# Why is the Laplacian of $1/r$ a Dirac delta? [duplicate]

How does one show that $\nabla^2 1/r$ (in spherical coords) is the Dirac delta function ? Intuitively, it would seem that the function undefined at the origin and I'm not able to construct a limiting argument that avoids this problem.

– user147263
Commented Apr 16, 2016 at 22:51
• Hint: you want to show that $\int_{\mathbb{R}^3} \nabla^2(1/r) v dV = c v(0)$ whenever $v$ is smooth with compact support, for some $c$ (which turns out to not be $1$). Write this as an integral over a ball $B$ whose boundary is outside the support of $v$. Then integration by parts gives $-\int_B \nabla(1/r) \cdot \nabla(v) dV$. Now write this integral in spherical coordinates. You should find that the $r^2$ cancels out, leaving behind a well-defined integral.
– Ian
Commented Apr 16, 2016 at 23:50

Delta function is zero everywhere except at origin, and integration over space is zero. The first property is easy to prove with vector identities. For the second property:

$$I = \iiint_V \nabla^2 \frac{1}{r} d V = \iiint_V \nabla \cdot \nabla \frac{1}{r} dV$$ With divergence theorem: $$I = \oint_S \nabla \frac{1}{r} \vec{dS} = \oint_S - \frac{1}{r^2} \vec{n} \vec{dS} = \int_{\Omega} - \frac{1}{r^2} r^2 {d \Omega} = -4 \pi = -4 \pi \iiint_V \delta(r) dV$$ That's all, your question missed the factor $-4 \pi$.

As to the test function (as response to comment below), the spherical surface in the second integral has to be infinitely small around the origin. In this way $f(r)$ in an integration like (The volume $V$ can be any size but only the infinitesimal region around the origin matters, as at outside laplacian of 1/r is zero): $$\iiint_V f \nabla^2 \frac{1}{r} dV$$ can be replaced by the constant $f(0)$. Of course $f$ must be continuous and finite around the origin.

• That's a bit deceptive, because you haven't included any test functions...
– Ian
Commented Apr 16, 2016 at 23:47
• @Ian Edited, I added discussion of test function. Commented Apr 17, 2016 at 0:01
• I don't think that really clarifies the situation. The point is that your first argument establishes that $\nabla^2(1/r)$ acts like a certain multiple of the Dirac delta when applied to $1$. It doesn't say how it acts when applied to nonconstant functions.
– Ian
Commented Apr 17, 2016 at 0:43
• @Ian I mentioned this. $\nabla^2 1/r$, after multiplied with a non-constant function $f$ and put inside an integral, acts like a filter to $f$. It is zero for the whole space except the infinitesimal volume around the origin. Within this volume $f$ can be replaced by $f(0)$. Commented Apr 17, 2016 at 3:23
• But why can't, say, $\nabla^2(1/r)$ be a derivative of a Dirac delta rather than just a Dirac delta?
– Ian
Commented Apr 17, 2016 at 3:44