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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.

Use $X_i=n_i/n_B$, so $X_D$ counts deuterium nuclei per baryon, not baryons bound into deuterium. For the radiative capture reaction, chemical equilibrium and the zero photon chemical potential give ${\mu }_D={\mu }_n+{\mu }_p$. Dividing the three nonrelativistic Maxwell--Boltzmann number densities removes all chemical potentials:Here $g_D=3$ and $g_n=g_p=2$. The binding convention in this PDF is a negative mass defect, $B_D=m_D-m_n-m_p<0$. The positive nuclear binding energy is ${\epsilon }_D=-B_D$, so the abundance is enhanced by $e^{-B_D/T}$, not by $e^{B_D/T}$.

Since $X_D/(X_n{X}_p)=n_B{n}_D/(n_n{n}_p)$, insert the baryon-to-photon ratio $n_B=\eta n_{\gamma }$ and the photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$. This givesAt equal nucleon mass $m_N$ in the prefactors, $m_D\simeq 2m_N$, the deuterium equilibrium abundance simplifies toThe dimensionless prefactor is about $8.14$. These expressions describe thermal equilibrium; residual primordial deuterium after nuclear reactions cease must instead be obtained from the reaction kinetics.

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.

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.

Because $X_i=n_i/n_B$,Using $n_B=\eta n_{\gamma }$ and the given photon density in the deuterium equilibrium abundance formula givesAlthough the exponential favors deuterium once $k_{B}T<B_D$, the very small factor $\eta$ expresses the enormous excess of photons. The high-energy tail therefore photodissociates newly formed deuterium until the temperature is well below $B_D/k_B$; this is the deuterium bottleneck.