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\begin{document}

\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions Midterm October 11.} \bigskip

\begin{enumerate}

\item {\bf Spacetrip.}

We only consider the first quarter (1~year) of the flight. The
other results for $t$ are the same due to symmetry. With $\beta
=v/c$ the acceleration in the rest frame is given by
$$ \frac{g}{c} = \frac{d\beta}{d\tau} = \frac{d\zeta}{d\tau}\,,$$
where $\zeta$ is the rapidity. As rapidities are additive, the 
following equation holds in the earth frame:
$$ d\zeta(\tau) = \frac{d\zeta}{d\tau}\,d\tau = \frac{g}{c}\,d\tau\,. $$
With the initial condition $\zeta(0)=0$ this integrates to
$$ \zeta = \int_0^{\zeta} d\zeta' = \frac{g}{c} \int_0^{\tau} d\tau' 
         = \frac{g}{c}\,\tau\,.$$ 
The age of the twin on earth follows from $dt'=\cosh(\zeta)\,d\tau'$:
$$ \int_0^t dt' = t = \int_0^{\tau} \cosh[\zeta(\tau')]\,d\tau'=
   \int_0^{\tau} \cosh\left[\frac{g}{c}\,\tau'\right]\,d\tau' =
   \frac{c}{g}\,\sinh\left[\frac{g}{c}\,\tau\right]\,.  $$
Inserting $\tau=1\,{\rm year} = 365\times 24\times 3600\,[s]$, 
$c=3\times 10^8\,[m/s]$ and $g=9.81\,[m/s^2]$, we find $t=1.187$~years 
for a quarter of the trip and 4.748~years for the entire trip.
\bigskip

\noindent
Distance traveled.

\noindent Seen from earth: ${\rm maximum\ distance} = 2\,x_1$ with
\begin{eqnarray} \nonumber
  x_1 &=& \int_0^{t_1}v(t)\,dt=\int_0^{\tau_1}\cosh\left[\zeta(\tau)
  \right]\,v(\tau)\,d\tau \\ \nonumber &=& c\,\int_0^{\tau_1} 
  \sinh\left(\frac{g\tau}{c} \right) \,d\tau\  =\ \frac{c^2}{g}\,
  \left[\cosh\left(\frac{g\tau_1}{c}\right)-1\right]\,.
\end{eqnarray}
Numerical values: $x_1=0.563$ light years, maximum distance $=1.126$
light years.
\bigskip


\item {\bf Lorentz invariance of anti-symmetry}

$$ F'^{\alpha\beta} = a^{\alpha}_{~\hat\alpha}\,a^{\beta}_{~\hat\beta}\,
   F^{\hat{\alpha}\hat{\beta}} $$
$$ F'^{\beta\alpha} = a^{\beta}_{~\hat\alpha}\,a^{\alpha}_{~\hat\beta}\,
   F^{\hat{\alpha}\hat{\beta}} =
   - a^{\beta}_{~\hat\alpha}\,a^{\alpha}_{~\hat\beta}\,
   F^{\hat{\beta}\hat{\alpha}} =
   - a^{\alpha}_{~\hat\alpha}\,a^{\beta}_{~\hat\beta}\,
   F^{\hat{\alpha}\hat{\beta}} = - F'^{\alpha\beta}\ . $$
\bigskip


\item {\bf Some electromagnetic invariants.} 

From the lecture notes (1.161), (1.162):

\begin{eqnarray} \nonumber
  (F^{\alpha\beta}) = \pmatrix{ 
         0   & -E^x & -E^y & -E^z \cr
         E^x &  0   & -B^z &  B^y \cr
         E^y &  B^z &  0   & -B^x \cr
         E^z & -B^y &  B^x &  0   }\,,~~
  (F_{\alpha\beta}) = \pmatrix{ 
         0   &  E^x &  E^y &  E^z \cr
        -E^x &  0   & -B^z &  B^y \cr
        -E^y &  B^z &  0   & -B^x \cr
        -E^z & -B^y &  B^x &  0   }
\end{eqnarray} % \vfil\eject
and, therefore, 
$$ F_{\alpha\beta}F^{\alpha\beta} =
   2\,\left(\vec{B}^{\,2}-\vec{E}^{\,2}\right)\,. $$
Next, from (1.164):
\begin{eqnarray} \nonumber
  ({^*\!F}^{\alpha\beta}) = \pmatrix{ 
         0   & -B^x & -B^y & -B^z \cr
         B^x &  0   &  E^z & -E^y \cr
         B^y & -E^z &  0   &  E^x \cr
         B^z &  E^y & -E^x &  0   }\,~~
  ({^*\!F}_{\alpha\beta}) = \pmatrix{ 
         0   &  B^x &  B^y &  B^z \cr
        -B^x &  0   &  E^z & -E^y \cr
        -B^y & -E^z &  0   &  E^x \cr
        -B^z &  E^y & -E^x &  0   }\,. 
\end{eqnarray} % \vfil\eject
and, therefore,
$$ F_{\alpha\beta}\,^*\!F^{\alpha\beta} = -4\,\vec{E}\cdot\vec{B}\,,
   ~~~ ^*\!F_{\alpha\beta}\,^*\!F^{\alpha\beta} = 
  2\,\left(\vec{E}^{\,2} - \vec{B}^{\,2}\right)\,.$$
\bigskip

\end{enumerate} 

\end{document}
