Talk:Virial theorem

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For the power-law relationship given is the number 'n' appearing in the exponent required to be a natural number? I suppose it is, otherwise any force should be described as satisfying a power-law. But then I've read numerous papers on 1/f noise discussing a 3/2 power law relation. Mandelbrot described fractional power-laws in a paper with John van Ness in the 1950's. So what is it that makes natural number power laws so different from real number power laws?

Reply: Why is it true that any force can satisfy a power-law with a non-natural exponent? Does this just happen to be the case for every force in existance? That would surprise me, though I cannot currently think of a counter example. Certainly arbitrary functions cannot be described as power laws.
Never heard of a Taylor expansion, then?
I think \overline{K} =  \frac{n+1}{2} \overline{V} holds also for fractional n. It assumes V = arn + 1, not a sum of powers.--Patrick 07:55, 13 October 2005 (UTC)
A Taylor expansion is a sum of power-laws, not a single power law. The strong force and weak force are not power laws, but drop precipitously at larger distances. The force between two magnetic dipoles is proportional to the inverse-cube of the distance, but also proportional to a factor determined by the relative orientations of the dipoles and the separation between them. 128.165.112.42 15:04, 25 May 2006 (UTC)

After the copyright situation is resolved, a possible good link to add is John Baez's explanation 128.165.112.42 15:04, 25 May 2006 (UTC)

Contents

[edit] Have you heard of Copyright?

I wonder who wrote this page, but I'd like to remind him that copyright should be respected.

If you compare this page with Goldstein's "Classical Mechanics" p.70, you'd understand by point.

Besides, if you intend to copy anyway, please state "if the force is central" just before "If V is a power-law function of r". This can't be omitted, because you are using the same r to represent the position vector and the radius vector ambiguously.

I put a notice at Wikipedia:Copyright problems but I don't know the proper procedure for this. 80.203.45.67 14:00, 8 April 2006 (UTC)
Which edition? Zarniwoot 00:20, 14 May 2006 (UTC)

[edit] Nomenclature

I don't like the nomenclature K for the kinetic energy. Remember that K is still often used for force, too, especially in Special Relativity, given that until the 40s, German was especially important in physics. In German, force is named Kraft and the relativistic 4-force is often written as K.

For the kinetic energy, the nomenclature KE (when not E(kin) or something like that) is much better. However, E would be very good to appear, since the official term for energy is E or W (of work). I think, an often used abbreviation for kinetic energy is T. N.M.B.R.Nbez 11:06, 7 May 2006 (UTC)

[edit] Copyvio

I've reverted to the revision (cur) (last) 03:08, 6 May 2004 158.193.210.45, as per concerns raised above that this is a copyright violation of "Classical Mechanics". As I don't have a copy of the book, I'm unable to check what exactly is a copyright violation.

Please see this for a list of differences between that version and the pre-copyright violation noted version. I've also restored the interwiki links as well as the external links to the version right before it was noted as a copyright violation. Jude (talk,contribs,email) 03:10, 15 May 2006 (UTC)

[edit] No copyvio; have fun!

Hi all, I re-wrote this article from scratch without looking at any outside sources (e.g., books, articles or websites), so there should be no copyright problem now. I did keep the G notation for the virial, but that's about the only hold-over. I'm clueless about how the virial theorem has been used in other settings (e.g., for gases or in quantum mechanics), though, so I'm looking forward to your additions! :D WillowW 10:51, 9 July 2006 (UTC)

[edit] Intro needed?

Something about its significance? —The preceding unsigned comment was added by HairyDan (talk • contribs) 22:08, 17 October 2006 (UTC).

How's that? Willow 23:06, 17 October 2006 (UTC)