Summary: Complete spatial fields determine a scalar interaction only when their full residual is scalar and equal-current points require the same law; unchanged Jacobi pairs cannot share the assigned simplex frequencies through such a law, while nonlinear dressed controls reconstruct both the common action and its coupling derivative.

Keep the causal kinetic equation and invert the interaction

The quadratic inverse construction matches the complete simplex frequencies, but its unrestricted polynomial dynamics fail the curved finite-propagation test. A different inverse problem keeps the original first-order spatial Dirac operator D\mathscr D and asks for one real local scalar law:

i∂τψ=Dψ+F(J)βψ,J=ψ†βψ.(1) \boxed{ i\partial_\tau\psi=\mathscr D\psi+F(J)\beta\psi,\qquad J=\psi^\dagger\beta\psi. } \tag{1}

Here β†=β\beta^\dagger=\beta, β2=I\beta^2=I and {D,β}=0\{\mathscr D,\beta\}=0. The physical geometry, connection, measure and canonical field normalization are fixed inputs. The scalar current is not identified with electric charge. All complex spinor components remain in (1).

For a smooth real FF, this is a semilinear first-order equation. As long as its smooth solutions exist and remain locally bounded, the difference of two solutions has the original Dirac principal part and a locally Lipschitz source. The usual exterior-cone energy estimate and Gronwall inequality preserve the Dirac domain of dependence. This does not prove global existence or stable particle solutions, but it avoids replacing the kinetic equation by an unrestricted sixth-order spatial dispersion.

Equation (1) is a candidate class of same-field laws. Its inverse conditions apply to arbitrary spatial polarizations. No dimensions are compactified, and the radius is not selected. Sector nesting does not require one particle field to be a restriction or constituent of another.

A complete pointwise test for a real scalar source

Suppose a proposed stationary field and frequency are given: ψ(τ,x)=e−iωτϕ(x)\psi(\tau,x)=e^{-i\omega\tau}\phi(x). Define the source that its full differential equation requires:

r(x)=(ω−D)ϕ(x),ρ(x)=ϕ†ϕ,v(x)=βϕ(x).(2) r(x)=(\omega-\mathscr D)\phi(x),\qquad \rho(x)=\phi^\dagger\phi,\qquad v(x)=\beta\phi(x). \tag{2}

At every point where ρ>0\rho>0, the best real scalar coefficient is uniquely determined:

Mϕ(x)=Re⁡[v(x)†r(x)]ρ(x),r⊥(x)=r(x)−Mϕ(x)v(x).(3) \boxed{ M_\phi(x)= \frac{\operatorname{Re}[v(x)^\dagger r(x)]}{\rho(x)},\qquad r_\perp(x)=r(x)-M_\phi(x)v(x). } \tag{3}

Indeed, for any real trial coefficient MM,

∥r−Mv∥2=∥r⊥∥2+ρ(M−Mϕ)2.(4) \|r-Mv\|^2 =\|r_\perp\|^2+\rho(M-M_\phi)^2. \tag{4}

The cross term vanishes by the definition in (3). Thus the full stationary equation admits a real scalar multiplication coefficient precisely when r⊥=0r_\perp=0. An imaginary coefficient or an orthogonal spinor source remains in that residual; a scalar expectation alone cannot remove it. At a zero of the complete field, stationarity requires r=0r=0, and that point does not determine a scalar coefficient.

These are pointwise statements on actual spatial fields, with all their covariant derivatives included in Dϕ\mathscr D\phi. They are invariant under a common change of spinor basis when the field, operator and β\beta are transformed consistently. A spatially varying change of frame also transforms the connection.

One action requires one function of the actual current

Passing (3) is only the first condition. For a collection of proposed particle fields ϕa\phi_a, one law (1) additionally needs

Mϕa(x)=F(Jϕa(x))(5) \boxed{M_{\phi_a}(x)=F(J_{\phi_a}(x))} \tag{5}

with the same smooth real function FF for every field and point. In particular,

Jϕa(x)=Jϕb(y)⟹Mϕa(x)=Mϕb(y).(6) J_{\phi_a}(x)=J_{\phi_b}(y) \quad\Longrightarrow\quad M_{\phi_a}(x)=M_{\phi_b}(y). \tag{6}

This compares independently occupied field structures; it does not impose inheritance between their spatial profiles. Equal-current compatibility is necessary. A smooth extension through critical current values and the boundaries of the sampled current ranges is required as well.

Where ∇J≠0\nabla J\ne0, a differentiable law must satisfy

∇Mϕ=F′(Jϕ)∇Jϕ,F′(Jϕ)=∇Mϕ⋅∇Jϕ∣∇Jϕ∣2.(7) \nabla M_\phi=F'(J_\phi)\nabla J_\phi,\qquad F'(J_\phi)= \frac{\nabla M_\phi\cdot\nabla J_\phi}{|\nabla J_\phi|^2}. \tag{7}

The derivative inferred at equal-current points must also agree. Physical norms and density units must be fixed before applying this test; rescaling each candidate afterward changes the argument of FF. Different depths additionally need their common master-field reduction and normalization.

If (3)–(5) hold with a smooth law on the required current range, put U(J)=∫0JF(s) dsU(J)=\int_0^J F(s)\,ds. The real action is

S[ψ]=∫dτ[i2(⟨ψ,ψ˙⟩−⟨ψ˙,ψ⟩)−E[ψ]],E[ψ]=⟨ψ,Dψ⟩+∫U(J) dV.(8) \begin{aligned} \mathcal S[\psi]&=\int d\tau\left[ \frac i2\bigl(\langle\psi,\dot\psi\rangle- \langle\dot\psi,\psi\rangle\bigr)-E[\psi]\right],\\ E[\psi]&=\langle\psi,\mathscr D\psi\rangle+\int U(J)\,dV. \end{aligned} \tag{8}

Its full complex variation gives exactly (1). The same scalar law fixes the complete local response:

δ ⁣(F(J)βψ)=F(J)βh+2F′(J)Re⁡(ψ†βh) βψ.(9) \boxed{ \delta\!\left(F(J)\beta\psi\right) =F(J)\beta h+ 2F'(J)\operatorname{Re}(\psi^\dagger\beta h)\,\beta\psi. } \tag{9}

The inferred law need not be unique outside the sampled current ranges. A background with constant current determines FF at one value but generally leaves F′F' undetermined there. Thus mass frequencies alone do not determine every coupling response or the complete master dynamics.

The excited spatial response calculation tests this constraint on nonlinearly deformed assigned modes. The same local interaction produces a calculable resonance in their full complex response, showing why stationary source compatibility and spatial stability must be assessed together.

Apply the test to unchanged spatial Dirac pairs

Let ϕn\phi_n be a normalized zero-form eigenfield and let ηn=Dϕn/λn\eta_n=\mathscr D\phi_n/\lambda_n be its normalized exact one-form partner:

Dϕn=λnηn,Dηn=λnϕn,λn=2n(n+5)R.(10) \mathscr D\phi_n=\lambda_n\eta_n,\qquad \mathscr D\eta_n=\lambda_n\phi_n,\qquad \lambda_n=\frac{2\sqrt{n(n+5)}}R. \tag{10}

Consider the unchanged paired spatial profile χn=cnϕn+snηn\chi_n=c_n\phi_n+s_n\eta_n, with real constant weights, cn2+sn2=1c_n^2+s_n^2=1, and both nonzero. The common overall phase is arbitrary. The two form equations require

λnsn+M(x)cn=ωncn,λncn−M(x)sn=ωnsn.(11) \begin{aligned} \lambda_ns_n+M(x)c_n&=\omega_nc_n,\\ \lambda_nc_n-M(x)s_n&=\omega_ns_n. \end{aligned} \tag{11}

Where the respective spatial components are nonzero, these give

M=ωn−λnsncn=λncnsn−ωn.(12) M=\omega_n-\lambda_n\frac{s_n}{c_n} =\lambda_n\frac{c_n}{s_n}-\omega_n. \tag{12}

Continuity extends this one constant coefficient over the remaining points of the explicit Jacobi profile. In particular,

ωn2=λn2+Mn2.(13) \boxed{\omega_n^2=\lambda_n^2+M_n^2.} \tag{13}

Changing FF does not alter this fixed-profile requirement. It follows from the full spatial equation, before taking any expectation or dropping a complementary source.

Up to its positive common normalization, the two local densities are

∣ϕn∣2∝t2pn2,∣ηn∣2∝t3(1−t)(pn′)2n(n+5).(14) |\phi_n|^2\propto t^2p_n^2,\qquad |\eta_n|^2\propto \frac{t^3(1-t)(p_n')^2}{n(n+5)}. \tag{14}

For every n≥1n\ge1, the polynomial has simple interior zeros. At such a zero the scalar current of χn\chi_n is negative, because the odd component is nonzero. At t=1t=1 the odd component vanishes and the scalar current is positive. Between the last polynomial zero and that pole, the current therefore crosses zero while the complete spinor is nonzero.

Every such unchanged pure pair consequently requires

F(0)=Mn.(15) \boxed{F(0)=M_n.} \tag{15}

Its complete current image is an interval containing zero. The same field forces FF to be constant on that interval, so F′=0F'=0 in its interior. Multiplying the physical norm by a positive factor preserves the zero-current argument.

The written massless scalar law F(J)=gJF(J)=gJ has F(0)=0F(0)=0 and therefore cannot support these unchanged profiles with Mn≠0M_n\ne0. A nonzero bare value F(0)F(0) would instead give the same constant coefficient to every such pair. Neither option supplies distinct simplex coefficients on the unchanged spatial profiles.

The assigned two-frequency target fails this pure-profile test

Keep the common six-dimensional geometry and its one fixed radius. The assigned target is ωn=μS(n,6)\omega_n=\mu S(n,6) at n=13,35n=13,35. Equations (13) and (15) would require one real constant MM. For xn=n(n+5)x_n=n(n+5), this gives

ω35ω13=x35+(MR/2)2x13+(MR/2)2≤x35x13.(16) \frac{\omega_{35}}{\omega_{13}} =\sqrt{\frac{x_{35}+(MR/2)^2}{x_{13}+(MR/2)^2}} \le\sqrt{\frac{x_{35}}{x_{13}}}. \tag{16}

The actual numbers are:

Quantity Value
Largest ratio from a shared real gap on this fixed geometry 1400234≈2.445998\sqrt{\frac{1400}{234}}\approx2.445998
Assigned simplex frequency ratio (406)(186)=351517≈206.764706\frac{\binom{40}{6}}{\binom{18}{6}}=\frac{3515}{17}\approx206.764706

The result does not cover nonlinear spatial dressing, other polarization textures or a self-consistently changing geometry. The round-space scalar branch, for example, has constant nonzero current and does not satisfy the current-crossing hypothesis used in (15). These distinctions are part of the theorem's scope, rather than reasons to reassign the particle depths or mass indices.

A dressed field can pass the same inverse test

They provide a positive control for the reconstruction, independently of the desired particle masses.

Use the radius-one, unit-volume convention and fix one local law F(J)=κJF(J)=\kappa J with κ=0.15\kappa=0.15. Take physical norms Q=1Q=1 and Q=3Q=3. The normalized branch solver then uses g=κQg=\kappa Q:

Physical norm QQ Normalized solver coefficient gg Stationary frequency
11 0.150.15 0.4173826624460.417382662446
33 0.450.45 1.1067127648741.106712764874

At successively larger paired cutoffs 8,14,228,14,22, independent uncompressed spatial derivatives reduce the full residual below 6×10−156\times10^{-15} in this numerical control. Both profiles retain nonzero odd spatial dressing. At equal physical currents J=0.25,1,3J=0.25,1,3, the inverse formula recovers the same M=κJM=\kappa J from both fields. Their spatial derivatives also recover the same F′(J)=0.15F'(J)=0.15.

This validates both parts of the inverse procedure on solutions of a common local action. It is not a mass prediction, a quantization of QQ or a stability proof for those branches. It demonstrates why solving the changed spatial field can succeed where the unchanged-profile test fails.

Frequency and rest energy still need separate identification

For a stationary solution of (1), (8) gives

ωQ−E=∫[JF(J)−U(J)] dV.(17) \boxed{\omega Q-E =\int\left[J F(J)-U(J)\right]\,dV.} \tag{17}

Along a smooth family of constrained stationary solutions, dE/dQ=ωdE/dQ=\omega. Neither identity makes E/QE/Q equal to ω\omega for a general nonlinear interaction. This agrees with the local scalar normalization calculation.

Thus the inverse frequency target, the total energy of a localized particle and an observed rest mass must be connected by the same physical normalization and dynamics. The current calculation reconstructs and tests a candidate scalar law; it does not establish that final physical identification.

The energy constraints from scalar response make the next comparison explicit. A common energy-frequency conversion can preserve the ratios, but its agreement across spatial branches must be checked. The same F′F' fixes the energy defect and full norm response; the latter also determines how a common change of action moves the energy and frequency targets.

Verification

inverse_scalar_law_checks.py checks the real scalar minimization and its complete complex residual at spinor ranks two, eight and thirty-two. General complex basis changes, imaginary-source controls and full action variations retain the ordinary and conjugate responses.

Actual curved derivatives verify the constant coefficient of the paired Jacobi fields and locate their nonvanishing-spinor current zeros. Exact integer comparisons check (16) at the unchanged assigned indices. Independent spatial derivatives, quadrature and increasing paired cutoffs reconstruct the common law and its coupling derivative on the dressed controls.