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For relativistic species in thermal equilibrium at a common temperature $T$, the total energy density can be writtenThe factor $7/8$ is the fermion-to-boson thermal integral ratio for energy density.

For a bosonic degree of freedom, the supplied integral givesFor a fermionic degree of freedom, useChanging variable $u=2y$ in the second integral givesThis is the energy-density case of the fermion-to-boson thermal integral ratio.

Summing the result of part (a) over all particle species therefore giveswhere the effective number of relativistic energy degrees of freedom is

The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.

For a relativistic species the supplied relation $P=ϵ/3$ makes its entropy density proportional to $T^3$. The fermion-to-boson thermal integral ratio gives the factor $7/8$ for each fermionic degree of freedom, so the electromagnetic plasma immediately before electron-positron annihilation in cosmology hasAfter annihilation only the two photon polarizations remain, so $g_{\ast s,\mathrm{after}}=2$.

The neutrinos have already undergone neutrino decoupling. They therefore receive none of the electron-positron entropy and cool as $T_{\nu }\propto a^{-1}$. In the still-coupled electromagnetic plasma, cosmological entropy conservation gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$. If $T_d$ and $a_d$ denote the common temperature and scale factor just before annihilation, then at a later scale factor $a$,Dividing gives the Cosmic neutrino background temperature ratio

For equal internal degeneracy, the number density of an isotropic ultrarelativistic species is proportional toWith $x=\beta pc$, compare the Fermi-Dirac distribution and Bose-Einstein distribution usingFor $s>0$, a change of variables in the second term gives the fermion-to-boson thermal integral ratioParticle number has $s=2$, so $n_f=(1-2^{-2})n_b=3n_b/4$. The energy density contains one further factor $E=pc$, so it has $s=3$ and