For solving problems involving elastic collisions it is useful to use the center of mass reference frame as to avoid quadratic equations. However, I am confused about why the total momentum in the center of mass reference frame is equal to zero. I understand that the change in momentum would be zero, but why is the total momentum equal to zero in this frame?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ That's how the reference frame is defined. $\endgroup$ – pfnuesel Feb 29 '16 at 4:22
-
$\begingroup$ But why is this the case? Is there some reason this holds true if you move with the center of mass? $\endgroup$ – foobar34 Feb 29 '16 at 4:25
-
$\begingroup$ By construction of the center of mass system in our mathematical analysis. $\endgroup$ – anna v Feb 29 '16 at 4:38
Add a comment
|
$\begingroup$
$\endgroup$
Just call it the center of momentum frame instead and it will seem less mysterious.
If you take the energy-momentum four-vector for each part of the system and add them up, you get a total energy-momentum four-vector. It will be timelike. And in the frame of that vector you get zero total momentum.
It's a useful frame. And calling it the center of mass frame just confuses the issue. Technically the center of mass frame is the center of momentum frame with a specific choice of origin, but an origin hardly matters for anything.
$\begingroup$
$\endgroup$
However, I am confused about why the total momentum in the center of mass reference frame is equal to zero.
The center of mass frame of a system of particles is defined as having two key qualities. Less important is that the center of mass at some time $t_0$ is located at the origin. Much more importantly, the time derivative of the center of mass at that time is the zero vector. It is always possible to define such a frame in both classical mechanics and in quantum mechanics. (Note well: This is not so well definable in general relativity, at least on the scale of the universe. I'll leave that aside.)
If no external forces act on a system of particles, and if conservation of linear momentum holds true, then by definition, the center of mass of that system of particles moves at a constant velocity. It is extremely convenient to choose a frame in which that constant velocity is zero. The center of mass frame, by definition, satisfies those constraints (assuming those two very big ifs).
answered Feb 29 '16 at 10:36
