# Why is linear momentum not conserved for a particle in a central force? [duplicate]

I am studying a two-body central force system in which the two particles, one of mass $m$ and one of mass $M$, experience a force directed along the line connecting the two particles.

We can reduce this to a system of just one fictitious particle with reduced mass and a central force. Why is linear momentum not conserved in the CM frame when it was conserved before shifting to the CM frame?

• Why should linear momentum be conserved in the CM frame with a fictitious particle? – user93237 Jun 21 '18 at 1:57

Suppose that I know that $$p$$ is conserved and $$q$$ is not. Now suppose I define the new variable $$p' = p + q.$$ If you weren't paying attention, you might conclude that $$p'$$ is also conserved, because it has the same letter as the conserved quantity $$p$$. But that's clearly wrong. The similarity is just superficial. Not every variable whose name looks like "$$p$$" has to be conserved.
Similarly, in the two-body problem, the conserved linear momentum is $$\mathbf{p} = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2.$$ To work in the reduced mass picture, we define, among other things $$m_{r} = \frac{m_1 m_2}{m_1 + m_2}, \quad \mathbf{v} = \mathbf{v}_1 - \mathbf{v}_2.$$ Your question is why the quantity $$\mathbf{p}' = m_{r} \mathbf{v}$$ is not conserved, while $$\mathbf{p}$$ is. But $$\mathbf{p}'$$ has nothing to do with the total linear momentum $$\mathbf{p}$$, the resemblance is totally superficial. There is no reason to expect that $$\mathbf{p}'$$ should be conserved.
Incidentally, the linear momentum is conserved in the reduced mass picture. The point of reduced mass is that the relative velocity between two bodies interacting by a central force, with masses $$m_1$$ and $$m_2$$ can also be computed by finding the relative velocity between two bodies with masses $$m_r$$ and infinity attracting by the same force; this second mass stays fixed at the origin. The total linear momentum in this one-body problem is the sum of $$\mathbf{p}'$$ and the momentum of the second mass, and is indeed conserved. However, it has nothing to do with the total linear momentum in the original two-body problem.