Recombination and Photon Decoupling

Derivation

Recombination and Photon Decoupling

D-345 Home PU-308 Threads light · matter · chance Depends on The Saha Ionization Equation, CMB Blackbody Spectrum and T ∝ 1/a

Statement

Applying the Saha ionization equation to a primordial hydrogen plasma in thermal equilibrium, and using the blackbody scaling \(T = T_0(1+z)\), fixes the free-electron fraction \(x_e(z)\) as a function of redshift. Setting \(x_e = \tfrac12\) locates the epoch of recombination at \(z_{\rm rec}\approx 1380\) (\(k_BT\approx 0.32\ \text{eV}\), \(T\approx 3760\ \text{K}\)); imposing that the Thomson scattering rate fall below the Hubble rate then locates the last-scattering surface, where photons decouple, at \(z_{\rm dec}\approx 1090\).

Why it matters

The cosmic microwave background is the oldest electromagnetic image of the Universe, and it exists only because free electrons disappeared: while hydrogen is ionized, photons Thomson-scatter off electrons on scales far shorter than the horizon, and the plasma is opaque. Recombination removes the electrons, the mean free path diverges, and the surviving photons free-stream to us. The Saha equation is the minimal quantitative tool that predicts when this happens.

The two redshifts differ, and the reason is itself instructive: \(z_{\rm rec}\) marks where the gas is half-neutral, but photons keep scattering off the residual ionized tail until the electron density is a hundred-fold lower still. That offset — recombination at \(z\approx 1380\), decoupling at \(z\approx 1090\) — is why last scattering happens in the tail of recombination and why an equilibrium calculation must be corrected by a rate calculation.

Assumptions

Local thermodynamic equilibrium at a single temperature \(T\).If matter and radiation temperatures decouple, or if the reaction \(p+e^-\leftrightarrow \mathrm{H}+\gamma\) is not fast compared with expansion, the Saha equation does not apply and one must integrate a rate (Boltzmann) equation instead.

Pure hydrogen; helium and metals neglected.Helium (\(\sim 24\%\) by mass) recombines earlier at \(z\sim 1800\)–\(6000\); ignoring it shifts the free-electron budget by up to \(\sim 8\%\) and mis-times the onset of hydrogen recombination slightly.

Ground-state recombination and a fixed ionization energy \(B=13.6\ \text{eV}\).Treating only the \(n=1\) level and its binding energy ignores the excited-state population and the partition function's temperature dependence; corrections are small near recombination because \(k_BT\ll B\).

Chemical equilibrium is maintained throughout recombination (Saha limit).This is the assumption that ultimately fails: as recombination proceeds the reaction rate cannot keep pace with the falling temperature, the ionization fraction "freezes out", and Saha over-predicts how fast \(x_e\) drops. A full treatment needs the Peebles effective three-level atom.

Non-relativistic, non-degenerate electrons and protons.At \(k_BT\sim 0.3\ \text{eV}\ll m_ec^2\) this is excellent; dropping it would require Fermi–Dirac statistics and a relativistic phase-space factor.

Derivation

\[ \frac{n_p\,n_e}{n_{\rm H}} = \left(\frac{m_e k_B T}{2\pi\hbar^2}\right)^{3/2} e^{-B/k_BT} \]

Saha equation for \(p+e^-\leftrightarrow \mathrm{H}+\gamma\), taken as a prior result. The right side is the electron quantum concentration times the Boltzmann factor for the binding energy \(B=13.6\ \text{eV}\); statistical weights of proton, electron and the hydrogen ground state combine to unity here. A

\[ n_e = n_p,\qquad n_B \equiv n_p + n_{\rm H},\qquad x_e \equiv \frac{n_e}{n_B} \]

Impose charge neutrality (\(n_e=n_p\)) and baryon conservation: every baryon is either a free proton or a bound hydrogen atom. This defines the free-electron fraction \(x_e\). A

\[ n_p = n_e = x_e\,n_B,\qquad n_{\rm H} = (1-x_e)\,n_B \]

Rewrite each number density in terms of \(x_e\) and the total baryon density \(n_B\) — pure algebra from the definitions in step 2. A

\[ \frac{x_e^2}{1-x_e}\,n_B = \left(\frac{m_e k_B T}{2\pi\hbar^2}\right)^{3/2} e^{-B/k_BT} \]

Substitute step 3 into step 1: \(n_p n_e/n_{\rm H} = (x_e n_B)^2/[(1-x_e)n_B]\). Symbols only, no numbers yet. A

\[ \frac{x_e^2}{1-x_e} = \frac{1}{n_B}\left(\frac{m_e k_B T}{2\pi\hbar^2}\right)^{3/2} e^{-B/k_BT} \]

Divide by \(n_B\). The large factor \(1/n_B\) — baryons are rare — is what keeps hydrogen ionized down to temperatures far below \(B\). B

\[ n_B = \eta\,n_\gamma = \eta\,\frac{2\zeta(3)}{\pi^2}\left(\frac{k_BT}{\hbar c}\right)^3 \]

Express the baryon density through the baryon-to-photon ratio \(\eta\) and the blackbody photon density \(n_\gamma=\frac{2\zeta(3)}{\pi^2}(k_BT/\hbar c)^3\). Since \(\eta\) is redshift-independent, this ties \(n_B\) to \(T\) alone. B

\[ \frac{x_e^2}{1-x_e} = \frac{\sqrt{\pi}}{4\sqrt{2}\,\zeta(3)\,\eta}\left(\frac{m_e c^2}{k_BT}\right)^{3/2} e^{-B/k_BT} \]

Insert step 6 into step 5 and simplify. The powers of \(\hbar\) cancel; \(m_e^{3/2}c^3=(m_ec^2)^{3/2}\), and the numerical prefactor collects as \(\pi^2/[2\zeta(3)(2\pi)^{3/2}]=\sqrt{\pi}/(4\sqrt2)\). Everything is now dimensionless and depends only on \(k_BT/m_ec^2\), \(B/k_BT\), and \(\eta\). C

\[ T = T_0(1+z),\qquad T_0 = 2.725\ \text{K} \]

Use the CMB blackbody temperature scaling (prior result): adiabatic expansion preserves the Planck spectrum with \(T\propto (1+z)\). This converts the temperature condition into a redshift. A

\[ x_e=\tfrac12 \;\Rightarrow\; \frac{x_e^2}{1-x_e}=\tfrac12 \;\Rightarrow\; \frac{B}{k_BT_{\rm rec}}\approx 42 \]

Define recombination by half-ionization. Because the right side of step 7 varies through the exponential \(e^{-B/k_BT}\), the condition is essentially a transcendental equation \(u^{3/2}e^{-u}=\text{const}\) with \(u=B/k_BT\); its root is \(u\approx 42\), giving \(k_BT_{\rm rec}\approx 0.32\ \text{eV}\), \(T_{\rm rec}\approx 3760\ \text{K}\), \(z_{\rm rec}\approx 1380\). B

\[ \Gamma_{\rm T}=n_e\sigma_{\rm T}c = x_e\,n_B\,\sigma_{\rm T}c \;\overset{!}{=}\; H(z) \]

Photon decoupling is a kinetic, not chemical, condition: it occurs when the Thomson scattering rate per photon drops below the Hubble expansion rate. With \(H\propto(1+z)^{3/2}\) (matter domination) and \(n_B\propto(1+z)^3\), and \(x_e\) from Saha, this crossing lies at \(z_{\rm dec}\approx 1090\), later than \(z_{\rm rec}\) because only \(x_e\sim 10^{-2}\) is needed to make the plasma transparent. C

Result

\[ \frac{x_e^2}{1-x_e} = \frac{\sqrt{\pi}}{4\sqrt{2}\,\zeta(3)\,\eta}\left(\frac{m_e c^2}{k_BT}\right)^{3/2} e^{-B/k_BT},\qquad x_e(z_{\rm rec})=\tfrac12 \;\Rightarrow\; z_{\rm rec}\approx 1380 \]

Reading. The free-electron fraction is set by a competition between the huge phase-space/dilution factor \((m_ec^2/k_BT)^{3/2}/\eta\sim 10^{15}\), which favours ionization, and the Boltzmann suppression \(e^{-B/k_BT}\), which favours neutrality. Neutrality wins only once \(k_BT\) has fallen to about \(B/42\approx 0.32\ \text{eV}\) — a temperature roughly forty times smaller than the binding energy, entirely because photons outnumber baryons by \(\sim 1.6\times10^{9}\). Half-ionization defines recombination at \(z\approx 1380\); the plasma becomes transparent slightly later, at \(z_{\rm dec}\approx 1090\), once \(x_e\) has fallen to \(\sim 10^{-2}\).

Units check. The left side \(x_e^2/(1-x_e)\) is dimensionless. On the right, \(\eta\), \(\zeta(3)\), and the numeric prefactor are pure numbers; \(m_ec^2/k_BT\) is (energy)/(energy), dimensionless, so its \(3/2\) power is too; and \(B/k_BT\) in the exponent is dimensionless. Both sides are pure numbers. In step 5, \([n_B^{-1}]=\text{m}^3\) times the quantum concentration \([(m_ek_BT/\hbar^2)^{3/2}]=\text{m}^{-3}\) cancels correctly.

Limiting cases

High temperature, \(k_BT\gg B\): the exponential \(\to 1\) and the prefactor is enormous, so \(x_e^2/(1-x_e)\to\infty\), forcing \(x_e\to 1\) — fully ionized plasma, as expected at early times.
Low temperature, \(k_BT\ll B\): \(e^{-B/k_BT}\to 0\), the right side vanishes, and \(x_e\to 0\) — complete recombination (in the equilibrium approximation).
Small-\(x_e\) branch: once \(x_e\ll 1\), \(1-x_e\to 1\) and \(x_e\simeq[\,\text{RHS}\,]^{1/2}\), so \(x_e\) falls as \(e^{-B/2k_BT}\) — half the exponential rate, the origin of the sharp but finite recombination width.
Large \(\eta\): more baryons dilute the ionizing radiation less effectively; \(1/\eta\) shrinks the right side, so recombination completes at higher \(T\) (earlier \(z\)). \(z_{\rm rec}\) rises only logarithmically with \(\eta\).

Breaks when

Chemical equilibrium fails (freeze-out). Direct recombination to the ground state emits an ionizing photon that re-ionizes a neighbour, so net recombination proceeds via the slow two-photon \(2s\to1s\) decay and Lyman-\(\alpha\) escape. Below \(z\sim 1100\) these rates fall behind the expansion, \(x_e\) freezes at a floor \(\sim 2\times10^{-4}\), and Saha (which assumes instantaneous equilibrium) badly under-predicts the residual ionization.
Helium is not negligible. Around \(z\sim 1800\)–\(6000\) helium recombines (He\(^{++}\to\)He\(^{+}\to\)He\(^0\)), changing the electron budget before hydrogen recombines; a pure-hydrogen Saha treatment mis-times the electron history at the \(\sim 5\)–\(8\%\) level.
Reionization at low redshift. After the first stars and quasars form (\(z\lesssim 10\)) the intergalactic medium is re-ionized; the Saha equilibrium curve, which assumes a cooling neutral gas, says nothing about this later ionized epoch.
Non-thermal or spectral-distortion conditions. If the radiation departs from a Planck spectrum (energy injection, decaying particles), the single-temperature Boltzmann factor is invalid and the ionization balance must be computed from the actual photon occupation.

Failure modes

Equating recombination with decoupling. Setting \(x_e=\tfrac12\) gives \(z\approx 1380\), not the last-scattering \(z\approx 1090\). Students who quote "\(z\approx 1100\) is where hydrogen becomes neutral" conflate the chemical and kinetic conditions.
Expecting recombination at \(k_BT\approx B\). Naively one guesses \(T\sim B/k_B\approx 1.6\times10^5\ \text{K}\). The factor-of-40 offset comes from the \(1/\eta\) dilution; forgetting it overestimates \(z_{\rm rec}\) by two orders of magnitude.
Dropping the \((m_ec^2/k_BT)^{3/2}\) prefactor. Treating recombination as purely \(e^{-B/k_BT}=\tfrac12\) omits the quantum-concentration factor and gives the wrong root of the transcendental equation.
Using \(n_e\propto(1+z)^3\) with fixed \(x_e\) in the decoupling condition. \(x_e\) itself plunges near recombination; ignoring its \(T\)-dependence makes \(\Gamma_{\rm T}\) fall far too slowly and pushes \(z_{\rm dec}\) to absurdly low values.
Sign/inversion errors in \(x_e^2/(1-x_e)\). Writing \((1-x_e)/x_e^2\) or forgetting that \(n_e=n_p\) (double-counting electrons) yields a quadratic with the wrong root.
Trusting Saha into the tail. Using the Saha \(x_e\) below \(z\sim 1000\) gives a residual ionization many orders of magnitude too small, because freeze-out is not an equilibrium effect.

Discussion

The central physical surprise is the factor of forty between \(B/k_B\) and the actual recombination temperature. Its origin is entropy: there are roughly \(1.6\times10^{9}\) photons per baryon, so even when the mean photon energy is far below \(13.6\ \text{eV}\), the exponentially rare high-energy tail of the Planck distribution still contains enough ionizing photons to keep hydrogen ionized. Recombination waits until \(k_BT\) drops to \(\sim B/\ln(1/\eta)\), a logarithm of the baryon-to-photon ratio. The same enormous photon number is why the Saha condition depends so weakly (logarithmically) on the precise value of \(\eta\) or \(B\): the exponent must simply run out to \(u\approx 42\).

The separation of recombination from decoupling is the second key idea. Thomson opacity depends on the absolute electron density \(n_e=x_e n_B\), not on the fraction \(x_e\). Because \(n_B\) is tiny, a fractional ionization of only \(\sim 1\%\) already makes the mean free path comparable to the Hubble length. So photons keep scattering well after half-recombination, decoupling only in the steep tail where \(x_e\sim 10^{-2}\). The last-scattering surface is therefore not a sharp shell but a layer of finite thickness \(\Delta z\approx 80\)–\(100\); the visibility function \(g(z)=e^{-\tau}\,d\tau/dz\), which weights where the CMB photons we see were last scattered, peaks at \(z\approx 1090\) with roughly this width. That thickness is what damps small-scale CMB anisotropies (Silk damping) and blurs the surface.

The result connects three threads. On the light side it sets the origin of the CMB and its blackbody temperature today. On the matter side it marks the moment baryons stop being dragged by radiation and can begin gravitational collapse, seeding the baryon acoustic oscillation scale imprinted at the sound horizon at \(z_{\rm dec}\). On the chance side it is a statistical-mechanics calculation par excellence: the entire timing is dictated by an equilibrium constant and a Boltzmann tail.

The deepest caveat is that the observed universe recombines out of equilibrium. Direct capture to the ground state is self-defeating — each such recombination emits a photon that ionizes another atom — so the net removal of electrons is throttled by two bottlenecks: the redshifting of Lyman-\(\alpha\) photons out of resonance and the forbidden two-photon \(2s\to1s\) transition (rate \(\sim 8.2\ \text{s}^{-1}\)). Peebles' effective three-level atom, and the modern multilevel codes RECFAST/CosmoRec/HyRec that refine it, track this competition and yield the frozen-out residual \(x_e\approx 2\times10^{-4}\). Saha is the correct equilibrium skeleton onto which this non-equilibrium kinetics is grafted; it gets \(z_{\rm rec}\) right to \(\sim 10\%\) but cannot predict the residual electrons, which matter for the reionization optical depth and for 21-cm cosmology.

Common misconceptions. Recombination is not the first time protons and electrons combine (the "re-" is historical); the CMB temperature today, \(2.725\ \text{K}\), is not the temperature at last scattering (that was \(\sim 2970\ \text{K}\), redshifted down by \(1+z_{\rm dec}\)); and "the universe became transparent" is a statement about the electron abundance falling, not about the neutral atoms — neutral hydrogen is nearly transparent to CMB-wavelength photons.

Worked examples

Recombination redshift from Saha with \(x_e=\tfrac12\). Symbols first, then numbers. B

\[ \frac{x_e^2}{1-x_e} = A\left(\frac{m_ec^2}{k_BT}\right)^{3/2}e^{-B/k_BT},\qquad A\equiv\frac{\sqrt\pi}{4\sqrt2\,\zeta(3)\,\eta} \]

\[ A = \frac{1.7725}{4\sqrt2\,(1.202)(6.1\times10^{-10})} = 4.27\times10^{8} \]

Set \(x_e=\tfrac12\Rightarrow x_e^2/(1-x_e)=\tfrac12\). Write \(u\equiv B/k_BT\) and \(m_ec^2/k_BT = (m_ec^2/B)\,u = 3.76\times10^{4}\,u\):

\[ \tfrac12 = A\,(3.76\times10^{4})^{3/2}\,u^{3/2}e^{-u} \;\Rightarrow\; u^{3/2}e^{-u} = 1.6\times10^{-16} \]

Take logarithms, \(\tfrac32\ln u - u = -36.4\), and iterate: \(u=42\) gives \(\tfrac32\ln 42 - 42 = 5.61-42=-36.4\). ✓

\[ k_BT_{\rm rec}=\frac{B}{u}=\frac{13.6\ \text{eV}}{42}=0.324\ \text{eV},\quad T_{\rm rec}=\frac{0.324}{8.617\times10^{-5}}=3.76\times10^{3}\ \text{K} \]

\[ z_{\rm rec}=\frac{T_{\rm rec}}{T_0}-1=\frac{3760}{2.725}-1 \]

\[ \boxed{\,z_{\rm rec}\approx 1.38\times10^{3},\quad T_{\rm rec}\approx 3760\ \text{K},\quad k_BT_{\rm rec}\approx 0.32\ \text{eV}\,} \]

Reading. Half-ionization occurs when \(k_BT\) is about \(B/42\), forty times below the binding energy — the entropy (photon-to-baryon) factor at work. Units: \(u\), \(A\), and \((m_ec^2/k_BT)^{3/2}\) are all dimensionless; \(k_BT\) comes out in eV, converted to kelvin via \(k_B=8.617\times10^{-5}\ \text{eV/K}\).

Ionization needed for decoupling: Thomson rate versus Hubble rate at \(z\approx 1100\). C

\[ \Gamma_{\rm T}=x_e\,n_{B,0}(1+z)^3\,\sigma_{\rm T}\,c,\qquad H(z)=H_0\sqrt{\Omega_m}\,(1+z)^{3/2} \]

Evaluate the baryon density today from \(n_{B,0}=\eta\,n_{\gamma,0}\) with \(n_{\gamma,0}=411\ \text{cm}^{-3}=4.11\times10^{8}\ \text{m}^{-3}\):

\[ n_{B,0}=(6.1\times10^{-10})(4.11\times10^{8}\ \text{m}^{-3})=0.251\ \text{m}^{-3} \]

At \(z=1100\), take \(x_e=1\) provisionally to compute the scattering rate of a fully ionized plasma; \(\sigma_{\rm T}=6.65\times10^{-29}\ \text{m}^2\):

\[ \Gamma_{\rm T}=(1)(0.251)(1101)^3(6.65\times10^{-29})(3\times10^{8})=6.7\times10^{-12}\ \text{s}^{-1} \]

Hubble rate with \(H_0=2.19\times10^{-18}\ \text{s}^{-1}\) (i.e. \(67.7\ \text{km s}^{-1}\text{Mpc}^{-1}\)) and \(\Omega_m=0.31\):

\[ H=(2.19\times10^{-18})\sqrt{0.31}\,(1101)^{3/2}=4.5\times10^{-14}\ \text{s}^{-1} \]

\[ \frac{\Gamma_{\rm T}}{H}\Big|_{x_e=1}=\frac{6.7\times10^{-12}}{4.5\times10^{-14}}\approx 1.5\times10^{2} \;\Rightarrow\; x_{e,\rm dec}\approx \frac{1}{150} \]

\[ \boxed{\,x_{e,\rm dec}\approx 7\times10^{-3}\ \text{ at } z_{\rm dec}\approx 1090\,} \]

Reading. Because a fully ionized plasma scatters \(\sim 150\) times faster than the universe expands at \(z\sim1100\), photons stay coupled until \(x_e\) has dropped to \(\sim 1/150\). Decoupling therefore occurs in the recombination tail, at \(z\approx 1090\) — well below \(z_{\rm rec}\approx1380\). Units: \(\Gamma_{\rm T}\) and \(H\) both carry \(\text{s}^{-1}\); their ratio and \(x_{e,\rm dec}\) are dimensionless.

Problems

(A) Compute the photon temperature and \(k_BT\) (in eV) at the last-scattering redshift \(z=1090\), given \(T_0=2.725\ \text{K}\).
Solution
\(T=T_0(1+z)=2.725\times1091=2.97\times10^{3}\ \text{K}\). Then \(k_BT=(8.617\times10^{-5}\ \text{eV/K})(2970\ \text{K})=0.256\ \text{eV}\). So the CMB was emitted at about \(2970\ \text{K}\) with characteristic thermal energy \(\approx 0.26\ \text{eV}\) — reddened to \(2.725\ \text{K}\) today by the factor \(1+z=1091\).
(B) Explain quantitatively why recombination occurs at \(k_BT\approx 0.3\ \text{eV}\), a factor \(\sim 40\) below \(B=13.6\ \text{eV}\). Estimate the required \(B/k_BT\) directly from the Saha balance.
Solution
Setting \(x_e^2/(1-x_e)=\tfrac12\) in \(A(m_ec^2/k_BT)^{3/2}e^{-B/k_BT}=\tfrac12\), the exponent must satisfy \(e^{-u}\sim \tfrac{1}{2A(m_ec^2/k_BT)^{3/2}}\). The prefactor \(A(m_ec^2/k_BT)^{3/2}\sim 10^{15}\) is dominated by \(1/\eta\sim1.6\times10^{9}\) (there are \(\sim1.6\times10^{9}\) photons per baryon, so the rare ionizing tail stays populated). Hence \(u\approx\ln(2A(\cdots))\approx\ln(10^{15})\approx 35\)–\(42\); the transcendental root is \(u=B/k_BT\approx 42\), i.e. \(k_BT\approx 13.6/42=0.32\ \text{eV}\). The factor of 40 is essentially \(\ln(1/\eta)\) plus the phase-space term.
(A/B) Using the Saha equation with \(\eta=6.1\times10^{-10}\), compute the free-electron fraction \(x_e\) at \(z=1200\).
Solution
\(T=2.725\times1201=3273\ \text{K}\Rightarrow k_BT=0.282\ \text{eV}\); \(u=B/k_BT=13.6/0.282=48.2\); \(m_ec^2/k_BT=5.11\times10^{5}/0.282=1.81\times10^{6}\), so \((\cdots)^{3/2}=2.44\times10^{9}\). Then RHS \(=A\,(2.44\times10^{9})\,e^{-48.2}=(4.27\times10^{8})(2.44\times10^{9})(1.2\times10^{-21})=1.2\times10^{-3}\). Solving \(x_e^2/(1-x_e)=1.2\times10^{-3}\) with \(x_e\ll1\): \(x_e\approx\sqrt{1.2\times10^{-3}}\approx 0.035\). So by \(z=1200\) (below \(z_{\rm rec}\)) the gas is already \(\sim96.5\%\) neutral — recombination is steep.
(B) If the baryon-to-photon ratio were ten times larger, \(\eta=6.1\times10^{-9}\), would recombination occur earlier or later, and by roughly how much in redshift?
Solution
The Saha right side scales as \(1/\eta\), so a \(10\times\) larger \(\eta\) lowers it by \(10\); to restore \(x_e=\tfrac12\) the \(u\)-factor \(u^{3/2}e^{-u}\) must rise by \(10\), i.e. \(u\) must decrease. Linearizing near \(u=42\), \(d[\tfrac32\ln u-u]/du=\tfrac{3}{2u}-1\approx-0.96\), so \(\Delta u=-\ln 10/0.96\approx-2.4\), giving \(u\approx 39.6\). Then \(k_BT=13.6/39.6=0.343\ \text{eV}\), \(T=3985\ \text{K}\), \(z_{\rm rec}=3985/2.725-1\approx 1460\). Recombination happens earlier (higher \(z\)), shifting up by \(\sim 80\) in redshift — a weak, logarithmic dependence on \(\eta\).
(C) The Saha equation assumes chemical equilibrium. Explain physically why it fails during recombination and what sets the frozen-out residual ionization. Estimate the residual \(x_e\).
Solution
As \(T\) falls, the volumetric recombination rate \(\alpha n_e^2\) (with \(\alpha\propto T^{-1/2}\) roughly) competes with the Hubble rate \(H\). Direct capture to \(n=1\) emits a Lyman-continuum photon that promptly re-ionizes another atom, giving no net recombination; the effective channels are the slow two-photon \(2s\to1s\) decay (\(\Lambda\approx 8.2\ \text{s}^{-1}\)) and the redshifting of Ly-\(\alpha\) photons out of resonance. Once \(\alpha n_e \lesssim H\), electrons can no longer find protons before the density dilutes, and \(x_e\) "freezes". Equating a rough recombination timescale to the expansion time near \(z\sim1000\) gives a floor \(x_e\sim H/(\alpha n_B)\). Numerically the Peebles/RECFAST result is \(x_e\to 2\)–\(3\times10^{-4}\), many orders above the (vanishing) equilibrium Saha value — which is exactly why an equilibrium calculation cannot be trusted into the tail, and why the residual electrons (relevant to the CMB optical depth and 21-cm signal) require the full kinetic treatment.