Chapter 12

Solutions of Maxwell’s Equations

We have noticed that ~∇. ~B = 0 allows us to write

~B = ~∇× ~A (12.1)

and then ~∇× ~E = −1c∂ ~B∂t

says:

~∇× ( ~E +1

c

∂ ~A

∂t) = 0 (12.2)

or

~E = −~∇φ− 1

c

∂ ~A

∂t(12.3)

Thus knowing 4 quantities ~A and φ we generate the six components ~E and ~B. But actually

there are only 3 components of ~A and φ are needed to generate ~E and ~B. Thus consider the

transformation

~A′ = ~A+ ~∇Λ; Λ(~r, t) = arbitrary (12.4)

Then

~∇× A′ = ~∇× A+ ~∇× ~∇Λ = ~B (12.5)

88

So ~A′ generate the same ~B fields as ~A does. if we simultaneously change φ

φ′ = φ− 1

c

∂Λ

∂t(12.6)

Then

−~∇φ′ − 1

c

∂ ~A′

∂t= −~∇φ− 1

c

∂ ~A

∂t= ~E (12.7)

Hence the set of potentials ~A′, φ gives rise to the same ~E and ~B as the the set ~A, φ.

Thus all the physics is invariant if we make the gauge transformation on the potentials:

~A→ ~A′ = ~A+ ~∇Λ; φ→ φ′ = φ− 1

c

∂Λ

∂t(12.8)

and one says that the theory is gauge invariant.

Let ~V (~r, t) be any vector that vanishes sufficient rapidly at |~r| → ∞. Then one can

always decompose ~V into a transverse part ~V T and a longitudinal part ~V L

~V = ~V T + ~V L (12.9)

where ∇.~V T = 0 and ~∇× ~V L = 0 We can also write

~V L = ~∇V L (12.10)

~V T and ~V L are orthogonal under an integral∫d3r~V L.~V T =

∫d3r~∇V L.~V T =

∫d3r[~∇.(V L.~V T )−∇.~V TV L =

∫d~S.~V TV L (12.11)

Provided ~V T and ~V L fall off sufficiently rapidly The decomposition of ~V is uniquie, i.e., if

~A = ~B then ~AT = ~BT and ~AL = ~BL Returning now to the gauge transformation, we can

decompose the equation ~A′ = ~A + ~∇Λ into its T and L parts. Clearly ~∇Λ is longitudinal

since ∇× (∇Λ) = 0. so we can write

~A′L = ~AL + ~∇Λ; ~A′T = ~AT (12.12)

89

Now ~A′L, ~AL can be written as ‘

~AL = ~∇AL; ~A′L = ~∇A′L (12.13)

so

A′L = AL + Λ (12.14)

(aside from an irrelevant constant). Since Λ is arbitray, AL can be changed arbitrarily.

without changing the physics. Thus AL carries no physical content. On the other hand ~AT

is unchanged under gauge transformation (i.e., gauge invariant) and it must be carrying the

physical content. What about φ?

~∇× (~∇× ~A)− 1

c

∂

∂t(−~∇φ− 1

c

∂ ~A

∂t) =

1

c~j (12.15)

and so

−∇2 ~A+1

c2

∂2

∂t2~A+ ~∇(

1

c

∂φ

∂t+ ~∇. ~A) =

1

c~j (12.16)

and using ~∇. ~E = ρ we have

−∇2φ− 1

c

∂

∂t~∇. ~A = ρ (12.17)

One can divide the ∇2A equation into its T and L parts. Since the ~∇ part of that equation

does not contribute to the T part

−∇2 ~AT +1

c2

∂2

∂t2~AT =

1

c~jT (12.18)

This equation is gauge invariant The φ equation is a Poisson’s equation to determine φ and

one can solve for it using ~∇. ~A = ~∇. ~AL = ∇2AL

φ = −∇−2ρ− 1

c

∂

∂tAL (12.19)

90

Thus we see that φ contains a physical part, the coulomb integral

−∇−2ρ(r, t) =

∫d3r′

ρ(r′, t)

4π|r − r′|(12.20)

plus unphysical part

−1

c

∂AL

∂t(12.21)

which can be changed arbitrarily by changing the gauge.

∇−2 means the solution of Poisson’s equation which vanishes at ∞, i.e., if

∇2f(~r) = ρ(~r) (12.22)

then

f(~r) = ∇−2ρ (12.23)

f(~r) = −∫d3r′

ρ(r′)

4π|~r − ~r′|(12.24)

ρ vanishes rapidly enough t at r → ∞ so that f(~r) vanishes. The operator notation is

convenient since

∇2(∇−2) = 1 (12.25)

∇2f(~r) = ∇2(∇−2ρ) = −∫d3r′∇2 1

4π|~r − ~r′|ρ(~r′) = −

∫d3r′δ3(~r − ~r′)ρ(~r′) = ρ(~r) (12.26)

Now since we can change AL(r, t) by an arbitrary function Λ(r, t) without changing the

physics it means that we ra efree to set AL to anything we wish. The most convenient choice

is to choose Λ so that AL vanishes, i.e.,

A′L = 0 = Λ + AL ⇒ Λ = −AL (12.27)

Thus in the primed gauge we have

~A′ = ~A′T = ~AT (12.28)

91

Since ~AT is gauge invariant. This means that in the prime gauge φ is just the Coulomb

integral

φ′ = −∇−2ρ, −∇2A′T +1

c2

∂2A′T

∂t2=

1

c~jT (12.29)

which are the content of Maxwell’s equations The choice of this gauge (Eq.12.27) implies

that we are imposing the condition

~∇. ~A′ = 0, radiation gauge or Coulomb gauge (12.30)

This gauge brings the physics out of Maxwell’s equation most clearly as it sets the irrelevant

gauge variant part AL to zero. The A′ field obeys the wave equation

−2A′ =1

c~jT , 2 ≡ ∇2 − 1

c2

∂2

∂t2= D′Alembertian (12.31)

and φ′ is the Coulomb integral of electrostatics. The dynamics is all in the ~A′ field.

However there can be other gauge choices that are mathematically more convenient. For

example, suppose we choose (instead of the Coulomb gauge):

~∇. ~A′ + 1

c

∂φ′

∂t= 0, Lorentz gauge (12.32)

This is convenient since it simplifies

−2A′ = (−∇2 +1

c2

∂2

∂t2) ~A′ =

1

c~j (12.33)

and eliminating ∇.A′ we have

−2φ′ = (−∇2 +1

c2

∂2

∂t2)φ′ = ρ (12.34)

Here both A′ and φ′ obey the same equation (with different sources). In the coulomb gauge

A eqn is dynamical but in the Lorentz gauge both A and φ equations are dynamical. This

added dynamical motion for φ makes the physics unclear sometimes.

92

Can we find a Λ to have A′L 6= 0 after satisfiying Lorentz gauge condition. The require-

ment is does there exist a Λ such that i.e., can we can we find a Λ which satisfies

0 = ~∇. ~A′ + 1

c

∂φ′

∂t= ~∇. ~A+

1

c

∂φ

∂t+∇2Λ− 1

c2

∂2Λ

∂t2(12.35)

, i.e., can we find a Λ such that

2Λ = −(~∇. ~A+1

c

∂φ

∂t) (12.36)

Thus suppose we are in Coulomb gauge (∇.A = 0, φ = −∇−2ρ), Then Λ that transformst

to the Lorentz gauge would be the solution of

2Λ =1

c∇−2(

∂ρ

∂t) (12.37)

Actually there are a infinite number of well behaved solution of the above equation which

means that the Lorentz gauge can be set up. and in fact it is not unique, but there are an

infinte number of Lorentz gauges.

12.0.2 Solution of Maxwell eqns. in Free Space

Let us consider the case without any charge and current density.

~E = −~∇φ− 1

c

∂ ~A

∂t; ~B = ~∇× ~A (12.38)

which leads to

−∇2 ~A+1

c2

∂2

∂t2~A = 0, ~∇. ~A = 0;φ = 0 (12.39)

and

~E = −1

c

∂ ~A

∂t; ~B = ~∇× ~A (12.40)

93

A convenient way to solve:

~A = a~εei(~k.~r−ωt) (12.41)

k is the wavenumber, ~ε; polarization vector, a: amplitude of wave. This corresponds to a

plane wave propagating in the direction ~k and all points ⊥ ~k have constant values at a fixed

plane ~k.~r = 0, ~A = a~εei(−ωt). So the wave is constant on this plane. We need

∂i ~A = ikia~εei(~k.~r−ωt) (12.42)

and take the second derivative

∇2 ~A = (i~k).(i~k) ~A = −k2 ~A (12.43)

Similarly

∂2

∂t2~A = −ω2 ~A (12.44)

(k2 − 1

c2ω2) ~A = 0 (12.45)

ω = ±|~k|c2, ωk = |~k|c = kc (12.46)

and there are two solutions for any ~k|, ω = ±ωk; e−iωkt, eiωkt We also need to implement the

gauge condition. Consider the positive frequency wave

0 = ~∇. ~A = ia~k.~εei(~k.~r−ωt) ⇒ ~k.~ε = 0, ~k ⊥ ~ε (12.47)

The polarization vector is transverse to the propagation vector. Now in general two

vector ⊥ any vector so can have two independent ~ε for each ~k vector. Here we label ~ελk ,

λ = 1, 2, ~ελk .~k = 0. There are a number of different possible choices for the ~ελk , but before

that let us look at ~E and ~B:

~E = −1

c

∂ ~A

∂t= i

ωkca~ελke

i(~k.~r−ωkt) = ika~ελkei(~k.~r−ωkt) (12.48)

94

E

B

k

Figure 12.1:

where k = |~k| So ~E is polarized in the same way as ~A is i.e., ⊥ to ~k. For ~B, we have

~B = ia(~k × ~ελk)ei(~k.~r−ωkt) (12.49)

Hence ~B is ⊥ to ~k and also to ~ελk . Note

( ~E × ~B) = ~ε× ~k × ~ε = ~ε.~ε~k − ~ε.~k~ε = |~ε|2~k (12.50)

The ( ~E × ~B) points in the directtion of ~k, i.e., ~E, ~B and ~k form a righthanded triad of

orthogonal vectors. We also have another constraint, ~A is a real field and so the physical

part of our solution is

~A = Re[aλk~ε

λkei(~k.~r−ωt)

]=

1

2

[aλk~ε

λkei(~k.~r−ωt) + a∗λk ~ε

∗λk e−i(~k.~r−ωt)

](12.51)

(since our eqn. for ~A is homogenous, the real part is also a solution for wave equation

separately) Similarly, the physical electric field and magnetic fields are

~E = Re[ikaλk~ε

λkei(~k.~r−ωt)

](12.52)

95

~B = Re[iaλk(

~k × ~ελk)ei(~k.~r−ωt)

](12.53)

Now the ~ελk can be chosen as any two vectors that form a basis ⊥ to ~k. A convenient choice

is

(i) ~ελk is real and orthogonal

~ελk .~ελ′

k = δλλ′ (12.54)

The general solution for a fixed ~k is then the sum of the two independent polarizations

~A = Re

[2∑

λ=1

aλk~ελkei(~k.~r−ωt)

](12.55)

and hence

~E = Re

[ik

2∑λ=1

aλk~ελkei(~k.~r−ωt)

]=

2∑λ=1

k~ελkIm[aλke

i(~k.~r−ωt)]

(12.56)

~B = Re

[i

2∑λ=1

aλk(~k × ~ελk)ei(

~k.~r−ωt)

]=

2∑λ=1

(~k × ~ελk)Im[aλke

i(~k.~r−ωt)]

(12.57)

, we get | ~E| = | ~B| A priori the amplitudes aλk are complex numbers, i.e., can have arbitrary

magnitues and phases. different possibilities correpond to different types of waves:

Linearly Polarized Wave:

aλk = |aλk |eiφk ;φk is same for eachλ (12.58)

Then

~E =2∑

λ=1

Re[ke

iπ2 aλk~ε

λkeiφkei(

~k.~r−ωt)]

(12.59)

or

~E =2∑

λ=1

k|aλk |~ελkcos(~k.~r − ωt+ φk +π

2) (12.60)

we have then The electric field oscillates in time along the line making angle θ with ~ε1k axis.

This is called linearly polarized wave.

96

θ

E2

E1

2kε

1kε

E

Figure 12.2:

Circularly Polarized Wave:

A second possibility could be wave has aλk with phase difference ±π2

for different λ

a1k = |a1

k|eiφk ; a2k = |a2

k|eiφk±iπ2 (12.61)

Then ~E has the form

~E = k|aλk |~ε1kcos(~k.~r − ωt+ φk +π

2) + k|aλk |~ε2kcos(~k.~r − ωt+ φk +

π

2± π

2) (12.62)

= Re k[|a1k|~ε1k ± i|a2

k|~ε2k]ei(

~k.~r−ωt+φk+π2

)

This is called elliptically polarized radiation A special case occurs when |a1k| = |a2

k| and then

~E = Re k|a1k|[~ε1k ± i~ε2k

]ei(

~k.~r−ωt+φk+π2

) (12.63)

The choice ~ε1k + i~ε2k: left circularly polarized wave or “positive helicity”

~ε1k− i~ε2k: right circularly polarized wave or “negative helicity”. One can see what this means

by calculating the components of ~E. E2 has an extra phase of ±π2

from the ±i factor, and

since cos(α± π2) = ∓sinα, we can write

E1 = E0cos(~k.~r − ωt+ φk +π

2) (12.64)

97

E2

E1

2kε

1kε

E

k.r-ωt+π/2+φk

Figure 12.3:

E2 = ∓E0sin(~k.~r − ωt+ φk +π

2) (12.65)

where E0 = k|a1k|.

For the left circularly polarized wave we see fig.12.3 As time goes on the phase decreases

and ~E moves counter-clockwise on a circle of radius E0. For the right circularly polarized

wave ~E will move clockwise. If |a1k| 6= |a2

k|, ~E traces out corresponding ellipses.

(ii) ~ελk complex While the choice real orthogonal basis is the simplest possibility, the

vectors that enter into the circulary polarized waves are complex and thus turn out to

be specially important in quantum electrodynamics. There the (postive/negative) helicites

correspond to photon whose spin is parallel(anti-parallel) to the momentum. The linearly

polarized states are not eigenstates of helicity (component of spin parallel to the momentum:

~S.p), and hence are less convenient.

To see the properties of the helicity basis we chose two complex vectors

~ελk ⊥ ~k, λ = ±1

+1

−1

positive helicity

negative helicity

(12.66)

98

normalized to

~ε∗λk .~ελ′

k = δλλ′ , λ, λ′ = ±1 (12.67)

One can introduce a third vector parallel to ~k

~ε0k × ~k = 0; λ = 0 (12.68)

To get the properties of the vector choose a special frame where ~k is along the positive z-axis.

Then our 3 vectors are chosen as (~k = (0, 0, |k|))

~ε0k = (0, 0, 1), ~ε±k =1√2

(1,±i, 0) (12.69)

± corresponds to left-right circularly polarization from these one can easily check

~ελ∗k .~ελk = δλλ′ ; ~ε

λ∗k = ~ε−λk ; λ = 0,±1, ~ε±k × k = ±i~ε±1

k (12.70)

Returning now back to the wave equation for ~A, ~B and ~E, we saw our plane wave solution

held for any value of ~k. Thus we can ge the general solution by linear superposition, i.e.,

by suumming over all ~k. This is just a fourier integral or fourier series representation of the

general solution. Before doing ths we note that these solutions are really unphysical as they

posess infinite energy. For example,

~E = k|aλk |~ελkcos(~k.~r − ωt+ φk +π

2) (12.71)

and so

E2 = k2|aλk |2cos2(~k.~r − ωt+ φk +π

2) (12.72)

Thus the electric energy is

∫d3r

1

2E2 =

∫d3r

1

2k2|aλk |2cos2(~k.~r − ωt+ φk +

π

2) =∞ (12.73)

99

That is the plane wave cannot extend out to infinity in all directions, but must fall off so that

the energy is finite. A physically acceptable solution would therefore be of the form of a wave

packet, rather than an infinite plane wave. One way to deal with this in an approximate

fashion is to imagine space to be a large box of volume

V = L3; −L/2 ≤ x ≤ L/2; −L/2 ≤ y ≤ L/2; −L/2 ≤ z ≤ L/2; (12.74)

which we will eventually let go to ∞. We then impose periodic boundary conditions on the

surface of the box,e.g., for surface ⊥ to x-axis we have

eik1L2 = eik1

−L2 (12.75)

or

eik1L = 1⇒ k1L = 2πn1, (12.76)

hence

k1 =2πn1

L, n1 = 0,±1,±2, ... (12.77)

In general the wave number is discrete, i.e.,

~kn =2π

L~n, ~n = (n1, n2, n3); ni = 0,±1,±2, ... (12.78)

As L gets large, the discreteness becomes a continuous, e.g., the difference between the

adjacent ~k values are

∆k =2π

L−−−→L→∞

0 (12.79)

with these b.c. our plane wave becomes orthogonal. Thus just consider just an x-wave

I =

∫ L/2

−L/2dxφ∗1φ2 =

1

L

∫ −L/2L/2

dxei(k′x−kx)x =

1

L(ei∆k

L2 − e−i∆kL2i∆k

) (12.80)

100

where ∆k = k′x − kx is an integer multiple 2π/L:

∆k = (n′1 − n1)2π

L= ∆n

2π

L(12.81)

Hence

I =e−i∆k

L2

iL∆k(e2πi∆n − 1) = 0, for∆n 6= 0 (12.82)

For ∆n = 0

I =

∫ L/2

−L/2

1

Ldx = 1 (12.83)

Thus

I = δkx,k′x (12.84)

In general one has:

~φλk =~ελk√Vei(k.r−ωkt) (12.85)

Then∫d3x~φλ∗k .

~φλ′

k′ =1

V~ελ∗k .~ε

λ′

k′

∫d3xei(k

′−k).re−i(ω′k−ωk)t = ~ελ∗k .~ε

λ′

k′δkke−i(ω′k−ωk)t (12.86)

or ∫d3x~φλ∗k .

~φλ′

k′ = δk′kδλ′λ (12.87)

We now take the general solution for ~A as a sum over all possible ~k and all λ

~A =∑k,λ

√c2

2ωkV(aλkε

λkei(k.r−ωkt) + aλ∗k ε

λ∗k e−i(k.r−ωkt)) (12.88)

(The front factor√c2/2ωk is for later convenience and we have taken the Re by adding in

the c.c.) The aλk are arbitrary amplitudes of each wave since ~k = 2π~nL

, the sum is overall

discrete integer in ~n We can write ~A as

~A =∑k,λ

√c2

2ωk(aλk

~φλk + aλ∗k~φλ∗k ) (12.89)

101

and the electric field is

~E =i

c

∑k,λ

√c2ωk

2(aλk

~φλk − aλ∗k ~φλ∗k ) (12.90)

The electric energy is

EE =1

2

∫d3r ~E. ~E = −1

2

∫1

c

∑k,λ

√c2ωk

2(aλk

~φλk − aλ∗k ~φλ∗k )1

c

∑k′,λ′

√c2ωk′

2(aλ

′

k′~φλ′

k′ − aλ′∗k′~φλ′∗k′ )

(12.91)

The a∗a cross-sterms are just right to give normalization integral and there are in addition

aa and a∗a∗ terms

EE =1

2c2

∑√c2ωk

2

c2ωk′

2((aλka

λ′∗k′ δkk′δλλ′ +aλ∗k a

λ′

k′δkk′δλλ′)− (aλkaλ′

k′

∫~φλk .

~φλ′

k′ + c.c.)) (12.92)

For the magnetic energy we have:

~B = i∑k,λ

√c2

2ωk(aλk

~k × ~φλk − aλ∗k ~k × ~φλ∗k ) (12.93)

and so

EB =1

2

∫d3r ~B. ~B = −1

2

∫1

c

∑k,λ

√c2

2ωk(aλk

~k×~φλk−aλ∗k ~k×~φλ∗k )1

c

∑k′,λ′

√c2

2ωk′(aλ

′

k′~k′×~φλ′k′−aλ

′∗k′~k′×~φλ′∗k′ )

(12.94)

In the cross-terms we get a factor of

(~k × ~ε∗λk ).(~k′ × ~ελ′k′)δkk′ = (~k × ~ελk).(~k × ~ε∗λ′

k )δkk′ = k2δkk′δλλ′ =ω2k

c2δkk′δλλ′ (12.95)

Thus

EB =1

2

∑√c2

2ωk

c2

2ωk′((aλka

λ′∗k′ δkk′δλλ′ + aλ∗k a

λ′

k′δkk′δλλ′)ω2k

c2− (aλka

λ′

k′

∫~k × ~φλk .~k′ × ~φλ

′

k′ + c.c.))

(12.96)

102

The first terms of EB and EE are identitical and just double. The last terms involve the

integral

1

V

∫eik.reik

′.r =1

V

∫ei(k+k′).r = δk,−k′ (12.97)

In serting ~k′ = ~k into the last terms cancel when we combine EE and EB, if we use ω2k = k2c2,

we have then for the total energy

EE+B =∑k,λ

ωkaλ∗k a

λk (12.98)

Thus each mode contributes a contribution to energy proportional to the frequency ωk and

square of amplitude |aλk |2. In the quantum theory aλk and aλ∗k → aλ†k become operator obeying

[a†λk , a

λ′

k′

]= /hδkk′δλλ′ (12.99)

[aλk , a

λ′

k′

]= 0 =

[aλ†k , a

λ′†k′

](12.100)

Thus aλ†k aλk has just integer quantum numbers

aλ†k aλk → nλk/h; nλk = 0, 1, 2, ... (12.101)

and the energy eigenvalues are

En =∑k,λ

/hωknλk (12.102)

where nλk=nos. of photons of energy /hωk, helicity λ = ±1 when our amplitude is normalized.

Then |akλ|2//h is the classical analogue of the number of photons.

We can take now the limit as our box grows to all space. We find

∆kx =2π

L−−−→L→∞

0 (12.103)

In the limit of L→∞, δkx becomes infintesimal,

∆kx∆ky∆kz =(2π)3

L3=

(2π)3

V→ 0 (12.104)

103

and

1

V=

d3k

(2π)3(12.105)

and the sum becomes integral

1

V

∑~kn

=

∫d3k

(2π)3(12.106)

Defining

aλk =1√Vaλ(~k) (12.107)

then using

~A =1

V

∑k,λ

√c2

2ωk(aλελke

i(k.r−ωt) + c.c.) (12.108)

and so as V →∞

~A =∑λ

∫d3k

(2π)3

√c2

2ωk(aλελke

i(k.r−ωt) + c.c.) (12.109)

The total energy of the field

EE+B =∑k,λ

ωk1

V|aλ|2 −−−→

V→∞

∑λ

∫d3k

(2π)3ωk|aλ|2 (12.110)

We see therefore if aλ(k) falls off sufficient rapidly as |~k| → ∞, the integral will converge

and we will have a situation of finite energy.

12.0.3 Solution of Maxwell eqns. with Sources

Let us generalize the solution of Maxwell eqns in free space. We had the general form of

Maxwell equations in the arbitrary gauge to be for the Vector and scalar potentials

−∇2 ~A+1

c2

∂2

∂t2~A+ ~∇(

1

c

∂φ

∂t+ ~∇. ~A) =

1

c~j (12.111)

−∇2φ− 1

c

∂

∂t~∇. ~A = ρ (12.112)

104

We are free to choose any gauge and this time we are going use the Lorentz gauge

(1

c

∂φ

∂t+ ~∇. ~A) = 0 (12.113)

Our question reduces to

−∇2 ~A+1

c2

∂2

∂t2~A =

1

c~j (12.114)

−∇2φ+1

c2

∂2

∂t2φ = ρ (12.115)

We can use the above two equations consistent with our gauge condition if we consider our

gauge condition, since if we consider (∇.(12.114) + 1c∂∂t

(12.115)) we get

(−∇2 +1

c2

∂2

∂t2)(

1

c

∂φ

∂t+ ~∇. ~A) =

1

c(∇.~j +

∂ρ

∂t) = 0 (12.116)

by equation of continuity (conservation of charges). Hence the choice of Eq.(12.113 is con-

sistent with Eq. (12.114) and Eq.(12.115). Both Eq.(12.114) and Eq.(12.115) are same type

of equations. Recall that the general solution of an inhomogenous differential equation is a

particular integral (P.I)+ general solution of the homogenous equation, e.g.,

~A = P.I.+ Homogenous eqn. soln. (12.117)

We have already noticed the general homogenous equation solution (as an arbitrary super-

sposition of plane waves) and so we need a particular integral. We do this by introducing

Green’s function. Let

(−∇2 +1

c2

∂2

∂t2)G(~r − ~r′, t− t′) = δ3(r − r′)δ(t− t′) (12.118)

Then we take

~AP.I =

∫G(~r − ~r′, t− t′)d3r′dt′~j(r′, t′) (12.119)

105

Because

−2 ~AP.I =

∫(−2G(~r−~r′, t−t′))d3r′dt′~j(r′, t′) =

∫δ3(r−r′)δ(t−t′)d3r′dt′~j(r′, t′) =

1

c~j(r, t)

(12.120)

and so Eq.(12.119) is a solution of Eq. (12.114) A convenient way solving for G is to fourier

analyze in time

G(~r − ~r′, t− t′) =1

2π

∫dωG(~r − ~r′, ω)e−iω(t−t′) (12.121)

and also we can write

δ(t− t′) =1

2π

∫dωe−iω(t−t′) (12.122)

Inserting into Eq.(12.118) gives

1

2π

∫dω(−∇2 − ω2

c2)G(r − r′, ω)e−iω(t−t′) =

1

2π

∫dωδ3e−iω(t−t′) (12.123)

and equating fourier coefficients

(−∇2 − ω2

c2)G(r − r′, ω) = δ3(r − r′) (12.124)

since δ3 depends only on

~R = ~r − ~r′ (12.125)

we can look for a solution which only depends on ~R Also δ3(R) is spherically symmetric in

~R so we can look for a solution that depends only on

R = |~R| (12.126)

Hence Eq.(12.124) reads

(1

R2

d

dR(R2 d

dR) +

ω2

c2)G(R,ω) = −δ3(R) (12.127)

106

For R 6= 0 the eqn. reads

(d2

dR2+

2

R

d

dR+ω2

c2)G(R,ω) = 0 (12.128)

we can get rid of the first derivative by writing

G =1

RG (12.129)

and one has

(d2

dR2+ω2

c2)G(R,ω) = 0; R 6= 0 (12.130)

or

G = Aeiω

Rc

R+B

e−iωRc

R(12.131)

we can determine the effect of the point charge at R = 0 by integrating over a small plane

round the origin

−∫∇2Gd3R− ω2

c2

∫Gd3R =

∫d3Rδ3(R) = 1 (12.132)

and since ∇2 = ∇.∇ one gets

−∫dS.∇G− ω2

c2

∫R2dRG(R,ω)dΩ = 1 (12.133)

we see that since from Eq. (12.131)

G ∼ 1

R(for R→ 0);

dG

dR∼ 1

R2(12.134)

The first term gives a finite contribution while the second term gives zero in the limit ε→ 0

Hence

−4π(R2dG

dR)R→0 = 1 (12.135)

Inserting E.(12.131) gives then

A+B =1

4π(12.136)

107

we have therefore two independent solutions

G±(R,ω) =e±iω

Rc

4πR(12.137)

and we get our G-function by insreting back into Eq.(12.121):

G±(r − r′, t− t′) =1

2π

∫ ∞−∞

dωe−iω(∓R

c+t−t′)

4πR(12.138)

and hence

G±(r − r′, t− t′) =δ(t− t′ ∓ R

c)

4πR(12.139)

one says

G+ = “retarted′′G− function ≡ Gret; G− = “advanced′′G− function ≡ Gadv; (12.140)

we can now insert our G functions into Eq.(12.119). Let us consider the retarded solution

~A = ~Arethom +

∫Gret(r − r′, t− t′)d3r′dt′

1

c~j(r′, t′) = ~Arethom +

∫δ(t− t′ − R

c)

4π|~r − ~r′|d3r′dt′

1

c~j(r′, t′)

(12.141)

or

~A = ~Arethom +

∫1

4π|~r − ~r′|d3r′

~j(r′, t− |~r−~r′|

c)

c(12.142)

φ is the same

φ = φrethom +

∫1

4π|~r − ~r′|d3r′ρ(r′, t− |~r − ~r

′|c

) (12.143)

The retarted time is defined tret is defined as:

tret = t− |~r − ~r′|

c≤ t (12.144)

what Eq.(12.142) or Eq.(12.143)) is saying for the P.I. part is the following: a source at

pt.~r′ emits radiation at time earlier t′ = tret. This radiation then, traveling with velocity c

108

arrives at point ~r at later time t. The total ~AP.I. is the sum(Integral) of all the radiation.

that can get to the point ~r at time t if emitted earlier and travel with velocity c. The factor

14π|r−r′| is the usual Coulomb or magnetostatic relating the source to the emitted potential.

~Ahom or φhom are determined by some determined by some boundary conditions. Suppose

we go backward in time t to earlier and earlier. The radiation seen at point ~r in API comes

from ~J at earlier times t′ = tret i.e., from the past of the observer at point ~r. As we go

towards t→ −∞. the beginning of the universe, this came from the earliest posible source.

One involves now “Sommerfield boundary condition”. All radiation observed at time t come

from sources radiating at an earlier time

~A(r, t), φ(r, t)→ 0 fort→ −∞, if currents started radiating at some finite timein the past

(12.145)

Since currents started radiating at some time in the past. With this boundary condition we

get

~Arethom = 0 (12.146)

Since there is no source with a homogenous soln. and AretPI already includes the radiation

from sources. Thus our solution obeying the Sommerfeld boundary condition is

~A =

∫1

4π|~r − ~r′|d3r′

~j(r′, tret)

c, φ =

∫1

4π|~r − ~r′|d3r′ρ(r′, tret) (12.147)

However, we could just as well used the advanced G-function. Then we would have gotten

~A = ~Aadvhom +

∫1

cGadv(r − r′, t− t′)d3r′dt′~j(r′, t′) (12.148)

and putting in Gadv = G−

~A = ~Aadvhom +

∫1

4π|~r − ~r′|d3r′

~j(r′, t+ |~r−~r′|c

)

c(12.149)

109

The particular integral now says that the radiation at time t comes from sources at later

times t′ = tadv = t + |~r−~r′|c

traveling backwards in time with velocity c. This would oppose

or violate causality with the future affecting the past. However this is not the case because

of the existence of aadvhom. Thus we can write

~A = ~Aadvhom +

∫1

cGret

~j +

∫1

c(Gadv −Gret)~j (12.150)

Now the last term is actually a homogenous solution i.e.,

−2 [Gadv(r − r′, t− t′)−Gret(r − r′, t− t′)]~j =

∫ [δ3(r − r′)δ(t− t′)− δ3(r − r′)δ(t− t′)

] 1

c~j = 0

(12.151)

Thus writtten in this form we see

Arethom = Aadvhom +1

c

∫(Gadv −Gret)~j (12.152)

If we now invoke the Sommerfeld b.c. we deremine Aadvhom:

~Aadvhom = −∫

1

c(Gadv −Gret)~j (12.153)

Thus we are free to use the advanced G-function in our solution but then the homogenous

solution is not zero to compensate The Sommerfeld bounday condition is not a consequence

of Maxwell’s theory, but verges on a statement about cosmology, i.e., what happened at the

beginning of the universe. We are saying tha the Universe started with no free radiation and

all the radiation we see was emitted subsequently by charges. In fact we do see radiation

incident on the Earth from the distant past, the cosmic microwave background (CMB)

radiation. But even here CMB arose from a hot plasma of charges at the time of the Big

Bang and again supporting the Sommerfeld consition.

110

As the final point we can insert into Eq.(12.147) our expression for ρ and ~j for point

charges:

φ =

∫d3r′

1

4π|r − r′|∑a

eaδ3(~r′ − ~ra(t)); t = t− |~r − ~ra(t)|

c(12.154)

We can use the δ3 to do the integral over d3r′. However the δ function is a complicated

function of ~r′ Recall now the theorem about δ-functions:

∫d3r′g(r′)δ3(~f(r′)) =

∫d3fg(r′)J(

∂~r′

∂ ~f)δ3(~f(r′)) =

∫d3fg(r′)

δ3(~f(r′))

J( ∂~f

∂~r′)

=g(r′)

J( ∂~f

∂~r′)

]f=0

(12.155)

For our case

~f = ~r′ − ~ra(t−|~r − ~r′|c

), g =1

4π|~r − ~r′|(12.156)

The Jacobian is

J =

∂fx∂x′

∂fx∂y′

∂fx∂z′

∂fy∂x′

∂fy∂y′

∂fy∂z′

∂fz∂x′

∂fz∂y′

∂fz∂z′

=

1− vxa

cx−x′r−r′

vxacy−y′r−r′

vxacz−z′r−r′

vyacx−x′r−r′ 1− vya

cy−y′r−r′

vyacz−z′r−r′

vzacx−x′r−r′

vzacy−y′r−r′ 1− vza

cz−z′r−r′

,∂fx∂x′

= 1−1

c

dxadt

x− x′

|~r − ~r′|

(12.157)

Now J is a scalar function of vector ~va and ~r − ~r′. We can evaluate it in any frame Let

us choose that ~va lies along x axis

~va = (vxa , 0, 0) (12.158)

Then

J =

1− vxa

cx−x′r−r′

vxacy−y′r−r′

vxacz−z′r−r′

0 1 0

0 0 1

= 1− vxac

x− x′

|~r − ~r′|(12.159)

111

and in that frame

J = 1− ~va.(~r − ~r′)c|~r − ~r′|

(12.160)

Since J is a scalar, this must be its value in any frame

We have then

φ(~r, t) =1

4π

∑a

ea1

|~r − ~r′| − va(t)c.(~r − ~r′)

(12.161)

From Eq. (12.156) f = 0 means ~r′ = ~ra(t(r′)) where t(r′) = t− |~r−~ra(t)|

cHence t(r′) = tret.

Then f = 0 means

~r′ = ~ra(tret), tret = t− |~r − ~ra(tret)|c

(12.162)

Hence

φ(~r, t) =1

4π

∑a

ea1

|~r − ~ra(tret)| − ~va(tret)c

.(~r − ~ra(tret))(12.163)

and similarly

~A(~r, t) =1

4π

∑a

eac~va(tret)

1

|~r − ~ra(tret)| − ~va(tret)c

.(~r − ~ra(tret))(12.164)

These are called Lienard-Wiechert Potentials.

Note that Eq.(12.162) is an implicit for tret. Thus ra(t) is the trajectory of the particle

and is some function of time and one is to solve Eq.(12.162) for tret. For example in a straight

line trajectory

~ra(t) = ~r0 + ~v0t (12.165)

Then Eq. (12.162) becomes

tret = t− ~r − ~r0 − ~v0tretc

≤ t (12.166)

which one can solve to get

tr = tr(~r, t) (12.167)

112

and this is to be inserted back into Eq. (12.163) and Eq. (12.164) This complexity just

gurantees that the signal from the source point trvales with velocity c to get to the field

point.

12.0.4 Radiation of Energy

Let us consider a current and charge density that is confined to be close to the origin. For

the electric and magnetic static situation, one knows that as one gets a distance r distant

from the origin that the fields drop off rapidly, i.e.,

~E ∼ 1

r2, ~B ∼ 1

r2, static electromagnetism (12.168)

If we are to calculate the power radiated across a sphere at a large distance then

P =

∫dS.P = power crossing surface = energy/time crossing surface (12.169)

then

P =

∫r2dΩr.(c ~E × ~B) ∼ r2(

1

r4)→ 0 (forr →∞) (12.170)

We will see now that in the electrodynamic situation this is not the case, i.e., then

~E ∼ 1

r, ~B ∼ 1

r, electrodynamics (12.171)

and P is non-zero even as r → ∞. We say that there is radiation of energy by the current

and charge distribution. To see this we need to calculate ~A and φ for large r and then from

this to calculate ~E and ~B. Now we can write

|~r − ~r′| =√

(~r − ~r′)2 =√r2 − 2~r.~r′ + r′2 = r

[1− 2

r.~r′

r+r′2

r2

]1/2

= r

[1− r.~r′

r

](12.172)

113

Hence for large r

|~r − ~r′| ' r − r.~r′ (12.173)

Aloso |~r − ~r′| appears in tret:

tret = t− |~r − ~r′|

c' t− r

c+r.~r′

c(12.174)

hence to leading order in 1/r

~A ' 1

4πr

∫d3r′

~j(~r′, t− rc

+ r.~r′

c)

c(12.175)

and similarly

φ ' 1

4πr

∫d3r′ρ(~r′, t− r

c+r.~r′

c) (12.176)

we next calculate B = ∇× A. We need

∇× (1

r~j(~r′, tret)) = ∇(

1

r)×~j +

1

r∇×~j (12.177)

Since

∇(1

r) = − r

r2(12.178)

the first term is negligible for large ~r. For the second term

1

r∇×~j =

1

r(∇tret)×

∂~j

∂tret' 1

r∇(t− r

c+r.~r′

c)× ∂~j

∂tret=

1

r

[− rc− 1

r2

~r.~r′

cr +

~r′

rc

]× ∂~j

∂tret

(12.179)

So the leading term is just the first term

1

r∇×~j ' −1

r

r

c× ∂~j

∂tret(12.180)

we have

~B ' − 1

4πrr ×

∫d3r′

1

c2

∂~j(r′, t− rc

+ r.~r′

c)

∂t(12.181)

114

One can similarly calculate

~E = ∇φ− 1

c

∂ ~A

∂t(12.182)

to find

~E ' −r × ~B (12.183)

There are several remarkable things about this result:

[1] Both ~E and ~B falls off 1/r asymptotically, which means as we saw energy will be radiated

over a large sphere at ∞.

[2] Last two equations implies for the electromagnetic fields

~E ⊥ ~B; ~E ⊥ ~r; ~B ⊥ ~r; (12.184)

and further

~E × ~B = −(r × ~B)× ~B = r ~B. ~B − ~Br. ~B = r ~B. ~B (12.185)

Hence ~E, ~B and r form a right handed triad of vectors or on the sphere with |E| = |B|

[3] Thus ~E and ~B precisely the relations that our plane wave did with r playing the role

of the direction of propagation vector. In fact we see that the Poyinting vector is in the

direction of r and so the energy is radiiated outwards in the direction of r.

[4] But these relations are just relations of a source free Maxwell eqn. solutions, i.e., of a

homogenous solution to Maxwell equations. In fact we can check that the leading 1/r part

of ~A and φ precisely obey the source free wave eqn. Thus we have

∂i ~A =1

4πr

∫d3r′∂i

~j(~r′, t− rc

+ r.~r′

c)

c− ri

4πr3

∫d3r′~j (12.186)

and since

∂i~j = (∂~j

∂t)∂tr∂xi

;∂tr∂xi

= −1

c

rir

+1

c

r′ir− 1

c

ri

r2r.~r′ (12.187)

115

so keeping term of O(1/r)

∂i~j = −1

c

rir

∂~j

∂t(12.188)

and hence

∇2~j = ∂i(∂i~j) = −1

c

rir

∂2~j

∂t2(−rirc

)− 1

c

3

r

∂~j

c∂t+riri

r3

∂~j

∂t(12.189)

Thus for the leading terms for large r

∇2 ~A =1

c2

∫d3r′

4πr

∂2~j(r′, tr)

∂t2=

1

c2

∂2

∂t2

∫d3r′

4πr~j(r′, tr) =

1

c2

∂2

∂t2~A (12.190)

and so Thus the 1/r term obeys the free field wave eqn. and act just like a source free

field. Of course the amplitude of the wave depends on the sources in the interior, but in the

asymptotic domain, the fields appear to be disconnected from the interior. As a consequence,

with the Sommerfeld boundary consition, we can think of any radiation field we may see as

having been emitted from a source in the distant past so that we are so far away from the

source that we see only 1/r part only and hence it gives the appearance of a homogenous

solution.

Let us now go back to our expression for power radiated across a large sphere of radius

r →∞. If we consider the power radiated over an element of surface d~S in the direction r,

we have

dP = cr2dΩr.( ~E × ~B) (12.191)

We define

dP

dΩ= cr2r.( ~E × ~B) = power radiated in all direction r (12.192)

. which gives us the angular distribution of energy radiated. The total powr radiated is

P =

∫dP

dΩdΩ =

total power radiated in direction r

solid angle(12.193)

116

Asymptotically we have

dP

dΩ= cr2r.(−(r × ~B)× ~B) (12.194)

and since ~B ⊥ ~r, we have for large ~r

dP

dΩ= cr2 ~B2 (12.195)

and then we have

dP

dΩ=

1

(4π)2

1

c3

[r ×

∫d3r′

∂~j(r′, tr)

∂t

]2

; tr ≡ t− r

c+r.~r′

c(12.196)

In general, the previous expression is difficult to evaluate. However, if our current and

charges are constrained to stay close to the origin, we can make approximation by expanding

the r.~r′/c in tret. Thus we can write

~j(r′, t− r

c− r.~r′

c) =

∞∑n=0

(r.~r′

c)n

1

n!

∂n

∂tn~j(r′, te); te ≡ t− r

c(12.197)

te is just the retarted time in the approximation that all the particles were located at the

origin. A a first approximation, let us keep just the first term in the above equation. Then

we can write

dP

dΩ=

1

(4π)2

1

c3

[r × ∂

∂t

∫d3r′~j(r′, te)

]2

(12.198)

where te does not depend on ~r′. We can use the equation of continuity to reqrite the integral.

One has

∇.~j +∂

∂tρ = 0 (12.199)

and hence ∫d3rxi(∂kjk +

∂

∂tρ) = 0 (12.200)

117

Integrating by parts in 1st term and setting the surface term to zero (since ~j = 0 at large ~r):

∫d3r

[−∂kxijk + xi

∂

∂tρ)

]= 0 (12.201)

Hence ∫d3rji(~r, t) =

∂

∂t

∫d3rxiρ(~r, t) (12.202)

One defines

~d(t) =

∫d3r~rρ(~r, t) = electric dipole moment (12.203)

(Similar to the electrostatic definition). We have then

dP

dΩ=

1

(4π)2

1

c3

[r × d2~d(te)

dt2

]2

(12.204)

In this approximation the radiation depens on the dipole moment at the retarded time

te in the past We can simplify a bit

(r × ~d).(r × ~d) = r.rd.d− (r.d)(r.d) (12.205)

If θ betwin d and ~r Then

(r × ~d)2 = (d)2 − (d)2cos2θ = (d)2sin2θ (12.206)

Thus

dP

dΩ=

1

(4π)2

1

c3d2sin2θ (12.207)

and the radiation comes out preferentially perpendicular to d (Fig.12.4)

For a distribution of point charges

~d(t) =∑

ea

∫d3r~rδ3(r − ra(t)) =

∑ea~ra(t) (12.208)

118

d

n

ΩddP

Figure 12.4:

so

~d(t) =∑

ea~ra(t) (12.209)

and

dP

dΩ=

1

(4π)2

1

c3|∑

ea~ra(t)|2sin2θ (12.210)

Thus electromagnetic power is radiated when the particles are accelerated. This is reasonable

if we look at our energy conservation law. Recall

−∑a

~Fa.~va =d

dt

∫V

d3rU +

∮s

d~S. ~P (12.211)

, where the term on L.H.S. is the rate of loss of energy by forces on particles and the second

term on the R.H.S. corresponds to the outward flux of energy. Thus suppose at some time

in past, te, forces on the particles cause it to accelerate. The particle then radiates and loses

energy. These waves enter into the volume V and cause increase of the E.M. energy in V ,

i.e., ddτ

∫U > 0. Eventually at time t that energy arrives at S and there is a flux of energy

over S.

Note also that one adds the contribution from each particle and then sqaures to get

119

d|P |/dΩ, i.e., there can be interference of the waves eminating from each particle.

If we consider just a single particle we have

dP

dΩ=

1

(4π)2

e2

c3|~r(t)|2sin2θ (12.212)

We can choose ~r as the polar axis and integrate

P =1

(4π)2

e2

c3|~r(t)|2

∫dΩsin2θ =

1

(4π)2

e2

c3|~r(t)|2 8π

3(12.213)

or

P =2

3

1

(4π)

e2

c3|~v(t)|2 (12.214)

note that we are using rationalized units so e2r/(4π) = e2

unrat]

120