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For a photon gas in thermal and chemical equilibrium, photon number is not a conserved thermodynamic charge. Number-changing processes can create and absorb photons without a corresponding conserved-number constraint, so the equilibrium photon chemical potential is zero.

At $T=5\times 10^9\mathrm{K}$, $k_{\mathrm{B}}T\simeq 0.431\mathrm{MeV}$. The other light species are electrons, Positrons, three neutrino species and their antiparticles. Neutrinos are ultrarelativistic; electrons and Positrons are only moderately relativistic at this particular temperature because $m_e{c}^2=0.511\mathrm{MeV}$. The later simple counting formulas use the intended ultrarelativistic approximation for them. Baryons, muons and heavier particles are not an abundant relativistic bath at this temperature.

Pair equilibrium implies ${\mu }_{e^{-}}+{\mu }_{e^{+}}=2{\mu }_{\gamma }=0$. Approximate charge symmetry, with the net electron excess tied to the much smaller baryon asymmetry, then makes both electron and Positron chemical potentials small compared with $k_{\mathrm{B}}T$. For neutrinos one additionally assumes negligible lepton asymmetry. Their near-zero chemical potentials are not implied merely by their statistics. After neutrino decoupling, a distribution initially having negligible chemical potential retains a redshifted Fermi-Dirac distribution, with its own temperature.

For zero chemical potential, each species contributes $S_i=(U_i+P_{i}V)/T=4U_i/(3T)$. ThusEntropy density scales as $T^3$ at fixed relativistic particle content; total entropy also includes the changing proper volume. An isolated adiabatically expanding comoving gas has conserved total entropy and $aT$ constant, rather than its total entropy falling as $T^3$.

At equal temperature, single electrons and photons each have two spin/polarization states. In the ultrarelativistic approximation,Positrons are a separate species with the same contribution, so $(S_{e^{-}}+S_{e^{+}})/S_{\gamma }=7/4$. This distinction is needed for the heating count in part d. At the exact supplied temperature the electron rest mass gives corrections to the ideal $7/8$ ratio; it is the limiting relativistic value requested by the problem.

There is also a sign convention issue in the introductory entropy formula. With the chemical-potential convention used in the supplied distribution, the Euler relation is $U=TS-PV+{\sum }_i{\mu }_i{N}_i$, so $S=(U+PV-{\sum }_i{\mu }_i{N}_i)/T$. Its chemical-potential term has the opposite sign to the printed one. For a concrete counterexample, a dilute monatomic ideal gas with one internal state and $n{\lambda }_T^3=e^{-1}$, where ${\lambda }_T$ is the thermal de Broglie wavelength, has $\mu =-k_{\mathrm{B}}T$, $U=3N{k}_{\mathrm{B}}T/2$ and $PV=N{k}_{\mathrm{B}}T$. Its entropy is $7N{k}_{\mathrm{B}}/2$; the printed plus sign would instead give $3N{k}_{\mathrm{B}}/2$. All chemical potentials have been neglected here, so this source error does not change any of the above results.

Use the observed present photon temperature $T_{\gamma 0}\simeq 2.725\mathrm{K}$. This is an additional observational input, consistent with ^[ref]. Standard neutrino decoupling and electron-positron entropy transfer giveFor nonrelativistic relic neutrinos this is the temperature parameter of the redshifted momentum distribution, not the temperature of a newly thermalized gas of massive particles.

The photon number density from part b isOne neutrino mass eigenstate and its antineutrino have two populated helicity states. ThusThe total over three species is about $336\mathrm{c}{\mathrm{m}}^{-3}$, including antineutrinos. Comoving number conservation preserves these abundances after the particles become nonrelativistic.

For the supplied $H_0=72\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}=2.33\times 10^{-18}{\mathrm{s}}^{-1}$,Assuming nonnegative other energy components, relic neutrino rest-mass density cannot exceed the near-critical total density. With the common per-species abundance $n_{\ast }=n_{{\nu }_i}+n_{{\bar{\nu }}_i}$,Using the total $336\mathrm{c}{\mathrm{m}}^{-3}$ would instead bound the mean mass, not the sum. This critical-density bound on the sum of relic neutrino masses is deliberately conservative: baryons, dark matter and dark energy leave less density available. It assumes the standard thermal abundance, negligible chemical potentials, no later dilution or extra populated states, and the idealized decoupling count. Small non-instantaneous-decoupling corrections do not matter for this estimate.

A quasi-Fermi level is the chemical potential parameter of an approximately thermalized carrier population within one energy band. Rapid intraband scattering can establish a Fermi-Dirac distribution in each energy band before slower interband recombination equilibrates their populations. Thus separate $E_{\mathrm{Fc}}$ and $E_{\mathrm{Fv}}$ describe a pumped population inversion even though the entire semiconductor is far from thermal equilibrium.

At a common positive temperature, the Fermi-Dirac distribution is strictly decreasing in $E-E_F$. ConsequentlyThe joint density of states is nonzero only above the band gap, soIn particular the quasi-Fermi-level splitting must exceed $E_g$. The upper boundary is transparency, where stimulated emission balances absorption; the lower boundary has zero joint density of states in this ideal three-dimensional parabolic model. The strict occupation comparison uses $T>0$; at exactly zero temperature occupations become step functions, and their supports should be checked explicitly, as in the next part.

For a mixture with phase compositions $x_1,x_2$, its molar Gibbs free energy is the straight-line combination $g=(1-f)g_1(x_1)+f{g}_2(x_2)$ at overall $x=(1-f)x_1+f{x}_2$. Minimizing over phase compositions gives equal slopes and equal tangent intercepts: the common-tangent construction expresses equality of both component chemical potentials. The equilibrium curve is the lower convex envelope of the candidate phase curves. Phase fractions follow the lever rule.

The Boltzmann distribution for two atomic levels at thermodynamic equilibrium iswhere $\omega$ is the statistical weight of an atomic level. For a resolved level of total angular momentum $J$, $\omega =2J+1$ when magnetic substates are unresolved.

To extend this counting to the continuum, assume an ideal, nondegenerate plasma with classical free electrons and neglect the small difference between the translational masses of the two heavy ionic species. Let $z_e=e^{{\mu }_e/kT}$ be the electron fugacity in the Saha equation, with the electron chemical potential measured relative to its rest energy. The two electron spin states and the phase-space cell volume $h^3$ giveChemical equilibrium for capture equates the bound species' chemical potential to the sum for its parent ion and free electron. Thus applying Boltzmann factors to a bound level at energy $-I_n$ gives $N_n/N_{+}=({\omega }_n/{\omega }_{+})z_e{e}^{I_n/kT}$. Eliminating the electron fugacity proves the level-specific Saha ionization equation:The factor two is the electron spin multiplicity; the factor with power $3/2$ is the inverse free-electron translational state density. A total-ionization Saha equation sums the bound states into an internal partition function, whereas this expression refers to one specified level.

For dielectronic recombination, write the entrance ion as $X^{(z+1)+}$ in ground state $i$, and the intermediate state of $X^{z+}$ as $d=(j,n\ell )$. Capture excites the core from $i$ to $j$ while binding the incident electron. The resonance energy relative to the entrance continuum is $\bar{E}=E_{ij}-I_{n\ell }>0$. The doubly excited state can eject the electron by autoionization, or undergo radiative stabilization to a bound state.

Let $A_{di}^a$ be the partial autoionization rate back to the entrance ion, $A_a$ the total autoionization rate, $A_r$ the total radiative decay rate, and $A_s$ the rate of radiative decays that stabilize into bound levels. For an isolated narrow resonance, the Saha-Boltzmann equilibrium ratio has the bound-state energy replaced by the positive resonance energy:At equilibrium, inverse dielectronic capture and autoionization fluxes balance: $N_e{N}_i{C}_d=N_d^{\mathrm{eq}}{A}_{di}^a$. This establishes the capture coefficient even when the actual plasma is not in local thermodynamic equilibrium. In the low-density, weak-radiation recombining plasma, the intermediate level instead has the stationary balanceMultiplying its population by the successful stabilization rate gives the actual recombination coefficientHere the equilibrium population was used only to determine the capture rate by detailed balance; it was not assumed to equal the intermediate population when photons escape. This is precisely why the competing decay probabilities appear.

Use the supplied core-transition relation between the Einstein coefficients and atomic oscillator strength,Here $\alpha$ is the fine-structure constant and $a_0$ the Bohr radius, not a recombination coefficient. Factor out $A_{ji}$ and defineSubstitution yieldsThis definition accommodates several entrance and decay channels. In the common spectator-electron approximation with one autoionization channel and stabilization by the core transition, $A_s=A_r=A_{ji}$ and $A_{di}^a=A_a$. If all coupled substates of the captured $n\ell$ electron are summed, $g_d=2(2\ell +1)g_j$, givingFor an intermediate state resolved by fine structure, retain its actual $g_d$ instead of that summed weight. The stabilization probability itself is $A_s/(A_a+A_r)$; the requested ${\beta }_{j,n\ell }$ is not just this branching probability, because the prefactor has already factored out the core radiative rate and statistical weights of atomic levels. Radiative transitions to another still-autoionizing state require that state's eventual survival probability, rather than automatically counting every emitted photon as stabilization.

The total dielectronic recombination coefficient sums all accessible positive-energy resonances. In this isolated-resonance approximation it has the formwith atomic factors included in $c_d$. The resonance-temperature dependence of dielectronic recombination follows by differentiating a single term's logarithm:Thus low temperatures suppress a positive-energy resonance, although resonances very near threshold can still matter; at sufficiently high temperature a fixed set of resonances falls as $T^{-3/2}$. Several core-excitation series can produce several peaks. When autoionization is much faster than radiation, the capture-and-survival product is limited by the stabilizing radiative rate; when capture is weak, autoionization is the limiting inverse-capture rate. For a fixed Rydberg angular channel, autoionization typically decreases at high principal quantum number while the core radiative rate changes little. Capture into very high levels is vulnerable to subsequent collisional or field ionization, so finite-density survival can suppress the total rate. dielectronic recombination is resonant and strongly temperature-selective; its magnitude can be important in coronal ionization balance.

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.

Use radiation-era $H=C_{\ast }{T}^2/m_{\mathrm{pl}}$, with $C_{\ast }=1.66\sqrt{g_{\ast }}$. The stipulated interaction rate is ${\alpha }_Y^4{T}^5/m_Y^4$, not ${\alpha }_Y^2{T}^5/m_Y^4$. Decoupling therefore givesAt the threshold $T_D=m_{Y\mathrm{nr}}$, the assumed $g_{\ast }\simeq 100$ is self-consistent with the stated high-temperature condition. Smaller masses decouple nonrelativistically, with $x_D=m_Y/T_D=(m_{Y\mathrm{nr}}/m_Y{)}^{1/3}\gg 1$.

For an thermal equilibrium species with zero chemical potential, $n_Y=2(m_Y{T}_D/2\pi {)}^{3/2}{e}^{-x_D}$, while $s=2{\pi }^2{g}_{\ast s}{T}_D^3/45$. ThusThe numerical coefficient before rounding is about $2.9\times 10^{-3}$ for $g_Y=2$ and $g_{\ast s}=100$. This is mass-dependent fermion freeze-out. It assumes the rate maintains chemical equilibrium until sudden cosmological particle freeze-out and that entropy is conserved afterwards. Elastic scattering alone would not justify the thermal equilibrium number-density estimate. For decoupling below the high-temperature range, changes in effective degrees of freedom modify the numerical estimate; the requested formula treats them as roughly constant.

Above the transition,for $g_{\ast }\sim 100$. The ratio increases during cooling, rather than at higher temperature. A GUT transition at $T_c\sim 10^{15}$–$10^{16}$ GeV occurs well above this scale, so the interaction is too slow to establish or maintain thermal equilibrium throughout $T>T_c$. In fact its integrated interaction probability from an arbitrarily high temperature down to $T_c$ is at most of order $T_{\mathrm{eq}}/T_c\ll 1$, since $dt=-dT/(HT)$.

An initial Fermi-Dirac distribution with zero chemical potential therefore cannot be assumed from this high-temperature rate alone. If imposed initially and collisionlessly redshifted while the particles are massless, its thermal-looking form can persist, but this is not dynamically maintained thermal equilibrium. Other initial states can remain nonthermal. This is the thermalization temperature for a massless gauge mediator.

The below-transition rate stipulated in the model has $\Gamma /H=(T/T_D{)}^3$. If $T_c\gg T_D$ and it includes efficient number-changing reactions, it can restore chemical equilibrium after the transition and erase the initial-state memory, making the estimates in a and b applicable. If it only equilibrates momenta, or if reheating never populates the relevant range, the thermal relic calculation must be reconsidered.

The baryon asymmetry is an excess of net baryon number over zero, conveniently measured by ${\eta }_B=(n_B-n_{\bar{B}})/n_{\gamma }$. Ordinary matter contains abundant baryons, whereas no comparable primordial antibaryon population is observed. Large nearby domains of matter and antimatter would produce radiation through particle-antiparticle annihilation at their boundaries. Antiparticles produced in energetic collisions are not evidence for a cosmological antimatter domain. The small nonzero baryon-to-photon ratio inferred from Big Bang nucleosynthesis is independently tested by baryon-density consistency between nucleosynthesis and the microwave background.

Starting with zero net baryon number, the Sakharov conditions require baryon-number violation, C and CP violation, and departure from thermal equilibrium. Conservation of baryon number would keep the initial value zero. C symmetry or CP symmetry would pair baryon production with equal antibaryon production; CP rate cancellation in baryogenesis explains why C violation alone does not suffice. In thermal equilibrium with zero imposed charge chemical potentials, CPT symmetry forbids the required net baryon number. An asymmetric rate in one reaction must not be mistaken for net production after all reverse reactions are included.

A simple example is heavy-boson decay baryogenesis. Take equally many heavy $X$ bosons and their antiparticles, with decay channelsThe symbols in parentheses are branching fractions. The two $X$ channels carry baryon numbers $2/3$ and $-1/3$, so both cannot conserve one assigned baryon number for $X$. Their mean baryon number is $r-1/3$, while the mean for $\bar{X}$ is $-\bar{r}+1/3$. Consequently,where $ϵ$ is the asymmetry per decaying particle. CP violation permits $r\ne \bar{r}$ without violating the equality of total decay widths required by CPT symmetry. Microscopically a direct CP asymmetry requires interference between decay amplitudes with both relative weak phases and absorptive phases; specifying complex couplings alone would not establish unequal partial decay widths.

For radiation domination, a sufficient decay condition is ${\Gamma }_X\lesssim H(T=M_X)=1.66\sqrt{g_{\ast }}{M}_X^2/m_P$. The heavy population can then survive as the equilibrium abundance drops, before decaying while inverse production is suppressed. Its actual abundance differs from its thermal equilibrium value. If subsequent baryon-number washout is negligible and cosmological entropy conservation holds, an initial total abundance $Y_X=(n_X+n_{\bar{X}})/s$ producesHere $s$ is the entropy density. Residual baryon-number washout multiplies this expression by an efficiency factor below one. The example realizes all three Sakharov conditions; it illustrates a mechanism, rather than identifying the actual cosmological source of the baryon asymmetry.

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.

Use natural units, assume negligible neutrino chemical potentials, and approximate neutrino decoupling as occurring before electron-positron annihilation in cosmology. At high temperature, weak interactions maintain a common temperature for neutrinos, antineutrinos, electrons, Positrons and photons. Since ${\Gamma }_{\nu }/H\propto T^3$, the reaction rate eventually loses against expansion. Decoupled relativistic neutrinos then free stream; their momenta redshift as $a^{-1}$, so their distribution retains a temperature parameter satisfying $a{T}_{\nu }=\mathrm{constant}$.

The still-coupled electromagnetic bath receives the entropy of annihilating electrons and Positrons. Its effective number of relativistic entropy degrees of freedom is initiallywhere the two photon polarizations count as bosonic states and the four electron-positron spin states have the fermion entropy weight $7/8$. After the pairs disappear, $g_{\ast s}^{\mathrm{EM}}=2$. Cosmological entropy conservation in this coupled bath gives $g_{\ast s}^{\mathrm{EM}}(a{T}_{\gamma }{)}^3=\mathrm{constant}$. The three already decoupled neutrino species are excluded from this bath's entropy count. At decoupling $T_{\nu }=T_{\gamma }$, henceSubsequent expansion preserves this ratio in the idealized history. Residual interactions during annihilation give small corrections to the instantaneous-decoupling approximation. If a relic species later becomes massive, $T_{\nu }$ here remains the redshifting momentum-distribution parameter; it need not be its nonrelativistic kinetic temperature.

A single ordinary neutrino species includes one populated neutrino helicity and one antineutrino helicity. For negligible chemical potential, integration of the Fermi-Dirac distribution givesTherefore the number density of a relativistic neutrino relative to photons, with the antiparticle included, isThe factor $3/4$ is the fermionic number-density weight, distinct from the energy and entropy weight $7/8$. This relation continues to hold after this species becomes nonrelativistic if its comoving particle number is conserved.

Under the proposed matter-dominated, spatially flat model, the Friedmann equation requires the present relic-neutrino energy density to be approximately the critical density. Using the unreduced Planck mass $M_P=G^{-1/2}$ givesThus the critical-density mass of a thermal relic neutrino isTaking the paper's rounded values, $T_{\gamma 0}=2\times 10^{-13}\mathrm{GeV}$, $H_0=h/(4.5\times 10^{41}{\mathrm{GeV}}^{-1})$ and $M_P\sim 10^{19}\mathrm{GeV}$, givesFor $0.5<h<1$, this is of order $30$--$100\mathrm{eV}$ with these approximate constants. More precise constants change the coefficient, not the requested mass scale. This mass is much larger than the present momentum scale $T_{\nu 0}\simeq 1.4\times 10^{-4}\mathrm{eV}$, validating the nonrelativistic energy density approximation. The result assumes that one thermal relic species dominates the matter density; it is conditional on that model rather than a measurement of a neutrino mass.

Let the uniform background number density be $n_{\infty }>0$ and set $\mu =U{n}_{\infty }$. The subtraction of this chemical potential requires a phase rotation as well as a rescaling:Substitution into the physical Gross–Pitaevskii equation givesConsequently the requested coefficients follow fromThe length unit is the specified healing length. Removing tildes produces the paper's dimensionless equation; in particular the coefficient of its time derivative fixes $t_0$, not its reciprocal. Without the phase rotation the linear $+\Psi$ term would not appear.

The dimensionless ground state is $\Psi =e^{i\chi }$ for any constant phase $\chi$, conventionally $\Psi =1$. Its subtracted energy is proportional to $\int [|\nabla \Psi {|}^2+(1-|\Psi {|}^2{)}^2/2]d^{3}x$, which is nonnegative and vanishes for this state. In physical variables it has density $n_{\infty }$ and phase rotating at $\mu /\hslash$.

The Hamiltonian functional isTreating the field and its complex conjugate as independent variables, and integrating the gradient term by parts with periodic, decaying or fixed-field boundary conditions, givesThus $i\hslash {\psi }_t=\delta H/\delta {\psi }^{\ast }$ is the Gross–Pitaevskii equation. The factor $1/2$ in the interaction energy avoids counting each pair twice.

At prescribed particle number, minimize $H$ subject to $N=\int |\psi {|}^2{d}^{3}x$. Introducing the chemical potential as a Lagrange multiplier, the variation of $H-\mu N$ vanishes at a ground-state profile ${\psi }_0$, provingThe corresponding time-dependent field is $\psi (\mathbf{x},t)=e^{-i\mu t/\hslash }{\psi }_0(\mathbf{x})$. A solution of this stationary equation is only a candidate ground state; the ground state is the minimizer among all fields satisfying the normalization. For the uniform repulsive gas in a periodic volume, the gradient energy is nonnegative and $\int n^2{d}^{3}x\ge N^2/\mathcal{V}$, so the constant-density, constant-phase state is indeed the minimum. The uniform-background solutions below assume $U>0$ and a positive background density.