1
$\begingroup$

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 $\nabla\cdot{F} = \operatorname{div} F$)?

$\endgroup$
1
  • 2
    $\begingroup$ It's worth pointing out that $\nabla \cdot F$ is not really the dot product of $\nabla$ and $F$ . . . $\endgroup$ – ruakh Oct 15 '14 at 8:36
6
$\begingroup$

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 $\nabla \cdot F = 1$; $F \cdot \nabla = x\cdot \frac{d}{dx}$.

$\endgroup$
1
  • 8
    $\begingroup$ Appropriate user image. $\endgroup$ – user856 Oct 15 '14 at 10:07
2
$\begingroup$

$\nabla \cdot F \in \mathbb{R}$, 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.

$\endgroup$
2
$\begingroup$

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.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.