Recombination and the Saha Equation

For its first 380,000 years the universe was an opaque, glowing plasma: every photon travelled a hair's breadth before Thomson-scattering off a free electron, so light and matter moved as one fluid. Recombination ends that. As the expanding universe cools through a few thousand kelvin, free protons and electrons lock into neutral hydrogen, the free-electron density collapses, and the photon mean free path explodes past the horizon. The relic light released at that instant is the cosmic microwave background — the oldest image of the universe we will ever have. The name is a misnomer inherited from the laboratory — the electrons and protons had never been "combined" before — but it has stuck. The quantitative backbone is the Saha ionization equation, written down by Meghnad Saha in 1920 for stellar atmospheres and pressed into cosmological service half a century later. Saha's insight was to treat ionization as a chemical reaction, $p+e^-\leftrightarrow H+\gamma$ , and apply the law of mass action in thermal equilibrium. The result pins the ionization fraction $X_e$ to just three inputs: temperature $T$ , baryon density, and binding energy $B=13.6\,\mathrm{eV}$ . Its most startling prediction is that hydrogen stays ionized until $T$ falls to about $0.3\,\mathrm{eV}$ — forty times below $B$ — because photons outnumber baryons a billion to one and the thin tail of the Planck spectrum keeps tearing atoms apart. Saha's picture is exact in spirit but wrong in the tail: real recombination cannot keep pace with expansion and freezes at a residual $X_e\approx2\times10^{-4}$ , the number that fixes the thickness of the last-scattering surface.

Key takeaways

Recombination is the epoch ( $z\approx1100$ , $T\approx0.3\,\mathrm{eV}\approx3000\,\mathrm{K}$ ) when free protons and electrons combined into neutral hydrogen, collapsing the free-electron density and releasing the CMB.
The Saha equation treats $p+e^-\leftrightarrow H+\gamma$ as a chemical reaction in equilibrium, giving the ionization fraction $X_e$ as a function of temperature alone (given the baryon density).
Recombination is delayed to $T\ll B=13.6\,\mathrm{eV}$ because $\eta\approx6\times10^{-10}$ is tiny: with $\sim1.6\times10^{9}$ photons per baryon, the Wien tail keeps hydrogen ionized until $B/T\approx42$ .
Recombination is not photon decoupling: the falling $X_e$ drives the Thomson rate below expansion, and only then does the CMB free-stream from the last-scattering surface.
Saha is only an equilibrium approximation; it over-predicts how completely hydrogen recombines. The true history needs the non-equilibrium Peebles treatment, which freezes $X_e$ at $\approx2\times10^{-4}$ .
Applied locally the same equation explains stellar spectral classification — its original 1920 purpose — and, with helium's ionization energies, the earlier helium recombination steps at $z\approx2500$ and $z\approx6000$ .
Key idea: the Saha equation applies the law of mass action to $p+e^-\leftrightarrow H+\gamma$ in thermal equilibrium, giving the ionization fraction as a function of temperature; the huge photon-to-baryon ratio delays recombination to $T\approx0.3\,\mathrm{eV}\ll13.6\,\mathrm{eV}$ .

Where this lives today

Cosmological recombination is an active precision frontier: the CMB power spectrum is now measured well enough that percent-level errors in $X_e(z)$ would bias the cosmological parameters. Planck's determinations of the baryon density, the Hubble constant, and the primordial spectrum all rest on a recombination history computed to sub-percent accuracy by modern codes — CosmoRec and HyRec — that track hundreds of hydrogen and helium levels and the radiative transfer of Lyman- $\alpha$ , far beyond Peebles's three-level atom. The thickness of the last-scattering surface, set by the visibility function peaking at $z\approx1090$ with width $\Delta z\approx80$ , controls the Silk-damping tail of the acoustic peaks that CMB-S4 and the Simons Observatory are built to measure. Two open frontiers stand out: the cosmological recombination radiation — faint, redshifted lines from the electron cascade down through the levels, a direct snapshot a future spectral-distortion mission could detect — and the residual $X_e\approx2\times10^{-4}$ , the reservoir that catalyses the first primordial molecules and, later, the 21-cm signal from the dark ages.

The physical picture

Picture an equilibrium tug-of-war over every hydrogen atom. Pulling one way, energetic photons photoionize atoms and strip their electrons; pulling back, free protons capture free electrons and radiate the excess energy away. The balance point is set not by the binding energy alone but by how many photons there are per atom. Because the universe carries about $1.6\times10^{9}$ photons per baryon, the fight is wildly lopsided: even when the mean photon energy has fallen far below $13.6\,\mathrm{eV}$ , the exponentially thin Wien tail of the Planck distribution still holds enough ionizing photons to win. Only when $T$ drops to $\approx B/42$ does that tail thin enough for neutral atoms to survive.

Figure

The equilibrium tug-of-war p + e⁻ ⇌ H + γ. Recombination (green) is opposed by photoionization (red). At recombination the mean photon energy is only ⟨E⟩ ≈ 2.7 T ≈ 0.9 eV, some 15 times below the binding energy B = 13.6 eV, yet with 1/η ≈ 1.6×10⁹ photons per baryon the exponentially rare Wien tail beyond B keeps hydrogen ionized until B/T ≈ 42

Now follow the same gas forward in time. As the universe expands, both $T$ and the density fall, so Saha drives $X_e$ smoothly from near unity toward zero — a rapid but continuous drop, not a phase transition. Reality, though, cannot follow the equilibrium curve all the way down: once the gas is dilute the reactions become too slow to track the falling temperature, and $X_e$ freezes out at a small residual value. The second figure is the punchline — the ionization history $X_e(z)$ , with the Saha prediction and the true frozen-out curve side by side.

Figure

The cosmic ionization history Xe(z). The Saha equilibrium curve (blue) plunges through half-ionization at z ≈ 1380, T ≈ 0.32 eV, and heads toward zero; the real non-equilibrium recombination (green) tracks it down but freezes out at Xe ≈ 2×10⁻⁴. Photon decoupling follows at z ≈ 1090, when the Thomson-scattering rate drops below expansion

One piece of vocabulary prevents the commonest confusion. Recombination is about atoms forming — the drop in $X_e$ . Photon decoupling is about light escaping — the moment the Thomson rate $n_e\sigma_T c$ falls below the Hubble rate. They are causally linked (the free electrons are the coupling agent) but are distinct events at distinct, if nearby, redshifts — and keeping them separate is essential for reading the CMB.

The core equations

The reaction that governs everything is the capture of an electron by a proton, balanced against its reverse, photoionization. Photons carry no conserved charge, so $\mu_\gamma=0$ — the fact that makes the whole equilibrium tractable.

p + e^- \leftrightarrow H + \gamma \qquad (1)

where: $p$ — free proton; $e^-$ — free electron; $H$ — neutral hydrogen (ground state); $\gamma$ — emitted or absorbed photon. Capture releases a photon; photoionization absorbs one. For any non-relativistic species in thermal equilibrium, Maxwell-Boltzmann statistics fix the number density in terms of temperature and chemical potential — the single input the Saha equation is built from.

n_i = g_i\left(\frac{m_i T}{2\pi}\right)^{3/2}\exp\!\left(\frac{\mu_i - m_i}{T}\right) \qquad (2)

where: $n_i$ — number density of species $i$ ; $g_i$ — internal degeneracy; $m_i$ — mass; $T$ — temperature in energy units ( $k_B=1$ ); $\mu_i$ — chemical potential. The rest-mass term is the exponential price of making a massive particle. Imposing chemical equilibrium on (1) and forming the ratio $n_pn_e/n_H$ cancels the chemical potentials, and the rest-mass energies collapse into the binding energy. This is the Saha equation proper.

\frac{n_p n_e}{n_H} = \frac{g_p g_e}{g_H}\left(\frac{m_e T}{2\pi}\right)^{3/2}\exp\!\left(-\frac{B}{T}\right) \qquad (3)

where: $n_p,n_e,n_H$ — densities of protons, electrons, neutral H; $g_p=g_e=2$ , $g_H=4$ so $g_pg_e/g_H=1$ ; $m_e$ — electron mass, $511\,\mathrm{keV}$ ; $B=m_p+m_e-m_H=13.6\,\mathrm{eV}$ . Only $m_e$ survives because the reduced mass $m_pm_e/m_H\approx m_e$ . To turn densities into a single history, define the ionization fraction; charge neutrality forces $n_e=n_p$ in pure hydrogen.

X_e \equiv \frac{n_e}{n_p+n_H}=\frac{n_e}{n_b} \qquad (4)

where: $X_e$ — ionization fraction, $0\le X_e\le1$ ; $n_b=n_p+n_H$ — total baryon density; $n_e=n_p$ — charge neutrality. $X_e=1$ is fully ionized, $X_e=0$ fully neutral. Substituting $n_e=n_p=X_en_b$ and $n_H=(1-X_e)n_b$ into (3) gives the working form — a quadratic in $X_e$ whose only external input is $n_b$ .

\frac{1-X_e}{X_e^2}=n_b\,\frac{g_H}{g_p g_e}\left(\frac{2\pi}{m_e T}\right)^{3/2}e^{B/T} \qquad (5)

where: the left side is the reaction quotient; $n_b$ carries the density; $e^{B/T}$ is the Boltzmann factor that switches recombination on sharply as $T$ falls. Solve for $X_e$ at each $T$ to draw the blue curve above. The baryon density is tied to the photon density by the baryon-to-photon ratio $\eta$ — the small number responsible for the whole delay below $B$ .

n_b=\eta\,n_\gamma=\eta\,\frac{2\zeta(3)}{\pi^2}T^3,\qquad \eta\approx6.1\times10^{-10} \qquad (6)

where: $\eta$ — baryon-to-photon ratio, fixed by BBN and the CMB; $n_\gamma$ — photon density; $\zeta(3)\approx1.202$ . Because $\eta$ is minuscule, there are $1/\eta\approx1.6\times10^{9}$ photons per baryon. Convert temperature to cosmic time via redshift; the CMB temperature scales as $1+z$ .

1+z=\frac{T}{T_0},\qquad T_0=2.725\,\mathrm{K}=2.35\times10^{-4}\,\mathrm{eV} \qquad (7)

where: $z$ — redshift; $T_0$ — present CMB temperature; $T$ — temperature at $z$ . Half-ionization at $T\approx0.32\,\mathrm{eV}$ maps to $z\approx1380$ . Saha breaks down in the dilute tail; the actual history obeys a rate equation that tracks how far $X_e$ lags equilibrium.

\dot X_e = -\left[\alpha^{(2)}\,X_e^2\,n_b - \beta\,(1-X_e)\,e^{-B_2/T}\right]C \qquad (8)

where: $\alpha^{(2)}$ — case-B recombination coefficient; $\beta$ — photoionization rate; $B_2=B/4=3.4\,\mathrm{eV}$ ; $C$ — Peebles factor for $2s\to1s$ two-photon decay and Lyman- $\alpha$ escape. When the bracket cannot keep up with expansion, $X_e$ freezes. Solving the full history yields the numbers that anchor observational cosmology.

z_{\mathrm{rec}}\approx1100,\qquad z_{\mathrm{dec}}\approx1090,\qquad X_e^{\mathrm{freeze}}\approx2\times10^{-4} \qquad (9)

where: $z_{\mathrm{rec}}$ — recombination ( $X_e\sim0.1$ ); $z_{\mathrm{dec}}$ — photon decoupling / last scattering; $X_e^{\mathrm{freeze}}$ — residual frozen ionization. These fix the geometry and thickness of the CMB surface.

Derivation ADeriving the Saha equation from chemical equilibrium

The goal is to build the Saha equation from a single physical input — the equilibrium number density of a non-relativistic gas — plus the law of mass action for $p+e^-\leftrightarrow H+\gamma$ . We will see how the chemical potentials cancel when we form the ratio $n_pn_e/n_H$ , how the three rest masses combine into the binding energy $B$ , and why the degeneracy factors give exactly unity — leaving an equation whose ionization balance depends only on $T$ , $n_b$ , and $B$ .

Step 1. Start from the non-relativistic equilibrium density, $n_i=g_i(m_iT/2\pi)^{3/2}e^{(\mu_i-m_i)/T}$ , valid here since $T\lesssim\mathrm{eV}\ll m_e=511\,\mathrm{keV}$ . Why: this Maxwell-Boltzmann form is the only statistical input needed — the reaction merely rearranges these densities.

Step 2. Impose chemical equilibrium: $\mu_p+\mu_e=\mu_H+\mu_\gamma$ with $\mu_\gamma=0$ , so $\mu_p+\mu_e=\mu_H$ . Why: this is the law of mass action, the hinge that lets the unknown chemical potentials drop out.

Step 3. Form the ratio using Step 1. The chemical-potential exponent becomes $e^0=1$ by Step 2, leaving $n_pn_e/n_H=(g_pg_e/g_H)(m_pm_e/m_H)^{3/2}(T/2\pi)^{3/2}e^{-(m_p+m_e-m_H)/T}$ . Why: this cancellation is the whole trick — it converts an intractable statement about $\mu$ 's into a pure function of $T$ .

Step 4. Simplify. Since $m_p\approx m_H\gg m_e$ the reduced mass $m_pm_e/m_H\approx m_e$ , and $g_pg_e/g_H=(2\cdot2)/4=1$ because hydrogen's four ground-state spin states match the free-particle count. Why: these give the clean prefactor $(m_eT/2\pi)^{3/2}$ with no leftover mass ratios or spin factors.

Step 5. Identify $B\equiv m_p+m_e-m_H=13.6\,\mathrm{eV}$ in the exponent, giving the boxed $n_pn_e/n_H=(m_eT/2\pi)^{3/2}e^{-B/T}$ . Why: the mass defect of the bound atom is exactly its binding energy — the rest-mass bookkeeping has delivered the physical energy scale.

Go deeper · Derivation B · 1 minWhy recombination is delayed to T about forty times below the binding energy

Derivation BWhy recombination is delayed to T about forty times below the binding energy

The goal is to explain the most surprising fact about recombination: hydrogen does not recombine at $T\sim B=13.6\,\mathrm{eV}$ but waits until $T\approx0.32\,\mathrm{eV}$ , forty times cooler. Writing Saha in its $X_e$ form and imposing half-ionization, we show the delay is set by the logarithm of the tiny baryon-to-photon ratio $\eta$ — a purely entropic effect, nothing to do with the atomic physics of hydrogen.

Step 1. Insert $n_b=\eta(2\zeta(3)/\pi^2)T^3$ into the $X_e$ -form equation: $(1-X_e)/X_e^2=\eta(2\zeta(3)/\pi^2)(2\pi T/m_e)^{3/2}e^{B/T}$ . Why: making $\eta$ explicit exposes the one small number that decides everything.

Step 2. Set $X_e=\tfrac12$ , giving $(1-X_e)/X_e^2=2$ . Why: half-ionization is a clean, convention-free marker of "recombining now"; the exact threshold barely matters because the exponential makes the transition sharp.

Step 3. Take logs. The exponential $e^{B/T}$ must balance the tiny prefactor $\eta(T/m_e)^{3/2}$ , so $B/T\approx\ln(1/\eta)+\tfrac32\ln(m_e/T)\approx42$ (with $\ln(1/\eta)\approx21$ ). Why: this is the delay laid bare — with $\sim10^{9}$ photons per baryon, $T$ must fall until $e^{-B/T}$ beats that number, and beating $10^{9}$ costs $\sim40$ in $B/T$ .

Step 4. Solve: $T_{\rm rec}\approx B/42\approx0.32\,\mathrm{eV}\approx3760\,\mathrm{K}$ , i.e. $z_{\rm rec}\approx1380$ (pure Saha). Why: this is the observable output; the full Peebles treatment shifts decoupling to $z\approx1090$ , but Saha already lands within tens of percent. *Direct numerical solution of the $X_e$ -form equation with $\eta=6.1\times10^{-10}$ gives $T_{\rm rec}=0.324\,\mathrm{eV}$ , $B/T_{\rm rec}=42.0$ , and $z=1379$ .*

Go deeper · Derivation C · 1 minWhy Saha fails in the tail and the ionization fraction freezes out

Derivation CWhy Saha fails in the tail and the ionization fraction freezes out

The goal is to understand why the equilibrium Saha curve, which drives $X_e$ toward zero, is wrong at late times, and why the true ionization fraction freezes at $X_e\approx2\times10^{-4}$ . Saha assumes chemical equilibrium — recombination and photoionization both fast compared to expansion. We will see that direct capture to the ground state is self-defeating, that the atom must recombine through slow bottleneck channels, and that once those fall behind the Hubble rate $X_e$ can no longer track equilibrium.

Step 1. The ground-state trap: a direct capture $p+e^-\to H(1s)+\gamma$ emits a photon of energy $\ge B=13.6\,\mathrm{eV}$ , a Lyman-continuum photon that promptly ionizes a neighbour, netting zero. Why: with $10^{9}$ photons per baryon, an emitted ionizing photon is reabsorbed almost immediately, so the obvious channel does nothing.

Step 2. The real channels dump the binding energy into photons that cannot re-ionize: the two-photon $2s\to1s$ decay (rate only $\sim8.2\,\mathrm{s^{-1}}$ ) or a redshifted, escaped Lyman- $\alpha$ photon. Why: these slow routes are the bottleneck; the Peebles factor $C$ is the fraction of excited atoms reaching the ground state through them.

Step 3. Compare rates to expansion: the recombination rate $\alpha^{(2)}X_e^2n_b$ falls with the density while $H$ falls more slowly, so when the reaction time exceeds the expansion time, $X_e$ leaves the equilibrium track. Why: this is a generic freeze-out — the same logic as neutrino or WIMP decoupling.

Step 4. Consequence: $X_e$ stalls at $\approx2\times10^{-4}$ , far above the near-zero Saha value at the same temperature. Why: this leftover ionization sets the post-recombination optical depth, the thickness of the last-scattering surface, and the electrons that seed primordial molecular chemistry.

Limiting cases

The Saha ionization fraction is most illuminating in its extremes, where the competition between the Boltzmann factor and the photon reservoir simplifies.

Case	Result	Meaning
High temperature $T\gg B$	$X_e\to1$	Full ionization: thermal energy dwarfs the binding energy and no neutral hydrogen survives.
Low temperature $T\ll B$	$X_e\to0$ (Saha)	Saha predicts near-complete recombination via $e^{-B/T}$ ; reality departs and freezes at $X_e\approx2\times10^{-4}$ .
Half-ionization $X_e=\tfrac12$	$T_{\rm rec}\approx0.32\,\mathrm{eV}\approx3760\,\mathrm{K}$	Defines recombination; occurs at $z\approx1380$ (Saha), far below $B/k_B\approx1.6\times10^{5}\,\mathrm{K}$ .
Larger $\eta$ (more baryons)	$T_{\rm rec}\uparrow$	Fewer ionizing photons per baryon, so recombination happens slightly hotter and earlier.
Helium (ionization $\chi>B$ )	$T\uparrow$ , earlier $z$	Higher ionization energy recombines first: He III→He II at $z\approx6000$ , He II→He I at $z\approx2500$ .

Common trapRecombination and photon decoupling are the same event

They are not, and conflating them is the commonest error here. Recombination is the formation of neutral atoms — the drop in $X_e$ governed by Saha and Peebles. Photon decoupling is a scattering statement — the moment the Thomson rate $n_e\sigma_Tc$ falls below the Hubble rate $H$ , so photons free-stream and the CMB is released. They are close in time ( $z\approx1090$ ) because decoupling is caused by the recombination-driven collapse of $n_e$ , but they are physically distinct and even the redshifts differ slightly. A related trap: Saha is only an equilibrium approximation. It over-predicts how fast and how completely hydrogen recombines, so the true residual ionization and exact decoupling redshift both require Peebles — never quote a freeze-out number from Saha alone.

Worked examples

1 · The recombination temperature from Saha (fully worked)

Using $\eta=6.1\times10^{-10}$ and $B=13.6\,\mathrm{eV}$ , find the temperature at which hydrogen is half-ionized, its redshift, and the ratio $B/T_{\rm rec}$ . Why is it so far below $B$ ?

Step 1. Write Saha in $X_e$ form with $n_b$ inserted: $(1-X_e)/X_e^2=\eta\,\dfrac{2\zeta(3)}{\pi^2}\left(\dfrac{2\pi T}{m_e}\right)^{3/2}e^{B/T}$ . At half-ionization the left side is $(1-\tfrac12)/(\tfrac12)^2=2$ .

Step 2. Solve $2=\eta(2\zeta(3)/\pi^2)(2\pi T/m_e)^{3/2}e^{B/T}$ numerically with $m_e=511\,\mathrm{keV}$ . The exponential dominates, so $B/T\approx\ln(1/\eta)+\tfrac32\ln(m_e/T)\approx42$ , giving $\mathbf{T_{\rm rec}=0.324\,\mathrm{eV}}$ , $\mathbf{B/T_{\rm rec}=42.0}$ .

Step 3. Convert to kelvin and redshift: $T_{\rm rec}=0.324/(8.617\times10^{-5})=\mathbf{3760\,K}$ , and $1+z=T_{\rm rec}/T_0$ , so $\mathbf{z_{\rm rec}\approx1380}$ (pure Saha; the full Peebles history gives decoupling at $z\approx1090$ ).

Step 4. Interpret. Recombination lands at $B/42$ , not $B$ , because there are $1/\eta\approx1.6\times10^{9}$ photons per baryon: the gas goes neutral only once the ionizing fraction $\sim e^{-B/T}$ drops below $\eta$ , and beating a billion costs $\ln(10^{9})\approx21$ in $B/T$ , doubled by the phase-space log to $\approx42$ . The delay is set by $\ln(1/\eta)$ , not by hydrogen's atomic physics.

2 · How fast is the drop? The width of recombination

The Saha curve is steep. Estimate the redshift range over which $X_e$ falls from $0.9$ to $0.1$ .

Step 1. Solving the $X_e$ -form equation at both endpoints gives $X_e=0.9$ at $T\approx0.348\,\mathrm{eV}$ and $X_e=0.1$ at $T\approx0.294\,\mathrm{eV}$ — the whole transition spans barely $\Delta T\approx0.05\,\mathrm{eV}$ , about $15\%$ in temperature.

Step 2. Convert with $1+z=T/T_0$ : $X_e=0.9$ at $z\approx1480$ , $X_e=0.1$ at $z\approx1250$ , so $\mathbf{\Delta z\approx230}$ .

Step 3. The steepness comes from $e^{B/T}$ with $B/T\approx42$ : a $15\%$ drop in $T$ shifts the exponent by $\Delta(B/T)\approx6$ , swinging the balance by $e^{6}\approx400$ . Recombination is rapid but continuous — a sharp crossover, not a phase transition — and this narrowness is why the last-scattering surface is thin ( $\Delta z\approx80$ with Peebles freeze-out).

3 · Helium recombines first

Neutral helium's first ionization energy is $\chi_{\rm HeI}=24.6\,\mathrm{eV}$ and He $^+$ 's is $\chi_{\rm HeII}=54.4\,\mathrm{eV}$ . Estimate the two helium recombination redshifts.

Step 1. The same balance holds with $B\to\chi$ ; since the delay is set by $\ln(1/\eta)$ , the half-ionization temperature scales as $T\approx\chi/42$ . For He I→neutral He, solving gives $\mathbf{T\approx0.60\,\mathrm{eV}}$ , $\mathbf{z\approx2550}$ ; for He $^{2+}$ →He $^+$ , $\mathbf{T\approx1.37\,\mathrm{eV}}$ , $\mathbf{z\approx5800}$ .

Step 2. Helium recombines before hydrogen — twice — because its electrons are more tightly bound, so their Boltzmann factors switch off hotter; by $z\approx1100$ both are long captured. Higher binding energy means earlier recombination, and these steps leave measurable imprints codes like RECFAST must include.

Physics of the portrait

The equation at the heart of cosmological recombination was never meant for cosmology. It was written to explain the colours of stars.

Origins · 1 minMeghnad Saha and how this was discovered

Meghnad Saha

1893–1956 · thermal ionization and stellar spectra

Meghnad Saha was born in 1893 near Dhaka, in what is now Bangladesh, the son of a small shopkeeper, and rose through a colonial education system that repeatedly tried to shut him out. In 1920, at the University of Calcutta, he published the thermal ionization equation that bears his name — the realization that the ionization state of a gas is a chemical equilibrium fixed by temperature and pressure, governed by the same law of mass action chemists apply to reacting molecules. Applied to stellar atmospheres it was revolutionary: it explained why spectral-line strengths change smoothly along the sequence O-B-A-F-G-K-M, showing this Harvard classification is a temperature sequence, not a compositional one. Within five years Cecilia Payne used Saha's equation to deduce that stars are overwhelmingly hydrogen and helium. A Fellow of the Royal Society from 1927, Saha founded the Institute of Nuclear Physics in Calcutta (now the Saha Institute of Nuclear Physics), drove India's river-planning and calendar reform, and served in Parliament. That his equilibrium formula for starlight would, decades later, become the foundation of how we compute the release of the cosmic microwave background is a testament to the reach of a single, deep idea.

History

1920

— Saha publishes the thermal ionization equation for stellar atmospheres

1925

— Cecilia Payne uses the Saha equation to show stars are mostly hydrogen

1948

— Alpher, Herman and Gamow predict a relic radiation from the hot early universe

1968

— Peebles, and independently Zel'dovich, Kurt and Sunyaev, compute non-equilibrium recombination

1999

— Seager, Sasselov and Scott release RECFAST, a fast effective recombination code

Going further

Texts. Ryden, Introduction to Cosmology, 2nd ed., ch. 8 · Dodelson & Schmidt, Modern Cosmology, 2nd ed., ch. 4 · Weinberg, Cosmology (2008), ch. 2 · Mo, van den Bosch & White, Galaxy Formation and Evolution, ch. 3 · Carroll & Ostlie, An Introduction to Modern Astrophysics, ch. 8 · Peebles, ApJ 153, 1 (1968). Key numerical checks: $g_pg_e/g_H=1$ , prefactor $(m_eT/2\pi)^{3/2}$ , $B=13.6\,\mathrm{eV}$ ; direct numerical solution of the $X_e$ -form equation with $\eta=6.1\times10^{-10}$ gives $T_{\rm rec}=0.324\,\mathrm{eV}$ , $B/T_{\rm rec}=42.0$ , $z\approx1380$ . The helium estimates $z\approx2550$ (He I) and $z\approx5800$ (He II) use the same equation with $\chi_{\rm HeI}=24.6\,\mathrm{eV}$ , $\chi_{\rm HeII}=54.4\,\mathrm{eV}$ . Standard values: $T_0=2.725\,\mathrm{K}$ , $z_{\rm dec}\approx1090$ , $\Delta z\approx80$ , $X_e^{\rm freeze}\approx2\times10^{-4}$ . The non-equilibrium freeze-out follows the Peebles three-level atom (Peebles 1968) and Zel'dovich, Kurt and Sunyaev (1968).

Prerequisites. Equilibrium Thermodynamics of the Early Universe · Thermal History of the Universe · Baryon Asymmetry and Baryogenesis. Leads to. The Last-Scattering Surface and Photon Decoupling · Decoupling and Freeze-Out.

Ask the tutor: Ask me to walk through solving the Saha equation numerically for $X_e(z)$ , or to show exactly how the Peebles rate equation bends the curve away from equilibrium into its frozen-out tail.

Why is it called "recombination" if the electrons and protons were never combined before?

It is a genuine misnomer, borrowed from laboratory plasma physics, where you ionize a neutral gas and then let it "recombine" back — so the prefix "re-" makes sense. In cosmology the plasma was primordial: protons and electrons had been free since the universe was seconds old and had never formed atoms; the first time they combined was at $z\approx1100$ . But by the time cosmologists borrowed the term in the 1960s it was standard, and it stuck. Some authors prefer "the recombination epoch" or "atom formation," but "recombination" is universal in the literature.

What exactly is the baryon-to-photon ratio η, and why does it matter so much?

$\eta=n_b/n_\gamma\approx6\times10^{-10}$ is the number of baryons per photon, a constant fixed by the tiny matter-antimatter asymmetry from baryogenesis. It is measured two independent ways — primordial deuterium (big-bang nucleosynthesis) and the CMB acoustic peaks — and they agree. It matters because its inverse, $1/\eta\approx1.6\times10^{9}$ , is the number of photons competing for each baryon. Even far below $13.6\,\mathrm{eV}$ , that billion-strong bath has an exponentially thin but numerically large ionizing tail, and hydrogen cannot stay neutral until that tail falls below $\eta$ — which is why recombination is delayed to $T\approx B/42$ .

Does the Saha equation give the correct decoupling redshift?

Not exactly. Pure Saha predicts half-ionization at $z\approx1380$ , but it assumes perfect chemical equilibrium, which fails as the universe dilutes. Because Saha lets recombination run to completion too eagerly, it gets the shape and residual of the ionization history wrong at low $X_e$ . The observed decoupling redshift $z\approx1090$ and residual $X_e\approx2\times10^{-4}$ both come from the non-equilibrium Peebles rate equation (and its successors HyRec and CosmoRec). Saha is the right conceptual, order-of-magnitude tool — it explains why recombination happens near $z\approx1000$ — but the precision numbers for fitting the CMB require the full kinetic treatment.

Why do the spin degeneracies g_p g_e / g_H combine to exactly one?

The free proton and electron each have two spin states, so $g_p=g_e=2$ and their product is $4$ . Ground-state hydrogen also has four: the electron and proton spins combine into a singlet (one state) plus a triplet (three), so $g_H=1+3=4$ . Thus $g_pg_e/g_H=4/4=1$ and the degeneracy prefactor drops out of the standard Saha form. This is not luck — forming the bound state does not create or destroy internal spin states, it only rearranges how they are counted. For other atoms and ions the analogous ratio is not unity and must be tracked explicitly.

Why can't a proton just capture an electron straight into the ground state?

It can — but it is nearly useless for net recombination. A direct capture to the $1s$ ground state releases a photon of at least the full binding energy, $13.6\,\mathrm{eV}$ : a Lyman-continuum photon that, in a bath of a billion photons per baryon, is reabsorbed almost instantly, ionizing a neighbour and undoing the capture. Net progress requires disposing of the binding energy in photons that cannot re-ionize hydrogen. There are only two routes: the forbidden two-photon decay $2s\to1s$ (rate $\approx8.2\,\mathrm{s^{-1}}$ ), which splits the energy into two sub-ionizing photons, and Lyman- $\alpha$ photons that redshift out of resonance before reabsorption. These slow channels are the true bottleneck.

How does the same equation apply to both stars and the early universe?

The Saha equation is a statement about local thermal and chemical equilibrium; it does not care whether the gas sits in a stellar photosphere or fills the cosmos. In a star you plug in the local temperature and electron pressure to find which ionization states dominate — that predicts the Balmer-line strengths behind the O-B-A-F-G-K-M classification. In cosmology you plug in the temperature and expansion-diluted baryon density to find $X_e(z)$ . The physics is identical; only the inputs differ. The one place cosmology needs more than Saha is the tail, where expansion drives the gas out of equilibrium — a complication a static atmosphere never faces.