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

Divide the Friedmann equation by $a^2$ and use $a=(1+z{)}^{-1}$. The Hubble parameter therefore satisfiesThe positive root describes the expanding branch. Spatial flatness and the neglect of radiation in cosmology imply ${\Omega }_{m,0}+{\Omega }_{\Lambda ,0}=1$.

The age of an FLRW universe at an epoch is the elapsed cosmic time since the Big Bang, not the lookback time from today. For positive matter and positive cosmological constant, integrating $dt=da/\dot{a}$ givesPut $u=\sqrt{{\Omega }_{\Lambda ,0}/{\Omega }_{m,0}}{a}^{\prime 3/2}$; then $du=(3/2)\sqrt{{\Omega }_{\Lambda ,0}/{\Omega }_{m,0}}\sqrt{a^{\prime }}d{a}^{\prime }$. The integral becomes the age of a flat matter-Lambda universe:Choose $0<\theta <\pi /2$ such that $\tan \theta =\sqrt{{\Omega }_{m,0}/{\Omega }_{\Lambda ,0}}{a}^{-3/2}$. The upper-limit value of $u$ is $\cot \theta$, andConsequentlyThe integration constant is fixed by $t\to 0$ as $z\to \infty$, where $\theta \to \pi /2$. Equivalently, differentiating $u=\cot \theta$ changes the last integral to $-\int d\theta /\sin \theta$, recovering the supplied trigonometric identity. The formula assumes positive ${\Omega }_{\Lambda ,0}$ and ${\Omega }_{m,0}$; its zero-vacuum limit is regular even though this parametrization is singular.

At $z=0$, spatial flatness gives $\tan {\theta }_0=\sqrt{(1-{\Omega }_{\Lambda ,0})/{\Omega }_{\Lambda ,0}}$, hence $\cos {\theta }_0=\sqrt{{\Omega }_{\Lambda ,0}}$ and $\sin {\theta }_0=\sqrt{1-{\Omega }_{\Lambda ,0}}$. ThereforeFor ${\Omega }_{\Lambda ,0}\to 0$, $\mathrm{arsinh}u\sim u$ in the previous integral gives $t_0\to 2/(3H_0)$, the Einstein-de Sitter universe result. For ${\Omega }_{m,0}=0$, the positive-vacuum flat slicing has no finite-age Big Bang, explaining the other endpoint divergence.

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.