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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.

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    $\begingroup$ Can you give more context? Like where you find this notation? $\endgroup$
    – Arctic Char
    Jun 12, 2020 at 2:33
  • $\begingroup$ I always thought that $\wedge$ in this context was just a synonym for $\times$, so $\nabla \wedge \mathbf F = \nabla \times \mathbf F$. $\endgroup$
    – Robert Lewis
    Jun 12, 2020 at 2:34
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    $\begingroup$ @RobertLewis that is a shorthand used in many physics books but it is not true. $\wedge$ denotes the wedge product. For example the wedge product of two vectors (or $1$ forms) $v \wedge w$ would be a $2$ form. However in $3$ dimensions the hodge dual of a $2$ form is a $1$ form, and that is the value assigned to the "cross product" $\endgroup$
    – Ninad Munshi
    Jun 12, 2020 at 2:52
  • $\begingroup$ The determinant formula given in answer of Rodrigo Dias is often used for $\nabla \times \mathbf F$. $\endgroup$
    – Robert Lewis
    Jun 12, 2020 at 2:56

1 Answer 1

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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

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

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

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