The Microcanonical Approach. (a) The volume of accessible phase space for a given total energy is proportional to. dq 1 dq 2 dq N dp 1 dp 2 dp N,

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1 8333: Statistical Mechanics I Problem Set # 6 Solutions Fall 003 Classical Harmonic Oscillators: The Microcanonical Approach a The volume of accessible phase space for a given total energy is proportional to Ω h H dq dq dq dp dp dp where the integration is carried out under the condition of constant energy H {q i p i } i [ p i m + mω qi ] ote that Planck s constant h is included as a measure of phase space volume so as to make the final result dimensionless The surface of constant energy is an ellipsoid in dimensions whose area is difficult to calculate However for the difference between volume and area is subleading in and we shall instead calculate the volume of the ellipsoid replacing the constraint H by H The ellipsoid can be distorted into a sphere by a canonical transformation changing coordinates to q i mωq i and p i p i mω The Hamiltonian in this coordinate system is H {q i p i } ω i p i + q i Since the canonical transformation preserves volume in phase space the Jacobian is unity we have Ω h H dq dq dp dp where the integral is now over the -dimensional hyper- sphere of radius R /ω As the volume of a d-dimensional sphere of radius R is S d R d /d we obtain Ω π! hω π hω!

2 The entropy is now given by S k B ln Ω k B ln b From the expression for temperature T S k B πe hω we obtain the energy and the heat capacity C k B c The single particle distribution function is calculated by summing over the undesired coordinates and momenta of the other particles Keeping track of the units of h used to make phase space dimensionless gives pp q dp dq H H dq h dq dp dp dp dq dq h dq dq dp dp dp h where p /m mω q / Using the results from part a pp q Ω hω π p hω! h π hω! ω p m + mω q π m mω q Using the approximation for and setting we have pp q ω π ω π exp p m + mω q p m + mω q

3 Let us denote p q by p q then pp q ω π exp p m mω q is a properly normalized product of two Gaussians The mean kinetic energy is p m while the mean potential energy is also mω q Quantum Harmonic Oscillators: pp q p m dqdp k BT pp q mω q dqdp k BT a The total energy of the set of oscillators is hω n i + Let us set the sum over the individual quantum numbers to i M n i Ē hω i The number of configurations {n i } for a given energy thus for a given value of M is equal to the possible number of ways of distributing M energy units into slots or of partitioning M particles using walls This argument gives to the number of states as Ω M +! M!! and a corresponding entropy M + S k B ln Ω k B [M + ln e b The temperature is calculated by T S k B M + hω ln M k B hω ln 3 M ln M ] e ln e Ē hω + hω k B hω ln + hω hω

4 By inverting this equation we get the energy hω exp hω/k BT + exp hω/ hω and a corresponding heat capacity C T hω k B [ ] + exp hω/ exp hω/ [exp hω/ ] c The probability that a particular oscillator is in its nth quantum level is given by summing the joint probability over states for all the other oscillators ie pn n+/ hω pn i Ω n + hω Ω {n i } [M n + ]! M!! M n!! M +! MM M n + M + M + M + n M + n M n where the approximations used are of the from I I for I Hence pn hω n + hω Ē hω n hω + hω + n which using T k B + hω hω ln hω Ē hω + hω n exp n hω leads to the probability d As found in part b pn exp C quantum k B hω n hω [ exp hω ] 4 exp [ exp hω hω ]

5 In the high temperature limit hω/ using the approximation e x + x for x gives C quantum k B C classical At low temperatures the quantized nature of the energy levels of the quantum oscillators becomes noticeable In the limit T 0 there is an energy gap between the ground state and the first excited state This results in a heat capacity that goes to zero exponentially as can be seen from the limit hω/ 3 Hard Sphere Gas: hω C quantum k B exp hω a The available phase space for identical particles is given by Ω!h 3 H d 3 q d 3 q d 3 p d 3 p where the integration is carried out under the condition H q i p i p i m or i i p i m The momentum integrals are now performed as in an ideal gas yielding Ω m3/!h 3 π 3 3! d 3 q d 3 q The joint integral over the spacial coordinates with excluded volume constraints is best performed by introducing particles one at a time The first particle can explore a volume V the second V ω the third V ω etc resulting in d 3 q d 3 q V V ωv ω V ω Using the approximation V aωv aω V ω/ we obtain d 3 q d 3 q V ω 5

6 Thus the entropy of the system is [ e S k B ln Ω k B ln V ω ] 3/ 4πme 3h b We can obtain the equation of state by calculating the expression for the pressure of the gas which is easily re-arranged to P T S V P k B V ω V ω ote that the joint effective excluded volume that appears in the above expressions is one half of the total volume excluded by particles c The isothermal compressibility is calculated from κ T V V P k BT T P V > 0 and is explicitly positive as required by stability constraints 4 Interacting Rod-Molecules: θ l excluded volume a Including both forms of entropy translational and rotational leads to [ S k B ln V Ωθ ] [ Aθ k B ln n Ωθ ] + + ln Aθ! b The extremum condition S/ θ 0 is equivalent to Ω n Ω A A 6

7 where primes indicate derivatives with respect to θ Solving for the density gives c lementary geometry yields n so that the equilibrium condition becomes with the function fθ plotted below: A ΩA + Ω A Ω l θ + sin θ n f θ l [θ + cos θ + sin θ] fθ n n c 0 θ θ c π d At high densities θ and the equilibrium condition reduces to V θl ; the angle θ is as open as allowed by the close packing The equilibrium value of θ increases as the density is decreased up to its optimal value θ c at n c and θ n < n c θ c The transition occurs at the minimum of f θ whence θ c satisfies d [θ + cos θ + sin θ] 0 dθ ie + cos θ c θ c sin θ c 7

8.333: Statistical Mechanics I Problem Set # 5 Due: 11/22/13 Interacting particles & Quantum ensembles

8.333: Statistical Mechanics I Problem Set # 5 Due: 11/22/13 Interacting particles & Quantum ensembles 1. Surfactant condensation: N surfactant molecules are added to the surface of water over an area