Timeline for How can we justify, in deriving quantum statistics, the use of Stirling approximation in the form $\ln(x!)\approx x \ln x - x$?
Current License: CC BY-SA 4.0
9 events
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| Jan 25, 2023 at 22:19 | comment | added | Silas | @JohnDonne shouldn‘t it be $\log N!=N(\log N-1)+ \mathcal{O}(\log N)$? | |
| Oct 15, 2018 at 12:40 | history | edited | John Donne | CC BY-SA 4.0 |
Added edit section following discussion in the comments
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| Aug 8, 2018 at 14:28 | comment | added | Ori | I think it is useful to emphasize that while (1) is correct, it is rather meaningless: the first term is itself $\mathcal{O}(N)$, so it's of the same order as the neglected term. Only (2) yields a sensible approximation for large $N$ (indeed, 5/2 might be much larger than the other term, depending on the values of $V/N$ and $\lambda$). | |
| Jul 30, 2018 at 22:09 | comment | added | John Donne | @Fausto Vezzaro My proof is the same as yours. Note that the constant $N$ in my proof corresponds to the $d_n$ in yours, which are given and nowhere one differentiates with respect to them. Generally you should take as many terms as needed. As explained in my answer there's no need to take terms $\mathcal{O}(1/N)$ and since we're differentiating it's also pointless to take the constant term. The only terms left are those of the form $x\log{x}-x$ | |
| Jul 30, 2018 at 21:46 | comment | added | Fausto Vezzaro | @JohnDonne In the proof I wrote above (you can find more details in Griffiths) there is no explicit mention of entropy and the logarithm only serves to break production in summation and to exploit Stirling approximation (even if the maximization of entropy is certainly a possible angle from which see this problem). In the proof we are searching the maximum with constraints of a $\mathbb{R}^\infty \to \mathbb{R}$ function (each point in the domain is a configuration). I find difficult what you wrote: shouldn't $N$ have a subscript $N_n$? And with respect to derive? Probably is the same proof? | |
| Jul 30, 2018 at 11:27 | history | edited | John Donne | CC BY-SA 4.0 |
added second example
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| Jul 28, 2018 at 16:03 | comment | added | John Donne | @Norbert Schuch Yes I was thinking about extensivity. Perhaps a better statement would be that it is as good depending on what you're doing. I will update later | |
| Jul 28, 2018 at 15:21 | comment | added | Norbert Schuch | Well, in some sense (2) is more correct than (1). (1) pretends that the $V/\lambda^3$ is meaningful, while in fact it isn't, since it is just another $O(N)$ term, so you should correctly write $S=-Nk\log N + O(N)$ (which seems odd, so I guess you want to use that $V$ scales in some way with $N$). | |
| Jul 28, 2018 at 8:59 | history | answered | John Donne | CC BY-SA 4.0 |