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Electric field in moving frame

  1. May 8, 2012 #1
    I'm trying to understand why the x-component of electric field is the same in the rest frame and the frame moving in the x direction. I thought I should just be able to write the force four vector in the rest frame and transform it. Symbols with arrows are four vectors.

    [tex]
    \vec{p}=\gamma(mv,mc)\\
    \vec{f}=\frac{d\vec{p}}{d\tau}=\gamma(\frac{dp}{dt},\frac{dmc}{dt})
    [/tex]
    For the charge in its rest frame, just looking at the x-component
    [tex]
    \gamma=1\\
    \vec{f}_{x}=\frac{dp_{x}}{dt}=qE_{x}
    [/tex]
    Now I transform to the frame moving with velocity v in the positive x direction
    [tex]
    \vec{f'}_{x}=\gamma'(qE_{x}-\frac{v}{c}\frac{dmc}{dt})
    [/tex]
    So, how does that prove that the x-component of x doesn't change? dmc/dt is not zero as far as I can tell.
     
  2. jcsd
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