University:
Massachusetts Institute of Technology
Course:
8.333 | Statistical Mechanics I: Statistical Mechanics of Particles
Academic year:

2021
Views:

98

Pages:

38
Author:

Mckenzie Hardy

8.333: Statistical Mechanics I Fall 2007 Test 3 Review Problems & Solutions The third in-class test will take place on Wednesday 11/28/07 from 2:30 to 4:00 pm. There will be a recitation with test review on Monday 11/26/07. The test is ‘closed book,’ but if you wish you may bring a one-sided sheet of formulas. The test will be composed entirely from a subset of the following problems. 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. Debye–Hückel theory and ring diagrams: The virial expansion gives the gas pressure as an analytic expansion in the density n = N/V . Long range interactions can result in non-analytic corrections to the ideal gas equation of state. A classic example is the Coulomb interaction in plasmas, whose treatment by Debye–Hückel theory is equivalent to summing all the ring diagrams in a cumulant expansion. For simplicity consider a gas of N electrons moving in a uniform background of positive charge density N e/V to ensure overall charge neutrality. The Coulomb interaction takes the form X e2 V(~qi − q~j ) , with V(~q ) = −c. UQ = 4π |~q | i < j The constant c results from the background and ensures that the ﬁrst order correction R vanishes, i.e. d3 ~q V(q~ ) = 0. (a) Show that the Fourier transform of V(~q ) takes the form 2 2 � 0 . Ṽ (~ω ) = e /ω for ~ω = 0 for ~ω = 0 • The Fourier transform of V(~q) is singular at the origin, and can be deﬁned explicitly as Z Ṽ(~ω ) = lim d3 q~V(~q)ei�ω·q�−εq . ε→0 The result at ~ω = 0 follows immediately from the deﬁnition of c. For ω ~ = � 0, 2 2 Z Z e e 3 i� ω ·q �−εq 3 −c e Ṽ(~ω) = lim d ~q = lim d ~q ei�ω·q�−εq ε→0 ε→0 4πq 4πq 2 Z π Z ∞ e iωq cos θ−εq 2 e sin θdθ q dq = lim 2π ε→0 4πq 0 0 Z e2 ∞ eiωq−εq − e−iωq−εq dq =− iω 2 0 e2 1 1 e2 e2 = lim + = lim = . ε→0 ω 2 + ε2 ε→0 2iω iω − ε iω + ε ω2 D E0 (b) In the cumulant expansion for U ℓQ , we shall retain only the diagrams forming a c ring; which are proportional to Z 3 d ~q1 d3 ~qℓ Rℓ = ··· V(~q1 − q~2 )V(~q2 − q~3 ) · · · V(~qℓ − q~1 ). V V Use properties of Fourier transforms to show that Z 1 d3 ~ω Ṽ(~ω)ℓ . Rℓ = ℓ−1 3 (2π) V 2 D E0 • In the cumulant expansion for U ℓQ , we retain only the diagrams forming a ring. The c contribution of these diagrams to the partition function is d3 ~q 1 d3 ~q 2 d3 ~q ℓ V(q~ 1 − ~q 2 )V(q~ 2 − q~ 3 ) · · · V(~q ℓ − ~q 1 ) ··· V V V Z Z 1 · · · d3 ~x1 d3 ~x2 · · · d3 ~xℓ−1 d3 q~ ℓ V(~x1 )V(~x2 ) · · · V(~xℓ−1 )V = ℓ V Rℓ = Z − ℓ−1 X i=1 ~xi ! , where we introduced the new set of variables {~xi ≡ ~q i − q~ i+1 }, for i = 1, 2, · · · , ℓ − 1. Note that since the integrand is independent of ~q ℓ , Rℓ = 1 V ℓ−1 Z ··· Z d3 ~x1 d3 ~x2 · · · d3 ~xℓ−1 V(~x1 )V(~x2 ) · · · V − ℓ−1 X ~xi i=1 ! . Using the inverse Fourier transform 1 V(~q ) = (2π )3 Z ˜ ω)e−i�q ·ω� , d3 ω ~ · V(~ the integral becomes 1 Rℓ = 3 (2π) ℓV ℓ−1 Z Z d3 ~x1 · · · d3 ~xℓ−1 Ṽ(~ω1 )e−i�ω1 ·�x1 Ṽ(~ω2 )e−i�ω2 ·�x2 ! ℓ−1 X · · · Ṽ(~ωℓ ) exp −i ~ ℓ · ~xk d3 ω ω ~ 1 · · · d3 ω ~ ℓ. ··· k=1 Since Z d3 ~q −i�ω ·�q e = δ 3 (~ω ), (2π )3 we have 1 Rℓ = 3 (2π) V ℓ−1 Z ··· ℓ−1 Y Z k=1 resulting ﬁnally in Rℓ = 1 V ℓ−1 Z δ(~ωk − ω ~ ℓ )Ṽ(ω ~ k )d3 ~ωk d3 ω ~ Ṽ(ω ~ )ℓ . 3 (2π) (c) Show that the number of ring graphs generated in Sℓ = D U ℓQ E0 c is (ℓ − 1)! (ℓ − 1)! ℓ N! × ≈ N . 2 2 (N − ℓ)! 3 ! d3 ω ~ ℓ, D U ℓQ • The number of rings graphs generated in of ways to choose ℓ particles out of a total of N , E0 c is given by the product of the number N! . (N − ℓ)! multiplied by the number of ways to arrange the ℓ particles in a ring ℓ! . 2ℓ The numerator is the number of ways of distributing the ℓ labels on the ℓ points of the ring. This overcounts by the number of equivalent arrangements that appear in the denom inator. The factor of 1/2 comes from the equivalence of clockwise and counterclockwise arrangements (reﬂection), and there are ℓ equivalent choices for the starting point of the ring (rotations). Hence Sℓ = N! ℓ! N! (ℓ − 1)! × × = . (N − ℓ)! 2 (N − ℓ)! 2ℓ For N ≫ ℓ, we can approximate N (N − 1) · · · (N − ℓ + 1) ≈ N ℓ , and Sℓ ≈ (ℓ − 1)! ℓ N . 2 Another way to justify this result is by induction: A ring of length ℓ + 1 can be created from a ring of ℓ links by inserting an additional point in between any of the existing ℓ nodes. Hence Sℓ+1 = Sℓ × (N − ℓ − 1) × ℓ, leading to the above result, when starting with S2 = N (N − 1)/2. (d) Show that the contribution of the ring diagrams can be summed as ln Zrings = ln Z0 + ∞ X (−β)ℓ ℓ! ℓ=2 V ≈ ln Z0 + 2 Z 0 ∞ Sℓ Rℓ 4πω 2 dω (2π)3 p κ 2 ω κ2 − ln 1 + 2 , ω βe2 N/V is the inverse Debye screening length. P∞ (Hint: Use ln(1 + x) = − ℓ=1 (−x)ℓ /ℓ.) where κ = 4 • The contribution of the ring diagrams is summed as ln Zrings = ln Z0 + ∞ X (−β)ℓ ℓ=2 ∞ X ℓ! Sℓ Rℓ Z (−β )ℓ (ℓ − 1)! ℓ 1 d3 ω ~ Ṽ(ω ~ )ℓ = ln Z0 + N ℓ−1 2 ℓ! V (2π )3 ℓ=2 ℓ Z ∞ ∞ 4πω 2 dω X 1 βN V − Ṽ(ω) = ln Z0 + (2π)3 ℓ V 2 0 ℓ=2 ℓ Z ∞ ∞ V 4πω 2 dω X 1 βN e2 − = ln Z0 + (2π)3 ℓ V ω2 2 0 ℓ=2 Z ∞ 4πω 2 dω βN e2 βN e2 V − ln 1 + , = ln Z0 + V ω2 V ω2 2 0 (2π)3 p P∞ where we have used ln(1 + x) = − ℓ=1 (−x)ℓ /ℓ. Finally, substituting κ = βe2 N/V , leads to Z ∞ V 4πω 2 dω κ 2 κ2 − ln 1 + 2 ln Zrings = ln Z0 + . (2π)3 ω ω 2 0 (e) The integral in the previous part can be simpliﬁed by changing variables to x = κ/ω, and performing integration by parts. Show that the ﬁnal result is ln Zrings = ln Z0 + V 3 κ 12π . • Changing variables to x = κ/ω, and integrating the integrand by parts, gives Z 0 ∞ Z ∞ κ2 dx 2 3 =κ ω dω − ln 1 + 2 x − ln(1 + x2 ) 4 ω ω x 0 3 Z ∞ 3 Z ∞ κ dx 2x 2κ dx πκ3 = = = , 2x − 3 0 1 + x2 3 3 0 x3 1 + x2 2 κ 2 resulting in ln Zrings = ln Z0 + V 3 V πκ3 · = ln Z + κ . 0 4π 2 3 12π (f) Calculate the correction to pressure from the above ring diagrams. 5 • The correction to the ideal gas pressure due to the Debye–Hückel approximation is ! ∂ ln Zrings P = kB T ∂V T,N ∂ V κ3 kB T 3 = P0 − = P0 + k B T κ ∂V 12π T,N 24π 2 3/2 kB T e N = P0 − . 24π kB T V Note that the correction to the ideal gas behavior is non-analytic, and cannot be expressed by a virial series. This is due to the long range nature of the Coulomb interaction. (g) We can introduce an eﬀective potential V (~q − ~q ′ ) between two particles by integrating over the coordinates of all the other particles. This is equivalent to an expectation value that can be calculated perturbatively in a cumulant expansion. If we include only the loop-less diagrams (the analog of the rings) between the particles, we have Z 3 ∞ X d3 ~qℓ d ~q1 ℓ ′ ′ V (~q − ~q ) = V (~q − ~q ) + (−βN ) ··· V(~q − ~q1 )V(~q1 − q~2 ) · · · V(~qℓ − q~ ′ ). V V ℓ=1 Show that this sum leads to the screened Coulomb interaction V (~q ) = exp(−κ|~ q |)/(4π|q~ |). ′ • Introducing the eﬀective potential V(~q − ~q ), and summing over the loop–less diagrams gives Z 3 ∞ X d q~ 1 d3 q~ ℓ ′ ′ ℓ ··· V(~q − q~ 1 )V(~q 1 − ~q 2 ) · · · V(~q ℓ − q~ ′ ) (−βN ) V(~q − q~ ) = V(~q − ~q ) + V V ℓ=1 Z 3 d ~q 1 = V(q~ − ~q ′ ) − βN V(q~ − ~q 1 )V(~q 1 − q~ ′ ) V Z 3 d q~ 1 d3 q~ 2 2 V(~q − q~ 1 )V(~q 1 − q~ 2 )V(~q 2 − q~ ′ ) − · · · . + (βN ) V V Using the changes of notation ~x1 ≡ ~q , ~x2 ≡ q~ ′ , V 12 ≡ V(~x1 − ~x2 ), ~x3 ≡ q~ 1 , and ~x4 ≡ q~ 2 , n ≡ N/V, ··· ~xℓ ≡ ~q ℓ , we can write V 12 = V 12 − βn Z 3 d ~x3 V 13 V 32 + (βn) 2 Z d3 ~x3 d3 ~x4 V 13 V 34 V 42 − · · · . Using the inverse Fourier transform (as in part (a)), and the notation ~xij ≡ ~xi − ~xj , Z 3 d ~x3 V 12 = V 12 − βn Ṽ(ω ~ 13 )Ṽ(ω ~ 32 )e−i(�x13 ·ω�13 +�x32 ·ω�32 ) d3 ~ω13 d3 ω ~ 32 + · · · , 6 (2π ) 6 and employing the delta function, as in part (a) Z 3 d ~ω13 d3 ω ~ 32 3 δ (~ω13 − ~ω32 ) Ṽ(~ω13 )Ṽ(ω ~ 32 ) exp[~x1 · ~ω13 − ~x2 · ~ω32 ] + · · · V 12 = V 12 − βn 3 (2π ) Z i2 d3 ~ω h = V 12 − βn Ṽ(ω ~ ) exp[~ω · ~x12 ] + · · · . (2π )3 Generalizing this result and dropping the subscript such that ~x ≡ ~x12 , V 12 Z h ∞ iℓ+1 X (−βn)ℓ ˜ ω) = V 12 + V(~ ei�x·ω� d3 ~ω . (2π)3 ℓ=1 Finally, including the Fourier transform of the direct potential (ﬁrst term), gives V 12 ∞ Z 2ℓ+2 X d3 ~ω (−1)ℓ e2 κ2ℓ ixω cos θ d3 ~ω ℓe i� x ·ω � (−βn) e = e = ω 2ℓ+2 (2π)3 ω 2ℓ+2 (2π)3 ℓ=0 ℓ=0 Z ∞ ∞ ℓ 2 2ℓ Z 1 X (−1) e κ = dω eixω cos θ d cos θ 2 2π ω 0 −1 ℓ=0 Z ∞ ∞ κ 2ℓ e2 2 sin xω X = dω 2 (−1)ℓ . xω ω 2π 0 ∞ Z X ℓ=0 Setting y ≡ ω/κ, gives V 12 1 = 2 Z ∞ −∞ e2 eixκy − e−ixκy −1 κ dy. 2π 2 2ixκy y2 + 1 Intergrating in the complex plane, via the residue theorem, gives −κx e2 e−κx e e−κx e2 · V 12 = + π = . 2x 2x 4πx 4π 2 Recalling our original notation, x = |~q − q~ ′ | ≡ |~q |, we obtain the screened Coulomb potential e2 e−κ|�q | V(~q ) = . 4π |~q | ******** 2. Virial coeﬃcients: Consider a gas of particles in d-dimensional space interacting through a pair-wise central potential, V(r), where ( +∞ for 0 < r < a, V(r) = −ε for a < r < b, 0 for b < r < ∞. 7 (a) Calculate the second virial coeﬃcient B2 (T ), and comment on its high and low tem perature behaviors. • The second virial coeﬃcient is obtained from Z 1 dd r12 {exp[−βV(r12 )] − 1} , B2 ≡ − 2 where r12 ≡ |~r1 − ~r2 |, as 1 B2 = − 2 "Z a dd r12 (−1) + 0 Z a b dd r12 eβε − 1 # 1 = − {Vd (a)(−1) + [Vd (b) − Vd (a)] · [exp(βε) − 1]} , 2 where Vd (r) = 2π d/2 Sd d r = rd , d d (d/2 − 1)! is the volume of a d–dimensional sphere of radius r. Thus, B2 (T ) = 1 1 Vd (b) − exp(βε) [Vd (b) − Vd (a)] . 2 2 For high temperatures exp(βε) ≈ 1 + βε, and B2 (T ) ≈ 1 βε [Vd (b) − Vd (a)] . Vd (a) − 2 2 At the highest temperatures, βε ≪ 1, the hard-core part of the potential is dominant, and B2 (T ) ≈ 1 Vd (a). 2 For low temperatures β ≫ 1, the attractive component takes over, and 1 B2 = − {Vd (a)(−1) + [Vd (b) − Vd (a)] · [exp(βε) − 1]} 2 1 ≈ − [Vd (b) − Vd (a)] exp(βε), 2 resulting in B2 < 0. (b) Calculate the ﬁrst correction to isothermal compressibility κT = − 1 ∂V V ∂P 8 . T,N • The isothermal compressibility is deﬁned by κT ≡ − From the expansion 1 ∂V V ∂P . T,N P N N2 = + 2 B2 , V V kB T for constant temperature and particle number, we get 1 N N2 dP = − 2 dV − 2B2 3 dV. kB T V V Thus and ∂V ∂P T,N 1 1 V2 =− = − kB T N/V 2 + 2B2 N 2 /V 3 N kB T V κT = N kB T 1 1 + 2B2 N/V V ≈ N kB T 1 1 + 2B2 N/V , N 1 − 2B2 . V (c) In the high temperature limit, reorganize the equation of state into the van der Waals form, and identify the van der Waals parameters. • Including the correction introduced by the second virial coeﬃcient, the equation of state becomes PV N = 1 + B2 (T ). V N kB T Using the expression for B2 in the high temperature limit, PV N =1+ {Vd (a) − βε[Vd (b) − Vd (a)]} , 2V N kB T and N2 N P+ ε[Vd (b) − Vd (a)] = kB T 2 V 2V N 1+ Vd (a) . 2V Using the variable n = N/V , and noting that for low concentrations −1 n n =V 1 + Vd (a) ≈ 1 − Vd (a) 2 2 −1 N V − Vd (a) , 2 the equation of state becomes N n2 ε P+ [Vd (b) − Vd (a)] · V − Vd (a) = N kB T. 2 2 9 This can be recast in the usual van der Waals form (P − an2 ) · (V − N b) = N kB T, with a= ε [Vd (b) − Vd (a)] , 2 and b= 1 Vd (a). 2 (d) For b = a (a hard sphere), and d = 1, calculate the third virial coeﬃcient B3 (T ). • By deﬁnition, the third virial coeﬃcient is 1 B3 = − 3 Z dd rdd r ′ f (r)f (r ′ )f (r − r ′ ), where, for a hard core gas ( −1 V(r) f (r) ≡ exp − −1= kB T 0 for 0 < r < a, for a < r < ∞ . In one–dimension, the only contributions come from 0 < r, and r ′ < a, where f (r) = f (r ′ ) = −1. Using the notations |x| ≡ r, |y| ≡ r ′ (i.e. −a < x, and y < a), 1 B3 = − 3 Z a a 1 dx dy · f (x − y) = 3 −a −a Z Z Z (−1) = −a < x,y < a,−a < x−y < a

Statistical Mechanics I: Exam Review 3 Solution

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