Question

what is the laplacian of 1/r ? (r being the magnitude of the position vector)

Answer 1

In polar or in cartesian coordinates? It's different

Answer 2

Cartesian !

Answer 3

(i know it should be zero but i can't prove it)

Answer 4

It's just the divergence of the gradient of 1/sqrt(x^2+y^2+z^2)

Answer 5

i know but it doesn't come out zero for me !

Answer 6

It's zero for all the points but the origin, where it is undefined. The procedure is a bit messy. Take the gradient, you can use Wolfram Alpha < http://www.wolframalpha.com/input/?i=gradient+of+1%2Fsqrt%28x%5E2%2By%5E2%2Bz%5E2%29 >. It's just a vector that contains all the partial derivatives of 1/sqrt(x^2+y^2+z^2). Then you take all the components of the gradient, derive the x component by x, and so on and sum all these derivatives together. It's algebraically ugly but you can also see the first of these derivatives with Wolfram Alpha < http://www.wolframalpha.com/input/?i=d%2Fdx+%28-x%2F%28x%5E2%2By%5E2%2Bz%5E2%29%5E%283%2F2%29%29 >. To get the others you cycle x,y,z. Then when you add them together it's clear they are zero

Answer 7

thanks man !

Answer 8

The laplacian of 1/r is \[-4\pi \delta ^{3}(r)\]
where

\[\delta ^{3}(r)\]

is a 3 dimensional Dirac delta function.