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Statistical Mechanics - Physics at Oregon State University

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5.1. FERMIONS IN A BOX. 95<br />

non-analytic behavior. The physics of such behavior is a phase transition! Hence<br />

if we do not discuss convergence issues properly, we will miss many important<br />

phase transitions! For example, if we replace sums by integrals for bosons, we<br />

see no Bose-Einstein condens<strong>at</strong>ion.<br />

Evalu<strong>at</strong>ing the integral.<br />

Our task is to evalu<strong>at</strong>e:<br />

1<br />

V Ω(T, µ, V ) = −(2S + 1)(2π)−3 <br />

kBT<br />

d 3 k log(1 + λe − 2 k 2<br />

2Mk B T ) (5.23)<br />

The integral is changed into a simple form by a coordin<strong>at</strong>e transform<strong>at</strong>ion.<br />

Define<br />

and we obtain<br />

2 <br />

x =<br />

2MkBT<br />

1<br />

V Ω(T, µ, V ) = −(2S + 1)(2π)−3 − 2 <br />

kBT<br />

2MkBT<br />

3<br />

2 <br />

After integr<strong>at</strong>ing over the angular variables we find<br />

1<br />

V Ω(T, µ, V ) = −(2S+1)(2π)−3 − 2 <br />

kBT<br />

2MkBT<br />

3<br />

2<br />

4π<br />

Next we introduce the thermal wavelength<br />

2 2π<br />

λT =<br />

MkBT<br />

1<br />

2<br />

k (5.24)<br />

1<br />

2<br />

∞<br />

0<br />

d 3 x log(1 + λe −x2<br />

)<br />

(5.25)<br />

x 2 dx log(1+λe −x2<br />

)<br />

(5.26)<br />

(5.27)<br />

which is approxim<strong>at</strong>ely the de-Broglie wavelength of a particle with energy<br />

kBT . Again, this combin<strong>at</strong>ion is introduced for historical reasons to simplify<br />

the formulas. The result is a small difference from the purist definition of a<br />

de-Broglie wavelength. Note th<strong>at</strong> nQ(T ) = λ −3<br />

T . At the quantum density a<br />

cube of dimension de-Broglie wavelength contains one particle. In other words,<br />

the average distance between particles is the de-Broglie wavelength, and it is no<br />

surprise th<strong>at</strong> quantum effects will be important!<br />

We also introduce a function f 5 (λ) by<br />

2<br />

∞<br />

4<br />

f 5 (λ) = √<br />

2 π<br />

0<br />

x 2 dx log(1 + λe −x2<br />

) (5.28)

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