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− e
2 = p √ 1 − e 2 = |L z | 2 |E|m , (5.39) 5.3 Motion in a central field 187 where r
m and r
M denote the distance from the apocentre and pericentre, respect- ively. The verification of equations (5.39) is immediate, while equation (5.38) is obtained from (5.27): E = V e
m ) =
− k p (1 + e) + L 2 z 2mp
2 (1 + e)
2 , E = V e (r M ) = − k p (1 − e) + L 2 z 2mp 2 (1 − e) 2 , yielding e 2 = 1 + 2EL
2 z k 2 m . (5.40) From Kepler’s second law it follows that the area swept by the radius over a period T of the motion of revolution, and hence the area of the ellipse with semi-axes a and b, is proportional to the period T , namely (see (5.5)) πab = |L
| 2m T. (5.41) On the other hand, it follows from equations (5.35) and (5.39) that |L z
2 ). (5.42) Substituting this into (5.41) and recalling that b = a √ 1 − e 2 , we find πa 2 1 − e 2 = 1 2 a(1 − e 2 ) k m T , and hence a 3 T 2 = k 4π 2 m , (5.43) which is the proportionality between the cube of the major semi-axis and the square of the period of revolution (Kepler’s third law ). 5.3 Potentials admitting closed orbits Among all closed orbits, the circular orbits are of particular relevance. These orbits are defined by the parametrisation r(t) = r c , ϕ(t) = ϕ(0) + ω c t, (5.44) where ω
2 c = f (r c )/mr
c (see (5.7)). By equation (5.9) r c is necessarily a critical point of the effective potential V e , i.e. it is a root of V e (r) = 0, (5.45) 188 Motion in a central field 5.3 and in particular it is a regular function of the angular momentum L z . The
energy E c of the orbit can be derived from equation (5.10): E c = V e (r c ) = V (r c ) + L 2 z 2mr 2 c . (5.46)
This is the critical value of V e corresponding to r c . Suppose that the point particle moves along a circular orbit and receives a small impulse δp. Since the resulting variation δL of the angular momentum satisfies δL = r × δp, the component of δp which is parallel to the angular momentum L has the effect of slightly altering |L| and the plane of the orbit. The component of δp orthogonal to L can be decomposed into a vector parallel to e r
ϕ , changing the absolute value but not the direction of the angular momentum. The orbit therefore always stays in a plane close to the plane of the initial reference circular orbit, and in studying stability we can as a first approximation neglect the variation of L. D efinition 5.2 Let L be fixed. A circular orbit (5.44) is called (Lagrange) stable if for every ε > 0 there exists δ > 0 such that, for any initial data r(0), ˙r(0) such that
|r(0) − r c | < δ, | ˙r(0)| < δ, it holds that |r(t) − r(0)| < ε, (5.47)
for every t ∈ R.
Remark 5.3 Note that ϕ(0) does not influence the computation of the orbit, while the value of ˙ ϕ(0) is fixed by the choice of r(0), after we fix the value of L z = mr(0) 2 ˙ ϕ(0). Remark 5.4 ‘Lagrange’ stability of a circular orbit is equivalent to ‘Lyapunov’ stability of the equilibrium position corresponding to r = r c for the corresponding one- dimensional motion (5.9). However, it is a weaker notion than that of Lyapunov stability in the phase space R 4 of the original problem. Indeed, we have neglected the change in polar angle, which, in general, differs linearly in time from that of the reference circular orbit. As an example, in Kepler’s problem the circular orbit r = r c = k/2 |E c | is clearly stable. However, if we consider a nearby orbit, of energy E = E c + δE with δE > 0, by Kepler’s third law (5.43) the period of the motion becomes T =
4π 2 m k a 3/2 = π 2 mk 2 2 |E| −3/2
, (5.48)
by equation (5.39). Hence, if T c is the period of the circular orbit, T = T c 1 (1 − δE/|E
c |) 3/2 = T c 1 + 3 2 δE |E c | + O((δE)
2 ) , (5.49) 5.3 Motion in a central field 189 and thus the difference of the polar angles along the two orbits grows linearly with time. From Remark 5.4 and applying Proposition 3.1 to the one-dimensional system (5.9) we obtain the following. T heorem 5.2 If the effective potential V e has an isolated relative minimum at r c ,
For the study of orbits near the circular orbits we linearise the equation of motion (5.9) following a procedure analogous to that of Section 3.4. Setting x =
L z √ mr , (5.50) since d dt = x 2 L z d dϕ and
d dr = − √ m L z x 2 d dx , it follows that equation (5.9) can be written as d 2
dϕ 2 = − dW dx , (5.51)
where W (x) = V L z
mx + x 2 2 = V e Lz √ mx . (5.52) Equation (5.51) can be identified with the equation of motion for a one- dimensional mechanical system with potential energy W (x) and total energy E = x
2 + W (x),
(5.53) where the polar angle ϕ replaces time and x denotes the derivative of x with respect to ϕ. The orbit equation (5.14) becomes dx dϕ = x = ± 2[E − W (x)], (5.54) and the angle Φ between a pericentre and an apocentre is given by Φ =
M x m dx 2[E
− W (x)] , (5.55) where x m = L z / √ mr M , x M = L
z / √ mr m . Equation (5.55) expresses the half- period of the one-dimensional motion (5.51). The circular orbits are obtained in 190 Motion in a central field 5.3 correspondence with the roots x c of W (x
c ) = 0, and if W (x c ) > 0 then the orbit is stable. Consider an orbit near a circular orbit. Setting x 1 = x − x c , the behaviour of the orbit is described by the equation obtained by linearising equation (5.51): d 2
1 dϕ 2 = −W (x
c ) − W (x c )x 1 + O(x
2 1 ). (5.56) From this it follows, by neglecting the nonlinear terms and setting W c =
2 W (x
c ), that d 2 x 1 dϕ 2 + 2W c x 1 = 0.
(5.57) This is the equation of a harmonic oscillator. The angle Φ c between the pericentre (the maximum of x 1 ) and apocentre (minimum of x 1 ) of an orbit close to a circular orbit of radius r c is to a first approximation equal to the half-period of oscillations of the system (5.57): Φ c = π √ 2W c = πL z r 2 c mV e (r c ) (5.58)
because d dx = − √ mr 2 L z d dr , which yields W c
mr 4 c 2L 2 z V e (r c ). From the expression (5.8) for the effective potential we find that V e (r) = V (r) + 3L 2 z mr 4 . However V e (r c ) = V (r
c ) − L 2 z /mr 3 c = 0, and hence mr 4 c /L 2 z = r c /V (r
c ), and
equation (5.58) becomes Φ c = π V (r
c ) r c V (r
c ) + 3V (r c )
(5.59) We can now give the proof of a theorem due to Bertrand (1873). T heorem 5.3 In a central field with analytic potential energy V (r), all bounded orbits are closed if and only if the potential energy V (r) has one of the following forms:
V (r) = ⎧ ⎨ ⎩ kr 2 , − k r , (5.60) where k > 0. 5.3 Motion in a central field 191 Hence the only central potentials for which all bounded orbits are closed are the elastic and Newtonian potentials; Bertrand commented that, ‘all attracting laws allow closed orbits but natural law is the only one dictating them.’ Before the proof we consider the following. L emma 5.1 If in a central field all orbits, close to a circular orbit, are closed, then the potential energy V (r) has the form V (r) =
ar b , b > −2, b =
/ 0, a log
r R , (5.61) where a, b and R are constants. For these potentials, the angle between a pericentre and an apocentre is given, respectively, by Φ c = π √ b + 2 , Φ c = π √ 2 . (5.62) Proof
Since r c depends continuously on L z , from equation (5.59) it follows that Φ c
z . The condition (5.31) that ensures that an orbit is closed is false for a dense set of values of Φ c . Hence the only way that it can be satisfied for varying L z is if
Φ c is a constant, independent of r c . In this case, by imposing V rV + 3V = c > 0,
(5.63) where c is a constant, it follows that V =
− 3c rc V , (5.64) from which, setting U = V , we get that U = ar (1−3c)/c . Integrating the last relation we obtain (5.61). Equation (5.62) follows from (5.59). Proof of Theorem 5.3 By Lemma 5.1 we can assume that V (r) has the form (5.61). Note that we must have ab > 0 (or a > 0 in the logarithmic case), otherwise V e
Let x = L z / √ mr as in (5.50). Then we can reduce the study of the variation of the angle between a pericentre and an apocentre (which must be a rational multiple of 2π for the orbits to be closed) to the study of the variation of the period (5.55) of the one-dimensional motion (5.53), (5.54), with W (x) =
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ x 2 2 + αx
−b , α = a
L z √ m b , b > −2, b =
/ 0, x 2 2 − a log
x X , X = L z √ mR . (5.65) 192 Motion in a central field 5.3 Each of these potentials has a stable equilibrium point, corresponding to a stable circular orbit, and obtained by imposing W (x c ) = 0: x c = (αb) 1/(2+b)
, b >
−2, b = / 0,
√ a. (5.66) To conclude the proof we use a result proved in Problem 4(ii) of Section 3.9, and in particular the formula (3.63) expressing the variation of the period of a one-dimensional motion with respect to the period of small oscillations, for orbits near the equilibrium position. Setting y = x − x
c , ε = E
− E c , ˆ W e (y) = W (x c + y) − W (x c ), we have ˆ W e (0) = ˆ W e (0) = 0, ˆ W e (0) =
b + 2, b >
−2, b = / 0,
2 (5.67)
and Φ = y M y m dy 2[ε − ˆ W e (y)] . (5.68) Use of the Taylor series expansion of ˆ W e yields ˆ W e (y) =
b + 2 2 y 2 1 − b + 1 3x c y + (b + 1)(b + 3) 12x 2
y 2 + O(y 5 ). (5.69) Note that equation (5.69) for b = 0 is precisely this Taylor expansion in the case that W (x) =
x 2 2 − a log x X . It follows then that Φ =
c + Φ 1 (ε), where Φ c
Φ 1 is obtained from (3.63) (by the substitution m = 1, ω = 2π/ Φ c , c 1 = (b + 1)/3x c , c
2 = (b + 1)(b + 3)/12x 2 c
Φ 1 (ε) = 3π 2 ε ω 3 5 2 c 2 1 − 2c
2 = 3ε Φ 3 c 16π 2 x 2 c 5 18 (b + 1)
2 − 1 6 (b + 1)(b + 3) = ε Φ
c 48π
2 x 2 c (b + 1)(b − 2). (5.70)
5.4 Motion in a central field 193 Hence
Φ 1 (ε) is independent of ε if and only if (b + 1)(b − 2) = 0. In all other cases, the angle between a pericentre and an apocentre varies continuously with ε, and hence not all orbits can be closed. Thus it must be that either b = −1 (Newtonian potential, Φ c = π) or b = 2 (elastic potential, Φ c = π/2). The interesting relation between the harmonic oscillator and Kepler’s problem is considered in Appendix 6, where we prove the existence of a transformation of coordinates and of time which maps the associated flows into one another. 5.4
Kepler’s equation In this section we derive the time dependence in Kepler’s problem. This problem can be addressed directly; this is done in the usual manner for the case of one-dimensional problems, by using the conservation of the total energy E =
m ˙r 2 2 + L 2 z 2mr
2 − k r , (5.71) from which it follows that t =
m 2 r (t) r (0) dr k r − L 2 z 2mr
2 + E
· (5.72)
Using equations (5.37) and (5.3) we also find t =
mp 2 L z ϕ (t) ϕ (0)
dϕ (1 + e cos(ϕ − ϕ(0))) 2 · (5.73) This integral can be solved in terms of elementary functions, noting that dϕ (1 + e cos ϕ) 2 = (1
− e 2 ) −1 − e sin ϕ 1 + e cos ϕ + dϕ 1 + e cos ϕ = (1
− e 2 ) −1 − e sin ϕ 1 + e cos ϕ + 2 √ 1 − e 2 arctan
√ 1 − e 2 tan (ϕ/2) 1 + e .
In addition, it is necessary to invert the relation t = t(ϕ). A simpler solution, due to Kepler, consists of introducing the so-called eccentric anomaly ξ (Fig. 5.4) and the mean anomaly (time normalised to an angle) l =
2πt T = L z mab t. (5.74)
194 Motion in a central field 5.4
Fig. 5.4
The substitution r − a = −ae cos ξ (see Problem 12) in the integral (5.72) yields Kepler’s equation ξ − e sin ξ = , (5.75) and a parametric representation of the relation between r and t: r = a(1 − e cos ξ), t = mab
L z (ξ − e sin ξ), (5.76)
where we assume that at time t = 0 the particle is at the pericentre. The variable ξ, the so-called eccentric anomaly, has a remarkable geometrical interpretation. Consider a Cartesian reference system with origin at the centre C of the ellipse traced by the particle along its orbit, and denote by X and Y the coordinates in the standard system, different from x = r cos ϕ and y = r sin ϕ which are referred to the system with a focus F at the origin, and axes parallel to X, Y . It then follows that x = X
− ae, y = Y,
(5.77) and in the new coordinate system the equation of the ellipse is given by X 2
2 + Y 2 b 2 = 1. (5.78)
5.4 Motion in a central field 195 The eccentric anomaly parametrises the ellipse in the form X = a cos ξ = r cos ϕ + ae, Y = b sin ξ = r sin ϕ, (5.79) and using this parametrisation the motion is described by the equation ξ = ξ(l). Trace a circle of centre C and radius equal to the major semi-axis a of the ellipse. If P denotes the point subject to the central force field, moving around the ellipse, denote by Q the point of the circle with the same x-coordinate as P . Let H be a point on the X-axis with the same x-coordinate X H as P and
Q. It then follows (see Fig. 5.4) that area (AF P ) = − F H
· HP 2 + a X H b a a 2 − X
2 dX.
On the other hand area (AF Q) = − F H
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