What will be value of $\vec{r} \cdot \nabla$

Question

I was studying on Nabla Operator and saw that $\nabla \cdot \vec{r} \neq \vec{r} \cdot \nabla$

So, if I were to find $\vec{r} \cdot \nabla$ how would I calculate it?

I know that $\vec{r} \cdot \nabla$ = $r_1 \cdot \frac{\partial}{\partial x} + r_2 \cdot \frac{\partial}{\partial y} + r_3 \cdot \frac{\partial}{\partial z}$

but what I'm confused is on what am i suppose to differentiate?

Thanks.

Answer 1

The expressions $\vec r\cdot \nabla$, $\vec r\times \nabla$, etc. are operators.

Thus, if $\Phi$ is a scalar field and $\vec F$ is a vector field, then

$$(\vec r\cdot \nabla)(\Phi)=\sum_i x_i \frac{\partial \Phi}{\partial x_i}$$

is a scalar

$$(\vec r\cdot \nabla)(\vec F)=\sum_i x_i \frac{\partial \vec F}{\partial x_i}$$

is a vector

$$(\vec r\times \nabla)(\Phi)=\sum_i \sum_j (x_i\hat x_i\times \hat x_j) \frac{\partial \Phi}{\partial x_j}$$

is a vector

$$(\vec r\times \nabla)(\vec F)=\sum_i \sum_j (x_i\hat x_i\times \hat x_j) \frac{\partial \vec F}{\partial x_j}$$

is a tensor.

Answer 2

The difference is that $\nabla \cdot \textbf{r}$ is a scalar, but $\textbf{r}\cdot \nabla$ is another operator. You're not going to get a scalar from this latter quantity unless you act it on a scalar field.