Some trouble with relativistic notations

  • Thread starter Amentia
  • Start date
  • #1
95
4
Hello, I am trying to solve a problem and I have trouble with almost every part of it.

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.
 

Answers and Replies

  • #2
178
1
For the first part you need to lower the indices as you stated and use the product rule.
 
  • #3
95
4
Hello, thank you for your answer, do you mean that this expression

[tex]\frac{\partial}{\partial(\partial_{\sigma}A_{\rho})} F^{\alpha\beta}[/tex]

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?
 
  • #4
178
1
Hello, thank you for your answer, do you mean that this expression

[tex]\frac{\partial}{\partial(\partial_{\sigma}A_{\rho})} F^{\alpha\beta}[/tex]

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?
[tex] F^{\alpha \beta} = \eta^{\mu \alpha} \eta^{\nu \beta} F_{\mu \nu}[/tex]

[tex]\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}[/tex]
 
  • #5
95
4
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.
 
  • #6
178
1
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.
 
  • #7
95
4
So in the second problem, when I rise nu,
[tex]\delta_{\nu}^{\mu}[/tex]
is directly replaced by
[tex]g^{\mu\nu}[/tex]
?
 
  • #9
95
4
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?
 
  • #10
95
4
Thank you anyway, I have to go to sleep, but I will take a look tomorrow to see if there are new answers.
 
  • #11
178
1
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.
 

Related Threads on Some trouble with relativistic notations

  • Last Post
Replies
2
Views
2K
Replies
8
Views
1K
  • Last Post
Replies
1
Views
3K
Replies
3
Views
726
  • Last Post
Replies
2
Views
2K
Replies
2
Views
2K
Replies
0
Views
1K
Replies
6
Views
473
Replies
7
Views
1K
Replies
2
Views
1K
Top