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Electromagnetism and Relativity

  1. Aug 22, 2011 #1
    Hello. Here is the problem:
    A charge q is moving at speed v0 to the right in the lab frame. At t=0 the particle passes at the origin (0,0,0). calculate the field at (-d,0,0) at the instant it passes the origin, both in the lab frame and in the charge frame. are they equal?

    My solution:

    In the lab frame, the field is
    [tex]
    \frac{q}{\gamma^2 d^2}
    [/tex]
    in the negative x direction. By the transformation of parallel fields, [tex]E_{||}'=E_{||} [/tex] so the result in the charge frame should be the same.
    However working in the charge frame,
    [tex]
    E=-\frac{q}{d'^2}, \qquad d'=d/\gamma
    [/tex]
    because of length contraction. therefore the field in the charge frame is
    [tex]
    E=-\frac{q}{(d/\gamma)^2}=\frac{\gamma^2 q}{d^2}
    [/tex]
    these are not equal, the gamma here is in the "wrong place".
    what is going on here?
    thanks.
     
    Last edited: Aug 22, 2011
  2. jcsd
  3. Aug 23, 2011 #2

    vela

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    The spacetime point [itex]x^\mu = (ct, x, y, z) = (0, -d, 0, 0)[/itex] doesn't correspond to the point [itex]x'^\mu = (ct', x', y', z') = (0, -d/γ, 0, 0)[/itex] as you've assumed. You need to use the Lorentz transformations to find the correct [itex]x'^\mu[/itex].
     
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