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Least-Action Principle Applied to the Kepler Problem

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Least-Action Principle Applied to the Kepler Problem Least-action Principle Applied to the Kepler Problem S. Olszewski Institute of Physical Chemistry of the Polish Academy of Sciences, Kasprzaka 44/52, 01-224 Warsaw, Poland Reprint requests to Prof. S. O.; e-mail: [email protected] Z. Naturforsch. 59a, 375 – 381 (2004); received March 1, 2004 The least-action principle is examined for the two-body Kepler problem. This examination allows one to couple the eccentricity parameter of the Kepler orbit with the size of the major semiaxis of that orbit. The obtained formula is applied to an estimate of eccentricities characteristic for a set of the planetary and satellitary tracks. Key words: Eccentricity of the Kepler Orbits; Least-action Principle. 1. Introduction: Dependence of the Eccentricity e the motion of the system. In the next step, the same on the Major Semiaxis a0 of the Kepler Orbit action variables J1,J2,... occur useful in the quantiza- tion process which takes an easy form according to the The Kepler problem is one of the best known in clas- well-known Bohr-Sommerfeld rules [9, 10]. sical mechanics. Being based partly on observations, it But beyond of J1,J2,... an elementary action func- accounts of the motion of a planet, or satellite, along an tion S0 can be defined, equal to double the kinetic en- elliptical trajectory extended about a gravitational cen- ergy of a body integrated over a time interval passed ter. The period T of the motion and the major semiaxis by that body along its trajectory. The least-action prin- a0 are coupled by the well-known Kepler law ciple states that – for a conservative system – this ac- 2 = π2 3, tion function should be at minimum [8, 11]. In the next GMST 4 a0 (1) step, the derivative of S0 calculated with respect to the energy E of the body at some time t provides us with where G is the gravitational constant, M is the mass p S the time interval of the motion of the body, beginning of the Sun in the planet case, or the mass of the planet from some time t : in the satellite case. In (1) it is assumed – for the sake 0 of simplicity – that the mass MS remains at rest. Si- ∂S 0 = t −t . (2) multaneously, the eccentricity e defining the elliptical ∂E 0 trajectory of a planet, or satellite, is considered as in- p dependent of a0 or T [1 – 6]. The aim of the present paper is to point out that (2) can In spite of the fundamental importance of the Kepler be satisfied only on the condition that the eccentricity problem its mechanics seems to be discussed insuffi- e of the planetary, or satellitary, track depends on the ciently. Here we have in mind the least-action principle major semiaxis a0 of the Kepler orbit. applied to a conservative system. The principle makes The Cartesian coordinates of a body on the Kepler reference to the variational foundations of mechanics orbit are and is based on the idea of the action function. In prac- = ( − ), tice, the action variables J1,J2,..., considered for the x a0 cosu e (3) Kepler problem, are parameters conjugated to the cor- 2 1/2 responding angle variables of the moving body [7, 8]. y = a0(1 − e ) sinu, (3a) Both kinds of the action and angle variables serve to where the coordinate u depends on the time t by the transform the Hamiltonian into a function dependent equation [8] solely on J1,J2,... This technique is useful because it enables one to calculate the frequencies of the pe- 2π t ≡ ωt = u − esinu. (3b) riodic motion without finding a complete solution of T 0932–0784 / 04 / 0600–0375 $ 06.00 c 2004 Verlag der Zeitschrift f¨ur Naturforschung, T¨ubingen · http://znaturforsch.com 376 S. Olszewski · Least-action Principle Applied to the Kepler Problem Here we assumed that the beginning of the motion is at ∂S0 = ∂ ∂ the perihelion of the Kepler orbit [8]. Obviously, t 0 = S0 = a0 = t2 at u 0; the frequency of the motion about the gravi- ∂Ep ∂Ep tation center has been put equal to ω. ∂a0 For a moment let us assume that e is constant in a ∂ω way similar to a . Because of (3), (3a) and (3b) we ω + 2 ( + ) 0 2a0 a0 ∂ u2 esinu2 (8) have = a0 GMS 2 2 2a0 2 + 2 = 2 2 +( − 2) 2 du x˙ y˙ a0 sin u 1 e cos u 3ω( + ) dt a0 u2 esinu2 1 = = (u2 + esinu2). 1 GM ω = a2(1 − e2 cos2 u)ω2 (4) S 0 (1 − e2 cosu)2 In the calculation of (8) we took into account + = 2ω2 1 ecosu. a0 1 − ecosu GMSmp Ep = − , (9) 2a0 The integral of the expression (4), extended over an which is the energy of the planet or satellite, as well as interval of t beginning with t = t = 0 and ending at 1 the relations (1) and (3b). Since from the third Kepler some t = t ,gives 2 law given in (1) we have t2 t2 ∂T 3 T 1 + ecosu = , (10) (x˙2 + y˙2)dt = a2ω2 dt ∂a 2 a 0 1 − ecosu 0 0 0 0 u 2ω 2 the derivative of the expression a0 entering (7) calcu- = 2ω ( + ) (5) a0 1 ecosu du lated with respect to a0 gives 0 = ∂(a2ω) ∂ω = 2ω( + )u u2 , 0 = 2a ω + a2 a0 u esinu = ∂a 0 0 ∂a u 0 0 0 1 3 T = 2a ω + 2πa2 − (11) 0 0 T 2 2 a since u(t1)=u(0)=0 and 0 1 = a0ω. du ω 2 = ; (6) dt 1 − ecosu Our task is now to compare the result of (8) with t2 = t2(u2) calculated from (3b) for an arbitrary fraction of see (3b). The integral giving the elementary action S 0, the period T of rotation of the celestial body about the equal to the integral of the double kinetic energy ex- gravitational center. This is done for different values of tended along the time interval (0,t2), is t2 = λT, (12) t2 where λ is a constant: 0 < λ ≤ 1. The mentioned com- S = m (x˙2 + y˙2)dt 0 p parison is presented in Table 1, where – for the sake 0 (7) of convenience – the variable u = u has been taken u = u 2 = m a2ω(u + esinu) 2 , as a parameter. It is evident that solely at λ = 1 and p 0 u = 0 2 λ = 1 there exists an agreement between t2 = t2(u2) taken from (3b) and t2 calculated from (8); the case where mp is the mass of a planet or satellite. λ = 0 has been omitted as being a trivial one. The fundamental equation (2), valid for any planet A discrepancy obtained between t2 in (3b) and t2 or satellite trajectory, becomes therefore given by Eq. (8), illustrated by the second and the third S. Olszewski · Least-action Principle Applied to the Kepler Problem 377 Table 1. A comparison of the time interval λT of the motion In the next step, because of the relation of a celestial body performed along the Kepler orbit with the time interval calculated from the elementary action function; Ep see (3b), (8) and (12). The orbit eccentricity e is assumed ∂ m GM 2π2a 1 equal to a constant number. T is the circulation period of the t p = t S = t 0 = a ω2t , (16) body about the gravitational center. The beginning point of 2 ∂ 2 2 2 2 0 2 a0 2a0 T 2 the motion is assumed in the perihelion when the position of the center of force is assumed in the Sun. the expression (15), substituted into the first part of (8), u2 λ obtained from t2 λ obtained from t2 gives see (3b) for a reference calculated calculated in (8) between t and u according to (3b) ∂ ∂ 2 2 √ √ 1 1 2 e 2 y 1 1 2 1 2 0 = a ωe + a ωy + a ω + a ω π − π e + π e 0 0 0 0 4 8 4 8 4 2 2 ∂a0 ∂a0 1 π 1 − 1 1 + 1 (17) 2 4 2π e 4 2π e ∂ √ √ = 1 ω( + )+ 2ω ( + ), 3 π 3 − 2 e 3 + 2 e a0 e y a0 e y 4 8 4π 8 4π 2 ∂a0 π 1 1 2 √ 2 √ or 5 π 5 + 2 5 − 2 4 8 4π e 8 4π e 3 3 1 3 1 ∂ π + π e − π e 2 4 √2 4 √2 e + y = −2a0 (e + y), (17a) 7 π 7 + 2 7 − 2 ∂a0 4 8 4π e 8 4π e π 2 11 where column of Table 1, implies that e entering S 0 should y = esin(ωt2 + e ). (18) depend on a0: The solution of (17a) is e = e(a0). (13) 1/2( + ) = ln a0 e y const (19) 2. Equation for e(a0) or 1/2 + (ω + ) = . ( ) This dependence can be examined when the Fourier a0 e esin t2 e const 19a expansion for u is applied [6, 12, 13]: It is evident from (19a) that e depends both on a 0 ∞ 1 and t2. For example an explicit differentiation of (19a) u = ωt + 2 ∑ J (ne)sin(nωt). (14) n with respect to a0 gives successively the equations n=1 n − / / ∂e Here J is the nth-order Bessel function of the first 1 2 + 1 2 = n ea0 2a0 ∂ 0 (20) kind. Because e is a small number, the expansion (14) a0 – as well as the power expansions for J – can be lim- π n for ωt = and ωt = 3 π, ited to either one or two first terms.
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