Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 74 / 1 / ii / Solution

For a radial null geodesic, the FLRW metric givesTwo successive wave crests traverse the same comoving path between comoving observers. Subtracting their path integrals, to first order in their short emission and reception periods, gives $\delta t_0/a(t_0)=\delta t_e/a(t_e)$. Since cosmic time is each observer's proper time and the frequency is the reciprocal period,This neglects peculiar velocities and local gravitational redshift contributions.

For B's light passing through A on its way to Earth, with the comparison event at A being that passage, cosmological redshift factors multiply:This is the intended numerical answer with that reception-event specification. The reception-event dependence of intergalactic redshift is essential: Earth's two redshifts by themselves neither specify the A–B separation nor identify the emission event at B whose light A receives at a chosen time. For arbitrary directions or a different reception time at A, the general answer is $a(t_{A,\mathrm{rec}})/a(t_{B,\mathrm{em}})-1$, with the emission time fixed by the A–B null geodesic; it cannot be deduced from the two supplied numbers alone.