Statistical Mechanics - Physics at Oregon State University

7.5. IDEAL GAS IN CLASSICAL STATISTICAL MECHANICS. 149
This example is studied in detail in a course on non-linear dynamics. An
important question, studied actively in that field, is the following. What are the
conditions for the Hamiltonian so that most of phase space for most energies
is covered by a chaotic orbital? In our words, can we construct a non-trivial
many body Hamiltonian for which statistical mechanics does not work? Note
that every Hamiltonian will have regular orbits, but the ergodic theorem is valid
only if these orbits have a measure zero.
7.5 Ideal gas in classical statistical mechanics.
The classical Hamiltonian for a system of N independent particles of mass m is
given by
Collisions are needed.
H =
N
i=1
p 2 i
2m
(7.32)
Such a system is clearly not ergodic. The solutions for the trajectories are
xi(t) = xi(0) + vit and sample only a small fraction of phase space. We have to
perform a spatial average in order to get some meaningful answers. In reality,
of course, there are collisions. These collisions are elastic and are assumed to be
instantaneous and they randomly redistribute the energy between all particles.
In this case, we can perform a time average to get a meaningful answer. The
latter procedure is more satisfying from an experimental point of view, and
allows us to compare experimental results with ensemble averages derived from
the Hamiltonian above.
If we assume that the atoms are hard spheres, instantaneous collisions are
the only possibility. But the interactions between atoms are more than that,
there are long range tails in the potential. This means that when the density of
the gas is too high, the interaction potential plays a role. Hence the ideal gas
is a first approximation of a description of a gas at low density, where we can
ignore the interaction terms.
Density of states.
We will calculate the entropy of the ideal gas in the micro-canonical ensemble.
We replace the delta functions by Gaussian exponentials, and use the limit
ɛ → 0. We have to use this ɛ-procedure in order to avoid problems with the
product of a delta-function and its logarithm. The entropy follows from
1
S(U, V, N) = −kB
N!h3N
d XC 1
ɛ √ (U−H)2
e− ɛ
π 2
log C 1
ɛ √ (U−H)2
e− ɛ
π 2
(7.33)