I'm studying thermodynamics I came across this definition

\begin{equation} p = \left( \frac{\partial u}{\partial x} \right)_{S} \end{equation}

where $p$ is a generalised pressure, $u$ is the energy of the system, $x$ is an extensive parameter and $S$ is the entropy. I know this page is not about physics but neither is my doubt, which is:

The text then proceeds to write this pressure in terms of a derivative of entropy and apply to it the chain rule

\begin{equation} p=\left( \frac{\partial u}{\partial x} \right)_{S}=-\left( \frac{\partial u}{\partial S} \right)_{x}\left( \frac{\partial S}{\partial x} \right)_{u} \end{equation}

Where does the minus sign come from? What is the rule here? If I was asked to write this pressure I would apply the chain rule without the minus sign, so what am I missing here?

Thank you very much.

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    $\begingroup$ I greatly suspect that negative comes from the $minus one$ rule for differentials, that is $\frac{dy}{dx}\frac{dx}{dz}\frac{dz}{dy}=-1$. It's in my calculus textbook for engineers, but unfortunately not in English. But this is the direction where you have to look into, therefore my answer as a mere comment... $\endgroup$ – imranfat May 16 '17 at 0:38
  • $\begingroup$ I will look into it. Thank you very much. $\endgroup$ – Gabu May 16 '17 at 0:40

Yes, thermodynamics people like to call this the cyclic rule for thermodynamics. If you have three variables, say, $x,u,S$, related by some functional equation $f(x,u,S)=0$, then the result is that $$\left(\frac{\partial u}{\partial x}\right)_S \left(\frac{\partial S}{\partial u}\right)_x \left(\frac{\partial x}{\partial S}\right)_u = -1,$$ along with the rule that $$\left(\frac{\partial u}{\partial x}\right)_S \left(\frac{\partial x}{\partial u}\right)_S = 1$$ (in the latter you're back to the usual single-variable calculus result that inverse functions have inverse derivatives, since $S$ is held constant in both).

To see where the surprising $-1$ comes from, say that $f(x,u,S)=0$ defines $x$ implicitly as a function of $u$ and $S$, and differentiate implicitly with respect to $u$ (fixing $S$, of course) to get that $$\frac{\partial f}{\partial x}\left(\frac{\partial x}{\partial u} \right)_S+ \frac{\partial f}{\partial u} = 0, \quad\text{and so}\quad \left(\frac{\partial x}{\partial u}\right)_S = -\frac{\frac{\partial f}{\partial u}}{\frac{\partial f}{\partial x}}.$$ Doing this, analogously, for the other two quantities, will give you the product of three $-1$s and everything else cancels!

REMARK: If you have four variables $x,y,z,w$ related by some functional equation $f(x,y,z,w)=0$, then you should be able to see that the corresponding cyclic rule is $$\left(\frac{\partial x}{\partial y}\right)_{z,w}\left(\frac{\partial y}{\partial z}\right)_{x,w}\left(\frac{\partial z}{\partial w}\right)_{x,y}\left(\frac{\partial w}{\partial x}\right)_{y,z}=+1.$$

  • $\begingroup$ Thank you very much. Just what I needed $\endgroup$ – Gabu May 16 '17 at 1:07
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    $\begingroup$ You're very welcome. It's a great question. $\endgroup$ – Ted Shifrin May 16 '17 at 1:09
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    $\begingroup$ @Gabu: You're relatively new here, so welcome. Once you're satisfied with an answer, please accept it so that the question will not stay on the "unanswered" list. If you're not satisfied, please follow up with a question. $\endgroup$ – Ted Shifrin May 16 '17 at 14:58
  • $\begingroup$ No, here are helpful instructions. $\endgroup$ – Ted Shifrin May 17 '17 at 15:08

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