For energetic fundamental relation $U=U(S,X_1,\ldots)$ where $X_k$ represent extensive parameters $V$ or $N_j$, let \begin{equation}P_k=\frac{\partial U}{\partial X_k}.\end{equation} For entropic relation $S=S(U,X_1,\ldots)$, let \begin{equation}F_k=\frac{\partial S}{\partial X_k}.\end{equation} By solving $dU=TdS\,+\,\Sigma\, P_k \,dX_k$ for $dS$ and comparing it with $dS=\frac{1}{T}dU\,+\,\Sigma\, F_k \,dX_k$, I can demonstrate that $F_k=-P_k\,/\,T$.

Alternatively, I want to get the same relation with following differentiation method:

Set $\bar{S}=S(U(S,X_1,\ldots),\, X_1,\, \ldots)$. Then by chain rule, we have \begin{equation} \left(\frac{\partial \bar{S}}{\partial X_k}\right)_{S,X_j,\,\ldots}=\left(\frac{\partial S}{\partial U}\right)_{X_j,\,\ldots}\left(\frac{\partial U}{\partial X_k}\right)_{S,X_k,\,\ldots} +\left(\frac{\partial S}{\partial X_k}\right)_{U, X_j,\,\ldots}.\end{equation} I can get the desired equation if the LHS vanishes. Then is it permissible to insist that the LHS equals zero since essentially $\bar{S}=S$? Is there any logical fallacy? I am sorry for my clumsy English. I appreciate your help.


1 Answer 1


When you wrote the first equation

$$U=U(S,X_1,\ldots) \tag 1$$

you were stating that $U$ is the dependent variable and the other variables on the right hand side are the independent variables.

So loose the bar on $\bar{S}$ and write

$$S=S(U(S,X_1,\ldots),\, X_1,\, \ldots) \tag 2$$

Then write down the total differentials for equations $(1)$ and $(2)$ and see if you can get to where ever it is you're going.

  • $\begingroup$ I think this method is same as the way I first mentioned. Could you examine the logic that $\frac{\partial \bar{S}}{\partial X_k}$ vanishes? $\endgroup$
    – asdf
    Commented Feb 20, 2019 at 6:07
  • $\begingroup$ The left hand side will be 0 because when you substitute dU into dS, then the dS will drop out of the equation, and you end up with the left hand side equal to 0. The left hand side you posted bothers me because you're taking a partial with respect to S when S is being held constant - which implies S is simultaneously a dependent and independent variable. $\endgroup$ Commented Feb 21, 2019 at 7:04
  • $\begingroup$ And what I mean by dS will drop out, is dS will show up on both sides of equal sign. $\endgroup$ Commented Feb 21, 2019 at 8:14

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