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I am having trouble understanding something very introductory in Chandler's statistical mechanics book regarding the derivation of $(\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f=0$

The argument laid out in the book goes as follows.

For thermodynamic equilibrium states, entropy is an extensive function of state: S(E,X) where E is the internal energy and X is a mechanical extensive variable. Computing the differential we get:

$dS=(\partial S/\partial E)_{X}dE+(\partial S/\partial X)_{E}dX$

If we assume we have a reversible process:

$dE=(dQ)_{rev}+fdX$,

where f is an applied force. Combining the previous two equations we get for a reversible process:

$dS=(\partial S/\partial E)_{X}(dQ)_{rev}+((\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f)dX$

For processes which are both adiabatic and reversible, $dS$ and $(dQ)_{rev}$ are zero. Hence we have

$(\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f=0$

Till here I have no trouble. Then it is stated that since all quantities involved in this equation are functions of state, that implies the equality holds for adiabatic as well as non-adiabatic processes. I do not see why this should hold for non-adiabatic processes .

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From what I understand, he's just saying that since state functions are path-independent, that is, they describe equilibrium states irrespectively of how has the system has arrived to them (either by an adiabatic or non-adiabatic process, for instance), then an equality involving uniquely state functions must always hold. That wouldn't be the case for the preceding equality, which involves $dQ$, a path-dependent function.

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  • $\begingroup$ I am trying to convince myself with that line of argument. However what stops one from saying for constant volume we get from $dE = TdS - PdV$ to $dE=TdS$, hence $dE=TdS$ holds in general. $\endgroup$ – hakeem Nov 4 '16 at 22:38
  • $\begingroup$ I'd say the difference might be that $dE=TdS$ doesn't describe an equilibrium state, but $(\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f=0$ does. $\endgroup$ – David Herrero Martí Nov 4 '16 at 22:50
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Here is another way of proceeding. Start with the equation $$dS=(\partial S/\partial E)_{X}dE+(\partial S/\partial X)_{E}dX$$So, if we set dS = 0, it follows from partial differential mathematics that, in general, $$\left(\frac{\partial E}{\partial X}\right)_S=-\frac{(\partial S/\partial X)_{E}}{(\partial S/\partial E)_{X}}$$ But, we know that $$dE=TdS+fdX$$ So, $$\left(\frac{\partial E}{\partial X}\right)_S=f$$ So,$$f=-\frac{(\partial S/\partial X)_{E}}{(\partial S/\partial E)_{X}}$$ This gives:$$(\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f=0$$ I don't know where the extra minus sign came from, but I'm confident in what I did.

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  • $\begingroup$ Thanks a lot for your explanation. However the text derives the result $dE=TdS+fdX$ as a result of the $(\partial S/\partial X)_{E}+(\partial S/\partial E)_{X}f=0$ and not the other way round. So it doesn't surprise me that it works out the way you wrote it. $\endgroup$ – hakeem Nov 4 '16 at 23:27

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