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Entropy in thermodynamics

  1. Feb 23, 2012 #1
    Entropy is in my book defined as:

    S = k ln[itex]\Omega[/itex] , where [itex]\Omega[/itex] is the multiplicity of the microstates.

    Now I have several questions regarding this definition (please try to answer them all! :) )

    1) Why do you take the log of the multiplicity? Is it just because additivity aswell as the fact that even for a lot of particles S is still a small number, just makes it a lot more convenient to work with this definition?
    At first I thought so, but then ln increases less per step the farther you are on the x axis. So wouldn't that make problems?

    2) I can understand that generally thermodynamics is more deeply described in statistical mechanics. Is entropy then on a deeper level still defined as the the above, or is that just something that also appears to be true? I'm asking this because, the k factor seems to indicate, that there's more to it than at first glance.
    Also I have seen a few videos, which discuss entropy in information theory, and here it seems that entropy is something more deep than just an expression for the multiplicity of a system.

    That covers all. Thanks :)
     
  2. jcsd
  3. Feb 23, 2012 #2
    (1) As you already said , the natural log grantees the additive property which is anticipated for an extensive quantity such as entropy. However, more generally you can define the entropy for any probability distribution function. In the microcanonical ensemble (Fixed, N,V,E) , the probability of any state of energy E is just 1/Ω (E). For this PDF , S has to be lnΩ. The extra factor of kB was added to match the thermodynamic dimensions of entropy (energy unit/degree kelvin).

    (2) Up to a constant , the thermodynamic entropy, Shanon entropy (information theory), statistical mechanics entropy, and H in Boltzmann H-theorem are equivalent.
    In statistical mechanics, the entropy in the microcanonical ensemble is given by the expression you wrote.
     
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