Some trouble with relativistic notations

[Amentia]

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}<br />
\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.

 

[mjordan2nd]

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

 

[Amentia]

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?

 

[mjordan2nd]

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}

 

[Amentia]

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.

 

[mjordan2nd]

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.

 

[Amentia]

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

 

[mjordan2nd]

Correct.

 

[Amentia]

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?

 

[Amentia]

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

 

[mjordan2nd]

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.