simple question about $\nabla r$

Question

In my physics notes, it says

$\nabla r = \underline{e_r} = \frac{\underline{r}}{r}$

and

$\nabla \frac{1}{r} = - \frac{\underline{r}}{r^3} = - \frac{1}{r^2} \underline{e_r}$

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I don't quite understand why this is?

I know that $\nabla \varphi = (\frac{\delta\varphi}{\delta x},\frac{\delta\varphi}{\delta y},\frac{\delta\varphi}{\delta z})$ but I do not understand the above.

Thanks.

Answer 1

Note that $r^{2} = x^{2} + y^{2} + z^{2}$. Taking gradients and using the chain rule, $$ 2r\, \nabla r = 2(x, y, z) = 2\vec{r},\quad\text{or}\quad \nabla r = \frac{\vec{r}}{r}. $$ For the second, apply the chain rule and use the preceding identity.

Answer 2

In rectangular coordinates, $$(\nabla r)_k = \frac{\partial r}{\partial x_k}.$$ Therefore, you must calculate $$\frac{\partial r}{\partial x_k} = \frac{\partial}{\partial x_k}\left(\sum_k (x_k)^2\right)^\frac{1}{2}.$$ You will see that the result is $$\frac{\partial r}{\partial x_k} = \frac{x_k}{r}.$$ Consequently $$\nabla r = \left(\frac{\partial r}{\partial x}, \, \frac{\partial r}{\partial y}, \, \frac{\partial r}{\partial z} \right) = \left(\frac{x}{r}, \, \frac{y}{r}, \, \frac{z}{r}\right) = \frac{1}{r}(x, \, y, \, z) = \frac{\mathbf{r}}{r}.$$ Similarly, you can prove the formula $$\nabla \left(\frac{1}{r}\right) = -\frac{\mathbf{r}}{r^3}$$ by showing that $$\left[\nabla\left(\frac{1}{r}\right)\right]_k = \frac{\partial}{\partial x_k}\left(\frac{1}{r}\right) = -\frac{x_k}{r^3}.$$