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

Set $u=Br/A$, retaining $r_0$ in $B$. Replacing only the lower integration limit by zero givesSubstitution of the constants yieldsTo quantify the approximation, write $ϵ=\delta M/M\ll 1$. The true lower limit is $u_0=B{r}_0/A=ϵ$, and an elementary integration by substitution, $u={\sin }^2\chi$, givesSince $\arcsin \sqrt{ϵ}-\sqrt{ϵ(1-ϵ)}=\frac{2}{3}{ϵ}^{3/2}+O({ϵ}^{5/2})$, the approximate time is too large by a relative fraction $\frac{4}{3\pi }{ϵ}^{3/2}+O({ϵ}^{5/2})$. This explains why changing the lower limit introduces little error for a small excess mass. The cosmological turnaround time diverges as $\delta M\to 0^{+}$, consistently with the absence of finite cosmological turnaround in an exactly critical Einstein-de Sitter universe.