Statistical Mechanics - Physics at Oregon State University

5.2. BOSONS IN A BOX. 107
with
Znx,ny,nz(T, µ, V ) =
1 − e
−1 µ−ɛ(nx,ny ,nz )
kB T
(5.98)
The same trick applies here as we used for fermions, and we replace the series
by an integral:
and
˜Ω(T,
1
µ, V ) = V kBT (2S + 1)
(2π) 3
d 3 µ−
k log(1 − e
2k 2
2M
kB T ) (5.99)
Ω(T, µ, V ) = ˜ Ω(T, µ, V ) + error(T, µ, V ) (5.100)
where the error is defined by making this equation exact. One can again show
that also for bosons we have limV →∞ error(T, µ, V ) = 0, but the big difference
with a system of fermions is that it is not possible to give an upper-bound to
the error which is only a function of V. The value of V we need in order to get
for example a 1% error becomes infinitely large when T → 0.
Note that µ < 0 for free particles, and hence 0 < λ < 1. Problems also occur
in the limit µ → 0, where the integrant diverges at the lower limit of integration.
These can be dealt with. The integral is of the form:
1
d 3 k log(
1 − λe − 2 k 2
2Mk B T
) (5.101)
and when k is small the argument of the logarithm is approximately log( 2MkBT
2k2 ).
Therefore, near the origin we need to evaluate d3k log(k), which behaves well
even though the logarithm diverges, since k2 log(k) goes to zero. Hence the
function ˜ Ω is well defined for 0 < λ 1, or for µ 0, and there are no problems
in this integral with the limit µ → 0. As we will see, this limit corresponds to
the limit T → 0.
Special functions for bosons.
In a manner similar to what we found for fermions, we define a set of special
functions by
gα(z) =
∞
n=1
zn , |z| < 1 (5.102)
nα and use analytic continuation to define these functions everywhere. The formula
for ˜ Ω is now manipulated in the same way we did for fermions, and we find
˜Ω(T, µ, V ) = −(2S + 1)V kBT nQ(T )g 5 (λ) (5.103)
2