Talk:Gas in a harmonic trap

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[edit] Error in formula

Potential error: in the expression for dg, the β3 should cancel. Perhaps this was miscopied? (I don't have the necessary information myself, it just seems like an error)

You are right, it does cancel, but its not cancelled so that its easier to integrate. βE is dimensionless, so its carried along without cancelling. PAR 05:42, 30 November 2005 (UTC)

Alright, that makes sense. New problem: For N = \frac{f}{(\hbar\omega\beta_c)^3}~\frac{1}{2}~\zeta(3), I think that factor of 1/2 should disappear. \int_0^\infty \frac{x^2}{e^x-1} dx = 2 \zeta (3), or so Mathematica tells me. --Keflavich 02:33, 1 December 2005 (UTC)

Thats right, and I have fixed it. Thank you for pointing that out. It's been a while since I wrote that article, so I fixed it by correcting the math, but that kind of thing worries me. Please, check the corrections I have made and, while reading this article, be on the lookout for any other errors. If you find them, please let me know. Thanks - PAR 05:01, 1 December 2005 (UTC)
That's all I noticed when I was going through it. However, it might be worth defining g0. In the text I'm using, Kittel/Kromer (I don't have the full cite handy b/c I'm using a photocopy), they use the equation N_0(\tau)=\frac{1}{\lambda^{-1}-1}=\frac{z}{1-z} (the third expression is my own, translating from lambda to z). They specify that it is for the orbital at ε = 0. Since g is defined as the number of states with energy less than E, should g0 just be defined to be 1? --Keflavich 21:05, 1 December 2005 (UTC)

It makes sense, but the "tacking on" of the ground state is an approximation. I'm not sure about g0=1, so I would rather leave it. With regard to defining it, yes, it needs a little explanation, and I will do that. Thanks - PAR 04:28, 2 December 2005 (UTC)

In discussing Massive Bose-Einstein particles in the sentence:

"Where Lis(z) is the polylogarithm function. The polylogarithm term must always be positive and real, which means its value will go from 0 to ζ(3 / 2) as z goes from 0 to 1."

Shouldn't Lis(z) go from 0 to ζ(3) not ζ(3 / 2)?

That's right. I fixed it, thank you. The 3/2 is for a gas in a box, and I believe I copied the section from that article, and then altered the value of the exponent, and that was one I missed. PAR 21:21, 17 January 2006 (UTC)
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