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Local construction of Maxwell's action

  1. Dec 10, 2014 #1
    What do we mean by "there is no local construction for an action in terms of [itex]F^{\mu\nu}[/itex], or E and B"?

    So, I understand the construction "on-shell", once we solve Maxwell's equations to find [itex]F^{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}[/itex], and how we can then write an action which is both gauge and Lorentz invariant.

    But before we know anything about gauge symmetry, which comes about once we solve for F and express it in terms of A, what tells us that our action in not good enough (i.e. there is no local action to be constructed)?

    Thanks
     
  2. jcsd
  3. Dec 15, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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