From quantum to classical statistical mechanics. Polyatomic ideal gas.

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1 From quantum to classical statistical mechanics. Polyatomic ideal gas. Peter Košovan Dept. of Physical and Macromolecular Chemistry Lecture 5, Statistical Thermodynamics, MC260P105, If you find a mistake, kindly report it to the author :-)

2 Newtonian approach to classical mechanics P. Košovan Lecture 5: Classical statistical mechanics 1/36 Basic equation of motion: dp dt ṗ = F = Ur) For m independent of time: dp dt = mdṙ = m r = ma dt Convenient in Cartesian coordinate system Cumbersome when Cartesian system is not natural to the problem

3 Newtonian approach to classical mechanics P. Košovan Lecture 5: Classical statistical mechanics 1/36 Basic equation of motion: dp dt ṗ = F = Ur) For m independent of time: dp dt = mdṙ = m r = ma dt Convenient in Cartesian coordinate system Cumbersome when Cartesian system is not natural to the problem Example 1: free fall Ux) = gx mẍ = mg xt) = 1 2 gt2 + v 0 t + x 0

4 Newtonian approach to classical mechanics P. Košovan Lecture 5: Classical statistical mechanics 1/36 Basic equation of motion: dp dt ṗ = F = Ur) For m independent of time: dp dt = mdṙ = m r = ma dt Convenient in Cartesian coordinate system Cumbersome when Cartesian system is not natural to the problem Example 1: free fall Ux) = gx mẍ = mg xt) = 1 2 gt2 + v 0 t + x 0 Example 2: harmonic oscillator Ux) = k 2 x x 0) 2 ; ξ = x x 0 ) 0 = ξt) + k m ξ ξt) = C sinωt + φ)

5 Newtonian approach to classical mechanics Example 3: centrosymmetric attractive potential in 2D Ur) = A/r, Fr) = Ar/r 3 Ax mẍ = F x = x 2 + y 2 ) 3/2, mÿ = F Ay y = x 2 + y 2 ) 3/2 Transformation to polar coordinates: x = r cosθ), y = r sinθ) 0 = m r θ 2 r) + Ar ) 2 cos θ mr θ + 2 θṙ) sin θ 0 = m r θ 2 r) + Ar ) 2 sin θ + mr θ + 2 θṙ) cos θ Multiply first equation by cos θ, the second by sin θ and add/subtract: 0 = m r + θ 2 r ) + A r 2, 0 = 1 r d dt mr2 θ) mr 2 θ = const. P. Košovan Lecture 5: Classical statistical mechanics 2/36

6 P. Košovan Lecture 5: Classical statistical mechanics 3/36 Newtonian approach to classical mechanics Example 3: centrosymmetric attractive potential in 2D contd.) On previous slide we obtained 0 = 1 r d dt mr2 θ) mr 2 θ = const. 0 = m r l2 mr 3 + A r 2 m r = A r 2 + l = mr 2 θ is the angular momentum l 2 /mr 3 is the centrifugal force. l2 mr 3 Newton is inconvenient for other than Cartesian coordinates. General case requires the introduction of additional forces to cast the equations in the form F η = m η

7 Lagrangian approach P. Košovan Lecture 5: Classical statistical mechanics 4/36 Define the Lagrangian: L{q}, { q}) K{ q}) U{q}) {q} arbitrary set of coordinates fully describing the system K{ q}) Kinetic energy U{q}) Potential energy as a function of position

8 Lagrangian approach P. Košovan Lecture 5: Classical statistical mechanics 4/36 Define the Lagrangian: L{q}, { q}) K{ q}) U{q}) {q} arbitrary set of coordinates fully describing the system K{ q}) Kinetic energy U{q}) Potential energy as a function of position Free fall problem with natural Cartesian coordinates {q} = {x, y}: L = m 2 ẋ2 + ẏ 2 ) Ux, y) L ẋ = K ẋ = mẋ, L x = U x

9 Lagrangian approach P. Košovan Lecture 5: Classical statistical mechanics 4/36 Define the Lagrangian: L{q}, { q}) K{ q}) U{q}) {q} arbitrary set of coordinates fully describing the system K{ q}) Kinetic energy U{q}) Potential energy as a function of position Free fall problem with natural Cartesian coordinates {q} = {x, y}: L = m 2 ẋ2 + ẏ 2 ) Ux, y) L ẋ = K ẋ = mẋ, m r = U L x = U x L t q j ) = L q j

10 Lagrangian approach P. Košovan Lecture 5: Classical statistical mechanics 4/36 Define the Lagrangian: L{q}, { q}) K{ q}) U{q}) {q} arbitrary set of coordinates fully describing the system K{ q}) Kinetic energy U{q}) Potential energy as a function of position Free fall problem with natural Cartesian coordinates {q} = {x, y}: L = m 2 ẋ2 + ẏ 2 ) Ux, y) L ẋ = K ẋ = mẋ, m r = U L x = U x L t q j ) = L q j Lagrange equations of motion

11 Lagrangian approach P. Košovan Lecture 5: Classical statistical mechanics 5/36 Centrosymmetric problem in polar coordinates {q} = {r, θ}: L = m 2 ṙ2 + r 2 θ2 ) + A r ; K = m 2 Apply Lagrange equations: ) L t θ = L mr 2 θ = 0 θ ) L = L t ṙ r m r = A r 2 + l2 mr 3 Same result as using Newton s equations Straightforward manipulations d dt r cos θ + r sin θ)2, U = A r

12 Lagrangian approach P. Košovan Lecture 5: Classical statistical mechanics 6/36 Irrespective of the coordinate system, Lagrange equations attain the form ) L = L t q j q j Generalized coordinates {q} With initial contitions {q0)}, { q0)} they fully and uniquely describe the evolution of a classical system Description equivalent with Newton s formulation No additional forces required In a suitable coordinate system Ux) can attain a simple form. Useful for solving problems in arbitrary geometries.

13 P. Košovan Lecture 5: Classical statistical mechanics 7/36 Hamiltonian approach Generalized momentum conjugate to generalized coordinate q j : p j = L q j Hamiltonian: H{p}, {q}) = j p j q j L{p}, {q}) For potential and kinetic energy of the form K = j a j {q}) q 2 j, U = U{q j }) p j = L q j = K q j = 2a j q j Which results in H = 2K K U) = K + U total energy of the system)

14 P. Košovan Lecture 5: Classical statistical mechanics 8/36 Hamiltonian approach If L is not an explicit function of time: using dh = j p j = L q j, we obtain q j dp j + j L t q j ) p j d q j j = L q j dh = q j dp j ṗ j dq j which we compare with dh = ) H dp j + H p j j q j j L q j d q j j ) dq j L q j dq j

15 P. Košovan Lecture 5: Classical statistical mechanics 8/36 Hamiltonian approach If L is not an explicit function of time: using dh = j p j = L q j, we obtain q j dp j + j L t q j ) p j d q j j = L q j dh = q j dp j ṗ j dq j which we compare with dh = ) H dp j + H p j j q j j L q j d q j j ) dq j L q j dq j to finally obtain H p j = q j, H q j = ṗ j

16 Summary of different formulations P. Košovan Lecture 5: Classical statistical mechanics 9/36 Newton: F x = mẍ Lagrange: ) L t q j Hamilton: H p j = q j, = L q j H q j = ṗ j Mathematically equivalent Generalized coordinates Generalized momenta H{p, q}) being the total energy is particularly important for statistical mechanics Just some illustrative examples without proofs!) If you want to learn more: Read Goldstein: Classical mechanics Attend the Theoretical mechanics course at MFF UK NOFY003)

17 The classical partition function conjecture) P. Košovan Lecture 5: Classical statistical mechanics 10/36 Quantum mechanics Sum over quantum states: q quant = j e βε j Classical mechanics Integral over classical states: q class =... e βhp,q) dp dq Translational partition function in the classical limit: H = 1 2m p2 x + p 2 y + p 2 z) q class = dq x dq y dq z exp βp2 x + p 2 y + p 2 ) z) dp x dp y dp z 2m ) ) 3 q class = V exp βp2 x) dp x = 2πmk B T ) 3/2 V 2m

18 The classical partition function conjecture) P. Košovan Lecture 5: Classical statistical mechanics 11/36 Quantum mechanics Translational partition function ) 2πmkB T 3/2 q trans = V h 2 They differ by a factor h 3 h stems from the QM treatment Classical mechanics Translational partition function q class = 2πmk B T ) 3/2 V Conjecture: in the high T limit, quantum and classical treatment yield the same expression for q, apart from the factor h s where s is the number of degrees of freedom DOF) per molecule

19 Classical partition function P. Košovan Lecture 5: Classical statistical mechanics 12/36 Rotation spherical top) Harmonic vibration H = 1 ) p 2 θ 2I + p2 φ sin 2 θ q rot... q rot 8π 2 Ik B T e βh dp θ dp φ dθdφ H = p2 2µ + k 2 x2 q vib... e βh dp x dx q vib k BT ν, ν = 1 ) k 1/2 2π µ

20 Classical partition function P. Košovan Lecture 5: Classical statistical mechanics 12/36 Rotation spherical top) H = 1 ) p 2 θ 2I + p2 φ sin 2 θ q rot... q rot 8π 2 Ik B T Conjecture: q = e βε j j e βh dp θ dp φ dθdφ... Harmonic vibration H = p2 2µ + k 2 x2 q vib... e βh dp x dx q vib k BT ν, ν = 1 ) k 1/2 2π µ e βh dp) s dq) s, where dp) s s dp j

21 Taking the conjecture further Identical non-interacting molecules: Q = qn N! = 1 N ) 1 N! h s... e βh dp) s dq) s Due to the symmetry with respect to permutation of molecules: Q = qn N! 1 h sn... e βh dp) sn dq) sn N! And further conjecture that this applies also to interacting molecules Q = e βe 1 j h sn... e βh dp) sn dq) sn N! j It turns out to be valid but we omit the proof To learn more, see Chapter 10-7 of McQuarrie P. Košovan Lecture 5: Classical statistical mechanics 13/36

22 P. Košovan Lecture 5: Classical statistical mechanics 14/36 Configuration integral Consider a classical Hamiltonian Hp, q) = 1 N p 2 2m x,j + p 2 y,j + p 2 ) z,j + Ux1, y 1, z 1,..., x N, y N, z N ) Very common case: Potential energy depends exclusively on positions Kinetic energy depends exclusively on momenta We can integrate over momenta Q = 1 ) 2πmkB T 3N/2 N! h 2 Z N = Z N Λ 3N N! where we introduced the classical configuration integral Z N =... e βu{q}) dq) 3N

23 Configuration integral P. Košovan Lecture 5: Classical statistical mechanics 15/36 Z N =... e βu{q}) dq) 3N For U = 0 we get Z N = V N and recover the ideal gas EOS. For U 0 interacting particles), Z N is N-dimensional integral monatomic species sn-dimensional integral polyatomic species Key quantity in statistical mechanics of fluids Very difficult to evaluate for N N A Cannot be easily computed or approximated, exceptions exist) Subject of research in equilibrium statistical mechanics from early days until present

24 Mixed classical and quantum treatment P. Košovan Lecture 5: Classical statistical mechanics 16/36 Some DOF cannot be treated classically electronic, vibrational) Others can be approximated classically translation, rotation, intermolecular interactions) H = H class + H quant, q = q class q quant, Q = Q class Q quant Q = Q quant h sn... N! e βh class dp class ) sn dq class ) sn Now s is the number of classically tractable DOF per molecule This approach can be used for linear polyatomics

25 Polyatomic ideal gas P. Košovan Lecture 5: Classical statistical mechanics 17/36 System consisting of N independent non-interacting molecules: QN, V, T ) = q trans q rot q vib q elec q nuc ) N N! qv, T ) = q trans q internal q trans q rot q vib q elec q nuc n atoms per molecule 3n coordinates degrees of freedom, DOF) describe each molecule Analogy with diatomics: rigid rotor harmonic oscillator approx. Generalized coordinates: 3 DOF for translation of centre of mass Linear molecules: 2 DOF for rotation, 3n 5) for vibrations Non-linear molecules: 3 DOF for rotation, 3n 6) for vibrations

26 Normal modes of vibration P. Košovan Lecture 5: Classical statistical mechanics 18/36 O C O O O O C O H H H H O C O O H H Suitable set of internal coordinates Normal modes are independent in the harmonic approximation. Frequencies determined from QM calculations or spectroscopically

27 Vibrations in polyatomics Normal coordinates, normal modes α = 3n 5) or α = 3n 6) independent harmonic oscillators ε = α n j )hν j ν j = 1 ) 1/2 kj 2π µ j Θ v,j = hν j k B q vib = α E vib = Nk B C v,vib = Nk B e Θ v,j/2t 1 e Θ v,j/t α Θv,j 2 + Θ v,je Θv,j/T 1 e Θ v,j/t α Θv,j T ) 2 e Θ v,j/t ) 1 e Θ v,j/t ) P. Košovan Lecture 5: Classical statistical mechanics 19/36

28 Rigid body rotation classical limit) Inertia tensor: Moments about Cartesian axes x, y, z): I xx = n m j y j y cm ) 2 + z j z cm ) 2), I yy =..., I zz =... Off-diagonal elements of the tensor: I xy = n m j x j x cm ) 2 y j y cm ) 2), I yz =..., I zx =..., I xz =..., I yx =..., I zy =... In a suitably chosen coordinate system X, Y, Z): principal moments: I XX = I A, I Y Y = I B, I ZZ = I C other moments vanish: I XY = I Y Z = I ZX = I XZ = I Y X = I ZY = 0 P. Košovan Lecture 5: Classical statistical mechanics 20/36

29 Rotations in polyatomics Linear polyatomics analogy with diatomics): q rot = 8π2 Ik B T σh 2 = T σθ r Non-linear polyatomics principal moments of inertia: I A, I B, I C. A = h2 8π 2, B = h2 I A 8π 2, C = h2 I B 8π 2 I C Θ A = 8π2 I A k B h 2, Θ B = 8π2 I B k B h 2, Θ C = 8π2 I C k B h 2, Special cases: Spherical top: I A = I B = I C Symmetric top: I A = I B I C Asymmetric top: I A I B I C P. Košovan Lecture 5: Classical statistical mechanics 21/36

30 P. Košovan Lecture 5: Classical statistical mechanics 22/36 Special cases Spherical top Θ A = Θ B = Θ C ): ε J = JJ + 1) 2 2I ω J = 2J + 1) 2 High T limit: q rot = 1 σ 1 σ 0 0 q rot = π1/2 σ 2J + 1) 2 e JJ+1) 2 /2Ik B T dj 4J 2 e J 2 2 /2Ik B T dj = π1/2 8π 2 Ik B T σ h 2 T 3 Θ A Θ B Θ C ) 1/2 ) 3/2

31 Special cases P. Košovan Lecture 5: Classical statistical mechanics 23/36 Symmetric top Θ A = Θ B Θ C ): ε JK = 2 2 JJ + 1) 1 + K 2 1 ) ) I A I C I A J = 0, 1, 2, ; ω JK = 2J + 1) High T limit: K = J, J 1,, J q rot = 1 +J 2J + 1) 2 e α AJJ+1) e α C α A )K 2, α j = 2 σ 2I j k B T J=0 K= J q rot = π1/2 8π 2 ) I A k B T 8π 2 ) I C k B T 1/2 σ h 2 h 2 = π1/2 T 3 ) 1/2 σ Θ A Θ B Θ C

32 P. Košovan Lecture 5: Classical statistical mechanics 24/36 Special cases Asymmetric top Θ A Θ B Θ C ): Very involved at quantum level. Can be solved numerically. Analytically solvable in the classical limit. High T limit: q rot = π1/2 σ 8π 2 ) I A k B T 1/2 8π 2 ) I B k B T 1/2 8π 2 I C k B T h 2 h 2 h 2 ) 1/2 Common formulation for all cases using rotational temperatures: q rot = π1/2 T 3 ) 1/2 σ Θ A Θ B Θ C E rot = 3 2 Nk BT, C rot V = 3 2 Nk B, S rot = Nk B ln π 1/2 σ T 3 e 3 Θ A Θ B Θ C ) 1/2 )

33 Thermodynamic functions of a non-linear polyatomic P. Košovan Lecture 5: Classical statistical mechanics 25/36 ) 2πMkB T 3/2 π q =V h 2 σ T 3 Θ A Θ B Θ C ) E Nk B T = n Θ v,j e Θ v,j/t 1 e Θ v,j/t C v = 3 Nk B n Θv,j T ) 1/2 3n 6 ) 2 e Θ v,j/t D e k B T 1 e Θ v,j/t ) ) e Θ v,j/2t 1 e Θ v,j/t ) ω e,1 e De/k BT

34 Thermodynamic functions of a non-linear polyatomic contd.) P. Košovan Lecture 5: Classical statistical mechanics 26/36 A Nk B T = ln 2πMkB T h 2 3n 6 S 2πMkB T = ln Nk B h 2 + 3n 6 ) 3/2 V e N π1/2 T 3 + ln σ Θ A Θ B Θ C ) Θ v,j 2T + ln1 e Θ v,j/t ) ) 3/2 V e 5/2 N + ln π1/2 e 3/2 σ Θ v,j /T e Θ v,j/t 1 ln1 e Θ v,j/t ) ) 1/2 + D e k B T + ln ω e,1 T 3 Θ A Θ B Θ C ) + ln ω e,1 ) 1/2

35 Molecular constants of selected polyatomics Table from McQuarrie, Statistical Mechanics, University Science Books 2000) P. Košovan Lecture 5: Classical statistical mechanics 27/36

36 Classical statistical mechanics: phase space P. Košovan Lecture 5: Classical statistical mechanics 28/36 Consider an N-body system: Configuration is fully described by l = sn spatial coordinates. Can be viewed as a single point in l-dimensional Euclidean configuration space {q} To describe its evolution we additionally need l momenta conjugate to {q} Phase point in 2l-dimensional Euclidean phase space {p, q} Time-evolution is a trajectory in phase space, governed by Hamilton s equations of motion H p j = q j, H q j = ṗ j, j = 1, 2,... l = sn

37 P. Košovan Lecture 5: Classical statistical mechanics 29/36 Density in phase space Microcanonical ensemble: A systems, same N, V, E Each system is a point in the same phase space Ensemble cloud of phase points Its evolution is described by a trajectory in phase space Trajectories are independent and deterministic. Each configuration conforming to the fixed N, V, E has a representative point in phase space Equal a priori probabilities: every classical state consistent with N, V, E is equally probable We can define density of phase points, fp, q, t):... fp, q, t) dp dq = A First postulate of statistical mechanics classical formulation): Φ = 1... Φp, q) fp, q, t) dp dq A

38 P. Košovan Lecture 5: Classical statistical mechanics 30/36 Density conservation Number of phase points N, V, E) inside a volume element δv = δp 1 δp 2... δp l 1 δp l δq 1... δq l 1 δq l δn = fp 1,..., p l, q 1,..., q l )δv Flux through face perpendicular to q 1 on a hypercube of volume δv : q 1 p 1,..., p l, q 1,..., q l ) fp 1,..., p l, q 1,..., q l )δq 2... δq l δp 1... δp l And through a parallel face at q 1 + δq 1 : q 1 p 1,..., p l, q 1 + δq 1,..., q l ) fp 1,..., p l, q 1 + δq 1,..., q l )δq 2... δq l δp 1... δp l Taylor-expanding f and q to linear terms yields f + f ) δq 1 q 1 + q ) 1 δq 1 δq 2... δq l δp 1... δp l q 1 q 1

39 P. Košovan Lecture 5: Classical statistical mechanics 31/36 Density conservation Subtracting fluxes through opposite faces gives in q 1 -direction f net flow = q 1 + f q ) 1 δq 1... δq l δp 1... δp l q 1 q 1 p and q have equal status in phase space, therefore in p 1 -direction f net flow = ṗ 1 + f ṗ ) 1 δq 1... δq l δp 1... δp l p 1 p 1 And the net flow through the whole hypercube of δv is l f net flow = q j + f q j + f ṗ j + f ṗ ) j δv q j q j p j p j Re-groupping the terms: dδn) l = f dt qj q j + ṗ j p j ) ) + f q j + f ṗ j δv q j p j

40 The Liouville equation P. Košovan Lecture 5: Classical statistical mechanics 32/36 In the equation dδn) dt = l qj f + ṗ ) ) j + f q j + f ṗ j δv q j p j q j p j the first term vanishes because H p j = q j, H q j = ṗ j q j q j + ṗ j p j = and dividing by δv we obtain the Liouville equation ) d δn = f dt δv t = l f q j + f ) ṗ j q j p j = 2 H p j q j l 2 H p j q j = 0 H p j f q j H q j f p j )

41 Implications of the Liouville equation P. Košovan Lecture 5: Classical statistical mechanics 33/36 Density of phase points is conserved in time f t = l f q j + f ) ṗ j q j p j df dt = 0 Cloud of phase points ensemble) behaves as an incompressible fluid Density in the neighbourhood of a phase point is constant in time Given initial conditions, the evolution is deterministic p = pp 0, q 0, t 0 ), q = qp 0, q 0, t 0 ) δpδq = δp 0 δq 0 Volume element around a given phase point is independent of the coordinate system δpδq = δp δq

42 P. Košovan Lecture 5: Classical statistical mechanics 34/36 Equipartition of energy Average energy per molecule: ε = He βh dpdq e βh dpdq For a Hamiltonian of the form m Hp 1,..., p l, q 1,... q l ) = a j p 2 j + n b j qj 2 + Hp m+1,... p l, q m+1... q l ) Equipartition: each of the quadratic terms in p and q contributes k B T/2 to the energy k B /2 to C v Can be generalized to include Hamiltonians of the form H = p2 x + p 2 y + p 2 z, H = 1 ) p 2 θ 2m 2I + p2 φ sin 2 θ

43 Equipartition of energy For the equipartition to hold, the contributing DOF have to be quadratic in p, q tractable classically, i. e., ε k B T Diatomic and polyatomic ideal gas: C v = 3 Nk B Θ v/t ) 2 e Θv/T e Θv/T 1) 2 C v = 3 Nk B n Θv,j T ) 2 e Θ v,j/t 1 e Θ v,j/t ) Vibrational levels are separated by ε > k B T Their contribution violates equipartition although U = kx x 0 ) 2 /2 It follows equipartition at Θ v T ), far above room temperature Violation of equipartition was a great concern in pre-quantum era P. Košovan Lecture 5: Classical statistical mechanics 35/36 )

44 Suggestions for individual projects P. Košovan Lecture 5: Classical statistical mechanics 36/36 Pen and paper: do some of the derivations which we omitted can be tedious!) Compute and plot populations of vibrational states of different molecules at a series of temperatures. Discuss the applicability of high-t approximation Compute the populations of rotational states for different molecules. Obtain the plot of the T-dependence of heat capacity of hydrogen, including ortho and para contribution Fig.6-9 from McQuarrie). Compare the applicability of various approximations direct sum, Euler-MacLaurin, integral). Select a suitable molecule and temperature can be several) where the approximations deviate. In any case, consult the topic with the lecturer before you start working on it

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