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%TCIDATA{Created=Mon Sep 16 18:15:19 1996}
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\markright{{\sf Power losses in cavities and waveguides}}
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\begin{document}


\section{8.5\quad Power losses in cavities and waveguides.}

\quad \quad (Covers also 8.6 and 8.8)

\subsection{8.5.1}

\quad \thinspace \textbf{a.} Power losses can occur for many reasons:

\quad \quad - currents flowing within the skin depth in the walls
(unavoidable ohmic losses);

\quad \quad - leaks in the walls (in an open structure, such as a
transmission line, deviations from cylindrical symmetry result in radiation
losses)

\quad \quad - absorption in the dielectric filling the cavity or guide, or
in samples deliberately placed in the cavity. Magnetic losses can occur too,
as in all kinds of magnetic resonance experiments

\smallskip\ 

\textbf{b}. In this section we concentrate on the power losses in the walls,
assuming that the effect of finite penetration on the field configuration in
the guide, or cavity, is negligible. This makes the problem tractable in a
simple and general way.

\smallskip\ 

\textbf{c. }According to formulae (8.1.6) and (8.1.7), the power loss per
unit area of the walls is 
\[
\left( \frac c{4\pi }\right) ^2\frac 1{2\sigma \delta }\left| \mathbf{H}%
_{/\!/}\right| ^2=\frac{\mu _c\omega \delta }{16\pi }\left| \mathbf{H}%
_{/\!/}\right| ^2\quad \quad \quad \text{with \quad }\delta =\frac c{\sqrt{%
2\pi \mu _c\omega \sigma }}\,. 
\]

\smallskip\ 

\textbf{d. }Note that we have expressed the power loss in terms of $H_{/\!/}$
at the wall, rather than in terms of $E_{/\!/},$ because $H_{/\!/}$ can be
well approximated by the value appropriate to the case of perfectly
conducting walls.

\smallskip\ 

\textbf{e. }The calculation of power losses is straightforward and will be
sketched in outline only.

\newpage 

\subsection{8.5.2}

\quad \thinspace \thinspace \textbf{a.} In a wave guide, the losses are
characterized by the imaginary part of the wave number, which is now written
as $k+i\alpha /2,$ for waves propagating in the $+z$ direction. In agreement
with the notation of Ch.7, $\alpha $ is an attenuation coefficient, which in
the case $k\gg \alpha $ (propagating waves well above cutoff) has the
meaning of an \textsl{absorption coefficient}. The fields behave as $\exp
(ikz)\exp (-\alpha z/2)$ and the energy density $U$ behaves as $\exp
(-\alpha z)$.

\smallskip\ 

\textbf{b.} The power transmitted in the guide, $P,$ is found by integrating
the $z$ component of the Poynting vector, $(c/8\pi )\func{Re}(\mathbf{%
E\times H}^{*})$ over the cross section. Being quadratic in the fields, $P$
also behaves as $\exp $(-$\alpha z).$ The most convenient way of determining 
$\alpha ,$ observationally, is from 
\[
\alpha =-\frac 1P\frac{dP}{dz}=-\frac{d\ln P}{dz} 
\]
(hence the name \textsl{logarithmic decrement}).

\smallskip\ 

\textbf{c. }From the general theory of wave propagation (see Jackson p.301) $%
P=Ud\omega /dk$. Differentiating the relation (3.5), $\gamma ^2=\mu
\varepsilon \left( \omega ^2/c^2\right) -k^2$ gives $\omega d\omega
=(c^2k/\mu \varepsilon )dk$, so that 
\[
P=U\frac k\omega \frac{c^2}{\mu \varepsilon }\,, 
\]
in the absence of losses.

\smallskip\ 

\textbf{d. }The power loss per unit length, $-dP/dz,$ is found by
integrating the power loss per unit area, from 1.c, over the perimeter of
the guide.

\smallskip\ 

\textbf{e. }From \textbf{b}., \textbf{c}. and \textbf{d}., 
\[
\alpha =\frac{\dfrac{\mu _c\omega \delta }{16\pi }}{\dfrac k\omega \dfrac{c^2%
}{\varepsilon \mu }\dfrac \mu {8\pi }}\frac{\int \left| \mathbf{H}%
_{/\!/}\right| ^2dl}{\int \left| \mathbf{H}\right| ^2dA} 
\]
and the ratio of integrals is of the order $G/A$, where $G$ is the girth and 
$A$ the cross section of the guide.

\newpage \ 

\subsection{8.5.3}

\quad \thinspace \thinspace \textbf{a.} From previous page, we conclude
that, up to a factor of order unity 
\begin{equation}
\frac \alpha k=\frac 12\left( \frac{\mu _c}\mu \right) \left( \frac{%
\varepsilon \mu \omega ^2}{c^2k^2}\right) \left( \frac{G\delta }A\right) .
\label{1}
\end{equation}
This is written in a way that emphasizes dimensionless ratios. It is
important to note that the theory holds only for $\alpha \ll k$ and fails
near cutoff (where $k\rightarrow 0).$

\smallskip\ 

\textbf{b.} For the mode characterized by the eigenvalue $\gamma _\lambda ^2$
and the cutoff frequency $\omega _\lambda =c\gamma _\lambda /\sqrt{\mu
\varepsilon },$ the dispersion relation is 
\[
k=k_0\sqrt{1-\left( \frac{\omega_\lambda}\omega \right) ^2}\quad \quad \quad 
\text{with\quad }k_0=\frac \omega c\sqrt{\mu \varepsilon }\,. 
\]
We can rewrite (\ref{1}) as 
\begin{equation}
\frac \alpha {k_0}=\frac 12\left( \frac{\mu _c}\mu \right) \frac 1{\sqrt{%
1-\left( \omega _\lambda /\omega \right) ^2}}\left( \frac{G\delta }A\right)
\label{1'}
\end{equation}

\smallskip\ 

\textbf{c.} An estimate of the missing ``factor of order unity'' in (\ref{1}%
) or (\ref{1'}) is given by Jackson. The factor does not depend on $\omega $
for TM modes and then $\alpha $ is minimum at $\omega =\sqrt{3}\omega
_\lambda .$ A plot of $\alpha $ vs. $\omega $ is in Fig. 8.6 of Jackson. The
divergence near cutoff just indicates that $\alpha $ becomes comparable to $%
k $ and then the simple theory fails. Typical attenuation lengths for copper
guides in the microwave region are a few hundred meters.

\smallskip\ 

\textbf{d.} There is also a shift in the real part of $k$ due to the finite
penetration in the walls. A rather straightforward application of
perturbation theory to the boundary value problem (Jackson, Sect. 8.6) shows
that 
\[
\left( k+\Delta k\right) ^2=k^2+(1+i)k\alpha 
\]
where $k$ and $\alpha $ are given in \textbf{b}. This formula is also valid
at or below cutoff, where (\ref{1'}) fails. Well above cutoff $\func{Re}%
(\Delta k)\simeq \func{Im}(\Delta k)\simeq \alpha /2.$

\newpage 

\subsection{8.5.4}

\quad \thinspace \textbf{a.} In a cavity, the losses are conveniently
characterized by 
\begin{equation}
Q=\frac{2\pi \,\,\text{Stored energy}}{\text{Energy loss per cycle}}
\label{2}
\end{equation}

\smallskip\ 

\textbf{b.} From 1c and the expression $(\mu /8\pi )\left| \mathbf{H}\right|
^2$ for the energy density 
\[
Q=\frac \mu {8\pi }\frac{16\pi }{\mu _c\omega \delta }\omega \frac{\int
\left| \mathbf{H}\right| ^2dV}{\int \left| \mathbf{H}_{/\!/}\right| ^2dS}\,. 
\]
The last ratio in this formula is about equal to the volume to surface ratio 
$V/S.$ Thus, up to a factor of order unity 
\begin{equation}
Q=\frac \mu {\mu _c}\frac V{S\delta }\,.  \label{2'}
\end{equation}

\smallskip\ 

\textbf{c. }This result, and the analogous result 2e for the ratio $\alpha
/k,$ are simply interpreted: $S\delta $ is the effective volume where losses
occur.

\smallskip\ 

\textbf{d.} A detailed calculation of the integrals in \textbf{b}. (see
Jackson) shows that for the lowest TM mode (with $m=0,n=1,p=0)$ in a
circular cylinder of length $d$, radius $R$%
\[
Q=\frac \mu {\mu _c}\frac d\delta \left( \frac 1{1+d/R}\right) . 
\]
For the lowest TE mode $(m=1,n=1,p=1)$ a slightly messier expression holds
(see Jackson).

\smallskip\ 

\textbf{e.} A cavity driven at a frequency near $\omega _\lambda $ will
respond as a harmonic oscillator of frequency $\omega _\lambda $ and damping
constant $\omega _\lambda /2Q$ (see next page).

\newpage 

\subsection{8.5.5}

\quad \thinspace \textbf{a.} This is a good time to review (once again) the
subject of linear response and resonance. A straightforward treatment starts
from the differential equation for a harmonic oscillator of unit mass,
frequency $\omega _0$ and damping constant $\gamma $ being driven by an
external force $F(t):$%
\begin{equation}
\frac{d^2x}{dt^2}+\omega _0^2x=-\gamma \frac{dx}{dt}+F(t).  \label{3}
\end{equation}

\smallskip\ 

\textbf{b.} If $F(t)=F(\omega )\exp (-i\omega t),$ the solution of (\ref{3})
gives $x=L(\omega )F(\omega )\exp (-i\omega t),$ with 
\begin{equation}
L(\omega )=\left[ \omega _0^2-\omega ^2-i\gamma \omega \right] ^{-1}\,.
\label{4}
\end{equation}

\smallskip\ 

\textbf{c.} For a general $F(t)=\int (d\omega /2\pi )F(\omega )\exp
(-i\omega t),$ the solution of (\ref{3}) is 
\[
x(t)=\int_{-\infty }^{+\infty }\frac{d\omega }{2\pi }L(\omega )F(\omega
)\exp (-i\omega t). 
\]

\smallskip\ 

\textbf{d.} For a real $F(t)$ we have $F(-\omega )=F^{*}(\omega ).$ Also, $%
L(-\omega )=L^{*}(\omega ).$

\smallskip\ 

\textbf{e.} The absorbed energy (integrated power absorption) is 
\[
\int_{-\infty }^{+\infty }F(t)(dx/dt)dt=\int_{-\infty }^{+\infty }\frac{%
d\omega }{2\pi }\omega \func{Im}L(\omega )\left| F(\omega )\right|
^2\,.\quad \quad \quad \text{(see 7.5.A5)} 
\]

\smallskip\ 

\textbf{f.} Multiplying each term in (\ref{3}) by $dx/dt$ and integrating,
it is seen that the power balance occurs between the input term $\int
F(dx/dt)dt$ and the dissipative or resistive term $\gamma \int (dx/dt)^2dt.$
The two reactive terms on the left hand side lead to the derivative of the
total energy of the unperturbed oscillator 
\[
\frac d{dt}\left[ \frac 12\left( \frac{dx}{dt}\right) ^2+\frac 12\omega
_0^2x^2\right] 
\]
and integrate to zero.

\newpage 

\subsection{8.5.6}

\quad \thinspace \textbf{a.} From 5.f, 
\begin{equation}
\gamma =\frac{\int F(dx/dt)dt}{\int (dx/dt)^2dt}=\frac{\int d\omega \omega 
\func{Im}L\left| F\right| ^2}{\int d\omega \omega ^2\left| L\right| ^2\left|
F\right| ^2}  \label{5}
\end{equation}
using 5.e. and 5.c.

With $L(\omega )$ given by (\ref{4}), $\func{Im}L=\gamma \omega \left|
L\right| ^2$ and everything checks, regardless of what $F(\omega )$ is. If $%
F $ is tuned to excite the resonance, even rather crudely, one can replace $%
\omega ^2$ by $(1/2)(\omega _0^2+\omega ^2)$ in the denominator of (\ref{5}%
). Then 
\begin{equation}
\gamma =\frac{\text{Power loss}}{\text{Stored energy}}  \label{5'}
\end{equation}
and $Q=\omega _0/\gamma $ by definition (\ref{2}).

\smallskip\ 

\textbf{b.} We emphasize that (\ref{5'}) gives a way of evaluating $\gamma $
by measuring the energy absorbed during a time $T\gg \omega _{0}^{-1}$ from
a source of constant power tuned to resonance (or at least such that $\left|
F(\omega _{0})\right| ^{2}$ is appreciable); under these conditions 
\begin{eqnarray*}
\int (dx/dt)^{2}dt &\simeq &(1/2)\int \left[ (dx/dt)^{2}+\omega
_{0}^{2}x^{2}\right] dt \\
&\simeq &T\times \text{Stored energy of resonance.}
\end{eqnarray*}

\smallskip\ 

\textbf{c. }Another way of evaluating (or defining) $\gamma $, hence $Q$, is
to measure the shape of the resonance absorption. We have 
\[
\omega \func{Im}L(\omega )=\frac{\gamma \omega ^2}{\left( \omega _0^2-\omega
^2\right) ^2+\gamma ^2\omega ^2}\simeq \frac{\gamma /4}{\left( \omega
-\omega _0\right) ^2+\gamma ^2/4} 
\]
where the last step gives the behavior near $\omega _0$ for $\omega _0\gg
\gamma .$ Thus the absorption has the maximum value $\gamma ^{-1}\left|
F\right| ^2$ for $\omega =\omega _0$ and falls to $(1/2)\gamma ^{-1}\left|
F\right| ^2$ for $\omega =\omega _0\pm (\gamma /2).$ Here the conditions
will be such that $\left| F(\omega )\right| ^2$ is appreciably constant over
the absorption line.

\smallskip\ 

\textbf{d.} From \textbf{c}., $\gamma $ can be identified as the \textsl{%
full width at half maximum} (FWHM) of the resonance.

\newpage 

\subsection{8.5.7}

\quad \thinspace \textbf{a.} Another physical meaning of $\gamma $ is that
of a \textsl{decay constant} for the stored energy after the external fields
are turned off. This can be seen as follows.

\smallskip\ 

\textbf{b.} If $F(t)=0$ for $t\geq 0,$ $F(\omega )$ is an analytic function
in the \textsl{lower} $\omega $-plane (see 7.10.2), while $L(\omega )$ is
analytic (in general) in the upper $\omega $-plane (see 7.10.2) and in our
case has poles at $\omega =\pm \omega _0-i(\gamma /2)$ (up to terms of order 
$(\gamma /\omega _0)^2.$ It follows that 
\[
\int L(\omega )F(\omega )\exp (-i\omega t)(d\omega /2\pi )\quad \text{%
for\quad }t\geq 0 
\]
can be evaluated by closing the contour down to give 
\[
x(t)=\frac 1{\omega _0}\func{Im}\left[ F(\omega _0)\exp (-i\omega
_0t)\right] \exp (-\gamma t/2). 
\]
Thus $x^2,$ $(dx/dt)^2$ and the energy all go as $\exp (-\gamma t),$ as can
be seen directly by solving (\ref{3}) with $F=0.$

\smallskip\ 

\textbf{c.} In carrying out the manipulations in \textbf{b.} it is
convenient to write 
\[
L(\omega )\simeq \frac 1{2\omega _0}\left[ \frac 1{\omega _0-\omega +i\gamma
/2}+\frac 1{\omega _0+\omega -i\gamma /2}\right] 
\]
valid for $\omega _0\gg \gamma .$ Compare 7.5A eqn.(9) and 7.5.2.

\smallskip

\textbf{d. }It is worth pointing out that $L(\omega )$ can also be viewed as
the retarded Green's function of eq.(\ref{3}), i.e. the solution with $%
F(t)=\delta (t).$

\smallskip

\textbf{e. }We have followed the easy way in obtaining all our results from
eq.(\ref{3}), but one could start almost at any point in the theory and
derive the rest. For instance, the behavior (decay) after switch-off (part 
\textbf{d}) gives the poles of $L(\omega ),$ which then can be approximated
by (\ref{4}). If $\omega \func{Im}L$ is as given in 6.c, the full $L$ can be
found by Kramers-Kroning and is (\ref{4}).

\newpage

\subsection{8.5.8}

\quad \thinspace \textbf{a.} To summarize the results of the last three
pages:

\[
\gamma =\frac{\omega _0}Q=\frac{\text{Power loss}}{\text{Stored energy }}=%
\text{FWHM}=\text{Decay constant.} 
\]

\smallskip

\textbf{b.} The physical nature of the variable $x$ in (\ref{3}) is totally
unimportant. Thus $x$ can be a field component in the case of the resonant
cavity. The important thing is that the energy is quadratic in $x.$

\smallskip

\textbf{c. }Comparing with the results of Sect. 7.5 and 7.10, we see that we
can regard atoms (or molecules) as tiny resonant cavities.

\smallskip

\textbf{d.} In analogy to the result for wave guides, there is also a shift
in the real part of the resonant frequency, given by 
\[
\left( \omega _0+\Delta \omega _0\right) ^2=\omega _0^2\,\left[ 1-\frac{(1+i)%
}Q\right] \,. 
\]
For $Q\gg 1,$ $\left| \Delta \omega _0\right| \ll \omega _0,$ this implies $%
\func{Re}(\Delta \omega _0)\simeq \func{Im}(\Delta \omega _0)\simeq \omega
_0/2Q$ (see Jackson).

\smallskip

\textbf{e.} There is actually a very elegant way to proof of 8.5.3d and 
\textbf{d}. here above. Simply note that $(k+\Delta k)$ must be an analytic
function of $\omega $ for $\func{Im}\omega >0,$ otherwise wave propagation
in a guide would violate causality with relativity (see 7.12). By 8.5.3b, or
Jackson's (8.63), $\alpha $ is proportional to $\omega \sqrt{\omega }%
f(\omega ^{2}),$ where $f(\omega ^{2})$ is a function analytic for $\func{Im}%
\omega >0.$ By the arguments of 7.11.3d, $\Delta k$ must be such that $\func{%
Re}\Delta k(-\omega )=-\func{Re}\Delta k(\omega )$ and $\alpha (-\omega
)=\alpha (\omega ),$ with $\alpha =2\func{Im}\Delta k.$ These conditions are
satisfied if $\Delta k=(1+i)\omega \sqrt{\omega }f(\omega ^{2}).$ Note that $%
\sqrt{\omega }$ comes from $\delta $ and always goes with a factor $(1+i).$

\end{document}
