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 "Boltzmann" 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.