\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{Waveguides and Cavities}
\author{John William Strutt\\also known as\\Lord Rayleigh\\(1842 - 1919)}
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\maketitle
\tableofcontents
\pagebreak
In this chapter we continue with the topic of solutions of the Maxwell
equations in the form of waves. This time we seek solutions in the presence
of bounding surfaces which may take a variety of forms. The basic
possibilities are to have boundaries in
\begin{enumerate}
\item one dimension only, such as a pair
of parallel planes;
\item two dimensions, such as several intersecting planes
forming a pipe or channel; and
\item three dimensions, such as a collection
of intersecting planes that completely bound some region of space.
\end{enumerate}
The materials employed to form the boundaries are usually\footnote{Dielectric
materials are also used, with conditions such that total internal
reflection takes place at the surfaces in order to keep the wave within the
channel or cavity.} conductors. The mathematical problem is a
boundary-value problem for solutions of the Maxwell equations. We shall
look at harmonic solutions within the cavity or channel and must match
these solutions onto appropriate ones within the walls or bounding
materials. If the walls are constructed from a ``good'' conductor, the
boundary conditions become simple and the boundary-value problem
itself is not too difficult. This point is explored in the following
sections.
\section{Reflection and Transmission at a Conducting Wall}
We consider the reflection and transmission of a harmonic plane wave
incident on a conducting material at a planar surface. We let the incident wave
have an arbitrary angle of incidence - which gives a hard problem to solve
in the general case - and then imagine that the conductivity is very large -
which simplifies the solution by allowing an expansion in a small parameter.
Physically, the central point is that if $\si>>\om$, then the skin depth
$\de=c/\sqrt{2\pi\si\om\mu}$
of the wave in the conductor is much smaller than the wavelength $\la$ of the
incident wave. The distance over which the fields vary in the conductor
depends on the direction. In the direction normal to the surface, this
distance is $\de$; in directions parallel to the surface, it is $\la$. Thus
by having $\de<<\la$, we can often ignore variations of the fields parallel to
the surface in comparison with variations normal to the surface; in effect,
the wave in the conductor travels normal to the surface no matter what the
angle of incidence.
\subsection{Boundary Conditions}
First, let us consider the boundary or continuity conditions at the
interface. We can find appropriate conditions by using the Maxwell
equations and either Stokes' theorem or Gauss's Law in the usual way. Let
us first do this by employing a rectangle or pillbox which has a size $t$
normal to the interface which is much larger than $\de$. At the
same time, the size $l$ of these constructs parallel to the interface must be
large compared to $t$ but small compared to $\la$, so we have the condition
\beq
\la>>l>>t>>\de
\eeq
\centerline{\psfig{figure=fig1.ps,height=2.5in,width=4.25in}}
{\narrower\em{Fig.1: Integration surfaces adjacent to a good conductor.}}
\noindent which can be satisfied by a metal with a large
enough conductivity (and an incident radiation with a small enough
frequency). Then, because the side of the rectangle, or face of the pillbox,
within the conductor is placed in a region where the transmitted fields
have been attenuated to very small values, compared to the incident
amplitudes, we can say that these fields are zero. The result is that the
continuity conditions become
\beq
\nn\cdot\D=4\pi\si_q\hsph\nn\times\H=\frac{4\pi}c\K\hsph\nn\cdot\B=0\andh
\nn\times\E=0
\eeq
where the fields are those just outside of the conductor, and $\si_q$ and $
\K$ are the charge and current density on the conductor's surface\footnote{
it is only appropriate to talk about surface currents or charges
in the limit of a perfect conductor; otherwise, these densities
will extend into the conductor to a finite extent}; $\nn$ is
the unit outward normal at the surface of the conductor. These relations
are only approximate because we have neglected in particular the term
$\partial\B/\partial t$ in Faraday's Law; it gives a correction of order
$\om\de/c$ times the incident field's amplitude to the statement that the
tangential electric field vanishes at the interface. To put it another way,
the tangential component of the reflected wave's electric field actually
differs from that of the incident wave's electric field by an amount of order
$\om\de/c$ times the amplitude of the incident wave. In lowest order we
ignore this difference. Outside of the conductor, $\partial\B/\partial t$
does not contribute to the integral since we assure the contour
has negligible area here.
So far we don't know the surface charge and current densities, but the
conditions that the tangential component of $\E$ and the normal component of
$\B$ are zero at the interface are already enough to allow us to determine
the reflected fields, given the incident ones. Hence we have at this point
all of the information we need to obtain, to lowest order in the small
parameter, the solution for the waves in the channel or cavity, i.e., the
solution to the boundary-value problem posed above.
\subsection{Power and Energy Loss}
Before going on to look at that problem, however, let's look at the
properties of the transmitted wave in the conductor. The reason for doing
this is that we want to know how much energy is lost in the reflection
process. To zero order in $\om\de/c$, none is lost, as is evident from the
boundary condition which says that the amplitude of the reflected wave is
the same as that of the incident wave. We must therefore look at the
first-order corrections to this result, and that is most easily done by
examining the transmitted wave.
From Faraday's Law and Amp\`ere's Law, assuming a harmonic wave, we find
that the fields in the conductor, which are identified by a subscript $c$,
obey the relations
\beq
\B_c=-i\frac c\om(\curl\E_c)\andh\E_c=\frac c{4\pi\si\mu}(\curl\B_c),
\eeq
where we have ignored the displacement current term because it is
of order $\om/\si$ relative to the real current term; $\mu$ is the permeability
of the conductor. Because the fields vary rapidly in the direction normal to
the interface (length scale $\de$) and slowly in directions parallel to the
interface (length scale $\la$), we may ignore spatial derivatives in all
directions except the normal one. The conditions\footnote{Of course, the
electric displacement has a non-zero divergence if $\rh\ne0$; as we saw in
Jackson 7.7, any initial non-zero $\rh$ dies out with some
characteristic lifetime and so when the steady-state is established, $\rh=
0$.} $\div\B_c=0$ and $\div\D_c=0$ tell us, to lowest order, that the fields
within the conductor have no components normal to the interface.
Taking the curl of the second of Eqs.~(3), and using the first of these
equations for the curl of $\E_c$, we find that
\beq
\lep i+\frac{c^2}{4\pi\si\mu\om}\lap\rip\B_c=0,
\eeq
or
\beq
\lep\lap+i\frac2{\de^2}\rip\B_c=0,
\eeq
which has the solution
\beq
\B_c(z,t)=\B_{c0}e^{\ka z}e^{-i\om t}
\eeq
where
\beq
\ka=\pm(1-i)/\de.
\eeq
Because the fields must vanish for $z\rta\infty$, we have to choose the
negative root and so find
\beq
\B_c=\B_{c0}e^{-z/\de}e^{iz/\de}e^{-i\om t}.
\eeq
Also, as is easily shown from this result and one of Eqs.~(3),
\beq
\E_c=-\frac{c}{4\pi\si\mu\de}(1-i)(\zh\times\B_c).
\eeq
These fields are the same in form as the ones that arise in the case of
normal incidence. The amplitudes are somewhat different from that case,
however.
The power per unit area entering the conductor is
\beqa
<\S\cdot\zh>|_{z=0}=\frac c{8\pi}\Re(\E_c\times\H_c^*)\cdot\zh|_{z=0}
\nonumber\\=-\frac c{8\pi}\Re\lec\frac{c}{4\pi\si\mu\de}(1-i)[(\zh\times\B_c)
\times(\B_c^*/\mu)]\cdot\zh|_{z=0}\ric\nonumber\\
=\frac{c^2}{32\pi^2\si\mu^2\de}\Re[(1-i)|\B_c|^2]|_{z=0}.
\eeqa
However, $c^2/2\pi\si\om\mu=\de^2$, so, writing the power per unit area as
${\cal P}$, we have
\beq
{\cal P}=\frac{\mu\om\de}{16\pi}|H_c|^2_{z=0}.
\eeq
We can relate $H_c$ at $z=0$ to the field at the interface on the outside
of the conductor by employing an appropriate continuity condition. It is
not the one derived above. This time, we use a value of $t$ which is much
smaller than $\de$ so that there is only a negligible amount of current
passing through the rectangle employed in applying Stokes' Theorem. Then
we find that $H_c$ at $z=0$ is the same as the tangential component of the
magnetic field on the outside. For definiteness, let the incident field
be polarized perpendicular to the plane of incidence. Then the reflected field
has an equal and opposite amplitude (to lowest order) and the sum of the
incident and reflected waves' magnetic field amplitudes parallel to the
interface is\footnote{We suppose that the
exterior medium is vacuum, or at least has $\mu=\ep=1$.} twice the amplitude of
the incident electric field times the cosine of the angle of incidence, or
$2E_0\cos\th$.
\centerline{\psfig{figure=fig2.ps,height=1.875in,width=4.25in}}
{\narrower\em{Fig.2: Wave polarized $\perp$ to the plane, between vacuum
and a good conductor.}}
\noindent Hence the power loss per unit area in the reflection process,
meaning the power per unit area entering the conductor and so not reflected,
is
\beq
{\cal P}=\frac{\mu\om\de}{4\pi}|E_0|^2\cos^2\th.
\eeq
The ratio of the lost to incident power, which is also the transmission
coefficient, is
\beq
T=\frac{\mu\om\de|E_0|^2\cos^2\th/4\pi}{c|E_0|^2\cos\th/8\pi}=\frac{2\mu
\om\de}c\cos\th.
\eeq
This agrees with the result we found earlier in the case of a good
conductor and for normal incidence, $\th=0$.
We will want to use the result for power loss later in connection with the
attenuation of waves travelling along a wave guide. First we shall obtain
the solution for the electromagnetic field within the waveguide in the
limit of perfectly conducting walls.
\section{Wave Guides}
A waveguide is a hollow conducting pipe, perhaps filled with dielectric. It
has a characteristic transverse size on the order of centimeters and is used to
transmit electromagnetic energy (waves) from one place to another.
\centerline{\psfig{figure=fig3.ps,height=1.5in,width=6.375in}}
{\narrower\em{Fig.3: Wave guide of arbitrary cross-section.}}
\noindent
The waves typically have frequencies such that the wavelength in vacuum would
be comparable to the size of the waveguide. Thus $\om=2\pi c/\la$ is of
order $10^{11}\,sec^{-1}$. If the walls of the guide are constructed of a
good conducting material, i.e., one with $\si\sim10^{17}\,sec^{-1}$, then
we are in the good conducting limit so that the treatment of the previous
section is valid. In particular, $T\sim10^{-3}$ which means that some
$10^3$ reflections can take place before the wave is seriously attenuated.
Also, we may adopt the boundary conditions that $\E_{tan}=0=B_n$ at the
conducting surfaces.
\subsection{Fundamental Equations}
Let the wave guide have its long axis parallel to the $z$-direction and let
its cross-section be invariant under translation along this direction. It
is useful to divide operators, such as $\grad$ and $\lap$, and also fields
into components parallel and perpendicular to the long axis. Thus we write
\beqa
\grad=\grad_t+\ept\pde{}{z},\hsph\lap=\lap_t+\spd{}{z},\nonumber\\
\E=\E_t+\ept E_z,\andh\B=\B_t+\ept B_z.
\eeqa
Further we shall assume that the fields' dependence on both $z$ and $t$ is
harmonic,
\beq
\Bxt=\B(x,y)\ekzot\andh\Ext=\E(x,y)\ekzot.
\eeq
Given $\om$, we need to find $k$ and the amplitudes $\B(x,y)$, $\E(x,y)$.
Letting the material within the guide have dielectric constant $\ep$ and
permeability $\mu$, and assuming no macroscopic sources in this region, we
can derive wave equations using familiar methods. Because of the
harmonic time dependences, the Maxwell equations read
\beq
\curl\E=i\frac\om c\B\hsph\div\B=0\hsph\div\E=0\hsph\curl\B=-i\frac\om c\mu
\ep\E.
\eeq
Taking the curl of each curl equation and using the forms of the fields as
well as the fact that both fields have zero divergence, we find that the
wave equations for all Cartesian components of $\E(x,y)$ and $\B(x,y)$ have the
same form; it is
\beq
\lep\lap_t-k^2+\mu\ep\frac{\om^2}{c^2}\rip\ps(x,y)=0.
\eeq
We can greatly simplify things by noting that if we find $E_z$ and $B_z$
first, then $\E_t$ and $\B_t$ follow. To demonstrate this statement, we
shall derive explicit expressions for the latter in terms of the former.
Consider the transverse components of the curl of the magnetic induction,
\beq
\leb\curl(B_z\ept+\B_t)\rib_t=(\grad_tB_z)\times\ept+\ept\times\lep\pde{
\B_t}z\rip=-i\mu\ep\frac\om c\E_t.
\eeq
Cross $\ept$ into this equation to find
\beq
\grad_tB_z-\pde{\B_t}z=-i\mu\ep\frac\om c(\ept\times\E_t).
\eeq
By similar means, one finds that the transverse part of Faraday's Law can
be written as
\beq
(\grad_tE_z)\times\ept+\ept\times\lep\pde{\E_t}z\rip=i\frac\om c\B_t.
\eeq
Take the derivative with respect to $z$ of the first of these equations and
substituted the result into the second equation; the result is
\beq
\lep\mu\ep\frac{\om^2}{c^2}+\spd{}z\rip\B_t=\grad_t\lep\pde{B_z}z\rip+
i\ep\mu\frac{\om}c\ept\times(\grad_tE_z).
\eeq
Because the fields' $z$-dependence is $\ekz$, we may take the derivatives
with respect to $z$ and express the result as an equation for $\B_t$:
\beq
\B_t=\frac1{\mu\ep\om^2/c^2-k^2}\leb\grad_t\lep\pde{B_z}z\rip+i\mu\ep
\frac\om c\ept\times(\grad_tE_z)\rib.
\eeq
In the same fashion, we can take the derivative with respect to $z$ of \eq{20}
and substitute into \eq{19} to find a relation for $\E_t$,
\beq
\E_t=\frac1{\mu\ep\om^2/c^2-k^2}\leb\grad_t\lep\pde{E_z}z\rip-i\frac\om c
\ept\times(\grad_tB_z)\rib.
\eeq
Now we need only solve for the $z$ components of the fields; from them and
the preceding relations, all components are determined.
\subsubsection{Boundary Conditions}
The boundary
condition on $E_z(x,y)$ is that it should vanish at the walls because the
tangential component of the electric field is zero there. The other
boundary condition, $B_n=0$, does not put any constraint on $B_z$; however,
there is a constraint on $B_z$ which can be extracted from the equation for
$\E_t$; one of the two components of $\E_t$ is tangential to the wall and this
one must vanish next to the wall. From \eq{23} we see that there is a
contribution to that component which is proportional to $\partial B_z/
\partial n$, so we conclude that the boundary condition on $B_z$ is
$\partial B_z/\partial n=0$. The other contribution to $\E_t$ is
proportional to the transverse gradient of $E_z$ at the boundary; since
$E_z$ is zero at all points on the boundary, it is clear that this term
will not give any tangential component of $\E_t$ at the boundary. Hence the
tangential components of $\E$ are zero on the boundary provided $E_z$
vanishes there along with the normal component of the gradient of $B_z$.
And what of the normal component of $\B$ itself? From \eq{22} for $\B_t$,
we see that the the normal component of $\B$ at the wall vanishes if, first,
the gradient of $B_z$ has zero normal component there, and, second, the
component of $\grad_tE_z$ parallel to the wall vanishes; these conditions are
met if $E_z=0$ and $\partial B_z/\partial n=0$ everywhere on the boundary.
Hence we are left with the following boundary-value, or eigenvalue, problem:
\beq
\lep\lap_t-k^2+\mu\ep\frac{\om^2}{c^2}\rip\lec\barr{c}E_z(x,y)\\B_z(x,y)
\earr\ric=0,
\eeq
with
\beq
E_z(x.y)=0\andh \partial B_z(x,y)/\partial n=0
\eeq
on the boundary.
\subsection{Transverse Modes}
Depending on the geometry, it may or may not be possible to find an
eigenvalue $k^2$ such that the conditions on $B_z$ and $E_z$ are both
satisfied. If it is not possible, then either $B_z\equiv0$ or $E_z\equiv0$.
In the former case, $\B$ is purely transverse and one speaks of a {\em
transverse magnetic mode}, often abbreviated as a TM mode; in the latter
case, $\E$ is purely transverse and the mode is called a {\em transverse
electric mode}, or a TE mode. For some geometries it is possible to have both
$E_z$ and $B_z$ identically zero although the transverse fields are finite;
then we have a {\em transverse electromagnetic mode} or a TEM mode.
\begin{center}\begin{tabular}{||l|l||} \hline\hline
MODE & CHARACTER \\ \hline\hline
TM (Transverse Magnetic) & $B_z=0$ \\ \hline
TE (Transverse Electric) & $E_z=0$ \\ \hline
TEM & $E_z=B_z=0$ \\ \hline\hline
\end{tabular}\end{center}
\subsubsection{TEM Mode}
Let's briefly discuss the TEM modes first. In order to see what are the
appropriate equations of motion of the fields, we have to go back to the
Maxwell equations. If we look just at the $z$-component of Faraday's Law
and of the generalized Amp\`ere's Law,
we find that
\beq
\ept\cdot(\grad_t\times\E_t)=0\andh\ept\cdot(\grad_t\times\B_t)=0.
\eeq
Since $\grad_t$ and $\E_t$ have only $x$ and $y$ components, the curls lie
entirely in the $z$ direction, so we can write
\beq
\grad_t\times\E_t=0\andh\grad_t\times\B_t=0.
\eeq
From the other two Maxwell equations we find that
\beq
\grad_t\cdot\E_t=0\andh\grad_t\cdot\B_t=0.
\eeq
The two transverse fields are solutions of problems identical to statics
problems in two dimensions. In particular, the transverse electric field,
which must have zero tangential component at the walls of the waveguide,
is found by solving an electrostatics problem. If the wave guide is
composed of a single conductor, which is an equipotential in the equivalent
electrostatics problem, then there is no nontrivial solution.
{\em{The conclusion is that a simple single-conductor waveguide cannot
have a TEM mode to lowest order in the parameter $\delta\om/c$}}. Where are there
TEM modes? These exist in guides which consist of a pair of parallel but
electrically unconnected conductors that can be at different potentials.
Such things are called {\em transmission lines}.
\subsubsection{TE and TM Modes}
Now let's turn to TE and TM modes. We will discuss here only TE modes; TM
modes can be treated by making a simple modification of the boundary
conditions when solving the eigenvalue problem. For TE modes, $E_z=0$ and so
the transverse fields are given in terms of $B_z$ by
\beq
\B_t=\frac1{\ga^2}\grad_t\lep\pde{B_z}z\rip=\frac{ik}{\ga^2}(\grad_tB_z)
\eeq
and
\beq
\E_t=-\frac i{\ga^2}\frac\om c\ept\times(\grad_tB_z)=-\frac\om{ck}(\ept
\times\B_t),
\eeq
where $\ga^2\equiv\mu\ep\om^2/c^2-k^2$. This parameter must be determined
by solving the eigenvalue equation
\beq
(\lap_t+\ga^2)B_z(x,y)=0,\hsph\mbox{with } \partial B_z/\partial n=0\mbox{ on
C};
\eeq
C is the boundary of the (cross-section of the) waveguide.
This problem will have solutions $B_z(x,y)\rta B_i(x,y)$ with eigenvalues
$\ga_i^2$, $i$=1,2,... . In terms of these eigenvalues, the wavenumber $k_i$
is given by
\beq
k_i^2(\om)=\mu\ep\om^2/c^2-\ga^2_i,
\eeq
which means that for given $\om$ and $i$, $k$ is determined.
The eigenvalues $\ga_i^2$ are always positive (otherwise the boundary
conditions cannot be satisfied), so we can see from the preceding equation
that for a given mode $i$, $\om^2$ must be larger than $\om_i^2\equiv\ga_i^2c^2
/\mu\ep$ in order for the squared wavenumber to be positive corresponding
to a real wavenumber $k$. If $\om^2$ is smaller than this cutoff value,
then the wavenumber is imaginary and the wave is attenuated as it moves in
the $z$-direction. As a particular consequence, one can choose $\om$ such
that only some fixed number of modes (one, for example) can propagate.
There are two velocities of interest in connection with any mode; these
are the phase velocity and the group velocity. The dispersion relation is
\beq
\om^2=\frac{c^2}{\mu\ep}(k^2+\ga_i^2),
\eeq
which can be expressed also as
\beq
\frac{\om^2}{k^2}=\frac{c^2/\mu\ep}{1-\om_i^2/\om^2}.
\eeq
From this form, one can see clearly that the phase velocity, $\om/k$, is
always larger than the phase velocity in the absence of walls,
$c/\sqrt{\ep\mu}$. Further, the phase velocity diverges as $\om$ approaches the
cutoff frequency. As for the group velocity, we have $v_g=d\om/dk$, and
\beq
\om\der\om k=\frac{c^2}{\mu\ep}k\orh\frac\om k\der\om k=\frac{c^2}{\mu\ep}
=v_pv_g.
\eeq
This equation tells us that the product of the group and phase velocities is a
constant, $c^2/\mu\ep$; the group velocity itself is
\beq
v_g=\frac{c^2}{\mu\ep}\frac k\om=\frac c{\sqrt{\ep\mu}}\sqrt{1-\frac{
\om_i^2}{\om^2}}
\eeq
which is always smaller than $c/\sqrt{\ep\mu}$.
\subsection{Energy Flow} The significance of the group velocity of the mode
becomes clear from a study of the energy flow in the guide. The time-average
of the Poynting vector's $z$ component is
\beq
<\S\cdot\ept>=\frac c{8\pi}\Re(\E_t\times\H_t^*)\cdot\ept=\frac c{8\pi}
\frac{\om k}{\mu c\ga^4}|\grad_tB_z|^2.
\eeq
This must be integrated over the cross-section of the guide to find the
power transmitted,
\beqa
{\cal P}=\ina<\S\cdot\ept>=\frac c{8\pi}\frac{\om k}{\mu c\ga^4}\ina|
\grad_tB_z|^2\nonumber \\=\frac{\om k}{8\pi\mu\ga^4}\ina[\grad_t\cdot(
B_z^*\grad_tB_z)-B_z^*\lap_tB_z].
\eeqa
The first term in the final expression converts to a surface integral which
is shown, from the boundary conditions on $B_z$, to be zero; the second term is
made simpler in appearance by using the fact that $\lap_tB_z=-\ga^2B_z$. Thus,
\beq
{\cal P}=\frac{\om k}{8\pi\mu\ga^2}\ina|B_z|^2.
\eeq
Compare this with the time-averaged energy per unit length in the guide
\beqa
U=\frac1{16\pi}\ina\lep\ep\E_t\cdot\E_t^*+\frac1\mu[\B_t\cdot\B_t^*+
B_zB_z^*]\rip\nonumber\\
=\frac1{16\pi}\ina\leb\ep\frac{\om^2}{c^2k^2}|\B_t|^2+\frac1\mu|\B_t|^2+
\frac1\mu|B_z|^2\rib\nonumber\\=\frac1{16\pi\mu}\ina\leb\lep\ep\mu\frac{
\om^2}{c^2\ga^4}+\frac{k^2}{\ga^4}\rip|\grad_tB_z|^2+|B_z|^2\rib\nonumber\\
=\frac1{16\pi\mu}\ina\leb\lep\ep\mu\frac{\om^2}{c^2}+k^2\rip\frac1{\ga^2}+1
\rib|B_z|^2=\frac1{8\pi}\lep\ep\frac{\om^2}{c^2}\rip\frac1{\ga^2}\ina|B_z|^2.
\eeqa
In arriving at the final result, we've used a whole collection of
identities related to the eigenvalue problem.
Comparison of $U$ and ${\cal P}$ shows that
\beq
\frac{{\cal P}}{U}=\frac k\om\frac{c^2}{\mu\ep}\equiv v_g.
\eeq
The obvious interpretation need not be stated.
\subsubsection{TE Modes in Rectangular and Circular Guides}
Before going on to other matters, let us look at the solutions of the
eigenvalue problem for some standard waveguide shapes, i.e., rectangles and
circles, assuming TE modes.
\centerline{\psfig{figure=fig4.ps,height=1.5in,width=6.375in}}
{\narrower\em{Fig.4: Geometry of Rectangular and Circular Wave Guides.}}
\noindent For the rectangle shown, the
solution\footnote{The eigenvalue problem for TE modes is formally equivalent to
that of a quantum mechanical particle in a box with somewhat unusual boundary
conditions; the case of the TM mode has the usual boundary conditions.} for
$B_z$ is
\beq
B_{mn}(x,y)=B_0\cos\lep\frac{m\pi x}a\rip\cos\lep\frac{n\pi y}b\rip
\eeq
with
\beq
\ga_{mn}^2=\pi^2\lep\frac{m^2}{a^2}+\frac{n^2}{b^2}\rip.
\eeq
For the circular guide of radius $a$, on the other hand, $B_z$ becomes
\beq
B_{mn}(\rh,\ph)=B_0e^{im\ph}J_m(y_{mn}\rh/a)
\eeq
with
\beq
\ga_{mn}^2=y^2_{mn}/a^2.
\eeq
Here, $y_{mn}$ is the $n^{th}$ zero of the derivative of the Bessel
function $J_m$, $J_m'(y)|_{y_{mn}}=0$.
The question of which modes will actually be excited in a waveguide for a
given source can be worked out (see below); one has to address the question of
how the source couples to the eigenfunctions for the different modes.
Different configurations of the source will produce different
superpositions of modes. A simple way to guarantee that just one
propagating mode will be present is to pick $\om$ smaller than the cutoff
frequencies of all modes but one.
\section{Attenuation of Modes in Waveguides}
Even modes for which the wave number is real will be somewhat damped
because of the finite conductivity of the walls. We have seen that the
power travelling down the pipe is given by \eq{39}; also, the power lost per
unit length is, from \eq{11},
\beq
\der{{\cal P}}z=-\frac{\mu\om\de}{16\pi}\inlc|\H\pll|^2
\eeq
where $\H\pll$ is the component of $\B/\mu$ which is parallel to the
boundary, and the integral is evaluated on the contour formed by the
cross-section of the guide.
We can evaluate $|\H\pll|^2$ up to a point. First, for TE modes
\beq
|\H\pll|^2=|\nn\times\H|^2=\frac1{\mu^2}(|B_z|^2+|\nn\times\B_t|^2)
=\frac1{\mu^2}\leb|B_z|^2+\frac{k^2}{\ga^4}|\nn\times(\grad_tB_z)|^2\rib.
\eeq
Further, at the surface $|\nn\times(\grad_tB_z)|^2=|\grad_tB_z|^2$
(since $\partial B_z/\partial n =0 $ there).
The latter can be expected to be comparable to $|B_z^*\lap_tB_z|=\ga^2
|B_z|^2$, so we write
\beq
|\grad_tB_z|^2=\xi\ga^2|B_z|^2,
\eeq
where $\xi$ is a mode-dependent dimension-free constant of order unity that
is independent of the frequency\footnote{For a given geometry of the guide,
and a particular mode therein, one may easily calculate $\xi$.}.
Hence,
\beq
|\nn\times\H|^2=\frac1{\mu^2}\lep1+\xi\frac{k^2}{\ga^2}\rip|B_z|^2
\eeq
and
\beq
\der{{\cal P}}z=-\frac{\mu\om\de}{16\pi}\frac1{\mu^2}\lep1+\xi\frac{k^2}{
\ga^2}\rip\inlc|B_z|^2
\eeq
This is to be compared with the power itself which is obtained by
integrating $|B_z|^2$ over a cross-section of the guide. We convert the
integral in the preceding equation into one over the cross-section by
noticing that $|B_z|$ behaves on the boundary in much the same way as in
the interior of the guide, so that
\beq
\inlc|H_z|^2=\frac CA\et\ina|H_z|^2
\eeq
where $A$ and $C$ are the cross-sectional area and circumference of the
guide and $\et$ is another dimension-free constant of order unity; it
depends on the shape of the guide and on the mode but not on the frequency.
Substituting this expression into \eq{50} and dividing by the power, given
by \eq{29}, we find an attenuation coefficient $\be$,
\beq
\be\equiv-\left.\der{{\cal P}}z\ris{\cal P}=\frac{\de}{2k}\et\frac CA
(\ga^2+\xi k^2).
\eeq
Substitute for $k$ using $k^2=\ep\mu\om^2/c^2-\ga^2$, and replace $\ga$
using $\ga^2=\ep\mu\om_i^2/c^2$; $\om_i$ is the cutoff frequency of the
mode. Then we find
\beq
\be=\frac{\sqrt{\ep\mu}}{2}\,\sqrt{\frac{\om_i}\om}\,\frac{\om_i}{
\sqrt{\om^2-\om_i^2}}\,\frac{\eta C}{A}\,\frac{\de_i\om_i}c
\leb1+\xi\lep\frac{\om^2-\om^2_i}{\om^2_i}\rip\rib;
\eeq
$\de_i$ is the penetration depth at cutoff. All of the frequency dependence
of $\be$ is explicit in this result. The damping becomes very large as
$\om\rta\om_i$; for $\om$ not too close to the cutoff for mode $i$, and for
$\si\sim10^{17}\,sec^{-1}$, we can see that the wave can travel some hundreds
of meters without disastrous attenuation. At very high frequencies,
$\om>>\om_i$, the attenuation increases once again.
The usefulness of our result is limited; in particular, it breaks down when the
frequency approaches the cutoff frequency for the mode. One can do a better
calculation by improving the treatment of the boundary conditions. The
solution (for the fields) that we have found satisfies boundary conditions
only slightly different from the correct ones; it turns out, not
surprisingly, that an improved solution can be obtained by looking for small
corrections to the fields we already have, with the corrections determined
by demanding that the exact boundary conditions be satisfied to one higher
order in $\om/\si$. This is a straightforward but somewhat technical
calculation and we shall not spend time on it.
\section{Resonating cavities}
The step from a waveguide to a resonating cavity is not a large one. We
need only think about what new constraints are placed on the
electromagnetic fields if end walls are placed on a waveguide. Suppose that
such walls are introduced at $z=0$ and $z=d$. Then we cannot have traveling
waves in the guide but must have instead standing waves; that is, in
addition to the fields varying as $\ekzot$, we must have ones that vary as
$\emkzot$ which means that the $z$ and $t$-dependent parts of the fields
can be expressed as some linear combination of $\sin(kz)\eot$ and
$\cos(kz)\eot$. Using such a combination, one may proceed as in the previous
sections and will find in particular that $E_z(x,y)$ and $B_z(x,y)$ satisfy the
same eigenequations as before.
\centerline{\psfig{figure=fig5.ps,height=1.5in,width=6.375in}}
{\narrower\em{Fig.5: Geometry of a Resonant Cavity.}}
\noindent
What is different in a cavity is that there are new boundary conditions or
constraints on the fields because of the presence of end walls. If
$B_z\ne0$, then one must add the condition that $B_z=0$ at $z=0$ and $z=d$.
Also, in order to guarantee that the tangential component of $\E$, (or $\E_t$)
vanish on the end walls, it must be the case that $\partial E_z/\partial z=0$
on the end walls; see \eq{23}. These are the additional conditions that
must be satisfied for a cavity.
Following tradition established in earlier sections, we examine only TE
modes. Then $B_z$ must vanish at the ends of the cavity, meaning that the
solution we want is the one having $z$-dependence $\sin(kz)$; this is zero
at $z=0$, and we make it zero at $z=d$ by choosing $k=p\pi/d$ with $p$ an
integer. It is also true, as for the waveguide, that $k$ must satisfy the
equation $k^2=\mu\ep\om^2/c^2-\ga_{mn}^2$. Both of these conditions on $k^2$
can be satisfied simultaneously only for certain discrete frequencies
$\om_{mnp}$ which are given by
\beq
\om_{mnp}^2=\frac{c^2}{\mu\ep}\lep\frac{p^2\pi^2}{d^2}+\ga_{mn}^2\rip.
\eeq
Hence a resonating cavity has a set of discrete natural ``resonant''
frequencies at which it can support a standing wave electromagnetic field.
These frequencies can be tuned by adjusting the size of the cavity, e.g.,
by changing $d$.
A resonant cavity is useful in that if excited in a single mode, it will
contain monochromatic radiation in the microwave frequency range
(ie.e a maser), or would do
so if it were perfect. However, the resonance is never perfectly sharp in
frequency, meaning that if one Fourier transforms the fields into frequency
space (instead of time) the result will not be a delta function at the
resonant frequency. There are several contributing factors to the width of the
resonance. One important factor is the power loss in the walls. This loss
is generally characterized by the ``$Q$'' of the cavity defined by
\beq
Q\equiv\om_0\lep\frac{\mbox{stored energy}}{\mbox{power loss}}\rip
\eeq
where $\om_0$ is the frequency of the mode in the cavity. In words, $Q$ is
$2\pi$ times the energy stored in the cavity divided by the energy loss per
cycle. From this
definition it follows that the connection between $Q$ and the rate at which the
energy stored in the cavity decays is $dU/dt=-U\om_0/Q$ when no new energy is
being pumped into the cavity. If $Q$ is independent of the amount of energy
in the cavity, then this differential equation has a simple exponential
solution,
\beq
U=U_0e^{-\om_0t/Q},
\eeq
and, under these circumstances, any field\footnote{That is, any component of
$\E$ or $\B$.} $\ps$ behaves in time as
\beq
\ps(t)=\ps_0e^{-i\om_0t}e^{-\om_0t/2Q}.
\eeq
Because of the decay, the frequency spectrum of the field contains
components in addition to $\om_0$. Specifically,
\beq
\ps(\om)\sim\int_0^\infty dte^{i(\om-\om_0)t}e^{-\om_0t/2Q}=\frac1{i(\om_0-
\om)+\om_0/2Q},
\eeq
or
\beq
|\ps(\om)|^2\sim\frac1{(\om_0-\om)^2+\om_0^2/4Q^2}
\eeq
which means that the resonance has a width in frequency space which is of
order $\om_0/2Q$.
And what is the value of $Q$? The energy (or fields) in the cavity decay in
time because of losses in the walls. We learned in section 1 how to compute
these losses. For any given mode in a particular cavity it is a
straightforward matter to do the calculations, provided one can solve for
the fields in that mode. One proceeds in much the same way as in the
previous section where we learned how to calculate the power loss in a
waveguide. Skipping over the details of the argument, which are much like
the calculation of $\be$ for a mode of a waveguide, we simply state the
conclusion which is that the energy lost per period is of the order of the
energy in the cavity times the ratio of the volume of the walls into which
the field penetrates to the volume of the cavity. If the area of the
walls is $A$ and the volume of the cavity is $V$, then
\beq
Q\sim\frac V{A\de}.
\eeq
Thus the relevant parameter for determining the $Q$ of the cavity is the ratio
of the penetration depth to the linear size of the cavity. For $\de$ on the
order of microns, or $10^{-4}\,cm$, and a cavity having a size on the order of
a centimeter, the $Q$ will be on the order of $10^3$ to $10^4$.
\edo