Ma432 Classical Field Theory

Notes by Chris Blair

These notes cover a lot of the 2008-2009 Ma432 Classical Field Theory course given by Dr Nigel
Buttimore (replaced by Ma3431 Classical Field Theory and Ma3432 Classical Electrodynamics,
the former corresponding to at least the rst four sections of these notes). The emphasis is
mostly on the Lagrangian formulation of classical electrodynamics and the solution of Maxwells
equations by Greens function methods. They are probably slightly suspect, particularly with
regard to indices and brackets (and no doubt contain other more unsettling errors).
I am told that Dr Buttimore has changed his units from those in these notes, so use at your
own discretion.
Contents
1 Simple eld theory 3
1.1 Introduction to eld theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Field theory as a continuum limit . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Euler-Lagrange equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Special relativity 6
2.1 Rapidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.2 Tensor notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3 Covariant eld theory 10
3.1 Relativistic free particle action . . . . . . . . . . . . . . . . . . . . . . . . . . 10
3.2 Relativistic interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.3 Electromagnetic eld tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.4 Maxwells equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.5 Four-current and charge conservation . . . . . . . . . . . . . . . . . . . . . . 15
4 Noethers theorem 17
4.1 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
4.3 Stress-energy tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.3.1 Stress-energy tensor for electromagnetic eld . . . . . . . . . . . . . . 20
5 Solving Maxwells equations 22
5.1 Time-independent solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5.1.1 Greens function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5.1.2 Magnetostatic and electrostatic potentials . . . . . . . . . . . . . . . 23
5.2 Time-dependent solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1
5.2.1 Greens function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
5.2.2 Lienard-Wiechart potentials . . . . . . . . . . . . . . . . . . . . . . . 26
5.2.3 Electromagnetic elds from Lienard-Wiechart potentials: method one 28
5.2.4 Electromagnetic elds from Lienard-Wiechart potentials: method two 29
5.2.5 Properties of the electromagnetic elds due to a moving charge . . . . 30
5.2.6 Local form of the electromagnetic elds due to a moving charge . . . 31
6 Power radiated by accelerating charge 33
7 Bibliography 34
Notes for Classical Field Theory Section 1: Simple eld theory
1 Simple eld theory
1.1 Introduction to eld theory
You are probably already familiar with the notion of electric and magnetic elds. Loosely
speaking, a eld in a physics is a physical quantity dened at every point of space and
time, which can be valued as a single number (scalar eld), a vector (vector eld, such as
electromagnetism and gravity), or as a tensor.
We will restrict ourselves to the study of the electric and magnetic elds, and do so in a
unied manner consistent with special relativity. The methods we use are based in the La-
grangian approach to classical mechanics. Recall that in classical mechanics the Lagrangian
L was dened as L = T V where T and V are the kinetic and potential energies of the
system in question. The action of the system was dened to be the quantity S =

Ldt, and
the equations of motion of the system were found from the principle of least action, which
states that the true time evolution of the system is such that the action is an extremum.
The equations of motion (known as the Euler-Lagrange equations) were thus derived from
the condition S =

Ldt = 0.
In studying elds which take on dierent values at dierent space points it is convenient
to express the Lagrangian itself as an integral, L =

d
3
x L, where L is called the Lagrangian
density. The full action is then S =

dtd
3
x L. Note that when we approach this from the
special relativistic point of view the separate time and space components will be unied into
a single package.
1.2 Field theory as a continuum limit
To begin, let us show how a simple eld theory may be derived by taking the contin-
uum limit of a system of N particles on a spring with spring constant k. Let the particles
have equilibrium positions a, 2a, . . . Na and denote the deviation of the i
th
particle from its
equilibrium by
i
. The force on the i
th
particle is
F
i
=

+k(
2

1
) i = 1
k(
i

i1
) + k(
i+1

i
) 1 < i < N
k(
N

N1
) i = N
The Lagrangian is
L = T V =
N

i=1
1
2
m

2
i

N

i=1
1
2
k(
i+1

i
)
2
and the equations of motion are
m

i
= k(
i

i1
) + k(
i1

i
) 1 < i < N
We now take the limit a 0 while keeping (N 1)a xed by letting N . If we write
x = ai as the position of the i
th
particle then we can regard
i
(x = ai, t), and using the
3
Notes for Classical Field Theory Section 1: Simple eld theory
equations of motion in the form
m
a

i
= ka
1
a
2

(
i1

i
) (
i

i1
)

we can apply the denition of the derivative

i
= lim
a0
([i + 1]a) (ai)
a
twice to obtain the equations of motion in the limit a 0:

t
=

x
2
where = lim
a0
ka and =
m
a
is the mass density which we keep xed. We see that our
simple eld obeys the wave equation.
If we dene
L =
N

i=1
aL
i
L
i
=
1
2

2
i

1
2
k
a
(
i+1

i
)
2
then in the limit we obtain the Lagrangian density
L =
1
2

1
2

2
such that L =

dxL.
1.3 Euler-Lagrange equations
A more general Lagrangian density would be of the form L(
t

i
,
x

i
,
i
, t, x). We can
use Hamiltons Principle of Least Action to nd the general form of the equations of motion.
Let us consider the simplest case where the eld is one-dimensional and the Lagrangian
density is invariant under time and space translation. Then we have that

Ldx dt = 0
and in full

L
(
t
)
(
t
) +
L
(
x
)
(
x
) +
L

dx dt = 0
Noting that (
t
) =
t
() and (
t
) =
t
() we rewrite this as

d
dt

L
(
t
)

t
L
(
t
)
+
d
dx

L
(
x
)

x
L
(
x
)
+
L

dx dt = 0
4
Notes for Classical Field Theory Section 1: Simple eld theory
We integrate out the total time and space derivatives and use the fact that the term must
vanish at the endpoints to then obtain
S =

t
L
(
t
)
+
x
L
(
x
)

L

dx dt = 0
hence we obtain the Euler-Lagrange equations for this eld:

t
L
(
t
)
+
x
L
(
x
)

L

= 0
For a vector eld just replace by A
i
.
5
Notes for Classical Field Theory Section 2: Special relativity
2 Special relativity
We will now introduce the machinery that allows us to express eld theory in a manner
consistent with the theory of special relativity. In particular, we seek to formulate the theory
of elds in a manner that is Lorentz covariant - that is, related from one frame to another
via Lorentz transformations. Note that we do not introduce special relativity systematically
but assume some prior knowledge of the subject. For completeness we note that the two
postulates of special relativity are that the laws of physics take the same form in all inertial
(non-accelerating) reference frames, and that the speed of light c in vacuum is an absolute
constant regardless of frame.
2.1 Rapidity
The basic Lorentz transformations in 1 + 1 dimensions are
t

t
vx
c
2

= (x vt) where =
1

1
v
2
c
2
for a frame S

moving with velocity v with respect to the frame S. We have that

ct

1 +
v
c
1
v
c
(ct x)
We dene the rapidity (v) as
(v) =
1
2
ln

1 +
v
c
1
v
c

so that
ct

= e

(ct x)
Note that rapidities add. We can then show that
v = tanh = cosh
which allows us to write the Lorentz transformations as
ct

= ct cosh x sinh x

= x cosh ct sinh
In the full 1 + 3 dimensions we can write this transformation in matrix form as
=

cosh sinh 0 0
sinh cosh 0 0
0 0 1 0
0 0 0 1

6
Notes for Classical Field Theory Section 2: Special relativity
called a boost in the x-direction. Note that the most general proper Lorentz transformation
can be written as a product of a 3-rotation to align the new x-axis with the direction of
motion, a boost along the new x-direction with velocity v and a second 3-space rotation.
2.2 Tensor notation
A basic invariant in special relativity is the interval ds separating two (innitesimally
close) events in four-dimensional space-time: ds
2
= c
2
dt
2
dx
2
1
dx
2
2
dx
2
3
. From this we
get the metric tensor:
g

= diag (1, 1, 1, 1) = g

which is used to raise and lower indices as follows:

x

= g

= g

Note that we sum over repeated indices. Upper indices are said to be contravariant, and
lower indices are said to be covariant. Note that (in this metric) raising a time-index has no
eect, x
0
= x
0
, while raising a space-index changes the sign, x
i
= x
i
. Note also that indices
with Greek letters can take any value in {0, 1, 2, 3} while indices with Roman letters refer to
spatial indices, {1, 2, 3}. Thus in our notation we have x

= (ct, x) and x
i
= x.
A Lorentz transformation relates events x

in the frame S

to events x in the frame S,

and is written as
x

=

The metric tensor is invariant under Lorentz transformations

g

= g

A four-dimensional vector, or four-vector, is written as A

and transforms like the coordinates

x

:
A

while a second rank tensor T

transforms like the product of (components of) two four-

vectors:
T

and similarly for tensors of higher rank.

We form the four-dimensional Kronecker delta by lowering the index of g

:
g

= g

= diag (1, 1, 1, 1)
We also have the four-dimensional Levi-Civita symbol,

which is +1 for even permu-

tations of and 1 for odd permutations, with
0123
= 1. Note that for the covariant
form,

, we have
0123
= 1. Both the Levi-Civita symbol and the Kronecker delta are
invariant under Lorentz transformations.
7
Notes for Classical Field Theory Section 2: Special relativity
We can form a scalar invariant under Lorentz transformations (a Lorentz scalar) by con-
tracting two four vectors
a

= a

The invariant time element d is given by

c
2
d
2
= dx

dx

It is related to the usual time element by

d =
dt

d
d
=
d
dt
This allows us to dene vectors of four-velocity V

=
dx

d
and four-momentum p

= m
dx

d
where m is the rest mass of the particle. The zero component of the four-momentum is related
to the energy E = mc
2
by cp
0
= E, so we can write the four-momentum as p

E
c
, p

.
Note that the contraction of the four-momentum with itself is p

= m
2
c
2
.
If this (or indeed the scalar formed by contracting any four-vector with itself) is equal
to zero we say that the four-vector is light-like, if it is greater than zero we say that it is
space-like, if it is less than zero it is timelike.
This can be related to the idea of light-cones:
ct
x
v < c
v > c
v = c
Past
Future
This picture can be understood as follows: events that occur inside the lightcone are
timelike; it is possible to nd a Lorentz transformation such that any two events occur at the
same point in space, but at dierent times. Similarly, events that occur outside the lightcone
are spacelike in that is possible to nd a Lorentz transformation to a frame such that any
two events occur at the same point in time, but are separated in space.
8
Notes for Classical Field Theory Section 2: Special relativity
The boundary of the light-cone is given by a line corresponding to the motion of a particle
of velocity c. The motion of a particle with velocity less than c lies within its light-cone.
Events that occur within the past light-cone of a particle can aect the particle in the
present, while events that occur outside it cannot.
Finally we must consider calculus in space-time. We will be integrating over the four-
dimensional volume element
d
4
x = dx
0
dx
1
dx
2
dx
3
= c dt dx
1
dx
2
dx
3
which is an invariant. Derivatives are denoted by

Note that dierentiating with respect to a lower index gives an upper index, while dieren-
tiating with respect to an upper index gives a lower index, so for instance
x

= g

= g

= g

)
(

)
=

9
Notes for Classical Field Theory Section 3: Covariant eld theory
3 Covariant eld theory
We now seek to formulate eld theories in a special relativistic context. We will be seeking
actions
S =

Ld
3
x dt =

Ldt =

L d
which are relativistically invariant. Recall that the rst postulate of special relativity is that
the laws of physics are the same in all (inertial) reference frames - these laws take the form
of the equations of motion derived from the condition S = 0, hence S must be Lorentz
invariant. As d is an invariant scalar we see that we must have L also a Lorentz scalar.
The simplest way to achieve this is to contract the available four-vectors.
Note that the Euler-Lagrange equations of motion for a eld A

are

L
(

L
A

= 0
3.1 Relativistic free particle action
For a free particle, the only scalar which preserves translational invariance is p

= (mc)
2
,
suggesting a Lagrangian of the form L = Cp

= Cm
2
c
2
, where C is some constant. Let us
look at the non-relativistic limit of this Lagrangian. We have
S =

Ldt =

L d
We want the non-relativistic limit of L to agree with the Lagrangian for a non-relativistic
free particle, L =
1
2
mv
2
. Consider the Taylor expansion of m
1
:
m
1
= m

1
v
2
c
2

= m
m
2
v
2
c
2
+ O

1
c
4

mc
2

1
= mc
2
+
1
2
mv
2
As the constant term mc
2
is unimportant, we see that we can take our Lagrangian to be
L =
mc
2

hence we have action

S = mc
2

dt

= mc
2

d
or
S =
1
m

d
10
Notes for Classical Field Theory Section 3: Covariant eld theory
3.2 Relativistic interactions
We now let our particle be acted on by some eld with potential A

(x

). Possible scalars
include A

and A

. Again we want to choose the Lagrangian so that the non-relativistic

limit gives us the interaction Lagrangian for a particle in the presence of an electric eld (in
the non-relativistic limit the particles velocity goes to zero, and so it will not interact with
the magnetic eld). This is L = e. This suggests we look at eA

= ep
0
A
0
ev

A. Now,
p
0
=
E
c
= mc, and we identify A
0
with . Then in the limit v 0 we have ep

= mce.
Thus we take our interaction term to be
L
int
=
e
mc
A

The total action is now

S =
1
m

+
e
c
A

d =

+
e
c
A

dx

using p

= m
dx

d
. The dynamics of the system can then be found by varying this action.
First, let us note that
(p

) = p

+ p

= p

+ p

= 2p

= 2m
dx

d
p

but also
(p

) = (m
2
c
2
) = 0
and hence if m = 0 we have
dx

d
p

= 0 dx

= 0
Now let us compute the variation:
S =

+
e
c
A

dx

+
e
c
A

(dx

+
e
c
A

dx

+
e
c
A

d(x

)
e
c

dx

where we have used the above result to eliminate the p

term. We now use

+
e
c
A

d(x

) =

+
e
c
A

(x

+
e
c
A

d
d

+
e
c
A

(x

+
e
c
A

11
Notes for Classical Field Theory Section 3: Covariant eld theory
and the fact that the variation x

vanishes at the end points to obtain

S =

+
e
c
A

1
. .. .
=0
+

dp

+
e
c
dA

e
c

dx

=

dp

+
e
c
A

dx

e
c

dx

=

dp

d
d +
e
c

dx

d
x

dx

d
x

Switching the dummy variables and in the second A

term we have:
S =

dx

dp

d
+
e
c

dx

d

A

dx

= 0
Hence we nd equations of motion
dp

d
=
e
c

dx

d
3.3 Electromagnetic eld tensor
The electric and magnetic elds can be expressed in terms of the 4-potential A

as

E =
1
c

t

A

(A
0
)

B =

A
Note in passing that

E has odd parity (i.e. transforms as

E

E under space reversal,

r r) and is even under time reversal, while

B has even parity and is odd under time
reversal.
We now introduce the electromagnetic eld tensor
F

where we identify
F
i0
= E
i
F
ij
=
ijk
B
k
as we have
E
i
=
0
A
i

i
A
0
=
0
A
i
+
i
A
0
= F
i0

ijk
B
k
=
ijk

klm

l
A
m
= (
i
l

j
m

i
m

j
l
)
l
A
m
=
i
A
j
+
j
A
i
=
i
A
j

j
A
i
= F
ij
12
Notes for Classical Field Theory Section 3: Covariant eld theory
Explicitly, the contravariant and covariant forms of the tensor are:
F

0 E
1
E
2
E
3
E
1
0 B
3
B
2
E
2
B
3
0 B
1
E
3
B
2
B
1
0

0 E
1
E
2
E
3
E
1
0 B
3
B
2
E
2
B
3
0 B
1
E
3
B
2
B
1
0

The equations of motion derived in the last section can then be written
dp

d
=
e
c
F

dx

d
To write these in three-dimensional form we rst set = 0 and sum over (noting that
d
d
=
d
dt
)
dp
0
dt
=
e
c
F
0i
dx
i
dt
=
e
c
F
i0
v
i

dE
dt
= e

E v
using p
0
=
E
c
and the fact that raising a spatial index changes the sign, as well as the
antisymmetry of F

. For = i
dp
i
dt
=
e
c
F
i0
dx
0
dt
+
e
c
F
ij
dx
j
dt

dp
i
dt
= eF
i0

e
c

ijk
v
k
B
k

d p
dt
= e

E +
e
c
v

B
Note that F

is invariant under an important class of transformations known as gauge

transformations. A gauge transformation of the electromagnetic four-potential is a trans-
formation of the form A

for some scalar eld , and under this we have

F

= F

The invariance of the electromagnetic eld tensor and hence the observable elds allows us to
simplify problems by choosing a particular gauge, i.e. a particular choice of the A

satisfying
certain conditions. For example we will later explicitly solve Maxwells equations (introduced
in the next section) in Lorenz gauge:

= 0.
3.4 Maxwells equations
In the moonlight opposite me were three young women, ladies
by their dress and manner. I thought at the time that I must be
dreaming when I saw them, they threw no shadow on the oor.
Bram Stoker, Dracula
The simplest choice of a Lagrangian density for the electromagnetic eld tensor is L =
CF

where C is some constant. We will now nd the equations of motion satised by

13
Notes for Classical Field Theory Section 3: Covariant eld theory
the eld using the Euler-Lagrange equations. We have
(F

)
(

)
= 2F

)
= 2F

)
(

)

(

)
(

= 2F

= 2F

2F

= 4F

and as
L
A
= 0 we have the equation of motion for a free eld

= 0
If we introduce a source of charge and 3-current J

= (c,

J) (see the next section) then we
have an interaction term
L
int
=
1
c
A

and we obtain

=
1
4Cc
J

and choosing C =
1
16
(i.e. Gaussian cgs units) gives

=
4
c
J

In three dimensions this becomes

E = 4
and

1
c

E +

B =
4
c

J
These are two of Maxwells equations. The other two may be expressed using the dual
tensor

F

dened by

=
1
2

The components of the dual tensor may be seen to be

F
i0
= B
i

F
ij
=
ijk
E
k
in words, replace

E with

B and

B with

E. Now consider

14
Notes for Classical Field Theory Section 3: Covariant eld theory
This is the contraction of an antisymmetric tensor with a symmetric tensor, and so is equal
to zero. Hence we have the equation

= 0
By comparing with the above Maxwells equations in the case that J

= 0 and interchang-
ing the electric and magnetic elds in the manner mentioned above, we nd the other two
Maxwells equations:

B = 0
and
1
c

B +

E = 0
Note that the equation

= 0 can also be written in the form

= 0
The various contractions arising from the eld tensor and its dual are
F

= 2

B

B

E

E

= 4

E

B

F

= 2(

E

E

B

B)
3.5 Four-current and charge conservation
The four-current density is given by
J

(t, x) = e
dx

dt

3
(x x
e
(t))
where x
e
(t) is the path of a particle of charge e generating the eld A

. The charge q and

current ev are then given by
q =

(t, x)d
3
x ev =

J(t, x)d
3
x
Consider the gauge transformation A

= A

. This gives an interaction action

S

=
1
c
2

(J

+ J

) d
4
x
We vary this with respect to :
S

=
1
c
2

(J

)d
4
x =
1
c
2

[J

]) d
4
x
The second term vanishes on the boundary, and we are left with
S

=
1
c
2

d
4
x = 0

= 0
15
Notes for Classical Field Theory Section 3: Covariant eld theory
so charge is conserved. This equation can also be written

t
+

J = 0
Let us integrate this over a surface ,

t
d
3
x +

Jd
3
x = 0

d
3
x +

J d

Ad
3
x = 0
or
q
t
+

J d

A = 0
We see that invariance under gauge transformations leads to charge conservation. In fact
there is a close relationship between certain types of transformational invariance and conser-
vation laws, which we treat in detail in the next section.
16
Notes for Classical Field Theory Section 4: Noethers theorem
4 Noethers theorem
Noethers theorem states for every continuous symmetry there is a conserved quantity.
4.1 Derivation
Let us suppose we have a Lagrangian density L(,

, ) invariant under a transformation

= + x

= x

+ x

parametrised by a small quantity such that

( = 0) = x

( = 0) = x

Now the change in x

means that the volume we integrate over will change; that is, we have
cS =

L(

, x

)d
4
x

L(, x

)d
4
x
cS =

[L(

, x

) L(, x

)]d
4
x +

Lx

where d

is a surface element (in the x

-direction) and x

can be thought of as giving

the change in the boundary caused by the transformation of the coordinates (for more
details see Classical Mechanics by Goldstein, 3rd edition, page 592). We have also switched
dummy variables in the rst integral from x

to x

.
Now the rst integrand is just the variation of L with respect to , that is

L

+
L
(

)
(

d
4
x =

L
(

L
(

. .. .
=0 (E-L)

d
4
x
while the second can be rewritten using the divergence theorem

Lx

(Lx

) d
4
x
hence
cS =

L
(

)
+Lx

d
4
x = 0
Now, consider
=

(x

) (x

)
=

(x

) (x

) [

(x

(x

)]
17
Notes for Classical Field Theory Section 4: Noethers theorem
We now Taylor expand

(x

) with respect to x

, obtaining

(x

(x

(x

) +

(x

)x

(x

) =

(x

)x

or, using

= +

(x

(x

(x

)x

and the term on the right is of second order and so can be neglected.
Similarly we expand

(x

) with respect to to nd

(x

) (x

(x

=0
+

=0
(x

) = (x

) +

=0
(x

) =

=0
and as we also have x

= x

|
=0
,
cS =

L
(

=0

L
(

=0

d
4
x = 0
and so we can conclude that the current
J

=
L
(

=0

L
(

=0
is conserved. In the case of a vector eld A

this becomes
J

=
L
(

)
A

=0

L
(

=0
Note that this derivation assumes has no indices (i.e. is a scalar). To be more precise we
should take into account the possibility that may be a vector or even matrix quantity, and
write it as

where stands for any possible index. In this case we should go back to the
second from last line of the proof and extract the conserved quantity
J

L
(

=0

L
(

=0

where there is now one conserved current J

for each

(see example iv) below).

4.2 Examples
Let us assume that the eld does not change, i.e.

(x) = (x).
18
Notes for Classical Field Theory Section 4: Noethers theorem
i) Let L be invariant under time translation, x
0
= x
0
+ , x
i
= x
i
, then
J
0
=
L
(
0
)

0
L H
is conserved (this is the Hamiltonian for the eld).
ii) Suppose L invariant under x

= x

+ , then
L
(

L
is conserved (this is the energy-momentum density for the eld).
iii) Suppose L is symmetric under rotations about the x
3
axis, that is,
x
1
= x
1
cos + x
2
sin x
2
= x
1
sin + x
2
cos x
3
= x
3
x
0
= x
0
then the angular momentum density of the eld is conserved,
L
(
3
)
(x
1

2
x
2

1
)
iv) Suppose that L is completely rotationally invariant in the space dimensions. An innites-
imal rotation can be written as
x
i
x
i
+
ij
x
j
where
ij
=
ji
is a three by three skew-symmetric real matrix, and so an element of so(3)
(i.e. a generator of rotations). We then have that
J
ijk

jk
=

L
(
i
)

l
g
il
L
x

jk

jk
=0

jk
is conserved. Now,
x

jk
=

ln
x
n

jk
=
j
l

k
n
x
n

j
n

k
l
x
n
=
j
l
x
k

k
l
x
j
giving
J
ijk

jk
=

L
(
i
)

j
x
k

k
x
j

g
ij
Lx
k
+ g
ik
Lx
j

jk
so we can pick out our conserved currents
J
ijk
=
L
(
i
)

j
x
k

k
x
j

g
ij
Lx
k
+ g
ik
Lx
j
19
Notes for Classical Field Theory Section 4: Noethers theorem
For example, consider rotations about the x
1
axis. Then we have the following conserved
currents:
J
123
=
L
(
1
)

2
x
3

3
x
2

J
112
=
L
(
1
)

1
x
2

2
x
1

Lx
2
J
113
=
L
(
1
)

1
x
3

3
x
1

Lx
3
J
132
= J
123
J
121
= J
112
J
131
= J
113
J
111
= J
112
= J
133
= 0
Note that J
132
corresponds to the angular momentum density about the x
1
axis.
4.3 Stress-energy tensor
If we assume that our system is invariant under the transformation x

then we
have that the tensor
T

=
L
(

L
is conserved (note we have changed the indices). This tensor is known as the stress-energy
tensor.
4.3.1 Stress-energy tensor for electromagnetic eld
For a free electromagnetic eld, L =
1
16
F

. The stress energy tensor is given by

T

=
1
16
(F

)
(

+
1
16
g

=
1
4
F

+
1
16
g

Although conserved, T

is not gauge invariant (as A

appears explicitly) or symmetric.

We can form a tensor with nicer properties as follows: rst, write F

= F

=
g

giving
T

=
1
4
g

+
1
16
g

and then substitute in

= F

to obtain
T

=
1
4
g

1
4
g

+
1
16
g

and now dene

= T

+
1
4
g

20
Notes for Classical Field Theory Section 4: Noethers theorem

=
1
4
g

+
1
16
g

(note interchange of and to remove the leading minus sign). This tensor is conserved as
the dierence between it and the stress-energy tensor is
1
4
g

=
1
4
F

and

(F

) = (

+ F

= 0
where the rst term vanishes due to the equations of motion and the second term vanishes
as it is the contraction of a symmetric and antisymmetric tensor.
The tensor

is gauge invariant, symmetric, traceless (

= 0) and can be
used to dene the angular momentum density M

.
But my very feelings changed to repulsion and terror when I
saw the whole man slowly emerge from the window and begin
to crawl down the castle wall over the dreadful abyss, face down
with his cloak spreading out around him like great wings.
Bram Stoker, Dracula
21
Notes for Classical Field Theory Section 5: Solving Maxwells equations
5 Solving Maxwells equations
We now wish to solve the equation

=
4
c
J

for a given J

. This equation can be

written

=
4
c
J

We impose the Lorenz gauge,

= 0 so that this becomes

=
4
c
J

This equation will be solved using Greens function methods. Recall that given some dif-
ferential equation Df(x) = g(x) the Greens function G solves DG(x) = (x); so that
f =

dx

G(x x

)g(x

) as then Df =

dx

DG(x x

)g(x

) =

dx

(x x

)g(x

) = g(x).
We will apply Fourier transform methods to obtain the Greens functions we need. The
four-dimensional Fourier transform and its inverse are
f(x

) =
1
(2)
2

d
4
k e
ikx

f(k

)

f(k

) =
1
(2)
2

d
4
x e
ikx

f(x

)
The Fourier representation of the delta-function is
(x) =
1
2

dk e
ikx
5.1 Time-independent solutions
First let us obtain solutions with no time dependence. This means we will be solving the
equation

2
A

=
4
c
J

5.1.1 Greens function

We need to nd a Greens function satisfying

2
G(x) =
3
(x). In terms of Fourier
transforms, this is

2
1
(2)
3/2

d
3
k e
i

kx

G(

k) =
1
(2)
3/2

d
3
k

k
2
e
i

kx

G(

k) =
1
(2)
3

d
3
k e
i

kx

G(

k) =
1
(2)
3/2
1

k
2
giving
G(x) =
1
(2)
3

d
3
k
1

k
2
e
i

kx
=

0
d

d
3
k e
i

kx

k
2
22
Notes for Classical Field Theory Section 5: Solving Maxwells equations
as

0
de
x
=
1

. Now the k integral can be expressed as a Gaussian:

d
3
k e
i

kx

k
2
=

d
3
k e

k+

ix
2

2
e

x
2
4
=

3/2
e

x
2
4
so we now have
G(x) =
1
(2)
3

3/2

0
d

3/2
e

x
2
4
Let u =
1/2
, then 2du =
3/2
d and we have
G(x) =
1
4
3/2

du e

x
2
4
u
2
=
1
4
3/2
1
2

4
x
2
using the fact that the Gaussian integral is symmetric, and hence
G(x) =
1
4|x|
5.1.2 Magnetostatic and electrostatic potentials
For an electrostatic system, we have

2
A
0
=

2
= 4(x) =
4
c

e
e
3
(x x
e
),
where the sum ranges over the dierent charges in the system, with x
e
signifying the position
of the charge e. Using the Greens function above we see that
(x) =

d
3
x

)
|x x

|
=

d
3
x

e
e
3
(x

x
e
)
|x x

|
=

e
e
|x x
e
|
giving

E =

e
e
x x
e
|x x
e
|
3
Similarly the vector potential is given by

A(x) =
1
c

d
3
x

J(x

)
|x x

|
5.2 Time-dependent solutions
We will now seek solutions of the full equation

=
4
c
J

23
Notes for Classical Field Theory Section 5: Solving Maxwells equations
5.2.1 Greens function
The Greens function we seek is a solution D(z

) of

D(z

) =
4
(z

) where z

=
x

x

. Appealing to Fourier transforms again, we have

1
(2)
2

d
4
k

D(k)e
ikz

=
1
(2)
2

d
4
k k

D(k

)e
ikz

=
1
(2)
4

d
4
k e
ikz

from which

D(k

) =
1
4
2
1
k

=
1
4
1
k
2
0
k
2
where k = |

k|. We now must solve

D(z

) =
1
(2)
4

d
4
k
1
k
2
0
k
2
e
ikz

=
1
(2)
4

d
3
k e
i

kz

dk
0
1
k
2
0
k
2
e
ik
0
z
0
Note the sign change in the rst exponential when converting to three-dimensions. Now,
to evaluate the k
0
integral we use contour integration, treating k
0
as a complex number,
k
0
= Re k
0
+iImk
0
. This gives e
i

kz
= e
Imk
0
z
0
e
iRe k
0
. So that the integral converges we must
choose Imk
0
< 0 for z
0
> 0. This condition is imposed as z
0
= x
0
x
0
= c(t t

), and so
positive z
0
ensures that contributions to the Greens function and hence to the potential A

only come from events that occur at times t

< t, i.e. events in the past. Thus causality is

ensured.
Now, the poles of the integrand are k and so lie on the real axis. To avoid them, we
displace our contour by an innitesimal amount i so that it lies just in the upper-half plane
(formally we should let 0 at the end). Our contour of integration then looks like:
Im k
0
Re k
0
R
k k

We than have

dk
0
e
ik
0
z
0
k
2
0
k
2
=

R
R
dk
0
e
ik
0
z
0
k
2
0
k
2
+

semi-circle
e
ik
0
z
0
k
2
0
k
2
Now on the semicircle we can write k
0
= Re
i
= Rcos iRsin , so that the integral
becomes an integral over from 0 to . So we have

semi-circle
e
ik
0
z
0
k
2
0
k
2
= i

0
dRe
i
e
Rz
0
sin iRz
0
cos
1
R
2
e
2i
k
2
24
Notes for Classical Field Theory Section 5: Solving Maxwells equations
and sin 0 for this range of , hence the integrand goes to zero as R , as required.
The value of the k
0
integral we are interested in will hence be given by the 2i times the
sum of the residues inside the contour. The residues are:
lim
k
0
k
(k
0
k)
e
ik
0
z
0
(k
0
k)(k
0
+ k)
=
e
ikz
0
2k
lim
k
0
k
(k
0
+ k)
e
ik
0
z
0
(k
0
k)(k
0
+ k)
=
e
ikz
0
2k
hence

dk
0
1
k
2
0
k
2
e
ik
0
z
0
=
i
k

e
ikz
0
e
ikz
0

(z
0
)
where the Heaviside function
(z
0
) =

1 z
0
> 0
0 z
0
< 0
is added as the residues only contribute for positive z
0
. So we now have
D(z

) =
(z
0
)
(2)
4
i

d
3
k e
i

kz
1
k

e
ikz
0
e
ikz
0

=
(z
0
)
16
3
i

0
dk k
2
1
k

e
ikz
0
e
ikz
0

2
0
d

0
d sin e
ikz cos
upon switching to polar coordinates and choosing the coordinate frame such that the k
3
axis
makes an angle of with z, and letting z = |z|. Integrating over the angles, we get
D(z

) =
(z
0
)
8
2
i

0
dk k

e
ikz
0
e
ikz
0

e
ikz
ikz

e
ikz
ikz

=
(z
0
)
8
2
z

0
dk

e
ik(z
0
+z)
+ e
ik(z
0
+z)
e
ik(z
0
z)
e
ik(z
0
z)

but if we let k k

0
dk e
ik(z
0
+z)
=

0
d(k)e
ik(z
0
+z)
=

dk e
ik(z
0
+z)
hence
D(z

) =
(z
0
)
8
2
z

dk

e
ik(z+z
0
)
e
ik(z
0
z)

=
(z
0
)
4z

(z
0
z) (z
0
+ z)

remembering the integral representation of the delta function. Owing to the Heaviside func-
tion only the rst delta function will contribute, so
D
ret
(z

) =
(z
0
)
4z
(z
0
z)
25
Notes for Classical Field Theory Section 5: Solving Maxwells equations
The subscript signies that this is the retarded Greens function (that is, the Greens function
resulting from the contribution of events in the past).
We can put the Greens function in covariant form using
(z

) = ([z
0
z][z
0
+ z]) =
(z
0
z)
|z
0
+ z|
+
(z
0
+ z)
|z
0
z|
as (ab) =
(a)
|b|
+
(b)
|a|
. Hence
(z
0
)(z

) = (z
0
)
(z
0
z)
|z
0
+ z|
= (z
0
)
(z
0
z)
|2z|
and so
D
ret
(z

) =
(z
0
)
2
(z

)
Recalling that z

= x

we can state our nal results for the Greens function as

D
ret
(x

) =
(x
0
x
0
)
4z
(x
0
x
0
|x

|)
and in covariant form,
D
ret
(x

) =
(x
0
x
0
)
2
([x

][x

])
Note that if we had taken z
0
< 0 and closed our contour in the upper half-plane (with poles
displaced upwards) we would have obtained the advanced Greens function
D
adv
(x

) =
(x
0
x
0
)
4z
(x
0
x
0
+|x

|)
5.2.2 Lienard-Wiechart potentials
Of night and light and the half-light.
W.B. Yeats, He Wishes For The Cloths Of Heaven
We can now work out the potentials A

that solve the Maxwell equation

=
4
c
J

.
They are given by
A

(x

) =
4
c

d
4
x

D
ret
(x x

)J

(x

)
where
J

(x

) = e
dx

dt

3
(x

e
(t))
or in covariant form
J

(x

) = ec

d
dx

d

4
(x

e
())
26
Notes for Classical Field Theory Section 5: Solving Maxwells equations
as integrating over and using that
(f()) =

i:f(
i
)=0
(
i
)
|f

(
i
)|
we have
J

(x

) = ec
dx

3
(x

e
(t))

dx
0
d

where x
0
x
0
e
(

) = 0. Now,
dx
0
d
|

=
dx
0
dt
dt
d
|

= c
dt
d
|

and
dx

d
|

=
dx

dt
dt
d
|

so this reduces
to the local form.
So substituting this in, we have
A

(x

) =
4
c

dd
4
x

(x
0
x
0
)
2
([x

][x

])ec
dx

d

4
(x

e
())
= 2e

d(x
0
x
0
e
)([x

x
e

()][x

e
()])V

()
where we have written V

dx

e
d
. Now, we need the roots of the argument of the delta
function:
[x

x
e

()][x

e
()] = 0 (x
0
x
e
0
())
2
|x x
e
()|
2
= 0
There are two possibilities:
x
0
x
e
0
() = |x x
e
()|
The Heaviside function constrains us to choose the positive option. We see that the unique
root of [x

x
e

()][x

e
()] = 0 which ensures causality is given by the retarded time

0
:
x
0
x
e
0
(
0
) = |x x
e
(
0
)|
Physically, the retarded time gives the unique time at which the charged particle intersects
the past light-cone of the observation point. Now, as
d
d
[x

x
e

()][x

e
()] = 2[x

x
e

()]
d
d
x

e
()
We then have that
(x
0
x
0
e
)([x

x
e

()][x

e
()]) =
(
0
)
2[x

x
e

(
0
)]V

(
0
)
so, writing R

= x

e
(
0
), we have
A

(x

) =
V

()
R

()V

()

0
Note that the retarded time is in this notation dened by R

(
0
)R

(
0
) = 0 and that then
R
0
= R = |

R| = |x x

e
(
0
)|.
27
Notes for Classical Field Theory Section 5: Solving Maxwells equations
5.2.3 Electromagnetic elds from Lienard-Wiechart potentials: method one
We wish to evaluate the electromagnetic elds F

arising from the

motion of a charged particle. Consider the integral expression for the potential
A

(x

) = 2e

d(x
0
x
0
e
)([x

x
e

()]
2
)V

()
We will dierentiate this with respect to x

. First, note that

(x
0
x
0
e
) = (x
0
x
0
e
)
and this will give a term (|x x
e
()]
2
) which only contributes for x = x
e
() and can be
neglected. Thus we have

= 2e

d(x
0
x
0
e
)

([x

x
e

()]
2
)V

()
Let us now write

([x

x
e

()]
2
) =

(R

()R

())
=

R

(R

()R

())

(R

)
=
d
d
(R

()R

())

R

(R

)
and as R

= x

e
() we have

(R

) = 2R

and
d
d
(R

) = 2R

d
dR

=
1
2R

V

so (using as our dummy variable in the denominator as is used in the numerator)

([x

x
e

()]
2
) =
R

V

d
d
(R

()R

())
Thus

= 2e

d(x
0
x
0
e
)
R

V

R

V

d
d
(R

()R

())
and we can integrate this by parts to obtain

= 2e

d(x
0
x
0
e
)(R

()R

())
d
d

V

R

Recalling from the derivation of the potentials that

(x
0
x
0
e
)(R

()R

()) =
(
0
)
2R

(
0
)V

(
0
)
28
Notes for Classical Field Theory Section 5: Solving Maxwells equations
we nd

=
e
R

V

d
d

V

R

with this evaluated at the retarded time. Carrying out the dierentiation, with a dot denoting
a derivative with respect to proper time,

=
e
R

V

V

+ R

V

R

V


R

V

(R

V

)
2

V

+ R

and hence
F

=
eV

V

(R

V

)
3

V

R

+
e
(R

V

)
2

V

R

eR

V

(R

V

)
3

V

R

Note that we can write F

as a sum of two parts, one a velocity eld containing terms

depending on V

and the other an acceleration or radiative eld containing derivatives of the
velocity,

V

:
F

= F

vel
+ F

rad
where
F

vel
=
eV

V

(R

V

)
3

V

R

and
F

rad
=
e
(R

V

)
2

V

R

eR

V

(R

V

)
3

V

R

5.2.4 Electromagnetic elds from Lienard-Wiechart potentials: method two

Consider the covariant form of the equation dening the retarded time:
R

(
0
)R

(
0
) = 0
This equation denes
0
as a function of the observation point x

. To nd how
0
changes
with x

we dierentiate

(R

(
0
)R

(
0
)) = 0 2R

= 0
Now,

x
e
() =

d
d
x

e
()

so at the retarded time

R

0
) = 0
and hence

0
=
R

0
=
R

V

29
Notes for Classical Field Theory Section 5: Solving Maxwells equations
with the expression on the right evaluated at the retarded time.
We will use this to now evaluate the electromagnetic eld tensor resulting from the four-
potential
A

(x

) =
eV

V

R

0
In what follows we will denote dierentiation with respect to proper time by a dot. Note
that our expressions must all be evaluated at the retarded time - so in eect we will work
out

treating R

and V

as functions of with the awareness that in reality everything

we do is evaluated at
0
. This allows us to substitute the above expression for

0
for

as we derive. So we have

= e

V

V

eV

(V

R

)
2

V

)R

+ V

(

= e

eV

(V

R

)
2

V

R

+ V

(

= e

V

R

(V

)
2

eV

(V

R

)
2

V

R

V

R

+ V

V

V

= e
R

V

V

V

(V

R

)
3

eV

V

(V

R

)
2
+
eR

V

(V

R

)
2

eR

V


V

R

(V

R

)
3
Hence we again nd we can write F

as a sum of two parts, one involving

terms containing the four-velocity of the charge V

and the other involving terms involving
the acceleration

V

, that is,
F

= F

vel
+ F

rad
where
F

vel
=
ec
2
(V

R

)
3

V

R

and
F

rad
=
e
(V

R

)
2

V

R

e

V

R

(V

R

)
3

V

R

where we have used that V

V

=
2
c
2

2
v
2
=
2
c
2
(1 v
2
/c
2
) = c
2
.
5.2.5 Properties of the electromagnetic elds due to a moving charge
We can immediately work out some important properties of the velocity and radiative
elds. Consider
R

vel
=
ec
2
(V

R

)
3
R

V

= 0
30
Notes for Classical Field Theory Section 5: Solving Maxwells equations
where we have used that R

= 0 at the retarded time (remember that all our expressions

for the elds are evaluated at
0
), and
R

rad
=
eR

V

(V

R

)
2
+
e

V

R

V

(V

R

)
3
= 0
upon making a single fraction and switching some of the dummy variables to agree. Setting
in R

acc
to zero and j successively we nd
R

F
0
rad
= 0 R
i
F
i0
rad
= 0 R
i
E
i
rad
= 0 n

E
rad
= 0
R

F
j
rad
= R
0
F
0j
rad
+ R
i
F
ij
rad
= RE
j
+ R
i

ijk
B
k
= 0 E
j
=
jik
R
i
R
B
k

E = n

B
where n is a unit direction in the direction of

R. Note that this implies n,

E
rad
and

B
rad
are
mutually orthogonal and have the same magnitude (in Gaussian/Heaviside-Lorentz units).
We can similarly consider the dual tensor,

F

. For both F

vel
and F

rad
we see
every term in R

will contain

or

and hence contract

to zero. Thus,
R

= 0
and hence
n

B = 0

B = n

E
5.2.6 Local form of the electromagnetic elds due to a moving charge
To transform our covariant expressions for F

vel
and F

rad
to our local frame we recall that
V

= (c, v) = (c, c

), where v = c

B. We need to work out

V

=
d
d
V

. Now,
d
d
=
d
dt
, so
we have
d
d
c = c
d
dt
1

2
= c

2
3
= c
4

where =
d
dt

, and
d
d
c = c
d
dt
= c

d
dt
+
d
dt

= c

3
(

) +

= c
4
(

+ c
2

We also have that V

R

= Rc

V

R = cR(1 n

). Hence we have
E
i
vel
= F
i0
vel
=
ec
2
(V

R

)
3

R
i
V
0
R
0
V
i

=
ec
2

3
c
3
R
3
(1 n

)
3

cR
i
Rc
i

31
Notes for Classical Field Theory Section 5: Solving Maxwells equations
so

E
vel
=
e(n

2
R
2
(1 n

)
3

t
0

B
vel
= n

E
vel
Turning to the radiative elds, we have
E
i
rad
= F
i0
rad
=
e
(V

R

)
2

R
i

V
0
R
0

V
i

e

V

R

(V

R

)
3

R
i
V
0
R
0
V
i

so, writing

R = Rn,

3
c
3
R
3
(1 n

)
3
e

E
rad
= cR

1 n

4
cRn

c
4
(

+ c
2

4
cR

4
cR(

)n

2
cRn

cRn cR

=
5
c
2
R
2

1 n

(n

(1 n

)
n

=
5
c
2
R
2

2
(1 n

) + (n

(1 n

(1 n

) +
1

2
n

=
3
c
2
R
2

(n

)n (1 n

hence

E
rad
=
e
cR
n (n

) (1 n

)
(1 n

)
3

t
0

B
rad
= n

E
rad
We can use the vector identity a (

b c) =

b(a c) c(a

b) with a = n,

b = n

and c =
to see that
n ([n

] ) = n (n

) (1 n

)
so we can write the radiative electric eld as

E
rad
=
e
Rc
n ([n

] )
(1 n

)
3

t
0
32
Notes for Classical Field Theory Section 6: Power radiated by accelerating charge
6 Power radiated by accelerating charge
The essential idea is that the energy ux per unit area in the direction n is given by

S n
where

S =
c
4

E
rad

B
rad
is the Poynting vector. We have that

E
rad
is perpendicular to

B
rad
and n is perpendicular to both with

B
rad
= n

E
rad
, so
|n

S| =
c
4
|

E
rad
|
2
Now the dierential power radiated into a solid angle element d in the direction n is
dP = R
2
|n

S|d
This expression is in terms of the time t at the observation point; it is often more convenient
to work with the time t

in the charges own frames. We can write the energy radiated

between times t
1
and t
2
as
E =

t
2
t
1
|n

S|dt dR
2
=

2
t

1
|n

S|
dt
dt

dt

dR
2
so we see that we should dene
dP(t

) = R
2
|n

S|d
dt
dt

= R
2
c
4
|

E
rad
|
2
d
dt
dt

To evaluate the derivative, we use that

ct ct

= R
from the denition of the retarded time. As R = |

R| = |x x
e
(t

) we nd
dt
dt

1 =

R v
Rc

dt
dt

= 1 n

and hence
dP(t

)
d
= R
2
c
4
|

E
rad
|
2
(1 n

)
and using the expression for

E
rad
evaluated in the preceding section we have
dP(t

)
d
=
e
2
4c

n ([n

] )

2
(1 n

)
5
with this expression evaluated at the retarded time.
An important consequence of the electromagnetic radiation of an accelerating charged
particle is that classically an electron turning in circular motion will lose energy due to
radiation, leading to a decay of its orbit.
The remainder of this section of the course is not covered by these notes.
Because I do not hope to turn again
Because I do not hope
Because I do not hope to turn
T.S. Eliot, Ash Wednesday
33
Notes for Classical Field Theory Section 7: Bibliography
7 Bibliography
Obviously most of the material above was taken from my notes from Dr Buttimores lec-
tures and shamelessly L
A
T
E
Xed out under my own name. However I can claim full credit
for any mistakes that have appeared, and would appreciate any corrections/suggestions
to cblair[at]maths.tcd.ie.
The covariant derivation of the Lienard-Wiechart potentials was borrowed from Elec-
trodynamics by Fulvio Melia and Classical Electrodynamics by Jackson (note that the
rst method for obtaining the electromagnetic elds from the potentials is taken from
Jackson while the second was contributed to the 432 course by former students).
The Classical Theory of Fields by Landau and Lifshitz.
There are some good exam-oriented notes by Eoin Curran at http://peelmeagrape.
net/eoin/notes/fields.pdf.
The quotations throughout these notes were all mentioned in lectures by Dr Butti-
more; their precise relevance is left as an exercise for the reader. The text of Drac-
ula by Bram Stoker is available online at http://classiclit.about.com/library/
bl-etexts/bstoker/bl-bsto-drac-1.htm. The poems by Yeats and Eliot can pre-
sumably also be found online, or in any major collection of their work.
34