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May
28
comment Helmholtz decomposition in the plane
Thanks for a very informative answer. I'm marking yours instead of mine as the answer.
May
28
answered Helmholtz decomposition in the plane
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May
28
accepted Uniqueness of Helmholtz decomposition?
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asked Uniqueness of Helmholtz decomposition?
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accepted $\nabla ^2\psi$ equals $\psi -$ average value of $\psi$ at neighboring points
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May
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answered $\nabla ^2\psi$ equals $\psi -$ average value of $\psi$ at neighboring points
May
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comment $\nabla ^2\psi$ equals $\psi -$ average value of $\psi$ at neighboring points
@Hans: I put the formula as it appears on page 7 of Morse & Feshbach Methods of Theoretical Physics (1953) vol.1.
May
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comment $\nabla ^2\psi$ equals $\psi -$ average value of $\psi$ at neighboring points
@Approximist: I think you're right. It must be a typo which has to be corrected to the formula you posted. Thanks.
May
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comment $\nabla ^2\psi$ equals $\psi -$ average value of $\psi$ at neighboring points
@Lubos: That's the exact formula that appears at Morse & Feshbach. I agree with you. I also think it can't be right since it is dimensionally incorrect. Perhaps it's a typo.
May
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comment $\nabla ^2\psi$ equals $\psi -$ average value of $\psi$ at neighboring points
@Approximist: I edited the post to explain what I had done. Why you say I could be missing a factor of $4\pi$? Thanks.