763620SS STATISTICAL PHYSICS Solutions 5 Autumn 2012

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1 7660SS STATISTICAL PHYSICS Solutions 5 Autumn 01 1 Classical Flow in Phase Space Determine the trajectories of classical Hamiltonian flow in -dim corresponding to a particle in a constant gravitational field phase space (q, p), H = p m + mgq Calculate how a phase space area moves in time, if at time t = 0 it is a triangle with corners at (q 0, p 0 ), (q 0 + a, p 0 ), (q 0, p 0 + b) Show that the surface area is conserved Solution: Hamilton s equations of motion imply q(t) = q 0 + p 0 m t 1 gt p(t) = p 0 mgt 5 C q 0, p 0 b t=0 4 q 0, p 0 q 0 a, p 0 mgt C' 1 A' a b t B' p 0 t 1 gt q Every point within the triangle moves downwards mgt Every point on the line AB moves the same distance p 0 m t 1 gt 1

2 Point C moves to the right more than the line AB, the distance being ( ) p0 + b t 1 m gt The both triangles have the same width a and height b, and thus, the surface area is conserved Distribution at constant p and T a) Derive the distribution of microstates for a system whose energy E and volume V are allowed to fluctuate b) Gibbs potential G(p, T ) is a Legendre trasform of the Helmholtz free energy A(V, T ), substituting the free variable V p Show that this corresponds to a Laplace transform e βg(p,t ) = What is the physical interpretation? 0 dv e βpv e βa(t,v ) Solution:a) We find the distribution by maximizing the entropy with constraints E r P r = constant r V r P r = constant r P r = 1 The method of Lagrange s multipliers gives r 0 = r,s δ(k B P r,s ln P r,s + λe r P r,s + λ V s P r,s + λ P r ) = r,s (k B ln P r,s + k B + λe r + λ V s + λ )δp r,s This implies where V s P r,s = e λer λ, = r,s e λer λ V s Let us identify λ and λ We know that S = k B P r,s ln P r,s = = k B (λ E + λ V + ln ) r,s

3 and ( ) 1 S T = E V = k B (λ + E λ E + V λ E + 1 [ λ λ E + ] λ λ E = k B λ λ = β Also, Because we have that p = k B T ln = E T S + λ β V ( ) A V T so λ /β is the pressure We obtain A = E T S, = λ β + V β λ V 1 λ λ V = λ β G(p, T ) = E T S + pv = k B T ln b) According to a) in the continuum limit of the volume = e βg(p,t ) = dv e β(er+pv ) V r = dv e βpv Z(T, V, N) V = dv e βpv e βa, V where Z(T, V, N) is the canonical partition function, and the free energy A = k B T ln Z Probabilities at different volumes are weighted with factors e βpv because energy is needed to increase the volume Statistics of DNA DNA-molecule chain can be described by a zipper with N links The energy of the link is 0 if it is closed and ɛ if it is open The chain can open from one direction only, so that link number s can open only if all links (1,,, s 1) are already open Show that the partition function is Z = 1 e β(n+1)ɛ 1 e βɛ Calculate the number of open links N open as a function of β number at limits T 0 and T? What happens to this

4 Solution: The possible energies are E s = sɛ where s = 0,, N Since the link s can be open only if the previous links are open, the energies E s are non-degenerate Now, the partition function is Z = N e βes = 1 e β(n+1)ɛ 1 e βɛ The number N open of open links is given by the expectation value N open = s = 1 Z se βsɛ = 1 Z (βɛ) Z = 1 x [ ] Nx N+ (N + 1)x N+1 + x, 1 x N+1 (1 x) where x = e βɛ Now, if T 0 we have that β and x 0 Accordingly, N open 0 If T, then β 0 and x 1 We obtain [ ] Nx N+ (N + 1)x N+1 + x N open = lim x 1 (1 x)(1 x N+1 ) = l H N(N + )x N+1 (N + 1) x N + 1 lim x 1 ( + N)x N+1 (N + 1)x N 1 = l H N(N + )(N + 1) N(N + 1) (N + )(N + 1) N(N + 1) = N where l H denotes l Hospital s rule for limits involving indeterminate forms We could calculate also Z directly by noticing that e βes 1 when β 0, since E s 0 This implies N Z = 1 = N + 1 Then, N open = s = 1 N + 1 N s = N(N + 1) (N + 1) = N Why is the N open = N/ in the high T limit, instead of N? 4

5 N open 5 4 N=10 1 N=6 N= k B T Ε 4 Heat Capacity of Anharmonic Oscillator The potential of a one-dimensional, anharmonic oscillator may be written as V (q) = cq gq fq 4, where c, g, f are positive constants; quite generally, g and f may be assumed to be very small in value Show that the leading contribution of anharmonic terms to the heat capacity of the oscillator, assumed classical, is given by ( f k B c + 5 ) g T 4 c and, to the same order, the mean value of the position coordinate q is given by gk B T 4 c (Hint: You might want to use Mathematica to make the relevant approximations) Solution: The partition function can be calculated in the phase space as We concentrate on the q-integral expand Z(β) = 1 e β m p dp e β(cq gq fq 4) dq h = 1 m e β(cq gq fq 4) dq h β e β(gq +fq 4) 1 + β(gq + fq 4 ) + 1 β (gq + fq 4 ) First acknowledging the smallness of g and f, we = 1 + β(gq + fq 4 ) + 1 β (g q 6 + fgq 7 + f q 8 ) 5

6 When the power of q is odd, its integral with the Gaussian vanishes due to antisymmetry of the product We just evaluate the integrals with even powers of q Generally for even powers we have 1 q n e q /σ dq = σ n+1 (n 1)!! where k!! is the double-factorial, ie product of odd integers from 1 to k In our case σ = 1/ βc and we obtain q 4 e q /σ dq = 4(βc) 5/ q 6 e q /σ dq = 15 8(βc) 7/ q 8 e q /σ dq = (βc) 9/ We obtain Z(β) = 1 [ m h β βc + β (fq βg q ] βf q 8 )e βcq dq = 1 [ m 1 c β + f 4β c + 15g 16β c + 105f ] β c 4 1 [ m 1 c β + A β + B ] β Now, recall that C V = E/ T and E = ln Z/ β Denote with x = 1/β = k B T 0, A = f/4c + 15g /16c, B = 105f /c 4, and expand ( ) 1 m ln Z = ln + ln(x + Ax + Bx ) c ( ) 1 m ln + ln x + Ax + ( A / + B)x c Thus, we obtain E = (1/x) ln Z = k B(T + Ak B T + ( A / + B)kBT ) and up to first order in T C V = E T = k B + [ ] f k BT c + 5g 4c Similarly, at small T limit Z 1 ( ) 1 1 m β c 6

7 and the expectation value for q is q = 1 β h e m p dp gq fq 4) qe β(cq dq Z ( 1 m c = k B T g 4c ) 1 β 1 h [ m βg ] β 4(βc) 5/ 5 Latent Heat in Phase Transitions Show that the latent heat must always be positive (heat is absorbed) when making a transition from a low-temperature phase to a high-temperature phase Solution: We are dealing with a transition at constant pressure and temperature, and thus the equilibrium state is that of minimum Gibbs energy We denote with I the high-temperature phase, and with II the low-temperature one Above the transition temperature G I < G II and below G II < G I Thus, the rate of change of the Gibbs energy of phase II has to be larger than that of phase I, on both sides of the transition temperature ( ) ( ) GI GII T T Thus, S I > S II and p,n j < S = S I S II = T H p,n j is always positive when going from the low-temperature phase to a high-temperature phase 7