Center of Mass Reference Frame

The Center of Mass frame is very convenient for analyzing certain events (collisions in free space). It is defined below, and its use exemplified.

In the discussion below, $\vec{x}_i$, $\vec{v}_i$ and $\vec{p}_i$ refer to the ``lab'' frame (where in general $\vec{P}_{\rm tot} = M_{\rm tot}\vec{V}_{\rm cm} \neq 0$) while $\vec{x}_i'$, $\vec{v}_i'$ and $\vec{p}_i'$ (with primes $'$) refer to the velocity and momentum are in the ``CM'' frame whose origin is at $\vec{X}_{\rm cm} = \vec{V}_{rm cm}t$ (starting the two origins at the same point at $t = 0$).

Recall that in this case the coordinate transformation between two frames (the new one with origin at $\vec{X}_{\rm cm}$) is:

$$\vec{x}_i' = \vec{x}_i - \vec{X}_{\rm cm} = \vec{x}_i - \vec{V}_{\rm cm} t$$ (4.29)

The reason that this is useful is that (differentiating with respect to time once as previously noted in the first few lectures)

$$\vec{v}_i' = \vec{v}_i - \vec{V}_{\rm cm}$$ (4.30)

provided $\vec{V}_{\rm cm}$ is constant in time. This is true provided only that the net external force is zero:

$$\vec{P}_{\rm tot} = M \vec{V}_{\rm cm} = {\rm constant vector}.$$ (4.31)

We thus see that a ``free'' system (one with no net external forces acting) indeed defines an inertial reference frame wherein Newton's laws are expected to hold.

To start, let's apply this idea for just two particles. The ideas are readily extended to $N$ particles, although the algebra and arithmetic can get pretty hairy for complicated problems. First we find the velocities of the particles in the CM frame

$$\vec{v}_1' = \vec{v}_1 - \vec{V}_{\rm cm},\quad \vec{v}_2' = \vec{v}_2 - \vec{V}_{\rm cm}$$ (4.32)

where

$$\vec{P}_{\rm tot} = (m_1+m_2)\vec{V}_{\rm cm} = \vec{p}_1 + \vec{p}_2 = m_1\vec{v}_1 + m_2\vec{v}_2$$ (4.33)

so that

$$\vec{V}_{\rm cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1+m_2}.$$ (4.34)

Now in the CM frame,

$$\vec{P}_{\rm tot}' = \vec{p}_1'+\vec{p}_2' = m_1\vec{v}_1' + m_2\vec{v}_2'$$

$$= m_1\vec{v}_1 + m_2\vec{v}_2 - \left( \frac{m_1+m_2}{m_1+m_2} \right) (m_1\vec{v}_1 + m_2\vec{v}_2)$$

$$= 0!$$ (4.35)

so the CM frame (by definition) is the frame where the total momentum is zero. (Note: See if you can generalize this proof to work for an arbitrary number of particles by summing $\sum_i$ instead of just 1 and 2.) From this we can readily conclude that:

$$\vec{P}_{\rm tot} = M_{\rm tot}\vec{V}_{\rm cm} + \vec{P}_{\rm tot}'$$ (4.36)

or the total momentum in the lab is the momentum of the ``system'' plus its internal momentum (relative to the CM), which sums to zero.

We can thus speak of the internal ``relative'' momentum $\vec{p}'$ in the CM frame as well as total momentum $\vec{P}_{\rm tot}$ of the CM frame. The former adds to zero. The latter describes the uniform motion of the whole frame with no pseudoforces.

A typical problem using the CM frame requires you to transform the problem from the lab frame (where it is difficult) into the CM frame where it is much easier), solve it in the CM frame, and transform the result back into the lab frame. We will see applications of this idea later.

The CM frame (and this problem-solving methodology) is also useful when a uniform force acts on all the masses in the frame. For example, the frame itself can freely ``fall'' under the influence of a uniform external force (like gravity) without destroying the internal relations within the frame! This is because

$$\sum \vec{F}_{\rm ext} = \frac{d \vec{P}_{\rm tot}}{dt} = M_{\rm tot} \frac{d \vec{V}_{\rm cm}}{dt}$$ (4.37)

can be integrated for the trajectory $\vec{X}_{\rm cm}(t)$ just as if it were a particle, in the cases where the total external force and the total mass $M_{\rm tot}$ are known. In this case, as well, one can solve for the motion of the CM as if it were a particle, evaluate the motion in the CM from a knowlege of its internal forces or conservation properties, and transform the combined result back into the lab to find the individual trajectories $\vec{x}_i(t)$ or anything else that might be desired.

This trick ``works'' but is considerably more complicated for rotating systems because of the non-uniform pseudoforces introduced into the rotating frame. We will examine only simple cases of it (for rigid objects, where the rigidity cancels most of the complexity) in a few chapters.