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

1. May 22, 2015

### beyondthemaths

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:
$$L=-\frac{1}{4}(ImZ)F_{\mu\nu}F^{\mu\nu}-\frac{1}{8}(ReZ)\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}$$

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. May 27, 2015

### Greg Bernhardt

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