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%% This document created by Scientific Word (R) Version 2.0


\documentclass[12pt]{article}
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%\usepackage{sw20jart}

%TCIDATA{TCIstyle=Article/ART4.LAT,jart,sw20jart}

%\input tcilatex
%\QQQ{Language}{
%American English
%}

\begin{document}

\section{Fields in a cavity}

Jackson works out in detail the fields in a waveguide, then uses the
waveguide solutions to work out the fields in a cavity. His procedure is to
break up the cavity fields into the $e^{ikz}$ part and the $e^{-ikz}$ part,
work on each separately, and recombine them. This can be cumbersome.

One can obtain formulas that work directly for fields proportional to any
combination of sinusoidal or exponential functions. The result shown here is
in cartesian coordinates, and thus it is especially suited for box-shaped
cavities.

Start from Maxwell's equations written out explicitly: 
\begin{eqnarray}
\partial _yH_z-\partial _zH_y &=&-i\left( \omega /c\right) \epsilon E_x
\label{dEx} \\
\partial _zH_x-\partial _xH_z &=&-i\left( \omega /c\right) \epsilon E_y
\label{dEy} \\
\partial _xH_y-\partial _yH_x &=&-i\left( \omega /c\right) \epsilon E_z
\label{dEz}
\end{eqnarray}
\begin{eqnarray}
\partial _yE_z-\partial _zE_y &=&i\left( \omega /c\right) \mu H_x
\label{dHx} \\
\partial _zE_x-\partial _xE_z &=&i\left( \omega /c\right) \mu H_y
\label{dHy} \\
\partial _xE_y-\partial _yE_x &=&i\left( \omega /c\right) \mu H_z
\label{dHz}
\end{eqnarray}
Assuming only that the each field component satisfies $\partial _z^2\phi
=-\gamma ^2\phi ,$ we can obtain all the other field components from the
``superpotentials'' $E_z$ and $H_z:$%
\begin{eqnarray}
\left( \epsilon \mu \left( \omega /c\right) ^2-\gamma ^2\right) E_x
&=&\partial _x\partial _zE_z+i(\omega /c)\mu \partial _yH_z  \label{Ex} \\
\left( \epsilon \mu \left( \omega /c\right) ^2-\gamma ^2\right) E_y
&=&\partial _y\partial _zE_z-i(\omega /c)\mu \partial _xH_z  \label{Ey} \\
\left( \epsilon \mu \left( \omega /c\right) ^2-\gamma ^2\right) H_x
&=&\partial _x\partial _zH_z-i(\omega /c)\epsilon \partial _yE_z  \label{Hx}
\\
\left( \epsilon \mu \left( \omega /c\right) ^2-\gamma ^2\right) H_y
&=&\partial _y\partial _zH_z+i(\omega /c)\epsilon \partial _xE_z  \label{Hy}
\end{eqnarray}
To derive (\ref{Ex}), for example, all we need is Eq. (\ref{dEx}) and the $z$
derivative of (\ref{dHy}): 
\begin{eqnarray*}
\partial _yH_z-\partial _zH_y &=&-i\left( \omega /c\right) \epsilon E_x \\
\partial _z^2E_x-\partial _z\partial _xE_z &=&i\left( \omega /c\right) \mu
\partial _zH_y
\end{eqnarray*}
Eliminate $\partial _zH_y$ from these two equations and replace $\partial
_z^2E_x$ with $-\gamma ^2E_x$.

\section{Boundary conditions on {\bf E} and {\bf H} }

Jackson is rather vague on this point. For a cavity with perfectly
conducting walls and end walls perpendicular to the $z$ axis: 
\newline\smallskip
\centerline
{$E_z=0$ \ on the side walls and \ $\partial _zE_z=0$ \ on the end walls}
\newline
\centerline
{$\partial _nH_z=0$ \ on the side walls and \ $H_z=0$ \ on the end walls}
\smallskip
To derive these boundary conditions, we note first that $E_{\parallel }$
must be inversely proportional to the conductivity, otherwise the current
would be infinite. This implies that on the side walls $E_z=0,$ and on the
end walls $E_x=0,\,E_y=0$ and their lateral derivatives also vanish. Then we
can use $\partial _xE_x=0$ and $\partial _yE_y=0$ in the divergence equation 
$\partial _xE_x+\partial _yE_y+\partial _zE_z=0$ to conclude that $\partial
_zE_z=0$ on the end walls$.$  We can also use $\partial _xE_y=0$ and $%
\partial _yE_x=0$ in the equation $i\left( \omega /c\right) \mu H_z=\partial
_xE_y-\partial _yE_x,$ from $-(1/c)\partial \mathbf{B}/\partial t=\nabla
\times \mathbf{E,}$ to conclude that $H_z=0$ on the end walls. In general
the normal component $H_n$ vanishes on a wall, and so do its lateral
derivatives.

Finally, we show that the normal derivative $\partial _nH_z$ vanishes on the
side walls. Consider a wall element and orient the $x$ axis normal to it.
Use $-i\left( \omega /c\right) \epsilon E_y=\partial _xH_z-\partial _zH_x,$
from $(1/c)\partial \mathbf{D}/\partial t=\nabla \times \mathbf{H,}$ and
note that $E_y=0,$ because it is a parallel component of $\mathbf{E,}$ and
that $\partial _zH_x=0,$ because it is a lateral derivative of the normal
component of $\mathbf{H}.$ It follows that $\partial _xH_z=0$.  In general, $%
\partial _n\mathbf{H}_{\parallel }=0$ on a wall.

\end{document}

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