Talk:Scalar potential

Latest comment: 8 years ago by Sławomir Biały in topic Calculating the scalar potential for n dimensions

Merge

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The articles on potential and [scalar potential]] seem to be substantially the same. Worse, the article on potential is filled with cruft and inaccurate remarks that should be deleted or heavily edited. Thus I propose a merge, and a disambiguation....Oh, never mind, I'll just do it. linas 17:30, 7 July 2006 (UTC)Reply

  • Please keep the scalar potential.

—Preceding unsigned comment added by 24.93.101.70 (talkcontribs)


  • I'm a phyisicist. I do think too that it is important to keep the scalar potential separated from the potential.

Thank you. —Preceding unsigned comment added by 193.204.167.87 (talkcontribs)


  • For. I'd be for merging Potential into scalar potential. There's no difference between potential and scalar potential (in this context at least, vector potential etc has it's own entry). --H2g2bob

Heading added by Smack (talk) 00:31, 8 June 2006 (UTC)Reply

Sub-Section Edit Button

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Why are the sub-section "Edit" buttons in Russian? —Preceding unsigned comment added by 216.54.92.149 (talk) 12:03, 10 June 2008 (UTC)Reply

Minus sign

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If this discussion is about the mathematical (scalar) potential shouldn't the - sign in the definition go away? In physics we write that, e.g. E = - grad(V), but this is the physical definition to get the signs there in the "natural" way. In math we usually define it as E = grad(V) only.

Ravn-hawk (talk) 15:58, 23 July 2009 (UTC) Agree. I asked a Professor in Engineering Mathematics and he told me that the mathematical concept doesn't contain the negative sign , but for a physical phenomenon , the negative sign is sometimes added for physical considerations Melnakeeb (talk) 09:55, 30 March 2010 (UTC)Reply

Is that true? In Helmholtz decomposition for instance there is a minus sign. (This article seems quite math-centric to me.) There is an advantage from a pure mathematical perspective in having the minus sign. The scalar potential yielding a vector field A is used when rotA=0 in which case divA=f(x,y,z) is the quantity of interest. If V is defined such that -gradV=A, the the solution for V in terms of f will have a positive sign which is symmetric with the equations for the solenoidal (divA=0) case.
As a physicist having the negative sign is strongly preferred. If there is a valid mathematical reason to not have it I will defer, but if there isn't or if there is some ambiguity in math then I want to use the negative. TStein (talk) 19:59, 11 October 2010 (UTC)Reply

Calculating the scalar potential

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Not all symbols are defined in the equations that are presented in the "Calculating the scalar potential" section. What is R(tau) for example? Also, no references are given. So, neither the equations nor the claim that "This formula is known to be correct if"... can be verified. This section really needs a reference to the information source or to further information. —Preceding unsigned comment added by Pms549 (talkcontribs) 23:36, 14 September 2010 (UTC)Reply

Calculating the scalar potential for n dimensions

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A formula is rigorously derived where the integrand is a multiple of   instead of  . — Preceding unsigned comment added by Math buff (talkcontribs) 08:14, 3 June 2016 (UTC)Reply

It follows by integration by parts. It does not require pages of calculation. Also, since I see you add long proofs to a lot of articles, you should familiarize yourself with the the manual of style for mathematics: "as a rule of thumb, include proofs when they expose or illuminate the concept or idea; don't include them when they serve only to establish the correctness of a result." There is general consensus that long proofs like the ones you are adding here (and elsewhere) are not appropriate for Wikipedia, so do not be surprised if they are removed. Sławomir Biały (talk) 11:29, 3 June 2016 (UTC)Reply

[Relatively Minor] Conservation conditions

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It should be made clear that the three conditions for a vector field to have a potential are one (you can think of them as being like different tests for the same thing). The second is a corollary of the first (this is obvious by single-valuedness – P(x)–P(x) = 0) , and the third is implied by the second (by Stokes' theorem).[1]

References

  1. ^ Corral, Michael. "Vector Calculus" (PDF). http://www.mecmath.net. Schoolcraft College. Retrieved 27 June 2014. {{cite web}}: External link in |website= (help)


Pressure as buoyant potential

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In the above mentioned chapter, the definition "The surfaces of constant pressure are planes parallel to the ground." is used. This might well be true for volumes with even ground, but regarding a lake, the definition should be "The surfaces of constant pressure are planes parallel to the surface, which represents the plane of zero pressure". Nevertheless, also this definition only holds true for undisturbed systems, in line with the example of the vertical vortex described a few lines further down. If I am not mistaken, a similar effect can be observed within any flowing liquid such as water in a river. If someone could justify to prefer to use of the "parallel to the ground." definition, it would be great. Otherwise, I'd like to alter the definition according to that described above. 132.187.199.137 (talk) 10:47, 17 December 2014 (UTC) TCSHReply