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The average and variance of distributions (thermodynamics)

  1. Oct 8, 2009 #1
    (Note: I'm not sure about international notations or terms, but I hope everything is comprehensible)

    Next Monday I will pass my exam in theoretical physics about thermodynamics.
    However, there's still one thing that I couldn't find explicitly described in my lecture notes or any additional literature.

    It's the average and variance of distributions. All I found was the formulas, but no further explications.

    Average:
    <x> = integral (x * f(x)) dx

    Variance:
    < (x - <x>)^2 > = integral (x * (x - <x>)^2) dx

    f(x) is in this case the Maxwell-Boltzmann distribution (such as f(x) = a * exp(-b*x^2) ).

    What I don't know is what interval do I have to choose?
    I thought about the whole set of real numbers, so from negative infinity to positive. However doing so, I don't get a sensible result, it's zero.
    I also have thought about [0; infinity] or [0; x], but I actually have no idea.

    Is the end result a term (including x) or a constant?


    Any hints are highly appreciated.
     
  2. jcsd
  3. Oct 8, 2009 #2

    mathman

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    Science Advisor
    Gold Member

    For the f(x) you described, the mean is 0.

    Your variance formula has an error. It should read:

    < (x - <x>)^2 > = integral (f(x) * (x - <x>)^2) dx
    The variance = 1/2b, a is needed so that the integral of f(x)=1.

    The integral is over the entire real line.
     
  4. Oct 9, 2009 #3
    Thank you!
    That helped me a lot!
     
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