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Is entropy the volume in phase space of energy E or LESS than E?

  1. Sep 26, 2011 #1
    Hello,

    I thought the statistical definition of entropy for an isolated system of energy E (i.e. microcanonical ensemble) was [itex]S=k \ln \Omega[/itex] where [itex]\Omega[/itex] is the volume in phase space of all the microstates with energy E.

    However, if you take a look here http://en.wikipedia.org/wiki/Equipartition_theorem#The_microcanonical_ensemble
    there is the line
    so they use the volume in phase space where energy < E instead of the surface where energy = E. Do these notions coincide? I would think they'd conflict. Why do they say "by the usual definitions", I'm confused.
     
  2. jcsd
  3. Sep 26, 2011 #2
    I'm finding a source (Huang, Statistical Mechanics, 2nd edition, p134) that states [itex]S = k \log \Omega[/itex] and [itex]S = k \log \Sigma[/itex] are indeed equivalent up to a constant dependent of N. The reason for that, I don't seem to get, as the text is a bit too advanced for me atm.

    In a way I'm willing to accept the equivalency (as it would clear up my problem), but there's one thing that bothers me: take for example a state of a certain system such that if you lower the energy, entropy goes up (think of a system with bounded energy), doesn't the [itex]\Sigma(E)[/itex] (= the volume in phase space where energy < E) definition make this behavior impossible, because surely (by definition) [itex]E_1 < E_2 \Rightarrow \Sigma(E_1) < \Sigma(E_2) \Rightarrow S(E_1) < S(E_2)[/itex]?

    What am I overlooking?
     
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