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What does it mean in thermodynamics when a derivative is computed "at constant $X$"? If I see

$\left.\frac{\partial S(E, N)}{\partial E}\middle| \right._N$

how is the derivation performed any differently? Other than forcing any derivatives of $N$ that happen to be present in the function $S$ to zero?

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    $\begingroup$ Why do you think it means something more than this? $\endgroup$ – G. Smith Sep 24 '20 at 19:34
  • $\begingroup$ @G.Smith Because my solution doesn't match the published solution, lol $\endgroup$ – Anonymous Sep 24 '20 at 19:59
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From a "physical" point of view, this means that we are only interested in the variation of $S$ as $E$ changes, even though probably changing $E$ might also (if $N$ depends on $E$ in some way) somewhat affect $N$ and therefore change $S$ not directly by the change of $E$ but rather through $N$. The partial derivative you wrote ignores this "implicit", "hidden" effect. What follows is a more detailed explanation.

The partial derivative you wrote, ${\partial S \over \partial E}|_N$, means you can treat $N$ as a constant. So for example if $$S=\alpha N^2E$$ its partial derivative with respect to to $E$ with $N$ constant is simply $\alpha N^2 $ because $N^2$ can be treated as a constant (I just used a random expression, there is no entropy I can think of with that expression..!).

This is usually not an issue when doing computations, you just regard $N$ as a constant and that's it, but it does avoid confusion when both $N$ and $E$ are functions of another parameter, say $w$ so that $E=E(w)$ and $N=N(w)$. In this case, you cannot claim to be just "slightly" changing $E$ keeping $N$ constant because they are intertwined by $w$. In this case, if what interested you is the total rate of change of $S$ then you would have to compute the total derivative of $S$ i.e. (by the chain rule): $${dS(E(w), N(w)) \over dw}={\partial S \over \partial E}|_N{dE\over dw}+{\partial S \over \partial N}|_E{dN\over dw}$$ You see that this expression takes into account not only the variation of $S$ with respect to $N$ but also the contribution due to the variation of $E$.

Also, in general, it could be that $N$ is a function of $E$, so that for example: $$S=\alpha N(E)^2E$$ In this case, when computing the partial derivative at constant $N$ we ignore this dependency and we are only concerned in the variation of $S$ as the part of it that explicitly contains $E$ varies, so that again we get $${\partial S \over \partial E}|_N=\alpha N(E)^2$$ because we treat $N$ as constant even though it contains an implicit dependency on $C$. Differently the total derivative would be (by the chain rule / by the formula above with $E$ instead of $w$):

$$dS/dE=\alpha N(E)^2+ 2\alpha N(E){dN\over dE} E$$

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  • $\begingroup$ Very complete, thanks! $\endgroup$ – Anonymous Sep 24 '20 at 19:59
  • $\begingroup$ thanks. I added a small edit at the beginning to give you more of physical sense of this if you missed that ;) $\endgroup$ – JalfredP Sep 24 '20 at 19:59

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