Statistical Mechanics - Physics at Oregon State University

5.1. FERMIONS IN A BOX. 91
Znx,ny,nz(T, µ, V ) = 1 + e µ−ɛ(nx,ny ,nz )
k B T (5.8)
We now introduce again λ = e µ
k B T , which is always positive, and if we are
allowed to replace the sum by an integral we obtain:
1
V Ω(T, µ, V ) = −(2S + 1)(2π)−3
kBT
d 3 k log(1 + λe − 2 k 2
2Mk B T ) (5.9)
Formulas like these make it clear that we really want to use kB for the Boltzmann
constant, in order to avoid confusion with the norm of the wave vector, k.
Free particles have a simple volume dependence!
One remarkable thing happened. The right hand side does not depend on
volume anymore! When we had a summation over discrete k-vectors, the volume
dependence was included in these vectors. But when we changed to a continuous
summation (or integration), this volume dependence disappeared, and the only
volume dependence is a simple pre-factor V , a linear behavior!
Those mathematical details!
What could go wrong here? There is an error associated with the change
from summation to integration, and in general we have:
1
Ω(T, µ, V ) = −(2S + 1)(2π)
V −3
kBT
d 3 k log(1 + λe − 2k 2
2MkB T ) [1 + E(T, V, µ)]
(5.10)
where we know that limV →∞ E(T, V, µ) = 0. But we would like to be able to
differentiate the result with respect to T and µ, and hence we need to require
uniform convergence! But that in turn also begs the question if the original
series is uniformly convergent! In order to discuss the convergence properties
we have to take two steps. First, we have to show that the series converges
uniformly. Second, we have to show that the series in the thermodynamic limit
converges uniformly to the integral. We will discuss the uniform convergence of
the series first. The important thing to notice is that both the integral and the
series converge.
For large values of the wave vector the terms in the series behave like:
tnx,ny,nz = log(1 + λe − ɛ(nx,ny ,nz )
k B T ) ≈ λe − ɛ(nx,ny ,nz )
k B T (5.11)
which approaches zero exponentially fast. Next we consider a domain in (T, µ, L)
space with 0 T Tm , −∞ µ µm , and 0 L Lm. It is clear that for
large values of the indices we have