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Integrating Maxwell-Boltzmann Distribution

  1. Feb 1, 2007 #1
    I don’t know how to integrate the Maxwell-Boltzmann distribution without approximation or help from Maple.

    Given the Maxwell-Boltzmann distribution:

    f(v) = 4*pi*[m/(2*pi*k*T)]^(3/2)*v^2*exp[(-m*v^2)/(2*k*T)]

    Assuming a fixed temperature and mass, one can simplify this equation:

    f(v) = a*v^2*exp[-bv^2]
    a = 4*pi*[m/(2*pi*k*T)]^(3/2)
    b = m/(2*k*T)

    In order to calculate the fraction of particles between two speeds v1 and v2, one should evaluate the definite integral:

    ∫f(v)dv

    Can this be done analytically?

    Also, this function predicts a non-zero (albeit very small) probability for particles to have a speed greater than the speed of light. Is there a correction to this distribution that takes this into account?
     
  2. jcsd
  3. Feb 1, 2007 #2
    Simplest way is 1) find it in a standard table of integrands and 2) then differentiate the integral just to convince yourself.
     
  4. Feb 1, 2007 #3
    Table of Integrals

    I haven't been able to find this integral in a table. I looked in Gradshteyn and Ryzhik's book and couldn't find it. Maybe I need to find a bigger book of integrals.
     
  5. Feb 1, 2007 #4
    There isn't one. Let's look at our good friend the Gaussian integral. If you're in a statistical mechanics course, you are going to become a master of these things by the time you're done.

    Okay, so
    [tex]I(\alpha) = \int_{-\infty}^\infty dx e^{- \alpha x^2}[/tex]
    Let's square it. Then we get that
    [tex]I(\alpha)^2 = \int_{-\infty}^\infty dx \int_{-\infty}^\infty dy e^{-\alpha(x^2 + y^2)}[/tex]
    After making a change of coordinates to polar and integrating over the entire plane, we get that the integral is given by
    [tex]I(\alpha) = \sqrt{\pi/\alpha}[/tex]
    Okay, but what about terms like x^2? Well...
    [tex] \int x^2 e^{-\alpha x^2} = \int - \frac{d}{d \alpha} e^{-\alpha x^2}[/tex]
    Which then means that
    [tex]\int x^2 e^{-\alpha x^2} = - \frac{\partial}{\partial \alpha} I(\alpha)[/tex]
    This works for all even powers. This is a rather handy trick to know, and is pretty standard.
     
    Last edited: Feb 1, 2007
  6. Feb 1, 2007 #5
    Integral

    Yes, the solution of that integral from -infinity to +infinity or zero to infinity is well known and can be used to show that the function normalizes. But what if you want to know the fraction of particles that are between two particular speeds? I did find some information at the following web site:

    http://user.mc.net/~buckeroo/MXDF.html

    It states below equation 35 that the integral "cannot be solved in closed form". I also found a computational chemistry class that had an assignment to numerically integrate the distribution, which implies that this might be necessary:

    http://www.glue.umd.edu/~nsw/ench250/boltzman.htm
     
  7. Feb 1, 2007 #6

    Dr Transport

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    Gold Member

    look for the function erf(c), ther error function and you should be able to integrate the Boltzmann function between two finite limits.
     
  8. Feb 2, 2007 #7
    Erf

    I see how "erf" would work. Thanks much.

    Does anybody have a reference for the relativistic correction to the speed distribution?
     
  9. Mar 14, 2009 #8
    error function

    Please describe what is mean by error function.Why we are introducing error function in Maxwell's function distribution?
     
  10. Mar 14, 2009 #9
    The error function is defined as
    [itex]\textrm{erf}(x) = \frac{2}{\sqrt{\pi}}\int^{x}_{0} e^{-t^{2}}~\textrm{d}t[/itex]
    and pops up a lot when one does integrals with exponentials, even though it doesn't always imply that there is some sort of "error" involved.


    cf. http://en.wikipedia.org/wiki/Error_function
     
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