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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
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!
sect 3 1 reduction to equivalent 1 body problem
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!
slide4
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

slide5
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,..)

cm relative motion
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!

slide7
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,..)

slide8
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!

slide9
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!

relative motion
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

slide11
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).
sect 3 2 eqtns of motion 1 st integrals
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.

angular momentum
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)!

motion in a plane
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.

summary so far
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)

slide16
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.

slide17
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.
kepler s 2 nd law
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
slide19
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).

lagrange s eqtn for r
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.

energy
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

r t t
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).
slide23
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

slide24
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

orbits
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) .
slide26
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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