Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 10D (Cosmology)

For dilute nonrelativistic nucleons in thermal equilibrium, their Maxwell-Boltzmann distribution gives $n_j=g_j(m_{j}kT/2\pi {\hslash }^2{)}^{3/2}\exp [({\mu }_j-m_j{c}^2)/kT]$. The neutron and proton spin degeneracies are equal, and their mass prefactor ratio is nearly one. ThusWeak reaction equilibrium imposes ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Charge neutrality in an electron-positron plasma, together with the small baryon asymmetry, makes ${\mu }_e/kT$ very small. Assuming negligible neutrino lepton asymmetry also makes ${\mu }_{{\nu }_e}/kT$ small. Hence the difference, rather than the absolute baryon chemical potentials, is negligible, yieldingA large neutrino chemical potential would invalidate this last approximation; it is an assumption about the lepton asymmetry, not a general rule that baryonic chemical potentials vanish.

The weak-conversion rate per expansion time falls as $T^3$. Below the stated threshold it becomes less than one, so neutron-proton interconversion cannot track the rapidly changing equilibrium ratio: cosmological weak freeze-out occurs. Free neutrons subsequently decay, reducing their abundance until nuclei can survive photodissociation. Once the deuterium bottleneck is passed, nuclear reactions rapidly assemble neutrons into the strongly bound helium-4 nucleus. The short available time, low density and missing stable mass-five and mass-eight nuclei impede extensive production of heavier nuclei, leaving nearly all surviving neutrons in helium.

If $r=n_n/n_p$ just before helium formation, each helium nucleus uses two neutrons and two protons, so $n_{\mathrm{He}}=n_n/2$. Neglecting binding-energy corrections to nuclear masses,The remaining protons mainly become hydrogen. Counting all four helium nucleons, rather than only the captured neutrons, is essential to this mass fraction.

If the weak-rate coefficient is reduced, equality with the expansion rate occurs at a higher temperature, since ${\Gamma }_W/H$ scales as that coefficient times $T^3$. The freeze-out temperature and interaction strength relation is $T_f\propto C_W^{-1/3}$. The frozen neutron-to-proton ratio is then closer to one, so the primordial helium mass fraction increases, potentially approaching unity if nearly equal numbers survive to nucleosynthesis. If the same weaker interaction also slows free-neutron decay, it reinforces this trend; changing conversion rates alone would instead require a separate treatment of the decay interval.

Less primordial hydrogen means less hydrogen fuel and, for otherwise comparable stellar physics, helium enrichment raises the mean molecular weight, tending to make a fixed-mass star hotter and more luminous and to shorten its hydrogen-burning lifetime. A genuinely weaker weak interaction also changes proton-proton fusion itself, so a quantitative lifetime cannot be inferred from composition alone. Much less hydrogen and fewer long-lived suitable stars would alter planet formation, water availability and the timescales needed for life on Earth. These are physical implications of the changed initial composition, not a numerical prediction of habitability from the helium fraction alone.