Section 2 Introduction to Statistical Mechanics

Transcription

1 Secton 2 Introducton to Statstcal Mechancs 2.1 Introducng entropy Boltzmann s formula A very mportant thermodynamc concept s that of entropy S. Entropy s a functon of state, lke the nternal energy. It measures the relatve degree of order (as opposed to dsorder) of the system when n ths state. An understandng of the meanng of entropy thus requres some apprecaton of the way systems can be descrbed mcroscopcally. Ths connecton between thermodynamcs and statstcal mechancs s enshrned n the formula due to Boltzmann and Planck: S = k ln Ω where Ω s the number of mcrostates accessble to the system (the meanng of that phrase to be explaned). We wll frst try to explan what a mcrostate s and how we count them. Ths leads to a statement of the second law of thermodynamcs, whch as we shall see has to do wth maxmsng the entropy of an solated system. As a thermodynamc functon of state, entropy s easy to understand. Entropy changes of a system are ntmately connected wth heat flow nto t. In an nfntesmal reversble process dq = TdS; the heat flowng nto the system s the product of the ncrement n entropy and the temperature. Thus whle heat s not a functon of state entropy s. 2.2 The spn 1/2 paramagnet as a model system Here we ntroduce the concepts of mcrostate, dstrbuton, average dstrbuton and the relatonshp to entropy usng a model system. Ths treatment s ntended to complement that of Guenault (chapter 1) whch s very clear and whch you must read. We choose as our man example a system that s very smple: one n whch each partcle has only two avalable quantum states. We use t to llustrate many of the prncples of statstcal mechancs. (Precsely the same arguments apply when consderng the number of atoms n two halves of a box.) Quantum states of a spn 1/2 paramagnet A spn S = 1/2has two possble orentatons. (These are the two possble values of the projecton of the spn on the z axs: m s = ±½.) Assocated wth each spn s a magnetc moment whch has the two possble values ±µ. An example of such a system s the nucleus of 3 He whch has S = ½ (ths s due to an unpared neutron; the other sotope of helum 4 He s non-magnetc). Another example s the electron. In a magnetc feld the two states are of dfferent energy, because magnetc moments prefer to lne up parallel (rather than antparallel) wth the feld. Thus snce E = µ.b, the energy of the two states s E = µb, E = µb. (Here the arrows refer to the drecton of the nuclear magnetc moment: means that the moment s parallel to the feld and means antparallel. There s potental confuson because for 3 He the moment ponts opposte to the spn!). PH261 BPC/JS 1997 Page 2.1

2 2.2.2 The noton of a mcrostate So much for a sngle partcle. But we are nterested n a system consstng of a large number of such partcles, N. A mcroscopc descrpton would necesstate specfyng the state of each partcle. In a localsed assembly of such partcles, each partcle has only two possble quantum states or. (By localsed we mean the partcles are fxed n poston, lke n a sold. We can label them f we lke by gvng a number to each poston. In ths way we can tell them apart; they are dstngushable). In a magnetc feld the energy of each partcle has only two possble values. (Ths s an example of a two level system). You can't get much smpler than that! Now we have set up the system, let's explan what we mean by a mcrostate and enumerate them. Frst consder the system n zero feld. Then the states and have the same energy. We specfy a mcrostate by gvng the quantum state of each partcle, whether t s or. For N = 10 spns That s t! are possble mcrostates Countng the mcrostates What s the total number of such mcrostates (accessble to the system). Ths s called Ω. Well for each partcle the spn can be or ( there s no restrcton on ths n zero feld). Two possbltes for each partcle gves 2 10 arrangements (we have merely gven three examples) for N = 10. For N partcles Ω = 2 N Dstrbuton of partcles among states Ths lst of possble mcrostates s far too detaled to handle. What's more we don t need all ths detal to calculate the propertes of the system. For example the total magnetc moment of all the partcles s M = (N N ) µ ; t just depends on the relatve number of up and down moments and not on the detal of whch are up or down. So we collect together all those mcrostates wth the same number of up moments and down moments. Snce the relatve number of ups and downs s constraned by N + N = N (the moments must be ether up or down), such a state can be charactersed by one number: m = N N. Ths number m tells us the dstrbuton of N partcles among possble states: N = (N m) /2, N = (N + m) /2. Now we ask the queston: how many mcrostates are there consstent wth a gven dstrbuton (gven value of m; gven values of N, N ). Call ths t (m). (Ths s Guenault s notaton). Look at N = 10. For m = 10 (all spns up) and m = 10 (all spns down), that's easy! t = 1. Now for m = 8 (one moment down) t (m = 8) = 10. There are ten ways of reversng one moment. PH261 BPC/JS 1997 Page 2.2

3 The general result s t (m) = N!/N! N!. Ths may be famlar to you as a bnomal coeffcent. It s the number of ways of dvdng N objects nto one set of N dentcal objects and N dfferent dentcal objects (red ball and blue balls or tossng an unbassed con and gettng heads or tals). In terms of m the functon t (m) may be wrtten N! t (m) = ( N 2 m )! ( N + 2 m )!. If you plot the functon t (m) t s peaked at m = 0, ( N = N = N /2). Here we llustrate that for N = 10. The sharpness of the dstrbuton ncreases wth the sze of the system N, snce the standard devaton goes as N. t (m) m = N N Plot of the functon t (m) for 10 spns When N = 100 the functon s already much narrower: Plot of the functon t (m) for 100 spns When N s large, then t (m) approaches the normal (Gaussan) functon t (m) 2 N exp m2 2N. Ths may be shown usng Strlng s approxmaton (Guenault, Appendx 2). Note that the RMS wdth of the functon s N. PH261 BPC/JS 1997 Page 2.3

4 2.2.5 The average dstrbuton and the most probable dstrbuton The physcal sgnfcance of ths result derves from the fundamental assumpton of statstcal physcs that each of these mcrostates s equally lkely. It follows that t (m) s the statstcal weght of the dstrbuton m (recall m determnes N and N ), that s the relatve probablty of that dstrbuton occurrng. Hence we can work out the average dstrbuton; n ths example ths s just the average value of m. The probablty of a partcular value of m s just t (m) / Ω.e. the number of mcrostates wth that value of m dvded by the total number of mcrostates ( Ω = m t (m)). So the average value of m s m mt(m) / Ω. In ths example because the dstrbuton functon t (m) s symmetrcal t s clear that m av = 0. Also the value of m for whch t (m) s maxmum s m = 0. So on average for a system of spns n zero feld m = 0: there are equal numbers of up and down spns. Ths average value s also the most probable value. For large N the functon t (m) s very strongly peaked at m = 0; there are far more mcrostates wth m = 0 than any other value. If we observe a system of spns as a functon of tme for most of the tme we wll fnd m to be at or near m = 0. The observed m wll fluctuate about m = 0, the (relatve) extent of these fluctuatons decreases wth N. States far away from m = 0 such as m = N (all spns up) are hghly unlkely; the probablty of observng that state s 1/Ω = 1/2 N snce there s only one such state out of a total 2 N. (Note: Accordng to the Gbbs method of ensembles we represent such a system by an ensemble of Ω systems, each n a defnte mcrostate and one for every mcrostate. The thermodynamc propertes are obtaned by an average over the ensemble. The equvalence of the ensemble average to the tme average for the system s a subtle pont and s the subject of the ergodc hypothess.) It s worthwhle to note that the model treated here s applcable to a range of dfferent phenomena. For example one can consder the number of partcles n two halves of a contaner, and examne the densty fluctuatons. Each partcle can be ether sde of the magnary dvson so that the dstrbuton of densty n ether half would follow the same dstrbuton as derved above. End of lecture Entropy and the second law of thermodynamcs Order and entropy Suppose now that somehow or other we could set up the spns, n zero magnetc feld, such that they were all n an up state m = N. Ths s not a state of nternal thermodynamc equlbrum; t s a hghly ordered state. We know the equlbrum state s random wth equal numbers of up and down spns. It also has a large number of mcrostates assocated wth t, so t s far the most probable state. The ntal state of the system wll therefore spontaneously evolve towards the more lkely, more dsordered m = 0 equlbrum state. Once there the fluctuatons about m = 0 wll be small, as we have seen, and the probablty of the system returnng fleetngly to the ntal ordered state s nfntesmally small. The tendency of solated systems to evolve n the drecton of ncreasng dsorder thus follows from () dsordered dstrbutons havng a larger number of mcrostates (mathematcal fact) () all mcrostates are equally lkely (physcal hypothess). The thermodynamc quantty ntmately lnked to ths dsorder s the entropy. Entropy s a functon of state and s defned for a system n thermodynamc equlbrum. (It s possble to ntroduce a PH261 BPC/JS 1997 Page 2.4

5 generalsed entropy for systems not n equlbrum but let's not complcate the ssue). Entropy was defned by Boltzmann and Planck as S = k ln Ω where Ω s the total number of mcrostates accessble to a system. Thus for our example of spns Ω = 2 N so that S = Nk ln 2. Here k s Boltzmann s constant The second law of thermodynamcs The second law of thermodynamcs s the law of ncreasng entropy. Durng any real process the entropy of an solated system always ncreases. In the state of equlbrum the entropy attans ts maxmum value. Ths law may be llustrated by askng what happens when a constrant s removed on an solated composte system. Is the number of mcrostates of the fnal equlbrum state be smaller, the same or bgger? We expect the system to evolve towards a more probable state. Hence the number of accessble mcrostates of the fnal state must be greater or the same and the entropy ncreases or stays the same. A nce smple example s gven by Guenault on p The mxng of two dfferent gases s another. For a composte system Ω = Ω 1 Ω 2. If the two systems are allowed to exchange energy, them n the fnal equlbrum state the total number of accessble mcrostates s always greater. The macroscopc descpton of ths s that heat flows untl the temperature of the two systems s the same. Ths leads us, n the next secton, to a defnton of statstcal temperature as 1 T = S. E V,N Thermal nteracton between systems and temperature Consder two solated systems of gven volume, number of partcles, total energy When separated and n equlbrum they wll ndvdually have Ω 1 = Ω 1 (E 1, V 1, N 1 ) and Ω 2 = Ω 2 (E 2, V 2, N 2 ) mcrostates. The total number of mcrostates for the combned system s Ω = Ω 1 Ω 2. Now suppose the two systems are brought nto contact through a dathermal wall, so they can now exchange energy. E 1, V 1, N 1 E 2, V 2, N 2 E 1, V 1, N 1 E 2, V 2, N 2 Thermal nteracton Ω fxed dathermal wall The composte system s solated, so ts total energy s constant. So whle the systems exchange energy ( E 1 and E 2 can vary) we must keep E 1 + E 2 = E 0 = const. And snce V 1, N 1, V 2, N 2 all PH261 BPC/JS 1997 Page 2.5

6 reman fxed, they can be gnored (and kept constant n any dfferentaton). Our problem s ths: after the two systems are brought together what wll be the equlbrum state? We know that they wll exchange energy, and they wll do ths so as to maxmse the total number of mcrostates for the composte system. The systems wll exchange energy so as to maxmse Ω. Wrtng Ω = Ω 1 (E)Ω 2 (E 0 E) we allow the systems to vary E so that Ω s a maxmum: or or Ω E = Ω 1 E Ω Ω 2 Ω 2 1 E = 0 1 Ω 1 Ω 1 E = 1 Ω 2 ln Ω 1 = ln Ω 2 E E. But from the defnton of entropy, S = k ln Ω, we see ths means that the equlbrum state s charactersed by S 1 E = S 2 E. In other words, when the systems have reached equlbrum the quantty S / E of system 1 s equal to S / E of system 2. Snce we have defned temperature to be that quantty whch s the same for two systems n thermal equlbrum, then t s clear that S / E (or ln Ω / E) must be related (somehow) to the temperature. We defne statstcal temperature as 1 T = S E V,N (recall that V and N are kept constant n the process) Wth ths defnton t turns out that statstcal temperature s dentcal to absolute temperature. It follows from ths defnton and the law of ncreasng entropy that heat must flow from hgh temperature to low temperature (ths s actually the Clausus statement of the second law of thermodynamcs. We wll dscuss all the varous formulatons of ths law a bt later on.) Let us prove ths. For a composte system Ω = Ω 1 Ω 2 so that S = S 1 + S 2 as we have seen. Accordng to the law of ncreasng entropy S 0, so that: Ω 2 E S = ( S 1 E S 2 E ) E 1 0 or, usng our defnton of statstcal temperature: S = ( 1 T 1 1 T 2) E 1 0. PH261 BPC/JS 1997 Page 2.6

7 ths means that E 1 ncreases f T 2 > T 1 E 1 decreases f T 2 < T 1 so energy flows from systems at hgher temperatures to systems at lower temperatures. End of lecture More on the S=1/2 paramagnet Energy dstrbuton of the S=1/2 paramagnet Havng motvated ths defnton of temperature we can now return to our smple model system and solve t to fnd N and N as a functon of magnetc feld B and temperature T. Prevously we had consdered the behavour n zero magnetc feld only and we had not dscussed temperature. As dscussed already n a magnetc feld the energes of the two possble spn states are E = µb and E = µb. We shall wrte these energes as ε. Now for an solated system, wth total energy E, number N, and volume V fxed, t turns out that N and N are unquely determned (snce the energy depends on N N ). Thus m = N N s fxed and therefore t exhbts no fluctuatons (as t dd n zero feld about the average value m = 0). Ths follows snce we must have both These two equatons are solved to gve E = (N N ) ε N = (N + N ) N = (N + E / ε) /2 N = (N E / ε) /2. All mcrostates have the same dstrbuton of partcles between the two energy levels N and N. But there are stll thngs we would lke to know about ths system; n partcular we know nothng about the temperature. We can approach ths from the entropy n our now-famlar way. The total number of mcrostates Ω s gven by N! Ω = N! N!. So we can calculate the entropy va S = k ln Ω: S = k {ln N! ln N! ln N!} Now we use Strlng s approxmaton (Guenault Appendx 2) whch says ln x! x ln x x; ths s mportant you should remember ths. PH261 BPC/JS 1997 Page 2.7

8 Hence S = k {N ln N N ln N N ln N }, where we have used N + N = N. Into ths we substtute for N and N snce we need S n terms of E for dfferentaton to get temperature. S = k N ln N 1 2 (N + E / ε) ln { 1 2 (N + E / ε) } 1 2 (N E / ε) ln { 1 2 (N E / ε) }. Now we use the defnton of statstcal temperature: 1/T = S / E N.V to obtan the temperature: 1 T = k ln { N E / ε 2ε N + E / ε}. (You can check ths dfferentaton wth Maple or Mathematca f you wsh!) Recallng the expressons for N and N, the temperature expresson can be wrtten: 1 T = k 2ε ln { N N }, whch can be nverted to gve the rato of the populatons as Or, snce N + N = N we fnd 1.0 N N = exp 2ε / kt. N N = exp ε / kt exp +ε / kt + exp ε / kt N N = exp +ε / kt exp +ε / kt + exp ε / kt 0.8 N / N N / N kt / ε Temperature varaton of up and sown populatons Ths s our fnal result. It tells us the fracton of partcles n each of the two energy states as a functon of temperature. Ths s the dstrbuton functon. On nspecton you see that t can be wrtten rather concsely as exp ε / kt n (ε) = N z PH261 BPC/JS 1997 Page 2.8

9 where the quantty z z = exp ε / kt + exp +ε / kt s called the (sngle partcle) partton functon. It s the sum over the possble states of the factor exp ε / kt. Ths dstrbuton among energy levels s an example of the Boltzmann dstrbuton. Ths dstrbuton apples generally(to dstngushable partcles) where there s a whole set of possble energes avalable and not merely two as we have consdered thus far. Ths problem we treat n the next secton. In general z = exp ε / kt, states where the sum s taken over all the possble energes of a sngle partcle. In the present example there are only two energy levels. In the general case n (ε) s the average number of partcles n the state of energy ε (In the present specal case of only two levels n (ε 1 ) and n (ε 2 ) are unquely determned). To remnd you, an example of countng the mcrostates and evaluatng the average dstrbuton for a system of a few partcles wth a large number of avalable energes s gven n Guenault chapter 1. The fluctuatons n n (ε) about ths average value are small for a suffcently large number of partcles. (The 1/ N factor) Magnetsaton of S = 1/2magnet Cure s law We can now obtan an expresson for the magnetsaton of the spn 1/2 paramagnet n terms of temperature and appled magnetc feld. The magnetsaton (total magnetc moment) s gven by: M = (N N ) µ. We have expressons for N and N, gvng the expresson for M as exp ε / kt exp +ε / kt M = Nµ exp +ε / kt + exp ε / kt = Nµ tanh ε / kt or, snce ε = µb, the magnetsaton n terms of the magnetc feld s M = Nµ tanh ( kt) µb. M / Nµ 1.0 saturaton lnear regon M Nµ2 B kt magnetsaton of paramagnet µb/ kt 3.0 PH261 BPC/JS 1997 Page 2.9

10 The general behavour of magnetsaton on magnetc feld s nonlnear. At low felds the magnetsaton starts lnearly but at hgher felds t saturates as all the moments become algned. The low feld, lnear behavour may be found by expandng the tanh: The magnetsaton has the general form M = Nµ2 kt B. M = C B T, proportonal to B and nversely proportonal to T. Ths behavour s referred to as Cure s law and the constant C = Nµ 2 / k s called the Cure constant Entropy of S = 1/2magnet We can see mmedately that for the spn system the entropy s a functon of the magnetc feld. In partcular at large B/ T all the moments are parallel to the feld. There s then only one possble mcrostate and the entropy s zero. But t s qute easy to determne the entropy at all B/ T from S = k ln Ω. As we have seen already S / k = N ln N N ln N N ln N. Snce N = N + N, ths can be wrtten S / k = N ln N / N N ln N / N. And substtutng for N and N we then get Nk Nk S = ln [exp ( 2µB/ kt) +1] + exp ( 2µB/ kt) +1 exp (2µB/ kt) +1 ln [exp (2µB/ kt) +1] S Nkln 2 lower B hgher B 0 0 Entropy of a spn 1/2 paramagnet T PH261 BPC/JS 1997 Page 2.10

11 1 At low temperatures kt 2µB. Then the frst term s neglgble and the 1s may be gnored. In ths case S Nk (2µB/ kt) exp ( 2µB/ kt). Clearly S 0 as T 0, n agreement wth our earler argument. 2 At hgh temperatures kt 2µB. Then both terms are equal and we fnd S = Nk ln 2. There are equal numbers of up and down spns: maxmum dsorder. 3 The entropy s a functon of B/ T and ncreasng B at constant T reduces S; the spns tend to lne up; the system becomes more dsordered. End of lecture The Boltzmann dstrbuton Thermal nteracton wth the rest of the world usng a Gbbs ensemble For an solated system all mcroststes are equally lkely; ths s our Fundamental Postulate. But what about a non-solated system? What can we say about the probabltes of mcrostates of such a system? Here the probablty of a mcrostate wll depend on ts energy and on propertes of the surroundngs. A real system wll have ts temperature determned by ts envronment. That s, t s not solated, ts energy s not fxed; t wll fluctuate (but the relatve fluctuatons are very small because of the 1/ N rule). All we really know about s solated systems; here all quantum states are equally probable. So to examne the propertes of a non-solated system we shall adopt a trck. We wll take many copes of our system. These wll be allowed to exchange (heat) energy wth each other, but the entre collecton wll be solated from the outsde world. Ths collecton of systems s called an ensemble. The pont now s that we now know that all mcrostates of the composte ensemble are equally lkely. Of the N ndvdual elements of the ensemble, there wll be n n the mcrostate of energy ε, so the probablty of an element beng n ths mcrostate wll be n / N. System n energy state of εj collecton of dentcal systems a Gbbs ensemble of systems used to calculate {n } There are many dstrbutons {n } whch are possble for ths ensemble. In partcular we know that because the whole ensemble s solated then the number of elements and the total energy are fxed. In other words any dstrbuton {n } must satsfy the requrements PH261 BPC/JS 1997 Page 2.11

12 n = N n ε = E. By analogy wth the case of the spn 1/2 paramagnet, we wll denote the number of mcrostates of the ensemble correspondng to a gven dstrbuton {n } by t ({n }). Then the most lkely dstrbuton s that for whch {n } s maxmsed. t ( ) Most lkely dstrbuton The value of {n } s gven by t ( ) t ({n }) = N! n! snce there are N elements all together and there are n n the th state. So the most probable dstrbuton s found by varyng the varous n to gve the maxmum value for t. It s actually more convenent (and mathematcally equvalent) to maxmse the logarthm of t. The maxmum n ln t corresponds to the place where ts dfferental s zero: dln t ({n = ln t dn 1 + ln t dn ln t }) dn + = 0 n 1 n 2 n If the n could be vared ndependently then we would have the followng set of equatons ln t ({n }) = 0, = 1, 2, 3,. n Unfortunately the n can not all be vared ndependently because all possble dstrbutons {n } must satsfy the two requrements there are two constrants on the dstrbuton Snce the constrants mply that and n = N n ε = E; {n }. These may be ncorporated n the followng way. d n = 0 d n ε = 0, we can consder the dfferental of the expresson ln t ({n }) α n β n ε where α and β are, at ths stage undetermned. Ths gves us two extra degrees of freedom so that we can maxmse ths by varyng all ndependently. In other words, the maxmum n ln t s also n PH261 BPC/JS 1997 Page 2.12

13 specfed by d ln t ({n }) α n β n ε = 0 Now we have recovered the two lost degrees of freedom so that the n can be vared ndependently. But then the multplers α and β are no longer free varables and ther values must be found from the constrants fxng N and E. (Ths s called the method of Lagrange s undetermned multplers.) The maxmsaton can now be performed by settng the N partal dervatves to zero ln t ({n }) α n n β We use Strlng s approxmaton for lnt, gven by ln t ({n }) = ( n ε = 0, = 1, 2, 3,. n ) ln ( so that upon evaluatng the N partal dervatves we obtan ) n n ln n. ln n ln n j α βε j = 0 whch has soluton n j = Ne α e βε j. These are the values of n j whch maxmse t subject to N and E beng fxed. Ths gves us the most probable dstrbuton {n }. The probablty that a system wll be n the state j s then found by dvdng by N: p j = e α e βε j. So the remanng step, then, s to fnd the constants α and β What are α and β? For the present we shall sdestep the queston of α by appealng to the normalsaton of the probablty dstrbuton. Snce we must have p j = 1 j t follows that we can express the probabltes as where the normalsaton constant Z s gven by p j = e βεj Z Z = e βε j. j The constant Z wll turn out to be of central mportance; from ths all thermodynamc propertes can be found. It s called the partton functon, and t s gven the symbol Z because of ts name n German: zustandsumme (sum over states). PH261 BPC/JS 1997 Page 2.13

14 But what s ths β? For the spn 1/2 paramagnet we had many smlar expressons wth 1/kT there, where we have β here. We shall now show that ths dentfcaton s correct. We wll approach ths from our defnton of temperature: 1 T = S. E V,N Now the fundamental expresson for entropy s S = k ln Ω, where Ω s the total number of mcrostates of the ensemble. Ths s a lttle dffcult to obtan. However we know t ({n }), the number of mcrostates correspondng to a gven dstrbuton {n }. And ths t s a very sharply peaked functon. Therefore we make neglgble error n approxmatng Ω by the value of t correspondng to the most probable dstrbuton. In other words, we take N! S = k ln ( n!) where n j = Ne α e βε j. Usng Strlng s approxmaton we have From ths we mmedately dentfy so that as we asserted. S = k N ln N n ln n = k {N ln N + αn + βe}. 1 T = S E V,N β = 1 kt = kβ The probablty of a (non-solated) system beng found n the th mcrostate s then p e ε /kt or /kt p = e ε Z Ths s known as the Boltzmann dstrbuton or the Boltzmann factor. It s a key result. Feynman says Ths fundamental law s the summt of statstcal mechancs, and the entre subject s ether a sldedown from the summt, as the prncple s appled to varous cases, or the clmb-up to where the fundamental law s derved and the concepts of thermal equlbrum and temperature clarfed. End of lecture 7 PH261 BPC/JS 1997 Page 2.14

15 2.5.4 Lnk between the partton functon and thermodynamc varables All the mportant thermodynamc varables for a system can be derved from Z and ts dervatves. We can see ths from the expresson for entropy. We have And snce so that e α S = k {N ln N + αn + βe}. = N / Z we have, on takng logarthms α = ln N + ln Z S = k {N ln N N ln N + N ln Z + E / kt} = Nk ln Z + E / T. Here both S and E refer to the ensemble of N elements. So for the entropy and nternal energy of a sngle system S = k ln Z + E / T, whch, upon rearrangement, can be wrtten E TS = k ln Z. Now the thermodynamc functon F = E TS s known as the Helmholtz free energy, or the Helmholtz potental. We then have the memorable result F = kt ln Z Fndng Thermodynamc Varables A host of thermodynamc varables can be obtaned from the partton functon. Ths s seen from the dfferental of the free energy. Snce de = TdS pdv t follows that df = SdT pdv. We can then dentfy the varous partal dervatves: S = F = kt ln Z + k ln Z T V p = F V T T V = kt ln Z V T Snce E = F + TS we can then express the nternal energy as E = kt 2 ln Z T V Thus we see that once the partton functon s evaluated by summng over the states, all relevant thermodynamc varables can be obtaned by dfferentatng Z. It s nstructve to examne ths expresson for the nternal energy further. Ths wll also confrm the dentfcaton of ths functon of state as the actual energy content of the system. If p j s the probablty of the system beng n the egenstate correspondng to energy ε j then the mean energy of the system may be expressed as PH261 BPC/JS 1997 Page

16 E = ε j p j j = 1 Z j ε j e βε j where Z s the prevously-defned partton functon, and t s convenent here to work n terms of β rather than convertng to 1/kT. In exammnng the sum we note that ε j e βε j = so that the expresson for E may be wrtten. after nterchangng the dfferentaton and the summaton, β e βε j E = 1 Z But the sum here s just the partton functon, so that E = 1 Z β j Z β e βε j. And snce β = β ln Z. = 1/kT, ths s equvalent to our prevous expresson E = kt 2 ln Z T V however the mathematcal manpulatons are often more convenent n terms of β Summary of methodology We have seen that the partton functon provdes a means of calculatng all thermodynamc nformaton from the mcroscopc descrpton of a system. We summarse ths procedure as follows:, 1 Wrte down the possble energes for the system. 2 Evaluate the partton functon for the system. 3 The Helmholtz free energy then follows from ths. 4 All thermodynamc varables follow from dfferentaton of the Helmholtz free energy. 2.6 Localsed systems Workng n terms of the partton functon of a sngle partcle The partton functon methodology descrbed n the prevous sectons s general and very powerful. However a consderable smplfcaton follows f one s consderng systems made up of nonnteractng dstngushable or localsed partcles. The pont about localsed partcles s that even though they may be ndstngushable, perhaps a sold made up of argon atoms, the partcles can be dstngushed by ther postons. PH261 BPC/JS 1997 Page 2.16

17 Snce the partton functon for dstngushable systems s the product of the partton functon for each system, f we have an assembly of N localsed dentcal partcles, and f the partton functon for a sngle such partcle s z, then the partton functon for the assembly s Z = z N. It then follows that the Helmholtz free energy for the assembly s F = kt ln Z = NkT ln z n other words the free energy s N tmes the free energy contrbuton of a sngle partcle; the free energy s an extensve quantty, as expected. Ths allows an mportant smplfcaton to the general methodology outlned above. For localsed systems we need only consder the energy levels of a sngle partcle. We then evaluate the partton functon z of a sngle partcle and then use the relaton F = NkT ln z n order to fnd the thermodynamc propertes of the system. We wll consder two examples of ths, one famlar and one new. End of lecture Usng the partton functon I the S = 1/2paramagnet (agan) Step 1) Wrte down the possble energes for the system The assembly of magnetc moments µ are placed n a magnetc feld B. The spn ½ has two quantum states, whch we label by and. The two energy levels are than ε = µb, ε = µb. Step 2) Evaluate the partton functon for the system We requre to fnd the partton functon for a sngle spn. Ths s z = e ε /kt states = e µb/kt + e µb/kt. Ths tme we shall obtan the results n terms of hyperbolc functons rather than exponentals for varety. The partton functon s expressed as the hyperbolc cosne: z = 2 cosh µb/ kt. Step 3) The Helmholtz free energy then follows from ths Here we use the relaton F = NkT ln z = NkT ln {2 cosh µb/ kt}. Step 4) All thermodynamc varables follow from dfferentaton of the Helmholtz free energy Before proceedng wth ths step we must pause to consder the performance of magnetc work. Here we don t have pressure and volume as our work varables; we have magnetsaton M and magnetc feld B. The expresson for magnetc work s PH261 BPC/JS 1997 Page 2.17

18 d W = MdB, so comparng ths wth our famlar pdv, we see that when dealng wth magnetc systems we must make the dentfcaton p M V B. The nternal energy dfferental of a magnetc system s then de = TdS MdB and, of more mmedate mportance, the Helmholtz free energy E TS has the dfferental df = SdT MdB. We can then dentfy the varous partal dervatves: S = F = NkT ln z + Nk ln z, T B M = F B T T B = NkT ln z B T Upon dfferentaton we then obtan S = NµB tanh µb/ kt + Nk ln {2 cosh µb/ kt}, T M = Nµ tanh µb/ kt. The nternal energy s best obtaned now from E = F + TS = NkT ln z + TS = NkT ln {2 cosh µb/ kt} + T NµB T tanh µb/ kt + Nk ln {2 cosh µb/ kt}, gvng E = NµB tanh µb/ kt. We note that the nternal energy can be wrtten as E = MB as expected. By dfferentatng the nternal energy wth respect to temperature we obtan the thermal capacty at constant feld C B = Nk ( kt) µb 2 sech 2 µb/ kt. Some of these expressons were derved before, from the mcrocanoncal approach; you should check that the exponental expressons there are equvalent to the hyperbolc expressons here. We plot the magnetsaton as a functon of nverse temperature. Recall that at hgh temperatures we have a lnear regon, where Cure s law holds. At low temperatures the magnetsaton saturates as all the moments become algned aganst the appled feld. Incdentally, we note that the expresson M = Nµ tanh µb/ kt PH261 BPC/JS 1997 Page 2.18.

19 s the equaton of state for the paramagnet. Ths s a nonlnear equaton. But recall that we found lnear behavour n the hgh T / B regon where M = Nµ2 B kt, just as t s n the hgh temperature regon that the deal gas equaton of state p = NkT / V s vald. M / Nµ 1.0 saturaton lnear regon M Nµ2 B kt µb/ kt magnetsaton of spn 1/2 paramagnet aganst nverse temperature Next we plot the entropy, nternal energy and the thermal capacty as a functon of temperature. Observe that they all go to zero as T 0. At low temperatures the entropy goes to zero, as expected. And at hgh temperatures the entropy goes to the classcal two-state value of k ln 2 per partcle S 2Nk ( kt) µb 2µB e kt T 0 S Nk ln 2 4Nk ( kt) µb 2 T. The thermal capacty s partcularly mportant as t s readly measurable. It exhbts a maxmum of C B 0.44Nk at a temperature of T 0.83µB/ k. Ths s known as a Schottky anomaly. Ordnarly the thermal capacty of a substance ncreases wth temperature, saturatng at a constant value at hgh temperatures. Spn systems are unusual n that the energy states of a spn are fnte and therefore bounded from above. The system then has a maxmum entropy. As the entropy ncreases towards ths maxmum value t becomes ncreasngly dffcult to pump n more heat energy. At both hgh and low temperatures the thermal capacty goes to zero, as may be seen by expandng the expresson for C B : C B 4Nk ( kt) µb 2 e 2µB kt T 0 C B 2Nk ( kt) µb 2 T. PH261 BPC/JS 1997 Page 2.19

20 S Nkln 2 lower B kt / µb entropy of spn ½ paramagnet E / NµB C B / Nk nternal energy of spn ½ paramagnet 5.0 kt / µb kt / µb thermal capacty of spn ½ paramagnet End of lecture 9 PH261 BPC/JS 1997 Page 2.20

21 2.6.3 Usng the partton functon II the Ensten model of a sold One of the challenges faced by Ensten was the explanaton of why the thermal capacty of solds tended to zero at low temperatures. He was concerned wth nonmagnetc nsulators, and he had an nklng that the explanaton was somethng to do wth quantum mechancs. The thermal exctatons n the sold are due to the vbratons of the atoms. Ensten constructed a smple model of ths whch was partally successful n explanng the thermal capacty. In Ensten s model each atom of the sold was regarded as a smple harmonc oscllator vbratng n the potental energy mnmum produced by ts neghbours. Each atom sees a smlar potental, so they all oscllate at the same frequency; let us call ths ω /2π. And snce each atom can vbrate n three ndependent drectons, the sold of N atoms s modelled as a collecton of 3N dentcal harmonc oscllators. We shall follow the procedures outlned above. Step 1) Wrte down the possble energes for the system The energy states of the harmonc oscllator form an nfnte ladder of equally spaced levels; you should be famlar wth ths from your quantum mechancs course j = ħω 7 2 ħω 5 2 ħω 3 2 ħω 1 2 ħω sngle partcle energes: εj = (j + 1 2) ħω j = 0, 1, 2, 3, Step 2) Evaluate the partton functon for the system We requre to fnd the partton functon for a sngle harmonc oscllator. Ths s z = e ε j/kt j = 0 = j = 0 exp { ( j + 1 2) ħω kt }. To proceed, let us defne a characterstc temperature θ, related to the oscllator frequency by θ = ħω / k. Then the partton functon may be wrtten as z = j = 0 = e θ/2t exp { ( j + 1 2) θ T } j = 0 (e θ/t ) j and we observe the sum here to be a (convergent) geometrc progresson. You should recall the result x n = 1 + x + x 2 + x = n = 0 1 x. PH261 BPC/JS 1997 Page 2.21

22 If you don t remember ths result then multply the power seres by 1 x and check the sum. The harmonc oscllator partton functon s then gven by z = e θ/2t 1 e θ/t. Step 3) The Helmholtz free energy then follows from ths Here we use the relaton F = 3NkT ln z = 3N 2 kθ + 3NkT ln {1 e θ/t }. Step 4) All thermodynamc varables follow from dfferentaton of the Helmholtz free energy Here we have no explct volume dependence, ndcatng that the equlbrum sold s at zero pressure. If an external pressure were appled then the vbraton frequency ω /2π would be a functon of the nterpartcle spacng or, equvalently, the volume. Ths would gve a volume dependence to the free energy from whch the pressure could be found. However we shall gnore ths. The thermodynamc varables we are nterested n are then entropy, nternal energy and thermal capacty. The entropy s S = F T V = 3Nk ln ( e θ/t e θ/t 1) + θ / T (e θ/t 1). In the low temperature lmt S goes to zero, as one would expect. The lmtng low temperature behavour s S 3Nk θ T e θ/t T 0. At hgh temperatures the entropy tends towards a logarthmc ncrease The nternal energy s found from where we wrte ln z as Then on dfferentaton we fnd S 3Nk ln ( T θ ) T. ln z = βħω 2 E = 3N β ln z ln {1 e βħω }. E = 3N { ħω 2 + ħω e βħω 1}. The frst term represents the contrbuton to the nternal energy from the zero pont oscllatons; t s present even at T = 0. PH261 BPC/JS 1997 Page 2.22

23 At hgh temperatures the varaton of the nternal energy s E 3NkT { ( θ T ) 2 }. The frst term s the classcal equpartton part, whch s ndependent of the vbraton frequency. Turnng now to the thermal capacty, we have C V = E Upon dfferentaton ths then gves T V, = 3Nkθ T { 1 e θ/t 1}. C V = 3Nk ( θ T ) 2 e θ/t (e θ/t 1) 2. At hgh temperatures the varaton of the thermal capacty s C V 3Nk 1 4 Nk ( θ T ) 2 + T θ where the frst term s the constant equpartton value. At low temperatures we fnd C V 3Nk ( θ T ) 2 e θ/t T θ. We see that ths model does ndeed predct the decrease of C V to zero as T 0, whch was Ensten s challenge. The decrease n heat capacty below the classcal temperature-ndependent value s seen to arse from the quantsaton of the energy levels. However the observed low temperature behavour of such thermal capactes s a smple T 3 varaton rather than the more complcated varaton predcted from ths model. The explanaton for ths s that the atoms do not all oscllate ndependently. The couplng of ther moton leads to normal mode waves propagatng n the sold wth a contnuous range of frequences. Ths was explaned by the Debye model of solds. The Ensten model ntroduces a sngle new parameter, the frequency ω /2π of the oscllatons or, equvalently, the characterstc temperature θ = ħω / k. And the predcton s that C V s a unversal functon of T / θ. To the extent that the model has some valdty, each substance should have ts own value of θ; as the C V fgure shows, the value for damond s 1300K. Ths s an example of a law of correspondng states: When the temperatures are scaled approprately all substances should exhbt the same behavour. PH261 BPC/JS 1997 Page 2.23

24 6 S / Nk entropy of Ensten sold 2.0 T / θ E / Nkθ 6 4 E cl = 2NkT 3 classcal lmt 2 E 0 = 3 2Nkθ zero pont moton nternal energy of Ensten sold 2.0 T / θ 3.0 C V / Nk 2.0 expermental ponts for damond ft to lne for θ = 1300K thermal capacty of Ensten sold T / θ 2.0 PH261 BPC/JS 1997 Page 2.24

25 Fnally we gve the calculated propertes of the Ensten model n terms of hyperbolc functons; you should check these. The partton functon for smple harmonc oscllator can be wrtten as z = 1 2 cosech ( θ 2T ), so that the Helmholtz free energy for the sold comprsng 3N such oscllators s F = 3NkT ln 2 snh ( θ 2T ). And from ths the varous propertes follow: S = 3Nk {( θ 2T ) coth ( θ 2T ) ln 2 snh ( θ E = 3 2 ( Nkθ coth θ 2T ) 2T ) } C V = 3Nk ( θ 2T ) 2 cosech ( 2 θ 2T ). End of lecture 10 Equaton of state for the Ensten sold We have no equaton of state for ths model so far. No p V relaton has been found, snce there was no volume-dependence n the energy levels, and thus n the partton functon and everythng whch followed from that. Ths defcency can be rectfed by recognsng that the Ensten frequency, or equvalently the temperature θ may vary wth volume. Then the pressure may be found from p = F V T = 3Nk coth 2 ( θ 2T ) dθ dv. We note that the equlbrum state of the sold when no pressure s appled, corresponds to the vanshng of dθ /dv. In fact θ wll be a mnmum at the equlbrum volume V 0 (why?). For small changes of volume from ths equlbrum we may then wrte θ = θ 0 + α ( V V 2 0 V 0 ) so that dθ dv = 2α (V V V0 2 0 ). Then the equaton of state for the sold s p = 3Nkα θ (V V0 2 0 V) coth ( 0 2T ). The hgh-temperature lmt of ths s p = 6Nkα (V V0 2 0 V) T θ and the low-temperature, temperature-ndependent lmt s p = (V 0 V) 3Nkα. V0 2 PH261 BPC/JS 1997 Page 2.25