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During radiation domination, the weak conversion rate scales as $T^5$ whereas the Hubble parameter scales as $\sqrt{G{g}_{\ast }}{T}^2$. Their crossing fixes the cosmological weak freeze-out temperature. A faster expansion raises it, leaving a larger neutron-to-proton ratio, and shortens the interval for free-neutron decay after freeze-out. Both effects increase the helium-4 mass fraction if the extra component does not independently change reaction rates or inject decay products.

Neglecting chemical potentials, the equilibrium abundance ratio is $n_n/n_p\simeq \exp [-(m_n-m_p)c^2/(k_{B}T)]$. Weak freeze-out fixes an approximate starting ratio, which is then reduced by free-neutron decay after freeze-out before most remaining neutrons bind into helium.

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.

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.

At a fixed temperature, the deuterium equilibrium abundance contains an explicit factor $n_B$ through $\eta$. Two nucleons have to encounter each other to make one bound nucleus. Their capture rate per volume is proportional to $n_n{n}_p$, whereas nuclear photodisintegration competes through the very numerous thermal photons. The small baryon-to-photon ratio therefore delays stable deuterium formation until $T\ll {\epsilon }_D$, producing the deuterium bottleneck. Changing the cosmological baryon density changes both its release time and the subsequent efficiency of nuclear burning. The residual primordial deuterium abundance is particularly sensitive to these rates: on the usual nucleosynthesis branch, increasing $n_B$ burns more deuterium into heavier nuclei and lowers the remaining D/H. The positive dependence of the instantaneous equilibrium expression and the negative dependence of the final relic D/H are different statements.

By contrast, once nuclear reactions are efficient, nearly every surviving neutron ends in helium-4. Two neutrons make each helium-4 nucleus, so $n_{\mathrm{He}}\simeq n_n/2$ andThis primordial helium mass fraction is mainly neutron limited: $r$ is set by weak conversion, cosmological weak freeze-out and free-neutron decay after freeze-out, rather than by the absolute baryon density. There is a weaker dependence because a higher $\eta$ releases the deuterium bottleneck earlier and allows fewer neutrons to decay. Thus $Y_P$ is comparatively insensitive, not exactly independent, of $n_B$. At such exceptionally small baryon densities that nuclei never form efficiently, the neutron-limited approximation would fail.

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.

The nucleons are nonrelativistic, so the Maxwell-Boltzmann distribution givesBoth spin degeneracies are two. Taking the ratio and using the specified negligible chemical potentials yields the neutron-to-proton ratioThe mass prefactor differs from unity only at the per-mille level. The dependence on $T_F$ is much stronger: $d\ln r_F/d\ln T_F=Q/T_F>0$. Thus earlier, hotter weak freeze-out leaves more neutrons relative to protons. For illustration, $T_F=1\mathrm{MeV}$ gives $r_F\simeq 0.28$; $T_F=0.8\mathrm{MeV}$ gives $r_F\simeq 0.20$. Free-neutron decay after freeze-out lowers the ratio further before most Big Bang nucleosynthesis takes place.

A larger baryon-to-photon ratio favours bound nuclei at a fixed temperature through the equilibrium factor ${\eta }^{A-1}$. In particular it helps deuterium survive photodissociation sooner: there are fewer energetic photons per baryon, and nuclear reaction rates per baryon also rise. A schematic deuterium bottleneck condition is $\eta e^{B_D/T_B}\sim \mathrm{constant}$, suppressing slowly varying powers of temperature. It givesThus higher $\eta$ permits burning at a higher temperature. During radiation domination that means earlier cosmic time, $t_B\propto T_B^{-2}$, leaving less time for free-neutron decay after freeze-out. More neutrons enter helium, so the final helium-4 mass fraction increases, ordinarily rather weakly. This is the baryon-density effect on primordial helium.

The equilibrium helium factor ${\eta }^3$ indicates the direction of the change but is not a prediction that the final helium abundance grows without bound as ${\eta }^3$. Once almost all surviving neutrons are captured, the final primordial helium mass fraction is limited by their supply, as the shared neutron budget shows.

Under the usual interpretation that the extra force simply slows otherwise adiabatic expansion, without changing reaction rates or injecting particles, the primordial hydrogen mass fraction would be larger. A smaller Hubble parameter makes cooling slower, delays the end of the deuterium bottleneck, and permits more free-neutron decay after freeze-out before nuclear assembly. If the reduced expansion also keeps weak conversions effective longer, the lower eventual cosmological weak freeze-out temperature further reduces the neutron-to-proton ratio. Both effects lower the primordial helium mass fraction and raise the remaining hydrogen mass fraction.

For fixed initial neutron fraction $x_{n,1}=n_n/(n_n+n_p)$, the post-freeze-out estimate is $x_n=x_{n,1}{e}^{-(t_N-t_1)/{\tau }_n}$, where $t_N$ is the onset of efficient nuclear assembly. Thus $Y_p\simeq 2x_n$ decreases when cooling delays $t_N$. The extra time is the time to reach the nuclear-assembly temperature, not an increase in the specified clock interval $300-1$ seconds. An unspecified force that also reheats the plasma, changes weak rates or causes recollapse would require additional information; the simple expansion-only interpretation is essential to this expectation.

Reducing the gravitational constant decreases $H$ at fixed temperature. The helium sensitivity to the expansion rate has two effects: weak interconversion persists to a lower freeze-out temperature, leaving fewer neutrons, and the universe takes longer to reach the deuterium-burning temperature, allowing more free-neutron decay after freeze-out. Both lower the helium-4 abundance. In particular $T_f\propto G^{1/6}$ and $t(T)\propto G^{-1/2}$.

At weak freeze-out, photons, electrons, positrons and all four neutrino flavors are relativistic and approximately at the same temperature. The effective number of relativistic energy degrees of freedom changes fromThe expansion rate at fixed temperature is larger by $\sqrt{50/43}\simeq 1.078$. Equating the supplied weak interaction time with the supplied age gives $T_f^3=\sqrt{g_{\ast }}$ when $T_f$ is measured in MeV. ThereforeEarlier cosmological weak freeze-out at a higher temperature leaves more neutrons. With negligible chemical potentials, the neutron-to-proton ratio in equilibrium is $r_f=\exp (-Q/T_f)$, where $Q=m_n-m_p\simeq 1.293\mathrm{MeV}$. If almost all surviving neutrons enter helium-4, the primordial helium mass fraction before neutron decay is $Y_f=2r_f/(1+r_f)$. Literal use of the given rate normalizations yieldsEquivalently, the leading response is $\delta Y_f/Y_f=[Q/(T_f(1+r_f))]\delta T_f/T_f$. This is a roughly one-percentage-point absolute increase in the helium fraction, rather than a one-percent relative increase.

The deuterium bottleneck delays nuclear burning, allowing free-neutron decay after freeze-out. If its end temperature is $T_N$, and ${\tau }_n$ is the neutron mean life, a sharper instantaneous-freeze-out estimate conserves total baryon number and givesThis formula evolves the neutron fraction, not just the neutron-to-proton ratio, because each neutron decay also creates a proton. After electron annihilation, let $r_T=(4/11{)}^{1/3}$. The radiation counts relevant to the age at fixed $T_N$ areConsequently $t_N^{(4)}/t_N^{(3)}\simeq \sqrt{3.36/3.82}\simeq 0.939$. Less time is available for neutron decay, adding to the helium increase. As an explicit illustrative completion, $T_N=0.07\mathrm{MeV}$ and ${\tau }_n\simeq 880\mathrm{s}$ give an increase of approximately $0.012$ in this toy model. The precise helium fraction requires the nuclear-burning temperature, baryon abundance and properly normalized weak rates; the question supplies only approximate time laws. In particular, the large no-decay fractions in the table must not be interpreted as precise predictions for the observed primordial abundance. The robust effect is extra relativistic energy raises the primordial helium fraction, with an absolute change of order $10^{-2}$ for one thermally populated flavor.

The effective number of relativistic energy degrees of freedom count two photon polarizations, four electron/positron spin states, and six neutrino/antineutrino states, with the Fermi-Dirac distribution factor $7/8$ for fermions. Treating these species as relativistic and at a common temperature gives $\boxed{g_{\ast }=2+(7/8)(4+6)=10.75}$.

The mass-equivalent radiation mass density is $\rho =({\pi }^2/30)g_{\ast }(k_{B}T{)}^4/({\hslash }^3{c}^5)$. Therefore the Friedmann equation givesAt $T_0=10^{10}\mathrm{K}$, $H_0\simeq 0.503{\mathrm{s}}^{-1}$. Equating the specified weak interaction rate to $H_0(T/T_0{)}^2$ gives $(T_{\ast }/T_0{)}^3=H_0/(1{\mathrm{s}}^{-1})$ and henceThe approximate equilibrium neutron-to-proton ratio, neglecting chemical potentials, is $r_{\ast }\simeq \exp [-(m_n-m_p)c^2/(k_B{T}_{\ast })]$, about $0.152$ using the mass difference $1.293\mathrm{MeV}$. This is a freeze-out estimate rather than a detailed reaction-network computation.

After weak cosmological weak freeze-out, free neutrons undergo beta-minus decay; neutrinos decouple, and electron-positron annihilation heats the photons relative to the neutrinos. Initially the abundant energetic photons destroy deuterium almost as soon as it forms. Once the deuterium bottleneck ends, nuclear reactions rapidly pass through deuterium, tritium and helium-three to helium-4. Reactions eventually become too slow compared with cosmic expansion; nearly all surviving neutrons are bound in helium-four. Its mass fraction is approximately $Y_4=2r/(1+r)$, where $r$ is evaluated when nuclear assembly begins.

Increasing $g_{\ast }$ increases the Hubble parameter, raises $T_{\ast }$, increases the freeze-out neutron-to-proton ratio, and leaves less time for free-neutron decay after freeze-out. Both effects increase $Y_4$; decreasing $g_{\ast }$ reverses them. Most deuterium is only an intermediate product and is burned into helium, so its residual abundance and its absolute change are tiny compared with the helium mass fraction. This does not imply a negligible fractional change in residual deuterium. Faster expansion also shortens the time available to burn it, and its relative abundance can be appreciably sensitive to extra radiation. The distinction matters when interpreting the final wording of the question; joint deuterium and helium abundances constrain extra relativistic species, as discussed in the Particle Data Group nucleosynthesis review.

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.