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How many frames of total 0-momentum in the Minkowski plane?

  1. Apr 2, 2014 #1
    Hello to everybody,

    the question seems trivial in my mind, yet, is it legal to say that there is not unique frame of 0 total momentum in the Minkowski spacetime plane?

    I am thinking of two non-accelerating equal masses on a horizontal plane, one is moving horizontally, the other perpendicularly as their respective 2-velocities in the Minkowski spacetime plane indicate. A quick way to find the frame of 0-momentum, one either moves along with the horizontal moving mass and Lorentz-transforms the velocity of the other mass, or does it by following the perpendicularly pointed mass.

    These two separate methods yield two different results as to the direction of the respected 2-velocities of the masses in the Minkowski plane, very much unlike the Newtonian case where they're the same regardless.
    So in this case, there must be two frames of 0-momentum. Is this legal to say, or they both cannot be told apart?
     
  2. jcsd
  3. Apr 2, 2014 #2

    stevendaryl

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    Unless the total energy is zero, there is a unique frame in which the total momentum is zero. Suppose that in one frame, the total momentum is in the x-direction. Then in another frame, the momentum in the x-direction will be given by:

    [itex]p' = \gamma (p - v E/c^2)[/itex]

    where [itex]v[/itex] is the relative speed between the two frames (assuming the velocity is in the x-direction), and where [itex]E[/itex] is the total energy in the original frame and [itex]p[/itex] is the momentum. That equation uniquely determines the value of [itex]v[/itex] to make [itex]p'=0[/itex]: [itex]v = p c^2/E[/itex].

    If the momentum is not in the x-direction, you can first perform a Lorentz transformation to get the momentum zero in the x-direction, then perform a second transformation to get it zero in the y-direction, then a third to get it zero in the z-direction. The resulting frame is uniquely determined by the momentum and energy in the initial frame.
     
  4. Apr 2, 2014 #3

    Meir Achuz

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    haruna: Both your methods are wrong.
    Add the two momenta vectorially so P=p1+p2, and E=e1+e2.
    Then Lorentz transform with v=-P/E.
     
  5. Apr 2, 2014 #4
    Yes, thank you both. I can see my mistake now, tho not for the reason you've just pointed out.

    You see, when i move along one of the two masses bringing its speed to a standstill, I add the momenta vectorially by transferring the one component of momentum to the other moving mass.

    My mistake was when I Lorentz-transformed with v=-P/E, I forgot to rotate the velocity 2 or (3-vectors in general case) to point both in the x-direction. I forgot to rotate the Minkowski plane in the direction θ of v where v=tanhθ, and was left with the same velocity directions as before the transformation.
     
  6. Apr 2, 2014 #5
    Oops. This is not right! The correct statement is: I forgot to rotate the velocity 3 or (4-vectors in the general case) to point both in the x-direction.
     
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