Introduction to Electrodynamics · Problem 1.64

Vector Analysis — David J. Griffiths

Physics · dirac-delta-function-3d

Problem

In case you're not persuaded that ∇2(1/r)=−4πδ3(r)\nabla^2(1/r) = -4\pi\delta^3(\mathbf{r}) (Eq. 1.102 with r′=0\mathbf{r}' = \mathbf{0} for simplicity), try replacing rr by r2+ϵ2\sqrt{r^2 + \epsilon^2}, and watching what happens as ϵ→0\epsilon \to 0. Specifically, let D(r,ϵ)≡−14π∇21r2+ϵ2.D(r, \epsilon) \equiv -\frac{1}{4\pi} \nabla^2 \frac{1}{\sqrt{r^2 + \epsilon^2}}. To demonstrate that this goes to δ3(r)\delta^3(\mathbf{r}) as ϵ→0\epsilon \to 0: (a) Show that D(r,ϵ)=(3ϵ2/4π)(r2+ϵ2)−5/2D(r, \epsilon) = (3\epsilon^2/4\pi)(r^2 + \epsilon^2)^{-5/2}. (b) Check that D(0,ϵ)→∞D(0, \epsilon) \to \infty, as ϵ→0\epsilon \to 0. (c) Check that D(r,ϵ)→0D(r, \epsilon) \to 0, as ϵ→0\epsilon \to 0, for all r≠0r \neq 0. (d) Check that the integral of D(r,ϵ)D(r, \epsilon) over all space is 1.
Intuitive explanation

The Laplacian of 1/r1/r is singular at the origin, but you can understand it by regularizing: replace rr with r2+ϵ2\sqrt{r^2 + \epsilon^2} (which smooths out the singularity), compute the Laplacian, then take ϵ→0\epsilon \to 0. The three properties you'll verify—infinite spike at origin, zero elsewhere, and unit integral—are exactly what define the Dirac delta function, so the limit recovers the famous result.

Intuitive explanation

The Laplacian of 1/r1/r is singular at the origin, but you can understand it by regularizing: replace rr with r2+ϵ2\sqrt{r^2 + \epsilon^2} (which smooths out the singularity), compute the Laplacian, then take ϵ→0\epsilon \to 0. The three properties you'll verify—infinite spike at origin, zero elsewhere, and unit integral—are exactly what define the Dirac delta function, so the limit recovers the famous result.

Related theory

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