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Particle´s acceleration respect two inertial frames

  1. Oct 1, 2016 #1
    1. The problem statement, all variables and given/known data
    System S' moves with constant speed v=(vx,0,0) respect to the system S. On the S' system a particle moves with a constant acceleration a=(ax,ay,az).
    What is the acceleration a'=(ax',ay',az') measured from the system S?.

    2. Relevant equations
    Lorentz transformation

    3. The attempt at a solution

    I use the velocity-addition formula and find the ecuations of the velocity transformation, I think maybe a derivation respect to proper time of the velocity transformation ux',uy',uz' give the aceleration of the particle, but im not really sure how to do it.


     
  2. jcsd
  3. Oct 1, 2016 #2

    TSny

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    The notation is confusing. You have the unprimed a for the acceleration as measured in the primed frame S, and you have the primed a' for the acceleration as measured in the unprimed frame S.
     
  4. Oct 1, 2016 #3
    I know, but I´m using the same notation that it´s used in the problem. Solving the problem I´m finding the unprimed acceleration as its seen from the system S.
     
  5. Oct 1, 2016 #4

    TSny

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    Well, it's confusing to me. But let's see how it goes.

    I agree that starting with the velocity transformation formula is a good idea. I'm not sure why you would want to differentiate with respect to proper time. The acceleration of the particle in frame S would be the derivative of the velocity of the particle in frame S with respect to time as measured in frame S. Similarly for frame S'.
     
  6. Oct 3, 2016 #5
    I talk with the teacher and he made some corrections:
    System S' moves with constant speed v=(vx,0,0) respect to the system S. On the S' system a particle moves with a constant acceleration a'=(ax',ay',az').
    What is the acceleration a'=(ax,ay,az) measured from the system S?.

    In this case I used the inverse lorentz transformation cause its from the S system and obtain something like
    Code (Text):
    [ tex ]ux=\frac{u_x'+v}{1+frac{u_x' v}{c^2}}[ /tex ]
    and
    Code (Text):
    uy=\frac{uy'}{ \gamma(1+ ux'v/c^2)}
    To find the acelleration we derive respect to dt, as you said, I was a bit confused, but it depents of proper time τ

    a=du/dt=du/dτ dτ/dt

    but how i use this equation with the velocitiy formulas we found before?
     
    Last edited: Oct 3, 2016
  7. Oct 3, 2016 #6

    TSny

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    When finding ##a_x = \frac{du_x}{dt}##, you can work with differentials. Thus ##a_x## equals the differential ##du_x## divided by the differential ##dt##.

    For ##du_x##, use your formula above. For ##dt##, use the appropriate Lorentz transformation equation.
     
  8. Oct 3, 2016 #7
    Thanks, I was complicating myself, I finally obtained the three acceleration transformations.

    Thanks for your help!!!
     
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