Statistical Mechanics - Physics at Oregon State University
Statistical Mechanics - Physics at Oregon State University
Statistical Mechanics - Physics at Oregon State University
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114 CHAPTER 5. FERMIONS AND BOSONS<br />
this is impossible! For such a combin<strong>at</strong>ion of density and temper<strong>at</strong>ure we need<br />
the first term in equ<strong>at</strong>ion (5.133). For large values of the volume , the value of<br />
λ will be close to one, and the value of the second term in equ<strong>at</strong>ion (5.133) can<br />
be replaced by (2S + 1)G. In the thermodynamic limit we have<br />
or<br />
lim<br />
V →∞<br />
2S + 1<br />
V<br />
λ = 1 −<br />
λ<br />
1 − λ = n − (2S + 1)nQ(T )G (5.136)<br />
2S + 1<br />
V (n − (2S + 1)nQ(T )G)<br />
(5.137)<br />
Since in most experiments the temper<strong>at</strong>ure is varied and not the density, it<br />
is customary to define a critical temper<strong>at</strong>ure TE, the Einstein temper<strong>at</strong>ure, by<br />
n = nQ(TE)(2S + 1)G, or<br />
2 2π n<br />
TE =<br />
MkB (2S + 1)G<br />
2<br />
3<br />
(5.138)<br />
The Einstein temper<strong>at</strong>ure for Helium is about 3K. The first term in the formula<br />
for n, equ<strong>at</strong>ion (5.133), gives the number of particles in the orbital with the<br />
lowest energy. If the temper<strong>at</strong>ure is higher than the Einstein temper<strong>at</strong>ure,<br />
there are essentially no particles in this orbital. If the temper<strong>at</strong>ure is below the<br />
Einstein temper<strong>at</strong>ure, the number of particles in this orbital becomes very large.<br />
More precisely, the Einstein temper<strong>at</strong>ure is the dividing point between having<br />
or not having a macroscopically large number of particles in the ground st<strong>at</strong>e<br />
orbital. In the thermodynamic limit this is the only orbital with a macroscopic<br />
number of particles, but for finite volumes other orbitals can be important, too.<br />
As we have shown before, for a volume of 1 m 3 we need a temper<strong>at</strong>ure less than<br />
10 −18 K for all particles to be in the ground st<strong>at</strong>e.<br />
The rel<strong>at</strong>ive number of particles in the ground st<strong>at</strong>e orbital, (2S+1)fBE(ɛ111)/N,<br />
for a temper<strong>at</strong>ure below the Einstein temper<strong>at</strong>ure follows from<br />
(2S + 1) fBE(ɛ111)<br />
V<br />
= n − (2S + 1)λ −3<br />
T G (5.139)<br />
and since the thermal wavelength is proportional to T − 1<br />
2 we can write this as<br />
the fraction of particles in the ground st<strong>at</strong>e orbital:<br />
(2S + 1) fBE(ɛ111)<br />
N<br />
3<br />
2 T<br />
= 1 −<br />
TE<br />
(5.140)<br />
This fraction is zero for T > TE. At T = 0, all particles are in the ground st<strong>at</strong>e<br />
orbital.<br />
Note th<strong>at</strong> there is some confusion in defining an orbital. If we define an<br />
orbital by the quantum numbers nx, ny, nz only, the total number of particles<br />
in th<strong>at</strong> orbital is given by (2S + 1)fBE(ɛ(nx, ny, nz)). If we define an orbital by<br />
the quantum numbers n plus a quantum number s for spin, the total number<br />
of particles in th<strong>at</strong> orbital is fBE(ɛ(nx, ny, nz, s)). But now we have 2S + 1