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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2018 
Solutions} \medskip

\centerline{\bf Set 4} \medskip

\begin{enumerate} \setcounter{enumi}{8}

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

\begin{enumerate}

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

\item The proper time of $B$ at $B_0$, the end of the first part 
of its travel, 
is $\sqrt{(5)^2-(4)^2}=3$. 
\smallskip

\item $2\times 5=10$ is the time on the clock of $A$ when $A$ meets
$B$ again.
\smallskip

\item $2\times 3=6$ is the time on the clock of $B$ when $A$ meets
$B$ again.
\smallskip

\item We have $\beta_A^2=-\beta_B^1=-4/5$ for the velocity of $A$
in the inertial frame $K_2$.
\smallskip

\item The coordinates of $B_0$ in $K_2$ are $\begin{pmatrix} 3\cr
0\end{pmatrix}$.
\smallskip

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

\item We may find the final meeting point by calculating where the
straight lines on which $A$ and $B$ travel in $K_2$ meet or by 
performing the Lorentz transformation from $K_1$ to $K_2$. For 
simplicity of the equations we denote the coordinates in $K_2$ 
just by $t$ and $x$ without subscripts indicating $K_2$.
\smallskip

First method: $x=-4\,t/5$ for $A$ and $x=-40\,(t-3)/41$ for $B$.
Equating these equations gives 
\begin{eqnarray} \nonumber
  4\,t/5 &=& 40\,t/41-120/41 \\ \nonumber
  41\times 4\,t &=& 5\times 40\,t - 5\times 120 \\ \nonumber
  164\,t &=& 200\,t - 600 
\end{eqnarray}
with the result $t=50/3$. Therefore, $x=-(4\times 50)/(3\times 5)
=-40/3$. Together: The final space-time point in $K_2$ is
$$ \begin{pmatrix} t  \cr  x    \end{pmatrix}\ =\
   \begin{pmatrix}50/3\cr -40/3 \end{pmatrix}\, . $$
\smallskip

Second method (Lorentz transformation):
$$ \gamma = \sqrt{\frac{1}{1-\beta^2}} = \sqrt{\frac{1}{1-(4/5)^2}}
   = \sqrt{\frac{5^2}{5^2-4^2}} = \frac{5}{3}~~{\rm and}~~
   \beta\gamma = \frac{4}{3}\,.  $$
So, we find for the position of the final point in $K_2$
$$ \begin{pmatrix}t\cr x\end{pmatrix} =  
   \begin{pmatrix}\gamma&-\beta\gamma \cr 
  -\beta\gamma & \gamma\end{pmatrix} 
   \begin{pmatrix} 10\cr 0\end{pmatrix} =  
   \begin{pmatrix}5/3 & -4/3 \cr -4/3 & 5/3\end{pmatrix} 
   \begin{pmatrix} 10\cr 0 \end{pmatrix} =
   \begin{pmatrix}50/3\cr -40/3 \end{pmatrix}\, . $$
\smallskip

\item See the figure. \medskip

\end{enumerate} \end{enumerate}

\begin{figure}[th] % See subfolder figs08plt.
\centerline{\hbox{\psfig{figure=figures/2mink,width=10cm} }}
\medskip

Figure: Travel of the inertial frame $K_1$ translated to the inertial 
frame $K_2$. \end{figure}

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