Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 310 / 2

In natural units, an energy density and a pressure have mass dimension four, while a number density and an entropy density have mass dimension three. Since $m$ and $T$ each have mass dimension one,

The reaction maintaining chemical equilibrium isFor $T<m_e$ and negligible chemical potentials, the electron and Positron are dilute and obey the nonrelativistic equilibrium number density. Including two spin states for each particle,The photon number density is $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, so the electron-positron equilibrium pair abundance below the electron mass isIf $n_e$ denotes only one of the charge-conjugate species, this expression is divided by two.

Let $Y_{\mathrm{He}}\simeq 0.25$ be the present primordial helium mass fraction. If $n_{\mathrm{He}}$ is the helium-nucleus density and $n_H$ the hydrogen-nucleus density, thenSince $Y_{\mathrm{He}}=4n_{\mathrm{He}}/n_b$,This retains the correction from neutrons bound in helium. Charge neutrality requires one electron per proton, including bound electrons, so with the baryon-to-photon ratio $\eta =n_b/n_{\gamma }\simeq 6\times 10^{-10}$,

Pair equilibrium with negligible chemical potentials cannot continue once its predicted electron abundance falls below the residual abundance required by charge neutrality. Put $x=m_e/T$ and equate part (b), to logarithmic accuracy, to the order-$10^{-10}$ result of part (c):Taking a logarithm and retaining the numerical factor singled out in the hint givesThe solution is $x\simeq 28$, soThus the zero-chemical-potential electron-positron equilibrium description must fail before roughly $20\mathrm{keV}$; thereafter the small asymmetric electron population survives.