Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 53 / 2

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.

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.

Since the species decouples while relativistic, its comoving particle number is conserved. Once it becomes nonrelativistic, its energy density is ${\rho }_{\nu ,0}\simeq m_{\nu }{n}_{\nu ,0}$. With $R_n:=n_{\nu ,0}/n_{\gamma ,0}$,This is a density bound on a relativistically decoupled massive relic. Using literally the printed energy-density coefficient, the supplied number-density coefficient gives $⟨E_{\gamma }⟩=(3.3/0.12)T_0=27.5T_0$. For $g_{\nu }=2$ and $g_d=100$,The estimate scales as $(2/g_{\nu })(g_d/100)$. At the bound the particles are safely nonrelativistic today, so neglecting their kinetic energy is consistent.

There is a physical normalization defect in the printed coefficient. Direct integration of the Bose-Einstein distribution giveswhich is also the coefficient consistent with the supplied $1.66$ expansion-rate formula. It implies $⟨E_{\gamma }⟩={\pi }^4{T}_0/(30\zeta (3))\simeq 2.70T_0$, rather than $27.5T_0$. With this physically normalized photon gas, the same density bound on a relativistically decoupled massive relic becomes $m_{\nu }\lesssim 3.7\times 10^1\mathrm{eV}$, or about $4\times 10^1\mathrm{eV}$ using the rounded coefficients with $0.33$ in place of $3.3$. The two estimates differ by the printed factor of ten; the entropy-dilution result is unaffected.

The preceding bound applies to a species that freezes out while relativistic, retaining an unsuppressed thermal number density apart from entropy dilution. It need not apply to a species that is already massive while reactions remain rapid.

For $m_{\nu }\gg T_{\mathrm{dec}}$, the interaction strength is sufficient to keep the species in thermal equilibrium through its relativistic-to-nonrelativistic transition. At $T\sim m_{\nu }$, the relativistic scaling would give $\Gamma /H\sim (m_{\nu }/T_{\mathrm{dec}}{)}^3\gg 1$. Subsequent annihilation removes most particle-antiparticle pairs while equilibrium remains possible. At negligible chemical potential, the nonrelativistic equilibrium number density isThis Boltzmann suppression of a nonrelativistic relic can make the eventual abundance much smaller than the relativistically decoupled value. Freeze-out now occurs when $n_{\mathrm{eq}}⟨\sigma v⟩\sim H$, with a nonrelativistic cross-section, rather than by extrapolating the massless $T^5$ rate indefinitely. A large mass can therefore have an acceptably small energy density: a hot-relic mass bound is not a universal bound on a heavy thermal relic.

Compatibility depends on the annihilation rate and any conserved particle-antiparticle asymmetry; the supplied relativistic rate alone does not determine the heavy relic's final abundance. Efficient annihilation is the mechanism that permits large masses within the standard expansion and thermal history.