= Reduced mass
{title2=$\mu$}
{wiki}
The <reduced mass> is the effective <mass> for the <two-particle relative coordinate> of two bodies with constant positive <masses> $m_1,m_2$:
$$
\mu=\frac{m_1m_2}{m_1+m_2},\qquad \frac1\mu=\frac1{m_1}+\frac1{m_2}.
$$
Both bodies can respond to their mutual <force>, so their relative <acceleration> is greater than if one were fixed. Consequently $0<\mu<\min(m_1,m_2)$. This explains why using either individual <mass>, or their sum, generally gives the wrong internal <oscillation> rate. The same <reduced mass> describes a <two-body problem>, a <free two-mass spring>, and the approximate <molecular vibration> of a <diatomic molecule>.<Newtonian Dynamics by Richard Fitzpatrick>[Reduced Mass, equations 322–328: equal-opposite forces, relative acceleration and reduced mass]{ref}
For equal <masses> $m_1=m_2=m$, $\mu=m/2$. Two $1\,\mathrm{kg}$ bodies therefore have $\mu=0.5\,\mathrm{kg}$: in their stationary <centre of mass> frame a relative <speed> $2\,\mathrm{m/s}$ means individual <speeds> $1\,\mathrm{m/s}$ and total <kinetic energy> $1\,\mathrm J$. With $m_1=1\,\mathrm{kg}$ and $m_2=100\,\mathrm{kg}$, instead $\mu=100/101\,\mathrm{kg}\approx0.990099\,\mathrm{kg}$. At the same relative <speed>, the individual <speeds> are $200/101$ and $2/101\,\mathrm{m/s}$, and the total <kinetic energy> is $200/101\,\mathrm J$. These are calculations for stipulated <masses> and <velocities>.
For fixed $m_1$ and $q=m_2/m_1>0$, $\mu/m_1=q/(1+q)$. The heavy-partner limit $q\to\infty$ gives $\mu\to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu)/\mu=1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.
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