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greswd

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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.

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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