Krdytkyu

It is often quoted in physics textbooks for finding the electric potential using Green's function that

$$nabla ^2 left(frac{1}{r}right)=-4pidelta^3({bf r}),$$

or more generally

$$nabla ^2 left(frac{1}{|| vec x - vec x'||}right)=-4pidelta^3(vec x - vec x'),$$

where $delta^3$ is the 3-dimensional Dirac delta distribution.
However I don't understand how/where this comes from. Would anyone mind explaining?

The gradient of $frac1r$ (noting that $r=sqrt{x^2+y^2+z^2}$) is

$$
nabla frac1r = -frac{mathbf{r}}{r^3}
$$
when $rneq 0$, where $mathbf{r}=xmathbf{i}+ymathbf{j}+zmathbf{k}$. Now, the divergence of this is

$$
nablacdot left(-frac{mathbf{r}}{r^3}right) = 0
$$
when $rneq 0$. Therefore, for all points for which $rneq 0$,

$$
nabla^2frac1r = 0
$$
However, if we integrate this function over a sphere, $S$, of radius $a$, then, applying Gauss's Theorem, we get

$$
iiint_S nabla^2frac1rdV = iint_{Delta S} -frac{mathbf{r}}{r^3}.dmathbf{S}
$$
where $Delta S$ is the surface of the sphere, and is outward-facing. Now, $dmathbf{S}=mathbf{hat r}dA$, where $dA=r^2sintheta dphi dtheta$. Therefore, we may write our surface integral as
$$begin{align}
iint_{Delta S} -frac{mathbf{r}}{r^3}.dmathbf{S}&=-int_0^piint_0^{2pi}frac{r}{r^3}r^2sintheta dphi dtheta\
&=-int_0^pisintheta dthetaint_0^{2pi}dphi\
&= -2cdot 2pi = -4pi
end{align}$$
Therefore, the value of the laplacian is zero everywhere except zero, and the integral over any volume containing the origin is equal to $-4pi$. Therefore, the laplacian is equal to $-4pi delta(mathbf{r})$.

EDIT: The general case is then obtained by replacing $r=|mathbf{r}|$ with $s=|mathbf{r}-mathbf{r_0}|$, in which case the function shifts to $-4pi delta(mathbf{r}-mathbf{r_0})$

I'm new around here (so suggestions about posting are welcome!) and want to give my contribution to this question, even though a bit old. I feel I need to because using the divergence theorem in this context is not quite rigorous. Strictly speaking $1/r$ is not even differentiable at the origin. So here's a proof using limits of distributions.

Let $mathbf{x}inmathbb{R}^{3}$ and $r=|mathbf{x}|=sqrt{x^2+y^2+z^2}$. It is evident from direct calculation that $nabla^{2}left(frac{1}{r}right)=0$ everywhere except in $mathbf{x}=0$, where it is in fact not defined. Thus, the integral of $nabla^{2}left(frac{1}{r}right)$ over any volume non containing the origin is zero.

So let $eta>0$ and $r_{eta}=sqrt{x^2+y^2+z^2+eta^2}$. Obviously $limlimits_{etarightarrow 0}r_{eta}=r$. Direct calculation brings
begin{equation}
nabla^{2}left(frac{1}{r_{eta}}right) = frac{-3eta^{2}}{r_{eta}^5}
end{equation}
Now let us consider the distribution represented by $nabla^{2}left(frac{1}{r_{eta}}right)$ and let $rho$ be a test function (for example in the Schwartz space). I use Dirac's bra-ket notation to express the action of a distribution over a test function. Let $S^{2}$ be the unit sphere. Thus we calculate
begin{align}
limlimits_{etarightarrow 0}left.leftlangle nabla^{2}left(frac{1}{r_{eta}}right)right|rhorightrangle &= limlimits_{etarightarrow 0}iiintlimits_{mathbb{R}^3}mathrm{d}^{3}x, nabla^{2}left(frac{1}{r_{eta}}right)rho(mathbf{x})\
&=limlimits_{etarightarrow 0}left{iiintlimits_{mathbb{R}^3setminus S^{2}}mathrm{d}^{3}x, frac{-3eta^{2}}{r_{eta}^5}rho(mathbf{x})+ iiintlimits_{S^{2}}mathrm{d}^{3}x, frac{-3eta^{2}}{r_{eta}^5}rho(mathbf{x})right}\
&=limlimits_{etarightarrow 0}iiintlimits_{S^{2}}mathrm{d}^{3}x, frac{-3eta^{2}}{r_{eta}^5}rho(mathbf{x})
end{align}
Where the limit of the first of the integrals in the curly braces is zero (easy to show, referring to the laplacian of $1/r$, no need for $eta$ in sets not containing the origin).
Now Taylor expand $rho$ at $mathbf{x}=0$ and integrate using spherical polar coordinates:
begin{equation}
limlimits_{etarightarrow 0}int_{0}^{pi}mathrm{d}theta,sinthetaint_{0}^{2pi}mathrm{d}varphi,int_{0}^{1}mathrm{d}t,frac{-3eta^{2}t^{2}}{(t^{2}+eta^{2})^{5/2}}(rho(0)+O(t^{2}))
end{equation}
Integrating you will get that all the terms contained in $O(t^2)$ vanish as $etarightarrow 0$, while the term with $rho(0)$ remains. In fact you get
begin{align}
limlimits_{etarightarrow 0}frac{-4pirho(0)}{sqrt{1+eta^{2}}}&=-4pirho(0)\
&=langle -4pidelta_{0}^{(3)}|rhorangle
end{align}

From this argument one defines the limit distribution
begin{equation}
nabla^{2}left(frac{1}{r}right):=limlimits_{etarightarrow 0}nabla^{2}left(frac{1}{r_{eta}}right)=-4pidelta_{0}^{(3)}
end{equation}
The generalization to $r=|mathbf{x}-mathbf{x}_{0}|$ is obvious.

Here is a back-alley derivation using Fourier transform properties.
Take the Fourier transform of $frac{1}{4 pi r}$ to get $frac{1}{k^2}$. Therefore,
$$
frac{1}{4pi r} = int frac{d^3k}{(2pi)^3} frac{e^{-ikcdot r}}{k^2}
$$
Now, apply $-nabla^2$ to get
$$
-nabla^2 frac{1}{4pi r} = int frac{d^3k}{(2pi)^3} e^{-ikcdot r}
$$
This is the delta function, but we can explicitly put in our test function:
begin{align*}
int d^3r left(-nabla^2 frac{1}{4 pi r}right) f(r) &=
int frac{d^3k}{(2pi)^3} int d^3r e^{-ikcdot r} f(r) \
&=
int frac{d^3k}{(2pi)^3} tilde{f}(k) \
& = f(r=0).
end{align*}