% Bernd Berg Oct 12 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 7} 

\begin{description} 
\item {\bf 22. Twin travel in 2D Minkowski space.}
\end{description}
\bigskip


\begin{figure}[th] % From EMscript/figures_plt/minkowski_space_sol.
\centerline{\hbox{\psfig{figure=figures/smid1mink,width=10cm} }}
IRS: Minkowski space in which observer~B is at rest and observer~A 
moves with speed $4c/5$ in negative $x'$ direction. \end{figure}

\begin{figure}[th] % See subfolder minkowski2Dtwin.
\centerline{\hbox{\psfig{figure=figures/2DtwinFRS,width=10cm} }}
FRS: Minkowski space in which observer~B is at rest and observer~A 
moves with speed $4c/5$ in positive $x''$ direction. \end{figure}

\begin{enumerate}

\item $\beta_1=60/75 = 4/5$.
\bigskip

\item The proper time of $B$ at the end of the first part of its travel 
is $\sqrt{(75)^2-(60)^2}=15\sqrt{5^2-4^2}=45$. This is also the time in 
the initial rest frame at which $B$ leaves the $x=0$ axis.
\bigskip

\item There are several ways to calculate the final meeting point. In
the $K$ frame the final position of B is $\begin{pmatrix}150\cr 
0\end{pmatrix}$. For the Lorentz transformation to this IRF $K'$ we
have to use
$$ \gamma = \sqrt{\frac{1}{1-\beta^2}} = \sqrt{\frac{5^2}{5^2-4^2}}
   = \frac{5}{3}~~{\rm and}~~\beta\gamma = \frac{4}{3}\,.  $$
So, we find for the position of the final point in $K'$
$$ \begin{pmatrix}x'^{\,0}\cr x'^{\,1}\end{pmatrix} =  
   \begin{pmatrix}\gamma&-\beta\gamma \cr 
  -\beta\gamma & \gamma\end{pmatrix} 
   \begin{pmatrix}x^0\cr x^1\end{pmatrix} =  
   \begin{pmatrix}5/3 & -4/3 \cr -4/3 & 5/3\end{pmatrix} 
   \begin{pmatrix}150\cr 0 \end{pmatrix} =
   \begin{pmatrix}250\cr - 200 \end{pmatrix}\, . $$
See the figure. \medskip

Note: $\sqrt{(250-45)^2-200^2}=5\sqrt{41^2-40^2}=45$ is the proper time 
for the second part of the travel of B.
\bigskip

\item From the previous result we see that $B$ travels with the velocity
$$ \beta'\ =\ - \frac{200}{205}\ =\ -\frac{40}{41} $$
to catch up with $A$, $B$. The same result follows from the addition
theorem of velocities:
$$ -\beta'\ =\ -\frac{\beta+\beta}{1+(\beta)^2}\ =\
   -\frac{8}{5}\,\frac{25}{41}\ =\ -\frac{40}{41}\,. $$ 
Calculating the velocity first, we coul have calculated the coordinates
of the last position of B in $K'$ as the meeting point of the tracetories
$x=-4\,t/5$ and $x=40\,(t-45)/41$.

Note: See the second figure for the same in the final rest frame (FRS).

\end{enumerate} \bigskip


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