Physics 562: Statistical Mechanics Spring 2003, James P. Sethna Homework 2, due Wednesday Feb. 12 Latest revision: February 10, 2003, 12:22 am

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1 Physics 562: Statistical Mechanics Spring 2003, James P. Sethna Homework 2, due Wednesday Feb. 12 Latest revision: February 10, 2003, 12:22 am Pathria, chapters 3, 4, Reading (2.1) Entropy of Glasses. Problems See Entropy of Glasses, Stephen A. Langer and James P. Sethna, Phys. Rev. Lett. 61, 570 (1988), although M. Goldstein noticed the bounds earlier. Glasses aren t really in equilibrium. In particular they do not obey the third law of thermodynamics, that the entropy S goes to zero at zero temperature. Experimentalists measure a residual entropy by subtracting the entropy change from the known entropy of the equilibrium liquid at a temperature T l at or above the crystalline melting temperature T c : Tl 1 dq S residual = S liquid (T l ) dt, (2.1.1) 0 T dt where Q is the net heat flow out of the bath into the glass. If you put a glass in an insulated box, it will warm up (very slowly) because of microscopic atomic rearrangements which lower the potential energy. So, glasses don t have a welldefined temperature or specific heat. In particular, the heat flow upon cooling and on heating dq dt (T ) won t precisely match (although their integrals will agree by conservation of energy). Thomas and Parks in the figure shown are making the approximation that the specific heat of the glass is dq/dt, the measured heat flow out of the glass divided by the temperature change of the heat bath. They find that the specific heat defined in this way measured on cooling and heating disagree. Consider the second cooling curve and the final heating curve, from 325 C to room temperature and back. Assume that the liquid at 325 Cis in equilibrium both before cooling and after heating (and so has the same liquid entropy S liquid. (a) Is the residual entropy, equation (2.1.1), larger on heating or on cooling? (Hint: Use the fact that the integrals under the curves, T l dq 0 dt dt give the heat flow, which by conservation of energy must be the same on heating and cooling. The heating curve shifts weight to higher temperatures: will that increase or decrease the integral in (2.1.1)?) (b) By using the second law (entropy can only increase), show that when cooling and then heating from an equilibrium liquid the residual entropy measured on cooling must always be less than the residual entropy measured on heating. 1

2 Specific heat of B 2 O 3 glass measured while heating and cooling. The glass was first rapidly cooled from the melt (500 C 50 C in a half hour), then heated from 33 C 345 C in 14 hours (solid curve with squares), cooled from 345 Cto room temperature in 18 hours (dotted curve with diamonds), and finally heated from 35 C 325 C (solid curve with crosses). Data from S. B. Thomas and G. S. Parks, J. Phys. Chem. 35, 2091 (1931); revisited in Nonequilibrium Entropy and Entropy Distributions, S. A. Langer, E. R. Grannan, and J. P. Sethna, Phys. Rev. B 41, 2261 (1990). The fact that the energy lags the temperature near the glass transition, in linear response, leads to the study of specific heat spectroscopy, (N. O. Birge and S. R. Nagel, Phys. Rev. Lett. 54, 2674 (1985).) The residual entropy of a typical glass is about k B per molecular unit. It s a measure of how many different glassy configurations of atoms the material can freeze into. (c) In a molecular dynamics simulation with one hundred atoms, and assuming that the residual entropy is k B log 2 per atom, what is the probability that two coolings to zero energy will arrive at equivalent atomic configurations? In a system with molecular units, with residual entropy k B log 2 per unit, how many coolings would be needed to arrive at the original configuration again, with probability 1/2? Up to, e.g., translations and rotations. 2

3 (2.2) Shannon entropy. See: Claude E. Shannon, A Mathematical Theory of Communication, from The Bell System Technical Journal, 27, pp , , July, October, Entropy can be viewed as a measure of the lack of information you have about a system. Claude Shannon realized, back in the 1940 s, that communication over telephone wires amounts to reducing the listener s uncertainty about the sender s message, and introduced a definition of an information entropy. Most natural languages (voice, written English) are highly redundant; the number of intelligible fifty-letter sentences is many fewer than 26 50, and the number of ten-second phone conversations is far smaller than the number of sound signals that could be generated with frequencies between 60 and 20,000 Hz. Shannon, knowing statistical mechanics, defined the entropy of an ensemble of messages: if there are N possible messages that can be sent in one package, and message m is being transmitted with probability p m, then Shannon s entropy is N S = k p m log p m (2.2.1) 1 where instead of Boltzmann s constant, Shannon picked k = 1/ log 2. This immediately suggests a theory for signal compression. If you can recode the alphabet so that common letters and common sequences of letters are abbreviated, while infrequent combinations are spelled out in lengthly fashion, you can dramatically reduce the channel capacity needed to send the data. (This is lossless compression, like zip and gz and gif). An obscure language A bç! for long-distance communication has only three sounds: a hoot represented by A, a slap represented by B, and a click represented by C. In a typical message, hoots and slaps occur equally often (p =1/4), but clicks are twice as common (p = 1/2). Assume the messages are otherwise random. (a) What is the Shannon entropy in this language? More specifically, what is the Shannon entropy rate (entropy per sound, or letter, transmitted)? (b) Show that a communication channel transmitting bits (ones and zeros) can transmit no more than one unit of Shannon entropy per bit. (Hint: this should follow by showing that, for N =2 m messages, equation (2.2.1) is maximized by p m =1/N. You needn t prove it s a global maximum: check that it is a local extremum. You ll need either a Lagrange multiplier or will need to explicitly set p N =1 N 1 m=1 p m.) (c) In general, argue that the Shannon entropy gives the minimum number of bits needed to transmit the ensemble of messages. (Hint: compare the Shannon entropy of the N original messages with the Shannon entropy of the N (shorter) encoded messages.) Calculate the minimum number of bits per letter on average needed to transmit messages for the particular case of an A bç! communication channel. (d) Find a compression scheme (a rule that converts a A bç! message to zeros and ones, that can be inverted to give back the original message) that is optimal, in the sense that it saturates the bound you derived in part (b). Shannon also developed a measure of the channel capacity of a noisy wire, and discussed error correction codes... 3

4 (2.3) Dyson s Time without End: Life and the Heat Death of the Universe. See Freeman J. Dyson, Time without end: Physics and biology in an open universe, Rev. Mod. Phys. 51, 447 (1979). Freeman Dyson discusses how living things might evolve to cope with the cooling and dimming we expect during the heat death of the universe. Dyson models an intelligent being as a heat engine that creates a fixed entropy S per thought. (a) Energy needed per thought. Assume that the being draws heat Q from a hot reservoir at T 1 and radiates it away to a cold reservoir at T 2. What is the minimum energy Q needed per thought, in terms of S and T 2?(YoumaytakeT 1 very large.) Time needed per thought to radiate energy. Dyson shows, using theory not important here, that the power radiated by our intelligent being as heat engine is no larger than CT 3 2, a constant times the cube of the cold temperature. (The constant scales with the number of electrons in the being, so we can think of our answer t as the time per thought per mole of electrons.) (b) Write an expression for the maximum rate of thoughts per unit time dh/dt (the inverse of the time t per thought), in terms of S, C, andt 2. Number of thoughts for an ecologically efficient being. Our universe is expanding: the radius R grows roughly linearly in time t. The microwave background radiation has a characteristic temperature Θ(t) R 1 which is getting lower as the universe expands: this red-shift is due to the Doppler effect. An ecologically efficient being would naturally try to use as little heat as possible, and so wants to choose T 2 as small as possible. It cannot radiate heat at a temperature below T 2 =Θ(t) =A/t. (c) How many thoughts H can an ecologically efficient being, radiating at a fixed multiple of Θ(t), have between now and time infinity, in terms of S, C, A, and the current time t 0? Time without end: Greedy beings. Dyson would like his beings to be able to think an infinite number of thoughts before the universe ends, but consume a finite amount of energy. He proposes that his beings need to be profligate in order to get their thoughts in before the world ends: he proposes that they radiate at a temperature T 2 (t) t 3/8 which falls with time, but not as fast as Θ(t) t 1. (d) Show that with Dyson s radiation schedule, the total number of thoughts H is infinite, but the total energy consumed U is finite. 4

5 (2.4) Thermodynamics Review II: Gibbs-Duhem relation and the Clausius- Clapeyron equation. (a) Using the fact that the entropy S(N,V,E) is extensive, show that N S N Show from this that in general + V S V,E V + E S N,E E = S. N,V and hence E = TS pv + µn. This is Euler s equation. S =(E + pv µn)/t (2.4.1) As a state function, S is supposed to depend only on E, V,andN. But equation (2.4.1) seems to show explicit dependence on T, p, andµ as well: how can this be? (c) One answer is to write the latter three as functions of E, V,andN. Do this explicitly for the ideal gas, using last week s equation (1.2.2) [ S(N,V,E) = 5 ( ) ] 3/2 2 Nk V 4πmE B + Nk B log Nh 3, (1.2.2) 3N and your (or the grader s) results for problem 1.2(c), and verify equation (2.4.1) in that case. Another answer is to consider a small shift of all six variables. We know that de = TdS pdv + µdn, but if we shift all six variables in Euler s equation we get de = TdS pdv + µdn + SdT Vdp+ Ndµ. This implies the Gibbs-Duhem relation 0=SdT Vdp+ Ndµ. (2.4.2) It means that the intensive variables T, p, andµ are not all independent. Generic phase diagram, from itl/2045 s99/lectures/lec f.html 5

6 Clausius-Clapeyron equation. Consider the phase diagram above. Along an equilibrium phase boundary, the temperatures, pressures, and chemical potentials of the two phases must agree: otherwise a flat interface between the two phases would transmit heat, shift sideways, or leak particles, respectively (violating the assumption of equilibrium). (d) Apply the Gibbs-Duham relation to both phases, for a small shift by T across the phase boundary. Let s 1, v 1, s 2,andv 2 be the molecular entropies and volumes (s = S/N, v = V/N for each phase); derive the Clausius-Clapeyron equation for the slope of the coexistence line on the phase diagram dp/dt =(s 1 s 2 )/(v 1 v 2 ). (2.4.3) (e) Latent Heat. It s hard to experimentally measure the entropies per particle: we don t have an entropy thermometer. But, as you will remember, the entropy difference upon a phase transformation S = Q/T is related to the heat flow Q needed to induce the phase change. Let the latent heat L be the heat flow per molecule: write a formula for dp/dt not involving the entropy. 6

7 (2.5) Excited Atoms, Negative Temperature, and Micro Canonical. AsystemofN atoms can be in the ground state or in an excited state. For convenience, we set the zero of energy exactly in between, so the energies of the two states of an atom are ±ɛ/2. The atoms are isolated from the outside world. There are only weak couplings between the atoms, sufficient to bring them into internal equilibrium but without other effects. Entropies and Energy Fluctuations Microcanonical and Canonical 30k B 0.1 Entropy S Microcanonical Entropy S micro (E) Canonical Entropy S c (T(E)) Probability ρ(e) Probability ρ(e) 0k B -20ε -10ε 0ε 10ε 20ε Energy E = -N/2 + M ε/3 Entropies and energy fluctuations for this problem with N = 50. The canonical probability distribution for the energy is for E = 10ɛ, andk B T =1.207ɛ. You may wish to check some of your answers against this plot. (a) Microcanonical Entropy. If the net energy is E (corresponding to a number of excited atoms m = E/ɛ + N/2), what is the microcanonical entropy S micro (E) ofour system? Simplify 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). (Why is it tricky to do it without approximation?) What happens to the temperature when E>0? Having the energy E>0is a kind of population inversion. Population inversion is the driving mechanism for lasers. Microcanonical simulations can lead also to states with negative specific heats. For many quantities, the thermodynamic derivatives have natural interpretations when viewed as sums over states. It s easiest to see this in small systems. (c) Canonical Ensemble: Explicit traces and thermodynamics. (i) Take one of our atoms and couple it to a heat bath of temperature k B T =1/β. Write explicit formulas for Z canon, E canon,ands canon in the canonical ensemble, as a trace (or sum) over the two states of the atom. (E should be the energy of each state multiplied by the probability ρ n of that state, S should be the trace of ρ n log ρ n.) (ii) Compare the results with what you get by using the thermodynamic relations. Using Z from the trace over states, calculate the Helmholtz free energy A, S as a derivative of A, ande from A = E TS. Do the thermodynamically derived formulas you get agree with the statistical traces? (iii) To remove some of the mystery from the thermodynamic relations, 7 0

8 consider the thermodynamically valid formula E = log Z/ β = (1/Z) Z/ β. Write out Z as a sum over energy states, and see that this formula follows naturally. (d) What happens to E in the canonical ensemble as T? regime discussed in part (b)? Can you get into the (e) Canonical-Microcanonical Correspondence. Find the entropy in the canonical distribution for N of our atoms coupled to the outside world, from your answer to part (c). How can you understand the value of S(T = ) S(T = 0) simply? Using the approximate form of the entropy from part (a) and the temperature from part (b), show that the canonical and microcanonical entropies agree, S micro (E) = S canon (T (E)). (Perhaps useful: arctanh(x) =(1/2) log ((1 + x)/(1 x)).) Notice that the two are not equal in the figure above: the form of Stirling s formula we used in part (a) is not very accurate for N = 50. In a simple way, explain why the microcanonical entropy is smaller than the canonical entropy. (f) Fluctuations. Show in general that the root-mean-squared fluctuations in the energy in the canonical distribution (E E ) 2 = E 2 E 2 is related to the specific heat C = E/ T. (I find it helpful to use the formula from part (c.iii), E = log(z)/ β.) Calculate the root-mean-square energy fluctuations for N of our atoms. Evaluate it at T (E) from part (b): it should have a particularly simple form. For large N, are the fluctuations in E small compared to E? 8

9 (2.6) Specific Heat: Strings and Defects. One-dimensional Phonons. A nano-string of length L with mass per unit length µ under tension τ has a vertical, transverse displacement u(x, t). The kinetic energy density is (µ/2)( u/ t) 2 and the potential energy density is (τ/2)( u/ x) 2. (a) Write the kinetic energy and the potential energy in new variables, changing from u(x, t) tonormalmodesq k (t) with u(x, t) = n q k n (t) sin(k n x), k n = nπ/l. Show in these variables that the system is a sum of decoupled harmonic oscillators. Calculate the density of states per unit frequency g(ω) = n δ(ω ω k n ) and the specific heat of the string c(t ) per unit length in the limit L, treating the oscillators quantum mechanically. What is the specific heat of the classical string? Defects in Crystals. A defect in a crystal has one on-center configuration with energy zero, and M off-center configurations with energy ɛ, with no significant quantum tunneling between the states. The Hamiltonian can be approximated by the (M +1) (M +1) matrix H = ɛ ɛ There are N defects in the crystal, which can be assumed stuck in position (and hence distinguishable) and assumed not to interact with one another. (b) Write the canonical partition function Z(T ), the mean energy E(T ), the fluctuations in the energy, the entropy S(T ), and the specific heat C(T ) as a function of temperature. Plot the specific heat per defect C(T )/N for M = 6; set the unit of energy equal to ɛ and k B = 1 for your plot. Derive a simple relation between M and the change in entropy between zero and infinite temperature. Check this relation using your formula for S(T ). The density of states here is the number of normal modes in a frequency range (ω, ω+ɛ) divided by ɛ. In the next homework, we ll study the density of energy eigenstates g(e) in a trap with precisely three frequencies (g(ω) will be = δ(ω ω 0 )+2δ(ω ω 1 )). Don t confuse the two. 9

10 (2.7) Laplace ( ), Lagrange ( ) and Legendre ( ). (a) Laplace Transform. Show (or note) that the canonical partition function can be written as the Laplace transform of the microcanonical partition function. I ve never noticed that this was useful. (b) Lagrange Multipliers. (See Pathria, problem 3.6). (i) Microcanonical: Show, using a Lagrange multiplier, that the probability distribution that extremizes the entropy S = k B Tr(ρlog ρ) = k B ρ(x, P)logρ(X, P) with the constraint that Energy Surface the density integrates to one, Tr(ρ) = ρ(x, P) = 1, is a constant Energy Surface (the microcanonical distribution). (ii) Canonical: Integrating over all X and P, use another Lagrange multiplier to fix the mean energy E = dxdp H(X, P)ρ(X, P). Show that maximizing the entropy given the two constraints gives you the canonical distribution. (iii) Grand Canonical: Summing over different numbers of particles N, adding the constaint that the average number is N = N dxdp NρN (X, P), show that you get the grand canonical distribution. (c) Legendre Transforms. What thermodynamics potentials correspond to the microcanonical, canonical, and grand canonical ensembles? What variables are they naturally functions of? Give the Legendre transformations which lead us from one to the other. As you showed in problem 2.2(b). 10

11 (2.8) Does Entropy Increase? The second law of thermodynamics says that entropy always increases. Perversely, it s easy to show that in an isolated system, no matter what non-equilibrium condition it starts in, entropy as precisely defined stays constant in time. Entropy is Constant: Classical. Liouville s theorem tells us that the total derivative of the probability density is zero: following the trajectory of a system, the local probability density never changes. The equilibrium states have probability densities that only depend on energy and number. Clearly something is wrong: if the density starts non-uniform, how can it become uniform? (a) Let ρ(p, Q) is the probability density in phase space (where P =(p 1,...,p 3N )are momenta and Q =(q 1,...,q 3N ) are the momenta. Show, for any function f(ρ) with f(0) = 0, that f(ρ)/ t = [f(ρ)v] = α / p α (f(ρ)ṗ α )+ / q α (f(ρ) q α ), where v = (Ṗ, Q) is the 6N dimensional velocity in phase space. Hence, (by Gauss s theorem in 6N dimensions), show f(ρ)/ t dpdq = 0, assuming that the probability density vanishes at large momenta and positions. Show, thus, that the entropy S = k B ρ log ρ is constant in time. (b) Entropy is Constant: Quantum. Prove that S = Trˆρ log ˆρ is time-independent, where ˆρ is any density matrix. The Arnol d Cat. Why do we think entropy increases? First, points in phase space don t just swirl in circles: they get stretched and twisted and folded back in complicated patterns especially in systems where statistical mechanics seems to hold! Arnol d, in a takeoff on Schrödinger s cat, suggested the following analogy. Instead of a continuous transformation of phase space onto itself preserving 6N-dimensional volume, let s think of an area-preserving mapping of the plane into itself. (You might think of the mapping as a Poincaré section, remembering last problem set.) Consider the mapping See the map illustrated below. Γ ( ) ( ) x x + y = mod n. y x +2y (c) Check that Γ preserves area (what is the determinant?). Show that it takes a square n n (or a picture of n n pixels) and maps it into itself with periodic boundary conditions. (With less cutting and pasting, you can view it as a map from the torus into itself.) As a linear map, find the eigenvalues and eigenvectors. Argue that a small neighborhood (say a circle in the center of the picture) will initially be stretched along an irrational direction into a thin strip. When this thin strip hits the boundary, it gets split into two; in the case of an n n square, further iterations stretch and chop our original circle into a line uniformly covering the square. In the pixel case, there are always exactly the same number of pixels that are black, white, and each shade of gray: they just get so kneaded together that everything looks 11

12 Arnol d Cat Transform, from see movie at sander/movies/arnold.html. Notice, however, that the cat comes back! This system exhibits Poincaré recurrance (which we haven t covered) much earlier than most systems. a uniform color. So, by putting a limit to the resolution of our measurement (rounding errors on the computer, for example), or by introducing any tiny coupling to the external world, the final state can be seen to rapidly approach equilibrium, proofs to the contrary notwithstanding! 12

13 (2.9) Quantum Stat Mech: Density Matrices and Density of States. (a) Density matrices for photons. Write the density matrix for a photon linearly traveling along z and linearly polarized along ˆx, in the basis where (1, 0) and (0, 1) are polarized along ˆx and ŷ. Write the density matrix for a right-handed polarized photon, (1/ 2,i/ 2), and the density matrix for unpolarized light. Calculate Tr(ˆρ), Tr(ˆρ 2 ), and S = k B Tr(ˆρ log ˆρ). Interpret the values of the three traces physically: one is a check for pure states, one is a measure of information, and one is a normalization. (b) Density matrices for a spin.(adapted from Halperin s course, 1976.) Let the Hamiltonian for a spin be H = h 2 B ˆ σ where ˆ σ =(σ x,σ y,σ z ) are the three Pauli spin matrices, and B may be interpreted as a magnetic field, in units where the gyromagnetic ratio is unity. Remember that σ i σ j σ j σ i =2iɛ ijk σ k. Show that any 2 2 density matrix may be written in the form ˆρ =(1/2)(ˆ1+p ˆ σ). Show that the equations of motion for the density matrix i h ˆρ/ t =[H, ˆρ] canbe written as dp/dt = B p. In classical mechanics, the entropy goes to minus infinity as the temperature is lowered to zero; in quantum mechanics the entropy goes to zero, because states are quantized. This is Nernst s theorem, the third law of thermodynamics. The classical phase-space volume has units of ((momentum) (distance)) 3N. It s a little perverse to take the logarithm of a quantity with units, and the obvious candidate with these dimensions is Planck s constant h 3N. Or should it be h 3N? (c) Arbitrary zero of the classical entropy. Show that the choice of units changes the entropy per particle, and hence that Nernst s theorem depends on getting this correct. (d) Fixing the zero of entropy: quantum Maxwell-Boltzmann ideal gas. Consider N particles in a L L L box with periodic boundary conditions. Show that the single-particle energy eigenstates exp(ik x) form a regular grid in k space: what is the spacing k? (Fixed boundary conditions would give a grid for only positive k, with a different spacing.) Show that the many-particle energy eigenstates form a regular grid in 3N-dimensional k space. Show that the constant-energy surfaces are spheres in this space. Assume that the particles are indistinguishable, but classical: only distinct fillings of single-particle eigenstates are different. Calculate the entropy per particle in the limit L, in analogy to the fixed boundary condition calculation in the text. Which version of Planck s constant is the right one to normalize the units of phase space? We ll see soon that real quantum mechanics gives symmetry constraints on the manybody wave function, which changes the statistics from Maxwell-Boltzmann to either Fermi or Bose-Einstein. This calculation, however, remains valid at low densities and high pressures, where multiply occupied states are rare. 13