The Two -Body Central Force Problem

The Two -Body Central Force Problem Physics W3003 March 6, 2015 1 The setup 1.1 Formulation of problem The two-body central potential problem is defined by the (conserved) total energy E = 1 2 m 1Ṙ2 1 + 1 2 m 1Ṙ2 1 + U( R 1 R 2 ) (1) This problem has 6 degrees of freedom (three components of position for each particle). The energy is conserved. The energy is invariant under translation in the three Cartesian directions, thus the the three components of total momentum are conserved. The energy is invariant under rotations about an arbitrary axis, so the three components of angular momentum are conserved. These conservation laws allow us to construct a complete formal solution. These notes take you through the solution, first in the general case (arbitrary potential) and then in the important specific case (gravitational or Coulomb) of U(r) 1/r. 1.2 Separation of motion into center of mass and relative coordinates Define the center of mass coordinate R COM and relative coordinate r via These equations imply R COM = m 1R 1 + m 2 R 2 m 1 + m 2 (2) r = R 1 R 2 (3) m 2 R 1 = R COM + r (4) m 1 + m 2 m 1 R 2 = R COM r (5) m 1 + m 2 1

Substituting Eqs. 4,5 into Eq. 1 gives with r = r and the reduced mass µ defined by E = 1 2 (m 1 + m 2 ) Ṙ2 COM + 1 2 µṙ2 + U(r) (6) µ = m 1m 2 m 1 + m 2 (7) Thus the total energy is the sum of a term corresponding to the free motion of a particle with mass equal to the total mass of the system and position equal to the position of the center of mass, and a second term corresponding to the motion of a fictitious second particle which represents the relative motion of the two particles (in other words, the internal degrees of freedom of the system). This fictitious particle feels a force arising from the interparticle potential U and has mass equal to the reduced mass. Conservation of center of mass momentum means that ṘCOM is independent of time so we may write down the solution for the motion of the center of mass as R COM = R init COM + V COM t (8) with initial position R init COM and velocity V COM determined by boundary conditions. From this point on we focus on the relative motion of the two particles 1.3 Simplifying the description of the relative motion Because the center of mass velocity is constant, the energy associated with the center of mass motion (the kinetic energy of the center of mass) is conserved. Therefore the energy of internal motion E int = E 1 2 (m 1 + m 2 ) Ṙ2 COM = 1 2 µṙ2 + U(r) (9) is conserved. The potential U(r) 0 in general so the momentum p = µṙ associated with internal motion is not conserved. However we assume that U(r) depends only on the magnitude of the coordinate r so that it is rotationally invariant. Thus the angular momentum L = r p associated with relative motion is conserved. Conservation of angular momentum has an important consequence. At any given time t the two vectors r and p define a plane and L is perpendicular to this plane. The conservation of L implies that L always remains parallel to its initial direction; thus r and p must also always lie in the same plane (if they did not, there would be a component of L perpendicular to its initial direction. Take the plane defined by r and p to be the x y plane; then the non-zero component of L is the z component L z. Thus the relative motion problem has only two degrees of 2

freedom, and has two conserved quantities, E int and L = L = L z, so the problem is integrable (the solution may be written down in terms of definite integrals). It is easiest to write down the solution in polar coordinates. Define Then the chain rule implies and Thus if r(t) is known, θ(t) may be found from implying x = r cos θ (10) y = r sin θ (11) ẋ = ṙ cos θ θ r sin θ (12) ẏ = ṙ sin θ + θ r sin θ (13) L = x(µẏ) y(µẋ) = µ r 2 θ (14) θ = θ(t) θ(t = 0) = L mr 2 (15) t 0 dt L µr 2 (t ) Now, substituting Eqs. 12 and 13 into Eq. 9 for the internal energy gives Finally using Eq. 15 gives (16) E internal = 1 2 µṙ2 + 1 2 µr2 θ2 + U(r) (17) E internal = 1 2 µṙ2 + L2 + U(r) (18) 2µr2 Thus the time dependence of the radial coordinate is determined by solving the problem of a particle of mass µ moving in the effective potential U eff = L2 + U(r) (19) 2µr2 given by the sum of the physical potential U(r) and the centrifugal potential L 2 /(2µr 2 ). The centrifugal potential blows up as r 0 and thus acts to keep the particle away from r = 0. It expresses the physics that as r decreases, the angular velocity θ must increase rapidly to keep L constant. Thus as r decreases more of the energy goes into rotational motion, leaving less for the radial motion. 3

1.4 Formal Solution From Eq. 18 we have (using the definition Eq. 19) 2 ṙ = ± E internal U eff (r) (20) µ Thus t(r) t(r init ) = µ r dr 2 r init Einternal U eff (r ) provides a formal solution for time as a function of position. Inverting this gives r(t); use of Eq. 16 then gives θ(t) completing the solution. Note also that use of Eq. 20 in Eq. 16 gives the explicit relation (21) θ(r) θ(r init ) = L r dr 2µ r init (r ) 2 E internal U eff (r) (22) between position and angle that specifies the orbit. The rest of these notes examine the physics contained in this formal solution. We begin by looking at the effective potential. 2 The effective potential The effective potential U eff (r) is constructed by adding a repulsive term L 2 /(2µr 2 ) to the original interparticle potential U(r). The result depends on the form of U(r). It is useful to distinguish three cases Repulsive U(r). If U(r) 0 for all r then U eff (r) 0 for all r and we simply have a repulsive potential which admits only unbound orbits. An example is sketched in panel a of Fig. 1. Attractive, short-ranged U(r). If U(r) 0 but is short ranged (goes to zero faster than 1/r 2 as r ) and is not divergent (does not increase as fast as 1/r 2 as r 0) then we have the interesting situation sketched in panel (b) of Fig. 1: for small L the potential has a minimum, and thus there can be both bound orbits and unbound orbits, but if L is too large, then there is no minimum and there are only unbound orbits. Gravitational: if U(r) = k/r then for any L at very large r the potential is negative, decreasing as r decreases, but at small r the potential is positive and increasing as r increases, so there is always a minimum and both bound and unbound orbits, as shown in Panel c of Fig. 1 4

2.0 1.5 - -1.5 - - -2.0 1.5 2.0 2.5 1.5 2.0 2.5 0 1 2 3 4 5 Figure 1: Left panel: effective potential (solid line) constructed from short ranged repulsive potential U(r) = e r (dashed line) and centrifugal potential computed with µ = 1 and L = 1. Central panel: effective potential (solid line) constructed from short ranged attractive potential U(r) = e r (dashed line) and centrifugal potential computed with µ1 and L = 0.25 (light solid line) and L = 1.1 (heavy line) showing cases with and without minimum. Right panel: effective potential (solid line) constructed from attractive 1/r potential U(r) = 1/r (dashed line) and centrifugal potential computed with µ = 1 and L = (light line) and 0.7 (heavy line). 3 Unbound Orbits The motion in an unbound orbit involves two particles, initially very far from each other. The two particles approach, have their motion affected by the interparticle potential, and then move far away, with the relative motion in a different direction than it was initially. When the particles are far from each other the relative motion is of a free particle r(t) = r init + vt (23) Careful!: these equations only describe the initial state of relative motion, when the particles are far apart. The angular momentum and internal energy are L = µr v = µr init v E internal = µ 2 v2 (24) Let us choose coordinates so that the initial velocity of relative motion is along x, so the initial motion is x(t) = x 0 vt (25) y = y 0 (26) The energy and angular momentum are E internal = µ 2 v2 and L = µy 0 v. If U(r) = 0 the motion would be described by Eqs. 25,26 at all times, and the distance of closest approach r min would be y 0. Let us now describe the motion in radial and angular coordinates. If the potential U(r) = 0 then U ef (r) = L 2 /(2µr 2 ) and the distance of closest approach r min would be 5

given by the solution of E internal = U eff (r min ) (27) or (using the expressions for E internal and L in terms of the initial velocity) or so that y 2.0 1.5 - -3-2 -1 0 1 2 3 Figure 2: Trajectories in xy plane computed as described in text for potentials used in panels (a) and (b) of Fig. 1 with E = 1.5 and L = 1. Upper line corresponds to repulsive interaction, lower to attractive and dashed to no interaction. Dot labels origin of relative coordinates µ 2 v2 = µ2 v 2 y0 2 2µrmin 2 + U(r min ) (28) r 2 min = y 2 0 + U(r min)r 2 min E initial (29) x r 2 min = y 2 0 1 U(r min) E initial (30) so that if the physical potential U(r) = 0 then r min = y 0, while if the potential is attractive the minimum distance is smaller than y 0 and if repulsive, the minimum distance is larger than y 0. Thus for an attractive potential the trajectory is bent towards the origin of relative motion and if the potential is repulsive the trajectory is bent away from the origin of relative motion. We can now use Eq. 21 to compute the relation between time and radial position (bearing in mind that we use the negative square root (rdecreases) to describe the approach phase and the positive to describe the separation, and then Eq. 16 to get the angle. From these we can get the x and y positions. I used the software package Mathematica to do the integrals for the short ranged attractive and repulsive potentials shown in panels (a) and (b) of Fig. 1. I took µ = 1 and said that at time t = 0 the particle was at position r = 3 with angular momentum L = 1 and energy E = 1.5. This implies v = 3 so y 0 = 1/ 3 implying that the initial angle is θ init = ArcSin1/(3 3). Using this initial data I performed the integrals. Results are presented in Fig. 2. 6

4 Bound orbits 4.1 General properties In a bound orbit the particle is confined between a minimum radius r min and a maximum radius r max which are two solutions of E internal = U eff (r) (31) A very important special case is when E internal is equal to the minimum value of U eff. In this case the minimum and maximum radius are equal: r min = r max = r circ. In this case the centripetal force L 2 /mr 3 arising from the derivative of the centrifugal potential just balances the attractive force du/dr. Because ṙ = 0 the radius is constant in time and we have a circular orbit. For a circular orbit the time dependence of the phase is easy to find: Eq. 16 becomes θ(t) θ(t = 0) = Lt (32) µr 2 circ which is just the statement that for a circular orbit the angular momentum is L = mr 2 theta. If E internal is greater than the minimum value of U eff the radial coordinate oscillates between r min amd r max according to Eq. 21 and the phase advances according to Eq. 16. Panel (a) of Fig. 3 shows the variation with time of the radial coordinate for the attractive potential discussed above. Here I have chosen L = 0.25 and E = 0.2. Notice that r varies much faster near its minimum point because the potential varies much more rapidly at small r, so the force is much greater (cf Fig. 1). Panel (b) of Fig. 3 shows the variation with time of the phase for the same case. We see again that the phase varies more rapidly when r is small, in this case because of conservation of angular momentum. We now ask a simple question: how much time, T r, does it take for the relative coordinate to go from r min to r max and back to r min? From Eq. 21 we find (the 2 is because the total time is twice the time it takes to go from r min to r max. µ T r = 2 2 rmax dr r min Einternal U eff (r ) (33) In the case shown in Fig. 3 the time is about 7. We may now ask another simple question: by how much does the phase advance in the time T r. From Eq. 16 we find Tr θ(t r ) = θ(t = 0) + 0 dt L µr 2 (t ) In the case shown in Fig. 3 the phase advances only by a bit more than π. In general, θ(t r ) 2π; in other words, the system does not come back to the same state after the 7 (34)

1.6 1.4 (a) (c) 1.2 0.8 0.6-0.4 0.2 0 5 10 15 20 25-1.5-3 2 (b) 1.5 (d) 1 0-1 -2 - -3 0 5 10 15 20 25-1.5-1.5-1.5 Figure 3: Panel (a) time dependence of relative coordinate computed using attractive potential of Fig 1 with E = 0.2 and L = 0.25. Panel (b) Time dependence of phase computed for same parameters. (c) and (d) trajectories in xy plane computed from r and θ for a short time (t = 15) and a longer time (t = 50). radial coordinate goes from its minimum to its maximum value and back to its minimum. The location of the point of closest approach changes with time. This is a generic feature of problems with multiple degrees of freedom: if the motion is bounded, the degrees of freedom are periodic in time, but with as many periods as there are degrees of freedom, and generically the periods are incommensurate (not rational multiples of each other). In this case the system explores the entire region energetically accessible to it, as you can see in panels c and d of Fig. 3, which show the trajectories in the xy plane. As we will see in the next section, the gravitational potential is a special case. For the gravitational potential the time needed for the radial coordinate to complete a cycle is the same as the time needed for the phase to advance by 2π so the position of the point of closest approach remains the same and the orbits have a simple geometric structure, in fact an ellipse. 8

5 The gravitational potential The gravitational case has two remarkable accidental features. First, for bound orbits the time required for the relative coordinate to move from its minimum to its maximum value is the same as the time required for the phase to advance by π. The orbits therefore have a simple structure. Second, the integrals can be done analytically, so simple expressions can be obtained in terms of known functions. It is interesting to speculate on how different the history of science might have been if this were not the case. In the gravitational case, the potential is k/r with k determined by Newton s gravitational constant and the masses of the two bodies. Thus U eff (r) = L2 2µr 2 k r (35) At very small r, U eff 1/r 2 so is large and positive, and increasing as r decreases. At very large r, U eff 1/r so is negative and gets more negative as r decreases. U eff must therefore have at least one minimum. Setting the r-derivative to zero gives du eff (r) dr implying that the extremum of the function is at = L2 µr 3 + k r 2 = 0 (36) r circ = L2 µk (37) The second derivative is positive so this is indeed a minimum. The value of U eff at the minimum is U circ = µk2 2L 2 (38) If the total energy of radial motion is greater than or equal to U circ but less than 0 the motion is confined between the two solutions of E internal = L2 2µr 2 k r The quantity 1/r appears repeatedly, so it is convenient to use this as a variable. Define η = r circ r Then using Eq. 37 we can write (39) (40) e int = η 2 2η (41) 9

with e int = E internal U circ The boundaries of the orbit are then given by (42) η ± = 1 ± 1 + e int (43) with η corresponding to r max and η + corresponding to r min. Instead of evaluating the time dependence we use Eq. 22 to obtain the angle as a function of the radial coordinate. Starting the integration from the minimum value we have θ(r) = θ(r min ) + L r 2µ Expressing this in terms of η and e int gives dr r min (r ) 2 E internal L2 + k 2µr 2 r (44) Now define η dη θ(r) = θ(r min ) η max eint η 2 + 2η (45) η = 1 + 1 + e int cosψ (46) and note that at r = r min, θ = 0 and at r = r max, θ = π. Eq. 45 becomes θ r circ r 1 θ(r) = dψ = ArcCos (47) θ rmin 1 + eint Note in particular that Eq. 47 implies that θ increases by π when r goes from r min to r max so the particle comes back to exactly the same point after each cycle. Rearranging: or 1 r = 1 ( 1 + 1 + eint cosθ ) (48) r circ r = r 0 (1 + ɛ cosθ) This is the equation of an ellipse with semimajor axis r 0 = r circ = L 2 /µk and eccentricity ɛ = 1 + 2E internall 2. µk 2 (49) 10