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Change independent variable from proper time to the scale factor. Then $\dot{\delta }=\dot{a}{\delta }_a$ and $\ddot{\delta }={\dot{a}}^2{\delta }_{aa}+\ddot{a}{\delta }_a$. Under radiation domination, $\ddot{a}=-{\dot{a}}^2/a$ and $H^2\simeq 8\pi G{\bar{\rho }}_r/3$. Since ${\bar{\rho }}_m/{\bar{\rho }}_r=a/a_{\mathrm{eq}}$, the linear matter perturbation growth equation becomesPutting $\delta =au$ givesThis keeps the matter self-gravity term in a radiation-dominated background. It is not an exact background equation through radiation-matter equality.

At $a/a_{\mathrm{eq}}\ll 1$, neglecting that small term gives ${\delta }_{aa}+{\delta }_a/a=0$. Integration yieldsThus matter has at most logarithmic growth during this leading radiation approximation, together with a constant independent mode. The constant is non-growing, not a mode that literally falls as $a^{-1}$; that power belongs to $u$ rather than ${\delta }_m$.

For an increasing/decreasing basis of the displayed equation with matter self-gravity retained, put $x=\sqrt{6a/a_{\mathrm{eq}}}$. Its equation becomes $x^2{\delta }_{xx}+x{\delta }_x-x^2\delta =0$. HenceThe Modified Bessel function of the first kind gives $I_0(x)=1+3a/(2a_{\mathrm{eq}})+O((a/a_{\mathrm{eq}}{)}^2)$, which increases slowly. The Modified Bessel function of the second kind gives $K_0(x)=-\frac{1}{2}\ln (a/a_{\mathrm{eq}})+\frac{1}{2}\ln (2/3)-{\gamma }_E+O((a/a_{\mathrm{eq}})|\ln (a/a_{\mathrm{eq}})|)$, where ${\gamma }_E$ is Euler's constant; this mode decreases as $a$ increases. Their leading span is precisely the constant/logarithmic pair above. Mode labels depend on the chosen basis and normalization; no rapid matter-era growth $\delta \propto a$ occurs here. Neglected background corrections can change subleading terms, so the Bessel basis should not be extrapolated through equality.

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, $\dot{a}=a^{\prime }/a$ and $H=a^{\prime }/a^2$. Thus the flat Friedmann equation isChoose the expanding branch and put the Big Bang at $\tau =0$. Integration givesInverting it supplies the radiation-matter scale factor in conformal time: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, $a_{\mathrm{eq}}={\Omega }_{r,0}/{\Omega }_{m,0}$. Since ${\Omega }_{r,0}+{\Omega }_{m,0}=1$,In units $c=1$, the comoving particle horizon at equality is ${\tau }_{\mathrm{eq}}$; its physical size is $a_{\mathrm{eq}}{\tau }_{\mathrm{eq}}$. The angular diameter distance to the equality surface is $a_{\mathrm{eq}}({\tau }_0-{\tau }_{\mathrm{eq}})$. Therefore the horizon angle at matter-radiation equality corresponding to one horizon length isFor ${\Omega }_{r,0}=10^{-4}$, this is $4.20\times 10^{-3}$ 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 $\sqrt{{\Omega }_{r,0}}\ll 1$ would give $4.14\times 10^{-3}$ 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.

Normalize the present scale factor to one. For $0<\Omega <1$, the parametric solution givesThus the age of a flat matter-coasting-fluid universe isA direct age integral proves the bound more transparently than manipulating inverse hyperbolic functions. Since $\dot{a}=H_0\sqrt{\Omega /a+b}$,For $0\le a\le 1$ and $0\le \Omega \le 1$, one has $a\le \Omega +ba\le 1$, henceIntegrating givesEquality on the left is the pure pressureless matter universe, $\Omega =1$; equality on the right is the pure coasting fluid, $\Omega =0$. These are all physically allowed density fractions for the nonnegative two-fluid model. The integrand decreases with $\Omega$, so the coasting-fluid age bound is approached continuously between those endpoints. In the closed expression the apparent $b=0$ singularities cancel using $\mathrm{arsinh}x=x-x^3/6+\cdots$; as $\Omega \to 0$, the term $\Omega \log (1/\Omega )$ vanishes. The limits are respectively $H_0{t}_0=2/3$ and $1$.