 Solution

On the expanding branch of the Friedmann equation, dropping the vacuum term gives the conformal timeThe lower limit sets $\tau =0$ at the Big Bang. The approximation requires ${\Omega }_{\Lambda }{a}^4\ll {\Omega }_{m}a$, and remains safe earlier when radiation in cosmology dominates. At matter-radiation equality, ${\rho }_{m,0}{a}^{-3}={\rho }_{r,0}{a}^{-4}$, so $a_{\mathrm{eq}}={\Omega }_r/{\Omega }_m$. Substitution yieldsIn conformal coordinates the radial FRW metric is $d{s}^2=a^2(-d{\tau }^2+d{r}^2)$. A radial light ray therefore travels a comoving distance $dr=d\tau$. The maximum distance traversed since the Big Bang is ${\int }_0^{t_{\mathrm{eq}}}dt/a={\tau }_{\mathrm{eq}}$: this is the comoving particle horizon, rather than the instantaneous comoving Hubble radius or the acoustic horizon. Restoring $c$ in the numerical conversion givesFor ${\Omega }_m=0.3$ and $h=0.7$, the comoving particle horizon at matter-radiation equality is about $110\mathrm{Mpc}$. The corresponding scale factor is $a_{\mathrm{eq}}\simeq 2.9\times 10^{-4}$.