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Why are probabilities distribution of thermodynamic variables tend to Gaussian?

  1. Dec 13, 2009 #1
    The probability distribution for some thermodynamic variable x is given by

    [tex]P = N e^{-A(x)/KT}[/tex]

    where A(x) is the availability, which can be replaced by Hemlholtz free energy F, Gibb's free energy G, etc depending on the conditions imposed. N is just some normalization constant. A(x) can be expanded in a taylor series about the equilibrium conditions,

    [tex] A(x) = A(x_{0}) + (x - x_{0})(\frac {\partial A} {\partial x})_{x = x_{0}} + \frac{1} {2} (x - x_{0})^{2} (\frac {\partial^2 A} {\partial x^2})_{x = x_{0}} + ... [/tex]

    The second term is 0 since dA/dx = 0 at equilibrium. If we truncate all the other terms, clearly we see that P will be a Gaussian distribution with mean of [tex] x_{0} [/tex] and standard deviation of

    [tex] \sqrt {\frac {K T} {(\frac {\partial^2 A} {\partial x^2})_{x = x_{0}}}} [/tex]

    What is the justification for truncating this series? This is justified if (x - x0) is small. But why will it be small for big N?
  2. jcsd
  3. Dec 14, 2009 #2


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    Science Advisor

    I am not familiar with the details of the physics. However such truncation would be based on the assumption |x-x0| is small.
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