 Solution

An adiabatic change of box size preserves the mode labels, whose momenta scale as $p\propto L^{-1}$. Since $V=L^3$, differentiation gives $dp/dV=-p/(3V)$. Occupation numbers are fixed under this slow mechanical change, so summing the changes in the individual particle energies givesComparison with $dE=-PdV$ at fixed entropy and particle number yieldsThe integration notation means summing over the original momentum-mode labels; it does not assume that the density in momentum space is unchanged after rescaling. In the ultrarelativistic limit $E(p)=cp$, so $p{E}^{\prime }(p)=E(p)$ and $P=E/(3V)=\rho c^2/3$.

Adiabatic radiation obeys $d(E{a}^3)=-Pd(a^3)$, giving energy density proportional to $a^{-4}$. Equilibrium photon energy density is proportional to $T^4$, so $T\propto a^{-1}$. Therefore the photon number in a comoving volume, $n_{\gamma }{a}^3\propto T^3{a}^3$, is constant. Its entropy is also constant: photon entropy density is $({ϵ}_{\gamma }+P)/T\propto T^3$. Photon number is not a microscopic conserved charge, but equilibrium processes maintaining the spectrum do not alter this comoving-number scaling in adiabatic expansion; entropy-producing processes would invalidate the assumptions.