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Lorentz Transformation

  1. Jul 15, 2009 #1
    How do we find d(gamma)/dt?:redface:
  2. jcsd
  3. Jul 15, 2009 #2


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    gamma is [itex]\sqrt{1- v^2/c^2}[/itex]?

    Differentiating that is fairly basic "Calculus I".
    [tex]\gamma= \left(1- \frac{v^2}{c^2}\right)^{\frac{1}{2}}[/tex]
    [tex]\frac{d\gamma}{dt}= \frac{1}{2}\left(1- \frac{v^2}{c^2}\right)^{-\frac{1}{2}}\left(2\frac{v}{c^2}\right)[/tex]

    [tex]= \frac{v}{c^2\sqrt{1- \frac{v^2}{c^2}}}= \frac{v}{\sqrt{c^2- v^2}}[/tex]
  4. Jul 15, 2009 #3


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    HallsOfIvy, I think you differentiated with respect to v, instead of t.
    For the t derivative, use the chain rule
    [tex]\frac{d\gamma}{dt} = \frac{d\gamma}{dv} \frac{dv}{dt}[/tex]

    And I didn't check, but you might have missed a minus sign (it's -v^2 giving -2v isn't it?)
  5. Jul 15, 2009 #4


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    Except [tex]\gamma=\frac{1}{\sqrt{1-v^2/c^2}}[/tex]
    which I'm sure you knew before you answered the question.
  6. Jul 15, 2009 #5
    the factor [tex]\frac{dv}{dt}[/tex]
    isn't missing in the equation?

    [tex]\frac{d\gamma}{dt}= \frac{1}{2}\left(1- \frac{v^2}{c^2}\right)^{-\frac{1}{2}}\left(2\frac{v}{c^2}\right)[/tex]
  7. Jul 15, 2009 #6


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    It is, along with a minus sign. See CompuChip's post.
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