Prove that #vec(nabla) xx vecA = curl(vecA)#?

1 Answer
Aritra G. · Truong-Son N.
Sep 4, 2017

Considering Cartesian coordinates, a vector #vec A = A_xhat x + A_yhat y + A_zhat z# has it's curl defined as,

#curl (vec A)#

#= 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,

#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)

#[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]#

#= hat x ((delA_z)/(dely) - (delA_y)/(delz)) + hat y((delA_x)/(delz) - (delA_z)/(delx)) + hat z((delA_y)/(delx) - (delA_x)/(dely))#

#implies vecnabla xx vec A = curl (vec A)#

Not sure if it was a proof, but I think this serves the purpose. Often, #vecnabla xx vec A# is used as a definition for #curl (vec A)#.

Also this holds for other coordinate systems as well.

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