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My intuition tells me that the divergence of the vector field

$$\vec{E} = \dfrac{\hat{r}}{r^3} $$

should be zero everywhere except at the origin. So I think it should be

$$ \vec{\nabla}\cdot \vec{E} = 4\pi \delta^3(\vec{r}).$$

However, if I use the polar version of divergence on this I get:

$$ \vec{\nabla}\cdot \vec{E} = \dfrac{1}{r^2}\dfrac{\partial r^2 \cdot 1/r^3}{\partial r} = -\dfrac{1}{r^4}, $$

which is quite upsetting. Any help would be greatly appreciated. I suppose I made some dumb misktake but I don't see where.

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    $\begingroup$ Intuition doesn't always work. You checked that it is not zero everywhere outside O. $\endgroup$ – FGSUZ Feb 17 '18 at 14:29
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    $\begingroup$ Did you mean $\dfrac{{\bf r}}{r^3}\equiv\dfrac{ \hat{\bf r}}{r^2}$ rather than $\dfrac{ \hat{\bf r}}{r^3}$? $\endgroup$ – Qmechanic Feb 17 '18 at 14:29
  • $\begingroup$ @Qmechanic, no, I did mean $\dfrac{\hat{r}}{r^3}$. A spherical symmetric field centered at $\vec{O}$ which drops with the distance cubed. $\endgroup$ – PaleBlueDot Feb 17 '18 at 14:40
  • $\begingroup$ I now see why my intuition was wrong. $\endgroup$ – PaleBlueDot Feb 17 '18 at 14:56
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If the divergence of your vector field was really zero everywhere, then it should be the case that the surface integral $$ \int_V (\vec{\nabla} \cdot \vec{E}) \, dV = \oint_{\partial V} \vec{E} \cdot d\vec{A} $$ should be zero for any volume (and its bounding surface) I care to name.

So let's try to calculate this for a thick spherical shell centered at the origin, with inner radius $a$ and outer radius $b$. In this case, the boundary of my volume has two parts: an inner boundary at $r = a$, with normal vector $\hat{n} = - \hat{r}$, and an outer boundary at $r = b$, with normal vector $\hat{n} = \hat{r}$. The surface integral of $\vec{E}$ over the boundary is then $$ \oint_{\partial V} \vec{E} \cdot d\vec{A} = \oint_{r = a} \left( \frac{\hat{r}}{r^3} \right) \cdot (-\hat{r}\, dA) + \oint_{r = b} \left( \frac{\hat{r}}{r^3} \right) \cdot (\hat{r}\, dA) \\= -\oint_{r = a} \left( \frac{1}{a^3} \right) dA + \oint_{r = b} \frac{1}{b^3} dA \\ = - \frac{4 \pi a^2}{a^3} + \frac{4 \pi b^2}{b^3} \\ = 4 \pi \left( \frac{1}{b} - \frac{1}{a} \right) \neq 0. $$ So as we can see, it cannot be the case that the divergence of the vector field $\vec{E} = \hat{r}/r^3$ is zero, because this integral never vanishes.

It should also be a bit clearer from this argument why the case of of $\vec{E} = \hat{r}/r^2$ is special: in that case, the $r^2$ behavior of the surface area exactly cancels out the $1/r^2$ behavior of the vector field, and so we have $$ \oint_{\partial V} \vec{E} \cdot d\vec{A} = \oint_{r = a} \left( \frac{\hat{r}}{r^3} \right) \cdot (-\hat{r}\, dA) + \oint_{r = b} \left( \frac{\hat{r}}{r^3} \right) \cdot (\hat{r}\, dA) \\ = - \frac{4 \pi a^2}{a^2} + \frac{4 \pi b^2}{b^2} =0. $$ The above argument then makes it plausible, at least, that the divergence of this vector field is zero at points other than the origin.

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