Finite nuclear-plane response

Physical premise (IDWT): Physical spacetime has infinitely many ordinary spatial dimensions plus time. Matter occupies up to ten spatial dimensions; all remaining spatial directions are unbound for every particle. Depth counts a pattern's bound spatial involvement within that space. “Observed” and “unobserved” describe an observer's access, not different kinds of dimension. Older projective and nested-geometry calculations below retain their additional hypotheses; they do not establish a separate physical manifold or compactified direction. See spatial meaning and mathematical scope.

Summary: An unobserved-localized control saddle cannot be stabilized by passive flexibility, while the selected cylindrical ontology instead requires a finite-size, finite-inertia plane rotor before molecular stability can be tested.

Question and measured input

The now-rejected unobserved-localized multicentre control found a trigonal cycle-averaged force-balance radius with positive radial force slope

$$F'(R_*)=0.0485176186.$$

Writing a displacement as $x$, its uncoupled local equation is $\ddot x=\sigma^2x$ with

$$\sigma=\sqrt{F'(R_*)}=0.2202671528.$$

This slope and every numerical threshold derived from it belong only to that control. They must not be transferred to the selected cylindrical bound-within/free-without kernel.

Passive relaxation is a no-go

Let $y$ be one small plane-orientation coordinate with positive stiffness $\kappa$ and reciprocal potential coupling $bxy$. The quadratic potential is

$$V(x,y)=-\frac{\sigma^2}{2}x^2+bxy+\frac\kappa2y^2.$$

Static elimination gives $y=-bx/\kappa$ and

$$V_{\rm eff}''=-\sigma^2-\frac{b^2}{\kappa}<0.$$

Equivalently, the Hessian determinant is $-\sigma^2\kappa-b^2<0$. Thus a passive positive-energy orientation mode makes the relaxed curvature more negative and the autonomous coupled system retains a real exponential eigenvalue. Runtime checks cover twelve $(\kappa,b)$ pairs and positive orientation inertias.

The driven susceptibility

$$\chi_N(\omega)=\frac1{\kappa-I\omega^2}$$

does reverse sign above resonance. That fact alone is not a stability proof: the autonomous static Hessian remains indefinite. Treating the sign reversal as stabilization would confuse forced response with the spectrum of the closed electron--nucleus system.

A plane coordinate alone is not a phase space

The configuration space of rank-three planes in six dimensions is

$$\operatorname{Gr}(3,6),\qquad \dim\operatorname{Gr}(3,6)=3(6-3)=9.$$

The tangent variations of $P_0=\operatorname{diag}(I_3,0)$ are $\delta P=[K,P_0]$ for $K\in\mathfrak{so}(6)$; the checked linear map has rank nine. Every antisymmetric bilinear form on a nine-dimensional real space is degenerate. Hence the plane coordinate by itself cannot carry a nondegenerate Hamiltonian/Berry two-form.

A complete flexible-plane model must use at least the cotangent phase space $T^*\operatorname{Gr}(3,6)$, of dimension eighteen, or supply additional field variables derived from the same master geometry. These are parameter counts for spatial orientation and angular momentum, not additional physical dimensions or separate sectors. Its conjugate momentum is not optional bookkeeping: it is where the plane stores angular momentum between electron encounters. A first-order precession equation written only for $P_a$ would necessarily be a reduced/presymplectic law and must identify the degrees of freedom already eliminated.

Gyroscopic stabilization is possible

A rotating nuclear three-plane has at least a two-component transverse normal mode. The minimal conservative gyroscopic normal form is

$$\ddot{\mathbf x}+2\Omega J\dot{\mathbf x}-\sigma^2\mathbf x=0, \qquad J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$

The antisymmetric term does no work, so

$$E=\frac12|\dot{\mathbf x}|^2-\frac{\sigma^2}{2}|\mathbf x|^2$$

is conserved. The characteristic equation is

$$(\lambda^2-\sigma^2)^2+4\Omega^2\lambda^2=0.$$

All four roots are imaginary exactly when

$$\boxed{|\Omega|\ge\sigma}.$$

Below threshold there is a real part. Above threshold the two positive oscillation frequencies are

$$\omega_\pm=|\Omega|\pm\sqrt{\Omega^2-\sigma^2}.$$

This is spectral gyroscopic stabilization of an indefinite-energy equilibrium, not a positive potential minimum.

Control-only comparison with the actual trigonal clock

The completed signed $L=3$ orbit has clock

$$\beta_3=0.2992305666.$$

Numerically,

$$\frac{\beta_3}{\sigma}=1.35849>1.$$

Therefore a reciprocal plane rotation acting with gyroscopic rate equal to the existing orbit clock would lie above the local threshold. A deterministic 100-turn integration of this normal form remains bounded, with maximum norm amplification $3.26186$ and energy error below $2.5\times10^{-15}$.

This comparison is only a scale observation inside the rejected unobserved-localized control; it is not evidence that the cylindrical molecular branch is stable. The full nuclear-plane equations must derive the coefficient $2\Omega J$ and identify its $\Omega$ with the orbit clock. Substituting $\Omega=\beta_3$ by hand would merely hide the missing lock in the linearized equation.

Rotating coordinates are not the mechanism

There is a decisive adversarial control. Expressing an ordinary unstable lab equation $\ddot q=\sigma^2q$ in coordinates rotating at rate $\Omega$ produces

$$\ddot x+2\Omega J\dot x-(\sigma^2+\Omega^2)x=0.$$

The Coriolis-looking term comes with the centrifugal $\Omega^2$ contribution. Its spectrum retains real growth $\sigma$ for every $\Omega$; changing frames does not stabilize anything. Runtime tests confirm this at four rates, including the trigonal clock.

Consequently the stabilizing normal form of the preceding section requires a physical antisymmetric term without the compensating coordinate artifact. Candidates include a Berry-curvature term on the dynamical orientation space, a magnetic bivector coupling, or a spinning-rotor relative equilibrium whose energy--momentum Hessian supplies the needed block. Merely declaring that the nuclear plane co-rotates with the electron is insufficient.

Next proof obligation

Introduce an orientation frame $U_a(t)\in SO(6)$, $P_a=U_aP_0U_a^T$, and a conjugate skew momentum $\Pi_a$. Derive both the torque and electron backreaction from one Hamiltonian. Linearization about the trigonal relative-periodic branch must then reproduce a transverse block of the form above with a calculated $\Omega$. The required checks are:

1. energy and the applicable $SO(3)_O\times SO(3)_H$ moment maps are conserved;
2. the gyroscopic coefficient appears from the rotating solution rather than as an independent coupling;
3. the complete Floquet spectrum, not only the radial block, has no growing multiplier;
4. the same action admits or excludes the tetrahedral continuation.

Reproduction

flexible_nuclear_response_checks.py checks the passive Schur-complement no-go, autonomous exponential modes, susceptibility sign reversal, exact gyroscopic threshold, long-time bounded integration, and energy conservation.

Related

• Multicentre orbit axes — prerequisite: supplies the angle-closure theorem, the provisional force kernel, and the measured trigonal radial instability.
• Nuclear-plane rotor — continuation: supplies the finite-inertia reciprocal phase space required by the dimensional obstruction found here.