Copyright Oxford University Press 2006 v2.0

10/27/21

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Lecture 19 PHYS 416 Thursday October 28 Fall 2021

1. More Entropic forces: Rubber bands! (Sethna 5.12)2. The nuclear Overhauser effect3. 6.4 What is thermodynamics?4. 6.5 Mechanics: friction and fluctuations

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Now consider Rubber Bands, Sethna 5.12 and Sethna 6.15

Copyright Oxford University Press 2006 v2.0 --

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and final states of the liquid are both in equilib-rium. See [100].)

100 200 300Temperature ( C)

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Fig. 5.18 Glass transition specific heat. Spe-cific heat of B2O3 glass measured while heating andcooling [181]. The glass was cooled from 345 ◦C toroom temperature in 18 hours (diamonds), and thenheated from 35 ◦C → 325 ◦C (crosses). See [98].

The residual entropy of a typical glass is aboutkB per molecular unit. It is a measure of howmany different glassy configurations of atoms thematerial can freeze into.(c) In a molecular dynamics simulation with onehundred indistinguishable atoms, and assumingthat the residual entropy is kB log 2 per atom,what is the probability that two coolings to zeroenergy will arrive at equivalent atomic config-urations (up to permutations)? In a systemwith 1023 molecular units, with residual entropykB log 2 per unit, about how many coolings wouldbe needed for one to duplicate the original con-figuration again at least once, with probability1/2?

(5.12) Rubber band. (Condensed matter) ©iFigure 5.19 shows a one-dimensional model forrubber. Rubber is formed from long polymericmolecules, which undergo random walks in theundeformed material. When we stretch the rub-ber, the molecules respond by rearranging theirrandom walk to elongate in the direction of theexternal stretch. In our model, the moleculeis represented by a set of N links of length d,which with equal energy point either parallel orantiparallel to the previous link. Let the totalchange in position to the right, from the begin-ning of the polymer to the end, be L.As the molecule extent L increases, the entropyof our rubber molecule decreases.

(a) Find an exact formula for the entropy of thissystem in terms of d, N , and L. (Hint: Howmany ways can one divide N links into M right-pointing links and N −M left-pointing links, sothat the total length is L?)

L

d

Fig. 5.19 Rubber band. Simple model of a rubberband with N = 100 segments. The beginning of thepolymer is at the top; the end is at the bottom; thevertical displacements are added for visualization.

The external world, in equilibrium at temper-ature T , exerts a force pulling the end of themolecule to the right. The molecule must exertan equal and opposite entropic force F .(b) Find an expression for the force F exertedby the molecule on the bath in terms of the bathentropy. (Hint: The bath temperature 1/T =∂Sbath/∂E, and force times distance is energy.)Using the fact that the length L must maximizethe entropy of the Universe, write a general ex-pression for F in terms of the internal entropyS of the molecule.(c) Take our model of the molecule from part (a),the general law of part (b), and Stirling’s formulalog(n!) ≈ n log n − n, write the force law F (L)for our molecule for large lengths N . What isthe spring constant K in Hooke’s law F = −KLfor our molecule, for small L?Our model has no internal energy; this force isentirely entropic. Note how magical this is –we never considered the mechanism of how thesegments would generate a force. Statistical me-chanics tells us that the force generated by oursegmented chain is independent of the mecha-nism. The joint angles in the chain may jigglefrom thermal motion, or the constituent poly-mer monomers may execute thermal motion – solong as the configuration space is segment orien-tations and the effective potential energy is zerothe force will be given by our calculation. Forthe same reason, the pressure due to compress-ing an ideal gas is independent of the mechanism.The kinetic energy of particle collisions for realdilute gases gives the same pressure as the com-


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plex water-solvent interactions give for osmoticpressure (Section 5.2.2).(d) If we increase the temperature of our rubberband while it is under tension, will it expand orcontract? Why?In a more realistic model of a rubber band, theentropy consists primarily of our configurationalrandom-walk entropy plus a vibrational entropyof the molecules. If we stretch the rubber bandwithout allowing heat to flow in or out of therubber, the total entropy should stay approxi-mately constant. (Rubber is designed to bouncewell; little irreversible entropy is generated in acycle of stretching and compression, so long asthe deformation is not too abrupt.)(e) True or false?(T) (F) When we stretch the rubber band, it willcool; the configurational entropy of the randomwalk will decrease, causing the entropy in the vi-brations to decrease, causing the temperature todecrease.(T) (F) When we stretch the rubber band, it willcool; the configurational entropy of the randomwalk will decrease, causing the entropy in the vi-brations to increase, causing the temperature todecrease.(T) (F) When we let the rubber band relax, itwill cool; the configurational entropy of the ran-dom walk will increase, causing the entropy inthe vibrations to decrease, causing the tempera-ture to decrease.(T) (F) When we let the rubber band relax, theremust be no temperature change, since the entropyis constant.This more realistic model is much like the idealgas, which also had no configurational energy.(T) (F) Like the ideal gas, the temperaturechanges because of the net work done on the sys-tem.(T) (F) Unlike the ideal gas, the work done onthe rubber band is positive when the rubber bandexpands.You should check your conclusions experimen-tally; find a rubber band (thick and stretchy isbest), touch it to your lips (which are very sen-sitive to temperature), and stretch and relax it.

(5.13) How many shuffles? (Mathematics) ©iFor this exercise, you will need a deck of cards,either a new box or sorted into a known order(conventionally with an ace at the top and a kingat the bottom).56

How many shuffles does it take to randomize adeck of 52 cards?The answer is a bit controversial; it dependson how one measures the information left inthe cards. Some suggest that seven shuffles areneeded; others say that six are enough.57 We willfollow reference [186], and measure the growingrandomness using the information entropy.We imagine the deck starts out in a known order(say, A♠, 2♠, . . . , K♣).(a) What is the information entropy of the deckbefore it is shuffled? After it is completely ran-domized?(b) Take a sorted deck of cards. Pay attentionto the order; (in particular, note the top and bot-tom cards). Riffle it exactly once, separating itinto two roughly equal portions and interleavingthe cards in the two portions.. Examine the cardsequence, paying particular attention to the topfew and bottom few cards. Can you tell whichcards came from the top portion and which camefrom the bottom?The mathematical definition of a riffle shuffle iseasiest to express if we look at it backward.58

Consider the deck after a riffle; each card inthe deck either came from the top portion orthe bottom portion of the original deck. A riffleshuffle makes each of the 252 patterns tbbtbttb . . .(denoting which card came from which portion)equally likely.It is clear that the pattern tbbtbttb . . . deter-mines the final card order: the number of t’stells us how many cards were in the top portion,and then the cards are deposited into the finalpile according to the pattern in order bottom totop. Let us first pretend the reverse is also true:that every pattern corresponds one-to-one witha unique final card ordering.(c) Ignoring the possibility that two different rif-fles could yield the same final sequence of cards,what is the information entropy after one riffle?You can convince yourself that the only way two

56Experience shows that those who do not do the experiment in part (b) find it challenging to solve part (c) by pure thought.57More substantively, as the number of cards N → ∞, some measures of information show an abrupt transition near 3/2 log2 N ,while by other measures the information vanishes smoothly and most of it is gone by log2 N shuffles.58The forward definition mimics an expert riffler by splitting the deck into two roughly equal portions by choosing n cards ontop with a binomial probability 2−52

(52n

). Then the expert drops cards with probability proportional to the number of cards

remaining in its portion. This makes each of the 252 choices we found in the backward definition equally likely.

What about internal energy?

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(6.15) Entropic forces: Gas vs. rubber band. ©3

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Fig. 6.19 Piston with rubber band. A piston invaccum is held closed by a rubber band. The gas inthe piston exerts an outward force. The rubber bandstretches across the piston, exerting an inward tensileforce. The piston adjusts to its thermal equilibriumdistance, with no external force applied.

In Fig. 6.19, we see a piston filled with a gaswhich exerts a pressure P . The volume outsidethe piston is empty (vacuum, Pext = 0), and noexternal forces are applied. A rubber band at-tached to the piston and the far wall resists theoutward motion with an inward force F . Thepiston moves to its equilibrium position.Assume the gas and the rubber band are mod-eled as both having zero potential energy Eint =0 for all accessible states. (So, for example, per-haps the gas was modeled as hard spheres, simi-lar to Exercise 3.5, and the rubber band as zero-energy random walk chain as in Exercise 5.12).What will happen to the position of the piston ifthe temperature is increased by a factor of two?

Fundamentally, why does this occur? Give anelegant and concise reason for your answer.

(6.16) Rubber band free energy. (Condensed mat-ter) ©iThis exercise illustrates the convenience ofchoosing the right ensemble to decouple systemsinto independent subunits. Consider again therandom-walk model of a rubber band in Exer-cise 5.12 – N segments of length d, connected byhinges that had zero energy both when straightand when bent by 180◦. We there calculated itsentropy S(L) as a function of the folded lengthL, used Stirling’s formula to simplify the com-binatorics, and found its spring constant K atL = 0.Here we shall do the same calculation, withoutthe combinatorics. Instead of calculating the en-tropy S(L) at fixed L,66 we shall work at fixedtemperature and fixed force, calculating an ap-propriate free energy χ(T, F ). View our modelrubber band as a collection of segments sn = ±1of length d, so L = d

∑Nn=1 sn. Let F be the

force exerted by the band on the world (nega-tive for positive L).(a) Write a Hamiltonian for our rubber band un-der an external force.67 Show that it can be writ-ten as a sum of uncoupled Hamiltonians Hn, onefor each link of the rubber band model. Solvefor the partition function Xn of one link, andthen use this to solve for the partition functionX(T, F ). (Hint: Just as for the canonical en-semble, the partition function for a sum of un-coupled Hamiltonians is the product of the par-tition functions of the individual Hamiltonians.)(b) Calculate the associated thermodynamic po-tential χ(T, F ). Derive the abstract formula for〈L〉 as a derivative of χ, in analogy with the cal-culation in eqn 6.11. Find the spring constantK = ∂F/∂L in terms of χ. Evaluate it for ourrubber band free energy at F = 0.The calculations of this exercise, and of the origi-nal rubber band Exercise 5.12, are closely relatedto Exercise 6.3 on negative temperatures.

66Usually we would write S(E,L), but our model rubber band is purely entropic – E = 0 for all states.67Our model rubber band has no internal energy. This Hamiltonian includes the energy exchanged with the outside worldwhen the length of the rubber band changes, just as the Gibbs free energy G(T, P ) = E−TS+PV includes the energy neededto borrow volume from the external world.


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Sbathα = Sbath(U0 )−

1T⎛

⎝⎜⎜⎜⎞

⎠⎟⎟⎟⎟εα

Large system at temperature T: “the bath” plus “the sample”

“the sample”

Substitute the definition of inverse temperature:

…and insert into: wα = e−(S0−Sbath

α )/kB

…to finally get:

wα = Ae−εα / kBT( )

1T= ∂S∂E V ,N

(from Lecture 12)

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ρ(s)∝ Ω2(Ε − Εs ) = exp(S2(E - Es )/kB )

2 2 2 B

2

2 2( ( ) ( ))/ ( )( / )/

( )/

( ) / ( ) ( ) / ( ) e e e .

B A B A B

A B B

B A B AS E E S E E k E E S E k

E E k T

s s E E E Er r- - - - ¶ ¶

-

=W - W -

= =

=

6.1 The canonical ensemble

We consider two weakly connected systems that can exchange energy. The temperature is fixed. The larger part is the reservoir (bath). What is the probability of the system having energy Es?

Apply this to 2 system states A & B:

(Boltzmann)

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Also coupled via conservation of angular

momentum

The ions in a metal, plus translational KE of conduction electrons

conduction electron

spins

nuclear spins

Let’s consider a metal as three systems. The underlying ion cores form the lattice. The lattice temperature is due to the vibrations of these ions. This is what a thermometer would measure. The conduction electrons are weakly coupled to the lattice and carry spin ½. Assume each ion core contains one nuclear spin, which also is weakly coupled to the lattice, and to the electron spin via spin-spin interactions.

Dynamic Nuclear Polarization

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What is the problem?

Nuclear Magnetic Resonance (NMR) is a powerful probe of many materials physics properties, and of course is the same thing as Magnetic Resonance Imaging (MRI). When a magnetic field is applied, the energy of a spin depends on whether it is parallel or antiparallel to the field. Resonant absorption of RF signals is impossible, however, if there are equal numbers of spin up and spin down (which is called “saturation”).

The energy splitting for nuclei is tiny compared to room temperature, so they are always saturated. This is even true at cryogenic temperatures, except for maybe a very small effect. (The splitting for electron spins, however, is thousands of times larger.)

So, the problem is, how can enough nuclei be flipped to make a big difference between spin up and spin down? (Cooling is not an option.)

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K&K Ch 3.5 The Overhauser effect

The importance of this effect was ignored in Kittel’s book, despite Kittel being Overhauser’sthesis advisor. It is the basis for Dynamical Nuclear Polarization, which has had an enormous impact (Google it!)

Here is the basic problem. If you want to study nuclear magnetism (NMR, MRI) it’s not possible if the magnetic moments are equally likely to be spin up or spin down. You must have a spin excess. The energy splitting for nuclear moments is so low, however that this is impossible for many materials.

If you add energy to a system, won’t that make it hotter, which would make the problem worse? Al Overhauser figured out a way to add energy and make the nuclear spins cooler!

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1. The lattice makes up the thermal bath that sets the temperature. 2. The electrons thermalize with the lattice. 3. The nuclear spins don’t interact much with the lattice, but they interact with

the conduction electron spins via the hyperfine interaction, and also thermalize.

Total angular momentum is conserved when there are no external torques.Assume the system is in equilibrium at a temperature where the electron spins are somewhat polarized, but there is no nuclear polarization.

What happens if you flip one electron spin? Due to the hyperfine coupling between electrons and nuclear spins, one nuclear spin must also flip to conserve angular momentum.

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Nuclear spins, no polarization at kT:

Electron spins, some polarization at kT:

Initially: Flip one nuclear spin:

One electron spin must also flip, to conserve angular momentum

Energy initially removed from bath:

DE=-2µIH~-2

DE=+2µBH~+2000

Later, electron scatters from lattice, spin flips, returns energy to bath:

DE=-2µBH~-2000

Net change to bath from flipping one nuclear spin: DE=-2µIH~-2

Hence, the Boltzmann factor for nuclear spin flips is: ⁄𝑃* 𝑃+ = 𝑒𝑥𝑝 −2𝜇,𝐻/𝑘𝑇 ~𝑒 ⁄+. /0

Normal process:

For accessible temperatures, this means P+ ~ P-, so no nuclear polarization.

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Normal flipping of one spin:

1. Reverse one nuclear spin, µ I. Then DE=-2µ IH.

2. Conserve angular momentum, reverse one electron spin, µB. Then DE=+2µBH.

3. This energy comes from the thermal bath, whose energy changes by DEbath=-2(µB – µ I)H

4. Some short time later, the electron flips back, so DE=-2µBH.

5. That energy is returned to the bath (first as electron translational kinetic energy).

6. At the end of this process, the change in the bath energy is just DEbath=2µ IH.

7. This energy goes into the Boltzmann factor:

⁄𝑃* 𝑃+ = 𝑒𝑥𝑝 −2𝜇,𝐻/𝑘𝑇

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Nuclear spins, no polarization at kT:

“Saturated” electron spins, forced to have zero polarization by rf field:

Initially: Flip one nuclear spin:

One electron spin must also flip, to conserve angular momentum

Energy initially removed from bath:

DE=-2µIH~-2

DE=+2µBH~+2000

But now the rf field provides the energy to reverse the electron spin flip, without changing the energy of the bath!

Net change to bath from flipping one nuclear spin: DE=-2(µI-µB)H~+2000

Hence, the Boltzmann factor for nuclear spin flips is:

⁄𝑃* 𝑃+ = 𝑒𝑥𝑝 2(𝜇3 − 𝜇,𝐻/𝑘𝑇 ~𝑒 ⁄*.444 /0

Dynamic nuclear polarization: saturate electrons with resonant rf field

This leads to an enormous polarization of the nuclear spin!

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Saturate all the conduction electron spins with a strong RF field at resonance. Then, repeat the previous spin flip process:

1. Reverse one nuclear spin, µ I. Then DE=-2µ IH.

2. Conserve angular momentum, reverse one electron spin, µB. Then DE=+2µBH.

3. This energy comes from the thermal bath, whose energy changes by DEbath=-2(µB – µ I)H

4. Some short time later, the electron is driven by the RF field back to its original orientation. The change in

energy comes from the RF field and is NOT returned to the bath.

5. At the end of this process, the change in the bath energy is still DEbath=-2(µB – µ I)H.

6. This energy goes into the Boltzmann factor:

⁄𝑃* 𝑃+ = 𝑒𝑥𝑝 2(𝜇3 − 𝜇,)𝐻/𝑘𝑇 ~ 𝑒𝑥𝑝 2𝜇3𝐻/𝑘𝑇

The nuclear spin distribution is now determined by the electron magnetic moment!

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⁄𝑃* 𝑃+ = 𝑒𝑥𝑝 +2(𝜇3 − 𝜇,)𝐻/𝑘𝑇 ~ 𝑒𝑥𝑝 +2𝝁𝑩𝐻/𝑘𝑇

⁄𝑃* 𝑃+ = 𝑒𝑥𝑝 −2𝝁𝑰𝐻/𝑘𝑇Normal:

Dynamical:

This emphasizes the central role of the entropy of the bath in determining the Boltzmann factor.This creates a thousand-fold increase in the NMR signal, and single-handedly opened up major new fields of research. (Google “NOESY spectrometer”.)

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Spin ratios (polarization)

el ectr on r at io nucl ear r at io Ser ie s3

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6.4 What is thermodynamics?

0. The theory of the relations between heat and mechanical energy, and of the conversion of one into the other.

1. Thermodynamics is the theory that emerged from statistical mechanics in the limit of large systems.

Thermodynamics is the statistical mechanics of near-equilibrium systems when one ignores the fluctuations.

We say letting N → infinity is the “thermodynamics” limit.

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2. Thermodynamics is a self-contained theory.Derivable completely from axioms:

(0) Transitivity of equilibria. If A and B are both in equilibrium with C, then A is in equilibrium with B.

(1) Conservation of energy. The total energy of an isolated system is constant.

(2) Entropy always increases. An isolated system may undergo irreversible changes, whose effect can be measured by a state function called entropy.

(3) Entropy goes to zero at absolute zero. The entropy per particle of all large systems approaches the same constant value at zero T, which is set to zero by choosing the appropriate phase space unit of volume.

We focus instead on justifying these from statistical mechanics.

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(3) Thermodynamics is a zoo of partial derivatives, transformations, and relations.

Why? So many free energies, thermodynamic potentials, enthalpy, etc.

Consider the transformation between free energies. Helmholtz:( , , ) ( , , )A T V N E TS E V N= -

/ 1 / 1 /s b s s b sA E T S E T T¶ ¶ = - ¶ ¶ = -

Why isn’t A a function of energy E, in the same way the entropy is? Let s= system, b= bath, take derivative with respect to Es:

Since A is an equilibrium property, of a system in equilibrium with the bath, Tb

must be equal to Ts, so the derivative is zero, meaning A is independent of E. Thus A is a transformation from independent variables (E,V,N) to (T,V,N).

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“Physically, energy is transferred until A is a minimum: E is no longer an independent variable.”

This follows from the Helmholtz free energy being a minimum at equilibrium.

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Thermodynamics is so cluttered because it is so powerful.

dE = TdS − PdV + µdN

∂E∂S

N ,V

= T , ∂E∂V

N ,S

= −P, and ∂E∂N

V ,S

= µ

From just the first derivatives we get:

Plus lots more from the second derivatives, such as the Maxwell relations.

On the other hand, statistical mechanics really isn’t much better…

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6.5 Mechanics: friction and fluctuations

Why does a mass on a spring stop oscillating? Isn’t energy conserved?

Think of the system being in thermal contact with a bath of internal atomic degrees of freedom. In equilibrium, the potential energy of the system is ½ kBT, which gives an equilibrium amplitude smaller than the size of an atom.

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Copyright Oxford University Press 2006 v2.0