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Ryder QFT, 2nd Edition

  1. Sep 15, 2007 #1
    We have the Lagrangian of EM field: [tex]L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}[/tex]

    Variation of Lagrangian give Maxwell's equations:

    [tex]\partial_{\mu} F^{\mu\nu}=0[/tex].


    [tex](g_{\mu\nu}\partial_{\mu}\partial^{\mu}-\partial_{\mu}\partial_{\nu})A^{\mu}=0.[/tex] (equation 7.3, p.241)

    Ryder, then, claims that after partial integration, and discarding of surface terms, Lagrangian may be written


    I simply can't figure out this last derivation. Which quantity has to be integrated to give such result? How do we obtain this particular form? I played with various quantities and integrals, but can't prove it..thanx in advance.
  2. jcsd
  3. Sep 15, 2007 #2


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    [tex] L=-\frac{1}{4}\left(\partial_{[\mu}A_{\nu]}\right)\left(\partial^{[\mu}A^{\nu]\right) =\frac{1}{2}\left[\left(\partial_{\nu}A_{\mu}\right)\left(\partial^{\mu}A^{\nu}\right)
    -\left(\partial_{\mu}A_{\nu}\right)\left(\partial^{\mu}A^{\nu}\right)\right] [/tex]

    Now move the derivatives and discard the 4-divergences.
  4. Sep 15, 2007 #3
    Thank you for your answer! Ok, here we go:





    I can see that discarding the first terms of each equation yields the correct result. But why one should discard these terms? I understand that when integrated, it can be converted to a surface integral and thus equal to zero. But why do we act so in this particular form, without integration?
  5. Sep 15, 2007 #4


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    Nope, discarding only takes place when both 4-divergences are under the integral sign . We act without integration and simply write

    [tex]L=L'+ 4div [/tex]

    And by integration we see that L and L' are physically equivalent, since the lagrangian action is the same.
  6. Sep 16, 2007 #5
    Very well then! Thanks again for your help!
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