Cosmology · Recombination, last scattering and the CMB
Neutral atoms and declining free-electron abundance
Track free and bound electron fractions in a supplied hydrogen-only equilibrium example, and distinguish those fractions from the scattering density.
Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.
Back to the chapterAfter reading, you can
- Define the free-electron fraction using total hydrogen nuclei as the denominator.
- Solve a supplied Saha-equilibrium relation and interpret its domain.
- Explain why an equilibrium curve alone does not determine the actual recombination history.
Recombination changes the charge state
Hydrogen nuclei already exist before neutral hydrogen atoms become abundant. An electron binding to a proton changes the ionization state; it does not create the proton or form a new nucleus.
Write n_H,tot = n_HI + n_p for total hydrogen nuclei, including both neutral atoms and free protons. In a hydrogen-only charge-neutral example, n_e = n_p. Define the free-electron fraction x_e = n_e/n_H,tot.
The neutral fraction is then 1 − x_e. The total physical nucleus density can fall through expansion while the fraction of electrons bound to nuclei rises. A fraction and a physical density therefore need separate accounts.
Helium and other populations would modify the electron account. They are omitted in the supplied example rather than silently included in the same hydrogen-only denominator.
Saha equilibrium is a conditional population relation
For a dilute nonrelativistic ground-state hydrogen equilibrium with the stated spin weights and approximate hydrogen/proton mass prefactor, the Saha relation is x_e²/(1 − x_e) = S(T)/n_H,tot. Here S(T) = [2πm_e k_B T/h²]^(3/2) exp[−E_H/(k_B T)].
E_H is the supplied hydrogen binding-energy symbol, m_e is electron mass and T is the shared temperature. This is a chemical-equilibrium relation, not a time-dependent atomic kinetics solution.
Let q = a/a_ref over 0.9 ≤ q ≤ 1.2, with T = T_ref/q and n_H,tot = n_H,ref/q³. Choose E_H/(k_B T_ref) = 40 and S(T_ref)/n_H,ref = 1. No physical reference temperature, density, date or redshift is assigned.
The right-hand side becomes R(q) = q^(3/2) exp[40(1 − q)]. A stable positive solution is x_e = 2/[1 + √(1 + 4/R)]. It lies between zero and one for positive R, with the other algebraic branch excluded.
Cooling and dilution compete within the equilibrium account
At q = 1, R = 1 and x_e² = 1 − x_e. The physical root is x_e = (√5 − 1)/2 ≈ 0.618034. The neutral fraction is about 0.381966.
At q = 1.2, the equilibrium free-electron fraction is about 0.0207801. The total nucleus density is also smaller, at q⁻³ ≈ 0.578704 of its reference value. Their product gives n_e/n_H,ref ≈ 0.0120255.
The number of nuclei in a comoving region has not vanished; electrons have become mostly bound in this equilibrium construction. For a fixed Thomson cross section, physical scattering opacity per unit path length n_eσ_T follows the free-electron density.
The curve’s rapid change results from the exponential binding-energy dependence under the supplied conditions. It is not a measured ionization record or an accurate calculation of the real recombination interval.
| q | T/T_ref | n_H,tot/n_H,ref | Free fraction x_e | Neutral fraction | n_e/n_H,ref |
|---|---|---|---|---|---|
| 0.9 | 1.111111 | 1.371742 | 0.979422 | 0.0205778 | 1.343515 |
| 0.95 | 1.052632 | 1.166351 | 0.885417 | 0.114583 | 1.032707 |
| 1 | 1 | 1 | 0.618034 | 0.381966 | 0.618034 |
| 1.05 | 0.952381 | 0.863838 | 0.315668 | 0.684332 | 0.272686 |
| 1.1 | 0.909091 | 0.751315 | 0.135182 | 0.864818 | 0.101564 |
| 1.2 | 0.833333 | 0.578704 | 0.0207801 | 0.979220 | 0.0120255 |
Atomic kinetics can lag behind equilibrium
Maintaining Saha equilibrium requires the relevant atomic and radiation processes to act effectively. Excited-state populations, radiative transfer and finite transition rates can prevent the real electron population from following the equilibrium curve.
A precision recombination history therefore solves a more complete kinetic account, including relevant levels and radiation processes. The equilibrium relation is useful for understanding a trend and checking limiting behavior, while its numerical timing needs a separate calculation.
The scattering history depends on n_e as well as the expansion and propagation convention. A small equilibrium x_e alone does not establish the actual last-scattering probability, which involves accumulated opacity and the chance of no later scattering.
The next lesson isolates that probability account using a supplied visibility profile. Neither toy replaces a precision coupled recombination and radiation-transport calculation.
Neutralization does not create the whole thermal background
The cosmic thermal radiation field already exists before the main neutralization interval in the standard thermal-history account. Atomic recombination also emits additional radiation through transitions, but those photons are not the origin of the whole thermal CMB bath.
As the free-electron abundance falls, the background can propagate with less Thomson scattering. This changes how radiation carries earlier spatial information to later observers, rather than turning every observed photon into a newly created atomic-transition photon.
Later scattering, gravitational effects and astrophysical foregrounds can still alter what is observed. Declining opacity should not be described as guaranteeing that every photon experiences no subsequent interaction or processing.
When interpreting a recombination plot, identify total nuclei, neutral atoms, free electrons, physical density and the equilibrium or kinetic assumptions. Those distinctions connect atomic physics to visibility without mixing their definitions.
Worked reasoning
Solve the reference equilibrium fraction
What is x_e at q = 1 in the supplied example?
- The chosen reference has R = S/n_H,tot = 1.
- Solve x_e² = 1 − x_e to obtain x_e = [−1 ± √5]/2.
- Choose the root between zero and one: x_e ≈ 0.618034, with neutral fraction about 0.381966.
About 61.8 percent of the hydrogen-associated electrons are free in this chosen equilibrium state.
Try the reasoning
Does n_H,tot count only neutral atoms?
Show the answer and reasoning
No.
It includes every hydrogen nucleus, whether a free proton or bound in neutral hydrogen.
At q = 1.2, is n_e/n_H,ref equal to x_e?
Show the answer and reasoning
No.
The physical total nucleus density has also fallen by q⁻³, so the density ratio is x_e q⁻³.
Does a small Saha-equilibrium x_e establish the actual recombination timing?
Show the answer and reasoning
No.
Atomic kinetics and radiative transfer can keep the real populations away from Saha equilibrium.
What to carry forward
- Recombination changes electron binding, not the number of hydrogen nuclei in a comoving region.
- Free-electron fraction and physical scattering density have different denominators.
- Saha equilibrium illustrates a trend while the actual recombination history requires kinetics.
References
Sources supporting the explanations, physical distinctions and examples in this lesson.
- Cosmology: 2 The Hot UniverseDavid TongCourse notes
- How exactly did the Universe become neutral?Sara Seager; Dimitar D. Sasselov; Douglas ScottResearch paper
- A Line of Sight Integration Approach to Cosmic Microwave Background AnisotropiesUroš Seljak; Matias ZaldarriagaResearch paper
- Cosmic Microwave Background AnisotropiesWayne Hu; Scott DodelsonTeaching resource