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%TCIDATA{Created=Tue Oct 15 12:19:57 1996}
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\markright{{\sf  Optical Theorem. Forward scattering}}
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\section{9.14\quad Optical Theorem. Forward Scattering.}

\subsection{9.14.1}

\quad \thinspace \textbf{a. }A very important and general result of
scattering theory is the optical theorem that relates the total cross
section to the imaginary part of the forward scattering amplitude 
\begin{equation}
\sigma _t=\frac{4\pi }k\func{Im}f_0.  \label{1}
\end{equation}

\smallskip\ 

\textbf{b}. For electromagnetic scattering, $f_0$ must be identical with $%
\mathbf{\epsilon }_0^{*}\cdot \mathbf{f}$ for $\mathbf{k=k}_0.$ (Here $%
\mathbf{k}$ and $\mathbf{k}_0$ are the final and initial wave vectors
formerly denoted by $k\mathbf{n}$ and $k_0\mathbf{n).}$ In words, not only
the momentum but also the spin orientation of the photon is unchanged. Also, 
$\sigma _t$ must be summed over all final polarizations. The result (\ref{1}%
) then holds for any $\mathbf{\epsilon }_0,$ and thus it is also true when
averaged over $\mathbf{\epsilon }_0,$ i.e. for mixed polarization.

\smallskip\ 

\textbf{c.} An elegant proof of (\ref{1}) is obtained by computing the
average power flow through a surface $S_1$ surrounding the scattering center
and comparing with the VKI expression of 9.2c.

\[
\]

\smallskip\ 

\textbf{d. }It is convenient to split the fields into an incident and a
scattered part: $\mathbf{E=E}_i\mathbf{+E}_s\mathbf{,}$ $\mathbf{B=B}_i%
\mathbf{+B}_s.$ The normal $\mathbf{n}$ is pointing outwards from $S_1.$

\smallskip\ 

\textbf{e. }The integral $\int_{S_1}\func{Re}(\mathbf{E}_i\times \mathbf{B}%
_i^{*}\cdot \mathbf{n)}da$ vanishes; the similar integral with $\mathbf{E}%
\times \mathbf{B}$ will also vanish if there are no powers losses (compare
potential scattering), while the integral of $\mathbf{E}_s\times \mathbf{B}%
_s $ gives the scattered flux. The total flux taken \textsl{out }of the
forward direction is then 
\begin{equation}
P=\frac c{8\pi }\int_{S_1}\left[ \func{Re}(\mathbf{E}_s\times \mathbf{B}%
_s^{*})\cdot \mathbf{n-}\func{Re}(\mathbf{E}\times \mathbf{B})\cdot \mathbf{n%
}\right] da.  \label{2}
\end{equation}

21 minutes

\newpage

\subsection{9.14.2}

\quad \quad \textbf{a. }The result (\ref{2}) can be rewritten as 
\[
P=-\frac c{8\pi }\int \func{Re}\left[ \mathbf{E}_s\times \mathbf{B}_i^{*}+%
\mathbf{E}_i^{*}\times \mathbf{B}_s\right] \cdot \mathbf{n}da. 
\]
Inserting 
\[
\mathbf{E}_i=E_0\mathbf{\epsilon }_0e^{i\mathbf{k}_0\cdot \mathbf{x}},\quad
\quad \mathbf{B}_i=\frac{\mathbf{k}_0}k\times \mathbf{E}_i 
\]
and comparing with 9.2c, one sees that 
\[
P=\frac c{2k}\left| E_0\right| ^2\func{Im}\left( \mathbf{\epsilon }%
_0^{*}\cdot \mathbf{f(k=k}_0)\right) . 
\]
This is identical with (\ref{1}), since $c\left| E_0\right| ^2/8\pi =I_0$
and $\sigma _t=P/I_0$ by definition.

\smallskip\ 

\textbf{b.} The derivation can be easily extended to cover not only elastic
(frequency unshifted) scattering and absorption, but also inelastic
(frequency shifted) scattering.

\smallskip\ 

\textbf{c. }A word of caution is necessary in using (\ref{1}). It is an
exact relation, but it may fail badly if both sides are computed
approximately in an apparently consistent way. Once this disconcerting fact
is accepted, it can be used to advantage by computing whichever side of (\ref
{1}) is easier.

\smallskip\ 

\textbf{d.} A striking example is given by the Born approximation for
elastic scattering. In this case $f_0$ is real to lowest order in the
perturbation, while the cross-section is finite. The point of course is that 
$\sigma $ is proportional to $\left| f\right| ^2,$ and thus is second order
in the perturbation, at least. Computing $f_0$ to second order, one finds
invariably that (\ref{1}) is satisfied, but this is usually harder than
computing the integral of $\left| f\right| ^2$ over angles.

\smallskip\ 

\textbf{e. }On the other hand, it may be trivial to compute $f_0$ when
absorption takes place. For instance, for the dielectric sphere we have,
from 6.3b, 
\[
f_0=k^2a^3\frac{\varepsilon -1}{\varepsilon +2}\,, 
\]
hence 
\[
\func{Im}f_0=2k^2a^3\frac{\func{Im}\varepsilon }{\left| \varepsilon
+2\right| ^2}\,. 
\]

\newpage

\subsection{9.14.3}

\quad \quad \textbf{a. }Another very important and general result is the
connection between the forward scattering amplitude and the dielectric
constant: 
\begin{equation}
\varepsilon (\omega )=1+\frac{4\pi N}{k^2}f_0(k)  \label{3}
\end{equation}
with $k=\omega /c$ and $N$ the number of scatterers per unit volume.

\smallskip\ 

\textbf{b. }This result has already been established in 9.7.3d within the
context of dipole scattering from an assembly of scatterers, with neglect of
the refraction and attenuation of the incident radiation. In that case, $%
f_0=k^2\chi ^{el}$ and $\varepsilon =1+4\pi N\chi ^{el},$ where $\chi 
^{el}$ is the atomic polarizability.

\smallskip\ 

\textbf{c. }A more general relation can be established by a rather
sophisticated treatment of multiple scattering that cannot be repeated here.
The propagation vector $k^{\prime }$ in the medium is connected to $k=\omega
/c$ by $k^{\prime 2}=k^2+4\pi Nf_0(\func{Re}k^{\prime }).$

\smallskip\ 

\textbf{d. }Since $\func{Im}k^{\prime }=\alpha /2,$ it follows from the
above result that the attenuation coefficient $\alpha $ is given by 
\begin{equation}
\alpha =\frac{4\pi N}{\func{Re}k^{\prime }}\func{Im}f_0(\func{Re}k^{\prime
}).  \label{4}
\end{equation}
This is consistent with the optical theorem (\ref{1}), with the
identification $\alpha =N\sigma _t$ and the replacement $k\rightarrow \func{%
Re}k^{\prime },$ as appropriate to the medium.

\end{document}
