Relative Velocity Derivation from Maxwell-Boltzmann

  • #1
Hi,
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:

\begin{eqnarray*}
c_r&=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}v_rP(v_{Ax},v_{Ay},v_{Az})dv_{Ax}dv_{Ay}dv_{Az}P(v_{Bx},v_{By},v_{Bz})dv_{Bx}dv_{By}dv_{Bz}\\
\ 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)}
\end{eqnarray*}

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

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

I have started like this ..
\begin{eqnarray*}
\ 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
\end{eqnarray*}

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!!
 

Answers and Replies

  • #2
38
0
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,

[tex]
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
[/tex]

so it looks like this coordinate change does not affect the volume element.
 
  • #3
Excellent! Thank you so much.
 

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