Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 67 / 2

Use natural units with $c=\hslash =k_B=1$ and the unreduced Planck mass $M_{\mathrm{Pl}}=G^{-1/2}$. In a radiation-dominated universe, the effective number of relativistic energy degrees of freedom $g_{\ast }=\mathcal{N}$ givesThe ratio of weak conversion rate to expansion rate is proportional to $T^3$, so it decreases as the universe cools. Cosmological weak freeze-out occurs approximately at $\Gamma =H$, givingFor $g_{\ast }=10.75$, $G_F=10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}=1.22\times 10^{19}\mathrm{GeV}$ this estimate is $1.65\mathrm{MeV}$. Equivalently $g_{\ast }^{1/6}\simeq 1.5$ is only a mild multiplicative correction. The rate itself was given only to order of magnitude, so this is a decoupling scale, not a precise prediction for the weak freeze-out abundance.

Both neutrons and protons are nonrelativistic at this temperature. The nonrelativistic Maxwell--Boltzmann number density follows by the Gaussian momentum integral:Their equal spin degeneracies are $g_n=g_p=2$. Chemical equilibrium of the weak reaction gives ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$; with negligible lepton asymmetry, ${\mu }_n\simeq {\mu }_p$. Taking the ratio and setting the nearly equal nucleon masses equal only in the prefactor gives the neutron-proton chemical equilibrium resultThe common baryon normalization cancels. Keeping the exact mass prefactor changes this by only order $Q/m_p$. Thus the printed simple exponential is the usual nonrelativistic equal-mass-prefactor approximation. Using $Q\simeq 1.29\mathrm{MeV}$ with the rough decoupling scale above gives a ratio of order one half; the rate's omitted numerical factors and gradual freeze-out must be included before using a more accurate neutron abundance.

Use $X_i=n_i/n_B$, so $X_D$ counts deuterium nuclei per baryon, not baryons bound into deuterium. For the radiative capture reaction, chemical equilibrium and the zero photon chemical potential give ${\mu }_D={\mu }_n+{\mu }_p$. Dividing the three nonrelativistic Maxwell--Boltzmann number densities removes all chemical potentials:Here $g_D=3$ and $g_n=g_p=2$. The binding convention in this PDF is a negative mass defect, $B_D=m_D-m_n-m_p<0$. The positive nuclear binding energy is ${\epsilon }_D=-B_D$, so the abundance is enhanced by $e^{-B_D/T}$, not by $e^{B_D/T}$.

Since $X_D/(X_n{X}_p)=n_B{n}_D/(n_n{n}_p)$, insert the baryon-to-photon ratio $n_B=\eta n_{\gamma }$ and the photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$. This givesAt equal nucleon mass $m_N$ in the prefactors, $m_D\simeq 2m_N$, the deuterium equilibrium abundance simplifies toThe dimensionless prefactor is about $8.14$. These expressions describe thermal equilibrium; residual primordial deuterium after nuclear reactions cease must instead be obtained from the reaction kinetics.

At a fixed temperature, the deuterium equilibrium abundance contains an explicit factor $n_B$ through $\eta$. Two nucleons have to encounter each other to make one bound nucleus. Their capture rate per volume is proportional to $n_n{n}_p$, whereas nuclear photodisintegration competes through the very numerous thermal photons. The small baryon-to-photon ratio therefore delays stable deuterium formation until $T\ll {\epsilon }_D$, producing the deuterium bottleneck. Changing the cosmological baryon density changes both its release time and the subsequent efficiency of nuclear burning. The residual primordial deuterium abundance is particularly sensitive to these rates: on the usual nucleosynthesis branch, increasing $n_B$ burns more deuterium into heavier nuclei and lowers the remaining D/H. The positive dependence of the instantaneous equilibrium expression and the negative dependence of the final relic D/H are different statements.

By contrast, once nuclear reactions are efficient, nearly every surviving neutron ends in helium-4. Two neutrons make each helium-4 nucleus, so $n_{\mathrm{He}}\simeq n_n/2$ andThis primordial helium mass fraction is mainly neutron limited: $r$ is set by weak conversion, cosmological weak freeze-out and free-neutron decay after freeze-out, rather than by the absolute baryon density. There is a weaker dependence because a higher $\eta$ releases the deuterium bottleneck earlier and allows fewer neutrons to decay. Thus $Y_P$ is comparatively insensitive, not exactly independent, of $n_B$. At such exceptionally small baryon densities that nuclei never form efficiently, the neutron-limited approximation would fail.