Can indistinguishable particles obey Boltzmann statistics

[DrClaude]

There were two interwoven discussions, so I split one off to https://www.physicsforums.com/threads/identical-and-indistinguishable-particles.939282/

 

[DrClaude]

I am not sure I understand the original problem. To quote from T. Guénault, Statistical Physics (Springer):

One type is a gaseous assembly, in which the identical particles are the gas molecules themselves. In quantum mechanics one recognizes that the molecules are not only identical, but they are also (in principle as well as in practice) indistinguishable. It is not possible to ‘put a blob of red paint’ on one particular molecule and to follow its history. Hence the microstate description must take full account of the indistinguishability of the particles. Gaseous assemblies will be introduced later in Chapter 4.

In this chapter we shall treat the other type of assembly, in which the particles are distinguishable. The physical example is that of a solid rather than that of a gas. Consider a simple solid which is made up of N identical atoms. It remains true that the atoms themselves are indistinguishable. However, a good description of our assembly is to think about the solid as a set of N lattice sites, in which each lattice site contains an atom. A ‘particle’ of the assembly then becomes ‘the atom at lattice site 4357 (or whatever)’. (Which of the atoms is at this site is not specified.) The particle is distinguished not by the identity of the atom, but by the distinct location of each lattice site. A solid is an assembly of localized particles, and it is this locality which makes the particles distinguishable.

This summarizes well how I always thought of the situation. So for me, Boltzmann follows from working with indistinguishable particles.

 

[Philip Koeck]

I am not sure I understand the original problem. To quote from T. Guénault, Statistical Physics (Springer):

This summarizes well how I always thought of the situation. So for me, Boltzmann follows from working with indistinguishable particles.

However, the derivations I've shared give Boltzmann only for distinguishable particles!

 

[Lord Jestocost]

For a system of identical, indistinguishable and independent molecules satisfying the condition that the number of available molecular states is much greater then N, the canonical ensemble partition function Q can be written

Q = 1/N! q^N .

This is the limiting form of Bose-Einstein and Fermi-Dirac statistics and is called classical or Boltzmann statistics. This equation is, for example, satisfied for a monoatomic gas at ordinary temperatures and densities.

For identical and independent molecules where a model can artificially introduce molecular distinguishability, the canonical ensemble partition function Q can be written as

Q = q^N .

The Einstein model of a crystal is an example where this relation can be applied.

These relations are unequivocally discussed in the textbook “An Introduction to Statistical Thermodynamics” by Terrell L. Hill.

 

[NFuller]

I also noticed the opinion that the factor 1/N! is required even for distinguishable particles if they are identical, but I'm not sure what to think of that and it's not what textbooks say.

We know that the $1/N!$ is needed for an ensemble of identical and distinguishable (a better word might be classical) particles to make the entropy extensive. The factorial is not needed if the particles are not identical. Such is the case if two different gasses are allowed to mix. The entropy should increase when two subsystems containing different gasses are combined since mixing the gasses is an irreversible process.

As far as agreeing with what your textbooks say, I think the confusion is again coming from differences in opinion of how to use the words identical and indistinguishable. That topic is for another thread though.

Textbooks claim that you need a factor 1/N! in the partition function to make sure that even "classical" particles are indistinguishable and the entropy becomes extensive.

Again, its not that classical particles of a pure gas are indistinguishable but that they are identical. This means that permuting their positions in phase space on a constant energy manifold leads to the same microstate.

On the other hand the mentioned derivations give Boltzmann only for distinguishable and Bose-Einstein only for indistinguishable particles.

Yes Boltzmann is for distinguishable particles and Bose-Einstein and Fermi-Dirac are for indistinguishable particles.

 

[Philip Koeck]

We know that the $1/N!$ is needed for an ensemble of identical and distinguishable (a better word might be classical) particles to make the entropy extensive. The factorial is not needed if the particles are not identical. Such is the case if two different gasses are allowed to mix. The entropy should increase when two subsystems containing different gasses are combined since mixing the gasses is an irreversible process.

As far as agreeing with what your textbooks say, I think the confusion is again coming from differences in opinion of how to use the words identical and indistinguishable. That topic is for another thread though.

Again, its not that classical particles of a pure gas are indistinguishable but that they are identical. This means that permuting their positions in phase space on a constant energy manifold leads to the same microstate.

Yes Boltzmann is for distinguishable particles and Bose-Einstein and Fermi-Dirac are for indistinguishable particles.

That would definitely solve my problem. So some textbooks simply got it wrong and the factor 1/N! does not account for industinguishability but only for identity.

 

[Lord Jestocost]

Yes Boltzmann is for distinguishable particles and Bose-Einstein and Fermi-Dirac are for indistinguishable particles.

The Boltzmann statistics is the limiting form of the Bose-Einstein and Fermi-Dirac statistics for a system of identical, indistinguishable “particles” in case the number of energy states available to anyone “particle” is very much larger than the number of “particles” in the system.

 

[Philip Koeck]

The Boltzmann statistics is the limiting form of the Bose-Einstein and Fermi-Dirac statistics for a system of identical, indistinguishable “particles” in case the number of energy states available to anyone “particle” is very much larger than the number of “particles” in the system.

If you have a system of identical or even slightly different, distinguishable particles, what distribution would you use to describe it?

 

[Stephen Tashi]

That would definitely solve my problem. So some textbooks simply got it wrong and the factor 1/N! does not account for industinguishability but only for identity.

Perhaps this is a topic for the other thread, but what deductive process is going on between premises concerning things being identical or indistinguishable and conclusions about formulae for physical quantities? ( Is it a Bayesian form of deduction - as according to Jaynes ?)

In purely mathematical problems about combinatorics, the given information about things being identical or indistinguishable is used to define what is meant by "ways". This is needed in order to interpret the inevitable question: "In how many different ways can...?".

In physics, in addition to showing we have counted the number of "ways" correctly, we need some justification that says: The following is the physically correct way to define a "way": ... .

A frequently seen deductive pattern is:
1. Provide formulae for the number of "ways" using combinatorics.
2. Deduce probability distributions from the combinatorial results - usually by assuming each "way" has the same probability.

In this thread there is a concern for:
3. Use the probability distributions to compute (Shannon) entropy
4. Check that the entropy computations resolve GIbb's paradox. - i.e. make sure entropy is an "extensive" quantity.

My interpretation of the Jaynes paper "The Gibbs Paradox" http://www.damtp.cam.ac.uk/user/tong/statphys/jaynes.pdf is that information about "identical" or "indistinguishable" particles is not an absolute form of information - i.e. it is not a property of Nature that is independent of who is performing experiments. If particles are "indistinguishable" to a certain experimenter then that experimenter doesn't know how to keep track of which one is which. An experimenter cannot perform any experiments that would require distinguishing among particles that are indistinguishable to that experimenter. From the Bayesian perspective, Entropy (when defined as a function of a probability distribution) is defined relative the experimenter's capabilities.

 

[Lord Jestocost]

If you have a system of identical or even slightly different, distinguishable particles, what distribution would you use to describe it?

There are either identical or non-identical “particles” (slightly different is no criterion for statistical considerations). To find the statistical distribution describing the system, one starts to set up the appropriate partition function for the system in question.

Identical, distinguishable particles:

That can be the case when modeling the solid state. By assuming, for example, that the identical atoms are confined to lattice sites and that each site is occupied at most once, one “artificially” introduces distinguishability as the positions in the lattice can be considered as distinguishable labels.

 

[Philip Koeck]

There are either identical or non-identical “particles” (slightly different is no criterion for statistical considerations). To find the statistical distribution describing the system, one starts to set up the appropriate partition function for the system in question.

Identical, distinguishable particles:

That can be the case when modeling the solid state. By assuming, for example, that the identical atoms are confined to lattice sites and that each site is occupied at most once, one “artificially” introduces distinguishability as the positions in the lattice can be considered as distinguishable labels.

I didn't want to consider a system with a maximum of one particle per state. Let's make it very specific: What would be the most appropriate distribution function for a gas of Argon atoms at room temperature with a density low enough to use the ideal gas law?

 

[Lord Jestocost]

What would be the most appropriate distribution function for a gas of Argon atoms at room temperature with a density low enough to use the ideal gas law?

The Maxwell-Boltzmann distribution.

 

[Stephen Tashi]

If you have a system of identical or even slightly different, distinguishable particles, what distribution would you use to describe it?

And what physics does "indistinguishable" vs "distinguishable" imply? For example, suppose I have a box with an open top that contains "sub boxes" inside it, also with open tops. I have a big bag of "indistinguishable" black marbles and I toss them into the big box using some physically implemented random process. Repeating this process estimates a joint probability distribution for the number of balls landing in each of the sub boxes. Now, I make tiny marks on each of the marbles that give each marble a unique identifier. I repeat the experiment using these now-distinguishable marbles. Does it necessarily follow that the tiny marks physically interact wtih the random process in such a way to change the joint probability distribution for the numbers of balls landing in sub-boxes?

When I use distinguishable marbles, the experimental data is more detailed. There are results like: The 5 marbles in box 1 were marbles A,B,C,F,E, The 3 marbles in box 2 were D,G,H ... etc. However, I can use detailed results to produce records with less information - like 5 marbles landed in box 1, 3 marbles landed in box 2. So the joint probability distribution for the number of marbles landing in the sub boxes can be estimated and compared to the joint distribution estimated from using indistinguishable marbles.

That example suggests that "ability ot make a distinction" may have no physical consequences.

So situations in physics where "distinguishable" vs "indistinguishable" particles behave differently are not (in spite of my preference for the Bayesian.outlook) due merely to the ability of an experimenter to make a distinction. For example, I can imagine that when distinguishable marbles are used in the experiment, someone might observe the results and say "Look! We can make a simple mathematical model that explains the joint probability distribution for numbers of balls landing in boxes, if we base it on the combinatorics of indistinguishable objects."

From that viewpoint, the correct deductive order isn't "The particles are indistinguishable therefore we calculate using the combinatorics of indistinguishable objects" Instead the deductive order is "The combinatorics of indistinguishable objects produces a correct model therefore we shall say these particles are indistinguishable".

 

[Philip Koeck]

The Boltzmann statistics is the limiting form of the Bose-Einstein and Fermi-Dirac statistics for a system of identical, indistinguishable “particles” in case the number of energy states available to anyone “particle” is very much larger than the number of “particles” in the system.

I think I can make a summary that sort of works for me. Boltzmann is for distinguishable particles, and atoms in ideal gas are distinguishable since they are sufficiently far apart (far more states than particles). At very high density particles become indistinguishable and Bose-Einstein applies (maybe easiest for Helium). This means of course that the factor 1/N! in the N-particle partition function for an ideal gas has nothing to do with indistinguishability. Some testbooks simply get it wrong. In other words indistinguishability is not required to resolve Gibbs' paradox.

 

[Lord Jestocost]

...and atoms in ideal gas are distinguishable since they are sufficiently far apart

Identical atoms in an ideal gas are indistinguishable and everything can be derived by using quantum statistics. The classical approach (distinguishability) works under certain circumstances as a good approximation to derive the distribution functions, but is in principle “wrong.” There exists nothing like a gas of identical classical “particles” where the “particles” become distinguishable at low densities and indistinguishable at high densities.

Regarding the “Gibbs’ paradox”: This is nothing else but a superficial paradox.

To recommend textbooks in this context:

“Thermodynamik und Statistik” by Arnold Sommerfeld (maybe, there exists an English translation)
“Theorie der Wärme” by Richard Becker (maybe, there exists an English translation)
“An Introduction to Statistical Thermodynamics” by Terrell L. Hill

 

[Philip Koeck]

Identical atoms in an ideal gas are indistinguishable and everything can be derived by using quantum statistics. The classical approach (distinguishability) works under certain circumstances as a good approximation to derive the distribution functions, but is in principle “wrong.” There exists nothing like a gas of identical classical “particles” where the “particles” become distinguishable at low densities and indistinguishable at high densities.

Regarding the “Gibbs’ paradox”: This is nothing else but a superficial paradox.

To recommend textbooks in this context:

“Thermodynamik und Statistik” by Arnold Sommerfeld (maybe, there exists an English translation)
“Theorie der Wärme” by Richard Becker (maybe, there exists an English translation)
“An Introduction to Statistical Thermodynamics” by Terrell L. Hill

German is fine. Thanks for the suggestions.
I'll just repeat my problem (same as original question):
You state that atoms in an ideal gas are indistinguishable and obey Boltzmann.
A combinatorial derivation gives Boltzmann only for distinguishable and Bose-Einstein only for indistinguishable particles.
How is this possible?
I can send my version of the derivation if you want.

 

[Lord Jestocost]

I'll just repeat my problem (same as original question):
You state that atoms in an ideal gas are indistinguishable and obey Boltzmann.
A combinatorial derivation gives Boltzmann only for distinguishable and Bose-Einstein only for indistinguishable particles.
How is this possible?

You can find the lengthy math in Chapter 3 "General Relations for Independent Distinguishable and Indistinguishable Molecules or Subsystems" of Terrell L. Hill's textbook “An Introduction to Statistical Thermodynamics”.

 

[Philip Koeck]

You can find the lengthy math in Chapter 3 "General Relations for Independent Distinguishable and Indistinguishable Molecules or Subsystems" of Terrell L. Hill's textbook “An Introduction to Statistical Thermodynamics”.

Thanks, I'll have a look at that.

 

[Philip Koeck]

You can find the lengthy math in Chapter 3 "General Relations for Independent Distinguishable and Indistinguishable Molecules or Subsystems" of Terrell L. Hill's textbook “An Introduction to Statistical Thermodynamics”.

What I've read in Hill so far seems to confirm the problem I see very nicely.

 

[Lord Jestocost]

What I've read in Hill so far seems to confirm the problem I see very nicely.

To my mind, there is no problem. According to Walter Guido Vincenti and Charles H. Krüger (see page 105 in “Introduction to Physical Gas Dynamics”, John Wiley and Sons, Inc., New York (1965)):

“The foregoing limiting results [for the BE and FD statistics, ed. LJ] can also be obtained directly on the basis of the so-called Boltzmann statistics. This approach, which was appropriate to the classical methods used before quantum statistical mechanics was known, assumes in counting up the microstates that the particles are truly distinguishable. This leads without difficulty to equation (6.1), but an approximate and much argued over correction to account for the actual indistinguishability is needed to arrive at equation (6.2). Because of this difficulty we have chosen to avoid the Boltzmann statistics here and obtain our results solely as the limit of the correct quantum statistics. For convenience we shall refer to these results as corresponding to the Boltzmann limit.”

 

[Philip Koeck]

To my mind, there is no problem. According to Walter Guido Vincenti and Charles H. Krüger (see page 105 in “Introduction to Physical Gas Dynamics”, John Wiley and Sons, Inc., New York (1965)):

“The foregoing limiting results [for the BE and FD statistics, ed. LJ] can also be obtained directly on the basis of the so-called Boltzmann statistics. This approach, which was appropriate to the classical methods used before quantum statistical mechanics was known, assumes in counting up the microstates that the particles are truly distinguishable. This leads without difficulty to equation (6.1), but an approximate and much argued over correction to account for the actual indistinguishability is needed to arrive at equation (6.2). Because of this difficulty we have chosen to avoid the Boltzmann statistics here and obtain our results solely as the limit of the correct quantum statistics. For convenience we shall refer to these results as corresponding to the Boltzmann limit.”

I like the part where they write "... much argued over...".
The Boltzmann distribution is derived for distinguishable particles, the correction (1/N!, I assume) accounts for their "actual indistinguishability".
I see a slight contradiction here.

 

[NFuller]

“The foregoing limiting results [for the BE and FD statistics, ed. LJ] can also be obtained directly on the basis of the so-called Boltzmann statistics. This approach, which was appropriate to the classical methods used before quantum statistical mechanics was known, assumes in counting up the microstates that the particles are truly distinguishable. This leads without difficulty to equation (6.1), but an approximate and much argued over correction to account for the actual indistinguishability is needed to arrive at equation (6.2). Because of this difficulty we have chosen to avoid the Boltzmann statistics here and obtain our results solely as the limit of the correct quantum statistics. For convenience we shall refer to these results as corresponding to the Boltzmann limit.”

The Boltzmann distribution is derived for distinguishable particles, the correction (1/N!, I assume) accounts for their "actual indistinguishability".

Remember, the $1/N!$ is not strictly a product of quantum mechanics. It is there because a permutation of the particles on a constant energy manifold in phase space does not alter the microstate. I think the text fumbles this point a bit and might be confusing you. Although all classical systems are in some way an approximation of a more fundamental quantum system, classical thermodynamics is still self consistent. Even if a system was truly classical, the $1/N!$ is still needed to give the correct counting.

 

[atyy]

I'm pretty sure I'm not. The argument in many/some textbooks is as follows: Distinguishable particles obeying Boltzmann give an entropy expression that is non-extensive and therefore wrong. If one includes a factor 1/N! to account for indistinguishability one arrives at the Sackur-Tetrode formula for entropy, which is extensive. Therefore particles obeying Boltzmann (such as atoms and molecules in a gas) have to be indistinguishable.
If this was only true in limiting cases it wouldn't be very helpful since the Gibbs paradox wouldn't be resolved in the general case.

The argument is ok as a heuristic argument, but it is wrong as an exact argument. The 1/N! for indistinguishability is made within the classical context within which the Boltzmann distribution is derived. However, classically, there are no indistinguishable particles in the quantum sense, because classical particles have trajectories, and are always distinguishable. Thus in classical physics, the 1/N! is a fudge factor. It comes from quantum mechanics, and the fact that reality is described by quantum physics, and the Boltzmann distribution is an extremely good approximation in some regime of the quantum description.

So indistinguishable particles must be quantum since quantum particles have no trajectories (in Copenhagen), whereas classical particles have trajectories. And the quantum distributions are Bose-Einstein or Fermi-Dirac.

See the the comment at the bottom of p86 of http://web.mit.edu/8.333/www/lectures/lec13.pdf.

 

[stevendaryl]

I'm pretty sure I'm not. The argument in many/some textbooks is as follows: Distinguishable particles obeying Boltzmann give an entropy expression that is non-extensive and therefore wrong. If one includes a factor 1/N! to account for indistinguishability one arrives at the Sackur-Tetrode formula for entropy, which is extensive. Therefore particles obeying Boltzmann (such as atoms and molecules in a gas) have to be indistinguishable.
If this was only true in limiting cases it wouldn't be very helpful since the Gibbs paradox wouldn't be resolved in the general case.

They are talking about an approximation here. Fixing the distribution by dividing by $N!$ is only approximately correct, in the low-density limit.

Let me work out an exact problem:

Suppose there are two possible states for a particle: $A, B$. Then the number of possible states of a two-particle system of distinguishable particles is 4:

1. Both in state A.
2. First particle in state A, second particle in state B.
3. First particle in state B, second particle in state A.
4. Both in state B.

So the number of states is 4 for distinguishable particles. If we divide by 2! to account for indistinguishability, we get: $4/(2!) = 2$. But the actual number of states for indistinguishable particles is 3:

1. Both in A.
2. One in A, one in B.
3. Both in B

So the heuristic of dividing by N! doesn't give the exact right answer.

 

[NFuller]

So the heuristic of dividing by N! doesn't give the exact right answer.

Strictly speaking this is true. The actual number of states valid for even small numbers of identical particles is
$$W=\frac{(N+g-1)!}{N!(g-1)!}$$
where $g$ is the number of states. For the case given above this means
$$W=\frac{(2+2-1)!}{2!(2-1)!}=3$$
In the high temperature limit where $g>>1$ and the low density limit where $g>>N$, the expression above can be simplified to give the standard Boltzmann counting
$$W=\frac{g^{N}}{N!}$$

 

[Stephen Tashi]

A micro-rant:

From the point of view of mathematical probability modeling, the language of statistical physics is a gosh-awful mess. Instead of clearly stated probability models ( e.g. clearly stated probability spaces) we have microstates, this-and-that kinds of ensembles, distinguishable identical particles, and identical particles that cannot be distinguished. We have "statistics" that do not satisfy the definition of "statistic" used in mathematical statistics and "Entropy" that seeks to be a property of matter instead of a property of a probability distribution.

A micro-reply to the micro-rant:

The origins of statistical physics preceded the modern development of probability theory and preceded the modern formulation of mathematical statistics. So it isn't surprising that presentations of statistical physics follow different traditions.

You can find my version of the mentioned derivations here:
https://www.researchgate.net/publication/322640913_A_microcanonical_derivation_gives_the_Boltzmann_for_distinguishable_and_the_Bose-Einstein_distribution_for_indistinguishable_particles

You say:

I will derive the most probable distribution of N particles among k energy levels from combinatorics and some classical thermodynamics.

We can survive the use of "distribution" to mean something different than a probability distribution. There is a probability distribution involved, but a particular "distribution of N particles among k energy levels" is an outcome of the probability distribution involved. For a probability distribution that assigns a probability for each possible "distribution" of N particles among k energy levels, you derive the most probable outcome of this probability distribution.

However, what physically is the interpretation of a particular outcome? After all, in a gas things are changing. So do we define an outcome with a reference to time? If we ignore time, do we think of realizing an outcome as picking a container of gas at random from a population of containers of gas sitting on a shelf , each in a static condition as far as the "distribution" of particles in energy levels goes?

I've read (somewhere, I can't remember) that Boltzman's original thinking did involve time - i.e that he thought of a point describing a specific container of gas moving around in phase space. In equilibrium, the probability of the gas have a given property was what fraction of the time the gas had that property within the confined set in phase space where its point moved around. So realizing an outcome physically amounted to picking a random time to observe the gas.

I also read that this concept of an outcome eventually caused problems - meaning problems within Boltzman's lifetime and before the advent of QM.

For a probability distribution whose outcomes give numbers of particles per energy level, what is the final (classical) verdict on the physical definition of such an outcome?

A microstate is a unique distribution of particles in phase space. Swapping the position and momentum of two identical particles will give the same configuration in phase space and the same microstate. If we didn't get the same microstate, that would imply that some microstates with many possible permutations are much more likely than others. The problem is that such a system cannot be at equilibrium. At equilibrium, the system must be in a maximum entropy configuration which occurs when each microstate comprising the equilibrium macrostate is equally likely.

The above quote alludes to a probability distribution whose outcome is a microstate - or some property of a microstate. So the question again arises: what physically is meant by realizing such an outcome? Observe the physical system at a random time? Pick a physical system at random from a set of physical systems where the property is not changing in time?

The passage deals with
1) The way a microstate is defined
2) The assertion that at equilbrium, the probability distribution whose outcomes are microstates is a maximum entropy distribution.

The definition of microstate can be justified by "the voice of authority" or tradition. I assume it can also be justified by arguments about practicality along the lines of "It wouldn't make physical sense to define a microstate so it depended on which particular particles are in an energy level because ..." It isn't clear (to me) which type of justification is being used for item 1)

It also isn't clear what justification is implied for item 2). It could be justified soley by empirical tests- or It might be justified by a Bayesian form of reasoning. If we take the Bayesian approach we face the usual criticism: You defined a particlar type of outcome (i.e. microstate) and as assumed a maximum entropy distribution for it, but why didn't you define a different type of outcome and assume a maximum entropy distribution for that type of outcome?

 

[NFuller]

The definition of microstate can be justified by "the voice of authority" or tradition. I assume it can also be justified by arguments about practicality along the lines of "It wouldn't make physical sense to define a microstate so it depended on which particular particles are in an energy level because ..." It isn't clear (to me) which type of justification is being used for item 1)

I agree with your rant that the definition of microstate, or whatever-state, is a bit sloppy. A microstate is generally interpreted to mean a specific configuration of the sub-units of the system. How to deal with those microstates, and what the physical meaning of those states is, depends on the type of ensemble used. In the micro-canonical ensemble, each microstate has an equal probability of being selected at equilibrium. In the canonical ensemble, the probability to select a microstate depends on the energy of the state and temperature of the system.

It also isn't clear what justification is implied for item 2). It could be justified soley by empirical tests- or It might be justified by a Bayesian form of reasoning. If we take the Bayesian approach we face the usual criticism: You defined a particlar type of outcome (i.e. microstate) and as assumed a maximum entropy distribution for it, but why didn't you define a different type of outcome and assume a maximum entropy distribution for that type of outcome?

The justification used is the second law of thermodynamics which states that when the system reaches equilibrium, that is a maximum entropy state. Since we want statistical mechanics to reproduce classical thermodynamics, we impose the requirement that the equilibrium probability distributions maximize the entropy.

 

[atyy]

Many textbooks claim that particles that obey Boltzmann statistics have to be indistinguishable in order to ensure an extensive expression for entropy. However, a first principle derivation using combinatorics gives the Boltzmann only for distinguishable and the Bose Einstein distribution for indistinguishable particles (see Beiser, Atkins or my own text on Research Gate). Is there any direct evidence that indistinguishable particles can obey Boltzmann statistics?

Reading Kardar's comments that I linked to in post #58, I think the answer is yes, classical indistinguishable particles can obey Boltzmann statistics. There is no derivation, simply a postulation. However, there seems to be no problem (in terms of consistency with thermodynamics and the other postulates of classical statistical mechanics) with postulating the 1/N! factor.

 

[Philip Koeck]

Strictly speaking this is true. The actual number of states valid for even small numbers of identical particles is
$$W=\frac{(N+g-1)!}{N!(g-1)!}$$
where $g$ is the number of states. For the case given above this means
$$W=\frac{(2+2-1)!}{2!(2-1)!}=3$$
In the high temperature limit where $g>>1$ and the low density limit where $g>>N$, the expression above can be simplified to give the standard Boltzmann counting
$$W=\frac{g^{N}}{N!}$$

Do you have a pdf or a link for the approximation you could send easily?
I can't quite get it right.

About $$W = \frac {g^N} {N!}$$ This is actually not the number of ways of distributing N distinguishable particles among g states. The correct expression for distinguishable particles is $$W = g^N$$
To me it seems that Bose-Einstein does not give Boltzmann for g>>N>>1, but something similar to Boltzmann, only for indistiguishable particles

 

[rubi]

One can also have indistinguishable particles in classical mechanics. Instead of using the configuration space $Q^N$ of distinguishable particles and its symplectic manifold $T^\ast Q^N$, one can use the configuration space $Q_N = (Q^N \setminus \Delta) / S_N$, where $\Delta$ is the set of coinciding points and $S_N$ is the group of permutations of $N$ elements. The phase space then becomes $T^\ast Q_N$. One can define Hamiltonians almost as usual, but one has to make sure that they obey permutation symmetry in order to be well defined on the equivalence classes: $H([p_1,\ldots, q_N]) = \sum_{i=1}^N \frac {p_i^2} {2m} + \sum_{i<j}^N V(\left|q_i - q_j\right|)$. The corresponding Liouville measure and the entropy will automatically get the correct $\frac 1 {N!}$ factors and the statistics is the usual Boltzmann statistics. So if one starts with a configuration space of indistinguishable particles, the Gibbs paradox is resolved naturally and no factors need to be smuggled in.

 

[NFuller]

Do you have a pdf or a link for the approximation you could send easily?

This wiki page actually goes through much of the derivation. https://en.wikipedia.org/wiki/Maxwell–Boltzmann_statistics

 

[Stephen Tashi]

Ieach microstate has an equal probability of being selected at equilibrium.

What is the physical interpretation of "being selected"? Are we talking about picking a random time and taking the microstate of the system at that time to be the one that is selected?

The justification used is the second law of thermodynamics which states that when the system reaches equilibrium, that is a maximum entropy state. Since we want statistical mechanics to reproduce classical thermodynamics, we impose the requirement that the equilibrium probability distributions maximize the entropy.

I can understand that as a purely empirical claim. What I don't understand is an attempt to justify the definition of microstate by deductive logic - if that's what's being attempted.

If we are discussing Shannon entropy then when a system is at equilibrium some probability distributions of its properties may be maximum entropy distributions and others may not. How do we explain why assuming a maximum entropy for microstates (as defined by occupancy numbers) is a good idea - as opposed to assuming a maximum entropy distribution for selecting the type of state whose description includes which particular particles occupy various energy levels.

 

[Stephen Tashi]

This wiki page actually goes through much of the derivation. https://en.wikipedia.org/wiki/Maxwell–Boltzmann_statistics

From that article we have:

Thus when we count the number of possible states of the system, we must count each and every microstate, and not just the possible sets of occupation numbers.

Which implies the author of the article is willing (at that stage of exposition) to define a microstate as set of information that includes more than just the occupancy numbers.

In post #24, in reply to

You seem to be saying that swapping two particles in different states does not lead to a different microsctate even if it's obvious that the particles have been swapped. My understanding was that swapping distinguishable particles in different states leads to a new micrstate even if the particles are identical.
You say in post #24

You wrote:

A microstate is a unique distribution of particles in phase space. Swapping the position and momentum of two identical particles will give the same configuration in phase space and the same microstate.

What is the consensus definition of a "microstate"?

 

[Philip Koeck]

This wiki page actually goes through much of the derivation. https://en.wikipedia.org/wiki/Maxwell–Boltzmann_statistics

I've tried to fill in the gaps of this derivation (see appended pdf) and I don't get the same result. Am I making a mistake?

 

[NFuller]

What is the physical interpretation of "being selected"? Are we talking about picking a random time and taking the microstate of the system at that time to be the one that is selected?

Yes

I can understand that as a purely empirical claim. What I don't understand is an attempt to justify the definition of microstate by deductive logic - if that's what's being attempted.

If we are discussing Shannon entropy then when a system is at equilibrium some probability distributions of its properties may be maximum entropy distributions and others may not. How do we explain why assuming a maximum entropy for microstates (as defined by occupancy numbers) is a good idea - as opposed to assuming a maximum entropy distribution for selecting the type of state whose description includes which particular particles occupy various energy levels.

I'm sorry but I don't think I understand what you are asking. Can you rephrase this?

What is the consensus definition of a "microstate"?

I think the closest I can give to a "consensus definition" is the one given in Kardar's statistical physics book. He says

At any time $t$, the microstate of a system of $N$ particles is described by specifying the positions $\vec{q}(t)$ and momenta $\vec{p}(t)$ of all of the particles. The microstate thus corresponds to a point $\mu(t)$, in the $6N$-dimensional phase space $\Gamma=\Pi_{i=1}^{N}\{\vec{q}_{i},\vec{p}_{i}\}$