= Solution
On the expanding branch of the <Friedmann equation>, dropping the vacuum term gives the <conformal time>
$$
\tau(a)\simeq\frac1{H_0}\int_0^a\frac{du}{\sqrt{\Omega_r+\Omega_m u}}
=\frac{2}{H_0\Omega_m}\left(\sqrt{\Omega_r+\Omega_m a}-\sqrt{\Omega_r}\right).
$$
The 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 $a_{\rm eq}=\Omega_r/\Omega_m$. Substitution yields
$$
\boxed{\tau_{\rm eq}=\frac{2\sqrt{\Omega_r}}{H_0\Omega_m}(\sqrt2-1).}
$$
In conformal coordinates the radial <FRW metric> is $ds^2=a^2(-d\tau^2+dr^2)$. A radial light ray therefore travels a comoving distance $d r=d\tau$. The maximum distance traversed since the <Big Bang> is $\int_0^{t_{\rm eq}}dt/a=\tau_{\rm eq}$: this is the <comoving particle horizon>, rather than the instantaneous <comoving Hubble radius> or the acoustic horizon. Restoring $c$ in the numerical conversion gives
$$
\tau_{\rm eq}\simeq
\frac{2(3000/h)\sqrt{4.2\times10^{-5}/h^2}(\sqrt2-1)}{\Omega_m}\,\mathrm{Mpc}
=\frac{16.1}{\Omega_m h^2}\,\mathrm{Mpc}.
$$
For $\Omega_m=0.3$ and $h=0.7$, the <comoving particle horizon> at <matter-radiation equality> is \b[about $110\,\mathrm{Mpc}$]. The corresponding <scale factor> is $a_{\rm eq}\simeq2.9\times10^{-4}$.
Back to article page
