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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 8.2 & 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)
• 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
• 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
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
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 particlelike 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
• 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)!
• 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)
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)μ (r2 + r2θ2) - 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
• 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
• 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