Dummy index question

Question

The Maxwell's Lagrangian density is given by the equation, $$\mathcal L = -\frac{1}{4} \space F_{\mu\nu} \space F^{\mu\nu},$$ where $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$.

Hence, one can rewrite the Lagrangian density into the following, $$\mathcal L= \frac{1}{4} (\partial_\mu A_\nu - \partial_\nu A_\mu) (\partial^\mu A^\nu - \partial^\nu A^\mu ). $$ This means that one can show that, $$ \frac{\partial \mathcal L}{\partial(\partial_\mu A_ \nu)} = -\partial^\mu A^\nu + (\partial_\rho A^\rho) \space \eta^{\mu\nu}. $$

But some references from page 9 of this page, and page 16 of this page, show that this expression can be written as, $$ \frac{\partial \mathcal L}{\partial(\partial_\mu A_ \nu)} = F^{\nu\mu} = (\partial^\nu A^\mu - \partial^\mu A^\nu ) = -\partial^\mu A^\nu + \partial^\nu A^\mu. $$ This means that, $$\partial^\nu A^\mu = (\partial_\rho A^\rho) \space \eta^{\mu\nu}. $$ But the only way that they are equal is that if I allow my dummy index $\rho$ to be equal to either $\nu$ or $\mu$. But it means that I allow my index to be repeated more than twice?

Answer 1

The Maxwell's Lagrangian density is given by the equation, $$\mathcal L = -\frac{1}{4} \space F_{\mu\nu} \space F^{\mu\nu},\tag{A}$$ where $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$.

...This means that one can show that, $$ \frac{\partial \mathcal L}{\partial(\partial_\mu A_ \nu)} = -\partial^\mu A^\nu + (\partial_\rho A^\rho) \space \eta^{\mu\nu}. \tag{B}$$

No, one can not show that. Your expression above (tagged above as equation B) is wrong, given your specified Lagrangian (tagged above as equation A).

The correct result is: $$ \frac{\partial \mathcal{L}}{\partial{\partial_\rho A_\sigma}} = F^{\sigma\rho}\;.\tag{C} $$

This means that...

No. It does not mean what you purport it to mean.

In the comments you provided a clarifying link to a different set of lecture notes that start from a different Lagrangian (Eq. 1.18 in those notes) and arrive at a different expression for the derivative of that different Lagrangian (the expression I tagged as equation B above). Because you start from a different Lagrangian, you end up with a different derivative of the Lagrangian. (Note, I ported the link from the comments into the body of the question as an edit)

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Regardless of which of the two Lagrangians you start with, in the end the equation of motion turns out to be the same: $$ \partial_\rho F^{\sigma\rho} = 0 $$

For example, from equation B, and from the usual Lagrangian equation of motion $\partial_\mu\frac{\partial\mathcal{L}}{\partial \partial_\mu A_\nu}=\frac{\partial \mathcal{L}}{\partial A_\nu}$, we have: $$ \partial_\mu \left(-\partial^\mu A^\nu\right) + \partial^\nu\left(\partial_\rho A^\rho\right) = 0 $$ $$ =\partial_\mu \left(-\partial^\mu A^\nu\right) + \partial^\nu\left(\partial_\mu A^\mu\right)=-\partial_\mu \left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)=-\partial_\mu F^{\mu\nu}\;. $$

And from equation C the same equation of motion results more directly as: $$ \partial_\rho F^{\nu\rho}=-\partial_\mu F^{\mu\nu}=0 $$