Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 346 / 1 / b / Solution

The shell feels only radial gravity and the radial force due to the cosmological constant, so its torque vanishes and its specific angular momentum $L=r^2\dot{\theta }$ is conserved. Multiplyingby $\dot{r}$ and integrating gives the conserved specific orbital energy

For a uniform sphere, assembling concentric shells gives its gravitational potential energyThe cosmological-constant potential per unit mass is $-\Lambda r^2/6$. Since $⟨r^2⟩=3r_{\mathrm{ta}}^2/5$ in a uniform sphere,

The scalar virial theorem weights a potential homogeneous of degree $n$ by $-nW$. Gravity has degree $-1$ and the $\Lambda$ potential degree $2$, so the final state obeysAt turnaround $T=0$, while the virial relation gives $E_f=W_{G,f}/2+2W_{\Lambda ,f}$. Conservation of energy, together with $M=4\pi {\rho }_{\mathrm{ta}}{r}_{\mathrm{ta}}^3/3$, then gives, for $x=r_f/r_{\mathrm{ta}}$ and $\eta =\Lambda /(4\pi G{\rho }_{\mathrm{ta}})$,The root connected continuously to the $\Lambda =0$ solution has $x=1/2-\eta /8+O({\eta }^2)$, equivalentlyto first order in $\eta$. With $\Lambda =0$, virialization occurs at half the turnaround radius. At fixed turnaround state, positive $\Lambda$ makes this equilibrium root slightly smaller because its repulsive quadratic potential enters both energy conservation and the virial relation; sufficiently strong repulsion instead prevents a bound virialized state. Negative $\Lambda$ shifts the root in the opposite direction.