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Time in the Lorentz transformation

  1. Aug 22, 2005 #1
    where does t`=(t-vx/c^2)/(1-v^2/c^2)^1/2 come from?
  2. jcsd
  3. Aug 22, 2005 #2


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    [tex] \left( \begin{array}{c} x' \\ t' \end{array} \right) = \left( \begin{array}{cc} cosh(\theta) & -sinh(\theta) \\ sinh(\theta) & cosh(\theta) \end{array} \right) \left( \begin{array}{c} x \\ t \end{array} \right) [/tex]

    where [tex]v = tanh(\theta)[/tex] and c is taken to be 1.
  4. Aug 22, 2005 #3
    There are many elegant ways to derive the Lorentz transformation:

    http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000043000005000434000001 [Broken]
    http://arxiv.org/PS_cache/physics/pdf/0302/0302045.pdf [Broken]
    Last edited by a moderator: May 2, 2017
  5. Aug 22, 2005 #4


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    It should be [tex]+\sinh(\theta)[/tex].
    (The determinant has to be 1.)
  6. Aug 22, 2005 #5


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    Or rather they should both be [tex]-\sinh(\theta)[/tex] I believe.
  7. Aug 22, 2005 #6


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    Yes, of course :tongue2: , considering the original post. Thanks.
  8. Aug 26, 2005 #7
    thanks! :)
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