% Bernd Berg feb 01 2017.
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\centerline{\bf Electrodynamics B (PHY 5347) Winter/Spring 2017 
Solutions} \bigskip

{\bf Set 5:} \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 11.\ Wave equation with point source.}
\smallskip

\noindent The wave is obtained by integrating the source with the
retarded Green function (the homogeneous solution is zero as the wave 
disappears for times $t<0$):
\begin{eqnarray} \nonumber
  \psi_r(x^0,\vec{x}) &=& 
  \int dy^0\,d^3y\,G_r(x-y)\,\theta(x^0)\,\delta(\vec{y}\,)\\ \nonumber 
  &=& \int dy^0\,d^3y\,\frac{\delta(x^0-y^0-|\vec{x}-\vec{y}\,|)}
   {4\pi\,|\vec{x}-\vec{y}\,|}\,\theta(y^0)\,\delta(\vec{y}\,)\\ \nonumber
  &=& \int_0^{\infty} dy^0\,\frac{\delta(x^0-y^0-|\vec{x}|)}
   {4\pi\,|\vec{x}|} \,=\, \begin{cases} 
   1/(4\pi\,|\vec{x}|)~~{\rm for}~~|\vec{x}| < x^0 = c\,t\,; \\
   0~~{\rm for}~~|\vec{x}| > x^0=c\,t\,. \end{cases}
\end{eqnarray}
Using the advanced Green function one gets
\begin{eqnarray} \nonumber
  \psi_a(x^0,\vec{x}) &=& \begin{cases}1/(4\pi\,|\vec{x}|)~~{\rm for}
  ~~|\vec{x}|<-x^0=-c\,t\,;\\ 0~~{\rm for}~~|\vec{x}|>-x^0=-c\,t\,.
  \end{cases}
\end{eqnarray}
\bigskip


\noindent
{\bf 12.\ Covariant retarded Green function.}
\smallskip

\noindent With $\tau=x^0-y^0$ the zeros of $\delta((x-y)^2)$ are at
$$ \tau^0=\pm\sqrt{(\vec{x}-\vec{y}\,)^2}=\pm|\vec{x}-\vec{y}\,|\,.$$
Let now
$$ f(\tau) = \tau^2 - \left(\vec{x}-\vec{y}\,\right)^2\,, $$
so that
$$ \left.\frac{d f}{d \tau}\right|_{\tau=\tau^0}=f'(\tau^0)=2\,\tau^0$$
holds. Therefore,
\begin{eqnarray} \nonumber
 \delta((x-y)^2)&=& \frac{1}{2\,|f'(\tau^0)|}\left[\delta(\tau-|\vec{x}
     -\vec{y}\,|) + \delta(\tau+|\vec{x}-\vec{y}\,|)\right] \\ \nonumber
 &=& \frac{1}{2\,|\vec{x}-\vec{y}\,|}\left[\delta(x^0-y^0-|\vec{x}
     -\vec{y}\,|)+\delta(x^0-y^0+|\vec{x}-\vec{y}\,|)\right] 
\end{eqnarray}
and the identity of the two expressions for the Green function follows:
$$ \frac{1}{2\pi}\,\theta(x^0-y^0)\,\delta((x-y)^2) = \frac{
   \delta(x^0-y^0-|\vec{x}-\vec{y}\,|)}{4\pi\,|\vec{x}-\vec{y}\,|}\,.$$
\bigskip


\noindent
{\bf 13.\ Four-potential of a moving point particle.} \smallskip

\begin{figure}[th] % From figprob2/. Spring 2014 phy5347.
\centerline{\hbox{ \psfig{figure=figures/4potential,width=10cm} }}
\caption{Backward light cone and worldline in Minkowski space.
\label{fig_prob3} } \end{figure}

\begin{enumerate} 
\item The distance of the point particle from the origin is given by
$$ r(t) = \int_0^t \beta(t')\,dt'\ =\ \int_0^t \tanh(t)'\,dt' = 
   \int_0^{\cosh(t)} \frac{d \cosh(t')}{\cosh(t')} = 
   \ln\left[\cosh(t)\right]\,.      $$
\item With $t_1=0.6931472$ the speed is $\beta(t_1)=0.6$ and the 
      position is $r(t_1)=0.2231436=R$.
\item The position $r(t_1)=0.2231436$ of the point particle is observed 
      at $\vec{x}=0$ when the time is $x^0=t_1+r(t_1)=0.9162908$.
\item We find the corresponding potentials from the equations
\begin{eqnarray} \nonumber
  \Phi\left(x^0,\vec{x}\right) = \left[\frac{q}{\left(1-
  \vec{\beta}\cdot\hat{n}\right)\,R}\right]_{\rm ret}\!\!,~~~
  \vec{A}\left(x^0,\vec{x}\right) = \left[\frac{q\,\vec\beta}{\left(
  1-\vec{\beta}\cdot\hat{n}\right)\,R}\right]_{\rm ret}\!\! .
\end{eqnarray}
We have $\vec{\beta}=\beta(t_1)\,\hat{x}^1=0.6\,\hat{x}^1$, $R=
0.2231436$ and $\vec{R}=-R\,\hat{x}^1$ because the point charge 
moves away from the origin. Therefore,
\begin{eqnarray} \nonumber
  \Phi &=& \frac{q_e}{(1+0.6)\,R}\ =\ \frac{q_e}{0.3570297}
  \ =\ 2.800888\,q_e\,, \\ \nonumber 
  \vec{A}&=&\frac{0.6\,q_e\,\hat{x}^1}{(1+0.6)\,R}\ =\ 
  \frac{0.6\,q_e\,\hat{x}^1}{0.3570297}\ =\ 
  1.680533\,\hat{x}^1\,q_e\,.
\end{eqnarray}
\end{enumerate}

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