- #1

feynman_freak

- 2

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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!!