\section{The entangling rate against the model's own dephasing}
\label{sec:rate}

\subsection{The first-order rate}

For equal masses write the rate~\eqref{eq:rate} in units of the Newtonian rate,
$\dot\lambda=-\kappa\,Gm^2/(\hbar d)$, with $d$ the separation and $\kappa\geq0$ dimensionless.
Eq.~\eqref{eq:rate} is also the initial rate when a local Hamiltonian is
present~\cite{TrilloNavascues2025}. The constant $\kappa$ is bounded, because the coupling, of weight
$\sqrt{1+C^2}$, must first pay for the dephasing of its own witness, of weight $C$.

\begin{theorem}[Rate bounds]
\label{thm:ratebounds}
Let $\theta=\min(1,\sth/\rmax)$.
\begin{enumerate}
\item[(i)] For every product state with closest approach $d$ and every $\sigma$,
$\kappa\leq(1-\theta)\sqrt{1+C^2}\,d f(d)\leq\sqrt{1+C^2}-C$.
\item[(ii)] For supports separated by a slab of width $d$, with the Coulomb kernel ($\sigma\to0$),
\begin{equation}
  \kappa\leq B(C)=\sqrt{1+C^2}-C-C\ln\frac{\sqrt{1+C^2}}{C},
  \label{eq:BC}
\end{equation}
which is $0.068$ at $C=1$ and falls as $1/(8C^3)$ for $C\to\infty$. For planar supports at finite $\sigma$
the corresponding bound is larger (Appendix~\ref{app:values}), so~\eqref{eq:BC} is a statement for
$\sigma\ll d$.
\end{enumerate}
\end{theorem}

The entangling rate is therefore a fraction of the Newtonian rate for every state, and a few per cent of
it for slab-separated states at $\sigma\ll d$. Explicit states show how much of the bound can be reached.
Give each mass an equal superposition of a planar sheet and a far copy of it, the two sheets facing each
other at gap $d$, in the formal limits $\sigma\to0$ and zero thickness. A Gaussian sheet of the best width
has $\kappa_G=0.027$ at $C=1$ (Eq.~\eqref{eq:kappaG} in Appendix~\ref{app:bounds}). So at $C=1$ and
$\sigma\ll d$ the best rate constant of slab-separated states lies between $0.027$ and $0.068$. The lower
end is a proven formula for one witness, in limits we do not control; at finite $d/\sigma$ smearing raises
both ends by corrections of first order in $\sigma/d$.

\subsection{The ratio to the dephasing rate}

The entangling rate has to be compared with the rate at which the same dynamics destroys the
superpositions it acts on. The coherence between two branches $a$, $a'$ of one body decays at the
Di\'{o}si--Penrose rate $C\,Gm^2[f(0)-f(|a-a'|)]/\hbar$. Because the body is rigid and only its centre is
delocalised, this saturates for $|a-a'|$ larger than a few $\seff$:
\begin{equation}
  \Gsat=\frac{C\,Gm^2K_0}{\hbar},\qquad K_0=\frac{1}{\sqrt\pi\,\seff},
  \label{eq:Gsat}
\end{equation}
with $\seff=\sigma$ for point masses and $K_0=6/(5R)$, $\seff=5R/(6\sqrt\pi)=0.470\,R$, for uniform spheres of
radius $R\gg\sigma$. A wide superposition does not dephase faster per pair. For extended bodies
Theorem~\ref{thm:criterion} and the bounds of this section hold verbatim with $f$ replaced by the Coulomb
energy $K(r)$ of two smeared bodies, which for $\sigma\ll R$ is $1/r$ beyond contact; at large separation
only the self-term $K_0$ changes, and the radius takes the place of $\sigma$ however small $\sigma$ is. With
$\Gent=|\dot\lambda|=\kappa\,Gm^2/(\hbar d)$,
\begin{equation}
  \Rr\equiv\frac{\Gent}{\Gsat}=\frac{\kappa}{CK_0d}=\frac{\sqrt\pi\,\kappa}{C}\,\frac{\seff}{d}.
  \label{eq:ratio}
\end{equation}
This is an identity, and its bounds are those of Theorem~\ref{thm:ratebounds}: the Newtonian rate falls as
$1/d$ while the dephasing rate is fixed by the size of the body, and $\kappa$ is small. At $C=1$ the ratio
is at most $0.12\,\seff/d$ for slab-separated supports at $\sigma\ll d$, and the best state known reaches
$0.048\,\seff/d$; at large $C$, $\Rr\leq(\sqrt\pi/8C^4)\,\seff/d$ (Fig.~\ref{fig:rate};
Table~\ref{tab:ratio} gives the dependence on $C$). As a function of the extent, $\Rr$ vanishes below a
threshold, peaks near an extent of a few gaps, and falls off like gap over extent beyond: a wider sheet is
not better.

\begin{figure}[t]
\centering
\includegraphics[width=0.64\textwidth]{fig-rate.pdf}
\caption{Entangling rate over single-body dephasing rate, in units of $\seff/d$, at $C=1$, for a Gaussian
sheet of rms width $b$ and a uniform disc of radius $a$ (each with a far copy, $\sigma\to0$), against the
extent in units of the gap. The dashed line is the ceiling~\eqref{eq:BC} for all slab-separated states.}
\label{fig:rate}
\end{figure}

The same dephasing is already inside the criterion, divided by the number of branches. The first term
of the numerator of~\eqref{eq:rate} is the Di\'{o}si--Penrose dephasing rate of the collective coherence
that the witness reads. For a smooth pattern over $n$ point-like branches its point self-energies $f_0\sum|q(a)|^2$ are
diluted as $1/n$, and to leading order the criterion $\rmax>\sth$ reads $n\,\Rr_{\rm coll}>1$, with
$\Rr_{\rm coll}$ the ratio of the collective rate to $\Gsat$.

\subsection{The extent law: how wide the superposition must be}
\label{sec:extent}

For sheets the criterion is the electrostatics of a plate condenser. For coaxial discs of radius $a$ at gap
$d\gg\sigma$, each with a
far copy, the criterion becomes scale-free (formal limit $\sigma\to0$, zero thickness). Writing the Hankel
transform of an axisymmetric charge as an Abel transform, $\hat q(k)=\int_0^1g(t)\cos(kt)\,dt$, turns the
canonical correlation into the top eigenvalue of Love's operator for the circular plate
condenser~\cite{Love1949},
\begin{equation}
  (T_\delta g)(t)=\frac1\pi\int_{-1}^{1}\frac{\delta\,g(s)}{\delta^2+(t-s)^2}\,ds,\qquad\delta=d/a,
  \label{eq:love}
\end{equation}
and the criterion is $\lambda_{\max}(T_{d/a})>\sth$ (Appendix~\ref{app:bounds}, which also gives the
condition~\eqref{eq:Phi} for the uniform pattern alone to be a witness). At $C=1$ the coherent
diameter must exceed about $5.5$ gaps, with the far copy. On a single connected sheet the best pattern
found is dipolar and needs a diameter of $14.8$ gaps. The required extent grows like $C^2$ at large $C$. In
a lattice with spacing larger than $\sigma$, reaching the threshold needs at least about $15\,d/\sigma$
point-like branches per mass. The thresholds, computed by converged Nystr\"om discretisation
of~\eqref{eq:love}, are in Table~\ref{tab:Cdep}, and the lattice counts in Appendix~\ref{app:values}.

\subsection{The size of the entanglement}
\label{sec:negativity}

The rate bounds limit how fast entanglement starts. The next theorem limits how large it becomes. Below,
the negativity is $|\lambda_{\min}(\rho_t^{T_1})|$, the quantity plotted in~\cite{TrilloNavascues2025}.

\begin{theorem}[Negativity bound]
\label{thm:negativity}
For $H=0$, equal masses, any product state with weights $p$, $r$ on finite branch sets, $\theta$ as in
Theorem~\ref{thm:ratebounds}, and all $t'=Gm^2t/(\hbar\sigma)\geq0$ (lengths in units of $\sigma$),
\begin{equation}
  |\lambda_{\min}(\rho_t^{T_1})|\leq(1-\theta)\,t'\Big\{\sum p_ap_{a'}r_br_{b'}\,(C^2\delta^2+\epsilon^2)\,
  e^{-2t'C\,(D_A+D_B+\min(\theta\delta,\delta))}\Big\}^{1/2},
  \label{eq:negbound}
\end{equation}
the sum over all $a,a',b,b'$, with $\epsilon=F(a,b)-F(a',b')$. If the branches of each mass are at least
$\ell$ apart, $D_\ell=f_0-f(\ell)>f(d)$ and $C\geq1/\sqrt3$, then
$\sup_t|\lambda_{\min}|\leq(1-\theta)(f(d)/e)\big[1/(D_\ell-f(d))+\sqrt{\sum p^2+\sum r^2}/(CD_\ell)\big]$; for
$\ell,d\gg\sigma$ and many branches this is at most $(\sqrt\pi/e)(1-\sth)\,\sigma/d$, which is
$0.19\,\sigma/d$ at $C=1$.
\end{theorem}

For many branches more than a few $\sigma$ apart the entanglement is thus at most of order $\sigma/d$ at
all times, whatever the state. The proof (Appendix~\ref{app:bounds}) compares the model with a copy of
itself whose cross kernel is weakened to $\theta F$, which is separable by Remark~\ref{rem:general}. For
states with structure below $\sigma$ the sum~\eqref{eq:negbound} has to be evaluated as it stands, and a
$\sigma/d$ bound for all product states is not proven.

\emph{A leading-order law.} The bound lies well above what explicit states reach. Expanding the exact
matrix elements to first order in the cross kernel, with the local dephasing kept exactly, gives an
estimate of the actual size. For $n$
branches per mass with uniform weights it is $n\lambda_{\min}\simeq e^{-2\tau}[e^\tau-1-\xi\tau]$ up to
terms of relative order $1/n$, with $\tau=\Gsat t$ and $\xi=n\Rr_{\rm coll}$ the quantity whose exceeding
one is the criterion. The factor $e^{-2\tau}$ is the survival of one coherence on each body, $e^\tau-1$ the
mixedness the dephasing creates in the witness subspace, $-\xi\tau$ the entangling term. For $\xi\gg1$,
\begin{equation}
  \lambda_{\min}(t)\simeq-\Gent\,t\,e^{-2\Gsat t},\qquad
  \max_t|\lambda_{\min}|\simeq\frac{\Rr}{2e}=\frac{\sqrt\pi}{2e}\,\frac{\kappa}{C}\,\frac{\seff}{d}
  \quad\text{at}\quad t_*=\frac{1}{2\Gsat}.
  \label{eq:law}
\end{equation}
The entanglement is what the entangling rate can build in half a dephasing time, so ``negativity of order
$10^{-2}\,\seff/d$'', the value for Gaussian sheets at $C=1$, is the statement that the entangling rate is
a fraction of the dephasing rate, not an extra suppression. The law is derived; its error is not bounded.
At $C=1$ its finite-$\xi$ form agrees to a few per cent with exact partial transposes of uniform lattice
sheets at separations of $4\,\sigma$ and $6\,\sigma$; at $C=2$ it overestimates the peak.

One explicit state shows both halves of the result (Appendix~\ref{app:values}). A sheet state with 38
branches per mass at slab separation $3.52\,\sigma$, beyond the two-branch ceiling of
Proposition~\ref{prop:ceilings}, is entangled, with an exact peak negativity of order $10^{-3}$, against
$0.41$ for unitary Newtonian gravity on the same state.