%% Bernd Berg sep 24 2016.
%  \documentclass{ws-jnmp}  
% \documentclass{sfp-ws}  
 \documentclass{ws-book9x6}  

\begin{document}

% \setcounter{page}{1}

\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
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 $z$.}

\begin{enumerate}

\setcounter{enumi}{15}

\item Exercise: Proper time under rotation.
\smallskip \noindent

$$ c^2\,d\tau^2 = c^2\,dt^2 - \omega^2 R^2\,dt^2 \Rightarrow
   \frac{d\tau}{dt} = \sqrt{1-\frac{\omega^2 R^2}{c^2}}\ .$$
Now $g=v^2/R=\omega^2\,R$, where $g=9.81\,[km/s^2]$ and $v$ is the 
velocity of the satellite.
For $R=(6,378+160)\,[km]$ (equatorial radius plus $160\,[km]$) we get 
$\omega=0.0012249\,[s^{-1}]$, which correpsonds to a period of $T=2\pi
/\omega=85.5\,[m]$. For the time dilation we find
$$ 1-\frac{d\tau}{dt} = \frac{1}{2}\,\left(\frac{\omega\,R}{c}\right)^2 
   = 0.356321\times 10^{-9}\,. $$
Compared to this the time dilation of a clock on the equator is with
$$ 1-\frac{d\tau}{dt} = \frac{1}{2}\,
   \left(\frac{\omega\,R}{c}\right)^2 = 0.119517\times 10^{-11} $$
negligle. However, gravity effects need to be taken into acount for
the final result.
\bigskip



\item Exercise: Homogeneous Maxwell equations
\smallskip \noindent

We choose $\beta =0$ in 
$\partial_{\alpha}\,^*F^{\alpha\beta}=0$ and obtain:
$\partial_{\alpha}\,^*F^{\alpha 0}=0$. Using the definition of the 
dual tensor, this reads
$$ \frac{1}{2}\,\left(\partial_{\alpha}\,
   \epsilon^{\alpha 0\gamma\delta}\,F_{\gamma\delta}\right)=0\,$$
or
\begin{equation} \label{2HMEQ_result}
  \partial_{\alpha}\,F_{\gamma\delta} +
  \partial_{\gamma}\,F_{\delta\alpha} +
  \partial_{\delta}\,F_{\alpha\gamma}\ =\ 0
\end{equation}
for all three indices different and not 0. Repeating this for 
$\beta=1,\,2$ and 3, we find that equation (\ref{2HMEQ_result})
hold for all combinations of distinct indices $\alpha,\,\beta$
and $\gamma$. Due to the antisymmetry of electromagnetic 
tensor, the equation still holds when some of the indices
agree. Finally, all indices can of course be raised. This
proves the equivalence of the two versions of the homogeneous
Maxwell equations. More by brute force, one could just write
out all 64 equations.

More elegantly: Use
$$ \frac{1}{2}\,\partial_{\alpha}\,
   \epsilon_{\hat\alpha\beta\hat\gamma\hat\delta}\,
   \epsilon^{\alpha\beta\gamma\delta}\,F_{\gamma\delta} = 0 $$
and expand the contraction in $\beta$ of the two Levi-Civita symbols 
into Kronecker deltas.
\bigskip


\item Exercise: Lorentz transformation of the electric and magnetic 
fields. 
\smallskip 

\noindent In matrix notation we have $F'=A\,F\,\tilde{A}$, where $F$ 
is the electromagnetic field tensor, $A$ the Lorentz matrix and
$\tilde{A}$ its transpose. We perform the calculation in two 
steps. First,
$$ A\,F\ =\ \left(\begin{matrix} 
        \gamma      & -\gamma\beta~&~0~&~0~\\
       -\gamma\beta &  \gamma     ~&~0~&~0~\\
                  0 &       0     ~&~1~&~0~\\
                  0 &       0     ~&~0~&~1~\end{matrix}\right)\,
\left(\begin{matrix} 
 0   & -E^1 & -E^2 & -E^3 \\
 E^1 &  0   & -B^3 &\ B^2 \\
 E^2 &\ B^3 &  0   & -B^1 \\
 E^3 & -B^2 &\ B^1 &  0   \end{matrix}\right) $$ 
$$=\ \left(\begin{matrix} 
  -\gamma\beta\,E^1 &~-\gamma\,E^1 &~-\gamma\,E^2+\gamma\beta\,B^3 & 
                                    ~-\gamma\,E^3 -\gamma\beta\,B^2\\
  ~\gamma\,E^1 &~~\gamma\beta\,E^1 &~~\gamma\beta\,E^2-\gamma\,B^3 & 
                                    ~~\gamma\,E^2 +\gamma\beta\,B^2\\
    E^2 & ~B^3 &  0  & -B^1 \\
    E^3 & -B^2 & B^1 &  0   \end{matrix}\right) $$ 
then (note $\tilde{A}=A$)
$$ F'\ =\ (A\,F)\,\tilde{A}\ =$$
$$ \left(\begin{matrix} 
   0   & -E^1 & -\gamma\,(E^2-\beta\,B^3) &~-\gamma\,(E^3+\beta\,B^2) \\
   E^1 &  0   & -\gamma\,(B^3-\beta\,E^2) &~~\gamma\,(B^2+\beta\,E^3) \\
   \gamma\,(E^2-\beta\,B^3) &~~\gamma\,(B^3-\beta\,E^2) & 0 & -B^1 \\
   \gamma\,(E^3+\beta\,B^2) &~-\gamma\,(B^2+\beta\,E^3)& B^1 & 0  
                            \end{matrix}\right) $$ 
Comparing this with the elements of $F'$ in its standard form, we
read off:
$$ \begin{matrix}  
   E'^1=E^1\,,\qquad~~~~~~~~~~~~~~& B'^1=B^1\,,\qquad~~~~~~~~\\
   E'^2=\gamma\,(E^2-\beta\,B^3)\,,~~~& ~B'^2=\gamma\,(B^2+\beta\,E^3)\,,\\
   E'^3=\gamma\,(E^3+\beta\,B^2)\,,~~~&~~B'^3=\gamma\,(B^3-\beta\,E^2)\,.
                            \end{matrix} $$ 
If $\vec{\beta}$ is no longer in $\hat{e}^1$ direction we decompose the
fields in components parallel and perpendicular to $\vec{\beta}$ and
these equation become
$$ \begin{matrix}  
   E'^{\,\|}=E^{\|}\,,\qquad~~~~~~~~~~~~& B'^{\,\|}=B^{\|}\,,
   \qquad~~~~~~~~\\
   \vec{E}^{\,'\perp}=\gamma\,\left(\vec{E}^{\perp}+
   \vec{\beta}\times\vec{B}^{\perp}\right)\,,~~~&~~
   \vec{B}^{\,'\perp}=\gamma\,\left(\vec{B}^{\perp}-
   \vec{\beta}\times\vec{E}^{\perp}\right)\,.  \end{matrix} $$ 
To transform this we first note that the equations
$$ \vec{\beta}\times\vec{E}^{\perp}=\vec{\beta}\times\vec{E}~~~{\rm and}
~~~\vec{\beta}\times\vec{B}^{\perp}=\vec{\beta}\times\vec{B}          $$
hold, so that
$$ \vec{E}^{\,'\perp}=\gamma\,\left(\vec{E}^{\perp}+
   \vec{\beta}\times\vec{B}\right)~~~{\rm and}~~~
\vec{B}^{\,'\perp}=\gamma\,\left(\vec{B}^{\perp}-
   \vec{\beta}\times\vec{E}\right) $$
are correct. To extend them to the parallel components, we have to
subtract a term that vanishes for the perpendicular components, i.e.,
it should be proportional to $\vec{\beta}\,\left(\vec{\beta}\cdot\vec{E}
\right)$ in the first and to $\vec{\beta}\,\left(\vec{\beta}\cdot\vec{B}
\right)$ in the second equation. The proportionality factor $\gamma^2/
(\gamma+1)$ follows then from the observation
$$ \gamma - \frac{\gamma^2\,\beta^2}{\gamma+1} = \frac{\gamma^2+\gamma
   -\gamma^2\,\beta^2}{\gamma+1} = \frac{\cosh^2\zeta+\cosh\zeta
   -\sinh^2\zeta}{\cosh\zeta+1} = 1\,.  $$
\medskip

\end{enumerate}
\end{document}
