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Probability distribution using the Maxwell Boltzmann formula

  1. Jan 31, 2013 #1
    According the the kinetic theory of gases, molecules moving along the x direction are given by Σx= (1/2) mv^2, where m = mass and vx is the velocity in the x direction
    The distribution of particles over velocities is given by the Boltzmann law p(x)=e^[(-mv^2)/ekT]
    where velocities range from -∞ to ∞

    Here, the probability distribution p(v), needs a constant, c, that will normalize the distribution so that
    c∫e^[(-mv^2)/2kT]dv=1 ( -∞ to ∞ )

    The publisher used a trick converting this equation into a double integral from
    I=∫e^(-ax^2)dx ( -∞ to ∞ ) where a = m/2kT
    I^2= ∫e^(-ax^2)dx ∫e^(-ay^2)dy -∞ to ∞
    and combined the exponentials from the double integral to get

    I^2=∫ ∫e^-a(x+y^)2 dxdy -∞ to ∞

    then converted to polar coordinates r and θ since r^2= x^2 + y^2

    to simplify to
    I^2= ∫rdr ∫e^-ar^2dθ from 0 to 2π

    simplifying further to exchange dθ for dr

    I^2= ∫dθ ∫e^-ar^2dr from 0 to 2π

    and integrated the first integrand to ∫dθ from 0 to 2π = 2π

    reducing the double integral to

    I^2=2π∫e^-ar^2dr 0 to 2π

    and finally used u substition u=-ar^2 & du = -2ardr to leave us with

    I^2= (-π/a)∫e^udu from 0 to ∞

    ==> I^2= (-π/a)e^u integrating from 0 to ∞ gives


    and I = (π/a)^-1/2

    where a= m/2kT gives us

    I= [(2πkT)/m]^1/2

    So when used as a constant, the normalized probability distribution equation becomes
    p(v)dv= [(m)/2kT)^1/2] * ∫e^[(-mv^2)/2kT]dv

    My questions is

    How are you able to take an Integral with respect to x and make it into a double integral with respect to x and y?
  2. jcsd
  3. Jan 31, 2013 #2
    Both sides were squared in that step. So on the right you have the integral times the integral. Remember that the variable you integrate over is a "dummy variable", that means it can be changed to a different character and it doesnt matter. http://mathworld.wolfram.com/DummyVariable.html

    By changing the dummy variable from "x" to "y" it is better illustrated how you are going from cartesian to polar coordinates.
  4. Feb 3, 2013 #3


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    Also, you should realize that the assumption was made that p(x,y)=p(x)p(y), which is not always the case. When p(x,y)=p(x)p(y), that means that the probability distribution for x has nothing to do with the value of y. In terms of velocities, it means that if I know the x-velocity is Vx, it tells me nothing about the y-velocity Vy. In probability theory, it says that the correlation between x and y is zero. It is also the assumption of isotropy, that there is no special direction in space. That means that p(x) and p(y) are the same function, different variables. If you pick an x-axis and a y-axis, you get the same result as if you choose two different axes, x' and y'. Those two assumptions are why you can build p(x,y) from p(x) by making it equal to p(x)p(y).
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