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\begin{document}

\centerline{\bf Special and General Relativity (PHZ 4601/560 
Fall 2017) Solutions Midterm October 20.} \bigskip 

\begin{figure}[th] % From EMscript/figures_plt/minkowski_space_sol.
\centerline{\hbox{ \psfig{figure=figures/1msol,width=12cm} }}
\caption{Minkowski space in which observer~B is at rest and
observer~A moves with speed $4c/5$ in negative $x$ direction. 
\label{fig_mid1mink} } \end{figure}

\begin{enumerate}

\item {\bf Light signals and travel in two inertial frames.}

We use natural units, $c=1$, and express everything in seconds.

\begin{enumerate}

\item In the rest frame $S$ of A the coordinates of $A_1$ are $(t,x)
=(15,0)$. In the rest frame $S'$ of B we have $15=\tau=\sqrt{t'^2-x'^2}$ 
for the proper time of A at this space-time point and $x'=-4t'/5$. 
Putting these equations together, we find:
$$ 15 = \sqrt{(t')^2-(4t'/5)^2} = 3\,t'/5~~\Rightarrow~~
   t'=25~~\Rightarrow~~(t',x')=(25,-20)\,.$$

\item In $S$ the coordinates of $B_0$ are the meeting point of the 
straight line $x=4t/5$ and $x=t-15$ for $t\ge 15$. The solution is 
$(t,x)=(75,60)$ and the proper time of B at this point is $\tau=
\sqrt{t^2-x^2}=45$. In the $S'$ frame we have by its definition 
$x'=0$ for B and the space-time point $B_0$ becomes $(t',x')=(45,0)$.

\item In $S$ the space-time position $A_2$ is at $(t,x)=(135,0)$. We
can deal with this like before with $(t,x)=(15,0)$: 
$$ 135 = 3\,t'/5~~\Rightarrow~~t'=225\,,~~x'= -4\,t'/5 = -180\,.$$
Alternatively, we may perform the Lorentz transformations
$$ t' = \gamma\,t - \beta\gamma\,x\,~~~
   x' = \gamma\,x - \beta\gamma\,t\,.$$
Now, $\beta=4/5$ and $\gamma=1/\sqrt{1-(4/5)^2}=5/3$, 
$\beta\gamma=4/3$. Hence, $(t,x)=(135,0)$ transforms into
$$ t' = 5t/3 = 225\,,~~x'=-4t/3=-180\,~~{\rm i.e.,}~~
   (t',x')=(225,-180)\,. $$

\item See the figure.

\end{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^{x_1} dx = \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}\cosh\left(\frac{g\tau}{c}\right)
  \,\tanh\left(\frac{g\tau}{c}\right)\,d\tau = 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

\end{enumerate} 
\end{document}
