Version 1 · earlier version; the current one is v4 · External review, round 1: major errors

# Inertia and weight of a pair bound through the field, full grip

- **Subproject:** 07-ilang-weight
- **Package:** 07-inertia-of-pair-full-grip
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-21

## Setup and assumptions

**Description (M8).** The field, the two bodies, the full grip and the field state of `question.md`: ring of even length $L\ge6$, $N_f\ge1$ field objects with own term $C_1$, $n=N_f/L$; Model D bodies $b_1,b_2$ of different types, own contract the motion term $A$ with $\tau_i$, rest and clock term only in the pair term $\gamma_i\sum_x\lvert x\rangle\langle x\rvert\otimes\sigma_3\otimes(m_i+D_i)\otimes\Phi_x$; no pair term between $b_1,b_2$; records at $\kappa\to0$. $D_i$ commutes with every term, so the clock level $k_i$ is conserved and on branch $\sigma$ the pair term is $\sigma\gamma_iM_i\Phi_x$; $g_i=\sigma_i\gamma_iM_i$ for the selected branches.

**Inputs** (quoted in `question.md`): (2.1), (2.10) of 07-ilang-weight/02-two-free-bodies; (4.2) of 06-ilang-source/04-profile; (4.2), (4.5), (4.7) of 07-ilang-weight/04-weight-of-bond (ring, $\mathcal X=L$); (5.1), (5.3), (5.4), (5.7), (5.14) of 07-ilang-weight/05-inertia-of-mark (FG: $\theta=0$, $K=T_-$, $v=\tau$); (6.1), (6.5), (6.6), (6.8), (6.9), (6.13) of 07-ilang-weight/06-inertia-of-pair. Their notation is used unchanged ($\ell,\ell_0,\ell^{(2)},\phi,B,\eta,\mu_i,\kappa_i,\bar\kappa_i,\bar t_0,d_i,U,f_0,b_0,\beta_0,m,\Delta E_0,K_0,\langle\cdot\rangle_0,W_{\rm pair}$). On $0\le x\le L$, $\ell(x)=\ell_0-\frac{x(L-x)}{2L}$, $\ell_0=\frac{L^2-1}{12L}$ (it solves $\Lambda\ell=\delta_0-\frac1L$, $\sum_x\ell=0$, by substitution). $F:=ng_1g_2/t_f$; $'$ is the central difference in the place.

**Assumptions** (M8, M9; each a choice of the package).
- **S1** Field state: the lowest state adapted to the bodies (A11): in the group $\mathcal G$ every field object is in the lowest one-object state of the held configuration.
- **S2** Band regime: $\max_i\tau_i\ll d$, with $d\simeq\min\big(2n\lvert g_1\rvert,2n\lvert g_2\rvert,\varepsilon_1-2n(g_1^++g_2^+)\big)$, $g^+:=\max(g,0)$, the distance of $\mathcal G$ to the other held levels ((5.4) for each body and for the joint flip, leading order in $\eta$). Omitted: relative $O(\tau^2/d^2)$.
- **S3** Field: $\eta=\max_i\lvert g_i\rvert\ell_0/t_f\ll1$ and $N_f\eta\ll1$. Omitted: relative $O(\eta)$ in mutual terms, $O(N_f\eta^2)$ in field overlaps.
- **S4** Binding: $w\ll L$, $w:=(4\lvert\bar t_0\rvert/\lvert F\rvert)^{1/3}$. Omitted: the flux term of (6.6) (exponentially small in $(L/w)^{3/2}$) and $\langle V\rangle_0/\lvert B(0)\rvert=O(w/L)$.
- **S5** Regime of item 3 (ii): $1-\beta_0\ll\eta$ (hence $w\gg1$).
- **S6** Source: a held source $g_s$ at $x_s$, weights to first order in $g_s$; pair centre $X$ at distance $s$ from $x_s$, $w\ll s\le L/2$, $u:=s/L$.
- Out of the class (tuned): $\lvert\tau_1^2g_2+\tau_2^2g_1\rvert$ small against $\max(\tau_1^2\lvert g_2\rvert,\tau_2^2\lvert g_1\rvert)$ (this is $\mu_1^{-1}+\mu_2^{-1}$ small, by Step 1); in item 4, $\lvert g_1+g_2\rvert$ small against $\max\lvert g_i\rvert$.

## Derivation

### Step 1. Held levels, half-gaps and odd parts

By S1, non-interacting objects in the lowest state all occupy the lowest one-object level, so the held level of $\mathcal G$ is $E(j,l)=N_f\varepsilon_0(g_1\,\text{at}\,j,\,g_2\,\text{at}\,l)$ and the mutual term $M(r)$, $r=l-j$, is (4.2) of 06-ilang-source/04-profile. Define, with the other coupling held fixed,
$$\sigma_iR_i:=\tfrac{N_f}2\big[\varepsilon_0(g_i)-\varepsilon_0(-g_i)\big],\quad c_i^2:=\frac{\sigma_iR_i}{\mu_i},\quad M_i^{o}(r):=\tfrac12\big[M(g_i)-M(-g_i)\big](r),\quad M_i^e:=M-M_i^o,\quad \bar f(\rho):=\tfrac14\big[f(\rho-1)+2f(\rho)+f(\rho+1)\big].\tag{7.1}$$
$\sigma_iR_i$ is the branch-odd part of the cost (5.3) ($\theta=0$). Rayleigh–Schrödinger perturbation of $\varepsilon_0$ in $\sum_ag_a\delta_{x_a}$, with $\varepsilon_q-\varepsilon_0=2t_f(1-\cos q)$ and $\frac1L\sum_{q\ne0}\frac{e^{iqx}}{2(1-\cos q)}=\ell(x)$, gives $N_f\varepsilon_0=-2N_ft_f+n\sum_ag_a-\frac n{t_f}\sum_{a,b}g_ag_b\ell(x_a-x_b)+O(n\lvert g\rvert\eta^2)$. Hence $\sigma_iR_i=ng_i[1+O(\eta^2)]$; $M=B(r)[1+O(\eta)]$ with $B$ of (4.2) of 04-weight-of-bond, and since $B\propto g_1g_2$ is odd in each coupling, $M_1^o=M_2^o=B[1+O(\eta)]$, $M_i^e=O(\eta B)$. In (5.3) the field state of the flipped branch differs at first order in $g_i$ by a vector of squared norm $O(\eta^2)$ per object, so $c_i^2=\tau_i^2[1+O(N_f\eta^2)]$ and $\mu_i=ng_i\tau_i^{-2}[1+O(N_f\eta^2)]$; in particular $\operatorname{sign}\mu_i=\operatorname{sign}g_i$.

### Step 2. Band contract and selected level (item 1)

The derivation of (2.10) uses only that $\mathcal G$ is isolated and that each motion term moves one body by one place and flips its branch; the full grip changes only the held levels ($\theta=0$ in (5.1)). So (2.10) holds under S2 with error $O(\tau^4/d^3)$, and with (2.1) the ansatz $\psi(j,l)=e^{iPj}g(l-j)$ gives (6.1) unchanged (the double step of $b_1$ carries the relative field coordinates: phase $e^{-2iP}$). Each motion term conserves $x_i$ plus the branch index mod 2, so the parity of $r$ is conserved in $\mathcal G$; each parity class has its own selected level and everything below holds for either.

*Amplitudes.* The double step of $b_i$ across the relative bond $\rho$ passes through the held level with $b_i$ flipped at relative place $\rho$. Since $M(-g_i)=-M_i^o+M_i^e$, the gap from the held level at $r=\rho\mp1$ is
$$\Delta_i(r;\rho)=2\sigma_iR_i+M_i^o(r)+M_i^o(\rho)+M_i^e(r)-M_i^e(\rho).\tag{7.2}$$
Second-order Schrieffer–Wolff with symmetrized denominators gives $\kappa_i(\rho)=-\frac{\tau_i^2a_i^2}8\big[\Delta_i(\rho-1;\rho)^{-1}+\Delta_i(\rho+1;\rho)^{-1}\big]$, up to the excited-state term of (5.3); for $M=0$ this is $\bar\kappa_i=-1/(8\mu_i)$ of (5.3). Expanding in $M/\sigma_iR_i=O(\eta)$ (the $M_i^e$ terms combine to a second difference, negligible), the **local inertia** of $b_i$ is
$$m_i(\rho):=-\frac1{8\kappa_i(\rho)}=\mu_i+\frac{\bar M_i^o(\rho)}{c_i^2}+O(\lvert\mu_i\rvert\eta^2)=\mu_i+\frac{\bar B(\rho)}{\tau_i^2}\big[1+O(\eta)\big].\tag{7.3}$$
The partner's mark changes each body's local inertia by the bond average of the mutual term over that body's own $\tau_i^2$. The same gaps give $d_i(r)=-2\bar\kappa_i[1-\bar M_i^o(r)/\sigma_iR_i]$. (6.5) and (6.6) hold: they use only the form (6.1), a real $f_0$ at $P=0$, and Hellmann–Feynman in the bond magnitudes and the flux.

*Bound level.* $U(r)-U(0)=V(r)[1+O(\eta,\tau^2/d^2)]$ with $V(r):=B(r)-B(0)=F\,r(L-r)/L$ on $0\le r\le L$. Near the edge $P=0$, $k=0$ of the relative band, (6.1) has relative hop $\bar t_0=-\frac18(\mu_1^{-1}+\mu_2^{-1})$: the selected level is the lowest if $\bar t_0<0$, the highest if $\bar t_0>0$. It is bound at $r\approx0$ iff the extremum of $V$ at $r=0$ pushes it out of the band: a minimum ($F>0$) for the lowest, a maximum ($F<0$) for the highest, i.e. $g_1g_2(\mu_1^{-1}+\mu_2^{-1})>0$. With Step 1:
$$\text{selected level bound}\iff\tau_1^2g_2+\tau_2^2g_1>0,\qquad g_i=\sigma_i\gamma_iM_i .\tag{7.4}$$
Both $g_i>0$: bound, lowest. Both $g_i<0$: not bound. Opposite signs: bound (highest) iff the inequality holds. Otherwise the extreme level sits at $r\approx L/2$, where $V$ has its other extremum (control: case D of the code). In the continuum limit of (6.1), $-4\lvert\bar t_0\rvert\partial_r^2+\lvert F\rvert\lvert r\rvert$, the width is $w$ of S4 and $1-\beta_0=O(w^{-2})$; $f_0>0$ (Perron–Frobenius), so $b_0>0$ and $0<\beta_0\le1$.

### Step 3. Inertia of the pair (item 2)

By (7.3), $m(\rho)=\mu_1+\mu_2+\delta m(\rho)$, $\delta m=\sum_i\bar M_i^o/c_i^2$. In (6.6), $1/\sum_\rho b_0/m$ is exact in the constant part of $\delta m$ and first order in its variation ($O(\lvert\mu\rvert\eta^2w^2/L^2)$ omitted):
$$D_0=\underbrace{(\mu_1+\mu_2)\frac{1-\beta_0}{\beta_0}}_{D^{\rm lat}}+\underbrace{\frac1{\beta_0^2}\sum_\rho b_0(\rho)\sum_i\frac{\bar M_i^o(\rho)}{c_i^2}}_{D^{\rm mark}}+O\Big((\lvert\mu_1\rvert+\lvert\mu_2\rvert)\big[\eta^2+\tfrac{\tau^2}{d^2}(\eta+1-\beta_0)\big]\Big),\quad D^{\rm mark}=\Big(\frac1{\tau_1^2}+\frac1{\tau_2^2}\Big)\frac{\sum_\rho b_0\bar B}{\beta_0^2}\big[1+O(\eta)\big].\tag{7.5}$$
This is (6.8) with $s_c=+1$ and $c_i=\tau_i$, the full-grip speeds of (5.7), plus the flux remainder of (6.6). $D^{\rm lat}$ comes from the lattice form of the bands, $D^{\rm mark}$ from the mutual term. Each body's inertia counts the part of $M$ odd in its own coupling, and at leading order this is all of $M$ for each body.

### Step 4. Cost of the bond and inertia against cost (item 3)

With $f_0$ real, $E_b(0)=\sum_rf_0^2U+2\sum_\rho b_0\,t(\rho;0)$, $E_i(0)=E_i^{\rm held}+\bar d_i+2\bar\kappa_i$, $E^{\rm held}_{\rm pair}(r)=E_1^{\rm held}+E_2^{\rm held}-E_\varnothing+M(r)$ (definition (4.2)). The partner shifts $d_i$ and $\kappa_i$ by the relative amount $-\bar M_i^o/\sigma_iR_i$ (Step 2), and $\bar d_i=-2\bar\kappa_i$, so these shifts cancel up to the factor $1-\beta_0$:
$$\Delta E_0=\langle M\rangle_0+K_0\big[1+O(\eta)\big],\qquad K_0=2\bar t_0(\beta_0-1)=\tfrac14(\mu_1^{-1}+\mu_2^{-1})(1-\beta_0).\tag{7.6}$$
**(i)** Since $(\mu_1+\mu_2)/(\mu_1^{-1}+\mu_2^{-1})=\mu_1\mu_2$, $D^{\rm lat}=4\mu_1\mu_2K_0/\beta_0$; and $\sum_\rho b_0\bar B/\beta_0^2=\langle M\rangle_0[1+O(\eta,1-\beta_0)]$. With $v=\tau_i$, the speed of (5.7) of either body, and $j\neq i$:
$$D_0-\frac{\Delta E_0}{\tau_i^2}=\frac{4\mu_1\mu_2}{\beta_0}K_0\Big[1-\frac{\beta_0}{4\mu_1\mu_2\tau_i^2}\Big]+\frac{\langle M\rangle_0}{\tau_j^2}\big[1+O(\eta,1-\beta_0)\big],\qquad\lvert4\mu_1\mu_2\tau_i^2\rvert=\frac{(2n\lvert g_1\rvert)(2n\lvert g_2\rvert)}{\tau_j^2}\ge\frac{d^2}{\tau_j^2}\gg1.\tag{7.7}$$
Both terms are nonzero: the lattice part exceeds $K_0/v^2$ by the factor $\ge d^2/\tau_j^2$, and the mark part exceeds $\langle M\rangle_0/v^2$ by $\langle M\rangle_0/\tau_j^2$. Both terms have the sign of $D^{\rm lat}$ when $g_1g_2<0$ ($\mu_1\mu_2K_0>0$, $\langle M\rangle_0>0$), so then (i) never holds; for $g_1,g_2>0$ it holds only on the surface where the two terms cancel, a tuned relation between $1-\beta_0$ and $\eta$. Elsewhere, no.

**(ii)** In S5, $D^{\rm lat}\ll\lvert D^{\rm mark}\rvert$, and by (7.7) $\lvert K_0\rvert\le\lvert D^{\rm lat}\rvert/\lvert4\mu_1\mu_2\rvert\ll2\tau_{\max}^2d^{-2}\lvert\langle M\rangle_0\rvert$, so $\Delta E_0=\langle M\rangle_0$ at leading order and
$$D_0=\Big(\frac1{\tau_1^2}+\frac1{\tau_2^2}\Big)\Delta E_0\Big[1+O\Big(\eta,\frac{1-\beta_0}\eta,\frac{\tau^2}{d^2}\Big)\Big];\qquad\text{equal speeds }\tau_1=\tau_2=v:\ D_0=\frac{2\Delta E_0}{v^2}.\tag{7.8}$$
The binding cost has inertia, but $(\tau_1^{-2}+\tau_2^{-2})\Delta E_0\ne\Delta E_0/\tau_i^2$: the mutual term is odd in both couplings, so it enters both half-gaps (7.3) but the cost only once. It approaches $\Delta E_0/\tau_i^2$ only for $\tau_i/\tau_j\to0$, with relative deviation $\tau_i^2/\tau_j^2$.

### Step 5. Weight and fall (item 4)

*Weight of the selected level.* The source enters (2.10) through the held levels at first order in $g_s$, so (6.13) carries over: $W_0'=\sum_rf_0^2(\partial_1+\partial_2)W_{\rm pair}$ plus the kinetic part. By (7.3) with the source in place of the partner, $\kappa_i,d_i$ change by the relative amount $-\phi$ ($M^o_{is}=ng_i\phi$). So the kinetic energies weigh $-(\text{kinetic energy})\,\phi'$, and the bond part of this is $-K_0\phi'$; all such terms are relative $O(\tau^2/d^2)$ against $ng\phi'$.

*Lattice identities.* For $1\le x\le L-1$: $\ell'(x)=-\frac12(1-\frac{2x}L)$; $(\ell^2)'=(2\ell+\frac1L)\ell'$ (as $\Lambda\ell=-\frac1L$ there); and since $\Lambda\ell^{(2)}=\ell$, $\ell^{(2)}$ is even about $0$ and $L/2$, the cubic $-\frac{Lx}{12}+\frac{x^2}4-\frac{x^3}{6L}=\frac{x(L-x)}{6}\ell'(x)$ is $(\ell^{(2)})'$. Inserting these into the derivative of $2\ell_0\ell+\ell^2-\frac2L\ell^{(2)}$ in (4.5) gives $4\ell_0\kappa_s\ell'$ with $\kappa_s$ below, and $\phi'=-\frac{2g_s}{t_f}\ell'$:
$$W_i'(X)=\phi'\big[ng_i+2\kappa_sC_i^{(2)}\big],\qquad C_i^{(2)}:=-\frac{ng_i^2\ell_0}{t_f},\qquad\kappa_s:=\frac{L^2(1-2u)^2+2}{L^2-1}=(1-2u)^2+O(L^{-2}),\tag{7.9}$$
up to relative $O(\eta^2)$. $C_i^{(2)}$ is the second-order even part of the cost; it is weighed with $2\kappa_s$ but has no inertia by (5.3).

*Bond.* In (4.7) at $x_{1,2}=X\mp r/2$: $f_0$ is even in $r$ at $P=0$ (as $V$ and $\kappa_i$ are), so $\langle r\rangle_0=0$; $W_b(X;r)-W_b(X;0)=V(r)\phi(X)$ up to $O(r^2/(Ls))$; and $W_b(X;0)$ is (4.5) with $g_i^2\to2g_1g_2$. So, with the kinetic part,
$$W_b':=W_0'-W_1'-W_2'=\phi'\big[2\kappa_sB(0)+\langle V\rangle_0-K_0\big]\big[1+O(\eta/\kappa_s)\big].\tag{7.10}$$
*Fall factors.* By (5.14), (6.13), $\Gamma_i:=E_i''(0)W_i'=W_i'/\mu_i$ and $\Gamma_{\rm pair}:=E_b''(0)W_0'$ set the acceleration from rest $-\Gamma$ of a smooth start. With $\Gamma_b:=W_b'/D_0$ and $1/E_b''=\mu_1+\mu_2+D_0$, exactly,
$$\Gamma_{\rm pair}=\frac{\mu_1\Gamma_1+\mu_2\Gamma_2+D_0\Gamma_b}{\mu_1+\mu_2+D_0},\qquad\Gamma_i=\tau_i^2\phi'\Big[1-\frac{2\kappa_sg_i\ell_0}{t_f}+O\Big(\eta^2,N_f\eta^2,\frac{\tau^2}{d^2}\Big)\Big].\tag{7.11}$$
*Leading order.* $W_b'$ and $D^{\rm mark}$ are $O(\eta)$; the lattice inertia has weight $-K_0\phi'=-\beta_0D^{\rm lat}\phi'/(4\mu_1\mu_2)$, which is $O(\tau^2/d^2)$ against $\tau^2\phi'D^{\rm lat}$. Hence $\Gamma_{\rm pair}=\beta_0\bar\Gamma[1+O(\eta,\tau^2/d^2)]$ with $\bar\Gamma:=\frac{\mu_1\Gamma_1+\mu_2\Gamma_2}{\mu_1+\mu_2}=\frac{\mu_1\tau_1^2+\mu_2\tau_2^2}{\mu_1+\mu_2}\phi'[1+O(\eta)]$. So $\Gamma_{\rm pair}=\Gamma_1=\Gamma_2$ at leading order (with $1-\beta_0\to0$) iff $\tau_1=\tau_2$. For $\tau_1\ne\tau_2$ the bodies already fall differently; $1-\beta_0$ lowers the pair's fall, since the internal kinetic energy has inertia but no weight at this order.

*Order of the mutual term (S5).* By (7.10), (7.5), (7.8),
$$\Gamma_b=\frac{2\kappa_s\,\phi'}{\tau_1^{-2}+\tau_2^{-2}}\Big[1+O\Big(\frac\eta{\kappa_s},\frac{1-\beta_0}\eta,\frac wL,\frac{\tau^2}{d^2}\Big)\Big],\qquad \Gamma_{\rm pair}-\bar\Gamma\overset{\tau_1=\tau_2=\tau}{=}4(1-\kappa_s)\frac{g_1g_2\ell_0}{t_f(g_1+g_2)}\tau^2\phi'.\tag{7.12}$$
For equal speeds the bond's weight and inertia both carry the factor 2 of (7.8). The weight also carries $\kappa_s=1-\frac{4s(L-s)-3}{L^2-1}<1$ for $1\le s\le L/2$, so $\Gamma_b=\kappa_s\tau^2\phi'$ against $\Gamma_i=\tau^2\phi'$ at leading order. The binding cost is weighed in the same proportion as its inertia only in the limit $u\to0$ ($w\ll s\ll L$), where $\Gamma_b/\Gamma_i=1-4u+O(u^2,w/L,\eta)$. For $\tau_1\ne\tau_2$, $\Gamma_b\ne\Gamma_i$ already because $2/(\tau_1^{-2}+\tau_2^{-2})\ne\tau_i^2$. The second equation of (7.12) is the first order of the difference, from (7.11) with $D_0\simeq2B(0)/\tau^2$, $\Gamma_b-\bar\Gamma\simeq(\kappa_s-1)\tau^2\phi'$, $\mu_1+\mu_2\simeq n(g_1+g_2)/\tau^2$. Besides this, $\bar\Gamma-\Gamma_i=O(\eta)$ for $g_1\ne g_2$ from the self terms of (7.11).

### Step 6. The hopping of the field (item 5)

At fixed $n,\gamma_i,M_i,\tau_i$ (so fixed $g_i$, $\mu_i$) and leading order: $B,\langle M\rangle_0,D^{\rm mark},W_b'/\phi'\propto t_f^{-1}$; $F\propto t_f^{-1}$, so $w\propto t_f^{1/3}$, and for $w\gg1$, $1-\beta_0\propto t_f^{-2/3}$, $K_0,\langle V\rangle_0,D^{\rm lat}\propto t_f^{-2/3}$; $\phi'\propto g_s/t_f$, $\Gamma_i/\phi'$ and $\mu_i$ are independent of $t_f$. Hence
$$\frac{D^{\rm lat}}{\lvert D^{\rm mark}\rvert}\propto t_f^{1/3},\qquad \frac{D^{\rm mark}}{\Delta E_0}\to\frac1{\tau_1^2}+\frac1{\tau_2^2},\quad \frac{\Gamma_b}{\Gamma_i}\to\frac{2\kappa_s}{1+\tau_i^2/\tau_j^2}\quad\text{independent of }t_f.\tag{7.13}$$
A stiffer field weakens the bond ($\propto t_f^{-1}$) and widens it, and S5 lies at the small-$t_f$ end, bounded by $\eta\propto t_f^{-1}\ll1$ and by $\varepsilon_1\propto t_f$ in S2. No value of $t_f$ restores the proportions of (7.8) and (7.12).

## Result

Class: full grip for both bodies, $N_f\ge1$, even $L\ge6$, S1–S4 (S5, S6 where stated), without the tuned cases.
- **(7.1)** Definitions: half-gap $\sigma_iR_i$, $c_i^2=\sigma_iR_i/\mu_i=\tau_i^2[1+O(N_f\eta^2)]$, the part $M_i^o$ of the mutual term odd in $g_i$, and the bond average $\bar f$.
- **(7.2), (7.3)** Under S2, S3 the band contract is (2.10)/(6.1) with gaps (7.2). Each body's local inertia is $m_i(\rho)=\mu_i+\bar M_i^o(\rho)/c_i^2+O(\lvert\mu_i\rvert\eta^2)=\mu_i+\bar B(\rho)/\tau_i^2\,[1+O(\eta)]$. (6.5), (6.6) hold for the full grip. Control: coupling (c) has the opposite sign of the mark part ($s_c=-1$ in (6.8)).
- **(7.4)** The selected level is bound at $r\approx0$ iff $\tau_1^2g_2+\tau_2^2g_1>0$: lowest if both $g_i>0$, highest for opposite signs, never for both $g_i<0$. Otherwise it sits at $r\approx L/2$ (control).
- **(7.5)** $D_0=1/E_b''(0)-\mu_1-\mu_2=D^{\rm lat}+D^{\rm mark}$ to $O((\lvert\mu_1\rvert+\lvert\mu_2\rvert)[\eta^2+\frac{\tau^2}{d^2}(\eta+1-\beta_0)])$ and the ring remainder. $D^{\rm lat}=(\mu_1+\mu_2)(1-\beta_0)/\beta_0$, and $D^{\rm mark}=(\tau_1^{-2}+\tau_2^{-2})\beta_0^{-2}\sum_\rho b_0\bar B\,[1+O(\eta)]$ is (6.8) with $s_c=+1$, $c_i=\tau_i$.
- **(7.6)** $\Delta E_0=\langle M\rangle_0+K_0[1+O(\eta)]$, $K_0=\frac14(\mu_1^{-1}+\mu_2^{-1})(1-\beta_0)$.
- **(7.7)** Item 3 (i), negative: $D_0\ne\Delta E_0/v^2$ for $v=\tau_1$ or $\tau_2$. The lattice part is $4\mu_1\mu_2K_0/\beta_0$, larger than $K_0/v^2$ by a factor $\ge d^2/\tau_j^2$; the mark part exceeds $\langle M\rangle_0/v^2$ by $\langle M\rangle_0/\tau_j^2$. Equality holds only on a tuned surface, which exists only for $g_1,g_2>0$.
- **(7.8)** Item 3 (ii), negative: in S5, $D_0=(\tau_1^{-2}+\tau_2^{-2})\Delta E_0[1+O(\eta,(1-\beta_0)/\eta,\tau^2/d^2)]$, i.e. $2\Delta E_0/v^2$ for $\tau_1=\tau_2=v$. It approaches $\Delta E_0/\tau_i^2$ only as $\tau_i/\tau_j\to0$.
- **(7.9), (7.10)** In S6: $W_i'=\phi'[ng_i+2\kappa_sC_i^{(2)}]$ up to relative $O(\eta^2)$, $\kappa_s=(L^2(1-2u)^2+2)/(L^2-1)$; $W_b'=\phi'[2\kappa_sB(0)+\langle V\rangle_0-K_0][1+O(\eta/\kappa_s)]$.
- **(7.11)** Exact: $\Gamma_{\rm pair}=(\mu_1\Gamma_1+\mu_2\Gamma_2+D_0\Gamma_b)/(\mu_1+\mu_2+D_0)$, and $\Gamma_i=\tau_i^2\phi'[1-2\kappa_sg_i\ell_0/t_f+O(\eta^2,N_f\eta^2,\tau^2/d^2)]$. At leading order $\Gamma_{\rm pair}=\beta_0\bar\Gamma$, equal to $\Gamma_1=\Gamma_2$ (for $1-\beta_0\to0$) iff $\tau_1=\tau_2$. Control: $\tau_1\ne\tau_2$.
- **(7.12)** In S5, S6: $\Gamma_b=2\kappa_s\phi'/(\tau_1^{-2}+\tau_2^{-2})$ to the orders stated. For $\tau_1=\tau_2$ the binding cost is weighed in the same proportion as its inertia only as $u\to0$; at finite $u$ it is weighed less, by $\kappa_s<1$, and $\Gamma_{\rm pair}-\bar\Gamma=4(1-\kappa_s)g_1g_2\ell_0\tau^2\phi'/(t_f(g_1+g_2))$. Control: any $u>0$ (e.g. $u=1/4$).
- **(7.13)** At fixed $n,\gamma_i,M_i,\tau_i$: mark quantities $\propto t_f^{-1}$, lattice quantities $\propto t_f^{-2/3}$ ($w\gg1$), and $D^{\rm mark}/\Delta E_0$, $\Gamma_b/\Gamma_i$ are independent of $t_f$ at leading order.

## Consistency checks

1. **Dimensions.** $g,\tau,t_f,M,K_0$ are energies (rates in $\lambda$); $P,r,u,\phi,\kappa_s$ are dimensionless. So $\mu,D$ are inverse energies, and (7.3), (7.5), (7.7), (7.8) are consistent ($M/\tau^2$, $\mu_1\mu_2K_0$). $\Gamma=E''W'$ and $\tau^2\phi'$ are both energy squared per place.
2. **$g_2\to0$** (no binding): $B,M_i^o,W_b',D^{\rm mark}\to0$ linearly, (7.3) gives $m_1\to\mu_1$, $F\to0$ so $w\to\infty$ (S4 fails, as it must), and (7.9) is the single-body weight of (4.5).
3. **Coincident held pair $r=0$** ($\langle V\rangle_0=K_0=0$). (7.9) plus (7.10) give $\phi'[n(g_1+g_2)+2\kappa_s(C_1^{(2)}+C_2^{(2)}+B(0))]=\phi'[nG+2\kappa_sC_G^{(2)}]$, $G=g_1+g_2$, as must hold: two point couplings at one place are one coupling $G$.

## Open issues

- The code check has $N_f=1$, so S5 is reached only moderately ($D^{\rm lat}/\lvert D^{\rm mark}\rvert=0.21$ at $L=128$, $\eta=0.15$). The $O(\eta)$ corrections have large coefficients ($\langle M_1^o+M_2^o\rangle_0/\langle M\rangle_0=2.11$ at $\eta=0.05$, $2.40$ at $\eta=0.15$).
- The kinetic weight $-K_0\phi'$ in (7.10) is derived, not checked: the code uses held weights.
- The fall itself (time evolution) is not simulated; $\Gamma$ is taken from (5.14), (6.13).
- $\kappa_s$ is a property of the ring (zero mode). Other geometries are out of scope (tori).
- $c_i^2=\tau_i^2[1+O(N_f\eta^2)]$ is an estimate of the overlap corrections of (5.3), not derived in detail.

## Code

- `code/v1/check_pair_full_grip.py` (role: check; $N_f=1$, $t_f=1$). It builds the exact sector contract of two full-grip bodies and one field object continued in $P$ (basis: branches, $r$, relative field place), finds the selected level by its overlap with the band contract, and takes $E_b''(0)$ by Richardson differences. It builds the band contract (2.10) numerically (Schrieffer–Wolff over the exact held blocks) and compares $D_0$ with (7.5), $m_i(\rho)$ with (7.3), and $\Delta E_0$ with (7.6). It computes weights from exact lowest eigenvalues of the field with held couplings and a held source ($\pm g_s$, odd part) against (7.9)–(7.12). Cases: A (equal bodies, $\eta=0.15$, $\tau/d=0.3$, $L=32,64$), A′ ($\eta=0.05$, $L=64$), A″ ($\eta=0.05$, $\tau/d=0.15$, $L=64$), B (unequal $g_i$ and $\tau_i$, $L=32,64$), R (S5, $L=128$), C (opposite signs), D (both negative).
- Outputs: `code/v1/output/check_log.txt` (readable log), `code/v1/output/check_results.json` (all numbers).
- Findings (all from `check_log.txt`):
  - Item 1: the selected levels have overlap $\ge0.984$ with the band contract at every $P$ used. $E_b''(0)$ of (2.10) and of (6.6) agree with each other to about $10^{-6}$, and with the exact value to 1.0–4.6%. At $L=64$ this difference falls from 2.9% (A) to 1.45% (A′) to 0.99% (A″), with $\eta$ and $\tau/d$. Case C is bound at the top ($\langle\lvert r\rvert\rangle=0.14$). In case D the extreme level sits at the antipode ($\langle\lvert r\rvert\rangle=21.4$, $L=48$), as (7.4) states.
  - Item 2: (7.3) holds near $r=0$ to 2.5% ($\eta=0.15$) and 0.26% ($\eta=0.05$) in the odd-part form, and to 9–25% and 4.8% in the form with $\bar B$. (7.5) reproduces $D_0$ of the band contract to 0.1–2% of $\lvert D^{\rm lat}\rvert+\lvert D^{\rm mark}\rvert$, and the exact $D_0$ to 3–11% of it (the band-versus-exact part).
  - Item 3: $\Delta E_0$ agrees with $\langle M\rangle_0+K_0$ to 0.7–3.2%. $(D_0-D^{\rm lat})/\Delta E_0=1.03\,(\tau_1^{-2}+\tau_2^{-2})=2.06/\tau^2$ in A′. In R (S5), $D_0/\Delta E_0=1.016\,(\tau_1^{-2}+\tau_2^{-2})$; there the corrections $D^{\rm lat}/D^{\rm mark}=-0.21$ and $O(\eta)$ partly cancel.
  - Item 4: $W_0'$ agrees with (7.9)+(7.10) (held part) to 0.3–7%, and $\Gamma_i/\phi'$ with (7.11) to 2–4% (R). $W_b'/[(D_0-D^{\rm lat})\phi']$ agrees with $2\kappa_s/(\tau_1^{-2}+\tau_2^{-2})$ to 7–20% (A′, R), within $O(\eta/\kappa_s,w/s)$. In R, $\Gamma_b/\Gamma_1=0.76$ at $u=1/8$ and $0.40$ at $u=1/4$ ($\kappa_s=0.56,0.25$): the bond is weighed below the single bodies' proportion, and more so at larger $u$. In case B ($\tau_1\neq\tau_2$) $\Gamma_1/\phi'$ and $\Gamma_2/\phi'$ differ by a factor 3.6–3.7, as (7.11) states.

## Methods used

- Rayleigh–Schrödinger and Schrieffer–Wolff perturbation theory
- Bloch sectors of the joint translation, band contract
- Hellmann–Feynman theorem, Perron–Frobenius
- Lattice Green function on the ring (pseudo-inverse of the Laplacian)
- Continuum limit of the relative motion (linear potential)
- Exact diagonalization, shift-invert Lanczos, Richardson extrapolation (code)

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