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During radiation domination, the effective number of relativistic energy degrees of freedom, denoted $g_{\ast }$, gives the energy densityThe paper's ${\mathcal{N}}_{\mathrm{eff}}$ is this $g_{\ast }$; it is not the conventionally normalized effective number of neutrino species. With the specified reduced Planck mass, the Friedmann equation reads $H^2={\rho }_R/(3M_{\mathrm{Pl}}^2)$, whenceWrite the weak interaction conversion rate as $\Gamma =A{G}_F^2{T}^5$. The ratio $\Gamma /H$ scales as $T^3$, so it falls during cooling: conversion eventually becomes too slow to maintain neutron-proton chemical equilibrium. Equating the conversion and expansion rates givesOn suppressing the numerical coefficient, the cosmological weak freeze-out temperature is consequentlyThe one-sixth power reflects a balance between a conversion rate proportional to $T^5$ and a Hubble parameter proportional to $\sqrt{g_{\ast }}{T}^2$.

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

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

The new neutrino species undergoes thermal decoupling in cosmology when its interaction rate becomes comparable to the Hubble parameter. Equating the given rates givesWith $g_{\ast }\simeq 100$, $G_A=10^{-12}{\mathrm{GeV}}^{-2}$ and $m_{\mathrm{Pl}}\simeq 10^{19}\mathrm{GeV}$,This is an order-of-magnitude electroweak-scale estimate under the given approximation. The mass in this formula is the unreduced Planck mass, not the reduced Planck mass.

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.

Assume the new neutrino is stable and retains the abundance established by relativistic thermal decoupling in cosmology. While it remains relativistic, its contribution to the effective number of neutrino species isThus its effect on the radiation density and expansion during Big Bang nucleosynthesis is reduced relative to a fully heated extra neutrino species.

After it becomes nonrelativistic, its energy density is approximately $m_X{n}_X$. A two-state relativistic fermion relic has $n_X/n_{\gamma }=(3/4)(T_X/T_{\gamma }{)}^3\simeq 0.0275$. The density bound on a relativistically decoupled massive relic then givesHere ${\Omega }_{X,\max }$ is the allowed present matter-density fraction and $⟨E_{\gamma }⟩={\pi }^4{T}_{\gamma ,0}/(30\zeta (3))$. Choosing the total available matter fraction yields the basic overclosure bound. Its mass cannot be raised freely at a fixed relic number density.

There is also a perturbation constraint. Neutrino free streaming suppresses clustering below the distance traveled before the species becomes nonrelativistic. The ratio $T_X/m_X$ controls this distance, and the relic matter fraction controls the strength of the suppression. A significant hot or warm relic component must therefore satisfy the observed structure-formation constraints in addition to the abundance bound. Its allowed dark-matter contribution is limited jointly by these two effects. Numerical observational bounds depend on the chosen matter budget and perturbation data; the temperature and abundance dilution derived here are the factors needed to apply them.

Write $n_{56}$ and $n_{54}$ for the number densities of nickel-56 and iron-54. Both nuclei contain 28 neutrons, while their proton numbers differ by two. Dividing the positive-exponential nuclear statistical equilibrium formulas printed in the PDF therefore gives the nickel–iron nuclear equilibrium ratioThe neutron-density factors cancel. $K$ contains the isotope-dependent statistical and translational mass factors suppressed in the supplied proportionality; it is constant in this simplified calculation. The TeX conversion incorrectly reverses the exponential sign, so the PDF sign is essential here.

In the specified restricted composition, equality of total proton and neutron numbers meansConsequentlyDifferentiation, with the suppressed prefactors held constant, givesUsing the paper's supplied signed energy difference producesWith that negative value, $F$ tends to zero at both temperature endpoints, increases below this temperature and decreases above it. At the stationary point $d^2\log F/d{T}^2=-3/T^2<0$, so it is indeed a maximum for the printed toy data.

There is a genuine physical-data qualification: the stated negative energy difference is inconsistent with interpreting $Q$ as the conventional positive total nuclear binding energy of these actual isotopes. Using atomic masses, whose electron masses cancel in this reaction,The masses used are $1.00782503223\mathrm{u}$, $53.93960899\mathrm{u}$ and $55.94212855\mathrm{u}$, from NIST hydrogen masses, NIST iron masses and NIST nickel masses. Small electronic binding corrections do not change the sign. With a positive $\Delta B$, the simplified $F(T)$ above is strictly decreasing for $T>0$ and has no finite maximum. Thus the numerical maximum follows from the supplied negative parameter, not from consistent physical binding energies; changing the TeX sign to recover a maximum would not repair the PDF's data.

Nuclear statistical equilibrium is appropriate when the relevant forward and inverse strong or electromagnetic reactions are fast compared with changes of temperature, density and composition. It is useful in sufficiently hot late silicon-burning material and in the hottest explosive burning regions of massive stars. Some burning zones reach only quasi-equilibrium, and cooling or expansion eventually freezes the abundances out; an equilibrium expression cannot by itself predict the full ashes when these timescale conditions fail. Temperature, density, conserved baryon number and the electron fraction in stellar matter must all be specified. Negligible free-particle contributions in bulk bookkeeping are not the same as exactly zero free densities in the equilibrium formula.

Strong and electromagnetic reactions preserve total charge and baryon number, so they conserve the proton-to-neutron ratio. Weak reactions need not: electron capture and beta-plus decay convert protons into neutrons and lower it. Such conversions occur in hydrogen-burning networks and in the nitrogen-to-neon chain during helium burning. At high density in advanced stellar cores and core collapse, electron capture becomes particularly important; neutrino emission also affects energy and lepton loss. Beta-minus decay changes the ratio in the opposite direction and can compete during later relaxation. Nuclear equilibrium for strong reactions does not automatically imply weak equilibrium or a fixed proton-to-neutron ratio throughout all stellar evolution.

Use $f=X_{22}$ for the neon-22 mass fraction, temporarily keeping it distinct from the earlier metallicity. Let $n_b=\rho /m_u$ be the baryon density, ignoring small nuclear mass defects. Equal masses of carbon and oxygen giveCarbon-12 and oxygen-16 each have equal proton and neutron numbers, while neon-22 has ten protons and twelve neutrons. The total proton and neutron numbers per baryon are consequentlyHenceThe equal carbon/oxygen mass split cancels from this result: any mixture of these two symmetric nuclei has the same electron fraction. Taking the paper's metallicity symbol to mean $f$ gives its requested expression.

The literal pre-helium-burning nitrogen interpretation, however, needs a correction. Each initial nitrogen-14 nucleus receives two alpha particles, including eight additional baryons, before becoming neon-22; the intervening beta-plus decay also emits a neutrino. Thus, if the entire original metallicity is nitrogen-14 with mass fraction $Z_{\mathrm{initial}}$, conservation of the number of seed nuclei givesThis is neon-22 neutron excess from nitrogen-14. The two expressions are compatible only when their arguments are distinguished: the displayed $(11-Z)/(11+Z)$ law uses the final neon mass fraction, not literally the earlier nitrogen mass fraction. Treating those fractions as equal is an extra simplifying approximation, not a consequence of the stated capture chain.

A radioactively powered Type Ia supernova light curve rises to a broad maximum over roughly weeks as energy deposited in expanding ejecta diffuses out, then declines over the following weeks and months. Its power comes predominantly from the nickel-56 decay chain, first nickel to cobalt and then cobalt to iron. Gamma rays and charged decay products deposit heat, which is reprocessed into optical and infrared radiation. At later times, declining cobalt decay power and increasing gamma-ray escape shape the tail. It is not sustained nuclear fusion in the expanding ejecta. The amount of radioactive nickel, photon diffusion time and deposition efficiency all matter, so an uncorrected fixed luminosity is only an idealization.

For the requested two-isotope yield calculation, assume the same total processed iron-group mass, namely $1M_{\odot }$ as inferred from the zero-neon reference. Treat the final freeze-out ashes as nickel-56 and iron-54, neglecting residual light particles and additional weak neutronization. If $w$ is the nickel mass fraction, conservation of electron fraction givesSince $26/54=1/2-1/54$, this reduces to the metallicity dependence of nickel-56 yieldUnder the paper's intended identification $f=Z=1/50$,If instead $Z=1/50$ denotes the earlier nitrogen mass fraction, retaining the actual capture stoichiometry gives $w=1-27Z/7=323/350$ and $M_{56}\simeq 0.92M_{\odot }$. The same temperature and density do not alone fix a total burned mass; equality of that mass is also implicit in the comparison. The two-isotope assumption is a bulk-ash approximation, not a complete equilibrium network with all free particles identically absent.

For otherwise comparable explosions, approximate the peak luminosity by a quantity proportional to the radioactive nickel mass. Let $d$ be the true distance to an extremely low-neon event, and use the solar calibration $L_{\odot }/L_0=523/550$. Its measured flux is $F=L_0/(4\pi d^2)$, but the inferred standard-candle distance isThe source is intrinsically brighter than the calibration, so the distance is underestimated by about $2.5\%$, conventionally described here as about $3\%$. Equivalently, $d/d_{\mathrm{est}}=\sqrt{550/523}\simeq 1.0255$, consistent with the supplied rounded factor $1.03$. Using the literal nitrogen metallicity instead gives about a $4\%$ underestimate. These are conditional composition effects in this simplified model; real Type Ia distance estimation also standardizes light-curve shape and color.

A thermally populated extra light neutrino species is relativistic during nucleosynthesis; its $0.001\mathrm{eV}$ mass is negligible at MeV temperatures. One neutrino-antineutrino species adds $(7/8)\times 2=7/4$ to $g_{\ast }$. Thus the expansion rate increases by $\sqrt{12.5/10.75}$ and the freeze-out temperature by $(12.5/10.75{)}^{1/6}$. Earlier freeze-out leaves more neutrons, while faster cooling leaves less time for decay. The helium-4 abundance increases. This assumes the new species was populated at the standard neutrino temperature; an unpopulated species would not contribute this extra density.

Put $s_W=\sin {\theta }_W$ and $c_W=\cos {\theta }_W$. The left-handed doublet haswhile the right-handed charged-lepton singlet has $T_R^3=0$ and $Q_R=-1$. There is no weakly coupled right-handed neutrino in the stated field content. For a fermion $f$, write $g_L^f=T_{3L}^f-Q_f{s}_W^2$ and $g_R^f=T_{3R}^f-Q_f{s}_W^2$, with $g_R^{\nu }=0$ here. Since ${\bar{f}}_L{\gamma }^{\mu }{f}_L=\bar{f}{\gamma }^{\mu }{P}_{L}f$ and similarly for $R$,Consequently the neutral-current vector and axial couplings in the specified overall normalization areThese coefficients accompany $g/c_W$. A convention with $g/(2c_W)$ uses twice these vector and axial coefficients; mixing the two conventions would make the widths wrong by a factor of four. For the neutrino, $c_V-c_A{\gamma }^5=\frac{1}{2}{P}_L$, so the chiral projection is already built into the vertex.

For either massless final pair with momenta $k_1,k_2$, $p=k_1+k_2$, the invariant amplitude is, up to an irrelevant overall phase,The spin-summed fermion tensor isThe symmetric part, using the four-gamma trace, isThe vector-axial interference is antisymmetric in $\mu ,\nu$, so it drops out of the symmetric $Z$ polarization sum. Also $p_{\mu }{T}_f^{\mu \nu }=0$ for massless external fermions, by their Dirac equations; hence the $p_{\mu }{p}_{\nu }/m_Z^2$ part contributes zero. Using $k_1\cdot k_2=m_Z^2/2$ givesThe massive-vector spin average supplies the factor $1/3$ for the three spin states of the massive $Z$. The massless two-body Lorentz-invariant phase space is $d{\Phi }_2=d\Omega /(32{\pi }^2)$, and so $\int d{\Phi }_2=1/(8\pi )$. The spin-averaged amplitude above is angle-independent in the rest frame, givingThe massless leptonic Z decay widths for one charged-lepton flavor and one active neutrino flavor respectively areThere is no color factor for leptons and no extra final-state identical-particle factor for a particle-antiparticle pair. The neutrino projectors already exclude sterile helicity states, so no additional factor of two is needed. These are tree-level massless widths; electroweak radiative corrections and charged-lepton masses have not been included. If desired, using $G_F=g^2/(4\sqrt{2}{m}_W^2)$ and $m_W=c_W{m}_Z$ converts the common prefactor to $G_F{m}_Z^3/(12\sqrt{2}\pi )$.

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.

At weak freeze-out, photons, electrons, positrons and all four neutrino flavors are relativistic and approximately at the same temperature. The effective number of relativistic energy degrees of freedom changes fromThe expansion rate at fixed temperature is larger by $\sqrt{50/43}\simeq 1.078$. Equating the supplied weak interaction time with the supplied age gives $T_f^3=\sqrt{g_{\ast }}$ when $T_f$ is measured in MeV. ThereforeEarlier cosmological weak freeze-out at a higher temperature leaves more neutrons. With negligible chemical potentials, the neutron-to-proton ratio in equilibrium is $r_f=\exp (-Q/T_f)$, where $Q=m_n-m_p\simeq 1.293\mathrm{MeV}$. If almost all surviving neutrons enter helium-4, the primordial helium mass fraction before neutron decay is $Y_f=2r_f/(1+r_f)$. Literal use of the given rate normalizations yieldsEquivalently, the leading response is $\delta Y_f/Y_f=[Q/(T_f(1+r_f))]\delta T_f/T_f$. This is a roughly one-percentage-point absolute increase in the helium fraction, rather than a one-percent relative increase.

The deuterium bottleneck delays nuclear burning, allowing free-neutron decay after freeze-out. If its end temperature is $T_N$, and ${\tau }_n$ is the neutron mean life, a sharper instantaneous-freeze-out estimate conserves total baryon number and givesThis formula evolves the neutron fraction, not just the neutron-to-proton ratio, because each neutron decay also creates a proton. After electron annihilation, let $r_T=(4/11{)}^{1/3}$. The radiation counts relevant to the age at fixed $T_N$ areConsequently $t_N^{(4)}/t_N^{(3)}\simeq \sqrt{3.36/3.82}\simeq 0.939$. Less time is available for neutron decay, adding to the helium increase. As an explicit illustrative completion, $T_N=0.07\mathrm{MeV}$ and ${\tau }_n\simeq 880\mathrm{s}$ give an increase of approximately $0.012$ in this toy model. The precise helium fraction requires the nuclear-burning temperature, baryon abundance and properly normalized weak rates; the question supplies only approximate time laws. In particular, the large no-decay fractions in the table must not be interpreted as precise predictions for the observed primordial abundance. The robust effect is extra relativistic energy raises the primordial helium fraction, with an absolute change of order $10^{-2}$ for one thermally populated flavor.

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.

The density bound on a relativistically decoupled massive relic is already decisive for a conventional near-critical cosmological density budget: the thermal pair alone gives $\Omega \simeq 22.4$. The same additional relativistic pair increases the effective number of neutrino species by one during nucleosynthesis and raises the primordial helium mass fraction, so its expansion effect supplies another constraint. Thus a stable, fully thermalized MeV-decoupling neutrino pair of this mass is incompatible with the usual cosmological density budget.

This conclusion is not a prohibition on every particle with that mass. Earlier decoupling followed by entropy dilution, a reduced production abundance, or interactions that change the assumed thermal history can alter both tests. At fixed mass, requiring this pair to contribute no more than the supplied critical density would demand a number density at most $1/22.4$ of the standard value, or a thermal temperature at most $(1/22.4{)}^{1/3}\simeq 0.355$ of the standard neutrino temperature. No numerical abundance or exclusion independent of these assumptions follows from the mass and equal particle-antiparticle counts alone.