A Prove a formula with Dirac Delta

[Salmone]

Why is the Laplacian of $1/r$ in spherical coordinates proportional to Dirac's Delta, namely:

$\left(\frac{\partial^2 }{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}\right)\left(\frac{1}{r}\right)=-\frac{\delta(r)}{r^2}$

I get that the result is zero.

 

[Charles Link]

I believe there should be a $4 \pi$ on the right side. (Edit: See post 4. What they have is correct.) I have seen this result in E&M books, e.g. Classical Dynamics by J.D. Jackson, but I have forgotten the details of the proof... I think it might be an application of Gauss' law to a small sphere centered at the origin. In addition, the gradient on $1/r$ gives $- (1/r^2) \hat{a}_r$, and when the $\nabla \cdot$ is done on this and then integrated over the small sphere (in $d \tau$ about the origin), the result is that it needs the delta function, because the volume integral by Gauss' law then integrates the function over the surface of the small sphere and gives a $4 \pi$ result.

See https://www.therightgate.com/deriving-divergence-in-cylindrical-and-spherical/
The divergence will be zero except for $r=0$ where it will be undefined. Gauss' law then tells what it needs to be at $r=0$.

 

[fresh_42]

Why is the Laplacian of $1/r$ in spherical coordinates proportional to Dirac's Delta, namely:

$\left(\frac{\partial^2 }{\partial r^2}+\frac{2}{r}\frac{\partial }{\partial r}\right)\left(\frac{1}{r}\right)=-\frac{\delta(r)}{r^2}$

I get that the result is zero.

If you read the LHS as differential operators, then the result is zero. The interesting part is the RHS. What does $\delta (r)$ stand for? If it is a distribution, then $\delta (r)=r(0)=0$ and everything is fine. But then you have to read the LHS as distribution, too. However, I cannot explain it without further information about the definition of $\delta (r).$

@Charles Link The $\pi$ factor is a convention for Fourier transforms according to Wikipedia.

 

[Charles Link]

Looking at it more closely, I remember $- 4 \pi \delta^3(\vec{r})$ for the right hand side.

Edit: and what they have I think is consistent with this. Integrating their RHS over a small volume will have $d \tau= 4 \pi r^2 \, dr$, and it will integrate to $- 4 \pi$, just as it should. They use a one dimensional delta function, where mine is 3D.