= Solution
The <cosmological perfect-fluid continuity equation> gives $\rho_r=\rho_{r,0}a^{-4}$ and $\rho_m=\rho_{m,0}a^{-3}$. In <conformal time>, $\dot a=a'/a$ and $H=a'/a^2$. Thus the flat <Friedmann equation> is
$$
\boxed{(a')^2=H_0^2(\Omega_{r,0}+\Omega_{m,0}a)}.
$$
Choose the expanding branch and put the <Big Bang> at $\tau=0$. Integration gives
$$
\tau(a)=\frac{2}{H_0\Omega_{m,0}}
\left[\sqrt{\Omega_{r,0}+\Omega_{m,0}a}-\sqrt{\Omega_{r,0}}\right].
$$
Inverting it supplies the <radiation-matter scale factor in conformal time>:
$$
\boxed{a(\tau)=\frac{H_0^2\Omega_{m,0}}4\tau^2
+H_0\sqrt{\Omega_{r,0}}\,\tau},\qquad
A=\frac{H_0^2\Omega_{m,0}}4,\quad B=H_0\sqrt{\Omega_{r,0}}.
$$
The early linear term gives <radiation domination>, while the late quadratic term gives <matter domination>. The solution presumes the stated radiation-plus-matter model, without a <cosmological constant>.
At <matter-radiation equality>, $a_{\rm eq}=\Omega_{r,0}/\Omega_{m,0}$. Since $\Omega_{r,0}+\Omega_{m,0}=1$,
$$
\tau_{\rm eq}=\frac{2\sqrt{\Omega_{r,0}}}{H_0\Omega_{m,0}}(\sqrt2-1),
\qquad
\tau_0=\frac{2}{H_0\Omega_{m,0}}(1-\sqrt{\Omega_{r,0}}).
$$
In units $c=1$, the comoving <particle horizon> at equality is $\tau_{\rm eq}$; its physical size is $a_{\rm eq}\tau_{\rm eq}$. The <angular diameter distance> to the equality surface is $a_{\rm eq}(\tau_0-\tau_{\rm eq})$. Therefore the <horizon angle at matter-radiation equality> corresponding to one horizon length is
$$
\boxed{\theta_{\rm hor}\simeq\frac{\tau_{\rm eq}}{\tau_0-\tau_{\rm eq}}
=\frac{(\sqrt2-1)\sqrt{\Omega_{r,0}}}{1-\sqrt{2\Omega_{r,0}}}}.
$$
For $\Omega_{r,0}=10^{-4}$, this is $4.20\times10^{-3}$ radians, or $0.241^\circ$, about $14.4$ arcminutes. If quoting the full diameter of a horizon-sized patch, the result is twice this, $0.481^\circ$. The small-angle approximation is excellent. Expanding in $\sqrt{\Omega_{r,0}}\ll1$ would give $4.14\times10^{-3}$ radians; retaining both cosmic components near equality avoids an inaccurate pure-matter extrapolation there. The <scale factor> cancels between physical size and <angular diameter distance>, and $H_0$ cancels from the ratio.
If <cosmological recombination> occurs only a modest expansion after equality, its causal patch still subtends a small part of the sky. Widely separated regions of the observed <Cosmic microwave background> have nearly the same temperature even though their past <particle horizons> did not overlap in the ordinary decelerating history. Local thermalization after the <Big Bang> therefore cannot explain this large-angle uniformity. This is the <Horizon problem>; inflation supplies an earlier connected region that can grow to encompass the observed sky.
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