Version 2 · earlier version; the current one is v4 · External review, round 1: major errors

# Formation of the made field: bound, conserved densities and oscillator cells

- **Subproject:** 08-ilang-propagation
- **Package:** 04-formation
- **Version:** v2
- **Mode:** internal regeneration
- **Date:** 2026-10-24

## Changes from previous version

- Step 6(b), (4.14): the finiteness of the caustic set is now a stated condition of the stationary-phase step and of (4.14), with a short argument that it holds for the nearest-neighbour contract and the control (4.16) (review I1).
- Step 7, (4.15): the arrival with $v_{\rm front}$ and $v_{\rm form}=c_0$ carry the same condition (review I1).
- Step 4, (4.9): within the class of the question's note (a), $\chi_B\le s_B/\lvert\mathbf q\rvert^2$, the answer of item 2 for $B\ne A$ is stated as the conclusion of (4.9) (no linear formation, $\lambda_{\rm form}\ge(1-\varepsilon)s_0/(\sqrt{c_As_B}\,\lvert\mathbf q\rvert^2)$); stronger sources are out of scope (review I2).
- Step 5, Result (4.14), Open issues: the square torus is out of scope by note (b); the two former open issues on stronger sources and on the square torus are removed (review I2).
- Code: the unchanged scripts were run again in `code/v2/`; every cited number is unchanged.

## Setup and assumptions

**Class F** (assumption, M8). A field of finitely many objects with finitely many places each (A1) on a torus $X_f=\mathbb Z_L^D$, contract $C_F$ a sum of pair terms of finite range (A5), invariant under the translations: unitaries $U_y$ with $U_yC_FU_y^\dagger=C_F$, $U_yA_xU_y^\dagger=A_{x+y}$, $U_yB_xU_y^\dagger=B_{x+y}$, where $A_x$ (read) and $B_x$ (coupled to the source) are self-adjoint translates acting near $x$. The source sits at $x_s=0$ (translation).

**Source and start** (assumption, M8, M9, A11). The contract is $C_F+g_sB_0$ at every $\lambda\ge0$; at $\lambda=0$ the state is the lowest eigenvector $\Omega$ of $C_F$, assumed nondegenerate, eigenvalue $E_0$. Since $U_y\Omega$ is again a lowest eigenvector, $U_y\Omega=e^{ip_0\cdot y}\Omega$. Eigenvectors $\lvert n\rangle$ of $C_F$ are chosen with $U_y\lvert n\rangle=e^{ip_n\cdot y}\lvert n\rangle$; $\omega_n=E_n-E_0>0$ for $n\neq0$.

**Made field.** $\delta a(x,\lambda)=\partial_{g_s}\langle A_x\rangle_\lambda\vert_{g_s=0}$ (first order: the limit $g_s\to0$ of the coefficient is taken first); $\delta a_{\rm st}(x)$ the same in the lowest eigenvector of $C_F+g_sB_0$ (simple for small $\lvert g_s\rvert$). Label components $Y_{\mathbf q}=\sum_xe^{-i\mathbf q\cdot x}Y_x$ for operators, $\delta a_{\mathbf q}=\sum_xe^{-i\mathbf q\cdot x}\delta a(x)$, $\mathbf q\in(2\pi/L)\mathbb Z^D$; $Y_{\mathbf q}^\dagger=Y_{-\mathbf q}$.

**Inputs** (quoted in the question, with their settings). (8.10), (8.14), (8.15) of 07-ilang-weight/08-field-speed; (5.11) of 04-ilang-time/05-moving-clock, $v_{\max}=\tau\kappa\sigma_X$; (4.15) of 06-ilang-source/04-profile, $d_\Gamma(\pi(m),\pi(m'))=k\lvert m-m'\rvert$.

**Mathematics used** (M5): first-order (Duhamel) perturbation of $e^{-i(C_F+g_sB_0)\lambda}$, Cauchy–Schwarz, Fourier series on tori, closed forms of explicit Fourier integrals (Bessel functions), stationary phase. All are applied to the stated operators; no field equation is used (M4).

## Derivation

### Step 1. First-order response in spectral form

With $\langle A_x\rangle_\lambda=\langle\Omega\vert e^{i(C_F+g_sB_0)\lambda}A_xe^{-i(C_F+g_sB_0)\lambda}\vert\Omega\rangle$, Duhamel's formula and $e^{-iC_F\lambda}\Omega=e^{-iE_0\lambda}\Omega$ give $\delta a(x,\lambda)=-i\int_0^\lambda\mathrm ds\,\langle\Omega\vert[A_x(s),B_0]\vert\Omega\rangle$, $A_x(s)=e^{iC_Fs}A_xe^{-iC_Fs}$. Inserting the eigenvectors (the $n=0$ terms cancel in the commutator) and integrating over $s$:

$$
\delta a(x,\lambda)=-\sum_{n\neq0}\Bigl[\langle\Omega\vert A_x\vert n\rangle\langle n\vert B_0\vert\Omega\rangle\,g_\lambda(\omega_n)+\langle\Omega\vert B_0\vert n\rangle\langle n\vert A_x\vert\Omega\rangle\,\overline{g_\lambda(\omega_n)}\Bigr],\qquad g_\lambda(\omega)=\frac{1-e^{-i\omega\lambda}}{\omega}.
\tag{4.1}
$$

The first-order change of the lowest eigenvector of $C_F+g_sB_0$ is $-\sum_{n\ne0}\lvert n\rangle\langle n\vert B_0\vert\Omega\rangle/\omega_n$, so $\delta a_{\rm st}$ is (4.1) with $g_\lambda(\omega)\to1/\omega$. Translation covariance gives $\langle\Omega\vert A_x\vert n\rangle=e^{i(p_0-p_n)\cdot x}\langle\Omega\vert A_0\vert n\rangle$ and $\langle n\vert B_x\vert\Omega\rangle=e^{i(p_n-p_0)\cdot x}\langle n\vert B_0\vert\Omega\rangle$. Summing over $x$, the first term of (4.1) survives only for $p_n=p_0-\mathbf q$, where $\langle n\vert B_0\vert\Omega\rangle=\lvert X_f\rvert^{-1}\langle n\vert B_{-\mathbf q}\vert\Omega\rangle$; the second only for $p_n=p_0+\mathbf q$. Hence

$$
\delta a_{\mathbf q}(\lambda)=-\frac1{\lvert X_f\rvert}\sum_{n\neq0}\Bigl[\langle\Omega\vert A_{\mathbf q}\vert n\rangle\langle n\vert B_{-\mathbf q}\vert\Omega\rangle\,g_\lambda(\omega_n)+\langle\Omega\vert B_{-\mathbf q}\vert n\rangle\langle n\vert A_{\mathbf q}\vert\Omega\rangle\,\overline{g_\lambda(\omega_n)}\Bigr],
\tag{4.2}
$$

and $\delta a^{\rm st}_{\mathbf q}$ is (4.2) with $g_\lambda,\overline{g_\lambda}\to1/\omega_n$.

### Step 2. The bound on formation (item 1)

Define, for the read operator and for any translate family $Y$,

$$
f_A(\mathbf q)=\frac{\langle\Omega\vert[A_{\mathbf q}^\dagger,[C_F,A_{\mathbf q}]]\vert\Omega\rangle}{\lvert X_f\rvert}=\frac1{\lvert X_f\rvert}\sum_n\omega_n\bigl(\lvert\langle n\vert A_{\mathbf q}\Omega\rangle\rvert^2+\lvert\langle n\vert A_{\mathbf q}^\dagger\Omega\rangle\rvert^2\bigr),\qquad
\chi_Y(\mathbf q)=\frac1{\lvert X_f\rvert}\sum_{n\neq0}\frac{\lvert\langle n\vert Y_{\mathbf q}\Omega\rangle\rvert^2+\lvert\langle n\vert Y_{\mathbf q}^\dagger\Omega\rangle\rvert^2}{\omega_n}.
\tag{4.3}
$$

The second form of $f_A$ follows by expanding the double commutator in the eigenvectors (sum rule). Both are $\ge0$, fixed by $C_F$, $A$ or $Y$, and $\Omega$, and unchanged by the common phase of $\Omega$ and by every replacement of contract terms that leaves $C_F$ unchanged (M6). By (4.2) with $B=A$, $\delta a^{\rm st}_{\mathbf q}=-\chi_A(\mathbf q)$: $\chi_Y$ is the label component of the static self-mark of $Y$.

Since $\lvert g_\lambda(\omega)\rvert=2\lvert\sin(\omega\lambda/2)\rvert/\omega\le\min(\lambda,2/\omega)$, bounding (4.2) termwise and applying Cauchy–Schwarz to the joint sum over $n$ and the two terms, with the weights $\omega_n^{\pm1/2}$ (for the bound $\lambda$) or $\omega_n^{-1/2}$ on both factors (for the bound $2/\omega$), gives for every $\mathbf q$, every $L$ and every $\lambda\ge0$

$$
\lvert\delta a_{\mathbf q}(\lambda)\rvert\le\min\Bigl\{\lambda\sqrt{f_A(\mathbf q)\,\chi_B(\mathbf q)}\,,\;2\sqrt{\chi_A(\mathbf q)\,\chi_B(\mathbf q)}\Bigr\},\qquad\text{for }B=A:\ \ \Bigl\lvert\frac{\delta a_{\mathbf q}(\lambda)}{\delta a^{\rm st}_{\mathbf q}}\Bigr\rvert\le\lambda\sqrt{\frac{f_A(\mathbf q)}{\chi_A(\mathbf q)}} .
\tag{4.4}
$$

*Highest eigenvector.* If $\Omega$ is the nondegenerate highest eigenvector, all $\omega_n<0$; Steps 1–2 hold with $\lvert\omega_n\rvert$, and (4.4) holds with $\lvert f_A\rvert$ and $\lvert\chi\rvert$, since the sum rule still equals $-\sum_n\lvert\omega_n\rvert(\dots)$. For an eigenvector inside the spectrum, $\sum_n\lvert\omega_n\rvert(\dots)$ is not a double commutator, and Step 3 does not apply.

### Step 3. Read operators that are conserved densities

Assumption (M8): $N=\sum_xA_x$ with $[A_x,A_y]=0$; $C_F=\sum_\gamma h_\gamma$, the translates of finitely many terms, $\nu$ terms per place, each with $[h_\gamma,N]=0$ (the terms that move the number have finite range). Let $x_\gamma$ be a reference place of $h_\gamma$, and $R_A$, $n_A$ the largest distance from $x_\gamma$ to, and the largest number of, places $x$ with $[h_\gamma,A_x]\neq0$.

Because $\sum_x[h_\gamma,A_x]=[h_\gamma,N]=0$, and $\sum_y[A_y,[h_\gamma,A_x]]=[[N,h_\gamma],A_x]+[h_\gamma,[N,A_x]]=0$, the phases may be replaced by their differences from $e^{\mp i\mathbf q\cdot x_\gamma}$:

$$
[A_{\mathbf q}^\dagger,[C_F,A_{\mathbf q}]]=\sum_\gamma\sum_{x,y}\bigl(e^{i\mathbf q\cdot y}-e^{i\mathbf q\cdot x_\gamma}\bigr)\bigl(e^{-i\mathbf q\cdot x}-e^{-i\mathbf q\cdot x_\gamma}\bigr)[A_y,[h_\gamma,A_x]].
$$

Only $x,y$ with $[h_\gamma,A_x]\ne0\ne[h_\gamma,A_y]$ contribute (with $[A_x,A_y]=0$, $[A_y,[h_\gamma,A_x]]=[[A_y,h_\gamma],A_x]$), each phase difference is at most $\lvert\mathbf q\rvert R_A$ in modulus, and there are $\nu\lvert X_f\rvert$ terms $\gamma$. Hence

$$
0\le f_A(\mathbf q)\le c_A\lvert\mathbf q\rvert^2,\qquad c_A=4\,\nu\,n_A^2R_A^2\,\max_\gamma\lVert h_\gamma\rVert\,\lVert A_0\rVert^2 ,
\tag{4.5}
$$

for every $L$; $c_A$ is fixed by $C_F$ and $A$. The same holds for the highest eigenvector.

### Step 4. No formation within a linear $\lambda$ for conserved densities (item 2)

**Label criterion** (stated before use). The label component $\mathbf q$ has formed to accuracy $\varepsilon\in(0,1)$ from $\lambda_f$ on if $\lvert\delta a_{\mathbf q}(\lambda)-\delta a^{\rm st}_{\mathbf q}\rvert\le\varepsilon\lvert\delta a^{\rm st}_{\mathbf q}\rvert$ for every $\lambda\ge\lambda_f$ (order: $g_s\to0$, then $L\to\infty$ at fixed $\mathbf q$, or $\lambda$ below the wrap time); $\lambda_{\rm form}(\mathbf q;\varepsilon)$ is the least such $\lambda_f$, and $1/\lvert\mathbf q\rvert$ the associated distance. It is not the arrival of a first change. **Long-range mark** in label form: $\lvert\delta a^{\rm st}_{\mathbf q}\rvert\ge s_0/\lvert\mathbf q\rvert^2$, $s_0>0$, for $\lvert\mathbf q\rvert\le q_1$; on the cubic torus an inverse-distance place dependence $s/\lvert x\rvert$ with a summable remainder has $\delta a^{\rm st}_{\mathbf q}=4\pi s/\lvert\mathbf q\rvert^2+O(1)$.

Formation at $\lambda$ requires $\lvert\delta a_{\mathbf q}(\lambda)\rvert\ge(1-\varepsilon)\lvert\delta a^{\rm st}_{\mathbf q}\rvert$; with (4.4) and (4.5), for every source operator $B$,

$$
\lambda_{\rm form}(\mathbf q;\varepsilon)\ \ge\ \frac{(1-\varepsilon)\,\lvert\delta a^{\rm st}_{\mathbf q}\rvert}{\lvert\mathbf q\rvert\sqrt{c_A\,\chi_B(\mathbf q)}} .
\tag{4.6}
$$

*$B=A$.* Then $\chi_B=\lvert\delta a^{\rm st}_{\mathbf q}\rvert$, and with the long-range mark

$$
\lambda_{\rm form}(\mathbf q;\varepsilon)\ \ge\ \frac{1-\varepsilon}{\lvert\mathbf q\rvert}\sqrt{\frac{\lvert\delta a^{\rm st}_{\mathbf q}\rvert}{c_A}}\ \ge\ \frac{1-\varepsilon}{\lvert\mathbf q\rvert^{2}}\sqrt{\frac{s_0}{c_A}}\qquad(\lvert\mathbf q\rvert\le q_1).
\tag{4.7}
$$

For every $\beta>0$, $\lambda_{\rm form}(\mathbf q;\varepsilon)>\beta/\lvert\mathbf q\rvert$ as soon as $\lvert\mathbf q\rvert<(1-\varepsilon)\sqrt{s_0/c_A}/\beta$: the long-range part cannot form within a $\lambda$ linear in the distance; it needs a $\lambda$ at least quadratic in it.

*Position form* ($B=A$, cubic torus). For real $\varphi$, Parseval gives $\sum_x\varphi(x)\delta a(x,\lambda)=\lvert X_f\rvert^{-1}\sum_{\mathbf q}\overline{\hat\varphi(\mathbf q)}\,\delta a_{\mathbf q}(\lambda)$. If also $\chi_A(\mathbf q)\le s_1/\lvert\mathbf q\rvert^2$ for all $\mathbf q$ (the mark is at most long-range), (4.4)–(4.5) give $\lvert\sum_x\varphi\,\delta a(\cdot,\lambda)\rvert\le\lambda\sqrt{c_As_1}\,\lvert X_f\rvert^{-1}\sum_{\mathbf q}\lvert\hat\varphi(\mathbf q)\rvert$. Take $\varphi(x)=\Phi(x/\ell)$ with $\Phi\ge0$ smooth on the shell $1\le\lvert y\rvert\le2$; as $L\to\infty$, $\lvert X_f\rvert^{-1}\sum_{\mathbf q}\lvert\hat\varphi\rvert\to\lVert\hat\Phi\rVert_1=\int\lvert\hat\Phi(\mathbf p)\rvert\,\mathrm d^3p/(2\pi)^3$, independent of $\ell$. If the made field has formed at every place of the shell $\ell\le\lvert x\rvert\le2\ell$ ($\lvert\delta a-\delta a_{\rm st}\rvert\le\varepsilon\lvert\delta a_{\rm st}\rvert$ there) and $\delta a_{\rm st}=s/\lvert x\rvert+o(\lvert x\rvert^{-1})$, then $\lvert\sum_x\varphi\,\delta a\rvert\ge(1-\varepsilon)\lvert s\rvert\ell^2\int\Phi(y)\lvert y\rvert^{-1}\mathrm d^3y\,(1+o(1))$, hence

$$
\lambda\ \ge\ \frac{(1-\varepsilon)\,\lvert s\rvert\int\Phi(y)\lvert y\rvert^{-1}\mathrm d^3y}{\sqrt{c_As_1}\;\lVert\hat\Phi\rVert_1}\;\ell^{2}\,(1+o(1))\qquad(\ell\to\infty).
\tag{4.8}
$$

*$B\neq A$* (any operator, fields of several kinds of objects included: only the conservation of $N$ and the locality of $A$ were used). By (4.6), formation within $\lambda\le\beta/\lvert\mathbf q\rvert$ at labels with a long-range mark requires the first relation below; for a source operator whose self-response is at most long-range, $\chi_B(\mathbf q)\le s_B/\lvert\mathbf q\rvert^2$ (the class of the question's note (a): a held body or a local source), (4.6) with $\lvert\mathbf q\rvert\sqrt{c_A\chi_B}\le\sqrt{c_As_B}$ gives the second:

$$
\chi_B(\mathbf q)\ \ge\ \frac{(1-\varepsilon)^2s_0^2}{\beta^2c_A}\,\frac1{\lvert\mathbf q\rvert^{4}};\qquad
\chi_B(\mathbf q)\le\frac{s_B}{\lvert\mathbf q\rvert^2}\ \Longrightarrow\ \lambda_{\rm form}(\mathbf q;\varepsilon)\ \ge\ \frac{(1-\varepsilon)\,s_0}{\sqrt{c_A s_B}}\,\frac1{\lvert\mathbf q\rvert^{2}}\quad(\lvert\mathbf q\rvert\le q_1).
\tag{4.9}
$$

The first relation is an exact necessary condition: the static self-mark of the source operator must grow at least as $\lvert\mathbf q\rvert^{-4}$ in label space (on the cubic torus, at least linearly in the distance); linear formation is excluded for every $B$ with $\lvert\mathbf q\rvert^4\chi_B(\mathbf q)\to0$ as $\mathbf q\to0$. In the class of note (a) the first relation fails for $\lvert\mathbf q\rvert<(1-\varepsilon)s_0/(\beta\sqrt{c_As_B})$ and the second gives the answer of item 2 for $B\ne A$: the long-range part of a conserved density cannot form within a $\lambda$ linear in the distance; it needs a $\lambda$ at least quadratic in it, as for $B=A$. Sources with a stronger self-response are out of scope by note (a).

The highest eigenvector gives the same, with $\lvert f_A\rvert$, $\lvert\chi\rvert$ (Step 2). **Control** (short argument): oscillator cells with $A=B=q_x$ in the limit $M\to\infty$ (Step 5). There $[q_x,[H,q_y]]=\delta_{xy}$, so $f_A(\mathbf q)=1$; $q_x$ is not the density of a conserved number, (4.5) fails, and Step 6 shows formation at the speed $c_0$.

### Step 5. Oscillator cells: the made field and the criterion (item 3)

**Class** (M8). $H=\tfrac12\sum_xp_x^2+\tfrac12\sum_{x,y}q_xK_{xy}q_y$, $A_x=B_x=q_x$, $K$ of finite range, invariant under the translations and the permutations and reflections of the axes, $\hat K(\mathbf k)=m_h^2+\hat K_0(\mathbf k)$ with $\hat K_0(\mathbf k)=c_0^2\lvert\mathbf k\rvert^2+O(\lvert\mathbf k\rvert^4)$ (isotropic at second order by the cubic symmetry), $\hat K_0(\mathbf k)>0$ for $\mathbf k\ne0$ (the pinning is the term $m_h^2$, assumption), $m_h,c_0>0$. $\hat K$ is real and even; $\Omega_{\mathbf k}=\hat K(\mathbf k)^{1/2}$ are the normal-mode frequencies. Limit: $M\to\infty$ first; the lowest eigenvector and the first-order response of the compression $C_M$ then become those of $H$ (stated limit, standard for compressions onto growing sets of low levels).

For $H+g_sq_0$, quadratic plus linear, the equations of the means are closed and exact: $\mathrm d\langle q\rangle/\mathrm d\lambda=\langle p\rangle$, $\mathrm d\langle p\rangle/\mathrm d\lambda=-K\langle q\rangle-g_se_0$. The ground state of $H$ is invariant under $(q,p)\to(-q,-p)$, so $\langle q\rangle=\langle p\rangle=0$ at $\lambda=0$, and $\langle q\rangle_\lambda=-g_sK^{-1}(1-\cos K^{1/2}\lambda)e_0$, exactly linear in $g_s$. The lowest eigenvector of $H+g_sq_0$ is the ground state displaced by $-g_sK^{-1}e_0$. Hence $\delta a_{\mathbf k}(\lambda)=-(1-\cos\Omega_{\mathbf k}\lambda)/\Omega_{\mathbf k}^2$, $\delta a^{\rm st}_{\mathbf k}=-1/\Omega_{\mathbf k}^2$, and for $L\to\infty$

$$
u(x,\lambda):=\delta a(x,\lambda)-\delta a_{\rm st}(x)=\int_{[-\pi,\pi]^D}\frac{\mathrm d^D k}{(2\pi)^D}\,e^{i\mathbf k\cdot x}\,\frac{\cos\Omega_{\mathbf k}\lambda}{\Omega_{\mathbf k}^2}.
\tag{4.10}
$$

(With (4.3): $f_A=1$, $\chi_A=\Omega_{\mathbf q}^{-2}$, and (4.4) reads $1-\cos\Omega_{\mathbf q}\lambda\le\lambda\Omega_{\mathbf q}$.)

**Criterion** (stated before use; places $\rho\mathbf e_1$, $\rho\in\mathbb N$, along an axis):

$$
\Delta_3(\rho,\lambda)=\frac{\delta a(\rho\mathbf e_1,\lambda)}{\delta a_{\rm st}(\rho\mathbf e_1)}-1\ \ (D=3),\qquad
\Delta_1(\rho,\lambda)=\frac{\nabla\delta a(\rho,\lambda)}{\nabla\delta a_{\rm st}(\rho)}-1,\ \ \nabla v(\rho)=v(\rho+1)-v(\rho)\ \ (D=1);
\tag{4.11}
$$

the made field at distance $\rho$ has formed to accuracy $\varepsilon\in(0,1)$ from $\lambda_f$ on if $\lvert\Delta_D(\rho,\lambda)\rvert\le\varepsilon$ for every $\lambda\ge\lambda_f$; $\lambda_{\rm form}(\rho;\varepsilon)$ is the least such $\lambda_f$. The first change arrives at $\lambda_{\rm arr}(\rho;\eta)$, the first $\lambda$ with $\lvert\Delta_D+1\rvert\ge\eta$. In $D=3$ the static mark at $1\ll\rho\ll c_0/m_h$ is dominated by its inverse-distance part; in $D=1$ it is a constant plus a part linear in $\rho$, and the difference removes the constant, as the gradient does in (8.10). The square torus is out of scope by note (b) of the question.

### Step 6. Scaling limit and the formation speed (item 3)

**Order of limits.** $g_s\to0$, $M\to\infty$, $L\to\infty$ (or $\lambda$ below the wrap time), then $\rho\to\infty$ along $\mathbf e_1$ at fixed $w=c_0\lambda/\rho$ and fixed $\mu=m_h\rho/c_0$ (so $m_h=\mu c_0/\rho\to0$ with $\hat K_0$ fixed: distances large against the lattice spacing and a fixed fraction $\mu$ of the pinning length $c_0/m_h$), then $\mu$ small.

*(a) Small labels*, $\lvert\mathbf k\rvert<\delta$. With $\mathbf k=\mathbf p/\rho$: $\Omega_{\mathbf k}\lambda=w\sqrt{\mu^2+p^2}\,(1+O(p^2/\rho^2))$ and $\Omega_{\mathbf k}^{-2}=\rho^2c_0^{-2}(\mu^2+p^2)^{-1}(1+O(p^2/\rho^2))$, so (4.10) and the static mark tend to $\rho^{2-D}c_0^{-2}$ times the integrals $\int\mathrm d^Dp\,(2\pi)^{-D}e^{ip_1}\cos(w\sqrt{\mu^2+p^2})/(\mu^2+p^2)$ at $w$ and at $w=0$. These are standard Fourier integrals of explicit functions (M5); in $D=3$ (value at unit distance) and in $D=1$ (derivative in the place at unit distance) they give the same ratio

$$
\Delta_\infty(w,\mu)=\begin{cases}-1, & w<1,\\[2pt] e^{\mu}-1-\mu e^{\mu}\,I_\mu\bigl(\mu\sqrt{w^2-1}\bigr), & w>1,\end{cases}\qquad I_\mu(Y)=\int_0^Y\frac{J_1(y)\,\mathrm dy}{\sqrt{y^2+\mu^2}} .
\tag{4.12}
$$

(In $D=3$, $\partial_w$ of the first integral is $-(4\pi)^{-1}[\delta(w-1)-\theta(w-1)\,\mu J_1(\mu\sqrt{w^2-1})/\sqrt{w^2-1}]$ and its value at $w=0$ is $e^{-\mu}/(4\pi)$; in $D=1$, $\partial_w$ of the place derivative is $\tfrac12[\delta(w-1)-\theta(w-1)\,\mu J_1(\mu\sqrt{w^2-1})/\sqrt{w^2-1}]$, starting from $-e^{-\mu}/2$; the substitution $y=\mu\sqrt{w'^2-1}$ gives $I_\mu$.) The lobes of $J_1$ between its zeros alternate in sign and decrease in weight against $1/\sqrt{y^2+\mu^2}$, so $0\le I_\mu(Y)\le I_\mu(j_{1,1})<1.1$ for all $Y\ge0$ (checked numerically for $10^{-3}\le\mu\le1$: $\min I_\mu=0$, $\max I_\mu\le1.092$). Writing $\Delta_\infty=(e^\mu-1-\mu e^\mu)+\mu e^\mu(1-I_\mu)$ and using $0\le1-e^{\mu}(1-\mu)\le\mu^2e^\mu/2$ and $\lvert1-I_\mu\rvert\le1$:

$$
\Delta_\infty(w,\mu)=-1\ \ (w<1),\qquad \lvert\Delta_\infty(w,\mu)\rvert\le\mu e^{\mu}\bigl(1+\tfrac{\mu}{2}\bigr)\ \ (w>1).
\tag{4.13}
$$

*(b) The rest of the zone.* For $\lvert\mathbf k\rvert\ge\delta$ the integrand is smooth in $\mathbf k$ and $m_h$, and the phases are $\rho\,(k_1\pm(w/c_0)\Omega_{\mathbf k})$. Let $v_{\rm front}=\sup_{\mathbf k}\partial_{k_1}\hat K_0(\mathbf k)^{1/2}$. For $\rho/\lambda>v_{\rm front}$ there is no stationary point and the contribution is $O(\rho^{-\infty})$; nondegenerate stationary points contribute $O(\rho^{-D/2})$ to $u$ and to $\nabla u$, i.e. $O(\rho^{-1/2})$ relative to the static value ($\propto\rho^{-1}$, $D=3$) or gradient ($\propto\rho^0$, $D=1$). **Caustic condition** (assumption, M8): the set of caustic values, the $w$ at which degenerate stationary points occur, is finite; each contributes $O(\rho^{-1/3})$ relative. It holds for the nearest-neighbour contract and for the control (4.16) (short argument). Both are separable, $\hat K_0=\sum_ig(k_i)$, with $g'$ vanishing only at $k\in\{0,\pi\}$ and $g''\ne0$ there ($g'=2\kappa_h\sin k$, resp. $g'=\sin k\,(14-12\cos k)$). Since $\partial_{k_i}\Omega_{\mathbf k}=g'(k_i)/(2\Omega_{\mathbf k})$, in $D=3$ the stationary points lie on the four branches $k_2,k_3\in\{0,\pi\}$, where the Hessian of $\Omega$ is diagonal with the nonzero entries $g''(k_{2,3})/(2\Omega)$ in $k_2,k_3$. They are degenerate only at the zeros of $\partial_{k_1}^2\Omega$ along a branch ($D=1$: the one branch). These are finitely many on $\lvert\mathbf k\rvert\ge\delta$, since there this function is analytic and not identically zero ($\Omega$ is periodic in $k_1$ and not constant). For the nearest-neighbour $K$ the caustic values are the front $w=c_0/v_{\rm front}$ and, in $D=3$, $w\approx2.41$ and $3.15$ (the group-velocity maxima of the branches). Near $w=1$ the $O(\lvert\mathbf k\rvert^4)$ term of $\hat K_0$ smooths the step of (4.12) over a window $\lvert c_0\lambda-\rho\rvert=O(\rho^{1/3})$, with a wake of relative size $O(\rho^{-1/2}(w-1)^{-3/4})$ beyond it (stationary phase near $\mathbf k=0$).

Hence, for every $K$ of the class that satisfies the caustic condition, under the stationary-phase estimates of (b), $\Delta_D(\rho,\rho w/c_0)\to\Delta_\infty(w,\mu)$ uniformly on $w\ge1+\eta$ and on $w\le1-\eta$ for every $\eta>0$, at the rate $\rho^{-1/3}$ at worst (for $w\to\infty$ the stationary-phase remainders decrease with $\lambda$, and $\Delta_\infty\to0$ since $I_\mu(\infty)=(1-e^{-\mu})/\mu$). With (4.13), for such $K$: for every $\varepsilon\in(0,1)$ and every $\mu$ with $\mu e^\mu(1+\mu/2)<\varepsilon$,

$$
\lim_{\rho\to\infty}\frac{c_0\,\lambda_{\rm form}(\rho;\varepsilon)}{\rho}=1,\qquad \lambda_{\rm form}(\rho;\varepsilon)-\frac{\rho}{c_0}=O(\rho^{1/3})\ \ \text{once the caustic wakes }O(\rho^{-1/3})\text{ are below }\varepsilon-\mu e^\mu(1+\tfrac{\mu}{2}) ,
\tag{4.14}
$$

on the ring ($D=1$) and on the cubic torus ($D=3$): the long-range part forms within a $\lambda$ linear in the distance, at the formation speed $v_{\rm form}=c_0$, independent of $\varepsilon$; for $\lambda<(1-\eta)\rho/c_0$ it has not formed ($\Delta_D\to-1$). The caustic condition is shown for the nearest-neighbour contract and for the control (4.16) (Step 6(b)); for other members of the class it is assumed.

### Step 7. Formation speed, slope and front

Let $v_{\rm slope}=\sup_{\mathbf k}\lvert\nabla\hat K_0(\mathbf k)^{1/2}\rvert$, the largest slope of the normal-mode frequencies (at $m_h\to0$, the regime of Step 6). Since $c_0=\lim_{k_1\to0^+}\partial_{k_1}\hat K_0(k_1\mathbf e_1)^{1/2}$,

$$
v_{\rm form}=c_0\ \le\ v_{\rm front}\ \le\ v_{\rm slope}.
\tag{4.15}
$$

The inequalities hold for every $K$ of the class; $v_{\rm form}=c_0$ holds under the conditions of (4.14) (caustic condition and the estimates of Step 6(b)). Under the same conditions the first change at $\rho\mathbf e_1$ arrives with $v_{\rm front}$ (Step 6(b): no stationary point for $\rho/\lambda>v_{\rm front}$); between $\rho/v_{\rm front}$ and $\rho/c_0$ the made field is a short-wavelength wake of relative size $O(\rho^{-1/3})$, not the long-range part. For the nearest-neighbour contract, $\hat K_0=2\kappa_h\sum_i(1-\cos k_i)$, $\lvert\nabla\hat K_0^{1/2}\rvert^2=\kappa_h\sum_i\sin^2\tfrac{k_i}2\cos^2\tfrac{k_i}2/\sum_i\sin^2\tfrac{k_i}2\le\kappa_h$, so $c_0=v_{\rm front}=v_{\rm slope}=\sqrt{\kappa_h}$: arrival and formation coincide, as in (8.15). **Control** (a member of the class with $c_0\ne v_{\rm slope}$):

$$
\hat K_0(\mathbf k)=\sum_{i}\bigl[14(1-\cos k_i)-3(1-\cos2k_i)\bigr]=\sum_i(1-\cos k_i)(8-6\cos k_i),\qquad c_0=1,\quad v_{\rm front}=v_{\rm slope}=2.475 ,
\tag{4.16}
$$

of range 2 along the axes, with the symmetries of the class and $\hat K_0\ge2\sum_i(1-\cos k_i)$; the speeds are computed by `code/v2/oscillator_formation.py`. There the first change runs ahead of the formation by the factor 2.475.

### Step 8. Speeds of bodies and field (item 4)

Let $\ell_0$ be the recorded distance of one step of the body's chain ($\ell_0=\kappa\sigma_X$ in the setting of (5.11)); by (5.11) $v_{\max}=\tau\ell_0$. By the coupling map (4.15) of 06-ilang-source/04-profile, one chain step is $k$ field steps along the ring (along a lattice axis of the tori, a stated assumption, M8), so one field step is the recorded distance $\ell_0/k$, and the formation speed $c_0$ (field steps per unit $\lambda$) is $c_0\ell_0/k$. Hence

$$
\frac{v_{\rm form}}{v_{\max}}=\frac{c_0}{k\,\tau},\qquad v_{\rm form}=v_{\max}\iff c_0=k\tau\quad(\text{nearest neighbour: }\sqrt{\kappa_h}=k\tau),
\tag{4.17}
$$

independent of $\ell_0$; the front moves at $v_{\rm front}\ell_0/k\ge v_{\rm form}$. Equality of the two speeds is a condition on the description (M7).

### Step 9. Numerical check (code)

Criterion (4.11) with $\varepsilon=0.1$; all numbers from `code/v2/output/`.

*(i) Oscillator cells* (`oscillator_formation.py`; $c_0=1$; ring $L=4096$, $m_h=5\cdot10^{-5}$; cubic torus $L=256$, $m_h=2.5\cdot10^{-4}$; so $\mu\le0.026$; windows end before the first image front). *Control (4.16):* $c_0\lambda_{\rm form}/\rho=1.266,\,1.162,\,1.089,\,1.053,\,1.031,\,1.018$ for $\rho=16,32,\dots,512$ ($D=1$) and $1.344,\,1.191,\,1.154,\,1.082$ for $\rho=8,16,24,48$ ($D=3$). The arrival ($\eta=0.01$) gives $v_{\rm front}\lambda_{\rm arr}/\rho$ from 0.890 to 0.991 ($D=1$) and from 0.665 to 0.872 ($D=3$). Between the arrival and $w=0.8$ the mean of $\Delta_D$ lies between $-1.18$ and $-0.99$: the change has arrived, the long-range part has not formed. *Nearest neighbour:* $c_0\lambda_{\rm form}/\rho=2.47,\,1.72,\,1.43,\,1.26,\,1.16,\,1.13$ ($D=1$, $\rho=16,\dots,512$), with $(\lambda_{\rm form}-\rho/c_0)/\rho^{1/3}$ between 6.4 and 9.3, bounded as (4.14) states. In $D=3$: $3.94,\,3.64,\,3.72,\,3.14$ for $\rho=8,16,24,48$, set by the caustic wakes at $w\approx2.41,\,3.15$. Their largest $\lvert\Delta_3\rvert$ on $1.5\le w\le3$ falls as $0.238,\,0.197,\,0.178,\,0.130$, i.e. as $\rho^{-1/3}$ (ratio 0.546 from $\rho=8$ to 48, against $6^{-1/3}=0.550$), and extrapolates below $\varepsilon=0.1$ at $\rho\approx10^2$. In short,

$$
\frac{c_0\lambda_{\rm form}}{\rho}\Big\vert_{\rm control}=1.018\ (D=1,\ \rho=512),\ \ 1.082\ (D=3,\ \rho=48);\qquad \frac{c_0\lambda_{\rm form}}{\rho}\Big\vert_{\rm n.n.}=1.13\ (D=1,\ \rho=512),\ \ 3.1\text{–}3.9\ (D=3,\ \rho\le48).
\tag{4.18}
$$ Static marks: relative deviation from the closed or Bessel form $10^{-9}$ ($D=1$) and $4\cdot10^{-13}$ ($D=3$). $I_\mu$: $\min=0$, $\max\le1.092$, $\sup\lvert1-I_\mu\rvert=1$ for $\mu=10^{-3},\dots,1$.

*(ii) Field of 06-ilang-source* (`conserved_density_formation.py`; $t_f=1$, infinite lattice). $\lambda_{\rm form}/[\rho^2/(4\zeta(0.1))]$ for $\rho=4,8,16,32$ ($D=1$): 1.69, 1.32, 1.15, 1.08; for $\rho=4,8,12,16$ ($D=3$): 0.965, 0.992, 0.997, 0.998; local exponents of $\lambda_{\rm form}$ in $\rho$: 1.90 ($D=1$, last two) and 2.006 ($D=3$). At $\lambda=2\rho$ the deviation is $\Delta\approx-1$ ($-0.96$ and $-1.33$ at the largest $\rho$): no linear formation, as (4.7)–(4.8) require.

## Result

Class F and its source and start as stated (Ω the nondegenerate lowest eigenvector of $C_F$), first order in $g_s$ throughout.

- **(4.1), (4.2)** (formula; every $L$, every $\lambda\ge0$): the made field and its label components in the eigenvectors of $C_F$; the static mark is the same with $g_\lambda(\omega)\to1/\omega$.
- **(4.3)** (definition): the sum-rule weight $f_A(\mathbf q)$ and the static self-response $\chi_Y(\mathbf q)$, fixed by $C_F$, the read or source operator and Ω; $\delta a^{\rm st}_{\mathbf q}=-\chi_A(\mathbf q)$ for $B=A$.
- **(4.4)** (theorem, item 1): $\lvert\delta a_{\mathbf q}(\lambda)\rvert\le\min\{\lambda\sqrt{f_A\chi_B},\,2\sqrt{\chi_A\chi_B}\}$ for every $\mathbf q$, $L$ and $\lambda\ge0$; for $B=A$, $\lvert\delta a_{\mathbf q}(\lambda)/\delta a^{\rm st}_{\mathbf q}\rvert\le\lambda\sqrt{f_A/\chi_A}$. It holds for the nondegenerate highest eigenvector with $\lvert f_A\rvert,\lvert\chi\rvert$.
- **(4.5)** (theorem): if $A$ is the density of a number $N$ conserved by every term of $C_F$, with $[A_x,A_y]=0$, then $f_A(\mathbf q)\le c_A\lvert\mathbf q\rvert^2$ for every $L$, with $c_A$ fixed by $C_F$ and $A$.
- **(4.6)** (theorem): under (4.5), for every source operator $B$, formation of the label component $\mathbf q$ to accuracy $\varepsilon$ needs $\lambda\ge(1-\varepsilon)\lvert\delta a^{\rm st}_{\mathbf q}\rvert/(\lvert\mathbf q\rvert\sqrt{c_A\chi_B(\mathbf q)})$.
- **(4.7)** (negative, item 2, $B=A$): with a long-range mark $\lvert\delta a^{\rm st}_{\mathbf q}\rvert\ge s_0/\lvert\mathbf q\rvert^2$ ($\lvert\mathbf q\rvert\le q_1$), $\lambda_{\rm form}(\mathbf q;\varepsilon)\ge(1-\varepsilon)\sqrt{s_0/c_A}\,\lvert\mathbf q\rvert^{-2}$: the long-range part of a conserved density cannot form within a $\lambda$ linear in the distance; also for the highest eigenvector.
- **(4.8)** (negative, position form, $B=A$, cubic torus, $L\to\infty$ then $\ell\to\infty$): if moreover $\chi_A\le s_1/\lvert\mathbf q\rvert^2$ and $\delta a_{\rm st}=s/\lvert x\rvert+o(\lvert x\rvert^{-1})$, formation at every place of the shell $\ell\le\lvert x\rvert\le2\ell$ needs $\lambda\ge C\ell^2(1+o(1))$, $C>0$ given in (4.8).
- **(4.9)** (negative, item 2, $B\ne A$, several kinds of objects included; under (4.5), with a long-range mark): formation within $\lambda\le\beta/\lvert\mathbf q\rvert$ at long-range labels requires $\chi_B(\mathbf q)\ge(1-\varepsilon)^2s_0^2/(\beta^2c_A\lvert\mathbf q\rvert^4)$, so it is excluded for every $B$ with $\lvert\mathbf q\rvert^4\chi_B(\mathbf q)\to0$. For every $B$ of the class of the question's note (a), $\chi_B\le s_B/\lvert\mathbf q\rvert^2$, $\lambda_{\rm form}(\mathbf q;\varepsilon)\ge(1-\varepsilon)s_0/(\sqrt{c_As_B}\,\lvert\mathbf q\rvert^2)$ for $\lvert\mathbf q\rvert\le q_1$: the long-range part of a conserved density cannot form within a $\lambda$ linear in the distance; also for the highest eigenvector. Sources with a stronger self-response are out of scope (note (a)). Control: oscillator cells with $A=B=q_x$, where $f_A=1$.
- **(4.10)** (formula, oscillator class, $M\to\infty$ then $L\to\infty$): exact made field $\delta a_{\mathbf k}(\lambda)=-(1-\cos\Omega_{\mathbf k}\lambda)/\Omega_{\mathbf k}^2$ and its deviation from the static mark.
- **(4.11)** (definition): the formation criterion along an axis (value ratio in $D=3$, difference ratio in $D=1$), $\lambda_{\rm form}(\rho;\varepsilon)$ and $\lambda_{\rm arr}(\rho;\eta)$.
- **(4.12), (4.13)** (formula, theorem; scaling limit $\rho\to\infty$ at fixed $w=c_0\lambda/\rho$, $\mu=m_h\rho/c_0$): $\Delta_D\to\Delta_\infty(w,\mu)$, equal to $-1$ for $w<1$ and bounded by $\mu e^\mu(1+\mu/2)$ for $w>1$, in $D=1$ and $D=3$.
- **(4.14)** (theorem, item 3, on the ring ($D=1$) and the cubic torus ($D=3$); the square torus is out of scope by note (b)): for every $K$ of the class with a finite caustic set (the caustic condition), under the stationary-phase estimates of Step 6(b), for every $\varepsilon\in(0,1)$ and $\mu e^\mu(1+\mu/2)<\varepsilon$, $c_0\lambda_{\rm form}(\rho;\varepsilon)/\rho\to1$: the long-range part forms within a $\lambda$ linear in the distance, at the speed $c_0$, independent of $\varepsilon$; the lattice correction is $O(\rho^{1/3})$ in $\lambda$ once the caustic wakes, $O(\rho^{-1/3})$ relative in $D=3$, are below $\varepsilon-\mu e^\mu(1+\mu/2)$. The caustic condition is shown for the nearest-neighbour contract (caustic values: the front and, in $D=3$, $w\approx2.41$ and $3.15$) and for the control (4.16), and assumed for the other members. Order: $g_s\to0$, $M\to\infty$, $L\to\infty$, $\rho\to\infty$, then $\mu$ small.
- **(4.15)** (theorem): $c_0\le v_{\rm front}\le v_{\rm slope}$ for every $K$ of the class; under the conditions of (4.14), $v_{\rm form}=c_0$ and the first change arrives with $v_{\rm front}$; for the nearest-neighbour contract all three equal $\sqrt{\kappa_h}$.
- **(4.16)** (example, control of item 3): $\hat K_0=\sum_i(1-\cos k_i)(8-6\cos k_i)$ is in the class with $c_0=1<v_{\rm front}=v_{\rm slope}=2.475$.
- **(4.17)** (formula, item 4, ring, or a lattice axis of the tori by assumption): $v_{\rm form}/v_{\max}=c_0/(k\tau)$; the two speeds are equal if and only if $c_0=k\tau$.
- **(4.18)** (numerical, Step 9, $\varepsilon=0.1$): for the control (4.16), $c_0\lambda_{\rm form}/\rho$ tends to 1 (1.018 at $\rho=512$, $D=1$; 1.082 at $\rho=48$, $D=3$), and the arrival tends to $\rho/v_{\rm front}$. For the nearest-neighbour $K$, $c_0\lambda_{\rm form}/\rho=1.13$ at $\rho=512$ ($D=1$), and between 3.1 and 3.9 in $D=3$ at $\rho\le48$, set by caustic wakes that decay as $\rho^{-1/3}$. For the field of 06-ilang-source, $4\zeta(0.1)\lambda_{\rm form}/\rho^2$ tends to 1 (1.08 at $\rho=32$, $D=1$; 0.998 at $\rho=16$, $D=3$).

## Consistency checks

1. **Dimensions.** $g_sB$ is a rate, so $\delta a$ per unit $g_s$ has the units $[A][B][\lambda]$; $f_A\sim[A]^2/[\lambda]$ and $\chi_B\sim[B]^2[\lambda]$, so both bounds of (4.4) and (4.6) are consistent; $c_0$ is in field steps per unit $\lambda$ and $\tau$ is a rate, so (4.17) is a pure number.
2. **(8.10), (8.14) as a special case.** For the field of 06-ilang-source with one field object ($n=1/\lvert X_f\rvert$) and $A=B=\Phi$, (4.2) gives one mode per label, $\delta a_{\mathbf q}(\lambda)/\delta a^{\rm st}_{\mathbf q}=1-\cos\omega_{\mathbf q}\lambda\le\lambda\omega_{\mathbf q}$, which is (4.4) with $f_A=2n\omega_{\mathbf q}\le2nt_f\lvert\mathbf q\rvert^2$ (4.5); (4.7) then gives a $\lambda$ quadratic in the distance, as (8.10) and (8.14). The code reproduces $\lambda_{\rm form}\to\rho^2/(4t_f\zeta)$ with $\zeta(0.1)=0.01555$, against $\zeta_1=\pi\varepsilon^2/2=0.01571$ of (8.10) (Step 9).
3. **(8.15) as a special case.** For the nearest-neighbour contract in $D=3$ with the value criterion, $c_0=\sqrt{\kappa_h}$, $w=c$ and $\mu=m_Y\rho$; (4.13) gives $\lvert\Delta\rvert\le\mu e^\mu(1+\mu/2)$ for $c>1$, the same order $O(m_Y\rho)$ as the bound $8.4\,m_Y\rho$ of (8.15), and the remainder $O(\rho^{-1/2})$ of nondegenerate stationary points matches $o(1)=O(\rho^{-1/2})$ for $1<c<2$, below the first caustic at $c\approx2.41$.

## Open issues

- Item 2, position form: derived for formation on a whole shell (4.8); for a single place, only for the field of 06-ilang-source ((8.14) and the code).
- Item 3: the remainder estimates of Step 6(b) are stationary-phase estimates; the caustic condition is shown here only for the nearest-neighbour contract and (4.16), and is an assumption of (4.14) for the other members of the class.
- Item 3: the approach to $v_{\rm form}=c_0$ is slow when the front coincides with $c_0$ (nearest neighbour), through the $O(\rho^{1/3})$ window and, in $D=3$, the caustic wakes; at the distances computed in $D=3$ ($\rho\le48$), $\lambda_{\rm form}$ is still 3 to 4 times $\rho/c_0$ there.
- Item 1 for eigenvectors inside the spectrum: (4.4) holds with $\sum_n\lvert\omega_n\rvert(\dots)$ in place of $f_A$, but (4.5) does not apply.
- Item 4 on the tori rests on the assumption that the coupling map runs along a lattice axis.

## Code

- `code/v2/oscillator_formation.py` (check, items 3 and 4's speeds; unchanged from v1, run again): exact made field (4.10) on a ring ($L=4096$) and a cubic torus ($L=256$), static mark on the infinite lattice by its Laplace representation (cross-checked against the closed form or Bessel form for the nearest-neighbour $K$), criterion (4.11) with $\varepsilon=0.1$, $\eta=0.01$, for the nearest-neighbour $K$ and the control (4.16); speeds $c_0$, $v_{\rm front}$, $v_{\rm slope}$; the bounds on $I_\mu$ used in (4.13). Output: `code/v2/output/oscillator_formation.json`.
- `code/v2/conserved_density_formation.py` (check, item 2; unchanged from v1, run again): the field of 06-ilang-source with one field object, $A=B=\Phi$, on the infinite chain and cubic lattice (Bessel-function forms of (4.1)), criterion (4.11) with $\varepsilon=0.1$, against the continuum prediction $\rho^2/(4t_f\zeta(\varepsilon))$. Output: `code/v2/output/conserved_density_formation.json`.

## Methods used

- first-order (Duhamel) perturbation, spectral representation
- sum rule of a double commutator, Cauchy–Schwarz inequality
- translation covariance, Fourier series on tori, Parseval
- Fourier integrals with Bessel functions, stationary phase, scaling limits
- exact equations of the means for a quadratic contract
- numerical lattice sums and Bessel-function integrals (Python)

Author: Ivan Honis · ivan.honis@ndot.io · License: CC BY 4.0 · Human–AI note