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Big Bang nucleosynthesis is a kinetic process in an expanding universe. A common particle temperature does not imply chemical equilibrium for every nuclear reaction: reaction rates must also exceed the expansion rate and connect the required species efficiently. Around $10^9\mathrm{K}$, nuclear abundances retain the history of earlier freeze-out and cannot be inferred from equilibrium Boltzmann factors alone.

In particular, weak reactions cease to maintain neutron-proton chemical equilibrium. Subsequent free-neutron decay after freeze-out changes the surviving neutron inventory. The large photon-to-baryon ratio delays deuterium survival through photodissociation; once the deuterium bottleneck opens, reactions rapidly bind many surviving neutrons into helium-4. Coulomb barriers, decreasing density and further freeze-out prevent unrestricted equilibration of all nuclei. The final helium abundance therefore requires a reaction network and expansion history, not just the temperature and nuclear binding energies.

For illustration, if essentially all available neutrons end in helium-4 and the neutron-to-proton ratio at that stage is $r$, simple particle counting gives $Y_{\mathrm{He}}\simeq 2r/(1+r)$. The crucial input $r$ is itself set by out-of-equilibrium history. Kinetic thermalization of the plasma can remain effective while these composition-changing processes are out of equilibrium.

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.

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.

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.

During radiation domination, the effective number of relativistic energy degrees of freedom, denoted $g_{\ast }$, gives the energy densityThe paper's ${\mathcal{N}}_{\mathrm{eff}}$ is this $g_{\ast }$; it is not the conventionally normalized effective number of neutrino species. With the specified reduced Planck mass, the Friedmann equation reads $H^2={\rho }_R/(3M_{\mathrm{Pl}}^2)$, whenceWrite the weak interaction conversion rate as $\Gamma =A{G}_F^2{T}^5$. The ratio $\Gamma /H$ scales as $T^3$, so it falls during cooling: conversion eventually becomes too slow to maintain neutron-proton chemical equilibrium. Equating the conversion and expansion rates givesOn suppressing the numerical coefficient, the cosmological weak freeze-out temperature is consequentlyThe one-sixth power reflects a balance between a conversion rate proportional to $T^5$ and a Hubble parameter proportional to $\sqrt{g_{\ast }}{T}^2$.

At temperatures above a few MeV, weak interactions such as $n+{\nu }_e\leftrightarrow p+e^{-}$ and $n+e^{+}\leftrightarrow p+{\bar{\nu }}_e$ maintain neutron-proton chemical equilibrium. Negligible lepton chemical potentials give the neutron-to-proton ratio $r_{\mathrm{eq}}=n_n/n_p=\exp [-\Delta /(k_{B}T)]$, where $\Delta =(m_n-m_p)c^2\simeq 1.293\mathrm{MeV}$. At high temperature this ratio is nearly one.

During radiation domination, the interaction time grows as $T^{-5}$, faster than the expansion time $T^{-2}$. Their crossing therefore produces cosmological weak freeze-out. If $\Theta =T/(10^{10}\mathrm{K})$, equating the displayed order-of-magnitude times gives ${\Theta }_f^{-5}\sim g_{\ast }^{-1/2}{\Theta }_f^{-2}$, soUsing $H^{-1}=2t$ instead of $t$ changes only the order-one coefficient. Before electron-positron annihilation, the usual effective relativistic count is $g_{\ast }=2+(7/8)(4+6)=10.75$. The factor $7/8$ weights fermionic relative to bosonic energy density.

After freeze-out, free neutrons decay while protons increase correspondingly. Define the neutron baryon fraction $X_n=n_n/(n_n+n_p)$. In the instantaneous-freeze-out estimate,where ${\tau }_n=t_{1/2}/\ln 2$ is the neutron mean lifetime. Hence the ratio at the onset of nuclear synthesis isThis expression accounts for the protons produced by neutron decay.

Nuclear burning initially waits at the deuterium bottleneck: although deuterium is bound by about $2.2\mathrm{MeV}$, the many photons per baryon can still photodissociate it. Roughly, ${\eta }_b^{-1}{e}^{-B_D/(k_B{T}_N)}\sim 1$ sets the bottleneck temperature, with refinements from reaction rates and phase space. Around $k_B{T}_N\sim 0.1\mathrm{MeV}$, deuterium survives, reactions through deuterium, tritium and helium-3 rapidly build tightly bound helium-4, and nearly all surviving neutrons are captured.

Thus $n_{\mathrm{He}}\simeq n_n/2$ and the primordial helium mass fraction isFor the standard estimate $r_N\simeq 1/7$, neutron and proton constituent fractions are $1/8$ and $7/8$, respectively, and $Y_{\mathrm{He}}\simeq 1/4$. Free neutrons are negligible after capture. The remaining hydrogen mass fraction is about $3/4$, and the helium-to-hydrogen number ratio is about $1/12$.

The printed numerical time coefficients are schematic, not a precision prediction of this quarter fraction: their literal crossing with $g_{\ast }=10.75$ gives $k_B{T}_f\simeq 1.28\mathrm{MeV}$. Keeping physical weak-rate coefficients and the gradual freeze-out instead gives an effective freeze-out around $0.7$–$0.8\mathrm{MeV}$, followed by decay over a few minutes. The abundance formulas above explain the standard quarter fraction and its parameter dependence without treating the rounded time formulas as exact nuclear-rate data.

At fixed cosmological weak freeze-out temperature $T_f$ and fixed neutron survival factor, neutron-proton chemical equilibrium gives $r_f=e^{-Q/T_f}$ and neutron-limited helium synthesis gives $Y_p\propto 2r_f/(1+r_f)$. Differentiating yields the displayed sensitivity. A finite change $Q\mapsto (1+\alpha )Q$ gives $Y_p^{\prime }/Y_p=(1+e^{Q/T_f})/(1+e^{(1+\alpha )Q/T_f})$. Increasing the positive mass difference reduces helium production. Changes to the weak conversion rates, neutron lifetime or nuclear history must be supplied separately for a full abundance prediction.