Dimensional Constriction and Hadronic Mass — Part 7

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31. Self-consistent Gaussian curvature closure

There is a stronger route to the $7/3$ coefficient than the raw reciprocal dimension trace.

The canonical normalized reciprocal $4\to3$ calculation gave the inverse symmetric weight

$$\boxed{ C_{34}=\frac76. }$$

Under the Gaussian profile and local harmonic self-consistency ansatz below, a factor of two appears. It is not a universal mean-field result and does not survive as an automatic doubling when the exact-profile comparisons later in this ledger replace that ansatz.

Take a normalized one-dimensional oscillator mode in the lost normal coordinate,

$$\psi(y) =\frac{1}{(\pi\ell^2)^{1/4}} \exp\left(-\frac{y^2}{2\ell^2}\right).$$

Its density is

$$\rho(y)=|\psi(y)|^2 =\frac{1}{\sqrt\pi\ell} \exp\left(-\frac{y^2}{\ell^2}\right).$$

Suppose an attractive local self-consistent interaction produces a normal mean-field well proportional to that density,

$$V(y)=-G\rho(y).$$

Define its central depth

$$A=G\rho(0).$$

Then

$$V(y) =-A\exp\left(-\frac{y^2}{\ell^2}\right).$$

Expanding around the minimum,

$$V(y) =-A +\frac{A}{\ell^2}y^2 +O(y^4).$$

Matching the quadratic term to a harmonic oscillator,

$$V(y) =-A+\frac12m_\perp\omega^2y^2+\cdots,$$

gives

$$\frac12m_\perp\omega^2 =\frac{A}{\ell^2}.$$

For the oscillator ground state written above,

$$\boxed{ \ell^2=\frac{1}{m_\perp\omega} }$$

in natural units. Substitution eliminates both the inertial parameter and the width:

$$\frac12m_\perp\omega^2 =A m_\perp\omega,$$

so for the nonzero localized solution,

$$\boxed{ \omega=2A. }$$

This factor of two follows within the stated field-intensity ansatz: the self-consistent potential is proportional to $|\psi|^2$, whose Gaussian exponent is twice that of the mode amplitude.

31.1 Combination with the canonical reciprocal normalization

If the normalized reciprocal $4\to3$ contact gives the normal well depth

$$\boxed{ A_{34} =\frac76\mu_3g_{33}, }$$

then Gaussian self-consistency yields

$$\hbar\omega_{\rm common} =2A_{34} =2\frac76\mu_3g_{33} =\boxed{\frac73\mu_3g_{33}}.$$

Thus the previously unexplained coefficient decomposes as

$$\boxed{ \frac73 =2\times\frac76. }$$

The two factors have distinct origins:

1. $7/6$ is the canonical inverse symmetric reciprocal normalization of the $4\to3$ shared phase-space carrier;
2. $2$ is the exact curvature factor generated when a Gaussian oscillator mode produces a mean field proportional to its squared amplitude.

Numerically,

$$\mu_3g_{33}=99.5195753\ {\rm MeV},$$

so

$$A_{34} =\frac76\mu_3g_{33} =116.106171\ {\rm MeV},$$

and therefore

$$\boxed{ \hbar\omega_{\rm common} =232.212342\ {\rm MeV}. }$$

Fixed-action dimensional constriction then gives

$$\boxed{ \hbar\omega_3 =\frac43\hbar\omega_{\rm common} =309.616456\ {\rm MeV}. }$$

This is exactly the previous nucleon-scale candidate, but the $7/3$ coefficient no longer has to be postulated as a raw dimension-incidence ratio.

31.2 What remains unproved

This calculation substantially reduces the closure problem, but one identification is still required from the V2 master action:

$$\boxed{ A_{34}=\frac76\mu_3g_{33}. }$$

The factor $7/6$ is already supplied by canonical reciprocal normalization, and the factor $2$ follows from self-consistent Gaussian curvature. What is not yet derived is that the interaction scale entering that normalized reciprocal well is specifically the product

$$\mu_3g_{33}.$$

Accordingly, the new proof target is narrower than before. One no longer needs to derive the full $7/3$ coefficient. One needs only to show that the normalized $4\to3$ normal mean-field depth is

$$\boxed{ A_{34}=C_{34}\mu_3g_{33}, \qquad C_{34}=\frac76. }$$

If the projected master action yields this relation, then the complete hadronic bulk-energy chain follows without any fitted coefficient:

$$A_{34} \xrightarrow{\text{Gaussian self-consistency}} 2A_{34} \xrightarrow{4\to3\text{ assumed frequency rescaling}} \frac83A_{34} =\frac{28}{9}\mu_3g_{33}.$$

32. Schur-complement derivation of the surviving $g_{33}$ factor

The previous question was whether a genuine $4\to3$ interaction should involve the cross-sector coupling $g_{34}$ rather than the surviving-sector self-coupling $g_{33}$. A two-channel quadratic model resolves this distinction.

Let $x$ denote the surviving depth-three normal amplitude and $y$ the depth-four amplitude that is removed by constriction. The most general positive quadratic interaction block is

$$E(x,y) =\frac12 \begin{pmatrix}x&y\end{pmatrix} \begin{pmatrix} g_{33} & g_{34}\\ g_{34} & g_{44} \end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix}.$$

Explicitly,

$$E(x,y) =\frac12g_{33}x^2+g_{34}xy+\frac12g_{44}y^2.$$

32.1 Relaxed higher-dimensional state

If the fourth-direction amplitude is allowed to relax freely, minimization gives

$$\frac{\partial E}{\partial y}=0 \quad\Longrightarrow\quad y_{\rm rel}=-\frac{g_{34}}{g_{44}}x.$$

Substitution gives the Schur-complement energy

$$E_{\rm rel}(x) =\frac12 \left( g_{33}-\frac{g_{34}^2}{g_{44}} \right)x^2.$$

Thus ordinary integration-out or adiabatic relaxation produces the effective coefficient

$$\boxed{ g_{33}^{\rm rel} =g_{33}-\frac{g_{34}^2}{g_{44}}. }$$

32.2 Constricted state

Dimensional constriction is different. The removed coordinate is not allowed to relax to $y_{\rm rel}$; it is forced onto the lower-dimensional subspace

$$y=0.$$

The constrained energy is therefore

$$E_{\rm con}(x) =\frac12g_{33}x^2.$$

The energetic cost of constriction relative to the relaxed higher-dimensional state is

$$\Delta E_{\rm con}(x) =E_{\rm con}(x)-E_{\rm rel}(x) =\frac12\frac{g_{34}^2}{g_{44}}x^2.$$

Hence the positive stiffness generated specifically by suppressing the fourth direction is

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

This explains why a $4\to3$ process can be governed by the surviving-sector self-coupling even though the transition itself is mediated by $g_{34}$.

32.3 Rank-one coupling closure

The V2 coupling hypothesis uses the rank-one relation

$$\boxed{ g_{34}^2=g_{33}g_{44}. }$$

Therefore

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

Even more strongly, the relaxed Schur complement becomes

$$g_{33}^{\rm rel} =g_{33}-\frac{g_{34}^2}{g_{44}} =0.$$

So under the rank-one condition the coupled higher-dimensional mode has a flat relaxed direction, while constraining the fourth component to zero restores the full depth-three stiffness $g_{33}$.

This gives a concrete geometric interpretation of the self-coupling:

$$\boxed{ g_{33}=\text{stiffness cost of forbidding the depth-four relaxation channel}. }$$

That is precisely the structure required by the hadronic constriction mechanism.

33. Combined closure chain

The previous sections supplied two other ingredients.

Canonical reciprocal normalization gives

$$C_{34}=\frac76.$$

The Schur-complement constriction calculation gives

$$g_{\rm con}=g_{33}.$$

Therefore, if the depth-three contact-energy unit $\mu_3$ converts this dimensionless constriction stiffness into the central normal mean-field depth, then

$$\boxed{ A_{34} =C_{34}\mu_3g_{\rm con} =\frac76\mu_3g_{33}. }$$

For a self-consistent Gaussian density well, the local curvature relation derived above gives

$$\boxed{ \hbar\omega_{\rm common}=2A_{34} =\frac73\mu_3g_{33}. }$$

Finally, assumed frequency-rescaling redistribution from four oscillator directions into three gives

$$\boxed{ \hbar\omega_3 =\frac43\hbar\omega_{\rm common} =\frac{28}{9}\mu_3g_{33}. }$$

Numerically,

$$\boxed{ \hbar\omega_3=309.616456\ {\rm MeV}. }$$

The logical chain is now

$$\boxed{ g_{34}^2=g_{33}g_{44} \Longrightarrow \frac{g_{34}^2}{g_{44}}=g_{33} }$$

from rank-one coupling and constrained-vs-relaxed quadratic energy,

$$\boxed{ \frac76 }$$

from canonical reciprocal normalization,

$$\boxed{ 2 }$$

from Gaussian density curvature self-consistency, and

$$\boxed{ \frac43 }$$

from assumed frequency-rescaling dimensional constriction.

Multiplying them gives

$$\boxed{ \frac76\times2\times\frac43 =\frac{28}{9}. }$$

33.1 Remaining assumption

The principal remaining assumption is now narrower still: that the dimensionless quadratic constriction stiffness is converted to physical normal-well depth by the same depth-three contact-energy unit $\mu_3$,

$$\boxed{ A_{34}=\frac76\mu_3g_{\rm con}. }$$

Unlike the earlier unexplained product $\mu_3g_{33}$, the factor $g_{33}$ now arises from an explicit constrained-versus-relaxed quadratic calculation. What remains to be derived from the master action is the normalization identifying $\mu_3$ as the energy unit of this normal quadratic block.

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