The Cosmic Recombination Epoch — Epoche C2
A surface that is not a surface The free-electron fraction of the cosmic plasma falls from near unity to near zero over a redshift interval of order a hundred, which gives the so-called surface of last scattering a comoving thickness of roughly fifteen megaparsecs; this essay works out that number, works out the comparable scale set by photon diffusion, and shows how the two together produce the exponential-looking fall-off of the microwave background power spectrum at multipoles above about a thousand. The point is not that the instantaneous-decoupling picture is wrong as a caricature — it gives the acoustic peak positions correctly — but that the small-angle tail is entirely a product of the features the caricature discards, and that the tail is now among the most information-rich parts of the measured spectrum. One correction to the standard compressed telling should be made at the outset, because it recurs and it matters for the physics. The falling ionisation does not reduce the Thomson cross-section. That cross-section, $\sigma_T = 6.652 \times 10^{-29}\,\mathrm{m^2}$, is a constant of nature and is what it is whatever the plasma is doing. What falls is the number density of free electrons $n_e$, and therefore the scattering rate per photon $n_e \sigma_T c$ and the optical depth built from it. Nothing about the electron changes; there are simply fewer of them. Why hydrogen stayed ionised down to a quarter of an electronvolt Recombination is often introduced by saying that photons cooled below the $13.6\,\mathrm{eV}$ ionisation potential of hydrogen. Taken literally that would place the event at $T \approx 1.6 \times 10^{5}\,\mathrm{K}$, redshift of order $6 \times 10^{4}$. It happened instead at $T \approx 3000\,\mathrm{K}$, where the mean photon energy is some fifty times smaller than the binding energy. The reason is the photon-to-baryon ratio, and the estimate is short enough to be worth doing. Write $\eta = n_b/n_\gamma$ for the baryon-to-photon number ratio; the Planck 2018 baryon density $\Omega_b h^2 = 0.02237$ corresponds to $\eta \approx 6.1 \times 10^{-10}$. In a blackbody at temperature $T$ the fraction of photons with energy above $B = 13.6\,\mathrm{eV}$ falls off as $e^{-B/k_BT}$ in the Wien tail. Ionising photons cease to outnumber hydrogen atoms only when $$ \frac{1}{\eta}\, e^{-B/k_BT} \sim 1 \qquad \Longrightarrow \qquad \frac{B}{k_BT} \sim \ln\frac{1}{\eta} = \ln(1.6\times 10^{9}) \approx 21 . $$ So the delay is by a factor of about twenty in temperature purely from the enormous photon excess: a plasma with $10^{9}$ photons per baryon stays ionised long after the typical photon has become harmless, because the rare hard photon is not rare enough. The full Saha calculation, which adds the phase-space factor $(m_e k_B T/2\pi\hbar^2)^{3/2}$ and asks for a specific residual ionisation rather than order unity, pushes the factor further, to about $53$: at the Planck value $z_* = 1089.92 \pm 0.25$ the temperature is $T = 2.7255 \times 1090.92 = 2973\,\mathrm{K}$, or $k_BT = 0.256\,\mathrm{eV}$, and $13.6/0.256 = 53$. The crude estimate and the exact answer differ by a factor of $2.5$, which is the correct level of agreement to expect from an argument that ignored phase space. The kinetic equation, and a term that is routinely misstated Saha equilibrium — chemical equilibrium between protons, electrons and neutral hydrogen with the photon bath — holds only while the reaction rates exceed the expansion rate, and it fails during recombination itself. Peebles showed in 1968, as did Zel'dovich, Kurt and Sunyaev independently in the same year, that the failure is not a small correction: recombination is bottlenecked, and the bottleneck is the disposal of the energy released. The bottleneck arises because direct recombination to the ground state is useless. Every such capture emits a Lyman-continuum photon energetic enough to ionise a neighbouring atom, so the net effect on the ionisation fraction is nil. This is why one uses the case B recombination coefficient $\alpha_B(T)$, which sums captures to all excited levels and excludes the ground state. An atom that has landed in $n = 2$ must then reach $n = 1$, and the two available routes are both slow: a Lyman-$\alpha$ photon from $2p \to 1s$, which is immediately reabsorbed by another atom unless cosmological redshifting has moved it out of the line before it finds one; or the forbidden two-photon decay $2s \to 1s$, with rate $\Lambda_{2s,1s} = 8.22\,\mathrm{s^{-1}}$ — slow by atomic standards but competitive here, because the alternative is almost completely blocked. The resulting evolution of the fractional ionisation $X_e = n_e/n_H$, with $n_e$ the free-electron and $n_H$ the total hydrogen number density, is $$ \frac{dX_e}{dt} = -C \left[\alpha_B(T)\, n_H X_e^2 - \beta_B(T)\,(1-X_e)\, e^{-B_2/k_BT}\right] , $$ and here a correction to the compressed version of this equation is needed. The exponent is not the ground-state binding energy. Because case-B recombination and the corresponding photoionisation both refer to the excited states, the Boltzmann factor is set by the binding energy of the $n = 2$ level, $B_2 = 13.6/2^2 = 3.4\,\mathrm{eV}$. Using $13.6\,\mathrm{eV}$ there would suppress the reverse term by $e^{-10.2/k_BT}$ too much — at $k_BT = 0.26\,\mathrm{eV}$ that is a factor $e^{-39} \approx 10^{-17}$ — and would abolish the equilibrium the equation is meant to describe. The prefactor $C$, the Peebles factor, is the fraction of excited atoms that succeed in reaching the ground state rather than being re-excited: $$ C = \frac{1 + K \Lambda_{2s,1s}\, n_H (1 - X_e)}{1 + K (\Lambda_{2s,1s} + \beta_B)\, n_H (1 - X_e)} , \qquad K = \frac{\lambda_\alpha^3}{8\pi H(z)} , $$ with $\lambda_\alpha = 121.57\,\mathrm{nm}$ the Lyman-$\alpha$ wavelength. The structure of $K$ carries the physics: it is the effective time a Lyman-$\alpha$ photon spends in the line before the expansion redshifts it out, so the escape probability, and he