Resolve Discrepancies in Boltzmann's Equilibrium Theory

[Bill Dreiss]

In Lectures on Gas Theory (Dover Books on Physics) (p. 74), Boltzmann states “In nature, the tendency of transformations is always to go from less probable to more probable states”, by which he means what are now called macrostates. Thus he claims that an ideal gas almost always evolves to the most probable macrostate, which he defines as equilibrium.

For example, the number of molecules found in one half of a box of ideal gas is given by the binomial distribution. However, it can easily be seen that if the number of molecules in one half of the box is within ± 0.6745 σ of the mean, the probability of the next macrostate being greater than the current macrostate is less than ½, contrary to Boltzmann’s statement.

Furthermore, by his definition of equilibrium, the probability of being in the most probable macrostate is approximately √(2/πN) for large N (where N is the total number of molecules in the gas). This means that the probability of being in equilibrium becomes vanishingly small for large systems, contradicting the common understanding of the meaning of equilibrium.

How can these discrepancies be resolved?

 

[bob012345]

In Lectures on Gas Theory (Dover Books on Physics) (p. 74), Boltzmann states “In nature, the tendency of transformations is always to go from less probable to more probable states”, by which he means what are now called macrostates. Thus he claims that an ideal gas almost always evolves to the most probable macrostate, which he defines as equilibrium.

For example, the number of molecules found in one half of a box of ideal gas is given by the binomial distribution. However, it can easily be seen that if the number of molecules in one half of the box is within ± 0.6745 σ of the mean, the probability of the next macrostate being greater than the current macrostate is less than ½, contrary to Boltzmann’s statement.

Furthermore, by his definition of equilibrium, the probability of being in the most probable macrostate is approximately √(2/πN) for large N (where N is the total number of molecules in the gas). This means that the probability of being in equilibrium becomes vanishingly small for large systems, contradicting the common understanding of the meaning of equilibrium.

How can these discrepancies be resolved?

Here is an article that discusses the topic and which hopefully will address your specific questions.

https://www.researchgate.net/publication/1769551_Boltzmann_Gibbs_and_the_Concept_of_Equilibrium

Abstract​

The Boltzmann and Gibbs approaches to statistical mechanics have very different definitions of equilibrium and entropy. The problems associated with this are discussed, and it is suggested that they can be resolved, to produce a version of statistical mechanics incorporating both approaches, by redefining equilibrium not as a binary property (being/not being in equilibrium) but as a continuous property (degrees of equilibrium) measured by the Boltzmann entropy and by introducing the idea of thermodynamic-like behavior for the Boltzmann entropy. The Kac ring model is used as an example to test the proposals.

 

[Bill Dreiss]

Thank you for the excellent reference. The paper is a welcome corroboration of my ideas, as indicated by the following quotes:

Page 2: This means that there will be no convergence to an attractor (which could in the dynamic sense be taken as an equilibrium state). [This negates Boltzmann’s explanation of the second law.]

Page 3: Along a trajectory sB will not be a monotonically increasing function of time. [For the binomial distribution representing the molecules in one half of a box of gas, a macrostate will be followed by a more probable macrostate less than half the time, as described in my previous post.]

Page 4: The proportion of ΓN contained within the band of macrostates B(N, k) decreases with N.4 [This means the proportion of ΓN outside the band of macrostates approaches ΓN, indicating that all microstates should be included in the equilibrium state.]

Page 4 footnote: 4For later reference we note that, in particular, this result applies to k = 0 with m(μ0)/m(ΓN) ≃ √(2/Nπ) for large N. The proportion of phase space in the largest macrostate decreases with N. [The attached paper by Jack Denur details the derivation for this equation.]

Page 12: It should be emphasized that the Gibbs entropy (20) is no longer taken as that of some (we would argue) non-existent equilibrium state, but as an approximation to the true thermodynamic entropy which is the time-average over macrostates of the Boltzmann entropy. [This is known as statistical stationarity (https://en.wikipedia.org/wiki/Stationary_process), which means that the mean and standard deviation of the probability/entropy distribution does not change with time. “White noise is the simplest example of a stationary process” and can be generated by tracking the position of a molecule over time.]

On page 11 coarse graining is portrayed as fundamental to statistical mechanics. However, one can see from the attached papers by Boltzmann and Brian Zhang that if the energy is assumed to be composed of discrete units, coarse graining is not needed for deriving the energy distribution. Also note that Zhang’s model does not use statistical macrostates.

Equation (23) on page 11 makes no sense since it is meaningless to take ratios of entropies, whose purpose is the creation of extensive variables that can be added and subtracted. Equation 23 implies that for large N, sB(ΓN) − sB(μMax) ≅ 0. But from footnote 4 above, we see that sB(ΓN) − sB(μMax) ≅ ½ ln (Nπ/2), which instead approaches ∞ for large N.

As this paper makes clear, the Boltzmann model fails in its primary purposes of defining equilibrium, providing a transfer function, and explaining the second law. The Markov model described by Brian Zhang accomplishes all three.

To the best of my knowledge, none of the above issues have been addressed by textbooks on statistical mechanics since the publication of Boltzmann’s Lectures on Gas Theory in 1895. I’ve traced this history through Fowler, Schrödinger, Tolman, Reif, Kittel and many others and have found no mention of any of the above issues. Boltzmann’s model has had a good run, but perhaps it’s time for a paradigm change.

 

[Demystifier]

Thus he claims that an ideal gas almost always evolves to the most probable macrostate, which he defines as equilibrium.

Not exactly, it evolves towards the most probable macrostate. It means that after a short time it will only be slightly closer to the most probable macrostate, it will not immediately arrive to that state.

the probability of the next macrostate being greater than the current macrostate is less than ½, contrary to Boltzmann’s statement.

Do you see now why it's consistent with the Boltzmann’s statement formulated as above?

Furthermore, by his definition of equilibrium, the probability of being in the most probable macrostate is approximately √(2/πN) for large N (where N is the total number of molecules in the gas). This means that the probability of being in equilibrium becomes vanishingly small for large systems, contradicting the common understanding of the meaning of equilibrium.

The precise statement is not that it will be exactly in the most probable state, but that it will be very close to the most probable state. It's similar to tossing coins; if you toss coin 1000 times the probability that there will be exactly 500 heads is very small, but the probability that there will be approximately 500 heads is very big.

 

[DrClaude]

There are also fluctuations that scale as $1/\sqrt{n}$, so in the thermodynamic limit they become too small to be measured.

 

[Bill Dreiss]

Not exactly, it evolves towards the most probable macrostate. It means that after a short time it will only be slightly closer to the most probable macrostate, it will not immediately arrive to that state.

Do you see now why it's consistent with the Boltzmann’s statement formulated as above?

The precise statement is not that it will be exactly in the most probable state, but that it will be very close to the most probable state. It's similar to tossing coins; if you toss coin 1000 times the probability that there will be exactly 500 heads is very small, but the probability that there will be approximately 500 heads is very big.

While the notion that Boltzmann’s equilibrium includes nearby macrostates is widespread, it is not supported by the literature. Boltzmann is both explicit and consistent throughout his papers and his book Lectures on Gas Theory that equilibrium consists of a single macrostate (singular). For instance, in the attached Boltzmann Energy 1877.pdf, he states on page 1975 that “The initial state in most cases is bound to be highly improbable and from it the system will always rapidly approach a more probable state until it finally reaches the most probable state, i.e., that of the heat equilibrium.” This couldn’t be clearer. All of the statistical mechanics textbooks I’ve reviewed follow Boltzmann’s lead in this regard. If you can find a counterexample, I would appreciate the reference.

My post was confusing in this regard, and should have read “if the number of molecules in one half of the box is within ± 0.6745 σ of the mean, the probability of the probability of the next macrostate being greater than the probability of the current macrostate is less than ½”. This should clarify what I meant.

Your ”precise statement” is true, but it’s not what Boltzmann says. Also, it is not “precise” since you don’t define what you mean by “very close”. The definition which best supports your argument is that “very close” means, say, within ±1% of the most probable macrostate. For the binomial distribution, 99% of the distribution will fall within this range for N ≅ 17,000. But we can save a lot of trouble by just recognizing a priori that all 2N microstates are equally accessible, according to Boltzmann’s assumption of independence, and that the entropy therefore remains constant at S = kN ln 2, once again because it is not defined as a mathematical function of time. So if the equilibrium entropy is not S = k ln Wmp, it must be S = k ln Wtotal = kN ln 2. Since this is correct, it eliminates the need for macrostates, which are not used to determine the entropy and are not relevant to the alleged progression toward equilibrium. For an alternative derivation of the energy distribution that makes no use of macrostates, see the attached paper by Brian Zhang entitled “Coconuts and Islanders”.

 

[Demystifier]

My post was confusing in this regard, and should have read ...

In other words, it matters what you meant, not what you said. I think the same applies to Boltzmann.

 

[Bill Dreiss]

My corrected sentence clarifies the original but doesn't contradict it. On the other hand, Boltzmann is consistent throughout his writings from 1872 through 1895 in identifying equilibrium as the most probable macrostate (singular). If you can find an exception, please let me know. Nevertheless, if one accepts your interpretation, it obviates the need for macrostates, rendering your interpretation meaningless.

For instance, the simulation in "Coconuts and Islanders" approaches equilibrium without utilizing macrostates and by incorporating all accesible microstates, as your interpretation implies. The simulation provides a numerical solution to the problem of the approach to equilibrium. Boltzmann (1877) also applies a numerical approach to determining the most probable macrostate, which he devotes most of section I to explaining. If he meant to include (almost) all 2^N microstates, why did he go to so much trouble to determine the most probable macrostate?

In the first paragraph of section II, he reiterates his earlier definition of equilibrium by saying "That state distribution which has the most complexions, we consider as the most likely, or corresponding to thermal equilibrium." He continues in section II to provide a continuous counterpart to the discrete derivation of the most probable macrostate provided in section I.