Timeline for What theorem Feynman is referring to?

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Oct 5, 2023 at 20:53	comment	added	S.G		@AndrewSteane Thank you for the first comment. For the second bit, since we already see that $\nabla \times \mathbf{A} = 0$, then via Helmholtz decomposition (en.wikipedia.org/wiki/…) doesn't it then follow that $\mathbf{A} = \nabla \psi$? In particular, I have shown that if $\nabla \times \mathbf{A} = 0$ then $\mathbf{A} = \nabla \psi$.
Oct 5, 2023 at 20:50	history	edited	S.G	CC BY-SA 4.0	added 4 characters in body
Oct 5, 2023 at 11:38	comment	added	Andrew Steane		1. Penultimate sentence should reach "since this holds for all closed paths, this means ...". 2. You have proved that if ${\bf A} = \nabla \psi$ then $\nabla \times {\bf A} = 0$ but not the converse. The question is whether $\nabla \times {\bf A}=0$ is enough to guarantee the existence of a $\psi$ of which $\bf A$ is the gradient.
S Oct 5, 2023 at 11:09	history	suggested	TAR86	CC BY-SA 4.0	clarity and typo
Oct 5, 2023 at 10:57	review	Suggested edits
S Oct 5, 2023 at 11:09
Oct 5, 2023 at 1:42	history	edited	S.G	CC BY-SA 4.0	added 18 characters in body
Oct 5, 2023 at 1:21	history	answered	S.G	CC BY-SA 4.0