# Boltzmann equation for photons in cosmology

I am trying to understand the derivation of the Boltzmann equation for photons given in Modern Cosmology (2nd Edition) by Scott Dodelson and Fabian Schmidt. In Eq. (5.4) they give the zeroth-order distribution function as the Bose– Einstein distribution with zero chemical potential, $$f^{(0)} \equiv\left[\exp \left\{\frac{p}{T}\right\}-1\right]^{-1}$$ where $$p$$ is the physical momentum of the photon and $$T$$ is the temperature. Later on the same page they evaluate the time derivative as follows $$\frac{\partial f^{(0)}}{\partial t}=\frac{\partial f^{(0)}}{\partial T} \frac{d T}{d t}\,.$$ But using the chain rule, shouldn't the above equation rather be the following? $$\frac{\partial f^{(0)}}{\partial t}=\frac{\partial f^{(0)}}{\partial T} \frac{d T}{d t}+\frac{\partial f^{(0)}}{\partial p} \frac{d p}{d t}\,.$$ I can't see any reason why they can neglect the $$\frac{\partial f^{(0)}}{\partial p} \frac{d p}{d t}$$ term.

This is because you are considering a partial derivative, not a total derivative. You should really have that \begin{align} \frac{\partial f^{(0)}}{\partial t} = \frac{\partial f^{(0)}}{\partial T}\frac{\partial T}{\partial t} \;, \end{align} but then they use implicitly the fact that $$\frac{\partial T}{\partial t}=\frac{d T}{d t}$$ since $$T$$ depends only on $$t$$.
• No, since $t$ and $p$ are both variables of $f$. You can see in the wiki example that you have no $\partial x/\partial y$ term. You could also see from the wiki example that taking $u=f$, $r=T$, $x=t$ and $y=p$ that you would obtain a $\partial p/\partial T$ term, which would be zero, and the remaining terms would lead you to the same result. May 18, 2021 at 11:07
• Thanks, so if I understand correctly when they write $\frac{\partial f^{(0)}}{\partial t}$ they actually mean $\left. \frac{\partial f^{(0)}}{\partial t}\right|_{p={constant}}$. May 18, 2021 at 18:54