Riddle of solar neutrinoes part 1,2

It is the copy of my relevant article from a site on Yandex (uCoz).

"Riddle" of solar neutrinos from the point of view of the field theory elementary particles. Part 1.

Contents:

1 . Stream of solar neutrinos

2 . Neutrino oscillations

3 . Energy transferred by solar neutrinos

1 . Stream of solar neutrinos

According to the field theory of the fundamental particles of a neutrino (as well as any other fundamental particle with the nonzero size of a rest-mass) possesses a variation electromagnetic field from a constant component, i.e. the neutrino possesses:

  • constant dipole electric field,

  • constant dipole magnetic field,

  • constant magnetic field of a moment of magnet

  • variation electromagnetic field.

On the other hand according to a classical electrodynamics constant electric, magnetic and variation electromagnetic fields of different fundamental particles interact. This interaction will lead to an exchange of energy between the fundamental particles. From this it follows that when passing relativistic neutrinos (such as are radiated by our sun) through substance of Earth, the neutrino will lose gradually the kinetic energy and to be slowed down. This power loss is caused by interaction of electronic neutrinos with electrons of substance of Earth (see S. S. Gerstein of "A riddle of solar neutrinos").

Having lost the considerable proportion of the kinetic energy, the neutrino will be able unnoticed to pass through detectors as energies will be insufficiently for exercise in the reaction detector with neutrino participation. Reactions on detection of electronic neutrinos have a threshold amount of energy below which reaction it is impossible as it will contradict the law of conservation of energy. Thus, rather delayed neutrinos will pass unnoticed through detectors and other similar inventory on their filing.

Now we need to remember that filing of a neutrino happens continuously a method of accumulation and day and night. I.e. and when the sun shines from above, and the neutrino separates from the detector only earth kilometers and when the sun shines from the opposite side of a planet, and the neutrino should pass through all planets. And then we still are surprised why in experiments of GALLEX, SAGE and GNO it is possible to catch solar neutrinos twice smaller than has to be. – Look, what part of time of day the sun shines from above, and everything becomes clear.

To be convinced of it, we will look at a few figures.

Most sensing of the types of neutrino detectors created so far is the Gallic detector with a threshold amount of energy of 0.233 MEV. With its help experiments on filing of a stream of electronic neutrinos of GALLEX, SAGE (most the long-lived) and GNO were made. As a result of these experiments the following data are obtained:

  • GALLEX (result during 1991-1997)...... 77.5 (SNU)

  • SAGE (result during 1990-2010)...... 65.4 (SNU) with an accuracy of 6%.

  • GNO (1998 - 2002) … …. 65.2

where SNU - solar neutrino unit.

And experiment of GNO was made in the same laboratory, as GALLEX. But its result coincided not with GALLEX experiment, and with SAGE experiment carried out in RHMDN OLVENA laboratories and GGNT BNO that gives the grounds to claim that data of experiment of SAGE are the most precise.

Thus, the stream of solar electronic neutrinos (SAGE measured by experiment and GNO confirmed with experiment) is equal 65.4(SNU) that makes 0.503 from value theoretically predicted by solar models 130±8(SNU). But the field theory of the fundamental particles, considering feature of operation of the Gallic detector, reduced for the last theoretical value twice and now it makes 65(SNU). When more sensing neutrino detectors will be developed (with lower threshold amount of energy), this number becomes another for these detectors (will be higher). And then it will be possible to find the "lost" solar electronic neutrinos. But for the Gallic detector it remains 65(SNU).

Thus: the deviation of the experimental value (65.4) from theoretical (65) makes 0,6%. It does not go beyond accuracy of experiment of SAGE (6%).

So, we dealt with a riddle of a trace amount of solar neutrinos. According to the field theory of the fundamental particles and feature of reaction of detection of electronic neutrinos, the Gallic neutrino detector will well catch those neutrinos which go from above (through the small thickness of the earth) and will be blind on the relation to many neutrinos passed through all planet. Thus, the result of experiment will depend, as on the district width where the neutrino detector is located, and from a season (when measurements were made).

Thus, becomes apparent that due to the lack of deficiency of solar electronic neutrinoes neutrino oscillations are not necessary to the nature.

2 . Neutrino oscillations

Without having managed to deal with unexpectedly small number of the solar neutrinos recorded on Earth, the new fairy tale under the name "neutrino oscillations" was composed. Its substance consists in the following: "the neutrino of any particular type at the driving in vacuum will periodically pass to a neutrino (or an antineutrino) other types and back". Well and as it is considered that in the nature there are three types of a neutrino, the reasonable explanation for observed deficiency of solar neutrinos turned out.

And now we will look at it from the point of view of the field theory of the fundamental particles.

  • Tau-lepton is an excited state of a muon - therefore, the tau-neutrino actually is the first excited state of a muonic neutrino.

  • Energy of the first excited state of a muonic neutrino it is considerable (several times) above an internal energy of a muonic neutrino – therefore, oscillations between a muonic neutrino and its first excited state will contradict the law of conservation of energy.

  • The size of a rest-mass of a muonic neutrino considerably differs from the size of a rest-mass of an electronic neutrino (at these fundamental particles quantum numbers do not coincide) – therefore, oscillations between a muonic neutrino and an electronic neutrino will contradict the law of conservation of energy.

Thus, neutrino oscillations will go with violation of the law of conservation of energy, such unloved reference model and a quantum theory. This law allows not transformation, and disintegration of heavier muonic neutrino. At products of such disintegration there will be electronic neutrinos together with other rather mild fundamental particles. Besides, the law of conservation of energy allows other reactions of the elementary particles in the presence of a sufficient kinetic energy. But wonderful transformations of one elementary particle into others are from the world of fairy tales.

Neutrino oscillations contradict also to electrodynamics laws. The spontaneous transformation of any neutrino in an antineutrino is impossible as these particles have a counter sign of electric and magnetic fields. Reaction of their annihilation, as well as for any other particle antiparticle vapors is possible. Also the spontaneous transformation of a neutrino of one type in a neutrino of other type as they have different linear dimensions and also various structure of their electric and magnetic fields contradicts electrodynamics laws.

And now we will look that happens actually to neutrino transformations.

The electronic neutrino can turn into a muonic neutrino, but as a result of collision with other electronic neutrino in the presence of a sufficient kinetic energy. Such collisions can happen on the sun, and the invented oscillations here not and. Besides on the sun there will be collisions of a neutrino of different reactions as a result of which particles will exchange also energies – neutrinos of higher energies will transfer part of the energy to other neutrinos.

The muonic neutrino as heavier (than an electronic neutrino) is an unstable fundamental particle and after a lifetime it will break up (most likely on one of the following channels of disintegration):

  • in an electronic neutrino steam plus electronic neutrino antineutrino

  • in an electronic neutrino plus of steam of photons.

The first excited state of a muonic neutrino is even more short-lived condition, than a muonic neutrino as it has even more opportunities for transition to conditions with lower energy. As a result of these transformations at the exit we will receive some number of electronic neutrinos.

Thus, to our planet as a result of neutrino reactions even more electronic neutrinos will reach. As we see, "neutrino oscillations" as a spontaneous transformation (of one neutrinos in others neutrinos) are the next fairy tale.

3 . Energy transferred by solar neutrinos

The stream of solar neutrinos through a surface of our planet is estimated today by physics as 0.66×1011 a neutrino / (cm2 × c).

The kinetic energy transferred by a neutrino depends on reaction in which it was formed:

p+ + p+2Н + e+ + νe - 0.4202 MEV (1)

p+ + е + p+2Н + νе - 1,4422 MEV (fix) (2)

e- + 7Be → 7Li + νe - 0.862 MEV (fix) (3)

3He + p+ 4He + e+ + νe - 18.77 MEV (4)

plus β+ disintegrations:

13N → 13C + e+ + νe - 1.198 MEV (5)

15O → 15N + e+ + νe - 1.732 MEV (6)

17F → 17O + e+ + νe - 1.738 MEV (7)

8B → 8Be + e+ + νe - 16.957(14.06?) MEV (8)

where it is specified maximal (or fixed) energy which is carried away by an electronic neutrino.

Fig. 1. The calculated range of solar electronic neutrinos

As we see, the majority makes a neutrino of the first reaction (1). To participants of experiment of Borexino, it was succeeded to register traces of the second reaction - having processed the data collected for two and a half years, they reported that they managed to register solar neutrinos with energy in the range of 1,0-1,5 MEV.

Well and as an overwhelming majority (> 91%) make a neutrino of the first reaction - we will consider only a neutrino of the first reaction.

Most sensing of the types of neutrino detectors created so far is the Gallic detector. Its threshold amount of energy makes 0,233 Mev. On the basis of this detector within 5 years Gallic experiment of "GALLEX" was made. To protect the detector from hindrances of space radiation, it was placed on depth of 3 300 m at the foot of a hill in Italy, in Gran Sass (to the east of Rome). As a result of this experiment about 50% from the stream of solar neutrinos predicted by solar models were registered. Well and as the average duration of light time of day when detectors can catch solar neutrinos, is equal to 12 clocks, we receive that the experimental data will be coordinated with the first section of this article.

But this experiment allows drawing still some conclusions. In order that neutrinos became invisible to the Gallic detector, it is necessary that when passing through substance of Earth they lowered the kinetic energy to level lower than 0,233 MEV. From here it is possible to make the conclusion that when passing through our planet at the width of Rome of a neutrino loses part of the kinetic energy (if to consider possible power loss of a neutrino at their driving through the sun, smaller size) will turn out. As a result it is possible to count percent of energy of the first reaction lost by solar neutrinos at their passing through 1 km of substance of Earth.

Average radius of our planet (r) makes 6371 km, the width of Rome - 41 ° 52', from here it is possible to define the maximal distance passed by solar neutrinos through substance of Earth:

l=2×6371×cos (41 ° 52') =12742×0.74466=9487.5 km. (5)

The average distance will be approximately twice less and will make approximately km l=4744.

Now we can estimate percent of energy lost by a solar neutrino when passing through substance of our planet.

Let when passing through 1 km of substance of a planet of a neutrino on the average loses some part of the kinetic energy,

E1 = E0 / (1 + αmed) (6)

where E0 – average initial energy at an entrance of a neutrino to substance of a planet, E1 – average energy after passing of 1 km of substance of a planet, αmed - parameter defining power loss by solar neutrinos.

Having passed l of kilometers through substance of a planet, the neutrino will lose more energy, and new energy of a neutrino will be equal:

El = E0 / (1 + αmed) l (7)

Average energy of solar neutrinos of the first reaction (E0) corresponds to a maximum in a range (fig. 1) – 0.3 MEV.

And now knowing E0=0.3 of MEV, El=0.233 MEV and l can solve an inverse problem and to define αmed. As a result we will receive:

αmed = 0,00005328.

I.e. when passing through 1 km of substance the solar neutrino of the first reaction loses on the average not less than 0,005352% of the kinetic energy. This size very small and therefore remained unnoticed, but our planet rather larger and for what – we will see soon.

For a start we will look, how many the kinetic energy loses a solar neutrino of the first reaction when passing through Earth middle. In this case l will be to equally doubled radius of our planet 12742км, and

1/ (1-αmed) l = 1 / (1 - 0,00005328)12742 =0,5145 (8)

As a result the solar neutrino of the first reaction when passing through the middle of Earth will lose not less than 48.5% of the kinetic energy that there correspond 0.1457 MEV.

Let's multiply a neutrino stream with their loss energy; we will receive an energy flux density passing through the center of a surface of our planet and remained in it not less:

0.66×1011 neutrino / (cm2 × c) × 0.1457×106ev = 0.9616×1016ev / (cm2 × c) = 0,154×10-2W/cm2 = 15,4 W/m2 (9)

For comparison fluency of sunlight makes 1360 W/m2. In spite of the fact that the energy transferred by a neutrino of the first reaction is about 100 times less than energy of sunlight and its influence on surface temperature of a planet can be neglected, this energy constantly warms the interior of our planet and it is necessary to reckon with this warming up.

Now we can estimate the complete stream of energy of solar neutrinos of the first reaction. Certainly, it is possible to take a definite integral and by means of the COMPUTER precisely to count it, but it is possible and is simple to take the maximal value to divide it into two (for obtaining mean value). And then to increase by the circle square with a radius is equal to average radius of Earth. As a result we will receive the following approximate value:

½ ×15.4 W/m2 ×πr2 = 7.7 × 3.14159 × 63710002 = 9.81872 × 1014 W. (10)

Average fluency of entering energy will be

9.81873 × 1014 W / (4 πr2) = 1.925 W/m2 (11)

The complete stream of neutrino energy passable through Earth, will be:

0.3 MEV/neutrino × 3.14159 ×63710002 m2 × 0.66×1015 a neutrino / (m2 × c) = 0.25248 ×1029 MEV / with = 4.045 ×1015 W. (12)

As we see, when passing through Earth, a neutrino loses not less than 24% of the kinetic energy. But not less is does not mean equally.

Let's look from the point of view of physics that will happen gradually to the cooled-down planet of the sizes as at Earth and placed at the same distance from the sun, little billions years ago.

  • At first the planet surface up to some average temperature will quickly get warm.

  • Heat from a surface gradually will get inside, thus the surface of a planet will continue to heat up already sluggishly.

  • Then temperature in a planet will reach average temperature on its surface and further will sluggishly increase, warmed by neutrino energy.

    • Then temperature in the center of a planet will reach level at which melting of low-melting substances will begin and process of transformation of a planet will be accelerated.

  • Then on the one hand the fusion zone of low-melting substances will extend, and in the planet center temperature will grow to level of melting of other breeds.

  • The fusion zone will increase gradually, process of movement of heavier breeds closer to the planet center will begin. This process will be the long-lived.

  • Then, when the fusion zone will come nearer to a surface of a planet and it will need to overcome bark thickness some tens kilometers, the vigorous volcanic activity and motions of bark of a planet will begin. As a result of it the exit of an internal energy to a surface of a planet and its warming up will sharply increase.

  • Gradually the planet will dump the excess energy saved up inside and will pass to more moderate course. Surface temperature of a planet will smoothly go down, there will be eruptions of volcanoes and the earthquakes dumping excess energy.

  • After a while average surface temperature of a planet is stabilized. There will be eruptions of volcanoes and an earthquake. The planet can be in such condition as much as long while with a row constantly with former activity the Sun shines.

In more detail and it can precisely describe mathematical models of Earth after their corresponding correction. It should be noted that this process depends on the planet sizes. With decrease of the sizes of a planet absorbed neutrino energy will decrease, and conditions for heat removal from the center of a planet will improve - thickness of bark of a planet as a result will increase and its tectonic activity will decrease, and volcanic activity in general can stop. Well and with an approximation to the Sun both streams of energy both light, and neutrino with all that it implies (temperature body height on surfaces, increase of volcanic activity, etc.) will increase.

Let's return to our planet. Average fluency of entering energy is equal 1.925 W/m2 therefore it was possible to expect, as the average density of thermal energy leaving a planet same. But by estimates of models of Earth the heat flux going from Earth several times less. If data of models of Earth are exact, there is a question: what did we do not consider?

  • First of all, the degree of approximation of calculations and also not the accounting of dependence αmed from a kinetic energy of a neutrino and inhomogeneity of substance of a planet takes place. Therefore the calculations given here should be considered as a preliminary estimate.

  • When passing through substance of the Sun, a neutrino (also as well as on Earth) has to lose some part of the kinetic energy. The size of these losses while is not established by physics.

  • When driving from the Sun of a neutrino can spend some part of the kinetic energy for destruction of the met connections of a molecular neutrino.

  • The part of energy will leave Earth together with thermal waters.

  • But most important: the part of energy entering in Earth will collect while it will not be able to escape in the form of eruptions of volcanoes, earthquakes, change of a surface of a planet and tectonic activity.

As we see, except disintegrations of long-lived radioactive isotopes (a potassium-40, uranium-238, uranium-235 and thorium-232) to which the part of almost single sources of internal heat of Earth earlier was assigned, there is one more considerable energy source - the Sun. And all statements about pervasive ability of a neutrino - are inexact. It is not necessary to put equality sign between reactions with participation of a neutrino and their interactions are different concepts.

Let's sum up: according to the field theory of the fundamental particles, one of natural energy sources of earthquakes, volcanic activity, geothermal activity, a heat flux, coming from Earth subsoil, the stream of a kinetic energy of the solar electronic neutrinos, resulting thermonuclear reactions to the Sun and passing through our planet is.

Vladimir Gorunovich

3.03.2013

"Riddle" of solar neutrinos from the point of view of the field theory elementary particles. Part 2.

Contents:

4. Range of solar electronic neutrinos

5. Electric fields of electronic neutrinos and their interaction

4. Range of solar electronic neutrinos

In the first part of article it was established that the size of the complete stream of solar electronic neutrinos of SAGE measured in neutrino experiments (most the long-lived) and GNO will be coordinated with reference solar model. There is the following question: and as affairs with a range are.

In Homestake experiment under the leadership of R. Davis with use chlorine - the argon detector with a threshold amount of energy of 0.814 MEV was measured a stream of solar electronic neutrinos equal 2.56 SNU (where by SNU - reference solar unit in neutrino astronomy).

Let's look that had to receive as a result of experiment. For this purpose again we will consider the range of solar electronic neutrinos calculated by John Bahcall and Pinsonneaul (Fig. 2).

Figure 2. The calculated range of solar electronic neutrinos

Some padding data necessary for definition of that had to register chlorine - the argon detector are presented in table 1.

Table 1. The calculated stream of energy of solar electronic neutrinos

Reactions: stream v % Eмакс(MEV) Еср(MEV) stream E(MEV)

pp 6,00×1010 90,955 0,4202 0,3 1,80×1010

pep 1,43×108 0,217 1,4422 1,4422 2,06×108

7Be 4,89×109 7,413 0,862 0,862 4,22×109

8B 5,69×106 0,009 16,957 6,5 3,70×107

13N 4,92×108 0,746 1,198 0,7 3,44×108

15O 4,26×108 0,646 1,732 1 4,26×108

17F 1,00×107 0,015 1,738 1,1 1,10×107

all: 6,60×1010 0,352 2,32×1010

Apparently from fig. 1 and table 1 chlorine - the argon detector has to fix:

  • all solar electronic neutrinos of reaction of pep – 0.217%,

  • all solar electronic neutrinos of reaction 7Be – 7.413%,

  • the little less than a half of solar electronic neutrinos β+ disintegration of an isotope 13N – 0.746×0.45 = 0.336%,

  • a little more a half of solar electronic neutrinos β+ disintegration of an isotope 15O – 0.646×0.55 = 0.355%,

  • more than a half of solar electronic neutrinos β+ disintegration of an isotope 17F – 0.015×0.6 = 0.009%,

  • about 90% of solar electronic neutrinos β+ disintegration of an isotope 8B – 0.009×0.9 = 0.008%,

As a result we will receive about 8.338% from all streams of solar electronic neutrinos taking into account the amendment from the field theory, i.e. from 65 SNU.

65 SNU×0.08338 = 5.4 SNU (18)

So, from expected 5.4 SNU units of a stream of solar electronic neutrinos it was registered less than a half (2.56 SNU). - Why?

About a range of solar electronic neutrinos Gallic neutrino experiments of SAGE (most the long-lived for today experiment) and GNO allow making one more conclusion. As a result of these experiments the stream of solar electronic neutrinos according to reference solar model was recorded. But apparently from fig. 1, thirty two percent of solar electronic neutrinos of reaction (pp) cannot be registered by the Gallic detector as their energy lies below a detector threshold amount of energy. And the Gallic detector, nevertheless, showed the exact value. - Why?

As we see, both neutrino detectors as chlorine - argon, and Gallic showed discrepancy of the calculated range (fig. 2) to a stream of the solar electronic neutrinos entering to Earth. Only total number of entering solar electronic neutrinos (on the Gallic detector) coincided, but there are considerable deviations in energies.

So from Gallic experiments follows that the maximum of energy of flow of solar electronic neutrinos of reaction (pp) cannot be equal neither 0.3 MEV, nor 0.352 MEV, and is much above: not less than 0.533 MEV. And from chlorine - argon experiment follows that the maximum of energy of flow of solar electronic neutrinos of reaction (pp) is not higher than 0.694 MEV.

Apparently from results of operation of both neutrino detectors the maximum of energy is somewhere between levels of 0.533 MEV and 0.694 MEV. Let's take mean value of 0,6135 MEV, in this case fluency of neutrino energy passing through our planet will be more:

6.60×1010 × 0.6135 MEV = 4 × 1010 MEV / (cm2 c) = 4 × 1020 ev / (m2 c) (19)

A half of solar electronic neutrinos with energies higher than 0.814 MEV (generally beryllium electronic neutrinos with energy of 0.862 MEV of making 7.4% of a common stream) lost part of the kinetic energy (not less than 0.048 MEV) and lower than 0.814 MEV moved to range – now these solar electronic neutrinos became invisible for chlorine - the argon detector.

It is necessary to find the answer to a question why at solar electronic neutrinos the size of a kinetic energy changed? The answer to the matter is covered in a structure of the elementary particles and structure of their electromagnetic fields. For this purpose it is necessary to address to the field theory of the elementary particles and a classical electrodynamics.

The matter is that the range of fig. 1 falls into to a place of formation of solar electronic neutrinos – instead of to a place of their detecting. Flying by from an education place (from the Sun core) through part of a solar core, a zone of beam transfer, a convective zone, a photosphere, the chromosphere and the Sun crown to a detecting place on Earth, the solar electronic neutrino met huge quantity of the elementary particles with which it is very weak on the way, but interacted. And a corollary of interactions is the exchange of part of energy. And if on Earth electronic neutrinos act only as energy sources - on the Sun these neutrinos can receive padding energy, for example, from photons.

5. Electric fields of electronic neutrinos and their interaction

Can seem that is in the name a mistake. What fields at a neutral elementary particle can be, after all its electric charge is equal to zero. But from is the zero size of electric charge lack of electric fields at the elementary particle does not follow. The charge and field is primary not. Electric fields of the elementary particles are created not by electric charges, and a constant component of an electromagnetic field, and this field dipole. We observe electric charge because the symmetry between the external and internal sites generating a constant electric field of the elementary particle is broken.

The elementary particles are arranged where it is more interesting, than the quantum theory and reference model, and quarks, gluons, gravitons, Higgs's bosons consider as it are not necessary.

So, we will remember structure of electromagnetic fields of a neutral elementary particle with a quantum number of L>0, for example, an electronic neutrino. According to the field theory of the elementary particles, the electronic neutrino possesses:

  • constant electric dipole field,

  • constant magnetic dipole field,

  • constant magnetic field of a moment of magnet,

  • variation electromagnetic field.

The zero size of boundary space electric charge does not speak about lack of a constant electric field, as well as (and even its zero size) does not tell absence of data on the size of a moment of magnet about lack of constants of magnetic fields.

Let's consider constant electric dipole fields of neutral elementary particles. Everything that will be told here, is treats and an electronic neutrino.

The dipole in physics is understood as system from two electric charges of +q and – q of s located apart between them. In case of the elementary particles of q=0.75e (a corollary of the field theory), and s decides by thickness of an annular domain generating a constant electric field and in inverse proportion to a rest-mass of the elementary particle (in case of an electronic neutrino on a rest-mass of 0.28 ev, s = 6.12×10-5 cm).

But the difference is that charges in the elementary particles are not present, and there are vector electric fields (and vector magnetic fields too). The physics did not face such fields earlier, and experience with them for the present is not present. But something can be made. Outside the elementary particle fields of vectors will be transformed in electric with which the classical electrodynamics is able to work. We can replace an actual vector electric field with a field of two self-contained ring parallel evenly the distributed electric charges of +q and – q of s located apart between their planes. At the distances several times exceeding s, this field will be few than to differ from an actual field, but this replacement will allow the computer to calculate their interactions, proceeding from laws of a classical electrodynamics. The made replacement is lawful as in the nature electric charges, and electric fields interact not. According to a classical electrodynamics, energy of such interaction is defined as a difference of the sum of energies of fields of each of interacting charged particles and energy of their resultant field (all in strict accordance with the law of conservation of energy). Well and with electric charges it is simpler to work, than to integrate fields on all space.

At distances (r) exceeding 10s these fields will be same, as well as electric dipole fields of point charges of +q and – q of s located apart between them. Problem of calculation of such fields of the physicist already solved, and there are corresponding equations.

In physics there is a vector size under the name electric electrical dipole moment (p)

p = q s (20)

In case of the elementary particles

s=0.85 ħ/m0~c (21)

and then:

p = 0.75es, |p | = 0.6375e ħ/m0~c (22)

where with a bold print are allocated a vector, ħ – a Dirac constant, e – the elementary electric charge, m0~ – a rest-mass concluded in a variation electromagnetic field of the elementary particle, with – light velocity, |p |-the size of the electrical dipole electric moment of p of a neutral elementary particle. The factor 0.85 is taken from the field theory because the average point of an arrangement of the equivalent electric charge has to lie not on a surface of area generating a field, and is about 15% deeper (a corollary of geometry of a form of a field of vectors).

The field gradient of an electric dipole (in a Si-system) is equal in a Si-system:

(23)

where the bold print allocated a vector, n - an unit vector from the dipole center in the direction of a measuring position, with a point (•) the scalar product, ε0 – a permittivity of vacuum, p – a vector of electric electrical dipole moment, r – distance from the dipole center to a measuring position is designated.

Force operating on electric charge (q) being in a field of an electric dipole of p, will be:

(24)

Potential (ϕ) fields of an electric dipole of p in a point and:

(25)

where α-a corner between a vector and the direction on a point of supervision (And).

Energy (W) of a point charge (q) of the electric dipole which was in a field of p (And) is equal in a point to work of size of electric charge on field potential:

(30)

where re – is the electric radius of the neutral elementary particle creating an electric dipole field.

In a formula (30) we replaced an electric dipole field of two point charges (size +0.75e and - 0.75e) in the field from two groups on three equal point charges (sizes +0.25e and - 0.25e), lying in the parallel planes, equidistant from each other and re which were apart from an axis of symmetry of the elementary particle. Such replacement of electric charge by three equal fractional charges is a minimum for reflection of internal structure of a field – it is not connected with imaginary quarks in any way.

Knowing energy size at r=0 and known orientation, it is possible to solve an inverse problem and to determine the r0 parameter for each elementary particle.

Similarly it is possible to finish the equation (26) for energy of the charged elementary particle 2 (is possessing mesh sizes, in comparison with the elementary particle 1) in an electric dipole field of a neutral elementary particle 1.

(29)

where 0.4064=0.63752, m0~1 and m0~2 – rest-masses prisoners in a variation electromagnetic field (m0~) the elementary particle 1 and 2 respectively.

In (29) we received a formula of energy of electric dipole interactions of couple of neutral elementary particles at distances (r>> s). In more short distances the size of energy can be calculated by means of the computer with application of a classical electrodynamics.

In order that the formula (29) could work approximately and at smaller distances we will add in a denominator to r3 still re and two items: ro13 and r023 respectively on one for the elementary particle 1 and 2.

(28)

where r - distance between points 1 and 2, θ12-a corner between vectors of the electrical dipole electric moments р1 and p2, θ1 and θ2-corners between vectors р1 and p2 and a vector of r, W12 - energy of a dipole р1 in the field of p2 dipole.

From the equations (27, 28) follows: W21=W12 (corollary of symmetry of a trigonometric function of cos).

Formulas (27, 28) are fair for long distances (r>> s), in this case energy of interactions of electric dipole fields decreases under the law 1/r3, even quicker, than for a point charge in the field of an electric dipole. And the size of this energy is proportional to work of vectors of electric electrical dipole moments of the elementary particles (p1 and p2) and depends on their relative orientation.

Now it is necessary to substitute in the equation (10) size of the electrical dipole electric moment (received by means of the field theory) p=0.6375e ħ/m0~c and we will receive:

(27)

or in a vector form:

(26)

As we see, the potential of a field of an electric dipole (unlike a field of electric charge) decreases under the law 1/r2.

Using a principle of superposition it is possible to count interaction energy of electric charges of a dipole of p1 with a field of an electric dipole of p2, and then to put them (considering a sign of each charge of a dipole of p1). As a result we will receive energy (W) of interaction of two dipoles р1 and p2 being in points 1 and 2 respectively:

(31)

where the r parameter for potential (it is similar to the r0 parameter for energy) is equal 1.5 L ħ/m0~1c.

Apparently, we had fractional degrees, but that is suitable for the equation (30) not absolutely is suitable for the equation (31). If to break an electric dipole charge of an electronic neutrino into four equal parts the equation will assume more convenient air:

(32)

where h - distance between the planes of both particles, he - the average height of an arrangement of electric charge in a neutral elementary particle 1 (r - distance between an axis (rotation of a variation electromagnetic field) elementary particle 1 is equal 0.5s from the equation 4), and the center of the elementary particle 2, - the electric radius of the neutral elementary particle creating an electric dipole field, roф1 - normalizing parameter is equal to re 1.5 L ħ/m0~1c, α-a corner between a vector of electrical dipole moment (p) of a particle 1 and the direction on a point of finding of a particle 2, m0~1 – a rest-mass concluded in a variation electromagnetic field (m0~) the elementary particle 1 creating an electric dipole field (for an electronic neutrino it is equal 0.97755m0 where m0 – a rest-mass), other designations - standard in physics.

Well and as in comparison with an electronic neutrino all other elementary particles with a quantum number of L>0; 0 can be considered as point wise objects, the equations (30) - (32) will work approximately and at small distances.

Now it is possible to look at some schedules.

So the schedule of the calculated potential energy of interactions of electric dipole fields of two electronic neutrinos lying in one plane with opposite spins looks.

Fig. 3 Interaction of Electric Fields of Two Electronic Neutrinos.

From the schedule existence of attractive forces between the elementary particles, operating in a near zone is visible.

And so the schedule of the calculated potential energy of interactions of electric dipole fields of an electronic neutrino with a neutron, lying in one plane with opposite spins looks.

Fig. 4 Interaction of Electric Fields of an Electronic Neutrino and Neutron.

From the schedule all is visible, and the answer to a question: "why" is in the equation (30). It is necessary to compare only sizes of rest-masses of a neutron and an electronic neutrino. It is easy to be convinced that the similar result will turn out and with any other neutral elementary particle, except a neutrino.

And now for the sake of what article was written: electric interactions of an electronic neutrino with an electron and with a proton.

The fig. 5 Interaction of an Electric Field of an Electronic Neutrino with an electron.

The fig. 6 Interaction of an Electric Field of an Electronic Neutrino with a proton.

In both calculations the charged particle lies in the plane, the parallel plane of an electronic neutrino and with shift in plus equal to the radius of an electronic neutrino.

We received that the electronic neutrino can exchange energy with the charged elementary particles. As you understand, continuation will follow.

But something left unfinished remained and here. In figures 5 and 6 the peak located apart equal to the electric radius (re) of an electronic neutrino is visible. As the elementary particle in the center empty, a maximum of an electric intensity of a neutral particle is the share of the distance equal to radius of the elementary particle from the field theory (L ħ/m0~c). Electric radius will be a little more, owing to asymmetry between the external and internal hemispheres generating constant electric a field of the elementary particle In process of body height of a quantum number the L this asymmetry will decrease owing to what the size of electric radius of the elementary particle will aspire to the size of field radius.

So:

  • both neutrinos electric radius have at 10.6% more field radius,

  • at a neutral pi meson, and other neutral mesons - for 3,54%,

  • at a neutron and other neutral baryons this difference makes already 1.77%,

  • at neutral vector mesons (one of which to us in every way try to give out for Higgs's imaginary boson) with a spin 1 difference makes 1,06%,

  • yet open neutral baryons with a spin 3/2 difference will have 0,7%.

The charged elementary particles will have a size of electric radius other.

In general attempts to adapt mathematics of point charges for actual fields is a task hard-hitting or nearly unattainable, it can be solved at the expense of accuracy loss.

Generally (for an electronic neutrino) we are faced by a problem of interaction of point electric charge with an electric dipole field of two parallel ring distributed charges (±0.75e) and here integrals and respectively computer calculation are required. If other neutral elementary particle acts as a source of an electric field - it is necessary to consider still structure and the charged elementary particle.

And with the equation (30) affairs are even more difficult. It is visible on interactions of electric fields of neutrons.

The fig. 7 Interaction of Electric Fields of Neutrons with opposite spins.

To be continued.

Vladimir Gorunovich

30.04.2013

The numbering of formulas and drawings is changed of 6.09.2013