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Velocity transformation

  1. Apr 22, 2007 #1
    When textbooks derive the transformation of the Y(Y') component of the velocity of a particle, take into account that y=y' and use the time transformation depending only on the X(X') component of the parfticle's velocity and on the relative velocity as well. Please tell me why?
  2. jcsd
  3. Apr 23, 2007 #2
    A more general choice of the two coordinate frames' relative orientation and a more general relative velocity (rather than along a common X,X' axis) will not change the physics (what happens where and when). Time dilation and length contraction still appear. But the computations that show this will be more complicated.
  4. Apr 23, 2007 #3
    Thanks. I have in mind only the case of the standard arrangement.
  5. Apr 24, 2007 #4
    In that case I don't understand what your question requires. Certainly not why the velocity is chosen along X, X' rather than Y, Y'. Please explain.
  6. Apr 24, 2007 #5
    Consider the standard arrangement of the involved inertial reference frames. The derivation of the LT starts with the transformation of the x(x') space coordinates and with the derivation of the time transformation
    t=g(t'+Vx'/cc). Extending the problem to two space dimensions we add y=y' and derive the transformation of the Y(Y') component of the speed using the time transformation derived in the one space dimension approach considering that the time transformation is not affected by that.
    Thanks for discussing with me the problem.
  7. Apr 24, 2007 #6

    Meir Achuz

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    Are you asking why y=y'?
  8. Apr 24, 2007 #7
    What I am asking for is:
    Starting from an one space dimension approach, textbooks derive the transformation for the x(x') and t(t') space and time coordinate . Going to two space dimensions it is stated that y=y' (a direct consequence of the principle of relativity) and in order to derive the transformation of the Y(Y') component of the velocity the time transformation derived in the case of one space dimension is used. Is there an explanation for that?
    Because all events characterized by the same x coordinate are simultaneous? If so, should that fact be mentioned?
  9. Apr 25, 2007 #8
    I'm not sure I understand Bernhard's question either, but will try.
    The velocity of the observer relative to light only occurs in one direction,
    and is arbitrarily selected as x because this choice simplifies the transformation equations. If a random vector other than x was selected,
    the observers velocity would have components in the y and z axes.
  10. Apr 25, 2007 #9
    Since Minkowski spacetime is an affine space it really does not matter how to describe parallel velocities. Any single direction or origin can be chosen.
    In case of velocities that are not parallel the situation is more complicated and not even associative and commutative but one can still select any principle direction and origin.
    Last edited: Apr 26, 2007
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