Proving $\nabla^2(\frac{1}{r})=-4\pi\delta^{(3)}(r)$

[Unkraut]

Homework Statement

Prove that $$\nabla^2(\frac{1}{r})=-4\pi\delta^{(3)}(r)$$, where $$\delta^{(3)}$$ is the three-dimensional Dirac delta function.

Homework Equations

1+1=2
$$\pi=3$$

The Attempt at a Solution

I am very confused. I don't see how this statement can be true. Deriving an ordinary nice differentiable function two times cannot yield a Dirac delta (which is actually not a function). Or do I misinterpret something here?

 

[kuruman]

Take the Laplacian of 1/r in spherical coordinates and see for yourself how "nice" this function is.

 

[dx]

I am very confused. I don't see how this statement can be true. Deriving an ordinary nice differentiable function two times cannot yield a Dirac delta (which is actually not a function). Or do I misinterpret something here?

Are you sure it's a nice differentiable function? What about at the origin? (r = 0)

 

[Unkraut]

$$\nabla^2(\frac{1}{r})=\frac{1}{r^2sin\theta}(sin\theta\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}(\frac{1}{r})+\left[\text{derivatives with respect to the angles = 0}\right])$$
$$=\frac{1}{r^2}\frac{\partial}{\partial r}(-r^2\frac{1}{r^2})$$
$$=\frac{1}{r^2}\frac{\partial}{\partial r}(-1)=0$$
Of course the function is not defined for r=0 and that's where the dirac delta takes on the value infinity.
Somehow now the statement looks plausible for me, but still I have no idea what kind of black magic a physicist would apply to prove it. I can only calculate the Laplacian for $$r\neq 0$$.

 

[dx]

Try a simpler problem. Can you show that (d/dx)f(x) = δ(x), where f(x) = 1 for x>0 and f(x) = -1 for x<0.

 

[gabbagabbahey]

$$\nabla^2(\frac{1}{r})=\frac{1}{r^2sin\theta}(sin\theta\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}(\frac{1}{r})+\left[\text{derivatives with respect to the angles = 0}\right])$$
$$=\frac{1}{r^2}\frac{\partial}{\partial r}(-r^2\frac{1}{r^2})$$
$$=\frac{1}{r^2}\frac{\partial}{\partial r}(-1)=0$$
Of course the function is not defined for r=0 and that's where the dirac delta takes on the value infinity.
Somehow now the statement looks plausible for me, but still I have no idea what kind of black magic a physicist would apply to prove it. I can only calculate the Laplacian for $$r\neq 0$$.

Okay, so now you have$\nabla^2\left(\frac{1}{r}\right)=0$ everywhere except at the origin, where it is undefined.

Now, take note that $\nabla^2\left(\frac{1}{r}\right)=\mathbf{\nabla}\cdot\mathbf{\nabla}\left(\frac{1}{r}\right)$ and use the divergence theorem to integrate $\nabla^2\left(\frac{1}{r}\right)$ over any volume that enclose the singularity at the origin (I'd recommend choosing a sphere of any radius for ease of calculation, but any volume will produce the same result)

 

[Unkraut]

Try a simpler problem. Can you show that (d/dx)f(x) = δ(x), where f(x) = 1 for x>0 and f(x) = -1 for x<0.

Hmm, no, I can't. But I can say that it's plausible...

Okay, so now you have$\nabla^2\left(\frac{1}{r}\right)=0$ everywhere except at the origin, where it is undefined.

Now, take note that $\nabla^2\left(\frac{1}{r}\right)=\mathbf{\nabla}\cdot\mathbf{\nabla}\left(\frac{1}{r}\right)$ and use the divergence theorem to integrate $\nabla^2\left(\frac{1}{r}\right)$ over any volume that enclose the singularity at the origin (I'd recommend choosing a sphere of any radius for ease of calculation, but any volume will produce the same result)

Hmm... probably I'm very stupid:
$$\int_V\nabla^2\frac{1}{r}dV=\int_{\partial V}\nabla(\frac{1}{r})\cdot \vec n dS=\int_{\partial V}(-\frac{1}{r^2})dS=\int\limits_0^{2\pi}\int_0^{\pi}(-\frac{1}{r^2})r^2sin\phi d\theta d\phi = -\pi\int\limits_0^{2\pi}sin\phi d\phi=0$$

Here I took $$\partial V$$ to be any sphere around the origin. The only non-zero component of $\nabla \frac{1}{r}$ in spherical coordinates is the radial component which is parallel to the normalized normal $\vec n$. So the second step should be correct? It looks like if I did something wrong, then probably there. But I don't see it.

 

[gabbagabbahey]

Hmm, no, I can't. But I can say that it's plausible...





Hmm... probably I'm very stupid:
$$\int_V\nabla^2\frac{1}{r}dV=\int_{\partial V}\nabla(\frac{1}{r})\cdot \vec n dS=\int_{\partial V}(-\frac{1}{r^2})dS=\int\limits_0^{2\pi}\int_0^{\pi}(-\frac{1}{r^2})r^2sin\phi d\theta d\phi = -\pi\int\limits_0^{2\pi}sin\phi d\phi=0$$

If $\phi$ is the polar angle (the angle between $\textbf{r}$ and the z-axis), then its limits are zero to $\pi$ and $\theta$'s limits are zero to $2\pi$, not the other way around.

 

[Unkraut]

Ah, yes, I confused the angles. Thank you very much!
Now it's even more plausible. But still somehow I'm not quite satisfied. Anyway if I wanted to have a mathematically rigorous understanding of it that would probably involve some work. I don't like work.
Thank you all for your answers.

 

[gabbagabbahey]

I'm not sure how much rigor you're looking for, but by definition of the 3D Dirac delta function, any function which is zero everywhere except at the origin, where it is undefined, yet when integrated over any volume enclosing the origin, produces a finite result ($-4\pi$ in this case) must be the product of that result (or equivalently, any function that takes on that value at the origin) with $\delta^3(\textbf{r})$. So, $\nabla^2(\frac{1}{r})=-4\pi\delta^{(3)}(\textbf{r})$ or equivalently $\nabla^2(\frac{1}{r})=f(\textbf{r})\delta^{(3)}(\textbf{r})$ for any function where $f(0)=-4\pi$.

 

[dx]

If you're concerned about physics rather than mathematics, then 'rigor' in the sense of a precise mathematical definition of the delta function is almost never useful. For instance, in the example I gave above, we don't think of the function f as being 'undefined' at the point 0, but rather that the function goes from -1 to 1 in a very very small interval.