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Lorentz transformation

  1. Apr 24, 2008 #1
    I need to show that:
    1. if E is [itex]\perp[/itex] to B in one Lorentz frame, it is [itex]\perp[/itex] in all Lorentz frames
    2. [itex]|E|>|cB|[/itex] in L. frame, [itex]|E|>|cB|[/itex] in all L. frame
    3. Angle b/t E and B is acute/obtuse in L. frame, it is acute/obtuse in all L. frame
    4. E is [itex]\perp[/itex] to B but [itex]|E|\neq|cB|[/itex], then there is a frame which the field is purely electric or magnetic

    Attempt:
    1. I believe I just show that [itex] \bar{E} \cdot \bar{B} =E \cdot B[/itex]
    2. I believe I just show [itex]\bar{E}^2-c^2 \bar{B}^2 =E^2-B^2c^2[/itex] so that if [itex]|E|>|cB|[/itex], then [itex]\bar{E}^2-c^2 \bar{B}^2[/itex] is positive and thus [itex]E^2-B^2c^2[/itex] has to be positive, thus, [itex]|E|>|cB|[/itex] in all frames.

    Not too sure where to start for 3 and 4. Open to suggestions, also it would be great if someone could check my approach for 1 and 2. thanks.
     
  2. jcsd
  3. Apr 26, 2008 #2

    tiny-tim

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    Hi LocationX! :smile:

    Your answers to 1 and 2 seem fine. :smile:

    3 should be similar to 1 … what is the sign of E.B if the angle between is acute?

    4: I suggest you start with the simple case of |E| > c|B|, and B and E along the x-direction and y-direction respectively, and see what happens when you transform parallel to the x-direction. :smile:
     
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