zotero/storage/PBW8LDCM/.zotero-ft-cache

arXiv:physics/0205057v3 [physics.gen-ph] 19 Feb 2006

Mechanical scenario for the reaction: neutron → proton + electron + antineutrino
Valery P. Dmitriyev Lomonosov University P.O.Box 160, Moscow 117574, Russia∗ (Dated: 1 February 2006)
Small perturbations of averaged ideal turbulence reproduce the electromagnetic ﬁeld. The luminiferous medium has the relatively low pressure and high energy. A vapor bubble in the ﬂuid models the neutron. The bubble stabilized via creating in the ﬂuid a ﬁeld of the positive perturbation of the turbulence energy serves as a model of the proton. An isle of the ﬂuid that produces the respective ﬁeld of the negative perturbation of the turbulence energy models the electron. The antineutrino corresponds to a local positive disturbance of the turbulence energy needed in order to compensate the diﬀerence in perturbations of the energy produced by the electron and proton.
Keywords: luminiferous medium, ideal ﬂuid, Reynolds turbulence, perturbation wave, electromagnetic ﬁeld, vapor bubble, particle, antiparticle, neutron, electron, neutrino, energy, mass

1. SUBSTRATUM FOR PHYSICS
The properties of physical space known as ﬁelds and particles can be modelled in terms of continuum mechanics1,2. The concept of substratum for physics is used to this end. The substratum serves as a medium to carry over the electromagnetic wave and to transmit interactions. On a large scale the substratum can be approximated by a turbulent medium.

2. PERTURBATIONS OF TURBULENCE

We consider the averaged turbulence of an ideal ﬂuid. The system is governed by linearized Reynolds equations3,4,5:

∂i ui = 0,

(1)

ς∂t ui + ς∂k u′iu′k + ∂i p = 0

(2)

where ς is the medium density, u the averaged velocity of the ﬂuid ﬂow, p the averaged pressure and u′ the turbulent ﬂuctuation of the velocity, so that

u = u + u′.

(3)

Here and further on we use denotations ∂t = ∂/∂t, ∂k = ∂/∂xk. Summation over recurrent index is implied throughout. We assume that in the unperturbed state the turbulence is homogeneous and isotropic, i.e.

u (0) = 0,

(4)

p (0) = p 0,

(5)

u′iu′k (0) = c2δik

(6)

where p 0, c = const and c is the speed of the turbulence perturbation wave in the medium. Integrating equation (2) for an isotropic turbulence

u′iu′k = u′1u′1 δik

(7)

and u = 0, we get a kind of Bernoulli equation

ς u′1u′1 + p = ςc2 + p 0.

(8)

Formally, by (8), any distribution of the turbulence energy density

1 2

ς

u′iu′i

(9)

2

may occur. The next linearized term in the chain of Reynolds equations looks as follows

∂t u′iu′k + c2 (∂i uk + ∂k ui ) + hik = 0

(10)

where

ςhik = u′i∂kp′ + u′k∂ip′ + ς∂j u′iu′ju′k .

(11)

The density (9) of the turbulence energy is in a way similar to distribution of the heat energy. However, the aether
ﬂuid is a true continuum, i.e. it does not consist of corpuscles and hence does not imply the dissipation of the mechanical energy. It can be shown4 that in the absence of diﬀusion

hii = 0.

(12)

Taking in (10) i = k and summing over the recurrent index we get with the account of (1) and (12)

∂t u′iu′k = 0.

(13)

By (13) the proﬁle of linear perturbations is conserved in the nondissipative incompressible medium. For uniformity with (2), it is expedient to diﬀerentiate equation (10) with respect to xk:

∂t∂k u′iu′k + c2∇2 ui + ∂khik = 0

(14)

where the incompressibility condition (1) was used.

3. MAXWELL’S EQUATIONS

With the deﬁnitions

Ai = κc ui − ui (0) ,

(15)

ςϕ = κ p − p (0) ,

(16)

Ei = κ∂k u′iu′k − u′iu′k (0) ,

ji

=

κ 4π

∂k

hik

(17) (18)

where κ is an arbitrary constant, (2), (14) and (1) take the form of Maxwell’s equations

1 c

∂tA

+

E

+

∇ϕ

=

0,

(19)

1 c

∂E ∂t

−

∇

×

(∇

×

A)

+

4π c

j

=

0

(20)

with the Coulomb gauge

∇·A=0

(21)

respectively, where in (20) the general vector relation

∇2A = ∇(∇ · A) − ∇ × (∇ × A)

(22)

and incompressibility condition (21) were used. So, small perturbations of an ideal turbulence reproduce the electromagnetic ﬁeld3.

3

4. THE NEUTRON AS A VAPOR BUBBLE

Discontinuities, or defects, of the medium model particles. Typically, a spherical cavity included into the medium represents a point-like discontinuity. A bubble ﬁlled with the vapor of the ﬂuid can be taken as a model of the neutron.
Let V ∗ be the volume of the turbulent ﬂuid evaporated into the bubble. The kinetic energy K∗ transferred with the ﬂuid into the gas phase can be found from (9). We get with the account of (6) for the unperturbed medium

K∗

=

V

∗

3 2

̺

u′1u′1

(0)

=

3 2

̺V

∗c2.

(23)

The vapor will be assumed to behave as an ideal gas. The equation of state of the ideal gas can be written in the form

pV

=

2 3

K.

(24)

In the mechanical equilibrium the gas pressure p must be equal to the ﬂuid pressure p . If V is the volume of the bubble then using (23) in (24) we get for a bubble in the unperturbed medium

p0V = ̺V ∗c2

(25)

where p0 is the background pressure (5). Insofar as V ∗ ≪ V we see from (25) that the occurrence in the ﬂuid of ideal vapor bubbles implies the turbulence of the high energy and low pressure:

p0 ≪ ςc2.

(26)

5. THE MASS AND SELF-ENERGY OF A PARTICLE

The mass of the bubble can be determined from the mass of the gas contained in it

m = ̺V ∗.

(27)

Substituting (27) into (25) we get

p0V = mc2.

(28)

The self-energy of a bubble can be deﬁned as the work needed in order to create the bubble in the unperturbed medium

E = p0V.

(29)

Using (29) in (28) gives

E = mc2.

(30)

In the phenomenological theory (see e.g. the textbook6) we can deduce for the increment of the kinetic energy of a particle

dE = c2dm.

(31)

In order to obtain from (31) the expression for the self-energy we must postulate that all the internal energy of the particle concerned with the mass has the kinetic origin. As you see from the microscopic theory considered the self-energy (30) of a particle is indeed immediately reduced to a part of the turbulence energy of the luminiferous medium.

6. CONVERSION TO THE PROTON
A turbulent ﬂuid may adjust itself to boundary conditions via the perturbation of the turbulence energy (Fig.1, left). A new stable conﬁguration is thus formed. The conversion of the neutron to the proton can be seen in the

4

following way. A microscopic restructuring takes place in the core of the particle. The gas in the bubble cools down and the pressure on the wall drops to some value p0 − ∆p, where ∆p > 0.
The ﬁeld of the turbulence perturbation produced by the point discontinuity was shown in5 to take the form 1/r. We have for the stable bubble of the radius Rp located at x′:

p

=

p

0

−

|x

a −

x′| ,

(32)

a = Rp∆p.

(33)

The respective ﬁeld of the turbulence energy can be found using (32) in (8):

ς u′1u′1

=

ς c2

+

|x

a −

x′| .

(34)

The bubble inclusion that produces in the turbulent ﬂuid the ﬁeld of the positive perturbation of the turbulence energy models the proton (Fig.1, left).

✻ 2 ❵♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ ςc

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ ς u′1u′1

✻
2
ςc ♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

ς u′1u′1 ♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

p

p0

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ 0

✲

x

p ♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣
p0

0

✲

x

FIG. 1: The proton (left) and antiproton (right). Dots on the pressure graph indicate the vapor in the bubble.

7. THE ELECTRON

The form (34) implies an inﬁnite quantity of the total energy perturbation

1 2

ς

Ω

u′iu′i − u′iu′i (0) d3x

(35)

where the medium volume Ω → ∞. So, the positive deviation (34) from the background (6) should be compensated by a negative deviation of the turbulence energy of the same form, yet with the opposite sign of the coeﬃcient (33):

a = −Rp∆p.

(36)

Supposing that the energy attains a reduced value ς u′1u′1 = ςc2 − ∆ε at the core |x − x′| = re of the negative disturbance center, we ﬁnd from (34):

a = − re∆ε.

(37)

The center of the negative perturbation of the turbulence energy serves as a model of the electron (Fig.2, right).

8. THE ANTINEUTRINO

Comparing energy conﬁgurations of the proton (Fig.1, left, top) and the electron (Fig.2, right, top) we see that the negative perturbation of the turbulence energy in the core of the electron should be compensated by a respective positive perturbation of the background energy (6). This surplus of the turbulence energy corresponds to the emission of a new particle. Assuming in (36) and (37) re ≃ Rp we get ∆p ≃ ∆ε. An isle of the ﬂuid with increased turbulence energy may serve as a model of the antineutrino (Fig.3, left).
Thus, we constructed a linear mesoscopic mechanical scenario of the reaction

n → p+ + e− + ν˜.

(38)

5

✻ 2 ❵♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ ςc
ς u′1u′1

✻
2

ς u′1u′1

ςc

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

p

p0

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ 0

✲

x

p ♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣
p0

0

✲

x

FIG. 2: The positron (left) and electron (right).

✻

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

♣

2

♣ ♣

♣ ♣ ♣

ςc ♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

ς u′1u′1

✻

2

ςc

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

♣ ♣ ♣

♣ ♣ ♣

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

ς

u′1 u′1

p

p0

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ ♣

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ ♣

♣

♣

♣

♣

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣ 0

✲

x

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

♣ ♣

♣ ♣

p

♣

♣

p 0 ♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣

0

✲

x

FIG. 3: The antineutrino (left) and neutrino (right).

9. PARTICLES AND WAVES
Regions of the turbulent ﬂuid such as in the models of the neutrino and inner part of the electron are referred to in hydrodynamics as the turbulence core. In the turbulence core the simple closure hik = 0 of the chain of governing equations (1), (2) and (10) is not valid. Using appropriate closure schemes for (10), perturbations of the energy in the turbulence core was shown4 to be in general nonstationary. So, the neutrino as well as the electron should be properly considered as waves of the turbulence energy.
The neutron and proton are associated with the inclusions into the medium of the empty space. This is the reason why respective perturbations of the turbulence energy are stationary, and the baryon number is conserved.
On the other hand, in the model of the electromagnetic wave there was shown5 to be no perturbation of the turbulence energy. Nondiagonal terms of correlations u′iu′k of turbulent ﬂuctuations are only perturbed.
Thus we come to the conclusion that a particle is concerned with the perturbation of the turbulence energy while the pure wave does not accompanied by a perturbation of the turbulence energy.

10. QUANTUM MECHANICS
There is considered a special kind of nonstationarity of the wave-particles. The core of the electron is an isle of the ﬂuid. So, the electron can be additively decomposed into the collection of fractions. Each fraction has an isle of the ﬂuid in the core. The size of the core is the same as in the whole particle. The turbulence energy in the core is decreased relative to the background level (6). Yet it is greater than in the core of the original localized electron. Each fraction generates in the ﬂuid the turbulence perturbation ﬁeld that corresponds to a part of the electric charge. The decomposition of the electron may proceed under the action of turbulent ﬂuctuations. This process underlies what is known as quantum phenomena.
The proton can not be decomposed into fractions in the same manner. It is a true particle.

11. ANTIPARTICLES
Models of antiparticles can be obtained mirroring graphs of respective particles in relation to the background levels of the pressure (5) and energy (6). Thus we come to a model of the positron as the center of the positive perturbation of the turbulence energy (Fig.2, left). The neutrino is the isle of the ﬂuid with the reduced turbulence energy (Fig.3, right).

6
In the present approach the proton was modelled by the bubble with vapor in it cooled as compared with the vapor in the neutron core. Then the bubble with the vapor heated in comparison with the vapor in the neutron core models the antiproton (Fig.1, right).
12. CONCLUSION
In the mesoscopic approach the luminiferous medium can be modelled by the averaged turbulence of an ideal ﬂuid. Particles of matter are concerned with inclusions into the medium of empty space. The neutron is modelled by the vapor bubble. The cooling of the vapor in the bubble forms the center of the positive perturbation of the turbulence energy that serves as a model of the proton. The center of the negative perturbation of the turbulence energy models the electron. The antineutrino is concerned with a local positive perturbation of the turbulence energy necessary in order to compensate the diﬀerence in perturbations of the turbulence energy between the proton and the electron.
∗ Electronic address: aether@yandex.ru 1 Valery P. Dmitriyev, “Electrodynamics and elasticity”,
2003, Am. J. Phys., 71, No 9, 952-953. 2 Valery P. Dmitriyev, “Mechanical analogy for the wave-particle: helix on a vortex ﬁlament”,
2002, J. of Applied Mathematics, 2, No 5, 241-263. 3 O.V.Troshkin, “On wave properties of an incompressible turbulent ﬂuid”,
1990, Physica A, 168, No 2, 881-898. 4 O.V.Troshkin, “Perturbation waves in turbulent meadia”,
1993, Comp. maths Math. Phys. , 33, No 12, 1613-1628. 5 V.P.Dmitriyev, “Towards an exact mechanical analogy of particles and ﬁelds”,
1998, Nuovo Cimento A, 111, No 5, 501-511. 6 I.E.Irodov, Fundamental laws of mechanics, §7.2 The basic equation of relativistic dynamics, Moscow, Mir Publishers, 1980.