For pressureless matter, $\rho_ma^3=\rho_{m,0}$ when $a_0=1$. Set
$$
D=\frac{8\pi G\rho_{m,0}}3;
$$
then $\dot a^2=D/a-k$. Substitution of $a=A(1-\cos\theta)$ makes this identity hold when
$$
\boxed{A=\frac{D}{2k}=\frac{4\pi G\rho_{m,0}}{3k}},
\qquad
\boxed{B=\frac{A}{\sqrt k}=\frac{D}{2k^{3/2}}}.
$$
Indeed $dt/d\theta=B(1-\cos\theta)=Ba/A$, and hence
$$
t=B(\theta-\sin\theta)
$$
after setting $t=0$ at $\theta=0$. This is the <closed matter-dominated Friedmann solution>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 310 / 1 / c / Solution

A positively curved pressureless Friedmann universe has the cycloidal parametrization $a=A(1-\cos\theta)$ and $t=B(\theta-\sin\theta)$, with $B=A/\sqrt{k}$. It expands from a Big Bang, reaches maximum size at $\theta=\pi$, and recollapses at $\theta=2\pi$.


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 Incoming links: Closed matter-dominated Friedmann solution