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Question about SR in Griffiths

  1. Aug 13, 2012 #1
    Hi, I'm reading the section about "How the fields transform" in Griffiths' Intro to Electrodynamics book, but I'm a little confused and want to understand this fully.

    He starts off with the example of a parallel plate capacitor where each plate has charge density [itex]\sigma _0[/itex]. So in the rest frame [itex]S_0[/itex] of the plates, the electric field in between the plates is [itex]E_0 = \sigma _0 /\epsilon_0[/itex]. Then he says, let's look at a frame (S) that's moving parallel to the plates (like, along them) at speed [itex]v_0[/itex]. Then, because the length of the plates is contracted in this frame but the charge is constant, the charge density increases in this frame and thus the E field increases to [itex]E = \gamma_0 E_0[/itex].

    So this makes sense. Then he says he wants to look at magnetic fields. So he says, in frame S, there's a magnetic field due to the surface currents on each plate: [itex]K = \sigma v_0[/itex]. I assume these "surface currents" are from the fact that in frame S, it look like the charge on the plates is moving at the speed S is moving with respect to [itex]S_0[/itex]. If this is true, that makes sense too.

    Then he introduces yet another frame, [itex]\overline{S}[/itex], that's moving at speed v relative to speed S. He derives the equations for the fields in this frame in terms of the fields in frame S. But this is what I'm confused about: He looks at two "special cases", when E = 0 and when B = 0, in frame S. But how can either equal zero? I guess B could be 0 if frame S isn't actually moving at all with respect to [itex]S_0[/itex], so it's basically the same as S, but I still can't see how E would be 0. I thought we basically chose to look at S because it necessarily has a nonzero magnetic field.

    If anyone has a copy handy, it's on pages 525-532. Can anyone help me?

    Thanks!
     
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  3. Aug 13, 2012 #2

    Matterwave

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    You are right, there cannot exist a frame where E=0 in this scenario since (E^2-B^2) is an Lorentz invariant and so if there exists some frame where E>B, then one cannot transform away E using a Lorentz transformation (and vice versa).

    As I don't have a copy of the book, I can't really help much more than that, sorry.
     
  4. Aug 13, 2012 #3

    George Jones

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    Griffiths uses two specific physical situations, a parallel plate capacitor and a solenoid, to motivate the transformation equations for general fields. After Griffiths writes down the general transformation equations, he then treats cases that don't necessarily have anything to do with the motivating cases.
     
  5. Aug 13, 2012 #4
    Ok, but he's still referring to some type of frame when he says "When B = 0 is frame S". So what is S here, just any type of frame now?
     
  6. Aug 13, 2012 #5

    George Jones

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    Yes, Griffiths means:

    "Suppose there is an electromagnetic field and an inertial frame S such that in S, B = 0. Then, in another inertial frame S' that moves with velocity v with respect to S, ..."
     
  7. Aug 13, 2012 #6
    Ahhh, ok, thanks!
     
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