Riddle of solar neutrinos part 3

"Riddle" of solar neutrinos from the point of view of the field theory elementary particles. Part 3

Contents:

6. The energy acquired by the charged particle at passing through an electric field of a moving electronic neutrino

7. Loss by an electronic neutrino of a kinetic energy when passing through substance

8. Interaction of electronic neutrinos with lava ions

9. Interaction of electronic neutrinos with ocean water

10. Interaction of electronic neutrinos with metals

6 . The energy acquired by the charged particle at passing through an electric field of a moving electronic neutrino

Let's ask a question: what kinetic energy will be got by the charged particle (not necessarily the partial) through which will pass a wave of a constant electric field strength of E and duration Δt, created by an electronic neutrino moving with near light speed.

According to electrodynamics laws, the particle with q charge in an electric field strength of E is affected by force f equal:

(33)

Under the influence of this force, according to the classical mechanics, the charged particle will get acceleration α, equal:

(34)

where m – the mass of the charged particle.

If the considered particle originally was based, in time Δt it will get the speed of V, equal:

(35)

If this speed not relativistic, the acquired kinetic energy of W (for stay during time Δt in the field strength of E) is equal:

(36)

In case of the inhomogeneous electric field for determination of the acquired speed it is necessary to take a definite integral:

(37)

In this case the acquired kinetic energy will be:

(38)

The factor of q²/2m depends on electric charge and weight of a particle. It accepts the greatest size at an electron – the easiest charged elementary particle. Thus, when passing through substance of Earth, solar electronic neutrinos have to warm, first of all, the free electrons of substance (where they are available). Except the free electrons, in much smaller degree, other free charges, for example, the free ions of substance, as in water (hydro’s, hydroxyl, salts …), and in the melted breeds (for example, in a lava in Earth) will be warmed.

The warming up of atoms electrically neutral substance will be interfered by an electric field of nuclear kernels. Under the influence of an external electric field of an electronic neutrino there will be a shift of average provisions of electrons in atoms therefore in atom there will be an electric field, opposite in direction to the external field, not allowing electrons to be displaced further. It will not give the chance to electrons to gather a kinetic energy from an electronic neutrino. There will be other mechanism and other levels of energy.

The size of an acquired kinetic energy is proportional to a square of an electric intensity and a square of time of finding of the charged particle in an electric field. In case of an electric dipole field of an electronic neutrino it depends on orientation of a spin of an electronic neutrino and a trajectory of passing of the charged particle through a neutrino (in a frame of reference bound to the center of an electronic neutrino). The greatest electric intensity and duration of interaction of two particles will be when passing the charged particle through the center of an electronic neutrino and orientation of a spin of an electronic neutrino in the direction, perpendicular to the driving direction. In this case, interaction time, according to the field theory of the elementary particles will be:

(39)

where m_0~ - the mass of a variation electromagnetic field of the electronic neutrino, Lħ/m_0~c – the radius of the elementary particle in the field theory, for an electronic neutrino equal 3.6044×10^-7м, L – the main quantum number (at leptons equal ½), ħ – a Dirac constant divided on 2π, с – light velocity.

To it is possible to add still interaction after an exit of the charged particle from an internal electric dipole field of an electronic neutrino that will add some more percent to the main size of a kinetic energy.

Well and an electric intensity we will take from a classical electrodynamics.

Strength of a constant electric field of a dipole on a perpendicular, regenerated to an axis from its middle (we also will consider this case) in a Si-system:

(40)

where p – is electric electrical dipole moment, r - distance from the dipole center to a supervision point, a factor 1/ (4πε₀) – from an electrodynamics, equal 8.98755×10⁹ m / ф, works in SI.

The size of electric electrical dipole moment of an electronic neutrino was determined by the field theory of the elementary particles. In case of distribution 1 it is equal:

Fig. of 8 Neutrinos

If the direction of a spin of an electronic neutrino is parallel to the driving direction, the maximal distance passable by the charged particle through an internal electric dipole field of an electronic neutrino will be reduced twice:

(50)

In this case the size of an acquired kinetic energy will decrease by 4 times and will be (for the free electron):

(51)

For spin orientations from 0 to π/3 it will be right (50), and, therefore, and (51).

But the acquired kinetic energy depends not only on a spin, but still, on what trajectory the charged particle flew by through an electronic neutrino. The maximum is reached when passing through the center of area generating a constant electric field of an electronic neutrino, and at shift from the center geometry laws come into effect.

Let's enter designation Δl – distance passable by the charged particle in an electronic neutrino within its homogeneous internal electric field. Then:

(52)

(53)

Now it is necessary to substitute in (36) for the free electron, we will receive:

(54)

In (54) it is necessary to substitute Δl which changes in limits:

0 - 2ħ/m_0~c, for the spin directions from π/3 to π/2,

0 - ħ/m_0~c, for the spin direction from 0 to π/3.

For other directions of a spin Δl undertakes symmetrically.

For the any charged particle (with z charge) we will receive the expression similar (54):

(55)

7. Loss by an electronic neutrino of a kinetic energy when passing through substance

Let's define what energy is lost by an electronic neutrino when passing through 1 meter of the homogeneous matter.

Let's consider a case of a parallel alignment of a spin (the least power loss).

The sectional area of an electronic neutrino in the plane passing through the center of a particle and perpendicular her back will be:

(56)

where r_v – the radius of an electronic neutrino in the field theory.

Having passed through 1 meter of substance, the electronic neutrino will meet on the way of Ni of the charged particles of i:

(57)

where n_i – concentration of the charged particles of i in cubic meter.

If the substance contains the unbound charged particles of different types (for example, ions of different chemical elements), it is necessary to take the sum on each type of the charged particles separately as interaction energy of an electronic neutrino with different charged particles will be different.

In this case the energy lost by an electronic neutrino at passing of 1 meter of substance, will be:

(58)

As

(59)

that

(60)

Considering that for leptons to which the electronic neutrino, the main quantum number in the field theory belongs:

(61)

Let's receive the average energy lost by an electronic neutrino at passing of 1 meter of substance:

(62)

Let's substitute in (62) initial data, as a result we will receive:

(63)

and for substance with the free electrons:

(64)

Now we will add in (63) fluence of solar electronic neutrinos on a surface of our planet and as a result we will receive a stream of the energy lost by solar electronic neutrinos , when passing through 1 meter of terrestrial substance:

(65)

For comparison fluence of sunlight makes 1360 W/m².

Even if in (65) we will substitute electronic mass (having lawfully assumed that will be warmed, first of all, the free electrons as the energy going on a warming up of other charged particles, is inversely proportional to the size of their weight), all of us will equally receive a small:

(66)

Now it is necessary to specify, how many the free electrons (it is possible the free charges) are available in one cubic meter of substance. It depends on substance and its temperature. In routine minerals, at averages for Earth surface temperatures, all electrons are bound in atoms and molecules, and there it is few free charges. The stream of solar electronic neutrinos with ease will pass through such substance, almost without loss energies. But, first, there are metals, and in them the free electrons are available also them much. And secondly, routine minerals can be warmed up to the temperatures (to melt) at which they will begin to carry out an electric current (to them there will be free carriers of electricity) and such temperatures exist in Earth subsoil. The minerals being dielectrics in solidity, in flux carry out an electric current, and under bark of Earth there is an ocean of the melted lava (in flux, the conductor of the electric current, possessing the free carriers of electricity).

And now we will ask a question: how many will be the free electrons in one cubic meter of the melted lava warmed up to the temperature at which 1 valence electron on one thousand molecules (it not such rigid conditions) is free at least. For this purpose it is necessary to count number of molecules in cubic meter of substance.

For example, we will take lava from 50% of dioxide of SiO₂ silicon, and 50% from Al₂O₃ adamant. Its density can be taken equal 2.8×10⁶ g/m³. The molar mass of SiO₂ is equal 60.05 g/mol, and Al₂O₃ – 101.96 g/mol. Now it is necessary to remember number of molecules in one mole of substance (Avogadro number) of N_A = 6,022 141 29×10²³ mole-1.

The number of molecules being in 1 cubic meter of lava will be:

(67)

From here number of the free electrons in 1 cubic meter of lava:

(68)

Let's set up him in (66), we will receive:

(69)

I.e. when passing through 6 km of such melted lava being in an Earth's mantle, solar electronic neutrinos lose almost all the kinetic energy. And after all we considered only electric interactions and only passing through an internal field of an electronic neutrino. But at an electronic neutrino has to be and the constant magnetic field, and magnetic fields too interact.

And at last, it is worth to remember that the radius of the melted Earth's mantle in one thousand times more.

Let's remember from the second part, fluence of neutrino energy passing through our planet:

(70)

that makes 4.8% from a sunlight energy flux density.

The complete stream of neutrino energy absorbed by our planet, (taking into account results of part 2) approximately will be:

0.6135 MEV/neutrino × 3.14159 ×6346000² m² × 0.66×10¹⁵ neutrino/(m² × s) = 0.5 ×10²⁹ MEV /s = 8 ×10¹⁵ W. (71)

Here earth radius and average radius of the melted cloak which is actively absorbing a kinetic energy of solar electronic neutrinos is taken not. And this energy warms our planet from within! Use of part of this energy can make eruptions of volcanoes by more infrequent as through eruptions there is a dumping of an excess stored energy.

Thus, the conclusion drawn in the first part of article can be specified: according to the field theory of the elementary particles, the main natural energy source of earthquakes, volcanic activity, tectonic 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. Now the physics can approve it.

It is possible to consider other orientation of a spin of an electronic neutrino in relation to the driving direction (when more energy is absorbed), but it already significantly will not change conclusions.

8. Interaction of electronic neutrinos with lava ions

For an example, at first we will consider a special case when practically all molecules of minerals of the lava considered earlier, (and Al₂O₃) are completely ionized by SiO₂ (100% ionization of molecules of substance take place). So, instead of each molecule of SiO₂ we will have 1 ion of silicon (Si) with a charge +4e and 2 ions of oxygen (O) with a charge - 2e, and instead of Al₂O₃ molecule – 2 ion of aluminum (Al) with a charge +3e and three ions of oxygen (O) with a charge - 2e. I.e. instead of 2 molecules we received 8 ions. So, it agrees (67), we have:

• Si⁴⁺ : n₁ = 1.0408×10²⁸, z₁ = +4, m₁ = 4.48×10^-26 kg,

• Al³⁺ : n₂ = 2.0816×10²⁸, z₂ = +3, m₂ = 4.663×10^-26 kg,

• O^2- : n₃ = 5.2040×10²⁸, z₃ = -2, m₃ = 2.657×10^-26 kg.

Having set up them in (63), we will receive the average energy lost by an electronic neutrino at passing of 1 meter of the melted lava (from 100% ionization):

(72)

And having set up them in (65) we will receive a stream of the energy lost by solar electronic neutrinos at passing of 1 meter of the melted lava from 100% by ionization:

Table 2. Ionic composition of ocean water.

Where: % - percentage in moles on kilogram of ocean water, z – electric charge of an ion, n_i – number of ions in cubic meter of ocean water, m – the mass of an ion.

If to substitute these data in (63) we will obtain an upper bound of the energy lost by an electronic neutrino at passing of 1 meter of ocean water:

(75)

Having added fluence of solar electronic neutrinos on a surface of our planet, and as a result we will receive an upper bound of a stream of the energy lost by solar electronic neutrinos, when passing through 1 meter of ocean water: