Induced cross-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: A positive-cost mixed-plane response generates the inverse orientation kernel in its static limit, while its dynamical source and physical normalization remain explicit assumptions.

Source audit and new premise

The source-level binding module still supplies a scalar nuclear density and an unevaluated cross-depth kernel. It does not supply a calculated mixed-plane tensor. The construction below is a candidate effective response channel, not a claim that the missing source has already been found in the master action.

Use the model's observer-relative coordinate grouping V6=O direct-sum H and electron sector coordinate J with mixed block Q. The grouping introduces no physical distinction among the electron's six dimensions. Introduce a real response matrix X in Hom(H,O), transforming as X -> R_O X R_H^T, just as Q does. This is relational cross-depth polarization at the electron–nucleus interaction. It is not a declaration that an isolated depth-three nucleus has acquired three additional spatial directions. Its microscopic current origin and its possible electromagnetic identification are open.

Assume its leading static energy is

$$E_X(J,X)=\frac1{2\chi}\|X\|_F^2-g\operatorname{tr}(X^TQ), \qquad \chi>0.$$

The positive susceptibility parameter chi makes the isolated response cost positive. The invariant bilinear coupling and a single isotropic chi are additional physical assumptions. This model preserves independent rotations of the two three-planes before a configuration is chosen.

Static elimination derives the inverse kernel

Complete the square:

$$E_X=\frac1{2\chi}\|X-\chi gQ\|_F^2 -\frac{\chi g^2}{2}\|Q\|_F^2.$$

The complex-structure identities give ||Q||²=3−2s. Consequently

$$X_* =\chi gQ,\qquad E_{\rm eff}(J)=-\frac32\chi g^2+\chi g^2s.$$

Thus the inverse quadratic orientation kernel is the static response energy, with positive coefficient

$$\boxed{\kappa=\chi g^2.}$$

This determines the coefficient's sign and factor within the response model, without fitting a precession frequency. It does not independently determine chi or g, nor prove that this channel is present. The sign is unchanged by reversing the coupling g. General anisotropic susceptibility would yield a tensor quadratic form, not necessarily a function of s alone.

The induced source produces the same base flow

For any X define the skew matrix

$$B_X=\begin{pmatrix}0&X\\-X^T&0\end{pmatrix},\qquad \langle B_X,J\rangle=\operatorname{tr}(X^TQ), \quad \langle U,V\rangle=-\tfrac12\operatorname{tr}(UV).$$

With the symplectic action scale I_e, the electron equation at fixed X is

$$\dot J=-\frac g{\mathcal I_e}[B_X,J].$$

Substitution of X_* yields

$$-\frac{\chi g^2}{\mathcal I_e}[B_Q,J] =\frac{2\kappa}{\mathcal I_e}[PJP,J].$$

The equality follows from J=B_Q+diag(A,C) and AQ+QC=0. This matches the previous precession equation, including its normalization. At s=0 the induced source is proportional to J and commutes with it, so this reduced base point is stationary despite a nonzero source. Source strength alone must not be mistaken for actual motion.

Dynamical response

Give X canonical momentum Y and a positive kinetic coefficient M_X. Use

$$E=\frac{\|Y\|^2}{2M_X}+\frac{\|X\|^2}{2\chi} -g\operatorname{tr}(X^TQ),\qquad \Omega_X^2=\frac1{M_X\chi}.$$

Together with the CP3 symplectic term, the equations are

$$\dot J=-\frac g{\mathcal I_e}[B_X,J],\qquad \dot X=Y/M_X,\qquad \dot Y=-X/\chi+gQ.$$

The static source is no longer imposed as a function of time: X and J respond to each other. Energy is conserved. Independent frame rotations also conserve the two skew charge matrices

$$\mathcal M_O=\mathcal I_e A-(XY^T-YX^T),\qquad \mathcal M_H=\mathcal I_e C-(X^TY-Y^TX).$$

Finite response can exchange the earlier invariant s with the response degrees of freedom; s need not remain constant. This is not dissipation or attraction to a settled orbit. The joint evolution remains Hamiltonian.

An exact interaction substep holds X fixed, conjugates J by exp(−g B_X dt/I_e), and integrates Q(t) to update Y. The oscillator substep is also exact. Their symmetric splitting preserves J²=−I and the rotation charges to roundoff, and its energy error converges quadratically. Independent DOP853 integration checks the full coupled equations.

Illustrative dimensionless runs use chi=1.2, g=0.7, I_e=1.3, initially X=chi g Q and Y=0, for time 0 through 6:

Halving dt from .02 to .01 reduces endpoint error against the independent integration by approximately four in each case. No parameters represent a calibrated nucleus; the table tests response dynamics, not atomic data.

With the response velocity prepared consistently with the static-limit motion, increasing Omega_X through 5,10,20,40 gives electron endpoint errors .0152191, .00371429, .000952156, .000240221 against that limit at time 3. This checks convergence to the reduced precession rather than assuming that eliminating X is valid at every response rate.

A tempting spectral match that does not close the orbit

In the static limit, the induced generator has mixed block proportional to Q. Its vector-plane strengths are proportional to {sqrt(1−s),sqrt(1−s),1}. At s=8/9 these have the visualization's 1:1:3 ratio. This is an algebraic compatibility worth recording, not a solved selection law: no equation here selects s=8/9.

More importantly, Q and therefore the induced generator rotate during the motion. Freezing that generator at its initial value gives a different trajectory. For the parameters above at time .8 the maximum entrywise difference in J is .074626. Thus matching instantaneous generator spectra does not reproduce the fixed-generator petal construction. The actual time-dependent evolution and its physical readout must be checked.

Verification and next calculation

cross_plane_response_checks.py currently completes 97 evaluations across 17 runtime conditions: static elimination, gradient equivalence, differential energy and charge balance, exact interaction substeps, time-step convergence, the fast-response limit, and the frozen-source counterexample. These are conditional-model checks, not evidence that the microscopic current supports X.

The next calculation is to seek self-consistent uniformly precessing solutions of the finite-response equations and test their linear stability. In parallel, the source audit must distinguish an allowed relational response from an already derived nuclear electromagnetic susceptibility.

Related

• Research status — current proof boundaries and investigation results.
• Inverse nuclear orientation kernel — target whose quadratic coefficient this response model supplies.
• Flat nucleus precession — reduced Hamiltonian flow recovered in the static response limit.
• Nuclear response feedback — contrasts a physical electric-dipole diagnostic with this additional relational tensor response.