Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 346 / 2 / Solution

A galaxy contains so many stars that its two-body relaxation time is generally much longer than its age. Individual encounters can therefore be neglected and each star moves in the smooth collective potential. Liouville conservation along these Hamiltonian trajectories gives the Collisionless Boltzmann equationIn spherical phase-space coordinates this is

The spherical line element is $d{s}^2=d{r}^2+r^{2}d{\theta }^2+r^2{\sin }^2\theta d{\phi }^2$, soFor a unit-mass star in a spherical potential,The Euler--Lagrange equations, followed by differentiating $v_{\theta }=r\dot{\theta }$ and $v_{\phi }=r\sin \theta \dot{\phi }$, give

Integrating the Boltzmann equation over velocity space, with vanishing velocity-space boundary terms, gives spherical mass conservation:Multiplication by $v_r$ and integration gives the radial Jeans equation. With isotropic dispersionthe geometric dispersion terms cancel, and use of continuity yieldsThe factor $\rho$ on the right is required dimensionally.

In a Lambda-CDM background, the local excess mass contributes $GM(r)/r^2$, homogeneous matter inside radius $r$ contributes $4\pi G{\rho }_{b}r/3$, and the cosmological constant contributes outward acceleration $\Lambda r/3$. HenceSince ${\Omega }_m=8\pi G{\rho }_b/(3H^2)$, ${\Omega }_{\Lambda }=\Lambda /(3H^2)$, and $q={\Omega }_m/2-{\Omega }_{\Lambda }$,

Write $⟨v_r⟩=Hr+v_p$, where $v_p$ is the peculiar velocity. ThenThe acceleration equation gives $\dot{H}+H^2=\ddot{a}/a=-q{H}^2$, which cancels the background term in ${\Phi }^{\prime }$. Therefore