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Topic: ever since 1842, the Doppler shift was assumed to exist for
lightwaves and never experimentally verified Chapt 8 #138; ATOM TOTALITY

Replies: 122   Last Post: Jul 10, 2010 2:38 PM

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 plutonium.archimedes@gmail.com Posts: 7,506 Registered: 3/31/08
Re: Schrodinger cubic set; deriving speed of light from pure math
Chapt 19 #219; ATOM TOTALITY

Posted: Jul 10, 2010 4:10 AM

Archimedes Plutonium wrote:
> Well, I think I can get it all from page 73 of Seaborg, Loveland's
> book The Elements Beyond
> Uranium, 1990.
>

There is a solution set that shows only one elliptical lobe. This is
intriguing
in that I could get the number 5300 from just one lobe rather than the
8 lobes
in the other solution set.

> I think I can use just the cube with the 8 lobes inside it, where I do
> not even have to transform
> the cube into a sphere and then transform the lobes into hyperbolic
> geometry making that
> of 4 pseudospheres. Although I could partake in that transformation.
> And, I see no reason
> that the Schrodinger Equation must always be *elliptic geometry
> solutions*? Why the solutions
> are always trigonometric ellipsoids of lobes or spheres? Why not
> solutions as pseudospheres
>
> But anyway, staying with my model of the Earth sphere as 40,000km x
> 40,000 of these strips
> when divided by the pseudosphere inside of Earth sphere of its maximum
> circle 5,300 seconds
> yields the speed of light as 300,000 km/sec.
>
> Now if one takes a globe of Earth in their home and measures with a
> plastic foldable tie string
> to a little more than 45 degree, about 48 degrees for 5,300/40,000
> x(360) which is the circumference of the largest circle of the
> enclosed pseudosphere. This circle is about the area
> of the western USA of California to Colorado to Montana in area. So
> envision Earth having 8 of
> these lobes inside of it whose largest circle on each of those lobes
> is 5,300 km circumference
> whilst Earth circumference is 40,000 km.
>
> Now I can remain with the Cubic Set and use the Euclidean geometry of
> the cube and use the
> lobes inside as shown of the Schrodinger Equation solutions on page
> 73. The strip geometry
> of 1 km wide strips is suitable in covering the cube, and perhaps
> works even better on the cube since the strips do not overlap as on
> the sphere.
>
> So we end up with 40,000 x 40,000/ 5,300 and thus the speed of light.
> Now if we vary the
> size of the cubes and their lobes inside, that constant of the speed
> of light remains. What I
> have done is made the speed of light a purely geometrical constant.

Now what I have to get is a reconciliation of the hyperbolic
pseudosphere
inside the sphere or the cube, either one. And I have to get these
numbers:

60,000 x 60,000/ 12,000 = 300,000

50,000 x 50,000 / 8,300 = 300,000

40,000 x 40,000/ 5,300 = 300,000

30,000 x 30,000/ 3,000 = 300,000

20,000 x 20,000 / 1,300 = 300,000

10,000 x 10,000 / 330 = 300,000

Notice that the 10,000 circumference with its 330 seconds of time is
the inverse
square of the 40,000 circumference with its 5,300 seconds of time.
This is very
promising.

Promising because the largest circle on the pseudosphere converted
from the
elliptic lobe is the seconds parameter of 330 for 10,000 and 5,300 for
the 40,000.

The physical meaning would be that a light wave travels 40,000 x
40,000 in a time
of 5,300, where the light wave constructs the sphere in the time of
of the pseudosphere's largest circle == 5,300 circumference.

Does that make sense? When the sphere size is smaller at 10,000, then
is the
largest pseudosphere circle that of merely 330? It makes sense because
it is the
inverse square.

Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies

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