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Iam Studying "Quantization of the electromagnetic field using Quantum Field Theory" by Lahiri and Pal.

But I don't get how they computed action in equation $8.23$.

$$A=-{1\over 4} \int d^4xF_{\mu \nu}F^{\mu \nu}=-{1\over2}\int d^4 x [(\partial _\mu A_ {\nu})(\partial ^\mu A^ {\nu})-(\partial _\mu A_ {\nu})(\partial ^\nu A^ {\mu}) ]~.$$

I don't get how they evaluated $~F_{\mu \nu}F^{\mu \nu}~$ and arrived at this result , can any one please help me?

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    $\begingroup$ Related : Squaring the E&M (Maxwell) field strength tensor. $\endgroup$
    – Frobenius
    Commented Feb 27, 2020 at 7:40
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    $\begingroup$ Start from the definition of $F^{\mu \nu}$ and show us how far you get, we should see an attempt at a solution. $\endgroup$
    – Triatticus
    Commented Feb 27, 2020 at 7:55
  • $\begingroup$ I know to evaluate second term in this integral, your link suggests an answer which only focuses on evaluation of this second term, my question is more fundamental than that @Frobenius $\endgroup$
    – ROBIN RAJ
    Commented Feb 27, 2020 at 12:59

2 Answers 2

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Expand $F_{\mu\nu}$ and $F^{\mu\nu}$ and multiply. Since $\mu,\nu$ are summed over, in the next step, they can be interchanged so that $$(\partial_\mu A_\nu)(\partial^\mu A^\nu)=(\partial_\nu A_\mu)(\partial^\nu A^\mu).$$ Hope this helps!

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  • $\begingroup$ Ok I get some ensight, I am forget about einstein's summation rule, any way thanks a lot. I get answer but it is multiplied by a minus sign , can you please define $F^{\mu\nu} $ and $F_{\mu \nu} $ @SRS $\endgroup$
    – ROBIN RAJ
    Commented Feb 27, 2020 at 17:44
  • $\begingroup$ @ROBINRAJ I am not sure what you are really asking. By definition, $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ and $F^{\rho\sigma}=\eta^{\rho\nu}\eta^{\sigma\mu}F_{\mu\nu}=\partial^\rho A^\sigma-\partial^\sigma A^\rho$. $\endgroup$
    – SRS
    Commented Feb 28, 2020 at 13:35
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I briefly state main steps:

1) Using antisymetry of $F_{\mu\nu}$ $$ F_{\mu\nu}\partial^\mu A^\nu = -F_{\mu\nu}\partial^\nu A^\mu $$ $$A=-{1\over 4} \int d^4xF_{\mu \nu}F^{\mu \nu}=-{1\over2} \int d^4 xF_{\mu\nu}\partial ^\mu A^\nu$$

2) You need use $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$

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  • $\begingroup$ I don't get your equation 1 , I know $F_{\mu \nu}$ is an antisymmetric tensor but $\partial _{\mu } A_{\nu } $ is not @Nikita $\endgroup$
    – ROBIN RAJ
    Commented Feb 27, 2020 at 14:10
  • $\begingroup$ Write $F_{\mu\nu} = \frac12(F_{\mu\nu}-F_{\nu\mu})$ $\endgroup$
    – MannyC
    Commented Feb 27, 2020 at 15:14
  • $\begingroup$ I edited answer. Is it clear now? $\endgroup$
    – Nikita
    Commented Feb 27, 2020 at 16:11
  • $\begingroup$ Can you define $F^{\mu \nu } $ $\endgroup$
    – ROBIN RAJ
    Commented Feb 27, 2020 at 17:01
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    $\begingroup$ It is as usual, but with upper indeces $\endgroup$
    – Nikita
    Commented Feb 27, 2020 at 18:14

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