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Lorrentz velocity transformation

  1. Sep 3, 2014 #1
    I'm being asked to derive the velocity transformation between vy and vy' and my result isn't exactly matching my goal but I don't know what I'm doing wrong. It's an introductory modern physics course and we're covering special relativity.

    Assume a reference frame S' moving in some constant velocity vx with respect to a stationary frame S. Since time isn't synchronized (dt ≠ dt') then the velocities vy and vy' aren't the same even though y and y' aren't being affected by length contraction (y = y'). The aforementioned statement isn't necessary for the following but I'm just wondering if the above is a correct statement.

    Anyways, this is my logic:
    Since y = y', dy = dy'. I want to know dy/dt'. From the chain rule:

    [tex] \frac{dy}{dt'} = \frac{dy}{dt}\frac{dt}{dt'} [/tex]

    Letting [tex] λ = \frac{1}{(1 - (v/c)^2)^{1/2}} [/tex], the Lorrentz transformation for time is:

    [tex] t' = λ(t - \frac{vx}{c^2}) [/tex]

    Since there is no fundamental difference between two inertial reference frames, the inverse transformation must be of the same form:

    [tex] t = λ(t' + \frac{vx'}{c^2}) [/tex]

    In differential form,

    [tex] dt = λ(dt' + \frac{vdx'}{c^2}) [/tex]

    Now differentiating the above wrt t',

    [tex] \frac{dt}{dt'} = λ(\frac{dt'}{dt'} + v\frac{dx'}{dt'}/c^2) [/tex]

    [tex] \frac{dt}{dt'} = λ(1 + vv_x'/c^2) [/tex]

    Going back to my original goal:

    [tex] \frac{dy}{dt'} = \frac{dy}{dt}\frac{dt}{dt'} = v_y\frac{dt}{dt'} [/tex]

    [tex] \frac{dy}{dt'} = v_y(λ(1 + vv_x'/c^2)) [/tex]

    ...which is wrong. Where am I going wrong with this? Would appreciate some clarity.
     
    Last edited: Sep 3, 2014
  2. jcsd
  3. Sep 4, 2014 #2

    Orodruin

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    Does the object have a velocity component also in the x-direction in S? What is ##v_x'## if ##v_x = v##?
     
  4. Sep 4, 2014 #3
    If I'm understanding correctly, yes, the object would have a velocity component in S. I just realized I should have been a little more clear in my first post.

    The origin O' in S' is moving along with some velocity vx and an observer at rest in S' is observing an object P in S' to be moving in a direction parallel to the direction of S' with velocity u.

    An observer in a stationary frame S would then see P's velocity to be given by the Lorrentz velocity transformation,

    [tex] v_x' = \frac{v_x - u}{1 - \frac{uv_x}{c^2}} [/tex]

    The inverse transform for vx in terms of vx would be obtained by swapping primes and replacing u by -u.

    The more I think about it, the more I get confused though. If P is moving only in the direction parallel to motion of S' and both observers in S and S' agree that y = y', I'm not seeing how it's possible for P to be observed to have a velocity perpendicular to vx in either frame S or S'. I think I'll just have to ask for clarification in lecture because I think I must have misheard or misunderstood something in the premise.

    The result I'm being asked should be

    [tex] v_y' = \frac{v_y(1 - (v/c)^2)^{1/2}}{1 - \frac{uv_x}{c^2}} [/tex]
     
    Last edited: Sep 4, 2014
  5. Sep 4, 2014 #4

    Doc Al

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    I think you have those mixed up. The origin O' is seen to move with velocity u in frame S (parallel to the x axis). The object moves with velocity v'x in S' and velocity vx in S.

    It's not. You're mixing yourself up. If P is moving only in the x' direction according to S', then it will be moving only in the x direction according to S.
     
  6. Sep 4, 2014 #5
    Yes, you're right. Thank you. I switched around vx and u.

    Yeah, I would think not. I'll just ask when I get the chance to clarify the premise. I must have misunderstood or misheard something.
     
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