Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 53 / 2 / i

At temperatures above a few MeV, weak interactions such as $n+{\nu }_e\leftrightarrow p+e^{-}$ and $n+e^{+}\leftrightarrow p+{\bar{\nu }}_e$ maintain neutron-proton chemical equilibrium. Negligible lepton chemical potentials give the neutron-to-proton ratio $r_{\mathrm{eq}}=n_n/n_p=\exp [-\Delta /(k_{B}T)]$, where $\Delta =(m_n-m_p)c^2\simeq 1.293\mathrm{MeV}$. At high temperature this ratio is nearly one.

During radiation domination, the interaction time grows as $T^{-5}$, faster than the expansion time $T^{-2}$. Their crossing therefore produces cosmological weak freeze-out. If $\Theta =T/(10^{10}\mathrm{K})$, equating the displayed order-of-magnitude times gives ${\Theta }_f^{-5}\sim g_{\ast }^{-1/2}{\Theta }_f^{-2}$, soUsing $H^{-1}=2t$ instead of $t$ changes only the order-one coefficient. Before electron-positron annihilation, the usual effective relativistic count is $g_{\ast }=2+(7/8)(4+6)=10.75$. The factor $7/8$ weights fermionic relative to bosonic energy density.

After freeze-out, free neutrons decay while protons increase correspondingly. Define the neutron baryon fraction $X_n=n_n/(n_n+n_p)$. In the instantaneous-freeze-out estimate,where ${\tau }_n=t_{1/2}/\ln 2$ is the neutron mean lifetime. Hence the ratio at the onset of nuclear synthesis isThis expression accounts for the protons produced by neutron decay.

Nuclear burning initially waits at the deuterium bottleneck: although deuterium is bound by about $2.2\mathrm{MeV}$, the many photons per baryon can still photodissociate it. Roughly, ${\eta }_b^{-1}{e}^{-B_D/(k_B{T}_N)}\sim 1$ sets the bottleneck temperature, with refinements from reaction rates and phase space. Around $k_B{T}_N\sim 0.1\mathrm{MeV}$, deuterium survives, reactions through deuterium, tritium and helium-3 rapidly build tightly bound helium-4, and nearly all surviving neutrons are captured.

Thus $n_{\mathrm{He}}\simeq n_n/2$ and the primordial helium mass fraction isFor the standard estimate $r_N\simeq 1/7$, neutron and proton constituent fractions are $1/8$ and $7/8$, respectively, and $Y_{\mathrm{He}}\simeq 1/4$. Free neutrons are negligible after capture. The remaining hydrogen mass fraction is about $3/4$, and the helium-to-hydrogen number ratio is about $1/12$.

The printed numerical time coefficients are schematic, not a precision prediction of this quarter fraction: their literal crossing with $g_{\ast }=10.75$ gives $k_B{T}_f\simeq 1.28\mathrm{MeV}$. Keeping physical weak-rate coefficients and the gradual freeze-out instead gives an effective freeze-out around $0.7$–$0.8\mathrm{MeV}$, followed by decay over a few minutes. The abundance formulas above explain the standard quarter fraction and its parameter dependence without treating the rounded time formulas as exact nuclear-rate data.

Reducing the gravitational constant decreases $H$ at fixed temperature. The helium sensitivity to the expansion rate has two effects: weak interconversion persists to a lower freeze-out temperature, leaving fewer neutrons, and the universe takes longer to reach the deuterium-burning temperature, allowing more free-neutron decay after freeze-out. Both lower the helium-4 abundance. In particular $T_f\propto G^{1/6}$ and $t(T)\propto G^{-1/2}$.

At fixed expansion and conversion rates, a smaller neutron half-life reduces ${\tau }_n$ and hence $X_n(t_N)=X_{n,f}{e}^{-(t_N-t_f)/{\tau }_n}$. Fewer neutrons survive, so less helium-4 forms. If the reduced lifetime arises from a stronger weak coupling, its accompanying faster neutron-proton conversion also delays freeze-out and reinforces this decrease.

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.

The equilibrium neutron-to-proton ratio is $e^{-\Delta /(k_{B}T)}$. Increasing the neutron-proton mass difference suppresses neutron abundance at the same freeze-out temperature. Therefore less helium-4 forms, with other inputs held fixed. Any associated shortening of neutron lifetime further reduces survival, rather than reversing this direct effect.

The helium yield is primarily neutron-limited, so small changes in cosmological baryon density have only a weak effect compared with their effect on residual deuterium. Increasing the baryon-to-photon ratio makes deuterium destruction less effective relative to its formation and releases the deuterium bottleneck at a slightly higher temperature. Nuclear reactions also proceed faster. Fewer neutrons decay before capture, so a small increase in baryon density slightly increases helium-4; a decrease slightly lowers it.

At nucleosynthesis temperatures, $50\mathrm{GeV}$ particles are nonrelativistic. Their equilibrium abundance is negligible because of Boltzmann suppression, so they do not behave as an extra relativistic species contributing its full degree count to $g_{\ast }$. A stable thermal weakly interacting massive particle generally freezes out much earlier; a surviving relic population with the usual dark-matter abundance remains negligible compared with radiation during nucleosynthesis. In that standard relic scenario the helium abundance is essentially unchanged.

The mass and weak coupling alone do not specify the relic density. If an unusually abundant stable population contributes appreciably to the expansion density, it raises $H$ at fixed radiation temperature and increases helium by the same freeze-out and survival mechanism as part (g). Decays or energetic annihilation products would require their own specified history. Thus a universal numerical change cannot be inferred from the mass alone.

A faster expansion at fixed temperature makes the weak reaction rate fall below $H$ sooner, at higher temperature, leaving a larger neutron-to-proton ratio. It also shortens the time before the deuterium bottleneck ends, reducing neutron decay. The helium-4 abundance increases, assuming nuclear rates and the temperature history of the particle bath are otherwise unmodified.