$\left( \nabla \cdot \overrightarrow{r} \right)=\sum_{i=1}^{3}\frac{\partial X_{i}}{\partial x_{i}}=3$

Question

Hi I have a silly question about a expression that I have found.

Let:

$$\nabla = e_{r}\frac{\partial}{\partial r}+e_{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}+e_{\phi}\frac{1}{r \sin(\theta)}\frac{\partial}{\partial \phi}$$ $$\overrightarrow{r}=r \, e_{r}$$ so $$\left( \nabla \cdot \overrightarrow{r} \right)=\sum_{i=1}^{3}\frac{\partial X_{i}}{\partial x_{i}}=3$$ I can not understand the argument of the $\sum$ and the following result. Thank you in advice.

Answer 1

In cartesian coordinates

$$ \nabla \cdot {\bf A} = {\partial A_x \over \partial x} + {\partial A_y \over \partial y} + {\partial A_z \over \partial z} $$

So if you take ${\bf A} = {\bf r} = x\hat{x} + y\hat{y} + z\hat{z}$, then

$$ \nabla \cdot {\bf r} = \frac{\partial r_x}{\partial x} + \frac{\partial r_y}{\partial y}+ \frac{\partial r_z}{\partial z} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1+ 1 = 3 \tag{1} $$

In spherical coordinates

$$ \nabla\cdot {\bf A} = {1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r} + {1 \over r\sin\theta}{\partial \over \partial \theta} \left( A_\theta\sin\theta \right) + {1 \over r\sin\theta}{\partial A_\varphi \over \partial \varphi} $$

So that for the field ${\bf A} = {\bf r} = r\hat{r}$ the divergence becomes

$$ \nabla \cdot {\bf r} = \frac{1}{r^2}\frac{\partial (r^3)}{\partial r} = 3 \tag{2} $$

The expression you have in your question is for the gradient, not the divergence