Quantum mechanics: a particle is a volumetric standing wave (the second part)

In [1], the concept of a particle having both the properties of a wave and the properties of a particle is considered. We will call such a particle a wave-AND-particle. Such a particle can explain those experiments in which (according to the concepts of quantum mechanics) a particle behaves either as a particle or as a wave, being (in contrast to a wave-AND-particle) a wave-OR-particle. How it behaves at a particular moment, the observer cannot know. So, the wave-OR-particle magically transforms from wave to particle and back, and wave-AND-particle constantly exhibits the properties of both waves and particles.

The wave-AND-particle (WAP) must be a standing wave and not propagate in space. It should not have its own speed (like the speed of a traveling wave), but it should have energy, momentum and mass. The quantum theory describes a wave in a vacuum in an open, unbounded space, which is associated with a material particle - the de Broglie wave. This is not an electromagnetic wave and its nature is unknown. But it is known that it is inextricably linked with the particle [2]. The definition of this wave includes its own speed. At zero own speed, this wave becomes a spatially unlimited standing wave. Moreover, of course, it cannot be identified with a particle.

So, below the WAP described, standing, existing in vacuum, limited in space, having no own velocity, having mass, momentum and energy. The first version of the mathematical description of such a wave was proposed in [3]. Here we will consider a more general case. In [3-6] various phenomena are considered that can be explained by the existence of such a wave.

Consider a certain volume V with magnetic permeability д and dielectric constant e . Let, as a result of some impact, an electromagnetic wave with energy W_o . has arisen in this volume V. There is no heat loss in volume V and no radiation from it. After a while, the wave parameters will take stationary values determined by the values д, e and W_o the size of the volume V. These parameters are electric field strength and magnetic field strength as a function of Cartesian coordinates and time, i.e. E(x, y, z,t) and H(x, y, z, t). Naturally, they satisfy the system of Maxwell equations of the form

9HZ	9H y	9Ex E — = 0	(1)
9У	9z	9t
9HX	9H z	9E y   n E—y = 0	(2)
9z	9x	9t
9H y	9HX —	E 9^ = 0	(3)
9x	9y	9t
9E z 9У	9Ey । -zz +	9HX ^^7 = 0 9t	(4)

ЭЕХ dz	—  + 11^ dx      dt	= 0	(5)
дЕ у dx	—   + ^81^ dy      dt	= 0	(6)
дЕХ dx	। dE y । dE z = dy    dz	0	(7)
9H x dx	++ di = dy    dz	0	(8)

Consider the following functions (proposed in [7]) that satisfy this system of equations:

E_x (x, У,z, t) = e_x cos(ax)sin(Pу)sin(yz) sin(tot) ,   (9)

Ey(x, у, z, t) = eysin(ax)cos(Pу)sin(yz) cos(tot), (10) Ez(x,у, z, t) = ezsin(ax)sin(Pу)cos(yz) sin(tot), (11) Hx(x,у,z, t) = hxsin(ax)cos(Pу)cos(yz) sin(tot), (12) Hy(x,у, z, t) = hycos(ax)sin(Pу)cos(yz) cos(tot), (13) Hz(x,у, z, t) = hzcos(ax)cos(Pу)sin(yz) cos(tot), (14) where ex, ey, ez, hx, hy, hz are constant amplitudes of functions,

• a, P, X, to are constants.

Differentiating (9-14) and substituting the obtained in (1-8), after reducing the common factors, we get:

hzp — hyy + exEto = 0

hxy — hza + eyEto = 0

hya — hxp + ezEto = 0

ezp - eyy — hxfito = 0

exy - eza — hyfito = 0

eya — exp — hzfito = 0

exa + eyP + ezy = 0

h xa + h yP + h zy = 0

Consider the solution of the resulting system of equations found in [8]. Since the system is symmetric, we take a = P = X.(23)

In this case, the system of equations (15-22) takes the form:

h z	— hy + exEto(a = 0	(24)
hx	— hz + eyEto(a = 0	(25)
hy	— hx + ezEto(a = 0	(26)
ez	— ey — hxp.to/a = 0	(27)

ex — ez — h y pxo/a = 0 ey — ex — hzp.a)/a = 0 e x + e y + e z 0 h x + h y + h z = 0

(28) (29) (30) (31)

In the system of equations (24-31), equations (30, 31) follow directly from the previous ones. Indeed, adding up equations (27-29), we obtain (31), and adding (24-26), we obtain (30).

The first 6 equations in system (24-31) with 6 unknowns are independent and the amplitudes of the functions e_x , e y , e_z , h_x , h_y , h_z can be found from them.

We will seek a solution to system (24-29) for

h z = 0 . Then this system will take the form:	(32)
ex£to/a — hy = 0,	(33)
ey£to/a + hx = 0,	(34)
ez£to/a — hx + hy = 0,	(35)
—ey + ez — hx.co/a = 0,	(36)
ex - ez — hy.o)/a = 0,	(37)
—e x + ey = 0-	(38)

The solution to system (33-38) has the form of equations (32) and

h y      h x ,	(39)
_  hxa x         Eto ’	(40)
ey = e x ,	(41)
ez      2e x .	(42)

3.    Energy of WAP

We write strengths (9-14) in the form

Г E x ⁽ x, y, z, t)! exi p \E_y (x, y,z, t)l = e_y L [E z (x, y, z, t)J ^ezl

cos(ax)sin(/?y)sin(yz)l

E =

sin(ax)cos(/?y)sin(yz) sin(tot) (43) sin(ax)sin(/?y)cos(yz)J

Г H x ⁽ x, y, z, t)l   r^xl rsin(ax)cos(/?y)cos(yz)l

H = H_y (x, y, z, t) = h y cos(ax)sin(/?y)cos(yz) cos(tot) (44) IH z (x, y, z, t)J   [hj [cos⁽ax⁾cos⁽Ay⁾sin⁽yz⁾J

Let us denote the parts of these expressions that are independent of time:

Г Ёх (х, у, z)j Ё = |Ёу (х, у, z) l = 1 Ё2 (х, у, z)	[ Кх1 б у 6 z	Г cos(aх)siп(/?у)siп(yz)] siп(aх)cos(/?у)siп(yz) [ siп(aх)siп(/?у)cos(yz)]
Г Н х ( х, у, z)l	[ h %	] Гsin(aх)cos(/?у)cos(yz)]
Н = \Ну(х, у,z) l =	| h у	cos(aх)siп(/?у )cos(yz)
1 Hz (х, у, z)	[ hz	] [cos(aх)cos(/?у)siп(yz)]

Let us now

find the square of the modulus of the total strengths: Ё2 = (Ё2 + Ё2 + Ё2),                       (47) Н2 = (Н2 + Н2 + H z 2).                     (48)

From (43-48) we find:

We denote:

Ё2 = ^( Ё ^ + Ё у + Ё 2 ) sin2 ( wt ))              (49) Н2 = ^( н 2 + Н у + n Z ) cos2 ( tot ) ^           (50) |Ё2| = ( Ё 2 _ + Ё У + Ё Z )                       (51) |Н21 = ( н 2 + н у + H Z )                    (52)

Then we get:

Ё2 = (|Е2 | sin2 ( tot ))                             (53) Н2 = (|Н2 | cos2 ( tot ))                            (54)

Let us write expressions (45, 46) using the solution obtained above (32, 39,

40, 41, 42):

Г Ё%(х, у, z)l         ri Ё = |Ёу(х,у,z) l = б х Ё 1                        (55) [ Ё z (х, у, z)J         L-2 Г Нх (х, у, z)!               i Н = 1йу(х,у,z)l = б х^^ Н -1                 (56) [ Hz (х, у, z)_l                 0

Let us find from (49, 50, 55, 56) the squares of the total strengths:

Е2 = (EX + Еу + Е^ sin2(ot) = бе2sin2(ot) • Е2(х, у, z)(57)

Н ² = ( н 2 + Н_у + Н^ cos²(cot ) = 2 (~) е 2 cos² (cot) Н²(х, у, z) (58)

Thus, the modules of these squares

|Е2| = бе2^Е2(х, у, z)(59)

|Н2| =2® ехН2(х,у, z)(60)

Let us consider a volume in which the integrals of sine and cosine are equal along any coordinate, and we will call such a volume agreed volume V. In such a volume, for any values of coordinates, the following conditions are satisfied:

/аХ . ,     \(PY . _     \/yZ          \ к2тт - integer), (2^ — teger), ^ — integerJ,

where X, Y, Z are the length of the segment of the coordinate axis located inside the volume V. In the agreed volume, for each such segment, conditions of the form

I cos2 (ах) dx = I sin2(ax)dx

J х                  'x

Consider, for an agreed volume, the ratio of modules (59, 60)

|Е2 |/|н2| = (6/2 (£r)2)q

where

q = Е2/Н2

Calculation using this formula gives a wonderful result: q = 1.5 .

This constant is NOT dependent on the size of the agreed volume. Therefore, for the agreed volume, the following condition can also be satisfied

U = £|Е2| = ^|Н21.

The energy density is

W = е Е ² + ц Н²

From (53, 54, 65) we get:

W = е |Е21 sin2(ot) + ц |Н2| cos2(cot).

Из (64, 66) следует, что

W = U(sin2(ot) + cos2(ot)) = U,

i.e. when condition (67) is satisfied, the energy density in the volume does not depend on time. In other words, a standing wave is created in the agreed volume, which does not radiate.

The U value is constant. Therefore, for a matched volume, the expression for the energy W_o in the entire volume V has the form:

W_o = U • V. (68)

Consequently, in a constant by magnitude of agreed volume, the energy of an electromagnetic wave does not depend on time, i.e. remains constant. This means that under the indicated conditions it is true

Statement 1.

WAP, like a standing electromagnetic wave, can exist in agreed volume.

The flow of energy can be thought of as a flow of energy quanta. The line of motion of a quantum of energy, obviously, should be such that the time derivative of energy remains zero. Let us find the equation of this derivative from (66)

^ = 2 w sin ( wt ) cos(cot ) • (e|E²| — рДН² |).          (69)

From here we find the equation of the line of the same energy density at a given time:

(_£| e² I— u| h² I) = 0.                                   (70)

4.    Energy flow of WAP

The energy flux densities by coordinates are determined by the formula

S = [^S y j = (E x H) =

Г E_yH_z— Е_гНД e_zh_x — e_xh_z\ = . [E_xH y — E y H_x\

where the functions E, H are determined from (9-14). The modulus of the total vector of the energy flux density has the form:

S o = (E y H z — E z H y )² + (E z H x — E x H z )² + (E x H y — E y H x ) 2 (72) This quantity is equal to zero if three terms in (34) are simultaneously equal to zero. We will write this condition as:

{(EyHz — EzHy), (EzHx — ExHz), (ExHy — EyHx)} = 0. (73) Consequently, the locus of points of zero energy flux density is described by the system of equations (73). In this equation, each parenthetical term has a common factor depending on time - see (43-44). If we are looking for a geometrical place that does not change in time, then we can cancel these factors. Then, taking into account the designations (45, 46), we obtain that the locus of points of zero energy flux density, which does not change in time, is described by a system of equations

{{E_yH_z - E_zH_y), (jUl - E_XH_Z), (E_xH_y - E_yH_x^ = 0. (74) This system of equations with three coordinates describes a surface where there is no energy flow. If this surface is closed and continuous, then it is the equation of the WAP boundary. It is such a surface that limits the WAP. It is important to emphasize that this is not a surface where the structure of the medium experiences a jump in any parameter of the medium. So,

Statement 2.

WAP can exist within a closed and continuous boundary , the equation of which has the form (74).

First of all, consider the cubic form proposed in [11]. In order for the cube not to radiate from the surface хоу , it is necessary that at all points of this surface the condition

ExHy = 0 и EyHx = 0.                           (75)

The function sin(a%) is present in the definition of one of the functions indicated in condition (75). If the origin of coordinates is in the center of the cube and α is the length of the half-edge of the cube, then the functions sin(a%) are equal to zero on the faces of the cube and, therefore, condition (75) is satisfied. Similar conditions are met on other faces of the cube. Therefore, Statement 2 is true. In addition, Statement 1 holds for such a cube is true. Thus, WAP can exist in the volume of the cube.

5.    The momentum and mass of WAP

The well-known Umov formula connects the energy density and energy flow with the speed of this flow:

V = -.                                              (76)

w

It is possible that this speed is different for different flow lines.

So, there is a flow of energy inside a constant volume with a certain speed V . As you know, an electromagnetic wave, together with a flow of energy, has momentum and mass. Consequently, the described standing wave in a agreed volume has momentum and mass even in the case when the volume is at rest.

The densities of these quantities are related to each other and to the velocity of propagation of electromagnetic energy by the known relations (76) and

Р = 7, m = ^

Consequently, m = ^.

This electromagnetic mass pulsates along with the flow of electromagnetic energy. However, the center of mass does not change position. Consequently, WAP can be considered both as a standing wave and as the volume of a pulsating mass with a constant center of mass.

6.    Wave frequency of WAP

From (61, 62) we find:

^61   aj31 \H\    ^V2      £to .                             (80)

Condition (64) is satisfied for the agreed volume. Consequently

(ф2|) (Jh^d = 1                              (81) or

Л 4H ■                  (82

From (80, 82) we find:

	aJ3q    /й V = /                                       (83) £Ш    ^ £
or	a^3q to =      = caJ3q .                  (84) Vy.£         n

Thus, the frequency of a standing wave in an agreed volume is determined

by the formula	to     ca f = 2^ = 2^ V3?                   (85)
or for q = 1.5	f « 0.34ca .                                 (86)

So, there is such a frequency of the electromagnetic wave, at which the energy of the electromagnetic wave in the agreed volume remains constant.

7.    Conclusion

We have established two conditions that must be satisfied by the region in which

WAP, like a standing electromagnetic wave, can exist in agreed volume.

WAP can exist within a closed and continuous boundary

The frequency of WAP standing wave is determined by the value of the coefficient Ct.

We have found that WAP forms a closed area, has a certain shape and volume. The size of WAP is determined by the value of the coefficient C . The results obtained can be applied for any arbitrarily small units of length measurement.

The shape of the WAP area is such that multiple WAPs can be adjacent to each other without gaps. Consequently, WAP groups can be of any size. Thus, WAP of any size and regions of WAP of any size can exist.

Energy, momentum and mass of WAP are also determined only by the value of the coefficient C . It can be assumed that the quantum of energy is also WAP and the minimum value of C is determined by the magnitude of the energy of the quantum.

WAP does not have its own speed and its mechanical energy is determined by its mass and the speed that it received when interacting with other masses (including other WAP).

The internal pressure on the WAP border is equal to the energy density at the border, although WAP does not have any shell. It can be assumed that WAP behaves like an absolutely elastic body and transmits the received impulse without changing its value. Then the WAP area also behaves as a conductor of the impulse.

Obviously, WAP can form elementary particles and larger structures. But we can assume that the vacuum is woven from WAP.