Cartesian Dirac Delta from divergence of gradient...?by Gan_HOPE326 Tags: cartesian, delta, dirac, divergence, gradient 

#1
Dec1809, 05:56 AM

P: 20

Hi, I've just found in an electrodynamics book a demonstration of Gauss' law involving a definition of Dirac's Delta I didn't know. Substantially, it states that:
[tex]\nabla^{2}(\frac{1}{\leftxx'\right})=4\pi\delta(xx')[/tex] (x and x' are vectors, of course). I can see it somewhat makes sense, since the singularity is the only place where the modulus of the laplacian is the sum of two infinites, but I can't find a real proof. Can someone help me? Thanks. 


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