Physics 562: Statistical Mechanics Spring 2003, James P. Sethna Final Exam, due Wednesday, May 14 Latest revision: May 11, 2003, 6:00 pm

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1 Physics 562: Statistical Mechanics Spring 2003, James P. Sethna Final Exam, due Wednesday, May 14 Latest revision: May 11, 2003, 6:00 pm Open Book Exam Workon your own for this exam. You may consult your notes, homeworks and answer keys, books and published work, or Web pages as you find useful. The problems have been designed to be doable given only material already presented in the course. If you find something in the literature or on the Web that is particularly helpful (e.g., solvesthe problem), feel free to use it: but, just as in a publication, give a citation. Problems (F.1) Connecting Two Macroscopic Systems. (30 points) An isolated system with energy E is composed of two macroscopic subsystems, each of fixed volume V and number of particles N. The subsystems are weakly coupled, so the sum of their energies is E 1 + E 2 = E. (a) Derive a formula for the volume of the energy surface of the whole system Ω(E) = dx 1 dp 1 dx 2 dp 2 δ (E (H 1 (X 1,P 1 )+H 2 (X 2,P 2 ))) (F.1.1) as an integral involving Ω 1 (E 1 ) and Ω 2 (E 2 ). (Hint: Insert δ(e 1 H 1 (X 1,P 1 )) de 1 = 1.) Consider a monatomic ideal gas (He) mixed with a diatomic ideal gas (H 2 ). We showed that a monatomic ideal gas of N atoms has Ω 1 (E 1 ) E 3N/2 1. A diatomic molecule has Ω 2 (E 2 ) E 5N/2 2. (b) For these two gases, which energy E1 max maximizes the integrand in your convolution answer to part (a)? (c) Use the saddle-point method to approximate the probability ρ(e 1 ) that the He has energy E 1 as a Gaussian. (That is, put the integrand into the form exp(f(e 1 )) and Taylor expand f(e 1 ) to second order in E 1 E1 max.) In this approximation, what are the energy fluctuations per particle (E 1 E1 max ) 2 /N? For subsystems with large numbers of particles N, temperature and energy density are well defined because Ω(E) for each subsystem grows. Notice that the temperatures T 1 and T 2 agree when E 1 = E1 max. In the range h 2 /2I k B T hω, whereω is the vibrational frequency of the stretch mode and I is the moment of inertia. The lower limit makes the rotations classical; the upper limit freezes out the vibrations, leaving us with three classical translation modes and two rotational modes a total of five degrees of freedom. See Pathria, 6.5(B). For example, see Matthews and Walker, Methods of Mathematical Physics, section

2 (F.2) Statistical Mechanics and Statistics. (20 points) Consider the problem of fitting a theoretical model to experimentally determined data. Let our model M predict a time-dependent function y (M) (t). Let there be N experimentally determined data points y i at times t i with errors of standard deviation σ. We assume that the experimental errors for the data points are independent and Gaussian distributed, so that the probability that our model actually generated the observed data points (the probability P (D M) of the data given the model) is P (D M) = N i=1 [ ( ) 1 y (M) 2 (t i ) y i exp 2πσ 2σ 2 ]. (F.2.1) (a) True or false: This probability density corresponds to a Boltzmann distribution with energy H and temperature T, with H = N i=1 (y(m) (t i ) y i ) 2 /2andk B T = σ 2. There are two schools of statistics. Among a family of models, the frequentists will pickthe model M with the largest value of P (D M). The Bayesians take a different point of view. They argue that there is no reason to believe that all models have the same likelihood: there is no analogue of Liouville s theorem in model space. Suppose the intrinsic probability of the model (the prior) isp (M). They use the simple theorem P (M D) =P (D M)P (M)/P (D) =P (D M)P (M) (F.2.2) where the last step notes that the probability that you measured the known data D is presumably one. The Bayesians often will pickthe maximum of P (M D) as their model for the experimental data. But, given their perspective, it s even more natural to consider the entire ensemble of models, weighted by P (M D), as the best description of the data. This ensemble average then naturally provides error bars as well as predictions for various quantities. 2

Consider the simple problem of fitting a line to two data points.

2 =2,wherebothy-values have uncorrelated Gaussian errors with

and two, and assumes that the probability density is otherwise

(b) (10 points) Which of the following contour plots accurately

3 Consider the simple problem of fitting a line to two data points. Suppose the experimental data points are at t 1 =0,y 1 =1andt 2 =1,y 2 =2,wherebothy-values have uncorrelated Gaussian errors with standard deviation σ = 1/2, as assumed in equation (F.2.1) above. Our model M(m, b) isy(t) =mt + b. Our Bayesian statistician knows that m and b both lie between zero and two, and assumes that the probability density is otherwise uniform: P (m, b) =1/4 for0<m<2 and 0 <b<2. (b) (10 points) Which of the following contour plots accurately represent the probability distribution P (M D) for the model, given the observed data? (The spacing between the contour lines is arbitrary.) (A) (B) (C) (D) (E) 3

On the right, you see the experimental setup used to apply a force F to the RNA: the two ends are attached, via complementary DNA strands, to two beads which are then pulled apart.

4 (F.3) Telegraph Noise and RNA Unfolding. (40 points) Figure F.3.1, Hairpins in RNA, from J. Liphardt et al., Science 292, 733 (2001). On the left, you see the hairpin in the RNA: a length of RNA attaches to an inverted, complementary strand immediately following. On the right, you see the experimental setup used to apply a force F to the RNA: the two ends are attached, via complementary DNA strands, to two beads which are then pulled apart. The RNA has two metastable states over a range of forces: the hairpin, zipped configuration shown and an unzipped configuration. Let the difference in length between the zipped and unzipped configuration be L. Figure F.3.2, Telegraph Noise in RNA unzipping, from Liphardt et al. above. As the force increases, the fraction of time spent in the zipped state decreases. RNA, ribonucleic acid, is a long polymer like DNA, with many functions in living cells. It has four monomer units (A, U, C, and G: DNA has T instead of U The sequence of Adenine, Uracil, Cytosine, and Guanine; DNA has Thymine. See 4

5 monomers can encode information for building proteins, and can also cause the RNA to fold into shapes that are important to its function. Michelle Wang s group here studies the strength of these hairpins by pulling on them (see data from a different group, above). Under a sufficiently large force, the hairpin will unzip. Near the threshold for unzipping, the RNA is found to jump between the zipped and unzipped states, giving telegraph noise, figure (F.3.2). Just as the current in a telegraph signal is either on or off, these systems are bistable and make transitions from one state to the other. When these systems are in equilibrium, they provide a prototype statistical mechanics problem. (Thinkof it as an Ising model with one spin.) The two RNA configurations presumably have different energies (E z,e u ) entropies (S z,s u ) and volumes (V z,v u ) for the local region around the zipped and unzipped states, respectively. The environment is at temperature T and pressure P.Letρ z be the fraction of the time our molecule is zipped at a given external force F,andρ u =1 ρ z be the unzipped fraction of time. Of the following statements, which are true, assuming that the pulled RNA is in equilibrium? (T) (F) ρ z /ρ u =exp((sz tot Su tot )/k B ), where Sz tot and Su tot are the total entropy of the universe when the RNA is in the zipped and unzipped states, respectively. (T) (F) ρ z /ρ u =exp( (E z E u )/k B T ). (T) (F) ρ z /ρ u =exp( (G z G u )/k B T ), where G z = E z TS z + PV z and G u = E u TS u + PV u are the Gibbs energies in the two states. (T) (F) ρ z /ρ u =exp( (G z G u +FL)/k B T ), where L is the extra length of the unzipped RNA and F is the applied force. 5

6 (F.4) Telegraph Noise in Nanojunctions: Continuous Time Markov Chains. (40 points) Many systems in physics exhibit telegraph noise. Dan Ralph, in an early part of his PhD with Bob Buhrman, worked with Kristen Ralls on telegraph noise in his nanojunctions. Figure F.4.1, Telegraph Noise in a metallic nanobridge. Resistance versus time R(t) for a copper constriction, from Kristen S. Ralls, Daniel C. Ralph, and Robert A. Buhrman, Phys.Rev.B (1989). We label A the state with low resistance R A,andB the state with high resistance R B. The two states probably represent a local shift of an atom or a small group of atoms in the constriction from one metastable state to another. The junction has two states, A and B. It makes transitions at random, with rate Γ B A = Γ BA from A to B and rate Γ AB from B to A. Master Equation. Consider an ensemble of many identical copies of this system. Let the state of this ensemble at time t be given by ρ(t) =(ρ A,ρ B ), a vector of probabilities that the system is in the two states. This vector thus evolves according to the master equation d ρ/dt = M ρ. (F.4.1) (a) What is the 2 2matrixM for our system? (It s different from the discrete-time case we studied earlier. Notice under Γ AB, ρ A increases and ρ B decreases.) At long times, what fraction of the time will our system be in the A state, ρ A = lim t ρ A (t)? (b) Find the eigenvalue eigenvector pairs for M. Which corresponds to the stationary state ρ( ) from part (a)? Suppose that at t = 0 the system is known to be in the A state, ρ(0) = (1, 0). Write this initial condition in the basis of eigenvectors, and hence give a formula for the subsequent time evolution ρ A (t). What is the rate of decay to the stationary state? Let s call your answer for ρ A (t) =P AA (t) to emphasize the fact that it is the probability of being in the A state at time t + t, given that it is in the A state at time t. You may wish to checkthat P AA (0) = 1, and that P AA (t) ρ A as t. (c) Correlation function. Let R(t) be the resistance as a function of time, hopping between R A and R B, as shown in figure F.4.1, and let R be the time average of the More specifically, the right eigenvectors M ρ λ = λ ρ λ 6

7 resistance. Write a formula for the correlation function C(t) = (R(t ) R)(R(t + t) R) in terms of P AA (t)? For ease of grading, please do not plug in your answer for the latter from part (b). (Hint: what is (R(t ) R B )(R(t + t) R B ), interms of P AA (t)? What is it in terms of C(t)?) You may wish to checkthat your C(t) 0ast,andthatC(0) = ρ A (R A R) 2 + ρ B (R B R) 2. Figure F.4.2, Telegraph Noise in a metallic nanobridge with three metastable states. From Kristen S. Ralls and Robert A. Buhrman, Phys. Rev. B (1991). Nanojunctions, especially at higher temperatures, often show more than two metastable states in the experimental bandwidth (fluctuating neither too fast or too slowly to measure). Usually these form independent two-level fluctuators (atomic rearrangements too far apart to interact substantially), but sometimes more complex behavior is seen. The figure above shows three resistance states, which we label A, B, andc from lowest resistance to highest. We notice from Figure F.4.2 that the rates Γ CB and Γ BC are the highest, followed by the rates Γ AC and Γ CA. There are no transitions seen going between states A and B, at least as far as I can see. There is a net current passing through the junction. This current can in principle cause lots of nonequilibrium effects: local heating, electromigration, different temperatures for electrons and phonons... On the other hand, the two-level fluctuators are described remarkably well by equilibrium statistical mechanics. Can we use three-level fluctuators to test the predictions of equilibrium statistical mechanics more sharply? At the time, Kristen Ralls and I discussed trying to test detailed balance. In an equilibrium system, for any two states i and j with equilibrium probabilities ρ i and ρ j, detailed balance states that Γ i j ρ j =Γ j i ρ i. When discussing systems that may not be in equilibrium, it s both possible and elegant to reformulate the condition for detailed balance so that it doesn t involve the equilibrium probabilities. (d) Assume that each of the three types of transitions in the three-state system satisfies detailed balance. Eliminate the equilibrium probability densities to write the unknown rate Γ AB in terms of the five other rates. If we view the three states A, B, andc to be around a circle, find a general relation implied by detailed balance between the rates going clockwise and the rates going counter-clockwise around the circle. 7

8 (F.5) Nucleation of Dislocation Pairs. (40 points) F/L = σ xy b y x b R F/L = σ xy Figure F.5.1, Dislocation Pair in a 2D Hexagonal Crystal, (NickBailey). The loop around the defect on the right shows an extra atom along the bottom row. By the conventions used in materials physics (assuming here that the dislocation points up out of the paper, see Hirth and Lothe, Theory of Dislocations, second edition, figure 1.20, p. 23). this edge dislocation has Burgers vector b = aˆx, where a is the distance between neighboring atoms. Similarly, the defect on the left has an extra atom in the top row, and has b = aˆx. The defects are centered at the same height, separated by a distance R. The crystal is under a shear stress σ xy = F/L,wheretheforceF = ±σ xy Lŷ is applied to the top and bottom as shown (but the crystal is kept from rotating). Consider the simple example of a two-dimensional crystal under shear shown in the figure above. The external force is being relieved by the motion of the upper half of the crystal to the left with respect to the bottom half of the crystal by one atomic spacing a. If the crystal is of length L, the energy released by this shuffle when it is complete will be F a = σ xy La. This shuffling has only partially been completed: only the span R between the two edge dislocations has been shifted (the dislocations are denoted by the conventional A similar problem, for superfluids, was studied by Vinay Ambegaokar, Halperin, Nelson, and Eric Siggia, PRL 40, 783 (1978) and PRB 21, 1806 (1980); see also Tang and Chen PRB 67, (2003). The complete solution is made more complex by the effects of other dislocation pairs renormalizing the elastic constants at high temperatures. 8

9 tee representing the end of the extra column of atoms). Thus the strain energy released by the dislocations so far is F ar/l = σ xy Ra. (F.5.1) This energy is analogous to the bulkfree energy gained for vapor droplet in a superheated fluid ( The dislocations, however, cost energy (analogous to the surface tension of the vapor droplet). They have a fixed core energy C that depends on the details of the interatomic interaction, and a long-range interaction energy which, for the geometry shown in the figure above, is µ 2π(1 ν) a2 log(r/a). (F.5.2) Here µ is the 2D shear elastic constant and ν is Poisson s ratio. For this problem, assume the temperature is low (so that the energies given by equations (F.5.1) and (F.5.2) are good approximations for the appropriate free energies). By subtracting the energy gained from the dislocation from the energy cost, one finds in analogy to other critical droplet problems a critical radius R c and a barrier height for thermally nucleated dislocation formation B. Of the following statements, which are true? (10 points each) (T) (F) The critical radius R c is proportional to 1/σ xy. (T) (F) The energy barrier to thermal nucleation is proportional to 1/σxy. 2 (T) (F) The rate Γ of thermal nucleation of dislocations predicted by our critical droplet calculation is of the form Γ = Γ 0 (T )(σ xy /µ) D/kBT, for a suitable materialdependent function Γ 0 (T ) and constant D. (T) (F) According to our calculation, the response of a two-dimensional crystal under stress is indistinguishable from that of a liquid: even at low temperatures, the strain rate due to an external shear force is proportional to the stress. The 2D elastic constants µ and ν can be related to their 3D values; in our notation µ has units of energy per unit area. 9

10 (F.6) Scaling Functions and Renormalization-Group Trajectories. (30 points) An Ising model near its critical temperature T c is described by two variables: the distance to the critical temperature t =(T T c )/T c, and the external field h = H/J. Under a coarse-graining of length x =(1 ɛ) x, the system is observed to be similar to itself at a shifted temperature t =(1+aɛ) t and a shifted external field h =(1+bɛ) h, with b>a>0 (so there are two relevant eigendirections, with the external field more relevant than the temperature). (a) (10 points) Which diagram has curves consistent with this flow, for b>a>0? h h h t t t (A) (B) (C) h h t t (D) (E) The magnetization M(t, h) is observed to rescale under this same coarse-graining operation to M =(1+cɛ) M, som ((1 + aɛ) t, (1 + bɛ) h) =(1+cɛ) M(t, h). (b) (20 points) Suppose M(t, h) isknownatt = t 1, the line of filled circles in the various figures in part (a). Give a formula for M(2t 1,h) (open circles) in terms of M(t 1,h). (Hint: Find the scaling variable in terms of t and h which is constant along the renormalization-group trajectories shown in (a). Write a scaling form for M(t, h) in terms of this scaling variable, and find the critical exponents in terms of a, b, andc. From there calculating M(t, h) att =2t 1 should be easy, given the values at t = t 1. By emphasizing self-similarity, we can view these flows in the two-dimensional space of the original model but the self-similarity is only approximate, valid sufficiently near to the critical point. The renormalization-group flows are in the infinite-dimensional space of Hamiltonians, and the irrelevant directions perpendicular to t and h cause corrections far from t =(T T c )/T c = h = 0. It s equivalent to view the flows in part (a) as the flows on the unstable manifold of the renormalizaiton-group fixed point, if you prefer. 10

Physics 562: Statistical Mechanics Spring 2002, James P. Sethna Prelim, due Wednesday, March 13 Latest revision: March 22, 2002, 10:9

Physics 562: Statistical Mechanics Spring 2002, James P. Sethna Prelim, due Wednesday, March 13 Latest revision: March 22, 2002, 10:9 Open Book Exam Work on your own for this exam. You may consult your