Orbits, binding and observed spatial patterns

The electron in IDWT is a six-dimensional field pattern following a literal deterministic trajectory around a nucleus. Its physical position $q(t)$, its internal classical field oscillations, coefficients describing that field and accumulated detector registration are different objects. A stationary field density, a circulating coefficient vector or an image resembling a conventional orbital cannot replace that trajectory.

The orbit-plus-precession construction supplies concrete paths and useful conditional force laws, and the precession rate now follows from the quark charges and the numbers of directions they act in, as the section on precession below shows. The open problem is to derive the physical attraction, any selected lock and the reciprocal response from the same field $\Psi_\infty$ on one spacetime with infinitely many ordinary spatial directions plus time, with matter occupying up to ten. None of these calculations introduces mediator particles, Kaluza–Klein towers, compactification or an internal configuration space in place of physical position.

Reviewed Connections as of 5 November 2026

The commuting seed tetrahedron resolves three normal two-planes inside an internal spatial field construction. It is neither a molecular ligand tetrahedron nor an electron's path around the nucleus. The centered-window translating profiles likewise describe internal field structure and do not derive the electron-nucleus force. The external baseline remains literal deterministic motion of the six-dimensional electron.

How the newer internal structure enters this question

In the current portion picture, an electron's internal count is $C=18564$, with all portions sharing the charged-lepton fundamental. Its registered mass-frequency is $C\Omega$. Neither that fundamental nor its multiplied rate has yet been identified with the electron's external revolution or precession rate. An internal simplex arrangement is not the path around the nucleus.

The nucleus may orient one or more of the electron's extra bound directions. Within the ten-dimensional matter domain, four directions complement the electron's six bound directions; infinitely many additional directions are also unbound. Changing the bound orientation constrains its orthogonal complement. Fixing one extra bound direction does not fix the full set; how much the attraction pins remains open. Translation along an exactly uniform free extension leaves the pattern unchanged, whereas changing its orientation changes the configuration.

The new compact-probe calculation also distinguishes the density centroid from the center governing its kinetic response. That distinction is useful when deriving a force on a structured object. Its supplied uniform scalar probe and one-portion shape response do not yet produce electron–nucleus attraction, the six-dimensional electron, or the external orbit below. A complete derivation must connect these notions of position through the same reciprocal field dynamics.

One exact external trajectory and its flat force law

Let $J$ be a real antisymmetric isoclinic rotation on four active ordinary spatial coordinates,

$$J^T=-J,\qquad J^2=-I.$$

Choose orthogonal real directions $e_1,e_3$, with their corresponding $Je_1,Je_3$, and set

$$q(t)=\rho e^{\beta Jt} [\cos(\delta t)e_1+\sin(\delta t)e_3].$$

The full radius is constant, $|q|=\rho$. Direct differentiation in flat coordinates gives

$$\boxed{\ddot q=2\beta J\dot q-(\delta^2-\beta^2)q.}$$

This follows from the ordinary collective Lagrangian

$$L=\frac m2|\dot q|^2+m\beta q^TJ\dot q -\frac m2(\delta^2-\beta^2)|q|^2, \qquad m>0.$$

The canonical momentum is $p=m\dot q-m\beta Jq$, so the sign of the antisymmetric force is fixed. That term does no instantaneous work. It has the mathematical form of a magnetic-curvature force on spatial rotation planes, not a particle-exchange mechanism. Its same-field source and the positive collective inertia still need their physical derivation.

In the frame rotating at $\beta$, the path is a circle at rate $\delta$. The signed normal frequencies are $\beta+\delta$ and $\beta-\delta$. The commensurability $\delta=L\beta$ requires

$$\omega_0^2=(L^2-1)\beta^2.$$

The action admits continuously variable ratios: $L=2$ needs $\omega_0^2=3\beta^2$, while $L=3$ needs $8\beta^2$. It does not select integers. These orbit ratios also are not the particle mass index $n$: the electron's maintained count is $S(13,6)=18564$, which supplies no automatic choice of this external lock or its absolute rate. The particle mass-frequency, sector unit, collective phase rate and revolution/precession rates require their actual dynamical relations.

The same curve fits in four directions and embeds in physical depths four, five, six and ten. This particular two-plane realization needs more than depths two or three, but other projected breathing trajectories exist already in three dimensions. It is a higher-dimensional compatibility result, not unique selection of depth six. The flat orbit derivation gives the equation, action and corrected physical interpretation.

An autonomous Coulomb-plus-source completion

A separate six-coordinate mechanical model has an autonomous Hamiltonian, rather than putting the desired path into the force at every time. With real positions $q_i$, momenta $p_i$ and a skew coadjoint source $C$, it uses

$$H=\sum_i\frac{|p_i-m\lambda Cq_i|^2}{2m} -\sum_i\frac{k}{|q_i|} +\sum_{i<j}\frac{\gamma}{|q_i-q_j|} +\frac{\langle C,C\rangle}{2I}, \qquad \langle A,B\rangle=-\frac12\operatorname{tr}(AB).$$

The completed kinetic square is essential. The attractive $-k/r$ is an effective mechanical comparison, not a derived six- or ten-dimensional nuclear field. The source is proposed as a collective description of the same field, not a mediator; its physical origin remains supplied.

Relative equilibria solve the Hamilton equations together with a simultaneous group rotation. The literal trajectory in its four active entries is

$$q(\theta)=\rho\bigl( \cos L\theta\cos\theta,\ \cos L\theta\sin\theta,\ \sin L\theta\cos\theta,\ \sin L\theta\sin\theta\bigr), \qquad \theta=\beta t,$$

For the recorded dimensionless one-electron preparation $m=k=I=1$, $\lambda=0.2$, representative branches are

The recorded equilibrium and independent integration checks support actual four- and eight-petal signed projections. Those tested one-electron branches have no resolved exponential linear instability within the reported calculation. This is not a theorem of nonlinear stability or a fit of atomic constants.

The source magnitude and spectrum are conserved; $C=0$ remains zero. Constraining a common total source norm still permits different lock branches, but their prepared spectra differ. Thus source preparation and integer selection remain unresolved. The tested antipodal two-electron preparation is not stabilized by the one-electron result. The completed Coulomb-petal article records the autonomous equations, exact projections and scoped numerical checks.

Cylinders, breathing and petals are different views of one path

Ordinary orthogonal changes of observation axes connect several appearances of the constant-radius trajectory. Its two orthogonal plane motions may be written, using a complex symbol only as notation for each pair of real spatial coordinates,

$$u_1+iu_2=\frac{\rho}{\sqrt2}e^{i(1-L)\theta},\qquad v_1+iv_2=\frac{\rho}{\sqrt2}e^{i(1+L)\theta}.$$

Mix the two planes by an observation angle $\chi$. The end-on readout is

$$X+iY=\frac{\rho}{\sqrt2} [\cos\chi\,e^{i(1-L)\theta} +\sin\chi\,e^{i(1+L)\theta}],$$ $$\boxed{r_{\rm end}^2=\frac{\rho^2}{2} [1+\sin(2\chi)\cos(2L\theta)].}$$

At $\chi=0$ the end-on image is circular while the third observed coordinate oscillates along a cylinder. At intermediate angles it breathes between nonzero inner and outer radii. At $\chi=\pi/4$ it is the signed projection $\rho\cos(L\theta)e^{i\theta}$. The full physical radius remains unchanged throughout.

The signed projection and the folded display $\rho|\cos(L\theta)|e^{i\theta}$ must not be conflated. For even integer $L$, their planar images agree. For odd integer $L$, the signed curve has $L$ petals rather than the folded display's $2L$: the signed $L=1$ case is an offset circle. Modulus folding is an additional readout operation, not ordinary linear spatial projection.

For uniform $\theta$, the projected squared radius $s$ has dwell-time density

$$f_s(s)=\frac1{\pi\sqrt{(s-s_{\min})(s_{\max}-s)}}, \qquad s_{\min}<s<s_{\max}.$$

Turning radii are enhanced because radial motion slows there. This is time spent at definite positions, not a position-probability cloud. A commensurate path reaches turning radii at finitely many angles; ring-like enhancements do not imply complete concentric circles. Irrational clock ratios have an annular infinite-time closure, while any finite trace remains a curve. Tuned long-time ring structure and untuned nearly uniform pictures must therefore be distinguished by their preparation, trace time and detector response.

The plain circular orbit is the working lowest-energy $s$-shape picture; tuned ringed cylindrical examples are higher-energy candidates. That qualitative ordering is not a derived atomic spectrum or identification of every displayed ring with a measured state. The projection analysis supplies the exact view transformation and occupied-depth controls. The interactive trajectory remains an illustration of specified preparations, not a replacement for a derived detector law.

Field coefficients and an apparatus map

The older observer construction gives useful mathematics when its additional premises are kept explicit. Let $z_j$ be paired real field coefficients, written in complex notation. Normalized shares and relative phases,

$$p_j=\frac{|z_j|^2}{\sum_k|z_k|^2},\qquad q=\frac{|z_1|^2}{|z_1|^2+|z_2|^2},\qquad e^{i\phi}=\frac{z_3\bar z_4}{|z_3z_4|},$$

are invariant under a common nonzero complex rescaling on their legal chart. They are algebraic coefficient data, not electron position probabilities.

If an aligned plane readout is restricted to $(q,\phi)$, rotational covariance gives

$$W(q,\phi)=R h(q)e^{i\phi}.$$

Reflection alignment makes $h$ real. Requiring a nonnegative, continuous, strictly increasing radial response $f$, with $f(0)=0$, is a further apparatus assumption. For a coefficient flow with $q=\cos^2(L\theta)$, this class preserves $2L$ separated radial maxima in the folded readout. It does not prove that the physical apparatus chooses $f$, that the coefficient flow is the position trajectory, or that a folded curve is the signed spatial projection above.

Given an actual trajectory and a justified sampling law, a finite detector kernel can describe accumulated registration:

$$\rho_{\rm obs}(x)=\frac1T\int_0^T K_\sigma(x-\Gamma(t))\,dt, \qquad K_\sigma\ge0,\quad\int K_\sigma=1.$$

The curve measure is one-dimensional before broadening. The kernel width, shape and sampling weights must come from the electron–apparatus interaction. They are not observations of an indeterminate electron position or a direct measurement of its internal shape. The observer-map module gives the restricted covariance theorem; the actual-orbit correction keeps physical position, coefficient descriptions and registration distinct.

Nuclear direction, frame orientation and uniform extension

The attraction must retain enough directional information to act on the actual relative position and the structures of both participants. A scalar rest-frequency term by itself does not generate a nontrivial spatial precession operator. A chosen observer-relative grouping into two three-planes establishes no electron $3+3$ split: the electron's six directions are the base four and one hidden pair (volume 2), and since a nucleus is uniform across the pairs, only the base four coordinates change in an atom.

In a conditional two-frame description, write a real orthogonal complex structure as

$$J=\begin{pmatrix}A&Q\\-Q^T&C\end{pmatrix}, \qquad J^2=-I, \qquad s=\tfrac12\|A\|_F^2.$$

Its exact algebra fixes $0\le s\le1$ and

$$\sigma(Q)=\{1,\sqrt{1-s},\sqrt{1-s}\}.$$

At $s=0$, $Q$ is orthogonal and, with the chosen orientation, lies in $SO(3)$. A potential preferring $s=0$ does not select one oriented $Q$ from this whole family. Under the specified reduced Hamiltonian, the conditional flow

$$\dot J=\frac{2V'(s)}{\mathcal I}[PJP,J]$$

conserves $s$; an energy minimum is not automatic capture. This older homogeneous-space reduction requires its stated geometry and symplectic normalization. It is not a derivation of the current flat trajectory from nuclear attraction. The binding module distinguishes the exact frame algebra from that physical bridge.

The minimum classical orientation source is smaller than a full-rank matrix. Relative phase-space data can provide the mixed orbital bivector

$$E=x_O p_H^T-p_O x_H^T.$$

If both pairs are noncollinear, $E$ has rank two. For a real rank-two tensor $E=U\operatorname{diag}(\sigma_1,\sigma_2,0)V^T$, $\sigma_i>0$, choose $U,V\in SO(3)$. The orientation-preserving Procrustes problem has the unique maximizer

$$\max_{Q\in SO(3)}\operatorname{tr}(Q^TE), \qquad Q=UV^T.$$

Two oriented directions determine the third. Rank-one or zero data leave a continuous ambiguity. This is useful orientation geometry; a complete interaction still has to produce and retain that mixed bivector rather than average it away. The rank-two orientation correction supplies the proof and degenerate cases.

Uniform extension through additional directions raises a separate invariance question. In a product-coordinate reduction with $\rho_e(x,y)=\rho_e(x)$, an overlap

$$U=g\int dx\,dy\,\rho_e(x)\rho_h(x,y) =g\int dx\,\rho_e(x)\bar\rho_h(x)$$

is unchanged by an outer translation when the full fibre and measure are translation invariant. Its corresponding outer force is zero. But full evolution must also avoid generating outer variation: for an appropriate projector $P_u$, a linear generator needs $(I-P_u)HP_u=0$, and a nonlinear equation needs $(I-P_u)F[P_uu]=0$. Finite overlap is not proof of finite whole-space energy on an infinite uniform fibre. The shared-manifold calculation separates these force, invariance, normalization and transport requirements.

Finite nuclear encounters and reciprocal dynamics

A nucleus that is cylindrically extended through directions it does not support is encountered when the supported coordinates coincide. A nonzero electron coordinate outside those directions does not automatically make a projected crossing a physical miss. This is a consequence of the stated extension convention, not evidence that an ideal point singularity is physical.

A finite supported-space nuclear profile regularizes the encounter while preserving cylindrical extension. In a reciprocal diagnostic, let nuclear centers $R_a$, momenta $P_a$, rank-three support projectors $\Pi_a$ and skew angular velocities $\Omega_a$ evolve. The source coupling must use the relative electron–central-nucleus coordinate $s=q-(R_0,0)$, so

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

One stated Hamiltonian is

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

Every force and torque is varied from that energy, including

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

This diagnostic conserves energy and its applicable total momentum/angular charges. The nuclear center-charge repulsion is a separately stated approximation, not the interaction of overlapping finite densities with arbitrary support planes. The full skew rotor also contains stabilizer angular momentum unless it is reduced to a horizontal plane rotor. The reciprocal molecular model gives the exact assumptions and checks; it does not establish a stable reciprocal periodic molecule.

Finite plane response needs both orientation and conjugate angular momentum. $\operatorname{Gr}(3,6)$ has nine configuration dimensions and cannot alone carry a nondegenerate symplectic two-form; its cotangent phase space has eighteen. These are orientation parameters, not extra physical spatial dimensions. A forced susceptibility above resonance is not a closed-system stability theorem. The finite-response analysis makes those distinctions without deriving a nuclear rotor from the field.

Molecular directions: exact geometry and actual force tests

Conservation can constrain angles once the interaction assumptions are specified. If equivalent nuclear passages exchange equal impulses along unit directions $n_a$, periodic momentum closure gives $\sum_a n_a=0$. If every distinct pair has the same inner product $c$, then

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

Thus three equivalent directions give $120^\circ$, while four give $\arccos(-1/3)=109.4712206\ldots^\circ$. Equal impulses and equivalence remain physical premises to derive. The same simplex angles occur in an external conventional hybrid-orbital comparison, but that is not the deterministic IDWT binding mechanism or independent evidence of depth six.

The signed $L=3$ trajectory directly has three outward maxima at $0,2\pi/3,4\pi/3$. A planar $L=4$ rose has eight directions, not a tetrahedron. A distinct constant-radius spatial curve,

$$q(\theta)=\bigl(\sqrt{2/3}\cos\theta, \sqrt{2/3}\sin\theta, (1/\sqrt3)\cos2\theta, (1/\sqrt3)\sin2\theta,0,0\bigr),$$

has four quarter-turn projected directions with pairwise dot product $-1/3$. This is kinematic availability; it does not show the completed interaction sustains that route. See the orbit-axis construction and simplex geometry.

Actual ligand force tests are more restrictive than those angles. With a prescribed electron route, fixed central nucleus and finite cylindrical nuclear profiles, the three-petal example gives a strict trigonal minimum against all nine ligand displacement coordinates. At profile radius $0.3$, diagnostic charges $(Z_0,Z_L)=(1,0.1)$ and common radius $0.937604839$, its Hessian eigenvalues lie between approximately $0.548468$ and $0.659033$.

The nominal four tetrahedral rays are not stationary after averaging the entire stated route: the transverse gradient is nonzero. Free ligand relaxation in that same approximation gives a disphenoid, with four angles $105.31619^\circ$, two $118.14528^\circ$, and positive ligand-Hessian eigenvalues approximately $0.183516$ to $0.608247$. These dimensionless model results are profile-dependent, not fitted molecular predictions. The electron route has not been relaxed reciprocally; the fixed-route surface is not a derived Born–Oppenheimer energy. The shape-stability calculation records the quantitative correction.

A radial force root, a positive fixed-route ligand Hessian and a recurrent full molecule are therefore different accomplishments. Stability of the latter must include electron, angular source, nuclear translations and plane-rotor perturbations under one conserved action. A frozen-path angle match cannot replace that test.

What the orbit program has established

The useful present content is concrete: actual deterministic curves, an exact flat force equation, autonomous Coulomb-plus-source relative equilibria, a precise relation between cylindrical and petal views, conditional radial persistence, a reduced minimum orientation source, and conservative reciprocal nuclear equations. These show that higher-dimensional literal motion is mathematically workable. The later sections add an exact precession ratio from the quark charges, the universality of that ratio across nuclei, a hydrogen line at one Bohr radius and molecular comparisons. A universal integer lock, hydrogen's binding energy on the same orbit as its line, multi-electron exclusion, stable molecular structures and the detector sampling law remain to be derived.

The same conclusions must be compared across occupied depths two, three, four, five, six and ten under one law. The four-direction orbit's embedding into higher depths is compatibility, not a six-only discriminator. None of the internal count-field or localization results supplies this external orbit merely by having the same frequency dimensions.

The sector case keeps geometric capability distinct from selective evidence. The neighboring accounts of same-field coupling, particle-pair encounters and localization describe further requirements on a common physical law. Binding must connect these to literal positions and reciprocal motion rather than replace them with conventional electron clouds.

A Scoped Orbit-Stiffness Control

For the separately supplied attractive scalar potential $-g|x-q(t)|^{-s}$, put $q=a n(t)$ with $|n|=1$ in the four active orbit directions and $Q_{ij}=\langle n_i n_j\rangle$. Differentiating the time-averaged potential at the centre gives

$$K_{ij}=gs\,a^{-s-2}\bigl[\delta_{ij}-(s+2)Q_{ij}\bigr].$$

The nonexceptional integer-ratio prepared orbit already has $Q=I_4/4$. Therefore this control has $K_{ww}=gs\,a^{-s-2}(2-s)/4$: restoring for an attractive tail with $0<s<2$, and zero at $s=2$. A positive entry along one direction would not establish a trap in all four directions. This is a calculation for the stated scalar law and prepared orbit, not the force law of the full multi-orientation particle patterns. Actual nuclear alignment requires their reciprocal field interaction and its complete spatial response. The independent assessment also corrects the author uniformity check's energy exponent; that check cannot validate the maintained density-law response as labelled.

At the zero-Hessian endpoint, the lock-three orbit has a cubic leading term, $-3g\epsilon^3/(4a^5)$ along the stated first coordinate and the opposite sign along the fourth. Locks two and four have inversion symmetry and no odd moments. A symmetric displacement average would miss that cubic. This correction concerns the specified scalar control.

Real Fourth-Direction Composite Escape

The real-spatial decay construction develops a conditional alpha-escape picture using the assigned quark depths. Up quarks involve four bound spatial directions, down quarks three. An alpha is the whole two-proton/two-neutron composite, with six up and six down valence constituents; this proposal does not describe individual free quarks escaping. The electron retains its full deterministic six-dimensional position and trajectory even where four coordinates supply the displayed nucleus-bound orbit.

The mathematical bridge is exact for the specified second-order scalar operator with matching coefficients and boundaries: its Euclidean continuation is the same elliptic operator as a static problem with one extra real spatial direction. Equivalence to the first-order nonlinear density-spinor dynamics and a physical decay rate remain to be established.

A chosen covariantly turned trial gives a conditional arclength contribution. Its exact reduced constant-tension graph equation is $T X''/(1+X'^2)^{3/2}=U'(X)$; $T X''=U'$ is the small-slope limit. Rotation covariance alone does not establish an exact relaxed curved solution, and the displayed Dirac trial is not a stable elastic minimum under all orientation changes. The spatial cost needs a field-action conversion and outgoing composite evolution before it becomes a dimensionless decay exponent. The tension itself is calculated in volume 12.

The alpha exponent and the prediction

Alpha decay, in which the quarks survive and the nucleus comes apart, needs no tunnelling or imaginary time: the fourth direction is the natural path for that motion. Across 142 even-even alpha emitters, spanning 30 decades of half-life, the exponent computed from the three-direction Coulomb barrier and the net electric charge has a slope of 0.956 against the expected 1, with an rms scatter of 0.98 decades. That is what this path gives when the barrier is unchanged along it. Replacing the charge by the up-quark count $(2Z+N)/3$ lowers the slope to 0.734 and raises the scatter to 1.63 decades; the down-quark count $(Z+2N)/3$ gives 0.559 and 2.43; a form linear in barrier height gives a scatter of 5.60 decades. The exponent follows net charge.

The up quarks are localized along the fourth direction, so the up-up part of the barrier varies along the path, and the exponent becomes the minimum over paths in $(r,w)$. This has not been calculated, and neither has the exit of a composite alpha from its daughter.

The prediction is that the same nuclide, stripped of its electrons, alpha-decays faster than conventional calculation allows. Electrons hold the fourth-direction position and so restrict the path. An order estimate of the electron hold is 0.16 eV. The measured rate of hydrogen-like francium-213 relative to the neutral atom, $1.004\pm0.18$, limits the change to about 18% so far. The size of the effect is not derived.

Precession from the quark charges

The electron's position relative to the nucleus has four coordinates, $x_1,x_2,x_3$ in the observed directions and $w$ in the fourth. An up quark is bound in all four directions and acts through $\rho=\sqrt{x_1^2+x_2^2+x_3^2+w^2}$. A down quark is bound in the observed three and spread evenly along the fourth, so it acts through $r_3=\sqrt{x_1^2+x_2^2+x_3^2}$ and has no grip along $w$. For a proton the up quarks pull with $a=4/3$ and the down quark pushes with $b=1/3$:

$$V=-\frac{a}{\rho}+\frac{b}{r_3},\qquad \ddot x_i=-\frac{a\,x_i}{\rho^3}+\frac{b\,x_i}{r_3^3},\qquad \ddot w=-\frac{a\,w}{\rho^3}.$$

At small tilt this is $V=-1/r+(2/3)w^2/r^3$. A circular orbit runs at $\Omega^2=(a-b)/r^3$, and a small fourth-direction displacement oscillates at $\omega_w^2=a/r^3$, so

$$\boxed{\frac{\omega_w^2}{\Omega^2}=\frac{a}{a-b}=\frac43 .}$$

The fourth-direction motion runs $\sqrt{4/3}=1.154701$ times as fast as the orbit, and the orbital plane turns by 0.154701 of a revolution per circuit, independent of the radius. Direct integration at a tilt of $0.5^\circ$ gives 0.154708. The drift grows with tilt: 0.1577 at $10^\circ$, 0.1671 at $20^\circ$ and 0.1847 at $30^\circ$. With no down-quark term the force is central, every one of 90 integrated orbits stays planar, and there is neither precession nor a second line. The non-central term is positive, so it resists tilt. The precession exists because the down quark is bound in one direction fewer than the up quark.

The line. The light of a spectral line has the frequency of this fourth-direction motion. The in-plane frequency matches no line.

The neutron. For a nucleus of $Z$ protons and $N$ neutrons, the orbital rate depends only on $a-b=Z$. The precession rate depends on $a$, so the spectra decide how neutrons act. The line relative to hydrogen's is $\sqrt{aZ^3/(4/3)}$:

The measured column is $Z^2$ times the reduced-mass factor. If a fraction $x$ of a neutron's up-quark pull acted, the deuterium line would be raised by $\sqrt{1+x/2}$ beyond the reduced mass, which the measured isotope shift excludes to better than a part per million. A neutron's quarks are in direct contact and its charge is negated completely, in the fourth direction as in the other three. Every nucleus therefore has $a=4Z/3$ and $b=Z/3$, and the ratio $\omega_w^2/\Omega^2=4/3$ holds for every atom. The precession is universal, the hydrogen-like lines scale as $Z^2$, and isotopes differ only by the reduced mass.

Positronium and muonium. These contain no quarks, so their attractor is a point, the orbit is planar and there is one frequency. They have no second line of this kind. Their spectra are a stated constraint on the account.

Zeeman. Larmor's theorem is exact for rotations in the observed directions. It gives the normal triplet on a line in the $(x_1,x_2)$ plane and none on the $w$ line, and a $(x_1,w)$ plane has no triplet.

The orbit with a nucleus of finite width

The proton pattern has finite width. Each charge softens over its own radius:

$$V=-\frac{4/3}{\sqrt{\rho^2+R_u^2}}+\frac{1/3}{\sqrt{r_3^2+R_d^2}},\qquad R_u=0.01524\,s,\quad R_d=0.00653\,s,$$

in Bohr radii, with one width parameter $s$. At $s=100$, $r=1$ and a tilt of $10^\circ$ the line is 0.4693522 hartree, 12.7717 eV, 97.08 nm. Lyman-γ is 97.20 nm for an infinitely heavy nucleus, a difference of 0.13%, and 97.25 nm as measured, a difference of 0.18%. An exact match needs $s=100.12$. The in-plane frequency at the same orbit is 0.1547 hartree, 294.6 nm, and matches no line. A point nucleus gives 0.99998 and 1.15761 hartree, which match no hydrogen line.

Tilt. The line moves only 1.4% as the tilt runs from $5^\circ$ to $85^\circ$, from 0.46923 to 0.47594 hartree. The force law fixes the line.

Binding energy. The circular orbit with Lyman-γ at one Bohr radius and the circular orbit with hydrogen's binding energy are different orbits. At $s=100$ and $r=1$ the energy is $-0.440$ hartree; the lowest circular energy is $-0.4536$ at $r=0.906$. Hydrogen's $-0.4997$ is reachable only for $s\lesssim90.5$, and at $s=90$ the orbit that has it lies at $r=0.841$ with line 0.566 hartree. The line at $r=1$ is 0.522, 0.511, 0.495 and 0.469 hartree at $s=90,92,95,100$. The angular momentum of the $s=100$ orbit is 0.156, below $\hbar/2$. Eccentric orbits with hydrogen's energy have not been examined.

Other hydrogen checks bound the width. The well depth is $1234.2/s$ eV, which fails 13.598 eV by about 10% at $s=100$ and needs $s\le90.8$. Muonic lines bound $s\le0.65$ at 1898 eV and $s\le0.49$ at 2528 eV, far below the width that gives the Lyman-γ match; this conflict is open. The isotope exponent at fixed angular momentum is $-0.84$ at $s=100$ and $1.000$ at $s=0.1$.

A thin slab. The zero-point energy of the fourth-direction motion, half its frequency, is 0.235 hartree, about twenty times the orbit's kinetic energy of 0.012. The electron's spread along the fourth direction at that frequency is 0.73 Bohr radii and the nucleus's is 0.10. Atoms are thin slabs in the fourth direction, thinnest at the nucleus.

What holds the orbit's plane in the observed directions. The uniform background couples to an electron-type pattern through a cross term. The coupling vector is exact and lies on exactly one pair of directions, the pair whose occupation differs from the background's, with the circular combination $(1,\pm i)$. For the two electron-type seeds the pair is $(3,4)$ or $(1,2)$, and both lie in the base four. A bound set containing the base four holds the vector in full; a bound set tilted out of it holds less. This is the first link found between the electron's bound set and the observed directions. The sign, restoring force and rate need the second-order energy.

The exact orbit family

The orbit family of the exact trajectory above has components at $|L-1|\beta$ and $(L+1)\beta$, with $H_L=mr^2\beta^2L^2$ and $k_L=(L^2-1)\beta^2$, and the potential share of the energy at fixed total is $(1-1/L^2)/2$. Finite-pulse absorption peaks exactly at $L\pm1$, with a width set by the pulse, and absorbed energy lowers the rates. The energy shares have the form of the Rydberg differences, and with hydrogen's radius and binding energy supplied one component lies 0.62% to 0.65% from Lyman-β. The scale is not derived.

Integer locking did not appear in any of seven nucleus and field models: point, fixed triangle, freely spinning triangle, radiation reaction, mutual feeding at four couplings, and finite size. The same holds for a second electron rotating in a hidden plane $(5,6)$ at weak coupling, scanned from 0.1 to 12 times the orbital rate over 300 orbits each. There is no lock at 1:1 and the orbit is undisturbed at equal rates. Two sharp resonances appear: near 3.6 times the orbital rate the orbit's energy changes by up to 2.4%, and near 5.9 times the fourth-direction amplitude rises from 0.059 to 0.096 with the energy undisturbed. At its true size a triangle of quarks is indistinguishable from a point.

Shells on the three-sphere

An orbit at fixed radius in four directions lies on a three-sphere. Harmonics of degree $m$ on the three-sphere have dimension $(m+1)^2$: 1, 4, 9, 16. Two electrons per pattern give the shell capacities 2, 8, 18, 32. The harmonics branch under rotations of the observed three directions into $l=0,\dots,m$, each once, which matches the subshell content of the periods. The mathematics is standard $SO(4)$ (Fock 1935, Pauli 1926, Bargmann 1936, Moser 1970, Kustaanheimo and Stiefel 1965). The new content is the physical reading: here the three-sphere is the shell of a real orbit in four directions.

The degree fixes which $l$ occur, not the principal number. Grouping by $n+l$ gives 2, 2, 8, 8, 18, 18, 32, 32, as the periodic table has it, but the doubling of the periods has no mechanism yet. The total energy differing between branches as $1/a^2-1/b^2$, with the scale derived, is still owed.

Equal circles in four directions generically do not meet. They are equidistant if and only if they are isoclinic, that is, Clifford parallel and members of one Hopf family; non-isoclinic circles approach twice per revolution. A shell's average fourth-direction fraction is exactly 1/4. Four mutually repelling pairs gain nothing by leaving the observed three directions, by the Gram bound $-1/3$.

Bonds and molecules

In a molecule the electron's orbit goes around all the nuclei involved. Sharing is literal. Multiple bonds are electrons on equal orbits kept apart by phase, so a multiple bond is one orbit with several phase slots and bond order is a count of those slots; this has not yet been computed. Two-centre bonding orbits are not circles about a common centre.

Halogens. With two nuclei at the measured bond length $R$ and the electron around both at $r=1+R/2$ Bohr radii:

The line moves into the visible with the bond length, in the order of the measured colours, and one factor of about 1.75 in energy carries it to all four maxima within 9%. That factor is open.

Polyenes. With the electron pair orbiting a chain of $N$ carbons spaced 1.40 Å apart:

Ethylene comes out at 170 nm against 165 nm measured. The line moves to longer wavelengths as the chain grows, but the orbit around the whole chain is too red from four carbons on, 3.4 times the measured wavelength at 22 carbons. A racetrack path along the chain holds the line at 80 to 100 nm at every length, too blue. The measured series lies between the two paths, and the path that gives it is open.

Angles. The pressing weights for water and ammonia are 0.803 and 0.799. A compensation tilt $s=1/3$ gives 104.48° for water. Ammonia follows through $|w|^2=2s^2$. Bends computed from the same picture give 1832 cm$^{-1}$ for NH$_3$, 1.93 times too high, and 1111, 1010 and 845 cm$^{-1}$ for PH$_3$, AsH$_3$ and SbH$_3$, 8% to 12% high. Gas-phase NO$_2^-$ gives 117.5°. The origin of the tilt is open and a two-centre calculation with free nuclei has not been run.

Fine-structure beats. The excess drift is approximately $0.10\sin^2(\text{tilt})$ for the proton and $0.25$ for $N=Z$ nuclei, whose non-central term is $(Z/2)w^2/r^3$. The tilts in that comparison are read from the lines, not derived.

Quarks in orbit

Colour attraction can put quarks into orbit about one another, given sideways motion, as electric attraction does for electrons. Opposite charges spread around the same circle cancel in the three observed directions. The up quark's pull along the fourth direction remains unless the negation is complete, and the spectra above require it to be complete. Contact provides that. A colour orbit is compatible with the neutral neutron as long as the negation stays complete.

Variable-Slope Contact Can Contribute to Four-Direction Holding

For a smooth central contribution $U<0$, set $s=-\mathrm d\log(-U)/\mathrm d\log R$. The maintained orbit second moments give

$$K=\frac14\left(U''+\frac{3U'}R\right) =\frac{-U}{4R^2}\bigl[s(2-s)+R s'\bigr].$$

This extends the fixed-power control: an increasing local slope adds positive stiffness, so a variable tail can stay restoring beyond its local inverse-square crossing. The independently assessed contact-curvature construction supplies the derivation and frozen-profile examples. Their root locations come from a twelve-sample spline and lack derivative convergence or physical length calibration. They support a conditional positional mechanism, not a complete bound nucleus, orientation theorem or calibrated atomic exclusion.