Find G_{\mu\nu} from Supergravity Lagrangian: Chapter 4, Section 4.3.2

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SUMMARY

The discussion focuses on deriving the expression for \( G^{\mu\nu} \) from the Supergravity Lagrangian as presented in Chapter 4, Section 4.3.2 of the book "Supergravity." The electromagnetic part of the Lagrangian is defined as \( L=-\frac{1}{4}(ImZ)F_{\mu\nu}F^{\mu\nu}-\frac{1}{8}(ReZ)\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma} \). The key equations discussed include the gauge Bianchi identity and the equations of motion, leading to the expression \( G^{\mu\nu}=\epsilon^{\mu\nu\rho\sigma}\frac{\delta S}{\delta F^{\rho \sigma}}=-i(ImZ)\tilde{F}^{\mu\nu}+(ReZ)F^{\mu\nu} \). The user expresses difficulty in deriving \( G^{\mu\nu} \) due to their novice status in tensor calculus and simultaneous study of General Relativity (GR) and Supergravity (SUGRA).

PREREQUISITES
  • Understanding of electromagnetic duality in field theory
  • Familiarity with tensor calculus and index notation
  • Knowledge of gauge fields and complex scalar fields
  • Basic principles of General Relativity (GR) and Supergravity (SUGRA)
NEXT STEPS
  • Study the derivation of the electromagnetic Lagrangian in detail
  • Learn about the Bianchi identity and its implications in gauge theories
  • Explore the contraction of indices in epsilon tensors and their physical significance
  • Investigate advanced topics in tensor calculus relevant to GR and SUGRA
USEFUL FOR

This discussion is beneficial for students and researchers in theoretical physics, particularly those focusing on General Relativity, Supergravity, and field theory. It is especially useful for individuals seeking to deepen their understanding of tensor calculus in the context of gauge theories.

beyondthemaths
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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.
 
beyondthemaths said:
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.
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?
 

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