If we have an isolated system $Sb$ with thermodynamic entropy $Eb=X$ (and growing by the 2nd law of thermodynamics),

we could define an abstract system $Sa$ (containing the system $Sb$) but define the possible microstates as a single bit, being 1 (one) when $Eb>=X$, and 0 (zero) when $Eb < X$.

The microstate will tend to be $1$, so the entropy of the system $Sa$ tends to zero.



Eb = - k [p0 * log (p0) + p1 * log (p1)] --> 0

How could it be possible that an isolated system have a decreasing entropy?

Does the growth (or lack of growth) of the entropy depend on the microstates choosen?

Finally, if there is a restriction on how to choose the definition of the microstates, then what is that restriction?


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    $\begingroup$ The way you're defining the word "microstate" is simply not what the word means in statistical mechanics. If you defined "microstate" to mean "a dollar in my bank account," then the number of microstates might, unfortunately, decline over time, but this would not be a violation of the second law of thermodynamics. $\endgroup$ – Ted Bunn Jul 4 '11 at 15:50
  • $\begingroup$ @Ted Bunn bank account is not an isolated system, I agree with you that there is some statistical mechanics microstate, and some that are not, but I can see what is that general restriction $\endgroup$ – HDE Jul 4 '11 at 16:11

I think the question you're really trying to ask is: what makes a set of microstates work for thermodynamics. In quantum mechanics, for a set of microstates, you can just take a set of orthogonal states of the system. For classical mechanics, you need to take a set of microstates which all have equal volume. Of course, what equal volume means depends on the measure you put on the system. This generally isn't covered in popular texts. (I don't know why; maybe because people who are popularizing physics think that measures are stupid, unnecessary, and overly complicated mathematical notions.) For thermodynamics of ideal gasses, you can choose a microstate to be one which contains equal volume in terms of the x and p (position and momentum) variables of the particles. If you chose an alternate measure, where the states had equal volumes in terms of x2 and p2 variables, I believe that thermodynamics would not work with this measure.

For other kinds of systems, if you know enough physics, it is fairly easy to see what the proper generalization of the variables x and p to these system (hint: position and momentum are complementary variables). I don't know what the proper way to explain this is, though.

  • $\begingroup$ Great for the answer to mentions measures, It seems central, I would like to ask you about the "volume". Are you talking about a phase state volume? I think so, then you mention "equal volume", do you mean equal in time? I've read that coarse grained state for particles is what finally leads to growing entropy, I don't know why in "classic" physics one have to assume coarse grained measurement, thanks in advance for any further info about this respect. $\endgroup$ – HDE Jul 27 '11 at 11:56
  • $\begingroup$ I do indeed mean phase space volume. You need course-grained states because physics is reversible, so in some sense, the infinitessimally-grained entropy doesn't increase. I don't know what you mean by "equal in time". I'm thinking about taking a snapshot of phase space at a single point in time. $\endgroup$ – Peter Shor Jul 27 '11 at 18:16

The entropy $E_b$ is a macroscopic thermodynamic quantity. Therefore, any quantity derived from it will also be a macroscopic quantity. Therefore, your states are not microstates.

The general restriction for a microstates is that once someone tells you what microstate a system is in, you know "everything" there is to know about the system. In classical mechanics, this means that a microstate specifies all positions and momenta of all particles in the system. In quantum mechanics, it means that you specify the complete many-body wavefunction or many-body state.

Your curious state does not tell you what state your system is in: There are many ways the system can be in state $1$, and many ways it can be in state $0$. Because of that, it is not a microstate.

So, the general restriction is that a microstate has a unique realization in the system.

  • $\begingroup$ I think your second paragraph is slightly wrong (perhaps that's why you put quotes), there is no need to know "everything" to specify a microstate, just the phase space, for ex. in a gas you don't need to know where all particles are, it's enough to know where they -could- be, so it's a definition about their posibilities without needing to know details about microstates, just the space of posibilities, so I think my curious state tell enough information to know the space of posibilities, has a unique realization in the system i.e. outcomes will be 1, with probability 1 (or very close). $\endgroup$ – HDE Jul 4 '11 at 18:07
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    $\begingroup$ This comment is very confusing to me. Lagerbaer's answer is quite right: a microstate, by definition, involves specifying everything that can be specified about a system. In particular, specifying the microstate for a gas does mean specifying where all the particles are. $\endgroup$ – Ted Bunn Jul 4 '11 at 21:00
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    $\begingroup$ @HDE "There is no need to know everything to specify a microsate [...] without needing to know details about microstates". That is contradictory. A macrostate is a set of possible microstates. $\endgroup$ – Lagerbaer Jul 4 '11 at 23:22
  • $\begingroup$ @Lagerbaer About your comment, you are right, I wanted to say "define" instead of "specify" : there is no need to know everything to define something as microstate, that's what I meant, thanks $\endgroup$ – HDE Jul 4 '11 at 23:57
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    $\begingroup$ Sorry, but I'm still utterly missing your original point. Coarse-graining works if there are well-defined degrees of freedom that can be ignored because they don't couple to the rest of the system. I can't see any sense in which your example looks like that. My original comment, about dollars in my bank account, was facetious, but in fact I meant something by it: as far as I can tell your definition of "microstate" in your question is just as unrelated to the usual meaning of that term as my silly example. I honestly can't see what point you're trying to make. Sorry. $\endgroup$ – Ted Bunn Jul 5 '11 at 17:25

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