Past exam of the general relativity course of the University of Oxford / 2022 / 4 / c / Solution

Let $t_0$ be the initial epoch, let $r(t)$ be the physical radius of the outer matter shell, and put $r(t_0)=r_0$. The background Einstein-de Sitter universe has critical density ${\rho }_b=3H_i^2/(8\pi G)$ at that epoch. Its mass inside $r_0$ is $M_b=M-\delta M$, where $M$ is the actual conserved mass. Hence the background Friedmann equation impliesFor spherical collapse of pressureless matter, with no shell crossing, the shell theorem gives $\ddot{r}=-GM/r^2$. Its conservation of energy first integral isConsequentlyFor $\delta M>0$, the initially expanding branch reaches cosmological turnaround at $r_{\mathrm{ta}}=A/B=M{r}_0/\delta M$. Since $dt=dr/\dot{r}$ on that branch, the elapsed cosmic time isThis is spherical turnaround from an unperturbed Hubble velocity. The integral measures the comoving matter observers' time interval in the pressureless model. The integration variable is the physical radius, not the radius divided by the background scale factor. If an absolute cosmic time is wanted, add $t_0$ to the displayed interval.