Centre of mass and reduced mass in classical mechanic

Question

Can we derive reduced mass from Cetre of mass equation if CM is not taken as origin? And Under what conditions can reduced mass concept be used or not?

Answer 1

In the two-body problem for two bodies whose position is represented by two position vectors ${\bf r}_1$ and ${\bf r}_2$, it is helpful to rewrite the equations of motion (it doesn't matter if classical or quantum), in terms of the two variables $$ \begin{align} {\bf r} &= {\bf r}_2 - {\bf r}_1\\ {\bf R} &= \frac{m_1 {\bf r}_1 + m_2 {\bf r}_2}{m_1+m_2}. \end{align} $$

The equation for ${\bf r}$ can be read as the equation of motion of a body of mass $\mu$, the reduced mass, defined through $$ \frac{1}{\mu}= \frac{1}{m_1} + \frac{1}{m_2}. $$ For example, in the case of Newtonian mechanics, in the presence of a central force, we have $$ \begin{align} m_1\ddot{\bf r}_1 &= ~~\Phi(r){\bf r}\tag{1}\\ m_2\ddot{\bf r}_2 &= -\Phi(r){\bf r}.\tag{2} \end{align} $$ By dividing equation ($1$) by $m_1$, and equation ($2$) by $m_2$ and subtracting , we get an equation of the form $$ \mu \ddot {\bf r} = -\Phi(r){\bf r}. $$

From these completely general formulas, it should be clear that the position of the center of mass does not matter. ${\bf R}$ may be the null vector or any other vector, even changing with time, without affecting the change of variables and the definition of the reduced mass.