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The cosmic microwave background consists of relic photons from the early hot, nearly thermal photon-baryon fluid. Before cosmological recombination, abundant free electrons repeatedly scatter photons by Thomson scattering. Their mean free path ${\lambda }_{\gamma }=(n_e{\sigma }_T{)}^{-1}$ is short and their Thomson scattering rate $n_e{\sigma }_{T}c$ is much greater than the Hubble parameter. An earlier photon therefore cannot stream directly to us: subsequent scattering repeatedly changes its direction.

As cosmic expansion cools the plasma to about $T\simeq 3000\mathrm{K}$, electrons and protons combine into neutral hydrogen. The free-electron abundance drops rapidly, the scattering optical depth to the observer becomes of order one, and photon decoupling occurs. Since the cosmic microwave background temperature scales as $T=T_0(1+z)$, with $T_0\simeq 2.725\mathrm{K}$, this occurs at $z\simeq 1100$. The small baryon-to-photon ratio delays neutralization to well below $13.6\mathrm{eV}/k_B$: the high-energy tail contains enough photons to ionize hydrogen until the temperature is much lower than the Hydrogen binding energy expressed as a temperature. The observed cosmic microwave background is primarily this pre-existing radiation undergoing its last scattering; its nearly blackbody spectrum is then preserved by cosmological redshift.

For the requested estimate take a flat matter-dominated universe, $a(t)=(t/t_0{)}^{2/3}$. Restore $c$ in distances. The comoving particle horizon at cosmological recombination and the comoving radial distance to the Cosmic microwave background last-scattering surface areThe proper particle horizon is $a_{\ast }{r}_{\mathrm{hor},\ast }=3c{t}_{\ast }$ and the angular diameter distance is $D_A=a_{\ast }{\chi }_{\ast }$. Therefore the angular particle-horizon size in a power-law universe, taking one horizon radius as the transverse length, isThe angular diameter of a patch of that radius is about $3.56^{\circ }$. The factor of two is a radius-versus-diameter convention; either gives a horizon scale of only a few degrees. This is a particle horizon, rather than the smaller sound horizon. Including early radiation in cosmology changes the numerical estimate but preserves its order of magnitude.

The horizon problem is that the cosmic microwave background has almost the same temperature across the whole sky, including pairs of regions whose past light cones do not overlap in this noninflating model. Local thermal equilibration cannot explain that observation within this history. A sufficiently long period of cosmic inflation makes the comoving Hubble radius decrease: a region once small enough to share a causal past can expand to encompass our entire observed Cosmic microwave background last-scattering surface. After reheating, it can be homogeneous across many later particle horizon patches. The matter-only estimate integrates from a decelerating Big Bang; extending the earlier history by cosmic inflation changes that causal conclusion.

Two stages of the thermal history. Big Bang nucleosynthesis is nuclear assembly in an almost fully ionized, radiation-dominated plasma. Around $t\sim 1\mathrm{s}$, at temperatures of order $1\mathrm{MeV}/k_B$, the weak neutron–proton conversion rate falls below the expansion rate. The neutron-to-proton abundance freezes out at order $1/5$ and then decreases by neutron decay. The deuterium bottleneck delays effective nuclear assembly until $k_{B}T\sim 0.07$–$0.1\mathrm{MeV}$, at ages of a few minutes: the high photon-to-baryon ratio permits photodissociation despite the deuteron binding energy exceeding the mean photon energy. Once deuterium survives, reactions rapidly put most remaining neutrons into helium-4. A neutron-to-proton ratio near $1/7$ gives a helium mass fraction $Y_p\simeq 2(n/p)/(1+n/p)\simeq 1/4$, with hydrogen dominant by number and small residual deuterium, helium-3 and lithium-7 abundances. Expansion and low density prevent substantial heavy-element production. Nuclear binding is involved; neutral atoms do not yet survive.

Photon decoupling occurs much later, near a few thousand kelvin and a few hundred thousand years, after matter has overtaken radiation in the expansion dynamics. Electrons and nuclei form neutral atoms during cosmological recombination. The resulting decline of free-electron density reduces Thomson opacity, and photons begin to free stream over cosmological distances. Those photons form the observed cosmic microwave background. Atomic binding and scattering opacity, rather than new nuclear reactions, control this stage. Helium forms neutral atoms earlier than hydrogen. Recombination, photon last scattering, baryon drag release and the later thermal decoupling of the gas are related but not identical events.

The temperature falls through expansion: after the relevant entropy releases have ended, $T_{\gamma }\propto a^{-1}$ and number densities scale as $a^{-3}$. Big Bang nucleosynthesis lies in the radiation-dominated universe; photon decoupling lies after matter–radiation equality, when a matter-dominated approximation is useful but radiation is not exactly negligible.

Photon density and baryon abundance. The cosmic microwave background has a black-body radiation spectrum obeying Planck's law:where $u_{\nu }$ is spectral energy density, not a photon-number density. The measured temperature is about $2.72548\mathrm{K}$.^[ref] The mean photon energy in a black body isUsing the supplied rounded energy density givesThe exact black-body number integral at $2.725\mathrm{K}$ gives $4.1\times 10^8{\mathrm{m}}^{-3}$; the difference reflects rounding of the stated energy density. Either estimate is adequate.

A numerical photon-to-baryon ratio additionally needs a baryon density, which is not printed in this question. Make that observational input explicit rather than infer baryons from radiation alone. For illustration take the standard rounded physical density ${\Omega }_b{h}_H^2\simeq 0.022$, with $H_0=100h_H\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$. This is consistent with the cosmological parameter fits available around the paper's date.^[ref] It givesCounting baryons, including those bound in helium, rather than counting nuclei then givesBoth populations dilute as $a^{-3}$ after electron–positron annihilation, so this ratio also applies at hydrogen recombination in the standard adiabatic history.

Estimating decoupling. The large photon excess explains why setting $k_{B}T$ equal to the Hydrogen binding energy ${\chi }_H=13.6\mathrm{eV}$ is wrong. Even a small ionizing tail can oppose neutral-atom formation. The crude exponential balance $e^{-{\chi }_H/(k_{B}T)}\sim \eta$ gives ${\chi }_H/(k_{B}T)\sim \ln (1/\eta )\sim 21$, or $T\sim 7\times 10^3\mathrm{K}$. This only establishes the scale and the large suppression below ${\chi }_H/k_B$; it neglects phase-space factors and the opacity criterion.

A more controlled elementary Saha-Thomson decoupling estimate uses the hydrogen-only Saha equation. With $X_e=n_e/n_H$ and ground-state statistical weights combined in the usual way,The electron translational phase-space factor is significant inside the logarithm. This equilibrium relation places the sharp neutralization interval around $T\sim (3\text{–}4)\times 10^3\mathrm{K}$. Combine it with the approximate Thomson decoupling rate criterionFor $Y_p\simeq 0.25$, $n_{b0}\simeq 0.25{\mathrm{m}}^{-3}$, $H_0\simeq 68\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, and ${\Omega }_m\simeq 0.31$, solving these two approximate relations gives $T\simeq 3.1\times 10^3\mathrm{K}$, with equilibrium $X_e$ of order $10^{-2}$. Thus the appropriate estimate isThis is not a precision recombination calculation. Hydrogen-only equilibrium omits the kinetic bottleneck in escaping Lyman photons, two-photon processes, the detailed helium history and freeze-out; $\Gamma /H\sim 1$ is not identical to the maximum of the last-scattering visibility function. These approximations must be improved to predict an accurate free-electron fraction or a percent-level last-scattering redshift. The high photon-to-baryon ratio alone does not fix that detailed history.

The electrons are highly nonrelativistic: $k_{B}T/(m_e{c}^2)\sim 5\times 10^{-7}$, and their thermal root-mean-square speed is only about $10^{-3}c$. Photon coupling does not make their kinetic energies relativistic.

Age–temperature relation. During a nearly flat pressureless matter-dominated phase, with radiation, curvature and vacuum energy neglected in the expansion equation,Here the Big Bang time is chosen as $t=0$. Conserved photon entropy gives $T\propto a^{-1}$, soTo normalize this relation, use the matter-era expansion rate expressed in present-day measured parameters, rather than assume the late universe is still matter dominated:The same illustrative parameters give $t_{\mathrm{dec}}\sim 4.5\text{–}5\times 10^5\mathrm{yr}$ for $T_{\mathrm{dec}}\sim 3.0\text{–}3.1\times 10^3\mathrm{K}$. Radiation was still a meaningful correction; retaining it reduces the age, and a more complete recombination calculation gives the familiar scale of roughly $4\times 10^5\mathrm{yr}$.^[ref] A numerical age cannot follow from a proportionality $t\propto T^{-3/2}$ alone: an expansion or density normalization is essential. The small amount of dark energy at decoupling can be neglected even though it affects the present age.

As the temperature falls, electrons and protons form neutral hydrogen through cosmological recombination. The free electron fraction drops sharply, and so does the scattering rate ${\Gamma }_{\gamma }=n_e{\sigma }_{T}c$. When this rate becomes comparable to or smaller than the Hubble parameter, the plasma becomes transparent and photon decoupling occurs. The Thomson scattering cross-section itself does not abruptly vanish. The rapid change is in the abundance of free targets.

The Hydrogen binding energy is $13.6\mathrm{eV}$, but recombination occurs at a much smaller thermal energy, of order $0.3\mathrm{eV}$, or a few thousand kelvin. There are roughly a billion photons per baryon, so the high-energy tail can keep hydrogen ionized long after $k_{B}T$ drops below the binding energy. The falling exponential ionizing tail then makes the transition comparatively sharp. Last scattering follows when sufficiently few electrons remain free.^[ref]

To estimate a horizon, specify an expansion model as well as the redshift. The proper particle horizon at emission isAs a transparent analytic estimate, approximate a spatially flat matter-dominated universe by $a(t)=(t/t_{\mathrm{now}}{)}^{2/3}$ throughout. ThenTaking the illustrative present value $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, this model has $t_{\mathrm{now}}=2/(3H_0)\simeq 9.3\times 10^9\mathrm{yr}$. At $z_{\ast }=10^3$,This is a proper horizon radius at emission; its corresponding present comoving length is about $270\mathrm{Mpc}$.

For the same model the comoving distance to the Cosmic microwave background last-scattering surface and its angular diameter distance areTherefore the angular extent associated with one radial horizon length isThis is an order-one-degree estimate. Including the earlier radiation domination and late dark energy changes the coefficient; it does not make the horizon span the sky. A horizon patch diameter has twice this angle, whereas the acoustic scale uses a sound horizon and differs again. The redshift by itself, without an expansion model or normalization, is insufficient for a unique horizon distance.

The horizon problem is that the observed cosmic microwave background has nearly the same mean temperature across angular separations vastly greater than this scale. In the purely decelerating Hot Big Bang model, sufficiently separated last-scattering regions had neither time to exchange signals nor a common post-Big-Bang causal region that could establish their observed thermal uniformity. The puzzle concerns the missing causal explanation of those initial correlations; it is not an inconsistency of specifying uniform initial conditions mathematically. An earlier period of cosmic inflation can enlarge a formerly causally connected region beyond the later observable horizon.

Work in natural units $k_B=\hslash =c=1$, initially neglecting helium and excited hydrogen states. The reaction $p+e^{-}\leftrightarrow H+\gamma$ is in chemical equilibrium when ${\mu }_p+{\mu }_e={\mu }_H$, since equilibrium photons have zero chemical potential. Substitution of the nonrelativistic Maxwell-Boltzmann distribution for the three densities givesFor ground-state hydrogen use $g_e=g_p=2$, $g_H=4$, and $m_H\simeq m_p$ in the translational prefactor. Writing its Hydrogen binding energy as $I$, this becomes $(m_{e}T/(2\pi ){)}^{3/2}{e}^{-I/T}$. Charge neutrality gives $n_p=n_e=X_e{n}_B$, while baryon conservation gives $n_H=(1-X_e)n_B$. Hence the hydrogen-only Saha equation isThe photon Bose-Einstein distribution supplies $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, where $\zeta (3)$ is the usual Riemann zeta function value in the thermal number integral. With baryon-to-photon ratio ${\eta }_{\gamma }=n_B/n_{\gamma }$, substitution yieldsThis reproduces the requested numerical coefficient, with the printed $\xi (3)$ interpreted as the thermal $\zeta (3)$ value. There is a normalization ambiguity in the PDF: it writes $\eta =n_B/s$ without defining $s$. To reproduce this coefficient, $s$ must mean photon number density. If $s$ has its usual entropy density meaning, the Saha baryon-abundance normalization instead givesThe two forms agree through ${\eta }_{\gamma }=[{\pi }^4{g}_{\ast s}/(45\zeta (3))]{\eta }_s$, with the entropy sector specified. A baryon-to-entropy ratio cannot simply be inserted into the photon-normalized expression.

As the universe cools, the Boltzmann factor favors neutral atoms and cosmological recombination sharply lowers $X_e$. For an order-one ionized fraction the photon-normalized equation gives the balanceup to an order-one ionization-fraction factor. The small baryon abundance means that there are very many photons per baryon: even a small high-energy photon tail can ionize most atoms. Thus the recombination temperature is far below the microscopic binding scale $I\simeq 13.6\mathrm{eV}$, typically a few tenths of an eV for the standard abundance, rather than of order $I$.

Photon decoupling is a related but distinct kinetic event. The Thomson scattering rate $n_e{\sigma }_T$ falls relative to $H$ as free electrons disappear, and photons begin free streaming. The precise last-scattering epoch depends on integrated optical depth, not only the local equality of rates. At still lower density the effective recombination rate per electron also drops below the expansion rate. Residual electron freeze-out leaves a small nonzero ionized fraction, larger than the vanishingly small equilibrium Saha prediction; radiative atomic bottlenecks mean the entire history cannot be obtained from equilibrium alone.

The temperature comparison in the last request needs an interpretation. If “reionisation temperature” means the microscopic thermal ionization scale, the large photon-to-baryon ratio explains why cosmological recombination requires $T\ll I$. If it means later cosmic reionization, there is no universal thermal threshold: stars or other sources supply ionizing photons and can reheat gas. The cosmological photon background is actually colder at that later epoch, because $T_{\gamma }\propto a^{-1}$. Its temperature therefore cannot literally be higher than the background temperature at recombination merely because hydrogen is reionized.

Chemical equilibrium for $p+e^{-}\leftrightarrow H+\gamma$ implies ${\mu }_H={\mu }_p+{\mu }_e$, because equilibrium photons have zero chemical potential. Dividing the three nonrelativistic Maxwell-Boltzmann number densities therefore givesThe Hydrogen binding energy is $I=m_p+m_e-m_H=13.6\mathrm{eV}$ in natural units. For ground-state hydrogen, with hyperfine splittings unresolved, $g_H=4$ and $g_p=g_e=2$, so the degeneracy ratio is one. Equivalently one can omit nuclear-spin degeneracy consistently from both the proton and atom. Excited hydrogen levels are exponentially suppressed at recombination temperatures. Also $m_H/m_p=1+O(m_e/m_p)$, so its effect in the prefactor is negligible. ThusThe sign of the exponential follows from the smaller rest mass of the bound atom: cooling favours neutral hydrogen once translational entropy no longer compensates the binding energy.

There are about ${\eta }^{-1}\sim 10^9$ photons per baryon. Even when the typical photon energy is below the Hydrogen binding energy, the high-energy tail can provide abundant ionizing photons. The nonrelativistic electron's translational phase space also favours the ionized state. Quantitatively, the Saha equation contains both suppressing factors $\eta$ and $(T/m_e{)}^{3/2}$. Recombination needs their tiny product to be overcome by $e^{I/T}$, so $I/T$ must be several tens, not of order one.

For an illustrative ${\Omega }_B{h}^2=0.02$, the supplied baryon-to-photon relation gives $\eta =5.36\times 10^{-10}$. With $m_e=5.11\times 10^5\mathrm{eV}$, direct evaluation givesThus the equilibrium ionization fraction reaches a few per cent just below $0.3\mathrm{eV}$, while at $13.6\mathrm{eV}$ it is essentially unity. The exact temperature depends weakly on the baryon abundance and on the fraction used to define cosmological recombination. This is small baryon abundance delays hydrogen recombination. Recombination, photon decoupling and residual electron freeze-out are related but distinct: the last two involve rates and cannot be determined by chemical equilibrium alone.

For $p+e\rightleftharpoons H+\gamma$, chemical equilibrium and the vanishing photon chemical potential imply ${\mu }_H={\mu }_p+{\mu }_e$. Taking the ratio of the three nonrelativistic Maxwell--Boltzmann number densities cancels these chemical potentials:The ground-state hydrogen atom has four combined electron and proton spin states if the small hyperfine splitting is ignored, whereas $g_p=g_e=2$. Hence $g_H/(g_p{g}_e)=1$. With Hydrogen binding energy $B=M_p+M_e-M_H$ and $M_H\simeq M_p$, the Saha ionization equation givesThe approximation in the prefactor neglects relative corrections of order $(M_e+B)/M_p$; the Hydrogen binding energy must be retained in the exponential.

Chemical equilibrium imposes ${\mu }_e+{\mu }_p={\mu }_H+{\mu }_{\gamma }$. For blackbody photons, ${\mu }_{\gamma }=0$. Define the positive Hydrogen binding energy by $I=(m_e+m_p-m_H)c^2$, approximately $13.6\mathrm{eV}$. Multiplying the two Maxwell-Boltzmann distribution number densities and dividing by the neutral-hydrogen density givesUse $g_e{g}_p/g_H=1$ and $m_p/m_H\simeq 1$ to obtain the Saha ionization equationThe assumptions are nonrelativistic, dilute, nondegenerate species at a common temperature, ground-state hydrogen with the stated degeneracy factors, negligible mass correction in the prefactor, and sufficiently rapid reactions to maintain chemical equilibrium.

In chemical equilibrium, the reaction $p+e\leftrightarrow H+\gamma$ obeys chemical-potential balance for ionizationbecause a photon gas in equilibrium has photon chemical potential ${\mu }_{\gamma }=0$. Applying the given nonrelativistic Maxwell-Boltzmann distribution to each massive species givesThe Hydrogen binding energy is $I=m_e+m_p-m_H$ in units with $c=1$. Thus Saha's equation isFor ground-state hydrogen, $g_e=g_p=2$ and $g_H=4$, so the degeneracy factor is one. Since $m_p\simeq m_H$, the translational factor is commonly written $(2\pi m_e{k}_{B}T/h^2{)}^{3/2}$.

Charge neutrality gives $n_e=n_p$. Write $n_B=n_e+n_H\simeq \eta n_{\gamma }$. Thenand henceCombining this with the stated photon density produces

In our universe the baryon-to-photon ratio is tiny, so there are roughly $10^9$ photons per baryon. Even when the mean photon energy is far below $I$, the high-energy blackbody radiation tail contains enough ionizing photons to suppress neutral hydrogen. The recombination temperature is reached only after the Boltzmann factor overwhelms this large photon-to-baryon ratio, at $k_{B}T\simeq 0.3\mathrm{eV}$.

For a universe with $\eta =1$, take the degeneracy factor as one, $m_H\simeq m_p$, and define $x=I/(k_{B}T)$. Neutral formation occurs when the boxed ratio is of order one, giving approximatelyWith $I=13.6\mathrm{eV}$ and $m_e=5.11\times 10^5\mathrm{eV}$, this gives $x\simeq 19$ and thereforeto order unity. This is much hotter than recombination in our universe because there is no enormous excess of photons, but it remains below $I$ because the electron's large translational phase space still favours the ionized state.

Chemical equilibrium for $p+e^{-}\rightleftharpoons H+\gamma$ gives ${\mu }_p+{\mu }_e={\mu }_H$, because the photon chemical potential vanishes. Applying the nonrelativistic number-density formula to all three massive species and taking the ratio givesUse charge neutrality $n_p=n_e$, the Hydrogen binding energy $E_{\mathrm{bind}}=m_p{c}^2+m_e{c}^2-m_H{c}^2$, $m_H\simeq m_p$, and the ground-state statistical weights $g_H=g_p{g}_e$. ThenThe approximations are a dilute, nondegenerate ideal gas in thermal and chemical equilibrium, nonrelativistic electrons, protons and hydrogen, negligible excited hydrogen states, and negligible corrections from the proton-to-hydrogen mass ratio. This relation is the Saha ionization equation.

In the nonrelativistic regime,The assumption $E-\mu \gg k_{B}T$ puts the Fermi-Dirac distribution in its Maxwell-Boltzmann limit, soTherefore the nonrelativistic Maxwell--Boltzmann number density is

For $p+e^{-}\leftrightarrow H+\gamma$, chemical equilibrium and the vanishing photon chemical potential implyApplying the number-density formula to each massive species and eliminating the chemical potentials givesUse charge neutrality, $n_p\simeq n_e$, the mass approximation $m_H\simeq m_p$, and the convention in the question that suppresses the order-one internal-degeneracy ratio $g_e{g}_p/g_H$. With the Hydrogen binding energywe obtain the Saha ionization equationThe assumptions are thermal equilibrium and chemical equilibrium before cosmological recombination, a nonrelativistic nondegenerate gas, zero photon chemical potential, charge neutrality, negligible proton-electron mass correction in the translational prefactor, and the stated degeneracy-factor approximation.

For hydrogen ionization fraction $X_e$ of order one, the hydrogen-only Saha equation gives $1\sim n_b(2\pi /(m_{e}T){)}^{3/2}{e}^{B/T}$. The baryon-to-photon ratio and photon number density imply $n_b\sim \eta T^3$. Therefore $1\sim \eta (T/m_e{)}^{3/2}{e}^{B/T}$ and the displayed logarithmic balance follows. A small $\eta$ supplies many ionizing photons per baryon, so neutral hydrogen forms at a temperature much smaller than its Hydrogen binding energy. The numerical temperature depends on the ionization-fraction convention and on whether equilibrium remains valid.