Version 1 · earlier version; the current one is v3 · Review of v1

# Fall from rest in curved level profiles: curvature terms and local slope

- **Subproject:** 05-ilang-free-fall
- **Package:** 15-curved-profile-fall
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-18

## Setup and assumptions

**Description (assumption of the package, M8).** A body $b$, the only instance of its type, with places $(x,s,a)$, recorded by a medium $c$ through $K_x=\kappa xX$. The clock level $k$ is conserved, and on it the body's contract is the level block of the question with $\zeta_{x,k}=1$ and the quadratic level profiles $M_{x,k}=M_k+G_ky+\tfrac12H_ky^2>0$, $V_{x,k}=-\mathcal F_ky-\tfrac12\mathcal H_ky^2$, $y=x-x_0$. In Steps 1–7 the level is fixed and its index is dropped: $\tau,M,G,\mathcal F,H,\mathcal H$ ($\mathcal H$ is the curvature of $V$, not a Hilbert space). $\varepsilon$ is the joint size of $G,\mathcal F,H,\mathcal H$.

**Regime and order (A13).** Weak records: $\kappa\to0$ at fixed $\lambda$, chain and start, with $\kappa x\langle X\rangle$ counted in $V_{x,k}$, so the body evolves with $C^{(k)}$. First order jointly in $G,\mathcal F,H,\mathcal H$ at fixed $\lambda$, start and chain, then a long chain: at fixed $\lambda$ the hopping is nearest-neighbour, so a start far inside a long chain does not feel its ends, and the computation is done on $x\in\mathbb Z$. $O(\varepsilon^2)$ is a second-order remainder at fixed $\lambda$ and start.

**Plane waves, branches, start.** $\lvert p\rangle:=\sum_xe^{ipx}\lvert x\rangle$. The block with all slopes and curvatures zero is $C_0=\int\frac{\mathrm dp}{2\pi}\lvert p\rangle\langle p\rvert\otimes h(p)$ with

$$
h(p)=M\sigma_3-\tau\sin p\,\sigma_1,\qquad h\beta_\pm=\pm\omega\beta_\pm,\qquad \omega(p)=\sqrt{M^2+\tau^2\sin^2p}\ \ge M>0 .
\tag{15.1}
$$

$h$ is real symmetric, so $\beta_+=(\cos\frac\theta2,\sin\frac\theta2)$, $\beta_-=(-\sin\frac\theta2,\cos\frac\theta2)$ with $\tan\theta=-\tau\sin p/M$, $\lvert\theta\rvert<\pi/2$, are real, smooth and $2\pi$-periodic. The start is $\psi_0=B_{k,+}[g]=\int\frac{\mathrm dp}{2\pi}g(p)\lvert p\rangle\otimes\beta_+(p)\otimes\lvert f_k\rangle$, $g$ smooth, $2\pi$-periodic, $\int\frac{\mathrm dp}{2\pi}\lvert g\rvert^2=1$; another phase convention for $\beta_+$ multiplies $g$ by a $p$-dependent phase, and all formulas refer to the real convention. $\langle F\rangle_g:=\int\frac{\mathrm dp}{2\pi}\lvert g\rvert^2F$. At rest: $\langle p\rangle_g=0$ on $(-\pi,\pi]$; $\delta p^2:=\langle p^2\rangle_g$, $\kappa_4:=\langle p^4\rangle_g/\delta p^4$. The start is not adapted to the profiles (M9).

**Inputs** (04-branch-fall@v3, quoted in the question; here $g_{\zeta,k}=0$, $\sigma=+$). (4.7): the coefficient of $\lambda$ in the first-order velocity of $B_{k,+}[g]$ is $\langle a_{k,+}\rangle_g$, $a_{k,+}(p)=-\frac{G_kM_k\tau_k^2\cos2p}{\omega_k^2}+\mathcal F_k\frac{\tau_k^2(M_k^2\cos2p-\tau_k^2\sin^4p)}{\omega_k^3}$. (4.11): $\lvert\langle a_{k,\sigma}\rangle_g-a_{k,\sigma}(p_0)\rvert\le\frac12\sup_p\lvert a_{k,\sigma}''\rvert\,\delta p^2$ for a start concentrated at $p_0$, and drift acceleration $=\lim_{\delta p\to0}\lim_{\Lambda\to\infty}\overline{a^{(1)}}=a_{k,\sigma}(p_0)$. (4.16): $a_{k,\sigma}(0)=\frac{\tau_k^2}{M_k}(\sigma\mathcal F_k-G_k)$. (4.19): for linear profiles, $a_{k,+}(0)=a_+\ \forall k\iff\frac{\tau_k^2}{M_k}(\mathcal F_k-G_k)=a_+\ \forall k$.

## Derivation

### Step 1. Branch representation

Write a state as $\int\frac{\mathrm dp}{2\pi}\lvert p\rangle\otimes\bigl(u_+(p)\beta_+(p)+u_-(p)\beta_-(p)\bigr)$, with Pauli matrices $\rho_{1,2,3}$ on $(u_+,u_-)$. Since $x\,e^{ipx}=-i\partial_pe^{ipx}$, integration by parts on the circle gives $\hat x=i\partial_p$ on the amplitude $u_+\beta_++u_-\beta_-$; with $\partial_p\beta_+=\alpha\beta_-$, $\partial_p\beta_-=-\alpha\beta_+$, $\alpha:=\theta'/2$, and the expansion of $\sigma_3$ in $\beta_\pm$:

$$
\hat x=\xi+\alpha\rho_2,\quad \xi:=i\partial_p,\quad \sigma_3=\mu\rho_3+\nu\rho_1,\quad C_0=\omega\rho_3,\qquad
\alpha=-\frac{\tau M\cos p}{2\omega^2},\ \ \mu=\frac M\omega,\ \ \nu=\frac{\tau\sin p}{\omega}.
\tag{15.2}
$$

Let $\eta:=\xi-x_0$, so $\hat y:=\hat x-x_0=\eta+\alpha\rho_2$. For a function $f(p)$: $[\eta,f]=if'$ and $e^{i\omega\lambda}\eta e^{-i\omega\lambda}=\eta+\omega'\lambda$. The start has $u_+=g$, $u_-=0$; centred means $\langle g\vert\eta\vert g\rangle=0$, and $W^2=\langle\psi_0\vert\hat y^2\vert\psi_0\rangle=\lVert\eta g\rVert^2+\langle\alpha^2\rangle_g\ge\lVert\eta g\rVert^2$. The position–momentum density of the start is

$$
j(p):=\operatorname{Re}\bigl[g^*(\eta g)\bigr]=-\operatorname{Im}(g^*\partial_pg)-x_0\lvert g\rvert^2,\qquad \int\tfrac{\mathrm dp}{2\pi}\,j=0 .
\tag{15.3}
$$

At zero order $u_-$ stays $0$ and $u_+(\lambda)=u_0(\lambda):=e^{-i\omega\lambda}g$, so $\bar x_{\rm lab}=x_0+\lambda\langle\omega'\rangle_g$ and the zero-order acceleration vanishes.

### Step 2. The profile term in the branch representation

$C_1:=C^{(k)}-C_0=\delta M(\hat y)\sigma_3+V(\hat y)$ with $\delta M(y)=Gy+\frac12Hy^2$, $V(y)=-\mathcal Fy-\frac12\mathcal Hy^2$ (diagonal in $x$, hence functions of $\hat x$). As $\hat y$ and $\sigma_3$ commute, $\hat y\sigma_3=\frac12\{\hat y,\sigma_3\}$ and $\hat y^2\sigma_3=\hat y\sigma_3\hat y$. Multiplying the blocks $\hat y_{\pm\pm}=\eta$, $\hat y_{+-}=-i\alpha$, $\hat y_{-+}=i\alpha$, $(\sigma_3)_{\pm\pm}=\pm\mu$, $(\sigma_3)_{\pm\mp}=\nu$, and using $[\eta,f]=if'$:

$$
\begin{aligned}
(C_1)_{++}&=\tfrac G2\{\eta,\mu\}+\tfrac H2\,\eta\mu\eta-\mathcal F\eta-\tfrac{\mathcal H}2\eta^2+Q,\qquad Q:=-\tfrac H2\bigl[(\nu\alpha)'+\mu\alpha^2\bigr]-\tfrac{\mathcal H}2\alpha^2,\\
(C_1)_{-+}&=\tfrac G2\{\eta,\nu\}+\tfrac H2\bigl[\eta\nu\eta-\nu\alpha^2+(\alpha\mu)'\bigr]-i\mathcal F\alpha-\tfrac{i\mathcal H}2\{\alpha,\eta\},\qquad (C_1)_{+-}=(C_1)_{-+}^\dagger .
\end{aligned}
\tag{15.4}
$$

### Step 3. Block diagonalization at first order and the oscillating terms

Define the anti-self-adjoint, branch-off-diagonal $S$ by

$$
S_{+-}:=\int_0^\infty\!\mathrm ds\;e^{-s\omega}(C_1)_{+-}e^{-s\omega},\qquad S_{-+}:=-S_{+-}^\dagger .
\tag{15.5}
$$

The integral converges since $\omega\ge M>0$, and integrating $\partial_s(e^{-s\omega}Xe^{-s\omega})$ gives $\omega S_{+-}+S_{+-}\omega=(C_1)_{+-}$. Hence $[S,C_0]$ cancels the off-diagonal part of $C_1$, and $C':=e^SC^{(k)}e^{-S}=C_0+C_1^{\rm d}+O(\varepsilon^2)$, $C_1^{\rm d}$ the diagonal blocks. Write $\bar x_{\rm lab}=\langle\Psi'(\lambda)\vert\hat x'\vert\Psi'(\lambda)\rangle$ with $\Psi'(\lambda)=e^{-iC'\lambda}e^S\psi_0$, $\hat x'=e^S\hat xe^{-S}$. At first order $\Psi'_+=e^{-iE\lambda}g$ with $E:=\omega+(C_1)_{++}$, $\Psi'_-=e^{i\omega\lambda}S_{-+}g$, $\hat x'_{++}=\xi+i(S_{+-}\alpha+\alpha S_{-+})$, $\hat x'_{+-}=-i\alpha$. Hence

$$
\bar x_{\rm lab}(\lambda)=\langle e^{-iE\lambda}g\vert\xi\vert e^{-iE\lambda}g\rangle+P(\lambda)+X_{\rm osc}(\lambda)+O(\varepsilon^2),\qquad
X_{\rm osc}:=2\operatorname{Re}\!\int\!\tfrac{\mathrm dp}{2\pi}\,e^{2i\omega(p)\lambda}f(p),\quad f:=-i\alpha\,g^*\,(S_{-+}g),
\tag{15.6}
$$

with $P(\lambda):=\langle u_0(\lambda)\vert i(S_{+-}\alpha+\alpha S_{-+})\vert u_0(\lambda)\rangle$. The only part of $(C_1)_{+-}$ of degree 2 in $\eta$ is $\frac H2\eta\nu\eta$; by $e^{-s\omega}\eta=(\eta+is\omega')e^{-s\omega}$ the degree-2 part of $S_{+-}$ is $\frac H4\eta\frac\nu\omega\eta$, which enters $S_{+-}\alpha-\alpha S_{+-}^\dagger$ as $\frac H4[\eta\frac\nu\omega\eta,\alpha]=\frac{iH}4\{\eta,\frac{\nu\alpha'}{\omega}\}$, of degree 1. An operator of degree $\le1$ in $\eta$ has, under $e^{i\omega\lambda}\eta e^{-i\omega\lambda}=\eta+\omega'\lambda$, an expectation of degree $\le1$ in $\lambda$: $P''=0$. Every term of $X_{\rm osc}$ carries $e^{2i\omega(p)\lambda}$ under the $p$-integral; these are the **oscillating terms**, with acceleration $A_{\rm osc}:=X_{\rm osc}''=-8\operatorname{Re}\int\frac{\mathrm dp}{2\pi}\omega^2e^{2i\omega\lambda}f$, bounded uniformly in $\lambda$ by $8\int\frac{\mathrm dp}{2\pi}\omega^2\lvert f\rvert$, and $\lvert X_{\rm osc}'\rvert\le v_{\rm osc}:=4\int\frac{\mathrm dp}{2\pi}\omega\lvert f\rvert$.

### Step 4. Acceleration operator of the upper branch

Under $E$, $\frac{\mathrm d^2}{\mathrm d\lambda^2}\langle\xi\rangle=\langle i[E,i[E,\xi]]\rangle$. With $i[f,\eta]=f'$, (15.4) gives $i[E,\xi]=\omega'+\frac G2\{\eta,\mu'\}+\frac H2\eta\mu'\eta+Q'$. Since $i[\omega,\omega']=0$, the first-order part is $i[\omega,\frac G2\{\eta,\mu'\}+\frac H2\eta\mu'\eta]+i[(C_1)_{++},\omega']=G\omega'\mu'+\frac H2\{\omega'\mu',\eta\}-G\mu\omega''-\frac H2\{\mu\omega'',\eta\}+\mathcal F\omega''+\frac{\mathcal H}2\{\omega'',\eta\}$ ($Q$ shifts the velocity by $Q'$ and drops out). Thus

$$
\hat A=a(p)+\tfrac12\{b(p),\eta\},\qquad a=\mathcal F\omega''+G(\omega'\mu'-\mu\omega''),\quad b=\mathcal H\omega''+H(\omega'\mu'-\mu\omega'').
\tag{15.7}
$$

With $\omega''=\tau^2(M^2\cos2p-\tau^2\sin^4p)/\omega^3$ and $\omega'\mu'-\mu\omega''=-M\tau^2\cos2p/\omega^2$:

$$
a(p)=-\frac{GM\tau^2\cos2p}{\omega^2}+\mathcal F\,\frac{\tau^2(M^2\cos2p-\tau^2\sin^4p)}{\omega^3},\qquad b(p)=-\frac{HM\tau^2\cos2p}{\omega^2}+\mathcal H\,\frac{\tau^2(M^2\cos2p-\tau^2\sin^4p)}{\omega^3}.
\tag{15.8}
$$

$a=a_{k,+}$ of (4.7) at $g_{\zeta,k}=0$, and $b$ is the same function of the curvatures; $\hat A$ is the acceleration of the linear profiles with the local slopes $G+H\eta$, $\mathcal F+\mathcal H\eta$, symmetrized.

### Step 5. First-order acceleration

$\hat A$ is of first order, so it is evaluated in $u_0(\lambda)$: $e^{i\omega\lambda}\eta e^{-i\omega\lambda}=\eta+\omega'\lambda$ and $\frac12\langle g\vert\{b,\eta\}\vert g\rangle=\operatorname{Re}\langle bg\vert\eta g\rangle=\int\frac{\mathrm dp}{2\pi}bj$. With (15.6) and $P''=0$:

$$
\frac{\mathrm d^2\bar x_{\rm lab}}{\mathrm d\lambda^2}=\underbrace{\langle a\rangle_g+c_g+\lambda\,\langle b\,\omega'\rangle_g}_{\bar A(\lambda)}+A_{\rm osc}(\lambda)+O(\varepsilon^2),\qquad c_g:=\int\tfrac{\mathrm dp}{2\pi}\,b\,j .
\tag{15.9}
$$

### Step 6. The curvature part

The curvatures contribute $c_g+\lambda\langle b\omega'\rangle_g$ to $\bar A$, and to $A_{\rm osc}$ through the $H,\mathcal H$ terms of $S_{-+}$. $b$ is even in $p$ and $\omega'$ odd, and $b(0)=\frac{\tau^2}{M}(\mathcal H-H)$ by (4.16) applied to $b$. Since $\int j=0$, $c_g=\operatorname{Re}\langle(b-b(0))g\vert\eta g\rangle$, $\lvert b-b(0)\rvert\le\frac12b_2p^2$ and Cauchy–Schwarz give the first bound; at rest, $\langle b\omega'\rangle_g=\langle b\omega'-(b\omega')'(0)p\rangle_g$, Taylor's remainder for the odd $b\omega'$ and $\langle\lvert p\rvert^3\rangle_g\le\langle p^2\rangle_g^{1/2}\langle p^4\rangle_g^{1/2}$ give the second:

$$
\lvert c_g\rvert\le\tfrac12\,b_2\,\kappa_4^{1/2}\,\delta p^2\,W,\qquad
\lvert\langle b\omega'\rangle_g\rvert\le\tfrac16\,b_3\,\kappa_4^{1/2}\,\delta p^3,\qquad b_2:=\sup_p\lvert b''\rvert,\ b_3:=\sup_p\lvert(b\omega')'''\rvert .
\tag{15.10}
$$

$b_2,b_3$ are linear in $(H,\mathcal H)$ with coefficients depending only on $\tau,M$. Exactly, $c_g=0$ if $j\equiv0$ (e.g. $g=re^{-ipx_0}$, $r$ real), and $\langle b\omega'\rangle_g=0$ if $\lvert g\rvert^2$ is even.

### Step 7. Local slope

Average (15.9) over a window $[\lambda_1,\lambda_1+\Lambda]$ with centre $\lambda_c$. The non-oscillating part, linear in $\lambda$, averages to $\bar A(\lambda_c)$; the oscillating part to $[X_{\rm osc}']/\Lambda$, at most $2v_{\rm osc}/\Lambda$. With (4.11) at $p_0=0$ (the mean of a start at rest), $a(0)=\frac{\tau^2}M(\mathcal F-G)$ by (4.16), and (15.10), the window average $\bar A_\Lambda$ of the first-order acceleration obeys

$$
\bigl\lvert\bar A_\Lambda-a(0)\bigr\rvert\le\tfrac12a_2\,\delta p^2+\tfrac12b_2\,\kappa_4^{1/2}\delta p^2\,W+\tfrac16b_3\,\kappa_4^{1/2}\delta p^3\lvert\lambda_c\rvert+\frac{2v_{\rm osc}}{\Lambda},\qquad a_2:=\sup_p\lvert a''\rvert .
\tag{15.11}
$$

A relative error $\epsilon$ ($a(0)\neq0$) holds if each term is at most $\epsilon\lvert a(0)\rvert/4$. At leading order in $\tau/M$ (with $\mathcal F\ne G$, $\mathcal H\ne H$), $a\simeq a(0)\cos2p$, $b\simeq\frac{\tau^2}M(\mathcal H-H)\cos2p$, $b\omega'\simeq\frac{\tau^4}{4M^2}(\mathcal H-H)\sin4p$, and the four relative terms become $2\delta p^2$, $2\kappa_4^{1/2}\delta p^2W\frac{\lvert\mathcal H-H\rvert}{\lvert\mathcal F-G\rvert}$, $\frac83\kappa_4^{1/2}\frac{\tau^2}M\delta p^3\lvert\lambda_c\rvert\frac{\lvert\mathcal H-H\rvert}{\lvert\mathcal F-G\rvert}$, $\frac{2v_{\rm osc}}{\Lambda\lvert a(0)\rvert}$.

Limits. For windows $[\lambda_1,\lambda_1+\Lambda]$ at fixed start, $\bar A_\Lambda=\langle a\rangle_g+c_g+(\lambda_1+\frac\Lambda2)\langle b\omega'\rangle_g+O(\Lambda^{-1})$, so $\lim_{\Lambda\to\infty}\bar A_\Lambda$ exists iff $\langle b\omega'\rangle_g=0$. Hence:

$$
\lvert g\rvert^2\ \text{even}:\quad \lim_{\delta p\to0}\ \lim_{\Lambda\to\infty}\bar A_\Lambda=\frac{\tau^2}{M}(\mathcal F-G)\quad\text{along families with }\kappa_4\ \text{bounded and}\ \delta p^2W\to0 ,
\tag{15.12}
$$

taken after first order and the long chain, in the order of (4.11); if also $j\equiv0$, $c_g=0$ and the curvatures drop out of $\bar A$ at every $\delta p$. If $\langle b\omega'\rangle_g\ne0$ the inner limit does not exist; a joint limit ($\lambda_1$ fixed) with $\delta p_n\to0$, $\kappa_4$ bounded, $\delta p_n^2W_n\to0$, $\Lambda_n\delta p_n^3\to0$ and $v_{\rm osc}(g_n)/\Lambda_n\to0$ gives the same value, and such $\Lambda_n$ exist whenever $v_{\rm osc}(g_n)\delta p_n^3\to0$ (e.g. $\Lambda_n=(v_{\rm osc}/\delta p_n^3)^{1/2}$ when $v_{\rm osc}>0$).

### Step 8. Level independence

To avoid a clash with the start function $g$, write $n_x=1+g_ny+\frac12h_ny^2$ (the question's $g,h$), and compare starts $B_{k,+}[g]$, $B_{l,+}[g]$ with the same $g$ ($j$ does not depend on the level). Let $\bar A_k$ be $\bar A$ of (15.9) on level $k$.

(a) Factor class: $M_k=m+d_k$, $G_k=g_nM_k$, $H_k=h_nM_k$, $\mathcal F_k=\mathcal H_k=0$, $\tau_k=\tau$. With $r_k:=\tau^2/M_k^2$, (15.8) gives $a_k=-g_n\tau^2\cos2p/(1+r_k\sin^2p)$, $b_k=-h_n\tau^2\cos2p/(1+r_k\sin^2p)$, $b_k\omega_k'=-\frac{h_n\tau^4}{M_k}\sin p\cos p\cos2p\,(1+r_k\sin^2p)^{-3/2}$, hence

$$
\bar A_k-\bar A_l=\tau^2(r_k-r_l)\Bigl[g_n\langle\Delta_{kl}\rangle_g+h_n\!\int\!\tfrac{\mathrm dp}{2\pi}\Delta_{kl}\,j\Bigr]+\lambda\bigl(\langle b_k\omega_k'\rangle_g-\langle b_l\omega_l'\rangle_g\bigr),\qquad \Delta_{kl}:=\frac{\sin^2p\,\cos2p}{(1+r_k\sin^2p)(1+r_l\sin^2p)} .
\tag{15.13}
$$

$r_k\ne r_l$ for $k\ne l$. For $g_n\ne0$, take $g=re^{-ipx_0}$ with $r$ real, even, smooth, supported in $\lvert p\rvert<\pi/4$, $\delta p>0$: then $j\equiv0$, the $\lambda$-term vanishes, $\Delta_{kl}>0$ for $0<\lvert p\rvert<\pi/4$, so $\bar A_k-\bar A_l=g_n\tau^2(r_k-r_l)\langle\Delta_{kl}\rangle_g\ne0$, and $\langle\Delta_{kl}\rangle_g=\delta p^2+O(\langle p^4\rangle_g)$ gives the leading difference $g_n\tau^4(M_k^{-2}-M_l^{-2})\,\delta p^2$. For $g_n=0\ne h_n$, take $g=re^{-ipx_0-i\gamma(p^3/3-\delta p^2p)}$, $\gamma\ne0$, with $r$ as before and supported where $\Delta_{kl}$ increases with $\lvert p\rvert$ (a neighbourhood of $0$): it is at rest and centred, $j=\gamma(p^2-\delta p^2)\lvert g\rvert^2$, and $\int\Delta_{kl}j=\gamma\,\mathrm{Cov}_g(\Delta_{kl},p^2)\ne0$, since both are increasing functions of $\lvert p\rvert$. The reason is that the hopping is not multiplied by $n_x$: $\omega_k/M_k=(1+r_k\sin^2p)^{1/2}$ depends on the level except at $p=0$. In the limit (15.12), $\bar A_k\to a_k(0)=-g_n\tau^2$ for every $k$, the case $a_+=-g_n\tau^2$ of (4.19).

(b) Control: $M_k=m+d_k$, $G_k=mg_n$, $H_k=mh_n$, rest as in (a). By (4.16), $a_k(0)=-\tau^2mg_n/(m+d_k)$, so already in the limit (15.12)

$$
a_k(0)-a_l(0)=\frac{\tau^2m\,g_n\,(d_k-d_l)}{(m+d_k)(m+d_l)}\ \ne0\quad(g_n\ne0,\ k\ne l),
\tag{15.14}
$$

of zeroth order in $\delta p$ ($\simeq\tau^2g_n(d_k-d_l)/m$ for $d_k,d_l\ll m$); $b_k(0)=-\tau^2mh_n/(m+d_k)$ makes the curvature terms level-dependent as well.

## Result

All results: one body in the weak-record regime, conserved level, first order jointly in the slopes and curvatures at fixed $\lambda$, start and chain, then a long chain.

- **(15.7)–(15.8)** (formula). For every level block with quadratic level profiles ($\zeta=1$), the first-order acceleration operator of the upper branch is $\hat A=a(p)+\frac12\{b(p),\eta\}$: $a$ is $a_{k,+}$ of (4.7) at $g_{\zeta,k}=0$, and $b$ is the same function with $(G,\mathcal F)\to(H,\mathcal H)$, i.e. the linear-profile acceleration with the local slopes $G+H\eta$, $\mathcal F+\mathcal H\eta$.
- **(15.9)** (formula). For every such block and every start $B_{k,+}[g]$ centred at $x_0$ with finite $W$: $\mathrm d^2\bar x_{\rm lab}/\mathrm d\lambda^2=\langle a\rangle_g+c_g+\lambda\langle b\omega'\rangle_g+A_{\rm osc}(\lambda)+O(\varepsilon^2)$, with $c_g=\int\frac{\mathrm dp}{2\pi}bj$, $j$ of (15.3), and $A_{\rm osc}=-8\operatorname{Re}\int\frac{\mathrm dp}{2\pi}\omega^2e^{2i\omega\lambda}f$, $f$ of (15.6), bounded in $\lambda$. The non-oscillating part is exactly linear in $\lambda$. No control inside the class (it is an identity at first order).
- **(15.10)** (theorem). The curvatures contribute $c_g+\lambda\langle b\omega'\rangle_g$ to the non-oscillating part, and a part of $A_{\rm osc}$ bounded in $\lambda$; there is no term in $W^2$. The contribution grows with $\lambda$, linearly (power 1), iff $\langle b\omega'\rangle_g\ne0$. For starts at rest, $\lvert c_g\rvert\le\frac12b_2\kappa_4^{1/2}\delta p^2W$ and $\lvert\langle b\omega'\rangle_g\rvert\le\frac16b_3\kappa_4^{1/2}\delta p^3$; $c_g=0$ if $j\equiv0$, $\langle b\omega'\rangle_g=0$ if $\lvert g\rvert^2$ is even. Control: a start at rest with uneven $\lvert g\rvert^2$ and $\langle b\omega'\rangle_g\ne0$ shows the growth.
- **(15.11)** (criterion). For starts at rest centred at $x_0$, the average of the first-order acceleration over a window of length $\Lambda$ and centre $\lambda_c$ differs from $a(0)=\frac{\tau^2}M(\mathcal F-G)$ by at most $\frac12a_2\delta p^2+\frac12b_2\kappa_4^{1/2}\delta p^2W+\frac16b_3\kappa_4^{1/2}\delta p^3\lvert\lambda_c\rvert+2v_{\rm osc}/\Lambda$; for $a(0)\ne0$ the relative error is at most $\epsilon$ if each term is $\le\epsilon\lvert a(0)\rvert/4$ (sufficient). At leading order in $\tau/M$, with $\mathcal F\ne G$ and $\mathcal H\ne H$, the curvature conditions read $\kappa_4^{1/2}\delta p^2W\lvert\mathcal H-H\rvert\ll\lvert\mathcal F-G\rvert$ and $\kappa_4^{1/2}\frac{\tau^2}M\delta p^3\lvert\lambda_c\rvert\lvert\mathcal H-H\rvert\ll\lvert\mathcal F-G\rvert$.
- **(15.12)** (theorem). For starts at rest centred at $x_0$ with even $\lvert g\rvert^2$: $\lim_{\delta p\to0}\lim_{\Lambda\to\infty}\bar A_\Lambda=\frac{\tau^2}M(\mathcal F-G)$ exactly at first order, along families with $\kappa_4$ bounded and $\delta p^2W\to0$, the limits taken after first order and the long chain, window before $\delta p$. For windows $[\lambda_1,\lambda_1+\Lambda]$ with fixed $\lambda_1$ and a fixed start, $\lim_{\Lambda\to\infty}\bar A_\Lambda$ exists iff $\langle b\omega'\rangle_g=0$; otherwise a joint limit that adds $\Lambda_n\delta p_n^3\to0$ and $v_{\rm osc}(g_n)/\Lambda_n\to0$ to the conditions above gives the same value.
- **(15.13)** (negative). Factor class, $N\ge2$: for $g_n\ne0$ or $h_n\ne0$ there are starts of the class for which the non-oscillating first-order acceleration differs between two levels; the difference is (15.13), with leading term $g_n\tau^4(M_k^{-2}-M_l^{-2})\delta p^2$ for a start with $j\equiv0$ and even $\lvert g\rvert^2$. Reason: $\tau_k=\tau$ is not multiplied by $n_x$. In the limit (15.12) it is level-independent, $-g_n\tau^2$.
- **(15.14)** (negative). Control $M_{x,k}=mn_x+d_k$: not the same for all levels for $g_n\ne0$, already in the limit (15.12), with difference $\tau^2mg_n(d_k-d_l)/((m+d_k)(m+d_l))$, of zeroth order in $\delta p$.

## Consistency checks

1. Dimensions: $\lambda\sim[C]^{-1}$, $x$ dimensionless, $\tau,M,G,\mathcal F,H,\mathcal H\sim[C]$; $a$, $c_g$ and $\lambda\langle b\omega'\rangle_g$ are all $[C]^2$, as is $\mathrm d^2\bar x_{\rm lab}/\mathrm d\lambda^2$. Passed.
2. Zero curvatures: $b=0$, so (15.9) gives a constant non-oscillating acceleration $\langle a_{k,+}\rangle_g$, i.e. the coefficient of $\lambda$ in the velocity of (4.7), and (15.12) reduces to (4.11) with $a(0)$ of (4.16). Passed.
3. $\tau\to0$: $\omega'=\omega''=\mu'=\alpha=0$, so $a=b=f=0$ and the first-order acceleration vanishes; exactly, $[C^{(k)},\hat x]=0$ at $\tau=0$, so $\bar x_{\rm lab}$ is constant. Passed.

## Open issues

- The oscillating amplitude $f$ is given through $S_{-+}$ (15.5), not as explicit averages; $v_{\rm osc}$ depends on the position spread of the start (through $\eta g$, $\eta^2g$) and sets the window length, but is not bounded explicitly.
- First order is not uniform in $\lambda$ (the non-oscillating $\bar x_{\rm lab}$ contains $\lambda^3\langle b\omega'\rangle_g/6$); long falls and second order are out of scope.
- In the factor class the level dependence at $\delta p>0$ comes from the unscaled hopping; whether a common factor on the hopping ($\zeta_{x,k}\ne1$) removes it is out of scope.

## Methods used

- plane waves and branch (band) representation
- first-order block diagonalization (Schrieffer–Wolff)
- Heisenberg equations and Duhamel expansion
- Taylor remainders and Cauchy–Schwarz bounds
- window averaging of oscillating terms

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