# I Differing first-principle models for Maxwell-Boltzmann statistics?

#### greswd

Let's consider a simple scenario in Maxwell-Boltzmann statistics: 6 identical but distinguishable particles, and 9 quanta of energy, 9 indivisible units, to be distributed among the particles.

The first model is like that of the wheels on a combination lock, or should I say "permutation lock".

There are 6 wheels, one for each particle, and the numbers run from 0 to 9, since there are 9 units of energy available.

The number on each wheel represents the amount of energy each particle possesses. Since there are 6 wheels, there are a total of a million permutations.

However, there are only 9 units of energy in total, so if the 6 numbers do not add up to 9 in total, it is an invalid permutation.

Considering only the valid permutations, the odds of a particle being in the ground state, having zero units of energy, is the highest of all, and it only decreases subsequently for each incremental energy level.

Now for the second model, 9 dice.

One die for each unit of energy, and of course 6 sides, one for each particle.

You can imagine rolling all 9 dice, with the number on each die indicating which particle the unit of energy the die represents belongs to.

This time, there's no need to remove invalid permutations, all are valid.

This model produces quite different results from the first one, as the odds of a particle having one unit of energy is the highest, and the odds of having two units of energy is even higher than the odds of having no energy at all.

Maxwell-Boltzmann statistics follows the first model, and I'm curious as to why nature is of the first model rather than of the second.
Because intuitively, it seems to me like the second model would be more likely to occur. It's simpler in a way, no assumptions need to be made about invalid permutations.

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#### BvU

Homework Helper
You can imagine rolling all 9 dice, with the number on each die indicating which particle the unit of energy the die represents belongs to.
So state 1 1 1 1 1 1 1 1 2 occurs 9 times (any of the other dice can be 2 instead of 1) ?

#### greswd

So state 1 1 1 1 1 1 1 1 2 occurs 9 times (any of the other dice can be 2 instead of 1) ?
yes, so maybe that state should have a statistical weight of 9?

#### BvU

Homework Helper
What would it get in the permutation lock scenario ?

#### greswd

What would it get in the permutation lock scenario ?
It is, as you've mentioned, a state, so I think it's weight is 1, out of 2002.

While for the dice it is 9 out of 10077696.

#### Dale

Mentor
You can imagine rolling all 9 dice, with the number on each die indicating which particle the unit of energy the die represents belongs to.

This time, there's no need to remove invalid permutations, all are valid.
This is not exactly correct. Although all permutations are valid, they are not all distinguishable. For example {1,1,1,1,1,1,1,2,2} refers to the same state as {2,2,1,1,1,1,1,1,1} and {2,1,1,1,1,1,1,1,2}. The particles are distinguishable, but the energy levels are not. If you go through and eliminate all of the invalid permutations of locks and all of the indistinguishable permutations of dice then you are left with 2002 sets in both cases.

The equipartition theorem basically states that all of these 2002 distinguishable/valid sets are equally probable. You could make an arbitrary number of models describing how many indistinguishable sets correspond to each distinguishable set, but that doesn't affect the physics. So physically both scenarios are the same since they both wind up with the same number of sets. In one case you have to remove invalid cases and in the other you have to remove indistinguishable cases.

It is, as you've mentioned, a state, so I think it's weight is 1, out of 2002.

While for the dice it is 9 out of 10077696.
In both cases it is 1 out of 2002.

#### BvU

Homework Helper
Conversely, all 9 energy bits for the first particle gets one out of 10077696 with the dice and 1 out of 2002 with the lock

The energy bits are indistinguishable, so you'd have to divide each of the 10077696 combinations by the number of ways such a combination can occur to get its weight.

(ah, Dale to the rescue while I waste my time trying to get an elegant expression for this dice combination weight )

#### greswd

The equipartition theorem basically states that all of these 2002 distinguishable/valid sets are equally probable. You could make an arbitrary number of models describing how many indistinguishable sets correspond to each distinguishable set, but that doesn't affect the physics.
thanks, so, what's the root, fundamental basis or effect that leads to equipartition probabilities and nature not following the dice distribution?

Is it like some fundamental rule/effect which describes distinguishability affecting probabilities?

Because it seems like the combination lock method isn't a direct, sole logical given, and that there is some fundamental physical principle or effect which makes it so.

#### BvU

Homework Helper
You can't expect nine dice to mimic nine units of energy ...

#### greswd

You can't expect nine dice to mimic nine units of energy ...
But six wheels of a lock demonstrate the mathematics of the phenomenon very well.

I think a more direct, clear and physical visual would be a beer-pong set-up.

Imagine 6 cups representing the 6 particles, and 9 ping-pong balls representing the units of energy, which are hurled at the cups.

Assuming the balls are ghostly and pass through each other instead of bouncing off each other, it will produce the same results as that of the dice.

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#### Dale

Mentor
thanks, so, what's the root, fundamental basis or effect that leads to equipartition probabilities and nature not following the dice distribution?
I think that it is called the identity of indiscernibles. Two states that are completely indistinguishable are the same state. I believe that it is simply an assumption that seems to work. Apparently it has a very powerful usage in quantum mechanics where it is the cornerstone of the statistical distributions used to represent particles.

The reason that the combination lock works and the dice does not is that you specified that the particles were distinguishable and the energy levels were not. The combination lock respects that distinction and the dice do not. If you had indistinguishable particles then the combination lock metaphor wouldn't work either.

#### greswd

I think that it is called the identity of indiscernibles. Two states that are completely indistinguishable are the same state.

I believe that it is simply an assumption that seems to work.
wow, that's pretty interesting. I think most would imagine a "beer-pong distribution" as the simplest method of probabilistically distributing energy among particles. but nature appears to have other ideas.

#### Dale

Mentor
I think most would imagine a "beer-pong distribution" as the simplest method of probabilistically distributing energy among particles. but nature appears to have other ideas.
As long as the balls are indistinguishable and the cups are distinguishable it will work fine.

#### greswd

As long as the balls are indistinguishable and the cups are distinguishable it will work fine.
Considering actual, macroscopic ping-pong balls, I'm not sure how indistinguishability would apply.

Whether all 9 balls are white, or each painted with a different color, it would have no effect on the overall distribution. A layer of paint can't change anything.

#### greswd

The energy bits are indistinguishable, so you'd have to divide each of the 10077696 combinations by the number of ways such a combination can occur to get its weight.
As long as the balls are indistinguishable and the cups are distinguishable it will work fine.
Considering actual, macroscopic ping-pong balls, I'm not sure how indistinguishability would apply.

Whether all 9 balls are white, or each painted with a different color, it would have no effect on the overall distribution. A layer of paint can't change anything.
Or, a reverse scenario, if each ping-pong ball is originally of a unique color, we could make them "indistinguishable" by painting all of them white.

And that layer of white paint also can't change anything, the kinematics will remain the same.

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#### BvU

Homework Helper
Does someone have a decent expression for the weight of a 9 dice throw ?

#### greswd

Does someone have a decent expression for the weight of a 9 dice throw ?
well, there are 10077696 possible permutations in total.

#### BvU

Homework Helper
I mean: What's the weight if I throw 1 2 3 4 5 6 6 6 6

#### greswd

I mean: What's the weight if I throw 1 2 3 4 5 6 6 6 6
it should be 15120

#### Dale

Mentor
Considering actual, macroscopic ping-pong balls, I'm not sure how indistinguishability would apply.
It doesn’t. But IF it did then the beer pong distribution would work. The observed fact that it doesn’t work is due to the violation of the indistinguishability assumption. Clearly, a layer of paint is not what makes a physical system indistinguishable.

#### BvU

Homework Helper
it should be 15120
So do you have an expression for that ? The number isn't interesting.

#### greswd

It doesn’t. But IF it did then the beer pong distribution would work. The observed fact that it doesn’t work is due to the violation of the indistinguishability assumption. Clearly, a layer of paint is not what makes a physical system indistinguishable.
what do you think makes a physical system indistinguishable?

each ball has tiny imperfections and flaws that distinguish it from the others.

however, even if they are somehow 100% perfect copies of each other, it wouldn't affect the kinematics and wouldn't produce different results.
Their trajectories through the air would not change much.

So energy quanta probably have some other properties in addition to their indistinguishability which make them behave differently and follow the lock model.

#### Dale

Mentor
So energy quanta probably have some other properties
I think it is the opposite. They have fewer properties, not additional ones. That is what makes them indistinguishable. Macroscopic systems have too many properties for the identity of indiscernible to play a role.

#### greswd

So do you have an expression for that ? The number isn't interesting.
$\operatorname{nCr}\left(9,5\right)\cdot5!$

#### greswd

I think it is the opposite. They have fewer properties, not additional ones. That is what makes them indistinguishable. Macroscopic systems have too many properties for the identity of indiscernible to play a role.
The concept of energy quanta is somewhat abstract, but if we replaced all of the ping-pong balls with classical electrons, which behave like tiny ping-pong balls somewhat, I think the result probably wouldn't change.

I think even real electrons in a beer-electron-pong set-up would result in the dice model.

"Differing first-principle models for Maxwell-Boltzmann statistics?"

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