Core and nuclear vacuum

The methodology, appears to be true at research of structure of space vacuum, is applicable to core vacuum. The core vacuum will mean medium with virtual pi-mesons, participating as exchange particles at core interactions. Using the already tested tool, we shall determine classical pion radius. Energy J and radius dipole according to an axiom 2 m. Classical charged pion radius is a little bit more than border strength of space vacuum. Ukava radius is equal m, at core distances it is a lot of less than this radius core forces are displayed to the greatest degree.

Radius of a proton is introduced into the formulas, as on smaller distances in core cannot and should not approach. For "penetrations" in meson medium of vacuum we shall take advantage of the phenomenon of a core photoeffect. It is known, that excitation of a core and subsequent emission from it a meson needs photon energy 140MeV or 140·1,6·10^-13J. If to assume, as well as in a case of photon field, that the meson field is formed by the tied charges (dipoles) from pion⁺ and pion- , the photon energy should surpass 280*1,6*10^-13J. The energy of rest for meson cluster with charges (+) and (- ) will be J. It is necessary to take into account defect of mass in meson cluster, i.e. its rest energy will be J. We find that J. By analogy of the formula (7) we shall determine arm of meson dipole:

Arm is less than classical pion radius in 2,0065 times. Let's find an alpha of the meson ether:

And limiting deformation

We carry out the control of the received results - J. It is possible to see a divergence with the previous result only in the fourth mark, i.e., that the counts are carried correctly. Thus, it is enough in a core to make by any way deformation of the tied charges more, than it is determined in (24), and that way from a core will be get as a minimum as one pion. Let's find factor of meson dipole elasticity by that manner , as well as in case of space dipole (see formula 15),

Elasticity of meson vacuum on 7 orders above the space vacuum. It is necessary to apply energy J to break off meson dipole and to receive two pions. It is almost in 300 times exceeds energy of ties of a photon field. As we do not find out a difference between Colombian and specific core forces, the following logic step is possible. The formula (25) gives an opportunity to enter concept of Newton interaction into a core (axiom by 1) and such opportunity it is necessary to take advantage. According to this "an arbitrariness ", the core vacuum should have a constant of gravitation, distinct from constants of gravitation of space vacuum. Let's find a meson constant of gravitation:

This value of a meson constant of gravitation equalizes Colombian and Newtonian ("nuclear") forces in a nucleus.

On fig.1 the diagrams Colombian and gravitational forces between protons are shown. Let's remind, that m - classical radius of a proton. Let us define its classical radius through the equation of energy:

Fig. 1

The deformation m, i.e. surpasses durability of vacuum. And only at m the destroying action of a proton will be stopped. Thus it is possible to consider, that between a proton and vacuum there is a zone "free" electron and positron, which as the exchange particles transfer in vacuum action of a proton.

Short distance of "core" forces are completely explained by meson vacuum spatially limited to the sizes of a core. Thus, photon vacuum and meson vacuum unite in the first case by usual gravitation and electromagnetism, in the second case by the core gravitation and core electromagnetism. The electromagnetism unites, probably, all interactions in a nature. A question on weak interaction here is not considered. Probably, it can be solved on the basis of structure of meson vacuum. It is possible to assume, that the weak interactions are displayed in spontaneous destruction of meson clusters on positrons, neutrino, radiation etc. On fig.1 it is possible to note one more fact, which should be related to amusing concurrence. The left slope of the diagram concerns to force of interaction proportional to a square of distance, instead of to its return size! At increase of distance between hypothetical quarks, taking place inside core - distance less than 10^-18 m, the force of "tension" hypothetical gluons is increased with increase of distance as the left slope of the diagram demonstrates. The force in peak gets infinite grate value, that guarantees of gluon force on durability and consequently "free" quarks are impossible.

On the basis of axioms we shall determine the electron classical radius, by taking advantage of the equation of electron energy: , where and - electron charge and mass. From the equation is determined, that m - there is the electron classical radius. Let's take advantage of results of a nuclear photoeffect in vacuum. Energy scale of quantum ( - mass of a proton), necessary for reception of a proton and antiproton from vacuum. Dipole distance of the tied charge of nuclear vacuum m. Electrical elasticity nuclear dipole kg/s². Strength of a proton or dipole deformation m, where - thin structure constant of nuclear vacuum media. Actually it means impossibility of deformation of a proton more than its radius, at which excess the proton collapses. Let's estimate nuclear gravitational constant, by taking advantage the equation for elastic force and force under the formula of Newton:

What the nuclear constant of gravitation means? Neither more, nor less as a condition of stability of a proton - powerful Colombian forces of repulsion away of a charge of a proton are equalized by Newtonian force of an attraction, i.e.

Let's make the equation of static forces of Newton and Coulomb for electron and we shall determine radius of electron mass:

Thus electron represents two-layer structure - nucleus electron mass has radius 1,534722*10^-18 m, the charged surface has classical radius 2,81794092*10^-15 m. Dividing classical radius on radius of electron mass is received: the relation of classical radius and mass electron radius is equal 1836,125. That is exact the mass number of a proton. The electron structure (positron structure) contains the information on mass number of a proton.

Cosmic Vacuum

Recent observational studies of remote supernovae have suggested the existence of cosmic vacuum with the density which is larger than the total density of all the other forms of cosmic energy in the Universe. Vacuum produces the field of anti-gravity and induces acceleration of the cosmological expansion. It is this acceleration of the expansion that has been discovered in the observations. The discovery of the cosmic vacuum leads to a drastic change in the current concepts concerning the present state of the Universe. It poses also a number of new important problems in both cosmology and fundamental physics. Why is the density of vacuum as that as it has been found in observations? Why are the components of the cosmic medium have different, but close to each other, on the order of magnitude, values of their densities? On the other hand, this discovery made at the large cosmological distances (hundreds and thousands megaparsecs) provides new insights into the dynamics of the nearby area in the Universe, the motions of galaxies in the local volume with the radius of 10-20 megaparsec where the cosmological expansion was first discovered.