1. Introduction

2. Electrical-Quantum Modelling of the Neutron as a Three-Phase Alternating Current Electrical Generator

3. Electrical-Quantum Modelling of the Proton as a Three-Phase Alternating Current Electrical Generator

4. Application OF the Model and Results

4.1. Proton Decay

We are going to work with the following figure to represent the decay of the proton.

Figure 20. Proton.

Figure 20. Proton.

In figure 21, we represent the decay of the proton, through interaction 1 and interaction 2. We call this particle Dproton. We observe that all the components are U quarks and we also observe that there are 9 U quark-antiquark dipoles, which are vector related. This proton decay configuration is represented by matter and antimatter.

We are going to make the vector diagram of the Dproton particle, of the interaction 1 and 2:

Figure 22. Vector diagram of the proton and Dproton particle.

Figure 22. Vector diagram of the proton and Dproton particle.

Figure 23. The vector diagram of the proton is represented in black. The vector diagram of Dproton particle - interaction 1 is represented in red. The vector diagram of Dproton - interaction 2 is represented in purple.

Figure 23. The vector diagram of the proton is represented in black. The vector diagram of Dproton particle - interaction 1 is represented in red. The vector diagram of Dproton - interaction 2 is represented in purple.

At first glance, if we add the three U quarks scalarly, we can affirm that the total charge is (2/3 + 2/3 + 2/3 = 2) 2, but we have to remember that we are adding scalarly and in this situation we must consider a vector sum and this gives us a net charge of zero. We can represent the U quarks as three vectors with the same module, 120 degrees out of phase, whose vector sum is zero, in analogy to electric generators.

The Dproton particle behaves analogously to a generator, whose current, voltage and charge are perfectly balanced, in equilibrium.

The scale factor is the following:

SF = (X / 48.28) 4

SF⁻¹ = (48.28 / 4) X

This scaling factor is unique and applies to all vectors in the diagram representing the proton decay.

We are going to describe the mass distribution of the dipoles of interaction 1 and 2, which represent the Dproton particle.

Figure 24. Mass distribution in the interaction 1 and 2 of the Dproton particle.

Figure 24. Mass distribution in the interaction 1 and 2 of the Dproton particle.

According to our vector calculation, the Dproton particle has a mass of 651.78 MeV/c².

We are going to make the vector diagram of the proton decay in the Dproton particle.

Let's analyse figure 25:

The dipoles of interaction 1 of the proton are represented in black.

The dipoles of interaction 2 of the proton are represented in orange.

The dipoles of interaction 1 of the Dproton are represented in red.

The dipoles of interaction 2 of the Dproton are represented in purple.

Represented in blue are the 3 vectors of interaction 1 and the 6 vectors of interaction 2 that correspond to the decay of the proton in the Dproton particle.

Now we are going to analyse the energy (mass) changes that occur when the proton decays in the Dproton particle. These changes in energy (mass) are represented in blue in figure 25.

Table 1. Calculation of the interactions in MeV/c², when the proton decays in the Dproton particle.

Table 1. Calculation of the interactions in MeV/c², when the proton decays in the Dproton particle.

PROTON DECAY
PROTON	DPROTON	INTERACTION	INTERACTION (MeV/c²)
R(DD)R	R(UU)R	E1 = 4.5	E1 = 54.31
B(UU)B	B(UU)B	E2 = 2.3	E2 = 27.76
G(UU)G	G(UU)G	E3 = 1.1	E3 = 13.27
R(DU)B	R(UU)B	E4 = 3.5	E4 = 42.24
R(DU)G	R(UU)G	E5 = 6.5	E5 = 78.45
B(UD)R	B(UU)R	E6 = 7.6	E6 = 91.73
B(UU)G	B(UU)G	E7 = 2.0	E7 = 24.14
G(UD)R	G(UU)R	E8 = 10.2	E8 = 123.62
G(UU)B	G(UU)B	E9 = 4.4	E9 = 53.10
TOTAL INTERACTION			IEtI = 508.72 MeV/c²
(W⁻)e INTERACTION			IEeI = W⁻ = 66.38 MeV/c²
(Z⁰)n INTERACTION			IEnI = 39.83 MeV/c²
(Z⁰)p INTERACTION			IEpI = 36.21 MeV/c²
(Z⁰) INTERACTION			IZ⁰I = 70.00 MeV/c²
Θ angle, Θ = arc cos W⁻/Z⁰			Θ = 16⁰

The vectors E1 to E9 are represented in blue in figure 25.

It is important to highlight that the decay of the proton in the Dproton particle produces a total interaction of Et = 508.72 MeV/c².

In figure 25, we divide these vectors into three parts, which we will explain below:

• En = E7 + E9

En = (3.3 x 48.28) / 4 = 39.83 MeV/c²

E7 = [B(UU)G]p → [B(UU)G]Dp

E9 = [G(UU)B]p → [G(UU)B]Dp

For E7, when the decay of [B(UU)G]p of the proton to [B(UU)G]Dp of the Dproton occurs; a neutral current is produced.

For E9, when the decay of [G(UU)B]p of the proton to [G(UU)B]Dp of the Dproton occurs; a neutral current is produced.

Here, we are going to hypothesize that the En energy is what gives origin to the neutrinos. Let us note that the BG and GB gluons are of same frequencies; B and G are associated with the U quark.

When E7 + E9 interacts, it is like the BG and GB gluons cancel each other and the UU and UU quarks also cancel and, in this way, the resulting energy gives origin to the neutrinos, whose mass is neutral.

• Ep = E2 + E3

Ep = (3 x 48.28) / 4 = 36.21 MeV/c²

E2 = [B(UU)B]p → [B(UU)B]Dp

E3 = [G(UU)G]p → [G(UU)G]Dp

For E2, when the decay of [B(UU)B]p of the proton, to [B(UU)B]Dp of the Dproton, occurs; a neutral current is produced.

For E3, when the decay of [G(UU)G]p of the proton, to [G(UU)G]p of the Dproton, occurs; a neutral current is produced.

Here, we are going to hypothesize that the Ep energy is what gives origin to the photons. Here it is important to highlight that the BB and GG gluons have the same frequency; BB and GG gluons, have the frequency of the U quark.

• Ee = E1 + E4 + E5 + E6 + E8

Ee = (5.5 x 48.28) / 4 = 66.38 MeV/c²

E1 = [R(DD)R]p → [R(UU)R]Dp

E4 = [R(DU)B]p → [R(UU)B]Dp

E5 = [R(DU)G]p → [R(UU)G]Dp

E6 = [B(UD)R]p → [B(UU)R]Dp

E8 = [G(UD)R]p → [G(UU)R]Dp

For the vectors E1, E4, E5, E6 and E8, when the decay of proton to Dproton occurs, it is observed that currents with positive and negative charges are generated.

The negative net energy of the vector Ee = 66.38 MeV/c²

Here, we are going to hypothesize that the Ee energy is what gives origin to the electron.

The emission of the energy vectors En, Ep and Ee are necessary in the βˉ decay that occurs when the proton decays into the Dproton particle, so that the Dproton stabilizes and reaches its corresponding energy.

In the following vector diagram, we are going to represent the vectors En, Ep and Ee.

Figure 26. - Calculation of En, Ep, Ee, W⁻, Z⁰ and Θ angle.

Figure 26. - Calculation of En, Ep, Ee, W⁻, Z⁰ and Θ angle.

calculation of W ⁻ , Z ⁰ and Θ angle:

Theoretical calculations that we obtain from figure 46, proton decay diagram:

theoretical W⁻ = Ee = (5.5 x 48.28) / 4 = 66.38 MeV/c²

theoretical W⁻ = 66.38 MeV/c²

theoretical Z⁰ = (5.8 x 48.28) / 4 = 70.00 MeV/c²

theoretical Z⁰ = 70.00 MeV/c²

W⁻/ Z⁰ = cos Θ

Θ = arc cos W⁻/Z⁰ = arc cos (66.38 / 70.00) = arc cos (0.948)

Θ = 18⁰

measured Θ = 16⁰

As analysed so far, the decay of the proton in the Dproton particle also includes a βˉ decay.

4.2. Dproton Decay

We are going to work with the following figure to represent the decay of the Dproton particle.

Figure 27. Dproton particle.

Figure 27. Dproton particle.

Figure 28. Protoniu particle, does not include antimatter.

Figure 28. Protoniu particle, does not include antimatter.

In figure 28, we represent the protoniu particle, through interaction 1 and interaction 2. We observe that all the components are U quarks and we also observe that the 9 dipoles are U quarks, which are related vectorially. This configuration is represented only by matter, there is no antimatter in this configuration.

We are going to make the vector diagram of the Protoniu particle, of the interaction 1 and 2:

Figure 29. Vector diagram of the proton and protoniu.

Figure 29. Vector diagram of the proton and protoniu.

Figure 30. The vector diagram of the proton is represented in black. The vector diagram of protonium - interaction 1 is represented in red. The vector diagram of protonium - interaction 2 is represented in blue.

Figure 30. The vector diagram of the proton is represented in black. The vector diagram of protonium - interaction 1 is represented in red. The vector diagram of protonium - interaction 2 is represented in blue.

At first glance, if we add the three U quarks scalarly, we can affirm that the total charge is (2/3 + 2/3 + 2/3 = 2) 2, but we have to remember that we are adding scalarly and in this situation we must consider a vector sum, this gives us a net charge of zero. We can represent the U quarks as three vectors with the same module, 120 degrees out of phase, whose vector sum is zero, in analogy to electric generators.

The protoniu particle behaves analogously to a generator, whose current, voltage and charge are perfectly balanced, in equilibrium.

If we compare figure 23 and figure 30, the vector diagrams for interaction 1 and interaction 2 are the same, they differ in the magnitude of the vectors. In figure 30, antimatter is not included. In figure 23, it includes the antimatter.

The scale factor is the following:

SF = (X / 48.28) 4

SF⁻¹ = (48.28 / 4) X

This scaling factor is unique and applies to all vectors in the diagram representing the proton decay.

We are going to describe the mass distribution of the dipoles of interaction 1 and 2, which represent the Protoniu particle.

Figure 31. Protoniu particle.

Figure 31. Protoniu particle.

According to our vector calculation, the Protoniu particle has a mass of 253.44 MeV/c².

We are going to make the vector diagram of the Dproton decay in the Protoniu particle.

Let's analyse figure 32:

The dipoles of the interaction 1 of the proton, are represented in black.

The dipoles of the interaction 1 of the Dproton, are represented in red.

The dipoles of the interaction 2 of the Dproton, are represented in purple.

The dipoles of the interaction 1 of the protoniu correspond to the point of intersection of all the vectors.

The dipoles of the interaction 2 of the protoniu, are represented in orange.

Represented in blue are the 3 vectors of interaction 1 and the 6 vectors of interaction 2 that correspond to the decay of the Dproton in the protoniu particle.

It can be seen in figure 31, interaction 1:

R(UU)R = 0

B(UU)B = 0

G(UU)G = 0

This is telling us that there is no direct interaction between the quarks (U,U). In analogy with an electric generator, it is as if the pair of quarks (U,U) were at the same electric potential.

Now we are going to analyse the energy (mass) changes that occur when the Dproton decays in the Protoniu particle. These changes in energy (mass) are represented in blue in figure 32.

Table 2. - Calculation of the interactions in MeV/c², when the Dproton decay in the protoniu particle.

Table 2. - Calculation of the interactions in MeV/c², when the Dproton decay in the protoniu particle.

DPROTON DECAY
DPROTON	PROTONIU	INTERACTION	INTERACTION (MeV/c²)
R(UU)R	R(UU)R	E1 = 4	E1 = 48.28
B(UU)B	B(UU)B	E2 = 4	E2 = 48.28
G(UU)G	G(UU)G	E3 = 4	E3 = 48.28
R(UU)B	R(UU)B	E4 = 3.5	E4 = 42.24
R(UU)G	R(UU)G	E5 = 3.5	E5 = 42.24
B(UU)R	B(UU)R	E6 = 3.5	E6 = 42.24
B(UU)G	B(UU)G	E7 = 3.5	E7 = 42.24
G(UU)R	G(UU)R	E8 = 3.5	E8 = 42.24
G(UU)B	G(UU)B	E9 = 3.5	E9 = 42.24
TOTAL INTERACTION			IEtI = 398.28 MeV/c²

The vectors E1 to E9 are represented in blue in figure 32.

It is important to highlight that the decay of the Dproton in the protoniu particle produces a total interaction of Et = 398.28 MeV/c².

In figure 32, these vectors in blue, we will explain below:

• E = E1 + E2 + E3 + E4 + E5 + E6 + E7 + E8 + E9

E1 = R(UU)R → R(UU)R

E2 = B(UU)B → B(UU)B

E3 = G(UU)G → G(UU)G

E4 = R(UU)B → R(UU)B

E5 = R(UU)G → R(UU)G

E6 = B(UU)R → B(UU)R

E7 = B(UU)G → B(UU)G

E8 = G(UU)R → G(UU)R

E9 = G(UU)B → G(UU)B

When the Dproton particle decays into the protoniu particle, it is observed that matter separates from antimatter, producing energy in the form of radiation.

The particle that results from the decay of the Dproton particle, protoniu, is formed only by U quark and does not have antimatter.

The configuration of the quarks inside the protoniu particle is formed by vectors that represent the quarks, which are displaced by 120 degrees, whose vector sum makes the net charge zero.

We are going to use the following equation to represent the decay of the particle Dproton into the particle protoniu:

Dproton → protonium + ∆E

4.3. Conservation Laws:

It is important to mention the conservation laws, which indicate that there are properties that should never change.

There are three conservation laws that we have to mention when we analyse the decay of the proton and Dproton.

• Electric Charge
• Energy
• Baryon and lepton number

We will analyse the decay of the proton in the Dproton particle, we will use figure 33, figure 34 and figure 35, for our analysis:

Proton → Dproton + e⁻ + ṽ + ∆E

(+1)       (+2)     (-1)

We see that the proton has an electric charge (+1), the Dproton particle has an electric charge (+2) and the electron has an electric charge (-1); We see that the electric charge is balanced, it is conserved.

Proton → Dproton + e⁻ + ṽ + ∆E

 (+1)      (+1)      (+1) (-1)

We see that the proton has a baryon number equal to 1 and the Dproton particle also has a baryon number equal to 1, the baryon number is conserved. We are going to analyze the lepton number, we see that the electron has a lepton number equal to (+1) and the antineutrino has a lepton number (-1), the lepton number is also conserved.

Proton → Dproton + e⁻ + ṽ + ∆E

938.27 MeV/c²  651.78 MeV/c² +…..

Considering the energy (mass), the proton has a mass of 938.27 MeV/c², the Dproton particle has a mass of 651.78 MeV/c², to this is added the mass of the electron, the neutrino and the energy of photons ∆E. Energy is also conserved.

In a simple analysis, we see that the electric charge, energy and baryon and lepton number are conserved for the decay of the proton in the Dproton particle.

We will analyse the decay of the Dproton in the protoniu particle, we will use figure 36, figure 37 and figure 38, for our analysis:

Dproton → protoniu + ∆E

 (+2)        (+2)

We see that the Dproton has an electric charge (+2), the protonium particle has an electric charge (+2); we see that the electric charge is balanced, it is conserved.

Dproton → protoniu + ∆E

 (+1)        (+1)

We see that the Dproton has a baryon number equal to 1 and the protoniu particle also has a baryon number equal to 1, the baryon number is conserved.

Dproton → protoniu + ∆E

651.78 MeV/c²  253.44 MeV/c² + ……

Considering the energy (mass), the Dproton has a mass of 651.78 MeV/c², the protoniu particle has a mass of 253.44 MeV/c², to this is added the energy of photons ∆E. Energy is also conserved.

Finally, summarizing we have:

In the following graph we are going to represent some fundamental parameters that describe the following particles:

Figure 39. Representation of characteristic parameters of the neutron, proton, Dproton and protoniu.

Figure 39. Representation of characteristic parameters of the neutron, proton, Dproton and protoniu.

4.4. Inverse Neutron Decay

We are going to work with the following figure to represent the inverse neutron decay.

Figure 40. Neutron.

Figure 40. Neutron.

Figure 41. Dneutron.

Figure 41. Dneutron.

In figure 41, we represent the inverse neutron decay through interaction 1 and interaction 2. We call this particle Dneutron. We observe that all the components are D quarks and we also observe that there are 9 D quark-antiquark dipoles, which are vector related. This inverse neutron decay configuration is represented by matter and antimatter.

We are going to make the vector diagram of the Dneutron particle, of the interaction 1 and 2:

Figure 42. Vector diagram of the neutron and Dneutron particle.

Figure 42. Vector diagram of the neutron and Dneutron particle.

Figure 43. The vector diagram of the neutron is represented in black. The vector diagram of Dneutron particle - interaction 1 is represented in red. The vector diagram of Dneutron - interaction 2 is represented in red.

Figure 43. The vector diagram of the neutron is represented in black. The vector diagram of Dneutron particle - interaction 1 is represented in red. The vector diagram of Dneutron - interaction 2 is represented in red.

At first glance, if we add the three D quarks scalarly, we can affirm that the total charge is (-1/3 - 1/3 - 1/3 = -1) -1, but we have to remember that we are adding scalarly and in this situation we must consider a vector sum, and this gives us a net charge of zero. We can represent the D quarks as three vectors with the same module, 120 degrees out of phase, whose vector sum is zero, in analogy to electric generators.

The Dneutron particle behaves analogously to a generator, whose current, voltage and charge are perfectly balanced, in equilibrium.

The scale factor is the following:

SF = (X / 39.39) 4

SF⁻¹ = (39.39 / 4) X

This scaling factor is unique and applies to all vectors in the diagram representing the proton decay.

We are going to describe the mass distribution of the dipoles of interaction 1 and 2, which represent the Dneutron particle.

Figure 44. Dneutron.

Figure 44. Dneutron.

According to our vector calculation, the Dneutron particle has a mass of 1140.33 MeV/c².

We are going to make the vector diagram of the neutron decay in the Dneutron particle.

Let's analyse figure 45:

The dipoles of the interaction 1 of the neutron, are represented in black.

The dipoles of the interaction 2 of the neutron, are represented in black.

The dipoles of the interaction 1 of the Dneutron, are represented in red.

The dipoles of the interaction 2 of the dneutron, are represented in red.

Represented in blue are the 3 vectors of interaction 1 and the 6 vectors of interaction 2 that correspond to the decay of the neutron in the Dneutron particle.

Now we are going to analyse the energy (mass) changes that occur when the inverse neutron decay in the Dneutron particle. These changes in energy (mass) are represented in blue in figure 45.

Table 3. Calculation of the interactions in MeV/c², when the neutron decay in the Dneutron particle.

Table 3. Calculation of the interactions in MeV/c², when the neutron decay in the Dneutron particle.

INVERSE NEUTRON DECAY
NEUTRON	DNEUTRON	INTERACTION	INTERACTION (MeV/c²)
R(DD)R	R(DD)R	E1 = 0	E1 = 0
B(DD)B	B(DD)B	E2 = 4.6	E2 = 45.29
G(UU)G	G(DD)G	E3 = 4.6	E3 = 45.29 (-)
R(DD)B	R(DD)B	E4 = 10	E4 = 98.47
R(DU)G	R(DD)G	E5 = 10	E5 = 98.47 (+)
B(DD)R	B(DD)R	E6 = 4.7	E6 = 46.28
B(DU)G	B(DD)G	E7 = 4.7	E7 = 46.28 (+)
G(UD)R	G(DD)R	E8 = 5.0	E8 = 49.23 (-)
G(UD)B	G(DD)B	E9 = 5.0	E9 = 49.23 (-)
TOTAL INTERACTION			IEtI = 478.54 MeV/c²
(W⁺)e⁺ INTERACTION			IEeI = W⁺ = 1.00 MeV/c²
(Z⁰)n INTERACTION			IEnI = 52.19 MeV/c²
(Z⁰)p INTERACTION			IEpI = 45.29 MeV/c²
(Z⁰) INTERACTION			IZ⁰I = 6.90 MeV/c²
Θ angle, Θ = arc cos W⁺/Z⁰			Θ = 82⁰

The vectors E1 to E9 are represented in blue in figure 45.

It is important to highlight that the decay of the neutron in the Dneutron particle produces a total interaction of Et = 478.54 MeV/c².

In figure 45, we divide these vectors into three parts, which we will explain below:

• En = E4 + E6

En = 98.47 – 46.28 = 52.19 MeV/c²

E4 = [R(DD)B]n → [R(DD)B]Dn

E6 = [B(DD)R]n → [B(DD)R]Dn

For E4, when the decay of [R(DD)B]n of the neutron, to [R(DD)B]Dn of the Dneutron occurs; a neutral current is produced.

For E6, when the decay of [B(DD)R]n of the neutron, to [B(DD)R]Dn of the Dneutron occurs; a neutral current is produced.

Here, we are going to hypothesize that the En energy is what gives origin to the antineutrinos. Let us note that the RB and BR gluons are of same frequencies; B and G are associated with the D quark.

When E4 + E6 interacts, it is like the RB and BR gluons cancel each other and the DD and DD quarks also cancel and, in this way, the resulting energy gives origin to the neutrinos, whose mass is neutral.

• Ep = E1 + E2

Ep = E2 = (4.6 x 39.39) / 4 = 45.29 MeV/c²

E1 = 0 MeV/c²

E1 = [R(DD)R]n → [R(DD)R]Dn

E2 = [B(DD)B]p → [B(DD)B]Dn

For E1, when the decay of [R(DD)R]n of the neutron, to [R(DD)R]Dn of the Dneutron occurs; no neutral current is produced.

For E2, when the decay of [B(DD)B]n of the neutron, to [B(DD)B]Dn of the Dneutron occurs; a neutral current is produced.

Here, we are going to hypothesize that the Ep energy is what gives origin to the photons. Here it is important to highlight that the RR and BB gluons have the same frequency; RR and BB gluons, have the frequency of the D quark.

• Ee = E3 + E5 + E7 + E8 + E9

Ee = 1 MeV/c²

E3 = [G(UU)G]n → [G(DD)G]Dn

E5 = [R(DU)G]n → [R(DD)G]Dn

E7 = [B(DU)G]n → [B(DD)G]Dn

E8 = [G(UD)R]n → [G(DD)R]Dn

E9 = [G(UD)B]n → [G(DD)B]Dn

E3 // E5 // E7 // E8 // E9

For the vectors E3, E5, E7, E8 and E9, when the decay of neutron to Dneutron occurs, it is observed that currents with positive and negative charges are generated.

The negative net energy of the vector Ee = 1 MeV/c²

Here, we are going to hypothesize that the Ee energy is what gives origin to the positron.

The emission of the energy vectors En, Ep and Ee are necessary in the β⁺ decay that occurs when the neutron decay into the Dneutron particle, so that the Dneutron stabilizes and reaches its corresponding energy.

In the following vector diagram, we are going to represent the vectors En, Ep and Ee.

It becomes difficult to graphically determine Ee, W⁺ and Θ angle.

calculation of W⁺, Z⁰ and Θ angle:

Theoretical calculations that we obtain from figure 46, inverse neutron decay diagram:

theoretical W⁺ = Ee = E3 + E5 + E7 + E8 + E9

E3 // E5 // E7 // E8 // E9

W⁺ = Ee = (- 45.29 + 98.47 + 46.28 – 49.23 -49.23 – 49.23) MeV/c²

theoretical W⁺ = 1.00 MeV/c²

theoretical Z⁰ = En + Ep

theoretical IZ⁰I = 52.19 MeV/c² - 45.29 MeV/c²

IZ⁰I = 6.9 MeV/c²

W⁺/ Z⁰ = cos Θ

Θ = arc cos IW⁺I/IZ⁰I = arc cos (1 / 6.9) = arc cos (0.144)

Θ = 82⁰

As analysed so far, the decay of the neutron in the Dneutron particle also includes a β⁺ decay.

4.5. Dneutron Decay

We are going to work with the following figures to represent the disintegration of the Dneutron particle into the neutroniumd particle

Figure 47. Dneutron particle.

Figure 47. Dneutron particle.

In figure 48, we represent the neutroniumd particle, through interaction 1 and interaction 2. We observe that all the components are D quarks and we also observe that the 9 dipoles are quarks, which are related vectorially. This configuration is represented only by matter, there is no antimatter in this configuration.

We are going to make the vector diagram of the neutroniumd particle, of the interaction 1 and 2:

In figure 49, the vector diagram of the neutron is represented in black.

In figure 49, the vector diagram of the Dneutron is represented in red.

In figure 49, the vector diagram of neutroniumd is represented in green.

The neutron and the Dneutron are particles made up of matter and antimatter.

Neutroniumd is a particle formed only by matter, it does not contain antimatter.

Figure 49. - Diagram of neutron particles, Dneutron and Neutronium.

Figure 49. - Diagram of neutron particles, Dneutron and Neutronium.

Figure 50. Dneutron Decay.

Figure 50. Dneutron Decay.

The vector diagram of the Dneutron particle, interaction 1, is represented in black.

The vector diagram of the Dneutron particle, interaction 2, is represented in black.

The vector diagram of Neutroniumd, interaction 2, is represented in red.

The vector diagram of Neutroniumd, interaction 1, is zero, there is no interaction between two D quarks of matter (DD).

Represented in blue are the 3 vectors of interaction 1 and the 6 vectors of interaction 2 that correspond to the decay of the Dneutron in the neutroniumd particle.

At first glance, if we add the three D quarks scalarly, we can affirm that the total charge is (-1/3 - 1/3 - 1/3 = -1) -1, but we have to remember that we are adding scalarly and in this situation we must consider a vector sum, this gives us a net charge of zero. We can represent the D quarks as three vectors with the same module, 120 degrees out of phase, whose vector sum is zero, in analogy to electric generators.

The neutronium particle behaves analogously to a generator, whose current, voltage and charge are perfectly balanced, in equilibrium.

The scale factor is the following:

SF = (X / 39.39) 4

SF⁻¹ = (39.39 / 4) X

This scaling factor is unique and applies to all vectors in the diagram representing the Dneutron decay.

We are going to describe the mass distribution of the dipoles of interaction 1 and 2, which represent the neutroniumd particle.

According to our vector calculation, the neutroniumd particle has a mass of 443.10 MeV/c².

It can be seen in figure 51, interaction 1:

R(DD)R = 0

B(DD)B = 0

G(DD)G = 0

This is telling us that there is no direct interaction between the quarks (DD). In analogy with an electric generator, it is as if the pair of quarks (DD) were at the same electric potential.

Now we are going to analyse the energy (mass) changes that occur when the Dneutron decay in the neutroniumd particle. These changes in energy (mass) are represented in blue in figure 50.

Table 4. Calculation of the interactions in MeV/c², when the Dneutron decay in the neutroniumd particle.

Table 4. Calculation of the interactions in MeV/c², when the Dneutron decay in the neutroniumd particle.

DNEUTRON DECAY
DNEUTRON	NEUTRONIUMD	INTERACTION	INTERACTION (MeV/c²)
R(DD)R	R(DD)R	E1 = 8.5	E1 = 84.69
B(DD)B	B(DD)B	E2 = 8.5	E2 = 84.69
G(DD)G	G(DD)G	E3 = 8.5	E3 = 84.69
R(DD)B	R(DD)B	E4 = 7.5	E4 = 73.85
R(DD)G	R(DD)G	E5 = 7.5	E5 = 73.85
B(DD)R	B(DD)R	E6 = 7.5	E6 = 73.85
B(DD)G	B(DD)G	E7 = 7.5	E7 = 73.85
G(DD)R	G(DD)R	E8 = 7.5	E8 = 73.85
G(DD)B	G(DD)B	E9 = 7.5	E9 = 73.85
TOTAL INTERACTION			IEtI = 697.17 MeV/c²

The vectors E1 to E9 are represented in blue in figure 50.

It is important to highlight that the decay of the Dneutron in the neutroniumd particle produces a total interaction of Et = 697.17 MeV/c².

In figure 50, these vectors in blue, we will explain below:

• E = E1 + E2 + E3 + E4 + E5 + E6 + E7 + E8 + E9

E1 = R(DD)R → R(DD)R

E2 = B(DD)B → B(DD)B

E3 = G(DD)G → G(DD)G

E4 = R(DD)B → R(DD)B

E5 = R(DD)G → R(DD)G

E6 = B(DD)R → B(DD)R

E7 = B(DD)G → B(DD)G

E8 = G(DD)R → G(DD)R

E9 = G(DD)B → G(DD)B

When the Dneutron particle decay into the neutroniumd particle, it is observed that matter separates from antimatter, producing energy in the form of radiation.

The particle that results from the decay of the Dneutrom particle, neutroniumd, is formed only by D quark and does not have antimatter.

The configuration of the quarks inside the neutroniumd particle is formed by vectors that represent the quarks, which are displaced by 120 degrees, whose vector sum makes the net charge zero.

We are going to represent the decay of the Dneutron particle into neutroniumd with the following equation:

Dneutron → neutronium + ∆E

4.6. Conservation laws

It is important to mention the conservation laws, which indicate that there are properties that should never change.

There are three conservation laws that we have to mention when we analyse the decay of the neutron and Dneutron.

• Electric Charge
• Energy
• Baryon number

We will analyse the decay of the neutron in the Dneutron particle, we will use figure 52 and figure 53, for our analysis:

Dneutron → neutron + e⁻ + ṽ + ∆E

  (-1)        (0)      (-1)

We see that the neutron has an electric charge (0), the Dneutron particle has an electric charge (-1) and the electron has an electric charge (-1); We see that the electric charge is balanced, it is conserved.

Dneutron → neutron + e⁻ + ṽ + ∆E

 (+1)         (+1)    (+1)(-1)

We see that the neutron has a baryon number equal to 1 and the Dneutron particle also has a baryon number equal to (+1), the baryon number is conserved. We are going to analyse the lepton number, we see that the electron has a lepton number equal to (+1) and the antineutrino has a lepton number (-1), the lepton number is also conserved.

Dneutron → neutron + e⁻ + ṽ + ∆E

1140.33 MeV/c² → 939.51 MeV/c² +…..

Considering the energy (mass), the Dneutron has a mass of 1140.33 MeV/c², the neutron particle has a mass of 939.51 MeV/c², to this is added the mass of the electron, the antineutrino and the energy of photons ∆E. Energy is also conserved.

In a simple analysis, we see that the electric charge, energy and baryon and lepton number are conserved for the decay of the neutron in the Dneutron particle.

We will analyse the decay of the Dneutron in the neutroniumd particle, we will use figure 54 and figure 55, for our analysis:

Dneutron → neutroniumd + ∆E

 (-1)        (-1)

We see that the Dneutron has an electric charge (-1), the neutroniumd particle has an electric charge (-1); We see that the electric charge is balanced, it is conserved.

Dneutron → neutroniumd + ∆E

 (+1)         (+1)

We see that the Dneutron has a baryon number equal to (+1) and the neutroniumd particle also has a baryon number equal to (+1), the baryon number is conserved.

Dneutron → neutroniumd + ∆E

1140.33 MeV/c² 443.10 MeV/c² + ……

Considering the energy (mass), the Dneutron has a mass of 1140.33 MeV/c², the neutronium particle has a mass of 443.10MeV/c², to this is added the energy of photons ∆E. Energy is also conserved.

Finally, summarizing we have:

In the following graph we are going to represent some fundamental parameters that describe the following particles:

Figure 56. Fundamental characteristics of particles.

Figure 56. Fundamental characteristics of particles.

Figure 57. Proton, Dproton and Protoniu.

Figure 57. Proton, Dproton and Protoniu.

Figure 58. Dneutron, Neutron and Neutroniumd.

Figure 58. Dneutron, Neutron and Neutroniumd.

5. Conclusions:

Figure 59. Characteristic parameters of the following particles: Neutron, Proton, Dproton, Protoniu, Dneutron and Neutroniumd.

Figure 59. Characteristic parameters of the following particles: Neutron, Proton, Dproton, Protoniu, Dneutron and Neutroniumd.

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