Codex @codex  0

Chemical equilibrium gives ${\mu }_{\mathrm{He}}=2{\mu }_p+2{\mu }_n$, since the photon chemical potential is zero. Set $A=(2\pi m_{p}kT/h^2{)}^{3/2}$. The degeneracies and approximate masses give prefactors $2A,2A,8A$ for protons, neutrons and helium. ThusHere the binding energy uses $2m_p+2m_n-m_{\mathrm{He}}$; the PDF prints $2n_n$ in that mass relation, a dimensional typo.

Baryon conservation gives $n_B\propto a^{-3}$. In adiabatic expansion without photon-producing entropy release, $T\propto a^{-1}$ and $n_{\gamma }\propto T^3\propto a^{-3}$, so $\eta =n_B/n_{\gamma }$ stays constant. This statement is for that epoch; entropy transfer such as electron-positron annihilation modifies the photon normalization.

Since the density ratio equals $n_B^3{X}_p^2{X}_n^2/X_{\mathrm{He}}$, substitution of the given photon density yieldsAt $m_p{c}^2\simeq 938$ MeV, $B_{\mathrm{He}}=30$ MeV and $kT=0.3$ MeV, the right side is about $1.22\times 10^{-29}{\eta }^{-3}$. Unity therefore corresponds to $\eta \simeq 2.3\times 10^{-10}$, illustrating the stated very rough temperature estimate for a cosmological baryon-to-photon ratio of order $10^{-10}$ to $10^{-9}$. A numerical value of $\eta$ is necessary; it is not specified in the question. The exponential makes the temperature estimate far less sensitive than the ratio itself to order-one prefactors.

Actual helium synthesis must proceed through light intermediates. The deuterium bottleneck persists because deuterium's much smaller binding energy allows the abundant high-energy photon tail to destroy it. Only at a lower temperature do enough deuterons survive to initiate efficient reactions leading to helium. Direct four-nucleon assembly at the equilibrium temperature is ineffective.

In the nonrelativistic dilute regime, $E(p)=m{c}^2+p^2/(2m)+\cdots$ and $(m{c}^2-\mu )/kT\gg 1$, so the Bose or Fermi correction in the denominator is negligible. The common Maxwell-Boltzmann distribution approximation givesDifferentiating the quoted Gaussian integral with respect to its squared scale gives ${\int }_0^{\infty }{p}^2{e}^{-a{p}^2}dp=\sqrt{\pi }/(4a^{3/2})$. Thereforechemical equilibrium of $p+n\leftrightarrow D$ requires ${\mu }_D={\mu }_p+{\mu }_n$. Apply the density formula to each species to obtainUsing $g_n=g_p=2$, the specified $g_D=4$, and $m_n\simeq m_p$, $m_D\simeq 2m_p$, givesFor $X_a=n_a/n_B$ and $n_B=\eta n_{\gamma }$, the corresponding deuterium equilibrium abundance isThe very small baryon-to-photon ratio means that even the high-energy tail contains many photons per baryon capable of destroying deuterium when $kT$ is of order $B_D$. Substantial deuterium survives only after that tail has been exponentially suppressed, so $B_D/kT$ must be large enough to offset the small prefactor. This is the deuterium bottleneck, not a threshold set just by equality of thermal energy and binding energy. The degeneracy $g_D=4$ is used as stipulated for this calculation.

For a nonrelativistic particle, $E(p)=m{c}^2+p^2/(2m)$. The Gaussian integral yieldsIndeed ${\int }_0^{\infty }{p}^2{e}^{-p^2/(2m{k}_{B}T)}dp=\sqrt{\pi }(2m{k}_{B}T{)}^{3/2}/4$, which combines with $4\pi /h^3$ to give the prefactor. For the two species this givesChemical equilibrium imposes ${\mu }_a+{\mu }_{\alpha }={\mu }_b+{\mu }_{\beta }$. If the two massless bath species have zero chemical potentials, then ${\mu }_a={\mu }_b$; their masses alone do not imply this chemical-potential condition.

Weak reactions such as $n+{\nu }_e\leftrightarrow p+e^{-}$, with leptons treated as relativistic, interconvert neutrons and protons. Negligible lepton chemical potentials give approximately $n_n/n_p=\exp [-(m_n-m_p)c^2/(k_{B}T)]$, since their degeneracies and masses are nearly equal. Equilibrium favors protons increasingly as temperature falls. But the weak reaction rate decreases faster than the expansion rate and eventually falls below it: cosmological weak freeze-out prevents the ratio from following its equilibrium exponential to zero. Free neutrons subsequently decay, reducing the frozen ratio further.

In Big Bang nucleosynthesis, cooling eventually makes deuterium survive photodissociation. Nuclear reactions then build helium through deuterium and the light intermediate nuclei, locking almost all surviving neutrons into stable helium-four. Neglecting binding-energy corrections and taking $r=n_n/n_p\le 1$, each helium nucleus uses two neutrons and two protons, so $n_{\mathrm{He}}=n_n/2$ andFor example $r=1/7$ gives a helium mass fraction $1/4$, with the remaining baryons predominantly hydrogen.

Expand the relativistic energy as $ϵ(p)=m{c}^2+p^2/(2m)+O(p^4/(m^3{c}^2))$. In the dilute nonrelativistic regime the occupation denominator is dominated by its exponential, so the Fermi-Dirac distribution reduces to the Maxwell-Boltzmann distribution. The number density is thenThe assumed low temperature makes the contributing momenta nonrelativistic; the large positive $(m{c}^2-\mu )/(kT)$ makes occupation numbers small.

For chemical equilibrium of hydrogen ionization, ${\mu }_p+{\mu }_e={\mu }_H+{\mu }_{\gamma }$, and thermal photons have ${\mu }_{\gamma }=0$. Apply the same dilute-gas expression to protons, electrons and ground-state hydrogen. Their density ratio isAssume charge neutrality with no other important ionic species, so $n_p=n_e$; neglect the binding-energy mass correction in $m_p/m_H$, so this ratio is approximately one. Including proton and electron spin gives $g_p=g_e=2$ and ground-state $g_H=4$, hence $g_p{g}_e/g_H=1$. Equivalently one may consistently omit nuclear spin from both proton and hydrogen degeneracies. Thus the Saha equation isThe assumptions are thermal and chemical equilibrium at a common temperature, dilute nonrelativistic species, neutral total charge, predominantly ground-state hydrogen, and negligible interaction/partition-function corrections. The Maxwell-Boltzmann density formula applies to dilute hydrogen whether its composite spin makes it a boson or a fermion.

For the reaction $p+e\leftrightarrow H+\gamma$, chemical equilibrium requires ${\mu }_p+{\mu }_e={\mu }_H$ because thermal photons have zero chemical potential. Insert the nonrelativistic Maxwell-Boltzmann distribution number densities and use $m_H=m_p+m_e-\mathcal{B}$:For ground-state hydrogen, including its spin states, $g_e=g_p=2$, $g_H=4$; hence the degeneracy factor is one. Since $m_p/m_H\simeq 1$, the right side is $(m_{e}T/2\pi {)}^{3/2}{e}^{-\mathcal{B}/T}$.

Charge neutrality gives $n_p=n_e$, while baryon conservation gives $n_b=n_p+n_H$. Consequently $n_e=n_p=X_e{n}_b$ and $n_H=(1-X_e)n_b$. Inverting the preceding ratio and inserting the photon number density with $n_b=\eta n_{\gamma }$ gives the hydrogen-only Saha equationWriting $y=\mathcal{B}/T$, its coefficient is $(2\zeta (3)/{\pi }^2)\eta (2\pi \mathcal{B}/m_e{)}^{3/2}\simeq 3.2\times 10^{-16}$, consistent with the rounded $3\times 10^{-16}$. The negligible mass-ratio correction and ground-state approximation are the assumptions behind this form.

Photon decoupling is a dynamical loss of frequent scattering, distinct from the chemical conversion of charged particles into neutral atoms. Before recombination, Thomson scattering on free electrons keeps photons coupled to the baryon plasma. The physical scattering rate isin units with $c=1$. Define the approximate decoupling temperature byEquivalently the photon mean scattering time becomes comparable to an expansion time. A refined last-scattering definition uses the optical depth and the maximum of the visibility function; it need not coincide exactly with the local rate criterion. As $X_e$ falls, scattering becomes inefficient, ordinarily after the half-ionized recombination stage, so $T_{\mathrm{dec}}<T_{\mathrm{rec}}$.

Residual electron freeze-out concerns the remaining ionized fraction. The Saha equation presumes sufficiently rapid forward and reverse reactions to maintain chemical equilibrium. Real recombination is slowed by radiative bottlenecks and expansion; the free-electron fraction can therefore exceed its equilibrium value. Eventually the effective recombination rate per free electron, roughly ${\alpha }_{\mathrm{eff}}(T)n_e$, falls below $H$. The remaining electrons cannot all find and recombine with protons within an expansion time. A small nonzero residual fraction survives while the Saha prediction decreases exponentially toward zero.

The sketch shows this lag and residual floor. Its kinetic curve and the position of the decoupling marker are schematic: the data supplied determine the equilibrium curve, but not a quantitative recombination history or an exact $T_{\mathrm{dec}}$. Determining those needs the expansion history and atomic transition rates. The horizontal axis decreases toward the right to follow cosmological cooling.

When the number-changing reaction is in chemical equilibrium, its chemical balance condition is $2{\mu }_{\varphi }=3{\mu }_{\varphi }$. Therefore zero chemical potential from number-changing reactions gives $\boxed{{\mu }_{\varphi }=0}$. Elastic scattering maintains kinetic equilibrium but imposes no independent condition on ${\mu }_{\varphi }$. The conclusion requires the number-changing rate to remain large compared with the Hubble parameter; merely listing an allowed reaction does not guarantee this after freeze-out.

Before thermal decoupling in cosmology, the sectors have a common temperature $T$. For relativistic species at that temperature their entropy densities aresoThese are the effective number of relativistic degrees of freedom for entropy. The paper's $g_{\ast }$ agrees with $g_{\ast s}$ for relativistic components at a common temperature. For one real scalar field, $g_{\ast s}^{\varphi }=1$. At temperatures where all Standard Model species are relativistic, both counts are constant, and so is $\xi$ before decoupling; thresholds can change this equilibrium ratio if they occur while the sectors remain coupled.

After decoupling, adiabatic isolation gives separately conserved visible and dark entropy: $s_{\mathrm{SM}}{a}^3$ and $s_{\varphi }{a}^3$ are individually constant. Thus $\boxed{\xi (T<T_d)=\xi (T_d)}$ even when either sector's entropy degrees of freedom change. Changes in particle content then alter their temperature ratio rather than the ratio of their comoving entropies. This assumes no late inter-sector transfer or entropy injection.

During the cannibal phase of a decoupled sector, cosmological entropy conservation and the leading result of part (c) giveDifferentiating and using $T_{\varphi }=m_{\varphi }/x$ yields logarithmic cooling during cannibalism:Here $a_{\ast }$ absorbs the entropy normalization, and the logarithmic expression is a large-$x$ asymptotic relation. With fixed visible entropy degrees of freedom the photon bath has $T_{\gamma }\propto a^{-1}$. Thus for $x\gg 1$ the scalar particles cool much more slowly than the photons and become relatively hotter.

The physical reason is that $3\varphi \to 2\varphi$ converts rest-mass energy into kinetic energy of the remaining particles. Elastic scattering distributes this energy, offsetting much of the cooling caused by expansion. The total comoving entropy remains constant; the number density does not merely dilute as $a^{-3}$. This explanation applies only while number-changing chemical equilibrium persists. If those reactions freeze out, an isolated number-conserving nonrelativistic gas instead has $T_{\varphi }\propto a^{-2}$ under adiabatic expansion, which is faster cooling than the photons. The paper supplies no rates, so that transition time cannot be determined.

Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting electron rest energy, isFor electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net electrons from thermally created electrons and Positrons. Chemical equilibrium with photons requires opposite electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

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.

Four mechanisms that can produce disequilibrium chemistry in an exoplanet atmosphere are:

The exoplanet transport and photochemical examples follow kinetic atmosphere calculations and models including horizontal transport; the condensate example is examined in cold-trap calculations.

Four possible molecular signatures of disequilibrium chemistry in an exoplanet atmosphere, measured through an exoplanet transmission spectrum or exoplanet emission spectrum, are:

The first two signatures can be consequences of the same process and are not independent evidence for two mechanisms. Changes in atmospheric carbon-to-oxygen ratio, atmospheric metallicity of a giant planet, exoplanet cloud coverage, and the atmospheric pressure-temperature profile can mimic abundance changes. A convincing inference compares several molecular bands with a chemically consistent equilibrium model rather than identifying one unusual band alone.

For solar elemental abundances, take a molecular hydrogen- and helium-dominated atmosphere with atmospheric carbon-to-oxygen ratio near $0.55$. The principal oxygen-bearing molecule is generally water; the carbon and nitrogen carriers depend on both temperature and pressure.

At the cool $500\mathrm{K}$ observable upper layers, methane is the expected main carbon reservoir under chemical equilibrium, with abundant water and much less carbon monoxide. At the hotter layers approaching $1400\mathrm{K}$, carbon monoxide becomes the main carbon carrier and water contains much of the oxygen not bound in it. The relevant law of mass action follows fromLower temperature favors the exothermic methane-forming direction, while increasing pressure favors the side with fewer molecules. This explains why the change of dominant carrier cannot be specified by temperature alone.

Nitrogen is distributed between molecular nitrogen and Ammonia. Cooling favors Ammonia, but low pressure favors molecular nitrogen; it is therefore unsafe to call Ammonia dominant throughout the $500\mathrm{K}$ low-pressure region. The exact partition requires the equilibrium constant for ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$.

Carbon dioxide is usually a minor constituent at solar composition; hydrogen cyanide and acetylene are much less abundant than the principal carbon carriers in this oxygen-rich equilibrium case. Condensation can remove refractory species where a condensation curve is crossed. The robust cool-atmosphere expectation is a molecular hydrogen–helium background with methane and water, changing toward carbon monoxide in hotter layers. Precise mixing ratios require thermodynamic data and an explicitly specified elemental inventory after rainout.

Assume all three planets retain approximately their original hydrogen-helium bulk composition. Stellar encounters change orbital energy and irradiation, not automatically the planet's elemental abundances. Their atmospheric structures can become approximately stationary long before their interiors finish cooling.

For the unperturbed Jupiter-like planet at $a\simeq 5.2\mathrm{au}$, the zero-albedo globally averaged planetary equilibrium temperature is about $122\mathrm{K}$. Stellar light heats the outer atmosphere, while internal cooling supports a deeper temperature gradient and convection. At cooler pressures, chemical equilibrium favors methane and ammonia; condensate clouds can include ammonia at high levels and water deeper down. Disequilibrium chemistry in an exoplanet atmosphere can preserve carbon monoxide or other species from deeper layers.

For the inward-migrated planet at $a\simeq 0.052\mathrm{au}$, the irradiation-only planetary equilibrium temperature is ten times higher, about $1220\mathrm{K}$, because $T_{\mathrm{eq}}\propto a^{-1/2}$. Its irradiated planetary atmosphere has a heated radiative exterior, potentially strong day-night differences, and a deep radiative-convective boundary. At suitable pressures, chemical equilibrium increasingly favors carbon monoxide over methane; water remains important, while alkali absorption and high-temperature condensates can matter. An atmospheric thermal inversion depends on absorbers, clouds, and irradiation and is not guaranteed simply by migration.

The ejected object is a rogue planet. Its irradiation-based planetary equilibrium temperature becomes very small, but its actual emitting temperature is set primarily by internal cooling and Kelvin-Helmholtz contraction. After only several million years it can remain warm and self-luminous, with deep convection and an outward-cooling radiative atmosphere. Its photospheric chemistry and clouds depend on that cooling temperature; they cannot be inferred from the absence of a host star alone. None of these cases fixes an exact pressure-temperature profile without a cooling model, opacities, and atmospheric composition.

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.

Chemical equilibrium for $e^{-}+p\leftrightarrow H+\gamma$ requires ${\mu }_e+{\mu }_p={\mu }_H$, since the photon chemical potential vanishes. Inserting the nonrelativistic Maxwell-Boltzmann distribution and using charge neutrality $n_p=n_e$ giveswhere the proton-to-hydrogen mass ratio and the stated degeneracy factors have been approximated by one. Thus

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