8.333: Statistical Mechanics I Re: 2007 Final Exam
Review Problems
The enclosed exams (and solutions) from the previous years are intended to help you review the material. ******** Note that the first parts of each problem are easier than its last parts. Therefore, make sure to proceed to the next problem when you get stuck. You may find the following information helpful: Physical Constants 31 27 Electron mass me 9.1 10− Kg Proton mass mp 1.7 10− Kg ≈ × 19 ≈ × 34 1 Electron Charge e 1.6 10− C Planck’s constant/2π ¯h 1.1 10− Js ≈ × 8 1 ≈ × 8 2 4 Speed of light c 3.0 10 ms− Stefan’s constant σ 5.7 10− W m− K− ≈ × 23 1 ≈ × 23 1 Boltzmann’s constant k 1.4 10− JK− Avogadro’s number N 6.0 10 mol− B ≈ × 0 ≈ × Conversion Factors 5 2 10 4 1atm 1.0 10 Nm− 1A˚ 10− m 1eV 1.1 10 K ≡ × ≡ ≡ × Thermodynamics dE = T dS+dW For a gas: dW = P dV For a film: dW = σdA − Mathematical Formulas
2x 4x 1 x 2 limx coth x = 1 + 2e− + e− limx 0 coth x = + + x →∞ O → x 3 O √π ∞ n αx n! 1 0 dx x e− = αn+1 ( 2)! = 2
R 2 2 2 ∞ x √ 2 σ k dx exp ikx 2σ2 = 2πσ exp 2 limN ln N! = N ln N N −∞ − − − →∞ − h i h i R n n ikx ( ik) n ikx ( ik) n e− = ∞ − x ln e− = ∞ − x n=1 n! � � n=1 n! � �c m−1 α η η P1 x α+1 z dfm 1 η P ∞ ∞ fm(z) = (m 1)! 0 dx z−1 ex η = α=1 η αm dz = z fm 1 − − −
R m 2 P 2 4 (ln z) π 2 π 7π limz fm−(z) = 1 + m(m 1)(ln z)− + f2−(1) = f4−(1) = →∞ m! 6 − · · · 12 720 h i 2 4 ζ f +(1) ζ 2.612 ζ = π ζ 1.341 ζ 1.202 ζ = π m ≡ m 3/2 ≈ 2 6 5/2 ≈ 3 ≈ 4 90
1 8.333: Statistical Mechanics I Fall 1998 Final Exam
1. Exciton dissociation in a semiconductor: By shining an intense laser beam on a semicon ductor, one can create a metastable collection of electrons (charge e, and effective mass − me) and holes (charge +e, and effective mass mh) in the bulk. The oppositely charged particles may pair up (as in a hydrogen atom) to form a gas of excitons, or they may dissociate into a plasma. We shall examine a much simplified model of this process.
(a) Calculate the free energy of a gas composed of Ne electrons and Nh holes, at temper
ature T , treating them as classical non-interacting particles of masses me and mh. The canonical partition function of gas of non-interacting electrons and holes is the • product of contributions from the electron gas, and from the hole gas, as
1 V Ne 1 V Nh Ze h =ZeZh = , − N ! λ3 · N ! λ3 e e h h
where λα = h/√2πmαkBT (α =e, h). Evaluating the factorials in Stirling’s approximation, we obtain the free energy
Ne 3 Nh 3 Fe h = kBT ln Ze h = NekBT ln λe + NhkBT ln λh . − − − eV eV
(b)By pairing into an excition, the electron hole pair lowers its energy by ε. [The binding energy of a hydrogen-like exciton is ε me4/(2¯h2ǫ2), where ǫ is the dielectric constant, 1 1 1 ≈ and m− = me− + mh− .] Calculate the free energy of a gas of Np excitons, treating them as classical non-interacting particles of mass m = me + mh. Similarly, the partition function of the exciton gas is calculated as •
Np 1 V β( Np ǫ) Z = e− − , p N ! λ3 p p leading to the free energy
N F = N k T ln p λ3 N ǫ, p p B eV p − p
where λp = h/ 2π (me + mh) kBT .
(c) Calculate thep chemical potentials µe, µh, and µp of the electron, hole, and exciton states, respectively. The chemical potentials are derived from the free energies, through •
∂Fe h 3 µ = − = k T ln n λ , e ∂N B e e e T,V
2
∂Fe h 3 µ = − = k T ln n λ , h ∂N B h h h T,V ∂F µ = p = k T ln n λ3 ǫ, p ∂N B p p − p T,V where nα = Nα/V (α =e, h, p). (d) Express the equilibrium condition between excitons and electron/holes in terms of their chemical potentials. The equilibrium condition is obtained by equating the chemical potentials of the electron • and hole gas with that of the exciton gas, since the exciton results from the pairing of an electron and a hole, electron + hole ↽⇀ exciton.
Thus, at equilibrium
µe (ne,T ) + µh (nh,T ) = µp (np,T ) , which is equivalent, after exponentiation, to
3 3 3 βǫ n λ n λ = n λ e− . e e · h h p p
(e) At a high temperature T , find the density np of excitons, as a function of the total density of excitations n n + n . ≈ e h The equilibrium condition yields • 3 3 λe λh βǫ np = nenh 3 e . λp
At high temperature, n n = n n/2, and p ≪ e h ≈
λ3λ3 n 2 h3 m + m 3/2 e h βǫ e h βǫ np = nenh 3 e = 3/2 e . λp 2 (2πk T ) memh B ******** 2. The Manning Transition: When ionic polymers (polyelectrolytes) such as DNA are immersed in water, the negatively charged counter-ions go into solution, leaving behind a positively charged polymer. Because of the electrostatic repulsion of the charges left behind, the polymer stretches out into a cylinder of radius a, as illustrated in the figure. While thermal fluctuations tend to make the ions wander about in the solvent, electrostatic attractions favor their return and condensation on the polymer. If the number of counter ions is N, they interact with the N positive charges left behind on the rod through the
3 potential U (r) = 2 (Ne/L) ln (r/L), where r is the radial coordinate in a cylindrical − geometry. If we ignore the Coulomb repulsion between counter-ions, they can be described by the classical Hamiltonian
N p2 r = i + 2e2 n ln , H 2m L i=1 X where n = N/L. R 2a + − + + − − z − − + r L − + − − + − + − − +
(a) For a cylindrical container of radius R, calculate the canonical partition function Z in terms of temperature T , density n, and radii R and a. The canonical partition function is •
3 3 N 2 i d pid qi pi 2 r Z = 3N exp β + 2e n ln N!h (− 2m L ) Z Q i=1 X N N R 2 2 2πLe N β2e n 2e n/kBT = L · rdr r− Nλ3 · "Za # 2 2 N N 2(1 e n/kBT ) 2(1 e n/kBT ) 2πe 2 R − a − = L2Ne nβ − . nλ3 2 (1 e2n/k T ) " − B #
(b) Calculate the probability distribution function p (r) for the radial position of a counter ion, and its first moment r , the average radial position of a counter-ion. � � 4 Integrating out the unspecified N momenta and N 1 positions from the canonical • − distribution, one obtains the distribution function
2 2 (2e n/kBT ) ln(r/L) 2 1 2e n/kBT re− e n r − p (r) = = 2 1 2 2 . R 2 2(1 e n/kBT ) 2(1 e n/kBT ) drre (2e n/kBT ) ln(r/L) − kBT R − a − a − −
R R (Note the normalization condition a drp(r) = 1.) The average position is then
R 2 2 R 2k T 2e2n R3 2e n/kBT a3 2e n/kBT B − − r = rp (r) dr = − 2 2 2e2n/k T − 2 2e2n/k T . � � 3kBT 2e n R B a B Za − − − − !
(c) The behavior of the results calculated above in the limit R a is very different at high ≫ and low temperatures. Identify the transition temperature, and characterize the nature of the two phases. In particular, how does r depend on R and a in each case? � � Consider first low temperatures, such that e2n/k T > 1. In the R a limit, the • B ≫ distribution function becomes
2 2 1 2e n/kBT e n r − p (r) = 2 1 2 , − k T a2(1 e n/kBT ) B − and r a. To see this, either examine the above calculated average r in the R a � � ∝ � � ≫ limit, or notice that 2 2 e n 1 2e n/kBT p (r) dr = 2 1 x − dx, − k T B 2 1 2e n/kBT 2 where x = r/a, immediately implying r a (as ∞ dxx − < if e n/k T > 1). � � ∝ 1 ∞ B On the other hand, at high temperatures (e2n/k T < 1), the distribution function reduces BR to 2 2 1 2e n/kBT e n r − p (r) = 2 1 2 , − k T R2(1 e n/kBT ) B − and r R, from similar arguments. Thus, at temperature T = e2n/k there is a � � ∝ c B transition from a “condensed” phase, in which the counter-ions are stuck on the polymer, to a “gas” phase, in which the counter-ions fluctuate in water at typical distances from the polymer which are determined by the container size. (d) Calculate the pressure exerted by the counter-ions on the wall of the container, at r = R, in the limit R a, at all temperatures. ≫ The work done by the counter-ions to expand the container from a radius R to a radius • R + dR is dW = dF = (force) dR = P (2πRL) dR, − 5 leading to 1 ∂F k T ∂lnZ P = = B . −2πRL ∂R 2πRL ∂R
At low temperatures, T < Tc, the pressure vanishes, since the partition function is inde pendent of R in the limit R a. At T > T , the above expression results in ≫ c
k T e2n 1 P =B 2N 1 , 2πRL − k TR B i.e. e2 n PV = Nk T 1 . B − k T B (e) The character of the transition examined in part (d) is modified if the Coulomb in teractions between counter-ions are taken into account. An approximate approach to the
interacting problem is to allow a fraction N1 of counter-ions to condense along the polymer rod, while the remaining N = N N fluctuate in the solvent. The free counter-ions are 2 − 1 again treated as non-interacting particles, governed by the Hamiltonian
N p2 r = i + 2e2 n ln , H 2m 2 L i=1 X
where n2 = N2/L. Guess the equilibrium number of non-interacting ions, N2∗, and justify your guess by discussing the response of the system to slight deviations from N2∗. (This is a qualitative question for which no new calculations are needed.) 2 Consider a deviation (n2) from n2∗ N2∗/V kBT/e , occuring at a temperature lower • 2 ≡ ≡ than Tc (i.e. e n/kBT > 1). If n2 > n2∗, the counter-ions have a tendency to condensate 2 (since e n/kBT > 1), thus decreasing n2. On the other hand, if n2 > n2∗, the counter-ions 2 tend to “evaporate” (since e n/kBT < 1). In both cases, the system drives the density n2 2 to the (equilibrium) value of n2∗ = kBT/e . If the temperature is higher than Tc, clearly n2∗ = n and there is no condensation. ******** 3. Bose gas in d dimensions: Consider a gas of non-interacting (spinless) bosons with an energy spectrum ǫ = p2/2m, contained in a box of “volume” V = Ld in d dimensions.
(a) Calculate the grand potential = kBT ln , and the density n = N/V , at a chemical G − Q + βµ potential µ. Express your answers in terms of d and fm (z), where z = e , and
m 1 1 ∞ x f + (z) = − dx. m Γ(m) z 1ex 1 Z0 − − (Hint: Use integration by parts on the expression for ln .) Q 6 We have • ni =N i ∞ = eNβµ exp β n ǫ P i i Q − ! N=0 ni i , X {X} X β(µ ǫi)ni 1 = e − = β(µ ǫi) 1 e − i ni i YX{ } Y −
β(µ ǫi) whence ln = ln 1 e − . Replacing the summation with a d dimensional Q − i − i d d d d 1 d/2 integration V d k/ (2π) = VS / (2π) k − dk, where S = 2π / (d/2 1)!, leads P d d P − to R h i R VSd d 1 βh¯ 2k2/2m ln = k − dk ln 1 ze− . Q − d − (2π) Z The change of variable x = βh¯2k2/2m ( k = 2mx/β/¯h and dk = dx 2m/βx/2¯h) ⇒ results in d/2 p p VSd 1 2m d/2 1 x ln = x − dx ln 1 ze− . Q − d 2 ¯h2β − (2π) Z Finally, integration by parts yields
d/2 x d/2 d/2 VSd 1 2m d/2 ze− Sd 2m x ln = d 2 x dx x = V 2 dx 1 x , Q (2π) d ¯h β 1 ze− d h β z− e 1 Z − Z − i.e. S 2m d/2 d = k T ln = V d k T Γ + 1 f + (z) , B 2 B d +1 G − Q − d h β 2 2 which can be simplified, using the property Γ (x + 1) = xΓ(x), to
V = k T f + (z) . d B d +1 G − λ 2 The average number of particles is calculated as
d/2 x ∂ Sd 2m d/2 1 ze− N = ln =V x − dx ∂ (βµ) Q d h2β 1 ze x Z − − , d/2 Sd 2m d + V + = V 2 Γ f d (z) = d f d (z) 2 h β 2 2 λ 2 i.e. 1 + n = d f d (z) . λ 2
(b) Calculate the ratio P V/E, and compare it to the classical value.
7 We have PV = , while • −G ∂ d ln d E = ln = + Q = . −∂β Q 2 β −2 G
Thus P V/E = 2/d, identical to the classical value.
(c) Find the critical temperature, Tc (n), for Bose-Einstein condensation. The critical temperature T (n) is given by • c
1+ 1 d n = d f d (1) = d ζ λ 2 λ 2 for d > 2, i.e. 2/d h2 n Tc = . 2mkB ζ d 2 !
(d) Calculate the heat capacity C (T ) for T < Tc (n). At T < T , z = 1 and • c
∂E d ∂ d d d d V C (T ) = = G = + 1 G = + 1 kBζ d +1. ∂T −2 ∂T −2 2 T 2 2 λd 2 z=1 z=1
(e) Sketch the heat capacity at all temperatures.
• .
(f) Find the ratio, C /C (T ), of the maximum heat capacity to its classical limit, max → ∞ and evaluate it in d = 3
8 As the maximum of the heat capacity occurs at the transition, •
d d V d d ζ d +1 C = C (T ) = + 1 k f + (1) = Nk + 1 2 . max c B d +1 B 2 2 2 2 2 ζ d ζ d /n 2 2 Thus C d ζ d +1 max = + 1 2 , C (T ) 2 ζ d → ∞ 2 which evaluates to 1.283 in d = 3. (g) How does the above calculated ratio behave as d 2? In what dimensions are your → results valid? Explain. The maximum heat capacity, as it stands above, vanishes as d 2! Since f + (x 1) • → m → → if m 2, the fugacuty z is always smaller than 1. Hence, there is no macroscopic ∞ ≤ occupation of the ground state, even at the lowest temperatures, i.e. no Bose-Einstein condensation in d 2. The above results are thus only valid for d 2. ≤ ≥ ********
9 8.333: Statistical Mechanics I Fall 1999 Final Exam
1. Electron Magnetism: The conduction electrons in a metal can be treated as a gas of fermions of spin 1/2 (with up/down degeneracy), and density n = N/V . (a) Ignoring the interactions between electrons, describe (in words) their ground state.
Calculate the fermi wave number kF, and the ground-state energy density E0/V in terms of the density n. In the ground state, the fermi sea is filled symmetrically by spin up and spin down • particles up to kF, where kF is related to the density through
N d3k kF 4π V k3 2 F = V 3 = V 3 k dk = 2 , 2 k The ground-state energy is calculated as 2 2 3 2 ¯h k d k ¯h 4π 5 E0 = 2V 3 = 2V 3 kF, k