Mechanics and Electromagnetism

D. Atkinson and O. Scholten ∗

(August 1991)

∗These lecture notes were originally written by D.Atkinson for the courses of 1988/1989 and1989/1990 and have been modified by O. Scholten for the course 1991/1992.

1

Chapter 1

Introduction

The purpose of this course is to explore mechanics and electromagnetism in theformalisms of Lagrange and of Hamilton. The unification that this gives to thesubject is very satisfying; and the pivotal role that Special Relativity has in thetreatment of electromagnetic phenomena is especially clear.

It is assumed that the student has some knowledge of classical mechanics in theformulation of Newton: from this springboard the method of Lagrange is developedin Chapter 2, and that of Hamilton in Chapter 3. Two classical problems are handledwith the new tools in Chapter 4: the Kepler problem of planetary motion, that wasfirst cracked by Newton, and the Rutherford scattering of α-particles by an atomicnucleus.

The reader probably knows the Lorentz transformation of Special Relativityalready: an interesting derivation is given in Chapter 5, and special importanceis attached to contravariant and covariant Lorentz vectors. The starting point ofChapter 6 is the Maxwell system of electromagnetic equations, and these are castinto manifestly relativistically covariant form. Lagrangian and Hamiltonian methodsare used consistently. Chapter 7 is devoted to the theorem of Emmy Noether, andthe conservation of electromagnetic energy, momentum, and angular momentum isshown to follow from invariance of the lagrangian under respectively translations intime and space, and under Lorentz transformations.

The effect of a field on a point charged particle, and the inverse effect of sucha charged particle on the field, are considered in Chapters 8 and 9. Some of thetechnical details of calculations are relegated to the decent obscurity of a half dozenappendices.

The main reference book used with the lectures is:L.D. Landau and E.M. Lifshits, Mechanics and Electrodynamics, Pergamon Press.In the notes at some places this book will be referred to. It is strongly recommendedto try to solve the problems given in the book. Other books that provide a moreextensive treatment of the material at approximately the same level are:K.R. Symon, Mechanics, Addison-Wesley,J.B. Marion, Classical Dynamics, Academic Press,J.R. Reitz, F.J. Milford and R.W. Christy, Foundations of Electromagnetic Theory,Addison-Wesey,R.K. Wangsness, Electromagnetic fields, Wiley,A much more indepth treatment of the material is given in the following two classics,

2

H. Goldstein, Classical Mechanics, Addison-Wesley,J.D. Jackson, Classical Electrodynamics, Wiley.

3

Chapter 2

Euler-Lagrange Equations

2.1 Point Particle

AIM: derive equations of motions without the explicit introduction of forces.REASON: Forces are often clumsy to deal with since it are vectors.

Consider a point-particle of mass,m, in a conservative force-field: that is, supposethat there exists a scalar potential energy, V (~r ), that depends only on the position,~r , of the particle, such that the force is given by

~F = −~∇V. (2.1)

Generally, it is convenient to express vectors in terms of Cartesian components,~r = (x1, x2, x3) , thus:

Fi = −∂V

∂xi

. (2.2)

Newton’s second law can then be written

d

dt[mxi] = −∂V

∂xi

, (2.3)

where dots mean time-derivatives.In the alternative approach the non-relativistic kinetic energy of the particle is

introduced,

T = 12mxixi, (2.4)

the convention being made that a repeated index (here i) is to be summed from 1to 3. Accordingly,

d

dt[∂T

∂xi

] = −∂V

∂xi

. (2.5)

The same equation (2.5) holds for N particles in a conservative force-field. Wewrite the kinetic energy in the form

T = 12mixixi, (2.6)

where we sum over the Cartesian coordinates of all N particles, with the conventionthat m3i−2 = m3i−1 = m3i is the mass of the i’th particle, the x, y, z coordinates

4

of which are x3i−2,x3i−1 and x3i respectively. The potential energy may includeparticle-particle interactions, so long as these are conservative:

V =3N∑i

Vi + 12

3N∑i6=j

Vij. (2.7)

2.2 Canonical Transformation

AIM: rewriting equation (2.5) to obtain an equation of motion that is not onlyvalid for a cartesian coordinate system but for any set of generalized coordinatesdescribing the system.

We shall now perform a (possibly explicitly time-dependent) transformation ofthe 3N Cartesian coordinates, xi, to a set of 3N independent coordinates, qn. A sim-ple example would be the polar coordinates of the N particles, but much more exoticpossibilities exist, as we will see. The purpose is eventually to obtain coordinate-independent equations of motion. By differentiating Eq. (2.6) with respect to qn

and to qn, we find

∂T

∂qn

= mixi∂xi

∂qn

, (2.8)

∂T

∂qn

= mixi∂xi

∂qn

. (2.9)

Please note that in general ∂T∂qn

6= dTdqn

and that care should be taken with thedifference between straight and partial derivatives.

Since the qn are functions of the xi and t, but not explicitly of the qn, it followsthat xi may be written as a function only of the qn and t, so that

dxi

dt≡ xi =

∂xi

∂qk

qk +∂xi

∂t. (2.10)

Now the only dependence on the variables, qk, here is in the explicit occurrence ofqk on the right-hand side, so that

∂xi

∂qn

=∂xi

∂qn

. (2.11)

Hence, combining eqs.(2.9,2.11),

∂T

∂qn

= mixi∂xi

∂qn

, (2.12)

so that

d

dt[∂T

∂qn

] =d

dt[mixi]

∂xi

∂qn

+ mixi∂xi

∂qn

. (2.13)

By using Eq. (2.3), we can write the first term on the right-hand side as

−∂V

∂xi

∂xi

∂qn

= − ∂V

∂qn

, (2.14)

5

since V does not depend on the xi explicitly. The second term on the right ofEq. (2.13) is equal to ∂T/∂qn, as can be seen from Eq. (2.8). Therefore

d

dt[∂T

∂qn

] = − ∂V

∂qn

+∂T

∂qn

=∂L

∂qn

, (2.15)

where the Lagrangian is defined by L = T − V . This is quite different from thetotal energy, E = T + V . Since V does not depend explicitly on xi, neither does itdepend explicitly on qn, which means that the partial qn-derivative of T is equal tothat of L. Hence Eq. (2.15) can be rewritten as follows:

d

dt[∂L

∂qn

]− ∂L

∂qn

= 0, (2.16)

which is the famous Euler-Lagrange equation, the derivation of which was the goalof this chapter. The Lagrangian L should be regarded as a function of q, q, and t;L(q, q, t).

2.3 Hamilton’s Variational Principle

AIM: use a variational principle to arrive at the Euler-Lagrange equation andintroduce the principle of ”least action”.

Finally, we introduce the action, as the time-integral of the Lagrangian:

S(t2, t1) =∫ t2

t1dtL(q, q, t). (2.17)

Hamilton’s variational principle states that, if we vary the functions q(t) and q(t),for t1 < t < t2, but in such a way that q(t1) and q(t2) remain fixed at their physicalvalues, then, of all the possible values that q(t) and q(t) can have, those values thatmake S extremal satisfy the Euler-Lagrange equations. We shall now prove thisstatement.

δS(t2, t1) =∫ t2

t1dt[L(q + δq, q + δq, t)− L(q, q, t)]

=∫ t2

t1dt[

∂L

∂qn

δqn +∂L

∂qn

δqn]. (2.18)

By an integration by parts, we find

δS(t2, t1) =∫ t2

t1dt[

∂L

∂qn

− d

dt(∂L

∂qn

)]δqn, (2.19)

where the integrated terms vanish, since δq is constrained to be zero at t1 and t2.Since δq is arbitrary, it follows that for δS to be zero and thus for S to be extremal,the expression between square brackets must vanish.

There now appear to be two different axioms from which the same Euler-Lagrangeequation can be derived;

• Newton: F = ma.

• Hamilton: The action S is minimal.

Since these give rise to the same equation of motion, which determines the physicalobservables, they are equivalent. For many generalizations used later, Hamiltonsprinciple is more versitile and it will therefore be used frequently.

6

Chapter 3

Hamilton’s Equations

3.1 Canonical Momentum

AIM: introduce momenta associated with the generalized coordinates, the gen-eralization of pi = mxi to non-cartesian coordinates.

In terms of Cartesian components, the Lagrangian of our system of N particles,interacting with one another, and with an external, conservative field, can be written

L(x, x, t) = 12mixixi − V (x, t), (3.1)

where V may depend explicitly on t (for example, a time-dependent external poten-tial), but not on x. Then clearly

∂L

∂xi

= m(i)x(i), (3.2)

where there is no summation over the bracketed index, (i). We recognize the right-hand side of Eq. (3.2) as a component of the linear momentum, and we now definethe generalized, or canonical momentum, by

pn =∂L

∂qn

. (3.3)

It is important to realize that the canonical momenta are not necessarily momentain the Cartesian sense: for example, in terms of spherical polar coordinates, the“momentum” conjugate to the angle, θ, is the angular momentum.

The rate of change of the Lagrangian can be written

dL

dt=

∂L

∂qn

qn +∂L

∂qn

qn +∂L

∂t. (3.4)

Now from the Euler-Lagrange equation, we know that

∂L

∂qn

=d

dt[∂L

∂qn

], (3.5)

so that Eq. (3.4) can be rewritten

dL

dt=

d

dt[∂L

∂qn

qn] +∂L

∂t. (3.6)

7

If the potential and thus L has no explicit time dependence, ∂L∂t

= 0, then fromEq. (3.6) we obtain

d

dt[∂L

∂qn

qn − L] = 0. (3.7)

Since the time derivative of qnpn−L is zero, it is associated with a conserved quantitywhich is the total energy (see Eq. (3.10)) of the system.

3.2 The Hamiltonian

AIM: derive first order equations of motion for the generalized coordinates andthe conjugate (canonical,generalized) momenta.

The Hamiltonian is defined by

H = pnqn − L, (3.8)

and we use Eq. (3.3) to rewrite Eq. (3.6) in the form

dH

dt+

∂L

∂t= 0. (3.9)

If L does not depend explicitly on the time, the second term above is absent, andhence the Hamiltonian is constant in time. If, moreover, the potential energy isconservative, Eq. (3.2) holds,so that

H = mnxnxn − L = 2T − [T − V ] = T + V, (3.10)

which means that the Hamiltonian is equal to the total energy, which is time-independent. Note that two conditions are necessary for these conclusions to betrue: (1) L must not explicitly depend on the time, and (2) V must not explicitlydepend on the q.

From the definition, Eq. (3.8), we deduce

dH = qndpn + pndqn −∂L

∂qn

dqn −∂L

∂qn

dqn −∂L

∂tdt. (3.11)

By using the definition (3.3), we see that the second and the fourth terms abovecancel. Moreover, the same definition, combined with the Euler-Lagrange equation,implies that

∂L

∂qn

= pn, (3.12)

so that

dH = qndpn − pndqn −∂L

∂tdt. (3.13)

The form of this increment suggests that we consider H to be a function of qn, pn

and t (and not independently of qn). Then the following partial derivatives can beread off from Eq. (3.13):

qn =∂H

∂pn

(3.14)

8

pn = −∂H

∂qn

(3.15)

∂L

∂t= −∂H

∂t= −dH

dt. (3.16)

These are the Hamilton equations, which constitute an alternative to the Euler-Lagrange system.

One of the advantages of Hamilton’s approach is that one can readily define moregeneral transformations of the variables. The set, qn,pn, may be replaced by anotherset, Qn,Pn, in which the Q’s and the P ’s can both be functions of all the q’s and thep’s, and possibly of t. If the Hamilton equations, Eq. (3.14)–Eq. (3.16), remain validin terms of the new variables (i.e. the transformation is such that these equations,with qn and pn replaced respectively by Qn and Pn,are true), then we speak of acanonical transformation. The new variables are just as acceptable as the old ones.Pn is called the momentum canonically conjugate to Qn.

3.3 Conservation Laws

AIM: relate the existence of conservation laws to presence of symmetries (invari-ants) in the system. Book: chapter 2

We have seen already that when L has no explicit time dependence i.e. L isinvariant under time translations, there is a conserved quantity, E, the total energyof the system, dE

dt= 0 (see also book, §6 ).

It can also be shown (see book, §7) that if the system is invariant under transla-tions, the operation ~r′i = ~ri+~ε with ~ε being a constant (space and time independent)vector, there is a conserved quantity, the momentum of the system. If the systemis invariant under rotations (see book §9) there is a conserved quantity, the angularmomentum.

In general it is the case that if the system has a symmetry, its properties areinvariant under a class of operations (space translations, rotations etc.) there will bean associated conserved quantity. Finding the conserved quantities greatly facilitatessolving the equations of motion. for this reason it is important to find the symmetriesof the system, the Lagrangian.

3.4 Poisson Brackets

AIM: derive the -so called- Poisson brackets which can be regarded as the classicalequivalent of quantum mechanical commutation relations.

Lastly, we introduce the Poisson bracket between two arbitrary functions of thecanonical variables ((q, p, t) and not q), say F and G:

F, G =∂F

∂qn

∂G

∂pn

− ∂F

∂pn

∂G

∂qn

(3.17)

The time-derivative of the arbitrary function, F , can be written

dF

dt=

∂F

∂qn

qn +∂F

∂pn

pn +∂F

∂t

9

=∂F

∂qn

∂H

∂pn

− ∂F

∂pn

∂H

∂qn

+∂F

∂t

= F, H+∂F

∂t(3.18)

In particular, the total time-derivative of a function that does not depend explic-itly on the time is given by the Poisson bracket of that function with the Hamiltonian:we say that the Hamiltonian generates time-translations. In particular,

qn = qn, H, (3.19)

pn = pn, H. (3.20)

These equations are impressively symmetrical between the q’s and the p’s, and theyreplace the first two of the Hamilton equations, (3.14) and (3.15). It is easy to seethat

qk, pl = δkl, (3.21)

and Dirac’s recipe for the intuitive leap from classical to quantum mechanics is torepresent dynamical variables by linear operators on a Hilbert space, and Poissonbrackets by commutators (multiplied by 2π/(ih)). With this interpretation, thegeneral equation (3.18), as well the “quantization condition”, Eq. (3.21), have thesame forms in classical and in quantum mechanics. Both Poisson brackets andcommutators satisfy the same algebra: in particular the Jacobi identity is satisfiedby both, namely

E, F, G+ F, G, E+ G, E, F = 0. (3.22)

10

Chapter 4

Central field and Two-BodyProblems

4.1 One Dimensional Problem

AIM: show application of E-L equation of motion.As a first problem we will consider the case of a potential depending on a single

coordinate only, U(x),

L = 12mx2 − U(x) (4.1)

Since there is no explicit time dependence the total energy E = 12mx2 + U(x) is a

constant of motion. Using this we arrive at

x =dx

dt=√

E − U(x)/√

12m (4.2)

which is straightforward to integrate yielding

t =√

12m

∫ dx√E − U(x)

+ constant. (4.3)

As a specific example the harmonic oscillator potential U(x) = 12kx2 will be

taken, giving

t =√

12m

∫ dx√E − 1

2kx2+ constant. (4.4)

With the standard substitution x =√

2E/k sin φ the integration can be performed.The full solution of the problem is thus

t =√

m/k(φ + φ0),

x =√

2E/k sin φ, (4.5)

or more familiarly,

x =√

2E/k sin√

k/mt + φ0, (4.6)

where φ0 is determined by the initial conditions.

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4.2 Two-Body Problem

AIM: show equivalence 2-body and central-field problems (see book §11).In this chapter, we shall consider a system of just two particles, at positions ~ra

and ~rb, with an interaction potential energy, V , that depends only on the relativepositions of the particles. The Lagrangian can be written

L = 12ma~ra

2 + 12mb~rb

2 − V (~ra − ~rb). (4.7)

We reduce this to an effective one-particle system by introducing the relative andthe centre-of-mass vectors:

~r = ~ra − ~rb, (4.8)

and

~R =ma~ra + mb~rb

M, (4.9)

where

M = ma + mb. (4.10)

In terms of these variables, the Lagrangian becomes

L = 12M

~R 2 + 12m~r 2 − V (~r), (4.11)

where m is defined by

1

m≡ 1

ma

+1

mb

. (4.12)

The motion thus separates into a trivial one for the center of mass and one that ismore interesting for the relative coordinate. The latter is equivalent to that of aparticle of mass m moving in a potential V (~r).

4.3 Central Field Problems

AIM: show the method of solution for a very general class of problems (see book§12).

A very common class of problems is that in which the force between two particlesdepends on their relative distance only. This problem is equivalent to that of aparticle moving in a central field, a potential V (r) depending only on r = |~r| andnot on its orientation, not on θ and φ in polar coordinates. In polar coordinates thekinetic energy of a particle with mass m is

T = 12m(r2 + r2θ2 + r2 sin2 θ φ2). (4.13)

The Lagrangian of the problem is

L = T − V (r) = 12m(r2 + r2θ2 + r2 sin2 θ φ2)− V (r). (4.14)

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The first observation to make is that this L has no explicit φ dependence (iscyclic in φ) and thus there should be an associated conserved quantity. From theEuler-Lagrange equation in the coordinate φ one deduces that

d

dt

∂L

∂φ=

∂L

∂φ= 0. (4.15)

The conserved quantity, a constant of motion, is therefore

Mz ≡∂L

∂φ= 1

2mr2 sin2 θ 2φ, (4.16)

The z-component of the angular momentum.By choosing a coordinate system in which initially φ = 0 and φ = 0 one obtains

Lz = 0 at t = 0 and at all later times since it it is a constant of motion. It is nowsimple to solve Eq. (4.16), giving

φ(t) = 0 (4.17)

and thus φ(t) = 0.Substituting Eq. (4.17) in the expression for the Lagrangian one observes that

(for this choice of coordinates) the lagrangian is cyclic in θ. From the Euler-Lagrangeequation one obtains the associated constant of motion,

M =∂L

∂θ= mr2θ (4.18)

which is the angular momentum. From the conservation of angular momentum onecan easily derive Keplers second law (book §12). The value of M is known once theinitial conditions, θ(t = 0) and r(t = 0) are specified.

Next the radial motion is solved, realizing that the problem is now reduced toone dimension. The energy of the system equals

E = 12m(r2 + r2θ2) + V (r) = 1

2mr2 + 12M

2/mr2 + V (r) (4.19)

Using the approach discussed in the first section one arrives at

t =∫ dr√

2m

(E − V (r))− M2

m2r2

+ constant (4.20)

Alternatively the angular motion can be expressed as a function of r by startingfrom

M = mr2dθ

dt= mr2dθ

dr

dr

dt, (4.21)

giving

dθ

dr=

1

rM/mr2. (4.22)

Integrating this gives the equation for the orbit of the particle,

θ(r) =∫ Mdr/r2√

2m(E − V (r))−M2/r2+ constant. (4.23)

13

The integration constant is determined by the boundary condition at t = 0.At the turning points the radial velocity r = 0. In general the kinetic energy

of the particle does not vanish since the angular velocity remains finite because ofangular momentum conservation. The position of the turning points can be obtainedfrom solving

E = M2/2mr2 + V (r). (4.24)

Please note that the direction of the angular motion (sign of θ) remains fixed.

4.3.1 Kepler Problem

A common problem (gravitation and coulomb) is one in which the potential isproportional to 1/r, a problem investigated extensively by Kepler in the study ofplanetary motion. The equation for the orbit of the particle, Eq. (4.23) can be solvedanalytically using V (r) = −α/r ,

θ(r) = cos−1

M/r −mα/M√2mE + m2α2/M2

+ constant (4.25)

In the book, §13, an extensive discussion of this problem and its special cases isgiven and will not be repeated here.

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4.3.2 Rutherford Scattering

book: §§14,15,16In the Kepler problem the interest was focussed on closed orbits or boundstates.

In Rutherford scattering one considers a charged probe coming in from a largedistance and scattering off a test charge (target). Since the size of each of theparticles is taken small compared to the typical size in the problem (the distance ofclosest approach), the particles are taken to be point charces.

In a typical experiment one observes the scattered particles, in particular theirdeviation in angle from a straight line trajectory. This deviation is the scatteringangle χ. In an imaginary experiment the scattering angle for each particle can bedetermined with infinite precision. In practice one aims a homogeneous beam ofparticles with n particles per unit area, and measures the number of particles dNscattered between the angles χ and χ′ = χ + dχ. The partial cross-section forscattering in this angle bin is defined as

dσ = dN/n. (4.26)

Please note that dσ carries the units of area.To calculate this we use Eq. (4.23) to obtain θ0, the angle at which the projectile

reaches the distance of closest approach rmin,

θ0 = θ(rmin) =∫ ∞

rmin

Mdr/r2√2m(E − V (r))−M2/r2

+ constant. (4.27)

where the constant in Eq. (4.23) is choosen such that the particle comes in at θ = 0.Instead of using the energy and angular momentum as constants of motion it provesto be easier to introduce instead the initial velocity of the particle,

E = 12mv2

∞, (4.28)

and the impact parameter ρ,

M = mρv∞. (4.29)

The scattering angle can be expressed in terms of θ0 as

χ = |π − 2θ0|, (4.30)

where time reversal symmetry has been used (incoming part of the orbit is same asout going part). Since θ0 depends on E and M or equivalently on ρ and v∞, alsothe scattering angle can be regarded as a function of these two, χ(ρ, v∞). At thispoint we can return to the problem of calculating the scattering cross-section.

Assume that a particle with impact parameter ρ (ρ′ = ρ + dρ) scatters at anangle χ (χ′ = χ + dχ), then dN equals the number of particles that were impingingat the scatterer with impact-parameters between ρ and ρ′,

dN = n2πρ|dρ|. (4.31)

The partial cross-section is

dσ = 2πρ

∣∣∣∣∣ dρ

dχ

∣∣∣∣∣ dχ. (4.32)

15

As a last step the solid angle, dω = 2π sin χdχ is introduced to obtain the differentialcross-section,

dσ

dω=

ρ

sin χ

∣∣∣∣∣ dρ

dχ

∣∣∣∣∣ (4.33)

which is a function of the scattering angle.For the case of the coulomb potential, V (r) = α/r the analytic solution can

be used for the orbits given in the previous section. After some manipulation thedetails of which are given in §16 of the book, the famous Rutherford equation forthe differential cross-section can be derived,

dσ

dω=

(α

2mv2∞

)21

sin4 12χ

. (4.34)

to describe the scattering of α particles off a heavy nucleus.The expression for the cross-section is the same for repulsive and attractive forces,

even though the orbits of the particles are very different.The expression for the cross-section diverges for χ = 0, even the total cross-

section diverges. The reason is that even at large distances an 1/r potential isnon-negligible. The large impact parameters give rise to small angle scattering.

4.4 Small Oscillations

book §§17,19,23Many physical systems are in a quasi-stationary state in the sense that they

are close to their lowest energy state but have not reached it yet. The system canbe described in terms of small oscillations around this equilibrium state. Also forinvestigating the stability of certain solutions one often considers the resonse of thesystem under small oscillations.

Assume that a system has a equilibrium state for the coordimate q = q0. Thepotential has a minimum at q0 which will be normalized to zero, V (q0) = 0. Intro-ducing the coordinate x = q − q0, for small values of x the potential can be writtenas

V (x) = 12kx2 +O(x3). (4.35)

In this expantion the linear term in x vanishes since the potential is required to havea minimum at x = 0. Similarly the kinetic energy can be expressed as

T = 12 a(q) q2 = 1

2 [a(q0) +O(x)] x2 ≈ 12mx2. (4.36)

To leading order the Lagrangian is

L = 12mx2 − 1

2kx2. (4.37)

The equation of motion is

mx + kx = 0, (4.38)

16

which has the well known solutions

x(t) = a cos(ωt + α),

ω =√

k/m (4.39)

or equivalently

x(t) = <(Ae−iωt),

A = ae−iα. (4.40)

From the Lagrangian the momentum can be constructed,

p =∂L

∂x= mx. (4.41)

As a next step the Hamiltonian is obtained,

H = px− L = 12mx2 + 1

2kx2 = 12p

2/m + 12kx2, (4.42)

where the r.h.s. should be used for obtaining the Hamilton equations of motion. Forthis system the energy, E = 1

2mω2a2 is a conserved quantity, a constant of motion.For a system with N degrees of freedom the approach is analogous, see book

§19. The eigen frequencies (in general many, but not more than N) of the systemare obtained from solving the characteristic equation,

|kik − ω2mik| = 0 (4.43)

with solutions

ωα ; α = 1, · · · , s ; s ≤ N. (4.44)

For each frequency there is at least one eigen-mode or normal-mode Qα in which thesystem can oscillate with frequency ω = ωα. The general solution of the problemcan be expressed in terms of these normal-modes,

xk =∑α

AkαQα. (4.45)

In problems like these it often will be advantageous to use the normal-modes as thegeneralized coordinates of the system.

perturbations As mentioned in the begining of this section, the harmonic oscil-lator Lagrangean represents only a lowest order (in the deviation from equilibrium)approximation to the full problem. As an example of the use of perturbation theorythe following Lagrangean is considered,

L = 12mx2 + 1

2m′xx2 − 1

2kx2 (4.46)

with a small (m′ m) anharmonic term in the kinetic energy. The equation ofmotion for this problem is

mx + kx + m′(12 x

2 + xx) = 0. (4.47)

17

Since m′ is small the solution is expanded in its powers, x = x0 +m′x1 +m′2x2 + · · ·.To zeroth order in m′ the equation of motion is:

mx0 + kx0 = 0, (4.48)

with the solution

x0(t) = A0e−iωt ; ω2 = k/m. (4.49)

The part of the equation of motion proportial to m′ reads

mx1 + kx1 + 12 x0

2 + xx0 = 0. (4.50)

Since the solution of the homogeneous equation is the solution for x0, only the specialsolution needs to be obtained,

x1(t) = −12A

20/m[1 + e−2iωt] . (4.51)

CHECK THIS EQN.where the explicit form of the solution for x0(t), Eq. (4.49), has been used. This

procedure can be extended to higher order terms, but beware of some pittfalls, seebook §23.

18

Chapter 5

Special Relativity

5.1 Lorentz Transformation

The transformation between one coordinate system, (t, x, y, z), and another,(t′, x′, y′, z′), such that the space axes of the two systems are coincident at t′ = 0 = t,and parallel thereafter, and such that the primed system has velocity v, along thex-direction, with respect to the unprimed system, can be written

x′ = γ(x− vt) y′ = y z′ = z . (5.1)

The classical assumption (Galileo, Newton) was that γ = 1, but if we only requirethat the origin of the primed system have velocity v with respect to the unprimedsystem, the above more general transformation is possible. What is γ? From theuniformity of space, we see that γ must not depend on the coordinates (in a non-uniform gravitational field, near a star, for example, the properties of space are notuniform, i.e. translation-invariant, but that is the domain of general, not of specialrelativity). In the absence of gravity, γ is independent of t, x, y, z, but it can dependon v. Space is also invariant under rotations: a rotation by 180 degrees about they-axis should have no measurable effect; but it changes the sign of the x and thez coordinates, and also of the relative velocity of the primed system. Rotationalinvariance therefore means that γ must be independent of the sign of v, i.e. it is afunction only of v2.

The unprimed system has a velocity −v, with respect to the primed system,along the x-direction, so the inverse relation between the two systems is

x = γ(x′ + vt′) y = y′ z = z′ , (5.2)

where γ is the same as in Eq. (5.1), since it is unaffected by the change in the signof v.

Note that we have not assumed that the coordinate, t, in the unprimed system, isthe same as t′, the coordinate in the primed system. Such an assumption was made,quite explicitly, by Newton, in his Principia: “Absolute, true, and mathematicaltime, of itself, and from its own nature, flows equably without relation to anythingexternal”. If we set t′ = t, with Newton, and do not enquire too closely what, ifanything, ‘equable flow of time’ means, then Eq. (5.1) and Eq. (5.2) together implyγ = 1. Now if light is emitted at the origin of space- and travels along the x-axis with

19

velocity c, then at t, the light will have reached the point x = ct. From Eq. (5.1),with γ = 1, we see that x′ = (c − v)t, which means that the speed of light, asmeasured in the primed system, should be c− v.

The experimental fact that the measured speed of light is however independentof the motion of the source, and of the observer (Michelson-Morley experiment andthe tests of ‘æther-drag’ in the solar-system), forces the conclusions t′ 6= t and γ 6= 1.Since the speed of light in the primed system is actually c, and not c− v, it followsthat x = ct in the unprimed system must correspond to x′ = ct′ in the primedsystem. Filling in these values into Eq. (5.1) and Eq. (5.2), we find

ct′ = γ(c− v)t (5.3)

ct = γ(c + v)t′ ;

and by multiplying these two equations together, we obtain

γ = 1/

√1− v2

c2. (5.4)

By eliminating x′ between Eq. (5.1) and Eq. (5.2), we see that

t′ = γ(t− v

c2x) . (5.5)

From now on, we choose units of distance and time in such a way that c = 1, sothat we may rewrite the Lorentz transformation, as it is called, in the form

t′ = γ(t− vx) x′ = γ(x− vt) . (5.6)

Further, by introducing the parametrization v = tanh u, so that γ = cosh u andγv = sinh u, we can rewrite Eq. (5.6) in the form

t′ = t cosh u− x sinh u x′ = −t sinh u + x cosh u . (5.7)

This form looks very much like a rotation: if the space axes are rotated by an angleθ about the z-axis, we have

y′ = y cos θ − x sin θ x′ = y sin θ + x cos θ . (5.8)

There are two important differences between Eq. (5.7) and Eq. (5.8), namely that,in the former, hyperbolic functions replace circular ones, and the signs of the twohyperbolic sines are the same, whereas those of the two circular sines are different.One can regard the Lorentz transformation as a ‘hyperbolic rotation’ between thet-axis and the x-axis.

20

5.2 Contravariant and Covariant Vectors

AIM: introduce a compact notation without which formulation of Electromag-netism will be very clumsy. The notation makes space-time equivalence manifest.

With c = 1, the relativistically invariant interval, ds, between two infinitesimallyseparated points is given by

ds2 = dt2 − dx2 − dy2 − dz2 (5.9)

In the following, we write x0 = t, x1 = x, x2 = y, x3 = z. In terms of the metrictensor, gµν , which is defined to be equal to be +1 if µ = 0 = ν, to be −1 if µ = i = ν,for i = 1, 2, or 3, and to be 0 if µ 6= ν, we can rewrite Eq. (5.9) compactly as follows:

ds2 = gµνxµxν . (5.10)

Here the Einstein summation convention has been used, that is, the repeated indices,µ and ν, are implicitly summed from 0 to 3 . If a new set of coordinates, x′µ ,arelinearly related to the old ones,

x′µ = Λµνx

ν , (5.11)

where the matrix, Λµν , is constant (i.e. independent of x), and is such that

gµνx′µx′

ν= gµνx

µxν , (5.12)

then we speak of a Lorentz transformation. In Eq. (5.11) and Eq. (5.12), there is animplicit summation over the repeated indices. This will be a general rule: if a Greekindex occurs above (a contravariant index), and below (a covariant index), in thesame term, then a summation over the values 0, 1, 2, 3 is implied. Any quadruple ofnumbers, aµ, together with the transformation law,

a′µ = Λµνa

ν , (5.13)

is said to be a contravariant Lorentz four-vector.Since the matrix, Λµ

ν , is independent of x, it follows, by differentiation ofEq. (5.11), that

Λµν =

∂x′µ

∂xν, (5.14)

so that the transformation law, Eq. (5.13), can also be written

a′µ =∂x′µ

∂xνaν . (5.15)

Suppose now that Φ is a Lorentz invariant function of x. Then, by the usualchain rule for partial derivatives,

∂Φ

∂x′µ=

∂xν

∂x′µ∂Φ

∂xν. (5.16)

Hence the partial differentiation operator does not have the same transformationlaw as the contravariant vector Eq. (5.15). Rather, it is an example of a covariantvector, bµ, which transforms as follows:

b′µ =∂xν

∂x′µbν . (5.17)

21

The product of any contravariant vector, aµ, and any covariant vector, bµ, which weoften write simply as ab, is Lorentz invariant:

a′b′ = a′µb′µ =

∂x′µ

∂xρ

∂xσ

∂x′µaρbσ = aρbρ = ab. (5.18)

If aν is a contravariant vector, and we define

aµ = gµνaν , (5.19)

then clearly

a′µ = gµνa′ν = gµν

∂x′ν

∂xρgρσaσ, (5.20)

where gρσ, the contravariant metric tensor, is equal to gρσ, the covariant tensor thatwe introduced above. However, by differentiating Eq. (5.12) with respect to xρ, weobtain

gµνx′µ ∂x′ν

∂xρ= xµgµρ. (5.21)

Next, differentiate both sides with respect to x′λ, and recall that the matrix Eq. (5.14)is independent of x:

gµν∂x′ν

∂xρ= gρτ

∂xτ

∂x′µ(5.22)

Multiply both sides by gρσ , and use the obvious fact that

gρτgρσ = δσ

τ , (5.23)

where the Kronecker δ is equal to 1 if both indices are equal, and to 0 if they arenot. Hence

gµν∂x′ν

∂xρgρσ =

∂xσ

∂x′µ. (5.24)

By using Eq. (5.20), we see that aµ transforms as a covariant vector. In a similarway, if bµ is a covariant vector, then

bµ = gµνbν , (5.25)

is a contravariant vector.

22

CONTRA- & CO-variant SummarryTake a point at time t and at space coordinates (x, y, z). In space-time coordi-

nates this point is specified as

Contravariant : xµ = (t, x, y, z) , by definition. (5.26)

Covariant : xµ = (t,−x,−y,−z) , µ = 0, 1, 2, 3. (5.27)

The relation between co- and contra- variant vectors is thus given by

x0 = x0 xi = −xi ; i = 1, 2, 3. (5.28)

The invariant quantity is s2 = t2 − x2 − y2 − z2. The metric tensor is thus given by

gµν = gµν =

1 µ = ν = 0−1 µ = ν = 1, 2, or 30 µ 6= ν

(5.29)

or

g =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

. (5.30)

The metric tensor is obviously symmetric, gµν = gνµ. The definition of a covariantvector is

xµ = gµνxν (5.31)

with an implicit sum over repeated indices. Please check that using Eq. (5.26) thisleads to Eq. (5.27). By definition of the metric tensor the invariant length can bewritten as

s2 = t2 − x2 − y2 − z2

= gµνxµxν = xµgµνx

ν = xνgρνxρ

= xnuxnu = xmux

mu

This last line is the reason for introducing the co- contra-variant notation. Thepermutations and relabelling are allowed since we are not dealing with quantumme-chanical operators but only with summations over real numbers, which commute.In other words, a summation over individual matrix elements is implied, not themultiplication of matrices. Always make sure that with a substitution of repeatedindices the correct number of implicit summations is implied;

(s2)2 = xµxµxνxν (5.32)

with the implicit summation convention µ = 0, · · · , 3 and ν = 0, · · · , 3, the summa-tion runs over 4× 4 = 16 terms, is thus not equal to

xµxµxµxµ (5.33)

since the summation convention is only µ = 0, · · · , 3 and the implicit sum containsa total of only 4 terms.

23

The contra/co variant notation has been introdeced because of the following rule:RULE: If a quantity contains an equal number of upper (contravariant) and lower(covariant) indices, and if each implicit summation runs over one upper and onelower index and all indices are summed over in this manner, the quantity is aninvariant.

Please note that the introduction of upper and lower indices is only a matter ofbookeeping. It does not contain any magic, the above rule only ensures that thecorrect metric tensors are included.

Lorentz transformationsContravariant vectors transform as:

x′µ = Λµν xν (5.34)

where Λµν are the matrixelements of a 4 × 4 matrix, or, in sloppy language, Λµ

ν isa 4× 4 matrix. The matrix elements can be expressed as

Λµν =

∂x′µ

∂xν(5.35)

Covariant vectors transform according to

x′µ = Λ νµ xν (5.36)

where

Λ νµ = gµρ gντ Λρ

τ 6= Λµν (5.37)

24

5.3 Mechanics of a Free Particle

In order to guess the correct Lagrangian for a free particle in relativity theory,and hence to infer the form of relativistic mechanics, we consider again Eq. (2.17),which we can rewrite in the rest-frame of the particle, in which the -variable is τ ,the proper of the particle, which is also the invariant interval:

(dτ)2 = (dx0)2 − (dx1)2 − (dx2)2 − (dx3)2. (5.38)

In the rest-system, we write Eq. (2.17) in the form

S =∫ b

adτL(xi, xi, τ). (5.39)

The integration is taken from a space- point point, a, to another one, b. The problemis that we do not know what L should be. It depends asymmetrically on space and, and certainly is not a relativistic invariant. However, the action only refers to thetwo space- points, a and b. This is true in all Lorentz frames, although of coursethe and space coordinates of a and b do depend on the frame that is chosen. TheHamilton variational principle simply says that the physical trajectory between aand b is the one for which δS = 0. This is a Lorentz invariant specification of thedynamics, and it is consistent with the principle of relativity to require that theaction be a Lorentz invariant. However, the only invariants available, out of whichthe action of a free particle could be made, are the mass, m, and the invariantinterval, τ . The action must be translationally invariant, so we cannot include τitself, but only the invariant measure, dτ . Hence Eq. (5.39) takes the form

S = κ∫ b

adτ, (5.40)

where κ is a function of the mass of the particle, m, only. Suppose that we nowtransform from the rest-system to any other Lorentz frame. From Eq. (5.38), we seethat

dτ = dt√

1− x21 − x2

2 − x23 = dt

√1− v2, (5.41)

where t ≡ x0, and v is the velocity of the new frame, with respect to the rest-system. By comparing this equation with Eq. (2.17), we can read off the form ofthe Lagrangian in a general Lorentz frame:

L = κ√

1− v2. (5.42)

The constant, κ, can be identified by expanding the above expression to first orderin v2:

L = κ− 12κv2 + O(v4), (5.43)

from which it follows that κ = −m, so that L reduces in the low velocity limit tothe correct non-relativistic kinetic energy,

T = 12mv2, (5.44)

25

aside from the constant, −m, which drops out of the Euler-Lagrange equation. TheLagrangian is thus

L = −m√

1− v2 . (5.45)

We next calculate the canonical momenta:

pi =∂L

∂xi= −m

∂

∂xi

√1− xjxj =

mxi

√1− v2

. (5.46)

The Hamiltonian is derived from the standard formula, Eq. (3.8), yielding

H = xi mxi

√1− v2

+ m√

1− v2 =m√

1− v2. (5.47)

According to the general argument of Chapter 3, H will be a constant in time, andwe now assume that it is equal to the total energy, E, as in the non-relativistic case.This assumption has far-reaching consequences, for we see that the energy of a freeparticle in its rest-frame (v = 0), is not zero, but is equal rather to the rest mass(in units for which c 6= 1, the rest-energy is mc2).

The form of Eq. (5.47) is not yet in canonical form, i.e. it is not expressed interms of the coordinates and momenta. To remedy this, we observe that

pipi

m2=

xixi

1− v2=

v2 − 1 + 1

1− v2= A2 − 1 , (5.48)

where A = 1/√

1− v2. Hence

A =

√1 +

pipi

m2; (5.49)

and by using the fact that H = mA, we obtain the Hamiltonian in canonical form:

H =√

m2 + pipi . (5.50)

Note that, for ~p 2 = pipi << m2,

H = m +~p 2

2m− ~p 4

8m3+ O(

~p 6

m5). (5.51)

We recognise the second term as the nonrelativistic kinetic energy. The first term,the rest-mass of the free particle, is the equivalent energy that is locked up ina particle at rest, and which can be liberated on the annihilation of matter andantimatter.

26

5.4 Four-Momentum

Next, we introduce the 4-velocity,

uµ =dxµ

dτ, (5.52)

which is clearly a contravariant 4-vector. From Eq. (5.41), we see that

uµ =1√

1− v2

dxµ

dx0(5.53)

so that

mu0 =m√

1− v2= E, (5.54)

and

mui =m√

1− v2

dxi

dx0= pi. (5.55)

Hence

pµ ≡ muµ = (E, ~p ), (5.56)

is a contravariant 4-vector. The invariant,

pµpµ = E2 − ~p 2, (5.57)

has the same value in any Lorentz frame. In the rest-frame, it is clearly equal tom2, so in general

E2 = ~p 2 + m2. (5.58)

As a simple application of this last formula, consider the decay at rest of aπ+ meson, of mass mπ, into a µ+ lepton, of mass mµ, and a neutrino, which hasmass zero. Since the 3-momentum is conserved, and it is zero before the decay, themomenta of the µ+ and of the neutrino must be equal and opposite. Suppose thatthe magnitude of the momentum of the µ+, which can be measured, is p. The zerothcomponent of the 4-momentum, the relativistic energy, is also conserved. Before thedecay, the energy of the π+ is just the pion mass (Eq. (5.58) with ~p = 0); and afterthe decay, it is the sum of the the neutrino energy, which is equal to p (Eq. (5.58)with m = 0), and the muon energy. That is,

mπ = p +√

p2 + mµ2 . (5.59)

This equation can be solved to give

p =mπ

2 −mµ2

2mπ

. (5.60)

27

Chapter 6

Maxwell’s Equations

AIM: Arrive at a lagrangian formulation of Electro-magnetism. This implies writingthe Maxwell equations as the equations of motion for the electromagnetic fields.

6.1 Electromagnetic Fields

AIM: write the Maxwell equations in a contravariant form.The Maxwell equations, in the presence of a charge-density, ρ(x), and a current-

density, ~ (x), are

~∇. ~E = ρ , (6.1)

~∇. ~B = 0 , (6.2)

~∇∧ ~B − ∂ ~E

∂t= ~ , (6.3)

~∇∧ ~E +∂ ~B

∂t= 0 . (6.4)

Note that no polarization or magnetization has been included: these are the equa-tions in vacuo, except in so far that charge distributions are taken into account. Ina polarizable and magnetizable medium, one distinguishes between ~D and ~E, andbetween ~B and ~H. When, however, one adopts the more fundamental view that allthe charges should be explicitly taken into account, this distinction need no longerbe made. As in Chapter 5, we choose units such that c = 1 ; moreover the ugly4π that often disfigures the right-hand sides of Eq. (6.2) and Eq. (6.4) has beenremoved by suitably redefining the unit of electric charge.

Since ~B is divergence-free (since there do not seem to be any magnetic monopoles),it follows that there is a vector field, A, whose curl it is, i.e.

~B = ~∇∧ ~A. (6.5)

A proof of this statement can be found in the Appendix A, from which it will beseen that ~B does not determine ~A uniquely. Define ~C = − ~E − ∂ ~A/∂t, so that

~∇∧ ~C = 0, (6.6)

28

and from Appendix A again, we know that this implies the existence of a scalarfield, Φ, such that

− ~E − ∂ ~A

∂t= ~C = ~∇Φ . (6.7)

Substituting Eq. (6.5) and Eq. (6.7) into Eq. (6.2) and Eq. (6.4), we find

∂2Φ

∂t2−∇2Φ− ∂

∂t[∂Φ

∂t+ ~∇. ~A ] = ρ ,

∂2 ~A

∂t2−∇2 ~A + ~∇[

∂Φ

∂t+ ~∇. ~A ] = ~ . (6.8)

The above equations can be cast into a more compact form by writing the operator∂2/∂t2 −∇2 = ∂µ∂µ = ∂2, and combining the scalar and the vector potentials intoone 4-potential:

Aµ = (Φ, ~A ) . (6.9)

We shall show in a moment that this 4-potential has the transformation propertiesof a contravariant Lorentz vector. We may write

∂2A0 − ∂0[∂µAµ] = ρ , (6.10)

∂2 ~A + ~∇[∂µAµ] = ~ .

These equations cry out to be combined into one covariant 4-dimensional equa-tion, do they not? However, there is an awkward sign difference in front of thesecond term on the left. If we suppose that electric charge is a relativistic invariant,so that the charge, e, of an electron is the same in any inertial system, then chargedensity is not invariant; rather, the product, e = ρ(x)d3x, is Lorentz invariant.Electric current is caused by the flow of electrons: it is given by the sum of theelectric charges, multiplied by their velocities. The current density is accordinglythe product of the charge density and the velocity:

~ = ρ~v , (6.11)

so that if we define the 4-current density by

jµ = ρdxµ

dt≡ (ρ,~ ) , (6.12)

then

edxµ = ρ dxµd3x = jµd4x . (6.13)

Now since e and d4x are Lorentz invariants, and dxµ is a contravariant 4-vector, itfollows that jµ must also be a contravariant vector.

We can rewrite Eq. (6.11) in component form as follows:

∂2A0 − ∂0[∂νAν ] = j0

∂2Ak + ∂k[∂νAν ] = jk . (6.14)

Now we can understand the apparently awkward sign difference, for the derivativeoperator is covariant, and if we cast it into the unnatural, contravariant form, ∂µ =

29

−∂µ for µ = 1, 2, 3 , we pick up a minus sign! Hence Eq. (6.14) can be written inthe beautiful form

∂2Aµ − ∂µ[∂νAν ] = jµ (6.15)

Since jµ is a contravariant vector, it follows that Aµ must also be a contravariantvector. In fact, Eq. (6.15), which is merely (!) a rewriting of Eq. (6.2)-Eq. (6.4),is in a relativistically covariant form. The equations knew more than their creator,Maxwell, did, when he invented them! To do Maxwell and Lorentz justice, theywere worried that the electromagnetic equations are not consistent with Galileancovariance; and they did their best to understand this fact.

6.2 Electromagnetic Field Tensor

The contravariant 4-potential changes under a Lorentz transformation as follows:

Aµ′(x′) =∂x′µ

∂xρAρ(x) ; (6.16)

that is, the transformed field, at the transformed point, is equal to the old field, atthe old point, multiplied by the Lorentz-transformation matrix. A covariant versionof the 4-potential can be defined:

Aµ(x) = gµνAν(x) ; (6.17)

and of course the transformation law for this is

Aµ′(x′) =

∂xρ

∂x′µAρ(x) . (6.18)

It is convenient to introduce the second-order tensor

Fµν = ∂µAν − ∂νAµ , (6.19)

which transforms as follows:

Fµν′(x′) =

∂xρ

∂x′µ∂xσ

∂x′νFρσ(x) . (6.20)

After these book-keeping preliminaries, we can write the Maxwell equationsEq. (6.15) in the still compacter form

∂µFµν = jν . (6.21)

The field tensor, which is manifestly antisymmetric, can be expressed directly interms of the electric field and the magnetic induction, for if i,j and k are restrictedto the values 1, 2, 3, then

F0k = ∂0Ak − ∂kA0 = −∂0Ak − ∂kA

0 = Ek , (6.22)

and

Fjk = ∂jAk − ∂kAj = −εjklBl . (6.23)

30

Thus the field tensor can be expressed wholly in terms of ~E and ~B, and vice-versa.For future reference it is helpful to give the matrix elements Fµν in matrix form,

Fµν :

−→ ν

µ ↓

0 E1 E2 E3

−E1 0 −B3 B2

−E2 B3 0 −B1

−E3 −B2 B1 0

. (6.24)

Please note that F µν is similar with some important sign changes,

F µν :

−→ ν

µ ↓

0 −E1 −E2 −E3

E1 0 −B3 B2

E2 B3 0 −B1

E3 −B2 B1 0

. (6.25)

Also, since ~E and ~B are not four vectors but three vectors, E1 = E1 = Ex.Despite the fact that the 4-potential, Aµ, is not uniquely determined by the field

tensor, it is an extremely useful quantity. If it is subjected to a gauge transformation,i.e.

Aµ −→ A′µ = Aµ + ∂µG , (6.26)

where G is any Lorentz scalar field, then clearly the field tensor is unchanged. Sucha gauge-transformation has no physical consequences; and so any interactions withthe electromagnetic 4-potential must respect this gauge invariance. The restrictionturns out to be very important, with ramifications far outside the field of electro-magnetism.

A particular restriction that is often made on the 4-potential is the so-calledLorentz condition, viz.

∂µAµ = 0 . (6.27)

By means of a gauge transformation, it is always possible to achieve the Lorentz con-dition, without changing the physics. For under the gauge transformation Eq. (6.26),

∂µA′µ = ∂µA

µ + ∂2G . (6.28)

and the right side can be made to vanish by choosing G such that ∂2G = −∂µAµ.

The solution of this partial differential equation can be readily performed by Fouriertransformation, as in Appendix B.

When the Lorentz condition, Eq. (6.27), is satisfied, the Maxwell equations,Eq. (6.21), become even simpler:

∂2Aν = jν . (6.29)

31

6.3 Lagrangian Density

Let us first examine the free electromagnetic field. We shall see how the Maxwellequation, Eq. (6.21), can be derived from a variational principle. In order to do this,we regard the field, Aν(t, ~r ), as a continuous infinity of canonical variables. For agiven time, t, the canonical variables are labelled by ν, and the continuous variable,~r. Since the expression for the Lagrangian will inevitably involve a summation overall space, it is convenient to introduce a Lagrangian density:

L =∫

d3xL(x). (6.30)

The action can accordingly be written

S =∫ tb

tadtL =

∫ tb

tadt∫

d3xL(x) =∫ b

ad4xL(x). (6.31)

Since the action, S, is a Lorentz invariant, and d4x is a Lorentz-invariant measure,it follows that the Lagrangian density, L , is Lorentz-invariant.

Consider now a variation in the fields, Aµ, such that the values stay fixed at thespace-time points a and b. The resultant change in the action is

δS =∫ b

ad4x[

∂L∂Aν

δAν +∂L

∂(∂µAν)δ(∂µAν)]

=∫ b

ad4x[

∂L∂Aν

− ∂µ∂L

∂(∂µAν)]δAν . (6.32)

Since δAν is arbitrary between the end-points, the Hamilton variation principle,δS = 0, implies

∂µ∂L

∂(∂µAν)− ∂L

∂Aν

= 0 (6.33)

This covariant expression is the continuum version of the Euler-Lagrange equation.We shall now show that the Lagrangian density,

L = −14F

µνFµν − Aµjµ = −12F

µν∂µAν − Aµjµ, (6.34)

when inserted into Eq. (6.33), yields the Maxwell equation, Eq. (6.21). To do this,we regard L as a function of the 4 variables Aν , and the 16 variables ∂µAν . We find

∂L∂Aν

= −jν , (6.35)

and

∂L∂(∂µAν)

= −12F

ρσ ∂Fρσ

∂(∂µAν)

= −12F

ρσ[δµρ δν

σ − δµσδν

ρ ]

= −F µν . (6.36)

It is clear now that the Euler-Lagrange field equations yield indeed the Maxwellequation, when the above Lagrangian density is used. However, the Lagrangian

32

density is not uniquely determined by the requirement that the equation of motionbe correct.

In the above account, the current density, jν , is simply treated as an externallyprescribed source of the electromagnetic field. In a more thoroughgoing theory, thissource is itself expressed in terms of canonical fields, for example those pertainingto the electron. This leads to quantum electrodynamics, which is a very beautifuland successful theory, which is, however, beyond the present scope.

6.4 Hamiltonian

The canonical momentum densities that correspond to the field variables, Aµ,are defined by

πν =∂L

∂(∂0Aν)= −F 0ν = −∂0Aν + ∂νA0. (6.37)

From this, it is clear that π0 vanishes identically. Moreover, from Eq. (6.7), we seethat

Ei = −∂0Ai + ∂iA0 = −F 0i = πi. (6.38)

Remember that ∂i = −∂i, so that the signs above are correct. We have written thecomponents of the electric field as subscripts: ~E = (E1, E2, E3), and of course ~Eand thus also ~π is emphatically not a Lorentz vector.

The Hamiltonian density is now defined by

H = πµAµ − L = πkAk − L= −F 0k∂0Ak + 1

4FµνFµν

= −πk[∂kA0 − πk] + 14F

j0Fj0 + 14F

0kF0k + 14F

jkFjk

= 12πkπk − πk∂kA0 + 1

4FjkFjk , (6.39)

in current free space. The last form is canonical, that is, the hamiltonian densityis expressed in terms of the fields, and their space derivatives, but not their timederivatives, and the momentum densities.

In terms of the electric field and the magnetic induction, the hamiltonian maybe written in the form

H =∫

d3xH(x) =∫

d3x [ ~E.~∇A0 + 12~E. ~E + 1

2~B. ~B] , (6.40)

which is not a lorentz scalar.

33

Chapter 7

Conservation Laws

7.1 Noether Theorem

THEOREM:For every continuous symmetry of the Lagrangian density, there is a conservedphysical quantity.

We shall illustrate this theorem by considering the invariance of the free elec-tromagnetic Lagrangian density, Eq. (6.34) without the current density term, undertime translations, space translations, and Lorentz transformations. These invari-ances lead respectively to the conservation of energy, momentum, and angular mo-mentum.

Suppose, in general, that L is unchanged under a transformation of the space-time points x → x′, and of the fields, Aµ(x) → A′µ(x′). That is

L(∂′µA′ν(x′)) = L(∂µAν(x)). (7.1)

Define

δAµ(x) = A′µ(x)− Aµ(x), (7.2)

and

δL(x) = L(∂µA′ν(x))− L(∂µAν(x)). (7.3)

Note that A′µ(x) and ∂µA′ν(x) occur in these definitions, and not A′µ(x′) and∂′µA′ν(x′) . We find therefore that

δL =∂L∂Aµ

δ(Aµ) +∂L

∂(∂µAν)δ(∂µAν), (7.4)

where the space-time argument, x, has been suppressed. Now

∂µ[∂L

∂(∂µAν)δAν ] = ∂µ[

∂L∂(∂µAν)

]δAν +∂L

∂(∂µAν)∂µ(δAν),

=∂L∂Aµ

δAµ +∂L

∂(∂µAν)δ(∂µAν), (7.5)

34

where the Euler-Lagrange equation, Eq. (6.33), has been used to obtain the last line.On comparing this with Eq. (7.4), we see that

δL = ∂µ[∂L

∂(∂µAν)δAν ], (7.6)

which is the general form of the Noether equation.To evaluate δL(x) in detail, we replace x′ in Eq. (7.1) by x, which means that

we must replace x by x, where the transformation x → x is the inverse of x → x′.So in place of Eq. (7.1), we have

L(∂µA′ν(x)) = L(∂µAν(x)). (7.7)

Thus Eq. (7.3) can be written

δL(x) = L(∂µAν(x))− L(∂µAν(x))

= L(x)− L(x)

= (x− x)ρ∂ρL(x) + O((x− x)2)

= −δxρ∂ρL(x) . (7.8)

In the last line use has been made of the fact that the transformation is infinetesimaland continuous and thus (x− x)ρ = (x− x′)ρ.

7.2 Energy Momentum Tensor

Under space-time translations, xµ → x′µ = xµ+aµ, we know that the transformedfield at the transformed point is just the original field at the original point, i.e.A′µ(x′) = Aµ(x), so that A′µ(x) = Aµ(x), where xµ = xµ − aµ . Thus

A′µ(x) = Aµ(x− a) = Aµ(x)− aν∂νAµ(x) + O(a2). (7.9)

Hence, for infinitesimal aµ,

δAµ(x) = −aλ∂λAµ(x), (7.10)

δL(x) = −aν∂νL(x). (7.11)

The Noether equation, Eq. (7.6), takes the form

aν∂νL = aλ∂µ

[∂L

∂(∂µAρ)∂λAρ

]. (7.12)

Since λ only denotes an index for an implicit summation it may also be relabelledby ν since this index has not been used on the right hand side. Bringing the termsto one side and flipping positions of ν indices one obtains

0 = aν∂µ

[∂L

∂(∂µAρ)∂νAρ − gµνL

]. (7.13)

Since this equation must hold for arbitrary aν , it follows that

∂µTµν = 0, (7.14)

35

where the energy-momentum tensor is defined by

T µν =∂L

∂(∂µAρ)∂νAρ − gµνL. (7.15)

By inserting the explicit form of the Lagrangian density for current free space,Eq. (6.34), we obtain

T µν = −F µρ∂νAρ + 14g

µνFρσFρσ. (7.16)

The four-momentum (check that it is a contravariant vector indeed!) of the electro-magnetic field is defined by

P ν =∫

d3xT 0ν , (7.17)

so that, by using Eq. (7.14), we find

P ν =∫

d3x∂0T0ν = −

∫d3x∂iT

iν = 0, (7.18)

on condition that the fields vanish at spatial infinity. The four quantities, P ν , arethe conserved quantities of the Noether theorem.

The zeroth component is just the Hamiltonian, since

P 0 =∫

d3xT 00 =∫

d3x[−F 0ρ∂0Aρ + 14FρσF

ρσ] =∫

d3xH = H, (7.19)

where we have used Eq. (6.39).The space components of the field four-momentum are

P i =∫

d3xT 0i = −∫

d3xF 0ρ∂iAρ

= −∫

d3x[∂0Aj − ∂jA0]∂iAj. (7.20)

However, from Appendix A, we see that

[ ~E ∧ ~B]i = [ ~E ∧ (~∇∧ ~A)]i = Ej[∂iAj − ∂jA

i]

= [∂0Aj − ∂jA0][∂iAj − ∂jAi]

= −[∂0Aj − ∂jA0]∂iAj − ∂j[∂0Aj − ∂jA0]Ai. (7.21)

where we have used the free Maxwell equation to obtain the last line. Hence

~P =∫

d3x~E ∧ ~B, (7.22)

where ~P = (P 1, P 2, P 3). The integrand of the above equation, the field momentumdensity, is called the Poynting vector.

36

7.3 Angular Momentum Tensor

The Lagrangian density is invariant under under a Lorentz transformation; butthe four potential is not, since it is a four vector. If we write

x′µ

= Λµνx

ν , (7.23)

then

A′µ(x′) = ΛµνA

ν(x) . (7.24)

In order to obtain δA, we need to calculate A′µ(x) rather than A′µ(x′). This is doneby replacing x′ by x, which means that x must be replaced by x = Λ−1x :

A′µ(x) = ΛµνA

ν(x). (7.25)

We consider an infinitesimal Lorentz transformation, and its inverse,

Λµν = δµ

ν + εµν , (Λ−1)ρ

σ = δρσ − ερ

σ. (7.26)

Hence

A′µ(x) = [δµν + εµ

ν ][1− ερσx

σ∂ρ]Aν(x), (7.27)

so that

δAµ = εµνA

ν − ερσx

σ∂ρAµ. (7.28)

Since the Lagrangian density is Lorentz invariant, we find from Eq. (7.8) that

δL(x) = −ερσx

σ∂ρL(x) . (7.29)

By inserting Eq. (7.28) and Eq. (7.29) into Eq. (7.6), we obtain the following:

ερσxσ∂ρL = ερσ∂µ[

∂L∂(∂µAσ)

Aρ +∂L

∂(∂µAν)xσ∂ρAν ]. (7.30)

Since εσρ is antisymmetric (see appendix C), but otherwise arbitrary, it follows thatthe odd part of its coefficient in the above equation must vanish. Thus

∂µLµρσ = 0, (7.31)

where

Lµρσ =∂L

∂(∂µAσ)Aρ +

∂L∂(∂µAν)

xσ∂ρAν − gµρxσL − [ρ ↔ σ]

= Jµρσ + Sµρσ, (7.32)

where the extrinsic (or orbital) angular momentum density is

Jµρσ = xσ[∂L

∂(∂µAν)∂ρAν − gµρL]− [ρ ↔ σ]

= T µρxσ − T µσxρ, (7.33)

37

and the intrinsic (or spin) angular momentum density is

Sµρσ =∂L

∂(∂µAσ)Aρ − (ρ ↔ σ)

= F µρAσ − F µσAρ. (7.34)

We define the angular momentum tensor by

Lρσ =∫

d3xL0ρσ. (7.35)

In view of Eq. (7.31), we see that

Lρσ =∫

d3x∂0L0ρσ = −

∫d3x∂iL

iρσ = 0, (7.36)

so that all the components of the angular momentum tensor are time independent(they are the Noether currents that correspond to the invariance of the Lagrangiandensity under Lorentz transformation). Consider

J2 1 =∫

d3xJ0 2 1 =∫

d3x[x1T 0 2 − x2T 0 1] . (7.37)

Since T 0 i are the momentum densities, we see that J2 1 is the third component ofthe orbital angular momentum, which we often write J3. Then S3 ≡ S2 1 is thethird component of the intrinsic, or spin angular momentum. We write L2 1 ≡L3 = J3 + S3. In a similar way, by permuting 1, 2, 3 cyclically, we define the othercomponents: L1 = J1 + S1 and L2 = J2 + S2 .

7.4 The Photon

We will now use the tensors, that we have introduced via the Noether theorem,to calculate the energy, momentum and angular momentum in a particular fieldconfiguration, namely that specified by the following 4-potential:

A1 = ∆ cos Ω, A2 = ∆ sin Ω, A0 = 0 = A3, (7.38)

where

Ω = kµxµ = k(x0 − x3). (7.39)

This corresponds to wave propagation along the x3 axis, with unit velocity (i.e. thespeed of light). In terms of the electric and magnetic fields, and the field tensor,wefind

E1 = B2 = F 10 = F 13 = k∆ sin Ω (7.40)

E2 = −B1 = F 20 = F 23 = −k∆ cos Ω (7.41)

E3 = B3 = F 03 = F 12 = 0. (7.42)

It is easy to see that ∂µFµν = 0, so that the fields correspond to regions of space-time

in which there is no charge nor current density. Moreover, the electric and magnetic

38

fields are perpendicular to the direction of propagation, and to each other. Bothvectors have constant magnitude, but they rotate around the x3 axis. We say thatthe radiation is circularly polarized.

Consider first the energy momentum tensor. We find

T 00 = 12 [(F

01)2 + (F 02)2 + (F 23)2 + (F 13)2] = k2∆2, (7.43)

so that the energy density is constant in space and time. Further,

T 0i = −∂0Aj∂iAj, (7.44)

so that T 01 = 0 = T 02 and

T 03 = −∂0A1∂3A

1 − ∂0A2∂3A

2 = k2∆2, (7.45)

which is the same as the energy density.Now suppose that a target is placed in this electromagnetic radiation, with cross-

sectional area α orthogonal to x3, and suppose that it absorbs all the radiation thatfalls on it, during a time T . The energy that is transferred to the target is obtainedby integrating T 00 over all the radiation that will fall on to the target, during thespecified period. This is, however, a volume of cross-sectional area α and length T(since the speed of light is unity). This absorbed energy is αTk2∆2. Similarly, wecan calculate the momentum that is transferred to the target, in the same time, byintegrating T 0i over the same volume. Clearly the momentum is purely in the x3

direction, and it is also equal to αTk2∆2.In the 17th century, Newton postulated that light consists of a stream of particles.

This idea fell into disrepute, largely because of the success of the wave-hypothesisof Huygens. The Maxwell equations are wave equations par excellence. At thebeginning of the 20th century, however, the particle theory of light was reintroducedby Planck, in order to deal with the ultra-violet catastrophe of black-body radiation,and by Einstein, in connection with the photo-electric effect, in which it was clearthat the particles of light, photons, each had an energy that was proportional to thefrequency.

Suppose that there are N photons, each of mass m, in the volume of the radiationthat falls on our target in the time, T . We calculated that both the energy and themomentum of these N photons is αTk2∆2. Hence the energy, E, and the momentum,p, per photon are equal to one another. However, from Eq. (5.58), we know thatm2 = E2 − p2, from which it follows that the mass of a photon is zero.

We will look lastly at the angular momentum transfer to the target. SinceT 01 = 0 = T 02, it follows that J3 vanishes. However,

S021 = A1F 02 − A2F 01 = k∆2. (7.46)

The total angular momentum that is contained in the volume of radiation, that fallson the target in the time T , is therefore wholly intrinsic, and it is equal to αTk∆2.The ratio of the energy to the angular momentum for the N photons, and thus foreach photon separately, is k = 2πν, where ν is the frequency of the radiation. Sincewe know, by studying the energy of photo-electrons, that the energy of a photon isproportional to the frequency, it follows that the intrinsic angular momentum, orspin, of a photon is independent of the frequency. This spin does not in fact dependon any of the accidental features of the field configuration. It is a fundamentalunit of angular momentum, usually written as h/(2π), where h is called Planck’sconstant.

39

Chapter 8

Point Charge

AIM: solve the dynamics of Electromagnetic fields interacting with massive particles.

8.1 Lagrangian

The force on a particle of charge e, at rest in an electric field ~E, is e ~E. The forceon a moving charge can be obtained by applying a Lorentz transformation to thisstatic situation, but some care is needed, since a three-dimensional force is not arelativistic 4-vector. Consider the infinitesimal quantity defined by

dpµ ≡ eF µνdxν , (8.1)

where e is the charge, and x the coordinate of an elementary charge, say an electron,in an electromagnetic field, F µν . This is manifestly a Lorentz 4-vector. If we nowdivide by dt, which is not a Lorentz scalar, but rather the zeroth component of a4-vector, we obtain the non-covariant relation,

dpµ

dt= eF µν dxν

dt, (8.2)

the spacelike components of which (i.e. µ = j = 1, 2, 3) are

dpj

dt= eF j0 + eF jkvk , (8.3)

where ~v is the velocity of the electron. By means of Eq. (6.22) and Eq. (6.23), wecan write this, in 3-vector notation, as

~p = e ~E + e~v ∧ ~B . (8.4)

If the electron is instantaneously at rest, ~v = 0, then the right-hand side reduces toe ~E, the static force, which implies that ~p is the 3-momentum of the electron. Weknow however that ~p constitutes the space components of the 4-momentum, whichis a contravariant 4-vector. Hence Eq. (8.1) is in fact the covariant relation betweenan increment of the 4-momentum of an electron and the electromagnetic field thatinteracts with it.

To summarize the sequence of the reasoning, Eq. (8.1) is the definition of acontravariant vector, which we identify with an increment of the 4-momentum of

40

the electron, for in the rest-frame of the electron the space components coincidewith the known static force exerted on the electron by the electric field. Sincewe know that the 4-momentum transforms under a Lorentz transformation as acontravariant vector, we can assert that Eq. (8.1) identifies the incremental changeof the 4-momentum in an arbitrary inertial frame. In a frame in which the electronhas velocity ~v , we see from Eq. (8.4) that the Lorentz force on a moving point chargein an electromagnetic field must be

~F = e ~E + e~v ∧ ~B. (8.5)

The zeroth component of Eq. (8.1), is also interesting, since it gives the incrementalchange in the energy of the electron, E. Since F 00 = 0, the zeroth component ofEq. (8.2) can be written

d

dtE = eF 0k dxk

dt

= e~v. ~E . (8.6)

The rate of change of the energy of the electron is equal to the rate at which theLorentz force does work, which is ~F .~v, and so we see that Eq. (8.5) and Eq. (8.6)are consistent with one another.

We shall now show that the Lorentz force, Eq. (8.5), can be derived from thefollowing action:

S = −m∫ b

a

√dxµdxµ − e

∫ b

adxµA

µ. (8.7)

This can be rewritten

S = −m∫ tb

tadt√

1− xixi − e∫ tb

tadt[A0 − xiAi], (8.8)

where the indices of x and A are contravariant ones, and where the repeated Latinindex, i, is to be summed from 1 to 3.

From Eq. (8.8),we can read off the Lagrangian as

L = −m√

1− xixi − eA0 + exiAi ; (8.9)

and so the canonical momentum is

pi ≡∂L

∂xi=

mxi

√1− v2

+ eAi . (8.10)

Note that this canonical momentum, pi , is not the same as the ordinary momentum,pi, the time derivative of which occurs in Eq. (8.4). Indeed pi = pi +eAi . L dependsexplicitly on xi via the potentials A0 and Ai . Thus we calculate

∂L

∂xi= −e

∂A0

∂xi+ exj ∂Aj

∂xi. (8.11)

The Euler Lagrange equation (2.16) reads then

d

dt[

mxi

√1− v2

+ eAi] + e∂A0

∂xi− exj ∂Aj

∂xi= 0. (8.12)

41

Now

dAi

dt=

∂Ai

∂t+

∂Ai

∂xjxj, (8.13)

so it follows from Eq. (8.12) that

d

dt[

mxi

√1− v2

] = −e∂Ai

∂t− e

∂A0

∂xi− e

∂Ai

∂xjxj + e

∂Aj

∂xixj. (8.14)

Since

Ei = −∂Ai

∂t− ∂A0

∂xi, (8.15)

and

(~v ∧ ~B)i = (~v ∧ (~∇∧ ~A))i =∂Aj

∂xixj − ∂Ai

∂xjxj, (8.16)

we see that the right-hand side of Eq. (8.14) is indeed the i’th component of theright-hand side of Eq. (8.5). This concludes the demonstration that the Lagrangian(8.9) does indeed yield the Lorentz force.

8.2 Hamilton’s Equations

It is convenient to make the transition from the Lagrangian to the Hamiltonianformalism. The Hamiltonian is

H = pixi − L =

mxixi

√1− v2

+ eAixi + m√

1− v2 + eA0 − exiAi

=m√

1− v2+ eA0. (8.17)

This is not yet in the canonical form, in which the Hamiltonian must be expressed asa function of the coordinates, xi, and the canonical momenta, pi. From Eq. (8.10),we have

m2 + (pi − eAi)(pi − eAi) = m2 +m2v2

1− v2=

m2

1− v2, (8.18)

so that

H = eA0 +√

m2 + (pk − eAk)(pk − eAk) , (8.19)

which is indeed canonical.The Hamilton equations (3.8) take the form

xi =∂H

∂pi

=pi − eAi√

m2 + (pk − eAk)(pk − eAk), (8.20)

˙pi = −∂H

∂xi= −e

∂A0

∂xi+

e(pj − eAj)√m2 + (pk − eAk)(pk − eAk)

∂Aj

∂xi. (8.21)

42

If the Lagrangian (8.9) does not depend explicitly on the time, the Hamiltonianis independent of time: it is one of the constants of the motion. In such a case,it is advantageous to use Eq. (8.19) to simplify the above two equations. FromEq. (8.20), we have immediately

[H − eA0]xi = pi − eAi, (8.22)

˙pi + e∂A0

∂xi= exj ∂Aj

∂xi. (8.23)

8.3 Constant Magnetic Field

In this section, we consider a charged point mass in a constant magnetic field.This is of relevance to the motion of elementary particles in a bubble chamber, orin an accelerator. Suppose that A0 = A2 = A3 = 0 but A1 = −Bx2, where B isconstant. Then all the components of ~E and ~B vanish, except B3 = −∂2A

1 = B.The Hamilton equations, Eq. (8.22) and Eq. (8.23), are

Hxi = pi + eBx2δi1, (8.24)

˙pi = −eBx1δi2. (8.25)

These equations reduce to

Hx1 = eBx2

Hx2 = −eBx1

Hx3 = 0. (8.26)

The velocity in the x3 direction is constant, and to solve the first two equations, weset ζ = x1 + ix2, so that

Hζ = eB(x2 − ix1) = −ieBζ. (8.27)

The solution is

ζ = α + βe−iωt, (8.28)

where ω = eB/H. However, H = m/√

1− v2, so that

ω =eB√

1− v2

m. (8.29)

From Eq. (8.28), we see that

ζ = −iωβe−iωt; (8.30)

and hence

v2⊥ ≡ (x1)2 + (x2)2 = |ζ|2 = ω2|β|2. (8.31)

Thus v⊥ is constant, and since

|ζ − α|2 = |β|2 =v2⊥

ω2, (8.32)

43

it follows that the motion in the (1, 2) plane is a circle, of radius R = v⊥/ω. In viewof the constant velocity in the x3 direction, it follows then that the trajectory of theparticle is a spiral, or helix, around the direction of the constant magnetic field.

In the case that x3 = 0, i.e. the motion is perpendicular to ~B, so that v⊥ = v,the particle travels with constant speed in a circle of radius

R =vm

eB√

1− v2. (8.33)

In a synchrotron, in order to keep the radius of the trajectory constant, one hasto increase the bending magnetic field, as the speed of the particle gets closer andcloser to that of light.

8.4 Constant Electric Field

Consider next a point charge in a constant electric field, E. For convenience, wechoose the x3 axis to be parallel to the field, and the origin to be such that xi = 0when t = 0, i = 1, 2, 3. Let the initial value of the particle velocity be (v1, v2, v3),

and choose the x2 axis such that v2 = 0. The fields can be specified by ~A = 0, andΦ = A0 = −Ex3. The Hamilton equations of motion are

(H + eEx3)xi = pi, (8.34)

˙pi = eEδi3. (8.35)

We can integrate Eq. (8.35) immediately and then combine it with Eq. (8.34):

(H + eEx3)xi = Hvi + eEtδi3, (8.36)

and the Hamiltonian,

H =m√

1− v2− eEx3, (8.37)

is a constant.For i = 2 this implies that x2 ≡ 0, i.e. that the motion is wholly in the (x1−x3)

plane. For i = 3, we rewrite Eq. (8.36) as follows:

d

dt[(H + eEx3)2] =

d

dt[(Hv3 + eEt)2], (8.38)

from which we deduce

(H + eEx3)2 − (Hv3 + eEt)2 = H2[1− v23]. (8.39)

By dividing the i = 1 and the i = 3 components of Eq. (8.34) , we obtain

dx3

dx1=

Hv3 + eEt

Hv1

. (8.40)

On using Eq. (8.39) to eliminate t in favour of x3, we find the solution

1 +eEx3

H=√

1− v23 cosh[

eEx1

Hv1

], (8.41)

44

which is a catenary, the same figure as that assumed by a uniform cord, hanging ina uniform gravitational field.

In the above work, we have relied heavily on the fact that H = 0. The actualvalue of the Hamiltonian can be found by computing Eq. (8.17) at t = 0:

H =m√

1− v21 − v2

3

, (8.42)

since at t = 0, x3 = 0, so that then also A0 = 0. It is important to realise thatthe velocities do change, and that although H is indeed time-independent, the twoterms in the last line of Eq. (8.17) separately do depend on the time.

At the risk of being tedious, but in the hope of avoiding future errors, let usnow treat the configuration A0 = A1 = A2 = 0 and A3 = −Et. Since again ~B = 0and Ei = Eδi3, this is the same physical problem, although at first sight it looksdifferent, since now

H =m√

1− v2, (8.43)

but here H is not time-independent, since L depends on t explicitly, through A3, soH = −∂L/∂t 6= 0.

The Hamilton equations of motion are

mxi

√1− v2

= pi + eEtδi3 , (8.44)

together with ˙pi = 0, so now all the pi are time-independent, but v is not. Since

m√1− v2

=

√m2 +

m2xixi

1− v2

=√

m2 + p21 + p2

2 + (p3 + eEt)2, (8.45)

we can cast Eq. (8.44), i = 3, into the form

x3 =p3 + eEt√

m2 + p21 + p2

2 + (p3 + eEt)2. (8.46)

This equation can be integrated to give

(C + eEx3)2 − (p3 + eEt)2 = C2 − p23 , (8.47)

where C is an integration constant. This equation has the same algebraic form asEq. (8.39), the only difference being that the constants have other names. We invitethe reader to complete the solution, and to show that the catenary, Eq. (8.41), canbe recovered.

It is instructive to observe that the two configurations are related by a gaugetransformation of the form Eq. (6.26). If ~A = 0, and A0 = −Ex3 (our first example),then it is sufficient to take G = Ex3t, in Eq. (6.26), to see that A′0 = 0 andA′3 = ∂3G = −∂3G = −Et (our second example).

Although the above problem is relatively straightforward, two lessons can belearned:

45

[1] The Hamiltonian method is easier if ∂L/∂t = 0 ,for then the Hamiltonian is a constant of the motion.

[2] One can sometimes use the gauge freedom to ensure that H = 0.

46

Chapter 9

Radiation from a Moving Charge

9.1 Solution of the Wave Equation

In a region of space-time in which there are no sources (i.e. where jν = 0), theMaxwell equations Eq. (6.29) in Lorentz gauge Eq. (6.27) reduce to

∂2Aµ = 0 . (9.1)

Solutions of this equation can be found by Fourier transformation:

Aµ(x) =∫

d3keikx3∑

λ=0

a(~k, λ)εµ(~k, λ), (9.2)

where kx = kνxν = k0t − ~k.~x. The four polarization vectors, ε(~k, λ), λ = 0, 1, 2, 3,

correspond to the four Lorentz components of Aµ. The spacelike components areas in Appendix B, that is, they are mutually orthogonal unit spacelike vectors,such that ε(~k, 3) is parallel to ~k. ε(~k, 0) is a unit vector in the timelike direction,(1, 0, 0, 0).

From Eq. (9.1) we have

∂2Aµ = −∫

d3keikxk23∑

λ=0

a(~k, λ)εµ(~k, λ) = 0 . (9.3)

Since the Fourier components are independent, k2 = (k0)2−~k.~k = 0, or equivalently,

k0 = |~k| ≡ k . (9.4)

This constraint is called the mass-shell condition; it corresponds to the fact thatphotons are massless.

The Lorentz condition Eq. (6.27) implies

∂µAµ =

∫d3keikxk

[a(~k, 0)− a(~k, 3)

]= 0 ; (9.5)

and this means simply that a(~k, 0) = a(~k, 3). Under this constraint, any solution ofthe Maxwell equations can be written in the form Eq. (9.2).

47

9.2 Retarded Potentials

The scalar potential engendered by a static charge, e, at a distance r is e/(4πr).If we have a static charge distribution, ρ(~x ′), then the scalar potential at the pointx is given by the integral

Φ(~x ) =1

4π

∫d3x′

ρ(~x ′)

|~x ′ − ~x |. (9.6)

If however the charge distribution is time-dependent, we expect the above integralto be modified, since the potential should not be determined by the instantaneouscharge-distribution, but rather by the retarded charge-distribution, i.e. the distrib-ution as it was when a light-signal was emitted from ~x ′, that has just arrived at ~x .Thus

Φ(t, ~x ) =1

4π

∫d3x′

ρ(t− |~x ′ − ~x |, ~x ′)

|~x ′ − ~x |. (9.7)

In Appendix E it is shown that Eq. (9.6) is indeed a solution of the differentialequation

∇2Φ(~x ) = −ρ(~x ) ; (9.8)

while Eq. (9.7) is a solution of the time-dependent wave-equation

∂2Φ(t, ~x ) = ρ(t, ~x ) , (9.9)

where ∂2 = ∂µ∂µ . The rather intuitive derivations given above are thus put on afirm foundation. Moreover, since the Maxwell equation for the 4-potential has theform

∂2Aµ = jµ , (9.10)

when the Lorentz gauge condition Eq. (6.27) is satisfied, it follows that we can usethe analysis of Appendix E to write the retarded solution to Eq. (9.10) in the form

Aµ(x) =1

4π

∫d3x′

jµ(t− |~x ′ − ~x |, ~x ′)

|~x ′ − ~x |. (9.11)

This is not the most general solution of the Maxwell equation, Eq. (9.10), since anysolution of the homogeneous equation, Eq. (9.1), may be added to Eq. (9.11). Theretarded solution is assumed, however, to give the physical, or causal solution of theMaxwell equation.

9.3 Lienard-Wiechert Fields

Let us specialize the general expression Eq. (9.11) for the retarded potential tothe case that the source is a single point charge, for example an electron, which isat a position ~X, that depends on the time. The corresponding charge density is

ρ(x′) = j0(x′) = eδ3(~x ′ − ~X) , (9.12)

48

where e is the electron’s charge and the δ-function is Dirac’s distribution, whichcan be thought of as the limit of an infinitely high, infinitely sharply-peaked func-tion, located at the point where its argument is zero. The corresponding currentdistribution is

~ (x′) = e~vδ3(~x ′ − ~X) , (9.13)

where ~v = ~X, the velocity of the electron.With these expressions for the charge and current densities, we can evaluate the

integral in Eq. (9.11). A difficulty, however, is that the time retardation means thatthe zeroth, or time component of the argument of jµ depends on ~x ′. The easiestway to keep things straight is to introduce a t′-integration, and a “time-retardation”delta-distribution. For the scalar potential, this yields

A0(x) =e

4π

∫d4x′δ3(~x ′ − ~X (t′))δ(t′ − t + |~x ′ − ~x |) 1

|~x ′ − ~x |

=e

4π

∫dt′δ(t′ − t + | ~X(t′)− ~x |) 1

| ~X(t′)− ~x |. (9.14)

Before the last integration is performed, the integration variable is changed fromt′ to t = t′ − t + | ~X(t′)− ~x |. However,

dt =

[1 +

∂

∂t′| ~X(t′)− ~x |

]dt′

=

1 +( ~X(t′)− ~x ). ~X(t′)

| ~X(t′)− ~x |

dt′ . (9.15)

In other words,

dt′ = dt/κ , (9.16)

where

κ = 1− ~n.~v , (9.17)

with

~n =~x− ~X(t′)

|~x− ~X(t′)|. (9.18)

Thus ~n is a unit vector pointing from the point charge, as it was at the retardedtime, to the field point, ~x . The integration over t can now be performed trivially,and we find

A0(x) =[

e

4πκR

]ret

. (9.19)

Here

R = |~x − ~X(t′)| , (9.20)

49

and the suffix ‘ret’ means that t′, which occurs implicitly in the definition of κ andof R, must be replaced by the retarded time, i.e. by the solution of the equation

t = t′ + R(t′) . (9.21)

Similarly, for the vector potential,

~A(x) =

[e~v

4πκR

]ret

. (9.22)

The expressions Eq. (9.19) and Eq. (9.22) are called the Lienard-Wiechert potentials.

They give the 4-potential at the point ~x , at time t, in terms of the position ~X ofthe point charge, as it was at the time t′ = t− | ~X(t′)− ~x |.

From the 4-potential, the Lienard-Wiechert fields can be calculated. They are

~E =e

4πκ3

~n ∧ [(~n− ~v) ∧ ~v]

R+

(1− v2)(~n− ~v)

R2

, (9.23)

~B = − e

4πκ3~n ∧

~v + ~n ∧ (~v ∧ ~v)

R+

(1− v2)~v

R2

. (9.24)

In these expressions, as in subsequent ones, the specification that the quantities onthe right are to be evaluated at the retarded time is implicit. The details of thiscalculation are to be found in Appendix F, as is the proof that the above Lienard-Wiechert fields are related by

~B = ~n ∧ ~E . (9.25)

As can be seen from the above expressions, both the electric field and the mag-netic induction have contributions that behave like R−2, and others that behave likeR−1. The former give the near-field, and are all that is left if the acceleration of thecharge, ~v , is zero:

~E =e(1− v2)

4πκ3R2(~n− ~v) ; (9.26)

while ~B can always be obtained from ~E by means of Eq. (9.25). These termscorrespond to electrostatics and magnetostatics, i.e. to a charge moving at constantspeed and so giving rise to a constant current. The fields fall off according to thefamous inverse square law of distance. Indeed, with v << 1, and to lowest order(O(1) for ~E and O(v) for ~B), and with ~v = 0, we find

~E =e~n

4πR2, (9.27)

~B = −e~n ∧ ~v

4πR2. (9.28)

These terms are not surprising: they arise essentially from the application of aLorentz transformation to the static situation.

50

The really amazing thing about the Lienard-Wiechert fields lies in the R−1 orradiation terms. These are present only if the charge undergoes acceleration. Forlarge R, we can neglect the R−2 terms, leaving

~E =e

4πκ3R~n ∧ [(~n− ~v) ∧ ~v] , (9.29)

together with Eq. (9.25). In the radiation zone (far from the accelerated source), theelectric field is at right angles to the direction of the source (i.e. orthogonal to ~n),and also to the magnetic induction, which is also at right angles to the direction ofthe source. In a word, the radiation is transverse. That we have to do with radiationis clear, since the energy and momentum densities are proportional to the squaresof the field strengths, so to R−2, and this means that there is a continual streamingof energy and momentum into space. We shall illustrate this in the nonrelativisticlimit.

For nonrelativistic motion of the electron, i.e. v << 1, the Lienard-Wiechertfields reduce to

~E =e

4πR~n ∧ (~n ∧ ~v) +

e

4πR2~n , (9.30)

~B = − e

4πR~n ∧ ~v . (9.31)

For large R, we neglect the static R−2 term, and we find

~E. ~E =

[e~n ∧ ~v

4πR

]2

= ~B. ~B . (9.32)

Now according to Eq. (6.40), the energy density of the field (the hamiltonian density)is

H = 12 [

~E. ~E + ~B. ~B] + ~E.~∇A0 . (9.33)

The last term here does not contribute in the radiation zone. Indeed, even withoutthe nonrelativistic approximation, we can calculate from Eq. (9.19) that

~E.~∇A0 =e(1− v2)

16π2κ5R4(~n− ~v).~v . (9.34)

Hence the hamiltonian density in the radiation zone is

H =e2v2sin2θ

16π2R2, (9.35)

where θ is the angle between ~n and ~v, and this means that, in a shell about thesource of radius R and infinitesimal thickness δR, the instantaneous radiant energyδE is

δE = 2π∫ R+δR

RR2dR

∫ π

0sin θdθ

e2v2sin2θ

16π2R2

=1

6πe2v2δR . (9.36)

51

The energy in the shell is independent of R: it propagates outwards, without loss,with the speed of light — indeed it is light!! (Or radio waves, or IR, UV, γ-radiation,or whatever!) Note that the R−2 terms in the fields do not contribute to this energyin the limit of large R: they fall off too rapidly.

Consider lastly a collection of point charges, en, located at ~Xn. We define thedipole moment of this collection of charges by

~D =∑n

en~Xn . (9.37)

In the nonrelativistic limit, and in the radiation zone (i.e. v << 1 and R large com-pared with the dimensions of the source), we have, from Eq. (9.30) and Eq. (9.31),

~E =~n ∧ (~n ∧ ~D)

4πR, (9.38)

and

~B = −~n ∧ ~D

4πR. (9.39)

As we have seen, the electric field and the magnetic induction are transverse to thepropagation direction, are orthogonal to each other, and are equal in amplitude. At

an angle θ from the direction of ~D, we have

E2 = B2 =D2sin2θ

16π2R2. (9.40)

Thus the radiation is most intense at right angles to the second derivative of thedipole moment, as anyone who has played with a radio transmission antenna knows!

52

Appendix A

Triple Vector Product

The vector product of two three-vectors can be written

(~a ∧~b )i = εijkajbk, (A.1)

where the repeated indices, j and k, are summed over 1, 2, 3. This is a general rule:if a Latin index is repeated in the same term, a summation is implied. The symbol,εijk , is defined to be +1 if i, j, k is a cyclic permutation of 1, 2, 3, and −1 for ananticyclic permutation. It is 0 if two or more of the indices are the same. FromEq. (A.1), we see that, for example, (~a ∧~b )1 = a2b3 − a3b2 .

A triple vector product can be calculated as follows:

[~a ∧ (~b ∧ ~c )]i = εijkajεklmblcm

= εijkεlmkajblcm

= (δilδjm − δimδjl)ajblcm

= ajbicj − ajbjci. (A.2)

The summation over j in the last line corresponds to a scalar product of two vectors.In vector notation, we have shown that

~a ∧ (~b ∧ ~c ) = ~b (~a .~c )− ~c (~a .~b ). (A.3)

From Eq. (A.2), we can immediately evaluate

[~v ∧ (~∇ ∧ ~A )]i = vj∂iAj − vj∂jAi, (A.4)

Note that the order of the factors is important. Further,

[~∇ ∧ (~∇ ∧ ~A )]i = ∂i∂jAj − ∂j∂jAi. (A.5)

In vector notation, this last equation reads

~∇ ∧ (~∇ ∧ ~A ) = ~∇ (~∇ . ~A )−∇2 ~A . (A.6)

53

Appendix B

Potentials

B.1 Vector Potential

If a vector field, ~B(x), satisfies ~∇. ~B = 0, then there exists a vector field, ~A(x),

such that ~B = ~∇∧ ~A.

ProofGiven ~B(x), we write a three-dimensional Fourier transform as follows,

~B(x) =∫

d3keikx3∑

λ=1

b(~k, λ)~ε (~k, λ). (B.1)

Here ~ε (~k, λ), λ = 1, 2, 3, are three orthonormal vectors, ~ε (~k, 3) being parallel to ~k,

the others being therefore orthogonal to it. kx means ~k.~x, in this, and in subsequentformulas. Clearly

~∇. ~B(x) = i∫

d3keikxkb(~k, 3), (B.2)

where k = |~k|. The necessary and sufficient condition that ~∇. ~B = 0 is accordingly

that b(~k, 3) = 0, k 6= 0. Define now ~A(x) as follows:

~A(x) =∫

d3keikx3∑

λ=1

a(~k, λ)~ε (~k, λ), (B.3)

where a(~k, 1) = −ib(~k, 2)/k, a(~k, 2) = ib(~k, 1)/k, for k 6= 0, while a(~k, 3) is arbitrary.Then

~∇∧ ~A(x) = i∫

d3keikx3∑

λ=1

a(~k, λ)~k ∧ ~ε (~k, λ)

= i∫

d3keikxk[a(~k, 1)~ε (~k, 2)− a(~k, 2)~ε (~k, 1)]

=∫

d3keikx[b(~k, 1)~ε (~k, 1) + b(~k, 2)~ε (~k, 2)]. (B.4)

The last expression can be seen to be equal to ~B(x), since b(~k, 3) = 0, k 6= 0.

54

B.2 Scalar Potential

If a vector field, ~C(x), satisfies ~∇∧ ~C = 0, then there exists a scalar field, Φ(x),

such that ~C = ~∇Φ.

ProofIn terms of the Fourier transform,

~C(x) =∫

d3keikx3∑

λ=1

c(~k, λ)~ε (~k, λ), (B.5)

we have

~∇∧ ~C(x) = i∫

d3keikxk[c(~k, 1)~ε (~k, 2)− c(~k, 2)~ε (~k, 1)]. (B.6)

For this to vanish, it is necessary and sufficient that c(~k, 1) = 0 = c(~k, 2), for k 6= 0.Define now the scalar field, Φ(x), by means of the Fourier transform

Φ(x) =∫

d3keikxφ(~k), (B.7)

where φ(~k) = −ic(~k, 3)/k, for k 6= 0. Then it follows that

~∇Φ(x) =∫

d3keikxc(~k, 3)~ε (~k, 3). (B.8)

We recognize the right-hand side of the above equation as ~C(x).

55

Appendix C

Invariant Measure

If x′ and x are related by a Lorentz transformation, and F (x) is a Lorentz scalar,then ∫

d4x′F (x′) =∫

d4xF (x) (C.1)

Proof

By the usual rules for changing variables in a multiple integral, we have∫d4x′F (x′) =

∫d4x

∂(x′0, x′1, x′2, x′3)

∂(x0, x1, x2, x3)F (x) , (C.2)

where the Jacobian can be written

∂(x′0, x′1, x′2, x′3)

∂(x0, x1, x2, x3)= det(Λµ

ν) , (C.3)

in which the matrix Λ is defined by

Λµν =

∂x′µ

∂xν, (C.4)

that is, it is the Lorentz transformation matrix itself. We shall show that thedeterminant of this matrix is always unity, thus completing the proof of Eq. (C.1).

For a rotation about one of the space axes by an angle θ,

det(Λµν) = cos2 θ + sin2 θ = 1 ; (C.5)

while a Lorentz transformation along one of the space axes by a hyperbolic boost,u, gives

det(Λµν) = cosh2 u− sinh2 u = 1 . (C.6)

Since a general Lorentz transformation can be built up by compounding rotationsabout the space axes (Euler angles), and a pure Lorentz boost along one space axis,and since the determinant of the product of a number of matrices is equal to theproduct of the determinants of each matrix, each of which is equal to unity, we seefinally that the determinant of the most general Lorentz transformation matrix isalso equal to unity.

56

Consider next an infinitesimal Lorentz transformation along the x1 axis, de-scribed by a hyperbolic angle, u. The transformation matrix is

Λµν = δµ

ν + εµν , (C.7)

where ε01 = −u = ε1

0, and the other elements of εµν are zero. For an infinitesimal

rotation by an angle, θ, about the x3-axis, on the other hand, Eq. (C.7) holds, withε1

2 = −θ = −ε21, and the other elements are zero.

Now consider the doubly contravariant form,

Λµν = Λµσg

σν = gµν + εµν . (C.8)

In the case of the Lorentz transformation,

ε01 = ε0σg

σ1 = u ε10 = ε1σg

σ0 = −u , (C.9)

so that εµν is antisymmetric. For the rotation,

ε12 = ε1σg

σ2 = θ ε21 = ε2σg

σ1 = −θ , (C.10)

which means that here, too, εµν is antisymmetric.Since any infinitesimal Lorentz transformation can be made by compounding

infinitesimal rotations around axes and a Lorentz transformation along an axis, itfollows that the general case can be written

Λµν = gµν + εµν , (C.11)

where the infinitesimal, ε, is antisymmetric, i.e.

εµν = −ενµ . (C.12)

It is important to note that this antisymmetry applies to the doubly contravariantform Λµν , or of course equally well for the doubly covariant form Λµν . It is not truefor the mixed form, Λµ

ν , in terms of which the Lorentz transformation was originallydefined.

57

Appendix D

Laplacian

Consider a transformation of Cartesian coordinates, x1, x2, x3, in three dimensions,to another set of orthogonal coordinates, u1, u2, u3, such that

(dx1)2 + (dx2)2 + (dx3)2 = (h1du1)2 + (h2du2)2 + (h3du3)2 , (D.1)

where the h’s are functions of the u’s. Let τ be an infinitesimal volume. Then thedivergence of a vector, ~A, can be written

~∇. ~A =1

τ

∫τdx1dx2dx3 ~∇. ~A =

1

τ

∫S

dS ~n. ~A , (D.2)

where S is the surface of the infinitesimal volume, and ~n is a unit vector normal tothis surface. In terms of the components in the u1, u2 and u3 directions, this can bewritten

1

τ

(∫u1+du1

−∫

u1

)h2du2h3du3A1 + permutations ,

where ‘permutations’ means the two terms obtained by replacing the indices 1, 2, 3by respectively 2, 3, 1 and 3, 1, 2 . This is, however,

1

τ

∫u1

du1du2du3 ∂

∂u1[h2h3A1] + permutations ;

and since du1du2du3 = (dx1dx2dx3)/(h1h2h3), it follows that

~∇. ~A =1

h1h2h3

∂

∂u1[h2h3A1] + permutations . (D.3)

In the case that ~A is the gradient of a scalar, ~A = ~∇Φ, we find

∇2Φ =1

h1h2h3

∂

∂u1

[h2h3

h1

∂Φ

∂u1

]+

∂

∂u2

[h3h1

h2

∂Φ

∂u2

]+

∂

∂u3

[h1h2

h3

∂Φ

∂u3

].(D.4)

The case of spherical polars is of particular interest. These are defined by

x1 = r sin θ cos φ

x2 = r sin θ sin φ

x3 = r cos θ , (D.5)

58

which yields

(dx1)2 + (dx2)2 + (dx3)2 = (dr)2 + (rdθ)2 + (r sin θdφ)2 , (D.6)

so that

h1 = 1 , h2 = r , h3 = r sin θ . (D.7)

Hence we can write

∇2Φ =1

r2

∂

∂rr2∂Φ

∂r+

1

r2 sin θ

∂

∂θsin θ

∂Φ

∂θ+

1

r2sin2θ

∂2Φ

∂φ2. (D.8)

A simple manipulation shows that the following operator identity holds:

∇2 =1

r

∂2

∂r2r +

L2

r2, (D.9)

where

L2 =1

sin θ

∂

∂θsin θ

∂

∂θ+

1

sin2θ

∂2

∂φ2. (D.10)

59

Appendix E

Charge Distributions

E.1 Static Potential

Suppose that ρ(~x ′) is a twice-differentiable function, such that∫ ∞

0r′dr′|ρ(~x ′)| < ∞, (E.1)

where r′ = |~x ′|. Define

Φ(~x ) =1

4π

∫d3x′

ρ(~x ′)

|~x ′ − ~x |. (E.2)

We shall show that

∇2Φ(~x ) = −ρ(~x ). (E.3)

E.1.1 Proof

Change the integration variable from ~x ′ to ~R = ~x − ~x ′, so that

Φ(~x ) =1

4π

∫ d3R

Rρ(~x − ~R ), (E.4)

where R = |~R |. Hence

∇2Φ(~x ) =1

4π

∫ d3R

R∇2ρ(~x − ~R ), (E.5)

where ∇2 on the RHS can be shifted from the ~x to the ~R variable. In terms of polarcoordinates,

∇2 =1

R

∂2

∂R2R +

L2

R2, (E.6)

where

L2 =1

sin θ

∂

∂θsin θ

∂

∂θ+

1

sin2θ

∂2

∂φ2. (E.7)

60

The integral in Eq. (E.5) can be separated into three parts:

∇2Φ(~x ) = (Φ1 + Φ2 + Φ3)/(4π), (E.8)

where

Φ1 =∫

dΩ∫ ∞

0dR

∂2

∂R2[Rρ(~x − ~R )]

= −∫

dΩ[R∂

∂Rρ(~x − ~R ) + ρ(~x − ~R )]R=0

= −4πρ(~x ), (E.9)

Φ2 =∫ ∞

0

dR

R

∫ 2π

0dφ∫ π

0dθ

∂

∂θsin θ

∂

∂θρ(~x − ~R ) = 0, (E.10)

Φ3 =∫ ∞

0

dR

R

∫ π

0

dθ

sin θ

∫ 2π

0dφ

∂2

∂φ2ρ(~x − ~R ) = 0. (E.11)

The result Eq. (E.3) follows on combining the last four equations.

There is a subtlety here that ought not to be glossed over. Suppose that we wereto apply the Laplacian to Eq. (E.2), without the change of integration variable:

∇2Φ(~x ) =1

4π

∫d3x′ρ(~x ′)∇2 1

|~x ′ − ~x |. (E.12)

Apparently nothing prevents us from rewriting the Laplacian in terms of R, θ andφ, as in Eq. (E.6), but without making the change in integration variable. However,

∇2 1

|~x ′ − ~x |=

1

R

∂2

∂R2R

1

R= 0, (E.13)

so that one is sorely tempted to the erroneous conclusion that ∇2Φ vanishes.Let us see what has gone wrong by considering the case that ρ(~x ′) = 1 if |~x ′| < ε

,and is zero otherwise. We can do an explicit calculation:

Φ(~x ) = 12

∫ ε

0r′2dr′

∫ 1

−1

d cos θ√r2 + r′2 − 2rr′ cos θ

=∫ ε

0dr′

r′2

max(r, r′)

= 12ε− 1

6r2, (E.14)

on condition that r < ε. For r > ε, we have clearly

Φ(~x ) =ε3

3r. (E.15)

It is now easy to check that ∇2Φ(~x ) = −1 if r < ε. For r > ε, the Laplacian of Φobviously vanishes.

The illegal step is Eq. (E.12), because if one performs the x−differentiationsunder the integral, the resulting integral is not absolutely convergent, and the for-mal vanishing of Eq. (E.13) does not guarantee the vanishing of the integral! Theobjection clearly does not apply to Eq. (E.5).

61

We need to separate the integration domain into a sphere of radius ε around|~x ′ − ~x | = 0, and the rest. However, it is necessary to do this very smoothly. We

define the test-function, fε(~R ), by the requirements that it be unity for |~R | < ε,

and zero for |~R | > 2ε. For ε < |~R | < 2ε, we wish fε(~R ) to go smoothly from 1 to 0.If we require it to be infinitely differentiable in (0 < R < ∞), then all its derivativesat R = ε and at at R = 2ε must vanish.

Let us write

Φε(~x ) =1

4π

∫d3x′fε(|~x ′ − ~x |) ρ(~x ′)

|~x ′ − ~x |, (E.16)

Φ⊥(~x ) =1

4π

∫d3x′[1− fε(|~x ′ − ~x |)] ρ(~x ′)

|~x ′ − ~x |. (E.17)

Clearly, from Eq. (E.2),

Φ(~x ) = Φε(~x ) + Φ⊥(~x ). (E.18)

Moreover, Φ⊥ really is divergenceless, for Eq. (E.12) is unexceptionable, when asphere around the point ~x ′ = ~x is excised from the domain of the integral. Sincewe have already demonstrated Eq. (E.3), by the legal proof Eq. (E.4) – Eq. (E.11),it follows that

∇2Φε(~x ) = −ρ(~x ), (E.19)

and this holds, no matter how small ε is.This result is summarized by the formula

∇2 1

|~x ′ − ~x |= −4πδ3(~x ′ − ~x ), (E.20)

where δ3 is the three-dimensional Dirac delta function (or more properly generalizedfunction, or distribution). This replaces Eq. (E.13). Within the more general frame-work of distribution theory, Eq. (E.12) is allowed, with the identification Eq. (E.20).

E.2 Retarded Potential

If the charge-distribution is time-dependent, we write ρ(x) ≡ ρ(t, ~x ), and wedefine the retarded potential by

Φ(t, ~x ) =1

4π

∫d3x′

ρ(t− |~x ′ − ~x |, ~x ′)

|~x ′ − ~x |. (E.21)

We shall show that

∂2Φ(t, ~x ) = ρ(t, ~x ). (E.22)

62

E.2.1 Proof

We first define Φε(t, ~x ) and Φ⊥(t, ~x ) , analogously to Eq. (E.16) and Eq. (E.17),i.e. in the former the integration domain of the integral in Eq. (E.21) is smoothlyrestricted to a sphere of radius 2ε about the point ~x ′ = ~x , while Φ⊥(t, ~x ) containsthe rest of the integral. In the latter,

∇2Φ⊥(t, ~x ) =1

4π

∫d3x′

1

R

∂2

∂R2R[1− fε(R)]

ρ(t−R, ~x ′)

R

=1

4π

∫d3x′[1− fε(R)]

1

R

∂2

∂t2ρ(t−R, ~x ′) + O(ε)

=∂2Φ⊥(t, ~x )

∂t2+ O(ε), (E.23)

i.e. ∂2Φ⊥(t, ~x ) → 0 as ε → 0. As to the contribution of the small sphere aboutR = 0, it is easy to see that ∂2Φε/∂t2 tends to zero as ε → 0. Moreover, just as inthe static case, we can apply the method of Eq. (E.8) to Eq. (E.11) to show that

∇2Φε = −ρ(t−R, ~x − ~R )|R=0 = −ρ(t, ~x ). (E.24)

This concludes the demonstration of Eq. (E.22).

In a completely analogous way, an advanced potential can be defined by

Φ(t, ~x ) =1

4π

∫d3x′

ρ(t + |~x ′ − ~x |, ~x ′)

|~x ′ − ~x |, (E.25)

and it is clear that this, too, is a solution of Eq. (E.22). That the retarded, andnot the advanced solution must be used, in order to describe the radiation from anaccelerating charge, is a prescription that is not contained in the Maxwell differentialequations: it is a deep question of boundary conditions.

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Appendix F

Retarded Fields

We will sketch the rather tedious calculation of the Lienard-Wiechert fields, Eq. (9.23)and Eq. (9.24), starting from the potentials, Eq. (9.19) and Eq. (9.22).

F.1 Lemmata

The following auxiliary theorems will first be proved:

α: dt′

dt= 1

κ,

β: ~∇t′ = −~nκ

,

γ: ∂~n∂t′

= ~n∧(~n∧~v)R

,

δ: [~∇(κR)]t′ fixed = ~n− ~v ,

ε: ∂(κR)∂t′

= v2 − ~n.~v −R~n.~v .

Here t = t′ + R(t′) , R(t′) = |~x− ~X(t′)| , ~n = ~x− ~X(t′)R

.

Further κ = 1− ~n.~v , and ~v = d ~X(t′)dt′

≡ ~X .

F.1.1 Proofs

α : dt = [1 + R]dt′ , and R ≡ ∂R(t′)∂t′

= −~x− ~X(t′)R

. ~X = −~n.~v .

β : 0 = ~∇t = ~∇t′ + ~n.~∇(~x− ~X) = ~∇t′ + ~n− (~n.~v)~∇t′

γ : ∂∂t′

~x− ~XR

= − ~XR− ~n

RR = (~n.~v)~n−~v

R

δ : ~∇(κR) = ~∇[R− (~x− ~X).~v ] = ~n− ~v

ε : ∂(κR)∂t′

= κR−[

∂~n∂t′

.~v + ~n.~v]R = −R(~n.~v)− (~n− ~v).~v .

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F.2 Theorems

The main theorems are:

I: ~∇A0 = − eκ3

[~n.~vR

+ 1−v2

R2

]~n + e

κ2R2~v ,

II: [~∇∧ ~A ]t′ fixed = − eκ2R2~n ∧ ~v ,

III:[

∂ ~A∂t′

]~xfixed = e

κ2R[~v + ~n ∧ (~v ∧ ~v )] + e

κ2R2 [~n.~v − v2]~v .

F.2.1 Proofs

With the help of the Lemmata, and the following decompositions, the desiredformulae for the Lienard-Wiechert fields can be readily obtained:

I: ~∇A0 = − eκ2R2

[~∇(κR)]t′ fixed + ∂(κR)

∂t′~∇t′

II: [~∇∧ ~A]t′ fixed = − e

κ2R2~∇(κR) ∧ ~v

III:[

∂ ~A∂t′

]~x fixed = e~v

κR− e~v

κ2R2∂(κR)

∂t′

In calculating ~B from Theorem II, one must remember that

* ~∇∧ ~A = [~∇∧ ~A ]t′ fixed + ~∇t′ ∧[

∂ ~A∂t′

]~x fixed .

Similarly, in using Theorem III to calculate ~E, one uses

* ∂ ~A∂t

=[

∂ ~A∂t′

]~x fixed

dt′

dt.

F.3 Corollary

From Eq. (9.23) we have

~n ∧ ~E = eκ3R

~n ∧ ~n ∧ [(~n− ~v) ∧ ~v ]+ e(1−v2)κ3R2 ~n ∧ (~n− ~v)

However~n ∧ ~n ∧ ~C = ~n~n. ~C − ~C.

So with~C = (~n− ~v) ∧ ~v

we find~n∧~n∧ [(~n−~v)∧ ~v ] = −~n~n.(~v∧ ~v )−~n∧ ~v +~v∧ ~v = −~n∧~v +~n∧ (~v∧ ~v ).

Hence~n ∧ ~E = − e

κ3R~n ∧ ~v + ~n ∧ (~v ∧ ~v ) − e(1−v2)

κ3R2 ~n ∧ ~v = ~B.As we see from Eq. (9.24).

65