Statistical Mechanics Solution Set #1 Instructor: Rigoberto Hernandez MoSE 2100L, , (Dated: September 4, 2014)

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1 CHEM 6481 TT 9:3-1:55 AM Fall 214 Statistical Mechanics Solution Set #1 Instructor: Rigoberto Hernandez MoSE 21L, , (Dated: September 4, Answered according to individual choice of paper 2. These essay questions were essentially discussed during the first lecture. See notes, or talk to me for further detail. 3. In this problem, we will solve the classical Harmonic Oscillator. Recall that for the harmonic oscillator, V (x = 1 2 kx2, and that the frequency ω is usually defined as k m. (a What is the (classical Hamiltonian? answer: H = p2 2m kx2 (b Write down Hamilton s Equations i.e., what are ẋ and ṗ as a function of x and p? answer: ẋ = H p = p m, ṗ = H x = kx (c Solve the two coupled equations in part (b, and explicitly provide the values of x(t and p(t. (Assume that the boundary conditions in time i.e. the requirements that the solution must satisfy are that x( = x and p( = p. answer: First, we set up the differential equation: p(t = d ( H dt x The solution of this is: and x(t = p mω sin(ωt+x cos(ωt (.1 = d dt V (x (.2 = kẋ (.3 = k H p = ω2 p (.4 p(t =p cos(ωt mωx sin(ωt (d Using your result from part (c, what is the value of the Hamiltonian as a function of t? Is it constant in time? Why or why not? answer: By substitution, H = p2 2m kx2 (.5 = p2 2m cos2 (ω(t 1 2 p x ω sin(2ωt+ 1 2 mω2 x 2 sin 2 (ωt (.6 + p2 2m sin2 (ω(t+ 1 2 p x ω sin(2ωt+ 1 2 mω2 x 2 cos 2 (ωt (.7 = p2 2m kx2 E (.8 which is a constant equal to the energy of the initial phase space point. The energy is a constant of the motion, and so should be constant throughout! 4. Now we solve the problem #3, using the more powerful machinery that we have learned.

2 (a Notice that the classical path for the Harmonic oscillator that you obtained in the previous problem is cyclic in time. Evaluate the action over one period. answer: Since the result in part (1c is cyclic, and the initial condition does not matter, we take x = x( =. Observe that the trig functions are cyclic over 2π. The period is T = 2π ω. The action over this period is: S(,T = T L(x, ẋdt (.9 ( T p 2 = p2 m = p2 2m = πp2 mω T T m E dt (.1 cos 2 (ωtdt ET (.11 (1 + cos(2ωt dt ET (.12 ET = (.13 NOTE: The boxed part is related to Hamilton s Principal action I. (b Use θ to denote the cyclic angle, i.e., identify a quantity in the classical path which you will set to take on the value of θ such that the solution takes on the same values at θ = and θ =2π. answer: Since the argument of the cosine is cyclic from to 2π, θ = ωt (.14 (c Evaluate W using the classical Harmonic Oscillator solution with x( = at t =, and arbitrary t 1. answer: We evaluate the integral by explicitly inserting the solutions for q(t and p(t from before, recognizing that dq =(p /m cos(ωtdt: W (,t 1 = = p2 m = p2 2m pdq (.15 cos 2 (ωtdt (.16 ( t ω sin(2ωt 1 (.17 (d Calculate the action S for the same conditions as in the previous part. By comparison with your result for S show that S and W are related by a Legendre transform over the variables E and t. (NOTE: It s because this simple relation holds in general, that they can both be properly called actions! answer: This calculation is similar to that of part (a, except that now, we allow the integral to go to arbitrary time t 1, instead of the period, T : S(,t 1 = L(x, ẋdt (.18 ( t1 p 2 m E dt (.19 = p2 cos 2 (ωtdt Et 1 (.2 m ( = p2 t m 4ω sin(2ωt 1 Et 1 (.21 = W (,t 1 Et 1 (.22 and this shows that for the harmonic oscillator, the actions are related by a Legendre transform.

3 (e Using your result in part (c, evaluate I explicitly. answer: Use of the definition of I provides the answer immediately: I = 1 π W (t <,t > (.23 = 1 W (,T (.24 2π p 2 = 1 2 mω (.25 (f In your solutions for x and p in the Harmonic oscillator, insert I and θ as obtained/defined in the previous parts as much as possible. (When your re done, there should be no reference to t in your expressions, nor should there be any x s or p s, as we will use these expressions to define a canonical transformation between the (x, p phase space, and the (I,θ vis-a-vis action-angle phase space. (g Calculate the energy, E i.e., the transformed Hamiltonian in terms of I and θ. answer: From part (c, p = 2mωI. Bysubstitution, p = 2mωI cos(θ (.26 2I x = sin(θ (.27 mω (h Assuming that the transformation in part (f is canonical, what are Hamilton s equations for I and θ? Notice that these equations are easy to solve I and θ are the action-angle variables that correspond to an integrated Hamiltonian! [No surprise, really, since we ve used the solution of problem 1 to obtain it. The correct way to solve this problem aprioriis to develop a systematic method for the construction of canonical transformations. This has been done for special cases, but it s likely that a general solution will never be found.] answer: Take I and θ to be the momentum and position, respectively, in the new phase space. The new Hamiltonian is obtained by inserting the formulas for x and p into the old Hamiltonian: H = 2mωI 2m In this coordinate system, Hamilton s equations are: cos2 (θ+ k2i 2mω sin2 (θ = ωi = E (.28 θ = H I = ω (.29 I = H θ = (.3 The solutions to these equations are θ = ωt and I is a constant. This is precisely what we got before. (i Bonus Question: Is the transformation of part (f canonical? answer: Yes. Here s why: Hamilton s equations can be written a: ż = J z H (.31 ( 1 ( where z =(x, p, J =, and 1 z = x, p. Suppose ζ (x ζ,p ζ = ζ(z is a transformation of the z phase space. The transformation is canonical if Hamilton s equations, ζ = J ζ H, (.32

4 are satisfied. Noting that ζ = z ζ ż and ζ H =( ζ z T z H,(wheretheT denotes the transpose then Hamilton s equations in ζ are equivalent to: z ζ ż = ζ z z H (.33 ż = ( z ζ 1 J( ζ z T z H. (.34 By comparison to the original Hamilton s equations, the requirement that the transformation be canonical is: In this case, the transformation is: J = ζ zj( ζ z T. (.35 x = 2I sin(θ (.36 mω p = 2mωIcos(θ (.37 and ζ z = ( θ I (x p = 2I mω cos(θ 1 2mωI sin(θ 2mωIsin(θ mω 2I cos(θ (.38 By substitution, Eq. (.35 is satisfied, and hence the transformation is canonical. Note: To answer this question correctly, you could have simply transformed the original Hamilton s equations by substitution using the relationships for x and p in terms of I and θ. However, the present demonstration also shows you that, in general, it may be easier to simply show that the relation in Eq. (.35 needs to be satisfied. 5. Chandler problem See below Chandler problem See below Chandler problem See below...

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