Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 73 / 2 / iv / Solution

In an Einstein-de Sitter universe, $H(a)=H_0{a}^{-3/2}$, so direct integration of $dt=da/(aH)$ gives $t(a)=2a^{3/2}/(3H_0)$. At $z=8$, $a=1/9$ and $a^{3/2}=1/27$. The fraction of the present age of an FLRW universe and the available formation interval are thereforeThis is only about $3.7\%$ of the present age. The available interval is $t(8)$ itself, not $t_0-t(8)$, which would instead be the lookback time.

For ${\Omega }_{m,0}=0.3$ and ${\Omega }_{\Lambda ,0}=0.7$, at the same Hubble constant, the exact integral givesThus the formation interval is longer, by about a factor of $1.83$. There is also a direct comparison: for $0<a<1$, $0.3a^{-3}+0.7<a^{-3}$, so $H(a)$ is smaller than its Einstein-de Sitter universe value at fixed $a$, and integrating $da/(aH)$ gives a greater age. At $z=8$ the leading matter-era result is $t\simeq 2/[3H_0\sqrt{0.3}(1+z{)}^{3/2}]$; the smaller matter density, with the same $H_0$, explains nearly all the increase.