Version 2 · earlier version; the current one is v3 · External review, round 2: major errors

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

- **Subproject:** 04-ilang-time
- **Package:** 06-inertia-internal-energy
- **Version:** v2
- **Mode:** external regeneration
- **Date:** 2026-10-10

## Changes from previous version

- **Step 6 rewritten, Step 7 revised (I1).** v1 answered items 3 and 4 for the decoupled body only when the on-site blocks are the same at every position, $T_{xx}=D'$. v2 covers the class of the question (arbitrary $T_{xx}$):
  - 6a: an exact operator identity for the whole class, (6.21);
  - 6b: uniform on-site blocks, where (6.17) holds as in v1;
  - 6c: position-dependent on-site blocks, with what holds and what fails, and a body E of the class, (6.22), whose $\mu_{\rm av}$ depends on the clock level and differs from $\mu$;
  - (6.18) of v1 stated $\mu=\mu_{\rm av}$ for the decoupled body without restriction. It is withdrawn and its tag is not reused. The comparison is now (6.23).
- **Step 5f (I2).** New (6.20): label inertia and background inertia. (6.15) and (6.16) are unchanged.
- **Setup and Step 5c (I3).** The chain limit is taken for compactly supported starts, and the order of the limits is stated.
- **Step 5e (I4).** (6.14) keeps its quantity, leading term and conditions; its error term is now given completely by the new (6.19). The average is obtained from the exact relation between the averaged acceleration and the change of the velocity. (6.15) is identified as the leading long-interval coefficient.
- Steps 1–4 and 5a–5d are unchanged in content, and so are the tags (6.1)–(6.13) and (6.15)–(6.17). Check 3 of v1 (cost sum rule for (6.9)) remains valid; because of the limit of three checks it is replaced by a check of the new (6.21).

## Response to verification

- **I1: Accepted and fixed in Step 6 (6a–6c) and Step 7.** The restriction $T_{xx}=D'$ was essential, as the verifier says. (6.21) gives the free acceleration for arbitrary $T_{xx}$; it vanishes exactly for uniform on-site blocks. For position-dependent blocks the result is not only "no universal formula": body E of (6.22) belongs to the class, and its $\mu_{\rm av}$ depends on the clock level and differs from $\mu$. The unrestricted claim (6.18) is withdrawn and replaced by (6.23).
- **I2: Accepted and fixed in Step 5f, (6.20).** $M_k/\tau^2$ is the inertia of the position label against the label force $\mathcal F$; the inertia of the background-space response is $M_k/(\tau^2\kappa^2\sigma_X^2)$. Both are stated; both are proportional to the rest cost.
- **I3: Accepted and fixed in "Setup and assumptions" (Chains and limits) and Step 5c.** Starts are compactly supported; the order of the chain-size limit and the concentration limit is stated.
- **I4: Accepted and fixed in Step 5e, (6.14) with (6.19), and Step 5f.** The remainder now contains the variation of the branch curvature along the drift and the record terms; (6.15) is stated as the coefficient at first order in $F$, leading order in $\kappa$, in the long-interval limit.

## 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 $e=(X-\langle X\rangle)\chi$, $\lVert e\rVert=\sigma_X>0$, $\bar x=\bar m\,e$ and $d_0(h_m,h_n)=\lvert m-n\rvert\sigma_X$.
- (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.

**Chains and limits (A1).**
- *Chain.* Model D, and the chain bodies of Step 6, are taken on $x\in\{-L,\dots,L\}$. The hopping terms act only between neighbours of the chain; every term diagonal in $x$ (on-site blocks, gradient, records) is unchanged.
- *Starts.* Compactly supported: $\phi_h=0$ for $\lvert x_h\rvert>W$, with $W<L$.
- *Propagation bound (standard).* The hopping part is nearest-neighbour with norm $\le\tau$ (Model D; $2\lvert t\rvert$ in Step 6), and every other term of $C$ is diagonal in $x$. Expand the evolution in powers of the hopping, in the interaction picture of the diagonal part. The terms of order $n<L-W$ are the same on the chain and on $\mathbb Z$, and the part of $\Psi(\lambda)$ at distance $>r$ from the support of the start has norm $\le\sum_{n>r}(\tau\lambda)^n/n!\le(e\tau\lambda/r)^r$.
- *Limit 1: $L\to\infty$*, at fixed start, $F$, $\kappa$ and $\lambda\le\lambda_{\max}$. The chain state converges in norm to the state on $\mathbb Z$, and the weighted tails $\sum_{\lvert x\rvert>W+r}\lvert x\rvert^jp_x$, $j=1,2$, are bounded by the same superexponentially small terms. Hence $\bar m$, $\dot{\bar m}$ and $\ddot{\bar m}$ converge uniformly on $[0,\lambda_{\max}]$. A vanishing tail weight alone would not give the convergence of $\bar m$; the compact support is what is used.
- *Limit 2: concentration*, $W\to\infty$ (Step 5c), taken after limit 1 at fixed $\lambda_1$, $\Lambda$, $F$, $\kappa$.
- *Then* first order in $F$ and leading order in $\kappa$, and last the long-interval limit $M_k\Lambda\to\infty$ (Steps 5e, 5f).

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: Step 5a, and the record terms of (6.19) in Steps 5e, 5f.
- (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) and (6.19).

## 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 and the bodies of Step 6 need both extensions: their diagonals are not uniform, and the starts used have 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** ($T_{xx'}=t_{xx'}\mathbb 1_N$ for $x\neq x'$, arbitrary $T_{xx}$, also position-dependent). 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 $\phi_X$. By (6.1) it holds for every choice of the $T_{xx}$, and 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
$$
\mathcal F_\xi:=\mathcal F-\kappa(\xi-\langle X\rangle),\qquad \sum_\xi w_\xi\mathcal F_\xi=\mathcal F,\qquad \sum_\xi w_\xi(\mathcal F_\xi-\mathcal F)^2=\kappa^2\sigma_X^2 .
\tag{6.11}
$$
The part of $\ddot{\bar m}$ that is linear in the sector force sums exactly to its value at $\mathcal F$, as for the single evolution by $C_b-\mathcal F\hat x$. The weak-record approximation drops the terms of higher order in the $\mathcal F_\xi$; their size is given in (6.19). 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 $\ddot{\bar m}(\lambda)=\int\frac{dq}{2\pi}\,w_q^\dagger\,\hat a_k(q+\mathcal F\lambda)\,w_q$.
- *Start.* $\phi^{(W)}=\phi_X^{(W)}\otimes\lvert1\rangle\otimes\lvert f_k\rangle$ with $\phi_X^{(W)}(x)=c_W\varphi(x/W)$, where $\varphi$ is smooth with support in $[-1,1]$ and $c_W$ normalizes. Then $w_q(0)=\tilde\phi_W(q)\lvert1\rangle$ and $\lvert\tilde\phi_W(q)\rvert^2\frac{dq}{2\pi}\to\delta(q)$ for $W\to\infty$: the start is concentrated at $p_0=0$.
- *Branch.* The real eigenvector of the positive branch is $\lvert+_k(q)\rangle=\lvert1\rangle+O(\tau q/M_k)$, by (6.7). The start therefore lies in the positive branch up to a part of norm $O(\tau/(M_kW))$. The start $\tilde\phi_W(q)\lvert+_k(q)\rangle$, exactly in the positive branch, differs from it by the same amount and has the same limit, because $\hat a_k$ is bounded.
- *Limit 2.* The integrand is $\lvert\tilde\phi_W(q)\rvert^2$ times a function of $q$ that is continuous and bounded, uniformly on bounded $\lambda$-intervals. For $W\to\infty$ the integral tends 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 leading order in $\kappa$: 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 $\bar x=\kappa\bar m e$:
$$
a(\lambda)=\frac{2\tau^2}{M_k}\,\sin^2(M_k\lambda)\;\kappa^2\,g(e,F)\,e+O(F^2)\qquad(\text{fixed }\lambda).
\tag{6.13}
$$
The solution of the two-level problem is analytic in $\mathcal F$ at fixed $\lambda$, so the coefficient of $F$ in (6.13) is exact for every fixed $\lambda\ge0$.

**5e. Average.** Let $T:=\lambda_1+\Lambda$ and $\overline{(\cdot)}:=\frac1\Lambda\int_{\lambda_1}^{T}(\cdot)\,d\lambda$.

*Coefficient of $F$ at a fixed interval.* By (6.13), at first order in $F$ and leading order in $\kappa$ the coefficient of $F$ in $\overline a$ is exactly $\frac{\tau^2}{M_k}(1-\rho)\,\kappa^2e\,g(e,\cdot)$, with $\rho$ as in (6.19), since $\overline{2\sin^2(M_k\lambda)}=1-\rho$.

*Remainder at finite $F$ and $\kappa$.* In each sector $\xi$ of 5a, exactly $\overline{\ddot{\bar m}}=[\dot{\bar m}(T)-\dot{\bar m}(\lambda_1)]/\Lambda$. Use the standard adiabatic response of a branch to a uniform gradient:
- for $\lvert\mathcal F_\xi\rvert\tau\ll M_k^2$, the state started in the positive branch at label $0$ follows this branch at the drifting label $\mathcal F_\xi\lambda$, with an admixture of the negative branch of amplitude $O(\lvert\mathcal F_\xi\rvert\tau/M_k^2)$;
- hence $\dot{\bar m}=\omega_k'(\mathcal F_\xi\lambda)+O(\lvert\mathcal F_\xi\rvert\tau^2/M_k^2)$, where the second term oscillates at the frequency $2\omega_k$. Its first order is the term $-\frac{\tau^2\mathcal F_\xi}{2M_k^2}\sin(2M_k\lambda)$ of the integral of (6.13). There is no further term, because the branch eigenvectors are real.

With the mean value theorem and $\omega_k''(q)=\frac{\tau^2}{M_k}\bigl[1+O\bigl(q^2(1+\tau^2/M_k^2)\bigr)\bigr]$, the sector average is $\mathcal F_\xi\frac{\tau^2}{M_k}\bigl[1-\rho+O\bigl((\mathcal F_\xi T)^2(1+\tau^2/M_k^2)\bigr)\bigr]$. Summing over the sectors with (6.11), the linear part gives $\mathcal F(1-\rho)$ exactly:
$$
\overline{a}=\frac{\tau^2}{M_k}\,\kappa^2\,g(e,F)\,e\,\bigl[1+r\bigr],\qquad
\frac{1}{2M_k}\ll\Lambda,\qquad \mathcal F_{\max}\,T\ll\min\Bigl(1,\frac{M_k}{\tau}\Bigr),
\tag{6.14}
$$
with $\mathcal F_{\max}:=\max_\xi\lvert\mathcal F_\xi\rvert$ and
$$
r=-\rho+\frac{R}{\mathcal F},\qquad
\rho=\frac{\sin(2M_kT)-\sin(2M_k\lambda_1)}{2M_k\Lambda},\quad\lvert\rho\rvert\le\frac{1}{M_k\Lambda},\qquad
R=O\Bigl(T^2\Bigl(1+\frac{\tau^2}{M_k^2}\Bigr)\sum_\xi w_\xi\lvert\mathcal F_\xi\rvert^3\Bigr).
\tag{6.19}
$$
Here $\sum_\xi w_\xi\lvert\mathcal F_\xi\rvert^3\le\mathcal F_{\max}(\mathcal F^2+\kappa^2\sigma_X^2)$, and $\sum_\xi w_\xi\mathcal F_\xi^3=\mathcal F^3+3\mathcal F\kappa^2\sigma_X^2-\kappa^3\langle(X-\langle X\rangle)^3\rangle$. So $R$ contains:
- terms of third order in $\mathcal F$, of relative size $O\bigl((\mathcal FT)^2(1+\tau^2/M_k^2)\bigr)$: the variation of the branch curvature along the drift, beyond first order in $F$;
- terms linear in $\mathcal F$, of relative size $O\bigl((\kappa\sigma_XT)^2(1+\tau^2/M_k^2)\bigr)$: records beyond leading order in $\kappa$;
- a term independent of $F$, of order $\kappa^3$: an acceleration at $F=0$ induced by the records, beyond leading order in $\kappa$.

About the conditions and the errors:
- The first condition of (6.14) says that $\Lambda$ is long compared with the inverse gap $1/(2M_k)$; it makes $\rho$ small.
- 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 the first-order treatment and for weak records; it makes $R$ small, but it does not bound $R$ by $\rho$. On intervals that grow as $F$ decreases, $R$ must be kept.
- Together the two conditions imply $\mathcal F_{\max}\tau\ll M_k^2$ (no branch transitions).
- Finite-packet corrections vanish in limit 2, which is taken before. The chain limit requires $L-W\gg\tau T$.

**5f. Averaged inverse inertia.** $\mu_{\rm av}(k)$ is the coefficient of $F$ in $\overline a$ at first order in $F$ and leading order in $\kappa$, in the long-interval limit $M_k\Lambda\to\infty$. At a finite interval this coefficient is $(1-\rho)$ times the following:
$$
\mu_{\rm av}(k)=\frac{\tau^2}{M_k}\,\kappa^2\,e\,g(e,\cdot\,)=\frac{\tau^2}{m+d_k}\,\kappa^2\,e\,g(e,\cdot\,).
\tag{6.15}
$$
$$
M_k\,\mu_{\rm av}(k)=\tau^2\kappa^2\,e\,g(e,\cdot\,)\ \ \text{for every }k,\qquad \frac{M_k}{\tau^2}=\frac{m}{\tau^2}+\frac{2\pi k}{N\lambda_0\,\tau^2}.
\tag{6.16}
$$
So $\mu_{\rm av}(k)$ decreases strictly with $d_k$, and it is inversely proportional to the rest cost (6.7).

*Two normalizations of the inertia.* Let $n:=e/\sigma_X$, a unit vector of $(\mathcal H_c,g)$, and $d_1:=\kappa\sigma_X$, the background distance of neighbouring points by (17.15) with $X\to\kappa X$. Then $\bar x=d_1\bar m\,n$ and $\mathcal F=d_1\,g(n,F)$, and
$$
\mu_{\rm av}(k)=\frac{1}{I_k}\,n\,g(n,\cdot\,),\qquad
I_k:=\frac{M_k}{\tau^2\kappa^2\sigma_X^2}=\frac{I_k^{\rm lab}}{d_1^{\,2}},\qquad
I_k^{\rm lab}:=\frac{M_k}{\tau^2}\quad\bigl(\overline{\ddot{\bar m}}=\mathcal F/I_k^{\rm lab}\bigr).
\tag{6.20}
$$
- $I_k$ is the **background inertia**: $\mu_{\rm av}(k)$ has the single nonzero eigenvalue $1/I_k$, on $n$, and vanishes on the $g$-orthogonal complement of $e$.
- $I_k^{\rm lab}$ is the **label inertia**: the response of the dimensionless position label $\bar m$ to the label force $\mathcal F$.
- Both are proportional to the rest cost $M_k=m+d_k$. The clock cost $d_k$ adds to each with the same coefficient as the rest term $m$.

### Step 6. Decoupled body on the chain

**6a. The whole class.** Take $x\in\mathbb Z$ (limits as in the Setup), $T_{x+1,x}=t\mathbb 1_N$ with $t\neq0$, and arbitrary self-adjoint on-site blocks $T_{xx}$; records and gradient as in (6.8). Let
$$\omega(p):=te^{-ip}+t^*e^{ip},\qquad V:=\sum_x\lvert x\rangle\langle x\rvert\otimes T_{xx},\qquad G_x:=T_{x+1,x+1}-T_{xx},$$
so that $C_b=\omega(\hat p)\otimes\mathbb 1_N+V$ with $\omega(\hat p)=t\hat S+t^*\hat S^\dagger$. Since $V$ commutes with $\hat x$, the steps of 5b give $\hat v=i[C_b-\mathcal F\hat x,\hat x]=\omega'(\hat p)=-it\hat S+it^*\hat S^\dagger$. With $[V,\hat S]=\sum_x\lvert x+1\rangle\langle x\rvert\otimes G_x$:
$$
\hat v=\omega'(\hat p)\otimes\mathbb 1_N,\qquad
\ddot{\hat x}=\mathcal F\,\omega''(\hat p)+\hat a_{\rm free},\qquad
\hat a_{\rm free}:=i[V,\hat v]=\sum_x\bigl(t\,\lvert x+1\rangle\langle x\rvert+t^*\lvert x\rangle\langle x+1\rvert\bigr)\otimes G_x .
\tag{6.21}
$$
Step 5a applies to the whole class, since it uses only that $c$ enters through $\hat x\otimes X$. For the start $\phi_X\otimes\eta$, (6.21) gives the following.
- **At $\lambda=0$.** The state does not depend on $\mathcal F$, and $\hat a_{\rm free}$ does not contain $\mathcal F$. The coefficient of $F$ in $a(0)$ is therefore $\kappa^2\langle\phi_X\vert\omega''(\hat p)\vert\phi_X\rangle\,e\,g(e,\cdot)$. By (6.2), $\langle\omega''(\hat p)\rangle_0=-2\sum_x\operatorname{Re}[t^*\phi_X(x)^*\phi_X(x+1)]=-\sum_xL_{x,x+1}(0)$, and $\Delta_{x,x+1}=-\kappa e$: this is $\mu$ of (6.4), for every choice of the $T_{xx}$. Only $a_0=\kappa\langle\hat a_{\rm free}\rangle_0\,e$ depends on $\eta$ and on the $T_{xx}$.
- **At $\lambda>0$.** The coefficient of $\mathcal F$ in $\ddot{\bar m}(\lambda)$ is $\langle\omega''(\hat p)\rangle_\lambda+\partial_{\mathcal F}\langle\hat a_{\rm free}\rangle_\lambda$, taken at $\mathcal F=0$.
- **Criterion.** Since $t\neq0$, $\hat a_{\rm free}=0$ iff $G_x=0$ for all $x$, i.e. iff the on-site blocks are the same at every position. Exactly then the velocity commutes with $C_b$.

**6b. Uniform on-site blocks.** Let $T_{xx}=D'$ for all $x$, with any self-adjoint clock term $D'$. Then $\hat a_{\rm free}=0$ and
$$C_b-\mathcal F\hat x=(\omega(\hat p)-\mathcal F\hat x)\otimes\mathbb 1_N+\mathbb 1\otimes D' .$$
The two parts commute, and $\bar m$ involves only the first. By (6.21), $\ddot{\hat x}=\mathcal F\omega''(\hat p)$ exactly, and $\frac{d}{d\lambda}\omega''(\hat p)=\mathcal F\omega'''(\hat p)$. Hence $a(\lambda)=a(0)+O(F^2\lambda)$: the response does not oscillate and equals its own average over every interval with $\mathcal F_{\max}T\ll1$. So
$$
\mu_{\rm av}=\mu=\kappa^2\,\langle\phi_X\vert\omega''(\hat p)\vert\phi_X\rangle\,e\,g(e,\cdot\,)\ \xrightarrow{\ \text{concentrated at }p_0\ }\ -2\operatorname{Re}\bigl(t^*e^{ip_0}\bigr)\,\kappa^2\,e\,g(e,\cdot\,).
\tag{6.17}
$$
This is independent of the clock level, of $\eta$ and of $D'$.

**6c. Position-dependent on-site blocks.** Now $G_x\neq0$ for some $x$.
- *What holds.* (6.4): $\mu$ is the same as in 6b and depends neither on $\eta$ nor on the $T_{xx}$. Also (6.21) and Step 5a hold.
- *What fails.* $[C_b,\hat v]=0$, the constancy of the response in $\lambda$, and with them the argument for $\mu_{\rm av}=\mu$.
- *Scalar differences*, $T_{xx}=D'+\nu_x\mathbb 1_N$. The factorization of 6b persists with $\omega(\hat p)$ replaced by $\omega(\hat p)+\nu(\hat x)$. The response at every $\lambda$ is that of the body without internal labels with site costs $\nu_x$. It is independent of the clock level, of $\eta$ and of $D'$, but it need not be constant in $\lambda$ or equal to $\mu$ (for $\nu_x=(-1)^xM$ it is body E below with $m\mathbb 1_N+D$ replaced by $M\mathbb 1_N$).
- *Otherwise* the internal state does not factor from the position. The term $\partial_{\mathcal F}\langle\hat a_{\rm free}\rangle_\lambda$ depends on the $T_{xx}$ and on $\eta$, and no formula in terms of $t$ alone exists. The following body of the class shows this.

**Body E.** Let $t=-\tfrac{i\tau}{2}$ and $T_{xx}=(-1)^x(m\mathbb 1_N+D)$, with records and gradient as in Model D. Define $J\lvert x\rangle\otimes\lvert a\rangle:=\lvert x\rangle\otimes\lvert s(x)\rangle\otimes\lvert a\rangle$, with $s(x)=1$ for even $x$ and $s(x)=2$ for odd $x$.
- $A$ of Model D maps $\lvert x,s\rangle$ to $-\tfrac{i\tau}{2}\lvert x+1,\bar s\rangle+\tfrac{i\tau}{2}\lvert x-1,\bar s\rangle$, where $\bar s$ is the other branch label, and $\overline{s(x)}=s(x\pm1)$. $B$ multiplies $\lvert x,s(x)\rangle$ by $(-1)^x(m\mathbb 1_N+D)$. Hence $C_b^{\rm D}J=JC_b^{\rm E}$.
- $J$ preserves $x$, so it also intertwines $\hat x$, the gradient and the records. It is a unitary from the places of E onto the places of Model D with $x+s$ odd, which span an invariant subspace of $C$.
- Therefore the place weights, $\bar x$ and $a(\lambda)$ of E with start $\phi$ are those of Model D with start $J\phi$, for all $\lambda$, $F$ and $\kappa$. (The places with $x+s$ even carry the body with $T_{xx}=-(-1)^x(m\mathbb 1_N+D)$: Model D is the direct sum of two bodies of this class.)

Take the start $\phi=\phi_X\otimes\lvert f_k\rangle$ with $\phi_X(x)\propto\varphi(x/W)$ on even $x$ and $\phi_X(x)=0$ on odd $x$. In the terms of E, it sits at the plane-wave label $0$ of the two-site cell, in the branch of positive eigenvalue $M_k$: at this label that branch lives on the even sites. Its image is $J\phi=\phi_X\otimes\lvert1\rangle\otimes\lvert f_k\rangle$, a start of item 2.
- *$\mu$.* Every edge between points joins an even and an odd site, so $L_{x,x+1}(0)=0$ in (6.4). This agrees with (6.10) for $J\phi$.
- *$\mu_{\rm av}$.* $\tilde\phi_X(q+\pi)=\tilde\phi_X(q)$, so $J\phi$ is concentrated at $q\in\{0,\pi\}$. Since $\sigma_3C_k(q)\sigma_3=C_k(q+\pi)$, $\sigma_3\hat a_k(q)\sigma_3=\hat a_k(q+\pi)$ and $\sigma_3\lvert1\rangle=\lvert1\rangle$, the integrand of 5c has period $\pi$ in $q$. The limit is the two-level problem of 5d, and Steps 5d–5f hold unchanged.

Hence
$$
\text{body E, level }k:\qquad \mu=0,\qquad \mu_{\rm av}(k)=\frac{\tau^2}{m+d_k}\,\kappa^2\,e\,g(e,\cdot\,)\neq\mu .
\tag{6.22}
$$
In the class of item 1 with position-dependent $T_{xx}$, $\mu_{\rm av}$ can thus depend on the clock level and differ from $\mu$, although $\mu$ itself depends on neither $\eta$ nor the $T_{xx}$.

### Step 7. Item 4: comparison

$$
\begin{aligned}
&\text{Model D (and body E):}&&\mu=0\neq\mu_{\rm av}(k)=\frac{\tau^2}{M_k}\,\kappa^2e\,g(e,\cdot\,);\\
&\text{decoupled body, }T_{xx}=D'\text{ at every }x:&&\mu=\mu_{\rm av}\quad\text{(6.17), independent of the clock};\\
&\text{decoupled body, }T_{xx}\text{ position-dependent:}&&\text{no general relation; }\mu\neq\mu_{\rm av}\text{ for body E (6.22)}.
\end{aligned}
\tag{6.23}
$$
Of the two bodies of the question, $\mu$ and $\mu_{\rm av}$ coincide for the decoupled body, provided its clock term is the same at every position; they differ for Model D.

**Reason.** $\mu$ is the coefficient of the term $-i\mathcal F[\hat x,\hat v]$ of the acceleration, evaluated in the start; it is fixed by the edge costs between points, which is the content of (20.10). The averaged response contains in addition the response of the free acceleration $i[C_b,\hat v]$ to the gradient.
- **Uniform on-site blocks.** The velocity commutes with $C_b$, so the free acceleration vanishes identically (6.21). The term $\mathcal F\omega''(\hat p)$ is the whole response, already at $\lambda=0$. The clock term commutes with position and hopping, and drops out.
- **Model D.** The velocity $-\tau\cos\hat p\,\sigma_1$ connects the two branches and does not commute with the term $\sigma_3\otimes(m\mathbb 1+D)$ that carries the clock.
  - At $\lambda=0$ the $s=1$ start has no edge cost between points, which gives (6.10).
  - The response comes from interference between 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 $2M_k$.
- **Position-dependent on-site blocks.** The same mechanism acts through $G_x\neq0$ in (6.21). Model D is itself of this kind in each of its two invariant subspaces (body E): the dividing line is whether the term that carries the clock commutes with the velocity, not whether the hopping blocks are multiples of $\mathbb 1_N$.

## 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 $T_{xx}$, uniform or position-dependent.
- **Item 2 (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): $a(\lambda)=\frac{2\tau^2}{M_k}\sin^2(M_k\lambda)\,\kappa^2g(e,F)e$ at first order in $F$.
  - (6.14) with (6.19): the average $\frac{\tau^2}{M_k}\kappa^2g(e,F)e\,[1+r]$ for $1/(2M_k)\ll\Lambda$ and $\mathcal F_{\max}T\ll\min(1,M_k/\tau)$; $r$ consists of the averaging error $\rho$, $\lvert\rho\rvert\le1/(M_k\Lambda)$, and of the higher orders in $F$ and $\kappa$.
  - (6.15): $\mu_{\rm av}(k)=\frac{\tau^2}{m+d_k}\kappa^2e\,g(e,\cdot)$, the coefficient at first order in $F$, leading order in $\kappa$, in the long-interval limit.
  - (6.16), (6.20): $\mu_{\rm av}(k)$ is inversely proportional to the rest cost. Background inertia $I_k=M_k/(\tau^2\kappa^2\sigma_X^2)$, label inertia $I_k^{\rm lab}=M_k/\tau^2$; the clock cost $d_k$ adds to each like $m$.
  - (6.21): for the decoupled chain body with arbitrary $T_{xx}$, $\ddot{\hat x}=\mathcal F\omega''(\hat p)+\hat a_{\rm free}$, and $\hat a_{\rm free}=0$ iff the on-site blocks are uniform.
  - (6.17): for uniform on-site blocks, $\mu_{\rm av}=\mu=\kappa^2\langle\omega''\rangle e\,g(e,\cdot)$, independent of the clock level.
  - (6.22): for position-dependent on-site blocks there is no such statement; for body E, $\mu=0$ and $\mu_{\rm av}(k)=\frac{\tau^2}{m+d_k}\kappa^2e\,g(e,\cdot)$.
- **Item 4, (6.23).** $\mu$ and $\mu_{\rm av}$ coincide for the decoupled body with uniform on-site blocks, whose velocity commutes with $C_b$. They differ for Model D, and for body E, where the term that carries the clock does not commute with the velocity.

## Consistency checks

1. **Dimensions and $\tau\to0$.** $C$, $\tau$, $m$, $d_k$ and $\mathcal F$ have the dimension $1/[\lambda]$. $I_k^{\rm lab}=M_k/\tau^2$ has the dimension $[\lambda]$, so $\mathcal F/I_k^{\rm lab}$ has the dimension $1/[\lambda]^2$ of $\ddot{\bar m}$. $\mu_{\rm av}F=\frac{\tau^2}{M_k}\,\kappa e\cdot\kappa g(e,F)$ has the dimension of $\kappa e/[\lambda]^2$, that of $\ddot{\bar x}$. In (6.9), $2\pi/(N^2\lambda_0)$ is a cost. As $\tau\to0$, $\hat v=0$, (6.13) and (6.15) vanish and $I_k\to\infty$. 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), obtained from (20.10) through edge costs, and with (6.13) at $\lambda=0$. For body E, (6.21) gives the same: $\langle\omega''(\hat p)\rangle_0$ and $\langle\hat a_{\rm free}\rangle_0$ are sums of bond terms $\phi_X(x)^*\phi_X(x+1)$, which vanish for $\phi_X$ on even sites. Passed.
3. **(6.21) against (6.12) through $J$.** For body E on level $k$, $G_x=-2(-1)^xM_k$ and (6.21) gives $\hat a_{\rm free}=i\tau M_k\sum_x(-1)^x\bigl(\lvert x+1\rangle\langle x\rvert+\lvert x-1\rangle\langle x\rvert\bigr)$. In Model D, $2\tau M_k\cos\hat p\,\sigma_2=\tau M_k(\hat S+\hat S^\dagger)\sigma_2$ and $\sigma_2\lvert s(x)\rangle=i(-1)^x\lvert\overline{s(x)}\rangle$, so it maps $J\lvert x\rangle$ to $i\tau M_k(-1)^x(J\lvert x+1\rangle+J\lvert x-1\rangle)$: the same operator. The force terms agree as well: $\omega''(p)=\tau\sin p$ for $t=-i\tau/2$, and $\tau\sin\hat p\,\sigma_1J=J\,\tau\sin\hat p$. For $G_x=0$, (6.21) reduces to 6b. 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.
- The remainder $R$ of (6.19) rests on the stated standard adiabatic response of a branch; it is an order estimate, without constants. Wide packets are not treated: the average is expected to be $\langle\omega_k''\rangle$, but this is not derived here.
- Position-dependent on-site blocks: only the scalar case and body E are solved. Which position-dependent blocks make $\mu_{\rm av}$ depend on the clock level is not classified, and a general member of the class has no gap that fixes the averaging interval.
- Second order in $\kappa$ (decoherence by the records, the acceleration of order $\kappa^3$ at $F=0$) and higher order in $F$ (Bloch oscillation, branch transitions) are only estimated in (6.19).
- The meaning of $\tau$ and of $d_1\tau$ as speeds, and of the inertias (6.20), 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 perturbation theory in the force; adiabatic response of a branch (stated)
- Averaged acceleration as velocity change over the interval; mean value theorem
- Conserved record observable: exact mixture decomposition of the view
- Invariant subspaces and unitary equivalence (staggered chain and two-branch chain)
- Propagation bound for nearest-neighbour hopping; limit of finite chains
- Geometric sums

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