Central Force Problem

Mechanics
Physics 151
Lecture 5
Central Force Problem
(Chapter 3)

What We Did Last Time
• Introduced Hamilton’s Principle
• Action integral is stationary for the actual path
• Derived Lagrange’s Equations
• Used calculus of variation
• Discussed conservation laws
• Generalized (conjugate) momentum
• Symmetry – Invariance – Momentum conservation
• We are almost done with the basic concepts
• One more thing to cover …

Goals for Today
• Energy conservation
• Define energy function
• Subtle difference from the Newtonian version
• Central force problem First application
• Motion of a particle under a central force
• Simplify the problem using angular momentum conservation
• Discuss qualitative behavior of the solution
• Use energy conservation
• Distinguish bounded/unbounded orbits
• Actual solution Thursday

Energy Conservation
• Consider time derivative of Lagrangian
dL( q, q
, t) ∂L dq
j ∂L dq
j ∂L
= ∑ + ∑ +
dt ∂q dt ∂q
dt ∂t
• Using Lagrange’s equation
one can derive
d ⎛ ∂L ⎞ ∂L
qj
− L + =
dt ⎜∑
j ∂q ⎟
⎝
j ⎠ ∂t
j
j
• Conserved if Lagrangian does not depend explicitly on t
j
0
j
∂L d ⎛ ∂L
⎞
= ∂qj
dt⎜
q ⎟
⎝∂
j ⎠
Define this as energy function hqqt ( , ,
)

“Energy” Function?
• Does energy function represent the total energy?
• Let’s try an easy example first
• Single particle moving along x axis
2
2
mx
L= −V( x)
h= mx
−L
2
2
mx
= + V( x)
= T + V
2
• How general is this?
∂L
hqqt ( , , ) ≡ ∑ qj
− L
∂q
j
Total
energy
j

Energy Function
• Suppose L can be written as
L( q, q, t) = L ( q, t) + L ( q, q, t) + L ( q, q, t)
• True in most cases
of interest
• Derivatives satisfy
∂L
∂q
0
0
j
=
∑ q
j
0 1 2
j
∂L
∂q
1
j
=
L
1
1 st order in q
∑ q
j
j
∂L
∂q
2
j
2 nd order in q
= 2L
∂L
hqqt ( , , ) ≡ ∑ qj
− L= L −L
j ∂q
j
2 0
∂L
hqqt ( , , ) ≡ ∑ qj
− L
∂q
2
j
Euler’s
theorem
j

Energy Function
hqqt ( , , ) = L − L L= T −V
2 0
• Energy function equals to the total energy T + V if
T L 2
and
= V = −L0
• 1 st condition is satisfied if transformation from r i to q j is
time-independent
• 2 nd condition holds if the potential is velocity-independent
• No frictions Friction would dissipate energy
• Let’s look into the 1 st condition

Kinetic Energy
mi
T = ∑
2
i
2
i
• Using the chain rule
r
• This wouldn’t work if r = r( q ,..., q , t)
because
dr
i
dt
r
= r(
q ,..., q )
i i 1 n
dr
∂
= ∑ ∂ q
j
dt
i
∂
= ∑ ∂ q
i i 1 n
ri
q
j
j
ri
q
j
j
∂ri
+ ∂ t
Time-independent
m m m
r = ∂r ⋅ ∂r qq = qq
∂r ⋅
∂r
∑ ∑ ∑ ∑ ∑
j
i 2
i i i i i i
i j k j k
i 2 i 2 jk , ∂qj ∂qk jk , i 2 ∂qj ∂qk
2 nd order homogeneous No q

Energy Conservation
∂L
hqqt ( , , ) ≡ ∑ qj
− L
j ∂q
j
• Energy function equals to the total energy if
• Constraints are time-independent
Kinetic energy T is 2 nd order homogeneous function of
the velocities
• Potential V is velocity-independent
• Energy function is conserved if
• Lagrangian does not depend explicitly on time
• These are restatement of the energy conservation
theorem in a more general framework
• Conditions are clearly defined

Central Force Problem
• Consider a particle under a central force
• Force F parallel to r
• Assume F is conservative
• V is function of |r| if F is central
• Such systems are quite common
• Planet around the Sun
• Satellite around the Earth
• Electron around a nucleus
F = −∇V()
r
• These examples assume the body at the center is heavy
and does not move
O
F
r
m

Two-Body Problem
• Consider two particles without external force
• r 1 and r 2 relative to center of mass
• Lagrangian is
L
Motion of CoM
( m + m ) R
m
∑ r
2 2 2
1 2
i i
= + −
2 i=
1 2
Motion of particles
around CoM
V()
r
m 1
O
R
r 1
CoM
r 2
m 2
Potential is function of
|r| = |r 2 – r 1 |
Strong law of action and reaction
m
m
( )
2
1
r1
=− r r2
=
( m1+
m2)
m1+
m2
r
2 2
mi i
1 mm
∑ r
1 2
=
i=
1 m1+
m2
2 2 ( )
r
2

Two-Body Central Force
L
( m + m ) R
1 mm
r
2 2 ( m + m )
2
1 2 1 2 2
= + −
• R is cyclic
1 2
V()
r
• CoM moves at a constant velocity
• Move O to CoM and forget about it
m 1
O
R
r
CoM
m 2
L
1 mm
r
2( m + m )
= 1 2 2
−
1 2
V()
r
Relative motion of two particles is identical to the motion of
one particle in a central-force potential
mm
1 2
• Reduced mass µ = or
( m m )
1 1 1
µ = m
+ m
1+ 2
1 2

Hydrogen and Positronium
• Positronium is a bound state of a positron
and an electron
• Similar to hydrogen except m(p) >> m(e + )
• Potential V(r) is identical
• Turn them into central force problem
µ = mm
e e
me
positronium
( me
+ me) = 2
µ = mm
p e
hydrogen
me
( m + m )
≈
p
• Spectrum of positronium identical to
hydrogen with m e m e /2
e
e −
e +
e −
p
q
V()
r = −
r
2

Spherical Symmetry
• Central-force system is spherically symmetric
• It can be rotated around any axis through the origin
2
• Lagrangian L= T( r ) −V( r)
doesn’t depend on the
direction
L= r× p=const
• Direction of L is fixed
• r ⊥ L by definition r is always in a plane
• Angular momentum is conserved
• Choose polar coordinates
• Polar axis = direction of L
r = r(, r θ , ψ) = r(, r θ)
Azimuth Zenith = 1/2π
L
O
r

More Formally
• Lagrangian in polar coordinates
• θ is cyclic, but ψ is not
• We can choose the polar axis so that the initial condition is
• Now ψ is constant. We can forget about it
r
=
r
(, r θ , ψ )
m
L= T − V = r
+ r + r −V r
2
d ⎛ ∂L ⎞ ∂L
dt ⎝∂ψ
⎠ ∂ψ
2 2 2 2 2 2
( sin ψθ ψ ) ( )
( sin cos ) 0
2 2
⎜ ⎟ − = mr ψ − ψ ψθ =
ψ = π , ψ
= 0
2
2 nd term vanishes
ψ = 0

Angular Momentum
m
L= T − V = r
+ r −V r
2
• θ is cyclic. Conjugate momentum p θ conserves
• Alternatively
2 2 2
( θ ) ( )
∂L
= = = ≡
∂ θ
2
pθ
mr θ const l
Areal velocity
• Kepler’s 2 nd law
dA
dt
1
r
2
2
= θ =
• True for any central force
const
Magnitude of
angular momentum
dr
dA

Radial Motion
m
L = T − V = r
+ r −V r
2
2 2 2
( θ ) ( )
• Lagrange’s equation for r
• Derivative of V is the force
2
mr = mr θ + f () r
Centrifugal force
• Using the angular momentum l
2
l
mr = + f () r
3
mr
d
dt
2 ∂V()
r
( mr
) − mr θ + = 0
∂r
Central force
∂V()
r
f()
r =− ∂ r
l
=
We know how to integrate this.
But we also know what we’ll
get by integrating this
2
mr θ

Energy Conservation
1
E = T + V = m r + r + V r = m r + l + V r =
2 2 2 mr
2
2 2 2 2
( θ ) ( )
( ) const
2
2
2 ⎛
l
r
= ⎜E−V()
r −
m⎝
2mr
• One can solve this (in principle) by
t r dr
t = ∫ dt = ∫
= t()
r
0 r0
2
2 ⎛
l ⎞
⎜E−V()
r −
2 ⎟
m⎝
2mr
⎠
• Then invert t(r) r(t)
• Then calculate θ(t) by integrating
2
⎞
⎟
⎠
1 st order differential
equation of r
θ =
l
mr
2
NB: This never
goes negative
Done! (?)

Degrees of Freedom
• A particle has 3 degrees of freedom
• Eqn of motion is 2 nd order differential 6 constants
• Each conservation law reduces one differentiation
• By saying “time-derivative equals zero”
• We used L and E 4 conserved quantities
• Left with 2 constants of integration = r 0 and θ 0
• We don’t have to use conservation laws
• It’s just easier than solving all of Lagrange’s equations

Qualitative Behavior
• Integrating the radial motion
isn’t always easy
• More often impossible…
• You can still tell general behavior by looking at
E
V′ () r ≡ V()
r +
2
l
2mr
2
• Energy E is conserved, and E – V’ must be positive
2
mr
= + V′
() r
2
mr
2
• Plot V’(r) and see how it intersects with E
2
2
2 ⎛
l
r
= ⎜E−V()
r −
m⎝
2mr
Quasi potential including
the centrifugal force
= E− V′
() r > 0
E
2
> V′
() r
⎞
⎟
⎠

Inverse-Square Force
• Consider an attractive 1/r 2 force
k
k
f()
r =− V()
r = −
2
r
r
• Gravity or electrostatic force
2
k l
V′ () r =− +
2
r 2mr
• 1/r 2 force dominates at large r
• Centrifugal force dominates at
small r
• A dip forms in the middle
V′
() r
2
l
2mr
−
2
k
r
r

Unbounded Motion
• Take V’ similar to 1/r 2 case
• Only general features are relevant
• E = E 1 r > r min
E
= V′
( r )
1 min
• Particle can go infinitely far
E 1
V′
() r
1
2 mr
2
Arrive from r = ∞
E 2
r
Turning point
E = V ′ r = 0
Go toward r = ∞
E 3
A 1/r 2 force would
make a hyperbola

Bounded Motion
• E = E 2 r min < r < r max
• Particle is confined between two
circles
E 1
V′
() r
Goes back and
forth between
two radii
E 2
1
2 mr
2
r
E 3
Orbit may or may not be
closed. (This one isn’t)
A 1/r 2 force would
make an ellipse

Circular Motion
• E = E 3 r = r 0 (fixed)
• Only one radius is allowed
Stays on a circle
• Classification into unbounded, bounded and circular motion
depends on the general shape of V’
• Not on the details (1/r 2 or otherwise)
E
= V′
( r )
r = 0
0
r = const = r
0
E 1
E 2
E 3
V′
() r
r 0
r

Another Example
V
a
r
=− f = −
3
4
• Attractive r -4 force
• V’ has a bump
3a
r
• Particle with energy E may be
either bounded or unbounded,
depending on the initial r
a
V ′ =− +
r
2
l
2mr
3 2
E
V ′
2
l
2mr
V
2
r

Stable Circular Orbit
• Circular orbit occurs at the bottom of a dip of V’
mr
2
2
= E− V′ =
• Top of a bump works in theory,
but it is unstable
• Initial condition must be exactly
r
= 0 and r = r
0
r = const
Stable circular orbit requires
dV ′
mr = − =
dr
0
0
2
dV
0
2
dr
′ >
E
E
r 0
r
stable
unstable
r

Power Law Force
V′ () r ≡ V()
r +
2
l
2mr
2
dV ′
dr
r=
r
0
2
l
=−f( r ) − = 0
0 3
mr0
2 2
dV′
df 3l
= − + >
dr dr mr
2 4
r=
r
r=
r0
0
0
0
df
dr
r=
r
0
0 for attractive force
• Condition for stable circular orbit is
n−1 n−1
− knr < 3kr
n >−3
0 0
f
=−kr
Power-law forces with n > –3 can make stable circular orbit
n

Summary
• Started discussing Central Force Problems
• Reduced 2-body problem into central force problem
2
l
mr = + f () r
3
mr
• Problem is reduced to one equation
• Used angular momentum conservation
2
l
V′ () r ≡ V()
r +
2mr
• Unbounded, bounded, and circular orbits
• Condition for stable circular orbits
• Qualitative behavior depends on
• Next step: Can we actually solve for the orbit?
2