Re: resolve to perpendicular components, because they are independent

On Fri, 20 Jan 2006, Ken S. Tucker wrote:

I find nonorthogonal axes easier than orthogonal, 

Then you must be some kind of bizarre freak of nature!!!

indeed a Curl
becomes A_u,v - A_v,u  (== &A_u/&x^v - &A_v/&x^u), because
manipulating equations in tensors is streamlined by notation.

Can't you just do that with orthogonal metrics too? (Mixing covariant and contravariant is just a naughty little trick to hide the metric tensor!)

--
Timo
.