 Solution

A radial null ray satisfies $dr/\sqrt{1-k{r}^2}=\pm cdt/a(t)$. Define the present-normalized radial comoving distanceFor $k<0$, write $\kappa =\sqrt{-k}=H_0\sqrt{{\Omega }_K}/c$. Integration gives $D_C=\mathrm{arsinh}(\kappa r_E)/\kappa$, so $r_E=\sinh (\kappa D_C)/\kappa$. Two neighboring pulses on the same comoving path obey $\Delta t_0/a_0=\Delta t_E/a_E$, hence $1+z=1/a_E$. For the angular diameter distance in an open dust universe, a transverse source size $\ell =a_E{r}_E\delta \theta$ defines the angular diameter distance $D_A=\ell /\delta \theta$. Thus

Since $dt=da/(aH)$, the dust-plus-curvature expansion givesAt a specified source redshift, the only independent cosmological parameters in this integral are ${\Omega }_M$ and $H_0$. The distance necessarily also has a source-position argument; it cannot be a constant for all sources.

For $0<{\Omega }_M<1$, the integral can be writtenThe apparently singular parameter endpoints have regular limits: ${\Omega }_M=0$ gives $D_C=(c/H_0)\ln (1+z)$, and ${\Omega }_M=1$ gives $D_C=(2c/H_0)[1-(1+z{)}^{-1/2}]$. The flat-curvature distance limit is $D_A=D_C/(1+z)$. For a negative ${\Omega }_K$, the transverse distance uses sine with $\sqrt{-{\Omega }_K}$ instead of hyperbolic sine; that closed case is outside the almost-empty open model, but explains the convention behind the square root.