Reciprocal cylindrical molecule

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 6D electron, relative angular source, moving finite cylindrical nuclei, and flexible nuclear-plane rotors form one translation-invariant reciprocal Hamiltonian with verified energy and Noether-charge conservation.

Why the first combination failed

The completed petal Hamiltonian was originally written about a fixed nuclear origin. Adding moving nuclei without changing that source makes its magnetic completion depend on the absolute electron coordinate and destroys visible translation invariance. A failed momentum-derivative check exposed this before the model was used for shooting.

The correction is to define the source coupling using the electron--central- nucleus relative coordinate

$$s=q-(R_0,0).$$

The completed kinetic term is then

$$H_e=\frac1{2m_e}|p-m_e\lambda Cs|^2 +\frac{\langle C,C\rangle}{2I_C}.$$

The central nucleus receives the opposite visible component of the source force. It receives no unobserved translational momentum because it remains a depth-three object.

Full diagnostic Hamiltonian

For nuclear centres $R_a\in\mathbb R^3$, momenta $P_a$, rank-three planes $\Pi_a$, and skew plane velocities $\Omega_a$, use

$$H=H_e+\sum_a\frac{|P_a|^2}{2M_a} +\sum_a\frac{I_a}{2}\langle\Omega_a,\Omega_a\rangle +\sum_a V_a\bigl(|\Pi_a(q-(R_a,0))|\bigr) +\sum_{a<b}\frac{Z_aZ_b}{|R_a-R_b|}.$$

Here $V_a$ is the finite uniform supported-space profile of the preceding note, extended cylindrically through unsupported coordinates. The source coadjoint equation uses $s\wedge p$ and $s\wedge Cs$, not $q\wedge p$ about an external origin. Each plane obeys

$$\dot\Pi_a=[\Omega_a,\Pi_a], \qquad I_a\dot\Omega_a=2[\Pi_a,\partial_{\Pi_a}V_a].$$

Every force and torque is obtained from this one energy.

The executable originally specialized the equations to $m_e=1$ while exposing a configurable mass. The September 7 audit restores the factors of $m_e$ in the completed square, velocity, magnetic force, and source gradient; the reported unit-mass run is unchanged. Nuclear mutual repulsion is a separately assumed centre-charge interaction here. It is not the mutual interaction of overlapping finite cylindrical charge profiles, particularly when their support planes differ. The full skew rotor also permits angular momentum in the plane stabilizer; a pure Grassmannian plane rotor requires horizontal initial angular velocity or an explicit reduction.

Conserved quantities

The visible total momentum is

$$\mathcal P_O=p_O+\sum_aP_a.$$

The visible rotational charge is

$$\mathcal J_O=q_O\wedge p_O+C_{OO} +\sum_aR_a\wedge P_a-\sum_a I_a(\Omega_a)_{OO}.$$

Ten random states independently check the differential conservation of $H$, $\mathcal P_O$, and $\mathcal J_O$. A nonlinear example contains one central cylinder plus three trigonal ligand cylinders with moving centres, four flexible planes, and a nonzero electron angular source. Over eight model time units its maximum errors are:

The differential energy residual is below $8.9\times10^{-10}$. Thus the finite profile, moving lower-depth centres, plane rotors, and completed source are mutually compatible at the Hamiltonian level.

Remaining periodic-orbit problem

This calculation establishes the system to shoot; it does not claim that the arbitrary nonlinear example is molecularly periodic. The next solve should start from the exact trigonal orbit and the restoring finite-profile radius, fix centre-of-mass and global-rotation gauge directions, and solve a relative-periodic boundary condition. The monodromy calculation must retain electron, source, nuclear-centre, and plane-rotor perturbations. Stable radial force balance alone is not sufficient.

The uniform-ball nuclear profile and dimensionless masses, charges, and inertias remain diagnostic inputs. A successful branch would establish existence within that controlled model, not yet derive those inputs from V2.

Fixed-Plane Canonical Response Comparison

A separate comparison replaces $C$ by a canonical skew response $B,\Pi$ with positive source kinetic and restoring terms. It freezes one nuclear support $P$ and uses $V=-1/\sqrt{q^TPq}$, without adding perpendicular localization. This is an infinite-inertia nuclear limit, not the full reciprocal Hamiltonian above. The proposed same-field collective reduction and its coefficients remain assumptions; no mediator particle is introduced.

For a relative equilibrium generated by $A$, in units $m=K=k=1$,

$$p=(A+gB)q,\quad \Pi=M[A,B],\quad B+M[A,[A,B]]=-g(q\wedge Aq).$$

One circle in the supported plane and one perpendicular circle, with clocks $\Omega,\nu$ and unit radii, give

$$\Omega^2(1+2g^2)-g^2F=1,\qquad \nu^2(1+2g^2)-g^2F=0,$$

$$F=\frac{-(\Omega^2+\nu^2)+M(\Omega^4+\nu^4+6\Omega^2\nu^2)} {\{1-M(\Omega+\nu)^2\}\{1-M(\Omega-\nu)^2\}}.$$

The finite response can therefore support perpendicular relative motion that the static response cannot. At $g=1,\Omega/\nu=2$, the exact solutions have $\Omega=2/3$, $\nu=1/3$ and $M=(71\pm\sqrt{4177})/6$. Both branches also belong to continuous noninteger-ratio families; no integer lock or simplex mass index is selected. Their visible projection is a circle rather than the full petal sequence, and their exact preparations have nonzero response amplitudes and momenta.

This particular two-plane geometry is available at occupied depths five, six and ten. Depths two, three and four lack its two perpendicular nuclear directions; this does not exclude other trajectories there. Full Jacobians at ratios $1.5,2,2.5,3,4$ test both branches at depths five, six and ten. All thirty have growing modes. At ratio two, the low/high growth rates are approximately $0.07087/0.16808$ at five, $0.61922/0.20473$ at six, and the same respective values at ten. These numerical results are not physical lifetimes or a theorem excluding every cylindrical trajectory.

The standalone calculation and recorded results check ten exact identities, all thirty equilibria and complete Jacobians, two nonlinear growth tests, and energy/applicable rotational charges across all six occupied depths. The fixed support preserves only the supported and perpendicular rotational-charge blocks; it receives the omitted mixed torque. The full moving-nucleus problem still needs the nuclear phase space above. This comparison supplies conditional orbit existence and a failed stability test, not new support for the assigned sector structure.

Reproduction

reciprocal_cylindrical_molecule_checks.py checks the relative-source energy gradient, visible translation and rotation charges, nonlinear integration, all four plane constraints, and a nonzero source preparation.

Related

• Finite cylindrical nuclear profile — prerequisite: supplies the finite encounter law and restoring candidate radii.
• Nuclear-plane rotor — prerequisite: supplies the finite-inertia projector dynamics and torque normalization.
• Cylindrical petal continuation — begins fixed-centre periodic continuation and diagnoses the soft shooting directions.