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PHY422/820: Classical Mechanics FS 2021 Worksheet #7 (Oct 11 – Oct 15)

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PHY422/820: Classical Mechanics FS 2021 Worksheet #7 (Oct 11 – Oct 15) H. Hergert Facility for Rare Isotope Beams and Department of Physics & Astronomy PHY422/820: Classical Mechanics FS 2021 Worksheet #7 (Oct 11 { Oct 15) October 8, 2021 1 Preparation • Lemos, Section 1.7 • Goldstein, Sections 3.1{3.3, 3.5{3.7 2 Central-Force Problems This week, we will start our discussion of central-force problems. An obvious example is the motion of a mass or particle in a spherically symmetric potential, but the importance of the following discussion extends far beyond this case. We have seen that the fundamental forces of nature like gravity or electromagnetism are pairwise interactions between objects that only depend on their relative distance, due to the fundamental symmetries of spacetime. For this reason, we will first demonstrate how motion under such pairwise interactions can be reduced to an equivalent one-body central-force problem in relative coordinates before discussing general strategies for analyzing such problems and the properties of their solutions. 2.1 Reducing Two-Body Problems to Equivalent One-Body Problems Let us consider a system of two masses m1 and m2 whose positions are given by ~r1 and ~r2, respec- tively. Their center of mass is given by 1 R~ = (m ~r + m ~r ) ; (1) M 1 1 2 2 where the total mass is M = m1 + m2. In the center-of-mass system, the coordinates of the masses are given by 0 ~ ~ri = ~ri − R; (2) and specifically m m m ~r0 = ~r − 1 ~r + 2 ~r = 2 (~r − ~r ) ; (3) 1 1 M 1 M 2 M 1 2 m m m ~r0 = ~r − 1 ~r + 2 ~r = 1 (~r − ~r ) : (4) 2 2 M 1 M 2 M 2 1 1 Introducing the relative distance vector ~r ≡ ~r2 − ~r1 ; (5) we see that m m m + m ~r0 − ~r0 = 1 ~r − 2 (−~r) = 1 2 ~r = ~r; (6) 2 1 M M M i.e., the relative distance of the two masses does not depend on our choice of coordinate system, which makes sense. We can use ~r to express the coordinates in the center-of-mass system as m m ~r0 = − 2 ~r; ~r0 = 1 ~r: (7) 1 M 2 M Now we can express the kinetic energy of the two masses in center-of-mass and relative coordi- nates. First, we have 1 1 T = m ~r_2 + m ~r_2 2 1 1 2 2 2 1 _ 2 1 _ 2 = m ~r_0 + R~ + m ~r_0 + R~ 2 1 1 2 2 2 1 _ _ 1 _ _ = m ~r_02 + 2~r_0 · R~ + R~ 2 + m ~r_02 + 2~r_0 · R~ + R~ 2 2 1 1 1 2 2 2 2 1 _ _ 1 1 = MR~ 2 + m ~r_0 + m ~r_0 · R~ + m ~r_02 + m ~r_02 2 1 1 2 2 2 1 1 2 2 2 1 _ m1m2 m2m1 _ 1 1 = MR~ 2 + ~r_ − ~r_ · R~ + m ~r_02 + m ~r_02 2 M M 2 1 1 2 2 2 1 _ 1 1 = MR~ 2 + m ~r_02 + m ~r_02 ; (8) 2 2 1 1 2 2 2 i.e., the kinetic energy is the sum of the center-of-mass kinetic energy and the kinetic energy of the masses as expressed in the center-of-mass system | the so-called intrinsic kinetic energy. This result extendes to general N-particle systems. For the two-body system, the intrinsic term can be rewritten using Eq. (7) as 1 m2 1 m2 1 m m 1 T = m 2 ~r_2 + m 1 ~r_2 = (m + m ) 1 2 ~r_2 ≡ µ~r_2 ; (9) intr 2 1 M 2 2 2 M 2 2 1 2 M 2 2 with the reduced mass m m m m µ ≡ 1 2 = 1 2 : (10) m1 + m2 M With a potential that only depends on the relative distance of the two masses, the Lagrangian now can be written as 1 _ 1 L = MR~ 2 + µ~r_2 − V (~r) = L + L (11) 2 2 com intr _ Since L only depends on R~ but not on R~, the center-of-mass motion decouples from the intrinsic motion, and the Lagrange equations imply @L d @L _ = 0 = ) MR~ = const. : (12) _ @R~ d @R~ Thus, the center-of-mass motion is uniform and linear. The intrinsic Lagrangian is equivalent to a one-body problem in the relative coordinate ~r. 2 2.2 Trajectories in the Central-Force Problem 2.2.1 General Solution of the Equations of Motion As we have seen in the previous section, the motion of a particle in a central force field can be modeled by the Lagrangian 1 L = m~r_ 2 − V (~r) ; (13) 2 where ~r can either indicate the coordinates of a single object of mass m in the external potential V (~r), or the relative coordinate of a two-body system with reduced mass m = µ. In general, a central force does not have to be spherically symmetric or conservative, but a conservative force will always be spherically symmetric. To see this, we first assume that the central force is conservative, i.e., that it can be written as @ 1 @ 1 @ F~ (~r) = −r~ V (~r) = − ~e + ~e + ~e V (~r) ; (14) r @r θ r @θ φ r sin θ @φ where we have used the gradient in spherical coordinates. Since F~ (~r) must be parallel to ~er, the partial derivatives of V (~r) with respect to the angles θ and φ must vanish, implying that V (~r) = V (j~rj) = V (r). Conversely, if the force is spherically symmetric, F~ (~r) = f(r)~er, then @ 1 @ 1 @ f(r) r~ × F~ (r) = ~e + ~e + ~e × ~r r @r θ r @θ φ r sin θ @φ r f 0(r)r − f(r) f(r) f(r) f(r) = ~e × ~e + ~e + ~e × ~e + ~e × r sin θ~e r r r r r θ r θ φ r2 sin θ φ = 0 ; (15) where we have used that @~r @~r @~r = ~e ; = r~e ; = r sin θ~e : (16) @r r @θ θ @φ φ Combining the proofs in both directions, we find that a central force is conservative if and only if it is spherically symmetric. Assuming a conservative, spherically symmetric central force going forward, we can write the Lagrangian in spherical coordinates as 1 L = m r_2 + r2θ_2 + r2 sin2 θ φ_2 − V (r) : (17) 2 The Lagrangian does not depend on φ, hence @L d @L @L 2 2 = 0 = ) = mr sin θ φ_ ≡ lz = const. ; (18) @φ dt @φ_ @φ_ where lz denotes the angular momentum around the z axis of our coordinate system. However, since the Lagrangian is spherically symmetric, we can orient our coordinate system at will, so all components of the angular momentum vector must be conserved. Since ~l = ~r × ~p = const:; (19) the motion of the mass(es) will always be confined to a plane that is perpendicular to ~l. Choosing our coordinate system such that ~l = l~ez ; (20) 3 π we have θ = 2 in spherical coordinates, and our Lagrangian (17) simplifies to ! 1 1 L = m r_2 + r2 θ_2 +r2 sin2 θ φ_2 − V (r) = m r_2 + r2φ_2 − V (r) : (21) 2 |{z} | {z } 2 =0 =1 The Lagrange equations now read d @L @L − = 0 ) mr¨ − mrφ_2 + V 0(r) = 0 ; dt @r_ @r d @L @L − = 0 ) mr2φ_ = l = const. (22) dt @φ_ @φ Expressing φ_ by the conserved angular momentum l, and introducing f(r) = −V 0(r), we can write the radial equation of motion as l2 mr¨ − − f(r) = 0 : (23) mr3 Since the force field we consider here is conservative, the total energy of the moving mass @L must be conserved. Alternatively, we can note that @t = 0 since L does not explicitly depend on time, which implies conservation of the total energy (cf. worksheet #4). Starting from the energy conservation law E = T + V 1 = m r_2 + r2φ_2 + V (r) 2 1 l2 = mr_2 + + V (r) = const. ; (24) 2 2mr2 we can solve forr _, s dr 2 l2 r_ = = ± E − − V (r) ; (25) dt m 2mr2 where the sign depends on the initial or boundary conditions, and must be chosen so that r ≥ 0 at all times. Separating the variables, we obtain dr dt = ±r (26) 2 l2 m E − 2mr2 − V (r) which can be integrated to give Z r(t) dr0 t − t = ± : (27) 0 r r0 2 l2 0 m E − 2mr02 − V (r ) Analogously, we can separate the variables in the angular momentum conservation law (22) and integrate to obtain Z t l 0 φ − φ0 = 0 2 dt : (28) t0 mr(t ) Together, Eqs. (27) and (28) constitute the general solution of any central force problem. As we might guess, these equations can be solved analytically only under very special circumstances. 4 However, we can continue the discussion of the general solutions qualitatively if we introduce the effective potential l2 V (r) = + V (r) (29) eff 2mr2 and write Eq. (27) as Z r(t) dr0 t − t0 = ± q : (30) r0 2 0 m (E − Veff(r )) In this form, the integral equation is equivalent to the general solution for one-dimensional motion of an object in the effective potential Veff - here, this is one-dimensional radial motion. The object is performing angular motion around the center of the potential due to the angular-momentum dependent term in Veff. Figure (1) shows the effective potential for the Kepler problem, κ V (r) = − ; κ > 0 ; (31) r as well as the main types of radial trajectories as a function of the energy E.
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