PHYS 3318 Section Notes
Joshua Berger
May 10, 2012
Note that the content of these notes refers heavily to the textbooks: Landau and Lifshits,
Mechanics; Goldstein, Classical Mechanics; Hand and Finch, Analytical Mechanics.
1
1.1
Lagrangians and Euler-Lagrange Equations
Overview and Context
In lecture, we learned an alternative way to formulate classical mechanics. Whereas in
previous courses, we started with an equation of motion which we could solve for x(t), we
are now working from something more fundamental. We start with the principle of least
action, from which we can derive the equations of motion. For the purposes of this course,
we will often consider the problem of classical mechnics (i.e. to find x(t)) to be solved once
we have determined the equations of motion. In this section, we wiill get some practice
determining the Lagrangian of a few simple systems and using these Lagrangians to find the
equations of motion.
1.2
Simple Pendulum with Sliding Pivot
Landau & Lifshitz, Chapter 1, Problem 2
Consider the system of a simple pendulum with a sliding pivot, as described in L & L.
The goal of this problem is to count the number of degrees of freedom, find a Lagrangian
describing them, and to derive the Euler-Lagrange equations of motion from this Lagrangian.
To determine the number of degrees of freedom, we note that there are two particles in the
system. Two particles in three dimensions means that there should be six degrees of freedom.
There are, however, some constraints on the system: both particles are constrained to move
in a plane (2 constraints), the pivot is constrained to move along a line (1 constraint), and
the length of the pendulum is fixed (1 constraint). We are left with two degrees of freedom
for the system.
Next, we would like to write down the constraints in Cartesian coordinates and, based
on the constraints, choose a more convenient set of coordinates. If the positions of the two
1
particles with masses m1 and m2 are denoted by q1 = (x1 , y1 ) and q2 = (x2 , y2 ) respectively,
then the sliding and pendulum constaints are as follows:
p
(x2 − x1 )2 + (y2 − y1 )2 = `.
(1.1)
y1 = 0,
Given these constraints, it is best to pick a convenient coordinate system. The natural
choice of coordinates is x1 and tan φ = −(x2 − x1 )/y2 as depicted in L & L. In terms of these
coordinates,
x2 = x1 + ` sin φ,
y2 = −` cos φ,
(1.2)
giving
ẋ2 = ẋ1 + `φ̇ cos φ,
ẏ2 = `φ̇ sin φ.
(1.3)
It is now time to determine the Lagrangian of the system. As we saw in lecture, L = T −V ,
where T is the kinetic energy and V is the potential energy. In Cartesian coordinates, it is
easy to write down the kinetic energy and the potential energy is due entirely to gravity:
1
1
T = m1 (ẋ21 + ẏ12 ) + m2 (ẋ22 + ẏ22 )
2
2
(1.4)
V = m1 gy1 + m2 gy2 .
(1.5)
and
Writin this in terms of our “smart” coordinates, we find:
1
1
T = m1 ẋ21 + m2 ẋ21 + 2ẋ1 `φ̇ cos φ + `2 φ̇2
2
2
(1.6)
and
V = −m2 g` cos φ.
(1.7)
1
1
L = (m1 + m2 )ẋ1 + m2 `ẋ1 φ̇ cos φ + m2 `2 φ̇2 + m2 g` cos φ.
2
2
(1.8)
The Lagrangian is given by
Finally, we want to find the Euler-Lagrange equations of motion:
d ∂L
∂L
=
dt ∂ q̇i
∂qi
(1.9)
To get there, we will start by calculating each side of the equation for our two generalized
2
coordinates:
d
d dL
=
{(m1 + m2 )ẋ1 + m2 `φ̇ cos φ}
dt dẋ1
dt
= (m1 + m2 )ẍ1 + m2 `φ̈ cos φ − m2 `φ̇2 sin φ
d dL
d
{m2 (ẋ1 ` cos φ + `2 φ̇)}
=
dt dφ̇
dt
= m2 ẍ1 ` cos φ − m2 ẋ1 `φ̇ sin φ + m2 `2 φ̈
dL
= 0
dx1
dL
= −m2 ẋ1 `φ̇ sin φ − m2 g` sin φ
dφ
Plugging these results into (1.9), we solve the problem:
1.3
(m1 + m2 )ẍ1 + m2 `φ̈ cos φ − m2 `φ̇2 sin φ = 0
(1.10)
m2 ẍ1 ` cos φ + m2 `2 φ̈ + m2 g` sin φ = 0
(1.11)
Simple Pendulum on Circle
Landau & Lifshitz, Chapter 1, Problem 3 (a)
We will work through the same exercise in this case.
The counting of degrees of freedom here is more difficult. We still have two particles
constrained to a plane. Naively, one might expect that there are four degrees of freedom,
with two for the pivot and two for the mass at the end of the pendulum. The motion of the
pivot, however, is completely constrained. It must follow the path q0 (t) = (a cos γt, a sin γt).
It does not matter that this constraint depends on time: there is nothing left to solve for to
determine the motion. It therefore has no degrees of freedom corresponding to its motion.
The mass at the end of the pendulum is constrained by the fixed length of the pendulum.
These constraints leave only one degree of freedom in this case, which we can take to be the
angle that the pendlum makes with the vertical.
We need only consider the Lagrangian for the bob at the end of the pendulum. It is
convenient to define q0 (t) = (x0 (t), y0 (t)) = (a cos(γt), a sin(γt)). Then the constraint on
motion of m can be written as
p
(x − x0 )2 + (y − y0 )2 = `
(1.12)
The convenient choice of coordinates is to use the angle of the pendulum respect to the
vertical:
x − x0
tan φ = −
(1.13)
y − y0
3
so that
y = y0 − ` cos φ.
x = x0 + ` sin φ,
(1.14)
Now we are ready to write down the Lagrangian way in the same way as for the previous
problem (neglecting the kinetic and potential energy for the pivot since it is not a degree of
freedom):
1
d
d
1
T = m(ẋ2 + ẏ 2 ) = m(x˙0 2 + y˙0 2 + 2x˙0 ` sin φ − 2y˙0 ` cos φ + `2 φ̇2
2
2
dt
dt
(1.15)
and
V = mg`y = mg`y0 − mg` cos φ
(1.16)
In writing down the Lagrangian, we can drop terms that do not depend on our coordinate
φ since they will not contribute to the Euler-Lagrange equations of motion. The resulting
Lagrangian, written in a suggestive way, is
d
d
1
L = m(−2aγ sin(γt)` sin φ − 2aγ cos(γt)` cos φ + `2 φ̇2 ) + mg` cos φ.
2
dt
dt
(1.17)
Now, inverting the product rule, f ġ = d(f g)/dt − f˙g. However, a term like d(f g)/dt will not
affect the equations of motion, as discussed in lecture, so we drop the total derivative with
respect to time. Applying this rule to the first two terms, we can put the time derivative
onto the part of these terms that is a funciton only of t, so that
1
L = m(2aγ 2 cos(γt)` sin φ − 2aγ 2 sin(γt)` cos φ + `2 φ̇2 ) + mg` cos φ.
2
(1.18)
Recall the trigonometric identity cos α sin β − sin α cos β = sin(β − α), so that
1
L = m`2 φ̇2 + ma`γ 2 sin(φ − γt) + mg` cos φ
2
(1.19)
This is the final, “nice” version of the Lagrangian. We could have kept the more complicated
version, but it will be much easier to find the equations of motion using this.
The equation of motion is then given by the Euler-Lagrange equation (1.9)
d ∂L
= m`2 φ̈
dt ∂ φ̇
∂L
= ma`γ 2 cos(φ − γt) − mg` sin φ
∂φ
m`2 φ̈ − ma`γ 2 cos(φ − γt) + mg` sin φ = 0
4
(1.20)
2
Lagrange Multipliers
The theory portion of this section is based on Goldstein, 1.3 and 2.4.
2.1
Motivation
When possible, we would like to write the Lagrangian of a system in terms of constrained
degrees of freedom. That is, given some constraints on the system, we would like to solve
the constraint equations and write the Lagrangian in terms of a smaller number of degrees of
freedom. In some cases, however, it is very difficult to go through with this procedure, so we
would like to find a way to impose the constraints without solving them directly beforehand.
In many interesting cases, we can do this using Lagrange multipliers.
Given constraints fα (q1 , . . . , qn , t) = 0, we can change to coordinates q 0 such that qα0 =
fα (q1 , . . . , qn , t) and remaining coordinates embody physical degrees of freedom. This kind of
constraint is called holonomic. Any other type of constraint is called nonholonomic (for e.g.
constraints with inequalities, constraints involving velocities, constraints involving friction).
In general, nonholonomic constraints need to implemented on a case-by-case basis in a nontrivial way. But if the constraint has the form
fα (q1 , . . . , qn , q̇1 , . . . , q̇n ) = 0,
(2.1)
then it is possible to implement the constraint using Lagrange multipliers. Also, Lagrange
multipliers give a nice way to solve problems with holonomic constraints when the constraint
equations are complicated and difficult to solve explicitly.
2.2
Theory
Suppose that equation (2.1) holds. Then, since each fα vanishes independently at all times,
any linear combination vanishes as well:
X
λα fα = 0,
(2.2)
α
for any functions of time λα (t). So such a term can be added to the Lagrangian without
affecting the unconstrained portion of the physics. Varying the action with respect to λα
enforces the constraints on the system. The number of independent equations of motion
should be reduced by the constraints and this is taken care of by additional terms in λα (t)
that then appear in the equations of motion. The resulting Euler-Lagrange equations of
motion are
∂L
d ∂L
−
= Qk
(2.3)
dt ∂ q̇k
∂qk
where
Qk =
X
α
{λα
∂fα
d
−
∂qk
dt
5
∂fα
∂ q̇k
− λ̇α
∂fα
}
∂ q̇k
(2.4)
as well as
fα = 0.
(2.5)
There is one equation per degree of freedom plus one equation per constraint. Although we
will not prove it formally, the coefficients λ are actually the generalized force required to
maintain the constraint (which is why they are a priori unknown).
2.3
Example 1: The Simple Pendulum
We have already studied the Lagrangian for the simple pendulum once, but we would like
to examine it once again using Lagrange multipliers. Rather than choosing coordinates such
that our one degree of freedom is explicit, we will impose the constraint that the length of
the pendulum is fixed. We will then derive the resulting equations of motion. Note that the
Lagrange multiplier in this case is the force required to maintain the length of the pendulum.
This force is of course known as the tension.
The kinetic energy, potential energy, and constraint equation in this case are:
mẋ2 mẏ 2
+
2
2
= mgy
= `
T =
V
p
2
x + y2
so that the Lagrangian in terms of Lagrange multipliers is:
L=
p
mẋ2 mẏ 2
+
− mgy + λ( x2 + y 2 − `)
2
2
(2.6)
We have already seen what the constraint equation is, so now let’s derive the other two
equations:
∂L
∂x
d ∂L
dt ∂ ẋ
mẍ
∂L
∂y
d ∂L
dt ∂ ẏ
mÿ
x
= λp
= λ sin θ
2
x + y2
= mẍ
= λ sin θ
(2.7)
y
= −mg + λ p
= −mg + λ cos θ
x2 + y 2
= mÿ
= −mg + λ cos θ
What is λ? It is just the tension of the string!
6
(2.8)
2.4
Example 2: Particle Sliding Down
A particle is sliding down a stationary quarter circle of radius R as shown below. At some
point the particle will detaches from the quarter circle. The goal of this problem is to
determine when the particle will detach. This will occur when the normal force maintaining
the constraint that the particle travel on the surface of the quarter circle goes to 0. This
normal force can be described by the Lagrange multiplier for the constraint. This is generally
true: when the Lagrange multiplier goes to an unphysical value, the constraint can no longer
be maintained. For simplicity, we will impose the intial conditionsθ(0) = θ0 and θ̇(0) = 0.
r
θ
The Lagrangian + multiplier is
L =
p
mẋ2 mẏ 2
+
− mgy + λ( x2 + y 2 − R).
2
2
In r, θ coordinates:
L =
mr2 θ̇2 mṙ2
+
− mgr cos θ + λ(r − R).
2
2
The EOMs are
mr2 θ̈ = mgr sin θ
mr̈ = mrθ̇2 − mg cos θ + λ.
Using the constraint r = R, we find that
mR2 θ̈ = mgR sin θ
v2
mar = m = mRθ̇2 = mg cos θ − λ
R
7
Solving the first equation and using θ̇(0) = 0, θ(0) = θ0 :
θ̈ =
dθ̇
=
dt
θ̇dθ̇ =
1 2t
θ̇ | =
2 0
⇒ θ̇2 =
g
sin θ
R
g1d
−
cos θ
R θ̇ dt
g
− d cos θ
R
g
− cos θ|t0
R
2g
2g
− cos θ +
cos θ0
R
R
Plugging this into the second equation,
−2gm cos θ + 2mg cos (θ0 ) = mg cos θ − λ
λ = −2mg cos θ0 + 3mg cos θ
The particle detaches when λ = 0:
cos θ =
2
cos θ0
3
θ = arccos
8
2
cos θ0
3
3
3.1
Symmetries
Energy Conservation
The goal of this problem is to explore energy conservation in a non-standard Lagrangian.
We would like to use conservation of energy to solve the problem of finding the maximum
velocity of a particle in this Lagrangian. . We will work with the Lagrangian
1
1
L = cv 6 − kx2
5
2
(3.1)
and initial conditions x(0) = x0 , v(0) = 0.
Since there is no explicit time dependence, energy is conserved. We can then calculate
the energy:
dL
6 5
=
cv
dv
5
dL
E =
v−L
dv
6 6 1 6 1 2
=
cv − v + kx
5
5
2
1
= cv 6 + kx2 .
2
(3.2)
When the potential energy decreases, the speed increases, so the maximum velocity occurs
when x = 0. We can then use energy conservation to say that
E(0) = Evmax =⇒
1 2
2
kx = cvmax
.
2 0
(3.3)
We can solve this trivially to find that vmax = (kx20 /2c)1/6 .
3.2
Rotation Symmetry
Rotation symmetry implies conservation of angular momentum. But how should we think
about rotation symmetry in the context of physics?
Let’s first start with the simplest case of 2D. Consider a point (x, y). First, we want to
describe a rotation counterclockwise by angle α about the origin. We write
(x, y) = (r cos φ, r sin φ)
(3.4)
in polar coordinates. We know that, under the rotation, φ → φ + α. Then,
(x, y) → (r cos(φ + α), r sin(φ + α)) =
(r [cos φ cos α − sin φ sin α] , r [sin φ cos α + sin α cos φ]) = (x cos α−y sin α, x sin α+y cos α).
(3.5)
9
x
Great! But we can do more. Let us write (x, y) as a column vector ~r =
. This
y
transformation can then be written as
x
x
cos α − sin α
→
≡ O~r.
(3.6)
y
sin α cos α
y
Notice the following property:
T
O =
cos θ sin θ
− sin θ cos θ
=⇒ OT O = 1.
(3.7)
Also note that
det O = cos2 α − (− sin2 α) = 1.
(3.8)
O is an example of what is called a (special) orthogonal matrix and all rotations are
defined to be orthogonal matrices (even in higher dimensions). We will not go through the
full description of a three dimensional rotation yet, but later in the course we will do this.
In physics language, we define a vector ~v to be any object that transforms under rotations
as ~v → O~v . So, in our example, (t, x)T is not a vector. Since O is independent of time, all
the time derivatives of ~r are also vectors. In particular, ~r˙ is a vector.
An object that is invariant under rotations is called a scalar. Very explicitly, a scalar s
transforms as s → s. How can we construct scalars out of vectors? One way (not quite the
only way–can you think of another?) is using the dot prouct. If we require our theory to
not violate mirror symmetry (parity), it is the only way. Notice that for vectors ~v1 and ~v2 ,
~v1 · ~v2 = ~v1T ~v2 . Then,
~v1T ~v2 → ~v1T OT O~v2 = ~v1T ~v2 .
(3.9)
Thus, the dot product of two vectors is a scalar.
Going back to physics, we have a Lagrangian L(q, q̇, t). Supposing for the minute that
the dot product were the only way to create scalars out of vectors, if the Lagrangian is to
be a scalar (that is, invariant under rotations), what must its form be? Well, L involves two
sets of vectors, qi and q̇i . In addition, there may be vectors related to the potential of the
system (like the location of masses/charges). It must be formed out of dot products of these
vectors.
All of this assumes that the system is completely rotationally invariant (spherical symmetry). What if there is only one rotational symmetry axis (cylindrical symmetry)? By
choosing our coordinates judiciously, we can pick that axis to be the z axis. Then, rotations
are given by


cos α − sin α 0
O =  sin α cos α 0 .
(3.10)
0
0
1
Scalars under these rotations are given by the dot product of the vectors (vx , vy )T .
10
3.3
Example of a Lagrangian
Let’s look at an example. Consider a particle with the Lagrangian
L=
m 2
v − mgr
2
(3.11)
What are the symmetries of this Lagrangian? The Lagrangian has no explicit time dependence, so energy is conserved.
It also has rotational symmetry about any axis through the ori√
gin: note that r = ~r · ~r. There is no translational or Gallilean symmetry.
To see this,
√ note
p
that under any translation ~a (time dependent or otherwise), r → (~r + ~a) · (~r + ~a) 6= ~r · ~r.
That means that momentum is not conserved! How can this be? It is simply a result of
the particle producing the potential absorbing the difference. How can this “problem” be
solved? Add a second particle so that the potential is V = mg|~r1 − ~r2 |. Then the total
momentum of the system is conserved and the relative separation of the particles obeys the
Lagrangian 3.11, with a mass given by the reduced mass.
3.4
Cylindrical Symmetry
Suppose that L = L(r, ṙ, θ̇, ż). What are the symmetries of this Lagrangian and what
conservation laws to they imply? Time translation invariance gives energy conservation.
z translation symmetry gives conservation of momentum in the z direction. Rotational
invariance about the z axis gives conservation of angular momentum about the z axis. Show
explicitly that L is not invariant under translations in the x direction.
L=
3.5
mv 2
mv 2
mv 2
− V (r) =
− V (x2 + y 2 ) →
− V ((x + c)2 + y 2 )
2
2
2
(3.12)
Energy in MOND
Recall the MOND theory from the homework.
L = T (v, a) − V (x).
(3.13)
Is energy conserved in this theory? Yes, since there is no explicit time dependence. What is
the conserved energy?
dL
∂L
∂L
∂L
=
ẋ +
v̇ +
ȧ
dt
∂x
∂v
∂a
d ∂L
d2 ∂L
∂L
=
− 2
v+
v̇
dt ∂v
dt ∂a
∂v
d ∂L
d d ∂L
d ∂L
=
v −
v +
v̇
dt ∂v
dt dt ∂a
dt ∂a
d ∂L
d ∂L
∂L
=
v−
v+
a
dt ∂v
dt ∂a
∂a
11
(3.14)
assuming ȧ = 0. Then, the conserved energy is given by
E=
∂ 2L
∂L
∂L
v−
va +
a − L.
∂v
∂v∂a
∂a
12
(3.15)
4
Relativity and the Pendulum
4.1
Special Relativity
In lecture, we saw the derivation of the Lagrangian for a free particle in Newtonian mechanics
beginning with 5 axioms, 4 of which defined the symmetries of the theory:
R
1. S = dtL(x, v, t);
2. t → t + T is a symmetry;
3. x → x + X is a symmetry;
4. x → Ox, where O is a rotation, is a symmetry;
5. x → x + Vt is a symmetry.
Using these axioms, we saw that the Lagrangian of a single particle is uniquely determined
to be
m
(4.1)
L(v2 ) = v2 .
2
Note that using the first 4 axioms alone, we know that L depends only on v2 and we write
L as L(v2 ). To go to special relativity, we only need to modify the last axiom. Gallilean
invariance breaks down in relativity and is replaced by Lorentz invariance. Recall that the
general Lorentz transformation is given by
ct → γ(ct + β · x),
β
x → x + (γ − 1)(β · x) 2 + γβct
β
(4.2)
(4.3)
This general transformation under which we want our Lagrangian to be invariant is rather
complicated. There is, however, a nice way to construct Lorentz invariants generally. We
define four-vectors v = (vt , vx , vy , vz ) as objects that transform like (ct, x, y, z) under (4.2).
We also define a dot product
v1 · v2 = v1t v2t − v1x v2x − v1y v2y − v1z v2z .
(4.4)
The dot product of two four-vectors is invariant under Lorentz transformations (though I
won’t show it here). This is (almost) the only way to construct Lorentz invariant quantities.
Thus, we have reduced our new fifth condition to the condition that S be made up of dot
products of four vectors. Note that we do not say L must be made up in this way: it is
important to note that dt transforms under Lorentz transformations now.
Next, we define the Lorentz-invariant “proper” time between two events:
r
r
2
∆x
v2
1√
(4.5)
∆τ = 2 ∆x · ∆x = ∆t2 − 2 = ∆t 1 − 2 .
c
c
c
13
The proper time elapsed between two events for a particle is Lorentz invariant. In its
infinitessimal form, it provides a natural (and unique) construction dtL(v 2 ) that is Lorentz
invariant, as we would want our action to be. Our Lagrangian can then be
r
v2
L = K 1− 2.
(4.6)
c
One can then show by applying a Lorentz transformation that it must have this form. How
do we figure out the constant K? Taylor series: the Lagrangian must reduce to that of
non-relativistic mechanics in the limit v/c 1. Taking the Taylor series,
1 v2 3 v4
(4.7)
L = K 1 − 2 + 4 + ...
2c
4c
The first term is just a constant and does not affect our equations of motion. The second
term should match the result from non-relativistic mechanics. It has the right form: constant
times v 2 . In non-relativistic mechanics, the constant is m/2. That means that K = −mc2 .
Note that the action principle just tells us that the path the particle takes is just the one
with the shortest four-length! This idea is crucial in making the jump to general relativity.
But I digress...
Is energy conserved in this case? Yes, since ∂L/∂t = 0. What is the conserved energy?
To find it, we use the general formula derived in lecture for any Lagrangian L(qi , q̇i ):
∂L
− L(qi , q̇i ).
∂ q̇i
(4.8)
∂L
v/c2
= mc2 p
∂v
1 − v 2 /c2
(4.9)
E = q̇i
In this case,
so that
E=p
mv 2
1 − v 2 /c2
+ mc2
p
1 − v 2 /c2 =
mc2
mv 2 + mc2 − mv 2
p
=p
.
1 − v 2 /c2
1 − v 2 /c2
(4.10)
E = mc2 , as we already knew. What is the conserved linear momentum?
p=
∂L
mv
=p
,
∂v
1 − v 2 /c2
(4.11)
which matches the definition of relativistic momentum. So what is the difference between
relativity and classical mechanics? The only difference is one set of symmetries: Gallilean
relativity becomes Einstein relativity.
14
4.2
The Pendulum in 1D
Consider the pendulum in 1D described by the Lagrangian
L=
m`2 φ̇2
+ mg` cos φ.
2
(4.12)
We would like to find the period of for an oscillation of this pendulum. First, let’s estimate!
What parameters do we have? m, `, φ0 , g only. Suppose the period contains α powers of m,
β powers of ` and γ powers of g. Then dimensions of the period would be
[T ] = M α Lβ Lγ T −2γ .
(4.13)
Clearly, γ = −1/2 or else [T ] cannot be T . Then, we want L to cancel since the period has no
units of length, so β = 1/2. Finally, the units of mass should cancel, so α = 1. What about
φ0 ? It’s dimensionless, so any function of φ0 could appear. Usually, it is safe to assume that
such a function is analytic. Furthermore, if it is analytic, it must involve even powers of φ0
only so that the period of starting at −φ0 is the same as the period starting at φ0 . Thus, we
expect the period can be expanded for small oscillations as follows:
s
`
T =
C0 + C2 φ20 + . . . .
(4.14)
g
We will explicitly calculate C0 and C2 . For small oscillations, we can make an expansion.
What is the expansion parameter? It is φ and/or φ0 . Why are we allowed to do this?
Because φ and φ0 are unitless and φ ≤ φ0 for all times. If we do this expansion, we get the
Lagrangian
m`2 φ̇2
φ2
4
L=
+ mg` 1 −
+ O(φ ) .
(4.15)
2
2
The first term in the potential part is a constant and can be dropped. What are we left
with? It is the Lagrangian for the Harmonic oscillator! Let’s solve it exactly. We will discuss
this example in detail in lecture, but for now, recall that
Z B
√
dφ
p
.
(4.16)
T (E) = 2m`
E − V (φ)
A
In this case, V (φ) = mg`φ2 /2, so that
T (E) =
√
Z
B
2m`
A
15
dφ
p
E − mg`φ2 /2
.
(4.17)
To do the integral, recall that
Z
Z
dx
√ p
a 1 − b/ax2
√
Z b/ax
dv
√ √
=
b 1 − v2
1
= √
q b
b sin−1
x
a
dx
√
=
a − bx2
(4.18)
(4.19)
(4.20)
Thus, we find that
√ "
2` m
sin−1
T (E) = √
mg`
r
mg`
B
2E
!
r
− sin−1
mg`
A
2E
!#
(4.21)
Now, what are B, A, and E in this case? Suppose the pendulum starts at some angle −φ0 .
By staring at the potential (see diagram), we notice that the pendulum will oscillate to an
angle φ0 . So A = −φ0 , B = φ0 . As for the energy, at the turning point there is no kinetic
energy, so E = mg`φ20 /2. Plugging these figures in,
s
s
s
s
!
`
2mg`
`
`
T (φ0 ) = 4
sin−1
φ =4
sin−1 (1) = 2π
,
(4.22)
2 0
g
2mg`φ0
g
g
recovering the well-known result. Suppose we wanted the full solution. That simply amounts
to a different choice of A, B, and E. Choose for example A = φ0 , B = φ and E =
m`2 ω02 /2 + mg`φ20 /2 to get a solution for φ with initial conditions φ(0) = φ0 , φ̇(0) = ω0 .
Then
s "
!
!#
r
r
`
mg`
mg`
t(φ) =
sin−1
φ − sin−1
φ0
.
(4.23)
g
2E
2E
Solving for φ,
s
φ(t) =
`2 ω02
φ20 +
sin
g`φ20
r
g
t + sin−1
`
"s
g`
φ0
2
g`φ0 + `2 ω02
#!
(4.24)
Let’s go back to the full system, before the approximation and see if we can get anywhere.
What dop
you think should be the next term in the expansion for the period of the pendulum?
Around `/gφ20 . Without making any approximations yet,
s Z
√
Z
2m` φ0
dφ
` φ0
dφ
√
q
(4.25)
T (φ0 ) = √
=
g −φ0 sin2 φ0 − sin2 φ
mg` −φ0 cos φ − cos φ0
2
16
2
Make a subsitution sin χ = sin(φ/2)/ sin(φ0 /2), cos χdχ = cos(φ/2)/[2 sin(φ0 /2)]dφ, giving
2dχ
dφ
p
=p 2
.
2
2
1 − sin (φ0 /2) sin χ
sin (φ0 /2) − sin2 (φ/2)
(4.26)
Note that by time reversal and reflection symmetry, the period is just twice the integral from
0 to φ0 . Equation (4.25) becomes
s Z
` π/2
dχ
q
sin2 χ.
(4.27)
T (φ0 ) = 4
g 0
2 φ0
1 − sin
2
For small oscillations, this gives
s Z
` π/2
1 φ20
2
T (φ0 ) = 4
dχ 1 −
sin χ + . . . .
g 0
2 4
(4.28)
s s 2
φ20
` π 1 φ0 π
`
−
+ . . . = 2π
+ ... .
1+
T (φ0 ) = 4
g 2 2 4 4
g
16
(4.29)
This gives
Matching up with our approximation above, we find C0 = 2π, C2 = π/8. The full integral
(4.27) is actually an eliptical integral and can be written as
s
`
2 φ0
T (E) = 4
K sin
,
(4.30)
g
2
where K is the complete elliptic integral of the first kind (EllipticK in Mathematica).
17
5
5.1
Central Potentials and the Kepler Problem
Modified Kepler Problem
Consider a two-body problem with the potential:
V (~r1 , ~r2 ) = −
β
α
+
,
|~r1 − ~r2 | |~r1 − ~r2 |2
α > 0,
(5.1)
where β is an arbitrary parameter. The particles 1 and 2 have masses m1 and m2 respectively.
First, let’s reduce this to an equivalent one body problem. What is the reduced mass? By
definition,
m1 m2
.
(5.2)
µ=
m1 + m2
How about the coordinates? We choose coordinates
~ = m1~r1 + m2~r2
R
m1 + m2
(5.3)
~r = ~r1 − ~r2 .
(5.4)
and
Choosing these coordinates, our potential depends only on ~r and our kinetic energy becomes
~˙ 2
m1 m2~r˙ 2
(m1 + m2 )R
+
.
T =
2
m1 + m2
(5.5)
~ is R
~¨ = 0, so we can choose an inertial frame where R
~ = 0.
The equation of motion for R
This is the center of mass frame. In this frame, we have the effective one-body problem
L=
α
β
µ~r˙ 2
+
− 2,
2
|~r| |~r|
(5.6)
E=
α
β
µ~r˙ 2
−
+ 2
2
|~r| |~r|
(5.7)
with a conserved energy T + V :
where µ is the reduced mass m1 m2 /(m1 + m2 ). Let’s reduce this to an effective one dimen~ is conserved as a vector since there is no angular dependence
sional problem. Noting that L
~ is the z-axis.
of the potential, we are free to rotate our coordinate such that the direction of L
This means that there is rotation about the z-axis, but not about any other axis. The only
way this is possible is if motion is confined to the plane z = 0. We can see this explicitly by
staring at the equations Lx = µ(y ż − z ẏ = 0 and Ly = µ(z ẋ − xż, which combine to give
ż(xẏ − y ẋ) = 0 and z(xẏ − y ẋ) = 0. So either z = ż = 0 for all times or Lz = 0. Assuming
L = Lz 6= 0 and using cylindrical coordinates,
E=
µṙ2 µr2 φ̇2 α
β
+
− + 2.
2
2
r
r
18
(5.8)
Now, φ is a cyclic coordinate and we still haven’t used up that fact. Conservation of angular
momentum gives
(5.9)
L = µr2 φ̇.
Plugging back into the energy,
L
1
µṙ2 α
− +
+β
E =
2
r
2µ
r2
2
µṙ
α L + 2µβ
=
− +
2
r
r2
02
2
α L
µṙ
− + 2,
=
2
r
r
(5.10)
(5.11)
(5.12)
where L02 = L2 + 2µβ. The energy has exactly the same form as the energy of the Kepler
problem. We can in principle solve for r(t) in exactly the same way. It is more usual to
solve for r(φ), but our angular momentum L0 corresponds to a different angular coordinate
φ0 . Let’s figure out what this coordinate is. By definition,
µr2 φ̇0 = L0
L 0
φ̇ = L0
φ̇
L0
φ̇0 =
φ̇.
L
(5.13)
(5.14)
(5.15)
Integrating gives φ0 = (L0 /L)φ + c, where c is any constant. We are free to choose c = 0 (a
different c just means that we start our two angle measurements at different values). Since
r(t) is the same regardless of which variables we choose, r0 = r. We can then just write down
the solution as in lecture in terms of the primed variables:
L02
1
r (φ ) =
,
µα 1 + 0 cos φ0
0
0
(5.16)
which can be trivially turned into a relation for r(φ):
r(φ) =
with L02 = L2 + 2µβ, 0 =
q
1+
2EL02
.
µα2
1
L02
0
µα 1 + cos
,
L0
φ
L
(5.17)
Now, we consider an interesting limit of this problem.
Suppose that µβ L2 . The result is a precessing ellipse. Let’s calculate the precession
of the perihelion (point of closest approach) during one period in this limit. Normally, the
period of the motion is 2π. Due to the fact that L0 /L 6= 0, the period is now 2πL/L0 < 2π.
The perihelion is then shifted by
!
L
L
µβ
µβ
∆φ = 2π 1 − 0 = 2π 1 − p
≈ 2π − 2π + 2 = 2
(5.18)
L
L
L
L2 + 2µβ
19
5.2
The Laplace-Runge-Lenz Vector
Consider the Kepler problem. The Laplace-Runge-Lenz vector is defined as
~ = ~v × L
~ − α ~r .
A
r
(5.19)
I claim that this vector is conserved. To check,
˙
~˙ = ~v˙ × L
~ − α ~r + α ~r ṙ
A
r
r2
˙
~˙ = (m~v˙ ) × (~r × ~r˙ − α ~r + α ~r ṙ,
A
r
r2
(5.20)
where we use the product rule to take the derivative in the first line and we use the fact that
~ = ~r × p~ = m~r × ~r˙ in the second line. The conseration laws depend on the equations of
L
motion, so we need to plug in the equation of motion:
m~v˙ = −
α
~r,
r3
(5.21)
that is, Newton’s law of Gravitation (recall that we are working with the Kepler problem).
Plugging in for ~v˙ ,
~˙ = − α ~r × (~r × ~r˙ ) + r2~r˙ − rṙ~r .
A
(5.22)
r3
Next,
~r · ~r˙
d√
~r · ~r =
ṙ =
(5.23)
dt
r
Plugging this in,
~˙ ∝ ~r × (~r × ~r˙ ) + r2~r˙ − ~r · ~r˙~r.
A
(5.24)
To finish, we need to use a vector identity on the cross product:
~ × (B
~ × C)
~ = B(
~ A
~ · C)
~ − (A
~ · B)
~ C.
~
A
(5.25)
~˙ ∝ ~r(~r · ~r˙ ) − ~r˙ r2 + r2~r˙ − (~r · ~r˙ )~r = 0
A
(5.26)
Using this relation,
~ also satisfies A
~·L
~ = 0. To see this,
The vector A
~
~·L
~ = (~v˙ × L)
~ ·L
~ − α ~r · L.
A
r
(5.27)
~ = m~r × ~v and the triple product satisfies the identity A
~ · (B
~ × C)
~ =C
~ · (A
~ × B),
~ so
Now, L
we have
~·L
~ = (L
~ × L)
~ · ~v˙ − mα ~v · (~r × ~r) = 0.
A
(5.28)
r
20
Finally, we want to show that
A2 = α2 +
2EL2
.
m
(5.29)
Let’s show this explicitly:
2
A
=
=
=
=
=
2
~
r
~ −α
~v × L
r
r2
α
2
2~
~
~
(~v × L) − 2 (~v × L) · ~r + α 2
r r
α
2 2
2
~
~ + α2
v L − (~v · L) + 2 (~v × ~r) · L
r
2T 2
2V 2
L −0 +
L + α2
m
m
2EL2
+ α2 .
m
(5.30)
(5.31)
(5.32)
(5.33)
(5.34)
~ useful for? Well, any conserved quantity is useful. But we can actually
What is the vector A
solve the Kepler problem in a different way that is perhaps a little easier. Consider the dot
product
~ · ~r = Ar cos θ = ~r · (~v × L)
~ − αr
A
(5.35)
~ =L
~ · (~r × ~v ) = L2 /m. Then, solving (5.35) for r,
Note that ~r · (~v × L)
r=
L2 /mα
.
1 + (A/α) cos θ
(5.36)
This is our solution for the Kepler problem with eccentricity given by e = A/α. Note that
~ ~v ⊥ ~r, that is at the perihelion. So A
~ points along the major axis of the ellipse.
when ~r k A,
21
6
Small Oscillations
6.1
Small Displacement Approximations
Consider the following potentials and determine whether there are points of stable equilibrium. If so, find them and determine the effective Lagrangian that describes oscillations
about the minima and the frequency of those oscillations. What is the condition such that
the oscillations are small?
1. V (x) = a sin(kx)/x
There are an inifinite number of stable equilibrium points at
kx cos kx = sin kx
(6.1)
whenever sin kx/x < 0. To see this, we expand the potential V (x) around its critical
points. Define a critical point x0 satisfying the equation (6.1). Then
V (x) = a
sin kx0
sin kx0
− ak 2
(x − x0 )2 .
x0
2x0
(6.2)
We see that if sin kx0 /x0 < 0, then small oscillations about x0 will have positive
frequency and will be well defined. The constant term can be dropped. We get the
Lagrangian
mẏ 2
2 sin kx0
− −ak
L=
y2 + . . . .
(6.3)
2
2x0
These oscillations have a frequency ω 2 = −ak 2 sin kx0 /mx0 . The only lenght scale in
the potential is 1/k. Therefore, the small oscillation approximation must be valid when
A 1/k =⇒ E ak,
(6.4)
Note that by “energy” here, I mean the energy above one of the minima of the potential.
If I went through the calculation technically, preferably with Mathematica, I would find
a small sublety: the position of the minimum about which we are expanding begins to
appear. It is important to remember that that this zero is at least as big as 1/k. The
bare minimum condition for the expansion to be valid is A 1/k. We need to go to
fourth order in the Taylor series to see this or else we would naively guess that A x0
is enough. It is not, but if A 1/k, then automatically A x0 .
6.2
The General 2D Problem Two-Ways
Consider a Lagrangian of the form:
1
1
L = mij q̇i q̇j − Kij qi qj
2
2
with two degrees of freedom. I would like to solve this problem in two different ways.
22
(6.5)
6.2.1
The Easy, Ugly Way
First, we find the E-L EOMs.
∂L
∂ q̇k
d ∂L
dt ∂ q̇k
∂L
∂qk
mki q̈i
= mkj q̇j
= mkj q̈j
= −Kkj qj
= −Kkj qj
(6.6)
How do we solve differential equations in general? We guess an answer and check that it
works. Let us guess an answer that qj (t) = Aj eiωt . Then, plugging into the equations of
motion, we find that
~ = −K A
~ =⇒ 0 = (K − ω 2 m)A.
~
−ω 2 mA
(6.7)
For this equation to have a non-trivial solution, the determinant of K − ω 2 m must be 0.
(Otherwise, the matrix is invertible and you can just multiply both sides on the left by its
inverse). The resulting equation gives the mode frequencies:
(k11 − ω 2 m11 )(k22 − ω 2 m22 ) − (k12 − m12 ω 2 )2 = 0.
This is a quadratic equation with the solution:
s
!
2
2
4(m
m
−
m
)(k
k
−
k
)
k
m
+
k
m
−
2k
m
11
22
11
22
11
22
22
11
12
12
12
12 2
1± 1−
)
ω2 =
2(m11 m22 − m212 )
(k11 m22 + k22 m11 − 2k12 m12
(6.8)
(6.9)
With these ω’s, each component of (6.7) should vanish independently. Picking, for example,
the first component, we can say that
2
±
(k11 − ω±
m11 )A±
1 + (k12 − ω± m12 )A2 = 0.
(6.10)
For each frequency, we have one corresponding amplitude initial condition giving us the
±
±
±
freedom to set A±
1 ≡ A . Then, A2 is determined in terms of A by (6.10):
A±
2
2
k11 − ω±
m11 ±
=−
A
2
k12 − ω± m12
(6.11)
Our final answer then, in terms of real variables, is:
q1 (t) = A+ cos(ω+ t + φ+ ) + A− cos(ω− t + φ− )
−
q2 (t) = A+
2 cos(ω+ t + φ+ ) + A2 cos(ω− t + φ− ),
(6.12)
where A± and φ± are aribtrary initial conditions, ω± are given by (6.9), and A±
2 are given
by (6.11). This solves the problem since we have written qi (t) (in principle) in terms of the
parameters of the problem and four initial conditions.
23
6.2.2
The Hard, Pretty Way
In order to solve the problem this way, I’m going to make some simplifying assumptions at
some point, so that the final answer is sane. In this case, we will postpone attempting to
find the equations of motion. We would like to apply three successive (linear) coordinate
transformations. The first is to diagonalize m. Let us make the transformation:
cos θ1 − sin θ1
0
q (t) =
q(t).
(6.13)
sin θ1 cos θ1
This is an orthogonal transformation. For now, I will not specify θ1 , but soon we will figure
out what it must be. What happens to our Lagrangian?
1 T
1
L = q˙0 OT mOq˙0 − q 0T OT KOq 0 .
2
2
(6.14)
But wait: in HW, we showed that you can find an O like this so that m0 = OT mO is diagonal.
Let us choose θ1 so that this is the case. We know from the homework that θ1 must be given
by
2m12
tan(2θ1 ) =
.
(6.15)
m22 − m11
So now our Lagrangian looks like
1 ˙0 T m1 0
1
L= q
q 0 − q 0T K 0 q 0 ,
(6.16)
0 m2
2
2
with
m1,2
1
=
2
q
m11 + m22 ± (m11 − m22 )2 + 4m212 .
(6.17)
and K 0 = OT KO. Now, the mass matrix is diagonal, but not proportional to the unit
matrix. To achieve that, we do a second coodinate transformation:
√
m1
0
00
√
q 0 (t).
(6.18)
q (t) =
0
m2
After this the Lagrangian becomes
1 T
1
L = q˙00 q˙00 − q 00T K 00 q 00 ,
2
2
where
K 00 =
√1
m1
0
0
√1
m2
!
K0
(6.19)
√1
m1
0
0
√1
m2
!
.
(6.20)
We need one more coordinate transformation to get this to a form that is trivial. We will
do another orthogonal transformation on the coordinates. Since there was a rescaling in
24
between, this is non-trivial. Clearly, we’re going to want to choose this one to diagonalize
K 00 . We are doing
cos θ2 − sin θ2 00
000
q (t) =
q (t).
(6.21)
sin θ2 cos θ2
Under this transformation,
L=
m1 ˙000 T T
1
q O2 O2 q˙000 − q 000T O2T K 00 O2 q 000
2
2
As before, we choose
tan(2θ2 ) =
00
2k12
,
00
00
− k11
k22
(6.22)
(6.23)
giving the Lagrangian
m1 ˙000 T ˙000 1 000T
L=
q q − q
2
2
with
k1,2
m1 ˙000 2 k1 000 2 m1 ˙000 2 k2 000 2
k1 0
q 000 =
q 1− q 1+
q 2 − q 2,
0 k2
2
2
2
2
1
=
2
00
k11
+
00
k22
q
00
00 2
00 2
± (k11 − k22 ) + 4k12 .
(6.24)
(6.25)
2
= k1,2 /m1 , we have the solution in terms of the non-trivial coordinates
Defining ω1,2
000 q1 (t)
A1 eiω1 t
=
,
q2000 (t)
A2 eiω2 t
(6.26)
where A1 , A2 are arbitrary initial conditions (complex constants). The last step is to relate
these back to the original (physical) coordinates q(t). But, keeping track of the transformations we did, we notice that q(t) = O1T p−1 O2T q 000 (t):
!
000 1
√
0
cos θ2 sin θ2
q1 (t)
q1 (t)
cos θ1 sin θ1
m1
(6.27)
=
√1
0
q2000 (t)
−
sin
θ
cos
θ
q2 (t)
− sin θ1 cos θ1
2
2
m2
25
7
7.1
Damped and Driven Oscillations
Asymmetric Double Spring
Consider the system we discussed in lecture with m1 6= m2 . Write down the Lagrangian. Go
to generalized coordinates where the two modes decouple. Determine the frequency of the
two modes. Find the solution in terms of the “physical” coordinates. Consider the limit of
m1 = m2 = m. Does the solution reduce to the one found in lecture?
The Lagrangian in terms of the variables x1 , x2 is given by
L=
m1 ẋ21 m2 ẋ22 kx21 kx22 α(x1 + x2 )2
+
−
−
−
2
2
2
2
2
(7.1)
First, let’s consider some limits. When m1 = m2 , we should get back the answer seen in
lecture. When m1 m2 , the two two masses should decouple. The light mode m2 should
have a frequency given by the (k + α)/m2 , since it is essentially stuck between two “walls.”
For the heavier mode, the mass m2 looks negligible, so it essentially sees a spring of constant k
on one side and two light springs in series on the other, with frequencies k and α. Recall that
the spring constants for spings in series add inversely. We find that the effective frequencyt
in this limit is
−1
1
kα
1
k(k + 2α)
2
m1 ω = k +
=k+
+
=
(7.2)
k α
k+α
k+α
When k α, the two masses vibrate independently with frequency k/mi . When k α, we
recover antisymmetric, symmetric modes with frequencies 2k/(m1 + m2 ), α/µ.
The masses being equal meant that any orthogonal transformation on (x1 , x2 ) left the
kinetic term untouched. Let’s rescale by:
q1 =
√
m1 x1 ,
q2 =
√
m2 x2 .
(7.3)
The Lagrangian in terms of q1 and q2 is:
kq 2
kq 2
α
q̇ 2 + q̇22
− 1 − 2 −
L= 1
2
2m1 2m2
2
q1
q2
+√
√
m1
m2
2
(7.4)
Next, we need to find another coordinate transformation that will allow us to seperate the
two modes of the system. Note that the potential can be written as
! k+α
√ α
1
q1
m1
m1 m2
q1 q 2
V =
.
(7.5)
α
k+α
√
√
q2
2
m1 m2
m1 m2
What we see from this is that finding the coordinates where the two modes decouple is the
same as the problem of diagonalizing the matrix in (7.5). The eigenvalues represent the
spring constant, while the eigenvectors give the combination of coordinates corresponding to
26
the modes. The frequencies of the modes are then given as the solutions of the eigenvalue
equation:
!
#
"
det
√ α
m1 m2
k+α
m2
k+α
m1
√ α
m1 m2
− ω2 = 0
Calculating the determinant explicitly, we see that
k+α
k+α
α2
2
2
−ω
−ω −
= 0.
m1
m2
m1 m2
This is a quadratic equation, which has a known solution:
s
2
k+α
k+α
α2
(k + α)2
2
ω =
±
+
.
−
2µ
2µ
m1 m2
m1 m2
(7.6)
(7.7)
(7.8)
Note that in the limit m1 = m2 = m, this reduces to
ω2 =
k+α
α
±
m
m
(7.9)
as seen in lecture. In the limit where α k, the leading order dependence of frequency is
ω12 =
α
µ
ω22 =
2k
m1 + m2
(7.10)
The other limits can be checked as well. As for the actual relation between the variables
corresponding to these modes and x1 , x2 , we can find the eigenvectors corresponding to these
eigenvalues. Since the matrix is symmetric, it can diagonalized by an orthogonal matrix.
A general formula for the rotation angle in the 2 × 2 case was given in lecture. Then the
solution to the problem is given by
!
0
1
√
0
x1
cos θ sin θ
q1
m1
=
(7.11)
√1
0
x2
−
sin
θ
cos
θ
q20
m2
and
q1,2 (t) = A1,2 cos(ω1,2 t + φ1,2 )
(7.12)
with ω1,2 given by (7.8). Now, let’s look at some limits on the angle θ. The solution for
tan(2θ) is
√
m1 m2
2α
(7.13)
tan(2θ) =
k + α m22 − m21
Consider all the limits of this angle. For m1 = m2 , we get tan(2θ) → inf. That means
θ = π/4, as we saw in lecture. For m1 m2 , θ → 0. That is the motion of the two masses
is independent (decoupling). For k α, θ → 0 as well. We get decoupling since it’s as if
the middle spring weren’t there.
27
7.2
Damped Oscillator Example
Consider the two block-three spring system with equal masses. First, without friction, let’s
just see how to apply some initial conditions. Recall that the Lagrangian was
k
k
α
m 2
(ẋ1 + ẋ22 ) − x21 − x22 − (x1 − x2 )2 .
(7.14)
2
2
2
2
Recall as well that there are two different ways to solve the problem. Method 1 is to
diagonalize the Lagrangian before finding the equations of motion. Let me briefly outline
how that would be done.
L=
• In this case, since the mass matrix is already proportional to the identity, we would
fist find the eigenvalues of the matrix K given by
k + α −α
K=
(7.15)
−α k + α
2
This tells us the frequencies of the modes of our system ω1,2
= k1,2 /m.
• Once we have the eigenvalues, we need to find the eigenvectors. This gives us the
orthogonal matrix O such that OT KO = diag(k1 , k2 ): the columns of O are the eigenvectors of K. Alternatively, in the 2 × 2 case, can use the formula from HW 6 for θ.
This matrix may be messy, but in principle we have an algorithm to solve.
• Next, we make a change of coordinates such that ~q = OT ~x. In these new coordinates,
the kinetic matrix is unchanged, but the spring constant matrix becomes diagonal. The
equations of motion in this case are for independent harmonic oscillators (easy) with
spring constants k1,2 . We can then solve. Our final answer for ~x(t) is then ~x(t) = O~q(t).
Method 2 is to find the equations of motion, guess a solution, and then find the constraints
such that our guessed solution works. Let’s do this explicitly. The equations of motion can
be written as
ẍ1
k + α −α
x1
m
=
(7.16)
ẍ2
−α k + α
x2
T
We guess a solution A1 A2 eiωt . Then, equation of motion becomes
mω 2
0
k + α −α
A1
(7.17)
2 −
0
mω
−α k + α
A2
This equation only has a non-trivial solution if det(mω 2 I − K) = 0 (think about this as
(mω 2 − k)a = 0: trivial if a = 0, nontrivial when ω 2 = k/m). The solution to that equation
is ω 2 = k/m or ω 2 = (k + 2α)/m. To finally solve the equation though, we need to plug
back in and get the proper relation between A1 and A2 :
k
m
k + 2α
for ω 2 =
m
A1 = A2 ≡ A for ω 2 =
(7.18)
A1 = −A2 ≡ B
(7.19)
28
Thus, the (general) solution to the equations of motion is
x1 (t)
A
B
=
cos(ω1 t + φ1 ) +
cos(ω2 t + φ2 )
x2 (t)
A
−B
(7.20)
Now, saw we want to apply the initial conditions x1 (0) = C1 , x2 (0) = C2 , ẋ1 (0) = 0, ẋ2 (0) =
0. Let’s start with the velocity conditions:
0 = −Aω1 sin(φ1 ) − Bω2 sin(φ2 )
(7.21)
0 = −Aω1 sin(φ1 ) + Bω2 sin(φ2 )
(7.22)
and
Taking the first equation plus the second tells us either A = 0 or φ1 = 0. Taking the
first equation minus the second tells us either B = 0 or φ2 = 0. Assuming that our answer
shouldn’t be “boring”, we can rule out the amplitudes being zero. Now, we apply the position
conditions:
C1 = A + B
(7.23)
and
C2 = A − B
(7.24)
Then, we can see that
A=
C1 + C2
,
2
B=
C1 − C2
.
2
(7.25)
Thus our final solution is
!
r !
r
C1 − C2 1
C1 + C2 1
k
k + 2α
x1 (t)
t +
t .
=
cos
cos
x2 (t)
1
−1
2
m
2
m
Now, let’s move the system into a viscous fluid. The equation of motion is then
m1 ẍ1
ẋ1
k + α −α
x1
+ 2λ
+
m2 ẍ2
ẋ2
−α k + α
x2
(7.26)
(7.27)
Notice that we cannot simultaneously diagonalize all three terms. We could guess an ansatz
~ rt . The issue here is that you get a quartic equation in r. In principle, we can
that ~x = Ae
solve it, but good luck! Let’s simplify to the case where the masses are equal and see if we
can do anything. Plugging in the ansatz, we get the equation
2
mr + 2λr + k + α
−α
A1
(7.28)
−α
mr2 + 2λr + k + α
A2
This only has a non-trivial solution when the determinant of the matrix vanishes:
(mr2 + 2λr + k + α)2 − α2 = 0
(mr2 + 2λr + k)(mr2 + 2λr + k + 2α) = 0
29
(7.29)
Now, we only need to solve two quadratic equations to get
!
r
λ
km
r1,2 = −
1± 1− 2
m
λ
!
r
λ
(k + 2α)m
1± 1−
r3,4 = −
m
λ2
(7.30)
Thus, the solution to the problem has the form
x1
A1 r1 t
B1 r2 t
C1 r3 t
D1 r4 t
=
e +
e +
e +
e
x2
A2
B2
C2
D2
(7.31)
A1 and A2 are related by plugging that solution for r back into the equation of motion
(similarly with the others, but I won’t do it explicitly):
A2 =
λ2
mm
2
q
1+ 1−
km
λ2
2
−
2
2 λm
α
q
1+ 1−
km
λ2
+k+α
A1
(7.32)
This in principle solves the general problem. It remains to plug in the initial conditions to
determine A1 , B1 , C1 and D1 .
30
8
8.1
Vectors and Intertia Tensors
Vectors and Cross Products
Recall when we discussed rotations that we defined a vector as an object v that transforms
under rotations as
v → Ov,
(8.1)
where O is an orthonormal (rotation) matrix. An orthonormal matrix is one that satisfies
as OT O = 1, det(O) = 1. That is a somewhat incomplete definition. We further break down
the classification of vectors into vectors and pseudo-vectors. There is an additonal spatial
symmetry it is convenient to define called parity. Under parity, a vector transforms as
v → −v,
(8.2)
v → v.
(8.3)
while a pseudo-vector transforms as
An example of a vector is the position vector r. It is a physical vector in space that
rotates if we rotate our coordinate system. Furthermore, by definition, parity transforms r
according to (8.2). Since time is a scalar (invariant under rotations), ṙ is also a vector. Since
m is a scalar as well, p is a vector. Now, we can see how to get a pseudo-vector. We take
two vectors, say r and p, and take their cross product. The resulting object the transforms
under parity as
L ≡ r × p → (−r) × (−p) = r × p = L.
(8.4)
That is, angular momentum is a pseudo-vector. Other vecors include acceleration, force, and
electric field. Other pseudo-vectors include angular velocity, torque, and magnetic field. Let
us briefly see why electric field is a vector while magnetic field is a pseudo-vector. All you
need to know to see this is the Lorentz force law:
F = q(E + v × B).
(8.5)
Charge is a scalar and force is a vector, so E must be a vector in order for this to be an
equation between objects with the same transformation properties. B on the other hand
must be a pseudo-vector since v is a vector.
Now, we would like to move to a more sophisticated notation for dealing with vectors,
pseudo-vectors, and cross products. This notation will allow you to efficiently derive all sorts
of cross-product identities. Let us introduce indices to label the components of our vectors.
We will now call a vector by its components vi . A matrix is then an object with two indices
Oij . Summation over an index is like taking a dot product or matrix product:
a·b=
3
X
ai b i ,
(Ov)i =
i=1
3
X
j=1
31
Oij vj .
(8.6)
Because we do such summations so often, we will suppress the summation symbols. It is
implied that we sum over repeated indices. For example,
ai b i ≡
3
X
ai b i .
(8.7)
i=1
How do we do cross-products in this notations? We will introduce an object called the LeviCivita symbol. It is also sometimes called the Levi-Civita tensor, though this is somewhat
sloppy for technical reasons. It is an object with three indices ijk . We define it to be
antisymmetric under any permutation of indices:
ijk = −jik = jki = −kji = kij = −ikj .
(8.8)
One important consequence is that if any two indices are equal, then the symbol vanishes.
For example, 112 = 0. That means that only if all three indices have different values does is
this symbol non-zero. Well, there are only three values an index can take, so there is only one
non-trivial component of the symbol to define. We therefore define 123 = 1. This completes
the definition of ijk : it is a three component object that is completely antisymmetric and
satisfies 123 = 1.
Notice that this object would not be well defined in the plane. This is essentially why
where is no cross product in the plane. We can only define a two component analogous
quantity ij which is an antisymmetric matrix. We can write it out explicitly:
0 1
=
.
(8.9)
−1 0
Let’s work with this a for a little bit to get a feel for this sort of object. First, let’s calculate
a few objects we can construct out of .
0 1
0 −1
1 0
T
T
ij ij = ij ji = Tr( ) = Tr
= Tr
= 2,
(8.10)
−1 0
1 0
0 1
0 1
0 −1
1 0
T
T
ik jk = ik kj = ( )ij =
=
= δij ,
(8.11)
−1 0
1 0
0 1 ij
ij
and
ij k` = δik δj` − δi` δjk .
(8.12)
Identities like these are what makes the object so convenient.
Let’s now put this object together with a vectors ai , bi . Let’s consider the object ij ai bj .
In matrix notation, this is
aT b.
(8.13)
Is this object a scalar? Let’s see how it transforms under a rotation
cos θ sin θ
O=
.
− sin θ cos θ
32
(8.14)
The transformation does:
aT b → aT OT Ob.
(8.15)
Let’s compute the matrix product in the middle:
cos θ − sin θ
0 1
cos θ sin θ
sin θ cos θ
cos θ sin θ
T
O O =
=
sin θ cos θ
−1 0
− sin θ cos θ
− cos θ sin θ
− sin θ cos θ
0 1
=
= . (8.16)
−1 0
That means,
aT b → aT b
(8.17)
under rotations. Clearly, under parity, aT b → aT b. Thus, this object is in fact a scalar.
Moving back to three dimensions, we want to consider a similar object
(a × b)i ≡ ijk aj bk .
(8.18)
You can check explicitly that this definition is equivalent to the usual one. Furthermore, we
have already seen that the result is a pseudo-vector. Well, we didn’t quite show that it has
the right properties under rotations. To really see that, we need a mathematical identity:
ijk Ojj 0 Okk0 = i0 j 0 k0 Oii0 .
(8.19)
To see where this comes from, notice first that if j 0 = k 0 , the left hand side vanishes since
it describes the cross product of identical vectors and the right hand side vanishes by the
definition of . Suppose then, WLOG, that j 0 = 2, k 0 = 3 (note that both sides are antisymmetric under exchange of j 0 , k 0 ). Then, the left hand side is the unit vector perpendicular to
Oj2 and Ok3 , with direction given by the right-hand rule from Oj2 to Ok3 . That is precisely
Oi1 . On the right hand side, j 0 = 2 and k 0 = 3 force i0 = 1, giving Oi1 on that side as well and
proving the identity. Then the fact that the cross-product of two vectors is a pseudo-vector
follows easily.
Now, the big advantage of Levi-Civita symbols (other than being able to generalize to
other dimensions) is that proving some identities becomes very straight forward. All we need
is a basic set of identities for combining Levi-Civita symbols:
ijk ijk = 6,
(8.20)
ijm ijn = 2δmn ,
(8.21)
ijk mnk = δim δjn − δin δjm .
(8.22)
and
Armed with these identities, it is trivial to prove various identities about cross products. For
example, consider the triple product. It is invariant under cyclic permutations. If we write
out the triple product in index notation,
a · (b × c) = ai ijk bj ck .
33
(8.23)
This has a cyclic property trivially, since is invariant under a cyclic permutation of its
indices. Let’s consider the triple cross product next:
[a × (b × c)]i = ijk aj kmn bm cn = (δim δjn − δin δjm )aj bm cn = aj cj bi − aj bj ci = (a · c)bi − (a · b)ci .
(8.24)
As a final example, consider
(a×b)·(c×d) = kij kmn ai bj cm dn = (δim δjn −δin δjm )ai bj cm dn = (a·c)(b·d)−(a·d)(b·c). (8.25)
Let me do a brief interlude to talk about 4 dimensions. We mentioned in class that there
was no analogue to the cross product in this case. That is because there is no Levi-Civita
symbol with three indices. The number of indices necessary to form a Levi-Civita symbol is
always the same as the number of dimensions of your space. Therefore, there is a four-index
Levi-Civita symbol in 4d. Applying this to two vectors, you get a two index object:
ijkl ak bl .
(8.26)
This defines a plane containing the i and j axes, which is the plane perpendicular to the kl
plane.
8.2
Intertia Tensor
Let us go through some examples of calculating the moment of inertia tensor. Recall that
for a rigid body, we define the moment of inertia tensor as
Z
X
2
Ijk =
m(x` δjk − xj xk ) ≈ dV ρ(x2` δjk − xj xk ).
(8.27)
As a first warm-up, let’s calculate the intertia tensor of a sphere with uniform density ρ and
radius R (in a frame centered at the center of the sphere). First, let’s get an idea of how
much work we need to do. By NDA, how big will the non-zero entries be roughly? Well, we
know that the units of I are M L2 . The only quantities that can go into this are ρ and R, so
it must be given by
I ∼ M R2 ∼ ρR5 .
(8.28)
Next, we notice the symmetries of the system. Since the sphere doesn’t care about direction,
Ixx = Iyy = Izz exactly. All the off-diagonal entries must be equal to one another as well for
the same reason (the matrix is symmetric and what’s the difference between xy and yz to a
34
sphere?). Let’s calculate these.
Z
drd(cos θ)dφρ(r2 − z 2 )
Z
= ρ drr2 d(cos θ)dφρr2 (1 − cos2 θ)
Z 1
Z R
4
drr
d(cos θ)(1 − cos2 θ)
= 2πρ
Izz =
−1
0
5
= 2πρ
R 4
5 3
2
M R2
5
=
(8.29)
and
Z
√
R
Z
dz
Ixz =
−R
Z
R
= ρ
dz
−R
√
R2 −z 2
Z
dx
√
− R2 −z 2
Z √R2 −z2
√
− R2 −z 2
R2 −z 2 −x2
√
− R2 −z 2 −x2
ρ(−xz)
√
dx2ρ R2 − z 2 − x2 xz.
(8.30)
This last integral must vanish since we integrating an odd integrand with symmetric limits.
Therefore, we see that


1 0 0
2
I = M R2 0 1 0 .
(8.31)
5
0 0 1
As our next example, let’s consider a cylinder. The cylinder is still a symmetric top, so
that Ixx = Iyy 6= Izz . If we choose our axes carefully, the off-diagonal elements should vanish.
We choose our origin to be at the center of the cylinder, with the z axis along the cylinder’s
axis. Then,
Z
Izz =
dzdrrdφρr2
Z
= 2πhρ drr3
= 2πhρ
35
R4
,
4
(8.32)
Z
Ixx =
=
=
=
=
dzdrrdφρr2 + z 2 − r2 cos2 φ
Z
h3
2
2
2
ρ drrdφ hr +
− hr cos φ
12
Z
h3
2
2
ρ drr 2πhr + 2π − πhr
12
4
3
2
R
h R
R4
ρ(2πh
+ 2π
− πh )
4
12 2
4
2
π 2
h
ρR h(R2 + )
4
3
36
(8.33)
9
9.1
Rigid Body Problems
Euler angles and Angular Velocity
We want to consider the motion of an arbitrary individual particle P on the rigid body.
First, we choose a frame. In this frame, we can describe the position of P by a vector r.
The position of the center of mass is R. We can then define the vector r as the vector that
points from the center of mass R to the particle P at r:
r = r − R =⇒ r = R + r.
(9.1)
In general, both R and r depend on time and we would like to parametrize this dependence. We know have already seen how to describe the center of mass motion. We have also
seen a way to describe the motion of r:
ṙ = Ω × r.
(9.2)
We would like to come at this description in a different way that will parametrize both r
and its derivative ṙ in terms of a set of Euler angles. This is contrasted to the case of (9.2),
where we can only really describe the angular velocity Ω.
To start with we notice that because the body is rigid, |r| is rigid. That is, the distance
between any two particles in the body is, by definition, constant. So far, we have not specified
our frame at all, so now let’s say that our frame is inertial. Next we have to pick an arbitrary
contant reference vector r0 with length |r|. This will be like choosing a coordinate systen:
you always have to pick an origin and an orientation for your axes, but these choice are
of course arbitrary. It will often by convenient to choose r0 so that its componets are the
components of r in the principal axis frame, but we could also choose it to be any other
constant vector of length |r|. It could be, for example, r(t = 0). Since r and r0 have the
same length, they must be related by a rotation O, that depends on time in general. That
is, the orientation of the body changes with time. We define
r = Or0 .
(9.3)
Inutitively, the matrix O spins the vector r0 by φ around the z axis, then swings it down by
θ, then twists the whole thing by ψ around the z axis again.
Since the body is rigid, the matrix O that implements this rotation is the same for any
particle in the body. We can then write the time dependence of r as
ṙ(t) = Ȯ(t)r0 ,
(9.4)
with the same O for every point on the rigid body. For this to be true, we must have
ṙ = Ω × r = Ȯr0 = ȮOT r.
(9.5)
Does this definition make any sense? In index notation, we are saying
ijk Ωj rk = Ȯij Okj rk .
37
(9.6)
What we need to show then is that we can write a relation like
Ȯij Okj = ijk Ωj .
(9.7)
We need to essentially prove two things in order to convince ourselves that (9.7) makes
sense: that an anti-symetric tensor Aij can always be written as ijk Vk , for some vector Vk
and that Ȯij Okj is anti-symmetric. First, we show that an anti-symmetric tensor is related
to a vector:
1
1
1
Aij = (Aij − Aji ) = (δim δjn − δin δjm )Amn = ijk mnk Amn = ijk Vk ,
2
2
2
1
Vk = mnk Amn .
2
(9.8)
Next, we need to show that Ȯij Okj is anti-symmetric. Start with
Oij Okj = δik =⇒ Ȯij Okj + Oij Ȯkj = 0 =⇒ Ȯij Okj = −Ȯkj Oij .
(9.9)
Combining these results, we see that
1
ij` Ȯik Ojk = Ω` .
2
(9.10)
Using this result, we can find that the components of the angular velocity are given by
~ = (cψ θ̇ + sθ sψ φ̇, cψ sθ φ̇ − sψ θ̇, ψ̇ + cθ φ̇),
Ω
(9.11)
by taking the derivative of the matrix (??).
We have already seen that the rotational kinetic energy of a rigid body can be written
as T = Iij Ωi Ωj /2, so we can now write the kinetic term explicitly in terms of Euler angles.
For example, in the case of the symmetric top, I1 = I2 . In this case, the kinetic energy is
given by
1
(I1 (θ̇2 + s2θ φ̇2 ) + I3 (ψ̇ + cθ φ̇)2 ).
(9.12)
2
It then remains to write the potential in terms of Euler angles and then the equations of
motion can be easily derived.
9.2
Rigid Body Kinematics
In this example, we calculate various quantities related to rigid body motion in a couple
of different systems. We calculate the center of mass velocity, the angular velocity, the
instantaneous axis of rotation, the total kinetic energy, and the anguar momentum. We
then determine which among these quantities is conserved.
Recall the following important formula before we begin. Given a rigid body with CM
~ the velocity a point on the body displaced by a vector ~a
velocity V~ and angular velocity Ω,
from the CM is given by
~ × ~a
~v = V~ + Ω
(9.13)
38
1. Homogeneous cone rolling on a plane.
dCM
α
The most important point about rolling bodies is that the instantaneous axis of rotation always goes through the point(s) touching the plane, assuming they roll without
slipping. In that case, note that the tip of the cone must have zero velocity at all times:
it is always touching the plane. So we choose our coordinate system so that the tip of
the cone is at the origin. Now, consider the motion of the center of mass. It has no
velocity due to rotation, by construction. The distance between the CM and the origin
is fixed, so the CM must undergo circular motion in the plane. It is a planar distance
dCM cos α from the origin, so it must have velocity dCM cos αθ̇. That is,
V~ = dCM θ̇ cos αθ̂.
(9.14)
Next, we want the angular velocity. To find it, consider the point directly below the
center of mass on the plane. This point is along the instantaneous axis of rotation, so
it has velocity 0. That is,
~ × ~a.
0 = V~ + Ω
(9.15)
â = −ẑ and V̂ = θ̂, so it must be that Ω̂ = −r̂. In this case, |~a| = dCM sin α so that
~ = dCM θ̇ cos α r̂ = θ̇ cot αr̂.
Ω
dCM sin α
(9.16)
In order to proceed, we need to know the inertia tensor in terms of the r, θ, z basis
used above. The inertia tensor for a cone in the body frame is
3

1 2
2
M
R
+
h
0
0
20
4
3
0
0 
M R2 + 14 h2
I0 = 
20
3
0
0
M R2
 102

0
0
tan α + 14
3
0
tan2 α + 14
0  . (9.17)
= M h2 
20
0
0
2 tan2 α
The two coordinate systems can be related by xˆ1 = sin αr̂ − cos αẑ, xˆ2 = −θ̂, and
xˆ3 = cos αr̂ + sin αẑ. They are related by a transformation
  
 
x1
sin α 0 − cos α
r
 x2  =  0


−1
0
θ ,
(9.18)
x3
cos α 0
sin α
z
39
so that



sin α
0 cos α
I1 0 0
sin α
−1
0   0 I1 0   0
I= 0
− cos α 0 sin α
0 0 I3
cos α
 2
sin αI1 + cos2 αI3
0
=
(I3 − I1 ) cos α sin α

0 − cos α
−1
0 
0
sin α

0 (I3 − I1 ) cos α sin α
 . (9.19)
I1
0
2
2
0 cos αI1 + sin αI3
Now, we can calculate the angular momentum and kinetic energy. The angular momentum is given by
~ = IΩ
~ = (sin2 αI1 + cos2 αI3 )θ̇ cot αr̂ + sin α cos α(I3 − I1 )θ̇ cot αẑ
L
1
1
3
2
2
2
2
sin α cos α tan α +
+ 2 sin α cos α r̂ + cos α tan α −
ẑ .
= Mh
20
4
4
(9.20)
We can also calculate the kinetic energy of the particle. The angular component is
given by
1~ T ~
1~ T ~
3 2
1
2
2
2
2
Ω I Ω = Ω L = θ̇ M h sin α + cos α + 2 cos α .
(9.21)
2
2
40
4
Putting this together with the CM component 1/2M V 2 = 1/2M a2 θ̇2 cos2 α =
3
5
3
15
2 2
2
2
T = M h θ̇
cos α + 1 + cos α = M h2 θ̇2 1 + 5 cos2 α
40
4
4
40
In this case, angular momentum and energy are both conserved.
2. An ellipsoid spinning about an arbitrary axis with no CM motion.
x3
θ̂
x1
α
40
9
M h2 θ̇2
32
(9.22)
cos2 α,
Let Ω̂ be the axis of rotation. Let x1 and x2 be the principle axes with the same moment
of intertia and let x3 be the third principle axis. Since the ellipsoid is symmetric in
rotations about x3 , the direction of x1 is arbitrary (while x2 is fixed relative to x1 since
they must be perpendicular). Thus, we are free to pick xˆ1 to be in the plane spanned by
xˆ3 and Ω̂ (supposing those two are not parallel, for if they are, the problem is trivial).
Then, the angular momentum must be given by
~ = I1 Ω(xˆ1 · Ω̂)xˆ1 + I3 Ω(xˆ3 · Ω̂)xˆ3
L
(9.23)
and the kinetic energy is given by
1 T = Ω2 I1 (xˆ1 · Ω̂)2 + I3 (xˆ3 · Ω̂)2 .
2
(9.24)
To make things more concrete, consider the case where the ellipsoid has two different
components to its rotation. Say the ellipsoid is tilted to an angle α as in the diagram.
Define the θ̇ rotation axis to be ẑ. Then, the φ̇ axis can be taken to be cos αx̂ + sin αẑ.
The two relevant principle axes are x3 = cos αx̂ + sin αẑ and x1 = sin αx̂ − cos αẑ. The
~ is given by
vector Ω
~ = θ̇ẑ + φ̇(cos αx̂ + sin αẑ).
Ω
(9.25)
The angular momentum is then
~ = I1 [(sin αx̂ − cos αẑ) · Ω](sin
~
~
L
αx̂ − cos αẑ) + I3 [(cos αx̂ + sin αẑ) · Ω](cos
αx̂ + sin αẑ)
= −I1 θ̇ cos α(sin αx̂ − cos αẑ) + I3 (φ̇ + θ̇ sin α)(cos αx̂ + sin αẑ)
= [I3 φ̇ cos α + (I3 − I1 )θ̇ cos α sin α]x̂ + [I3 φ̇ sin α + (I1 cos2 α + I3 sin2 α)θ̇]ẑ
(9.26)
Let’s examine a few limits. When I1 = I3 , we get a spherical top. In that case, we
~ = I1 Ω.
~ We see that this is indeed the case. When α = π/2, there is only
know that L
one axis of rotation, the ẑ axis. The two angular velocities simply add up and we are
rotating about the principle axis xˆ3 . We see this case realized as well. Finally, let’s
write the kinetic energy explicitly:
i
1h
I1 cos2 αθ̇2 + I3 (θ̇ sin α + φ̇)2 .
(9.27)
T =
2
9.3
The Heavy Symmetric Top
Consider a heavy symmetric top with one point, the lowest point, fixed. We want to write
down the Lagrangian in terms of a single dynamical variable. This is possible because we
have 6 degrees of freedom with 3 constraints (lowest point fixed: specifies x, y, and z) and
2 conserved quantities (two components of the angular momentum of the body. First, let’s
write the full Lagrangian, imposing only the constraints, in terms of Euler angles. Recall
that we said the rotation kinetic energy is
1
1
Trot = I1 (θ̇2 + s2θ φ̇2 ) + I3 (ψ̇ + cθ φ̇)2 .
2
2
41
(9.28)
The position of the center of mass is independent of ψ (the spin about the z axis). It is is
given by `(cφ sθ , sφ sθ , cθ ), so that
V = (−sφ sθ φ̇ + cφ cθ θ̇, cφ sθ φ̇ + sφ cθ θ̇, −sθ θ̇).
(9.29)
Thus, the kinetic energy is
1
Tcm = M (s2θ φ̇2 + θ̇2 )
2
Gravity is uniform, so it acts at the center of mass and pulls down, so that
V = M g`cθ .
(9.30)
(9.31)
Now, we notice the cool part: both ψ and φ are cyclic and so have conserved components of
angular momentum associated with that:
∂L
= I3 (ψ̇ + φ̇cθ ) = const.
∂ ψ̇
(9.32)
∂L
= (I10 s2θ + I3 c2θ )φ̇ + I3 ψ̇cθ = const.,
∂ φ̇
(9.33)
pψ =
and
pφ =
for I10 = I1 + M `2 . We can then plug these two equations in to the Lagrangian to get a
Lagrangian that is independent of both φ and ψ and only depends on θ.
42
10
The “Everything” Problem
ϕ
M
R
m
θ
Consider the problem of a bead on a ring in gravity, where the ring is held in place, but
free to rotate about the vertical axis (see figure). We will consider doing everything in this
problem!
1. How many degrees of freedom are there? Two. We have a rigid body whose center
of mass is fixed and which is free to rotate only about one axis, leaving one degree of
freedom. The bead is free to move around the ring, but its distance from the center is
fixed and its angle in the horizontal plane is determined by the angle of the ring.
2. Write down the Lagrangian for this system. Working in spherical coodinates so that
θ = 0 is the bottom for simplicity, we find that the Lagrangian is
1 2 1
IΩ + m(ṙ2 + r2 θ̇2 + r2 sin2 θϕ̇2 ) + mgr cos θ
2
2
1
1
=
M R2 θ̇2 + mR2 (θ̇2 + sin2 θϕ̇2 ) + mgR cos θ
4
2
L =
(10.1)
3. Are there any cyclic coordinates? Identify them. ϕ is cyclic and so its conjugate
momentum will be conserved.
4. Find Hamiltonian for the system. The conjugate momenta are
pϕ =
∂L
= (I + I˜ sin2 θ)ϕ̇,
∂ ϕ̇
pθ = I˜θ̇,
(10.2)
where I use I = M R2 /2 and define I˜ = mR2 . We can invert these relations to find
ϕ̇ =
pϕ
,
I + I˜ sin2 θ
43
θ̇ =
pθ
.
I˜
(10.3)
Next, we write down the Hamiltonian as
H = pϕ ϕ̇ + pθ θ̇ − L
(10.4)
Plugging in (10.3), we find
H=
p2ϕ
p2θ
+
− mgR cos θ
2(I + I˜ sin2 θ) 2I˜
(10.5)
5. Show explicitly using Poisson brackets that pϕ is conserved. Does this make sense based
on symmetries and based on forces/torques? We can plug into the Poisson bracket:
[pϕ , H] = −
∂pϕ ∂H
= 0,
∂pϕ ∂ϕ
(10.6)
since H does not depend on ϕ. This makes sense since there is a rotational symmetry
of the system in the horizontal plane (but note that any particular state of the system
could break this symmetry). Another way to see this is that the force is vertical, so
that the torque about the origin is always perpendicular to the ring and therefore
perpendicular to the angular momentum vector corresponding to rotations about the
vertical axis.
6. Find the Hamiltonian equations of motion for θ. We find that
˜ 2 sin θ cos θ
Ip
∂H
ϕ
ṗθ = −
=
− mgR sin θ.
∂θ
(I + I˜ sin2 θ)2
pθ
∂H
= ,
θ̇ =
∂pθ
I˜
(10.7)
7. How many equilibrium points does the effective potential in terms of the conserved
momentum have? Which are stable? The answer actually depends on the size of the
conserved angular momentum. The effective potential is
Veff =
p2ϕ
− mgR cos θ.
2(I + I˜ sin2 θ)
(10.8)
We have already seen what the first derivative is: it’s the negative of theRHS of the
Hamilton equation for pθ . Explicitly,
˜ 2ϕ sin θ cos θ
Ip
∂Veff
=−
+ mgR sin θ.
∂θ
(I + I˜ sin2 θ)2
(10.9)
There are at least θ for which this vansihes, so that the potential has either a maximum
or a minimum: θ = 0 and θ = π. Intuitively, this is when the bead is at the top or
bottom of the ring. In this case, the second derivative gives
˜2
Ip
∂ 2 Veff
ϕ
|
=
−
± mgR,
θ=0,π
2
2
∂θ
I
44
(10.10)
where we find + (−) for θ = 0 (θ = π). For θ = π, the second derivative is always
negative and the equilibrium is always unstable. This is as we expect. For θ = 0, it
depends. For small pϕ , the second term is larger, gravity dominates, and there is stable
equilibrium at the bottom. For large pϕ , the first term is larger and the point at the
bottom is an point of unstable equilibrium. The critical value of angular momentum
at which we transition from stable to unstable is
p2ϕ = mgR
I2
≡ p2crit .
˜
I
(10.11)
Now, the effective potential is a smooth function of cos θ, which must lie between −1
and 1, so it must have both a maximum and a minimum. We have found for large
enough pϕ , that the minimum is neither at 0 nor at π, so there must be another
equilibrium point. Let’s see how this comes about. Suppose θ 6= 0, π so that sin θ 6= 0.
Then, the equation for the equilibrium points reduces to
˜ 2ϕ cos θ = mgR(I + I˜ − I˜ cos2 θ)2 .
Ip
(10.12)
Then the LHS is monotonically increasing and the RHS is monotonically decreasing as
˜ 2 and
we move away from θ = 0 toward θ = π. The maximum value of the LHS is Ip
ϕ
˜ 2 > mgRI 2 , the two never
the minimum value of the RHS is mgRI 2 . So long as Ip
ϕ
meet and there is no solution. This is exactly when pϕ < pcrit . Once pϕ > pcrit , there
is a new critical point on either side of ring, elevated from the bottom. This must be
a minimum in this case, since it is a critical point that lies between two maxima.
8. Expand in small deviations away from the equilibrium point in each scenario and
determine the frequency of oscillations. The frequency is given by
1 ∂ 2 Veff
|θ=θc .
I˜ ∂θ2
ω2 =
(10.13)
Below the critical angular momentum, this gives
ω2 =
mgR p2ϕ
− 2
I
I˜
(10.14)
Above the critical angular momentum, this gives
˜ 2ϕ sin2 θc (I + 4I˜ cos2 θc + I˜ sin2 θc )
Ip
ω =
(I + I˜ sin2 θc )3
2
(10.15)
9. Next, we would like to solve for ϕ. To do this we use seperation of scales. Assume that
I˜ ≈ I, or in other words that the mass of the ring is roughly the same as the mass of
the bead. The relevant physical quantities in this problem are then I, pϕ , mgR, and θ0 .
By θ0 we mean the amplitude of the motion of the bead (the maximum displacement
45
away from whichever equilibrium point is relevant). Out of these, we can form two
time scales (neglecting for now the angular dependence),
r
mgR
pϕ
ω∼
(10.16)
Ω∼ ,
I
I
First suppose that Ω ω. This corresponds to the limit pϕ pcrit . In this limit, the
equilibrium should be at θ = 0. To find the frequency of rotation in φ, we note that
this motion is much slower than the motion of the bead, so we can average over the
motion of the bead. The frequency is then given by using the averaged total moment
of inertia
Z θ0
3 sin 2θ0
1
2
dθI + I sin θ =
−
I.
(10.17)
I=
2θ0 −θ0
2
4θ0
The resulting appoximate frequency of motion of the ring is then
Ω≈
pϕ
4θ0
I 6θ0 − sin 2θ0
(10.18)
In the small angle approximation, we find
pϕ
Ω≈
I
θ02
1−
3
(10.19)
Now, consider the case ω Ω. In this case, we can the bead is to good approximation
at some fixed θ(t) over the time of rotation of the ring. That is the frequency while
the bead is at height θ(t) is,
Ω(t) =
pϕ
,
I(1 + sin2 θ(t))
(10.20)
so that the solution for ϕ is
ϕ=
pϕ
t + ϕ0
I(1 + sin2 θ(t))
(10.21)
Note that in the small oscillation approximation, we can solve for ϕ explicitly, but the
results we found so far are completely general. In the small oscillation approximation,
the solution is given by
Z
pϕ
pϕ t 0
t−
dt [θ0 cos(ωt0 ) + θc ]2 + φ0
ϕ(t) ≈
I
I 0
θ02
pϕ
pϕ
1 2
2
=
1−
− θc
t−
2θ0 θc sin ωt + θ0 sin 2ωt .
(10.22)
2
I
Iω
4
46
11
11.1
Hamiltonian-Jacobi Equations
Solution of H-J Equations
We would like to solve some problems using the H-J equations, but this requires a small
extra piece of information that we didn’t derive in lecture. I will not derive it here, but it
can be found in any textbook. The point of the H-J equation is that it defines a canonical
transformation under which we get a new Hamiltonian H(Q, P, t) = 0, where Q(q, p) and
P (q, p) are new coordinates. If such a transformation is possible, then clearly the equations
of motion become Q̇ = Ṗ = 0.
Suppose now that S(q, t) solves the H-J equation. Since it is the solution to a first order
equation in n degrees of freedom, it has n integration constants, which we define to be P .
It turns out that to get the conjugate Q, the procedure is to take
∂S
= Q.
∂P
(11.1)
See the textbooks for more details if you would like. Another useful relation that we mentioned in lecture, but which will come in handy is:
∂S
= p.
∂q
(11.2)
Thus, we now have a procedure for solving mechanics using the H-J equation:
1. Solve the H-J equation for S(q, P, t) (the hardest part)
2. The solution will be parametrized by n constants, which we call P ≡ P0 .
3. Take ∂S/∂P0 = Q0 , a constant. These algebraic equations could also be difficult to
solve.
At this stage, we have an implicit equation for q in terms of constants Q0 and P0 , as well as
t. Thus, in principle, we have solved mechanics!
One last note. As we saw in lecture, recall that if dH/dt = 0, which also means that
∂H/∂t = 0, then ∂ 2 S/∂t2 = 0, so that ∂S/∂t = const. ≡ −E. This allows us to rewrite the
H-J EOM in terms of a new “time-independent” action W as follows:
∂W (q, E)
= E,
(11.3)
H q,
∂q
where S = W (q, E) − Et. In this way, we can eliminate one of the momenta P0 in favor of
E if we would like to. We then know that ∂W/∂E − t = −t0 , a constant. This can be done
with any conserved quantity: it can be chosen to be one of the momenta P and separated
out from S. For example, if there is a conserved momentum p associated with coordinate q,
we can make the trivial choice P = p, with S = Sno q + pq.
47
11.2
Harmonic Oscillator
Consider a one dimensional Harmonic oscillator. The Hamilton-Jacobi equation is:
2
1
∂W
mω 2 2
∂W
=
+
x = E.
H x,
∂x
2m ∂x
2
(11.4)
Trivially, this is a ordinary differential equation for W , as long as we hold E constant, which
we can solve be doing:
Z √
W =
2mE − m2 ω 2 x2 dx.
(11.5)
Now, we just have to do this integral! To make this easier on ourselves, note that we are
only interested in ∂W/∂E. This is given by:
Z
m
∂W
√
=
dx.
(11.6)
∂E
2mE − m2 ω 2 x2
p
If we make the substitution u = mω 2 /2Ex, then we get
r
Z
∂W
2E
m
√
√
=
du.
(11.7)
∂E
2mE 1 − u2 mω 2
The integral here is just sin−1 , as seen in any calculus class, so that
!
r
2
mω
∂W
1
= sin−1
x .
∂E
ω
2E
(11.8)
We can then take the partial with respect to E, so that
∂W
= t − t0 ,
∂E
so that we recover exactly our usual solution for the Harmonic oscillator:
r
2E
x=
sin[ω(t − t0 )].
mω 2
Note that we can also solve for the action by integrating (11.8):
r
1 √
E
mω 2
−1
W = x 2mE − m2 ω 2 x2 + sin
x,
2
ω
2E
(11.9)
(11.10)
(11.11)
up to a constant which we set to 0. We can show that this is the action, up to a term like
const. × t by plugging in the solution for x to both this and our expression for the action
itself. We find that, up to the term we just discussed,
S=
E
sin[2ω(t − t0 )].
2ω
48
(11.12)
For our system to be a classical Harmonic oscillator, we need
ω~
1
E sin(2ω∆t)
(11.13)
If ∆t ω −1 , then classical mechanics breaks down. But this just what we expect from the
energy-time uncertainty relation! Suppose now that ∆t ω −1 . This just means that we are
looking at time scales much shorter than the period. Then, (11.14) simplifies to
~
1 =⇒ E∆t ~,
E∆t
(11.14)
exactly as expected!
11.3
The Kepler Problem
Consider pair of particles with a 1/r interaction. Since this system does not have any explicit
time dependence, we can write the Hamilton-Jacobi equation as
"
"
2 #
2 #
∂W ∂W
1
∂W
1
∂W
α
H x1 , x2 ,
,
= E. (11.15)
=
+
−
∂x1 ∂x2
2m1
∂x1,i
2m2
∂x2,i
|x1 − x2 |
If we switch to our usual coordinates where one is the center of mass and one is the displacement from one particle to the other, then this becomes:
"
"
2 #
2 #
1
∂W
1
∂W
α
+
−
= E.
(11.16)
2M
∂Ri
2µ
∂ri
|r|
Finally, we can switch the ri into spherical coordinates. This gives:
"
"
2 #
2
2
2 #
∂W
1
∂W
1 ∂W
1
α
1
∂W
+
+ 2
+ 2 2
− = E.
2M
∂Ri
2µ
∂r
r
∂θ
r
r sin θ ∂φ
(11.17)
In order to solve this, we make an ansatz that this equation can be separated in the following
sense. We guess that
X
W (Ri , r, θ, φ, E, Pi , pθ , pφ , t) =
WRi (Ri , E, Pi , pθ , pφ ) + Wr (r, E, Pi , pθ , pφ )(11.18)
i
+Wθ (θ, E, Pi , pθ , pφ ) + Wφ (φ, E, Pi , pθ , pφ ).
(11.19)
We already separated out the last term to get (11.15). Now, we assume separation for the
other variables. This is just the usual technique for solving EOMs: we guess!
49
12
Perturbation Theory
12.1
Frequency Shifts in Perturbation Theory
Consider the Lagrangian:
mv 2 mω 2 x2 mλx4
−
−
(12.1)
2
2
4
We would like to treat the λ term as a perturbation on the harmonic Lagrangian. In what
limit can we do this? Suppose that the maximal displacement in x is A. Then we need
L=
λA2 ω 2 .
(12.2)
Let’s proceed in the naive way and try to do perturbation theory. We define = λA2 /ω 2 .
The equation of motion is then
ẍ + ω 2 x2 = −
ω2 3
x.
A2
(12.3)
We plug in the ansatz x = x(0) + x(1) . Then, we break up the equation by order in .
ẍ
(0)
+ ẍ
(1)
2 (0)
+ω x
2 (1)
+ ω x
ω 2 (0) 3
= − 2 (x ) + O(2 )
A
(12.4)
The terms of order 0 give
ẍ(0) + ω 2 x(0) = 0
(12.5)
which has the solution x(0) (t) = A cos(ωt), choosing initial conditions such that x(0) = A
and ẋ(0) = 0 for convenience. Then, the terms of order give (canceling out the ’s)
ẍ(1) + ω 2 x(1) = −
ω 2 (0) 3
ω2A
2
3
(x
)
=
−ω
A
cos
(ωt)
=
−
(cos(3ωt) + 3 cos(ωt)) .
A2
4
(12.6)
What is wrong with this equation? Why is it pathological? The effective driving force is
at the resonant frequency! The amplitude of x(1) diverges! Why does this happen? Well,
we assumed implicitly that the frequency of the system is the same as the frequency of the
unperturbed case. We know this is not the case, but often you can get away with that
assumption at lower orders. Not so here. Let’s start over. Let’s redefine ω → ω0 , the
frequency of the harmonic part of the Lagrangian. The “exact” frequency of the system for
a given amplitude will now be called ω. We expand ω, like we expanded x: ω 2 = ω02 + ∆ω 2 =
ω02 + ω12 , plus higher order terms in principle, but that won’t be necessary here. Now, we
pull a slight trick. We rewrite our equations of motion as
ẍ + ω 2 x = −
ω2 3
x + ∆ω 2 x.
A2
(12.7)
To do this, notice that we have to redefine our small parameter in terms of ω 2 as opposed
to ω02 . This is arbitrary, since the factors of either ω 2 or ω02 will cancel out anyway. Given
50
that caveat, we have not changed our equation of motion at all. All we did was add the same
term ∆ω 2 x to both sides of the equation. We still have not determined what we would like
∆ω to be, but that’s coming. Patience! It’s value for now is arbitrary: the choice does not
affect the equations of motion so long as we modify ω 2 accordingly. Now, our zeroth order
solution is the same, but in terms of the shifted frequency ω: x(0) (t) = A cos(ωt). The first
order equation, writing out ∆ω = ω12 , is
ẍ(1) + ω 2 x(1) = −
ω2A
(cos(3ωt) + 3 cos(ωt)) + ω12 A cos(ωt),
4
(12.8)
keeping only terms up to order . Naively, this equation still seems pathological. The trick
is to choose ω1 to exactly the case where this equation is no longer pathological. That is, we
choose
3ω 2
ω12 = 0
(12.9)
4
With this choice, our equation of motion reduces to
ẍ(1) + ω 2 x(1) = −
ω02 A
cos(3ωt),
4
(12.10)
which has a solution
A
cos(3ωt).
32
Explicitly, to first order in = λA2 /ω02 , the solution to the equations of motion is
λA2
3λA2
2
2
x(t) = A cos(ωt) +
cos(3ωt)
,
ω
=
ω
+
.
0
32ω 2
4
x(1) (t) =
(12.11)
(12.12)
Note that the frequency is now amplitude-dependent as we expect! The frequency increases,
as we expect. The freedom to choose whatever ∆ω 2 we wanted has allowed us to render our
perturbation theory meaningful again. We have explicitly found an x(t) that satisfies the
equations of motion up to terms proportional to 2 .
12.2
Second-Order Perturbation Theory
Armed with the tools we learned about frequency shifts, we are ready to deal with a second
order perturbation expansion for the problem of the Lagrangian
L=
mv 2 mω02 x2 mλx3
−
−
.
2
2
3
(12.13)
The equation of motion, as seen in lecture, can be written as
ẍ + ω02 x = −
ω02 2
x,
A
51
=
λA
.
ω02
(12.14)
We make the following definitions for our perturbation expansion:
x = x(0) + x(1) + 2 x(2) ,
ω 2 = ω02 + ω12 + 2 ω22
(12.15)
We rewrite our equation as
ẍ + ω 2 x = −
ω2 2
x + ω12 x + 2 ω22 x
A
(12.16)
To zeroth order in , we have the usual solution seen in lecture: x(0) = A cos(ωt). To first
order in , the equation looks like
ẍ
(1)
2 (1)
+ω x
ω 2 (0) 2
1 + cos(2ωt)
= − (x ) + ω12 x(0) = −ω02 A
+ ω12 A cos(ωt).
A
2
(12.17)
In order to avoid problematic resonance, we must set ω1 = 0, as seen in lecture: to first order
there is no frequency shift. Then, as in lecture, we find that
A
1
(1)
(12.18)
x (t) =
−1 + cos(2ωt) .
2
3
Finally, we write out the equation to second order:
ω 2 (0) (1)
Aω 2
x x +ω12 x(1) +ω22 x(0) = −
(−5 cos(ωt) + cos(3ωt))+ω22 A cos(ωt).
A
6
(12.19)
In order to avoid resonance, we choose ω22 = − 65 ω02 . Then, the equation reduces to
ẍ(2) +ω 2 x(2) = −2
ẍ(2) + ω 2 x(2) = −
Aω 2
cos(3ωt),
6
(12.20)
which has the solution
A
cos(3ωt).
48
Thus, the full solution to second order in perturbation theory is
λA
1
λ2 A2
x(t) = A cos(ωt) + 2 −1 + cos(2ωt) +
cos(3ωt) ,
2ω
3
48ω 4
x(2) (t) =
52
(12.21)
ω 2 = ω02 −
5λ2 A2
.
6ω02
(12.22)