n=0 We cannot compute this exactly, but we know [the maximum term approximation ]

Transcription

1 Homework [Explicit esemble equivalece for the Frekel defects] The caoical partitio fuctio for Frekel defects reads ZT ) = mi{n,m} = N ) ) M e βɛ. HW5.1) We caot compute this exactly, but we kow [the maximum term approximatio ] [ ) ] N M log max )e βɛ = log Z, HW5.2) if we igore the O[log N] terms. Therefore, we ca compute A from Z. Compute the Helmholtz free eergy A assumig /Nad /M may be igored relative to 1. Warig: However, do ot itroduce this approximatio too quickly you may get Y = ). Sol. Notice that max ca be located through the logarithmic differetiatio, or recall max x [log fx)] = log[max x fx)]. Therefore, we maximize Y : Y log [ N ) M )e βɛ ] [ = N N log N + 1 N M ) log [ M log M + 1 M 1 N ) log )] 1 M )] βɛ. Let us differetiate Y wrt recall that we eed ot differetiate log terms, because their derivatives kill each other, so almost immediately we get): That is, Y = log /N 1 /N log /M 1 /M βɛ = e βɛ = This is already i Y may be rewritte as max Y = log 2 NM N )M ) + log NM 2 2 N )M ). HW5.3) N log ) 1 N ) M log 1 M 1 ) βɛ M ) βɛ = log 1 N )M ) N log M log N = N log 1 ) M log 1 ) = 2. HW5.4) N M We have used HW5.3). Now, we may get from the same equatio as = NMe βɛ/2. 1

2 Thus, we get log Z = N log 1 ) ) M N N e βɛ/2 M log 1 M e βɛ/2 2 NMe βɛ/2. Or A = 2k B T NMe βɛ/ [Determiig N A ] 1 By usig particles of radius a =.55 µm suspeded i water at T = 23.6 C shear viscosity η =.919 mpa s), the mea square displacemets were measured. The results o a 2D observatio table are summarized as 2 R 2 = t m 2, HW5.5) where time t is measured i secods. Assumig that we kow R = 8.31 J/mol K, estimate Avogadro s costat N A. Sol. We kow R 2 = 2dDt = 4Dt, which maybe obtaied from the diffusio equatio as Eistei did see 9.9). Therefore, D = /4 = m 2 /s. We use Eistei s formula: D = k B T/ζ. Therefore, ζ = 6π = k B = Dζ/T = /296.6 = This correspods to N A = , but gettig a accurate value with this method is ot so trivial. 5.3 [Electric polarizatio of ideal gas] Cosider a ideal gas cosistig of N diatomic molecules with a electric dipole momet µ. We impose a exteral electric field E. The thermodyamic coordiates for this system are E, V, P ), where P is the electric polarizatio. The stadard Gibbs relatio thus reads de = T ds P dv + E dp, HW5.6) where E is the electric field. We wish to study this system uder costat T ad costat E. 1 I forgot to ask this questio two weeks ago. 2 A example of such experimets may be foud i Marco A. Catipovic, Paul M. Tyler, Josef G. Trapai, ad Ashley R. Carter, Improvig the quatificatio of Browia motio, Am. J. Phys ). 2

3 1) Itroduce a coveiet thermodyamic potetial F ad write dow df, Sol. The thermodyamic space is spaed by E, V ad P as you ca see from the Gibbs relatio for E. Now we wish to put this i a thermostat, ad also set E costat. Therefore, the coveiet thermodyamics potetial F ) is give by This Legedre trasformatio meas F = sup[t S + P E E]. S,P df = SdT P dv P de. 2) We wish to obtai F directly with the aid of a particular esemble theory i.e., a particular statistics) as F = k B T log Y. Fid a appropriate formula for Y ad cofirm F = k B T log Y. Sol. Let us follow our stadard guess work : max S,P eβt S+P E E] = e βf. Note that e βst = we, V, P ) w is a fuctio of thermodyamic coordiates of our system), so max E,P we, V, P )eβ[p E E] = e βf, we expect Y = microstates e β[p E E]. This may be iterpreted as the max term approximatio of the sum over all the microscopic cofiguratios. Note that P = µ i Therefore, F = k B T log Y with Y = dγ e βh+β µ i E. If you are familiar with the esemble formalism, you ca immediately write the above formula dow. To cofirm the esemble equivalece, we must show up to O[N] factor C Or to show with a O[N] width E Y = C max E,P we, V, P )eβ[p E E]. max we, V, P E,P )eβ[p E E] we, V, P )e β[p E E] E max we, V, P E,P )eβ[p E E] + O[1], E,P 3

4 where O[1] is a lit. I oly require you to recall this geeral logic, sice to show the possibility of E = O[N] is ot quite trivial. I will probably come to this with the grad caoical esemble. 3) The system Hamiltoia cosists of the usual traslatioal kietic eergy part ad the molecular rotatioal portio. Sice we are ot iterested i the traslatioal part, ad sice we kow that portio ca be factored out, here we cocetrate o our attetio to the rotatioal portio of the molecules. The rotatioal kietic Hamiltoia reads for each molecule H r = p2 θ 2I + 2I si θ. where the polar coordiates are defied aroud E see the figure). p2 φ HW5.7) directio of φ E θ μ Remark Perhaps you forgot aalytical mechaics. θ ad φ are chose as the caoical coordiates, ad their cojugate mometa i.e., agular mometa) are writte as p θ ad p φ, respectively. 4) Compute the rotatioal portio of) Y ad get P as a fuctio of T ad E. 3 Sol. It is a well-kow questio, so you ca fid the aswer somewhere easily, BUT at least you should try to say it i your words. Y or its rotatioal portio) reads Y = Y1 N, where Y 1 is the partitio fuctio for a sigle molecule: The caoical volume elemet for oe molecule rotatioal part) reads dγ 1 = dθdp θ dφdp φ ever write si θdθ): Y 1 = π dθ 2π dφ dp θ dp φ e β[p2 θ /2I+p2 φ /2I si θ]+βµe cos θ. The agular mometum itegrals are Gaussia itegrals: 2π π Y 1 = dφ dθ 2πI/β 2πI si 2 θ/βe βµe cos θ = 2πIk B T 2π π βµe cos θ dφ si θdθ e 3 You eed ot write the cotributio of the traslatioal motio, i.e., the ordiary ideal gas caoical partitio fuctio. 4

5 1 = 4π 2 Ik B T dxe βµex = 8π 2 sih βµe Ik B T 1 βµe. Notice that the correct volume elemet si θdθ is automatically obtaied. Therefore, igorig irrelevat additive costats, we get F T, V, E) = Nk B T log sih βµe βµe. Usig the Gibbs relatio, we get the E- or i our choice the z)-compoet of P as ) F P = = NµLβEµ), E T,V where Lx) = d sih x log = coth x 1 dx x x. Lx) is called the Lagevi fuctio. O the average o other compoet perpedicular to E survives, so if you wish, you ca write P = NµLβEµ) E E. 5) Now a hard for may ucritical professors) coceptual questio. It is ofte the case that the rotatioal portio of the Hamiltoia is give as H r = p2 θ 2I + p2 φ µe cos θ, 2I si θ HW5.8) ad the usual caoical formalism is used: Z = dγe βk βh r, HW5.9) where K is the total traslatioal kietic eergy. The, A = k B T log Z is used. Perhaps, eve without actually usig da = SdT P dv + E dp, you may have realized somethig very basic is fudametally screwed up. What is it? [The aswer may be uderstood by everyoe who uderstads that the iteral eergy is the average of the system Hamiltoia/eergy. 4 ] Sol. Let us first stop thikig, ad follow the usual path: We ca calculate Z, which is of course exactly the same as Y we computed. The, the use A = k B T log Z forces upo us the followig Gibbs relatio: da = SdT + E dp. 4 If you do ot see the key poit of this questio, proceed to compute the rotatioal portio of Z, which is actually Y, ad the try to get thermodyamics. 5

6 However, A thus obtaied is exactly the same as our F, which is a fuctio of T or β) ad E. There is o hadle P, so this Gibbs relatio caot be used. Of course, we kow A actually F ) is ot the Helmholtz free eergy but a couterpart of the Gibbs free eergy called a 5 geeralized Gibbs free eergy), so df = SdT P de must be used. As is already hited at, the fudametal coceptual mistake is to idetify H as the true itrisic Hamiltoia, whose average is the iteral eergy. H is cotamiated with the iteractio betwee the system ad the device supplyig E or the field itself). Thus, its average is ot the iteral eergy but a sort of ethalpy. Cosequetly, its Legedre trasformatio wrt S is ot the Helmholtz free eergy but a sort of Gibbs free eergy that is why it is called a geeralized Gibbs free eergy). 5 sice there are may kids. 6