Two Particle System

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The two-body problem is one of the most important problems in Classical Mechanics and describes the interaction of two particles. It is solvable analytically in closed form.

Reduced Mass

The two-body problem can be simplified by splitting their motion into the motion of the bodies relative to each other and the motion of their center of mass. The potential of the Lagrangian of the two-body system only depends on the distance $|\mathbf{r}_1-\mathbf{r}_2|$ and is given by

\begin{align} L=\frac{m_1 \dot{\mathbf{r}}_1^2}{2}+\frac{m_2 \dot{\mathbf{r}}_2^2}{2}-U(|\mathbf{r}_1-\mathbf{r}_2|) \end{align}

We define the $\mathbf{r}$ as the distance between the two particles

\begin{align} \mathbf{r}=\mathbf{r}_1-\mathbf{r}_2 \end{align}

If we lay the origin into the center of mass we have \begin{align} m_1 \mathbf{r}_1 + m_2\mathbf{r}_2=0 \end{align} These two equations together give \begin{align} m_1 (\mathbf{r}+\mathbf{r}_2) + m_2\mathbf{r}_2&=0\\ m_1\mathbf{r}&= -(m_1+m_2) \mathbf{r}_2\\ \mathbf{r}_2&=-\frac{m_1}{m_1+m_2}\mathbf{r} \end{align} and similarly \begin{align} \mathbf{r}_1=\frac{m_2}{m_1+m_2}\mathbf{r} \end{align} If we plug in this expressions for $\mathbf{r}_1$ and $\mathbf{r}_2$ into the Lagrangian we get

\begin{align} L= \left( \frac{m_1 \cdot m_2^2}{2 \cdot (m_1+m_2)^2}+\frac{m_2 \cdot m_1^2}{2 \cdot (m_1+m_2)^2} \right) \dot{\mathbf{r}}^2-U(|\mathbf{r}|) \end{align} \begin{align} L= \left( \frac{m_1 \cdot m_2 \cdot(m_1+m_2) }{2 \cdot (m_1+m_2)^2}\right) \dot{\mathbf{r}}^2-U(|\mathbf{r}|) \end{align} The Lagrangian for the reduced mass system simply becomes \begin{align} L= \frac{m \cdot \dot{\mathbf{r}}^2}{2}-U(r) \end{align} with \begin{align} m=\frac{m_1 \cdot m_2}{m_1+m_2} \end{align}