Transcription

1 Statistical Mechanics Dr Alfred Huan

2 Contents What is Statistical Mechanics? 3. Key Ideas in the module Why do we study Statistical Mechanics? Denitions Microstate Macrostate Distribution Law 6. Distribution Law & Ensemble Approaches Computation of Most Probable Macrostate Distribution, n Problems in Derivation of Maxwell-Boltzmann Distribution n not being always large What about the other macrostates? What happens when we have many energy levels? Time Evolution of system of microstates Alternative Derivation for Maxwell-Boltzmann Distribution Examples of Counting in Statistical Mechanics Indistinguishable Particles 3. Introduction Bose-Einstein Distribution for Bosons Fermi-Dirac Distribution for Fermions Alternative Derivation for Bose-Einstein and Fermi-Dirac Statistics Statistical Mechanics and Thermodynamic Laws 6 4. First Law Second Law Third Law Calculation of Other Thermodynamic Variables Applications of Maxwell-Boltzmann Statistics Classical Perfect Gas Harmonic Oscillator Equipartition Theorem Paramagnetic Systems 4 6. Introduction Spin Quantum System Spin J Quantum System

3 CONTENTS 6.4 Adiabatic Cooling Applications of Fermi-Dirac Statistics 5 7. Fermi-Dirac Distribution Free Electron Model of Metals Applications of Bose-Einstein Statistics Introduction to Black Body Radiation Bose-Einstein Distribution Planck's Radiation Formula Radiation emitted by blackbody The Classical Limit The Classical Regime Monatomic Gas Gibbs Paradox Kinetic Theory of Gases 7. Introduction Number of impacts on Area A ofawall Eusion of molecules Mean Free Path, Collision cross-section,

4 Chapter What is Statistical Mechanics?. Key Ideas in the module We begin with the introduction of some important ideas which are fundamental in our understanding of statistical mechanics. The list of important terms are as follows :. Microstate. Macrostate 3. Partition Function 4. Derivation of Thermodynamic laws by Statistical Mechanics 5. The idea of interaction (a) particle (system) - external environment (b) particle (within a system of many particles) 6. Entropy : Disorder Number 7. Equipartition Theorem : the energies are distributed equally in dierent degrees of freedom. 8. Temperature 9. Distributions : Maxwell-Boltzmann, Fermi-Dirac, Bose-Einstein In the 3rd year course, we shall restrict to equilibrium distributions, and static macroscopic variables. We will deal with single particle states and no ensembles, weakly interacting particles and the behaviour for distinguishable (degeneracies) and indistinguishable particles. 3

5 CHAPTER. WHAT IS STATISTICAL MECHANICS? 4. Why do we study Statistical Mechanics? We apply statistical mechanics to solve for real systems (a system for many particles). We can easily solve the Schrodinger's equation for particle/atom/molecule. For many particles, the solution will take the form : total = linear combination of a () b () c (3)::: where a means a particle in state a with an energy E a. For example, we consider particles conned in a cubic box of length L. From quantum mechanics, the possible energies for each particle is : E = ~ m L (n + x n + y n) z For example in an ideal gas, We assume that the molecules are non-interacting, i.e. they do not aect each other's energy levels. Each particle contains a certain energy. At T >, the system possesses a total energy,e. So, how is E distributed among the particles? Another question would be, how the particles are distributed over the energy levels. We apply statistical mechanics to provide the answer and thermodynamics demands that the entropy be maximum at equilibrium. Thermodynamics is concerned about heat and the direction of heat ow, whereas statistical mechanics gives a microscopic perspective of heat in terms of the structure of matter and provides a way of evaluating the thermal properties of matter, for e.g., heat capacity..3 Denitions.3. Microstate A Microstate is dened as a state of the system where all the parameters of the constituents (particles) are specied Many microstates exist for each state of the system specied in macroscopic variables (E, V, N,...) and there are many parameters for each state. We have perspectives to approach in looking at a microstate :. Classical Mechanics The position (x,y,z) and momentum (P x, P y, P z ) will give 6N degrees of freedom and this is put in a phase space representation.. Quantum Mechanics The energy levels and the state of particles in terms of quantum numbers are used to specify the parameters of a microstate.

6 CHAPTER. WHAT IS STATISTICAL MECHANICS? 5.3. Macrostate A Macrostate is dened as a state of the system where the distribution of particles over the energy levels is specied.. The macrostate includes what are the dierent energy levels and the number of particles having particular energies. It contains many microstates. In the equilibrium thermodynamic state, we only need to specify 3 macroscopic variables (P,V,T) or (P,V,N) or (E,V,N), where P : pressure, V : Volume, T : Temperature, N : Number of particles and E : Energy. The equation of state for the system relates the 3variables to a fourth, for e.g. for an ideal gas. PV = NkT We also have examples of other systems like magnetic systems (where we include M, the magnetization). However, the equilibrium macrostate is unknown from thermodynamics. In statistical mechanics, the equilibrium tends towards a macrostate which is the most stable. The stability of the macrostate depends on the perspective of microstates. We adopt the a priori assumption which states the macrostate that is the most stable contains the overwhelming maority of microstates. This also gives the probabilistic approach in physics.

7 Chapter Distribution Law. Distribution Law & Ensemble Approaches The distribution law is dependent on how we dene the system. There are three approaches in dening the system and we adopt the three ensemble approaches.. Microcanonical Ensemble : Isolated system with xed energy, E and number of particles, N.. Canonical Ensemble : The system is in a heat bath,at constant temperature with xed N. 3. Grand Canonical Ensemble : Nothing is xed. The disorder number, is dened as the number of microstates available to a macrostate, and is very large for the equilibrium macrostate. The disorder number is related to entropy, which is maximum at equilibrium. To enumerate, we need to decide whether the particles are distinguishable or indistinguishable. If they are indistinguishable, every microstate is a macrostate. Let us now consider independent particles, with energy levels xed and the particles don't inuence each other's occupation. The energy levels are xed by each particle's interaction with its surroundings but not particle-particle interactions. Mutual interaction exist only for energy exchange (during collisions). Each level has energy ", degeneracy g, and occupation n. Distinguishability : small and identical particles cannot be distinguished during collisions, i.e. non-localised. However, localised particles are distinguishable because they are xed in space, the sites being distinguishable e.g. in a paramagnetic solid. For an isolated system, the volume V, energy E and total number of particles, N are xed. 6

8 CHAPTER. DISTRIBUTION LAW 7 N = n Total E = n " The A Priori assumption states that all microstates of a given E are equally probable thus the equilibrium macrostate must have overwhelming. Ω Equilibrium state Macrostates The disorder number is proportional to thermodynamic probability, i.e. how probable for a particular macrostate to be the one corresponding to equilibrium.. Computation of Most Probable Macrostate Distribution, n To nd the n at equilibrium, i.e. we must nd the most probable macrostate distribution. Considering the case for distinguishable particles :. Enumerate. We distribute N particles over the energy levels: For example, For the rst level, For the second level, The number of ways to assign n particles = N! n!(n, n )! (.) The number of ways to assign n particles = N, n! n!(n, n, n )! (.)

9 CHAPTER. DISTRIBUTION LAW 8 N! Numberofways = n!(n, n!) N, n! n!(n, n, n )! :::::: N! = n!n!::::n!::::: = Q N! n! (.3) Level has degeneracy g so any of the n particles can enter into any of the g levels. For all levels, Number of ways = g n (.4) Number of ways = Y g n (.5) Total = N! Y g n n! (.6). Maximise by keeping xed E, N. These constraints are introduced through Lagrange multipliers. For a xed V, it means that the energy levels," are unchanged. Maximise subect to n = N and n " = E At maximum, d = (.7) ) d(ln ) = d = Hence we can choose to maximise ln instead of ln = ln N!+ n ln g, (.8) ln n! (.9) By Stirling's approximation for large N : ln N! =Nln N, N (.)

10 CHAPTER. DISTRIBUTION LAW 9 Proof : Or N ln n 4 n = ln N! = Z N = N Z N ln n ln ndn =[nln n, n] N = N ln N, N + (N ln n N + N ln N)d( n N ) = N[x ln x, x] N + N ln N(, N ) =,N +lnn++nln N, ln N = N ln N, N + (.) (.) where 4n = d( n N )=dx. Now we come back to ln ln N ln N, N + (n ln g, n ln n + n ) = N ln N + = N ln N + (n ln g, n lnn ) n ln( g n ) (.3) by assuming large n also, which is not always true. d(ln ) = (ln g n dn, dn )= (.4) It is subected to the constraints : dn = (for xed N) " dn =(for xed E) Note " does not vary. Therefore ln( g n )dn = (.5)

11 CHAPTER. DISTRIBUTION LAW Now we introduce the constraints : so we get We choose and such that, The equilibrium macrostate becomes : n = g e e," dn = " dn = (ln( g n )+," )dn = (.6) ln( g n )+," = (.7) = P Ng g e," e," = Ng Z e," where z is the partition function (Maxwell-Boltzmann Distribution) (.8) We will prove later that is equal to where k is the Boltzmann constant (k = kt :38,3 JK, mol, ) and T is the temperature. For " >" i,weget n <n i for g = g i and T > For T =,n =, except for n, i.e. all the particles are at ground state. For T >, the upper levels get piled by particles provided Otherwise, the population of upper levels remain small. In lasers, population inversion leads to negative temperature, 4" kt (.9) for g = g, because k>. Since " >",then T < " kt n = g e, n g e, " kt = e, " kt e, " kt > ) e, " kt >e, " kt, " kt >, " kt, " T >," T (.) (.)

12 CHAPTER. DISTRIBUTION LAW.3 Problems in Derivation of Maxwell-Boltzmann Distribution There are inherent problems present in the derivation of Maxwell-Boltzmann distribution :. n is not always large.. The other macrostates of the same xed energy E. What are their s? Are they small enough to be neglected? The equilibrium distribution ~n should be weighted against of all possible macrostates i, so ~n = P P n i (n ) i(n ) The most probable n (Maxwell-Boltzmann Distribution) (.) n = Ng e," kt P g e," kt (.3).3. n not being always large P If the total number of levels is small and T is large enough, then every level has many particles so that for large n. Stirling's approximation can be used. If many levels exist, the lower levels will have large n but higher levels will have small n. Many of the higher levels will give ln n!. (Note that! =! = ). However, the higher levels contribute very negligibly to P ln n! so that we can approximate again by P (n ln n, n ) because many of the lower levels have large n..3. What about the other macrostates? How does the change when we consider other macrostates? Consider the changes from the most probable macrostate n to another distribution n. n = n + n (.4) Since ln (n )=Nln N, N, (n ln n, n ) (.5) = N ln N, n ln n

13 CHAPTER. DISTRIBUTION LAW Expand about n, ln (n )=Nln N, (n +n )ln(n +n ) = N ln N, = N ln N, = N ln N, =ln(n ), (n +n )(ln n +ln( + n n ) (n +n )(ln n + n n, ((n ) n )+::::) n ln n + n + n ln n + ((n ) n, n ln n, (n ) n )+::: n + :::: (.6) where n = N Z e," kt neglecting g for the purpose of simplifying the discusion. Now as before, n = and n " = Thus, ln (n )=ln( n ), = ln( n ), ln N Z = ln( n ), n (ln N Z, " kt ), (n ) + ::: n (n ) n + kt n + :::::: " n, (n ) n + ::: (.7) It follows that (n ) will be small, if n p n. (n ) = ( n )e P (n ) n (.8) Since n are large, the uctuations allowed are very small, by comparison, e.g. if n, n. for (n ) to be large. Thus n n, so the uctuation is minor.. Ultimately, in deriving properties of the system, minor uctuations will not matter very much.

14 CHAPTER. DISTRIBUTION LAW 3 Ω (i) (ii) (ii) Macrostates where (i) represents n n and (ii) n large. Alternatively, we can also argue the problem out by the choice of a fractional deviation = (n ) n, and we assume is the same for all. Then, Hence (n ) n = = N n (.9) (n )=(n )e, N (.3) which is small unless. N, i.e, n. n p < (.3) N so n needs to be small in order for (n ) to be signicantly large. In either case, we see that (n ) will be small when the uctuation n is large Now we come back to the total number of macrostates : The total number of macrostates consistent with constraints : Z,n ( n )Z,n e n dn e n dn ::::::: =(n ) Y (n ) (.3)

15 CHAPTER. DISTRIBUTION LAW 4 Stopping the product when n = If total number of energy levels, G is very much less than N, then n N G..., G. and =,, ( n )( N G ) G (.33) ln ln ( n )+ G ln N, G lng (.34) Now ln ( n ) N ln N ln N ln G (.35) so so the most probable microstate has overwhelming. ln ln ( n ) (.36).3.3 What happens when we have many energy levels? If n for most levels, then we have many energy levels G N. This means that all macrostates look roughly the same. The total number of microstates are not mainly due to the maximum term ( n ), i.e. P 6= (~n ). Hence Stirling's approximation is invalid in this case. We resort to the following trick : Energy levels are closely spaced in groups because G is large. They are grouped into bunches of g i with m i occupation. g i and m i can be made large in this case. i = i = first bundle of g bundles nd bundle of g bundles The energy levels are so closely spaced such that they are grouped into bundles of g levels i Using the previous derivation by counting ( m i energy states), m i = Ng ie,"i kt Pi g ie," kt = Ng ie,"i kt P," e kt (.37)

16 CHAPTER. DISTRIBUTION LAW 5 We take the time average value of n over a large number of dierent macrostates. m i = m i g i = Ne," kt P e," kt (.38) ( m i )=N! Y i g m i i m i! (.39) ln ( n )=ln (.4) By grouping closely spaced levels, the time-averaged variations are small and the Maxwell-Boltzmann distribution describes n very well. Otherwise with n, the variation at each level appears large. The treatment tries to obtain the average n i over many dierent but similar macrostates. After a series of attack on our fundamental assumptions, we come to establish the basis of statistical mechanics : All macrostates of signicant probability (i.e. large ) have roughly the same properties and distribution close to the Maxwell- Boltzmann distribution ignoring minor changes..4 Time Evolution of system of microstates How many microstates does the system explore within the time for the experiment? For example, we consider paramagnetic system (where the atomic spins interact with the external magnetic eld - Zeeman Eect). The particles inside this system (= 3 particles) are assumed to be non-interacting, and each electron have a spin of. We also x E and T for a xed macrostate i.e. n. Note in Quantum Mechanics, the energy levels are discretized. k k :B k E f(t )k- each atom The change in microstates will mean that the spin energy of the electron is exchanged between neighbouring atoms so the atoms interchange energy levels. The typical time for spin ip is, s. So for 3 electrons : Now, the number of microstates : The Transition Speed = 3 = 35 s, e N ln N >e 35 o 35 So it takes forever to go through all the microstates. Hence, for every experiment, one can only explore a small proportion of microstates. So we need a statistical assumption that what we observe corresponds to the real average over all microstates.

17 CHAPTER. DISTRIBUTION LAW 6.5 Alternative Derivation for Maxwell-Boltzmann Distribution We present an alternative method to derive Maxwell-Boltzmann distribution from the perspective of equilibrium. Suppose we change from macrostate to another macrostate, keeping E, V and N xed. We need a transition from i to k to be accompanied by another transition from to l. E l k E i Both the transitions from to l or i to k requires the same amount of energy. Therefore, the rate R i!kl = N i N T i!kl (.4) where T i!kl is the transition probability. The total rate out from leveli: R iout = N i N T i!kl (.4) The total rate in to leveli: ;k;l R iin = ;k;l N k N l T kl!i (.43) At equilibrium, R i out = R i in (.44) Therefore it follows that : N i ;k;l N T i!kl = ;k;l N k N l T kl!i (.45) Now we invoke the time-reversal invariance, which is true for almost all transitions, T kl!i = T i!kl (.46)

18 CHAPTER. DISTRIBUTION LAW 7 Thus for every combination i,, k and l, we require N i N = N k N l (Principle of Detailed Balancing) (.47) or ln N i +lnn =lnn k +lnn l (.48) and " i + " = " k + " l (.49) is required for energy conservation. Since N i N(" i ), both equations will lead to ln N i =," i + (.5) which is the Maxwell-Boltzmann Distribution. N i = e e," i (.5).6 Examples of Counting in Statistical Mechanics Problem : Given 6 distinguishable particles, energy levels (one with a degeneracy of and the other degeneracy of 5). We want to calculate the number of macrostates and microstates in this system. First of all, we present a diagramatic scheme of the problem : g = 5 g = 6 particles We also note importantly that there are no equilibrium macrostate in this problem because all the macrostates have dierent total energies. There are 7 macrostates namely : (6,), (5,), (4,), (3,3), (,4), (,5) and (,6). Now we come to calculate the microstates : For (6,) macrostate, = 6 ways

19 CHAPTER. DISTRIBUTION LAW 8 For (5,) macrostate, = N! n!n! gn gn = 6! 5!! 5 5ways Subsequently, we can calculate for the remaining macrostates, using the example above. Problem : Given 5 energy levels, each separated by E. We consider the particles to be weakly interacting, i.e. they exchange energy but do not aect each other's energy levels. Let us look at the rst 3 levels : (i) a particle moves from the 3rd level down to the nd level and (ii) a particle moves from st level to the nd level. In this case, the change in energy is balanced by exchange of particles for the processes. Hence we can talk about a xed macrostate. Now we try to calculate Now we consider the schematic picture of this problem : E E E E Macrostate (5,,,,), = 6 C 5 C C C C =6ways Macrostate (4,,,,), = 6 C 4 C C C C =5ways At equilibrium, we apply the weighted average for microstates, because the number of microstates in either macrostate is not overwhelming enough. ~n = P n i (n ) P i (.5) Therefore, ~n = = 9

20 CHAPTER. DISTRIBUTION LAW 9 ~n = = 3 ~n 3 = = 6 ~n 4 = Note : ~n 5 = ~n =6 which is the same as P n for any macrostate.

21 Chapter 3 Indistinguishable Particles 3. Introduction Why are particles indistinguishable? This is a quantum mechanical eect. The exchange of quantum particles cannot be detected. The 's of the individual particles combine to form a total, and they are non-localised. We can write a many particle wavefunction as a linear combination of a () b ():::::: plus all the possible permutations. For example, for three particle permutations, we get six possibilities : a() b () c (3)::::: a() b (3) c ()::::: a() b (3) c ()::::: All particles are identical so all permutations are solutions to Schrodinger's equations, and so are all linear combination of the permutations. There are two physically meaningful solutions.. Symmetric : etc. s = p P [ (; ; ::::; N)] (3.) E.g. for 3 particles, we get s = a () b () c (3) + a () b () c (3) + a (3) b () c () + a () b (3) c () + a () b (3) c () + a (3) b () c (). Antisymmetric a = p (,) p P [ (; ; :::::; N)] (3.)

22 CHAPTER 3. INDISTINGUISHABLE PARTICLES where P is the permutation operator; (,) P two particle states. E.g., for 3 particles, we get means (,) every time we exchange s = a () b () c (3), a () b () c (3), a (3) b () c (), a () b (3) c () + a () b (3) c () + a (3) b () c () If we exchange and, then a (; ; 3) =, a (; ; 3) There are two kinds of quantum particles in nature:. Bosons : Symmetric wavefunctions with integral spins.. Fermions : Anti-symmetric wavefunctions with half-integral spins. Note when fermions are in the same state, total is zero. For example, a = b, meaning all permutations cancel in pairs, and hence we obtain Pauli's Exclusion Principle. 3. Bose-Einstein Distribution for Bosons There is no limit to the occupation number at each level of quantum state. In fact, the more the better. Note that = for all macrostates because of the indistinguishability of particles. (Macrostates Microstates) Unless g>> for some levels, for example, by the clumping of levels together to make g i large. We can then count microstates ( >> ) because the states in i are distinguishable. To put m i particles in g i states, imagine arranging m i particles into g i states, by arranging m i identical particles and g i, identical barriers on a straight line. etc. The number of ways = (m i + g i, )! m i!(g i, )! (3.3) Therefore, = (m i+g i,)! m i!(g i, )! (3.4) To maximise subect to the constraints : i i m i = N m i " i = E

23 CHAPTER 3. INDISTINGUISHABLE PARTICLES Therefore, ln = i i i [ln(m i + g i, )!, ln m i!, ln(g i, )!] [(m i + g i, ) ln(m i + g i, ), (m i + g i, ), m i ln m i + m i, (g i, ) ln(g i, ), (g i, ) [(m i + g i )ln(m i +g i ),m i ln m i, g i ln g i ] (3.5) for g i ;m i d(ln ) = i dm i ln(m i + g i )+dm i, dm i ln m i, dm i = (3.6) i (ln m i + g i m i )dm i = (3.7) Add in : So i, i i dm i = " i dm i = (ln( m i + g i m i )+," i )dm i = (3.8) Then, m i + g i m i = e, e " i (3.9) Therefore, m i = g i e, e " i, (3.) where = kt (to be proved later). Averaging over the g i levels in each clump, n = which is the Bose-Einstein Distribution. We will nd later that = kt e, e ", (3.) where is the chemical potential. Two systems at thermodynamic equilibrium have the same temperature T. The systems at chemical (particle number) equilibrium have the same.

24 CHAPTER 3. INDISTINGUISHABLE PARTICLES Fermi-Dirac Distribution for Fermions In this case, we allow or particle to each quantum state. Now we clump the energy levels to produce g i states and m i (<< g i ). etc... Either there is a particle or no particle in each quantum state. Therefore Number of ways to ll m i boxes out of g i = g i m i!(g i, m i )! (3.) = Y i g i m i!(g i, m i )! (3.3) To maximise subect to the constraints : Therefore, i i m i = N m i " i = E ln i g i ln g i, m i ln m i, (g i, m i )ln(g i,m i ) (3.4) where g i >> m i >> d ln = i = i =,dm i lnm i, dm i + dm i ln(g i, m i )+dm i ln( g i, m i )dm i m i (3.5) Now we add in : So i, i i dm i = " i dm i = (ln( g i, m i m i )+," i )dm i = (3.6)

25 CHAPTER 3. INDISTINGUISHABLE PARTICLES 4 Then, Therefore, g i, m i m i = e, e " i (3.7) m i = Averaging over the g i levels in each clump, n = which is the Fermi-Dirac Distribution. g i e, e " i + e, e " + (3.8) (3.9) Note : ) The positive sign in Fermi-Dirac distribution implies that n which is Pauli Exclusion principle (which states that no fermions can share the same quantum state). ) If P n = N is not a constraint, is zero and this is true for open systems. For isolated systems with xed N, has avalue. 3.4 Alternative Derivation for Bose-Einstein and Fermi- Dirac Statistics For equilibrium consideration, T i!kl is the transition rate from i! k accompanied by! l to maintain energy conservation. Suppose the rate depends on the occupation at k and l, Ri out = N i N T i!kl ( N k )( N l ) (3.) ;k;l Ri in = ;k;l N k N l T kl!i ( N i )( N ) =(N i ) ;k;l N k N l T kl!i ( N ) (3.) At equilibrium, N i N T i!kl ( N k )( N l )=(N i ) ;k;l ;k;l N k N l T kl!i ( N ) (3.) Applying time-reversal invariance, T i!kl = T kl!i (3.3)

26 CHAPTER 3. INDISTINGUISHABLE PARTICLES 5 Then for every combination i,, k, l, we need So N i N ( N k )( N l )=N k N l ( N i )( N ) (3.4) N i N i N N = N k N k N l N l (3.5) ln N i N i +ln N N =ln N k N l +ln (3.6) N k N l with which is the conservation of energy. Therefore then, ln " i + " = " k + " l (3.7) N i N i =, " i (3.8) N i N i = e e," i (3.9) N i N i e e," i = e e," i (3.3) N i = So starting with the positive (+) sign, we obtain e, e " i (3.3) N i = e, e " i + Fermi-Dirac Distribution (3.3) and with the negative (-) sign, we obtain N i = e, e " i, Bose-Einstein Distribution (3.33) For fermions, the rate Ri out / (, N k )(, N l ) which demostrates Pauli's Exclusion Principle, and for bosons, the rate Ri out / ( + N k )(, N l ), i.e. the more particles in k and l, the more likely the transition. (The more you have, the more you receive).

27 Chapter 4 Statistical Mechanics and Thermodynamic Laws 4. First Law The First law of Thermodynamics : du = dq + dw (4.) In Statistical Mechanics : Total Energy;E =U = n " (4.) Therefore du = " dn + n d" (4.3) Now, we have dw =,PdV or Hdm etc. This occurs when there is a change in the boundary conditions (V, M, etc). When the boundary conditions change, the energy levels change. For example : particle in a box Hence " = ~ m V 3 (n x + n y + n z ) (4.4) dw = n d" (4.5) So, e.g.,p (4.6) or H (4.7) 6

28 CHAPTER 4. STATISTICAL MECHANICS AND THERMODYNAMIC LAWS 7 dw Original dq Then dq= P " dn which takes place by rearrangement in the distribution. In kinetic theory, the particles may collide with the chamber walls and acquire more energy where the walls are hotter (with more heat added). 4. Second Law We now search for a microscopic explanation for S in terms of. We start from the TdS equation : so Also TdS = du + PdV,dN (4.8) S f(e;v;n) (4.9) g(e;v;n) (4.) Note : In adiabatic changes, for example, volume increase with no heat added (dq = ). ds =i.e. S is constant.also dn = sois also constant. During irreversible changes, the total S in the Universe increases. The total of the Universe also increases because the equilibrium macrostate must be the maximum. Now we consider systems with entropy S and S and disorder number and, Total entropy,s = S + S (4.) Total = (4.) Now S f() (4.3) so S = f( ) and S = f( ), Now f( )=f( )+f( ) (4.4)

29 CHAPTER 4. STATISTICAL MECHANICS AND THERMODYNAMIC LAWS 8 Then So ) @ @ (4.7) ) where k is a constant by separation of variables. = k (4.9) S = f( )=kln + C (4.) S = f( )=kln + C (4.) which implies that C =. Therefore S = k ln + k ln + C = S + S (4.) where k is the Boltzmann constant R N A. (To be proven later) S = k ln (4.3) Hence we have a microscopic meaning of entropy. The system is perfectly ordered when = and S = k ln =. When more microstates are availiable > so S >. The lesser the knowledge about a system, the more disordered it gets. From the formula which we obtained for the relation of entropy with, we are unable to deduce the value of k. The value of k comes out from the ideal gas equation.

30 CHAPTER 4. STATISTICAL MECHANICS AND THERMODYNAMIC LAWS Third Law As the temperature T tends to zero, the contribution to the entropy S from the system components also tend to zero. When all the particles take the ground state, (E ) =. When the macrostate has only one microstate, it implies that there are no further permutations for the Maxwell-Boltzmann distribution. S = k log (E )= Practically, some small interactions (e.g. nuclear paramagnetism) may lift the ground state degeneracy. If kt > " int, then S 6= at a given T. Hence we can only say S! S as T! where S is the entropy associated with the remaining small interactions for which " int <kt and independent of all the entropy for all the other interactions (degrees of freedom). The atoms occupying in the ground state are placed in fixed position in a lattice like this. Note that it is meaningless to talk about the exchange of particles in the same state because they do not exchange their positions. We know from the nd law of thermodynamics that heat passes from hot (high temperature) to cold (low temperature) if no work is done. The internal energy U(T) increases when T increases. Now we come to the formulation of the third law : The Third Law of Thermodynamics states that as the temperature T!, the entropy S!, it is approaching ground state as U(T) also approaches, ) that the disorder number tends to. From the third law, T= is a theoretically correct statement, but it is an impossibility. Why is this so? The answer : There exists very small interactions in the nucleus, made up of magnetic moments, quadrapole moments and octapole moments etc. In order to make T =, the magnetic moments are required to point in the same direction. However, experimentally, this is not achievable. The third law of thermodynamics can also be understood in a quantum mechanical

31 CHAPTER 4. STATISTICAL MECHANICS AND THERMODYNAMIC LAWS 3 perspective as the freezing out of interactions. By the Schrodinger equation :, + V ( ) = H The V( ) consists of many interaction terms. Hence we can approach the total entropy, S as a combination of entropies of many sub-systems : S = S total + S total + S total Calculation of Other Thermodynamic Variables From chapter, the Partition function Z is dened as : Z = g e," kt (4.4) where represents all levels. Now we try to nd the relation between P and T, when systems are in thermal equilibrium, where will be the same. The depends on the nature of the particles and volume V etc. For the system of same identical particles, free exchange of particles will allow to reach the same value. Now we proceed to calculate the other thermodynamic g e," V;N = where = kt which is proven in tutorial. We also dene the internal energy, U (or E) U = n " = N Z =, N Z g V;N (4.6) Since, d dt =, kt

32 CHAPTER 4. STATISTICAL MECHANICS AND THERMODYNAMIC LAWS 3 and therefore, We know Z) Z U ln V;N = ln V;N (4.7) n = N Z g e," Then we work out ln, ) g n = Z N e" ln = N ln N!+ (n ln g, ln n!) = N ln N, N + =Nln N + ( (n ln( g n )+n ) n " )+Nln Z, N ln N (4.8) = U + N ln Z Therefore S = k ln = ku + Nkln Z (4.9) = U T + Nkln Z The Helmholtz free energy, F is dened as : F = U, TS (4.3) Therefore F =,NkT ln Z (4.3) Since df = du, TdS,SdT = TdS,PdV,TdS,SdT =,SdT, PdV (4.3)

33 CHAPTER 4. STATISTICAL MECHANICS AND THERMODYNAMIC LAWS 3 From the chain rule Therefore, we can calculate : Pressure : ) ) V dt Entropy : ) T = ln ) T (4.33) ) V = Nkln Z + NkT Z = Nkln Z + (4.34) Then we can use the results above to compute the heat capacity at constant volume : C ) V (4.35)

34 Chapter 5 Applications of Maxwell-Boltzmann Statistics 5. Classical Perfect Gas To apply the Maxwell-Boltzmann Statistics to the classical perfect gas, we require a few conditions :. The gas is monatomic, hence we consider only the translational kinetic energy (KE).. There is no internal structure for the particles, i.e. no excited electronic states and the paricles are treated as single particles. 3. There is no interaction with external elds, e.g. gravity, magnetic elds. 4. The energy is quantized as in the case of particle in a box (in QM). 5. There is a weak mutual interaction - only for exchanging energy between particles and establishing thermal equilibrium, but there is no interference on each other's ". 6. The discrete levels can be replaced by a continuous one if the spacing is small as compared to T, i.e. " kt which is the classical limit. 7. The atoms are pointlike - occupying no volume. From quantum mechanics, " = ~ ml (n x + n y + n z ) for V = L 3 and where n x, n y and n z are integers. 33

35 CHAPTER 5. APPLICATIONS OF MAWELL-BOLTZMANN STATISTICS 34 Now we compute the partition function, Z and in this case g states are summed over. Z = = nx = nx e, " kt ny nz e,~ mkt L n x e, ~ mkt L (n x +n y +n z ) e,~ mkt L n y e,~ mkt L n z =, because all the (5.) = Z To evaluate the integral : ny dn x e,~ mkt L n x Z nz dn y e,~ mkt L n y Let x =(,~ ml kt ) nx Z dn z e,~ mkt L n z where Therefore Z Z e ~ n x ml kt dn x =( ml kt ) ~, Z e,x dx =( =( =( = L( mkt h ), Z Z, Z =( ) = p Z, Z, Z e,x dx,z rdr Z = L 3 ( mkt h ) 3, e,x dx e,x dx e,y dy) e,(x +y ) dxdy) = V ( mkt h ) 3 de,r ) (5.) (5.3) (5.4) So, ln Z = lnv + 3 ln T + 3 ln mk h (5.5) P = ln ) T = NkT V (5.6)

36 CHAPTER 5. APPLICATIONS OF MAWELL-BOLTZMANN STATISTICS 35 since i.e. PV = NkT PV = nrt Nk = nr k = n N R k = R N A (5.7) because N n = N A molecules per mole U = NkT ln = NkT 3 T = 3 NkT ) V (5.8) For each atom, U = 3 kt Also the entropy is given by : S = U T + Nkln Z Therefore, the distribution becomes : = 3 Nk + 3 Nkln(mk h )+Nk(ln V + 3 ln T ) n = N Z e," kt = Nh 3 V(mkT) 3 e," kt (5.9) (5.) where represent all n x, n y and n z states. The number of particles having energies between " and " +": n= Nh 3 V (mkt) 3 e," dn kt " (5.) d" where dn d" is the density of states, i.e degeneracy dn d" ". To calculate the density of states, we consider the quantum numbers n x, n y making up the grid points with each grid point representing one state. and n z

37 CHAPTER 5. APPLICATIONS OF MAWELL-BOLTZMANN STATISTICS 36 By writing R = n x + n y + n z The number of points (states) with energy up to " R lie within a sphere of volume 4 3 R3. Only positive n's are allowed so we take the positive octant. The number of states lying between energy " and " +"will become : Using we obtain, Therefore, Therefore, dn d" " = 8 (4R )R = 8 (4R ) dr d" " " = ~ ml R d" dr = ~ ml R dn d" " = RmL ~ " = (m) 3 L 3 " 4 ~ 3 3 " =V (m 3 ) " h 3 " n = VNh3 e, kt " v (m) 3 " (mkt) 3 h 3 " e, " = N" kt p " 4 (kt)3 (5.) (5.3) (5.4) For the momentum distribution, we use : and " = p m d" dp = p m

38 CHAPTER 5. APPLICATIONS OF MAWELL-BOLTZMANN STATISTICS 37 So, the number of particles have momentum between p and p+p is : Note that ZZZ all p space n = N p 4 (kt)3 p p e, m p d" mkt dp p N = 4e, p (mkt) 3 mkt p p dp x dp y dp z = = Z Z Therefore we can also express in p x, p y and p z as Z Z d p sin p d p p dp 4p dp (5.5) (5.6) n = = = N (mkt) 3 N (mkt) 3 N (mkt) 3 e, p mkt px p y p z x e +p y +p z,p mkt p x p y p z p x p y e, mkt px e, mkt py e, p z mkt pz (5.7) For velocity distribution, we use p = mv, n = f(v x ;v y ;v z )dv x dv y dv z = Nm 3 (mkt) 3 = g(v)dv = Nm3 4 (mkt) 3 x e,m(v +v y +v z ) kt v x v y v z e,m(v ) kt v dv (5.8) which are the Maxwell distribution laws for velocities and speeds. We invoke a few integral identities : Z I n = e,x x n dx p I = I = I = I I 3 = I I 4 = 3 I

39 CHAPTER 5. APPLICATIONS OF MAWELL-BOLTZMANN STATISTICS 38 So, Z g(v)dv = Nm3 4 (mkt) 3 = N ( kt m ) 3 Z e,x x dx (5.9) The mean speed, v v = N Z vg(v)dv Z = m3 4 ( kt (mkt) 3 m ) e,x x 3 dx r 8kT = m (5.) and mean square speed is v = N Z = m3 4 (mkt) 5 = 3kT m v g(v)dv Z e,x x 4 dx (5.) Average energy, m v = 3 kt (5.) Root mean square (RMS) speed, p v = r 3kT m > v (5.3) The most probable speed : dg(v) dv =)v p e,mv kt, mv3 p kt e,mv r kt )v p = m kt = (5.4)

40 CHAPTER 5. APPLICATIONS OF MAWELL-BOLTZMANN STATISTICS 39 Maxwell Distribution of Molecular Speeds V Vp V V T 5. Harmonic Oscillator v p < v <p v We now consider the simple -D harmonic oscillator in equilibrium at T, with the Hamiltonian, H = p m + k x where =,,,... " =(+ )~!! = r k m Assume the oscillators are independent and distinguishable, n = N Z e," kt Z = = = e, ~! kt = e, ~! kt = e, ~! e, (+ )~! kt = e, ~! kt, e, kt ~!, e,~! (5.5)

41 CHAPTER 5. APPLICATIONS OF MAWELL-BOLTZMANN STATISTICS 4 Note : for a series of a, ar, ar,... n= ar n = a, r ln Z =, ~!, ln(, e,~! ) (5.6) = N~! + N ~!e,~!, e,~! = N~![ + e ~!, ] = N~![ + e ~! kt, ] (5.7) At low T,U N~!, it means that the oscillators are in ground state. At high T, U = N~![ + + :::, ] + ~! kt = N~![ + kt ~! ] = NkT (5.8) which is the classical result. The high energy states are occupied. Quantization is not important for kt >> "(= ~!) 5.3 Equipartition Theorem The Equipartition theorem states that all terms in the Hamiltonian with squared coordinates (x or p x etc) will contribute kt to the average energy, provided quantization is not important. This is reected in the classical regime at high T (where kt >> " int. Example : Ideal Gas, H = kinetic energy only Therefore = p x m + p y m + p z m U = 3 kt N atoms = 3 NkT

42 CHAPTER 5. APPLICATIONS OF MAWELL-BOLTZMANN STATISTICS 4 Example : -D Harmonic Oscillator, Therefore H = Kinetic energy + Potential energy = p x m + k x U = kt N atoms = NkT With quantization setting in at low T, the Hamiltonian is frozen out and the contribution is not kt. Example 3 : Diatomic gas, For translational motion, For rotational motion, H trans = p x m + p y m + p z m neglecting rotation about z-axis. For vibrational motion, H rot = L x + L y I x I y H vib = p z m + k z where z' is the relative coordinate between the atoms. Therefore, the total internal energy, U should be 7 NkT. However, the experimental results show that U = 5 NkT. The reason is because the vibrational motion is frozen out, i.e. ~! vib >> kt. So in terms of H vib, the atoms are in ground state, i.e. atoms don't appear to vibrate much at normal temperature. Total H = H vib + H trans + H rot (5.9)

43 Chapter 6 Paramagnetic Systems 6. Introduction For a paramagnetic spin system, the atoms have unpaired electrons or partially lled orbitals. The net angular momentum ~ J for each atom is equal to both the linear orbital angular momentum, ~ L and spin, ~ S. It also acts like the magnetic moment. When the system interacts with an external magnetic eld, J =,g J B J (6.) Hamiltonian, H =,~ ~ B = g J B ~ B ~ J = g J B B z J z (Quantization) (6.) The J z and J are observed from atomic physics. B = e~ mc (Bohr Magneton) In a solid, the outer electrons are involved in valency for chemical bonds, which are ultimately paired up so that the net J of outer electrons is zero. The inner core, particularly 3d, 4d and 4f orbitals of many elements are not involved in valency, so magnetism arises from their S, L and J. In many solids of transition elements (3d), the orbital quenching occurs, so that the net L = for atoms in the ground electronic state. Note that the excited electronic states are unoccupied at room temperature because " >>kt. Hence in the ground electronic state of 3d elements, there is a degeneracy of (S +) states. Note that J = S. 4

44 CHAPTER 6. PARAMAGNETIC SYSTEMS 43 Applying B-eld to a spin 3 system : For example, S = 3/ system B Hamiltonian, H = g s B BS z (6.3) and g s for electron spin. The distribution in the spin states is associated with the spin temperature T S which may not be the same as the lattice temperature T L. A mechanism must exist for T S and T L to establish thermal equilibrium - spin-phonon interaction. This is important inspin resonance experiments such as electron spin resonance and nuclear magnetic resonance, and determines the relaxation time. 6. Spin Quantum System We consider the system of N magnetic atoms of spin due to unpaired electron. The magnetic moment, =,g s B S z = g (6.4) s B The Zeeman interaction with external eld gives energy levels g s B B z S z = g s B B (6.5) where g s and B is the Bohr magneton. The energies correspond to spin # ( antiparallel to B) or spin " ( parallel to B).

45 CHAPTER 6. PARAMAGNETIC SYSTEMS 44 So Using the Maxwell-Boltzmann distribution, " " =, g s B B (6.6) " # =+ g s B B (6.7) n " = N Z e+ gs B B kt (6.8) n # = N Z e, gs B B kt (6.9) where Therefore, the total energy, E : Z = e + gs B B kt + e, gs B B kt = cosh( g s B B kt ) (6.) The entropy S: E = N Z (, g s B Be +gs B B kt =, Ng s B B (sinh( g s B B Z kt )) =, Ng s B B (tanh( g s B B Z kt )) + g s B Be,gs B B kt ) (6.) The magnetization M : where " = B S= U T +Nkln Z =, Ng s B B T M = V = N VZ (tanh( g s B B kt )) + Nkln( cosh(g s B B kt )) n e + B kt = N VZ [ g s B e gsbb kt, g s B e, gs B B kt = N VZ g s B sinh( g s B B kt ) = N V g s B tanh( g s B B kt ) =, E BV (6.) (6.3)

46 CHAPTER 6. PARAMAGNETIC SYSTEMS Spin J Quantum System Now we consider a system with total angular momentum J in an external B-eld. +J +(J-) -(J-) -J B= B Therefore there are (J + )states. m J =,J;,(J, ); ::::::; ; ; ::::; (J, );J where g J is the Lande g-factor (an experimental quantity). Let " J = g J B Bm J (6.4) y = g J B B Therefore, the partition function, Z = = J m J =,J J m J =,J e g J B B e ym J = e,yj J m J = e ym J =(e,yj ) e(j+)y, e y, = e(j+)y, e,jy e y, )y, e,(j+ )y = e(j+ e y, e, y = sinh(j + )y sinh y (G.P sum = a(rn, ) r, ) (6.5)

47 CHAPTER 6. PARAMAGNETIC SYSTEMS 46 Therefore, n J = N Z e," J = N Z e,g J B Bm J (6.6) Then Magnetization, M = N g J B m J e g J B Bm J VZ = VZ@B @B = N VZ J = N V ln Z J (6.7) ln Z =ln(sinh(j + )y), ln(sinh y ) (6.8) M = N V g J B ( (J + ) cosh(j + )y sinh(j + )y, cosh y ) sinh y = N V g J B ((J + ) coth(j + )y, coth y ) (6.9) = N V g J B JB J (y) where B J (y) = J [(J + ) coth(j + )y, coth y ] Brillouin function (6.) We proceed to evaluate other thermodynamic quantities : Helmholtz free energy, F =,NkT ln Z =,NkT[ln(sinh(J + )y), ln(sinh y )] (6.) Internal energy, U ln @ =,Ng J B BJB J (y) (6.)

48 CHAPTER 6. PARAMAGNETIC SYSTEMS 47 Entropy, S = U T + NK ln Z =, Ng J B BJB J (y) T At high T, y<<, and for small x, + Nk[ln(sinh(J + )y), ln(sinh y )] (6.3) coth x = ex + e,x e x, e,x = ex + e x, =(+x+ 4x 4x +:::: + )( + x + + 8x3 + :::::, ), 3! = x (+x+x +::::)( + x + x 3 + :::) = x ( + x + x + :::)(, (x + x 3 + :::)+(x+x 3 :::::) ::::) = x ( + x, x + x, x, x 3 +::::) = x ( + x 3 + :::::) (6.4) = x + x 3 So, B J (y) = J [(J + )( (J + )y + (J + )y ), 3 ( y = J [ y + (J + ) y, 3 y, y ] = 3J [(J + J)y] = y (J +) 3 + y 3 )] (6.5) Therefore, so, M = N V g J B J(J +) y 3 = N g J B J(J+)B V 3kT ) T = N V / T g J B J(J +) 3kT (Curie's law) (6.6) (6.7)

49 CHAPTER 6. PARAMAGNETIC SYSTEMS 48 Since, U =, Ng J B BJ(J +)y 3! Since, ln Z ln(j +) (6.8) Proof : ln Z =ln e(j+ )y,e,(j+ )y =ln[e,(j+ )ye,(j+ )y,, ln[e,y e y, ] =,(J + )y + ln(+(j+ )y + :::, )+ y,ln(+y+::::, =,Jy + ln[(j + )y + ::::], ln[y + ::] (6.9) =,Jy + ln(j +) ln(j +) or ln Z = ln[ (+(J+ )y+::::), (, (J + + :::) = ln(j + )y, ln y ], ln[ ( + y + :::), (, y + :::) ] = ln(j +) (6.3) Then S = Nkln(J +) (6.3) for (J + ) levels. So, At low T,y>>, and for large x, Therefore coth x = ex + e,x = e x, e,x B J (y) = J [J+, ]= (6.3) M = N V g J B J (6.33) where all spins are align in the same direction (parallel to B), hence!.

50 CHAPTER 6. PARAMAGNETIC SYSTEMS 49 ln Z =ln( e(j+ )y,e,(j+ )y =(J+ )y+ln,y,ln ),ln( e y, e,y ) (6.34) = Jy Internal energy, U =,Ng J B BJ (6.35) Entropy, S =,Ng J B BJ T 6.4 Adiabatic Cooling + NkJy = (6.36) From the third law, S! as T!, so cooling reduces the entropy of system, so we need magnetic dipoles to align in one direction. But alignment can also be produced with a B-eld, so we need the alignment to stay as we reduce the B-eld to zero. So adiabatic cooling is two-staged.. Reduce the entropy by turning on the B-eld isothermally. The paramagnetic system is in contact with the lattice and the heat bath. As B-eld increases, dipoles align more and energy and entropy is taken out from the system. Ts Spin System T L Lattice Heat Bath at constant T Heat flow Entropy passed to lattice and heat bath = dq Ts. Reduce the temperature by turning o the B-eld adiabatically, i.e. isolated from heat bath.

51 CHAPTER 6. PARAMAGNETIC SYSTEMS 5 Ts Spin System T L Lattice Remove the B-field adiabatically to keep spins align. As B decreases, S spin stays constant, so T S decreases. The lattice passes heat and entropy back to spin. But this entropy is so small because of low T L, S Lat is small, so the heat passing is very small. (dq = TdS). Hence both T S and T L will drop drastically. S Entropy Curves obeying the third law P B = () R () Q B = B T Note : They obey the 3rd law Stage () : From P to Q, Isothemal Magnetization Stage () : From Q to R, Adiabatic Demagnetization

52 Chapter 7 Applications of Fermi-Dirac Statistics 7. Fermi-Dirac Distribution From chapter 3, we obtain two quantum distributions :. Fermi-Dirac Distribution n = g e, e " + (7.). Bose-Einstein Distribution n = g e, e ", (7.) For either systems, =. This can be proven by having the quantum system in kt contact with a classical perfect gas where =. The two s will reach equilibrium at kt same temperature. Now to identify for Fermi-Dirac distribution, we consider fermion systems in thermal and diusive contact so to allow particle exchange. We also assume that the same type of fermions for both systems. With the same constraints : n i + n = N n i " i + n " = E = i Y g i = n i!(g i, n i )! i Y g n!(g, n )! (7.3) 5

53 CHAPTER 7. APPLICATIONS OF FERMI-DIRAC STATISTICS 5 ln = i g i ln g i, g i, (g i, n i )ln(g i,n i )+(g i,n i ),n i ln n i + n i + g ln g, g, (g, n )ln(g,n )+(g,n ),n ln n + n = i g i ln g i, (g i, n i )ln(g i,n i ),n i ln n i + g ln g, (g, n )ln(g,n ),n ln n (7.4) dln = i ln(g i, n i )dn i + dn i, ln n i dn i, dn i + ln(g, n )dn + dn, ln n dn, dn = i (ln g i, n i )dn i + n i (ln g, n )dn n = Using the method of Lagrange multipliers, (7.5) ( i dn i + dn )=,( i " i dn i + " dn )= So, i (ln g i, n i n i Then Choose and, sothat, + " i )dn i + (ln g, n n, + " )dn = (7.6) Then ln g i, n i n i, + " i = ln g, n n, + " = g i n i = e, e " i + g n = e, e " + So the two systems in thermal contact have the same and the two systems in diusive contact will have the same, which is related to. The diusive contact gives rise to chemical equilibrium, so i = = (chemical potential) Suppose we take one of the systems : S = k ln = g ln g, (g, n )ln(g,n ),n ln n (7.7)

54 CHAPTER 7. APPLICATIONS OF FERMI-DIRAC STATISTICS 53 Then ds = k = k = k =,k ln(g, n )dn + dn, n ln n, dn ln g, n dn n (, + " )dn dn + k =,kdn + kdu, kdw [d(n " ), n d" ] (7.8) because TdS =,ktdn + du, dw =,dn + du, dw du = TdS +dw + dn (7.9) Therefore So = kt (same derivation for bosons) (7.) n = g e ", kt + (7.) where dn is the work done in moving the fermions from one place to another, e.g electrical work. 7. Free Electron Model of Metals The valence electrons (in the outermost shell) are loosely bound, i.e. takes little energy to detach them, hence they have low ionisation potential. In metal, the weakly attached conducting electrons are \free" to move in the lattice. The electrons interact with the electrostatic potential of the positive ions in the lattice. If we consider the periodicity of the potential, we will obtain the solution to the Schrodinger equation in the form of Bloch waves. The energy will form a band. For perfect stationary lattice, there is free propagation and innite conductivity of electrons. However lattice imperfections, lattice vibrations (phonons) and the scattering by the other electrons leads to resistance in the lattice. In the free electron model, we apply Fermi-Dirac Statistics - electrons of spin and =. We neglect the lattice potential, the electrons are conned in a box (body of

55 CHAPTER 7. APPLICATIONS OF FERMI-DIRAC STATISTICS 54 metal) and we neglect the mutual interactions(electron-electron scattering). For single particle state (n x, n y, n z, ), the energy is given by : " n x;ny;nz; = ~ ml (n x + n y + n z ) = ~ m k (7.) where k is the wavevector and is dened by the following relation : k = L (n x + n y + n z ) (7.3) n z n R n y n x Number of states from k= to k=k = n3 R (7.4) (Note : we multiply the relation by because of the spin states and 8 only take the positive octant, i.e. n x, n y and n z are all positive integers.) because we Number of states between k and k+dk = 8 4k dk ( L )3 = V k dk (7.5) Number of states between " and "+d" is g(")d" = V m" ~ dk d" d" = V m" m ~ ~ r k d" = V m " ~ ~ 4 m" d" 3 = V (m) " ~ 3 d" = VC" d" (7.6)

56 CHAPTER 7. APPLICATIONS OF FERMI-DIRAC STATISTICS 55 where Therefore, C = (m) ~ 3 3 N = n = g e (", + (7.7) = Z g(")d" e (",) + E = n " = " g e (",) + (7.8) = Z "g(")d" e (",) + We solve equation (7.7) for (N; T; V ) and use in (7.8) to obtain ". This is dicult to evaluate except at low T.Weshall try to evaluate this in cases. At T = Probability of occupation of state of energy is f F,D (") = e (",) + (7.9) f F-D (E) E F E For (T =)=" F Fermi energy (7.) and f F,D (") = f F,D (")= ">" F "<" F

57 CHAPTER 7. APPLICATIONS OF FERMI-DIRAC STATISTICS 56 Hence and N = = Z Z "F = VC g(")f F,D (")d" g(")d" Z "F = VC 3 "3 F " F =[ 3N V " d" ~ 3 (m) 3 =( 3N V ) 3 = ~ m k F ] 3 ~ m (7.) (7.) k F =( 3N V ) 3 (7.3) The meaning of Fermi energy is that all states below " F are occupied and all states above " F are unoccupied at T =. At T=, all electrons occupy the lowest possible state subect to the Exclusion Principle, so the states are lled until " F. Total Energy, E = Z g(")f F,D (")d" Z "F = VC " 3 d" (7.4) = VC 5 "5 = 3 5 N" F. At low T, kt " F <<, we want to nd C v and we know that the experimental value is proportional to T. Therefore, U = = = Z Z Z N = Z g(")f F,D (")d" "g(")d" e (",) + ("," F )g(")d" e (",) + ("," F )g(")d" e (",) + + " Fg(")d" e (",) + +N" F (7.5)

58 CHAPTER 7. APPLICATIONS OF FERMI-DIRAC STATISTICS 57 Then C ) V =, Z (", " F )g(")e (",) (e (",) +) [(", )(, kt, kt Since does not change much, so =and = " F, Now, C v = )]d" (7.6) (", " F ) e ("," F ) VC" d" (e ("," F ) +) (7.7) e ("," F) (e ("," F) +) =,@f (7.8) The maximum occurs at " = " F. The value drops to of the maximum value when e ", " F kt. So the function is sharp for small kt (<< " F ). df F-D (E) de kt E F E The contribution to the C v " " F integral comes mainly from values of " near " F, i.e. C v = kt VC" F =VC" F k T Z Z ("," F )e ("," F)d" d" (e ("," F ) +) xex dx where x = (", " " F (e x +) F ) The lower limit can be extended to, because kt<< " F. Therefore Z, x e x dx (e x +) = 3 (7.9) (7.3) C v = 3 k Tg(" F ) (7.3) where g(" F ) is the density of states of Fermi level. The physical picture :

59 CHAPTER 7. APPLICATIONS OF FERMI-DIRAC STATISTICS 58 g(e)f F-D (E) T = T> E F E For T =: All states are occupied within " = " F is a truncated parabola. For T> : We will expect electrons to gain energy kt. However, most electrons cannot gain kt because the states above them are already occupied. Only those electrons near " F can gain kt to move into unoccupied states. This results in a rounded-o cli at " F. An alternative estimate : Number of electrons involved is g(" F )kt and they gain kt. Therefore Energy of excitation = g(" F )k T (7.3) The number of electrons participating energy Contribution to C v, heat capacity = T = g(" F )kt kt T = f(" F )k T (7.33) Experimental result : C v / T for a fermi gas. The theory is in agreement but does not give the right constant of proportionality, and need to include eective mass of electron m e >m ebecause electrons are not completely free but the lattice potential constrains them.

60 Chapter 8 Applications of Bose-Einstein Statistics 8. Introduction to Black Body Radiation First, we discuss radiation in thermal equilibrium. The radiation is enclosed in vessel, and all the walls are at temperature T. The atoms in the wall will absorb and emit electromagnetic radiation. At equilibrium, the distribution of photons will follow the temperature T. Second, in order to study the radiation, we assume a hole is cut in the chamber wall to emit some radiation. The hole is very small so that the equilibrium is not disturbed. This will approximate to a blackbody. The black body will also absorb all radiation falling on it. The radiation bounces inside the chamber until nally being absorbed. In this chapter, we will study black body radiation using statistical mechanics. We note that the photons have the following characteristics :. The photons have spin (bosons) and they obey Bose-Einstein statistics.. The photons do not interact with one another. 3. The photons have zero rest mass and travel at c, the speed of light. 4. The photons satisfy the dispersion relation,! = ck, so, " = ~! = ~ck 59

61 CHAPTER 8. APPLICATIONS OF BOSE-EINSTEIN STATISTICS 6 8. Bose-Einstein Distribution Small Blackbody Perfect Insulating Walls We consider a system of a small blackbody placed inside a chamber with perfect insulating walls. The features of the system are as follows :. The blackbody absorbs and emits radiation until thermal equilibrium is established.. The number of photons are not conserved because of absorption and re-emission. This implies that and =. 3. The blackbody is so small that heat capacity is negligible. The energy content of photons are not changed. E = n " = constant (for the photons) For bosons, = Y (n +g,)! n!(g, )! (8.) Maximise subect to E = constant only. ln = [ln(n + g, )!, ln n!, ln(g, )!] = = (n + g, ) ln(n + g, ), (n + g, ), n ln n + n, (g, ) ln(g, )+(g,) (n + g, ) ln(n + g, ), (g, ) ln(g, ), n ln n d(ln) = (8.) ln(n + g, )dn + dn, n ln n, dn = (8.3) ln( n + g, n )dn =

62 CHAPTER 8. APPLICATIONS OF BOSE-EINSTEIN STATISTICS 6 Therefore So, for large g and n. dn " = [ln( n + g, n ), " ]dn = (8.4) n = g e ", (8.5) Bose-Einstein function = e ", (8.6) Prove = byhaving the photon system in thermal contact with another system at kt temperature T and use walls that can exchange energy. 8.3 Planck's Radiation Formula Consider photons enclosed in a box L 3, standing electromagnetic waves result such that n = L Therefore k x = x k y = y k z = z = n x L = n y L = n z L For any photon with ~ k = (k x, k y, k z ), For any ~ k, there are polarizations - linear or circular.! c = k = k x + k y + k z (8.7) n z n R n y n x

63 CHAPTER 8. APPLICATIONS OF BOSE-EINSTEIN STATISTICS 6 Therefore, since Number of photon states for k= to k=k = n3 R n R = L k = 3 L 3 3 k3 (8.8) Number of photon states between k and k+dk = V k dk (8.9) Number of photon states between frequency range! and!+d! = V! d! c c = V! d! c 3 (8.) The number of photons in frequency range! and!+d! is N(!)d! = (Density of states) (Bose-Einstein function) = V! d! c 3 (e ~!, ) (8.) Thus the energy of the photons in the frequency range! and!+d! is ~!N(!)d! = V ~ c 3! 3 d! e ~!, (8.) The energy density (per unit volume) in the frequency range is u(!; T)d! = ~ c 3! 3 d! e ~!, (Planck's formula) (8.3) u(w, T) T T>T T w For T >T,the maximum of u(!, T)is given by : e ~! 3!,! ~e ~! (e ~!, ) =

64 CHAPTER 8. APPLICATIONS OF BOSE-EINSTEIN STATISTICS 63 ) 3(e ~!, ) = ~e ~! ) 3(e x, ) = xe x where x = ~! Solving numerically, x3, by neglecting (-), or more accurately, x =.8. Therefore, ~! max kt =:8 (8.4) or! max = :8k T (Wien's displacementlaw) (8.5) ~ c = :8k T max ~ (8.6) max = ~c :8k T (8.7) So, as T rises, the colour of the blackbody changes from red to blue : Therefore where Total Energy Intensity, u(t) = Z u(!; T)d! Z = ~! 3 d! c 3 e ~!, Z = ~ x 3 dx c 3 4 ~ 4 e x, = k4 T 4 4 c 3 ~ 3 5 (8.8) u(t )=at 4 (Stefan-Boltzmann law) (8.9) a = k 4 5c 3 ~ Radiation emitted by blackbody Consider a tiny hole, A, with photons coming from direction (! + d,! + d), with a subtended solid angle d, where rsin drd d = r = sin dd

65 CHAPTER 8. APPLICATIONS OF BOSE-EINSTEIN STATISTICS 64 The total solid angle = 4. The photons that will pass through in time t are located in a parallelpiped of length ct. The volume of the parallelpied is da ct cos. θ θ θ c t da Therefore, No. of photons passing per unit time from direction (! + d,! + d) with a frequency range! to! + d! Volume of Parallelpiped = Fractional solid angle Number density in frequency interval t = dac cos d 4 N(!)d! V Therefore, the total energy radiated per unit time in frequency interval! to! + d! Therefore = Z d Z = 4 c~!dan(!)d! V = cu(!; T)dAd! 4 dac cos sin d ~! N(!)d! 4 V Power radiated per unit area in frequency interval from! to! + d! = cu(!; T)d! 4

66 CHAPTER 8. APPLICATIONS OF BOSE-EINSTEIN STATISTICS 65 where a= k 4 5c 3 ~ 3. Therefore, Z Total power radiated = 4 c u(!; T)d! = 4 cat 4 = T 4 = k 4 6c ~ 3 =5:67,8 W m, K,4

67 Chapter 9 The Classical Limit 9. The Classical Regime In the classical regime, there is a lack of discreteness, we consider the continuous variable " rather than the discrete energy " r. The energy spacing is " <<kt, and <n r ><< which means that there are many unoccupied states, which are averaged over time. i.e. In the Fermi-Dirac or Bose Einstein statistics, the classical regime is only satised if so that <n r > F-D,B-E e,("r,) e ("r,) << e ("r,) >> = e e,"r M-B distribution So all the 3 distributions(maxwell-boltzmann, Fermi-Dirac and Bose-Einstein) distributions approximate to Maxwell-Boltzmann distributions in the classical regime. According to the Maxwell-Boltzmann statistics, for all r, including r =. <n r > M-B = N Z e,"r << (9.) So we require N Z <<, since "! for ground state. 66

68 CHAPTER 9. THE CLASSICAL LIMIT Monatomic Gas For monatomic gas, so, Z = V 3 mkt h 3 Nh 3 V mkt 3 << So, N T >> V 3 h mk The conditions are high temperature, T and low density N, and high mass, m. V Then and i.e. Assume the energy is equipartitioned, " = 3 kt (for each atom) = p m p = p m" = mkt h db (in QM terms) (9.) N 3 V db << 3 3 = m" 3 3 = p = h db 3 3 db << V 3 = N 3 where l is the average seperation between atoms. Examples : 3 (9.3) l

69 CHAPTER 9. THE CLASSICAL LIMIT 68. He gas : so l =,7 cm At T = 73 K, and N V = per cm 3 m = g so db =8,9 cm << l. Hence Maxwell-Boltzmann distribution is applicable. For liquid He at T =4K, N V = per cm 3 so l =4,8 cm. Since db >l,so Bose Einstein statistics apply.. For conduction electrons : Assume electron per atom, hence N V 3 per cm 3 l,8 cm For m = 9:,3 kg and T = 73 K, db > l, so Fermi-Dirac statistics apply. g(e)f F-D (E) T = Approximately M-B T> E F E f F-D (E) T= T> E F E

70 CHAPTER 9. THE CLASSICAL LIMIT 69 Note that for M-B distributions. n(") / " e," 9.3 Gibbs Paradox For classical ideal gas, we assume the atoms to be distinguishable. Now we mix the volumes of identical atoms. V N T S V N T S S = U T + Nkln Z = 3 Nk + Nk ln V + 3 ln T + 3 ln mk h (9.4) Suppose now N! N, V! V, we expect S!S. However, S = 3 (N)k +(N)k =S old +Nkln ln V + 3 ln T + 3 mk ln h (9.5) The problem lies with distinguishability and overcounting. Hence we use a corrected. Corrected = old N! Y g n = n! (9.6) Therefore, Corrected S = k ln Corrected, k(n ln N, N) = 5 V Nk + Nk ln + 3 N ln T + 3 mk ln h (9.7)

71 CHAPTER 9. THE CLASSICAL LIMIT 7 Then for N! N, V! V, S = 5 V (N)k +Nk ln N + 3 ln T + 3 mk ln h (9.8) =S old

72 Chapter Kinetic Theory of Gases. Introduction Consider the distribution (for e.g. velocity, momentum, position, energy, etc) of system of gas molecules in thermal equilibrium. At thermal equilibrium, the momentum and spatial coordinates are independent of each other, i.e. a molecule in any position (x,y,z) can have any momentum (p x, p y, p z ). For the Hamiltonian :. The kinetic energy (KE) terms are determined by p x, p y and p z.. The potential energy (PE) terms are determined by x, y and z. The Boltzmann law applies to every coordinate in the Hamiltonian. For example,. The probability that x-momentum which lies between p x and p x +dp x is : P [p x ]dp x = Ae,p x mkt dp x (.) where Z A e,p x mkt dp x = (.),. The probability that x-velocity lies between v x and v x +dv x is : P [v x ]dv x = A e,mv x kt dv x (.3) where A Z, e,mv x kt dvx = (.4) 7

73 CHAPTER. KINETIC THEORY OF GASES 7 3. The probability that the molecule is found between and x and x+dx is :,V (x) P [x]dx = Be kt dx (.5) where B Z,,V (x) e kt dx = (.6) We concentrate on the momentum and velocity distributions, hence we do not require the interaction potential except for establishing thermal equilibrium. For graphs of P[p x ] or P[v x ]: the gaussian curves centred at zero. P[p ], P[v x x ] p x, v x For speed distribution : P[c] dc where c = v x + v y + v z. dφ dc dθ P [v x ;v y ;v z ]dv x dv y dv z = P [v x ]dv x P [v y ]dv y P [v z ]dv z 3 m = e,mv x kt kt m = kt Integrating over all angles : e,mv y kt e,mv z kt dv x dv y dv z 3 e,mc kt dc(cd)(csind) (.7) Hence, we can calculate : m P [c]dc =4 kt 3 e,mc kt c dc (.8)

74 CHAPTER. KINETIC THEORY OF GASES 73. Most probable speed, c max : For c max, put dp [c] dc kt =)c max = m (.9). Mean speed c : c = Z 8kT cp [c]dc = m (.) 3. Mean square speed, c : c = Z c P [c]dc = 3kT m (.) 4. Root mean square speed, p c : p c = 3kT m (.) 5. Mean kinetic energy (KE) : Mean KE = m c = 3 kt (.3) Note : so c = v x + v y + v z =3 v x v x = kt m (.4) (.5). Number of impacts on Area A of a wall θ A

75 CHAPTER. KINETIC THEORY OF GASES 74 The distribution of speeds : P[c] dc The fraction of molecules travelling in direction (! + d,! + d) is cdcsind 4c = sindd 4 The fraction that will reach A in time t for above direction where Visthe volume of the gas. Therefore, Number of impacts on area A in time t = ZZZ = Ac cos t V NP[c]dc sindd Ac cos t 4 V Number of impacts on unit area per unit time = N 4V Z = N 4V c () = 4 nc cp [c]dc Z sin cos d Z d (.6) (.7) (.8) (.9) where the number density, n = N V..3 Eusion of molecules The number of molecules with speeds c!c+dc from direction (! + d,! + d) imping a area in time t is N cp [c]dcsinddat 4V The distribution is proportional to c P[c] dc. The mean speed of escape, c e : c e = R c P [c]dc R cp [c]dc 3kT m = ( 8kT ) m 9kT = 8m (.)

76 CHAPTER. KINETIC THEORY OF GASES 75 Number of molecules escaping per unit time per unit area 4 mc = p 8kT kt m / m (.) So, the smaller molecules escapes(euses) faster..4 Mean Free Path, The free path is dened as the distance between collisions. It is not all equal, for some are longer and some are shorter. The more probable the collision, the shorter is the free path. The distribution of free path : Probability that free path in between x and x+dx / e, x dx. We can prove this by a self consistent argument : Probablity that collision occurs between x and x+dx = (Probability of no collision from to x) (Probability of collisions in dx) Comparing with the above equation, we write Probability ofnocollision in dx = e,dx (.) Probability of collision in dx = dx (.3) According to (.), the probability of no collision in dx = e, dx (.4) Then, Probability of collision in dx =, e,dx =,, dx + :::: (.5) dx which agrees with (.3).

77 CHAPTER. KINETIC THEORY OF GASES 76 So the argument is self consistent. Also, we see that Z e, x dx = (.6) which is as required for probability. Therefore, Z Mean free path = xe, x dx Z =[,xe, x ] + e, x dx =[,e, x ] = (.7).5 Collision cross-section, d d The moving gas molecules sweeps out a cylinder of cross section : where d is the diameter of the each molecule. = d (.8) Average volume occupied by each molecule = V N = n (.9) So, Mean free path, = V N d = nd (.3) Taking in account of the motion of gas molecules, = p n (.3) At stp, 7,8 m. At,6 mbar, 5m.