# Dummy index 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?

• Could you show how you get $\frac{\partial L}{\partial\partial_\mu A_\nu}=-\partial^\mu A^\nu+\eta^{\mu\nu}\partial_\rho A^\rho$? It doesn't look right to me.
– J.G.
Commented Jan 8 at 18:02
• +1 to above. It is definitely not correct. Commented Jan 8 at 19:34
• But it is correct, It is actually the same expression in the david tong's QFT notes. Look at this link page 10 equation 1.19, damtp.cam.ac.uk/user/tong/qft/qft.pdf Commented Jan 8 at 23:15
• That's not the same Lagrangian. Tong is starting from the Lagrangian (1.18) which is related to the more common $-(1/4) F^{\mu\nu} F_{\mu\nu}$ by integration by parts. Commented Jan 9 at 0:01
• "But it is correct..." No, it's not correct. The derivative of the Lagrangian that you presented in your question post is just $F^{\mu\nu}$. The different Lagrangian (from Eq. 1.18 of the notes you linked in your comments) will result in a different derivative... But the EOM is still the same...
– hft
Commented Jan 9 at 0:39

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)

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: $$\partial_\rho F^{\nu\rho}=-\partial_\mu F^{\mu\nu}=0$$

• Yes and I really thought that you can actually directly derive one Lagrangian density from the other. Actually the complete derivation (using integration by parts) from one Lagrangian density to the other is from this page, physics.stackexchange.com/questions/481903/… Commented Jan 9 at 6:17
• @KingMeruem But what IBP does is obtain a new Lagrangian with the same EOMs, not an equivalent form of the same Lagrangian. So derivatives change.
– J.G.
Commented Jan 9 at 8:52
• yes yes, the two lagrangian densities are not equivalent but have the same EOM. Commented Jan 9 at 9:06