The general rank-2 linear approximated field equation is (2-1) [4] .

∂ λ ∂ λ h μ ν + a ∂ μ ∂ ν h + b ( ∂ λ ∂ ν h μ λ + ∂ λ ∂ μ h ν λ ) + c η μ ν ∂ λ ∂ λ h + d η μ ν ∂ λ ∂ σ h λ σ   + e h μ ν + f η μ ν h = − k T μ ν (2-1)

where a, b, c, d, e, and f are unknown coefficients and k = 8 π G / c 4 .

Here, through Nöether’s theorem [2] [3] , it was found that the linear approximated gravitational equation has two gauges as follows. [1]

h μ ν → h μ ν + ∂ ν Λ μ + ∂ μ Λ ν + η ′ μ ν Λ (2-2)

where Λ is an arbitrary scalar function and η ′ μ ν is a metric tensor with a trace value of 2 [1] . We have named ∂ ν Λ μ + ∂ μ Λ ν as a “vector gauge” and η ′ μ ν Λ as a “scalar gauge”. With a vector gauge, it was shown that gravity and electromagnetic fields have been unified [5] [6] . By finding a new gauge, the scalar gauge [1] , we have acquired a cornerstone to step to a new potential. Nöether’s theorem plays a role in the process of finding all coefficients in (2-1) [1] .

In the first equation, both gauges (vector and scalar) satisfy the symmetries, which also include conservation law, all coefficients are determined as follows. [1]

a = 1 b = − 1 c = − 1 d = 1 e = f = 0 (2-3)

The field equation with these coefficients corresponds to the equation of gravity. Here, when describing gravitational waves (gravitational particles), the following equation can be used [4] (pp. 241-294).

Φ μ ν = h μ ν − 1 2 η μ ν h (2-4)

Using the above equation, the linear approximated gravitational equation can be expressed as follows [4] (pp. 241-294).

∂ λ ∂ λ Φ μ ν = − κ T μ ν (2-5)

∂ μ Φ μ ν = 0 (2-6)

Equation (2-4) is called the Hilbert gauge condition, and it allows for the formation of a wave equation and description of gravitational waves. The reason for focusing only on the linear equation is that we want to deal with local gauges that are applicable in linear space-time. On the other hand, global gauges apply to curved space-time [1] .

The second equation that we can do further is to adopt the symmetry of only a scalar gauge while keeping the conservation law. We will examine what happens to the newly created potential and its meaning in the following chapters.

As mentioned in the previous chapter, gravity is a case where both gauges are satisfied. This chapter addresses the case where a scalar gauge maintains symmetry but a vector gauge does not require symmetry. We summarize this in Table 1 [1] .

When Case (1) and Case (2) were both satisfied (blue box), then Case (3) was automatically satisfied [4] . Now let us consider the cases where Case (2) and Case (3) are satisfied (red box). This is a case where the conservation law (Case (3)) is satisfied and only one of the two symmetries (Case (2)) is satisfied. In this case, the values of the coefficients (a, b, c, d, e, and f) can be fixed through the equations in Table 1 [1] (green box).

a = 1 b = − 1 c = − 1 d = 1 e + 2 f = 0 (2-7)

Except for e and f, all coefficients are equal to (2-3). When the coefficients determined in (2-7) are put into Equation (2-1), we have:

∂ λ ∂ λ h μ ν + ∂ μ ∂ ν h − ( ∂ λ ∂ ν h μ λ + ∂ λ ∂ μ h ν λ ) − η ′ μ ν ∂ λ ∂ λ h + η ′ μ ν ∂ λ ∂ σ h λ σ   + e ( h μ ν − 1 2 η ′ μ ν h ) = − κ T μ ν (2-8)

According to the symmetry of h μ ν → h μ ν + η ′ μ ν Λ [1] , so all η μ ν are modified to η ′ μ ν in (2-8) [1] . When we use the conservation law, we have:

∂ μ ( ∂ λ ∂ λ h μ ν + ∂ μ ∂ ν h − ( ∂ λ ∂ ν h μ λ + ∂ λ ∂ μ h ν λ ) − η ′ μ ν ∂ λ ∂ λ h + η ′ μ ν ∂ λ ∂ σ h λ σ )   + e ∂ μ ( h μ ν − 1 2 η ′ μ ν h ) = − κ ∂ μ T μ ν = 0 (2-9)

The blue terms in Equation (2-9) are self-terminated and can also be seen in the conservation law for the original linear gravity equation. Then, the last black terms on the left side in Equation (2-9) satisfy the following condition:

∂ μ ( h μ ν − 1 2 η ′ μ ν h ) = 0   or   ∂ ν ( h μ ν − 1 2 η ′ μ ν h ) = 0 (2-10)

Although Equation (2-10) is similar to the existing Hilbert gauge condition [4] , it is not the same. Be careful that the metric tensor in Equation (2-10) is not the conventional η μ ν , but a metric tensor η ′ μ ν with a trace value of 2 [1] . We may call it a “new gauge conditional expression” or “Hilbert-like gauge condition”. When the trace of η ′ μ ν is 2, there can be three cases, as shown in Table 2, depicted in four dimensions of space-time.

Because the trace of all the tensors in Table 2 is 2, whose components are one space element and the other time element, the field looks to operate only in one space dimension. This one-dimensional field can explain the force line that binds quarks together, which will be interpreted mathematically in a later

chapter. The meaning of Equation (2-9) is that h μ ν − 1 2 η ′ μ ν h makes the new

conserved physical quantity. h μ ν − 1 2 η ′ μ ν h can be one of the above three metric

tensors in Table 2. When (2-10) is applied to (2-8), we have (2-11).

∂ ν h μ ν = 1 2 ∂ ν h ∂ λ ∂ λ h μ ν + ∂ μ ∂ ν h − ( 1 2 ∂ μ ∂ ν h + 1 2 ∂ μ ∂ ν h ) − η ′ μ ν ∂ λ ∂ λ h   + η ′ μ ν 1 2 ∂ σ ∂ σ h + e ( h μ ν − 1 2 η ′ μ ν h ) = − κ T μ ν (2-11)

The red terms are erased, and the blue terms are left. Rearranging this, we have (2-12).

∂ λ ∂ λ h μ ν − 1 2 η ′ μ ν ∂ σ ∂ σ h + e ( h μ ν − 1 2 η ′ μ ν h ) = − κ T μ ν (2-12)

or

∂ λ ∂ λ ( h μ ν − 1 2 η ′ μ ν h ) + e ( h μ ν − 1 2 η ′ μ ν h ) = − κ T μ ν (2-13)

(2-13) is a new field equation different from the gravitational equation, and the 2nd term, including the coefficient e in (2-13), is the new term, which can be explained as a mass term. We note that the mass term is obtained by breaking the symmetry of case (1) in Table 1. We emphasize that the conservation law is satisfied even with this mass because case (3) in Table 1 is still valid. We attempt to find the solution of this new field equation in the next chapter.

From the above equations, we can find the exact mass of the particle. When we put Φ μ ν = C μ ν exp ( i ( ∓ E t ± P r ) ) into Φ μ ν = h μ ν − 1 2 η ′ μ ν h (3-1), we have

( 0 − 0 + m 2 ) Φ μ ν = 0 (4-1)

Therefore, we have

m = 0 (4-2)

According to Equation (3-12), the speed of this particle is that of light, as shown in (4-3). Thus, the rest mass m = 0 is a correct result.

d r d t = E P = P c P = c (4-3)

In this chapter, we will determine the spin value of the new particle. The spin of the gravitational particle (graviton) is 2. We can also infer the spin of the new particle by reasoning from the spin of the graviton. Let’s consider a gravitational wave traveling along the Z-axis [4] (pp. 241-294) [9] .

h μ ν = ( − 1 0 0 0 0 h + h x 0 0 − h x h − 0 0 0 0 1 ) (4-4)

The time elements of the metric perturbation h 00 and h 0 i in (4-4) remain invariant when rotated about the Z-axis by any angle ϕ , while the spatial element h i j transforms as follows:

h i j = R ( ϕ ) i a h a b R ( ϕ ) j b (4-5)

where R ( ϕ ) i j is a rotation matrix given by:

R ( ϕ ) i j = ( cos ϕ sin ϕ 0 − sin ϕ cos ϕ 0 0 0 1 ) (4-6)

By working through the algebra and using various trigonometric identities, we can show that h i j is invariant under a rotation by π , which is a characteristic feature of a spin-2 particle. In general, after a rotation, we find that the components of h i j are multiplied by functions of 2 ϕ , another characteristic sign of a spin-2 particle. However, in the case of h μ ν 3 (Z-axis) when the trace of the metric tensor is 2, performing the calculation as in (4-5) results in invariance in terms of ϕ , while in Equation (3-8), it demonstrates symmetry under 2 π rotations. This implies the spin becoming 1, akin to the characteristics of an electromagnetic wave. Table 4 shows this explanation.

If we apply ∂ μ to both sides in (3-2), we get:

∂ λ ∂ λ ∂ μ Φ μ ν + e ∂ μ Φ μ ν = 0 (4-7)

According to (2-10), since ∂ μ Φ μ ν = 0 , both terms on the left side are 0. The first term becomes:

∂ λ ∂ λ ∂ μ Φ μ ν = 0 (4-8)

This always holds since ∂ μ Φ μ ν = 0 . According to (3-1),

∂ λ ∂ λ ∂ μ ( h μ ν − 1 2 η ′ μ ν h ) = 0 (4-9)

The two terms on the left side of (4-9) can be defined as j ν , respectively, as shown in (4-10):

∂ λ ∂ λ ∂ μ h μ ν = ∂ λ ∂ λ ∂ μ ( 1 2 η ′ μ ν h ) ≡ j ν (4-10)

The difference of the two terms becomes 0 because they are equal, which implies the pair annihilation of particle and antiparticle.

Be careful that ∂ λ ∂ λ is not four-dimensional but two-dimensional, and it is not strictly an electromagnetic equation. However, it is analogous to the four-vector potential of an electromagnetic field. Let us define A ν = ( Φ , A ) , which is similar to the electromagnetic 4-potential [10] , and we adopt this formalism in 2-dimensions (4-11), (4-12), and (4-13):

□ A = μ 0 J (4-11)

□ Φ = − ρ ε 0 (4-12)

∂ λ ∂ λ A ν =   □ ( Φ , A ) = ( − ρ ε 0 , μ 0 J ) = 0 (4-13)

where □   = ∂ λ ∂ λ = 1 c 2 ∂ 2 ∂ t 2 − ∇ 2 .

Here, □ is the 2-dimensional d’Alembertian operator. Comparing Equation (4-8) with (4-13), it is as if there is no charge source.

If we know Φ μ ν as in Equation (3-15), we can obtain h μ ν in the linearly approximated space-time, and we can also obtain the metric tensor g μ ν as given in Equation (4-14) [4] (pp. 176-179).

g μ ν = η μ ν + k h μ ν (4-14)

where k is an infinitesimally small value. Assuming h μ ν as in Equation (4-15), we can obtain Φ μ ν as shown in Equation (3-15).

h μ ν = ( a 0 0 b ) (4-15)

where a and b are certain constants. According to Equation (3-1),

Φ μ ν = ( a 0 0 b ) − 1 2 ( − 1 0 0 1 ) ( − a + b ) (4-16)

= ( a + b 2 0 0 a + b 2 ) = 1 k ( E 0 0 E ) (4-17)

The parameter α in Equation (3-15) and (3-16) is 1/k [4] (pp. 176-179). That is, (3-15) and (4-17) are the same because ( a + b ) / 2 is constant. Therefore, a choice such as Equation (4-15) is appropriate. Using (4-15), we can obtain (4-18).

g μ ν = ( − 1 0 0 1 ) + k ( a 0 0 b ) = ( − 1 + k a 0 0 1 + k b ) (4-18)

where k ≪ 1 . g μ ν is the metric tensor for the potential well. Although we have used a 2-dimensional representation (2 × 2 matrix) in (4-15) to (4-18) for convenience, they are part of the 4-dimensional representation and can have 3 variants in a 4 × 4 representation, as shown in Table 5.

In the previous chapters, we have identified the mass, charge, and spin of the new particle, which suggests that it could be either a gluon or a photon in the standard model, as depicted in Figure 2 [11] .

Figure 3 shows the force lines of gluons (up) and photons (down) [12] .

Among the two candidates, as shown in Figure 3, it is reasonable to interpret the new particle as a gluon, rather than a photon with radial force, since the trace of the tensor is 2, implying that the force acting is one-dimensional spatial force, as depicted in the upper part of Figure 3.

We have obtained a local gauge using linear approximated gravitational equations [1] and differentiated the strong force from it. The differentiation can be justified by comparing it with cosmological phenomena, which suggest that there was differentiation of force during the period of linear expansion after cosmic inflation following the Big Bang. Nature is made up of four-dimensional space-time, and there is an action to create three-dimensional symmetry spatially. However, as the trace value remains at 2, a new pattern emerges. In the next chapter, we will give a qualitative interpretation of this, followed by a quantitative one.

In this section, we qualitatively discuss how and why three quarks can be stably gathered when the new potential is considered. Figure 4 and Figure 5 illustrate why the case with three quarks is stable. According to Table 5, force lines can be created from any angle in 3D, allowing many quarks to stick together. However, we observe that only three quarks always come together. The reason for this is related to the default space-time being trace 2.

In Figure 4 (left), there are initially 3 quarks. When 2 quarks are far enough to reach the limit of confinement, a force line (red line) is generated. It should be noted that, according to (9-1), only one force line can exist to satisfy the condition of trace 2. Even though the quarks connected by the red force line become closer, no quarks move farther apart than the initial distance. In Figure 4 (right), a becomes a', and b becomes b', but the distance between each quark becomes closer.

Now, let’s consider the case where four quarks are inside and one tries to go out. It shows that the outgoing quark cannot be captured because only one force line is allowed. Figure 5 shows the transition from 4 quarks to 3. We assume that initially, 4 quarks exist within the range of confinement. The force line is a single strand, so the 2nd quark cannot simultaneously attract the 1st and 3rd quarks. Consequently, if the 2nd quark is pulling the 3rd quark (Figure 4, left), it cannot pull the 1st quark. As the 2nd and 3rd quarks move into a stable position (Figure 4, right), the distance between the 1st and 2nd quarks is already beyond the confinement. So it has no force between them. In this way, the 1st quark becomes disconnected from the rest of the three-quark ensemble. New force lines can only be generated according to the distances between pairs of the 2nd, 3rd, and 4th quarks. The 1st quark might disappear after meeting an anti-quark or might form a new nucleon if it acquires sufficient energy. Thus, when quarks are collected in groups of 4, 5, or more, they eventually fall off and result in 3 quarks.

Previously, this phenomenon was explained using the color charge model to explain why the three quarks gather, as shown in Figure 6 (top). Each quark has one charge among R, G, and B, and when they come together, they neutralize and become stable [13] [14] . This is not to say that the existing model is incorrect, but both the existing model and our model independently explain the same phenomenon. According to the existing theory, gluons themselves have two color charges. They are composed of particles and antiparticles, where one has a color among R, G, B, and the other has a color charge with its complementary color. In our case, as shown in Figure 6 (bottom) and according to (2-13),

Φ μ ν = h μ ν − 1 2 η ′ μ ν h , the two potentials are always bundled together. From

(4-13), the charges of the two are opposite, and the mass is the same, indicating a particle-antiparticle relationship. This result aligns with the existing analysis.

Now, we would like to quantitatively describe the reason for the gathering of three quarks. The existing Gluon eight-fold way is shown on the left of Figure 7 [15] . Similarly, you can see the eight-fold way claimed in this paper on the right.

In Chapter 10, we said that h μ ν and − 1 2 η ′ μ ν h are particle and antiparticle relationships. Therefore, the logic shown in Figure 7 can be made.

The potential of each position at the eight-fold way in Figure 7 is as follows. h μ ν and − 1 2 η ′ μ ν h can be expressed for each component through Table 2 and Table 4.

Φ 12 μ ν = h 1 μ ν − 1 2 η ′ 2 μ ν h = ( a 0 0 0 0 b 0 0 0 0 0 0 0 0 0 0 ) − 1 2 ( a − b 0 0 0 0 0 0 0 0 0 − a + b 0 0 0 0 0 ) = ( a + b 2 0 0 0 0 b 0 0 0 0 a − b 2 0 0 0 0 0 ) (5-1)

Φ 13 μ ν = h 1 μ ν − 1 2 η ′ 3 μ ν h = ( a 0 0 0 0 b 0 0 0 0 0 0 0 0 0 0 ) − 1 2 ( a − b 0 0 0 0 0 0 0 0 0 0 0 0 0 0 − a + b ) = ( a + b 2 0 0 0 0 b 0 0 0 0 0 0 0 0 0 a − b 2 ) (5-2)

Φ 21 μ ν = h 2 μ ν − 1 2 η ′ 1 μ λ h = ( a 0 0 0 0 0 0 0 0 0 b 0 0 0 0 0 ) − 1 2 ( a − b 0 0 0 0 − a + b 0 0 0 0 0 0 0 0 0 0 ) = ( a + b 2 0 0 0 0 a − b 2 0 0 0 0 b 0 b 0 0 0 ) (5-3)

Φ 23 μ ν = h 2 μ ν − 1 2 η ′ 3 μ ν h = ( a 0 0 0 0 0 0 0 0 0 b 0 0 0 0 0 ) − 1 2 ( a − b 0 0 0 0 0 0 0 0 0 0 0 0 0 0 − a + b ) = ( a + b 2 0 0 0 0 0 0 0 0 0 b 0 b 0 0 a − b 2 ) (5-4)

Φ 31 μ ν = h 3 μ ν − 1 2 η ′ 1 μ ν h = ( a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b ) − 1 2 ( a − b 0 0 0 0 − a + b 0 0 0 0 0 0 0 0 0 0 ) = ( a + b 2 0 0 0 0 a − b 2 0 0 0 0 0 0 0 0 0 b ) (5-5)

Φ 32 μ ν = h 3 μ ν − 1 2 η ′ 2 μ ν h = ( a 0 0 0 0 0 0 0 0 0 0 0 0 0 0 b ) − 1 2 ( a − b 0 0 0 0 0 0 0 0 0 − a + b 0 0 0 0 0 ) = ( a + b 2 0 0 0 0 0 0 0 0 0 a − b 2 0 0 0 0 b ) (5-6)

Φ 11 μ ν = ( a 0 0 0 0 b 0 0 0 0 0 0 0 0 0 0 ) − 1 2 ( a − b 0 0 0 0 − a + b 0 0 0 0 0 0 0 0 0 0 ) = ( a + b 2 0 0 0 0 a + b 2 0 0 0 0 0 0 0 0 0 0 ) (5-7)

Φ 22 μ ν = ( a 0 0 0 0 0 0 0 0 0 b 0 0 0 0 0 ) − 1 2 ( a − b 0 0 0 0 0 0 0 0 0 − a + b 0 0 0 0 0 ) = ( a + b 2 0 0 0 0 0 0 0 0 0 a + b 2 0 0 0 0 0 ) (5-8)

Φ 33 μ ν = ( a 0 0 0 0 0 0 0 0 0 b 0 0 0 0 0 ) − 1 2 ( a − b 0 0 0 0 0 0 0 0 0 0 0 0 0 0 − a + b ) = ( a + b 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 a + b 2 ) (5-9)

Φ c 1 μ ν = Φ 11 μ ν − Φ 22 μ ν = ( 0 0 0 0 0 a + b 0 0 0 0 − a − b 0 0 0 0 0 ) (5-10)

Φ c 2 μ ν = 1 3 ( Φ 11 μ ν + Φ 22 μ ν − 2 Φ 33 μ ν ) = 1 3 ( 0 0 0 0 0 a + b 2 0 0 0 0 a + b 2 0 0 0 0 − a − b ) (5-11)

At the eight-fold way, the outer 6 ( Φ 12 μ ν , Φ 13 μ ν , Φ 21 μ ν , Φ 23 μ ν , Φ 31 μ ν , Φ 32 μ ν ) have a diagonal sum of a + b , and the middle two ( Φ c 1 μ ν , Φ c 2 μ ν ) are 0. The six outer potentials represent the attraction between quarks when they move away from each other by a certain distance. When there is nothing to attract, the two potentials in the middle hide the force but retain energy by setting the sum of the diagonals to 0.