Competing Ranges from the Shape Energy of a Geometric Guide

A geometric guide can produce long-range attraction and a shorter-range outward response without postulating two independent force laws. In the existing neutral-frame action, guide tension supplies the first contribution; the cost of matching different endpoint shapes supplies the second. The calculation is conditional on a supported commuting guide and prescribed small endpoint deformations. It also reveals why a radial force balance alone does not establish a stable composite: freely moving endpoint shapes can rotate toward agreement.

This uses the independently varied connection version of the mixed neutral-frame action. Its reduction from the single field remains a physical premise. The later canonical J-only connection and determinant orientation actions have different variations and cannot inherit this guide by identification of notation.

1. Two Shape Channels Already Present in the Frame

On the active ordinary four-carrier retain two real spatial deformation vectors a,b with common canonical gradient and restoring energy, and

$$\epsilon=\kappa_2(|a|^2+|b|^2)+\kappa_4|a\wedge b|^2, \qquad \kappa_2,\kappa_4,\Omega>0.$$

Let u_0(y) be a balanced fixed-flux guide profile on the transverse space R^(d-1), with electric field E_0=Q/I_0 and tension T>0. Set mu^2=Omega^2-kappa_2 E_0^2>0. In its fixed ordinary spatial two-plane write

$$(a\ b)=\frac1{\sqrt2} \begin{pmatrix}u_0+v&w\\w&u_0-v\end{pmatrix}, \qquad r^2=v^2+w^2. \tag{1}$$

The two omitted four-carrier rows are zero in this subspace. Direct Gram identities give

$$|a|^2+|b|^2=u_0^2+r^2,\qquad |a\wedge b|^2=\frac{(u_0^2-r^2)^2}{4},$$ $$|\partial a|^2+|\partial b|^2 =|\partial u_0|^2+|\partial v|^2+|\partial w|^2. \tag{2}$$

Thus these are actual retained frame deformations, not introduced mediators. Their first dielectric variation vanishes. Expanding the complete fixed-flux electric energy Q^2/(2I) therefore contributes -E_0^2 delta I/2 at quadratic order, with no squared first-variation term. The full shape energy is

$$E^{(2)}=\frac12\int\left( |\nabla v|^2+|\nabla w|^2+ \left[\mu^2+\frac{\kappa_4 E_0^2}{2}u_0^2\right](v^2+w^2) \right). \tag{3}$$

At this order the electric displacement need not change: its first source variation is zero. The same local quadratic expression follows by retaining and minimizing the electric variation in the complete Hamiltonian. Relaxing the stationary background at second order does not change its second variation. This does not solve the nonlinear finite guide or its end caps.

Both channels have the same transverse operator

$$\mathcal A=-\Delta_y+\mu^2+\frac{\kappa_4E_0^2}{2}u_0(y)^2 \ \ge\ \mu^2 I>0. \tag{4}$$

For the actual balanced guide u_0=A v_n(mu y), its profile equation fixes A^2=2mu^2/(kappa_4 E_0^2), hence

$$\mathcal A=\mu^2[-\Delta_{\mu y}+1+v_n(\mu y)^2]. \tag{5}$$

The short response scale is tied to the existing guide width mu^(-1). Identifying Omega with the sector fundamental remains an additional hypothesis. No isolated transverse mode below the gap exists, because the profile potential is positive. The following calculation retains the complete transverse operator and its continuum.

2. Exact Finite-Segment Energy

Let x run along an ordinary spatial segment of length L. Write z=(v,w), with endpoint shapes g_1(y),g_2(y). These are slow shape coordinates, which can be constructed from phase-invariant field bilinears. Static shape data do not impose equal rapid carrier frequencies on distinct particles. The quasistatic boundary approximation and endpoint-source work remain explicit.

The conditional quadratic segment energy is

$$U[z,L]=TL+\frac12\int_0^L \left(\|z_x\|^2+\langle z,\mathcal A z\rangle\right)dx. \tag{6}$$

Let K=sqrt(A) by positive self-adjoint functional calculus. For sufficiently regular finite-energy endpoint data, strict convexity gives the unique minimizer

$$z(x)=\frac{\sinh[K(L-x)]}{\sinh(KL)}g_1 +\frac{\sinh(Kx)}{\sinh(KL)}g_2. \tag{7}$$

Each ratio is a bounded spectral function; no separate exponentially unbounded operator is applied to the endpoint data. Integration by parts using z_xx=A z gives the exact on-shell result

$$U(L)=TL+\langle g_+,K\tanh(KL/2)g_+\rangle +\langle g_-,K\coth(KL/2)g_-\rangle, \qquad g_\pm=(g_1\pm g_2)/2. \tag{8}$$

The independent longitudinal stress is constant:

$$P=\frac12(\|z_x\|^2-\langle z,\mathcal A z\rangle), \qquad U'(L)=T-P. \tag{9}$$

Thus the inward force is F_in=U'(L); the two endpoints feel opposite forces. It is derived from the same energy as the shape torque. This is a force within a prescribed guide preparation, not already an observed quark force.

3. When Attraction Increases with Separation

For g_i=q_i f with a common transverse shape f and real two-component shape polarizations q_i, equation (8) gives

$$F_{\rm in}=T+ \frac{|q_+|^2}{2}\langle f,\mathcal A\operatorname{sech}^2(KL/2)f\rangle -\frac{|q_-|^2}{2}\langle f,\mathcal A\operatorname{csch}^2(KL/2)f\rangle. \tag{10}$$

The last term is outward. If q_1 dot q_2<=0, then |q_-|^2>=|q_+|^2. For every spectral k>0,

$$\frac{\operatorname{csch}^2 z\coth z} {\operatorname{sech}^2 z\tanh z}=\coth^4z\ge1, \qquad z=kL/2.$$

Differentiating (10) under the spectral integral proves that F_in increases with L and tends to T. It can be outward at small separation; where it is inward it strengthens as the shorter-range outward response fades. The long-range force saturates rather than growing without bound. The energy still grows linearly. These are distinct statements.

For opposite shapes g_2=-g_1=g up to the sign convention,

$$F_{\rm in}=T-R(L),\qquad R(L)=\frac12\langle g,\mathcal A\operatorname{csch}^2(KL/2)g\rangle>0,$$ $$R(L)\le\frac{\mu^2}{2}\operatorname{csch}^2(\mu L/2)\|g\|^2 \le\frac{2\|g\|^2}{L^2}. \tag{11}$$

The first bound uses the decreasing function k^2 csch^2(kL/2) and K>=mu. It bounds the entire continuum response by an exponentially decreasing large-L envelope. There is no finite transverse box or invented bound mode.

4. The Same Energy Permits Alignment

Take equal endpoint norms and let theta be the angle between q_1 and q_2. With |q_i|=q, the angular derivatives are

$$U_\theta=q^2\sin\theta\,\langle f,K\operatorname{csch}(KL)f\rangle,$$ $$U_{\theta\theta}|_{\theta=\pi} =-q^2\langle f,K\operatorname{csch}(KL)f\rangle<0. \tag{12}$$

Opposite shapes are an angular maximum even when their separation sits at a radial energy minimum. An orthogonal pair generally has nonzero alignment torque. Allowing unconstrained orientations therefore removes the inference of a stable composite from the radial balance alone. Actual cores, conserved data or continuing rotation could change the encounter, but they must be derived rather than used as an implicit angular clamp.

The polarization angle here describes the traceless symmetric frame deformation in (1). Under ordinary simultaneous rotation of its physical two-plane and mode basis, (v,w) rotates by twice the spatial angle. Opposite shape coordinates need not mean opposite spatial vectors. They do not identify a particle with an antiparticle, electric charge or a color label.

The L->0 radial divergence is a property of the quadratic boundary problem. Large gradients or overlapping end cores can leave its valid regime. A formal finite radial minimum need not lie in the small-deformation domain.

There is a quantitative control on that last issue. Suppose opposite endpoint deformations obey ||g||^2<=delta^2 M_0, where M_0=int u_0^2 and delta is a small relative shape amplitude. At a formal force balance, equation (11) gives L_*^2<=2 delta^2 M_0/T. The actual mixed-frame guide moments give, with n=d-1 and t=kappa_2 E_0^2/Omega^2,

$$\frac{T}{M_0}=\Omega^2\frac{3+(1-n)t}{4-n},\qquad (\mu L_*)^2\le 2\delta^2\frac{(1-t)(4-n)}{3+(1-n)t} \le\begin{cases}4\delta^2/3,&d=3,\\2\delta^2/3,&d=4.\end{cases} \tag{13}$$

Thus small endpoint distortions cannot balance tension at separations much larger than the guide width. Their formal radial balance lies where a long uniform guide with independent end caps is not controlled. The increasing force correction at larger separations remains derived; a controlled hadron-sized equilibrium is a further problem, not an outcome of this linearization.

The dynamic endpoint continuation retains each drive frequency, source work and the actual longitudinal stress. Distinct count-harmonic cross terms average exactly to zero over a fundamental period, while individual response remains. Direct high-frequency forcing can open radiation channels; phase-invariant slow shapes can remain static without locking unequal mass clocks. These alternatives must not be interchanged.

5. Occupied-Sector Control and Physical Scope

No depth-specific deformation coefficient was introduced. The boundary mathematics would also work in another dimension if a supported positive-gap guide were supplied there. Its three/four restriction is inherited from the existing Hodge and guide premises; it is not an independent new color selection rule or an explanation of every possible higher-sector force.

The full non-Abelian compact-source alternatives still prevent extending the commuting tension bound to unrestricted classical sources. This result does not select observed particle indices or replace the simplex clock law. It supplies a concrete, reciprocal realization of a competing-range force within the existing geometric action, together with the orientation instability and endpoint assumptions that its physical use must address.

Reproducibility

The companion finite-guide-imbalance.py checks the exact frame identities, fixed-flux quadratic variation, scalar boundary solution, stress, force and torque. Independent numerical boundary-value solutions test the operator formula on positive matrix controls. These finite matrices verify the functional-calculus reduction; they are not claimed transverse guide modes. The continuum and force conclusions rely on the displayed operator proof.

Verification script and recorded results.