The Friedmann equation for the scale factor $a(t)$ of a homogeneous and isotropic universe of mass density $\rho$ is

$$\left(\frac{\dot{a}}{a}\right)^{2}=\frac{8 \pi}{3} G \rho-\frac{k c^{2}}{a^{2}}$$

where $\dot{a}=d a / d t$. Explain how the value of the constant $k$ affects the late-time $(t \rightarrow \infty)$ behaviour of $a$.

Explain briefly why $\rho \propto 1 / a^{3}$ in a matter-dominated (zero-pressure) universe. By considering the scale factor $a$ of a closed universe as a function of conformal time $\tau$, defined by $d \tau=a^{-1} d t$, show that

$$a(\tau)=\frac{\Omega_{0}}{2\left(\Omega_{0}-1\right)}[1-\cos (\sqrt{k} c \tau)]$$

where $\Omega_{0}$ is the present $\left(\tau=\tau_{0}\right)$ density parameter, with $a\left(\tau_{0}\right)=1$. Use this result to show that

$$t(\tau)=\frac{\Omega_{0}}{2 H_{0}\left(\Omega_{0}-1\right)^{3 / 2}}[\sqrt{k} c \tau-\sin (\sqrt{k} c \tau)],$$

where $H_{0}$ is the present Hubble parameter. Find the time $t_{B C}$ at which this model universe ends in a "big crunch".

Given that $\sqrt{k} c \tau_{0} \ll 1$, obtain an expression for the present age of the universe in terms of $H_{0}$ and $\Omega_{0}$, according to this model. How does it compare with the age of a flat universe?