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Relative Velocity Derivation from Maxwell-Boltzmann

  1. Sep 30, 2012 #1
    I have a specific question about the derivation of the relative velocity between two molecules A and B from the Maxwell Boltzmann distribution (which should equal square root of 2 times the average velocity).
    In equations, I have been evaluating this integral:

    \ c_r&=&\bigg(\frac{m_A}{2\pi kT}\bigg)^{3/2}\bigg(\frac{m_B}{2\pi kT}\bigg)^{3/2}\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty}v_rdv_{Ax}...dv{_Bz}e^{-(m_A(v_{Ax}^2+v_{Ay}^2+v_{Az}^2)+m_B(v_{Bx}^2+v_{By}^2+v_{Bz}^2))/(2k_BT)}

    I have transformed to relative and center-of-mass velocity coordinates using these relations:
    \ v_{ri}&=&v_{Ai}-v_{Bi}\\
    \ v_{ci}&=&\frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}

    But now I cannot seem to prove that
    \ dv_{Ai}dv_{Bi} = dv_{ri}dv_{ci}\
    (to transform the Cartesian differentials to center-of-mass differentials).

    I have started like this ..
    \ dv_{Ai}dv_{Bi}&=&\bigg(dv_{ci}+\frac{m_B}{m_A+m_B}dv_{ri}\bigg)\bigg(dv_{ci}-\frac{m_A}{m_A+m_B}dv_{ri}\bigg)\\
    \ dv_{ri}dv_{ci}&\stackrel{?}{=}& dv_{ci}^2+\frac{m_B-m_A}{m_A+m_B}dv_{ri}dv_{ci}-\frac{m_Am_B}{(m_A+m_B)^2}dv_{ri}^2

    But do not end up with the final line actually being equal. If anyone has done this derivation before and has advice for converting the differentials I would be very, very grateful!!
  2. jcsd
  3. Sep 30, 2012 #2
    When doing a change of coordinates with multiple variables, the way to find the general volume element [itex]dv_A dv_B[/itex] in terms of the new variables is the Jacobian determinant; you can look at what I am referring to here. What you are really looking for is,

    dv_{r}dv_{c}= \begin{vmatrix}\frac{\partial v_r}{\partial v_A} & \frac{\partial v_r}{\partial v_B}\\ \frac{\partial v_c}{\partial v_A} & \frac{\partial v_c}{\partial v_B}\end{vmatrix}=\begin{vmatrix} 1 & -1 \\ \frac{m_A}{m_A +m_B} & \frac{m_B}{m_A+m_B}\end{vmatrix} = 1

    so it looks like this coordinate change does not affect the volume element.
  4. Sep 30, 2012 #3
    Excellent! Thank you so much.
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