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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.

firdaus
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    See this answer –  Apr 16 '16 at 22:51
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    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 Apr 16 '16 at 23:50

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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.

Taozi
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  • That's a bit deceptive, because you haven't included any test functions... – Ian Apr 16 '16 at 23:47
  • @Ian Edited, I added discussion of test function. – Taozi Apr 17 '16 at 00:01
  • 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 Apr 17 '16 at 00:43
  • @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 $f(0)$. – Taozi Apr 17 '16 at 03:23
  • But why can't, say, $\nabla^2(1/r)$ be a derivative of a Dirac delta rather than just a Dirac delta? – Ian Apr 17 '16 at 03:44
  • The derivative of $\delta$ function has different property. At least the integration of it over space is $0$. Besides, I see no way in which $\delta'(r)$ can emerge. – Taozi Apr 17 '16 at 04:25
  • You're right, there needs to be a regular delta function in there so that you get the right result when you apply it to $1$. But now all you've said is that you have a distribution whose support is ${ 0 }$ which gives $-4 \pi$ when applied to $1$. This could include additional derivatives of the Dirac delta. You haven't shown that it doesn't. – Ian Apr 17 '16 at 13:03
  • Let us continue this discussion in chat. – Taozi Apr 17 '16 at 14:58