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Consider $F_{\mu\nu}=\partial_{\mu}A_\nu-\partial_\nu A_\mu$, I am trying to understand how to fast calculate $$\frac{\partial(F_{\mu\nu}F^{\mu\nu})}{\partial (\partial_\alpha A_\beta)}$$

without expanding the multiplication in terms of $A$. Clearly if we do not have indices, this can be done very quickly with a glance, yet when there are indices multiplication factors come into play we would, in this case get a factor of $4$ before usual derivative.

Is there a way to glance at it and get correct answer immediately?

Qmechanic
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Rescy_
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1 Answers1

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The product rule (Leibniz rule) applies to functional derivatives, so we have $$ \frac{\partial(F_{\mu\nu}F^{\mu\nu})}{\partial (\partial_\alpha A_\beta)} = 2 F^{\mu \nu} \frac{\partial(F_{\mu\nu})}{\partial (\partial_\alpha A_\beta)} = 2 F^{\mu \nu}\left( \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} - \delta^{\alpha}_{\nu} \delta^{\beta}_{\mu}\right) = 4 F^{\alpha \beta}. $$ The step where I actually take the derivative of $F_{\mu \nu}$ with respect to $\partial_\alpha A_\beta$ does involve writing out the field strength in terms of the derivatives, but it's less complicated in this version.

Michael Seifert
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  • Aren’t functional derivatives usually written with $\delta$ rather than $\partial$? Or is that old-fashioned? Are you just following the OP’s notation? – Ghoster Nov 10 '22 at 21:33
  • @Ghoster: There are different schools of thought, I think. My own style is to use $\delta$ for small variations when I'm doing calculus of variations problems ab initio and to use partial derivatives for functional derivatives. It does overload the notation a bit but it's not usually a problem. – Michael Seifert Nov 11 '22 at 02:12