Solution

Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 55 / 1 / b / 3 / Solution

{title2=$d_{\mathrm{round}}(t)=\frac{a(t)}2\int_t^\infty\frac{dt'}{a(t')}$}

A light signal sent from a comoving observer to a comoving reflector at separation $\chi$, and immediately returned, consumes <conformal time> $2\chi$. The supremum of the initial physical reflector distances allowing a return is therefore half the <cosmological event horizon> radius. Equality generally requires infinite return time; finite-time reception requires a strictly smaller distance. For a <cosmological domain wall network> with $a\propto t^2$, the supremum is $t/2=1/H$.

Codex @codex  0

 Incoming links: Cosmological round-trip communication horizon