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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106 CHAPTER 5. FERMIONS AND BOSONS<br />
Coulomb <strong>at</strong>traction between the electrons and the ion cores in a solid. A similar<br />
large pressure occurs in white dwarf stars. In th<strong>at</strong> case gravity is the force th<strong>at</strong><br />
keeps the particles together.<br />
Large temper<strong>at</strong>ures.<br />
The other limit of interest is T → ∞. In this case we expect, of course,<br />
to find the results for an ideal gas. Like we discussed before, since n = (2S +<br />
1)nQ(T )f 3 (λ) the function f 3 (λ) has to approach zero and hence λ → 0. There-<br />
2 2<br />
fore this function is domin<strong>at</strong>ed by the first term in the power series and we find<br />
N(T, µ, V ) ≈ (2S + 1)V nQ(T )λ (5.92)<br />
Apart from the factor 2S + 1 this is the result we had before. This last factor<br />
is a result of the additional degeneracy we introduced by including spin and<br />
reduces the chemical potential. In this high temper<strong>at</strong>ure limit we find<br />
Ω(T, µ, V ) ≈ −(2S + 1)V kBT nQ(T )λ (5.93)<br />
Together with the formula for N this yields:<br />
and using p = − Ω<br />
V :<br />
Ω(T, µ, V ) ≈ −NkBT (5.94)<br />
p = NkBT<br />
(5.95)<br />
V<br />
Hence the ideal gas law is not influenced by the extra factor 2S+1. The pressure<br />
does not change due to the spin degeneracy, unlike the chemical potential,<br />
which is now equal to (found by inverting the equ<strong>at</strong>ion for N):<br />
5.2 Bosons in a box.<br />
Integral form.<br />
n<br />
µ(T, V, N) = kBT log(<br />
) (5.96)<br />
(2S + 1)nQ(T )<br />
The discussion in the previous chapter for bosons was again general for all<br />
types of bosons, and in order to derive some analytical results we have again<br />
to choose a specific model for the orbital energies. Of course we will again take<br />
free particles in a box, using ɛo = 2<br />
2M k2 with k = π<br />
L (nx, ny, nz). This leads to<br />
Ω(T, µ, V ) = −(2S + 1)kBT <br />
nx,ny,nz<br />
log(Znx,ny,nz(T, µ, V )) (5.97)