I have an expression:

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

Where $\eta^{\mu \nu}$ is the Minkowski metric, F is an antisymmetric tensor, and the comma on the middle tensor denotes a derivative. I am not going to pretend like I have a lot of experience manipulating tensors, but I am trying to raise the co-variant derivative so it is a contravariant derivative, and change it from $\nu$ to $\mu$. Here is what I've accomplished:

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

I am unsure where to proceed from here, but essentially what I'd like to do is be able to get rid of the $\eta's$. Any help is appreciated


I will start again, the trouble I spotted right away with your initial calculation is that you have used $\mu$ as a dummy index of summation whereas it is apparently free from the initial expression: $$ \eta^{\mu \nu} F_{\alpha \beta, \nu} F^{\alpha \beta} $$ Ok, so, to raise the derivative index, you can just use the existing metric in the expression above: poof it's gone and we have: $$ F_{\alpha \beta}^{ \ \ \ , \mu} F^{\alpha \beta} $$ Now, as the metric here is constant we can just as well write: $$ F_{\alpha \beta} F^{\alpha \beta, \mu} $$ To see why the above is true, notice, $$ F_{\alpha \beta} = \eta_{\alpha \gamma}\eta_{\beta \sigma} F^{\gamma \sigma}$$ and $$ F_{\alpha \beta}^{ \ \ \ , \mu} = \partial^{\mu}F_{\alpha \beta} = \partial^{\mu} \left( \eta_{\alpha \gamma}\eta_{\beta \sigma} F^{\gamma \sigma}\right) = \eta_{\alpha \gamma}\eta_{\beta \sigma}\partial^{\mu} F^{\gamma \sigma}$$ hence, $$ F_{\alpha \beta}^{ \ \ \ , \mu} F^{\alpha \beta} = \eta_{\alpha \gamma}\eta_{\beta \sigma}\partial^{\mu} (F^{\gamma \sigma}) F^{\alpha \beta} = \partial^{\mu} (F^{\gamma \sigma}) \eta_{\gamma \alpha}\eta_{ \sigma\beta}F^{\alpha \beta} = \partial^{\mu} (F^{\gamma \sigma}) F_{\gamma \sigma} = F^{\alpha \beta, \mu} F_{\alpha \beta}.$$ In the last step I switched the dummy variables of summation back to $\alpha,\beta$. Note I also used the symmetry of the minkowski metric in the middle step.

Well, I hope this helps. Let me know if you need further elaboration.

  • $\begingroup$ Okay, so the metric doesn't have to be on the right to raise the derivative? The book I am using defines raising of the derivative to be done by the action of the metric on the right. If it doesn't matter that's perfectly okay with me, it just seemed weird that the book would specifically put it on the right. $\endgroup$ – mphy Nov 4 '14 at 11:21
  • $\begingroup$ Also, in your third equation you commuted the tensors, but that's not what you ended up with in your calculation. $\endgroup$ – mphy Nov 4 '14 at 11:57
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    $\begingroup$ so the expression $v_{\mu} = \eta_{\mu \nu} v^{\nu}$ can just as well be written $v^{\nu}\eta_{\mu \nu} $ since the quantities involved for a specific choice of indices are just scalar functions which commute. So the choice to put it on the right is just one of style. Likewise, $v^{\mu}v_{\mu} = v_{\mu}v^{\mu}$ because these are just scalars which commute... or for $a_{\mu}b^{\mu} = b_{\mu}a^{\mu}$ because the metric is symmetric. For that reason the third equation is what I claimed (but, perhaps I should add another step to emphasize this equivalence) $\endgroup$ – James S. Cook Nov 4 '14 at 13:33

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