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Adding a total derivative to the Lagrangian

  1. Dec 16, 2013 #1
    I recently posted another thread on the General Physics sub forum, but didn't get as much feedback as I was hoping for, regarding this issue. Let's say I have two Lagrangians:
    $$ \mathcal{L}_1 = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\mu A^\mu)^2 $$
    $$ \mathcal{L}_2 = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\nu A_\mu)(\partial^\mu A^\nu) $$
    They refer to the same Maxwell in a vacuum field theory, so they give rise to the same equations of motion. As such they should differ only by a total derivative. I can test if something is a total derivative by plugging it into the Euler-Lagrange equations, and see if I get something that can be reduced to [itex]0\,=\,0[/itex]. In this case, there should be some vector [itex]B^\mu[/itex] such that the difference between the two Lagrangians, is equal to [itex]\partial_\mu B^\mu[/itex], right? My question is, how can you find [itex]B^\mu[/itex] (up to the addition of a divergence-less term)?
     
  2. jcsd
  3. Dec 16, 2013 #2

    dextercioby

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    Since the first 2 terms are identical, you need to show that the second term of L1 = L2 + 4 div, thing which I find trivial.
     
  4. Dec 16, 2013 #3
    I'm sorry, but I'm just not finding it trivial. Could you write it out for me, please?
     
  5. Dec 16, 2013 #4

    ChrisVer

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    or you can just try the Lorentz gauge... for the 1st
    For the second you just need to do a partial derivative, and use again the Lorentz gauge...
    You'll have the same lagrangians, differing by a total derivative term
     
    Last edited: Dec 16, 2013
  6. Dec 17, 2013 #5

    Bill_K

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    No, you cannot impose gauge conditions on the Lagrangian!


    Aμ,νAν,μ = (Aμ,νAν) - Aμ,νμAν = (Aν,μAμ) - Aμ,νμAν = (Aν,μAμ - Aμ,μAν) + Aμ,μAν,ν
     
    Last edited: Dec 17, 2013
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