= Two-particle relative coordinate
{title2=$\mathbf r=\mathbf r_2-\mathbf r_1$}
Choose <positions> in one <inertial frame> and let $M=m_1+m_2$. The <centre of mass> and <two-particle relative coordinate> are
$$
\mathbf R=\frac{m_1\mathbf r_1+m_2\mathbf r_2}{M},\qquad \mathbf r=\mathbf r_2-\mathbf r_1,\qquad
\mathbf r_1=\mathbf R-\frac{m_2}{M}\mathbf r,\quad \mathbf r_2=\mathbf R+\frac{m_1}{M}\mathbf r.
$$
The lighter body is farther from the <centre of mass>. Reconstructing both <positions> from these equations is part of solving the <two-body problem>; $\mathbf r$ alone describes their separation, not either body's laboratory <position>.<Newtonian Dynamics by Richard Fitzpatrick>[Reduced Mass, equations 324–326: centre and reconstruction of both particle positions]{ref}
\Image[/reduced-mass-center-of-mass-relative-motion.png]
{title=<Centre of mass> translation and internal <two-particle relative coordinate> for a pair with mass ratio three}
{id=image-reduced-mass-center-of-mass-relative-motion}
{description=Original illustrative <free two-mass spring> model, with $m_2=3m_1$, equilibrium separation $\ell$, relative extension $s/\ell=0.3\cos\tau$, $\tau=\omega t$, and $R/\ell=0.08\tau$. The right plot subtracts both the translating <centre of mass> and each body's equilibrium offset. It shows internal <displacements>, not full distances from the <centre of mass>.}
Substitute the reconstructed <velocities> into the <kinetic energy>. The result is the exact two-particle instance of <kinetic-energy decomposition about the centre of mass>:
$$
T=\frac12m_1|\dot{\mathbf r}_1|^2+\frac12m_2|\dot{\mathbf r}_2|^2
=\frac12M|\dot{\mathbf R}|^2+\frac12\mu|\dot{\mathbf r}|^2.
$$
The mixed <dot product> has coefficient $-m_1m_2/M+m_2m_1/M=0$; the coefficient of $|\dot{\mathbf r}|^2$ is $(m_1m_2^2+m_2m_1^2)/M^2=\mu$. This algebra needs constant positive <masses> but no particular <force> law. In the stationary <centre of mass> frame, the fraction of internal <kinetic energy> carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.
For a closed pair with no external <forces>, write $\mathbf F$ for the <force> on body 2 and $-\mathbf F$ for that on body 1, as required by <Newton's third law>. Adding and subtracting their <Newton's second law> equations yields
$$
M\ddot{\mathbf R}=\mathbf0,\qquad
\ddot{\mathbf r}=\left(\frac1{m_2}+\frac1{m_1}\right)\mathbf F,\qquad
\mu\ddot{\mathbf r}=\mathbf F.
$$
Thus the <centre of mass> moves uniformly while the <two-particle relative coordinate> behaves like one body of <reduced mass>. Equal-opposite internal <forces> suffice for these equations. For an internal <central force>, $\mathbf F=f(r)\widehat{\mathbf r}$ with $r=|\mathbf r|>0$, the internal <torque> vanishes and <conservation of angular momentum> gives constant $\mathbf L_{\rm rel}=\mu\mathbf r\times\dot{\mathbf r}$. If the <central force> comes from a time-independent <potential energy> $U(r)$, then $f(r)=-dU/dr$ and the internal <mechanical energy> $E_{\rm rel}=\mu|\dot{\mathbf r}|^2/2+U(r)$ is conserved. The <kinetic energy> split alone does not establish either conservation law.
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