advanced20 min studyLesson 858 in Physics

Recombination and photon decoupling

Distinguish equilibrium ionization, radiative recombination kinetics, and the probability of last scattering.

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What you’ll understand

  • Apply the hydrogen-only Saha equation with a defined ionization fraction.
  • Explain the radiative bottlenecks that invalidate instantaneous equilibrium.
  • Separate an instantaneous scattering rate from integrated optical depth.

Hydrogen recombination reduces the free-electron density, enabling cosmic photons to travel much farther between scatterings. Atomic neutralization and photon decoupling are related but distinct processes. One concerns populations of atomic and ionized states; the other concerns the probability that a photon scatters along its remaining path. A thermal equilibrium formula provides an instructive baseline, but accurate recombination requires radiative kinetics, and accurate last-scattering predictions require an optical-depth history. We first use a hydrogen-only, electrically neutral gas, then identify the missing physics needed before comparing with a precision cosmic microwave background calculation.

Saha predicts an equilibrium population, not a reaction speed

Let nHn_H denote the total number density of hydrogen nuclei, whether bound or ionized. Define the free-electron fraction by xe=ne/nHx_e=n_e/n_H; electrical neutrality gives the same free-proton fraction. In chemical equilibrium, the reaction linking an electron, proton, hydrogen atom, and photon fixes the chemical-potential relation. Nondegenerate nonrelativistic matter distributions and a thermal photon bath then give the Saha equation. The ground-state spin multiplicities are included in the conventional prefactor displayed here.

The hydrogen binding energy alone does not set the recombination temperature. Because photons outnumber hydrogen nuclei, the exponential photon tail can keep the gas ionized at temperatures far below the binding energy. The explicit inverse density in the Saha expression matters: at fixed temperature, a denser hydrogen gas favors neutralization. In a real expanding history, both temperature and density change, so one must specify their relation before solving for an epoch.

xe21−xe=1nH(meT2π)3/2e−χH/T,χH≃13.6 eV\frac{x_e^2}{1-x_e}=\frac1{n_H}\left(\frac{m_eT}{2\pi}\right)^{3/2}e^{-\chi_H/T},\qquad \chi_H\simeq13.6\,\mathrm{eV}
Natural units are used; mem_e is the electron mass and TT the common matter-radiation temperature. This hydrogen-only ground-state approximation neglects helium, reduced-mass corrections, excited-state populations, and nonequilibrium radiative transfer.↔ Scroll wide equations horizontally.

Atomic radiative bottlenecks leave a nonequilibrium history

Direct capture into the ground state emits a photon energetic enough to ionize another atom, so it need not yield a lasting increase in neutral hydrogen. Population must instead pass through excited states and lose radiation without immediately reversing the net recombination. Two important escape routes are the metastable two-photon transition from the second s state and redshifting of resonant Lyman-alpha photons out of the line. These processes introduce timescales absent from an equilibrium population formula.

As expansion outpaces effective atomic relaxation, the electron fraction departs from the Saha value. A simplified effective-level treatment captures the bottleneck qualitatively; precision work follows more levels and radiative transfer. Helium recombination and later astrophysical reionization also alter the electron history. Thus using Saha alone all the way through last scattering cannot justify a precise visibility peak or residual ionization fraction, even if it gives a reasonable initial scale for when atoms begin forming.

Transparency depends on accumulated scattering probability

At low photon energy, free electrons scatter photons with the Thomson cross section. The rate per photon is free-electron density times cross section times light speed. Compare it with expansion for a local diagnostic, but integrate the rate along a trajectory to obtain optical depth. The probability of no further scattering is an exponential of minus that integrated depth. A particular ionized fraction therefore cannot universally define decoupling independently of density and expansion.

For optical depth defined from an emission epoch to the present, the last-scattering visibility combines the chance to scatter near that epoch with the chance of no scattering afterward. In conformal time it is g(η)=−τ′(η)e−τg(\eta)=-\tau'(\eta)e^{-\tau}, since the remaining optical depth decreases toward today. Recombination, the peak of this visibility, and baryon drag decoupling answer different questions. A later reionized population can add optical depth without undoing the earlier atomic history.

Γγ=neσTc,τ(z)=∫0zcσTne(z′)(1+z′)H(z′) dz′,Pno scatter=e−τ\Gamma_\gamma=n_e\sigma_Tc,\qquad \tau(z)=\int_0^z\frac{c\sigma_Tn_e(z')}{(1+z')H(z')}\,dz',\qquad P_{\rm no\ scatter}=e^{-\tau}
Light speed is restored here. nen_e is proper free-electron density, zz is redshift, and σT\sigma_T is the Thomson cross section. The integral uses dt=−dz/[(1+z)H]dt=-dz/[(1+z)H] and a specified ionization history.↔ Scroll wide equations horizontally.

Half neutral need not mean transparent

At one stipulated instant, a hydrogen-only equilibrium toy model has Saha right-hand side 1/21/2. Take nH=300 cm−3n_H=300\,\mathrm{cm}^{-3}, σTc=2×10−14 cm3 s−1\sigma_Tc=2\times10^{-14}\,\mathrm{cm^3\,s^{-1}}, and H=5×10−14 s−1H=5\times10^{-14}\,\mathrm{s^{-1}}. Find the physical electron fraction and scattering-to-expansion ratio. These rounded inputs do not specify a fitted recombination epoch.

  1. Solve xe2/(1−xe)=1/2x_e^2/(1-x_e)=1/2, giving 2xe2+xe−1=02x_e^2+x_e-1=0. The roots are 1/21/2 and −1-1; only the first lies in the allowed interval from zero to one. The toy gas is half ionized.

  2. Multiply the total hydrogen-nucleus density by the free fraction to obtain ne=150 cm−3n_e=150\,\mathrm{cm}^{-3}. Bound electrons do not enter the free-electron Thomson rate used here.

  3. Compute Γγ=150(2×10−14)=3×10−12 s−1\Gamma_\gamma=150(2\times10^{-14})=3\times10^{-12}\,\mathrm{s^{-1}}. Division by the stipulated expansion rate gives Γγ/H=60\Gamma_\gamma/H=60. Many scatterings fit into one expansion time despite the substantial neutral fraction.

The result

The equilibrium fraction is xe=0.5x_e=0.5, but the instantaneous scattering-to-expansion ratio is 6060. Determining a last-scattering probability still requires the future density and ionization history, rather than this single instant.

Explore the model

Interactive lab

Interactive lab · heat

An exponential attenuation analogy

Reset and set both controls to mode n=1n=1, diffusivity κ=0.2\kappa=0.2, then Step once: the amplitude is e−0.1≃0.9048e^{-0.1}\simeq0.9048. Reset and set both controls to n=1n=1, κ=1\kappa=1, then Step once: it is e−0.5≃0.6065e^{-0.5}\simeq0.6065. This is decay of a periodic diffusion mode; its exponential illustrates accumulated attenuation only. It does not model photon paths, atomic populations, Thomson scattering, or cosmological optical depth.

An exponential attenuation analogyThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn exponential depends on the accumulated decay parameter; a real transparency calculation must supply the correct physical integral.

Check your understanding

What is missing when the Saha equation is used as a complete description of cosmic hydrogen recombination?

Try it yourself

In a separate constant-rate segment along a photon path, let Γγ=2×10−13 s−1\Gamma_\gamma=2\times10^{-13}\,\mathrm{s^{-1}} for Δt=5×1012 s\Delta t=5\times10^{12}\,\mathrm{s}. Find the optical depth and probability of zero scatterings in that segment. Repeat for twice the duration, with the same rate.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Integrate the rate over time before exponentiating; this segment model is independent of Saha equilibrium.

Compare with the solution

The first optical depth is τ=ΓγΔt=1\tau=\Gamma_\gamma\Delta t=1, so the no-scattering probability is e−1≃0.3679e^{-1}\simeq0.3679. Doubling the duration gives τ=2\tau=2 and probability e−2≃0.1353e^{-2}\simeq0.1353. The rate is unchanged while the survival probability changes substantially. A real cosmological calculation integrates the time-dependent free-electron density and does not assume a constant rate over the whole recombination history.

Take it with you

  • Saha supplies equilibrium populations under specified thermodynamic assumptions.
  • Radiative bottlenecks make recombination a kinetic problem.
  • Photon decoupling is governed by an optical-depth history, not a universal ionized fraction.

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