Lect21 - web.ics.purdue.edu

11/3/21

1

Lecture 21 PHYS 416 Thursday November 4 Fall 2021

1. HW problems 6.2 Two states, 6.11 Barrier crossing2. Minimize free energy in equilibrium3. HW problem 6.4 Molecular motors4. Sethna 6.7 Free energy density for ideal gas5. Quiz 7

1

Copyright Oxford University Press 2006 v2.0 --

Exercises 159

at time t. Their probability density (per unitvertical velocity) is ρ(vz, z, t)A∆z. After a time∆t, this slab will have accelerated to vz − g∆t,and risen a distance h+ vz∆t, so

ρ(vz, z, t) = ρ(vz−g∆t, z+vz∆t, t+∆t). (6.68)

(d) Using the fact that ρ(vz, z, t) is time in-dependent in equilibrium, write a relation be-tween ∂ρ/∂vz and ∂ρ/∂z. Using your result frompart (c), derive the equilibrium velocity distribu-tion for the ideal gas.Feynman then argues that interactions and col-lisions will not change the velocity distribution.

(e) Simulate an interacting gas in a box with re-flecting walls, under the influence of gravity. Usea temperature and a density for which there is alayer of liquid at the bottom (just like water in aglass). Plot the height distribution (which shouldshow clear interaction effects) and the momen-tum distribution. Use the latter to determine thetemperature; do the interactions indeed not dis-tort the momentum distribution?What about the atoms which evaporate from thefluid? Only the very most energetic atoms canleave the liquid to become gas molecules. Theymust, however, use up every bit of their extraenergy (on average) to depart; their kinetic en-ergy distribution is precisely the same as that ofthe liquid.48

Feynman concludes his chapter by pointing outthat the predictions resulting from the classicalBoltzmann distribution, although they describemany properties well, do not match experimentson the specific heats of gases, foreshadowing theneed for quantum mechanics.49

(6.2) Two-state system. ©iConsider the statistical mechanics of a tiny ob-ject with only two discrete states:50 one of en-ergy E1 and the other of higher energy E2 > E1.(a) Boltzmann probability ratio. Find the ra-tio of the equilibrium probabilities ρ2/ρ1 to findour system in the two states, when weakly cou-pled to a heat bath of temperature T . What isthe limiting probability as T → ∞? As T →

0? Related formula: Boltzmann probability= Z(T ) exp(−E/kT ) ∝ exp(−E/kT ).(b) Probabilities and averages. Use the normal-ization of the probability distribution (the systemmust be in one or the other state) to find ρ1 andρ2 separately. (That is, solve for Z(T ) in the ‘re-lated formula’ for part (a).) What is the averagevalue of the energy E?

(6.3) Negative temperature. ©3A system of N atoms each can be in the groundstate or in an excited state. For convenience,we set the zero of energy exactly in between, sothe energies of the two states of an atom are±ε/2. The atoms are isolated from the outsideworld. There are only weak couplings betweenthe atoms, sufficient to bring them into internalequilibrium but without other effects.(a) Microcanonical entropy. If the net energy isE (corresponding to a number of excited atomsm = E/ε + N/2), what is the microcanon-ical entropy Smicro(E) of our system? Sim-plify your expression using Stirling’s formula,log n! ∼ n log n− n.(b) Negative temperature. Find the temperature,using your simplified expression from part (a).What happens to the temperature when E > 0?Having the energy E > 0 is a kind of popula-tion inversion. Population inversion is the driv-ing mechanism for lasers.For many quantities, the thermodynamic deriva-tives have natural interpretations when viewedas sums over states. It is easiest to see this insmall systems.(c) Canonical ensemble. (i) Take one of ouratoms and couple it to a heat bath of temperaturekBT = 1/β. Write explicit formulæ for Zcanon,Ecanon, and Scanon in the canonical ensemble, asa trace (or sum) over the two states of the atom.(E should be the energy of each state multipliedby the probability ρn of that state, and S shouldbe the trace of −kBρn log ρn.) (ii) Compare theresults with what you get by using the thermo-dynamic relations. Using Z from the trace overstates, calculate the Helmholtz free energy A, Sas a derivative of A, and E from A = E − TS.

48Ignoring quantum mechanics.49Quantum mechanics is important for the internal vibrations within molecules, which absorb energy as the gas is heated.Quantum effects are not so important for the pressure and other properties of gases, which are dominated by the molecularcenter-of-mass motions.50Visualize this as a tiny biased coin, which can be in the ‘heads’ or ‘tails’ state but has no other internal vibrations or centerof mass degrees of freedom. Many systems are well described by large numbers of these two-state systems: some paramagnets,carbon monoxide on surfaces, glasses at low temperatures, . . .


Exercises 159

at time t. Their probability density (per unitvertical velocity) is ρ(vz, z, t)A∆z. After a time∆t, this slab will have accelerated to vz − g∆t,and risen a distance h+ vz∆t, so

ρ(vz, z, t) = ρ(vz−g∆t, z+vz∆t, t+∆t). (6.68)

(d) Using the fact that ρ(vz, z, t) is time in-dependent in equilibrium, write a relation be-tween ∂ρ/∂vz and ∂ρ/∂z. Using your result frompart (c), derive the equilibrium velocity distribu-tion for the ideal gas.Feynman then argues that interactions and col-lisions will not change the velocity distribution.

(e) Simulate an interacting gas in a box with re-flecting walls, under the influence of gravity. Usea temperature and a density for which there is alayer of liquid at the bottom (just like water in aglass). Plot the height distribution (which shouldshow clear interaction effects) and the momen-tum distribution. Use the latter to determine thetemperature; do the interactions indeed not dis-tort the momentum distribution?What about the atoms which evaporate from thefluid? Only the very most energetic atoms canleave the liquid to become gas molecules. Theymust, however, use up every bit of their extraenergy (on average) to depart; their kinetic en-ergy distribution is precisely the same as that ofthe liquid.48

Feynman concludes his chapter by pointing outthat the predictions resulting from the classicalBoltzmann distribution, although they describemany properties well, do not match experimentson the specific heats of gases, foreshadowing theneed for quantum mechanics.49

(6.2) Two-state system. ©iConsider the statistical mechanics of a tiny ob-ject with only two discrete states:50 one of en-ergy E1 and the other of higher energy E2 > E1.(a) Boltzmann probability ratio. Find the ra-tio of the equilibrium probabilities ρ2/ρ1 to findour system in the two states, when weakly cou-pled to a heat bath of temperature T . What isthe limiting probability as T → ∞? As T →

0? Related formula: Boltzmann probability= Z(T ) exp(−E/kT ) ∝ exp(−E/kT ).(b) Probabilities and averages. Use the normal-ization of the probability distribution (the systemmust be in one or the other state) to find ρ1 andρ2 separately. (That is, solve for Z(T ) in the ‘re-lated formula’ for part (a).) What is the averagevalue of the energy E?

(6.3) Negative temperature. ©3A system of N atoms each can be in the groundstate or in an excited state. For convenience,we set the zero of energy exactly in between, sothe energies of the two states of an atom are±ε/2. The atoms are isolated from the outsideworld. There are only weak couplings betweenthe atoms, sufficient to bring them into internalequilibrium but without other effects.(a) Microcanonical entropy. If the net energy isE (corresponding to a number of excited atomsm = E/ε + N/2), what is the microcanon-ical entropy Smicro(E) of our system? Sim-plify your expression using Stirling’s formula,log n! ∼ n log n− n.(b) Negative temperature. Find the temperature,using your simplified expression from part (a).What happens to the temperature when E > 0?Having the energy E > 0 is a kind of popula-tion inversion. Population inversion is the driv-ing mechanism for lasers.For many quantities, the thermodynamic deriva-tives have natural interpretations when viewedas sums over states. It is easiest to see this insmall systems.(c) Canonical ensemble. (i) Take one of ouratoms and couple it to a heat bath of temperaturekBT = 1/β. Write explicit formulæ for Zcanon,Ecanon, and Scanon in the canonical ensemble, asa trace (or sum) over the two states of the atom.(E should be the energy of each state multipliedby the probability ρn of that state, and S shouldbe the trace of −kBρn log ρn.) (ii) Compare theresults with what you get by using the thermo-dynamic relations. Using Z from the trace overstates, calculate the Helmholtz free energy A, Sas a derivative of A, and E from A = E − TS.

48Ignoring quantum mechanics.49Quantum mechanics is important for the internal vibrations within molecules, which absorb energy as the gas is heated.Quantum effects are not so important for the pressure and other properties of gases, which are dominated by the molecularcenter-of-mass motions.50Visualize this as a tiny biased coin, which can be in the ‘heads’ or ‘tails’ state but has no other internal vibrations or centerof mass degrees of freedom. Many systems are well described by large numbers of these two-state systems: some paramagnets,carbon monoxide on surfaces, glasses at low temperatures, . . .

2

Γ = Γ0 exp(−B / kBT )

The previous discussion was all for the system in equilibrium. When it is out of equilibrium, the individual forward and backward reaction rates are involved. Consider an initial situation with lots of reactants. What determines the rate at which the reaction proceeds?

This is described in terms of a barrier crossing process. The Boltzmann-like factor is traditionally called the Arrhenius Law:

3


164 Free energies

(6.11) Barrier crossing. (Chemistry) ©2In this exercise, we will derive the Arrhenius law(eqn 6.60)

Γ = Γ0 exp(−E/kBT ), (6.79)

giving the rate at which chemical reactions crossenergy barriers. The important exponential de-pendence on the barrier height E is the relativeBoltzmann probability that a particle is near thetop of the barrier (and hence able to escape).Here we will do a relatively careful job of calcu-lating the prefactor Γ0.Consider a system having an energy U(X), withan energy well with a local minimum at X = X0

having energy U(X0) = 0. Assume there isan energy barrier of height U(XB) = B acrosswhich particles can escape.59 Let the tempera-ture of the system be much smaller than B/kB .To do our calculation, we will make some ap-proximations. (1) We assume that the atomsescaping across the barrier to the right do notscatter back into the well. (2) We assume thatthe atoms deep inside the well are in local equi-librium. (3) We assume that the particles cross-ing to the right across the barrier are given bythe equilibrium distribution inside the well.(a) Let the probability density that a particlehas position X be ρ(X). What is the ratio ofprobability densities ρ(XB)/ρ(X0) if the parti-cles near the top of the barrier are assumedto be in equilibrium with those deep inside thewell? Related formula: Boltzmann distributionρ ∝ exp(−E/kBT ).

Fig. 6.15Well probability distribution. The ap-proximate probability distribution for the atoms stilltrapped inside the well.

If the barrier height B $ kBT , then most of theparticles in the well stay near the bottom of thewell. Often, the potential near the bottom is ac-curately described by a quadratic approximationU(X) ≈ 1/2Mω2(X −X0)

2, where M is the massof our system and ω is the frequency of smalloscillations in the well.(b) In this approximation, what is the probabilitydensity ρ(X) near the bottom of the well? (SeeFig. 6.15.) What is ρ(X0), the probability den-sity of being precisely at the bottom of the well?Related formula: Gaussian probability distribu-tion (1/

√2πσ2) exp(−x2/2σ2).

Knowing the answers from (a) and (b), we knowthe probability density ρ(XB) at the top of thebarrier.60 We also need to know the proba-bility that particles near the top of the bar-rier have velocity V , because the faster-movingparts of the distribution of velocities contributemore to the flux of probability over the bar-rier (see Fig. 6.16). As usual, because the to-tal energy is the sum of the kinetic energy andpotential energy, the total Boltzmann proba-bility factors; in equilibrium the particles willalways have a velocity probability distributionρ(V ) = 1/

√2πkBT/M exp(−1/2MV 2/kBT ).

Δv t

Fig. 6.16 Crossing the barrier. The range of po-sitions for which atoms moving to the right with ve-locity v will cross the barrier top in time ∆t.

(c) First give a formula for the decay rate Γ(the probability per unit time that a given par-ticle crosses the barrier towards the right), foran unknown probability density ρ(XB)ρ(V ) as anintegral over the velocity V . Then, using your

59This potential could describe a chemical reaction, with X being a reaction coordinate. At the other extreme, it could describethe escape of gas from a moon of Jupiter, with X being the distance from the moon in Jupiter’s direction.60Or rather, we have calculated ρ(XB) in equilibrium, half of which (the right movers) we assume will also be crossing thebarrier in the non-equilibrium reaction.


164 Free energies


Γ = Γ0 exp(−E/kBT ), (6.79)






√2πσ2) exp(−x2/2σ2).



Δv t





164 Free energies


Γ = Γ0 exp(−E/kBT ), (6.79)






√2πσ2) exp(−x2/2σ2).



Δv t





164 Free energies


Γ = Γ0 exp(−E/kBT ), (6.79)






√2πσ2) exp(−x2/2σ2).



Δv t





Exercises 165

formulæ from parts (a) and (b), give your esti-mate of the decay rate for our system. Relatedformula:

∫∞0

x exp(−x2/2σ2) dx = σ2.How could we go beyond this one-dimensionalcalculation? In the olden days, Kramers stud-ied other one-dimensional models, changing theways in which the system was coupled to the ex-ternal bath. On the computer, one can avoid aseparate heat bath and directly work with thefull multidimensional configuration space, lead-ing to transition-state theory. The transition-state theory formula is very similar to the oneyou derived in part (c), except that the prefac-tor involves the product of all the frequenciesat the bottom of the well and all the positivefrequencies at the saddlepoint at the top of thebarrier (see [72]). Other generalizations arisewhen crossing multiple barriers [81] or in non-equilibrium systems [109].

(6.12) Michaelis–Menten and Hill. (Biology, Com-putation) ©3Biological reaction rates are often saturable; thecell needs to respond sensitively to the introduc-tion of a new chemical S, but the response shouldnot keep growing indefinitely as the new chem-ical concentration [S] grows.61 Other biologi-cal reactions act as switches; a switch not onlysaturates, but its rate or state changes sharplyfrom one value to another as the concentrationof a chemical S is varied. These reactions givetangible examples of how one develops effectivedynamical theories by removing degrees of free-dom; here, instead of coarse-graining some largestatistical mechanics system, we remove a singleenzyme E from the equations to get an effectivereaction rate.The rate of a chemical reaction,

NS +B → C, (6.80)

where N substrate molecules S combine with a Bmolecule to make a C molecule, will occur witha reaction rate given by the law of mass-action:

d[C]dt

= k[S]N [B]. (6.81)

Saturation and the Michaelis–Menten equation.Saturation is not seen in ordinary chemical reac-tion kinetics. Notice that the reaction rate goesas the Nth power of the concentration [S]; far

from saturating, the reaction rate grows linearlyor faster with concentration.The archetypal example of saturation in biolog-ical systems is the Michaelis–Menten reactionform. A reaction of this form converting a chem-ical S (the substrate) into P (the product) has arate given by the formula

d[P ]dt

=Vmax[S]KM + [S]

, (6.82)

where KM is called the Michaelis constant(Fig. 6.17). This reaction at small concentra-tions acts like an ordinary chemical reaction withN = 1 and k = Vmax/KM , but the rate saturatesat Vmax as [S]→∞. The Michaelis constantKM

is the concentration [S] at which the rate is equalto half of its saturation rate (Fig. 6.17).

0 KM, KHSubstrate concentration [S]

0

Vmax

Rate

d[P

]/dt

Michaelis-MentenHill, n = 4

Fig. 6.17 Michaelis–Menten and Hill equationforms.

We can derive the Michaelis–Menten form by hy-pothesizing the existence of a catalyst or enzymeE, which is in short supply. The enzyme is pre-sumed to be partly free and available for bind-ing (concentration [E]) and partly bound to thesubstrate (concentration62 [E : S]), helping it toturn into the product. The total concentration[E] + [E : S] = Etot is fixed. The reactions areas follows:

E + S !k−1k1

E : Skcat→ E + P. (6.83)

We must then assume that the supply of sub-strate is large, so its concentration changesslowly with time. We can then assume that theconcentration [E : S] is in steady state, and re-move it as a degree of freedom.

61[S] is the concentration of S (number per unit volume). S stands for substrate.62The colon denotes the bound state of the two, called a dimer.

4

Example of minimizing free energy at equilibrium, taken from:

https://scholar.harvard.edu/files/schwartz/files/8-freeenergy.pdfMatthew Schwartz 2019

(On blackboard; will post the notes.)

5


160 Free energies

Do the thermodynamically derived formulæ youget agree with the statistical traces?(d) What happens to E in the canonical ensem-ble as T → ∞? Can you get into the negative-temperature regime discussed in part (b)?

-20ε -10ε 0ε 10ε 20εEnergy E = ε(-N/2 + m)

0kB

40kB

Entro

py S

0

0.1

Prob

abili

ty ρ

(E)

Microcanonical Smicro(E)Canonical Sc(T(E))Probability ρ(E)

Fig. 6.10 Negative temperature. Entropies andenergy fluctuations for this problem with N = 50.The canonical probability distribution for the energyis for 〈E〉 = −10ε, and kBT = 1.207ε. You may wishto check some of your answers against this plot.

(e) Canonical–microcanonical correspondence.Find the entropy in the canonical distributionfor N of our atoms coupled to the outside world,from your answer to part (c). Explain the valueof S(T = ∞) − S(T = 0) by counting states.Using the approximate form of the entropy frompart (a) and the temperature from part (b), showthat the canonical and microcanonical entropiesagree, Smicro(E) = Scanon(T (E)). (Perhaps use-ful: arctanh(x) = 1/2 log ((1 + x)/(1− x)) .) No-tice that the two are not equal in Fig. 6.10; theform of Stirling’s formula we used in part (a) isnot very accurate for N = 50. Explain in wordswhy the microcanonical entropy is smaller thanthe canonical entropy.(f) Fluctuations. Calculate the root-mean-squareenergy fluctuations in our system in the canoni-cal ensemble. Evaluate it at T (E) from part (b).For large N , are the fluctuations in E small com-pared to E?

(6.4) Molecular motors and free energies.51 (Bi-ology) ©iFigure 6.11 shows the set-up of an experimenton the molecular motor RNA polymerase thattranscribes DNA into RNA.52 Choosing a goodensemble for this system is a bit involved. Itis under two constant forces (F and pressure),and involves complicated chemistry and biology.Nonetheless, you know some things based on fun-damental principles. Let us consider the opticaltrap and the distant fluid as being part of theexternal environment, and define the ‘system’ asthe local region of DNA, the RNA, motor, andthe fluid and local molecules in a region immedi-ately enclosing the region, as shown in Fig. 6.11.

F

DNA

RNA

laserFocused

Bead

x

System

Fig. 6.11 RNA polymerase molecular motorattached to a glass slide is pulling along a DNAmolecule (transcribing it into RNA). The oppositeend of the DNA molecule is attached to a bead whichis being pulled by an optical trap with a constant ex-ternal force F . Let the distance from the motor tothe bead be x; thus the motor is trying to move todecrease x and the force is trying to increase x.

(a) Without knowing anything further about thechemistry or biology in the system, which of thefollowing must be true on average, in all cases?(T) (F) The total entropy of the Universe (thesystem, bead, trap, laser beam, . . . ) must in-crease or stay unchanged with time.(T) (F) The entropy Ss of the system cannot de-crease with time.

51This exercise was developed with the help of Michelle Wang.52RNA, ribonucleic acid, is a long polymer like DNA, with many functions in living cells. It has four monomer units (A, U, C,and G: Adenine, Uracil, Cytosine, and Guanine); DNA has T (Thymine) instead of Uracil. Transcription just copies the DNAsequence letter for letter into RNA, except for this substitution.


160 Free energies



0kB

40kB

Entro

py S

0

0.1

Prob

abili

ty ρ

(E)





F

DNA

RNA

laserFocused

Bead

x

System





160 Free energies



0kB

40kB

Entro

py S

0

0.1

Prob

abili

ty ρ

(E)





F

DNA

RNA

laserFocused

Bead

x

System





Exercises 161

(T) (F) The total energy ET of the Universemust decrease with time.(T) (F) The energy Es of the system cannot in-crease with time.(T) (F) Gs−Fx = Es−TSs+PVs−Fx cannotincrease with time, where Gs is the Gibbs freeenergy of the system.Related formula: G = E − TS + PV .(Hint: Precisely two of the answers are correct.)The sequence of monomers on the RNA canencode information for building proteins, andcan also cause the RNA to fold into shapesthat are important to its function. One of themost important such structures is the hairpin(Fig. 6.12). Experimentalists study the strengthof these hairpins by pulling on them (also withlaser tweezers). Under a sufficiently large force,the hairpin will unzip. Near the threshold forunzipping, the RNA is found to jump betweenthe zipped and unzipped states, giving telegraphnoise53 (Fig. 6.13). Just as the current in a tele-graph signal is either on or off, these systems arebistable and make transitions from one state tothe other; they are a two-state system.

Fig. 6.12 Hairpins in RNA. (Reprinted with per-mission from Liphardt et al. [105], c©2001 AAAS.) Alength of RNA attaches to an inverted, complemen-tary strand immediately following, forming a hairpinfold.

The two RNA configurations presumably havedifferent energies (Ez, Eu), entropies (Sz, Su),and volumes (Vz, Vu) for the local region aroundthe zipped and unzipped states, respectively.The environment is at temperature T and pres-

sure P . Let L = Lu − Lz be the extra lengthof RNA in the unzipped state. Let ρz be thefraction of the time our molecule is zipped at agiven external force F , and ρu = 1 − ρz be theunzipped fraction of time.(b) Of the following statements, which are true,assuming that the pulled RNA is in equilibrium?(T) (F) ρz/ρu = exp((Stot

z − Stotu )/kB), where

Stotz and Stot

u are the total entropy of the Uni-verse when the RNA is in the zipped and un-zipped states, respectively.(T) (F) ρz/ρu = exp(−(Ez − Eu)/kBT ).(T) (F) ρz/ρu = exp(−(Gz − Gu)/kBT ), whereGz = Ez−TSz+PVz and Gu = Eu−TSu+PVu

are the Gibbs energies in the two states.(T) (F) ρz/ρu = exp(−(Gz − Gu + FL)/kBT ),where L is the extra length of the unzipped RNAand F is the applied force.

Fig. 6.13 Telegraph noise in RNA unzip-ping. (Reprinted with permission from Liphardt etal. [105], c©2001 AAAS.) As the force increases, thefraction of time spent in the zipped state decreases.

(6.5) Laplace.54 (Thermodynamics) ©iThe Laplace transform of a function f(t) is afunction of x:

L{f}(x) =∫ ∞

0

f(t)e−xt dt. (6.69)

53Like a telegraph key going on and off at different intervals to send dots and dashes, a system showing telegraph noise jumpsbetween two states at random intervals.54Pierre-Simon Laplace (1749–1827). See [115, section 4.3].


Exercises 161






Stotz and Stot





L{f}(x) =∫ ∞

0

f(t)e−xt dt. (6.69)



Exercises 161






Stotz and Stot





L{f}(x) =∫ ∞

0

f(t)e−xt dt. (6.69)


6

https://scholar.harvard.edu/files/schwartz/files/8-freeenergy.pdf

11/3/21

2

A(N ,V ,T ) = NkBT log(ρλ3)−1⎡⎣ ⎤⎦

F ideal (ρ(x j ),T ) =A(nj ,ΔV ,T )

ΔV= ρ(x j )kBT log(ρ(x j ) λ

3)−1⎡⎣ ⎤⎦


6.7 Free energy density for the ideal gas 155

scopic phenomena like nucleation rates (Section 11.3).37 37There are basically three ways inwhich slow processes arise in physics.(1) Large systems can respond slowlyto external changes because communi-cation from one end of the system to theother is sluggish; examples are the slowdecay at long wavelengths in the diffu-sion equation (Section 2.2) and Gold-stone modes (Section 9.3). (2) Sys-tems like radioactive nuclei can respondslowly—decaying with lifetimes of bil-lions of years—because of the slow rateof quantum tunneling through barriers.(3) Systems can be slow because theymust thermally activate over barriers(with the Arrhenius rate of eqn 6.60).

6.7 Free energy density for the ideal gas

We began our text (Section 2.2) studying the diffusion equation. How isit connected with free energies and ensembles? Broadly speaking, inho-mogeneous systems out of equilibrium can also be described by statisticalmechanics, if the gradients in space and time are small enough that thesystem is close to a local equilibrium. We can then represent the localstate of the system by order parameter fields, one field for each property(density, temperature, magnetization) needed to characterize the stateof a uniform, macroscopic body. We can describe a spatially-varying,inhomogeneous system that is nearly in equilibrium using a free energydensity, typically depending on the order parameter fields and their gra-dients. The free energy of the inhomogeneous system will be given byintegrating the free energy density.38 38Properly, given an order parameter

field s(x) there is a functional F{s}which gives the system free energy. (Afunctional is a mapping from a space offunctions into the real numbers.) Writ-ing this functional as an integral overa free energy density (as we do) canbe subtle, not only due to long-rangefields, but also due to total divergenceterms (Exercise 9.3).

We will be discussing order parameter fields and free energy densitiesfor a wide variety of complex systems in Chapter 9. There we will usesymmetries and gradient expansions to derive the form of the free energydensity, because it will often be too complex to compute directly. In thissection, we will directly derive the free energy density for an inhomoge-neous ideal gas, to give a tangible example of the general case.39

39We will also use the free energy den-sity of the ideal gas when we study cor-relation functions in Section 10.3.

Fig. 6.8 Density fluctuations inspace. If nj is the number of points ina box of size ∆V at position xj , thenρ(xj ) = nj/∆V .

Remember that the Helmholtz free energy of an ideal gas is nicelywritten (eqn 6.24) in terms of the density ρ = N/V and the thermaldeBroglie wavelength λ:

A(N,V, T ) = NkBT[log(ρλ3)− 1

]. (6.61)

Hence the free energy density for nj = ρ(xj)∆V atoms in a small volume∆V is

F ideal(ρ(xj), T ) =A(nj ,∆V, T )

∆V= ρ(xj)kBT

[log(ρ(xj)λ

3)− 1].

(6.62)The probability for a given particle density ρ(x) is

P{ρ} = e−β∫dV F ideal(ρ(x))/Z. (6.63)

As usual, the free energy F{ρ} =∫F(ρ(x)) dx acts just like the energy

in the Boltzmann distribution. We have integrated out the microscopicdegrees of freedom (positions and velocities of the individual particles)and replaced them with a coarse-grained field ρ(x). The free energydensity of eqn 6.62 can be used to determine any equilibrium propertyof the system that can be written in terms of the density ρ(x).40

40In Chapter 10, for example, we willuse it to calculate correlation functions〈ρ(x)ρ(x′)〉, and will discuss their rela-tionship with susceptibilities and dissi-pation.

The free energy density also provides a framework for discussing theevolution laws for non-uniform densities. A system prepared with somenon-uniform density will evolve in time ρ(x, t). If in each small volume∆V the system is close to equilibrium, then one may expect that itstime evolution can be described by equilibrium statistical mechanics


156 Free energies

even though it is not globally in equilibrium. A non-uniform density willhave a force which pushes it towards uniformity; the total free energywill decrease when particles flow from regions of high particle density tolow density. We can use our free energy density to calculate this force,and then use the force to derive laws (depending on the system) for thetime evolution.

Fig. 6.9 Coarse-grained density inspace. The density field ρ(x) repre-sents all configurations of points Q con-sistent with the average. Its free energydensity F ideal(ρ(x)) contains contribu-tions from all microstates (P,Q) withthe correct number of particles per box.The probability of finding this particu-lar density is proportional to the inte-gral of all ways Q that the particles inFig. 6.8 can be arranged to give thisdensity.

The chemical potential for a uniform system is

µ =∂A

∂N=∂A/V

∂N/V=∂F∂ρ

,

i.e., the change in free energy for a change in the average density ρ. Fora non-uniform system, the local chemical potential at a point x is

µ(x) =δ

δρ(x)

∫F(y)dy (6.64)

the variational derivative41 of the free energy density with respect to

41That is, the derivative in functionspace, giving the linear approximationto∫(F{ρ+ δρ}−F{ρ}) (see note 7 on

p. 280). We will seldom touch uponvariational derivatives in this text.

ρ(x). Because our ideal gas free energy has no terms involving gradientsof ρ, the variational derivative δ

∫F/δρ equals the partial derivative

∂F/∂ρ:

µ(x) =δ∫F ideal

δρ=

∂

∂ρ

(ρkBT

[log(ρλ3)− 1

])

= kBT[log(ρλ3)− 1

]+ ρkBT/ρ

= kBT log(ρλ3). (6.65)

The chemical potential is like a number pressure for particles: a particlecan lower the free energy by moving from regions of high chemical po-tential to low chemical potential. The gradient of the chemical potential−∂µ/∂x is thus a pressure gradient, effectively the statistical mechanicalforce on a particle.How will the particle density ρ evolve in response to this force µ(x)?

This depends upon the problem. If our density were the density ofthe entire gas, the atoms would accelerate under the force—leading tosound waves.42 There momentum is conserved as well as particle density.

42In that case, we would need to addthe local velocity field into our descrip-tion of the local environment. If our particles could be created and destroyed, the density evolution

would include a term ∂ρ/∂t = −ηµ not involving a current. In systemsthat conserve (or nearly conserve) energy, the evolution will depend onHamilton’s equations of motion for the free energy density; in magnets,the magnetization responds to an external force by precessing; in super-fluids, gradients in the chemical potential are associated with windingand unwinding the phase of the order parameter field (vortex motion). . .Let us focus on the case of a small amount of perfume in a large body

of still air. Here particle density is locally conserved, but momentumis strongly damped (since the perfume particles can scatter off of theair molecules). The velocity of our particles will be proportional to theeffective force on them v = −γ(∂µ/∂x), with γ the mobility.43 Hence

43This is linear response. Systemsthat are nearly in equilibrium typi-cally have currents proportional to gra-dients in their properties. Examplesinclude Ohm’s law where the electri-cal current is proportional to the gra-dient of the electromagnetic potentialI = V/R = (1/R)(dφ/dx), thermalconductivity where the heat flow is pro-portional to the gradient in temper-ature J = κ∇T , and viscous fluids,where the shear rate is proportional tothe shear stress. We will study linearresponse with more rigor in Chapter 10.

6.7 Free energy density for the ideal gasSpatially inhomogeneous systems can be described in terms of their free energy density (involves coarse-graining).

Recall the Helmholtz free energy of the ideal gas:

Number of atoms in a small volume:

Hence the free energy density is:

nj = ρ(x j )ΔV

7












]. (6.61)



∆V= ρ(xj)kBT

[log(ρ(xj)λ

3)− 1].








156 Free energies




µ =∂A

∂N=∂A/V

∂N/V=∂F∂ρ

,


µ(x) =δ

δρ(x)

∫F(y)dy (6.64)






∂F/∂ρ:

µ(x) =δ∫F ideal

δρ=

∂

∂ρ

(ρkBT

[log(ρλ3)− 1

])


]+ ρkBT/ρ

= kBT log(ρλ3). (6.65)







F ideal (ρ(x j ),T ) =A(nj ,ΔV ,T )

ΔV= ρ(x j )kBT log(ρ(x j ) λ

3)−1⎡⎣ ⎤⎦

P ρ{ } = e−β ∫ dVF ideal (ρ (x )) / Z

The probability of a given particle density is:

What we are doing is replacing the detailed discrete particle picture with more of a continuum density picture. This is what “coarse-graining” means.

F ρ{ } = F ρ(x)( )∫ dxThe free energy acts just like the energy in the usual Boltzmann distribution.

8

µ = ∂A∂N

= ∂A /V∂N /V

= ∂F∂ρ

µ(x) = δFδρ

µ(x) = δF ideal

δρ= ∂∂ρ

(ρkBT log(ρλ3)−1⎡⎣ ⎤⎦)

= kBT log(ρλ3)−1⎡⎣ ⎤⎦ + ρkBT / ρ

= kBT log(ρλ3)⎡⎣ ⎤⎦












]. (6.61)



∆V= ρ(xj)kBT

[log(ρ(xj)λ

3)− 1].








156 Free energies




µ =∂A

∂N=∂A/V

∂N/V=∂F∂ρ

,


µ(x) =δ

δρ(x)

∫F(y)dy (6.64)






∂F/∂ρ:

µ(x) =δ∫F ideal

δρ=

∂

∂ρ

(ρkBT

[log(ρλ3)− 1

])


]+ ρkBT/ρ

= kBT log(ρλ3). (6.65)







The chemical potential for a uniform system is:

For a non-uniform system, the local chemical potential is:

For the ideal gas, the variational derivative is just the partial derivative:

“The chemical potential is like a number pressure for particles: a particle can lower the free energy by moving from regions of high chemical potential to low chemical potential. The gradient of the chemical potential is thus a pressure gradient, effectively a statistical mechanical force on a particle.”

9

J = γρ(x) − ∂µ∂x

⎛⎝⎜

⎞⎠⎟= −γρ(x)

∂kBT log(ρλ3)∂x

= −γρ(x)kBTρ

∂ρ∂x

= −γ kBT∂ρ∂x












]. (6.61)



∆V= ρ(xj)kBT

[log(ρ(xj)λ

3)− 1].








156 Free energies




µ =∂A

∂N=∂A/V

∂N/V=∂F∂ρ

,


µ(x) =δ

δρ(x)

∫F(y)dy (6.64)






∂F/∂ρ:

µ(x) =δ∫F ideal

δρ=

∂

∂ρ

(ρkBT

[log(ρλ3)− 1

])


]+ ρkBT/ρ

= kBT log(ρλ3). (6.65)







For particles in a fluid (like perfume particles in air), the velocity is proportional to the effective force (not F=ma).

v=−γ(∂µ / ∂x)

Hence the current will be: J = ρv

10












]. (6.61)



∆V= ρ(xj)kBT

[log(ρ(xj)λ

3)− 1].








156 Free energies




µ =∂A

∂N=∂A/V

∂N/V=∂F∂ρ

,


µ(x) =δ

δρ(x)

∫F(y)dy (6.64)






∂F/∂ρ:

µ(x) =δ∫F ideal

δρ=

∂

∂ρ

(ρkBT

[log(ρλ3)− 1

])


]+ ρkBT/ρ

= kBT log(ρλ3). (6.65)







∂ρ∂t

= −∇⋅ J = γ kBT∂2ρ∂x2

J = −γ kBT∂ρ∂x

1. We have again derived the diffusion equation, starting with an inhomogeneous free energy density and velocity-dependent force.

2. We have derived the diffusion constant D (the Einstein relation).

3. We claim the chemical potential gradient acts like a force, even though in the ideal gas there is no potential energy.

11

Lect21 - web.ics.purdue.edu

Documents

Transcript of Lect21 - web.ics.purdue.edu