= Solution
Write the <photon energy density> as $\epsilon=a_RT^4$, where $a_R=4\sigma/c$ by the <Stefan–Boltzmann law>. The <photon gas> has <pressure> $P=\epsilon/3$ and zero <photon chemical potential>. The <first law of thermodynamics> therefore gives
$$
TdS=d(\epsilon V)+P\,dV=4a_RT^3V\,dT+\frac43a_RT^4\,dV.
$$
Dividing by $T$ and integrating gives the <entropy density of a photon gas>:
$$
\boxed{S=\frac{4a_R}{3}T^3V,\qquad s=\frac SV=\frac{16\sigma}{3c}T^3.}
$$
The integration constant is fixed to zero at zero <temperature>, consistently with the <Third law of thermodynamics>.
The inequality $\Gamma\gg H$ says interactions act much faster than the expansion time, maintaining local <thermal equilibrium>. For adiabatic reversible expansion with no additional source of <entropy>, <cosmological entropy conservation> fixes the entropy in a comoving volume. Since that volume scales as $a^3$, and the relativistic degrees of freedom are fixed here, $T^3a^3$ is constant. Thus
$$
\boxed{T\propto a^{-1}.}
$$
During <radiation domination>, the <Friedmann equation> gives $H^2\propto\rho\propto T^4$, and hence $\boxed{H\propto T^2}$. Fast interactions alone would not exclude an additional entropy-producing process; changing relativistic species likewise changes the simple temperature scaling.
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