Multicentre orbit axes

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: Three-petal motion gives exact trigonal axes and a distinct 6D rotation gives tetrahedral axes; under bound-within/free-without, projected nuclear crossing is an encounter rather than an unobserved-depth miss.

The corrected molecular question

Molecular nuclei should not be inserted as generic fixed perturbations. The stronger proposal is that they settle along directions selected by a shared electron trajectory. The electron follows Hamiltonian motion in six coordinates; in a weak-force interval this approaches a straight line, while nuclear encounters bend it. Molecular shape may then be a closure and force-balance condition constrained by conservation of total momentum, energy, and the six-dimensional angular-momentum bivector.

There are two different apparent encounters:

1. the orbit can project through a central nucleus while carrying nonzero coordinates unsupported by that nucleus;
2. an outer nucleus can lie along a petal axis without lying at the electron's instantaneous radius.

Three trigonal directions are exact

For the checked signed relative equilibrium

$$(x,y)=\cos(3\theta)(\cos\theta,\sin\theta),$$

the three distinct outward maxima occur at $0,2\pi/3,4\pi/3$. Their unit vectors have off-diagonal Gram entries $-1/2$, so their mutual angle is exactly $120^\circ$. The actual signed three-petal branch, rather than a modulus-folded six-lobe image, therefore supplies the trigonal axes directly.

Four directions require a three-dimensional projection

The planar signed $L=4$ rose has eight directions and is not tetrahedral merely because its integer is four. A different one-parameter rotation in six coordinates does produce four tetrahedral directions. Set

$$a=\sqrt{\frac23},\qquad b=\frac1{\sqrt3},$$

and

$$q(\theta)=(a\cos\theta,a\sin\theta,b\cos2\theta, b\sin2\theta,0,0).$$

This is a genuine constant-radius group orbit, $|q|=1$. At the four quarter turns its visible positions are

$$(a,0,b),\quad(0,a,-b),\quad(-a,0,b),\quad(0,-a,-b).$$

Every distinct pair has inner product $-1/3$, giving $\arccos(-1/3)=109.4712206\ldots^\circ$. Equiangularity forces the amplitude ratio: $-b^2=-a^2+b^2$ gives $a^2=2b^2$, and unit radius fixes both amplitudes. This is an orbit-level counterpart of the $s+3p$ simplex geometry, but the completed interaction has not yet been shown to sustain this $(1,2)$ rotation.

Conservation closes the angle calculation

Suppose the electron follows near-straight segments between $n$ equivalent interactions and each passage exchanges the same impulse magnitude along a unit nuclear direction $\mathbf n_a$. Periodic momentum closure requires

$$\sum_{a=1}^n\mathbf n_a=0.$$

If equivalence also makes every distinct pairwise product equal to $c$, dotting the closure equation with any $\mathbf n_b$ gives

$$1+(n-1)c=0, \qquad c=-\frac1{n-1}.$$

Consequently $n=3$ forces $c=-1/2$ and the trigonal angle $120^\circ$, while $n=4$ forces $c=-1/3$ and the tetrahedral angle $109.471\ldots^\circ$. Conversely, both direction sets checked above sum to zero, so equal impulses close exactly.

This supplies a candidate dynamical link that the earlier hybrid-state proof did not: straightest-path tendency supplies the segments, conservation closes their impulse polygon, and equivalent nuclear interactions make it a regular simplex. The conclusion is conditional on equal impulse magnitudes and equivalent passages. It is not by itself evidence that the presently proposed interaction kernel produces those impulses, and the visible simplex theorem is not uniquely six-dimensional; the six-dimensional content is the continuous six-dimensional orbit that realizes the passages.

Exact passage through the cylindrically extended nucleus

For every checked signed-rose branch $L=2,3,4$, the visible coordinates vanish at

$$\theta_k=\frac{(k+1/2)\pi}{L}.$$

Direct evaluation gives

$$|q_{\rm vis}(\theta_k)|=0, \qquad |q_{\rm hid}(\theta_k)|=1.$$

Under "bound within, free without," the first equation is the physically relevant one. A depth-three nucleus extends cylindrically through the three directions it does not support, so nonzero $q_{\rm hid}$ does not provide separation. The trajectory encounters the nuclear support whenever its three supported coordinates coincide. The earlier "full-depth miss" interpretation is withdrawn as inconsistent with that premise.

What the first fixed-ligand test says

A provisional embedded-point kernel

$$V(q)=-\sum_a\frac{Z_a}{|q-(R_a,0)|}$$

was used only to perturb the completed one-centre solutions. With ligands at radius 2.5 and charge 0.01, the nearest visible central approaches were about 0.00215 for three trigonal ligands and 0.00120 for four tetrahedral ligands; unsupported-coordinate norms were about 0.9972 and 0.9978. Energy was conserved, but the old orbit did not recur exactly. This embedded-point calculation belongs to the alternative unobserved-localized model; it is retained only as a rejected-control comparison.

Cycle-averaged force-balance verdict

The provisional kernel was next tested against the reciprocal condition

$$\frac1T\int_0^T F_a(t)\,dt=0$$

for every symmetry-related ligand. A root at $R=1$ is inadmissible: the ideal axis-aligned trajectory reaches the outer nucleus with zero unobserved separation, so the point kernel is singular there.

For a unit central source, three trigonal ligands of effective charge $0.1$, and the signed $L=3$ candidate, one noncollision root occurs at $R=1.89676096$. Its radial force derivative is $+0.0485176$, so it is unstable rather than restoring. For the orbit-aligned tetrahedral curve with central and ligand effective charges $0.1$, no noncollision root occurs on the tested ranges $R\in[0.2,0.95]$ and $R\in[1.05,4]$.

Thus the alternative embedded-point $1/r$ kernel does not turn the available conservation geometry into a stable molecule. Conservation and equivalence derive the angles, but a derived lower-depth interaction/response kernel—or a materially retuned electron trajectory—is still required for stable reciprocal closure.

Flexible nuclear three-planes

Let a nucleus carry a rank-three orthoprojector $P_a$ in the electron's six-coordinate space. A minimal anisotropic distance family is

$$d_{\eta,P}^2=\Delta q^T\bigl(P+\eta(I-P)\bigr)\Delta q.$$

The unobserved-localized point-kernel control is $\eta=1$, while bound-within/free-without is the cylindrical limit $\eta=0$. Allowing $P$ to relax instantaneously with no orientation cost does not retain an anisotropic force. A rank-three plane in six dimensions can always be chosen either to contain the one separation vector $\Delta q$ or to lie wholly perpendicular to it. Therefore

$$\min_{\operatorname{rank}P=3}d_{\eta,P}^2 =\min(1,\eta)|\Delta q|^2.$$

For $\eta>0$ the relaxed interaction is only a constant rescaling of $1/|\Delta q|$. In the selected cylindrical limit $\eta=0$, the minimum is zero and the ideal Coulomb kernel becomes singular: a freely reorienting plane can turn until it intersects the electron. Complete cost-free flexibility is therefore unusable in either case. Finite inertia or stiffness is essential.

The physically useful interpretation of flexibility must be finite response. The nuclear plane needs rotational inertia, an orientation stiffness relative to its lower-depth structure, or a nonzero response time. It can then lag the electron, store angular momentum, and return it through backreaction. This is the same kind of dynamical susceptibility already isolated by the response modules, now applied to molecular closure. A fixed projector and an instantaneously minimized projector are the two controls on either side of that finite-response problem.

Correct next solve: reciprocal closure

The nuclei must now be dynamical variables rather than fixed scenery. For a candidate period $T$, solve simultaneously for electron and nuclear initial data, the response source, and the nuclear scale, subject to

$$X(T)=gX(0),\qquad P(T)=gP(0),$$

where $g$ is the appropriate trigonal or tetrahedral permutation-rotation, together with energy and total bivector conservation. Equivalently, in a Born--Oppenheimer reduction require zero cycle-averaged force on every nucleus,

$$\frac1T\int_0^T F_a(t)\,dt=0.$$

This formulation tests the proposed spatial mechanism: near-straight electron motion, deflection at the lower-depth nuclear structures, and a molecular shape selected partly by closure and conservation in six dimensions. It also makes the proposal falsifiable: if no stable reciprocal periodic branch exists for the derived kernel, geometrical availability alone is insufficient.

Reproduction

multicentre_visible_crossing_checks.py checks the exact central coincidences and unsupported-coordinate values, trigonal and tetrahedral Gram matrices, constant six-radius, energy conservation, and the first fixed-ligand continuations.

Related

• Completed Coulomb petals — prerequisite: supplies the autonomous one-centre signed-rose equilibria and completed Hamiltonian.
• Finite nuclear-plane response — continuation: proves passive flexibility cannot stabilize the measured saddle and isolates the gyroscopic threshold.
• Finite cylindrical nuclear profile — correction: replaces the rejected unobserved-localized force kernel and finds restoring trigonal and tetrahedral radial branches.