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Extra factor of u in the expression for relativistic momentum?

  1. Oct 27, 2014 #1
    PROBLEM SOLVED - the worked example I was referring too was wrong :/
    ----------------------------

    Hello, I've been stuck on a question in one of my SR problem sets for some time now, and managed to find a worked solution to a similar problem online. I've attached an image of the problem (the missing text just says that two particles of equal mass are accelerated to a speed u).

    It begins by showing that in frame S, both particles have the same energy, which is

    [itex]E = \gamma m u[/itex]

    It then uses the 4-momentum transformations for the second particle (the one that is moving in both frames), noting that

    [itex](p^0)' = \gamma (p^0 - \frac{v}{c} (p^1)')[/itex]

    But then it goes on to define [itex]p^1[/itex] as

    [itex]p^1 = \gamma m u (-u)[/itex]

    This is where I made the mistake in my working, I used

    [itex]p^1 = \gamma m (-u)[/itex]

    I don't understand why the proper momentum of the second particle has an additional factor of u associated with it. Any explanation would be greatly appreciated.

    Cheers.
     

    Attached Files:

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    Last edited: Oct 27, 2014
  2. jcsd
  3. Oct 27, 2014 #2

    Simon Bridge

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