Physics 550 Problem Set 1: Classical Mechanics

Physics 550 Problem Set 1: Classical Mechanics

Problem Set 1: Classical Mechanics Physics 550 Physics 550 Problem Set 1: Classical Mechanics Name: Instructions / Notes / Suggestions: • Each problem is worth five points. • In order to receive credit, you must show your work. • Circle your final answer. • Unless otherwise specified, answers should be given in terms of the variables in the problem and/or physical constants and/or cartesian unit vectors (e^x; e^x; e^z) or radial unit vector ^r. Problem Set 1: Classical Mechanics Physics 550 Problem #1: Consider the pendulum in Fig. 1. Find the equation(s) of motion using: (a) Newtonian mechanics (b) Lagrangian mechanics (c) Hamiltonian mechanics Answer: Not included. Problem Set 1: Classical Mechanics Physics 550 Problem #2: Consider the double pendulum in Fig. 2. Find the equations of motion. Note that you may use any formulation of classical mechanics that you want. Answer: Multiple answers are possible. Problem Set 1: Classical Mechanics Physics 550 Problem #3: In Problem #1, you probably found the Lagrangian: 1 L = ml2θ_2 + mgl cos θ 2 Using the Legendre transformation (explicitly), find the Hamiltonian. Answer: Recall the Legendre transform: X H = q_ipi − L i where: @L pi = @q_i _ In the present problem, q = θ andq _ = θ. Using the expression for pi: 2 _ pi = ml θ We can therefore write: _ 2 _ 1 2 _2 H =θml θ − 2 ml θ − mgl cos θ 1 2 _2 = 2 ml θ − mgl cos θ Problem Set 1: Classical Mechanics Physics 550 Problem #4: Consider a dynamical system with canonical coordinate q and momentum p vectors. Verify the Poisson brackets: fqi; qjg = 0 fpi; pjg = 0 fqi; pjg = δij Answer #4: Using the definition of the Poisson brackets: X @f @g @f @g ff; gg = − @qi @pk @pk @qk k For the first two Poisson brackets: X @qi @qj @qi @qj fqi; qjg = − @qk @pk @pk @qk k X @pi @pj @pi @pj fpi; pjg = − @qk @pk @pk @qk k Since, for any i and k: @q i = 0 @pk @p i = 0 @qk those Poisson brackets always vanish. For the last Poisson bracket: X @qi @pj @qi @pj fqi; pjg = − @qk @pk @pk @qk k For the reasons above, this reduces to: P @qi @pj fqi; pjg = k @qk @pk P = k (δikδjk) = δij Problem Set 1: Classical Mechanics Physics 550 Problem #5: Consider a dynamical system with canonical position r and momentum p vectors, and L = r × p (angular momentum). Evaluate the Poisson brackets: fLx;Lxg fLx;Lyg fLy;Lxg Answer: Recall the definition of angular momentum (for a point particle): L = q×p The components of L are: Lx =qypz − qzpy Ly =qzpx − qxpz Lz =qxpy − qypx The Poisson brackets for this problem are easily calculated by using the properties of Poisson brackets: Recall the distributivity rule: ff + g; hg = ff; hg + fg; hg and anticommutivity: ff; gg = −{g; fg and the product rule: ffg; hg = ff; hgg + ffg; hg Problem Set 1: Classical Mechanics Physics 550 Consider the first relation: fLx;Lxg = fqypz − qzpy; qypz − qzpyg = fqypz; qypz − qzpyg − fqzpy; qypz − qzpyg =fqypz; qypzg − fqypz; qzpyg − fqzpy; qypzg + fqzpy; qzpyg = fqypz; qypzg + fqzpy; qzpyg = 0 Doing the same for fLx;Lxg: fLx;Lyg = Lz (1) and from the properties: fLy;Lxg = −{Lx;Lyg = −Lz (2) NOTE: One could also apply the definition of the Poisson bracket directly..

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