Cyclic co-ordinates implying the constant velocity motion of center of mass of a system of particles

Question

I'm reading the section on Central Force in my textbook (Goldstein's Classical Mechanics has a similar argument in the chapter titled "The Central Force Problem", first section), where we have the following:

The Lagrangian for the system of two particles is found to be $$L=\frac{1}{2}M \dot R^2+\frac{1}{2}\mu \dot r^2-V(r)$$ where $R$ is the position vector of the center of mass of the particles.

The textbook says that since the three components of $R$ do not appear in the Lagrangian, they are cyclic.

(My first question is : Is it referring to the fact that $L$ is not a function of $(x,y,z)$? What about the $V(r)$ term. This introduces a position-dependence, doesn't it?)

We continue "..(therefore) the center of mass is either at rest or moving at a constant velocity, and we can drop the first term of the Lagrangian in our discussion. The effective Lagrangian is now given by $$L=\frac{1}{2}\mu \dot r^2-V(r)$$

"(end quote)

I don't quite see how we conclude that the center of mass is either at rest or moving at a constant velocity based on the fact that $L$ is not a function of ($x,y,z$).

Answer 1

The three components of $R$ indeed do not appear in the Lagrangian; $V(r)$ is a function of only $r$ (i.e. the distance between the particles). If $V$ were a function of $R$ it would imply the presence of some external field and you wouldn't be dealing with the same two-body problem anymore.

That the center of mass is either at rest or moving at a constant velocity can easily be seen from the Euler-Lagrange equations for the original Lagrangian. For $R$ E-L equation reads: $$ \ddot{R} = 0 . $$ In fact, in the absence of external forces, the center of mass of a system is always at rest or moving at a constant velocity.