= Solution
The <particle horizon> is the greatest distance from which a light signal could have reached the observer since the initial cosmic time $t_i$. Its comoving radius and its radial proper distance at time $t$ are
$$
\chi_h(t)=\int_{t_i}^t\frac{dt'}{a(t')},\qquad
\boxed{d_h(t)=a(t)\int_{t_i}^t\frac{dt'}{a(t')}.}
$$
Here proper distance is measured along the spatial slice, rather than by the transverse area of a sphere. The <Hubble parameter>, <deceleration parameter> and <cosmological density parameter> are
$$
\boxed{H=\frac{\dot a}{a},\qquad q=-\frac{a\ddot a}{\dot a^2},\qquad
\Omega=\frac{\rho}{\rho_{\rm cr}},\qquad\rho_{\rm cr}=\frac{3H^2}{\kappa}.}
$$
The last expression defines the <critical density>. For <blackbody radiation>, $p=\rho/3$, so the <Friedmann acceleration equation> gives $\ddot a/a=-\kappa\rho/3$. Consequently
$$
\boxed{q=\frac{\kappa\rho}{3H^2}=\Omega.}
$$
This is the <radiation relation between deceleration and density>. It does not require spatial flatness.
For the <open radiation universe particle horizon>, write $x=a/a_0$. The <cosmological continuity equation> gives $\rho=\rho_0x^{-4}$, while the present <Friedmann equation> gives $-k/a_0^2=H_0^2(1-\Omega_0)$. Thus
$$
H^2=H_0^2\left[\Omega_0x^{-4}+(1-\Omega_0)x^{-2}\right],\qquad
\dot x=\frac{H_0}{x}\sqrt{\Omega_0+(1-\Omega_0)x^2}.
$$
Taking the big-bang endpoint $x=0$ and using $dt=dx/\dot x$, the present radial proper distance becomes
$$
\begin{aligned}
d_{h0}&=\int_0^1\frac{dx}{x\dot x}
=\frac1{H_0}\int_0^1\frac{dx}{\sqrt{\Omega_0+(1-\Omega_0)x^2}}\\
&=\frac{\operatorname{arsinh}\sqrt{(1-\Omega_0)/\Omega_0}}
{H_0\sqrt{1-\Omega_0}}.
\end{aligned}
$$
Therefore
$$
\boxed{d_{h0}=\frac1{H_0\sqrt{1-\Omega_0}}
\ln\left[\sqrt{\frac{1-\Omega_0}{\Omega_0}}+
\sqrt{1+\frac{1-\Omega_0}{\Omega_0}}\right].}
$$
The formula applies to $0<\Omega_0<1$. Its flat-radiation limit is $H_0d_{h0}\to1$, agreeing with $a\propto t^{1/2}$ and $d_h=2t=H^{-1}$. The divergence as $\Omega_0\to0$ reflects the unbounded past conformal interval of the limiting empty open model; that endpoint is not a radiation-filled universe.
The <horizon problem> concerns the nearly uniform temperature of widely separated parts of the <cosmic microwave background>. In a purely decelerating <Hot Big Bang model>, their past light cones at last scattering do not overlap far enough to explain this agreement by thermal contact. During <cosmic inflation>, accelerated expansion shrinks the <comoving Hubble radius>. A patch initially small enough for causal communication can be stretched to encompass the later observable universe. <Reheating> converts the inflationary energy into a hot plasma with correlated initial conditions across that patch. Enough inflation must occur before the observable scales leave the <Hubble radius>; inflation cannot establish contact between regions that were never initially causally related.
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