Boltzmann particle distinguishability

[rabbed]

What justifies the use of multinomial coefficient in the combinatorics used by Boltzmann?
The particles are distinct but counts as identical when they are in the same energy state?

 

[DrClaude]

Classically, the particles have to be considered simply because otherwise it leads to incorrect results, see the Gibbs paradox.

From quantum mechanics, we know that fundamentally, identical particles are indistinguishable, and this property remains when we take the classical limit.

 

[rabbed]

Without prior knowledge of quantum mechanics, would my wording above be the correct assumption?

 

[DrClaude]

No, it has nothing to do with the energy state the particles are in. Actually, in the classical limit, the occupancy of a single state must be << 1.

 

[rabbed]

So the multinomial coefficient is justified because experimental results agree with the mathematical results following from using it?
But because its multinomial, the particles must be considered identical (otherwise we would only use a factorial for all permutations), so why are they considered distinct?

 

[DrClaude]

So the multinomial coefficient is justified because experimental results agree with the mathematical results following from using it?
But because its multinomial, the particles must be considered identical (otherwise we would only use a factorial for all permutations), so why are they considered distinct?

Now I am confused about what case you are considering. Can you give some equations please?

 

[vanhees71]

Well, classical statistical physics is much more complicated than quantum statistical physics, the reason being that there's no consistent classical theory of matter ;-)). You are not even able to define the phase-space measure from classical theory alone. Boltzmann was very aware of all this trouble, but had no quantum theory but the right physics intuition to still get the theory correct.

 

[rabbed]

Now I am confused about what case you are considering. Can you give some equations please?

I see my error now, we can arrive at a multinomial coefficient expression both for sets of purely distinct objects and for sets of objects having duplicates (identical objects). There will be a conceptual difference in what the denominators stand for in the two expressions.

Thanks for making me check :)

 

[rabbed]

Number (11) and (12) here is what my thread is about:
https://staff.imsa.edu/~fogel/APS/PDF/List%20of%20Standard%20Counting%20Problems.pdf

I agree with (12) after listing all the possible permutations of [1, 2 and 2] and count all distinct cases/microstates. First treating them as the distinct symbols 1A, 2A and 2B to get 3!, and then divide by the number of permutations for 1 indistinct one's and 2 indistinct two's to get W = 3!/(1!*2!).

I am not able to list all possible permutations to see the cases of (11), using objects [1A, 2A and 2B] by for example first filling box1 with 1 symbol, then filling box 2 with 2 symbols. How is this done in W = 3! / (1!*2!) = 3 ways?

Could you say that the number of ways are identical in this case?

 

[DrClaude]

I am not able to list all possible permutations to see the cases of (11), by for example first filling box1 with 1 symbol 1A, then filling box 2 with 2 symbols, 2A and 2B. How is this done in W = 3! / (1!*2!) = 3 ways?

Start by listing all possible permutations:
1 2A 2B
1 2B 2A
2A 1 2B
2A 2B 1
2B 1 2A
2B 2A 1
Total number 3! = 6. Now you say that box 1 contains one particle and box 2, two. So you put the first particle on the list in the first box, and the two others in the second box. Some combinations repeat, so you need to consider permutations within a box to be identical configurations, so you divide by 1! for the suffering of box 1 and 2! for the shuffling of box 2. Therefore, W = 3! / (1!*2!) = 3.

I am having difficulty understanding the link you are making with statistical physics, because the above is true for distinguishable particles, which is almost never the case in stat. phys.

 

[rabbed]

Thanks
But if we draw a line after each first symbol/particle in each permutation to signify that what's to the left is in box1 and what's to the right is in box 2, do we not then have 6 ways of filling box1 with 1 symbol AND box 2 with 2 symbols?

The link, isn't this the combinatorical case that Boltzmann used?

 

[rabbed]

So it's some permutations per box that are identical.. But we still have distinct objects and distinct boxes.
Tricky to get intuitively..

So this is one way:
1 2A 2B
1 2B 2A

This is another:
2A 1 2B
2A 2B 1

And this is the third?
2B 1 2A
2B 2A 1

 

[DrClaude]

That's correct.

 

[rabbed]

I am having difficulty understanding the link you are making with statistical physics, because the above is true for distinguishable particles, which is almost never the case in stat. phys.

Didn't Boltzmann get experimental results that agreed with mathematical results following from this combinatorical case?

 

[DrClaude]

Didn't Boltzmann get experimental results that agreed with mathematical results following from this combinatorical case?

I don't know about this.

 

[rabbed]

Can you say that considering distinct objects into distinct boxes leads to correct Boltzmann distribution but incorrect thermodynamical entropy (not extensive)? Although correct information theoretical entropy.
And considering identical objects into distinct boxes leads to correct Boltzmann distribution and correct thermodynamical entropy?

 

[DrClaude]

Can you say that considering distinct objects into distinct boxes leads to correct Boltzmann distribution but incorrect thermodynamical entropy (not extensive)? Although correct information theoretical entropy.
And considering identical objects into distinct boxes leads to correct Boltzmann distribution and correct thermodynamical entropy?

I don't really understand what you are saying here.

Remember that statistical physics is not about objects in boxes. These can be useful analogies to help understand concepts, but in the end we are dealing with things like gas particles or a crystal of paramagnets.

 

[rabbed]

I mean, using the analogy of distinct objects into distinct boxes (with specified energy level/box occupancies) and applying the constraint sum(p) = 1 will give you a valid information theoretical entropy. Also applying the constraint that sum(energy of level) = E (where E is the total energy) will not give you a valid thermodynamical entropy (it's not extensive) because of the analogy used. But it will lead to the (correct) Boltzmann distribution?

 

[rabbed]

One thought that popped up, using the analogy of distinct->distinct we're calculating the number of ways to realize a particular state of occupations, in other words - the unnormalized probability of the state, W.
In order to get the probability of that state, we would need the total number of ways, T, of realizing all possible states of occupancies, so that p = W/T?
How do i get T?
If the number of particles/objects = 4 and the number of energy levels/boxes = 3, is it the number of different ways to combine the occupancy numbers?
4!/(0!*0!*4), 4!/(0!*1!*3), 4!/(0!*2!*2!), etc..
Do you understand what I mean?

 

[DrClaude]

The probability is always the Boltzmann factor divided by the partition function. What will change in the case of distinguishable vs indistinguishable particle is the partition function, as it involves a sum over all states. These states are not the same in both cases.

 

[vanhees71]

You find my 2 cents about the approach to Bose-Eintein and Fermi-Dirac statistics by the simple counting rules (as already Planck did it for Bose-Einstein statistics of course without knowing about it when doing the statistics for em-field modes when "deriving" his before found empirical Planck-spectrum formula for black-body radiation) are here (adopted from Landau&Lifshitz vol. V by the way):

https://th.physik.uni-frankfurt.de/~hees/publ/kolkata.pdf

 

[rabbed]

Start by listing all possible permutations:
1 2A 2B
1 2B 2A
2A 1 2B
2A 2B 1
2B 1 2A
2B 2A 1
Total number 3! = 6. Now you say that box 1 contains one particle and box 2, two. So you put the first particle on the list in the first box, and the two others in the second box. Some combinations repeat, so you need to consider permutations within a box to be identical configurations, so you divide by 1! for the suffering of box 1 and 2! for the shuffling of box 2. Therefore, W = 3! / (1!*2!) = 3.

Still having some issues with this.
Could you say something more about the reasoning why (1 | 2A 2B) and (1 | 2B 2A) should count as just one "number of ways" instead of two?
Is there any way to practically get the number of ways? Would it help to make some trials manually and create a PMF?

 

[rabbed]

Nevermind..
It helped seeing indexing (treat identical as distinct) and faculty division (of the number of objects considered identical) as a means of adding and removing order into account of the number of ways.
We can distribute N distinct balls into one box in 1 way (if we don’t care about their order in the box) or in N! ways (if we care about their order in the box).

 

[vanhees71]

It's pretty useless to think about these questions in classical terms. It was Boltzmann's ingenious insight into physics making him write the famous $1/N!$ factor to get rid of Gibbs's paradoxon. Nowadays if follows from the indistinguishability of particles in quantum theory. Classical statistical physics is most convincingly derived as an approximation of quantum statistics for small occupation numbers (i.e., less than 1 particle occupation number for each one-particle state on average).