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

\centerline{\bf Set 5} 

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\item{\bf 15. Spacetrip Elapsed time on earth.}
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We only consider the first quarter (five years) of the flight. The
other results for $t$ are the same by symmetry reasons. 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
$$ d\zeta(\tau) = \frac{d\zeta}{d\tau}\,d\tau = \frac{g}{c}\,d\tau $$
holds. 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.
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