Prove that $- \to \nabla \times \to A = c u r l (\to A)$ $vec(nabla) xx vecA = curl(vecA)$ ?

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Prove that $- \to \nabla \times \to A = c u r l (\to A)$ $vec(nabla) xx vecA = curl(vecA)$ ?

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Answer 1 · 2017-09-04T10:26:25

Considering Cartesian coordinates, a vector $\to A = A_{x} ˆ x + A_{y} ˆ y + A_{z} ˆ z$ $vec A = A_xhat x + A_yhat y + A_zhat z$ has it's curl defined as,

$c u r l (\to A)$ $curl (vec A)$

$= ˆ x (\frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z}) + ˆ y (\frac{\partial A_{x}}{\partial z} - \frac{\partial A_{z}}{\partial x}) + ˆ z (\frac{\partial A_{y}}{\partial x} - \frac{\partial A_{x}}{\partial y})$ $= hat x ((delA_z)/(dely) - (delA_y)/(delz)) + hat y((delA_x)/(delz) - (delA_z)/(delx)) + hat z((delA_y)/(delx) - (delA_x)/(dely))$

Then in terms of the vector differential operator,

$- \to \nabla = ˆ x \frac{\partial}{\partial x} + ˆ y \frac{\partial}{\partial y} + ˆ z \frac{\partial}{\partial z}$ $vecnabla = hat x del/(delx) + hat y del/(dely) + hat z del/(del z)$

It may be represented as a cross product such that, (verify this yourself)

$[ˆ x \frac{\partial}{\partial x} + ˆ y \frac{\partial}{\partial y} + ˆ z \frac{\partial}{\partial z}] \times [A_{x} ˆ x + A_{y} ˆ y + A_{z} ˆ z]$ $[hat x del/(delx) + hat y del/(dely) + hat z del/(del z)] xx [A_x hat x + A_y hat y + A_z hat z]$

$= ˆ x (\frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z}) + ˆ y (\frac{\partial A_{x}}{\partial z} - \frac{\partial A_{z}}{\partial x}) + ˆ z (\frac{\partial A_{y}}{\partial x} - \frac{\partial A_{x}}{\partial y})$ $= hat x ((delA_z)/(dely) - (delA_y)/(delz)) + hat y((delA_x)/(delz) - (delA_z)/(delx)) + hat z((delA_y)/(delx) - (delA_x)/(dely))$

$\Rightarrow - \to \nabla \times \to A = c u r l (\to A)$ $implies vecnabla xx vec A = curl (vec A)$

Not sure if it was a proof, but I think this serves the purpose. Often, $- \to \nabla \times \to A$ $vecnabla xx vec A$ is used as a definition for $c u r l (\to A)$ $curl (vec A)$ .

Also this holds for other coordinate systems as well.

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Prove that $- \to \nabla \times \to A = c u r l (\to A)$ $vec(nabla) xx vecA = curl(vecA)$ ?

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