Elements of Ensemble Theory

Transcription

1 Chapte Elements of Ensemble Theoy In 19 Gibbs, one of the foundes of Statistical Mechanics, intoduced the concept of ensemble a Fench wod meaning assembly of systems. An ensemble is a collection of essentially independent systems, which ae macoscopically identical but micoscopically diffeent. Thee types of ensemble ae impotant in statistical mechanics. The classification depends on the matte in which the systems inteact..1 MICROCANONICAL ENSEMBLE The micocanonical ensemble is a collection of independent systems, having the same numbe of paticles N, volume V and an enegy between E and E + δe. So the individual systems of a micocanonical ensemble ae sepaated by igid, impemeable and insulated walls, such that the values of E,V and N fo a paticula system ae not affected by the pesence of othe systems. Thus, in micocanonical ensemble, neithe enegy no matte is exchanged. N,V, E N,V, E N,V, E N,V, E Fig..1. Micocanonical ensemble. 57

2 6 An Intoduction to Equilibium Statistical Mechanics Now E = 1 m me = N i=1 N 3 i=1 α=1 p i = 1 m N 3 p iα i=1 α=1 p iα.1.9 whee p i = p i1 + p i + p i3, denoting x,y,z components by 1,, 3 espectively. Obviously, the sum contains 3N = f squae tems f = degees of feedom. Fo E = constant, it descibes a sphee of adius ςe=me 1 in the f -dimensional space of the momentum coodinates. Hence ΩE is popotional to the volume of phase space contained in the spheical shell lying between the sphee of adius ςe and that of slightly lage adius ςe + δe. We now calculate the volume of the N-dimensional sphee. Fom the dimensional analysis, the volume of a N-dimensional sphee is W N = C N ς N.1.1 whee C N is a constant and is given by C N = Γ N see Appendix II. Using.1.11 in.1.1, the volume of the N-dimensional sphee becomes W N = Hence, the volume of the 3N-dimensional sphee of adius me is π N π N Γ N + 1ςN.1.1 Neglecting 1 compaed to N W 3N E= ΩE= V N h 3N π 3N Γ 3N 3N me W N E δe E 3N mπ V N ΩE= h Γ E 3N 3N δe.1.14 Since 3N = f, we see that 3N mπ V N ΩE=Ω E δe = h Γ E 3N 3N δe.1.15 ΩE E f.1.16

3 Elements of Ensemble Theoy 61 N is of the ode of Avogado s numbe and so vey lage. Thus, ΩE is an extemely apidly inceasing function of the enegy E of the system. The entopy of the ideal gas is S = k lnω E [ ] 3 mπe 3N = Nkln + NklnV k lnγ h 3N 3N Now Γ = 1! = 3N!, since N 1. Again by Stiling s appoximation.1.17 Using.1.18 in.1.17 lnγ ln 3N! = 3N ln 3N 3N 3 3N 3N = N ln 3 3N 3N 3N = N ln.1.18 [ ] 3 mπe 3 3N 3 SE,V =Nkln + NklnV Nkln + h Nk [ 3 ] 4mπ SE,V =Nkln V 3h E N + 3 Nk.1.19 Solving fo E in tems of S and V, we obtain the intenal enegy given by 3 h N S E S,V = exp 4π m 3 Nk 1 Again fom themodynamics V 3.1. TdS = de + PdV.1.1 Then the tempeatue is given by E T = = E.1. S V 3 Nk Hence, the specific heat at constant volume is given by E C V = = 3 Nk.1.3 T Fom.1.1, the pessue P is given by E P = = E V S 3 V V.1.4

4 6 An Intoduction to Equilibium Statistical Mechanics Using.1. in.1.4, we finally get P = N V This is the equation of state of the ideal gas. PV = N PROBLEM.1: Conside an isolated system consisting of a lage numbe N of weakly inteacting localized paticles of spin 1/ in an extenal magnetic field H. Each paticle has a magnetic moment µ, which can point eithe paallel o anti-paallel to the field H. Conside the enegy ange between E and E + δe whee δe is vey small compaed to E but micoscopically lage so that δe µh. What is the total numbe of states ΩE lying in this enegy ange? Hence, find the tempeatue T. Let n be the numbe of spins aligned paallel to H. So N n is the numbe of spins aligned anti-paallel to H. The total enegy of the system is E = nµh +N nµh = NµH nµh n = N E µh Then, if the enegy of the system is exactly E, the numbe of accessible micostates is Ω E = N N! C n = n!n n! = N! N E N! µh + E ii! µh whee we use equation i. Fom equation i, δn = δe consideing magnitude only. µh Hence, the total numbe of states in the ange E to E + δe is N! ΩE=Ω E δn = N E N! µh + E δe iii µh! µh Fom equation 1..1 the tempeatue T is defined though the elation 1 = lnω iv E Taking logaithms on both sides of equation ii N lnω = lnn! ln E N! ln µh + E!. v µh Using Stiling s appoximation N lnω = N lnn N E µh N + E l1n µh N ln N + E µh E µh + N + N + E µh E µh i

5 N = N lnn E µh lnω E = 1 N µh ln E µh 1 = 1 N = 1 µh ln E µh N + E µh N µh ln E µh N + E µh N E µh N + E µh N E µh NµH E = e µh = e µh +1 e µh 1 = e µh µh +e e µh e µh N ln E µh + 1 µh 1 µh ln = coth µh µh E = NµH tanh This is the elation between enegy E and tempeatue T. N Elements of Ensemble Theoy 63 + E µh N + E µh N ln + E µh 1 µh PROBLEM.: Conside an exteme elativistic gas, chaacteized by the single paticle enegy states εn x,n y,n z = hc n l x + n y + n 1 z. Calculate Ω and hence show that the atio γ = C p = 4 C v 3. The single paticle enegy states ae given by εn x,n y,n z = hc l n x + n y + n z = 4ε L Then the total enegy of the system is given by E = ε = hc L N =1 n x + n y + n 1 z n x + n y + n 1 z i h c 3 = 4ε V h c n 1x + n 1y + n 1z + n x + n y + n z + + n Nx + n Ny + n Nz = 4L E h c = R say ii iii

6 64 An Intoduction to Equilibium Statistical Mechanics This is the equation of a sphee of adius R in 3N-dimensional space; the suface of the sphee coesponds to a constant enegy E. The numbe of micostates ΩN,V,E is equal to the numbe of lattice points on the positive segment of this spheical suface. If the enegy of the system is specified in between E and E + δe, the numbe of micostates ΩN,V,E E + δe within this ange is equal to the numbe of lattice points lying within the positive pat of the spheical shell fomed between the sufaces of adii R and R + δr in the 3N-dimensional space. If ΦN,V,E is the volume of the positive pat of the 3N-dimensional sphee of adius R, then ΩN,V,E E + δe=ω = ΦN,V,E + δe ΦN,V,E By Taylo s expansion, ΦN,V,E + δe = ΦN,V,E+ Φ E δe + 1! = ΦN,V,E+ Φ E δe Φ E δe +... Ω = ΦN,V,E + δe ΦN,V,E = Φ E δe Now fom appendix II, the volume of the 3N-dimensional sphee of adius R is iv Hence, Thus, Taking log on both sides ΦN,V,E= But fom vi, we have [ πle lnφ = N ln hc Φ 3N E = V 3N R= π 3N R 3N 3N! 1 3N π 3N 3N! πl hc 3N LE 3N = hc 3N E 3N 1! Ω = Φ E δe = 3NΦN,V,E δe E lnω = lnφ + ln3n + ln δe E ] ln 3N! = N ln πl hc 3N 3N E 3N! v vi = 3N E Φ vii [ πle hc ] 3N ln 3N + 3N viii ix x

7 Elements of Ensemble Theoy 65 whee we use Stiling s appoximation. [ ] πle lnω = N ln 3N 3N hc ln + 3N + ln3n + ln δe E [ = 3 ] N ln πle 3 + 3N 3N hc + ln3n + ln δe E δe Fo macoscopic systems, N 1 and so ln3n 3N. Again since δe E, ln is negligible E and so [ lnω = 3 /3 ] N ln πle + 3N xi 3N hc Then the entopy of the system is [ S = k lnω = 3 /3 ] Nkln πle + 3Nk 3N hc Solving fo E, The tempeatue of the gas is given by 1 T = E = The specific heat at constant volume is S E hc πl 3N C V = Fo the pessue of the system, we obtain P = The specific heat at constant pessue is E V C P = + P = T N,P T N,P γ = C P V = 4 3 Poved N,V 3 e s Nk 3 = Nk E E = N xii xiii xiv E = Nk xv T N,S E = E V N,S 3V E + 1 T N,P 3 E = 4 T N,P 3 xvi E = 4 Nk xvii T N,P 3 xviii PROBLEM.3: The quantity ΩN,V, E is called the micocanonical patition function. Show that lnωn,v,e=lnω 1 + Ω + = lnω max, whee Ω max is the lagest Ω i in the seies.

8 66 An Intoduction to Equilibium Statistical Mechanics Let us assume that thee ae as many N systems with Ω i s compaable to Ω max, that is, as many such systems as thee ae paticles in any one system. Thus, Ω = N Ω max lnω = lnω max + lnn But lnn N. lnωn,v,e = lnω max Poved ii i PROBLEM.4: A one-dimensional chain is made up of N identical elements of length l. The angle between successive elements can be eithe o 18, but thee is no diffeence in intenal enegy between these two possibilities. Fo the sake of counting, one can think of each element as eithe pointing to the ight + o to the left. Then one has N = n + + n L = l n + n =ln + N. a Use the micocanonical ensemble to find the entopy as a function of N and n +, SN,n +. b Find an expession fo the tension in the chain as a function of T, N and n +, IT,N,n +. c Reaange b to give the length as a function of N, T and I. N! a The numbe of ways of choosing n + elements fom a total of N is. It follows N n +!n +! that N! ΩN,n + = i N n +!n +! The entopy of the system is SN,n + =klnω = k [lnn! lnn n +! lnn +!] Using Stiling s appoximation, the above equation becomes SN,n + = k [N lnn N N n + lnn n + n + lnn + + n + ] b = k [N lnn N n + lnn n + n + lnn + ] ii I S T = = S n + L N,E n + L = S 1 n + l = k [ N n+ + lnn n + n ] + lnn + l N n + n + = k N l ln n+ n + N,T,n + = N l ln n+ n + iii

9 Elements of Ensemble Theoy 67 c Fom iii, we have 1 n + = N 1 + exp li iv We now use this expession fo n + to find L: L=l n + N=Nl 1 + exp /I 1 =Nl 1 + exp /I 1 + exp /I 1 + exp /I 1 exp /I exp 1/I exp 1/I = Nl 1 + exp /I = Nl exp 1/I + exp 1/I li L = Nl tanh v Fo high tempeatues, whee li,tanhx x fo x, so L = Nl I vi The fact that the length is popotional to the tension I shows that Hooke s law applies to this system, at least fo high tempeatues..1.3 A two-level system Conside a system of N non-inteacting and distinguishable paticles. Let the only accessible states fo the system be the gound state of zeo enegy and an excited state of enegy ε. ε Fig... Two non-degeneate enegy levels sepaated by enegy. Let the numbe of paticles in the gound state of zeo enegy be N and that in the excited state with enegy ε be N ε. N + N ε = N.1.5 The total enegy of the system is E =N +N ε ε N ε = E ε.1.6

10 68 An Intoduction to Equilibium Statistical Mechanics If the paticles ae distinguishable, the numbe of micostates accessible to the system is equal to the numbe of ways choosing N ε paticles fom N paticles and is given by Ω = N N! C Nε = N ε!n N ε! = N! E ε! N E ε!.1.7 Assuming that N, N ε and N N ε ae all lage numbes compaed to 1, we use Stiling s appoximation fo the factoials of lage numbes and then the entopy is S = k lnω = k [N lnn N ε lnn ε N N ε lnn N ε ] [ E = k ε N ln 1 E E Nε ε ln E ] Nε The tempeatue of the system is given by 1 S T = E N,V fom which the enegy E of the system is given by E = = k Nε ε ln E 1 Nε 1 + e ε N ε = E ε = N 1 + e ε ε Now fo T, i.e., 1,N ε, that is, all paticles ae fozen in the gound state. As T ε inceases, both E and N ε incease and when T, i.e., 1,N ε; N, that is, half the paticles ae in the gound state and half in the excited state and the enegy of the system attains its maximum value of E = Nε. The specific heat at constant volume is given by E ε ε e C V = = Nk.1.3 T 1 + e ε V C V.4 ε Fig..3. The specific heat at constant volume of a two-state system, showing Schottky hump.

11 Elements of Ensemble Theoy 69 The specific heat C V is zeo at both vey low and vey high tempeatues and is maximum at T =.4 ε k. The hump, obseved in the C V vs T plot, is called Schottky anomaly. Such a peak, when obseved expeimentally, is theefoe an indication of a gap in its enegy states.. CANONICAL ENSEMBLE This ensemble is a collection of independent systems having same tempeatue T, volume V and numbe N of paticles. So the systems of this ensemble ae sepaated by igid, impemeable, but conducting walls. Since the systems ae sepaated by conducting walls, heat can be exchanged between the systems till they each a common tempeatue T. Thus, in canonical ensemble, the systems can exchange enegy but not paticles. N,V,T N,V,T N,V,T N,V,T Fig..4. Canonical ensemble...1 Pobability distibution and canonical patition function Conside a small system A in themal inteaction with a heat esevoi A. What is the pobability P of finding the system A in any one paticula micostate of enegy E? A A Fig..5. A small system A in themal inteaction with a heat esevoi A. We assume weak inteaction between A and A, so that thei enegies ae additive. The enegy of A is, of couse, not fixed. Only the total enegy of the combined system, A = A + A which is isolated has a constant value in some ange between E and E + δe: E + E = E = constant..1

12 7 An Intoduction to Equilibium Statistical Mechanics whee E denotes the enegy of the esevoi A. Thus, when A has an enegy E, the esevoi A must have an enegy E = E E. Let Ω E denote the numbe of accessible micostates of A, when it has an enegy in the ange nea E. If the system A is in the paticula state, the numbe of states accessible to the combined system A is 1 Ω E =Ω E E. Then the pobability of finding the system A in this state is P = P E = Ω E Ω = Ω E E total Ω, total whee Ω total denotes the total numbe of states accessible to the combined system A. But Ω total = constant = c say. P = c Ω E E.. lnp = lnc + lnω E E..3 Since A is vey small compaed to A,E E. Expanding lnω about E = E, lnω E E = lnω E [ ] lnω E E... E Since A acts as a esevoi, E E and so we can neglect highe ode tems. lnω E E = lnω E [ ] lnω E E..4 E [ ] lnω The deivative E is evaluated at the fixed enegy E = E and is thus a constant, independent of the enegy E and lnω E [ ] E = β = 1, whee T is the tempeatue of the esevoi. E The equation..4 then becomes lnω E E = lnω E βe Ω E E = Ω E e βe Since Ω E is just a constant, independent of, equation.. becomes P = ce βe..5 whee c is a constant, independent of. The pobability..5 is a vey geneal esult and is of fundamental impotance in statistical mechanics. The exponential facto e βe is called the Boltzmann facto ; the coesponding pobability distibution..5 is known as the canonical distibution. Now accoding to the nomalization conditions fo pobabilities, P = 1 c 1 = e βe

13 Elements of Ensemble Theoy 71 P = e βe e βe = e βe Z..6 whee Z = e βe..6a Z is called the canonical patition function. The lette Z is used because the Geman name is Zustandsumme. It is the sum ove all accessible micostates of the system A. It should be noted that the expession..6a is coect, if the enegy levels ae discete and non-degeneate. In the geneal case, an enegy level E consists of a goup of states g in numbe all of the same enegy E and of equal pobability. Thus in summing ove all accessible micostates, it is necessay to epeat equal tems involving E g times. Instead, it is simple to use a modified fom of the patition function Z = g e βe..6b whee the label goes ove all enegy levels. If the level is non-degeneate, then g = 1. Equation..6b is the geneal fom of the patition function. Accodingly, P = g e βe g e βe..6c Once the pobability distibution is known, vaious mean values can be computed. Fo example, let x be any quantity assuming the value x in state of the system A. Then the mean value of x is x P x = P x e βe = x P = e βe..7.. Enegy fluctuations in the canonical ensemble: coespondence with the micocanonical ensemble Accoding to equation..7, the mean enegy value is E e βe E = e βe..8 Now the canonical patition function is Using..9 in..8 Z = e βe Z β = E e βe..9 E = 1 Z Z β = lnz β...1

14 Elements of Ensemble Theoy To calculate the canonical patition function 1. Classical ideal gas We conside a classical monatomic gas at an absolute tempeatue T, contained in a volume V. Let us denote the position vecto of the ith molecule by i, its momentum by p i. Then the total enegy of the gas is given by E = N i=1 p i m +U 1,... N..17 The fist tem on the ight-hand side of the above equation epesents the total kinetic enegy of all the molecules and the second tem epesents the potential enegy of inteaction between the molecules. The classical canonical patition is given by [ Z = exp β = 1 h 3N { 1 p m p N +U 1,..., d N }] d 3 N d 3 p 1...d 3 p N h 3N e βp 1 m d 3 p 1... e βp N m d 3 p N e βu 1... N d d 3 N Since the kinetic enegy is a sum of tems, one fo each molecule, the coesponding pat of the patition function beaks into a poduct of N integals, each identical except fo the ielevant vaiable of integation and equal to e βp m d 3 p. Since U 1,..., N is not in the fom of a simple sum of tems fo individual molecules, the integation ove the coodinates 1,..., N is vey difficult to cay out. This is why the teatment of non-ideal gases is complicated Chapte 7. Fo an ideal gas, we can set U = and so the integal becomes tivial: d d 3 N = V N..18 since each integation extends ove the volume V of the containe. Z = ξ N..19 whee ξ = V h 3 is the patition function fo a single paticle. e βp m d 3 p.. Equation..19 is the elation between the single paticle canonical patition function and that fo the whole system, if the paticles of the system ae identical, non-inteacting and distinguishable.

15 74 An Intoduction to Equilibium Statistical Mechanics Now e βp m d 3 p = = whee we use the standad integal e β m p x + p y + p z dp x dp y dp z e βp x m dpx e αx dx = e βp y m dpy π α e βp z m dpz = [ ] 3 πm, β ξ = V 3 πm V = h 3 β h 3 πm 3..1 Using..1 in..19 Z N,V,T = V N πm 3N.. h 3N This is the canonical patition function of a classical ideal gas. What is going to happen, if we would use the same fomula, i.e., equation..19 to calculate the patition function fo a system of N indistinguishable, identical, non-inteacting paticles. To illustate that this is wong, we conside two indistinguishable paticles: Z N =,V,T =ξ = 1 e βε 1 e βε = e βε + e βε 1+ε = 1 = }{{ } 1 }{{} paticles both paticles in diffeent in the same state states Howeve, the paticles in the second tem ae counted twice, but since the paticles ae indistinguishable this does not geneate a new state, because e βε 1 +ε = e βε +ε 1. Hence, Z N =,V,T =ξ counts the paticles in diffeent states falsely twice! We can coect fo that by Z N =,V,T = e βε + 1 = 1 = 1 e βε 1+ε = = 1 = e βε + 1! 1 e βε 1 +ε.

16 Elements of Ensemble Theoy 75 Similaly, fo a system of 3 paticles Z N = 3,V,T = = 1 = = 3 paticles in same state = = 1 = = 3 paticles in same state e 3βε + some paticles in same state e 3βε + some paticles in same state In geneal, fo a system of N indistinguishable paticles, Z N,V,T = = 1 = = 3 = = N all paticles in same state e Nβε + some paticles in same state We now use semi-classical agument accoding to which + 1 N! e βε +ε 1 +ε all paticles in diffeent states + 1 3! e βε +ε 1 +ε all paticles in diffeent states e βε +ε 1 +ε 3 + +ε N all paticles in 1 diffeent states gas paticles ae identical. only vey weak o no inteaction between the paticles exists 3. thee ae much moe available enegy levels than paticles classical continuum of quantum mechanics. Thus, the pobability that any single paticle state is occupied by moe than one paticle is vey small. This agument geatly simplifies..3 because only the last tem counts. Thus, fo a system of N indistinguishable, identical, non-inteacting paticles Z N,V,T = ξn N!..3a. Classical non-ideal gas Fo a non-ideal monatomic gas, the inteaction tem U. Theefoe, the canonical patition function fo a classical non-ideal gas is given by Z N,V,T = exp = 1 h 3N Z N,V,T = 1 h 3N whee [ β { 1 m e βp 1 m d 3 p N πm Q β p p N Q = +U 1,..., d N }] d 3 N d 3 p 1...d 3 p N h 3N e βp N m d 3 p N e βu 1... N d d 3 N e βu 1... N d d 3 N..4 known as configuational patition function chapte 7. The calculation of Q is quite difficult. Fo the sake of simplicity, we apply mean field appoximation. Then equation..4 becomes Z N,V,T = 1 3N πm [ V ] N h 3N e UMβ d 3 β

17 76 An Intoduction to Equilibium Statistical Mechanics whee U M is the mean field. We take it as follows: {, < U M = U, U M U Fig..6. U M vs. plot. Theefoe, Z N,V,T = 1 h 3N 3N πm β V V e UMβ d 3 + V e UMβ d 3 N..5 whee V is the volume coesponding to =. Now in the ange to V,U M and in the ange V to V, U M = U. Theefoe, equation..5 becomes Z N,V,T = 1 3N πm [ V ] N h 3N e Uβ d 3 β V Z N,V,T = 1 h 3N πm 3N [ V V e U ] N..6 This is the canonical patition function fo a classical non-ideal gas in the mean field appoximation. This topic will be futhe discussed in Chapte Themodynamic quantities in tems of patition function All the physical quantities can be expessed completely in tems of the patition function Z. Some of them ae shown below.

18 Elements of Ensemble Theoy Entopy Accoding to Boltzmann s entopy elation, the entopy S is elated to the themodynamic pobability Ω though the elation S = k lnω..7 Fo a classical system, the total numbe of molecules is N = n i..8 i and Ω = N! i n i! Taking logaithm and then using Stiling s appoximation g n i i..9 lnω = N lnn N +n i lng i n i lnn i + n i..3 Accoding to classical statistics, fo equilibium state [see equation 3.1.7] n i = g i e α e βe i = g i Ae βe i..31 whee A is a new constant. Using..31 in..3 n i lng i n i lng i Ae βe i + n i lnω = N lnn N + i = N lnn N + i n i lng i i = N lnn N N lna + βe + N = N lnn N lna + βe lnω = N ln N A + βe n i lng i lna i n i + β i n i E i +n i i..3 whee we put n i = N = total numbe of paticles and n i E i = E = total enegy of the system. i i Again using..8 in..31 i n i = Ag i e βe i N = AZ Z = N i A..33 whee Z = g i e βe i patition function. i Afte using..33, equation..3 becomes lnω = N lnz + βe..34 Finally, using..34 in..7, the expession fo entopy becomes S = NklnZ + kβe 1 = NklnZ + k E S = NklnZ + E..35 T

19 8 An Intoduction to Equilibium Statistical Mechanics Taking logaithms on both sides lnz = N lnv + 3N Diffeentiating patially with espect to V Using..49 in..48 This is equation of state fo an ideal gas. lnz V [ ln πmk = N V h ] + lnt...49 pv = N..5. Repoduction of van de Waals equation of state Fom equation..6, the canonical patition function fo a non-ideal gas in the mean field appoximation is Z N,V,T = 1 [ ] N πm 3N/ V V e U Taking logaithms on both sides h 3N lnz = 3N ln πm h lnz = 3N πm ln h Diffeentiating patially with espect to V + N lnv V + NU..51 Now, V N V = bn, whee b is a constant and U N V U = an, whee a is also a constant. V Then equation..51 becomes + N lnv bn+ an V lnz V = a N + N V V bn Using..5 in..48 N p = a + N V V bn p = [ ] [V p + a ] N b = V N V N a + V N b..5 Let V N = v, known as specific volume. p + a v v b=..53 This is the van de Waals equation of state fo a eal gas.

20 Elements of Ensemble Theoy Genealized equipatition theoem Let q i be the genealized co-odinate. We want to calculate the ensemble aveage of the quantity H q i, whee H is the Hamiltonian of the system unde consideation. Using dτ d 3N qd 3N p, q j we can wite in the canonical ensemble H H q i q j e βh dτ q i = q j e βh..54 dτ Let us now conside the integal in the numeato. Integating ove q j by pats, it becomes [ 1 β q ie βh q j + 1 ] qi e βh dq q j j dτ 1 β q j..55 j whee q j 1 and q j ae the exteme values of the coodinate q j. If q j is a space coodinate, then its exteme values will coespond to positions of the walls of the containe and so the potential enegy would be infinite The infinite potential enegy immediately implies [ that the Hamiltonian H of the system would be infinite. Theefoe, the integated pat, i.e., 1 ] β q ie βh q j vanishes. q j 1 The integal pat contains the tem q i / q j, which is equal to δ ij. Finally,..55 becomes 1 β δ ij e βh dτ..55a Substituting this in..54 H q i = δ ij..56 q j independent of H. Similaly H p i p j = δ ij..57 which is independent of the fom of the Hamiltonian H. Equations..55 and..56 ae known as genealized equipatition theoem. This topic will be futhe discussed in Chapte Viial theoem Accoding to the Hamilton s equations of motion, ṗ i = H..58 q i Using this in..57 q i ṗ i =..59 Adding ove all i fom i = 1to3N, we obtain 3N i ṗ i i=1q = 3N..6

21 8 An Intoduction to Equilibium Statistical Mechanics This is known as the viial theoem, because q i ṗ i is known as viial in classical mechanics. i PROBLEM.5: Find the classical canonical patition function of a one-dimensional linea hamonic oscillato. Also find its aveage enegy E. Since this is a one-dimensional case, the classical canonical patition function is βe dp dx Z = e. h whee E is the total enegy of the oscillato and is given by E = p m + 1 mω x. Z = 1 e βp m dp e 1 mω x dx. h Using the standad integal π e αx dx = α, ][ Z = 1 h [ πm β This is the canonical patition function. The aveage enegy of the oscillato is π mβω E = lnz β ] = 1 β =. = π h βω = π h ω. PROBLEM.6: Find the canonical patition function of a elativistic classical ideal gas with enegy-momentum elationship E = pc. Fo a classical ideal gas, the canonical patition function is Z = e βe d3n pd 3N x h 3N = ξ N, whee ξ is the single paticle patition function and is given by ξ = e βpc d3 pd 3 x h 3 = V π π h 3 d pe βpc p dθsinθ dφ = 4πV h 3 = 4πV 1 h 3 βc 3 d βpce βpc βpc = 4πV 3 h 3 Γ3 c d pe βpc p

22 Elements of Ensemble Theoy 85 Also Z can be witten as Now the mean value of n is n N C n e βnw n = n N C n e βnw n n N C n e βnw = n Z Z = N C n e βnw n = 1 lnz w β = N w [ ln 1 + e βw] β iii = N e βw 1 + e βw n + ne βw = Ne βw n N n = e w When n N, n Ne w iv PROBLEM.1: Show that the patition function of an exteme elativistic gas consisting of 3N indistinguishable paticles, with enegy-momentum elationship ε = pc, c being the speed of light and moving in one dimension is given by Z N,L,T= 1 [ 3N! L ] 3N, hc whee L being the length of the space available. The single paticle canonical patition function is ξ = 1 e βε dqd p = L e βpc dp= L h h hc Theefoe, the canonical patition function of the system of N non-inteacting indistinguishable paticles is Z N,L,T = ξ3n 3N! = 1 [ ] 3N L ii 3N! hc PROBLEM.11: Conside a system in themal equilibium at tempeatue T its two states with enegy diffeence eg occu with elative pobability e eg / deg. Calculate the tempeatue. Boltzmann s constant k = eg / deg. Let P 1 and P be espectively the pobabilities of the state of enegies E 1 and E. Then P 1 = ce E 1 and P = ce E, whee c is a constant of popotionality. P 1 = e E E1 P T = E E 1 P1 k ln P = lne = K i

23 86 An Intoduction to Equilibium Statistical Mechanics PROBLEM.1: The fist vibational enegy of a diatomic molecule is 6 cm 1 above the gound state. Calculate the elative population of molecules in these two levels at T = 4 K. Let N and E o be espectively the numbe of molecules and the enegy of the gound state and N 1 and E 1 be the coesponding values fo the fist excited state. N 1 = N e E 1 E Now i E 1 E =hv = hλ c = hcv = = eg N 1 = e E 1 E = e N =.1 N 1 =.1 N PROBLEM.13: Conside a system which can take only thee diffeent enegy states E 1 =, E = egs,e 3 = egs. Find the pobability that at tempeatue 1 K the system may be a in one of the micostates of enegy E 3, b in the gound state E 1. Let P 1, P and P 3 be espectively the pobabilities with which the micostates can occu in thee enegy values E 1, E and E 3. Then P 1 = ce E 1 = c P = ce E = ce = ce 1 and P 3 = ce E 3 = ce = ce whee c is a constant of popotionality. Again these thee states can occu in, 5 and 4 diffeent ways espectively. Then P 1 = c, P = 5ce 1 and P 3 = 4ce. But Since e =.7. 3 i=1 P i = 1 P 1 + P + P 3 =1 c + 5ce 1 + 4ce = 1 c = c = a The pobability fo the system to be in one of the micostates of enegy E 3 is P 3 = 4ce = =.1.

24 Elements of Ensemble Theoy 89 Again Compaing v and vi Finally, using vii in iii E = lnz β Fβ = lnz F = lnz P lnp = E + F = 1 1 F E= TS= S k S = kp lnp [Poved] viii In case of micocanonical ensemble, we have a goup of Ω states which ae equally likely to occu fo each membe. The value of P is then 1 Ω fo each of these states and fo all othes. Consequently, S = k Ω =1 vi vii { } 1 1 Ω ln = k lnω [Poved] ix Ω PROBLEM.17: The states of a system ae i a goup of g 1 equally likely states with a common enegy ε 1, and ii a goup of g equally likely states with a common enegy ε ε 1. [ ] P1 P a Show that the entopy of the system is given by S = k P 1 ln + P ln [ { } g 1 ] g g b Also show that S = k lng 1 + ln 1 + e x x + g 1 1 +g 1 /g e x, whee x = ε ε 1, assumed positive. c Check that at T,S k lng 1. a The canonical patition function is and P ln P g Z = g e βe P = g e βe Z P P = lnz βe i ii = e βe ln g Z g = lnz βe iii

25 9 An Intoduction to Equilibium Statistical Mechanics The Helmholtz fee enegy is Hence, Again Compaing v and vi F = E TS df = de TdS SdT = de de + PdV SdT = PdV SdT S = F T E = F + TS= F T Using vii in iii, we obtain P ln b We know that Using ix in viii P g P = F V V T [ ] F F = F + β = T V β V β Fβ V g 1 E = lnz β F = lnz = F E = S k S = k P ln [ ] P1 P S = k P 1 ln + P ln g P 1 = g 1e βε 1 Z P = g e βε Z S = k [P 1 βε 1 lnz+p βε lnz] = k [βp 1 ε 1 + P ε +P 1 + P lnz] P g Poved = k [βp 1 ε 1 + P ε +lnz] P 1 + P = 1 [ ] g 1 ε 1 e βε 1 + g ε e βε = k β + lnz Z [ ] g 1 ε 1 e βε 1 + g ε e βε = k β g 1 e βε 1 + g e βε + lng 1 e βε 1 + g e βε [ g 1 ε 1 e βε 1 + g ε e βε { = k β g 1 e βε 1 + g e βε + lng 1 e βε 1 +ln 1 + g } ] e βε ε 1 g 1 iv v vi vii viii ix

26 9 An Intoduction to Equilibium Statistical Mechanics If the tempeatue T is so high that the themal enegy is lage compaed to the sepaation hν between enegy levels, i.e., βhν = hν 1, 1 E hν = hν hν + hν = classical esult. hν hν PROBLEM.19: A classical point paticle is moving in a 3-dimensional hamonic oscillato potential well, V =1/K = 1/Kx + y + z at absolute tempeatue T. Obtain a fomula fo the pobability that the paticle is in between and + d fom the cente of attaction. Also obtain a fomula fo the mean squae distance of the paticle fom the cente of attaction and check you esult by compaison with the equipatition pinciple. The pobability that the paticle will be in phase space volume element d 3 d 3 p βp e m β 1 K d 3 pd 3 βp e m β 1 K d 3 p sinθ dθ dφ d = = e βp m β 1 K d 3 pd 3 e βp m β 1 K d 3 p sinθ dθ dφ d Theefoe, the pobability that the paticle will lie in the ange and + d is θ φ p Pd = = θ φ p e βk sinθ dθ dφ d e βp m d 3 p e βk d e βk sinθ dθ dφ d e βp m d 3 p e βk d = e βk d 1 4 π Kβ Pd = 4 3 Kβ e 1 βk d i π The mean squae distance is = Pd Pd = 4 e βk d = e βk d π βk 1 4 π βk Theefoe, the aveage potential enegy = 1 K = 1 K 3 K equipatition law. 3 = 3 = 3 βk K ii = 3, which is consistent with the PROBLEM.: Suppose a system of N non-inteacting paticles at tempeatue T in an extenal magnetic field H pointing along the z-diection. Assume that each atom has spin 1/ and an intinsic magnetic moment µ.

27 Elements of Ensemble Theoy If the magnetic moment of each atom can point eithe paallel o anti-paallel to the extenal field H, what is the mean magnetic moment µ H in the diection of H, i.e., z-diection of such atom?. Calculate the patition function of one atom, the mean magnetic moment along the field diection and hence the susceptibility of the system, fist by teating the system classically and then quantum mechanically. Each atom can be in two possible states: the state + whee its spin points up i.e., paallel to H and the state whee its spin points down i.e. anti-paallel to H. In the + state, µ is paallel to H and so µ H = µ. The coesponding magnetic enegy of the atom is E + = µh. The pobability of finding the atom in this state is P + = Ce βe + = Ce βµh whee C is a constant of popotionality and β = 1. Similaly, in the state, µ is anti-paallel to H and so µh = µ. The coesponding magnetic enegy of the atom is E =+µh. The pobability of finding the atom in this state is The mean value of µ H is µ H = P + µ+p µ P + + P P = Ce βe = Ce βµh = µ eβµh e βµh µh e βµh = µtanhβµh=µtanh. + e βµh When the system is teated classically The enegy of a magnetic dipole in the pesence of the extenal magnetic field is E = µ H = µh cosθ The canonical patition function of the system is Z = ξ N whee ξ is the single paticle canonical patition function and is given by π π ξ = expβµh cosθ= e βµh cosθ sinθ dθ dφ = 4π sinhβµh θ φ= θ= βµh [ Z = 4π sinhβµh ] N βµh The aveage magnetic moment along the field diection is given by µcosθexpβµh cosθ θ µ H = µcosθ = = 1 [ expβµh cosθ β H lnξ = µ cothβµh 1 ] βµh θ = µl βµh i ii iii iv v vi vii

28 94 An Intoduction to Equilibium Statistical Mechanics whee Lx is the so-called Langevin function: Lx=cothx 1 x viii Theefoe, the magnetization of the system is given by M = Nµ H = NµL x ix Fo magnetic fields so stong o tempeatues so low that the paamete x 1, Lx = 1 and so M = Nµ satuation. On the othe hand, fo high tempeatues o the magnetic fields so weak that the paamete x 1,Lx = x and so 3 M = Nµ 3 H Theefoe, the high tempeatue susceptibility of the system is given by M χt =Limit = Nµ H H 3 = C T This is Cuie law of paamagnetism, the constant C being the Cuie constant of the system. x xi When the system is teated quantum mechanically In this case, the majo modification aises fom the fact that the magnetic dipole moment µ and its component µ H along the field diection cannot have abitay values. The magnetic moment µ is elated to the total angula momentum ħ J though the elation µ = gµb J xii whee µ B = eħ mc, known as Boh magneton and g is the so-called g facto of the atom, e = chage of an electon, m = mass of an electon and c = velocity of light in vacuum. Using xii in iii, we have E = gµ B J H = gµb HJ z xiii since H is along the z-diection. In quantum mechanical desciption, the values which J z can assume ae discete and ae given by J z = m, whee m = J, J 1,...,J 1,J. The possible enegies of the atom ae E m = gµ B Hm xiv Then the patition function of one atom is Z = m e βe m = J e βgµ BHm m= J xv

29 Elements of Ensemble Theoy 95 Let x = βgµ B H = gµ BH Z = J e xm m= J a dimensionless paamete. Theefoe, = [e xj + e xj e xj] = e xj e xj+1 1 e x = e xj+ 1 e xj+ 1 e x e x = sinh J + 1 x sinh x Z = sinh J + 1 βgµb H sinh 1 βgµ BH Now the pobability of finding the atom in a state labelled m is given by P m = Ce βe m xvi xvii whee C is a constant of popotionality, detemined fom the nomalization condition, m P m = 1. Theefoe, the mean value of the magnetic moment along the field diection i.e. z-diection is µ H = m µ H P m m P m = J m= J µ H e βgµ BHm J e βgµ BHm m= J Since the field H is along the z-diection, µ H = µ z = gµ B J z = gµ B m, fom iv. µ H = J m= J gµ B me βgµ BHm J e βgµ BHm m= J = 1 lnz β x = gµ B [ J + 1 = 1 β J m= J βgµ B me βgµ BmH J m= J x H = gµ lnz B = gµ B x x cosh J + 1 x sinh J + 1 x e βgµ BmH [ ln sinh 1 cosh x sinh x ] = 1 1 Z β Z H = 1 lnz β H J + 1 x lnsinh x ] xviii µ H = gµ B JB J x whee B J x= 1 [ J + 1 coth J + 1 x 1 J coth x ], known as Billouin function. Theefoe, the magnetization of the system is given by M = N µ H = Ngµ B JB J x Fo stong fields and low tempeatues x 1, the function B J x = 1 fo all J and so M = Ngµ B J magnetic satuation. xix

30 96 An Intoduction to Equilibium Statistical Mechanics On the othe hand, fo high tempeatues and weak fields x 1,B J x = x, so that J M = N gµ BJ H = N g µ BJ J + 1 J 3 Theefoe, the susceptibility is given by χt = Ng µ BJ J + 1 1, which shows that the Cuie law is again valid. 3 T H xx PROBLEM.1: Show that entopy inceases when two ideal gases at the same tempeatue and pessue diffuse into each othe. Discuss Gibbs paadox in this connection. Accoding to Boltzmann s elation, the entopy S of a system is elated to the themodynamic pobability Ω and is given by Fom Maxwell-Boltzmann statistical count Ω = N! i g n i i n i! lnω = lnn! + i = lnn! + i = lnn! + i Ω = lnn! + i S = k lnω n i lng i lnn i! n i lng i n i lnn i + n i n i lng i lnn i + 1 n i ln g i + 1 n i Again fo most pobable state, Ω o lnω is maximum and then we have i ii [see equation 3.1.7]. Using iv in ii n i = e g i n i = e α g i e E i α+ E i ln g i n i = α + E i Ei lnω = lnn! +n i i + α + 1 = lnn! + 1 i n i E i + α i n i +n i i iii iv

31 98 An Intoduction to Equilibium Statistical Mechanics S B = 3 3 ] πm [V Nk+ Nkln. Hence, the total entopy is simply [ 3 ] πm S A + S B = 3Nk+ Nkln V We now emove the patition vey slowly so that the gases diffuse into each othe. Theefoe, the total numbe of molecules N of the two gases occupy a volume V. Since thee is no change in pessue and tempeatue, the entopy of the system afte the patition is emoved is [ 3 ] πm S = 3Nk+ Nkln V x Now, accoding to the additive popety of entopy But fom ix and x, we see that S = S A + S B h h h S [S A + S B ]=Nkln >. which is not equal to zeo as equied by xi. This paadox was fist discussed by Gibbs and is commonly efeed to as the Gibbs paadox. The oot of the difficulty embodied in the Gibbs paadox is that we teated the gas molecules as distinguishable, as though intechanging the positions of two molecules would lead to a physically distinct state of the gas. This is not so. Indeed, if we teated the gas by quantum mechanics, the molecules would have to be egaded as indistinguishable see Chapte 3. In this case, N! pemutations of the molecules among themselves do not lead to physically distinct situations, so that we should subtact the tem k lnn! fom vi, so that [ S = 5 3 ] Nk+ Nkln V πm N h xiii This has been veified expeimentally at high tempeatues and is known as Sacku-Tetode equation. Equation xii can be witten as S = Nkln [ V πm N h 3 ] 5 e Using this expession, it can be easily veified that S A + S B S = and hence the paadox is emoved. PROBLEM.: Imagine that a system R 1 has pobability P 1 of being found in a state of and a system R has pobability P s of being found in a state s. Then one has S 1 = k P 1 lnp 1 and ix xi xii xiv

32 Elements of Ensemble Theoy 99 S = k P s lnp s. Each state of the composite system R= R 1 + R can be labelled by a pai of numbes, s. Let the pobability of R being found in this state be denoted by P s. Then the entopy is defined by S = k s P s lnp s. If R 1 and R ae weakly inteacting so that they ae statistically independent, then P s = P 1 P s. Show that unde these cicumstances, the entopy is simply additive, i.e., S = S 1 + S. We have S = k = Ps 1 s s P s lnp s = k kp 1 lnp 1 = kp 1 lnp 1 + S = S 1 + S Poved P 1 Ps ln P 1 P s Ps = P 1 P s s + kps lnps s P 1 kps lnps s P 1 = 1 = Ps s PROBLEM.3: Suppose that a system R 1 has pobability P 1 of being found in a state and a system R has pobability P s of being found in a state s. The entopy of R 1 is S 1 = k P 1 lnp 1 and that of R is S = k s P s lnp s. Each state of the composite system R= R 1 +R is then labelled by the pai numbes, s. Let the pobability of R being found in this state be P s ; its entopy is defined by s = k s P s lnp s. Assume that R 1 and R ae not weakly inteacting, so that P s P 1 P s. Of couse P 1 = P s and P s = P s. Moeove, all the pobabilities ae popely nomalized so that P 1 = 1, P s = 1, s P s = 1. Show that a S s 1 + S =k,s P s ln P 1 P s P s, and b S S 1 + S. a We have P 1 = P s p s = P s S 1 = kp 1 lnp 1 S = kp lnps s i ii iii iv

33 1 An Intoduction to Equilibium Statistical Mechanics Using i in iii Again using ii in iv Adding v and vi Again accoding to the question [viii - vii] gives S 1 = k S = k S 1 + S = k S = k S S 1 + S =k 1 b Using the inequality, ln x x 1 ln P1 P s P s P s ln P1 P s P s k P 1 Ps P s lnp 1 s P s lnps s lnp 1 Ps s P s lnp s s v vi vii viii +P s ln P1 Ps Poved ix s P s 1 P s P s P s ln P1 Ps s P s [S S 1 + S ] P 1 Ps P s 1 =P 1 Ps P s S S 1 + S x PROBLEM.4: Conside a system of N vey weakly inteacting paticles at a tempeatue T sufficiently high so that classical statistical mechanics is applicable. Each paticle has mass m and is fee to pefom one-dimensional oscillation about its equilibium position. Calculate the heat capacity of this system, if the estoing foce is popotional to x 3. The estoing foce is given by whee K is the constant of popotionality. Hence, the potential enegy of a paticle is F = Kx 3 i Theefoe, the total enegy of a paticle is V = 1 4 Kx3 ε = p m kx 3 ii iii

34 1 An Intoduction to Equilibium Statistical Mechanics Z L L L Y X Now and L e mgz dz = mg 1 e mgl L ze mgz dz = mg mgz = mg 1 e mgl mg 1 e mgl mg L mgl e mg L mgl e mg = mgl e 1 e mgl mgl 1 e mgl mgz = + mgl 1 e mgl 1 Ans. ii The aveage kinetic enegy of a molecule is given by [ πm p p p = m e β m d +mgz p 3 pd 3 βp e m d 3 3 ] p m e β p = m β β = m d +mgz 3 pd 3 e βp m d 3 3 = 3 p πm β = 3 iii β PROBLEM.6: Calculate the otational patition function fo heteonuclea and homonuclea molecules sepaately.

35 Elements of Ensemble Theoy 13 The otation of a diatomic molecule is specified by the angles θ,φ and the coesponding momenta P θ,p φ and its kinetic enegy assumes the fom ε ot = P θ I + P φ I sin θ. Fo heteonuclea diatomic molecule: ξ ot heteonuclea = 1 h e 3/ ldp θ dp φ dθdφ = 1 π h φ= ξ ot heteonuclea = π I h dφ pφ= e p I dpθ = 1 h π 1 πi π θ= π θ= dθ e p φ I sin θ dpφ pφ= dθ 1 πi sinθ Fo homonuclea diatomic molecule: Fo homonuclea diatomic molecule, φ anges fom to π see Section ξ ot homonuclea = π I h PROBLEM.7: Study the themodynamics of a system of N non-inteacting diatomic molecules, each having an electic dipole moment µ, placed in an extenal electic field E. Assume that i the system is classical and ii µe, whee T is the absolute tempeatue of the system. i The enegy of a diatomic molecule in the pesence of the electic field E is given by { } ε = P m + Pθ I + P φ I sin µe cosθ θ whee I is the moment of inetia of the molecule. Now the single paticle canonical patition function is given by ξ = 1 h 3 e βε d dθ dφ dpdp θ dp φ V 1 3 ξ = 1 h 3 = d π φ= dφ P= d pe βp m P θ = dp θ e βpθ m i ii

36 14 An Intoduction to Equilibium Statistical Mechanics βµe cosθ dθe dpφe βp φ I sin θ θ= Pφ= = mπ 1 Iπ h 3V 3 π βµe dθe cosθ 1 Iπ β β β sinθ θ= 5 miπ = h 3 1 π βµe cosθ dθsinθe β 3 5 miπ = h e β 3 βµe βµe e βµe 5 miπ 1 ξ = µeh 3 sinhβµe iii β 5 Theefoe, the canonical patition function of the system is [ ] 5 N [ ] N miπ Z = ξ N sinh βµe = µeh 3 β 5 iv 5 miπ lnz = N ln µeh 3 + N [lnsinhβµe 5 ] lnβ v The mean enegy of the system is given by ε = lnz β = 5 N µe β NµE cothβµe=5 N NµE sinh The specific heat at constant volume is given by ε C V = T Fo T, i.e., x = µe V C v 5 = Nk Now fo µe/ 1, i.e., fo high tempeatue cosec h µe = 5 Nk Nk µe = 1 µe cosec h Lt cosec hx = x µe vi vii viii µe ix

37 16 An Intoduction to Equilibium Statistical Mechanics and 3 = n= n e un+ 1 = e u n= n e ν = e [ u e u + e u + 3 e 3u + ] = e u e u [ 1 + e u + 3 e u + ] = e 3u 1 + e u 1 e u 3 iv Substituting ii, iii and iv in i [ { }] ξ = e u e u 1 + e u 1 e u 1 + xu 1 e u + e u 1 e u = e u [ { e u 1 e u 1 + xu 1 e u + 1 }] v 4 Theefoe, the canonical patition function fo the system is given by Z = ξ N lnz = N lnξ = N [ u ln 1 e u + ln Then the intenal enegy of the system is whee { 1 xu E = lnz lnz = β T = hv lnz u { x e u = Nhv 1 e u 1 e u + 1 e u { 1 + xu e u 1 e u }] ue u ue u 1 e u 1 e u e u 1 e u } } = E + E coection vii E = 1 Nhv Nhv + e u 1 = 1 Nhv + The coesponding specific heat at constant volume is C V = E T V = Nk hv { e u E coection = xnhv 1 e u 1 xu e hv [ e hv 1 Nhv e hv 1 vi viii ] ix ue u ue u 1 e u 1 e u { } e u 1 e u } x

38 Elements of Ensemble Theoy 17 Upto the fist ode in x, x educes to [ E coection = xnhv The coesponding specific heat is e u 1 e u ue u 1 e u 4uee u 1 e u C v coection = xnku e u 1 e u + 1ue u 1 e u 4 4e u 1 e u 3 4e u ] 1 e u 3 This is the coection to the specific heat upto the fist ode in x. Comment: Fo u = hv 1, i.e., fo high tempeatue, C V coection = xnk 4 4 4Nk = xnk u hv = x T T, hv which shows that the coection due to the anhamonicity inceases with tempeatue, T. PROBLEM.9: Calculate the canonical patition function and hence the specific heat at constant volume fo a classical system of N non-inteacting diatomic molecules enclosed in a box of volume V at tempeatue T. The enegy of a diatomic molecule can be witten as E p 1, p, 1, = 1 p m 1 + p 1 + K 1 i whee p 1, p, 1 and ae the momenta and position coodinates of the two atoms in a diatomic molecule. Let R = 1 + and ρ = 1. Then the Jacobian is 1 J = R R 1 = = 1 J = 1 ii ρ ρ d 3 1 d 3 = d 3 Rd 3 ρ Now the single paticle canonical patition function is ξ = e βh d 3 1 d 3 d 3 p 1 d 3 p = e βh d 3 1 d 3 d 3 Rd 3 ρ [ ][ ][ ][ ] = e βp 1 d 3 p 1 e βp d 3 p d 3 R e βkp d 3 ρ 3 πm πm = β β 8π 3 m = V K 3 1 β π V βk xi xii iii

39 18 An Intoduction to Equilibium Statistical Mechanics Theefoe, the canonical patition function fo the system is 8π Z = ξ N = V N 3 m 3N 1 K β 9N [ 8π lnz = ln V N 3 m 3N ] 9N K lnβ The mean enegy of the system is E = lnz β Hence, the specific heat at constant volume is E C V = T = 9N β = 9 N V iv v vi = 9 Nk vii PROBLEM.3: In Poblem.31, calculate the mean squae molecula diamete, 1. The mean squae molecula diamete is 1 1 e βe d 3 1 d 3 d 3 p 1 d 3 p = e βe d 3 1 d 3 d 3 p 1 d 3 p = = ρ e β[ mp 1 1 +p + 1 Kρ ] d 3 1 d 3 d 3 p 1 d 3 p e β[ mp 1 1 +p + 1 Kρ ] d 3 1 d 3 d 3 p 1 d 3 p ρ e 1 βkρ d 3 ρ e 1 βkρ d 3 ρ = 1 T. ρ 4 e 1 βkρ dρ ρ e 1 βkρ dρ = K x 3 e x dx x 3 e x dx = K Γ 5 Γ 3 = 3 K PROBLEM.31: Conside a classical system of N non-inteacting diatomic molecules enclosed in a box of volume V and tempeatue T. The enegy fo a single molecule is E p 1, p, 1, = 1 m p 1 + p + ε 1, whee p 1, p, 1 and ae the momenta and position coodinates of the two atoms in a molecule. ε and ae given positive constants and 1 = 1. Find the specific heat of the system at a constant volume as a function of tempeatue.

40 Elements of Ensemble Theoy 19 The enegy of a molecule is E p 1, p, 1, = 1 m Let R = 1 + and ρ = 1. Then the Jacobian is 1 R R J = = = 1 J = 1 ii ρ ρ d 3 1 d 3 = d 3 Rd 3 ρ p 1 + p + ε 1 i The single paticle canonical patition function is ξ = e βh d 3 1 d 3 d 3 p 1 d 3 p = e βh d 3 1 d 3 d 3 Rd 3 ρ [ ] [ ] [ ] [ ] = e βp 1 m d 3 p 1 e βp m d 3 p d 3 R ρ βε ρ d 3 ρ [ 3 m ] [ m = 4π β τ ] 1 3 4π β τ V 4π βε 3 y e y x dy whee y = βερ and x = βε. Now y e y dy = y e y ye y + e y dy = y e y ye y + e y Similaly, y e y dy = y e y ye y e y Again y e y x dy = x x y e y dy = x e x + xe y e x y e y x dy+ y e y x dy = e x x y e y dy+ e x x y e y dy iii iv v vi vii = e x [ x e x xe x + e x ] + e x [ x e x + xe x + e x] viii y e y x dy = x + 4 e x ix

41 Elements of Ensemble Theoy 111 Now, when the cylinde is otated, it becomes enegetically favouable fo the molecules to move towads the edge of the cylinde, so we expect the density to incease with adius. In a fame of efeence otating with a gas, a molecule of mass m at a distance fom the axis expeiences a centifugal foce F c = mω ii away fom the axis. The centifugal potential enegy is V c = 1 mω Fom the Boltzmann distibution law, we expect the pobability of a molecule to be popotional to exp V c and so we can wite mω σ=aexp iv whee A is a constant of popotionality. Now the total mass of the gas within the cylinde is given by a a mw M = πl σd = πla exp The integal can be easily evaluated by noting that [ d mω ] exp = mω mω d exp M = πla [ mω a ] mω exp 1 Equating i and vii, we have the expession fo A: A = σ mω a 1 mω a Using viii, we get the expession fo σ as σ= σ mω a [ exp [ exp mω a exp [ mω exp d iii v vi vii ] viii 1 mω a a 1] 1 ] ix Let R = a and Ω = mω a. σr σ = Ω expωr [expω 1] x

42 11 An Intoduction to Equilibium Statistical Mechanics Witing the function as a seies expansion, we have σr 1 + ΩR + 1! = Ω Ω R σ Ω + 1 =! Ω Ω + 1! Ω Ω +... xi Fo small ω i.e., fo small Ω, we can wite xi appoximately as σr 1 + ΩR = xii σ 1 + Ω which shows that σr=σ, i.e., the density emains unchanged when R = 1 a = 1 = a. Thus fo small ω, the density emains unchanged at a distance a fom the axis of the cylinde. PROBLEM.33: Conside an ideal gas at absolute tempeatue T in a unifom gavitational field descibed by acceleation g. By witing the condition of hydostatic equilibium fo a slice of the gas located between heights z and z+dz, deive an expession fo nz, the numbe of molecules pe cm 3 at height z. Also show that this esult is identical with that deived fom statistical mechanics. Let us conside a slice between heights z and z + dz. Let the pessue at z be p and that at z + dz be p + dp. z + dz, p + dp z, p Then fo hydostatic equilibium, whee, A = coss-sectional aea of the slice n = numbe density m = paticle mass g = acceleation due to gavity. pa p + dpa = mngadz dp= mngdz ii Again teating the gas as an ideal gas, we can wite Using iii in ii, we have p = n dp= dn dn n = mg dz i iii

43 Elements of Ensemble Theoy 113 On integation Let at z =,n = n. Using ideal gas law: p = n, p= n lnn = mg z + constant n = ne mgz p = pe mgz Now the pobability P, pd 3 d 3 p that the molecule has position lying in the ange between and + d and momentum in the ange between p and p + d p is given by P, pd 3 d 3 p d3 d 3 p h 3 e β p m +mgz Theefoe, the pobability Pzdz that a molecule is located at a height between z and z + dz is Pzdz 1 h 3 x,y dx dy π π P= θ= φ= P β e m P sinθ dpdθ dφ e βmgz dz But x,y dx dt = A, coss-sectional aea of the slice. Pzdz = constant A 3 πm / e mg h 3 z β Let at z =,Pz z= = P. P=constant A πm h 3 β Pz=Pe mg z Thus, the pobability of finding a molecule at height z deceases exponentially with height. Now nz dz pz dz = constant Pe mgz dz xii At z =,n = n constant P=n iv v vi vii viii ix x xi which is same as v. nz=ne mgz xiii PROBLEM.34: Conside a system consists of N weakly inteacting paticles, each of which can be in eithe of two states with espective enegies ε and ε ε > ε 1. Calculate explicitly the mean enegy E and heat capacity C V of this system.

44 Elements of Ensemble Theoy 115 Conside a pai of one +ve and one ve paticle. a a + + a + + The adjoining figue shows that thee ae 4 possible states of this pai. Since the potential enegy of a chaged paticle of chage in an extenal electic field is V x= eεx, the aveage sepaation between a +ve ion and a ve ion is given by x = a eβ a εe a e β a εe e β a εe + e β a εe = a βeεa tanh If thee ae N numbe of pais pe unit volume, the electic polaization o aveage dipole moment pe unit volume is given by µ = Ne x = Ne a βeεa tanh PROBLEM.36: Calculate the canonical patition function and hence the specific heat of a classical ideal two-dimensional gas. i ii The enegy of a single paticle is ε = p x + p y m Then the single paticle canonical patition function is ξ = 1 h e mp β x+p y dx dy dpx dp y = A [ ] [ h e βp x ] m dpx e βp y m dpy = A πm h β whee A is the aea. i ii

45 116 An Intoduction to Equilibium Statistical Mechanics Hence, the canonical patition function of the system is given by Z = AN mπ N h 3N iii β The mean enegy of the system is E = The specific heat at constant volume is C V = lnz = N β V E = Nk iv T V PROBLEM.37: Suppose that a wie of adius is placed along the axis of a metal cylinde of adius R and length L. The wie is maintained at a positive potential V with espect to the cylinde and the whole system is at some high absolute tempeatue T. As a esult, electons emitted fom the hot metals fom a dilute gas filling the cylindical containe and in equilibium with it. The density of these electons is so low that thei mutual electostatic inteaction can be neglected. Usine Gauss theoem, obtain an expession fo the electostatic potential at a adial distance fom the wie < < R. [Assume the cylinde to be infinitely long so that end effects ae neglected.] Also find the density of the electon gas filling the space between the wie and cylinde, as a function of the adial distance in equilibium. V V = R L