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The dual field strength tensor is defined as $$\hat{F}^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta},$$ where $F_{\alpha\beta} = \partial_\alpha A_\beta-\partial_\beta A_\alpha$ is the electromagnetic field strength tensor and $A_\gamma$ is the four-potential. We know that $\partial_\mu\hat{F}^{\mu\nu} = 0$. I now want to rewrite the expression to get $$\partial_\sigma F_{\mu\nu}+\partial_\mu F_{\nu\sigma}+\partial_\nu F_{\sigma\mu} = 0.$$

Unfortunately, I don't quite know how to do this. My first thought was to multiply $\partial_\mu\hat{F}^{\mu\nu} = 0$ by $\epsilon_{\mu\rho\sigma\gamma}$ which yields $$\begin{align}\frac{1}{2}\partial_\mu \epsilon_{\mu\rho\sigma\gamma}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta} &= \frac{1}{2}\partial_\mu\left(\delta_\rho^\nu\delta_\sigma^\alpha\delta_\gamma^\beta+\delta_\rho^\alpha\delta_\sigma^\beta\delta_\gamma^\nu+\delta_\rho^\beta\delta_\sigma^\nu\delta_\gamma^\alpha\right)F_{\alpha\beta}\\ &= \frac{1}{2}\partial_\mu\left(\delta_\rho^\nu F_{\sigma\gamma}+\delta_\gamma^\nu F_{\rho\sigma}+\delta_\sigma^\nu F_{\sigma\rho} \right) \\ &= 0.\end{align}$$

I don't really see what to do next. So any help is appreciated.

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closed as off-topic by AccidentalFourierTransform, John Rennie, heather, Gert, JamalS Nov 11 '16 at 10:52

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    $\begingroup$ $\frac{1}{2}\partial_{\color{red}{\mu}} \epsilon_{{\color{red}{\mu}}\rho\sigma\gamma}\epsilon^{{\color{red}{\mu}}\nu\alpha\beta}$ makes no sense (three $\mu$'s) $\endgroup$ – AccidentalFourierTransform Nov 7 '16 at 19:33
  • $\begingroup$ I think that's exactly my problem. I'm not really familiar with tensor notation yet, so maybe you could explain to me how to do it better. $\endgroup$ – MeMeansMe Nov 7 '16 at 19:49
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    $\begingroup$ Maybe you should do a review of Einstein summation notation before tackling this. Not to be mean, but "does $\partial_\mu A_\mu^\mu$ make sense?" will be chapter 1 problem 1 of any book on tensors. $\endgroup$ – user12029 Nov 7 '16 at 20:16
  • $\begingroup$ I do see now that $\partial_\mu\epsilon_{\mu\rho\sigma\gamma}\epsilon^{\mu\nu\alpha\beta}$ does not make sense, but I don't know how to get rid of that problem. $\endgroup$ – MeMeansMe Nov 7 '16 at 20:20
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The expression that you want to get automatically follows from the definition of the electromagnetic tensor as $F_{\mu\nu} = \partial_{\mu}A_{\nu}- \partial_{\nu}A_{\mu}$. In particular one has $$ \partial_{\sigma}F_{\mu\nu} + \partial_{\mu}F_{\nu\sigma} + \partial_{\nu}F_{\sigma\mu} =\partial_{\sigma}\left(\partial_{\mu}A_{\nu}- \partial_{\nu}A_{\mu}\right) + \partial_{\mu}\left(\partial_{\nu}A_{\sigma}- \partial_{\sigma}A_{\mu}\right) + \partial_{\nu}\left(\partial_{\sigma}A_{\mu}- \partial_{\mu}A_{\sigma}\right) = 0 $$ under the assumptions that $A\in C^2(\mathbb{R}^4)$.

The result is nothing but the Bianchi identity: since $F= \textrm{d}\,A$, then $\mathcal{D}F=\mathcal{D}\,\textrm{d}\,A=0$.

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    $\begingroup$ Yes, this is another possibility of proving the identity, but I would like to prove it by rewriting the expression in my first post. $\endgroup$ – MeMeansMe Nov 7 '16 at 19:57
  • $\begingroup$ @MeMeansMe More than applying the derivatives on both sides of your initial expression there's not really much to do (bring the derivatives inside the $\epsilon$ symbol and you're done). $\endgroup$ – gented Nov 7 '16 at 20:02

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