Past exam of the astrophysics course of the University of Oxford / 2022 / 6 / c / Solution

For $r_s\gg d_H$, a regular adiabatic perturbation follows the large-scale density contrast growth in b; this does not require causal communication across the perturbation. Since $d_H=c/H$, the Friedmann equation gives $d_H\propto a^2$ during radiation domination and $d_H\propto a^{3/2}$ during matter domination. A fixed comoving perturbation has physical size $r_s=aR$, so its ratio to $d_H$ decreases as $a^{-1}$ or $a^{-1/2}$ until cosmological horizon entry.

The numerical “exactly at equality” assertion is inconsistent with the supplied parameters and $d_H=c/H$. Normalize $a_0=1$ and use the supplied ${\Omega }_{\gamma }$ as the whole radiation component. ThenBoth terms must be retained at matter-radiation equality. The present comoving size corresponding to its Hubble radius isIts physical size then is $0.0355\mathrm{Mpc}$. A mode with $R=79\mathrm{Mpc}$ has physical size only $0.0263\mathrm{Mpc}$ at equality, already inside the Hubble radius. More explicitly, putting $q=R{H}_0/c$ in $aR=c/H(a)$ givesThis is a direct numerical counterexample using the stated data. Neglecting one component exactly at equality instead gives $150.5\mathrm{Mpc}$, which also does not produce $79\mathrm{Mpc}$. Late-time dark energy is negligible at these early times.