Talk:Exact differential

Just asking why they would use Q and dQ since this can cause confusion in a thermodynamic context —Preceding unsigned comment added by 70.29.38.49 (talk) 02:02, 5 October 2009 (UTC)[reply]

The introduction to this article states that "It is always possible to calculate the differential dQ of a given function Q(x, y, z)." But don't we need to require that the total derivative of Q exists? 128.193.11.102 10:17, 23 June 2007 (UTC)[reply]

I'm not good enough with Wikipedia's formatting to fix it, but the notation used with the subscript outside the parentheses looks way too much like the vertical bar notation for partial derivative. It looks like it is saying to differentiate first in terms of x and then in terms of y, for instance, rather than to differentiate in terms of x while holding y constant. I really don't see how that notation is necessary at all, since we are assuming anyone who is reading the article knows how to partially-differentiate. 212.29.216.114 (talk) 18:35, 9 July 2008 (UTC)LFStokols[reply]

.

[W]e are assuming anyone who is reading the article knows how to partially-differentiate.

.We should not be assuming that readers of this article know calculus; in fact, it can be argued that people may read this page to learn about partial differentiation. A common occurance with Wikipedia is that articles dealing with mathematics or technical subjects often become wordy, unorganized, or written by professionals for a professional audience, not the general public. KraziTV 23:30, 22 June 2009 (UTC)[reply]

Why is this in "thermodynamics relations"? 22:48, 16 July 2008 (UTC)

Is this mathematics or physics

“A mathematical differential dQ is said to ...”

Contrary to what this sentence suggests, the adopted viewpoint is not mathematical. Something called dQ in maths is necessarly exact. In maths "d" should always represents the differential operator applied to a function. Moreover, the question of exactness apply to a differential form. A differential means the differential of a certain function, which once more is always exact ...

Wanna talk about some differential form ? Call it $\omega$ or at least $\delta Q$ .--213.144.210.105 (talk) 08:16, 9 June 2010 (UTC)[reply]

(newbie warning; apologies in advance.) I don't find the differential notation to be clear. In fact I believe, and have seen in another place, that the differential notation is really shorthand for an integral equation. The following link illustrates the point.

http://en.wikipedia.org/wiki/Stochastic_differential_equation#Note_on_.22the_Langevin_equation.22

If the notational shortcut is in common use in physics/thermodynamics, then I think a citation to a definition is in order. (pondhockey) — Preceding unsigned comment added by Pondhockey (talk • contribs) 21:05, 24 January 2012 (UTC)[reply]

In fact in modern maths you can give a definition to dx as a object on its own and not a shortcut for an integral. Sure thermodynamicians generally do not know or or do not care about that. But my whole point is: what is the viewpoint adopted in this article, if it's really maths, modifications are needed to clarify. --94.245.127.15 (talk) 11:40, 18 April 2012 (UTC)[reply]

<< Note: The subscripts outside the parenthesis indicate which variables are being held constant during differentiation. Due to the definition of the partial derivative, these subscripts are not required, but they are included as a reminder. >>

Once again, I disagree with the text. Subscripts can be needed. It is not if functions are defined on R^n or if the coordinate system/map is obvious. But in the general case, they really convey information.--94.245.127.15 (talk) 11:51, 18 April 2012 (UTC)[reply]

Exactness of one-dimensional forms

Resolved

I believe that it is not necessary for a one-dimensional form, $A(x)\,dx$ , to have an anti-derivative to be exact as the page currently states. A function $Q(x)$ that satisfies the exactness condition is any integral of $A(x)$ , even if there is no closed form version. Let us choose $Q(x)=\int _{x^{*}}^{x}A(t)\,dt$ for any fixed $x^{*}$ in the domain on which $A(x)$ is defined. This is a function such that that $dQ=A(x)\,dx$ and therefore $A(x)\,dx$ is exact. Notice that requiring an exact form to have an elementary anti-derivative begs us to ask which operations and functions we consider elementary. Since the set of "elementary" operations is somewhat arbitrary, it would render the notion of an exact form arbitrary too. Jason Quinn (talk) 22:35, 11 March 2016 (UTC)[reply]

I now think the section is okay. When it said "is exact as long as A has an antiderivative", I was assuming that it meant in closed form but that was probably not what was meant. Regardless, the text could use some clarification to prevent such a misinterpretation, which I will do. Jason Quinn (talk) 19:55, 12 March 2016 (UTC)[reply]