I’m currently working on the practice problems in Introduction to Electrodynamics by Griffiths. I got confused by the solution to this problem.

enter image description here

What does “ill-defined divergence” even mean? I understand how and when to use delta function, but I don’t understand how divergence is not defined.

  • $\begingroup$ is the delta-function well defined at the origin? $\endgroup$ Nov 26, 2022 at 0:02
  • $\begingroup$ I suggest that you evaluate the divergence of the ${\bf v}$ field. Is there any delta function? And what is the value of the divergence at the origin? $\endgroup$ Nov 26, 2022 at 0:15
  • $\begingroup$ @GiorgioP I think I got it. When $n<-2$, there’s a zero in the denominator if we want to calculate the divergence at the origin $\endgroup$
    – Irene
    Nov 26, 2022 at 0:22

1 Answer 1


I think we can use

$$ \nabla \cdot (\psi \vec{a}) = \vec{a} \cdot \nabla \psi + \psi \nabla \cdot \vec{a} $$

to see what's happened for $n \lt -2$.

$$\begin{align*} \nabla \cdot (r^{-3} \hat{r}) &= \left( \frac{1}{r^2} \hat{r} \right) \cdot \nabla \frac{1}{r} + \frac{1}{r} \nabla \cdot \frac{1}{r^2} \hat{r} \\ &= - \frac{1}{r^4} + \frac{1}{r} \nabla \cdot \frac{1}{r^2} \hat{r} && \left( \nabla \frac{1}{r} = - \frac{1}{r^2} \right) \\ &= - \frac{1}{r^4} + \frac{4 \pi}{r} \delta^3(\vec{r}) \end{align*}$$

When $r$ toward $0$ (below misuses the delta function, delta function is meaning less outside of integral),

$$\begin{align*} \nabla \cdot (r^{-3} \hat{r}) &= - \frac{1}{r^4} + \frac{4 \pi}{r} \delta^3(\vec{r}) \\ &= -\infty + \infty \cdot \infty \end{align*}$$

We can not assign a meaningful value to $\nabla \cdot (r^{-3} \hat{r})$, so it is called "ill-defined".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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