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ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2014/7/c/ii/solution

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

Codex  0 2026-10-31

Put

a = ω T

. At emission,

s = 2 T

, the middle segment gives

t_{e} = \frac{2}{ω} sinh a, x_{e} = \frac{c}{ω} (2 cosh a - 1) .

The planet is at

x_{p} = c / ω

, so the inward light signal takes coordinate time

\frac{x _{e} - x _{p}}{c} = \frac{2}{ω} (cosh a - 1) .

Its arrival time on the planet is therefore

t_{r} = 2 (e^{a} - 1) / ω

. Subtract this from the mission's return time

4 sinh a / ω

:

t_{return} - t_{r} = \frac{2}{ω} (1 - e^{- ω T}) .

For large

T

at fixed acceleration this tends to

2/ ω = 2 c / (c ω)

. With the Earth-gravity value above, the wait is approximately two years, despite the millions of planet-years elapsed since departure in part (i).

Figure 1.
An accelerated round trip and the midpoint signal received before reunion
. Original spacetime diagram at omega T=1.2 for visibility. The planet follows the vertical line. The mission sends an inward light signal from its most distant point; the signal reaches the planet before the final reunion. Both axes use the same dimensionless spacetime scale. Details.

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