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How did my professor get this integral

  1. May 6, 2017 #1

    grandpa2390

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    Gold Member

    1. The problem statement, all variables and given/known data
    derive maxwell distribution function in case of 1-d and 2-d classical gas

    2. Relevant equations


    3. The attempt at a solution

    The constant K can be solved from normalization.
    ##\int_{-∞}^{∞} F(V_x)dV_x = 1##

    substituting ##F(V_x)=Ke^{+/- kV_x^2}##

    ##1 = K\int_{-∞}^{∞} e^{-kV_x^2}dV_x##
    which equals:
    ##K(\frac{π}{k})^{1/2}##

    how did that integral become that result?
     
  2. jcsd
  3. May 6, 2017 #2

    gneill

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    Staff: Mentor

    I suspect that he used the definition of the Gauss error function and its limiting values at +/- ∞. At least, that's what I would do :smile:
     
  4. May 6, 2017 #3
    This is the integral of a Gaussian (a bell curve essentially). There are plenty of resources online that give you the general form of this integral. One such resource is http://www.umich.edu/~chem461/Gaussian Integrals.pdf. They also include a very handy derivation.
     
  5. May 6, 2017 #4

    grandpa2390

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    Gold Member

    I followed that. And it assuming I did it correctly... I got the same result.
    ##K = \frac{k}{π}^\frac{1}{2}## in 1-D
    and
    ##K^2 = \frac{k}{π}## which is the same thing for 2-D

    is that right?
     
    Last edited: May 6, 2017
  6. May 6, 2017 #5
    For 1D, make sure to put some parenthesis around you fraction so we can tell you're setting the whole fraction the the 1/2 power.

    For 2D, I believe your formula generalizes to [tex]1 = K\int_{-∞}^{∞}\int_{-∞}^{∞} e^{-k(V_x^2+V_y^2)}dV_x \, dV_y[/tex]
     
  7. May 6, 2017 #6

    grandpa2390

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    Gold Member

    well I thought it was ##F(V_x)F(V_y) = Ke^{(..)}Ke^{(...)}## and so K^2
     
  8. May 7, 2017 #7
    A one-dimensional Gaussian has the form [tex]F(x)=Ke^{-\alpha x^2}[/tex] A two-dimensional Gaussian has the form [tex]F(x,y)=Ke^{-\alpha (x^2+y^2)}[/tex] For an ideal gas in two dimensions, the Maxwell velocity distribution is a two-dimensional Gaussian (I'm assuming [itex]V_i[/itex] represents the speed in the [itex]i[/itex]th direction). This is a consequence of the equipartition theorem.
     
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