Electrodynamics: Dirac Delta function and Gauss' divergence theorem

[lys04]

Homework Statement

Show that $$ \nabla ^ {2} (\frac{1}{|\vec{x}|}) = -4\pi \delta(\vec{x}) $$

Relevant Equations

Gauss's divergence theorem, Laplacian in spherical coordinates

Computing the Laplacian of $(\frac{1}{|\vec{x}|})$ in spherical coordinates I get 0 everywhere except at 0.
Now I want integrate the Laplacian over a spherical surface that encloses the origin. I do this using Gauss's divergence theorem, i.e $$ \iint_S \vec{F}.\vec{dA} = \iiint_V \nabla . F dV $$

Do I need to use ∇F here in substitution for F so that when I take the divergence of the gradient of F I get the Laplacian of F which is what I need?
need?

 

[TSny]

Now I want integrate the Laplacian over a spherical surface that encloses the origin. I do this using Gauss's divergence theorem, i.e $$ \iint_S \vec{F}.\vec{dA} = \iiint_V \nabla . F dV $$

Do I need to use ∇F here in substitution for F so that when I take the divergence of the gradient of F I get the Laplacian of F which is what I need?
need?

Yes. Well, you need to sub $\nabla \dfrac 1 {|\vec x|}$ for $\vec F$ in your integral equation.

 

[lys04]

Yes. Well, you need to sub $\nabla \dfrac 1 {|\vec x|}$ for $\vec F$ in your integral equation.

Yes. That's what I did.
And to derive the more general result, i.e $$ \nabla (\frac{1}{|\vec{x}-\vec{x'}|}) = -4 \pi \delta (\vec{x}-\vec{x'})$$, do I just let $\vec{x}-\vec{x'}$ equal to some other vector say $\vec{r}$ and then using my previous result it gives me $$ \nabla (\frac{1}{|\vec{r}|}) = -4 \pi \delta (\vec{r})$$ so $$ \nabla (\frac{1}{|\vec{x}-\vec{x'}|}) = -4 \pi \delta (\vec{x}-\vec{x'})$$ is correct?

 

[PhDeezNutz]

OP check out this thread. Particularly post#11

https://www.physicsforums.com/threads/divergence-of-the-electric-field-of-a-point-charge.1051807/

 

[TSny]

And to derive the more general result, i.e $$ \nabla (\frac{1}{|\vec{x}-\vec{x'}|}) = -4 \pi \delta (\vec{x}-\vec{x'})$$
do I just let $\vec{x}-\vec{x'}$ equal to some other vector say $\vec{r}$ and then using my previous result it gives me $$ \nabla (\frac{1}{|\vec{r}|}) = -4 \pi \delta (\vec{r})$$ so $$ \nabla (\frac{1}{|\vec{x}-\vec{x'}|}) = -4 \pi \delta (\vec{x}-\vec{x'})$$ is correct?

Typos: $\nabla$ should be $\nabla^2$ in the expressions above. Otherwise, yes.

One way to think about it is to imagine shifting the origin of the coordinate system to the location of the charge, $\vec x'$. Then, $\vec x - \vec x'$ in the original system becomes $\vec x$ in the new system. After evaluating $\nabla^2 \dfrac 1 { |\vec x|}$ in the new system you can shift back to the original coordinate system.

 

[Albertus Magnus]

You can use "\cdot" to write your dot products correctly.

 

[lys04]

Typos: ∇ should be ∇2 in the expressions above. Otherwise, yes.

Oh yes, sorry about that.

One way to think about it is to imagine shifting the origin of the coordinate system to the location of the charge, x→′. Then, x→−x→′ in the original system becomes x→ in the new system. After evaluating ∇21|x→| in the new system you can shift back to the original coordinate system.

Ah, thats a good way of thinking about it.