Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 310 / 3 / b / Solution

When the number-changing reaction is in chemical equilibrium, its chemical balance condition is $2{\mu }_{\varphi }=3{\mu }_{\varphi }$. Therefore zero chemical potential from number-changing reactions gives $\boxed{{\mu }_{\varphi }=0}$. Elastic scattering maintains kinetic equilibrium but imposes no independent condition on ${\mu }_{\varphi }$. The conclusion requires the number-changing rate to remain large compared with the Hubble parameter; merely listing an allowed reaction does not guarantee this after freeze-out.

Before thermal decoupling in cosmology, the sectors have a common temperature $T$. For relativistic species at that temperature their entropy densities aresoThese are the effective number of relativistic degrees of freedom for entropy. The paper's $g_{\ast }$ agrees with $g_{\ast s}$ for relativistic components at a common temperature. For one real scalar field, $g_{\ast s}^{\varphi }=1$. At temperatures where all Standard Model species are relativistic, both counts are constant, and so is $\xi$ before decoupling; thresholds can change this equilibrium ratio if they occur while the sectors remain coupled.

After decoupling, adiabatic isolation gives separately conserved visible and dark entropy: $s_{\mathrm{SM}}{a}^3$ and $s_{\varphi }{a}^3$ are individually constant. Thus $\boxed{\xi (T<T_d)=\xi (T_d)}$ even when either sector's entropy degrees of freedom change. Changes in particle content then alter their temperature ratio rather than the ratio of their comoving entropies. This assumes no late inter-sector transfer or entropy injection.