Remark 12.31. Derivation Algebra.
Recall the definitions of center of mass position \(\vec R\) and the relative position \(\vec r\) in terms of the positions of the two particles.
\begin{align*}
\amp m_1 \vec r_1 + m_2 \vec r_2 = M \vec R\\
\amp \vec r_1 - \vec r_2 = \vec r
\end{align*}
From these, we can get
\begin{align*}
\amp \vec r_1 = \frac{m_2}{M}\,\vec r + \vec R\\
\amp \vec r_2 = -\frac{m_1}{M}\,\vec r + \vec R
\end{align*}
Taking one time derivative with respect to time we get (write \(V\) for \(V_\text{cm}\text{.}\))
\begin{align*}
\amp \vec v_1 = \frac{m_2}{M}\,\vec v + \vec V \\
\amp \vec v_2 = -\frac{m_1}{M}\,\vec v + \vec V
\end{align*}
Now, take dot product of each with itself; do the left side with the left side of the same equation; you will get
\begin{align*}
\amp v_1^2 = \left(\frac{m_2}{M}\right)^2\,v^2+ V^2 + 2 \frac{m_2}{M} \vec v \cdot \vec V\\
\amp v_2^2 = \left(\frac{m_1}{M}\right)^2\,v^2+ V^2 - 2 \frac{m_1}{M} \vec v \cdot \vec V
\end{align*}
Now, we multiply the first one by \(m_1\) and the second one by \(m_2\text{,}\) and add. The last term will cancel out. Then multiply everything by \(1/2\) and then simpifying further, we get
\begin{equation*}
\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2} \mu v^2 + \frac{1}{2} M V^2.
\end{equation*}
