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Relativistic Action

  1. Nov 16, 2006 #1
    I am at rest. I see an object with action A. Someone moves by with
    velocity v. How should I use the Lorentz transform to get the action A' in his frame?
  2. jcsd
  3. Nov 16, 2006 #2
    What do you mean by "I see an object with action A"? This doesn't make much sense to me.
  4. Nov 16, 2006 #3
    Do you mean an "I see an object which follows a path that minimises the action integral A"?
  5. Nov 16, 2006 #4
    Not really - I am trying to treat action on it's own merit. Let's keep the path (dt) or (dx) infinitesimally small.
  6. Nov 16, 2006 #5
    I'm still confused. The action is just a number formed from an integral of some quantity:

    [tex] S = \int L(q,\dot{q}).[/tex]

    If you're interested in an action which is appropriate for, say, a relativistic point particle of mass [itex]m[/itex], then the obvious candidate is

    [tex] S = -m\int d\tau[/tex]

    where [itex]d\tau[/itex] is the infinitesimal proper time along the world-line. The point here is that Lorentz invariance is built in to the action from the start since [itex]d\tau[/itex] is a Lorentz invariant. Because of this you're guaranteed that the equation of motion that you get from varying this action will automatically be Lorentz invariant also.

    Apologies if I'm missing something here, but the question seems trivial.
  7. Nov 16, 2006 #6
    What I am missing is the following: I am at rest. I see an object with energy E. I watch it for time dt. I calculate E*T. A moving person will see the same object with energy E' and the time interval in question is dt'. It is not obvious to me that Edt = E'dt' but I will try to convince myself of that.
  8. Nov 16, 2006 #7
    Is it possible, that you are confusing the coordinate time t and the proper time [tex]\tau[/tex], which is by definition Lorentz invariant (as coalquay404 explained) as it is the square root of the spacetime interval?
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