# Manipulation of Tensors

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.
• 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) – James S. Cook Nov 4 '14 at 13:33