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This is a problem concerning covariant formulation of electromagnetism.

Given $$\partial_{[\alpha} F_{\beta\gamma]}~=~ 0 $$ how does one prove that $F$ can be obtained from a 4-potential $A$ such that $$F_{\alpha \beta}~=~\partial_{\alpha} A_{\beta} - \partial_{\beta} A_{\alpha}~? $$

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1 Answer 1

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The local existence of a one-form $A$ such that the closed two-form $F$ is exact $F=\mathrm{d}A$ is a consequence of the Poincare Lemma. There might be global obstructions.

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  • $\begingroup$ Great answer! Do you know if there is an extension of the Poincare Lemma for the case where the space where $F$ is defined is not an open subset of $\mathbb{R}^{n}$? I mean, you may have a non-compact manifold that cannot be embedded. Just curious $\endgroup$ May 20, 2014 at 23:12
  • $\begingroup$ Well, a sufficient condition for the global existence of $A\in\Gamma(T^{*}M)$ on a manifold $M$ is if the second cohomology group $H^2(M)=0$ is trivial, but you probably know that already. $\endgroup$
    – Qmechanic
    May 21, 2014 at 18:38

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