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Why is the logarithm of the number of all possible states of

  1. Dec 25, 2015 #1
    Temperature of a system is defined as
    $$\left( \frac{\partial \ln(\Omega)}{ \partial E} \right)_{N, X_i} = \frac{1}{kT}$$Where Ω is the number of all accessible states (ways) for the system. Ω can only take discrete values. What does this mean from a mathematical perspective? Many people say we have 10^23 particles so Ω is almost continuous function of energy. Why is 10^23 a nice number but 1000 is not? When can one be sure they can differentiate ln(Omega)?

    If you agree with me, do you know an alternative accurate definition for temperature?
     
  2. jcsd
  3. Dec 25, 2015 #2

    Fredrik

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    There's no better definition of temperature. Ω is only approximately equal to the number of accessible states. I don't see why that bothers you.

    When two systems are in contact with each other, energy will flow in the direction that increases the number of accessible states of the combined system. A fairly simple argument shows that if we define temperature this way, then energy is flowing from higher temperature to lower. We define Ω and T this way to be able to make this argument.
     
  4. Dec 25, 2015 #3
    @Fredrik Thank you for your reply. log[Ω] can only take discrete values; so it is ambiguous what is its derivative. I mean there is infinitely many functions that have the same values as Ω has, but they have different derivative. What does it mean when we differentiate log[Ω]; it is not trivial.
     
  5. Dec 25, 2015 #4

    Fredrik

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    OK, so the main issue is that there's more than one way to approximate a function defined on a set of integers by a function defined on an interval. I suppose the answer has to be that all reasonable ways to define what we mean by a "best" approximation (e.g. for some n, the nth-degree polynomial that's the best approximation in the least squares sense) give us functions that are empirically indistinguishable from each other.

    I haven't tried to verify this, but I also don't think it's necessary. There's obviously some ambiguity in a word like "reasonable", but this is to expected in arguments that are used to find the theory that we're going to use. Such arguments don't have to be rigorous, since the goal is to guess what definitions will be useful, not to prove that the theory is right.
     
  6. Dec 25, 2015 #5

    jtbell

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    Physicists are well-known for doing things that make mathematicians tear their hair out, but that somehow "work", anyway. :-p
     
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