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

A photon loses energy by $1+z=a_0/a_e$ between emission and reception. The interval between successive photon arrivals is also longer by $1+z$, so the received energy per unit time is reduced by $(1+z{)}^2$. An isotropic wavefront at reception has area $4\pi (a_{0}r{)}^2$: the astronomical radial coordinate is an areal coordinate, even in curved space. HenceThis defines the luminosity distance. Photon absorption or conversion would require a further transmission factor; the relation assumes freely propagating conserved photons and bolometric luminosity. The radial cosmological proper distance is generally different from the areal distance $a_{0}r$.

Take zero cosmological constant and an open geometry, which means $k<0$ in the metric convention. Evaluating the Friedmann equations today gives $H_0^2=H_0^2{\Omega }_0-k/a_0^2$, soOn a constant-time slice the radial line element is $d\ell =a_{0}dr/\sqrt{1-k{r}^2}$. ThereforeThis is a cosmological proper distance on today's slice, not the distance traveled during the photon flight. The printed condition $k<1$ is insufficient to specify an open model; the derivation uses its intended $k<0$.

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.