By S. G. Rajeev

This path might be in most cases approximately structures that can't be solved during this manner in order that approximation equipment are worthwhile.

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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. Change variables to χ = θ − Ωt to get 1 1 1−ν ν L = r˙ 2 + r2 [χ˙ + Ω]2 + G[M + m] + 2 2 r1 r2 where r1 = √ r2 + ν 2 R2 + 2νrR cos χ , r2 = √ r2 + (1 − ν)2 R2 − 2(1 − ν)rR cos χ are now independent of time. 5 Now the hamiltonian in the rotating frame, H = r˙ ∂L ∂L 1 1 r2 1−ν ν + χ˙ − L = r˙ 2 + r2 χ˙ 2 − G[M + m] + + ∂ r˙ ∂ χ˙ 2 2 2R3 r1 r2 is a constant of the motion.

Is it possible to reconstruct the potential V (x) knowing the time delay T (E) of scattering for all energies? That is, can we solve the above integral equation for V (x) given T (E) ?. •This is a nonlinear integral equation which is hard to solve. But a little trick will reduce it to a linear integral equation: we try to find the inverse function of V (x) . That is we look for x(V ) . But this is a multiple-valued function: since V (x) goes from 0 at x = −∞ to a negative value to 0 again at x = ∞ , its inverse is at least double valued.

The choice of this rule will depend on two constants, which are determined by the boundary conditions on the equation of motion. •Let us now consider a more general mechanical system with Lagrangian L= 1 2 gij (q)q˙i q˙j − V (q). ij Practically any mechanical system (except those with velocity dependent forces) is of this form. The equation of motion is d dt gij (q)q˙j + j ∂V = 0. ∂q i 30 PHY411 S. G. Rajeev ∂V Any point at which ∂q is an equilibrium point: there is a constant i = 0 solution to the equation of motion at that point.