Prove that vec(nabla) xx vecA = curl(vecA)?
1 Answer
Considering Cartesian coordinates, a vector
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,
Also this holds for other coordinate systems as well.