Now it’s time for a more general description of what Boltzmann and Planck did with their different versions of discrete energy elements distributed in equilibrium among their respective systems of molecules (Boltz) or resonators (Planck). This description goes by the name “energy partitions”. I'm not sure who invented this generic terminology, but it was put to good use in 1922 by Darwin and Fowler in their paper "On the partition of energy".
The reason for the invention of energy partitioning by Boltzmann and its later use by Planck was to calculate entropy.
Energy partitions
describe the number of ways in which discrete energy elements can be
distributed among a collection of particles capable of absorbing and emitting
energy. You may not have encountered the name “energy partitions” before, but
if you’ve studied thermodynamics and kinetic theory beyond the freshman level,
you’ve encountered the equipartition theorem and the partition function. Since
I’m providing links to those terms, I won’t describe them further except to say
they are based on the idea of energy partitions.
Occupancy numbers and
occupation numbers
There are two equivalent ways energy partitions can be expressed: the number of energy elements per particle, or the number of particles per energy level. The first use of energy partitions in a way that later became standard was by Boltzmann in 1877, in his paper “On the Relationship between the Second Fundamental Theorem of the Mechanical Theory of Heat and Probability Calculations Regarding the Conditions for Thermal Equilibrium.”
In this paper, Boltzmann described how to calculate the number of different ways kinetic energy elements could be partitioned among the molecules of a gas, and he considered his “molecules” to be structureless particles possessing only translational kinetic energy. In his model, portions of kinetic energy that are integer multiples of a very small indivisible element of energy, ε, are transferred between the molecules during elastic collisions. (Check on Goodstein or d’Abro or somebody and 1/3 of molecules traveling in each x, y, z direction, no angles of collision needed in calculations.)
A more rigorous mathematical
description of Boltzmann’s energy partitions was developed by Alexander Bach in
the late 1980s. (Bach claims that Boltzmann used the idea of indistinguishable
classical particles, and also introduced Bose-Einstein statistics, which was supposedly
not discovered until 1924 by Bose and Einstein. I’ll discuss both of these controversial
claims later.) Bach’s scheme describes energy partitions in terms of cells
that contain a certain number of particles, where the cells and particles
can be either energy elements or physical particles (molecules or resonators in
our case).
Bach uses the term
“occupancy number” for the number of energy elements per physical particle, and
“occupation number” for the number of physical particles per energy level. (Bach also has a category called "configurations," but that would get us into the problem of particle distinguishability, which I'm not ready to do yet.)
Let’s look at occupancy
numbers first. In Boltzmann’s notation the total kinetic energy of his gas molecules
is L = λε and the number of molecules is n. The overall occupancy number for his gas is therefore
λ energy elements (particles) distributed among the n molecules (cells). Individual-cell occupancy numbers are needed to calculate probabilities, and those are dealt with in the following section.
It may seem that just
inverting the overall occupancy number description would give the overall occupation number as n molecules
(particles) distributed over λ energy levels (cells), but here we run
into a little counting problem. Or maybe I should say an accounting
problem.
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One part of the problem is that Boltzmann
introduced another constant in addition to λ to count the number of
energy elements, which, remember, are necessarily kinetic energy elements. He
says “we assume that the kinetic energies of each molecule form an arithmetic
progression, such as the following: 0,
ε, 2ε, 3ε,…, pε. We call P the largest value of the kinetic energy,
But this little p seems to me to be the same as λ, because the largest possible kinetic energy any one molecule can have is the total kinetic energy. Then all others would have zero energy, and p would equal λ.
(This would never happen
in a realistic setting with many molecules, with “never” meaning an
infinitesimal probability of something like 1/n. But we can imagine a video-game (not much of
a game, but never mind that) in which n-1 molecules are at rest at the
bottom of the screen, and we send in the nth molecule from
the top of the screen with kinetic energy pε = λε. Eventually, this total energy would be
distributed evenly, in an average way, among the molecules. Equilibrium would be established, with the usual mean-square fluctuations about the equilibrium value or average value of the kinetic-energy-per-molecule.)
But Boltzmann was no
dummy. He introduced p and λ
for a reason. He wanted “the highest
possible value of the kinetic energy” to be different from the constant total
energy. Apparently he meant this to be interpreted as the highest possible kinetic energy
of any single distribution, with p = λ being only one such p,
and a highly unlikely one at that.
But the strangest thing
is that Boltzmann later gives an example in which he allows p to go to infinity while λ = 7 (see below), so there’s some kind of madness in his method (rather than method in his
madness), but it’s too much for me to juggle at the moment, and I’ll just continue
with occupancy numbers and occupation numbers now.
--------
I have to backtrack
a little bit, because of the accounting problem. Boltzmann’s states the main calculational purpose of his 1877 paper this way: “I want to establish
the probability of a state distribution regardless of how it is created, or
more specifically, I want to find all possible combinations of the p+1
kinetic energy values allowed to each of the n molecules and then establish
how many of these combinations correspond to each state distribution.”
But nobody
counts zero as a value when we’re talking about occupancy numbers. There are λ energy elements. There are
n molecules. No matter how many
of them have zero energy, there are still λ energy elements distributed
over n molecules. Zero as an
energy element is not counted as an occupancy number. However, again, Boltz was no dummy, and his (English-translated) statement is certainly correct: each molecule when considered independently of the others, has equal access to
But this is not very significant, since nobody uses occupancy numbers in calculations of probability anyway, except me (see below). Occupation numbers are used instead, and since occupation numbers have cells that are energy levels, the zero-energy level must be counted.
For occupation numbers, then, we have p+1 cells, and the overall occupation number is n molecules (particles) distributed over p+1 energy levels (cells). The counting problem I mentioned above is now accounted for.
Occupancy versus occupation calculations for n=3 and λ=p=5
Before any calculations are done, the
total-number-of-particles constraint and the total-energy constraint must be
expressed in equations. An occupancy-number variable and an occupation-number
variable are needed in order to do this.
Occupancy number variable: kj
= the number of energy elements in molecule j. The sum of the kj’s
from j = 1 to j = n is the total number of energy elements
λ. Let’s say there are three molecules and five energy elements. Then
k1 + k2
+ k3 = 5.
This equation is both a
constraint on the energy (the sum = λ = 5) and a constraint on the
number of molecules (the upper limit of the sum = n = 3)
Occupation number variable: wi is the number
of molecules in energy level i. The sum of the wi’s
from i = 0 to i = λ is the total number of molecules n.
Using three molecules and five energy elements tambien, we have
w0 + w1 + w2 + w3 +
w4 + w5 = 3
0·w0 + 1·w1
+ 2·w2 + 3·w3 + 4·w4 + 5·w5
= 5.
Now let’s make a table. It's helpful to show occupancy and occupation number tables together. So, here’s my two-tables-in-one, with occupancy numbers on the left, occupation numbers on the right, using n=3 and λ=5:
occupancy #’s
occupation #’s
|
k1 |
k2 |
k3 |
--- |
w0 |
w1 |
w2 |
w3 |
w4 |
w5 |
|
5 |
0 |
0 |
|
2 |
0 |
0 |
0 |
0 |
1 |
|
0 |
5 |
0 |
|
2 |
0 |
0 |
0 |
0 |
1 |
|
0 |
0 |
5 |
|
2 |
0 |
0 |
0 |
0 |
1 |
|
|
|
|
|
|
|
|
|
|
|
|
4 |
1 |
0 |
|
1 |
1 |
0 |
0 |
1 |
0 |
|
4 |
0 |
1 |
|
1 |
1 |
0 |
0 |
1 |
0 |
|
0 |
4 |
1 |
|
1 |
1 |
0 |
0 |
1 |
0 |
|
1 |
4 |
0 |
|
1 |
1 |
0 |
0 |
1 |
0 |
|
1 |
0 |
4 |
|
1 |
1 |
0 |
0 |
1 |
0 |
|
0 |
1 |
4 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
3 |
2 |
0 |
|
1 |
0 |
1 |
1 |
0 |
0 |
|
3 |
0 |
2 |
|
1 |
0 |
1 |
1 |
0 |
0 |
|
0 |
3 |
2 |
|
1 |
0 |
1 |
1 |
0 |
0 |
|
2 |
3 |
0 |
|
1 |
0 |
1 |
1 |
0 |
0 |
|
2 |
0 |
3 |
|
1 |
0 |
1 |
1 |
0 |
0 |
|
0 |
2 |
3 |
|
1 |
0 |
1 |
1 |
0 |
0 |
|
|
|
|
|
|
|
|
|
|
|
|
2 |
2 |
1 |
|
0 |
1 |
2 |
0 |
0 |
0 |
|
2 |
1 |
2 |
|
0 |
1 |
2 |
0 |
0 |
0 |
|
1 |
2 |
2 |
|
0 |
1 |
2 |
0 |
0 |
0 |
|
|
|
|
|
|
|
|
|
|
|
|
3 |
1 |
1 |
|
0 |
2 |
0 |
1 |
0 |
0 |
|
1 |
3 |
1 |
|
0 |
2 |
0 |
1 |
0 |
0 |
|
1 |
1 |
3 |
|
0 |
2 |
0 |
1 |
0 |
0 |
(to be continued with permutations, combinations, particle distinguishability (or not), and the multinomial theorem)
