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Find G_{\mu\nu} from this lagrangian.

  1. May 22, 2015 #1
    I am reading the book Supergravity.

    In chapter 4, section 4.3.2

    - Duality for gauge fields and complex scalar:

    The simplest case of electromagnetic duality in an interacting field theeory occurs with one abelian gauge field ##A_{\mu}(x)## and a complex scalar field ##Z(x)##. The electromagnetic part of the Lagrangian is:

    The author said:

    The gauge Bianci identity and E.OM of our theory are:
    $$\partial_{\mu}\tilde{F}^{\mu\nu}=0, \hspace{2cm} \partial_{\mu}[(ImZ)F^{\mu\nu} +i(ReZ)\tilde{F}^{\mu\nu}]=0$$

    He continued to say:
    $$G^{\mu\nu}=\epsilon^{\mu\nu\rho\sigma}\frac{\delta S}{\delta F^{\rho \sigma}}=-i(ImZ)\tilde{F}^{\mu\nu}+(ReZ)F^{\mu\nu}$$

    My question is I tried to carry on the calculations as I moved with the reading, but I failed in deriving the ##G^{\mu\nu}## and I am not getting this final result because I am new to tensors and am trying to learn GR and SUGRA simultaneously and want to get a better picture of how to work this out before I move to more advanced levels.. Any sort of help is appreciated.
  2. jcsd
  3. May 27, 2015 #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?
  4. May 29, 2015 #3


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    First, do you know what we get when we contract the indices in [itex] \epsilon^{{\mu\nu\rho\sigma}} F_{\mu \nu} [/itex] ? And do you know what we get when contracting two epsilon tensors?
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