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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 16. End of Spacetrip.}
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We use energy-momentum conservation
$$ 0 = dp = \left(
      \begin{matrix}dp^0_r\cr dp^1_r\cr 0\cr 0\end{matrix}\right) +
\left(\begin{matrix}dp^0_e\cr dp^1_e\cr 0\cr 0\end{matrix}\right)\,,$$
where the subscript $r$ stands for rocket and $e$ for exhaust. In 
the temporary rest frame of the rocket we have 
$$  dp^1_e=-v\,dm\,,\qquad dp^1_r=m\,du=m\,g\,d\tau\,,$$
where $v$ is the velocity of the exhaust, $dm$ the infinitesimal 
change of rest mass of the rocket (not the exhaust!) and $du$ the 
infinitesimal velocity of the rocket. Note that the exhaust may have 
no rest mass (v=c). The equation for $dp^1_e$ (no $\gamma$ in front
of $v$!) follows from
$$dp^0_e=-dp^0_r=c\,dm~~{\rm and}~~-\beta=\frac{dp^1_e}{dp^0_e}:
  \quad dp^1_e = -\beta\,dp^0_e = v\,dm\,.                            $$
With this definition $\beta=v/c$ is positive, because $dp^1_e$ is
negative when we choose $dp^1_r$ positive. Now, $0=-v\,dm+\,g\,d\tau$
and separation of variables gives
$$ \frac{dm}{m} = - \frac{g}{v}\,d\tau\ \Rightarrow\ 
   \int_{m_0}^{m(\tau)} \frac{dm}{m} = 
   - \frac{g}{v} \int_0^{\tau}d\tau'\ \Rightarrow
   \ \ln\left(\frac{m(\tau)}{m_0}\right) = -\frac{g\,\tau}{v}\,. $$
(1) The mass of the spaceship decreases with its proper time according to
$$ m(\tau) = m_0\,\exp\left(-\frac{g\tau}{v}\right)\,.$$
(2) With $v=c$ and after a trip of twenty years the remaining 
fraction of the mass is $m(20\,{\rm years})/m_0=1.10\times 10^{-9}$.

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