2. 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!

3. 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).

4. 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)

5. 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

6. 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

7. 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)

8.  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)

9.  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!

10. 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!

11.  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!

12. 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)

13. 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).

14. 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.

15. 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)!

16.  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)

17. 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.

18. 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.

19. 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

20.  (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)

21. 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