Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 9E (Cosmology)

Since $a(\tau )=(\tau /{\tau }_0{)}^2$, the conformal Hubble parameter is $\mathcal{H}=2/\tau$. With ${\Omega }_M=1$, the mode equation becomesSubstitution of $\delta ={\tau }^{\beta }$ givesThe two independent solutions are therefore ${\tau }^2$ and ${\tau }^{-3}$, so

The physical horizon size at conformal time $\tau$ is $a(\tau )c\tau$. Horizon crossing occurs when it equals the physical wavelength:and hence

At late times the decaying mode is neglected, so the growing solution in part (a) obeys $\delta \propto {\tau }^2\propto a$. For the equality-crossing mode this gives

A mode that crosses during matter domination subsequently grows byIts present amplitude is consequentlyNowSquaring therefore gives

Since ${\tau }_H\propto k^{-1}$, modes crossing chronologically between equality and today satisfy $k_0\le k\le k_{\mathrm{eq}}$. The reverse ordering printed with the target formula is inconsistent with the stated definitions of $k_0$ and $k_{\mathrm{eq}}$; it does not affect the spectrum calculation.