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I personally have not seen how Boltzmann's entropy (S being proportional to Ln of W) is not valid when taking into account quantum mechanics and all its effects on microstates. In the article where the review is meant to be taken from they only talk about the H theorem yielding incorrect results. If I am wrong please correct me. —Preceding unsigned comment added by 86.132.199.201 ( talk • contribs) 21:51, 9 March 2007
- This article isn't about the S = k ln W formula, strictly speaking. It's about the assumption that W can be factorised into separate identical terms from each particle, W = W1 × W2 × W3 ...
- For a discussion of S = k ln W see Boltzmann entropy formula. -- Jheald 22:54, 9 March 2007 (UTC)
Then my question would be be: Would S = k ln W yield incorrect results in a system where there are interactions among the particles? (e.g any other than the ideal gas where PE due to conservative forces might have to be taken into account. Thank you for the tip above. —Preceding
unsigned comment added by
86.132.199.201 (
talk •
contribs) 13:45, 10 March 2007
- S = k ln W can give the right answer, but only if each 'way' counted in W is a way of arranging the entire universe -- ie each 'way' is including all N particles, and all their inter-particle interactions/energies. If there are interparticle interactions, you cannot just calculate the single-particle states available to one particle, the way Boltzmann did in his gas lectures, and then multiply up -- to do so is to make a serious approximation. Jheald 14:00, 10 March 2007 (UTC)
The comment(s) below were originally left at Talk:Boltzmann entropy/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.
| The main expression contains errors:
Erase N, in which case Gibbs entropy is well expressed, or Erase de Sum symbol and the subindexes i, giving an equivalent to k.lnW the Boltzmann entropy. -- Afaigon ( talk) 21:51, 6 February 2008 (UTC) |
Last edited at 21:51, 6 February 2008 (UTC). Substituted at 10:00, 29 April 2016 (UTC)
|
| This redirect does not require a rating on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||
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I personally have not seen how Boltzmann's entropy (S being proportional to Ln of W) is not valid when taking into account quantum mechanics and all its effects on microstates. In the article where the review is meant to be taken from they only talk about the H theorem yielding incorrect results. If I am wrong please correct me. —Preceding unsigned comment added by 86.132.199.201 ( talk • contribs) 21:51, 9 March 2007
- This article isn't about the S = k ln W formula, strictly speaking. It's about the assumption that W can be factorised into separate identical terms from each particle, W = W1 × W2 × W3 ...
- For a discussion of S = k ln W see Boltzmann entropy formula. -- Jheald 22:54, 9 March 2007 (UTC)
Then my question would be be: Would S = k ln W yield incorrect results in a system where there are interactions among the particles? (e.g any other than the ideal gas where PE due to conservative forces might have to be taken into account. Thank you for the tip above. —Preceding
unsigned comment added by
86.132.199.201 (
talk •
contribs) 13:45, 10 March 2007
- S = k ln W can give the right answer, but only if each 'way' counted in W is a way of arranging the entire universe -- ie each 'way' is including all N particles, and all their inter-particle interactions/energies. If there are interparticle interactions, you cannot just calculate the single-particle states available to one particle, the way Boltzmann did in his gas lectures, and then multiply up -- to do so is to make a serious approximation. Jheald 14:00, 10 March 2007 (UTC)
The comment(s) below were originally left at Talk:Boltzmann entropy/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.
| The main expression contains errors:
Erase N, in which case Gibbs entropy is well expressed, or Erase de Sum symbol and the subindexes i, giving an equivalent to k.lnW the Boltzmann entropy. -- Afaigon ( talk) 21:51, 6 February 2008 (UTC) |
Last edited at 21:51, 6 February 2008 (UTC). Substituted at 10:00, 29 April 2016 (UTC)