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Proton 6: 2nd Degree (P6-30)

A second degree solution set for P6. The first three images are side views, the fourth & fifth are views of the polar triangles, and the sixth is shown without the balloon inserted to demonstrate the symmetry of the shell. As you can see, this model for P6 is more spherically symmetric in its structure than the polar star shells shown below, all of the structural circlets also qualify as equatorial circlets. These spherical shells are extremely difficult to model from scratch using the glue-gun and wire techniques I am forced to use. The 6-particle spherical shells presented here are the result of a great deal of frustrating trial and error. Look at the pictures of the shell and note that all the filler circlets can be replaced by a single sinusoidal string wrapped once around the particle. This sinusoidal string can be calculated exactly by virtue of the fact that the peaks of the sine correspond to the fixed position of a quark on the shells surface. (Figures for the sinusoidal solution are not given.)

Stereograms of P6-30 Shell

Required Snap Points: 60
Available Snap Points: 60
Pinned Vector Bosons: 0
L1 = 32.60432506" (Structural)
L2 = 32.60432506" (Structural)
L3 = 18.39567494" (Filler)
Model Diameter: 10.378279"
Loop Ratios:
(1:2): 1
(1:3): 1.772390802
(2:3): 1.772390802
Loop Equation:
3(L1) + 3(L2) + 6(L2) = 306"
Snap Point Equation:
3(6) + 3(8) + 6(3) = 60 R.S.P.

Proton 6: 3rd Degree, Polar Star (P6-30PA)

This Proton Shell of 6 is built from 13 hadron strings. Six of the strings are structural, six are filler circlets, and one is a polar circlet. The polar circlet has six snap points, and a different loop value than the filler circlets. All the down quarks are found at one pole of the shell, making for a highly polarized charge structure.

Stereograms of P6-30PA Shell

Required Snap Points: 60
Available Snap Points: 60
Pinned Vector Bosons: 0
L1 = 29.75263657" (Structural)
L2 = 17.94151492" (Filler)
L3 = 19.83509105" (Polar)
Model Diameter: 10.93566267"
Loop Ratios:
(1:2): 1.658312395
(1:3): 1.5
(2:3): 0.904534034
Loop Equation:
6(L1) + 6(L2) + 1(L3) = 306"
Snap Point Equation:
6(6) + 6(3) + 1(6) = 60 R.S.P.

Proton 6: 3rd Degree, Polar Star (P6-30PB)

A third degree solution set for P6. This one places all six down quarks on an equatorial circlet. The P6-30PB shell stacks with N6-30P to form an extremely stable complementary shell structure, and I believe it represents the most likely solution for the principal nuclear isomer of carbon 12. This complementary shell structure for C12 has highlighted a facet of chemistry which has always mystified me, and finding this unexpected solution for C12 has caused me to revisit my assumptions on the matter, resulting in something of an epiphany for me. It is a puzzle that I now intend to share with you.

I once thought only inert gases would be characterized by complementary shells with no P.V.B's, so I was quite surprised when I stumbled onto this rather mysterious solution for C12. With this proton shell stacked on top of N6-30P the charge structure of the resulting cation would be "dumbell" or "peanut" shaped. According to the chemistry texts, C12 contributes four electrons to bonding. In the free state, one pair of electrons would reside at either pole. In the bound state, such as in a buckyball, three electron pairs (one of which is contributed by an adjacent C12) would locate at one pole 120 degrees apart, while the opposite pole might harbor the third native pair, or be capable of accepting an additional electron pair, by virtue of the resulting highly polarized state of this nucleus. It all kind of depends on where the "s" orbital (non-valence) pair are truly located. Since this is an extremely homogeneous charge structure for C12 by virtue of the complementary nuclear shells, the two electrons presumed to form the "s" orbital may actually be nestled in the nuclear interior. There should be plenty of room in there for them, as well as strong positive charge currents off the underside of the neutron shell. On the other hand, and more in line with conventional wisdom, the lack of P.V.B.s may actually encourage electron pairs to hover more closely to the nuclear surface by permitting them to avoid direct interaction with the charge jets emanating from the quarks on the nuclear surface.

I used to suspect electron pairs residing on P.V.B.s would form stronger bonds by virtue of the P.V.B.'s themselves, but the action of the P.V.B.'s would tend to place the pinned electrons within the direct influence of the negative nuclear charge jets, essentially pushing them farther from the nuclear surface, resulting in a longer bond geometry and consequently a weaker charge exchange between electrons and nuclei. Now this seems slightly mysterious, in that the stacked shells for helium 4 should form a strong bond as well, but the answer may lie in the mutual repulsion of the electron pairs. In Helium 4, there is only one electron pair, and that pair may hover so close to the He4 nucleus, even encircling it, so that the nuclear charge and the electron charge cancel quickly and very near the nuclear surface, making helium 4 a very slippery atom. (Even more so if the electron pair takes up residence in the nuclear interior, as then it would present no opportunity to form any sort of electron-mediated bonds.) In the buckyball example of bound carbon 12, with three pairs of electrons mutually attracted to one of the C12 poles, the mutual repulsion of the electron pairs may act to force the electron pairs to hover higher above the nuclear surface, extending the charge structure of the atom to more readily facilitate covalent bonds. But if that is the case, it should force the "s" orbital pair to the opposite pole, permitting another covalent bond off of that pole as well. I don't think that's what is happening though, which is one reason why I suspect the "s" orbital pair may be hiding out in the nuclear interior.

If this turns out to be true, then the strong nuclear force is considerably weaker than old-style calculations would lead us to believe. That is because those calculations are based on the electrostatic repulsion of the protons, which were assumed to be discrete entities (like a cluster of grapes) repulsing each other from their surfaces. (Note 2008: Well, um, at least I assume that's what they based them on, if not, they were based on the binding energies, like how many million electron volts of energy it takes to knock a proton or neutron out of the nucleus. I'll have to look into that. Sorry for the interruption.) We can see with these shell structures that this "cluster of grapes" is not the case, a proton shell is a composite of all the protons in that shell, and the electrostatic repulsion from surface to surface of discrete protons simply does not apply. Nuclear shells are more like delicately webbed soap bubbles nestled in minimum energy configurations. This configuration of nuclear structure means that there is, or can be a great deal of empty space in the interior of a nucleus, and if one or more electron pairs take up residence there, then there is a significant reduction in the electrostatic repulsion between opposite sides of the proton shells as the internal electrons would cancel much of that effect. This would result in a reduction in the required strength of the strong nuclear force to bind protons in the nucleus. Even the stacking of neutron shells and proton shells is done in such a way so as to achieve the greatest neutralization of the charge currents between adjacent quarks in upper and lower shells, and this too will reduce the electrostatic repulsion between the combined protons in the proton shells. Additionally, the theory underlying this string model suggests that the underside of both up and down quarks act as positive charge sources, making the hollow nuclear interior a very attractive environment for electrons, and it is almost inconceivable that electrons would not find their way into that enormous, hollow, positively charged space. Another factor to consider is that leptons and hadrons have structure in different domains in this theory... an electron does not have to pass through the string geometry of a nuclear shell to get inside of it, as those nuclear strings do not reside in lepton space where the electrons display their structure. So yes, I think the non-valence electrons will often be found hanging out in the nuclear interior.

Of course, at this point in time the idea of electrons locating themselves in the nuclear interior is highly conjectural. To my knowledge I am currently the only supporter of this belief. (For what its worth, I am currently the only supporter of nuclear geometry.) If electrons do reside in nuclear interiors, it is quite possible that these internalized electrons will construct shell geometries identical to the proton shells, but residing in lepton space instead of hadron space. It is something to ponder. When all of these contributing factors are taken into consideration, it seems as if the strong nuclear force may not amount to much more than the familiar electromagnetic force and the electrostatic interplay of charged surfaces. You can see, simply by looking at the stacked shell models, that these shells are maintained by electrostatic forces, and if they turn out to be an accurate description of the nucleus, then there is no strong nuclear force. For now I will call this the "mystery of the "s" orbitals," and reserve further judgement. (For reference to casual readers, "s" orbitals are typically characterized as tightly bound spherical orbitals, encompassing the nucleus, whereas "p" orbitals are teardrop shaped lobes, and overlapping "sp" orbitals are broad fan shaped orbitals, also referred to as hybridized orbitals. An "orbital" is essentially a region in which a pair of electrons are located.)

NOTE, 03 MAR 2008: I'm fairly well convinced at this point in time that there is no strong nuclear force, per se. I believe three classical influences are at work in binding the nucleus; the electromagnetic force, represented by the strings, the electrostatic forces of charge courier exchanges, and the Casimir effect operating between quarks stacked on top of each other in adjacent nuclear shells. It is becoming increasingly obvious to me that all non-valence electrons are building interstitial shells within the hadron shells. The repulsive effects of the positive charges of the quarks in the nuclear shells are greatly cancelled by interleaving electron shells. However, this raises an interesting question... do these electron shells (composed of lepton string) pin strings, or is the geometry more restrictive for electron shells interleaved with proton and neutron shells? (The rules for geometering electron shells from lepton string are almost identical to the rules for constructing proton shells, the only real difference being in the way external strings are pinned.) I now suspect that "partons" found in nuclear collision data may represent collisions with leptonic down quarks in intelaced nuclear electron shells. (I previously attributed partons to string-string collisions, however discovering that my model describes another class of hard point-like objects swimming in the nuclear sea, this seems the more likely candidate.)

Stereograms of P6-30PB Shell

Required Snap Points: 60
Available Snap Points: 60
Pinned Vector Bosons: 0
L1 = 28.40477376" (Structural)
L2 = 32.79901906" (Equatorial)
L3 = 17.12872306" (Filler)
Model Diameter: 10.44025202"
Loop Ratios:
(1:2): 0.8660251
(1:3): 1.658312395
(2:3): 1.914854887
Loop Equation:
6(L1) + 1(L2) + 6(L3) = 306"
Snap Point Equation:
6(6) + 1(6) + 6(3)= 60 R.S.P.

Proton 6: 4th Degree, Polar Star (P6-40PS)

A fourth degree sinusoidal solution set for P6. Because there are an even number of facets to this shell, the sinusoidal string only wraps once around the shell before closing on itself. A more conventional equatorial string cohabits the equator with the sinusoidal string. The sinusoidal string is estimated, and the length of this string will affect the lengths of the other strings in the model, but the loop ratios for the non-sinusoidal strings will remain unchanged. Knowing that, it is possible to substitute values for the sinusoidal string and recalculate the dimensions of the other loops so as to satisfy the loop equation. I have not yet been able to determine any concrete rules or laws which govern the amplitude of the sinusoidal strings, making it difficult to get concrete values for most shells containing sinusoidal strings.

Stereograms of P6-40PS Shell

Required Snap Points: 60
Available Snap Points: 60
Pinned Vector Bosons: 0
L1 = 29.9410405" (Structural)
L2 = 51.8594216" (Sinusoidal)
L3 = 34.5729478" (Equatorial)
L4 = 19.9606937" (Polar)
Model Diameter: 11.00491108"
Loop Ratios:
(1:2): 0.577350067
(1:3): 0.8660251
(1:4): 1.5
(2:3): 1.5
(2:4): 2.598077123
(3:4): 1.732051415
Loop Equation:
6(L1) + 1(L2) + 1(L3) + 2(L4) = 306"
Snap Point Equation:
6(6) + 1(6) + 1(6) + 2(6) = 60 R.S.P.

Copyright 1997 by Arnold J. Barzydlo
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