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A small increase in temperature can produce a large increase in a reaction rate. In the simplest Arrhenius equation model, warming from $300$ to $310\mathrm{K}$ increases a rate constant with $E_a=80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ by about $2.81$ times, even though the absolute thermodynamic temperature rises by only about $3.33\%$. This sensitivity helps explain why cooling can slow unwanted chemical reactions and why a hot process can become difficult to control. The comparison assumes the same rate law and reactant concentrations; the activation energy characterizes $k(T)$, rather than every change in the measured reaction rate.

The physical motivation is the distribution of molecular energies. Warming changes the Boltzmann distribution, increasing the relative population of energetically costly configurations. A transition state can therefore become much more accessible than the modest change in average kinetic energy might suggest. The exponential function in an Arrhenius equation captures such sensitivity in a simple model; it is not a universal formula for the fraction above every barrier in a Maxwell-Boltzmann distribution.^[ref]

Activation energy quantifies how strongly a rate constant depends on thermodynamic temperature. For a positive rate constant $k(T)$, the empirical definition iswhere $R$ is the molar gas constant and $k_{\mathrm{ref}}>0$ is a fixed reference with the same units as $k$. The ratio is a dimensionless quantity, keeping the natural logarithm meaningful without changing its derivative. Thus $E_a$ is a molar energy, commonly in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ or $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The Arrhenius equation makes $E_a$ constant over its assumed range. This rate-based definition, and the need for care when identifying it with a threshold on a potential energy surface, are stated in the activation energy.

In a simple barrier picture, reaching a transition state costs energy relative to the starting reactants. The height of that barrier is different from the difference between the reactants and reaction products. An energetically favourable endpoint can still be reached slowly. An energy profile makes the distinction visible, while an Arrhenius plot tests temperature sensitivity from rate constants. A measured $E_a$ need not equal a barrier read from an arbitrarily chosen diagram.

For a computed example, assume a first-order reaction has $k(300\mathrm{K})=0.100{\mathrm{s}}^{-1}$, constant $E_a=80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, and a temperature-independent pre-exponential factor. The Arrhenius equation givesHence $k(310\mathrm{K})\approx 0.281398{\mathrm{s}}^{-1}$. Its half-life falls from $\ln 2/0.100\approx 6.93147\mathrm{s}$ to $\ln 2/0.281398\approx 2.46322\mathrm{s}$. These are chosen model parameters, not measurements of an identified chemical reaction. The exponent uses kelvin, not unconverted Celsius temperatures, and $80000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ must match $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

A microscopic reaction energy barrier belongs to a specified path on a potential energy surface. An experimental activation energy belongs to a specified rate constant and its temperature dependence. A multi-step reaction mechanism can change which step controls the observed conversion, and a temperature-dependent pre-exponential factor contributes to the observed activation energy. Even the Gibbs energy of activation of transition state theory is a different quantity: it incorporates entropy as well as enthalpy and depends on the standard states. These distinctions prevent a fitted Arrhenius plot from being mistaken for a direct microscopic energy measurement.^[ref]

Catalysis illustrates the separation between a path and its endpoints. At the same temperature and with the same reactants, reaction products, and standard states, a regenerated catalyst changes the available reaction mechanism and rate constants, while the overall standard change in Gibbs free energy and the equilibrium constant remain the same:For a reversible first-order reaction pair $\mathrm{A}\rightleftharpoons \mathrm{B}$, consistency at chemical equilibrium requires $k_f/k_r=K$. A model that multiplies only $k_f$ by a catalytic speedup while holding $k_r$ fixed would change this ratio and misrepresent catalysis. This simple ratio applies to that specified rate law; a multi-step catalytic reaction mechanism needs its own rate equations. A potential-energy profile alone cannot determine $K$, because its endpoint difference is not generally the standard change in Gibbs free energy.^[ref]

During collisionless effusion, selective loss changes the distribution of surviving particle velocities without collision-driven relaxation to a new Maxwell-Boltzmann distribution. The loss law depends on flight paths and on the wall-reflection rule. Speed-dependent depletion and, for specular walls, directional structure can remain.

A semiconductor whose mobile charge carrier population requires Fermi-Dirac statistics rather than a dilute Maxwell-Boltzmann distribution. In a strongly degenerate n-type semiconductor, the Fermi level lies inside the conduction band; in a strongly degenerate p-type semiconductor, it lies inside the valence band. The occupied and empty quantum states on opposite sides of a tunnel diode can then overlap in energy. Heavy semiconductor doping both changes these occupations and narrows the depletion region; a narrow barrier and accessible final quantum states serve different roles.

A broad electron Maxwell-Boltzmann distribution of width ${\sigma }_p$ plus a narrow beam of width ${\sigma }_b$ centred at speed $V$ can have a positive resonant velocity slope. The beam slope is largest at $V-{\sigma }_b$. Treating the background slope as constant over the beam width gives the displayed density threshold. Replacing the Maxwellian tail at $V-{\sigma }_b$ by its value at $V$ also requires $V{\sigma }_b/{\sigma }_p^2\ll 1$; width separation alone does not ensure that exponentially sensitive approximation. A positive slope is the local resonant-drive criterion, while an exact unstable mode also requires the kinetic dispersion relation.

A gas is nondegenerate when its occupation numbers are low enough that quantum statistics reduce to the Maxwell-Boltzmann distribution. A common criterion is $n{\lambda }_{\mathrm{th}}^3\ll 1$, where ${\lambda }_{\mathrm{th}}$ is the thermal de Broglie wavelength.

For an ideal gas, this is the classical statistical regime; nondegeneracy alone does not assert the absence of interactions.

For a dilute nonrelativistic gas with $x=m/T\gg 1$ and zero chemical potential, the Bose-Einstein distribution approaches the Maxwell-Boltzmann distribution. Expanding $E=m+p^2/(2m)+\cdots$ in the number density integral gives $n\sim g(mT/(2\pi ){)}^{3/2}{e}^{-x}$. Since $\rho =mn+3nT/2+\cdots$ and $P=nT$, the entropy density at zero chemical potential gives $s=n(x+5/2+\cdots )$ and hence the displayed leading expression. This includes the internal multiplicity $g$; a single real scalar field has $g=1$.

The claimed equality does not hold for the printed joint distribution. The original PDF specifies a centered Maxwell-Boltzmann distribution of $\mathbf{v}=\dot{\mathbf{R}}$ independent of $\mathbf{R}$, withSince the double average includes number density $n$, evaluating the Gaussian integrals givesThe cross term vanishes because the conditional mean of $\mathbf{v}$ is zero. In particular, set $K=0$ and take the equilibrium configuration probability density ${\psi }_{\mathrm{eq}}\propto e^{-H{R}^2/(2k_{B}T)}$. The left side is $3n\zeta k_{B}T/m>0$, while $nHC:K=0$. This is a counterexample even at thermal equilibrium, so no rearrangement of the earlier conformation tensor equation can prove the printed statement.

The discrepancy is thermal bead drag and heat-bath power: gross viscous loss of microscopic thermal motion is replenished by the thermal bath. It is not identical to mechanical polymer stress power. Moreover, $\dot{\mathbf{R}}$ in the earlier configuration continuity equation is a probability current velocity, including a distribution-dependent osmotic drift; it cannot simultaneously be identified with an independently sampled centered Maxwellian velocity.

A consistent overdamped result is the free-energy balance of a Hookean dumbbell. Define $\mathbf{q}=H\mathbf{R}+k_{B}T{\nabla }_{\mathbf{R}}\log \psi$ andDifferentiate the relative entropy and use the configuration Fokker-Planck probability current $\psi \mathbf{V}$. After integration by parts,The affine osmotic term is $n{k}_{B}T\int \nabla \psi \cdot K\mathbf{R}d\mathbf{R}=-n{k}_{B}T\mathrm{tr}K=0$. HenceThe polymer stress tensor ${\sigma }_p=n(HC-k_{B}TI)$ has ${\sigma }_p:K=nHC:K$ in an incompressible flow. It stores or releases free energy as well as dissipating it, so the total stress-work term in part a need not itself be a purely dissipative quantity for this viscoelastic material. Stress power equals $\mathcal{D}$ only for stationary free energy, and $\mathcal{D}$ uses the probability-current slip. Neither condition turns the printed independent-Maxwellian average into that expression.

Identify the transition-state separation with contact, $R=r_A+r_B$, take distinct atoms so $\sigma =1$, and let $E_A=N_A\Delta {\epsilon }_0^{‡}$. Then $\pi R^2=\rho$ and $\Delta {\epsilon }_0^{‡}/kT=E_A/RT$, yieldingThe velocity factor is the mean relative speed from the Maxwell-Boltzmann distribution. Thus this simple transition state theory model reproduces the stated collision theory expression. Molecular orientation constraints, quantum transmission and reaction-surface recrossing can invalidate the equality outside these simplifying assumptions.

Use Gaussian units and the kinetic magnetohydrodynamics equationstogether with species continuity equations, quasineutrality and the drift-kinetic equation. Here $p_{\perp },p_{\parallel }$ sum the species pressures. Interpret the thermal speed denominator in the source as the species mass $m_{\alpha }$, so $v_{\mathrm{th}\alpha }^2=2T_{0\alpha }/m_{\alpha }$ and the first adiabatic invariant is $\mu =v_{\perp }^2/(2B)$.

The specified Maxwell-Boltzmann distribution has zero mean motion and isotropic pressure, $p_{\perp 0}=p_{\parallel 0}=p_0$. All equilibrium scalars depend only on $x$, while ${\mathbf{b}}_0\cdot \nabla x=0$ and ${\mathbf{b}}_0\cdot \nabla B_0=0$. Therefore the equilibrium streaming and mirror terms of the drift-kinetic equation vanish. The remaining terms vanish because ${\mathbf{u}}_0=0$ and $E_{\parallel 0}=0$; the parallel electric constraint is also satisfied because the equilibrium parallel pressure gradient is zero. Species continuity equations and the induction equation are stationary, and quasineutrality holds by assumption.

It remains to check the perpendicular force balance. The radial magnetic pressure and magnetic tension giveThis is preciselyThus the Z-pinch is a steady solution of the stated kinetic magnetohydrodynamics system. Its equilibrium electric current is supplied by the small species drifts implicit in this approximation; zero leading common flow does not require zero electric current.

Choose the drift-kinetic equation in $(\mathbf{r},\mu ,v_{\parallel })$ with $v_{\parallel }$ the laboratory parallel velocity along the exact $\mathbf{b}$. To first order, before discarding terms small in the stated steep-gradient ordering, it isThe last term is the linear acceleration associated with the moving curved field. Terms quadratic in the perturbations have already been removed. A formulation using parallel velocity relative to the common flow gives the same physical result after transforming back to these variables.

The two mirror terms and the curvature acceleration are subleading here. Their relative sizes are controlled by $h\ll \delta p/p_0$ and $L_p/L_B,L_p/R\ll 1$. The field-direction perturbation multiplying the steep equilibrium gradient must remain:Using $u_x=-i\omega {\xi }_x$ and $\delta b_x=i{k}_{\parallel }{\xi }_x$ therefore gives the leading equationThe displaced branch satisfying quasineutrality has $\delta E_{\parallel }=0$ to this order: its species number densities are $\delta n_{\alpha }=-{\xi }_x{n}_{0\alpha }^{\prime }$, so ${\sum }_{\alpha }{q}_{\alpha }\delta n_{\alpha }=-{\xi }_x{\partial }_x{\sum }_{\alpha }{q}_{\alpha }{n}_{0\alpha }=0$. More explicitly, putting $\mathcal{D}=-i\omega +i{k}_{\parallel }{v}_{\parallel }$ givesAway from an independent longitudinal eigenfrequency, the summed susceptibility is nonzero, giving $\delta E_{\parallel }=0$. An independent longitudinal eigenmode is not forced by the interchange displacement. The required solution is thusThe apparent streaming denominator cancels identically, including at $\omega =k_{\parallel }{v}_{\parallel }$; this response has no leading Landau damping singularity.

To see the temperature and density contributions explicitly, differentiate at fixed $\mu ,v_{\parallel }$ and neglect $B_0^{\prime }$ at this order:The gyrotropic velocity-space measure is $d^{3}v=2\pi B_{0}d\mu d{v}_{\parallel }$. The species perpendicular plasma pressure and parallel plasma pressure are the momentsExplicit perturbations of the measure and of $\mu B$ produce terms proportional to $h$, which are subleading in these pressure moments; derivatives of the equilibrium measure produce the discarded $B_0^{\prime }$ terms. The remaining Maxwell-Boltzmann distribution integrals giveInsertion into b proves

The result is a fluid Eulerian perturbation generated by translating the entire equilibrium phase-space distribution function through ${\xi }_x$:Here $\Delta$ follows the displaced flux tube. Magnetic flux freezing carries the tube and its particles across the equilibrium pressure gradient. To leading order the tube changes neither its field strength nor its length by enough to alter the particle phase-space distribution function: these changes are smaller by plasma beta or $L_p/R$. Parallel streaming remains within the same displaced tube, on which the original equilibrium phase-space distribution function was uniform. It therefore cannot erase the displacement-generated contrast between adjacent tubes. This is why the term $v_{\parallel }\delta b_x{\partial }_x{f}_{0\alpha }$ in c exactly cancels the apparent parallel relaxation of $-{\xi }_x{\partial }_x{f}_{0\alpha }$. No collisional restoration of a Maxwell-Boltzmann distribution is needed.

The event rate counts unordered pairs of reacting nuclei. For unlike species there are $n_i{n}_j$ pairs per unit-volume normalization; for identical species there are approximately $n_i^2/2$ pairs, since counting ordered pairs counts each collision twice. The Kronecker delta therefore supplies exactly the required divisor $1+{\delta }_{ij}$. It is the event rate, not the destruction rate of identical nuclei: one identical-pair reaction destroys two nuclei.

The temperature dependence combines the high-energy tail of the Maxwell-Boltzmann distribution with quantum tunnelling through the Coulomb barrier. The principal exponential factor in the energy integral is $\exp [-E/(k_{B}T)-b/\sqrt{E}]$. Its stationary point, the Gamow peak, has $E_0\propto T^{2/3}$ and exponent $\eta =3E_0/(k_{B}T)\propto T^{-1/3}$. Integrating around this peak gives a prefactor proportional to $T^{-2/3}$, or ${\eta }^2$, when the nuclear factor varies slowly. Thus the supplied thermonuclear reaction rate is locally proportional to ${\eta }^2{e}^{-\eta }$. The slow weak interaction in the proton-proton reaction also makes its overall coefficient very small; its local thermal slope follows the supplied formula.

At fixed number densities, define the local temperature exponent of a thermonuclear reaction by a logarithmic derivative. Since $d\eta /d\log T=-\eta /3$,For proton-proton reactions, $A=1/2$ and $Z_1=1$. At $T_6=15$, ${\eta }_{11}=13.67$ and $\alpha =3.89$. For two helium-3 nuclei, $A=3/2$ and $Z_3^2{Z}_3^2=16$, giving ${\eta }_{33}=49.68$ and $\beta =15.89$. For helium-3 with helium-4, $A=12/7$ and the same charge product gives ${\eta }_{34}=51.95$ and $\gamma =16.65$. HenceThese are local powers near the specified temperature, not constant exponents over an arbitrarily large range. The greater reduced nuclear mass in the $34$ reaction increases its tunnelling exponent and its thermal sensitivity.

For the effective proton-proton reaction network, denote the event rates byThe short-lived beryllium intermediate can be eliminated, so production through $R_{34}$ feeds $n_7$, representing predominantly lithium-7. This rapid conversion is electron capture. Count the nuclei consumed and produced in each reaction: $11$ consumes two protons and makes one deuterium; $21$ consumes a proton and a deuterium and makes one helium-3; $33$ consumes two helium-3 and makes one helium-4 and two protons; $34$ consumes one of each helium isotope; $17$ consumes a proton and lithium-7 and makes two helium-4. It follows thatFor closure, ${\dot{n}}_7={\lambda }_{34}{n}_3{n}_4-{\lambda }_{17}{n}_1{n}_7$. These nuclear terms conserve $n_1+2n_2+3n_3+4n_4+7n_7$, providing an independent stoichiometric check. They describe a fixed local volume; compression or mixing would add transport terms.

The roughly one-second destruction time of deuterium is negligible compared with the proton-burning and helium-3 evolution times. Its abundance rapidly adjusts to the slowly evolving production rate. Set ${\dot{n}}_2\simeq 0$, giving the deuterium quasi-equilibrium in proton-proton burningSubstitution, without setting ${\dot{n}}_1=0$, givesThe helium-3 relaxation time of about $6\times 10^5$ years is likewise short compared with the central proton-burning time and the age of the Sun. On this relaxation time the proton and helium-4 abundances, and the background temperature, can be treated as fixed. Put $A={\lambda }_{33}$, $B={\lambda }_{34}{n}_4$, $C={\lambda }_{11}{n}_1^2/2$. Then ${\dot{n}}_3=C-A{n}_3^2-B{n}_3$, and its nonnegative equilibrium isThe other root is negative and unphysical. Write $n_3=n_{3e}+x$ and subtract the equilibrium equation. Exactly, for a fixed background,Dropping the quadratic displacement givesThus small displacements decay exponentially. A slowly moving equilibrium introduces an additional $-{\dot{n}}_{3e}$ term, negligible to leading order when $\tau$ is short compared with background evolution.

To estimate the temperature for helium-3 freeze-out, use the cooler pp-I-dominated region and neglect $B$ compared with $2A{n}_{3e}$. Then $n_{3e}\simeq n_1\sqrt{{\lambda }_{11}/(2{\lambda }_{33})}$ and ${\tau }^{-1}\simeq n_1\sqrt{2{\lambda }_{11}{\lambda }_{33}}$. With the local powers just obtained and fixed proton number density, $n_{3e}\propto T^{-6}$ and $\tau \propto T^{-10}$. Adopting $t_{\odot }\simeq 4.6\times 10^9\mathrm{yr}$ givesThis is an order-of-magnitude extrapolation of local exponents. Keeping the supplied exponential factors gives the helium-3 freeze-out estimate with Gamow factorswhich yields about $6.8\times 10^6\mathrm{K}$ with the same normalization and fixed-density approximation. Thus the robust estimate is several million kelvin. The supplied data do not fix the local density, composition or pp-II fraction away from the centre, so they cannot determine an exact transition temperature or radius. In particular, using $\tau \propto T^{-16}$ alone would omit the temperature dependence of the equilibrium helium-3 abundance.

The solar hydrogen and helium-3 abundance profiles follow from burning, finite equilibration time and convection. Central hydrogen has been consumed, so $X_1$ is lowest in the centre and increases toward its much less processed outer value. In the hot centre, rapid destruction keeps helium-3 small. Moving outward initially raises its equilibrium abundance because ${\lambda }_{11}/{\lambda }_{33}$ rises strongly as temperature falls. Further out, equilibration is no longer reached within the solar age, and still farther out even production is slow. Hence $X_3$ has an off-centre maximum and then declines toward its small outer abundance. The outer convective envelope mixes each abundance toward a constant value.

The sketch shows these qualitative shapes; the helium-3 ordinate is scaled to its own maximum. The illustrated amplitudes, peak radius and envelope boundary are schematic, not a numerical solar-model solution.

Work in natural units $k_B=\hslash =c=1$, initially neglecting helium and excited hydrogen states. The reaction $p+e^{-}\leftrightarrow H+\gamma$ is in chemical equilibrium when ${\mu }_p+{\mu }_e={\mu }_H$, since equilibrium photons have zero chemical potential. Substitution of the nonrelativistic Maxwell-Boltzmann distribution for the three densities givesFor ground-state hydrogen use $g_e=g_p=2$, $g_H=4$, and $m_H\simeq m_p$ in the translational prefactor. Writing its Hydrogen binding energy as $I$, this becomes $(m_{e}T/(2\pi ){)}^{3/2}{e}^{-I/T}$. Charge neutrality gives $n_p=n_e=X_e{n}_B$, while baryon conservation gives $n_H=(1-X_e)n_B$. Hence the hydrogen-only Saha equation isThe photon Bose-Einstein distribution supplies $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, where $\zeta (3)$ is the usual Riemann zeta function value in the thermal number integral. With baryon-to-photon ratio ${\eta }_{\gamma }=n_B/n_{\gamma }$, substitution yieldsThis reproduces the requested numerical coefficient, with the printed $\xi (3)$ interpreted as the thermal $\zeta (3)$ value. There is a normalization ambiguity in the PDF: it writes $\eta =n_B/s$ without defining $s$. To reproduce this coefficient, $s$ must mean photon number density. If $s$ has its usual entropy density meaning, the Saha baryon-abundance normalization instead givesThe two forms agree through ${\eta }_{\gamma }=[{\pi }^4{g}_{\ast s}/(45\zeta (3))]{\eta }_s$, with the entropy sector specified. A baryon-to-entropy ratio cannot simply be inserted into the photon-normalized expression.

As the universe cools, the Boltzmann factor favors neutral atoms and cosmological recombination sharply lowers $X_e$. For an order-one ionized fraction the photon-normalized equation gives the balanceup to an order-one ionization-fraction factor. The small baryon abundance means that there are very many photons per baryon: even a small high-energy photon tail can ionize most atoms. Thus the recombination temperature is far below the microscopic binding scale $I\simeq 13.6\mathrm{eV}$, typically a few tenths of an eV for the standard abundance, rather than of order $I$.

Photon decoupling is a related but distinct kinetic event. The Thomson scattering rate $n_e{\sigma }_T$ falls relative to $H$ as free electrons disappear, and photons begin free streaming. The precise last-scattering epoch depends on integrated optical depth, not only the local equality of rates. At still lower density the effective recombination rate per electron also drops below the expansion rate. Residual electron freeze-out leaves a small nonzero ionized fraction, larger than the vanishingly small equilibrium Saha prediction; radiative atomic bottlenecks mean the entire history cannot be obtained from equilibrium alone.

The temperature comparison in the last request needs an interpretation. If “reionisation temperature” means the microscopic thermal ionization scale, the large photon-to-baryon ratio explains why cosmological recombination requires $T\ll I$. If it means later cosmic reionization, there is no universal thermal threshold: stars or other sources supply ionizing photons and can reheat gas. The cosmological photon background is actually colder at that later epoch, because $T_{\gamma }\propto a^{-1}$. Its temperature therefore cannot literally be higher than the background temperature at recombination merely because hydrogen is reionized.

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 dilute nonrelativistic nucleons in thermal equilibrium, their Maxwell-Boltzmann distribution gives $n_j=g_j(m_{j}kT/2\pi {\hslash }^2{)}^{3/2}\exp [({\mu }_j-m_j{c}^2)/kT]$. The neutron and proton spin degeneracies are equal, and their mass prefactor ratio is nearly one. ThusWeak reaction equilibrium imposes ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Charge neutrality in an electron-positron plasma, together with the small baryon asymmetry, makes ${\mu }_e/kT$ very small. Assuming negligible neutrino lepton asymmetry also makes ${\mu }_{{\nu }_e}/kT$ small. Hence the difference, rather than the absolute baryon chemical potentials, is negligible, yieldingA large neutrino chemical potential would invalidate this last approximation; it is an assumption about the lepton asymmetry, not a general rule that baryonic chemical potentials vanish.

In the nonrelativistic dilute regime, $E(p)=m{c}^2+p^2/(2m)+\cdots$ and the Fermi-Dirac distribution reduces to the Maxwell-Boltzmann distribution. HenceDifferentiating the Gaussian integral with respect to its coefficient gives ${\int }_0^{\infty }{p}^2{e}^{-a{p}^2}dp=\sqrt{\pi }/(4a^{3/2})$. ThereforeThe degeneracy factor here is the one in the supplied integral, namely one. Substituting $kT=m{c}^2/\alpha$, $\mu =0$ and dividing by the supplied photon density givesAt $\alpha =20$ this is about $4.8\times 10^{-8}$. The relic is already nonrelativistic at decoupling and becomes still colder as its momenta redshift, making it a possible cold dark matter component if it is stable and interacts sufficiently weakly with light.

For orientation, write ${\eta }_b=n_b/n_{\gamma }$. Ignoring subsequent entropy transfer, its density relative to baryons would be ${\rho }_X/{\rho }_b=(1/20)(n/n_{\gamma })/{\eta }_b$. With the cosmological baryon-to-photon ratio of order $6\times 10^{-10}$, this is of order four, demonstrating the right cosmological scale. The measured dark-to-baryonic density ratio is likewise of order five; an observational reference is arxiv.org/abs/1807.06209 . This is a candidate abundance argument, not an exact relic-density calculation: photon heating after decoupling changes $n/n_{\gamma }$, and stability and interaction properties were not specified. The supplied mass and freeze-out parameter alone therefore do not establish a precise fit.

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.

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.