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Lecture 21. Boltzmann Statistics (Ch. 6)

We have followed the following logic: 1. Statistical treatment of isolated systems: ⇒ entropy ⇒ the 2nd Law. 2. Thermodynamic treatment of systems in contact with the heat reservoir ⇒ the minimum free principle. However, the link between G and the process of counting of accessible microstates was not straightforward. Now we want to learn how to “statistically” treat a system in contact with a heat bath. The fundamental assumption states that a closed (isolated) system visits every one of its microstates with equal : all allowed states of the system are equally probable. This statement applies to the combined system (the system of interest + the reservoir). We wish to translate this statement into a statement that applies to the system of interest only. Thus, the question: how often does the system visit each of its microstates being in the thermal equilibrium with the reservoir? The only information we have about the reservoir is that it is at the T.

Reservoir R U - ε Combined system 0

U0 = const System S ε

a combined (isolated) system = a heat reservoir and a system in thermal contact The Fundamental Assumption for an Isolated System Isolated ⇒ the energy is conserved. The ensemble of all σ i equi-energetic states ⇒ a mirocanonical ensemble The ergodic hypothesis: an isolated system in thermal equilibrium, evolving in time, will pass through all the accessible σ 2 microstates states at the same recurrence rate, i.e. all accessible σ 1 microstates are equally probable. The average over long times will equal the average over the ensemble of all equi- energetic microstates: if we take a snapshot of a system with N microstates, we will find the system in any of these microstates with the same probability.

Probability of a particular microstate of a microcanonical ensemble = 1 / (# of all accessible microstates) The probability of a certain macrostate is determined by how many microstates correspond to this macrostate – the multiplicity of a given macrostate macrostate Ω

Probability of a particular macrostate = (Ω of a particular macrostate) / (# of all accessible microstates)

Note that the assumption that a system is isolated is important. If a system is coupled to a heat reservoir and is able to exchange energy, in order to replace the system’s trajectory by an ensemble, we must determine the relative occurrence of states with different . Systems in Contact Reservoir R System S ε with the Reservoir U0 - ε The system – any small macroscopic or R microscopic object. If the interactions between the system and the reservoir are S weak, we can assume that the of ε2 energy levels of a weakly-interacting system is

ε1 the same as that for an isolated system. Example: a two-level system in thermal contact with a heat bath.

We ask the following question: under conditions of equilibrium between the system and reservoir, what is the probability P(εk) of finding the system S in a particular state k of energy εk? We assume weak interaction between R and S so that their energies are additive. The in the isolated system “system+reservoir”:

U0 = UR+ US = const According to the fundamental assumption of thermodynamics, all the states of the combined (isolated) system “R+S”are equally probable. By specifying the microstate of the system k, we have reduced ΩS to 1 and SS to 0. Thus, the probability of occurrence of a situation where the system is in state k is proportional to the number of states accessible to the reservoir R . The total multiplicity:

Ω()ε k ,U 0 −ε k = ΩS (ε k )×ΩR (U 0 −ε k )(=1×ΩR U 0 −ε k )= ΩR (U 0 −ε k ) Systems in Contact with the Reservoir (cont.)

The ratio of the probability that the system is in 1 at energy ε1 to the probability that the system is in quantum state 2 at energy ε2 is:

PS ()ε1 R ()U −Ω ε10 exp[ R ( −ε10 )/ kUS B ] ⎡ R ()ε10 R (USUS −−− ε 20 )⎤ ⎛ ΔSR ⎞ = = = exp⎢ ⎥ = exp⎜ ⎟ PS ()ε 2 R ()U −Ω ε 20 exp[]R ()−ε 20 / kUS B ⎣ kB ⎦ ⎝ kB ⎠

SR

S(U0) Let’s now use the fact that S is much smaller than R (US=εk << UR).

S(U0- ε1) Also, we’ll consider the case of fixed volume and number of (the latter limitation will be removed later, when we’ll S(U - ε ) 0 2 allow the system to exchange particles with the heat bath 1 2 ε 0 ε UR 1 (Pr. 6.9 addresses the case - - U 0 0 dSR ()R R −+= μdNPdVdU R nd U

U when the 2 is not negligible) TR 0

⎡ ∂SR ⎤ ⎡ dSR ⎤ R ()R USUS 0 (+≈ ⎢ ) ()0 ⎥ (−UUU 0 ) R ()()0 ε Ri USUS 0 −≈− ε i ⎢ ()U 0 ⎥ ⎣∂U R ⎦ ,NV ⎣dU R ⎦ ,NV

⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎡⎛ ∂S ⎞ 1 ⎤ −εε S =Δ ⎜ R ()U ⎟ ε − ⎜ R ()U ⎟ ε = ⎢⎜ R ()U ⎟ = ⎥ −= 21 R ⎜ ∂U 0 ⎟ 2 ⎜ ∂U 0 ⎟ 1 ⎜ ∂U 0 ⎟ TT ⎝ R ⎠ ,NV ⎝ R ⎠ ,NV ⎣⎢⎝ R ⎠ ,NV R ⎦⎥ R Boltzmann Factor

PS ()ε1 ⎛ ΔSR ⎞ ⎛ −εε21 ⎞ ⎛ ⎞ = exp⎜ ⎟ exp⎜−= ⎟ ε1 ⎜ ⎟ ⎜ ⎟ exp⎜− ⎟ PS ()ε 2 ⎝ kB ⎠ ⎝ BTk ⎠ P ()ε Tk S 1 = ⎝ B ⎠ T is the characteristic of the heat reservoir PS ()ε 2 ⎛ ε 2 ⎞ exp⎜− ⎟ ⎝ BTk ⎠ exp(- εk/kBT) is called the Boltzmann factor

This result shows that we do not have to know anything about the reservoir except that it maintains a constant temperature T !

The corresponding probability distribution is known as the reservoir canonical distribution. An ensemble of identical systems all of which are in contact with the same heat reservoir and T distributed over states in accordance with the Boltzmann distribution is called a canonical ensemble. The fundamental assumption for an isolated system has been transformed into the following statement for the system of interest which is in thermal equilibrium with the thermal reservoir: the system visits each microstate with a frequency proportional to the Boltzmann factor. Apparently, this is what the system actually does, but from the macroscopic point of view of thermodynamics, it appears as if the system is trying to minimize its free energy. Or conversely, this is what the system has to do in order to minimize its free energy. One of the most useful equations in TD+SP... Firstly, notice that only the energy difference Δε = ⎛ ⎞ ⎛ ⎞ P()ε1 ΔSR −εε21 ε - ε comes into the result so that provided that = exp⎜ ⎟ exp⎜−= ⎟ i j P()ε 2 ⎝ kB ⎠ ⎝ BTk ⎠ both energies are measured from the same origin it does not matter what that origin is. Secondly, what matters in determining the ratio of the occupation P(ε ) numbers is the ratio of the energy difference Δε to k T. Suppose i e9 ≈= 8000 B P()ε that εi = kBT and εj = 10kBT . Then (εi - εj ) / kBT = -9, and j

The lowest ε0 available to a system (e.g., a ) is referred to as the “”. If we measure all n(ε ) ⎛ ε ⎞ exp⎜−= ⎟ energies relative to ε0 and n0 is the number of in this ⎜ ⎟ n0 ⎝ BTk ⎠ state, than the number molecules with energy ε > ε0

Problem 6.13. At very high temperature (as in the very early universe), the and the can be thought of as two different states of the same , called the “nucleon”. Since the neutron’s mass is higher than the proton’s by Δm = 2.3·10-30 kg, its energy is higher by Δmc2. What was the ratio of the number of to the number of at T=1011 K?

2 ⎛ ncm ⎞ exp⎜− ⎟ 2 ()nP ⎜ Tk ⎟ ⎛ Δmc2 ⎞ ⎛ 103.2 −30 kg ()⋅×⋅ 103 8 m/s ⎞ = ⎝ B ⎠ exp⎜−= ⎟ exp⎜−= ⎟ = 86.0 ()pP ⎛ cm 2 ⎞ ⎜ Tk ⎟ ⎜ 1038.1 −23 ×⋅ 10J/K 11 K ⎟ ⎜ p ⎟ ⎝ B ⎠ ⎝ ⎠ exp⎜− ⎟ ⎝ BTk ⎠ More problems Problem 6.14. Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere. The system is a single air molecule, with two states: 1 at the sea level (z = 0), 2 – at a height z. The energies of these two states differ only by the mgz (the temperature T does not vary with z): ⎛ ε ⎞ ⎛ ε + mgz ⎞ exp⎜− 2 ⎟ exp⎜− 1 ⎟ P()ε ⎜ Tk ⎟ ⎜ Tk ⎟ ⎛ mgz ⎞ ⎛ mgz ⎞ 2 ⎝ B ⎠ ⎝ B ⎠ ⎜ ⎟ ⎜ ⎟ = = exp⎜−= ⎟ ()z ρρ0 exp⎜−= ⎟ P()ε1 ⎛ ε1 ⎞ ⎛ ε1 ⎞ ⎝ BTk ⎠ ⎝ BTk ⎠ exp⎜− ⎟ exp⎜− ⎟ ⎝ BTk ⎠ ⎝ BTk ⎠

At home: A system of particles are placed in a uniform field at T=280K. The particle concentrations measured at two points along the field line 3 cm apart differ by a factor of 2. Find the force F acting on the particles. Answer: F = 0.9·10-19 N

A mixture of two with the molecular masses m1 and m2 (m2 >m1) is placed in a very tall container. The container is in a uniform gravitational field, the acceleration of free fall, g, is given. Near the bottom of the container, the concentrations of these two types of molecules are n1 and n2 respectively (n2 >n1) . Find the height from the bottom where these two concentrations become equal. ( /ln nnTk ) h = B 12 Answer: ()− 12 gmm The Partition Function

For the absolute values of probability (rather 1 ⎛ ε ⎞ 1 ⎜ i ⎟ than the ratio of probabilities), we need an P()ε i exp⎜ −= ⎟ exp()−= βε i explicit formula for the normalizing factor 1/Z: ⎝ B ⎠ ZTkZ 1 β ≡ - we will often use this notation BTk The quantity Z, the partition function, can be found from the normalization condition - the total probability to find the system in all allowed quantum states is 1: 1 ( ,, NVTZ ) exp(−= βε ) The Zustandsumme 1 P()ε i == ∑∑ exp()− βε i or ∑ i i i Z i in German

The partition function Z is called “function” because it depends on T, the spectrum (thus, V), etc.

∞ ⎛ ε ⎞ ∞ Example: a single particle, ()TZ exp⎜−= ⎟ ε = exp()=− TkdxxTkd ∫ ⎜ ⎟ B ∫ B continuous spectrum. 0 ⎝ BTk ⎠ 0 -1 (kBT1) ⎛ ε ⎞ The areas under these curves must be exp⎜− ⎟ the same (=1). Thus, with increasing T, ⎜ Tk ⎟ T1 ⎝ B ⎠ 1/Z decreases, and Z increases. At T = < P()ε = -1 Tk 0, the system is in its ground state (ε=0) (k T ) T2 B B 2 with the probability =1.

0 ε Average Values in a Canonical Ensemble We have developed the tools that permit us to calculate the average value of different physical quantities for a canonical ensemble of identical systems. In general, if the systems in an ensemble are distributed over their accessible states in accordance with

the distribution P(εi), the average value of some quantity x (εi) can be found as: ⎛ ⎞ ()ε == ()Pxxx (εε) In particular, for a 1 ε i i ∑ iii x()ε i = x ()ε ii exp⎜− ⎟ i ∑ ⎜ ⎟ canonical ensemble: (),, NVTZ i ⎝ BTk ⎠ Let’s apply this result to the average (mean) energy of the systems in a canonical ensemble (the average of the energies of the visited microstates according to the frequency of visits): 1 1 = PU ()ii = ∑∑ εεεi exp()βε i =− ∑ i exp()− βεεi i i Z Z i

The average values are additive. The average total tot = UNU energy Utot of N identical systems is:

⎛ ε ⎞ ⎡ 1 ⎤ ∂ ∂ Another useful representation for ⎜ i ⎟ ∑ε i exp⎜− ⎟ ⎢ ≡= β ⎥ −= ∑exp()i βε −=− Z the average energy (Pr. 6.16): i ⎝ B ⎠ ⎣ BTkTk ⎦ ∂β i ∂β

1 ∂ Z ∂ ∂ thus, if we know Z=Z(T,V,N), U −= −= ln = TkZ 2 ()ln Z Z ∂ ∂ββ B ∂T we know the average energy! Example: energy and heat capacity of a two-level system

⎛ 0 ⎞ ⎛ ε ⎞ ⎛ ε ⎞ the slope ~ T The partition Z exp⎜−= ⎟ exp⎜−+ ⎟ exp1 ⎜−+= ⎟ Ei ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ function: ⎝ BTk ⎠ ⎝ BTk ⎠ ⎝ BTk ⎠ ε = ε 2 The average energy:

ε = 0 1 ⎡ ⎛ 0 ⎞ ⎛ ε ⎞⎤ ε 1 U ⎢ exp0 ⎜−×= ⎟ ε exp⎜−×+ ⎟⎥ = ()TZ ⎣ ⎝ BTk ⎠ ⎝ BTk ⎠⎦ ⎛ ε ⎞ -lnni + exp1 ⎜ ⎟ ⎝ BTk ⎠ U 1 ∂Z (check that the same result follows from U −= ) ε /2 Z ∂β

CV

The heat capacity at constant volume: 0 T

2 ⎛ ∂U ⎞ ∂ ⎡ ε ⎤ (ε BB ) (ε /exp/ BTkTkk ) CV = ⎜ ⎟ = ⎢ ⎥ = 2 ⎝ ∂ ⎠V TT ⎣ +∂ ()ε /exp1 BTk ⎦ []+ ()ε /exp1 BTk Partition Function and Helmholtz Free Energy

∂ F ⎛ F 1 ∂ F ⎞ ∂ ⎛ − F ⎞ Now we can relate F to Z: −=+= TFTSFU T 2 −= = Tk 2 ⎜ ⎟ ⎜ 2 ⎟ B ⎜ ⎟ ∂T ⎝ TT ∂T ⎠ ∂T ⎝ BTk ⎠ ⎛ F ⎞ Comparing this with the expression for Z exp⎜−= ⎟ exp()−= β F the average energy: ⎜ ⎟ ⎝ BTk ⎠ This equation provides the connection between the microscopic world which we specified with microstates and the macroscopic world which we described with F.

If we know Z=Z(T,V,N), we know everything we want to know about the thermal behavior of a system. We can compute all the thermodynamic properties:

⎛ ∂ F ⎞ ⎛ ∂ ln Z ⎞ ⎜ ⎟ ⎜ ⎟ S −= ⎜ ⎟ B ln += BTkZk ⎜ ⎟ ⎝ ∂T ⎠ ,NV ⎝ ∂T ⎠ ,NV

⎛ ∂ F ⎞ ⎛ ∂ ln Z ⎞ ⎜ ⎟ ⎜ ⎟ P −= ⎜ ⎟ = BTk ⎜ ⎟ ⎝ ∂V ⎠ ,NT ⎝ ∂V ⎠ ,NT

⎛ ∂ F ⎞ ⎛ ∂ ln Z ⎞ ⎜ ⎟ ⎜ ⎟ μ −= ⎜ ⎟ = BTk ⎜ ⎟ ⎝ ∂ N ⎠ ,VT ⎝ ∂ N ⎠ ,VT Microcanonical ⇔ Canonical Our description of the microcanonical and canonical ensembles was based on counting the number of accessible microstates. Let’s compare these two cases: microcanonical ensemble canonical ensemble For an isolated system, the multiplicity For a system in thermal contact with Ω provides the number of accessible reservoir, the partition function Z provides microstates. The constraint in the # of accessible microstates. The calculating the states: U, V, N – const constraint: T, V, N – const For a fixed U, the mean temperature T For a fixed T, the mean energy U is is specified, but T can fluctuate. specified, but U can fluctuate. E 1 - the probability of finding − n - the probability of finding 1 B Tk Pn = P = e Ω a system in one of the n Z a system in one of these accessible states states

()= kNVUS B ln,, Ω ( ,, )= − B ln ZTkNVTF - in equilibrium, S reaches a maximum - in equilibrium, F reaches a minimum For the canonical ensemble, the role of Z is similar to that of the multiplicity Ω for the microcanonical ensemble. This equation gives the fundamental relation between statistical and thermodynamics for given values of T, V, and N, just as

S = kB lnΩ gives the fundamental relation between statistical mechanics and thermodynamics for given values of U, V, and N.

1 ⎛ ∂ S ⎞ P ⎛ ∂ S ⎞ μ ⎛ ∂ S ⎞ ⎛ ∂ F ⎞ ⎛ ∂ F ⎞ ⎛ ∂ F ⎞ = ⎜ ⎟ = ⎜ ⎟ −= ⎜ ⎟ S −= ⎜ ⎟ P −= ⎜ ⎟ μ −= ⎜ ⎟ T ⎜ ∂U ⎟ T ⎜ ∂ ⎟ TV ⎜ ∂ N ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ,NV ⎝ ⎠ ,NU ⎝ ⎠ ,VU ⎝ ∂T ⎠ ,NV ⎝ ∂V ⎠ ,NT ⎝ ∂ N ⎠ ,VT Boltzmann Statistics: classical (low-density) limit We have developed the formalism for calculating the thermodynamic properties of the systems whose particles can occupy particular quantum states regardless of the other particles (classical systems). In other words, we ignored all kind of interactions between the particles. However, the occupation numbers are not free from the rule of . In quantum mechanics, even if the particles do not interact through forces, they still might participate in the so-called exchange interaction which is dependent on the of interacting particles (this is related to the principle of indistinguishability of elementary particles, we’ll consider bosons and fermions in Lecture 23). This type of interactions becomes important if the particles are in the same quantum state (the same set of quantum numbers), and their wave functions overlap in space:

strong exchange interaction weak exchange interaction

the de Broglie λ ↔ (n) 3/1 the average distance btw particles h h (nh ) 3/1 Boltzmann λ = ~ (for N molecule, λ ~ 10-11 m at RT) <<1 statistics p mTk 2 B B mTk applies Violations of the Boltzmann statistics are observed if the the density of particles is very large (neutron stars), or particles are very ( in metals, ), or they are at very low (liquid ), However, in the limit of small density of particles, the distinctions between Boltzmann, Fermi, and Bose-Einstein statistics vanish. If several quantum states of the system (different sets of quantum numbers) correspond to the same energy level, this level is called degenerate. The probability to find the system in one of these degenerate states is the same for all the degenerate

states. Thus, the total probability to find the system in a state with energy εi is ⎛ ε ⎞ ()dP ⎜ i ⎟ ε ii exp⎜−= ⎟ where di is the degree of . ⎝ BTk ⎠ ⎛ ε ⎞ Taking the degeneracy of energy levels into ⎜ i ⎟ = ∑ dZ i exp⎜− ⎟ account, the partition function should be modified: i ⎝ BTk ⎠ 6.13 eV Example: The energy levels of an in the : ε ()n −= i n 2 ε i r where ni = 1,2,... is the principle (these levels εi 2 are obtained by solving the Schrödinger equation for the di =2ni Coulomb potential). In addition to ni, the states of the electron ε2 d2 =8 in the H atom are characterized with three other quantum

numbers: the orbital quantum number li max = 0,1, ..., ni – 1, the projection of the orbital momentum m = - l , - l +1,...0, l -1,l , ε1 d1 =2 li i i i i and the projection of spin si = ±1/2. In the absence of the (for a continuous external magnetic field, all electron states with the same ni are spectrum, we need degenerate (the property of Coulomb potential). The degree of another approach) degeneracy in this case: max nl i −= 1 (ni )+−+ 1121 2 di = ∑[]i 122 =+ nl i = 2ni lmin =0 2 Problem (final 2005, partition function) Consider a system of distinguishable particles with five microstates with energies 0, ε, ε, ε, and 2ε ( ε = 1 eV ) in equilibrium with a reservoir at temperature T =0.5 eV. 1. Find the partition function of the system. 2. Find the average energy of a particle. 3. What is the average energy of 10 such particles?

⎛ ε ⎞ ⎛ 2ε ⎞ ⎜ ⎟ ⎜ ⎟ Z1 exp31 ⎜−+= ⎟ exp⎜−+ ⎟ =++= 424.1018.0406.01 ⎝ BTk ⎠ ⎝ BTk ⎠ the average energy of a single particle: ⎛ ε ⎞ ⎛ 2ε ⎞ ε exp3 ⎜−× ⎟ ε exp2 ⎜−×+ ⎟ ⎝ BTk ⎠ ⎝ BTk ⎠ 0.406 + 0.036 = ∑ P()εεεii = 1 eV ×= = 310.0 eV i ⎛ ε ⎞ ⎛ 2ε ⎞ 1.424 exp31 ⎜−+ ⎟ exp⎜−+ ⎟ ⎝ BTk ⎠ ⎝ BTk ⎠ the same ∂Z ×exp31 (− βε )×(−ε )+ (− βε )×(− 22exp ε ) result you’d ε −= −= get from this: Z ∂β exp31 ()βε −+−+ 2exp βε ( )

the average energy of N = 10 such particles: 10 UNU ×== eV = 1.3310.010 eV Problem 1 (partition function, average energy) Consider a system of N particles with only 3 possible energy levels separated by ε (let the ground state energy be 0). The system occupies a fixed volume V and is in thermal equilibrium with a reservoir at temperature T. Ignore interactions between particles and assume that Boltzmann statistics applies. (a) (2) What is the partition function for a single particle in the system? (b) (5) What is the average energy per particle?

(c) (5) What is probability that the 2ε level is occupied in the high temperature limit, kBT >> ε? Explain your answer on physical grounds.

(d) (5) What is the average energy per particle in the high temperature limit, kBT >> ε? (e) (3) At what temperature is the ground state 1.1 times as likely to be occupied as the 2ε level?

(f) (25) Find the heat capacity of the system, CV, analyze the low-T (kBT<<ε) and high-T (kBT >> ε) limits, and sketch CV as a function of T. Explain your answer on physical grounds.

βε −− 2βε (a) = ∑ dZ i exp( βε i ) 1 ++=− ee i (b) 1 ∂ ( ε )eZ −βε (−+− 2ε )e−2βε −βε + 2ee −2βε ε −= −= = ε Z ∂β 1 βε ++ ee −− 2βε 1 βε ++ ee −− 2βε e−2βε − 21 βε 1 (c) P = ≈ ≈ all 3 levels are populated with 1 βε ++ ee −− 2βε −+−+ 2111 βεβε3 the same probability

(d) −βε + 2ee −2βε + 21 = εε ≈ = εε 1 βε ++ ee −− 2βε ++ 111 Problem 1 (partition function, average energy) 1 2ε (e) ()2exp βε =− βε = 1.1ln2 T = 1.1 kB 1.1ln dU d d εεdβ (f) C N === N V dT dT dβ dT ⎛ 1 ⎞⎪⎧()ε e−βε −+− (ε 22 e−2βε )()−βε 2 −2βε []()ε eee −βε −+−+ 2 (ε e−2βε )⎪⎫ Nε⎜−= ⎟ − ⎜ 2 ⎟⎨ βε −− 2βε βε −− 2βε 2 ⎬ ⎝ BTk ⎠⎩⎪ 1 ++ ee []1 ++ ee ⎭⎪ ⎛ Nε 2 ⎞⎪⎧[]−βε −2βε []14 βε −− 2βε ()−βε +−+++ 2 2 ()−− βεβε+ 2eeeeeeee −2βε ⎪⎫ = ⎜ ⎟ ⎜ 2 ⎟⎨ βε −− 2βε 2 ⎬ ⎝ BTk ⎠⎩⎪ []1 ++ ee ⎭⎪ Nε 2 ⎪⎧ −βε 4 −2βε −2βε 4 −3βε −3βε 4eeeeee −4βε −2βε −3βε −−−+++++ 44 eee −4βε ⎪⎫ = 2 ⎨ βε −− 2βε 2 ⎬ BTk ⎩⎪ []1 ++ ee ⎭⎪ Nε 2 −βε 4 −2βε ++ eee −3βε = 2 βε −− 2βε 2 C BTk []1 ++ ee V ε Nε 2 −βε 4 −2βε ++ eee −3βε Nε 2 − Low T (β>>ε): C = ≈ e BTk V 2 βε −− 2βε 2 2 BTk []1 ++ ee BTk Nε 2 −βε 4 −2βε ++ eee −3βε 2 Nε 2 high T (β<<ε): C = ≈ V 2 βε −− 2βε 2 2 BTk []1 ++ ee 3 BTk T H Problem (the average values) A is placed in a very tall container at the temperature T. The container is in a uniform gravitational field, the acceleration of free fall, g, is given. Find the average potential energy of the molecules. h

⎛ mgh ⎞ # of molecules within dh: ⎜ ⎟ ()ndhhhdN 0 exp⎜−=+↔ ⎟ )( dharea ⎝ BTk ⎠

0 mgH mgH H ⎛ mgh ⎞ BTk Tk BTk nmgh exp⎜−× ⎟ )( dharea ()exp()yyTk ×−× B dy exp()−× dyyy ∫ 0 ⎜ ⎟ ∫ B ∫ 0 ⎝ BTk ⎠ ⎡mgh ⎤ 0 mg 0 U = H ⎢ == y⎥ = mgH BTk ×= mgH ⎛ mgh ⎞ BTk ⎣ ⎦ BTk BTk n0 exp⎜− ⎟ )( dharea Tk ∫ ⎜ Tk ⎟ exp()y ×− B dy exp()− dyy 0 ⎝ B ⎠ ∫ ∫ 0 mg 0

A d A ∫ exp()dyyy []exp()yy exp()exp −×−−=−×→−× ()∫ dyyyy 0 dy 0 A A A []exp()yy A exp()exp −×−−=−× ()exp()AAdyyydyyydyy ()exp1exp −−+−×−=−× A () 0 ∫∫ ∫ 0 0 0 A () () For a very tall container (mgH/k T ): = TkU ∫ exp dyy A +−−=− 1exp B →∞ B 0 Partition Function for a (Pr. 6.9)

Any reference energy can be chosen. Let’s choose ε = 0 in the ground state: ε1=0, ε2=10.2 eV, ε3=12.1 eV, etc. The partition function: ε ⎛ ε ⎞ ⎡ ( − /116.13 neV 2 )⎤ ⎜ i ⎟ 2 i r = ∑ dZ i exp⎜− ⎟ = ∑ ni exp2 ⎢− ⎥ i ⎝ BTk ⎠ i ⎣ BTk ⎦

ε3= -1.5 eV (a) Estimate the partition function for a hydrogen atom ε = -3.4 eV 2 at T = 5800K (= 0.5 eV) by taking into account only three lowest energy states.

ε1= -13.6 eV - we can forget about the spin degeneracy – it is the same for all the levels – the only factor that matters is n2

⎛ ε1 ⎞ ⎛ ε 2 ⎞ ⎛ ε 3 ⎞ 0 − 4.20 − 2.24 −9 −10 Z exp⎜−= ⎟ exp4 ⎜−×+ ⎟ exp9 ⎜−×+ ⎟ eee ≈⋅+⋅+=×+×+= 1108.2105.5194 ⎝ BTk ⎠ ⎝ BTk ⎠ ⎝ BTk ⎠

However, if we take into account all discrete levels, the full partition function diverges: ⎡ ⎛ 1 ⎞⎤ ∞ ⎢ ⎜16.13 − 2 ⎟⎥ ∞ 2 ⎝ n ⎠ ⎛ 6.13 ⎞ 2 = ∑ nZ exp⎢− ⎥ exp⎜−> ⎟ ∑ n ∞= n=1 ⎢ BTk ⎥ ⎝ BTk ⎠ n=1 ⎣⎢ ⎦⎥ Partition Function for a Hydrogen Atom (cont.)

Intuitively, only the lowest levels should matter at ε >> kBT . To resolve this paradox, let’s go back to our assumptions: we neglected the term PdV in 1 dS ()+= PdVdU R T R R

⎛ PVE ⎞ If we keep this term, then + Boltzmann New Boltzmann expfactor ⎜−= ⎟ ⎝ BTk ⎠

For a H atom in its ground state, V~(0.1 nm)3 , and at the atmospheric pressure, PV~ 10-6 eV (negligible correction). However, this volume increases as n3 (the grows as n), and for n=100, PV is already ~1 eV. The PV terms cause the Boltzmann factors to decrease exponentially, and this rehabilitates our physical intuition: the correct partition function will be dominated by just a few lowest energy levels.