Atomic Displacements in the RS

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bperet
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Atomic Displacements in the RS

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 (31.81 KiB) Viewed 5900 times
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bperet
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Displacement as Spherical and Polar coordinates

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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Sun
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Re: Atomic Displacements in the RS

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?

bperet
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Re: Atomic Displacements in the RS

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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Sun
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Re: Atomic Displacements in the RS

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>.
Conventional science doesn't recognize the speed of light actually don't have a direction. Therefore a location of an object in space can be expressed by a quaternion R=ct+r. The vector algebra is wrong, should back to the quaternion algebra. When this kind of location is applied to photons, it is R=1+r, the same as <1 i j k>, except <1 i j k> is used for rotation. I think it means by a gravitate observer, the SOLID time, the aether is actually fixed in space, it is not dynamic. Space instead is flying apart. So it is possible to alternate time by vectorial motion, but not possible for vectorial motion to alternate space since space is not fixed. I'm always wondering does vectorial motion producing scalar effect? Like sound can be produced by oscillation, but once it forms a sound, it is independent of its source. It seems it is the oscillation vibrating time but the effect is compression wave in space. I think in coordinate space, the neutral axis NN in Nehru's paper "The Nature of Scalar Rotation" is no more than an EM barrier, implied by figure2 that rotation penetrate NN, should not be an ordinary line in coordinate space. I notice both Larson and Nehru think about scalar motion as motion of a massless point, splitting it into space and time aspect to see which part take effect. Then in coordinate space, the A-B is compression from space and time respectively. Linear motion, rotation and oscillation are all possible to penetrate NN. Why can't we think about light is produced by something oscillating fast enough to penetrate NN? (Not sure if i get the same conclusion implied by this post sonoluminescence)
Frequency of light we used is not the actual frequency of "aether", while acoustic frequency is. It is also possible turning sound waves into a rotation.

bperet
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Re: Atomic Displacements in the RS

Can you describe your understanding of the term "scalar?"

For me, a "scalar" is nothing more than a single variable, with a magnitude of 1..n, where "n" is a finite number. (Vector requires more than one variable/dimension.)
Sun wrote:
Sat May 05, 2018 4:37 am
Frequency of light we used is not the actual frequency of "aether", while acoustic frequency is. It is also possible turning sound waves into a rotation.
This is an interesting conclusion... how did you determine that acoustic frequencies are the frequency of aether?
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Sun
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Re: Atomic Displacements in the RS

I've posted my discussion on this thread Katirai on Photon frequency as a pressure

bperet
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Atomic models

I think there is a "design flaw" in the way Larson models the atom. First, let's take a look at the conventional system, which consists of three particles: the proton, neutron and electron, which are the observed decay products.

Proton: since everything in conventional physics is charge-based, this is a charged proton, which means that protons cannot stick together in a nucleus, unless some magical force exists to override the charge repulsion. The number of protons defines the atomic number and half of the primary mass.

Neutron: neutral particle that, having NO charge, just decides to hang around with the protons in the nucleus for no apparent reason--they should just wander off, unless another magical force is holding them in place. Neutrons account for the other half of primary mass, but can also keep going to add additional mass (RS isotopic mass).

Electron: tiny wave/particle thingies that float around the nucleus to get rid of all the positive charge the protons are producing, except a tiny amount to cause ionization, electrical conduction and chemical bonding.

Larson's model is described in the initial topic post, which consists of two, double-rotating systems that share a common point of rotation (nucleus) with the entire structure spinning in the inverse aspect to produce a surrounding electrical effect, much like the electron cloud of the conventional model.

In the material sector, the double-rotating (magnetic) systems are temporal rotations, and the single-rotating (electric) system (spinning structure) is spatial. The conventional view just keeps adding protons to build atomic number; Larson does it like a staircase, where the magnetic rotations are the landings, with the electric rotation defining the atomic number; the steps between the landings.

We can connect the two by taking a proton-neutron pair as one set of two, double-rotating systems. But in the RS, the double-rotating systems are indistinguishable from each other, so that electron that changes the proton into a neutron is missing. (This also indicates that hydrogen is a subatomic particle, having no neutron, and it is deuterium that is atomic number 1.) The RS double-rotating systems should be either 2 protons or 2 neutrons, not a mix, but because it maps as a mix, that extra electron must be part of the "C" displacement, the electric rotation. So there is a bit of a twist that exists between the models... which indicates we are looking at two different aspects of something else.

Let's apply this to my "rotational dimensions" concept that was introduced in the "Photon model as a Quaternion" discussion. Nehru, in one of his papers on quantum phenomenon, noticed that the nucleus of the atom could be described as a quaternion... what if we atoms have exactly the same structure as the quaternion photon model I have proposed?

That changes things, as we no longer need particles or magnetic/electric rotations--all we have is a compound rotatio system, the quaternion, with THREE SPEED RANGES.

A speed range starts with the progression, +1. Then it rotates to the second dimension, crossing the "real" axis at 0, then continues into a third dimension, ending up at -1. A change in polarity is tantamount to the switching of aspects, so if +1 were in space, -1 (in space) would be +1 in time. That creates an interesting sequence (see Photon diagram for reference):
• 1-x, the blue rotational plane, are nuclear protons, with a charge of +1 (one end of the real axis).
• 2-x, the green rotational plane, are the neutrons, with a charge of 0 (neutral, orthogonal to the real axis).
• 3-x, the red rotational plane, are the electrons, with a charge of -1 (the other end of the axis).
Now we have a common denominator to both conventional and RS models... Larson's "magnetic rotation" being a measurement of the +1→0 "real axis" that includes the green rotational plane (since displacement in that plane always has a real component of zero), making it 2D, indicating that the magnetic rotation of the atom is nothing more than a motion in the intermediate speed (2-x) range. The remaining dimension, the "k" of i.j=k, is the remaining, 1D component in the inverse, ultra-high speed (3-x) range, the electric displacement.

This makes the atom just a "scaled up" photon... nice to have in a system based on "scalar motion."
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Sun
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Re: Atomic Displacements in the RS

If i get it correctly, Larson's conversion of magnetic displacement to electric has some problems.
Gopi wrote:
Tue Jan 08, 2019 9:43 am
bperet wrote: 1. c-magnetic rotation is observed as m-electric rotation, with twice as many units as m-magnetic rotation, doubling the maximum magnitude (electric rotation).
2. m-magnetic rotation interacts as c-electric rotation, needing twice as many units--but appearing as half-steps, rather than doubling (specific rotation).
So that means (1) is the reason that we have the 2n2 relation in the periodic table (electric to magnetic), and (2) is the reason for spin ½ and specific rotation ½.
2n2 relation is not resulted from (2n)2/2 but 2×n2, which means the electric displacement is not equal to two natural units but one. There are TWO 2n2 since there are two double-rotating system in an atom. Nehru call it "folds" in his paper The Nature of Scalar Rotation.
What i got it wrong for a long long time was that i thought it had gravitated after the two photons was neutralized by positive displacements. However, after being neutralized, you still need TWO -1(positive) displacement opposing +1 progression to gravitate.

Applying new understanding (1) stated above, 2-1-0 is actually built up from 1-0-0 with +1 magnetic displacement, which is equal to +2 electric displacements, neutralizes negative displacements of photons brings the whole system to speed 1/1, still at +1, moving in speed of light. Adding another positive displacement we get 1-1-0, one positive displacement opposing +1 progression, total of ZERO, a roaming particle but not gravitate. Finally, adding another 1 positive displacement we reach 2-1-0, which has 1 net displacement in both a-b, TWO -1 opposing +1 progression results of total of -1 displacement, gravitates. That's why we need TWO photons (TWO double-rotating system) to composes an atom. Since net displacement has reached 1 in each of the TWO double-rotating system, onion 2-2-0 will have no choice to add another layer of 2 net magnetic displacement in ONE of its double-rotating system.
In a single double-rotating system, if we use notation a-b-c instead of M a-b-c, rotational base is 1-1-0. You have to imagine atom has two of this thing, not indicated in this notation.
2-1-0 has speed 1/3 in a and 1/2 in b, displacement 2 in a and 1 in b, total net displacement -1 (which is positive). See, it is a very bad number system raising a lot of confusions.
Still don't understand the quaternion model and specific rotation...

bperet
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Re: Atomic Displacements in the RS

Sun wrote:
Fri May 31, 2019 8:47 pm
2n2 relation is not resulted from (2n)2/2 but 2×n2, which means the electric displacement is not equal to two natural units but one. There are TWO 2n2 since there are two double-rotating system in an atom.
If you haven't already, I recommend you read "Structure of the Physical Universe" (Larson's first book), which is much clearer on the nature of these relations. (He doesn't waste a lot of time quoting other scientists).

The Periodic Table is broken into three, 4n2 sections, where each of the double-rotating systems comprises 2(2n2) electric rotations (4n2/2).

I'm not following your deductive reasoning here; one electric unit is equivalent to 8 natural units and one magnetic unit is equivalent to 4 natural units (the ratio of electric to magnetic is 2:1).
Sun wrote:
Fri May 31, 2019 8:47 pm
However, after being neutralized, you still need TWO -1(positive) displacement opposing +1 progression to gravitate.
Did you account for the two -1(positive) displacements that are used to create the "direction reversal" of the photon? Larson omits them in his notation; the rotational base was originally 1-0-0 (SPU), not 0-0-0 as in later works.

To add to the confusion, Larson uses up to 6 different notational systems, which I have documented elsewhere. There are three forms of the atomic notation, the first was in SPU where the photon displacement was included, and there are two others, the particle notation and the atomic notation that appear later. This is why hydrogen has two different notation forms: 1½-1½-(2) and 2-1-(1). The former shows the combination of two particles (proton + neutrino), whereas the latter shows the atomic displacement. And yes, they are off by 1 unit.
Every dogma has its day...