Re: resolve to perpendicular components, because they are independent

"Ken S. Tucker" <dynamics@xxxxxxxxxxxx> wrote in message
news:1137800854.404588.75250@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
|
| 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?

Hmm... I wonder if that would apply to what Lisa Randall is calling
"Warped Passages"?

FrediFizzx

.