Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
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?
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.
