Divergence of curl in spherical coordinates

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
member 428835
Hey pf!

I was thinking about how div(curl(f)) = 0 for any vector field f. However, is this true for div and curl in spherical coordinates? It doesn't seem to be.

If not, what needs to happen for this to be true in spherical coordinates??

Thanks all!
 
Physics news on Phys.org
div and curl do not depend on coordinates so the result holds for all coordinates including spherical

Keep in mind you omitted some conditions, f must be well behaved for that to be true
 
joshmccraney said:
Hey pf!

I was thinking about how div(curl(f)) = 0 for any vector field f. However, is this true for div and curl in spherical coordinates? It doesn't seem to be.

If not, what needs to happen for this to be true in spherical coordinates??

Thanks all!

If you don't get [itex]\nabla \cdot (\nabla \times F) = 0[/itex] for well-behaved [itex]F[/itex] in spherical coordinates then you are making an error in your calculations, such as forgetting that the basis vectors are functions of position and not constant as in the Cartesian case.