Laplacian and Dirac function gives contradictory result.

Asked Nov 16 '19 at 13:59

Active Nov 16 '19 at 13:59

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The following equation is correct for all non-negative real numbers:

$$4\pi\delta^{(3)}(\mathbf{r})=\nabla\cdot\frac{\mathbf{r}}{r^{3}},$$ $$r\in[0,+\infty)$$

especially, when r=0, both sides give infinity.

However, the right handside can further give:

$$\nabla\cdot\frac{\mathbf{r}}{r^{3}}=\frac{\nabla\cdot\mathbf{r}}{r^{3}}+\mathbf{r}\cdot\nabla\frac{1}{r^{3}}=\frac{3}{r^{3}}-\frac{3}{r^{3}}=0$$

which is very strange, since $$4\pi\delta^{(3)}(\mathbf{r})=0,$$

I have read the proof for $r\neq0$ at Laplacians and Dirac delta functions enter link description here, but there, this issue is not addressed.

Can anyone explain what happens here?

asked Nov 16 '19 at 13:59

ZHANG Juenjie

2

The equality holds in the sense of distributions. This is not an equality of functions. Note that $\frac{{\bf{r}}}{r^{3}}$ is not even defined at $r = 0$, so it cannot be differentiable at this point. Thus, your conclusion $\nabla \cdot \frac{{\bf{r}}}{r^{3}}=0$ holds if $r\neq 0$. – IamWill Nov 16 '19 at 14:24
Crossposted to https://physics.stackexchange.com/q/514339/2451 – Qmechanic Nov 17 '19 at 06:27
Related : Divergence of r^r2 , what is the 'paradox'? – Frobenius Nov 17 '19 at 08:04

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