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After cosmological weak freeze-out, the surviving neutron fraction is $x_n=r_f(1+r_f{)}^{-1}{e}^{-\Delta t/{\tau }_n}$, where $r_f$ is the initial neutron-to-proton ratio and ${\tau }_n=t_{1/2}/\ln 2$ is the mean lifetime. Baryon number is conserved because a decaying neutron becomes a proton. If almost all surviving neutrons enter helium-4, neutron-limited helium synthesis gives the displayed primordial helium mass fraction. Increasing the half-life increases survival. If the longer lifetime also represents uniformly weaker weak interactions, it raises the temperature of cosmological weak freeze-out and the initial neutron fraction, reinforcing the helium increase. That last conclusion requires specifying how the weak rates change, rather than treating lifetime alone as a complete microscopic model.

A moderately enhanced Hubble parameter raises the primordial helium mass fraction through earlier cosmological weak freeze-out and less free-neutron decay after freeze-out, provided nuclear burning still incorporates nearly all surviving neutrons into helium-4. This provision matters: if expansion carries the gas through the nuclear-burning temperature range faster than the reaction times, neutron-limited helium synthesis no longer applies. More neutrons surviving the bottleneck alone does not prove more final helium in that limit.

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.

For a moderately faster cosmic expansion that still permits efficient nuclear burning, the hydrogen mass fraction is smaller, since cooling from cosmological weak freeze-out to helium formation takes less time, fewer free neutrons decay into protons, and more surviving neutrons enter helium-4. If the change also affects cosmological weak freeze-out, its higher freeze-out temperature leaves a larger initial neutron-to-proton ratio and reinforces the helium increase. The conclusion assumes efficient neutron-limited helium synthesis: an unspecified arbitrarily rapid expansion could instead outrun the nuclear reactions, so the nuclear-burning qualification of the helium expansion-rate response prevents an unconditional assertion.

In the hydrogen–helium approximation, take helium to be helium-4, and assume essentially all neutrons remaining at the onset of efficient nuclear burning enter it. Each helium-4 nucleus contains two neutrons and two protons, henceHere $n_p,n_n$ count constituent nucleons, not just free particles after burning. Neglecting the small nucleon mass difference and nuclear binding-energy correction, the helium mass is $4m_u{n}_{\mathrm{He}}$ and the total baryon mass is $m_u{n}_B$. Thus neutron-limited helium synthesis givesFor $n_n/n_p\simeq 1/7$, this yields $Y_p\simeq 1/4$ and the hydrogen mass fraction is $X_p\simeq 3/4$. The assumptions require $n_p\ge n_n$ and negligible free neutrons, deuterium and helium-3: the phrase hydrogen and helium must here mean predominantly hydrogen-1 and helium-4, not an arbitrary mixture of helium isotopes.

The small baryon-to-photon ratio suppresses nuclear equilibrium abundances through the factor ${\eta }^{A-1}$. Cooling eventually makes the binding-energy exponential favourable, but the deuterium bottleneck delays substantial nuclear burning until well after cosmological weak freeze-out. In ordinary Big Bang nucleosynthesis, nearly every neutron surviving that delay ends up in helium-4. This makes the final primordial helium mass fraction chiefly a neutron budget rather than an unrestricted equilibrium abundance.

Let $r_F$ be the neutron-to-proton ratio at freeze-out, let $\Delta t$ be the delay until effective nuclear burning, and let ${\tau }_n$ be the neutron mean lifetime. The initial fraction of baryons which are neutrons is $x_{n,F}=r_F/(1+r_F)$. Because neutron decay converts a neutron into a proton, total baryon number remains constant, while the neutron fraction becomes $x_{n,B}=x_{n,F}{e}^{-\Delta t/{\tau }_n}$. Each helium nucleus uses two neutrons and four baryons. Hence neutron-limited helium synthesis givesEquivalently, $Y_4=2r_B/(1+r_B)$ with $r_B=x_{n,B}/(1-x_{n,B})$. This accounts for the extra protons produced by neutron decay; multiplying $r_F$ alone by the survival factor would miss their contribution. The following responses refer to ordinary neutron-limited burning, without exotic particle injection or a change to the nuclear network.

Let $n_n$ and $n_p$ count the available neutrons and protons immediately before their incorporation into nuclei. In the usual neutron-limited helium synthesis approximation, all surviving neutrons go into ${}^4\mathrm{He}$ and the remaining protons form ${}^1\mathrm{H}$. Each helium nucleus contains two neutrons and two protons, soUsing the equal-nucleon-mass approximation and neglecting trace nuclei and binding-energy corrections, the primordial helium mass fraction isThe ratio is the pre-assembly nucleon ratio, not the number ratio of free hydrogen nuclei to helium nuclei afterward. The derivation assumes enough protons to bind all the neutrons, as in standard Big Bang nucleosynthesis. The familiar $n_p/n_n\simeq 7$ gives $Y_p\simeq 1/4$ and hydrogen mass fraction $X_p\simeq 3/4$.

Two effects must be distinguished in Big Bang nucleosynthesis. The positive extra neutrino energy density increases the Hubble parameter through the Friedmann equation. Holding weak rates fixed, faster expansion causes earlier neutron-proton cosmological weak freeze-out, leaving more neutrons, and leaves less time for neutron decay before nuclear reactions begin. Both tendencies increase the primordial helium mass fraction.

Electron-flavor degeneracy also changes the charged-current balance directly. Chemical equilibrium for $n+{\nu }_e\leftrightarrow p+e^{-}$ gives, with negligible electron chemical potential,This is the primordial helium response to electron-neutrino degeneracy. Positive ${\xi }_{{\nu }_e}$ favors conversion of neutrons to protons and reduces the neutron abundance; negative ${\xi }_{{\nu }_e}$ favors the opposite balance. When neutrons are the limiting ingredient and nearly all surviving neutrons enter helium-4, the neutron-limited helium synthesis estimate isIn this usual proton-rich regime, the direct electron-neutrino effect lowers $Y_p$ for positive degeneracy and raises it for negative degeneracy, other conditions held fixed. At sufficiently large negative degeneracy the equilibrium ratio can instead exceed one; protons then become limiting. The corresponding ideal complete-capture estimate is $Y_p\simeq 2\min (r,1)/(1+r)$, so the neutron-limited trend is not a universal monotonic law at arbitrary negative chemical potential. Degeneracy in the other flavors primarily affects helium through the expansion rate in this simplified discussion.

For very large degeneracy the weak reaction rates themselves also change, so a quantitative freeze-out calculation must include both the altered distributions and the altered expansion rate. The two effects can compete for a positive electron-neutrino asymmetry. The primordial helium mass fraction change is not determined by $|{\mu }_{\nu }|$ alone: its flavor and sign matter as well as its energy density.

If the Friedmann equation remains approximately $H^2=8\pi G\rho /3$, radiation domination and fixed weak microphysics give $H\propto \sqrt{G}{T}^2$ and a neutron-proton conversion rate proportional to $T^5$. Cosmological weak freeze-out then occurs at $T_f\propto G^{1/6}$. Increasing $G$ during this era raises the neutron-to-proton ratio at freeze-out and shortens the time for free-neutron decay after freeze-out. Both increase the primordial helium mass fraction when neutron-limited helium synthesis remains efficient. Arbitrary variation of $G$ over time requires its history and the full gravitational field equations; this scaling is a controlled background approximation, rather than a complete model of gravity with arbitrary time dependence.

Chemical equilibrium of $n+{\nu }_e\leftrightarrow p+e^{-}$, with negligible electron chemical potential, gives the stated neutron-to-proton ratio. Positive electron-neutrino degeneracy reduces the neutron fraction, whereas negative degeneracy increases it. The primordial helium mass fraction then responds through neutron-limited helium synthesis while protons outnumber neutrons; a sufficiently neutron-rich mixture instead becomes proton limited, so that trend must not be extrapolated indefinitely. Extra neutrino energy density also accelerates cosmic expansion, changing weak freeze-out and the time available for neutron decay. These effects can compete; a quantitative Big Bang nucleosynthesis calculation must account for the altered reaction rates as well as the altered expansion rate.

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.