In this section, we give a synthesis of the representative didactic ideas of quarks. We found between Compton’s wavelength (Δ_pl) and (Δ_p) there is a golden relation [1] , at less than a scale s-factor (10)¹⁹: [n_(pl,p) = ()²s/2], with () the “aureus” (golden). We find golden relations in a pentagon between the side (base) and apothem. Recall the protons are composed of three quarks: three centres of elastic diffusion which could indicate an internal “geometric structure” of the proton. Then, we formulate the Didactic Idea (1): “A proton has a ‘pentagonal’ geometric structure Ψ_1p where its three component quarks are coincident with three constituent triangles of a pentagon”. The quarks (u,d) are represented by “golden triangles”, see Figure 1 :

This conjecture has physical sense if we admit the Didactic Idea (2): “some quantum oscillators can couple to build a geometric figure” [2] . Then, we place three quantum oscillators (I_A, I_B, I_C) at the vertices (A, B, D) and three junction oscillators (I_AB, I_BC, I_CA), see Figure 2 :

Weindicate by the acronym “IQuO” (Intrinsic Quantum Oscillator) the oscillators of coupling building a geometric structure. Note that the oscillations of IQuO are longitudinal along the sides. The structure of coupled oscillators (IQuO) is possible only if the IQuO oscillators [3] [4] are decomposed into several oscillating or “sub-oscillators” with energy at “semi-quanta” (sq()): 1 quantum () = (,) 2 sq().

From oscillations’ theory, for any n, one has:

$\begin{array}{c}\left[H_{\left(n\right)}\right]=\left[U_{\left(n\right)}+K_{\left(n\right)}\right]=\left[{\left(U_{\left(n\right)}\right)}_{el}+\left(K_{\left(n\right)}\right)\right]\\ =\left[\left(2n+1\right){\left(\frac{1}{4}\hslash \omega \right)}_{el}+\left(2n+1\right){\left(\frac{1}{4}\hslash \omega \right)}_{in}\right]\\ =\left[\left(2n\right){\left(\frac{1}{4}\hslash \omega \right)}_{el}+\left(2n\right){\left(\frac{1}{4}\hslash \omega \right)}_{in}\right]+\left[{\left(\frac{1}{4}\hslash \omega \right)}_{el}+{\left(\frac{1}{4}\hslash \omega \right)}_{in}\right]\end{array}$ (1)

From this equation, it derives conjecture that the energy value of [( = 1/4h)], indicated as an “empty” semi-quantum with the symbol (o). Note IQuO_(n=1) is [(o, ) + (o, )].

In Figure 2 there are two “golden” triangles, with IQuO in eigenstate n = 1 (I_AB) and n = 2 (all other IQuO). Recall the IQuO_(n=2) has three sub-oscillators. We used the following representation, Figure 3 , where we highlight the presence of [sq(o,), sq (o, o)] in each sub-oscillator of vertex:

Note the sub-oscillators (S_A2, S _B2, S_C2) are very important: we conjecture that they can originate the “gluons” from which emerge the hadronic jets (two or three jets) in the experiments to the CERN or the Fermilab. We stated that to have a “compact” triangular structure (golden) the junction oscillators and vertex ones need to be “wedged” one inside the other (superpositions of IQuO). In this case, on each oblique side we must have two junction oscillators, with n = 2, (I_BF, I_DC; I_AE, I_CG) and on the base side another junction oscillator (I_AB), see Figure 4(a) , Figure 4(b) :

At each junction, [(H, D, F) (H, G, E)], there will be a phase shift during the couplings between the sub-oscillators, but to each phase shift follows a phase adaptation for realize the coupling. Each phase adaptation of the overlapping IQuOs identifies a degree of freedom in the structure to which we assign a “color”, see Figure 4 . It is understood that the degrees of freedom are associated with transformations operating on the phase of the oscillation, that is they are “gauge” transformations. The overlapping of IQuO origins a double coupling along the sides. Note a double sub-oscillator in the state of an “excited vacuum” (one only sq () in one only IQuO) constitute a “gluon”. In conclusion, a quark is a structure of coupled gluons. A last note, the sub-oscillators (S_A2, S_C2) will perform a very important function: they will be available to be hooked by a line (guide rail) of B-type quantum oscillators (Bosons) which will allow the triangular structure of “roto-translate” along it. The “gluon boson” becomes so the gauge field that adapts the phase shifts in the junction for maintain the structure and to allow to this (in moment eigenstate) to propagate along an axis.

Didactic Idea (3): “A structure-particle”, in an eigenstate of impulse (p_x), can propagate along a line (X-axis) or a “guiderail” given by line of oscillators of a gauge base field (waveguide)” [6] .

Didactic Idea (4): “We indicate in the triangle-structure the particle the sign ± of the electric charge (q), the possible proper rotation of the structure or ‘spin’ (s) and the direction of the () ‘flow’ vectorof the internal quanta sq () to the oscillators of the structure”, Figure 5 :

The direction v of propagation of the particle coincides with the direction of the flux along the propagating side (the one lying along the X axis: v).

Didactic Idea (5): “The value of the electric charge will be given by the probability P() of detecting the quantum () along the propagating side” [1] .

By the number of quanta () in Figure 5 , we find:

$q\left(d\right)=P{(•)}_d=-1/3$, $q\left(u\right)=P{(•)}_u=+2/3$.

The doubling of the energy in a quantum oscillator with sub-oscillators determines the doubling of the components of each quantization operator (a, a⁺) [3] , that is $\left[\left(a_{el},a_{el}^{+}\right),\left(a_{in},a_{in}^{+}\right)\right]$. Let us then correspond to each component of the operators(a, a⁺) a sq(o, ) with their respective elastic and inertial characteristics; we will have the following representation of IQuO, Figure 6 :

The IQuO with phase rotation opposites is said: B-IQuO (Boson type). The aspect 2-dim of the oscillation determines an oscillator with an internal degree of freedom extra whit two eigenvalues (the two directions of phase rotation) [3] [5] . Coupling two B-IQuO, we obtain an IQuO with mono-direction of phase rotation, Figure 7 :

These IQuO are F-IQuO, different from those of B-type (Boson).

Didactic Idea (6): “We associate the ± sign of electric charge to the directions of the phase rotation: Direction clockwise (−e), Direction anticlockwise (+e)”.

A B-IQuO coupling with a F-IQuO reads the direction of rotation of the phase of F-IQuO: this boson is the photon. An IQuO_(n=0) in the vacuum state has an only sub-oscillator. We associate [5] to the configurations of the sq(o, ) a “color” and an anti-color, and a “gluon” to them superpositions (Didactic Idea (7)), see Figure 8 :

So, the gluon has two colors. The gluons of the guideline have two functions:

Didactic Idea (8): “There is the possibility that two quarks can bind along the propagation side and generate a ‘real’ particle, the pion” [1] , see Figure 10 :

We pose the masses of quarks bound in a pion:

$\left[m{\left(d\right)}_{\pi }=\left(86.26\right)\text{MeV}/c^2,m{\left(u\right)}_{\pi }=\left(\text{53}.\text{21}\right)\text{MeV}/c^2\right]$

Finally, note, in Figure 10 , that Φ_π v_x. We can describe a hadron by a “structure equation” built on the component quarks. We have:

${\pi }^{\pm }=\left(u\underset{\_}{\otimes }d\right),{\pi }^0=\left[\left({\pi }^{+}\right)\underset{\_}{\otimes }\left({\pi }^{-}\right)\right]$ (2)

The operation () is (here, it is expressed in intuitive way) a combination of two operations [ (, )], where -operation describes the property of “interpenetration” of the quarks (structures with oscillations), instead, the-operation describes “dynamics interactions” with the exchange of sq(). The neutral pion is a unique elementary particle: its components [(u, d) and (u, d)] are reciprocally “interpenetrating” (Didactic Idea (9)). As in literature [7] [8] , now the “spin” is:

$s\left({\pi }^{\pm }\right)=\left[s\left(u\right)+s\left(d\right)\right]=0$ and $s\left({\pi }^0\right)=\left[s\left(u_1\right)+s\left({\underset{\_}{d}}_1\right)\right]+\left[s\left(u_2\right)+s\left({\underset{\_}{d}}_2\right)\right]=0$

From its structure equation, we have, see Figure 11 :

The structure of Figure 11 predicts the decay of neutral pion in two photons caused by annihilation of two quark pairs [(u, u), (d, d)] along the side BD or axis X. The annihilation process defines the “electromagnetic mass” of quarks:

$m_{em}\left(u\right)=\left(3.51\right)\text{MeV}/c^2$, $m_{em}\left(d\right)=\left(5.67\right)\text{MeV}/c^2$.

From structure equation of charged pion one has the parity of pions [9] :

$P\left({\pi }^{\pm }\right)=-1$, $P\left({\pi }^0\right)=-1$

Didactic Idea (10) “Light mesons are structures composed of pion molecules”.

The -meson has the structure following [10] :

$\eta =\left[{\left({\pi }^{+}\oplus {\pi }^{-}\right)}_1\otimes {\left({\pi }^{+}\oplus {\pi }^{-}\right)}_2\right]=\left[{\left({\pi }^0\right)}_{r1}\underset{\_}{\otimes }{\left({\pi }^0\right)}_{r2}\right]$

where ${\left({\pi }^0\right)}_r=\left({\pi }^{+}\oplus {\pi }^{-}\right)$ is a “molecule” of pions. The structure is, Figure 12 :

There are four possible particles (π⁺, π⁻, π⁰, γ) with the following combination or channels:(π⁺, π⁻, π⁰)_(23%), (π⁰, π⁰, π⁰)_(33%), with a virtual lattice π⁰, (γ, γ)_(39%), (π⁺, π⁻ ,γ) _(5%), ((e⁻ + e⁺), γ) (see Figure 12 ).

Other mesons are [11] ${\eta }^{\prime }={\left({\pi }^0\right)}_r\otimes \left(\eta \right)$, and the -meson and -meson (with added the pair (d, d)_π)

$\rho =\left\{{\left(d,\underset{\_}{d}\right)}_{\pi }\otimes \left[2\eta \oplus {\left({\pi }^0\right)}_r\right]\right\}=\left\{{\left(d,\underset{\_}{d}\right)}_p\otimes \left[\left(\eta \oplus \eta \right)\oplus {\left({\pi }^0\right)}_r\right]\right\}$

$\omega =\left[2{\left(d,\underset{\_}{d}\right)}_p\right]\otimes \left[\left({\pi }^{+}\oplus {\pi }_r^0\right)\otimes \left({\pi }^{-}\oplus {\pi }_r^0\right)\right]$

Recall the geometric rule of golden triangles: in a golden triangle (u, d) it is possible to draw two other triangles (sub-triangles) inside, always of the type (u, d), which are indicated as (u, d) [2] [12] ; then we have: Didactic Idea (11) “In each quark (u*, d*) one can insert another pair of golden triangles of type (u, d) which give origin to sub-quarks (u, d) inside the ‘tank’ quark (u*, d*)”.

So, we will have two possible configurations, see Figure 13 :

We see the composed quarks have two forms: [d*, (u,d); u*, (u,d)].

The quarks (u*, d*) are called “tank” quarks because contain the “sub-quarks” (u, d). The first combination will give us the “strange” quark or s-quark (s) [9] . The second will give us the “charm” quark or c-quark (c) [13] .

The electric charge is: q(s) = q([u*, (u, d)]) = −1/3, q(s) = q([u*, (u, d)]) = +1/3.

Recall the “weak Isospin” in quarks: the s-quark is of d-type. Then, we will have, see Figure 14 :

Figure 13(b) represents the c-quark, with the structure: c = [d*, (u, d)]. The electric charge will be given by q(c) = q([d*, (u, d)] = q(d*) + q(u) + q(d)] = q(u) = +2/3.

Recall the “weak Isospin”. Note T_z(c) = +1/2, the c-quark is type up (u).

The other possibility of combinations gives the other quarks. The b-Quark:

$q\left(b\right)=q\left(\left\{d*\left[\underset{\_}{u}*\left[\left(u_1,d_1\right)\right],\underset{\_}{d}\right]\right\}\right)=-1/3$, $q\left(\underset{\_}{b}\right)=q\left(\left\{\underset{\_}{d}*\left[u*\left[\left({\underset{\_}{u}}_1,{\underset{\_}{d}}_1\right)\right],d\right]\right\}\right)=+1/3$

Note T_z(b) = −1/2 , the b-quark is type up (d).

The t-quark: q(t) = q(u*[u*[(u₁, d₁)], d]) = +2/3 and T_z(t) = +1/2, the c-quark is type up (u).

In summary, we will only have “six” types of quarks [22] [23] , see Figure 15 :

The diffusion experiments (e + p) determine the quarks structure of the proton [14] , in “parton” model, Figure 16 :

We indicate by Ψ_1p this structure state, which corresponds to the structure of Figure 1 , with the three diffusing centres coinciding with the three vertices (ABD):

Didactic Idea (12): “A proton belonging to a nucleus assumes the state configuration of the Figure 1 , called ‘static’ proton, with the u-quarks in the ‘periphery’respect to d-quark”.

(Physical Aspect) This because the u-quarks are attracted to other quarks from other nucleons. However, since the parton model was formulated in experiments with electrons (bullets) and hydrogen H (target), we could conjecture that also in a hydrogen atom the proton is arranged in the Ψ_1p configuration. The “static” term says to us that the “indivisible” proton having two propagation axes, see Figure 1 , cannot move in any direction. This aspect determines that the proton can become an aggregation centre gravitational and nuclear of matter. For a moving “free” proton, instead, we formulate the following idea [15] , Figure 17 .

Didactic Idea (13): “We indicate by Ψ_2p the following structure representation of a free proton”.

Here, a proton propagates along the X-direction coinciding with one of the sides of the geometric structure (we speak of “kinetic proton”). Also, in this structure we highlight three diffusing centres, the vertices (A, C, D) or (A, C, E). The presence of two configurations of quarks associated with a proton was highlighted in the “ECM” effect [16] , found in a series of experiments with nuclear protons, see ref. [17] . The representation Ψ_2p could correspond to an eigenstate of the momentum (p_x) of the proton along the X-axis, in the Reference Frame (RF) S_lab, with uncertainty of x-position (see ΔxΔp_x h) and wavelength (λ_p = h/p_x). In RF at rest S˚ the λ_p→λ_p that is the Compton wavelength of the proton. Note that the structure-particle of IQuO, in a different RF (S), can have of the “relativistic deformations” but it remains invariant in geometric form. The movement along the X-axis of the particle-structure can be associated with a rotation of the quarks around the same X-axis. From the configurations of the charged pions and the neutral pion, see also Equation 2, we deduce, considering the spins, that the two u-quarks are paired, [s(u₁) + s(u₂)] = ± 1. This is confirmed by some recent experiments [18] [19] . Since the proton is a fermion then the following spin relation is admitted:

$s\left(p\right)=s\left(d\right)+\left[s\left(u_1\right)+s\left(u_2\right)\right]=s\left(u\right)$ with $\left[s\left(u_i\right)+s\left(d\right)\right]=0$

Looking at the structure of the proton Ψ_2p, we can draw the structure of the neutron, see Figure 18 :

Since the neutron is a fermion then we find [19] that s(n) = s(u) + [s(d₁) + s(d₂)]. Having [s(d_i) + s(u)] = 0 → s(n) = s(d₂), with the two d-quarks are paired, that is: [s(d₁) + s(d₂)] = ± 1. Since the neutron spin is coincident to that of d-quark then the neutron magnetic moment cannot be zero; instead, its electric dipole moment is zero since along the diagonal AC the electric charge is zero. So, in the PGM one shows in simple way the values of the magnetic moment and electric dipole of neutron. We represent the Deuteron and Tritium, Figure 19 :

Recall that the neutron binds to the proton thanks to the strong force represented by overall “neutral” virtual pions. However, we note that the mass of the pions is about seven times smaller than that of the nucleon. In Compton lengths will instead be about seven times larger. It follows:

Didactic Idea (14) “The long-distance interaction between proton and neutron mediated by a neutral pion $\left[n+\left({\pi }_v^0\right)+p\right]\to \left[n+p\right]$ can be represented in the following way, see Figure 20 ”:

(Physical Aspect) This aspect is coherent with the Yucawa’s theory [20] . As they get closer, they always overlap within a neutral pion, assuming a larger size due to the loss of mass as binding energy. When the (n + p) system becomes bound but it is free of propagate then it could assume the configuration Ψ_2p of Figure 17 , where the pion of bind will have the same size as the two nucleons (the pion is not represented in Figure 20 ). Note the configuration is compatible with relativity and QM. In PGM we represent all the strong interactions as for example [π⁺ + p⁺ → Δ⁺⁺], Figure 21 :

We can also construct the freely moving alpha particle, see Figure 22 :

The proton in shape Ψ_1p will be called “static proton” and will participate in the formation of nuclei and atoms. So, in a nucleus the alpha particle is built with a “static” proton, see Figure 23 :

The structure equations of nucleons are:

$p={\kappa }_p\left(u_1\underset{\_}{\otimes }d\underset{\_}{\otimes }{u}_2\right)$ (3a)

$n={\kappa }_n\left\{{\left[\left(d\otimes \underset{\_}{d}\right)\right]}_A\underset{\_}{\otimes }{\left[N_a\oplus N_b\oplus N_c\right]}_B\right\}$ (3b)

(Physical Aspect) The _i coefficient are connected to the elastic tensions between (IQuO) building the geometric structure [15] and the N_i are coefficients relative to (d, u, d) quarks combinations by two operations (, ). From these equations it is possible to derive them mass [15] . The pair (d, d) in the structure equation of the is the proof of a d, d background lattice [G] which mediates the interactions and allows the propagation of free neutron and its decay, Figure 24 :

Note that the energy of the couple [ = (d, d)] is a bond energy that concerns the gluons (g), that is “gluon” energy [ = (d, d) = g _g]. This aspect is present also in the structures of mesons [10] .

Through the property of golden triangles it is possible to explain one of the current puzzles that have emerged from experimental studies on the structure of the proton, see experiments at CERN, ref. [21] , the experimental discovery of quarks’ “sea” (s,c) inside the proton, Figure 25 :

Didactic Idea (15) “There are two forms of structure in neutral pions and nucleons: one form is called “kinetic” (ordinary and interacting) while the other is “static” (not interacting)”, see Figure 26 :

Also, the neutron has two configurations: the one anomalous is, Figure 27 :

(Physical Aspect) The physical behavior of these forms may be different since there are not all quarks on the propagation side, see Figure 26 and Figure 27 . In the static shape of neutron its decay could be different from the ordinary one: in some experiments an anomaly in the decay of the free neutron was detected [22] [23] . Note that the anomalous shape has the following characteristics: the particle no propagates, no decay, no interaction (only gravitational). These characteristics could push us to conjecture the existence of some relationship with dark matter [24] [25] . Besides, anomalous particles and ordinary antiparticles could not annihilate, thus it is possible that the antimatter hides in galactic halos. This aspect could solve the problem of the asymmetry between matter-antimatter.

We take into consideration the scalar equation describing massive particles with zero spin:

$\left\{\frac{{\partial }^2\Psi \left(x,t\right)}{\partial x^2}-\frac{{\partial }^2\Psi \left(x,t\right)}{c^2\partial t^2}={\left(\frac{mc}{\hslash }\right)}^2\Psi \left(x,t\right)\right\}\iff \left\{{\nabla }^2\Psi \left(x,t\right)={\left(\frac{mc}{\hslash }\right)}^2\Psi \left(x,t\right)\right\}$ (4)

$\left\{E^2=m^2{c}^4+p^2{c}^2\iff {\omega }^2={\omega }_0^2+k^2{c}^2\right\}$ (5)

This equation also describes oscillations in a set of “pendulums” coupled through springs [26] with frequency . In analogy with this system, we connect the ₀ toa particular elastic coupling (T₀), see Figure 28 , which is in addition to the one already existing between the oscillators of a massless scalar field (), as the springs. This “additional coupling” [1] [5] is referred to as a “massive coupling”:

In this way, we introduce a “new paradigm” in physical field theory: “the m mass is the oscillation frequency ₀ (relative toa transversal elastic tension T₀) of a structure of coupled quantum oscillators of field” (Didactic Idea (16)). The model that we develop considering the geometric aspect of a particle will be referred to as the “Particles’ Geometric Model”, with the acronym (PGM).

Let’s remember the grid in Figure 28 . The grid could represent a virtual lattice Wof bosons electrically charged, W^±, with a sq() -flow along the sides, with phase rotation indicating an electric charge and a rotation, indicating the spin s, around the diagonals A_iB_i, lying along the propagation axes X_i of quanta, see Figure 29 :

Didactic Idea (17) “W^±-bosons are golden rectangle and contain the golden d-quarks”. In the pion decay, we insert (u, d) quarks in a lattice W^±, see Figure 30 :

The sq() of the d-quark (ACE) runs along the AC_d side. The AC_d side belong also W-boson: [AC AC(W⁺)_ABCD]. In C the sq()_d enters the AC diagonal of (W⁺)_ICLG, and, thanks to this, it transforms into a sq()⁺ becoming a sq()_u⁺ of the u-quark (ACG). Having used the boson (W⁺)_ABCD of the W^± lattice, the W⁻ boson remains unpaired, and, thus, after decays. Didactic Idea (18). “The Feynman diagrams can be replaced in the -decay by lattice of bosons (W⁺, W⁻) W”.

Then, we consider the reactions [(W u)→d, (W d)→u], see Figure 31 :

The W-boson falls into a local state (in the W lattice the W⁻ was in a non-local or virtual state) and, being no longer connected to the lattice but in a free “excited” state (it has absorbed the photon γ = (u, u)), thus it decays. The virtual photon, incorporated in W⁻, participates in the formation of the muon in energy and spin. We consider that W = (W⁺ W⁻) and π^± = (u d); using the matrix form of (W, π), the spin (,) conservation tells us that:

$\begin{array}{c}\left(\begin{array}{c}{W}_{\uparrow \uparrow }^{-}\\ W_{\downarrow \downarrow }^{+}\end{array}\right)\underset{\_}{\otimes }\left(\begin{array}{c}{\underset{\_}{u}}_{\downarrow }\\ d_{\uparrow }\end{array}\right)=\left(\begin{array}{c}{W}_{\uparrow \uparrow }^{-}\underset{\_}{\otimes }{\underset{\_}{u}}_{\downarrow }\\ W_{\downarrow \downarrow }^{+}\underset{\_}{\otimes }{d}_{\uparrow }\end{array}\right)=\left(\begin{array}{c}{W}_{\uparrow \uparrow }^{-}\underset{\_}{\otimes }{\underset{\_}{u}}_{\downarrow }\\ u_{\downarrow }\end{array}\right)\equiv \left[\left(W_{\uparrow \uparrow }^{-}\underset{\_}{\otimes }{\underset{\_}{u}}_{\downarrow }\right)\underset{\_}{\otimes }{u}_{\downarrow }\right]\\ =\left[\left(u_{\downarrow }\underset{\_}{\otimes }{\underset{\_}{u}}_{\downarrow }\right)\underset{\_}{\otimes }{W}_{\uparrow \uparrow }^{-}\right]=\left[{\gamma }_{\downarrow \downarrow }\underset{\_}{\otimes }{W}_{\uparrow \uparrow }^{-}\right]=\left[{\gamma }_{\downarrow \downarrow }\underset{\_}{\otimes }\left({\mu }_{\uparrow }^{-}\underset{\_}{\otimes }{\nu }_{\uparrow }\right)\right]\\ =\left[\left({\gamma }_{\downarrow \downarrow }\underset{\_}{\otimes }{\mu }_{\uparrow }^{-}\right)\underset{\_}{\otimes }\left({\nu }_{\uparrow }\right)\right]\to \left(\begin{array}{c}{\underset{\_}{\nu }}_{\uparrow }\\ {\mu }_{\downarrow }^{-}\end{array}\right)\end{array}$ (6)

The Kaon structures will be: K⁰ = (s, d), K⁰ = (s, d) ; K⁺ = (s, u), K⁻ = (s, u). The charge is:

$q\left(K^{+}\right)=q\left(\underset{\_}{s},u\right)=q\left\{\left[u*,\left(\underset{\_}{u},\underset{\_}{d}\right)\right]\underset{\_}{\otimes }\left(u\right)\right\}=q\left(u*\right)+q\left(\underset{\_}{u}\right)+q\left(\underset{\_}{d}\right)+q\left(u\right)=+1$ (7a)

$q\left(K^{-}\right)=q\left(s,\underset{\_}{u}\right)=q\{\left\{\left[\underset{\_}{u}*,\left(u,d\right)\right]\underset{\_}{\otimes }\left(\underset{\_}{u}\right)\right\}=q\left(\underset{\_}{u}*\right)+q\left(u\right)+q\left(d\right)+q\left(\underset{\_}{u}\right)=-1$ (7b)

$q\left(K^0\right)=q\left(\underset{\_}{s},d\right)=q\{\left\{\left[u*,\left(\underset{\_}{u},d\right)\right]\underset{\_}{\otimes }\left(d\right)\right\}=q\left(u*\right)+q\left(\underset{\_}{u}\right)+q\left(\underset{\_}{d}\right)+q\left(d\right)=0$ (7c)

$q\left(K^0\right)=q\left(s,\underset{\_}{d}\right)=q\left\{\left[\underset{\_}{u}*,\left(u,d\right)\right]\underset{\_}{\otimes }\left(\underset{\_}{d}\right)\right\}=q\left(\underset{\_}{u}*\right)+q\left(u\right)+q\left(d\right)+q\left(\underset{\_}{d}\right)=0$ (7d)

The structures of K-mesons are, Figure 32 :

Since the sub-quark are attached to its tank u*, it follows that the spin of s-quark is:

$\sigma \left(s\right)=\sigma \left(\underset{\_}{u}*\right)+\sigma \left(u+d\right)=\sigma \left(\underset{\_}{u}*\right)$

Then, the Kaon spin is: $\sigma \left(K\right)=\sigma \left(s\right)+\sigma \left(q\right)=\sigma \left(\underset{\_}{u}*\right)+\sigma \left(d\right)=\left[\pm 1,0\right]$

If we consider the interpenetration of quark (u, s) as that of pion then it follows:

$\sigma \left(K\right)=\sigma \left(s\right)+\sigma \left(q\right)=\sigma \left(\underset{\_}{u}*\right)+\sigma \left(d\right)=0$

As it happens in pions (u*) and (q = u, d) have opposite spins. The structure equations are:

$K^{+}=\left(\underset{\_}{s}\underset{\_}{\otimes }u\right)=\left\{\left[u*,\left(\underset{\_}{u},\underset{\_}{d}\right)\right]\underset{\_}{\otimes }\left(u\right)\right\};K^{-}=\left(s\underset{\_}{\otimes }u\right)=\left\{\left[\underset{\_}{u}*,\left(u,d\right)\right]\underset{\_}{\otimes }\left(u\right)\right\}$ (8a)

$K^0=\left(\underset{\_}{s}\underset{\_}{\otimes }d\right)=\left\{\left[u*,\left(\underset{\_}{u},\underset{\_}{d}\right)\right]\underset{\_}{\otimes }\left(d\right)\right\};K^0=\left(s\underset{\_}{\otimes }d\right)=\left\{\left[\underset{\_}{u}*,\left(u,d\right)\right]\underset{\_}{\otimes }\left(\underset{\_}{d}\right)\right\}$ (8b)

We calculate the parity P(K⁰); we will have:

$\begin{array}{c}P\left(K^0\right)=P\left\{\left[u*,\left(\underset{\_}{u},\underset{\_}{d}\right)\right]\underset{\_}{\otimes }\left(d\right)\right\}=P\left\{\left[u*\otimes \left(\underset{\_}{u}\oplus \underset{\_}{d}\right)\right]\oplus \left(d\right)\right\}\\ =\left\{P\left[u*\underset{\_}{\otimes }\left(\underset{\_}{u}\oplus \underset{\_}{d}\right)\right]P\left(d\right)\right\}=P\left(\underset{\_}{u}\oplus \underset{\_}{d}\right)P\left(\underset{\_}{d}\right)\\ =P\left(\underset{\_}{u}\right)P\left(\underset{\_}{d}\right)P\left(d\right)=\left(+1\right)\left(+1\right)\left(-1\right)=-1\end{array}$

Recall the connection between u* and (u, d), and no interpenetration, therefore we have: $P\left[u*\otimes \left(\underset{\_}{u}\oplus \underset{\_}{d}\right)\right]=P\left(\underset{\_}{u}\oplus \underset{\_}{d}\right)$

Here the parity of K-meson would be negative. However, we can also have the following possibility:

$\begin{array}{c}P\left(K^0\right)=P\left\{\left[u*,\left(\underset{\_}{u},\underset{\_}{d}\right)\right]\underset{\_}{\otimes }\left(d\right)\right\}=P\left\{\left[u*\oplus \left(\underset{\_}{u}\oplus \underset{\_}{d}\right)\right]\otimes \left(d\right)\right\}\\ =\left\{P\left[u*\oplus \left(\underset{\_}{u}\oplus \underset{\_}{d}\right)\right]P\left(d\right)\right\}=P\left(u*\right)P\left(\underset{\_}{u}\oplus \underset{\_}{d}\right)P\left(d\right)\\ =P\left(u*\right)P\left(\underset{\_}{u}\right)P\left(\underset{\_}{d}\right)P\left(d\right)=\left(-1\right)\left(+1\right)\left(+1\right)\left(-1\right)=+1\end{array}$

Here the parity of K-meson would be positive. There are two possibilities due to divalent behaviour of the s-quark in relation to operation : the two component operations (, ) determinate that the s-quark can induce two different ways of decay. In fact, the K meson can decay in two pions (K→ππ) with negative parity or in three pions (K→πππ) with positive parity [27] [28] . In PGM it is easy to demonstrate that K⁰ K⁰. The oscillations of the K-meson affirm the existence of lattice u, d.

In the literature [27] [28] , it is assumed that the decays of the Kaon are weak (about 10⁻⁸ sec), that is mediated by W bosons as happens to the pion. The theory of the K decay describes that, Figure 33 :

The description, in truth, would be rather problematic. In the interactions’ literature [29] [30] it is not physically explained why a point particle “decays” (or transform itself) in two other particles as

$\left[\left(s\to W^{+}+\underset{\_}{u}\right),\left(W^{+}\to \underset{\_}{d}+u\right),\left(g\to \underset{\_}{d}+d\right),\cdots ,\left(\pi \to \mu +\nu \right)\right]$

and others. Now, we change the representation of interactions: the intermediary “agent” presents itself in lattice form. We admit the existence of W-bosons’ lattice to which the quarks couple. We now describe the transformation of s-quark operated by the background lattice W^±, using the matrices of the Equation (6) and introducing the W-matrix and s-matrix with three terms:

$\begin{array}{l}{\left(\begin{array}{c}{W}^{+}\\ I\\ W^{-}\end{array}\right)}_{\left\{W\right\}}\underset{\_}{\otimes }{\left(\begin{array}{c}\underset{\_}{u}*\\ \sout{u}\\ \sout{d}\end{array}\right)}_s=\left(\begin{array}{c}{W}^{+}\underset{\_}{\otimes }\underset{\_}{u}*\\ I\underset{\_}{\otimes }\sout{u}\\ W^{-}\underset{\_}{\otimes }\sout{d}\end{array}\right)=\left(\begin{array}{c}\underset{\_}{d}*\\ \sout{u}\\ W^{-}\underset{\_}{\otimes }\sout{d}\end{array}\right)\\ \to \left\{\begin{array}{c}{W}^{-}\underset{\_}{\otimes }\left[\left(\underset{\_}{d}*\underset{\_}{\otimes }\sout{d}\right)\underset{\_}{\otimes }\sout{u}\right]\\ W^{-}\underset{\_}{\otimes }\left[\underset{\_}{d}*\underset{\_}{\otimes }\left(\sout{d}\underset{\_}{\otimes }\sout{u}\right)\right]\end{array}\right\}=\left\{\begin{array}{c}{W}^{-}\underset{\_}{\otimes }\left[\gamma \oplus \sout{u}\right]\\ W^{-}\underset{\_}{\otimes }c\end{array}\right\}\Rightarrow \left\{\begin{array}{c}{W}^{-}+{\sout{u}}_{\gamma }\\ W^{-}+c\end{array}\right\}\end{array}$ (9)

There are so two possibilities:

$\begin{array}{l}\left[\left\{W\right\}\underset{\_}{\otimes }s\to \left(c+W^{-}\right),\left\{W\right\}\underset{\_}{\otimes }\underset{\_}{s}\to \left(\underset{\_}{c}+W^{+}\right)\right]\\ \left[\left\{W\right\}\underset{\_}{\otimes }s\to \left(u+W^{-}\right),\left\{W\right\}\underset{\_}{\otimes }\underset{\_}{s}\to {\left(\underset{\_}{u}+W\right)}^{+}\right]\end{array}$ (10)

We obtained the same result as the interaction theory applied to SM: there is a physical equivalence between a W-bosons’ lattice and Feynman diagrams with bosons. An ongoing study would demonstrate that by applying a W-boson lattice the weak interaction would appear “renormalizable”. This would be a remarkable achievement. By using lattices, the descriptive perspective of the particle’s changes: we have a new the descriptive paradigm of the particles and their interactions. In place of the gluons, we will put the quarks-gluons lattice u,d_γg = [(u, u)_γ (d, d)_γg]_γg and in place of the reactions (s → W + u) we will put (s + W). From the structure equation we could obtain the decays of the Kaons.

We analyse the neutral K and its weak decays using lattices I_g=u, d, W.

See Figure 30 and Figure 32 , the lattice W interpenetrating the quarks (u*, d) and lattice W interpenetrating the quarks (u, d). In these cases, more W-bosons can contain the quark pairs [(u, u), (d, d)], see in the Figure 30 the bosons [(W)_ICLG, (W)_ABCD, (W)_IAMGN,]. This determines the following equivalences: [W (q_i, q_i, q_i)]. We can have also:

$\boxed{\begin{array}{c}{K}^0\left(\underset{\_}{s},d\right)=\left(\begin{array}{c}\underset{\_}{s}\\ d\end{array}\right)\underset{\_}{\otimes }\left[{\left(\begin{array}{c}\left\{W\right\}\\ \left\{W\right\}\end{array}\right)}_{\left\{W\right\}}\right]=\left(\begin{array}{c}\underset{\_}{s}\underset{\_}{\otimes }\left\{W\right\}\\ d\underset{\_}{\otimes }\left\{W\right\}\end{array}\right)=\left(\begin{array}{c}\underset{\_}{s}\underset{\_}{\otimes }\left\{W^{+}\underset{\_}{\otimes }{W}^{-}\right\}\\ d\underset{\_}{\otimes }\left\{W^{+}\underset{\_}{\otimes }{W}^{-}\right\}\end{array}\right)\\ =\left(\begin{array}{c}\underset{\_}{u}\underset{\_}{\otimes }\left\{W^{+}\right\}\\ u\underset{\_}{\otimes }\left\{W^{-}\right\}\end{array}\right)=\left(\begin{array}{c}\underset{\_}{u}\underset{\_}{\otimes }\left\{W^{+}\right\}\\ u\underset{\_}{\otimes }\left\{W^{-}\right\}\end{array}\right)=\left(\begin{array}{c}\underset{\_}{u}\\ I_g\\ I_g\\ u\end{array}\right)\underset{\_}{\otimes }\left(\begin{array}{c}\left\{W^{+}\right\}\\ W^{-}\\ {\sout{W}}^{+}\\ \left\{W^{-}\right\}\end{array}\right)\\ =\left(\begin{array}{c}\underset{\_}{u}\\ I_g\\ I_g\\ u\end{array}\right)\underset{\_}{\otimes }\left(\begin{array}{c}{\left({\sout{u}}_1\underset{\_}{\otimes }{\sout{u}}_2\underset{\_}{\otimes }{\sout{u}}_3\right)}_{\left(W^{+},W^{+}\right)}\underset{\_}{\otimes }{W}^{-}\\ {\left(d_1\underset{\_}{\otimes }{d}_2\underset{\_}{\otimes }{d}_3\right)}_{W^{-}}\\ {\left({\underset{\_}{d}}_1\underset{\_}{\otimes }{\underset{\_}{d}}_2\underset{\_}{\otimes }{\underset{\_}{d}}_3\right)}_{W^{+}}\\ {\left({\underset{\_}{u}}_1\underset{\_}{\otimes }{\underset{\_}{u}}_2\underset{\_}{\otimes }{\underset{\_}{u}}_3\right)}_{\left(W^{-},W^{-}\right)}\underset{\_}{\otimes }{W}^{+}\end{array}\right)\end{array}}$ (11)

Here, there are two possibilities: if u-quark annihilates u₁-quark then we will have two neutral pions, while in the opposite case we will have three neutral pions. We consider the first case:

(12)

The second case is:

(13)

The coupling between (K, W) determines the weak decays of K-meson and the oscillation between two states [_(π,π), _(π,π,π)], see the literature [28] . Note that these decays make it clear why the decay into three pions is longer than that into two pions. The literature has considered this aspect as two possible states of the neutral Kaon (K_Short), K_Longe) that is τ_dec(3π⁰) >τ_dec(3π⁰). Using the two lattices [W, u,d], we can demonstrate transformation K⁰(s, d) K⁰(s, d):

(14)

The phenomenology of the Kaons validates so the Geometric Model and the existence of intermediary lattices.