= Solution
Use the flat matter-only Einstein-de Sitter approximation for the distance-redshift relation, neglecting radiation corrections at <cosmological recombination>. Let $z_*$ be the <cosmological recombination> <redshift>. Adiabatic <photon> cooling gives $1+z_*=T_*/T_0$. The Hubble rate is $H(z)=H_0(1+z)^{3/2}$, so the <comoving radial distance> to last scattering is
$$
\chi_* =\int_0^{z_*}\frac{dz}{H(z)}=\frac2{H_0}\left(1-\frac1{\sqrt{1+z_*}}\right).
$$
The <angular diameter distance> is $D_A=\chi_* /(1+z_*)$. A proper <Hubble radius> $H_*^{-1}=H_0^{-1}(1+z_*)^{-3/2}$ therefore subtends the small angle
$$
\boxed{\theta_H=\frac{H_*^{-1}}{D_A}=\frac1{2(\sqrt{1+z_*}-1)}\quad\text{radians}.}
$$
This is the <angular Hubble-radius scale in an Einstein-de Sitter universe>; the present <Hubble constant> cancels.
Taking $T_*=0.3\,\mathrm{eV}$ and the literal printed $T_0=2.35\,\mathrm{meV}$ gives $1+z_*=127.7$ and $\theta_H\simeq0.0486$ radians, or $2.8$ degrees. As noted above, that present <CMB> <energy> is too large by ten. With the corrected $T_0=0.235\,\mathrm{meV}$, $1+z_*=1277$ and
$$
\boxed{\theta_H\simeq0.0144\ \mathrm{rad}\simeq0.83^\circ,}
$$
which gives the familiar order-one-degree scale within the assumed cosmology. Using $0.28\,\mathrm{eV}$ to mark a few-per-cent ionization fraction changes this only modestly. These numbers refer to the Hubble-radius scale, not its diameter. In a matter-only universe the particle-horizon radius is $2H^{-1}$ and would give twice the angle; the acoustic <sound horizon> is a different length again.
Back to article page
