= Solution
Normalize the <scale factor> to $a(t_0)=1$, use $c=1$, and define the instantaneous <critical density> by $\rho_{\rm cr}(t)=3H(t)^2/(8\pi G)$. Separately conserved <pressureless matter> and <radiation in cosmology> obey the <cosmological continuity equation>,
$$
\dot\rho+3H(\rho+p)=0,\qquad
\rho_M=\rho_{M0}a^{-3},\qquad
\rho_R=\rho_{R0}a^{-4}.
$$
The present <cosmological density parameters> are
$$
\Omega_{M0}=\frac{8\pi G\rho_{M0}}{3H_0^2},\qquad
\Omega_{R0}=\frac{8\pi G\rho_{R0}}{3H_0^2},\qquad
\Omega_{\Lambda0}=\frac{\Lambda}{3H_0^2},\qquad
\Omega_0=\Omega_{M0}+\Omega_{R0}+\Omega_{\Lambda0}.
$$
The <cosmological constant> has equivalent <energy density> $\rho_\Lambda=\Lambda/(8\pi G)$. Substituting into the <Friedmann equation> yields
$$
H^2=\frac{8\pi G}{3}(\rho_M+\rho_R)+\frac{\Lambda}{3}-\frac{k}{a^2}.
$$
At the present epoch this gives $k=H_0^2(\Omega_0-1)$, or curvature contribution $\Omega_{K0}=-k/H_0^2=1-\Omega_0$. Dividing each component by $H_0^2$ proves
$$
\boxed{\frac{H^2}{H_0^2}
=\Omega_{R0}a^{-4}+\Omega_{M0}a^{-3}
+\Omega_{\Lambda0}+(1-\Omega_0)a^{-2}.}
$$
Spatial curvature is not included in the total physical-density parameter $\Omega_0$. In particular, positive vacuum density does not have the radiation or matter scaling.
Back to article page
