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Probability of a distrubition

  1. Aug 18, 2015 #1
    in stat mechanics we derive most probable distribution .but this does not say any thing about existence of other less probable distributions. is there a way to find out how probable is the most probable
     
  2. jcsd
  3. Aug 18, 2015 #2
    I come to question about these things a lot and I totally agree with what you say and there is a way to find the most propable propability, even though we aren't aware of where stuff is and we don't know their state for sure, we don't know that no matter what they wouldn't (and shouldn't) smash a bunch of laws of physics and the distribution that applies this MIGHT bd the best propability distribution, but for the best not so sure because we cannot know pricesly the state and properly describe the microstates of each little particle (example) this is an artefact of our ignorance !
     
  4. Aug 18, 2015 #3

    mathman

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    The distribution is derived from the laws of physics. To get a different distribution you need to get a different set of physics laws.
     
  5. Aug 20, 2015 #4

    Jano L.

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    It is a probability question. The probability is proportional to number of ways distribution can be realized.
     
  6. Aug 21, 2015 #5

    Philip Wood

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    Most introductory SM books state without much back-up that the most probable distribution of energy states in a large ensemble of systems is so overwhelmingly the most probable that we can forget about the others. More precisely, they say that for a very large number of systems the logarithm of the number of ways of achieving the most probable distribution is the same as the logarithm of the sum of the numbers of ways of achieving every distribution!

    Usually they refer to the binomial distribution for back-up, without properly explaining why the binomial distribution is related to the matter in hand.

    I've found the following elementary example useful for clarifying what's going on…

    Take an ensemble of N 3-level systems. Let the levels be non-degenerate with energies 0, E, 2E. Let the total ensemble energy be [itex]\frac{4}{7}NE[/itex]. Suppose n1 systems are on the lowest level, n2 systems are on the middle level, n3 systems are on the top level. It's easy to express n1 and n2 in terms of n3 (and the constant, N). So there's only one free variable, n3.
    For this simple system it's possible to show by quite elementary means (using Stirling's formula and second order Taylor expansion) that, as N approaches infinity, the most probable distribution is the only one that carries any weight.

    There's nothing special about the choice of [itex]\frac{4}{7}NE[/itex] for the total ensemble energy; it just makes the arithmetic slightly neater then many other choices. For example, in the most probable distribution n2 turns out be half n1 and twice n3.
     
    Last edited: Aug 21, 2015
  7. Aug 24, 2015 #6

    Philip Wood

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    Gayle. It would be nice to know if any of the responses to your question were of any use. If any were impossible to understand, or didn't go far enough towards answering your question, you can ask for clarification.
     
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