= Solution
For $r_s\gg d_H$, a regular adiabatic perturbation follows the large-scale <density contrast> growth in <6/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>.
\b[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. Then
$$
a_{\rm eq}=\frac{\Omega_\gamma}{\Omega_m}=\frac1{3000},\qquad
H^2=H_0^2\left(\Omega_\gamma a^{-4}+\Omega_ma^{-3}\right).
$$
Both terms must be retained at <matter-radiation equality>. The present comoving size corresponding to its <Hubble radius> is
$$
\boxed{R_{H,\rm eq}=\frac{c}{a_{\rm eq}H_{\rm eq}}
=\frac{c}{H_0}\frac{\sqrt{\Omega_\gamma}}{\sqrt2\,\Omega_m}=106.4\ \mathrm{Mpc}.}
$$
Its 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 $q=RH_0/c$ in $aR=c/H(a)$ gives
$$
a_{\rm ent}=\frac{q^2\Omega_m+\sqrt{q^4\Omega_m^2+4q^2\Omega_\gamma}}2
=2.27\times10^{-4}=0.681a_{\rm eq}.
$$
This 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.
Back to article page
