Atomic Displacements in the RS

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bperet
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Atomic Displacements in the RS

Post by bperet » Mon Oct 28, 2013 3:38 pm

Larson represents particles and atoms with "displacements," which is a measure of the deviation from unit speed. For example, a speed of 1/3 would have a displacement of 2 (1-1 = 0 in space, 3-1 = 2 in time). A spatial displacement is indicated in parenthesis, such as a speed of 3/1 would have a displacement of (2).

The rotational system that forms the basis of particles and atoms has two, magnetic (2D) rotations and a single electric (1D) rotation.

A magnetic rotation can be visualized as a disk with a radial measure. Larson's two, magnetic rotations form two, interpenetrating disks to produce the two, magnetic rotations that are designated "a" and "b" in the a-b-c notation. Technically, the "a" rotation is a×a and the "b" rotation is a×b, as the second, or subordinate rotation shares a common axis with the primary rotation. "a" therefore has the dimensions of a2 and "b" is "ab". Larson only uses the "a" and "b" values to represent magnetic rotations.

The electric rotation is a spinning of the magnetic rotating system, in a single dimension, like spinning a ball.

The system can be visualized like this (the rotating system of a particle; atoms have two of these rotating systems, of the same displacements):
Larson-Atom.png
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Displacement as Spherical and Polar coordinates

Post by bperet » Sun Mar 11, 2018 11:37 am

I noticed something regarding Larson's displacement notation, regarding rotating systems. Larson assumed his "a" and "b" dimensions were radial, producing an atomic shape that was a sphere (a=b) or ellipsoid (a<>b). But in a universe of motion, these variables are speeds, not radial distances.

In the A-B-C notation, A-B form the 2-dimensional, magnetic rotating system that, in RS2, is expressed as a quaternion, <1 i j k>. C is the 1-dimensional, electric rotating system, the complex number, <1 i>.

Why does Larson only have two values, A-B (like i,j) instead of three, and no "real" component?

The real component is always normalized to unity; since unity is the natural datum, it is ignored. As Hamilton discovered, there is no real solution to i.j, and in order to normalize the real component out of the system, a real solution is required. Hence the quaternion, i.j.k=-1.

So what do A, B and C represent? The A-B notation is that of spherical coordinates, and the C notation, that of a polar coordinate--with the exception that the magnitudes involved are angular velocities, not angles.

Spherical notation has the form: (r,θ,φ); polar notation: (r,θ), where "r" is the radius (always 1.0 in the RS). So you end up with A-B-C actually being spherical-polar, θ-φ-θ, describing a sphere (θ-φ, A-B) spinning on its equator (θ, C)--just like the Earth, with its longitude-latitude and axial rotation.

Hopefully, this parallel with our homeworld gives a way to visualize what Larson was thinking about when he described his atomic model in The Structure of the Physical Universe--the Earth is just a gigantic atom.
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Re: Atomic Displacements in the RS

Post by Sun » Sun Apr 15, 2018 9:01 pm

I have difficulties understanding Larson's triplet, getting stuck in the calculation of mass for years.
Logic seems not consistent in NBM. Magnetic displacement 2n is equal to 2n2 electric displacement, no more than two areas, easy to understand.
Larson's logic is 2-1-0 has 1 net displacement in both a and b, but i don't see why.
Adding another magnetic displacement would add up to 2 in a, still 1 net in b, so we need 8 electric displacement. Adding another 8 electric displacement arrive at 3-2-0 has 2 net displacement in both a and b. What he's doing is building areas by 1D electric displacement. This logic consistent with 8-x speed range, in which volume built up by 1D displacement. It doesn't matter it's a cube or a sphere, since it's a π problem. But in 5-4-0 he just added two 4 units 2D displacement up to 8, adding two areas to equate a volume which is confusing me.
Would you clarify these?
Would you elaborate how a×a, a×b are involved in equivalent electric displacement calculation?
RS2 treats all scalar to be rotation is really smart. But i don't understand what is the scalar part, the radius 1.0 means in physics? Should this 1 turns to -1 for c-atoms?

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Re: Atomic Displacements in the RS

Post by bperet » Tue Apr 17, 2018 12:25 pm

Might want to take a look at a couple of Nehru's papers...

The Nature of Scalar Rotation, where he explains the electric-to-magnetic conversion process.

Relative Abundance of the Elements, where he shows the atomic building process and why the elements occur in the proportions in which they do.
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