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

# Moving clock: a ring clock carried by a body in the record background

- **Subproject:** 04-ilang-time
- **Package:** 05-moving-clock
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-10

## Setup and assumptions

- **System.** The body $b$ and the medium $c$ are each the only instance of their type, so no swap condition (1.11) applies. The joint space is $\mathcal H=\mathcal H_b\otimes\mathcal H_c$, with $\mathcal H_c$ finite-dimensional (A1).
  - Item 1: $\mathcal H_b=\mathcal H_X\otimes\mathbb C^N$, with places $(x,a)$.
  - Items 2–4: $\mathcal H_b=\ell^2(\mathbb Z)\otimes\mathbb C^2\otimes\mathbb C^N$ (position, branch, clock label), with places $(x,s,a)$, obtained as the limit of Step 2.
  - Contract: $C=C_b\otimes\mathbb 1_c+\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$ with $K_h=K_x$.
  - Start: $\lvert\phi\rangle\otimes\lvert\chi\rangle$. We take $\lambda\ge0$ (A9 restricted).
- **Clock reading** (A8). It is applied at a stated $\lambda$. Its projectors are $\Pi_a=\mathbb 1\otimes\lvert a\rangle\langle a\rvert\otimes\mathbb 1_c$, with $\lvert a\rangle\langle a\rvert$ on the clock-label factor. By (1.3), $q(a;\lambda)=\lVert\Pi_a\Psi(\lambda)\rVert^2$ with $\Psi(\lambda)=e^{-iC\lambda}(\phi\otimes\chi)$.
- **Branch label.** $\sigma_3\lvert1\rangle=\lvert1\rangle$, $\sigma_3\lvert2\rangle=-\lvert2\rangle$, $\sigma_1\lvert1\rangle=\lvert2\rangle$. Spinors are written as columns $(\beta_1,\beta_2)^T$.
- **Clock inputs (02-clock@v2).** Take (2.9) with $\alpha_m=0$, $\varepsilon=0$, $\ell_k=k$. Then $D=\sum_{k=0}^{N-1}d_k\lvert f_k\rangle\langle f_k\rvert$, $d_k=\frac{2\pi k}{N\lambda_0}$, satisfies $e^{-iD\lambda_0}=S$. By (2.12) with $c=0$, it has the minimal spread for the start $\lvert0\rangle$.
  - (2.13) is its reading distribution. It is written below with the place called $a$, since $m$ denotes the mass parameter.
  - Since $\langle a\vert f_k\rangle=N^{-1/2}e^{2\pi ika/N}$, (2.13) is a geometric sum:
$$
P_N(a;\vartheta):=\Bigl\lvert\frac1N\sum_{k=0}^{N-1}e^{2\pi ik(a-\vartheta)/N}\Bigr\rvert^2=\frac{\sin^2(\pi\vartheta)}{N^2\sin^2\bigl(\pi(\vartheta-a)/N\bigr)},\qquad p(a;\lambda)=P_N(a;\lambda/\lambda_0)\ \text{in (2.13)}.
\tag{5.1}
$$
- **Space inputs (03-ilang-space).**
  - By (5.10) and (5.11) of 03-ilang-space/05-recording-contract, the background point of a place $h$ is $u_h=(K_h-\langle K_h\rangle)\chi$, and the distance is $d_0(h,h')=\lVert u_h-u_{h'}\rVert$. Both depend only on $K_h$ and $\chi$. (5.10) of 03-ilang-space/05-recording-contract is stated for $\phi_h\neq0$; only $u_h$ and $d_0$ are used below.
  - From 03-ilang-space/17-motion: the definitions $\bar x=\sum_hp_hu_h$ and $v=\mathrm d\bar x/\mathrm d\lambda$; the bound (17.7), $\lVert v\rVert\le v_{\max}:=\varrho(a)\le\max_h\sum_{h'}\lvert t_{hh'}\rvert d_0(h,h')$ with $a_{hh'}=\lvert t_{hh'}\rvert d_0(h,h')$; the chain relations (17.15); and the chain value (17.20), $v_{\max}=2t\sigma_X$.
- **Weak-record regime [W]** (items 2–4, as fixed in item 2(b)).
  - Everything is computed at $\kappa=0$, except that positions are the background points $u_h=O(\kappa)$. Then $\Psi(\lambda)=(e^{-iC_b\lambda}\phi)\otimes\chi$, and the view of $b$ is that of $C_b$ alone.
  - By Duhamel's formula, the exact state differs from this one by at most $\int_0^\lambda\lVert\sum_x\lvert x\rangle\langle x\rvert\otimes K_x\Psi_0(\lambda')\rVert\mathrm d\lambda'\le\kappa\lVert X\rVert\lambda\sup_{\lambda'\le\lambda}\langle x^2\rangle_{\lambda'}^{1/2}$. [W] is valid while this is $\ll1$.
- **Starts near $p_0$.** $g$ is smooth and $2\pi$-periodic, with $\int_{-\pi}^{\pi}\frac{\mathrm dp}{2\pi}\lvert g\rvert^2=1$ and $\lvert g\rvert^2$ concentrated within $\delta p$ of $p_0$.
- **Clock parameter.** $\epsilon:=\frac{2\pi}{N\lambda_0m}$, so $d_k=km\epsilon\le(N-1)m\epsilon$. Item 3 needs $N\epsilon\ll1$.

## Derivation

### Step 1. Decoupled body (item 1)

Since $K_{(x,a)}=K_x$, the record term is $\sum_x\lvert x\rangle\langle x\rvert\otimes\mathbb 1_N\otimes K_x$. Order the factors as $(\mathcal H_X\otimes\mathcal H_c)\otimes\mathbb C^N$; this relabelling of tensor factors changes no reading. Then $C=G\otimes\mathbb 1_N+\mathbb 1\otimes D'$, with $G:=H\otimes\mathbb 1_c+\sum_x\lvert x\rangle\langle x\rvert\otimes K_x$.

The two terms act on different factors and commute. Hence $e^{-iC\lambda}=e^{-iG\lambda}\otimes e^{-iD'\lambda}$ and $\Psi(\lambda)=e^{-iG\lambda}(\phi_X\otimes\chi)\otimes e^{-iD'\lambda}\eta$. $\Pi_a$ acts on the last factor only, and $e^{-iG\lambda}$ is unitary, so
$$
q(a;\lambda)=\bigl\lvert\langle a\vert e^{-iD'\lambda}\vert\eta\rangle\bigr\rvert^2\qquad\text{for all }\lambda .
\tag{5.2}
$$
This holds exactly, for every $\chi$, every $K_x$ and every $\kappa$. The answer to each of the three questions is no: $q$ depends neither on $H$, nor on $\phi_X$, nor on the $K_x$ (nor on $\chi$). For $D'=D$ and $\eta=\lvert0\rangle$, (5.2) is (2.13).

The model of items 2–4 is not of this form: $B$ contains $\sigma_3\otimes D$, which does not commute with $A\propto\sigma_1$.

### Step 2. The infinite chain as a limit (item 2)

**Finite chains.** Take $\Lambda_L=\{-L,\dots,L\}$, keep in $A$ only the bonds inside $\Lambda_L$, and restrict $B$ and the records to $\Lambda_L$. This gives $C^{(L)}$; the same expressions on $\ell^2(\mathbb Z)$ give $C^{(\infty)}$.

**Locality bound.** Write $C^{(\cdot)}=A^{(\cdot)}\otimes\mathbb 1_c+Q$, where $Q$ (the $B$ term and the records) is diagonal in $x$. $A$ is $\frac\tau2$ times a sum of two operators of norm 1, so $\lVert A\rVert\le\tau$. Expand $e^{-iC\lambda}$ in powers of $A$ around $e^{-iQ\lambda}$ (Dyson series). The term of order $n$ moves $x$ by at most $n$ and has norm at most $(\tau\lambda)^n/n!$. For a start supported in $\lvert x\rvert\le R$, therefore, for both $C^{(L)}$ and $C^{(\infty)}$:
$$
\bigl\lVert\Pi_{\lvert x\rvert\ge L}\,e^{-iC^{(\cdot)}\lambda}(\phi\otimes\chi)\bigr\rVert\le\sum_{n\ge L-R}\frac{(\tau\lambda)^n}{n!} .
\tag{5.3}
$$
On vectors supported in $\Lambda_L$, $C^{(\infty)}-C^{(L)}$ is the hopping out of the sites $x=\pm L$. Its norm is at most $\tau\lVert\Pi_{\lvert x\rvert\ge L}\,\cdot\,\rVert$. By Duhamel's formula, the two evolved states therefore differ by at most $\tau\lambda\sum_{n\ge L-R}(\tau\lambda)^n/n!\to0$.

**How the limit is taken.**
- First $L\to\infty$, at fixed $\lambda$, $\kappa$ and start. Then the order $\kappa\to0$ of [W], the leading order in the spread, and the expansion in $\epsilon$.
- Starts with rapidly decaying tails are approximated in norm by finitely supported ones; unitarity preserves the error.
- By (5.3) the tails stay superexponentially small. Hence $q$, $p_h$, $\bar x$ and $v$ converge.

**Plane waves.** On $\ell^2(\mathbb Z)$ we use the standard Fourier facts.
- The plane waves are $\lvert p\rangle=\sum_xe^{ipx}\lvert x\rangle$, $p\in(-\pi,\pi]$, with $\langle p\vert p'\rangle=2\pi\delta(p-p')$.
- With $T:=\sum_x\lvert x+1\rangle\langle x\rvert$, $T\lvert p\rangle=e^{-ip}\lvert p\rangle$.
- A vector $\psi=\int\frac{\mathrm dp}{2\pi}\lvert p\rangle\otimes G(p)$ has $\lVert\psi\rVert^2=\int\frac{\mathrm dp}{2\pi}\lVert G\rVert^2$.
- The position label acts as $G\mapsto i\partial_pG$, for smooth periodic $G$ (integration by parts).

### Step 3. Anticommutation and spectrum (item 2(a))

In terms of $T$, $A=\frac\tau2\bigl(-i(T-T^\dagger)\bigr)\otimes\sigma_1\otimes\mathbb 1_N$ and $B=\mathbb 1\otimes\sigma_3\otimes M$, with $M:=m\mathbb 1_N+D$. On the position and clock factors the factors of $A$ and $B$ commute, and $\sigma_1\sigma_3+\sigma_3\sigma_1=0$. Hence
$$
AB+BA=0,\qquad C_b^2=A^2+B^2 .
\tag{5.4}
$$
Further, $-i(T-T^\dagger)\lvert p\rangle=-2\sin p\,\lvert p\rangle$ and $M\lvert f_k\rangle=\mu_k\lvert f_k\rangle$. So
$$
C_b\bigl(\lvert p\rangle\otimes\beta\otimes\lvert f_k\rangle\bigr)=\lvert p\rangle\otimes C_k(p)\beta\otimes\lvert f_k\rangle,\qquad C_k(p):=-\tau\sin p\,\sigma_1+\mu_k\sigma_3,\qquad\mu_k:=m+d_k .
\tag{5.5}
$$
By (5.4), $C_k(p)^2=(\mu_k^2+\tau^2\sin^2p)\mathbb 1$; moreover $\operatorname{Tr}C_k=0$. Write $C_k=\omega_k(\sin\theta_k\sigma_1+\cos\theta_k\sigma_3)$, with $\cos\theta_k=\mu_k/\omega_k>0$. The half-angle formulas then give the generalized eigenvectors $\lvert p\rangle\otimes\beta_{k,\pm}(p)\otimes\lvert f_k\rangle$ of $C_b$, with eigenvalues $\pm\omega_k(p)$:
$$
\omega_k(p)=\sqrt{\mu_k^2+\tau^2\sin^2p},\quad
\beta_{k,+}=\begin{pmatrix}\cos\frac{\theta_k}2\\ \sin\frac{\theta_k}2\end{pmatrix},\quad
\beta_{k,-}=\begin{pmatrix}-\sin\frac{\theta_k}2\\ \cos\frac{\theta_k}2\end{pmatrix},\quad
\tan\theta_k(p)=-\frac{\tau\sin p}{\mu_k},\ \ \lvert\theta_k\rvert<\frac\pi2 .
\tag{5.6}
$$
The two branches are separated by the gap $2\mu_k\ge2m>0$. The spinors $\beta_{k,\pm}$ are real, of unit length, and smooth and $2\pi$-periodic in $p$.

### Step 4. Background positions and velocity (item 2(b))

Every place has $u_{(x,s,a)}=\kappa x(X-\langle X\rangle)\chi=\kappa x\,e$, with $\lVert e\rVert=\sigma_X$. Hence $d_0(h,h')=\kappa\sigma_X\lvert x-x'\rvert$: this is (17.15) of 03-ilang-space/17-motion with $X\to\kappa X$. By (5.11) of 03-ilang-space/05-recording-contract, all places with the same $x$ are one background point. With $P(x;\lambda):=\sum_{s,a}p_{(x,s,a)}(\lambda)$,
$$
\bar x(\lambda)=\kappa\,\bar x_{\rm lab}(\lambda)\,e,\qquad\bar x_{\rm lab}:=\sum_xx\,P(x;\lambda),\qquad v=\kappa\,\frac{\mathrm d\bar x_{\rm lab}}{\mathrm d\lambda}\,e .
\tag{5.7}
$$
**Start.** $\phi=\psi_{k,+}\otimes\lvert f_k\rangle$, with $\psi_{k,+}:=\int\frac{\mathrm dp}{2\pi}g(p)\lvert p\rangle\otimes\beta_{k,+}(p)$. Under [W], (5.6) gives the evolved amplitude $G(p,\lambda)=g(p)e^{-i\omega_k(p)\lambda}\beta_{k,+}(p)$.

**Mean position.** By Step 2,
$$\bar x_{\rm lab}=\int\frac{\mathrm dp}{2\pi}G^\dagger i\partial_pG=\int\frac{\mathrm dp}{2\pi}\bigl(ig^*\partial_pg+\lambda\lvert g\rvert^2\partial_p\omega_k+i\lvert g\rvert^2\beta_{k,+}^T\partial_p\beta_{k,+}\bigr).$$
The last term vanishes, because $\beta_{k,+}$ is real and of unit length. So $\bar x_{\rm lab}$ is exactly linear in $\lambda$, with the group velocity as slope:
$$
\frac{\mathrm d\bar x_{\rm lab}}{\mathrm d\lambda}=\int_{-\pi}^{\pi}\frac{\mathrm dp}{2\pi}\lvert g(p)\rvert^2\,\partial_p\omega_k(p)=\partial_p\omega_k(p_0)+O(\delta p).
\tag{5.8}
$$
With $\hat e:=e/\sigma_X$:
$$
v=\kappa\sigma_X\,w_k(p_0)\,\hat e,\qquad w_k(p):=\partial_p\omega_k(p)=\frac{\tau^2\sin p\cos p}{\sqrt{\mu_k^2+\tau^2\sin^2p}} .
\tag{5.9}
$$

### Step 5. Largest velocity and the bound (17.7) of 03-ilang-space/17-motion (item 2(b))

**Largest velocity.** With $u=\sin^2p_0$, $w_k^2=\tau^4u(1-u)/(\mu_k^2+\tau^2u)$. It vanishes at $u=0$ and $u=1$, and its $u$-derivative vanishes iff $\tau^2u^2+2\mu_k^2u-\mu_k^2=0$. The root in $[0,1]$ is $u_*=\mu_k/(\mu_k+W_k)$, with $W_k:=\sqrt{\mu_k^2+\tau^2}$. There $\mu_k^2+\tau^2u_*=\mu_kW_k$ and $1-u_*=W_k/(\mu_k+W_k)$, so $w_k^2=\tau^4/(\mu_k+W_k)^2$:
$$
\max_{p_0}\lVert v\rVert=\kappa\sigma_X\,(W_k-\mu_k)=\kappa\sigma_X\frac{\tau^2}{\sqrt{\mu_k^2+\tau^2}+\mu_k},\qquad\text{attained at }\sin^2p_0=\frac{\mu_k}{\mu_k+W_k}.
\tag{5.10}
$$
**The bound.**
- $A$ has $\lvert t_{hh'}\rvert=\tau/2$ between $(x,s,a)$ and $(x\pm1,\bar s,a)$, $\bar s\ne s$.
- The off-diagonal entries of $B$ come from $D$, which is not diagonal on the clock places. They join places with equal $x$, where $d_0=0$.
- Hence $a_{hh'}=\frac\tau2\kappa\sigma_X$ on the edges $(x,s,a)$–$(x\pm1,\bar s,a)$, and $a_{hh'}=0$ otherwise.
- This graph is the disjoint union of $2N$ chains $\{(x,s,a):x+s\equiv c\ (\mathrm{mod}\ 2)\}$, along which $d_0=\kappa\sigma_X\lvert x-x'\rvert$.
- $a$ depends only on $\lvert t_{hh'}\rvert$ and $d_0$. Each chain is therefore that of (17.20) of 03-ilang-space/17-motion, with $t=\tau/2$ and $X\to\kappa X$.

$\varrho(a)$ is the maximum over the components, so
$$
v_{\max}=\tau\,\kappa\sigma_X=\max_h\sum_{h'}\lvert t_{hh'}\rvert\,d_0(h,h') .
\tag{5.11}
$$
On $\Lambda_L$, $\varrho(a)=\tau\kappa\sigma_X\cos\frac{\pi}{2L+2}\to\tau\kappa\sigma_X$.

In units of $\kappa\sigma_X$: $v$ is $w_k(p_0)$, its largest value is $W_k-\mu_k$, and the bound is $\tau$. The bound is not reached, since $(W_k-\mu_k)/\tau=\sqrt{(W_k-\mu_k)/(W_k+\mu_k)}<1$.

### Step 6. Clock reading of the moving clock (item 3)

**Start.** $\phi=\psi_{0,+}\otimes\lvert0\rangle$. Write $\beta_0:=\beta_{0,+}$, $\beta_{0,-}$ for the other spinor of level $0$, and $\omega:=\omega_0$. Since $\lvert0\rangle=N^{-1/2}\sum_k\lvert f_k\rangle$, (5.5), [W] and Step 2 give
$$
q(a;\lambda)=\int\frac{\mathrm dp}{2\pi}\lvert g(p)\rvert^2\,\lVert\Phi_a(p,\lambda)\rVert^2,\qquad
\Phi_a:=\frac1N\sum_{k=0}^{N-1}e^{2\pi ika/N}\,e^{-iC_k(p)\lambda}\beta_0(p).
\tag{5.12}
$$
**Meaning of "first order in $\epsilon$".** Amplitudes are kept to first order in $d_k/m\le(N-1)\epsilon$. Phases $\omega_k\lambda$ are kept to first order in $d_k$, but $d_k\lambda=2\pi k\lambda/(N\lambda_0)$ is kept to all orders, since it is $O(1)$ at the $\lambda$ of interest.

*(i) Branch mixing.*
- $\beta_{k,\pm}$ is $\beta_{0,\pm}$ rotated by $-\delta_k$, with $\delta_k:=(\theta_0-\theta_k)/2$.
- $\lvert\partial_\mu\theta\rvert=\tau\lvert\sin p\rvert/(\mu^2+\tau^2\sin^2p)\le1/(2\mu)$, so $\lvert\delta_k\rvert\le d_k/(4m)$.
- Expanding $\beta_0=\cos\delta_k\,\beta_{k,+}+\sin\delta_k\,\beta_{k,-}$ and rotating back gives $e^{-iC_k\lambda}\beta_0=\bigl(e^{-i\omega_k\lambda}+2i\sin^2\delta_k\sin\omega_k\lambda\bigr)\beta_0+i\sin2\delta_k\sin\omega_k\lambda\,\beta_{0,-}$.
- Since $\beta_0\perp\beta_{0,-}$, the first-order term enters $\lVert\Phi_a\rVert^2$ only squared: $\lVert\Phi_a\rVert^2=\bigl\lvert\frac1N\sum_ke^{2\pi ika/N-i\omega_k\lambda}\bigr\rvert^2+O((N\epsilon)^2)$.
- There is thus no first-order correction. The same argument holds for any other reference level used to define "the positive branch".

*(ii) Dispersion.* Expand $\omega_k$ in $d_k$ by Taylor's theorem, using $\partial_d^2\sqrt{(m+d)^2+\tau^2\sin^2p}=\tau^2\sin^2p/\omega_k^3\in[0,1/m]$:
$$
\omega_k(p)=\omega(p)+\gamma(p)\,d_k+r_k(p),\qquad\gamma(p):=\frac{m}{\omega(p)}=\frac{m}{\sqrt{m^2+\tau^2\sin^2p}},\qquad0\le r_k\le\frac{d_k^2}{2m}.
\tag{5.13}
$$
Equivalently, by Hellmann–Feynman, $\gamma=\cos\theta_0=\beta_0^T\sigma_3\beta_0$: the clock term $\sigma_3\otimes D$ averaged in the positive branch.
- The common phase $e^{-i\omega\lambda}$ drops out of the modulus.
- Dropping $r_k\lambda$ changes $\lVert\Phi_a\rVert^2$ by at most $2\max_kr_k\lambda\le2\pi\frac{(N-1)^2}{N}\epsilon\frac\lambda{\lambda_0}$.
- With $\gamma d_k\lambda=2\pi k\gamma\lambda/(N\lambda_0)$, (5.1) gives $\lVert\Phi_a\rVert^2=P_N(a;\gamma(p)\lambda/\lambda_0)$.

*(iii) Spread.* Replacing $\gamma(p)$ by $\gamma(p_0)$ shifts the argument by $O(\delta p\,\lvert\gamma'(p_0)\rvert\,\lambda/\lambda_0)$. To leading order in $\delta p$:
$$
q(a;\lambda)=P_N\Bigl(a;\gamma(p_0)\frac{\lambda}{\lambda_0}\Bigr)=\frac{\sin^2\bigl(\pi\gamma(p_0)\lambda/\lambda_0\bigr)}{N^2\sin^2\bigl(\pi(\gamma(p_0)\lambda/\lambda_0-a)/N\bigr)},\qquad\gamma(p_0)=\frac{m}{\sqrt{m^2+\tau^2\sin^2p_0}} .
\tag{5.14}
$$
This is (2.13) with $\lambda$ replaced by $\gamma(p_0)\lambda$. The neglected terms are:
- $O((N\epsilon)^2)$ from (i);
- $O(N\epsilon\,\lambda/\lambda_0)$ from (ii);
- $O(\delta p\,\lvert\gamma'(p_0)\rvert\,\lambda/\lambda_0)$ from (iii);
- $O(\kappa)$ from [W].

### Step 7. Tick and ratio (item 3)

By (5.1), $P_N(\cdot\,;\vartheta)$ is concentrated on a single place iff $\vartheta\in\mathbb Z$, and then on $a\equiv\vartheta$. The reading starts at $\vartheta=0$. The first $\lambda>0$ at which it gives $a=1$ with certainty is therefore $\gamma(p_0)\lambda=\lambda_0$. Since $\gamma(0)=1$,
$$
\lambda_0(p_0)=\frac{\lambda_0}{\gamma(p_0)}=\lambda_0\sqrt{1+\frac{\tau^2}{m^2}\sin^2p_0},\qquad
R(p_0):=\frac{\lambda_0(p_0)}{\lambda_0(0)}=\frac{\omega(p_0)}{m}=\sqrt{1+\frac{\tau^2}{m^2}\sin^2p_0}\ \ge1 .
\tag{5.15}
$$
Relative to its rest rate, the carried clock advances at the rate $\gamma(p_0)=1/R\le1$. The rate equals $1$ only for $\sin p_0=0$. It is smallest at $p_0=\pm\pi/2$, where it equals $m/W$, with $W:=\sqrt{m^2+\tau^2}$.

### Step 8. Rate and velocity (item 4)

**The velocity used.** To the order of item 3 we may set $\mu_k\to m$ in (5.9). Then all levels of the item-3 start move, up to $O(N\epsilon)$, with $\lVert v\rVert=\kappa\sigma_X\,w$, where $w:=\lvert w_0(p_0)\rvert$. By (5.11), $\lVert v\rVert/v_{\max}=w/\tau$.

**Exact relation.** By (5.15), $\tau^2\sin^2p_0=m^2(R^2-1)$ and $\tau^2\cos^2p_0=W^2-m^2R^2$. Inserting these in $w^2=\tau^4\sin^2p_0\cos^2p_0/\omega^2$ gives
$$
m^2R^4-(W^2+m^2-w^2)R^2+W^2=0 .
\tag{5.16}
$$
The discriminant is $\bigl((W-m)^2-w^2\bigr)\bigl((W+m)^2-w^2\bigr)$. It is nonnegative iff $w\le W-m$, consistent with (5.10). The roots are
$$
R=\frac{\sqrt{(W+m)^2-w^2}\mp\sqrt{(W-m)^2-w^2}}{2m},\qquad\text{sign }-\text{ for }\sin^2p_0\le\frac{m}{m+W},\quad\text{sign }+\text{ for }\sin^2p_0\ge\frac{m}{m+W}.
\tag{5.17}
$$
- Squaring confirms the roots; their product is $W/m$.
- $R$ increases with $\sin^2p_0$, while $w$ increases up to $u_*$ and then decreases; this fixes the sign.
- At $w=0$: $R=1$ ($p_0=0,\pi$) and $R=W/m$ ($p_0=\pm\pi/2$). At $w=W-m$ both roots equal $\sqrt{W/m}$.
- So $R$ is a two-valued function of $\lVert v\rVert$. The branch through $p_0=0$ has the sign $-$.

**Small $p_0$.** $\omega=m+\frac{\tau^2p_0^2}{2m}+O(p_0^4)$ and $w=\frac{\tau^2\lvert p_0\rvert}{m}+O(\lvert p_0\rvert^3)$. Let $v_{\rm top}:=\kappa\sigma_X(W-m)$ be the largest velocity (5.10) at $\mu_k=m$. Then
$$
R=1+\frac{\lVert v\rVert^2}{2v_{\max}^2}+O(p_0^4)=1+\frac{W-m}{2(W+m)}\,\frac{\lVert v\rVert^2}{v_{\rm top}^2}+O(p_0^4).
\tag{5.18}
$$
**Comparison.** Exactly, $1-w^2/\tau^2=(m^2+\tau^2\sin^4p_0)/\omega^2$. Hence
$$
R^2\Bigl(1-\frac{\lVert v\rVert^2}{v_{\max}^2}\Bigr)=1+\frac{\tau^2}{m^2}\sin^4p_0 .
\tag{5.19}
$$
- In units of the bound $v_{\max}$ of (17.7) of 03-ilang-space/17-motion, the coefficient in (5.18) is universal: $\frac12$ for all $\tau,m$. In units of the largest velocity $v_{\rm top}$ it depends on $\tau/m$.
- By (5.19), $R=(1-\lVert v\rVert^2/v_{\max}^2)^{-1/2}\sqrt{1+\tau^2\sin^4p_0/m^2}$. The correction factor is $1+O(p_0^4)$.
  - At fixed $\lVert v\rVert/v_{\max}<1$ on the branch through $p_0=0$, the factor tends to $1$ as $m/\tau\to0$, since there $\sin p_0=O(m/\tau)$.
- Always $\lVert v\rVert/v_{\max}\le(W-m)/\tau<1$: the bound is never reached, and $R$ stays finite. On the branch through $p_0=0$, $R\le\sqrt{W/m}$, with equality at $v_{\rm top}$.

## Result

- **(5.2)** Decoupled body: $q(a;\lambda)=\lvert\langle a\vert e^{-iD'\lambda}\vert\eta\rangle\rvert^2$, exactly, for all $\lambda$. It does not depend on $H$, $\phi_X$, the $K_x$ or $\chi$.
- **(5.3)** The infinite chain is the limit $L\to\infty$, taken first at fixed $\lambda$. It is controlled by the locality bound (5.3).
- **(5.4)** $AB+BA=0$ and $C_b^2=A^2+B^2$. **(5.6)** For each level $d_k$, the eigenvalues are $\pm\omega_k(p)=\pm\sqrt{(m+d_k)^2+\tau^2\sin^2p}$, with eigenvectors $\lvert p\rangle\otimes\beta_{k,\pm}(p)\otimes\lvert f_k\rangle$.
- **(5.9)** $v=\kappa\sigma_X\,w_k(p_0)\hat e$, with $w_k=\tau^2\sin p_0\cos p_0/\omega_k(p_0)$. **(5.10)** Largest velocity: $\kappa\sigma_X(\sqrt{\mu_k^2+\tau^2}-\mu_k)$. **(5.11)** Bound (17.7) of 03-ilang-space/17-motion: $v_{\max}=\tau\kappa\sigma_X$, strictly larger.
- **(5.14)** $q(a;\lambda)=P_N(a;\gamma(p_0)\lambda/\lambda_0)$, i.e. (2.13) with $\lambda\to\gamma(p_0)\lambda$, where $\gamma(p_0)=m/\sqrt{m^2+\tau^2\sin^2p_0}$.
- **(5.15)** $\lambda_0(p_0)=\lambda_0\sqrt{1+(\tau/m)^2\sin^2p_0}$, and $R=\lambda_0(p_0)/\lambda_0(0)=\omega(p_0)/m\ge1$.
- **(5.17)** Exact relation: $R=\bigl[\sqrt{(W+m)^2-w^2}\mp\sqrt{(W-m)^2-w^2}\bigr]/(2m)$, a two-valued function of $w=\lVert v\rVert/(\kappa\sigma_X)$.
- **(5.18)** Small $p_0$: $R=1+\lVert v\rVert^2/(2v_{\max}^2)+O(p_0^4)$; the coefficient is universal only in units of the bound $v_{\max}$.
- **(5.19)** $R^2(1-\lVert v\rVert^2/v_{\max}^2)=1+(\tau/m)^2\sin^4p_0$.

## Consistency checks

1. **Dimensions.** $[\tau]=[m]=[d_k]=[C]=[\lambda]^{-1}$, so $\epsilon$, $\gamma$, $R$ and $w/\tau$ are dimensionless. $[v]=[\kappa\sigma_X][C]$, consistent with (5.9)–(5.11) and (5.17)–(5.19). Passed.
2. **$p_0=0$ and $\tau\to0$.**
   - At $p=0$, $C_k(0)=\mu_k\sigma_3$ and $\beta_0(0)=(1,0)^T$ is an eigenvector of every $C_k(0)$. (5.12) then gives exactly $P_N(a;\lambda/\lambda_0)$, i.e. (2.13), with no $\epsilon$ expansion. This agrees with (5.14) at $\gamma(0)=1$ and with (5.2).
   - Also $v=0$ by (5.9), and $R=1$ by (5.17) with sign $-$.
   - For $\tau\to0$ the same holds for every $p_0$, and (5.10) and (5.11) tend to $0$. Passed.
3. **$N=2$.** Directly, $\lVert\Phi_0\rVert^2=\frac14\lvert1+e^{-i\pi\gamma\lambda/\lambda_0}\rvert^2=\cos^2(\pi\gamma\lambda/(2\lambda_0))$. (5.14) gives $\sin^2(\pi\vartheta)/(4\sin^2(\pi\vartheta/2))=\cos^2(\pi\vartheta/2)$ with $\vartheta=\gamma\lambda/\lambda_0$. Passed.

## Open issues

- The curvature $r_k$ of $\omega_k$ in $d_k$ gives corrections $O(N\epsilon\lambda/\lambda_0)$ to (5.14) and relative corrections $O(N\epsilon)$ to $\lambda_0(p_0)$. A treatment consistent at fixed $\lambda/\lambda_0$ would replace $\gamma(p_0)$ by a level-weighted rate between $\gamma$ at $\mu=m$ and at $\mu=m+d_{N-1}$; this is not computed.
- The interbranch admixture of the item-3 start, of order $N\epsilon$, adds oscillating velocity terms of order $N\epsilon$; these are not computed.
- Effects of the records beyond leading order in $\kappa$ are out of scope: the drift from $\langle X\rangle$, and the decoherence of the view.
- (5.10) of 03-ilang-space/05-recording-contract assumes $\phi_h\ne0$ for all $h$, but the item-3 start has $\phi_h=0$ for $a\ne0$. Only $u_h$ and $d_0$ are used here, and they do not depend on $\phi$.
- The comparison of two carried clocks between two meetings is left to a later package.

## Methods used

- tensor-product factorization of commuting generators
- Dyson series and Duhamel formula (locality bound)
- Fourier analysis on $\mathbb Z$, plane waves, position as $i\partial_p$
- anticommuting (Clifford) terms, $2\times2$ diagonalization, half-angle formulas
- group velocity
- Taylor expansion with remainder bounds, Hellmann–Feynman
- geometric sums
- spectral radius of path adjacency matrices

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