IDOS Energy, Coherence, and Supporting Dynamics — Part 2

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It is a stable center of the reduced fixed-action two-mode dynamics at every maintained depth. This projected model has nonzero leakage under the full unprojected local quartic law, so it is not a solved full spatial particle. Its lock also fails equal bare modal energy: it requires $I_1/J>1/2$, whereas equal energies with $\nu_1>\nu_0$ require $I_1/J<1/2$. The complete energy identity is $\Omega J=H_2+2H_4$, not $H=\Omega J$. Thus locking and equal energy must be solved together in a future action.

The Kernel Is a Trace, With a Precise Spatial Profile

For the ball weight $w=\sqrt{1-r^2}$, put $W_d=\int w\,dx$, $\lambda=(d+1)/2$, and let $K_N$ be the reproducing kernel in the unnormalized measure $w\,dx$. The exact identity is

$$w^2K_N(x,x)=\frac{D_{N+1}+D_N- C_{N+1}^{\lambda}(2r^2-1)-C_N^{\lambda}(2r^2-1)}{2(d+1)W_d}.$$

Here $C_n^\lambda$ denotes a Gegenbauer polynomial. Integrating the auxiliary spherical addition kernel over its three-coordinate fibre derives the identity; those auxiliary coordinates are not physical dimensions. Independent separated Jacobi sums verify it. With $c_d=[(d+1)W_d]^{-1}$,

$$\frac{w^2K_N}{D_Nc_d}=1+\frac d{2(N+1)}- \frac{C_{N+1}^{\lambda}(2r^2-1)+C_N^{\lambda}(2r^2-1)}{2D_N}.$$

At fixed interior radius this tends to one. At the center it is exactly one for odd $N$, and $(N+d+1)/(N+1)$ for even $N$. At every finite cutoff it vanishes at the rim. The limiting inverse-weight trace density therefore has a boundary layer; it cannot be imposed uniformly to the edge. If $u_{\rm trace}=\varepsilon wK_N$, then $u_{\rm trace}w$ tends to an interior constant after normalization by $D_N$. Energy divided by material density instead scales as $1/w^2$, not a constant.

A coherent field does not automatically realize this trace. For coefficients $a(t)$, its averaged quadratic density is $u(x)^T\langle aa^T\rangle u(x)$. Equal diagonal modal energies do not fix the cross correlations. One common-frequency scalar field has covariance rank at most two; isotropic covariance in an arbitrary larger band does not follow. Distinct deterministic integer frequencies can produce an exactly diagonal cycle average while retaining instantaneous beats. This concerns real field motion, not random particle positions.

Nor is the trace automatically the complete local energy. For the positive local derivative-energy representative of the ball action, writing $P(L)=\sum b_mL^m$, $Q_{\rm odd}(z)=\sum_{m\text{ odd}}b_mz^{m-1}$, and degree-shell kernels $K_{[k]}$, diagonal averaging gives

$$u_{\rm deriv}=\varepsilon w\left[K_N- \frac14L\sum_{k=0}^N\frac{Q_{\rm odd}(\lambda_k)}{P(\lambda_k)}K_{[k]}\right].$$

The correction integrates to zero but can change the local density. Material backreaction must be varied from the action, including its kinetic measure.

Supporting Energy and the Static Load

For generalized modes $Ka_j=\omega_j^2Ma_j$, $a_i^TMa_j=\delta_{ij}$, a slow conservative support variation at fixed actions gives

$$\delta E_{\rm support}+\sum_j\frac{I_j}{2\omega_j} a_j^T(\delta K-\omega_j^2\delta M)a_j=0.$$

At an equal-energy reference this wave term is $\varepsilon\,\delta\log\prod_j\omega_j$. The complete material pullback is statically isospectral, so this load vanishes for its admitted static deformations. A nonconstant trace density alone does not contradict that cancellation. Dynamical carrier beats remain a separate source.

A conditional scale model illustrates complete energy accounting. Let $U(L)=aL^p$, $\omega_j(L)=v_jL^{-q}$, with positive $a,p,q$. At fixed actions, $H=aL^p+BL^{-q}$, $B=\sum I_jv_j$, so equilibrium requires $pU=qE_{\rm wave}$. If each wave mode has energy $\varepsilon$ there, then

$$L_*^p=\frac{qD\varepsilon}{pa},\quad E_{\rm total}=(1+q/p)D\varepsilon,\quad L_*^2H''(L_*)=p(p+q)aL_*^p>0.$$

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