Central Force Motion Chapter 8 Introduction The “2 body” Central Force problem! We want to describe the motion of 2 bodies interacting through a central force. Central Force  A force between 2 bodies which is directed along the line between them. A very important problem! Can it solve exactly! Planetary motion & Kepler’s Laws. Nuclear forces Atomic physics (H atom), but we need the quantum mechanical version for this!

Center of Mass & Relative Coordinates, Reduced Mass Section 8 Center of Mass & Relative Coordinates, Reduced Mass Section 8.2 & Outside Sources Consider the general 3 dimensional, 2 body problem. 2 masses m1 & m2  Need 6 coordinates to describe the system. Use the components of the 2 position vectors r1 & r2 (with respect to an arbitrary origin, as in the figure).

Now, specialize to the 2 body problem with conservative Central Forces only. The 2 bodies interact with a force F = F(r), which depends only on the distance r = |r1 - r2| between m1 & m2 (no angular dependences!). F = F(r) is a conservative force  A PE exists: U = U(r)  F(r) = -U(r) r = -(dU/dr) r  The Lagrangian is: L = (½)[m1|r1|2 + m2|r2|2] - U(r)

CENTER OF MASS COORDINATE is defined as: Instead of the 6 components of the 2 vectors r1 & r2, Its usually much more convenient to transform to the (6 components of) the Center of Mass (CM) & Relative Coordinate systems. CENTER OF MASS COORDINATE is defined as: R  (m1r1 +m2r2)(m1+m2) Or: R = (m1r1+m2r2)(M) M  (m1+m2) = total mass RELATIVE COORDINATE is defined: r  r1 - r2 Its also convenient to define the Reduced Mass: μ  (m1m2)(m1+m2) A useful relation is: (1/μ)  (1/m1) +(1/m2) Algebra (student exercise!) gives inverse coordinate relations: r1 = R + (μ/m1)r; r2 = R - (μ/m2)r

Center of Mass (CM) & Relative Coordinates CENTER OF MASS (CM) COORDINATE: R = (m1r1+m2r2)(M) (see figure) RELATIVE COORDINATE: r  r1 - r2 Inverse relations: r1 = R + (μ/m1)r; r2 = R - (μ/m2)r

v1 = V + (μ/m1)v; v2 = V - (μ /m2)v (1) The Lagrangian is The velocities [vi = ri, V = R, v = r] are related by v1 = V + (μ/m1)v; v2 = V - (μ /m2)v (1) The Lagrangian is L = (½)[m1|v1|2 + m2|v2|2] - U(r) (2) Combining (1) & (2) + algebra (student exercise!) gives the Lagrangian in terms of V, r, v: L = (½)M|V|2 + (½)μ|v|2 - U(r) Or: L = LCM + Lrel Where: LCM  (½)M|V|2 & Lrel  (½)μ|v|2 - U(r)

 For the 2 body, Central Force. Problem, the motion separates into 2  For the 2 body, Central Force Problem, the motion separates into 2 distinct parts! 1. The Center of Mass Motion, governed by: LCM  (½)M|V|2 2. The Relative Motion, governed by Lrel  (½)μ|v|2 - U(r)

 For the 2 body, Central Force Problem, the  For the 2 body, Central Force Problem, the motion separates into 2 distinct parts! 1. The center of mass motion: LCM  (½)M|V|2 2. The relative motion: Lrel  (½)μ|v|2 - U(r)  Lagrange’s Equations for the 3 components of the CM coordinate vector R clearly gives equations of motion independent of r. Lagrange’s Equations for the 3 components of the relative coordinate vector r clearly gives equations of motion independent of R. By transforming coordinates from (r1, r2) to (R,r): The 2 body problem has been separated into 2 INDEPENDENT one body problems!

Center of Mass (CM) Motion The motion of the CM is governed by LCM  (½)M|V|2 (assuming no external forces). Let R = (X,Y,Z)  3 Lagrange Equations. Each looks like: ([LCM]/X) - (d/dt)([LCM]/X) = 0 [LCM]/X = 0  (d/dt)([LCM]/X) = 0  X = 0, The CM acts like a free particle! Solution: X = Vx0 = constant. Determined by initial conditions!  X(t) = X0 + Vx0t , exactly like a free particle! Similar eqtns for Y, Z:  R(t) = R0 + V0t , like a free particle! The motion of the CM is identical to the trivial motion of a free particle. It corresponds to a uniform translation of the CM through space. Trivial & uninteresting!

 We’ve transformed the 2 body, Central Force Problem, motion separates into 2 one body problems, ONE OF WHICH IS TRIVIAL! 1. The center of mass motion is governed by: LCM  (½)M|V|2 As we’ve just seen, this motion is trivial! 2. The relative motion is governed by Lrel  (½)μ|v|2 - U(r) Clearly, all of the interesting physics is in the relative motion part!  We now focus on it exclusively!

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

The 2 body, central force problem has been formally reduced to an EQUIVALENT ONE BODY PROBLEM in which the motion of a “particle” of mass μ in a potential U(r) is what is to be determined! The full solution superimposes the uniform, free particle-like translation of the CM onto the relative motion solution! If desired, if we get r(t), we can get r1(t) & r2(t) from the relations on the previous page. Usually, the relative motion (or orbit) only is wanted & we stop at r(t).

Conservation Theorems “First Integrals of the Motion” Section 8.3 Our System is effectively a “particle” of mass μ moving in a central force field described by a potential U(r). Note: U(r) depends only on r = |r1 - r2| = distance of the “particle” from the force center. There is no orientation dependence!  The system has spherical symmetry  Rotation about any fixed axis cannot affect the equations of motion.

Angular Momentum Conservation! In Section 7.9, it was shown: Spherical Symmetry Total Angular Momentum is conserved. That is: L = r  p = const (magnitude & direction!) Angular Momentum Conservation!  r & p always lie in a plane  L, which is fixed in space (figure).  The problem has now effectively been reduced from a 3d to a 2d problem (motion in a plane)!

 L = (½)μ (r2 + r2θ2) - U(r) Remarkable enough to emphasize again! We started with a 6d, 2 body problem. We reduced it to 2, 3d 1 body problems, one (the CM motion) of which is trivial. Angular momentum conservation effectively reduces second 3d problem (relative motion) from 3d to 2d (motion in a plane)! The Lagrangian is: L = (½)μ|v|2 - U(r) Motion in a plane  Use plane polar coordinates to do the problem:  L = (½)μ (r2 + r2θ2) - U(r)

pθ  A “First Integral” of the motion. L = (½)μ (r2 + r2θ2) - U(r) The Lagrangian is cyclic in θ  The corresponding generalized momentum pθ is conserved: pθ  (L/θ) = μr2θ; (L/θ) - (d/dt)[(L/θ)]= 0  pθ = 0, pθ = constant = μr2θ PHYSICS: pθ = μr2θ = The (magnitude of the) angular momentum about an axis  to the plane of the motion. Conservation of angular momentum! The problem symmetry has allowed us to integrate one equation of motion (the θ equation). pθ  A “First Integral” of the motion. It is convenient to define:   pθ = μr2θ = constant.

L = (½)μr2 + [(2)/(2μr2)] - U(r) Using the angular momentum  = μr2θ = const, the Lagrangian is: L = (½)μr2 + [(2)/(2μr2)] - U(r) 2nd Term: The KE due to the θ degree of freedom! Symmetry & the resulting conservation of angular momentum has reduced the effective 2d problem (2 degrees of freedom) to an effective (almost) 1d problem! The only non-trivial equation of motion is for the single generalized coordinate r! Could set up & solve the problem using the above Lagrangian. Instead, follow the authors & do it with energy.

Kepler’s 2nd Law First, more discussion of the consequences of the constant angular momentum  = μr2θ Note that  could be < 0 or > 0. Geometric interpretation:  = constant. See figure: In describing the “particle” path r(t), in time dt, the radius vector sweeps out an area dA = (½)r2dθ

 (dA/dt) = (½)r2(dθ/dt) = (½)r2θ (1) But  = μr2θ = constant In dt, the radius vector sweeps out an area dA = (½)r2dθ Define AREAL VELOCITY  (dA/dt)  (dA/dt) = (½)r2(dθ/dt) = (½)r2θ (1) But  = μr2θ = constant  θ = (/μr2) (2) Combine (1) & (2):  (dA/dt) = (½)(/μ) = constant!  The areal velocity is constant in time! The radius vector from the origin sweeps out equal areas in equal times  Kepler’s 2nd Law Derived empirically by Kepler for planetary motion. A general result for central forces! Not limited to gravitational forces (r-2)

Momentum & Energy E = T + U = const = (½)μ(r2 + r2θ2) + U(r) The linear momentum of the system is conserved: Linear momentum of the CM.  Uninteresting free particle motion! The total mechanical energy is E also conserved since we assumed that the central force is conservative: E = T + U = const = (½)μ(r2 + r2θ2) + U(r) Recall that the angular momentum is:   μr2θ= const  θ = [/(μr2)]  E = (½)μr2 + [2(2μr2)] + U(r) = const A “second integral” of the motion