Register to reply

Cartesian Dirac Delta from divergence of gradient...?

by Gan_HOPE326
Tags: cartesian, delta, dirac, divergence, gradient
Share this thread:
Gan_HOPE326
#1
Dec18-09, 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}{\left|x-x'\right|})=4\pi\delta(x-x')[/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.
Phys.Org News Partner Science news on Phys.org
An interesting glimpse into how future state-of-the-art electronics might work
Tissue regeneration using anti-inflammatory nanomolecules
C2D2 fighting corrosion

Register to reply

Related Discussions
Kronecker delta and Dirac delta Calculus 3
Find gradient in spherical and cartesian coordinates Calculus & Beyond Homework 6
Purpose of each of the operators , divergence, gradient and curl? Calculus 12
Dirac delta and divergence Classical Physics 1
Vecctor analysis and got the mathematical formulae for gradient General Physics 4