Version 1 · earlier version; the current one is v2 · External review, round 1: minor issues

# Fall of the bound pair one order beyond leading order

- **Subproject:** 05-ilang-free-fall
- **Package:** 16-pair-fall-next-order
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-18

## Setup and assumptions

**Description (assumption of the package, M8, M9).** As in 13-pair-fall: two Model D bodies $b_1,b_2$ of different types, one medium, and a sharp stationary field with a linear profile that multiplies the rest and clock term of each part, $(1+g_ix)\,G_i$ with $G_i=M_i\sigma^{(i)}_3$ on the fixed clock levels $k_1,k_2$, and not the binding; operator (13.2) and gradient terms (13.3) of 13-pair-fall. Diagonal on-site binding with $\hat w_{11}\neq0$, of either sign $\varsigma=\pm1$; the bound level (11.12) at the band $\mathcal B_{++}$; regime $\mathsf R$ at fixed hatted data, hence fixed $u\in(0,\infty)$; packets at $P_0=0$; first order in $(g_1,g_2,\kappa\langle X\rangle)$, records at leading order in $\kappa$. Equal parameters of the two types are never assumed (M7). The controls are the ones fixed by the question: items 2 and 3 compare the pair with its cost and with its free parts.

**Definitions of the question.** $\mathcal W:=\langle\mathcal K\rangle_0=g_1M_1\langle\sigma^{(1)}_3\rangle_0+g_2M_2\langle\sigma^{(2)}_3\rangle_0+2\kappa\langle X\rangle$ and $a_{\rm pair}=-E_b''(0)\,\mathcal W=-\mathcal W/\mathfrak m_b$, so $\mathfrak m_b=1/E_b''(0)$; $E_b'(0)=0$ exactly (Step 6.1 of 13-pair-fall). $a^{(i)}=-(g_iM_i+\kappa\langle X\rangle)/\mathfrak m_i$; comparison with (4.16) of 04-branch-fall ($G_k=g_iM_i$, $\mathcal F_k=-\kappa\langle X\rangle$, $\sigma=+$) gives $\mathfrak m_i=M_i/\tau_i^2$. $\bar a=(\mathfrak m_1a^{(1)}+\mathfrak m_2a^{(2)})/(\mathfrak m_1+\mathfrak m_2)$.

**Inputs, as quoted in the question.**
- (13.11) of 13-pair-fall at $P_0=0$: $a_{\rm pair}=g_1\alpha_1+g_2\alpha_2+\kappa\langle X\rangle\,\alpha_X$, $\alpha_i=-M_i\bigl[E_b''\,\partial_{M_i}E_b-E_b'\,\partial_{M_i}E_b'\bigr]$, $\alpha_X=-2E_b''$.
- (13.13) of 13-pair-fall, which defines $r_1,r_2,r_X$: $a_{\rm pair}=\frac{\varsigma}{\bar s}\,\frac{g_1M_1(1+r_1)+g_2M_2(1+r_2)+2\kappa\langle X\rangle(1+r_X)}{\mathfrak m_1+\mathfrak m_2}$.
- (11.5) of 11-bound-pair: $\partial E_b/\partial M_i=\langle\sigma^{(i)}_3\rangle_P$ (at fixed $\tau_1,\tau_2$ and binding).
- (14.7)–(14.10) and (14.12) of 14-pair-level-next-order: $E_{\rm bind}=a(1+\varsigma\bar s)+\mathcal E_4+O(\epsilon^6)$ with $\mathcal E_4,\phi_A,\phi_B$; $\langle\sigma^{(i)}_3\rangle_0=1-\beta_i\Lambda+\beta_i\Xi_i+O(\epsilon^6)$; $E_b''(0)=-4\varsigma\nu a/\bar s+\nu\,\mathcal C+O(\epsilon^6)$; $\mathfrak m_b=-\varsigma(\mathfrak m_1+\mathfrak m_2)\bar s\bigl[1+\frac{\varsigma\bar s}{4a}\mathcal C+O(\epsilon^4)\bigr]$; the expansions at $\varsigma=-1$, $u\to0$.

**Notation of 11-bound-pair and 14-pair-level-next-order not spelled out in the question.** $a_i=\tau_i^2/(4M_i)$, so $a_i=1/(4\mathfrak m_i)$; $a=a_1+a_2$; $u=\lvert w_{11}\rvert/a$ (only $u\propto1/a$ at fixed binding is used); $\varsigma=\operatorname{sgn}w_{11}$; $\beta_i,\beta,\nu,\bar s,\gamma_s,\gamma_c,\gamma_m$ as in the question. Step 1 shows that this reading makes the quoted inputs consistent and fixes the symbol $\Lambda$ of (14.8). Elementary identities used below:
$$
M_i\beta_i=a_i,\qquad \sum_ia_i=a,\qquad \sum_ia_i\beta_i=\gamma_s,\qquad a\beta=\gamma_s+\gamma_m .
$$

**Orders (A13).** At fixed hatted data, with costs in the units of (14.7): $a_i,a=O(\epsilon^2)$; $\gamma_s,\gamma_c,\gamma_m,\mathcal E_4,\mathcal C=O(\epsilon^4)$; $\beta_i=O(\epsilon^2)$. Every $O(\epsilon^n)$ holds at fixed $u\in(0,\infty)$, with constants that depend on the hatted data. In item 4, $u\to0$ is taken in the coefficients after $\epsilon\to0$; no statement is uniform in $u$.

## Derivation

### Step 1. Inertia and $\Lambda$ in usable form

(a) Since $\mathfrak m_b=1/E_b''(0)$, (14.10) and $\varsigma^2=1$ give
$$
-\varsigma\bar s\,(\mathfrak m_1+\mathfrak m_2)\,E_b''(0)=\frac{-\varsigma\bar s\,(\mathfrak m_1+\mathfrak m_2)}{\mathfrak m_b}=1-\mu+O(\epsilon^4),\qquad \mu:=\frac{\varsigma\bar s}{4a}\,\mathcal C=O(\epsilon^2).
\tag{16.1}
$$
(14.9) gives the same with the leading factor $4\nu a(\mathfrak m_1+\mathfrak m_2)=4\frac{a_1a_2}{a}\bigl(\frac1{4a_1}+\frac1{4a_2}\bigr)=1$.

(b) By (11.5), $\langle\sigma^{(i)}_3\rangle_0=\partial E_b(0)/\partial M_i$ at fixed $\tau_1,\tau_2$ and binding, with $E_b(0)=M_1+M_2+E_{\rm bind}$ and $E_{\rm bind}$ of (14.7). In the notation above, $\partial_{M_i}a=-\beta_i$, $\partial_{M_i}u=u\beta_i/a$, $\partial_{M_i}\gamma_s=-3\beta_i^2$ and $\partial_{M_i}(\gamma_s+\gamma_m)=\partial_{M_i}(a\beta)=-\beta_i(\beta+2a/M_i)$. Inserted in $\partial_{M_i}\mathcal E_4$, these reproduce $\beta_i\Xi_i$ of (14.8) term by term, which confirms the reading. The order-$\epsilon^2$ term is $\partial_{M_i}[a(1+\varsigma\bar s)]=-\beta_i(1+\varsigma\bar s)+\varsigma\frac{u}{\bar s}\,u\beta_i=-\beta_i(1+\varsigma/\bar s)$, by $\bar s^2-u^2=1$. Hence
$$
\Lambda=1+\frac{\varsigma}{\bar s}.
\tag{16.2}
$$

### Step 2. Item 1: the remainders $r_1,r_2,r_X$

At $P_0=0$, $E_b'(0)=0$ and (11.5) reduce (13.11) to $\alpha_i=-M_iE_b''(0)\langle\sigma^{(i)}_3\rangle_0$ and $\alpha_X=-2E_b''(0)$. Both (13.11) and (13.13) are linear in the three independent gradients $g_1,g_2,\kappa\langle X\rangle$. Equating their coefficients gives, exactly,
$$
1+r_i=-\varsigma\bar s\,(\mathfrak m_1+\mathfrak m_2)\,E_b''(0)\,\langle\sigma^{(i)}_3\rangle_0\quad(i=1,2),\qquad 1+r_X=-\varsigma\bar s\,(\mathfrak m_1+\mathfrak m_2)\,E_b''(0).
\tag{16.3}
$$
Insert (16.1), and (14.8) with (16.2), and keep the order $\epsilon^2$:
$$
r_i=-\beta_i\Bigl(1+\frac{\varsigma}{\bar s}\Bigr)-\frac{\varsigma\bar s}{4a}\,\mathcal C+O(\epsilon^4),\qquad r_X=-\frac{\varsigma\bar s}{4a}\,\mathcal C+O(\epsilon^4),
\tag{16.4}
$$
with $\mathcal C$ of (14.9). The first term of $r_i$ is the shift of $\langle\sigma^{(i)}_3\rangle_0$ from 1. The common term $-\mu$ is the relative change of $1/\mathfrak m_b$ beyond its leading value.

### Step 3. Item 2: weight against cost

For $g_1=g_2=g_{\rm c}\neq0$ and $\langle X\rangle=0$, $\mathcal W/g_{\rm c}=\sum_iM_i\langle\sigma^{(i)}_3\rangle_0$. With (14.8), (16.2), $M_i\beta_i=a_i$ and $\sum_ia_i=a$:
$$
\frac{\mathcal W}{g_{\rm c}}=M_1+M_2-a\Bigl(1+\frac{\varsigma}{\bar s}\Bigr)+\sum_ia_i\Xi_i+O(\epsilon^6).
\tag{16.5}
$$
Summing $\Xi_i$ of (14.8) with the elementary identities gives $\sum_ia_i\Xi_i=\frac{\gamma_s}{2}(3\phi_A-u\phi_A')+\frac{\gamma_s+\gamma_m}{2}(3\phi_B-u\phi_B')$. Now subtract $E_b(0)$ of (14.7). At order $\epsilon^2$ the difference is $-a(1+\varsigma/\bar s)-a(1+\varsigma\bar s)=-\frac{a}{\bar s}(\varsigma\bar s^2+2\bar s+\varsigma)$. At order $\epsilon^4$ it is $\sum_ia_i\Xi_i-\mathcal E_4$. Hence
$$
\frac{\mathcal W}{g_{\rm c}}-E_b(0)=-\frac{\varsigma a(\bar s+\varsigma)^2}{\bar s}+\mathcal V_4+O(\epsilon^6),\qquad \mathcal V_4:=\Bigl(u\frac{\partial}{\partial u}-4\Bigr)\mathcal E_4\Big|_{\gamma_s,\gamma_m}=\frac{\gamma_s}{2}\,\psi_A+\frac{\gamma_s+\gamma_m}{2}\,\psi_B,
\tag{16.6}
$$
where $\psi=4\phi-u\phi'$. With $\bar s'=u/\bar s$ this is
$$
\psi_A=8-2u^2+\frac{8\varsigma+10\varsigma u^2+u^3+2u^5}{\bar s^3},\qquad \psi_B=2u^2+\frac{2\varsigma u^2-u^3+3\varsigma u^4-2u^5}{\bar s^3}.
\tag{16.7}
$$
Reading of the order $\epsilon^2$: with $E^{(2)}_{\rm bind}:=a(1+\varsigma\bar s)$, (16.5) reads $\mathcal W/g_{\rm c}=M_1+M_2-\frac{\varsigma}{\bar s}E^{(2)}_{\rm bind}+O(\epsilon^4)$. The rest costs enter the weight with the factor 1, the binding cost with the factor $-\varsigma/\bar s$. This factor is $1/\bar s<1$ for $\varsigma=-1$ and $-1/\bar s<0$ for $\varsigma=+1$. The order-$\epsilon^2$ coefficient of (16.6) is nonzero for every $u\in(0,\infty)$ and both signs, since $a>0$ and $\bar s+\varsigma\ge\bar s-1>0$. Its sign is $-\varsigma$.

### Step 4. Item 3: pair against its parts

For $g_{\rm c}\neq0$ and $\langle X\rangle=0$, $\bar a=-g_{\rm c}(M_1+M_2)/(\mathfrak m_1+\mathfrak m_2)\neq0$ and $a_{\rm pair}=-\mathcal W/\mathfrak m_b$. Define the relative changes
$$
\delta_W:=\frac{\mathcal W}{g_{\rm c}(M_1+M_2)}-1,\qquad \delta_{\mathfrak m}:=\frac{\mathfrak m_b}{\mathfrak m_1+\mathfrak m_2}-1 .
\tag{16.8}
$$
Then, exactly,
$$
\frac{a_{\rm pair}}{\bar a}-1=\frac{1+\delta_W}{1+\delta_{\mathfrak m}}-1=\frac{\delta_W-\delta_{\mathfrak m}}{1+\delta_{\mathfrak m}} .
\tag{16.9}
$$
By (16.5) and (14.10),
$$
\delta_W=-\frac{a}{M_1+M_2}\Bigl(1+\frac{\varsigma}{\bar s}\Bigr)+O(\epsilon^4),\qquad 1+\delta_{\mathfrak m}=-\varsigma\bar s\,(1+\mu)+O(\epsilon^4).
\tag{16.10}
$$
Here $\delta_W=O(\epsilon^2)$, while $\delta_{\mathfrak m}=-1-\varsigma\bar s+O(\epsilon^2)$ is of order one. Hence $(1+\delta_W)/(1+\delta_{\mathfrak m})=-\frac{\varsigma}{\bar s}(1+\delta_W-\mu)+O(\epsilon^4)$, and with $\frac{\varsigma}{\bar s}\mu=\frac{\mathcal C}{4a}$:
$$
\frac{a_{\rm pair}}{\bar a}-1=-1-\frac{\varsigma}{\bar s}+c_2+O(\epsilon^4),\qquad c_2:=\frac{\varsigma}{\bar s}\,\frac{a}{M_1+M_2}\Bigl(1+\frac{\varsigma}{\bar s}\Bigr)+\frac{\mathcal C}{4a}.
\tag{16.11}
$$
The leading term $-1-\varsigma/\bar s$ is nonzero for every $u\in(0,\infty)$ and both signs. It tends to 0 only for $\varsigma=-1$, $u\to0$. $c_2=O(\epsilon^2)$ is the relative correction.

### Step 5. Item 4: wide pair

Take $\varsigma=-1$. First $\epsilon\to0$ at fixed $u$ (Steps 3 and 4), then $u\to0$ in each coefficient, with $\bar s^{-1}=1-\frac{u^2}{2}+\frac{3u^4}{8}+O(u^6)$.

*Item 2.* $\phi_A=u^3-\frac34u^4-\frac12u^5+O(u^6)$ and $\phi_B=-u^3+\frac12u^4+\frac12u^5+O(u^6)$. The operator $u\partial_u-4$ multiplies $u^n$ by $n-4$. So $\psi_A=u^3+\frac12u^5+O(u^6)$ and $\psi_B=-u^3-\frac12u^5+O(u^6)$, and the $\gamma_s$ terms of $\mathcal V_4$ cancel through $u^5$. The order-$\epsilon^2$ coefficient is $-\varsigma a(\bar s+\varsigma)^2/\bar s=a(\bar s-1)^2/\bar s$. Hence
$$
\frac{a(\bar s-1)^2}{\bar s}=\frac a4\,u^4+O(u^6),\qquad \mathcal V_4=-\frac{\gamma_m}{2}\,u^3+O(u^5),\qquad \gamma_m=a_1a_2\Bigl(\frac1{M_1}+\frac1{M_2}\Bigr)>0 .
\tag{16.12}
$$
$\gamma_m>0$ for all data ($\tau_i>0$, $M_i\ge m_i>0$), so the leading term of $\mathcal V_4$ vanishes on no data.

*Item 3.* At $\varsigma=-1$, $c_2=-\frac{a}{\bar s(M_1+M_2)}\bigl(1-\frac1{\bar s}\bigr)+\frac{\mathcal C}{4a}$. Insert $1-\frac1{\bar s}=\frac{u^2}{2}+O(u^4)$ and $\mathcal C=2(\gamma_c-2\gamma_m)u^2+O(u^3)$ from (14.12). Then write $\gamma_c-2\gamma_m=\frac{a_2(a_2-2a_1)}{M_1}+\frac{a_1(a_1-2a_2)}{M_2}$ and use $x_2=1-x_1$, $y_1=1-y_2$:
$$
c_2=K_2\,u^2+O(u^3),\qquad K_2=\frac{\gamma_c-2\gamma_m}{2a}-\frac{a}{2(M_1+M_2)}=\frac{a(M_1+M_2)}{2M_1M_2}\Bigl[(x_1-y_2)^2-2x_1x_2\Bigr],
\tag{16.13}
$$
with $x_i:=a_i/a$ and $y_i:=M_i/(M_1+M_2)$. By $a_i=1/(4\mathfrak m_i)$, $x_1=\mathfrak m_2/(\mathfrak m_1+\mathfrak m_2)$: $x_1-y_2$ is the inertia share minus the rest-cost share of body 2. The prefactor is positive, so $K_2=0$ if and only if $(x_1-y_2)^2=2x_1x_2$, that is $3x_1^2-2(1+y_2)x_1+y_2^2=0$, or
$$
x_1=\tfrac13\bigl(1+y_2\pm\sqrt{1+2y_1y_2}\bigr).
$$
For every $y_2\in(0,1)$ both roots lie in $(0,1)$: $x_1^+<1\iff 2y_1y_2<2y_1+y_1^2$, and $x_1^->0\iff 3y_2^2>0$. So $K_2$ vanishes exactly on two surfaces of the data, and each surface meets every ratio $M_1/M_2$. Equal data of the two types ($x_1=y_2=\frac12$) give $K_2=-a/(2M_1)\neq0$. The correction taken relative to the leading value $1/\bar s$ of $a_{\rm pair}/\bar a$, namely $\bar s\,c_2$, has the same leading term $K_2u^2$.

## Result

All results hold for the setting under "Setup and assumptions": both signs $\varsigma=\pm1$, fixed hatted data (fixed $u\in(0,\infty)$), first order in the gradients, records at leading order in $\kappa$, packets at $P_0=0$. Every $O(\epsilon^n)$ holds at fixed $u$, with constants that depend on the hatted data.

1. **Item 1 (formula; every diagonal on-site binding with $\hat w_{11}\neq0$, all three gradients).** Exactly, $1+r_i=-\varsigma\bar s(\mathfrak m_1+\mathfrak m_2)E_b''(0)\langle\sigma^{(i)}_3\rangle_0$ and $1+r_X=-\varsigma\bar s(\mathfrak m_1+\mathfrak m_2)E_b''(0)$ (16.3). At leading order, $r_i=-\beta_i(1+\varsigma/\bar s)-\frac{\varsigma\bar s}{4a}\mathcal C+O(\epsilon^4)$ and $r_X=-\frac{\varsigma\bar s}{4a}\mathcal C+O(\epsilon^4)$ (16.4), with $\mathcal C$ of (14.9). The symbol $\Lambda$ of (14.8) is $\Lambda=1+\varsigma/\bar s$ (16.2).
2. **Item 2 (formula; common gradient $g_{\rm c}\neq0$, $\langle X\rangle=0$).** $\mathcal W/g_{\rm c}-E_b(0)=-\varsigma a(\bar s+\varsigma)^2/\bar s+\mathcal V_4+O(\epsilon^6)$, with $\mathcal V_4=(u\partial_u-4)\mathcal E_4|_{\gamma_s,\gamma_m}=\frac{\gamma_s}2\psi_A+\frac{\gamma_s+\gamma_m}2\psi_B$ (16.6) and $\psi_A,\psi_B$ of (16.7). At order $\epsilon^2$ the weight per unit gradient is $M_1+M_2-\frac{\varsigma}{\bar s}\,a(1+\varsigma\bar s)$. The binding cost is weighed with the factor $-\varsigma/\bar s$, not 1. The order-$\epsilon^2$ difference is nonzero for every $u\in(0,\infty)$ and both signs, and has the sign $-\varsigma$.
3. **Item 3 (formula; common gradient $g_{\rm c}\neq0$, $\langle X\rangle=0$).** Exactly, $a_{\rm pair}/\bar a-1=(\delta_W-\delta_{\mathfrak m})/(1+\delta_{\mathfrak m})$ (16.9). Here $\delta_W$ and $\delta_{\mathfrak m}$ (16.8) are the relative changes of the weight against $g_{\rm c}(M_1+M_2)$ and of the inertia against $\mathfrak m_1+\mathfrak m_2$, given by (16.10). Through relative order $\epsilon^2$, $a_{\rm pair}/\bar a-1=-1-\varsigma/\bar s+c_2+O(\epsilon^4)$, with $c_2=\frac{\varsigma}{\bar s}\frac{a}{M_1+M_2}(1+\frac{\varsigma}{\bar s})+\frac{\mathcal C}{4a}$ (16.11).
4. **Item 4 (formula; $w_{11}<0$, ordered limit: first $\epsilon\to0$ at fixed $u$, then $u\to0$ in each coefficient).** The order-$\epsilon^2$ coefficient of item 2 is $\frac a4u^4+O(u^6)$. The order-$\epsilon^4$ coefficient is $-\frac12\gamma_mu^3+O(u^5)$, and $\gamma_m>0$ for all data, so its leading term vanishes on no data (16.12). The order-$\epsilon^2$ correction of item 3 is $c_2=K_2u^2+O(u^3)$, with $K_2=\frac{a(M_1+M_2)}{2M_1M_2}[(x_1-y_2)^2-2x_1x_2]$, $x_i=a_i/a$, $y_i=M_i/(M_1+M_2)$ (16.13). $K_2=0$ if and only if $a_1/a=\frac13\bigl(1+y_2\pm\sqrt{1+2y_1y_2}\bigr)$. These are two surfaces of the data, each meeting every ratio $M_1/M_2$; equal data of the two types are not on them.

## Consistency checks

1. **Dimensions.** $\beta_i$, $\mathcal C/a$, $\gamma/a$ and $a/M_i$ are dimensionless, and so are $r_i,r_X,\delta_W,\delta_{\mathfrak m},c_2,K_2$. $a$, $\gamma_s,\gamma_m$ and $\mathcal V_4$ are costs, as is $\mathcal W/g_{\rm c}-E_b(0)$. Passed.
2. **Ordered limit $u\to0$ at $\varsigma=-1$ against the free parts.** $\Lambda=1-1/\bar s\to0$, and $\mu=-\frac{\gamma_c-2\gamma_m}{2a}u^2+O(u^3)\to0$ by (14.12). Therefore $r_1,r_2,r_X\to0$, $1+\delta_{\mathfrak m}\to1$, and $-1+1/\bar s\to0$ and $c_2\to0$. Both coefficients of item 2 also tend to 0. Coefficient by coefficient, the wide pair falls with $\bar a$ and weighs $E_b(0)$ per unit gradient. Passed.
3. **Independent route to (16.12).** From the $X_k$ of (14.12), with $X^{(2)}_k$ obtained by exchanging the labels: $\sum_ia_iX^{(i)}_3=0$, $\sum_ia_iX^{(i)}_4=\frac{\gamma_s}8-\frac{\gamma_m}4$ and $\sum_ia_iX^{(i)}_5=-\frac{\gamma_m}2$. From (14.7), $\mathcal E_4=\frac{\gamma_m}2u^3+(\frac{\gamma_s}8-\frac{\gamma_m}4)u^4-\frac{\gamma_m}4u^5+O(u^6)$. Then $\mathcal V_4=\sum_ia_i\Xi_i-\mathcal E_4=-\frac{\gamma_m}2u^3+0\cdot u^4-\frac{\gamma_m}4u^5+O(u^6)$, the same as from (16.7). Passed.

## Open issues

- The question does not define the notation $a_i$, $a$, $u$, $\varsigma$, $\epsilon$ or the symbol $\Lambda$ of (14.8). Step 1 reads them off from the consistency of (11.5), (14.7)–(14.10) and (4.16); the reading should be checked against 14-pair-level-next-order.
- Joint limits in $\epsilon$ and $u$ are out of scope. At $\varsigma=-1$ the order-$\epsilon^2$ coefficient of item 2 is $O(u^4)$ and the order-$\epsilon^4$ one is $O(u^3)$, so which one dominates depends on how $\epsilon$ and $u$ go to zero together.
- On the two surfaces where $K_2=0$, the leading term of $c_2$ is not determined (restricted by the question).
- Item 2 compares $\mathcal W/g_{\rm c}$ with $E_b(0)$ in the zero of the cost fixed by $G_i=M_i\sigma^{(i)}_3$. Couplings to other operators (criterion 6) are not treated.

## Methods used

- Hellmann–Feynman relation (11.5) summed over the rest terms
- Matching coefficients of expressions linear in the gradients
- Expansion in $\epsilon$ at fixed $u$; Euler-type operator $u\partial_u-4$
- Ordered limit and Taylor expansion of coefficients in $u$
- Zero set of a quadratic form in the data

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