What is $\nabla\wedge\vec{\mathbf F}(x,y,z)$

Question

What is $\nabla\wedge\vec{\mathbf F}(x,y,z) \quad$ where F is a vector field?

I assume its the dual of the curl but I need an equation. And what is it called? Google is getting me nowhere.

Answer 1

You can start with the formula for the curl and work your way backwards. We have that

$$\operatorname{curl} \vec{F} = (\partial_yF_z - \partial_z F_y)dx + (\partial_z F_x - \partial_x F_z)dy + (\partial_xF_y - \partial_yF_x)dz$$

We have that in $3$ dimensions the formula for the Hodge dual of a $1$ form by applying the Hodge star operator (the buzzwords to Google) is given by

$$\vec{v} = a dx + b dy + c dz \implies \star \vec{v} = \begin{pmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \\ \end{pmatrix}$$

This means that the proper value to assign would be

$$\nabla \wedge \vec{F} = \begin{pmatrix} 0 & \partial_xF_y - \partial_yF_x & \partial_x F_z - \partial_z F_x \\ \partial_yF_x - \partial_xF_y & 0 & \partial_yF_z - \partial_z F_y \\ \partial_z F_x - \partial_x F_z & \partial_z F_y - \partial_yF_z & 0 \\ \end{pmatrix}$$

$$ = \begin{pmatrix} 0 & \partial_xF_y & - \partial_z F_x \\ - \partial_xF_y & 0 & \partial_yF_z \\ \partial_z F_x & - \partial_yF_z & 0 \\ \end{pmatrix} - \begin{pmatrix} 0 & \partial_yF_x & - \partial_x F_z \\ - \partial_yF_x & 0 & \partial_z F_y \\ \partial_x F_z & - \partial_zF_y & 0 \\ \end{pmatrix}$$