# Some trouble with relativistic notations

Hello, I am trying to solve a problem and I have trouble with almost every part of it.

## Homework Statement

For the Lagrangian
$$L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$
show that
$$\Pi^{\sigma\rho} = \frac{\partial L}{\partial(\partial_{\sigma}A_{\rho})}=-F^{\sigma\rho}$$
Hence show that the energy-momentum tensor
$$T_{\nu}^{\mu} = \Pi^{\mu\sigma}\partial_{\nu}A_{\sigma}-\delta_{\nu}^{\mu}L$$
can be written as
$$T^{\mu\nu} = -F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\frac{1}{4}g^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}$$
We symmetrize the tensor by adding
$$\partial_{\lambda}X^{\lambda\mu\nu} \text{ with } X^{\lambda\mu\nu}=F^{\mu\lambda}A^{\nu}$$
Show that the new tensor is:
$$\hat{T}^{\mu\nu}=F^{\mu\sigma}F_{\sigma}^{\nu} + \frac{1}{4}g^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}$$

## Homework Equations

$$F_{\alpha\beta} = \partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}$$
$$F^{\alpha\beta} = \partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha}$$

## The Attempt at a Solution

$$\Pi^{\sigma\rho} = \frac{\partial L}{\partial(\partial_{\sigma}A_{\rho})}$$
$$\Pi^{\sigma\rho}=-\frac{1}{4}\frac{\partial}{\partial(\partial_{\sigma}A_{\rho})}\left([\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}]F^{\alpha\beta}\right)$$
$$\Pi^{\sigma\rho}=-\frac{1}{4}\left([\frac{\partial(\partial_{\alpha}A_{\beta})}{\partial(\partial_{\sigma}A_{\rho})}-\frac{\partial(\partial_{\beta}A_{\alpha})}{\partial(\partial_{\sigma}A_{\rho})}]F^{\alpha\beta}\right)$$
$$\Pi^{\sigma\rho}=-\frac{1}{4}\left(\delta_{\sigma}^{\alpha}\delta_{\rho}^{\beta}-\delta_{\sigma}^{\beta}\delta_{\rho}^{\alpha}\right)F^{\alpha\beta}$$
$$\Pi^{\sigma\rho}=-\frac{1}{4}(F^{\sigma\rho}-F^{\rho\sigma})$$
$$\Pi^{\sigma\rho}=-\frac{1}{4}(F^{\sigma\rho}+F^{\sigma\rho})$$
$$\Pi^{\sigma\rho}=-\frac{1}{2}F^{\sigma\rho}$$

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...
$$T_{\nu}^{\mu} = \Pi^{\mu\sigma}\partial_{\nu}A_{\sigma}-\delta_{\nu}^{\mu}L$$
$$T_{\nu}^{\mu} = -F^{\mu\sigma}\partial_{\nu}A_{\sigma}+\frac{1}{4}\delta_{\nu}^{\mu}F_{\alpha\beta}F^{\alpha\beta}$$

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:
$$-F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda}A^{\nu})=-F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda})A^{\nu}+F^{\mu\lambda}\partial_{\lambda}(A^{\nu})$$
$$-F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda}A^{\nu})=-F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\sigma}(F^{\mu\sigma})A^{\nu}+F^{\mu\sigma}\partial_{\sigma}A^{\nu}$$
$$-F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda}A^{\nu})=F^{\mu\sigma}(-\partial^{\nu}A_{\sigma}+\partial_{\sigma}A^{\nu}) +\partial_{\sigma}(F^{\mu\sigma})A^{\nu}$$
$$-F^{\mu\sigma}\partial^{\nu}A_{\sigma}+\partial_{\lambda}(F^{\mu\lambda}A^{\nu})=F^{\mu\sigma}F_{\sigma}^{\nu} +\partial_{\sigma}(F^{\mu\sigma})A^{\nu}$$

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.

For the first part you need to lower the indices as you stated and use the product rule.

Hello, thank you for your answer, do you mean that this expression

$$\frac{\partial}{\partial(\partial_{\sigma}A_{\rho})} F^{\alpha\beta}$$

is different from zero?

In that case how does that work when I perform a derivative with lower indices on an object with upper indices?

Hello, thank you for your answer, do you mean that this expression

$$\frac{\partial}{\partial(\partial_{\sigma}A_{\rho})} F^{\alpha\beta}$$

is different from zero?

In that case how does that work when I perform a derivative with lower indices on an object with upper indices?

$$F^{\alpha \beta} = \eta^{\mu \alpha} \eta^{\nu \beta} F_{\mu \nu}$$

$$\frac{\partial}{\partial(\partial_{\sigma}A_{\rho})} F^{\alpha\beta}= \eta^{\mu \alpha} \eta^{\nu \beta} \frac{\partial}{\partial(\partial_{\sigma}A_{\rho})} F_{\mu \nu}$$

Wow thank you for this equation, I believe it may solve all my factors 2 problem, and not only in this calculation but many other, when I calculate conjugate momenta from Lagrangians!! Can you just remember me what eta is? I think I have seen somewhere that this is related to the metric tensor g but it is not defined in the book I am reading. I think they assume more knowledge of special relativity and tensor notations than I have.

Sorry, should have read your post more carefully. Eta is what the flat spacetime metric is typically denoted by. Simply replace it with g and it's consistent with your notation. You can raise the index on the second part of your problem in the same way.

So in the second problem, when I rise nu,
$$\delta_{\nu}^{\mu}$$
is directly replaced by
$$g^{\mu\nu}$$
?

Correct.

Do you know if there is a special reason why in the beginning the tensor is defined with low nu and then we rise it instead of having the tensor with all the indices up directly?

Thank you anyway, I have to go to sleep, but I will take a look tomorrow to see if there are new answers.

I'm not entirely sure. From what I recall, when we derive the energy-momentum tensor from the Lagrangian it's derived as a mixed tensor as in 1.41 here: http://www.damtp.cam.ac.uk/user/tong/qft/one.pdf. At the same time, I know a lot of identities regarding the tensor and equations using the tensor are given in terms of the energy-momentum tensor with both upper (or lower) indices. Perhaps that's why they're having you raise and lower it. Honestly, I'm just guessing and I don't think I have a satisfactory answer to this.