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Accessible States



The "particle position" thread has been interesting, but not what I had in
mind when I asked the original question -- it has warmed me toward a
derivation that I still have a lot of trouble with but ...

Consider a box of volume V which contains N atoms of an ideal gas which
have a total energy E.

Now I start to worry: Assume that each atom is contained in one of a
number of
6-dimensional cells of

V(6) = DxDyDzDpxDpyDpz = h^3

The total number of states in the box for each atom is

Omega(1) = (total 6-d volume)/cell volume
= V(real)V(momentum)/DxDyDzDpxDpyDpz
= VrVp/h^3

Now Vp = (4/3)pi x p(max)^3 & p(max) = sqrt(2mE)

So Omega(1) = (4/3)pi(2m)^(3/2) x V E^(3/2)

Now Omega(N) =Prod[Omega(i)]

So Omega(N) = Const x V^N x E^(3N/2) Cf big Rief Eqn 2.5.19

The result seems OK (assuming I typed it correctly), but I have problems:

1) Is it ok to say that the atom fills a Heisenberg volume? ie that the
possible 6-d positions of the atoms are this size? (This is what the
"Particle Position" thread has warmed me to.) Is it ok to find Omega by
just dividing the total 6-d volume by the cell volume? This seems to be in
accord with the similar approach of Stowe, but alas I think that he makes
mistakes elsewhere. I would feel a lot better if there were some
concurrence by the list here. At least this approach arrives at the usual
answer.

2) Is there a way to get this result via some method not using modern
concepts -- none of the usual texts do. What was Boltzman's thinking here
for S=k ln Omega?

TX for your help

Jim Green
mailto:JMGreen@sisna.com
http://users.sisna.com/jmgreen