Thank you for your reply. I was familiar with the whole theory about Dirac delta function, but whole your text especially that example with sinus cleared things out for me perfectly now
But if you would be kind enough I would need some more help in proving that equation:
[math]
\nabla \frac{ 1}{ | \vec{r} - \vec{r_0} |} = - 4 \pi \delta ( \vec{r} - \vec{r_0})
[/math]
I have just figured out all that stuff from the link I have posted above and I think the method to achieve this:
[math] \delta( \vec{r} - \vec{r_0})= \frac{1}{4 \pi r^2 }\delta ( r - r_0) [/math]
is quite understandable now for me.
But according to the excersie I have:
[math]
\nabla \frac{ 1}{ | \vec{r} - \vec{r_0} |} = - \frac{1}{r^2} \delta ( r - r_0)
[/math]
And I have just thought that if I counted this in spherical coordinates, gradient of my function 1/r should be also presented in that form. If thats ok I would have:
[math]-\frac{1}{ | \vec{r} - \vec{r_0} |^2}= - \frac{1}{r^2} \delta ( r - r_0) [/math]
And if I integrate both sides I get:
[math]\frac{1}{ | \vec{r_0} - \vec{r} |}= \frac{1}{r_0^2} [/math]
So rather it is not equal? :/
Uh I am confused with this...
Merged post follows:
Consecutive posts mergedUh I am sorry, dunno why but I write everywhere gradient, where there should be laplasian. As far as I see it will change my last equation so that on the left side we will have 1/(r0-r)^2... Sorry one more time.
