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Maxwell distribution

  1. Feb 25, 2015 #1
    1. The problem statement, all variables and given/known data

    In thermal equilibrium, the particle in a gas are distributed in velocity space according to the Maxwell distribution

    f(v) = A*exp(-mv^2/(2KT))

    What is the average velocity ? What is the most probable velocity ?

    2. Relevant equations

    <v> = ∫∫∫vf(v)d3v = (0 to infinty) ∫∫∫vxf(v)d3vx^ + ∫∫∫vyf(v)d3vy^ + ∫∫∫vzf(v)d3vz^


    3. The attempt at a solution

    I started with the vy coordinate

    [0 to inf] ∫∫∫vyAexp(-mv^2/(2KT))dy = ∫[0 to 2pi]∫[0 to pi][0 to inf]vyA*exp(-mv2/(2KT))*v2y sin(θ)dθdvdφ = 4piA∫vy3exp(-mv^2/(2KT)) dv = Here is the problem, if i evaluate the integral i got a non-zero value. I know the 3 integrals should be zero since they are mult. of odd times even. And the most probable velocity is just the derivative of the f(v) function by resp. coordinate right ?
     
  2. jcsd
  3. Feb 25, 2015 #2

    TSny

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    If you are going to use spherical coordinates, then you will need to express vy in terms of v, θ, and φ before you integrate over θ and φ.

    Or, you can use Cartesian coordinates and write the volume element as dvxdvydvz. Then you would integrate over the variables vx, vy, and vz.
     
  4. Feb 26, 2015 #3
    Yes, i forgot to include that in my latex code, but i have done that. If i integrate with cartesian coordinates i got:
    [0 to inf]∫∫∫Ae-m/(2KT)*(Vx2+vy2+vz2dvxdvydvz =[0 to inf] ∫Ae-m/(2KT)*vx2 +--- = non-zero value which is wrong. What goes wrong ?
     
    Last edited: Feb 26, 2015
  5. Feb 26, 2015 #4

    TSny

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    If you are trying to get the average y-component of velocity, then you need to include a factor of vy in the integrand.

    Also, what are the correct limits in the integral for vx, vy, and vz? (Note that the components of velocity can be negative.)
     
  6. Feb 27, 2015 #5
    Did you mean the integral = 4piA∫vy3exp(-mv^2/(2KT)) dv or [0 to inf] ∫Ae-m/(2KT)*vx2 (x component) ?
     
  7. Feb 27, 2015 #6
    If i do the following integral

    v- = [inf to -inf]∫vƒm^d3v = and put in spherical coordinates and simplify i got 4π(πvth2)-3/2*vth4 [inf to -inf] ∫y3 e-y2dy = [Part. int.] =...= 2*(2KT/mπ)1/2

    It should be zero
     
  8. Feb 27, 2015 #7

    TSny

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    The Maxwell probability distribution is ##Ce^{-m(v_x^2+v_y^2+v_z^2)/(2kT)}## where ##C## is a constant.

    This means that the average value of any function ##f(v_x, v_y, v_z)## is given by $$C\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(v_x, v_y, v_z)e^{-m(v_x^2+v_y^2+v_z^2)/(2kT)}dv_xdv_ydv_z$$

    So, to find the average value of ##v_x##, you would let ##f(v_x, v_y, v_z) = v_x##. Try this and see what you get.

    It's important not to confuse the probability distribution for the variables ##v_x, v_y, v_z## (as given above) with the probability distribution for the single variable ##v## (the speed of the particle). The distribution function for ##v## is ##Bv^2e^{-mv^2/(2kT)}##, where ##B## is a constant.
     
  9. Mar 1, 2015 #8
    If i integrate with resp. cartesian coordinates i got it to be equal to C*((2πKT)/m)3/2*(KT/m). Since the Gaussian integral is equal to (2πKT/m)1/2. Is something missing here ?
     
  10. Mar 1, 2015 #9

    TSny

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    Suppose you are trying to find the average value of ##v_x##. What does the integrand look like when you are integrating over all ##v_x##? Is the integrand an even function or an odd function of ##v_x##?
     
  11. Mar 1, 2015 #10
    If i integrate over vx: [-inf to inf]C∫∫[-KTvx/m*exp(-m(vx^2+vy^2+vz^2)/(2KT))](-inf to inf] + KT/m [-inf to inf]∫exp(-m((vx^2+vy^2+vz^2)/(2KT)) dvy dvz. The integral [-inf to inf] Is odd right ?. But i dont know if vx is even or odd
     
  12. Mar 1, 2015 #11

    TSny

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    $$<v_x> = C\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}v_xe^{-m(v_x^2+v_y^2+v_z^2)/(2kT)}dv_xdv_ydv_z$$

    $$<v_x> = C\int_{-\infty}^{\infty}v_xe^{-mv_x^2/(2kT)}dv_x\int_{-\infty}^{\infty}e^{-mv_y^2/(2kT)}dv_y\int_{-\infty}^{\infty}e^{-mv_z^2/(2kT)}dv_z$$

    What do you get when you evaluate the first integral on the right?
     
  13. Mar 1, 2015 #12
    This was the integral i just evaluated above which i got to be equal to C*((2πKT)/m)^3/2*(KT/m). One with part. integration(and gaussian) and the last two integrals is the gaussian integral.
     
  14. Mar 1, 2015 #13

    TSny

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    Did you integrate the first integral from -∞ to +∞? The integrand is $$v_xe^{-mv_x^2/(2kT)}$$ If you replace ##v_x## by ##-v_x## what happens to the overall sign of the integrand?
     
    Last edited: Mar 1, 2015
  15. Mar 1, 2015 #14

    vela

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    Since you're having problems evaluating the integral, it would help if you showed what you're doing rather than just saying you got the wrong answer.

    I don't know if it's just a typo, but what's up with the red part?

    [-inf to inf]C∫∫[-KTvx/m*exp(-m(vx^2+vy^2+vz^2)/(2KT))](-inf to inf] + KT/m [-inf to inf]∫exp(-m((vx^2+vy^2+vz^2)/(2KT)) dvy dvz
     
  16. Mar 2, 2015 #15
    I got the right integral to be equal to C[-inf to inf on all integrals]∫vxe((-m*vx2))/(2KT)dvx∫e((-m*vy2))/(2KT)dvy*((2πKT)/(m)) = C((2πKT)/(m))^(1/2)∫vxe((-m*vx2))/(2KT)dvx*((2πKT)/(m)) = C((2πKT)/(m))^(1/2) ([-/KTvx)/m*e-mvx2/(2KT)]+ KT/m∫e-mvx2/(2KT) dvx = the first is equal to zero and the last is (2πKT/m)^(1/2) =C(2πKT/m)^(3/2)*(KT/m)

    If i replace vx with -vx in the integral then i got it to be equal to C[inf to -inf]vxe-(mvx2)/(2KT), when i switch -inf to inf
     
  17. Mar 2, 2015 #16

    TSny

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    The integral over ##v_x## has an odd integrand, so the integral must be zero. There is then no need to evaluate the other integrals over ##v_y## and ##v_z##.
     
  18. Mar 4, 2015 #17

    vela

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    The problem is that you're not integrating by parts correctly. In fact, integration by parts doesn't work with this integral; the substitution ##u=v_x^2## would work. The easiest route, of course, is what TSny has said. The integrand is odd, so the integral must vanish.
     
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