Timeline for Thermodynamics Chain Rule And Independent Variables
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 3, 2022 at 10:43 | comment | added | Tobias Fünke | @Themis I understand your point of view and I mostly agree. But IMHO, one should first write down everything explicitly and only then one can abuse the notation in order to ease it! Many books/lectures fail here, which quite often leads to confusion, in my experience. | |
| Nov 3, 2022 at 10:33 | comment | added | Themis | @TobiasFünke As you say, it is cumbersome. Personally, I don't find the standard notation bad. And when it comes to thermodynamics, the simpler the notation the easier is to convey the concepts. | |
| Nov 2, 2022 at 19:03 | history | edited | Tobias Fünke |
edited tags
|
|
| Nov 2, 2022 at 16:20 | comment | added | Tobias Fünke | @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. | |
| Nov 2, 2022 at 15:57 | comment | added | Tobias Fünke | 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". | |
| Nov 2, 2022 at 15:53 | comment | added | Tobias Fünke | @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. | |
| Nov 2, 2022 at 15:38 | comment | added | Themis | @TobiasFünke I agree with your explanation, but what notation would you propose instead? | |
| Nov 1, 2022 at 7:45 | comment | added | Tobias Fünke | 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. | |
| Nov 1, 2022 at 0:10 | history | edited | Buzz♦ | CC BY-SA 4.0 |
added 104 characters in body
|
| S Oct 31, 2022 at 23:58 | review | First questions | |||
| Nov 1, 2022 at 2:46 | |||||
| S Oct 31, 2022 at 23:58 | history | asked | Abe | CC BY-SA 4.0 |