Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 14D (Cosmology)

Writing $r=a(q+\psi )$ to first order gives $x^{\prime \prime }+\mathcal{H}{x}^{\prime }\simeq {\psi }^{\prime \prime }+\mathcal{H}{\psi }^{\prime }=-{\nabla }_q\phi$. The perturbation of Poisson's equation is ${\nabla }_q^2\phi =(4\pi G/c^2)a^2{\bar{\rho }}_M{\delta }_M$.

The Jacobian is $a^3\det (I+\partial \psi /\partial q)\simeq a^3(1+\nabla \cdot \psi )$, so mass conservation gives ${\delta }_M=-\nabla \cdot \psi$. Taking the divergence of (i) and using Poisson yields ${\delta }_M^{\prime \prime }+\mathcal{H}{\delta }_M^{\prime }-(4\pi G/c^2){\bar{\rho }}_M{a}^2{\delta }_M=0$.

Set ${\delta }_R=e^{-S}u$ and neglect $S^{\prime \prime }$, $(S^{\prime }{)}^2$, and slow variation relative to $k^2$. Cancelling the first derivative gives $S^{\prime }=\alpha k/(2\mathcal{H})$, hence $S={\int }^{\tau }\alpha k/(2\mathcal{H})d{\tau }^{\prime }$, and $u^{\prime \prime }+(c^2{k}^2/3)u\simeq 0$. Thus acoustic oscillations are exponentially damped, unlike pressureless matter perturbations, which have a growing mode.