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\centerline{\bf Electrodynamics A (PHY 5346) Fall 2016 
Solutions} \bigskip

{\bf Set 3:} \bigskip

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% \end{figure} \centerline{Potential as function of $z$.}

\begin{enumerate}

\setcounter{enumi}{6}

\item Exercise (CW 5 points): Lorentz group Lie matrix generator.
\smallskip

The Matrix $L$ is:
$$L=\left( 
\begin{array}{cccc}
l_{\ 0}^{0} & l_{\ 1}^{0} & l_{\ 2}^{0} & l_{\ 3}^{0} \\ 
l_{\ 0}^{1} & l_{\ 1}^{1} & l_{\ 2}^{1} & l_{\ 3}^{1} \\ 
l_{\ 0}^{2} & l_{\ 1}^{2} & l_{\ 2}^{2} & l_{\ 3}^{2} \\ 
l_{\ 0}^{3} & l_{\ 1}^{3} & l_{\ 2}^{3} & l_{\ 3}^{3}
\end{array} \right) .$$
(1)~The matrix g is:
$$ g=\left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & -1 & 0 & 0 \\ 
0 & 0 & -1 & 0 \\ 
0 & 0 & 0 & -1 \end{array} \right) .$$
By matrix multiplication:
$$-gL=\left( \begin{array}{cccc}
-l_{\ 0}^{0} & -l_{\ 1}^{0} & -l_{\ 2}^{0} & -l_{\ 3}^{0} \\ 
l_{\ 0}^{1} & l_{\ 1}^{1} & l_{\ 2}^{1} & l_{\ 3}^{1} \\ 
l_{\ 0}^{2} & l_{\ 1}^{2} & l_{\ 2}^{2} & l_{\ 3}^{2} \\ 
l_{\ 0}^{3} & l_{\ 1}^{3} & l_{\ 2}^{3} & l_{\ 3}^{3}
\end{array} \right) .$$
(2)~The transpose of $L$ is:
$$\widetilde{L} = 
\left( \begin{array}{cccc}
\widetilde{l}^{\ 0}_{0} & \widetilde{l}^{\ 1}_{0} & 
\widetilde{l}^{\ 2}_{0} & \widetilde{l}^{\ 3}_{0} \\ 
\widetilde{l}^{\ 0}_{1} & \widetilde{l}^{\ 1}_{1} & 
\widetilde{l}^{\ 2}_{1} & \widetilde{l}^{\ 3}_{1} \\ 
\widetilde{l}^{\ 0}_{2} & \widetilde{l}^{\ 1}_{2} & 
\widetilde{l}^{\ 2}_{2} & \widetilde{l}^{\ 3}_{2} \\ 
\widetilde{l}^{\ 0}_{3} & \widetilde{l}^{\ 1}_{3} & 
\widetilde{l}^{\ 2}_{3} & \widetilde{l}^{\ 3}_{3}
\end{array} \right)
=
\left( \begin{array}{cccc}
l_{\ 0}^{0} & l_{\ 0}^{1} & l_{\ 0}^{2} & l_{\ 0}^{3} \\ 
l_{\ 1}^{0} & l_{\ 1}^{1} & l_{\ 1}^{2} & l_{\ 1}^{3} \\ 
l_{\ 2}^{0} & l_{\ 2}^{1} & l_{\ 2}^{2} & l_{\ 2}^{3} \\ 
l_{\ 3}^{0} & l_{\ 3}^{1} & l_{\ 3}^{2} & l_{\ 3}^{3}
\end{array} \right) \ .$$
(3)~By matrix multiplication:
$$\widetilde{L}g=\left( 
\begin{array}{cccc}
l_{\ 0}^{0} & -l_{\ 0}^{1} & -l_{\ 0}^{2} & -l_{\ 0}^{3} \\ 
l_{\ 1}^{0} & -l_{\ 1}^{1} & -l_{\ 1}^{2} & -l_{\ 1}^{3} \\ 
l_{\ 2}^{0} & -l_{\ 2}^{1} & -l_{\ 2}^{2} & -l_{\ 2}^{3} \\ 
l_{\ 3}^{0} & -l_{\ 3}^{1} & -l_{\ 3}^{2} & -l_{\ 3}^{3}
\end{array} \right) .$$
(4)~Set $-gL=\widetilde{L}g$ and compare the components:
$$
\left( \begin{array}{cccc}
-l_{\ 0}^{0}=l_{\ 0}^{0} & -l_{\ 1}^{0}=-l_{\ 0}^{1} & -l_{\ 2}^{0}=-l_{\
0}^{2} & -l_{\ 3}^{0}=-l_{\ 0}^{3} \\ 
l_{\ 0}^{1}=l_{\ 1}^{0} & l_{\ 1}^{1}=-l_{\ 1}^{1} & l_{\ 2}^{1}=-l_{\ 1}^{2}
& l_{\ 3}^{1}=-l_{\ 1}^{3} \\ 
l_{\ 0}^{2}=l_{\ 2}^{0} & l_{\ 1}^{2}=-l_{\ 2}^{1} & l_{\ 2}^{2}=-l_{\ 2}^{2}
& l_{\ 3}^{2}=-l_{\ 2}^{3} \\ 
l_{\ 0}^{3}=l_{\ 3}^{0} & l_{\ 1}^{3}=-l_{\ 3}^{1} & l_{\ 2}^{3}=-l_{\ 3}^{2}
& l_{\ 3}^{3}=-l_{\ 3}^{3}
\end{array} \right)\ .
$$
The resulting matrix is:
$$
\left( \begin{array}{cccc}
0 & l_{\ 1}^{0} & l_{\ 2}^{0} & l_{\ 3}^{0} \\ 
l_{\ 1}^{0} & 0 & l_{\ 2}^{1} & l_{\ 3}^{1} \\ 
l_{\ 2}^{0} & -l_{\ 2}^{1} & 0 & l_{\ 3}^{2} \\ 
l_{\ 3}^{0} & -l_{\ 3}^{1} & -l_{\ 3}^{2} & 0
\end{array} \right)\ .
$$


\item Exercise (5 points): Obtain the result of the previous Exercise
discussing the elements of
$ g^{\alpha\beta}\, \tilde{l}_{\beta}^{\ \gamma}\,
   g_{\gamma\delta} = - l^{\alpha}_{\ \delta}$.
Performing the contractions we get
$ \tilde{l}^{\alpha}_{\ \delta} = - l^{\alpha}_{\ \delta} $. Using now 
the definition $\tilde{l}^{\alpha}_{\ \delta} = l_{\delta}^{\ \alpha}$,
the following equations are obtained (no summations, $a=0,1,2,3$,
$i=1,2,3$ and $j=1,2,3$):
$$ \tilde{l}^a_{\ a} = l_a^{\ a} = + l^a_{\ a} \Rightarrow
   l^a_{\ a} = - l^a_{\ a} = 0\ , $$
$$ \tilde{l}^i_{\ 0} = l_0^{\ i} = - l^0_{\ i} \Rightarrow
   l^0_{\ i} = + l^i_{\ 0}\ , $$
$$ \tilde{l}^i_{\ j} = l_j^{\ i} = + l^j_{\ i} \Rightarrow
   l^j_{\ i} = - l^i_{\ j} $$
and the resulting matrix is the same as before.


\item Exercise: A rotation and a boost generator.
\smallskip

For $S_3$ we have
$$ S_3 = \left(\begin{matrix} 
                         0 & 0 & 0 & 0 \\
                         0 & 0 &-1 & 0 \\
                         0 & 1 & 0 & 0 \\
                         0 & 0 & 0 & 0 \end{matrix}\right)
  ~~{\rm and}~~ (S_3)^2 = - {\bf 1}_{2,3} ~~{\rm with}~~ 
    {\bf 1}_{2,3} = \left(\begin{matrix} 
                     0 & 0 & 0 & 0 \\
                     0 & 1 & 0 & 0 \\
                     0 & 0 & 1 & 0 \\
                     0 & 0 & 0 & 0 \end{matrix}\right)\,.$$
Therefore, for $n=1,2,3,\dots$
$$ (S_3)^{2n} = (-1)^n\, {\bf 1}_{2,3} ~~{\rm and}~~
   (S_3)^{2n+1} = (-1)^n\, S_3 $$
and the result follows:
\begin{eqnarray} \nonumber
  e^{-\omega S_3} &=& {\bf 1} + 
  \sum_{n=1}^{\infty} (-1)^n\,\frac{\omega^{2n}}{(2n)!}\,{\bf 1}_{2,3} 
- \sum_{n=0}^{\infty} (-1)^n\,\frac{\omega^{2n+1}}{(2n+1)!}\,S_3 \\
 \nonumber &=& \left(\begin{matrix} 
           1 & 0          & 0          & 0 \\
           0 & \cos\omega & \sin\omega & 0 \\
           0 &-\sin\omega & \cos\omega & 0 \\
           0 & 0 & 0 & 1 \end{matrix}\right)\,.
\end{eqnarray}
Similarly we deal with $K_1$:
$$ K_1 = \left(\begin{matrix} 
         0 & 1 & 0 & 0 \\
         1 & 0 & 0 & 0 \\
         0 & 0 & 0 & 0 \\
         0 & 0 & 0 & 0 \end{matrix}\right) 
  ~~{\rm and}~~ (K_1)^2 = {\bf 1}_{1,2} ~~{\rm with}~~
    {\bf 1}_{1,2} = \left(\begin{matrix} 
                     1 & 0 & 0 & 0 \\
                     0 & 1 & 0 & 0 \\
                     0 & 0 & 0 & 0 \\
                     0 & 0 & 0 & 0 \end{matrix}\right)\,.$$
Therefore, for $n=1,2,3,\dots$
$$ (K_1)^{2n} = {\bf 1}_{1,2} ~~{\rm and}~~ (K_1)^{2n+1} = K_1 $$
and
$$ e^{-\zeta K_1} = {\bf 1} + 
  \sum_{n=1}^{\infty}\frac{\omega^{2n}}{(2n)!}\,{\bf 1}_{1,2} -
  \sum_{n=0}^{\infty}\frac{\omega^{2n+1}}{(2n+1)!}\,K_1 
  = \left(\begin{matrix} 
    \cosh\zeta &-\sinh\zeta & 0 & 0 \\
   -\sinh\zeta & \cosh\zeta & 0 & 0 \\
             0 & 0          & 1 & 0 \\
             0 & 0          & 0 & 1 \end{matrix}\right)\,.$$
% \bigskip % \eject

\item Exercise: Four-dimensional Levi-Civita tensor.
\smallskip

\noindent (1) Using the Einstein summation convention of the left-hand
side, but not on the right-hand side, of the following equation we
have
$$ g_{\alpha_1\alpha_2} g_{\beta_1\beta_2} g_{\gamma_1\gamma_2} 
   g_{\delta_1\delta_2} \epsilon^{\alpha_2\beta_2\gamma_2\delta_2}
   = g_{\alpha_1\alpha_1} g_{\beta_1\beta_1} g_{\gamma_1\gamma_1} 
   g_{\delta_1\delta_1} \epsilon^{\alpha_1\beta_1\gamma_1\delta_1}\,.$$
The right-hand side is zero unless $\alpha_1\beta_1\gamma_1\delta_1$ are
a permutation $\pi_0\pi_1\pi_2\pi_3$ of the numbers $0,1,2,3$ and in that 
case
$$ \epsilon_{\pi_0\pi_1\pi_2\pi_3} = g_{\pi_0\pi_0} g_{\pi_1\pi_1} 
   g_{\pi_1\pi_2} g_{\pi_1\pi_3} \epsilon^{\pi_0\pi_1\pi_2\pi_3} = 
   (-1)^3\,\epsilon^{\pi_0\pi_1\pi_2\pi_3}\,, $$
which proves $\epsilon_{\alpha\beta\gamma\delta}=-
\epsilon^{\alpha\beta\gamma\delta}$.
\smallskip

\noindent (2) We have $\epsilon_{\alpha\beta\gamma_1\delta_1}
\epsilon^{\alpha\beta\gamma_2\delta_2}=0$ unless either $\gamma_1=
\gamma_2$, $\delta_1=\delta_2$ or $\gamma_1=\delta_2$, $\delta_1=
\gamma_2$ or (otherwise one of the already used numbers will be 
repeated by $\alpha$ or $\beta$ of the sums). Therefore,
$$\epsilon_{\alpha\beta\gamma_1\delta_1} 
  \epsilon^{\alpha\beta\gamma_2\delta_2} = 
  a\,\delta_{\gamma_1}^{\ \gamma_2}\,\delta_{\delta_1}^{\ \delta_2}+
  b\,\delta_{\gamma_1}^{\ \delta_2}\,\delta_{\delta_1}^{\ \gamma_2}$$
holds. With no summation in the permutation indices $\pi_0$ to $\pi_3$
the constants follow from
\begin{eqnarray} \nonumber
 \epsilon_{\alpha\beta\pi_2\pi_3}\epsilon^{\alpha\beta\pi_2\pi_3} =
 \epsilon_{\pi_0\pi_1\pi_2\pi_3}\epsilon^{\pi_0\pi_1\pi_2\pi_3} +
 \epsilon_{\pi_1\pi_0\pi_2\pi_3}\epsilon^{\pi_1\pi_0\pi_2\pi_3} =
 -2 = a\,, \\ \nonumber
 \epsilon_{\alpha\beta\pi_2\pi_3}\epsilon^{\alpha\beta\pi_3\pi_2} =
 \epsilon_{\pi_0\pi_1\pi_2\pi_3}\epsilon^{\pi_0\pi_1\pi_3\pi_2} +
 \epsilon_{\pi_1\pi_0\pi_2\pi_3}\epsilon^{\pi_1\pi_0\pi_3\pi_2} =
 +2 = b\,. 
\end{eqnarray}
\smallskip

\noindent (3) We have $\epsilon_{\alpha\beta_1\gamma_1\delta_1}
\epsilon^{\alpha\beta_2\gamma_2\delta_2}=0$ unless $\beta_2\gamma_2
\delta_2$ is a permutation of $\beta_1\gamma_1\delta_1$. There are
six such permutations, so that the results is a sum of the form
\begin{eqnarray} \nonumber
   \epsilon_{\alpha\beta_1\gamma_1\delta_1} 
   \epsilon^{\alpha\beta_2\gamma_2\delta_2}\, &=&
    a_1\,\delta_{\beta_1}^{~\beta_2}\delta_{\gamma_1}^{~\gamma_2}
         \delta_{\delta_1}^{~\delta_2}
  + a_2\,\delta_{\beta_1}^{~\gamma_2}\delta_{\gamma1}^{~\delta_2}
         \delta_{\delta_1}^{~\beta_2}
  + a_3\,\delta_{\beta_1}^{~\delta_2}\delta_{\gamma_1}^{~\beta_2}
         \delta_{\delta_1}^{~\gamma_2} \\ \nonumber &+&
    b_1\,\delta_{\beta_1}^{~\beta_2}\delta_{\gamma_1}^{~\delta_2}
         \delta_{\delta_1}^{~\gamma_2}
  + b_2\,\delta_{\beta_1}^{~\delta_2}\delta_{\gamma_1}^{~\gamma_2}
         \delta_{\delta_1}^{~\beta_2}
  + b_3\,\delta_{\beta_1}^{~\gamma_2}\delta_{\gamma_1}^{~\beta_2}
         \delta_{\delta_1}^{~\delta_2} \,,
\end{eqnarray}
where it follows from (no summation)
$\epsilon_{\pi_0\pi_1\pi_2\pi_3}\epsilon^{\pi_0\pi_1\pi_2\pi_3}=-1$
that $a_i=-1$ and $b_i=1$ holds for $i=1,2,3$.
\bigskip % \vfil\eject

\item Exercise: Addition theorem for transverse velocity components.
\smallskip 

\noindent In $K'$ the motion with velocity $\vec{u'}$ is
$$ x'^i\ =\ c^{-1}\,u'^i\,x'^0\,.$$ 
With respect to frame $K$ the origin of $K'$ frame moves with 
speed $v$ along the $x^1$ axis of $K$. For $i=2,3$ the Lorentz 
transformations give then
$$ x^i\ =\ x'^i\ =\ c^{-1}\,u'^i\,\gamma\,(x^0-\beta\,x^1)\,. $$ 
Dividing by $x^0$ gives the velocity components in $K$:
\begin{eqnarray} \nonumber
  c^{-1}\,u^i&=&c^{-1}\,u'^i\,\gamma\,(1-\beta\,c^{-1}\,u^1)\
  =\ c^{-1}\,u'^i\,\gamma\,\left(1-\beta\,c^{-1}\,\frac{u'^1+v}{1+
     u'^1v/c^2}\right)\,,
\end{eqnarray}
where in the second step the already calculated equation for $u^1$
has been inserted. Bringing everything to the common denominator
gives
\begin{eqnarray} \nonumber
 u^i &=& u'^i\,\gamma\,\frac{(1+\beta\,u'^1/c)-\beta\,c^{-1}\,(u'^1+v)}
      {1+u'^1v/c^2} \\ \nonumber &=& u'^i\,\gamma\,\frac{1-\beta^2}
      {1+u'^1v/c^2}\ =\ \frac{u'^i}{\gamma\,(1+u'^1v/c^2)}\,,
\end{eqnarray}
where $1-\beta^2=1/\gamma^2$ was used in the last step.
\bigskip

\end{enumerate}
\end{document}

\item Exercise: Spacetrip Elapsed time on earth.

We only consider the first quarter (five years) of the flight. The
other results for $t$ are the same by symmetry reasons. 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
$$ d\zeta(\tau) = \frac{d\zeta}{d\tau}\,d\tau = \frac{g}{c}\,d\tau $$
holds. 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 $d\tau=dt/\cosh(\zeta)$:
$$ \int_0^t dt' = t = \int_0^{\tau} \cosh[\zeta(\tau')]\,d\tau'=
   \int_0^{\tau} \cosh\left[\frac{g}{c}\,\tau'\right]\,d\tau' =
   \frac{c}{g}\,\sinh\left[\frac{g}{c}\,\tau\right]\,.  $$
Inserting $\tau=5\,{\rm years} = 5\times 365\times 24\times
3600\,[s]$, $c=3\times 10^8\,[m/s]$ and $g=9.81\,[m/s^2]$, we find 
$t=84.1$~years for a quarter of the trip and 337~years for the
entire trip.
\bigskip
