Undergrad Why Does div[grad(1/r)] Equal -4πδ(r vec)?

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The discussion centers on the mathematical derivation of the gradient and divergence of the function 1/r, particularly at the origin where it is undefined. It clarifies that the gradient of 1/r is indeed (-1/r^2)ê_r for r > 0, and addresses the challenge of defining this at r = 0. The divergence of the gradient, div[grad(1/r)], is shown to equal -4πδ(r vec), which relates to the behavior of the function in the context of distributions. The conversation emphasizes the importance of using limits and integrals to handle the singularity at the origin. Understanding these concepts is crucial for grasping the implications in physics and mathematics.
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I don't understand why grad(1/r)= (-1/r^2)ê_r. I know its derivate, but in cero the function is not definite and you can't find its derivative in cero.And, why div [grad(1/r)]=-4πδ(r vec) ?
 
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cristianbahena said:
I don't understand why grad(1/r)= (-1/r^2)ê_r. I know its derivate, but in cero the function is not definite and you can't find its derivative in cero.And, why div [grad(1/r)]=-4πδ(r vec) ?

This is simply applying definitions. Also,

##\frac{d|x|}{dx} = \frac{|x|}{x} = \begin{cases} 1 \quad \mathrm{if} \quad x >0\\-1 \quad \mathrm{if} \quad x <0 \end{cases}##
 
You're probably supposed to show that

##\lim_{\epsilon \rightarrow 0+}\int_{\mathbb{R}^3}\nabla \cdot \left(\nabla \left(\frac{1}{|\mathbf{r}|+\epsilon}\right)\right) f(\mathbf{r})dV \propto f(\mathbf{0})##

for a relatively arbitrary function ##f( \mathbf{r})##. Note that ##\mathbf{r}## is a position vector, ##|\mathbf{r}|## its norm and ##dV## a volume element.
 
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