In the post Chemical potential of particles with zero mass it's explained that for any species such that in equilibrium satisfies entropy and volume constant, the chemical potential of said species must be zero if particle number is not conserved.
Now, I'm trying to apply this to Cosmology. According to this subject, the Universe is in expansion so the volume in which you find radiation is not constant. Cosmic expansion is adiabatic so,
$$ 0 = dE + pdV - \mu dN \Rightarrow \mu = p\frac{dV}{dN} + \frac{dE}{dN} $$
Where $p$ is the pressure, $\mu$ the chemical potential and $N$ the particle number.
Now, as it's usual in Cosmology, we can deal with radiation as if it was a barotropic fluid with equation of state
$$ p = \frac{1}{3}\rho = \frac{1}{3}\frac{dE}{dV} $$
With $E$ the energy of the radiation which we now is given by Statiscal Physics as
$$ E \propto gT^4 $$
$g$ is the degrees of freedom of the species and $T$ its equilibrium temperature. I understand the thermal equilibrium condition (which is always applied in Cosmology) as an approximation valid while $\Gamma \gg H$, being $\Gamma$ the rate for the interations that are supposed to be in equilibrium; and $H$ the Hubble parameter. In other words, equilibrium as long as transition rates are greater than Universe expansion, so reactions don't feel the expansion.
So all together renders,
$$ \mu = (p/3 + 1)\frac{dE}{dN} \propto (p/3 + 1)gT^3 \frac{dT}{dN} $$
So, can we now assert $dT/dN \equiv 0$? Maybe this whole post is non-sense since $\mu$ is defined as $\mu = \left(\frac{\partial U}{\partial N}\right)\Big|_{S, V}$, i.e., $V$ must be taken constant at derivation even if it is not really constant. Or maybe since we know that $T \propto R^{\alpha}$ being $\alpha$ a number, then $dT/dN \propto R^{\alpha - 1}dR/dN = 0$ due to $R$, the expansion scale of the universe, is only time dependent. What do you think?
