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Lecture 7 – Phase Space, Part 1 1 Phase Portraits

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Lecture 7 – Phase Space, Part 1 1 Phase Portraits Lecture 7 { Phase Space, Part 1 MATH-GA 2710.001 Mechanics 1 Phase portraits 1.1 One dimensional system Consider the generic one dimensional case of a point mass m described by a generalized coordinate q and subject to a time independent potential V (q) such that a Lagrangian for the point mass m is m L(q; q_) = q_2 − V (q) 2 The Hamiltonian of the system is 1 p2 H(q; p; t) = + V (q) 2 m Energy is conserved, so we can write 1 p2 = E − V (q) , p = ±p2m(E − V (q)) 2 m Since the first canonical equation is dq p r 2 = = ± (E − V (q)) dt m m the solution q(t) can be expressed as a definite integral and obtained as follows Z dq 2 1=2 Z = ± dt pE − V (q) m Even if so, the definite integral on the left can be hard to evaluate without a computer for certain V (q), and may not yield as many insights as looking at the phase portrait of the system. What exactly is a phase portrait? As Hamilton's equations make clear, the dynamical state of a system is completely specified by the phase state (q; p; t) given either the Lagrangian or the Hamiltonian to have the map between generalized velocities and momenta. A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase space (q; p). For the one dimensional system studied above, the phase space is two dimensional: q gives one dimension, and p the other dimension. Now, since the energy is conserved, and we could write p2 = E − V (q) 2m the orbit in phase space corresponding to a certain energy is a curve in phase space. All orbits in phase space (i.e. the phase portrait of the system) are just level curves of the Hamiltonian. In Lecture 1, we plotted the phase portrait for the simple pendulum, which we reproduce in Figure 1 as an illustration of our discussion. The time evolution of the system in phase space is given by Hamilton's equations. Specifically, since (@1H; @2H) is the gradient of the Hamiltonian, we see that Hamilton's equations tells the system to move perpendicular to the gradient of H. This is equivalent to saying that the system follows contours of constant energy, and is intuitive. Now, there are two points in the phase portrait of the pendulum where the gradient of H vanishes: at the center of the bowl, corresponding to (θ; pθ) = (0; 0), and at the saddle point (θ; pθ) = (π; 0). Phase space trajectories in the neighborhood of the minimum point (θ; pθ) = (0; 0) remain close to that minimum point: it is a stable equilibrium, corresponding to the situation in which the bob is simply hanging. Phase space trajectories in the neighborhood of the saddle point (θ; pθ) = (π; 0) leave the vicinity of that equilibrium point: it is an unstable equilibrium, corresponding to the situation in which the bob is standing upright. The contour that crosses the saddle point is called a separatrix, as it separates two regions with vastly different behavior. Inside the separatrix, the contours describe bounded oscillations about the stable equi- librium, and outside the separatrix the contours correspond to situations in which the pendulum circulates, with θ continuously increasing or decreasing without changing sign. Note finally that along a trajectory on the separatrix, the pendulum takes an infinite time to approach the saddle point. This can however not be deduced from the phase portrait. 1 8 6 4 2 θ p 0 −2 −4 −6 −8 −3 −2 −1 0 1 2 3 θ Figure 1: Phase portrait for the simple pendulum with g = 9:8 and l = 1. 1.2 Phase space reduction Let H(q1; : : : ; qN ; p1; : : : ; pN ; t) be a Hamiltonian for an N-dimensional problem, with a 2N-dimensional phase space. Let us suppose that H does not depend explicitly on qi. Hamilton's equations then tell us that pi is conserved. Therefore, Hamilton's equations for the 2N − 2 quantities do not depend on qi and involves the constant pi that only depends on initial conditions. The dimension of the problem that is left to solve has been reduced by 2. This is quite powerful. Of course, one may still want to solve for the time evolution of qi, but once the rest of the problem is solved, all one has to do to compute qi(t) is to evaluate a definite integral. Note that in principle, all this is also possible in the Lagrangian formulation. However, that requires replacing theq _i with their associated pi wherever they appear, which can be cumbersome. In the Hamiltonian formulation, that happens automatically. We already saw an example of automatic phase space reduction when presenting the Hamiltonian representation of the point mass in a central potential in Section 2.3.2 of Lecture 6. Let us give another example with the axisymmetric top we studied in Lecture 4. After a bit of algebra, we had obtained the following expression for the energy, which we can call the Hamiltonian function since it only involves generalized coordinates and conjugate momenta: 2 2 p2 pθ (pφ − cos θp ) H(θ; pθ; pφ; p ) = + 2 + + mgR cos θ (1) 2I 2I sin θ 2I33 We see that the Hamiltonian does not depend explicitly on φ and , so pφ and p are conserved quantity, as we had already proved in Lecture 4. Let us focus on the special case when pφ = p = p. By definition of the Euler angles, this equality must hold when the top is vertical. However, it is not limited to this situation, as we will now see. The Hamiltonian becomes 2 2 2 2 pθ 2 (1 − cos θ) p pθ H(θ; pθ) = + p 2 + + mgR cos θ = + Veff (θ) (2) 2I 2I sin θ 2I33 2I with 2 2 2 (1 − cos θ) p Veff (θ) = p 2 + + mgR cos θ 2I sin θ 2I33 As one can see in Figure 2, for p large, Veff has a unique minimum at θ = 0. For p small, there is a maximum at θ = 0, and two symmetric minima on each side of θ = 0. The transition occurs for p = pc such that the minimum of Veff at θ = 0 becomes a maximum. Computing the second derivative of Veff , we easily find the condition 2 p pc = 4ImgR ) pc = 4ImgR At θ = 0, we have p = I33!, where ! is the rotation rate of the spinning motion of the top. pc corresponds to the following critical rotation: p !c = 4IgmR=I33 2 0.7 0.65 0.6 0.55 ) θ ( 0.5 eff V p>p 0.45 crit p=p crit 0.4 p<p crit 0.35 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 θ Figure 2: Veff (θ) as a function of θ for different values of p For ! > !c, the top is stable standing vertically, as can be seen from the phase portrait shown in Figure 3, obtained for ! > !c. For ! < !c, the top is unstable standing vertically, as can be seen from the phase portrait shown in Figure 4, obtained for ! < !c. 2 Phase space evolution and Liouville's theorem Most problems do not have enough symmetries and associated conserved quantities to be reducible to quadra- tures or to situations in which the phase portraits are just given by contour lines corresponding the value of a conserved quantity such as the energy. In general, one thus has to solve the problem numerically by integrating the canonical equations in time. Even if so, phase portraits can be constructed numerically that yield plenty of insight on the dynamics of the system, as we will soon see when we discuss Poincar´eplots. Before we do so, we present some important properties of the evolution of a Hamiltonian system in phase space, and introduce the simple mechanical system we will use to illustrate these properties, namely the periodically driven pendulum. 2.1 The periodically driven pendulum Consider a pendulum of length l and mass m supported by a pivot that is driven in the vertical direction by a given function of time yD(t), as shown in Figure 5. The position of the bob in the Cartesian coordinate system x − y shown in the figure is x = l sin θ y = yD(t) − l cos θ The velocity of the bob therefore is dy v = lθ_ cos θ v = D + lθ_ sin θ x y dt We thus see that the kinetic energy T of the bob can be expressed in terms of θ, θ_ and t as " # m dy 2 dy T (θ; θ;_ t) = l2θ_2 + D + 2lθ_ D sin θ 2 dt dt The potential energy can also be written in terms of θ, θ_ and t: _ V (θ; θ; t) = mg (yD(t) − l cos θ) 3 x 10−3 2 1.5 1 0.5 θ p 0 −0.5 −1 −1.5 −2 −3 −2 −1 0 1 2 3 θ −5 2 −5 Figure 3: Phase portrait (θ; pθ) for the axisymmetric top with pφ = p , I = 3:28×10 kg m , I33 = 6:6×10 2 −2 2 −2 kg m , mgR = 4:56 × 10 kg m s and ! = 130 rad/s> !c ≈ rad/s x 10−3 5 4 3 2 1 θ p 0 −1 −2 −3 −4 −5 −3 −2 −1 0 1 2 3 θ −5 2 −5 Figure 4: Phase portrait (θ; pθ) for the axisymmetric top with pφ = p , I = 3:28×10 kg m , I33 = 6:6×10 2 −2 2 −2 kg m , mgR = 4:56 × 10 kg m s and ! = 90 rad/s< !c ≈ rad/s 4 Figure 5: Periodically driven pendulum.
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