

Leon Foucault (1819 - 1868)

Leon Foucault was born in Paris and lived theremost of his life. He attended the College Stanis-las, where he did not do well. Fellow student J.A. Lissajous wrote: “Nothing about the boy an-nounced that he would be illustrious some day;he health was delicate, his character mild, timid,and not expansive. The frailty of his constitutionand the slow way he worked made it impossible forhim to study at college. He was only able to studysuccessfully thanks to the help of dedicated tutorswatched over by his mother.” However, Foucaultbecame good friends with A. H. L. Fizeau, who became the best of colleagues.

After first studying medicine, Foucault’s fear of blood led him to physicsinstead. He then worked with Fizeau to study light, measuring its intensity invarious situations, and its interference and polarization properties. They alsomeasured the speed of light, and showed that light travels more slowly in waterthan in glass, which directly contradicted Newton’s corpuscular theory.

Foucault became most famous for the Foucault pendulum, which demon-strated Earth’s rotation. His pendulum was hung from the ceiling of the Pan-theon in Paris; huge crowds came to observe the pendulum’s plane of oscillationslowly rotate as the Earth turned. Foucault likewise invented the gyroscope,which he also used to demonstrate Earth’s rotation. He built superb telescopesfor the Imperial Observatory, and traveled with the astronomer Urbain Le Verrierto Spain to photograph a solar eclipse.

Foucault’s career was aided by Napoleon III, who made him an Officier Legiond’Honneur; he was also elected a fellow of the French Academy and of the RoyalSociety of London. His name is one of the 72 inscribed on the Eiffel Tower.

355

8.7. CONTACT FORCES

356

chapter 8 electromagnetism - harvey mudd collegesaeta/courses/p111/uploads/y2013/chap08.pdfchapter 8...

Documents