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by oh_my_goodness
1520 days ago
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Sure. A fine-grained macro-state contains fewer micro-states. A coarse-grained macro-state contains more micro-states. Say I flip 8 coins and I don't look at the results. A fine-grained macro state is TTTT TTTT. A coarser-grained macro state is TTTT xxxx. The one has 4 bits more entropy than the other. It works the same way in statistical mechanics. Call them spins. We're just talking about some ensemble of micro states, and then we divide the ensemble up into macro-states. To do statistical mechanics at all, I think I have to define some macro-states according to which micro-states they contain. That doesn't mean I necessarily have any information about which macro-state the system is actually in. |
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This macrostate seems fine enough to be a microstate, but sure. For the macrostates with at least one 'x' in it, that 'x' seems to be a placeholder for the concept of subjective ignorance.
> That doesn't mean I necessarily have any information about which macro-state the system is actually in.
But the entire purpose of the exercise of assigning microstates to macrostates is so that you can match up a description of some system to a microstate ensemble and calculate its entropy! Otherwise there's no point to arbitrarily labelling various groups of microstates.
To follow your example more practically, let's say you have an 8-spin system, whose net spin is zero. (You know because you've measured its overall magnetic moment or something). I've just described a system that is in one of the following possible microstates:
TTTT HHHH, TTTH HHHT, ..., HHHH TTTT
Now you can go ahead and define the macrostates as fine-grained as you want, where TTTT HHHH and HHHH TTTT are in different macrostates, but to calculate the entropy of this system, you're going to have to sum up all of those macrostates anyway to get the one that's consistent with the described system.