A Can indistinguishable particles obey Boltzmann statistics

[Philip Koeck]

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?

 

[mfb]

The Boltzmann statistics is the high-temperature (or low-density) limit of both the Bose-Einstein and the Fermi-Dirac statistics.
It is an approximation only, but a very good one in many cases.

 

[Philip Koeck]

I forgot to say that I don't want to consider limiting cases such as high temperature. The combinatorics derivation I mention gives Boltzmann for distinguishable and Bose-Einstein for indistinguishable particles at any temperature. On the other hand, according to books, even particles obeying Boltzmann have to be indistinguishable in order to resolve the Gibbs paradox, independent of the temperature. There seems to be a contradiction, I would say. So my question is: Does anything else point to the possibility of indistinguishable particles obeying Boltzmann at any temperature and density?

 

[Philip Koeck]

I have to add: For the model system treated in the mentioned derivation the Boltzmann distribution is an exact result, not an approximation. I don't know, however, whether there is a real system that is described sufficiently well by this model.

 

[mfb]

Indistinguishable particles cannot follow Boltzmann exactly at finite temperature. So what?
Distinguishable particles are like many individual distributions with single particles summed, for these particles all three distributions are the same.

 

[Philip Koeck]

Indistinguishable particles cannot follow Boltzmann exactly at finite temperature. So what?

According to many textbooks they can and do. Thus my question.

 

[Philip Koeck]

Distinguishable particles are like many individual distributions with single particles summed, for these particles all three distributions are the same.

I don't understand what you are saying here. Could you rephrase?

 

[mfb]

According to many textbooks they can and do. Thus my question.

I'm sure you are missing the part where they call it an approximation.

I don't understand what you are saying here. Could you rephrase?

I don't understand what is unclear.

 

[Philip Koeck]

I'm sure you are missing the part where they call it an approximation.

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.

 

[Philip Koeck]

I don't understand what is unclear.

The following sentence is unclear: "Distinguishable particles are like many individual distributions with single particles summed,..."

 

[ZapperZ]

The Drude model for conduction electrons in a conductor is based on Boltzmann statistics. Isn't this a clear example of what you are looking for?

Zz.

 

[Philip Koeck]

The Drude model for conduction electrons in a conductor is based on Boltzmann statistics. Isn't this a clear example of what you are looking for?Zz.

Good point! However, in 1900 Drude had no idea that electrons were actually fermions and should obey Fermi-Dirac statistics. He simply described them as an ideal gas because that's the only thing he could do at the time. I'm not sure if that qualifies as evidence that indistinguishable particles can obey Boltzmann.

 

[ZapperZ]

Good point! However, in 1900 Drude had no idea that electrons were actually fermions and should obey Fermi-Dirac statistics. He simply described them as an ideal gas because that's the only thing he could do at the time. I'm not sure if that qualifies as evidence that indistinguishable particles can obey Boltzmann.

But what does history have anything to do with it? The Drude model is STILL being used, and in fact, it is the foundation on how we got Ohm's law! Many of the basic properties of conductors are based on such a model, i.e. single-particle, non-interacting model.

In fact, in electron particle accelerators, we typically model charge particles using such classical statistics as well! Particle beam modeling packages do not use quantum statistics to get the beam dynamics.

Zz.

 

[NFuller]

According to many textbooks they can and do. Thus my question.

I haven't heard of this. Could you provide a reference?

 

[Philip Koeck]

I haven't heard of this. Could you provide a reference?

For example Blundell and Blundell: Concepts in thermal physics, 2nd edition, chapter 21, sections 21.3 to 21.5 and exercise 21.2.

 

[Philip Koeck]

But what does history have anything to do with it? The Drude model is STILL being used, and in fact, it is the foundation on how we got Ohm's law! Many of the basic properties of conductors are based on such a model, i.e. single-particle, non-interacting model.

In fact, in electron particle accelerators, we typically model charge particles using such classical statistics as well! Particle beam modeling packages do not use quantum statistics to get the beam dynamics.

Zz.

You have a very good point there. One could say that particles that are actually indistinguishable are described in a classical way (like distinguishable atoms in an ideal gas), and the results match experimental findings. That might be one answer to my question. Thanks.

 

[NFuller]

For example Blundell and Blundell: Concepts in thermal physics, 2nd edition, chapter 21, sections 21.3 to 21.5 and exercise 21.2.

Reading this, I think I understand the confusion here. This book uses the terms distinguishable and indistinguishable more loosely than I assumed. It appears to assume distinguishable means something like, the particles have a different color or shape, and assumes indistinguishable means they are identical. The issue I have with this wording is that identical is not the same thing as indistinguishable. If I carefully watch an ensemble of classical identical particles, I can keep track of were each particle moved and what it is doing. Therefore, they are distinguishable. For quantum particles, I can not keep track of each particle as they interact with each other. Quantum particles are truly indistinguishable.

When discussing Boltzmann statistics of a pure gas, the particles are assumed identical but distinguishable. It is really the fact that they are identical which leads to the need to prevent the over counting of states.

 

[Stephen Tashi]

On the other hand, according to books, even particles obeying Boltzmann have to be indistinguishable in order to resolve the Gibbs paradox, independent of the temperature.

One could say that particles that are actually indistinguishable are described in a classical way (like distinguishable atoms in an ideal gas), and the results match experimental findings. That might be one answer to my question.

In another thread, @Andy Resnick mentioned a paper by Jaynes (e.g. http://www.damtp.cam.ac.uk/user/tong/statphys/jaynes.pdf ). Jaynes says that (Gibbs said that) whether to distinguish or not distinguish microstates is a choice made by the experimenter.

 

[Vanadium 50]

The Boltzmann statistics is the high-temperature (or low-density) limit of both the Bose-Einstein and the Fermi-Dirac statistics.

I forgot to say that I don't want to consider limiting cases such as high temperature.

So I'm confused then. As mfb points out, Boltzmann is always an approximation/limiting case.

 

[Philip Koeck]

Reading this, I think I understand the confusion here. This book uses the terms distinguishable and indistinguishable more loosely than I assumed. It appears to assume distinguishable means something like, the particles have a different color or shape, and assumes indistinguishable means they are identical. The issue I have with this wording is that identical is not the same thing as indistinguishable. If I carefully watch an ensemble of classical identical particles, I can keep track of were each particle moved and what it is doing. Therefore, they are distinguishable. For quantum particles, I can not keep track of each particle as they interact with each other. Quantum particles are truly indistinguishable.

When discussing Boltzmann statistics of a pure gas, the particles are assumed identical but distinguishable. It is really the fact that they are identical which leads to the need to prevent the over counting of states.

I agree that identical particles can be distinguishable (in the sense of trackable). I actually thought Blundell saw it that way too.
About your last sentence:
I don't understand why the fact that particles are identical but distinguishable would necessitate a factor 1/N! in the partition function to avoid overcounting.
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.

 

[Philip Koeck]

So I'm confused then. As mfb points out, Boltzmann is always an approximation/limiting case.

That's an interesting statement. The derivations I referred to earlier give Boltzmann for distinguishable particles at any temperature, not just for very high temperatures.

 

[bhobba]

I am totally confused as well.

In QM you exchange particles and exchange them back and you get the same wave-function. This means under exchange the wave-function must change by +1 or - 1. That's the elementary argument anyway - something in the back of my mind is it has a flaw - but its still an experimental fact.

If it doesn't change its called a boson.

The rest is math rather than physics - its just a probability modelling exercise - you will find it in for example Ross - Introduction To Probability Models.
https://www.amazon.com/dp/0123756863/?tag=pfamazon01-20

That's all there is to it really - of course I may be missing something.

Thanks
Bill

 

[Philip Koeck]

I am totally confused as well.

In QM you exchange particles and exchange them back and you get the same wave-function. This means under exchange the wave-function must change by +1 or - 1. That's the elementary argument anyway - something in the back of my mind is it has a flaw - but its still an experimental fact.

If it doesn't change its called a boson.

The rest is math rather than physics - its just a probability modelling exercise - you will find it in for example Ross - Introduction To Probability Models.
https://www.amazon.com/dp/0123756863/?tag=pfamazon01-20

That's all there is to it really - of course I may be missing something.

Thanks
Bill

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?

 

[NFuller]

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.

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.

 

[Stephen Tashi]

I agree that identical particles can be distinguishable (in the sense of trackable). I actually thought Blundell saw it that way too.

It would be helpful to know if the terms "indistinguishable" and "identical" have technical definitions in thermodynamics. For example, from the point of view of common speech it is paradoxical to refer to "two indistinguishable cups" or "two identical cups" if this terminology is taken to imply a set with cardinality 2. A set of two "indistinguishable things" or "two identical things" is not a set of two things. It is a set of 1 thing. The "two" things are the same thing.

Likewise, in mathematics , if I say "Let S be the set {13,x} whose members are two identical real numbers", this would usually be interpreted to mean that I have defined a set with cardinality 1.

So it seems to me that the adjectives "indistinguishable" or "identical" as used in physics must have some qualification like "indistinguishable with respect to ..." or "identical with respect to..." and there should be list of properties the adjectives apply to.

Of course, I'm thinking in terms of classical physics. Perhaps someone can explain whether the concept of "N particles"in the setting of QM differs from the ordinary concept of cardinality in mathematics. Perhaps there is some concept like "You can know a system is in a state with the property "There are 13 particles", but , you can't perform any process that will, in a manner of speaking, lay them all out on a table in a distinguishable way so you can count them.

 

[ZapperZ]

It would be helpful to know if the terms "indistinguishable" and "identical" have technical definitions in thermodynamics. For example, from the point of view of common speech it is paradoxical to refer to "two indistinguishable cups" or "two identical cups" if this terminology is taken to imply a set with cardinality 2. A set of two "indistinguishable things" or "two identical things" is not a set of two things. It is a set of 1 thing. The "two" things are the same thing.

Likewise, in mathematics , if I say "Let S be the set {13,x} whose members are two identical real numbers", this would usually be interpreted to mean that I have defined a set with cardinality 1.

So it seems to me that the adjectives "indistinguishable" or "identical" as used in physics must have some qualification like "indistinguishable with respect to ..." or "identical with respect to..." and there should be list of properties the adjectives apply to.

Of course, I'm thinking in terms of classical physics. Perhaps someone can explain whether the concept of "N particles"in the setting of QM differs from the ordinary concept of cardinality in mathematics. Perhaps there is some concept like "You can know a system is in a state with the property "There are 13 particles", but , you can't perform any process that will, in a manner of speaking, lay them all out on a table in a distinguishable way so you can count them.

Two electrons are identical to each other. They have exactly the same properties and characteristics.

But these two electrons become INDISTINGUISHABLE if they are so close to one another that their wavefuctions significantly overlap, so much so that you can no longer distinguish which is which.

Is that clear enough?

Zz.

 

[Vanadium 50]

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?

That is correct.

shouldn't a classical description become appropriate at some point?

That point is when the error induced by this approximation becomes small when compared to the precision you care about.

In short "Boltzman" and "no approximations" together seem contradictory.

 

[Philip Koeck]

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.

By definition, or because we don't really know, all microstates are equally likely. The macrostate with the largest number of microstates has the highest entropy and is the equilibrium state. A system of distinguishable particles can be in equlibrium, just like a system of indistinguishable particles can.

 

[NFuller]

By definition, or because we don't really know, all microstates are equally likely. The macrostate with the largest number of microstates has the highest entropy and is the equilibrium state.

It is a consequence of the Gibbs entropy formula. For a system with $\Omega$ accessible microstates the entropy is
$$S=-k\sum_{i=1}^{\Omega}p_{i}\ln p_{i}$$
The second law states that the equilibrium state is the maximum entropy state. The probability distribution which maximizes $S$ is the uniform distrabution $p_{i}=1/\Omega$. Plugging this in leads to the Boltzmann expression for the entropy:
$$S=-k\sum_{i=1}^{\Omega}\frac{1}{\Omega}\ln \frac{1}{\Omega}=k\ln \Omega$$
So Boltzmann statistics inherently has the assertion that all microstates are equally likely built in.

 

[Philip Koeck]

It is a consequence of the Gibbs entropy formula. For a system with $\Omega$ accessible microstates the entropy is
$$S=-k\sum_{i=1}^{\Omega}p_{i}\ln p_{i}$$
The second law states that the equilibrium state is the maximum entropy state. The probability distribution which maximizes $S$ is the uniform distrabution $p_{i}=1/\Omega$. Plugging this in leads to the Boltzmann expression for the entropy:
$$S=-k\sum_{i=1}^{\Omega}\frac{1}{\Omega}\ln \frac{1}{\Omega}=k\ln \Omega$$
So Boltzmann statistics inherently has the assertion that all microstates are equally likely built in.

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.

 

[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