Can you take the Laplacian of a vector field with cylindrical coordinates?

[hunt_mat]

Tell me I'm not going mad. If I have a vector field of the form \mathbf{A}=(0,A(x,y,z),0) and I want to take the Laplacian of it, \nabla^{2}\mathbf{A}, can I take the Laplacian of the co-ordinate function A(x,y,z)? Will this be the same for the case of cylindrical co-ordinates?

Mat

 

[arildno]

That's true for Cartesian coordinates.
For cylindrical coordinates, you must remember that a planar unit vector is dependent upon the angular variable, so take care to differentiate that unit vector as well.

For a vector field \vec{A}=A(r,\theta,z)\vec{i}_{\theta} we'll get, for example,
\nabla^{2}\vec{A}=(A_{rr}+\frac{A_{r}}{r}-\frac{A}{r^{2}}+\frac{A_{\theta\theta}}{r^{2}}+A_{zz})\vec{i}_{\theta}-\frac{2A_{\theta}}{r^{2}}\vec{i}_{r}
where lower case on the scalar function indicates differentiation with respect to that variable.
Note that this does NOT equal a naive application of the Laplacian operator merely to the scalar component A.