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Velocities as seen by different observers.

  1. Aug 20, 2012 #1
    You have two frames (A & B) with each having velocity v with respect to one another. To make things easier, A is at rest with you, and B is moving to the right at v. Also, in frame B, it observes a particle moving to the right at speed u'. We can then find the velocity of this particle as observed by A, u via u = (u'+v)/(1+u'v/c^2).

    From this, we can then calculate the velocity difference of the particle and B as observed by A. This velocity is not what is observed by B, that would be u'.

    Now have u' to the left in B (it is now implicitly "negative").

    Once again calculating the velocity difference in A to find the velocity that A observes the particle to have with respect to B, I find that its magnitude is not equal to that of the first case where the particle was moving to the right in B.

    I can see why this is the case mathematically, but it doesn't seem intuitive. That is, there is only a change in direction of u' in B, but A observes two different velocities with respect to B.


    Can anyone offer me some intuition on this? I have a feeling it has to do with clocks not being synchronized from frame to frame.

    Thanks.
     
  2. jcsd
  3. Aug 20, 2012 #2
    I think your concern ultimately results from this statement. While you can freely calculate the relative velocities between two objects according to some arbitrary frame, this is not a quantity that observers in all frames will agree on (and hence it is not meaningful) while all observers will agree on the relative velocities between the two objects according to their own frames.
     
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