# Chapter 12 Statistical Thermodynamics 1 Introduction to statistical Chapter 12 Statistical Thermodynamics 1 Introduction to statistical mechanics Statistical mechanics was developed alongside macroscopic thermodynamics. Macroscopic thermodynamics has great generality, but does not explain, in any fundamental way, why certain processes occur. As our understanding of the molecular nature of matter developed this knowledge was used to obtain a deeper understanding of thermal processes. Some uses: 1) ideal gas:- very successful 2) real gases:- more difficult, but some success 3) liquids:- very difficult, not much success 4) crystalline solids:- since they are highly organized they can be treated successfully 5) electron gas:- electrical properties of solids 6) photon gas:- radiation 7) plasmas:- very important As the results of kinetic theory can be obtained from statistical mechanics, we will not discuss kinetic theory. 2 Stat mech adds something very useful to thermodynamics, but does not replace it. Can we use our knowledge of the microscopic nature of a gas to, say, violate the 2nd law? Maxwell investigated this possibility and invented an intelligent being, now called a Maxwell demon, who does just that. As an example, imagine a container with a partition at the center which has a small trapdoor.

The demon opens, momentarily, the trapdoor when a fast molecule approaches the trapdoor adiabatic from the right. She also opens it when a slow walls molecule approaches it from the left. As a demon result the gas on the left becomes hotter and the gas on the right becomes cooler. One can then consider operating a heat engine between the two sides to produce work, violating the second law. 3 H adiabatic wall demon QC QH C The demon, clever lady, can keep the energy content of the cold reservoir constant (for a time). The net result is that energy is removed from a single reservoir (H) and is used to do some work. This violates the 2nd law. QC

E work Of course(?) no demon exists, but could some clever mechanical device be used?The demon must have information about the molecules if she is to operate successfully. Is there a connection between information and entropy? Yes! 4 The subject of information theory uses the concept of entropy. Let us consider another example:- free expansion. The demon removes the partition, free expansion occurs and the entropy of the system increases. gas vacuum Because of random motion of the molecules, there is some probability that, at some instant, they will all be in the region initially occupied by the gas. For demon this to occur, you will probably have to wait 1010 yrs The demon could, at this instant, slide in the partition and we would have a decrease in entropy of the universe. Again the demon must have some information about the location of the molecules. No such demon has been sighted. Before starting Ch. 12 I should warn you that there are two different types of statistics that have some similarities and have similar names. These two types of statistics are easily confused. (1) Maxwell-Boltzmann Statistics:-classical limit applies to dilute gases. The particles are indistinguishable. 5 (2) Boltzmann Statistics: particles are distinguishable 10 Some jargon: assembly (or system): N identical submicroscopic entities, such

as molecules. macrostate (or configuration): number of particles in each of the energy levels. microstate: number of particles in each energy state. thermodynamic probability: number of different microstates leading to a given macrostate. k th macrostate wk is the thermodynamic probability Basic postulate: All possible microstates are equally probable. 6 RECALL: In statistics, probabilities are multiplicative. As an example, consider a true die. The probability of throwing a one is 1/6. Now if there are two dies, the probability of one coming up on both dies is 1 1 1 6 6 36 7 Elementary Statistics We begin by considering 3 distinguishable coins (N D Q) The possible macrostates are HHH HHT HTT TTT Let us consider the microstates for the macrostate HHT H

H T N D Q D N Q N Q D Q N D D Q

N Q D N The table shows the possible selection of coins. There are 6 possibilities. However the pairs shown are not different microstates (the order does not matter). Hence we have 3 microstates. 8 More generally, suppose that we have N distinct coins and we wish to select N1 heads (a particular macrostate). There are N choices for the first head. There are (N-1) choices for the second head. There are [ N ( N1 1)] [ N N1 1] choices for the N1th head The thermodynamic probability (w) is the number of microstates for a given macrostate. We are then tempted to write N! w ( N )(N 1) ( N N1 1) ( N N1 )! However permuting the N1 heads results in the same microstate, so N! w N1!( N N1 )! 3!

3 For the above simple example with 3 coins: w 2!(3 2)! This is the number of microstates for the HHT macrostate. 9 Suppose we plot w as a function of N1 a number of cases (Thermocoin.mws). for a given N. We plot (N1 is the number of heads.) 10 Notice that the peak occurs at N1 N / 2 wmax N! N N ! ! 2 2 For large N we can use Stirlings formula ln(N !) N ln(N ) N N

ln(wmax ) ln( N !) 2 ln ! N ln( N ) N 2 N N N 2 ln 2 2 2 N N ln(wmax ) N ln( N ) 2 ln N ln(2) ln(2 N ) 2 2 For N=1000 wmax 2 N 300 wmax 10 This is the number of distinct microstates for the most probable macrostate (N1=500). Note that it is a very large number! 11 For large N the plot of w versus N1 is very sharp (see next slide) The total thermodynamic probability is obtained by summing over all macrostates. Let k indicate a particular macrostate: wk k Since, for large N, the peak is very sharp: wk wmax k

12 13 Now we consider N distinguishable particles placed in n boxes with N1 in the first box, N2 in the second box, etc. We wish to calculate w( N1, N2, N n ) (a particular macrostate) Before doing a general calculation, we consider the case of 4 particles (ABCD) with 3 boxes and N1 2 N2 1 N 3 1 We begin by indicating the possibilities for the first box. Since the order is irrelevant, there are A B B C 6 possible microstates. B A C B Now suppose A and B were selected for the first box. This leaves C and D when A C B D we consider filling the second box. We C A D B obviously have only two possibilities, C or D. Suppose that C was selected. A D C D That leaves only one possibility (D) for D A D C the third box. The total number of

14 possibilities for this macrostate is (6)(2)(1)=12 Now we consider the general problem (macrostate(N1,N2,N3,..)): Consider placing N1 of the N distinguishable particles in the first box. 1st box N ( N 1)(N 2)( N N1 1) N! N1! N1!( N N1 )! ( N N1 )! N2!( N N1 N2 )! 2nd box ( N N1 N2 )! N 3!( N N1 N2 N 3 )! 3rd box The thermodynamic probability for this macrostate is: w ( N )! ( N N1 )! ( N N1 N2 )! N1!( N N1 )! N2!( N N1 N2 )! N 3!( N N1 N2 N 3 )!

We have been considering distinguishable particles, such as atoms rigidly set in the lattice of a solid. For a gas, the statistics will be different. w N! n Nk ! k 1 15 Example (Problem 12.6) We will do an example illustrating the use of the formula on the previous slide. We have 4 distinguishable particles (ABCD). We wish to place them in 4 energy levels (boxes) 0, , 2 , 3 subject to the constraint that the total energy is U 6 A macrostate will be labeled by k and wk is the thermodynamic probability for the kth macrostate. k 1 2 3 3 2 1 0

1 wk 2 1 1 1 3 4 5 4! 4! 4! 4! w1 w2 w3 w4 2!2! 1!1!1!1! 1!3! 2!2! 2 3

The most probable state, k=2, is the most random distribution. 1 {Students should explicitly display one of the macrostates.} 2 2 1 6 24 4 6 4 16 Now consider an isolated system of volume V containing N distinguishable particles. The internal energy U is then fixed and the macrostate will be characterized by (N,V,U). There are n energy levels (like boxes) available and we wish to know the set N i at equilibrium. There are the following restrictions: n N i

N i 1 n N i i U Conservation of particles Conservation of energy i 1 The central problem is then to determine the most probable distribution. Since the system is isolated the total entropy must be a maximum with respect to all possible variations within the ensemble. The actual distribution of particles amongst the energy levels will be the one that maximizes the entropy of the system. Can we make a connection between the entropy and some specification of the macrostate? 17 A study of simple systems suggests that there is a connection between entropy and disorder. For example if one considers the free expansion of a gas, the entropy of the gas increases and so does the disorder. We know less about the distribution of the molecules after the expansion. The thermodynamic probability is also a measure of disorder. The larger the value of w, the greater the disorder. A simple example

is as follows: Suppose we distribute 5 distinguishable particles among 4 boxes. We can use the equation developed to determine w. N1 N 2 N 3 N 4 wk 5 0 0 0 1 4 1 0 0 5 3 2 0 0 10

3 1 1 0 20 2 2 1 0 30 2 1 1 1 60 5!

30 2!2!1!0! 18 The most ordered state, that is with all the particles in a single box, has the lowest w. The most disordered state, that is with the particles distributed amongst all the boxes, has the largest thermodynamic probability. As a system approaches equilibrium not only does the entropy approach a maximum, but the thermodynamic probability also approaches a maximum. Is there a relationship between entropy and thermodynamic probability? If so we would expect that S would be a monotonically increasing function of w: as the probability increases, so does S. 19 Thermodynamic Probability and Entropy Ludwig Boltzmann made many important contributions to thermodynamics. His most important contribution to physics is the relationship between w and the classical concept of entropy. His argument was as follows. Consider an isolated assembly which undergoes a spontaneous, irreversible process. At equilibrium S has its maximum value consistent with U and V. But w also increases and approaches a maximum when equilibrium is achieved. Boltzmann therefore assumed that there must be some connection between w and S. He therefore wrote S=f(w), and S and w are state variables. To be physically meaningful f(w) must be a single-valued monotonically increasing function. Now consider two systems, A and B, in thermal contact. (Such a system of two or more assemblies is called a canonical ensemble.) Entropy is an extensive property and so S for

the composite system is the sum of the individual entropies: S S A SB Hence f (w) S A S B or 20 f (w) f (wA ) f (wB ) (1) On the other hand independent probabilities are multiplicative so w wA wB Hence f (w) f (wA wB )(2) From (1) and (2) we obtain: f (wA wB ) f (wA ) f (wB ) The only appropriate function for which this relationship is true is a logarithm. Hence Boltzmann wrote S k ln w The constant k has the units of entropy and is, in fact, the Boltzmann constant that we have previously introduced. This celebrated equation provides the connecting link between statistical and classical thermodynamics. (One can begin with statistical mechanics and define S by the above equation.) 21 Quantum States and Energy Levels We consider a closed system containing a monatomic ideal gas of N particles. They are in some macroscopic volume V. According to quantum mechanics only

certain discrete energy levels are permitted for the particles. These allowed energy states are given by h2 8mV2 / 3 n 2 x 2 n y n2z where the nj are integers commencing with 1. The symbol h represents Plancks constant, which is a fundamental constant. The symbol m is the mass of a molecule. The symbol n is called a quantum number. 22 At ordinary temperatures the s of the particles are such that the n - values are extremely large (109 is a

typical value). When n changes by 1, the change in is so small that may be treated as a continuous variable. This will later permit us to replace sums by integrals. Example: A Hg atom moves in a cubical box whose edges are 1m long. Its kinetic energy is equal to the average kinetic energy of an atom of an ideal gas at 1000K. If the quantum numbers in the three directions are all equal to n, calculate n. Hg atom: m 201amu (201)(1.66 10 27)kg 2 x 2 y 2 z n n n 3n 2 h2 2 ( 3 n )

2 8mL 1 n 2 (4mL2kT ) h 2 m 3.34 10 25 kg 3 3h2n2 kT 2 8mL2 2L n mkT h 23 n 2(1.00m) 25 23 J 3 ( 3 .

34 10 kg )( 1 . 38 10 )( 10 K) 34 K 6.63 10 J s n 2.05 1011 24 Each different represents a quantum level. Each specification of (nx, ny, nz) represents a quantum state. The energy levels are degenerate in that a number of different states have the same energy. The degree of degeneracy of level i will be specified by gi. There is only one way to form the level 1 (nx = ny = nz = 1) so g1 = 1

that is, the ground state is not degenerate. The next level 2 occurs when one of the ns assumes the value 2 so g2 = 3 and so forth. As one goes to higher energy levels gi increases very rapidly. 25 In the terminology of statistical mechanics a number N of identical particles is called an assembly or a system. Let us now consider an assembly of N indistinguishable particles. A macrostate is a given distribution of particles in the various energy levels. A microstate is a given distribution of particles in the energy states. Basic Postulate of statistical mechanics: All accessible microstates of an isolated system are equally probable of occurring. We are interested in the macrostates N i In particular, what is the macrostate when the system is in equilibrium? We address this problem in succeeding chapters. 26 Density of Quantum States. A concept that is important for later work is that of the density of states. Under conditions in which the ns are large and the energy levels close together, we regard n, as continuous variables. From h2 8mV2 / 3 n

2 x 2 2 z ny n h2 8mV2 / 3 n2 2/ 3 8 m V 2 n h2 We consider a quantum- number space, (n x , n y , nz ) Each point in this space represents an energy state. Each unit volume in this space will contain one state. All the states are in the first quadrant. We then consider a radius R (which is n) in this space and

a second radius (R+dR). The volume between these two surfaces is 1 (4 R2dR) This gives the number of states between and d 8 1 We represent this number by g ( )d g ( )d (4 R2dR) 8 2/ 3 2/ 3 8mV 4mV 2 2 Substitute in RdR d But R n 2 2 h h 27 2/ 3 2/ 3 8mV 4mV g( )d

2 h2 h2 4 d 3 h 3 2 2 Vm d 4 2 V g ( )d m 3 h 3 2 d This result is correct for only certain particles. We have assumed that a state is uniquely specified by the quantum numbers (n x , n y , nz ) In many cases other quantum numbers play a role in the unique specification of a state. Particles fall into two categories which are radically different. Bosons: have integral spin quantum number

Fermions: have odd half-integral spin quantum number Examples are: Bosons photons, gravitons, pi mesons Fermions electrons, muons, nucleons, quarks 28 For electrons, two spin states are possible for each translational state. Thus each point in space represents two distinctly different states. This leads to a multiplicative factor of 2 in the density of states formula. To be completely general we write 3 4 2 V 2 g ( )d s m d 3 h For s=(1/2) fermions, s =2 The density of states replaces the degeneracy when we go from discrete energy levels to a continuum of energy levels. Notice that g depends on V, but not on N. {Students: Show that the unit of g is J-1.} 29 Problem 12.1 Consider N honest coins.

(a) How many microstates are possible? Consider the coins lined up in a row. Each coin has two possibilities N (H or T). For the N coins w 2 As an example consider 3 coins, so 3 w 2 8 We will show these microstates explicitly by considering the possibilities for the 2nd and 3rd coins and then adding H or T for the first coin. The possibilities are displayed in the next slide. 30 COIN 1 COIN 2 COIN 3 H H H T H

H H H T T H T H T H T T H H T T

T T We use MAPLE to calculate the factorials. N 50 w 250 w 1.13 1015 (b) How many microstates for the most probable macrostate? The most probable macrostate has the same number of heads and tails. (slide 9) T N! 50! wmax wmax 1.26 1014 N N 25! 25! ! ! 2 2 14 w 1 .

26 10 (c) True probability: P max Pmax 0.112 max 15 w 1.13 10 31 Problem 12.2 This is the same problem as 12.1 except that N=1000 The results are: w 1.07 10301 wmax 2.70 10299 Pmax 0.0252 {Students: Consider 4 identical coins in a row. Display all the possible microstates and indicate the various macrostates.} 32 Problem 12.5 We have N distinguishable coins. The thermodynamic probability for a particular microstate is (slide 9) N! w

(a) ln(w) ln( N !) ln( N1!) ln(( N N1 )!) (Stirlings Formula) N1!( N N1 )! ln( w) N ln( N ) N N1 ln( N1 ) N1 ( N N1 ) ln( N N1 ) ( N N1 ) ln(w) N ln( N ) N1 ln( N1 ) ( N N1 ) ln( N N1 ) d ln(w) ln( N1 ) 1 ln( N N1 ) 1 ln( N N1 ) ln( N1 ) dN1 N N1 N N1 N {Maximum} 0 ln 1 N1 N1 2 N1 (b) Now for the number of microstates at the maximum 33 wmax

N! NN ! ! 2 2 N ln(wmax ) ln( N !) 2 ln ! 2 N N N ln( wmax ) N ln( N ) N 2 ln 2 2 2 N ln(wmax ) N ln( N ) N ln N ln 2 2 wmax e N ln 2 34 Problem 12.8. In this problem we show explicity the microstates associated with each macrostate. There are two distinguishable particles and three energy levels, with a total energy of U 2 (a) A macrostate is labeled k. k 0 1 2

1 1 0 1 A B wk 2 0 2 0 2 0 2 0 1 2 1

1 2 1 0 w=3 S=k ln(w) S=k ln(3) (b) Now we have 3 particles with the restriction that at least one particle is in the ground state. (This is obviously necessary.) 35 k 0 1 2 A B C w 1 2

1 2 0 0 3 0 2 0 0 0 0 1 2 1 1 2

1 0 2 0 1 0 1 1 For this case, 3 0 S=k ln(6) k ln(6) 1.63 k ln(3) 36 What have we accomplished in this chapter? We have started to consider the statistics of the microscopic particles (atoms, molecules,.) of a system. The thermodynamic probability, w, was introduced. For a given macrostate k, wk is the number of different microstates that give rise to this particular macrostate. A larger value of w for a macrostate means that the macrostate is more likely to occur. We also saw the link between macroscopic thermodynamics (S)

and statistical mechanics (w). S k ln w The basic postulate of statistical mechanics was also introduced: Basic Postulate of statistical mechanics: All accessible microstates of an isolated system are equally probable of occurring. 37 We will be considering situations for which the energy levels are so closely spaced that they may be considered to form a continuum. The degeneracy of isolated states is then replaced by the density of states: 3 4 2 V 2 g ( )d s m d 3 h We now apply what we have developed in this chapter to different situations. 38