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 

The Thermal Reservoir Oceans

From the free dictionary here are two terms that are often confused.

Heat sink
n.
1. An environment capable of absorbing heat from an object with which it is in thermal contact without a phase change or an appreciable change in temperature.
2. A protective device that absorbs and dissipates the excess heat generated by a system. (my bold)

Heat reservoir
A thermal reservoir, a short-form of thermal energy reservoir, or thermal bath is a thermodynamic system with a heat capacity that is large enough that when it is in thermal contact with another system of interest or its environment, its temperature remains effectively constant.[1] It is an effectively infinite pool of thermal energy at a given, constant temperature. The temperature of the reservoir does not change irrespective of whether heat is added or extracted. As it can act as a source and sink of heat it is often also referred to a heat reservoir or heat bath.

Lakes, oceans and rivers often serve as thermal reservoirs in geophysical processes, such as the weather. In atmospheric science, large air masses in the atmosphere often function as thermal reservoirs.


The oceans are often incorrectly called heat sinks by many involved in the climate change debate.   They are more appropriately called a heat reservoir since the heat capacity of the oceans is about 1000 times greater than the atmosphere and the average temperature of the upper ocean is greater than the atmosphere.  There are also two phase changes that can be expected with the oceans, fusion or freezing near the poles and evaporation away from the poles.   I tend to cringe every time the term Heat Sink is used knowing any observations that follow are most likely flawed.  

The large thermal capacity of the oceans and lakes is not contained at their surface but rather at considerable depth.  The most stable temperature in the oceans for example is the 4 C thermocline at an average depth of around 1000 meters.  This is below the majority of surface generated mixing which is the reason for its stability.  Properties of water including the maximum density of fresh water at 4C degrees and the temperature of fusion ranging from 0C to around -2 C for salt water with around 35 grams of salt per liter of volume help maintain a fairly stable reference temperature.   In deeper fresh water lakes the 4 C density prevents them from freezing solid by allowing less dense ice to form on the surface insulating the more dense 4 C sub-surface liquid water.

Depending on the source, the oceans contain about 1.5 x 10^9 cubic kilometers of water.   Each cubic kilometer of water contains one billion cubic meters (10^9), with each cubic meter containing 1000 (10^3) liters each containing 1000 (10^3) grams.   That would be a total of approximately 1.5 x 10^24 grams of water with an average temperature of about 4 C degrees.  

How much heat is contained in that volume is a bit confusing due to the phase change potential.   If it all evaporates it would have to gain 2260 Joules per gram plus about 4.2 joules per degree increase in temperature to the evaporation point which could be lower than the standard boiling point.  For all of it to freeze it would have to release 334 Joules per gram plus about 4.2 Joules per gram as the temperature reduced to possibly -18 C for a saturated salt brine.   A fair SWAG for the effective total heat capacity for the oceans at 4 C average would be about 334 Joules per gram which happens to be equal to the effective radiant energy of a surface at 4 C degrees using the Stefan-Boltzman relationship for an ideal black body.

Typically, the specific heat capacity is used to determine how much energy would be required to change a volume of a substance by 1 C degree which for a gram of water would be about 4.2 Joules.   To raise the entire global oceans by 1 C would require about 4.2(Joules/gram)*1.5 x 10^24 grams or 6.3 x 10^24 Joules assuming no change in the rate of evaporation or freezing takes place in the process, which is unlikely for larger changes in temperature.

The SWAG estimate of total heat capacity may be relevant  to basic Greenhouse Effect considerations since the ocean reservoir effective heat capacity of 334 Joules per gram happens to equal the best estimate of the effective Down Welling Long Wave Radiation (DWLR) felt at the "surface" of the Earth.  DWLR is a bit unique to atmospheric physics.  Most other branches of physics consider the net heat flux not the difficult to manage gross heat fluxes.   While the SWAG may not have an obvious physical meaning, it does at least provide a stable, as in thermal reservoir, frame of reference.  


Hopefully some will find this bit of minutia useful.  If you happen to notice any calculation errors, please do not hesitate to comment.
  

Using CO2 as a Climate Reference

There is still a battle between the extreme pessimist and different degrees of optimist/realists over the impact of Greenhouse Gases, primarily CO2 - Carbon Dioxide "Climate".  Since atmospheric CO2 is one of the factors that we have reasonable good data on, it makes a reasonable good proxy or reference to relate changes in our atmosphere to changes in our "climate".  There are many other factors that can impact "climate", but most have much less recorded data.  There is another issue with what is "climate"?

UPDATE: the axis titles are swapped.

Here is a simple illustration of CO2 being used as a reference.  Since the current data should be the most reliable, the baseline for this graph is the Mauna Loa CO2 sampling period from 1959 to 2013 or all the full years, J-D, converted to a temperature anomaly form using the simple ln(CO2/CO2ref) times scaling factors of 1.6 and 3.0.  Since the natural log (ln) of a doubling of CO2 concentration would be about 0.69, the "sensitivity" would be about 1.1 C for the 1.6 scale which is about the no feedback value for CO2 all by itself and 2.1 C for the 3.0 scaling factor which is the lower end of the IPCC estimates for CO2 only.  I could have divided by the ln(2) for a direct relationship to temperature per doubling, but this way people will actually have to read in order to see what the curves mean.  Along with the two CO2 reference curves is the normally ice free ocean area surface temperature from 60S to 60N from the ERSSTv3b SST data set available at KNMI Climate Explorer.  From 1958 to Mauna Loa CO2 data is used and prior to that the Berkeley interpolated data is used.  The ERSSTv3b is shown both as monthly anomaly and in a 60 year moving average assuming 60 year averages can be considered Climate.  This assumption should reduce the tenancy to cherry pick shorter periods that might not be long enough to be called "Climate".

The 1.6lnCO2 appears to match the ocean SST very well for the 60 YMA and the 3.0lnCO2 better match the 1970 to present portion of the SST curve.  Now remember that the CO2 references would also include all the "other" factors, anthropogenic and natural.  The object of this post is to see how pessimists, optimists and realists might interpret this simple comparison.