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Lorentz transformations

  1. Jun 21, 2005 #1

    What is the signifigance of the first derivative of the Lorentz transformation gamma function with respect to [tex]dv[/tex]?

    What type of system does this derivative represent?

    [tex]\gamma'(v) = \frac{d}{dv} \left( \frac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} \right) = \frac{v}{c^2 \left[ 1 - \left( \frac{v}{c} \right)^2 \right]^\frac{3}{2}}[/tex]

    [tex]\boxed{\gamma'(v) = \frac{v}{c^2 \left[ 1 - \left( \frac{v}{c} \right)^2 \right]^\frac{3}{2}}}[/tex]
  2. jcsd
  3. Jun 22, 2005 #2
    I don't see any usefulness for this particular derivative. Why do you ask?

    might be meaningful. It represents the change of the time-velocity [itex]cd\tau/dt}=c/\gamma[/itex] (see e.g. Brian Greene's "The elegant universe") as a function of the change of the spatial velocity [itex]v[/itex]. The function is goniometric.
    Last edited: Jun 22, 2005
  4. Jun 23, 2005 #3
    Reletive Relation...

    [tex]\gamma'(v)^{-1} = \frac{d}{dv} \left( \frac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} \right)^{-1} = \frac{d}{dv} \left( \sqrt{1 - \left( \frac{v}{c} \right)} \right) = - \frac{v}{c^2 \sqrt{1 - \left( \frac{v}{c} \right)^2 }}[/tex]

    [tex]\boxed{\gamma'(v)^{-1} = - \frac{v}{c^2 \sqrt{1 - \left( \frac{v}{c} \right)^2 }}}[/tex]

    [tex]\gamma'(v)^{-1} = - \frac{v \gamma}{c^2} = - \frac{ds}{dt} \left( \frac{\gamma}{c^2} \right)[/tex]

    [tex]\boxed{\gamma'(v)^{-1} = - \frac{ds}{dt} \left( \frac{\gamma}{c^2} \right)}[/tex]

    Are these equation solutions correct?
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