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## Part 2 Statistical Mechanics

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**Part 2Statistical Mechanics**PHYS 4315 R. S. Rubins, Spring 2011**Partition Function Z**Starting from the fundamental postulate ofequal a priori probabilities, the following are obtained: • the results of classical thermodynamics, plus their statistical underpinnings; • the means of calculating the thermodynamic parameters (U, H, F, G, S ) from a single statistical parameter, the partition functionZ (or Q), which may be obtained from the energy-level scheme for a quantum system. The partition function for a quantum system in contact with a heat bath is Z = i exp(– εi/kT), where εi is the energy of the i’th state.**Statistical Weight ω (or Ω)**The partition function for a quantum system in contact with a heat bath is Z = i exp(– εi/kT), where εi is the energy of the i’th state. The connection to the macroscopic thermodynamic function S is through the microscopic parameter Ω (or ω), which is known as thermodynamic degeneracy or statistical weight, and gives the number of microstates in a given macrostate. The connection between them, known as Boltzmann’s principle, is S = k lnω. (S = k lnΩis carved on Boltzmann’s tombstone).**The Fundamental Postulate 1**The foundation of statistical mechanics is theFundamental Postulate: an isolated system is equally likely to be in any of its accessible states. Quantum case Consider a system of N weakly interacting spin ½ particles of magnetic moment m in a strong uniform magnetic field B. Each particle has a magnetic energy of + mB or – mB. If n particles are in the lower state, then the statistical weight is ω = N!/n!(N – n)! .**The Fundamental Postulate 2**Classical case For non-discrete states, an artificial counting method is used. Let ω(U,V,N,α) be the number of microstates in the macrostate (V,N,α), with the energy lying in the range U to U + U, with U >> U >> U, where U is the separation of quantum levels. α represents one or more variables, for example the energy of one part of an isolated system separated by a partition. For each particle of an ideal gas, counting is achieved by dividing phase space into elemental volumes, given by pxx ~ ћ.**Sharpness of Distribution 1**The sharpness of the peak of the distribution ω(α) is given by *α/<α>, where *α is the rms deviation <(α)2>. We show later that *α/<α> ~ 1/N, where N is the number of particles. For N ~ 1024, *α/<α> ~ 10–12, so the probability of obtaining a value of α other than the mean is negligibly small. If the particle starts in a state with an accessible value of α for which ω(α) is negligibly small, the fundamental postulate states that it will be found with essentially 100% probability of being at the value of α corresponding to ωmax. Thusωωmax , so thatω 0andω/α 0. Letting S = k lnω, we obtainS Smax , S 0 andS/α 0.**Thermal Equilibrium**Equilibrium between two parts of an isolated system with a diathermal wall U = U1 + U2, V = V1 + V2, where U, V, V1, and V2 are constants. The variable parameter α is U1, withU1 = – U2. Statistical weights are multiplicative, so thatω = ω1ω2. Since S = k lnω, S = S1(U1,V1) + S2(U2,V2). The equilibrium condition is given by ω/α = 0, S/α = 0, where α = U1. Thus S/U1 = 0, so that S1/U1 + S2/U1 = 0, S1/U1 = S2/U2. The absolute temperature is defined as1/T S/U. Note:The thermodynamic identity is dU = T dS – P dV; so that 1/T S/U.**Thermal and Mechanical Equilibrium**Thermal and Mechanical Equilibrium in a two-part system U = U1 + U2, V = V1 + V2, where just U and V are constants. The variable parameters are U1and V1, with U1 = – U2 andV1 = – V2. The equilibrium condition may be written as dS = 0, where dS = (S1/U1 + S2/U1)dU1 + (S1/V1 + S2/V1)dV1 = 0. So that dS = 0 only if S1/U1 = S2/U2and S1/V1 = S2/V2. The equilibrium condition givesT1 = T2, where 1/T = S/U. The pressure is defined asP T S/V, so thatP1 = P2, Note:P is defined to agree with the standard definition of pressure obtained from TdS = dU + PdV.**Diffusional Equilibrium in a one-component system**U = U1 + U2, V = V1 + V2, N = N1 + N2, where U, V and N are constants. The variables are U1, V1and N1, with U1 = – U2, V1 = – V2and N1 = – N2. The equilibrium condition may be written as dS = 0, so that (S1/U1 + S2/U1)dU1 + (S1/V1 + S2/V1)dV1+ (S1/N1 + S2/N1)dN1 = 0. Thus dS = 0 only if S1/U1 = S2/U2, S1/V1 = S2/V2and S1/N1 = S2/N2. Diffusional Equilibrium 1**Diffusional Equilibrium 2**Maximizing S requires S1/U1 = S2/U2, S1/V1 = S2/V2, S1/N1 = S2/N2. Thermal equilibrium requiresT1 = T2, where1/T = (S/U)V,N. Mechanical equilibrium requiresP1 = P2, whereP = T (S/V)U,N. Diffusional equilibrium requires μ1 = μ2, whereμ – T (S/N)U,V μ is known as thechemical potential. The thermodynamic identity becomes dU = T dS – P dV + μ dN. Likewise dG(T,P,N) = – S dT + V dP + μ dN, so thatμ = (G/N)T,P = g(T,P), and dF(T,V,N) = – S dT – P dV + μ dN, so thatμ = (F/N)T,V = f(T,V).**The Boltzmann Distribution 1**Single-particle system • Consider a quantum system with energies Er (r =1,2 …) in contact with a heat bath of energy Ub at temp.T0. • Assume that the system is in the specific quantum state r, so that the total energy U of the system & bathis U = Ub+ Er Ub= U – Er . • Sinceωs = 1, the total statistical weight is ωtot = ωsωb = 1.ωb(Ub) = ωb(U – Er).**The probability of finding the system in the state r is**proportional to the total statistical weight; i.e. prω. Now pr ωtot = ωb(U – Er). Thus Stot = k ln ωtot, pr exp(Sb/k), where Sb = Sb(U – Er). SinceU >> Er, a Taylor expansion may be used; i.e. f(x) = f(xo) + (x – xo)f’. Thus, Sb(U – Er) Sb(U) – Er(S/Ub) = Sb(U) – Er/T0, since 1/T0 = S/Ub The Boltzmann Distribution 2**The Boltzmann Distribution 3**• Now pr = K exp[Sb(U) – Er/T0))/k] = K’exp(– Er/kT0), where K and K’ are constants. • If the probabilities are normalized,Σrpr = 1 . • The normalized Boltzmann distribution is, pr = exp(– Er/kT0)/Z, where thepartition function Z is given by Z = rexp (– Er/kT0). • Although introduced here only to normalize pr, the partition function Z may be considered as the central concept of statistical mechanics, since it links the microscopic and macroscopicviewpoints. a**The Boltzmann Distribution 4**• If the energy states UR of a quantum system is known, then the partition function Z may be determined. • For a sufficiently low temperature T, the factor exp(–UR/kT) becomes negligible, so that often only the lowest energy levels need to be considered in calculating Z. • For a system of N very weakly interacting, like particles, each with a partition function z, the partition function of the total system is given by Z = zNfor distinguishable particles (e.g. solids); Z = zN/Nfor indistinguishable particles (e.g.fluids). • For very large N, all the thermodynamic parameters (U, H, F, G, S ) may be easily obtained from Z.**Relation of Z to Classical Parameters**Summary of results to be obtained in this section <U> = – ∂(lnZ)/∂β = – (1/Z)(∂Z/∂β), CV = <(ΔU)2>/kT2, where β = 1/kT, with k = Boltzmann’s constant. S = kβ<U> + k lnZ , where <U> = U for a very large system. F = U – TS = – kT lnZ, From dF = S dT – PdV, we obtain S = – (∂F/∂T)Vand P = – (∂F/∂V)T . Also, G = F + PV = PV – kT lnZ. H = U + PV = PV – ∂(lnZ)/∂β.**Z = zNfor distinguishable particles (e.g. solids);**Z = zN/Nfor indistinguishable particles (e.g.fluids). <u> = – ∂(lnz)/∂β = – (1/z)(∂z/∂β), U = N<u>. cV = <(Δu)2>/kT2, CV =NcV, CP =NcP. Distinguishable particles: F = Nf = – kT ln zN = – NkT lnz. Since F = U – TS, so that S = (U – F)/T or S = – (∂F/∂T)V. Indistinguishable particles: F = – kT ln(zN/N) = – kT [ln(zN)– ln N] = – NkT [ln(z/N)– 1], Since for very large N, Stirling’s theorem gives ln N! = N lnN – N. Also, S = – (∂F/∂T)Vand P = (∂F/∂V)Tas before. Systems of N Particles of the Same Species**Mean Energies and Heat Capacities**• Equations obtained from Z = rexp (– Er), where = 1/kT. • U = rprEr/rpr= – (ln Z)/ = – (1/Z) Z/ . • U2 = rprEr2/rpr = (1/Z) 2Z/2. • Un = rprErn/rpr= (–1)n(1/Z) nZ/n. • (ΔU)2 = U2 – (U)2 = 2lnZ/2 or – U/ . • CV = U/T = U/ . d/dT = – k2. U/, or CV = k2 (ΔU)2 = (ΔU)2/kT2; i.e. (ΔU)2 = kT2CV. Notes Since (ΔU)2 ≥ 0, (i) CV≥ 0, (ii) U/T ≥ 0.**Sharpness of Distribution 2**Replacing α by U, the sharpness of the distribution is given by *U/<U>, where *U =<(U)2>, so that *U/<U> = (kT2CV)/<U>. For a system of N identical particles, <U> ~ f kT, where f = 3N is the number of degrees of freedom of the system (equipartition theorem). CV = dU/dT ~ f k, so that *U/<U> ~ (f k2T2)/<f kT> ~ f –1/2. Thus for a very large system, the peak value of U is the only value that a measurement could give, allowing the mean value <U> of statistical mechanics to be equated to the internal energy U of classical thermodynamics.**Estimating Mean Energies and Entropies 1**Basic Rules • Very low temperatures As T approaches 0 K, only the ground level of a quantum system is occupied, so that for a single-particle system, the mean energy <ε> = ε1 (the energy of the ground-level), the statistical weight ω = g1 (the degeneracy of the ground level), so that s = k ln g1. • Very high temperatures As T becomes so large that kT becomes much larger than the relative energy (ε – ε1) of the state of highest energy, all states approach equal occupancy, so that <ε> is just the mean energy calculated by assuming all states to be equally occupied, and ω is simply the number of total number of states of the system.**Estimating Mean Energies and Entropies 2**εigi Example ε32 ε23 ε11**Use of Equivalent Temperatures 1**• It is sometimes convenient to write energies in terms of equivalent temperatures; i.e. εi = kθi, where k is Boltzmann’s constant and θi is a temperature in K. Diatomic molecule**Use of Equivalent Temperatures 2**θigi Example 500 K2 10 K3 0.1K1**Relationship between Entropy and Probability**• Consider an ensemble of n replicas of a system. • If the probability of finding a member in the state r is pr, the number of systems that would be found in the r’th state is nr= n pr, if n is large. • The statistical weight of the ensemble Ωn(n1 systems are in state 1, etc.), is Ωn= n/(n1 n2…nr..), so thatSn= k ln n – k r ln nr. • From Stirling’s theorem, ln n ≈ n ln n – n, r ln nr ≈ rnr ln nr – n. Thus Sn= k {n ln n – rnr ln nr} = k {n ln n – rnr ln n – r nr ln pr}, so thatSn= – k r nr ln pr= – kn rpr ln pr. For a single system, S = Sn/n ; i.e.S = – k r pr ln pr.**Ensembles 1**A microcanonical ensembleis a large number of identical isolated systems. The thermodynamic degeneracy may be written asω(U, V, N). From the fundamental postulate, the probability of finding the system in the state r ispr = 1/ω. Thus, S = – k rpr ln pr = k r(1/ω) ln ω = (k/ω) ln ωr1 = k ln ω. Statistical parameter: ω(U, V, N). Thermodynamic parameter: S(U, V, N) [T dS = dU – PdV + μdN]. Connection: S = k ln ω. Equilibrium condition: S Smax.**Ensembles 2**Acanonical ensembleconsists of a large number of identically prepared systems, which are in thermal contact with a heat reservoir at temperature T. The probability pr of finding the system in the state r is given by the Boltzmann distribution: pr = exp(– Er)/Z, where Z = rexp(–Er), and = 1/kT. Now S = – k rpr ln pr =– k r [exp(–Er)/Z] ln[exp(–Er)/Z] = – (k/Z) rexp(–Er) {ln exp(–Er) – ln Z} = (k/Z) rErexp(–Er) + (k lnZ)/Z . rexp(–Er), so that S = k U + k lnZ = k lnZ + kU. Thus, S(T, V, N) = k lnZ + U/T and F = U – TS = – kT lnZ.**Ensembles 3**S(T, V, N) = k lnZ + U/T , F = U – TS = – kT lnZ. Statistical parameter: Z(T, V, N). Thermodynamic parameter: F(T, V, N). Connection: F = – kT ln Z. Equilibrium condition: F Fmin. A grand canonical ensembleis a large number of identical systems, which interact diffusively with a particle reservoir. Each system is described by agrand partition function, G(T, V, μ) = N{r(μN – EN,r)}, where N refers to the number of particles and r to the set of states associated with a given value of N.**Entropy of an Ideal Monatomic Gas 1**We wish to find a general expression ω(U,V,N) for a system of N weakly-interacting particles of an ideal monatomic gas, confined to a volume V, with the total energy in the range U to U + U. Since U = p2/2m, the total momentum lies in the ranges ± p to ± (p + p). In order to count microstates, we imagine the phase-space to be divided into cells of “area” ΔpiΔqi ≈ h, per degree of freedom. N particles moving in 3 dimensions is represented by a point in 6N-dimensional phase space, while the number of degrees of freedom of the system is f = 3N. Note that f may be more than 3N if a particle is vibrating in simple harmonic motion, since degrees of freedom are associated both with kinetic energy and with vibrational potential energy.**Entropy of an Ideal Monatomic Gas 2**A state describing N particles moving in 3 dimensions is a point in 6N-dimensional phase space, The number of degrees of freedom of the system is f = 3N. For f = 1 (one particle moving in one dimension), the relation U = p2/2m means that p lies between the values ± p to ± (p + p). Thus ω(U,V,N) = 2 p L/h, where is the length of the container, so that ω L = V1/3and ω p0 U0 .**Entropy of an Ideal Monatomic Gas 3**• f = 1 (1 atom in 1 dimension) ω is the number of squares contained in the two rectangles; i.e. ω= 2 p L/h p0L U0V1/3. • f = 2 (1 atom in 2 dimensions or 2 atoms in 1 dimension) ω is the number of squares contained in the area between the circles; i.e. ω= 2πp p L2/h2 p1L2 U1/2V2/3. • f = 3 (1 atom in 3 dimensions or 3 atoms in 1 dimension) ω is the number of squares contained in the area between the spheres; i.e. ω= 4πp2p L3/h3 p2L3 U1V.**Entropy of an Ideal Monatomic Gas 4**• General result for f degrees of freedom ω p(f – 1)Lf or p(f – 1)Vf/3. ω (mU)(f – 1)/2Vf/3 = (mU)(3N – 1)/2VN ≈(mU)3N/2VN for large N. Thus ω(U,V,N) = A(N) m3N/2 U3N/2 VN For a single species of mass m, we may write B(N) = A(N) m3N/2. Thusω(U,V,N) = B(N) VN U3N/2, so that S(U,V,N) = k [C(N) + N lnV + (3N/2) lnU]. Assuming that S is an extensive function, we have S(U,V,N) = N s(U,V) = Nk [K + lnv + (3/2) ln u]. Using the results u =(3/2)kT and v = V/N, we obtain S(U,V,N) = Nk [D + (3/2)lnT + ln(V/N)].**Mixing of gases 1: Gibbs’ Paradox**If N molecules of an ideal monatomic gas make a free expansion from a V to 2V, ΔU = 0, so that ΔS = Sf – Si = Nk ln(2V/V) = Nk ln 2. If the container of volume 2V is now redivided into equal parts, so that (N/2) molecules are in each half, then ΔS’ = Sf’ – Si’ = 2(N/2)k ln(V/2V) = – Nk ln 2. This is Gibbs’ paradox, which is removed by assuming that the molecules are indistiguishable; i.e. that ω(U,V,N) = B(N) U3N/2 VN/N!. The term Nk lnV in the expression for S is replaced by Nk ln(V/N!), where (by Stirling’s theorem), Nk ln(V/N!) ≈ Nk [ln(V/N) + 1].**Entropy of an Ideal Monatomic Gas 5**• If S is assumed to be an extensive quantity, Gibbs’ paradox may be avoided. • Letting S = Ns, with s(U,V) = k [lnC + lnv +(3/2) lnu], where C = B(N)/N, v = V/N, and u = U/N. • Thus, S(U,V,N) = Nk [lnC + (3/2)ln(U/N) + ln(V/N)]. • Temperature is defined as 1/T (S/U)V,N = (3/2)Nk/U, since U = (3/2)NkT, so that S(U,V,N) = Nk [D + (3/2)lnT + ln(V/N)]. • Note that P/T = (S/V)T,N = Nk/V, so that PV = NkT. Mixing gases • Gas A in the left compartment, a different gas B in the right compartment. ΔStot = ΔS1 + ΔS2. b. Gas A in the left compartment, the same gas B in the right compartment. ΔStot = Sfinal– S1initial– S2initial.**Some Useful Integrals**In polar coordinates, r varies from 0 to . In Cartesian coordinates x, y and z vary from – to +.**Maxwell Velocity Distribution 1**• One molecule of a monatomic ideal gas has internal energy = ½ mv2, so that the probability of a molecule lying in therangevtov + dvis P(v)d3v = exp(– /kT) d3v /d3v exp(– /kT), since its energy does not involve its position coordinates. • TheMaxwell velocity distributionfunctionf(v)d3v is the number of molecules per unit volume with velocities from vto v + dv. • Thedistribution function g(vx)dvxfor the velocity component vxis the number of molecules per unit volume with components in the rangevxtovx + dvx. • The Maxwell Speed Distribution F(v) dv is the number of molecules per unit volume with speeds between vand v + dv, regardless of direction.**Maxwell Velocity Distribution 2**• For an ideal monatomic gas in uniform surroundings, theMaxwell velocity distributionfunctionf(v) d3v is the number of molecules per unit volume with velocities between vand v + dv. • Now f(v)d3v = (N/V) P(v)d3v, where P(v)d3v = exp(– mv2/2kT) d3v / d3v exp(– mv2/2kT). • Thus f(v)d3v =(N/V) exp(– mv2/2kT) d3v / d3v exp(– mv2/2kT) • Since d3v = dvxdvydvz, f(v)d3v = (N/V)exp(– mv2/2kT) d3v /[–+dvx exp(– mvx2/2kT)]3. • Now I0(a) = ½ (π/a)1/2, so that f(v)d3v = (N/V)(m/2πkT)3/2exp(–mv2/2kT) d3v.**Distributions of Velocity Components 1**• Thedistribution function g(vx)dvxfor the velocity component vxis the number of molecules per unit volume with components in the rangevxtovx + dvx. • To obtain g(vx)dvx, the Maxwell velocity distribution f(v)d3v is integrated over all values ofvy and vz.**Distributions of Velocity Components 2**• The distribution of velocity components is, g(vx)dvx = (N/V)(m/2πkT)1/2exp(–mvx2/2kT) dvx. • This is a Gaussian distribution, symmetrical about the mean value, <vx>= 0, with a root-mean-square(rms)deviation, √(Δvx2) = √(vx2) = √(kT/m).**Maxwell Speed Distribution 1**• TheMaxwell Speed Distribution F(v) dv is the number of molecules per unit volume with speeds between vand v + dv, regardless of direction. • F(v) dv is obtained by working in polar coordinates and integrating f(v) d3v over all orientations (θ, Φ), so that the volume element d3v is replaced by v2sinθ dΦ dθ dv. Integration over all anglesgives F(v)dv = 4πv2f(v)dv. The factor v2 present in F(v) dv, makes the distribution unsymmetrical. F(v)dv = 4π(N/V)(m/2πkT)3/2 v2 exp(– mv2/2kT) dv.**Maxwell Speed Distribution 2**(V/N) vpeak= 1.41 √(kT/m), <v> = 1.60 √(kT/m), vrms = 1.73 √(kT/m).**Partition Function of an Ideal Monatomic Gas 1**z = iexp(– i) → d3x d3p exp(– )/h3(classical system).**Entropy of an Ideal Monatomic Gas 6**Z =zN/N!, where z = V(2πmkT/h2)3/2, and U = (3/2)N kT. Also S = k(lnZ + U) = k [ln zN – lnN! + (3/2)N], = Nk [lnz – lnN + 1 + 3/2]. Thus S = Nk [K + ln(V/N) + (3/2) ln T], with K = ln (2πmk/h2)3/2+ 5/2.**Degrees of Freedom**• Applying the equipartition theorem is simplified by using the concept of degrees of freedom f,which is the minimum number of independent coordinates needed to the specify the motion of a system of particles.* • For a single atom (assumed to be a point), there are 3 directions of motion so that f =3. • For a molecule consisting of n point-atoms, it is necessary to separate translational, rotational and vibrational motions; however the total number of degrees of freedom must be 3f. • For linear molecules, ftrans= 3, frot= 2 (since there can be no rotation about the axis for point-atoms), so that fvib= 3n – 5. • For non-linear molecules, ftrans= 3, frot= 3, so that fvib= 3n – 6. * Beware! Some texts, such as Carter, define f differently.**Equipartition Theorem : Examples**Harmonic oscillator in one dimension = p2/2m + ½ k’ x2, where k’ is the spring constant. • = ½ kT + ½ kT = kT. Ideal diatomic gas 50 Kvib,rotT trans, so that = (3/2)kT and CV = (3/2)nR. 500 KvibT trans,rot, so that = (5/2)kT and CV = (5/2)nR. CP = CV + nR = (7/2)nR(Mayer’s equation), so thatγ= 7/5 = 1.2. 5000 K(assuming no dissociation)T trans,rot,vib, so thatCV = (7/2)nR.**Harmonic Oscillator in One Dimension 1**For a 1D harmonic oscillator, n = (n + ½) ħω, where ω = √(k/m).