Curl Operator on Vector Space is Cross Product of Del Operator

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Theorem

Let $R$ be a region of Cartesian $3$-space $\R^3$.

Let $\map {\mathbf V} {x, y, z}$ be a vector field acting over $R$.


Then

$\curl \mathbf V = \nabla \times \mathbf V$

where:

$\curl \mathbf V $ denotes the curl of $\mathbf V$
$\nabla$ denotes the del operator.


Proof

Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.

We have by definition of curl of $\mathbf V$:

$\curl \mathbf V = \paren {\dfrac {\partial V_z} {\partial y} - \dfrac {\partial V_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial V_x} {\partial z} - \dfrac {\partial V_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} } \mathbf k$

Now:

\(\ds \nabla \times \mathbf V\) \(=\) \(\ds \paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } \times \paren {V_x \mathbf i + V_y \mathbf j + V_z \mathbf k}\) Definition of Del Operator
\(\ds \) \(=\) \(\ds \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ \dfrac \partial {\partial x} & \dfrac \partial {\partial y} & \dfrac \partial {\partial z} \\ V_x & V_y & V_z \end {vmatrix}\) Definition of Vector Cross Product
\(\ds \) \(=\) \(\ds \paren {\dfrac {\partial V_z} {\partial y} - \dfrac {\partial V_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial V_x} {\partial z} - \dfrac {\partial V_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} } \mathbf k\) Determinant Form of Curl Operator

$\blacksquare$


Sources

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