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

# Inertia and internal energy: response at λ = 0 and averaged response

- **Subproject:** 04-ilang-time
- **Package:** 06-inertia-internal-energy
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-10

## Setup and assumptions

**Inputs** (as quoted in question.md).
- (2.9), (2.12) of 02-clock: a uniform ring clock with tick $\lambda_0$ and $\theta_m=0$ has $e^{-iC_K\lambda_0}=S$ and $C_K=\sum_k\frac{2\pi\ell_k}{N\lambda_0}\lvert f_k\rangle\langle f_k\rvert$ with $\ell_k\equiv k$. Its spread is minimal iff the $\ell_k$ are $N$ consecutive integers.
- (5.10) of 03-ilang-space/05-recording-contract: $\lvert u_h\rangle=(K_h-\langle K_h\rangle)\lvert\chi\rangle$, and $g(v,w)=\operatorname{Re}\langle v\vert w\rangle$.
- (17.15) of 03-ilang-space/17-motion: for $K_{h_m}=mX$, $u_{h_m}=m\,e$ with $e=(X-\langle X\rangle)\chi$, $\lVert e\rVert=\sigma_X>0$, and $\bar x=\bar m\,e$.
- (20.10) of 03-ilang-space/20-cost-and-mass: for a uniform gradient, $a(0)=a_0+\mu F$ with $\mu=-\sum_{\{h,h'\}\in E}L_{hh'}(0)\,\Delta_{hh'}\,g(\Delta_{hh'},\cdot)$, where $L_{hh'}(0)=2\operatorname{Re}[t_{hh'}\phi_h^*\phi_{h'}]$ and $\Delta_{hh'}=u_h-u_{h'}$.

**Setting.** As in question.md:
- a body $b$, the only instance of its type, and a medium $c$;
- the contract $C=C_b\otimes\mathbb 1_c+\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$, with $K_h$ depending only on the position label $x_h$;
- a product start $\phi\otimes\chi$, $\lambda\ge0$, and contracts that do not depend on $\lambda$.

Notation: $\hat x:=\sum_hx_h\lvert h\rangle\langle h\rvert$ on $\mathcal H_b$ and $\bar m:=\sum_hx_h\,p_h$. Clock places are labelled $a\in\mathbb Z_N$; these are the $\lvert m\rangle$ of 02-clock.

**Model D on $\mathbb Z$ (A1).**
- Model D is taken on the finite chains $x\in\{-L,\dots,L\}$. The terms of $A$ act only between neighbours of the chain; $B$ and $K_x$ are unchanged.
- The start has its position weight in $\lvert x\rvert\le W$, up to tails that vanish in the limit.
- $\lambda$ stays in a bounded range $[0,\lambda_{\max}]$, and the limit $L\to\infty$ is taken first.
- The hopping part of $C_b$ is nearest-neighbour with norm $\le\tau$. By the standard propagation bound, weight therefore spreads over at most $O(\tau\lambda)$ sites, up to tails. Hence $p_x$, $\bar m$ and their $\lambda$-derivatives converge to their values on $\mathbb Z$ once $L\gg W+\tau\lambda_{\max}$.

On $\mathbb Z$, the plane waves are $\lvert p\rangle=\sum_xe^{ipx}\lvert x\rangle$ with $p\in(-\pi,\pi]$, and $\Phi(x)=\int\frac{dp}{2\pi}e^{ipx}\tilde\Phi(p)$. In this representation $\hat S:=\sum_x\lvert x+1\rangle\langle x\rvert$ acts as $e^{-ip}$ and $\hat x$ acts as $i\partial_p$.

**Approximations, used in item 3 and Step 6 only.**
- (W) Weak records: used once, in Step 5a.
- (F) First order in the force: Steps 5d–5f and 6.
- (Cst) Start concentrated at $p_0=0$: Step 5c.

Their ranges of validity are stated in (6.14).

## Derivation

### Step 1. The coefficient (20.10) for any diagonal and any start

Let $\Pi_h:=\lvert h\rangle\langle h\rvert\otimes\mathbb 1_c$. Then $p_h=\langle\Psi\vert\Pi_h\vert\Psi\rangle$, and (1.9) gives $\ddot p_h(0)=-\langle[C,[C,\Pi_h]]\rangle_{\Psi(0)}$.

Split $C=C_{\rm d}+C_{\rm o}$ with $C_{\rm d}:=\sum_h\Pi_hC\Pi_h$. Since $[C_{\rm d},\Pi_h]=0$,
$$a(0)=-\sum_hu_h\langle[C_{\rm o},[C_{\rm o},\Pi_h]]+[C_{\rm d},[C_{\rm o},\Pi_h]]\rangle .$$
This is affine in $C_{\rm d}$, hence in the entries $t_{hh}$. Its linear part depends only on $C_{\rm o}$ and on the start, not on the base values of the $t_{hh}$.

Change $t_{hh}\to t_{hh}+\delta_h$ with $\delta_h-\delta_{h'}=-g(F,\Delta_{hh'})$. From a base with all $\Omega_h$ equal this is the uniform gradient of 20, and by (20.10) $a(0)$ changes by $\mu F$. By affinity the change is the same from any base:
$$
a(0)\big|_{t_{hh}+\delta_h}=a(0)\big|_{t_{hh}}+\mu F,\qquad \mu\ \text{as in (20.10) of 03-ilang-space/20-cost-and-mass}.
\tag{6.1}
$$
Both sides of (6.1) are polynomials in the start amplitudes. The identity therefore extends by continuity from starts with all $\phi_h\neq0$ to every start.

Model D needs both extensions: its diagonal is not uniform (the $\sigma_3$ term), and the start of item 2 has zero amplitudes.

### Step 2. Item 1: a body with internal labels

For $h=(x,a)$ and $h'=(x',a')$ we have $t_{hh'}=\langle a\vert T_{xx'}\vert a'\rangle$ and $\phi_h=(\phi_x)_a$. Since $K_h$ depends only on $x$, (5.10) gives $u_h=u_x$. Hence $\Delta_{hh'}=\Delta_{xx'}:=u_x-u_{x'}$, which vanishes for $x=x'$.

Inserting this into (20.10):
$$
L_{hh'}(0)=2\operatorname{Re}\bigl[(\phi_x)_a^*\langle a\vert T_{xx'}\vert a'\rangle(\phi_{x'})_{a'}\bigr],\qquad
L_{xx'}(0):=\sum_{a,a'}L_{(x,a)(x',a')}(0)=2\operatorname{Re}\langle\phi_x\vert T_{xx'}\vert\phi_{x'}\rangle .
\tag{6.2}
$$
Pairs with $t_{hh'}=0$ add zero, so the sum may run over all $a,a'$. Then
$$
\mu=-\sum_{\{x,x'\},\,x\neq x'}L_{xx'}(0)\,\Delta_{xx'}\,g(\Delta_{xx'},\cdot\,).
\tag{6.3}
$$
Edges between places with the same $x$ carry edge cost, but they have $\Delta_{hh'}=0$ and do not contribute.

**Decoupled body.** For the start $\phi_X\otimes\eta$ with $\lVert\eta\rVert=1$, we have $\phi_x=\phi_X(x)\eta$ and $L_{xx'}(0)=2\operatorname{Re}[t_{xx'}\phi_X(x)^*\phi_X(x')]$. Hence
$$
\mu=-\sum_{\{x,x'\},\,x\neq x'}2\operatorname{Re}\bigl[t_{xx'}\phi_X(x)^*\phi_X(x')\bigr]\,\Delta_{xx'}\,g(\Delta_{xx'},\cdot\,).
\tag{6.4}
$$
This is the value of (20.10) for the body without internal labels, with hopping $t_{xx'}$ and start $\phi_X$. It depends neither on $\eta$ nor on the $T_{xx}$.

### Step 3. Model D: clock, branches, rest cost, positions, gradient

**Clock.** $S\lvert f_k\rangle=e^{-2\pi ik/N}\lvert f_k\rangle=e^{-id_k\lambda_0}\lvert f_k\rangle$, so $e^{-iD\lambda_0}=S$. By (2.9), $D$ is the uniform ring clock with $\theta_m=0$ and $\ell_k=k$; by (2.12) with $c=0$, its spread is minimal. On branch $s$ it enters as $(\sigma_3)_{ss}D$.

In the place basis, use $\sum_{j=0}^{N-1}jz^j=N/(z-1)$ for $z^N=1\neq z$ (the derivative of the geometric sum):
$$
D_{aa}=\frac{\pi(N-1)}{N\lambda_0},\qquad D_{aa'}=\frac{2\pi}{N\lambda_0}\,\frac{1}{e^{2\pi i(a-a')/N}-1}\neq0\quad(a\neq a').
\tag{6.5}
$$

**Branches.** Every term of $C_b$, and also $\hat x$, commutes with $\mathbb 1\otimes\mathbb 1\otimes\lvert f_k\rangle\langle f_k\rvert$, so the clock level $k$ is conserved. On level $k$, $A$ acts on $\lvert p\rangle\otimes v$ as $\frac\tau2(-ie^{-ip}+ie^{ip})\sigma_1=-\tau\sin p\,\sigma_1$:
$$
C_b\bigl(\lvert p\rangle\otimes v\otimes\lvert f_k\rangle\bigr)=\lvert p\rangle\otimes C_k(p)v\otimes\lvert f_k\rangle,\qquad
C_k(p)=-\tau\sin p\,\sigma_1+M_k\sigma_3,
\tag{6.6}
$$
with eigenvalues $\pm\omega_k(p)$, $\omega_k(p)=\sqrt{M_k^2+\tau^2\sin^2p}$.

**Rest cost.** $C_k(0)=M_k\sigma_3$, so
$$
M_k:=m+d_k=m+\frac{2\pi k}{N\lambda_0}\ \ (\text{rest cost of level }k,\ \text{eigenvector }\lvert1\rangle),\qquad\text{gap }2M_k,\qquad \omega_k''(0)=\frac{\tau^2}{M_k}.
\tag{6.7}
$$

**Positions and gradient.**
- $K_x=\kappa xX$ is (17.15) with $X$ replaced by $\kappa X$. Hence $u_{(x,s,a)}=\kappa x\,e$, $\bar x=\kappa\bar m\,e$ and $\Delta_{hh'}=\kappa(x-x')e$.
- The record means $\epsilon_h=\kappa x\langle X\rangle$ depend on position. Let $F$ be the force of the uniform gradient of $\delta_h+\epsilon_h$, where $\delta_h$ is the change of $t_{hh}$.
- $F=0$ thus means that the $\delta_h$ cancel the record means, the analogue of "all $\Omega_h$ equal" in 20. The $B$-diagonal is the same at every $x$ and does not enter, by Step 1.

Then
$$
\delta_h+\epsilon_h=-\mathcal F\,x+\text{const},\qquad \mathcal F:=\kappa\,g(e,F),\qquad \bar x=\kappa\,\bar m\,e .
\tag{6.8}
$$
Only $g(e,F)$ enters, as in (20.10). If $F$ is meant as the force of the $\delta_h$ alone, replace $\mathcal F$ by $\mathcal F-\kappa\langle X\rangle$; the coefficients of $F$ below are unchanged.

### Step 4. Item 2: response at $\lambda=0$

The start has the amplitudes $\phi_{(x,s,a)}=\phi_X(x)\,\delta_{s1}\,N^{-1/2}e^{2\pi ika/N}$. The edges of $C_b$ are as follows; the records, $m\mathbb 1$ and the gradient are diagonal.
- **(A)** $\{(x,s,a),(x+1,s',a)\}$ with $s\neq s'$, and $t_{(x+1,s',a),(x,s,a)}=-i\tau/2$.
- **(B)** $\{(x,s,a),(x,s,a')\}$ with $a\neq a'$, and $t=(\sigma_3)_{ss}D_{aa'}$, which is nonzero by (6.5).

Every (A)-edge, and every (B)-edge with $s=2$, has an end with $s=2$, where $\phi=0$; there $L_{hh'}(0)=0$.

On the (B)-edges with $s=1$, let $n:=a-a'$ and $\vartheta:=2\pi n/N$. Since $\operatorname{Re}[e^{-ik\vartheta}/(e^{i\vartheta}-1)]=-\sin((k+\tfrac12)\vartheta)/(2\sin\tfrac\vartheta2)$,
$$
L_{(x,1,a)(x,1,a')}(0)=-\frac{2\pi}{N^2\lambda_0}\,\lvert\phi_X(x)\rvert^2\,\frac{\sin\bigl((2k+1)\pi n/N\bigr)}{\sin(\pi n/N)},\qquad L_{hh'}(0)=0\ \text{on all other edges.}
\tag{6.9}
$$
The (B)-edges lie inside one point and do not enter (6.3). On the (A)-blocks, $L_{x,x+1}(0)=2\operatorname{Re}[\phi_X(x)^*\phi_X(x+1)\,\tfrac{i\tau}{2}\langle1\vert\sigma_1\vert1\rangle]=0$. Hence
$$
\mu=0\qquad\text{for every }\phi_X,\ k,\ N,\ \tau,\ m,\ \kappa .
\tag{6.10}
$$

### Step 5. Item 3: averaged response

**5a. Records (W).** $C$ contains $c$ only through $\hat x\otimes X$, so $X$ is conserved. Decompose $\chi=\sum_\xi\Pi_\xi\chi$ over the eigenspaces of $X$, with weights $w_\xi:=\lVert\Pi_\xi\chi\rVert^2$. The view of $b$ is then exactly the $w_\xi$-mixture of the pure evolutions by $C_b-\mathcal F_\xi\hat x$ (+const), with
$$
\mathcal F_\xi:=\mathcal F-\kappa(\xi-\langle X\rangle),\qquad \sum_\xi w_\xi\mathcal F_\xi=\mathcal F,\qquad \sum_\xi w_\xi(\mathcal F_\xi-\mathcal F)^2=\kappa^2\sigma_X^2 .
\tag{6.11}
$$
To first order in the sector force, $\ddot{\bar m}$ is linear in $\mathcal F_\xi$. The mixture therefore gives the single evolution by $C_b-\mathcal F\hat x$, up to $O(\mathcal F^2+\kappa^2\sigma_X^2)$. Dropping the $\kappa^2\sigma_X^2$ term is the weak-record approximation, and it is used nowhere else.

**5b. Equations of motion.** By (1.9), $\frac{d}{d\lambda}\langle O\rangle=\langle i[C,O]\rangle$. In the $p$-representation on level $k$, $[f(p),\hat x]=-if'(p)$. Hence $\hat v:=i[C_k-\mathcal F\hat x,\hat x]=\partial_pC_k$ and $[\hat x,\hat v]=i\partial_p\hat v$, so that
$$
\dot{\bar m}=\langle\hat v\rangle,\quad \hat v=-\tau\cos\hat p\,\sigma_1;\qquad
\ddot{\bar m}=\langle\hat a_k\rangle,\quad \hat a_k=i[C_k,\hat v]-i\mathcal F[\hat x,\hat v]=2\tau M_k\cos\hat p\,\sigma_2+\mathcal F\tau\sin\hat p\,\sigma_1 .
\tag{6.12}
$$

**5c. Drift and concentrated start (Cst).**
- With $\hat x=i\partial_p$, (1.9) reads $(\partial_\lambda+\mathcal F\partial_p)\tilde\Phi=-iC_k(p)\tilde\Phi$.
- Along $p=q+\mathcal F\lambda$ this gives $\tilde\Phi(q+\mathcal F\lambda,\lambda)=w_q(\lambda)$ with $i\dot w_q=C_k(q+\mathcal F\lambda)w_q$. This is the standard drift $\dot p=\mathcal F$ of the plane-wave label.
- Hence $\ddot{\bar m}(\lambda)=\int\frac{dq}{2\pi}\,w_q^\dagger\,\hat a_k(q+\mathcal F\lambda)\,w_q$.
- The start is $w_q(0)=\tilde\phi(q)\lvert+_k(q)\rangle$ with $\int\frac{dq}{2\pi}\lvert\tilde\phi\rvert^2=1$. Here $\lvert+_k(q)\rangle$ is the real eigenvector of the positive branch, and $\lvert+_k(0)\rangle=\lvert1\rangle$ by (6.7).
- The integrand is continuous in $q$. For a start concentrated at $q=0$, the integral therefore tends, on bounded $\lambda$-intervals, to the value of the two-level problem $i\dot w=C_k(\mathcal F\lambda)w$ with $w(0)=\lvert1\rangle$.

**5d. First order (F).** $C_k(\mathcal F\lambda)=M_k\sigma_3-\tau\mathcal F\lambda\,\sigma_1+O(\mathcal F^3)$. To first order, $w_1=e^{-iM_k\lambda}$ and $w_2=e^{iM_k\lambda}\gamma$, with $\dot\gamma=i\tau\mathcal F\lambda\,e^{-2iM_k\lambda}$ and $\gamma(0)=0$. Then
$$\langle\sigma_2\rangle=2\operatorname{Im}(w_1^*w_2)=2\tau\mathcal F\int_0^\lambda\lambda'\cos\bigl(2M_k(\lambda-\lambda')\bigr)d\lambda'=\tau\mathcal F\sin^2(M_k\lambda)/M_k^2 .$$
The term $\mathcal F\tau\sin(\mathcal F\lambda)\langle\sigma_1\rangle$ is $O(\mathcal F^2)$.

Also $a_0(\lambda)=0$: at $F=0$ the state stays in the positive branch, and $\langle+\vert\sigma_2\vert+\rangle=0$ for real vectors. Using (6.12) and $\bar x=\kappa\bar m e$:
$$
a(\lambda)=\frac{2\tau^2}{M_k}\,\sin^2(M_k\lambda)\;\kappa^2\,g(e,F)\,e+O(F^2)\qquad(\text{fixed }\lambda).
\tag{6.13}
$$

**5e. Average.** $\frac1\Lambda\int_{\lambda_1}^{\lambda_1+\Lambda}2\sin^2(M_k\lambda)\,d\lambda=1-\rho$ with $\lvert\rho\rvert\le1/(M_k\Lambda)$.

Averaging needs $\Lambda\gg1/M_k$, which goes beyond fixed $\lambda$. For this, use the standard first-order adiabatic response of a branch to a uniform gradient. It holds when there are no branch transitions, $\lvert\mathcal F\rvert\tau\ll M_k^2$. It has two parts:
- a part $\mathcal F\omega_k''(\mathcal F\lambda)$ that does not oscillate;
- a part of amplitude $\approx\lvert\mathcal F\rvert\tau^2/M_k$ that oscillates at frequency $2\omega_k\approx2M_k$.

At fixed $\lambda$, its first order is (6.13). Since $\omega_k''(q)=\frac{\tau^2}{M_k}\bigl[1+O\bigl(q^2(1+\tau^2/M_k^2)\bigr)\bigr]$,
$$
\overline{a}=\frac{\tau^2}{M_k}\,\kappa^2\,g(e,F)\,e\,\Bigl[1+O\Bigl(\frac{1}{M_k\Lambda}\Bigr)\Bigr],\qquad
\frac{1}{2M_k}\ll\Lambda,\qquad \mathcal F_{\max}\,(\lambda_1+\Lambda)\ll\min\Bigl(1,\frac{M_k}{\tau}\Bigr),
\tag{6.14}
$$
with $\mathcal F_{\max}:=\max_\xi\lvert\mathcal F_\xi\rvert$.
- The first condition says that $\Lambda$ is long compared with the inverse gap.
- The second keeps the drift $\mathcal F_\xi\lambda$ of every sector inside the region where $\omega_k''\approx\omega_k''(0)$. This is the condition for first order in $F$ and for weak records.
- Together the two conditions imply $\mathcal F_{\max}\tau\ll M_k^2$.
- The chain limit requires $L\gg W+\tau(\lambda_1+\Lambda)$.

**5f. Averaged inverse inertia.** The coefficient of $F$ in (6.14) is
$$
\mu_{\rm av}(k)=\frac{\tau^2}{M_k}\,\kappa^2\,e\,g(e,\cdot\,)=\frac{\tau^2}{m+d_k}\,\kappa^2\,e\,g(e,\cdot\,).
\tag{6.15}
$$
$$
M_k\,\mu_{\rm av}(k)=\tau^2\kappa^2\,e\,g(e,\cdot\,)\ \ \text{for every }k,\qquad \frac{M_k}{\tau^2}=\frac{m}{\tau^2}+\frac{2\pi k}{N\lambda_0\,\tau^2}.
\tag{6.16}
$$
So $\mu_{\rm av}(k)$ decreases strictly with $d_k$. The scalar inertia $M_k/\tau^2$ is the rest cost (6.7) divided by $\tau^2$. The clock cost $d_k$ adds to it with the same coefficient $1/\tau^2$ as the rest term $m$.

### Step 6. Decoupled body on the chain

Take $T_{x+1,x}=t\mathbb 1_N$ with $t\neq0$, and $T_{xx}=D'$, an arbitrary self-adjoint clock term that is the same for all $x$. Then
$$C_b-\mathcal F\hat x=(\omega(\hat p)-\mathcal F\hat x)\otimes\mathbb 1_N+\mathbb 1\otimes D',\qquad \omega(p)=te^{-ip}+t^*e^{ip} .$$
The two parts commute, and $\bar m$ involves only the first. Hence $\hat v=\omega'(\hat p)$ and $i[C_b,\hat v]=0$, so $\ddot{\hat x}=-i\mathcal F[\hat x,\omega'(\hat p)]=\mathcal F\omega''(\hat p)$ exactly.

Since $\frac{d}{d\lambda}\langle\omega''(\hat p)\rangle=O(\mathcal F)$, $a(\lambda)=a(0)+O(F^2\lambda)$: the response does not oscillate and equals its own average. By (6.2), $\langle\omega''(\hat p)\rangle_0=-2\sum_x\operatorname{Re}[t^*\phi_X(x)^*\phi_X(x+1)]=-\sum_xL_{x,x+1}(0)$. With $\Delta_{x,x+1}=-\kappa e$ this gives
$$
\mu_{\rm av}=\mu=\kappa^2\,\langle\phi_X\vert\omega''(\hat p)\vert\phi_X\rangle\,e\,g(e,\cdot\,)\ \xrightarrow{\ \text{concentrated at }p_0\ }\ -2\operatorname{Re}\bigl(t^*e^{ip_0}\bigr)\,\kappa^2\,e\,g(e,\cdot\,).
\tag{6.17}
$$
This is independent of the clock level, of $\eta$ and of $D'$; Step 5a applies unchanged.

### Step 7. Item 4: comparison

$$
\text{Model D: }\ \mu=0\neq\mu_{\rm av}(k)=\frac{\tau^2}{M_k}\,\kappa^2e\,g(e,\cdot\,);\qquad
\text{decoupled body: }\ \mu=\mu_{\rm av}.
\tag{6.18}
$$
**Decoupled body.** Its velocity commutes with $C_b$, so the free acceleration vanishes. The gradient term $\mathcal F\omega''(\hat p)$ is then the whole response, already at $\lambda=0$. Its value is fixed by the edge costs between points, which is the content of (20.10).

**Model D.** The velocity $-\tau\cos\hat p\,\sigma_1$ connects the two branches.
- At $\lambda=0$ the $s=1$ start has no edge cost between points, which gives (6.10).
- The response comes from interference between two terms: the admixture of the negative branch that the gradient induces, of amplitude $\propto\mathcal F\tau/M_k^2$, and the free acceleration $2\tau M_k\cos\hat p\,\sigma_2$.
- It builds up as $\sin^2(M_k\lambda)$, and its average is the branch curvature $\tau^2/M_k$.
- The clock cost enters through the gap, because $D$ sits in the branch term $\sigma_3\otimes(m\mathbb 1+D)$. In the decoupled body the clock term commutes with position and hopping, and drops out.

## Result

- **(6.1)** The inverse inertia $\mu$ of (20.10) of 03-ilang-space/20-cost-and-mass is the coefficient of $F$ for a uniform gradient added to any diagonal, and for every product start.
- **Item 1.**
  - (6.2): $L_{hh'}(0)$ in terms of the blocks, and $L_{xx'}(0)=2\operatorname{Re}\langle\phi_x\vert T_{xx'}\vert\phi_{x'}\rangle$.
  - (6.3): $\mu=-\sum_{x\neq x'}L_{xx'}(0)\Delta_{xx'}g(\Delta_{xx'},\cdot)$. Edges inside one point do not contribute.
  - (6.4): for the decoupled body, $\mu$ depends neither on $\eta$ nor on the $T_{xx}$.
- **Model D.**
  - (6.7): rest cost $M_k=m+d_k$.
  - (6.9): edge costs at $\lambda=0$, nonzero only on the clock edges of branch $s=1$.
  - (6.10): $\mu=0$.
- **Item 3.**
  - (6.13): $a(\lambda)=\frac{2\tau^2}{M_k}\sin^2(M_k\lambda)\,\kappa^2g(e,F)e$.
  - (6.14): its average $\frac{\tau^2}{M_k}\kappa^2g(e,F)e$, for $1/(2M_k)\ll\Lambda$ and $\mathcal F_{\max}(\lambda_1+\Lambda)\ll\min(1,M_k/\tau)$.
  - (6.15): $\mu_{\rm av}(k)=\frac{\tau^2}{m+d_k}\kappa^2e\,g(e,\cdot)$.
  - (6.16): the inertia is the rest cost over $\tau^2$; the clock cost $d_k$ adds to it like $m$.
  - (6.17): for the decoupled chain body, $\mu_{\rm av}=\mu=\kappa^2\langle\omega''\rangle e\,g(e,\cdot)$, independent of the clock level.
- **Item 4, (6.18).** $\mu$ and $\mu_{\rm av}$ coincide for the decoupled body, whose velocity commutes with $C_b$. They differ for Model D, whose velocity mixes the branches.

## Consistency checks

1. **Dimensions and $\tau\to0$.** $C$, $\tau$, $m$, $d_k$ and $\mathcal F$ have the dimension $1/[\lambda]$, and $\ddot{\bar m}\sim\mathcal F\tau^2/M_k$ has the dimension $1/[\lambda]^2$. In (6.9), $2\pi/(N^2\lambda_0)$ is a cost. As $\tau\to0$, $\hat v=0$, and (6.13) and (6.15) vanish. Passed.
2. **$\lambda=0$ by an independent route.** For the item-2 start with any $\phi_X$, $\langle1\vert\sigma_2\vert1\rangle=\langle1\vert\sigma_1\vert1\rangle=0$ in (6.12) gives $a(0)=0$ for every $F$. This agrees with (6.10), which was obtained from (20.10) through edge costs, and with $a_0=0$. Also, (6.13) vanishes at $\lambda=0$. Passed.
3. **Cost sum rule at one point.** The site costs of $D$ on the $s=1$ places of $x$ sum to $\lvert\phi_X(x)\rvert^2\pi(N-1)/(N\lambda_0)$. The edge costs (6.9) sum, by the Dirichlet-kernel sum $\sum_{n=1}^{N-1}\sin((2k+1)\pi n/N)/\sin(\pi n/N)=N-1-2k$, to $-\lvert\phi_X(x)\rvert^2\pi(N-1-2k)/(N\lambda_0)$. The total is $d_k\lvert\phi_X(x)\rvert^2=\langle f_k\vert D\vert f_k\rangle\lvert\phi_X(x)\rvert^2$. Passed.

## Open issues

- A ticking clock, i.e. a superposition of levels, is not treated. Since the level is conserved, the averaged response is the level-weighted mean of (6.15); whether this mixture is the inertia of one body is left open.
- Beyond fixed $\lambda$, Model D uses the stated standard adiabatic response of a branch, and the concentration limit. For wide packets the average is expected to be $\langle\omega_k''\rangle$, but this is not derived here.
- Second order in $\kappa$ (decoherence by the records, $\kappa^2\sigma_X^2$) and in $F$ (drift beyond the quadratic region, Bloch oscillation, branch transitions) is not treated.
- The meaning of $\tau$ as a maximal speed, and of $M_k/\tau^2$, belongs to criteria 4 and 6; it is outside this package (M1, M3).

## Methods used

- Heisenberg equations of motion (double commutators)
- Affine dependence of the acceleration on the diagonal of the contract
- Fourier transform on $\mathbb Z$, plane waves, two-branch dispersion
- Method of characteristics (drift of the plane-wave label in a uniform gradient)
- First-order time-dependent perturbation theory; adiabatic perturbation theory (stated)
- Conserved record observable: exact mixture decomposition of the view
- Geometric and Dirichlet-kernel sums
- Limit of finite chains with a propagation bound

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