= Solution
Take zero <cosmological constant> and an open geometry, which means $k<0$ in the metric convention. Evaluating the <Friedmann equations> today gives $H_0^2=H_0^2\Omega_0-k/a_0^2$, so
$$
\boxed{-k=a_0^2H_0^2(1-\Omega_0)},\qquad 0<\Omega_0<1.
$$
On a constant-time slice the radial line element is $d\ell=a_0dr/\sqrt{1-kr^2}$. Therefore
$$
d_S=a_0\int_0^r\frac{dr'}{\sqrt{1+(-k)r'^2}}=\boxed{\frac{\operatorname{arsinh}[a_0H_0\sqrt{1-\Omega_0}\,r]}{H_0\sqrt{1-\Omega_0}}}.
$$
This is a <cosmological proper distance> on today's slice, not the distance traveled during the <photon> flight. The printed condition $k<1$ is insufficient to specify an open model; the derivation uses its intended $k<0$.
Back to article page
