Statistical Mechanics - Physics at Oregon State University

5.2. BOSONS IN A BOX. 111
where we wrote N again for the number of particles, since the function N did not
appear anymore. Because the series converges in a uniform manner, we could
interchange the summation and the limit T → 0. For all terms beyond the first
we have
2 π
(
2M L )2 (n 2 x + n 2 y + n 2 z) − µ 2 π
(
2M L )2 (n 2 x + n 2 y + n 2 z − 3) > 0 (5.119)
and hence in the limit T → 0 the exponent goes to infinity and the term goes
to zero. As a consequence, we must have
or
N = (2S + 1) lim
T →0
lim
T →0
e
2
2M ( π L )23−µ(T,N,V )
kB T − 1
2
π
2M ( L )23 − µ(T, N, V )
= log(1 +
kBT
−1
(5.120)
2S + 1
) (5.121)
N
Hence we find the low temperature expansion for the chemical potential and
for the absolute activity:
µ(T, N, V ) ≈ 2 π
(
2M L )23 − kBT log(
What is a low temperature?
N + 2S + 1
) + O(T
N
2 ) (5.122)
In the treatment above we call the temperature low if the first term is dominant,
or if (comparing the first and the second term):
which is the case (using the limit for µ) if
2 π
(
2M L )26 − µ ≫ kBT (5.123)
2 π
(
2M L )23 ≫ kBT (5.124)
For L = 1 m and He atoms this gives kBT ≪ 4 × 10 −41 J, or T ≪ 4 × 10 −18 K,
which is a very low temperature.
But this is not the right answer! We asked the question when are all particles
in the ground state. A more important question is when is the number of
particles in the ground state comparable to N, say one percent of N. Now we
will get a much larger value for the temperature, as we will see a bit later.
Limits cannot be interchanged.