Download Green's Function and Central Forces, Lecture Notes - Physics and more Study notes Engineering Physics in PDF only on Docsity!
Green’s Function and Central Forces
Adrian Down
October 06, 2005
1 Green’s Function
1.1 Recap
Last time, we found the response of an underdamped oscillator to an impulse,
z(t) =
I
mω 1
e
−βt
sin (ω 1 t)
Note. This solution assumes that z(0) = z˙(0) = 0. For different initial
conditions, the derivation would have to be redone.
1.2 Multiple impulses
1.2.1 Motivation
Our goal is to move towards a solution for the response of the oscillator
to an arbitrary number of impulses. We then argue that any force can be
decomposed into an large number of infinitesimally short impulses.
1.2.2 Two impulses
Consider two impulses at times t 1
and t 2
with magnitudes I 1
and I 2
, respec-
tively. We know that the solutions are
z¨ 1
2
0
z 1
= F
1
(t)
z¨ 2
2
0
z 2
= F
1
Since the solutions are linear, they add by superposition.
Region I
z = 0 t < t 1
Region II
z(t) =
I 1
mω 1
e
−β(t−t 1 )
sin (ω 1 (t − t 1 )) t 1 ≤ t < t 2
Region III
z(t) =
I
1
mω 1
e
−β(t−t 1 )
sin (ω 1
(t − t 1
I
2
mω 1
e
−β(t−t 2 )
sin (ω 1
(t − t 2
1.2.3 Any number of impulses
By linearity, we can add solutions for an arbitrary number of impulses,
z(t) =
tn<t
I
n
mω 1
e
−β(t−tn)
sin (ω 1
(t − t n
where I n
= F
n
(t)dt
We can consider an arbitrary force to be composed of an infinite number
of infinitesimally short impulses. The sum then becomes an integral.
z(t) =
t
0
F (t
′
)dt
′
mω 1
e
−β(t−t
′ )
sin (ω 1
(t − t
′
))
The integral is usually written,
z(t) =
t
0
dt
′
G(t, t
′
)F (t
′
)
where =
e
−β(t−t
′ )
mω 1
sin (ω 1
(t − t
′
))
Note. • Remember that this solution is only valid for an underdamped
oscillator with our assumed initial conditions. Any other case would
require the derivation to be re-performed.
- The general method is
- Solve the response of the system to an impulse
- Use linearity to superpose solutions
- Consider the forcing function as a series of contiguous impulses
Let ˆs be a particular direction. We saw from Noether’s theorem,
F · ˆs = 0 ⇒
d
dt
~p · ˆs = 0 ⇒ ~ps = consntant
Thus if there is no force in a particular direction, momentum in that direction
is conserved.
In the case of torque,
~τ = ~r ×
F
~τ =
d
L
dt
L = ~r × p~
The same results for linear momentum hold in the angular case,
~τ · sˆ = 0
d
dt
L · sˆ = 0
L · sˆ = constant
Definition (Central force). A central force is one such that
F (~r) = f (r)ˆr
In the case of central forces,
~τ = ~r ×
F (r) = 0
d
L
dt
L = constant
Thus central forces conserve angular momentum.
2.2 Recasting the kinetic energy of a two-body prob-
lem
2.2.1 Motivation
Our goal is to write the Lagrangian for two interacting bodies such that it is
a function only of motion of the center of mass,
R, and the relative motion
of the two bodies, ~r,
L = T (
R, r˙) − U (r, r˙)
2.2.2 Definitions
Definition. We use the vectors
~r 1
from 0 to m 1
~r 2
from 0 to m 2
~r from m 1
to m 2
R from 0 to cm
r
′
1
from cm to m 1
r
′
2
from cm to m 2
We have the relations,
~r 1
~r 1 − ~r 2 = ~r
R +
r
′
1
= ~r 1
R +
r
′
2
= ~r 2
Computing time derivatives,
~r 1
R +
r
′
1
~r 2 =
R +
r
′
2
2.2.3 Kinetic energy
By linearity, the kinetic energies of the individual masses adds.
T =
m 1
r˙
2
1
m 2
r˙
2
2
m 1
R +
r
′
1
2
m 2
R +
r
′
2
2
m 1
R
2
m 1
r
′
2
1
R ·
r
′
1
m 2
R
2
m 2
r
′
2
2
R ·
r
′
2
Cross terms can be eliminated using the definition of the center of mass,
i
m i
~r i
i
m i
R
Definition (Reduced mass).
m 1 m 2
m 1
Using these, we can write,
L =
M
R
2
m r˙
2
− U (~r,
~r)
2.3 Center of mass motion
The first term in L is the kinetic energy of the center of mass, and the second
term represents the relative kinetic energy We get that
∂L
∂R
∂L
R
= const = p cm
Thus the momentum of the center of mass is conserved. We can consider
the motion of the system in the CM frame to get rid of the
R term. Since
the CM is translating with fixed velocity, we can translate back to the space
system with a Galilean transformation.
2.3.1 Reducing to 2D
The force is conservative, which means by definition,
F = −
∇V (r)
V (r) = −
~r
~r 0
F (
r
′ ) · d~r
′
Because the force is purely radial, angular momentum is conserved
~τ = ~r ×
F = 0
L = ~r × ~p = constant
Thus
L is always perpendicular to the motion, so there is always a single
plane in which the particle moves. In this plane, the motion is a 2D problem.
2.4 Kepler’s second law
Assume we are in the center of mass system, and for convenience, write
μ = m. Then,
L =
m( ˙r
2
2 ˙ θ
2
) − V (r)
∂L
∂θ
d
dt
∂L
θ
∂L
θ
= l θ
= constant = mr
2 ˙ θ
The area swept out by the radius vector to the orbiting body per unit
time is
dA =
r(rdθ)
dA
dt
r
2
dθ
dt
= constant
Law (Kepler’s Second). The radius vector sweeps out equal area in equal
time. really just a statement of conservation of angular momentum.
2.5 Equation of motion
2.5.1 Euler-Lagrange equations
d
dt
∂L
∂ r˙
∂L
∂r
d
dt
(m r˙) − mr
θ
2
∂V
∂r
m¨r − mr
θ
2
= f (r)
Plug in for
θ
2 to get the equation of motion,
m¨r −
l
2
mr
3
= f (r)
