* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Example of Boltzmann distribution.
Survey
Document related concepts
Chemical thermodynamics wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
Double layer forces wikipedia , lookup
Thermodynamics wikipedia , lookup
Colloidal crystal wikipedia , lookup
Depletion force wikipedia , lookup
Brownian motion wikipedia , lookup
Transition state theory wikipedia , lookup
Freeze-casting wikipedia , lookup
Double-slit experiment wikipedia , lookup
Nanoparticle wikipedia , lookup
Stöber process wikipedia , lookup
Molecular dynamics wikipedia , lookup
Self-assembly of nanoparticles wikipedia , lookup
Electron scattering wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Atomic theory wikipedia , lookup
Transcript
Chapter 13 Classical and Quantum Statistics 1 So far, with the exception of the previous chapter, we have dealt with the 1st and 2nd laws of thermodynamics. In using these laws to make numerical calculations it is usually necessary to appeal to experimental measurements. We would like to calculate all thermodynamic properties of a system from a microscopic model of that system. We have made a start in the last chapter and we will make significant progress in the present chapter. 2 Lagrange Undetermined Multipliers I would like to give you a trivial example of the use of Lagrange Undetermined Multipliers. The term is somewhat misleading because the multipliers can, in fact, be determined. In the following example it is not necessary to use this sophisticated method and you should solve the problem in a simple fashion. Consider the equation ax+by=0……………..(1) If y=y(x) then y=-(a/b)x However this is not true if both x and y are independent. The only solution is then a=b=0 Suppose, however, that x and y are not completely independent but satisfy a constraint condition such as x+2y=0………(2) What can we say about the coefficients a and b? The procedure, using Lagrange Multipliers, is to multiply each constraint condition by a multiplier, which is initially unknown. In the present case we have one constraint condition and let multiplier. giving (x 2 y) 0 (1) (2) 0 (a )x (b 2) y 0.....(3) 3 Now that the constraints have been implicitly taken into consideration we treat x and y as two independent variables. We must have ( a) 0 (b 2 ) 0 so so a b 2 This yields b 2a Placing this in equation (1) shows that the constraint condition is satisfied. 4 EXAMPLE: Lagrange Undetermined Multiplier A cylindrical nuclear reactor has radius R and height H We wish to minimize the volume of the reactor. V R2 H There is a constraint supplied by neutron diffusion theory 2 2 constraint 2.4048 dV 2 RHdR R2 dH For an extremum dV=0 2 RHdR R2 dH 0 (1) 2(2.4048 )2 2 2 dR 3 dH 0 3 R H (1) (2) 0 R constant H differentiating the constraint (2.4048 )2 2 dR 3 dH 0 (2) 3 R H 2 2 ( 2 . 4048 ) 2 RHdR R2 dH dR 3 dH 0 3 R H 2 (2.4048 )2 2 2 RH dR R 3 dH 0 3 R H 5 Now we consider R and H to be independent 2 (2.4048 )2 2 2 RH 0 .....( 3 ) R 3 0.....(4) 3 R H 1 2 3 (multiplier is determined) R H From equation (4) Substituting into equation (3) 2 2 2 R H2 (2.4048 )2 (2.4048 )2 R2 H 3 2 RH R3 H 2R (2.4048 ) 6 Boltzmann Statistics. We consider N distinguishable particles and we can place any number into a particular state. We wish to determine the thermodynamic probability for a particular macrostate. To guide us we consider a simple example: N=3 (A B C) and take the macrostate in which N1 2 g1 3 We begin with a box labeled N1 and we wish to throw 2 particles into the box. 1st particle: 3 choices ( N) 2nd particle 2 choices (N-1) The total number of choices is (3)(2) (N)(N-1) These choices are shown on the next slide 7 Choices However we can permute the particles in the box without changing the contents of the box. The number 1st 2nd of permutations is 2! ( N1 !) A B A C Therefore the number of distinct choices is: B A 32 N ( N 1) B C 2! N1! C A They are AB AC BC C B Now we arrange these particles into the 3 degenerate states. g1 1st particle has 3 possibilities 2nd particle has 3 possibilities g1 2 The total number of possibilities is 3 N1 1 g We will list these possibilities for the case where the box has AB 8 STATES 1 2 AB AB 3 The total number of possibilities is therefore AB A A B B B A A B We see that there are 9 possibilities. 32 We also have the AC and BC possibilities, each with 9 possible distributions among the states B B A A 32 2 3 2! N ( N 1) N1 g1 N1! Now we concentrate on the general case of N particles and consider placing N1 particles in energy level 1 which has a degeneracy g1 As above selecting the N1 particles for the N ( N 1)( N 2)( N N1 1) N1! 1 box, we have possibilities We can write this as N ( N 1)( N 2)( N N1 1)( N N1 )( N N1 1)1 N1! ( N N1 )( N N1 1)1 9 N! and this is evidently N !( N N )! 1 1 Now we consider the possibilities for distributing these particles into the degenerate states of 1 N The total number of possibilities is then N! g1 1 N1!( N N1 )! Now we go to the 2 level. The procedure is obviously the same except that we no longer have N particles. The number of available particles is ( N N1 ) N ( N N1 )! g2 2 As above we have, for this level, N2 ! ( N N1 N2 )! For the 3 possibilities ( N N1 N2 )! g 3 3 N 3! ( N N1 N2 N 3 )! N level 10 Hence the thermodynamic probability is ( N )! g1 1 ( N N1 )! g2 2 ( N N1 N2 )! g3 3 w( N1, N2 , N n ) N1!( N N1 )! N2!( N N1 N2 )! N 3!( N N1 N2 N 3 )! N N N N N N ( N )! g1 1 g2 2 g3 3 w( N1, N2 , N n ) N1! N2! N 3! n N gi i w( N1, N2 , N n ) N! i 1 N i ! Boltzmann statistics (distinguishable particles) This is the total number of accessible microstates for a particular macrostate. The value of w will be different for each particular macrostate. The greater the value of w, the greater the probability of occurrence. Remember: The equilibrium macrostate is the one for which w is a maximum. 11 Example: Before continuing with our discussion of Boltzmann statistics, let us consider a simple situation. Consider 6 distinguishable objects (A,B,C,D,E,F) which can be placed in 6 boxes (1,2,3,4,5,6). We will calculate w for several macrostates. The degeneracy will be unity for the boxes. 6 6 g iNi 1N i 1 w( N 1 , N 2 ,N n ) N! 6! 6! i 1 N i ! i 1 N i ! i 1 N i ! 1 w(1,1,1,1,1,1) 6! 6! w(1,1,1,1,1,1) 720 1!1!1!1!1!1! (most disorderd) 1 w(0,0,0,0,0,6) 6! 1 w(0,0,0,0,0,6) 1 0!0!0!0!0!6! 1 6! w(1,1,1,1,2,0) 6! w(1,1,1,1,2,0) 360 1!1!1!1!2!0! 2! 1 6 w(5,1,0,0,0,0, ) 6! w(5,1,0,0,0,0, ) 6 12 5!1!0!0!0!0! 1! n Now continuing with Boltzmann Statistics: We vary the N i to find the maximum value for w. This will give the N i in terms of g i and i For convenience we will work with ln(w) instead of w. (The range of values is much smaller.) This will also permit us to use Stirling’s fromula, valid for large x: ln(x!)=xln(x)-x Ni n n n g ln( w) ln N ! i ln( N ! ) N i ln( g i ) ln( N i ! ) i 1 N ! i 1 i 1 i Apply Stirling’s formula to the last term : n n n i 1 n i 1 i 1 ln( w) ln( N! ) Ni ln( gi ) Ni ln( Ni ) Ni Ni ln( w) ln( N! ) Ni ln N i 1 gi constants Now we maximize to obtain the equilibrium distribution. n Ni gi 1 dNi d (ln( w)) ln dNi Ni Ni gi i 1 i 1 gi n 13 n Ni d (ln( w)) ln dNi d Ni i 1 i 1 gi n gi d (ln( w)) ln dNi (1) i 1 Ni n We have the constraints: n N i 1 Differentiating i N constant n N i i i 1 n dN i 1 i 0 n i 1 U constant i dNi 0 We multiply the constraint conditions by the Lagrange multipliers ln( ), ( ) (This form of the multipliers is for convenience.) We also have the condition for a maximum that d(ln(w))=0 n n gi ln dNi ln( ) dNi i dNi 0 i 1 i 1 i 1 Ni n 14 The N i are now taken as independent, so the coefficient of each dN i must vanish. This gives gi ln ln( ) i 0 (2) Ni Ni gi e Solving for N i yields i Hence N i varies with the degeneracy and with the energy of the level. Now we need to determine the Lagrange multipliers. N n N i 1 N Z n i gi e with i We write this in the form i 1 Z n g e i 1 i i N i Ni gi e (3) Z Z comes from zustandsumme, which means a sum over states. It is usually referred to as a partition function. As we shall see, partition functions are very useful. It turns out that Z depends only on T and the parameters that determine the energy levels. 15 Once Z is known it is straightforward to calculate many thermodynamic properties, such as U, S and P. Of course there are other ways of calculating these quantities, but the simplest way is to first calculate the appropriate ln(Z). We still need to determine β. 16 Now we bring T into statistical mechanics. For a process taking place between two equilibrium states: đQ = dU+ PdV and dS = đQ/T so dS T1 dU TP dV Considering S as a function of U and V, dS US V dU VS U dV and comparing we see that 1 T US V (reciprocity relation) The state variables S and U may be calculated by statistical methods and so this equation brings the macroscopic concept of T into statistical thermodynamics. 17 Now we are in a position to discuss the Lagrange multiplier We have the following two equations (equations 1 and 2 above) gi ln ln( ) i 0 Ni gi d (ln( w)) ln dNi i 1 Ni n Substituting the second equation into the first one n n n n i 1 i 1 i 1 i 1 d (ln( w)) i dNi ln( )dNi d i Ni ln( )d Ni We are dealing with a closed system with constant V so n n d Ni 0 and This gives d(ln( w)) dU i 1 N i 1 i i U Using the Boltzmann relationship S=kln(w) 1 k dS dU or S U V dS=k d(ln(w)) k 18 From the reciprocity relationship: 1 1 k or T kT The temperature turns out to be a Lagrange multiplier! Now that we have determined the Lagrange multipliers, we write down the following two fundamental equations (see equation (3)): i Ni N kT Boltzmann distribution fi e (See note on next slide) g Z i The Boltzmann distribution gives the probability of occupation of a single state belonging to the ith energy level. Z n ge i 1 i i kT The partition function, an explicit function of T and an implicit function of V (through the energy levels) contains the statistical information about the system. 19 NOTE: There is some confusion regarding terminology. The Ni N term Boltzmann distribution for the equation fi gi Z e i kT is unfortunate because, among other things, it is then easily confused with the Maxwell-Boltzmann Distribution. The Boltzmann distribution applies to systems which have distinguishable particles and N, V and U are fixed. The Maxwell-Boltzmann Distribution is applicable only to dilute gases. 20 Consider the case in which the energy levels are very closely spaced. Then, instead of considering discrete levels with degeneracies g i , we consider a continuum and replace the g i by g( )d , the number of states in the energy range between and d The Boltzmann distribution is then written in the form: f ( ) Ne g( )e kT kT Distinguishable particles d 21 Example of Boltzmann distribution. Suppose that we have 1000 particles and T=10000K. The available states and their multiplicities are shown below. The partition function and the distribution of particles amongst the states are i n calculated using i N Z ge i 1 (eV) gi 3 4 2 3 1 2 0 1 kT i Ni Z g ie kT Z=2.045 Ε(eV) Ni 0 1 2 3 489 307 144 60 22 23 Canonical ensemble This consists of a set of systems in contact with a large thermal reservoir. For the Boltzmann distribution V, N and U are fixed. We now consider a system in which V, N and T are fixed. The energy of a system is not fixed, but will fluctuate about some average value due to the continuous interchange of energy between the system and reservoir. For large N these fluctuations are small. The formula giving the probability distribution is one of the most important in statistical mechanics. It applies to any system with a fixed number of particles in thermal equilibrium (V, N, T fixed). We will not provide a derivation of this probability formula, which is essentially the same as the Boltzmann formula {See section 13.9 of the textbook.} Ni gi e N Z i kT 24 e i kT is called the Boltzmann Factor We see that at low temperatures only low energy states are populated. As the temperature increases, the population shifts to higher energy states. 25 Example: An atmosphere at a temperature of 6000K has, as one constituent, neutral atoms with the first excited state 1.20eV above the ground state. The ground state (taken as the reference energy level of 0eV) is doubly degenerate while the 1st excited state is 6-fold degenerate. Assuming that there is negligible population of higher excited states, what fraction of these molecules are in the 1st excited state? (Use Canonical Distribution.) 1.2eV 0eV 6 2 N N i g ie Z Z 2e 0 eV kT i kT 6e (1) Z 1.20eV eV ( 8.6210 5 ( 6103 K )) K n gi e i kT i 1 2.59 i i Ni 1 1 Placing in (1) g i e kT g i e kT N Z 2.59 N2 1 (6)e N 2.59 1.20eV eV ( 8.6210 5 ( 6103 K )) K N2 0.227 N 26 Now we are going to discuss two other important distributions. In preparation we review some aspects of fundamental particles. These particles are divided into two classes, depending on their quantum number called the intrinsic angular momentum or “spin” Spin quantum number an integer: Bosons (examples are photons, gravitons, pi mesons,……..) Spin quantum number an odd number of (1/2): Fermions (examples are electrons, quarks, muons,……….) Fermions obey the Pauli Exclusion Principle: In an isolated system, no two fermions can occupy the same state. 27 Fermi-Dirac Distribution. This distribution is for indistinguishable fermions. There can be no more than one particle in any state. This places the following restriction on any macrostate: N i g i Consider the ith energy level. We wish to place N i particles into the g i states. For the 1st particle there are gi possibilities. For the 2nd particle there are ( g i 1) possibilities. For the ith particle there are ( g i N i 1) possibilities. The total number of possibilities is then ( g i )( g i 1)( g i N i 1) Since the particles are indistinguishable, we can permute them in a particular distribution without obtaining a different distribution. 28 The total number of distinct possibilities is then gi ! ( gi )( gi 1)( gi Ni 1) or Ni !( gi Ni )! Ni ! The total thermodynamic probability for a particular macrostate is n gi ! w( N1, N2 N n ) {What are the Ni N ! ( g N )! i 1 i i i at equilibrium?} Again we will work with ln(w) n n n i 1 i 1 i 1 ln( w) ln( gi ! ) ln( Ni ! ) ln(( gi Ni )! ) The first term is constant. We use Stirling’s approximation for the n n n other terms. ln( w) ln( g i ! ) N i ln( N i ) N i (gi constant!) i 1 n i 1 i 1 n ( g i N i ) ln( g i N i ) ( g i N i ) i 1 i 1 29 n n n Ni d (ln( w)) dN i ln( N i ) dN i (dN i ) ln( g i N i ) i 1 i 1 N i i 1 ( gi Ni ) (dN i ) i 1 ( g i N i ) n n n i 1 i 1 d (ln( w)) dNi ln( Ni ) (dNi ) ln( gi Ni ) gi Ni (dNi ) d (ln( w)) ln i 1 Ni As before, we have the constraints: n N i 1 i N constant n n N i 1 i i We introduce the Lagrange multipliers U constant and n n gi Ni (dNi ) (dNi ) i (dNi ) 0 ln i 1 i 1 i 1 Ni n 30 gi Ni i (dN i ) 0 ln Ni i 1 n With the constraints included in the equation, the N i are independent The coefficient of each dN i 0 gi Ni gi Ni i 0 ln i ln Ni Ni gi i 1 e Ni Ni 1 i gi e 1 It is not trivial to determine the Lagrange multipliers. It turns out, as 1 before, that kT The other multiplier is related to the chemical potential kT With these assignments we have 31 Ni 1 fi ( ) / kT gi e i 1 Fermi-Dirac distribution For a continuous energy distribution, this becomes f ( ) 1 e ( ) / kT 1 Later in the course we will use the Fermi-Dirac Distribution. 32 Bose-Einstein Statistics The particles are again indistinguishable, but now any number of particles can be in a particular energy state. (The Pauli Exclusion Principle does not apply.) Example: 13 particles in the ith energy level which has a degeneracy of 8 partition xxx x xxx x xx xx x 1 2 xx xx 3 4 5 x xx x 6 7 8 State x x xx x We can obtain different distributions by moving particles and partitions around. How many ways can we arrange N i particles and ( g i 1) partitions to form different distributions given that the particles and partitions are identical? The answer is Ni gi 1! (Students: convince yourselves.) Number of ways= Ni ! gi 1! 33 6! 20 Example: 3 particles in 4 degenerate states 3! 3! Students: show these 20 distributions (BEEx.ppt) The thermodynamic probability for a macrostate is then: n w( N1, N2 , N n ) i 1 N g i 1! N i ! g i 1! i We proceed as before: n n n i 1 i 1 i 1 ln( w) ln(( Ni gi 1)! ) ln( Ni ! ) ln(( gi 1)! ) Using Stirling’s formula ln( w) n (N i 1 i n n i 1 i 1 g i 1) ln( N i g i 1) ( N i g i 1) N i ln( N i ) n n n i 1 i 1 i 1 N i ( g i 1) ln( g i 1) ( g i 1) 34 n n n ln( w) ( Ni gi 1) ln(Ni gi 1) Ni ln( Ni ) ( gi 1) ln( gi 1) i 1 i 1 n n i 1 n n i 1 i 1 d (ln( w)) (dNi ) ln(Ni gi 1) dNi (dNi ) ln( Ni ) dNi i 1 i 1 n n i 1 i 1 d (ln( w)) (dNi ) ln(Ni gi 1) (dNi ) ln( Ni ) N gi 1 dNi d (ln( w)) ln i Ni i 1 n Introducing Lagrange multipliers with the constraint conditions: n n n Ni gi 1 dNi dNi i dNi 0 ln Ni i 1 i 1 i 1 Proceeding as before: ln Ni gi 1 i N i Neglecting the 1 (gi large!) Ni gi i ln Ni 35 gi i ln1 Ni gi i e 1 Ni gi i 1 e Ni Ni 1 i gi e 1 Again, a more detailed analysis gives the same values as before for the Lagrange multipliers Ni 1 fi ( ) / kT gi e i 1 Bose-Einstein Distribution This distribution will be used later in the course. 36 Maxwell-Boltzmann Statistics and Review Let us consider an assembly (system) 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 have considered some general assembly with: N1 particles in any of the g1 states of 1 N 2 particles in any of the g2 states of 2 N i particles in any of the g i states of i We now impose the restriction that g i N i for all i. This gives the Maxwell-Boltzmann Statistics. This condition holds for all real gases except at very low temperatures. At low temperatures one must use either Bose-Einstein or Fermi-Dirac Statistics depending on the 37 spin of the molecules. This restriction means that it is very unlikely that more than one particle will be in a given state. Subject to this restriction we first consider the number of ways that N i distinguishable particles can be distributed among the g i states. The first particle can be placed in any one of the g i states. The second particle can then be placed in any of the remaining ( g i 1) states, and so forth. The total number of different ways is then: g i ( g i 1)( g i 2)( g i N i1) Since g i N i this is approximately g iN i At present we are interested in indistinguishable particles and so many of these distributions will be the same when the condition of distinguishability is removed. We can start with one particular distribution and then obtain identical distributions by permuting the indistinguishable particles among themselves. The number of such permutations is N i ! . Hence the number of ways, subject to the restriction, that the N i indistinguishable particles can be distributed among the states is (g iNi / Ni !)(1) 38 A fundamental problem in statistical mechanics is to determine the particular macrostate of a system when at equilibrium. The U of the isolated system is fixed and so each microstate has this value. The laws of mechanics do not lead one to expect that the system will be found in one microstate rather than in any other. This is consistent with with our postulate that all microstates are equally probable. Of course, all such postulates must be verified by comparing the calculations based on the postulates with experimental results. Now let us return to a consideration of a particular macrostate, that is, a particular set ( i , N i ) As we stated earlier, the number of ways this particular macrostate can be achieved is called the thermodynamic probability w. We have calculated this probability (unnormalized) above for one energy level. The total thermodynamic probability is the product of the individual probabilities for all the N n accessible levels. From equation (1): gi i w( N1, N2 , N n ) i 1 N i ! 39 As shown in the textbook, the Fermi-Dirac and Bose-Einstein probabilities reduce to this equation when one makes the approximation g i N i The thermodynamic probability for Boltzmann Statistics and Maxwell-Boltzmann Statistics is w(Boltzmann ) N! w(Maxwell - Boltzmann ) Distinguishable Indistinguishable These two probabilities differ by a constant and hence their derivatives (dw) will be the same. In addition the constraint conditions (constant N and U) are the same. Hence, in using the method of Lagrange Undetermined Multipliers, we will obtain the same result. Z n gi e i 1 i kT Ni N fi e gi Z g i N i i kT Maxwell-Boltzmann so fi<<1 40 Statistical Interpretation of Heat and Work For concreteness we return to the quantum mechanical formula for the energy levels available for molecules of a gas in a container 2 of volume V. h2 2 / 3 2 h 2 2 2 / 3 2 i 8m V (n x n y n z ) 8m V n For a given quantum level (given n) the energy depends only upon the volume. d i i i (V ) d i dV dV The internal energy of a gas is U n N i 1 i i n d i dU i dNi Ni d i i dNi Ni dV dV i 1 i 1 i 1 i 1 n n d i dU i dNi Ni dV dV i 1 i 1 n n Macroscopically n dU= đ Q-PdV 41 Now suppose that the volume of the gas does not change. (No work is done on the gas.) Comparison of the two equations yields n đ Q i dN i i 1 When energy is added by the heating process, the energies do not change, but the distribution of particles among the energy levels changes. (See Fig. 13.4). In heating the gas you are promoting molecules from lower levels to higher levels. Now suppose that, instead of heating the gas, one does work on the gas, so that its volume decreases. In this case, the above formulae shows that: n d i n dV N i d i đ W N i dV i 1 i 1 In this case the distribution of particles among the energy levels does not change, but the energy of the levels change. Doing work on the gas raises the energy levels. (See Fig. 13.5). 42 W Q 43 The entropy, Helmholtz function and the chemical potential in terms of the partition function. We obtain formulae using the Maxwell-Boltzmann distribution. n Ni i g w( N1, N2 , N n ) i 1 N i ! Ni N e gi Z n n S k ln w k N i ln g i ln( N i ! ) i 1 i 1 i kT Using the Stirling approx. n n n n Ni S k Ni ln gi Ni ln Ni Ni k N Ni ln gi i 1 i 1 i 1 i 1 Ni i But ln ln N ln Z n gi kT S k N N ln N ln Z i i 1 1 S k N (ln N )( N ) (ln Z )( N ) U kT i kT 44 U S Nk ln Z ln N 1 T (M-B Dist.) Since we can write the Helmholtz function in terms of U and S we can easily obtain an expression for F in terms of Z U F=U-TS F U T Nk ln Z ln N 1 T F NkTln Z ln N 1 {Note: Z=Z(T,V) so F(T,V)} We have already determined the chemical potential in terms of F: F N T ,V Differentiating F with respect to N 1 kT ln Z ln N 1 NkT N N kT ln (M-B Dist.) Z 45 Finally we write the Maxwell-Boltzmann distribution in terms of the chemical potential. This will permit a comparison of the three i distributions. N N fi i gi Z e kT Using the above expression for the chemical potential: Substituting into the distribution: N e kT Z i Ni fi e kT e kT gi Ni 1 fi ( ) / kT gi e i 46 Comparison of the distributions The chemical potential, which enters these distributions, will be discussed when we use the distributions. At this stage we will simply plot the distributions, which can all be represented by fi a= +1 Fermi-Dirac -1 Bose-Einstein 0 Maxwell-Boltzman 1 e ( i ) / kT We plot these distributions as a function of (MAPLE plot distrib.mws) i x 0 i x 0 a x ( i ) kT i x 0 47 limit of validity (M-B) 48 Consider the BE distribution: For i the distribution is infinite and for i the distribution is negative and hence meaningless. The particles tend to condense into the lower energy states. Consider the FD distribution. For x=0 f i 1 / 2 For x fi 1 The low-energy levels are very nearly uniformly populated with one particle per state. Consider the MB distribution This distribution is only valid for f i 1 In this limit this distribution is an approximation for the BE and FD distributions. 49 We have introduced four distributions. These give the distribution of particles in the accessible states. Boltzmann: Distinguishable particles. Any number can go into an energy state. Fermi-Dirac: Indistinguishable particles which obey the PEP. Bose-Einstein: Indistinguishable particles. Any number can go into an energy state. Maxwell-Boltzmann: Indistinguishable particles. Any number can go into an energy state. Valid when g i N i An extremely useful function was introduced: The partition function. n Z(T, V ) g i e i kT {Boltzmann Factor} i 1 Many thermodynamic variables and functions can be written in terms of this function. This function contains all the statistical information about the system. 50 We also introduced the canonical distribution, a probability i distribution, N i 1 N Z e kT which has wide applicability. 51