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!

• 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

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• 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