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.
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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 $t$.
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$\begingroup$ I thought my question would also hold in the partial differential case. At least that is what is implied by en.wikipedia.org/wiki/Chain_rule#Example . $\endgroup$– VirgoCommented May 18, 2021 at 10:07
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1$\begingroup$ 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. $\endgroup$– Free_ionCommented May 18, 2021 at 11:07
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$\begingroup$ Thanks, so if I understand correctly when they write $\frac{\partial f^{(0)}}{\partial t}$ they actually mean $\left. \frac{\partial f^{(0)}}{\partial t}\right|_{p={constant}}$. $\endgroup$– VirgoCommented May 18, 2021 at 18:54
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$\begingroup$ Yes, this is exactly what they mean. $\endgroup$– Free_ionCommented May 19, 2021 at 6:36