How can we do this tensor product $F_{\mu \nu}F^{\mu \nu}$?

Question

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?

Start from the definition of $F^{\mu \nu}$ and show us how far you get, we should see an attempt at a solution. — Triatticus, Feb 27 '20 at 07:55
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 — ROBIN RAJ, Feb 27 '20 at 12:59

score 2 · Accepted Answer · answered Feb 27 '20 at 14:41

2

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!

answered Feb 27 '20 at 14:41

SRS

26,333

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 – ROBIN RAJ Feb 27 '20 at 17:44
@ROBINRAJ I am not sure what you are really asking. By definition, $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ and $F^{\rho\sigma}=\eta^{\rho\nu}\eta^{\sigma\mu}F_{\mu\nu}=\partial^\rho A^\sigma-\partial^\sigma A^\rho$. – SRS Feb 28 '20 at 13:35

Nikita · Answer 2 · 2020-02-27T16:10:59.410

1

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 $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$

edited Feb 27 '20 at 16:10

answered Feb 27 '20 at 13:26

Nikita

5,657

I don't get your equation 1 , I know $F_{\mu \nu}$ is an antisymmetric tensor but $\partial {\mu } A{\nu } $ is not @Nikita – ROBIN RAJ Feb 27 '20 at 14:10
Write $F_{\mu\nu} = \frac12(F_{\mu\nu}-F_{\nu\mu})$ – MannyC Feb 27 '20 at 15:14
I edited answer. Is it clear now? – Nikita Feb 27 '20 at 16:11
Can you define $F^{\mu \nu } $ – ROBIN RAJ Feb 27 '20 at 17:01
1

It is as usual, but with upper indeces – Nikita Feb 27 '20 at 18:14
Which one is correct $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ or $F^{\mu\nu} = \partial^\nu A^\mu - \partial^\mu A^\nu$ ? different sourses give different definition @Nikita – ROBIN RAJ Feb 27 '20 at 19:06
Of course first – Nikita Feb 27 '20 at 19:41

How can we do this tensor product $F_{\mu \nu}F^{\mu \nu}$?

2 Answers2