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Eliminating x in Lorentz Transformation

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    At the bottom of this page:
    http://www.cv.nrao.edu/course/astr534/LorentzTransform.html

    It states;
    Lorentz factor:
    * y =gamma

    How did they eliminate x?





    3. The attempt at a solution


    I'm taking the x value which is y(x'+vt') and substituting this into the x value in the x' equation.

    So I'm getting;

    x' = y^2(x'+vt')-yvt

    Then I've separated the equation so that t is on LHS;

    t = [y^2(x' + vt')] / yv

    If anyone knows how this is done, I'd be very interested in learning! Thank you. :smile:
     
  2. jcsd
  3. Oct 2, 2011 #2

    cepheid

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    Staff Emeritus
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    Gold Member

    Your last expression is wrong, it should be:

    [tex]t = \frac{(\gamma^2 - 1)x^\prime + \gamma^2 vt^\prime }{\gamma v}[/tex]

    Now:

    [tex] \gamma^2 = \frac{1}{1-v^2/c^2} [/tex]

    [tex] = \frac{c^2}{c^2 - v^2} [/tex]

    so that:

    [tex] \gamma^2 - 1 = \frac{c^2}{c^2 - v^2} - \frac{c^2 - v^2}{c^2 - v^2} [/tex]

    [tex] = \frac{v^2}{c^2 - v^2} = v^2 \frac{\gamma^2}{c^2} [/tex]

    Therefore, the above expression becomes:

    [tex]t = \frac{v^2\gamma^2 x^\prime}{\gamma vc^2} + \frac{\gamma^2 vt^\prime}{\gamma v}[/tex]

    [tex]= \frac{\gamma v x^\prime}{c^2} + \gamma t^\prime[/tex]

    [tex] = \gamma(t^\prime + vx^\prime / c^2) [/tex]
     
  4. Oct 4, 2011 #3
    Cheers very much for that.

    I understand it now!

    Thank you :smile:
     
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