Brownian Motion as a Free Energy Minimizing Process#

Consider a suspension of a large number of ideal beads:

Figure

Goal:#

Construct a phenomenological model for the dynamics of the particle density c(x,t) at position x and time t that leads to the equilibrium (Boltzmann) distribution

c(x,t)ceq(x)eU(x)kBT.

Based on simple and natural assumptions, we will be led to the generalized diffusion equation. One nice (and strange) feature of these dynamics is that they continuously lower the free energy.


Conservation of Particle Number#

The conservation of particle number requires that the dynamical equation of motion is described by the continuity equation:

(1)c(x,t)t=j(x,t)x

where j(x,t) is the probability current, i.e., the number of particles crossing x per unit time.

The probability current j(x,t) has two contributions:#

  1. Deterministic motion (drift due to potential U(x)):

    jdet=cv¯=cξUx

    where we introduced the mean velocity v¯=xU/ξ of an overdamped particle in a potential U. The friction coefficient ξ for a spherical particle of diameter σ in a fluid of viscosity η is given by:

    ξ=6πησ
  2. Random motion (diffusion):

    jdiff=Dcx(Fick’s Law)

Figure

Total current:#

Combining both contributions yields

j=jdet+jdiff,

which can be rewritten as j=cvfl in terms of a net velocity vfl,

vfl=ξ1x(U+Dξlnc)=ξ1μx,

where μ~(x) is defined as $μ~(x)=Dξlnc+U(x);.$

Note that μ~(x) is the local chemical potential for non-interacting particles if we could replace Dξ by kBT. We will soon see that this has to be the case to reproduce the Bolzmann distribution in equilibrium.

Generalized Diffusion Equation#

Substituting this into the continuity equation gives

ct=xj=x(cξμ~x)=x(1ξ(Dξcx+cUx)).

Equilibrium (t):#

At equilibrium, ct=0 implies j=0, so:

μ~=U+Dξlnceq=const.

Solving for ceq:

ceqexp(U(x)Dξ).

Since this must reproduce the Boltzmann distribution:

ceqexp(U(x)kBT),

we require the Stokes-Einstein relation:

D=kBTξ.

This is a key example of a fluctuation-dissipation theorem. So, indeed, μ~=μ, the local chemical potential.

Example values:#

  • For a spherical particle with radius R: ξ=6πηR.

  • Proteins (R2 nm): D100μm2/s.

  • Bacteria: Dbac0.5μm2/s.

(For physical intuition, it’s helpful to anticipate Δx2(t)=2Dt.)


Generalization: Multiple degrees of freedom#

Knowing μ=const. in equilibrium suggests that, close to equilibrium, gradients of the chemical potential should act as driving forces towards equilibrium. This insight can be used to construct Brownian motion for a complex system with potentially very many degrees of freedom {xi}. Here, xi could be the position of a segments of a polymer or a membrane, or just coordinates of interacting particles. The gradient Fi=iμ is the thermodynamic driving force.

As before, we demand a continuity equation,

tΨ({xi}i)=iiji,

where ji=Ψvi and

vi=jHijFj

with mobility matrix Hij and forces:

Fj=jμ,μ=U+kBTlnΨ.

This is known as the Smoluchowski equation.


Free Energy Minimization#

Define the free energy functional:

F[c]=dxc(x,t)μ(x).

The free energy is monotonically decreasing under Brownian motion:

dFdt=dx[μct+cμt].

Using integration by parts and boundary conditions at infinity, we obtain:

dFdt=dx[c(xμ)2+kBTtc],

which is negative unless μ=const., i.e., we have reached equilibrium.

Note: The dynamics are time-irreversible due to coarse-graining and underlying microscopic chaos (see next lecture on ergodicity).


Diffusion Equation – Basic Solution:#

For U=0:

(2)c(x,t)t=Dx2c(x,t)

Diffusion

Assume c(x,0)=δ(x) (point source at x=0).

Fourier transform of (2):

c(x,t)=dq2πeiqxc(q,t),c(q,t)t=Dq2c(q,t).

Solution:

c(q,t)=eDq2t.

Inverse Fourier transform:

c(x,t)=ex2/(4Dt)4πDt=ex2/(2σt2)2πσt2

with σt2=2Dt.

Note: This is the Green’s function solution. For arbitrary initial conditions, convolve with this Green’s function.

Since particles are non-interacting:

c(x,t)=Np(x,t),

where p(x,t) is the probability density for a single particle.


Practical Example: Nerve Cell Transport#

Is diffusion fast enough to transport vesicles?

Given D0.5μm2/s:

x22D12days!

Nerve cells require active transport (e.g., via motor proteins).

Example: Bacteria#

For L1μm and D100μm2/s:

tL22D2ms,

which is much shorter than the 20 min lifecycle of E. coli diffusion is effective for bacteria.


Application: Diffusion-limited Reaction Rates#

Ligand capture rate:#

How many ligands are captured per sec?

Use diffusion equation to model the ligands: $tc=D2c=D(x2c+y2c+z2c)$

tc=D1r2r(r2rc);c=c(r,t)

Steady state: 0=tcs, so, rcs= const r2 for r>R

cs(r)=Ar+B=Ar+c0cs()=c0

Perfect absorber limit: cs(R)=0.

cs(r)=c0(1Rr).

The particle current is j(R)=|j(R)|=Drc(r,t)=Dc0RR2.

Thus, the total flux of captured ligands is given by the expression

Φ=j(R)4πR2=4πDc0R

(Check dimensions!)


Universal Bound on Reaction Rates#

Key question: How fast is A+BkC if it is limited by AB collision rates?

In the extreme case, where each A-B collision leads to a C particle we have k=cAk~AB, where where k~AB is the rate of which a focal A particle is hit by B particles.

Since xAxB diffuses w/ twice the diffusion constant, we have that the number of collisions per unit time of B particles with a focal A particle is given by the above rate of ligand capture with twice the diffusivity and with twice the radius,

k~AB=4π(2D)(2R)CB
CCt=kdiff CACB

where

kdiff =16πDR=16πRkBT6πnR=8kBT3η

for spherical particles. Note that this is a nice universal bound, dependent only on temperature and viscosity. For water at room temperature, one estimates

kdiff 71091Ms

Notes:

  • Most enzymes are 100-1000 times slower they require more than just one collision to form a product.

  • … could be modeled by assuming a finite absorption rate.

  • Relation to Michaelis-Menten kinetics: Rate of product formation =kmax[E][S]/(KM+[S]).

Comparing with the above equation, we see that the enzyme efficience is bounded by kdiff $[kmaxKμ]=1 Concentration time $

For A+BC:

k~AB=4π(2D)(2R)CB=16πDRCB.

Thus,

CCt=kdiffCACB

with:

kdiff=16πDR=8kBT3η.

Estimate:

kdiff7×1091Ms.
  • Most enzymes operate 100–1000x slower.

  • Enzyme efficiency is thus bounded by kdiff.

Relation to Michaelis-Menten:

Rate=kmax[E][S]KM+[S],

where kmax/KM is upper bounded by kdiff:

[kmaxKM]=1concentrationtime.