\documentstyle[12pt,psfig]{article}
\message{
 	Copyright, 1993, all rights reserved, Charlie Ebner (Dept.of Physics, 
 The Ohio State University, Columbus OH 43210) and Mark Jarrell (Dept.of 
 Physics, The University of Cincinnati, Cincinnati, OH 45221-0011).  This 
 material may not be reproduced for profit, modified or published in any 
 form (this includes electronic redistribution) without the prior written 
 permission of the authors listed above.	
}
\title{Maxwell's Equations}
\author{James Clerk Maxwell\\(1831 - 1879)}
\input{../defs.tex}
\def\baselinestretch{1.4}
\bdo
\maketitle

{\large{
\[
\div\Bxt=0, \;\;\;\;\;
\curl\Hxt=\frac{4\pi}c\Jxt+\frac1c\pde{\Dxt}t
\]
\[
\curl\Ext=-\frac1c\pde{\Bxt}t,\;\;\;\;\; \div\Dxt=4\pi\rhxt
\]
}}

\tableofcontents

\pagebreak

Our first task in this chapter is to put time into the equations of 
electromagnetism. There are traditionally two steps in this process. The 
first of them is to develop Faraday's Law of Induction which is the culmination 
of a series of experiments performed by Michael Faraday (1791 - 1867) around 
1830. Faraday studied the current ``induced'' in one closed circuit when a 
second nearby current-carrying circuit either was moved or had its current 
varied as a function of time. He also did experiments in which the second 
circuit was 
replaced by a permanent magnet in motion. The general conclusion of these 
experiments is that if the ``magnetic flux'' through a closed loop or 
circuit changes with time, an induced voltage or electromotive force 
(abbreviated by ${\cal E}mf$) appears in the circuit.
\section{Faraday's Law of Induction}
Define the {\em magnetic flux} through, or ``linking'' a closed loop C as
\beq
F=\ina\B\cdot\nn
\eeq 
\noindent where S is an open surface that ends on the curve C and $\nn$ is the usual 
unit right-hand normal (see below) to the surface. So long as $\div\B=0$, this 
integral 
is the same for all such surfaces. 

\centerline{\psfig{figure=fig1.ps,height=1.625in,width=4.25in}}

\noindent Define next the ${\cal E}mf$ in the loop as
\beq
{\cal E}=\invlc\cdot\E'
\eeq 
where $\E'$ is the electric field in that ``frame of reference'' in which 
the loop is at rest.\footnote{Notice that this definition cannot cover the 
case of a rotating loop since such a loop is not at rest in any one 
inertial (unaccelerated) frame.} The path C is traversed in such a 
direction that the unit normal $\nn$ is the right-hand normal relative to 
this direction. We may now write {\em Faraday's Law of Induction} as
\beq
{\cal E}=-k\der Ft
\eeq
where $t$ is the time and $k$ is a positive constant. 

Let us make two points in relation to this equation. 

{\bf{First}}, the minus sign 
has the consequence, when taken along with our definitions of the direction 
of $\nn$, etc., that the induced electromotive force will try to drive a 
current through the loop in such a direction as to produce a flux through 
the loop that will be opposite in sign to the change in $F$ that gave rise to 
the ${\cal E}mf$ in the first place. Thus if one tries to increase the flux $F$ 
through the loop by manipulating some external currents or magnets, the 
resulting induced current will produce a $\Bx$ which acts to counter the 
externally applied magnetic induction. This particular aspect of Faraday's 
Law is also known as {\em Lenz's Law}. 

\centerline{\psfig{figure=fig2.ps,height=1.25in,width=4.25in}}

{\bf{Second}}, in Gaussian units, the constant 
$k$ has dimensions of $T/L$ or inverse speed. It is in fact $1/c$ where $c$ 
is the constant that appears in Amp\`ere's Law and in the Lorentz force 
law. This is not a separate experimental fact but may be deduced from 
classical notions of relativity and the laws of electromagnetism as we 
currently understand them. We may demonstrate this claim; consider the time 
rate of change of flux through a loop C that is moving with some constant 
velocity $\v$ relative to the (lab) frame in which we are 
measuring $\x$ and $t$. 
We have
\beq
\der Ft=\frac{d}{dt}\lep\ina\Bxt\cdot\nn\rip=\ina\pde{\Bxt}{t}\cdot\nn+\lep
\der Ft\rip_2
\eeq

\centerline{\psfig{figure=fig3.ps,height=1.875in,width=6.375in}}

\noindent where the final term on the right accounts for the fact that 
the surface S over which we must integrate changes with time (it moves). 
To evaluate this term, we 
note that the distance the loop moves in time $dt$ is $\v dt$ and that S is 
displaced by the same amount. A point on S initially at $\x$ goes to 
$\x+\v dt$ in time $dt$, so if $\Bx$ is sampled before $dt$ elapses, then 
$\B(\x+\v dt)$ is sampled afterwards. We can expand the latter as

\beq
\B(\x+\v dt)=\Bx+dt(\v\cdot\grad)\Bx+...
\eeq
For an infinitesimal time element, we can ignore the higher-order terms. 
Hence, $dF_2$, which is the integral of the change $[\B(\x+\v dt)-\Bx]\cdot
\nn$ over the surface S (at time $t$) is easy to formulate:
\beq
\lep\der Ft\rip_2=\ina(\v\cdot\grad)[\Bx\cdot\nn].
\eeq
Now recall the vector identity $\curl(\v\times\B)=\v(\div\B)-(\v\cdot\grad)\B$.
If we use the law $\div\Bx=0$, we can use this identity to 
find that
\beq
\lep\der Ft\rip_2=-\ina\nn\cdot[\curl(\v\times\B)]=-\invlc\cdot(\v\times\B),
\eeq
the last step following from Stokes theorem. Using \eqs{4}{7} in 
Faraday's Law of Induction\footnote{$\der{F}{t}=-1/k\invlc\cdot\E'$}, 
we find
\beq
\invlc\cdot[\E'-k(\v\times\B)]=-k\ina\lep\pde{\B}{t}\cdot\nn\rip
\eeq
where $\E'$ is the electric field in the frame where the loop is at rest; 
this frame moves at velocity $\v$ relative to the one where $\B$ is 
measured (along with $\x$ and $t$).

Now consider a second circuit which is at rest in the frame where 
$\B$, $\x$, and $t$ are measured and which coincides with the first loop 
at the particular time $t$.

\centerline{\psfig{figure=fig3b.ps,height=1.875in,width=6.375in}}
 
\noindent For this loop Faraday's Law says that
\beq
\invlc\cdot\E=-k\ina\lep\pde{\B}{t}\cdot\nn\rip
\eeq 
where $\E$ is the electric field in the lab frame. Comparing 
\eqs{8}{9}, we see that
\beq 
\invlc\cdot[\E'-k(\v\times\B)]=\invlc\cdot\E.
\eeq
This relation tells us that the integrands are equal, give or take a vector 
field $\V$ which has the property that the line integral of its component 
along the line is zero when taken around the loop C. This condition plus 
the arbitrariness of C tells us the $\curl\V=0$. This field can depend on $
\v$, so we shall write it as $\V(\v)$. Taking these statements together, 
we have the relation
\beq
\E'=\E+k(\v\times\B)+\V(\v).
\eeq
This relation gives us the transformation of the electric field from the 
lab frame to the rest frame of the moving circuit.

There is a second way to get this transformation. Consider a charge $q$ 
following some trajectory $\x(t)$ under the influence of $\E$ and $\B$, as 
seen from the ``lab'' frame (the one where the unprimed quantities are 
measured). 

\centerline{\psfig{figure=fig4.ps,height=1.875in,width=6.375in}}
 
\noindent If at time $t$ the particle is at some point $\x$ and has velocity 
$\v$, the same as the velocity of the moving circuit, then if feels a force 
$\F$ given by
\beq
\F=q[\E+\frac1c(\v\times\B)].
\eeq
But in its rest frame, where the electric field is $\E'$, it feels a force
\beq
\F'=q\E'.
\eeq
Now, according to classical, or Galilean, relativity, $\F'$ is the same as 
$\F$, so, upon comparing the expressions for the force, we see that
\beq
\E'=\E+\frac1c(\v\times\B).
\eeq
Comparison with \eq{11} gives $k=1/c$ and $\V(\v)=0$. Using this result for 
$k$ in Faraday's Law, we find that it is fully specified by 
\beq
{\cal E}=-\frac1c\der Ft \hbox{ or }\invlc\cdot\Ex=-\frac1c\ina\lep
\pde{\B}{t}\cdot\nn\rip
\eeq
for a stationary path.\footnote{It should be remarked that all of this 
makes sense only to order $v/c$ since in the next order, $v^2/c^2$, 
Galilean relativity fails. However, the conclusion that $k=1/c$ must 
remain valid since $k$ is a constant (according to experiments), independent 
of the relative size of any velocities.} If we take the integral relation 
and apply Stokes theorem, we find
\beq
\ina\lep\curl\E+\frac1c\pde{\B}{t}\rip\cdot\nn=0.
\eeq
Because S is an arbitrary open surface, this relation must be true for all 
such surfaces. That can only be if the integrand is everywhere zero, 
leading us to the differential equation
\beq
\curl\Ext=-\frac1c\pde\Bxt t
\eeq
which is the differential equation statement of Faraday's Law.
\section{Energy in the Magnetic Field}
Given Faraday's Law, we are in a position to calculate the energy required 
to produce a certain current distribution $\J$ starting from a state with 
$\J=0$ even though we do not as yet know all of the time-dependent terms in 
the field equations. In this section, we shall determine what is this 
energy. The mechanism that requires work to be done is as follows: If we 
attempt to make a change in any existing current distribution, there will 
be time-dependent sources (the current) with an accompanying time-dependent 
magnetic induction. The latter must in turn produce electromagnetic forces, or 
electric fields, against which work must be done not only in order to 
change the currents but also simply in order to maintain them. By examining 
this work, we can determine the change in the ``magnetic energy'' of the 
system. 

To get started, consider a single loop or circuit carrying current $I_1$. 
Given a changing flux $dF_1/dt$ through this loop, there is an
${\cal E}_1=-c^{-1}dF_1/dt$. If we wish to maintain the current in the face of 
this electromotive force, we must counter the latter by introducing an 

\centerline{\psfig{figure=fig5.ps,height=3.25in,width=4.25in}}
 
\noindent external agent which maintains a voltage 
$V_1=-{\cal E}_1=\frac{1}{c}\der{F_1}{t}$ 
around the loop. 
This agent thus does work at a rate $V_1I_1=-{\cal E}_1I_1$. The total work 
that it does is
\beq
\de W_1=-\int dtI_1{\cal E}_1=\frac{I_1}{c}\int dt\der{F_1}t=\frac{I_1
\de F_1}{c} 
\eeq
where $\de F_1$ is the total change of the flux through the loop. We may 
also express our result as
\beq
\de W_1=\frac{I_1}c\ina(\de \B\cdot\nn)
\eeq
where S and $\nn$ are related to the loop in the usual way, and $\de \B$ is 
the change in the magnetic induction. If we write $\de\B=\curl\de\A$, which 
is possible so long as $\div\de\B=0$, we have
\beq
\de W_1=\frac{I_1}c\ina(\curl\de\A)\cdot\nn=\frac{I_1}c\invlc\cdot\de\A
\eeq
where Stokes theorem has been invoked.

Now let us generalize. A current distribution $\J$ can be thought of as the 
sum of many infinitesimal loops. Given many such loops, the total work 
done by the external agent(s) will be the sum of the works done on each of the 
loops. The relation between the current in a loop, $I$, and $\J$ is 
$|\J|(d\si)$ where $d\si$ is the (infinitesimal) cross-sectional area of 
the loop. Thus $Id\l\rightarrow\J d\si dl$ or $\J d^3x$. Integrating over 
individual loops and summing over all loops is equivalent to integrating over 
all space, so we find that the change in energy accompanying a change $\de\Ax$ 
in the vector potential (reflecting an infinitesimal change $\de\Bx$ in the 
magnetic induction) is
\beq
\de W=\frac1c\iniv[\Jx\cdot\de\Ax].
\eeq
The change in the magnetic induction has to be infinitesimal for this 
expression to be valid because we did not ask how the current density must 
be changed to produce it.  Note that this form indicates that $W$ is a 
natural thermodynamic function function of the potential or magnetic
flux, rather than the sources

Now let us write the current density in terms of the fields. If we think we 
are doing macroscopic electromagnetism, then\footnote{This relation is only 
true for static phenomena; since we are changing the fields with time, it 
is not valid. However, if the change is accomplished sufficiently slowly
that $\pde{\D}{t}$ may be neglected, 
then the corrections to Amp\`ere's Law are so small as to have negligible 
influence on our argument. Notice too that use of this equation 
demands that changes in $\Jx$ be done in such a way that $\div\J=0$.} 
$\J=c(\curl\H)/4\pi$ and we can proceed as follows:
\beqa
\de W=\frac1{4\pi}\inv[(\curl\H)\cdot\de\A]=\frac1{4\pi}\inv[\div(\H\times
\de\A)+\H\cdot(\curl\de\A)]\nonumber\\
=\frac1{4\pi}\inac(\H\times\de\A)\cdot\nn+\frac1{4\pi}\inv(\H\cdot\de\B)
\nonumber\\=\frac1{4\pi}\iniv(\H\cdot\de\B).
\eeqa
The final step in this argument is achieved by letting the domain of 
integration be all space and assuming that the fields $\H$ and $\de\A$ fall 
off fast enough far away that the surface integral vanishes. This is in 
fact true for a set of localized sources in the limit that changes are made 
very slowly.

In the final step of our derivation we want to integrate $\de\B$ up to some 
final $\B$ starting from zero magnetic induction or zero current density. 
We can only do this functional integral if we know how $\H$ depends on 
$\B$. For a {\bf linear} medium or set of media, we know that
\beq
\H\cdot\de\B=\frac12\de(\H\cdot\B)
\eeq
and so 
\beq
\de W=\frac1{8\pi}\iniv\de(\B\cdot\H)
\eeq
which integrates to
\beq
W=\frac1{8\pi}\iniv\Bx\cdot\Hx
\eeq
provided we define $W\equiv0$ in the state with $\B\equiv0$.

Our result, \eq{25}, can be written in other forms; a particularly useful 
one is obtained by writing $\B=\curl\A$ and doing a parts integration in 
the familiar way. Assuming that one can discard the resulting surface term
(valid for localized sources), we find
\beq
W=\frac1{2c}\iniv\Jx\cdot\Ax.
\eeq
In analogy with the electrostatic case, one conventionally defines the magnetic
energy density to be
\beq
w\equiv\frac1{8\pi}(\Hx\cdot\Bx). 
\eeq

\subsection{Example: Motion of a permeable Bit in a Fixed $\J$}

Let us look at a specific example of the use of the expression(s) for 
the energy. Suppose that there is some initial current distribution $\J_0$ 
which produces fields $\H_0$ and $\B_0$ and energy $W_0$. Then we have
\beq
W_0=\frac1{8\pi}\iniv\B_0\cdot\H_0=\frac1{2c}\iniv\J_0\cdot\A_o.
\eeq
Now move some permeable materials around without changing the macroscopic 
current density $J_0$. 

\centerline{\psfig{figure=fig6.ps,height=1.5in,width=4.25in}}
 
\noindent After doing so we have a new energy which is
\beq
W_1=\frac1{8\pi}\iniv\B_1\cdot\H_1=\frac1{2c}\iniv\J_0\cdot\A_1
\eeq
where the new fields, energy, etc., are designated by the subscript `1'.
The change in energy may be written as
\beqa
W_1-W_0=\frac1{2c}\iniv\J_0\cdot(\A_1-\A_0)=\frac1{8\pi}\iniv
[(\curl\H_0)\cdot\A_1-(\curl\H_1)\cdot\A_0]\nonumber\\
=\lep\frac1{8\pi}\iniv\div[\H_0\times(\A_1-\A_0)]+\iniv
[\H_0\cdot(\curl\A_1)-\H_1\cdot(\curl\A_0)]\rip\nonumber\\
=\frac1{8\pi}\lep\inac\nn\cdot[\H_0\times(\A_1-\A_0)]+\iniv[\H_0
\cdot\B_1-\H_1\cdot\B_0]\rip
\eeqa
The surface term may be discarded for localized sources. Assuming further 
that all materials are isotropic so that $\B=\mu\H$, we find
\beq
W_1-W_0=\frac1{8\pi}\iniv(\mu_1-\mu_0)(\H_0\cdot\H_1)
\eeq
where $\mu_1$ is the final value of the permeability (a function of 
position) and $\mu_0$ is the initial value. If in addition $\mu_0=1$, a value 
appropriate for non-permeable materials or empty space, and $\mu_1\ne1$ only 
in some particular domain $V$, then
\beq
W_1-W_0=\frac12\inv(\M_1\cdot\B_0)
\eeq
where we have made use of the facts that $\M_1=(\mu_1-1)\H_1/4\pi$ and
$\B_0=\H_0$.

It is instructive (maybe it's just confusing) to compare the answer with 
other things that we have seen. First, we calculated previously the change 
in the electrostatic energy when a piece of dielectric is introduced into a 
previously empty space in the presence of some fixed charges. We found 
\beq
W_1-W_0=-\frac12\inv\P_1\cdot\E_0.
\eeq 
The difference in sign between the electrostatic and magnetostatic energies 
is a reflection of the fact that in the magnetic system we maintained fixed 
currents and in the electrostatic system, fixed charges. In the former case 
external agents must do work to maintain the currents, and in the latter one, 
no work need be done to maintain the fixed charges. Hence in the latter 
case, the force on the dielectric may be found by using the argument that 
in a conservative system, the force is the negative gradient of the energy  
(which is given above) with respect to the displacement of the dielectric. 
{\bf If}, in the magnetic system, the work done on 
the system by the current-maintaining external agents turns out to be 
precisely twice as large as the change in the system's energy, then the 
force on the permeable material will be the (positive) gradient of the 
energy with respect to the displacement of the material. This is, in fact, 
the case, as we shall see below.

\subsection{Energy of a Current distribution in an External Field}
As a second example, consider the energy of a current distribution in 
an external field.
\beq
W=\frac1{c}\iniv\Jx\cdot\Ax.
\eeq
where $\Ax$ is due to sources other than $\Jx$, which do not overlap the
region where $\Jx$ is finite.  (note the lack of a factor of $\frac12$, why?)

\centerline{\psfig{figure=fig7.ps,height=1.25in,width=4.25in}}
 
	Now assume that $\Ax$ changes little over the region where $\Jx$
is finite, and thus expand $\Ax$ about the origin of the current distribution.
\[
\Ax=\A(0) + \left.\x\cdot\grad \Ax\rib_{\x=0}  +\cdots
\]
through the now familiar manipulations, we end up with
\[
W=\mdm\cdot\B(0) + \cdots
\]
where $\mdm$ is the dipole moment of the current distribution

	This appears to make no sense at all!!  Recall last quarter, we
found that the force on a permanent dipole is
\[
\F=\grad(\M\cdot\B_0)
\]
which acts to minimize the potential $-\mdm\cdot\B$, rather than $+\mdm\cdot\B$!

\centerline{\psfig{figure=fig8.ps,height=1.25in,width=4.25in}}
 
Consider a current loop
Clearly the force tries to make $\mdm$ and $\B$ parallel.  Have we misplaced
a ``-''-sign?  Our expression for $W$ is correct; however, we do not
have a conservative situation, since in calculating the force on the loop,
we assumed that the current $I$ is constant.  However, rotating the
loop changes the flux through it which induces an ${\cal E}mf$
which opposes the changing flux.  Thus $I$ will not be constant unless
external work is done to make it so.  In such a non-conservative
situation, the force is not the negative gradient of the energy.  For this
particular situation, it must be that the the force is given by the
{\em{positive}} gradient of the energy.

To demonstrate that the last statement above is correct, we consider a set of 
circuits $C_i$, $i=1,2,...,n$, with currents $I_i$. The energy of this system is
\beqa
W=\frac1{2c}\iniv\J\cdot\A=\frac1{2c}\sum_iI_i\oint_{C_i}d\l\cdot\A
=\frac1{2c}\sum_i\int_{S_i}d^2x\,\nn\cdot(\curl\A)\nonumber\\
=\frac1{2c}\sum_iI_i\int_{S_i}d^2x\,(\nn\cdot\B)=\frac1{2c}\sum_iI_iF_i
\eeqa
where $F_i$ is the flux linking the $i^{th}$ circuit.

Now suppose that a piece of permeable material moves through the system at 
velocity $\v$. 

\centerline{\psfig{figure=fig9.ps,height=1.875in,width=6.375in}}
 
\noindent It will experience a force which we may calculate by demanding 
total energy conservation. The various energies that must be considered are
as follows: 
\begin{enumerate}
\item The energy transferred to the moving object from the magnetic 
field. In time $dt$, this energy is
\beq
dW_m=\F\cdot\v dt=F_\et vdt
\eeq 
given that $\v$ is in the $\et$-direction. 
\item The field energy; for fixed 
currents in the circuits, this energy changes by 
\beq
dW=\frac1{2c}\sum_iI_idF_i
\eeq
in time $dt$. 
\item The energy transferred to the magnetic field (or 
circuits) by some external agents whose business it is to maintain the 
currents. The ${\cal E}mf$'s in the circuits are ${\cal E}_i=-c^{-1}dF_i/dt$ 
and so the external agents do work on the $i^{th}$ circuit at a rate
$-{\cal E}_iI_i$ in order to maintain the currents. The work done on these 
external agents in time $dt$ is therefore
\beq
dW_e=\sum_i{\cal E}_iI_idt=-\frac1c\sum_iI_idF_i=-2dW
\eeq
\end{enumerate}

Now invoke conservation of energy which demands that the sum of all of the 
preceding infinitesimal energy changes must be zero
\beq
dW_m+dW+dW_e=0
\eeq
or, using \eqs{36}{38}, 
\beq
F_\et vdt=dW.
\eeq
Further, $vdt=d\et$, so we find that
\beq
F_\et=+\lep\pde W\et\rip_\J
\eeq
as suggested earlier. 

What would the force be if the flux $F$ was held constant?

\section{Maxwell's Displacement Current; Maxwell's Equations}
Let us summarize the equations of electromagnetism as we now have them:
\beqa
\div\B=0\nonumber\\
\div\D=4\pi\rh\nonumber\\
\curl\H=\frac{4\pi}c\J\nonumber\\
\curl\E=-\frac1c\pde\B t.
\eeqa
These are not internally consistent. Consider the divergence of Amp\`ere's 
Law:
\beq
\div(\curl\H)=0=\frac{4\pi}c(\div\J)
\eeq
which implies that $\div\J=0$. We know, however, that for general 
time-dependent phenomena the divergence of the current density is not 
necessarily zero; indeed, we have seen that charge conservation requires
\beq
\div\J+\pde\rh t=0.
\eeq
In consequence of this requirement, we must, at the very least, add a
(time-dependent) term to Amp\`ere's Law which will give, in distinction to 
\eq{43},
\beq
\div(\curl\H)=\frac{4\pi}c\leb\div\J+\pde\rh t\rib.
\eeq
This new term which must be a vector field $\X$, has to be such that
\beq
\div\X=\frac{4\pi}c\pde{\rh}t.
\eeq
We need not look beyond the things we have already learned to find a 
plausible candidate for this term. We have a (static) equation which reads
\beq
\rh=\frac1{4\pi}\div\D;
\eeq
if we accept this as correct, we have 
\beq
\frac{4\pi}c\pde\rh t=\frac1c\frac{\partial}{\partial t}\div\D=\div\lep\frac1c
\pde\D t\rip.
\eeq

The simplest possible resolution of the inconsistency in the field 
equations is thus to choose $\X$ to be $c^{-1}\partial\D/\partial t$, which 
would turn Amp\`ere's Law into the relation
\beq
\curl\H=\frac{4\pi}c\J+\frac1c\pde\D t.
\eeq
This adjustment was made\footnote{He delivered a paper containing this 
statement in 1864, but he'd had the idea at least as early as 1861.} by 
J.~C.~Maxwell in 1864, and the resulting set of differential field 
equations has since become known as {\em Maxwell's Equations}:
\beq
\div\Bxt=0,
\eeq
\beq
\div\Dxt=4\pi\rhxt,
\eeq
\beq
\curl\Ext=-\frac1c\pde{\Bxt}t,
\eeq
and
\beq
\curl\Hxt=\frac{4\pi}c\Jxt+\frac1c\pde{\Dxt}t.
\eeq
The term that Maxwell added was called by him the {\em displacement 
current} $\J_D$:
\beq
\J_D\equiv\frac1{4\pi}\pde{\D}t. 
\eeq
This term has the appearance of an additional current entering 
Amp\`ere's Law so that the latter reads
\beq
\curl\H=\frac{4\pi}c(\J+\J_D).
\eeq
Notice that $\div(\J+\J_D)=0$. We shall not emphasize the ``current'' 
interpretation of $\J_D$ because it is misleading; the displacement current 
does not describe a flow of charge and is not a true current density.

We close this section with two comments. {\bf{First}}, the Maxwell 
equations must 
be regarded as empirically justified. In the years since Maxwell's final 
adjustment of the field equations of electromagnetism, they have been 
subjected to extensive experimental tests and have been found to be correct 
for classical phenomena (no quantum effects or general relativistic 
effects); with proper interpretation, they even have considerable validity 
within the realm of quantum phenomena. {\bf{Second}}, the equations as we have 
written them are for macroscopic electromagnetism. The more fundamental 
version of these equations has $\B$ in place of $\H$ and $\E$ in place of 
$\D$; then the sources $\rh$ and $\J$ are the total charge and current 
densities.
\section{Vector and Scalar Potentials}
For time-dependent phenomena, which are fully described by the Maxwell 
equations, one can still write the fields $\E$ and $\B$ in terms of a 
scalar potential and a vector potential. Because the divergence of the 
magnetic induction is still zero, one continues to be able to find a vector 
potential $\Axt$ which has the property that
\beq
\Bxt=\curl\Axt;
\eeq
this potential is not unique. Further, since the curl of the electric field 
is not zero in general, we cannot write $\Ext$ as the gradient of a scalar; 
however, from Faraday's Law, \eq{52}, and from \eq{56}, we have
\beq
\curl\lep\Ext+\frac1c\pde{\Axt}t\rip=0
\eeq
and so we can write the combination of fields in parentheses as the 
gradient of a scalar function $\Phxt$:
\beq
\Ext+\frac1c\pde{\Axt}t=-\grad\Phxt
\eeq
or
\beq
\Ext=-\grad\Phxt-\frac1c\pde{\Axt}t.
\eeq
Equations (56) and (59) tell us how to find $\B$ and $\E$ from potentials; 
these potentials must themselves satisfy certain field equations that can 
be derived from the Maxwell equations involving the sources $\rh$ and $\J$; 
the ``homogeneous'' or source-free equations $\div\B=0$ and $\curl\E
=c^{-1}\partial\B/\partial t$ are then automatically satisfied. Letting $\H=
\B$ and $\D=\E$ for simplicity, we have, upon substituting \eqs{56}{59} 
into \eqs{51}{53}, 
\beq
\lap\Ph+\frac1c\frac\partial {\partial t}(\div\A)=-4\pi\rh
\eeq
and
\beq
\curl(\curl\A)+\frac1{c^2}\spd{\A}t+\grad\lep\frac1c\pde{\Ph}t\rip=\frac
{4\pi}c\J.
\eeq
These equations clearly do not have particularly simple pleasing or 
symmetric forms. However, we have some flexibility left in the choice of 
the potentials because we can choose the vector potential's divergence in 
an arbitrary fashion.

\section{Gauge Transformations}

Suppose that we have some $\A$ and $\Ph$ which give $\E$ and $\B$ 
correctly. Let us add to $\A$ the gradient of a scalar function $\chxt$, 
thereby obtaining $\A'$:
\beq
\A'\equiv\A+\grad\ch.
\eeq
The field $\A'$ has the same curl as $\A$, and hence $\B=\curl\A'$. 
However, the electric field is {\bf not} given by $-\grad\Ph-c^{-1}\partial
\A'/\partial t$; we must therefore change $\Ph$ to $\Ph'$ where $\Ph'$ is 
chosen to that
\beq
\E=-\grad\Ph'-\frac1c\pde{\A'}t.
\eeq
To this end we consider $\Ph'=\Ph+\ps$ where $\ps$ is some scalar function 
of $\x$ and $t$. Our requirement, \eq{63}, is that
\beq
\E=-\grad\Ph-\grad\ps-\frac1c\pde{\A}t-\frac1c\pde{(\grad\ch)}t.
\eeq
However, we know that $\E$ is given according to \eq{59}, so, combining 
this relation and \eq{64}, we find that $\ps$ must satisfy the equation
\beq
\grad\ps=-\frac1c\pde{(\grad\ch)}t.
\eeq
A clear possible choice of $\ps$ is $\ps=-c^{-1}\partial\ch/\partial 
t$.

What we have learned is that, given potentials $\A$ and $\Ph$, we may make 
a {\em gauge transformation} to equally acceptable potentials $\A'$ and 
$\Ph'$ given by
\beqa
\A'=\A+\grad\ch\nonumber\\
\Ph'=\Ph-\frac1c\pde{\ch}t
\eeqa
where $\chxt$ is an arbitrary scalar function of position and time.

\subsection{Lorentz Gauge}

Now let's look again at the differential equations, \eqs{60}{61}, for the 
potentials; these may be written as 
\beqa
\lap\Ph+\frac1c\frac{\partial}{\partial t}(\div\A)=-4\pi\rh\nonumber\\
\lap\A-\frac1{c^2}\spd{\A}t-\grad\lep\div\A+\frac1c\pde{\Ph}t\rip
=-\frac{4\pi}c\J
\eeqa
where we have used the identity $\curl(\curl\A)=\grad(\div\A)-\lap\A$. 
These have a much more pleasing form if $\A$ and $\Ph$ satisfy the {\em 
Lorentz condition}
\beq
\div\A+\frac1c\pde{\Ph}t=0.
\eeq
Supposing for the moment that such a choice is possible, we make use of 
\eq{68} in \eq{67} and find
\beqa
\lap\Ph-\frac1{c^2}\spd{\Ph}t=-4\pi\rh\nonumber\\
\lap\A-\frac1{c^2}\spd{\A}t=-\frac{4\pi}c\J.
\eeqa
These are very pleasing in that there are distinct equations for $\Ph$ and
$\A$, driven by $\rh$ and $\J$, respectively; furthermore, all equations 
have the form of the classical wave equation,
\beq
\dal\psxt=-4\pi\fxt
\eeq
where
\beq
\dal\equiv\lap-\frac1{c^2}\spd{}t
\eeq
is known as the {\em D'Alembertian} operator. 

Consider next whether it is generally possible to find potentials that 
satisfy the Lorentz condition. Suppose we have some potentials $\A_0$ and $
\Ph_0$ for a given set of sources $\J$ and $\rh$. We make a gauge 
transformation to new potentials $\A$ and $\Ph$,
\beqa
\A=\A_0+\grad\ch\nonumber\\
\Ph=\Ph_0-\frac1c\pde{\ch}t,
\eeqa
where the gauge function $\ch$ is to be chosen so that the Lorentz 
condition is satisfied by the new potentials. The condition on $\ch$ is 
thus
\beq
0=\div\A+\frac1c\pde\Ph t=\div\A_0+\lap\ch+\frac1c\pde{\Ph_0}t
-\frac1{c^2}\spd\ch t,
\eeq
or
\beq
\lep\lap-\frac1{c^2}\frac{\partial^2}{\partial t^2}\rip\ch=-\div\A_0
-\frac1c\pde{\Ph_0}t.
\eeq
The function that we seek is thus itself a solution of the classical wave 
equation with a ``source'' which is, aside from some constant factor,
just $\div\A_0+c^{-1}(\partial\Ph_0/\partial t)$. Such a function always 
exists. In fact, there are many solutions to this wave equation which means 
that there are many sets of potentials $\Ph$ and $\A$ which satisfy the 
Lorentz condition. Potentials satisfying the Lorentz condition are said to 
be in the {\em Lorentz gauge}.

\subsection{Coulomb Transverse Gauge}

Another gauge which can be useful is the {\em Coulomb} or {\em transverse 
gauge}. It is defined by the condition that $\div\A=0$. The beauty of this 
gauge is that in it the scalar potential satisfies the Poisson equation,
\beq
\lap\Phxt=-4\pi\rhxt.
\eeq
We know the solution of this equation:
\beq
\Phxt=\inivp\frac{\rh(\xp,t)}{\xxpa}.
\eeq
Notice that the time is the same at both the source point $\xp$ and the 
field point $\x$. There is nothing unacceptable about this because the 
scalar potential is not a measurable quantity.

The vector potential in the Coulomb gauge is less satisfying; it obeys the 
wave equation
\beq
\dal\A=-\frac{4\pi}c\J+\frac1c\frac{\partial}{\partial t}(\grad\Ph).
\eeq

In practice one may solve for the potentials in the transverse gauge, given 
the sources $\rh$ and $\J$, by first finding the scalar potential from the 
integral \eq{76} and then using the result in \eq{77} and solving the wave 
equation (see the following section). One should not be surprised to learn 
that the ``source'' term in \eq{77} involving the scalar potential can be 
made to look like a current; consider
\beqa
\frac1c\frac{\partial}{\partial t}(\grad\Ph)=\frac1c\frac{\partial}{
\partial t}\leb\grad\lep\inivp\frac{\rh(\xp,t)}{\xxpa}\rip\rib=
-\grad\inivp\frac{\divp\J(\xp,t)}{\xxpa}\nonumber\\
=-\frac{4\pi}c\lep\frac1{4\pi}\grad\inivp\frac{\divp\J(\xp,t)}{\xxpa}\rip
\eeqa
The negative of the quantity within the parentheses (...) is called the 
{\em longitudinal current density}. More generally, the {\em longitudinal} and 
{\em transverse} components $\J_l$ and $\J_t$ of a vector field such as $\J$ 
are defined by the conditions\footnote{In the appendix it is shown that
such a decomposition, into longitudinal and transverse parts, of a vector 
function of position is always possible}
\beq
\J_l+\J_t=\J,\hbox{ }\curl\J_l=0,\hbox{ and }\div\J_t=0.
\eeq
In other words, $J_l$ and $J_t$ satisfy the equations
\beq
\barr{cc}
\div\J_l=\div\J & \div\J_t=0\\
\curl\J_l=0 & \curl\J_t=\curl\J,\ear
\eeq
which means that $\J_l$ carries all of the divergence of the current 
density and $\J_t$ has all of the curl.

From these equations and our knowledge of electrostatics
we can see immediately that
\beq
\J_l(\x,t)=-\frac1{4\pi}\grad\lep\inivp\frac{\divp\J(\xp,t)}{\xxpa}\rip
\eeq
which may also be written as
\beq
\J_l(\x,t)=-\frac1{4\pi}\grad\lep\div\iniv\frac{\J(\xp,t)}{\xxpa}\rip,
\eeq
and from that of magnetostatics we see that
\beq
\J_t(\x,t)=\curl\lep\frac1{4\pi}\inivp\frac{\curlp\J(\xp,t)}{\xxpa}\rip
\eeq
which may also be written as
\beq
\J_t(\x,t)=\curl\leb\frac1{4\pi}\curl\lep\inivp\frac{\J(\xp,t)}{\xxpa}\rip
\rib.
\eeq
Comparison of \eq{81} with \eqs{77}{76} shows that 
the wave equation for $\A$ in the Coulomb or transverse gauge is
\beq
\dal\A=-\frac{4\pi}c(\J-\J_l)=-\frac{4\pi}c\J_t.
\eeq
This equation lends justification to the term ``transverse gauge.'' In this 
gauge, the vector potential is driven by the transverse part of the current 
in the same sense that the vector potential is driven by the entire current 
in the Lorentz gauge.

\section{Green's Functions for the Wave Equation}

We're going to spend quite a lot of time looking for solutions of the 
classical wave equation
\beq
\dal\psxt=-4\pi\fxt;
\eeq
Therefore, it could be useful to have Green's functions for the 
D'Alembertian operator, meaning functions $\gxt$ which satisfy the equation
\beq
\dal\gxt=-4\pi\dexxp\dettp
\eeq
subject to some boundary conditions.

Compare this equation with that for the Green's function for the Laplacian 
operator, $\lap\gxxp=-4\pi\dexxp$. The solution for the latter is, as we 
know, $\gxxp=1/\xxpa$ in an infinite space; physically, it is the scalar 
potential at $\x$ produced by a unit point charge located at $\xp$. The 
function $\gxt$, by contrast has dimensions $1/LT$ and may be thought of as 
the ``response'' at the space-time point $(\x,t)$ to a unit ``source'' at 
the space-time point $(\xp,t')$, that is, to a source which exists only for 
a moment at a single space point. If this source is a pulse of current or 
an evanescent point charge, then the Green's function would be the vector 
or scalar potential in the Lorentz gauge produced by that pulse.

As for the question of boundary conditions, we shall keep things as simple 
as possible by considering an infinite system and applying the boundary 
condition that $\gxt\rightarrow0$ for $|\x|\rightarrow\infty$. There is 
also a boundary condition on the behavior of $\gxt$ as a function of $t$; 
this is usually called an initial condition. We shall not make any specific 
statement now of the initial conditions but will keep them in mind as our 
derivation progresses.

We shall solve the wave equation for $\gxt$ by appealing to Fourier 
analysis. Assuming that the Fourier transform $q(\k,\om)$ of a function $\fxt$ 
exists, then one can write that function in terms of the transform as
\beq
\fxt=\inko\ekxot\qko.
\eeq
This expression may be thought of as an expansion of $\fxt$ using basis 
functions $\ekxot$. These form a complete orthonormal set; the 
orthonormality condition is
\beq
\inxt e^{-i(\kp\cdot\x-\om't)}\ekxot=(2\pi)^4\dekkp\deoop
\eeq
and the completeness relation is much the same,
\beq
\inko e^{-i(\k\cdot\xp-\om t')}\ekxot=(2\pi)^4\dexxp\dettp.
\eeq
If we substitute this last relation into the right-hand side of \eq{87} 
and also expand $\gxt$ on the  left-hand side as\footnote{From \eq{85} one may i
infer that $\gxt$ can be written as a function of $\x-\xp$ and of $t-t'$ in 
an infinite space.}
\beq
\gxt=\inko\gko\ekdod,
\eeq
then one finds that
\beqa
\dal\gxt=\inko\lep-k^2+\frac{\om^2}{c^2}\rip\gko\ekdod\nonumber\\
=-4\pi\dexxp\dettp=-\frac{4\pi}{(2\pi)^4}\inko\ekdod.
\eeqa
The orthonormality of the basis functions may now be used to argue that the 
integrands on the two sides of this equation must be equal; hence,
\beq
\gko=\frac{4\pi}{(2\pi)^4}\frac1{k^2-\om^2/c^2}.
\eeq
Now we have only to do the Fourier transform to find $\gxt$. Consider first 
the frequency integral,
\beq
I(\k,t-t')=\int_{-\infty}^\infty d\om\lep\frac{e^{-i\om(t-t')}}{k^2-\om^2/
c^2}\rip.
\eeq
This integral can have different values depending on how we handle the 
poles in the integrand at $\om=\pm ck$. Let's see what are the 
possibilities: If $t<t'$, we can extend the path of integration at
$|\om|\rightarrow\infty$ around a semi-circle in the upper half-plane where 
$\Re\;\om>0$ without getting an additional non-zero contribution to the 
integral. 

\centerline{\psfig{figure=fig10.ps,height=1.875in,width=6.375in}}
 
\noindent Hence, in this case the integral in \eq{94} is just $2\pi 
i$ times the sum of the residues of the poles of the integrand in the upper 
half-plane. The only poles in the integrand as it stands are the ones at
$\om=\pm ck$. The path specified in the integral in fact runs right across 
both of them.
However, if we make infinitesimal deformations of the path so 
that the poles are either inside of or outside of the contour, we will 
still obtain a function $\gxt$ which is a solution of \eq{87}. Thus the 
solution that we are obtaining is not unique.

Why not? The answer lies in the fact that we are dealing with a 
differential equation which is second-order in time, and there are two 
independent solutions which we can discriminate by specifying some initial 
condition. Consider what happens if we deform the contour so as to go just 
above the poles on the real-$\om$ axis. Then, for $t<t'$, we close the 
contour in the upper half-plane and find that the integral around the 
closed contour is zero (there are no poles inside of it). 
This means that 
$\gxt=0$ for $t<t'$. But if $t>t'$, then we must close the contour in the 
lower half-plane and will find that $\gxt\ne0$. This Green's function is 
called the {\em retarded Green's function} and its essential character is 
that it is non-zero only for times $t$ which are {\bf later} than the time 
$t'$ of the source pulse which ``produces'' it.
Instead of deforming the 
contour, we may change the integrand in such a way as to put the poles an 
infinitesimal distance below the real-$\om$ axis and so have
\beq
I(\k,t-t')=\int_{-\infty}^\infty d\om\lep\frac{e^{-i\om(t-t')}}{k^2-(\om+i
\ep)^2/c^2}\rip
\eeq
where the integral is to be evaluated in the limit that $\ep$ is a positive 
infinitesimal constant. In this way we obtain a Green's function that 
vanishes at all times $t$ earlier than the time $t'$.

\centerline{\psfig{figure=fig11.ps,height=1.875in,width=6.375in}}
 
The second possibility is that we displace the poles so that they lie 
slightly above the real-$\om$ axis. Then we will get a Green's function 
that will vanish for all $t>t'$ but not for times $t$ earlier than $t'$; 
this one is called the {\em advanced Green's function} and to have an 
expression for it, we need only change $+\ep$ to $-\ep$ in \eq{95}.

Now that we have determined how to handle the poles in the integrand, let us 
proceed with the evaluation of the retarded Green's 
function in particular. For $t<t'$, it is identically zero, and for $t>t'$, 
we close the contour in the lower half-plane and find
\beqa
I(\k,t-t')=-c^2\oint_Cd\om\frac{e^{-i\om(t-t')}}{[\om-(ck-i\ep)][\om
-(-ck-i\ep)]}\nonumber\\=-c^2(-2\pi i)\leb\frac{e^{-ick(t-t')}}{2ck}
+\frac{e^{ick(t-t')}}{-2ck}\rib=\frac{2\pi c}k\sin[ck(t-t')].
\eeqa
Now we may complete the determination of $\gxt$ by doing the integration 
over the wavevector:
\beq
G(R,\ta)=\frac1{4\pi^3}\inik\frac{2\pi c}k\sin(ck\ta)e^{i\k\cdot\R}
\eeq
where $\ta\equiv t-t'$ and $\R\equiv\xxp$.
Thus,
\beqa
G(R,\ta)=\frac c{2\pi^2}\int_0^\infty kdk\int_{-1}^1du\int_0^{2\pi}d\ph
\sin(ck\ta)e^{ikRu}\nonumber\\ =\frac c\pi\int_0^\infty 
kdk\lep\frac{e^{ikR}-e^{-ikR}}
{ikR}\rip\sin(ck\ta)\nonumber\\=-\frac c{4\pi R}\int_{-\infty}^\infty dk
\lep e^{ikR}-e^{-ikR}\rip\lep e^{ick\ta}-e^{-ick\ta}\rip\nonumber\\
=-\frac c{2R}[\de(R+c\ta)+\de(-R-c\ta)-\de(R-c\ta)-\de(-R+c\ta)]=\frac{\de(\ta
-R/c)}R;
\eeqa
the final step here follows from the fact that this expression is only 
correct for $\ta>0$; for $\ta<0$ $G\equiv0$.
In more detail, our result for the retarded Green's function is
\beq
G^{(+)}(\x,t;\xp,t')=\frac{\de(t-t'-\xxpa/c)}\xxpa.
\eeq

By similar manipulations one may show that the advanced Green's function is
\beq
G^{(-)}(\x,t;\xp,t')=\frac{\de(t'-t-\xxpa/c)}\xxpa.
\eeq

Given the appropriate Green's function, we can write down a solution to the 
classical wave equation with matching initial conditions and the 
appropriate boundary conditions as $|\x|\rightarrow\infty$. Suppose that 
the wave equation is
\beq
\dal\psxt=-4\pi\fxt;
\eeq
the solution is
\beq
\psxt=\inxtp\gxt\fxtp
\eeq
as may be seen by operating on this equation with $\dal$:
\beqa
\dal\psxt=\inxtp\dal\gxt\fxtp\nonumber\\=\inxtp[-4\pi\dexxp\dettp]\fxtp
=-4\pi\fxt.
\eeqa

The fact that $\gxt$ is proportional to a delta function means that one can 
always complete the integration over time trivially. For the retarded 
Green's function in particular, one finds that
\beq
\psxt=\inivp\frac{f(\xp,t-\xxpa/c)}\xxpa
\eeq
which has a fairly obvious interpretation.

Finally, it is worth pointing out that one may always add to $\psxt$ a 
solution of the homogeneous wave equation.
\section{Derivation of Macroscopic Electromagnetism}
Regretfully omitted because of time constraints.
\section{Poynting's Theorem; Energy and Momentum Conservation}
A conservation law is a statement to the effect that some quantity $q$ (such as 
charge) is a constant for an isolated system. Often, it is possible to express 
the law mathematically in the form
\beq
\div\J_q+\pde{\rh_q}{t}=\lep\der{\rh_q}t\rip_e.
\eeq
Here, the density of the conserved quantity is $\rh_q$, its current density 
is $\J_q$, and the term on the right-hand side of the equation represents 
the contribution to the density at a particular space-time point arising 
from external sources or sinks. As we have already seen for the particular case
of electrical charge, the divergence term represents the flow of the 
conserved quantity away from a point in space while the partial time 
derivative represents the rate at which its density is changing at that 
point. For electrical charge, there are no (known) sources or sinks and so 
the term on the right-hand side is zero.

\subsection{Energy Conservation}

It is possible to derive several such conservation laws from the Maxwell 
equations. These include the charge conservation law\footnote{Not surprising 
since the equations were explicitly constructed so as to be consistent with 
charge conservation.} and also ones that are interpreted as energy, 
momentum, and angular momentum conservation. To get started, consider the 
rate per unit volume at which the fields transfer energy to the charged 
particles, or sources. The magnetic induction does no work since it is 
directed normal to the velocity of a particle, and so we have only the 
electric field which does work a the rate $\J\cdot\E$ per unit volume. This 
is, in the context of \eq{105}, the term $-(d\rh/dt)$ representing the 
transfer of energy to external (to the fields) sources or sinks.

The total power ${\cal P}_m$ transferred to the sources within some domain V 
is found by integrating over that domain,
\beq
{\cal P}_m=\inv(\J\cdot\E).
\eeq

Now what we would like to do is perform some manipulations on $\J\cdot\E$ 
designed to remove all reference to the current density, leaving only 
electromagnetic fields; further, we want to make the expression look like 
the left-hand side of \eq{105}. That is not so hard to do, starting with the 
generalization of Amp\`ere's law:
\beqa
\J\cdot\E=\frac1{4\pi}\leb c\E\cdot(\curl\H)-\E\cdot\pde\D t\rib\nonumber\\
=\frac1{4\pi}\leb-c\div(\E\times\H)+c\H\cdot(\curl\E)-\E\cdot\pde\D t\rib
\nonumber\\=-\frac c{4\pi}\div(\E\times\H)-\frac1{4\pi}\lep\H\cdot\pde\B t+
\E\cdot\pde\D t\rip.
\eeqa
In the last step, Faraday's law has been employed. The result is promising; 
there is a divergence and the remainder is almost a time derivative. It 
becomes a time derivative if the materials in the system have linear 
properties so that 
\beq
\H\cdot\pde\B t=\frac12\frac{\partial}{\partial t}(\B\cdot\H)\hbox{ and }
\E\cdot\pde\D t=\frac12\frac{\partial}{\partial t}(\E\cdot\D).
\eeq
Then \eq{107} becomes
\beq
-\J\cdot\E=\frac c{4\pi}\div(\E\times\H)+\frac1{8\pi}\frac{\partial}{
\partial t}(\E\cdot\D+\B\cdot\H)
\eeq
which is of the very form we seek. The interpretation of this equation is 
that the {\em Poynting vector} $\S$, defined by 
\beq
\S\equiv\frac c{4\pi}(\E\times\H)
\eeq
is the energy current density of the electromagnetic field and that the 
{\em energy density} $u$, defined by 
\beq
u\equiv\frac1{8\pi}(\E\cdot\D+\B\cdot\H)
\eeq
is, indeed, the energy density of the electromagnetic field. In terms of 
these quantities the conservation law is
\beq
-\J\cdot\E=\div\S+\pde ut.
\eeq
Notice, however, that the energy current density is not unique. Because only 
its divergence enters into the conservation law, we may add to $\S$ the curl of
any vector field and would still have \eq{112}. The distinction is not 
important for measurable quantities because one always measures the rate 
of energy flow through a closed surface (the surface of the detector),
\beq
\inac(\nn\cdot\S)\equiv\inv(\div\S), 
\eeq
so that only the divergence of $\S$ is important. Finally, note that if the 
conservation law is integrated over some volume V, it can be expressed as
\beq
\inv\pde ut=-\inac(\nn\cdot\S)-\inv(\J\cdot\E)
\eeq
with the interpretation that rate of change of field energy within the 
domain V is equal to the rate at which field energy flows in through the 
surface of the domain plus the rate at which the sources within V transfer 
energy to the field.

\subsection{Momentum Conservation}

We can find a similar-looking momentum conservation law. We start from an 
expression for the rate per unit volume at which the fields transfer 
momentum to the sources; this is just the force density $\f$,
\beq
\f=\rh\E+\frac1c(\J\times\B).
\eeq

\centerline{\psfig{figure=fig12.ps,height=1.875in,width=6.375in}}
 
\noindent Now use the Maxwell equations to remove all appearance of 
the sources $\rh$ and $\J$:
\beqa
\f=\lep\frac{\div\D}{4\pi}\rip\E+\frac1c\lep\frac c{4\pi}\curl\H-\frac1{4
\pi}\pde\D t\rip\times\B\nonumber\\
=\frac1{4\pi}\leb\E(\div\D)+(\curl\H)\times\B-\frac1c\lep\pde\D t\times
\B\rip\rib\nonumber\\=\frac1{4\pi}\leb\E(\div\D)-\B\times(\curl\H)-\frac1c
\frac{\partial}{\partial t}(\D\times\B)+\frac1c\lep\D\times\pde\B t\rip\rib
\hsph\\=-\frac1{4\pi}\lec\frac1c\pde{(\D\times\B)}{t}
+\leb\B\times(\curl\H)-\H(\div\B)+\D\times(\curl\E)-\E(\div\D)
\rib\ric\nonumber
\eeqa
where we've used Faraday's law and the field equation $\div\B=0$. Now we 
have to get all of the terms in the square brackets [...] in the final 
expression to look like a divergence. There is one complication which is 
that the equation is a vector equation, so we need a divergence which 
yields a vector, not a scalar. 

Alternatively, we can look at \eq{116} as 
three scalar equations, one for each Cartesian component of the force 
density. Then the term in brackets contains three components $U_i$ and we 
want to write each component as the divergence of some vector field $\V_i$,
$U_i=\div\V_i$. If we expand the vector fields as
\beq
\V_i=\sum_{j=1}^3V_{ji}\epj,
\eeq
with
\beq
U_i=\div\V_i=\sum_j\pde{V_{ji}}{x_j},
\eeq
then we have nine scalar functions $V_{ji}$ which can be put into a $3\times3$ 
matrix. Let us go a step further and define a {\em dyadic} $\dV$ by its 
inner products with the complete set of basis vectors $\ep_i$:
\beq
\epj\cdot\dV\cdot\epi\equiv V_{ji}.
\eeq
A convenient way to write $\dV$ in terms of its components is
\beq
\dV=\sum_{i,j=1}^3\epj V_{ji}\epi,
\eeq
with the understanding that when an inner product is taken of $\dV$ with a 
vector, the dot product is taken with the left or right $\epi$ depending on 
whether the other vector lies to the left or right of $\dV$. In particular,
\beq
\div\dV=\sum_k\epk\frac{\partial}{\partial x_k}\sum_{i,j}\epj V_{ji}\epi=
\sum_{i,j}\pde{V_{ji}}{x_j}\epi=\sum_iU_i\epi.
\eeq

Now what we need, referring back to \eq{116}, is some $\dV$ such that
\beq
\div\dV=\B\times(\curl\H)-\H(\div\B)+\D\times(\curl\E)-\E(\div\D)
\eeq
where we shall restrict ourselves to the case of vacuum, or at least 
constant $\mu$ and $\ep$. First, we have an identity,
\beq
\B\times(\curl\B)=\frac12\grad(\B\cdot\B)-(\B\cdot\grad)\B
\eeq
which allows us to write
\beqa
\B\times(\curl\H)-\H(\div\B)=\frac1\mu\leb\frac12\grad(\B\cdot\B)-(\B\cdot
\grad)\B-\B(\div\B)\rib\nonumber\\=\frac1\mu\sum_{i,j}\leb\frac12\de_{ij}
\epj\pde{B^2}{x_i}-B_i\pde{B_j}{x_i}\epj-B_j\epj\pde{B_i}{x_i}\rib\nonumber
\\=\frac1\mu\sum_k\leb\epk\frac{\partial}{\partial x_k}\rib\cdot\leb
\sum_{i,j}\epi\lep\frac12\de_{ij}B^2-B_iB_j\rip\epj\rib\nonumber\\
=\div\leb\sum_{i,j}\frac1\mu\epi\lep\frac12\de_{ij}B^2-B_iB_j\rip\epj\rib
\eeqa
which is indeed the divergence of a dyadic. By similar means one may 
demonstrate that
\beq
\D\times(\curl\E)-\E(\div\D)=\div\leb\sum_{i,j}\ep\epi\lep\frac12
\de_{ij}E^2-E_iE_j\rip\epj\rib.
\eeq
Putting these back into \eq{124}, we may write
\beq
-\f=\frac\ep{4\pi c}\frac{\partial}{\partial t}\leb(\E\times\B)\rib-
\div\dT
\eeq
where the components $T_{ij}$ of the {\em Maxwell stress tensor} are
\beq
T_{ij}\equiv\frac1{4\pi}\leb\ep E_iE_j+\frac1\mu B_iB_j-\frac12\de_{ij}\lep\ep
E^2+\frac1\mu B^2\rip\rib.
\eeq
The Maxwell stress tensor is a symmetric tensor (in a uniform, isotropic, 
linear medium) and has the physical interpretation that $-T_{ij}$ is the 
$j$-component of the current density of the $i$-component of momentum. 
Note, however, that $\dT$ is not unique in that if we redefine it to 
include the curl of another dyadic, \eq{116} would still hold. 

Let us define also a vector field $\g$,
\beq
\g\equiv\frac1{4\pi c}(\D\times\B)
\eeq
which is interpreted as the momentum density of the electromagnetic field, 
or the momentum per unit volume. Notice that for $\D=\ep\E$ and $\B=\mu\H$, 
\beq
\g=\frac{\ep\mu}{4\pi c}(\E\times\H)=\frac{\mu\ep}{c^2}\S;
\eeq
there is thus a simple connection between the energy current density 
and the momentum density of the field. Our conservation law is now simply 
written as 
\beq
-\f=\pde\g t-\div\dT; 
\eeq
if integrated over some domain V, it may also be expressed as
\beq
-\inv\f=\inv\pde\g t-\ina(\nn\cdot\dT).
\eeq
Finally, let us define
\beq
\der{\P_m}t\equiv\inv\f\hbox{\hsph and\hsph}\der{\P_f}t\equiv\inv\pde\g t;
\eeq
these are rather obviously meant to be the time rate of change of the 
mechanical momentum\footnote{Note, however, that it includes only the rate 
of change of mechanical momentum as a consequence of the forces applied to 
the particles by the electromagnetic field and does not include the change 
in mechanical momentum which comes about because particles are entering and 
leaving the domain V.} and of the field momentum in V. Now we can write the 
conservation law as
\beq
\der{\P_m}t+\der{\P_f}t=\ina\nn\cdot\dT.
\eeq

\subsection{Need for Field Momentum}

	We defined our fields, the electric field and the magnetic induction,
in terms of the force and torque, respectively, that they inflict upon 
an elementary source.  However, as we have seen above, this definition 
is incomplete since the fields also carry energy and momentum.  From a quantum
point of view, this is obvious since we can think of the fields as
resulting from the exchange of real and virtual photons each with
momentum $\hbar\k$ and energy $\hbar\om$.  However, from a 19'th
century perspective the need of the field to carry energy and momentum
(especially momentum) is less obvious.

	However, Newton's law of action and reaction requires that the
field carry momentum.  First consider two completely isolated particles 
in free space.  If the only force exerted on either particle is from
its counterpart, then the net momentum $\P$ is conserved when
the forces are equal and opposite.
\[
\der{\P}{t}=\der{}{t}\lep \P_1+\P_2\rip = 
m_1\der{\v_1}{t}+m_2\der{\v_2}{t}=\F_1+\F_2
\]
I.e. the momentum is conserved when $\F_1=-\F_2$

	Now consider the situation where the particles are charged, and have 
trajectories as shown in the figure below.

\centerline{\psfig{figure=fig13.ps,height=1.875in,width=6.375in}}

\noindent Clearly the forces due to the electric field are equal
and opposite, but those due to the magnetic fields are not.  In fact,
charge 1 exerts a magnetic force on charge 2, but charge 2 does not
exert a magnetic force on charge 1.  Momentum is not conserved by the
particles (and the forces on the particles) alone.  The excess momentum
must be carried by the field!  Thus, the field is required to have a momentum
density by Newtonian law.


\subsection{Example: Force on a Conductor}

	As an example, consider a conductor in the presence of an external
electric field ($\B=0$).

\centerline{\psfig{figure=fig14.ps,height=1.875in,width=6.375in}}
 
\noindent Since $\B=0$ everywhere, $\E\times\B=0$, and there is no
electromagnetic momentum density.  By conservation of momentum, the force 
on the conducting surface is then given
\beq
\lep\der{\P}{t}\rip_i=F_i=\sum_j\ina n_i T_{ij}
\eeq
If we take the surface of integration as shown in the figure, then
we need only consider the surface in the xy plane for which $\nn$ is in the
$-\zh$ direction.  Thus we only need $T_{ij}$ when $i=z$.  Since
$\E=\zh E$, and since 
\beq
T_{ij}=\frac{1}{4\pi}\lep E_iE_j-\frac12 \delta_{ij}E^2\rip
\eeq
when $\B=0$, we have
\beq
T_{zx}=T_{zy}=0\;\;\;\;\;\;\;T_{zz}=\frac{1}{8\pi}E^2\,.
\eeq
Thus,
\beq
F_z=-\ina \frac{1}{8\pi}E^2=-\frac{A}{8\pi}E^2
\eeq
where $A$ is the area of the conductor presented to the E-field.  Thus
the force on the conductor is downward


\section{Conservation Laws for Macroscopic Systems}
Regretfully omitted.
\section{Poynting's Theorem for Harmonic Fields; Impedance, Admittance, 
etc.}
Poynting's theorem has many important applications at the practical 
(electrical engineering) level. In this section we briefly make a foray 
into such an application. We shall restrict attention to harmonic fields 
which means ones that are harmonic (sine) functions of time. To this end we 
write the time-dependent part of the fields, as well as the sources, as
$\emot$ where $\om$ is the angular frequency, and the physical field or 
source so represented is the real part of the complex mathematical function 
that we are using. For example,
\beq
\Ext=\Re\leb\Ex\emot\rib\equiv\frac12\leb\Ex\emot+\E^*(\x)\eot\rib.
\eeq
Similarly,
\beqa
\Jxt\cdot\Ext=\frac14\leb\Jx\emot+\J^*(\x)\eot\rib\cdot\leb\Ex\emot+\E^*(\x)
\eot\rib\\ \nonumber
=\frac12\Re\leb\J^*(\x)\cdot\Ex+\Jx\cdot\Ex\etot\rib.
\eeqa
The time-average of the product is particularly simple; the term which 
varies as $\etot$ has a zero time-average and so we are left with
\beq
<\Jxt\cdot\Ext>=\frac12\Re\leb\J^*(\x)\cdot\Ex\rib.
\eeq

One can easily see that there is a general rule for the time-average of 
the product of two harmonic functions. Given $Q(\x,t)$ and $R(\x,t)$,
\beq
Q(\x,t)=\Re\leb Q(\x)\emot\rib\hsph R(\x,t)=\Re\leb R(\x)\emot\rib,
\eeq
the time average of the product may be expressed as
\beq
<Q(\x,t)R(\x,t)>=\frac12\Re\leb R^*(\x)Q(\x)\rib=\frac12\Re\leb R(\x)Q^*(\x).
\rib
\eeq

We start by supposing that all fields are harmonic. Then they all have the 
complex form $\Fx\emot$ and they all have time derivatives give by
\beq
\frac{\partial(\Fx\emot)}{\partial t}=-i\om\Fx\emot
\eeq
so that the Maxwell equations for the complex amplitudes (just the 
position-dependent parts of the fields) are
\beqa
\div\Bx=0\hsp{1.0} \curl\Ex-i\frac\om c\Bx\nonumber\\
\div\Dx=4\pi\rhx\hsp{0.5}\curl\Hx+i\frac\om c\Dx=\frac{4\pi}c\Jx.
\eeqa
Notice that there is no problem in generalizing the Maxwell equations to 
complex fields because the equations involve linear combinations, with real 
coefficients, of complex objects. One thus has two sets of equations, one for 
the real parts of these objects and one for the imaginary parts. The set for 
the real parts comprises the ``true'' Maxwell equations.

For the remainder of this section, the symbols $\E$, $\B$, etc, stand for 
the complex amplitudes $\Ex$, $\Bx$, ...~. We can rederive the Poynting 
theorem for these by starting from the inner product $\J^*\cdot\E$ and 
proceeding as in the original derivation. The result is 
\beq
\J^*\cdot\E=\frac{c}{4\pi}\leb-\div(\E\times\H^*)-i\frac\om c(\E\cdot\D^*
-\B\cdot\H^*)\rib. 
\eeq
Define now the (complex) Poynting vector and energy densities
\beq
\S\equiv\frac c{8\pi}(\E\times\H^*),
\eeq
and
\beq
w_e\equiv\frac1{16\pi}(\E\cdot\D^*)\hbox{\hspace{0.5in}}w_m\equiv\frac1{16\pi}
(\B\cdot\H^*).
\eeq
Notice that the real part of $\S$ is just the time-averaged real Poynting 
vector while the real parts of the energy densities are the time-averaged 
energy densities. More generally, the energy densities can be complex 
functions, 
depending on the relations between $\D$ and $\E$ and between $\B$ and $\H$. 
If the two members of each pair of fields are in phase with one another, then 
the corresponding energy density is real. Similarly, if $\E$ and $\H$ are 
in phase, then $\S$ is also real.

In terms of our densities, Poynting's theorem for harmonic fields becomes
\beq
\frac12\J^*\cdot\E=-\div\S-2i\om(w_e-w_m).
\eeq
If we integrate this expression over some domain V and apply the divergence 
theorem to the term involving the Poynting vector, we find the relation
\beq
\frac12\inv\J^*\cdot\E+2i\om\inv(w_e-w_m)+\inac\S\cdot\nn=0.
\eeq
The interpretation of this equation is that the real part expresses the 
time-averaged conservation of energy. The imaginary part also has a meaning 
in connection with energy and its flow. 

Consider first the simplest case 
of real $w_e$ and $w_m$. Then the energy densities drop out of the real 
part of this equation and what it (the real part) tells us is that the 
time-average rate of doing work on the sources in V is equal to the 
time-averaged flow of energy (expressed by the Poynting vector) into V 
through the surface S. If the energy densities are not real, then there is 
an additional real piece in \eq{149} so that the work done on the sources in 
V is not equal to the energy that comes in through S; this case corresponds 
to having ``lossy'' materials within V which dissipate additional energy.

Now let's suppose that there is some electromagnetic device within V, i.e., 
surrounded by S. Let it have two input terminals which are its only 
material communication with the rest of the world. At these terminals there 
are some input current $I_i$ and voltage $V_i$ which we suppose are 
harmonic and which may also be written in the form \eq{138}. 

\centerline{\psfig{figure=fig16.ps,height=2.25in,width=6.375in}}
 
\noindent Then the 
(complex) input power is $I^*_iV_i/2$, meaning that the time-averaged input 
power is the real part of this quantity. Using our interpretation of the 
Poynting vector, we can express the input power in terms of a surface 
integral of the normal component of $\S$,
\beq
\frac12I^*_iV_i=-\int_{S_i}d^2x\,\S\cdot\nn
\eeq
where the surface integral is done just over the cross-section of the 
(presumed) coaxial cable feeding power into the device; it is assumed that 
for such a cable, the input fields are confined to the region within 
the shield on the cable and so the integral over the remainder of the 
surface S surrounding the device has no contribution from the incident 
fields.

If we now combine this equation with \eq{149}, we find that we can write
\beq
\frac12I^*_iV_i=\frac12\inv\J^*\cdot\E+2i\om\inv(w_e-w_m)+\int_{S-S_i}
d^2x\,\S\cdot\nn.
\eeq
The surface integral in this expression gives the power passing through the 
surface S, excluding the part through which the input power comes. The real 
part of this integral is the power radiated by the device.

Now let us define the {\em input impedance} $Z$ of the device,
\beq
V_i\equiv ZI_i;
\eeq
the impedance is complex and so can be written as
\beq
Z\equiv R-iX
\eeq
where the {\em resistance} $R$ and the {\em reactance} $X$ are real. 
From \eq{151} we find expressions for these:
\beq
R=\frac1{|I_i|^2}\lec Re\leb\inv\J^*\cdot\E+2\int_{S-S_i}d^2x\,\S\cdot\nn
\rib+4\om Im\leb\inv(w_m-w_e)\rib\ric
\eeq
and
\beq
X=\frac1{|I_i|^2}\lec4\om Re\leb\inv(w_m-w_e)\rib-Im\leb\inv\J^*\cdot\E
+2\int_{S-S_i}d^2x\,\S\cdot\nn\rib\ric.
\eeq
By deforming the surface so that it lies far away from the device, one may 
make the integral over $\S\cdot\nn$ purely real so that it does not 
contribute to the reactance; then it is only a part of the resistance and 
is the so-called ``radiation resistance'' which will be present if the 
device radiates a significant amount of power.

Our result has a simple and pleasing form at low frequencies. Then 
radiation is negligible and so the contributions of the surface integral 
may be ignored. Also, we may drop the term in the resistance 
proportional to $\om$. Then, assuming the current density and electric field 
are related by $\J=\si\E$ where $\si$ is the (real) {\em electrical 
conductivity}, and assuming real energy densities, we find
\beq
R=\frac1{|I_i|^2}\inv\si|\E|^2
\eeq
and 
\beq
X=\frac{4\om}{|I_i|^2}\inv(w_m-w_e)
\eeq
The last equation may be used to established contact between our 
expressions, based on the electromagnetic field equations, and some 
standard and fundamental relations in elementary circuit theory. If there 
is an inductance (magnetic energy-storing device) in the ``black-box,'' 
then the integral of the magnetic energy may be expressed (see the first 
two or three problems at the end of Chap. 6 of Jackson) as $\L|I_i|^2/4$, 
and so we find the familiar (if one knows anything about circuits) result that 
$X=L\om$. But if there is a capacitor, the energy becomes $|Q_i|^2/4C$ where 
the charge $Q_i$ is obtained by integrating the current over time; that 
gives $|Q_i|^2=|I_i|^2/\om^2$ and so $X=-1/\om C$, another familiar tenet 
of elementary circuit theory.

\section{Transformations: Reflection, Rotation, and Time Reversal}

Before entering into discussion of the specific transformations of 
interest, we give a brief review of orthogonal transformations. Introduce a 
$3\times3$ matrix ${\em a}$ with components $a_{ij}$ and use it to 
transform a position vector $\x=(x_1,x_2,x_3)=(x,y,z)$ into a new vector
$\xp$:
\beq
x_i'=\sum_ja_{ij}x_j.
\eeq
An orthogonal transformation is one that leaves the length of the vector 
unchanged,
\beq
\sum_i(x_i')^2=\sum_ix_i^2.
\eeq
Using this condition, one may show that ${\em a}$ must satisfy the 
conditions
\beq
\sum_ia_{ij}a_{ik}=\de_{jk}
\eeq
and
\beq
det({\em a})=\pm1.
\eeq
Orthogonal transformations with $det({\em a})=+1$ are simple rotations. The 
other ones are combinations of a rotation and an inversion\footnote{An 
inversion is a transformation $\xp=-\x$.}; these are called {\em improper 
rotations}.

It is common to refer to a collection of three objects $\ps_i,\;i=1,2,3$, 
which transform, under orthogonal transformations, in the same way as the 
components of $\x$, as a {\em vector}, a {\em polar vector},  or a 
{\em rank-one tensor}. A 
collection of nine objects $q_{ij},\;i,j=1,2,3$, which transform in the 
same way as the nine objects $x_ix_j$ is called a {\em rank-two tensor}. 
And so on. An object which is {\em invariant}, that is, which is unchanged 
under an orthogonal transformation, is called a {\em scalar} or {\em 
rank-zero tensor}; the length of a vector is such an object. 

One also defines {\em pseudotensors} of each rank. A {\em rank-p 
pseudotensor} comprises a set of $3^p$ objects which transform in the same way 
as a rank-p tensor under ordinary rotations but which transform with an extra 
change of sign relative to a rank-p tensor under improper rotations. One 
also uses the terms {\em pseudoscalar} for a rank-0 pseudotensor and {\em 
pseudovector} or {\em axial vector} for a rank-1 pseudotensor. Notice that 
under inversion, for which ${\em a}$ is just the negative of the unit 
$3\times3$ matrix, a vector changes sign, $\xp=-\x$, while a pseudovector 
is invariant. This statement can be generalized: Under inversion, a tensor
$T$ of rank n transforms to $T'$ with 
\beq
T'=(-1)^nT.
\eeq
A pseudotensor $P$ of the same rank, on the other hand, transforms 
according to 
\beq
P'=(-1)^{(n+1)}P
\eeq
under inversion.
 
\subsection{Transformation Properties of Physical Quantities}
It is important to realize that objects which we are accustomed to referring to 
as ``vectors,'' such as $\B$, are not necessarily vectors in the sense 
introduced here; indeed, it is one of our tasks in this section to find out 
just what sorts of tensor are the various physical quantities we have been 
studying. Consider for example the charge density. Suppose that we 
have a system with a certain $\rhx$ and that we rotate it; then $\rh$ 
becomes $\rh'$ and $\x$ becomes $\xp$. 

\centerline{\psfig{figure=fig15.ps,height=1.875in,width=6.375in}}
 
\noindent  The question is, how is $\rh'(\xp)$ 
related to $\rhx$? It is easy to see, since $\xp$ is what $\x$ becomes as 
a consequence of the rotation, that $\rh'(\xp)=\rhx$. Under an inversion 
also, this relation is true. Hence we conclude that the charge density is a 
scalar or rank-0 tensor.

An example of a vector or rank-1 tensor is, of course, $\x$. Similarly, 
from this fact one may show that the operator $\grad$ is a rank-1 tensor
(Differential operators can also be tensors or components of tensors);
\beq
\frac\partial{\partial x_i'}=\sum_ja_{ij}\frac\partial{\partial x_j}. 
\eeq
What then is $\grad\rh$? From the (known) transformation properties of $\rh$ 
and of $\grad$, it is easy to show that it is a rank-1 tensor. The gradient of 
any scalar function is a rank-1 tensor. Similarly, one may show that the 
inner product of two rank-1 tensors, or vectors, is a scalar as is the 
inner product of two rank-1 pseudotensors; the inner product of a rank-1 
tensor and a rank-1 pseudotensor is a pseudoscalar; and the gradient of a 
pseudoscalar is a rank-1 pseudotensor. 

All of the foregoing are quite easy to demonstrate. A little harder is the 
crossproduct of two vectors (rank-1 tensors).
Consider that $\b$ and $\c$ are rank-1 tensors. Their cross product may be 
written as
\beq
\u=\b\times\c
\eeq 
with a Cartesian component given by
\beq
u_i=\sum_{j,k}\ep_{ijk}b_jc_k
\eeq
where
\beq
\ep_{ijk}\equiv\lec\barr{cc}
+1 & \hbox{if }(i,j,k)=(1,2,3),(2,3,1),(3,1,2)\\
-1 & \hbox{if }(i,j,k)=(2,1,3),(1,3,2),(3,2,1)\\
0 & \hbox{otherwise}.\ear\right.
\eeq
If we {\bf define} $\ep_{ijk}$ to be given by this equation in all frames, 
then we can {\bf show} that it is a rank-3 pseudotensor. Alternatively, we 
can use \eq{164} to specify it in a single frame, {\bf define} it to be a 
rank-3 pseudotensor, and then {\bf show} that it is given by \eq{164} in 
{\bf any} frame. However one chooses to do it, one can use this object, called 
the {\em completely antisymmetric unit rank-3 pseudotensor}, and the assumed 
transformation properties of $\b$ and $\c$ (rank-1 tensors) to determine 
the transformation properties of the crossproduct. What one finds is that
\beq
u_i'=det({\em a})\sum_ja_{ij}u_j
\eeq
which means that $\u$ is a pseudovector or a rank-1 pseudotensor.

The transformations considered so far have all dealt with space; to them we 
wish to add the time-reversal transformation. The question to ask of a 
given entity is how it changes if time is reversed. Imagine making a 
videotape of the entity's behavior and then running the tape backwards. If, 
in this viewing, the quantity is the same at a given point on the tape as 
when the tape is running forward, then the quantity is {\em even} or {\em
invariant} under time reversal. If its sign has been reversed, then it is {\em
odd} under time reversal. For example, the position $\x(t)$ of an object is 
even under time reversal; the velocity of the object, however, is odd.

In Table 1, we catalog some familiar mechanical functions according to their 
rotation, inversion, and time-reversal symmetries.
\begin{table}
\caption{Rotation, inversion, and time-reversal properties of some common 
mechanical quantities.}
\btab{||l|c|l|c||} \hline
Function & Rank & Inversion Symmetry & Time-reversal Symmetry \\ \hline
$\x$     & 1    & $-$ (vector)         & +                      \\
$\v=d\x/dt$ & 1 & $-$ (vector)         & $-$                      \\
$\p=m\v$ & 1    & $-$ (vector)         & $-$                      \\
${\bf{L}}=\x\times m\v$ & 1 & + (pseudovector)     & $-$           \\
$\F=d\p/dt$ & 1 & $-$ (vector)         & +                      \\
$\N=\x\times\F$ & 1 & + (pseudovector) & +                    \\
$T=p^2/2m$ & 0  & + (scalar)           & +                      \\
$V$      & 0    & + (scalar)           & +                       \\ \hline
\etab
\end{table}

We may make the same sort of table for various electromagnetic quantities, 
basing our analysis on the Maxwell equations, which we assume to be the 
correct equations of electromagnetism. Given that $\rh$ is a scalar and 
that $\grad$ is a vector, the equation $\div\E=4\pi\rh$ tells us that the 
electric field is a vector; further, it is even under time reversal (since 
$\rh$ and $\grad$ are both even). Similarly, $\D$ and $\P$ must be vectors 
and even under time reversal. 
Moving on to Faraday's Law, $\curl\E=-c^{-1}\partial\B/\partial t$, from 
our knowledge of the properties of the gradient, the cross product, and the 
electric field, we see that $\B$ is a pseudovector and that it is odd under 
time reversal; $\H$ and $\M$ have the same properties. Finally, Amp\`ere's Law,
$\curl\B=(4\pi/c)\J+c^{-1}\partial\E/\partial t$ is consistent with these 
determinations and with the statement that $\J$ is a vector, odd under 
time reversal, which follows from the fact that $\J=\rh\v$. Finally, $\S$ 
and $\g$ are vectors with odd time-reversal symmetry while the Maxwell 
stress tensor is a rank-2 tensor, even under time reversal. These 
properties are summarized in Table 2.
\begin{table}
\caption{Rotation, inversion and time-reversal properties of some common 
electromagnetic quantities.}
\btab{||l|c|l|c||} \hline
function   & rank  & inversion symmetry & time-reversal symmetry \\ \hline
$\rh$      &   0   &   + (scalar)       &   +   \\
$\J$       &   1   &   $-$ (vector)       &   $-$   \\
$\E,\D,\P$ & 1     &   $-$ (vector)       &   +   \\
$\B,\H,\M$ & 1     &   + (pseudovector) &   $-$   \\
$\S,\g$    &   1   &   $-$ (vector)       &   $-$   \\
$\dT$      &   2   &   + (tensor)       &   +   \\ \hline
\etab
\end{table}

The usefulness of these expressions lies in the belief that acceptable 
equations of physics should be invariant under various symmetry operations. 
The Maxwell equations and the classical equations of mechanics (Newton's 
Laws), for example, are invariant under time reversal and under orthogonal 
transformations, meaning that each term in any given equation transforms in 
the same way as all of the other terms in that equation\footnote{Of course
there are some equations, like Ohm's Law, which describe truly irreversible
processes for which time reversal invariance does not hold.}. If we 
believe that 
this should be true of all elementary equations of classical physics, then 
there are certain implied constraints on the form of the equations. 
Consider as an example the relation between $\P$ and $\E$. Supposing that 
one can make an expansion of a component of $\P$ in terms of the components 
of $\E$, we have
\beq
P_i=\sum_j\al_{ij}E_j+\sum_{jk}\be_{ijk}E_jE_k+\sum_{jkl}
\ga_{ijkl}E_jE_kE_l+...
\eeq
where, since $\P$ and $\E$ are both rank-1 tensors, invariant under time 
reversal, it follows, using the invariance argument, that the coefficients
$\al_{ij}$ are the components of a rank-2 tensor, invariant under time 
reversal; the $\be_{ijk}$ are components of a rank-3 tensor, invariant 
under time reversal; and the $\ga_{ijkl}$ are components of a rank-4 
tensor, also invariant under time reversal. 

If we now add some statement 
about the properties of the medium, we can get further conditions. In the 
simplest case of an isotropic material, it must be the case that each of 
these tensors is invariant under orthogonal transformations. This condition 
severely limits their forms; in particular, it means that $\al_{ij}=
\al\de_{ij}$.  We can see this by appealing to the transformation
properties of second rank tensors.  Thus, $\al_{ij}$ must transform like
$x_ix_j$, or
\beq
\al'_{nm}=a_{ni}a_{mj}\al_{ij}
\eeq
Since the medium is isotropic, we require that $\al'_{ij}=\al_{ij}$.
The only way to satisfy both of these conditions of transformation
is if
\beq
\al'_{nm}=\al a_{ni} a_{mi}=\al\de_{nm}
\eeq
The same type of thing cannot be done with $\be$, so 
that $\be_{ijk}=0$, bu we can perform similar manipulations on $\ga$
so the coefficients $\ga_{ijkl}$ are such as to produce 
\beq
\sum_{jkl}\ga_{ijkl}E_iE_jE_l=\ga(\E\cdot\E)\E
\eeq
where $\ga$ is a constant. Thus, through third-order terms, the expansion 
of $\P$ in terms of $\E$ must have the form
\beq
\P=\al\E+\ga E^2\E.
\eeq
The general forms of many other less obvious relations may be determined by 
similar considerations.

\section{Do Maxwell's Equations Allow Magnetic Monopoles?}
The answer is yes, but only in a restricted, and trivial, sense. If there 
were magnetic charges of density $\rh_m$ and an associated magnetic current 
density $\J_m$, with a corresponding conservation law
\beq
\pde{\rh_m}t+\div\J_m=0,
\eeq
then the field equations would read
\beqa
\div\B=4\pi\rh_m\hspace{1.0in}\curl\H=\frac{4\pi}c\J+\frac1c\pde\D t
\nonumber\\ \div\D=4\pi\rh\hspace{1.0in}\curl\E=-\frac{4\pi}c\J_m-\frac1c
\pde\B t.
\eeqa
In fact, the Maxwell equations as we understand them can be put into this 
form by making  a particular kind of transformation, called a {\em duality 
transformation} of the fields and sources. Introduce
\beqa \E=\E'\cos\et+\H'\sin\et\hsph\D=\D'\cos\et+\B'\sin\et\nonumber\\
\H=-\E'\sin\et+\H'\cos\et\hsph\B=-\D'\sin\et+\B'\cos\et\nonumber\\
\rh=\rh'\cos\et+\rh_m'\sin\et\hsp{0.75}\J=\J'\cos\et+\J_m'\sin\et\nonumber\\
\rh_m=-\rh'\sin\et+\rh_m'\cos\et\hsph\J_m=-\J'\sin\et+\J_m'\cos\et.
\eeqa
where $\et$ is an arbitrary real constant.

If one now substitutes these into the generalized field equations, one 
finds, upon separating the coefficients of $\sin\et$ from those of 
$\cos\et$ (These must be independent because $\et$ is arbitrary), that the 
primed fields and sources obey an identical set of field equations. What 
this means is that the Maxwell equations (with no magnetic sources) may be 
thought of as a special case of the generalized field equations, one in 
which $\et$ is chosen so that $\rh_m$ and $\J_m$ are equal to zero. From 
the form of the transformations for the sources, we see that this is 
possible if the ratio of $\rh$ to $\rh_m$ for each source (particle)is the same
as that for all of the other sources (particles). Hence it is meaningless 
to say that there are no magnetic monopoles; the real question is whether 
all elementary particles have the same ratio of electric to magnetic 
charge. If they do, then Maxwell's equations are correct and correspond, as 
stated above, to a particular choice of $\et$ in the more general field 
equations.

If one subjects the electron and proton to scrutiny regarding the question 
of whether they have the same ratio of electric to magnetic charge, one 
finds that if one defines (by choice of $\et$) the magnetic charge of the 
electron to be zero, then experimentally the magnetic charge of the proton 
is known to be smaller than $10^{-24}$ of its electric charge. That's 
pretty good evidence for its being zero. 

But there remains the question whether there are other kinds of particles, not 
yet discovered, which have a different ratio $\rh/\rh_m$ than do electrons and 
protons. Dirac, for example, has given a simple and clever argument which 
shows that the quantization of the electric charge follows from the mere 
existence of an electrically uncharged magnetic monopole. Moreover, the 
argument gives the magnetic charge $g$ of the monopole as $g=nhc/4\pi e$ 
where $n$ is any integer and $h$ is Planck's constant. This is, in 
comparison with the electric charge, very large so that it ought to be in 
principle easy to detect a ``Dirac monopole'' should there by any of them 
around. So far, none has been reliably detected. 

\appendix
\section{Helmholtz' Theorem}
Any vector function of position $\C(x)$ can be written as the sum of two vector
functions such that the divergence vanishes for one and the curl vanishes
for the other.  In other words, the decomposition
\beq
\Cx=\Dx+\Fx
\eeq
is always possible, where
\beq
\div\D=0\;\;\;\; \curl\F=0
\eeq

	{\bf{Proof.}}  We may satisfy the two conditions for $\F$ and $\D$,
by writing
\beq
\D=\curl\A\;\;\;\;\;\F=-\grad\Phi\,.
\eeq
Then taking the curl and divergence of these equations respectively,
we can write
\beq
\lap\Phi=-\div\C\;\;\;\;\;\curl\lep\curl\A\rip=\curl\C\,.
\eeq
We already know how to solve these solutions (at least in Cartesian
coordinates).
\beq
\Phxp=\frac{1}{4\pi}\inv\frac{\div\Cx}{|\x-\xp|}\;\;\;\;\;\;
\A(\xp)=\frac{1}{4\pi}\inv\frac{\curl\Cx}{|\x-\xp|}
\eeq
Since $\D$ and $\F$ can now be found from these potentials, we have
demonstrated the decomposition claimed by Helmholtz' Theorem, and
thus proved it.  

An interesting corollary of this theorem is that a vector function
is completely determined if its curl and divergence are known everywhere.
The field $\F=-\grad\Phi$, where produced by a point
source, is longitudinal to the vector from the source to the point
where the field is evaluated.  The field $\D=\curl\A$ is transverse
to the vector from the source to the field point.  Thus $\F$ is typically
called the longitudinal, and $\D$ the transverse, part of $\C$.

\edo


