# The center of mass frame and lab frame

I have a very basic question on lab frame and COM frame. I assume that the lab frame is the frame where we perform experiments. The definition says

"The frame of reference in which particle 2 is initially at rest is known as the Lab Frame."

But why is that so? Why cannot both the particles be moving? So does it mean if both of them are moving initially, we cannot perform any mathematical treatments in lab frame?

Also, I would like to quote a line from wikipedia's article on Center of Momentum Frame. While walking through the 2 body system numerically, it says:

"The situation is analyzed using Galilean transformations and conservation of momentum (for generality, rather than kinetic energies alone), for two particles of mass m1 and m2, moving at initial velocities (before collision) u1 and u2 respectively. The transformations are applied to take the velocity of the frame from the velocity of each particle from the lab frame (unprimed quantities) to the COM frame (primed quantities):

u1' = u1 - V and u2' = u2 - V

where V is the velocity of the COM frame.

What does it mean by "velocity of COM frame"?

## 2 Answers

Both particles can be moving but often there is a stationary target particle at which "probing" particle are sent e.g. as in the Rutherford scattering experiment where alpha particles are fired at gold atoms.

"velocity of COM frame" is the velocity of the centre of mass of the system under consideration relative to the laboratory frame.
Relative to the centre of mass frame the total momentum of the system under consideration is zero.

• I got your point, but still what is bothering me is the definition of the lab frame, which says the frame in which the particle 2 is at rest. It seems as if it is required that the particle 2 be at rest and if it is not, lab frame is not a valid idea. – Patrick Sep 20 '17 at 17:09
• @Patrick Imagine that the particle at rest is on a laboratory bench. On the bench there are three axes which ar at rest relative to the bench and stay at rest relative to the bench. That is the lab frame. If particle 2 subsequently starts to move it will be moving relative to the lab frame (the bench with the axes on it). – Farcher Sep 20 '17 at 17:23
• Oh that helps a bit. So, does it make sense to ask "with what velocity does one frame move relative to the other?" – Patrick Sep 20 '17 at 17:55
• Yes. The velocity of he CoM frame relative to the lab frame is often the first calculation to be made. – Farcher Sep 20 '17 at 18:14
• Maybe I am just confused, but why are we talking about velocity of the frames themselves? Shouldn't we just be concerned about particles' motion in the frames? – Patrick Sep 20 '17 at 18:24

A picture might help you understand the concept of the center of mass frame and the velocities of the particles relative to the CM frame. In this 2d picture, the two frames are parallel to each other... that doesn't have to be the case and the equations generalize to the case of non-parallel 3d coordinate systems. ${\vec r'_1}$ is the position vector of particle 1 in the CM frame (in other words, if the center of mass is located at the origin of the center of mass frame, ${\vec r'_1}$ is the position vector of particle 1 relative to the center of mass). ${\vec r_1}$ is the position vector of particle one relative to the origin of the lab frame.

From the image... $${{\vec r_1}} = {{\vec R}_{cm}} + {{\vec r'_1}}$$ $${{\vec r'_1}} = {{\vec r_1}} - {{\vec R}_{cm}}$$

Taking the derivative of both sides of the bottom equation: $$\underbrace {\frac{{d{{\vec r'_1}}}}{{dt}}}_{{{\vec u'_1}}} = \underbrace {\frac{{d{{\vec r_1}}}}{{dt}}}_{{{\vec u_1}}} - \underbrace {\frac{{d{{\vec R}_{CM}}}}{{dt}}}_{{{\vec V_{CM}}}}$$

$${{\vec u'_1}} = {{\vec u_1}} - {{\vec V_{CM}}}$$ Using the same steps... we could get the velocity, $\vec{u'_2}$, of particle 2 relative to the CM frame...

If you're still a little bit confused about what $V_{CM}$ represents, this video might be helpful for you. MIT Classical Mechanics: 4.3 - Reference Frames.