%% Bernd Berg sep 24 2016.
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\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions} \bigskip

{\bf Set 6:} \bigskip

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% \caption{for assignment~14.} \label{fig_lader} \vskip -10pt
% \end{figure} \centerline{Potential as function of $z$.}

\begin{enumerate}

\setcounter{enumi}{18}

\item Exercise: Lorentz transformation for the field of a point charge.
\smallskip \noindent

\noindent In $K'$ the field is
$$ \vec{E}^{\,'} = \frac{q\,\left(x'\,\hat{e}^1+y'\,\hat{e}^2+
z'\,\hat{e}^3\right)}{\left(x'^{\,2}+y'^{\,2}+z'^{\,2}\right)^{3/2}}$$
This field transforms from $K'$ to $K$ with the inverse transformations
% of equations~(\ref{LTE}) and (\ref{LTB}):
$$ \vec{E} = \gamma\,\vec{E}^{\,'} - \frac{\gamma^2}{\gamma+1}\,
             \vec{\beta}\,\left(\vec{\beta}\cdot\vec{E}^{\,'}\right)
~~~{\rm and}~~~\vec{B} = \gamma\,\vec{\beta}\times\vec{E}^{\,'}\,.$$
Therefore, with $\vec{\beta}=\beta\,\hat{e}^1$ we get
\begin{eqnarray} \nonumber
  E^1 &=& \gamma\,E'^{\,1}-\frac{\gamma^2}{\gamma+1}\,\beta^2\,E'^{\,1}
  \ =\ \frac{\gamma^2+\gamma-\gamma^2\beta^2}{\gamma+1}\,E'^{\,1}\\ 
  \nonumber &=& \frac{\cosh^2\zeta+\gamma-\sinh^2\zeta}{\gamma+1}\,
  E'^{\,1}\ =\ \frac{\gamma+1}{\gamma+1}\,E'^{\,1} = E'^{\,1}\,,\\
  \nonumber E^2&=&\gamma\,E'^{\,2}\,,~~~E^3\ =\ \gamma\,E'^{\,3}\,,\\
  \nonumber B^1&=&0\,,~~~B^2=-\gamma\,\beta\,E'^{\,3}\,,~~~
                         B^3=+\gamma\,\beta\,E'^{\,2}\,.
\end{eqnarray}
It remains to express the arguments of the $\vec{E}^{\,'}$ components 
in the coordinates of $K$:
\begin{eqnarray} \nonumber
  c\,t'&=& +\gamma\,c\,t - \gamma\,\beta\,x\,,\\ \nonumber
    x' &=& -\gamma\,\beta\,c\,t + \gamma\,x\,,\\ \nonumber
    y' &=& y\,,~~~ z'\ =\ z\,, 
\end{eqnarray}
resulting in 
\begin{eqnarray} \nonumber
  E^1 &=& \frac{q\,(-\gamma\,\beta\,c\,t+\gamma\,x)}{
  \left[(-\gamma\,\beta\,c\,t+\gamma\,x)^2+y^2+z^2\right]^{3/2}}\,,
  \\ \nonumber E^2 &=& \frac{q\,\gamma\,y}{
  \left[(-\gamma\,\beta\,c\,t+\gamma\,x)^2+y^2+z^2\right]^{3/2}}\,,
  \\ \nonumber E^3 &=& \frac{q\,\gamma\,z}{
  \left[(-\gamma\,\beta\,c\,t+\gamma\,x)^2+y^2+z^2\right]^{3/2}}\,,
  \\ \nonumber B^1 &=& 0\,,
  \\ \nonumber B^2 &=& \frac{-q\,\gamma\,\beta\,z}{
  \left[(-\gamma\,\beta\,c\,t+\gamma\,x)^2+y^2+z^2\right]^{3/2}}\,,
  \\ \nonumber B^3 &=& \frac{+q\,\gamma\,\beta\,y}{
  \left[(-\gamma\,\beta\,c\,t+\gamma\,x)^2+y^2+z^2\right]^{3/2}}\,.
\end{eqnarray}
\bigskip


\item Classwork: Divergence and Laplace operator.

\begin{enumerate}

\item 
$$ \nabla = \sum_{i=1}^3 \hat{e}^i\,\frac{\partial~}{\partial x^i}\,, 
   ~~~ \nabla\cdot\vec{r}\ =\ 
   \sum_{i=1}^3 \frac{\partial x^i}{\partial x^i}\ =\ 3\,. $$

\item 
\begin{eqnarray} \nonumber
  \nabla &=& \hat{r}\,\frac{\partial\ }{\partial r} + 
  \frac{\hat\theta}{r}\,\frac{\partial~}{\partial\theta} +
  \frac{\hat\phi}{r\sin\theta}\,\frac{\partial~}{\partial\phi}\,,
  \\ \nonumber
  \nabla\cdot\,\vec{r}\,f(r) &=& f(r)\,\nabla\cdot\vec{r} +
  \vec{r}\cdot\nabla f =\ 3\,f(r)+\vec{r}\cdot\hat{r}\,
  \frac{df}{dr} = 3\,f(r)+r\,\frac{df}{dr}\,. 
\end{eqnarray}

\item Example: Calculate (b) for $f(r)=r^{n-1}$. Solution:
$$\nabla\cdot\,\vec{r}\,r^{n-1} = 3\,r^{n-1}+r\,\frac{dr^{n-1}}{d\,r}
  = 3\,r^{n-1}+(n-1)\,r^{n-1} = (n+2)\,r^{n-1}\,. $$

\item Solution for $n=-2$, $r\ne 0$:
$$ \nabla\cdot\vec{E} = \nabla\cdot\frac{q\,\vec{r}}{r^3} = 
    (-2+2)\,\frac{q}{r^3} = 0\,. $$

\item Solution for $r\ne 0$:
$$ \nabla^2\,\frac{q}{r} = \nabla\cdot\left(\nabla\,\frac{q}{r}
   \right) = - \nabla\cdot\left(\frac{q\,\hat{r}}{r^2}\right) 
   = - \nabla\cdot\left(\frac{q\,\vec{r}}{r^3}\,\right) = 0\,. $$
\end{enumerate}


\end{enumerate}
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