Two bodies interacting through a central force

We'll consider two particles interacting with each other through a conservative force, where the force is directed along the line that connects them: a central force. For example...

• two charged particles interacting via the electric force, for example an exciton: an electron and its hole circling each other in a semiconductor.
• A planet and one of its moons--we'll treat each body as a "point particle"--for example Enceladus and Saturn in this image from the Cassini mission.

[Skip to the conclusions]

Gravitational (central) force

The gravitational force is one kind of such a central force. In terms of the positions of the two particles $\myv{r}_1$ and $\myv{r}_2$ the gravitational energy can be written $$U(\myv{r}_1,\myv{r}_2) = -\frac{Gm_1m_2}{|\myv{r}_1-\myv{r}_2|} = U(r).$$

We already discussed conservative, central forces, where we found that they must be spherically symmetric. This central force is a function of the magnitude of the relative position between the two particles $$\myv{r} = \myv{r}_1-\myv{r}_2$$

The Lagrangian for this system is then $${\cal L}=\frac{1}{2}m_1\dot{r}_1^2 + \frac{1}{2}m_2\dot{r}_2^2 - U(r).$$

Center of mass / relative coordinates

We saw previously that the center of mass $\myv{R}$ of two particles moves under the influence of external forces as if it were one particle with the total mass $M=m_1+m_2$ concentrated at $\myv{R}$. The center of mass lies on the line connecting the two masses: $$\myv{R} = \frac{m_1 \myv{r}_1+m_2 \myv{r}_2}{m_1+m_2} = \frac{m_1 \myv{r}_1+m_2 \myv{r}_2}{M}.$$

If there are no external forces,

• $\dot{\myv{P}} = M \ddot{\myv{R}} = 0$.
• $\Rightarrow$ a reference frame in which $\myv{R}$ is at rest, is an inertial reference frame.

For our other generalized coordinate we'll use the relative position vector $\myv{r} = \myv{r}_1-\myv{r}_2$. Using equation (4) we can show $$\myv{r}_1=\myv{R}+\frac{m_2}{M}\myv{r};\ \myv{r}_2=\myv{R}-\frac{m_1}{M}\myv{r}.$$

Re-write the kinetic energy in terms of $\myv{R}$ and $\myv{r}$: $$\begineq T =& \frac{1}{2}(m_1\dot{\myv{r}}_1^2 + m_2\dot{\myv{r}}_2^2)\\ =& \frac{1}{2}\left(M\dot{\myv{R}}^2 + \frac{m_1 m_2}{m_1+m_2}\dot{\myv{r}}^2\right)\\ =&\frac{1}{2}\left(M\dot{\myv{R}}^2 + \mu \dot{\myv{r}}^2\right) \endeq$$

Where the reduced mass, $\mu$, is defined as

$$\mu \equiv \frac{m_1m_2}{M}=\frac{m_1m_2}{m_1+m_2}$$

The Lagrangian is: $$\begineq{\cal L} = T - U =& \frac{1}{2}M\dot{\myv{R}}^2 + \left[\frac{1}{2}\mu \dot{\myv{r}}^2 - U(r)\right]\\ &\equiv&{\cal L}_{CM} + {\cal L}_{r e l}\endeq $$

Applying the Euler-Lagrange equation, $\frac{\del {\cal L}}{\del R_i}=\frac{d}{dt}\frac{\del {\cal L}}{\del \dot{R}_i}$ for for the 3 components of $\myv R$ leads to: $$M \ddot{\myv{R}} = 0.$$ This is the equation of motion of a free particle not subject to any potential energy. The solution is: $$\dot{\myv{R}}= C o n s t,$$ that is, the velocity of the center of mass is constant.

By the way, this is an example of a system with an ignorable coordinate:
• The Lagrangian does not depend explicitly on any of the 3 coordinates $R_i$. Such coordinates are called "ignorable".
• The "generalized force" associate with each of the 3 $R_i$ is $-\frac{\del {\cal L}}{\del R_i}=0$.
• The "generalized momentum" associated with $R_i$ is $\frac{\del {\cal L}}{\del \dot{R}_i}$.
• This generalized momentum, $M\dot{R}_i$ in this case, is conserved when the generalized force is 0.

Shift to a reference frame moving at a constant velocity (same as the CM), which is an inertial fram. Now, our system's Lagrangian is just $${\cal L} = {\cal L}_{r e l} = \frac{1}{2}\mu\dot{\myv{r}}^2 - U(r).$$

This is the same as the Lagrangian for a single particle...

• of mass $\mu$
• with coordinate $r$
• moving in the field of a fixed (at the origin) central force $U(r)$.

So, it's as if we've exchanged our two-body problem for a one-body problem.

In this CM ref frame, the center of mass is not moving, so $\myv P_{\text{tot}}(t)=0$ which implies that $$m_1v_1 = -m_2v_2.$$

Conservation of angular momentum

Since $\myv{R}=0$ in our center-of-mass frame, the expressions for the positions of the particles are now

$$\myv{r}_1=\frac{m_2}{M}\myv{r};\ \ \myv{r}_2= -\frac{m_1}{M}\myv{r}$$

We'll use these in evaluating the total angular momentum, which must be constant (no external forces $\Rightarrow$ no external torques), and can be expressed as $$\begineq \myv{L} =&\myv L_1+\myv L_2= \myv{r}_1 \times \myv{p}_1 + \myv{r}_2 \times \myv{p}_2 \\ & =&m_1 \myv{r}_1 \times \dot{\myv{r}}_1 + m_2 \myv{r}_2 \times \dot{\myv{r}}_2 \\ =& m_1\frac{m_2^2}{M^2}\myv{r} \times \dot{\myv{r}} + m_2\frac{m_1^2}{M^2}\myv{r} \times \dot{\myv{r}} \\ =& \frac{m_1m_2}{M^2} \myv{r} \times \dot{\myv{r}}[m_1+m_2]\\ =&\mu \myv{r} \times \dot{\myv{r}} = \myv{r} \times \mu \dot{\myv{r}}\endeq$$

This is the same as the angular momentum of a single particle of mass $\mu$.

And since $\myv L$ is conserved, the direction of $\myv{r} \times \dot{\myv{r}}$ is not changing, so $\myv{r}$ and $\dot{\myv{r}}$ will remain co-planer.

Therefore our single-body problem in 3 dimensions is now just a single-body problem in 2 dimensions.

Conclusions

Consider a system of two particles with positions $\myv r_1$ and $\myv r_2$, and masses $m_1$ and $m_2$, we've shown that we can describe the system instead in terms of the position of the center of mass, $\myv R$, and the relative coordinate $\myv r\equiv\myv r_1-\myv r_2$:

If there are no external forces, then:

• the equation of motion for the center of mass is $\ddot{\myv R}=0$,
• so the center of mass moves with a constant velocity,
• A reference frame attached to the CM is an inertial reference frame.

Transforming to the CM reference frame, we find that the Lagrangian of the system, in terms of the relative coordinate, is $${\cal L}_\text{rel} = \frac{1}{2}\mu \dot{\myv{r}}^2 - U(r)$$ that is to say, the same as a particle with the "reduced mass", $\mu$ $$\mu=\frac{m_1m_2}{m_1+m_2}$$ subject to a central force, $U(r)$, relative to a fixed point in space.