Meeting a Terrific Challenge

Discussion of Larson Research Center work.

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Re: Meeting a Terrific Challenge

Post by dbundy » Fri May 18, 2018 4:22 am

It's certainly understandable, if the ideas presented in the previous post above seem bizarre to the reader. Nevertheless, the development is logical and follows from the system's fundamental postulate. The basic idea follows from the broken symmetry of the quarks, which is quite apparent.

Unlike the leptons and neutrino, which are fully symmetrical, so that rotating their schematic symbols makes no change to their appearance, rotating the symbols of the quarks does change their appearance, due to the broken symmetry of their constituent S|T (T|S) units, and even though this schematic difference is meaningless from a physical standpoint, it does illustrate that there exists a lack of symmetry in the analogous physical attributes of the particles.

What I am asserting is that the broken symmetry in the physical attributes of the quarks may be due to the combinations of different unit sizes. It seems clear that a given LC can be based on any of the three units, √1, √2 or √3. Regardless of which unit size is considered, there are always the three radii associated with it, the radius of the inner, the middle and the outer balls, defined by the LC based on that unit size.

Each of the three sets of dimensional magnitudes, or values generated by these three possible LCs, contains one integer magnitude: √1, √4, √9, as shown below

For unit size = √1:
1) √((√1)2+(√1)2+(√1)2) = √(1+1+1) = √3;
2) √((√1)2+(√1)2+(√0)2) = √(1+1+0) = √2;
3) √((√1)2+(√0)2+(√0)2) = √(1+0+0) = √1;

For unit size = √2:
1) √((√2)2+(√2)2+(√2)2) = √(2+2+2) = √6;
2) √((√2)2+(√2)2+(√0)2) = √(2+2+0) = √4;
3) √((√2)2+(√0)2+(√0)2) = √(2+0+0) = √2;

For unit size = √3:
1) √((√3)2+(√3)2+(√3)2) = √(3+3+3) = √9;
2) √((√3)2+(√3)2+(√0)2) = √(3+3+0) = √6;
3) √((√3)2+(√0)2+(√0)2) = √(3+0+0) = √3;

The conjecture is that if one quark of a nucleon's three quarks, whose constituent S|T units are combos of LCs that are based on a √3 unit size, then the S|T units of the other two quarks would necessarily have to consist of √1 and √2 based LCs, in order to dimensionally balance the opposing scalar motions, and when a neutron under goes beta decay, the unit size of the transformed down quark's LCs would have to change size accordingly, forcing the others to change as well, in order to maintain dimensional balance.

While this may seem bizarre, I would submit that it's no more so than the observed physical properties and behavior of the nucleons themselves.

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