Size variation of a common positive response

A shared length among the harmonics of one depth must follow from the same spatial action as their standing frequencies. The fixed-norm response result gives a frequency shift, but it does not yet determine a length. This note varies that length for a minimal local response whose source, stiffness and kinetic normalization all scale together. The calculation supplies a necessary test for that class of candidate actions, not a general theorem about every same-field support mechanism.

One scale-covariant sector model

Work at one physical depth $d\geq2$. Let a common reference region or profile family have length $L$. It need not be a hard-walled cavity or a compactified dimension. For a normalized spatial pattern write

$$\psi_{i,L}(x)=\sqrt{Q_i}\,L^{-d/2}u_i(x/L),\qquad \int |u_i(y)|^2\,d^dy=1. \tag{1}$$

The canonical norm $Q_i$ can differ between solutions. At this stage $u_i$ is held fixed during the dilation; allowing it to relax can change the result. Assume that the positive-frequency bare branch has

$$E_{0,i}(Q_i,L)=Q_i\frac{c\mu_i}{L},\qquad \mu_i>0, \tag{2}$$

as for a scale-covariant first-order spatial operator. The dimensionless $\mu_i$ is an output of that operator, not automatically a simplex count. For the proposed mass law, one would need $\mu_i=\gamma_d\mathcal S(i,d)$ with a single factor $\gamma_d$ at that depth.

Let $z$ be a retained, relaxing component of the same field. Consider its quadratic energy and a local quadratic source,

$$E_z=\frac12\langle z,A_Lz\rangle -g\langle z,S[\psi]\rangle, \quad A_L=L^{-2}U_L\widehat A U_L^{-1}>0, \quad S[\psi_{i,L}]\bigl(x\bigr)=Q_iL^{-d}\sigma_i(x/L). \tag{3}$$

Here $U_L f(x)=L^{-d/2}f(x/L)$ is unitary between the reference and physical $L^2$ spaces. The reference operator $\widehat A$ may be a positive dimensionless Laplacian plus a positive gap. The gap in (3) scales with $L^{-2}$; a fixed external gap introduces another length and is outside this one-scale test. The source shape $\sigma_i$ includes the actual spatial polarization contractions if the action supplies them. Its arbitrary normalization cannot be absorbed separately for each mode.

Eliminating $z$ gives the exact fixed-profile coefficient

$$K_i=\langle\sigma_i,\widehat A^{-1}\sigma_i\rangle>0, \qquad E_i(Q_i,L)=\frac{Q_ic\mu_i}{L} -\frac{g^2Q_i^2K_i}{2L^{d-2}}. \tag{4}$$

The power $L^{2-d}$ follows from the physical $d$-volume, the two factors $L^{-d}$ in the source and the inverse stiffness $L^2$. It does not depend on a chosen radial ansatz. At fixed norm the standing frequency is the canonical multiplier

$$\boxed{\omega_i=\left.\frac{\partial E_i}{\partial Q_i}\right|_L =\frac{c\mu_i}{L}-\frac{g^2Q_iK_i}{L^{d-2}}.} \tag{5}$$

Mass is read from this frequency in equivalent units. The second term can depend on amplitude and profile without making amplitude itself mass.

The size extremum

Put $A_i=Q_ic\mu_i>0$, $B_i=g^2Q_i^2K_i/2>0$, and $p=d-2$. Along the specified norm-preserving dilation,

$$E_i(L)=\frac{A_i}{L}-\frac{B_i}{L^p},\qquad E_i'(L)=-\frac{A_i}{L^2}+\frac{pB_i}{L^{p+1}}. \tag{6}$$

This one-dimensional variation is a necessary direction for spatial stability. It gives the following exhaustive cases for the maintained depths:

For $d>3$, its position obeys

$$L_*^{d-3}=\frac{(d-2)B_i}{A_i} =\frac{(d-2)g^2Q_iK_i}{2c\mu_i}. \tag{7}$$

The energy tends to $-\infty$ under contraction in this restricted model. Thus the extremum is a maximum even before other field and geometric perturbations are considered. A stationary common length across harmonics would also require the right side of (7) to be mode-independent. Equal $Q_i$ makes that a condition on $K_i/\mu_i$. The existing equal-norm source controls fail its simplex version at every multi-mode maintained depth. Adjusting $Q_i$ after seeing the masses or lengths would not derive a selection law.

Substituting the stationary-size condition into the multiplier (5) gives

$$\omega_i(L_*)=\left(1-\frac{2}{d-2}\right)\frac{c\mu_i}{L_*} =\frac{d-4}{d-2}\frac{c\mu_i}{L_*}. \tag{8}$$

At depth three the flat condition gives $\omega_i=-c\mu_i/L<0$. The depth-four extremum has $\omega_i=0$, so it cannot represent a positive mass-frequency branch. The depth-five, six and ten extrema have positive multipliers, respectively one-third, one-half and three-quarters of the bare frequency, but remain size maxima. Energy per norm and the canonical frequency differ because the eliminated response is quadratic in $Q_i$; the positive energy at those maxima does not repair their negative size curvature.

For the positive response alone, unconstrained amplitude does not repair the problem. At fixed $L$, $\partial_{Q_i}^2E_i=-g^2K_i/L^{d-2}<0$. A specified conserved norm removes that radial direction from the constrained tangent space; it does not select its value or supply a positive size curvature.

What an additional stabilizer must accomplish

If the complete same-field energy contains a further term $G_i(Q_i,L)$, a stable common length $L_d$ must satisfy, for every occupied harmonic at that depth,

$$\left.\partial_LG_i\right|_{L_d} =\frac{A_i}{L_d^2}-\frac{pB_i}{L_d^{p+1}}, \qquad \left.\partial_L^2(E_i+G_i)\right|_{L_d}>0. \tag{9}$$

The same $G$ must also enter the canonical multiplier and the complete energy-momentum response. A mode-independent geometric cost is not automatically sufficient: its one size derivative must equal the mode-dependent right side of (8) for all occupied harmonics. A coupling or stiffness coefficient independently chosen for each harmonic would hide the missing mechanism. A different spatial order, a fixed physical gap, nonlinear profile relaxation or a constrained off-diagonal field equation can change the scaling argument; each must be calculated from its own full action. The localized current construction, for example, uses an auxiliary spatial fixed-point functional whose value is not additional particle energy, so (4) is not its size law.

The next constructive task is to derive a common stabilizing term and solve its profile and norm equations together with (5) and (9), then check the full Hessian and reciprocal pair response. The present result shows exactly why a positive quadratic response and a common length symbol alone cannot complete that task.