Electromagnetism: Magnetostatics and Electrodynamics

Simon Ross

Centre for Particle Theory, Department of Mathematical SciencesDurham University, South Road, Durham DH1 3LE

February 26, 2015

Contents

1 Magnetostatics: Lorentz force law 2

2 Biot-Savart law 4

3 Divergence and curl of ~B 63.1 Integral form and applications of Ampere’s law . . . . . . . . . . . . . . . . . . 7

4 Vector potential 84.1 Multipole expansion for the vector potential . . . . . . . . . . . . . . . . . . . . 9

5 Time-dependent fields 115.1 Faraday’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135.2 Maxwell’s modification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145.3 Magnetic charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

6 Energy 166.1 Mutual inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166.2 Energy in the magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.3 Energy conservation: Poynting’s theorem . . . . . . . . . . . . . . . . . . . . . . 18

7 Wave solutions, light 207.1 Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207.2 Electromagnetic waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

8 Sources: Potentials and gauge transformations 238.1 Gauge transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248.2 Retarded potential solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268.3 Dipole fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

9 Electrodynamics and special relativity 309.1 Lorentz transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319.2 Four-vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329.3 Electrodynamics in four-vector notation . . . . . . . . . . . . . . . . . . . . . . 34

1

1 Magnetostatics: Lorentz force law

Study of magnetic fields has a much longer history than electric fields. The story becomesinteresting to us with the realization in the 19th century that currents, that is, moving charges,create and feel a force from magnetic fields.

We will first consider situations where the magnetic fields are time-independent: mag-netostatics. Magnetic fields are produced and felt by currents, that is moving charges, soin magnetostatics the charges cannot all be stationary, but we can consider steady currents,situations where ρ(~r) and ~j(~r) are independent of time.

In general, we had a continuity equation, expressing conservation of charge, which said

∂ρ

∂t+∇ ·~j = 0.

for steady currents, this simplifies to ∇ ·~j = 0. This divergencelessness expresses the conserva-tion of current; it must circulate, if charge density is time-independent there are no sources orsinks of current. [Picture]

This means we can’t start with point charges; a point charge in motion has a time-dependentcharge and current density. The elementary situation will be a line current flowing in a wire.[Picture].

Take the net charge density on the wire to vanish, the current density is ~j = Iδ2(r). Notethe line need not be straight.

Orsted (1819) found that a wire produces a magnetic field, which can be measured bycompasses placed near the wire:

~B =2I

c

1

r~eφ. (1.1)

But the basic observation (analogue of Coulomb’s law) is that there is a force between currents.Ampere (1825): If I have two parallel wires carrying steady currents, there is a force betweenthem. [Picture]. Force per unit length

~F ∝ I1I2~err

(1.2)

Similar in functional form to the force between long linear charge distributions, but dependingon the currents, not charge density.

Units: usually one has a constant µ0 here. In the units where 4πε0 = 1 that I used lastterm, this turns out to be related to the speed of light: ε0µ0 = 1

c2, so in my units µ0 = 4π

c2.

I will write this in terms of c from the outset. Relative to Griffiths, I have also changed thedefinition of ~B by scaling it by a factor of c so that ~E and ~B have the same units (C/L2). NoteI has units of C/T .

We can then think of the force on the other wire as due to a velocity-dependent force on thecharges moving in the wire, in the presence of the magnetic field ~B created by the first wire.The force is orthogonal to the velocity and ~B, so it goes as ~v ∧ ~B. Experimentally, the force ona single point charge moving in a magnetic field ~B is

~F = q~v

c∧ ~B.

This is the Lorentz force law. We could imagine (ahistorically) that we introduce the mag-netic field to explain a force on moving charges, just as we introduced the electric field to

2

explain a force on static charges. In the presence of both kinds of fields, the total force on apoint charge is

~F = q( ~E +1

c~v ∧ ~B). (1.3)

Remarks:

• We derived this for static fields, but the above expression is true in general, for pointparticles moving in any electric and magnetic fields.

• For charge and current distributions, the force on a volume element is

d~F = ρE +1

c~j ∧ ~B,

where ~j = ~vρ is the current density.

• Because the force is orthogonal to the velocity, the magnetic field does no work.

• Exercise (for problems class) motion in a constant field ~B0.

• There is a factor of 1/c here relative to the expression in Griffiths. This is because I definethe magnetic field differently.

Returning to the wires, we can work out the force on a current-carrying wire from the aboveexpression. Suppose the wire carries no net charge density and a current I. Then for a smalllength element,

d~F =1

c~j ∧ ~B =

1

cId~l ∧ ~B,

as the direction of the current density is in the direction of the wire. [Picture]So

~F =1

cI

∫d~l ∧ ~B. (1.4)

Example: For the situation we had before with two parallel long straight wires, the mag-netic field was

~B =2I1

c

1

r~eφ,

which is constant along the second wire, so

~F = −2I1I2

c2

1

r~er

∫d~l,

reproducing the force per unit length I had before.For a closed loop of wire in a constant field ~B0,

~F =I

c

∮d~l ∧ ~B0.

This vanishes because∮d~l = 0.

3

2 Biot-Savart law

Experimentally, we see that currents produce magnetic fields. We want to write down anintegral expression for the field in general, analogous to the one we had for the electric field.In this case, however, we can’t build it up by superposition of point charges. Consider theexpression for a straight line current:

~B(~r) =2I

c

1

r~eφ.

This is orthogonal to the direction of the current and to the separation vector ~r. Motivates usto propose that for a current element Id~l at ~r′,

d ~B(~r) =I

c

d~l ∧ (~r − ~r′)|~r − ~r′|3

.

Note that unlike in the electric case, I can’t consider this in isolation. The conservation law∇ ·~j = 0 means there can’t be isolated current elements.

For a single element, we take ~B to fall off as the inverse square of the distance, as in theelectric case. This will reproduce the inverse distance for a long straight wire once we integrateover the source.

Integrating, we have the Biot-Savart law

~B(~r) =I

c

∫d~l ∧ (~r − ~r′)|~r − ~r′|3

. (2.1)

For a continuous current density,

~B(~r) =1

c

∫ ~j(~r′) ∧ (~r − ~r′)|~r − ~r′|3

dV ′. (2.2)

Compare to

~E(~r) =

∫ρ(~r′)(~r − ~r′)|~r − ~r′|3

dV ′.

Similar distance dependence, local source dependence, but different vector character.Example: Long straight wire, carrying current I. Work in cylindrical polar coordinates,

so ~r′ = z′~ez, ~r = z~ez + r~er. d~l = dz~ez. So

d~l ∧ (~r − ~r′) = rdz~eφ,

and~B =

I

cr

∫ ∞−∞

dz′

((z − z′)2 + r2)3/2~eφ

Set z′ = z + rx; dz′ = rdx

~B =I

c

1

r

∫ ∞−∞

dx

(1 + x2)3/2~eφ

So see there’s no z dependence, 1/r falloff as claimed before. Now∫ ∞−∞

dx

(1 + x2)3/2= [

x√1 + x2

]∞−∞ = 2,

4

so~B =

2I

c

1

r~eφ.

Example: Circular current loop of radius R, carrying current I: calculate ~B at a point onthe axis of the loop a distance z from its center. Work in cylindrical polar coordinates again:by symmetry, ~B = B~ez. ~r = z~ez, ~r

′ = R~er, d~l = Rdφ~eφ, so

[d~l ∧ (~r − ~r′)] · ~ez = R2dφ,

and

B =I

cR2

∫ 2π

0

dφ

(R2 + z2)3/2=I

c

2πR2

(R2 + z2)3/2.

Can’t work out field off axis in closed form, but will discuss analogue of the electric dipoleapproximation later.

Example: Solenoid. Consider a long cylinder with current I flowing in a wire with N turnsper unit length, that is current density NI per unit length, radius R. Field on axis: add upcontributions from many loops as above. At one end,

B =NI

c2πR2

∫ L

0

dz

(R2 + z2)3/2=

2πNI

c[

z√R2 + z2

]L0 =2πNI

c

L√R2 + L2

.

In the middle? Use superposition, it’s like two such cylinders.

B = 22πNI

c

L/2√R2 + L2/4

=4πNI

c

L√4R2 + L2

.

Inside a very long solenoid, B = 4πNIc

, which we will recover in a different way shortly.Example: Force between two closed circuitsThe magnetic field due to a closed loop carrying current I2 is

~B(~r1) =I

c

∮2

d~l2 ∧ (~r1 − ~r2)

|~r1 − ~r2|3,

so total force on loop 1 due to loop 2 is

~F1 =I1

c

∮d~l1 ∧ ~B =

I1I2

c

∮1

∮2

d~l1 ∧ (d~l2 ∧ (~r1 − ~r2))

|~r1 − ~r2|3.

This is not obviously antisymmetric. Would have expected ~F2 = −~F1.Use ~A ∧ ( ~B ∧ ~C) = ~B( ~A · ~C)− ~C( ~A · ~B) to write as

~F1 =I1I2

c

[∮d~l2

∮d~l1 · (~r1 − ~r2)

|~r1 − ~r2|3−∮

1

∮2

(~r1 − ~r2)d~l1 · d~l2|~r1 − ~r2|3

].

The second term has the expected symmetry. See that the first term vanishes: note that for agradient,

∮d~l · ~∇f = 0 for any f . But

(~r1 − ~r2)

|~r1 − ~r2|3= −~∇1

1

|~r1 − ~r2|,

so the first term is of gradient form and hence vanishes.

5

3 Divergence and curl of ~B

We now want to derive the (time-independent) Maxwell equations for the magnetic field fromthe Biot-Savart law. I will do this in a slightly different way from Griffiths, anticipating thenext section. We observe that the Biot-Savart law involves

~r − ~r′

|~r − ~r′|3.

Now, in the electric case, we used

~∇ 1

|~r − ~r′|= − ~r − ~r′

|~r − ~r′|3

in writing ~E = −~∇φ. Let’s use the same observation here to write the Biot-Savart law as

~B = −1

c

∫~j(~r′) ∧ ~∇ 1

|~r − ~r′|dV ′ =

1

c~∇∧

∫ ~j(~r′)

|~r − ~r′|dV ′,

where in the second step we used that the derivative is with respect to ~r, so it doesn’t act onthe ~j, and we can take it outside the integral over ~r′. Note the sign change from reversing theorder of arguments in the cross product.

Now we can easily evaluate the divergence and curl of ~B:

~∇ · ~B = ~∇ · (~∇∧ . . .) = 0,

as ~∇ · (~∇∧ ~A) = 0 for any vector field ~A. Thus, the divergence of the magnetic field vanishes,

~∇ · ~B = 0. (3.1)

For the curl, we need to use the vector identity

~∇∧ (~∇∧ ~A) = ~∇(~∇ · ~A)− ~∇2 ~A

to write~∇∧ ~B =

1

c~∇(

∫~j(~r′) · ~∇ 1

|~r − ~r′|)− 1

c

∫~j(~r′)~∇2 1

|~r − ~r′|.

In the first term, we note that

~∇ 1

|~r − ~r′|= −~∇′ 1

|~r − ~r′|,

and having written it in terms of a derivative with respect to ~r′, we can integrate by parts,assuming the boundary contribution vanishes, to rewrite∫

~j(~r′) · ~∇ 1

|~r − ~r′|= −

∫~j(~r′) · ~∇′ 1

|~r − ~r′|=

∫(~∇′ ·~j(~r′)) 1

|~r − ~r′|= 0

by the time-independent continuity equation, ~∇ · ~j = 0. For the second term, we can recallfrom electrostatics that

~∇2 1

|~r − ~r′|= −4πδ3(~r − ~r′),

6

so~∇∧ ~B =

4π

c~j(~r). (3.2)

This is known as Ampere’s law. As with Gauss’ law in electrostatics, it gives us a local relationbetween the magnetic field and its source. We now have derived all of the time-independentMaxwell equations: we have that

~∇ · ~E = 4πρ, ~∇∧ ~E = 0,

so electric field lines diverge from a point source, and

~∇ · ~B = 0, ~∇∧ ~B =4π

c~j,

magnetic field lines curl around a current. [Picture]. In the static situation, the charge density

ρ is any function, but the current density ~j is divergenceless, ~∇ ·~j = 0. In the time-dependentcase, the divergence equations stay the same but the curl equations get modified.

3.1 Integral form and applications of Ampere’s law

As with Gauss’ law, we can obtain an integral form of Ampere’s law, which is useful to determinethe magnetic field in simple situations. Since Ampere’s law involves a curl, we should integrateover a surface, so we can use Stokes’ theorem.[Picture] We integrate over a closed surface Sbounded by some curve C. The flux of current through the surface S is defined by

Ienc =

∫S

~j · ~ndS =c

4π

∫S

(~∇∧ ~B) · ~ndS =c

4π

∮C

~B · d~l,

where we used Stokes’ theorem in the last step. Thus, the integral of ~B along a closed loopis related to the current enclosed by the loop, as the integral of the electric field over a closedsurface was related to the charge inside.∮

C

~B · d~l =4π

cIenc. (3.3)

We can use this to find the magnetic field in cases with enough symmetry: when we onlyneed to find one component, and we can take the ~B outside the integral.

Example: Solenoid. Consider an infinitely long solenoid of radius R, with n turns per unitlength carrying current I - that is, a current density per unit length nI flowing in the angulardirection around the solenoid. [Picture].

Symmetry: work in cylindrical polar coordinates. The current distribution has translationalsymmetry in z and rotational symmetry in φ, so we assume the magnetic field is a functiononly of r, ~B(r). The current distribution is symmetric under I → −I, z → −z; under I → −I,~B → − ~B by the Biot-Savart law, while z → −z only reverses the sign of Bz. Hence Bφ =−Bφ = 0, Br = −Br = 0, and the only non-zero component is Bz.

We could also see Bφ = 0 by applying Ampere’s law to a loop of radius r0:∮C

~B · d~l = 2πr0Bφ(r0),

but Ienc = 0 as the current flows in the φ direction, not the z direction. We could also seeBr = 0 using ~∇ · ~B = 0.

7

Now let’s use Ampere’s law to determine Bz(r). Consider a rectangular loop of length L,between r = a and r = b, b > a [Picture].∮

C

~B · d~l = L(Bz(a)−Bz(b)).

If b > a > R, Ienc = 0, so Bz(a) = Bz(b). Assuming ~B vanishes far from the solenoid, Bz = 0for all r > R. If b > R > a, Ienc = nIL, so Bz(a)−Bz(b) = 4π

cnI. That is,

Bz(r) =

{4πnIc

r < R,0 r > R.

This is consistent with the on-axis result obtained by explicit integration; the advantage ofAmpere’s law is it enables us to extend it off the axis for an infinitely long solenoid, whichwould be a hard calculation using the explicit integral.

From a practical point of view, we learn long solenoids are a useful way to generate approx-imately constant fields.

4 Vector potential

In electrostatics, the vanishing of the curl, ~∇ ∧ ~E = 0 implied that we could write ~E = −~∇φ.Similarly in magnetostatics, the fact that ~∇ · ~B = 0 implies that we can write

~B = ~∇∧ ~A (4.1)

where ~A is the vector potential. Clearly, if we can find such an ~A, ~∇·(~∇∧ ~A) = 0 would imply

the divergence equation is satisfied. But can we always find such an ~A? This is a system ofthree equations for the three components of ~A, so we would expect it to have a solution. Sinceit’s a differential equation, there’s a consistency condition coming from the fact that secondderivatives commute: writing the equation in components,

∂yAz − ∂zAy = Bx, ∂zAx − ∂xAz = By, ∂xAy − ∂yAx = Bz,

but taking derivatives,

∂x(∂yAz − ∂zAy) + ∂y(∂zAx − ∂xAz) + ∂z(∂xAy − ∂yAx) = 0

for any ~A as partial derivatives commute. The LHS is ~∇ · ~B, so this consistency condition issatisfied if the divergence equation is satisfied. Thus there is generically a solution for ~A. Infact, we already saw a solution for ~A previously, when we observed that we could rewrite theBiot-Savart law as

~B =1

c~∇∧

∫ ~j(~r′)

|~r − ~r′|dV ′.

We’ll shortly rederive this solution.The equation does not determine ~A uniquely. In the electrostatic case, there was a similar

non-uniqueness, of an overall constant φ0 we could add to φ; here it’s bigger. As ~∇ ∧ ~∇λ = 0for any function λ, we can shift

~A→ ~A+ ~∇λ (4.2)

and still have a solution. This is called a gauge freedom, and the function λ is called agauge transformation. The choice of λ has no effect on the physics. We can make use of

8

this freedom to try to pick a simple form of ~A (this is called gauge choice or gauge fixing). Aconvenient choice is the Coulomb gauge

~∇ · ~A = 0. (4.3)

If we have a vector field ~A0 not in Coulomb gauge, and we want ~A = ~A0 + ~∇λ to be inCoulomb gauge, this requires

~∇ · ~A = ~∇ · ~A0 +∇2λ = 0,

that is, the required gauge transformation λ satisfies a Poisson equation,

∇2λ = −~∇ · ~A0.

This will generically have a solution for λ. However, it doesn’t fix λ uniquely: gauge transfor-mations that satisfy Laplace’s equation ∇2λ = 0 preserve the Coulomb gauge.

One might think that we haven’t gained much in writing the magnetic field in terms of avector potential: we’ve traded a vector for another vector, and we seem to have gained a newcomplication in the form of gauge freedom.

But gauge freedom gives us the freedom to simplify things, so the solution for ~A for a givencurrent is simpler than for ~B. Consider Ampere’s law in terms of ~A:

4π

c~j = ~∇∧ ~B = ~∇∧ (~∇∧ ~A) = ~∇(~∇ · ~A)−∇2 ~A.

So far, not much simpler. But if we choose the Coulomb gauge ~∇ · ~A = 0, then the equationbecomes simply

∇2 ~A = −4π

c~j (4.4)

which is an independent Poisson equation for each component of ~A, and we can write a solutionjust as in electrostatics,

~A(~r) =1

c

∫ ~j(~r′)

|~r − ~r′|dV ′. (4.5)

Thus, the previous solution for ~B amounted to writing the vector potential in Coulomb gauge.We see that the problem for the vector potential in this gauge is essentially the same as theelectrostatic problem. For a line current,

~A(~r) =I

c

∫d~l

|~r − ~r′|. (4.6)

The vector potential will also play an important role in the relativistic formulation of elec-trodynamics, and in quantum treatments of the fields (although we won’t get to the latter).

4.1 Multipole expansion for the vector potential

As in the electrostatic case, it’s still hard to do the integral for ~A for general current distribu-tions. Therefore, we develop a multipole expansion for distant fields. This works only whenwe consider currents which circulate around inside of some finite region. For simplicity, we’llconsider a closed wire loop of arbitrary shape, carrying current I. [Picture]

We make use of the same Fourier series expansion,

1

|~r − ~r′|=

1

r+~r · ~r′

r3+ . . .

9

For simplicity, we’re only going to keep up to the dipole term, but in principle there’s an infiniteseries of corrections. Then

~A(~r) =I

c

∮d~l

|~r − ~r′|≈ I

cr

∮d~l +

I

c

1

r3·∮

(~r · ~r′)d~l + . . .

The first term vanishes:∮d~l = 0. That is, there is no monopole contribution to the magnetic

field. Not surprising.We would like to separate the dependence on ~r from the dependence on the source in the

second term, but the vector structure makes that harder here than in electrostatics. We can avariant of Stokes’ theorem on the vector calculus handout to write

~A(~r) ≈ I

c

1

r3·∮

(~r · ~r′)d~l =I

c

1

r3

∫~n ∧ ~∇′(~r · ~r′)dS,

where the integral is over a surface S spanning the closed loop (it doesn’t matter which one).

Now the derivative is with respect to position on the surface, so ~∇′(~r · ~r′) = ~r, and

~A(~r) ≈ I

c

1

r3

∫~n ∧ ~rdS =

I

c

1

r3(

∫~ndS) ∧ ~r =

~µ ∧ ~rr3

,

so the dipole field is

~A =~µ ∧ ~rr3

, (4.7)

where we have introduced the magnetic dipole moment

~µ =I

c

∫~ndS. (4.8)

We did this discussion for a closed loop carrying current I, but it extends to any currentdensity ~j which circulates within some finite region. We can see this hueristically by thinking of~j as some linear superposition of closed loops. The monopole moment term will always vanish.

The simplest case to consider is a circular loop of radius a. The monopole moment is~µ = I

cπa2~n, where ~n is the normal to the loop. We can define an ideal dipole by taking a→ 0

and I →∞ holding ~µ fixed.The dipole term in the magnetic field can be worked out by evaluating ~B = ~∇∧ ~A. So

~B = ~∇∧ ~µ ∧ ~rr3

.

This is easiest to evaluate in components:

Bi = εijk∂jεklmµlxm

r3.

Now ∂jxm = δjm, and εijkεklj = 2δil, so the derivative acting on xm gives 2µi/r3. On the second

term, ∂jr−3 = −3xjr

−5, so

Bi = 2µir3− 3r−5εijkεklmµlxjxm,

that is

~B = 2~µ

r3− 3

~r ∧ (~µ ∧ ~r)r5

.

10

Using the triple product identity, ~A∧ ( ~B∧ ~C) = ~B( ~A · ~C)− ~C( ~A · ~B), ~r∧ (~µ∧~r) = ~µr2−~r(~µ ·~r),so

~B = − ~µr3

+ 3~r(~µ · ~r)r5

. (4.9)

Remarkably, this is identical in form to the electric dipole,

~E = − ~p

r3+ 3

~r(~p · ~r)r5

.

[Picture] Different for physical dipoles. [Picture]On the axis of a circular loop, we previously worked out that

~B =I

c

2πR2

(R2 + z2)3/2~ez.

At large z,

~B ≈ I

c

2πR2

z3~ez,

which agrees with the above dipole expression, given that ~µ = µ~ez = IcπR2~ez and ~r = z~ez, so

~B = − µz3~ez + 3

z~ez(µz)

z5= 2

µ

z3~ez =

I

c

2πR2

z3~ez.

This concludes our discussion of static fields.

5 Time-dependent fields

Let’s turn to time-dependent fields. We start by considering a wire loop moving in a staticfield. [Picture]

For a static loop, the velocity of the electrons in the wire is along the wire, so the Lorentzforce

~F = q ~E +q

c~v ∧ ~B

does not lead to any net work done on the electrons, ~F · d~l = 0. However, for a moving loop, itcould lead to non-zero work done along the wire: as ~F · d~l = q

∮(~vcc ∧ ~B) · d~l = q

∮(~v ∧ ~B) · d~l,

where ~v is the velocity of the loop. We define the electromotive force (emf)

E =1

q

∮~F · d~l. (5.1)

This is a bad name; it’s not a force, it’s work done per unit charge. But we’re stuck with it.That’s life. For the moving loop,

E =1

c

∮(~v ∧ ~B) · d~l.

This is called a motional emf. Note that if there is a static electric field, it makes no con-tribution to the emf, as ~∇ ∧ ~E = 0 implies

∮~E · d~l = 0. The component of the velocity of

the electrons due to the current is along the wire, so it drops out, and we can take ~v in thisexpression to be the velocity of the loop.

A non-zero emf means work is being done on the charges in the wire, moving them aroundthe wire, producing a current. Note that it is not the magnetic field that is doing work; its

11

force is always orthogonal to velocity, so it can’t do any work. The work is done by whateveris moving the loop. The amount of work you have to do to move the loop depends on themagnetic field.

Example: a square loop with a constant ~B field. [Picture] Suppose the ~B field is ~B = B~ezfor x < 0, ~B = 0 for x > 0. Consider a square loop of side length 2a in the x − y plane, withits center between −a, a. Then

~v ∧ ~B = vyB~ex − vxB~ey

for x < 0. The contributions to E from the two horizontal components cancel, as ~dl points inopposite directions on the two sides, but for the vertical components, one is at x < 0 and oneis at x > 0 where ~B = 0. Thus

E =1

c

∫x<0

(~v ∧ ~B) · d~l =1

c

∫ a

−avxBdy =

2a

cvxB.

Why does only velocity in the x direction contribute to the emf? This is because this motionis changing the amount of magnetic field threading the loop. Indeed, we can recognise theright-hand side as −1

cdΦB

dt, where we define the magnetic flux through a surface S spanning the

loop as [Picture]

ΦB =

∫S

~B · ~ndS, (5.2)

where ~n is the unit normal to the surface. Using ~B = ~∇∧ ~A, we can rewrite this as

ΦB =

∮C

~A · d~l.

Note in passing that this quantity is invariant under the gauge transformation ~A → ~A + ~∇λ,as∮C~∇λ · d~l =

∮C∂φλdφ = 0 as it’s a total derivative.

Remarkably, we can write the emf in terms of the change in flux in general. As we move anarbitrary loop, ΦB may change. [Picture]

dΦb(t) = ΦB(t+ dt)− ΦB(t) = ΦB(ribbon) =

∫ribbon

~B · ~ndS.

where we have used ~∇· ~B = 0 to say that the total flux through a closed surface vanishes. Now[Picture]

~ndS = ~v⊥dt ∧ d~l = ~vdt ∧ d~l,where ~v⊥ is the component of the velocity orthogonal to the loop, and we can add the parallelcomponent in without changing anything. So

dΦb(t) =

∫~B · (dt~v ∧ d~l).

Using ~B · ( ~A ∧ ~C) = ~C · ( ~B ∧ ~A),

dΦB

dt=

∫( ~B ∧ ~v) · d~l = −

∫(~v ∧ ~B) · d~l = −cE ,

so the motional emf is simply

E = −1

c

dΦB

dt. (5.3)

12

We see that a changing ΦB generates an emf, and hence a current. There’s a rule of thumb,called Lenz’s law, which helps to remember the sign of the emf: “Nature abhors a change influx”. So the emf always generates a current which produces a ~B flux in the opposite directionto the change. If ΦB is decreasing, the emf will produce positive flux through the loop, andvice-versa. [Picture].

5.1 Faraday’s law

So we see that changing the magnetic flux through a loop by moving the loop produces a currentin the loop. What about other ways of changing the magnetic flux? Faraday (1831) found thatby moving the source of the field (i.e. moving the magnet) or changing the strength of the field(e.g. electromagnet) also produced an emf. Empirically,

E = −1

c

dΦB

dt

in all cases. This is not a surprise in light of relativity (whether you’re moving the loop ormoving the magnet depends on what frame you’re working in - this experimental observationin part inspired Einstein’s interest in relativity), but it was a big surprise at the time!

Why was it surprising? Because it contradicts the laws we’ve learned so far. Assume there’sno current initially, so the Lorentz force law says ~F = q ~E. But we didn’t have any electric fieldin the problem to start with (before we moved the magnet), and anyway the emf is

E =

∮~E · d~l,

and our Maxwell equation ~∇∧ ~E = 0 says this should always vanish!Faraday’s resolution was that a changing magnetic field induces an electric field. To

reproduce the experiment,

E =

∮~E · d~l = −1

c

dΦB

dt= −1

c

∫∂ ~B

∂t· ~ndS,

if we assume that only the ~B is time-dependent (the loop isn’t changing). Using Stokes,∮~E · d~l =

∫(~∇∧ ~E) · ~ndS,

so we obtain

~∇∧ ~E = −1

c

∂ ~B

∂t. (5.4)

This is called Faraday’s law. It’s one of our time-dependent Maxwell equations.In time-varying fields, we see ~E and ~B are coupled.Example: A square loop of wire of side length a lies at a distance d from a long straight

wire carrying current I(t). [Picture] The magnetic field is ~B = 2 Ic1/r~eφ, so the flux through

the loop is

ΦB =

∫Bφdrdz = 2

I

ca

∫ d+a

d

dr

r= 2

I

ca ln

d+ a

d.

Thus the emf is E = −1cdΦB

dt= 2 a

c2ln d+a

ddIdt

.

13

What is the electric field that creates this emf? By symmetry, ~E(r, t). We want ~∇∧ ~E ∝ ~eφ,

so ~E = Ez(r, t)~ez. (Assuming no net charge on the wire, a radial component of ~E is forbidden

by ~∇ · ~E = 0.) The problem is then simple enough to fix Ez directly from the emf:

E =

∮~E · d~l = −

∫r=d+a

Ezdz +

∫r=d

Ezdz = a(Ez(d+ a, t)− Ez(d, t)) = −2a

c2lnd+ a

d

dI

dt,

so~E =

2

c2ln r

dI

dt~ez.

Verify

~∇∧ ~E = −∂Ez∂r

~eφ = − 2

c2r

dI

dt~eφ = −1

c

∂ ~B

∂t.

5.2 Maxwell’s modification

Faraday’s experiments demonstrated the need to modify the curl of ~E. One of Maxwell’simportant contributions was to note a purely theoretical problem with the curl of ~B. Inmagnetostatics, we had

~∇∧ ~B =4π

c~j.

The divergence of a curl is always zero, but in a time-dependent situation, the divergence ofthe right-hand side is not zero:

~∇ · (~∇∧ ~B) =4π

c~∇ ·~j = −4π

c

∂ρ

∂t6= 0,

where in the last step we used the continuity equation. So for consistency with the continuityequation, we need to modify Ampere’s law. Since

~∇ · ~E = 4πρ,

a natural modification is an analogue of Faraday’s law:

~∇∧ ~B =4π

c~j +

1

c

∂ ~E

∂t, (5.5)

so that

0 = ~∇ · (~∇∧ ~B) =4π

c~∇ ·~j +

1

c

∂

∂t~∇ · ~E =

4π

c~∇ ·~j +

4π

c

∂ρ

∂t,

which is satisfied by virtue of the continuity equation. This seems relatively obvious now,but was a real breakthrough in Maxwell’s time: part of his brilliance was to recognise theimportance of this combination of derivatives.

By analogy to Faraday’s statement that changing ~B produces ~E, we see that changing ~Eproduces ~B.

This problem with Ampere’s equation can be seen clearly by considering a capacitor [Pic-ture]. The integral form of Ampere’s equation says∮

C

~B · d~l =4π

cIencl =

4π

c

∫~j · ~ndS.

14

But the enclosed current depends on which surface we use to evaluate it. This is because~∇ ·~j 6= 0. If we add in the ~E contribution, we have∮

C

~B · d~l =1

c

∫(4π~j +

∂ ~E

∂t) · ~ndS. (5.6)

Now if we consider the surface between the plates, ~E = 4πσ~n, so∫~E · ~ndS = 4πσA = 4πQ = 4πIt+Q0,

and ∮C

~B · d~l =1

c

∫(4π~j +

∂ ~E

∂t) · ~ndS =

4π

cI

on either surface.We now have the full set of dynamical Maxwell equations:

~∇ · ~B = 0, ~∇ · ~E = 4πρ, (5.7)

~∇∧ ~B =4π

c~j +

1

c

∂ ~E

∂t, ~∇∧ ~E = −1

c

∂ ~B

∂t. (5.8)

Along with the Lorentz force law, these describe all of classical electrodynamics. Note thatthe continuity equation

~∇ ·~j +∂ρ

∂t= 0

can be obtained as a consequence of Maxwell’s equations.

5.3 Magnetic charge

It is interesting to note that in the absence of sources, there is an interesting symmetry inMaxwell’s equations between ~E and ~B:

~∇ · ~B = 0, ~∇ · ~E = 0,

~∇∧ ~B =1

c

∂ ~E

∂t, ~∇∧ ~E = −1

c

∂ ~B

∂t,

so if we replaced ~B → ~E, ~E → − ~B, we get back the same equations.This is broken by sources, but would be restored if we allowed for magnetic sources. Imagine

there was a type of particle that acted as a point source for the magnetic field, as charges dofor the electric field, so

~∇ · ~B = 4πρm,

where ρm is the magnetic charge density. Now we would have problems with the divergence ofcurl of ~E:

~∇ · (~∇∧ ~E) = −1

c

∂

∂t~∇ · ~B = −4π

c

∂ρm∂t6= 0.

But assuming a continuity equation is satisfied, we could resolve this by assuming the magneticcurrent sources ~E: if

~∇∧ ~E = −4π

c~jm −

1

c

∂ ~B

∂t,

15

~∇ · (~∇∧ ~E) = −4π

c~∇ ·~jm −

4π

c

∂ρm∂t

= 0

by the continuity equation. The two sets of charges would appear symmetrically.Nature is not that symmetrical, as far as we know: no magnetic monopoles have been

observed in nature. But such symmetries between different fields, known as dualities, play animportant role in contemporary theoretical physics.

6 Energy

Now return to the subject we skipped, and determine the energy stored in a static magneticfield, analogous to that worked out in electrostatics for the electric field. In our magnetostaticsdiscussion, we saw that the magnetic forces do no work, so we might expect it costs no energyto arrange a charge distribution to produce a magnetic field, so there would be no energy inthe magnetic field. But we’ve seen that changing magnetic fields produce electric fields, andwe need to do work against those.

6.1 Mutual inductance

If we consider the flux through one loop produced by the current in another loop, there’s asimple relation. [Picture].

Current I1 in C1 produces a field ~B1, the flux through C2 is

ΦB2 =

∫S2

~B1 · ~n2dS =

∮C2

~A1 · d~l2.

But

A1 =I1

c

∮C1

d~l1|~r1 − ~r2|

,

so

ΦB2 =I1

c

∮C1

∮C2

d~l1 · d~l2|~r1 − ~r2|

.

This has a nice symmetry between 1 and 2. Define the mutual inductance

M12 =1

c2

∮C1

∮C2

d~l1 · d~l2|~r1 − ~r2|

. (6.1)

Then ΦB2 = cI1M12, and M21 = M12, so ΦB1 = cI2M12. Beautiful symmetry in the flux. Notethe mutual inductance is a purely geometrical quantity, specified by the shape and relativeposition of the loops.

Example: Flux due to a short solenoid. Consider a solenoid of length l, radius r, n turnsper unit length contained in an infinitely long one of radius R > r, N turns per unit length. Ifa current I flows in the short one, what is the flux through the long one?

Can’t calculate mutual inductance directly, but we can infer it by calculating the flux inthe short one due to a current in the long one. A current I in the long solenoid produces

~B =4πNI

c~ez

so

ΦB =4πNI

cπr2nl

16

so mutual inductance is

M =4πNn

c2πr2l.

Proportional to the volume of the smaller solenoid. Flux in larger one due to a current in thesmaller one is the same.

We considered the field through C1 due to the current in C2, but there is also a field in C1

due to current in C1. This is determined by the self-inductance. Formally

L =1

c2

∮C1

∮C1

d~l1 · d~l2|~r1 − ~r2|

,

but this is divergent, because we’ve idealised the wire as infinitesimally thin. In practice wecan’t do the integral anyway, so we just define the self-inductance as the coefficient in Φ = cLI.Important point is this depends only on the geometry of the loop.

If we vary the current in one loop, this produces an emf in the other:

E = −1

c

dΦB

dt= −MdI

dt.

(Assuming the change in the current is slow enough for the magnetostatic calculation of ~B tobe valid.)

Similarly, changing the current in the loop produces an emf in the loop itself:

E = −1

c

dΦB

dt= −LdI

dt.

this emf acts against the change in the current; it wants to set up a flux opposing the changein flux. It’s therefore called a “back emf”, and we have to do work against it to set up thecurrent.

6.2 Energy in the magnetic field

To find the energy in the magnetic field, look at the work done against the back emf in settingup the field. The emf E is work per unit charge, so −E is the work per unit charge done againstthe back emf. The current I measures charge per unit time, so

dW

dt= −IE = IL

dI

dt

is the work done per unit time in changing the current. Note that this is not counting energydissipated in resistance; this is useful energy, which can be recovered by running the currentback down. For a fixed circuit, the self-inductance is a constant, so

W =1

2LI2. (6.2)

This looks simple, but L is hard to calculate, and we would like an expression for more generalsources than just wire loops. Rewrite using ΦB = cLI:

LI =1

c

∫~B · ~ndS =

1

c

∮~A · d~l,

so

W =1

2c

∮~A · Id~l.

17

this is easy to generalise to arbitrary current distributions ~j(~r):

W =1

2c

∫~A ·~jdV.

Now using Ampere’s law we can go further:

~∇∧ ~B =4π

c~j,

so

W =1

8π

∫( ~A · (∇∧ ~B)dV )

and ~∇ · ( ~A ∧ ~B) = ~B · (∇∧ ~A)− ~A · (~∇∧ ~B), so

W =1

8π

∫( ~B · (∇∧ ~A)− ~∇ · ( ~A ∧ ~B))dV =

1

8π

∫B2dV,

neglecting the surface term from the last term. Thus

W =1

8π

∫B2dV (6.3)

Energy stored in the magnetic field, just as for the energy stored in the electric field we foundearlier. Again, there is an attractive symmetry between ~E and ~B.

We derived this using magnetostatic formulas for the relation of ~A and ~B, so the derivationassumes the fields are set up sufficiently slowly that Maxwell’s correction to Ampere’s law isnegligible. But the resulting expression provides a valid notion of the energy stored in anyconfiguration of the magnetic fields at a moment in time.

6.3 Energy conservation: Poynting’s theorem

In electrostatics, we derived the expression

W =1

8π

∫E2dV

for the energy stored in a static electric field by assuming that the charge distribution wasassembled by moving the charges in slowly (adiabatically) from infinity so that the static

formulas relating ~E and ρ remained valid. Similarly here, we derived

W =1

8π

∫B2dV

assuming that the field ~B was set up slowly so that the magnetostatic formulas for the relationof ~A and ~B remain valid. One might therefore wonder if these formulas really remain validwhen ~E and ~B are rapidly changing, and it doesn’t make sense to think about setting up thefields slowly.

Here we will give an independent argument for these formulas, by showing that if we takethe energy to have this form, the total energy in a region is conserved, in the sense that itstime derivative can be related to a surface integral representing energy flux out of the region.

dU

dt+

∫~S · ~ndS = 0.

18

This can compared to the conservation of electric charge:

dQ

dt+

∫~j · ~ndS = 0.

We take the total energy to consist of the energy in the fields, plus whatever kinetic energythe charged particles have. (Recall the field energy already takes into account potential energyof the particles in these fields). So we take

U =1

8π

∫(E2 +B2)dV + UKE.

Taking the time derivative,

dU

dt=

1

4π(

∫~E · ∂

~E

∂t+ ~B · ∂

~B

∂t)dV +

dUKEdt

.

We will assume that the only forces acting on the charges are electromagnetic; then for a single(non-relativistic) particle, the rate of change of the kinetic energy is

d

dt

1

2mv2 = m~v · ~a = ~v · ~F = q~v · ~E.

Generalising to a smooth distribution, we set q~v = ~j, so

dU

dt=

1

4π

∫( ~E · ∂

~E

∂t+ ~B · ∂

~B

∂t)dV +

∫~E ·~jdV.

The term dotted with ~E is familiar: from our corrected Ampere’s law,

~j +1

4π

∂ ~E

∂t=

c

4π~∇∧ ~B.

Let us similarly replace using Faraday’s law

∂ ~B

∂t= −c~∇∧ ~E.

so we havedU

dt=

c

4π

∫[ ~E · (~∇∧ ~B)− ~B · (~∇∧ ~E)]dV.

Now a general vector identity tells us

~∇ · ( ~E ∧ ~B) = ~B · (~∇∧ ~E)− ~E · (~∇∧ ~B).

ThusdU

dt= − c

4π

∫~∇ · ( ~E ∧ ~B)dV = − c

4π

∫( ~E ∧ ~B) · ~ndS,

where in the last step we used the divergence theorem. Thus, using the Lorentz force law andMaxwell’s equations, we can rewrite the time-derivative of the energy as a surface integral,which we identify as energy flux out of the region. This is Poynting’s theorem, and theintegrand is called the Poynting vector

~S =c

4π~E ∧ ~B; (6.4)

19

this is the energy flux density.This derivation used no adiabaticity assumption; it applies to general ~E, ~B fields (but only

to non-relativistic particles), and as a result we are now entitled to view

Uem =1

8π

∫(E2 +B2)dV (6.5)

as the energy stored in the electric and magnetic fields quite generally.Example: for constant fields ~E = E0~ex, ~B = B0~ey, the Poynting vector is S = c

4πE0B0~ez. It

might seem quite strange to have a non-zero energy flux for constant fields, but note that fora constant vector ~S, the integral over any closed surface vanishes∮

~S · ~ndS ∝∮~ndS = 0,

so there is no net flux of energy into any closed region.

7 Wave solutions, light

We now turn to the problem of solving the time-dependent Maxwell equations. We will firstconsider the source-free equations, and relate the wave solutions to light. The source-freeequations are

~∇ · ~B = 0, ~∇ · ~E = 0, (7.1)

~∇∧ ~B =1

c

∂ ~E

∂t, ~∇∧ ~E = −1

c

∂ ~B

∂t. (7.2)

Through the second set of equations, changing ~E sources ~B and changing ~B sources ~E, so wecan have non-trivial solutions with no charge sources. The two fields can sustain each other.

7.1 Waves

We want to look for wave solutions of these equations - the idea is that light is an electromagneticwave. But what is a wave? A wave is a disturbance that propagates at some constant speed,maintaining its shape [picture]. In the simplest case, the disturbance is a function only of onedirection, and we can have left- or right-moving waves,

f(x+ vt), f(x− vt).

Mathematically, waves usually arise as solutions of the wave equation,

∂2f

∂x2− 1

v2

∂2f

∂t2= 0.

We can write the differential operator here as(∂

∂x− 1

v

∂

∂t

)(∂

∂x+

1

v

∂

∂t

)f = 0,

making it clear that arbitrary left- or right-moving waves are solutions of this equation. Infact the general solution is a linear combination of a left- and a right-moving wave, f =fL(x+ vt) + fR(x− vt). The three-dimensional generalization is

∇2f −− 1

v2

∂2f

∂t2= 0.

20

If the solution is a function of just one of the coordinates, this reduces to the previous case. Wecall this a plane wave, as the solution is uniform over planes perpendicular to the direction ofpropagation. This is a linear equation, and we can take the different plane wave solutions as abasis for the solutions of this equation; that is, any solution is a linear superposition of planewaves.

7.2 Electromagnetic waves

The source-free Maxwell equations are a coupled system of equations for ~E and ~B. We canobtain decoupled equations, which include some of the information in the Maxwell equations.We take the curl of the second set of equations, and use

~∇∧ (~∇∧ ~F ) = ~∇(~∇ · ~F )−∇2 ~F ,

and use the fact that ~∇ · ~F = 0 by the first equation for ~F = ~E, ~B, so

~∇∧ (~∇∧ ~E) = −∇2 ~E = −~∇∧ 1

c

∂ ~B

∂t= −1

c

∂

∂t~∇∧ ~B = −1

c

∂

∂t

1

c

∂ ~E

∂t= − 1

c2

∂2 ~E

∂t2,

and similarly

~∇∧ (~∇∧ ~B) = −∇2 ~B = ~∇∧ 1

c

∂ ~E

∂t=

1

c

∂

∂t~∇∧ ~E = −1

c

∂

∂t

1

c

∂ ~B

∂t= − 1

c2

∂2 ~B

∂t2.

So Maxwell’s equations imply that the components of ~E and ~B individually satisfy the waveequation,

∇2 ~E − 1

c2

∂2 ~E

∂t2= 0, ∇2 ~B − 1

c2

∂2 ~B

∂t2= 0. (7.3)

Note that this follows from Maxwell’s equations, but is not equivalent to it. We had 8 Maxwell’sequations, and there are 6 equations here.

Solutions of the source-free Maxwell’s equations are thus solutions of the wave equation,and we call such solutions electromagnetic waves. We see that the velocity of these wavesis c. In arbitrary units, the factor appearing here is

c =1

√ε0µ0

(recall in my units, 4πε0 = 1 and µ0 = 4π/c2.) Maxwell’s remarkable discovery was that thisspeed of electromagnetic waves – determined by constants appearing the physics of static fields– was equal to the experimentally measured speed of light. Thus, we guess that light is anelectromagnetic wave.

We have a special notation for the differential operator in this case,

� = ∇2 − 1

c2

∂2

∂t2,

so equations are � ~E = 0, � ~B = 0.Let us examine the electromagnetic waves in more detail. We consider plane waves, where

the fields depend on only one direction (in Cartesian coordinates), which we will take to bez. We will consider sinusoidal waves, of a given frequency ω. This is of obvious interest for

21

light, as it corresponds to monochromatic light, and by Fourier analysis any plane wave can bewritten as a superposition of sine waves of different frequencies. Then

~E = ~E0 cos(kz − ωt+ δ).

Here ~E0 is a constant vector, the polarization vector, δ is the phase, k is the wavenumberand ω is the frequency. This is a solution of the wave equation if ω = ±ck, so that this is afunction of z ± ct. Likewise for the ~B field,

~B = ~B0 cos(kz − ωt+ δ′).

Now let’s consider the further constraints on the solution from Maxwell’s equations. Gauss’law gives

0 = ~∇ · ~E = ~E0 · ~∇ cos(kz − ωt+ δ) = − ~E0 · ~ezk sin(kz − ωt+ δ),

so ~E0 has no z component; the polarization is transverse to the direction of propagation.Similarly for the ~B field,

0 = ~∇ · ~B = ~B0 · ~∇ cos(kz − ωt+ δ′) = − ~B0 · ~ezk sin(kz − ωt+ δ),

so ~B0 also has no z component, although this turns out not to be an independent conditionbecause of the coupling of ~E and ~B. Faraday’s law gives

~∇∧ ~E = −1

c

∂B

∂t.

first of all, this implies that if ~E only involves frequency ω and wavevector k, so does ~B (as wehave been assuming). Then

∂B

∂t= ω ~B0 sin(kz − ωt+ δ′),

while~∇∧ ~E = − ~E0 ∧ ~∇ cos(kz − ωt+ δ) = ~E0 ∧ ~ezk sin(kz − ωt+ δ),

so δ′ = δ, and

k ~E0 ∧ ~ez = −ωc~B0,

using ω = ±ck,~B0 = ∓ ~E0 ∧ ~ez,

so the electric and magnetic fields are orthogonal (and ~B0 is automatically transverse by

this condition). [Picture]. There is a right-hand rule, where ~E ∧ ~B points in the direction ofpropagation.

This is called a linearly polarized wave; the electric field vector at a point lies in somedirection (along a line), and oscillates up and down as a function of time. The magnetic fielddoes the same in the orthogonal direction.

By taking a superposition of linearly polarised waves, we can construct the circularlypolarised wave. Take

~E = E0(cos(kz − ωt+ δ)~ex + sin(kz − ωt+ δ)~ey),

which implies~B = ±E0(− sin(kz − ωt+ δ)~ex + cos(kz − ωt+ δ)~ey).

22

Then | ~E| = E0, so the polarisation vector has constant length, and rotates in the x − y planeas a function of time, with angular frequency ω.

The energy density in the linearly polarised wave is

u =1

8π(E2 +B2) =

1

4πE2

0 cos2(kz − ωt+ δ),

and the Poynting vector (energy flux density) is

~S =c

4π~E ∧ ~B = ∓ c

4π~E ∧ ( ~E ∧ ~ez) = ± c

4πE2~ez = ±cu~ez.

Poynting vector points in the direction of travel. So there’s an energy density u in the field,which is travelling along with the wave at speed c. The energy density fluctuates along thewave, but extremely rapidly; generally we would measure the averaged energy density and flux,

〈u〉 =1

4πE2

0〈cos2(kz − ωt+ δ)〉 =1

8πE2

0 , 〈~S〉 = ±c〈u〉~ez.

As with my trivial constant example, energy isn’t building up anywhere in the plane wavesolutions; it’s just flowing through.

For a circularly polarised wave,

u =1

8π(E2 +B2) =

1

4πE2

0 ,

~S =c

4π~E ∧ ~B = ±cu~ez.

This is two linearly polarised waves with a phase shift of π/2, so it’s basically just cos2 + sin2 =1.

8 Sources: Potentials and gauge transformations

The solutions of the source-free equations are electromagnetic waves. How are these related tosources? If we consider Maxwell’s equations with sources,

~∇ · ~B = 0, ~∇ · ~E = 4πρ, (8.1)

~∇∧ ~B =4π

c~j +

1

c

∂ ~E

∂t, ~∇∧ ~E = −1

c

∂ ~B

∂t, (8.2)

we can see that taking the curl of the curl equations, we will have some sources in our waveequations. But it turns out to be much simpler if we first introduce potentials.

For the static case, we had a scalar potential φ and a vector potential ~A:

~∇∧ ~E = 0 ⇒ ~E = −~∇φ,

~∇ · ~B = 0 ⇒ ~B = ~∇∧ ~A.

The first no longer works for time-dependent fields, but it’s still true that ~∇ · ~B = 0, so ingeneral we can write

~B = ~∇∧ ~A. (8.3)

23

Now the curl of ~E becomes

~∇∧ ~E = −1

c

∂

∂t~∇∧ ~A = −~∇∧ 1

c

∂A

∂t,

So

~∇∧(~E +

1

c

∂A

∂t

)= 0,⇒ ~E +

1

c

∂A

∂t= −~∇φ.

That is, I can write the electric field as

~E = −~∇φ− 1

c

∂A

∂t. (8.4)

Thus, there is again a representation of the electric and magnetic fields in terms of potentials,but the coupling of the fields has implied that I have potentials for ~E and ~B, not separatepotentials for ~E and ~B individually.

We can plug these forms for ~E and ~B into the equations with sources:

~∇ · ~E = ~∇ ·(−~∇φ− 1

c

∂A

∂t

)= 4πρ,

so

∇2φ+1

c

∂

∂t(~∇ · ~A) = −4πρ,

and

~∇∧ ~B = ~∇∧ (~∇∧ A) =4π

c~j +

1

c

∂ ~E

∂t.

so

~∇∧ (~∇∧ A) = ~∇(~∇ · ~A)−∇2 ~A =4π

c~j +

1

c

∂

∂t

(−~∇φ− 1

c

∂A

∂t

).

That is,

∇2 ~A− 1

c2

∂2A

∂t2− ~∇

(~∇ · ~A+

1

c

∂φ

∂t

)= −4π

c~j.

This looks a right mess, but we can simplify it by cunning choices.

8.1 Gauge transformations

Just as in the static case, ~A and φ are not uniquely determined by the defining equations

~B = ~∇∧ ~A, ~E = −~∇φ− 1

c

∂A

∂t.

The first is still invariant under~A→ ~A′ = ~A+ ~∇λ

for some arbitrary function λ. This could change ~E, but if I assume φ changes at the sametime,

φ→ φ′ = φ− 1

c

∂λ

∂t,

then

~E ′ = −~∇φ′ − 1

c

∂A′

∂t= −~∇φ+

1

c~∇∂λ∂t− 1

c

∂A

∂t− 1

c

∂

∂t~∇λ = −~∇φ− 1

c

∂A

∂t= ~E.

24

so the ambiguity in the definition of ~A, φ is the freedom to make gauge transformations

~A→ ~A+ ~∇λ φ→ φ− 1

c

∂λ

∂t(8.5)

for an arbitrary function λ. This includes the gauge invariance of the static case as a specialcase.

As in the static case, this freedom can be used to simplify the form of the equations writtenin terms of ~A, φ. There are two obvious choices:

Coulomb gauge: Choose λ such that

~∇ · ~A = 0. (8.6)

This is just what we did in the static case. We know such a choice of λ exists. This simplifiesthe φ equation: it’s just

∇2φ = −4πρ,

as in electrostatics. The ~A equation is not so good, though, as it depends on φ:

∇2 ~A− 1

c2

∂2A

∂t2− ~∇

(1

c

∂φ

∂t

)= −4π

c~j.

This is a particularly convenient choice for considering the vacuum solutions in terms of thepotentials. ρ = 0, so ∇2φ = 0, and the only solution which vanishes at infinity is φ = 0. Thenthe vector potential satisfies the wave equation,

∇2 ~A− 1

c2

∂2 ~A

∂t2= 0,

and the Coulomb gauge condition ~∇ · ~A = 0. If we consider a sinusoidal wave as before,

~A = ~A0 sin(kz − ωt+ δ),

then the Coulomb gauge condition says

0 = ~∇ · ~A = ~A0 · ~ezk cos(kz − ωt+ δ),

which tells us the wave must be transverse: ~A0 · ~ez = 0. Taking derivatives, we recover thesolution we had before for the electric and magnetic fields:

~E = −1

c

∂A

∂t=ω

c~A0 cos(kz − ωt+ δ) = ±k ~A0 cos(kz − ωt+ δ),

~B = ~∇∧ ~A = k~ez ∧ ~A0 cos(kz − ωt+ δ).

But to consider sources it’s better to make a different gauge choice.Lorentz gauge: Choose λ such that

~∇ · ~A+1

c

∂φ

∂t= 0. (8.7)

Is this always possible? Suppose it wasn’t true for ~A, φ. To make it true for ~A′, φ′, need

0 = ~∇ · ~A′ + 1

c

∂φ′

∂t= ~∇ · ~A+∇2λ+

1

c

∂φ

∂t− 1

c2

∂2λ

∂t2.

25

That is, the gauge transformation must satisfy

�λ = ∇2λ− 1

c2

∂2λ

∂t2= −~∇ · ~A− 1

c

∂φ

∂t.

This inhomogeneous wave equation has solutions for generic sources, as we’ll see more explicitlylater. So it is possible to put the potentials in Lorentz gauge. Note that if I’m in Lorentz gauge,doing a gauge transformation where λ satisfies �λ = 0 will leave me still in Lorentz gauge. Sothere is some residual freedom.

The value of this gauge is that it simplifies the ~A equation, which becomes an inhomogeneouswave equation

∇2 ~A− 1

c2

∂2A

∂t2= −4π

c~j. (8.8)

Furthemore, the φ equation becomes

∇2φ+1

c

∂

∂t

(−1

c

∂φ

∂t

)= ∇2φ− 1

c2

∂2φ

∂t2= −4πρ. (8.9)

so in Lorentz gauge, the potentials satisfy decoupled wave equations with sources. This is thenicest way to solve for the fields in presence of sources.

8.2 Retarded potential solutions

Having reduced the problem to wave equations with sources, we would now like to determinethe solutions of these equations. Let’s consider the equation for φ in detail; the equation for ~Aworks similarly.

In the special case with no time dependence, the wave equation reduces to Poisson’s equa-tion,

∇2φ = −4πρ,

and in the electrostatics section we learned that a solution which goes to zero at large r is

φ(~r) =

∫ρ(~r′)

|~r − ~r′|dV ′.

We want to see how this solution is modified when we include time dependence. We know thatin a region with no sources, the solution is some superposition of waves, so we would expectthat far from the sources, φ(~r, t) should be a function of r − ct, representing a wave travellingoutward from the source [Picture]. More precisely, if we consider the homogeneous equationand take a spherically symmetric ansatz φ(~r, t) = ψ(r, t)/r, then

�φ =1

r∂2r (rφ)− 1

c2∂2t φ =

1

r(∂2r −

1

c2∂2t )ψ = 0,

so ψ = ψ(r − ct) satisfies the equation.Physically, the idea is that “information” or “news” about changes in the sources moves at

the speed of light (the wave speed) so the field φ(~r, t) at a given point depends on the behaviourof the sources not at the same time, but at the retarded time

tr = t− |~r − ~r′|

c. (8.10)

[Picture] The solution depends on the behaviour of more distant sources at earlier times.

26

We propose that the solution is obtained by simply setting

φ(~r, t) =

∫ρ(~r′, tr)

|~r − ~r′|dV ′. (8.11)

The integral is over all space as before, but the solution at time t depends on the sources at theretarded time tr defined above. This solution can be derived using Green’s function methods,but we will just verify that it satisfies the wave equation. Consider the laplacian:

∇2φ =

∫∇2ρ(~r′, tr)

|~r − ~r′|dV ′

Previously, the derivative acted only on the denominator, but now the source ρ also dependson ~r through the retarded time, so

∇2ρ(~r′, tr)

|~r − ~r′|= ρ(~r′, tr)∇2 1

|~r − ~r′|+

1

|~r − ~r′|∇2ρ(~r′, tr) + 2~∇ 1

|~r − ~r′|· ~∇ρ(~r′, tr).

Now recall~∇ 1

|~r − ~r′|= − 1

|~r − ~r′|3(~r − ~r′), ~∇2 1

|~r − ~r′|= −4πδ3(~r − ~r′).

For the derivatives of ρ, we have

~∇ρ(~r′, tr) = ∂trρ(~r′, tr)~∇tr = −1

c∂tρ(~r′, tr)

1

|~r − ~r′|(~r − ~r′),

where in the second step we have noted that derivatives with respect to tr are equivalent toderivatives with respect to t, and and

∇2ρ(~r′, tr) = −1

c~∇·[∂tρ(~r′, tr)

1

|~r − ~r′|(~r−~r′)] = −1

c~∇∂tρ(~r′, tr)·

1

|~r − ~r′|(~r−~r′)−1

c∂tρ(~r′, tr)~∇·

1

|~r − ~r′|(~r−~r′)

where~∇ · 1

|~r − ~r′|(~r − ~r′) = − 1

|~r − ~r′|3(~r − ~r′) · (~r − ~r′) +

3

|~r − ~r′|=

2

|~r − ~r′|,

and

−1

c~∇∂tρ(~r′, tr) ·

1

|~r − ~r′|(~r − ~r′) =

1

c2∂2t ρ(~r′, tr)

1

|~r − ~r′|(~r − ~r′) · 1

|~r − ~r′|(~r − ~r′) =

1

c2∂2t ρ(~r′, tr).

Thus

∇2ρ(~r′, tr) =1

c2∂2t ρ(~r′, tr)−

1

c∂tρ(~r′, tr)

2

|~r − ~r′|,

so

∇2ρ(~r′, tr)

|~r − ~r′|= −4πδ3(~r − ~r′)ρ(~r′, tr) +

1

c2

∂2t ρ(~r′, tr)

|~r − ~r′|− 1

c∂tρ(~r′, tr)

2

|~r − ~r′|2

+21

|~r − ~r′|3(~r − ~r′) · 1

c∂tρ(~r′, tr)

1

|~r − ~r′|(~r − ~r′)

= −4πδ3(~r − ~r′)ρ(~r′, tr) +1

c2∂2t ρ(~r′, tr),

so finally

∇2φ =

∫∇2ρ(~r′, tr)

|~r − ~r′|dV ′ = −4πρ(~r, t) +

1

c2

∫∂2t

ρ(~r′, tr)

|~r − ~r′|dV ′ = −4πρ(~r, t) +

1

c2∂2t φ,

27

so this does indeed solve the wave equation

�φ = ∇2φ− 1

c2∂2t φ = −4πρ(~r, t).

Note that there is another solution, where the retarded time is replaced by the advanced time,

tadv = t+|~r − ~r′|c

.

but this solution is of little physical interest. It represents a wave propagating towards (ratherthan away from) the source. Its existence is required by the time-reversal invariance of theequations, which we break by choosing boundary conditions.

Similarly setting

~A(~r, t) =1

c

∫ ~j(~r′, tr)

|~r − ~r′|dV ′ (8.12)

will satisfy

� ~A = −4π

c~j.

These solutions in terms of the sources evaluated at retarded times occur whenever we need tosolve inhomogeneous wave equations.

Are we done? No, we derived these equations in Lorentz gauge, so we also need to checkthat our proposed form for φ, ~A satisfy the gauge condition

~∇ · ~A+1

c

∂φ

∂t= 0.

Now

c~∇ · ~A =

∫ ~∇ ·~j(~r′, tr)|~r − ~r′|

dV ′ +

∫~j(~r′, tr) · ~∇

1

|~r − ~r′|dV ′.

In the second term, we can trade the derivative with respect to r for a derivative with respectto r′, and integrating by parts,

c~∇ · ~A =

∫(~∇+ ~∇′) ·~j(~r′, tr)

|~r − ~r′|dV ′.

The derivatives are at fixed t, so they act on both the ~r′ dependence and the tr dependence in~j. But

(~∇+ ~∇′)tr = (~∇+ ~∇′)[t− |~r − ~r′|

c] = 0,

-shifting both the source and the field point doesn’t change tr. So this is equivalent to takingthe derivative along r′ at fixed tr,

c~∇ · ~A =

∫ ~∇′fixedtr ·~j(~r′, tr)

|~r − ~r′|dV ′

But by the continuity equation,

~∇ ·~j(~r, t) = −∂tρ(~r, t),

where the derivative is at constant values of the time argument, so

c~∇ · ~A = −∫∂tρ(~r′, tr)

|~r − ~r′|dV ′ = −∂φ

∂t

as required.

28

8.3 Dipole fields

As in the electrostatic case, when we are far from the sources we can expand the general solutionin a Taylor series. [Picture] Taking the origin in the source region,

|~r − ~r′| ≈ r

(1− ~r′ · ~r

r2+ . . .

).

We now need to insert this expansion both in the denominator and in tr, so

φ ≈∫ρ(~r′, t− r

c+~r′ · ~rcr

)

(1

r+~r′ · ~r′

2r3

)dV ′

That is, we have

φ ≈∫ (

ρ(~r′, t− r

c) +

~r′ · ~rcr

∂tρ(~r′, t− r

c)

)(1

r+~r′ · ~r2r3

)dV ′

There are two contributions here from the dipole order in the Taylor expansion: the familiarone from the expansion of the denominator, and a new one from the expansion of tr. Thefirst gives a field falling off like 1/r2, while the new one falls off like 1/r, like the leading term!Keeping only the 1/r terms,

φ ≈ 1

r

∫ρ(~r′, t− r

c)dV ′ +

~r

cr2·∫~r′∂tρ(~r′, t− r

c)dV ′ ≈ Q(t− r/c)

r+~r · ∂t~p(t− r/c)

cr2

The first term is the total charge of the source distribution (measured at time t − r/c). Thisis time-independent by charge conservation, and this contribution is the usual Coulomb lawcontribution. The second term involves the time-derivative of the dipole moment of the source.This is the leading contribution to the time-dependence of the field at large distances. Hencethe radiation (the time-dependent part of the field) is determined at leading order by the timederivative of the dipole moment.

For the vector potential, the relevant term is just the leading order:

~A ≈ 1

c

∫ ~j(~r′, t− rc)

rdV ≈

~jav(t− r/c)cr

,

this term is again related to the dipole moment. This is easy to see if we have a single charge:then ρ = qδ3(~r−~r0(t)), so ~p = q~r0, and ~j = q~vδ3(~r−~r0(t)) = q∂t~rδ

3(~r−~r0(t)), so ~jav = q∂t~r0 =

∂t~p. In general, we can see the relation by observing that ~A and φ satisfy the Lorentz gaugecondition:

~∇ · ~A+1

c

∂φ

∂t= 0.

The leading contribution in ~∇· ~A is when the derivative acts on jav rather than the denominator,so

~∇ · ~A ≈ − 1

c2

∂t~jav · ~∇rr

= − 1

c2

∂t~jav · ~rr2

,

while∂φ

∂t=~r · ∂2

t ~p

cr2,

so the continuity equation is satisfied if ~jav = ∂t~p in general. This argument only fixes ∂tjav,but we know it has no constant part as there is no monopole contribution to static magneticfields.

29

We can then calculate the leading contributions to the electric and magnetic fields. Again,the leading contributions come from letting the derivatives act on the t − r/c dependence ofthe sources.

~B = ~∇∧ ~A ≈ ∂t~jav ∧ ~∇rc2r

=∂2t ~p ∧ ~rc2r2

,

~E = −~∇φ− 1

c

∂ ~A

∂t≈ Q

r3~r +

(~r · ∂2t ~p)~r

c2r3− ∂2

t ~p

c2r,

so both electric and magnetic fields depend on the second time derivative of the dipole moment(apart from the constant Coulomb part in the electric field) . They are both orthogonal to ~r,the direction of propagation of the wave.

Setting Q = 0, the Poynting vector is then

~S =1

4π~E∧ ~B =

1

4πc2r2~E∧(∂2

t ~p∧~r) =1

4πc2r2(∂2t ~p( ~E·~r)−~r( ~E·∂2

t ~p)) =1

4πc2r3~r

(|∂2t ~p|2 −

(~r · ∂2t ~p)

2

r2

).

The energy flow is radially outwards away from the source, as expected. But there is a non-trivial antenna pattern: the flux is strongest where ~r is orthogonal to ∂2

t ~p, and vanishes inthe directions where ~r is parallel to ∂2

t ~p. [Picture] This is called dipole radiation.Do explicit dipole example if time permits.

9 Electrodynamics and special relativity

We have seen that light is an electromagnetic wave, moving with speed c. This speed appearsin Maxwell’s equations, so it is a fundamental feature of our physical laws. This was the basisof the first great revolution in twentieth century physics, special relativity.

The issue is that under the existing understanding of the relation between different observersmeasurements, two observers in relative motion will measure different speeds. [Picture] If wedefine coordinates t, x for my rest frame and t, x′ for your rest frame, moving at a relative speedv, then we have the Galilean transformation

x′ = x− vt

between my coordinates and your coordinates, which implies that if x = Wt in my coordinates,x′ = Wt− vt = (W − v)t in your coordinates. So if I see something moving with speed W , andyou move relative to me at some speed v, you will say it has speed W − v.

As a result, we won’t agree on the speed of light. In the nineteenth century, this wasn’tseen as a major problem. Light is an electromagnetic wave, and it was assumed that this wavewas an excitation of some medium, called the ether – like water waves travel through water,and sound waves travel through air. This ether presumably is at rest in some intertial frame,and Maxwell’s equations are valid in that frame.

But nobody managed to detect the ether! If it’s there, we ought to be able to feel it in someway. This led Einstein to propose the famous postulates of special relativity

• The laws of physics are the same in all inertial frames (moving with constant velocitywith respect to each other).

• The speed of light is the same for all inertail observers, independent of the motion of thesource (there is no ether).

30

9.1 Lorentz transformations

This implies that the Galilean transformation above is simply wrong! What is then the trans-formation between inertial frames?

• The transformation must be a linear relation between ~x, t and ~x′, t′, or constant (linear)motion in one frame would not correspond to constant motion in another frame

• We can choose the origin in spacetime such that ~x = 0, t = 0 implies ~x′ = 0, t′ = 0.

• We can choose the orientation of the coordinates such that the relative motion is alongthe x axis in the original coordinates, and argue that y′ = y, z′ = z.

• The line x = vt passes through the spacetime origin, and is the worldline of an object atrest in the new coordinates, so it should correspond to x′ = 0.

Thus, we should havex′ = γ(x− vt),

for some constant γ, possibly a function of v. Very little difference from the Galilean transfor-mation is possible! By reversing the direction of x, we interchange the role of the two coordinatesystems, so the transformation the other way must be the same,

x = γ(x′ + vt′).

Now the essential novelty is that we want x = ct to correspond to x′ = ct′. Substituting thisinto the above relations,

ct′ = γ(c− v)t, ct = γ(c+ v)t′,

soc2t′ = γ2(c− v)(c+ v)t′,

giving us

γ =1√

1− v2/c2.

We can then invert the second relation to solve for t′:

t′ = −x′

v+

x

vγ= −γx

v+ γt+

x

vγ= γ

(t− (1− γ−2)

x

v

)= γ(t− v

c2x).

This is the revolutionary change: events at t = 0 do not all lie at t′ = 0; relativity of simultaneity.Note that the key point is that constancy of the speed of light implies γ 6= 1. This gives us theLorentz transformation, also called a boost,

x′ = γ(x− vt), y′ = y, z′ = z, t′ = γ(t− v

c2x). (9.1)

The factors of γ lead to length contraction and time dilation, so the spatial distance andtime between events in different coordinate systems are not invariant. But the spacetimeinterval

∆s2 = −c2∆t2 + ∆~x2

is. (Note some people will use the opposite sign in defining this.) This is constructed so that∆s2 = 0 for events which are separated by travelling at the speed of light, so the relativityprinciple implies directly that ∆s2 vanishing in one coordinate system implies that it vanishes

31

in another coordinate system. One can show that given the linearity of the coordinate trans-formation, it will follow that ∆s2 is invariant for any two events. The Lorentz transformationscan then be alternatively defined as the transformations which leave this interval invariant.Explicitly,

∆s′2 = −c2∆t

′2 + ∆~x′2

= −c2γ2(∆t2 − 2v

c2∆t∆x+

v2

c4∆x2) + γ2(∆x2 − 2v∆x∆t+ v2∆t2) + ∆y2 + ∆z2

= −c2γ2(1− v2/c2)∆t2 + γ2(1− v2/c2)∆x2 + +∆y2 + ∆z2 = ∆s2.

9.2 Four-vectors

To write the Lorentz transformations in a compact notation, it is convenient to introducefour-vectors. The defining example is the coordinate position: we define the four-vector xµ,µ = 0, 1, 2, 3 by

x0 = ct, x1 = x, x2 = y, x3 = z.

Then the Lorentz transformation can be written as a matrix multiplication:x0′

x1′

x2′

x3′

=

γ −γ v

c0 0

−γ vc

γ 0 00 0 1 00 0 0 1

x0

x1

x2

x3

or more compactly xµ

′= Λµ

νxν , where as usual a sum on the repeated index is implied and Λµ

ν

are the components of the above matrix. The invariant spacetime interval is

∆s2 = xµxνηµν ,

where we introduce the metric

η =

−1 0 0 00 1 0 00 0 1 00 0 0 1

. (9.2)

The invariance of the interval is the statement that

ηµνx′µx

′ν = ηµνΛµρΛν

σxρxσ = ηρσx

ρxσ.

For this to be true for arbitrary xρ, we must have

ηµνΛµρΛν

σ = ηρσ. (9.3)

We can think of this equation as defining the Lorentz transformations: the Lorentz transfor-mations are the transformations which preserve the metric ηµν . These form a group, calledSO(3, 1). In addition to the boosts defined above, this includes spatial rotations, which aretransformations Ri

j in SO(3), just acting on the spatial coordinates xi, i = 1, 2, 3 by x′i = Ri

jxj

such thatδijR

ikR

jl = δkl.

The Lorentz transformations extend this spatial symmetry to include the time direction in asymmetric way.

32

We can now define a (contravariant) four-vector Aµ as any combination of a spatial vector~A with a number A0 which transforms in the same way as the coordinates under Lorentztransformations:

A′µ = Λµ

νAν .

The metric ηµν defines an invariant inner product on four-vectors, which generalizes the usualEuclidean inner product of the spatial vectors, so

A ·B ≡ ηµνAµBν

is invariant asηµνA

′µB′ν = ηµνΛ

µρΛν

σAρBσ = ηρσA

ρBσ.

A simple physical example of a four-vector is the four-momentum of a particle, pµ = (E/c, ~p).I won’t repeat the derivation of the transformation properties here, but just observe that in theregime where the Galilean transformation is approximately valid, E ≈ mc2, and ~p → ~p + mv,so it’s natural that these two quantities get combined into a four-vector.

The derivatives ∂t, ∂~x form a four-component object, but they don’t transform in the sameway as the coordinates: we define

∂µ =

(1

c

∂

∂t,∂

∂x,∂

∂y,∂

∂z

)Then

∂µ = Λνµ∂′ν ,

the transformation is inverted. These provide an example of a covariant four-vector Aµ. Ingeneral a covariant vector can be obtained from a contravariant vector by contraction with themetric:

Aµ = ηµνAν .

This satisfies the covariant transformation:

ΛνµA′ν = Λν

µηνσA′σ = Λν

µηνσΛσρA

ρ = ηµρAρ = Aµ.

If Aµ has components (A0, Ai), Aµ has components (−A0, Ai), so the difference between thecontravariant and the covariant vectors is just the sign of the time component.

The inverse metric ηµν , which has the same components, defines an invariant inner producton the covariant vectors, A ·B = ηµνAµBν .

For a four-vector, the contracted derivative

∂µAµ

is invariant under Lorentz transformations: we say it’s a scalar. This is analogous to the factthat ~∇ · ~A is a scalar under rotations. Our wave operator can be written as

� = − 1

c2

∂2

∂t2+∇2 = ηµν∂µ∂ν = ∂ν∂

ν ,

so it’s invariant under Lorentz transformations. This is analogous to ∇2 being a scalar operatorunder rotations.

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9.3 Electrodynamics in four-vector notation

We now want to determine how the objects in electromagnetism transform under Lorentztransformations, and write them in terms of the four-vector notation introduced above whichmakes the transformation properties explicit. We won’t derive the transformation propertiesin detail (it’s somewhat tedious, see Griffiths for the details) but simply identify objects thatcan naturally be combined into four-vectors.

A first example is the charge density and current. Under Galilean transformations,

~j′ = ~j − ρ~v,

so it’s natural for them to be combined into a four-vector under Lorentz transformations,

jµ = (cρ,~j). (9.4)

The first simplification we see is that

∂µjµ =

1

c

∂

∂t(cρ) + ~∇ ·~j =

∂ρ

∂t+ ~∇ ·~j,

so the continuity equation in four-vector notation is simply

∂µjµ = 0. (9.5)

The equations for the potentials in Lorentz gauge are

�φ = −4πρ, � ~A = −4π

c~j.

Given that cρ,~j form a four-vector, and that � is a scalar operator, we conclude that φ, ~A alsoform a four-vector:

Aµ = (φ, ~A). (9.6)

the dynamical equations are

�Aµ = −4π

cjµ,

and the Lorentz gauge condition is

∂µAµ =

1

c

∂φ

∂t+ ~∇ · ~A = 0.

The gauge transformation is

(φ, ~A)→ (φ, ~A) + (−1

c

∂

∂t, ~∇)λ,

which in four-vector notation isAµ → Aµ + ∂µλ

(recalling that changing the index position changes the sign of the time component).

What about the fields ~E, ~B? One might guess these should be the spatial parts of four-vectors, but no suitable scalars are available, and it’s not consistent with their relations to thepotentials.

~E = −~∇φ− 1

c

∂A

∂t,

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~B = ~∇∧ A.Inspired by the form of ~E, let’s define an object with two indices, a tensor

Fµν = ∂µAν − ∂νAµ.

This is antisymmetric, Fµν = −Fνµ, so it has 6 independent components, the off-diagonalelements. The antisymmetric structure is to make it gauge invariant, like the electric andmagnetic fields: Under Aµ → Aµ + ∂µλ,

Fµν → Fµν + ∂µ∂νλ− ∂ν∂µλ = Fµν .

The covariant vector Aµ has components (−φ, ~A), so

F0i = ∂0Ai − ∂iA0 =1

c

∂

∂tAi +

∂

∂xiφ = −Ei.

the purely spatial components are

Fij = ∂iAj − ∂jAi,

soBi = εijkFjk.

Thus,

Fµν =

0 −Ex −Ey −EzEx 0 Bz −By

Ey −Bz 0 Bx

Ez By −Bx 0

.

Also useful to have it with indices raised:

F µν =

0 Ex Ey Ez−Ex 0 Bz −By

−Ey −Bz 0 Bx

−Ez By −Bx 0

.

Since F is constructed by combining two objects that are Lorentz vectors, we know how ittransforms under Lorentz transformations:

Fµν = ΛρµΛσ

νF′ρσ.

We can read off the transformations of the electric and magnetic fields from this. Note thatthe transformation will mix the electric and magnetic fields.

Do an example: field of a wire.We can now write Maxwell’s equations in four-vector notation. Let’s derive the equations

from the potential form in four-vector notation, and then check they agree with Maxwell’sequations. There are two interesting derivatives. First, a suitably symmetric combinationvanishes by the definition of F :

∂λFµν + ∂µFνλ + ∂νFλµ = ∂λ∂µAν − ∂λ∂νAµ + ∂µ∂νAλ − ∂µ∂λAν + ∂ν∂λAµ − ∂ν∂µAλ = 0.

Thus, F satisfies the Bianchi identity

∂λFµν + ∂µFνλ + ∂νFλµ = 0. (9.7)

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Given the antisymmetry of Fµν , the LHS is actually completely antisymmetric in the threeindices λ, µ, ν. Second, there’s a contracted derivative:

∂µFµν = ∂µ∂

µAν − ∂µ∂νAµ.

If we are in Lorentz gauge, the second term vanishes, and

∂µFµν = ∂µ∂

µAν = −4π

cjν . (9.8)

Check these are the same as Maxwell’s equations:

∂µFµ0 = −~∇ · ~E = −4π

ccρ.

∂µFµi = ∂jF

ji + ∂0F0i = −(~∇∧ ~B)i +

1

c

∂Ei∂t

= −4π

cji.

For the ∂µFνρ equation, there is one index that does not occur: if it is 0, we have

0 = εijk∂iFjk = ~∇ · ~B;

if it is one of the spatial indices, we have

0 = εijk(∂0Fjk + ∂jF0k + ∂kFj0) =1

c

∂Bi

∂t+ (~∇∧ ~E)i.

Thus, these two four-vector equations have the same content as Maxwell’s equations. Thisrelativistic perspective makes the precise way time derivatives enter into Maxwell’s equationsinevitable: it’s just the relativistic extension of the spatial derivatives. The theory can beelegantly developed further in the language of differential forms - a subject for anothercourse.

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