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 $$F_{\alpha \beta}~=~\partial_{\alpha} A_{\beta} - \partial_{\beta} A_{\alpha}~? $$


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.

  • $\begingroup$ 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 $\endgroup$ – Arthur Suvorov May 20 '14 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 '14 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.