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Prove square of four-momentum is relativistic invariant

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi everyone,

    I have a physics assignment that asks: Prove that the square of relativistic four-momentum for a massive particle is a relativistic invariant under Lorentz transformations.

    Can anyone help me to work on the problem? I'm always lost in the class ever since my professor starts teaching modern physics. Any help would be greatly appreciated!

    2. Relevant equations

    p = <γmc, γmv>

    3. The attempt at a solution

    From what I can gather from the question and my understanding, relativistic invariant means that the four-momentum from two different observer should be the same. My best shot on this question is through example from my class notes:

    suppose p1 (four-momentum from observer 1) is <γmc, γmv> and p2 four-momentum from observer 2, travelling at speed v) is <mc, 0>. Therefore when we calculate p^2 for both observer we find it to be (mc)^2

    Again I'm really grateful for any help you could give me
     
  2. jcsd
  3. Jun 6, 2012 #2
    Hi Falken_47. If you want to derive it explicitly, take the square of the 4-momentum [itex]p^i=({\gamma}mc,{\gamma}m{\mathbf{v}})[/itex], apply the Lorentz transformations for a contravariant 4-vector to obtain [itex]p^i{'}[/itex] which moves in a frame at an arbitrary speed V with respect to the original frame. So you should have [itex]p^i{'}[/itex] in terms of non-primes, take the square of this and after some algebra you should arrive at the same solution as you had for the square of [itex]p^i[/itex]. Lorentz Invariant just means that 2 observers in 2 different inertial frames will agree on the quantity which is invariant.
     
  4. Jun 6, 2012 #3
    I think that your method is both elegant and correct.
     
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