A Positive Gradient Gauss Law Separates Static Confinement from Signal Propagation

The fourth-order source window distinguishes depths three and four if the complete constrained source energy is positive and proportional to the inverse of the spatial bi-Laplacian. That dimensional result does not require a tensor Gauss law. An ordinary gauge law can produce the same static energy and ordinary charge continuity. The explicit candidate below does, but its localized physical response has a spacelike tail despite finite-speed homogeneous waves.

These calculations concern a prospective collective connection of the one field, linearized in one color direction. They introduce no exchanged particle. A full non-Abelian normal-frame action and an actual source reduction have not been derived. The linear result identifies a limitation that a completion would have to address.

An Ordinary Constraint with Fourth-Order Static Energy

On a flat ordinary spatial sector of dimension $d$, define

$$E_i=-\dot A_i-\partial_i\phi,\qquad F_{ij}=\partial_iA_j-\partial_jA_i,\qquad M=-\ell^2\Delta.$$

Consider the local action

$$L=\frac12\langle E,ME\rangle -\frac{c^2}{4}\langle F,MF\rangle +\langle j,A\rangle-\langle\rho,\phi\rangle.$$

After integration by parts, its electric energy is $\ell^2\|\nabla E\|^2/2$. With decay or another stated boundary domain, nonzero spatial modes have positive electric and magnetic energy. The zero spatial mode is degenerate and requires a domain prescription.

The ordinary transformations $A\mapsto A+\nabla\chi$, $\phi\mapsto\phi-\dot\chi$ require

$$\dot\rho+\nabla\cdot j=0.$$

They do not require conservation of dipole moment. Variation in $\phi$ gives $\nabla\cdot ME=\rho$. In a static field, $E=-\nabla\phi$, hence

$$\ell^2k^4\widehat\phi=\widehat\rho,\qquad E_{\rm source}=\frac1{2\ell^2}\int\frac{d^dk}{(2\pi)^d} \frac{|\widehat\rho(k)|^2}{k^4}.$$

The positive sign follows from the constrained electric energy. It is not the negative effective energy obtained by minimizing an elastic displacement with an imposed linear force. Thus this candidate supplies the desired static sign and ordinary source conservation simultaneously.

The transverse quadratic action at nonzero $k$ is

$$L_T(k)=\frac{\ell^2k^2}{2} \left(|\dot A_T|^2-c^2k^2|A_T|^2\right).$$

Every homogeneous transverse plane wave has $\omega=c|k|$. The factor $k^2$ multiplying its time-derivative term is nevertheless decisive for a localized source.

The Gauge-Invariant Retarded Response Has an Exterior Tail

Set $c=\ell=1$ temporarily. In Lorenz gauge the potential response has scalar kernel

$$\widehat K_d(t,k)=\Theta(t)\frac{\sin(|k|t)}{|k|^3},\qquad K_d=(-\Delta)^{-1}G_{{\rm wave},d}.$$

For three spatial dimensions its exact form is

$$K_3(t,r)=\Theta(t)\frac{\min(r,t)}{4\pi r}.$$

It is continuous at $r=t$; its radial derivative has exactly the Poisson jump corresponding to $G_{{\rm wave},3}=\delta(t-r)/(4\pi r)$. Independent Abel-damped radial Fourier integrals verify the same kernel. Outside the wave cone, $K_3=t/(4\pi r)$ remains nonzero.

This fact alone concerns potentials. The physical field must also be checked, because instantaneous potential terms cancel in ordinary Maxwell theory. Here they do not cancel completely.

Let $f(x)$ be a normalized smooth radial core supported inside radius $a$. Use a conserved quadrupole source

$$\rho(t,x)=Q(t)\partial_1^2 f(x),\qquad j_1(t,x)=-\dot Q(t)\partial_1 f(x),\qquad j_{i\ne1}=0.$$

Both total charge and dipole vanish. Its Fourier amplitude is $O(k^2)$, so its fourth-order source energy is infrared finite at every occupied depth, including two. Smoothness of the core makes the ultraviolet energy finite. It avoids reliance on the infinite-energy isolated or dipole preparations at low depth.

For $r>t+a$, the exterior radial kernel has

$$K_d(t,r)=tG_{\Delta,d}(r),\qquad G_{\Delta,d}(r)=\frac{r^{2-d}}{(d-2)S_{d-1}}\quad(d>2),$$

where $S_{d-1}=2\pi^{d/2}/\Gamma(d/2)$. This follows from $-\Delta K_d=G_{{\rm wave},d}$, radial symmetry and the integrated wave source $\int G_{{\rm wave},d}dx=t$. Radial averaging over the normalized compact core leaves the exterior harmonic kernel unchanged. At two, use $G_{\Delta,2}=-\log r/(2\pi)$; its additive infrared ambiguity drops out of the spatial derivatives below.

The conserved-source potentials give $E=-\dot A-\nabla\phi$. On the positive first axis, outside every source wave cone and for zero initial $Q,\dot Q$,

$$E_{1,\mathrm{out}}(t,r) =Q(t)G'_{\Delta,d}(r) -P_2(t)G'''_{\Delta,d}(r),\qquad P_2(t)=\int_0^t(t-\tau)Q(\tau)d\tau.$$

Choose $Q(t)=t^3/6$ during the initial switching interval, with $Q=0$ before zero. Then $P_2=t^5/120$ and

$$\boxed{ E_{1,\mathrm{out}} =-\frac{t^3}{6S_{d-1}r^{d-1}} \left(1-\frac{d(d-1)t^2}{20r^2}\right). }$$

At $r=4t$, with $a<t$ so the observation lies outside the complete source cone, the bracket equals:

Every value is nonzero. This is a gauge-invariant field response to an admissible conserved source, rather than an uncancelled gauge potential or singular point-source artifact. The degenerate spatial factor in the kinetic operator requires an inverse spatial Laplacian when solving for the response. Finite homogeneous wave velocity is therefore insufficient to establish a finite source-response cone.

A detector coupled reciprocally to $A$ measures this field. Calling $ME$ the observable would change the readout and its reciprocal source vertex; it cannot preserve the original fourth-order interaction by a change of name. Whether another complete same-field realization can repair the response while preserving the color window remains open. This limitation is not evidence for an instantaneous color interaction.

A Complementary Positive Spectral Bound

Consider a finite collection of ordinary quadratic gauge channels with positive-definite kinetic matrix $K$, positive-semidefinite squared-mass matrix $H$ and a common source vector $u$. Its static response is

$$R(k)=u^T(k^2K+H)^{-1}u =\sum_a\frac{w_a}{k^2+m_a^2},\qquad w_a\ge0.$$

Whitening $K$ and diagonalizing $K^{-1/2}HK^{-1/2}$ gives this identity directly. Consequently

$$0\le k^2R(k)\le u^TK^{-1}u.$$

A fourth-order infrared pole cannot arise inside this finite positive ordinary-channel class. Adding more positive ordinary modes does not by itself supply it. The same bound holds for a positive spectral measure with finite total source norm. This is not a theorem about nonlinear confinement, singular infinite measures, tensor constraints, spatially degenerate kinetic operators or all possible one-field actions.

A conventional covariant factor instead has

$$\frac1{z(z+m^2)}=\frac1{m^2}\left(\frac1z-\frac1{z+m^2}\right).$$

Its residues have opposite signs in the usual quadratic canonical realization, and its lower ordinary term restores a simple infrared pole. The pure double-pole limit requires separate treatment of the higher-time-derivative degrees of freedom. One cannot infer a positive physical completion from the static denominator alone.

Neutral Motion in the Tensor-Gauss Alternative

An Abelian double-divergence Gauss law conserves dipole moment. A neutral composite can translate rigidly because its dipole changes by translation times total charge, which is zero. That removes one overly strong mobility objection.

It does not automatically allow arbitrary internal motion. For the retained determinant color singlet,

$$\left\langle\left|\sum_i r_iT_i\right|^2\right\rangle =\frac23\sum_{i<j}|r_i-r_j|^2.$$

Rigid translation preserves this norm; dilation multiplies it by the squared dilation factor. Zero expectation of an individual color dipole operator is weaker than its annihilating the state. A direct collection of Abelian tensor constraints would restrict these internal changes unless other same-field degrees of freedom carry the compensating moments. A true non-Abelian covariant constraint can have different conservation laws and must be derived independently.

Verification

The checker and report preserve 149 checks across 50 cases: the conserved quadrupole, all occupied-depth exterior tails, independent three-dimensional Fourier integrals, exact shell jumps, positive multichannel spectral diagonalization and neutral determinant translations versus dilations. The calculation uses no particle simulation or independently fitted sector coefficient.