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\centerline{\bf Electrodynamics B (PHY 5347) Winter/Spring 2017 
Solutions} \bigskip

{\bf Set 10} \bigskip

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\noindent
{\bf 27.\ Principal value integral (E.93).} \bigskip

A.

\begin{enumerate} \setcounter{enumi}{3}

\item Calculation of the principal value integral using the
      first contour becomes:
\begin{eqnarray} \nonumber
  f(\omega_R)&=&{1\over 2\pi\,i}\,\lim_{\eta\to 0} 
  \left[ \int_{-\infty+i\eta}^{\omega_R-\eta+i\eta} 
  \frac{f(z)\,dz}{z-\omega_R+i\eta} +
  \int_{\omega_R+\eta+i\eta}^{+\infty+i\eta}
  \frac{f(z)\,dz}{z-\omega_R+i\eta} \right. \\ \nonumber
  &+& \left. \int_{C_+(\omega_R,\eta)}
  \frac{f(z)\,dz}{z-\omega_R+i\eta} \right] \ =\
  \frac{1}{2\pi i}\,P \int_{-\infty}^{\infty} 
  \frac{f(x)\,dx}{x-\omega_R} + \frac{1}{2}\,f(\omega_R)
  \\ \nonumber &\Rightarrow& 
  \frac{1}{2\pi i}\,P \int_{-\infty}^{\infty} 
  \frac{f(x)\,dx}{x-\omega_R}\ =\ \frac{1}{2}\,f(\omega_R)\,.
\end{eqnarray}
\item Calculation of the principal value integral using the
      second contour: 
\begin{eqnarray} \nonumber
  0 &=& {1\over 2\pi\,i}\,\lim_{\eta\to 0} 
  \left[ \int_{-\infty+i\eta}^{\omega_R-\eta+i\eta} 
  \frac{f(z)\,dz}{z-\omega_R+i\eta} +
  \int_{\omega_R+\eta+i\eta}^{+\infty+i\eta}
  \frac{f(z)\,dz}{z-\omega_R+i\eta} \right. \\ \nonumber
  &+& \left. \int_{C_-(\omega_R,\eta)}
  \frac{f(z)\,dz}{z-\omega_R+i\eta} \right] \ =\
  \frac{1}{2\pi i}\,P \int_{-\infty}^{\infty} 
  \frac{f(x)\,dx}{x-\omega_R} - \frac{1}{2}\,f(\omega_R)
  \\ \nonumber &\Rightarrow& 
  \frac{1}{2\pi i}\,P \int_{-\infty}^{\infty} 
  \frac{f(x)\,dx}{x-\omega_R}\ =\ \frac{1}{2}\,f(\omega_R)\,.
\end{eqnarray}
As it has to be, the results for the principal value integral agree.
\end{enumerate}

B. We assume that $g(\tau )$ is for small $\tau >0$ 
defined by the Taylor expansion
\begin{equation} \label{g_Taylor}
  g(\tau) = \sum_{n=1}^{\infty} {\tau^n \over n!}\, g^{(n)}(0) 
\end{equation}
to find the asymptotic expansion of $\epsilon_0(\omega)$ for
large $|\omega|$
$$ \epsilon_0 (\omega) - 1 =
  \sum_{n=1}^{\infty} \int_0^{\infty} d\tau\, {\tau^n \over n!}\,
  g^{(n)}(0)\, e^{i\,\omega\,\tau} = \sum_{n=1}^{\infty}
  {g^{(n)}(0)\over n!} \left( {\ d\ \over i\,d\omega} \right)^n
  \int_0^{\infty} d\tau\, e^{i\,\omega\,\tau} ,  $$
where all integrals are for $\omega$ in the upper complex plane. 
Hence, we get
$$ \epsilon_0 (\omega) - 1 = \sum_{n=1}^{\infty} {g^{(n)}(0)\over n!}
 \left( {\ d\ \over i\,d\omega} \right)^n\, \left( { -1 \over i\,
 \omega} \right) = \sum_{n=1}^{\infty}
 g^{(n)}(0)\, \left( { -1 \over i\,\omega} \right)^{n+1} . $$
So, we have the fall-off behavior ${\cal O}(|\omega|^{-2})$ for
$|\omega|\to\infty$. In particular, in the limit $\omega_I\to 0^+$ 
we get
$${\rm Re}\,[\epsilon_0(\omega_R)-1]={\cal O}(\omega_R^{-2})~~
  {\rm and}~~ 
  {\rm Im}\,[\epsilon_0 (\omega_R)-1] = {\cal O} (\omega_R^{-3})\,.$$


\noindent
{\bf 28.\ A relation for the group velocity (E.92).}
\smallskip

\noindent Using the index of refraction, the dispersion relation reads 
$$ \omega(k)\ =\ \frac{c\,k}{n(k)}\,.$$
Now, for the group velocity
\begin{eqnarray} \nonumber
  v_g = {d\omega\over dk} = \frac{c\,n}{n^2} - \frac{c\,k}{n^2}\,
  \frac{dn}{d\omega}\,\frac{d\omega}{dk}
  &\Rightarrow& n^2\,v_g + c\,k\,n'\,v_g = c\,n ~~{\rm with}~~
  n'=\frac{dn}{d\omega}\,. \\ \nonumber
  v_g &=& \frac{c\,n}{n^2+c\,k\,n'}\,.
\end{eqnarray} 
\bigskip


\noindent
{\bf 29.\ Kramers-Kronig relation in a model (E.94). }
\smallskip

\noindent (1) The poles of $\epsilon(\omega)$ are at
$$ \omega_{1/2} = -\frac{i\gamma_j}{2}\pm\sqrt{-\left(
   \frac{\gamma_j}{2}\right)^2+\omega_j^2}\,.          $$
As $\omega_j^2>0$ we have for the imaginary parts the inequality
$$ {\rm Im}\,\omega_{1/2}< -\frac{\gamma_j}{2}+\frac{\gamma_j}{2}=0\,.$$
The poles are in the lower half-plane and the function is analytical in
the upper half-plane.

\noindent (2) Let $x=\omega^2$ (real $\omega$). We have to 
differentiate the expression
$$ {f\over g} = {\omega_j^2-x \over (x-\omega_j^2)^2+\gamma_j^2\,x} $$
with respect to $x$. We use the formula
$$ \left( {f\over g} \right)' = {f'g-g'f\over g^2} $$
the denominator $g^2$ is positive (already $g$ is positive) and the
numerator is
$$ - [(x-\omega_j^2)^2+\gamma_j^2\,x] -[2(x-\omega_j^2)+\gamma_j^2]\,
   (\omega_j^2-x)=x^2-2\omega_j^2\,x-\omega_j^2\gamma_j^2+\omega^4_j\,.$$
For the maximum the numerator has to be zero and the solutions
$$ x_0 = \omega_0^2 =  \omega_j^2 \pm \gamma_j\,\omega_j $$
are found. For $\gamma_j<\omega_j$ and $0\le\omega<\omega_{j,\max}
=\sqrt{\omega_j^2-\gamma_j\,\omega_j}$ the numerator $f'g-g'f=(\omega^2-
\omega_j)^2-\gamma_j^2\omega_j^2$ as well as the denominator $g^2$ 
are positive. So, $f/g$ is increasing and we have a maximum. For 
$\omega_{j,\max}< \omega < \sqrt{\omega_j^2+\gamma_j\omega_j}$ the ratio 
$f/g$ is then decreasing with a minimum at the second solution of $f'g-g'f
=0$. In between the imaginary part takes its maximum at $\omega=\omega_j$.
\bigskip % \vfill \eject

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