- #1

- 95

- 4

## Homework Statement

For the Lagrangian

[tex]L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}[/tex]

show that

[tex]\Pi^{\sigma\rho} = \frac{\partial L}{\partial(\partial_{\sigma}A_{\rho})}=-F^{\sigma\rho}[/tex]

Hence show that the energy-momentum tensor

[tex]T_{\nu}^{\mu} = \Pi^{\mu\sigma}\partial_{\nu}A_{\sigma}-\delta_{\nu}^{\mu}L[/tex]

can be written as

[tex]T^{\mu\nu} = -F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\frac{1}{4}g^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}[/tex]

We symmetrize the tensor by adding

[tex]\partial_{\lambda}X^{\lambda\mu\nu} \text{ with } X^{\lambda\mu\nu}=F^{\mu\lambda}A^{\nu}[/tex]

Show that the new tensor is:

[tex]\hat{T}^{\mu\nu}=F^{\mu\sigma}F_{\sigma}^{\nu} + \frac{1}{4}g^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}[/tex]

## Homework Equations

[tex]F_{\alpha\beta} = \partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}[/tex]

[tex]F^{\alpha\beta} = \partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha}[/tex]

## The Attempt at a Solution

[tex]\Pi^{\sigma\rho} = \frac{\partial L}{\partial(\partial_{\sigma}A_{\rho})}[/tex]

[tex]\Pi^{\sigma\rho}=-\frac{1}{4}\frac{\partial}{\partial(\partial_{\sigma}A_{\rho})}\left([\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}]F^{\alpha\beta}\right)[/tex]

[tex]\Pi^{\sigma\rho}=-\frac{1}{4}\left([\frac{\partial(\partial_{\alpha}A_{\beta})}{\partial(\partial_{\sigma}A_{\rho})}-\frac{\partial(\partial_{\beta}A_{\alpha})}{\partial(\partial_{\sigma}A_{\rho})}]F^{\alpha\beta}\right)[/tex]

[tex]\Pi^{\sigma\rho}=-\frac{1}{4}\left(\delta_{\sigma}^{\alpha}\delta_{\rho}^{\beta}-\delta_{\sigma}^{\beta}\delta_{\rho}^{\alpha}\right)F^{\alpha\beta}

[/tex]

[tex]\Pi^{\sigma\rho}=-\frac{1}{4}(F^{\sigma\rho}-F^{\rho\sigma})[/tex]

[tex]\Pi^{\sigma\rho}=-\frac{1}{4}(F^{\sigma\rho}+F^{\sigma\rho})[/tex]

[tex]\Pi^{\sigma\rho}=-\frac{1}{2}F^{\sigma\rho}[/tex]

So I don't know where the missing factor 2 should appear in my result. I am not used to taking such derivatives but it seems obvious to me that it must give a delta function because only one of the terms does not vanish. And I don't see where is my mistake. Unless the tensor with upper indices is not a constant for this operator of derivation.

Next, let's assume I have the correct result...

[tex]T_{\nu}^{\mu} = \Pi^{\mu\sigma}\partial_{\nu}A_{\sigma}-\delta_{\nu}^{\mu}L[/tex]

[tex]T_{\nu}^{\mu} = -F^{\mu\sigma}\partial_{\nu}A_{\sigma}+\frac{1}{4}\delta_{\nu}^{\mu}F_{\alpha\beta}F^{\alpha\beta}[/tex]

Here it looks close to the expected result but I don't understand how the indices are rised. And what is the point of moving nu from a lower index to the top?

For the last calculation, one term is the same, so I work the other one:

[tex] -F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda}A^{\nu})=-F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda})A^{\nu}+F^{\mu\lambda}\partial_{\lambda}(A^{\nu})[/tex]

[tex] -F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda}A^{\nu})=-F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\sigma}(F^{\mu\sigma})A^{\nu}+F^{\mu\sigma}\partial_{\sigma}A^{\nu}[/tex]

[tex] -F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda}A^{\nu})=F^{\mu\sigma}(-\partial^{\nu}A_{\sigma}+\partial_{\sigma}A^{\nu}) +\partial_{\sigma}(F^{\mu\sigma})A^{\nu}[/tex]

[tex] -F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda}A^{\nu})=F^{\mu\sigma}F_{\sigma}^{\nu} +\partial_{\sigma}(F^{\mu\sigma})A^{\nu}[/tex]

This time nu seems to naturally come as an upper index. But what about my second term? It is not supposed to exist? I simply derived a product and I don't think this term vanishes because I remember it should define a current and generate Maxwell's equations from another problem I did.

I appreciate any comment and help.