% Bernd Berg sep 21 2017.
 \documentclass{ws-book9x6}  
\begin{document}

\setcounter{page}{1}

\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 5} 

% \begin{figure}[tb] % \vspace{5pc}
% \centerline{\hbox{ \psfig{figure=hsphere,width=12.5cm} }}
% \caption{for assignment~14.} \label{fig_lader} \vskip -10pt
% \end{figure} \centerline{Potential as function of $z$.}

\begin{description} 
\item{\bf 12. Spacetrip Elapsed time on earth.}
\end{description}
% smallskip

\noindent 
We only consider the first quarter (five years) of the flight. The
other results for the time are the same by symmetry reasons. 

\noindent 
Let us denote the inertial frame (IF) ``earth'' by $S_0$ and the 
speed of the spaceship in $S_0$ by $\beta(\tau)$ (this is equivalent
to $\beta(t)$, but more convenient in the problem at hand). We consider
a fixed proper time $\tau_1$, $\beta_1=\beta(\tau_1)$. By $S_1$ we
denote the IF that moves with $\beta_1$ away from $S$ and has the
spaceship at its origin at proper time $\tau_1$. At $\tau_1$ we have
then in $S_1$
$$ (d\beta)_{S_1} = (d\xi)_{S_1} = \frac{g}{c}\,d\tau\,.$$
This translates in $S_0$ to
$$ (d\beta)_{S_0} = \frac{\beta_1+(d\beta)_{S_1}}{1+\beta_1\,(d\beta)_{
   S_1}}-\beta_1 = (\beta_1+(d\beta)_{S_1})\,(1+\beta_1\,(d\beta)_{S_1}) 
   - \beta_1 $$
$$ = (\beta_1+(d\beta)_{S_1})\,(1-\beta_1\,(d\beta)_{S_1}) - \beta_1 
   = \beta_1+(d\beta)_{S_1}-(\beta_1)^2\,(d\beta)_{S_1} - \beta_1 $$
$$ = (d\beta)_{S_1}-(\beta_1)^2\,(d\beta)_{S_1} 
   = (d\beta)_{S_1}\,(1-(\beta_1)^2)\,.  $$
It is more convenient to work with the rapidity. Because the rapidity
is simply additive, we have in $S_0$
$$ (d\xi)_{S_0} = \xi_1+(d\xi)_{S_1}-\xi_1 = (d\xi)_{S_1} = d\xi\,,$$
where we jsut dropped the label $S_1$ at the last equal sign. 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

\noindent 
{\bf The maximum distance traveled from earth} is $2\,x$ with
$$ x=\int_0^{t}v(t)\,dt=\int_0^{\tau}\cosh\left[\zeta(\tau)\right]
   \,v(\tau)\,d\tau = c\,\int_0^{\tau} \sinh\left(\frac{g\tau}{c}
   \right) \,d\tau  $$
$$  = \frac{c^2}{g}\,\left[\cosh\left(\frac{g\tau}{c}
   \right) -1\right]\,.$$
Numerical values: $x = 83.4$ light years, maximum distance $=166.8$
light years.

\end{document}

