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Why is the electrodynamic Lagrangian E^2 - B^2?

  1. Jun 11, 2009 #1
    Is there a simple way to understand why the Lagrangian of the classical electrodynamic field is (in SI units)

    E^2/2 e0 - B^2/2 mu0 ?

    Why is there a minus in it? Is there some simple, intuitive explanation for it?

  2. jcsd
  3. Jun 11, 2009 #2


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    One reason: this Lagrangian density has to be a Lorentz-invariant scalar. A plus-sign there would not be Lorentz-invariant.

    It has a resemblance to T-V...
    "E involves time-derivatives of A" is similar to "v involves time-derivatives of x".

    Why this minus sign? A common answer is that it is what gives the correct equations of motion.
  4. Jun 13, 2009 #3
    http://www.hep.caltech.edu/~peck/lecture_EMRelativisticSymmetry.pdf" [Broken]

    This link has a little more detail about what robphy was saying.
    Last edited by a moderator: May 4, 2017
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