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Feb 22, 2023 at 12:11 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 11:58 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 11:52 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 5:45 comment added Iosif Pinelis @WillieWong : Thank you for your comments. I have been learning about this just while working, gradually, on this answer.
Feb 22, 2023 at 5:40 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 5:39 vote accept MrPie
Feb 22, 2023 at 5:32 comment added Willie Wong Though in the usual case, if $F$ is of moderate decrease then there are integral formulas for $\sigma$ and $\Gamma$, in which case both decay to zero at infinity. // The OP should look up Hodge theory, where the harmonic factor is emphasized. (There are some additional technical problems with $\mathbb{R}^3$ being non-compact. But the ideas are the same.)
Feb 22, 2023 at 5:28 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 5:26 comment added Willie Wong Basically, a harmonic vector field (including a constant one) is both divergence and curl free, and so if $F = \nabla \sigma + \nabla\times \Gamma$, given any harmonic vector field $\omega$, there exists $\tau$ such that $\nabla \tau = \omega$ and there exists $\Delta$ such that $\nabla \times \Delta = \omega$. So $F = \nabla (\sigma + \tau) + \nabla\times (\Gamma - \Delta)$. But then both the divergence free and curl free parts contain a factor of $\omega$ and so are not independent.
Feb 22, 2023 at 5:14 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 3:46 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 3:20 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 3:01 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 2:55 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 2:36 history edited Iosif Pinelis CC BY-SA 4.0
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Feb 22, 2023 at 2:24 history answered Iosif Pinelis CC BY-SA 4.0