Talk:Gradient theorem

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Mathematics Mid‑priority

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Physics Mid‑importance

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Why did the sum over i disappear in
$d\phi =\sum _{i}{\frac {\partial \phi }{\partial x_{i}}}dx_{i}=\left({\frac {\partial }{\partial x_{i}}}\right)\phi \cdot \left(dx_{i}\right)$ ? Randomblue (talk) 18:36, 2 February 2008 (UTC)[reply]

To say "No matter which path you take up a hill, when you reach the summit you will be X meter higher than the starting point" a valid "layman's interpretation" of this theorem? —Preceding unsigned comment added by 193.11.220.249 (talk) 19:37, 28 January 2011 (UTC)[reply]

Converse needs to be rephrased

The converse should not assume that the field is conservative, as conservative vector field is defined as the one having a gradient field. The proof, as far as I can tell, only assumes path independent line integrals. This inconsistency runs throughout the article, starting with the intro, but only in the relation to the converse. The subj seems to be stated correctly. melikamp (talk) 13:55, 11 December 2013 (UTC)[reply]

I don't think the converse does. I saw a couple of things in the article that could have been misinterpreted, and I changed them. However, I'm not sure this addresses your concerns. Could you be more specific? Ozob (talk) 02:16, 12 December 2013 (UTC)[reply]

Melikamp initially posted that comment in June 2013. At that time, the converse was indeed phrased incorrectly. However, I fixed it on 23 August 2013. Hence by the time Ozob saw the page (i.e, 12 December), all of the inconsistencies should already have been removed. I apologize, as I should have posted this on the talk page in August when I made these fixes. On an unrelated note, where specifically is the axiom of choice used when choosing integration paths (I'm referring to a previous comment by Melikamp)? 06:57, 19 December 2013 (UTC) — Preceding unsigned comment added by 69.114.94.117 (talk)

I see, it seems that there's some confusion because the date on the signature was edited. But there's no harm done.

I agree that I don't think the axiom of choice is necessary for choosing integration paths. I do think that the article glosses over this, and maybe it ought to say something. The point seems to be that when constructing the primitive of f, one needs to choose for each x a path from a to x. This can be done using AC, as there's a set whose elements are indexed by points x and whose members are sets of paths from a to x. But it should be possible to avoid AC as follows: U, being open and contained in Rⁿ, is a union of balls. If there is a smooth path from a to a point y in one of these balls, then we can draw a smooth path from a to x by drawing a straight-line path from y to x and interpolating the paths from a to y and y to x using a bump function. It ought to be possible to use this construction to give an explicit, distinguished path from a to x, and that should avoid the use of AC. But one would have to be careful in writing this out. Ozob (talk) 14:45, 19 December 2013 (UTC)[reply]

I agree fully. I removed my comment about the AC a while ago because I also realized that it is not needed. At least on the open regions (which is the way the theorem stated here and anywhere else I looked), the path can indeed be picked without AC. melikamp (talk) 17:09, 21 December 2013 (UTC)[reply]

It looks like my original concern has been addressed, so there's probably no need for this thread anymore. melikamp (talk) 03:40, 24 December 2013 (UTC)[reply]