#  Supporting Information:
The Mechanics of Nucleation and Growth and the Surface Tensions of Active
Matter

Luke Langford1 Ahmad K. Omar1,2†

(1Department of Materials Science and Engineering, University of California,
Berkeley, CA, USA 94720
2Materials Sciences Division, Lawrence Berkeley National Laboratory, Berkeley,
California 94720, USA )

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##  1 Mechanics of the Critical Nucleus

In this Section, we detail the theory and numerical procedures used to solve
for the stationary density profile of the critical nucles of a system
arbitrarily far from equilibrium. We first derive the finite-sized phase
coexistence criteria for active particles by extending the coexistence theory
used to obtain nonequilibrium binodals [1]. We then apply this theory to
active Brownian particles (ABPs). In addition, we provide a step by step walk
through of the numerical scheme used to translate phase coexistence criteria
into density profiles and critical radii. Lastly, we demonstrate how the
generalized Gibbs-Duhem relation found in Ref. [2] can be used to rewrite the
coexistence criteria in terms of the generalized chemical potential discussed
in Ref. [3]. Throughout this section, as well as the rest of the Supporting
Information, we assume three dimensions ($d=3$) unless specified otherwise.

###  1.1 Nonequilibrium Critical Nucleus Coexistence Criteria

We aim to describe the structure of a critical nucleus with density $\rho^{\rm
in}$ in a system that has been prepared at a supersaturation density
$\rho^{\rm sat}$. We assume that the critical nucleus is sufficiently small
compared to the system size such that the density outside the nucleus is well
approximated by the supersaturation density $\rho^{\rm sat}$. In addition, we
approximate the geometry of the critical nucleus as a perfect sphere with
radius $R_{C}$.

When the critical nucleus is an unstable fixed point, a nucleus smaller than
the critical nucleus will experience a flux normal to its interface such that
the nucleus shrinks. A nucleus larger than the critical nucleus will
experience the opposite flux normal to the interface such that the nucleus
grows. Therefore, in order to solve for the critical nucleus, we seek states
which have zero flux normal to the interface. Even if the critical nucleus is
a stable fixed point, as recently found in some active systems [4], similar
arguments may be used to still conclude the critical nucleus will be flux-
free. Two flux-free states, the macroscopically phase separated and metastable
homogeneous states, are already known. The critical nucleus will correspond to
a state with a finite-size minority phase and zero flux normal to the
interface. The evolution of the density flux
$\mathbf{J}=\rho(\mathbf{r},t)\mathbf{u}(\mathbf{r},t)$, where $\mathbf{u}$ is
the number-averaged velocity of particles, can be expressed as a balance of
linear momentum:

$\frac{\partial\left(m\mathbf{J}\right)}{\partial
t}=-\bm{\nabla}\cdot\left(\frac{m\mathbf{J}\mathbf{J}}{\rho}\right)+\bm{\nabla}\cdot\bm{\sigma}+\mathbf{b},$
(1.1)

where $m$ is the mass of each particle, $\bm{\sigma}$ is the stress tensor and
$\mathbf{b}$ are the body forces acting on the particles. Additionally, we
assume that $\bm{\sigma}$ and $\mathbf{b}$ can be written as local gradient
expansions, which is typically valid when interactions are short-ranged. For
an overdamped system with spherical symmetry and zero flux in the radial
direction, Eq. (1.1) reduces to
$\bm{\nabla}\cdot\bm{\sigma}+\mathbf{b}=\bm{0}$. Exact microscopic expressions
for both $\bm{\sigma}$ and $\mathbf{b}$ can be determined through an Irving-
Kirkwood-Noll procedure [5, 6, 7]. These microscopic expressions can include
terms that appear to be non-local, requiring integrals over all space to, for
instance, compute the stress at a particular point in space arising from
particle interactions. However, these terms are often well-approximated by a
local gradient expansion, the coefficients of which can be systematically
determined [8]. We thus focus on systems in which $\bm{\sigma}$ and
$\mathbf{b}$ can be fully expressed as locally.

Up to a divergence-free term, we may absorb the body force into a stress-like
contribution such that:

$\bm{\nabla}\cdot\bm{\sigma}+\mathbf{b}=\bm{\nabla}\cdot\left[\bm{\sigma}+\mathbf{F}\right]\equiv\bm{\nabla}\cdot\bm{\Sigma}=\bm{0},$
(1.2a)
$\mathbf{F}=\frac{1}{4\pi}\int\text{d}\mathbf{r}^{\prime}\mathbf{b}(\mathbf{r}^{\prime})\frac{\mathbf{r}-\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}},$
(1.2b)

where we have used the Green’s function identity
$\bm{\nabla}\cdot\left((\mathbf{r}-\mathbf{r}^{\prime})/|\mathbf{r}-\mathbf{r}^{\prime}|^{3}\right)=4\pi\delta(\mathbf{r}-\mathbf{r}^{\prime})$
in three dimensions. Details on evaluating the integral Eq. (1.2b) are
provided in Section 1.5. In general, $\mathbf{F}$ will contain terms which may
be expressed as local density gradient expansions as well as non-local terms
which depend on spatial integrals over the entire density field. Due to
spherical symmetry, $\bm{\Sigma}$ has only two independent components and is
given by:

$\bm{\Sigma}=\begin{bmatrix}\Sigma_{rr}&0&0\\\ 0&\Sigma_{tt}&0\\\
0&0&\Sigma_{tt}\end{bmatrix}=\Sigma_{rr}\mathbf{e}_{r}\otimes\mathbf{e}_{r}+\Sigma_{tt}\left(\mathbf{I}-\mathbf{e}_{r}\otimes\mathbf{e}_{r}\right).$
(1.3)

We now find the flux-free steady states by demanding the dynamic stress tensor
has zero divergence, and we expand the stress tensor to second order in
density gradients:

$\displaystyle\bm{\nabla}\cdot\bm{\Sigma}=\mathbf{0},$ (1.4a)
$\displaystyle\Sigma_{rr}=-\mathcal{P}(\rho)+\kappa_{1}(\rho)\left(\frac{\partial}{\partial
r}+\frac{2}{r}\right)\frac{\partial\rho}{\partial r}$
$\displaystyle+\left(\kappa_{2}(\rho)+\kappa_{3}(\rho)\right)\left(\frac{\partial\rho}{\partial
r}\right)^{2}+\kappa_{4}(\rho)\frac{\partial^{2}\rho}{\partial
r^{2}}+\mathcal{K}_{r},$ (1.4b)
$\displaystyle\Sigma_{tt}=-\mathcal{P}(\rho)+\kappa_{1}(\rho)$
$\displaystyle\left(\frac{\partial}{\partial
r}+\frac{2}{r}\right)\frac{\partial\rho}{\partial
r}+\kappa_{2}(\rho)\left(\frac{\partial\rho}{\partial
r}\right)^{2}+\mathcal{K}_{t}.$ (1.4c)

Here $\mathcal{P}(\rho)$ is the bulk isotropic contribution to $\bm{\Sigma}$,
which is the dynamic pressure, $\\{\kappa_{i}\\}$ are density gradient
expansion coefficients, and $\mathcal{K}_{t/r}$ are the non-local terms
obtained from evaluating Eq. (1.2b).

In spherical coordinates, the divergence of $\bm{\Sigma}$ is given by:

$0=\frac{\partial\Sigma_{rr}}{\partial r}-\frac{2}{r}\mathcal{G}(\rho),$ (1.5)

where we have defined the “excess stress”:

$\mathcal{G}(\rho)=\Sigma_{tt}-\Sigma_{rr}.$ (1.6)

The definite integral of $\mathcal{G}(\rho)$ across the interface is precisely
the mechanical surface found by Kirkwood and Buff [9]:

$\upgamma_{\rm
mech}=\int_{0}^{\infty}\left[\Sigma_{tt}-\Sigma_{rr}\right]\text{d}r.$ (1.7)

For active particles with hard-sphere interactions, this mechanical surface
tension was found to be strongly negative [10].

Let us consider the definite integral of Eq. (1.5) from $r=0$ to $\infty$:

$\mathcal{P}(\rho^{\rm in})-\mathcal{P}(\rho^{\rm
sat})-\mathcal{K}_{r}\bigr{|}_{r=0}+\mathcal{K}_{r}\bigr{|}_{r=\infty}=\int_{0}^{\infty}\left[\frac{2}{r}\mathcal{G}(\rho)\right]\text{d}r.$
(1.8)

Within a region of homogeneous density, $\bm{\Sigma}$ becomes isotropic.
Therefore $\mathcal{G}$ is only non-zero in the vicinity of the interface. If
the width of the interfacial region is small compared to the radius of the
critical nucleus, then we may approximate Eq. (1.8) as:

$\mathcal{P}(\rho^{\rm in})-\mathcal{P}(\rho^{\rm
sat})-\mathcal{K}_{r}\bigr{|}_{r=0}+\mathcal{K}_{r}\bigr{|}_{r=\infty}=\frac{2\upgamma_{\rm
mech}}{R_{C}}+\mathcal{O}\left(\frac{1}{R_{C}^{2}}\right),$ (1.9)

which demonstrates that the Laplace pressure difference holds in active
systems only without non-local stress components and is broken by systems with
non-local stress components. A derivation of an active Laplace pressure
difference was discussed in Ref. [3], however this work did not consider the
contribution of non-local stress terms.

Suppose we had taken the indefinite integral of Eq. (1.5) rather than the
definite integral. Doing so results in:

$\Sigma_{rr}(r)=-\mathcal{P}(\rho)+\kappa_{1}(\rho)\left[\frac{\partial}{\partial
r}+\frac{2}{r}\right]\frac{\partial\rho}{\partial
r}+\bigl{(}\kappa_{2}(\rho)+\kappa_{3}(\rho)\bigr{)}\left(\frac{\partial\rho}{\partial
r}\right)^{2}+\kappa_{4}(\rho)\frac{\partial^{2}\rho}{\partial
r^{2}}+\mathcal{K}_{r}=W(r)+C,$ (1.10)

where $C$ is a constant of integration and we have defined $W$ as:

$W(r)=\int_{0}^{r}\left[\frac{2\mathcal{G}}{r}\right]\text{d}r.$ (1.11)

While we do not know the value of $W(r)$ for any arbitrary $r$, we have
already approximated $W(\infty)$ as $2\upgamma_{\rm mech}/R_{C}$ when $R_{C}$
is much larger than the interfacial width. In addition, L’Hôpital’s rule can
be used to show that the integrand of $W(r)$ is zero at $r=0$, which implies
that $W(0)=0$. In order to determine the precise value of $C$, we recall that
the density as $r\rightarrow\infty$ is $\rho^{\rm sat}$:

$\Sigma_{rr}(\infty)=W(\infty)+C=\frac{2\upgamma_{\rm
mech}}{R_{C}}+C=-\mathcal{P}(\rho^{\rm
sat})+\mathcal{K}_{r}\bigr{|}_{r=\infty}$ (1.12)

Therefore

$C=-\mathcal{P}(\rho^{\rm sat})-\frac{2\upgamma_{\rm
mech}}{R_{C}}+\mathcal{K}_{r}\bigr{|}_{r=\infty}$ (1.13)

We are then left with the equation:

$\kappa_{1}(\rho)\left[\frac{\partial}{\partial
r}+\frac{2}{r}\right]\frac{\partial\rho}{\partial
r}+\bigl{(}\kappa_{2}(\rho)+\kappa_{3}(\rho)\bigr{)}\left(\frac{\partial\rho}{\partial
r}\right)^{2}+\kappa_{4}(\rho)\frac{\partial^{2}\rho}{\partial
r^{2}}+\mathcal{K}_{r}=W(r)+\mathcal{K}_{r}\bigr{|}_{r=\infty}-\frac{2\upgamma_{\rm
mech}}{R_{C}}+\left(\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm sat})\right).$
(1.14)

We now introduce the generalized Maxwell construction pseudovariable [11, 12,
1] $\mathcal{E}$ defined by the differential equation:

$\bigl{(}\kappa_{1}(\rho)+\kappa_{4}(\rho)\bigr{)}\frac{\partial^{2}\mathcal{E}}{\partial\rho^{2}}=\left(2\left(\kappa_{2}(\rho)+\kappa_{3}(\rho)\right)-\frac{\partial}{\partial\rho}\left(\kappa_{1}(\rho)+\kappa_{4}(\rho)\right)\right)\frac{\partial\mathcal{E}}{\partial\rho}$
(1.15)

For passive systems, $\\{\kappa_{i}(\rho)\\}$ are dictated by the Korteweg
expansion [13], resulting in $\mathcal{E}\sim 1/\rho$. This pseudovariable was
found to eliminate the need to evaluate interfacial terms when solving for the
macroscopic binodals, thus providing coexistence criteria in terms of bulk
equations of state. In our case multiplying by the derivative of the
pseudovariable with respect to $r$ and integrating from $r=0$ to $r=\infty$
will eliminate some, but not all, of the interfacial terms, resulting in:

$\displaystyle\int_{0}^{\infty}\left[\frac{2\kappa_{1}(\rho)}{r}\frac{\partial\mathcal{E}}{\partial\rho}\left(\frac{\partial\rho}{\partial
r}\right)^{2}+\mathcal{K}_{r}\frac{\partial\mathcal{E}}{\partial\rho}\frac{\partial\rho}{\partial
r}\right]\text{d}r=$
$\displaystyle\mathcal{K}_{r}\bigr{|}_{r=\infty}\left(\mathcal{E}^{\rm
sat}-\mathcal{E}^{\rm in}\right)-\frac{2\upgamma_{\rm
mech}}{R_{C}}\left(\mathcal{E}^{\rm sat}-\mathcal{E}^{\rm in}\right)$
$\displaystyle+\int_{0}^{\infty}\frac{\partial\mathcal{E}}{\partial
r}W(r)\text{d}r+\int_{\mathcal{E}^{\rm in}}^{\mathcal{E}^{\rm
sat}}\left[\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm
sat})\right]\text{d}\mathcal{E}.$ (1.16)

We now integrate the first integral on the right hand side of Eq. (1.16) by
parts to find:

$\int_{0}^{\infty}\frac{\partial\mathcal{E}}{\partial
r}W(r)\text{d}r=\mathcal{E}^{\rm sat}\frac{2\upgamma_{\rm
mech}}{R_{C}}-\int_{0}^{\infty}\frac{2}{r}\mathcal{E}\mathcal{G}\text{d}r.$
(1.17)

We also integrate the second integral on the left hand side of Eq. (1.16) by
parts to find:

$\int_{0}^{\infty}\mathcal{K}_{r}\frac{\partial\mathcal{E}}{\partial
r}\text{d}r=\mathcal{E}^{\rm
sat}\mathcal{K}_{r}\bigr{|}_{r=\infty}-\mathcal{E}^{\rm
in}\mathcal{K}_{r}\bigr{|}_{r=0}-\int_{0}^{\infty}\frac{\partial\mathcal{K}_{r}}{\partial
r}\mathcal{E}\text{d}r.$ (1.18)

Substitution of Eqs. (1.17) and (1.18) into Eq. (1.16) results in:

$\displaystyle\int_{0}^{\infty}\left[\frac{2\kappa_{1}(\rho)}{r}\frac{\partial\mathcal{E}}{\partial\rho}\left(\frac{\partial\rho}{\partial
r}\right)^{2}-\frac{\partial\mathcal{K}_{r}}{\partial
r}\mathcal{E}\right]\text{d}r$ $\displaystyle=\mathcal{E}^{\rm
in}\left(\mathcal{K}_{r}\bigr{|}_{r=0}-\mathcal{K}_{r}\bigr{|}_{r=\infty}\right)-\int_{0}^{\infty}\frac{2}{r}\mathcal{E}\mathcal{G}\text{d}r$
$\displaystyle+\frac{2\upgamma_{\rm mech}}{R_{C}}\mathcal{E}^{\rm
in}+\int_{\mathcal{E}^{\rm in}}^{\mathcal{E}^{\rm
sat}}\left[\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm
sat})\right]\text{d}\mathcal{E}.$ (1.19)

Finally, we apply our approximation that $R_{C}$ is much larger than the
interfacial width to find:

$\int_{\mathcal{E}^{\rm in}}^{\mathcal{E}^{\rm
sat}}\left[\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm
sat})\right]\text{d}\mathcal{E}=\frac{2}{R_{C}}\int_{0}^{\infty}\left[\kappa_{1}(\rho)\frac{\partial\mathcal{E}}{\partial\rho}\left(\frac{\partial\rho}{\partial
r}\right)^{2}+\mathcal{G}\left(\mathcal{E}-\mathcal{E}^{\rm
in}\right)+\frac{R_{C}}{2}\frac{\partial\mathcal{K}_{r}}{\partial
r}\left(\mathcal{E}^{\rm in}-\mathcal{E}\right)\right],$ (1.20)

where by making this approximation we have introduced error
$\mathcal{O}\left(R_{C}^{-2}\right)$. We now define the “Ostwald” tension [4]
as:

$\upgamma_{\rm ost}=\frac{\rho^{\rm sat}-\rho^{\rm in}}{\rho^{\rm
sat}\left(\mathcal{E}^{\rm sat}-\mathcal{E}^{\rm
in}\right)}\left[\int_{0}^{\infty}\left(\kappa_{1}(\rho)\frac{\partial\mathcal{E}}{\partial\rho}\left(\frac{\partial\rho}{\partial
r}\right)^{2}+\mathcal{G}\left(\mathcal{E}-\mathcal{E}^{\rm
in}\right)+\frac{R_{C}}{2}\frac{\partial\mathcal{K}_{r}}{\partial
r}\left(\mathcal{E}^{\rm in}-\mathcal{E}\right)\right)\text{d}r\right],$
(1.21)

where $\upgamma_{\rm ost}$ has the same units of mechanical surface tension.
We show in Section 3.2 that $\upgamma_{\rm ost}$ becomes identical to
$\upgamma_{\rm mech}$ in a passive system where $\mathcal{E}\sim 1/\rho$.
Using the definition of the Ostwald tension, Eq. (1.20) can be rewritten as:

$\int_{\mathcal{E}^{\rm in}}^{\mathcal{E}^{\rm
sat}}\left[\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm
sat})\right]\text{d}\mathcal{E}=\frac{2\upgamma_{\rm
ost}}{R_{C}}\frac{\rho^{\rm sat}\left(\mathcal{E}^{\rm sat}-\mathcal{E}^{\rm
in}\right)}{\rho^{\rm sat}-\rho^{\rm
in}}+\mathcal{O}\left(\frac{1}{R_{C}^{2}}\right).$ (1.22)

In addition, we restate the Laplace pressure difference found above:

$\mathcal{P}(\rho^{\rm in})-\mathcal{P}(\rho^{\rm
sat})-\mathcal{K}_{r}\bigr{|}_{r=0}+\mathcal{K}_{r}\bigr{|}_{r=\infty}=\frac{2\upgamma_{\rm
mech}}{R_{C}}+\mathcal{O}\left(\frac{1}{R_{C}^{2}}\right).$ (1.23)

Any flux-free state will satisfy Eqs. (1.23) and (1.22) simultaneously. The
solution with $R_{C}=\infty$ corresponds to macroscopic phase separation and
reduces to the phase coexistence criteria found in Ref. [1]. If a system is
prepared at a saturation density $\rho^{\rm sat}$ not equal to the binodal,
simultaneous solution of Eqs. (1.23) Eq. (1.20) allows for the calculation of
the critical nucleus density $\rho^{\rm in}$ and finite critical radius
$R_{C}$.

Simultaneous solution of Eqs. (1.23) and (1.20) involves a serious
complication. The right hand sides of both equation can only be evaluated if
the density profile of the critical nucleus is already known, which in turn
depends on $R_{C}$ and $\rho^{\rm in}$. Therefore a self-consistent approach
must be employed, which we outline in Section 1.3.

###  1.2 Critical Nucleus of Phase-Separating Active Brownian Particles

We consider a system of $N$ interacting ABPs in which the time variation of
the position $\mathbf{r}_{i}$ and orientation $\mathbf{q}_{i}$ of the $i$th
particle follow overdamped Langevin equations:

$\displaystyle\dot{\mathbf{r}}_{i}$
$\displaystyle=U_{o}\mathbf{q}_{i}+\frac{1}{\zeta}\sum_{j\neq
i}^{N}\mathbf{F}_{ij},$ (1.24a) $\displaystyle\dot{\mathbf{q}}_{i}$
$\displaystyle=\mathbf{q}_{i}\times\bm{\Omega}_{i},$ (1.24b)

where $\mathbf{F}_{ij}$ is the (pairwise) interparticle force, $U_{o}$ is the
intrinsic active speed, $\zeta$ is the translational drag coefficient, and
$\bm{\Omega}_{i}$ is a stochastic angular velocity with zero mean and variance
$\langle\bm{\Omega}_{i}(t)\bm{\Omega}_{j}(t^{\prime})\rangle=2D_{R}\delta_{ij}\delta(t-t^{\prime})\mathbf{I}$
(where $\mathbf{I}$ is the identity tensor). Here, $D_{R}$ is the rotational
diffusivity, which may be athermal in origin, and is inversely related to the
orientational relaxation time $\tau_{R}\equiv D_{R}^{-1}$. One can then define
an intrinsic run length, $\ell_{o}=U_{o}\tau_{R}$, the typical distance an
ideal particle travels before reorienting. We consider a purely repulsive form
of $\mathbf{F}_{ij}$, corresponding to a sufficiently stiff interaction
potential such that hard-sphere statistics are closely approximated with an
effective hard-sphere diameter of $d_{\rm hs}$ [14, 15]. As demonstrated in
Refs. [1, 16], the form of $\mathcal{P}(\rho)$ and and
$\\{\kappa_{i}(\rho)\\}$ for ABPs is given by:

$\displaystyle\mathcal{P}(\rho)$ $\displaystyle=p_{C}(\rho)+\frac{\rho\zeta
U_{o}\ell_{o}\overline{U}(\rho)}{d(d-1)},$ (1.25a)
$\displaystyle\kappa_{1}(\rho)$
$\displaystyle=\frac{3\ell_{o}^{2}}{2d(d-1)(d+2)}\left(\overline{U}(\rho)\right)^{2}\frac{\partial
p_{C}}{\partial\rho},$ (1.25b) $\displaystyle\kappa_{2}(\rho)$
$\displaystyle=0,$ (1.25c) $\displaystyle\kappa_{3}(\rho)$
$\displaystyle=\frac{3\ell_{o}^{2}}{2d(d-1)(d+2)}\overline{U}(\rho)\frac{\partial}{\partial\rho}\left[\overline{U}(\rho)\frac{\partial
p_{C}}{\partial\rho}\right],$ (1.25d) $\displaystyle\kappa_{4}(\rho)$
$\displaystyle=0,$ (1.25e) $\displaystyle\mathcal{K}_{r}$ $\displaystyle=0,$
(1.25f) $\displaystyle\mathcal{K}_{t}$ $\displaystyle=0,$ (1.25g)

where $p_{C}(\rho)$ is the conservative pressure arising from particle
interactions, $\overline{U}(\rho)$ is the dimensionless effective active
speed, and $d(>1)$ is the system dimensionality. In Eqs. (1.25) we have
ignored the gradient contributions of the interaction potential to
$\bm{\Sigma}$, which is justified when $\ell_{o}>>1$ [3, 17, 2]. Given our
expressions for $\\{\kappa_{i}(\rho)\\}$, we can solve for the generalized
Maxwell construction pseudovariable as $\mathcal{E}\sim p_{C}(\rho)$. We note
that the theory derived in Section 1.1 is valid for small supersaturations,
i.e. the overall density of the system is close to the binodals. For
reference, we include the phase diagram of ABPs as solved for by Ref. [1] in
Fig. S.1 and include a blue star in the figure representing a point near the
binodal where our theory may hold. Finally, we remark that ABPs were found to
have no non-local terms in the stress tensor, so the familiar Laplace pressure
difference holds.

Figure S.1: Mechanically constructed phase diagram as calculated by Ref. [1].
Dashed line indicates spinodals while solid line indicates binodals. Blue star
included as an example point where supersaturation is small.

###  1.3 Numerical Scheme for Density Profile of Active Brownian Particles

Before outlining the self-consistent scheme to solve Eqs. (1.23) and (1.20)
simultaneously, we describe the numerical construction of the density profile
of the critical nucleus. Solving for this density profile is a key step of the
self-consistent scheme we will later outline.

While constructing a profile, we will assume that $R$, $\rho^{\rm in}$, and
$\upgamma_{\rm mech}$ are already known. Going from Eq. (1.14) to Eq. (1.16)
involved integrating from $r=0$ to $r=\infty$. Suppose we had instead
integrated to some arbitrary radius between $0$ and $\infty$. Doing so would
have resulted in:

$\displaystyle\frac{1}{2}\frac{\partial\mathcal{E}}{\partial\rho}\kappa_{1}(\rho)\left(\frac{\partial\rho}{\partial
r}\right)^{2}\biggl{|}^{r}_{0}=$ $\displaystyle\int_{\mathcal{E}^{\rm
in}}^{\mathcal{E}(r)}\left[\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm
in})\right]\text{d}\mathcal{E}$
$\displaystyle-\frac{2}{R_{C}}\left[\int_{0}^{r}\left(\kappa_{1}(\rho)\frac{\partial\mathcal{E}}{\partial\rho}\left(\frac{\partial\rho}{\partial
r}\right)^{2}+\mathcal{E}\mathcal{G}\right)\text{d}r-\mathcal{E}(r)\int_{0}^{r}\mathcal{G}\text{d}r\right].$
(1.26)

We now convert the integrals to be over $\rho$ instead of $\mathcal{E}$ or
$r$:

$\displaystyle\frac{1}{2}\frac{\partial\mathcal{E}}{\partial\rho}\kappa_{1}(\rho)\left(\frac{\partial\rho}{\partial
r}\right)^{2}=$ $\displaystyle\int_{\rho^{\rm
in}}^{\rho(r)}\left[\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm
in})\right]\frac{\partial\mathcal{E}}{\partial\rho}\text{d}\rho$
$\displaystyle-\frac{2}{R_{C}}\left[\int_{\rho^{\rm
in}}^{\rho(r)}\left(\kappa_{1}(\rho)\frac{\partial\mathcal{E}}{\partial\rho}\left(\frac{\partial\rho}{\partial
r}\right)+\mathcal{E}\mathcal{G}\left(\frac{\partial\rho}{\partial
r}\right)^{-1}\right)\text{d}\rho-\mathcal{E}(\rho)\int_{\rho^{\rm
in}}^{\rho(r)}\mathcal{G}\left(\frac{\partial\rho}{\partial
r}\right)^{-1}\text{d}\rho\right],$ (1.27)

where we have used the radial symmetry condition $\partial\rho/\partial r=0$
at $r=0$. We can then isolate the density gradient by rearranging and taking
the square root of both sides:

$\displaystyle\frac{\partial\rho}{\partial r}=\pm f(\rho)=$
$\displaystyle\pm\Biggl{(}\frac{2}{\kappa_{1}(\rho)}\left(\frac{\partial\mathcal{E}}{\partial\rho}\right)^{-1}\int_{\rho^{\rm
in}}^{\rho(r)}\left[\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm
in})\right]\frac{\partial\mathcal{E}}{\partial\rho}\text{d}\rho$
$\displaystyle-\frac{2}{R_{C}}\left[\int_{\rho^{\rm
in}}^{\rho(r)}\left(\kappa_{1}(\rho)\frac{\partial\mathcal{E}}{\partial\rho}\left(\frac{\partial\rho}{\partial
r}\right)+\mathcal{E}\mathcal{G}\left(\frac{\partial\rho}{\partial
r}\right)^{-1}\right)\text{d}\rho-\mathcal{E}(\rho)\int_{\rho^{\rm
in}}^{\rho(r)}\mathcal{G}\left(\frac{\partial\rho}{\partial
r}\right)^{-1}\text{d}\rho\right]\Biggr{)}^{1/2}.$ (1.28)

Equation (1.28) maps any value of density between $\rho^{\rm in}$ and
$\rho^{\rm sat}$ to its derivative in the $r$ direction. Using $\rho^{\rm in}$
as an initial value, this mapping between density values and $r$ derivatives
allows for the construction of another mapping between density values and $r$
coordinate in the density profile:

$r(\rho)=\int_{\rho^{\rm
in}}^{\rho}\frac{1}{f(\rho^{\prime})}\text{d}\rho^{\prime}.$ (1.29)

We identify this mapping as $\upvarphi_{R}(r)$, the average density profile of
the critical nucleus.

The above procedure which requires $R_{C}$, $\upgamma_{\rm mech}$, and
$\rho^{\rm in}$ can be used with the following self-consistent scheme for
simultaneous solution of Eqs. (1.23) and (1.20):

  1. 1.

Begin with an initial guess for the density profile, $\rho(r)$.

  2. 2.

Using $\rho(r)$, evaluate $\upgamma_{\rm mech}$ from Eq. (1.7).

  3. 3.

Using a given value of $\rho^{\rm sat}$, $\rho(r)$, and $\upgamma_{\rm mech}$,
find the optimal value of $R_{C}$ and $\rho^{\rm in}$ that solve Eqs. (1.23)
and (1.20).

  4. 4.

If on the first iteration of the scheme, record the Laplace pressure
difference, $2\upgamma_{\rm mech}/R_{C}$.

  5. 5.

Solve for a new density profile, $\rho^{*}(r)$ using the $\upgamma_{\rm mech}$
and the newly calculated $R_{C}$ and $\rho^{\rm in}$.

  6. 6.

Using $\rho^{*}(r)$, update $\upgamma_{\rm mech}$ from Eq. (1.7).

  7. 7.

Record the updated value of the Laplace pressure difference, $2\upgamma_{\rm
mech}/R_{C}$.

  8. 8.

If the new value of the Laplace pressure is within tolerance of the old one,
the procedure has converged and set $\upvarphi_{R}(r)=\rho^{*}(r)$. Return the
current values of $R_{c}$, $\rho^{\rm in}$, and the newest density profile.
Otherwise, repeat the process starting from Step 1 using $\rho^{*}(r)$ as the
updated $\rho(r)$.

The above numerical scheme was used to calculate the data plotted in Figs. 1
and 2 of the main text. The density profile corresponding to macroscopic phase
separation was used as the initial guess of the above numerical scheme for the
closest value of $\rho^{\rm sat}$ to the binodal, and the density profile
obtained there was used as the initial guess for the next closest value of
$\rho^{\rm sat}$ to the binodal, etc.

###  1.4 Connection to Nonequilibrium Chemical Potential

As recently proposed in Ref. [2], we define the nonequilibrium chemical
potential $g$ (equivalent to that defined by Solon et al. [3]) through a
generalized Gibbs-Duhem relation:

$\text{d}g=\mathcal{E}\text{d}\mathcal{P}.$ (1.30)

This definition ensures that $g$ will be equal between two macroscopic
coexisting phases, a coexistence criterion that is identical to the a
generalized equal-area Maxwell construction on $\mathcal{P}$ with respect to
$\mathcal{E}$. The product rule results in:

$\text{d}g=\text{d}\left(\mathcal{P}\mathcal{E}\right)-\mathcal{P}\text{d}\mathcal{E}.$
(1.31)

We may then integrate Eq. (1.31) from the inside of the nucleus to the outside
to find:

$g^{\rm sat}-g^{\rm in}=\mathcal{P}^{\rm sat}\mathcal{E}^{\rm
sat}-\mathcal{P}^{\rm in}\mathcal{E}^{\rm in}-\int_{\mathcal{E}^{\rm
in}}^{\mathcal{E}^{\rm sat}}\mathcal{P}\text{d}\mathcal{E}.$ (1.32)

Substitution of Eq. (1.9) into Eq. (1.32) results in:

$g^{\rm sat}-g^{\rm in}=-\frac{\mathcal{E}^{\rm in}2\upgamma_{\rm
mech}}{R_{C}}-\int_{\mathcal{E}^{\rm in}}^{\mathcal{E}^{\rm
out}}\left[\mathcal{P}+\mathcal{E}^{\rm
in}\left(\mathcal{K}_{r}\bigr{|}_{r=\infty}-\mathcal{K}_{r}\bigr{|}_{r=0}\right)-\mathcal{P}^{\rm
sat}\right]\text{d}\mathcal{E}.$ (1.33)

Substitution of Eq. (1.22) into Eq. (1.33) results in:

$g^{\rm sat}-g^{\rm in}=-\Delta g=-\frac{2}{R_{C}}\left[\mathcal{E}^{\rm
in}\upgamma_{\rm mech}-\frac{\upgamma_{\rm ost}\rho^{\rm
sat}\Delta\mathcal{E}}{\Delta\rho}\right]+\mathcal{E}^{\rm
in}\left(\mathcal{K}_{r}\bigr{|}_{r=\infty}-\mathcal{K}_{r}\bigr{|}_{r=0}\right).$
(1.34)

In general, the chemical potential within and outside the droplet will be
different. However, as we demonstrate in Section 3, this difference vanishes
for a passive system. We may also define a pseudopotential as
$\Phi\equiv\int\mathcal{P}\text{d}\mathcal{E}$ as the antiderivative of the
pressure with respect to $\mathcal{E}$, allowing indefinite integration of Eq.
(1.31) to result in:

$g=\mathcal{P}\mathcal{E}-\Phi.$ (1.35)

Defining $\Phi$ allows us to express the critical radius from Eq. (1.22) as:

$R_{C}=\frac{2\upgamma_{\rm ost}\rho^{\rm
sat}\Delta\mathcal{E}}{\left(-\Delta\Phi+\Delta\mathcal{E}\mathcal{P}^{\rm
sat}\right)\Delta\rho}.$ (1.36)

This form for $R_{C}$ will become convenient in Section 2.3.

###  1.5 Local and Non-local Stress Tensors

In Section 1.1 we introduced the dynamic stress tensor
$\bm{\Sigma}=\bm{\sigma}+\mathbf{F}$, where $\bm{\sigma}$ is the conservative
stress tensor due to interaction forces and ideal gas pressure while
$\mathbf{F}$ is defined as the Coulomb integral of the body force:

$\mathbf{F}=\frac{1}{4\pi}\int\text{d}\mathbf{r}^{\prime}\mathbf{b}(\mathbf{r}^{\prime})\frac{\mathbf{r}-\mathbf{r}^{\prime}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}.$
(1.37)

It is straightforward to show that:

$\frac{\partial}{\partial\mathbf{r}^{\prime}}\frac{1}{|\mathbf{r}-\mathbf{r}^{\prime}|}=-\frac{\mathbf{r}-\mathbf{r}}{|\mathbf{r}-\mathbf{r}^{\prime}|^{3}}.$
(1.38)

Then, following integration by parts, Eq. (1.37) becomes:

$\mathbf{F}=\frac{1}{4\pi}\int\text{d}\mathbf{r}^{\prime}\frac{\partial}{\partial\mathbf{r}^{\prime}}\mathbf{b}(\mathbf{r}^{\prime})\frac{1}{|\mathbf{r}-\mathbf{r}^{\prime}|}.$
(1.39)

Note that $|\mathbf{r}-\mathbf{r}^{\prime}|^{-1}$ is the Green’s function of
the Laplace operator in three dimensions. Therefore solving for $\mathbf{F}$
is equivalent to finding the inverse Laplacian of $\bm{\nabla}\mathbf{b}$.
Under conditions of spherical symmetry this can be solved for as:

$\displaystyle\mathbf{F}=$
$\displaystyle-\int_{r}^{\infty}\text{d}r_{2}\frac{1}{r_{2}^{2}}\int_{0}^{r_{2}}\text{d}r_{1}r_{1}^{2}\frac{\partial}{\partial
r_{1}}\left[\text{b}_{r}\right]\mathbf{e}_{r}\mathbf{e}_{r}+\int_{r}^{\infty}\text{d}r_{2}\frac{1}{r_{2}^{2}}\int_{0}^{r_{2}}\text{d}r_{1}r_{1}\left(\text{b}_{r}\right)\mathbf{e}_{\theta}\mathbf{e}_{\theta}$
$\displaystyle+\int_{r}^{\infty}\text{d}r_{2}\frac{1}{r_{2}^{2}}\int_{0}^{r_{2}}\text{d}r_{1}r_{1}\left(\text{b}_{r}\right)\mathbf{e}_{\phi}\mathbf{e}_{\phi}$
(1.40)

While the divergence of $\mathbf{F}$ is equal to $\mathbf{b}$, we are not
guaranteed that all components of $\mathbf{F}$ are local. Depending on the
form of $\mathbf{b}$ one may find components of $\mathbf{F}$ that depend on
integrals over the entire spatial profile, conceptually similar to the non-
local chemical potential found in Ref. [4]. In order to obey the required
symmetries of the flux, an expansion of $\mathbf{b}$ in gradients of the
density may only contain odd gradients in the density. Any first order term
will have the form $b_{o}(\rho)\bm{\nabla}\rho$, which is trivially the
divergence of $\left[\int b_{o}(\rho)\text{d}\rho\right]\mathbf{I}$. Thus
first order terms in $\mathbf{b}$ will always result in bulk isotropic terms
of $\mathbf{F}$, which can be expressed locally. However, the third order
terms in $\mathbf{b}$ will result in second order terms of $\mathbf{F}$ that
may or may not have non-local contributions. The exact form of these stress
components in spherical coordinates may be computed by substituting the third
order terms of an expansion of $\mathbf{b}$ into Eq. (1.40). In certain cases,
such as ABPs, the body force can naturally be found as the divergence of a
stress tensor, and in these cases no non-local terms appear in $\mathbf{F}$
[17, 1].

##  2 Dynamics of an Active Nucleus

In this Section, we start from the fluctuating hydrodynamics derived in our
recent work [16] and derive the Langevin dynamics of a nucleus using the route
outlined by Cates and Nardini [18]. We then present a detailed demonstration
of how the noise statistics associated with the Langevin dynamics were
calculated. Third, we discuss and numerically justify the existence of an
effective potential barrier governing the nucleation dynamics. Finally, we
discuss the implication of this effective potential for predicting the rate of
nucleation.

###  2.1 Langevin Dynamics of a Spherical Nucleus

We aim to derive the Langevin equation for the radius of a nucleus of active
Brownian particles in the spherically symmetric geometry described by Section
1.1. We assume the dynamics of the nucleus to proceed quasistatically: any
given nucleus with radius $R$ (not necessarily equal to $R_{C}$) will be in a
local mechanical balance with its surroundings. This implies that the density
immediately outside of the nucleus $\rho^{\rm out}\neq\rho^{\rm sat}$, with
diffusive transport in the majority phase driven by this inequality. Then, as
diffusion will cause $\rho^{\rm out}$ to change in time, the local mechanical
balance will demand $R$ to change in time as well. The starting point to
derive the dynamics described above is the fluctuating hydrodynamics which we
recently derived [16]:

$\displaystyle\frac{\partial\rho}{\partial t}$
$\displaystyle=-\bm{\nabla}\cdot\mathbf{J},$ (2.1a) $\displaystyle\mathbf{J}$
$\displaystyle=\frac{1}{\zeta}\bm{\nabla}\cdot\bm{\Sigma}+\bm{\eta}^{\rm
act},$ (2.1b) where $\bm{\Sigma}$ is defined identically as the dynamic stress
tensor in Eqs. (1.3) and (1.25), while $\bm{\eta^{\rm act}}$ is an entirely
athermal stochastic contribution to the flux with zero mean and a variance of:
$\displaystyle\langle\bm{\eta}^{\rm act}(\mathbf{r},t)\bm{\eta}^{\rm
act}(\mathbf{r}^{\prime},t^{\prime})\rangle=2\frac{k_{B}T^{\rm
act}}{\zeta}\left(\rho\mathbf{I}-\frac{d}{d-1}\mathbf{Q}^{\prime}\right)\delta(t-t^{\prime})\delta(\mathbf{r}-\mathbf{r}^{\prime})\
,$ (2.1c)

Here ${k_{B}T^{\rm act}\equiv\zeta U_{o}\ell_{o}/d(d-1)}$ is the energy scale
of the athermal active fluctuations and $\mathbf{Q}^{\prime}$ is the traceless
nematic order tensor. Substitution of Eqs. (1.3) and Eq. (2.1b) into Eq.
(2.1a) results in:

$\zeta\frac{\partial\rho}{\partial
t}=-\nabla^{2}\Sigma_{rr}+\frac{2}{r}\frac{\partial\mathcal{G}}{\partial
r}+\frac{2}{r^{2}}\mathcal{G}-\zeta\bm{\nabla}\cdot\bm{\eta}^{\rm act}.$ (2.2)

From here, we proceed in a similar fashion as Cates and Nardini [18]. We
recently established a surface-area minimizing principle for liquid-gas
interfaces of ABPs resulting from a positive capillary-wave tension,
$\upgamma_{\rm cw}$, which resists fluctuations to the area-minimizing
interfacial interfacial shape. The strength of the fluctuations are set by
$k_{B}T^{\rm act}$ [16], while the ratio of the capillary tension to
$k_{B}T^{\rm act}$ – the active interfacial stiffness – controls the amplitude
of long-wavelength interfacial fluctuations. A fluctuating nucleus can be
well-approximated as a sphere when the average radius $R$ satisfies
$(R^{2}\upgamma_{\rm cw}/k_{B}T^{\rm act})^{1/2}>>1$. Therefore we introduce
the radialy symmetric ansatz connecting the density field and the radius of a
nucleus $R$ at time $t$, which was proposed by Bray [19]:

$\rho=\upvarphi_{R}\left[r-R(t)\right],$ (2.3)

where $\upvarphi_{R}$ is the density profile calculated in Section 1.3.
Substitution of Eq. (2.3) into Eq. (2.2) results in:

$-\zeta\upvarphi_{R}^{\prime}\frac{\partial R}{\partial
t}=-\nabla^{2}\Sigma_{rr}+\frac{2}{r}\frac{\partial\mathcal{G}}{\partial
r}+\frac{2}{r^{2}}\mathcal{G}-\zeta\bm{\nabla}\cdot\bm{\eta}^{\rm act},$ (2.4)

where $\upvarphi_{R}^{\prime}$ is the $r$ derivative of $\upvarphi_{R}$. We
now define the inverse Laplace operator as:

$\nabla^{-2}f(r)=-\int_{r}^{\infty}\int_{0}^{r_{2}}\text{d}r_{2}\text{d}r_{1}\left(\frac{r_{1}}{r_{2}}\right)^{2}f(r_{1}),$
(2.5)

where $f(r)$ is a function that vanishes in the limit of $r\to\infty$. We
apply the inverse Laplace operator to both sides of Eq. (2.4). The effect of
the inverse Laplace operator on the second term on the left hand side is:

$\nabla^{-2}\left[\frac{2}{r}\frac{\partial\mathcal{G}}{\partial
r}\right]=-\int_{r}^{\infty}\text{d}r_{2}\frac{1}{r_{2}^{2}}\int_{0}^{r_{2}}\text{d}r_{1}2r_{1}\frac{\partial\mathcal{G}}{\partial
r_{1}}.$ (2.6)

We now integrate Eq. (2.6) by parts while choosing:

$\displaystyle
u=\int_{0}^{r_{2}}\text{d}r_{1}2r_{1}\frac{\partial\mathcal{G}}{\partial
r_{1}}$ $\displaystyle,\quad du=2r_{2}\frac{\partial\mathcal{G}}{\partial
r_{2}}\text{d}r_{2},$ $\displaystyle dv=-\frac{1}{r_{2}^{2}}\text{d}r_{2}$
$\displaystyle,\quad v=\frac{1}{r_{2}}.$

Then Eq. (2.6) can equivalently be expressed as

$\nabla^{-2}\left[\frac{2}{r}\frac{\partial\mathcal{G}}{\partial
r}\right]=-\frac{2}{r}\int_{0}^{r}\text{d}r_{1}r_{1}\frac{\partial\mathcal{G}}{\partial
r_{1}}+2\mathcal{G}(r).$ (2.7)

A second integration by parts results in:

$\displaystyle\nabla^{-2}\left[\frac{2}{r}\frac{\partial\mathcal{G}}{\partial
r}\right]$
$\displaystyle=-\frac{2}{r}\left(r\mathcal{G}(r)-\int_{0}^{r}\mathcal{G}\text{d}r_{1}\right)+2\mathcal{G}(r),$
$\displaystyle=\frac{2}{r}\int_{0}^{r}\mathcal{G}\text{d}r_{1}.$ (2.8)

Applying a similar procedure to the other terms in Eq. (2.4) results in:

$\nabla^{-2}\left[\frac{2}{r^{2}}\mathcal{G}\right]=-\frac{2}{r}\int_{0}^{r}\mathcal{G}\text{d}r_{1}-2\int_{r}^{\infty}\frac{\mathcal{G}}{r_{1}}\text{d}r_{1},$
(2.9) $\nabla^{-2}\left[-\zeta\upvarphi^{\prime}_{R}\frac{\partial R}{\partial
t}\right]=\zeta R\frac{\partial R}{\partial t}\left(\rho^{\rm sat}-\rho^{\rm
in}\right)+\mathcal{O}\left(\frac{1}{R}\right),$ (2.10)
$\nabla^{-2}\left[-\nabla^{2}\Sigma_{rr}\right]=-\Sigma_{rr}-\mathcal{P}(\rho^{\rm
sat}),$ (2.11)

where the $\mathcal{P}(\rho^{\rm sat})$ in Eq. (2.11) comes from the boundary
term of $\Sigma_{rr}$ as $r\to\infty$. From the above equations one can see
that applying the inverse Laplace operator to both sides of Eq. (2.4) results
in:

$\zeta R\frac{\partial R}{\partial
t}\Delta\rho+\mathcal{O}\left(\frac{1}{R}\right)=\Sigma_{rr}+\mathcal{P}(\rho^{\rm
sat})+\int_{r}^{\infty}\text{d}r_{1}\frac{2}{r_{1}}\mathcal{G}-\nabla^{-2}\left[\zeta\bm{\nabla}\cdot\bm{\eta}^{\rm
act}\right],$ (2.12)

where we have defined the difference between an arbitrary property $a$ at the
interior and exterior of the bubble/droplet as $\Delta a\equiv a^{\rm
in}-a^{\rm sat}$. We now multiply both sides of the equation by
$\partial\mathcal{E}/\partial r$ and integrate from $r=0$ to $r=\infty$:

$\displaystyle-\zeta R\frac{\partial R}{\partial
t}\Delta\rho\Delta\mathcal{E}+\mathcal{O}\left(\frac{1}{R}\right)=$
$\displaystyle\int_{0}^{\infty}\Sigma_{rr}\frac{\partial\mathcal{E}}{\partial
r}\text{d}r+\mathcal{P}(\rho^{\rm sat})\left(\mathcal{E}^{\rm
sat}-\mathcal{E}^{\rm in}\right)$
$\displaystyle+\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\int_{r}^{\infty}\text{d}r_{1}\frac{2\mathcal{G}}{r_{1}}-\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\nabla^{-2}\left[\zeta\bm{\nabla}\cdot\bm{\eta}^{\rm act}\right].$ (2.13)

The double integral can be integrating by parts with:

$\displaystyle u=\int_{r}^{\infty}\frac{2\mathcal{G}}{r_{2}}\text{d}r_{1}$
$\displaystyle,\quad du=-\frac{2\mathcal{G}}{r}\text{d}r,$ $\displaystyle
dv=\frac{\partial\mathcal{E}}{\partial r}\text{d}r$ $\displaystyle,\quad
v=\mathcal{E}$ $\displaystyle-\zeta R\frac{\partial R}{\partial
t}\Delta\rho\Delta\mathcal{E}+\mathcal{O}\left(\frac{1}{R}\right)=$
$\displaystyle\int_{0}^{\infty}\Sigma_{rr}\frac{\partial\mathcal{E}}{\partial
r}\text{d}r+\mathcal{P}(\rho^{\rm sat})\left(\mathcal{E}^{\rm
sat}-\mathcal{E}^{\rm in}\right)-\frac{2\upgamma_{\rm
mech}}{R}\mathcal{E}^{\rm in}$
$\displaystyle+\frac{2}{R}\int_{0}^{\infty}\mathcal{E}\mathcal{G}\text{d}r-\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\nabla^{-2}\left[\zeta\bm{\nabla}\cdot\bm{\eta}^{\rm act}\right].$ (2.14)

Now consider the first integral on the right hand side. Substituting in the
constitutive expression for $\Sigma_{rr}$ from Eq. (1.4) and integrating by
parts results in

$\int_{0}^{\infty}\Sigma_{rr}\frac{\partial\mathcal{E}}{\partial
r}\text{d}r=-\int_{\mathcal{E}^{\rm in}}^{\mathcal{E}^{\rm
sat}}\text{d}\mathcal{E}\mathcal{P}(\rho)+\int_{0}^{\infty}\frac{2}{r}a(\rho)\frac{\partial\mathcal{E}}{\partial\rho}\left(\upvarphi^{\prime}_{R}\right)^{2}.$
(2.15)

Substitution of Eq. (2.15) into Eq. (2.14) gives:

$\displaystyle-\zeta R\frac{\partial R}{\partial
t}\Delta\rho\Delta\mathcal{E}+\mathcal{O}\left(\frac{1}{R}\right)=$
$\displaystyle-\int_{\mathcal{E}^{\rm in}}^{\mathcal{E}^{\rm
sat}}\text{d}\mathcal{E}\mathcal{P}(\rho)+\mathcal{P}(\rho^{\rm
sat})\left(\mathcal{E}^{\rm sat}-\mathcal{E}^{\rm
in}\right)-\frac{2\upgamma_{\rm mech}}{R}\mathcal{E}^{\rm in}$
$\displaystyle+\frac{2}{R}\int_{0}^{\infty}\left(\mathcal{E}\mathcal{G}+a(\rho)\frac{\partial\mathcal{E}}{\partial\rho}\left(\frac{\partial\upvarphi_{R}}{\partial
r}\right)^{2}\right)\text{d}r$
$\displaystyle-\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\nabla^{-2}\left[\zeta\bm{\nabla}\cdot\bm{\eta}^{\rm act}\right].$ (2.16)

From Eq. (1.21) we can rewrite Eq. (2.16) as:

$\displaystyle-\zeta R\frac{\partial R}{\partial
t}\Delta\rho\Delta\mathcal{E}+\mathcal{O}\left(\frac{1}{R}\right)=$
$\displaystyle-\int_{\mathcal{E}^{\rm in}}^{\mathcal{E}^{\rm
sat}}\text{d}\mathcal{E}\left[\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm
sat})\right]+\frac{2\upgamma_{\rm ost}}{R}\frac{\rho^{\rm
sat}\Delta\mathcal{E}}{\Delta\rho}$
$\displaystyle-\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\nabla^{-2}\left[\zeta\bm{\nabla}\cdot\bm{\eta}^{\rm act}\right].$ (2.17)

Recognizing that the first integral on the right-hand-side of Eq. (2.17) is a
constant, we can rearrange:

$\displaystyle\zeta R\frac{\partial R}{\partial
t}\Delta\rho\Delta\mathcal{E}+\mathcal{O}\left(\frac{1}{R}\right)=\frac{2\upgamma_{\rm
ost}\rho^{\rm
sat}\Delta\mathcal{E}}{\Delta\rho}\left[\frac{1}{R_{C}}-\frac{1}{R}\right]+\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\nabla^{-2}\left[\zeta\bm{\nabla}\cdot\bm{\eta}^{\rm act}\right],$ (2.18)

where the critical radius is defined as:

$R_{C}=\frac{2\upgamma_{\rm ost}\rho^{\rm
sat}\Delta\mathcal{E}}{\Delta\rho\int_{\mathcal{E}^{\rm in}}^{\mathcal{E}^{\rm
sat}}\left[\mathcal{P}(\rho)-\mathcal{P}(\rho^{\rm
sat})\right]\text{d}\mathcal{E}},$ (2.19)

which is identical to the critical radius found in Section 1.1 [see Eq.
(1.22)], highlighting the consistency of the two approaches.

The final evolution equation for the radial growth can be compactly written
as:

$\zeta_{\rm eff}\frac{\partial R}{\partial t}=F^{D}(R)+\tilde{\chi}(t),$
(2.20)

where we have defined an effective drag force:

$\zeta_{\rm eff}=\frac{4\pi\left(\rho^{\rm sat}-\rho^{\rm
in}\right)^{2}R^{3}\zeta}{\rho^{\rm sat}},$ (2.21)

as well as a deterministic force:

$F^{D}(R)=8\pi R^{2}\upgamma_{\rm
ost}\left[\frac{1}{R_{C}}-\frac{1}{R}\right].$ (2.22)

$\tilde{\chi}(t)$ represents the stochastic force with:

$\tilde{\chi}(t)=\frac{\zeta_{\rm
eff}}{R\Delta\rho\Delta\mathcal{E}}\nabla^{-2}\left[\bm{\nabla}\cdot\bm{\eta}^{\rm
act}\right].$ (2.23)

In Section 2.2, we derive the statistics of $\tilde{\chi}(t)$ finding that it
has zero mean and a variance given by:

$\langle\tilde{\chi}(t)\tilde{\chi}(t^{\prime})\rangle=2k_{B}T^{\rm
act}\zeta_{\rm eff}\delta(t-t^{\prime}).$ (2.24)

From Eqs. (2.20) and (2.24) we can see that the dynamics of the nucleus obey
the fluctuation dissipation theorem. In addition, the deterministic term can
be integrated to define an effective potential governing nucleation as further
discussed in Section 2.3.

###  2.2 Noise Statistics

We now seek to derive the statistics of $\tilde{\chi}(t)$ [Eq. (2.24)]. We
start by rewriting the statistics of $\bm{\eta}^{\rm act}$ [Eq. (2.1c)] to be
compatible with our spherical symmetry conditions:

$\langle\bm{\eta}^{\rm act}(\mathbf{r},t)\bm{\eta}^{\rm
act}(\mathbf{r}^{\prime},t^{\prime})\rangle=2\frac{k_{B}T^{\rm
act}}{4\pi\zeta}\left(\rho\mathbf{I}-\frac{d}{d-1}\mathbf{Q}^{\prime}\right)\delta(t-t^{\prime})\left(\frac{\delta(r-r^{\prime})}{r^{2}}\right).$
(2.25)

Before finding the statistics of $\tilde{\chi}$, we first find the statistics
of $\xi_{R}(\mathbf{r},t)$, which we define as

$\xi_{R}(\mathbf{r},t)=-\bm{\nabla}\cdot\bm{\eta}^{\rm act}.$ (2.26)

We split the correlations of $\xi_{R}$ into three contributions,

$\langle\xi_{R}(\mathbf{r},t)\xi_{R}(\mathbf{r}^{\prime},t^{\prime})\rangle=\langle\xi_{R}(\mathbf{r},t)\xi_{R}(\mathbf{r}^{\prime},t^{\prime})\rangle^{\rm
iso}+\langle\xi_{R}(\mathbf{r},t)\xi_{R}(\mathbf{r}^{\prime},t^{\prime})\rangle^{\rm
aniso}_{1}+\langle\xi_{R}(\mathbf{r},t)\xi_{R}(\mathbf{r}^{\prime},t^{\prime})\rangle^{\rm
aniso}_{2},$ (2.27)

where each contribution is given by:

$\langle\xi_{R}(\mathbf{r},t)\xi_{R}(\mathbf{r}^{\prime},t^{\prime})\rangle^{\rm
iso}=-\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\left(\frac{\partial}{\partial
r}+\frac{2}{r}\right)\left[\rho\frac{\partial}{\partial
r}\left(\frac{\delta(r-r^{\prime})}{r^{2}}\right)\right],$ (2.28)
$\displaystyle\langle\xi_{R}(\mathbf{r},t)\xi_{R}(\mathbf{r}^{\prime},t^{\prime})\rangle^{\rm
aniso}_{1}=$ $\displaystyle-\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\left(\frac{\partial}{\partial
r}+\frac{2}{r}\right)$
$\displaystyle\left[\left(\frac{1}{\ell_{o}^{2}D_{R}\zeta\overline{U}}\left[\kappa_{1}(\rho)\left(\frac{\partial}{\partial
r}+\frac{2}{r}\right)\frac{\partial\rho}{\partial
r}-\kappa_{3}(\rho)\left(\frac{\partial\rho}{\partial
r}\right)^{2}\right]\right)\frac{\partial}{\partial
r}\left(\frac{\delta(r-r^{\prime})}{r^{2}}\right)\right],$ (2.29)

and

$\langle\xi_{R}(\mathbf{r},t)\xi_{R}(\mathbf{r}^{\prime},t^{\prime})\rangle^{\rm
aniso}_{2}=\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\left(\frac{\partial}{\partial
r}+\frac{2}{r}\right)\left[\left(\frac{1}{r^{\prime}}\frac{1}{\ell_{o}^{2}D_{R}\zeta\overline{U}}\left[\kappa_{3}(\rho)\left(\frac{\partial\rho}{\partial
r}\right)^{2}\right]\right)\frac{\delta(r-r^{\prime})}{r^{2}}\right].$ (2.30)

We now define $\chi(t)$ as:

$\chi(t)=-\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\int_{r}^{\infty}\text{d}r_{2}\frac{1}{r_{2}^{2}}\int_{0}^{r_{2}}r_{1}^{2}\xi_{R}(\mathbf{r}_{1},t).$
(2.31)

The correlations of $\chi(t)$ are then given by

$\langle\chi(t)\chi(t^{\prime})\rangle=\left\langle\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\int_{r}^{\infty}\text{d}r_{2}\frac{1}{r_{2}^{2}}\int_{0}^{r_{2}}r_{1}^{2}\xi_{R}(\mathbf{r}_{1},t)\int_{0}^{\infty}\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{r}^{\infty}\text{d}r_{4}\frac{1}{r_{4}^{2}}\int_{0}^{r_{4}}r_{3}^{2}\xi_{R}(\mathbf{r}_{3},t^{\prime})\right\rangle,$
(2.32)

which can be rewritten as:

$\langle\chi(t)\chi(t^{\prime})\rangle=\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{2}}\int_{0}^{r_{4}}\text{d}r_{1}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{1}r_{3}}{r_{2}r_{4}}\right)^{2}\langle\xi_{R}(\mathbf{r}_{1},t)\xi_{R}(\mathbf{r}_{3},t^{\prime})\rangle.$
(2.33)

Now we similarly divide $\chi(t)$ into contributions:

$\langle\chi(t)\chi(t^{\prime})\rangle=\langle\chi(t)\chi(t^{\prime})\rangle^{\rm
iso}+\langle\chi(t)\chi(t^{\prime})\rangle^{\rm
aniso}_{1}+\langle\chi(t)\chi(t^{\prime})\rangle^{\rm aniso}_{2},$ (2.34)

where

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{2}}\int_{0}^{r_{4}}\text{d}r_{1}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{1}r_{3}}{r_{2}r_{4}}\right)^{2}\langle\xi_{R}(\mathbf{r}_{1},t)\xi_{R}(\mathbf{r}_{3},t^{\prime})\rangle^{\rm
iso},$ (2.35) $\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm
aniso}_{1}=$
$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}dr\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{2}}\int_{0}^{r_{4}}\text{d}r_{1}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{1}r_{3}}{r_{2}r_{4}}\right)^{2}\langle\xi_{R}(\mathbf{r}_{1},t)\xi_{R}(\mathbf{r}_{3},t^{\prime})\rangle^{\rm
aniso}_{1},$ (2.36)

and

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm aniso}_{2}=$
$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{2}}\int_{0}^{r_{4}}\text{d}r_{1}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{1}r_{3}}{r_{2}r_{4}}\right)^{2}\langle\xi_{R}(\mathbf{r}_{1},t)\xi_{R}(\mathbf{r}_{3},t^{\prime})\rangle^{\rm
aniso}_{2}.$ (2.37)

We will first solve for $\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}$.
Substitution of Eq. (2.28) into Eq. (2.35) results in:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{2}}\int_{0}^{r_{4}}\text{d}r_{1}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{1}r_{3}}{r_{2}r_{4}}\right)^{2}\left(\frac{\partial}{\partial
r_{1}}+\frac{2}{r_{1}}\right)\left[\rho(r_{1})\frac{\partial}{\partial
r_{1}}\left(\frac{\delta(r_{1}-r_{3})}{r_{1}^{2}}\right)\right].$ (2.38)

We now distribute the divergence in $r_{1}$ to find:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{2}}\int_{0}^{r_{4}}\text{d}r_{1}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{1}r_{3}}{r_{2}r_{4}}\right)^{2}\frac{\partial}{\partial
r_{1}}\left[\rho(r_{1})\frac{\partial}{\partial
r_{1}}\left(\frac{\delta(r_{1}-r_{3})}{r_{1}^{2}}\right)\right]$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}dr\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{2}}\int_{0}^{r_{4}}\text{d}r_{1}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{3}}{r_{2}r_{4}}\right)^{2}2r_{1}\left[\rho(r_{1})\frac{\partial}{\partial
r_{1}}\left(\frac{\delta(r_{1}-r_{3})}{r_{1}^{2}}\right)\right].$ (2.39)

We now integrate the top integral by parts, which results in a boundary term
as well as an integral which cancels out with the bottom integral. We are then
left with:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle-\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{4}}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{1}r_{3}}{r_{2}r_{4}}\right)^{2}\left(\rho(r_{1})\frac{\partial}{\partial
r_{1}}\left(\frac{\delta(r_{1}-r_{3})}{r_{1}^{2}}\right)\right)\Biggr{|}^{r_{1}=r_{2}}_{r_{1}=0}$
(2.40)

Evaluation of the limits must be done with care, particularly for the
$r_{1}=0$ term. Before evaluating these limits, we rearrange such that there
are no derivatives in $r_{1}$:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{4}}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{3}}{r_{2}r_{4}}\right)^{2}\rho(r_{1})\frac{2}{r_{1}}\delta(r_{1}-r_{3})\Biggr{|}^{r_{1}=r_{2}}_{r_{1}=0}$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{4}}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{3}}{r_{2}r_{4}}\right)^{2}\left(\rho(r_{1})\frac{\partial}{\partial
r_{3}}\delta(r_{1}-r_{3})\right)\Biggr{|}^{r_{1}=r_{2}}_{r_{1}=0}.$ (2.41)

We now evaluate the limits, noting that the $r_{1}\to 0$ limits will vanish:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{4}}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{3}}{r_{2}r_{4}}\right)^{2}\rho(r_{2})\frac{2}{r_{2}}\delta(r_{3}-r_{2})$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{4}}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{3}}{r_{2}r_{4}}\right)^{2}\left(\rho(r_{2})\frac{\partial}{\partial
r_{3}}\delta(r_{2}-r_{3})\right).$ (2.42)

We now integrate the first integral with respect to $r_{3}$. Because the
integral with respect to $r_{3}$ goes from $r_{3}=0$ to $r_{3}=r_{4}$, the
integral will be zero if $r_{4}<r_{2}$. However, if $r_{4}>r_{2}$, we can
apply the sifting property of the delta function. We can handle both cases by
applying the sifting function while applying up a factor of $H(r_{4}-r_{2})$,

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\text{d}r_{2}\text{d}r_{4}\left(\frac{1}{r_{4}}\right)^{2}\rho(r_{2})\frac{2}{r_{2}}H(r_{4}-r_{2})$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{4}}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{r_{3}}{r_{2}r_{4}}\right)^{2}\left(\rho(r_{2})\frac{\partial}{\partial
r_{3}}\delta(r_{2}-r_{3})\right),$ (2.43)

where $H$ is the Heaviside step function. We can now integrate the bottom
integral by parts which results in:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\text{d}r_{2}\text{d}r_{4}\left(\frac{1}{r_{4}}\right)^{2}\rho(r_{2})\frac{2}{r_{2}}H(r_{4}-r_{2})$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\text{d}r_{2}\text{d}r_{4}\left(\frac{r_{3}}{r_{2}r_{4}}\right)^{2}\left(\rho(r_{2})\delta(r_{2}-r_{3})\right)\Biggr{|}^{r_{3}=r_{4}}_{r_{3}=0}$
$\displaystyle-\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\int_{0}^{r_{4}}\text{d}r_{2}\text{d}r_{3}\text{d}r_{4}\left(\frac{1}{r_{2}r_{4}}\right)^{2}\left(2r_{3}\rho(r_{2})\delta(r_{2}-r_{3})\right).$
(2.44)

Evaluate the integral in $r_{3}$ of the bottom integral leads to a term which
cancels with the top integral, resulting in:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial r^{\prime}}$
$\displaystyle\times\int_{r}^{\infty}\int_{r^{\prime}}^{\infty}\text{d}r_{2}\text{d}r_{4}\left(\frac{r_{3}}{r_{2}r_{4}}\right)^{2}\left(\rho(r_{2})\delta(r_{2}-r_{3})\right)\Biggr{|}^{r_{3}=r_{4}}_{r_{3}=0}.$
(2.45)

We can now evaluate the limits,

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{r}^{\infty}\text{d}r_{2}\left(\frac{1}{r_{2}}\right)^{2}\rho(r_{2})\int_{r^{\prime}}^{\infty}\text{d}r_{4}\delta(r_{2}-r_{4}).$
(2.46)

Evaluating the integral in $r_{4}$ now depends on $r^{\prime}$. If
$r^{\prime}<r_{2}$ we may apply the sifting property of the delta function. If
$r^{\prime}>r_{2}$ the integral vanishes. Therefore we apply the sifting
property and gain a factor of the Heaviside step function:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{r}^{\infty}\text{d}r_{2}\left(\frac{1}{r_{2}}\right)^{2}\rho(r_{2})H(r_{2}-r^{\prime}).$
(2.47)

If $r<r^{\prime}$ then the integral will be zero until $r_{2}$ reaches
$r^{\prime}$. Then we may absorb the step function into the limits of
integration:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{\text{max}(r,r^{\prime})}^{\infty}\text{d}r_{2}\left(\frac{1}{r_{2}}\right)^{2}\rho(r_{2}).$
(2.48)

Integration by parts results in:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\frac{\rho(\text{max}(r,r^{\prime}))}{\text{max}(r,r^{\prime})}$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{\text{max}(r,r^{\prime})}^{\infty}\text{d}r_{2}\left(\frac{1}{r_{2}}\right)\frac{\partial\rho}{\partial
r_{2}}.$ (2.49)

The $\text{max}(r,r^{\prime})$ functions are cumbersome to deal with as is. We
therefore divide the integrals over $r^{\prime}$ into sections where
$r^{\prime}<r$ or $r^{\prime}>r$:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\left(\int_{0}^{\infty}\int_{0}^{r}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\frac{\rho(r)}{r}+\int_{0}^{\infty}\int_{r}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\frac{\rho(r^{\prime})}{r^{\prime}}\right)$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{0}^{r}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{r}^{\infty}\text{d}r_{2}\left(\frac{1}{r_{2}}\right)\frac{\partial\rho}{\partial
r_{2}}$ $\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\int_{r}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\int_{r^{\prime}}^{\infty}\text{d}r_{2}\left(\frac{1}{r_{2}}\right)\frac{\partial\rho}{\partial
r_{2}}.$ (2.50)

We now integrate the last line by parts in $r^{\prime}$, directly evaluate the
integral over $r^{\prime}$ in the second line, and combine like terms. This
results in:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\left(\int_{0}^{\infty}\int_{0}^{r}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\frac{\rho(r)}{r}+\int_{0}^{\infty}\int_{r}^{\infty}\text{d}r\text{d}r^{\prime}\frac{\partial\mathcal{E}}{\partial
r}\frac{\partial\mathcal{E}}{\partial
r^{\prime}}\frac{\rho(r^{\prime})}{r^{\prime}}\right)$
$\displaystyle-\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\int_{r}^{\infty}\text{d}r_{2}\left(\frac{1}{r_{2}}\right)\frac{\partial\rho}{\partial
r_{2}}$ $\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\int_{r}^{\infty}\text{d}r^{\prime}\mathcal{E}\left(\frac{1}{r^{\prime}}\right)\frac{\partial\rho}{\partial
r^{\prime}}.$ (2.51)

We now integrate the second integral of the first line by parts and directly
evaluate the $r^{\prime}$ integral in the first term of the first line, which
results in:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{4k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\mathcal{E}\frac{\rho(r)}{r}-\frac{4k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\frac{\rho(r)}{r}$ $\displaystyle-\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\int_{r}^{\infty}\text{d}r_{2}\left(\frac{1}{r_{2}}\right)\frac{\partial\rho}{\partial
r_{2}}$ $\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\int_{r}^{\infty}\text{d}r^{\prime}\mathcal{E}\left(\frac{1}{r^{\prime}}\right)\frac{\partial\rho}{\partial
r^{\prime}}.$ (2.52)

We now integrate both the first and second line by parts, then combine like
terms, which results in:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{4k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\mathcal{E}\frac{\rho(r)}{r}-\frac{4k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\frac{\rho(r)}{r}$ $\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\left(\mathcal{E}^{\rm
in}\right)^{2}\int_{0}^{\infty}\text{d}r\left(\frac{1}{r}\right)\frac{\partial\rho}{\partial
r}-\frac{4k_{B}T^{\rm act}}{4\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\mathcal{E}\left(\frac{1}{r}\right)\frac{\partial\rho}{\partial
r}$ $\displaystyle+\frac{2k_{B}T^{\rm
act}}{4\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\left(\mathcal{E}(r)\right)^{2}\left(\frac{1}{r}\right)\frac{\partial\rho}{\partial
r}.$ (2.53)

Now, as was assumed before, we note that each of the above integrals is non-
zero only near the interface. Then if $R$ is much larger than the width of the
interface, we may rewrite each integral as:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{4k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial
r}\mathcal{E}\rho(r)-\frac{4k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial r}\rho(r)$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\left(\mathcal{E}^{\rm
in}\right)^{2}\int_{0}^{\infty}\text{d}r\frac{\partial\rho}{\partial
r}-\frac{4k_{B}T^{\rm act}}{4R\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\mathcal{E}\frac{\partial\rho}{\partial r}$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\left(\mathcal{E}(r)\right)^{2}\frac{\partial\rho}{\partial
r}+\mathcal{O}(1/R^{2}).$ (2.54)

The first integral may be rewritten as:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\frac{\partial}{\partial
r}\left[\left(\mathcal{E}\right)^{2}\right]\rho(r)-\frac{4k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\frac{\partial\mathcal{E}}{\partial r}\rho(r)$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\left(\mathcal{E}^{\rm
in}\right)^{2}\int_{0}^{\infty}\text{d}r\frac{\partial\rho}{\partial
r}-\frac{4k_{B}T^{\rm act}}{4R\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\mathcal{E}\frac{\partial\rho}{\partial r}$
$\displaystyle+\frac{2k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\left(\mathcal{E}(r)\right)^{2}\frac{\partial\rho}{\partial
r}+\mathcal{O}(1/R^{2}).$ (2.55)

Then the above integrals may be compactly rewritten as:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\int_{0}^{\infty}\text{d}r\frac{\partial}{\partial
r}\left[\left(\mathcal{E}\right)^{2}\rho(r)\right]-\frac{4k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\mathcal{E}^{\rm
in}\int_{0}^{\infty}\text{d}r\frac{\partial}{\partial
r}\left[\mathcal{E}\rho\right]$ $\displaystyle+\frac{2k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\left(\mathcal{E}^{\rm
in}\right)^{2}\int_{0}^{\infty}\text{d}r\frac{\partial\rho}{\partial
r}+\mathcal{O}(1/R^{2}).$ (2.56)

All integrals may now be evaluated directly and simplify to:

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\rho^{\rm sat}\left(\mathcal{E}^{\rm
sat}-\mathcal{E}^{\rm in}\right)^{2}+\mathcal{O}(1/R^{2}).$ (2.57)

$\langle\chi(t)\chi(t^{\prime})\rangle^{\rm aniso}_{1}$ and
$\langle\chi(t)\chi(t^{\prime})\rangle^{\rm aniso}_{2}$ can be solved for in a
similar manner as $\langle\chi(t)\chi(t^{\prime})\rangle^{\rm iso}$, and we
find that $\langle\chi(t)\chi(t^{\prime})\rangle^{\rm aniso}_{1}$ is zero
while $\langle\chi(t)\chi(t^{\prime})\rangle^{\rm aniso}_{2}$ is:

$\langle\chi(t)\chi(t^{\prime})\rangle^{\rm aniso}_{2}=\frac{4k_{B}T^{\rm
act}}{4\pi
D_{R}\zeta\ell_{o}^{2}R^{2}}\Delta\mathcal{E}\delta(t-t^{\prime})\left[\int_{0}^{\infty}dr\frac{\kappa_{3}(\rho)}{\overline{U}(\rho)}\left(\frac{\partial\rho}{\partial
r}\right)^{2}\left(\mathcal{E}^{\rm
in}-\mathcal{E}\right)\right]+O\left(\frac{1}{R^{3}}\right),$

which implies that the most significant contributions to the anisotropic noise
correlations are $\mathcal{O}(1/R^{2})$ or smaller. Therefore,

$\displaystyle\langle\chi(t)\chi(t^{\prime})\rangle=$
$\displaystyle\frac{2k_{B}T^{\rm
act}}{4R\pi\zeta}\delta(t-t^{\prime})\rho^{\rm sat}\left(\mathcal{E}^{\rm
sat}-\mathcal{E}^{\rm in}\right)^{2}+\mathcal{O}(1/R^{2}).$ (2.58)

We now define $\tilde{\chi}$ as:

$\tilde{\chi}(t)=-\frac{\zeta_{\rm eff}}{R\left(\rho^{\rm sat}-\rho^{\rm
in}\right)\left(\mathcal{E}^{\rm sat}-\mathcal{E}^{\rm in}\right)}\chi(t),$
(2.59)

which will have zero mean and correlations:

$\langle\tilde{\chi}(t)\tilde{\chi}(t^{\prime})\rangle=2k_{B}T^{\rm
act}\zeta_{\rm eff}\delta(t-t^{\prime}).$ (2.60)

###  2.3 Effective Potential

Cates and Nardini [18] have argued that the nucleation of active matter must
have a defineable potential barrier because Eqs. (2.20) and (2.24) satisfy an
effective fluctuation dissipation theorem. This nucleation potential is
defined such that $F^{D}(R)=-\partial U(R)/\partial R$:

$U(R)=\int\text{d}R8\pi R^{2}\upgamma_{\rm
ost}\left[\frac{1}{R}-\frac{1}{R_{C}}\right],$ (2.61)

where $F^{D}(R)$ is defined by Eq. (2.22). In general, evaluating this
integral analytically is difficult because $\upgamma_{\rm ost}$ is implicitly
a function of $R$. Cates and Nardini proposed neglecting the $R$ dependence of
$\upgamma_{\rm ost}$ and treating it as a constant:

$U(R)=8\pi\upgamma_{\rm
ost}\left[\frac{R^{2}}{2}-\frac{R^{3}}{3R_{C}}\right].$ (2.62)

We verify the validity of this approximation by evaluating the integrand of
Eq. (2.61) both numerically and via the approximation proposed by Cates and
Nardini. We follow the same numerical scheme as discussed in Section 1.3,
except now $R$ is taken as an input and the two unknown values solved for are
$\rho^{\rm in}$ and $\rho^{\rm out}$. The result of calculating $-\partial
U(R)/\partial R$ both from our numerical scheme and the approximation of Cates
and Nardini is plotted in Fig. S.2.

Figure S.2: Deterministic force as defined by the negative derivative in $R$
of Eq. (2.61) as evaluated via the numerical scheme described above or
treating coefficients as constant in $R$. The bubble forces were evaluated for
a system at $\ell_{o}/d_{\rm hs}=70.7$ and $R_{C}/d_{\rm hs}=67$. The droplet
forces were evaluated for a system at $\ell_{o}/d_{\rm hs}=70.7$ and
$R_{C}/d_{\rm hs}=70$.

The difference between ignoring and accounting for the $R$ dependence of
$\upgamma_{\rm ost}$ is marginal for both bubbles and droplets, as shown by
Fig. S.2. Therefore evaluating coefficients at $R=R_{C}$ and treating them as
constant is a robust approximation, justifying the use of Eq. (2.62). We can
then define a potential barrier as $U(R_{C})$, which is given by:

$U(R_{C})=\frac{4}{3}\pi\upgamma_{\rm ost}R_{C}^{2}.$ (2.63)

We use Eq. (2.63) to calculate the nucleation barriers presented in the main
text.

We may also substitute $R_{C}$ as defined by Eq. (1.36) into Eq. (2.62) to
find a clean separation between terms proportional to surface area and volume:

$U(R)=4\pi\upgamma_{\rm ost}R^{2}-\frac{4}{3}\pi R^{3}\mathcal{V},$ (2.64)

where $\mathcal{V}$ has units of energy per volume and has form:

$\mathcal{V}=\frac{\mathcal{P}(\rho^{\rm sat})\Delta\rho}{\rho^{\rm
sat}}-\frac{\Delta\Phi\Delta\rho}{\rho^{\rm sat}\Delta\mathcal{E}}.$ (2.65)

###  2.4 Becker-Döring Nucleation Rate

The existence of an effective potential and the establishment of nuclei
obeying an effective detailed balance allows for one to extend equilibrium
Becker-Dp̈ring theory [20, 21, 22] to active particles in order to predict the
rate of droplet nucleation. The nucleation event is characterized by the
formation of a cluster with a critical number of particles, so we would like
to determine the rate at which such critical clusters form. We will denote the
number of clusters with $\ell$ particles per unit volume as $c_{\ell}$.
Additionally, we will assume that clusters are formed and destroyed by monomer
attachment and detachment. In other words, a size $(\ell+1)$ cluster is formed
by the merging of a size $\ell$ cluster and a size $1$ cluster. The opposite
process would be a size $\ell$ cluster being destroyed by splitting into a
size $\ell-1$ cluster and a size $1$ cluster. The flux, $J_{\ell}$, in
cluster-size space of $\ell$ clusters is then given by the difference between
the rate at which droplets containing $\ell$ particles are formed and the rate
at which they are destroyed:

$J_{\ell}=c_{\ell-1}S_{\ell-1}\beta_{\ell-1}-c_{\ell}S_{\ell}\alpha_{\ell};\ell\geq
2.$ (2.66)

Here $S_{\ell}$ is the surface area of an $\ell$ cluster, $\beta_{\ell}$ is
the flux of single particles arriving on the surface of the cluster, and
$\alpha_{\ell}$ is the flux of single particles leaving the surface of the
cluster. The evolution of $c_{\ell}$ is then given by:

$\frac{\partial c_{\ell}}{\partial t}=J_{\ell}-J_{\ell+1}.$ (2.67)

We now need to distinguish between two useful limits. One limit, which we call
steady state requires that all $\partial_{t}c_{\ell}=0$ but the individual
$J_{\ell}$ may remain finite. The other limit, which we call quasi-equilibrium
requires that all $\partial_{t}c_{\ell}=0$ but additionally requires that all
individual $J_{\ell}=0$. The effective detailed balance established for
nucleation dynamics implies that the dynamics which govern quasi-equilibrium
must also lead to quasi-equilibrium, as Eq. (2.20) obeys an effective time-
reversal symmetry. This implies that the values of $\beta_{\ell}$ and
$\alpha_{\ell}$ which are valid at quasi-equilibrium must also be valid in the
steady-state. Cates and Nardini [18] established that, to linear
approximation, the principle of detailed balance is recovered by nucleating
clusters in active system even if no such principle exists microscopically.
Then $\alpha_{\ell}$ is:

$\alpha_{\ell}=\frac{c_{\ell-1}^{\rm
eq}S_{\ell-1}\beta_{\ell-1}}{c_{\ell}^{\rm eq}S_{\ell}},$ (2.68)

and the flux $J_{\ell}$ at steady-state may be expressed as:

$J_{\ell}=c_{\ell-1}S_{\ell-1}\beta_{\ell-1}-c_{\ell}S_{\ell}\frac{c_{\ell-1}^{\rm
eq}S_{\ell-1}\beta_{\ell-1}}{c_{\ell}^{\rm eq}S_{\ell}}.$ (2.69)

This flux may be rewritten as:

$J_{\ell}=\beta_{\ell-1}S_{\ell-1}c_{\ell-1}^{\rm
eq}\left[\frac{c_{\ell-1}}{c_{\ell-1}^{\rm eq}}-\frac{c_{\ell}}{c_{\ell}^{\rm
eq}}\right].$ (2.70)

Note that in the steady state all $J_{\ell}$ with $\ell\geq 2$ are equal to
$J_{\ell+1}$. Then we can label $J_{\ell}=\mathcal{R}$ and write:

$\frac{\mathcal{R}}{\beta_{\ell-1}S_{\ell-1}c^{\rm
eq}_{\ell-1}}=\left[\frac{c_{\ell-1}}{c_{\ell-1}^{\rm
eq}}-\frac{c_{\ell}}{c_{\ell}^{\rm eq}}\right].$ (2.71)

We now sum both sides of the equation from $\ell=2$ to $\ell=\Lambda+1$, where
$\Lambda$ is some very large number (larger than the critical nucleus size).
After summing, the right hand side of Eq. (2.71) will vanish except for the
first term where $\ell=2$ and the last term where $\ell=\Lambda+1$:

$\mathcal{R}=\left[\frac{c_{1}}{c_{1}^{\rm
eq}}-\frac{c_{\Lambda+1}}{c_{\Lambda+1}^{\rm
eq}}\right]\left[\sum_{\ell=1}^{\Lambda}\frac{1}{\beta_{\ell}S_{\ell}c_{\ell}^{\rm
eq}}\right]^{-1}.$ (2.72)

We now argue that since the bulk metastable phase (a dilute gas) exists
overwhelmingly as single particles, so the quasi-equilibrium and steady-state
concentration of single particles are indistinuishable and $c_{1}/c_{1}^{\rm
eq}\approx 1$. Additionally, $\Lambda$ is larger than the size of the critical
nucleus, so $c_{\Lambda+1}$ should be zero in the metastable steady state but
finite at quasi-equilibrium. Then Eq. (2.72) reduces to:

$\mathcal{R}=\left[\sum_{\ell=1}^{\Lambda}\frac{1}{\beta_{\ell}S_{\ell}c_{\ell}^{\rm
eq}}\right]^{-1}.$ (2.73)

Using the effective nucleation potential [18], we know the relative
probability of observing a nucleus with radius $R$, and (given a spherical
geometry), we therefore we have an expression for $c_{\ell}^{\rm eq}$ as:

$c_{\ell}^{\rm eq}=\rho^{\rm sat}\text{exp}\left[\left(k_{B}T^{\rm
act}\right)^{-1}U(R_{\ell})\right],$ (2.74)

where $U(R_{\ell})$ is the effective potential evaluated at a radius
corresponding to a sphere of $\ell$ particles. Switching from a summation to
an integral, we can then approximate the nucleation rate as:

$\mathcal{R}=\rho^{\rm
sat}\left[\int_{1}^{\Lambda}\text{d}\ell\frac{1}{\beta_{\rm\ell}S_{\ell}}\text{exp}\left[\frac{U(R_{\ell})}{k_{B}T^{\rm
act}}\right]\right]^{-1}.$ (2.75)

We now would like to make the connection between radius and cluster size more
explicit:

$\ell=\frac{4}{3}\pi R^{3}\rho^{\rm in}.$ (2.76)

Alternatively, one may rearrange Eq. (2.76) and obtain an expression for the
radius in terms of size:

$R=\left[\frac{3\ell}{4\pi\rho^{\rm in}}\right]^{1/3}.$ (2.77)

We now consider the derived form of the nucleation barrier, Eq. (2.64), and
express it in terms of cluster size:

$U(\ell)=4\pi\upgamma_{\rm ost}\left[\frac{3\ell}{4\pi\rho^{\rm
in}}\right]^{2/3}-\frac{\ell\mathcal{V}}{\rho^{\rm in}}.$ (2.78)

From this form the critical cluster size $\ell^{*}$ can be solved for by
extremizing $U(\ell)$, resulting in:

$\ell^{*}=\frac{32}{3}\pi\rho^{\rm in}\left(\frac{\upgamma_{\rm
ost}}{\mathcal{V}}\right)^{3}.$ (2.79)

We now Taylor expand the the barrier about this maximum to second order,

$U(\ell)=U(\ell^{*})-\frac{1}{2}\eta\left(\ell-\ell^{*}\right)^{2},$ (2.80)

where we have defined $\eta$ as the second derivative of the barrier with
respect to $\ell$ and evaluated at $\ell^{*}$. Explicitly, this is solved for
as:

$\eta=\frac{1}{32\pi\left(\rho^{\rm
in}\right)^{2}}\frac{\mathcal{V}^{4}}{\upgamma_{\rm ost}^{3}}.$ (2.81)

We now consider the integral from Eq. (2.75). Substitution of Eq. (2.80) into
this integral results in:

$\int_{1}^{\Lambda}\text{d}\ell\frac{1}{\beta_{\rm\ell}S_{\ell}}\text{exp}\left[\frac{U(R_{\ell})}{k_{B}T^{\rm
act}}\right]=\int_{1}^{\Lambda}\text{d}\ell\frac{1}{\beta_{\rm\ell}S_{\ell}}\text{exp}\left[\frac{U(\ell^{*})}{k_{B}T^{\rm
act}}\right]\text{exp}\left[-\frac{\eta(\ell-\ell^{*})^{2}}{2k_{B}T^{\rm
act}}\right].$ (2.82)

The exponential of the barrier can be pulled out of the integral, which leads
to:

$\int_{1}^{\Lambda}\text{d}\ell\frac{1}{\beta_{\rm\ell}S_{\ell}}\text{exp}\left[\frac{U(R_{\ell})}{k_{B}T^{\rm
act}}\right]=\text{exp}\left[\frac{U(\ell^{*})}{k_{B}T^{\rm
act}}\right]\int_{1}^{\Lambda}\text{d}\ell\frac{1}{\beta_{\rm\ell}S_{\ell}}\text{exp}\left[-\frac{\eta(\ell-\ell^{*})^{2}}{2k_{B}T^{\rm
act}}\right].$ (2.83)

We now change integration variables from $\ell$ to $\delta\ell=\ell-\ell^{*}$:

$\int_{1}^{\Lambda}\text{d}\ell\frac{1}{\beta_{\rm\ell}S_{\ell}}\text{exp}\left[\frac{U(R_{\ell})}{k_{B}T^{\rm
act}}\right]=\text{exp}\left[\frac{U(\ell^{*})}{k_{B}T^{\rm
act}}\right]\int_{1-\ell^{*}}^{\Lambda-\ell^{*}}\text{d}\delta\ell\frac{1}{\beta_{\rm\ell}S_{\ell}}\text{exp}\left[-\frac{\eta\delta\ell^{2}}{2k_{B}T^{\rm
act}}\right].$ (2.84)

The integral will be dominated by value at which it is peaked, which is where
$\delta\ell=0$, as long as $1/\eta$ is large. Then we approximate
$\beta_{\ell}$, $S_{\ell}$ as their values when $\ell=\ell^{*}$, resulting in:

$\int_{1}^{\Lambda}\text{d}\ell\frac{1}{\beta_{\rm\ell}S_{\ell}}\text{exp}\left[\frac{U(R_{\ell})}{k_{B}T^{\rm
act}}\right]=\frac{1}{\beta_{\ell^{*}}4\pi\left(\frac{3\ell^{*}}{4\pi\rho^{\rm
in}}\right)^{2/3}}\text{exp}\left[\frac{U(\ell^{*})}{k_{B}T^{\rm
act}}\right]\int_{-\infty}^{\infty}\text{d}\delta\ell\text{exp}\left[-\frac{\eta\delta\ell^{2}}{2k_{B}T^{\rm
act}}\right].$ (2.85)

Finally, we can evaluate the Gaussian integral:

$\int_{1}^{\Lambda}\text{d}\ell\frac{1}{\beta_{\rm\ell}S_{\ell}}\text{exp}\left[\frac{U(R_{\ell})}{k_{B}T^{\rm
act}}\right]=\frac{1}{\beta_{\ell^{*}}4\pi\left(\frac{3\ell^{*}}{4\pi\rho^{\rm
in}}\right)^{2/3}}\text{exp}\left[\frac{U(\ell^{*})}{k_{B}T^{\rm
act}}\right]\sqrt{\frac{2\pi k_{B}T^{\rm act}}{\eta}}.$ (2.86)

Substitution of Eq. (2.86) into Eq. (2.75) results in an expression for the
rate of nucleation:

$\mathcal{R}=\rho^{\rm
sat}\beta_{\ell^{*}}4\pi\left(\frac{3\ell^{*}}{4\pi\rho^{\rm
in}}\right)^{2/3}\sqrt{\frac{\eta}{2\pi k_{B}T^{\rm
act}}}\text{exp}\left[-\frac{U(\ell^{*})}{k_{B}T^{\rm act}}\right].$ (2.87)

Everything in Eq. (2.87) can now be evaluated except for the rate of single
particle attachment on the critical nucleus $\beta_{\ell^{*}}$, for which we
use the simple kinetic model proposed Redner [23, 24]. $\beta_{\ell}$ is the
rate of absorption of particles onto the surface of a cluster. This rate
should be given by the density of the gas phase times the speed at which the
particles swim times the probability that a given particle will absorb onto
the surface:

$\beta_{\ell}=\rho^{\rm sat}U_{o}P^{\rm abs},$ (2.88)

where $P^{\rm abs}$ is the probability a particle absorbs onto the surface.
Assuming that the surface of the nucleus has an inward pointing normal vector
$\mathbf{n}$, we model $P^{\rm abs}$ as zero if a particle points away from
the interface and proportional to $\mathbf{n}\cdot\mathbf{q}$ if pointing in
the hemisphere towards interface. In other words $P^{\rm abs}$, is given by:

$P^{\rm
abs}=\begin{cases}\mathbf{n}\cdot\mathbf{q}&\mathbf{n}\cdot\mathbf{q}>0\\\
0&\mathbf{n}\cdot\mathbf{q}<0\end{cases}$ (2.89)

We pick a point on the sphere with unit normal vector $-\mathbf{e}_{x}$, and
then represent the orientation of a given particle as:

$\mathbf{q}=\sin{\theta}\cos{\phi}\mathbf{e}_{x}+\sin{\theta}\sin{\phi}\mathbf{e}_{y}+\cos{\theta}\mathbf{e}_{z}.$
(2.90)

Then, assuming that the orientations of the relevant particles are uniformly
distributed, the average of $\beta_{\ell}$ can be expressed as:

$\langle\beta_{\ell}\rangle=\rho^{\rm sat}U_{o}\frac{1}{4\pi}\int_{0}^{\rm
2\pi}\int_{0}^{\pi}\sin{\theta}P^{\rm abs}\text{d}\theta\text{d}\phi.$ (2.91)

$P^{\rm abs}$ is only nonzero when $\phi$ is between $0$ and $\pi/2$ as well
as when $\phi$ is between $3\pi/2$ and $2\pi$, for all values of $\theta$.
Then Eq. (2.91) reduces to:

$\langle\beta_{\ell}\rangle=\rho^{\rm
sat}U_{o}\frac{1}{2\pi^{2}}\left[\int_{0}^{\pi/2}\int_{0}^{\pi}\sin^{2}{\theta}\cos{\phi}\text{d}\theta\text{d}\phi+\int_{3\pi/2}^{2\pi}\int_{0}^{\pi}\sin^{2}{\theta}\cos{\phi}\text{d}\theta\text{d}\phi\right]=\frac{\rho^{\rm
sat}U_{o}}{4}.$ (2.92)

Substitution of Eq. (2.92) into Eq. (2.87) results in:

$\mathcal{R}=\left(\rho^{\rm sat}\right)^{2}\pi
U_{o}\left(\frac{3\ell^{*}}{4\pi\rho^{\rm
in}}\right)^{2/3}\sqrt{\frac{\eta}{2\pi k_{B}T^{\rm
act}}}\text{exp}\left[-\frac{U(\ell^{*})}{k_{B}T^{\rm act}}\right].$ (2.93)

The expected time to observe a nucleation event is given by $\tau_{\rm
nuc}=(\mathcal{R}V)^{-1}$, where $V$ is the total volume of the system. We
calculate $\mathcal{R}$ for ABP droplets and plot the results in Fig. S.3.
Following straightforward manipulations, it can be shown that $\eta$ is
alternatively expressed as:

$\eta=\frac{\upgamma_{\rm ost}}{2\pi(\rho^{\rm in})^{2}R_{C}^{4}}.$ (2.94)

Substitution of Eq. (2.94) into Eq. (2.93) results in:

$\mathcal{R}=\left(\rho^{\rm sat}\right)^{2}\pi U_{o}\sqrt{\frac{\upgamma_{\rm
ost}}{2\pi(\rho^{\rm in})^{2}k_{B}T^{\rm
act}}}\text{exp}\left[-\frac{U(\ell^{*})}{k_{B}T^{\rm act}}\right].$ (2.95)

At constant $\nu^{\rm sat}$, while $\upgamma_{\rm ost}/k_{B}T^{\rm act}$
scales roughly linearly with run length, the absolute value of $\rho^{\rm
sat}$ also decreases. At constant $\ell_{o}$, $\rho^{\rm sat}$ increases
linearly with $\nu^{\rm sat}$. These dependencies on activity and saturation
within the prefactor are far weaker than the exponential dependence of a
barrier which depends strongly on both activity and saturation. Thus the
qualitative behavior of Fig. S.3 is dominated by the strength of the
nucleation barrier.

Figure S.3: Nucleation rate of droplets as a function of activity and
supersaturation, as predicted by Eq. (2.93)

##  3 Equilibrium Nucleus Coexistence

In this Section we first derive the finite-sized coexistence of an equilibrium
system through a constrained free energy minimization. We then show that a
mechanical theory constructed in the same manner as Section 1 is equivalent to
equilibrium nucleation theory when the underlying dynamics are passive. For
brevity we will skip many steps of the derivation as this procedure will be
entirely analogous to those outlined in the previous two Sections.

###  3.1 Thermodynamics of Classical Nucleation

We consider a nucleus of radius $R$ and density $\rho^{\rm in}$ coexisting
with a parent phase of density $\rho^{\rm out}$. The system has an overall
density of $\rho^{\rm sat}$ and volume $V$. The Helmholtz free energy of this
system is then given by:

$\mathcal{F}=\frac{4}{3}\pi R^{3}f(\rho^{\rm in})+\left(V-\frac{4}{3}\pi
R^{3}\right)f(\rho^{\rm out})+4\pi R^{2}\upgamma,$ (3.1)

where $\upgamma$ is the surface tension and $f(\rho)$ is a bulk Helmholtz free
energy density of a system with uniform density $\rho$. The total number of
particles in the system is conserved with:

$\rho^{\rm in}\frac{4}{3}\pi R^{3}+\rho^{\rm out}\left(V-\frac{4}{3}\pi
R^{3}\right)=\rho^{\rm sat}V.$ (3.2)

Our free energy has dependencies on three adjustable internal variables,
$\rho^{\rm in}$, $\rho^{\rm out}$, and $R$. The properties associated with a
critical nucleus can be found by performing a constrained extremization with
respect to $\rho^{\rm in}$, $\rho^{\rm out}$, and $R$ and looking for
solutions where the radius is finite and not macroscopic. We construct the
following Lagrangian:

$\mathcal{W}=\mathcal{F}(\rho^{\rm in},\rho^{\rm
out},R)-\lambda\left(\rho^{\rm in}\frac{4}{3}\pi R^{3}+\rho^{\rm
out}\left(V-\frac{4}{3}\pi R^{3}\right)-\rho^{\rm sat}V\right),$ (3.3)

where $\lambda$ is a Lagrange multiplier. We extremize $\mathcal{W}$ by
setting its partial derivatives with respect to $\rho^{\rm in}$, $\rho^{\rm
out}$, $\lambda$, and $R$ (and evaluating these variables at their unknown
critical values) to zero:

$\displaystyle 0$ $\displaystyle=\frac{\partial\mathcal{W}}{\partial\rho^{\rm
in}}\Bigr{|}_{\rho^{\rm in}=\rho^{\rm in*}}=\frac{\partial
f}{\partial\rho}\Bigr{|}_{\rho=\rho^{\rm in^{*}}}-\lambda^{*},$ (3.4a)
$\displaystyle 0$ $\displaystyle=\frac{\partial\mathcal{W}}{\partial\rho^{\rm
out}}\Bigr{|}_{\rho^{\rm out}=\rho^{\rm out*}}=\frac{\partial
f}{\partial\rho}\Bigr{|}_{\rho=\rho^{\rm out^{*}}}-\lambda^{*},$ (3.4b)
$\displaystyle 0$ $\displaystyle=\frac{\partial\mathcal{W}}{\partial
R}\Bigr{|}_{R=R_{C}}=f(\rho^{\rm in})-f(\rho^{\rm
out})-\lambda^{*}\left(\rho^{\rm in}-\rho^{\rm
out}\right)+\frac{2\upgamma}{R_{C}},$ (3.4c) $\displaystyle 0$
$\displaystyle=\frac{\partial\mathcal{W}}{\partial\lambda}\Bigr{|}_{\lambda=\lambda^{*}}=(\rho^{\rm
in}-\rho^{\rm out})\frac{4}{3}\pi R^{3}+(\rho^{\rm out}-\rho^{\rm sat})V.$
(3.4d)

Here $\rho^{\rm in*}$, $\rho^{\rm out*}$, $\lambda^{*}$, and $R_{C}$ denote
the extremized values of $\rho^{\rm in}$, $\rho^{\rm out}$, $\lambda$, and
$R$, respectively. Defining the chemical potential as $\mu(\rho)\equiv\partial
f/\partial\rho$ allows us to appreciate that the extremized value of the
Lagrange multiplier, $\lambda^{*}$, is simply the coexistence chemical
potential with $\lambda^{*}\equiv\mu^{\rm coexist}=\mu(\rho^{\rm
in*})=\mu(\rho^{\rm out*})$ by combining Eqs. (3.4a) and (3.4b). By defining
the thermodynamic pressure, $p(\rho)\equiv-f(\rho)+\rho\mu(\rho)$, we can
express the coexistence criteria with the following three equations:

$\displaystyle\mu(\rho^{\rm in*})$ $\displaystyle=\mu(\rho^{\rm out*}),$
(3.5a) $\displaystyle p(\rho^{\rm in*})$ $\displaystyle=p(\rho^{\rm
out*})+\frac{2\upgamma}{R_{C}},$ (3.5b) $\displaystyle\rho^{\rm sat}V$
$\displaystyle=\rho^{\rm in*}\frac{4}{3}\pi R_{C}^{3}+\rho^{\rm
out*}\left(V-\frac{4}{3}\pi R_{C}^{3}\right).$ (3.5c)

Equations (3.5) allow for the determination of the unknown densities and
critical radius.

As we are interested in solutions in which the radius takes on large but non-
macroscopic values, we consider the limit of $R^{3}\ll V$. In this limit, the
particle conservation equation [Eq. (3.2)] reduces to $\rho^{\rm
out}\approx\rho^{\rm sat}$ and the nucleus in contact with a infinite particle
reservoir with chemical potential $\mu(\rho^{\rm sat})\equiv\mu^{\rm sat}$.
The Lagrangian then takes the following form:

$\mathcal{W}\approx\frac{4}{3}\pi R^{3}\left(w(\rho^{\rm in})-w(\rho^{\rm
sat})\right)+4\pi R^{2}\upgamma,$ (3.6)

where $w(\rho)\equiv f(\rho)-\mu^{\rm sat}\rho$ is the grand potential energy
density (also equivalent to the negative pressure, $-p$) of a system with
density $\rho$ in contact with a particle reservoir with chemical potential
$\mu^{\rm sat}$. We can plainly identify that the simplified Lagrangian
contains two familiar terms in classical nucleation theory: a negative
volumetric term proportional to the grand potential energy density difference
between the nucleus and the parent phase and positive areal term proportional
to the surface tension. The volumetric driving term defined in the main text
[i.e., ${\mathcal{W}=-\frac{4}{3}\pi R^{3}\mathcal{V}+4\pi R^{2}\upgamma}$]
has several equivalent forms ${\mathcal{V}\equiv-\Delta w\equiv\Delta
p\equiv-\Delta f+\mu^{\rm sat}\Delta\rho}$ where we have again used the
notation of ${\Delta a=a(\rho^{\rm in})-a(\rho^{\rm sat})}$. Extremization of
$\mathcal{W}$ with respect to $R$ and $\rho^{\rm in}$ results in the following
coexistence criteria:

$\displaystyle\mu(\rho^{\rm in*})$ $\displaystyle=\mu^{\rm sat},$ (3.7a)
$\displaystyle p(\rho^{\rm in*})$ $\displaystyle=p(\rho^{\rm
sat})+\frac{2\upgamma}{R_{C}}.$ (3.7b) Note that these criteria could also
have been obtained by directly taking the limit of $R_{C}\ll V$ in Eqs. (3.5).
The value of the Lagrangian evaluated at the critical radius and interior
density is the nucleation barrier, $F^{{\dagger}}\equiv\mathcal{W}(\rho^{\rm
in}=\rho^{\rm in*},R=R_{C})$ Finally, Eqs. (3.7) can be trivially rearranged
to isolate $R_{C}$ with: $\displaystyle R_{C}$
$\displaystyle=\frac{2\upgamma}{w(\rho^{\rm sat})-w(\rho^{\rm in*})}.$ (3.7c)

We note that in the main text and for most of this SI we denote $\rho^{\rm
in*}$ simply as $\rho^{\rm in}$ for notational convenience. The stability of
the solutions to Eqs. (3.7) can be determined by computing the Hessian matrix
of $\mathcal{W}$ and evaluating it at the solutions. A positive surface energy
and volumetric driving force for growth will generally result in the critical
nucleus representing a unstable fixed point.

###  3.2 Equilibrium Density Field

We consider particles which follow an overdamped Langevin equation with an
additional translational Brownian stochastic noise that satisfies the
fluctuation-dissipation theorem. The overdamped Langevin equation for the
$i$th particle of this system is:

$\displaystyle\dot{\mathbf{r}}_{i}=\frac{1}{\zeta}\sum_{j\neq
i}^{N}\mathbf{F}_{ij}+\bm{\eta}_{i}(t),$ (3.8)

where $\bm{\eta}_{i}$ is the translational Brownian noise with zero average
and variance
$\langle\bm{\eta}_{i}(t)\bm{\eta}_{j}(t^{\prime})=2(k_{B}T/\zeta)\delta(t-t^{\prime})\delta_{ij}\mathbf{I}\rangle$.
We assume that the interparticle forces, overall system density, and
temperature are such that the system is phase separated. Following the Irving-
Kirkwood-Noll [5, 7] procedure for a system described by Eq. (3.8) results in
the following density field dynamics at the mean-field level:

$\displaystyle\frac{\partial\rho}{\partial t}$
$\displaystyle=-\bm{\nabla}\cdot\mathbf{J},\ $ (3.9a)
$\displaystyle\mathbf{J}$
$\displaystyle=\frac{1}{\zeta}\bm{\nabla}\cdot\left(\bm{\sigma}^{C}+\bm{\sigma}^{B}\right),\
$ (3.9b) $\displaystyle\bm{\sigma}^{C}$
$\displaystyle=\left[-p_{C}(\rho)+\rho\kappa(\rho)\nabla^{2}\rho+\frac{\left(\kappa(\rho)+\rho\frac{\partial\kappa}{\partial\rho}\right)}{2}|\bm{\nabla}\rho|^{2}\right]\mathbf{I}-\kappa(\rho)\bm{\nabla}\rho\bm{\nabla}\rho,\
$ (3.9c) $\displaystyle\bm{\sigma}^{B}$ $\displaystyle=-k_{B}T\rho\mathbf{I},\
$ (3.9d)

where $\bm{\sigma}^{B}$ is the “ideal gas” stress generated by the stochastic
translational Brownian noise (absent in our treatment of athermal active
matter) and we define $p(\rho)=p_{C}(\rho)+k_{B}T\rho$. We also note that in
this model of equilibrium dynamics the stress tensor is fully local.

Within these passive dynamics we may identify the analogue to $a(\rho)$ as
$\rho\kappa(\rho)$ and the analogue to $b(\rho)$ as
$1/2\left(\rho\partial\kappa/\partial\rho-\kappa(\rho)\right)$. In this case
the identity of the pseudovariable becomes $\mathcal{E}\sim 1/\rho=v$, the
molecular volume. We may simplify the mechanical surface tension to:

$\upgamma_{\rm
mech}=\int_{0}^{\infty}dr\mathcal{G}(\rho)=\int_{0}^{\infty}dr\kappa(\rho)\left(\frac{\partial\rho}{\partial
r}\right)^{2},$ (3.10)

and we may simplify the Ostwald tension to:

$\displaystyle\upgamma_{\rm ost}$ $\displaystyle=\frac{v^{\rm
sat}\left(\rho^{\rm sat}-\rho^{\rm in}\right)}{\left(v^{\rm sat}-v^{\rm
in}\right)}\left[\int_{0}^{\infty}\left(-\kappa(\rho)\frac{1}{\rho}\left(\frac{\partial\rho}{\partial
r}\right)^{2}+\kappa(\rho)\left(\frac{\partial\rho}{\partial
r}\right)^{2}\left(\frac{1}{\rho}-v^{\rm in}\right)\right)\text{d}r\right],$
$\displaystyle=-\frac{v^{\rm sat}v^{\rm in}\left(\rho^{\rm sat}-\rho^{\rm
in}\right)}{\left(v^{\rm sat}-v^{\rm
in}\right)}\left[\int_{0}^{\infty}\kappa(\rho)\left(\frac{\partial\rho}{\partial
r}\right)^{2}\text{d}r\right],$
$\displaystyle=\int_{0}^{\infty}\kappa(\rho)\left(\frac{\partial\rho}{\partial
r}\right)^{2}\text{d}r=\upgamma_{\rm mech}.$ (3.11)

Therefore the equilibrium Ostwald tension is identical to the equilibrium
mechanical surface tension.

###  3.3 Equivalence of Thermodynamic and Mechanical Perspectives

We now demonstrate that the mechanically-derived generalized critical nucleus
criteria [Eqs. (1.23) and (1.22)] applied to passive systems recover the
criteria obtained through the variational method described in Section 3.1. The
pseudovariable variable for passive systems [11, 1] is simply $\mathcal{E}\sim
v$ where $v=1/\rho$ is the specific volume and the equivalence of the Ostwald
and mechanical tensions $\upgamma_{\rm ost}=\upgamma_{\rm mech}\equiv\upgamma$
results in the generalized criteria now taking the following form for passive
systems:

$p(\rho^{\rm in})-p(\rho^{\rm sat})=\frac{2\upgamma}{R_{C}},$ (3.12a)
$\int_{v^{\rm in}}^{v^{\rm sat}}\left[p(v)-p(v^{\rm
sat})\right]\text{d}v+\frac{2\upgamma v^{\rm in}}{R_{C}}=0,$ (3.12b)

where we have also recognized that the dynamic pressure is simply the static
thermodynamic pressure. Eq. (3.12a) is the familiar Laplace pressure
difference criterion that was recovered in Section 3.1 using thermodynamics.
All that is required to establish the equivalence of the criteria obtained
from our mechanical approach [Eqs. (3.12)] and the thermodynamically obtained
criteria [Eqs. (3.7)] is to demonstrate the equivalence of Eq. (3.12b) with
equality of chemical potential. We first eliminate the surface tension from
Eq. (3.12b) by using the Laplace pressure difference, resulting in:

$\int_{v^{\rm in}}^{v^{\rm sat}}p(v)dv-\left(p(v^{\rm sat})v^{\rm
sat}-p(v^{\rm in})v^{\rm in}\right)=0.$ (3.13)

Integration by parts demonstrates that the left-hand-side of the above
equation is simply $\int_{p^{\rm in}}^{p^{\rm sat}}vdp$, which is equivalent
to $\int_{\mu^{\rm in}}^{\mu^{\rm sat}}d\mu$ via the Gibbs-Duhem relation,
$vdp=d\mu$. We thus arrive at the expected:

$\mu(\rho^{\rm in})=\mu^{\rm sat},$ (3.14)

demonstrating the equivalence of our mechanically derived criteria with those
derived from thermodynamic considerations.

In Section 2.3 we identified a general expression for the volumetric
contribution to the effective nucleation barrier:

$\mathcal{V}=\frac{\mathcal{P}(\rho^{\rm sat})\Delta\rho}{\rho^{\rm
sat}}-\frac{\Delta\Phi\Delta\rho}{\rho^{\rm sat}\Delta\mathcal{E}}.$ (3.15)

With $\Phi=\int pd\mathcal{E}$ and again noting $\upgamma_{\rm
ost}=\upgamma_{\rm mech}\equiv\upgamma$ and $\mathcal{E}\sim v$, one can
straightforwardly show:

$\mathcal{V}=\mu^{\rm sat}\Delta\rho-\Delta f=-\Delta w.$ (3.16)

Therefore the mechanical theory recovers the expected thermodynamic driving
force for nucleation found in Eq. (3.6).

###  3.4 Application to van der Waals Fluid

Let us consider an equilibrium system where $\kappa(\rho)$ is a constant and
$p(\rho)$ is given by the van der Waals equation of state:

$\left(p+\rho^{2}a\right)\left(\frac{1}{\rho}-b\right)=k_{B}T,$ (3.17)

where $a$ is a measure of the average attraction between molecules and $b$ is
the volume excluded by a single particle. The critical point of this system is
given by:

$p^{C}=\frac{a}{27b^{2}},\quad
k_{B}T^{C}=\frac{8a}{27b},\quad\rho^{C}=\frac{1}{3b}.$ (3.18)

We now non-dimensionalize variables by their value at the critical point:

$p^{\prime}=\frac{p}{p^{C}},\quad
T^{\prime}=\frac{T}{T^{C}},\quad\rho^{\prime}=\frac{\rho}{\rho^{C}}.$ (3.19)

Then the dimensionless van der Waals equation of state is given by:

$p^{\prime}=\frac{8T^{\prime}}{\frac{3}{\rho^{\prime}}-1}-3(\rho^{\prime})^{2}.$
(3.20)

Equation (3.19) results in the following dimensionless form of $\kappa$:

$\kappa^{\prime}=\kappa\frac{\left(\rho^{C}\right)^{8/3}}{p^{C}}$ (3.21)

We note that this non-dimensionalized form of the van der Waals equation of
state leads to two different natural energy scales: $k_{B}T^{C}$ and
$p^{C}/\rho^{C}$. When calculating the mechanical surface tension of a phase
separated Van der Waals fluid via Eq. (1.7), our above non-dimensionalization
results in the latter energy scale. In order to convert it to have units of
$k_{B}T^{C}$ one can simply multiply the final result by a factor of $3/8$.

We now set $\kappa^{\prime}=1$ and calculate the coexisting density and radius
of the critical nuclei for van der Waals fluids. This calculation was carried
out similarly to those done for the active nuclei, only now with the van der
Waals equation of state. The results of this calculation are plotted in Fig.
S.4.

Figure S.4: Critical radius and relative change in nucleus density ($\nu^{\rm
in}\equiv|\rho^{\rm in}-\rho^{\rm gas,\rm liq}|/\rho^{\rm gas,\rm liq}$) as a
function of supersaturation ($\nu^{\rm sat}\equiv|\rho^{\rm sat}-\rho^{\rm
gas,\rm liq}|/\rho^{\rm gas,\rm liq}$).

In addition, we may calculate the nucleation barrier as a function of
saturation and temperature using Eq. (2.63), noting that $\upgamma_{\rm
ost}=\upgamma_{\rm mech}$ for this passive system. The results of this
calculation are plotted in Fig. S.5.

Figure S.5: Barrier $U(R_{C})$ relative to thermal energy as a function of
temperature and supersaturation.

##  4 Active Coarsening

In this Section we demonstrate Eqs. (2.20), (1.22) and (1.23) may be applied
to the late-stage coarsening dynamics of Active Brownian Particles. We first
demonstrate that a collection of noninteracting active nuclei obeys the
expected Lifshitz-Slyozov-Wagner (LSW) [25] growth rate. We then consider a
system where nuclei have appreciable volume fraction (and therefore nucleus
interactions must be considered), and summarize the derivation of the
necessary matrix equations to solve for the growth rate of a periodic array of
$N$ nuclei. This derivation was first shown by Voorhees and Glicksman [26, 27]
for passive alloys and requires minimal modification to be applicable to ABPs.
We then report numerical results for a highly dilute ABP system, demonstrating
the recovery of LSW theory [25].

###  4.1 Active Lifshitz-Slyozov-Wagner Theory

We consider the growth of many nuclei in a supersaturated solution. However,
we will also assume that the saturation is sufficiently small that the overall
volume fraction of the final phase is effectively zero. In this limit the
nuclei will evolve independently of one another, and we may use the single
nucleus Langevin equation [Eq. (2.20)] as the dynamics for all nuclei. For
convenience, we will rewrite Eq. (2.20) as:

$\frac{\partial R}{\partial t}=\frac{1}{\zeta_{\rm eff}}F^{D}(R)+g(R)\eta(t),$
(4.1)

where $\eta(t)$ is a unit Gaussian white noise with zero mean and variance
$\delta(t-t^{\prime})$ and $g(R)$ is defined as:

$g(R)=\sqrt{\frac{2k_{B}T^{\rm act}\rho^{\rm
sat}}{4\pi\zeta\Delta\rho^{2}R^{3}}}.$ (4.2)

and $F^{D}(R)$ remains defined as:

$F^{D}(R)=8\pi R^{2}\upgamma_{\rm
ost}\left[\frac{1}{R_{C}}-\frac{1}{R}\right].$ (4.3)

Although Eq. (2.20) contains multiplicative noise, the choice of time
discretization scheme (e.g. Itô, Stratonovich, etc.) is inconsequential as the
drift term [28] $\frac{1}{2}g(R)g^{\prime}(R)$ is
$\mathcal{O}\left(R^{-4}\right)$. We now wish to solve for how the
distribution of particle sizes evolves with time. This can be accomplished by
first determining the Fokker-Planck equation associated with Eq. (4.1):

$\frac{\partial P}{\partial t}=-\frac{\partial}{\partial
R}\left[\frac{1}{\zeta_{\rm
eff}}F^{D}(R)P(R)+\frac{1}{2}\frac{\partial}{\partial
R}\left(g^{2}(R)P(R)\right)\right].$ (4.4)

From Eq. (4.2), we can see that the diffusive term will be negligible in
comparison to the drift term for large $R$, resulting in:

$\frac{\partial P}{\partial t}=-\frac{\partial}{\partial
R}\left[\frac{1}{\zeta_{\rm eff}}F^{D}(R)P(R;t)\right].$ (4.5)

We can now proceed in the same fashion as Lifshitz and Slyozov [25], where it
was assumed that, at long times, the asymptotic form of the distribution takes
the following form:

$P(R;t)=\frac{1}{R_{C}(t)}\mathcal{N}\left(x\right),$ (4.6)

where we have defined the dimensionless radius $x\equiv R/R_{C}$ and
$\mathcal{N}(x)$ is the dimensionless distribution of radii scaled by $R_{C}$.
The time-dependence of $P(R;t)$ comes from that of the critical radius.
Substitution of Eq. (4.6) into Eq. (4.5) results in:

$\frac{1}{R_{C}^{2}}\frac{\partial R_{C}}{\partial
t}\left[\mathcal{N}(x)+x\frac{d\mathcal{N}}{dx}\right]=\frac{2\rho^{\rm
sat}\upgamma_{\rm ost}}{\zeta
R_{C}^{4}\Delta\rho^{2}}\left[\frac{d\mathcal{N}}{dx}\left(x^{-1}-x^{-2}\right)+\mathcal{N}(x)\left(2x^{-3}-x^{-2}\right)\right]$
(4.7)

In order for the ansatz Eq. (4.6) to be consistent with Eq. (4.7) (i.e.,
$\mathcal{N}(x)$ only depends on x), we require:

$R_{C}^{2}\frac{\partial R_{C}}{\partial t}=\frac{2\rho^{\rm
sat}}{\zeta\Delta\rho^{2}}\upgamma_{\rm ost}\mathcal{C},$ (4.8)

where $\mathcal{C}$ is some constant. This implies that the critical radius
evolves in time according to:

$R_{C}(t)=\left[R^{3}_{C}(t_{o})+\frac{6\rho^{\rm
sat}}{\zeta\Delta\rho^{2}}\upgamma_{\rm ost}\mathcal{C}t\right]^{1/3}.$ (4.9)

By equating the bracketed terms in Eq. (4.7) (within a dimensionless constant)
one obtains a differential equation for $\mathcal{N}(x)$ which one can solve
to obtain the asymptotic LSW distribution for nucleus size [25]. Here, we are
simply interested in how the average nucleus size evolves in time.
Straightforward insertion of Eq. (4.6) into ${\langle
R\rangle=\int_{0}^{\infty}P(R;t)RdR}$ reveals that the average radial dynamics
are identical to those of $R_{C}$ [Eq. (4.9)] within a multiplicative
dimensionless constant.

###  4.2 Coarsening Beyond the Zero Saturation Limit

We consider a system of phase separated active particles where the nucleation
of minority phase has completed and the the final phase composition has been
reached. In general, the system may contain many domains of minority phase.
The overall volume fraction of this system will be finite and thus nuclei may
interact with one another. When $\upgamma_{\rm ost}>0$, we expect the forward
Ostwald ripening process to occur and large domains of minority phase to grow
at the expense of small domains until the system reaches a final steady state
with a single domain. Rather than fixing the system to be finite size and
considering the influence of boundaries, we consider an infinite tiling of
periodic cells. Due to the large system sizes and long timescales necessary to
probe the dynamics of this process which evolves towards larger and larger
domains, direct integration of Eq. (2.1) may be impractical. The problem may
be greatly simplified by noting that majority phase will relax on timescales
much faster than the domains of minority phase grow or shrink. Ignoring
stochastic contributions, the density field dynamics (2.1) take the form:

$\frac{\partial\rho}{\partial
t}=-\frac{1}{\zeta}\bm{\nabla}\cdot\bm{\nabla}\cdot\bm{\Sigma}.$ (4.10)

By restricting our perspective to only describe the majority phase, we may
drop the time dependence from Eq. (4.10) by taking the quasi-static limit.
From the perspective of the majority phase, the particles injected/adsorbed
from the nuclei can be represented with finite-flux (and moving) boundary
conditions at the interface between nuclei and the parent phase. Here, we take
another approach and approximate majority phase particle addition/removal
through spatially varying volumetric sources and sinks. This approach results
in the quasi-static continuity equation now taking the following form:

$0=-\bm{\nabla}\cdot\bm{\nabla}\cdot\bm{\Sigma}+\zeta\sum_{u,v,w}\sum_{i=1}^{N}S_{i}(\mathbf{r}),$
(4.11)

where we gain the source or sink term $S_{i}$ in lieu of explicit
consideration of the $i$th minority domain. Here the sum over $(u,v,w)$ is a
summation over all periodic translations of the cell. Additionally, coarse-
graining the effect of the minority phase on the density field into
sources/sinks renders the interface between the nuclei and parent phase beyond
our level of description. We may therefore disregard the higher order gradient
terms from the dynamic stress tensor resulting in:

$0=\bm{\nabla}^{2}\mathcal{P}(\mathbf{r})+\zeta\sum_{u,v,w}\sum_{i=1}^{N}S_{i}(\mathbf{r}).$
(4.12)

Globally, the number of particles must remain conserved, imposing the
following constraint on $S_{i}$:

$\sum_{i=1}^{N}\int_{V}S_{i}(\mathbf{r})\text{d}\mathbf{r}=0.$ (4.13)

From Eq. (4.12), a mathematical analogue between the pressure field and the
electrostatic potential due to $N$ charge densities may be drawn. By
approximating the minority domains as spherically symmetric we may safely
collapse the “charge densities” $S_{i}$ into point charges at their centers by
defining $S_{i}(\mathbf{r})\equiv 2\pi
B_{i}\delta(\mathbf{r}-\mathbf{r}_{i})/\zeta$. Equation (4.12) can now be
expressed as:

$0=\bm{\nabla}^{2}\mathcal{P}(\mathbf{r})+4\pi\sum_{u,v,w}\sum_{i=1}^{N}B_{i}\delta(\mathbf{r}-\mathbf{r}_{i}-u\mathbf{a}-v\mathbf{b}-w\mathbf{c}),$
(4.14)

where $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are the lattice vectors and the
particle conservation constraint [Eq. (4.13)] becomes $\sum B_{i}=0$. The
solution to Eq. (4.14) is well-known and given by:

$\mathcal{P}(\mathbf{r})=\mathcal{P}_{o}+\sum_{u,v,w}\sum_{i=1}^{N}\frac{B_{i}}{|\mathbf{r}-\mathbf{r}_{i}-u\mathbf{a}-v\mathbf{b}-w\mathbf{c}|},$
(4.15)

where $\mathcal{P}_{o}$ is the unknown pressure of the majority phase far from
the nuclei. We now assume that every individual nucleus is in mechanical
equilibrium with its immediate surroundings. As outlined in Section 1, Eqs.
(1.22) and (1.23) allow us to solve for the coexistence densities of a finite
sized nucleus with radius $R$. We define $\rho^{R}(R)$ as a function mapping
the size of a nucleus with size $R$ to the outside coexistence density. Then,
if $\mathbf{r}^{s}_{j}$ is a point on the surface of a nucleus of size
$R_{j}$, we have:

$\mathcal{P}(\mathbf{r}^{s}_{j})=\mathcal{P}(\rho^{R}(R_{j}))$ (4.16)

Substitution of the boundary condition [Eq. (4.16)] into Eq. (4.15) gives:

$\mathcal{P}(\mathbf{r}^{s}_{j})=\mathcal{P}(\rho^{R}(R_{j}))=\mathcal{P}_{o}+\frac{B_{j}}{R_{j}}+\sum_{u,v,w}\sum_{i=1}^{N}\frac{B_{i}}{|\mathbf{r}^{s}_{j}-\mathbf{r}_{i}-u\mathbf{a}-v\mathbf{b}-w\mathbf{c}|},$
(4.17)

where the sum ignores the $i=j$ term when $u,v,w=0$. A key subtlety of Eq.
(4.17) is that $\mathbf{r}^{s}_{j}$ is a point on the surface of nucleus $j$
while $\mathbf{r}_{i}$ is the center of nucleus $i$. By expanding the
summation in terms of Legendre polynomials and averaging over all possible
choices of $\mathbf{r}^{s}_{j}$ on the surface of the $j$th nucleus, one may
show that

$\mathcal{P}(\rho^{R}(R_{j}))=\mathcal{P}_{o}+\frac{B_{j}}{R_{j}}+\sum_{u,v,w}\sum_{i=1}^{N}\frac{B_{i}}{|\mathbf{r}_{j}-\mathbf{r}_{i}-u\mathbf{a}-v\mathbf{b}-w\mathbf{c}|},$
(4.18)

where $\mathbf{r}_{j}$ is the center of nucleus $j$. Evaluation of the
summation over an infinite periodic lattice can be accomplished via standard
techniques of Ewald summation, although with the added complication that the
sum ignores the $i=j$ case when $u,v,w=0$, resulting in:

$\displaystyle\mathcal{P}(\rho^{R}(R_{j}))=$
$\displaystyle\mathcal{P}(\rho^{\rm
mat})+B_{j}\left(\frac{1}{R_{j}}-\sqrt{\frac{1}{\eta\pi}}\right)+\frac{4\pi}{\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)}\sum_{\mathbf{k}\neq\bm{0}}\sum_{i=1}^{N}\frac{B_{i}}{|\mathbf{k}|^{2}}\text{exp}\left[-\eta|\mathbf{k}|^{2}-i\mathbf{k}\cdot\left(\mathbf{r}_{j}-\mathbf{r}_{i}\right)\right]$
$\displaystyle+\sum_{u,v,w}\sum_{i=1}^{N}\frac{B_{i}}{|\mathbf{r}_{j}-\mathbf{r}_{i}-u\mathbf{a}-v\mathbf{b}-w\mathbf{c}|}\text{erfc}\left(\frac{|\mathbf{r}_{j}-\mathbf{r}_{i}-u\mathbf{a}-v\mathbf{b}-w\mathbf{c}|}{2\sqrt{\eta}}\right),$
(4.19)

where $\mathbf{k}$ are the reciprocal lattice vectors corresponding to our
periodic cell and $\eta$ is free parameter that controls how quickly the
summation converges in real space as opposed to reciprocal space. Equation
(4.19) can be expressed as a matrix equation
$\mathbf{b}=\mathbf{A}\cdot\mathbf{x}$, where

$\mathbf{b}=\begin{bmatrix}\mathcal{P}(\rho^{R}(R_{1}))\\\
\mathcal{P}(\rho^{R}(R_{2}))\\\ \vdots\\\ \mathcal{P}(\rho^{R}(R_{N}))\\\
0\end{bmatrix},$ (4.20a) $\mathbf{x}=\begin{bmatrix}B_{1}\\\ B_{2}\\\
\vdots\\\ B_{N}\\\ \mathcal{P}(\rho^{\rm mat})\end{bmatrix},$ (4.20b) and
$\mathbf{A}=\begin{bmatrix}\frac{1}{R_{1}}+D&\mathcal{O}_{12}&\mathcal{O}_{13}&\ldots&1\\\
\mathcal{O}_{21}&\frac{1}{R_{2}}+D&\mathcal{O}_{23}&\ldots&1\\\
\mathcal{O}_{31}&\mathcal{O}_{32}&\frac{1}{R_{3}}+D&\ldots&1\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\ 1&1&1&\ldots&0\end{bmatrix}.$ (4.20c)

Here $D$ is given by:

$\displaystyle D=$
$\displaystyle-\sqrt{\frac{1}{\eta\pi}}+\frac{4\pi}{L_{x}L_{y}L_{z}}\sum_{\mathbf{k}\neq
0}\frac{1}{|\mathbf{k}|^{2}}\text{exp}\left(-\eta|\mathbf{k}|^{2}\right)$
$\displaystyle+\sum_{u,v,w\neq
0}\frac{1}{\left(|u\mathbf{a}|^{2}+|v\mathbf{b}|^{2}+|w\mathbf{c}|^{2}\right)^{1/2}}\text{erfc}\left(\frac{1}{2\sqrt{\eta}}\left(|u\mathbf{a}|^{2}+|v\mathbf{b}|^{2}+|w\mathbf{c}|^{2}\right)^{1/2}\right),$
(4.21)

and $\mathcal{O}_{ij}$ is given by:

$\displaystyle\mathcal{O}_{ij}=$
$\displaystyle\frac{4\pi}{\mathbf{a}\cdot\left(\mathbf{b}\times\mathbf{c}\right)}\sum_{\mathbf{k}\neq\bm{0}}\frac{1}{|\mathbf{k}|^{2}}\text{exp}\left(-\eta|\mathbf{k}|^{2}-i\mathbf{k}\cdot\left(\mathbf{r}_{j}-\mathbf{r}_{i}\right)\right)$
$\displaystyle+\sum_{u,v,w}\frac{1}{|\mathbf{r}_{j}-\mathbf{r}_{i}-u\mathbf{a}-v\mathbf{b}-w\mathbf{c}|}\text{erfc}\left(\frac{|\mathbf{r}_{j}-\mathbf{r}_{i}-u\mathbf{a}-v\mathbf{b}-w\mathbf{c}|}{2\sqrt{\eta}}\right).$
(4.22)

Given the size and locations of each nucleus in the cell, one may then solve
Eqs. (4.20) in order to find all $B_{i}$ and the background pressure
$\mathcal{P}_{o}$. Let’s now consider the rate of change of the radius of a
particular nucleus and how it translates to the nucleus source/sink strength
$B_{i}$. If nucleus $i$ changes its radius $R_{i}$ by an amount $dR_{i}$ over
a time period $dt$, then the rate of mass injected into the majority phase by
particle $i$ is given by:

$4\pi R_{i}^{2}\frac{dR_{i}}{dt}\left(\rho^{\rm out}-\rho^{\rm in}\right)$
(4.23)

By the definition of $S_{i}$, the rate of mass injected into the majority
phase by nucleus $i$ is also equal to the integral of $S_{i}$ across the
entire system of volume. Then we have:

$4\pi R_{i}^{2}\frac{dR_{i}}{dt}\left(\rho^{\rm out}-\rho^{\rm
in}\right)=\int_{V}d\mathbf{r}S_{i}$ (4.24)

From the definition of $B_{i}$ we then have:

$4\pi R_{i}^{2}\frac{dR_{i}}{dt}\left(\rho^{\rm out}-\rho^{\rm
in}\right)=\frac{4\pi}{\zeta}B_{i}$ (4.25)

Then the rate of change of the radius of nucleus $i$ is given by:

$\frac{dR_{i}}{dt}=\frac{B_{i}}{\zeta R_{i}^{2}(\rho^{\rm out}-\rho^{\rm
in})}$ (4.26)

Therefore solution of the matrix equation defined by Eqs. (4.20) allows for
the extraction of the coarsening rates.

###  4.3 Numerical Results

Given the periodic cell dimensions, nucleus locations, and nucleus radii, it
is straightforward to set up the necessary matrix equations to find the
coarsening rates and numerically integrate. We conduct numerical simulation of
periodic cells of 250 nuclei radii randomly drawn from a distribution. The
PACKMOL software package [29] was used to give random locations within the
periodic cells such that all nuclei had a separation of at least two times the
largest radius. Particle sizes were sampled such that they had an average
radius of $750d_{\rm hs}$ and a cubic cell of width $a_{o}=2.1\times
10^{5}d_{\rm hs}$, corresponding to an overall volume fraction of $2\times
10^{-7}$. For the overall phase fractions simulated, we expect the results of
LSW theory [25] to be recovered. The main predictions of LSW theory are the
average nucleus size growing with the cube root of time and a steady-state
distribution of nucleus sizes $P(R)\propto\mathcal{N}(x)$ at long times, where
$x$ is the ratio of the nucleus radius to the average nucleus radius and
$\mathcal{N}$ is proportional to:

$\begin{cases}x^{2}(3+x)^{-7/3}\left(\frac{3}{2}-x\right)^{-11/3}\text{exp}\left(-\frac{3}{3-2x}\right)&x<3/2\\\
0&x\geq 3/2.\end{cases}$

We therefore initialized all simulations with radii sampled from a Gaussian
distribution as well as initial radii sampled from the distribution predicted
by LSW. Simulations were integrated in time for a duration of $6\times
10^{8}d_{hs}/U_{o}$. Simulation results were averaged over 10 independent
randomly generated initial positions and radii for both of the intial
distributions. We plot the average nucleus radius cubed as a function of time
in Figs. S.6 and S.7.

Figure S.6: Average radius growth as a function of time for nuclei intialized
with a Gaussian distribution with mean $750d_{\rm hs}$. Figure S.7: Average
radius growth as a function of time for nuclei initialized with an LSW
distribution with mean $750d_{\rm hs}$.

Growth rates demonstrate excellent agreement with the scaling predicted by LSW
theory. Droplets coarsen significantly faster than bubbles due to the higher
$\upgamma_{\rm ost}$ and therefore larger driving force for coarsening. The
distribution of nucleus sizes was also measured in the final third of the
simulation time (i.e. times from $4-6\times 10^{8}$) and are given by Figs.
S.8 and S.9.

Figure S.8: Nucleus size distribution sampled from last third of simulation
steps for particles initialized with Gaussian distribution. Figure S.9:
Nucleus size distribution sampled from last third of simulation steps for
particles initialized with LSW distribution.

The nuclei initialized from Gaussian distribution veer from their initial
sampling significantly by the end of the simulation time while nuclei
initialized from the LSW distribution stay close to the LSW distribution,
although with some error due to the finite amount of sampling.

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