Re: resolve to perpendicular components, because they are independent


Timo Nieminen wrote:
> 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!!!

Not really, as in Chess, solving problems in mathematical
physics consists of keeping your options open, to be
closed by physical principle, and certainly not by an aprior
preceived convenience. It's well known "orthogonality" is
at best an approximation in a g-field, but Reimann and his
"gang" evolved quite a nice "tensor" analysis notation that
is easier to use than clunky "ijk" unit vectors.

> > 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!)

If your intrinsic dimensionality differs from an integer, i.e
let n= intrinsic dimensionality =2.9, then how the heck do
you expect to squeeze 3 orthogonals into that?
Ken

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