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I want a convincing proof of this...

\nabla^2 (1/|r-r'|) = -4\pi \delta^3 (r-r')

(where r and r' are vectors and \delta^3 (r-r') represents the 3-D delta function...)

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    that approach is not completely appropriate: in spherical coordinates you easily see that $\nabla^2 1/|r|$ is 0 at every point with $r \neq 0$. But at $r = 0$ your coordinate system is degenerate; but all the "action" happens at $r=0$.2011-02-02

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