# FeaturedA Can indistinguishable particles obey Boltzmann statistics

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1. Feb 7, 2018

### Philip Koeck

Many text books 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?

2. Feb 7, 2018

### Staff: Mentor

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.

3. Feb 7, 2018

### 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?

4. Feb 7, 2018

### 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.

5. Feb 7, 2018

### Staff: Mentor

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.

6. Feb 7, 2018

### Philip Koeck

According to many text books they can and do. Thus my question.

7. Feb 7, 2018

### Philip Koeck

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

8. Feb 7, 2018

### Staff: Mentor

I'm sure you are missing the part where they call it an approximation.
I don't understand what is unclear.

9. Feb 7, 2018

### Philip Koeck

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.

10. Feb 7, 2018

### Philip Koeck

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

11. Feb 7, 2018

### ZapperZ

Staff Emeritus
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.

12. Feb 7, 2018

### Philip Koeck

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.

13. Feb 7, 2018

### ZapperZ

Staff Emeritus
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.

14. Feb 7, 2018

### NFuller

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

15. Feb 7, 2018

### Philip Koeck

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

16. Feb 7, 2018

### Philip Koeck

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.

17. Feb 7, 2018

### NFuller

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.

18. Feb 7, 2018

### Stephen Tashi

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.

19. Feb 7, 2018

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

20. Feb 8, 2018

### Philip Koeck

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

21. Feb 8, 2018

### Philip Koeck

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.

22. Feb 8, 2018

### Staff: Mentor

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/Introduction-Probability-Models-Tenth-Sheldon/dp/0123756863

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

Thanks
Bill

23. Feb 8, 2018

### Philip Koeck

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?

24. Feb 8, 2018

### NFuller

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.

25. Feb 8, 2018

### Stephen Tashi

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.