1. The Two-Body Problem

2. The two-body problem The two-body problem: two point objects in 3D interacting with each other (closed system) Interaction between the objects depends only on the distance between them The number of degrees of freedom: 6 Phase space dimensions: 12 3.1

3. The two-body problem The Lagrangian of the system in Cartesian coordinates: It is a very non-trivial problem if we try to deal with the Lagrangian in this format: all the 6 independent coordinates are entangled in the potential function Let us look for a different configuration space 3.1

4. New generalized coordinates Let us introduce new system of coordinates: R – center of mass vector Then And 3.1

5. New generalized coordinates The Lagrangian in the new coordinates: 3.1

6. New generalized coordinates The Lagrangian in the new coordinates: The center of mass coordinates are cyclic! Three Euler-Lagrange equations for them can be solved immediately Total momentum of the system is conserved: three integrals of motion 3.1

7. New generalized coordinates The Lagrangian in the new coordinates: Let’s re-gauge the Lagrangian Constant term 3.1

8. New generalized coordinates The re-gauged Lagrangian: We reduced the two-body problem to a one-body problem in a central potential (potential that depends only on the distance from the origin) m: reduced mass The number of degrees of freedom: 3 Phase space dimensions: 6 3.1

9. Spherical coordinates Central potential is spherically symmetric It is convenient to work in spherical coordinates Then 8.1

10. Spherical coordinates The Euler-Lagrange equation for φ The φ coordinate is cyclic Since the system is spherically symmetric, we have a freedom of choosing the reference frame We chose it as follows: the initial velocity vector belongs to a plane φ = const Then 3.2

11. Spherical coordinates The Euler-Lagrange equation for θ The θ coordinate is also cyclic Momentum conjugate to the θ coordinate Angular momentum in the plane of motion relative to the origin is conserved 3.2

12. Spherical coordinates The Euler-Lagrange equation for r Momentum conjugate to the r coordinate Now we can write a Hamiltonian 3.2 8.1

13. The effective potential The Hamiltonian effectively depends only on 1 coordinate now We reduced the two-body problem to a 1D problem of a particle with a reduced mass m in the effective potential The number of degrees of freedom: 1 Phase space dimensions: 2 3.2

14. The effective potential Hamilton equations of motion: 3.2

15. The orbit equation On the other hand Orbit equation 3.2 3.5

16. The orbit equation The orbit equation can be integrated for potentials with the power dependence on the distance If n = 2, - 1, - 2, the integral can be expressed in trigonometric functions If n = 6, 4, 1, - 3, - 4, - 6, the integral can be expressed in elliptic functions 3.5

17. The orbit equation From Hamilton’s equations of motion: If the orbit is known, the potential can be calculated 3.5

18. Example Restore a potential for a spiral orbit:

19. Stable circular orbits For a circular orbit: On the other hand For the extremum of the effective potential Extremum of the of the effective potential corresponds to a circular orbit 3.6

20. Stable circular orbits For a stable circular orbit, the second derivative of the of the effective potential should be positive: For potentials with the power dependence on the distance 3.6

21. Application of the Hamilton-Jacobi theory The problem of spherically symmetric potential can be neatly treated employing Hamilton-Jacobi theory Then equation for Hamilton’s characteristic function 10.5

22. Application of the Hamilton-Jacobi theory Let us assume that the variables can be separated Then The φ coordinate is cyclic, therefore 10.5

23. Application of the Hamilton-Jacobi theory The circled part should be constant, since it contains only the θ dependence Then, finally The variables are completely separated! The resulting equation can be integrated in quadratures and is equivalent to the orbit equation 10.5

24. The Kepler problem Kepler potential: Mediating gravitational and electrostatic interactions Attraction: Repulsion: Integral orbit equation: 3.7 Johannes Kepler (1571-1630)

25. The Kepler problem Let us consider an attractive potential: Table integral: 3.7

26. The Kepler problem 3.7

27. The Kepler problem We obtained an explicit expression for the orbit Depending on the values of C and e, the orbits can assume qualitatively different shapes For a positive C (attraction), the shapes of the orbits represent all possible conic sections 3.7

28. Classification of Kepler’s orbits Effective potential for the attractive Kepler case: 3.3 3.7

29. Classification of Kepler’s orbits Effective potential for the attractive Kepler case: Minimum point of the effective potential 3.3 3.7

30. Classification of Kepler’s orbits Effective potential for the attractive Kepler case: The simplest case Circular orbit 3.3 3.7

31. Classification of Kepler’s orbits Circular orbit Circle is one of conic sections 3.3 3.7

32. Classification of Kepler’s orbits If 3.3 3.7

33. Classification of Kepler’s orbits If Then is real and is positive The orbit is an ellipse with its center shifted from the origin by and two semi-major axes and Ellipse is also a conic section 3.3 3.7

34. Classification of Kepler’s orbits Elliptic motion is limited by two values of r 3.3 3.7 Perihelion Aphelion

35. Classification of Kepler’s orbits This parameter is known as an eccentricity of an ellipse For a constant energy, perihelion is decreasing with increasing eccentricity 3.3 3.7

36. 3.3 3.7 Classification of Kepler’s orbits If Then is imaginary and is negative The orbit is a hyperbola Hyperbola is a conic section as well

37. Classification of Kepler’s orbits Hyperbolic motion is limited by one value of r - perihelion 3.3 3.7

38. Classification of Kepler’s orbits Finally, if Then The orbit is a parabola with its center shifted from the origin by Parabola is a conic section 3.3 3.7

39. Classification of Kepler’s orbits Parabolic motion is also limited by one value of r - perihelion 3.3 3.7

40. Synopsys for orbit classification

41. Motion in time In Kepler’s case: Substitution: For ψ = 2π (one period): 3.8

42. A bit of history: Kepler’s laws First law: “The planets move in elliptical orbits with the sun at one focus” 3.8 Johannes Kepler (1571-1630) Tycho Brahe/ Tyge Ottesen Brahe de Knudstrup (1546-1601)

43. A bit of history: Kepler’s laws Second law: “The radius vector to a planet sweeps out area at a rate that is independent of its position in the orbit” Third law: “The square of the period of an orbit is proportional to the cube of the semi-major axis length” 3.8

44. The Kepler problem in action-angle variables We consider periodic motion in the case of attraction By definition, the action variables are 10.8

45. The Kepler problem in action-angle variables We found earlier for the two-body problem Therefore 10.8

46. The Kepler problem in action-angle variables Frequencies 10.8

47. The Kepler problem in action-angle variables Degenerate case: the frequencies for all three variables coincide Hence it takes the same time for all three variables to return to the same value – the same point on the orbit Therefore, a completely degenerate solution corresponds to a closed orbit We did not have to obtain an explicit expression for the orbit to realize that it is closed 10.8

48. The Kepler problem in action-angle variables To lift the degeneracy, we can introduce another canonical transformation employing the following generating function: Then And the Hamiltonian 10.8

49. The Kepler problem in action-angle variables This Hamiltonian is cyclic in 5 variables, therefore their 5 corresponding conjugates are conserved: We obtained 5 constants of motion for a system with 6 degrees of freedom (the last two can be shown to be related to certain orbit parameters) 10.8

50. Repulsive Kepler potential Let us consider a repulsive potential: Total mechanical energy: Orbit equation 3.10

51. Repulsive Kepler potential This is a hyperbola Therefore, hyperbolic orbits correspond to the case of a positive total energy for both the attractive and the repulsive interactions Scattering – the orbit is never closed for E > 0 3.10

52. Scattering Scattering angle: Impact parameter: On the other hand: 3.10

53. Scattering For a beam of (noninteracting) particles incident on the scattering center, intensity (flux density) is the number of particles crossing unit area normal to the beam in unit time Scattering cross-section in a given direction is the ratio of the number of particles scattered into a solid angle per unit time to the incident intensity Differential scattering cross-section: 3.10

54. Scattering cross-section Conservation of the number of particles: 3.10

55. Scattering cross-section Rutherford scattering cross-section It is independent of the sign of k! 3.10 Ernest Rutherford (1871 – 1937)

56. Total scattering cross-section It diverges because of the long-range nature of Kepler’s potential All the particles in an incident beam of infinite lateral extent will be scattered to some extent and must be included in the total scattering cross-section 3.10

57. Laboratory coordinates Let us recall the initial transformation of coordinates: Let us recall the re-gauged Lagrangian: All the results obtained so far are in the re-gauged center-of-mass system, in which the center of mass is at rest: 3.11

58. Laboratory coordinates In the center-of-mass system, the scattering process of two particles will look like this: Often, while the incident particle is moving, the second one is initially at rest We introduce the laboratory system of coordinates, in which the center of mass is moving with a constant velocity 3.11

59. Laboratory coordinates In the laboratory system, the scattering process of two particles will look like this: Let us introduce notations: Then Initially 3.11

60. Laboratory coordinates Taking the ratio of these two equations: 3.11

61. Laboratory coordinates Now we can write the differential scattering cross section expressed in laboratory system Conservation of the number of particles: 3.11

62. The three-body problem The Lagrangian of the system in Cartesian coordinates: This problem has 9 independent coordinates entangled by the 3 potential functions This Lagrangian cannot be re-gauged to a one- particle Lagrangian No general explicit solution is known 3.12

63. The three-body problem Constants of motion (not independent): total energy, three components of the center of mass linear and angular momenta If the three objects are allowed to move freely in 3D the orbits become very complex and sensitive to initial conditions Even after fixing positions of two particles and letting the third particle move in a plane, the orbit still can not be found explicitly 3.12