%% Bernd Berg sep 12 2016.
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\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions} \bigskip

\begin{figure}[tb]
\centerline{\hbox{ \psfig{figure=figures/1mink,width=9cm} }}
{Minkowski space in which observer~A is at rest and
flashes a light signal at observer~B, who moves with speed $4c/5$
and flashes the signal back.  \label{fig_exmink1} }
% \end{figure} \begin{figure}[tb]
\centerline{\hbox{ \psfig{figure=figures/2mink,width=9cm} }}
{Minkowski space in which observer~A is at rest and
observer~B moves with speed $4c/5$. Observer~B flashes a light
signal at observer~A, who flashes it back.  \label{fig_exmink2} }
\end{figure}

{\bf Set 2:} \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}{4}

\item Exercise: Time in Minkowski space.
\smallskip

We use natural units, $c=1$, and give all results in units of seconds.

\begin{enumerate}
\item It follows from $x^1=4\,t/5=t-15$ that the coordinates of 
      B$_0$ are $(t,x)=(75,60)$. 
\item The proper time of B is then $\tau=\sqrt{t^2-x^2}=45$.
\item At position A$_2$ the time on the clock of A is $75+60=135$.
\item The coordinates of position B$_1$ are $(t,x)=(t,4t/5)=(25,20)$ 
      so that $\tau=\sqrt{t^2-x^2}=15$ holds.
\item The light signal from B reaches A at time $25+20=45$ at position 
      A$_0$, which agrees with the time of B at B$_0$.
\item It follows from $x^1=4\,t/5=t-45$ that the coordinates of 
      B$_2$ are $(t,x)=(225,180)$ and the clock of B shows $\tau
      =\sqrt{t^2-x^2}=135$.
\item The times agree. That has to be the case, because the travel of 
A in the rest frame of B does just mirror (opposite velocity signs)
the travel of B in the rest frame of B.
\end{enumerate}
\bigskip

\item Exercise: Euclidean and hyperbolic rotations.
\smallskip

Transformation of the 2D Euclidean rotation.
$$ \begin{pmatrix} x'^{\,1}\cr i\,x'^{\,0}\cr \end{pmatrix} = 
   \begin{pmatrix} \cosh(\zeta)& i\,\sinh(\zeta)\cr 
   -i\,\sinh(\zeta)&    \cosh(\zeta)\cr \end{pmatrix}\, 
   \begin{pmatrix} x^1\cr i\,x^0\cr \end{pmatrix} $$
and in components
\begin{eqnarray} \nonumber
     x'^{\,1} &=& \cosh(\zeta)\,x^1 - \sinh(\zeta)\,x^0\,,\\ \nonumber
  i\,x'^{\,0} &=& -i\,\sinh(\zeta)\,x^1 + i\,\cosh(\zeta)\,x^0\,,
\end{eqnarray}
or, equivalently,
\begin{eqnarray} \nonumber
  x'^{\,0} &=& +\cosh(\zeta)\,x^0 - \sinh(\zeta)\,x^1\,,\\ \nonumber
  x'^{\,1} &=& -\sinh(\zeta)\,x^0 + \cosh(\zeta)\,x^1\,,\\ \nonumber 
  \\ \nonumber
  \begin{pmatrix} x'^{\,0}\cr x'^{\,1}\cr \end{pmatrix} &=&
  \begin{pmatrix} \cosh(\zeta)& - \sinh(\zeta)\cr
                - \sinh(\zeta)&   \cosh(\zeta)\cr \end{pmatrix}\, 
  \begin{pmatrix} x^0\cr x^1\cr \end{pmatrix} 
\end{eqnarray} 
\bigskip


\end{enumerate}
\end{document}

