Motion in a Central-Force Field

Transcription

1 Alain J. Brizard Saint Michael's College Motion in a Central-Force Field 1 Motion inthecenter-of-massframe The Lagrangian for an isolated two-particle system is L = m 1 j_r 1j + m j_r j U(r 1 r ); where r 1 and r represent the positions of the particles of mass m 1 and m,respectively, and U(r 1 ; r )=U(jr 1 r j)isthepotential energy for an isolated two-particle system (see Figure below). Let us now de ne the position R of the center-of-mass (CM) R = m 1 r 1 + m r m 1 + m ; and de ne the inter-particle vector r = r 1 r,sothatthe particle positions can be expressed as r 1 = R + m M r and r = R m 1 M r; where M = m 1 + m is the total mass of the two-particle system. The Lagrangian of the isolated two-particle system thus becomes L = M j _ Rj + ¹ j_rj U(r); 1

2 where ¹ = m µ 1 m 1 = m 1 + m m1 m denotes thereduced mass of the two-particle system. For an isolatedsystem,the CM canonical momentum R _ = M R _ is a constant of the motion. The CM reference frame is de ned by the condition R =, i.e., we move the origin of our coordinate system to the CM position (the Figure below shows the case where m 1 >m ). In this case, the Lagrangian for an isolated two-particle system in the CM reference frame is L(r; _r) = ¹ j_rj U (r); where r 1 = m M r and r = m 1 M r: Hence, once the Euler-Lagrange euation for r Ã! ¹ Är = is solved for r = r(t), the motion of m 1 and m are determined through r 1 (t) =(m =M ) r(t) and r (t) = (m 1 =M) r(t). Motion in a Central-Force Field Aparticle moves under the in uence of a central-force eld F(r) =F(r) br if the force on the particle is independent of the angular position of the particle about the center of force

3 and depends only on its distance r from the center of force. Here, the magnitude F(r) (which is positive for a repulsive force and negative for an attractive force) is de ned in terms of the central potential U (r) asf(r) = U (r)..1 Lagrangian Formalism The motion of two particles in an isolated system takes place on a two-dimensional plane. When these particles move in a central-force eld, the Lagrangian is simply L = ¹ ³ _r + r µ _ U(r); (1) where polar coordinates (r; µ) aremostconveniently used. Since the potential U is independent of µ, thecanonical momentum p µ _ = ¹r µ _ ` () is a constant of motion(here, labeled `). The Euler-Lagrange euation for r, therefore, becomes the radial force euation ¹ ³ Är r µ _ ` = ¹ Är = F(r): (3) ¹r3 In this description, the planar orbit is parametrized by time, i.e., once r(t) and µ(t) are obtained, apathr(µ) ontotheplane is de ned. Since _ µ does not change sign on its path along the orbit, we may replace _r and Är with r (µ) andr (µ) asfollows.first, Next, using E. () we nd Lastly, using the identity _r = _ µr = `r ¹r = ` µ 1 : ¹ r ĵ = ` _r ¹r 3 = ` r ¹ r 5 ; Är = Ä µr + _ µ r = ` µ 1 r we nd an expression for Är: = à ¹ r 5 h (r ) rr i :! r = 1 h (r ) rr i ; r r 3 Är = ` µ 1 ¹ r r 3 ;

4 and the radial force euation (3) becomes s + s = ¹ du (s) F(1=s) = ` s ds ; (4) where s(µ) =1=r(µ) andu(s) =(¹=`) U(1=s). Note that the form of the potential can be calculated from the solution s(µ) =1=r(µ) as follows. For example, consider the particle trajectory described in terms of the solution r(µ) =r sec( µ), where r and are constants, then s + s = ³ 1 s = du(s) ds ; and thus U (s) = 1 ³ 1 s ` ³! U(r) = ¹r 1 : Note also that the function µ(t) isdetermined from the relation _µ = ` ¹r (µ) Returning to our example, we nd t(µ) = ¹r ` Z µ! t(µ) = ¹` Z µ r (µ) dµ: sec ÁdÁ = ¹r ` tan( µ)! r(t) = r v u t 1+ Ã ` t ¹r! and the total energy E = ` ¹r ; is determined from the initial conditions r() = r and _r() =.. Hamiltonian Formalism The Hamiltonian for the central-force problem is H = p r ` ¹ + ¹r + U(r); where p r = ¹ _r is the radial canonical momentum. Since energy is also conserved, we solve as E = ¹ _r + ` ¹ _r + U(r) = ¹r _r = + V (r); s [ E V (r) ]; (5) ¹ 4

5 where V (r) isknownasthee ective potential and the sign depends on initial conditions. This euation can then be used with E.()to yield dµ = ` ¹r dt = ` dr ¹r _r = ds ² U(s) s ; (6) where ² =¹E=`, or s (µ) = ² U (s) s : (7) We readily check that this euation is a proper solution of the radial force euation (4) since s s = [du=ds + s] ² U(s) s = du ds s is indeed identical to E. (4). Hence, for a given central-force potential U (r), we can solve for r(µ)=1=s(µ) byintegrating Z s d¾ µ(s) = ; (8) ² U(¾) ¾ where s de nes µ(s )=,andperformingtheinversionµ(s)! s(µ). s.3 Turning Points E. (7) yields the following energy euation E = ¹ ` ` h _r + + U(r) = ¹r (s ) + s +U(s) i ; ¹ where s = ¹ _r=` = p r =p µ. Turning points are those special values of r n (or s n )(n = 1; ;:::)forwhich E = U(r n )+ ` = ` " # U (s ¹r n ¹ n )+ s n ; i.e., _r (or s )vanishesatthesepoints. Iftwonon-vanishing turning points r <r 1 < 1 (or <s 1 <s )exist,themotion is said to be bounded in the interval r <r<r 1 (or s 1 <s<s ), otherwise the motion is unbounded. 3 Kepler Problem We now solve the Kepler problem where U(r) = k=r, wherek is a constant, so that U(s) = s s,wheres = ¹k=`. Theturning points for the Kepler problem are solutions of the uadratic euation s s s ² = ; 5

6 which can be written as s = s s + ² s 1 = s (1 e) and s = s (1 + e); where e = 1+²=s = 1 + E`=¹k : We note that motion is bounded when E<(< e < 1) and unbounded when E (e > 1). 3.1 Bounded Keplerian Orbits We will now look at thebounded case (e < 1). We de ne µ(s )=,sothatforthekepler problem, E. (8) becomes µ(s) = Z s s (1+e) d¾ s e (¾ s ) ; (9) which can easily be integrated by using the identity dx p a x = d arccos µ x a ; so that E. (9) yields µ(s) = arccos This euation can easily be inverted to yield µ s s : s e s(µ) = s (1 + e cos µ): (1) We can readily check that this solution also satis es the radial force euation (4) Kepler's First Law The solution for r(µ) is now trivially obtained from s(µ) as r(µ) = r 1+e cos µ ; (11) where r =1=s denotes the position of the minimum of the e ective potential à V (r ) = 1 r k `! ¹r = : 6

7 E. (11) generates an ellipse ofmajor radius a = r 1 e = k jej and minor radius b = a p 1 e = and, therefore, yields Kepler's First Law. v u t ` ¹ jej 3.1. Kepler's Second Law Using E. (), we nd dt = ¹` r dµ = ¹` da(µ); where da(µ) = R rdrdµ= 1 [r(µ)] dµ denotes an in nitesimal area swept by dµ. When integrated, this relation yields Kepler's Second law t = ¹` A; (1) i.e., eual areas are swept in eual timessince¹ and ` are constants Kepler's Third Law The orbital period T of a bound system is de ned as T = Z ¼ dµ _µ = ¹` Z ¼ r dµ = ¹` A = ¼¹ ` ab 7

8 where A = ¼ab denotes the area of an ellipse with major radius a and minor radius b. Using the expressions for a and b found above, we nd T = ¼¹ ` k jej v u t ` ¹ jej = ¼ v u t ¹k ( jej) 3 : If we now substitute the expression for a = k=jej and suare both sides of this euation, we obtain Kepler's Third Law T = (¼) ¹ a 3 : (13) k Note that in Newtonian gravitational theory, k=¹ = G (m 1 + m ); although Kepler's Third Law states that T =a 3 is a constant for all planets in the solar system, we nd that this is only an approximation that holds for m 1 À m. 3. Unbounded Keplerian Orbits We now look at the case where the total energy is positive or zero (i.e., e 1). E. (11) yields r (1 + e cos µ) =r or à pe 1 x e r p e 1! y = r e 1 : For e =1,theparticle orbit is aparabola x =(r y )=r,withdistance of closest approach at x() = r =, while for e > 1, the particle orbit is ahyperbola. 3.3 Laplace-Runge-Lenz Vector Let us now investigate an additional constant of the motion for the Kepler problem. First, we consider the time derivative of the vector p L, wherethelinear momentum p and angular momentum L are p = ¹ ³ _r br + r µ _ µ b and L = ` bz = ¹r µ _ bz: The time derivative of the linear momentum is _p = ru(r)= U (r) br while the angular momentum L = r p is itself a constant of themotion so that d dp (p L) = L = ¹_r ru r + ¹r ru _r dt dt = d (¹U r) +¹ (r ru + U)_r; dt 8

9 hence the vector A = p L + ¹U(r) r is a constant of the motion if the potential U(r) satis es the condition r ru(r)+u(r)=. For thekepler problem, with central potential U(r)= k=r, thelaplace-runge-lenz (LRL) vector Ã` A = p L k¹ br = r k¹ br `¹_r µ b is, therefore, a constant of the motion since r ru = U. Since the vector A is constant in both magnitude and direction, we choose its direction to be along the x-axis and its amplitudeis determined at the distance of closest approach r min = r =(1 + e) andwecaneasily show that A br = A cos µ leads to the Kepler solution r r(µ) = 1+e cos µ ; where r = `=k¹ and e = A=k¹. Note that if the Keplerian orbital motion is perturbed by the introduction of an additional potential term ±U(r), we can show that the LRL vector is no longer conserved (i.e., da=dt 6= )andthatthe direction of the Keplerian elliptical orbit precesses with a precession freuency! p (µ) = bz A A da dt ; where the unperturbed Kepler solution r(µ) is to be used.! 4 Isotropic Simple Harmonic Oscillator We now investigate the case when the central potential is of the form U (r) = k r! U (s) = ¹k ` s : (14) The turning points for this problem are expressed as r 1 µ 1 e 1 4 = r 1+e = 1 µ 1+e 1 4 and r s = r 1 e = 1 s 1 ; where r =(`=¹k) 1=4 =1=s is the radial position at which the e ective potential has a minimum V = kr and v Ã! u e = kr 1 : E Here, we see that orbits exist and are always bounded for E>V (and thus e 1). Next, using the change of coordinate = s in E. (8), we obtain µ = 1 Z d (¹=`)[E k] ; (15) 9

10 where =(1+e)¹E=`. We now substitute (') =(1+e cos ') ¹E=` in E. (15) to obtain µ = 1 " Ã!# 1 ` arccos e ¹E 1 or r(µ) = r p 1 e p : (16) 1+e cos µ This euation describes the ellipse x (1 e) + y (1 + e) = r of semi-major axis a = r p 1+e and semi-minor axis b = r p 1 e. Lastly,wenote that one revolution along the orbit r(µ) corresponds to an angular period of ¼, i.e., r(µ + ¼) = r(µ), and not ¼ as found in thekeplerproblem. The area of the ellipse A = ¼ab = ¼ (`=¹E) while the period is T (E; `) = Z ¼ dµ _µ = ¹À = ¼` E : If we introduce the angularfreuency! = ¼=T, then we nd the important relation between energy and angular momentum E = `!. 1