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The cosmic neutrino background last-scattering surface is the locus from which relic neutrinos that reach an observer underwent their final appreciable scattering. Although neutrino decoupling occurs before photon decoupling, a massive neutrino travels slower than light after becoming nonrelativistic, so its last-scattering surface need not be farther away than the cosmic microwave background last-scattering surface.

Combining the degenerate neutrino and antineutrino energy density with a critical density bound constrains the neutrino degeneracy parameter through the fourth root of the available energy budget. The numerical bound depends on $T_{\nu ,0}/T_{\gamma ,0}$ and the number of degenerate flavors. The standard ratio $(4/11{)}^{1/3}$ assumes neutrino decoupling before electron-positron annihilation in cosmology; it should be stated as a thermal-history assumption if the degeneracy is very large. The massless formula is inappropriate for a species that is nonrelativistic today.

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.

Assume standard radiation-dominated expansion, negligible chemical potential, one neutrino helicity and one antineutrino helicity, and a stable species $X$ with no later entropy injection. Use natural units and unreduced Planck mass $m_{\mathrm{pl}}\simeq 1.22\times 10^{19}\mathrm{GeV}$. The relativistic energy density is ${\rho }_R={\pi }^2{g}_{\ast }{T}^4/30$. The Friedmann equation then givesFor thermal decoupling in cosmology, equate the interaction rate $G_A^2{T}_D^5$ to $H(T_D)$:This order-of-magnitude estimate assumes the supplied rate law remains applicable there.

A decoupled relativistic species cools as $T_X\propto a^{-1}$, while annihilations heat the coupled bath. Treat the supplied high-temperature count $100$ as the remaining bath's entropy count, approximately equal to its energy count. Cosmological entropy conservation down to ordinary neutrino decoupling givesSubsequent electron-positron annihilation in cosmology heats photons but neither neutrino population. Therefore the two-stage entropy dilution of an early-decoupled neutrino givesThe ordinary Cosmic neutrino background has $T_{\nu }/T_{\gamma }=(4/11{)}^{1/3}$. If the rounded count $100$ instead includes the new two helicities, the remaining bath count is $98.25$; this changes the answer only slightly, to about $0.341$. Keeping the extra species' entropy separate is essential in either convention.

For the two relativistic fermion states, $g_{\ast s,X}=(7/8)2=7/4$. Photon and three standard neutrino species have present entropy degrees of freedom, referred to $T_{\gamma }$,Writing $\theta =T_X/T_{\gamma }$, the fractional entropy of a relativistically decoupled species isThus approximately 1.7% of the thermal particle entropy is in the new neutrinos and antineutrinos. If $100$ counts all states before the new decoupling, the fraction is $1.75\%$. These are the same rounded estimate. The question concerns the radiation/relic thermal entropy budget; it does not supply a model for entropy in later black holes or other gravitational structures.

If $X$ has a small mass but decoupled relativistically, its comoving particle number is still conserved. The relativistic-neutrino number-density ratio givesOnce nonrelativistic, ${\rho }_X\simeq m_X{n}_X$. The density bound on a relativistically decoupled massive relic is thereforeUsing the standard present photon abundance and critical-density normalization, this is approximately ${\Omega }_X{h}^2\simeq m_X/(9\times 10^2\mathrm{eV})$, where $h=H_0/(100\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1})$ is the present Hubble normalization, not a magnetic field. A stable relic must not exceed the permitted matter density; for example, if ${\Omega }_M{h}^2=0.15$, even assigning all matter to it gives only an order-$10^2\mathrm{eV}$ mass ceiling.

A stronger restriction can follow from neutrino free streaming and the observed formation of large-scale structure of the universe: a light thermal population acts as hot rather than cold dark matter, erasing structure below its streaming scale. If at most a fraction $f_{\max }$ of the matter may be in this population, replace the overclosure ceiling by $m_X\lesssim 9\times 10^2{f}_{\max }{\Omega }_M{h}^2\mathrm{eV}$. The allowed fraction is observational input and is not supplied, so the paper does not determine a unique tighter mass limit. A species still relativistic today is radiation rather than matter; its contribution to the effective number of neutrino species is $\Delta N_{\mathrm{eff}}=(T_X/T_{\nu }{)}^4\simeq 0.051$, and must instead satisfy the radiation-density constraints. These conclusions assume stable thermally populated states; decay or nonstandard entropy histories change the bounds.

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.

Before electron-positron annihilation in cosmology, photons contribute two helicity states; Electrons and positrons contribute two spin states each. Each of the three light neutrino flavours contributes one active neutrino helicity and one antineutrino helicity. At a common temperature the effective number of relativistic energy degrees of freedom is thereforeUsing $G_F=1.17\times 10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}\simeq 2.435\times 10^{18}\mathrm{GeV}$ in the scaling formula alone gives $T_F\sim 2\mathrm{MeV}$. Including the explicit expansion coefficient from (ii), but setting the unknown rate coefficient $A$ to one, gives $T_F\simeq 1.5\mathrm{MeV}$. Thus the warranted conclusion is a freeze-out temperature of order $1\mathrm{MeV}$, above the $0.511\mathrm{MeV}$ Electron mass. The omitted weak-rate coefficient prevents a more precise prediction from this estimate.

Neutrino decoupling occurs before most electron-positron pairs annihilate. Afterwards neutrinos redshift freely, $T_{\nu }a=\mathrm{constant}$, while the photons receive the pair entropy. For the still-coupled photon-pair gas, cosmological entropy conservation gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$. Its entropy degrees of freedom drop from $2+(7/8)4=11/2$ to $2$. Comparing this with the freely cooling neutrinos givesAfter annihilation, photon and neutrino temperatures differ. To write the total energy density as $({\pi }^2/30)g_{\ast }{T}_{\gamma }^4$, the neutrino contribution must include that temperature ratio:This is the paper's post-annihilation ${\mathcal{N}}_{\mathrm{eff}}$. Merely deleting the Electron term while keeping all species at the same temperature would be incorrect. For comparison the entropy count, normalized to $T_{\gamma }^3$, is $g_{\ast s}=2+(7/8)6(4/11)=43/11\simeq 3.91$. These results assume instantaneous neutrino decoupling and relativistic neutrinos; the two counts represent different thermal moments.

Count one neutrino species as the usual two thermally populated particle-antiparticle helicity states, with negligible chemical potential. Its entropy weight is $g_{\ast s,X}=(7/8)2=7/4$. In the ideal high-temperature Standard Model, the bosonic count is $16+8+4=28$ from gluons, electroweak gauge bosons and the Higgs field. The fermionic count is $72+12+6=90$ from quarks, charged leptons and the three ordinary neutrino species. Hence the relativistic entropy content of the Standard Model is $g_{\ast s,D}=28+(7/8)90=106.75$ for the remaining bath. Before the new species leaves, the total count is $108.5$; its separately cooling states are excluded from the subsequent bath count.

Near ordinary neutrino decoupling, the still-coupled bath hasfrom photons, electron-positron pairs and the three ordinary neutrinos. After those neutrinos decouple, the electromagnetic bath falls from $2+(7/8)4=11/2$ to two when the pairs annihilate. Cosmological entropy conservation in that bath gives $T_{\nu }/T_{\gamma }=(4/11{)}^{1/3}$.

The new species has cooled as $T_X\propto a^{-1}$ since its earlier decoupling, whereas annihilation of Standard Model species heated the remaining bath. The temperature ratio of an early-decoupled neutrino is thusThese are ideal-gas estimates with the high-temperature count implicit in the question; precise electroweak thresholds change them slightly.

Entropy degrees of freedom weight temperature ratios with their cube, while energy degrees of freedom use their fourth power. The remaining Standard Model entropy today, in photon-temperature units, is $g_{\ast s,0}^{\mathrm{SM}}=2+(7/8)6(4/11)=43/11$. The new contribution is $(7/4)(T_X/T_{\gamma }{)}^3$. Therefore the fractional entropy of a relativistically decoupled species isAbout 1.6% of the total entropy is in the new species. Separate comoving entropy conservation also derives this fraction directly at decoupling. A model populating four helicity states instead would require the corresponding doubled species count.

For a quantitative thermal-relic prediction, assume one ordinary weakly interacting neutrino flavor: one populated neutrino helicity and one antineutrino helicity, negligible chemical potential, early thermal equilibrium, and relativistic decoupling around the MeV epoch. Equal neutrino and antineutrino populations imply no new lepton asymmetry. The Fermi-Dirac distribution is maintained while the weak interaction rate exceeds the expansion rate. After neutrino decoupling, the distribution free streams, its momentum scale falls as $a^{-1}$, and its comoving number is conserved. A mass of $1\mathrm{keV}$ is negligible at Big Bang nucleosynthesis temperatures, but the relic later becomes nonrelativistic. Its energy density then changes from radiation-like $a^{-4}$ scaling to matter-like $a^{-3}$ scaling. Stability prevents its removal by decay; decoupling while relativistic prevents a subsequent equilibrium Boltzmann suppression of its number.

Before electron-positron annihilation in cosmology, the pair has two spin states, just as photons do, and the fermionic number integral gives $n_{\nu +\bar{\nu }}/n_{\gamma }=3/4$ at equal temperature. The factor follows by expanding $(e^x+1{)}^{-1}$ into an alternating exponential series: its $x^2$ integral is $3/4$ of the bosonic one.

After neutrino decoupling, cosmological entropy conservation in the photon-electron plasma changes its relativistic entropy count from $2+(7/8)4=11/2$ to $2$. The decoupled neutrinos do not share this heating. HenceThese counts include both the new neutrino and its antineutrino; another factor of two would double count them. This is the number density of a relativistic neutrino relative to photons, evaluated after photon heating.

They are nonrelativistic today, so the relic-neutrino energy density is dominated by rest mass:The number follows from the supplied critical density, without a Hubble-unit convention. It is conditional on the thermal history stated above; a sterile, colder or nonthermally produced state need not have this abundance.

Let $r_{\nu }=T_{\nu ,0}/T_0$, and retain the assumption that these neutrinos remain massless. Constancy of the neutrino degeneracy parameter gives $|{\mu }_{\nu ,0}|=|{\xi }_{\nu }|r_{\nu }{T}_0$. The condition on the critical density therefore yields the critical-density bound on massless neutrino degeneracyFor the standard instantaneous-neutrino decoupling approximation before electron-positron annihilation in cosmology, cosmological entropy conservation heats photons relative to the decoupled neutrinos. The electromagnetic entropy degrees of freedom change from $2+(7/8)4=11/2$ to $2$, givingThis numerical value requires that thermal-history assumption; the bound in terms of $r_{\nu }$ is the general answer if large degeneracy changes decoupling or subsequent entropy transfer. Setting $T_{\nu ,0}=T_0$ would instead give $26.2$. For $N$ equally degenerate flavors with the same temperature, the leading bound becomes $36.8N^{-1/4}$ under the standard temperature assumption. The neglected thermal corrections make a small change near this loose energy-density limit; this is not a detailed Big Bang nucleosynthesis bound.

A collisionless relativistic species redshifts with $T_{\nu }\propto a^{-1}$. The coupled plasma instead obeys cosmological entropy conservation, $g_{\ast s}{T}^3{a}^3=\mathrm{constant}$, and is heated relative to a decoupled species as massive particles annihilate. The effective number of relativistic energy degrees of freedom $g_{\ast }$ and effective number of relativistic entropy degrees of freedom $g_{\ast s}$ coincide while all species share a common relativistic temperature. Use $g_d:=g_{\ast s}(T_{\mathrm{dec}})\simeq 100$ for the Standard Model bath, excluding the already decoupled extra species.

Just before ordinary neutrino decoupling, the coupled bath contains two photon states, four Electron/Positron states and six ordinary neutrino/antineutrino states. ThereforeThe extra-neutrino temperature relative to that bath is then $(43/(4g_d){)}^{1/3}$. After the ordinary neutrinos decouple, Electron/Positron annihilation heats only the photon bath: its entropy count changes from $2+(7/8)4=11/2$ to $2$. Both neutrino populations miss that last heating, soThis is the two-stage entropy dilution of an early-decoupled neutrino. Numerically $T_{\nu ,0}\simeq 6.8\times 10^{-14}\mathrm{GeV}$. Relative to ordinary relic neutrinos it is cooler by $(43/(4g_d){)}^{1/3}\simeq 0.48$.

Let $g_{\nu }$ count the extra neutrino's populated helicity and antiparticle states. For one conventional neutrino plus antineutrino species $g_{\nu }=2$, the same state count as photons. The Fermi-Dirac distribution supplies the $3/4$ number-density factor, soThus a two-state species has about $2.9\%$ of the photon number density. Stating $g_{\nu }$ explicitly also covers a different choice of populated states.

Neutrino decoupling occurs when the interaction time ceases to be short compared with the expansion time, or $\Gamma \simeq H$. In natural units, this givesUsing the supplied high-temperature estimate $g_{\ast }\simeq 100$,Here $M_{\mathrm{Pl}}$ is the unreduced Planck mass. The scaling $\Gamma /H\propto T^3$ explains the direction of the transition: interactions maintain thermal equilibrium at high temperature, but become ineffective as the universe cools. The extra species changes $g_{\ast }$ only by a small amount compared with the precision of the estimate.

For a relativistic equilibrium species with temperature $T_i$, the entropy density is $s_i=({\rho }_i+P_i)/T_i=4{\rho }_i/(3T_i)$. In units $k_B=\hslash =c=1$, bosons have ${\rho }_i={\pi }^2{g}_i{T}_i^4/30$, while fermions have $7/8$ times this expression. Relative to a reference bath temperature $T$, define the effective number of relativistic degrees of freedom for entropy byThe sums here concern relativistic species, or their effective entropy contributions. Since a physical comoving patch has volume proportional to $a^3$, cosmological entropy conservation givesfor an isolated adiabatic bath. After thermal decoupling in cosmology, one must apply this law separately to sectors that no longer exchange entropy.

Before electron-positron annihilation in cosmology, the electromagnetic bath contains two photon polarizations and four electron/positron spin states. After neutrino decoupling, its entropy degrees of freedom are initiallyand after annihilation only the two photon polarizations remain. Electromagnetic entropy conservation therefore givesThe decoupled Cosmic neutrino background receives none of this entropy; its distribution temperature obeys $a{T}_{\nu }=\text{constant}$, and $T_{\nu }=T_{\gamma }$ before annihilation. ThusThis is the temperature of a decoupled relativistic relic in the instantaneous-decoupling approximation. The ratio remains the same afterwards in the absence of additional photon heating. At late times $T_{\nu }$ denotes the redshifted relic distribution temperature even when neutrino masses become relevant; it need not be a kinetic-equilibrium temperature.

For a relativistic species the supplied relation $P=ϵ/3$ makes its entropy density proportional to $T^3$. The fermion-to-boson thermal integral ratio gives the factor $7/8$ for each fermionic degree of freedom, so the electromagnetic plasma immediately before electron-positron annihilation in cosmology hasAfter annihilation only the two photon polarizations remain, so $g_{\ast s,\mathrm{after}}=2$.

The neutrinos have already undergone neutrino decoupling. They therefore receive none of the electron-positron entropy and cool as $T_{\nu }\propto a^{-1}$. In the still-coupled electromagnetic plasma, cosmological entropy conservation gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$. If $T_d$ and $a_d$ denote the common temperature and scale factor just before annihilation, then at a later scale factor $a$,Dividing gives the Cosmic neutrino background temperature ratio

At neutrino decoupling, neutrinos, electrons, positrons, and photons share one temperature. The neutrinos subsequently free stream, so $T_{\nu }\propto a^{-1}$. In the still-coupled electromagnetic plasma, the effective entropy degrees of freedom change during electron-positron annihilation in cosmology fromto $g_{\ast s,\mathrm{after}}=2$. Separate cosmological entropy conservation in that plasma gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$, while $T_{\nu }a$ remains constant. Consequently the Cosmic neutrino background temperature obeys

The Cosmic neutrino background began free streaming at neutrino decoupling, long before cosmological recombination, so its directional flux can retain information about density fluctuations from epochs inaccessible to the cosmic microwave background. Earlier decoupling does not by itself guarantee a larger comoving radial distance. A sufficiently massive neutrino eventually becomes nonrelativistic, and its travelled distance iswhich may be smaller than the photon distance because $v(a)<c$. Hence the cosmic neutrino background last-scattering surface need not lie beyond the Cosmic microwave background last-scattering surface; it generally does for neutrinos that remain sufficiently relativistic for sufficiently long.