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Reading through the proof on the Helmholtz decomposition of a vector field, I came across the following identity: $$\delta(x-x')=-1/4\pi*\nabla^2*(|x-x'|)^{-1}$$

Does anyone have any insight on how to prove/derive this identity? Thanks in advance. Here's the article for reference:

http://faculty.uml.edu/cbaird/95.657(2013)/Helmholtz_Decomposition.pdf

Mr. Vubio
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1 Answers1

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There are a couple of ways to derive this, Ill sketch out what I think is the most intuitive.

If you write out the algebra for $\nabla \cdot \nabla (1/\vert \vec{x}-\vec{x}'\vert) $ you will find formally that it is zero. However the algebraic manipulations you perform are only valid in a neighborhood where the functions is continuous and differentiable. There is one glaring place where that isn't true and that is when $\vec{x}=\vec{x}'$.

The question is how can we define the divergence of the gradient at a point where it technically doesn't exist. The key is by thinking of the divergence of a vector field at a point as the flux per unit volume of the vector field at the given point. This is motivated by Gauss' Theorem,

$$ \int_{\partial V} \vec{A} \cdot d\vec{a} = \int_V \nabla \cdot V dV$$

So to find the divergence at $\vec{x}=\vec{x}'$ we calculate the flux of the gradient over the surface of a sphere centered at $\vec{x}'$ and then evaluate the limit of the flux as the radius of the sphere goes to $0$.

$$ Flux = \int_S \nabla \frac{1}{\vert \vec{x} - \vec{x}'\vert } \cdot \hat{n} da = -\int_S \frac{\vec{x} - \vec{x}'}{\vert \vec{x} - \vec{x}' \vert^3} \cdot \frac{\vec{x} - \vec{x}'}{\vert \vec{x} - \vec{x}' \vert } da \int_S \frac{\vec{R}}{R^3} \cdot \frac{\vec{R}}{R} R^2 d\Omega = \int_S d\Omega = -4\pi$$

So the flux per unit volume is $\lim_{R\rightarrow 0 } -4\pi/V = \infty$. We have that the lapacian of our function is zero everywhere $ \vec{x} \neq \vec{x}'$ and is infinite at $\vec{x}=\vec{x'}$ but integrates to a finite value. This is clearly a delta function so we conclude that,

$$ \nabla^2 \frac{1}{\vert \vec{x} - \vec{x}' \vert} = -4\pi \delta^{(3)}(\vec{x}-\vec{x}') $$

Thought I would add an explanation about what a delta function is:

A Dirac delta function can be thought of as a normalized distribution which is nonzero only at one point.

So we write,

$$ \delta^{(3)}(\vec{x}) = \begin{cases} 0 \quad \text{when } \vec{x} \neq 0 \\ \infty \quad \text{when } \vec{x} = 0 \end{cases} $$

with the understanding that

$$ \int_{V} \delta^{(3)}(\vec{x}) dV = \begin{cases} 1 \quad \text{if } 0 \in V \\ 0 \quad \text{if } 0 \notin V \end{cases}$$

You can construct such a "function" by taking the limit of some appropriate distribution. In 1 dimension we could write,

$$ \int \delta(x) f(x) dx = \lim_{n \rightarrow \infty} \int \delta_n(x) f(x) dx; \quad \text{where} \quad \delta_n(x) = \sqrt{\frac{n}{\pi}} e^{-n x^2}$$

Notice the importance of keeping the limit outside of the integral otherwise we wouldn't have something we could integrate.

Spencer
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  • For a more formal derivation see J.D. Jackson's Classical Electrodynamics book. – Spencer Nov 26 '13 at 04:31
  • Thanks Spencer, there's still one small thing that isn't clicking for me; since the function isn't defined at x=x', and the value of the function approaches infinity at the undefined point, does this imply that the laplacian of the function at that point will be infinity as well? – Mr. Vubio Nov 26 '13 at 21:37
  • Informally yes, a quick and loose definition of the delta function is a "function" which is zero when $x\neq x'$, infinity when $x=x'$, integrates to a finite value when the region of integration includes the singularity. – Spencer Nov 26 '13 at 21:44
  • Sorry to keep bothering you, but looking back over the derivation is Gauss' Theorem applicable to this scenario due to the singularity? Or is this the part I'd need to look at in the formal derivation of the identity? – Mr. Vubio Nov 26 '13 at 22:38
  • No problem, it depends on how you define divergence. Some people prefer to define it as the flux per unit volume in a limiting process and then derive the differential operator from that. There is no doubt that the flux is $-4\pi$ if you include $x'$ in the volume and $0$ if you don't include it so it at the very least tells you that something is up. But yes there are more formal proofs which involve removing the singularity computing the divergence and then putting it back. No matter what you do it though you will always have a singularity involved somewhere. – Spencer Nov 27 '13 at 07:12
  • I haven't read it myself but I here Lighthill's book on Generalised functions is quite good if you want to study more of this, http://www.amazon.com/Introduction-Generalised-Functions-Cambridge-Monographs/dp/0521091284/ref=sr_1_1?ie=UTF8&qid=1385536390&sr=8-1&keywords=lighthill+generalised. – Spencer Nov 27 '13 at 07:14