Statistical Mechanics - Physics at Oregon State University

5.4. PROBLEMS FOR CHAPTER 5 115
degenerate orbitals! Hence at T = 0 the number of particles in a ground state
orbital (1, 1, 1, s) is N
2S+1 and the total number of particles in all ground state
orbitals (1, 1, 1, s) together is N. In other words, all particles are in the ground
state orbital (1, 1, 1) defined without spin. In a number of books one changes
the actual definition of the distribution function by
fBE(ɛ) =
2S + 1
e ɛ−µ
k B T − 1
(5.141)
and similar for fermions. Keep this in mind.
The number of particles in the ground state orbital(s) has a sharp transition
according to our formulas. This is because we took the thermodynamic limit.
If we keep both terms in the expression for n, however, this transition is smooth
for finite volumes. Even for values above the Einstein temperature there are
some particles in the ground state orbital. For most experimental situations,
however, the differences are small and one is allowed to use the 3
2
power law
formula. Only right at the transition does one see finite size effects. The process
in which a macroscopic number of particles go into the ground state orbital
is called Bose-Einstein condensation. It is observed experimentally in liquid
Helium. Below 2.17 K liquid Helium has no viscosity and a very large thermal
conductivity. One can think of these effects as a manifestation of a macroscopic
quantum state. Since many particles are in the same orbital, there are no
interference effects between their wave functions and they all act in a coherent
way.
Note, however, that our calculations are performed for a gas. In a real
experiment the temperature is very low and the material is a liquid. Corrections
due to particle interactions are needed and the actual calculations are much more
difficult. Also, our calculations only show that there is a transition. In order
to describe the state with Bose-Einstein condensation we need to take particle
interactions into account! Again, these correlation effects are very difficult to
incorporate.
5.4 Problems for chapter 5
Problem 1.
An ideal gas consists of H2 molecules. We need six degrees of freedom to
describe such a molecule. Three pertain to the motion of the center of mass
of the molecule and these degrees of freedom follow the statistical mechanical
description of an ideal gas. There are three internal degrees of freedom, two
pertaining to rotations and one to vibrations of each molecule. The values for
the energy levels of the rotational degrees of freedom are kBTrj(j + 1) and for
the vibrational states kBTv(n + 1
2 ). Quantum effects start to play a role below
a temperature Tq, defined by nQ(Tq) = n. The H2 gas in enclosed in a fixed