Intermediate Classical Mechanics
Charles B. Thorn1
Institute for Fundamental Theory
Department of Physics, University of Florida, Gainesville FL 32611
Abstract
1
E-mail address: thorn@phys.ufl.edu
1
c
2013
by Charles Thorn
Contents
1 Introduction
1.1 Newton’s Three Laws of motion . . . . . . . . . . . . . . . . . . . . . . . . .
2 Vector notation, algebra, and calculus
2.1 Vector notation . . . . . . . . . . . . .
2.2 Addition of two vectors . . . . . . . . .
2.3 Scalar product of two vectors . . . . .
2.4 Vector product of two vectors . . . . .
2.5 Some force laws in vector notation . . .
2.6 Differentiating vectors . . . . . . . . .
2.7 Momentum Conservation . . . . . . . .
2.8 Inertial Frames of Reference . . . . . .
2.9 Coordinate Systems . . . . . . . . . . .
2.10 Polar coordinates in a plane . . . . . .
2.11 Newtonian Dynamics . . . . . . . . . .
2.12 The Value of New Formulations . . . .
4
4
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
5
5
5
6
6
7
8
8
9
9
10
12
12
3 Simple motions in one and two dimensions
3.1 Projectile motion without air resistance . . . . .
3.2 Air resistance . . . . . . . . . . . . . . . . . . .
3.3 Linear air resistance . . . . . . . . . . . . . . .
3.4 Millikan oil drop measurement of electric charge.
3.5 Quadratic air resistance . . . . . . . . . . . . .
3.6 A charged particle in a uniform magnetic field .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
13
13
14
14
15
16
18
4 Momentum and Angular Momentum
4.1 Momentum Conservation . . . . . . .
4.2 Rockets . . . . . . . . . . . . . . . .
4.3 The Center of Mass Coordinate . . .
4.4 Angular Momentum . . . . . . . . .
4.5 Systems of several particles . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
20
20
20
21
22
22
5 Energy
5.1 Conservative forces and potential energy
5.2 Stoke’s Theorem . . . . . . . . . . . . .
5.3 Potential energy examples . . . . . . . .
5.4 One dimensional systems . . . . . . . . .
5.5 Central forces . . . . . . . . . . . . . . .
5.6 Energy of systems of particles . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
24
24
25
26
27
29
32
2
.
.
.
.
.
c
2013
by Charles Thorn
6 Oscillations
6.1 Descriptions of Simple Harmonic Motion
6.2 2 and 3 dimensional oscillations . . . . .
6.3 Damped Oscillations . . . . . . . . . . .
6.4 Driving oscillations . . . . . . . . . . . .
6.5 Resonance . . . . . . . . . . . . . . . . .
6.6 Fourier Series . . . . . . . . . . . . . . .
6.7 RMS Displacement, Parseval’s Theorem
.
.
.
.
.
.
.
35
36
37
38
39
41
42
43
7 Variational Principles
7.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Calculus of Variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3 Nonmechanics applications of the calculus of variations. . . . . . . . . . . . .
44
44
45
46
8 Hamilton’s Principle of Least (Stationary)
8.1 Generalized coordinates . . . . . . . . . .
8.2 The Action and Hamilton’s Principle . . .
8.3 Changing Coordinates in the Lagrangian .
8.4 The energy from the Lagrangian . . . . . .
8.5 The simple pendulum . . . . . . . . . . . .
8.6 Constraints in general . . . . . . . . . . .
8.7 Examples . . . . . . . . . . . . . . . . . .
48
48
48
49
50
51
51
53
3
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Action
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
. . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
c
2013
by Charles Thorn
1
Introduction
The subject of mechanics deals with the motion of material bodies in three dimensional
space. The bodies might be fundamental particles like the electron, or extended objects
like baseballs, planets, or stars. We usually distinguish material bodies from other physical
entities such as electromagnetic or gravitational fields, whose description goes beyond the
mechanics we study in this course.
The subject of this course is classical mechanics which neglects quantum effects. In
other words we pretend that Planck’s constant ~ = h/2π is zero. This is a very good
approximation for all of the everyday physics we encounter around us. Quantum effects
become important only when we delve into atomic and molecular scales. For the most part,
in this first semester, we also assume that all speeds are much smaller than c the speed of
light, so that the modifications of mechanics required by Einstein’s theory of relativity can
also be neglected. Relativistic physics will be deferred to the second semester of the course.
Thus in this first semester we will be studying the application of Newton’s laws of motion,
which have been in place since the seventeenth century.
1.1
Newton’s Three Laws of motion
We start with an idealization: we imagine that we can isolate a material body from all
external influences (we say that no forces act on the body). In the real physical world
this idealization is never exactly realized but it can be approached by gradually reducing
frictional forces and for example restricting motion near the earth’s surface to a horizontal
plane. Newton’s first law then describes the body’s motion in this idealized situation:
1st Law: In the absence of forces a body moves with uniform velocity.
It is important to note that the velocity is a vector quantity having a direction as well as
magnitude (speed). Thus the isolated body is either at rest or moves in a straight line at
constant speed. The first law is the least specific of the three laws, and in fact is valid
for relativistic mechanics and even in quantum mechanics, to the extent allowed by the
uncertainty principle (i.e. the isolated particle’s wave function is a plane wave!).
The second law begins to pin down the effect of forces on a body:
2nd Law: ma = F .
Here a is the vector acceleration of the body, which has a clear meaning. The mass m is
an attribute of the body that quantifies its resistance to the action of the force. A heavier
body accelerates less than a lighter one if subjected to the same force. The force is not
an attribute of the body, but quantifies the interaction of the body with its environment.
Although not explicit in the second law there is the promise that the force is a cogent way
of characterizing the body’s interaction with its environment.
The third law puts an important constraint on the nature of forces.
4
c
2013
by Charles Thorn
3rd Law: Every action has an equal and opposite reaction.
Quantitatively it means that if body 1 exerts a force F on body 2, then of necessity body 2
exerts a force −F on body 1. It is important to appreciate that Newton’s formulation of the
constraint is specific to Newtonian mechanics which includes forces which act instantaneously
at a distance. Indeed such forces violate the causality of special relativity which requires that
influences can not travel faster than the speed of light. As we shall see, the third law implies
the conservation of momentum which will hold in general, if all the momentum is accounted
for. For example, the electromagnetic field can carry momentum, and two charged bodies
can give up some of their momentum to the field. Thus the third law applied to the two
bodies would fail but momentum would still be conserved.
2
Vector notation, algebra, and calculus
One of the objectives of this intermediate course on mechanics is to get beyond the oversimplifications that were made in the introductory course (PHY2048). We want to be able to
describe more complicated motions that involve all three dimensions of space. The concept
of vectors is a valuable tool for this task.
The motion of a point mass can be specified by giving its location at each time t. To
do this we establish a Cartesian coordinate system with three perpendicular coordinate axes
intersecting at the origin O of the system. Any point in space can be located by its three
coordinates x, y, z. The trajectory of a point particle is then specified by the three functions
x(t), y(t), z(t).
2.1
Vector notation
Next we introduce vector notation. We are familiar with geometric notion of a vector as an
arrow that points in the direction of the vector and whose length is the magnitude of the
vector. We denote a vector in the text by a bold face v. On the blackboard I will frequently
use instead an over arrow ~v . We can multiply vectors by numbers cv. Multiplying by a
negative number reverses the direction of the vector.
2.2
Addition of two vectors
The sum of two vectors v 1 + v 2 is given geometrically by the parallelogram rule. But a
more efficient procedure, especially when several vectors are to be added, is to resolve the
vector into its components along the coordinate axes. To do this we introduce unit vectors
i = x̂, j = ŷ, k = ẑ, pointing in the positive x, y, and z directions respectively. Then the
components of v are such that
v ≡ vx i + vy j + vz k = (vx , vy , vz )
(1)
Using component notation we then have for the sum v 1 + v 2 = (v1x + v2x , v1y + v2y , v1z + v2z )
5
c
2013
by Charles Thorn
With this notation we can now think of the coordinates of a point in space as the position
vector:
r ≡ xi + yj + zk = (x, y, z)
(2)
The position vector differs from a true vector in that it depends on the origin of coordinates.
But the notation is nonetheless extremely useful. Examples of true vectors are the velocity
and acceleration of a particle as well as the force. Thus Newton’s second law F = ma is an
equation between two true vectors. If we have two points specified by position vectors r 1
and r 2 the displacement d12 = r 2 − r 1 is a true vector, since it doesn’t depend on the origin
of the coordinates system.
2.3
Scalar product of two vectors
Geometrically the scalar product of two vectors A, B is defined as A · B = |A||B| cos θ
where θ is the angle between the two vectors. In terms of components it is given by
A · B = Ax Bx + Ay By + Az Bz =
3
X
Ai Bi = Ai Bi
(3)
i=1
where the last equality invokes the summation convention that repeated indices are always
summed. In the summation we identify x → 1, y → 2, z → 3. Along with this labeling, it
is also convenient to similarly label the basis vectors:
i = (1, 0, 0) ≡ e1 ,
j = (0, 1, 0) ≡ e2 ,
(
0 j 6= k
ej · ek = δkj ≡
1 j=k
i = (0, 0, 1) ≡ e3
(4)
(5)
Then we can write any vector
A =
X
Ai ei = Ai ei ,
i
Ai = ei · A
(6)
Notice that the scalar product is distributive (A + B) · C = A · C + B · C. Note also that
A · A = |A|2 .
2.4
Vector product of two vectors
We have seen that the scalar product of two vectors is a number. In contrast the vector
product of two vectors A, B is a vector. Its magnitude is |A||B| sin θ, and its direction is
perpendicular to both A and B, in the sense of the right hand rule. In terms of components
we have
(A × B)z = Ax By − Ay Bx ,
(A × B)y = Az Bx − Ax Bz ,
6
(A × B)x = Ay Bz − Az By
c
2013
by Charles Thorn
By introducing the antisymmetric symbol ǫijk , the relation of components can be more
succinctly written
(A × B)i = ǫijk Aj Bk
(7)
where the summation convention is in force. ǫijk is antisymmetric under the interchange of
any pair of indices: ǫijk = −ǫjik = −ǫikj = −ǫkji , and therefore vanishes unless i, j, k are all
different. By convention ǫ123 = +1. Using ǫijk we see that
A · (B × C) = ǫijk Ai Bj Ck = ǫkij Ai Bj Ck = (A × B) · C
(8)
The fundamental identity
ǫijk ǫklm = δil δjm − δim δjl
(9)
is easily proven: just put (i, j) = (1, 2), (2, 3), (3, 1) in turn. This identity implies the triple
vector product identity
[A × (B × C)]i = ǫijk Aj (B × C)k = ǫijk Aj ǫklm Bl Cm = (δil δjm − δim δjl )Aj Bl Cm
= Aj Bi Cj − A · BCi = A · CBi − A · BCi
(10)
where the last form is true if B and C commute.
2.5
Some force laws in vector notation
Newton’s inverse square force law for gravitational systems can be written:
F Grav
= −mk
k
X Gmj (r k − r j )
j6=k
|r k − r j |3
(11)
where we have several particles present. The vector notation makes Newton’s third law
evident:
F kj = −
Gmk mj (r k − r j )
= −F jk
|r k − r j |3
(12)
which has no dependence on the velocities of the particles. Newton didn’t know about
magnetic forces, which were understood much later. A particle of charge q moving in an
electromagnetic field experiences a force
F EM = q [E(r, t) + v × B(r, t)]
(13)
The existence of magnetic forces means that we have to allow for velocity dependent forces
to hope to describe all phenomena. Note that the 3rd law has no obvious interpretation in
this formula since the fields which exert the force on the charged particle are not themselves
material bodies. When the sources of the fields are included in the dynamics, however,
momentum conservation will be valid with an appropriate identification of the momentum
carried by the fields.
7
c
2013
by Charles Thorn
2.6
Differentiating vectors
When we describe the trajectory of a particle as a time dependent position vector r(t), we
can obtain the velocity and acceleration of the particle as first and second derivatives:
dr
dv
d2 r
,
a = v̇ =
= r̈ = 2
(14)
dt
dt
dt
where we have introduced the dot notation for time derivatives.
The time derivative of a vector is defined in parallel to the time derivative of a scalar
function:
v = ṙ =
A(t + ǫ) − A(t)
dA
≡ lim
(15)
ǫ→0
dt
ǫ
Since the Cartesian unit vectors i, j, k are independent of time, it is cleat that the components of dA/dt are simply (Ȧx , Ȧy , Ȧz ). In other words the Cartesian components of the
time derivative of a vector are simply the time derivatives of the Cartesian components of
the vector.
Spatial derivatives of vector functions of the coordinates can have several manifestations.
The gradient operator ∇ is defined by
∂f ∂f ∂f
∇f ≡
(16)
,
,
∂x ∂y ∂z
For example the divergence and curl of a vector function A(r) are given by
X ∂Ai
∂Ai
∂Ak
=
,
(∇ × A)i = ǫijk
∇·A =
∂xi
∂xi
∂xj
i
(17)
where the summation convention has been used. Notice that the meaning of the dot and
cross products here are exactly those already defined.
For completeness we mention the integral theorems of Gauss and Stokes:
Z
I
Z
I
dV ∇ · E =
dS · E,
dS · ∇ × E = dl · E
(18)
familiar from electromagnetic theory.
2.7
Momentum Conservation
We have mentioned that momentum conservation is a consequence of Newton’s third law.
Let us first consider two particles. According to the 3rd law if particle 2 exerts a force F 12
on particle 1, then particle 1 exerts a force F 21 = −F 12 on particle 2. Then the second law
reads
dp1
dp2
= F 12 ,
= F 21 = −F 12
(19)
dt
dt
d
(p + p2 ) = F 12 + F 21 = 0
(20)
dt 1
8
c
2013
by Charles Thorn
That is, P = p1 + p2 is a constant. If we have more particles exerting pairwise forces on
each other, the second law reads for particle k
X
dpk
=
F kl
(21)
dt
l6=k
Summing over all particles gives
XX
X
X
X
X
dP
d X
=
pk =
F kl =
F kl +
F kl =
F kl +
F lk = 0
dt
dt k
k l6=k
k<l
k>l
k<l
l>k
(22)
by the third law.
2.8
Inertial Frames of Reference
Inertial frames are those in which the first law holds. By implication this implicitly carries
the assumption that we know all of the forces. If all the known forces are zero, but an
unknown force is not zero we would judge the frame to be non-inertial. Once we have found
one inertial frame, any frame of reference moving at constant velocity relative to it will also
be an inertial frame.
2.9
Coordinate Systems
We have used vector notation to express the second law concisely ṗ = F . But to deal
with the equations mathematically it is usually best to write out the components of the
vector equation. In Cartesian coordinates, the basis vectors are fixed once and for all so
that applying time derivatives to the position vector r = (x, y, z) simply means applying the
derivatives to the components: r̈ = (ẍ, ÿ, z̈). For the case where F is a constant vector the
differential equation of motion is solved by two integrations. A familiar example is gravity
near the surface of the earth, which exerts a force −mgẑ on a particle of mass m. Then the
force only enters the z component:
1
z̈ = −g,
ż = −gt + v0 ,
z(t) = − gt2 + v0z t + z0
(23)
2
meanwhile x(t) = v0x t + x0 and y(t) = v0y t + y0 . In these solutions the 6 integration constants
(v0x , v0y , v0z ), (x0 , y0 , z0 ) are given by the initial conditions.
A more complicated uniform force problem is the classic freshman physics problem of
a block sliding down an inclined plane in the presence of friction. In this problem it is
convenient to pick the x-axis of the Cartesian system parallel to the inclined plane, so
motion is only in the x coordinate. Then the three forces are gravity that points vertically
down mg(sin θ, − cos θ, 0), the normal force N (0, 1, 0) exerted by the plane, and the frictional
force (−µN, 0, 0). Since there is no motion in the y coordinate, Fy = N − mg cos θ = 0.
Then Fx = mg sin θ −µN = mg(sin θ −µ cos θ) determines the acceleration down the inclined
plane. then x(t) is given by
g
x(t) =
(sin θ − µ cos θ)t2 + v0 t + x0
(24)
2
9
c
2013
by Charles Thorn
2.10
Polar coordinates in a plane
For something a little different let’s consider how the equations look in polar coordinates on
the xy-plane (z = 0).
x = r cos ϕ,
y = r sin ϕ,
r = (r cos ϕ, r sin ϕ, 0)
(25)
Now take derivatives of r to obtain the velocity and acceleration:
v = ṙ = ṙ(cos ϕ, sin ϕ, 0) + rϕ̇(− sin ϕ, cos ϕ, 0)
(26)
2
a = v̇ = r̈(cos ϕ, sin ϕ, 0) + (2ṙϕ̇ + rϕ̈)(− sin ϕ, cos ϕ, 0) − rϕ̇ (cos ϕ, sin ϕ, 0)
= (r̈ − rϕ̇2 )(cos ϕ, sin ϕ, 0) + (2ṙϕ̇ + rϕ̈)(− sin ϕ, cos ϕ, 0)
(27)
Now we can recognize r̂ = (cos ϕ, sin ϕ, 0) as the unit vector pointing in the direction of r,
and ϕ̂ as the unit vector pointing in the direction of increasing ϕ at fixed r. So we can write
more compactly:
r = rr̂,
a = (r̈ − rϕ̇2 )r̂ + (2ṙϕ̇ + rϕ̈)ϕ̂
v = ṙr̂ + rϕ̇ϕ̂,
(28)
The reason for these complicated formulas is that the unit basis vectors r̂, ϕ̂ of the polar
coordinate system are not constant in time:
dr̂
= ϕ̇ϕ̂,
dt
dϕ̂
= −ϕ̇r̂
dt
(29)
As a consequence, Newton’s second law in polar coordinates is a bit complicated: write
F = Fr r̂ + Fϕ ϕ̂, and find
m(r̈ − rϕ̇2 ) = Fr ,
m(2ṙϕ̇ + rϕ̈) = Fϕ
(30)
The oscillating skateboard on a half pipe in the text is essentially the same as a pendulum
consisting of a point mass attached to a massless rod of length R. The mass point feels
the downward force of gravity mg = mg(cos ϕr̂ − sin ϕϕ̂) and the normal force N = −N r̂.
Since the rod is a constant length ṙ = r̈ = 0, so the second law reads:
−mRϕ̇2 = −N + mg cos ϕ,
mRϕ̈ = −mg sin ϕ
(31)
the first equation determines N = m(Rϕ̇2 +g cos ϕ), which is precisely the force of constraint
keeping the bob moving in a circle. The important difference between the pendulum and
the skateboard on a half pipe is that the normal force exerted by the rod can be in either
direction, whereas the half pipe requires N > 0. We see that if ϕ > π/2 and ϕ̇ is small
enough, N has to be negative to keep the mass moving in a circle! The second equation
determines the motion of the pendulum
ϕ̈ = −
g
sin ϕ
R
10
(32)
c
2013
by Charles Thorn
This differential equation determines ϕ(t), however not in terms of elementary functions.
when ϕ ≪ 1 the right side can be approximated sin ϕ ≈ ϕ and the equation linearizes,
reducing to the equation of motion for a simple harmonic oscillator:
g
ϕ̈ = − ϕ
R
(33)
Since the diff eq is linear with constant coefficients we can always find solutions of the form
ert , becausepthen each derivative gives a factor of r and the equationpreduces to r2 = −g/R
or r = ±i g/R. The angular frequency of the oscillator is ω ≡ g/R and the general
solution is
ϕ(t) = Ae−iωt + Beiωt = (A + B) cos ωt − i(A − B) sin ωt
(34)
Since ϕ(t) is a real function, we must require B = A∗ . A can be determined by initial
conditions on ϕ:
ϕ̇(0)
sin ωt.
(35)
ω
p
The motion is periodic with period T = 2π/ω = 2π R/g.
What can we say if ϕ is not small? We can find an energy conservation law by multiplying
the equation by ϕ̇:
d 1 2 g
g
ϕ̇ − cos ϕ
(36)
0 = ϕ̇ϕ̈ + ϕ̇ sin ϕ =
R
dt 2
R
1 2 g
ϕ̇ − cos ϕ = C
(37)
2
R
ϕ(t) = ϕ(0) cos ωt +
which we can solve for ϕ̇:
p
dϕ
dϕ
dt = p
=
2C + 2ω 2 cos ϕ,
dt
2C + 2ω 2 cos ϕ
Z ϕ(t)
dϕ′
p
t =
2C + 2ω 2 cos ϕ′
ϕ(0)
(38)
The integral on the right is known as an elliptic integral which can not be expressed in
terms of elementary functions. We will return to its study later on when we come to a more
systematic discussion of energy. For small oscillations the integral can be done
t ≈
Z
0
ϕ(t)
dϕ′
p
2C +
2ω 2
−
ω 2 ϕ′2
=
ωϕ(t)
1
arcsin √
ω
2C + 2ω 2
(39)
recovering simple harmonic oscillations.
11
c
2013
by Charles Thorn
2.11
Newtonian Dynamics
Classical mechanics has not really changed, in substance, since the days of Isaac Newton.
The essence of Newton’s insight, encoded in his second law F = ma, is that the motion of
a particle r(t) is determined once its initial position and velocity are known. His famous
equation relates the acceleration d2 r/dt2 to the force on the particle, which is implicitly
assumed to depend only on the positions, and possible the velocities of the particles in the
system. Consider a system of N particles, whose trajectories are described by 3N coordinates
r k (t), k = 1, . . . , N . Then Newton’s laws of motion take the mathematical form of 3N second
order differential equations in time:
mk
d2 r k
= F k ({r i }, {ṙ i })
dt2
(40)
This is the general framework, but of course for each dynamical system we also need to know
the force law. Newton also didn’t know about Einstein’s relativity, but his law of motion,
appropriately modified carries over to this domain as well. The necessary modification is to
replace ma with dp/dt, where p is the relativistic momentum of the particle:
p ≡ p
mv
1 − v 2 /c2
(41)
We see that for particles moving slowly compared to the speed of light v ≪ c, the momentum
goes approximately to its Newtonian expression p ≈ mv, so ṗ ≈ ma. But even for fully
relativistic motion the basic structure of Newtonian dynamics holds.
2.12
The Value of New Formulations
The previous subsection contains everything you need to know about Newtonian dynamics–
once you solve the equations there is nothing more to say. However, there are other ways
to look at the dynamics that reveal features of the motion that brute force solution of the
equations might leave obscure. For example in elementary physics we learn how to exploit
energy conservation when the force is the gradient of a potential F = −∇V . Then instead
of working with the second order differential equations we can use energy conservation
E =
1
mv 2 + V (r) = Constant
2
(42)
to immediately read off the speed of the particle in terms of its location and total energy E.
In one dimensional motion the potential energy curve tells us a lot about the types of
motion that will occur. A horizontal line of height E intersects V (x) at the “turning points”
where the particle comes to rest. Points where dV /dx = 0 tell where a particle feels zero
force. A particle placed there at rest will stay at rest: we spot the static solutions by looking
for the maxima and minima of the potential. A minimum is stable equilibrium, whereas a
maximum is unstable equilibrium.
12
c
2013
by Charles Thorn
For motion in one dimension, energy conservation implies Newton’s equations
∂V
∂V
dE
= mẋẍ + ẋ
= ẋ mẍ +
=0
dt
∂x
∂x
(43)
so as long as ẋ 6= 0, Newton’s equation holds. But in two or more dimensions energy
conservation only tells us
ẋ · (mẍ + ∇V ) = 0
(44)
which only implies that mẍ + ∇V is perpendicular to ẋ. For example the magnetic force
might or might not be present:
mẍ = −∇V + q ẋ × B
(45)
In the second semester of this course the ideas of Lagrange, Hamilton, and Jacobi will be
used to interpret general nonstatic solutions in terms of maxima or minima of an energy-like
quantity called the action. Since a nonstatic solution is a curve in space rather than simply
a point, we have to study the action as a function of curves, which requires the concepts of
the calculus of variations, which we will develop near the end of this semester.
A great advantage of the action is that by construction it is an invariant under the
symmetries of the dynamical system. It is a scalar functional of the coordinates qk (t) and
velocities q̇k (t). It therefore summarizes the dynamical content of a system in a compact
and transparent way. As we shall see, it also greatly simplifies the problem of imposing
constraints on the dynamical variables.
3
Simple motions in one and two dimensions
To gain some experience with solving Newton’s equations we go beyond uniform forces by
considering velocity dependent forces that are independent of position. In such situations,
Newton’s laws are first order differential equations for the velocity, which makes them relatively easy to solve. Examples of velocity dependent forces are air resistance in projectile
motion and magnetic forces in a uniform magnetic field.
3.1
Projectile motion without air resistance
When we ignore air resistance, projectile motion near the earth’s surface is governed purely
by a uniform gravitational force −mgẑ. The trajectory stays in the plane determined by the
z-axis and the initial horizontal velocity. We choose coordinates so that this is the xz-plane.
Take the initial time to be t = 0, the initial location of the particle to be x = y = 0 and
z = h, and the initial velocity to be v 0 = Vx x̂ + Vz ẑ. Then we have
1
z(t) = h + Vz t − gt2
(46)
2
For example the range of the projectile is the horizontal distance traveled when z returns to
its initial value. This happens at the time t = 2Vz /g, so the range is 2Vx Vz /g.
x(t) = Vx t,
13
c
2013
by Charles Thorn
3.2
Air resistance
To add a little realism to projectile motion (and make the equations of motion a little less
trivial!) Let’s consider adding the effect of air resistance. We will assume the force of air
resistance depends only on the velocity, and is directed oppositely to the velocity (i.e. it
must tend to slow the speed!). We also make the simplifying assumption that its magnitude
is at most quadratic in v:
f ≈ −bv − cvv
(47)
This should be a good approximation if v is not too large. The coefficients b, c depend on the
properties of the medium as well as the geometry of the projectile. Assuming the projectile
is a sphere of diameter D, then b = βD and c = γD2 , and β, γ depend on the medium. For
air at STP β = 1.6 × 10−4 Ns/m2 and γ = 0.25Ns2 /m4 . What this means for the relative size
of the two terms is
fquad
Dv
∼ 1600 2
flin
1m /s
(48)
which means that for normal projectiles the quadratic term dominates. This is unfortunate,
because the linear term is so much easier to deal with!
3.3
Linear air resistance
The equation of motion for a projectile with linear air resistance is
m
dv
= −mgẑ − bv,
dt
m
dv ′
= −bv ′ ,
dt
v ′ = v ′0 e−bt/m
(49)
where we temporarily defined v ′ ≡ v + (mg/b)ẑ. Going back
v(t) = −(mg/b)ẑ(1 − e−bt/m ) + v 0 e−bt/m
(50)
We see that in characteristic time τ = m/b, v(t) approaches the terminal velocity v ter =
−gτ ẑ. It is interesting to note that the time τ and hence the terminal velocity depends on
the mass of the object, since b depends only on the shape of the object. So unlike motion
in a gravitational field in vacuum, the effects of air resistance are different for varying mass.
Clearly heavier particles are less affected by air resistance than lighter ones. For objects of
the same composition (density ρ) and different size (diameter D) the mass is proportional
to D3 . For linear air resistance b ∝ D, so τ ∝ D2 : larger drops fall faster.
By a simple integration of v(t), we find the trajectory, with initial position at the origin
(0, 0, 0),
r(t) = −(gτ )ẑ(t − τ (1 − e−t/τ )) + v 0 τ (1 − e−t/τ )
x(t) = v0x τ (1 − e−t/τ ),
z(t) = −(gτ )(t − τ (1 − e−t/τ )) + v0z τ (1 − e−t/τ )
14
(51)
(52)
c
2013
by Charles Thorn
To get the shape of the trajectory, we need to eliminate t in favor of x in the expression for
z:
x
x
−t/τ
e
= 1− x ,
(53)
t = −τ ln 1 − x
v0 τ
v0 τ
x
v0z + gτ
v0z + gτ
2
=
gτ
ln
1
−
+
x
(54)
z(x) = −gτ t + x
v0x
v0x τ
v0x
Notice that z → −∞ when x → v0x τ , meaning that x will never reach that value. Thus v0x τ
is an upper limit on the possible range of motion.
If air resistance is a small effect τ = m/b is large and it is reasonable to expand the
logarithm in a Taylor series:
ln(1 − a) = −a −
a2 a3
−
− ···
2
3
(55)
Putting a = x/(v0x τ ), we see that the first term cancels and we get
2x2
gx2
gx3
v0z
gx
2v0x v0z
z ≈ − x2 −
+ x x = − x2 x +
−
2v0
3τ v0x3
v0
2v0
3τ v0x
g
(56)
The range R is the value of x where the right side is 0:
2
2v0x v0z
2v0x v0z 8v0x v0z2
2v0x v0z
2
−
−
=
R ≈
R ≈
g
3τ v0x
g
3τ g 2
g
4v0z
1−
3vter
(57)
The factor in parentheses multiplies the range in vacuum, i.e. no air resistance. This factor
is clearly smaller than one, but close to one in the approximation used to obtain the result.
3.4
Millikan oil drop measurement of electric charge.
The formula for the terminal velocity vter = mg/b gets modified in the presence of an electric
field to vter = (mg − qE)/b in the presence of an electric field. Thus one can turn a careful
measurement of the terminal velocity a charged particle into a measurement of the particle’s
charge q. Looking back to the value of b = βD, with β ≈ 1.6 × 10−4 , the terminal velocity
of a spherical drop of oil with density ρ at zero electric field is
vter =
ρV g
4π D2
π 9.8 · 840 × 10−12 m
m
=
≈
≈ 2.7 × 10−5
−4
βD
3 8β
6
1.6 × 10
s
s
(58)
for a one micron oil drop of density 840kg/m3 = 0.840g/cm3 . This is so slow that it can
be directly measured by time of flight with the aid of a microscope. For example one can
fine-tune the electric field to cancel the gravitational force, and then get q = −mg/E. By
measuring lots of different drops of the same size one learns that q is an integer multiple of
a fundamental unit of charge. Incidentally the time it takes to reach this terminal speed is
τ = vver /g ≈ 3 × 10−6 s, which is virtually instantaneous!.
15
c
2013
by Charles Thorn
3.5
Quadratic air resistance
Now let’s consider the case where the linear resistance force is utterly negligible, so the
quadratic force must be used. Then we need to deal with the 2nd law
dv
c
cq 2
vx + vy2 + vz2 v
(59)
= −gẑ − vv = −gẑ −
dt
m
m
For projectile motion we can assume vy = 0, but then we still have a hard mathematical
problem. To see this resolve into components:
dvx
cp 2
= −
vx + vz2 vx ,
dt
m
cp 2
dvz
= −g −
vx + vz2 vz
dt
m
(60)
We could for example use the first equation to find vz in terms of vx , v̇x and plug into the
second equation, leading to a nonlinear second order differential equation!
However, if we restrict the motion to either purely horizontal or purely vertical we get
soluble equations. Let’s start with horizontal motion so gravity plays no role. So set vz =
vy = 0, after which the equation for vx ≡ v becomes
c
dv
= − v2,
dt
m
m
m
−
,
t(v) =
cv cv0
dt
m
=− 2
dv
cv
v0
mv0
=
v(t) =
cv0 t + m
1 + t/τ2
(61)
(62)
where τ2 = m/(cv0 ), which unlike τ depends on the initial conditions. We see that the speed
slows down toward 0, but only as 1/t. As it slows down, it will eventually have so low a
speed that the quadratic approximation is no longer valid, and linear air resistance takes
over. After that the slowing will become exponential. By a single integration we obtain x(t),
assuming the quadratic approximation:
x(t) = x0 + v0 τ2 ln(1 + t/τ2 )
(63)
The fact that x → ∞ as t → ∞ is unrealistic and is due to the assumption that the quadratic
approximation is valid down to arbitrarily low speeds. If we put both terms into the equation
of motion we get a more reasonable (though more complicated) result:
dv
c
dt
m
m
m
mc/b
= −bv − v 2 ,
=−
=
−
=
−
+
dt
m
dv
bv + cv 2
c(v + b/2c)2 − b2 /4c
bv b + cv
Z v
m
m v(cv0 + b)
c
v(cv0 + b)
1
t = −
= − ln
,
e−bt/m =
dv ′ ′ −
′
b v0
v
b + cv
b
v0 (cv + b)
v0 (cv + b)
v0
v0 e−bt/m
∼
,
−bt/m
1 + (v0 c/b)(1 − e
)
1 + cv0 t/m
m
x(t) = x0 + ln 1 + (v0 c/b)(1 − e−bt/m )
c
v(t) =
for
bt
≪1
m
(64)
(65)
We see that after a very long time the total distance traveled x(∞)−x0 = (m/c) ln(1+v0 c/b)
is finite as long as b is finite.
16
c
2013
by Charles Thorn
Next we consider vertical motion, so that gravity plays a role;
Z v
dv
dv ′
m
= −mg − c|v|v,
t = −m
′ ′
dt
v0 mg + c|v |v
(66)
Notice that we have to be careful with the sign of the resistive term. If v > 0 (upward
motion) it should be negative so it tends to slow the projectile. But if v < 0 (downward
motion) it must be positive to reduce the speed (magnitude of v). If v0 > 0 at the beginning
of the motion both gravity and air resistance tend to slow the projectile. But then after
it reaches its high point and starts to descend, gravity and air resistance work in opposite
directions. Let us first take v0 ≤ 0, so that v will stay negative thereafter. Then the equation
is solved by doing the integral
r
Z
Z v
dv ′
1 v
dv ′
mg
=−
,
vter =
t = −m
2
′2
′2
g v0 1 − v /vter
c
v mg − cv
Z v 0 1
vter (1 + v/vter )(1 − v0 /vter )
1
1
−gt =
dv ′
=
ln
(67)
+
′
′
2 1 + v /vter 1 − v /vter
2
(1 − v/vter )(1 + v0 /vter )
v0
To simplify life let’s assume that v0 = 0, i.e. the ball is dropped from rest. Then inverting
the last result gives
gt
1 − e−2gt/vter
1 + v/vter
=
−v
tanh
v(t) = −vter
= e−2gt/vter ,
ter
1 − v/vter
1 + e−2gt/vter
vter
v2
gt
v2
v2
z(t) = z0 − ter ln cosh
= z0 + ter ln 2 − ter ln(egt/vter + e−gt/vter )
g
vter
g
g
2
v
∼ z0 + ter ln 2 − vter t
g
(68)
(69)
where the last line shows the asymptotic behavior at large t, which simply reflects the
approach to a terminal velocity.
If v0 > 0, i.e. the projectile is thrown upward, the upward part of the journey is described
by
v
Z v
Z
dv ′
dv ′
1 v
v ′ vter
t = −m
=−
arctan
=−
2
′2
g v0 1 + v ′2 /vter
g
vter v0
v0 mg + cv
2
v0
vter
v0
gt
gt
d
v(t) = vter tan arctan
=
(70)
ln cos arctan
−
−
vter vter
g dt
vter vter
2
2
vter
vter
v0
v0
gt
z(t) = z(0) +
−
(71)
ln cos arctan
ln cos arctan
−
g
vter vter
g
vter
The upward journey ends when v(t) = 0por gt = vter arctan(v0 /vter ). At that point it has
2
2
/vter ). Then it starts falling as described
risen an amount zmax −z(0) = (vter
/g) ln( v02 + vter
in the previous paragraph.
17
c
2013
by Charles Thorn
Finally we briefly consider general projectile motion. As we already mentioned it is
hopeless to find an explicit solution to the equations of motion.
dvz
dvx
cp 2
cp 2
(72)
vx + vz2 vx ,
vx + vz2 vz
= −
= −g −
dt
m
dt
m
We of course know what to expect qualitatively, because either type of air resistance qualitatively hinders the motion relative to that in the vacuum. The projectile will not rise as high
nor go as far as in vacuum; and of course there is a terminal velocity in both cases. When
we considered pure horizontal motion with quadratic air resistance we found the unreasonable result that although the velocity drops to zero, it does so slowly enough that there is
no upper limit to the horizontal displacement. However this doesn’t happen in combined
vertical and horizontal motion.
To see this we look at the equations at very late times, when we know that v x → 0 and
v z → vter . In this regime the equations become
dvx
cvter
dvz
cvter
≈ −
vx ,
≈ −g −
vz
(73)
dt
m
dt
m
which are the equations for linear air resistance with the parameter b = cvter . So eventually
v x approaches zero exponentially, which means that x(t) approaches a finite value at large t:
Z ∞
x(t) → x∞ ≡
dtv x (t)
(74)
0
3.6
A charged particle in a uniform magnetic field
The velocity dependent forces we have considered thus far are not fundamental: they only
take into account the fundamental interactions of air molecules with the projectile in an
average and over-simplified way. In contrast the magnetic force F = qv ×B is a fundamental
force of nature. Unlike the force of air resistance which dissipates energy, the magnetic force
acts in a way that conserves kinetic energy. It is fairly easy to see this2
dK
mq
=
2v · (v × B) = 0
(76)
dt
2
because v is orthogonal to v × B! So the magnetic force is non-dissipative.
Since v 2 is a constant, only the direction of v changes with time. To investigate the
motion, let us choose coordinates so that B, which we shall assume is uniform, is parallel
to the z axis. Then the magnetic force is parallel to the xy plane. Newton’s equation is
m
2
find
dv
= q(v × B ẑ)
dt
(77)
The complete electromagnetic force includes an electric component F = q(E + v × B). In this case we
dK
dt
= qv · E → −
d
(qφ)
dt
(75)
for an electrostatic field E = −∇φ. Then E = K + qφ is conserved.
18
c
2013
by Charles Thorn
Using the definition of the cross product we easily resolve this equation into components:
dv x
dv y
dv z
= 0,
m
= qBv y ,
m
= −qBv x
(78)
m
dt
dt
dt
The first equation is trivial to solve: it just says v z =constant, so z(t) = z0 + v z t. There are
at least two ways to solve remaining equations.the most straightforward way is to use the
second equation to eliminate v y from the third equation:
m x
qB
vy =
v̇ ,
v̈ x = −ω 2 v x ,
ω=
(79)
qB
m
We recognize the second equation as the equation for trig functions cos ωt or sin ωt. Thus
the general solution for v x is
m x
1
v x = C sin ωt + S cos ωt ≡ A sin(ωt − δ),
vy =
v̇ = v̇ x = A cos(ωt − δ) (80)
qB
ω
√
where C = A cos δ, S = −A sin δ or A = C 2 + S 2 and δ = − arctan(S/C). Then a single
integration gives x(t), y(t):
A
A
x(t) = x0 − cos(ωt − δ),
y(t) = y0 + sin(ωt − δ)
(81)
ω
ω
r(t) = r 0 + (−R cos(ωt − δ), R sin(ωt − δ), v z t)
(82)
The projection of the trajectory onto the xy plane is a circle of radius R = v0 /ω = mv0 /qB
centered on the point (x0 , y0 , 0). The particle traces this circle in a clockwise direction (If
we flip the sign of B the particle will move in the counter-clockwise direction). If v z 6= 0
the three dimensional trajectory traces a circular helix with axis parallel to the z axis. Note
that we have 6 arbitrary constants of integration x0 , y0 , z0 , v z , R, δ, so that this is in fact the
most general solution.
There is an interesting mathematical representation of the motion of a magnetic field in
terms of complex exponentials. Let us go back to the equations of motion, and define the
complex quantity η ≡ vx + ivy . Then
dη
dvx
dvy
=
+i
= ω(vy − ivx ) = −iω(vx + ivy ) = −iωη
(83)
dt
dt
dt
This is just the differential equation satisfied by e−iωt ! Thus we can write the solution as
q
2
2
η(t) = vx (t) + ivy (t) = (v0x + iv0y )e−iωt ≡ iAeiδ−iωt ,
A = v0x
+ v0y
(84)
A iδ−iωt
e
(85)
ω
By taking the real and imaginary parts of both sides we recover the previous solution:
A
A
y(t) = y0 + sin(ωt − δ)
(86)
x(t) = x0 − cos(ωt − δ),
ω
ω
We see that the complex exponential e−iωt gives a very efficient representation of uniform
circular motion. Looking down from positive z toward negative z, the motion is clockwise
when Bz > 0 and counterclockwise when Bz < 0. This is the direction of motion that
guarantees that the magnetic force points radially toward the center of the circle.
x(t) + iy(t) = x0 + iy0 −
19
c
2013
by Charles Thorn
4
4.1
Momentum and Angular Momentum
Momentum Conservation
In our discussion of Newton’s 3rd law we noted that in a closed system with no external
forces F ext = 0, the 3rd law guarantees momentum conservation:
N
d X
mk v k = F ext = 0
dt k=1
(87)
In a collision of two particles, momentum conservation reads
m1 v 1 + m2 v 2 = m1 v ′1 + m2 v ′2
(88)
Suppose, for example, that particle 2 starts at rest, v 2 = 0. Then the particle 2’s final
velocity is determined:
v ′2 =
m1
(v 1 − v ′1 )
m2
(89)
The description of collisions is much simpler in the inertial frame where the total momentum vanishes m1 v 1 = −m2 v 2 . (This is called the center of mass system) Then conservation
of momentum simply says that m1 v ′1 = −m2 v ′2 . It is then very easy to calculate the change
in kinetic energy in the collision process:
m21
1
v 21 − v ′2
(90)
m1 +
∆K.E. =
1
2
m2
From this we immediately see that the maximum loss of kinetic energy (i.e. the collision is
maximally inelastic) occurs when v ′1 = 0 = v ′2 . That is when both particles in the final state
are at rest (carrying no kinetic energy). In this frame the condition for an elastic collision
∆K.E. = 0 is simply that the final particles have the same speeds as the initial particles.
The Lab frame, in which particle 2 is at rest, moves relative to the CM frame with velocity
V = v 2 CM = −(m1 /m2 )v 1 CM . In this Lab frame v 1 Lab = v 1 CM − V = (1 + m1 /m2 )v 1 CM ,
v 2 Lab = 0, v 1 ′Lab = v 1 ′CM − V and v 2 ′Lab = −(m1 /m2 )v 1 ′CM − V . The condition for maximal
inelasticity, v 1 ′CM = 0, then becomes the statement that the particles in the final state both
travel at the same velocity, −V = (m1 /m2 )v 1CM = m1 v 1Lab /(m1 + m2 ).
4.2
Rockets
Rockets are propelled by expelling momentum in the form of its fuel exhaust, giving itself
momentum equal to the opposite of the exhaust momentum. For simplicity consider only
motion in a straight line. Let the particles in the exhaust have a constant velocity vex relative
to the rocket. At the moment when rocket plus fuel have mass m and velocity v, suppose
20
c
2013
by Charles Thorn
a mass −dm is expelled. Its momentum relative to an observer at rest is dm(vex − v). The
change in the system momentum
∆Psys = (m + dm)(v + dv) − dm(v − vex ) − mv = mdv + vex dm
(91)
Newton’s second law gives ∆Psys = mdv + vex dm = Fext dt, but in the absence of external
forces we have simply (assuming vex is constant
dv
vex
=− ,
dm
m
v(t) − v0 = −vex ln
m(t)
m0
(92)
The smallest the final mass can be is the mass with all fuel spent. The log dependence shows
that a huge fraction of the initial mass must be fuel an discardable fuel tanks (stages) to get
a substantial velocity change.
4.3
The Center of Mass Coordinate
Let’s look again at the statement of momentum conservation for an N particle system:
Ṗ =
d2 X
d X
mk r k = 0,
mk v k = 2
dt k
dt k
M≡
X
mk
(93)
k
P
This equation says that the center of mass coordinate defined by R ≡ (1/M ) k mk r k
has zero acceleration R̈ = 0. It therefore has time dependence R(t) = R0 + V t, where
V = P /M . If there is an external force on the system so Ṗ = F ext , Then R satisfies
Newton’s second law M R̈ = F ext .
For a two particle system the point R lies on a straight line joining the particles and is
closest to the heavier of the two particles. For a continuous body the center of mass is a
volume integral
Z
1
R =
dV rρ(r)
(94)
M
R
where ρ is the mass density M = dV ρ. Notice that if a large system is broken into two
subsystems we have
R =
1 X
1
1 X
(M1 R1 + M2 R2 )
mk r k +
mk r k =
M k∈1
M k∈2
M
(95)
P
P
where M1 = k∈1 mk , M2 = k∈2 mk are the total masses of the subsystems. For example
the center of mass of a uniform sphere of radius a and mass m1 with a point mass m2 glued
somewhere on its surface is the point on the line joining the point mass to the center of the
sphere at radius m2 R/(m1 + m2 ).
21
c
2013
by Charles Thorn
4.4
Angular Momentum
For a single particle angular momentum about the point with position vector R is defined
by
L = (r − R) × p → r × p
(96)
L = mr2 ϕ̇ẑ
(99)
when it is about the origin of coordinates. We can use the 2nd law to find its time derivative
dL
= ṙ × p + (r − R) × F = (r − R) × F ≡ N
(97)
dt
where N is called the torque about R. The first term vanished because p is parallel to ṙ.
We see that angular momentum about R is conserved if the torque about R vanishes. For
simplicity we assume that R = 0, namely that we compute angular momentum and torque
about the origin of coordinates.
An important situation in which the torque vanishes is if the vector F (r) is parallel to
r. Such a force is called a central force. Examples include the gravitational and Coulomb
forces exerted by a mass or charge fixed to the origin of coordinates.
Conservation of angular momentum provides an elegant explanation of Kepler’s 2nd law
of planetary motion, which states that a single planet revolving about the sun sweeps out
equal areas in equal times. first notice that if L is a constant vector, both r and p, which
at all times are perpendicular to the constant direction of L, lie in a fixed plane, and so the
entire trajectory stays in that plane, which we take to be the xy-plane.
The area dA swept out in an infinitesimal time dt is simply the area of the triangle
determined by r and vdt. From our work with vectors, we have found that the area of a
parallelogram spanned by any two vectors is |v 1 × v 2 |, so the area of the triangle is just half
of this:
1
|L|
dA =
|r × v|dt =
dt
(98)
2
2m
Hence Ȧ = |L|/2m. Clearly, if L is conserved Ȧ is a constant, and Kepler’s second law
follows. Notice that Kepler’s law is true for any central force, not just inverse square ones.
In polar coordinates, we found that ṙ = ṙr̂ + rϕ̇ϕ̂ so that
When L is conserved, the angular velocity of the planet varies as 1/r2 .
4.5
Systems of several particles
Let us consider a system of N particles and define the angular momentum about the origin
of coordinates as simply the sum of the angular momenta of the individual particles:
L =
dL
=
dt
N
X
k=1
N
X
k=1
r k × pk
(100)
r k × F k = N total
(101)
22
c
2013
by Charles Thorn
When we considered the time derivative of the total momentum, the 3rd law guaranteed that
the internal forces canceled out so we had Ṗ = F ext , so that total momentum is conserved
in the absence of external forces. The internal force contributions to the total torque do not
automatically cancel out, but they do simplify:
X
X
X
X
rk ×
F kl =
r k × F kl +
r k × F kl
k
l6=k
k<l
k>l
X
X
=
(r k × F kl + r l × F lk ) =
(r k − r l ) × F kl
k<l
(102)
k<l
If all of the internal forces are central forces, which means F kl is parallel to (r k − r l ) for
each pair k, l then the internal force contributions to the total torque do cancel.
X
dL
=
r k × F k ext = N ext ,
Central Internal Forces
(103)
dt
k
Then in the absence of external forces the total angular momentum of the N particle system
will be conserved. [Discuss Putty-turntable example]
The angular momentum of a system of particles about its center of mass has special
interest. We might call this the “spin” of the system: think of a spinning top. When
external forces are present, the center of mass coordinate R(t) will be accelerating according
to M R̈ = F ext , so the spin is not simply the angular momentum in another inertial system.
With r k the coordinates relative to the origin of an inertial system, we define the coordinates
relative to the center of mass as r ′k ≡ r k − R(t), and the momenta relative to the center of
mass are p′k = mk ṙ ′k . Now write out L in terms of the primed variables:
X
L =
(r ′k + R(t)) × (p′k + mk Ṙ(t)
k
=
X
k
=
X
k
r ′k × p′k + R(t) ×
X
p′k +
X
k
k
mk R(t) × Ṙ(t) +
r ′k × p′k + M R(t) × Ṙ ≡ S + R × P
X
k
mk r ′k × Ṙ
(104)
P
P
P
where M = k mk is the total mass, and k mk r ′k = k mk r k − M R = 0 by the definition
P ′
of center of mass. Then also
k pk = P − M Ṙ = 0. This result shows that we can
always decompose the total angular momentum of a system into its “spin” plus the angular
momentum of the system as a whole R × P .
If all the internal forces are central and obey the 3rd law, we then have
dL
dS
N ext =
=
+ R × F ext
dt
dt
X
dS
= N ext − R × F ext =
(r k − R) × F k ext
(105)
dt
k
In other words the angular momentum of a system of particles about its center of mass has
its time derivative equal to the torque about the center of mass, even if the center of mass
is accelerating. [Discuss dumbbell motion after sharp impulse]
23
c
2013
by Charles Thorn
5
Energy
The last great conservation law is that of energy. In mechanics we start with a precise
notion of Kinetic energy (KE for short). the kinetic energy of a particle of mass m is simply
T = mv 2 /2 = p2 /2m. We can use Newton’s 2nd law to calculate its time derivative
dT
dt
= v·F
(106)
In general F can depend on position r, velocity ṙ and time t. When it depends only on
position, we can formulate the concept of work and the work energy theorem. Integrate both
sides of the last equation w.r.t. time:
∆T = T (t) − T (0) =
t
Z
0
dr
dt′ ′
dt
′
· F (r(t ) =
Z
r(t)
r(0)
dr ′ · F (r ′ )
(107)
The integral on the right is an example of a line integral and can be defined for any path
connecting the two points r 1 = r(0) and r 2 = r(t). To see this describe the chosen path
parametrically r(λ) = (x(λ, y(λ), z(λ), with r(0) = r 1 and r(1) = r 2 . Then dr/dλ is a
vector tangent to the curve, and the line integral is defined as the ordinary integral
Z r2
Z 1
dr
′
′
W (r 1 , r 2 ; P ) =
· F (r(λ))
(108)
dλ
dr · F (r ) ≡
dλ
P,r1
0
(In mechanics we can think of the parameter λ as time and the path as some trajectory
of the particle.) It is called the work W (r 1 , r 2 ; P ) done by the force on the particle as it
moves from r 1 to r 2 along the path P . The previous equation shows that this work gives
the change in kinetic energy of the particle as it follows this motion.
The dependence of the work on the path used to define it limits its utility. Random
choices for F are very likely to give path dependent work functions. Suppose for example
F = (ay, bx, 0), and we calculate the work from (0, 0, 0) to (R, R, 0) in two ways.
W1 =
Z
R
dxFx (x, 0, 0) +
0
W2 =
Z
0
R
dyFy (0, y, 0) +
Z
Z
R
dyFy (R, y, 0) = 0 + bR2 = bR2
0
R
0
dxFx (x, R, 0) = 0 + aR2 = aR2 6= W1
(109)
Of course for the special case b = a the two paths give the same answer!
5.1
Conservative forces and potential energy
However, for the important class of forces, called conservative forces the work turns out to
be independent of the path. In that case, fixing r 0 as some reference point, we can define a
function of all of space r by U (r) ≡ −W (r 0 , r). The reference point is the point at which
24
c
2013
by Charles Thorn
U (r 0 ) = 0. If W depended on the path, there would be no way to consistently define such
a single valued function. When it can be defined this function is the potential energy.
When the force is conservative, the work done from r 1 to r 2 can be written
Z r2
Z r0
Z r2
W (r 1 , r 2 ) =
dr · F =
dr · F +
dr · F
r1
r1
r0
= U (r 1 ) − U (r 2 ) = −∆U
(110)
Combining this with the work energy theorem then gives energy conservation ∆T + ∆U = 0.
Now let’s consider more closely the relation between U (r) and F (r). By definition
Z r
U (r) = U (x, y, z) = −
(dxFx + dyFy + dzFz )
r1
∂U
∂x
∂U
= −Fy ,
∂y
= −Fx ,
∂U
= −Fz
∂z
(111)
That is, the components of the vector F are the corresponding derivatives of U : Fi =
−∂U/∂xi . Since there are three coordinates there are three derivatives that can be taken,
each holding the other 2 constant. This is what is meant by the partial derivatives ∂/∂xi .
These partial derivatives can be denoted by the vector gradient ∇:
∂
∂
∂
+ ŷ
+ ẑ
∂x
∂y
∂z
F = −∇U
∇ = x̂
(112)
(113)
Forces which can be derived from a potential energy this way are precisely the conservative
forces.
It is a fundamental property of partial derivatives that the order in which two partials
are applied doesn’t matter:
∂2f
∂ ∂f
∂ ∂f
∂2f
≡
(114)
=
=
∂x∂y
∂x ∂y
∂y ∂x
∂y∂x
and this is true for each pair, x, y, y, z, z, x. This means that
∂2U
∂2U
∂Fj
∂Fi
=−
= −
=
∂xj
∂xi ∂xj
∂xj ∂xi
∂xi
∇i Fj − ∇j Fi = 0
(115)
We can identify the combinations ∇i Fj −∇j Fi as the various components of the cross product
∇ × F , which is called the curl of F . Thus conservative forces have zero curl: ∇ × F = 0.
5.2
Stoke’s Theorem
Another way to say that the work function is independent of the path is to say that the work
done along any closed path is zero:
I
dr · F (r) = 0,
All closed paths P
(116)
P
25
c
2013
by Charles Thorn
We can relate the line integral on the right to a surface integral of the curl of F . To do this
let’s parameterize a surface by the three function rk (λ1 , λ2 ). Where the parameters λ1 , λ2
range over the unit square 0 ≤ λ1 , λ2 ≤ 1. Then consider the integral
i
i j
Z 1
Z 1
∂r ∂F i
∂r ∂r
∂ri ∂rj ∂F i
∂ri ∂F i
−
=
dλ1 dλ2
−
dλ1 dλ2
∂λ1 ∂λ2 ∂λ2 ∂λ1 ∂rj
∂λ1 ∂λ2
∂λ2 ∂λ1
0
0
Z 1
∂r
∂r
∂
∂
=
·F −
·F
dλ1 dλ2
∂λ2 ∂λ1
∂λ1 ∂λ2
0
λ2 =1 Z 1
λ1 =1
Z 1
I
∂r
∂r
·F ·F =
dλ1
−
dλ2
= − dr · F (117)
∂λ1
∂λ2
0
0
P
λ2 =0
λ1 =0
where P is the closed loop described by r(λ1 , 0), 0 < λ1 < 1; r(1, λ2 ), 0 < λ1 < 1; r(λ1 , 1),
1 > λ1 > 0; r(0, λ2 ), 1 > λ2 > 0. Clearly this closed loop is the boundary of the surface
parameterized by r(λ1 , λ2 ). this is known as Stoke’s theorem. Our argument actually applies
in any number of spatial dimensions. In 3 dimensions we can write
k
i j
m
l
∂ri ∂rj ∂F i
∂r
∂r ∂r
∂r
i ∂r ∂r
(118)
−
= ǫijk ǫklm ∇j F
= −(∇ × F ) ·
×
∂λ1 ∂λ2 ∂λ2 ∂λ1 ∂rj
∂λ1 ∂λ2
∂λ1 ∂λ2
∂r
∂r
The infinitesimal vector dA = ∂λ
× ∂λ
dλ1 dλ2 has magnitude equal to the element of area of
1
2
∂r
∂r
the parallelogram spanned by the two vectors ∂λ
dλ1 and ∂λ
dλ2 , and direction perpendicular
1
2
to the surface with sign given by the right hand rule. So in 3 dimensions Stoke’s theorem
can be expressed as
Z
I
dA · ∇ × F =
dr · F
(119)
S
∂S
where ∂S is a standard notation for the boundary of the surface S. The sign is consistent
with the right hand rule: if the fingers of the right hand are pointed in the sense of the
contour on the right, the thumb points in the direction of dA.
With the aid of Stoke’s theorem, we see immediately that the work integral is independent
of the path if and only if ∇ × F = 0 throughout all of space. Equivalently any force with
zero curl can be expressed as minus the gradient of a potential.
5.3
Potential energy examples
• Central forces: U (r): F = −∇U = −U ′ (r) rr ≡ f (r)r. Checking the curl:
f ′ (r)
r × r + f (r)∇ × r = 0
(120)
r
We see from the above the relation between U and F for central forces. We can also
do the work integral explicitly
Z r
Z
Z r
1 r
′
′
′
′ 2
′
U (r) = −
dr · r f (r ) = −
d(r ) f (r ) = −
r′ f (r′ )dr′
(121)
2 r0
r0
r0
∇ × (f (r)r) =
from which we see that U depends only on r = |r| and that U ′ = −rf (r).
26
c
2013
by Charles Thorn
• Uniform force (F = constant). U = −r · F .
−∇i (−rj F j ) = δij F j = F i
Z r
U (r) = −
dr ′ · F = −F · (r − r 0 )
(122)
(123)
r0
• Several distinct forces. EachPconservative force can be expressed F k = −∇Uk . then the
total potential energy U = k Uk makes sense. This can’t be done for nonconservative
forces F nc . Then energy conservation is replaced by ∆(T + U ) = Wnc . Example:
Inclined plane with friction.
5.4
One dimensional systems
With one dimensional systems, one only needs to require that the force is a function only of
x, F (x), not of t nor of v. Then the path independence of the work function is automatic,
and one simply defines the potential energy by
Z x
dU
U (x) = −
dx′ F (x′ ),
F (x) = −
(124)
dx
x0
The force is zero whenever the curve U (x) has zero slope, i.e. at relative maxima or minima.
At those points the particle can be in static equilibrium. At a maximum the equilibrium is
unstable. A particle in motion has energy larger than U (x).
Energy conservation allows us to express the solution of a general one dimensional conservative force problem in terms of an explicit integral. We start with the statement of energy
conservation:
1 2
mv + U (x) = E
2
(125)
with E the total (kinetic + potential) energy, which is a constant. We then solve for v as a
function of x:
r
r Z x
Z t
2(E − U (x))
m
dx′
dx
′
p
dt =
v=
= ±
,
±
(126)
dt
m
2 x0 E − U (x′ )
0
The sign is fixed by initial conditions and will change at turning points. Let’s choose the +
sign from now on:
r Z x(t)
m
dx′
p
(127)
t =
2 x0
E − U (x′ )
Here we get t(x) as an explicit function of x, rather than x as an explicit function of t. But
it nevertheless tells us everything about the motion! To go further, we have to specialize to
special potentials.
27
c
2013
by Charles Thorn
Let’s do the integral for a uniform force, such as gravity near the earth’s surface. Then
F = −mg or U (z) = mgz and
r h
r Z z(t)
i
p
2
dz ′
m
m p
√
(128)
=
∓
t = ±
E
−
mgz(t)
−
E
−
mgz
0
2 z0
mg 2
E − mgz ′
If the particle starts out at rest, E = 0 + mgz0 = mgz0 , so the second term is zero we should
choose the − sign because the subsequent velocity will be negative. Then
r
1
2p
t=
(129)
z0 − z(t), z(t) = z0 − gt2
g
2
a result we are of course very familiar with.
Systems can be effectively one-dimensional without literally involving motion along a
straight line. This can happen if three dimensional motion is suitably constrained–think
of a roller coaster. Or consider a simple pendulum consisting of a massless rod of length L
connected to a bob mass m. The motion is constrained to a circle of radius L. In the absence
of friction energy is conserved. The position of the bob is given in terms of the angle θ the
rod makes with the vertical: v = Lθ̇ and z = L(1 − cos θ)
E=
m 2
mL2 2
mL2 2
θ
v + mgz =
θ̇ + mgL(1 − cos θ) =
θ̇ + 2mgL sin2
2
2
2
2
(130)
Note that we can quickly recover the e.o.m. by setting Ė = 0: mL2 θ̈ + mgL sin θ = 0. The
we apply the 1-d formula
r
Z
mL2 θ
dθ′
p
t=
(131)
2
E − mgL(1 − cos θ′ )
θ0
Taking θ0 = 0 and noting that the maximum angle occurs when θ̇ = 0, E = 2mgL sin2 θm
this simplifies to
s Z
L θ
dθ′
p 2
(132)
t=
4g 0
sin (θm /2) − sin2 (θ′ /2)
the period T of the pendulum is 4 times the time it takes θ to rise from 0 to θm :
s Z
s Z
θm
′
du
dθ
L
L 1
√
p 2
√
T =2
=4
2
2
′
g 0
g 0
1 − u 1 − k 2 u2
sin (θm /2) − sin (θ /2)
(133)
where k = sin(θm /2).
Other examples: (1) Cube balanced on a cylinder; (2) Atwood machine. Constraints
make the motion effectively one dimensional. The key ingredient is that the constraints do
no work: normal constraining forces are perpendicular to the motion, and friction is either
absent or does no work as with rolling without slipping.
28
c
2013
by Charles Thorn
5.5
Central forces
Central conservative forces are derived from a potential U (r) which depends only on the
distance from the center of force, which we choose to be the origin of coordinates. The
conserved energy is then
E =
m 2
v + U (r)
2
(134)
With motion in three dimensions energy conservation is less powerful than in one dimension,
because the direction of v is not constrained by it. It is convenient to use spherical polar
coordinates to describe the motion in central potentials. These are
r = r(sin θ cos ϕ, sin θ sin ϕ, cos θ)
(135)
Then the velocity is fairly complicated:
v = ṙr̂ + r sin θϕ̇(− sin ϕ, cos ϕ, 0) + rθ̇(cos θ cos ϕ, cos θ sin ϕ, − sin θ)
= ṙr̂ + r sin θϕ̇ϕ̂ + rθ̇θ̂
2
v = ṙ2 + r2 sin2 θϕ̇2 + r2 θ̇2
(136)
(137)
Here we have introduced the mutually orthogonal unit vectors r̂, θ̂, and ϕ̂. It is easy to
check that r̂ × θ̂ = ϕ̂, r̂ × ϕ̂ = −θ̂,and θ̂ × ϕ̂ = r̂.
Incidentally, the gradient operator in spherical coordinates is complicated by the fact
that the unit vectors are not constant: We can infer from
∂
∂
∂
+ dθ
+ dϕ
∂r
∂θ
∂ϕ
(138)
1
1 ∂
∂
∂
+ ϕ̂
+ θ̂
∂r
r sin θ ∂ϕ
r ∂θ
(139)
dr · ∇ = dr
and dr = drr̂ + r sin θdϕϕ̂ + rdθθ̂ that
∇ = r̂
which makes it obvious that −∇U can be a central force only if U is independent of angles!
Returning to energy conservation we have in terms of spherical coordinates
E =
m 2
(ṙ + r2 sin2 θϕ̇2 + r2 θ̇2 ) + U (r)
2
(140)
The angular velocities are not constrained by energy conservation, but remember that we
also have angular momentum conservation!
L = mr × v = mr × (r sin θϕ̇ϕ̂ + rθ̇θ̂) = mr2 (− sin θϕ̇θ̂ + θ̇ϕ̂)
L2 = m2 r4 (sin2 θϕ̇2 + θ̇2 )
29
(141)
(142)
c
2013
by Charles Thorn
We notice that the angular velocities in E occur exactly in the same combination as they do
in L2 ≡ L2 . Since L2 is a constant, we can safely eliminate the angular velocities in E:
L2
m
m 2
ṙ +
+ U (r) ≡ ṙ2 + UL (r)
2
2
2mr
2
L2
UL (r) = U (r) +
2mr2
(143)
E =
(144)
The radial coordinate can now be treated as in one dimensional motion. In particular we
can solve for t as a function of r as an explicit integral. By studying the graph of UL (r) we
can also map out all of the qualitative motions.
Since the direction of L is a constant, it is convenient to choose our polar axis (z-axis)
parallel to L, which means that θ = π/2 and L = mr2 ϕ̇ẑ. The particle moves in the xy
plane (θ = π/2) and its angular velocity is related to L and r:
L
mr2
r
2(E − UL (r))
ṙ =
,
m
ϕ̇ =
(145)
dϕ
L
= 2
dr
r
s
1
2m(E − UL (r))
We can integrate the last equation to get ϕ(r):
Z 1/r0 s
Z r ′s
dr
1
1
=L
du
ϕ = L
′2
′
2m(E − UL (r ))
2m(E − UL (1/u))
1/r
r0 r
(146)
(147)
For the −C/r potential UL (1/u) = L2 u2 /2m − Cu and the integral is an elementary one
involving an inverse trig function. The turning points are given by
p
C ± C 2 + 2EL2 /m
1±e
2 2
0 = L u /2m − Cu − E,
u± =
≡
(148)
2
L /m
p
r
L2
2EL2
p =
,
e= 1+
(149)
mC
mC 2
For an actual turning point r = 1/u must be positive. We see that u± are both positive
when C > 0 (attraction) and E < 0. In this case r oscillates between a minimum rp = 1/u+
(perihelion) and a maximum (aphelion) ra = 1/u− . When C > 0 and E > 0, only u+ is
positive, and the particle motion is unbounded. When C < 0 (repulsion), then necessarily
E > 0 and again only u+ is positive and the motion is unbounded. For gravity C = Gm1 m2 >
0. For the Coulomb potential C = −q1 q2 /4πǫ0 is negative for like sign charges and positive
for opposite sign charges.
30
c
2013
by Charles Thorn
To do the integral, we write
2 !
L 2 u2
L2
L2 e2
1
E + Cu −
(150)
= −
(u − u+ )(u − u− ) =
− u−
2m
2m
2m p2
p
Z u+ s
1
du
ϕ =
e2 /p2 − [u − 1/p]2
1/r
pu+ 1
1
1
π
p
p
= arcsin
−
−
−
− arcsin
= − arcsin
e
e
re e
2
re e
p
1
p
cos ϕ =
− ,
r=
(151)
re e
1 + e cos ϕ
When the eccentricity e < 1 (so E < 0), the radial coordinate r varies between a minimum
p/(1 + e) and a maximum p/(1 − e). The orbit is then an ellipse with semi-major axis
a = p/(1 − e2 ). Clearly when e = 0 the radius is constant so the orbit is a circle. When e > 1
(so E > 0) there is a minimum p/(1 + e) but no maximum: r → ∞ when cos ϕ → −1/e.
To find the time dependence of the motion we need
r r
L2
1 √
C
2
=
ṙ =
E+ −
2mEr2 + 2mCr − L2
m
r
2mr2
mr
√
−2mE p
(rmax − r)(r − rmin )
(152)
=
mr
When the orbit is bounded (E < 0), it has a period given by
Z rmax
Z rmax
rdr
m
dr
p
= 2√
T = 2
ṙ
−2mE rmin
(rmax − r)(r − rmin )
rmin
Z 1
m
(rmax − rmin )xdx + rmin dx
p
= 2√
−2mE 0
x(1 − x)
Z 1
dx
mπ(rmax + rmin )
m(rmax + rmin )
√
p
√
=
=
−2mE
−2mE
x(1 − x)
0
(153)
Now
C
L2 mC 2
=
mC (−2EL2 )
−E
4π 2 m
T2
=
a3
C
2a ≡ rmax + rmin = 2p/(1 − e2 ) = 2
which is Kepler’s third law.
2πa3/2
T = p
,
C/m
31
(154)
c
2013
by Charles Thorn
5.6
Energy of systems of particles
Now we would like to extend the concepts of kinetic energy and potential energy to more
than one particle. The total kinetic energy of several particles is simply the sum of the
kinetic energies of each particle:
T =
N
X
1
k=1
2
mk ṙ 2k
(155)
The definition of potential energy is less obvious. If the particles do not interact with each
other but each particle k moves independently in an external force field F k (r) then the force
on particle k is just F k (r k ). If the force is conservative F k (r) = −∇Uk (r) and the force on
particle k is
F (r k ) = −∇k Uk (r k ) = −∇k
N
X
l=1
Ul (r l ) ≡ −∇k UT ,
UT =
N
X
Ul (r l )
(156)
l=1
where we understand that ∇k involves only derivatives w.r.t. the components of r k , holding
constant the components of all r l with l 6= k. Thus all terms in the sum with l 6= k contribute
nothing to the force on particle k. With these definitions of total kinetic and potential energy,
we have energy conservation:
E =
dE
=
dt
=
N X
1
k=1
N
X
k=1
N
X
k=1
2
mk ṙ 2k
+ Uk (r k )
(157)
(mk r̈ k · ṙ k + ṙ k · ∇k Uk (r k )) =
N
X
k=1
ṙ k · (mk r̈ k + ∇k Uk (r k ))
ṙ k · (mk r̈ k − F k (r k )) = 0
(158)
by Newton’s second law for each particle.
Next we want to include interactions between particles. Consider first two particles with
position vectors r 1 , r 2 . By Newton’s third law the mutual forces are equal and opposite
F 12 = −F 21 . We assume that the mutual forces are translationally invariant. This means
that they are the same when the particles are at positions r 1 + a, r 2 + a as they are at
positions r 1 , r 2 . This means that the mutual force depends only on the difference r 1 − r 2 .
F 12 = F 12 (r 1 − r 2 ) = −F 21 = −F 12 (r 2 − r 1 )
(159)
The mutual gravitational and Coulomb forces are important examples. For example Newton’s law of mutual gravity can be written
F grav
= −Gm1 m2
12
32
r1 − r2
= −F 21
|r 1 − r 2 |3
(160)
c
2013
by Charles Thorn
The crucial feature here is that the third law is included in the statement by the oddness of
F grav as a function of its argument.
If the mutual two-body force is conservative F 12 (r) = −∇U12 (r) where r = r 1 − r 2 . The
oddness of F 12 implies that U12 (r) = U12 (−r). Interestingly, we can write F 12 in three ways
F 12 = −∇U12 = −∇1 U12 = +∇2 U12 = −F 21
(161)
Thus we have the intuitive results F 12 = −∇1 U12 and F 21 = −∇2 U12 . It is important to
appreciate that the single potential energy U12 accounts for both the force of particle 2 on
particle 1 and the force of particle 1 on particle 2. For this reason the total potential energy
describing external forces F k and the mutual forces F 12 = −F 21 is written
UT = U1 (r 1 ) + U2 (r 2 ) + U12 (r 1 − r 2 )
dE
= ṙ 1 · (m1 r̈ 1 + ∇1 U1 ) + ṙ 2 · (m2 r̈ 2 + ∇2 U2 ) + (ṙ1 − ṙ2 ) · ∇1 U12
dt
= ṙ 1 · (m1 r̈ 1 + ∇1 U1 + ∇1 U12 ) + ṙ 2 · (m2 r̈ 2 + ∇2 U2 + ∇2 U12 )
= ṙ 1 · (m1 r̈ 1 − F 1 − F 12 ) + ṙ 2 · (m2 r̈ 2 − F 2 − F 21 ) = 0
(162)
(163)
by Newton’s second law applied to each particle.
The iconic application of two body energy conservation is elastic collisions. It is based
on the fact that physical potential energies vanish at large separation. Thus the total energy
E = T + UT → T when r 1 , r 2 , r 1 − r 2 all become large. In a scattering process, the two
particles are initially very far from each other and from the center of external forces, so
initially E ≈ Ti . After scattering the two particles are again very far from each other and
from the center of external forces, so E ≈ Tf . But E is constant through out the motion,
so we have Tf = Ti for an elastic two body scattering process. If the external forces are all
zero, then Newton’s third law implies that total momentum is also conserved. In this latter
case the initial and final variables are subject to the constraints
p1 + p2 = p′1 + p′2
1 2
1 2
1 ′2
1 ′2
p1 +
p2 =
p1 +
p
m1
m2
m1
m2 2
(164)
(165)
The most efficient way to impose these constraints is to work in the center of mass system
for which p1 + p2 = 0. Then momentum conservation simple states that p′2 = −p′1 . Plugging
2
these relations into the energy conservation equation then shows that p′2
1 = p1 . That is p1
and p′1 have the same length. Thus p′1 ·p1 = p21 cos θ, and the scattering process is completely
determined by the scattering angle θ. If the two particles have the same mass m1 = m2 ,
these statements apply to the velocities and speeds of all the particles. Then the lab frame,
in which v 2 = 0, moves with velocity −v 1 relative to the center of mass. Then the final state
particles have velocities v ′1 + v 1 and −v ′1 + v 1 and their scalar product is
(v ′1 + v 1 ) · (−v ′1 + v 1 ) = v 1 2 − v 1 ′2 = 0
(166)
So in the equal mass case the final state particles make tracks perpendicular to each other.
33
c
2013
by Charles Thorn
Finally, we want to consider a general N particle system. We shall assume that the
mutual forces are all two body forces. This means that the force on each particle k is a sum
of forces from each of the other particles plus an external force
X
F k = F k (r k ) +
F kl (r k − r l )
(167)
l6=k
where we also assume that the mutual forces are translationally invariant. One could imagine
a more complicated situation with multi-body forces, but this will not be necessary. In this
notation Newton’s third law is the statement that F lk (r l − r k ) = −F kl (r k − r l ). Newton’s
law of universal gravity
F grav
kl (r k − r l ) = −Gmk ml
rk − rl
|r k − r l |3
(168)
clearly satisfies these requirements. The meaning of conservative forces immediately extends
from our two particle considerations. Namely,
F kl = −∇k Ukl (r k − r l ) = +∇l Ukl (r k − r l )
(169)
where the third law requires that Ulk (r l − r k ) = Ukl (r k − r l ). Just as in the two particle
case, the total potential energy will contain only one contribution for each pair:
X
X
Utot =
Uk (r k ) +
Ukl (r k − r l )
(170)
k
k<l
The justification for this choice is that it is the one that guarantees energy conservation:
X
X
dE
=
ṙ k · [mk r̈ k + ∇k Uk (r k )] +
(ṙ k − ṙ l ) · ∇k Ukl (r k − r l )
dt
k
k<l
X
X
X
=
ṙ k · [mk r̈ k + ∇k Uk (r k )] +
ṙ k · ∇k Ukl (r k − r l ) −
ṙ l · ∇k Ukl (r k − r l )
=
k
k<l
k<l
X
X
X
k
=
X
k
=
X
k
=
X
k
=
X
k
ṙ k · [mk r̈ k + ∇k Uk (r k )] +
ṙ k · [mk r̈ k + ∇k Uk (r k )] +
ṙ k · [mk r̈ k + ∇k Uk (r k )] +
k<l
X
k<l
X
l6=k
ṙ k · [mk r̈ k + ∇k Uk (r k ) + ∇k
ṙ k · ∇k Ukl (r k − r l ) +
ṙ k · ∇k Ukl (r k − r l ) +
ṙ k · ∇k Ukl (r k − r l )
X
l6=k
ṙ k · [mk r̈ k − F k ] = 0
k<l
X
l<k
ṙ l · ∇l Ukl (r k − r l )
ṙ k · ∇k Ukl (r k − r l )
Ukl (r k − r l )]
(171)
by Newton’s second law for each particle.
34
c
2013
by Charles Thorn
It is important to appreciate that energy conservation will follow if the potential energy is
an arbitrary function of the coordinates of the system, as long as it has no explicit dependence
on time. This is just a consequence of the chain rule:
dU
dt
=
N
X
k=1
ṙ k · ∇k U (r 1 , r 2 , · · · , r N ) = −
X
dE
=
ṙ k · (mk r̈ k − F k ) = 0
dt
k
N
X
k=1
ṙ k · F k
(173)
by Newton’s second law. Furthermore, we will also have momentum conservation if
X
X
0 =
Fk = −
∇k U (r 1 , r 2 , · · · , r N )
k
(172)
(174)
k
which is true if U (r 1 + a, r 2 + a, · · · , r N + a) = U (r 1 , r 2 , · · · , r N ). In other words, momentum conservation holds if the potential energy is invariant under a uniform translation of
coordinates.
6
Oscillations
Our next subject is to study the physics of harmonic oscillations. Recall Hooke’s law for a
one dimensional oscillator F (x) = −kx. This force can be derived from the potential energy
U (x) = kx2 /2. We will spend a lot of time studying the physics of this system, but first it is
important to appreciate the universality of Hooke’s law. It will come into play whenever one
has a point of stable equilibrium. Equilibrium requires that the force is zero or U ′ (x0 ) = 0.
Stability requires that U ′′ (x0 ) > 0. whatever function U (x) is, we can always do a Taylor
expansion about a point x0 :
1
U (x) = U (x0 ) + (x − x0 )U ′ (x0 ) + (x − x0 )2 U ′′ (x0 ) + O(x − x0 )3
2
k
→ U (x0 ) + (x − x0 )2 + O(x − x0 )3
2
(175)
for an equilibrium point and where k ≡ U ′′ (x0 ) > 0 for stability. As long as the motion stays
sufficiently close to x0 , the dynamics can be well approximated by Hooke’s law! A simple
example that we have already discussed is the motion of a pendulum in small oscillations:
U (θ) = mgL(1 − cos θ) ≈
mgL 2
θ
2
(176)
For small oscillations the energy is then
E ≈
1 2 2 mgL 2
ml θ̇ +
θ
2
2
(177)
From which energy conservation gives Hooke’s law θ̈ = −(g/L)θ.
35
c
2013
by Charles Thorn
6.1
Descriptions of Simple Harmonic Motion
We can put the equation of motion for harmonic oscillations in the generic form
ẍ = −ω 2 x
(178)
p
where
the angular frequency ω depends on the system: e.g. k/m for a spring constant k,
p
g/L for a pendulum.
We already know the general solution is a linear combination of trig functions cos ωt and
sin ωt, but there are several convenient ways to “package the solutions. One popular way is
using complex exponentials e±iωt :
x(t) = Ceiωt + C ∗ e−iωt
(179)
where C is any complex number and C ∗ is the complex conjugate of C, so that x(t) is a real
number. We can easily relate C to the initial conditions:
ẋ(0)
1
∗
∗
x(0) − i
(180)
x(0) = C + C ,
ẋ(0) = iω(C − C ),
C=
2
ω
We can of course use Euler’s formula e±iωt = cos ωt ± i sin ωt to rewrite the solution in terms
of trig functions
ẋ(0)
sin ωt
(181)
ω
However we write the solution, it is clear that the motion is periodic with period T = 2π/ω.
There is one more useful way to describe the general motion in terms of amplitude and
phase:
x(t) = x(0) cos ωt +
x(t) = A cos(ωt − δ),
x(0) = A cos δ,
ẋ(0) = Aω sin δ
p
ẋ(0)
x(0)2 + ẋ(0)2 /ω 2 ,
tan δ =
A =
ωx(0)
(182)
This last representation has a neat connection with the complex exponential representation:
put C = (A/2)eiδ . Then
A i(δ−ωt
e
+ e−i(δ−ωt = ReAeiδ e−iωt = A cos(ωt − δ).
(183)
2
If we understand that we will always take the real part of any complex solution at the end,
there is no harm in working with complex solutions from the beginning. This will be an
extremely useful point of view when we consider damping.
Finally it is useful to relate the various parameterizations of simple harmonic motion to
the conserved energy of the system.
x(t) =
1
1
1
1
1
E = mẋ(t)2 + kx(t)2 = mω 2 A2 sin2 (ωt − δ) + kA2 cos2 (ωt − δ) = kA2
(184)
2
2
2
2
2
So the energy is proportional to the square of the amplitude of motion. The total energy is
conserved as it oscillates between kinetic and potential energy with angular frequency 2ω.
36
c
2013
by Charles Thorn
6.2
2 and 3 dimensional oscillations
We can look for stable equilibrium points in higher dimensional potentials U (r). Equilibrium
requires that ∇U = 0. The multidimensional Taylor expansion reads
1
U (r) = U (r 0 ) + (r − r 0 ) · ∇U (r 0 ) + (ri − r0i )(rj − r0j )∇i ∇j U (r 0 ) + · · ·
2
1
→ U (r 0 ) + kij (ri − r0i )(rj − r0j ) + · · ·
(185)
2
where kij = ∇i ∇j U (r 0 ) is a 3 × 3 spring constant matrix. For now we will assume that this
matrix is diagonal with positive diagonal elements when the equilibrium is stable:


kx 0 0
(186)
k =  0 ky 0 
0 0 kz
Taking the equilibrium point as our origin of coordinates and the zero of potential energy to
be the equilibrium value, we then have
1
U (x, y, z) =
(187)
kx x2 + ky y 2 + kz z 2 + · · ·
2
We see that the total energy will then be a sum of three independently conserved terms
1
1
1
E = (mẋ2 + kx x2 ) + (mẏ 2 + ky y 2 ) + (mż 2 + kz z 2 )
2
2
2
(188)
p
In principle the angular frequency of each coordinate will be different ωx,y,z = kx,y,z /m.
If all spring constants are equal we have the isotropic oscillator, with potential energy
U = (k/2)(x2 + y 2 + z 2 ) = kr2 /2. In that case the force will be central F = −kr and
angular momentum will be conserved. As we know this implies that the motion is in a plane
perpendicular to L, and we can assume this plane is the xy-plane. then the general solution
can be written
x(t) = Ax cos ωt,
y(t) = Ay cos(ωt − δ) = Ay cos δ cos ωt + Ay sin δ sin ωt (189)
(y(t) − x(t)Ay cos δ/Ax )2 = A2y sin2 δ sin2 ωt
(190)
1 =
x2
(y(t) − x(t)Ay cos δ/Ax )2
+
A2x
A2y
(191)
which is the equation for a rotated ellipse. When δ = π/2, the equation reduces to the
standard form x2 /a2 + y 2 /b2 = 1 with a = Ax and b = Ay . If δ = 0, x = (Ax /Ay )y and the
elliptical orbit. collapses to a straight line.
The motion gets quite interesting when the spring constants in different directions are
different. If ωx /ωy is rational the resulting motion will eventually be periodic although x
and y may have to go through many periods for the motion to repeat. For example if
ωx = 2ωy , x executes 2 periods as y executes only one. However if the ratio of frequencies is
irrational the two dimensional motion will never repeat: this non-repeating motion is termed
quasi-periodic.
37
c
2013
by Charles Thorn
6.3
Damped Oscillations
In the real world no system will oscillate forever because of inevitable frictional forces.
Fortunately it is easy to incorporate damping and keep the equations of motion linear. We
simply take the damping force proportional to the velocity F damp = −bẋ ≡ −2mβ ẋ, with
β > 0 so that it impedes motion. The equation of motion for a single damped oscillator is
then
ẍ + 2β ẋ + ω02 x = 0.
(192)
This e.o.m. is a special case of a linear differential equation with constant coefficients.
Let’s take a brief mathematical excursion to consider how any such equation can be
handled. The key ingredient is to find a complete set of solutions, each of which diagonalizes
the derivative operator d/dt. We are of course very familiar with the functions that do this:
The exponential functions ert . Let us denote a general differential operator by the symbol
D:
n
X
d
dk
d2
dn
Dx = a0 + a1 + a2 2 + · · · + an n x =
ak k
(193)
dt
dt
dt
dt
k=0
Then
Dert =
n
X
k=0
ak rk ert = P(r)ert
(194)
In other words the differential equation Dx = 0 is solved by x(t) = Cert provided r is
a root of the polynomial P. We have converted a problem in differential equations to a
problem in algebra! To get the complete solution to the differential equation we need n
independent solutions. Fortunately the fundamental theorem of algebra guarantees that
there are precisely n complex roots of any nth order polynomial. It is of course crucial that
we allow complex roots for this theorem to hold. It is also important that some of the n
roots can be repeated. For example the two roots of the equation (r − 1)2 = 0 are both
equal to 1. When tire are multiple roots, we can get several solutions by multiplying ert by
a polynomial of t. For example if r is a double root, it means that D contains the factors
2
d
−r
(195)
dt
applying these operators to (a + bt)ert we find
2
d
d
rt
− r (a + bt)e =
− r bert = 0
dt
dt
(196)
More generally
d
−r
dt
k
(b0 + b1 t + · · · + bk−1 tk−1 )ert = 0
38
(197)
c
2013
by Charles Thorn
because the polynomial factor can prevent only k − 1 of the factors from giving zero.
With this background we return to our damped oscillator e.o.m. for which the polynomial
is quadratic
q
2
2
r + 2rβ + ω0 = 0,
r± = −β ± β 2 − ω02
(198)
and the general solution of the equation is
√ 2 2
√ 2 2
x(t) = e−βt C1 et β −ω0 + C2 e−t β −ω0
(199)
The qualitative nature of these solutions depends dramatically
on the
p
prelationship between
β and ω0 . If β < ω0 (underdamping), we can write β 2 − ω02 = −i ω02 − β 2 ≡ −iω1 and
the solution reads
x(t) = e−βt C1 e−itω1 + C1∗ eitω1
(200)
and we see that we have oscillations with frequency ω1 < ω0 and amplitudes that decay
exponentially in time. Note that in this case we choose C2 = C1∗ to make the solution real.
In contrast when β > ω0 (overdamping) the square root is real and both terms are
exponentially damped at different rates
x(t) = C1 e−β1 t + C2 e−β2 t
q
β1 = β − β 2 − ω02 ,
(201)
q
β2 = β + β 2 − ω02
(202)
with the first term dying off least slowly.
For β = ω0 (critical damping), the two roots are the same (the polynomial has a double
root) so the two independent solutions are e−βt and te−βt . Notice that in overdamped
and critically damped cases, damped oscillation is a misnomer since not even one cycle
of oscillation is completed. There can be at most one time when the amplitude vanishes,
and that is possible only when the two solutions occur with opposite signs in the linear
superposition.
6.4
Driving oscillations
Given the reality of damping forces, in order to keep the system oscillating, a driving force
must be applied. For example, a pendulum clock receives a “kick” with each tick, powered
by gravity (via weights) or perhaps by a spring, or perhaps by a battery. Of course the power
source must be regularly renewed.
Sinusoidal driving forces are the easiest to solve. We can put F = mf0 cos ωt = mRe f0 e−iωt .
It is in fact most efficient to use the time dependence e−iωt , taking the real part at the end
of the calculation. Then the e.o.m. to solve is simply
ẍ + 2β ẋ + ω02 x ≡ Dx = f0 e−iωt
39
(203)
c
2013
by Charles Thorn
This is what we call an inhomogeneous linear differential equation. This is because the
driving term does not involve x. The equation with f0 = 0 is called a homogeneous differential
equation. Suppose we have found some particular solution xp (t) to the inhomogeneous
equation. This solution usually not obey the desired initial conditions. But we can add to
xp any solution xh to the homogeneous equation Dxh = 0 and get another general solution:
D(xp + xh ) = Dxp + Dxh = Dxp + 0 = f0 e−iωt .
(204)
With the two free parameters in xh we have the flexibility to impose any initial conditions
on x(0) and ẋ(0).
So the problem is to find a particular solution. Recall that the differential operator D
applied to an exponential ert just multiplies the exponential by a polynomial in r:
DAert = P(r)Aert
(205)
Thus, by choosing r = −iω, the time dependence of both sides of the equation matches and
can be cancelled out leaving the equation
P(−iω)A = f0 ,
A=
f0
P(−iω)
x(t) = xp (t) + xh (t) = xh (t) +
= xh (t) + Re
ω02
(206)
f0
e−iωt
P(−iω)
f0
e−iωt
− ω 2 − 2iβω
(207)
In order for this construction to work it is important that P(−iω) 6= 0. When there is
nonzero damping β > 0, P is never zero for real ω. At the same time the solution xh is
exponentially damped e−βt , so that after a sufficiently long time βt ≫ 1 the xh term dies
away and the system settles down to a steady state of motion
x(t) ∼ Re
ω02
f0
e−iωt
− ω 2 − 2iβω
(208)
Before discussing this damped driven solution, let us briefly consider what happens in the
absence of damping β = 0. For ω 6= ω0 , the solution is much the same except the xh term no
longer dies out with time. However when ω = ω0 in the undamped case this method breaks
down, and one has to go back to the original diff eq to get xp .
ẍ + ω02 x = f0 e−iω0 t
(209)
The problem is that xp = Ae−iω0 t gives 0 on the left side. However, simply multiplying by t
fixes the problem:
d2
(te−iω0 t ) = −ω02 te−iω0 t − 2iω0 e−iω0 t
dt2
40
(210)
c
2013
by Charles Thorn
so we solve the inhomogeneous equation with xp = Ate−iω0 t and A = if0 /(2ω0 ). This is a
similar phenomenon to our discussion of critical damping. The xh term doesn’t die away,
but the xp term grows relative to the xh term linearly in t. So it is still true that the xp term
will eventually dominate the solution.
We now return to the damped case where this subtlety is avoided. It is interesting that
after the xh term (“transients”) dies away the solution is uniquely determined by the driving
term. Let’s assume that f0 is real and positive. Then we can write
ω02
f0
f0
= p 2
eiδ ≡ Aeiδ
2
− ω − 2iβω
(ω0 − ω 2 )2 + 4β 2 ω 2
tan δ =
2βω
− ω2
ω02
(211)
In summary the driving force F = mf0 cos ωt uniquely produces the steady state solution
(after transients have died away)
xsteady = A cos(ωt − δ)
(212)
We sometimes call xsteady the response to the driving force. Its time dependence is out of
phase with the driving force, lagging by the time δ/ω. Both A and δ depend sensitively on
the relation of ω to ω0 . When ω = ω0 tan δ blows up implying that δ → π/2. For small
damping A can get very large at that frequency.
6.5
Resonance
As we have seen, the driven damped harmonic oscillator is, for all practical purposes, described by its steady state motion after transients have died away. For sinusoidal driving
force, the response is sinusoidal with the same frequency but with a phase shift. The strength
of the response, as measured by the squared amplitude (which is proportional to the power
delivered to the system) is completely fixed:
A2 (ω) =
f02
(ω02 − ω 2 )2 + 4β 2 ω 2
(213)
The power delivered to the system is the rate at which the driving force does work:
P = ẋF = −ωAmf0 sin(ωt − δ) cos ωt = −ωAmf0 (sin ωt cos ωt cos δ − cos2 ωt sin δ
1
1
2βω
mβω 2 f02
hP i =
ωAmf0 sin δ = ωAmf0 p 2
(214)
= 2
2
2
(ω0 − ω 2 )2 + 4β 2 ω 2
(ω0 − ω 2 )2 + 4β 2 ω 2
The actual motion of the oscillator is less interesting that the the way the response depends
on the frequency. In spection of the preceding formula shows that the power delivered to the
oscillator is a maximum when ω = ω0 . It is perhaps not surprising that the biggest response
comes when driven at the natural frequency: we say that the system is in resonance. The
resonance phenomenon is most dramatic for weak damping, β ≪ ω0 . Then the resonance
response gets very large:
A2 (ω0 ) =
f02
f02
≫
4β 2 ω02
ω04
41
(215)
c
2013
by Charles Thorn
The resonance peak stands very tall compared to the non resonant amplitude. (When
ω 2 > 2ω02 , A2 is smaller than f02 /ω04 .)
With weak damping, the resonance peak is not only very high, but it is also very narrow.
A convenient measure of the width of the resonance peak is the value of ω − ω0 where the
response is half of the maximum value. Taking the power delivered as a good measure of
the response, this condition gives
q
2
2
|ω − ω0 | = 2βω,
ω = β + ω02 − β 2 ≈ ω0 + β
(216)
So for weak damping the width of the peak at half maximum is 2β ≪ ω0 . if the name of the
game is to have the sharpest possible peak (say to tune in a desired radio station) then the
ratio Q = ω0 /(2β) (the Quality factor) should be as large as possible, Q ≫ 1. Notice that
Q is a parameter characteristic of the damped oscillator: the properties of the driving force
don’t enter into it. Q is simply proportional to the ratio of the decay time of the unforced
oscillator to its period: high Q means that the oscillations will last many periods before
dying away. high Q is synonymous with a long lifetime.
So far we have studied the strength of the response at resonance. The other feature of
the response is the phase shift δ determining the lag time δ/ω of the response compared to
the driving force. We have noted that tan δ blows up at ω = ω0 implying that δ → π/2
at resonance. This means the response and driving force are 90◦ out of phase: when one
is a maximum the other is zero. When ω < ω0 and β is small δ is also small. It rises
rapidly to π/2 at ω = ω0 . For ω > ω0 , tan δ becomes small and negative, implying that δ is
approximately π. So the behavior of the phase shift is a rapid rise through π/2 from near 0
to π as ω increases past resonance.
6.6
Fourier Series
We have seen that sinusoidal driving forces are particularly easy to handle. Because the
equations are linear we can solve any linear combination of sinusoidal driving forces by
simply taking the corresponding linear combination of solutions:
f (t) =
X
fk e−iωk t ,
x=
k
X
k
ω02
−
fk
e−iωk t
− 2iβωk
ωk2
(217)
For example the periodic kicks given to the pendulum of a clock are by no means sinusoidal.
Let’s find the condition on the ωk for f to be periodic of period τ , f (t + τ ) = f (t).
e−iωk (t+τ ) = e−iωk t ,
e−iωk τ = 1,
ωk =
2nπ
≡ nω
τ
(218)
where n is any integer. We might as well use n as our summation index and so we write
f (t) =
∞
X
an e−inωt ,
x=
n=−∞
42
∞
X
ω2
n=−∞ 0
−
an
2
n ω2 −
2iβnω
e−inωt
(219)
c
2013
by Charles Thorn
This equation displays the Fourier series representation of f (t). The powerful fact is that
such a Fourier series can described practically any periodic function of t. Even better we
can get a formula for each fn as an integral. This is based on the following property:
Z τ
ei(m−n)ωτ − 1
= 0,
for m 6= n
(220)
dtei(m−n)ωt =
i(m − n)ω
0
Clearly when m = n this integral just gives τ . Then we find
Z
τ
imωt
dtf (t)e
=
0
am =
∞
X
n=−∞
Z τ
1
τ
an
Z
τ
ei(m−n)ωt = τ am
0
dtf (t)eimωt
(221)
0
As an important example, we find the coefficients for a square pulse f (t) = f0 for 0 < t <
ǫ < τ and f (t) = 0 for ǫ < t < τ . Then a0 = f0 ǫ/τ , and for m 6= 0,
Z
f0
f0
f0 imωǫ/2
mπǫ
f0 ǫ
dteimωt =
(eimωǫ − 1) =
(eimωǫ − 1) =
e
sin
(222)
am =
τ 0
imωτ
2πim
mπ
τ
Notice that if we pretend m is continuous we have a0 = lim am as m → 0. Then
f (t) =
nπǫ −inω(t−ǫ/2)
ǫf0 X f0
+
sin
e
τ
nπ
τ
n6=0
∞
=
ǫf0 X 2f0
nπǫ
+
sin
cos nω(t − ǫ/2)
τ
nπ
τ
n=1
(223)
The particular solution describing the response to this force is simply the sum of responses
to each term:
X f0
ǫf0
nπǫ
1
xp (t) =
e−inω(t−ǫ/2)
+
sin
2
2
τ ω0 n6=0 nπ
τ ω0 − n2 ω 2 − 2iβnω
∞
X
1
nπǫ
f0
ǫf0
p
sin
+2
cos[nω(t − ǫ/2) − δn ](224)
=
2
2
2
2
τ ω0
nπ
τ
(ω0 − n ω )2 + 4n2 β 2 ω 2
n=1
For weak damping β ≪ ω0 , the nth denominator gets small when ω = ω0 /n or τ = nτ0 .
when this happens we can say that the nth term is in resonance, and we can expect that
single term to be dominant.
6.7
RMS Displacement, Parseval’s Theorem
Sometimes it is good to have a single time independent measure of the response to a driving
force. If we average x over a period we simply pick out the 0th term: hxi = ǫf0 /(τ ω02 ). This
43
c
2013
by Charles Thorn
shows none of the resonance behavior. A better measure is the mean squared displacement:
Z
X
am
an
1 τ
2
hx i =
dte−i(n+m)ωt
2
2 ω 2 − 2iβmω ω 2 − n2 ω 2 − 2iβnω τ
ω
−
m
0
0
m,n 0
=
X
n
X f2
sin2 (nπǫ/τ )
an a−n
0
→
(ω02 − n2 ω 2 )2 + 4n2 β 2 ω 2
n2 π 2 (ω02 − n2 ω 2 )2 + 4n2 β 2 ω 2
n
(225)
for the square pulse example. The RMS displacement is just the square root of this result,
defined to give a measure with the dimensions of length instead of length squared. The
strength of the nth resonance peak to the RMS Displacement is
decreasing like 1/n.
7
p
hx2 in ∼
ǫω0
f0
sin
2βω0 nπ
2
(226)
Variational Principles
Up to now we have formulated classical mechanics in terms of Newton’s three laws. When
we began to understand conservation laws, we learned that there were dramatic shortcuts to
solving the equations of motion. The special case of finding static solutions, we learned that
they were simply the stationary points of the potential energy function U ′ (x) = 0. Thought
about this way, it is clear that a minimum is a minimum, regardless of the coordinate choice.
What we would now like to understand is how general solutions can be determined by a
similar principle, known as Hamilton’s principle of least action. The solutions are required
to be stationary points of the action, defined for dynamics determined by a potential energy
as
Z t2
I =
dt(T − V )
(227)
t1
One can calculate I for any trajectory with initial and final conditions x(t1 ) = x1 and
x(t2 ) = x2 . But the actual solution is precisely the one which minimizes I. We need a new
way of testing for minima w.r.t. small changes in the function x(t). This is the so-called
calculus of variations.
7.1
Examples
Consider the question: what path gives the shortest distance between two points? To answer
this question we have to compare the distances given by different paths. We first need a
formula for the distance on a fixed path P : To describe a path it is enough to give three
functions r(λ) where λ marks a unique point on the path, say with r(0) = r 1 and r(1) = r 2 .
44
c
2013
by Charles Thorn
This is just a parametric description of a curve in√space. To calculate the distance we start
with the differential dr = dλr ′ (λ). Then ds = dλ r ′2 and
Z 1 √
(228)
dλ r ′2
D12 =
0
The calculus of variations will then tell us how to minimize this distance.
Another example of a variational principle is Fermat’s principle of optics: A light ray
follows a path that minimizes the time to complete the journey. The speed of light c/n
depends on the index of refraction n(r) which can vary in space. Then the time of the
journey is
Z 1
Z
p
|r ′ |
1 1
T12 =
dλ
=
dλn(r(λ)) r ′2 (λ)
(229)
c/n
c 0
0
Hamilton’s Principle is exactly the same type of mathematical problem. We have to find a
function that minimizes a quantity that is expressed as the integral over some parameter of
a function of the path r and its derivative r ′ over the extent of the path. For Hamilton’s
principle the parameter is the time t, the path is the particle’s trajectory and the velocity is
the derivative.
So in complete generality we need to minimize an integral of the form
Z λ2
I =
dλL(qk (λ), q̇k (λ), λ)
(230)
λ1
Here we use a dot to denote derivative w.r.t. λ, which will be time for mechanics. For
pedagogical purposes we will just consider one coordinate in the following discussion. The
generalization to any number is easy.
7.2
Calculus of Variations
So we have to study how I changes under a small distortion of the path. Suppose we have
a trial path qk (λ). Then we can write a nearby path as qk (λ) + δqk (λ). Here δq is small but
arbitrary in shape. We can even choose δq to be zero over most of the path, departing from
the path only in a tiny interval of λ:
Z λ2
∆I =
dλ [L(qk (λ) + δqk (λ), q̇k (λ) + δ q̇k (λ), λ) − L(qk (λ), q̇k (λ), λ)]
λ1
Z λ2 X ∂L
∂L
δql
+ O(δq 2 )
dλ
+ δ q̇l (λ)
=
∂ql
∂ q̇l
λ1
l
Z
λ2
X ∂L λ2
X
d
∂L
∂L
+
+ O(δq 2 )
(231)
−
=
δql
dλ
δql
∂
q̇
∂q
dλ
∂
q̇
l λ1
l
l
λ
1
l
l
45
c
2013
by Charles Thorn
Now since the ends of the trajectory are fixed, δqk (λ1 ) = δqk (λ2 ) = 0, so qk (λ) will make I
stationary if
d ∂L
∂L
=
,
dλ ∂ q̇l
∂ql
for all l
(232)
These are the Euler Lagrange equations Without saying so, we have just applied what is
known as the calculus of variations!
7.3
Nonmechanics applications of the calculus of variations.
Let us apply Lagrange’s equations to a couple of examples, starting with the geodesic problem
in flat space. Recall that the distance between two points in 3 dimensional space is given by
Z 1 p
dλ ṙ 2
D12 =
(233)
0
where now we use the dot to indicate derivative w.r.t. λ. So in this case L =
need
ṙk
∂L
ṙk
=√ 2 =
∂ ṙk
|ṙ|
ṙ
∂L
= 0,
∂rk
√
ṙ 2 so we
(234)
The E-L equations just say that |ṙṙ| , which is just the unit tangent vector to the path, is a
constant over the path. This means that the path is a straight line!
Another famous example is the brachistochrone problem: What is the shape of the roller
coaster track between two points such that the frictionless car takes the minimum time to
complete the journey? To answer this we need to know the speed of the car at any point
on the track. Take point 1, where the car starts at rest, to be the origin of Cartesian
coordinates and√the gravitational potential energy to be U = mgz.√ Then by conservation
of energy v = −2gz. Recall the element of arc length ds = dλ ẋ2 + ż 2 so we need to
minimize
Z λ2 √ 2
ẋ + ż 2
dλ √
T12 =
(235)
−2gz
λ1
∂L
∂L
∂L
L
∂L
ẋ
ż
√
√
,
(236)
= 0,
=− ,
=√
=√
2
2
∂x
∂z
2z
∂ ẋ
∂ ż
−2gz ẋ + ż
−2gz ẋ2 + ż 2
We are free to choose λ in a way to simplify the√equations. In the text the choice λ = z is
made. For variety, let’s try choosing λ so that ẋ2 + ż 2 = 1. Then the E-L equations tell
us that
ẋ
∂L
= C,
= √
∂ ẋ
−2gz
46
p
ż = − 1 + 2C 2 gz
(237)
c
2013
by Charles Thorn
where we chose the negative square root because initially z decreases with λ. Integrating
the second equation gives
Z
dz
1
1 p
λ = − p
1 + 2C 2 gz,
z=
(C 4 g 2 λ2 − 1)
(238)
= 2
2
C g
2C 2 g
1 + 2C gz
Since we want z(λ1 ) = 0 we see we should choose λ1 = −1/(gC 2 ). z reaches a minimum of
−1/(2C 2 g) at λ = 0 and then increases to 0 at λ = 1/(C 2 g). We also need x, obtained from
p
p
dx
= C −2gz = 1 − C 4 g 2 λ2
dλ
(239)
It is convenient to put λ = −(1/C 2 g) cos θ so dλ/dθ = (1/C 2 g) sin θ so that
dx
1
1
=
sin2 θ =
(1 − cos 2θ)
2
dθ
C g
2C 2 g
1
1
x(θ) =
(θ − (1/2) sin 2θ) =
(2θ − sin 2θ) ≡ a(2θ − sin 2θ),
2
2C g
4C 2 g
1
1
1
(C 4 g 2 λ2 − 1) = − 2 sin2 θ = − 2 (1 − cos 2θ) ≡ −a(1 − cos 2θ)
z(θ) =
2
2C g
2C g
4C g
We have chosen the release point to be x = z = 0 or θ = 0. The constant a = 1/(4C 2 g)
must be chosen so that the curve passes through the final point x2 , z2 .
An interesting feature of the cycloid roller coaster track is that wherever the car is released
from rest on the track, the period of oscillations about the midpoint of the track is the same.
This is the subject of the 3 star problem 6.25, for which we sketch the solution. Let us write
the cycloid solution parametrically as follows
x(θ) = a(θ − sin θ),
y(θ) = −a(1 − cos θ))
then the lowest point is θ = π. Then
q
√
ds = adθ (1 − cos θ)2 + sin2 θ = adθ 2 − 2 cos θ
(240)
(241)
p
and the speed v = 2ga(cos θ0 − cos θ) where θ0 marks the point the cart is released from
rest. Then a quarter period is the time the cart takes to go from θ0 to π:
r
r Z π r
r Z u0
a
1 − cos θ
a
1−u
du
T
√
=
=
dθ
4
g θ
cos θ0 − cos θ
g −1 1 − u2 u0 − u
r Z 0u0
r Z 1
r
a
a
a
1
1
du p
du p
(242)
=
=
=π
g −1
g 0
g
(1 + u)(u0 − u)
u(1 − u)
So the period is just the small oscillation period of a pendulum of length 2a, but for any size
of amplitude!
47
c
2013
by Charles Thorn
8
8.1
Hamilton’s Principle of Least (Stationary) Action
Generalized coordinates
Cartesian coordinates are just fine for describing particles that can move unconstrained
throughout space. But when the motion is constrained in some way, another choice of
coordinates may be preferable. As a simple example suppose a particle is constrained to
move in a circle in the xy-plane. Then we have the constraints
x2 (t) + y 2 (t) = R2
z(t) = 0,
(243)
p
We could solve the constraints to eliminate the coordinate y(t) = ± R2 − x2 (t), but the
sign ambiguity is a nuisance. But in passing to polar coordinates x = ρ cos ϕ, y = ρ sin ϕ, we
see that the constraint is simply ρ = R, and ϕ gives a perfectly natural and unambiguous
description of the particle’s location. Thus in this situation it would be nice to use ϕ and
ϕ̇ as coordinate and velocity. It is standard to use qk and q̇k to denote such generalized
coordinates. There is no need to commit to a particular choice of coordinates in advance.
For example a system of N particles can be described by 3N Cartesian coordinates. If there
are k constraints, we can choose s = 3N − k independent generalized coordinates in any way
that is convenient.
8.2
The Action and Hamilton’s Principle
The action is defined as a time integral
Z
I =
t2
dtL(qk (t), q̇k (t), t)
(244)
t1
where L is called the Lagrangian of the system. For the moment we don’t specify it in
detail. It is a single scalar function of the generalized coordinates and their velocities, that
determines the equations of motion according to Hamilton’s principle: The trajectory qk (t)
of the system which starts at the point qk1 at time t1 and ends up at the point qk2 at time t2
is that trajectory which minimizes the action I.
This means that if we evaluate I for a trajectory qk (t) + δqk (t) infinitesimally different
from the solution, the change in the action will be of order δqk2 . So calculate
Z t2
dt [L(qk (t) + δqk (t), q̇k (t) + δ q̇k (t), t) − L(qk (t), q̇k (t), t)]
∆I =
t1
Z t2 X ∂L
∂L
δql
=
dt
+ O(δq 2 )
+ δ q̇l (t)
∂q
∂
q̇
l
l
t1
l
X ∂L t2 Z t2 X
d
∂L
∂L
+
=
δql
+ O(δq 2 )
(245)
−
dt
δql
∂
q̇
∂q
dt
∂
q̇
l t1
l
l
t
1
l
l
48
c
2013
by Charles Thorn
Now since the ends of the trajectory are fixed, δqk (t1 ) = δqk (t2 ) = 0, so qk (t) will satisfy
Hamilton’s principle if
d ∂L
∂L
=
,
dt ∂ q̇l
∂ql
for all l
(246)
Without saying so, we have just applied what is known as the calculus of variations!
If the qk ’s are Cartesian coordinates, this will be the form of Newton’s equations if
∂L
= ml q̇l (t),
∂ q̇l
∂L
∂V
=−
∂ql
∂ql
(247)
which tells us that the Lagrangian in that case can be taken to be
L =
1X
ml q̇l2 − V (q) = T (q̇l ) − V (ql )
2 l
(248)
where T is the kinetic energy of the system and V is the potential energy. Note carefully
the difference of L from the total energy T + V ! there is an all important sign difference in
the second term.
The Lagrangian is not uniquely determined, because Hamilton’s principle requires that
δq(t1 ) = δq(t2 ) = 0. For this reason a different Lagrangian
L′ = L +
d
f (q(t), t),
dt
I ′ = I + f (q(t2 ), t2 ) − f (q(t1 ), t1 )
(249)
will imply the same equations of motion. In other words, two Lagrangians, that differ by the
total time derivative of a function of coordinates and time, will imply the same equations of
motion.
8.3
Changing Coordinates in the Lagrangian
One of the virtues of Hamilton’s principle is that the points of stationary action do not
depend on the coordinate choice. Thus if you change coordinates in the action, Lagrange’s
equations in the new coordinates still determine the stationary points. It is much easier
to change coordinates in the Lagrangian than in the equations of motion! As an example
suppose we want to use spherical polar coordinates
r = r(cos ϕ sin θ, sin ϕ sin θ, cos θ)
ṙ = ṙ(cos ϕ sin θ, sin ϕ sin θ, cos θ) + r sin θϕ̇(− sin ϕ, cos ϕ, 0)
+rθ̇(cos ϕ cos θ, sin ϕ cos θ, − sin θ)
= ṙr̂ + r sin θϕ̇ϕ̂ + rθ̇θ̂
ṙ 2 = ṙ2 + r2 θ̇2 + r2 sin θ2 ϕ̇2
m 2
L =
(ṙ + r2 θ̇2 + r2 sin θ2 ϕ̇2 ) − U (r, θ, ϕ)
2
49
(250)
(251)
(252)
c
2013
by Charles Thorn
Notice that if U is independent of ϕ, Lagrange’s equations for ϕ simplify to the conservation
of ∂L/∂ ϕ̇ = mr2 sin2 θϕ̇ which is just Lz the z-component of angular momentum.
For motion in the xy plane, we can simply put θ = π/2 and the coordinate change is just
to polar coordinates r, ϕ:
L=
m 2
(ṙ + r2 ϕ̇2 ) − U (r, ϕ).
2
(253)
We calculate
∂L
= mṙ,
∂ ṙ
∂L
= mr2 ϕ̇,
∂ ϕ̇
∂L
∂U
∂U
= mrϕ̇2 −
,
mr̈ = mrϕ̇2 −
∂r
∂r
∂r
∂L
∂U
d
∂U
=−
,
(mr2 ϕ̇) = −
= rFϕ
∂ϕ
∂ϕ
dt
∂ϕ
(254)
It is important to realize that when one changes from Cartesian to other coordinates, the
kinetic energy in the new coordinate system can acquire coordinate dependence in addition
to velocity dependence, as spherical and polar coordinates demonstrate. Thus in a general
coordinate system we have T (q, q̇) but still U (q, t). To incorporate magnetic forces, we will
have to allow velocities to appear in a part of the Lagrangian that is neither kinetic nor
potential energy. In other words there will be systems for which the Lagrangian is not of the
form T − U !
8.4
The energy from the Lagrangian
We are all familiar with the conservation of energy by Newton’s equations when the forces are
conservative. In the Lagrange formulation we can generally identify an energy conservation
law when the Lagrangian has no explicit time dependence. Consider the Hamiltonian defined
by
∂L
−L
∂ q̇i
i
X ∂L X d ∂L X ∂L X ∂L ∂L
=
q̈i
+
q̇i
−
q̇i
−
q̈i
−
∂
q̇
dt
∂
q̇
∂q
∂
q̇
∂t
i
i
i
i
i
i
i
i
X
d ∂L ∂L
∂L
∂L
=
q̇i
−
=−
−
dt ∂ q̇i ∂qi
∂t
∂t
i
H ≡
dH
dt
X
q̇i
(255)
(256)
by Lagrange’s equations. Thus H is conserved provided ∂L/∂t = 0. For standard Newtonian
systems where T is quadratic in the velocities and L = T − V
X
i
q̇i
∂L
= 2T
∂ q̇i
(257)
so H = T + V as we expect.
50
c
2013
by Charles Thorn
8.5
The simple pendulum
As a familiar example of a problem with constraints consider the simple frictionless pendulum
with massless rod of length l, swinging in the xz-plane with the pivot at the origin of
coordinates. Let ϕ be the angle from the vertical. Then x = l sin ϕ, z = −l cos ϕ), and
T = (m/2)(ẋ2 + ẏ 2 ) = (ml2 /2)ϕ̇2 and V = −mgl cos ϕ. Hence
ml2 2
ϕ̇ + mgl cos ϕ
(258)
2
and Lagrange’s equation gives the familiar ϕ̈ = −(g/l) sin ϕ. Here we have solved the
constraint x2 + z 2 = l2 by going to polar coordinates and setting ρ = l.
Let’s consider the same problem in terms of Cartesian coordinates. Then the unconstrained Lagrangian is
m
(259)
L = (ẋ2 + ż 2 ) − mgz
2
L =
8.6
Constraints in general
When we consider constraints on a system’s motion, a useful first step is to find the Lagrangian for the unconstrained system. This might be a system with several particles with
positions r k (t) moving in a potential U (r 1 , r 2 , · · · , r N , t). In this case we immediately know
the unconstrained Lagrangian L = T − U , and there are 3N degrees of freedom. A typical
constraint would be to confine the motion to a surface in space. For example, a spherical
pendulum can swing in any direction, with the bob confined to the surface of a sphere. To
described the constrained motion it is useful to pick convenient coordinates of the constraint
surface. These generalized coordinates are generically called qk (t) with k = 1, · · · n < 3N .
When the positions r k satisfy the constraints they can be expressed as functions of the qk ’s:
X ∂r k ∂r k
r k (q1 , · · · qn , t),
ṙ k =
q̇l
(260)
+
∂ql
∂t
l
Plugging these relations into the unconstrained Lagrangian gives the Lagrangian for the
constrained motion L(q, q̇, t).
As an example consider the spherical pendulum. Introducing spherical coordinates, the
position of the bob can be expressed
r(t) = R(cos ϕ sin θ, sin ϕ sin θ, cos θ)
ṙ(t) = R sin θϕ̇(− sin ϕ, cos ϕ, 0) + Rθ̇(cos ϕ cos θ, sin ϕ cos θ, − sin θ)
m 2 2 2 m 2 2
R sin θϕ̇ + R θ̇ − mgR cos θ
L =
2
2
The constrained equations of motion are just the E-L equations:
∂L
∂L
= mR2 sin2 θϕ̇,
= mR2 θ̇
∂ ϕ̇
∂ θ̇
∂L
∂L
= 0,
= mR2 sin θ cos θϕ̇2 + mgR sin θ
∂ϕ
∂θ
51
(261)
(262)
(263)
(264)
(265)
c
2013
by Charles Thorn
Since the Lagrangian is independent of ϕ, mR2 sin2 θϕ̇ = M is a constant and then
θ̈ = sin θ cos θϕ̇2 +
M 2 cos θ
g
g
sin θ = 2 4 3 + sin θ
R
m R sin θ R
(266)
this gives a rather complicated equation of motion for M 6= 0! However, one can use energy
conservation, as usual, to express t in terms of an integral over θ:
m 2 2 2 m 2 2
R sin θϕ̇ + R θ̇ + mgR cos θ
2
2
2
M
m
=
+ R2 θ̇2 + mgR cos θ = E
2
2
2
2mR
r sin θ
dt
m
1
p
= R
dθ
2 E − mgR cos θ − M 2 /(2mR2 sin2 θ)
H =
(267)
(268)
Or consider the double pendulum swinging in a fixed plane.
r 1 = l1 (sin θ1 , cos θ1 ),
r 2 = r 1 + l2 (sin θ2 , cos θ2 )
(269)
or the pendulum on an accelerating cart:
r = (l sin θ + at2 /2, l cos θ)
(270)
All of these constraints are so-called holonomic constraints, in which the number of coordinates needed to describe the system is equal to the number of degrees of freedom: nonholonomic constraints are beyond the scope of this course. (Example: rubber ball rolling
without slipping on a plane).
Although we won’t be considering non-holonomic constraints further, there is a method of
handling constraints that avoids eliminating coordinates explicitly, and is a preferred method
for dealing with non-holonomic constraints.
As an example of this alternative treatment of constraints, let’s return to the simple
pendulum where we have to impose the holonomic constraint x2 + z 2 = l2 . The method of
Lagrange multipliers adds a term λ(t)(x2 + z 2 − l2 ) to the Lagrangian:
L→
m 2
(ẋ + ż 2 ) − mgz + λ(t)(x2 + z 2 − l2 )
2
(271)
We now regard λ as a generalized coordinate. Since λ̇ doesn’t appear in the Lagrangian, the
e.o.m. for λ is just
∂L
= x2 + y 2 − l 2 = 0
∂λ
(272)
which is seen to be precisely the constraint we wish to impose. The e.o.m’s for x, z now
involve λ:
mẍ = 2λx,
mz̈ = −mg + 2λz
52
(273)
c
2013
by Charles Thorn
From this we see that the force exerted by the constraint is F c = 2λ(x, 0, z) ≡ 2λρ. Passing
to polar coordinates (x, 0, z) = l(sin ϕ, 0, − cos ϕ) at this point, the e.o.m’s become
ml(ϕ̈ cos ϕ − ϕ̇2 sin ϕ) = 2λl sin ϕ,
2λ = m(ϕ̈ cot ϕ − ϕ̇2 )
ml(ϕ̈ sin ϕ + ϕ̇2 cos ϕ) = −mg − 2λl cos ϕ = −mg − ml(ϕ̈ cot ϕ − ϕ̇2 ) cos ϕ
g
g
2λ = −m
cos ϕ + ϕ̇2
(274)
ϕ̈ = − sin ϕ,
l
l
(275)
F c = −m g cos ϕ + lϕ̇2 (sin ϕ, 0, − cos ϕ)
For example, at the bottom ϕ = 0, F c = m(g + lϕ̇2 )ẑ to compensate gravity and match m×
the centripetal acceleration. Notice that if the rod is replaced by a rope, it can only pull on
the particle, which means that it forces the constraint only when λ < 0. This is always true
if ϕ < π/2. But if ϕ > π/2 the rope will only do its job if ϕ̇2 > −(g/l) cos ϕ!
8.7
Examples
Let’s go through some of the examples in the textbook.
Particle Moving on a Cylinder. In this example the force law is taken to be F = −kr,
corresponding to potential energy U = kr2 /2 = k(z 2 + R2 )/2, where the cylinder axis
coincides with the z-axis and R is the radius of the cylinder. The kinetic energy of the
particle is (m/2)(ż 2 + R2 ϕ̇2 ). thus the Lagrangian is
m 2
k
(ż + R2 ϕ̇2 ) − (z 2 + R2 )
2
2
∂L
∂L
∂L
∂L
= mR2 ϕ̇,
= 0,
= mż,
= −kz
∂ ϕ̇
∂ϕ
∂ ż
∂ ż
p
Clearly z undergoes simple harmonic motion with ω = k/m and ϕ̇ is a constant.
L =
(276)
(277)
Block sliding on a wedge free to move on a horizontal surface: Let q1 be the
displacement along the wedge of the block from the top of the wedge, and q2 be the horizontal
displacement of the left edge of the wedge. The kinetic energy of the wedge is just m2 q̇22 /2.
The Cartesian coordinates of the block are (q2 + q1 cos α, 0, q1 sin α) where α is the angle the
wedge makes with the horizontal. the block’s kinetic energy is therefore m1 [(q̇2 + q̇1 cos α)2 +
q̇12 sin2 α]/2. The block’s potential energy is −mgq1 sin α. Putting all this together
m2 2 m1 2
q̇ +
(q̇ + q̇12 + 2q̇2 q̇1 cos α) + m1 gq1 sin α
(278)
L =
2 2
2 2
∂L
∂L
= m1 q̇1 + m1 q̇2 cos α,
= (m1 + m2 )q̇2 + m1 q̇1 cos α
∂ q̇1
∂ q̇2
∂L
∂L
= m1 g sin α,
=0
∂q1
∂q2
m1 cos α
q̈2 = −
q̈1 ,
m1 q̈1 + m1 q̈2 cos α = m1 g sin α
m1 + m2
g sin α
(279)
q̈1 =
1 − m1 cos2 α/(m1 + m2 )
53
c
2013
by Charles Thorn
Note that this goes to the answer for a fixed wedge when m2 /m1 → ∞. It also goes to free
fall when α → π/2.
Bead on a spinning frictionless circular hoop The bead’s position vector is
r = R(cos ωt sin θ, sin ωt sin θ, − cos θ)
ṙ = R sin θω(− sin ωt, cos ωt, 0) + Rθ̇(cos ωt cos θ, sin ωt cos θ, sin θ)
m 2 2 2
(ω R sin θ + R2 θ̇2 ) + mgR cos θ
L =
2
(280)
Notice that in spite of the forced rotation, L does not depend explicitly on the time. Therefore
the Hamiltonian is a constant of the motion:
H = θ̇(mR2 θ̇) − L =
m 2 2
(R θ̇ − ω 2 R2 sin2 θ) − mgR cos θ
2
(281)
This is not the total energy of the bead which is
T +U =
m 2 2 2
(ω R sin θ + R2 θ̇2 ) − mgR cos θ = H + mω 2 R2 sin2 θ
2
(282)
which is not conserved. The E-L equations are:
g
mR2 θ̈ = −mgR sin θ + mω 2 R2 sin θ cos θ = mR2 sin θ ω 2 cos θ −
(283)
R
p
Static solutions are possible for θ = 0, π p
and for cos θ = g/(Rω 2 ) if ω > g/R.pθ = π is
always unstable; θ = 0 is stable for ω < g/R, but becomes unstable for ω > g/R. In
the latter case, put θ0 = arccos(g/Rω 2 ) and expand
cos θ =
g
− (θ − θ0 ) sin θ0 + O(θ − θ0 )2
Rω 2
(284)
so the right side behaves near θ = θ0 as
2
2
2
−mR ω sin θ0 (θ − θ0 ) = −mR
2
g2
ω − 2 2
R ω
2
(θ − θ0 )
(285)
Thus the solution
p θ = θ0 is stable, and harmonic small oscillations about it have angular
frequency ω0 = ω 2 − g 2 /(R2 ω 2 ).
A nice way to think about the different behaviors of this system is the effective potential
energy
1
Ueff (θ) = −mgR cos θ − mω 2 R2 sin2 θ
2
(286)
plotted as a function of θ in the following figure (a = ω 2 R/g):
54
c
2013
by Charles Thorn
55
c
2013
by Charles Thorn