Fight Night - Max and Boltz versus the Bose and 'Steiner - Be there or be indistinguishable

A mind is a terrible thing to waste they say.  With that in mind, I wander around a few "sciency" blogs,  websites and fora in an attempt to knock some of the cobwebs out of the corners of my mind.  Sometimes visits lead me to believe that there may not be any intelligent life left on Earth.  One recent example is a wild, completely irrelevant discussion on a suggestion in a new textbook.  The suggestion was that Bose-Einstein Statistics might be useful when attempting to describe cloud formation at low temperatures.  Clouds tend to be a beautiful PITA at times since they appear to have their own personal drummer.  On Earth most clouds are made of water in its various stages, solid, liquid and gas with the solid and liquid being the parts we generally see.  When water vapor becomes liquid, it is considered to condense.  When water vapor becomes ice it is considered to nucleate.  When ice directly changes to water vapor it is considered to sublimate.  When water is well below its freezing point it is consider super cooled and when water vapor is present in greater amount that it should be based on saturation pressure it is considered super saturated.  Predicting the path that water vapor will take in a cloud based on just temperature and/or energy gets to be a bit difficult.  If you can't predict what path it will follow, you may be able to determine a probability that it will follow a particular path which will give you some idea how likely you can pick the right path.  If the way you try to predict tends to be useless or close to useless, you might want to consider a different approach.

To better understand why you might want to try a different approach, it would be nice to understand a few approaches.  So I went looking for an fairly simple description of applicable statistics and found this:

I don't really see the answer in the other answer so let me do the calculation here. Your general Boltzmann Ansatz says that the probability of a state n depends on its energy as

pn=Cexp(−βEn)

where β=1/kT. Fermions are identical particles that, for each "box" or one-particle state they can occupy (given e.g. by nlms in the case of the Hydrogen atom-like states), admit either N=0 or N=1particles in it. Higher numbers are forbidden by the Pauli exclusion principle. The energies of the multi-particle state with N=1 and N=0 in a particular one-particle state nlms differ by ϵ. Consequently,

p1p0=Cexp(−β(E+ϵ))exp(−βE)=exp(−βϵ)

where I used the Boltzmann distribution. However, the probabilities that the number of particles in the given one-particle state is equal to N=0 or N=1 must add to one,

p0+p1=1.

These conditions are obviously solved by

p0=11+exp(−βϵ),p1=exp(−βϵ)1+exp(−βϵ),

which implies that the expectation value of n is equal to the right formula for the Fermi-Dirac distribution:

⟨N⟩=p0×0+p1×1=p1=1exp(βϵ)+1

The calculation for bosons is analogous except that the Pauli exclusion principle doesn't restrict N. So the number of particles (indistinguishable bosons) in the given one-particle state may be N=0,1,2,…. For each such number N, we have exactly one distinct state (because we can't distinguish the particles). The probability of each such state is called pn where n=0,1,2,….

We still have

pn+1pn=exp(−βϵ)

and

p0+p1+p2+⋯=1

These conditions are solved by

pn=exp(−nβϵ)1+exp(−βϵ)+exp(−2βϵ)+…

Note that the ratio of the adjacent pn is what it should be and the denominator was chosen so that all the pn from n=0,1,2… sum up to one.

The expectation value of the number of particles is

⟨N⟩=p0×0+p1×1+p2×2+…

because the number of particles, an integer, must be weighted by the probability of each such possibility. The denominator is still inherited from the denominator of pn above; it is equal to a geometric series that sums up to

11−q=11−exp(−βϵ)

Don't forget that 1−exp(−βϵ) is in the denominator of the denominator, so it is effectively in the numerator.

However, the numerator of ⟨N⟩ is tougher and contains the extra factor of n in each term. Nevertheless, the sum is analytically calculable:

∑n=0∞nexp(−nβϵ)=−∂∂(βϵ)∑n=0∞exp(−nβϵ)=…

⋯=−∂∂(βϵ)11−exp(−βϵ)=exp(−βϵ)(1−exp(−βϵ))2

This result's denominator has a second power. One of the copies gets cancelled with the denominator before and the result is therefore

⟨N⟩=exp(−βϵ)1−exp(−βϵ)=1exp(βϵ)−1

which is the Bose-Einstein distribution.

You could also obtain another version of the Boltzmann distribution for distinguishable particles by a similar calculation. For such particles, N could take the same values as it did for bosons. However, the multiparticle state with N particles in the one-particle state would be degenerate because the particles are distinguishable. There would actually be N! multiparticle states with N particles in them. The sum would yield a Taylor expansion for the same exponential.

Note added later: the derivation above was for μ=0. When the chemical potential is nonzero, all appearances of ϵ have to be replaced by (ϵ−μ). Of course that one may only talk about a well-defined value of μ when we deal with a grand canonical potential; it is impossible to derive a formula depending on μ from one that contains no μ and assumes it's ill-defined. The derivation above was meant to show that the difficult 1/(exp±1) structures do appear from a simpler exp(−βE) Ansatz because I think it's the only nontrivial thing to be shown while discussing the relations between the Boltzmann and BE/FD distributions. If that derivation proves the same link as the textbook does, then I apologize but I think there is "nothing else" of a similar kind to be proven.

I made the whole explanation a link to the comment of Luboš Motl who could be described as a quantum physics junkies go to dealer.  

Given the tools Lubos describe, how could one possibly use Bose-Einstein Statistics for water vapor?

One simple example would be to consider the limits of T as they relate to water.  At -43 C degrees, water vapor is super cooled to the point that it will change to ice, but there is a fairly large time window of hours.  So if you replace T with (T-T(-43C)) you would be indicating that the probability that water will be in its solid state is 100% at -43C degrees.  However, since that isn't really true because of the time window, you could pick a lower temperature that makes the probability more likely for a smaller time window.  Say minus 50 C for example.  That could be useful even though it is not completely true.  

The suggestion boils down to if the "accepted" methods don't provide enough accuracy consider options like Bose-Einstein, which created a tempest in a teacup.  Personally, I found the suggestion interesting and the controversy inspired me to review tidbits lurking behind some of the cobwebs. i