8.333: Statistical Mechanics I Re: 2007 Final Exam. Review Problems

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1 8.333: Statistical Mechanics I Re: 007 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 Electron mass m e Kg Proton mass m p Kg Electron Charge e C Planck s constant/π h Js 1 Speed of light c ms 1 Stefan s constant σ Wm K 4 Boltzmann s constant k B JK 1 Avogadro s number N mol 1 Conversion Factors 1atm Nm 1 A m 1eV K Thermodynamics de = TdS+dW For a gas: dw = PdV For a film: dw = σda Mathematical Formulas ( lim x coth x = 1 + e x + O e 4x 1 x lim x 0 cothx = x O ) x e αx n! dx x n = π α ( 1 )! = n+1 0 [ ] [ ] x σ k dx exp ikx = πσ exp lim N lnn! = N ln N N σ e ikx ( ik) = n x n ln e ikx = n=1 n! m 1 α f η 1 (z) = dx η α+1 z df m x1 = η = 1 f η η m (m 1)! 0 z e x α=1 α m dz z m 1 n=1 ( ik n [ ] lim z f (ln z) (z) = m π 1 + m(m 1)(lnz) + f π (1) = f 7π (1) = 4 m m! !) n x n c ζ m f + π m (1) ζ 3/.61 ζ = 6 ζ 5/ ζ ζ 4 = π

2 8.333: Statistical Mechanics I Fall 1998 Final Exam 1. Exciton dissociation in a semiconductor: By shining an intense laser beam on a semiconductor, one can create a metastable collection of electrons (charge e, and effective mass m e ) and holes (charge +e, and effective mass m h ) 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 N e electrons and N h holes, at temperature T, treating them as classical non-interacting particles of masses m e and m h. 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 Ne Nh 1 V 1 V Z e h = Z e Z h =, N e! λ 3 e N h! where λ α = h/ πm α k B T (α =e, h). Evaluating the factorials in Stirling s approximation, we obtain the free energy N e 3 N h 3 F e h = k B T lnz e h = N e k B T ln λ e + N h k B T ln λ h. ev ev λ 3 h (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 me 4 /( h ǫ ), where ǫ is the dielectric constant, and m 1 = m 1 e + m 1 h.] Calculate the free energy of a gas of N p excitons, treating them as classical non-interacting particles of mass m = m e + m h. Similarly, the partition function of the exciton gas is calculated as Z p = Np 1 V e β( N p ǫ), Np! λ3 p leading to the free energy N p F p = N p k B T ln λ 3 p N p ǫ, ev where λ p = h/ π (m e + m h ) k B T. (c) Calculate the chemical potentials µ e, µ h, and µ p of the electron, hole, and exciton states, respectively. The chemical potentials are derived from the free energies, through F µ e = e h = k B T ln ne λ 3 e, N e T,V

3 where n α = N α /V (α =e, h, p). F e h ( µ h = = k B T ln nh λ 3), N h h T,V F p ( µ p = = k B T ln np λ 3 ) p ǫ, N p T,V (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 µ e (n e, T ) + µ h (n h, T ) = µ p (n p, T ), which is equivalent, after exponentiation, to n e λ 3 3 e n h λ h = n p λ 3 p e βǫ. (e) At a high temperature T, find the density n p of excitons, as a function of the total density of excitations n n e + n h. The equilibrium condition yields λe 3 λ 3 h βǫ n p = n e n h e. At high temperature, n p n e = n h n/, and λ 3 p ( λ 3 3 e λ ( h βǫ n ) ) h 3 3/ m e + m h βǫ n p = n e n h e = e. λ 3 p (πk B T ) 3/ m e m h ********. 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 counterions is N, they interact with the N positive charges left behind on the rod through the 3

4 potential U (r) = (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 [ ] p i r H = m + e n ln, L i=1 where n = N/L. R z a 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 { } N [ ] d 3 p i d 3 q i p ( r ) i i Z = N!h 3N exp β m + e n ln L i=1 [ ] N N πle R = L N βe n rdr r e n/k B T Nλ 3 a [ ] N πe R (1 e n/k B T) a (1 e N n/k B T) = L Ne nβ. nλ 3 (1 e n/k B T ) (b) Calculate the probability distribution function p (r) for the radial position of a counterion, and its first moment r, the average radial position of a counter-ion. 4

5 Integrating out the unspecified N momenta and N 1 positions from the canonical distribution, one obtains the distribution function p (r) = re (e n/k B T)ln(r/L) ( e n ) r 1 e n/k B T = 1. kb T R (1 e n/k B T) a (1 e n/k B T) R drre a (e n/k B T)ln(r/L) R (Note the normalization condition a drp(r) = 1.) The average position is then R k B T e n R 3 e n/k B T a 3 e n/k B T r = rp (r)dr = 3kB T e. n R e n/k B T a e n/k B T a (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 e n/k B T > 1. In the R a limit, the distribution function becomes e 1 e n n/k B T r p (r) = 1 kb T a, (1 e n/k B T) and r a. To see this, either examine the above calculated average r in the R a limit, or notice that e n p (r)dr = 1 x 1 e n/k B T dx, k BT where x = r/a, immediately implying r a (as dxx 1 e n/k B T < if e n/k B T > 1). 1 On the other hand, at high temperatures (e n/k B T < 1), the distribution function reduces to e 1 e n n/k B T r p (r) = 1 kb T R, (1 e n/k B T) and r R, from similar arguments. Thus, at temperature T c = e n/k B there is a 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 (πrl)dr, 5

6 leading to 1 F k B T lnz P = =. πrl R πrl R At low temperatures, T < T c, the pressure vanishes, since the partition function is independent of R in the limit R a. At T > T c, the above expression results in k B T e n 1 P = N 1, πrl k B T R i.e. e n PV = Nk B T 1 kb T. (e) The character of the transition examined in part (d) is modified if the Coulomb interactions between counter-ions are taken into account. An approximate approach to the interacting problem is to allow a fraction N 1 of counter-ions to condense along the polymer rod, while the remaining N = N N 1 fluctuate in the solvent. The free counter-ions are again treated as non-interacting particles, governed by the Hamiltonian N [ ] p = i + e r H n ln, m L i=1 where n = N /L. Guess the equilibrium number of non-interacting ions, N, and justify your guess by discussing the response of the system to slight deviations from N. (This is a qualitative question for which no new calculations are needed.) Consider a deviation (n ) from n N /V k B T/e, occuring at a temperature lower than T c (i.e. e n/k B T > 1). If n > n, the counter-ions have a tendency to condensate (since e n/k B T > 1), thus decreasing n. On the other hand, if n > n, the counter-ions tend to evaporate (since e n/k B T < 1). In both cases, the system drives the density n to the (equilibrium) value of n = k B T/e. If the temperature is higher than T c, clearly n = 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 = p /m, contained in a box of volume V = L d in d dimensions. (a) Calculate the grand potential G = k B T ln Q, and the density n = N/V, at a chemical potential µ. Express your answers in terms of d and f + (z), where z = e βµ, and m 1 x m 1 f m + (z) = dx. Γ (m) 0 z 1 e x 1 (Hint: Use integration by parts on the expression for ln Q.) 6

7 We have n i =N i Q = e Nβµ exp β n i ǫ i N=0 {n i } i, β(µ ǫi )n i 1 = e = 1 e β(µ ǫ i) i {n i } ( whence ln Q = i ln ) 1 e β(µ ǫ i ) [. Replacing ] the summation i with a d dimensional d d integration V d d k/ (π) = V S d / (π) k d 1 dk, where S d = π d/ / (d/ 1)!, leads to V S d h k /m ln Q = k d 1 dk ln 1 ze β. d (π) i The change of variable x = βh k /m ( k = mx/β/ h and dk = dx m/βx/ h) results in d/ V S d 1 m ln Q = d (π) h β Finally, integration by parts yields i.e. d/ V S d 1 m ln Q = d (π) d h β x d/ dx x d/ 1 dx ln ( 1 ze x ). ze x d/ S d m 1 ze x = V d h β G = k B T ln Q = V S d/ d m d k B T Γ + 1 f + (z), d h β d +1 which can be simplified, using the property Γ (x + 1) = xγ (x), to V G = k B Tf + (z). λ d d +1 The average number of particles is calculated as d/ S d m d/ 1 ze x N = dx (βµ) ln Q =V x d h β 1 ze x (, S d m ) d/ d V = V Γ f + (z) = f + (z) h d β λ d d x d/ dx z 1 e x 1, i.e. 1 n = f + d (z). λ d (b) Calculate the ratio PV/E, and compare it to the classical value. 7

8 We have PV = G, while d ln Q d E = β ln Q = + β = G. Thus PV/E = /d, identical to the classical value. (c) Find the critical temperature, T c (n), for Bose-Einstein condensation. The critical temperature T c (n) is given by n = 1 1 (1) = ζ λ d f + d λ d d for d >, i.e. /d h n T c =. mk B ζ d (d) Calculate the heat capacity C (T ) for T < T c (n). At T < T c, z = 1 and E d G d d G d d V C (T) = T = = + 1 = + 1 z=1 T z=1 T λ d k Bζ d +1. (e) Sketch the heat capacity at all temperatures.. (f) Find the ratio, C max /C (T ), of the maximum heat capacity to its classical limit, and evaluate it in d = 3 8

9 As the maximum of the heat capacity occurs at the transition, Thus d d V d d ζ k B f + d C +1 max = C (T c ) = + 1 d (1) = Nk B ζ /n ζd d which evaluates to 1.83 in d = 3. C max d ζ d = + 1 C (T ) ζ +1, (g) How does the above calculated ratio behave as d? In what dimensions are your results valid? Explain. The maximum heat capacity, as it stands above, vanishes as d! Since f m + (x 1) if m, 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. The above results are thus only valid for d. ******** d 9

10 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/ (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 k F, and the ground-state energy density E 0 /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 k F, where k F is related to the density through N d 3 k kf 4π V k 3 = V = V k F dk =, k<k F (π) 3 0 (π) 3 6π i.e. 1/3 k F = 3π n. The ground-state energy is calculated as h k d 3 k h 4π 5 E 0 = V = V k F, k<k F m (π) 3 m 5 (π) 3 and the energy density is E 0 3 ( 3π ) /3 h 5/3 = n. V 5 m Electrons also interact via the Coulomb repulsion, which favors a wave function which is antisymmetric in position space, thus keeping them apart. Because of the full (position and spin) antisymmetry of fermionic wave functions, this interaction may be described as an effective spin-spin coupling which favors states with parallel spins. In a simple approximation, the effect of this interaction is represented by adding a potential U = α N +N, V to the Hamiltonian, where N + and N = N N + are the numbers of electrons with up and down spins, and V is the volume. (The parameter α is related to the scattering length a by α = 4π h a/m.) We would like to find out if the unmagnetized gas with N + = N = N/ still minimizes the energy, or if the gas is spontaneously magnetized. (b) Express the modified Fermi wave numbers k F+ and k F, in terms of the densities n + = N + /V and n = N /V. From the solution to part (a), we can read off 1/3 k F ± = 6π n ±. 10

11 (c) Assuming small deviations n + = n/ + δ and n = n/ δ from the symmetric state, calculate the change in the kinetic energy of the system to second order in δ. We can repeat the calculation of energy in part (a), now for two gases of spin up and spin down fermions, to get E kin = 1 h ( k 5 V 10π m + k 5 ) 3 ( = 6π ) /3 h n 5/3 + n 5/ m F+ F Using n ± = n/ ± δ, and expanding the above result to second order in δ, gives E kin E 0 4 /3 h n 1/3 = + 3π δ + O δ 4. V V 3 m (d) Express the spin-spin interaction density in terms of δ. Find the critical value of α c, such that for α > α c the electron gas can lower its total energy by spontaneously developing a magnetization. (This is known as the Stoner instability.) The interaction energy density is U ( n )( n ) n V = αn +n = α + δ δ = α αδ. 4 The total energy density is now given by [ ] E E 0 + αn /4 4 ( = + 3π ) /3 h n 1/3 α δ ( + O δ 4 ). V V 3 m When the second order term in δ is negative, the electron gas has lower energy for finite δ, i.e. it acquires a spontaneous magnetization. This occurs for α > α c = 4 ( 3π ) /3 h n 1/3. 3 m (e) Explain qualitatively, and sketch the behavior of the spontaneous magnetization as a function of α. For α > α c, the optimal value of δ is obtained by expanding the energy density to fourth order in δ. The coefficient of the fourth order term is positive, and the minimum energy is obtained for a value of δ (α α c ). The magnetization is proportional to δ, and hence grows in the vicinity of α c as α αc, as sketched below 11

12 ********. Boson magnetism: Consider a gas of non-interacting spin 1 bosons, each subject to a Hamiltonian p H 1 (p, s z ) = µ 0 s z B, m where µ 0 = e h/mc, and s z takes three possible values of (-1, 0, +1). (The orbital effect, p p ea, has been ignored.) (a) In a grand { canonical ensemble of } chemical potential µ, what are the average occupation numbers n + ( k), k), n ( n0 ( k), of one-particle states of wavenumber k = p/ h? Average occupation numbers of the one-particle states in the grand canonical ensemble of chemical potential µ, are given by the Bose-Einstein distribution n s ( 1 k) =, (for s = 1, 0, 1) e β[h(s) µ] 1 1 = [ ] h exp β k µ 0 sb βµ 1 m (b) Calculate the average total numbers {N +, N 0, N }, of bosons with the three possible values of s z in terms of the functions f m + (z). Total numbers of particles with spin s are given by N s = ns ( V k), = N s = d 3 1 k [ ]. (π) 3 h k exp β m µ 0 sb βµ 1 { k} After a change of variables, k x 1/ mkb T/h, we get where f + (z) m N s = V ( + f ) λ3 3 / ze βµ 0 sb, 1 dx x m 1 h βµ,, e. Γ(m) 0 z 1 e x λ z 1 πmkb T 1

13 (c) Write down the expression for the magnetization M(T, µ) = µ 0 (N + N ), and by expanding the result for small B find the zero field susceptibility χ(t, µ) = M/ B B=0. Magnetization is obtained from M(T, µ) = µ 0 (N + N ) V [ ( ] + = µ 0 f λ3 3 / ze βµ 0 B ) ( + f 3 / ze βµ 0 sb ). Expanding the result for small B gives f + ( ze ±βµ 0 B ) f + (z[1 ± βµ 0 B]) f + (z) ± z βµ 0 B f + 3/ 3/ 3/ 3/(z). z Using zdf + (z)/dz f + (z), we obtain and m = m 1 V M = µ 0 λ (βµ + 3 0B) f 1 / (z) = µ 0 V M χ B = B=0 k B T λ 3 µ 0 V + k B T λ 3 f 1 / (z). B f 1 + / (z), To find the behavior of χ(t, n), where n = N/V is the total density, proceed as follows: (d) For B = 0, find the high temperature expansion for z(β, n) = e βµ, correct to second order in n. Hence obtain the first correction from quantum statistics to χ(t, n) at high temperatures. In the high temperature limit, z is small. Use the Taylor expansion for f m + (z) to write the total density n(b = 0), as N + + N 0 + N n(b = 0) = = 3 + f V 3 / (z) B=0 λ 3 3 z z 3 z λ 3 3/ 3 3/ Inverting the above equation gives nλ 3 1 nλ 3 z = / 3 The susceptibility is then calculated as µ 0 V + χ = f kbt λ3 1 / (z), µ 0 1 z χ/n = z + + k B T nλ 3 1/ [ ] µ nλ 3 = O n. 3k B T 3/ 1/ 3 13

14 (e) Find the temperature T c (n, B = 0), of Bose-Einstein condensation. What happens to χ(t, n) on approaching T c (n) from the high temperature side? Bose-Einstein condensation occurs when z = 1, at a density or a temperature 3 + n = f λ 3 3 / (1), h /3 n T c (n) =, πmk B 3 ζ 3/ + where ζ 3/ f 3 / (1).61. Since lim + z 1 f 1 /(z) =, the susceptibility χ(t, n) diverges on approaching T c (n) from the high temperature side. (f) What is the chemical potential µ for T < T c (n), at a small but finite value of B? Which one-particle state has a macroscopic occupation number? [ ] Chemical potential for T < T c : Since n s ( k, B) = z 1 e βe s( k,b) 1 1 is a positive number for all k and sz, µ is bounded above by the minimum possible energy, i.e. for T < T c, and B finite, ze βµ 0 B = 1, = µ = µ 0 B. Hence the macroscopically occupied one particle state has k = 0, and sz = +1. (g) Using the result in (f), find the spontaneous magnetization, M(T, n) = lim M(T, n, B). B 0 Spontaneous magnetization: Contribution of the excited states to the magnetization vanishes as B 0. Therefore the total magnetization for T < T c is due to the macroscopic occupation of the (k = 0, s z = +1) state, and M(T, n) = µ 0 V n + (k = 0) 3 V = µ 0 V n n excited = µ 0 N λ 3 ζ 3/. ******** 3. The virial theorem is a consequence of the invariance of the phase space for a system of N (classical or quantum) particles under canonical transformations, such as a change of scale. In the following, consider N particles with coordinates { q i }, and conjugate momenta {p i } (with i = 1,, N), and subject to a Hamiltonian H ({p i }, { q i }). (a) Classical version: Write down the expression for classical partition function, Z Z [H]. Show that it is invariant under the rescaling q 1 λ q 1, p 1 p 1 /λ of a pair of conjugate 14

15 variables, i.e. Z [H λ ] is independent of λ, where H λ is the Hamiltonian obtained after the above rescaling. The classical partition function is obtained by appropriate integrations over phase space as 1 Z = d 3 p i d 3 q i e βh. N!h 3N i The rescaled Hamiltonian H λ = H (p 1 /λ, { p i=1 },λ q 1, { q i=1 }) leads to a rescaled partition function 1 Z [H λ ] = d 3 p i d 3 q i e βh λ, N!h 3N which reduces to Z [H λ ] = 1 N!h3N under the change of variables q 1 = λ q 1, p 1 = p 1 /λ. i ( λ 3 d 3 )( p 1 λ 3 d 3 ) q 1 d 3 p i d 3 q i e βh = Z, (b) Quantum mechanical version: Write down the expression for quantum partition function. Show that it is also invariant under the rescalings q 1 λ q 1, p 1 p 1 /λ, where p i and q i are now quantum mechanical operators. (Hint: start with the time-independent Schrödinger equation.) Using the energy basis Z = tr e βh = e βe n, where E n are the energy eigenstates of the system, obtained from the Schrödinger equation H ({ p i }, {q i }) ψ n = E n ψ n, where ψ n are the eigenstates. After the rescaling transformation, the corresponding equation is (λ) (λ) (λ) H (p 1 /λ, {p i=1 },λ q 1, {q i=1 }) ψ n = E n ψ n. In the coordinate representation, the momentum operator is p i = ih / q i, and therefore (λ) ψ λ ({ q i }) = ψ ({λ q i }) is a solution of the rescaled equation with eigenvalue E n = E n. Since the eigen-energies are invariant under the transformation, so is the partition function which is simply the sum of corresponding exponentials. (c) Now assume a Hamiltonian of the form p i H = m + V ({ q i}). i 15 n i

16 Use the result that Z [H λ ] is independent of λ to prove the virial relation p 1 V = q 1, m q 1 where the brackets denote thermal averages. (You may formulate your answer in the classical language, as a possible quantum derivation is similar.) Differentiating the free energy with respect to λ at λ = 1, we obtain lnz λ H λ p 1 V 0 = λ = β λ=1 λ = β + q 1, λ=1 m q 1 i.e., p 1 V = q 1. m q 1 (d) The above relation is sometimes used to estimate the mass of distant galaxies. The stars on the outer boundary of the G galaxy have been measured to move with velocity v 00 km/s. Give a numerical estimate of the ratio of the G s mass to its size. The virial relation applied to a gravitational system gives GMm mv =. R Assuming that the kinetic and potential energies of the starts in the galaxy have reached some form of equilibrium gives M R G kg/m. v ******** 16

17 8.333: Statistical Mechanics I Fall 000 Final Exam 1. Freezing of He 3 : At low temperatures He 3 can be converted from liquid to solid by application of pressure. A peculiar feature of its phase boundary is that (dp/dt ) melting is negative at temperatures below 0.3 o K [(dp/dt ) m 30atm o K 1 at T 0.1 o K]. We will use a simple model of liquid and solid phases of He 3 to account for this feature. (a) In the solid phase, the He 3 atoms form a crystal lattice. Each atom has nuclear spin of 1/. Ignoring the interaction between spins, what is the entropy per particle s s, due to the spin degrees of freedom? Entropy of solid He 3 comes from the nuclear spin degeneracies, and is given by S s k B ln( N ) s s = = = k B ln. N N (b) Liquid He 3 is modelled as an ideal Fermi gas, with a volume of 46 A 3 per atom. What is its Fermi temperature T F, in degrees Kelvin? The Fermi temperature for liquid 3 He may be obtained from its density as ε F h /3 3N T F = = k B mk B 8πV ( ) ( ) / o K. ( )( ) 8π (c) How does the heat capacity of liquid He 3 behave at low temperatures? Write down an expression for C V in terms of N, T, k B, T F, up to a numerical constant, that is valid for T T F. The heat capacity comes from the excited states at the fermi surface, and is given by π π 3N π T C V = k k B T D(ε F ) = k B B T = Nk B. 6 6 k B T F 4 T F (d) Using the result in (c), calculate the entropy per particle s l, in the liquid at low temperatures. For T T F, which phase (solid or liquid) has the higher entropy? The entropy can be obtained from the heat capacity as TdS 1 T C V dt π T C V =, s l = = k B. dt N 0 T 4 T F As T 0, s l 0, while s s remains finite. This is an unusual situation in which the solid has more entropy than the liquid! (The finite entropy is due to treating the nuclear spins 17

18 as independent. There is actually a weak coupling between spins which causes magnetic ordering at a much lower temperature, removing the finite entropy.) (e) By equating chemical potentials, or by any other technique, prove the Clausius Clapeyron equation (dp/dt ) melting = (s l s s )/(v l v s ), where v l and v s are the volumes per particle in the liquid and solid phases respectively. The Clausius-Clapeyron equation can be obtained by equating the chemical potentials at the phase boundary, µ l (T, P ) = µ s (T, P ), and µ l (T + ΔT, P + ΔP ) = µ s (T + ΔT, P + ΔP ). Expanding the second equation, and using the thermodynamic identities µ µ = S, and = V, T P P T results in P s l s s =. T melting v l v s (f) It is found experimentally that v l v s = 3 A 3 per atom. Using this information, plus the results obtained in previous parts, estimate (dp/dt) melting at T T F. The negative slope of the phase boundary results from the solid having more entropy than the liquid, and can be calculated from the Clausius-Clapeyron relation π T P s l s s 4 T F ln = k B. T melting v l v s v l v s Using the values, T = 0.1 o K, T F = 9. J o K, and v l v s = 3 A 3, we estimate P T melting in reasonable agreement with the observations. ******** Pa K 1,. Non-interacting bosons: Consider a grand canonical ensemble of non-interacting bosons with chemical potential µ. The one particle states are labelled by a wavevector q, and have energies E( q). (a) What is the joint probability P ({n q }), of finding a set of occupation numbers {n q }, of the one particle states, in terms of the fugacities z q exp[β(µ E( q))]? 18

19 In the grand canonical ensemble with chemical potential µ, the joint probability of finding a set of occupation numbers {n q }, for one particle states of energies E( q) is given by the normalized bose distribution P ({n q }) = {1 exp [β(µ E(q ))]} exp [β(µ E( q))n q ] q = (1 z q ) z n q q, with n q = 0, 1,,, for each q. q (b) For a particular q, calculate the characteristic function exp [ikn q ]. ( Summing the geometric series with terms growing as z q e ik) n q, gives exp[ikn = 1 exp[β(µ E( q))] = 1 z q q]. 1 exp [β(µ E( q)) + ik] 1 z q e ik (c) Using the result of part (b), or otherwise, give expressions for the mean and variance of n q. occupation number n q. Cumulnats can be generated by expanding the logarithm of the characteristic function in powers of k. Using the expansion formula for ln(1 + x), we obtain [ ( ln exp [ikn q ] = ln (1 z q ) ln 1 z q 1 + ik k / + )] [ ] z q k z q = ln 1 ik z q 1 z q [ ] z q k z q z q = ik z q 1 z q 1 z q z q k z q = ik +. 1 z q (1 z q ) From the coefficients in the expansion, we can read off the mean and variance z q z q n q =, and n 1 q = zq c (1 z q ). (d) Express the variance in part (c) in terms of the mean occupation number n q. Inverting the relation relating n q to z q, we obtain n q z q =. 1 + n q 19

20 Substituting this value in the expression for the variance gives z q n = c (1 z ) = n q (1 + n q ). q q (e) Express your answer to part (a) in terms of the occupation numbers { n q }. Using the relation between z q and n q, the joint probability can be reexpressed as [ ] q P ({n q }) = ( n q ) n q (1 + n q ) 1 n. q (f) Calculate the entropy of the probability distribution for bosons, in terms of { n q }, and comment on its zero temperature limit. Quite generally, the entropy of a probability distribution P is given by S = k B lnp. Since the occupation numbers of different one-particle states are independent, the corresponding entropies are additive, and given by S = k B [ n q ln n q (1 + n q )ln (1 + n q )]. q In the zero temperature limit all occupation numbers are either 0 (for excited states) or infinity (for the ground states). In either case the contribution to entropy is zero, and the system at T = 0 has zero entropy. ******** 3. Hard rods: A collection of N asymmetric molecules in two dimensions may be modeled as a gas of rods, each of length l and lying in a plane. A rod can move by translation of its center of mass and rotation about latter, as long as it does not encounter another rod. Without treating the hard-core interaction exactly, we can incorporate it approximately by assuming that the rotational motion of each rod is restricted (by the other rods) to an angle θ, which in turn introduces an excluded volume Ω (θ) (associated with each rod). The value of θ is then calculated self consistently by maximizing the entropy at a given density n = N/V, where V is the total accessible area. θ l excluded volume 0

21 (a) Write down the entropy of such a collection of rods in terms of N, n, Ω, and A (θ), the entropy associated to the rotational freedom of a single rod. (You may ignore the momentum contributions throughout, and consider the large N limit.) Including both forms of entropy, translational and rotational, leads to [ ] N [ ] 1 NΩ(θ) Ω(θ) S = k B ln V A(θ) N Nk B ln n lna(θ). N! (b) Extremizing the entropy as a function of θ, relate the density to Ω, A, and their derivatives Ω, A ; express your result in the form n = f (Ω, A, Ω, A ). The extremum condition S/ θ = 0 is equivalent to Ω A n 1 Ω = A, where primes indicate derivatives with respect to θ. Solving for the density gives A n =. ΩA + Ω A (c) Express the excluded volume Ω in terms of θ and sketch f as a function of θ [0, π], assuming A θ. Elementary geometry yields Ω = l (θ + sin θ), so that the equilibrium condition becomes n = f (θ) = with the function f(θ) plotted below: f(θ) [θ ( + cos θ) + sin θ] 1, l n n c 0 θ θ c π 1

22 (d) Describe the equilibrium state at high densities. Can you identify a phase transition as the density is decreased? Draw the corresponding critical density n c on your sketch. What is the critical angle θ c at the transition? You don t need to calculate θ c explicitly, but give an (implicit) relation defining it. What value does θ adopt at n < n c? At high densities, θ 1 and the equilibrium condition reduces to V N ; θl the angle θ is as open as allowed by the close packing. The equilibrium value of θ increases as the density is decreased, up to its optimal value θ c at n c, and θ (n < n c ) = θ c. The transition occurs at the minimum of f (θ), whence θ c satisfies d [θ ( + cos θ) + sin θ] = 0, dθ i.e. (1 + cos θ c ) = θ c sin θ c. Actually, the above argument tracks the stability of a local maximum in entropy (as density is varied) which becomes unstable at θ c. There is another entropy maximum at θ = π, corresponding to freely rotating rods, which becomes more advantageous (i.e. the global equilibrium state) at a density slightly below θ c. ********

23 8.333: Statistical Mechanics I Fall 003 Final Exam 1. Helium 4: 4 He at low temperatures can be converted from liquid to solid by application of pressure. An interesting feature of the phase boundary is that the melting pressure is reduced slightly from its T = 0K value, by approximately 0Nm at its minimum at T = 0.8K. We will use a simple model of liquid and solid phases of 4 He to account for this feature. (a) By equating chemical potentials, or by any other technique, prove the Clausius Clapeyron equation (dp/dt) melting = (s l s s )/(v l v s ), where (v l, s l ) and (v s, s s ) are the volumes and entropies per atom in the liquid and solid phases respectively. Clausius-Clapeyron equation can be obtained by equating the chemical potentials at the phase boundary, µ l (T, P ) = µ s (T, P ), and µ l (T + ΔT, P + ΔP ) = µ s (T + ΔT, P + ΔP ). Expanding the second equation, and using the thermodynamic identities µ µ = S, and = V, T P P T results in P s l s s =. T melting v l v s (b) The important excitations in liquid 4 He at T < 1 K are phonons of velocity c. Calculate the contribution of these modes to the heat capacity per particle C V /N, of the l liquid. The important excitations in liquid 4 He at T < 1 K are phonons of velocity c. The corresponding dispersion relation is ε(k) = hck. From the average number of phonons in mode k, given by n( k) = [exp(β hck) 1] 1, we obtain the net excitation energy as hck E phonons = exp(β hck) 1 k 4πk dk hck = V (change variables to x = βhck ) (π) 3 exp(β hck) 1 ( V kb T ) 4 6 x 3 π ( kb T ) 4 = hc dx = V hc, π hc 3! 0 e x 1 30 hc where we have used 1 ζ 4 3! 0 x 3 π 4 dx =. e x

24 The corresponding heat capacity is now obtained as de π ( kb T ) 3 C V = = V k B, dt 15 hc resulting in a heat capacity per particle for the liquid of ( C l π k B T ) 3 V = k B v l. N 15 hc s (c) Calculate the low temperature heat capacity per particle C V /N, of solid 4 He in terms of longitudinal and transverse sound velocities c L, and c T. The elementary excitations of the solid are also phonons, but there are now two transverse sound modes of velocity c T, and one longitudinal sound mode of velocity c L. The contributions of these modes are additive, each similar inform to the liquid result calculated above, resulting in the final expression for solid heat capacity of s C V π ( kb T ) 3 1 = k B v s +. N 15 h c 3 c 3 T L (d) Using the above results calculate the entropy difference (s l s s ), assuming a single sound velocity c c L c T, and approximately equal volumes per particle v l v s v. Which phase (solid or liquid) has the higher entropy? The entropies can be calculated from the heat capacities as T l 3 C s l (T) = V (T )dt = π k B T k B v l, 0 T 45 hc T 3 s C V (T )dt π k B T 1 s s (T) = = k B v s T 45 h c c 0 T L Assuming approximately equal sound speeds c c L c T 300ms 1, and specific volumes v l v s v = 46 A 3, we obtain the entropy difference 4π ( kb T ) 3 s l s s k B v. 45 hc The solid phase has more entropy than the liquid because it has two more phonon excitation bands. (e) Assuming a small (temperature independent) volume difference δv = v l v s, calculate the form of the melting curve. To explain the anomaly described at the beginning, which phase (solid or liquid) must have the higher density? 4

25 Using the Clausius-Clapeyron equation, and the above calculation of the entropy difference, we get 3 P s l s s = = 4π v k B T k B. T melting v l v s 45 δv hc Integrating the above equation gives the melting curve π v ( k B T ) 3 P melt (T) = P (0) k B T. 45 δv hc To explain the reduction in pressure, we need δv = v l v s > 0, i.e. the solid phase has the higher density, which is expected. ********. Surfactant Condensation: N surfactant molecules are added to the surface of water over an area A. They are subject to a Hamiltonian N p i 1 H = m + V( r i r j ), i=1 i,j where r i and p i are two dimensional vectors indicating the position and momentum of particle i. (a) Write down the expression for the partition function Z(N, T, A) in terms of integrals over r i and p i, and perform the integrals over the momenta. The partition function is obtained by integrating the Boltzmann weight over phase space, as N N i=1 d p i d q i p Z(N, T, A) = exp β i N!h N m β V(q i q j ), i=1 i<j with β = 1/(k B T ). The integrals over momenta are simple Gaussians, yielding N 1 1 Z(N, T, A) = d q i exp β V( q i q j ), N! λ N i=1 i<j where as usual λ = h/ πmk B T denotes the thermal wavelength. The inter particle potential V( r) is infinite for separations r < a, and attractive for r > a such that πrdrv(r) = u0. a (b) Estimate the total non excluded area available in the positional phase space of the system of N particles. To estimate the joint phase space of particles with excluded areas, add them to the system one by one. The first one can occupy the whole area A, while the second can 5

26 explore only A Ω, where Ω = πa. Neglecting three body effects (i.e. in the dilute limit), the area available to the third particle is (A Ω), and similarly (A nω) for the n-th particle. Hence the joint excluded volume in this dilute limit is A(A Ω)(A Ω) (A (N 1)Ω) (A NΩ/) N, where the last approximation is obtained by pairing terms m and (N m), and ignoring order of Ω contributions to their product. (c) Estimate the total potential energy of the system, assuming a constant density n = N/A. Assuming this potential energy for all configurations allowed in the previous part, write down an approximation for Z. Assuming a uniform density n = N/A, an average attractive potential energy, U, is estimated as Ū = 1 V attr. ( q i q j ) = 1 d r 1 d r n( r 1 )n( r )V attr. ( r 1 r ) i,j n N A d r V attr. ( r ) A u 0. Combining the previous results gives [ ] 1 1 Z(N, T, A) (A NΩ/) N βu 0 N exp. N! λ N A (d) The surface tension of water without surfactants is σ 0, approximately independent of temperature. Calculate the surface tension σ(n, T ) in the presence of surfactants. Since the work done is changing the surface area is dw = σda, we have df = TdS + σda + µdn, where F = k B T lnz is the free energy. Hence, the contribution of the surfactants to the surface tension of the film is lnz Nk B T u 0 N σ s = = +, A A NΩ/ A T,N which is a two-dimensional variant of the familiar van der Waals equation. Adding the (constant) contribution in the absence of surfactants gives σ(n, T ) = σ 0 lnz Nk B T u 0 N A = +. A NΩ/ A T,N (e) Show that below a certain temperature, T c, the expression for σ is manifestly incorrect. What do you think happens at low temperatures? 6

27 Thermodynamic stability requires δσδa 0, i.e. σ must be a monotonically increasing function of A at any temperature. This is the case at high temperatures where the first term in the equation for σ s dominates, but breaks down at low temperatures when the term from the attractive interactions becomes significant. The critical temperature is obtained by the usual conditions of σ s / A = σ s / A = 0, i.e. from σ s Nk B T u 0 N = = 0 A (A NΩ/) T A 3, σ s Nk B T 3u 0 N = + = 0 A (A NΩ/) 3 A 4 dq E C A = = = Nk B, dt T and dq E A C σ = =. dt T σ T T The two equations are simultaneously satisfied for A c = 3NΩ/, at a temperature 8u 0 T c =. 7k B Ω As in the van der Waals gas, at temperatures below T c, the surfactants separate into a high density (liquid) and a low density (gas) phase. (f) Compute the heat capacities, C A and write down an expression for C σ without explicit evaluation, due to thesurfactants. The contribution of the surfactants to the energy of the film is given by ln Z k B T u 0 N E s = = N. β A The first term is due to the kinetic energy of the surfactants, while the second arises from their (mean-field) attraction. The heat capacities are then calculated as A E ± ( k) = ± m c 4 + h k c, A σ σ σ ******** 3. Dirac Fermions are non-interacting particles of spin 1/. The one-particle states come in pairs of positive and negative energies, independent of spin. 7

28 (a) For any fermionic system of chemical potential µ, show that the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ δ. (δ is any constant energy.) According to Fermi statistics, the probability of occupation of a state of of energy E is For a state of energy µ + δ, e β(µ E)n p [n(e)] =, for n = 0, e β(µ E) e βδn e βδ 1 p [n(µ + δ)] = 1 + e, = p [n(µ + δ) = 1] = 1 + e βδ = 1 + e βδ. βδ Similarly, for a state of energy µ δ, e βδn 1 p [n(µ δ)] =, = p [n(µ δ) = 0] = 1 + e βδ 1 + e βδ E(T ) E(0) = 4V = p [n(µ + δ) = 1], i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ δ. (b) At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(t = 0) = 0. Using the result in (a) find the chemical potential at finite temperatures T. The above result implies that for µ = 0, n(e) + n E) is unchanged for an temperature; any particle leaving an occupied negative energy state goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle hole symmetry enfrces µ(t ) = 0. (c) Show that the mean excitation energy of this system at finite temperature satisfies d3 k E+ ( k ) (π) 3 exp βe ( k ) Using the label +(-) for the positive (energy) states, the excitation energy is calculated as E(T ) E(0) = [ n + (k) E + (k) + (1 n (k) ) E (k)] k,s z d 3 k E+ ( k) = n + (k) E + (k) = 4V. (π) 3 exp βe + ( k) + 1 k 8

29 (d) Evaluate the integral in part (c) for massless Dirac particles (i.e. for m = 0). For m = 0, E + (k) = hc k, and 4πk dk hck E(T ) E(0) = 4V = (set β hck = x) 8π 3 e β hck ( V kb T ) 3 x 3 = k B T dx π hc 0 e x + 1 7π ( kb T ) 3 = V k B T. 60 hc For the final expression, we have noted that the needed integral is 3!f 4 (1), and used the given value of f 4 (1) = 7π 4 /70. (e) Calculate the heat capacity, C V, of such massless Dirac particles. The heat capacity can now be evaluated as E 7π ( kb T ) 3 C V = = V k B. T V 15 hc (f) Describe the qualitative dependence of the heat capacity at low temperature if the particles are massive. When m = 0, there is an energy gap between occupied and empty states, and we thus expect an exponentially activated energy, and hence heat capacity. For the low energy excitations, h k E + (k) mc + +, m and thus V E(T ) E(0) mc e 4π π βmc dxx e x π λ V = mc e βmc. π λ3 The corresponding heat capacity, to leading order thus behaves as V C(T ) k B λ 3 βmc e βmc. ******** 9

30 8.333: Statistical Mechanics I Fall 004 Final Exam 1. Neutron star core: Professor Rajagopal s group has proposed that a new phase of QCD matter may exist in the core of neutron stars. This phase can be viewed as a condensate of quarks in which the low energy excitations are approximately E( k k F k) ± = ± h. M The excitations are fermionic, with a degeneracy of g = from spin. (a) At zero temperature all negative energy states are occupied and all positive energy ones are empty, i.e. µ(t = 0) = 0. By relating occupation numbers of states of energies µ + δ and µ δ, or otherwise, find the chemical potential at finite temperatures T. According to Fermi statistics, the probability of occupation of a state of of energy E is For a state of energy µ + δ, e β(µ E)n p [n(e)] =, for n = 0, e β(µ E) e βδn e βδ 1 p [n(µ + δ)] =, = p [n(µ + δ) = 1] = =. 1 + e βδ 1 + e βδ 1 + e βδ Similarly, for a state of energy µ δ, e βδn 1 p [n(µ δ)] = 1 + e βδ, = p [n(µ δ) = 0] = 1 + e βδ = p [n(µ + δ) = 1], i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ δ. This implies that for µ = 0, n(e) + n( E) is unchanged for an temperature; for every particle leaving an occupied negative energy state a particle goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle hole symmetry enforces µ(t ) = 0. (b) Assuming a constant density of states near k = k F, i.e. setting d 3 k 4πk dq with F q = k k F, show that the mean excitation energy of this system at finite temperature is E(T ) E(0) gv k πf 0 E+ ( q) dq exp (β E ( q))

31 Using the label +(-) for the positive (energy) states, the excitation energy is calculated as E(T ) E(0) = [ n + (k) E + (k) + (1 n (k) ) E (k)] k,s d 3 k E+ ( k) = g n + (k) E + (k) = gv. (π) 3 exp βe + ( k) + 1 k The largest contribution to the integral comes for k k F. and setting q = ( k k F ) and using d 3 k 4πkF dq, we obtain 4πk E + (q) k F F E + (q) E(T ) E(0) gv dq = gv dq. 8π 3 exp(βe+ (q)) + 1 π exp (βe+ (q)) (c) Give a closed form answer for the excitation energy by evaluating the above integral. For E + (q) = h q /(M), we have E(T ) E(0) = gv k F h q /M dq = (set βh q /M = x) π e βh q /M gv k ( MkB T ) 1/ F x 1/ = k B T dx π h 0 e x + 1 ( gv k F MkB T ) 1/ π 1 1 ζ3/ V k = F π k B T h 1 ζ 3/ = 1 π λ k BT. For the final expression, we have used the value of f (1), and introduced the thermal m wavelength λ = h/ πmk B T. (d) Calculate the heat capacity, C V, of this system, and comment on its behavior at low temperature. Since E T 3/, E C V = = 3 E = 3ζ 3/ 1 V k F T. T 1 V T π λ k B This is similar to the behavior of a one dimensional system of bosons (since the density of states is constant in q as in d = 1). Of course, for any fermionic system the density of states close to the Fermi surface has this character. The difference with the usual Fermi systems is the quadratic nature of the excitations above the Fermi surface. ********. Critical point behavior: The pressure P of a gas is related to its density n = N/V, and temperature T by the truncated expansion b c 3 P = k B Tn n + n, 6 31

32 where b and c are assumed to be positive temperature independent constants. (a) Locate the critical temperature T c below which this equation must be invalid, and the corresponding density n c and pressure P c of the critical point. Hence find the ratio k B T c n c /P c. Mechanical stability of the gas requires that any spontaneous change in volume should be opposed by a compensating change in pressure. This corresponds to δpδv < 0, and since δn = (N/V )δv, any equation of state must have a pressure that is an increasing function of density. The transition point between pressure isotherms that are monotonically increasing functions of n, and those that are not (hence manifestly incorrect) is obtained by the usual conditions of dp/dn = 0 and d P/dn = 0. Starting from the cubic equation of state, we thus obtain dp c =k B T c bn c + n c = 0 dn d. P dn = b + cn c = 0 From the second equation we obtain n c = b/c, which substituted in the first equation gives k B T c = b /(c). From the equation of state we then find P c = b 3 /(6c ), and the dimensionless ratio of k B T c n c = 3. P c 1 V (b) Calculate the isothermal compressibility κ T = V P, and sketch its behavior as a T function of T for n = n c. Using V = N/n, we get κ T (n) = 1 V V P T = 1 P 1 = n n T For n = n c, κ T (n c ) (T T c ) 1, and diverges at T c. [ ] n kb T bn + cn 1 /. (c) On the critical isotherm give an expression for (P P c ) as a function of (n n c ). Using the coordinates of the critical point computed above, we find b 3 b b c P P c = + n n + n 3 6c ( c 6 ) c b b b 3 = n c n + 3 c n c 3 c 3 = (n n c ). 6 (d) The instability in the isotherms for T < T c is avoided by phase separation into a liquid of density n + and gas of density n. For temperatures close to T c, these densities behave 3

33 as n ± n c (1 ± δ). Using a Maxwell construction, or otherwise, find an implicit equation for δ(t ), and indicate its behavior for (T c T ) 0. (Hint: Along an isotherm, variations of chemical potential obey dµ = dp/n.) According to the Gibbs Duhem relation, the variations of the intensive variables are related by SdT V dp + Ndµ = 0, and thus along an isotherm (dt = 0) dµ = dp/n = P/ n T dn/n. Since the liquid and gas states are in coexistence they should have the same chemical potential. Integrating the above expression for dµ from n to n + leads to the so-called Maxwell construction, which reads n+ nc (1+δ) dp k B T bn + cn / 0 = µ(n + ) µ(n ) = = dn. n n n c (1 δ) n Performing the integrals gives the equation 1 + δ c [ 0 = k B T ln bn c (δ) + nc (1 + δ) (1 δ) ] 1 + δ = k B T ln k B T c δ, 1 δ 4 1 δ where for the find expression, we have used n c = b/c and k B T c = b /(c). The implicit equation for δ is thus T 1 + δ T δ = ln δ δ 3 +. T c 1 δ T The leading behavior as (T c T ) 0 is obtained by keeping up to the cubic term, and given by T c δ 1. T c ******** 3. Relativistic Bose gas in d dimensions: Consider a gas of non-interacting (spinless) bosons with energy ǫ = c p, contained in a box of volume V = L d in d dimensions. (a) Calculate the grand potential G = k B T ln Q, and the density n = N/V, at a chemical potential µ. Express your answers in terms of d and f m + (z), where z = e βµ, and 1 x m 1 f m + (z) = dx. (m 1)! 0 z 1 e x 1 (Hint: Use integration by parts on the expression for ln Q.) We have n i =N i Nβµ Q = e exp β n i ǫ i N=0 {n i } i, β(µ ǫ i )n i 1 = e = 1 e β(µ ǫ i) i {n i } 33 i

34 ( whence ln Q = ln e ) β(µ ǫ i) i 1 [. Replacing ] the summation i with a d dimensional integration V d d k/ (π) d = V S d / (π) d k d 1 dk, where S d = π d/ / (d/ 1)!, 0 0 leads to V S d hck ln Q = k d 1 dk ln 1 ze β. d (π) 0 The change of variable x = β hck results in Finally, integration by parts yields ( V S d kb T ) d ln Q = x d 1 dx ln 1 ze x. d (π) hc 0 ( V S d 1 kb T ) d ze x ( S d kb T ) d x d ln Q = x d dx = V dx d (π) d hc 0 1 ze x d hc 0 z 1 e x 1, leading to d S d k B T + G = k B T ln Q = V k B Td!f d hc d +1 (z), which can be somewhat simplified to V π d/ d! + G = k B T f λc d (d/)! d +1 (z), where λ c hc/(k B T ). The average number of particles is calculated as G G V π d/ N = d! = βz = + f λd d (z), µ z c (d/)! where we have used z f d+1 (z)/ z = f d (z). Dividing by volume, the density is obtained as n = 1 π d/ d! + λ d f (d/)! d (z). c (b) Calculate the gas pressure P, its energy E, and compare the ratio E/(PV ) to the classical value. We have PV = G, while ln Q ln Q E = = +d = β dg. β Thus E/(PV ) = d, identical to the classical value for a relativistic gas. z 34

35 (c) Find the critical temperature, T c (n), for Bose-Einstein condensation, indicating the dimensions where there is a transition. The critical temperature T c (n) is given by This leads to 1 π d/ d! + 1 π d/ d! n = f d (z = 1) = ζ d. λ d c (d/)! λ d c (d/)! ) 1/d hc ( n(d/)! T c =. k B π d/ d!ζ d However, ζ d is finite only for d > 1, and thus a transition exists for all d > 1. (d) What is the temperature dependence of the heat capacity C (T) for T < T c (n)? At T < T c, z = 1 and E = dg T d+1, resulting in E E V π d/ d! C (T) = T = (d + 1) T = d(d + 1) G = d(d + 1) T λ dk B ζ d+1 T d. (d/)! z=1 c (e) Evaluate the dimensionless heat capacity C(T )/(Nk B ) at the critical temperature T = T c, and compare its value to the classical (high temperature) limit. We can divide the above formula of C(T Tc), and the one obtained earlier for N(T Tc), and evaluate the result at T = T c (z = 1) to obtain C(T c ) d(d + 1)ζ d+1 =. Nk B ζ d In the absence of quantum effects, the heat capacity of a relativistic gas is C/(Nk B ) = d; this is the limiting value for the quantum gas at infinite temperature. ******** 35

36 8.333: Statistical Mechanics I Fall 005 Final Exam 1. Graphene is a single sheet of carbon atoms bonded into a two dimensional hexagonal lattice. It can be obtained by exfoliation (repeated peeling) of graphite. The band structure of graphene is such that the single particles excitations behave as relativistic Dirac fermions, with a spectrum that at low energies can be approximated by E ± ( k) = ± hv k. There is spin degeneracy of g =, and v 10 6 ms 1. Recent experiments on unusual transport properties of graphene were reported in Nature 438, 197 (005). In this problem, you shall calculate the heat capacity of this material. (a) If at zero temperature all negative energy states are occupied and all positive energy ones are empty, find the chemical potential µ(t ). According to Fermi statistics, the probability of occupation of a state of of energy E is For a state of energy µ + δ, e β(µ E)n p [n(e)] =, for n = 0, e β(µ E) e βδn e βδ 1 p [n(µ + δ)] =, = p [n(µ + δ) = 1] = =. 1 + e βδ 1 + e βδ 1 + e βδ Similarly, for a state of energy µ δ, e βδn 1 p [n(µ δ)] = 1 + e βδ, = p [n(µ δ) = 0] = 1 + e βδ = p [n(µ + δ) = 1], i.e. the probability of finding an occupied state of energy µ + δ is the same as that of finding an unoccupied state of energy µ δ. At zero temperature all negative energy Dirac states are occupied and all positive energy ones are empty, i.e. µ(t = 0) = 0. The above result implies that for µ = 0, n(e) + n E) is unchanged for all temperatures; any particle leaving an occupied negative energy state goes to the corresponding unoccupied positive energy state. Adding up all such energies, we conclude that the total particle number is unchanged if µ stays at zero. Thus, the particle hole symmetry enforces µ(t ) = 0. (b) Show that the mean excitation energy of this system at finite temperature satisfies E(T ) E(0) = 4A d k E+ ( k ). (π) exp βe ( k)