# Why electromagnetic tensor (Faraday 2-form) is exact? (and not closed)

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• phoenix95
Gold Member
Following from Wikipedia, the covariant formulation of electromagnetic field involves postulating an electromagnetic field tensor(Faraday 2-form) F such that
F=dA
where A is a 1-form, which makes F an exact differential form. However, is there any specific reason for expecting F to be exact? Could it be the case that in general, F is a closed differential form, but by virtue of the Poincare lemma we define F to be this way?

## Answers and Replies

Gold Member
2022 Award
That's just the homogeneous Maxwell equations, ##\mathrm{d} F=0##. In Ricci-calculus notation that's
$$\partial_{\mu} ^{\dagger} F^{\mu \nu}=\partial_{\mu} \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}=0.$$
The Poincare lemma tells you that (at least locally) ##F=\mathrm{d} A## or, in Ricci notation,
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$

phoenix95
Gold Member
That's just the homogeneous Maxwell equations, ##\mathrm{d} F=0##. In Ricci-calculus notation that's
$$\partial_{\mu} ^{\dagger} F^{\mu \nu}=\partial_{\mu} \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma}=0.$$
The Poincare lemma tells you that (at least locally) ##F=\mathrm{d} A## or, in Ricci notation,
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
Thanks for the reply. I understood that. But as much as I know, not all closed forms are exact (although all exact forms are closed). So is there a specific reason why we always write F=dA? In other words, just because it is closed why do we expect it to be exact?

In your answer, you wrote F=dA at least locally right? So am I right in saying that the differential 2-form F, in general, is not exact globally (although we both agree that F has to be closed globally)?

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$$\int_{K} \mathrm{d} \vec{r} \vec{B}=2 \pi C N$$