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

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In summary, the book discusses electromagnetic duality in an interacting field theory with one abelian gauge field and a complex scalar field. The gauge Bianci identity and Euler-Lagrange equations of motion are given, and the author shows that \epsilon^{ {\mu\nu\rho\sigma} } \frac{\delta S}{\delta F^{\rho \sigma}} = -i(ImZ)\tilde{F}^{\mu\nu}+(ReZ)F^{\mu\nu}. The reader asks for help understanding this result.
  • #1
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.
 
  • #3
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 [itex] \epsilon^{{\mu\nu\rho\sigma}} F_{\mu \nu} [/itex] ? And do you know what we get when contracting two epsilon tensors?
 

1. What is the Supergravity Lagrangian?

The Supergravity Lagrangian is a mathematical framework that describes the interactions between gravity and other fundamental forces, such as electromagnetism and the strong and weak nuclear forces. It is an extension of Einstein's theory of general relativity and incorporates principles from supersymmetry, a theoretical framework that proposes a symmetry between fermions and bosons.

2. What is G_{\mu\nu} in the context of Supergravity?

G_{\mu\nu} is a mathematical symbol used to represent the Einstein tensor, which is a mathematical object that describes the curvature of spacetime in Einstein's theory of general relativity. In the context of Supergravity, G_{\mu\nu} is used to describe the curvature of the superspace, which is the space that includes both the ordinary space and the space of supersymmetry.

3. How is G_{\mu\nu} derived from the Supergravity Lagrangian?

G_{\mu\nu} is derived from the Supergravity Lagrangian using the principle of least action, which states that the actual path taken by a system between two points in spacetime is the one that minimizes the action, a mathematical quantity that describes the energy of the system. In the case of Supergravity, the action is calculated by varying the Lagrangian with respect to G_{\mu\nu} and then setting the resulting equations of motion to zero.

4. What is the significance of finding G_{\mu\nu} from the Supergravity Lagrangian?

Finding G_{\mu\nu} from the Supergravity Lagrangian is significant because it allows us to understand the behavior of the gravitational field in the context of supersymmetry. This can provide insights into the unification of all fundamental forces and contribute to the development of a more complete theory of physics.

5. Are there any limitations to using the Supergravity Lagrangian to find G_{\mu\nu}?

Yes, there are limitations to using the Supergravity Lagrangian. One limitation is that it is a purely classical theory and does not incorporate quantum effects. Additionally, the Supergravity Lagrangian is a highly complex mathematical framework and may not accurately describe certain physical phenomena, such as the behavior of black holes. It is important to continue researching and refining the Supergravity Lagrangian to overcome these limitations.

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