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Brownian Ratchets and Diffusion Pumps.

Chen Lin Department of Physics, UCLA, Los Angeles, CA 90095-1596    Alex J. Levine, UCLA Center For Biological Physics Department of Physics, UCLA, Los Angeles, CA 90095-1596 Department of Chemistry & Biochemistry and The California Nanosystems Institute
UCLA, Los Angeles, CA 90095-1596
   Robijn Bruinsma Department of Physics, UCLA, Los Angeles, CA 90095-1596 bruinsma@physics.ucla.edu
Abstract

ABSTRACT

The steady-state dynamics of non-equilibrium systems that consume free energy and transform some of that energy into work under steady-state conditions is currently of intense interest for biological physics. Molecular motors and Brownian ratchets are classic examples. The focus of this paper is on a rather unusual non-equilibrium system where thermal fluctuations cause the system to appear and disappear. As an example, consider the familiar problem of the heterogeneous nucleation and growth of a droplet of the liquid phase of certain solute particles in an, initially, homogeneous supersaturated solution. Assume that there is only a single heterogeneous nucleation site in the solution. Also assume that the chemical potential of the dispersed solute particles at a distance RmR_{m} far from the droplet is kept fixed at a value μRm\mu_{R_{m}} that is larger than the chemical potential μRl\mu_{R_{l}} of the solute particles in the liquid state. There is thus a fixed chemical potential difference Δμ=μRmμRl\Delta\mu=\mu_{R_{m}}-\mu_{R_{l}} between solute particles that are part of a (large) droplet and solute particles far from the droplet. At least in principle, Δμ\Delta\mu can function as a source of free energy for active processes. The free energy of the droplet has, as a function of its radius, a maximum at the “critical radius” RcR_{c}, which represents an activation-free energy barrier that a droplet needs to overcome in order to continue growing. The value of RcR_{c} value is determined by the surface energy per unit area γ\gamma of the condensed phase, which promotes droplet shrinkage, and the chemical potential difference Δμ\Delta\mu, which promotes droplet growth. A droplet with an initial radius less than RcR_{c} will tend to shrink and disappear. During the shrinkage process, a diffusion current flows outwards from the droplet. If, on the other hand, the droplet has an initial radius larger than the critical radius then the droplet will tend to grow. Now, a diffusion current flows inward to the droplet. The droplet radius R(t)R(t) is proportional to D(tt0)\sqrt{D(t-t_{0})} with DD the solute diffusion coefficient and tt0t-t_{0} the time elapsed after the droplet crossed the activation energy barrier. The final state is a large, uniform condensate filling the system volume.

Droplets are subject to thermal fluctuations. Some droplets with an initial radius less than the critical radius still may succeed in crossing the activation energy barrier while some droplets with a radius larger than the critical radius still may shrink and disappear. In general, there will be a statistically variable number of nucleation attempts before the droplet succeeds to cross the energy activation barrier and continue on growing. During these initial nucleation attempts, diffusion currents flow to and from the nucleation site. The droplet acts as a kind of pump that consumes free energy provided by a time-dependent chemical potential difference between the outer radius RmR_{m} and the nucleation site. This interesting state is only a transient stage in the example discussed so far. Recently, there has been intense interest in the nucleation and growth of condensed, macromolecular droplets in an aqueous solvent that also contains a (bio)polymer matrix, where the polymers are not soluble in the condensate. A striking experimental observation is that in this case, the final state is not a uniform condensate but a static dispersion of droplets trapped inside the matrix. This is attributed to the elastic deformation of the polymer matrix produced by a droplet growing in the matrix. Assume that the biopolymer matrix has a distribution of mesh sizes and a distribution of heterogeneous nucleation sites. If the mesh size is small compared to the critical radius of the droplet then droplet nucleation and growth will be inhibited. If, on the other hand, the mesh size is large compared to the critical radius, then – after the droplet has crossed the energy activation barrier – it will grow until the reduction in free energy produced by the transport of solute particle from infinity to the droplet is balanced by the increased in elastic deformation energy of the polymer matrix. This droplet is then in mechanical equilibrium. Finally, consider mesh sizes comparable to, but somewhat larger than, the critical radius. The free energy of the droplet has, as a function of the radius the shape of a double-well potential with a minimum corresponding to the absence of a droplet while the other minimum corresponds to a droplet with a radius somewhat larger than the critical radius. This metastable droplet is in mechanical but not in thermodynamic equilibrium. Thermal fluctuations will produce transitions between the two states. This would be a non-equilibrium steady-state system that consumes free energy, driving diffusion currents back and forth from infinity.

Current density JJ chemical potential difference Δμ\Delta\mu. Steady-state consumption of free energy. One of the poles is a system with multiple states with certain transition rates between the states. If Δμ=0\Delta\mu=0 Boltzmann distribution rates obey detailed balance. If Δμ0\Delta\mu\neq 0, the system may cycle through the states in a unidirectional way, breaking time-reversal symmetry and the principle of detailed balance. Operates like a motor. Examples: Brownian ratchet, molecular motor.

An interesting special case is where the system is allowed to disappear. This is the case for one of the earliest examples in non-equilibrium statistical physics where free energy consumption

then the free energy consumption. can produce a cyclic. Motor: can a molecular motor consume free energy consumed by a

Chemical potential gradient. Poles of a battery.

Current

One (or both) of the poles is subject to statistical fluctuations: appears and disappears.

Example: Metastable, supersaturated solution. Chemical potential at Heterogeneous nucleation.

Then the radius change rate can be obtained by reorganizing 39:

R˙=DRnnlnρR\dot{R}=\frac{D}{R}\frac{n_{\infty}-n^{*}}{ln\frac{\rho}{R}} (1)

then the number density of minority phase can be expressed by the equilibrium condition we mentioned above: n=n0eμinkBTn^{*}=n_{0}e^{\frac{\mu_{in}}{k_{B}T}}. Since ρ>>R\rho>>R is always true in our assumption, so we could further set lnρRln\frac{\rho}{R} as a constant D~\tilde{D}. Thus the radius change rate can be reorganized as:

R˙=D~R(nn0eμinkBT)\dot{R}=\frac{\tilde{D}}{R}(n_{\infty}-n_{0}e^{\frac{\mu_{in}}{k_{B}T}}) (2)

here the heart of our model is the chemical potential μin\mu_{in} is dependent on the interaction energy:

μin=μin+αU(x,R)R2\mu_{in}^{{}^{\prime}}=\mu_{in}+\alpha\frac{U(\vec{x},R)}{R^{2}} (3)

one can regard the last term as the excess Gibbs free energy offered by interaction potential up to a parameter α\alpha. 1R2\frac{1}{R^{2}} here represent unit area, alpha combining 1R2\frac{1}{R^{2}} represent the number of molecules. So α\alpha is the density of minority phase, which is the intrinsic property of different material. In this equation it represents how sensitive the droplet against interactions, thus we name α\alpha as the ”dewetting parameter”. Noticed that, to make equation 42 reasonable, we assume that in the condensed phase the Gibbs free energy is nearly identical to Helmholtz free energy. We also drop the entropy term.

I Introduction

The Brownian ratchet is a device composed of a saw-tooth wheel connected to a pawl mechanism that allows it rotate in only one direction. Introduced by Smoluchowski and popularized by Feynman, it is striking because a sufficiently small version of the device would seem to transform the kinetic energy of random thermal fluctuations into unidirectional rotational motion of the wheel that can do work. This would violate the Second Law of Thermodynamics and careful analysis indeed shows that in actuality the device cannot do work. The – by now – numerous experimental realizations of Brownian ratchets invariably involve some form of irreversible consumption of free energy. For example, In this letter we focus on a Brownian ratchet type diffusion pump that seemingly operates without free energy consumption.

The system is a polymer gel with a solvent that is a binary fluid. The majority component of the fluid is a good solvent for the gel while the minority component is a poor solvent. The solvent is initially in a homogeneous equilibrium state that is transformed into a supersaturated solution by a temperature jump. Minority phase droplets start to form by thermal activation. If the radius of such droplet exceeds the critical radius, which is proportional to the surface tension and inversely proportional to the degree of supersaturation, then the droplet size grows by diffusive transport while droplets with radii less than the critical radius disappear, thereby releasing minority phase molecules that will diffuse away. Liquid-liquid phase separation in binary liquids wei2020modeling normally is characterized by Oswald ripening, or “coarsening”, which refers to the growth of larger droplets of the minority phase and a corresponding shrinkage of smaller droplets, which is driven by capillary action. Oswald ripening is not observed in liquid-liquid phase-separation in gels. Instead, a stationary dispersion is observed composed of droplets of various sizes. This is attributed to the fact that both the cytoplasm of cells and the interior of the cell are permeated by a network of filamentous proteins such as actin, microtubules or DNA/nucleosome fibers that prevent the droplets from growing.

Cellular phase separation events are complex but the same phenomenon has been observed in a well-characterized model system composed of a polymeric gel in a solvent that is a binary liquid. The majority phase of the binary liquid is a good solvent for the gel polymers and the minority phase is a poor solvent. The solvent is initially in a homogeneous mixed state that is thermodynamically stable. After a discontinuous drop in temperature (or composition?), the homogeneous mixed state turns metastable. Minority phase droplets can form by thermal activation. If the radius of the droplet exceeds a critical radius, which is proportional to the surface tension and inversely proportional to the degree of supersaturation, then the droplet size can grow by diffusive transport while droplets with radii less than the critical radius disappear, thereby releasing minority phase molecules that will diffuse away. Assume the gel is placed in a solvent reservoir that is in a homogeneous metastable phase. If the mesh-size of the gel exceeds the size of the critical radius then thermally activated minority phase droplets can grow to at least the mesh-size. Further growth generates elastic deformation energy of the gel that eventually stops the growth of the droplet. The chemical potential of these minority phase droplets is the same as that of the reservoir while the free energy of the drop in the stressed polymer mesh can be lower or higher than that of the relaxed mesh without the droplet. Thermal fluctuations of such a “restrained” drop can cause the drop the change location to an adjacent mesh but it can also cause it to disappear. Attach an Ising variable to each mesh of the gel with “spin down” indicating the absence of a drop inside the mesh location and “spin up” the presence of a drop. Diffusive transport inside the gel drives minority phase transport from mesh locations where a drop has just disappeared to a mesh site where the drop is growing. The statistical physics of such a system could be reasonably described by an Ising model in a magnetic field and does not appear to be exceptional. But now assume that the gel is a sufficiently small system so that it contains, on average, either no drop or only one drop. When a droplet appears somewhere in that system, droplet growth requires diffusive transport of minority phase molecules from the reservoir outside the gel to the gel. When a droplet disappears, the diffusive transport would be reversed. It would seem that there is an alternating diffusion current to and from the gel system, somewhat like a diffusive form of breathing. Moreover, as discussed below, the disappearance of a drop involves momentum exchange between the drop and the reservoir, which might seem to allow for active Brownian motion of the gel system in the reservoir. The metastable reservoir could be viewed as a store of free energy so the First Law of Thermodynamics is not violated if this stored free energy is used to power the alternating diffusion current and active Brownian motion. In that case, however, the dynamics of the system should violate the fluctuation-dissipation theorem. In the next section we develop a simplified pair of coupled equations that describes the thermal fluctuations of the radius and location of the minority phase droplet. It allows us to check whether or not the fluctuation-dissipation theorem is violated.

II Model

First consider the formation of minority-phase droplets without the gel. The minority phase molecules are assumed to form an ideal solution in a solvent of majority phase molecules. They have a chemical potential

μ=kBTln(cc0)\mu=k_{B}T\ln\left(\frac{c}{c_{0}}\right) (4)

with c=N/Ac_{\infty}=N/A the area density of the minority phase (A is the area of the system and NN is the total number of minority phase molecules). The solvation concentration csc_{s} is defined as the solution concentration at which the minority phase molecules would be in thermodynamic equilibrium with a pure liquid of minority phase molecules. The chemical potential μs\mu_{s} of minority phase molecules in the liquid state is then

μ==kBTln(csc0)\mu==k_{B}T\ln\left(\frac{c_{s}}{c_{0}}\right) (5)

The Gibbs free energy of a two-dimensional circular droplet of radius RR is then

ΔG(R)=πR2cl(μμs)+2πRτ=πR2clkBTln(ccs)+2πRτ\begin{split}\Delta G(R)=&\pi R^{2}c_{l}(\mu-\mu_{s})+2\pi R\tau\\ &=\pi R^{2}c_{l}k_{B}T\ln\left(\frac{c_{\infty}}{c_{s}}\right)+2\pi R\tau\end{split} (6)

with clc_{l} the area density of liquid state molecules and with τ\tau the line tension of the drop. The critical radius of the drop corresponds to the maximum of ΔG(R)\Delta G(R) at

R=τ/(clkBTln(ccs))R^{*}=\tau/\left(c_{l}k_{B}T\ln\left(\frac{c_{\infty}}{c_{s}}\right)\right) (7)

The gel is represented as a two-dimensional potential energy landscape u(x)u(\vec{x}) with a distribution of maxima. The potential energy of a drop with center at r\vec{r} is then

U(r)=|ρ|<Rd2ρu(r+ρ)U(\vec{r})=\int_{|\rho|<R}{d^{2}}\vec{\rho}\medspace u(\vec{r}+\vec{\rho}) (8)

Transport is by diffusion where the diffusion flux is J\vec{J} equals nνμ-n\nu\nabla\mu with ν\nu the mobility. This relation is actually the diffusion equation:

tn=J=D2n\partial_{t}n=-\nabla\vec{J}=D\nabla^{2}n (9)

in two-dimensions. The solution will be:

n(r)=Alnr+Bn(r)=Alnr+B (10)

where AA and BB are undetermined constants.

Now we focus on the equilibrium condition of the diffusion. In this condition the chemical potential of minority phase is balanced with that of mixed phase (ideal solution as we assumed): μin=μ=kBTln(nn)\mu_{in}=\mu=k_{B}Tln(\frac{n^{*}}{n}), in which nn^{*} is an arbitrary number density of minority phase. Here we further set the number density at infinite region is a fixed value nn_{\infty}, which plays a role of a reservoir of minority phase. If those two regions relaxes separately as steady currents, we can get an equation series by plug them in equation 35:

n(R)=AlnR+B=n;n(ρ)=Alnρ+B=nn(R)=AlnR+B=n^{*};n(\rho)=Aln\rho+B=n_{\infty} (11)

here RR is the radius of the droplet and ρ\rho is the location at infinity. Then by solving this, the undetermined constants can be expressed by the constants we have:

A=nnlnρR;B=nlnρRlnρnlnRlnρRA=\frac{n_{\infty}-n^{*}}{ln\frac{\rho}{R}};B=\frac{n^{*}}{ln\frac{\rho}{R}}ln\rho-\frac{n_{\infty}lnR}{ln\frac{\rho}{R}} (12)

now we plug them back to the equation 36, then take gradient respect to rr. The constant BB directly vanished and the current on the droplet RR will be a simple expression:

J(R)=DRnnlnρRJ(R)=-\frac{D}{R}\frac{n_{\infty}-n^{*}}{ln\frac{\rho}{R}} (13)

Now we can determine the dynamics of radius RR by assuming the incoming current, the flux equation 38 times 2πR2\pi R, equals to the area change rate:

dadt=2πRDRnnlnρR=2πRR˙\frac{da}{dt}=2\pi R\frac{D}{R}\frac{n_{\infty}-n^{*}}{ln\frac{\rho}{R}}=2\pi R\dot{R} (14)

where we applied a=πR2a=\pi R^{2}. Then the radius change rate can be obtained by reorganizing 39:

R˙=DRnnlnρR\dot{R}=\frac{D}{R}\frac{n_{\infty}-n^{*}}{ln\frac{\rho}{R}} (15)

then the number density of minority phase can be expressed by the equilibrium condition we mentioned above: n=n0eμinkBTn^{*}=n_{0}e^{\frac{\mu_{in}}{k_{B}T}}. Since ρ>>R\rho>>R is always true in our assumption, so we could further set lnρRln\frac{\rho}{R} as a constant D~\tilde{D}. Thus the radius change rate can be reorganized as:

R˙=D~R(nn0eμinkBT)\dot{R}=\frac{\tilde{D}}{R}(n_{\infty}-n_{0}e^{\frac{\mu_{in}}{k_{B}T}}) (16)

here the heart of our model is the chemical potential μin\mu_{in} is dependent on the interaction energy:

μin=μin+αU(x,R)R2\mu_{in}^{{}^{\prime}}=\mu_{in}+\alpha\frac{U(\vec{x},R)}{R^{2}} (17)

one can regard the last term as the excess Gibbs free energy offered by interaction potential up to a parameter α\alpha. 1R2\frac{1}{R^{2}} here represent unit area, alpha combining 1R2\frac{1}{R^{2}} represent the number of molecules. So α\alpha is the density of minority phase, which is the intrinsic property of different material. In this equation it represents how sensitive the droplet against interactions, thus we name α\alpha as the ”dewetting parameter”. Noticed that, to make equation 42 reasonable, we assume that in the condensed phase the Gibbs free energy is nearly identical to Helmholtz free energy. We also drop the entropy term.

III Two dimensional model of the growing droplet

III.1 Basic Settings

Different from past works we reviewed above, we create a two-dimensional model dynamical model for the numerical simulation. In this model, a hexagon lattice of deformable Gaussian potentials is used to represent either filaments or a coarse-grained representation of a higher filament concentration region. This allows one to simulate different elastic networks by changing parameters of the Gaussian potentials. Harmonic spring connecting the Gaussians provides mechanical stability of the network. Next, a disk is used to represent a droplet of minority phase. The motion of the Gaussian bumps and the droplet are both governed by Brownian dynamics with external potentials:

Xd˙(t)=DdkBTUint+2Ddσ(t)\dot{X_{d}}(t)=-\frac{D_{d}}{k_{B}T}\nabla U_{int}+\sqrt{2D_{d}}\sigma(t) (18)
Xg˙(t)=DgkBT(Uint+Uh)+2Dgσ(t)\dot{X_{g}}(t)=-\frac{D_{g}}{k_{B}T}(-\nabla U_{int}+\nabla U_{h})+\sqrt{2D_{g}}\sigma(t) (19)

Here, XdX_{d} and XgX_{g} are locations, DdD_{d} and DgD_{g} are the diffusion constants of the droplet and Gaussians respectively while σ\sigma again is the Wiener noise we used previously. UhU_{h} is the harmonic potentials between each pair of Gaussians:

Uh=12k(xgixgj)2U_{h}=\frac{1}{2}k(\vec{x_{gi}}-\vec{x_{gj}})^{2} (20)

where xgi\vec{x_{gi}} and xgj\vec{x_{gj}} are locations of the Gaussian ii and jj.

Importantly, the droplet has an interaction potential UintU_{int} that depends on the overlapping area between the droplet and Gaussians. Conversely, a Gaussian will interact with the droplet by Newton’s third law, for which explains the negative sign in equation 31. Importantly, the size of the droplet is a variable that depends on UintU_{int}, surface tension, chemical potential, and the dewetting parameter. The dynamics of radius (size of the droplet) will couple with the motions of the droplet and Gaussians, which leads to intriguing properties. I will explain this in the next section.

Refer to caption
Figure 1: 2D lattice of Gaussian (red dots) with a droplet (blue dot)

IV Interaction between Gaussians and growing droplet

IV.1 Overlapping & Interaction Energy

As we just mentioned, the interaction energy has a key impact on the dynamics of the system. We now formally define how it depends on the overlap between droplet and Gaussian and then how to calculate it use a ”soft droplet” trick.

First we define the magnitude of Gaussian potential density. For an arbitrary Gaussian with its center locate at xg\vec{x_{g}}, the potential UgU_{g} is:

Ug(X)=U0e12σ2(Xxg)2U_{g}(\vec{X})=U_{0}e^{-\frac{1}{2\sigma^{2}}(\vec{X}-\vec{x_{g}})^{2}} (21)

Here, U0U_{0} is the maximum of Gaussian and σ\sigma is the standard deviation. Those two parameters represent the height and width of the bump, it allows one to mimic different strength & density of filaments. We further set a disk with radius RR at x\vec{x}, and y\vec{y} is the vector locate within the disk, it is the substitution of an arbitrary vector respect to the center of Gaussian Xxg\vec{X}-\vec{x_{g}}. The interaction energy is the integral of the potential density over the area of the disk. A naive cartoon is shown below:

Refer to caption
Figure 2: Cartoon representation of the interaction between the Gaussian and the droplet

Intuitively, we use this interaction potential to simulation the forces generated by filaments’ deformation. Mathematically it follows:

Uint=|yx<R|Ug(y)d2yU_{int}=\int\limits_{|\vec{y}-\vec{x}<R|}U_{g}(\vec{y})d^{2}\vec{y} (22)

Normally we set the disk as a ”hard disk” with sharp edge, i.e. the vector y\vec{y} are limited within the edge. Then we can rewrite the equation 22 into a polar coordinate form:

Uint=02π𝑑θ0Rr𝑑rex2+2xrcosθ+r22σ2U_{int}=\int_{0}^{2\pi}d\theta\int_{0}^{R}rdre^{-\frac{x^{2}+2xrcos\theta+r^{2}}{2\sigma^{2}}} (23)

Here we extend equation 22 by equation 21. The x\vec{x} would still be the location vector of the disk, however the vector Xxg\vec{X}-\vec{x_{g}} in equation 21 be expressed in the polar coordinates as a vector z\vec{z}:

z=x+x^rcosθ+y^rsinθ,0<r<R\vec{z}=\vec{x}+\hat{x}rcos\theta+\hat{y}rsin\theta,0<r<R (24)

A simple illustration is helpful for understanding the model:

Refer to caption
Figure 3: Toy representation of integral in polar coordinate

However, it is hard to truncate this integral analytically which requires Bessel function. We will put the derivation into the appendix session. Here, we regard the droplet as a ”soft disk” to simplify the integral. In the ”soft disk” version, we regard the disk itself as a Gaussian density:

ρDsoft=ρ0er22R2\rho_{D}^{soft}=\rho_{0}e^{-\frac{r^{2}}{2R^{2}}} (25)

To be clear, we can also formulate the ’hard disk’ we derived above, its density should be a step function:

ρDhard=H(R|r|)\rho_{D}^{hard}=H(R-|r|) (26)

and in equation 22, we drop the constant density ρDhard=1\rho_{D}^{hard}=1 within the integral range. Differently from the soft version, the part be integrated is fixed to be one which fulfills the condition that the minority phase (droplet) has a fixed amount. With this setting, the interaction energy will be:

Uint=d2yUg(y)ρ(y)U_{int}=\int d^{2}\vec{y}U_{g}(\vec{y})\rho(\vec{y}) (27)

Along with equation 21, the integral would be:

Uint=U0d2yey22σ2e|yx|22R2U_{int}=U_{0}\int d^{2}\vec{y}e^{-\frac{\vec{y}^{2}}{2\sigma^{2}}}e^{-\frac{|\vec{y}-\vec{x}|^{2}}{2R^{2}}} (28)

A simple illustration would be exactly identical as before. Here we first focus on the argument of exponential term, it can be collected and then reorganized by completing the square:

12σ2y212R2(y22xy+x2)=x22(σ2+R2)+R2+σ22σ2R2(yσ2R2+σ2x)2-\frac{1}{2\sigma^{2}}y^{2}-\frac{1}{2R^{2}}(y^{2}-2\vec{x}\vec{y}+\vec{x}^{2})=-\frac{x^{2}}{2(\sigma^{2}+R^{2})}+\frac{R^{2}+\sigma^{2}}{2\sigma^{2}R^{2}}(\vec{y}-\frac{\sigma^{2}}{R^{2}+\sigma^{2}}\vec{x})^{2} (29)

plug it back with putting y\vec{y} irrelevant term in front of the integral:

Uint=ex22(σ2+R2)d2yeR2+σ22σ2R2(yσ2R2+σ2x)2U_{int}=e^{-\frac{x^{2}}{2(\sigma^{2}+R^{2})}}\int d^{2}\vec{y}e^{\frac{R^{2}+\sigma^{2}}{2\sigma^{2}R^{2}}(\vec{y}-\frac{\sigma^{2}}{R^{2}+\sigma^{2}}\vec{x})^{2}} (30)

which is a Gaussian integral and the result is straightforward:

Uint=π(2σ2R2+σ2)U0ex22(σ2+R2)U_{int}=\pi(\frac{2\sigma^{2}}{R^{2}+\sigma^{2}})U_{0}e^{-\frac{x^{2}}{2(\sigma^{2}+R^{2})}} (31)

The resulting expression shows that the interaction strength grows with increasing RR, which is consistent with intuition.

IV.2 Chemical Interaction

We now start to formulate the dynamics of RR. We first focus on how a Gaussian interacts with er droplet chemically. For this purpose, we start from fundamental thermodynamics, where we assume the minority phase act like an ideal solution and its Gibbs free energy is:

F=NkBTln(nn0e)F=Nk_{B}Tln(\frac{n}{n_{0}e}) (32)

in which nn is the number density of the minority phase, and n0=(mkBT2π2)32n_{0}=(\frac{mk_{B}T}{2\pi\hbar^{2}})^{\frac{3}{2}} is a reference density. Then we can obtain the chemical potential:

μ=FN|T,V=kBTln(nn0)\mu=\frac{\partial F}{\partial N}|_{T,V}=k_{B}Tln(\frac{n}{n_{0}}) (33)

whew we applied the chain rule by using relation n=NVn=\frac{N}{V}. The gradient of the chemical potential will generate a current of particles J=νμn\vec{J}=-\nu\nabla\mu n, where ν\nu is the mobility. This relation is actually the diffusion equation:

tn=J=D2n\partial_{t}n=-\nabla\vec{J}=D\nabla^{2}n (34)

in two-dimensions. The solution will be:

n(r)=Alnr+Bn(r)=Alnr+B (35)

where AA and BB are undetermined constants.

Now we focus on the equilibrium condition of the diffusion. In this condition the chemical potential of minority phase is balanced with that of mixed phase (ideal solution as we assumed): μin=μ=kBTln(nn)\mu_{in}=\mu=k_{B}Tln(\frac{n^{*}}{n}), in which nn^{*} is an arbitrary number density of minority phase. Here we further set the number density at infinite region is a fixed value nn_{\infty}, which plays a role of a reservoir of minority phase. If those two regions relaxes separately as steady currents, we can get an equation series by plug them in equation 35:

n(R)=AlnR+B=n;n(ρ)=Alnρ+B=nn(R)=AlnR+B=n^{*};n(\rho)=Aln\rho+B=n_{\infty} (36)

here RR is the radius of the droplet and ρ\rho is the location at infinity. Then by solving this, the undetermined constants can be expressed by the constants we have:

A=nnlnρR;B=nlnρRlnρnlnRlnρRA=\frac{n_{\infty}-n^{*}}{ln\frac{\rho}{R}};B=\frac{n^{*}}{ln\frac{\rho}{R}}ln\rho-\frac{n_{\infty}lnR}{ln\frac{\rho}{R}} (37)

now we plug them back to the equation 36, then take gradient respect to rr. The constant BB directly vanished and the current on the droplet RR will be a simple expression:

J(R)=DRnnlnρRJ(R)=-\frac{D}{R}\frac{n_{\infty}-n^{*}}{ln\frac{\rho}{R}} (38)

Now we can determine the dynamics of radius RR by assuming the incoming current, the flux equation 38 times 2πR2\pi R, equals to the area change rate:

dadt=2πRDRnnlnρR=2πRR˙\frac{da}{dt}=2\pi R\frac{D}{R}\frac{n_{\infty}-n^{*}}{ln\frac{\rho}{R}}=2\pi R\dot{R} (39)

where we applied a=πR2a=\pi R^{2}. Then the radius change rate can be obtained by reorganizing 39:

R˙=DRnnlnρR\dot{R}=\frac{D}{R}\frac{n_{\infty}-n^{*}}{ln\frac{\rho}{R}} (40)

then the number density of minority phase can be expressed by the equilibrium condition we mentioned above: n=n0eμinkBTn^{*}=n_{0}e^{\frac{\mu_{in}}{k_{B}T}}. Since ρ>>R\rho>>R is always true in our assumption, so we could further set lnρRln\frac{\rho}{R} as a constant D~\tilde{D}. Thus the radius change rate can be reorganized as:

R˙=D~R(nn0eμinkBT)\dot{R}=\frac{\tilde{D}}{R}(n_{\infty}-n_{0}e^{\frac{\mu_{in}}{k_{B}T}}) (41)

here the heart of our model is the chemical potential μin\mu_{in} is dependent on the interaction energy:

μin=μin+αU(x,R)R2\mu_{in}^{{}^{\prime}}=\mu_{in}+\alpha\frac{U(\vec{x},R)}{R^{2}} (42)

one can regard the last term as the excess Gibbs free energy offered by interaction potential up to a parameter α\alpha. 1R2\frac{1}{R^{2}} here represent unit area, alpha combining 1R2\frac{1}{R^{2}} represent the number of molecules. So α\alpha is the density of the minority phase, which is the intrinsic property of different materials. This equation it represents how sensitive the droplet is against interactions, thus we name α\alpha as the ”dewetting parameter”. Noticed that, to make equation 42 reasonable, we assume that in the condensed phase, the Gibbs free energy is nearly identical to Helmholtz free energy. We also drop the entropy term.

This equation gives us good intuition about the dynamics of radius. It is controlled by a constant flow from infinity proportional to nn_{\infty}. When a droplet is located at the local minimum of the potential surface, the present chemical potential of the droplet μ\mu^{^{\prime}} is smaller than that at infinity, thus particles flow in so the droplet will grow. Conversely, the droplet’s free energy increases when it overlaps with the Gaussian, leading to a larger chemical potential than that in the reservoir, thus particles flow out and the droplet shrinks. Below in figure 4 we again use a cartoon to illustrate these dynamics.

Refer to caption
Figure 4: Cartoon of the size change of the droplet

Finally, we could absorb all constant terms to a single parameter for a clean look, which we assume:

R0˙=D~n;R1˙=n0D~eμinkBT\dot{R_{0}}=\tilde{D}n_{\infty};\dot{R_{1}}=n_{0}\tilde{D}e^{\frac{\mu_{in}}{k_{B}T}} (43)

then equation 41 can be written as:

R˙=1R(R0˙R1˙eαU(x,R)kBTR2)\dot{R}=\frac{1}{R}(\dot{R_{0}}-\dot{R_{1}}e^{\frac{\alpha U(\vec{x},R)}{k_{B}TR^{2}}}) (44)

IV.3 Mechanical Interaction

Besides the chemical interaction via excess Gibbs free energy, now we also consider the mechanical interaction with the droplet. For this, we consider the surface/line tension of the droplet. We already know the chemical potential of the minority phase by equation 42, now if we add a small particle to the droplet, the updated chemical potential should be:

μ=μ+ΓaR\mu^{{}^{\prime}}=\mu+\frac{\Gamma a}{R} (45)

in which a=1n0a=\frac{1}{n_{0}} is the area per molecule of the incoming cluster.

To prove equation 45, one just needs to image a simple case: add a small particle to the drop, which results in a perturbation of radius, R+dRR+dR. Then we know the perimeter change should equal to 2πdR2\pi dR. Thus the area work (analogy of volume work in 2 dimensions) should be:

Δμ=Γ(2πdR)=Γ2πa2πR\Delta\mu=\Gamma(2\pi dR)=\Gamma 2\pi\frac{a}{2\pi R} (46)

where we use the relation that area change 2πRdR2\pi RdR equals to the parameter aa. As we assume the area aa is the area/molecule, area work is the Gibbs free energy change per molecule, that is the chemical potential change. Then the chemical potential change from thermodynamic and mechanical interactions can be collected together, and the resulting expression of R˙\dot{R} is:

R˙=1R(R0˙R1˙eαU(x,R)kBTR2+ΓaRkBT)\dot{R}=\frac{1}{R}(\dot{R_{0}}-\dot{R_{1}}e^{\frac{\alpha U(\vec{x},R)}{k_{B}TR^{2}}+\frac{\Gamma a}{Rk_{B}T}}) (47)

V Summary.

We found that the parameter α\alpha controls the dynamical properties of the droplet. For α\alpha between six and ten, the droplet interacts strongly with the filament. They may evaporate shortly after the simulation start of the simulation or undergo a number of shrink-and-growth cycles, producing the characteristic ”shark fin shape” dependence of radius on time. Eventually, the droplet will evaporate when there is a sufficiently large thermal fluctuation. For α\alpha between 1.3 and 6, the system is in a quasi-equilibrium steady-state regime with the radius fluctuating around a mean value. Finally, for α\alpha smaller than 1.3, the droplet interacts only weakly with the surroundings. The radius will grow in time proportional to t12t^{\frac{1}{2}}, permeating the surrounding filaments. The changes that take place in the growth characteristics also affect the mobility of droplets. In both the evaporation regime and the steady-state regime, the droplet will be localized most of the time in a minimum energy location, with rare jumps to neighboring minima. In the diffusive regime of small α\alpha where the droplet permeates the surrounding filaments, the droplet will be much more mobile. Below is fig.5 are displacement of the droplet over time with different α\alpha.

Refer to caption
Figure 5: Different regimes of droplet’s mobility with α=1,1.3,6,10\alpha=1,1.3,6,10

It is interesting to compare these regimes with the equilibrium phase diagram that was recently obtained by the Kosmrlj group for this problem ronceray2021liquid. The phase diagram obtained in that paper is shown in figure 8.

{subfigure}

[b]0.4 Refer to caption {subfigure}[b]0.4 Refer to caption according to ref.ronceray2021liquid

Figure 6: A. LLPS without interplaying with polymer networks; B. LLPS in an elastic network has three scenarios: (i) Cavitation; (ii) Micro-droplets; (iii) Permeation
Figure 7: Phase diagram of LLPS in an elastic network as a function of elasto-capillary hh and permeo-elastic number pp
Figure 8: Phase diagram and possible scenarios of LLPS

The regime of micro-droplets would correspond to the regime of large α\alpha parameter while the regime permeation corresponds to the regime of small α\alpha. The so-called permeo-elastic number is roughly inversely proportional to α\alpha. There is no analog in our theory for cavitation where the droplet pushes the filaments out of the way and it would be interesting to pursue that.

VI Radius fluctuations and concentration field

With our current settings, the fluxes are driven by the gradient of chemical potential, and the growth rate is identical to the incoming flux. This setting is under the assumption that the diffusion is large enough such that the concentration field along this transportation can be soon vanished, 2c=0;ct=0\nabla^{2}c=0;\frac{\partial c}{\partial t}=0.

However, in actuality, the low and high-density/concentration domains are separated by an interface. The relaxation of the concentration field is not instantaneous as we assumed. Instead of analytically solving the Cahn-Hilliard equation yeung1992phase for this diffusive system, we use Einstein relation costigliola2019revisiting to add a radius fluctuation to the dynamics of RR as a proxy of the concentration field. Specifically, we add a noise term to the flux term which corresponds to the time evolution of the concentration profile:

ct=J;J=νcμ+Jn\frac{\partial c}{\partial t}=-\vec{\nabla}\vec{J};\vec{J}=-\nu c\vec{\nabla}\mu+\vec{J_{n}} (48)

Again, here ν\nu is the mobility, and Jn\vec{J_{n}} is the noise term we add to the flux. Specifically, a path integral around the perimeter of the sphere was added to integrate fluxes around it:

R˙=1R(R0˙R1˙eαU(x,R)kBTR2+ΓaRkBT)+a2π02πR𝑑sJrn\dot{R}=\frac{1}{R}(\dot{R_{0}}-\dot{R_{1}}e^{\frac{\alpha U(\vec{x},R)}{k_{B}TR^{2}}+\frac{\Gamma a}{Rk_{B}T}})+\frac{a}{2\pi}\int_{0}^{2\pi R}dsJ_{r}^{n} (49)

where s is the path along the perimeter and aa again is the molecular area, and this integral can be represented as a cartoon in figure 9:

Refer to caption
Figure 9: Toy representation of flux integral around the surface

We now consider this correction term together as ζ(t)=a2π02πR𝑑sJrn\zeta(t)=\frac{a}{2\pi}\int_{0}^{2\pi R}dsJ_{r}^{n}. Then the autocorrelation function of ζ(t)\zeta(t) is:

<ζ(t)ζ(t)>=(a2π)202πR𝑑s02πR𝑑s<Jrn(s,t)Jrn(s,t)><\zeta(t)\zeta(t^{\prime})>=(\frac{a}{2\pi})^{2}\int_{0}^{2\pi R}ds\int_{0}^{2\pi R}ds^{\prime}<J_{r}^{n}(s,t)J_{r}^{n}(s^{\prime},t^{\prime})> (50)

it can be easily estimated as 2πR(a2π)2Δ1lδ(tt)2\pi R(\frac{a}{2\pi})^{2}\Delta\frac{1}{l}\delta(t-t^{\prime}) up to an undetermined constant Δ\Delta. That relies on equal time correlation of Helmholtz free energy and kinetics of the concentration change: δct=D2δcJn\frac{\partial\delta c}{\partial t}=D\nabla^{2}\delta c-\vec{\nabla}\vec{J_{n}}. Here we only want audiences to notice the radius would fluctuate up to a noise due to such relaxation of the matter transfer. As we are not turning on this noise term in our later investigation, here We only leave the result as Δ=2Dc(R)\Delta=2Dc(R).

VII Non-linear 1D system and Bifurcation

Before we turn on the fluctuations of dynamics of radius and displacement, we first simplified the system into a 1-dimensional model by coarse-graining the net force around the droplet.

This simplification can be done by assuming several Gaussian potentials around the droplet exert an effective harmonic force on it, thus we could regard all point mass on the droplet as feeling such potential. For a location center of mass of the droplet r\vec{r} and an arbitrary point ρ\vec{\rho} deviates from the center, this integrated potential is:

Ueff=ρ<|R|d2ρ12u0(r+ρ)2=12u0(r2πR2+π2R4)U_{eff}=\int_{\vec{\rho}<|R|}d^{2}\vec{\rho}\frac{1}{2}u_{0}(\vec{r}+\vec{\rho})^{2}=\frac{1}{2}u_{0}(r^{2}\pi R^{2}+\frac{\pi}{2}R^{4}) (51)

here u0u_{0} is the effective spring constant of the harmonic, we measure this constant by simulating the 2D model with a small deviation, to make it comparable with the original model.

After obtaining the effective potential, we just need to replace the UintU_{int} in equation 44 by this. The resulting expression has a clean look:

R˙=1R(R0˙R1˙eαu0(r2π+π2R2)kBT+ΓaRkBT)\dot{R}=\frac{1}{R}(\dot{R_{0}}-\dot{R_{1}}e^{\frac{\alpha u_{0}(r^{2}\pi+\frac{\pi}{2}R^{2})}{k_{B}T}+\frac{\Gamma a}{Rk_{B}T}}) (52)

Together with the equation of motion γdrdt=Ur\gamma\frac{d\vec{r}}{dt}=-\frac{\partial U}{\partial r} of the droplet, and with turning off both noise terms, the result differential equation series can be regarded as a nonlinear dynamical system. This deterministic description offers us insight into the dynamics. Here we use the resulting flow lines of the vector field to illustrate it.

{subfigure}

[b]0.3 Refer to caption {subfigure}[b]0.3 Refer to caption {subfigure}[b]0.3 Refer to caption

Figure 10: α=0\alpha=0
Figure 11: α=5\alpha=5
Figure 12: α=9.2\alpha=9.2
Figure 13: Three regimes of flow lines

When we turn off the interaction energy, all vectors point toward the center location dominated by a regular harmonic potential. With α=5\alpha=5, a stable fixed point emerges at r=0,R=Rr=0,R=R^{*} in which RR^{*} is the saturated radius, the horizontal vectors indicate at almost all finite displacements the radius tends to converge to the saturated radius. Two unstable fixed points result from surface tension when the size of the droplet is small enough, i.e. R0R\approx 0 the mechanical interaction forces the evaporation at a finite displacement. However, at a later stage α=9.2\alpha=9.2, almost all vectors strictly rain down towards negative RR, indicating the droplet will constantly evaporate even with a finite interaction with the potential.

VIII The effective dynamics and theory by a perturbation

VIII.1 Stochastic dynamics

Now we turn on the thermal noise to use the stochastic dynamics to illustrate the difference between those two regimes. The resulting dynamics of radius and displacement with two different α\alpha are shown at the top of this page:

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[b]0.7 Refer to caption {subfigure}[b]0.7 Refer to caption

Figure 14: α=5\alpha=5
Figure 15: α=9.2\alpha=9.2
Figure 16: Two regimes of dynamics

With α=5\alpha=5, the radius rarely fluctuates but converge to the saturated radius. But with α=9.2\alpha=9.2, the evaporation events frequently happened. To mimic the real biophysical process, we couple a Poisson process as the nucleation time after the droplet evaporated. In this case, the dissipation by interacting with potential cannot balanced by the thermal fluctuation, the fluctuation-evaporation-nucleation process keeps consuming free energy and a steady state non-equilibrium condition has developed, in which, the droplet cannot forget its initial condition with finite lifetime before it reaches the saturated radius state.

VIII.2 A theoretical explanation by a small perturbation

The dynamics presented above are hard to truncate analytically. However, here we offer a theory to explain it by assuming the radius change around saturated value up to a small perturbation: R2=R¯2+x(t)R^{2}=\bar{R}^{2}+x(t), where x(t)x(t) is the time-dependent perturbation. We further collect all prefactors as two constants in equation 51 in a cleaner form: Ueff=12Cr2R2+DR4U_{eff}=\frac{1}{2}Cr^{2}R^{2}+DR^{4}. Thus at steady state, we have both differentiation equals zero, noticed here we also ignore the surface tension γ\gamma as the nonlinear dynamics showed, it just has light influence when the droplet is small enough:

CR2r=0;R¯=1αDln(R0˙R1˙)-CR^{2}r=0;\bar{R}=\sqrt{\frac{1}{\alpha D}ln(\frac{\dot{R_{0}}}{\dot{R_{1}}})} (53)

then the effective potential and the dynamics of the displacement can be written as:

Ueff=12C(R¯2+x)r2+D(R¯2+x)2;ζdrdt=C(R¯2+x)+η(t)U_{eff}^{\prime}=\frac{1}{2}C(\bar{R}^{2}+x)r^{2}+D(\bar{R}^{2}+x)^{2};\zeta\frac{dr}{dt}=-C(\bar{R}^{2}+x)+\eta(t) (54)

here η(t)\eta(t) is the white noise of the 1D dynamics. For x=0x=0, this is a trivial over-damped harmonic oscillator and the time correlation of the displacement has the form <r(t)r(0)>=<r2>etτr<r(t)r(0)>=<r^{2}>e^{-\frac{t}{\tau_{r}}} which up to a constant τr=ηCR¯2\tau_{r}=\frac{\eta}{C\bar{R}^{2}}. then the dynamics of xx can be captured by plugging in the perturbation expression of RR:

12dxdt=R0˙R1˙eα(DR¯2+12Cr2+Dx)\frac{1}{2}\frac{dx}{dt}=\dot{R_{0}}-\dot{R_{1}}e^{\alpha(D\bar{R}^{2}+\frac{1}{2}Cr^{2}+Dx)} (55)

by extracting constant term with R¯\bar{R} and DD, one can Taylor expand the remaining exponential term as both rr and xx are small values at steady state. The resulting expression is:

12dxdt=R0˙R1˙eαDR¯2(1+12αCr2+αDx)\frac{1}{2}\frac{dx}{dt}=\dot{R_{0}}-\dot{R_{1}}e^{\alpha D\bar{R}^{2}}(1+\frac{1}{2}\alpha Cr^{2}+\alpha Dx) (56)

By applying the saturated radius, equation 53, the above expression was cleaned up:

12dxdt=R0˙(12αCr2+αDx)\frac{1}{2}\frac{dx}{dt}=-\dot{R_{0}}(\frac{1}{2}\alpha Cr^{2}+\alpha Dx) (57)

Next, we defined the time scale τR=12R0˙αD\tau_{R}=\frac{1}{2\dot{R_{0}}\alpha D}, then we again rewrite the dynamics as:

dxdt=1τRxR0˙αCr2(t)\frac{dx}{dt}=-\frac{1}{\tau_{R}}x-\dot{R_{0}}\alpha Cr^{2}(t) (58)

The dynamics of xx can be solved analytically up to an integral:

x(t)=R0¯αC0t𝑑tettτRr2(t)+x(0)etτRx(t)=-\bar{R_{0}}\alpha C\int_{0}^{t}dt^{\prime}e^{-\frac{t-t^{\prime}}{\tau_{R}}}r^{2}(t^{\prime})+x(0)e^{-\frac{t}{\tau_{R}}} (59)

then with a long timescale t>>τRt>>\tau_{R}, the last term can be ignored. Then we plug in equation 59 into equation 54:

ζdrdt=CR¯2r(t)+R0¯αC0t𝑑tettτRr2(t)+η(t)\zeta\frac{dr}{dt}=-C\bar{R}^{2}r(t)+\bar{R_{0}}\alpha C\int_{0}^{t}dt^{\prime}e^{-\frac{t-t^{\prime}}{\tau_{R}}}r^{2}(t^{\prime})+\eta(t) (60)

it is clear that the second term introduces a ”memory kernel” to the dynamics of rr, which is consistent with the stochastic dynamics result, in which the droplet cannot forget its initial condition with a finite lifetime before reaching the equilibrium steady state. This non-uniformity in time is a piece of strong evidence for a non-equilibrium condition.