Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 60 / 1 / ii / Solution

Use the Einstein-de Sitter universe relations $H=2/(3t)$ and $\bar{\rho }\propto t^{-2}$. Write $H_{\mathrm{ta}}=H(t_{\mathrm{ta}})$ and $R_{\mathrm{ta}}=R(t_{\mathrm{ta}})$. Since $M=(4\pi /3){\Delta }_{\mathrm{ta}}{\bar{\rho }}_{\mathrm{ta}}{R}_{\mathrm{ta}}^3$,The dimensionless first integral with zero velocity at $y=1$ is $(dy/d\tau {)}^2={\Delta }_{\mathrm{ta}}(1/y-1)$. On the expanding branch this givesEvaluating the integral gives the inverse-trigonometric expression in the source. Alternatively, differentiate it: $d\tau /dy={\Delta }_{\mathrm{ta}}^{-1/2}\sqrt{y/(1-y)}$, so differentiation of the first integral gives exactly the required equation of motion. At $y=1$, the integral equals $\pi /2$. Thus ${\tau }_{\mathrm{ta}}=2/3$ fixes $\sqrt{{\Delta }_{\mathrm{ta}}}=3\pi /4$. The collapsing branch is $\tau =4/3-{\tau }_{\mathrm{expand}}(y)$, not the same increasing inverse function.

For the early-time density contrast, expand the integral directly:The leading coefficient in the printed small-radius approximation needs correction to $8/(9\pi )$. The printed coefficient is inconsistent with both the exact integral and approach to the homogeneous background. Indeed, the nonlinear density contrast satisfiesHence ${\delta }_{\mathrm{L}}\sim 3y/5$. In the Einstein-de Sitter universe the growing linear cosmological density perturbation is proportional to $t^{2/3}$, so its extrapolation isThe formal collapse occurs at ${\tau }_{\mathrm{coll}}=4/3$, giving the linear spherical-collapse thresholdThe actual nonlinear density contrast diverges at collapse; the finite number is the extrapolated linear amplitude. Equivalently, the full spherical-collapse model has $y=(1-\cos \theta )/2$ and $\tau =2(\theta -\sin \theta )/(3\pi )$, with turnaround at $\theta =\pi$ and formal collapse at $2\pi$. Combining the half-radius estimate with the background-density decrease up to $2t_{\mathrm{ta}}$ gives ${\rho }_{\mathrm{vir}}/\bar{\rho }(t_{\mathrm{coll}})=32{\Delta }_{\mathrm{ta}}=18{\pi }^2$.