Memory, Entropy and the Arrow of Time

[Nugatory]

Okay. So we're basically making both the holes and the marbles anonymous (i.e. interchangeable, lacking particular identity)?

My description is based on the assumption that the marbles are distinguishable, so that marble one in hole one and marble two in hole two (and the other 998 in the same places) is a different configuration than marble two in hole one and marble one in hole two (and the other 998 in the same places). If you drop this assumption, then the statistics become interestingly different. The problem changes from "How many different ways are there to arrange 1000 different marbles in 100 holes?" to "Suppose I toss 1000 identical marbles at random onto a surface with 100 holes. How likely is it that they'll all end up in the same hole?". The answer is still unimaginably small, but it's an interestingly different unimaginably small number/.

 

[David Carroll]

Okay. I see now. I was imagining that Barbour was suggesting that second quoted question, when he was really suggesting the first. Thanks.

 

[PeterDonis]

So we're basically making both the holes and the marbles anonymous (i.e. interchangeable, lacking particular identity)?

From the standpoint of defining macroscopic states, yes. It's worth noting, though, that the details of the statistics involved actually do depend on whether or not the "marbles" and "holes" are distinguishable or not at a microscopic level. Distinguishable particles give Boltzmann statistics, which is what is standardly assumed classically. Quantum mechanically, particles of the same type are considered indistinguishable, so you get either Bose-Einstein or Fermi-Dirac statistics, depending on whether the particles have integer or half-integer spin. (In quantum field theories, there are even more kinds of statistics possible.)

 

[David Carroll]

so you get either Bose-Einstein or Fermi-Dirac statistics, depending on whether the particles have integer or half-integer spin. (In quantum field theories, there are even more kinds of statistics possible.)

Respectively?

So if we have a closed system, where one atom of each of the first 105 elements, and each of which has integer spin, is bouncing around off the other atoms, any arbitrary thermal state of this closed system has lower entropy than some other closed system where 105 atoms, all of which are, say, hydrogen, are bouncing around...according to Bose-Einstein statistics?

 

[PeterDonis]

Respectively?

Yes.

if we have a closed system, where one atom of each of the first 105 elements, and each of which has integer spin, is bouncing around off the other atoms, any arbitrary thermal state of this closed system has lower entropy than some other closed system where 105 atoms, all of which are, say, hydrogen, are bouncing around...according to Bose-Einstein statistics?

I'm not sure how you're imagining these two scenarios. If the 105 atoms are all of different elements, they're distinguishable, so you would use Boltzmann statistics. If they're all hydrogen atoms, they're not, so you would use Bose-Einstein statistics. This would result in a different count of microstates for the two cases, yes. Is that what you mean?

(I believe the count of microstates would be lower for the Bose-Einstein case, i.e., the 105 hydrogen atoms. But I haven't done the calculation to confirm that.)

 

[David Carroll]

Yeah. That's what I meant. Thanks.