Solution to the exam in TFY4230 STATISTICAL PHYSICS Wednesday december 1, 2010

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1 NTNU Page of 6 Institutt for fysikk Fakultet for fysikk, informatikk og matematikk This solution consists of 6 pages. Solution to the exam in TFY423 STATISTICAL PHYSICS Wenesay ecember, 2 Problem. Particles in a spherical volume A system of N classical non-relativistic particles is confine to a spherical 3-imensional) volume with soft walls, escribe by the Hamiltonian N x 2 ) n H = 2m p2 i + ε i r 2, ) i= where ε is a positive constant, r is a length characterizing the raius of the sphere, an n is a positive integer. a) Write own the canonical partition function Z for this system at temperature T. Since all particles have the same finite mass m it is very likely that they are ientical. Hence Z = N! i 3 p i 3 [ x i h 3 e βh = h 3N N! ] N 3 p e βp2 /2m 3 x e βεx2 /r) 2 n. 2) b) Calculate the internal energy U = H an heat capacity C for this system. Since we have H = Z β Z = ln Z, β we only nee to factor out the β-epenence of the integrals. As simple way to o this is by introucting new integration variables, π = β /2 p an ξ = β /2n x. Since 3 p = β 3/2 3 π an 3 x = β 3/2n 3 ξ this gives Z = β 3N/2 3N/2n) Z, where Z oes not epen on β. It follows that H = 3 + ) N 2 n C = T H = 3 2 β ln β = n + ) Nk B T, 3) n ) Nk B. 4) c) Does your result for C agree with the equipartition theorem when n = or n =? The case n = correspons to N three-imensional oscillators, each contributing 3k B to the heat capacity accoring to the equipartition theorem. The case n = correspons to N particles in a volume with har walls, each contributing 3 2 k B to the heat capacity accoring to the equipartition theorem. The result 4) agrees with these statements.

2 Solution TFY423 Statistical Physics, Page 2 of 6 ) Calculate the mean particle ensity, efine as N ρx) = δx x i ). 5) We have ρx) = Z N i= N! i= N j= 3 p j 3 x j h 3 δx x i )e βh. Due to the factorize form of the integran most factors of the integral cancels against ientical factors in Z, leaving N ientical contributions, ρx) = N 3 x δx x ) e βεx2 /r2 ) n = N /r) 2 n Z Z e βεx2. 6) Here the normalization factor Z is the single uncancelle factor of Z, ) /2n Z = 3 x e βεx2 /r2 ) n 4π t = βε 2n r3 t t3/2n e t = βε Note that βε ) /2n, 2n Γ ) 3 2n 3, an Z 4π 3 r3 when n. ) /2n 4π 3 2n r3 Γ 2n ). 7) Next assume the particles to have charge Q measure in units of the positron charge e, an that the system is expose to a magnetic fiel B = A. This implies that we must make the substitution p i p i + QeAx i ) 8) in the Hamiltonian ). e) What is the effect of this magnetic fiel on the classical partition function Z? There is no effect of a magnetic fiel in classical statistical mechanics. This is known as the Bohr van Leuween theorem pointe out by Niels Bohr in his octoral issertation of 9 before the avent of quantum mechanics). A simple proof is that we may introuce new momentum integration variables, π i = p i + QeAx i ) in the partition function integrals, thereby removing every trace of the magnetic fiel from the integran. The Gamma function: Γν) = t t tν e t, Γ + ν) = ν Γν), 9) Γ) =, Γ 2 ) = π, Γν) = ν + when ν. ) Problem 2. Monte-Carlo simulation of a thermal system Here you shoul to prepare for a numerical simulation of the system iscusse in the previous problem, for the case of N = an n = 2. We further simplify the system to be one-imensional. a) Write own the classical equations of motion ictate by the Hamiltonian ). After reuction to one space imension one obtains ẋ = H p = p m, ) ṗ = H ) x = 2nε x 2 n x. 2) r Remark: The three-imensional version of these equations is not much ifferent, r 2 ẋ = p m, 3) ṗ = 2nε r x 2 ) n x. 4) r 2

3 Solution TFY423 Statistical Physics, Page 3 of 6 b) Fin suitable units for time an length so that the equations of motion can be written in terms of imensionless variables. It seems obvious that r must be a suitable unit of time. I.e., we write x = r ξ with ξ imensionless. It follows from equation ) that ε has imension energy, i.e. that ) ε/mr 2 /2 has imension time an coul serve as a suitable unit of time. However, it seems that mr t = 5) 2nε is a slightly better choice. We write t = t τ with τ imensionless, so that t = t τ. A natural unit of momentum then is p = mr t. Hence we write p = p η with η imensionsless. This leas to the imensionless equations ξ = η, τ 6) τ η = ξ2n. 7) Remark: The fastest way to solve this problem, completely acceptable in fact the recommene one when time is scarce), is to say that we may choose units for length so that r =, for mass so that m =, an for energy so that 2nε =. c) How woul you iscretize the ifferential equations for a numerical solution of the problem? We sample the function at iscrete times τ k = k τ, an approximate the time erivative with the iscrete ifference, t ξτ) = ξ k+ ξ k, with ξ k ξk τ), 8) τ=k τ τ an similar for ητ). This leas to the ifference equations which can be solve iteratively. ξ k+ = ξ k + τ η k, 9) η k+ = η k τ ξ 2n k. 2) ) To simulate temperature one has to introuce aitional fluctuating an a amping forces. Inicate how this shoul be one. In aition to the force H x we shoul a a issipative amping) force Γ p an a completely ranom fluctuating) force F. In imensionless form this changes the ifference equations to ξ k+ = ξ k + τ η k, 2) η k+ = η k τ ξ 2n k γ τ η k + τf k, 22) where the f k s are ranom numbers generate inepenently for each k, an γ is a imensionless parameter. Hamilton s equations: ẋ α = H p α, ṗ α = H x α. 23)

4 Solution TFY423 Statistical Physics, Page 4 of 6 Problem 3. Quantum statistics of thermal raiation The eigen-energies for the free raiation fiel can be written E = ω k Nk, r) 24) k,r where ω k = c k, an where Nk, r) =,,... is the occupation number of the state with wavevector k an polarization r. We have subtracte the zero-point energy. With av volume V an perioic bounary conitions the allowe values for k lie on a lattice, k = 2π nx, ny, nz) V /3 with all n s integer. 25) a) Show that the partition function for this system can be written ln Z = ) ln e β ω k. 26) k,r Backgroun: The following general backgroun was not expecte as part of the solution; it is inclue as a review of the concepts involve. The quantum partition function can in general be written Z = E e βe, where the sum runs over all possible eigenenergies E. You shoul be familiar with the fact that the eigenstates are usually labele by several quantum numbers, like n the principal quantum number), l the total angular momentum quantum number) an m the magnetic quantum number labels the z-component of the angular momentum vector) in atomic physics. Likewise the states of a 3-imensional harmonic oscillator may be labelle by nonnegative integer quantum numbers N x, N y, an N z escribing excitations of the oscillator in respectively the x-, y-, an z-irections. In the latter case the eigen-energies of the system is E = E Nx,N y,n z = ω x N x + ω y N y + ω z N z ) + E = ω α N α + E α=x,y,x where the secon term of each line is the zero-point energy E = 2 α=x,y,x ω α. Ignoring the zero-point energy, the partition function can be written Z = e βenx,ny,nz = e β ωxnx+ωyny+ωznz) N x,n y,n z N x,n y,n z = e β ωxnx e β ωyny e β ωznz N x= = α=x,y,z N α= N y= e β ωαnα = α=x,y,z N z= e β ωα ). I.e., since the logarithm of a prouct is the sum of logarithms of its factors, ln Z = ln e β ωα). α=x,y,z The eigenstates of the raiation fiel is like those of the 3-imensional oscillator, except that we on t have 3 but infinitely many irections each irection labele by a wavenumber k an a polarization r which together specifies a possible propagation moe of the electromagnetic fiel). The occupation number Nk, r) then specifies the excitation of that moe.

5 Solution TFY423 Statistical Physics, Page 5 of 6 Solution: We have Z = e β ωk = e β ω, 27) k ) k,r Nk,r)= k,r so that ln Z = k,r ln e β ω k). 28) b) Explain why the the average occupations numbers can be written as Nk, r) = ln Z. 29) 2 β ω k Since the occupation probabilities of ifferents moes are inepenent we have Nk, r) = Zk, r) Zk, r) = Nk,r)= Nk,r)= Nk, r) e β ω knk,r), e β ω knk,r). with I.e., Nk, r) = Zk, r) = ln Zk, r) = ln Z. 3) Zk, r) β ω k β ω k 2 β ω k There are two terms in ln Z which epens on ω k, one for each value of r. The factor 2 compensates for this. c) Fin an explicit expression for Nk, r). We fin Nk, r) = ln e β ω ) k = β ω k e β ω. 3) k ) ) To evaluate many physical quantities explicitly in the limit V one makes the substitution F k, r) V N 3 k F k, r), 32) k,r r vali for continuous functions F k, r). Explain the origin of this substitution. What is the imensionless number N? The vector k runs over the points of a cubic lattice, with volume k,r v = 2π)3 V of each elementary cell. As V becomes large the points becomes very close together, an we may approximate the sum by an integral, F k, r) = V 2π) 3 v F k, r) V 2π) 3 3 k F k, r), 34) r k where we in the last step have interprete the sum over k as a Riemann approximation to the integral. As V becomes very large this approximation becomes very goo. We have foun that r 33) N = 2π) 3. 35)

6 Solution TFY423 Statistical Physics, Page 6 of 6 e) In most of the universe the photons have a temperature T = K. How many photons N = k,r Nk, r) are there on average per m3? We fin N = 2 V 2π) 3 = π 2 V kb T c 3 k e β ω = 2 k ) 3 V 2π) 3 4π x 2 x e x = 2 π 2 ζ3) V k 2 k e β ck ) 3 kb T = ) c Some physical constants, an an integral: = Js, k B = J/K, c = m/s 37) x 2 x e x = 2ζ3) )