Conservation of linear momentum at relativistic speeds

  1. The question I ask is linear momentum conserved in
    in instance cited below?


    You place a particle at the origin on a x-y axis
    and accelerate it to 61% of c in the y direction.
    Then you accelerate it to 61% of c in the x direction.
    The net velocity of the particle will be
    86% of c at 45 degrees.The key here is that it
    takes approximately 3 times the energy to
    accelerate the particle in the x direction than
    the y direction, due to the fact that the net
    velocity change in the y direction is 0%-61% of c
    and in x direction the net velocity change is
    61%-86% of c.If the rate of acceleration,distance
    of acceleration and time of acceleration are the
    same on the x and y axis, then force of acceleration on
    the x axis has to be greater than
    on the y axis, since the energy of acceleration
    on the x axis is 3 times that of the y axis.
    Therefore the momentum on the x axis is greater
    the y axis.

    If the particle's final velocity is 86% of c at
    45 degrees then the momentum of acceleration
    should be equal on both the x and y axis.
    Is there a discrepancy in momentum here?
     
  2. jcsd
  3. Labguy

    Labguy 734
    Science Advisor

    I don't think so. No acceleration above 0.61c is used or provides momentum. The 0.86c is the apparent VECTORED V (in your example). But, I don't think that your example is correct since you seem to be using two-dimensional (graph-paper) trig here, while with excluding time. In V calculations, a Y velocity at 0.61c and an X velocity at 0.61c takes time to reach the "end-point" from where you measure the Hypotenuse. The "actual V" of the vectored triangle cannot exceed the V of the greater of the XV or the YV if time is included. The "apparent" V can though, as is seen in apparent superluminal expansion around some supernova remnants.
     
    Last edited: Oct 7, 2003
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