Talk:Statistical mechanics

Out-of-place quotation ?

"Ludwig Boltzmann, who spent much of his life studying statistical mechanics, died in 1906, by his own hand. Paul Ehrenfest, carrying on the work, died similarly in 1933. Now it is our turn to study statistical mechanics". -- David L. Goodman "States of Matter"

It is a little weird to start an article with a quotation, like that. — Miguel 07:33, 2004 May 28 (UTC)

I know, but it is such a great quote from a standard reference, I thought it was worth including. Michael L. Kaufman 22:29, May 28, 2004 (UTC)

It is pretty interesting, maybe inclusion at the end of the article. Edsanville 05:40, 19 Aug 2004 (UTC)

Link

Perhaps the page should link to Maxwell-Boltzmann distribution (AC)

Structure of statistical mechanics articles

I'd like to make some more contributions to the Stat. Mech. subject, so I was trying to figure out the present structure. I've made up the following "summary" and I suppose it ought to go on the Stat. Mech. page, but I don't know where. Any suggestions or modifications? Paul Reiser 21:32, 11 Dec 2004 (UTC)

		Maxwell Boltzmann		Bose-Einstein		Fermi-Dirac
Particle				Boson		Fermion
Statistics		Maxwell-Boltzmann statistics Derivation of the partition function Gibbs paradox		Bose-Einstein statistics		Fermi-Dirac statistics
Gas		Ideal gas		Bose gas Bose-Einstein condensate Planck's law of black body radiation		Fermi gas Fermion condensate

General:

• Partition_function_(statistical_mechanics)
• Statistical_ensemble
  • Microcanonical_ensemble
  • Canonical_ensemble
  • Grand_canonical_ensemble
• Entropy
  • Principle of maximum entropy

Factorising Z: is it correct?

$Z=\sum _{i}\exp \left(-\beta (E_{ti}+E_{ci}+E_{ni}+E_{ei}+E_{ri}+E_{vi})\right)$

$=\sum _{i}\exp \left(-\beta E_{ti}\right)\exp \left(-\beta E_{ci}\right)\exp \left(-\beta E_{ni}\right)\exp \left(-\beta E_{ei}\right)\exp \left(-\beta E_{ri}\right)\exp \left(-\beta E_{vi}\right)$

$=Z_{t}Z_{c}Z_{n}Z_{e}Z_{r}Z_{v}\,$

Is sum of the products equal to product of sums, or I am missing something here?(Igny 02:55, 8 September 2005 (UTC))

If only it were that easy :-) It is generally false, unless all the energies are completely independent (completely uncoupled).

Consider an example:

Zf = Zf (pr(f1) + pr(f2)),

Zg = Zg ((pr(g1) + pr(g2)).

(Zf)(Zg) = Zf (pr(f1) + pr(f2)) Zg ((pr(g1) + pr(g2))

= Zf Zg (pr(f1)pr(g1) + pr(f1)pr(g2) + pr(f2)pr(g1) + pr(f2)pr(g2))

Does this equal Zfg? -- Only if the probabilities for f and g are completely independent, so pr(f1)pr(g1) = pr(f1,g1) etc; ie pr(f1|g1) = pr(f1), etc.

Assuming you can neglect correlations between f and g may or may not be a reasonable first guess. Jheald 22:29, 6 November 2005 (UTC)

Line added to article, to reflect this Jheald 22:46, 6 November 2005 (UTC)

:: You're right, this is linked to the idea of independent degrees of freedom There are still quite some cases where it works, obvious examples are the monoatomic ideal gas and paramagnetic systems. Actually, any system that has N quadratic degrees of freedom can be reduced into a set of N or less independent and quadratic degrees of freedom. For example, this is what happens with the diatomic ideal gas, and even with some quite complex systems like phonons in a solid. The degrees of freedom (physics and chemistry) article will explain this when I'll have finished with it. Actually, in all the examples I give here, the partition function breaks down into a product involving only individual independent dofs. ThorinMuglindir 23:02, 6 November 2005 (UTC)