First, use the divergence theorem to show that [tex]\int_{\mathcal{V}}( \vec{\nabla}\cdot\frac{\hat{r}}{r^2})d\tau=4\pi[/tex] for any surface enclosing the origin (use a spherical surface centered at the origin for simplicity).Griffiths' section 1.5.3 states that the divergence of the vector function r/r^2 = 4*Pi*δ^3(r). Can someone show me how this is derived and what it means physically? Thanks in advance.
The fact that the divergence of [tex]\frac{\hat{r}}{r^2}[/tex] is zero everywhere, except at the origin where it is infinite, but when integrated gives a constant, means that it must be some unknown normal function [itex]f(\vec{r})[/itex] times a delta function so, you call it [itex]f(\vec{r})\delta^3(\vec{r})[/itex] and solve for [itex]f[/itex] by setting the integral (eq. 1.98) equal to the value you calculated using the diverergence theorem ([itex]4\pi[/itex])Can't use latex. Why does f(R)δ^3(R) = the divergence of R/r^2?