= Solution
Radiation conservation gives
$$
\rho_R=\rho_{R0}a^{-4}.
$$
Consequently
$$
H^2=H_0^2\Omega_{R0}a^{-4}
+\frac{\Lambda c^2}{3}.
$$
Evaluating at $t_0$, where $a=1$, shows that
$$
\frac{\Lambda c^2}{3}=H_0^2(1-\Omega_{R0}).
$$
Put
$$
A=H_0^2\Omega_{R0},
\qquad
B=H_0^2(1-\Omega_{R0}),
\qquad
b=a^2.
$$
Then $\dot b=2\sqrt{A+Bb^2}$. Choosing the big bang as $t=0$ and integrating gives
$$
b=\sqrt{\frac AB}\sinh(2\sqrt B\,t).
$$
Therefore
$$
\boxed{
a(t)=\alpha[\sinh(\beta t)]^{1/2}},
$$
where
$$
\boxed{
\alpha=\left(\frac{\Omega_{R0}}{1-\Omega_{R0}}\right)^{1/4},
\qquad
\beta=2H_0\sqrt{1-\Omega_{R0}}}.
$$
At early times, $\sinh(\beta t)\sim\beta t$, so $a\propto t^{1/2}$, as in a radiation-dominated universe. At late times,
$$
a\sim\frac{\alpha}{\sqrt2}e^{\beta t/2},
$$
the expected de Sitter expansion with $H\to\sqrt{\Lambda c^2/3}$.
<Acceleration> begins when the radiation deceleration and cosmological-constant <acceleration> balance:
$$
H_0^2\Omega_{R0}a^{-4}
=H_0^2(1-\Omega_{R0}).
$$
Thus $a=\alpha$, which in the exact solution means $\sinh(\beta t_\Lambda)=1$. Hence
$$
\boxed{
t_\Lambda=\frac{\operatorname{arsinh}1}
{2H_0\sqrt{1-\Omega_{R0}}}}.
$$
This is the <radiation-to-cosmological-constant transition>.
Back to article page
