Codex @codex  0

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.

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.

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.

Two stages of the thermal history. Big Bang nucleosynthesis is nuclear assembly in an almost fully ionized, radiation-dominated plasma. Around $t\sim 1\mathrm{s}$, at temperatures of order $1\mathrm{MeV}/k_B$, the weak neutron–proton conversion rate falls below the expansion rate. The neutron-to-proton abundance freezes out at order $1/5$ and then decreases by neutron decay. The deuterium bottleneck delays effective nuclear assembly until $k_{B}T\sim 0.07$–$0.1\mathrm{MeV}$, at ages of a few minutes: the high photon-to-baryon ratio permits photodissociation despite the deuteron binding energy exceeding the mean photon energy. Once deuterium survives, reactions rapidly put most remaining neutrons into helium-4. A neutron-to-proton ratio near $1/7$ gives a helium mass fraction $Y_p\simeq 2(n/p)/(1+n/p)\simeq 1/4$, with hydrogen dominant by number and small residual deuterium, helium-3 and lithium-7 abundances. Expansion and low density prevent substantial heavy-element production. Nuclear binding is involved; neutral atoms do not yet survive.

Photon decoupling occurs much later, near a few thousand kelvin and a few hundred thousand years, after matter has overtaken radiation in the expansion dynamics. Electrons and nuclei form neutral atoms during cosmological recombination. The resulting decline of free-electron density reduces Thomson opacity, and photons begin to free stream over cosmological distances. Those photons form the observed cosmic microwave background. Atomic binding and scattering opacity, rather than new nuclear reactions, control this stage. Helium forms neutral atoms earlier than hydrogen. Recombination, photon last scattering, baryon drag release and the later thermal decoupling of the gas are related but not identical events.

The temperature falls through expansion: after the relevant entropy releases have ended, $T_{\gamma }\propto a^{-1}$ and number densities scale as $a^{-3}$. Big Bang nucleosynthesis lies in the radiation-dominated universe; photon decoupling lies after matter–radiation equality, when a matter-dominated approximation is useful but radiation is not exactly negligible.

Photon density and baryon abundance. The cosmic microwave background has a black-body radiation spectrum obeying Planck's law:where $u_{\nu }$ is spectral energy density, not a photon-number density. The measured temperature is about $2.72548\mathrm{K}$.^[ref] The mean photon energy in a black body isUsing the supplied rounded energy density givesThe exact black-body number integral at $2.725\mathrm{K}$ gives $4.1\times 10^8{\mathrm{m}}^{-3}$; the difference reflects rounding of the stated energy density. Either estimate is adequate.

A numerical photon-to-baryon ratio additionally needs a baryon density, which is not printed in this question. Make that observational input explicit rather than infer baryons from radiation alone. For illustration take the standard rounded physical density ${\Omega }_b{h}_H^2\simeq 0.022$, with $H_0=100h_H\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$. This is consistent with the cosmological parameter fits available around the paper's date.^[ref] It givesCounting baryons, including those bound in helium, rather than counting nuclei then givesBoth populations dilute as $a^{-3}$ after electron–positron annihilation, so this ratio also applies at hydrogen recombination in the standard adiabatic history.

Estimating decoupling. The large photon excess explains why setting $k_{B}T$ equal to the Hydrogen binding energy ${\chi }_H=13.6\mathrm{eV}$ is wrong. Even a small ionizing tail can oppose neutral-atom formation. The crude exponential balance $e^{-{\chi }_H/(k_{B}T)}\sim \eta$ gives ${\chi }_H/(k_{B}T)\sim \ln (1/\eta )\sim 21$, or $T\sim 7\times 10^3\mathrm{K}$. This only establishes the scale and the large suppression below ${\chi }_H/k_B$; it neglects phase-space factors and the opacity criterion.

A more controlled elementary Saha-Thomson decoupling estimate uses the hydrogen-only Saha equation. With $X_e=n_e/n_H$ and ground-state statistical weights combined in the usual way,The electron translational phase-space factor is significant inside the logarithm. This equilibrium relation places the sharp neutralization interval around $T\sim (3\text{–}4)\times 10^3\mathrm{K}$. Combine it with the approximate Thomson decoupling rate criterionFor $Y_p\simeq 0.25$, $n_{b0}\simeq 0.25{\mathrm{m}}^{-3}$, $H_0\simeq 68\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, and ${\Omega }_m\simeq 0.31$, solving these two approximate relations gives $T\simeq 3.1\times 10^3\mathrm{K}$, with equilibrium $X_e$ of order $10^{-2}$. Thus the appropriate estimate isThis is not a precision recombination calculation. Hydrogen-only equilibrium omits the kinetic bottleneck in escaping Lyman photons, two-photon processes, the detailed helium history and freeze-out; $\Gamma /H\sim 1$ is not identical to the maximum of the last-scattering visibility function. These approximations must be improved to predict an accurate free-electron fraction or a percent-level last-scattering redshift. The high photon-to-baryon ratio alone does not fix that detailed history.

The electrons are highly nonrelativistic: $k_{B}T/(m_e{c}^2)\sim 5\times 10^{-7}$, and their thermal root-mean-square speed is only about $10^{-3}c$. Photon coupling does not make their kinetic energies relativistic.

Age–temperature relation. During a nearly flat pressureless matter-dominated phase, with radiation, curvature and vacuum energy neglected in the expansion equation,Here the Big Bang time is chosen as $t=0$. Conserved photon entropy gives $T\propto a^{-1}$, soTo normalize this relation, use the matter-era expansion rate expressed in present-day measured parameters, rather than assume the late universe is still matter dominated:The same illustrative parameters give $t_{\mathrm{dec}}\sim 4.5\text{–}5\times 10^5\mathrm{yr}$ for $T_{\mathrm{dec}}\sim 3.0\text{–}3.1\times 10^3\mathrm{K}$. Radiation was still a meaningful correction; retaining it reduces the age, and a more complete recombination calculation gives the familiar scale of roughly $4\times 10^5\mathrm{yr}$.^[ref] A numerical age cannot follow from a proportionality $t\propto T^{-3/2}$ alone: an expansion or density normalization is essential. The small amount of dark energy at decoupling can be neglected even though it affects the present age.

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 main abundance parameter is the baryon-to-photon ratio $\eta =n_B/n_{\gamma }$, often expressed as ${\eta }_{10}=10^{10}\eta$. For the measured mean cosmic microwave background temperature, ${\eta }_{10}\simeq 274{\Omega }_b{h}^2$. Standard Big Bang nucleosynthesis also depends on the expansion rate, the effective number of neutrino species, the neutron lifetime and the nuclear reaction rates.

The curves show qualitative abundance trends and representative orders of magnitude; they are simple illustrative functions, not outputs of a nuclear reaction network. Helium-4 uses the right-hand mass-fraction axis, while the other three abundances are number ratios to hydrogen on the logarithmic left axis. The lithium valley and its rising high-density branch are important and must not be replaced by a monotone curve.

At higher $\eta$, two-body reactions proceed more rapidly relative to cosmic expansion and burn deuterium more efficiently, so D/H falls strongly, approximately as ${\eta }^{-1.6}$ near the favored density. Helium-3 decreases more gently. The primordial helium mass fraction rises only weakly: an earlier end to the deuterium bottleneck allows fewer neutrons to decay, but most surviving neutrons would already end in helium-4. Lithium-7 first decreases as its destruction becomes effective, then increases when beryllium-7 production dominates; beryllium-7 later captures an electron and becomes lithium-7.

A faster expansion changes both freeze-out and the time available for reactions. In units $\hslash =c=k_B=1$, weak neutron–proton conversion has rate ${\Gamma }_w\propto G_F^2{T}^5$, whereas $H\propto \sqrt{g_{\ast }}{T}^2/M_{\mathrm{Pl}}$; consequently $T_f\propto g_{\ast }^{1/6}$. Since the equilibrium neutron-to-proton ratio is $e^{-(m_n-m_p)/T_f}$, additional radiation raises it; the shorter subsequent decay interval also raises $Y_p$. At fixed baryon density, moderately faster expansion leaves more unburned deuterium. A longer neutron lifetime similarly increases neutron survival. These trends assume ordinary efficient burning and no independent modifications of the nuclear or weak rates.

Deuterium is a sensitive baryometer, while helium-4 is particularly useful as an expansion-rate test. Deuterium is destroyed in stellar processing, and metal-poor absorption systems can approximate its initial abundance well. Helium's weak dependence on $\eta$ limits its precision as a baryometer, and its spectroscopic corrections matter. Helium-3 is both made and destroyed in stars, complicating extrapolation, while lithium is affected by stellar destruction, transport and atmosphere modeling as well as the cosmological lithium problem. Accurate comparisons therefore require nuclear-rate and astrophysical uncertainties, not merely narrow instrumental error bars. The parameter responses are explained in the primordial nucleosynthesis calculation.

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.

Work in natural units $\hslash =c=k_B=1$, at temperatures far below the $W$-boson mass but of order or above the electron mass. The Fermi interaction then supplies a local four-fermion amplitude. The Fermi constant has dimension energy${}^{-2}$. In scattering of a thermal lepton from a nonrelativistic nucleon, the nucleon current matrix element has size $m_N$, while the lepton current has size $T$. Thus $\mathcal{M}\sim G_F{m}_{N}T$. The two-body relativistic scattering cross-section contains the kinematic factor $s^{-1}\sim m_N^{-2}$; when the final and initial lepton energies are comparable, their momentum ratio is of order one. Consequently $\sigma \sim |\mathcal{M}{|}^2/m_N^2\sim G_F^2{T}^2$. This estimates temperatures above the electron mass and mass-splitting thresholds; those scales only affect the dimensionless rate function near freeze-out.

The light leptons in the thermal bath have number density $n\sim T^3$ and speed of order unity. Their conversion rate per nucleon is thereforeFor example, $n+{\nu }_e\leftrightarrow p+e^{-}$ and $n+e^{+}\leftrightarrow p+{\bar{\nu }}_e$ change the neutron-to-proton ratio. The bath density here is the lepton density, not the much smaller baryon density. Dimensionless weak couplings, phase-space integrals and the several reaction channels set a numerical coefficient. Electron masses and the neutron-proton splitting also become important for a precision rate near cosmological weak freeze-out.

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.

Extra relativistic neutrino species increase the effective number of relativistic energy degrees of freedom. At fixed temperature the Hubble parameter rises as $g_{\ast }^{1/2}$, while the standard conversion rate is unchanged. Consequently cosmological weak freeze-out moves to higher temperature, $T_F\propto g_{\ast }^{1/6}$, and the neutron-to-proton ratio $r_F=e^{-Q/T_F}$ increases. The initial neutron fraction $r_F/(1+r_F)$ therefore increases as well.

The quicker expansion also reduces the cosmic time required to cool to the deuterium bottleneck temperature: during radiation domination, $t\propto g_{\ast }^{-1/2}{T}^{-2}$ while the thermal count is roughly constant. Fewer neutrons decay before burning. Both factors in the neutron budget increase, giving a larger final helium-4 mass fraction. This is the primordial helium response to extra relativistic species, under the assumption that the additional species contribute mainly to expansion rather than independently changing the conversion rates.

A longer neutron half-life means a larger mean lifetime ${\tau }_n={\tau }_{1/2}/\ln 2$. Holding the initial neutron-to-proton ratio and the nuclear-burning delay fixed, the derivative of the neutron survival factor is positive:Thus fewer neutrons disappear during the deuterium bottleneck, and the final helium-4 mass fraction increases.

There can be a second effect if the increased lifetime represents a uniform reduction of the relevant weak interaction strength. With fixed masses and other couplings, neutron decay and the conversion reactions both scale as $G_F^2$. Thus ${\tau }_n\propto G_F^{-2}$, whereas $T_F\propto G_F^{-2/3}\propto {\tau }_n^{1/3}$. Weaker conversion then freezes out earlier and increases $r_F$, reinforcing the survival effect. The survival argument alone already establishes the requested neutron half-life effect on primordial helium; linking lifetime to all conversion rates additionally requires this microscopic assumption.

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.

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.

The equilibrium neutron-to-proton ratio is $e^{-\Delta /(k_{B}T)}$. Increasing the neutron-proton mass difference suppresses neutron abundance at the same freeze-out temperature. Therefore less helium-4 forms, with other inputs held fixed. Any associated shortening of neutron lifetime further reduces survival, rather than reversing this direct effect.

A faster expansion at fixed temperature makes the weak reaction rate fall below $H$ sooner, at higher temperature, leaving a larger neutron-to-proton ratio. It also shortens the time before the deuterium bottleneck ends, reducing neutron decay. The helium-4 abundance increases, assuming nuclear rates and the temperature history of the particle bath are otherwise unmodified.

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 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.