## Universality of the Theory

## The riddle of protons and neutrons

Some of the mysteries of the diagram obtained in the CERN experiments are solved by the Unified Theory of Nature (Why are there fewer protons than neutrons? see). In addition, the Unified Theory makes it possible to solve one more riddle: why does the band of stable nuclei on the indicated diagram remain narrow for all values of Z?

Let's start with the fact that the narrow band in the middle of the diagram is explained by the fact that the stability of the nuclei located in this band is conditioned by the strongest linking of protons and neutrons - etheric tori (How a neutron star has been organized - new see)

But this also means that the entanglement in nucleon collisions can be of different strength. However, according to the Unified Theory, the strength of the engagement of protons and neutrons is determined by the strength of the ether torus - a proton and a neutron, but nucleons, as you know, are stable in size. Then this different strength indicates only that sáveral tori are engaged with one torus. This is where the answer to the diagram on the narrowness of the band of stable nuclei lies.

The Unified Theory has established that the ethereal microvortex - a proton and a neutron is very strong (Why the nuclear interaction is the strong see). Then this strength with mutual engagement of microvortices - tori can be ensured only with a small number of engagement of such tori with one torus. This means that the ratio of the thickness of the jet of one torus to the size of the aperture of the second in the engagement of the torus will be expressed in a large fraction.

Cross sections of three tori - nucleons located inside the fourth torus - nucleon

This means that this fraction will be limited to the first terms of the simple fraction: 1/2, 1/3 or 1/4. Let us estimate, on the basis of the theory of stability of systems, the physical viability of a torus connection corresponding to each of these fractions. Let's start with extreme fractions. The first fraction will lead to the instability of the existence of a microvortex - a proton. This follows from the fact that a system of two material circles (sections of two microvortices), located inside a material circle (inner surface of the third microvortex), will be an oscillatory, and therefore unstable system [27]. When such a connection arises, that is, when the system starts up, the two indicated micro-vortices, which are engaged in the third, will begin to vibrate relative to the third micro-vortex with an increasing amplitude. As a result, the system will go into a "runaway" and break. But it is known that the nuclei (clusters) located in the middle band of the diagram are stable. This means that when a microvortex - a proton is entangled with several microvortices, not 2 vortices are entangled in it.

In accordance with the theory of oscillatory systems, the dynamics is very similar to this oscillatory system of the 2nd order and a system of engagement of 4 microvortices in one. Accordingly, it also cannot be viable. The fraction 1/4 also disappears. The ratio remains 1/3. Let's rate it.

A gearing system of three material circles within a material circle (see the picture above) is the most viable. This follows from the fact that it is geometrically the most compact (most dense) of all three possible. Therefore, in comparison with systems of 2 and 4 material circles, it has fewer degrees of freedom inside the material circle limiting them, which means that there are fewer opportunities for the occurrence of beats. Hence it follows that the maximum number of microvortices in the engagement is 3.

This, in turn, implies the reason for the narrowness of the band of stable nuclei in the diagram. This is the band of the most durable bond: 3 in 1.. Thus, the stable nucleon linking has the following form

Accordingly, hooks with a smaller number of entangled vortices (nucleons) in one: 1, 2 will be less tenacious due to beats (see above about the 1/2 fraction) and a smaller number of breaking hooks. Such hooks will tear faster, and nuclei with such hooks will not live long enough. This can be seen in the diagram outside the specified narrow band.