In-class exercises. Day 1

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1 Physics 4488/6562: Statistical Mechanics Material for Week 13 Exercises due Mon Apr 30 Last correction at April 24, 2018, 12:59 pm c 2018, James Sethna, all rights reserved Pre-class Preparation All exercises are from Version 2.0 of the text: sethna/ StatMech/v2EntropyOrderParametersComplexity.pdf Wednesday Read: Chapter 12, Introduction Pre-class question: 12.16: The Gutenberg Richter law. (Submit electronically by 9:30 Tuesday evening.) Friday Read: Chapter 12, Sec (Universality) Pre-class question: 12.7: Renormalization-group trajectories. (Submit electronically by 9:30 Thursday evening.) Monday Read: Chapter 12, Sec (Scale Invariance) Pre-class question: 12.3: Scaling and coarsening. (Submit electronically by 9:30 Sunday evening.) Exercises Those in 4488 may choose two of the four exercises. 11.9: Snowflakes and linear stability : The renormalization group and the central limit theorem: long. 12.9: Period doubling and the renormalization group. (Hints are available in Python and Mathematica: sethna/statmech/computerexercises.html.)

2 In-class exercises Day Origami microstructure. 1 (Mathematics, Engineering) i Figure shows the domain structure in a thin sheet of material that has undergone a martensitic phase transition. These phase transitions change the shape of the crystalline unit cell; for example, the high-temperature phase might be cubic, and the low-temperature phase might be stretched along one of the three axes and contracted along the other two. These three possibilities are called variants. A large single crystal at high temperatures thus can transform locally into any one of the three variants at low temperatures. The order parameter for the martensitic transition is a deformation field y(x), representing the final position y in the martensite of an original position x in the undeformed, unrotated austenite. The variants differ by their deformation gradients y representing the stretch, shear, and rotation of the unit cells during the crystalline shape transition. In this exercise, we develop an analogy between martensites and paper folding. Consider a piece of graph paper, white on one side and gray on the other, lying flat on a table. This piece of paper has two distinct low-energy states, one variant with white side up and one variant with gray side up. θ l θ c θ w SO(2) SO(2) P Fig. 1 Paper order parameter space. The allowed zero-energy deformation gradients for a piece of paper lying flat on a table. Let θ be the angle between the x axis of the graph paper and the near edge of the table. The paper can be rotated by any angle θ l (so the deformation gradient is a pure rotation in the group 2 SO(2)). Or, it can be flipped over horizontally ((x, y) (x, y), multiplying by P = ( ) ) and then rotated by θw (deformation gradient in the set SO(2) P ). Our two variants are hence given by the identity rotation I and the reflection P ; the ground states rotate the two variants. An interface between two of these ground states is a straight crease at angle θ c (Fig. 2). 1 This exercise was developed in collaboration with Richard D. James. A one-sided patterned sheet suitable for creasing and a printable version of Fig. 3 (both the full version and the foldable version with two levels) can be found at the book Web site [2].

3 The (free) energy density for the paper is independent of rotations, but grows quickly when the paper is stretched or sheared. The paper, like martensites, can be represented as a deformation field y(x), representing the final position y of a point x of the paper placed horizontally on the table with the gray side up. Naturally y(x) must be a continuous function to avoid ripping the paper. Since the energy is independent of an overall translation of the paper on the table, it can depend only on gradients of the deformation field. To lowest order, 3 the energy density can be written in terms of the deformation gradient y = j y i : F = α ( y) y I 2 = α( i y j i y k δ jk ) 2. (1) The constant α is large, since paper is hard to stretch. In this problem, we will be interested in the zero-energy ground states for the free energy. (a) Show that the zero-energy ground states of the paper free energy density (eqn 1) include the two variants and rotations thereof, as shown in Fig. 1. Specifically, show (1) that any rotation y i (x j ) = R ij x j of the gray-side-up position is a ground state, where R ij = ( cos θ l sin θ l ) sin θ l cos θ l, and (2) that flipping the paper to the white-side-up and then rotating, y i (x j ) = R ik P kj x j = ( cos θ w sin θ w )( 1 0 x sin θ w cos θ w 0 1)( y) also gives a ground state. (Hint: If y(x) = Mx = M ij x j for a linear transformation M, what is y = j y i?) ) and P = ( 1 0 Hence our two variants are the rotations I = ( ). In the real martensite, there are definite rules (or compatibility conditions for boundaries between variants: given one variant, only certain special orientations are allowed for the boundary and the other variant. A boundary in our piece of paper between a gray-up and white-up variant lying flat on the table is simply a crease (Fig. 2). θ w θ c Fig. 2 Paper crease. An interface between two ground states θ l = 0 and θ w for our paper on the table is a straight crease with angle θ c. (b) Place a piece of paper long-edge downward on the table. Holding the left end fixed θ l = 0, try folding it along crease lines at different angles θ c. Find a definite relation between the crease angle θ c and the angle θ w of the right-hand portion of the paper. Finding the displacement y(x) for the right-hand portion of the paper demands that 3 Including higher derivatives of the deformation field into the energy density would lead to an energy per unit length for the creases.

4 we know where the crease is. The right-hand portion has been flipped and rotated. Find y w = R θw P in terms of θ c. Suppose the crease is along an axis ĉ. We can derive the compatibility condition governing a crease by noting that y along the crease must agree for the white and the gray faces, so the directional derivative y c = (ĉ )y must agree. (c) Given the relation you deduced for the geometry in part (b), show that the difference in the directional derivatives ( y l y w ) is zero along c, ( y l y w ) c = ( j y l i j y w i )c j = 0. (Hints: y l is the identity. cos(2θ) = cos 2 θ sin 2 θ, sin(2θ) = 2 sin θ cos θ.) In general, two variants with deformation gradients A and B of a martensite can be connected together along a flat boundary perpendicular to n if there are rotation matrices R (1) and R (2) such that 4 R (1) B R (2) A = a n, k R (1) ik B kj k R (2) ik A kj = a i n j, (2) n. where a n is the outer product of a and n. This compatibility condition ensures that the directional derivatives of y along a boundary direction c (perpendicular to n) will be the same for the two variants, ( y 1 y 2 )c = (R (1) B R (2) A)c = a(n c) = 0 and hence that the deformation field is continuous at the boundary. For our folded paper, y is either RI or RP for some proper rotation R, and hence eqn 2 is just what you proved in part (c). As can be seen in Fig , the real martensite did not transform by stretching uniformly along one axis. Instead, it formed multiple thin layers of two of the variants. It can do so for a modest energy cost because the surface energy of the boundary between two variants is low. The martensite is driven to this laminated structure to satisfy boundary conditions. Steels go through a martensitic transition; as the blacksmith cools the horseshoe, local crystalline regions of the iron stretch along one of several possible axes. The red-hot horseshoe does not change shape overall as it is plunged into the water, though. This is for two reasons. First, if part of the horseshoe started stretching before the rest, there would be big stresses at the boundary between the transformed and untransformed regions. Second, a horseshoe is made up of many different crystalline grains, and the stretching is along different axes in different grains. Instead, the horseshoe, to a good approximation, picks a local mixture between the different variants that overall produces no net average stretch. 4 That is, the difference is a rank one matrix, with zero eigenvalues along all directions perpendicular to

5 Fig. 3 Origami microstructure. formation, by Richard D. James. Two-dimensional origami example of microstructure This is done by creating finely-divided structures, like the laminated structure seen in Fig At the boundaries of the square region, the martensite must not stretch, so it produces a fine laminated structure where the stretching in one domain cancels the contraction for its neighbors. Our paper folding example forms a similar microstructure when we insist that the boundary lie along a curve other than the natural one. (d) Print out a full-sized simplified version of Fig. 3. Cut out the hexagon, and fold along the edges. Where does the boundary go? 6 The mathematicians and engineers who study these problems take the convenient limit where the energy of the boundaries between the variants (the crease energy in our exercise) goes to zero. In that limit, the microstructures can become infinitely fine, and only quantities like the relative mixtures between variants are well defined. It is a 5 The laminated microstructure of the real martensite is mathematically even more strange than that of the paper. The martensite, in the limit where the boundary energy is ignored, has a deformation gradient which is discontinuous everywhere in the region; our folded paper has a deformation gradient which is discontinuous only everywhere along the boundary. See Exercise Deducing the final shape of the boundary can be done by considering how the triangles along the edge overlap after being folded. Note that the full structure of Fig. 3 cannot be folded in three dimensions, because the paper must pass through itself (Chen Wang, private communication, see [?]). The analogous problem does not arise in martensites.

6 wonderful example where the pathological functions of real analysis describe important physical phenomena. Day Ising self-similarity. 7 (Computation) i Start up the Ising model. Run a large system at zero external field and T = T c = 2/ log(1 + 2) Set the refresh rate low enough that graphics is not the bottle-neck, and run for at least a few hundred sweeps to equilibrate. You should see a fairly self-similar structure, with fractal-looking up-spin clusters inside larger down-spin structures inside... Can you find a nested chain of three clusters? Four? Day Hearing chaos. 8 (Dynamical systems) i This exercise listens to a system as it transitions from regular to chaotic behavior. Chaos scrambles our knowledge of the initial state of a system (Exercise 5.9), allowing entropy to increase and mediating equilibration. Also, the onset of chaos (Exercises and 12.9) is studied using the same renormalization-group methods we use to study continuous phase transitions (Chapter 12). The English word chaos makes us think of billiard balls, bumble bees, bumper cars, or turbulence the realm of statistical mechanics. The mathematical term chaos in dynamical systems often reflects a more subtle irregularity of motion. Think of an unbalanced dryer, or a faucet that drips at irregular intervals. One of the characteristics of chaotic motion is a continuous spectrum of frequencies. (Or, perhaps more precisely, systems that are non-chaotic but non-stationary usually have a spectrum that consists of sums and differences of a few frequencies.) Our ears are spectacularly sensitive to frequencies. Listen to the audio files that take the time series for the logistic map, slowly cross through a sequence of period-doubling bifurcations, and end in a chaotic noise. (a) If the period of a sound wave doubles, how much does the pitch change? (One note? A perfect fourth? A perfect fifth? An octave?) Listen to the two audio files. (One starts closer to the onset of chaos.) (b) Does the pitch start sounding like a pure tone? Can you hear the first period doubling? Does the new pitch agree with your answer to part (a)? Shortly after the first period doubling, you should begin to hear a chaotic signal. (c) Describe the new sound. Is it musical, or noisy? 7 A link to the software can be found at the book Web site [2]. 8 The book Web site [2] provides links to the audio files for this exercise, created by Erich Mueller.

7 Figure shows the states visited at long times by the logistic map, as the parameter µ is varied along the horizontal axis. It does not show the dynamics as the system hops between one point and another along the curves, but you can reconstruct that. For example, between µ 1 and µ 2 the system hops between the upper and lower branches of the curve, giving a tone with period equal to one iteration of the map. Between µ 2 and µ 3 you heard a second tone one octave lower. You probably did not hear the next period doubling, but after µ you began to hear noise. (d) Note the windows of non-chaotic motion deep inside the chaotic region. In the audio file, do you hear periods of tonal sounds interspersed in the noise? Crackling noises. 9 i Day 2 Listen to the sound file for the earthquakes in 1995 (shown graphically in Fig Each click represents an earthquake, with the sound energy proportional to the energy release and with the year compressed into a couple of seconds. (a) Roughly how many clicks can you hear? Consider the histogram in Fig. 12.3(b), down to what magnitude do you estimate you can perceive? Now listen to paper crumpling, Rice Krispies c, fire, and to our model of magnetic noise. All of these sound files share common features they are composed of brief pulses or avalanches with a broad range of sizes. Just as for earthquakes, the largest avalanches are rare, and the smaller avalanches are more common, with avalanches of size S happening with probability proportional to S τ. These sound files differ, however, in their values for the exponent τ. For earthquakes, τ can vary from one fault type to another; our data set (Fig. 12.3(b)) shows τ between 1.5 and 1.8 (see Exercise 12.16). Milk invading bubbles in puffed rice is likely a special case of the well-studied problem of fluid invasion into porous media (also known as imbibition), where in three dimensions estimates of τ range between around 1.3 and 1.5. Paper crumpling and fire are not well understood, but we have an excellent understanding of our model of magnetic noise, 10 where τ 2. Changing τ should change the sound of the crackling noise changing the predominance of the loudest crackles over the rest. (b) Should larger τ correspond to fewer very large avalanches, or more? Does that correspond roughly to your perception of the difference between the sounds from earthquakes and magnets? What about Rice Krispies c? Day 3 9 The computer exercises link on the book Web site [2] provides links to audio files of various types of crackling noise for this exercise. 10 Barkhausen noise in magnets is usually studied not using sound emission, but by measuring jumps in the magnetization as the external field is increased. Also, real magnets are not described by our model, but are in the same universality class as fluid invasion.

8 12.17 Random walks and critical exponents. 3 Self-similar behavior also emerges without proximity to any obvious transition. One might say that some phases naturally have self-similarity and power laws. Mathematicians have a technical term generic which roughly translates to without tuning a parameter to a special value, and so this is termed generic scale invariance. The simplest example of generic scale invariance is that of a random walk. Figure 2.2 shows that a random walk appears statistically self-similar. Let X(T ) = T t=1 ξ t be a random walk of length T, where ξ t are independent random variables chosen from a distribution of mean zero and finite standard deviation. Derive the exponent ν governing the growth of the root-mean-square end-to-end distance d(t ) = (X(T ) X(0))2 with T. Explain the connection between this and the formula from freshman lab courses for the way the standard deviation of the mean scales with the number of measurements. Day Mean-field theory. (Condensed matter) i In Chapter 11 and Exercise 9.5, we make reference to mean-field theories, a term which is often loosely used for any theory which absorbs the fluctuations of the order parameter field into a single degree of freedom in an effective free energy. The original mean-field theory actually used the mean value of the field on neighboring sites to approximate their effects. In the Ising model on a square lattice, this amounts to assuming each spin s j = ±1 has four neighbors which are magnetized with the average magnetization m = s j, leading to a one-spin mean-field Hamiltonian H = 4Jms j. (3) (a) At temperature k B T, what is the value for s j in eqn 3, given m? At what temperature T c is the phase transition, in mean field theory? (Hint: At what temperature is a non-zero m = s self-consistent?) Argue as in Exercise 12.4 part (c) that m (T c T ) β near T c. Is this value for the critical exponent β correct for the Ising model in either two dimensions (β = 1/8) or three dimensions (β 0.325)? (b) Show that the mean-field solution you found in part (a) is the minimum in an effective temperature-dependent free energy ( m 2 V (m) = k B T 2 log (cosh(4jm/k B T )) k BT 4J ). (4) On a single graph, plot V (m) for 1/(k B T ) = 0.1, 0.25, and 0.5, for 2 < m < 2, showing the continuous phase transition. Compare with Fig

9 (c) What would the mean-field Hamiltonian be for the square-lattice Ising model in an external field H? Show that the mean-field magnetization is given by the minima in 11 ( m 2 V (m) = k B T 2 log (cosh((h + 4Jm)/k B T )) k BT 4J ). (5) On a single graph, plot V (m, H) for β = 0.5 and H = 0, 0.5, 1.0, and 1.5, showing metastability and an abrupt transition. At what value of H does the metastable state become completely unstable? Compare with Fig. 11.2(a). References [1] Mézard, M. and Montanari, A. (To be published (2006)). Constraint satisfaction networks in Physics and Computation. Clarendon Press, Oxford. [2] Sethna, J. P. and Myers, C. R. (2004). Entropy, Order Parameters, and Complexity computer exercises: Hints and software. StatMech/ComputerExercises.html. 11 One must admit that it is a bit weird to have the external field H inside the effective potential, rather than coupled linearly to m outside.

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