In connection with introduction to electro dynamics, we deduced that $$\frac{\vec r}{r^3}=-\nabla\left(\frac{1}{r}\right)$$ This is fairly obvious for three dimensions, but we did it in $n$ dimensions.

In the process of derivation, there was the following step included: $$-\nabla\left(\frac{1}{r}\right)=-\left(\frac{d}{dr}\frac{1}{r}\right)\cdot(\nabla r)$$ Once again, the correctness of this can be easily checked in three dimensions, but it seemed kind of obvious that this is generally true. Is this a commonly known rule? And how does one derive it?


Let $$\vec{v}=\operatorname{grad}\left(\frac{1}{r}\right)$$ Then we have that $$\begin{align} v_i&=\partial_i \frac{1}{r}\\ &= \partial_i r^{-1}\\ &= -r^{-2} (\partial_i r)\\ &= -r^{-2} (\partial_i \sqrt{x_jx_j})\\ &= -r^{-2} \frac{1}{2\sqrt{x_jx_j}} \partial_i (x_jx_j)\\ &= -r^{-2} \frac{1}{2 r} 2(\partial_ix_j)x_j\\ &=-r^{-3} \delta_{ij}x_j\\ &=-\frac{x_i}{r^3} \end{align}$$ i.e. $$\vec{v}=-\frac{\vec{r}}{r^3}=\operatorname{grad}\left(\frac{1}{r}\right)$$

  • $\begingroup$ But you've implied the part I don't understand again - why is $\partial_ir^{-1}=-r^{-2}(\partial_ir)$? $\endgroup$ – MetaColon Apr 2 at 21:12
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    $\begingroup$ @MetaColon It's the chain rule: $\frac{\partial r^{-1}}{\partial x_i}=\frac{\partial r^{-1}}{\partial r}\frac{\partial r}{\partial x_i}$ $\endgroup$ – Botond Apr 2 at 21:12
  • $\begingroup$ Oh I see, it's because you firstly derive the outer and then the inner part. Thanks! $\endgroup$ – MetaColon Apr 2 at 21:13
  • $\begingroup$ @MetaColon Yes. You're welcome! $\endgroup$ – Botond Apr 2 at 21:14
  • $\begingroup$ @Botond very nice answer (+1) $\endgroup$ – the_candyman Apr 2 at 21:28

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