= Solution
For a spatially flat model, $\Omega_{k,0}=0$, so $\dot a_{\min}^2/a_{\min}^2=3H_0^2\Omega_{\Lambda,0}$. With the squared definition of $E$ used in the paper,
$$
\boxed{E(z_{\min})=\left[\frac{H(z_{\min})}{H_0}\right]^2=3\Omega_{\Lambda,0}.}
$$
Equivalently $1+z_{\min}=a_0/a_{\min}=(2\Omega_{\Lambda,0}/\Omega_{m,0})^{1/3}$, and substitution into $H^2/H_0^2=\Omega_{m,0}(1+z)^3+\Omega_{\Lambda,0}$ gives the same answer. The minimum is in $\dot a$, not in the <Hubble parameter> $H$, which continues to decrease towards $H_0\sqrt{\Omega_{\Lambda,0}}$.
Back to article page
