7
$\begingroup$

As far as i understand it, total differentials are linear maps that map vectors to numbers. In thermodynamics we encounter statements that a we have reached equilibrium when a total differential of a certain function is zero. For example equilibrium is reached when the Helmholtz free energy function differential at constant temperature is zero, $$ \mathrm{d}A = 0. $$

But that seems strange to me. $A$ map=the total differential=$\mathrm{d}A$ is a different "object" than the value of a map evaluated at a certain point. So what exactly does a statement like $\mathrm{d}A=0$ mean? I guess what is meant is the value of the total differential evaluated at some point with some vector?

So what is the vector and the point that we are feeding in?

EDIT:

Perhaps i should not have used the case $\mathrm{d}A=0$ since that seems to be unambiguous. But what is meant when we say for example the total differential is positive ? I've also read that in textbooks. A statement like that doesn't make sense unless i refer to an assumption/convention about the input vector, or does it ? I mean the value "positive" does vary in general with the input vector unless i have exactly the zero map. Please include a discussion about that also in an answer.

Optionally, I would also like a better understanding what the actual vectors are in the context of thermodynamics.

$\endgroup$
4
  • $\begingroup$ I just want to point out that the context can be important. I have already seem the abuse of notation by which $df$ (which is a one-form) was really meaning $df(v)$ (which is a number) for some $v$ implicit in the context. Apart from that, most often $df = 0$ means that at some point, probably also implicitly understood, given any vector $v$ at the point $df(v) =0$. $\endgroup$
    – Gold
    Feb 14, 2021 at 13:59
  • $\begingroup$ Why should be any difficulty with $dA>0$? Should be treated differently from $f>0$? $\endgroup$
    – GiorgioP
    Feb 14, 2021 at 16:16
  • $\begingroup$ The differential is linear in its argument, that means i can simply take the argument multiplied with minus 1 to obtain a negative value, with the very same differential. Saying that a differential is positive is therefore only qualified if i actually define the input but i have never seen any mention of that with such statements. The only case where the input doesn't matter is the special case of the zero map which was already mentioned in some answers. So i am either misunderstanding something or statements like the "differential is positive" are unqualified. I'd like to clear that up. $\endgroup$
    – Hans Wurst
    Feb 14, 2021 at 19:58
  • $\begingroup$ @GiorgioP A linear map which takes values in $\mathbb{R}$ either vanishes or is surjective. $\endgroup$ Feb 14, 2021 at 21:38

2 Answers 2

8
$\begingroup$

In all cases in which I met with the expression $\mathrm{d}A=0$ it meant that the map associated to it was identically null on its codomain. Then it's only an "abuse of notation", an abbreviation to mean that the differential maps all the points in which is evaluated to the null element of the codomain.

$\mathrm{d}A=0 \equiv$ $\mathrm{d}A(\vec{v})=0$ $\forall$ $\vec{v} \in D(\mathrm{d}A)$

where $D(\mathrm{d}A)$ is the domain of $\mathrm{d}A$ (a set in which its application is well defined).

[EDIT] - Answer to the second part of the question

With the differential, saying that it's positive means that $\mathrm{d}A$ is such that the direction of growth of the function that has $\mathrm{d}A$ as differential is coherent whit the direction of growth of the coordinates. For example for $f: \mathbb{R} \to \mathbb{R}$ if $\mathrm{d}f$ is said positive then the value of $f$ increases while we increase the value of the input variable. In particular the case $\mathbb{R} \to \mathbb{R}$ is the one in which I met with the expression "the differential is positive" more often, because there is only one variable. Trying to extend the expression in the case of multivariable functions, $f: \mathbb{R^n} \to \mathbb{R}$, the total differential takes in input a vector and gives in output a real number, and if it's positive in that point means that the function grows up in the direction coherent whit the one of the vector. So the direction intended should to be specified to be more precise, otherwise I agree it could sound so ambiguous, saying just that the total differential is positive.

$\endgroup$
3
  • 1
    $\begingroup$ Posts where notation is important are better for using the mathrm tag: $\mathrm{d}A=0$, which emphasise $\mathrm{d}$ as an operator, rather than $d$. Just a thought... $\endgroup$
    – Gert
    Feb 14, 2021 at 16:17
  • $\begingroup$ @Gert Yes sorry, you're right, I edit it. Thank you! $\endgroup$
    – annAB
    Feb 14, 2021 at 17:23
  • $\begingroup$ You're welcome, Ta. $\endgroup$
    – Gert
    Feb 14, 2021 at 17:26
7
$\begingroup$

An expression like $$ dA = 0 $$ means that $dA$ is the zero map, i.e. the map that returns $0$ for all inputs. It is not hard to see that this is also the zero vector of the space of linear maps to which $dA$ belongs when treated as a vector space in its own right.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.