# Is $\nabla\cdot{F} = F\cdot\nabla$?

According to the vector dot product, $a\cdot{}b = b\cdot{a}$ for all $a, b.$

However, is $\nabla\cdot{F} = F\cdot\nabla$
(where $\nabla\cdot{F} = \operatorname{div} F$)?

• It's worth pointing out that $\nabla \cdot F$ is not really the dot product of $\nabla$ and $F$ . . . – ruakh Oct 15 '14 at 8:36

## 3 Answers

No. $\nabla \cdot F$ is a function, whereas $F \cdot \nabla$ is an operator (a function of a function). It is easily seen with an example: let $F = x$. Then $\nabla \cdot F = 1$; $F \cdot \nabla = x\cdot \frac{d}{dx}$.

• Appropriate user image. – user856 Oct 15 '14 at 10:07

$\nabla \cdot F \in \mathbb{R}$, whereas $F \cdot \nabla = F_1 \frac{\partial}{\partial x_1} + F_2 \frac{\partial}{\partial x_2} + \ldots + F_n \frac{\partial}{\partial x_n}$ which makes it an operator.

The commutativity of vector terms in a vector dot product requires commutativity of the members of the vectors.   The members of the nabla are partial differentiation operators, they don't commute.

$$\frac{\partial }{\partial x} f(x) \not\equiv f(x) \frac{\partial }{\partial x} \implies \nabla\cdot \vec F(\vec x) \not\equiv \vec F(\vec x)\cdot \nabla$$