Version 1 · earlier version; the current one is v3 · External review, round 1: major errors

# The field made by a held source as a common local scale for a test body

- **Subproject:** 06-ilang-source
- **Package:** 05-made-field-scale
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-19

## Setup and assumptions

**Inputs** (as quoted in question.md).
- (2.1) of 02-what-a-body-presents@v3: $C_{bf}=\gamma\sum_xQ_x\otimes\Phi_x$, $Q^{(a)}_x=\Pi_xO$, $Q^{(c)}_x=\Pi_xO+\tfrac12(A_x+A_{x-1})$, with $\Pi_x=\lvert x\rangle\langle x\rvert\otimes\mathbb 1_2\otimes\mathbb 1_N$, $O=\sigma_3\otimes(m\mathbb 1_N+D)$, $A_x$ the bond $(x,x+1)$ of the motion term, $A=\sum_xA_x$, $B=\sum_x\Pi_xO$.
- (3.6), (3.9) of 03-held-sources@v4: $\chi_n(x,y)=\sum_{m:\,e_m\neq e_n}2\operatorname{Re}\bigl(\langle\varphi_n\vert\Phi_x\vert\varphi_m\rangle\langle\varphi_m\vert\Phi_y\vert\varphi_n\rangle\bigr)/(e_m-e_n)$ and $\langle\varphi_n(g)\vert\Phi_y\vert\varphi_n(g)\rangle=\bar\Phi_n(y)-\sum_ig_i\chi_n(y,x_i)+O(\lVert\Phi_y\rVert v^2/\Delta_n^2)$.
- (4.4) of 04-profile@v4: $M^{(2)}(x_1,x_2)=-\frac{2ng_1g_2}{t_f}\Lambda^+(x_1,x_2)$, $\Lambda^+(x,y)=\lvert X_f\rvert^{-1}\sum_{\mathbf k\neq0}e^{i\mathbf k\cdot(x-y)}/\omega(\mathbf k)$, $\omega(\mathbf k)=\sum_{\mu=1}^{D}4\sin^2\frac{k_\mu}2$.
- (17.11) of 05-ilang-free-fall/17-typical-or-forced@v3: for linear level profiles $M_{x,k}=M_k+G_kx$, $V_{x,k}=-\mathcal F_kx$: R3 $\iff G_k=\gamma'M_k,\ \mathcal F_k=\phi M_k\ \forall k\iff\boldsymbol g\in\mathcal U_4^{\pm}$ (fall from rest level independent on both branches). Here $\gamma'$ is the $\gamma$ of (17.11), not the coupling constant $\gamma$.
- (4.16), (4.24) of 05-ilang-free-fall/04-branch-fall@v3: $a_{k,\sigma}(0)=\frac{\tau_k^2}{M_k}(\sigma\mathcal F_k-G_k)$; $\frac{\mathrm d^2\bar x}{\mathrm d\lambda^2}\big\vert_{\rm drift}=d_0\,a_{k,+}(0)\,\hat e=-\frac{v_{\max}^2}{d_0}\frac{r(x+1)-r(x)}{r(x)}\hat e$, with $v_{\max}=\tau d_0$, $d_0=\kappa\sigma_X$, $r(x)$ the clock rate at $x$.

**Assumptions** (the description chosen for study, M8, M9).
- **S1 Field.** $N_f\ge1$ identical field objects ($c=+1$), no pair terms among them; places $X_f$: a ring of $L\ge5$ places ($D=1$), an $L\times L$ torus ($D=2$) or an $L^3$ torus ($D=3$); own term $C_1=-t_f\sum_{\{x,y\}}(\lvert x\rangle\langle y\rvert+\lvert y\rangle\langle x\rvert)$, $t_f>0$; $\Phi_x$ the number of field objects at $x$; $n=N_f/\lvert X_f\rvert$. With the source the field operator is $C_f^s:=\sum_{j=1}^{N_f}C_1^{(j)}+g_s\Phi_{x_s}$ on the symmetric subspace $\mathcal H_f$.
- **S2 Source.** The held source is in a common eigenvector of its own term and of its coupled operator (eigenvalue $g_s$). $C$ then keeps the source in that vector, and the source enters only through $g_s\Phi_{x_s}$ and a constant; it is suppressed below.
- **S3 Test body.** Model D with positions $x\in\ell$, a line of $X_f$ (the whole ring for $D=1$), recorded by a medium $c$. $C_{bc}$ is the part of $C$ not involving the field: $C_{\rm D}=A+B$, the record term and the medium's own term. Coupling (2.1), form (a) or (c). $d_k=2\pi k/(N\lambda_0)$, $M_k=m+d_k$, $\lVert O\rVert=M_{N-1}$.
- **S4 Start.** $\Psi(0)=\xi\otimes\varphi$ with any body–medium state $\xi\in\mathcal H_{bc}=\mathcal H_b\otimes\mathcal H_c$ and a field state $\varphi$; $\varphi(\lambda):=e^{-iC_f^s\lambda}\varphi$. Items 2–3 use a stationary $\varphi$ (an eigenvector of $C_f^s$), item 4 the lowest adapted state (Step 5).
- **S5 Positivity** (items 3–4). $\nu(x):=1+\gamma N_\varphi(x)>0$ on $\ell$, and $1+\gamma n>0$.
- **S6 Regime of item 3.** The conditions of (4.16) of 05-ilang-free-fall/04-branch-fall (branch start at rest, first order in the gradients), and the local-slope approximation: on the support of the start (width $w$ around $x_0$) $N_\varphi(x)=N_\varphi(x_0)+N'(x_0)(x-x_0)$ up to a curvature term, with $\lvert N''\rvert w\ll\lvert N'\rvert$.

Notation: $N_\lambda(x):=\langle\varphi(\lambda)\vert\Phi_x\vert\varphi(\lambda)\rangle$, $\delta\Phi_x:=\Phi_x-N_\lambda(x)$, $\sigma_\Phi(\lambda):=\max_{x\in\ell}\bigl(\operatorname{Var}_{\varphi(\lambda)}\Phi_x\bigr)^{1/2}$. For stationary $\varphi$, $N_\lambda=N_\varphi$ and $\sigma_\Phi$ are constant.

## Derivation

### Step 1. The operator by which the test body evolves

By S2, on $\mathcal H_{bc}\otimes\mathcal H_f$ and up to a constant (a common phase, (1.2)), $C=C_{bc}\otimes\mathbb 1+\mathbb 1\otimes C_f^s+\gamma\sum_{x\in\ell}Q_x\otimes\Phi_x$. For a unit vector $\varphi$ define the partial expectation $\bar C_\varphi$ on $\mathcal H_{bc}$ by $\langle\xi'\vert\bar C_\varphi\vert\xi\rangle:=\langle\xi'\otimes\varphi\vert C\vert\xi\otimes\varphi\rangle$. Inserting $C$:

$$
\bar C_\varphi=C_{bc}+\gamma\sum_{x\in\ell}N_\varphi(x)\,Q_x+E_\varphi\mathbb 1,\qquad N_\varphi(x)=\langle\varphi\vert\Phi_x\vert\varphi\rangle,\quad E_\varphi=\langle\varphi\vert C_f^s\vert\varphi\rangle .
\tag{5.1}
$$

Let $\xi(\lambda)$ solve $i\,\mathrm d\xi/\mathrm d\lambda=(\bar C_{\varphi(\lambda)}-E_\varphi)\xi$ ($E_{\varphi(\lambda)}=E_\varphi$ is conserved) and $\Psi_{\rm eff}:=\xi(\lambda)\otimes\varphi(\lambda)$. The product rule gives $i\dot\Psi_{\rm eff}=C\Psi_{\rm eff}-\gamma W_\lambda\Psi_{\rm eff}$ with $W_\lambda:=\sum_xQ_x\otimes\delta\Phi_x$. Hence $D:=\Psi-\Psi_{\rm eff}$ obeys $i\dot D=CD+\gamma W_\lambda\Psi_{\rm eff}$, $D(0)=0$, and Duhamel's formula with the unitarity of $e^{-iC\lambda}$ gives $\lVert D(\lambda)\rVert\le\lvert\gamma\rvert\int_0^{\lvert\lambda\rvert}\lVert W_\mu\Psi_{\rm eff}(\mu)\rVert\,\mathrm d\mu$. Write $W\Psi_{\rm eff}=\sum_xv_x$, $v_x=Q_x\xi\otimes\delta\Phi_x\varphi$, $\lVert\delta\Phi_x\varphi\rVert^2=\operatorname{Var}\Phi_x$.
- (a): $Q_xQ_y=\delta_{xy}\Pi_xO^2$, so $\lVert\sum_xv_x\rVert^2=\sum_x\langle\xi\vert\Pi_xO^2\vert\xi\rangle\operatorname{Var}\Phi_x\le\lVert O\rVert^2\sigma_\Phi^2$.
- (c): $Q_x\xi$ lies on the places $x-1,x,x+1$, so $\langle v_x,v_y\rangle=0$ for $\lvert x-y\rvert\ge3$ and $\lVert\sum_xv_x\rVert^2\le5\sum_x\lVert v_x\rVert^2$. With $\lVert A_x\rVert=\tau/2$, $\lVert Q_x\xi\rVert\le(\lVert O\rVert+\frac\tau2)\lVert(\Pi_{x-1}+\Pi_x+\Pi_{x+1})\xi\rVert$, whose squares sum to $3$.

Therefore

$$
\bigl\lVert\Psi(\lambda)-\xi(\lambda)\otimes\varphi(\lambda)\bigr\rVert\le\lvert\gamma\rvert\,c_Q\int_0^{\lvert\lambda\rvert}\sigma_\Phi(\mu)\,\mathrm d\mu,\qquad c_{Q^{(a)}}=\lVert O\rVert,\quad c_{Q^{(c)}}=\sqrt{15}\,\bigl(\lVert O\rVert+\tfrac\tau2\bigr).
\tag{5.2}
$$

For stationary $\varphi$, $\xi(\lambda)=e^{-i(\bar C_\varphi-E_\varphi)\lambda}\xi$: up to a common phase the test body evolves by the $\lambda$-independent $\bar C_\varphi$ and the field stays in $\varphi$, with remainder at most $\lvert\gamma\rvert c_Q\sigma_\Phi\lvert\lambda\rvert$; reading probabilities change by at most twice that. The neglected term $\gamma W$ is the change of the field by the test body, so (5.2) is the order of the test-body neglect (A15): first order in $\gamma$, proportional to the field fluctuation $\sigma_\Phi$ and to $\lvert\lambda\rvert$. The description holds for $\lvert\lambda\rvert\ll(\lvert\gamma\rvert c_Q\sigma_\Phi)^{-1}$.

*M6.* $\bar C_\varphi$ is computed from $C$ and $\varphi$ alone. It is therefore unchanged by every relisting that leaves $C$ unchanged, e.g. $\Phi_x\to\Phi_x-c_x$ together with $C_{\rm D}\to C_{\rm D}+\gamma\sum_xc_xQ_x$, place-dependent $c_x$ included. It does not depend on the labels of the field objects ($\Phi_x$ is symmetric) or on the phase of $\varphi$, and its multiple of $\mathbb 1$ is a common phase. The split of $\bar C_\varphi$ into $C_{bc}$ and $\gamma\sum_xN_\varphi(x)Q_x$ is not fixed: the separate values of the bare blocks, of $\gamma$ and of $N_\varphi(x)$ depend on the listing.

### Step 2. Level blocks

Let $\hat\nu:=\sum_x\nu(x)\Pi_x$ with $\nu(x)=1+\gamma N_\varphi(x)$. On level $k$, $O=M_k\sigma_3$. For (a), (5.1) gives $\gamma\sum_xN_\varphi(x)\Pi_xO=(\hat\nu-1)B$. For (c) it gives in addition $\gamma\sum_xN_\varphi(x)\tfrac12(A_x+A_{x-1})=\gamma\sum_x\tfrac12(N_\varphi(x)+N_\varphi(x+1))A_x$. Since $A_x$ connects only $x$ and $x+1$, $\tfrac12\{\hat\nu,A_x\}=\tfrac12(\nu(x)+\nu(x+1))A_x$. The record term and the medium do not involve the field and stay the background of 05-ilang-free-fall. The level-$k$ blocks of $\bar C_\varphi$ are

$$
\begin{aligned}
&B_{x,k}=M_{x,k}\sigma_3+V_{x,k}\mathbb 1_2,\qquad M_{x,k}=\nu(x)\,M_k,\qquad V_{x,k}=E_\varphi,\\
&\text{hopping on }(x,x+1):\ \tfrac{\tau_{x,x+1}}2(-i\sigma_1),\qquad \tau^{(a)}_{x,x+1}=\tau,\qquad \tau^{(c)}_{x,x+1}=\tau\,\tfrac12\bigl(\nu(x)+\nu(x+1)\bigr),
\end{aligned}
\tag{5.3}
$$

i.e. the Model D part of $\bar C_\varphi$ is $\hat\nu B+A$ for (a) and $\tfrac12\{\hat\nu,C_{\rm D}\}$ for (c). Fixed by $C$ and $\varphi$: the blocks of $\bar C_\varphi$, i.e. $M_{x,k}$, the hopping amplitudes and $V_{x,k}$ up to the common constant. Hence the level-independent ratios $\nu(y)/\nu(x)=M_{y,k}/M_{x,k}$ are fixed. Listing-dependent: the factorization $M_{x,k}=\nu(x)M_k$ (a constant relisting $c_x=c$ multiplies $M_k$, and $\tau$ for (c), by $1+\gamma c$ and divides $\nu$ by it) and the place-independent constant $E_\varphi$ in $V_{x,k}$.

### Step 3. R3 place by place, the hopping, and a control

From (5.3), for both couplings and every $\varphi$ (at every $\lambda$ if $\varphi$ is not stationary):

$$
B_{x+1,k}-B_{x,k}=M_k\,\Gamma(x),\qquad \Gamma(x)=\gamma\bigl(N_\varphi(x+1)-N_\varphi(x)\bigr)\sigma_3 ;\qquad\text{for }\nu(x)\neq0:\ B_{x+1,k}-B_{x,k}=M_{x,k}\,\frac{\nu(x+1)-\nu(x)}{\nu(x)}\,\sigma_3 .
\tag{5.4}
$$

$\Gamma(x)$ does not depend on $k$, so R3 holds at every place. It has no $\mathbb 1_2$ part ($\phi=0$ in the parametrization of (17.11) of 05-ilang-free-fall/17-typical-or-forced). The second form uses only fixed quantities.

*Hopping.* (a): the hopping is $\tau$ at every bond, while the on-site blocks change by $\nu(x+1)/\nu(x)$. The two factors differ at every place with $\gamma(N_\varphi(x+1)-N_\varphi(x))\neq0$. (c): by (5.3), every matrix element of the Model D part of $\bar C_\varphi$ between places $x,x'$ ($x'=x$ or a neighbour) equals that of $C_{\rm D}$ times $\tfrac12(\nu(x)+\nu(x'))$. The hopping of every level thus carries the same local factor as the on-site blocks, evaluated as the mean over the two ends of the bond.

*Control.* Take the pair term of the same form with $Q^{(r)}_x=\Pi_x\,\sigma_3\otimes m\mathbb 1_N$ (rest term only). Then $B^{(r)}_{x,k}=(M_k+\gamma mN_\varphi(x))\sigma_3+E_\varphi\mathbb 1_2$. R3 at $x$ would require $M_k\Gamma=\gamma m\Delta N\sigma_3$ for $k=0,1$, $\Delta N:=N_\varphi(x+1)-N_\varphi(x)$. With $M_0=m$ this forces $\Gamma=\gamma\Delta N\sigma_3$ and then $d_1\gamma\Delta N=0$, where $d_1=2\pi/(N\lambda_0)>0$. Under (4.16) of 05-ilang-free-fall/04-branch-fall ($G_k=\gamma mN'$, $\mathcal F_k=0$, $\tau_k=\tau$):

$$
B^{(r)}_{x+1,k}-B^{(r)}_{x,k}=\gamma m\,\Delta N\,\sigma_3,\qquad a^{(r)}_{k,\sigma}(0)=-\tau^2\,\frac{\gamma m\,N'(x_0)}{M_k+\gamma mN_\varphi(x_0)} .
\tag{5.5}
$$

So R3 fails at every place with $\gamma\Delta N\neq0$, and there the fall from rest depends on the level.

### Step 4. Clock rate and fall in a stationary field state

The on-site eigenvalues are $E_\varphi\pm\nu(x)M_k$. By S5 the upper one is $E_\varphi+\nu(x)M_k$, and its spacing between levels $k,l$ is $\nu(x)(d_l-d_k)$. The clock-rate ratio of two places is therefore

$$
\frac{r(y)}{r(x)}=\frac{\nu(y)}{\nu(x)}=\frac{M_{y,k}}{M_{x,k}},
\tag{5.6}
$$

the same for every pair of levels. It is exact in $\bar C_\varphi$ (no gradient expansion), and the lower branch gives the same ratio.

*Fall.* Under S6 the profiles across the start are those of 05-ilang-free-fall with base half-gap $\nu(x_0)M_k$, $G_k=\gamma N'(x_0)M_k$, $\mathcal F_k=0$, and hopping $\tau_k=\tau_{\rm loc}(x_0)$: $\tau$ for (a), $\tau\nu(x_0)$ for (c) up to first order in the gradient. The variation of the hopping across the start (c) does not enter at first order. In the band energy of a branch, $E_\sigma(x,p)=E_\varphi+\sigma\sqrt{M_{x,k}^2+\tau_{\rm loc}(x)^2\sin^2p}$, the hopping multiplies $\sin^2p$, whose $x$-derivative vanishes at rest. With (4.16) of 05-ilang-free-fall/04-branch-fall and (5.6):

$$
a_{k,\sigma}(0)=-\tau_{\rm loc}(x_0)^2\,\frac{\gamma N'(x_0)}{\nu(x_0)}=-\tau_{\rm loc}(x_0)^2\,\frac{r(x_0+1)-r(x_0)}{r(x_0)},\qquad v_{\max}(x_0):=\tau_{\rm loc}(x_0)\,d_0 .
\tag{5.7}
$$

The second equality holds at first order in the gradient. $a_{k,\sigma}$ is the same for all levels and both branches, as (17.11) of 05-ilang-free-fall/17-typical-or-forced requires for R3 with $\phi=0$. Multiplying by $d_0$, $d_0a_{k,+}(0)=-\frac{v_{\max}(x_0)^2}{d_0}\frac{r(x_0+1)-r(x_0)}{r(x_0)}$: (4.24) holds with the maximal speed $v_{\max}(x_0)$ of the effective contract at the start. For (a), $v_{\max}(x_0)=\tau d_0$ at every place. For (c), $v_{\max}(x_0)=\tau\nu(x_0)d_0$ changes from place to place by the factor of $r$; with the value at a fixed reference place $y$ instead, (4.24) acquires the factor $(r(x_0)/r(y))^2$. $\tau_{\rm loc}$ and $\gamma N'/\nu=(M_{x_0+1,k}-M_{x_0,k})/M_{x_0,k}$ are fixed by $C$ and $\varphi$.

### Step 5. The made field in the lowest adapted state

Without pair terms among the field objects, $C_f^s=\sum_jh^{(j)}$ with $h=C_1+g_s\lvert x_s\rangle\langle x_s\rvert$. $h$ has off-diagonal entries $-t_f\le0$ on a connected graph, so by Perron–Frobenius its lowest eigenvector $u_g$ is unique and positive, for every real $g_s$. The lowest state of $C_f^s$ is then $u_g^{\otimes N_f}$, which is symmetric and unique. In it the $N_f$ objects are independent: $N(y)=N_fu_g(y)^2$ and $\operatorname{Var}\Phi_y=N_fu_g(y)^2(1-u_g(y)^2)\le N(y)$. In (5.2), $\sigma_\Phi^2\le\max_yN(y)$.

Apply (3.9) to one field object ($\Phi_y\to\lvert y\rangle\langle y\rvert$, $v=\lvert g_s\rvert$). The eigenvectors of $C_1$ are the plane waves $\lvert X_f\rvert^{-1/2}e^{i\mathbf k\cdot x}$ with eigenvalues $-2Dt_f+t_f\omega(\mathbf k)$; the lowest is uniform, with gap $t_f\omega_{\min}$, $\omega_{\min}=4\sin^2(\pi/L)$. In (3.6) each term is $\lvert X_f\rvert^{-2}e^{i\mathbf k\cdot(y-x_s)}/(t_f\omega(\mathbf k))$, so $\chi_0=\frac{2}{\lvert X_f\rvert t_f}\Lambda^+$. Multiplying by $N_f$:

$$
N(y)=n-\frac{2n\,g_s}{t_f}\,\Lambda^+(y,x_s)+O\Bigl(\frac{N_fg_s^2}{t_f^2\omega_{\min}^2}\Bigr),\qquad \delta N(y):=N(y)-n,
\tag{5.8}
$$

with the kernel $\frac{2n}{t_f}\Lambda^+$ of (4.4). On the ring, with $\varrho\in[0,L/2]$ the ring distance from $x_s$,

$$
\Lambda^+(x,x_s)=\frac{L^2-1}{12L}-\frac{\varrho(L-\varrho)}{2L}.
\tag{5.9}
$$

Check: the second difference of $-\varrho(L-\varrho)/(2L)$ is $-1/L$ for $\varrho\neq0$ and $1-1/L$ at $\varrho=0$, so $\Lambda G=\delta_{x_s}-1/L$; the constant makes $\sum_xG=0$, which fixes $\Lambda^+\delta_{x_s}$. On the tori, with $\vec\varrho=x-x_s$ (nearest image) and $\varrho=\lvert\vec\varrho\rvert$, the standard asymptotics of the lattice Green function give, for $1\ll\varrho\ll L$,

$$
\Lambda^+(x,x_s)=c_L-\frac{\ln\varrho}{2\pi}+O(\varrho^{-2})+O\Bigl(\frac{\varrho^2}{L^2}\Bigr)\ (D=2),\qquad \Lambda^+(x,x_s)=\frac{1}{4\pi\varrho}-c_L'+O(\varrho^{-3})+O\Bigl(\frac{\varrho^2}{L^3}\Bigr)\ (D=3),
\tag{5.10}
$$

with constants $c_L,c'_L$ and gradients given by the term-by-term derivatives.

### Step 6. Dependence on the distance from the source

Insert (5.8) into (5.6) and (5.7). With $\bar\gamma:=\gamma/(1+\gamma n)$, which has the sign of $\gamma$ by S5, $\gamma/\nu(x)=\bar\gamma+O(g_s)$ and $\nu(y)/\nu(x)=1+\bar\gamma(\delta N(y)-\delta N(x))+O(g_s^2)$. Also $\tau_{\rm loc}=\tau_0+O(g_s)$, with $\tau_0=\tau$ (a) and $\tau_0=\tau(1+\gamma n)$ (c). Then

$$
\frac{r(y)}{r(x)}=1+\frac{2n\bar\gamma g_s}{t_f}\bigl[\Lambda^+(x,x_s)-\Lambda^+(y,x_s)\bigr]+O(g_s^2),
\tag{5.11}
$$

$$
a_{k,\sigma}(0)\big\vert_{x_0}=\frac{2n\bar\gamma g_s}{t_f}\,\tau_0^2\;\partial_\ell\Lambda^+(x_0,x_s)+O(g_s^2),
\tag{5.12}
$$

where $\partial_\ell$ is the slope along $\ell$. With (5.9) on the ring ($\varrho_x$ the ring distance of $x$; $a$ counted positive away from the source):

$$
\frac{r(y)}{r(x)}=1+\frac{n\bar\gamma g_s}{t_fL}\bigl[\varrho_y(L-\varrho_y)-\varrho_x(L-\varrho_x)\bigr]+O(g_s^2),\qquad a_{k,\sigma}(0)=-\frac{n\bar\gamma g_s}{t_f}\,\tau_0^2\Bigl(1-\frac{2\varrho}{L}\Bigr)+O(g_s^2).
\tag{5.13}
$$

With (5.10) on the tori, to leading order in $1\ll\varrho\ll L$:

$$
\frac{r(y)}{r(x)}=1+\frac{n\bar\gamma g_s}{\pi t_f}\ln\frac{\varrho_y}{\varrho_x},\quad a=-\frac{n\bar\gamma g_s\tau_0^2}{\pi t_f}\frac{\hat e_\ell\cdot\vec\varrho}{\varrho^2}\ (D=2);\qquad \frac{r(y)}{r(x)}=1+\frac{n\bar\gamma g_s}{2\pi t_f}\Bigl(\frac1{\varrho_x}-\frac1{\varrho_y}\Bigr),\quad a=-\frac{n\bar\gamma g_s\tau_0^2}{2\pi t_f}\frac{\hat e_\ell\cdot\vec\varrho}{\varrho^3}\ (D=3).
\tag{5.14}
$$

*Range.* S6 fails where the made field has a kink or a vanishing slope. On the ring, $\Lambda^+$ has a kink at $x_s$ and zero slope at the antipode, and its curvature is $-1/L$; the acceleration formulas need $w\ll\varrho$ and $w\ll L/2-\varrho$. On the tori the curvature is $\sim\varrho^{-2}$ against a slope $\sim\varrho^{-1}$, so they need $w\ll\varrho$.

*Signs.* $\Lambda^+(\cdot,x_s)$ decreases with $\varrho$: exactly on the ring for $0\le\varrho\le L/2$ by (5.9), and in the leading forms (5.10) on the tori. Hence, at first order in $g_s$, for $g_s\gamma>0$ the clock runs slower nearer the source and the drift acceleration points towards it: on the ring towards $x_s$ along the shorter arc, on the tori with the component along $\ell$ that decreases $\varrho$. For $g_s\gamma<0$ both reverse: the clock runs faster nearer the source and the fall is away from it.

## Result

- **(5.1), (5.2)** [theorem]. Class: couplings (a) and (c); every field state $\varphi$; product starts $\xi\otimes\varphi$ with every body–medium state $\xi$ (S1–S4). The joint state equals $\xi(\lambda)\otimes\varphi(\lambda)$, with $\xi$ evolving by the partial expectation $\bar C_{\varphi(\lambda)}$ of the total contract in the field state (5.1), up to a common phase and a remainder of norm at most $\lvert\gamma\rvert c_Q\int_0^{\lvert\lambda\rvert}\sigma_\Phi$ (5.2). This is first order in $\gamma$, valid for $\lvert\lambda\rvert\ll(\lvert\gamma\rvert c_Q\sigma_\Phi)^{-1}$; the neglected term is the test body's own change of the field. For stationary $\varphi$ the operator is $\lambda$-independent, and in the lowest adapted state $\sigma_\Phi^2\le\max_yN(y)$. $\bar C_\varphi$ is fixed by $C$ and $\varphi$; its split into own and field-made parts is not. The bound holds for every member of the class.
- **(5.3)** [formula]. Same class. The level-$k$ blocks of $\bar C_\varphi$: on-site $\nu(x)M_k\sigma_3+E_\varphi\mathbb 1_2$; hopping $\tau$ for (a) and $\tau\frac12(\nu(x)+\nu(x+1))$ for (c). Fixed by $C$ and $\varphi$: $M_{x,k}$, the hopping amplitudes and $V_{x,k}$ up to a common constant, hence the ratios $M_{y,k}/M_{x,k}=\nu(y)/\nu(x)$. Listing-dependent: the factorization $M_{x,k}=\nu(x)M_k$ and the place-independent constant in $V_{x,k}$.
- **(5.4)** [theorem]. Class: (a) and (c), every field state, every place (at every $\lambda$). R3 holds with $\Gamma(x)=\gamma(N_\varphi(x+1)-N_\varphi(x))\sigma_3$, without a $\mathbb 1_2$ part. In fixed quantities, $B_{x+1,k}-B_{x,k}=M_{x,k}\frac{\nu(x+1)-\nu(x)}{\nu(x)}\sigma_3$ for $\nu(x)\neq0$. The hopping changes by the same factor as the on-site blocks for (c), with the bond carrying the mean $\frac12(\nu(x)+\nu(x+1))$ of the factors of its ends. For (a) it does not: the hopping stays $\tau$ at every place, and the factors differ wherever $\gamma(N_\varphi(x+1)-N_\varphi(x))\neq0$. No field state violates R3 for (a) or (c); the control is (5.5).
- **(5.5)** [negative]. Control: the pair term $\gamma\sum_x\Pi_x\sigma_3\otimes m\mathbb 1_N\otimes\Phi_x$ (rest term only). R3 fails at every place with $\gamma(N_\varphi(x+1)-N_\varphi(x))\neq0$, and there the fall from rest (under S6) depends on the level.
- **(5.6)** [formula]. Class: (a), (c), stationary $\varphi$, S5. The clock-rate ratio of two places is $r(y)/r(x)=\nu(y)/\nu(x)=M_{y,k}/M_{x,k}$, the same for every pair of levels and for both branches. It is exact in $\bar C_\varphi$.
- **(5.7)** [theorem]. Class: (a), (c), stationary $\varphi$, branch starts at rest, S5–S6; first order in the gradients. $a_{k,\sigma}(0)=-\tau_{\rm loc}(x_0)^2\,\gamma N'(x_0)/\nu(x_0)$, the same for all levels and both branches. (4.24) of 05-ilang-free-fall/04-branch-fall holds with $v_{\max}(x_0)=\tau_{\rm loc}(x_0)d_0$. For (a) this is $\tau d_0$ at every place. For (c) it is $\tau\nu(x_0)d_0$, which changes from place to place by the factor of $r$; with a place-independent $v_{\max}$ taken at a reference place $y$, (4.24) carries the factor $(r(x_0)/r(y))^2$.
- **(5.8), (5.9)** [formula]. The lowest adapted state is $u_g^{\otimes N_f}$, unique for every real $g_s$. In it $N(y)=n-\frac{2ng_s}{t_f}\Lambda^+(y,x_s)$ up to $O(N_fg_s^2/(t_f\omega_{\min})^2)$, with $\Lambda^+$ on the ring given exactly by (5.9).
- **(5.11), (5.12)** [formula]. Class: (a), (c), lowest adapted state, S5 (and S6 for the acceleration); first order in $g_s$, with $\bar\gamma=\gamma/(1+\gamma n)$ and $\tau_0$ the source-free hopping of $\bar C_\varphi$. Clock-rate ratio $1+\frac{2n\bar\gamma g_s}{t_f}[\Lambda^+(x,x_s)-\Lambda^+(y,x_s)]$; acceleration $\frac{2n\bar\gamma g_s}{t_f}\tau_0^2\partial_\ell\Lambda^+(x_0,x_s)$, level and branch independent.
- **(5.13), (5.14)** [formula, criterion]. Ring, exact in $L$ at first order in $g_s$: (5.13), for the acceleration with $w\ll\varrho$ and $w\ll L/2-\varrho$. Square and cubic tori, leading forms for $1\ll\varrho\ll L$, $w\ll\varrho$: (5.14), with the logarithm ($D=2$) and $1/\varrho$ ($D=3$) in the clock ratio, and the acceleration along $\ell$ the projection on $\hat e_\ell$ of a radial vector of size $\propto\varrho^{-1}$ ($D=2$), $\propto\varrho^{-2}$ ($D=3$). Signs, at first order in $g_s$, with S5 and in these ranges: for $g_s\gamma>0$ the clock is slower nearer the source and the fall is towards it; for $g_s\gamma<0$ the clock is faster nearer the source and the fall is away from it.

## Consistency checks

- **Dimensions.** $\gamma$ and $n,\Lambda^+$ are numbers; $g_s,t_f,\tau,M_k$ have the dimension of $C$, i.e. $1/\lambda$. Then $n\bar\gamma g_s/t_f$ and $\lvert\gamma\rvert c_Q\sigma_\Phi\lvert\lambda\rvert$ are dimensionless, and $a\propto\tau_0^2$ has the dimension of label$/\lambda^2$. Passed.
- **$g_s=0$ or $\gamma=0$.** For $g_s=0$, (5.8) gives $N=n$ uniform, hence $\Gamma=0$, $r(y)/r(x)=1$ and $a=0$ in (5.11)–(5.14). For $\gamma=0$, $\bar C_\varphi=C_{bc}+E_\varphi$ and the bound (5.2) vanishes. Passed.
- **Sharp field.** Take $t_f=0$ and $\varphi$ a joint eigenvector of all $\Phi_x$. Then $\varphi$ is stationary, $\sigma_\Phi=0$, (5.2) is exact, and $C(\xi\otimes\varphi)=(\bar C_\varphi\xi)\otimes\varphi$ directly: the given lapse field of 05-ilang-free-fall with on-site factor $1+\gamma N(x)$. Passed.

## Open issues

- The bound (5.2) grows linearly in $\lambda$. A bound that uses the field gap $t_f\omega_{\min}$ (a non-growing first-order remainder plus a second-order shift, which is the test body's own mark of 06-own-mark) is not derived.
- The remainder of (5.8), taken from (3.9) per field object, is crude ($N_fg_s^2/(t_f\omega_{\min})^2$); a sharp estimate of the second-order profile is not derived.
- Non-product starts, and non-stationary field states for items 3–4 (a $\lambda$-dependent profile), are not treated.
- Regions with $\nu(x)\le0$ and the neighbourhoods $\varrho\lesssim w$ of the source are outside the formulas, as are the ring antipode and $\varrho\sim L$ on the tori.

## Methods used

- partial expectation (mean field) over the field state
- Duhamel formula, unitarity
- Rayleigh–Schrödinger perturbation theory (via (3.9))
- Perron–Frobenius theorem, product states of independent identical objects
- plane-wave diagonalization, graph Laplacian pseudo-inverse, lattice Green function asymptotics
- semiclassical band dynamics, local-slope (Taylor) expansion

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