---
title: "Saha balance and the recombination bottleneck"
description: "Saha equilibrium puts half of hydrogen in atoms by redshift 1370. A slow last step leaves 30 times more free electrons at redshift 1100."
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language: "en"
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last_modified: "2026-11-12"
---

[How the universe became transparent](https://sandboxphysics.com/stories/cosmic-transparency/)

# Saha balance and the recombination bottleneck

Sandbox Physics · 2026-11-12

If every reaction kept pace with the cooling universe, half of all hydrogen would have formed atoms by redshift 1370, at 3,737 kelvin. It did not keep pace. An atom that reaches its first excited state is usually broken apart again before it can drop to the ground state, so at redshift 1100 about 14 per cent of hydrogen is still ionized instead of the 0.5 per cent that equilibrium predicts.

This chapter follows [What can tiny departures from a perfect blackbody record?](https://sandboxphysics.com/stories/cosmic-transparency/a-nearly-perfect-blackbody/). You need exponentials, logarithms and number densities in particles per cubic metre. The aims are to derive the Saha balance, solve it at one redshift, then see which step of atom formation is slow and how a three-level model of hydrogen describes it.

## Quantities, units and what is left out

Redshift $z$ labels the epoch. The photon temperature is $T=T_0(1+z)$ with $T_0=2.7255\,\mathrm K$, and the hydrogen number density is $n_{\mathrm H}=n_{\mathrm H,0}(1+z)^3$ with $n_{\mathrm H,0}=0.190\,\mathrm{m^{-3}}$. These are the background values in the saved [model inputs](https://sandboxphysics.com/stories/cosmic-transparency/materials/inputs.json), taken from the HyRec input file shipped with CLASS v3.3.4.

The free-electron fraction is $x=n_{\mathrm e}/n_{\mathrm H}$. Treat the gas as hydrogen only. Helium finishes becoming neutral between redshifts 3500 and 1600, so below 1600 it adds almost no free electrons, though it does add mass. [Lucca and colleagues, 2020, section 2.4.3](https://arxiv.org/abs/1910.04619v3) The ground-state binding energy is $I=13.6\,\mathrm{eV}$, and $k_{\mathrm B}T$ is the typical thermal energy. The model also lets matter share the photon temperature throughout. That holds during recombination but fails late in the dark ages, when the gas cools faster than the light, so values near redshift 200 inherit that error.

## Deriving the Saha balance

Picture a box in which the reaction $\mathrm p+\mathrm e\rightleftharpoons\mathrm H+\gamma$ runs both ways so fast that it settles. For each species, statistical physics assigns a chemical potential: the energy cost of adding one more particle while keeping the temperature fixed. Photons can be created and destroyed freely, so theirs is zero. When the reaction has settled, the costs on both sides match:

$$
\mu_{\mathrm p}+\mu_{\mathrm e}=\mu_{\mathrm H}.
$$

For a dilute, non-relativistic gas, the density of species $i$ with $g_i$ internal states and mass $m_i$ is

$$
n_i=g_i\left(\frac{2\pi m_ik_{\mathrm B}T}{h^2}\right)^{3/2}\exp\!\left(\frac{\mu_i-m_ic^2}{k_{\mathrm B}T}\right).
$$

Form $n_{\mathrm p}n_{\mathrm e}/n_{\mathrm H}$. The chemical potentials cancel by the balance above. The rest energies leave $\mathrm e^{-I/k_{\mathrm B}T}$, because the atom is lighter than a proton plus an electron by $I/c^2$. The state counts give $g_{\mathrm p}g_{\mathrm e}/g_{\mathrm H}=2\times2/4=1$, and $m_{\mathrm p}\simeq m_{\mathrm H}$ leaves only the electron mass. With $n_{\mathrm e}=n_{\mathrm p}=xn_{\mathrm H}$ and $n_{\mathrm{HI}}=(1-x)n_{\mathrm H}$,

$$
\frac{x^2}{1-x}=S,\qquad S=\frac{1}{n_{\mathrm H}}\left(\frac{2\pi m_{\mathrm e}k_{\mathrm B}T}{h^2}\right)^{3/2}\mathrm e^{-I/k_{\mathrm B}T}.
$$

It is a quadratic in $x$ with one root between zero and one:

$$
x=\frac{-S+\sqrt{S^2+4S}}{2}.
$$

The assumptions are that every atom sits in its ground state and that formation and break-up are much faster than the expansion. The second assumption is the one that fails.

## Worked example: half recombined

Try $z=1370$. Then $T=3737\,\mathrm K$ and $k_{\mathrm B}T=0.322\,\mathrm{eV}$, so $I/k_{\mathrm B}T=42.3$ and $\mathrm e^{-42.3}=4.46\times10^{-19}$. The bracket raised to the power three halves is $5.52\times10^{26}\,\mathrm{m^{-3}}$, and $n_{\mathrm H}=4.90\times10^{8}\,\mathrm{m^{-3}}$. Together $S=0.503$, so $x=0.501$.

The exponential is tiny, yet it is offset by a ratio of about $10^{18}$ between the bracket and the hydrogen density. That ratio is large because the gas is so thin. Ionization needs only one photon to hit an atom; recombination needs an electron to meet a proton. Thin gas favours ionization.

A popular shortcut says atoms form once fewer than one photon per atom can still ionize hydrogen. At $z=1370$ the photons above $13.6\,\mathrm{eV}$ number only about $7.5\times10^{-7}$ per hydrogen nucleus, far below one, yet equilibrium is only halfway. What sets the balance is a ratio of rates, and a photon count gets its scale badly wrong.

## Why the last step is slow

Recombining straight into the ground state emits a photon above $13.6\,\mathrm{eV}$. That photon ionizes the next atom it meets, so the step achieves nothing overall. Only captures into excited states count; their total coefficient is called the case-B coefficient $\alpha_{\mathrm B}$. The captured electron cascades down to the second level, $n=2$.

From $n=2$ the atom must still reach the ground state, and both routes are slow. A Lyman-alpha photon from $2\mathrm p\to1\mathrm s$ is absorbed by the next ground-state atom, so it only escapes by redshifting out of the line as space expands. The $2\mathrm s$ level can decay by emitting two photons at once, at the rate $\Lambda_{2\mathrm s}=8.22\,\mathrm{s^{-1}}$. Meanwhile, ordinary photons of $3.4\,\mathrm{eV}$ ionize atoms in $n=2$ at a rate $\beta_{\mathrm B}$.

Let $C$ be the probability that an atom in $n=2$ reaches the ground state before it is ionized. With $s=\ln(1+z)$, the effective three-level atom gives

$$
\frac{\mathrm dx}{\mathrm ds}=\frac{C}{H}\left[x^2n_{\mathrm H}\alpha_{\mathrm B}-\beta_{\mathrm B}(1-x)\mathrm e^{-E_\alpha/k_{\mathrm B}T}\right],
$$

$$
C=\frac{1+K\Lambda_{2\mathrm s}n_{1\mathrm s}}{1+K\left(\Lambda_{2\mathrm s}+\beta_{\mathrm B}\right)n_{1\mathrm s}},\qquad K=\frac{\lambda_\alpha^3}{8\pi H}.
$$

$H$ is the expansion rate in $\mathrm{s^{-1}}$, $n_{1\mathrm s}=(1-x)n_{\mathrm H}$, $E_\alpha=10.2\,\mathrm{eV}$ and $\lambda_\alpha=121.6\,\mathrm{nm}$. The first bracketed term counts captures; the second counts ionizations from $n=2$, whose population follows the ground state through the Boltzmann factor. Read the sign with care: lowering $z$ lowers $s$, so a positive rate means $x$ falls as the universe ages. The coefficient $\alpha_{\mathrm B}$ uses the fitted formula and the factor $F=1.14$ given by [Seager, Sasselov and Scott, 1999](https://arxiv.org/abs/astro-ph/9909275v2); $F$ is tuned so this simple atom follows their multi-level calculation.

## Worked example: the bottleneck at redshift 1100

At $z=1100$ the solved history gives $x=0.142$. The expansion rate is $H=5.13\times10^{-14}\,\mathrm{s^{-1}}$, so $K=1.39\times10^{-9}\,\mathrm{m^3\,s}$. With $n_{1\mathrm s}=2.18\times10^{8}\,\mathrm{m^{-3}}$ and $\beta_{\mathrm B}=586\,\mathrm{s^{-1}}$,

$$
K\Lambda_{2\mathrm s}n_{1\mathrm s}=2.49,\qquad K\beta_{\mathrm B}n_{1\mathrm s}=178,\qquad C=\frac{3.49}{181}\simeq0.019.
$$

Fewer than two atoms in a hundred that reach $n=2$ go on to the ground state. The term $1/(Kn_{1\mathrm s})=3.3\,\mathrm{s^{-1}}$ is the effective Lyman-alpha escape rate, so at this epoch the two-photon decay at $8.22\,\mathrm{s^{-1}}$ carries more atoms down than redshifting does.

**Table 2.1** Free-electron fraction from three calculations.

| Redshift | Saha | Three-level atom | HyRec-2 in CLASS v3.3.4 |
| --- | --- | --- | --- |
| $1400$ | $0.649$ | $0.805$ | $0.803$ |
| $1200$ | $0.0386$ | $0.322$ | $0.322$ |
| $1100$ | $0.00469$ | $0.142$ | $0.145$ |
| $1000$ | $0.000364$ | $0.0472$ | $0.0487$ |

The three-level atom stays within 4 per cent of the frozen HyRec-2 output at these redshifts, while Saha falls 30 times too low by redshift 1100. HyRec-2 is a much more complete atomic calculation; here we read its output file and did not run it. The comparison and its errors are worked through in [Comparing Saha, a three-level atom and a frozen CLASS reference](https://sandboxphysics.com/stories/cosmic-transparency/compare-recombination-codes/).

## Lab tasks

Open the [recombination Lab](https://sandboxphysics.com/labs/recombination-spectral-distortions/). The ionization chart shows Saha as a dashed grey line and the three-level atom in cyan, with gold dots from the HyRec-2 file, which also counts helium electrons.

1. Predict where the two curves separate, then check: they agree above redshift 1500 and part between 1500 and 1400.
2. Choose the Dark-age ionization glow preset. Before looking, predict whether the free-electron fraction at redshift 200 rises or falls. The model gives $1.10\times10^{-3}$, against $3.39\times10^{-4}$ with no extra energy.
3. Explain from the bottleneck why a small energy release after recombination changes $x$ by a large factor while Saha would not change at all.

## Exercises

1. When $S\gg1$, show that $1-x\simeq1/S$. Check it at $z=1800$, where $S=8032$.
2. Show that $x=1/2$ exactly when $S=1/2$.
3. Double the baryon density at fixed temperature history. Does the Saha half-point move to higher or lower redshift, and by about how much?
4. At $z=1000$ the model gives $C=0.075$. What does this number mean for an atom in the second level, and why is it larger than at $z=1100$?

Answers

1. Write $\epsilon=1-x$. Then $(1-\epsilon)^2/\epsilon=S$, so $\epsilon=1/(S+2-\epsilon)\simeq1/S-2/S^2$. At $S=8032$ this gives $1.2447\times10^{-4}$, the same as the exact root. The first term alone is $1.2450\times10^{-4}$. This limit is checked in the repository tests independently of the quadratic root.
2. Put $x=1/2$ into the left side: $(1/4)/(1/2)=1/2$.
3. $S$ is proportional to $1/n_{\mathrm H}$, so doubling the density halves $S$ at a given temperature. To restore $S=1/2$ the gas must be hotter, which means earlier. Near the half-point $\mathrm d\ln S/\mathrm d\ln T=I/k_{\mathrm B}T-3/2\simeq40.8$, so a change $\ln2$ in $S$ needs $\Delta\ln T\simeq0.017$, or $\Delta z\simeq23$. The exact half-point moves from $1369.8$ to $1393.5$.
4. Seven or eight atoms in a hundred that reach $n=2$ now make it to the ground state. Fewer photons can ionize the second level as the gas cools, so $\beta_{\mathrm B}$ falls quickly and the bottleneck eases.

## Sources

1. [Lucca and colleagues, 2020, section 2.4.3](https://arxiv.org/abs/1910.04619v3)
2. [Seager, Sasselov and Scott, 1999](https://arxiv.org/abs/astro-ph/9909275v2)

## Keep reading

How the universe became transparent · Reading path

### From a blackbody to limits on hidden energy

1. [Article **What can tiny departures from a perfect blackbody record?**](https://sandboxphysics.com/stories/cosmic-transparency/a-nearly-perfect-blackbody/)
2. [You are here **Saha balance and the recombination bottleneck**](https://sandboxphysics.com/stories/cosmic-transparency/saha-and-the-recombination-bottleneck/)
3. [Article **Optical depth and the visibility function**](https://sandboxphysics.com/stories/cosmic-transparency/optical-depth-and-the-visibility-function/)
4. [Article **Thermalization, chemical-potential and Compton distortions**](https://sandboxphysics.com/stories/cosmic-transparency/thermalization-mu-and-y/)
5. [Article **Comparing Saha, a three-level atom and a frozen CLASS reference**](https://sandboxphysics.com/stories/cosmic-transparency/compare-recombination-codes/)
6. [Article **How to read the decaying dark matter limits of Lucca and colleagues**](https://sandboxphysics.com/stories/cosmic-transparency/reading-energy-injection-limits/)