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suppose we have the free field lagrangian: $$L=-\frac{1}{4}F_{\mu \nu} ^2$$ then its just $$L=-\frac{1}{4}(\partial_\mu A_\nu -\partial_\nu A_\mu)^2$$ what I don't understand is how its equal to: $$L=-\frac{1}{2}(\partial_\mu A_\nu)^2 +\frac{1}{2}(\partial_\mu A_\mu)^2$$

I appreciate any help. Thank you.

MSB
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  • I don't even get why we can add up $$(\partial_\mu A_\nu)^2 $$ and $$(\partial_\nu A_\mu)^2 $$ because if we take two different values of mu and nu these will be different. – MSB Oct 16 '16 at 16:48
  • say mu=0 nu=1 then these terms will give us $$(\partial_0 A_1)^2 +(\partial_1 A_0)^2 $$ which are not the same and we cant add them up – MSB Oct 16 '16 at 16:51
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    Do you understand why the expression $$\int_0^1 x dx + \int_0^1 y dy$$ makes sense? It doesn't matter that $x$ and $y$ are different letters, because they're both integrated over. Same here, the $\mu$ and $\nu$ are summed over. I could even write $(\partial_\mu A_\nu)^2 + (\partial_{\widetilde{Q}} A_{\mathfrak{G}})^2$. – knzhou Oct 16 '16 at 16:53
  • oh my God! I totally forgot about the square i didnt see the repeating indices, thank you so much knzhou! – MSB Oct 16 '16 at 16:55
  • Use integration by parts, for the boundary condition in which $A_{\mu}$ is finite at infinity you get that result. – Hossein Oct 16 '16 at 16:56

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