Editing Reduced mass (section)

=== Newtonian mechanics ===
{{see also|Newtonian mechanics}}

Using [[Newton's second law]], the force exerted by a body (particle 2) on another body (particle 1) is:
<math display="block">\mathbf{F}_{12} = m_1 \mathbf{a}_1</math>

The force exerted by particle 1 on particle 2 is:
<math display="block">\mathbf{F}_{21} = m_2 \mathbf{a}_2</math>

According to [[Newton's third law]], the force that particle 2 exerts on particle 1 is equal and opposite to the force that particle 1 exerts on particle 2:
<math display="block">\mathbf{F}_{12} = - \mathbf{F}_{21}</math>

Therefore:
<math display="block">m_1 \mathbf{a}_1 = - m_2 \mathbf{a}_2 \;\; \Rightarrow \;\; \mathbf{a}_2=-{m_1 \over m_2} \mathbf{a}_1</math>

The relative acceleration '''a'''<sub>rel</sub> between the two bodies is given by:
<math display="block">\mathbf{a}_\text{rel} := \mathbf{a}_1-\mathbf{a}_2 = \left(1+\frac{m_1}{m_2}\right) \mathbf{a}_1 = \frac{m_2+m_1}{m_1 m_2} m_1 \mathbf{a}_1 = \frac{\mathbf{F}_{12}}{\mu}</math>

Note that (since the derivative is a linear operator) the relative acceleration <math>\mathbf{a}_\text{rel}</math> is equal to the acceleration of the separation <math>\mathbf{x}_\text{rel}</math> between the two particles.
<math display="block">\mathbf{a}_\text{rel} = \mathbf{a}_1-\mathbf{a}_2 = \frac{d^2\mathbf{x}_1}{dt^2} - \frac{d^2\mathbf{x}_2}{dt^2} = \frac{d^2}{dt^2}\left(\mathbf{x}_1 - \mathbf{x}_2\right) = \frac{d^2\mathbf{x}_\text{rel}}{dt^2}</math>

This simplifies the description of the system to one force (since <math>\mathbf{F}_{12} = - \mathbf{F}_{21}</math>), one coordinate <math>\mathbf{x}_\text{rel}</math>, and one mass <math>\mu</math>. Thus we have reduced our problem to a single degree of freedom, and we can conclude that particle 1 moves with respect to the position of particle 2 as a single particle of mass equal to the reduced mass, <math>\mu</math>.