Divergence proof for a vector field

Question

I have to demonstrate that $\textrm{div}\left(\frac{\mathbf{r}}{r^3}\right)=0. $

I know that $\text{div}(\mathbf{r}\frac{1}{r^3})= \frac{1}{r^3}\text{div}\mathbf{r}+\nabla\frac{1}{r^3}\mathbf{r}$. Since $\nabla\frac{1}{r^3}\mathbf{r}=0$ Then how $(\frac{1}{r^3}div\mathbf{r})=0?$

Answer 1

The first part of your reasoning is not correct. Indeed the gradient of $1/r^3$ is $$\vec{\nabla} \frac{1}{r^3} = -3\frac{1}{r^4}\,\frac{\vec{r}}{r}$$

Then $$\vec{\nabla} \frac{1}{r^3}\cdot\vec{r} = -\frac{3}{r^3}$$

Furthermore, $\mathrm{div}(\vec{r}) = 3$, which proves the equality (Indeed if $\vec{r} = \vec{x}+\vec{y}+\vec{z}$, you have $\mathrm{div}(\vec{r}) = \frac{\partial \vec{x}}{\partial x}+\frac{\partial \vec{y}}{\partial y}+\frac{\partial \vec{z}}{\partial z}=1+1+1=3$)

Answer 2

Use divergence operator in spherical coordinates. Here you have $F_r = \frac{1}{r^2}$, (why?). Then you get $$ \nabla .F = \frac{1}{r^2} \frac{\partial (r^2 F_r)}{\partial r} = \frac{1}{r^2} \frac{\partial (1)}{\partial r} =0$$

Notice that this requires $r\ne 0$.

Answer 3

$$\nabla\cdot f\vec{A}=\nabla f\cdot\vec{A}+f\nabla\cdot\vec{A}$$ $$\nabla\cdot\left(\frac{\vec{r}}{r^3}\right)=\nabla\left(\frac{1}{r^3}\right)\cdot\vec{r}+\frac{1}{r^3}\nabla\cdot\vec{r}$$ $$LHS=\frac{-3}{r^4}\hat{r}\cdot r\hat{r}+\frac{1}{r^3}\nabla\cdot\vec{r}$$

$$\nabla\cdot\vec{r}=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2.r\right)=3$$ Hence, zero