By S. G. Rajeev

This path could be commonly approximately platforms that can't be solved during this approach in order that approximation tools are invaluable.

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Extra resources for Advanced classical mechanics: chaos

Sample text

We can choose choose co-ordinates so that this point is at the origin. 1 1 L = gij q˙i q˙j − Kij q i q j − V (0) + O(q 3 ). 2 2 2 V Here Kij = ∂q∂i ∂q is the matrix of second derivatives of the potential: j (0) also called the Hessian. , K must be positive. An equilibrium point is stable if the linearized equations of motion have bounded solutions for any initial condition. Now we get the equation of motion (using matrix notation) gij j d2 q j + dt2 Kij q j = 0. j We can look for solutions of the form q j = Aj eiωt .

The whole question is then whether such co-ordinates exist and if so how to find them. •The key to understanding these issues is, surprisingly, geometry. Let us first look at the hamrmoic oscillator: H= p2 1 + kq 2 . 2m 2 We saw that the orbits are ellipses in the phase plane: these are just the curves of constant energy: p2 1 + kq 2 = E. 2m 2 52 PHY411 S. G. Rajeev The particle moves along this ellipse as time evolves. In this case we know explicit answer: 1 √ 2E 2 q(t) = cos[ω(t − t0 )], p(t) = [2mE] sin[ω(t − t0 )] k Thus the change of variables q = A cos θ, p = B sin θ suggests itself, where A, B are the axes of the ellipse.

Then the position of the Sun is at (νR, π − Ωt) and the Moon is at ((1 − ν)R, Ωt) , where R is the distance between them, and ν = Mm . The distance from the asteroid to the Sun is +m ρ1 (t) = √ r2 + ν 2 R2 + 2νrR cos[θ − Ωt] 34 PHY411 S. G. Rajeev 35 and to Juptiter is ρ2 (t) = √ r2 + (1 − ν)2 R2 − 2(1 − ν)rR cos[θ − Ωt] . The lagrangian for the motion of the asteroid is 1 1 1−ν ν L = r˙ 2 + r2 θ˙2 + G(M + m) + . 4 In this co-ordinate system the Lagrangian has an explicit time dependence: the energy is not conserved.