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

Use the dark-sector temperature in every dimensionless integration variable: $x=m_{\varphi }/T_{\varphi }$ and $y=p/T_{\varphi }$. Assume a single real scalar field, $g=1$, and fast number-changing reactions keeping ${\mu }_{\varphi }=0$. For $x\gg 1$, the populated momenta satisfy $y=O(\sqrt{x})$, soThis is the nonrelativistic Maxwell-Boltzmann distribution limit of the Bose-Einstein distribution. The given Gaussian Gamma function integral givesSince ${\rho }_{\varphi }=m_{\varphi }n+3n{T}_{\varphi }/2+\cdots$ and $P_{\varphi }=n{T}_{\varphi }$, the entropy density at zero chemical potential is $s_{\varphi }=n(x+5/2+\cdots )$. Its leading nonrelativistic entropy at zero chemical potential is thereforeAn internal multiplicity $g$ would multiply this expression; the absence of $g$ in the printed result assumes a real one-state scalar.

Now use the constant separately conserved visible and dark entropy ratio and the relativistic visible-bath entropy law, $s_{\mathrm{SM}}=(2{\pi }^2/45)g_{\ast s}^{\mathrm{SM}}{T}_{\gamma }^3=\xi s_{\varphi }$. Since $T_{\varphi }=m_{\varphi }/x$,Thus the cannibal-sector temperature ratio hasThe numerical coefficient is approximately $0.525$. The visible-bath temperature is the photon temperature while that bath is internally thermalized; after distinct visible species acquire different temperatures, its total entropy must instead use the appropriate temperature-weighted $g_{\ast s}^{\mathrm{SM}}$.

The displayed relativistic scalar entropy formula must not be continued into this regime with $g_{\ast s}^{\varphi }=1$ held constant. If it is used as an effective definition at arbitrary temperature, thenand $T_{\gamma }/T_{\varphi }=(\xi g_{\ast s}^{\varphi }/g_{\ast s}^{\mathrm{SM}}{)}^{1/3}$ gives the same result. This specifies how the relativistic expression can legitimately be combined with the nonrelativistic one.