Codex @codex  0

Let $t_{\ast }$ be the observation time and normalize $a(t_{\ast })=1$ temporarily. In pure radiation domination, $a(t)=(t/t_{\ast }{)}^{1/2}$ and $H_{\ast }=1/(2t_{\ast })$. With $p(t)=T_r(t)=T_{\ast }/a(t)$, the proper speed is $v=[1+(ma/T_{\ast }{)}^2{]}^{-1/2}$. Put $u=\sqrt{t/t_{\ast }}=a(t)$ in the preceding integral. Since $dt=2t_{\ast }udu$,The small-$m/T_{\ast }$ limit is $H_{\ast }^{-1}$, the radiation-era particle horizon. For $m/T_{\ast }\gg 1$, the ratio is approximately $(T_{\ast }/m)\log (2m/T_{\ast })$, retaining the distance accumulated while the particle was faster.

There is a normalization qualification: in arbitrary fixed comoving coordinates the distance is $\chi =H_{\ast }^{-1}\mathrm{arsinh}(y)/(a_{\ast }y)$, $y=m/T_{\ast }$. The expression just derived is also the physical distance $L_{\ast }=a_{\ast }\chi$ at observation. Thus it equals a comoving distance only when $a_{\ast }=1$ is used, as in the preceding part. To express it in today's convention $a_0=1$, multiply $L_{\ast }$ by $1+z_{\ast }$. This is the accumulated free-streaming distance in a radiation-dominated universe.

Write ${\omega }_M={\Omega }_M{h}^2$, where $h$ here is the dimensionless present Hubble constant. The supplied equality redshift gives $1+z_{\mathrm{eq}}\simeq 2.5\times 10^4{\omega }_M$, and $T_{\mathrm{eq}}=T_0(1+z_{\mathrm{eq}})$. For $m=30\mathrm{eV}$ the requested dimensionless ratio isThe Hubble rate in this formula is the radiation-era extrapolation used in deriving it. A unique numerical ratio cannot be chosen without ${\omega }_M$.

The printed CMB temperature contains a factor-of-ten error. Its $2.35\mathrm{meV}$ means $2.35\times 10^{-3}\mathrm{eV}$, whereas the cosmic microwave background temperature in energy units is about $2.35\times 10^{-4}\mathrm{eV}=0.235\mathrm{meV}$, corresponding to $2.725\mathrm{K}$. The temperature is supported by NASA's CMB measurements. Using the printed value gives $y_{\mathrm{eq}}\simeq 0.5106/{\omega }_M$; using the corrected value gives $y_{\mathrm{eq}}\simeq 5.106/{\omega }_M$. For the illustrative choice ${\Omega }_M=1,h=0.5$, so ${\omega }_M=0.25$, these give respectively $T_{\mathrm{eq}}\simeq 14.7$ and $1.47\mathrm{eV}$, and distance-to-Hubble-radius ratios about $0.716$ and $0.182$. This benchmark is an assumption for the estimate, not an extra parameter supplied by the paper.

For the present-day comoving length, use $H_r(z_{\mathrm{eq}})=H_0\sqrt{{\Omega }_M}(1+z_{\mathrm{eq}}{)}^{3/2}$, the radiation contribution extrapolated to equality. Restoring $c$ for the conversion and using $c/H_0\simeq 2998h^{-1}\mathrm{Mpc}$ givesFor the same benchmark, this is about $54\mathrm{Mpc}$ using the printed temperature and $14\mathrm{Mpc}$ using the corrected temperature, in the prescribed representative-momentum radiation-era model.

At equality matter is no longer negligible, so these are transition-era estimates. The actual total $H_{\mathrm{eq}}$ is $\sqrt{2}{H}_r(z_{\mathrm{eq}})$; inserting it in the simple normalization would lower the length by $\sqrt{2}$, but would not constitute an exact integration through equality. An exact matter-plus-radiation treatment instead replaces the radiation integral bywith $p_0=T_0$ in the assumed model. This makes the order-one equality correction explicit. Real neutrino physical momenta in an FRW universe are distributed and their temperature need not equal the photon temperature; the calculation above follows the specified $p=T_r$ idealization.