Applications of non-equilibrium statistical mechanics

Differences and similarities with equilibrium theories and applications to problems in several areas.

Birger BergersenDept. Of Physics and Astronomy

University of British Columbiabirger@phas.ubc.ca

Overview• Overview of equilibrium systems: Ensembles, fluctuations , detailed balance,

equations of state, phase transitions, broken symmetry, information theory.• Steady states vs. Equilibrium: Linear response, Onsager coefficients, variational

principles, equilibrium “economics”, approach to equilibrium, Le Chatelier’s principle.

• Oscillatory reactions: Belousov –Zhabotinsky reaction, brusselerator.• Effects of delay: Logistic equation with delayed carrying capacity• Inertia: Transition to turbulence• Bachelier-Wiener processes, Master equation, Birth and death, Fokker Planck

equation, Brownian motion, escape probability, stochastic resonance, Brownian motors, Parrondo games.

• Other processes: Material failure, multiplicative noise, income distribution, Pareto tail.

• Levy distributions, colored noise,

Equilibrium: ensembles

• Microcanonical ensemble:

• Canonical ensemble:

T,V,N specified

Insulating walls

Entropy S maximum subject to constraintsTemperature T, pressure P, chemical potential μ fluctuating

Heat bath at temperature T

N,E,V specified

Helmholtz free energy A=E-TS minimum subject to constraints. Competition between energy and entropy.Entropy S , energy E, pressure, chemical potential fluctuating

• Gibbs ensembleG=E-TS+PV =μN minimumGibbs-Duhem relationE=TS+ μN-PV valid ifE,N,V,S extensiveμ,T,P intensiveExtensive= proportional to system sizeIntensive= independent of system size

• Grand canonical ensemble

P

Heat bath

P,T,N specified

μ, V, T specified

Heat bath

Porous walls

N,P,S fluctuating Grand potential Ω=-PV minimum

Canonical ensemble

dA =-SdT -PdV +μdNExample isothermal athmosphereIf atmospheric pressure at height h=0 is p(0) and m the mass of a gas molecule

Canonical fluctuations

I assume that what we done so far are things you have seen before, but to refresh your memory let us do some simple problems:Problem 1:In the micro-canonical ensemble the change in energy E, when there is an infinitesimal change in the control parameters S,V, N is

dE= TdS-PdV+μdNSo that T=∂E/ ∂S, P=- ∂E/ ∂V, μ= ∂E/ ∂N

a. Show that if two systems are put in contact by a conducting wall their temperatures a equilibrium will be equal.b. Show that, if the two systems are separated by a frictionless piston, their pressures will be equal at equilibrium.c. Show that if they are separated by a porous wall the chemical potential will be equal on the two sides at equilibrium.

Solution:Instead of considering the state variable E as E(S,V,N) we can treat the entropy as S(E,V,N)

dS=E/T dE +P/T dV –μ/ T dNOr 1/T=∂S/ ∂ELet E=E₁+E₂ be the energy of the compound system and the entropy is S=S ₁+S₂. At equilibrium the entropy is maximum or

Since ∂E₁ / ∂E₂=-1 we find 1/T₁ =1/T₂ or T₁ =T₂ .Similar arguments for the case of the piston tells us that at equilibrium P₁/T₁ = P₂/T₂. But, since T₁ =T₂, P₁= P₂.The same argument tells us that μ₁= μ₂.Note that if the two systems are kept at different temperatures, the entropy is not maximum, when the pressure is equal as required by mechanical equilibrium!

∂

Problem 2:If q and p are canonical coordinated and momenta the number of quantum states between q and q+dq, p and p+dp are dp/dq/h, where h is Planck’s constant. If a particle om mass m is in a 3D box of volume V its canonical partition function is

The partition function for an ideal gas of N identical particles is The factor N! comes about because we cannot tell which particle is in which state.a. Find expressions for the Helmholz energy A, the mean entropy

S and the mean energy E and the pressure Pb. Two ideal gases with N₁=N₂=N molecules occupy separate

volumes V₁=V₂=V at the same temperature. They are then mixed in a volume 2V. Find the entropy of mixing. Show that there is no entropy change if the molecules are the same!

Solution:We have

For large N we can use Stirlings formula ln N!=N ln N-N and find

The entropy can be obtained from S=-∂A/ ∂T, and

Similarly E=A+TS=3nkT/2 and P=-∂A/ ∂V=NkT

If the two gases are different the change in entropy comes about because V→2V for both gases. Hence, ∆S=2Nk ln 2If the two gases are the same there is no change in the total entropy.

Problem 3. Maxwell speed distribution

a. Assuming that the number of states with velocity components betweenis proportional to and that the energy of a molecule with speed is find theprobability distribution for the speed.

b. Find the average speed

c. The mean kinetic energy

d. The most probable speed i.e. the speed for which

Solution:Going to polar coordinate we find that To find the proportionalityconstant we use

after some algebra:

Integrating we find

Principle of detailed balanceThe canonical ensemble is based on the assumption that theaccessible states j have a stationary probability distribution

π(j)=exp(-βE(j))/Z(β)This distribution is maintained in a fluctuating environmentin which the system undergoes a series of small changes with probabilities P(m←j). For the distribution to be stationary it must not change by transitions in and out of state

π(m)=∑ P(j←m) π(m)= ∑ P(m←j) π(j)This is satisfied by requiring detailed balance:

P(j ←m)/P(m ←j)= π(j)/ π(m)=exp(-β(E(j)-E(m))In a equilibrium simulation we get the correct equilibrium properties, by satisfying detailed balance and making sure there is a path between all allowed states. We can ignore the actual dynamics (Monte Carlo method). Very convenient!

jj

Equations of stateThe equilibrium state of ordinary matter can be specified by a small number of control variables. For a one component system the Helmholtz free energy is A=A(N,V,T). Other variables such as the pressureP, the chemical potential μ, the entropy S are given by equations of state μ=∂A/ ∂N, P= - ∂A/ ∂V, S=- ∂A/ ∂TAn approximate equation of state describing non-ideal gasesIs the van der Waals equation

which obtains from the free energy

Comparing with the monatomic ideal gas expression

We interpret the constant a as a measure of the attraction between molecules outside a hard core volume b

Equilibrium phase transitionsFor special values of the control parameters (P,N,T in the Gibbs ensemble) the order Parameters(V,μ,S) changes abruptly. P and T for a single component system when solids/liquid and gas/liquid coexists at equilibrium is shown at the top while the specific volume v=V/N for a given P is below.The gas/liquid and solid/liquidtransition are 1st order (or dis-continuous). The critical endpoint of the latter is a 2nd order (or continuous ). Specific heatand fluctuations diverge at criticalpoint.

Combining the P-V and P-T plots into a single 3D plot.

Problem 4: Van der Waals equation.

a. Express the equation as a cubic equation in v=V/N.

b. Find the values for which the roots coincide.

c. Rewrite the equation in terms of the reduced variables.

the resulting equation is called the law of corresponding states.

d. Discuss the behaviour of the isotherms of the the law of corresponding states.

Solution:a. Multiplying out and dividing by a common factor we find

b. When the roots are equal this expression must be on the form

comparing term by term we find after some algebra

c. Again, after some algebra we find in terms of the new variables the law of corresponding states

d. We next plot p(v,t) for different values of t

←t=1.1

t=1 →

←t=0.9

PFor t<1 there is a range of pressures p₁<p<p₂ with 3 solutions . The points p₁ and p₂ are spinodals. In the region between them the compressibility -∂P/∂V is negative and the system is unstable. The equilibrium state has the lowest chemical potential μ We have μ=(A+PV)/N so that dμ=(SdT+VdP)/N

μ₂- μ₁=∫²₁=∫VdP/NThe equilibrium vapor pressure is when the areas between the horizontal line and the isotherms are equal. Between the equilibrium pressure and the spinodal there is a metastable region. Dust can nucleate rain drops. If particles too small surface tension prevents nucleation. VdW theory gualitatively but not quantitatively ok.

Symmetry breaking: Alben model.A large class of equilibrium systems characterized by a high temperature symmetric phase and low temperature broken symmetry phase. Some examples:Magnetic models: Selection of spin orientation of “spins”, Ising model, Heisenberg model, Potts model....Miscibility models: High temperature solution phase separates at lower temperaturesMany other models in cosmology, particle and condensed with analogies in other fields.

The Alben model is a simple mechanical example.An airtight piston of mass M separates two regions, each containing N ideal gas molecules at temperatureT. The piston is in a semicircular tube of cross section a.The “order parameter” of the problem is the angle Φ of the piston.

The free energy of an ideal gas is A=-NkT ln (V/N)+terms independent of V, N

yielding the equation of state PV=-∂A/ ∂V=NkT

We add the gravitational potential energy to obtainA=M gR cos φ- NkT(ln [aR(π/2+ φ)/N]+ln [aR(π/2+-φ)/N]

Minimizing with respect to φ gives0= ∂A/ ∂ φ=-MgR sin φ-8NkT φ/(π²-4 φ²)

This equation has a nonzero solution corresponding to minimum forT<T₀=MgR π²/(8Nk)

The resulting temperature dependence of the order parameter is shown below to the left. To the right we show the behavior if the amount of gas on the two sides is N(1± 0.05)

At equilibrium:• Second law: Variational principles apply: entropy

maximum subject to constraints → free energy minimum . System stable with reversible, Gaussian fluctuations (except at phase transitions, continuous or discontinuous) → No arrow of time, detailed balance.

• Zeroth law: Pressure, temperature chemical potential constant throughout system.

• No currents (of heat, charge, particles)• Thermodynamic variables: Extensive (proportional to

system size, N,V,E,S), or intensive (independent of size, μ,T,P). Law of existence of matter.

• Typical fluctuations proportional to square root of size

Information theory approach

Steady state vs. equilibrium:

• Steady states are time independent, but not in equilibrium.

J T+Δ TQ

Heat currentElectric current

Salt water Fresh waterOsmosis No zeroth law! Gradients of

temperature, chemical potential and pressure allowed.

Energy

Labour

Raw materials

Capital

Economy

ProfitsWagesProductsWaste

Economy

If no throughput there is no economy!Economic “equilibrium” an oxymoron.

Equilibrium “economics”

• N = # of citizen >>1w = W/N= average incomeMost probable income distributionp(w)=exp(-x/w)/wAverage income analogous to thermodynamic temperature.Problem: Income inequality varies from society to society! More parameters needed!

In non-equilibrium steady state entropy not maximized (subject to constraint).Example: Two gas containers of equal size are connected byA capillary. One is kept at a higher temperature than the otherby contact with heat baths.

If the capillary is not too thin (compared to mean free path of molecules ) system will reach mechanical equilibrium with constant pressure. The cold container will then have more molecules than the hot one.

P, V Tc P, V Th

The entropy per molecule is higher in hot gas. If entropy maximum subjectto temperature constraint molecules would move from cold to hot. This does not happen!

Chemical potential will be different on the two sides in steady state.

Steady states not governed by variational principle!Zeroth law does not apply!

Systems near equilibrium• Ohm’s law: Electrical current

V=R I V=voltage, R= resistance, I=current

• Fick’s law of diffusion

• Stokes law for settling velocity

Many similar laws!

More subtle effects:Temperature gradient in bimetallic strip produces electromotive force → Seebeck effectElectric current through bimetallic strip leads to heating at one endAnd cooling at the other → Peltier effectDust particle moves away from hot surface towards cooler region.→Thermophoresis. Particles in mixture may selectively move Towards hotter or cooler regions → Ludwig-Soret effectIn rarefied gases a temperature gradient may a pressuregradient → thermomolecular effect

Economic analogies? Wealth gradient may cause migration, emigration and ghetto formation.

Le Chatelier’s principleAny change in concentrations, volume or pressure is counteracted by shift in equilibrium that opposes change.

If a system is disturbed from equilibrium , it will return monotonically towards equilibrium as disturbance is removed (no overshoot).

In neoclassical economics it has been suggested that if supply exceeds demand, prices will fall, production cut and equilibrium restored. A similar process should occur if demand exceedssupply

But is this true?

Belousov-Zhabotinsky reaction

In the 1950s Boris Pavlovich Belousov discovered an oscillatory chemical reaction in which Cerium IV ions in solution is reduced by one chemical to Ce III and subsequently oxidized to Ce IV causing a periodic color change. Belousov was unable to publish his results, but years later Anatol Zhabotinsky repeated the experiments and produced a theory . By now many other chemical oscillators have been found.

A linear stability analysis of the fixed point inparameter space is shown in the picture to the right. Trajectories of the concentrations X and Yare shown below.

Le Chatelier’s principle does not apply because the system is driven, A,B must be kept constant. Reactions are irreversible.

Problem 5: Lotka Volterra modelBig fish eat little fishes to survive. The little fish reproduce at a certain net rate, and die by predation. The rate model rate eqs are

a. Simplify the model by introducing new variables b. Find the steady states of the populations and their stability.

Discuss the solutions to the rate equations.c. In the model, the populations of little fish grows without limits

if the big fish disappear. Modify the little fish equation to add a logistic term, with K the carrying capacity

How does this change things?d. If the population of big fish becomes too small it may become

extinct. Modify model to include small immigration rates. Neglect the effect of finite carrying capacity.

Solution:a. With the substitutions the equations become

dy/dτ=-ry(1-x);dx/dτ=x(1-y)b. There are two steady states x=0,y=0, and x=1,y=1. The former is a saddle point (exponential growth in the x-direction, and exponential decay of the big fishes). Linearizing around the second point x=1+ ξ, y=1+η yields

dη/dτ=r ξ, dξ/dτ=-ηwhich are the equations for harmonic motion around a center. The full differential equations can be solved exactly.Dividing the two equations and crossmultiplying

This equation can easily be integrated.

r=1

↓ ↓

c. In reduced units we write for the rate equationsdy/dτ=-ry(1-x); dx/dτ= x(1-y-x/K)

There is a new fixed point y=0,x=K. Linearizing about this pointy=η; x=K+ξdη/dτ=-rη(1-1/K); dξ/dτ=-K(η+ξ/K)

We find that the fixed point is a saddle for K>1, stable focus for K<1.

The old fixed point at 1,1 is now shifted to 1,1-1/K, it is unphysical for K<1. Linearizing for K>1 with

y=1-1/K+η; x=1+ξdη/dτ=-r(1-K)ξ; dξ/dτ=-η-ξ /K

The eigenvalues are

i.e. If 4K²r(K-1)>1 it is a stable focus otherwise a stable node.

Here are some typical situations:

r=0.1, K=1.6

↓↓

↓

r=1, K=0.5

↓

↓ ↓r=1, K=2

In the left plot there is not enough little fish habitat to support a big fish population.

In the middle plot the big fish can survive a long time without food.

In the plot to the right the big fish are more dependent on a steady supply of little fish.

d. With λ the big fish immigration rate in reduced units, we havedy/dτ= -r x (1-y)+λ; dx/dτ=x (1-y)

The x=0,y=0 fixed point shifts to x=0, y= λ/r. Linearizing around this point with x=ξ, y= λ/r+η gives

d ξ/dτ= ξ(1-λ/r); dη /dτ= λ ξ-r ηAssuming r>λ we see that the point remains unstable. There is a second fixed point at x=1-λ/r, y=1 . With

x=1-λ/r + ξ; y=1+ η dη/dτ= rξ- η λ, d ξ/dτ= (1-λ/r)η

The eigenvalues are

The system will thus undergo damped oscillations towards a stable fixed point. If λ is small the damping is weak.If the carrying capacity is finite immigration will not change things much except, when K<1, there will be a small, number of starving big fish. If they have any sense they will emigrate again!

The logistic equationdN/dτ=λN(1-N/K)is important in ecology. It gives an idealized description of limitation to growth K= carrying capacity, λ= growth rate. Put t= λ τ, y=N/K

dy/dt=y(1-y)

Equation is easy to solve. Solution assumes instantaneous response to changes in population size.In practice there is often a time delay in how changes in population affects growth rate.In economics time delay between investment and results.“Pig-farmers dilemma”

Modify logistic equation to account for delays between changes in the environment and their effect. A celebrated example is the Nicholson equation which describes population fluctuations in sheep blowflies, a known pest on Australian sheep farms:

dN/dt=r N(t-τ) exp(-N(t-τ)/K)-mNHere τ is a time delay due to hatching and maturing, r is the birth rate when the population is small compared to the carrying capacity K and m is the death rate.

The Nicholson equation is said to describe the data quite well. We will instead consider the simpler Hutchinson’s equation

dN/dt’=r N 1-N(t’-τ)/KWith y=N/K, t=t’/ τ we get

dy/dt=r1-y(t-1)The solution y=1 is a fixed point we need to study its stability. Assume that r is positive and

y=1+δ exp(λt)We obtain the eigenvalue equation λ=-r exp(- λ) or

r=- λ exp(λ)(i) λ is realWe find 2 negative roots for r<1/eNo real roots for r>1/e Negative real roots→ stability

- λ exp(λ)

(ii) λ = λ ₁ + i λ₂ complexWe have

r=- λ ₁ cos( λ ₁) = λ₂ sin(λ₂)We see that λ ₁ changes sign when r=(2n+1)π/2, n=0,1,2...For large values of r the Hutchinson equation is probably not very realistic.To solve dy/dt=r1-y(t-1) numerically for t>0 we need to specify initial conditions in the range -1..0. We plot belowNumerical solutions for r=1.4 and r=1.7

r=1.4r=1.7

In the previous example the stability of the system depends on the ratio between the timescale for growth and the time delay.If the delay is too large the system becomes unstable. There are numerous applications in Ecology (Delay due to maturation, regrowth)Medicine (Regulatory mechanism require production of hormones etc.)Economics (Investments takes time to produce results)Politics (Decisions take time)Mechanics (Regulatory feedback involves sensory input that must be processed) There is evidence that our brains automatically tries to calculate a trajectory to compensate for delay when catching or avoiding a flying object.The effect of inertia:Physically the effect of inertia isSimilar to that of delay. It takes time for a force to take effect!

Flow in a pipeAt low flow velociy flow is laminar and constant. Inertia plays no role. At higher velocity irregularities in pipe wall and external noise cause a disturbance. Competition between inertia and viscosity determines if laminar flow is reestablised.

Discrete Markov processesThere are a number of situations in which the system evolution can be described by a series of stochastic transitions from one state to another, but detailed balance can not be assumed. Suppose the system is in state i and let the P(j←i) be the probability that the next state is j. In a Markov process these probabilities only depends on the states and not on the history. P(j←i) can be represented by a matrix in which i is a column and j a row. This matrix is regular if all elements of (P(j←i)) ⁿ are positive and nonzero for some n (it must be possible to go from any state to any other state in a finite number of steps). There will then be a steady state probability distribution π(i) given by the eigenvector of P(j←i) with eigenvalue 1 normalized so that

∑ π(i) =1in analogy with ensembles in equilibrium theory.

1

2 3

Problem 6:A random walker moves back and forth between three different rooms as shown. The walker exits through any available door with equal probability, but on average stays twice as long in room 1 than in any other room.a. Set up the transition matrix P(j←i).

b. Is the matrix regular?

c. Find the steady state probability distribution π(i).

Solution:a. We find for the transition matrix i→ j ↓

Note that the column entries have to add up to one, but there is no such restriction on the rows.b. The matrix is regular. It is possible to get from any room to any other room in two steps. c. To find the steady state probabilities we must solve the eigenvalue problem

π(j) = λ Σ P( j←i) π (i)We find that the normalized eigenvector with eigenvalue λ=1 is π(1)= π(2)=2/5, π(3)=1/5.

Master equation:Transition events need not be evenly spaced in time.

To obtain the time evolution of the probability we need to consider the transition rate W(q←q’) which for a continuous process is governed by the master equation

∂P(q,t)/ ∂t=∫ dq’[W(q←q’)P(q’,t)-W(q’←q)P(q,t)]

One step birth and deathThe extension to discrete processes is straightforward. Consider the case in which the populations changes because individuals are born or die. We put

W(n+1←n)= b(n) birth rateW(n-1 ←n)=d(n) death rateW(n’ ←n)=0 otherwise

The master equation then becomes∂P(n,t)/ ∂t=d(n+1)P(n+1,t)+b(n-1)P(n-1,t)-[d(n)+b(n)]P(n,t)To obtain the steady state distribution π(n) put the r.h.s. to zero. We must have d(0)=π(-1)=0, allowing us to solve the problem by recursion d(1) π(1)=b(0)π(0)

d(n) π(n)=b(n-1)π(n-1)

We can determine π(0) by normalizing the probabilities and thus have an exact solution!

Example: Spruce budworm modelThese insects are a serious pest killing and damaging coniferous trees in many parts of Canada and USA.

They have a 12 month life cycle eggs → larvae, which hibernate in the winter → wake up in spring → feed on (preferably fresh) needles → pupate to become moths → mate → produce eggs . They damage trees by eating the needles and are in turn predated on by birds.

Ludwig and coworkers [12] presented a very simplified model in which the number of insects is represented by a single number N.The birth rate is

b(N)=βN+∆The immigration ∆ assumed to be very small. The death rate is

d(N)= (β-r₀)N -r₀N²/K+BN²/(A²+N²)r₀ is an effective growth rate, K is a logistic carrying capacity, proportional to the number of trees and assumed constant.B is proportional to the number of birds (assumed constant).If the number of insects is small, the birds will feed on something else, while A represents a saturation effect. We first consider a rate equation approach similar to what we did for the Brusselerator

dN/dt= r₀N(1-N/K)-BN²/(A²+N²)+ ∆

A very similar situation arises in the case of the gas-liquid transition as described by the van der Waals equation. When there are two stable solutions one will have a higher probability than the other which is said to be metastable.

Returning to the master equation approach: When the system size parameter A is very large we can replace a sum over N by an integral and use some other reference number than N=0 to start the recursion. We find

Near the maxima the distribution will be Gaussian except at the critical point and typical fluctuations proportional to sqrt(A). Intermediate states have low probability for large A and metastable states can live long.State with highest prob. analogue to equilibrium state!

The Fokker –Planck equationWe had for the master equation of a continuous system

∂P(q)/ ∂t=∫ dq’[W(q←q’)P(q’,t)-W(q’←q)P(q,t)]The equation can be rewritten in terms the “jump” r=q-q’ and the jump rate w(q’,r)=W(q ←q’). We find

∂P(q,t)/ ∂t=∫dr w(q,r)P(q-r,t)-P(q,t) ∫dr w(q,-r)We expand the integrand in the first integral in powers of r. The leading term cancels with the second integral and we are left with

The expansion is called the Kramers-Moyal expansion.

If we truncate the Kramers-Moyal expansion after the second term and use the notation

A(q)=a₁(q) , D=½ a₂we obtain the Fokker-Planck equation

It turns out that there is no point in truncating the series at any finite order beyond second. Pawula’s theorem states that truncating the expansion at any finite order beyond second leads to probability distributions that are not positive definite. The full expansion is equivalent to the master equation and we are no further ahead. The basic assumption of the Fokker-Planck equation is that the jumps are short range. It can then also be derived from a system size expansion.

The structure of the Fokker=Planck equation becomes more transparent if we rewrite it as

∂P(q,t)/ ∂t=- ∂J/ ∂qWhere

J=A(q)P(q,t)- ∂/ ∂x [D(q)P(q,t)]is the probability current. The situation is analogous to the customary treatment of macroscopic currents of particles or charges

Here μ is the mobility, f a force, and D the diffusion constant. The Fokker-Planck equation is just a statement of the conservation of probability. The equation also goes under the names of Smoluchowski equation.

Mesoscopic particle in a stationary fluidOrders of magnitude:Particle size: r= nm to micronTime between collisions with molecules in fluid: τ< nsTimes of interest: t =s to days“Infinitesimal “ time: δt>> τOn these time-scales mean velocity: <v>=μfForce= f, mobility= μ Jump moments, The Fokker-Planck equation then describes the motion of a Brownian particle:

If μf = A₀+A₁x is linear and D constant the Fokker Planck equation is said to be linear. The process it describes is a Ornstein-Uhlenbeck process. If D is not constant the diffusion is heterogeneous.

Example: A settling dust particleForce due to gravity: mg, assume D constant. Fokker-Planck equation now

To solve this go to moving framey=x+MgμtP(x,t)= Π(x+Mgμt,t)

If at t=0 the particle was located at y=0 the solution for t>0 is

A random walk of this type is called a Bachelier-Wiener process.The motion is continuous, but the instantaneous velocity nowhere defined. Π is a Gaussian distribution with variance σ²=2Dt, standard deviation σ

The Gaussian distribution

is stable, meaning that the sum of two Gaussians is another Gaussian with σ²=σ₁²+σ₂². This means that we can generate an instance of a Brownian random walk by considering the motion as made up of a series of small incremental steps of length δy and duration δt generated from a Gaussian random distribution of variance

σ²=2Dδt Gaussian random number generators are easily available, for an example see [13].The probability distribution is self-affine

Meanin it is invariant under a rescaling

Reflection at a boundary:The falling dust particle solution does not admit a steady state.Eventually, though it will reach ground. If it then diffuses back into the atmosphere when hitting ground at x=0 we require that the probability current

is zero for x=0. In the steady state J=0 everywhere. This gives us the ordinary d.e.

With normalized solution

The equilibrium probability distribution is

In the case of thermal Brownian motion the mobility and diffusion constant must thus satisfy the Einstein relation

D=μkT

Absorbing boundaries:We wish to find the rate at which a diffusing particle leaves a region. We handle this by putting particle in a ”limbo state *” at boundary, and requiring the probability to be zero at the actual boundary. Consider diffusion on a square with corners at (0,0), (0,L), (L,0), (L,L). The Fokker-Planck equation is

With the condition that P(x,y,t) is zero at the boundary we find

In the long time limit the smallest eigenvalue dominates and the survival probability is

Kramer’s escape rate:Consider a Brownian particle moving in an energy well as in the figure. Initially the particle is located near b in the figure. We wish to calculate the rate at which it reaches an absorbing boundary at e. The force on the particle is f(x)=-dU/dx and we assume μ=DkT. We also assume that near b there is approximate thermal equilibrum so that the probability that the particle is near b is

The probability current is given by

We now assume that near b

While near d

We finally assume that the probability current J(x,t) is independent of x.Multiplying both sides of the bottom expression on the previous slideWith exp([U(x)-U(b)]/kT) and integrating from b to e, noting that since e is absorbing P(e,t)=0, we get

Most of the contributions from this integrals comes from the rgion near d, while in the formula for the probability p that the particle is in the well comes from the region near b. We find for the escape rate r

Our next approximation is to extend the integrations to . Performing the Gaussian integrations we get the Kramers escape rate

The Kramers’ escape rate formula only depends on the behavior near d and b, what happens at c is of no consequence!Landauer’s blow-torch paradox [14]:If the path is heated near c, the diffusion is speeded up there, but this is not reflected in the formula!Probability current with heterogeneous diffusion: J=μf(x)P(x,t)- ∂/∂x [D(x)P(x,t)]=

=[μf(x)P(x,t)- d/dx D(x)]P(x,t)-D(x)∂P(x,t)/∂xThe x-dependence of D leads to an effective force, causing the effective barrier height to change. Heterogeneous diffusion is responsible for thermophoresis and the Ludwig-Soret effect and the difference between electromotive and electric field in the thermoelectric effects. We also assume Gaussian noise, which may not apply in non-thermal cases.

Over the barrier hopping:A particle is moving in a periodic potential with period a and is also subject to an external force F. Most of the time it will be trapped by barriers V aF/2.Near one of the maxima at x=d

near the minimum at b=d+a/2

F

V+Fa/2

Faa

The net jump rate to the right and left is then

The mean velocity towards the left is

Double well with periodic forcing

Bachelier developed the theory of Brownian motion not by considering motion of mesoscopic particles, but from price fluctuations on the market. In a large market there is again a clear separation between the time between individual transaction and the time horizon of agents.

References and further readingThe material on equilibrium statistical mechanics is standard and can be found in many textbooks. A possible choice for a graduate level text is:[1] M. Plischke and B. Bergersen: Equilibrium Statistical Physics, 3rd edition, World Scientific 2005.A more elementary text is[2] D. V. Schroeder: An introduction to Thermal Physics, Addison Wesley 2000.Both[1] and [2] covers the information theoretic approach.Alben’s original paper is[3] R. Alben. Amer. J. Phys. 40, 3 (1972)An excellent reference for condensed matter steady states can be found in[4] N.W. Ashcroft and N.D. Mermin: Solid state Physics, Holt Rinehart and Winston 1976.The authors argue in[5] A. Dragulescu and V.M. Yakovenko, European Physical J., B17, 723 (2000) and other articles that “equilibrium economics” can describe the income distribution in countries like USA, but this ignores the fact that income inequality changed a great deal during the 20th century.

The pictures of the Belousov Zhabotinsky reaction were taken from the Wikipedia article on the subject.The Brusselerator is described in more detail in[6] R.H. Enns and G. C. McGuire: Nonlinear physics with Maple for Scientists and Engineers, 2nd edition Birkhauser 2000. See also the Wikipedia article on the subjectThe picture of the Australian sheep blow-fly Lucilla cuprina was downloaded from the Wikipedia. The standard reference to the Nicholson equation is[7] W.S.C. Gurney, S.P. Blythe and R.M. Nisbet, Nicholson's blowflies revisited, Nature 287, 17-21 (1980).A readable introduction to delay differential equations is [8] Thomas Erneux, Applied differential equations, Springer 2009Hutchinson’s original paper is.[9] G.E. Hutchinson, Circular causal systems in ecology, Ann. N.Y. Acad. Sci. 50 221 (1948).The dimensional analysis approach to the physics of fluids is discussed in my lecture notes Fluids that can be downloaded fromhttp://www.physics.ubc.ca/~birger/314.pdf where you also can find further references.

The discussion of steady state probability distribution is taken from Section 9.2.1 in ref. [1], while the material which follows on stochastic processes borrows heavily from Chapter 8 of the same book. A classic reference on stochastic processes is[10] N.G. Van Kampen, Stochastic processes in Physics and Chemistry, North Holland (1981).The spruce budworm model originated with[11] D. Ludwig, D.D. Jones and C.S. Holling, J. Animal Ecology 47, 315.The rate equation version of the model is described in Section 4.5 of [1], while the stochastic version is in Section 8.2. The pictures of the insect and the description of the life cycle comes from the Wikipedia article on the subject.The classic text on the Fokker-Planck equation is[12] H. Risken, The Fokker-Planck equation, 2nd ed. Springer 1989, The book contains a proof of Pawula’s theorem. In [10] and [1] the Fokker-Planck equation is derived from a system size expansion, with the moment expansion presented as an after-thought.[13]W. H. Press, S.A. Teukolsky, W.T. Vetterling, B.P. Flannery. Numerical recipies in C, 2nd edition, Cambridge Univeristy Press 1992.[14] R. Landauer, J. of Stat. Phys. 53, 233 (1988)