Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 15D / a

A comoving material cell contains the same particles in its perturbed and background descriptions, so its mass is $\bar{\rho }{a}^3{d}^{3}q=\rho d^{3}r$. The Jacobian determinant is $a^3\det (I+{\nabla }_q\psi )$. Expanding the determinant givesSplit the potential into homogeneous and perturbed parts. There is a sign error in the printed hint: ${\nabla }^2{\Phi }_0=4\pi G\bar{\rho }$ requires $\nabla {\Phi }_0=+4\pi G\bar{\rho }r/3$, not the printed negative sign. Its acceleration $-\nabla {\Phi }_0$ agrees with $\ddot{a}/a=-4\pi G\bar{\rho }/3$.

For the perturbation, ${\nabla }_q^2\varphi =4\pi G\bar{\rho }{a}^2\delta =-4\pi G\bar{\rho }{a}^2{\nabla }_q\cdot \psi$. Under the usual homogeneous boundary convention the curl-free force therefore satisfieswhere ${\psi }_L$ is the longitudinal part of the displacement. Expanding the particle acceleration gives $\ddot{r}=\ddot{a}(q+\psi )+2\dot{a}\dot{\psi }+a\ddot{\psi }$. The background acceleration cancels the first term, leavingFor a longitudinal displacement this proves the displayed longitudinal displacement in linear dust perturbations equation with ${\psi }_L=\psi$. For unrestricted displacement, a transverse component has no density contrast and instead obeys ${\ddot{\psi }}_T+2H{\dot{\psi }}_T=0$; the full-vector equation printed in the question thus needs the longitudinal assumption. Taking the divergence removes this qualification and gives