Center-of-mass reduction

Center-of-mass reduction separates translational motion from internal coordinates by using the conserved or externally forced center coordinate.

For fixed particle masses and positive total mass, write ${\mathbf{x}}_i=\mathbf{R}+{\mathbf{y}}_i$ with $\mathbf{R}$ the centre of mass. The weighted relative velocities sum to zero, so expanding the kinetic energy eliminates the mixed term and yields the displayed decomposition. For a rigid body, the relative contribution is the rotational kinetic energy $I{\omega }^2/2$ about the chosen centre-of-mass axis.

For three particles, choose the relative position of particles one and two, the position of their center of mass relative to particle three, and the total center of mass. The kinetic energy diagonalizes with masses

For Jacobi coordinates $\mathbf{a},\mathbf{b},\mathbf{c}$ and their corresponding reduced masses $\alpha ,\beta ,\gamma$, total angular momentum is

For three ordered angular positions $q_1,q_2,q_3$, define consecutive gapsand the collective angle $\Phi =(q_1+q_2+q_3)/3$. Then $\alpha +\beta +\gamma =2\pi$. For three equal masses on a circle of radius $r$,Thus the ignorable coordinate $\Phi$ decouples from the relative gap dynamics whenever the potential depends only on the gaps.