8.333 Statistical Mechanics I: Exam Review 2 Solution

8.333: Statistical Mechanics I Fall 2007 Test 2 Review Problems The second in-class test will take place on Wednesday 10/24/07 from 2:30 to 4:00 pm. There will be a recitation with test review on Monday 10/22/07. The test is ‘closed book,’ and composed entirely from a subset of the following prob­ lems. Thus if you are familiar and comfortable with these problems, there will be no surprises! ******** You may find the following information helpful: Physical Constants Electron mass Electron Charge Speed of light Boltzmann’s const. me ≈ 9.1 × 10−31 kg e ≈ 1.6 × 10−19 C c ≈ 3.0 × 108 ms−1 kB ≈ 1.4 × 10−23 JK −1 Proton mass Planck’s const./2π Stefan’s const. Avogadro’s number mp ≈ 1.7 × 10−27 kg h̄ ≈ 1.1 × 10−34 Js−1 σ ≈ 5.7 × 10−8 W m−2 K −4 N0 ≈ 6.0 × 1023 mol−1 Conversion Factors ˚ ≡ 10−10 m 1A 1atm ≡ 1.0 × 105 N m−2 1eV ≡ 1.1 × 104 K Thermodynamics dE = T dS+dW ¯ R∞ 0 For a gas: dW ¯ = −P dV dx xn e−αx = n! αn+1 h dx exp −ikx − −∞ R∞ e−ikx = P∞ n=0 cosh(x) = 1 + x2 2! x2 2σ 2 (−ik)n n! + x4 4! i For a wire: dW ¯ = Jdx Mathematical Formulas
1 != 2 h 2 2i √ 2 = 2πσ exp − σ 2k �xn � √ π 2 limN→∞ ln N ! = N ln N − N ln e−ikx = +··· P∞ sinh(x) = x + Surface area of a unit sphere in d dimensions 1 Sd = 2π d/2 (d/2−1)! n=1 x3 3! + (−ik)n n! x5 5! �xn �c +··· 1. One dimensional gas: A thermalized gas particle is suddenly confined to a one– dimensional trap. The corresponding mixed state is described by an initial density function √ ρ(q, p, t = 0) = δ(q)f (p), where f (p) = exp(−p2 /2mkB T )/ 2πmkB T . (a) Starting from Liouville’s equation, derive ρ(q, p, t) and sketch it in the (q, p) plane. • Liouville’s equation, describing the incompressible nature of phase space density, is ∂ρ ∂ρ ∂ρ ∂H ∂ρ ∂H ∂ρ = −q̇ − ṗ =− + ≡ −{ρ, H}. ∂t ∂q ∂p ∂p ∂q ∂q ∂p For the gas particle confined to a 1–dimensional trap, the Hamiltonian can be written as H= p2 p2 , + V (qx ) = 2m 2m since Vqx = 0, and there is no motion in the y and z directions. With this Hamiltonian, Liouville’s equation becomes ∂ρ p ∂ρ =− , ∂t m ∂q whose solution, subject to the specified initial conditions, is    p p  ρ(q, p, t) = ρ q − t, p, 0 = δ q − t f (p). m m p t=0 q slope m/ t (b) Derive the expressions for the averages q 2 and p2 at t > 0. • The expectation value for any observable O is Z �O� = dΓOρ(Γ, t), 2 and hence 2 p Z  p  = p f (p)δ q − t dp dq = p2 f (p)dp m   Z ∞ 1 p2 2 = = mkB T. dp p √ exp − 2mkB T 2πmkB T −∞ Z 2 Likewise, we obtain  2 Z Z Z  2  p  p kB T 2 t 2 2 q = q f (p)δ q − t dp dq = p2 f (p)dp = t f (p)dp = t . m m m m (c) Suppose that hard walls are placed at q = ±Q. Describe ρ(q, p, t ≫ τ ), where τ is an appropriately large relaxation time. • Now suppose that hard walls are placed at q = ±Q. The appropriate relaxation time τ , is the characteristic length between the containing walls divided by the characteristic velocity of the particle, i.e. r 2Q 2Qm m τ∼ =p . = 2Q 2 |q̇ | kB T �p � Initially ρ(q, p, t) resembles the distribution shown in part (a), but each time the particle hits the barrier, reflection changes p to −p. As time goes on, the slopes become less, and ρ(q, p, t) becomes a set of closely spaced lines whose separation vanishes as 2mQ/t. p q -Q +Q (d) A “coarse–grained” density ρ̃, is obtained by ignoring variations of ρ below some small resolution in the (q, p) plane; e.g., by averaging ρ over cells of the resolution area. Find ρ̃(q, p) for the situation in part (c), and show that it is stationary. 3 • We can choose any resolution ε in (p, q) space, subdividing the plane into an array of pixels of this area. For any ε, after sufficiently long time many lines will pass through this area. Averaging over them leads to ρ̃(q, p, t ≫ τ ) = 1 f (p), 2Q as (i) the density f (p) at each p is always the same, and (ii) all points along q ∈ [−Q, +Q] are equally likely. For the time variation of this coarse-grained density, we find ∂ρ̃ p ∂ρ̃ =− = 0, i.e. ρ̃ is stationary. ∂t m ∂q ******** 2. Evolution of entropy: The normalized ensemble density is a probability in the phase R space Γ. This probability has an associated entropy S(t) = − dΓρ(Γ, t) ln ρ(Γ, t). (a) Show that if ρ(Γ, t) satisfies Liouville’s equation for a Hamiltonian H, dS/dt = 0. • A candidate “entropy” is defined by Z S(t) = − dΓρ(Γ, t) ln ρ(Γ, t) = − �ln ρ(Γ, t)� . Taking the derivative with respect to time gives   Z Z dS ∂ρ 1 ∂ρ ∂ρ = − dΓ ln ρ + ρ = − dΓ (ln ρ + 1) . dt ∂t ρ ∂t ∂t Substituting the expression for ∂ρ/∂t obtained from Liouville’s theorem gives  Z 3N  X dS ∂ρ ∂H ∂ρ ∂H = − dΓ − (ln ρ + 1) . dt ∂p ∂q ∂p ∂q i i i i i=1 (Here the index i is used to label the 3 coordinates, as well as the N particles, and hence runs from 1 to 3N .) Integrating the above expression by parts yields† Ra Ra a † This is standard integration by parts, i.e. b F dG = F G|b − b GdF . Looking explicitly at one term in the expression to be integrated in this problem, Z Y Z 3N ∂ρ ∂H ∂ρ ∂H dVi = dq1 dp1 · · · dqi dpi · · · dq3N dp3N , ∂qi ∂pi ∂qi ∂pi i=1 ∂ρ , and F with the remainder of the expression. Noting that we identify dG = dqi ∂q i ρ(qi ) = 0 at the boundaries of the box, we get Z Y Z Y 3N 3N ∂ρ ∂H ∂ ∂H =− dVi ρ . dVi ∂qi ∂pi ∂qi ∂pi i=1 i=1 4     3N  X ∂H ∂H ∂ ∂ dΓ ρ (ln ρ + 1) − ρ (ln ρ + 1) ∂q ∂p ∂p ∂q i i i i i=1  Z 3N  X ∂2H ∂H 1 ∂ρ ∂2H ∂H 1 ∂ρ = dΓ ρ −ρ (ln ρ + 1) + ρ (ln ρ + 1) − ρ ∂pi ∂qi ∂qi ρ ∂pi ∂pi ρ ∂qi ∂qi ∂pi i=1  Z 3N  X ∂H ∂ρ ∂H ∂ρ − = dΓ . ∂pi ∂qi ∂qi ∂pi dS = dt Z i=1 Integrating the final expression by parts gives dS =− dt  3N  X ∂ 2H ∂ 2H dΓ −ρ +ρ = 0. ∂pi ∂qi ∂qi ∂pi i=1 Z (b) Using the method of Lagrange multipliers, find the function ρmax (Γ) which maximizes R the functional S[ρ], subject to the constraint of fixed average energy, �H� = dΓρH = E. • There are two constraints, normalization and constant average energy, written respec­ tively as Z Z dΓρ(Γ) = 1, and �H� = dΓρ(Γ)H = E. Rewriting the expression for entropy, Z S(t) = dΓρ(Γ) [− ln ρ(Γ) − α − βH] + α + βE, where α and β are Lagrange multipliers used to enforce the two constraints. Extremizing the above expression with respect to the function ρ(Γ), results in ∂S ∂ρ(Γ) ρ=ρmax = − ln ρmax (Γ) − α − βH(Γ) − 1 = 0. The solution to this equation is ln ρmax = −(α + 1) − βH, which can be rewritten as ρmax = C exp (−βH) , where 5 C = e−(α+1) . (c) Show that the solution to part (b) is stationary, i.e. ∂ρmax /∂t = 0. • The density obtained in part (b) is stationary, as can be easily checked from n o ∂ρmax −β H = − {ρmax , H} = − Ce ,H ∂t ∂H ∂H −β H ∂H ∂H −β H = C(−β) e − C(−β) e = 0. ∂p ∂q ∂q ∂p (d) How can one reconcile the result in (a), with the observed increase in entropy as the system approaches the equilibrium density in (b)? (Hint: Think of the situation encountered in the previous problem.) • Liouville’s equation preserves the information content of the PDF ρ(Γ, t), and hence S(t) does not increase in time. However, as illustrated in the example in problem 1, the density becomes more finely dispersed in phase space. In the presence of any coarse-graining of phase space, information disappears. The maximum entropy, corresponding to ρ̃, describes equilibrium in this sense. ******** 3. The Vlasov equation is obtained in the limit of high particle density n = N/V , or large inter-particle interaction range λ, such that nλ3 ≫ 1. In this limit, the collision terms are dropped from the left hand side of the equations in the BBGKY hierarchy. The BBGKY hierarchy " ! # s s X X X ∂V(q�n − �ql ) ∂ ∂U ∂ ∂ �n p + · − + · fs ∂t n=1 m ∂�qn n=1 ∂�qn ∂�qn ∂�pn l s Z X ∂V(�qn − �qs+1 ) ∂fs+1 = dVs+1 · , ∂�qn ∂�pn n=1 has the characteristic time scales  1 ∂U ∂ v   ∼ · ∼ ,   τU ∂�q ∂�p L    1 ∂V ∂ v · ∼ , ∼  τc ∂�q ∂�p λ   Z   ∂V ∂ fs+1 1  1  ∼ dx · ∼ · nλ3 , τX ∂�q ∂�p fs τc where nλ3 is the number of particles within the interaction range λ, and v is a typical velocity. The Boltzmann equation is obtained in the dilute limit, nλ3 ≪ 1, by disregarding terms of order 1/τX ≪ 1/τc . The Vlasov equation is obtained in the dense limit of nλ3 ≫ 1 by ignoring terms of order 1/τc ≪ 1/τX . 6 (a) Assume that the N body density is a product of one particle densities, i.e. ρ = QN �i , �qi ). Calculate the densities fs , and their normalizations. i=1 ρ1 (xi , t), where xi ≡ (p • Let bf xi denote the coordinates and momenta for particle i. Starting from the joint QN probability ρN = i=1 ρ1 (xi , t), for independent particles, we find N! fs = (N − s)! Z N Y The normalizations follow from Z dΓρ = 1, and Z Y s dVn fs = n=1 dVα ρN α=s+1 Z =⇒ s Y N! = ρ1 (xn , t). (N − s)! n=1 dV1 ρ1 (x, t) = 1, N! ≈ Ns (N − s)! for s ≪ N. (b) Show that once the collision terms are eliminated, all the equations in the BBGKY hierarchy are equivalent to the single equation   ∂ � ∂ p ∂Ueff ∂ + · − · f1 (� p, �q, t) = 0, ∂t m ∂�q ∂�q ∂�p where Ueff (� q, t) = U (q� ) + • Noting that Z dx′ V(�q − q� ′ )f1 (x′ , t). fs+1 (N − s)! = ρ1 (xs+1 ), (N − s − 1)! fs the reduced BBGKY hierarchy is " ≈ s Z X s X ∂ + ∂t n=1  p� n ∂ ∂U ∂ · − · m ∂�qn ∂�qn ∂�pn # fs ∂V(�qn − q�s+1 ) ∂ · [(N − s)fs ρ1 (xs+1 )] ∂�qn ∂�pn n=1 Z  s X ∂ ∂ dVs+1 ρ1 (xs+1 )V(�qn − �qs+1 ) · N fs , ≈ ∂�pn ∂�qn n=1 dVs+1 where we have used the approximation (N − s) ≈ N for N ≫ s. Rewriting the above expression, " # s  X ∂ ∂Uef f ∂ ∂ p�n + · − · fs = 0, ∂t n=1 m ∂�qn ∂�qn ∂�pn 7 where Uef f = U (�q ) + N Z dV ′ V(q� − �q ′ )ρ1 (x′ , t). (c) Now consider N particles confined to a box of volume V , with no additional potential. Show that f1 (� q, p� ) = g(p� )/V is a stationary solution to the Vlasov equation for any g(p� ). Why is there no relaxation towards equilibrium for g(p� )? • Starting from ρ1 = g(p� )/V, we obtain Hef f = with N  X p�i 2 i=1 2m  + Uef f (�qi ) , Z 1 N Uef f = 0 + N dV V(q� − �q ) g(p� ) = d3 qV(�q ). V V R (We have taken advantage of the normalization d3 pg(p� ) = 1.) Substituting into the Vlasov equation yields   ∂ p� ∂ + · ρ1 = 0. ∂t m ∂�q Z ′ ′ There is no relaxation towards equilibrium because there are no collisions which allow g(p� ) to relax. The momentum of each particle is conserved by Hef f ; i.e. {ρ1 , Hef f } = 0, preventing its change. ******** 4. Two component plasma: Consider a neutral mixture of N ions of charge +e and mass m+ , and N electrons of charge −e and mass m− , in a volume V = N/n0 . (a) Show that the Vlasov equations for this two component system are   ∂ p� ∂ ∂Φeff ∂   f+ (� + · +e · p, �q, t) = 0  ∂t m+ ∂�q ∂�q ∂�p    ∂ �p ∂ ∂Φeff ∂   · −e · + f− (� p, �q, t) = 0 ∂�q ∂�p ∂t m− ∂�q , where the effective Coulomb potential is given by Φeff (�q, t) = Φext (q� ) + e Z dx′ C(�q − q� ′ ) [f+ (x′ , t) − f− (x′ , t)] . q) Here, Φext is the potential set up by the external charges, and the Coulomb potential C(� 2 3 satisfies the differential equation ∇ C = 4πδ (q� ). 8 • The Hamiltonian for the two component mixture is  X 2N 2N N  X X 1 p�i 2 �pi 2 + + + ei ej ei Φext (�qi ), H= 2m |q � − � q | 2m + − i j i,j=1 i=1 i=1 where C(q�i − �qj ) = 1/|q�i − �qj |, resulting in X ∂H ∂Φext ∂ + ei ej = ei C(�qi − q�j ). ∂�qi ∂�qi ∂�qi j=i 6 Substituting this into the Vlasov equation, we obtain   ∂ p� ∂ ∂Φef f ∂   p, �q, t) = 0,  ∂t + m · ∂�q + e ∂�q · ∂�p f+ (� +    ∂ p� ∂ ∂Φef f ∂   + · −e · f− (� p, �q, t) = 0. ∂t m− ∂�q ∂�q ∂�p (b) Assume that the one particle densities have the stationary forms f± = g± (p� )n± (q� ). Show that the effective potential satisfies the equation ∇2 Φeff = 4πρext + 4πe (n+ (�q ) − n− (�q )) , where ρext is the external charge density. R • Setting f± (� p, �q ) = g± (� p )n± (�q ), and using d3 pg± (p� ) = 1, the integrals in the effective potential simplify to Z Φef f (�q, t) = Φext (�q ) + e d3 q ′ C(� q − q� ′ ) [n+ (q� ′ ) − n− (q� ′ )] . Apply ∇2 to the above equation, and use ∇2 Φext = 4πρext and ∇2 C(�q−q� ′ ) = 4πδ 3 (q�−� q ′ ), to obtain ∇2 φef f = 4πρext + 4πe [n+ (q� ) − n− (q� )] . (c) Further assuming that the densities relax to the equilibrium Boltzmann weights n± (�q ) = n0 exp [±βeΦeff (q�)], leads to the self-consistency condition
  ∇2 Φeff = 4π ρext + n0 e eβeΦeff − e−βeΦeff , known as the Poisson–Boltzmann equation. Due to its nonlinear form, it is generally not possible to solve the Poisson–Boltzmann equation. By linearizing the exponentials, one obtains the simpler Debye equation ∇2 Φeff = 4πρext + Φeff /λ2 . 9 Give the expression for the Debye screening length λ. • Linearizing the Boltzmann weights gives n± = no exp[∓βeΦef f (�q )] ≈ no [1 ∓ βeΦef f ] , resulting in ∇2 Φef f = 4πρext + 1 Φef f , λ2 with the screening length given by λ2 = kB T . 8πno e2 (d) Show that the Debye equation has the general solution Φeff (�q ) = Z d3 �q′ G(�q − q� ′ )ρext (q� ′ ), where G(�q ) = exp(−|�q |/λ)/|q� | is the screened Coulomb potential. • We want to show that the Debye equation has the general solution Φef f (�q ) = Z d3 qG � (�q − q� ′ )ρext (q� ′ ), where G(�q ) = exp(−|q|/λ) . |q| Effectively, we want to show that ∇2 G = G/λ2 for �q = � 0. In spherical coordinates, 2 G = exp(−r/λ)/r. Evaluating ∇ in spherical coordinates gives     1 ∂ 1 ∂ 2 1 e−r/λ e−r/λ 2 ∂G ∇ G= 2 r − = 2 r − r ∂r ∂r r ∂r λ r r2   G 1 1 −r/λ 1 −r/λ 1 −r/λ 1 e−r/λ − e + e = = 2. = 2 re 2 2 r λ r λ λ λ λ 2 (e) Give the condition for the self-consistency of the Vlasov approximation, and interpret it in terms of the inter-particle spacing? • The Vlasov equation assumes the limit no λ3 ≫ 1, which requires that (kB T )3/2 1/2 no e3 ≫ 1, =⇒ 10 e2 ≪ n−1/3 ∼ ℓ, o kB T where ℓ is the interparticle spacing. consistency condition is In terms of the interparticle spacing, the self– e2 ≪ kB T, ℓ i.e. the interaction energy is much less than the kinetic (thermal) energy. (f) Show that the characteristic relaxation time (τ ≈ λ/c) is temperature independent. What property of the plasma is it related to? • A characteristic time is obtained from r r r λ kB T m m 1 ∼ · ∼ τ∼ ∼ , 2 2 no e kB T ωp c no e where ωp is the plasma frequency. ******** 5. Two dimensional electron gas in a magnetic field: When donor atoms (such as P or As) are added to a semiconductor (e.g. Si or Ge), their conduction electrons can be thermally excited to move freely in the host lattice. By growing layers of different materials, it is possible to generate a spatially varying potential (work–function) which traps electrons at the boundaries between layers. In the following, we shall treat the trapped electrons as a gas of classical particles in two dimensions. If the layer of electrons is sufficiently separated from the donors, the main source of scattering is from electron–electron collisions. (a) The Hamiltonian for non–interacting free electrons in a magnetic field has the form   2 � X  �pi − eA  � | . H= ± µB |B  2m i (The two signs correspond to electron spins parallel or anti-parallel to the field.) The �=B � × �q/2 describes a uniform magnetic field B � . Obtain the classical vector potential A equations of motion, and show that they describe rotation of electrons in cyclotron orbits �. in a plane orthogonal to B • The Hamiltonian for non–interacting free electrons in a magnetic field has the form   2 � �i + eA X p � | ± µB |B H= ,  2m i or in expanded form H= 2 p2 e � |. �+ e A � 2 ± µB |B + p� · A 2m m 2m 11 �=B � × �q/2, results in Substituting A p2 e � × �q + H= + � ·B p 2m 2m p2 e � · �q + = + �p × B 2m 2m 2 e2  � � B × q� ± µB |B| 8m  e2  2 2 � · �q )2 ± µB |B � |. B q − (B 8m Using the canonical equations, �q̇ = ∂H/� p and �ṗ = −∂H/�q , we find  ∂H �p e � e�  ˙  × �q , =⇒ p� = m�q˙ − B × �q ,  q� = ∂�p = m + 2m B 2  2 2   e ∂H p � · �q B. � � − e B 2 �q + e  �˙ = − =− p� × B B ∂�q 2m 4m 4m Differentiating the first expression obtained for p�, and setting it equal to the second ex­ pression for p�˙ above, gives   e� ˙ e2  � e  ˙ e� e2 � 2 ¨ � � mq� − B × �q = − mq� − B × q� × B − |B| �q + B · �q B. 2 2m 2 4m 4m   � · �q B � , leads to � × �q × B � = B 2 �q − B Simplifying the above expression, using B � × �q˙ . m¨q� = eB This describes the rotation of electrons in cyclotron orbits, ¨�q = ω � c × q, �˙ � �. i.e. rotations are in the plane perpendicular to B where �ωc = eB/m; (b) Write down heuristically (i.e. not through a step by step derivation), the Boltzmann equations for the densities f↑ (� p, �q, t) and f↓ (� p, �q, t) of electrons with up and down spins, in terms of the two cross-sections σ ≡ σ↑↑ = σ↓↓ , and σ× ≡ σ↑↓ , of spin conserving collisions. • Consider the classes of collisions described by cross-sections σ ≡ σ↑↑ = σ↓↓ , and σ× ≡ σ↑↓ . We can write the Boltzmann equations for the densities as  Z ∂f↑ dσ 2 − {H↑ , f↑ } = d p2 dΩ|v1 − v2 | [f↑ (p�1 ′ )f↑ (p�2 ′ ) − f↑ (p�1 )f↑ (p�2 )] + ∂t dΩ  dσ× ′ ′ [f↑ (p�1 )f↓ (p�2 ) − f↑ (p�1 )f↓ (p�2 )] , dΩ and ∂f↓ − {H↓ , f↓ } = ∂t Z 2 d p2 dΩ|v1 − v2 |  dσ [f↓ (p�1 ′ )f↓ (p�2 ′ ) − f↓ (p�1 )f↓ (p�2 )] + dΩ 12  dσ× ′ ′ [f↓ (p�1 )f↑ (p�2 ) − f↓ (p�1 )f↑ (p�2 )] . dΩ (c) Show that dH/dt ≤ 0, where H = H↑ + H↓ is the sum of the corresponding H functions. • The usual Boltzmann H–Theorem states that dH/dt ≤ 0, where H= Z d2 qd2 pf (�q, �p, t) ln f (�q, �p, t). For the electron gas in a magnetic field, the H function can be generalized to H= Z d2 qd2 p [f↑ ln f↑ + f↓ ln f↓ ] , where the condition dH/dt ≤ 0 is proved as follows: dH = dt = Z Z  ∂f↑ ∂f↓ d qd p (ln f↑ + 1) + (ln f↓ + 1) ∂t ∂t 2 2  d2 qd2 p [(ln f↑ + 1) ({f↑ , H↑ } + C↑↑ + C↑↓ ) + (ln f↓ + 1) ({f↓ , H↓ } + C↓↓ + C↓↑ )] , with C↑↑ , etc., defined via the right hand side of the equations in part (b). Hence dH = dt = ≡ Z Z d2 qd2 p (ln f↑ + 1) (C↑↑ + C↑↓ ) + (ln f↓ + 1) (C↓↓ + C↓↑ ) d2 qd2 p (ln f↑ + 1) C↑↑ + (ln f↓ + 1) C↓↓ + (ln f↑ + 1) C↑↓ + (ln f↓ + 1) C↓↑ dH↑↑ dH↓↓ d + + (H↑↓ + H↓↑ ) , dt dt dt where the H′ s are in correspondence to the integrals for the collisions. We have also made R R use of the fact that d2 pd2 q {f↑ , H↑ } = d2 pd2 q {f↓ , H↓ } = 0. Dealing with each of the terms in the final equation individually, dH↑↑ = dt Z d2 qd2 p1 d2 p2 dΩ|v1 − v2 | (ln f↑ + 1) dσ [f↑ (p�1 ′ )f↑ (p�2 ′ ) − f↑ (p�1 )f↑ (p�2 )] . dΩ After symmetrizing this equation, as done in the text, dH↑↑ 1 =− dt 4 Z d2 qd2 p1 d2 p2 dΩ|v1 − v2 | dσ [ln f↑ (p�1 )f↑ (p�2 ) − ln f↑ (p�1 ′ )f↑ (p�2 ′ )] dΩ · [f↑ (p�1 )f↑ (p�2 ) − f↑ (p�1 ′ )f↑ (p�2 ′ )] ≤ 0. 13 Similarly, dH↓↓ /dt ≤ 0. Dealing with the two remaining terms, dH↑↓ = dt dσ× [f↑ (p�1 ′ )f↓ (p�2 ′ ) − f↑ (p�1 )f↓ (p�2 )] dΩ Z dσ× [f↑ (p�1 )f↓ (p�2 ) − f↑ (p�1 ′ )f↓ (p�2 ′ )] , = d2 qd2 p1 d2 p2 dΩ|v1 − v2 | [ln f↑ (p�1 ′ ) + 1] dΩ Z d2 qd2 p1 d2 p2 dΩ|v1 − v2 | [ln f↑ (p�1 ) + 1] where we have exchanged (p�1 , �p2 ↔ �p1 ′ , �p2 ′ ). Averaging these two expressions together, dH↑↓ 1 =− dt 2 Z d2 qd2 p1 d2 p2 dΩ|v1 − v2 | dσ× [ln f↑ (p�1 ) − ln f↑ (p�1 ′ )] dΩ · [f↑ (p�1 )f↓ (p�2 ) − f↑ (p�1 ′ )f↓ (p�2 ′ )] . dH↓↑ 1 =− dt 2 Z d2 qd2 p1 d2 p2 dΩ|v1 − v2 | Likewise dσ× [ln f↓ (p�2 ) − ln f↓ (p�2 ′ )] dΩ · [f↓ (p�2 )f↑ (p�1 ) − f↓ (p�2 ′ )f↑ (p�1 ′ )] . Combining these two expressions, Z d 1 dσ× (H↑↓ + H↓↑ ) = − d2 qd2 p1 d2 p2 dΩ|v1 − v2 | dt 4 dΩ ′ ′ [ln f↑ (p�1 )f↓ (p�2 ) − ln f↑ (p�1 )f↓ (p�2 )] [f↑ (p�1 )f↓ (p�2 ) − f↑ (p�1 ′ )f↓ (p�2 ′ )] ≤ 0. Since each contribution is separately negative, we have dH dH↑↑ dH↓↓ d + + (H↑↓ + H↓↑ ) ≤ 0. = dt dt dt dt (d) Show that dH/dt = 0 for any ln f which is, at each location, a linear combination of quantities conserved in the collisions. • For dH/dt = 0 we need each of the three square brackets in the previous derivation to be zero. The first two contributions, from dH↑↓ /dt and dH↓↓ /dt, are similar to those discussed in the notes for a single particle, and vanish for any ln f which is a linear combination of quantities conserved in collisions ln fα = X aα �)χi (p�), i (q i where α = (↑ or ↓) . Clearly at each location �q, for such fα , ln fα (p�1 ) + ln fα (p�2 ) = ln fα (p�1 ′ ) + ln fα (p�2 ′ ). 14 If we consider only the first two terms of dH/dt = 0, the coefficients aα �) can vary i (q with both �q and α = (↑ or ↓). This changes when we consider the third term d (H↑↓ + H↓↑ ) /dt. The conservations of momentum and kinetic energy constrain the cor­ responding four functions to be the same, i.e. they require a↑i (�q) = a↓i (�q). There is, however, no similar constraint for the overall constant that comes from particle number conservation, as the numbers of spin-up and spin-down particles is separately conserved, i.e. a↑0 (�q) = a↓0 (q�). This implies that the densities of up and down spins can be different in the final equilibrium, while the two systems must share the same velocity and temperature. (e) Show that the streaming terms in the Boltzmann equation are zero for any function that depends only on the quantities conserved by the one body Hamiltonians. • The Boltzmann equation is ∂fα = − {fα , Hα } + Cαα + Cαβ , ∂t where the right hand side consists of streaming terms {fα , Hα }, and collision terms C. Let Ii denote any quantity conserved by the one body Hamiltonian, i.e. {Ii , Hα } = 0. Consider fα which is a function only of the Ii′ s fα ≡ fα (I1 , I2 , · · ·) . Then {fα , Hα } = X ∂fα j ∂Ij {Ij , Hα } = 0. � = �q × p�, is conserved during, and away from collisions. (f) Show that angular momentum L • Conservation of momentum for a collision at �q (p�1 + p�2 ) = (p�1 ′ + p�2 ′ ), implies �q × (� p1 + p�2 ) = �q × (� p1 ′ + p�2 ′ ), or �1 + L �2 = L �1 ′ + L � 2 ′, L � i = �q × p�i . Hence angular momentum L � is conserved during colli­ where we have used L sions. Note that only the z–component Lz is present for electrons moving in 2–dimensions, �q ≡ (x1 , x2 ), as is the case for the electron gas studied in this problem. Consider the Hamil­ tonian discussed in (a) H=  2  p2 e � · �q + e � · �q )2 ± µB |B|. � + p� × B B 2 q 2 − (B 2m 2m 8m 15 Let us evaluate the Poisson brackets of the individual terms with Lz = q� × p� |z . The first term is |p� |2 , �q × �p  = εijk {pl pl , xj pk } = εijk 2pl ∂ (xj pk ) = 2εilk pl pk = 0, ∂xl where we have used εijk pj pk = 0 since pj pk = pk pj is symmetric. The second term is proportional to Lz , n o � · q, p� × B � Lz = {Bz Lz , Lz } = 0.  The final terms are proportional to q 2 , and q 2 , �q × p�  2 p , �q × p� = 0, leading to {H, �q × p� } = 0. = 0 for the same reason that Hence angular momentum is conserved away from collisions as well. (g) Write down the most general form for the equilibrium distribution functions for particles confined to a circularly symmetric potential. • The most general form of the equilibrium distribution functions must set both the collision terms, and the streaming terms to zero. Based on the results of the previous parts, we thus obtain fα = Aα exp [−βHα − γLz ] . The collision terms allow for the possibility of a term −�u · p� in the exponent, corresponding to an average velocity. Such a term will not commute with the potential set up by a stationary box, and is thus ruled out by the streaming terms. On the other hand, the angular momentum does commute with a circular potential {V (�q), L} = 0, and is allowed by the streaming terms. A non–zero γ describes the electron gas rotating in a circular box. (h) How is the result in part (g) modified by including scattering from magnetic and non-magnetic impurities? � , in collisions. • Scattering from any impurity removes the conservation of p�, and hence L The γ term will no longer be needed. Scattering from magnetic impurities mixes popula­ tions of up and down spins, necessitating A↑ = A↓ ; non–magnetic impurities do not have this effect. (i) Do conservation of spin and angular momentum lead to new hydrodynamic equations? • Conservation of angular momentum is related to conservation of p�, as shown in (f), and hence does not lead to any new equation. In contrast, conservation of spin leads to an additional hydrodynamic equation involving the magnetization, which is proportional to (n↑ − n↓ ). ******** 16 6. The Lorentz gas describes non-interacting particles colliding with a fixed set of scat­ terers. It is a good model for scattering of electrons from donor impurities. Consider a uniform two dimensional density n0 of fixed impurities, which are hard circles of radius a. (a) Show that the differential cross section of a hard circle scattering through an angle θ is dσ = a θ sin dθ, 2 2 and calculate the total cross section. • Let b denote the impact parameter, which (see figure) is related to the angle θ between � by p� ′ and p π−θ θ = a cos . 2 2 The differential cross section is then given by b(θ) = a sin θ dσ = 2|db| = a sin dθ. 2 Hence the total cross section σtot = Z π 0  π θ θ dθa sin = 2a − cos = 2a. 2 2 0 (b) Write down the Boltzmann equation for the one particle density f (�q, �p, t) of the Lorentz gas (including only collisions with the fixed impurities). (Ignore the electron spin.) • The corresponding Boltzmann equation is ∂f p� ∂f � · ∂f + · +F ∂t m ∂�q ∂�p Z Z no |p� | dσ dσ |p� | ′ dθ [f (p� ′ ) − f (p� )] ≡ C [f (p� )] . = dθ n0 [−f (p� ) ) + f (p� )] = dθ m m dθ 17 � ≡ −∂U/∂�q, and (c) Using the definitions F Z � (�q, p, n(�q, t) = d2 pf � t), and �g(�q, t)� = 1 n(� q, t) Z d2 pf � (� q, �p, t)g(� q, t), show that for any function χ(|p�|), we have       ∂ ∂ p� ∂χ � (n �χ�) + · n χ =F · n . ∂t ∂�q m ∂�p • Using the definitions F� ≡ −∂U/∂�q, Z � (�q, p, � t), n(�q, t) = d2 pf and 1 �g(�q, t)� = n(�q, t) Z d2 pf � (�q, p, � t)g(�q, t), we can write   Z p� ∂f ∂f dσ |p� | ′ � −F · + dθ d pχ(|p� |) − · no (f (p�) − f (p� )) m ∂�q ∂�p dθ m       ∂ p� ∂χ =− · n χ + F� · n . ∂�q m ∂�p d (n �χ(|p� |)�) = dt Z 2 Rewriting this final expression gives the hydrodynamic equation       ∂ ∂ p� ∂χ (n �χ�) + · n χ = F� · n . ∂t ∂�q m ∂�p (d) Derive the conservation equation for local density ρ ≡ mn(�q, t), in terms of the local velocity �u ≡ �p/m � �. • Using χ = 1 in the above expression    ∂ ∂ p� n+ · n = 0. ∂t ∂�q m In terms of the local density ρ = mn, and velocity �u ≡ �� p/m�, we have ∂ ∂ ρ+ · (ρ�u ) = 0. ∂t ∂�q (e) Since the magnitude of particle momentum is unchanged by impurity scattering, the Lorentz gas has an infinity of conserved quantities |p�|m . This unrealistic feature is removed 18 upon inclusion of particle–particle collisions. For the rest of this problem focus only on p2 /2m as a conserved quantity. Derive the conservation equation for the energy density ǫ(�q, t) ≡ ρ 2 c , 2 where �c ≡ p� − �u, m in terms of the energy flux �h ≡ ρ �c c2 /2, and the pressure tensor Pαβ ≡ ρ �cα cβ �. • With the kinetic energy χ = p2 /2m as a conserved quantity, the equation found in (c) gives       ∂ n p� p2 �p� � ∂  n 2 � + |p� | · =F· n . ∂t 2m ∂�q 2 mm m Substituting �p/m = �u + �c , where ��c � = 0, and using ρ = nm, i ∂ hρ 2 ρ 2 i ∂ hρ ρ c + · (�u + �c )(u2 + c2 + 2�u · �c ) = F� · �u. u + ∂t 2 2 ∂�q 2 m From the definition ε = ρ c2 /2, we have i
i ρ ∂ hρ 2 ∂ hρ u +ε + · �uu2 + �u c2 + �cc2 + 2�u · ��c �c � = F� · �u. ∂t 2 ∂�q 2 m Finally, by substituting �h ≡ ρ �c c2 /2 and Pαβ = ρ �cα cβ � , we get i  i ∂ h ρ 2 ∂ ρ ∂ hρ 2 � u +ε + u +ε +h + (uβ Pαβ ) = F� · �u. · �u m ∂t 2 ∂�q 2 ∂qα (f) Starting with a one particle density  1 p2 f (� p, �q, t) = n(�q, t) exp − , 2mkB T (�q, t) 2πmkB T (�q, t)  0 reflecting local equilibrium conditions, calculate �u, �h, and Pαβ . Hence obtain the zeroth order hydrodynamic equations. • There are only two quantities, 1 and p2 /2m, conserved in collisions. Let us start with the one particle density  1 p2 . f (� p, �q, t) = n(�q, t) exp − 2mkB T (�q, t) 2πmkB T (�q, t)  0 Then �u =  p� m  = 0, and 0 19 �h =  p� p2 mm  0 ρ = 0, 2 since both are odd functions of p�, while f o is an even function of p�, while Pαβ = ρ �cα cβ � = n n �pα pβ � = δαβ · mkB T = nkB T δαβ . m m Substituting these expressions into the results for (c) and (d), we obtain the zeroth–order hydrodynamic equations  ∂ρ   = 0, ∂t  ∂ ∂ ρ 2  ε= c = 0. ∂t ∂t 2 The above equations imply that ρ and ε are independent of time, i.e. ρ = n(�q ), or and ε = kB T (q� ),   n(�q ) p2 exp − . f = 2mkB T (q� ) 2πmkB T (�q ) 0 (g) Show that in the single collision time approximation to the collision term in the Bolz­ mann equation, the first order solution is " p� p, �q, t) = f 0 (� p, �q, t) 1 − τ · f 1 (� m � ∂ln ρ ∂ln T p2 ∂T F − + − ∂�q ∂�q 2mkB T 2 ∂�q kB T • The single collision time approximation is C [f ] = f0 − f . τ The first order solution to Boltzmann equation f = f 0 (1 + g) , is obtained from  L f as Noting that 0  f 0g =− , τ   1  0 ∂ p� ∂ ∂ 0 0 0 g = −τ 0 L f = −τ ln f + · ln f + F� · ln f . f ∂t m ∂�q ∂�p ln f 0 = − p2 + ln n − ln T − ln(2πmkB ), 2mkB T 20 !# . where n and T are independent of t, we have ∂ ln f 0 /∂t = 0, and     ∂T � p 1 ∂n 1 ∂T p2 −p� � + · − + g = −τ F · m n ∂�q T ∂�q 2mkB T 2 ∂�q mkB T ( !) � p� 1 ∂ρ 1 ∂T p2 ∂T F = −τ · − + − . m ρ ∂�q T ∂�q 2mkB T 2 ∂�q kB T  (h) Show that using the first order expression for f , we obtain h i ρ�u = nτ F� − kB T ∇ ln (ρT ) . • Clearly R d2 qf 0 (1 + g) = Dp E R d2 qf 0 = n, and Z 1 pα = d2 p f 0 (1 + g) uα = m n m    Z pβ ∂ ln ρ ∂ ln T Fβ ∂T p2 1 2 pα f 0. −τ − − + = d p 2mkB T 2 ∂qβ n m m ∂qβ ∂qβ kB T α Wick’s theorem can be used to check that �pα pβ �0 = δαβ mkB T, p2 pα pβ 0 = (mkB T )2 [2δαβ + 2δαβ ] = 4δαβ (mkB T )2 , resulting in     ρ nτ ∂ 1 ∂T − δαβ kB T δαβ . ln Fβ + 2kB uα = − T kB T ∂qβ ρ ∂qβ Rearranging these terms yields   ∂ ln (ρT ) . ρuα = nτ Fα − kB T ∂qα (i) From the above equation, calculate the velocity response function χαβ = ∂uα /∂Fβ . • The velocity response function is now calculated easily as χαβ = ∂uα nτ = δαβ . ρ ∂Fβ (j) Calculate Pαβ , and �h, and hence write down the first order hydrodynamic equations. 21 • The first order expressions for pressure tensor and heat flux are ρ �pα pβ � = δαβ nkB T, and δ 1 Pαβ = 0, m2
E τρ D ρ 2 2 pi 2 pα p = − 3 pα p a i + bi p . hα = 2m3 2m m 0 Pαβ = The latter is calculated from Wick’s theorem results 2 pi pα p2 = 4δαi (mkB T ) , and 3 pi pα p2 p2 = (mkB T ) [δαi (4 + 2) + 4 × 2δαi + 4 × 2δαi ] = 22δαi , as     3 ∂ ρ 22(mk T ) ∂ ρτ B 2 ln − F� + ln T hα = − 3 (mkB T ) ∂qα T 2mkb T ∂qα 2m ∂T 2 = −11nkB Tτ . ∂qα Substitute these expressions for Pαβ and hα into the equation obtained in (e)    i  ∂ ρ 2 ρ ∂ hρ 2 ∂T ∂ 2 · �u u + ǫ − 11nkB T τ (�u nkB T ) = F� · �u. u +ǫ + + ∂t 2 ∂�q 2 ∂�q ∂�q m ******** 7. Thermal conductivity: Consider a classical gas between two plates separated by a distance w. One plate at y = 0 is maintained at a temperature T1 , while the other plate at y = w is at a different temperature T2 . The gas velocity is zero, so that the initial zeroth order approximation to the one particle density is, f10 (p, � x, y, z ) = n(y) 3/2 [2πmkB T (y)]   p� · p� . exp − 2mkB T (y) (a) What is the necessary relation between n(y) and T (y) to ensure that the gas velocity �u remains zero? (Use this relation between n(y) and T (y) in the remainder of this problem.) • Since there is no external force acting on the gas between plates, the gas can only flow locally if there are variations in pressure. Since the local pressure is P (y) = n(y)kB T (y), the condition for the fluid to be stationary is n(y)T (y) = constant. 22 (b) Using Wick’s theorem, or otherwise, show that 0 p2 0 ≡ �pα pα � = 3 (mkB T ) , 0 p4 and 0 2 ≡ �pα pα pβ pβ � = 15 (mkB T ) , 0 where �O� indicates local averages with the Gaussian weight f10 . Use the result p6 105(mkB T )3 in conjunction with symmetry arguments to conclude p2y p4 0 0 = 3 = 35 (mkB T ) . 0 • The Gaussian weight has a covariance �pα pβ � = δαβ (mkB T ). Using Wick’s theorem gives 0 0 p2 = �pα pα � = (mkB T ) δαα = 3 (mkB T ) . Similarly p4 0 0 2 2 = �pα pα pβ pβ � = (mkB T ) (δαα + 2δαβ δαβ ) = 15 (mkB T ) . The symmetry along the three directions implies p2x p4 0 = py2 p4 0 = pz2 p4 0 = 1 2 4 p p 3 0 = 1 3 3 × 105 (mkB T ) = 35 (mkB T ) . 3 (c) The zeroth order approximation does not lead to relaxation of temperature/density variations related as in part (a). Find a better (time independent) approximation f11 (� p, y), by linearizing the Boltzmann equation in the single collision time approximation, to    1 f 1 − f10 ∂ py ∂ , L f1 ≈ + f10 ≈ − 1 τK ∂t m ∂y where τK is of the order of the mean time between collisions. • Since there are only variations in y, we have     3 p2 ∂ py ∂ 3 0 py 0 py 0 0 − ln (2πmkB ) + f = f1 ∂y ln f1 = f1 ∂y ln n − ln T − ∂t m ∂y 1 m m 2 2mkB T 2     5 p2 3 ∂y T p2 ∂T ∂y T 0 py ∂y n 0 py − + = f1 − + , = f1 m n 2 T 2mkB T T m 2 2mkB T T where in the last equality we have used nT = constant to get ∂y n/n = −∂y T /T . Hence the first order result is     py p2 5 ∂y T 0 1 f1 (� p, y) = f1 (� − . p, y) 1 − τK m 2mkB T 2 T 23 (d) Use f11 , along with the averages obtained in part (b), to calculate hy , the y component of the heat transfer vector, and hence find K, the coefficient of thermal conductivity. • Since the velocity �u is zero, the heat transfer vector is  1 mc2 n py p2 hy = n cy = 2 2m2 1 . In the zeroth order Gaussian weight all odd moments of p have zero average. The correc­ tions in f11 , however, give a non-zero heat transfer n ∂y T hy = −τK 2m2 T  py m  p2 5 − 2mkB T 2  2 py p 0 0 . 0 Note that we need the Gaussian averages of p2y p4 and p2y p2 . From the results of part (b), these averages are equal to 35(mkB T )3 and 5(mkB T )2 , respectively. Hence n ∂y T 2 hy = −τK (mkB T ) 3 2m T  35 5 × 5 − 2 2  =− 2 5 nτK kB T ∂y T. 2 m The coefficient of thermal conductivity relates the heat transferred to the temperature gradient by �h = −K∇T , and hence we can identify K= 2 T 5 nτK kB . 2 m (e) What is the temperature profile, T (y), of the gas in steady state? • Since ∂t T is proportional to −∂y hy , there will be no time variation if hy is a constant. But hy = −K∂y T , where K, which is proportional to the product nT , is a constant in the situation under investigation. Hence ∂y T must be constant, and T (y) varies linearly between the two plates. Subject to the boundary conditions of T (0) = T1 , and T (w) = T2 , this gives T2 − T1 T (y) = T1 + y. w ******** 24