Two-Body Problem. Central Potential. D Motion The simplest non-trivial dynamical problem is the problem of two particles. The equations of motion read. m r = F 2, () We already know that the center of mass m 2 r 2 = F 2. (2) R = m r + m 2 r 2 m + m 2 (3) moves with a constant velocity: R(t) = V 0 t, (4) V 0 = m v (0) + m 2 v (0) 2, (5) m + m 2 where v (0) and v (0) 2 are the initial velocities of the particles; without loss of generality, we choose R(t = 0) = 0. Another important variable is the vector of relative distance Given r and R, one immediately finds r and r 2 : r = r r 2. (6) r = R + r 2 = R m 2 m + m 2 r, (7) m m + m 2 r. (8) Our goal thus is to set up a problem for finding r. Multiplying Eq. () by m 2 and Eq. (2) by m, and subtracting one from another, with Eqs. (4)-(8) of the section Newtonian Dynamics taken into account, we readily get m r = F(r), (9) m = m m 2 m + m 2, (0) F(r) = U 2(r) r = r r du 2 (r) dr. () L.h.s. of (9) formally looks like l.h.s. of the equation of motion of a particle with the mass m (this quantity is called reduced mass). The form of r.h.s. is slightly different from what we have considered so far. It depends on the coordinates of the particle, rather than on some relative distance to other particles. In this connection, the quantities F and U are called external force and external potential, respectively.
Particle in a Central Potential We have reduced a problem of two particles to a single-particle problem with an external central potential, U(r) U(r). What are the general properties of the motion in an external central potential? As is easily seen the momentum of particle is not conserved, however the energy, E = mv2 2 + U(r), (2) and the angular momentum with respect to the center (origin of the coordinates), are conserved. Problem 5. Make sure that E and L are constants of motion. L = m r v, (3) From the conservation of momentum we can deduce one very important property of the motion in a central potential. By definition (3), we have r L = 0. (4) In combination with conservation of L it means that there exists such a constant vector L that r(t) L, (5) i.e. the point with the radius-vector r(t) is in the plane that contains the origin and is perpendicular to L. We thus arrive at a generic conclusion about the motion in a central potential. The motion is actually two-dimensional; the z-component of the radius-vector is identically equal to zero, provided the z axis is chosen parallel to L. In view of the two-dimensional character of the motion and the fact that U depends only on absolute value of r, it turns out that the most convenient coordinates are the polar coordinates r and θ in the xy-plane: x = r cos θ, (6) y = r sin θ. (7) In these polar coordinates, the conservation of angular momentum implies θ = l mr 2 (l L = const). (8) Problem 6. Prove Eq. (8). Also, argue that Eq. (8) has the following geometric meaning, known as the Kepler s Second law: The area per unit time swept out by a radius-vector is constant. From Eq. (8) we draw yet another generic qualitative conclusion. The angular velocity θ does not change its sign, so that the particle permanently rotates around the center in one and the same direction, depending on the initial condition. (The case l = 0 is a special one and corresponds to θ = const.) The pair of equations of motion in polar coordinates can be obtained by just plugging (6) and (7) into the two Newton s equations for x and y, respectively. It is more wise, however, to take advantage of the conservations laws (2) and (8), since from the point of view of the theory of differential equations the two conservation laws are what is known as first integrals. First integrals are the 2
differential equations that are equivalent to the original ones, but of the lower order. First integrals correspond to partially integrated original equations and thus contain free integration constants, to be fixed by initial conditions. In the case of Eq. (2), the free constant is E, while in Eq. (8) the free constant is l. With (6)-(7) we readily find (note that v 2 ṙ 2 ) and conclude that Eq. (2) in polar coordinates reads v 2 = ṙ 2 + r 2 θ2, (9) 2 m(ṙ2 + r 2 θ2 ) + U(r) = E. (20) Eqs. (8), (20) form a system of two first-order ordinary differential equations for finding r(t) and θ(t), the constants E and l, as well as r 0 r(0) and θ 0 θ(0) being fixed by the initial condition at t = 0. The next good idea is to utilize Eq. (8) to exclude θ from Eq. (20), with the result 2 mṙ2 + l2 + U(r) = E. (2) 2mr2 We get a first-order differential equation for finding r(t). Mathematically, this is not a difficult problem, since one can always rely on a computer. Once r(t) is found, one then employs (8) to obtain θ(t) by integration: θ(t) = θ 0 + l t dt m r 2 (t ). (22) Eq. (2) thus plays the most crucial part in the theory of motion in the central potential. In what follows we concentrate on the general properties of the motion obeying this equation. First we notice that the term l 2 2mr 2 (23) has qualitatively the same structure as U(r) it is just some function of r which can be formally treated as a potential. That is why the term (23) is called centrifugal potential, the gradient of which is called centrifugal force. It makes then a perfect sense to combine the centrifugal potential with the genuine potential U(r) into what is known as the effective potential U eff (r) = U(r) + 0 l2 2mr 2. (24) Eq. (2) then says mṙ 2 + U eff (r) = E. (25) 2 If we differentiate Eq. (25) with respect to time, we get an elegant and physically transparent result m r = F eff (r), F eff (r) = d dr U eff(r). (26) That is r(t) evolves like a coordinate of a one-dimensional particle of the mass m moving in an external potential U eff. Hence, to further proceed with our analysis, we need to consider general properties of motion of a one-dimensional particle in an external potential. 3
One-Dimensional Particle in an External Potential Consider the equation of motion mẍ = F (x), F (x) = d U(x). (27) dx Conservation of energy yields the first integral of the differential equation (27) in the form mẋ 2 2 + U(x) = E. (28) The standard mathematical procedure of integrating differential equation (28) is as follows. dx dt = ± (2/m) [E U(x)]. (29) dx (2/m) [E U(x)] = ± dt. (30) m/2 dx [E U(x)] = ± t. (3) If the integral in the l.h.s. of Eq. (3) can be done analytically, then the solution comes in the form of a function t t(x). If the analytic integration is problematic, then one can do the integration numerically, or just numerically solve the differential equation (29). Time-reversal symmetry of the solution. The time-reversal symmetry of the equation of motion is explicitly seen due to the ± symbol. The time-translation symmetry is also explicit, since the integral in (3) is defined up to a free constant. In D, however, it is quite typical to have a timereversal symmetry of the solution, which is a stronger constraint than just the time-reversal symmetry of the equation of motion. In terms of the equation of motion, the time-reversal and time-translation symmetries require that if x(t) is a solution to the equation, then the function x (t), such that x (t) = x(t 0 t), (32) is also a solution to the equation, but, speaking generally, different from x(t), because of the difference in the initial condition. Now in D it is quite natural to have turning points at which the velocity is equal to zero (see Fig. ). Let such a turning event happen at t = t and consider the time-reversed and time-shifted solution x (t) = x(2t t). (33) By construction, x (t ) = x(t ) and ẋ (t ) = ẋ(t ) = 0 = ẋ(t ). We see that at t = t the values of the two functions, x (t) and x(t), coincide, and the same is true for the values of their derivatives. By uniqueness of the solution of a second-order differential equation with given initial conditions for the function and its derivative, x (t) x(t), and from Eq. (33) we get x(t) = x(2t t). (34) Since t here is an arbitrary number, we can shift it by t and present Eq. (34) in a more symmetric form: x(t + t) = x(t t). (35) Hence, whenever the motion involves a turning event at some time moment t, the solution x(t) of the equation of motion features time-reversal symmetry with respect to this point. 4
.0 0.8 U(x) E 0.6 0.4 0.2 0.0-0.2-0.4 x a x b x Figure : Motion in a trapping potential. Conservation of energy implies that the regions x < x a and x > x b are unaccessible for the particle. Hence the velocity of the particle is supposed to become zero and change its sign whenever x(t) = x a, or x(t) = x b. With respect to the given energy E, the points x a and x b are called turning points. x(t) x b 5 4 3 2 "+" "-" x a 0 -.0-0.8-0.6-0.4-0.2 0.0 0.2 0.4 0.6 0.8.0 t * t *2 t *3 t Figure 2: Evolution of a trapped D particle. The function x(t) is a train of + and - elements, the two being related to each other by time-reversal transformation. 5
Oscillatory motion. The situation of Fig. corresponds to the oscillatory motion of D particle. Qualitatively, the character of motion is always similar to that of Fig. 2. The particle oscillates between two turning points, x a and x b, the motion from x a to x b corresponding to the sign + in Egs. (29)-(3) being related to the motion from x b to x a corresponding to the sign - in Egs. (29)- (3) by time-reversal transformation. The period of oscillations is easily found from Eq. (3). If we integrate in the l.h.s. of (3) from x a to x b, then in the r.h.s. we get the time of motion from x a to x b, which, by time reversal symmetry, is equal to the time of motion from x b to x a, and thus is a half of the period. The period then is T = 2m xb x a dx [E U(x)]. (36) Harmonic oscillations in the vicinity of the potential minimum. The D oscillatory motion becomes especially simple if the energy E is close enough to its minimum value with x corresponding to a (local) minimum of the potential U(x): E min = U(x ), (37) U (x ) = 0, U (x ) = κ > 0. (38) The closeness of E to E min implies closeness of x to x and allows one to approximately replace U(x) with the first two non-zero terms of its Taylor expansion: U(x) U(x ) + (κ/2)(x x ) 2. (39) The equation of motion (27) then reduces to ẍ = ω 2 (x x ), ω = κ/m, (40) and is immediately solved: x(t) = x + A sin(ωt ϕ 0 ). (4) Here the amplitude A and the phase ϕ 0 are two constants fixed by the initial condition. The system (40) is known as harmonic oscillator. Unstable equilibrium at the potential maximum. (local) maximum at the point x : Suppose now that potential U(x) has a U (x ) = 0, U (x ) = κ < 0. (42) Independently of the sign of the second derivative, the fact that U (x ) = 0 guarantees that there is a special equilibrium solution of the equation of motion x(t) x. (43) However, the sign of the second derivative is crucial for the stability of the equilibrium solution. In the above-discussed case of U (x ) > 0, small deviations from equilibrium lead just to small-amplitude harmonic oscillations of x(t) around x. To see what happens at U (x ) < 0, we Taylor-expand the potential in the vicinity of x, and get the equation of motion ẍ = λ 2 (x x ), λ = κ/m. (44) 6
The solution of Eq. (44) is x(t) = x + Ae λt + Be λt. (45) In a general case, A 0, and we conclude that in the limit of large t the perturbed solution x(t) exponentially deviates from the equilibrium solution x = x. Hence, the solution x = x is unstable with respect to arbitrary small perturbations. Back to the Central Potential Now we are in a position to apply the facts we learned of the D motion to the central-potential problem. By the direct analogy between r(t) and x(t), we have dr (2/m) [E Ueff (r)] = ± dt. (46) m/2 dr [E Ueff (r)] = ± t. (47) Then, combining differential relation (46) with the integral (22) we get θ = ± l 2m dr/r 2 [E Ueff (r)]. (48) As it is seen from (46), the sign plus corresponds to the part of trajectory where ṙ > 0 while the sign minus corresponds to the part of trajectory where ṙ < 0. Eq. (48) relates the polar angle θ to the radius r, thus solving the problem of the geometric shape of the trajectory. This defines the curve, called orbit, in the xy-plane that the radius-vector r(t) draws with changing time. Note that the notion of orbit is not the same as the notion of trajectory: Orbit is a static geometric curve which is not sufficient to completely describe the temporal evolution of r. Since Eq. (47) directly relates the radius r to the time t, the pair of equations (47)-(48) completely solves the dynamical problem. Apogee and perigee. Reflectional symmetry of orbits. Let at some time moment t = t the function r(t) reach its local maximum (minimum), r = r. The corresponding point (r, θ ) in the particle s orbit is then called apogee (perigee). By definition, at t = t the function ṙ changes its sign, and from Eq. (48) we have: θ(r) = { θ + (l/ 2m) r r dr /r 2 E U eff (r ), θ > θ, θ (l/ 2m) r r dr /r 2 E U eff (r ), θ < θ, (perigee) (49) θ(r) = { θ + (l/ 2m) r r dr /r 2 E U eff (r ), θ > θ, θ (l/ 2m) r r dr /r 2 E U eff (r ), θ < θ, (apogee) (50) From (49) and (50) we can readily see that the orbit features a reflectional symmetry with respect to any axis defined by the origin and a perigee/apogee (below θ is an arbitrary angle): r(θ θ ) = r(θ + θ ). (5) To make sure that Eq. (5) holds true, one just needs to utilize the general structure of Egs. (49) and (50). Namely, { θ + f(r), θ > θ θ(r) =, (52) θ f(r), θ < θ, 7
the particular form of the function f(r) being irrelevant. Eq. (5) is trivially equivalent to f[r(θ θ )] = f[r(θ + θ )], and the latter relation is explicitly seen with (52). Bounded central-potential motion. If an attractive potential U(r) leads to an effective potential U eff (r) like the one plotted in Fig. 3, then there is a range of energies where the motion is bounded: The radius r(t) oscillates between perigee and apogee values, r p and r a, while the angle θ monotonously increases/decreases. By the symmetry properties, Eq. (35) and Eq. (5), the whole trajectory can be reconstructed from its one piece starting from one of the perigees and ending at the next apogee. In particular, for the time interval between two adjacent perigee and apogee, T p a (apogee and perigee, T a p ), we have: T p a = T a p = ra m/2 r p dr E Ueff (r). (53) Similarly, for the angle between two adjacent perigee and apogee, θ p a (apogee and perigee, θ a p ), we get θ p a = θ a p = (l/ ra dr/r 2 2m) r p E Ueff (r). (54) In a general case, the integral in (54) is some real number that depends on the particle mass, the form of the potential, and the initial conditions. The angle θ p a thus does not have to be a rational fraction of π, and we conclude that in a general case the orbit of a bounded motion has an infinite number of perigees and apogees, each next perigee (apogee) being obtained from the previous perigee (apogee) by rotation by the angle θ p p = θ a a = 2θ p a = 2θ a p. Nevertheless, there are two very important exceptions from this general rule. The first one is the potential U(r) /r, for which θ p p = 2π and the orbit is a closed curve with one perigee and one apogee. The second case is the harmonic potential U(r) r 2, for which θ p p = π and the orbit is a closed curve with two perigees and two apogees. [It is interesting to note that in both cases the closed orbit is an ellipse, but with different orientation with respect to the origin. In the case of U(r) /r the origin coincides with one of the focuses of the ellipse, while in the case of U(r) r 2 the center of the ellipse coincides with the origin. Both examples will be considered in detail later on.] Circular orbits. In any central potential having an attractive region U (r) > 0, for any radius r = r 0 of this attractive region there exists a circular orbit: r(t) r 0, θ(t) = θ 0 + lt mr 2 0 The angular momentum of the circular orbit is given by (l l(r 0 )). (55) U eff(r 0 ) = 0, (56) which means that the centrifugal force completely compensates the attractive force of the bare potential, and implies l 2 = mr 3 0U (r 0 ). (57) Problem 7. Prove Eqs. (55)-(57). Translate Eq. (57) into the initial condition for the velocity. The period, T circ, of a circular motion is the time required for θ(t) to change by ±2π. That is by definition θ(t + T circ ) = θ(t) ± 2π, and from (55), and then (57), we get T circ = 2πmr2 0 l = 2π mr0 U (r 0 ). (58) 8
U eff (r).0 0.8 0.6 0.4 0.2 0 0.0-0.2-0.4-0.6 E min -0.8 r 0 -.0 0 2 3 4 5 r Figure 3: Effective potential for the case U(r) = γ/r α, with γ > 0 and 0 < α < 2. From our D analysis, we conclude that a circular orbit is stable when U eff(r 0 ) = κ > 0, (59) in which case a small perturbation results in harmonic oscillations of r around the value r 0 with the frequency ω = κ/m. The case U eff(r 0 ) = κ < 0, (60) is unstable. At t, a perturbed orbit exponentially departs from the equilibrium, r r 0 e λt, with the exponent λ = κ/m. Let us explore the stability of circular trajectories in the family of power-law attractive potentials of the form U(r) = γ r α, γ > 0, α > 0. (6) Qualitatively, there are two generic cases: (i) α < 2 and (ii) α > 2 (see Figs. 3-4), and one special case α = 2. As is seen from Fig. 3, at α < 2 the circular motion is stable. From Fig. 4 we conclude that the circular trajectory is unstable. Moreover, for a given angular momentum l, any trajectory with r < r 0 (l) and E < U eff (r 0 ) will lead to a collapse at some finite t = t, when r(t ) = 0 and E kin (t ) =. Problem 8. Show that any α < 2 leads to the situation of Fig. 3, while any α > 2 leads to the situation of Fig. 4. Also, explore the stability of circular motion in the potentials U(r) = γr β, with γ, β > 0. Problem 9. Analyze the special case of α = 2: Is circular motion stable or unstable? Is it possible to have a collapse? If yes, then when; if no, then why. Motion in power-law potentials: Analysis of dimensions. Similarity Consider the bounded motion in the potentials (6) with α < 2, Fig. 3. In contrast to the bare potential U(r) that has no special scales of distance and energy in view of its scale-invariant form, the effective potential U eff (r) has both special distance, r 0, corresponding to its minimum, and special energy E min = U eff (r 0 ). That is whenever we fix some angular momentum l (for a particle of the given mass m), then we immediately fix the scale of characteristic distance and the scale of characteristic 9
.0 0.8 U eff (r) 0.6 0.4 0.2 0 0.0-0.2-0.4-0.6-0.8 r 0 -.0 0 2 3 4 5 r Figure 4: Effective potential for the case U(r) = γ/r α, with γ > 0 and α > 2. energy, given by r 0 and E min, respectively. Moreover, since we have already discussed that the two quantities characterize the circular motion with a given angular momentum, we conclude that there is also a characteristic time given by the period of the circular motion, Eq. (58). It makes then a perfect sense to re-scale all the physical quantities by introducing the following dimensionless quantities. ρ = r/r 0, (62) ε = E/ E min, (63) τ = t/t circ (r 0 ). (64) By this definition, ρ is the radius measured in units of r 0 ; ε is the energy measured in units of E min ; and τ is the time measured in units of T circ (r 0 ). Since r 0 is defined by U eff (r 0) = 0, we have (r 0 ) 2 α = l2 αγm, (65) E min = 2 α α and after some straightforward algebra arrive at θ = ± τ = ±(/2π) l 2 2mr 2 0, (66) dρ/ρ 2 (2 α)ε/α /ρ 2 + (2/α)/ρ α, (67) dρ (2 α)ε/α /ρ 2 + (2/α)/ρ α, (68) and, in particular, τ p a θ p a = ρa ρa = (/2π) ρ p dρ/ρ 2 ρ p (2 α)ε/α /ρ 2 + (2/α)/ρ, (69) α dρ (2 α)ε/α /ρ 2 + (2/α)/ρ α. (70) 0
At the turning points ρ p and ρ a the square-root expression becomes zero. Hence, ρ p and ρ a can be found by solving the equation (2 α)ε/α /ρ 2 + (2/α)/ρ α = 0 (ρ ρ p, ρ a ). (7) Problem 0. Do all the algebra leading to Eqs. (65)-(7). Our re-scaling procedure has revealed a remarkable similarity of the dynamics. There are only two dimensionless parameters, ɛ and α, that completely characterize the motion, up to changing the units of physical quantities. Moreover, for a given potential α is fixed, and the only variable that can depend on the initial conditions is ε. In a given potential, there is just one dimensionless variable, ɛ, that controls all the properties of the trajectory (up to changing the physical units)! Potential U(r) = γ/r The potential U(r) = γ/r describes the motion of celestial bodies and thus is of special interest. In accordance with generic results of the previous section, we chose convenient units of distance, time, and energy: the radius of the circular orbit (at a given angular momentum) the period of the circular motion with r = r 0 r 0 = l2 γm, (72) the absolute value of minimal energy T circ = 2πmr2 0 l = 2πl3 mγ 2, (73) Eq. (67) now reads E min = l2 2mr 2 0 θ = ± = mγ2 2l 2. (74) dρ/ρ 2 ε /ρ 2 + 2/ρ. (75) The integral can be done analytically with the result [ θ = ± arccos (/ρ )/ ] + ε + θ 0. (76) Without loss of generality, we can set the constant θ 0 to be equal to zero, since this corresponds to just a rotation of the coordinate system in the xy-plane by the angle θ 0. We then rewrite (76) as ρ = + + ε cos θ. (77) Eq. (77) is known to describe a conic section with the focus at the origin. The conventional geometric form of this equation is ρ = + λ e cos θ, (78) where the non-negative parameter λ e is called eccentricity. If λ e <, then the conic section is an ellipse (λ e = 0 corresponding to a circle); if λ e >, then the conic section is a hyperbola; and λ e = corresponds to a parabola. With the relation λ e = + ε (79)
3.0 2.5 2.0.5.0 0.5 0.0-0.5 -.0 -.5-2.0-2.5-3.0 y B D b A P O' C O a -3.5-3.0-2.5-2.0 -.5 -.0-0.5 0.0 0.5.0 x Figure 5: Elliptic orbit: The origin of coordinates (center of the potential) O coincides with one of the two focuses of the ellipse; O is the other focus. The point C is the center of symmetry of the ellipse. The points P and A are the perigee and apogee. we translate this as follows. (i) If ε < 0, then the orbit is an ellipse (ε = corresponding to a circle). (ii) For ε > 0, the conic section is a hyperbola. (iii) The case ε = 0 corresponds to a parabola. For all the three types of orbits, the perigee is at θ = 0, and ρ p = + λ e = + + ε. (80) The apogee is defined only for ε < 0, and corresponds to θ = π: ρ a = λ e = ε (ε < 0). (8) In Fig. 5 we show the elliptic orbit of bounded motion. Its geometric characteristics can be found from Eq. (78). OD =. (82) a CP = ε b CB = ε = =. (83) λ 2 e = a. (84) λ 2 e CO = ε ε = λ e λ 2 e = aλ e. (85) Problem. Derive Eqs. (82)-(85). Hint. Eq. (84) is readily obtained by observing that the point B corresponds to the maximal y. Note that to restore the genuine units of length we just need to multiply all the distances in Eqs. (80)- (85) by r 0. 2
2 y D 0 O P x -2-3 -2-0 Figure 6: Hyperbolic orbit. A hyperbolic orbit is shown in Fig. 6. Eqs. (80) and (82) are generic for any orbit, and thus are true for the hyperbolic one. A specific feature of the hyperbolic orbit is the maximal value of the cos θ that corresponds to the straight-line asymptotic motion of the particle in the limit ρ. From Eq. (78) we see that this value is cos θ m = λ e = + ε. (86) Yet Another Constant of Motion Eq. (79) says that any orbit of the bounded motion is closed. However, it does not give any idea of what is so special, if any, about the potential U /r that any bounded trajectory is closed. Meanwhile, it turns out that the closedness of the bounded orbits in our potential is a consequence of a very deep circumstance the existence of yet another constant of motion. This constant of motion is a vector given by the following expression. I = γ (v L) r r m γ [v2 r (r v) v] r r. (87) Problem 2. Prove that I is conserved by the equations of motion. You may use the previously-established fact that L is a constant of motion. As is easily seen from Eq. (87), the vector I is perpendicular to L. Hence, I always lies in the plain of motion (xy-plane in our notation), and this is crucial for the structure of the orbits. Indeed, each orbit, either bounded or non-bounded is supposed to have a perigee, and be reflection-symmetric with respect to the axis connecting the origin to the perigee, see Eq. (5). As far as the whole motion is concerned, the combination of the reflection symmetry with the time-reversal symmetry of the 3
function r(t) at perigees and apogees amounts to the statement that in any central potential the trajectory features the following symmetry property r(t) = R r( t), r(t = 0) r p/a. (88) Here r p/a means perigee (apogee), while R stands for the operation of reflection with respect to the axis connecting the origin to the perigee (apogee). For the quantity I this symmetry means (note that v(t) = R v( t), but I is quadratic in v and changing the sign of v does not affect the final expression) I(t) = R I( t), (89) and immediately leads to I = R I, (90) given that I is a constant of motion: I( t) = I(t) = const. Eq. (90) says that I is invariant with respect to the reflections about any axis connecting the origin with a perigee or apogee. Geometrically, it is possible only if there is just one perigee and either zero or only one apogee. The direction of the vector I then has to be parallel to the direction of the axis OP connecting the origin with the perigee; moreover, the apogee, if any, is supposed to lie on this axis. As far as the absolute value of I is concerned, it is easy to show that I = λ e. (9) Note that from Eq. (9) it follows that I is zero only for the circular orbits. It is also easy to check that the direction of I along the axis OP is from the origin to perigee. Problem 3. Check Eq. (9) and the sign of the direction of I by evaluating the x-component of I at the perigee. We conclude that the constant of motion (87) contains all the information about the geometry of the orbit: the direction of the main axis and the eccentricity. Time evolution To find the time evolution, we need to integrate Eq. (68) that in our case of α = reads τ = ±(/2π) dρ ε /ρ 2 + 2/ρ ±(/2π) ρ dρ ερ 2 + 2ρ. (92) This integral is well known in calculus. Its evaluation essentially depends on the sign of ε (with ε = 0 being a special case). Assuming ε 0, one first gets rid of linear in ρ terms in the argument of the square root. This is done by just shifting ρ: ρ = ρ + c, c = /ε, (93) which leads to τ = ±(/2π) ( ρ + c) d ρ ε ρ 2 εc 2. (94) It is then convenient to re-scale the integration variable: ρ = Λξ, Λ = + ε ε > 0. (95) 4
This yields: τ = ± (Λξ + c) dξ 2π ε ξ 2 τ = ± 2π ε (Λξ + c) dξ ξ 2 The integrals (96) and (97) are done by standard substitutions: (ε < 0), (96) (ε > 0). (97) ξ = cos ψ (ε < 0), (98) ξ = cosh ψ (ε > 0), (99) the idea of which is to cancel the denominator with the pre-factor of the differential: dξ = sin ψ dψ ξ 2 cos 2 ψ dξ ξ 2 = sinh ψ dψ cosh 2 ψ = dψ (ε < 0), (00) = dψ (ε > 0). (0) The integration becomes elementary and readily yields the final results { τ = ± (ψ λe sin ψ) /2π ε 3/2 (ε < 0), (02) ρ = ( λ e cos ψ) / ε { τ = ± (λe sinh ψ ψ) /2πε 3/2 ρ = (λ e cosh ψ ) /ε (ε > 0). (03) From the equations for ρ we see that ψ = 0 corresponds to the perigee. We also see that in the case of ε < 0 there is an apogee at ψ = π. The above equations solve the problem of finding the function ρ(τ) in the parametric form: the pair of values (τ, ρ) comes in the form of (τ(ψ), ρ(ψ)). In the case ε < 0 the motion is periodic. It is seen from the second line in Eq. (02) that the period corresponds to ψ changing by 2π, and thus τ period = 2τ p a = 2τ p a = ε 3/2. (04) It is interesting to note that this result can be easily obtained without explicitly integrating the equations of motion. Indeed, from Eq. (8) it follows (cf. Problem 6) that in any central potential, there is a relation t = 2mA(t), (05) l between the time t and the area A(t) swept out by the radius-vector (starting from t = 0). Problem 4. Prove Eq. (05) and show that it immediately leads to Eq. (04), provided one utilizes the known geometric formula for the area of an ellipse: A = πab. Eq. (04) suggests an alternative scaling of the temporal variable, that is especially convenient in the case of closed orbits. Introduce a new dimensionless temporal variable The equation for τ then is τ = τ ε 3/2. (06) τ = ± (ψ λ e sin ψ) /2π (ε < 0), (07) 5
and the new period is always unity by construction: τ period =, τ p a = τ p a = /2. (08) That is we simply measure time in units of the period of the orbital motion. Problem 5. Being trapped by an attractive potential U(r) /r, a particle moves along an elliptical orbit with the eccentricity / 2. Describe the time evolution of the position of the particle in the following format. For the 2 values of the parameter ψ namely, ψ j = πj/6 (j = 0,, 2,..., ) show corresponding positions of the particle in the xy-plane (with x axis pointing to the perigee). At each position, specify corresponding time moment in the units of the period of the orbital motion (with τ = 0 corresponding to the perigee). The sufficient accuracy is 3 significant digits. In the special case ε = 0 we have τ = ±(/2π) ρ dρ 2ρ (ε = 0). (09) The integral is easily done with the substitution 2ρ = ξ, and the result is τ = ± 2ρ (ρ + ) (ε = 0). (0) 6π Harmonic potential So far we were considering power-law potentials that vanish away from the center. Consider now attractive power-law potentials of the form U(r) = γ r α, γ > 0, α > 0. () The effective potential in this case looks like the one plotted in Fig. 7. The motion is always bounded and, at the qualitative level, is not different from the bounded motion in the potentials U /r α, discussed earlier. In a general case, the trajectory has infinite number of perigees and apogees. The only special case here is the so-called harmonic potential U(r) = γ r 2, γ > 0. (2) What make the harmonic potential special? If we write potential (2) in terms of Cartesian coordinates, we see that it separates into three terms: U(r) = γ r 2 = γ (x 2 + y 2 + z 2 ). (3) The potentials of this structure are known as separable. The separability of the potential has a crucial implication for the dynamics. Now each component of the force depends exclusively on corresponding coordinate, and the three-dimensional motion separates into three independent one-dimensional motions: mẍ = F x (x), mÿ = F y (y), m z = F z (z). (4) In our case we deal with just three independent D harmonic oscillators of one and the same frequency: ẍ = ω 2 x, ÿ = ω 2 y, z = ω 2 z, (5) 6
U 5 eff (r) 4 3 2 E min 0 0 0.0 0.2 0.4 0.6 0.8r.0.2.4.6.8 2.0 2.2 2.4 0 r Figure 7: Effective potential for the case U(r) = γr α, with γ > 0 and α > 0. The solution of the problem thus is: ω = 2γ/m. (6) x(t) = A x cos(ωt ϕ x ), y(t) = A y cos(ωt ϕ y ), z(t) = A z cos(ωt ϕ z ). (7) Since we know that in any central potential the motion is actually two-dimensional, we can always select the direction of the z axis parallel to the angular momentum, so that A z = 0 and z(t) 0. The trajectory in the xy-plane then is x(t) = A x cos(ωt ϕ x ), y(t) = A y cos(ωt ϕ y ). (8) In the analytic geometry, these two equations are known to describe (in the parametric form in terms of the parameter ψ = ωt) an ellipse centered at the origin. The orientation of the ellipse depends on the initial conditions. Note that the central symmetry and periodicity of the trajectory are explicitly seen from (8): if t t + π/ω, then r r. Without loss of generality, we can direct the x-axis to one of the two perigees (one of the two apogees); and t = 0 is conveniently associated with this perigee (apogee). Such a trajectory satisfies the generic symmetry requirement Eq. (88). That is and we see that this corresponds to Finally, we can exclude ωt by noticing that x(t) = x( t), (9) y(t) = y( t), (20) x(t) = A x cos ωt, y(t) = A y sin ωt. (2) x 2 A 2 x + y2 A 2 y =. (22) In analytic geometry, this is a canonic equation for an ellipse centered at the origin, the major and minor axis corresponding to x and y axis, or vice versa, depending on whether A x > A y, or A x < A y. The special case A x = A y corresponds to a circle. 7