Re: resolve to perpendicular components, because they are independent


FrediFizzx wrote:
> "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"?

LOL, ok, how about a link, Randall is super-pop, so
I know your not jokin...

> FrediFizzx
Ken

.

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