Statistical Mechanics - Physics at Oregon State University

94 CHAPTER 5. FERMIONS AND BOSONS
with
Ω(0, µ, V ) = −(2S + 1)
lim
T →0
nx,ny,nz
kBT log(Znx,ny,nz(T, µ, V )) (5.17)
Znx,ny,nz(T, µ, V ) = 1 + e µ−ɛ(nx,ny ,nz )
k B T (5.18)
The limit of the argument is easy, since µ(0) = ɛF (N) > 0. Therefore the
argument is µ − ɛ(nx, ny, nz) if µ > ɛ(nx, ny, nz), because in that case we have
and
Znx,ny,nz(T, µ, V ) ≈ e µ−ɛ(nx,ny ,nz )
k B T (5.19)
kBT log(Znx,ny,nz (T, µ, V )) ≈ kBT µ − ɛ(nx, ny, nz)
kBT
(5.20)
The argument is 0 if µ < ɛ(nx, ny, nz), because now the function Znx,ny,nz(T, µ, V )
approaches 1 rapidly. As a result we have
Ω(0, µ, V )
V
2S + 1
= −
(2π) 3
d 3 k(µ − ɛ( k))Θ(µ − ɛ( k)) + error (5.21)
with Θ(x) being the well-known step function: Θ(x) = 0 for x < 0 and 1 for
x > 0. In this case the error also goes to zero for V → ∞, only in a different
algebraic manner. This shows that we can use
Ω(T, µ, V )
V
2S + 1
= −
(2π) 3
d 3 µ−
kkBT log(1 + e
2k 2
2M
kB T ) (5.22)
for all values of T on [0, Tm] and also in the limit T → 0 with an error less than
a small number which only depends on V and not on T. The limit for small
temperatures in the last general form of the integral is the same step function
we need for the integral that is equal to the sum for T = 0. Therefore, we
have shown that in the limit T → 0 the series also converges to an integral,
and that this is the same integral we obtain by taking the limit T → 0 of the
general form. Hence we can use the general form of the integral for all our
calculations and derive low temperature series expansions, for example. This
situation will be very different for bosons, where these two forms are not the
same, and Bose-Einstein condensation occurs!
Why bother?
So we went through all this math stuff and found that there was nothing
special. Why bother? Why would there be anything special? The answer is very
simple and important. Every time a convergence fails, we have a singularity and