Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 35D (General Relativity)

For $k=1$, put $r=\sin \chi$. The spatial metric tensor becomesThe global closed spatial slice is a 3-sphere with $0\le \chi \le \pi$; the original $r$ chart covers only a hemisphere, so two charts are needed. A circle at fixed $\chi$ in an equatorial angular plane has circumference $\boxed{2\pi a\sin \chi }$. The full spatial volume isIntegrating a single $r\in [0,1]$ chart without its second copy would give only half this volume.

For the Einstein static universe with $P=0$ and ${\rho }_0>0$, the acceleration equation gives $\Lambda =4\pi {\rho }_0$. The first Friedmann equation then gives $k/a_0^2=4\pi {\rho }_0$, soKeep $\Lambda ,k$ fixed and use the specified perturbation closure $\delta P=\delta \rho /3$. Linearization of the constraint gives $\delta \rho =-3{\rho }_0\delta a/a_0$. In the acceleration equation the background terms cancel, leavingThusThe growing mode establishes linear instability of the static universe; a specially chosen decaying perturbation does not remove instability. This uses the pressure-density relation explicitly supplied for the perturbation. If zero density is allowed, there is also the degenerate vacuum static case $k=0$, $\Lambda =0$, which is not the positive-density static universe considered here.