Problem with taking the divergence

[lavster]

hi, can someone tell me how $\nabla\dot(\frac{\widehat{r}}{r^2})=4\pi\delta^3(r)$

thanks

 

[berkeman]

hi, can someone tell me how $\nabla\dot(\frac{\widehat{r}}{r^2})=4\pi\delta^3(r)$

thanks

How would you normally proceed with taking the divergence? What coordinate system is this in?

 

[lavster]

i would use spherical coord system. using the corresponding form of div. but this gives 0 as the r^2 in the formula is canceled but the r^2 in the div eqn

 

[gabbagabbahey]

i would use spherical coord system. using the corresponding form of div. but this gives 0 as the r^2 in the formula is canceled but the r^2 in the div eqn

Sure, it gives zero everywhere, except at r=0. Remember, $\frac{1}{r^2}$ is undefined at $r=0$

What does the divergence theorem tell you about the volume integral

$$\int_{\mathcal{V}}\left(\mathbf{\nabla}\cdot\frac{\hat{\mathbf{r}}}{r^2}\right)dV$$

in two cases:

(1) When $\mathcal{V}$ is any volume enclosing the origin?
(2) When $\mathcal{V}$ is any volume not enclosing the origin?

Compare these results, along with the fact that [tex]\mathbf{\nabla}\cdot\frac{\hat{\mathbf{r}}}{r^2}[/itex] is zero everywhere except at $r=0$, where it is undefined, to the properties defining the 3D Dirac Delta function.