Graduate Can indistinguishable particles obey Boltzmann statistics

[NFuller]

Are you saying that the only correct distributions are Bose-Einstein and Fermi-Dirac (since all particles are either Bosons or Fermions), and Boltzmann can only be an approximation? I think there are many authors who believe that there are "classical" systems that are correctly described even at low temperatures by Boltzmann statistics. You could think of colloids and aerosols, but there's also the gray-zone of gases of atoms, I would say. If you think of He, Ne, Ar etc., shouldn't a classical description become appropriate at some point?

The assumption is that at sufficiently high energies and low densities, the region of phase space excluded by the Pauli exclusion principle is a very small fraction of the total phase space, so we can safely ignore it. Under this assumption the over counting is just a permutation of the particles $N!$. In the case where this assumption is not valid, we also need to start excluding regions of the phase space. If there are $G$ levels then I believe the over counting will be by a factor of $N!(G-N)!$.

 

[NFuller]

What you've shown is that Boltzmann's definition of entropy follows from Gibb's if all microstates are equally likely, but I don't see what that says about the Boltzmann distribution.

When the position of two particles in phase space are swapped, the total energy of the system is the same. So we should be looking at these particular groups of microstates in the Boltzmann picture (the microcanonical ensemble).

For microstates which are not the same energy, the probabilities are not necessarily the same but this is not related to the question at hand. The $N!$ term is specifically dealing with particle arrangements of the same energy.

 

[Philip Koeck]

When the position of two particles in phase space are swapped, the total energy of the system is the same. So we should be looking at these particular groups of microstates in the Boltzmann picture (the microcanonical ensemble).

Are you saying that the Boltzmann distribution is the only one possible in the microcanonical ensemble? The derivations I mentioned all assume constant energy and give all three distributions depending on the properties of the system.

 

[NFuller]

Are you saying that the Boltzmann distribution is the only one possible in the microcanonical ensemble?

No, but if you are looking at a constant energy system (assuming $N$ and $V$ are also fixed) then the microcanonical ensemble is needed.

The derivations I mentioned all assume constant energy and give all three distributions depending on the properties of the system.

Are you taking about non-classical distributions here? I thought the question was about where the $N!$ came from in the classical case.

 

[Philip Koeck]

Are you taking about non-classical distributions here? I thought the question was about where the $N!$ came from in the classical case.

That's right. The original question was: 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. On the other hand the mentioned derivations give Boltzmann only for distinguishable and Bose-Einstein only for indistinguishable particles. So I'm wondering how this is possible and whether there is any other evidence that indistinguishable particles can obey Boltzmann.
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.
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

 

[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
$$W=\frac{g^{N}}{N!}$$