Dimensional Constriction and Hadronic Mass — Part 10

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43. Factor-of-two correction to the reciprocal zero-point trace

The earlier interpretation of the reciprocal zero-point trace contained a hidden factor-of-two ambiguity that becomes clear once the coordinate harmonic kernel is quantized directly.

For a common oscillator gap

$$\varepsilon=\hbar\omega,$$

the zero-point energies are

$$E_{0,4}=\frac42\varepsilon=2\varepsilon,$$

and

$$E_{0,3}=\frac32\varepsilon.$$

Their sum is

$$E_{0,4}+E_{0,3} =\frac72\varepsilon.$$

The dimensionless ratio

$$\frac{E_{0,4}+E_{0,3}}{E_{0,3}} =\frac73$$

is correct, but it compares the total reciprocal zero-point energy to the entire three-dimensional zero-point energy $3\varepsilon/2$. It does not say that the oscillator gap itself is multiplied by $7/3$.

Distributed over the three surviving directions, the reciprocal zero-point energy per direction is instead

$$\boxed{ \frac{E_{0,4}+E_{0,3}}{3} =\frac76\varepsilon. }$$

This is exactly the same factor obtained independently from the canonical inverse symmetric reciprocal normalization.

Therefore the consistent reciprocal energy factor is

$$\boxed{\frac76,}$$

not $7/3$, when applied to an oscillator energy quantum.

43.1 Direct harmonic quantization

The V2 oscillator coordinates are already written in canonical dimensionless form,

$$a =\frac1{\sqrt2}(X+iP), \qquad X=\frac{x}{\ell}, \qquad P=\frac{\ell p}{\hbar},$$

with

$$[X,P]=i.$$

A band-blind harmonic normal block with energy coefficient $\varepsilon_0$ is therefore

$$H_\perp =\frac{\varepsilon_0}{2}(X^2+P^2) =\varepsilon_0\left(a^\dagger a+\frac12\right).$$

Its level spacing is exactly

$$\boxed{\Delta E=\varepsilon_0.}$$

There is no additional universal factor of two from quantization.

Using the Schur-complement stiffness

$$g_{\rm con} =\frac{g_{34}^2}{g_{44}} =g_{33}$$

and the depth-three projected energy unit $\mu_3$ gives the base harmonic gap

$$\varepsilon_0=\mu_3g_{33}.$$

Canonical reciprocal normalization then gives

$$\varepsilon_{34} =\frac76\mu_3g_{33}.$$

Fixed-action $4\to3$ redistribution gives

$$\boxed{ \varepsilon_3 =\frac43\varepsilon_{34} =\frac{14}{9}\mu_3g_{33}. }$$

Numerically,

$$\boxed{ \varepsilon_3 =154.808228\ {\rm MeV}. }$$

This is exactly one half of

$$\frac{28}{9}\mu_3g_{33} =309.616456\ {\rm MeV}.$$

Hence the successful $309.6$ MeV formula requires an additional factor of two not supplied by

• canonical reciprocal normalization;
• direct harmonic quantization;
• the exact local cubic Schrödinger self-contact;
• the exact scalar-Soler nonlinear Dirac profile; or
• the finite-band Fubini--Study coherent-contact curvature.

44. Revised hard result

The mathematically supported constriction chain presently reaches

$$\boxed{ \Delta E_{\rm bulk}^{\rm derived\,candidate} =\frac{14}{9}\mu_3g_{33} \approx154.81\ {\rm MeV}, }$$

not the observed nucleon-scale value near $309.5$ MeV.

The empirical target is almost exactly twice this amount.

Therefore the remaining V2 problem is exceptionally sharp:

$$\boxed{ \text{find or rule out a second independent contribution equal to } \frac{14}{9}\mu_3g_{33}. }$$

Any such doubling must arise from a genuinely distinct degree of freedom, sector contribution, particle/antiparticle component, chiral block, reciprocal branch, or other operator multiplicity. It cannot be obtained by counting the same harmonic zero-point or reciprocal interaction twice.

45. No-go for obtaining the missing factor two from reciprocity or Dirac doubling

The remaining discrepancy is exactly a factor of two, so the obvious possibilities are two reciprocal variations, two chiral halves, or particle/antiparticle doubling. The written V2 action does not support any of these as an independent second copy of the same scalar confinement energy.

45.1 The scalar quartic variation already removes the combinatorial factor two

The leading scalar self-interaction is

$$\mathcal L_{\rm sc} =-\frac{\kappa_d}{2}J_d^2, \qquad J_d=\bar\Psi_d\Psi_d.$$

Varying with respect to $\bar\Psi_d$ gives

$$\frac{\partial\mathcal L_{\rm sc}}{\partial\bar\Psi_d} =-\frac{\kappa_d}{2}\,2J_d\Psi_d =-\kappa_dJ_d\Psi_d.$$

Thus the two identical factors in $J_d^2$ are already compensated by the conventional $1/2$ in the action.

There is therefore no additional factor of two available from saying that the self-contact acts in two reciprocal directions. Counting both variations of the same quadratic interaction would double-count one action term.

45.2 Cross-depth reciprocity is also one action term

The V2 cross-depth interaction is written as

$$S_{34}=g_{34}\int dt\,I_{34}.$$

Functional differentiation with respect to the depth-three and depth-four densities gives two equations of motion, but both are variations of this same interaction functional. They do not represent two additive copies of its energy.

Hence

$$\boxed{ \text{reciprocity does not provide the missing factor }2. }$$

45.3 Chiral halves are not two independent scalar energies

A Dirac spinor may decompose into two chiral halves, but the scalar bilinear

$$\bar\Psi\Psi$$

is a single Lorentz-scalar contraction of the full field. The V2 scalar contact normalization $\mu_d$ is obtained after the scalar spinor contraction $B_d$ has already been performed.

Consequently, multiplying the resulting scalar energy again by two merely because the underlying Dirac fibre has two chiral components would count spinor components already included in $B_d$.

Likewise, positive- and negative-frequency solutions of one Dirac operator are not simultaneous additive rest-energy contributions to one one-particle state.

Therefore neither chirality nor particle/antiparticle structure supplies a justified universal doubling.

46. Strongest present conclusion

After the explicit tests, the supported band-blind harmonic constriction scale is

$$\boxed{ \Delta E_3 =\frac{14}{9}\mu_3g_{33} =154.808228\ {\rm MeV}, }$$

subject to the scalar-channel normalization assumptions described above.

The observed nucleon bulk contribution per valence quark is approximately

$$309.5\ {\rm MeV},$$

which is almost exactly

$$2\times154.8\ {\rm MeV}.$$

But the factor two is not supplied by

• the scalar quartic variation;
• cross-depth reciprocity;
• Dirac chirality;
• particle/antiparticle doubling;
• canonical harmonic quantization;
• exact cubic self-localization;
• exact scalar-Soler localization; or
• finite-band projective coherent-contact curvature.

Thus a valid V2 derivation of the full nucleon bulk mass now requires a genuinely distinct second physical contribution, not a normalization reinterpretation of the same contact term.

One especially sharp possibility to test next is whether a confined valence quark carries two independent band-blind quadratic energies, one from coordinate constriction and one from its conjugate normal momentum/connection sector, each equal to

$$\frac{14}{9}\mu_3g_{33}.$$

That possibility is not established by the ordinary harmonic Hamiltonian, where coordinate and momentum terms are already the two halves of a single oscillator energy. It would require the V2 master geometry to supply a second independent normal operator rather than the canonical conjugate term of the first.

47. Baryon incidence closure candidate

The remaining factor of two has one non-double-counting geometric realization: pairwise incidence in a three-quark baryon.

A baryon with three valence quarks has pair graph

$$K_3.$$

There are three unordered pair contacts,

$$\{12,23,31\},$$

and every quark has graph degree

$$\boxed{\deg(i)=2.}$$

Let the one-incidence constriction quantum be the conditional band-blind trial value from the rank-one coupling, reciprocal normalization, and assumed frequency-rescaling reduction,

$$\boxed{ \varepsilon_{\rm inc} =\frac{14}{9}\mu_3g_{33} =154.808228\ {\rm MeV}. }$$

If each pairwise constraint contributes this one-particle rest-energy shift to each of its two endpoints, then the constriction energy assigned to quark $i$ is

$$E_i^{\rm constr} =\deg(i)\varepsilon_{\rm inc} =2\varepsilon_{\rm inc}.$$

Hence

$$\boxed{ E_i^{\rm constr} =\frac{28}{9}\mu_3g_{33} =309.616456\ {\rm MeV}. }$$

The total baryon constriction energy is

$$E_{K_3}^{\rm constr} =\varepsilon_{\rm inc}\sum_{i=1}^3\deg(i) =6\varepsilon_{\rm inc}.$$

Equivalently, each of the three physical pair constraints contributes to two one-particle endpoint operators.

For the proton this gives

$$m_p^{\rm cand} =2m_u+m_d+6\varepsilon_{\rm inc} =937.869369\ {\rm MeV},$$

and for the neutron

$$m_n^{\rm cand} =m_u+2m_d+6\varepsilon_{\rm inc} =940.409369\ {\rm MeV}.$$

The agreement is at the $10^{-3}$ level, while strange and higher-spin baryons require additional state-dependent contributions. Therefore the incidence mechanism is a candidate for the light-baryon bulk term, not a universal hadron mass law.

47.1 Decisive bookkeeping question

The graph factor two is physically legitimate only if a symmetric pair-contact action produces one induced one-particle rest-energy operator on each endpoint, while the underlying interaction action itself remains counted once.

The required distinction is

$$\boxed{ \text{one pair energy functional} \neq \text{one endpoint self-energy only}. }$$

For a symmetric pair functional $S_{ij}$, variation with respect to particle $i$ and particle $j$ may produce two endpoint equations of motion without implying that $S_{ij}$ itself should be summed twice in the total Hamiltonian.

The next test is to derive this endpoint-versus-pair distinction explicitly for the V2 cross-depth contact functional.

48. Exact $K_3$ collective-mode test: endpoint incidence is not an independent doubling

The baryon-incidence idea must be checked against the exact three-body harmonic Hamiltonian rather than by summing endpoint self-energies.

For one spatial direction, take three equal-mass quarks with symmetric pairwise quadratic constraints

$$V =\frac{k}{2} \left[(x_1-x_2)^2+(x_2-x_3)^2+(x_3-x_1)^2\right].$$

This can be written

$$V=\frac{k}{2}x^TL_{K_3}x,$$

with graph Laplacian

$$L_{K_3} =\begin{pmatrix} 2&-1&-1\\ -1&2&-1\\ -1&-1&2 \end{pmatrix}.$$

Its spectrum is

$$\boxed{\operatorname{spec}(L_{K_3})=\{0,3,3\}.}$$

The zero eigenvalue is the center-of-mass translation. The two nonzero eigenvalues are the two independent relative Jacobi modes.

Therefore, per spatial axis, a three-quark baryon has two independent relative spatial harmonic modes, not six independent endpoint energies.

For equal quark mass parameter $m$, their common frequency is

$$\boxed{\omega_{\rm rel}=\sqrt{\frac{3k}{m}}.}$$

Across three spatial dimensions there are six relative oscillator coordinates, but their ground-state energy is

$$E_0^{\rm rel} =6\left(\frac12\hbar\omega_{\rm rel}\right) =3\hbar\omega_{\rm rel}.$$

Thus the six coordinate components do not supply six full oscillator gaps; they supply three gaps worth of zero-point energy.

The graph identity

$$\sum_i\deg(i)=6$$

is therefore not, by itself, a six-fold energy multiplicity. It counts endpoint incidences of the three pair interactions.

Equivalently,

$$\sum_{i<j}E_{ij}$$

contains each physical pair once, while summing the corresponding one-particle mean-field contributions over both endpoints gives twice that interaction energy.

Hence

$$\boxed{ \text{the }K_3\text{ incidence factor }2\text{ cannot be used as an independent baryon-mass doubling without a separate many-body mass prescription.} }$$

This preserves the useful geometric fact that each quark participates in two constraints, but it removes the claim that this alone explains the missing factor two in the total nucleon energy.

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