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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1 the Basic Set-Up 2 Poisson Brackets
MATHEMATICS 7302 (Analytical Dynamics) YEAR 2016–2017, TERM 2 HANDOUT #12: THE HAMILTONIAN APPROACH TO MECHANICS These notes are intended to be read as a supplement to the handout from Gregory, Classical Mechanics, Chapter 14. 1 The basic set-up I assume that you have already studied Gregory, Sections 14.1–14.4. The following is intended only as a succinct summary. We are considering a system whose equations of motion are written in Hamiltonian form. This means that: 1. The phase space of the system is parametrized by canonical coordinates q =(q1,...,qn) and p =(p1,...,pn). 2. We are given a Hamiltonian function H(q, p, t). 3. The dynamics of the system is given by Hamilton’s equations of motion ∂H q˙i = (1a) ∂pi ∂H p˙i = − (1b) ∂qi for i =1,...,n. In these notes we will consider some deeper aspects of Hamiltonian dynamics. 2 Poisson brackets Let us start by considering an arbitrary function f(q, p, t). Then its time evolution is given by n df ∂f ∂f ∂f = q˙ + p˙ + (2a) dt ∂q i ∂p i ∂t i=1 i i X n ∂f ∂H ∂f ∂H ∂f = − + (2b) ∂q ∂p ∂p ∂q ∂t i=1 i i i i X 1 where the first equality used the definition of total time derivative together with the chain rule, and the second equality used Hamilton’s equations of motion. The formula (2b) suggests that we make a more general definition. Let f(q, p, t) and g(q, p, t) be any two functions; we then define their Poisson bracket {f,g} to be n def ∂f ∂g ∂f ∂g {f,g} = − .
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Poisson Structures and Integrability
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Hamilton-Poisson Formulation for the Rotational Motion of a Rigid Body In
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2.3. Classical Mechanics in Hamiltonian Formulation
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PHY411 Lecture Notes Part 2
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Hamilton's Principle and Symmetries
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Hamiltonian Dynamics CDS 140B
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