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Problem proving the uniqueness of the center of momentum frame

  1. Apr 4, 2014 #1
    Hello again,

    Two days ago, I started a thread asking about the same question more or less, and I was thinking that the matter was clear now in my mind, because I had made an error in my calculations...

    Before I begin, I want to admit that my English is not very good, and my exposition to physics is an elementary one, that's what I've learned from high school and one or two courses at the university.

    Wikipedia, http://en.wikipedia.org/wiki/Center-of-momentum_frame says that:
    "In physics, the center-of-momentum frame (zero-momentum frame, or COM frame) of a system is the unique inertial frame in which the center of mass of the system is at rest."

    Unique inertial frame in which the center of mass of the system is at rest, in my mind translates (please correct me if I am wrong) as:
    the constituent velocities of the individual masses of the system, vectorially summing up to the 0-total momentum of the system cannot be re-arranged by not any Lorentzian transformation (rotations of the Minkowski plane) thus resulting in 0-total momentum again, other than the identity transformation.
    I think, this is always true for Galilean transformations.

    I am aware that for the relativistic case, finding the center of momentum is no different than Newtonian mechanics, that is just use the transformation with v=P/E (in geometric units).

    The pesky question which is troubling me is: is this center unique? That is, is there only one inertial frame where the velocities of the masses can be re-arranged (transformed) summing up to 0-momentum?

    For any given system of free moving masses, if we can find any two distinct transformations (other than trivial rotations of the axes with no velocity), both giving us a 0-momentum frame, then this frame is not unique.

    If in relativity this frame by definition, is taken to be the one with the v=P/E transformation, not taking into account any other 0-momentum frames that may exist, or if I have made any mistake in what I have assumed before, then I think there is no point in bothering you and trying to read the rest of my calculations, but if I am not so wrong, would you care to point out any errors in them that I might have done?

    So, I assume geometric units throughout, and start with two masses, one is m moving at speed u in the x-positive direction, the other is M at speed v in the y-positive direction.

    I am trying to find two clearly distinct composite Lorentz transformations, both leading into two-clearly distinct 0-momentum frames.

    First, use the transformation with u. Mass m gets immediately at rest in new frame, M has velocity x-direction component to be -u and y-component to be: v[itex]\sqrt{1-u^{2}}[/itex]. Calculate new total P and E as if the two masses were lying on the line:y=tanθx=-[itex]\frac{v\sqrt{1-u^{2}}}{u}[/itex]x and find the center of momentum. Clearly, the two equal opposite velocities in this 0-momentum frame are tilted at angle θ to the horizontal.

    Second, use transformation with v. Mass M gets immediately at rest in new frame, m has velocity x-direction component to be: u[itex]\sqrt{1-v^{2}}[/itex] and y-component to be: -v.
    Similarly, we can find a new center of momentum and now the two equal opposite velocities are tilted at a different angle φ to the horizontal with tanφ=-[itex]\frac{v}{u\sqrt{1-v^{2}}}[/itex].
  2. jcsd
  3. Apr 4, 2014 #2


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    1. The CoM frame is unique up to rotations. I am still free to rotate my x, y, z, axes as I wish.

    2. Do not forget that the way you've constructed the CoM frame requires two Lorentz boosts in two different directions, first to the rest frame of one particle, and then to the CoM frame itself. Now the tricky part is to remember that two Lorentz boosts in different directions will no longer be a pure Lorentz boost (this is why the boosts don't form a group). The product of two Lorentz boosts not in the same direction is the product of a Lorentz boost plus a rotation. If you take this rotation into account, you will see that you get the same CoM frame (but they might be rotated with respect to each other).
  4. Apr 4, 2014 #3
    Thanks a million! This is it!
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