Physics 562: Statistical Mechanics Spring 2003, James P. Sethna Homework 5, due Wednesday, April 2 Latest revision: April 4, 2003, 8:53 am

Transcription

1 Physics 562: Statistical Mechanics Spring 2003, James P. Sethna Homework 5, due Wednesday, April 2 Latest revision: April 4, 2003, 8:53 am Reading David Chandler, Introduction to Modern Statistical Mechanics, chapter 8. Problems (5.1) Pollen and Hard Squares. Q B b Square pollen grain in fluid of oriented square molecules, next to a wall. The thin lines represents the exclusion region around the pollen grain and away from the wall. We model the entropic attraction between a pollen grain and a wall using a two-dimensional ideal gas of classical indistinguishable particles as the fluid. For convenience, we imagine that the pollen grain and the fluid are formed from square particles lined up with the axes of the box, of length B and b, respectively. We assume no interaction between the ideal gas molecules (unlike in problem set 1); the potential energy is infinite, though, if the gas molecules overlap with the pollen grain or with the wall. The container as a whole has one pollen grain, N gas molecules, and total area L L. Assume the pollen grain is close to only one wall. Let the distance from the surface of the wall to the closest face of the pollen grain be Q. (A similar square-particle problem with interacting small molecules was studied by D. Frenkel and A. A. Louis, Phase separation in a binary hard-core mixture. An exact result, Phys.Rev.Lett.68, 3363 (1992).) 1

2 (a) What is the area A(Q 0) available for the gas molecules, in units of (length) 2, when the pollen grain is far from the wall? What is the overlap of the excluded regions, A(0) A( ), when the pollen grain touches the wall, Q = 0? Give formulas for A(Q) as a function of Q for the two relevant regions, Q<band Q>b. (b) What is the configuration-space volume Ω(Q) for the gas, in units of (length) 2N? What is the configurational entropy of the ideal gas, S(Q)? (Write your answers here in terms of A(Q) to simplify grading.) Your answers to part (b) can be viewed as giving a free energy for the pollen grain after integrating over the gas degrees of freedom (also known as a partial trace, or coarse-grained free energy). (c) What is the resulting coarse-grained free energy of the pollen grain, F(Q) = E TS(Q), in the two regions Q>band Q<b?UseF(Q) to calculate the force on the pollen grain for Q<b. Is the force positive (away from the wall) or negative? Why? (d) Directly calculate the force due to the ideal-gas pressure on the far side of the pollen grain, in terms of A(Q). Compare it to the force from the partial trace in part (c). Why is there no balancing force from the other side? Effectively how long is the far side of the pollen grain? 2

3 (5.2) Entropy Increases: Diffusion. We saw that entropy technically doesn t increase for a closed system, for any Hamiltonian, either classical or quantum. However, we can show that entropy increases for most of the coarse-grained effective theories that we use in practice: when we integrate out degrees of freedom, we provide a means for the information about the initial condition to be destroyed. Here you ll show that entropy increases for the diffusion equation. Diffusion Equation Entropy. Let ρ(x, t) obey the one-dimensional diffusion equation ρ/ t = D 2 ρ/ x 2. Assume that the density ρ and all its gradients die away rapidly at x = ±. Derive a formula for the time derivative of the entropy S = k B ρ(x)logρ(x) dx and show that it strictly increases in time. (Hint: integrate by parts. You should get an integral of a positive definite quantity.) (5.3) Solving the Diffusion Equation. (Optional: for those for whom Fourier and Greens function methods are unfamiliar.) If needed: Matthews and Walker, Mathematical Methods of Physics, Chapter 8.4 p (Diffusion Equation). Consider a one-dimensional diffusion equation ρ/ t = D 2 ρ/ x 2, with initial condition periodic in space with period L, consisting of a δ function at every x n = nl: ρ(x, 0) = n= δ(x nl). (a) Using the Greens function method, give an approximate expression for the the density, valid at short times and for L/2 <x<l/2, involving only one term (not an infinite sum). (Hint: how many of the Gaussians are important in this region at early times?) (b) Using the Fourier method, give an approximate expression for the density, valid at long times, involving only two terms (not an infinite sum). (Hint: how many of the wavelengths are important at late times?) (c) Give a characteristic time τ in terms of L and D, such that your answer in (a) is valid for t τ and your answer in (b) is valid for t τ. Also, you may assume n ρ/ x n log ρ goes to zero at x = ±, even though log ρ goes to. If you use a Fourier transform of ρ(x, 0), you ll need to sum over n to get δ-function contributions at discrete values of k = 2πm/L. If you use a Fourier series, you ll need to unfold the sum over n of partial Gaussians into a single integral over an unbounded Gaussian. 3

4 (5.4) Coarse-Grained Magnetic Dynamics. If needed: Matthews and Walker, Mathematical Methods of Physics, Section 4.2 (Fourier Transforms) and Chapter 12 (Calculus of Variations). A one-dimensional classical magnet above its critical point is described by a free energy density F[M] =(C/2)( M) 2 +(B/2)M 2 (5.4.1) where M(x) is the variation of the magnetization with position along the single coordinate x. The average magnetization is zero, and the total free energy of the configuration M(x) is F [M] = F[M]dx. The methods we developed in class to find the correlation functions and susceptibilities for the diffusion equation can be applied with small modifications to this (mathematically more challenging) magnetic system. Assume for simplicity that the magnet is of length L, and that it has periodic boundary conditions. We can then write M(x) in a Fourier series M(x) = n= M n exp(ik n x) (5.4.2) with k n =2πn/L and L M n =(1/L) 0 M(x)exp( ik n x). (5.4.3) (a) Show that (as always, for linear systems with translation invariance) the free energy F [M], when written in terms of Mn, becomes an independent sum over modes, with a quadratic energy in each mode. What is M n 2, by equipartition? Argue that M m Mn = k B T L(Ck 2 n + B) δ mn (5.4.4) (b) Calculate the equal-time correlation function for the magnetization in equilibrium, C(x, 0) = M(x, 0)M(0, 0). (First, find the formula for the magnet of length L, in terms of a sum over n. Then convert the sum to an integral: dk k δk = 2π/L k.) You ll want to know that the Fourier transform e ikx /(1 + a 2 k 2 ) dk =(π/a)exp( x /a). Assume the magnetic order parameter is not conserved, and is overdamped, so the time derivative of M is given by the mobility γ times the variational derivative of the free I d call them harmonic oscillators, except that we don t have any kinetic energy in our model, so they don t really oscillate. 4

5 energy: M / t = γδf/δm. M evolves in the direction of the total force on it. The average M is over all future evolutions given the initial condition. The term δf/δm is the variational derivative: δf = F [M + δm] F [M] = F[M + δm] F[M] dx = (δf/δm)δm dx. (5.4.5) (c) Calculate δf/δm. As in the derivation of the Euler-Lagrange equations, you ll need to integrate one term by parts to pull things out of the integral. (d) From your answer to part (c), calculate the Greens function for M, G(x, t) giving the time evolution of an initial condition M(x, 0) = G(x, 0) = δ(x). (Hint: You can solve this with Fourier transforms.) The Onsager regression hypothesis tells us that the time evolution of a spontaneous fluctuation (like those giving C(r, 0) in part (b)) is given by the same formula as the evolution of an imposed initial condition (given by the Greens function of part (d)). C(x, t) = M(x, t)m(0, 0) = M(x, 0)G(x x,t) dx M(0, 0) = C(x, 0)G(x x,t) dx. (5.4.6) (e) Using the Onsager regression hypothesis calculate the space-time correlation function C(x, t) = M(x, t)m(0, 0). (This part is a challenge: your answer will involve the error function.) If it s convenient, plot it for short times and for long times: does it look like exp( y ) in one limit and exp( y 2 ) in another? The fluctuation dissipation theorem can be used to relate the susceptibility χ(x, t) to ta time dependent impulse to the correlation function C(x, t) (see Chandler, p. 257). (f) Calculate the susceptibility χ(x, t) from C(x, t). Start by giving the abstract formula (so we can grade it), and then plug in your answer from part (e). This formula is analogous to taking the gradient of a scalar function of a vector, f( y + δ) f( y) f δ, with the dot product for vector gradient replaced by the integral over x for derivative in function space. 5

6 (5.5) The Ising Model: Correlations, Response, and the Fluctuation-Dissipation Theorem. This problem again uses the program ising, downloadable from This time we ll consider the Ising Hamiltonian in a time-dependent external field H(t), H = J S i S j H(t) ij i S i, 3.7.1, and look at the fluctuations and response of the time-dependent magnetization M(t) = i S i(t). Our Ising model simulation outputs both the time-dependent magnetization per spin m(t) =(1/N ) i S i and the time-time correlation function of the magnetization per spin, g(t) = (m(0) m )(m(t) m ). (5.5.1) (select autocorrelation in the menu above the plot). We ll be working above T c,so m 0. Note, as before, that most of our formulas in class are in terms of the total magnetization M = Nm and its correlation function G = N 2 g. The time-time correlation function will start non-zero, and should die to zero over time. Suppose we start with a non-zero small external field, and turn it off at t =0: H(t) = H 0 Θ( t)? The magnetization M(t) will be non-zero at t = 0, but will decrease to zero over time. By the Onsager regression hypothesis, these two time decays should look the same. Run the Ising model, changing the size to Equilibrate at T = 3 and H = 0, reset, do a good measurement of the time-time autocorrelation function, and copy graph. Rescale the graph using configure to focus on the short times where things are interesting. Now equilibrate at T =3,H =0.05, set H =0andreset, and run for a short time. (a) Does the shape and the time-scale of the magnetization decay look the same as that of the autocorrelation function? Note down the values for g(0), G(0), m(0), and M(0). Response Functions and the Fluctuation-Dissipation Theorem. The response function χ(t) gives the change in magnetization due to an infinitesimal impulse in the external field H. By superposition, we can use χ(t) to generate the linear response to any external perturbation. If we impose a small time-dependent external field H(t), the average magnetization M(t) M = t dt χ(t t )H(t ), (5.5.2) where M is the equilibrium magnetization without the extra field H(t) (zero for us, above T c ). (b) Using equation (5.5.2), write M(t) for the step down H(t) =H 0 Θ( t), in terms of χ(t). 6

7 The fluctuation-dissipation theorem states that this response χ(t) = βdg(t)/dt. (5.5.3) where G(t) = (M(0) M )(M(t) M ) is the time-time correlation function for the total magnetization. (c) Use equation (5.5.3) and your answer to part (b) to predict the relationship between the demagnetization M(t) and the correlation G(t) you measured in part (a). How does your analytical ratio compare with the t =0ratioyounoteddowninpart(a)? 7