Relaxed Frame Twists Link Guide Force and Longitudinal Stiffness

The balanced neutral frame has continuous orientation modes as well as its gapped imbalance channels. A common rotation of its two vectors leaves the dielectric unchanged. Its longitudinal gradient costs energy, and the transverse guide can respond by changing its profile. This response can be calculated from the existing guide action rather than choosing a new potential or holding the profile fixed.

The construction remains in the independently varied commuting connection action. Its neutral vectors are retained deformation modes of the proposed single field, not a derived master-field reduction. End cores, full non-Abelian response and physical normalizations remain open.

1. A Uniform Frame Twist Solves the Bulk Equations

In an ordinary spatial two-plane let e_1(theta),e_2(theta) be its rotating orthonormal basis. Set

$$a=\frac{u(y)}{\sqrt2}e_1(\theta),\qquad b=\frac{u(y)}{\sqrt2}e_2(\theta),\qquad \theta=\varpi t+kx. \tag{1}$$

Here x is the longitudinal direction and y the transverse coordinates. The rotation varpi is the neutral frame's rotation frequency; it has not been identified with a particle mass-frequency. Both vectors satisfy partial_theta^2 e_i=-e_i, and the dielectric remains

$$\epsilon=\kappa_2u^2+\frac{\kappa_4}{4}u^4.$$

Direct substitution into the full neutral equations gives

$$-\Delta_y u+ (\Omega^2+k^2-\varpi^2-\kappa_2E^2)u =\frac{\kappa_4 E^2}{2}u^3. \tag{2}$$

The angular equation partial_mu(u^2 partial^mu theta)=0 is also satisfied. The action retains a and b as physical deformation variables. Their global mode-rotation symmetry has not been declared a local gauge redundancy. Theta rotates these vectors within their occupied plane; it does not tilt that plane, rotate the sector or change the rapid particle carrier phase. The displacement is independent of x and t for this uniform preparation, so ordinary commuting Gauss and electric evolution remain satisfied in the source-free bulk. Consequently the established scalar guide profile applies with the single substitution

$$s=\Omega^2+k^2-\varpi^2>0,\qquad\Omega_{\rm eff}=\sqrt s. \tag{3}$$

This is an exact bulk family within the stated frame action, not a complete finite-source solution. The area-preserving rotation exists in ordinary three- and four-dimensional directions; it does not add a spatial dimension. It retains the existing odd-sector source-carrier premise where required.

2. Eliminate the Profile at Fixed Flux

Let T(Q,sqrt(s)) be the established globally minimizing scalar guide tension on a smooth finite-width bound branch. Write tau(s)=T(Q,sqrt(s)) and M(s)=int u_s^2. The envelope derivative is

$$\tau'(s)=\frac{M(s)}2>0. \tag{4}$$

The infimum is concave in s: before minimization its energy is an affine function of s for every transverse profile. Thus tau''<=0 on a smooth global branch, or M_s<=0. A branch crossing or loss of a finite profile requires separate treatment.

For a finite static uniform twist with endpoint angle Theta, k=Theta/L. Allowing the profile to relax gives the bulk energy

$$U(L,\Theta)=L\tau(\Omega^2+\Theta^2/L^2). \tag{5}$$

Its inward force and endpoint torque follow from this same energy:

$$F_{\rm in}=\tau-k^2M,\qquad \partial_\Theta U=kM. \tag{6}$$

The longitudinal stress computed directly from the neutral gradients and electric energy gives the same force. There is no separate imposed repulsive potential. For large L, with M_0=M(Omega^2),

$$U=T_0L+\frac{M_0\Theta^2}{2L}+O(L^{-3}),\qquad F_{\rm in}=T_0-\frac{M_0\Theta^2}{2L^2}+O(L^{-4}). \tag{7}$$

The orientation mismatch supplies an algebraically decreasing outward correction to tension. It is distinct from the exponential correction of the gapped imbalance channel. Both occur in the same retained frame action.

3. Profile Relaxation Removes the Frozen-Profile Radial Balance

The actual mixed-guide moment identities give, with n=d-1 and t=kappa_2 E^2/s in(0,1),

$$\frac{\tau}{M}=s\,C_n(t),\qquad C_n(t)=\frac{3+(1-n)t}{4-n}>1,\qquad n=2,3. \tag{8}$$

Therefore

$$\boxed{F_{\rm in}=M[\Omega^2C_n(t)+k^2(C_n(t)-1)]>0.} \tag{9}$$

The relaxed uniform guide is inward-pulling at every supported twist in this family. Extrapolating the frozen-profile 1/L correction to small L would incorrectly create a radial equilibrium. Equation (9) retains all electric, restoring and transverse gradient energy. Finite end-cap work still prevents identifying (5) with the exact energy of arbitrary particles.

4. The Force Slope and Phase Stiffness Are the Same Quantity

The slowly varying longitudinal reduction has fixed-flux action

$$L_{\rm eff}=-\tau(\Omega^2+\theta_x^2-\theta_t^2). \tag{10}$$

The electric term uses its fixed-flux Routh reduction, not the unvaried positive electric Lagrangian. Profile derivatives and the transverse fields required by nonuniform Gauss data produce additional higher-gradient corrections. Uniform jets (1) are exact; (10) is a leading long-wavelength reduction for their modulation, not an exact action for arbitrary theta.

At a static uniform twist its phase time coefficient is M>0 and its longitudinal spatial stiffness is

$$G=M+2k^2M_s. \tag{11}$$

Differentiating the actual force (6) at fixed Theta gives the exact identity

$$\boxed{\frac{dF_{\rm in}}{dL}=\frac{k^2}{L}G.} \tag{12}$$

Thus increasing inward force with separation is precisely the positive phase-stiffness regime of this reduction. Since M_s<=0, that regime has phase speed squared G/M in(0,1]. If G<0 the reduced phase channel has long-wavelength negative stiffness. This does not establish full frame, connection or endpoint stability; other orientation modes are retained in the original theory and need their own test.

For a homogeneous frame rotation k=0, the time coefficient is M-2varpi^2 M_s>=M, while the spatial coefficient is M. Its reduced phase speed is at most one wherever the smooth supported branch exists. These are propagation properties of the retained-component action; the original neutral spatial vectors' full spacetime tensor completion remains open.

5. The Pure Quartic Guide Gives a Closed Example

With kappa_2=0, the established sharp tension scales as

$$\tau(s)=T_0(s/\Omega^2)^a,\qquad a=\frac{4-n}{6}.$$

Thus a=1/3 at depth three and a=1/6 at four. For y=k^2/Omega^2,

$$\frac{F_{\rm in}}{T_0}=(1+y)^{a-1}[1+(1-2a)y]>0, \qquad \frac GM=\frac{1-(1-2a)y}{1+y}. \tag{13}$$

The force has a minimum at y=1/(1-2a), exactly where this phase stiffness changes sign. At larger separations it increases toward T_0 from below. At very small formal L the relaxed bulk energy behaves as L^(1-2a), rather than the frozen profile's 1/L divergence. This is profile concentration, not a predicted short-distance hadron potential.

The pure quartic example reuses the existing guide law. The mixed positive quadratic correction has a different tau but retains equations (4), (6), (8), (9) and (12) on its supported global branch. No particle-specific coefficient was added to obtain the distinction.

6. An Exact Full-Frame Tilt Test

The phase stiffness is not the entire orientation test. Tilt one vector into an ordinary third direction e_3, while preserving its norm and its orthogonality to the other vector:

$$a=\frac{u}{\sqrt2} [\sqrt{1-\zeta^2}\,e_1(\theta)+\zeta e_3],\qquad b=\frac{u}{\sqrt2}e_2(\theta).$$

The dielectric is unchanged exactly. At quadratic order the actual frame action for a transversely constant tilt is

$$L_\zeta^{(2)}=\frac{M}{4}\int [\zeta_t^2-\zeta_x^2+(k^2-\varpi^2)\zeta^2]dx, \qquad \zeta_{tt}-\zeta_{xx}+(\varpi^2-k^2)\zeta=0. \tag{14}$$

This also follows directly by applying the full linearized vector equation to delta a=u zeta e_3/sqrt(2), with no profile elimination. The electric source has zero first variation, so no connection perturbation is needed in this channel. Uniform rotation k=0 gives positive tilt frequency squared varpi^2. An infinitely extended static twist instead has growing long-wave tilts with longitudinal wave number below |k|.

For fixed tilt endpoints on a finite segment the lowest squared frequency is (pi/L)^2+varpi^2-k^2. In a static shortest-angle preparation |Theta|<pi it is positive, so the infinite-guide instability cannot simply be assigned to every finite endpoint configuration. This is one exact tilt family, not a classification of all coupled frame and connection modes.

7. Rotating Guide Energy Has a Complete Variational Bound

At k=0 choose a fixed skew generator J of unit-rate rotation in the occupied two-plane, acting trivially on its orthogonal complement. Its conserved retained-frame rotation charge is

$$\mathcal J=\int(\dot a\cdot Ja+\dot b\cdot Jb).$$

It is not electric charge. Completing the velocity squares in the full positive commuting Hamiltonian gives

$$H-\varpi\mathcal J= \frac12\int\sum_{c=a,b} \left[|\dot c-\varpi Jc|^2+|\nabla c|^2+ (\Omega^2-\varpi^2)|c|^2+ \varpi^2(|c|^2-|Jc|^2)\right]+H_{\rm connection}. \tag{15}$$

Every last norm deficit is nonnegative. The established frame-area and gradient inequalities then give the same global static guide bound with restoring coefficient sqrt(Omega^2-varpi^2), where it has a supported global finite-width profile. The rotating balanced frame attains it. Thus this branch is a global energetic minimum of H-varpi J at fixed commuting flux, and also of H among configurations with its same rotation charge, within the stated source and boundary domain. The inequality includes tilt, imbalance and the positive remaining commuting connection energy. The attained uniform-guide statements use transverse integrals per unit length. A finite source-free segment requires compatible boundary conditions for the comparison; no finite-total-energy rotating endpoint solution has been constructed by this bound.

This is stronger than checking a single perturbation. It remains an energetic statement: the degenerate vacuum's evolution and coercivity modulo all symmetries have not been solved, so no general nonlinear stability theorem is claimed. The guide has finite energy per length rather than a particle's finite total energy.

For a general uniform rotating twist, the complete bulk energy, longitudinal momentum and tension are

$$\mathcal E=\tau+\varpi^2M,\qquad \mathcal P_x=-\varpi kM,\qquad \mathcal T=\tau-k^2M. \tag{16}$$

They follow both from the original stress and the reduced uniform action. In the retained-component action a longitudinal boost of the k=0 rotating guide has varpi=gamma varpi_0 and k=-gamma v varpi_0, leaving s unchanged. Its momentum is gamma^2 v varpi_0^2 M per laboratory length. The linear longitudinal momentum response is therefore positive and supplied by the internal frame motion. It differs from the complete energy per length because the extended guide also carries tension. Finite caps and complete supporting stresses are needed before a particle's inertial mass can be identified. The original vector variables' physical spacetime tensor completion remains an additional requirement.

8. Sector Scope and Reproducibility

The common existing Hodge rule supplies this guide at occupied three/four, not at two/five/six/ten. Its force and stiffness mathematics would work elsewhere if a supported guide were supplied; it is not an independent selection of color sectors. The full non-Abelian compact alternatives, the master-field reduction, canonical physical units, dynamical endpoint cores and occupied particle frequencies remain unresolved.

The companion relaxed-frame-twist.py checks the full rotating-frame bulk substitution, fixed-flux envelope and stress, exact pure-quartic formulas, the force-stiffness identity, the exact tilt channel, complete rotating Hamiltonian bound, and independently relaxed mixed-guide branch controls at three/four with common coefficients. All occupied sectors retain their original gate controls. No particle simulation or mass fit enters the calculation.

Verification script and recorded results.