Introduction to Statistical Physics (2SP)

Transcription

1 Introducton to Statstcal Physcs (2SP) Rchard Sear March 5, 20 Contents What s the entropy (aka the uncertanty)? 2. One macroscopc state s the result of many many mcroscopc states States wth equal energes are equally probable States wth dfferent energes: the role of Temperature The two-level system 6 3 Fluctuatons and the Central Lmt Theorem of Statstcs 7 4 States wth dfferent numbers of partcles 9 5 Ferm-Drac and Bose-Ensten Statstcs 0 5. A sngle level wth Ferm-Drac statstcs A sngle level wth Bose-Ensten statstcs Averages and the relatonshp between the partton functon and thermodynamc quanttes 3 6. Averages Average quanttes at constant T The partton functon Z and the Helmholtz free energy A Classcal Statstcal Mechancs 5 7. Example The equpartton theorem When s a gas classcal and when are quantum mechancal effects mportant Recommended textbooks I recommend four:. Statstcal Mechancs: A Survval Gude by A. M. Glazer and J. S. Wark (Oxford). 2. Cencepts n Thermal Physcs by S. J. Blundell and K. M. Blundell (Oxford).

2 r p r p 2 2 r p 3 3 N V U Macroscopc state Thermodynamcs Mcroscopc state Statstcal Mechancs Fgure : Schematcs of a gas or argon atoms from the vewpont of thermodynamcs (left-hand sde) and from the vewpont of statstcal mechancs (rght-hand sde). In thermodynamcs, we just deal wth a handful of, macroscopc, propertes, such as the total number of partcles, N, volume V and total energy U. In statstcal mechancs, we deal wth mcroscopc states. For the gas of argon atoms the mcroscopc state s specfed by the postons and momenta (or veloctes) of all N argon atoms. We have only shown 3 atoms here. r s the poston vector of atom, p s the momentum vector of atom, r 2 s the poston vector of atom 2, and so on. The arrows ndcate the momentum vectors, and the dots are the postons of the atoms. 3. Introducton to Modern Statstcal Mechancs by D. Chandler (Oxford). 4. Introducton to Statstcal Physcs by K. Huang (Oxford). 3 goes a bt faster than ether, 2 or 4. All four are n the lbrary. What s the entropy (aka the uncertanty)? We wll start off by lookng at the most basc functon n thermodynamcs, and n statstcal mechancs, the entropy, usng a statstcal descrpton. You have already encountered the entropy n thermodynamcs. The second law of thermodynamcs s that at thermodynamc equlbrum the entropy s at a maxmum. Note: a) thermodynamcs does not tell you how to calculate the entropy of say, a gven amount of steam. As we wll see, statstcal physcs does allow you calculate the entropy. Also, b) recall that thermodynamcs deals wth large amounts of ar, steam, ron etc,.e., Avogadro s number of atoms or molecules, 0 24, not one or a handful of molecules.. One macroscopc state s the result of many many mcroscopc states Consder a gas, say, of N molecules, n a volume V and wth an energy U. The gas has just one macroscopc, or thermodynamc, state but a huge number of possble states of ts molecules. If a gas has N = 0 24 molecules then the mcroscopc states of the gas are specfed by the postons and momenta of all these N molecules. Whenever a molecule moves t changes ts poston and so the mcroscopc state of the system. Also, whenever t colldes wth another molecule, ts velocty and 2

3 hence momentum wll change and ths too changes the mcroscopc state of the gas. See Fg. for an llustraton of the macroscopc and mcroscopc states of a gas. Ths s for a gas of molecules that can be treated as classcal objects,.e., lke bllard balls. For a quantum mechancal system, the mcroscopc state of a system s specfed by specfyng the quantum mechancal state the system s n. The possble states you get from solvng Schrödnger s equaton. For example, a hydrogen atom wll have two ground states, the s state wth up and down spns, excted states, 2s, 2p etc. If we have two atoms then one could be n the s state, the other n a 2p state etc., whle wth three atoms we could have one n the s state, another n the 2p state and the thrd n the 2s state etc. Thus agan f we have 0 24 H atoms there are huge number of possble mcroscopc states of these atoms, correspondng to all the possble dfferent electronc states the atoms can be n. So, we have a huge number of states and typcally we don t know what the mcroscopc state of the system s. Ths s because to know ths we would need to know the ndvdual states of a huge number of molecules, and we would have to measure all these states very quckly, as as soon as a molecule moves or colldes then the mcroscopc state changes. Ths s completely mpractcal so we need to resort to statstcs. Instead of tryng to work out what the mcroscopc state s we wll just assume each mcroscopc state occurs wth a gven probablty,.e., that the probablty that we observe the system n the mcroscopc state number, s p. Here, we wll calculate the entropy from Shannon s equaton. Ths was derved by Claude Shannon a lttle after the second world war. Shannon sad that f we knew all probabltes p of all S = N = p ln p entropy of one macroscopc state sum over N mcroscopc states () where p s the probablty of the th mcroscopc state, and the sum s over all the possble mcroscopc states. Note that here a macroscopc quantty, the entropy on the left-hand sde, s gven by a sum over mcroscopc quanttes, the probabltes p, of the mcroscopc states of the system. Statstcs s requred to calculate thermodynamc propertes such as the entropy, energy etc because each macroscopc or thermodynamc state s the result of many many mcroscopc states and we don t know whch mcroscopc states the system s an any nstant, all we know s the probablty that the system s n a gven mcroscopc state (below we wll see how to calculate these probabltes). So we need to work wth probabltes, whch requres statstcs. For example, f I tell you that there s say a volume cm 3 of steam wth an energy of MJ, and a total number of water molecules of 0 20, whch s all that s known and consdered n thermodynamcs, then you would not know what mcroscopc state the system s n. In order for you to know the mcroscopc state of the system I would have to tell you the poston and momentum etc., of every molecule n the vapour. Ths s llustrated n Fg.. As wth the nformaton that thermodynamcs provdes, N, V and U, you do not know what the mcroscopc state s, then you have to make a guess and ths nherently nvolves probabltes. We have to use probabltes because we do not know the exact mcroscopc state. Indeed the number of mcroscopc states s enormous and the system goes from one to another very rapdly as the partcles move around, as the momenta are changed by collsons etc.. Ths lack of knowledge of the exact mcroscopc state s why ths course s called statstcal physcs. Note on unts As defned by Eq. () the entropy S s just a pure number: t has no unts. In thermodynamcs n partcular, entropy s often defned so that t does have unts, generally JK. In ths course we wll always use entropes defned so that they are dmensonless, as n Eq. (). The entropes used n your thermal physcs course, wth unts of JK are obtaned by multplyng the entropes here by Boltzmann s constant k = JK. 3

4 .2 States wth equal energes are equally probable If dfferent states (wth the same number of partcles) have the same energy, then there s no reason for them to have dfferent probabltes. Thus, we assume that states wth the same energy have the same probablty. For example, consder a system wth two states, call them states and 2. If they have the same energy then they are equally probable,.e., the probablty that the system s n state, p, equals the probablty that the system s n state 2, p 2. As probabltes must add to one, p + p 2 = and so as p = p 2, then both p and p 2 are equal to /2. Thus the entropy s of the smple system s S = p ln p p 2 ln p 2 = ( ) 2 ln ( ) ( ) 2 2 ln = ln = ln 2 (2) 2 2 Here the entropy s just equal to the log of the number of states, 2. In fact n general f a system has not 2 mcroscopc states but Ω states wth equal energes then all these Ω states wll be equally probable. The probablty of each state s then just /Ω, and the the entropy s S = Ω p ln p = = Ω = Ω ln ( ) = Ω ( ) Ω Ω ln = ln Ω ( ) Ω S = ln Ω (4) where we used the fact that the sum s of Ω terms all of whch are the same. So, we see that n general when all the mcroscopc states are equally probable, that the entropy s just the log of the number of states. As a gas has somethng lke exp(0 23 ) states not 2. The equaton S = ln Ω s Boltzmann s equaton for the entropy, for states all wth the same energy. It s n fact on hs tombstone n Venna. Boltzmann s equaton predates that of Shannon, but s less general. The entropy s sometmes called the uncertanty, and we can see why from Eq. (4). As the number of states, Ω, ncreases, our uncertanty as to whch state the system s n ncreases. If Ω = 0 then the system could be n of 0 states, but f Ω = 0, 000, then the system could be n any of 0, 000 states. Thus, as Ω ncreases so does our uncertanty as to whch state the system s actually n..3 States wth dfferent energes: the role of Temperature Most often the mcroscopc states of a system do not all have the same energy. Then the system wll have some average energy U, whch depends on the probabltes of the system beng n the dfferent mcroscopc states wth dfferent energes. By defnton the average energy U s the sum over all states wth each state contrbutng ts energy tmes the probablty that the system s actually n that state. So, U = p ǫ defnton (5) where ǫ s the energy n state. So, f all states have the same energy then they are all equally probable,.e., all p are equal. However, f the states have dfferent energes, then the probabltes p are dfferent. They are proportonal to what s called the Boltzmann weght. We wll derve ths now. To do so we need the defnton of the temperature. Ths s S U = (6) kt (3) 4

5 .e., the temperature s, by defnton, the nverse of the dervatve of the entropy wth respect to the energy U. The proportonalty constant k s Boltzmann s constant, approxmately 0 23 JK. To keep thngs smple, let us consder a system that has only 2 mcroscopc states, or just states for short. Label these states and 2, and let the energy n state be 0, ǫ = 0, whle the energy n state 2 s ǫ,.e., ǫ 2 = ǫ. Also, let the probablty of beng n state 2 p 2 = p. Then the probablty of beng n state s then p = p 2 = p. The entropy and energy are then S = p ln p ( p) ln( p) As U equals just p tmes ǫ, and ǫ s a constant, then U = pǫ (7) S U = S (pǫ) = S p ǫ = kt We can easly take the dervatve of S, and get S p S p p p = ln p + ln( p) = ǫ kt [ ] p = ln = ǫ p kt S p = ǫ kt = exp( ǫ/kt) or p = exp( ǫ/kt) + exp( ǫ/kt) = p 2 (9) So, for a system wth just two levels (the smplest nontrval system) we have derved the Boltzmann expresson for the probablty p = p 2 of beng n state 2, the state wth energy ǫ 2 = ǫ. The factor exp( ǫ/kt) s very common, so t has a specal name: t s the Boltzmann weght of a state wth energy ǫ. The probablty of beng n the ground state, state s just p = p, whch s p = + exp( ǫ/kt) Note that the denomnator s the same n both cases. It s what s called the partton functon and s denoted by Z,.e., for the system wth two mcroscopc states the partton functon (8) (0) Z = + exp( ǫ/kt) () So, we have determned the probabltes of the states n a system wth two mcroscopc states. I wll not derve the general expresson as t s a lttle harder. See any one of the recommended textbooks for the dervaton of the general expresson. However, we need the general result. Ths s that the partton functon Z, the denomnator n the expresson for the probablty, s n general just the sum over all the mcroscopc states, wth each state contrbutng ts Boltzmann weght. The Boltzmann weght s exp( ǫ /kt) for state. So, n general the partton functon s Z = exp( ǫ /kt) (2) Then the probablty of beng n state, p, s the Boltzmann weght of that state, dvded by Z p = exp( ǫ /kt) Z = exp( ǫ /kt) exp( ǫ /kt (3) 5

6 kt/ ε Fgure 2: The energy < ǫ > and heat capacty C of a two-level system, plotted as a functon of temperature. The sold curve s the mean energy dvded by ǫ, u/ǫ, and the dashed curve s the heat capacty over Boltzmann s constant, C/k. 2 The two-level system The two-level system s pretty much the smplest system there s. It has two states: state wth an energy 0 and state 2 wth an energy ǫ. Its partton functon s defned by Z = exp ( ǫ /kt) = + exp ( ǫ/kt). (4) =,2 Now that we know the partton functon, we know the Helmholtz free energy, t s A = kt ln Z = kt ln ( + exp( ǫ/kt)). (5) The mean energy u at a temperature T s easly found from ts defnton u = p ǫ = p 2 ǫ, (6) =,2 as the energy of state s zero and so t does not contrbute to the mean energy. Now, p 2 = exp( ǫ/kt)/z and so u = exp( ǫ/kt)ǫ + exp ( ǫ/kt). (7) We can smplfy ths expresson slghtly by multplyng top and bottom of the rght-hand sde by exp(ǫ/kt). Then ǫ u = + exp(ǫ/kt). (8) Now that we have the mean energy, we can calculate the heat capacty, C. Ths s defned by ( ) u C = (9) T 6

7 left rght Fgure 3: Schematcs of a box contanng four molecules, all of whch are movng due to ther thermal knetc energy. The box s dvded n two, and there are two molecules n each half of the box, although one molecule s about to leave the left-sde half and enter the rght-sde half. If we take the temperature dervatve of Eq. (8) C = ǫ2 kt 2 exp(ǫ/kt) ( + exp(ǫ/kt)) 2 (20) Both the mean energy u and the heat capacty C are plotted as functons of temperature T n Fg. 2. The energy s zero at zero temperature; the exponental n the denomnator of Eq. (8) blows up as T 0 takng the energy to zero. At the other extreme, when the temperature s very large, then the exponental tends to one, as ts argument ǫ/kt tends to zero. Then the denomnator of Eq. (8) tends to two and the energy tends to ǫ/2 per system. At nfnte temperature half of the systems are n the hgher energy state and half are n the lower energy state. Ths s easy to see from the probablty of beng n the ground state probablty of beng n state wth energy 0 = + exp( ǫ/kt) = Z. (2) As T tends to nfnty exp( ǫ/kt) tends to and Z tends to 2. At nfnte temperature the Boltzmann weghts of both states are both, so each state s equally lkely. Conversely, at T = 0, all the systems are n the ground state and so u = 0. In other words at T = 0, Z = and so from Eq. (2) the probablty of beng n the lower energy state s. 3 Fluctuatons and the Central Lmt Theorem of Statstcs Consder a box that contans a gas of N molecules. A schematc of a box wth N = 4 molecules s shown n Fg. 3. It s dvded nto two equal halves so the probablty that a sngle one of these N 7

8 molecules s n the left half s p l = /2. The probablty that t s n the rght half s the same, p r = /2. The entropy of a sngle molecule s then S = p l ln p l p r ln p r = (/2) ln(/2) (/2) ln(/2) = ln 2 As the molecule has two states and they are equally lkely, the entropy s just the log of the number of states, whch s 2 here. So, for N = t s easy: we have a probablty of /2 of t beng n each half, thus leavng the other half empty. But often N s large and we want to know how many of the molecules are n each half. On average we wll always get N/2 molecules n each half. But ths leaves open the queston: Say N = 00, what are the probabltes that say 30 or 7 or 83 of the 00 molecules are n the left-hand sde of the box? We wll address ths queston here, wth the ad of the fundamental theorem n statstcs: The Central Lmt Theorem. To use the Central Lmt Theorem we need to ntroduce the dea of random numbers. A random number s smply a number whose value we don t ntally know, but we do know the probablty whch wth t takes each possble value. For example, the random number generators of computers typcally produce a number anywhere between 0 and, wth each possble number n that range beng equally lkely. Thus f ρ s a random number that wll be produced by such a computer random-number generator, then ρ wll be any number between 0 and, wth equal probablty of beng any number n that range. Another example would be f the random number ρ s the outcome of the throw of a dce. Then ρ would be one of, 2, 3, 4, 5 or 6, wth each of the 6 numbers beng equally lkely. For the case of the gas of molecules we defne a random number ρ for the th molecule that s equal to f the molecule s n the left-hand half and equal to 0 f the molecule s n the rght-hand half. The th molecule s one of the N, so =, 2, 3, 4,..., N. Also note that the mean value of ρ, denoted by ρ, equals one half because on average half the ρ are, and the other half are zero. So, ρ ρ = /2. Wth ths defnton of ρ, the total number of molecules n the left-hand half s the sum over the number on left-hand sde of the box = S l = because for each molecule that s on the left, ρ =. Frst let us consder the average of S l, S l. Ths s S l = N ρ = = N ρ = = N = 2 = N 2 Now the Central Lmt Theorem (CLT) s a theorem for sums over random numbers. It concerns the varatons or fluctuatons of a sum, n ths case S l, about ts mean value S l. The pont s that the sum of a set of random numbers s tself a random number, and f we know the probabltes wth whch the ndvdual random numbers take partcular values, then the CLT tells us about the probabltes wth whch the sum takes partcular values. So, here S l s tself a random number and the CLT tells about how broad the dstrbuton of possble values of S l s. In fact t clarfes thngs f we subtract off the mean value of S l, to get a new random number R defned by R = S l S l R = 0 N = ρ 8

9 Note that by ts defnton the average or mean value of R must be zero. Then R s the random fluctuaton or varaton of the the sum S l from ts average value. For S l, the CLT states that, for large values of N, the root-mean-square of R s gven by R 2 N or ( R 2 ) /2 N /2 The crucal feature (whch you must remember) s that the root-mean-square of R,.e., the standard devaton of S l about ts mean value, scales as the square root of N, whle S l tself scales lnearly wth N. Thus, for N = 00 molecules n the box, the number n the left-hand sde at any nstant wll typcally be somewhere n the range /2 to /2,.e., from 40 to 60. At any gven nstant t s unlkely that there wll be exactly 50 molecules n one half, but t s very lkely that that there wll be between 40 and 60 molecules. Ths corresponds to a fracton on the left of between 0.4 and 0.6. However, f there are N = 0 6 molecules n the box then n one half there wll be between 500,000-,000 and 500,000+,000,.e., between 499,999 and 50,000 molecules on the left. Ths corresponds to a fracton on the left n the range to We see that for N = 0 6 we have very close to half the molecules n each half at all tmes. For N = a reasonable value for a volume of around ltre - the fracton n one half les n the range 0.5 (0 23 ) /2 to (0 23 ) /2, whch s to Here we have essentally exactly half the molecules n each half of the box at all tmes. The CLT s used n pretty much all of statstcs. As an example, consder opnon pollng. Pollsters typcally poll a few thousand people, n order to use the CLT and get relable results. For example, say they are nterested n knowng what percentage of people wll vote Conservatve or Labour. Let us assume that n fact a fracton 0.54 wll vote Conservatve and 0.46 wll vote Labour. If they ask 00 people then the number that wll say they wll vote Labour wll be n the range /2 to /2,.e., from 36 to 56. Thus the pollsters could fnd that anywhere between 36% and 56% would vote Labour and between 44% and 64% would vote Conservatve. These estmates are not accurate enough to tell whether the Conservatves or Labour are more popular. However, f they asked 000 people then between /2 and /2,.e., from 428 to 492 would vote Labour. Ths gves a % votng Labour somewhere between 43% and 49%, and a % votng Conservatve of 5% to 57%. Thus, a poll of 000 people wll relably show that the Conservatves are more popular whle a poll of 00 wll not. Ths s why Mor etc poll large numbers of people. 4 States wth dfferent numbers of partcles So far we have dealt wth states that only dffer n ther energes,.e., state has energy ǫ, state 2 has energy ǫ 2 ǫ, etc. Here, we wll consder states that also have varyng numbers of partcles,.e., state has energy ǫ and a number of partcles n, state 2 has energy ǫ 2 ǫ and a number of partcles n 2 n, etc. Dealng wth states wth dfferent numbers of partcles s completely analogous to dealng wth states wth dfferent energes. So we won t derve any expressons, we wll just wrte them down. See any of the recommended textbooks for dervatons. So, recall that for states that dffer only by ther energy the probablty of beng n a state wth an energy ǫ s p (ǫ ) exp( ǫ /kt) p (ǫ ) = exp( ǫ /kt) Z for Z = exp( ǫ /kt) 9

10 When the number of partcles vares as well the probablty also depends on the number of partcles. We denote the number of partcles n state, as n. Then the probablty of state s p(ǫ,n ). The dependence of the probablty on n s exponental, just as ts dependence on ǫ. The coeffcent n the exponental s µ/kt, where µ s the chemcal potental. Just as the temperature T controls the probabltes of states wth dfferent energes, the chemcal potental µ controls the probabltes of states wth dfferent numbers of partcles. So we have p (ǫ,n ) exp( ǫ /kt + n µ/kt) p (ǫ,n ) = exp( ǫ /kt + n µ/kt) and wth = exp( ǫ /kt + n µ/kt) where s the partton functon when the number of partcles can vary as well as the energy. s the sum over all possble states of the system, just as Z s, but now each state contrbutes a factor that depends on n as well as ǫ. The probablty of a state s proportonal to exp(µ number of partcles n state/kt) and so f the chemcal potental µ s ncreased, states wth large numbers of partcles n them become more probable, more lkely. Also, unlke the temperature, the chemcal potental can n general be ether postve or negatve. If t s postve, states wth large numbers of partcles are favoured, whereas f t s negatve, states wth small numbers of partcles are favoured. As an example, consder a system that has just two states: state one wth 0 partcles, n = 0, and 0 energy, ǫ = 0, and state two wth partcle, n 2 =, and energy ǫ 2 /kt =,.e., the energy s n unts of kt. The chemcal potental µ/kt = 2,.e., s 2 n unts of kt. To calculate the probabltes of beng n each of the states, p and p 2, we frst need to calculate the partton functon = 2 exp( ǫ /kt + n µ/kt) = exp( ) + exp( + 2) = + e = 3.72 = Then the probabltes are p = exp( ǫ /kt + n µ/kt) p = exp(0) 3.72 = p 2 = exp() 3.72 = 0.73 the state wth the partcle n t s about three tmes as probable as the state wthout. 5 Ferm-Drac and Bose-Ensten Statstcs Now we wll do statstcal mechancs on partcles that behave quantum mechancally. Of course, quantum-mechancal partcles come n two varetes: fermons and bosons. Partcles wth half ntegral spn, electrons, 3 He etc., are fermons and partcles wth zero or ntegral spn, photons, 4 He etc., are bosons. The crucal dfference between fermons and bosons s that fermons obey Paul s excluson prncple whch states that no more than one fermon can occupy an energy level. An energy level can be empty,.e., contanng 0 fermons, or t can contan fermon, no more. An energy level can contan any number of bosons, ndeed very cold 4 He undergoes what s known as Bose-Ensten condensaton and then a sgnfcant fracton of all the atoms (there could be 0 23 atoms) n the helum are all n one level. 0

11 5. A sngle level wth Ferm-Drac statstcs Consder a sngle energy level (don t worry about where ths level comes from for now) wth an energy ǫ. Ths level can accommodate a fermon (no more than one of course). Ths level has two states: state wthout a fermon n t, energy = 0, and state 2 wth a fermon, energy = ǫ. Thus ǫ = 0, n = 0, and ǫ 2 = ǫ, n 2 =. We want to work at fxed chemcal potental µ and temperature T. Then the approprate weght of a state wth energy ǫ = n ǫ and number of partcles n s exp[ n ǫ/kt + n µ/kt]. The canoncal partton functon s just the sum of these weghts over the two possble states,.e., = exp [ n ǫ/kt + n µ/kt] + exp [ n 2 ǫ/kt + n 2 µ/kt], (22) or = + exp[(µ ǫ)/kt]. (23) Now that we have the partton functon we can calculate the probabltes of beng n the each of the two states, p and p 2, p = = + exp[(µ ǫ)/kt] p 2 = exp[(µ ǫ)/kt] = exp(µ ǫ)/kt] + exp[(µ ǫ)/kt] We can also work out the average number of fermons n the energy level, n. It s just a sum over the two states, wth each state contrbutng the probablty of t beng occuped tmes the number of fermons n t. For our system wth two states n = p n + p 2 n 2 = 0 + exp[(µ ǫ)/kt] + exp[(µ ǫ)/kt] = exp[(ǫ µ)/kt] +. (24) Because no more than one fermon can occupy the level n can never be more than. In Eq. (24) the exponental s never negatve so n equals, dvded by plus a nonnegatve quantty, and so les between 0 and. A plot of n as a functon of µ ǫ s shown n Fg A sngle level wth Bose-Ensten statstcs Agan consder a sngle energy level wth an energy ǫ. Ths level now accommodates bosons, any number of them. The level has an nfnte number of states: one wthout a boson n t, energy= 0; one wth boson, energy ǫ; one wth 2 bosons, energy 2ǫ; one wth 3 bosons, energy 3ǫ; etc. Agan we want to work at fxed chemcal potental µ and temperature T. Then the approprate weght of a state wth energy nǫ and number of partcles n s exp[ nǫ/kt + nµ/kt]. The partton functon s just the sum of these weghts over all possble states. So, = + exp[(µ ǫ)/kt] + exp[2(µ ǫ)/kt] + exp[3(µ ǫ)/kt] + = + exp[(µ ǫ)/kt] + (exp[(µ ǫ)/kt]) 2 + (exp[(µ ǫ)/kt]) 3 + (25) because for any x, exp(2x) = [exp(x)] 2 etc. Here the st term s for an empty level, the 2nd for a level wth boson, the 3rd for two bosons n the level, etc. If we defne y = exp[(µ ǫ)/kt], then we have = + y + y 2 + y 3 + y = y + y 2 + y 3 + y 4 + y = = y, (26)

12 n ( µ ε ) / kt 3 4 Fgure 4: Plots of the mean number of partcles, n, n a sngle energy level of energy ǫ, as a functon of the chemcal potental mnus ths energy, over kt. The sold black curve s for fermons and the dashed grey curve s for bosons. Note that the curve for bosons dverges at µ = ǫ and s not defned for µ > ǫ. and = exp[(µ ǫ)/kt]. (27) The probablty of the state wth n bosons n t s then just p n = exp [n(µ ǫ)/kt] so p 0 = p = exp [(µ ǫ)/kt] We can also work out the average number of bosons n the level, n. It s just a sum over each state, wth each state contrbutng p n n. So, n = p p + p exp [(µ ǫ)/kt] n = n = y + 2y2 + 3y 3 + exp [2(µ ǫ)/kt] 2 + exp [3(µ ǫ)/kt] 3 + The sum on top s y + 2y 2 + 3y 3 +. Now, f we take the dervatve of = + y + y 2 + y 3 + we have d dy = + 2y + 3y2 + and then f we multply by y y d dy = y + 2y2 + 3y 3 + whch s the sum we need. So the top half of the expresson for n s equal to yd/dy. As we know = /( y) ths dervatve s easy to calculate d dy = d( y) dy 2 = ( y) 2 (28)

13 and so and fnally n = y + 2y2 + 3y 3 + = n = y d dy = y( y) 2 y( y) 2 ( y) = y( y) = exp[(ǫ µ)/kt] exp [(µ ǫ)/kt] exp[(µ ǫ)/kt] Because many bosons can occupy the level n ranges from 0 to. n = when exp[(ǫ µ)/kt] =..e., when ǫ = µ. Note that µ cannot be larger than ǫ, f t were larger than ǫ, exp[(ǫ µ)/kt] < and n would be negatve. n s an average number of partcles n a level, t must be nonnegatve, you can t have a mnus number of partcles. If µ s ncreased so that t equals ǫ then Bose-Ensten condensaton takes place. Ths phenomenon s beyond the scope of ths course. Here we wll only consder the case ǫ µ > 0. A plot of n as a functon of µ ǫ s shown n Fg Averages and the relatonshp between the partton functon and thermodynamc quanttes 6. Averages The average of a quantty x s denoted by x. By defnton t s gven by x = p x, (29) where the sum s over all possble states, and x s the value of the quantty x n state. Ths works for any quantty that has a defnte value n a gven state. Here p s the probablty that the system s n state. Thus, f we can calculate the value of x n each state and can fnd the probabltes p we can fnd the average value of x. The above expresson s always true, but for the rest of ths secton we wll consder only averages at constant T. 6.2 Average quanttes at constant T We have already found that the probablty p that a system s n a state wth an energy ǫ s p = exp( ǫ /kt)/z, where Z s the partton functon Z = exp( ǫ /kt) Knowng the probabltes p, we can calculate averages: the average of x s x = p x = Z x exp( ǫ /kt), 3

14 For example, when x s the energy ǫ, we have an expresson for the average energy ǫ. Ths s gven by just the above equaton wth x replaced by ǫ ǫ = ǫ exp( ǫ /kt). Z For example, for a smple system wth only 2 states. State wth energy ǫ and state 2 wth energy ǫ 2 we have that the partton functon Z s and the average energy s Z = exp( ǫ /kt) + exp( ǫ 2 /kt) ǫ = Z [ǫ exp( ǫ /kt) + ǫ 2 exp( ǫ 2 /kt)] = ǫ exp( ǫ /kt) + ǫ 2 exp( ǫ 2 /kt). exp( ǫ /kt) + exp( ǫ 2 /kt) 6.2. The partton functon Z and the Helmholtz free energy A Here we wll relate the statstcal-physcs quantty, Z, and the thermodynamcs quantty, A, the Helmholtz free energy. Recall from your thermodynamcs lectures that the Helmholtz free energy s defned by A = U kts where U s the thermodynamc energy. Note that S was gven n dfferent unts n your thermodynamcs course so n that course there was no factor of k n ths equaton. Earler, we found that the typcal sze of fluctuatons about the mean was equal to the mean tself dvded by N /2. Thus for macroscopc systems (N 0 22 or more, whch s what thermodynamcs consders) these fluctuatons are only a fracton 0 of the mean and so are neglgble. Thus the energy s always ndstngushable from the average, ǫ, so, U = ǫ Ths s the, trval, relatonshp between the average energy n statstcal physcs and the thermodynamc energy. To see how we can calculate the Helmholtz free energy n statstcal physcs we wll start from Shannon s equaton for the entropy. Ths s S = p ln p = ( ) exp( ǫ /kt) exp( ǫ /kt) ln = [ exp( ǫ /kt) ǫ ] Z Z Z kt ln Z where we used p = exp( βǫ )/Z. We can smplfy the above expresson S = ZkT S = ǫ kt + lnz exp( ǫ /kt)ǫ + ln Z Z exp( ǫ /kt) because the frst sum over Z s smply the defnton of the average energy ǫ (over kt), and the second sum s just Z. At the moment the last equaton just relates the entropy S, the average energy 4

15 ǫ and the log of the partton functon. But f we take the thermodynamc equaton A = U kts and rearrange t we have S = U kt A kt. If we compare ths equaton wth the one above, clearly A = kt ln Z. So, from ths equaton f we can calculate the partton functon Z, then we can fnd the Helmholtz free energy A smply by takng ts logarthm and multplyng t by kt. Fnally, we note that f we take the dervatve of the partton functon Z wth respect to temperature, we obtan or ( ) Z T = T exp ( ǫ /kt) = ǫ kt 2 exp ( ǫ /kt) = ǫ Z kt. 2 Z ( ) Z = ǫ T kt 2 But ( ) ln Z = ( ) Z T Z T so ln Z T = ǫ or ǫ = kt 2 ln Z kt 2 T So, we can calculate the average energy by calculatng Z and then takng the dervatve of ts logarthm. 7 Classcal Statstcal Mechancs So far we have defned the partton functon Z as a sum over all the states, wth each state contrbutng ts Boltzmann weght, Z = exp ( ǫ /kt). (30) But we would now lke to consder classcal systems such as gases at relatvely hgh temperatures. There the state s specfed by specfyng the postons and veloctes (or equvalently the momenta) of all the partcles. These postons and veloctes are contnuous varables,.e., they are not quantsed and can therefore take any value, not just a dscrete set of values. By gases at relatvely hgh temperatures we mean gases where quantum mechancal effects are small and so we can treat them as beng composed of partcles whch can be treated usng classcal mechancs. Ths s true of the atmosphere, but not of low temperature helum. In low temperature helum we need to take quantum mechancs nto account. A gas has a very large number of molecules movng n three dmensons but to keep thngs smple we wll just consder a sngle partcle movng n one dmenson. The generalsaton to many partcles s easy. So consder a sngle partcle n a one-dmensonal box so that t can only move back and fore along the x axs. The mcroscopc state of the system of one partcle s specfed by specfyng the x 5

16 coordnate of the partcle and the x-component of the momentum of the partcle, p x. So, f we take one mcroscopc confguraton specfed by x and p x, and move the partcle from x to x + dx then we have a new mcroscopc state. We can also generate a new confguraton by changng the momentum from p x to p x +dp x. So, n order to sum over all the states we must ntegrate over all possble postons and momenta of the partcle. The sum becomes an ntegral dxdp x. (3) Also, the energy ǫ, whch dd depend on the state, hence the subscrpt on ǫ, s now a functon of x and p: ǫ(x,p). Puttng all ths together, Z exp( ǫ(x,p)/kt)dxdp x. (32) The proportonal sgn,, s there because we expect the partton functon to be dmensonless as t s a sum of pure numbers, the Boltzmann weghts. Ths s clear from Eq. (30). The ntegral on the rght hand sde of Eq. (32) has dmensons of length tmes momentum (from the factor dxdp x ). So, we need a proportonalty constant wth dmensons of one over length tmes momentum to put on the rght hand sde of Eq. (32). In fact, the correct proportonalty constant s one over Planck s constant, whch has the correct dmensons. So, Z = h exp( ǫ(x,p)/kt)dxdp x (33) I am not gong to prove ths. It s also not very mportant as changng the proportonalty constant does not actually change propertes we can measure such as the heat capacty etc. Indeed n the very next secton I am gong to drop the constant n front of the ntegral. The mportant thng to remember s that n classcal systems you just replace the sums over states by ntegrals over all possble values of the varables lke poston and momentum. 7. Example For example, consder a partcle restrcted to a box of length L, stretchng from x = 0 to x = L, and a momentum range from p x = 0 to p x = m. If the energy s zero nsde the box and nfnte nsde t the classcal partton functon s x=l px=m Z = h dx exp( ǫ(x,p)/kt)dxdp x = h dx dp x exp(0) = h [x] x=l x=0 [p x ] px=m p x=0 = h L m = Lm h Compare ths wth the partton functon Z for a quantum mechancal system wth Ω states all wth zero energy Ω Z = exp ( 0/kT) = Ω = Note that for the quantum mechancal partton functon doublng the number of states doubles the partton functon whle the classcal partton functon doubles f ether the length or the momentum range are doubled. Essentally, the longer the length of the box the more possble postons of the partcle there are and ths s equvalent to ncreasng the number of possble states. x=0 p x=0 6

17 7.2 The equpartton theorem Ths theorem apples to classcal systems n whch the energy vares quadratcally wth a coordnate,.e., the energy vares as some constant tmes the square of the coordnate. It does not matter what the coordnate s. For example, n an deal gas there s a knetc energy of (p 2 x + p 2 y + p 2 z)/(2m) per partcle whch s quadratc. Also, the energy of a Hookean sprng, = (/2)kx 2, s quadratc. We can derve the theorem usng any quadratc energy. For example, consder the quadratc energy ǫ(p) = bp 2, (34) where b s a constant. If p s the momentum of a partcle along say the x axs, then b would be /2m, where m s the mass of the partcle. The partton functon for the coordnate p s obtaned by ntegratng over all possble values of p. Z = dp exp( bp 2 /kt), (35) We have dropped the factor of h as t s a constant and wll not affect the energy, whch s a dervatve of the logarthm of the partton functon. The ntegral s a smple Gaussan ntegral lke So, ( π ) /2 exp( az 2 )dz =. (36) a ( ) /2 πkt Z =, (37) b the quadratc energy has contrbuted a factor whch s proportonal to T /2 to the partton functon. Ths s always true of any coordnate whch has a contrbuton to the energy whch depends on the square of the coordnate. We would lke the average energy U = ǫ, whch s obtaned from the dervatve of the partton functon (see prevous secton) U = kt 2 ln Z T = 2 kt 2 T = kt 2 [ ln ( πkt ( ) /2 πkt T ln b )] = b 2 kt 2 T [ ln T + ln ( )] πk = 2 [ ] T b kt = kt (38) 2 The average energy U of a partcle wth an energy whch depends quadratcally on ts momentum s just (/2)kT. Note that t does not depend on the value of b, changng b has no effect on the mean energy ǫ. Ths s general for any quadratc term, and s called the equpartton theorem Equpartton theorem = For a classcal system any quadratc term n the energy contrbutes (/2)kT to the energy. For example, n an classcal gas, each of the N partcles has a knetc energy (p 2 x + p 2 y + p 2 z)/2m three quadratc terms, one each from the x, y and z drectons therefore each gas partcle has an energy (3/2)kT and N partcles have an energy (3/2)NkT. 7

18 The heat capacty s obtaned from C = U T = k. (39) 2 The quadratc term has contrbuted k/2 to the heat capacty C; ths s a general result. The heat capacty C = (/2)k the number of quadratc terms n the energy. 7.3 When s a gas classcal and when are quantum mechancal effects mportant We would lke to know when we can treat the molecules n a gas usng classcal mechancs and when quantum mechancs s requred. For example, the above smple treatment of a gas s only vald, only gves the rght answer, when the molecules can be treated usng classcal mechancs. So to reassure ourselves that our answer s rght, we would lke to satsfy ourself that classcal mechancs s a vald descrpton of the gas molecules. Now, classcal mechancs of course treats molecules as pont-lke partcles wth a defnte poston, whereas wthn quantum mechancs (QM) we have the wave/partcle dualty. In QM we have Hesenberg s uncertanty prncple whch states that x p, where x s the uncertanty or blurrng of the poston x, and p s the uncertanty n momentum p. Now, f the typcal separaton between the molecules n a gas, call t s, s much larger than the blurrng of the postons of the molecules due to QM, x, then ths blurrng can be neglected and classcal mechancs s a good approxmaton. However, f s < x then the uncertanty n the poston of the partcles s greater than ther dstance apart,.e., ther wavefunctons are overlappng. Then classcal mechancs s not a good approxmaton, we have to use QM and calculate the wavefuncton etc. We estmate the uncertanty n x by fndng the uncertanty n p, p, and then usng Hesenberg s uncertanty prncple: x = / p. To fnd p, we start wth the equpartton theorem. If a gas s classcal then the average knetc energy of a partcle for moton along one of the axes, x, y or z, s smply (/2)kT. Ths average knetc energy s just p 2 /2m, where p 2 s just the average of the square of the momentum along the x axs, p. So, we have mean K.E. = p2 2m = 2 kt p2 = mkt. (40) and p2 = (mkt) /2 (4) So the root-mean-square momentum p 2 = (mkt) /2. Ths gves the typcal magntude of the momentum and ths momentum can be n ether drecton so the uncertanty n the momentum s approxmately the same as ths magntude p p 2 = (mkt) /2. Combnng ths wth Hesenberg s uncertanty prncple we have x p (mkt) /2 (42) Ths uncertanty n, x, x s often called the thermal or de Brogle wavelength. For a classcal-mechancs descrpton of a gas to be vald x must be smaller than s. Now, f the gas s at a densty of ρ molecules per cubc metre, then the volume per molecule s /ρ. The separaton 8

19 s s roughly the cube root of the volume per partcle so s (/ρ) /3. So fnally, for a gas of molecules to be classcal we requre that x s (mkt) /2 ρ /3 (43) Let us put some numbers nto ths equaton. Atmospherc pressure s about p = 0 5 Pa, and p = ρkt. Ths means that at room temperature, kt = J, ρ = 0 5 /(4 0 2 ) 0 26 molecules/m 3. So the average dstance between molecules s (/0 26 ) /3 0 9 m or about nm. Now for the QM uncertanty n the poston of the molecule, x, at room temperature. The mass of a proton s kg and the most abundant gas n the atmosphere s 4 N 2 whch has a mass kg. Thus x /(mkt) 0 34 /( ) /2 0 0 m. So the blurrng of the poston of the ntrogen molecules due to QM effects, at room temperature, s about 0.nm, or tenth of ther average spacng n Earth s atmosphere. Therefore classcal mechancs works for the gas molecules n the atmosphere. 9