a slightly different view

A slightly different view PhL 20.21.16 I was going to put this at the end of Section 5.5, but decided not to. I will explain why below. First here is the section I was going to add: _______________________ A slightly different view In the above discussion, the functions N(v) and g(v) are densities in "speed space" v and that is the usual path one takes with the Maxwell distribution. One can alternatively define N(v) to be the density of particles in velocity space so that Ni = N(v)d3v. The degeneracy in this case is just gi = (d3v/V) where V is the same tiny lattice volume in v-space we used before. Quantum theory tells us that V = (π/mL)3 for our box of helium gas of edge L. Then Ni = A gi e-βε N(v)d3v = A (1/V) d3v exp(-βmv2/2) N(v) = (A/V) exp(-βmv2/2) To eliminate the derived Lagrange multiplier A we write M = ∫first octant N(v)d3v = ∫N(v) v2dv dΩ = ∫dΩ ∫N(v) v2dv = (1/8) 4π ∫ (A/V) exp(-βmv2/2) v2dv = (π/2) (A/V) { (1/2)Γ(3/2) (βm/2)-3/2} // from *** = (π/2) (A/V) (1/2)(/2) (βm/2)-3/2 = (π3/2/8) (A/V) (βm/2)-3/2. Then (A/V) = M (8 π-3/2) (βm/2)3/2 and so N(v) = M (8 π-3/2) (βm/2)3/2 exp(-βmv2/2) Thus N(v) is really just a function of speed v, but we write it as N(v) as a reminder that it is the particle density in 3D velocity space. Recall now from *** that N(v) = M (4π-1/2) (βm/2)3/2 v2 exp(-βmv2/2) Comparison shows that N(v) = (1/8) (4πv2)N(v) = (πv2/2) N(v) so N(v) = (2/πv2)N(v) We can interpret this to say that N(v) is N(v) times the area of the octant shell of radius v in v-space. The result is in retrospect obvious if one writes d3v = v2dvdΩ so N(v)dv = ∫octant shell volume N(v) v2dvdΩ = (1/8)4π N(v) v2dv We are now in a better position to answer the following question: How may helium atoms in the 1 cm3 box have the specific "quantum numbers" (nx,ny,nz) ? This set of integers is associated with a tiny volume V = (π/mL)3 so the answer to the question is , Nnnn = N(vnnn) V = [ M (8 π-3/2) (βm/2)3/2 exp(-βm v2nnn/2)] * (π/mL)3 = (2/πvnnn2)N(vnnn) * (π/mL)3 where recall from ** that v2nnn = (π/mL)2 ( nx2+ny2+nz2) This count Nnnn is then analogous to the count Ni in a discrete-energy-level system where the energy levels are labeled by the single index i. Now here is some Maple code (continues previous Maple code) _______________________ When I wrote the above, I was assuming that perhaps N_(1,2,5) would be maybe 26 particles. But what I learned is that the quantum numbers ni have to be on the order of 108 to get a speed v which is out in the middle of the Maxwell distribution. But then you find that there is only 10-6 of one particle in the state corresponding to (nx,ny,nz) = (108,108,108). Basically here is what you have: Nnnn = N(v)V = M (8 π-3/2) (βm/2)3/2 exp(-βmv2/2)V The right side has a max value at v = 0 which I see is around 0.3 x 10-5 and for any larger v the result is just going to be smaller. So you never get a number like 26! The cube V is so small that you have to take a million cubes just to get a single particle being in that region of v-space! How many V cubes are in a sphere whose radius is the mean speed? (4/3)πv3/V ~ 4 * (1243)3 * 1016 = 1026 By showing g(v) >> N(v) I have already shown that there are many more states (think cubes) than there are particles in a shell near any v. Think of a large sphere in velocity space where you can see all the tiny cubes. Most of the 1026 cubes are empty, but 1019 of the cubes are occupied and appear as black cubes in my picture. The unoccupied cubes are clear. So if you ask about the probability of there being a particle in a particular cube (nx,ny,nz), that probability is going to be 10-7 or some small number like that. That is why my number Nnnn is coming out so small. So I conclude that computing Nnnn is just not a very interesting thing to do.