Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 45 / 1 / b

Use $c=1$ and signature $+---$. The positively curved FLRW metric in the usual radial chart isSet $r=\sin \chi$ and introduce conformal time by $dt=ad\eta$. Then $d{r}^2/(1-r^2)=d{\chi }^2$ in the chart, and the spherical polar coordinate $\chi \in [0,\pi ]$ extends the spatial description to the full three-sphere. ThusThe $r$ coordinate itself is not one-to-one across the equator, which is why the new polar coordinate is useful.

For $\Lambda =0$ and $k=1$, the Friedmann equation is ${\dot{a}}^2=(8\pi G/3)\rho a^2-1$. Since $a^{\prime }=a\dot{a}$, it becomesFor pressureless matter, the background continuity equation gives $\rho a^3={\rho }_0{a}_0^3$. Define the conserved normalization $M=(4\pi /3){\rho }_0{a}_0^3$. It makes the equation $(a^{\prime }{)}^2=2GMa-a^2$. The closed matter-dominated Friedmann solution is the cycloidIndeed $a^{\prime }=GM\sin \eta$ and $2GMa-a^2=(GM{)}^2{\sin }^2\eta$. Integrating $dt/d\eta =a$ gives the second expression, with the Big Bang at $t=\eta =0$. It expands to $a=2GM$ at $\eta =\pi$ and recollapses at $\eta =2\pi$. Here $M$ is the constant appearing in the Friedmann equation; it is not the total mass of the entire closed spatial slice, whose volume is $2{\pi }^2{a}^3$.