π e ax2 dx = x 2 e ax2 dx or x 3 e ax2 dx = 1 x 4 e ax2 dx = 3 π 8a 5/2 (a) We are considering the Maxwell velocity distribution function: 2πτ/m

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1 Homework Solutons Problem In solvng ths problem, we wll need to calculate some moments of the Gaussan dstrbuton. The brute-force method s to ntegrate by parts but there s a nce trck. The followng ntegrals should be known: e ax dx = π a xe ax dx = a Suppose we now want to calculate x e ax dx or x 3 e ax dx. We can evaluate these ntegrals by dfferentatng the above expressons. For example: d e ax dx = x e ax dx = d π π = da da a a 3/ d xe ax dx = x 3 e ax dx = d = da da a a d x e ax dx = x 4 e ax dx = d π π 3 = da da a 3/ 4 a 5/ Whch panlessly mples: π x e ax dx = a 3/ x 3 e ax dx = a x 4 e ax dx = 3 π 8a 5/ a We are consderng the Maxwell velocty dstrbuton functon: 3 pvdv = 4π e mv /τ v dv πτ/m where C = 4π πτ/m 3 = Ce av v dv and a = m τ. The most probable speed, v prob, s the speed whch maxmzes pv. dp = constant ve av + v e av av = dv τ v prob = a = m The average speed s calculated usng our Gaussan ntegrals: < v > = vpvdv = C v 3 e av dv = C a = 4π πτ/m 8τ < v >= πm 3 τ m

2 Smlarly: < v > = v pvdv = C v 4 e av dv = C 3 π 8a = 4π 5/ πτ/m v rms = 3τ < v > = m 3 3 π 8 b For N, the molecular weght s approxmately m N = 8m p where m p =.67 7 kg. Puttng n the numbers gves: τ m 5 = 3τ m v prob = τ m = [.38 3 J K 93K kg ] / = 45 m s < v > = 468 m s v rms = 57 m s Problem V partcle wth velocty v,v,v x y z vdt θ A dv Fgure : The fgure shows the system at tme t. Partcles wth velocty v x, v y, v z must be n volume dv n order to ht area A by the tme t + dt. See fgure. We frst determne the contrbuton of those partcles havng velocty v = v x, v y, v z known henceforth as v-partcles to the flux. If our hole has area A, then n a tme nterval dt, only those v-partcles n the volume dv wll leave the hole. We may calculate dv : dv = Adtv cosθ where v = v. The number of v-partcles n the volume dv, whch we call nvdv, s gven by the number of v-partcles n the entre volume tmes the rato dv/v. The number of v-partcles actually the number of partcles wth veloctes between v and v + dv n the entre volume V s proportonal to the Maxwell velocty dstrbuton functon. nvdv = C dv V exp mv dv τ = C Adtexp mv v 3 dv cosθ snθdθdφ τ

3 where C and C are constants. The prevous expresson gves the number of v-partcles leavng the hole of area A n tme dt. The flux of v-partcles, whch we call Fv, s defned as ths expresson dvded by Adt: Fvdv = C exp mv v 3 dv cosθ sn θdθdφ τ So we have calculated the velocty dstrbuton of the flux. In order to get a speed dstrbuton, we must ntegrate over sold angle. Fvdv = π dφ π/ dθfv = D exp mv v 3 dv τ where D s a constant. Note that the θ ntegraton was only over [, π ] because the range [ π, π] corresponds to partcles movng away from the hole. We can get D by requrng our functon to be normalzed. Fvdv = D v 3 exp mv dv = D τ = τ m m D = τ Problem 3 A sngle spn can be n one of two energy states: +E and E. The sngle partcle partton functon s gven by: Zτ = e E/τ + e E/τ The partton functon for N partcles s a product of the sngle partcle functons: Z N τ = Zτ N = e E/τ + e E/τ N In partcular, there s no factor of N! n the denomnator because the spns are dstngushable, beng labeled by ther lattce postons. If the spns were not fxed n space but nstead could move from ste to ste, then we would need the N!. Whle ths does not effect the energy, t wll effect the entropy. The free energy s obtaned n the usual way: and the entropy: Sτ and the energy: Fτ = τ lnzτ = Nτ lne E/τ + e E/τ F = τ V = N lne E/τ + e E/τ τ E τ e E/τ + E τ e E/τ e E/τ + e E/τ = N lne E/τ + e E/τ E τ tanh E τ U = F + τs = NE tanh E τ 3

4 Problem 4 Suppose we have two systems whch nteract suffcently weakly that ther nteracton energy may be gnored. The fact that they nteract at all mples they should be at the same temperature. A state of the combned system s specfed by gvng the sngle system states of ndvdual systems. The combned partton functon s: Z tot = e E+E/τ states of,states of = e E/τ e E/τ states of states of = Z Z Problem 5 a The states of our system are unquely specfed by the number of open lnks. The state havng n open lnks corresponds to an energy of nɛ. Thus: N Z = e ɛ/τn = n= exp N + ɛ/τ exp ɛ/τ b Let x = ɛ/τ. In the lmt x >> : N n= ne ɛ/τn = d dx Z = e N+x N + e x e N+x e x + e x e x e x e x Problem 6 a Ths problem s the same as havng a one-dmensonal chan of spns. The spn on each lattce ste can pont left or rght. If s s the rght excess, then: N gn, s = N + s s the number of confguratons wth that rght excess. Because the energy of the system only depends on the magntude of s, we can fnd the total number of states wth a gven s : gn, s + gn, s = N! N + s! N s! b To fnd the entropy, we make use of the Strlng approxmaton and also the fact that s << N here we wrte s but mean s : σl = lngn, s + gn, s = ln N! ln N + s! ln N s! 4

5 N lnn N = gn, N = gn, N = gn, N gn, s N = gn, l Nρ s + N ln N s + N N s N ln N s + N s s + ln + N N N s s ln N N s + N s N s N N s N s N s N s + N s N s N + N s N s N + s N where n the last lne, we used the defnton l = sρ. c σ f = τ l = lτ U Nρ Problem 7 We model a sngle partcle as a free partcle n a one-dmensonal box. The energy levels of ths system are gven by: E n = h π ML n where n s a postve nteger. The partton functon s gven by: Z = e En/τ = dne α n π Mτ = n= α = π h L h where α = π MτL. The N partcle partton functon s gven by: Z N = ZN N! Here the N! s needed because the partcles are ndstngushable. We get the free energy by the usual method: Mτ F = τ lnz N = τ N ln π h L N lnn + N = τ N ln And fnally, the entropy: σ F = τ = N ln Mτ π h L N + N Mτ = N ln L L π h N + N + N Mτ L π h N + 3 5

6 Problem 8 The probabltes whch maxmzed the entropy n the canoncal ensemble were gven by p = e λ λe. We note that the probabltes must sum to by defnton: p = e λ e λe = e λ = = e λe Zλ p = e λe Zλ where the notaton Zλ dentfes the sum wth a partton functon. The average energy and entropy of ths system are gven by: U = p E σ = p lnp The parameter λ may be vared, as t s somethng mposed on the system from the outsde to mantan the average energy constrant. Varyng λ mples the followng varatons n the energy and entropy: δu = E δp δσ = δp lnp + δp Because the p s are probabltes, they must always sum to meanng that the sum of ther varatons must gve : δp =. Ths mples: δσ = = δp lnp + δp δp lnp = δp λ E lnzλ = λ E δp Fnally, we use the statstcal defnton of temperature: τ = σ U V to conclude that λ = τ. 6