Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 71 / 1 / c / Solution

For radial null propagation, $dr/\sqrt{1-k{r}^2}=|dt|/a$. Since $dz/dt=-(1+z)H(z)$ and $a=a_0/(1+z)$, its present-distance integral isThus the printed extra $a_0$ before the redshift integral is inappropriate when $H$ is the physical Hubble parameter; it disappears only if $a_0$ has been set to one.

Dust conservation and the curvature relation giveTo evaluate it, put $s=\sqrt{1-{\Omega }_0}$ and $v=\sqrt{1+{\Omega }_{0}z}$. The dimensionless integral $I=H_0{d}_S$ isInverting the proper-distance expression in part b gives $r=\sinh (sI)/(a_0{H}_{0}s)$. Substituting the logarithm, writing the hyperbolic sine as half the difference of its exponential and inverse, and using $v^2-s^2={\Omega }_0(1+z)$ yieldsThis is the open dust distance-redshift relation. The printed square root lacks the factor $z$; its literal form would give a negative, nonzero coordinate at $z=0$. The corrected expression has $r\sim z/(a_0{H}_0)$ near zero. At fixed positive ${\Omega }_0$ and $z\gg {\Omega }_0^{-1}$, the term linear in $z$ dominates, givingThe zero-density limit is nonuniform: a pure Milne model has no such finite limiting radial coordinate.