I'm trying to integrate $(1)$, regarding $T$:

$$\left( \frac{\partial f_s}{\partial T}\right)_{E/T}= -\frac{1}{HT} f_\alpha \tag{1}$$

where $ \displaystyle f_\alpha= \frac{1}{e^{E/T}+1}$ .

  • $f_\alpha$ and $f_s$ are distribution functions.
  • $E$ represents energy/momentum;
  • $T$ represents temperature
  • The subscript $E/T$ represents a constant ration of these quantities.


  • Even though $E/T$ is constant, the individual quantities, $E$ and $T$ are still variables and therefore the $T$ in $f_\alpha$ must still be integrated. Is this statement correct? Or should I ignore the $f_\alpha$ term?
  • If the statement is correct, how do I integrate the $(1)$? I've attempted integration by parts but I keep on getting extra $T$ terms, I might have done something wrong.


When trying to integrate $(1)$ I get:

$$dv = -\frac{1}{HT} \hspace{5mm} \hspace{5mm}; u = \frac{1}{e^{E/T}+1}$$ so $$ v= -\frac{1}{H} \ln(T) \hspace{5mm}; \hspace{5mm} \frac{du}{dT}= \frac{e^{E/T} E}{(e^{E/T}+1)^2 T^2}$$

and using the formula for integration by parts $\int u dv = uv - \int v du$ , I obtain:

$$f_s= -\frac{1}{H} \ln(T) \frac{1}{e^{E/T}+1} + \frac{1}{H} \int \ln(T)\frac{e^{E/T}E}{(e^{E/T}+1)^2 T^2}$$

This seems to have made the integration more difficult instead of simplifying it.


1 Answer 1


If $T$ is a variable and $E/T$ is constant (say $k$), it means $E$ is a variable as well. Therefore $f_{\alpha}$ and $f_s$ are functions of two variables: $(T,E)$. Your derivative $\left( \frac{\partial f_s}{\partial T} \right)_{E/T}$ is thus actually the directional derivative of $f_s$ along the direction given by the vector $(T,kT)$ (for a certain $k$) in the $TE$ plane.

All that to say that, in this direction, $E/T$ is constant and therefore you can replace it by $k$ in your expression for $f_{\alpha}$ and only consider the $T$ in the fraction in front of $f_{\alpha}$ when you integrate $f_s$.

edit: typo correction


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