2
$\begingroup$

I was reading my textbook and I came up across the entropy $S(T,V,N)$ where temperature $T$, volume $V$, and number of particles $N$ are the independent variables. According to the chain rule the total differential of the entropy in this case is : $$dS=\left(\frac{\partial S}{\partial T}\right)_{V,N}dT + \left(\frac{\partial S}{\partial V}\right)_{T,N}dV + \left(\frac{\partial S}{\partial N}\right)_{T,V}dN.$$ Then my textbook proceeds to evaluate $\left(\frac{\partial S}{\partial T}\right)_{V,N}$ as $$\left(\frac{\partial S}{\partial T}\right)_{V,N}=\left(\frac{\partial S}{\partial E}\right)_{V,N}\left(\frac{\partial E}{\partial T}\right)_{V,N}$$ and then also inserts the result $\left(\frac{\partial S}{\partial E}\right)_{V,N}=\frac{1}{T}$ from when $E$, $N$, and $V$ are the independent variables not $T$, $N$ and $V$.

Is this mathematically correct/legal?

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Yes, it is correct (but as usual in thermodynamics, this is the abuse of notation hell). An easy way to see this is, as $V,N$ are held fixed in all expressions, to consider $S$ on the LHS only as a function of $T$, on the RHS only as a function of $E$ and $E$ only as a function of $T$. Doing so then is just the application of the "ordinary" chain rule, as all partial derivatives are "ordinary" derivatives. $\endgroup$
    – Tobias Fünke
    Commented Nov 1, 2022 at 7:45
  • $\begingroup$ @TobiasFünke I agree with your explanation, but what notation would you propose instead? $\endgroup$
    – Themis
    Commented Nov 2, 2022 at 15:38
  • $\begingroup$ @Themis Well, for example we could write $\mathcal S:\mathbb R^3\longrightarrow \mathbb R$ with $(E,V,N) \mapsto \mathcal S(E,V,N)=S$ and $\mathfrak S: \mathbb R^3\longrightarrow \mathbb R$ with $(T,V,N)\mapsto \mathfrak S(T,V,N) = S$ for the entropy. In other words, we obtain the entropy of the system $S$ through the two functions $\mathcal S$ and $\mathfrak S$. Obviously, if $\mathcal E:\mathbb R^3\longrightarrow \mathbb R^3$ with $(T,V,N) \mapsto (E,V,N)$ then $\mathfrak S = \mathcal S \circ \mathcal E$. Now the notation is not as ambiguous as before, but much more cumbersome. $\endgroup$
    – Tobias Fünke
    Commented Nov 2, 2022 at 15:53
  • $\begingroup$ Moreover, we shouldn't write something like $\partial_T$ but e.g. $\partial_1$ instead, i.e. we take the derivative with respect to the first "entry" of the function. In my experience something like $\partial_T$ also causes some confusion at some points. See also the [related] question by the OP and the discussion below an answer. It is important to distinguish the several functions which all yield the same physical quantity (e.g. the entropy). As I tried to explain, this reduces the potential confusion regarding derivatives etc., but of course is more "notational work". $\endgroup$
    – Tobias Fünke
    Commented Nov 2, 2022 at 15:57
  • $\begingroup$ @Themis Regarding the issue in the question. The derivative in the question can then be written as: $(\partial_1 \mathfrak S)(T,V,N) = \partial_1 (\mathcal S \circ \mathcal E)(T,V,N) = \partial_1 \mathcal S(\mathcal E(T,V,N)) \, \partial_1 \mathcal E(T,V,N)$, where we've used the chain rule. See also excellent answer to a related question. $\endgroup$
    – Tobias Fünke
    Commented Nov 2, 2022 at 16:20

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.