Past exam of the short options course of the University of Oxford / 2014 / 7 / c

At $s=0$ and $s=4T$, the mission is at the planet's position $x=c/\omega$. The final coordinate time from the last segment isThe planet is at rest in these coordinates, so its proper time elapsed equals this coordinate-time difference. The astronauts' proper time is the parameter interval, $4T$. ThereforeFor proper acceleration $c\omega =10\mathrm{m}{\mathrm{s}}^{-2}$ and $c=3\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, ${\omega }^{-1}=3\times 10^7$ seconds, approximately one year. Thus $\omega T=16$ and $e^{16}\approx 3000^2=9\times 10^6$. Neglecting $e^{-16}$ in the hyperbolic sine,The clocks compare two different timelike paths between the same departure and reunion events.

Put $a=\omega T$. At emission, $s=2T$, the middle segment givesThe planet is at $x_p=c/\omega$, so the inward light signal takes coordinate timeIts arrival time on the planet is therefore $t_r=2(e^a-1)/\omega$. Subtract this from the mission's return time $4\sinh a/\omega$:For large $T$ at fixed acceleration this tends to $2/\omega =2c/(c\omega )$. With the Earth-gravity value above, the wait is approximately two years, despite the millions of planet-years elapsed since departure in part (i).