---
title: Recombination and the Cosmic Microwave Background
draft: false
module: The Hot Big Bang
moduleNumber: 12
lessonNumber: 3
order: 1203
summary: >
  As the universe cooled through a few thousand kelvin the free electrons bound to
  protons, and the Saha equation tracks the falling ionization fraction. Once the
  plasma neutralized, photons stopped scattering and streamed freely from a
  spherical surface of last scattering at redshift about 1100. Those photons are
  the cosmic microwave background, an almost perfect blackbody at 2.725 kelvin with
  a dipole from our motion through it.
topics: [The Hot Big Bang]
sources:
  - book: Ryden
    ref: "Ch. 8 — The Cosmic Microwave Background; §8.1–8.3"
  - book: Carroll & Ostlie
    ref: "Ch. 30 — The Early Universe; §30.3"
  - book: WMAP
    ref: "NASA WMAP mission — COBE/FIRAS blackbody spectrum and the CMB dipole"
---

For its first 380,000 years the universe was an opaque plasma: free electrons
scattered photons so efficiently that light could not travel a meaningful
distance before being deflected. When the temperature fell far enough for
electrons to bind to protons, the free-electron density collapsed, the scattering
stopped, and the photons that had last scattered streamed to us across the entire
subsequent history of the universe. Those photons are the cosmic microwave
background (CMB), the oldest electromagnetic image available and a near-perfect
blackbody. This lesson derives the ionization history through the Saha equation,
locates the surface of last scattering, and characterizes the CMB spectrum and
its dipole.

## Recombination and the Saha equation

The relevant reaction is the capture of an electron by a proton to form neutral
hydrogen, balanced by photoionization,

$$
p + e^- \leftrightarrow \text{H} + \gamma ,
$$

with binding energy $B = 13.6\ \text{eV}$. Define the **ionization fraction** as
the fraction of baryons in free protons,

$$
X = \frac{n_p}{n_p + n_{\text{H}}} = \frac{n_p}{n_b},
$$

where charge neutrality sets $n_e = n_p$. In chemical equilibrium the relative
abundances of the three species follow the **Saha equation**, obtained by
equating chemical potentials with each equilibrium density given by its
non-relativistic phase-space integral:

$$
\frac{n_p\,n_e}{n_{\text{H}}}
= \left(\frac{m_e k_B T}{2\pi\hbar^2}\right)^{3/2}
  \exp\!\left(-\frac{B}{k_B T}\right).
$$

Dividing by $n_b$ and using $n_e = n_p = X n_b$ and $n_{\text{H}} = (1-X)n_b$
gives the Saha equation in the form that determines $X(T)$:

$$
\frac{X^2}{1 - X}
= \frac{1}{n_b}\left(\frac{m_e k_B T}{2\pi\hbar^2}\right)^{3/2}
  \exp\!\left(-\frac{B}{k_B T}\right).
$$

The baryon density $n_b = \eta n_\gamma \propto T^3$ is known from the
baryon-to-photon ratio. The right side is a steep function of temperature through
the exponential, so $X$ falls sharply once $k_B T$ drops well below $B$.

The transition happens at a temperature well below the binding energy for the
same reason as the deuterium bottleneck: photons outnumber baryons by $\sim
10^9$, so the high-energy tail of the blackbody spectrum keeps hydrogen ionized
until $k_B T \ll B$. Setting $X = 1/2$ in the Saha equation gives the temperature
of half-recombination,

$$
k_B T_{\text{rec}} \approx 0.32\ \text{eV},
\qquad T_{\text{rec}} \approx 3740\ \text{K},
\qquad 1 + z_{\text{rec}} \approx 1370 ,
$$

roughly forty times cooler than $B/k_B$. The ionization fraction plunges from
near unity to below $10^{-3}$ over a redshift interval of only a few hundred.

> **Worked example.** Evaluate the Saha ratio at $z = 1500$, where $T \approx
> 4090\ \text{K}$ and $k_B T \approx 0.35\ \text{eV}$. The exponential factor is
> $\exp(-B/k_B T) = \exp(-13.6/0.35) \approx e^{-39} \approx 10^{-17}$. The
> prefactor $n_b^{-1}(m_e k_B T/2\pi\hbar^2)^{3/2}$ is of order $10^{15}$ at this
> density, so $X^2/(1-X) \approx 10^{-2}$, giving $X \approx 0.1$: the plasma is
> already 90% neutral. A hundred kelvin higher, at $z = 1600$, the same estimate
> gives $X$ close to unity. The steepness of the exponential compresses
> recombination into a narrow shell, which is why the surface of last scattering
> is thin.

$$
% caption: The ionization fraction from the Saha equation falls sharply near
% redshift 1300 as the plasma neutralizes; the true history freezes out at a
% small residual ionization once recombination cannot keep pace with expansion.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (9.6,0) node[right, black!70] {redshift, decreasing to the right};
\draw[->, black] (0,0) -- (0,4.4) node[above, black!70] {ionization fraction};
% Saha curve: high, then sharp drop
\draw[acc, very thick] (0.4,4.0) -- (3.6,3.85)
  .. controls (4.4,3.6) and (4.8,0.6) .. (5.6,0.45) -- (9.0,0.35);
\node[acc, anchor=west] at (0.7,3.7) {fully ionized};
\node[acc, anchor=west] at (6.0,0.55) {neutral};
% residual freeze-out (dashed)
\draw[black, very thick, dashed] (5.6,0.45) -- (9.0,0.55);
\node[black, anchor=south] at (7.4,0.6) {residual freeze-out};
% recombination marker
\draw[black, densely dotted] (4.8,0) node[below, black!70] {recombination} -- (4.8,2.4);
\end{tikzpicture}
$$

The Saha equation assumes equilibrium, which fails at the end of recombination.
As $X$ drops, the recombination rate $\propto n_e n_p \propto X^2$ falls faster
than the expansion, so the reaction cannot keep the ionization at its equilibrium
value. A full treatment with the recombination rate equations (the Peebles
analysis) shows the true ionization freezes out at a residual $X \approx 10^{-4}$
rather than continuing to zero. Recombination also proceeds not directly to the
ground state — each such capture emits a photon that immediately reionizes
another atom — but through the two-photon decay of the metastable 2s level and
the redshifting of Lyman-$\alpha$ photons, both slow. These delays push the
completion of recombination and photon decoupling to slightly lower redshift than
the Saha estimate.

## The surface of last scattering

Photons in the plasma scatter off free electrons by Thomson scattering, with
cross section $\sigma_T = 6.65 \times 10^{-29}\ \text{m}^2$. The scattering rate
per photon is $\Gamma = n_e \sigma_T c$, and the mean free path is $1/(n_e
\sigma_T)$. While $X \approx 1$, the free-electron density is high and the mean
free path is tiny; the universe is optically thick and photons random-walk. As
$X$ collapses during recombination, $n_e$ drops by orders of magnitude, the mean
free path grows past the Hubble length, and photons **decouple** — they travel
freely thereafter.

The optical depth from an observer back to redshift $z$ is

$$
\tau(z) = \int_0^z n_e(z')\,\sigma_T\,\frac{c\,\d z'}{(1+z')H(z')} ,
$$

and photon decoupling occurs where $\tau \sim 1$. This happens at

$$
1 + z_{\text{dec}} \approx 1090,
\qquad T_{\text{dec}} \approx 2970\ \text{K},
\qquad t_{\text{dec}} \approx 380{,}000\ \text{yr}.
$$

Because decoupling is not instantaneous, the last-scattering event of a given
photon is spread over a range of redshift; the **visibility function** $g(z) = e^{-\tau}\,\d\tau/\d z$,
the probability that a photon last scattered at $z$, is a peaked distribution of
width $\Delta z \approx 80$ centered on $z_{\text{dec}}$. The **surface of last
scattering** is therefore a shell of finite thickness surrounding every observer,
at a comoving distance equal to the present particle horizon minus a negligible
correction.

$$
% caption: The observer sits at the center of a spherical shell of last
% scattering; CMB photons arriving now last scattered on that shell at redshift
% about 1090, and the shell has a finite thickness set by the visibility function.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% observer
\fill[black!70] (0,0) circle (2.2pt);
\node[black!70, anchor=north] at (0,-0.15) {observer};
% last-scattering shell (annulus)
\draw[acc, very thick] (0,0) circle (3.4);
\draw[thick, dashed] (0,0) circle (3.05);
\draw[thick, dashed] (0,0) circle (3.75);
\node[acc, anchor=south] at (0,3.5) {surface of last scattering};
% incoming photon rays
\foreach \a in {20,70,130,200,250,310}
  \draw[->, black] (\a:3.3) -- (\a:0.35);
\node[black, anchor=west] at (3.8,-1.6) {redshift about 1090};
\node[black, anchor=west] at (3.05,-2.5) {thick shell};
\end{tikzpicture}
$$

Before decoupling the photons were tightly coupled to the baryons in a single
photon-baryon fluid; after decoupling they free-stream, preserving the
temperature they had at last scattering, redshifted by the expansion. The CMB is
thus a snapshot of the universe at $z \approx 1090$, and its tiny temperature
variations, treated in the next lesson, are a direct image of the density field
at that epoch.

## The blackbody spectrum

Before decoupling, frequent interactions — Thomson scattering, together with
double Compton and bremsstrahlung that create and destroy photons — drove the
radiation to a blackbody spectrum. Free expansion after decoupling preserves the
blackbody form, because redshifting a Planck spectrum of temperature $T$ produces
a Planck spectrum of temperature $T/(1+z)$. The present CMB is therefore predicted
to be a blackbody, and it is: the COBE/FIRAS instrument measured

$$
T_0 = 2.725\ \text{K},
$$

with deviations from a pure Planck spectrum below one part in $10^4$, making the
CMB the most perfect blackbody known. The spectral radiance follows the Planck
function

$$
B_\nu(T_0) = \frac{2h\nu^3}{c^2}\,\frac{1}{e^{h\nu/k_B T_0} - 1},
$$

peaking near $\nu \approx 160\ \text{GHz}$ (wavelength $\approx 2\ \text{mm}$), in
the microwave band. The energy density is $\varepsilon_\gamma = a_{\text{rad}}
T_0^4$ and the number density $n_{\gamma,0} \approx 411\ \text{cm}^{-3}$, the same
photons whose abundance relative to baryons fixed $\eta$ in nucleosynthesis.

$$
% caption: The measured CMB spectrum matches a 2.725 K Planck blackbody with
% error bars far smaller than the line width; the intensity peaks near 160 GHz in
% the microwave band.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (9.6,0) node[right, black!70] {frequency};
\draw[->, black] (0,0) -- (0,4.4) node[above, black!70] {intensity};
% Planck curve
\draw[acc, very thick] (0.3,0.2)
  .. controls (1.6,1.2) and (2.6,3.9) .. (3.6,3.95)
  .. controls (4.8,4.0) and (6.4,1.6) .. (9.0,0.4);
% peak marker
\draw[black, densely dotted] (3.6,0) node[below, black!70] {about 160 GHz} -- (3.6,3.9);
\node[acc, anchor=west] at (4.2,3.4) {blackbody spectrum};
% a few data points on the curve
\foreach \x/\y in {1.4/1.0,2.4/3.2,3.6/3.95,5.0/3.2,6.6/1.4}
  \fill[black!70] (\x,\y) circle (1.6pt);
\end{tikzpicture}
$$

The blackbody form is itself a stringent test. A universe that was merely cold
and dilute, without a hot dense past, would have no mechanism to thermalize
starlight into so perfect a Planck spectrum across the whole sky. The FIRAS
spectrum is among the most direct pieces of evidence for the hot Big Bang.[^wmap-firas]

## The dipole anisotropy

The CMB is isotropic to about one part in $10^3$, but at that level it shows a
**dipole**: one hemisphere of the sky is slightly hotter, the opposite slightly
cooler. This is a Doppler effect from the observer's motion through the CMB rest
frame. An observer moving with speed $v = \beta c$ sees the temperature vary with
angle $\theta$ from the direction of motion as

$$
T(\theta) = T_0\,\frac{\sqrt{1 - \beta^2}}{1 - \beta\cos\theta}
\approx T_0\left(1 + \beta\cos\theta + \dots\right),
$$

a dipole of amplitude $\Delta T = \beta T_0$ to first order. The measured dipole
amplitude is $\Delta T \approx 3.4\ \text{mK}$, corresponding to

$$
\beta = \frac{\Delta T}{T_0} \approx 1.2 \times 10^{-3},
\qquad v \approx 370\ \text{km s}^{-1}.
$$

This is the velocity of the Solar System with respect to the CMB rest frame,
compounded of the Sun's orbit in the Galaxy, the Galaxy's motion in the Local
Group, and the Local Group's infall toward the Great Attractor. The dipole is a
kinematic foreground, subtracted before the primordial anisotropies are analyzed.

$$
% caption: The observer's motion through the CMB rest frame Doppler-shifts the
% radiation to a slightly hotter pole ahead and a cooler pole behind, producing
% the dipole anisotropy at the millikelvin level.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
% sphere outline
\draw[black, very thick] (0,0) circle (2.6);
% hotter pole (denser tint) on the right
\begin{scope}
\clip (0,0) circle (2.6);
\fill[acc!22] (2.6,0) ellipse (1.7 and 2.6);
\fill[acc!8] (-2.6,0) ellipse (1.7 and 2.6);
\end{scope}
\draw[black, very thick] (0,0) circle (2.6);
% motion arrow
\draw[->, very thick] (0,0) -- (2.3,0);
\node[anchor=south] at (1.4,0.1) {motion};
\node[black!70, anchor=west] at (2.7,0.9) {hotter pole};
\node[black!70, anchor=east] at (-2.7,0.9) {cooler pole};
\node[black, anchor=north] at (0,-2.9) {dipole anisotropy};
\end{tikzpicture}
$$

Recombination is not the last time the gas changes ionization state. When the
first stars and galaxies formed, at $z \sim 6$–$10$, their ultraviolet light
**reionized** the intergalactic hydrogen. Reionization scatters a small fraction
of CMB photons on their way to us, quantified by the **optical depth to
reionization** $\tau \approx 0.054$. This scattering slightly damps the
small-scale anisotropies and generates a large-scale polarization signal, from
which $\tau$ is measured; it is one of the six parameters of the concordance
model. Reionization does not erase the last-scattering surface — the universe is
far too dilute at $z \sim 8$ to become opaque again — but it is a second, partial
scattering screen between us and the CMB.

Once the dipole is removed, the residual temperature fluctuations are at the
$10^{-5}$ level and are primordial — the density perturbations at last scattering.
Their statistics, encoded in the angular power spectrum, are the subject of the
next lesson on [CMB anisotropies and cosmological
parameters](/astrophysics-cosmology/the-hot-big-bang/cmb-anisotropies-and-cosmological-parameters),
where the positions and heights of the acoustic peaks measure the geometry and
contents of the universe.

[^wmap-firas]: NASA WMAP mission and the COBE/FIRAS results — the blackbody spectrum at $T_0 = 2.725\ \text{K}$, the surface of last scattering at $z \approx 1090$, and the CMB dipole from the Solar System's motion. https://map.gsfc.nasa.gov