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Feynman field transformations

  1. Mar 26, 2013 #1
    I'm trying to follow Feynman's explanation on page 26-10 of Volume 2 of The Feynman Lectures on Physics. He describes the electric and magnetic fields in a FoR S' moving between the plates of a condenser. Feynman writes that we see a reduced E and an added transverse B in S'. I've attached a copy of the figure in question.

    The transformation formula provided in the text is E' = γE, because B = 0 in S. My thinking is that, according to the formula, E' increases. This effect is also consistent with the S length contraction in S'.

    I am confused. Thank you.
     

    Attached Files:

  2. jcsd
  3. Mar 27, 2013 #2
    I have attached additional information in the hope that someone will kindly point out the error in my thinking.
     

    Attached Files:

  4. Mar 27, 2013 #3

    PeterDonis

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    2016 Award

    Staff: Mentor

    You're right, it does. The "reduced" in the text is in error; possibly a typo or a mis-transcription (since the text was originally transcribed from Feynman's actual lectures).

    Another way of seeing that E' has to increase is that [itex]E^2 - B^2[/itex] is an invariant; it's the same in all frames. In frame S, B = 0, so the invariant is just [itex]E^2[/itex], the square of the electric field. In frame S', moving relative to the plates, B' is nonzero, so E' must increase to keep the invariant the same.
     
  5. Mar 27, 2013 #4
    I appreciate your response. Thank you.
     
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