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Following the Projection-Gell-Mann Standard Model obtained from 5D homogeneous space-time, we extended to the definition of Strange and Charm quarks from the SU(3) fractional charged generators superposition representations, and calculated all the well-known baryon masses. The results indicate how the Orthogonality between the SU(2) × L and SU(3) × L manifolds is broken in the hadrons, such that the superposition of the time projection P
_{o}, and the conformal space projection P
_{1} are established, the basic requirement for General Relativity and the establishment of the Riemannian curvature in the presence of mass.

In the Standard Model, baryons are made of products from 3 quarks, selected from 3 pairs of quarks. Each pair is quarks with (2/3)e, and (−1/3)e fractional charges. These pairs are named (u, d) up and down quarks, (t, b) top and bottom quarks and (c, s) charm and strange quarks. According to the Projection-Gell-Mann Standard Model, there are however, only 3 fractional charged quarks generators, namely (u, t, d), where both u and t have charge of (2/3)e, while d has charge (−1/3)e, plus their antiparticles. These quarks form the SU(3) generators. (See Ref. [

Based on the above we define the proton as (uud) and the neutron as (udd). And since these two baryons only contain u and d quarks, their gluon potential is also generated by u and d intermediate quark currents. Hence the gluon potential U is generated by (uud) + (udd) + (ddd). As the gluon potential is a product of the 3 vector potentials generated from charge currents, the change of u and d by c, t and s, b will only modify through the perturbation correction on the potential energy which then depends on the mass change, which must be small. None the less no matter how small, it breaks the SU(3) symmetry. Lastly, this perturbative change of U, due to s, c, etc. will not change the baryon SU(3) representations as given by (1) × (8) × (8) × (10). Thus the missing heavy baryon octet contention [

In the recent papers [_{0} already determined [^{0} should be replaced by π^{+}, in order the gauge loop radius is reduced by 1/2, needed for the binding between p and π^{+} to be 33.8 GeV.

It is one of our goals in this paper to investigate and calculate the rest of the known baryon masses [

As mentioned in the introduction the replacement of d quark by s quark in the baryon signature will result in the alpha invariant requirement of the conformal projection, so that the orthogonality between SU(2) × L and SU(3) × L is broken. Let us first review the proton, neutron states, where the Orthogonality holds. The signatures of proton is (uud), while that for the neutron is (udd), giving us the gluon potential proportional to:

where v is the equal valued intermediate quark velocity.

Since the proton and neutron are only given by u and d quarks, there are no projection-gauge violations, and thus no perturbative correction to the vector potentials generated. Based on the determined gauge loop r_{0} from the mesons gluon potential, U_{0} is calculated equal to 934.6 MeV. The terms in Equation (2.1) can be divided into the sum of 21 equal values E_{o}, each then is of 44.5 MeV [_{0} roughly equal to 178 MeV, which is clearly in agreement with data. If no perturbative corrections are introduced due to u, d replacement by c, s, b, then the fine mass splitting are within each gluon level and completely due to the total inter quark energies from each baryon. As shown in Ref. [^{0} and Σ^{0}, these particles signatures are given as (udd) and (uds) respectively, and they also are within the same general gluon potential of 1112.6 MeV. Yet the Σ^{0} mass is heavier than that of Λ^{0} by nearly 80 MeV. In fact, Σ^{0} decays into Λ^{0} and emits photons. This is a clear indication, that their gluon potentials composed of vector potential fields only must be different due to s quark replacement for d quark in the intermediate states that generate the potential and must decay by photon emission.

It also means strangeness is not a conserved quantum. To show this quantitatively, let us start with the proton U(0). We obtain the next gluon level:

The total energy T.E. for Λ^{0} (udd),

where the total mass m_{c} = (4/3)m, and the reduced mass^{2}/r is that of the revolving mass m', and for this semion limit γ = 0.18 [_{c}. And via orbital quantization we get γ = 0.486. Leading to T.E. = 86 MeV.

While the Σ^{0}, is (uds), with m_{c} = 2m, and

Should U(d) represent the gluon potential for Λ^{0}, then it's mass is given by (see Ch. 8 of Ref. [

Although the T.E. for Σ^{0} is larger than that of Λ^{0}, it is insufficient to produce a mass difference of nearly 80 MeV. In order that Σ^{0} is much heavier, U(d) must be replaced by U(s). By U(s), we mean one of the d quark in the intermediate states must be replaced by the s quark. Since all terms in U(d) involves at least one d quark. That means the correction due to such a replacement is linearly scaled.

where f > 1, and is due to the perturbative energy correction caused by breaking of the α constant constrain required by P_{1}, to obtain the SU(3) generators.

Should we assume

The factor (1/3) comes from the ratio of d mass to s mass. Using the Semion limit γ = 0.18, while the factor ^{3} dependence for the U correction, such that the effective corrective

term m’ γ, where m’ represents the mass ratio modified as a result of the change in quark rest energy due to the replacement of the d quark by the s quark, remains and represents the spinor particle m’ reduced effective binding energy contribution. With the above, we get f = 1.063 and hence

Finally, we get

From this numerical calculation, we can conclude, that indeed the s quark when replacing d in the intermediate quark states will cause a SU(3) gluon potential symmetry breaking, and is likely also observed in other hadron masses.

With the gluon potential U(s) as given by Equation (2.7), we now proceed to calculate the T.E. for Σ^{+} = (uus). It is obvious that the net Coulomb potential is zero, and according to the net zero current rule for the pairs, in the trinity structure as discussed in Refs. [

Thus

Similarly, Σ^{−} = (dds), has a net repulsive Coulomb potential of (1/3)e^{2}/r. And according to the net pair zero current rule for the trinity structure, it is static. Thus

Here we assumed that the value of the repulsive Coulomb potential can be approximated the same as that of the attractive potential giving the Chern-Simons Ground State [

This quantity is less than the T.E. of Σ^{0}, which implies if Σ^{−} is heavier, its U value must be larger. As we have shown the increase in U due to s replacement of d is a perturbative correction. The fact that Σ^{−} signature is (dds), means s has 3! choices of replacing any one of the d quark. Making f modified to

The factor (1/3)^{3} comes from the potential strength arising from the product of the (1/3)e charges giving U'(s) = 1188.9 MeV, and

Moreover this extra U correction implies the SU(3) symmetry breaking of U can be a continuous probabilistic function of the alpha change in quark mass due to superposition. Our question is then: Is our fitting a purely numerical game or is the principle demonstrated in other baryons? In the Σ mass analysis, we were led by T.E. of each state being different because of their signatures. In the baryon decuplet, we have 4 Δ; (ddd), (ddu), (duu) and (uuu). None has s or c quark replacement, thus their T.E. are well defined and independent of the definition chosen for s and c. Irrespective of the U value, each T.E. can be computed with accuracy, and are of course different. Yet data shows their masses are close to the same. This can happen if U changes just so much as to cancel the change in T.E. in support of the U adjustment of Σ^{−} from Σ^{0}, while s, c substitutions in U are present, at least in small fractions.

The T.E. values of any of the Δ baryons are within 80 MeV and would only affect the resultant mass by^{−}.

Another such an example of the U adjustment due to s replacement of d, is shown by Ξ^{−}. The Ξ particles are given by (uss) and (dss). Here we have the signature replacement of two d quarks with two s quarks. Obviously, the U potential expressed in u,d,s, must be modified by a factor f, as follows:

for f = 1.063 as given by Equation (2.6).

Again here the term 1, in the bracket in Equation (3.5) comes from the state (ddd), and thus can also be subjected to s replacement.

The net Coulomb potential for the (uss) Ξ^{0}, is −(1/3)e^{2}/r; while the Coulomb potential for (dss) Ξ^{−}, is repulsive, and equal to (1/3)e^{2}/r. Considering Ξ^{0}, we have m_{c} = 2.67m = 90.64 MeV and m* = 1.3m = 44.2 MeV. Thus its

Taking U(Ξ^{0}) as given by Equation (3.5), we obtain M(Ξ^{0}) = 1319 MeV, which is in agreement with data [

However the T.E. for Ξ^{−} gives 139.8 MeV if we assume the same rule for the repulsive potential e^{2}/r as equal to m_{c}. This T.E. value is less than the T.E. of Ξ^{0}, and therefore would need for further f correction, since Ξ^{−} is heavier. The fact that the signature of Ξ^{−} is (dss) implies that the last term in U(Ξ^{−}) needs to be weighted higher, giving.

The last term in the bracket comes from the extra probability of replacing 2 d in d^{3} by s^{2}.

Thus

This serves as another example of U not being the same for both states, even though we have s in the U generation.

Lastly, we have Ω^{−}, which is (sss). Hence according to our treatment of U(Ξ),

with m_{c} = 3m, m* = (3/2)m and

Finally

Among the known baryons we have only one with c charm quark replacement for the u quark, which is Λ^{+}, (ucd). U(c) is then generated by (ucd + 2cdd + ddd). The c quark is given by (s + d)*, thus its mass is (4/3)m. Like the proton, the Coulomb net V = 0, hence

as T.E. value here compared to U(c) is unimportant.

The Λ^{+} mass is 2281 MeV meaning U(c) is close to 2275 MeV. This means U(c) must be scaled by f", such that^{0} which is 1112.6 MeV. While U(ddd) is 44.5 MeV.

Suppose f" = 1 + f the increased factor 1 is because c actually is composed of 2 (1/3)e charge as compared to s, then

Combining with T.E., we get

This mass value is low compared to data. Thus there must be an error coming likely from f" approximation. By calculating backward, we get

This increased value actually came from the mass ratio between u and c, which is

Thus

This number then give the Λ^{+} mass equal to 2284 MeV, a little high, but still in line with data.

On the whole the Projection-Standard-Model agrees very well with all data [

This paper is not solely intended for showing how we can calculate the baryon masses from the Projection-Gell- Mann Standard Model. Although one might feel that the data fits could be artificially fixed. In reality, it is quite difficult, as the model started with the fixed U(0) proton gluon potential of 934.6 MeV plus the bare quark mass m of 34 MeV. For example, if we replace the signature of Λ^{0} from (udd) to (uds), there is no way to obtain its mass as compared to data to within 5 MeV which is beyond any experimental error bar. In fact it is not just difficult, but when they fit, the value is nearly precise to the decimal place of masses of baryons over 1000 MeV [

It was through these same procedures that we managed to calculate masses for all the known baryons shown in Section 3 above. In the process, we found that the 3 pairs of quarks, from the Standard Model, namely (u, d); (t, b) and (c, s), only (u, d, t) were the basic SU(3) generators as shown in Ref. [_{o} + P_{1}, and the realization of a Riemannian space-time, where the curvature is dependent on the presence of mass, such that Perelman mappings [

Kai-WaiWong,Gisela A. M.Dreschhoff,Högne J. N.Jungner, (2015) Breaking SU(3) Symmetry and Baryon Masses. Journal of Modern Physics,06,1492-1497. doi: 10.4236/jmp.2015.611153