UNIVERSITY OF OSLO FACULTY OF MATHEMATICS AND NATURAL SCIENCES

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1 UNIVERSITY OF OSLO FCULTY OF MTHEMTICS ND NTURL SCIENCES Exam in: FYS430, Statistical Mechanics Day of exam: Jun Problem :. The relative fluctuations in an extensive quantity, like the energy, depends on particle number as / N. quantity like temperature? How is the N-dependence in the relative fluctuations of an extensive Solution: For a small subsystem, the energy fluctuates around its equilibrium value E with a mean square fluctuation given by where k is the Boltzmann s constant and C V E 2 = kt 2 C V, () is the heat capacity which is proportional to N. Thus the mean square fluctuations of energy are proportional to the number of particles. Since the mean total energy is also E N, it follows that the relative energy fluctuations E2 / E / N. n instantaneous fluctuation in energy of a subsystem, that has fixed N and V and that may exchange energy with a larger system, induces an instantaneous temperature fluctuation about its equilibrium value given as T = Hence, the mean square fluctuations in temperature are ( ) T E = E. (2) E V,N C V T 2 = kt 2 C V. (3) Thus, the mean square temperature fluctuations scale inversely proportional with N, unlike E 2 which are proportional to N. However, since the average temperature is fixed by the thermodynamic equilibrium and is thus N-independent, the relative temperature fluctuations scale as T 2 /T / N same as for energy. Eq. 3 describes the accuracy

2 with which the temperature of an isolated system can be defined. Since T 2 /N, the uncertainty in defining a temperature increases as the number of particles is reduced; temperature has a well defined meaning in the thermodynamic limit..2 What is the equipartition theorem? Solution: The equipartition theorem states that, in thermal equilibrium, any degree of freedom which appears only quadratically in the energy has an average energy of kt/2. For example, the Hamiltonian of a classical ideal gas contains only the kinetic energy H = m(vx 2 + vy 2 + vz)/2. 2 Thus, the average kinetic energy per particle is e = 3kT/2..3 Show that Gibbs entropy formula S = k s P s ln P s (4) is correct for both the microcanonical and canonical ensemble. Start from the expressions for P s. Solution: In the canonical ensemble, the probability P s is P s = Z e βes. (5) Thus, S = k s P s ( ln Z βe s ) = k ln Z + β E s, (6) where we used the normalization condition s P s =. Using the definition of the Helmholtz free energy F = kt ln Z, we arrive at F = E s ST. (7) In the microcanonical ensemble, P s is constant by virtue of Liouville s theorem and equal to P s = /W, (8) where W is the number of accessible microstates. Thus, the Boltzmann s formula for the configurational entropy S = k s W ln W = kw W ln W 2 = k ln W. (9)

3 .4 What does Liouville s theorem express, and how does it relate to the fundamental hypothesis that all microstates may be taken as equally probable? Solution: Liouville s theorem applies to all Hamiltonian systems and states the conservation of the phase space volume (or equivalently the probability density of states). For a Hamiltonian system given by H = 3 α= 2 p2 α + U(q,, q 3N ), (0) with p α and q α being the generalized momentum and position coordinates of a system of N particles in 3D, the probability density in phase space ρ(q,, q 3N, p, p 3N ) is locally conserved, namely satisfies the continuity equation Dρ Dt = ρ 3 ( (ρ t + qα ) + (ρṗ ) α) = 0, () q α p α α= where Dρ/Dt is the material or total derivative with respect to time; it is the evolution of ρ seen by a particle moving with flow. Hence by virtue of the Liouville s theorem, Dρ/Dt = 0, it follows that the flow in phase space is conservative (a small volume in the phase space can only change its shape) and that the microcanonical ensembles are time independent (an initial uniform density in phase space will remain uniform). Thus ρ being a constant (equal probability for all microstates) is a particular solution of the Liouville s theorem. Problem 2: 2. classical model for diffusive processes is the random walker where the position x i at an instant i is updated as x i+ = x i + δx i. The steps δx i = ±a are uncorrelated, δx i δx j = a 2 δ ij. If time is defined as t i = i t, x 0 = 0, show that x 2 behaves diffusively (linear in time) and derive the diffusion constant as a function of a and t. Solution: Given that x 0 = 0, the position of the walker after N steps is x N = n δx i, (2) i= 3

4 thus x 2 N = = i= i= = Na 2 δx i δx j j= a 2 δ ij j= Thus, the diffusivity is D = a2 2 t and x2 t = 2Dt. = a2 t t N. (3) 2.2 State the central limit theorem and explain how the result in 2. may be generalized to variable a values. Solution: The central limit theorem states that the distribution P (s N ) of the sum s N = N i= x i of independent, identically distributed variables {x i } drawn from an arbitrary distribution p(x) that has zero mean and a finite second moment x 2 < converges to the Gaussian distribution P (s N ) = 2πσ 2 N e s2 N /2σ2 N (4) where σ 2 N = s2 = N x 2 = Nσ 2. In 2., the steps are assumed to be of constant length a, thus the parent distribution p( x) = /2 (δ( x + a) + δ( x a)), where δ(x) is the Dirac delta function. similar expression as in Eq. 3 holds for a random walker with an arbitrary length a of the step. In this case, we have that x 2 = N a Explain, at least in one way, how the Langevin equation mdv/dt = αv + F (t), where F (t) is a random force of zero mean and α is a friction constant, extends the description of a random walker. From this equation it is possible to show that x 2 (t) = 2kT α ( t m α ( e αt/m )) (5) Derive the short and long time limits of this equation and discuss the result. Solution: The Langevin equation as written above describes the Brownian motion of a particle in a fluid. There are two forces that drive the motion of the random walker: F (t) is due to random collisions with the molecules of the fluid, and αv is the viscous drag force due to the surrounding fluid. The random walk in 2. is obtained in the limit where the 4

5 inertial effects (md 2 x/dt 2 ) are ignored. The inertial effect is important on short timescales, i.e. t m/α. In the time limit t m/α, Eq. 5 simplifies to ballistic scaling x 2 (t) 2kT ( t m ( + αt )) α α m α2 t 2 2m 2 kt m t2. (6) However in the long time t m/α, we recover the diffusive behavior x 2 (t) 2kT α t. (7) Problem 3: non-relativistic gas of N particles with mass m in the gravitational field of acceleration g is contained in a cylinder of area and height L = L 2 L. The axis of the cylinder is the z axis which points upwards and is parallel to the force of gravity. The bottom of the cylinder is at z = L and the top at z = L 2. The energy of a particle with momentum p is ɛ = p2 + mgz. (8) 2m 3. Calculate the classical particle partition function Z as well as Z N, the corresponding N particle partition function. Solution: The one particle partition function follows from the phase-space integration d 3 p L2 Z = /2m (2π h) 3 e βp2 dze βmgz L ( ) 3/2 m ( = e βmgl 2π h 2 e ) βmgl 2. (9) β βmg The N particle partition function for independent and indistinguishable particles is then Z N = Z N /N!. 3.2 Find the g 0 limit of Z and interpret the result. Show that there is a typical length l that defines when we can neglect gravity if only L, L 2 l. 5

6 Solution: In the g 0 limit, the exponentials can be Taylor expanded to give Z ( ) 3/2 m 2π h 2 β βmg (βmgl 2 βmgl ) ( ) 3/2 m 2π h 2 (L 2 L ) β V Λ 3, (20) where Λ = 2π h 2 β/m is the thermal wavelength. Thus, in the g 0 limit we obtain the ideal gas partition function. The characteristic length-scale is l = βmg. Thus, g 0 corresponds to L, L 2 l. 3.3 Write down the general formula that relates the pressure and Z N. Explain why the pressure at the bottom is while Calculate P and P 2. P = NkT P 2 = NkT Solution: Pressure follows from the Helmholtz free energy as ( ) F P = V ln Z L (2) ln Z L 2. (22) where the free energy is related to the N particle partition function by Using that V = (L 2 L ) we have that T (23) F = kt ln Z N = kt ln ZN N!. (24) P = NkT ln Z V = NkT ln Z L = NkT ln ( ) e βmgl e βmgl 2 L = NkT βmge βmgl = Nmg e βmgl e βmgl 2 (25) e βmgl e βmgl e βmgl 2, (26) 6

7 and P 2 = NkT ln Z V = NkT ln Z L 2 = NkT ln ( ) e βmgl e βmgl 2 L 2 = Nmg e βmgl 2 e βmgl e βmgl 2 (27) 3.4 Obtain the g 0 limit of P i and comment on the result. Solution: For g 0 and to the leading order we have, P i Nmg βmgl 2 βmgl N β(l 2 L ) (28) thus P i V = NkT for both i = and i = 2. Thus, no pressure difference. 3.5 Calculate the net force on the cylinder from the gas from the pressure difference in the system and explain how this result could have been calculated in a simpler way. Solution: The pressure difference is calculated from P i as and equal P = P P 2 = Nmg e βmgl e βmgl2 e βmgl e βmgl2 (29) P = Nmg. (30) This is equivalently to having that the net force exerted by a column of particles under gravity is F = p = Nmg. 7