5 Home exercise sheet 5.1 The central force problem and Scattering

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Home exercise sheet
5.1
The central force problem and Scattering
Exercise 5.1: Kepler’s First Law
Proof that the only 1r gives proper ellipse orbits by doing the following.
We said in class that there are two type of closed orbit potentials. We now want to check
kr2 and so our Lagrangian will be
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L = m(ṙ2 + r2 φ̇2 ) − kr2
2
(1)
1. Write down the Lagrangian in Cartesian coordinates. What does this look like?
2. We can parametrize the trajectory as follows
ω(t) = |ω|eiφ = a cos t + ib sin t
(2)
Notice that the parameter t is time and must not be confused with the angle φ between
the radius vector and the x - axis. This is called a Hooke Ellipse.
Draw point for different times in this parametrization and check to see what a Hooke
Ellipse is.
3. Check to see how Kepeler’s second law looks for this transformation.
4. In order to change this Hooke Ellipse to a Kepler one we must make the coordinate
transformation
Z = ω2
(3)
Again draw point for different times in this parametrization to check if this is true.
5. Define a new time τ which relates to t in such a way that Kepler’s second law still
holds for this transformation. Remember that for Z the phase is twice the phase of φ.
6. Use items three and four to find Z(τ ). This now describes a motion in a Kepler’s
ellipse which complies both the first and second Kepler’s laws.
7. Find the EOM that Z satisfies. (Hint: write down
the force)
1
d2 Z
dτ 2
and use the EOM for ω to find
Solution:
1. We write the Lagrangian in Cartesian coordinates
L=
1 2
ẋ + ẏ 2 − k x2 + y 2
2
(4)
This is a two dimensional oscillator in the x, y plane.
2. Using the parametrization we can draw points for different times and we’ll find that we
have an ellipse around the center. This is called a Hooke ellipse.
3. We know that Kepler’s law looks like |r|2 dφ
= Const. Since the parametrization we
dt
used is very similar to normal polar coordinates then Kepler’s law will be
|ω|2
dφ
= Const
dt
(5)
4. We now write Z explicitly
Z = ω 2 = |ω|2 e−2iφ =
a2 − b 2 a2 + b 2
+
cos 2t + iab sin 2t
2
2
(6)
Again we can draw points for different times, this time we will find that we have a
Kepler’s Ellipse, which is a two focal points ellipse
5. We now define a new time τ which will make the second law hold for this transformation. Remembering that the phase of Z is 2φ we get
|Z|2
d2φ
= Const
dτ
(7)
6. A suitable choice of the two constants gives the following relation
dτ |Z|2
= |ω|2
dt |ω|2
⇔
d
1 d
=
dτ
|ω|2 dt
(8)
We now have a function Z(τ ) that describes the motion in accordance with Kepler’s
First and Second Laws.
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7. We just need to find what equation Z satisfies
1 dω 2
2 d ω̇
d2 Z
1 d
=
=
dτ 2
|ω|2 dt |ω|2 dt
|ω|2 dt ω
(9)
This is solved by using the force law equation for ω which is
ω̈ = −ω
|ω̇|2 + |ω|2 = 2
and
(10)
And we receive
d2 Z
Z
= −4 3
2
dτ
|Z|
(11)
This is precisely Newton’s force law for gravity. So we conclude that Kepler’s First and
Second laws together imply the inverse square law, with the potential
V (r) = −
k
r
(12)
As requested.
Exercise 5.2: Spherically symmetric potential well
Find the motion and characterize possible trajectories of a particle of mass m in a spherically
symmetric potential well V (r) where
V (r) =
−V
0
for r < R
for r > R
(13)
(V > 0)) for different values of angular momentum E and total energy E
a. Find the particle velocity inside and outside the potential well.
b. What is the minimum distance to the center of the potential rmin for the particle with
angular momentum L and total energy E which is moving inside the well?
c. What is the relationship between L and E for a particle that is permanently bound
inside the well?
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d. What is the trajectory motion of the particle with E > 0 and angular momentum
L2
which satisfies 2mR
2 > E (large angular momentum)?
e. Describe qualitatively the trajectory motion of a particle with E > 0 and small angular
L2
momentum 2mR
2 < E. Note that even though the particle can move outside of the
well (E > 0) its minimum distance to the center of the potential rmin < R. This means
that if the particle is moving towards the center it will traverse the well.
f. Describe qualitatively (make a sketch) of trajectories of particle that is permanently
bound in the potential well.
Solution:
It is good to always draw your effective potential on a graph (makes seeing everything easier).
a. Since the Lagrangian is time independent we know that the energy is a constant of
motion and so
r
1 2
2E
Outside E =
mv
⇒ v=
(14)
2
mr
1 2
2(E + V )
Inside E =
mv − V
⇒ v=
(15)
2
m
2
L
b. Let us look at the effective potential Vef f = 2mr
2 − V , we know that the radius is
minimal when Vef f = E (this is easily deduced from the graph we drew at the beginning).
L2
Assuming that E > 2mR
2 we get
L
rmin = p
2m(E + V )
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(16)
c. A particle that is permanently bound inside the well has an energy that wont allow it
L2
to escape meaning E < 2mR
2
d. There are two possible scenarios:
– The particle comes from infinity and completely misses the spherical potential.
– The particle starts inside the potential well and is stuck in there.
e. This time we assume that the particle can enter the potential well and so it will go
through the potential well and head out.
Notice that since the angular momentum is conserved we get mbvout − mrmin vin and
since vin > vout this means that b > rmin which is why the particle moves towards to
the center and not the other way.
f. The bound particle will move between rmin ≤ r ≤ R
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r
Notice that the condition for closed orbits is given by
rational number.
arccos( min
R )
2π
= m where m is a
Exercise 5.3: Interacting particles
Two point particles of mass m1 and m2 interact via the central potential
r2
U (r) = U0 log 2
r + b2
(17)
where b is a constant with dimensions of length.
1. For what values of the relative angular momentum l does a circular orbit exist? Find
the radius r0 of the circular orbit. Is it stable or unstable?
2. Suppose the orbit is nearly circular, with r = r0 + η, where |η| << r0 . Find the
equation for the shape η(φ) of the perturbation.
3. What is the angle ∆φ through which periapsis changes each cycle? For which value(s)
of l does the perturbed orbit not precess?
Solution:
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1. We first need to write the effective potential
l2
Uef f (r) =
+ U0 log
2µr2
r2
r 2 + b2
(18)
To find the circular orbit we need to find the minimum of the potential
l2
2rU0 b2
+
µr3 r2 (r2 + b2 )
b2 l 2
=
2µb2 U0 − l2
U 0 (r) = 0 = −
⇒
r02
⇒
(19)
(20)
The condition on l is born from the condition on r0 which is r02 > 0 which means that
p
l < lc ≡ 2µb2 U0
(21)
To check if the orbit is stable or unstable we just need to check if the point is a minimum
or maximum. There are two ways:
(a) Either take the second derivative and check to see if it’s positive or negative.
(b) Draw the effective potential and check if we need a maximum or minimum.
The second way is easier so that’s what we’ll do. Since Uef f (0+ ) = ∞ and Uef f (∞) = 0,
this means that if there’s a circular orbit it will be stable.
2. To find the shape of η(φ) we will start in the same way that we would for a regular
orbit.
2
l2
dr
1 2
+ Uef f (r)
(22)
E = µṙ + Uef f (r) =
4
2
2µr
dφ
The transition is done by using l = mr2 φ̇. We can now plug in r = r0 + η and
differentiate E with respect to φ , which will give us
η 00 = −β 2 η
where β 2 =
µr04 00
U (r0 )
l2 ef f
(23)
For our potential we get
l2
β =2 1− 2
lc
2
(24)
Which has the trivial solution
η(φ) = A cos (βφ + δ)
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(25)
3. The amount by which periapsis changes each cycle is given by
∆φ = φn+1 − φn − 2π
(26)
We now need to check what is φn . Setting η = η0 we obtain the sequence of φ values
φn = δ0 +
2πn
β
(27)
This we plug into ∆φ to receive
∆φ = 2π β − 1 − 1
(28)
If β > 1 then ∆φ < 0 and the periapsis advances each cycle (i.e. it comes sooner with
every cycle).
p If β < 1 then ∆φ > 0 and the periapsis recedes. For β = 1 which means
that l = µb2 U0 there is no precession and ∆φ = 0.
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