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In Griffiths Intro Electrodynamics, it is shown that $$\nabla \cdot( \frac{\hat{r} }{r^2}) = 4 \pi \delta^3 (r) \tag{1}$$

Eqtn discussed in this post and linked.

Now in page 72 of the international 4th ed of the book, the following equation is shown:

$$ \nabla \cdot ( \frac{\hat{\hat{q}} }{|\vec{q}|^2}) = 4 \pi \delta^3 (q) \tag{2}$$

Where $\vec{q}$ is defined as $ \vec{q}= \vec{r} - \vec{r'}$ , in which, $\vec{r'}$ is a vector from a 'source' to a $\vec{r}$ which is a fixed point in space, and, $\hat{q} =\frac{\vec{q} }{|\vec{q} |}$. In application, we do integrals where we integrate quantites multiplied by equation (2) over all sources. I want to know how to extrapolate eqtn (2) from eqtn (1).

One possibility to prove $(2)$ is to expand everything out in cartesian coordinates and take derivatives but that is a bit messy. Can we explain (2) through (1) .. somehow?

  • Translate by $r'$ ? – SolubleFish May 26 '21 at 09:38
  • Hmm I had thought that, but I wasn't sure how to write it. How do I write about translating my coordinates from the $(r,\theta,\phi)$ to be centered at $r'$ and then taking derivative in this new coordinates. I guess once that is done I can argue that gradient must be invariant (i.e: depend on geometric points instead of coordinates) – tryst with freedom May 26 '21 at 09:44
  • Don't do this in spherical coordinates ! Do the translation in euclidean coordinates. – SolubleFish May 26 '21 at 09:52

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