Timeline for Irreversibility and organisation of trajectories in the phase space
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 17 at 22:09 | vote | accept | Chevallier | ||
| Jan 17 at 22:08 | comment | added | Chevallier | @YvanVelenik, thanks again, I believe that I get it now. Essentially in my reasoning I didn't to take into account the fact that high entropy values correspond to larger volumes in the phase space. | |
| Jan 17 at 21:05 | comment | added | FlatterMann | @TobiasFünke The phase space of a gravitating n-body problem is infinite. The system can never fill it, hence the assumptions of statistical mechanics don't apply. As soon as such a system kicks out the first body there is one trajectory that goes towards infinity in the radial coordinate and it never stops. The naive ergodic hypothesis for such systems is surely false. Maybe the mathematicians will one day rescue statistical mechanics from itself by finding a properly defined measure that can actually "fill" the phase space under that re-definition, but I am not aware of such a proposal. | |
| Jan 17 at 20:48 | comment | added | Tobias Fünke | @FlatterMann I don't understand. You have a phase space, you can divide it into regions of different macroscopic parameters (hence macro states) and define an entropy function on these macrostates. I don't see any problem. The second law then states that for every experimentally reproducible process, the entropy cannot decrease (roughly speaking). | |
| Jan 17 at 20:42 | comment | added | FlatterMann | @TobiasFünke Hamiltonian systems don't obey the second law. That the second law wins is because Hamiltonian systems are a poor poor approximation in the infinite time limit. Or, to put it less mildly, the infinite time limit is simply unphysical, even in most trivial cases. Newtonian gravity, for instance, is unstable for all but the most simple examples. Gravitating n-body systems evaporate long before they explore all of their phase space. | |
| Jan 17 at 20:03 | comment | added | Yvan Velenik | Because the equilibrium macrostate occupies almost all of the phase space, so that a generic trajectory has little choice but to stay there (or go there if it starts out of equilibrium). The papers I linked probably go into quantitative estimates (I read them very long ago, so I don't remember). | |
| Jan 17 at 19:50 | comment | added | Chevallier | @Yvan Velenik, thank you very much for your answer. I need to meditate on it. I should probably think longer before writing but here is what comes to my mind. You say :'However, if you sample a random point in the equilibrium macrostate 𝐵, then it will almost certainly not belong to $A_t$ and would remain at equilibrium.', but how do we know the part 'and would remain at equilibrium.' ? | |
| Jan 17 at 17:14 | comment | added | Tobias Fünke | +1 A nice discussion about these matters is also given in Jaynes' "The second law as physical fact and as human inference", in the section "What does 'reversibility' mean?". In particular, there he emphasizes the connection to experiments. | |
| Jan 17 at 16:32 | history | edited | Yvan Velenik | CC BY-SA 4.0 |
added 89 characters in body
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| Jan 17 at 16:26 | history | edited | Yvan Velenik | CC BY-SA 4.0 |
added 537 characters in body
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| Jan 17 at 14:38 | history | edited | Yvan Velenik | CC BY-SA 4.0 |
Better formulation
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| Jan 17 at 13:25 | history | answered | Yvan Velenik | CC BY-SA 4.0 |