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Let ${\epsilon }_s$ be the energy of one adsorbed atom, using the same energy reference as the gas-phase value ${\epsilon }_0$. For equivalent, noninteracting adsorption sites, choosing the occupied positions givesEvery configuration has energy $N_1{\epsilon }_s$, so the canonical lattice-gas partition function isThere is no further $1/N_1!$ factor: the combinatorial choice already counts indistinguishable adsorbates correctly. The Helmholtz free energy, using Stirling's approximation for a macroscopic number of occupied and empty sites, isHence the adsorbed-phase chemical potential at fixed $T,M$ isThe logarithm comes from configurational entropy: adding an atom is less favorable when vacancies are scarce. In the gas, the stipulated expression can be writtenwhere $\Lambda$ is the thermal de Broglie wavelength. Equality of chemical potentials at chemical equilibrium givesThereforeThe binding energy $E_b={\epsilon }_0-{\epsilon }_s$ is positive for favorable binding. The units of $b$ are inverse pressure, making $bp$ dimensionless. Choosing the adsorbed energy as zero simply sets ${\epsilon }_s=0$ and leaves the same physics; it must not remove the relative binding energy. This is the Langmuir isotherm, with low-pressure linear uptake and high-pressure monolayer saturation.

At chemical equilibrium, adsorption and desorption balance:Canceling $M$ and rearranging givesThus $K_{\mathrm{ads}}=b$ when the kinetic and thermodynamic descriptions refer to the same ideal surface and energy convention. In particular,The ratio fixes the adsorption isotherm, while the individual rate constants also determine relaxation toward equilibrium. Here $k_a$ has units of inverse pressure per time and $k_d$ of inverse time, so $K_{\mathrm{ads}}$ is dimensional. This equivalent-site arrival/departure argument is the simple case of the Langmuir adsorption model.^[ref]

At chemical equilibrium, ${\mu }_{\mathrm{AB}}={\mu }_{\mathrm{A}}+{\mu }_{\mathrm{B}}$. Using the number-concentration form givesHere $\Delta {\epsilon }_0$ is the reaction energy between the chosen molecular reference states. Rearranging,Define the dimensionless concentration equilibrium constant byThusThe standard concentration $c^{\circ }$ has the same units as $c_i$. It converts concentration ratios to ideal thermodynamic activities; changing it changes the numerical dimensionless $K_c$ for this dissociation, which creates an extra particle. The unnormalized quotient $c_{\mathrm{A}}{c}_{\mathrm{B}}/c_{\mathrm{AB}}$ instead has concentration units.^[ref]

The exponential function makes modest Gibbs free energy differences correspond to extreme equilibrium ratios. At 298 K, $RT\simeq 2.48\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; for example,The minor equilibrium component may therefore lie below detection, or a species' required thermodynamic activity may be experimentally inaccessible. Apparently complete conversion supplies a bound on $K$, rather than an accurate finite value. Additionally, some reactions do not reach chemical equilibrium within a practicable observation time because of kinetic barriers. The thermodynamic identity still holds even when direct room-temperature determination of both sides of the equilibrium ratio fails.

At chemical equilibrium, the forward and reverse event rates are equal, rather than both reactions stopping:For the dilute-solution concentration model,The ratio is dimensionless because both concentrations and both rate constants have matching units. An exact thermodynamic equilibrium constant uses activities; here the common standard concentration and ideal-solution activity coefficients cancel.

This labelled part introduces the reversible reaction pair without a separate calculation to perform. For the following derivations, define its forward and reverse elementary reaction rates asand the initiating rate as $r_1=k_1[\mathrm{N}{\mathrm{O}}_2][{\mathrm{O}}_3]$. The law of mass action gives $k_2$ bimolecular units and $k_{-2}$ first-order units. Reversibility does not require $r_2=r_{-2}$ while the complete network is reacting: that equality would impose chemical equilibrium on the pair, an extra assumption not needed here.

Assume an ideal nondegenerate hydrogen gas in thermal equilibrium and chemical equilibrium, with ground-state atoms and charge neutrality. Let $n_p=n_e=X{n}_B$ and $n_H=(1-X)n_B$. The ground-state Saha equation, with the usual electron/proton spin factors cancelling the atomic ground-state degeneracy, isInvert it and use $n_B=n_{\gamma }/s=(0.24/s)(k_{B}T/(\hslash c){)}^3$. This givessoHere $m_e$ is the electron mass denoted $M_e$ in the source, and $0.24(2\pi {)}^{3/2}=3.78$. At $X=1/2$, the left side is two. At $k_{B}T$ of order $B$, the right side is extremely small because $1/s\sim 10^{-9}$ and $B/(m_e{c}^2)\ll 1$. A large exponential is needed to compensate, requiring $B/(k_{B}T)\gg 1$ rather than a temperature of order the binding energy.

Physically, there are enormously many photons per baryon, so even a small high-energy tail of the thermal photon distribution can maintain ionization. Recombination is delayed until that tail is sufficiently depleted. This explains the low recombination temperature without solving the equilibrium equation; actual cosmological recombination also eventually requires rate equations once equilibrium fails.

For a photon gas in thermal and chemical equilibrium, photon number is not a conserved thermodynamic charge. Number-changing processes can create and absorb photons without a corresponding conserved-number constraint, so the equilibrium photon chemical potential is zero.

At $T=5\times 10^9\mathrm{K}$, $k_{\mathrm{B}}T\simeq 0.431\mathrm{MeV}$. The other light species are electrons, Positrons, three neutrino species and their antiparticles. Neutrinos are ultrarelativistic; electrons and Positrons are only moderately relativistic at this particular temperature because $m_e{c}^2=0.511\mathrm{MeV}$. The later simple counting formulas use the intended ultrarelativistic approximation for them. Baryons, muons and heavier particles are not an abundant relativistic bath at this temperature.

Pair equilibrium implies ${\mu }_{e^{-}}+{\mu }_{e^{+}}=2{\mu }_{\gamma }=0$. Approximate charge symmetry, with the net electron excess tied to the much smaller baryon asymmetry, then makes both electron and Positron chemical potentials small compared with $k_{\mathrm{B}}T$. For neutrinos one additionally assumes negligible lepton asymmetry. Their near-zero chemical potentials are not implied merely by their statistics. After neutrino decoupling, a distribution initially having negligible chemical potential retains a redshifted Fermi-Dirac distribution, with its own temperature.

Use $X_i=n_i/n_B$, so $X_D$ counts deuterium nuclei per baryon, not baryons bound into deuterium. For the radiative capture reaction, chemical equilibrium and the zero photon chemical potential give ${\mu }_D={\mu }_n+{\mu }_p$. Dividing the three nonrelativistic Maxwell--Boltzmann number densities removes all chemical potentials:Here $g_D=3$ and $g_n=g_p=2$. The binding convention in this PDF is a negative mass defect, $B_D=m_D-m_n-m_p<0$. The positive nuclear binding energy is ${\epsilon }_D=-B_D$, so the abundance is enhanced by $e^{-B_D/T}$, not by $e^{B_D/T}$.

Since $X_D/(X_n{X}_p)=n_B{n}_D/(n_n{n}_p)$, insert the baryon-to-photon ratio $n_B=\eta n_{\gamma }$ and the photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$. This givesAt equal nucleon mass $m_N$ in the prefactors, $m_D\simeq 2m_N$, the deuterium equilibrium abundance simplifies toThe dimensionless prefactor is about $8.14$. These expressions describe thermal equilibrium; residual primordial deuterium after nuclear reactions cease must instead be obtained from the reaction kinetics.

Use natural units with $c=\hslash =k_B=1$ and the unreduced Planck mass $M_{\mathrm{Pl}}=G^{-1/2}$. In a radiation-dominated universe, the effective number of relativistic energy degrees of freedom $g_{\ast }=\mathcal{N}$ givesThe ratio of weak conversion rate to expansion rate is proportional to $T^3$, so it decreases as the universe cools. Cosmological weak freeze-out occurs approximately at $\Gamma =H$, givingFor $g_{\ast }=10.75$, $G_F=10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}=1.22\times 10^{19}\mathrm{GeV}$ this estimate is $1.65\mathrm{MeV}$. Equivalently $g_{\ast }^{1/6}\simeq 1.5$ is only a mild multiplicative correction. The rate itself was given only to order of magnitude, so this is a decoupling scale, not a precise prediction for the weak freeze-out abundance.

Both neutrons and protons are nonrelativistic at this temperature. The nonrelativistic Maxwell--Boltzmann number density follows by the Gaussian momentum integral:Their equal spin degeneracies are $g_n=g_p=2$. Chemical equilibrium of the weak reaction gives ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$; with negligible lepton asymmetry, ${\mu }_n\simeq {\mu }_p$. Taking the ratio and setting the nearly equal nucleon masses equal only in the prefactor gives the neutron-proton chemical equilibrium resultThe common baryon normalization cancels. Keeping the exact mass prefactor changes this by only order $Q/m_p$. Thus the printed simple exponential is the usual nonrelativistic equal-mass-prefactor approximation. Using $Q\simeq 1.29\mathrm{MeV}$ with the rough decoupling scale above gives a ratio of order one half; the rate's omitted numerical factors and gradual freeze-out must be included before using a more accurate neutron abundance.

Use radiation-era $H=C_{\ast }{T}^2/m_{\mathrm{pl}}$, with $C_{\ast }=1.66\sqrt{g_{\ast }}$. The stipulated interaction rate is ${\alpha }_Y^4{T}^5/m_Y^4$, not ${\alpha }_Y^2{T}^5/m_Y^4$. Decoupling therefore givesAt the threshold $T_D=m_{Y\mathrm{nr}}$, the assumed $g_{\ast }\simeq 100$ is self-consistent with the stated high-temperature condition. Smaller masses decouple nonrelativistically, with $x_D=m_Y/T_D=(m_{Y\mathrm{nr}}/m_Y{)}^{1/3}\gg 1$.

For an thermal equilibrium species with zero chemical potential, $n_Y=2(m_Y{T}_D/2\pi {)}^{3/2}{e}^{-x_D}$, while $s=2{\pi }^2{g}_{\ast s}{T}_D^3/45$. ThusThe numerical coefficient before rounding is about $2.9\times 10^{-3}$ for $g_Y=2$ and $g_{\ast s}=100$. This is mass-dependent fermion freeze-out. It assumes the rate maintains chemical equilibrium until sudden cosmological particle freeze-out and that entropy is conserved afterwards. Elastic scattering alone would not justify the thermal equilibrium number-density estimate. For decoupling below the high-temperature range, changes in effective degrees of freedom modify the numerical estimate; the requested formula treats them as roughly constant.

Above the transition,for $g_{\ast }\sim 100$. The ratio increases during cooling, rather than at higher temperature. A GUT transition at $T_c\sim 10^{15}$–$10^{16}$ GeV occurs well above this scale, so the interaction is too slow to establish or maintain thermal equilibrium throughout $T>T_c$. In fact its integrated interaction probability from an arbitrarily high temperature down to $T_c$ is at most of order $T_{\mathrm{eq}}/T_c\ll 1$, since $dt=-dT/(HT)$.

An initial Fermi-Dirac distribution with zero chemical potential therefore cannot be assumed from this high-temperature rate alone. If imposed initially and collisionlessly redshifted while the particles are massless, its thermal-looking form can persist, but this is not dynamically maintained thermal equilibrium. Other initial states can remain nonthermal. This is the thermalization temperature for a massless gauge mediator.

The below-transition rate stipulated in the model has $\Gamma /H=(T/T_D{)}^3$. If $T_c\gg T_D$ and it includes efficient number-changing reactions, it can restore chemical equilibrium after the transition and erase the initial-state memory, making the estimates in a and b applicable. If it only equilibrates momenta, or if reheating never populates the relevant range, the thermal relic calculation must be reconsidered.

At neutrino decoupling, the interaction rate is comparable with the Hubble parameter. Equating the supplied rates givesHere the Planck mass is the unreduced $m_P=G^{-1/2}\simeq 1.22\times 10^{19}\mathrm{GeV}$ in natural units, $G_F\simeq 10^{-5}{\mathrm{GeV}}^{-2}$, and the effective number of relativistic energy degrees of freedom is $g_{\ast }=10.75$. The supplied rate gives an order-of-magnitude neutrino decoupling estimate; a precise neutron-proton conversion temperature uses its own weak-rate coefficients.

The chemical equilibrium condition for $n+{\nu }_e\leftrightarrow p+e^{-}$ is ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. With negligible lepton chemical potentials, ${\mu }_n={\mu }_p$. The nonrelativistic Maxwell--Boltzmann number density therefore givesbecause $g_n=g_p=2$ and the small mass difference can be neglected in the prefactor. This is the neutron-proton chemical equilibrium relation.

For the deuterium equilibrium abundance, ${\mu }_D={\mu }_n+{\mu }_p$ because photons have zero chemical potential. The sign convention in the paper is important: $B=m_D-m_n-m_p<0$, while the positive nuclear binding energy is ${\Delta }_D=-B$. Using spin multiplicities $g_D=3$, $g_n=g_p=2$ givesWith $n_B=\eta n_{\gamma }$, $m_n\simeq m_p\simeq m_N$, $m_D\simeq 2m_N$ and photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, this becomesHere $X_D=n_D/n_B$ is a number fraction; the fraction of baryons bound in deuterium is $2X_D$. The formula estimates the onset of the deuterium bottleneck opening while depletion of free neutrons and protons remains small. Because $\eta \ll 1$, stable deuterium becomes abundant only far below the positive nuclear binding energy, not simply at $T={\Delta }_D$.

Let $T_f$ denote cosmological weak freeze-out of the neutron-to-proton ratio, $r_f=e^{-Q/T_f}$, and let $t_N$ mark efficient nuclear burning. Free-neutron decay after freeze-out givesAlmost every surviving neutron enters helium-4, whose atomic nucleus contains two neutrons and four baryons. Thus neutron-limited helium synthesis givesA longer neutron half-life leaves more neutrons available, increasing the primordial helium mass fraction. If the longer lifetime reflects weaker weak interactions, the accompanying earlier cosmological weak freeze-out can reinforce that effect.

During radiation domination, $H\propto \sqrt{G{g}_{\ast }}{T}^2$ whereas the weak conversion rate is proportional to $T^5$. Hence $T_f\propto (G{g}_{\ast }{)}^{1/6}$. Increasing the effective number of relativistic energy degrees of freedom raises $T_f$, increasing $r_f$, and shortens the interval for free-neutron decay after freeze-out. Both increase the primordial helium mass fraction while nuclear burning remains efficient. The same reasoning gives the primordial helium response to a varying gravitational constant: larger $G$ over this interval produces more helium, with fixed microscopic rates and the approximate Friedmann equation unchanged. Smaller $G$ has the opposite effect. A particular varying-$G$ model must specify its background expansion and time history; the simple trend must not be extrapolated to changes so large that nuclear burning itself fails.

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.

Use natural units $\hslash =c=k_B=1$ and the unreduced Planck mass $M_P=G^{-1/2}$. Neutrino decoupling occurs when the weak interaction rate falls below the Hubble parameter. The weak-decoupling temperature estimate equates the given rate with the radiation-era expansion rate:Here $g_{\ast }$ counts the effective number of relativistic energy degrees of freedom, and the Fermi constant is $G_F\simeq 10^{-5}{\mathrm{GeV}}^{-2}$. The omitted reaction-rate coefficients make this an order-of-magnitude estimate: decoupling occurs at a temperature of order one MeV.

The charge-conserving channel is $n+{\nu }_e\leftrightarrow p+e^{-}$, together with $p+{\bar{\nu }}_e\leftrightarrow n+e^{+}$. The positron sign in the printed neutron-capture channel violates electric charge conservation. Chemical equilibrium requires ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Negligible lepton chemical potentials therefore give ${\mu }_n\simeq {\mu }_p$. Since neutrons and protons are nonrelativistic particles, their Maxwell-Boltzmann distribution givesBoth nucleons have two spin states, and their masses agree closely in the prefactor. Thus neutron-proton chemical equilibrium impliesThe heavier neutron is suppressed by a Boltzmann factor.

For the deuterium equilibrium abundance, retain the paper's convention $B=m_D-m_n-m_p<0$; the positive nuclear binding energy is ${\epsilon }_D=-B\simeq 2.22\mathrm{MeV}$. Chemical equilibrium of $n+p\leftrightarrow D+\gamma$ gives ${\mu }_D={\mu }_n+{\mu }_p$ because the equilibrium photon chemical potential vanishes. Using the nonrelativistic Maxwell--Boltzmann number density for each species and dividing givesHere $g_D=3$ and $g_n=g_p=2$. With the baryon-to-photon ratio $\eta =n_B/n_{\gamma }$, the photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, and $m_D\simeq 2m_N$, $m_n\simeq m_p\simeq m_N$, this becomesIn this convention $X_D=n_D/n_B$ counts nuclei, so its contribution to the fraction of baryons in deuterium is $2X_D$.

The deuterium bottleneck explains why helium-4 formation waits until $T$ is far below ${\epsilon }_D$. At higher temperature, there are vastly more photons than baryons; even the dilute energetic tail can destroy newly formed deuterium. A crude estimate balances ${\eta }^{-1}{e}^{-{\epsilon }_D/T}$ against one, giving $T\sim {\epsilon }_D/\log ({\eta }^{-1})\sim 0.1\mathrm{MeV}$ for $\eta$ of order $10^{-9}$. The power-law factors in the deuterium equilibrium abundance and the nuclear reaction rates refine this estimate. Once stable deuterium accumulates, further reactions can efficiently place nearly all surviving neutrons into helium-4.

Write $r_N=n_n/n_p$ at the onset of nuclear burning. Each helium-4 nucleus requires two neutrons and contains four baryons. Neutron-limited helium synthesis therefore gives the primordial helium mass fractionFor a realistic rough cosmological weak freeze-out estimate, $r_f$ is about $1/6$ and free-neutron decay after freeze-out reduces it to about $1/7$, giving roughly one quarter of the baryonic mass in helium-4. The survival relation that conserves baryon number is $X_n(T_N)=r_f(1+r_f{)}^{-1}{e}^{-\Delta t/{\tau }_n}$; the corresponding $r_N$ is $X_n/(1-X_n)$.

The rough neutrino decoupling rate by itself does not determine that one-quarter value. Substituting $T_d=1.8\mathrm{MeV}$ literally into $e^{-Q/T_d}$ would give $r\simeq 0.49$ and $Y_4\simeq 0.65$ before neutron decay. Neutron-proton chemical equilibrium persists to a lower temperature, approximately $0.7$--$0.8\mathrm{MeV}$ once the actual conversion rates are included. The rate scaling fixes the MeV scale; accurate abundances need the conversion coefficients and the delay through the deuterium bottleneck. This sequence is described in David Tong's discussion of nucleosynthesis.

Increasing $Q$ reduces the equilibrium neutron-to-proton ratio, so the primordial helium abundance decreases. Holding the freeze-out temperature $T_f$ and the decay survival factor fixed, the primordial helium response to the neutron-proton mass difference isFor $Q\simeq 1.29\mathrm{MeV}$ and $T_f=0.7$--$0.8\mathrm{MeV}$ this is about $0.85$--$0.87$, a reduction of roughly fifteen percent. A larger $Q$ also changes neutron decay and detailed weak rates; their quantitative effects require more microscopic information. The sign from the Boltzmann factor is already clear.

In a large volume, one one-particle state occupies momentum-space volume $h^3/V$. A spherical shell of radius $p$ has volume $4\pi p^{2}dp$, so for spin degeneracy $g$ the density of states is $4\pi gV{p}^{2}dp/h^3$. With ultra-relativistic dispersion $E=cp$ and zero chemical potential, the Bose-Einstein distribution givesThe convergent dimensionless integral is ${\pi }^4/15$, obtained by expansion into exponentials and summing $\sum n^{-4}$. Therefore $ϵ\propto T^4$ and $\alpha =4$. Photons are massless bosons with two polarizations; equilibrium emission and absorption do not conserve photon number and force ${\mu }_{\gamma }=0$.

For recombination, chemical equilibrium imposes ${\mu }_p+{\mu }_e={\mu }_H+{\mu }_{\gamma }={\mu }_H$, including rest energies in the chemical potentials. Divide the three nonrelativistic number-density expressions to obtainHere $m_p{c}^2+m_e{c}^2-m_H{c}^2=I$. Neglect the relative proton/hydrogen mass difference and take the ground-state degeneracy ratio $g_p{g}_e/g_H=1$ (electron and proton spins give $2\cdot 2$ and hydrogen gives $4$). Charge neutrality gives $n_p=n_e$. This yields the stated Saha equation

With $n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$, the ratio asked for isEquivalently the prefactor is $16\pi \zeta (3)\eta (2\pi {)}^{-3/2}[kT/(m_e{c}^2){]}^{3/2}$. Because photons vastly outnumber baryons, even a small high-energy tail can keep hydrogen ionized. Neutralization therefore requires exponential suppression of that tail far below $kT=I$, explaining the much lower recombination temperature. The equilibrium formula is an estimate; the actual expanding-universe recombination kinetics eventually depart from Saha equilibrium.

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.