\documentclass{ws-book9x6}  % Bernd Berg Sep 09 2018.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 5} \smallskip

\begin{enumerate} \setcounter{enumi}{10}

{\bf \item Twin travel in 2D Minkowski space continued.}
\medskip

\begin{enumerate}

\item $\beta_A^3 = \beta_B^1 = 4/5$.
\smallskip

\item From the addition theorem of velocities we find
      $$\beta_B^3=\left(\frac{\beta_B^1+\beta_B^1}{1+(\beta_B^1)^2}
        \right) = \frac{8/5}{1+16/25} = \frac{40}{41}\,.$$
\smallskip

\item We give two solutions. For simplicity of the notation we denote 
time and space coordinates in $K_3$ by $t$ and $x$ without subscripts
for $K_3$.
\smallskip

First, we may use that the proper time is invariant. So, from
$$ t^2-x^2=3^2 = 9~~{\rm and}~~x=\beta_B^3\,t=40\,t/41 $$ 
we get 
$$ t^2-(40\,t/41)^2 = 9~~\Rightarrow~~t=\pm 41/3 $$ 
of which we 
have to discard the negative solution. The corresponding $x$ coordinate 
follows from $x=\beta_B^3\,t=(40/41)\times(41/3)=40/3$, i.e., the 
position of $B_0$ is given by 
$$\begin{pmatrix} t\cr x\end{pmatrix} =
  \begin{pmatrix} 41/3\cr 40/3\end{pmatrix}\,.$$
\smallskip

Second, we work out the Lorentz transformation from $K_1$ to $K_3$:
$$ \gamma = \sqrt{\frac{1}{1-\beta^2}} = \sqrt{\frac{5^2}{5^2-4^2}}
   = \frac{5}{3}~~{\rm and}~~\beta\,\gamma = - \frac{4}{3}\,.  $$
So, we find for the position of of $B_0$ in $K_3$
$$ \begin{pmatrix} t\cr x\end{pmatrix} =  
   \begin{pmatrix}\gamma&-\beta\,\gamma \cr 
  -\beta\,\gamma & \gamma\end{pmatrix} 
   \begin{pmatrix} 5\cr 4\end{pmatrix} =  
   \begin{pmatrix}5/3 & 4/3 \cr 4/3 & 5/3\end{pmatrix} 
   \begin{pmatrix} 5\cr 4\end{pmatrix} =
   \begin{pmatrix}41/3\cr 40/3 \end{pmatrix}\, . $$
\smallskip
\smallskip

\item At $B_0$ the elapsed time on the clock of $B$ is $3$.
\smallskip

\item We may find the coordinates of the space-time point where 
$A$ meets $B$ again in $K_3$ by Lorentz transformation from $K_1$. 
Shorter is using the observation that $B$ performs the travel 
from $B_0$ to the final point in its rest frame in 3 time units.
So, we only have to add 3 to the previously calculated position of 
$B_0$ and find
$$ \begin{pmatrix}3+41/3\cr 40/3 \end{pmatrix} =
   \begin{pmatrix}  50/3\cr 40/3 \end{pmatrix}\, . $$
\smallskip

\item See the first figure of this solution.

\begin{figure}[th] % See subfolder figs08plt.
\centerline{\hbox{\psfig{figure=figures/3mink,width=10cm} }}

Figure: Travel of the inertial frame $K_1$ translated to the inertial 
frame $K_3$. \end{figure}
\smallskip 

\item The inertial frames $K_4$ and $K_3$ differ by the translation 
$\begin{pmatrix} \triangle t\cr \triangle x\end{pmatrix}$: 
$\begin{pmatrix} t_4\cr x_4 \end{pmatrix} = 
\begin{pmatrix} t_3\cr x_3 \end{pmatrix} + \begin{pmatrix} \triangle t
\cr \triangle x\end{pmatrix}$, where
$$\begin{pmatrix} \triangle t \cr \triangle x\end{pmatrix} = 
\begin{pmatrix} t_2(B_0)\cr x_2(B_0) \end{pmatrix} - 
\begin{pmatrix} t_3(B_0)\cr x_3(B_0) \end{pmatrix} = 
\begin{pmatrix} 3\cr 0 \end{pmatrix} - 
\begin{pmatrix} 41/3\cr 40/3 \end{pmatrix} = -
\begin{pmatrix} 32/3\cr 40/3 \end{pmatrix}\,. $$

\item See the second figure of this solution.

\begin{figure}[th] % See subfolder figs08plt. 
\vskip -50pt
\centerline{\hbox{\psfig{figure=figures/4mink,width=10cm} }}

Figure: Travel of the inertial frame $K_3$ shifted to $K_4$ by the 
translation
$$\begin{pmatrix} \triangle t \cr \triangle x\end{pmatrix} = 
 - \begin{pmatrix} 32/3\cr 40/3 \end{pmatrix}\,,~~~
\begin{pmatrix} t_4\cr x_4 \end{pmatrix} = 
\begin{pmatrix} t_3\cr x_3 \end{pmatrix} + \begin{pmatrix} \triangle t
\cr \triangle x\end{pmatrix}\,.$$ 
\end{figure}
\smallskip 

\item See the third and last figure of this solution.
\begin{figure}[th] % See subfolder figs08plt. 
\vskip -20pt
\centerline{\hbox{\psfig{figure=figures/5mink,width=10cm} }}
Figure: Inertial frames $K_2$ and $K_4$ patched together, so that
the travel of $B$ is always in its rest frame. The lines for $A$
meet at 
$\begin{pmatrix} t \cr x \end{pmatrix} =
  \begin{pmatrix} 3 \cr -2.4 \end{pmatrix}\,.$
\end{figure}
\smallskip 
\end{enumerate} \end{enumerate}

\vfil\eject

\begin{figure}[th] % See subfolder figs08plt. 
\noindent


(j) $A$ travels to the patch point in proper time $3^2-(2.4)^2=
3.24$, in $K_2$ counted from the starting point, and in $K_4$
counted backward from the terminal point. In $K_1$ these positions 
are on the $x=0$ axis, at $t=3.24$ for the space-time point from
$K_2$ and at $t=10-3.24=6.76$ for the space-time point from
$K_4$.
\end{figure}


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