Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Why div [grad(1/r)]=-4πδ(r vec)

Tags:
  1. Jul 3, 2017 #1
    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) ?
     
  2. jcsd
  3. Jul 3, 2017 #2

    Math_QED

    User Avatar
    Homework Helper

    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}##
     
  4. Jul 3, 2017 #3

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Why div [grad(1/r)]=-4πδ(r vec)
  1. Integral 1/r grad 1/r (Replies: 2)

Loading...