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Sect. 3.3: Equivalent “1d” Problem

Sect. 3.3: Equivalent “1d” Problem. Formally, the 2 body Central Force problem has been reduced to evaluation of 2 integrals , which will give r(t) & θ (t) : (Given V(r) can do them, in principle.) t(r) =  ∫ dr ({2/ m}[E - V(r)] - [  2  (m 2 r 2 )]) -½ (1)

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Sect. 3.3: Equivalent “1d” Problem

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  1. Sect. 3.3: Equivalent “1d” Problem • Formally, the 2 body Central Force problem has been reduced to evaluation of 2 integrals, which will give r(t) & θ(t) : (Given V(r) can do them, in principle.) t(r) = ∫dr({2/m}[E - V(r)] - [2(m2r2)])-½ (1) • Limits r0  r, r0determined by initial conditions • Invert this to get r(t) & use that in θ(t) (below) θ(t) = (/m) ∫(dt/[r2(t)]) + θ0 (2) • Limits 0  t, θ0determined by initial condition • Need 4 integration constants:E, , r0, θ0 • Most cases: (1), (2) can’t be done except numerically • Before looking at cases where they can be done: Discuss the PHYSICS of motion obtained from conservation theorems.

  2. Assume the system has known energyE & angular momentum ( mr2θ). • Find the magnitude & direction of velocity v in terms of r: • Conservation of Mechanical Energy:  E = (½)mv2 + V(r) = const (1) Or: E = (½)m(r2 + r2θ2) + V(r) = const (2) v2 = square of total (2d) velocity: v2 r2 + r2θ2 (3) (1)  Magnitude of v: v =  ({2/m}[E - V(r)])½ (4) (2)  r =  ({2/m}[E - V(r)] - [2(m2r2)])½ (5) Combining (3), (4), (5) gives the direction of v • Alternatively,  = mr2θ = const, gives θ. Combined with (5) gives both magnitude & direction of v.

  3. Lagrangian :L= (½)m(r2 + r2θ2) - V(r) • In terms of   mr2θ= const, this is: L= (½)mr2 + [2(2mr2)] - V(r) • Lagrange Eqtn for r: (d/dt)[(∂L/∂r)] - (∂L/∂r) = 0  mr -[2(mr3)] = - (∂V/∂r)  f(r) (f(r) force along r) Or: mr = f(r) +[2(mr3)] (1) • (1) involves only r & r.  Same Eqtn of motion (Newton’s 2nd Law) as for a fictitious (or effective) 1d (r) problem of mass m subject to a force: f´(r) = f(r) +[2(mr3)]

  4. Centrifugal “Force” & Potential • Effective 1d (r) problem: m subject to a force: f´(r) = f(r) +[2(mr3)] • PHYSICS:Using   mr2θ: [2(mr3)]  mrθ2 m(vθ)2/r  “Centrifugal Force” • Return to this in a minute. • Equivalently, energy: E = (½)m(r2 + r2θ2) + V(r) = (½)mr2 + (½)[2(mr2)] + V(r) =const • Same energy Eqtn as for afictitious(or effective) 1d (r) problem of mass m subject to a potential: V´(r) =V(r) + (½)[2(mr2)] • Easy to show that f ´(r) = - (∂V´/∂r) • Can clearly write E = (½)mr2 + V´(r) = const

  5. Comments on Centrifugal “Force” & Potential: • Consider: E = (½)mr2 + (½)[2(mr2)] + V(r) • Physicsof [2 (2mr2)]. Conservation of angular momentum:  = mr2θ [2(2mr2)]  (½)mr2θ2  Angular part of kinetic energyof mass m. • Because of the form [2 (2mr2)], this contribution to the energy depends only on r: When analyzing the r part of the motion, can treat this as an additional part of the potential energy.  It’s often convenient to call it another potential energy term “Centrifugal” Potential Energy

  6. [2 (2mr2)]“Centrifugal” PE Vc(r) • As just discussed, this is really the angular part of the Kinetic Energy!  “Force” associated with Vc(r): fc(r)  - (∂Vc/∂r) = [2 (mr3)] Or, using  = mr2θ: fc(r) = [2(mr3)] = mrθ2 m(vθ)2/r  “Centrifugal Force”

  7. fc(r) = [2 (mr3)]  “Centrifugal Force” • fc(r) = Fictitious “force” arising due to fact that the reference frame of the relative coordinate r (of “particle” of mass m) is not an inertial frame! • NOT (!!)a force in the Newtonian sense! A part of the “ma” of Newton’s 2nd Law, rewritten to appear on the “F” side. For more discussion: See Marion, Ch. 10. • “Centrifugal Force”:Unfortunate terminology! Confusing to elementary physics students! Direction of fc :Outward from the force center! • I always tell such students that there is no such thing as centrifugal force! • Particle moving in a circular arc: Force in an Inertial Frame is directed INWARD TOWARDS THE CIRCLE CENTER  Centripetal Force

  8. Effective Potential • For both qualitative & quantitative analysis of the RADIAL motion for “particle” of mass m in a central potential V(r), Vc(r) = [2(2mr2)] acts as an additional potential & we can treat it as such! • But recall that physically, it comes from the Kinetic Energy of the particle!  As already said, lump V(r) &Vc(r) together into an Effective Potential V´(r)  V(r) + Vc(r)  V(r) + [2(2mr2)]

  9. Effective Potential V´(r)  V(r) + Vc(r)  V(r) + [2(2mr2)] • Consider now: E = (½)mr2 + (½)[2(mr2)] + V(r)= (½)mr2 +V´(r) = const r =  ({2/m}[E - V´(r)])½ (1) • Given U(r), can use (1) to qualitatively analyze the RADIAL motion for the “particle”. Get turning points, oscillations, etc. Gives r vs.r phase diagram. • Similar to analysis of 1 d motion where one analyzes particle motion for various E using E = (½)mx2 + V(x) = const

  10. Importantspecial case:Inverse square law central force:f(r) = -(k/r2)  V(r) = -(k/r) • Taking V(r  )  0 • k > 0: Attractive force. k < 0: Repulsive force. • Gravity: k = GmM. Always attractive! • Coulomb (SI Units):k = (q1q2)/(4πε0). Could be attractive or repulsive! • For inverse square law force, effective potential is: V´(r)  V(r) + [2(2mr2)] = -(k/r) + [2(2mr2)]

  11. V´(r) for Attractive r-2 Forces • Qualitatively analyze motion for different energies E in effective potential for inverse square law force. (Figure): V´(r) = -(k/r) + [2(2mr2)] E = (½)mr2 + V´(r)  E - V´(r) = (½)mr2 0  r = 0 at turning points (E = V´(r)) NOTE: This analysis is for the r part of the motion only. To get the particle orbit r(θ), must superimpose θ motion on this!

  12. Motion of particle with energy E1 >0 (figure): E1 - V´(r) = (½)mr2  0  r = 0 at turning point r1 (E1 = V´(r1)) • r1 = min distance of approach • No max r:  Unbounded orbit! • Particle from r   comes in towards r=0. At r = r1, it “strikes” the “repulsive centrifugal barrier”, is repelled (turns around) & travels back out towards r   . It speeds up until r = r0 = min of V´(r). Then, slows down as it approaches r1. After it turns around, it speeds up to r0& then slows down to r   .

  13. Motion of particle with energy E1 >0 (continued): E1 -V´(r) = (½)mr2  0 • Also: E1 - V(r) = (½)mv2  0  V (r) - V´(r) = (½)mv2- (½)mr2 = (½)mr2θ2 = [2 (2mr2)] = Vc(r)  From this analysis, can, for any r, get the magnitude of the velocity v +its r & θ components.  Can use this info to get anapproximate picture of the particle orbitr(θ).

  14. Motion of particle with energy E1 >0 (continued): E1 -V´(r) = (½)mr2  0. Qualitative orbit r(θ). center of force  

  15. Motion of particle with energy E2 = 0: E2 -V´(r) = (½)mr2  0.  -V´(r) = (½)mr2  0 Qualitative motion is ~ the same as for E1, except the turning point is at r0(figure): r0 

  16. “oscillatory” in r • Motion of particle with energy E3 < 0: E3 -V´(r) = (½)mr2  0. Qualitative motion: 2 turning points, min & max r: (r1& r2). Turning points given by solutions to E3 = V´(r).Orbit is bounded.r1& r2 “apsidal’ distances.

  17. Motion of particle with energy E3 < 0 (continued): E3 -V´(r) = (½)mr2  0. Qualitative orbitr(θ). Turning points, r1& r2. Orbit is bounded; but not necessarily closed!Bounded: Contained in the plane between the 2 circles of radii r1& r2. Only closed if eventually comes back to itself & retraces the same path over. More on this later.

  18. Motion of particle with energy E4 < 0:E4 -V´(r) = 0 (r = 0)  r = r1 (min r ofV´(r)) = constant  Circular orbit (& bounded, of course!) r(θ) = r1! Effective potential:V´(r) = V(r) + (½)[2(mr2)] Effective force:f´(r) = f(r) +[2(mr3)] = - (∂V´/∂r). Atr = r1 (min ofV´(r)) f´(r) = - (∂V´/∂r) = 0 or f(r)= -[2(mr3)]= - mrθ2. Appl. force = centripetal force

  19. Energy E < E4? E -V´(r) = (½)mr2 < 0  Unphysical! Requires r = imaginary. • All discussion has been for one value of angular momentum. Clearly changing  changes V´(r) quantitatively, but not qualitatively (except for  = 0 for which the centrifugal barrier goes away.)  Orbit types will be the same for similar energies.

  20. V´(r) for Attractive r-2 Forces • Will analyze orbits in detail later. Will find: • Energy E1 > 0: Hyperbolic Orbit • Energy E2 = 0: Parabolic Orbit • Energy E3 < 0: Elliptic Orbit • Energy E4 = [V´(r)]min: Circular Orbit

  21. Other Attractive Forces • For other types of Forces: Orbits aren’t so simple. • For any attractive V(r) still have the same qualitative division into open, bounded, & circular orbits if: 1.V(r)falls off slower thanr-2 as r   Ensures that V(r) > (½)[2(mr2)] as r    V(r) dominates the Centrifugal Potential at large r. 2.V(r) slower thanr-2 as r  0 Ensures that V(r) < (½)[2(mr2)] as r  0  The centrifugal Potential dominates V(r) at small r.

  22. If the attractivepotential V(r) doesn’t satisfy these conditions, the qualitative nature of the orbits will be altered from our discussion. • However, we can still use same method to examine the orbits. • Example: V(r) = -(a/r3) (a = constant)  Force: f(r) = - (∂V/∂r) = -(3a/r4).

  23. V´(r) for Attractive r-4 Forces • Example:V(r) = -(a/r3);  f(r) = -(3a/r4). (Fig): Eff. potential: V´(r) = -(a/r3) + (½)[2(mr2)] • Energy E, 2 motion types, depending on r: • r < r1, bounded orbit. r < r1 always. Particle passes through center of force (r = 0). • r > r2, unbounded orbit. r > r2 always. Particle can never get to the center force (r = 0). • r1 < r < r2: Not possible physically, since would require E -V´(r) = (½)mr2 < 0  Unphysical!  r imaginary!

  24. V´(r): Isotropic Simple Harmonic Oscillator • Example: Isotropic Simple Harmonic Oscillator: f(r) = - kr, V(r) = (½)kr2 Effective potential: V´(r) =(½)kr2+(½)[2(mr2)] •  = 0  V´(r) = V(r) =(½)kr2 (figure): Any E >0: Motion is straight line in “r” direction. Simple harmonic. Passes through r = 0. Turning point at r1 = motion amplitude. E -V(r) = (½)mr2 > 0  Speeds up as heads towards r = 0, slows down as heads away from r = 0. Stops at r1, turns around.

  25. Isotropic Simple Harmonic Oscillator: f(r) = - kr, V´(r) =(½)kr2+(½)[2(mr2)] •   0 (fig): All E: Bounded orbit. Turning points r1& r2. E -V´(r) = (½) mr2 > 0 Does not pass through r = 0  Oscillates in r between r1& r2. Motion in plane (r(θ))is elliptic.Proof: Take x & y components of force: fx = -kx, fy = -ky. r(θ)= Superposition of 2, 1d SHO’s, same frequency, moving at right angles to each other

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