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.

  • 2
    $\begingroup$ Intuition doesn't always work. You checked that it is not zero everywhere outside O. $\endgroup$ – FGSUZ Feb 17 '18 at 14:29
  • 2
    $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.