central force motion chapter 8 introduction the “2 body” central force problem! –we want to...
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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!
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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).
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• 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)
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• 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
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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
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• 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)
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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)
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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!
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• 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!
Center of Mass (CM) Motion
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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!
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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)
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• 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).
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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.
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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)!
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• 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)
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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.
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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.
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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
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• 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)
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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