Sects. 8.1,8.2 & 8.3
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Transcript Sects. 8.1,8.2 & 8.3
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