IDOS Energy, Coherence, and Supporting Dynamics — Part 24

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to preserve down. With $r(t)=0$ for $t\le0$ and $r(t)=e^{-1/t}$ otherwise, define $H(N)=r(N-1/2)/[r(N-1/2)+r(1-N)]$ and

$$m(q)=[1-H(N)]\omega_0+H(N)F(a,N)\quad(q\ne0), \qquad m(0)=\omega_0.$$

This is positive and smooth on the physical domain. It is constant near the origin because $N\le2z_1$, and it equals $F$ at $N\ge1$. All fourteen maintained clocks and their first five mass jets survive. The formula also assigns the count law at every positive integer cutoff supported by the finite carrier. This is an imposed constitutive extension, not a derivation of count feedback or dynamical selection of those cutoffs. Six depth knots replace fourteen target knots; both versions already use six sector frequency scales.

Replacing only $m$ in SF-DW preserves the fourteen isolated first variations, quadratic branch checks, and positive-chart principal cones. The optional $\kappa_a=8/m,b=m/128$ retains $b\kappa_a=1/16$ and the same causal bound. Extending the supporting-state positivity claim to every new low cutoff requires a common conservative frequency floor, such as $\min(\omega_0,\mu_2,\mu_3,\mu_4,\mu_5,\mu_6,\mu_{10})$, or direct verification of its support-gradient bound. The fourteen-target floor alone does not establish it.

For the actual same-family pair preparations, $a=d$ exactly and

$$N=N_a+(N_b-N_a)\frac{x^2}{1+x^2+2Cx\cos\Delta}.$$

Their mass uses only the same family's scale, including down/strange because $\omega_0=\mu_3$. Near the smaller positive cutoff, $m-\omega_a=O(x^{14})$, while the phase current still shifts at first order. Consequently the fractional log-density slope remains $128(r-1)/\sqrt r$ under variable $b$, where $r=D_b/D_a$. The destructive-phase maximum is exactly

$$N_{\max}=N_a+\frac{N_b-N_a}{1-D_a/D_b}, \qquad x=1/C,\quad\Delta=\pi.$$

For electron/muon it is $5423/159$, above the larger cutoff 34. At equal constructive envelopes, $N\simeq22.28475$ and the clock-only quantity $128(\chi-m)/m\simeq521.70981$. Alignment and occupation contribute additional nonpositive terms to log density; this quantity is not the full density, its magnitude bound, or an evolved encounter. The simpler law therefore fixes one interpolation defect while leaving the severe density response and reciprocal encounter dynamics unresolved.

The report sector-count-mass-law/report.md, its principal review and 2,325 assertions give the general proof, fourteen targets, all six depths through cutoff 94, eleven actual pair comparisons and source provenance. No second observed tau or integer bottom assignment is invented. The law does not distinguish depth five by a special coefficient and does not satisfy the outstanding requirement of spatially uniform participating energy.

Charge-Constrained Homogeneous Dynamics

The exponential preparation diagnostic is not automatically a density runaway. Its interpretation depends on which canonical data are fixed. In a homogeneous control with fixed positive $n_0$, rest flow and fixed frame, write the normalized carrier as

$$e=e^{J\theta}[\sqrt{1-z}\,u_A+\sqrt z\,e^{J\phi}u_B], \qquad f=z(1-z),\qquad I=\epsilon n_0.$$

Here $u_A,u_B$ are orthonormal real-pair modes uniformly supported on the smaller prefix and its disjoint extra labels. Their sizes are $D_a$ and $L=D_b-D_a$. Then $a_0=\dot\theta+z\dot\phi$ and $\|K\|^2=\dot z^2/(4f)+f\dot\phi^2$. The horizontal inertia is retained; neither shape velocity is discarded.

A genuinely spatially constant carrier has $R_i=-J(T_i-q_i)e/\ell$, so its potential also contains

$$W=\frac{I}{8\ell^2}\sum_i\operatorname{Var}(T_i).$$

This is not the isolated centered texture with $R_i=0$. Cancelling that residual by a spatial phase while changing mean labels generates additional time derivatives and is not the same homogeneous control. With the exact occupation/phase potential $V$ above, the restricted Lagrangian is

$$\mathcal L=\frac I2\left[\frac{\dot z^2}{4f}+f\dot\phi^2\right] +n[\dot\theta+z\dot\phi-m(z)-\gamma V] -b(z)[n\log(n/n_0)-n+n_0]-W(z).$$

Since $\theta$ is cyclic, its momentum is exactly $Q=n$ and is constant. This denotes charge per unit volume; an infinite homogeneous field does not

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