% Bernd Berg jan 31 2017.
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\centerline{\bf Electrodynamics B (PHY 5347) Winter/Spring 2017 
Solutions} \bigskip

{\bf Set 4:} \bigskip

% \begin{figure}[tb] % \vspace{5pc}
% \centerline{\hbox{ \psfig{figure=hsphere,width=12.5cm} }}
% \caption{for assignment~14.} \label{fig_lader} \vskip -10pt
% \end{figure} \centerline{Potential as function of $z$.}

\noindent
{\bf 8.\ Generalized Residue Theorem.}
\smallskip

\noindent (a) See appendix D.

\noindent (b) As $f(z)$ is analytical in $S$, we can transform the 
integration contour $C$ into a small circle about $z_0$ without
changing the value of the integral. Using the Taylor expansion
we have
$$ \frac{1}{2\pi i} \oint_C \frac{dz\,f(z)}{(z-z_0)^n} = 
   \frac{1}{2\pi i} \sum_{k=0}^{\infty} \oint_C 
   \frac{dz\,(z-z_0)^k\,f^k(z_0)}{k!\,(z-z_0)^n} $$
$$ = \frac{1}{2\pi i} \sum_{k=0}^{\infty} \frac{1}{k!} 
   \oint_C\frac{dz\,f^k(z_0)}{(z-z_0)^{n-k}}~~{\rm with}~~
  f^k(z_0)=\left.\frac{d^k f(z)}{dz^k}\right|_{z=z_0} .$$
Using the previously derived result
$$ \oint_C \frac{dz}{z^n} = \begin{cases} 2\pi i~~{\rm for}~~n=1\,,\\ 
   0~~{\rm otherwise}\,,\end{cases}  $$
we find
$$ \oint_C \frac{dz\,f^k(z_0)}{(z-z_0)^{n-k}} = \begin{cases}2\pi i\,
   f^k(z_0)~~{\rm for}~~n-k=1\,,\\ 0~~{\rm otherwise}\,.\end{cases} $$
Therefore, only $k=n-1$ contributes to the sum for $n=1,\,2,\,3,\,\dots$ 
and gives
$$ \frac{1}{2\pi i} \oint_C \frac{dz\,f(z)}{(z-z_0)^n} =
   \frac{f^{n-1}(z_0)}{(n-1)!}\,. $$
\bigskip


\noindent
{\bf 9.\ Green function by Fourier transformation.}
\smallskip

\noindent We use the notation $\vec{\xi}=\vec{x}-\vec{y}$ and 
introduce the Fourier transformation by
$$ G(\vec{x}-\vec{y}\,) = G(\vec{\xi}\,) = \frac{1}{(2\pi)^3}
   \int d^3p\,e^{-i\,\vec{p}\cdot\vec{\xi}}\,\widetilde{G}(\vec{p}\,)$$
and
$$ \delta(\vec{\xi}\,)=\nabla^2_x G(\vec{\xi}\,)=\frac{1}{(2\pi)^3}
   \int d^3p\,(-\vec{p}\,)^2\,e^{-i\,\vec{p}\cdot\vec{\xi}}\,
   \widetilde{G}(\vec{p}\,)\ \Rightarrow\ 
   \widetilde{G}(\vec{p}\,) = - \frac{1}{\vec{p}^{\,2}}\ .$$
We use spherical coordinates with $d^3p=p^2\,d\phi\,dx,\, x =
\cos(\theta)$, where $\theta$ is the angle between $\vec{p}$ and 
$\vec{\xi}$, to evaluate the integral
$$ G(\vec{\xi}\,) = - \frac{1}{(2\pi)^3} \int d^3p\,
   \frac{e^{-i\,\vec{p}\cdot\vec{\xi}}}{\vec{p}^{\ 2}\,} = 
  -\frac{1}{(2\pi)^2} \int_0^{\infty} dp \int_{-1}^{+1} dx\,
   e^{-i\,(p\,\xi)\,x}\,,~~\xi = |\vec{\xi}\,|\,,  $$
where we performed the $d\phi$ integration. Now the $dx$ integration
is also elementary
$$ G(\vec{\xi}\,) = \frac{i}{(2\pi)^2\xi}\int_0^{\infty} \frac{dp}{p}
   \left[ e^{-i\,(p\,\xi)\,x}\right]_{-1}^{+1} =
- \frac{1}{2\pi^2\xi}\int_0^{\infty} dp\,\frac{\sin(p\xi)}{p} $$
Using 
$$ \int_0^{\infty} dp\,\frac{\sin(p\xi)}{p}\ =\ \frac{\pi}{2} $$
the result is (as expected)
$$ G(\vec{\xi}\,) = - \frac{1}{4\pi\,\xi} =
   - \frac{1}{4\pi\,|\vec{x}-\vec{y}\,|}\,. $$
It remains to calculate the integral
$$ I\ =\ \int_0^{\infty} dx\,\frac{\sin(a\,x)}{x}\ =\ 
   \frac{1}{2} \int_{-\infty}^{+\infty} dx\,\frac{\sin(x)}{x} $$
where the last equality follows from the substitution $x'=ax$ and 
the fact that the integrand is symmetric under $x\to -x$. Although 
the integrand is regular at $x=0$, $\lim_{x\to 0} [\sin(x)/x]=1$, the 
integral can be evaluated using the residue theorem by writing
$$\frac{\sin(x)}{x}\ =\ \frac{1}{2\,i}\,\lim_{\epsilon\to 0^+} \left[
\frac{e^{+ix}}{x-i\,\epsilon}-\frac{e^{-ix}}{x-i\,\epsilon}\right]\,,$$
where the notation $\epsilon\to 0^+$ stands for $\epsilon >0$, $\epsilon
to 0$. The integral becomes
$$ I\ =\ \frac{1}{4\,i} \lim_{\epsilon\to 0^+}\, \left[ 
   \int_{-\infty}^{+\infty} dx\,\frac{e^{+ix}}{x-i\,\epsilon}
 - \int_{-\infty}^{+\infty} dx\,\frac{e^{-ix}}{x-i\,\epsilon}\right]\ 
 =\ \lim_{\epsilon\to 0^+}\,\left[I_1(\epsilon)+I_2(\epsilon)\right)\ .$$
We can evaluate $I_1(\epsilon)$ by closing on a staple in the upper 
and $I_2(\epsilon)$ on a staple in the lower complex planes, so
that the staples to not contribute. E.g.,\
\begin{eqnarray} \nonumber
  I_1(\epsilon)\ =\ \frac{1}{4\,i} \lim_{L\to\infty} &~& 
  \left[ \int_{-L}^{+L} dx\,\frac{e^{+ix}}{x-i\,\epsilon} +
  \int_0^{+L} i\,dy\,\frac{e^{-y}\,e^{+iL}}{L-i\,\epsilon}\,\right.
  \\ \nonumber + &~& \left. \,
  \int_{+L}^{-L} dx\,\frac{e^{-L}\,e^{+ix}}{x-i\,\epsilon} +
  \int_{+L}^0 i\,dy\,\frac{e^{-y}\,e^{-iL}}{-L-i\,\epsilon}\,\right] 
\end{eqnarray}
where the staple integral disappear for $L\to\infty$. Note that the
sign in $\exp(ix)$ determines whether we have to close in the upper
or lower plane. Now we are free to choose the $\epsilon$ limit 
to be from $\epsilon >0$. The residue theorem gives then
$$ \lim_{\epsilon\to 0^+}\,I_1(\epsilon) = \frac{2\pi\,i}{4\,i} =
   \frac{\pi}{2}~~{\rm and}~~I_2(\epsilon) = 0\,.              $$
The latter equality holds because there is no pole in the lower
plane (with the choice $\epsilon<0$ it is the other way round,
leading to the same result). This proves $I=\pi/2$.
\bigskip

{\bf 10.\ Current density of a point charge.}
\smallskip

\noindent Let $x^0$ be given. Because $r^0(\tau)$ is a monotonically 
increasing function, there will be a unique solution
$$ x^0-r^0(\tau^0)=0 ~~~{\rm with}~~~ U^0(\tau^0) =
   \left. \frac{d r^0}{d \tau} \right|_{\tau=\tau^0} \ne 0\ .$$
Then
$$ \delta\left[x^0-r^0(\tau)\right] =
   \frac{1}{\left|U^0(\tau^0)\right|}\,\delta(\tau^0-\tau) $$ 
and the components of
$$ j^{\alpha}(x)\ =\ q \int c\, d\tau\, U^{\alpha}(\tau)\,
  \delta^{(4)} \left[x-r(\tau)\right]  $$
become (note $U^0(\tau^0)=\gamma(\tau^0)\,c>0$)
$$ j^{\alpha}(x)\ =\ q\,c\,\frac{U^{\alpha}(\tau^0)}{U^0(\tau^0)}
   \,\delta^{(3)}\left[\vec{x}-\vec{r}(\tau^0)\right] $$
where $\tau^0=\tau^0(x^0)$. As $U^{\alpha}(\tau^0)=\gamma(\tau^0)\,
u^{\alpha}(\tau^0)$, $u^0=c$. The final result is
$$ j^0(x) = q\,c\,\delta^{(3)}\left[\vec{x}-\vec{r}(\tau^0)\right]\,,
   \qquad \vec{j}(x) = q\,\vec{u}\,
   \delta^{(3)}\left[\vec{x}-\vec{r}(\tau^0)\right]\ . $$
% Remark: $\tau^0$ is {\bf not} the light cone $\tau^0$ from the
% Li\'enard-Wiechert potential.
\bigskip

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