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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.

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.