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Chapter 3: Central Forces

Chapter 3: Central Forces. Introduction Interested in the “2 body” problem! Start out generally, but eventually restrict to motion of 2 bodies interacting through a central force . Central Force  Force between 2 bodies which is directed along the line between them.

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Chapter 3: Central Forces

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  1. Chapter 3: Central Forces Introduction • Interested in the “2 body”problem! Start out generally, but eventually restrict to motion of 2 bodies interacting through a central force. • Central Force Force between 2 bodies which is directed along the line between them. • Importantphysical problem! Solvable exactly! • Planetary motion & Kepler’s Laws. • Nuclear forces • Atomic physics (H atom). Needs quantum version!

  2. Sect. 3.1: Reduction to Equivalent 1-Body Problem • My treatment differs slightly from text’s. Same results, of course! • General 3d, 2 body problem. 2 massesm1& m2: Need 6 coordinates: For example, components of 2 position vectors r1& r2(arbitrary origin). • Assumeonly forces are due to an interaction potential U. At first, U = any function of the vector between 2 particles, r = r1 - r2, of their relative velocity r = r1 - r2, & possibly of higher derivatives of r = r1 - r2: U = U(r,r,..) • Very soon, will restrict to central forces!

  3. Lagrangian: L = (½)m1|r1|2 + (½)m2|r2|2 - U(r,r, ..) • Instead of 6 components of 2 vectors r1& r2, usually transform to (6 components of) Center of Mass (CM) & Relative Coordinates. • Center of Mass Coordinate:(M  (m1+m2)) R  (m1r1 +m2r2)(M) • Relative Coordinate: r  r1 - r2 • Define:Reduced Mass:μ (m1m2)(m1+m2) Useful relation: μ-1 (m1)-1 + (m2)-1 • Algebra  Inverse coordinate relations: r1 = R + (μ/m1)r; r2 = R - (μ/m2)r

  4. Lagrangian:L = (½)m1|r1|2 + (½)m2|r2|2 - U(r,r, ..) (1) • Velocities related by r1 = R + (μ /m1)r; r2 = R - (μ /m2)r (2) • Combining (1) & (2) + algebra gives Lagrangian in terms of R,r,r: L = (½)M|R|2 + (½)μ|r|2 - U Or: L = LCM + Lrel . Where: LCM  (½)M|R|2 Lrel  (½)μ|r|2 - U  Motion separates into 2 parts: 1.CM motion, governed by LCM  (½)M|R|2 2.Relative motion, governed by Lrel  (½)μ|r|2 - U(r,r,..)

  5. CM & Relative Motion • Lagrangian for 2 body problem:L = LCM + Lrel LCM  (½)M|R|2 ; Lrel  (½)μ|r|2 - U(r,r,..)  Motion separates into 2 parts: 1. Lagrange’s Eqtns for 3 components of CM coordinate vector R clearly gives eqtns of motion independent of r. 2. Lagrange Eqtns for 3 components of relative coordinate vector r clearly gives eqtns of motion independent of R. • By transforming from (r1, r2) to (R,r): The 2 body problem has been separated into 2 one body problems!

  6. Lagrangian for 2 body problem L = LCM + Lrel  Have transformed the 2 body problem into 2 one body problems! 1.Motion of the CM, governed by LCM  (½)M|R|2 2. Relative Motion,governed by Lrel  (½)μ|r|2 - U(r,r,..)

  7. Motion of CM is governed by LCM  (½)M|R|2 • Assuming no external forces. • R = (X,Y,Z) 3 Lagrange Eqtns; each like: (d/dt)(∂[LCM]/∂X) - (∂[LCM]/∂X) = 0 (∂[LCM]/∂X) = 0  (d/dt)(∂[LCM]/∂X) = 0  X = 0, CM acts like a free particle! • Solution: X = Vx0 = constant • Determined by initial conditions! X(t) = X0 + Vx0t , exactly like a free particle! • Same eqtns for Y, Z: R(t) = R0 + V0t , exactly like a free particle! CM Motion is identical to trivial motion of a free particle.Uniform translation of CM. Trivial & uninteresting!

  8. 2 bodyLagrangian: L = LCM + Lrel 2 body problem is transformed to 2 one body problems! 1.Motion of the CM,governedbyLCM  (½)M|R|2 Trivialfree particle-like motion! 2. Relative Motion, governed by Lrel  (½)μ|r|2 - U(r,r,..)  2 body problem is transformed to 2 one body problems, one of which is trivial! All interesting physics is in relative motion part!  Focus on it exclusively!

  9. Relative Motion • Relative Motion is governed by Lrel  (½)μ|r|2 - U(r,r,..) • Assuming no external forces. • Henceforth: Lrel  L(Drop subscript) • For convenience, take origin of coordinates at CM:  R = 0 r1 = (μ/m1)r; r2 = - (μ/m2)r μ (m1m2)(m1+m2) (μ)-1 (m1)-1 +(m2)-1

  10. The 2 body, central force problemhas been formally reduced to an EQUIVALENT ONE BODY PROBLEM in which the motion of a “particle” of mass μin U(r,r,..) is what is to be determined! • Superimpose the uniform, free particle-like translation of CM onto the relative motion solution! • If desired, if get r(t), can get r1(t) & r2(t) from above. Usually, the relative motion (orbits) only is wanted & we stop atr(t).

  11. Sect. 3.2: Eqtns of Motion & 1st Integrals • System: “Particle” of mass μ(μ m in what follows) moving in a force field described by potential U(r,r,..). • Now, restrict to conservative Central Forces: U  V where V = V(r) • Note:V(r) depends only on r = |r1 - r2| = distance of particle from force center. No orientation dependence. System has spherical symmetry  Rotation about any fixed axis can’t affect eqtns of motion.  Expect the angle representing such a rotation to be cyclic &the corresponding generalized momentum (angular momentum)to be conserved.

  12. Angular Momentum • By the discussion in Ch. 2: Spherical symmetry  The Angular Momentum of the system is conserved: L = r  p = constant(magnitude & direction!) Angular momentum conservation!  r & p(& thus the particle motion!) always lie in a plane  L, which is fixed in space. Figure: (See text discussion for L = 0)  The problem is effectively reduced from 3d to 2d (particle motion in a plane)!

  13. Motion in a Plane • Describe 3d motion in spherical coordinates(Goldstein notation!):(r,θ,). θ = angle in the plane (plane polar coordinates). = azimuthal angle. • Lis fixed,as we saw.The motion is in a plane.Effectively reducing the 3d problem to a 2d one! • Choose the polar (z) axis along L.  = (½)π& drops out of the problem. • Conservation of angular momentumL 3 independent constants of the motion (1st integrals of the motion): Effectively we’ve used 2 of these to limit the motion to a plane. The third (|L| = constant) will be used to complete the solution to the problem.

  14. Summary So Far • Started with 6d, 2 body problem. Reduced it to 2, 3d 1 body problems, one (CM motion) of which is trivial. Angular momentum conservation reduces 2nd 3d problem (relative motion)from 3d to 2d(motion in a plane)! • Lagrangian(μ m , conservative, central forces): L= (½)m|r|2 - V(r) • Motion in a plane  Choose plane polar coordinates to do the problem:  L= (½)m(r2 + r2θ2) - V(r)

  15. L= (½)m(r2 + r2θ2) - V(r) • The Lagrangian is cyclic in θ  The generalized momentum pθis conserved: p (∂L/∂θ) = mr2 θ Lagrange’s Eqtn: (d/dt)[(∂L/∂θ)] - (∂L/∂θ) = 0  pθ= 0, pθ= constant = mr2θ • Physics: pθ = mr2θ = angular momentum about an axis  the plane of motion. Conservation of angular momentum,as we already said! • The problem symmetry has allowed us to integrate one eqtn of motion. pθ a “1st Integral” of motion. Convenient to define:   pθ mr2θ= constant.

  16. L= (½)m(r2 + r2θ2) - V(r) • In terms of   mr2θ= constant, the Lagrangian is: L= (½)mr2 + [2(2mr2)] - V(r) • Symmetry & the resulting conservation of angular momentum has reduced the effective 2d problem (2 degrees of freedom) to an effective1d problem! 1 degree of freedom, one generalized coordinate r! • Now: Set up & solve the problem using the above Lagrangian. Also, follow authors & do with energy conservation. However, first, a side issue.

  17. Kepler’s 2nd Law • Const. angular momentum   mr2θ • Note that  could be < 0 or > 0. • Geometric interpretation: = const: See figure: • In describing the path r(t), in time dt, the radius vector sweeps out an area: dA = (½)r2dθ

  18. In dt, radius vector sweeps out area dA = (½)r2dθ • Define Areal Velocity (dA/dt)  (dA/dt) = (½)r2(dθ/dt) = dA = (½)r2θ(1) But   mr2 θ = constant  θ = (/mr2) (2) • Combine (1) & (2):  (dA/dt) = (½)(/m) = constant!  Areal velocity is constant in time!  the Radius vector from the origin sweeps out equal areas in equal times  Kepler’s 2nd Law • First derived by empirically by Kepler for planetary motion.General result for central forces! Not limited to the gravitational force law (r-2).

  19. Lagrange’s Eqtn for r • In terms of   mr2θ= const, the Lagrangian is: L= (½)mr2 + [2(2mr2)] - V(r) • Lagrange’s Eqtn for r: (d/dt)[(∂L/∂r)] - (∂L/∂r) = 0  mr -[2(mr3)] = - (∂V/∂r)  f(r) (f(r) force along r) Rather than solve this directly, its easier to use Energy Conservation. Come back to this later.

  20. Energy • Note: Linear momentum is conserved also: • Linear momentum of CM.  Uninteresting free particle motion • Total mechanical energy is also conserved since the central force is conservative: E = T + V = constant E = (½)m(r2 + r2θ2) + V(r) • Recall, angular momentum is:   mr2θ= const  θ = [(mr2)]  E = (½)mr2 + (½)[2(mr2)] + V(r) =const Another “1st integral” of the motion

  21. r(t) & θ(t) E = (½)mr2 + [2(2mr2)] + V(r) = const • Energy Conservation allows us to get solutions to the eqtns of motion in terms of r(t) & θ(t) and r(θ) or θ(r)The orbit of the particle! • Eqtn of motion to get r(t): One degree of freedom  Very similar to a 1 d problem! • Solve for r = (dr/dt) : r =  ({2/m}[E - V(r)] - [2(m2r2)])½ • Note: This gives r(r), the phase diagram for the relative coordinate & velocity. Can qualitatively analyze (r part of) motion using it, just as in 1d. • Solve for dt & formally integrate to get t(r). In principle, invert to get r(t).

  22. r =  ({2/m}[E - V(r)] - [2(m2r2)])½ • Solve for dt & formally integrate to get t(r): t(r) = ∫dr({2/m}[E - V(r)] - [2(m2r2)])-½ • Limits r0  r, r0determined by initial condition • Note the square root in denominator! • Get θ(t) in terms of r(t) using conservation of angular momentum again:   mr2θ= const  (dθ/dt) = [(mr2)]  θ(t) = (/m)∫(dt[r(t)]-2) + θ0 • Limits 0  t θ0determined by initial condition

  23. Formally,the 2 body Central Force problem has been reduced to the evaluation of 2 integrals: (Given V(r) can do them, in principle.) t(r) = ∫dr({2/m}[E - V(r)] - [2(m2r2)])-½ • Limits r0  r, r0determined by initial condition θ(t) = (/m)∫(dt[r(t)]-2) + θ0 • Limits 0  t, θ0determined by initial condition • To solve the problem, need 4 integration constants: E, , r0 , θ0

  24. Orbits • Often, we are much more interested in the path in the r, θ plane: r(θ) or θ(r)The orbit. • Note that (chain rule): (dθ/dr) = (dθ/dt)(dt/dr) = (dθ/dt)(dr/dt) Or: (dθ/dr) = (θ/r) Also,   mr2θ= const  θ = [/(mr2)] Use r =  ({2/m}[E - V(r)] - [2(m2r2)])½  (dθ/dr) =  [/(mr2)]({2/m}[E - V(r)] - [2(m2r2)])-½ Or: (dθ/dr) =  (/r2)(2m)-½[E - V(r) -{2(2mr2)}]-½ • Integrating this will giveθ(r) .

  25. Formally: (dθ/dr) =  (/r2)(2m)-½[E - V(r) -{2(2mr2)}]-½ • Integrating this gives a formal eqtn for the orbit: θ(r) = ∫ (/r2)(2m)-½[E - V(r) - {2(2mr2)}]-½ dr • Once the central force is specified, we know V(r) & can, in principle, do the integral & get the orbit θ(r), or, (if this can be inverted!) r(θ).  This is quite remarkable! Assuming only a central force law & nothing else: We have reduced the original 6 d problem of 2 particles to a 2 d problem with only 1 degree of freedom. The solution for the orbit can be obtained simply by doing the above (1d) integral!

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