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Solitonic Model of the Electron, Proton and Neutron | OMICS International
ISSN: 2090-0902
Journal of Physical Mathematics
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Solitonic Model of the Electron, Proton and Neutron

Sladkov P*

Independent Researcher, Russia

*Corresponding Author:
Sladkov P
Independent Researcher, Russia
Tel: 086958945
E-mail: [email protected]

Received Date: June 09, 2016; Accepted Date: August 19, 2016; Published Date: August 23, 2016

Citation: Sladkov P (2016) Solitonic Model of the Electron, Proton and Neutron. J Phys Math 7: 193. doi: 10.4172/2090-0902.1000193

Copyright: © 2016 Sladkov P. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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Abstract

In paper, which is submitted, electron, proton and neutron are considered as spherical areas, inside which monochromatic electromagnetic wave of corresponding frequency spread along parallels, at that along each parallel exactly half of wave length for electron and proton and exactly one wave length for neutron is kept within, thus this is rotating soliton. This is caused by presence of spatial dispersion and anisotropy of strictly defined type inside the particles. Electric field has only radial component, and magnetic field - only meridional component. By solution of corresponding edge task, functions of distribution of electromagnetic field inside the particles and on their boundary surfaces were obtained. Integration of distribution functions of electromagnetic field through volume of the particles lead to system of algebraic equations, solution of which give all basic parameters of particles: charge, rest energy, mass, radius, magnetic moment and spin.

Keywords

Structure of elementary particles; Structure of matter; Theory of elementary particles; Electron; Proton; Neutron; Nuclei; Electromagnetic field; Atom; Microcosm; Elementary particles; Fundamental interactions; New theory; New physical theory

Introduction

In present article alternative (to Standard Model) hypothesis of structure of electron, proton and neutron is suggested. The others elementary particles (except photon and neutrino) are not stable and they are considered as unsteady soliton-similar formations. In series of experiments indirect confirmations of existence of quarks were obtained, for instance in experiments by scattering of electrons at nuclei, performed at Stanford linear accelerator by R. Hofshtadter, look for instance [1]. At that, experiments by elastic and deeply inelastic scattering gave quite different results: in first case take place pattern of scattering at lengthy object, in second case is pattern of scattering at "point" centers, that is interpreted as confirmations of existence of quarks. However what "point" formations appear only in deeply inelastic scattering don’t may be an evidence of quarks existence, because to above-mentioned fact may be given and another explanations: in moment of birth of new particles, which take place in deeply inelastic scattering, structure of nucleon change, it sharply diminish in volume, but after appearance of new particles nucleon return to initial state. Or process of birth of new particles occur in "point" volume inside nucleon and these energy "point" centers disappear after completion of process particles birth. And fact that experiments by elastic scattering gave pattern of scattering at lengthy object prove inexistence of quarks in nucleus. In theory of Standard (quarkual) Model come into at least 20 parameters artificially introduced from outside, such as "colour" of particles, "aroma" etc., that is its fundamental demerit. Theoretical work, which is present here, has no demerits of Standard Model, it completely describe structure of elementary particles therefore it can help in discovery new ways of making energy, elaboration perfectly new devices for its production and to achieve progress in such fields as nuclear power engineering, nanotechnology, high-powerful lasers, clean energy and others.

Rotating Monochromatic Electromagnetic Wave

Let us write down Maxwell’s equations in spherical coordinates supposing that:

1) There are no losses;

2) Only image are not equal to zero.

image(1)

image(2)

image(3)

image(4)

image(5)

image(6)

Here r,θ, ? spherical coordinates of the observation point; imageimage - components of the electromagnetic field, image - density of electric current, image - volume charge density; ω-circular frequency of field alteration i - imaginary unit ε-dielectric permittivity - magnetic permeability Figure 1.

physical-mathematics-spherical

Figure 1:Representing the spherical coordinates of the observation point.

Substituting the expression for Hθ ? from (2) in (4), we obtain:

image(7)

image(7‘)

This is Helmholtz homogeneous equation. Let us designate.

image(7‘‘)

Wave number - General solution of Helmholtz equation:

image(8)

This expression describes two waves, moving to meet one another by circular trajectories, along the parallels. Pointing’s vector in each point is directed at tangent to the corresponding parallel [2,3].

Let us consider a wave, moving in positive direction ?.

image(9)

Here

image

Wave phase;

K1 - Dimensionless analog of the wave number. If to introduce a wave number of traditional dimension image

image

The wave phase will be written down as

image

Where

image

Arc length along the corresponding parallel. In the considered case the wave number is a function of coordinates and frequency. Thus, the wave, which is described, can exist only at availability of spatial and frequency dispersion [4-6]. Dispersion equations will be obtained below, apart from the already found expression (7′′).

From expression (2), taking into account (7′′) and (9), we have:

image(9‘)

For actual amplitudes:

image(10)

image(10‘)

Here

image

Means characteristic impedance.

The last expressions describe an electromagnetic wave, rotating around axis Z in positive direction ?.Conditions of self-consistency:

1) z=constant

2) Along each parallel on the circle length, the integer number of half-waves must be kept within.

image(11)

Here image

Wave length, v - phase velocity of wave, f - frequency, n=1,2,3…

Let us consider the case when n=1,

image

image(11‘)

Along each parallel, exactly half of wave length is kept within.

Phase velocity of wave is the function of frequency and distance up to the axis of rotation.

image

image(11‘)

image

image

We are substituting in (11′′):

image(12)

image(12‘)

image

We are substituting in (12′).

image(12‘‘)

image(12‘‘‘)

Taking into account (8) and (11′′)

image

Then

image(13)

image(13‘)

Function sin image is on valued in angles interval 0 ≤φ ≤ 2π

This situation can be interpreted as rotation of spherical coordinate system around axis z in positive direction ? with angular velocity image

Let us find it from the condition:

image

Having differentiated this expression on t, we receive,

image

At the same time the electromagnetic field, about spherical coordinate system, is determined by expressions (13) and (13′). Further from (3): as H?=0

image

image(14)

From equation (6)

image

Follows

image

image(14‘)

To receive field dependence from image , let us find solution of three-dimensional Helmholtz equation in spherical coordinates.

image(15)

Er does not depend from θ, look (14), therefore three-dimensional Helmholtz equation transfers into two-dimensional one.

image(15‘)

Let us suppose that

image

now

image(15‘‘)

This equation can be satisfied, if

image(16),(17)

Thus, initial Helmholtz equation has split into the system of two equations. We substitute in these equations instead of image (i.e. we are searching the solution as the product of two functions) and divide the first equation byf(r), and the second – by g(?). We receive

image(18), (19)

Equations (16) and (18) are equivalent to equations (7) ? (7′), which were received earlier from Maxwell’s equations, and

image

image

The solution of equation (18) was found earlier, look (13).

image

Let us copy (19) as:

image(19′)

Where

Where

image

19’ - centrally symmetric Helmholtz equation. Let us suggest, k4=k3 r

image

Where vr- phase velocity of electromagnetic wave in radial direction. As in the central symmetric equation angular dependence is absent, it is logical to assume that

vr=v=2ωrsinθ

at image i.e.

v=2ωr

image(21)

image(21′)

Instead of (19′), we are having

image(19′′)

This is Euler equation, it has the solution

image(22)

Let us converse expression (22).

image

Here image a-value of radius r, at which the rotating monochromatic electromagnetic wave ceases to exist, and Er=E0; f=1 hence

image

image

image(22′′)

image

In view of this,

image

Let us designate C=p now

image

Thus, for Er we are having

image

image

Really

image

So that at alteration of r within the interval from 0 to a, Er would not change its sign, observance of the following requirement is necessary P≤0

image

image

System of Equations for Electron

Basing on results of the previous section, let us write down expressions for electromagnetic field inside the electron, assuming that it is concentrated inside the orb of radius a

image(23′)

image(23′′)

Here a is electron radius, E0 amplitude of electric field intensity at r=a; z=const characteristic impedance inside the electron, P- unknown coefficient and P≤0

At that the internal electron medium possesses frequent and spatial dispersion, as well as anisotropy. Dispersion equations have the following appearance [7-11].

image(24)

image(24′)

image(24′′)

Here vr,vθ,v? - phase velocity of rotating monochromatic electromagnetic wave in corresponding direction. In viewed case, the electromagnetic wave is being spread only in the direction ?, and we shall need expressions vr and vθ for searching the formulas of dielectric and magnetic permeability, as well as wave numbers of corresponding directions; zr, zθ, z?, z characteristic impedances inside the electron; εφ, ?, μφ were found before, see (12′) (12′′)

image

image

In view of (24)(24′)(24′′), let us write down expressions for εr, εθ,μr,μθ

image

image

image

image(24′′′)

From considerations and formulas adduced, it follows that dielectric and magnetic permeability are tensor values.

image

image

Let us find dimensionless wave numbers.

image

image

image

Thus

image

Let us remind that in the viewed case, the electromagnetic wave is spread only in the direction of ?

At r=0 we are having a special point:

image

Despite of this, all basic electrons’ parameters – charge q rest energy W, magnetic moment M-expressed through integrals by volume from the functions specified above, prove to be finite quantities [12-15]. Look further.

From (5), we find volume charge density inside electron

image

image

Integrating ρ on electron’s volume, we shall receive this expression for its charge q.

image

image(26)

On the other hand, from the third integral Maxwell’s equation, it is possible to find electron’s charge as a stream of vector electric induction D through the surface of the orb of radius a

image(26′)

As we can see, expressions (26) ? (26′) are equivalent to each other.

From (1), we obtain expression for current density j?

image (27)

From expressions (25), (27) it is visible that in the interval of change of r from 0 to a,ρ and j? once change the sign. It can be explained by the fact that in the viewed structure, the substantial role is played by the rotating monochromatic electromagnetic wave, and the space charge density and electric current density – are auxiliary or even fictitious quantities in the sense that inside the particle there is neither any charged substance nor its motion [16]. Inside the electron, it is not the charge that is the source of electric field, but electric field is the source of the charge. In its turn, it is not the electric current that is the source of magnetic field, but magnetic field is the source of the electric current [17-22]. Thus, a deduction about vector nature of elementary charge can be made.

Now we shall determine electron’s rest energy as electromagnetic wave energy inside a particle.

image

Here w- is volume density of electromagnetic wave energy,

image

? – Pointing vector,

image

vφ -phase velocity of electromagnetic wave in direction of v?.

v?=2ωrsinθ

image

image

image(28)

image(28′)

Here image is Planck’s constant.

We shall be searching electron’s magnetic moment in the form of a sum. M=Mm+ML

M=Mm+ML

Where Mm- is magnetic moment, created by volumetric current; ML -magnetic moment, attributed to impulse moment, i.e. to rotation.

ML=γL

Where γ-gyromagnetic ratio; L-impulse moment of electron.

Basing on Barnett effect, we are making a supposition, that the impulse moment, attributed to rotation, creates additional magnetic moment [21,23].

Being aware of the fact that electron’s impulse moment is equal image from (28′) we find expression for L.

image

image

image

Let us calculate Mm as electric current magnetic moment in volume V, relating to axis z by the formula:

image

See for instance [3], page 111, where rz - distance to axis z,

rz=rsinθ

image(29)

image

image(29′)

Or

image(29′′)

Thus, we have received the system of algebraic equations for electron.

image

Here e - charge of electron, m- it’s mass.

Three equations contain five unknown quantities: E0, a, z, p, γ Let us add this system with equations, which we shall receive from boundary conditions.

At r=a, R=a

image(33)

In the exterior area, the same as and in the interior area, electric field intensity possesses only radial component. Here R - distance from electron’s center to the observation point in the exterior area, ε0- vacuum dielectric permeability [24].

Further. image(34)

In the exterior area, the same as and in the interior area, magnetic field intensity possesses only meridional component.

It is obvious that

image(33′)

then from (33) follows:

image(33′′)

On the other hand it is known that the electric field, having passed through dielectric layer, cannot increase, therefore

image(33′′′)

In other words, correlations (33′) (33′′) (33′′′) will be simultaneously executed only in one case, if

εr0;(35)

image(36)

Now under Biot-Savart’s law, we are finding magnetic field in the exterior area.

image

In last expression we substitute (12′′) and (27).

image(37)

image

image(38)

At r=a R=a

image

image

image(39)

On the other hand, from (24′′′)

image

At r=a

image

We substitute in (39).

image

image(40)

Thus, at r=a

image

image(41)

image

Here c - velocity of light, ω=7,7634421*1020 Hz- Compton circular frequency of electron.

image(42)

As it is known, atom’s radius approximately equals to 10-10 m, volume of atom - 4,18879*10-30 m3. We found, that radius of electron equals to 1,930796*10-13 m, volume of electron –3,0150724*10-38 m3. That is one electron occupies 0,7197955*10-8 from atom’s volume and, for example, 100 electrons (as in atoms located at the end of the periodic system) occupy 0,7197955*10-6 from atom’s volume [8,25].

We substitute (42) ? (39).

image(43)

Let us solve the system (30), (31), (32), taking into account (42) and (43).

image

image(30′)

image(31′)

image(32′)

We substitute (30′) in (32′).

image

image

image

P must be negative, therefore we select

image

We substitute (30′) in (31′)

image

We substitute p meaning in (31′′) and find γ

image

From solution of equation (31), it is visible that two components of magnetic moment of electron Mm ? ML are directed to opposite sides and image

Let us also calculate numerical value of E0 by formula (30′)

image

"Dimensions" of electron for the present are not discovered by experimental way, though precision of measuring is led to 10-18 m. Within the framework of the model considered it may be explained by the next way: electron is not hard particle with this quantity of vector E, which exist inside it, unlike from proton and neutron, quantity of vector E inside which approximately 107 times as much [26-31].

For positron, the system of equations will take a somewhat different view.

image

Boundary conditions are the same as for electron. Hence

image

image

The system of equations (44), (45), (46) with exactness to a sign, has the same solutions, as the system (30), (31), (32).

image

image

image

System of Equations for Proton

By applying reasoning and mathematical calculations of the previous section in relation to proton, we shall receive the relevant system of equations.

image

Here corresponding letters mean parameters of proton.

Boundary conditions: at r=a

image

image

image

image

image

Here:ω=1,425486*1024 Hz- Compton circular frequency of proton [32,33].

Solving the system (47), (48), (49), we shall receive

image

image

image

From the solution of equation (48) it is visible that two components of proton’s magnetic moment Mm ? ML have identical direction, and image

Let us write down the system of equations for antiproton explained in [34].

image(50)

image(51)

image(52)

Boundary conditions: at r=a

image

image

Hence

image

image

System of equations (50), (51), (52) with exactness to a sign has the same solutions, as system (47), (48), (49).

image

image

image

System of equations for neutron

image(53)

image(53′)

Along each parallel, exactly one wave length is kept within. In this case:

image

image(54)

image

image

image(54′)

image

image

image(54′′)

image

In other words, anisotropy is taking place, ε and μ are tensor quantities.

image

image

Here and further, corresponding letters mean parameters of neutron.

Let us find rest energy of neutron.

image

image

image

Further. Charge of neutron is equal to zero.

image.

Really,

image

It is obvious that

image

It is logical to assume that

image

Then

image(56)

Magnetic moment for neutron will be searched as the sum:

M=Mm + ML

Where Mm - magnetic moment created by volume current; ML - magnetic moment, attributed to impulse moment, i.e. to rotation.

image

as

image

image

image(57)

Now we shall write down the system of equations for neutron.

image

Boundary conditions: at r=a

image

image

Hence

image

From (54) ? (54′) follows that

image

and from (54) ? (54′′)that

image

image

So

image

image

Here ω=1,4274508*1024 Hz- Compton circular frequency of neutron.

Let us solve system (56)(55′)(57′)

image(56′′)

image

We substitute (56′′) B (55′)

image

image

image

P must be negative, therefore we select

image

From (57′) we find γ

image

Let us write down the system of equations for antineutron.

image

Boundary conditions are the same, as at neutron, hence

image

image

The last system with exactness to a sign has the same solutions, as system (56)(55′)(57)

image

image

image

Conclusion

Within the framework of the model, which is considered, electron, proton and neutron represent a monochromatic electromagnetic wave of corresponding frequency spread along parallels inside the spherical area, i.e. a wave, rotating around some axis. At that along each parallel, exactly half of wave length for electron and proton and exactly one wave length for neutron, is kept within, thus this is rotating soliton. This is caused by presence of spatial dispersion and anisotropy of a strictly defined type inside the particles. In electron vector E is directed to center of particle, that correspond to negative charge, and in proton vector E is directed from center of particle, that correspond to positive charge. Thus, by natural way, all basic parameters of particles are obtained: charge, rest energy, mass, radius, magnetic moment and spin, that is confirmed by mathematical expressions, which are discovered.

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