TITLE: Intuition on the curl formula
QUESTION [17 upvotes]: I'm trying to understand the formula for the curl:  $$\operatorname{Curl}(F) = \nabla \times F $$
Why the vector product? What means the vector product of a vector and a operator? What is the meaning of each of the resulting terms $  \frac { \partial F_y }{ \partial x } -\frac { \partial F_x }{ \partial y } $  ?
I think that I understand the concept of the curl, I recommend this if anyone is interested. The problem is in the formula.

REPLY [2 votes]: let $\mathbf{F}=(L(x,y),M(x,y))$
from Green's theorem
$\oint_{C} (\bf F\, \cdot d \bf r) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dx\, dy $
now define $\operatorname{Curl}(F)$ = $\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}$
now assume you want to find the curl formula for space.
from Green's theorem we know that the curl of gradient field must be zero . we will use that to get formula for curl in space
so let $f=f(x,y,z)$
$$\nabla f=\mathbf{F}=(\frac { \partial f }{ \partial x },\frac { \partial f }{ \partial y },\frac { \partial f }{ \partial z })$$
$\frac { \partial f }{ \partial x }=P$ , $\frac { \partial f }{ \partial y }=Q$ , $\frac { \partial f }{ \partial z }=R$
and since we can interchange the order of taking partial derivatives

$\frac { \partial^2 f }{ \partial x \partial y}=\frac { \partial^2 f }{ \partial y \partial x}$ $\Rightarrow$ $\frac { \partial P }{ \partial y }=\frac { \partial Q }{ \partial x }$
$\frac { \partial^2 f }{ \partial x \partial z}=\frac { \partial^2 f }{ \partial z \partial x}$ $\Rightarrow$ $\frac { \partial P }{ \partial z }=\frac { \partial R }{ \partial x }$
$\frac { \partial^2 f }{ \partial y \partial z}=\frac { \partial^2 f }{ \partial z \partial y}$ $\Rightarrow$ $\frac { \partial Q }{ \partial z }=\frac { \partial R }{ \partial y }$

so for any vector field $\mathbf{F}=(P(x,y,z),Q(x,y,z),R(x,y,z))$ to be gradient field, it must satisfy those conditions

$\frac { \partial Q }{ \partial z }=\frac { \partial R }{ \partial y }$
$\frac { \partial P }{ \partial z }=\frac { \partial R }{ \partial x }$
$\frac { \partial P }{ \partial y }=\frac { \partial Q }{ \partial x }$

we combine them in one formula so that when the vector field is gradient field ,thus satisfying the above conditions, the curl would be zero
$\operatorname{Curl}(F)$=($\frac { \partial Q }{ \partial z }-\frac { \partial R }{ \partial y }$)+($\frac { \partial P }{ \partial z }-\frac { \partial R }{ \partial x }$)+($\frac { \partial Q }{ \partial x }-\frac { \partial P }{ \partial y }$)
then I'll rewrite it in differently to make it easier to remember (in determinant form)
$\operatorname{Curl}(F)$=($\frac { \partial Q }{ \partial z }-\frac { \partial R }{ \partial y }$)-($\frac { \partial R }{ \partial x }-\frac { \partial P }{ \partial z }$)+($\frac { \partial Q }{ \partial x }-\frac { \partial P }{ \partial y }$)
now after we reached this step we notice that this almost exactly the same as the expansion of cross product
$\begin{align} \mathbf{u}\times\mathbf{v}\ =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ u_1&u_2&u_3\\ v_1&v_2&v_3\\ \end{vmatrix} =&(u_2v_3-u_3v_2)\mathbf{i}-(u_1v_3-u_3v_1)\mathbf{j}+(u_1v_2-u_2v_1)\mathbf{k}\\ \end{align}$
let $\nabla=\mathbf{u}$ , $\mathbf{F}=\mathbf{v}$
$\begin{align} \nabla\times\mathbf{F}\ =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ {\frac{\partial}{\partial x}}&{\frac{\partial}{\partial y}}&{\frac{\partial}{\partial z}}\\ P&Q&R\\ \end{vmatrix} =&(\frac { \partial R }{ \partial y }-\frac { \partial Q }{ \partial z })\mathbf{i}-(\frac { \partial R }{ \partial x }-\frac { \partial P }{ \partial z })\mathbf{j}+(\frac { \partial Q }{ \partial x }-\frac { \partial P }{ \partial y })\mathbf{k}\\ \end{align}$
which is precisely the curl.
$\operatorname{Curl}(F)=\nabla\times\mathbf{F}\ $
and as Hayden mentioned that doesn't mean that we take the cross product of any thing it's just abuse of the notation because it's convenient that way.