Re: resolve to perpendicular components, because they are independent


Hero.van.Jindelt@xxxxxx wrote:
> Ken S. Tucker wrote about
> > > > ....an orthogonal 4D....
> To be honest, i feel like a woman, being told by her husband "Yes dear,
> You are right." while he is thinking "'...and i got my peace."', so my
> words got labelled and stored away in the back part of the wrong
> drawer.
> I'm not a physicist, having just basic knowledge, but i notice, that
> You avoid a direct answer to:
> "Can You explain, how a fourth dimension can be orthogonal to the three
> of space, measured in cm³?,
> which arose from Your mentioning of "an orthogonal 4D".

3 spatial dimensions and 1 time dimension, is commonly
called 4D. An *orthogonal* 4D implies 3 orthogonal spatial
axes x,y,z, and 1 time axis perpendicular to x,y,z.

> The mathematical ideas and speculations of Riemann, which can be found
> here
> http://www.maths.tcd.ie/pub/HistMath
> led to best science fiction and some interesting math and he could
> differ between both. This is testified and described in the last pages
> of
> James Clark Maxwell's "A treatise of electricity and magnetism".
> (Actually i always thought the "Preliminary" of the first 31 pages
> would be all i could achieve understanding in my lifetime, but reading
> here so much about aether, so i wanted to know, what Maxwell thought
> about this. Just read for Yourself the last sentences of his work and
> what was his "constant aim in this treatise" of about a thousand pages.
> That intrigues me very much, so i hope, i can achieve some
> understanding of this too in my life.)
> Now, one hears a lot of loud talking about Riemann, but i never met
> even one, who didn't mix the math with the math-fiction to an
> undistinguishing mud, something Riemann always tried to avoid. All what
> they are doing, is:
> Expanding a 2D-space twice. Some do this on purpose, like Hilbert, and
> some out of not-knowing better - but all skip the third dimension.
> Take a simple sphere and consider a diameter - that are three points on
> a straight line, two are on the surface and one is the center. The line
> is the shortest connection between the two outer points with extension
> ( the two points mark an intervall on this line). Where do You find
> this with Hilbert? How can one proceed to "an orthogonal 4D" and
> skipping this 3D-shortest-path? I don't grasp.
> Just one step further is a step back to Thales: From every point of a
> circle a diameter appears under a right angle. This is true for every
> point on a sphere too ( the proof is left to the reader).
> And that's it in 3D. Now please do this in 4D and try to explain.
> Or- if You preferr analytical structure over geometrical - please
> explain, how the square of a density can be on equal footing with the
> square of a length of a distance and at the same time doesn't coincide
> somehow with three orthogonal extensions (and this for every set of
> two points possible)? To illustrate density in 3D i only know the use
> of different shades of grey, or "translated" into different colours or
> into "frames"
> (the discret form of displaying time with films and computers). Please
> show me "an orthogonal 4D".

Ok, maybe you could inform me/us of your background
in math & physics, especially geometry. A term like
"orthogonal" can be easy to explain but difficult to define.
Ken





> Regards
> Hero

.

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