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123
An Introduction
Dieter Strauch
Classical
Mechanics
Classical Mechanics
Dieter Strauch
Classical Mechanics
An Introduction
With 149 Figures
ABC
Professor Dr. Dieter Strauch
Universität Regensburg
Institut Theoretische Physik
Universitätsstr. 31, 93053 Regensburg, Germany
E-mail: dieter.strauch@physik.uni-regensburg.de
ISBN 978-3-540-73615-8 e-ISBN 978-3-540-73616-5
DOI 10.1007/978-3-540-73616-5
Library of Congress Control Number: 2009929359
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Preface
On this Textbook
This book has evolved from the series of lecture notes that I had handed out
to the physics students at the University of Regensburg in Bavaria, Germany.
Over the years, various bits and pieces had been added to the contents of these
lecture notes, and others had to be left out for reasons of time limitations.
These notes differed from the common textbooks, and as the students seemed
to like them, I have collected all those pieces in this book.
The Scope of this Book
The scope of this book is twofold. The reader can learn that alternative sets
of the very few principles of Classical Mechanics carry on very far. Thus, the
book contains an amount of applications of varying degrees of sophistication.
Also, different physical problems require different methods for their solutions
with varying degrees of mathematical sophistication.
The Organization of this Book
In order not to blur the physics with mathematical intricacies, the necessary
mathematical techniques are transferred to appendices.
The largest difference of this textbook from other books on Classical Me-
chanics may be that I have tried to make a particularly strong separation
between axioms and fundamental experiences, on the one hand, and between
claims, their proofs, various comments, on the other, rather than telling a
more or less continuous story. Also, frequent references are made to other
parts of the book or to other physical disciplines. If needed, the reader can
skip proofs, comments, applications, and footnotes and thus follow only the
main ideas.
An asterisk has been attached to the more advanced topics of this book,
which might also be skipped upon first reading.
vi Preface
The Readership
On the one hand, the book aims at the beginners by being very explicit on
going through the proofs of statements being made and of the examples being
given. On the other hand, the applications of the principles are extended to a
number of topics which might be of interest to the more advanced students.
The Contents of this Book
For the beginners I have often shown different paths to solutions of a problem,
so they might estimate the (dis)advantage of one method over another. Thus,
the reader should appreciate the employment of symmetry at an early stage.
It is equally important to realise that physics, unlike mathematics, is not
an exact science, but an approximation to the mathematical description of
processes in nature, usually relying on a number of simplifying assumptions.
In selecting the various assumptions the physicist is trained to choose the
important over the unimportant effects and to set up a sensibly simple model.
In fact, classical mechanics has its limitations, as astronomers found out a
long time ago, and nowadays these limitations have been overcome by taking
account of relativistic effects.1 Also, the mechanics of the microscopic world
is governed by quantum theory.2
Thus, I have tried to concentrate on the practical applications of the prin-
ciples and on the calculational techniques rather than on the construction of a
theory as such. Even though the structure and method of theoretical physics
is an exciting topic (in particular for the advanced physicist), I believe that
it is indispensable to have a hands-on understanding of the theory with its
basic results and of the techniques if one wants to handle and apply them
in practice. Therefore, this book is an introduction to theoretical mechanics
rather than an introduction to formal theoretical physics.
By no means have I tried to be complete. Thus, some topics are missing
which a physicist may need to know, like non-linear dynamics and (determin-
istic) chaos, and possibly others.
Relativistic mechanics is only touched upon.3 Also the statistical methods
of the many-body systems is a subject of its own.4
Even though skipped in most textbooks on theoretical mechanics, I have
included a short chapter on continuum mechanics, since the mechanics of de-
formable bodies presents a first example of a (classical) field theory. Here,
1 See the Course on Special Relativity.
2 See the Course on Quantum Mechanics.
3 More details can be found in the separate Course on Special Relativity where
the principle of relativity is applied to problems in mechanics, optics, electromag-
netism, and quantum theory.
4 See the Course on Thermodynamics and Statistics.
Preface vii
numerous concepts are introduced which will be found again in electromag-
netism,5 as, for example, the energy-momentum tensor, a lesser liked detail
from electromagnetism, and other (classical and quantum) field theories.
Material for Electromagnetism and Optics, Quantum Mechanics, Thermo-
dynamics, Statistics, and Special Relativity are planned to be published in the
near future.
What is Theoretical Mechanics?
Physics is the mathematical description of events occurring in nature.
(Comment : Nature offers numerous additional facets which are recognised
by man: esthetic, emotional, economical, etc. Let us approach all these with
wonder and responsibility!)
Theoretical physics is the classification of these physical events (respec-
tively of the corresponding mathematical ways of description) under a mini-
mum number of axiomatic basic assumptions. (A detailed exposition on the
nature of a physical theory can be found in the book by Ludwig [4].) These
physical axioms cannot be proved (but can be falsified by just one experi-
ment); they can only turn out to be useful.
For the application of a physical theory to natural phenomena one pre-
sumes in general an idealization of nature, in that the considered aspect of
nature is approximated by a model (point masses, neglect of the friction by
the air in the case of free motion or by the support in the case of rolling
cylinders, etc.).
The topic of classical mechanics is the description of the motion of massive
macroscopic bodies, namely
the kinematics: the geometrical form of the trajectories (e.g. the elliptical
trajectories in the Kepler problem) and
the dynamics: the time development of the particle position and/or orientation
(e.g. the orientation of the children’s top).
One can treat only the motion of just one or two particles (e.g. the pen-
dulum or a central star and a satellite, respectively) in an exact way. The
motion of three particles cannot be treated analytically. For the treatment of
the properties of very many particles (like fluids and gases) statistical methods
are needed.6
Ais not conservative; however, the integral then
does depend upon the path. This is true in particular for frictional forces.
20 1 Newtonian Mechanics: General Properties
Also in thermodynamics it turns out that in general the result of the integral∫
dA depends upon the path. In this case an (infinitesimal) work element is
not denoted by dA but by δA or d−A.
1.7.3 Energy Conservation Theorem
In a conservative force field the total energy is a conserved quantity,
F is conservative ⇔ Ė = 0 ⇔ E = const. (1.33)
Proof:
Ė =
d
dt
(T + V ) =
d
dt
[m
2
ṙ · ṙ + V (r)
]
= m r̈ · ṙ + ∇V · ṙ
= (m r̈ + ∇V ) · ṙ (1.3,1.26)
= (F − F ) · ṙ = 0.
�	
1.8 Invariances and Conservation Laws
1.8.1 Translations and Rotations
In the following section we will investigate the invariance of a system un-
der infinitesimal coordinate transformations. To this end we start with some
preparatory remarks:
Translation in Time
An infinitesimally small time translation from t to t′ by a time interval dt is
described by
t′ = t+ dt.
Translation in Space
The translation of the coordinate from r to r′ by an infinitesimally small
space-translation vector dr is described by
r′ = r + dr. (1.34)
Rotation in Space
Let the vector ϕ be directed along the direction of the axis of rotation, and
let the modulus have the value of the rotational angle. The transformation of
1.8 Invariances and Conservation Laws 21
Fig. 1.11. Axis of rotation Δϕ (as a vector) and
angle of rotation Δϕ (as its modulus)
the angle ϕ in an infinitesimal rotation by an angle dϕ along the direction eϕ
is described by24
r′
i = ri + dr = ri + dϕ × ri (1.35)
with the rotation vector
dϕ = ϕeϕ, (1.36)
see Fig. 1.11.
Comment: For a time-independent axis one has ėϕ = 0.
1.8.2 Invariances and Conservation Laws: Noether’s Theorem (I)
The invariance of a system under any one of the continuous transformations
leads to a conservation law (Noether’s theorem).25 This theorem plays an
important role in the modern physics, in particular in elementary-particle
physics. In classical mechanics one deals primarily with three possible in-
variances of the potential energy and thus with conserved quantities of the
system:
(i) Space homogeneity of the system, space translational symmetry: The in-
variance under continuous space translations leads to the conservation of
the total momentum.
(ii) Space isotropy of the system, rotational symmetry: The invariance under
continuous space rotations leads to the conservation of the total angular
momentum.
(iii) Time homogeneity of the system: The invariance under continuous time
translations leads to the conservation of the total energy.26
24 See also Sect. 7.2.2 further below.
25 This is considered in more detail in the context with the Lagrangian formulation
of mechanics, Sect. 3.11.
26 This one has only in the case of the motion without rheonomous constraints, see
Sect. 3.11.3.
22 1 Newtonian Mechanics: General Properties
Proof: (i) Invariance under continuous space translations means
V (r1, . . . , rN ) = V (r1 + r, . . . , rN + r) for arbitrary r.
If one expands the right side into a Taylor series and compares with the left
side, then one sees that the terms of all orders in r vanish except for that
of zeroth order, since the different powers of r are linearly independent. In
particular one finds for the first-order term
0 = V (r1 + r, . . . , rN + r) − V (r1, . . . , rN )
=
N∑
i=1
(r · ∇i)V (r1, . . . , rN ) = −
∑
i
r · F i = −r · F = −r · Ṗ . (1.37)
Here ∇i is the gradient with respect to the coordinate ri, F i the force on the
ith particle and F the total force, which is responsible for the motion of the
center of mass. Since r is arbitrary with respect to modulus and direction,
the total force and the time change of the total momentum must vanish,
F = 0 ⇔ P = const.
For the proofs to (ii) and (iii) see problem 1.8. �	
Comments:
• If the invariance applies only to one or two components in (i) or (ii), then
only the respective components of the conserved quantities are conserved.
• If a closed system has certain invariance properties, then these are gener-
ally destroyed by additional external fields. For example, a homogeneous
gravitational field destroys the invariance of a free point mass under an
(arbitrary) translation in vertical direction. For the potential energy of a
point mass in a gravitational field one namely has
V (x, y, z) = mgz �= V (x, y, z + h) = mg (z + h) .
In contrast, the translation invariance is conserved in the horizontal direc-
tions,
V (x, y, z) = mgz = V (x, y + y0, z) = V (x+ x0, y, z).
In contrast to the momentum components in horizontal direction the ver-
tical component of the momentum is not a conserved quantity.
• Even though the translation of a crystal by a lattice vector, which maps ev-
ery point of the crystal onto an equivalent one (neglecting surface effects),
is an invariance property, this translation is discrete and not continuous,
since it is not infinitesimally small and cannot be chosen arbitrarily.27
27 However, in solid-state physics this lattice translation invariance is connected to
the conservation of the so-called quasi-momentum, which leads, among others, to
the Laue spots of X-ray structure determination. This, in turn is related to the
“quantized” structure of the Fourier coefficients of periodic structures.
1.9 * Virial Theorem 23
1.8.3 The Usefulness of the Conservation Laws
The energy is a first so-called integral of the motion. Therefore, if one knows
that the force is conservative and that thus the energy is conserved, one can
immediately start from an equation like
1
2mṙ2 + V (r) = E
rather than the equation of motion
mṙ = −∇V (r).
For a given system, it is therefore of practical importance to know the constants
of the motion, since with their help a first integration of the equations of
motion is performed more or less automatically.
1.8.4 The Usefulness of Symmetry Analysis
The various conservation laws are interconnected with different invariances,
i.e., with different symmetries of the systems. It is, therefore of great practical
advantage to investigate the symmetry of the system. Obeying this advice
reduces the labor by an amount which is the more extensive the more complex
the system is. (None of the problems in modern physics is as simple as the
harmonic oscillator or the Kepler problem.)
1.9 * Virial Theorem
In a closed conservative system the total energy is a conserved quantity; the
energy is distributed over kinetic and potential energy in a time dependent
way. In the virial theorem a statement is made on the time-average values of
kinetic (and potential) energy for finite motions (|ri| �= ∞ and |pi| �= ∞) and
holds in particular for one-particle systems.
Time-average Value
Definition 15. (Time Average): The time-average value 〈A〉t of a time-
dependent quantity A(t) is defined as
〈A〉t = lim
τ→∞
1
τ
∫ τ
0
dt′A(t+ t′). (1.38)
This makes sense only if this mean value is independent of the arbitrarily
chosen time t.
24 1 Newtonian Mechanics: General Properties
Comment: For a periodic quantity A with the period28 T it suffices to
average over one single period,
〈A〉t =
1
T
∫ T
0
dt′A(t+ t′).
The Virial Theorem
Virial Theorem: Let the positions ri and momenta pi be finite, and let F i
be the force acting on the particle i; then one has for the time average value
〈T 〉t of the kinetic energy29 T
2〈T 〉t = −
∑
i
〈ri · F i〉t. (1.39)
For conservative systems one has
2〈T 〉t =
∑
i
〈ri · ∇iV 〉t.
For a single particle one has in particular
2〈T 〉t
(1.39)
= −〈r · F 〉t generally
(1.3)
= 〈r · ∇V 〉t for a conservative force
= 〈rdV (r)
dr
〉t for a central potential
= n〈V 〉t for V (r) = arn .
Proof (for a one-particle system): One obtains by partial integration
−〈r · F 〉t
(1.38)
= − lim
τ→∞
1
τ
∫ τ
0
dt′ r(t+ t′) · F (t+ t′)
(1.3)
= − lim
τ→∞
1
τ
∫ τ
0
dt′ r(t+ t′) ·mr̈(t+ t′)
= lim
τ→∞
1
τ
∫ τ
0
dt′mṙ(t+ t′) · ṙ(t+ t′)
− lim
τ→∞
1
τ
m [r(t+ τ) · ṙ(t+ τ) − r(t)· ṙ(t)]
= mṙ2(t) + 0 = 2〈T 〉t.
Since the quantities r and ṙ are finite by supposition, also the difference of the
scalar products remains finite, and the second (integrated) term vanishes in
the limit τ → ∞. The first term is just the average value of twice the kinetic
energy. �	
28 Both, the period as well as the kinetic energy are denoted by T .
29 See footnote 28.
Summary: Newtonian Mechanics – General Properties 25
1.10 The Basic Problem of the Mechanics
of a Point Mass
The basic problem of the mechanics of a point mass thus consists of the
following: Given a force F (r, ṙ, t); to be found is the trajectory r(t) of a point
mass under the influence of the force F , i.e., the solution r(t) of the differential
equation
d
dt
(m ṙ) = F (r, ṙ, t).
Sometimes also reversely: Given a trajectory; to be determined is the force
and/or the potential. This is the object of the dynamics.
In the kinematics one determines only the form of the trajectory from the
knowledge of the forces or of the potential and not the time dependence with
which the trajectory is traced out (the time dependence would be but one of
various possible parametric representations of the trajectory).
Reversely one can determine the potential from the form of the trajectory,
see, e.g., the consideration of the planet trajectories in Sect. 5.1.3.
Summary: Newtonian Mechanics – General Properties
Newton’s Axioms
(I) Law of inertia: postulate of inertial systems:
F i
(1.2)
= 0 ⇔ ṗi = 0 ⇔ pi = const.
(II) Law of motion: dynamics in inertial systems
F i
(1.3)
= ṗi.
(III) Reaction law:
F ji
(1.4)
= −F ij .
(IV) Superposition principle:
F i
(1.5)
=
∑
j
F ij .
Einstein’s Equivalence Principle
mi
(1.8)
= mg
26 1 Newtonian Mechanics: General Properties
Equations of Motion
Momentum:
Ṗ
(1.21)
= F ext
Angular momentum:
L̇
(1.22)
= N ext
Energy:
Ė
(1.23)
= P ext
Special Symmetries Lead to Conservation Laws (and vice versa)
Conservation of momentum:
Ṗ = 0
(1.24)⇔ F = 0
Angular-momentum conservation:
L̇ = 0
(1.25)⇐ F = F (r)er
Energy conservation:
Ė = 0
(1.33)⇔ F conservative.
⇔ ∇× F
(1.27)
= 0,
∮
F · dr
(1.29)
= 0, F
(1.26)
= −∇V
Work, Potential Energy
dA
(1.31)
= +F · dr, dV
(1.30)
= −F · dr
Power
P = Ȧ
(1.32)
= F · ṙ
Virial Theorem
2〈T 〉t
(1.39)
= −〈r · F 〉t generally
= 〈r · ∇V 〉t for conservative force
=
〈
r
dV (r)
dr
〉
t
for central potential
= n〈V 〉t for V (r) = arn .
Problems 27
Problems
1.1. Mechanics of the Many-Body System. Given be a system of n point
masses, {m1, m2, . . . , mn}. Acting between these masses are internal forces
F ik of the form
F ik = Fik(rik)
rik
rik
, rik = |rik| .
In addition, the system shall be subject to external forces F i.
(a) Which restriction for the internal forces follows from the ansatz for F ik?
(b) Set up the equation of motion for the particle i in such a system.
(c) Prove under these conditions
(i) the center-of-mass theorem,
M R̈ =
n∑
i=1
Ki with M =
n∑
i=1
mi R =
1
M
n∑
i=1
miri;
(ii) the angular momentum theorem,
L̇ =
n∑
i=1
ri × Ki with L =
n∑
i=1
li;
(iii) the energy-conservation theorem,
d
dt
(T + U) =
n∑
i=1
(vi · Ki) with
T =
n∑
i=1
Ti, U =
n∑
i=1
n∑
k=i+1
Uik(rik).
(d) Consider now a closed system. Which conservation theorems apply?
1.2. Torque on a Spool. A nearly empty spool rests on a horizontal plane. A
horizontal force F pulls on the free end of the thread, see Fig. 1.12. Determine
the direction of the torque on the spool by the force with respect to the spool
axis and to the axis through the point of contact (A) of the spool with the
plane.
1.3. Conservative Forces. Examine (by investigating ∇× F ) whether the
following forces are conservative or not, and, if so, determine the potential
energy V ; then calculate −∇V , and convince yourself that the original force
is obtained.
F (α)(x, y, z) = α
(
2xyzex + x2zey + x2yez
)
F (β)(x, y, z) = β
(
y e−γrex + z e−γrey + xyez
)
with r2 = x2 + y2 + z2 and the constants α, β, and γ.
28 1 Newtonian Mechanics: General Properties
F
A Fig. 1.12. The spool and the force F due to the thread
1.4. Central Potentials. Prove that energy and angular momentum is con-
served for a particle in a central potential V (r) = V (r).
1.5. Central Force and Central Potential. In problem 1.4 it is concluded
that the force corresponding to a central potential is a central force. Determine
the condition for the reverse case.
1.6. Rotational Motion. Prove that for a purely rotational motion the
power P = N · ω is compatible with P
(1.32)
= F · v. (ω is the vector of
the angular velocity ω = dϕ/dt.)
1.7. Particle in a Static Electric and Magnetic Field.
(a) Prove that the static electric field E = −∇φ represents a conservative force
field (F = qE for a particle with the charge q). Determine the potential
energy.
(b) Prove that the kinetic energy of a particle in a magnetic induction field
B is a conserved quantity.
1.8. Invariances and Conservation Laws. Prove that the invariances of a
system under time translations and under space rotations lead to the conser-
vation of the energy and of the angular momentum, respectively.
1.9. The Virial Theorem. Prove the virial theorem
〈T 〉 =
1
2
〈
r
dV
dr
〉
for a central potential V (r). Discuss the case V (r) = a · rn+1, in particular
for n = 2 and n = −1. (Give examples for these latter cases.)
2
Newtonian Mechanics: First Applications
In this chapter we shall treat some examples, useful in the following, for the
motion of one and two particles. Thereby the separation into center-of-mass
and relative coordinates shall be performed and some techniques shall be
learned, mostly for the solution of typical equations of motion.
2.1 One-Dimensional Motion
The equation of motion
mẍ = F (x, ẋ, t) (2.1)
is a differential equation of second order. The general solution contains two
integration constants. These integration constants are to be fixed by two initial
conditions (boundary conditions). Even though there are no general solution
methods, the solution in special cases can be given.
2.1.1 Constant Force
Let the force be
F (x, ẋ, t) = F0.
Then one can integrate once immediately,
m
∫ t
t0
ẍdt′ = F0
∫ t
t0
dt′ = m [ẋ(t) − ẋ(t0)] = F0 (t− t0)
or
ẋ(t) = v0 +
F0
m
(t− t0)
with v0 = ẋ(t0). A second integration leads to
30 2 Newtonian Mechanics: First Applications
x(t) = x0 + v0 (t− t0) +
F0
2m
(t− t0)
2
with x0 = x(t0). The two integration constants are v0 and x0.
Example: Particle in the homogeneous gravitational field. A particularly
simple example for an equation of motion, which can be integrated directly,
is the motion in a homogeneous gravitational field. In this case the force is
constant,
mz̈ = −mg.
Details will be treated in problem 2.2.
2.1.2 Time-Dependent Force
Let the force be
F (x, ẋ, t) = F (t).
Then a first integration (over t′) yields
m
∫ t
t0
dt′ẍ(t′) =
∫ t
t0
dt′F (t′) = m [ẋ(t) − v0] =
∫ t
t0
dt′F (t′)
and from this
ẋ(t) = v0 +
1
m
∫ t
t0
dt′F (t′).
A second integration (over t′′) leads to
x(t) = x0 + v0 (t− t0) +
1
m
∫ t
t0
dt′′
∫ t′′
t0
dt′F (t′)
with x0 = x(t0) and v0 = ẋ(t0).
Example: Motion of a particle with charge q in an alternating electric field
E(t) = E0 sin(ωt) of a condenser. The force is
F (t) = qE(t) = qE0 sin(ωt).
The first integration yields
ẋ(t) = ẋ(t0) +
qE0
m
∫ t
t0
dt′ sin(ωt′) = v0 −
qE0
mω
[cos(ωt) − cos(ωt0)]
with v0 = ẋ(t0). If one chooses t0 = 0 for simplicity one obtains
ẋ(t) = v0 −
qE0
mω
[cos(ωt) − 1] .
2.1 One-Dimensional Motion 31
An additional integration leads with x0 = x(0) to
x(t) = x0 + v0t−
qE0
mω
∫ t
0
dt′ cos(ωt′) +
qE0
mω
t
= x0 + v0t−
qE0
mω2
[ωt− sin(ωt)] .
2.1.3 Velocity-Dependent Force
Let the force be
F (x, ẋ, t) = F (ẋ).
This can be integrated directly once with the method of the separation of
variables. With ẋ(t) = v(t) one obtains
mv̇ = m
dv
dt
= F (v)
or
m
dv
F (v)
= dt
and from this with v = v(t) and v0 = v(t0)
m
∫ v
v0
dv′
F (v′)
=∫ t
t0
dt′ = t− t0.
This has the form
t = t0 + f(v, v0)
and in most cases can be solved for
dx
dt
= v = g(t, t0, v0),
from which the solution of the equation of motion is obtained by a second
integration,
x(t) − x(t0) =
∫ t
t0
dt′g(t′, t0, v0).
Example: For a motion under frictional forces see, e.g., problems 2.6
and 2.7.
32 2 Newtonian Mechanics: First Applications
2.1.4 Coordinate-Dependent Force
Let the force be
F (x, ẋ, t) = F (x).
Every one-dimensional force F (x) is in general a conservative force, since the
potential V (x) can be calculated from the integral
V (x) = −
∫
dxF (x). (2.2)
With the so-called “integration of the energy conservation theorem” one
obtains
mẍ = F (x)
⇒ mẍẋ =
m
2
dẋ2
dt
= F (x)ẋ = F (x)
dx
dt
⇒ m
2
d(ẋ2) = F (x) dx
⇒ m
2
ẋ2(t) − m
2
ẋ2(t0) =
∫ x(t)
x(t0)
dxF (x) = −V (x) + V (x(t0))
or
m
2
ẋ2 + V (x) =
m
2
ẋ2(t0) + V (x(t0)) = E. (2.3)
Comments:
• This procedure, namely the multiplication with the velocity, leads in many
cases to a first integral, which is easy to obtain.
The next integration can be performed again via the separation of variables:
From (2.3) one obtains
dx
dt
= ẋ = ±
√
2
m
√
E − V (x) (2.4)
or
dt = ±
√
m
2
dx√
E − V (x)
.
If the particle was at the position x0 = x(t0) a time t0, it will be at the
position x at a time t, given by
t− t0 =
∫ t
t0
dt = ±
√
m
2
∫ x
x0
dx′√
E − V (x′)
. (2.5)
This equation yields – in principle – the time t(x) as function of position x
and, if one can invert this relation, also reversely the position x(t) is given as
a function of time t.
2.1 One-Dimensional Motion 33
V
E
x1 x2
x
forbidden allowed forbidden
x3
Fig. 2.1. Allowed and forbidden regions of motion in a one-dimensional potential
V (x) at a given energy E
With a given energyE those regions of x are allowed, in which the radicand
(namely ẋ2) in (2.5) is positive, E − V (x) ≥ 0.
If one assumes for example a potential as in Fig. 2.1, i.e., a potential of the
form
V (x) > V (x1) x V (x3) x2 x3
with
V (x1) = V (x2) = V (x3) = E,
then there are three types of motion:
(i) Bound Motion: The energy lies between the local minimum and the local
maximum of the potential energy. The motion consists of an (anharmonic)
oscillation with turning points x1 and x2, at which the kinetic energy vanishes
and at which the total energy is thus equal to the potential energy.
(ii) Free (Unbound) Motion: The energy lies above of the local maximum. The
motion consists of an approach towards the center (x = 0) up to the point x1
and/or towards infinitely large values of x. This is the process of scattering
or of a collision.
(iii) The energy is equal to the potential energy of the local maximum. The
motion ends at the x-value of the local maximum.
34 2 Newtonian Mechanics: First Applications
(iv) A motion with an energy below that of the local minimum is again a
scattering or a collision, however with an approach up to x′3.
In the case (i) of the oscillation the regime x1 ≤ x ≤ x2 is traced once in
half a period, and one finds from (2.5) for the period of the oscillations
T = 2
√
m
2
∫ x2
x1
dx√
E − V (x)
. (2.6)
Notice that the turning points x1 and x2 depend upon the energy E!
Comments:
• Also in the case that the kinetic energy has the form
T (x, ẋ) = a(x) ẋ2 + b(x) ẋ+ c(x),
there is a solution ẋ(E, V (x)) from the energy conservation theorem,
E = T (x, ẋ) + V (x) ⇒ ẋ = ẋ(E, V (x))
⇒
∫
dt =
∫
dx
ẋ(E, V (x))
(2.7)
in analogy to (2.4), see for example the pendulum in the following section
or the Kepler motion in Sect. 5.
• With the energy as sketched in Fig. 2.1 the regions x 0) is that (oriented towards the center, the hinge) force of
constraint, which (in addition to the external forces) lets the pendulum execute its
motion in such a way that the constraint is fulfilled. Thus the constraint consists
of the fact that the motion is restricted (constrained) to a circular arc.
3 For the general treatment of constraints see Chap. 3.
2.1 One-Dimensional Motion 35
ϕ
Fig. 2.2. The definition of the (mathematically) positive angle ϕ and of the positive
force Fϕ (in the direction of increasing angle!) as well as of the positive force Fr (in
the direction of increasing distance r)
ϕ l
S
m
mg
mg cos ϕ
mg sin ϕ
Fig. 2.3. The decomposition of the forces acting on the plane pendulum
system. Also, so far we have not presented a criterion according to which one
could decide whether or not the system is conservative.
The motion can be described by the coordinate ϕ (Fig. 2.2). The forces
acting on the pendulum mass are the (known) gravitational force F = mg
and the (so far unknown) tension S = −Ser (in the direction towards the
hinge) (Fig. 2.3).
There is no motion in radial direction,
r = l = const,
such that the tension S = −Ser, the fugal force mrϕ̇2er, and the radial
component of the gravitational force add to zero,4
0 = mr̈ = Fr = −S +mrϕ̇2 +mg cosϕ.
4 See also Sect. 2.2.4.
36 2 Newtonian Mechanics: First Applications
The motion along the circular arc is forced by the tangential component Fϕ
of the gravitational force,
Fϕ = −mg sinϕ.
(Fϕ points into the direction of decreasing angle ϕ!) The equation of motion
is then (with the line element ds = l dϕ of the circular arc)
m (r̈)ϕ = ms̈ = ml ϕ̈ = Fϕ = −mg sinϕ
or
ϕ̈ = −g
l
sinϕ. (2.8)
First Integration, First Alternative
The integration of this equation of motion is done implicitly by the use of the
energy conservation theorem (for conservative forces). It is always the same
trick: One multiplies the equation of motion with ϕ̇ (in the general case with
a first derivative with respect to time),
ϕ̇ϕ̈ = −g
l
ϕ̇ sinϕ,
what one writes in the form
1
2
d
dt
ϕ̇2 =
g
l
d
dt
cosϕ,
and can integrate directly once,
1
2
[
ϕ̇2(t) − ϕ̇2(t0)
]
=
g
l
[cosϕ(t) − cosϕ(t0)]. (2.9)
The additional integration is not elementary;5 the methodological procedure
has been presented in Sect. 2.1.4.
First Integration, Second Alternative
If one multiplies (2.9) with ml2 and reorders, one obtains the energy conser-
vation theorem,
E(t) = 1
2ml
2ϕ̇2(t) −mgl cosϕ(t) = 1
2ml
2ϕ̇2(t0) −mgl cosϕ(t0) = E(t0),
from which one could have started right away.
5 See (2.6) in Sect. 2.1.6.
2.1 One-Dimensional Motion 37
φ/π
−1
0
1
f
−1 0 1 −2 −1 0 1 2
φ/π
0
w
E
φ0 / π
Fig. 2.4. The (reduced) force f = Fϕ/mg acting on the mathematical pendulum
(left) and the (reduced) potential energy w = V/mgl (right) as a function of the
angle ϕ ≡ φ; the approximation of small amplitudes is shown by the broken lines
Second Integration in the Case of Small Amplitudes
To begin with, small displacements6 shall be considered in the following. Here,
the one-dimensional path is the path x = lϕ on the circular arc.
One has (Fig. 2.4)
line element ds = l dϕ
kinetic energy T =
m
2
ṡ2 =
m
2
l2ϕ̇2
potential energy V = mgl (1 − cosϕ) ≈ m
2
glϕ2 (2.10)
total energy E = T + V.
The (total) energy and the turning points are interconnected; atthe turning
points ±ϕ0 the kinetic energy vanishes,
ϕ̇ = 0, E = V (ϕ0) = mgl (1 − cosϕ0) ≈
m
2
glϕ2
0.
One obtains7 from (2.5) by choosing the initial condition8 as ϕ(t) = 0 for
t = t0 = 0
t =
∫ ϕ
0
l dϕ′
√
2
m
m
2 gl (ϕ
2
0 − ϕ′2)
=
√
l
g
∫ ϕ
0
dϕ′
√
ϕ2
0 − ϕ′2
GR 2.261=
1
ω0
arcsin
ϕ
ϕ0
6 Large displacements are treated in Sect. 2.1.6.
7 With the integral # 2.261 with c = −1 and Δ = −4ϕ2
0 from the extremely
voluminous tables in [11].
8 In order to perform one integration one needs one initial condition.
38 2 Newtonian Mechanics: First Applications
K(z)
p/2
0
0 z 1
Fig. 2.5. Graphical representation of K(z); see also [10], p. 592
with ω2
0 = l/g. From this one obtains by solving for ϕ
ϕ = ϕ0 sin(ω0t).
Comment: The solution of the linear differential equation ϕ̈ = − g
lϕ is simple
within the standard procedure; here, however, the method of the integration
of the energy conservation theorem according to (2.7) was to be demonstrated.
2.1.6 * Example: Plane Pendulum with Large Amplitude
As an example, the plane pendulum shall be considered again, but now for
displacements which are not necessarily small; the pendulum is assumed to
oscillate, but not to rotate. The kinetic energy T is thus smaller than the
potential energy for the upright standing pendulum, T ≤ 2mgl. The one-
dimensional path x is as in Sect. 2.1.5 again the path on the circular arc.
The evaluation of the integral in (2.5) for arbitrary times t and angles ϕ
poses difficulties, but the period can be given. The (total) energy and the
turning points are related to each other; at the turning points ±ϕ0 one has,
cf. Fig. 2.4,
ϕ̇ = 0 , E = V (±ϕ0) = mgl (1 − cosϕ0) . (2.11)
The regime 0 0 for
0 ≤ ϕ ≤ ϕ0 ≤ π.) Then, with (2.13) and (2.14) one has for the square root in
the denominator of (2.12)
√
cosϕ− cosϕ0 =
√
2
√
sin2 ϕ0
2
− sin2 ϕ
2
=
√
2 sin
ϕ0
2
√
1 − sin2 α
=
√
2 sin
ϕ0
2
cosα,
9 An extensive collection of special functions, their properties, tabulated values,
and graphical representations can be found in [10].
40 2 Newtonian Mechanics: First Applications
and for the period one obtains
T =
√
8l
g
∫ ϕ0
0
dϕ
1√
cosϕ− cosϕ0
=
√
8l
g
∫ π/2
0
2 sin ϕ0
2 cosα dα√
1 − sin2 ϕ0
2 sin2 α
· 1√
2 sin ϕ0
2 cosα
= 4
√
l
g
∫ π/2
0
dα√
1 − sin2 ϕ0
2 sin2 α
.
�	
Small Amplitudes
For small amplitudes one obtains from K(z) → π
2 for z → 0 the known result
ω =
√
g
l
,
and the period diverges with the amplitude ϕ0 approaching π.
Proof: One expands for small amplitudes in (2.12),
cosϕ− cosϕ0 ≈ 1
2
(
ϕ2
0 − ϕ2
)
and obtains for the period
T ≈ 4
√
l
g
∫ ϕ0
0
dϕ√
ϕ2
0 − ϕ2
= 4
√
l
g
∫ ϕ0
0
dϕ/ϕ0√
1 − (ϕ/ϕ0)2
.
With the substitution
ϕ = ϕ0 sinβ
one obtains
T = 4
√
l
g
∫ π/2
0
cosβ dβ
cosβ
= 4
√
l
g
π
2
⇒ ω =
2π
T
=
√
g
l
.
Who ever looks for exceptional integrals should consult alternatively the in-
tegral tables in [11], see footnote 7. There one finds
T ≈ 4
√
l
g
∫ ϕ0
0
dϕ√
ϕ2
0 − ϕ2
GR 2.261= 4
√
l
g
[
arcsin
ϕ
ϕ0
]ϕ0
0
= 4
√
l
g
π
2
.
�	
2.2 Motion in a Plane 41
Intermediate Amplitudes
One can determine corrections of the harmonic result by expansion of the
elliptic integral K in terms of small ϕ0. With
1√
1 − sin2 ϕ0
2 sin2 α
= 1 +
1
2
sin2 ϕ0
2
sin2 α+ . . .
one obtains
K
(
sin
ϕ0
2
)
=
∫ π/2
0
[
1 +
1
2
sin2 ϕ0
2
sin2 α+ . . .
]
dα
=
π
2
[
1 +
1
2
sin2 ϕ0
2
· 1
2
+ . . .
]
≈ π
2
[
1 +
ϕ2
0
16
+ . . .
]
,
and from this one obtains for the period
T = 2π
√
l
g
[
1 +
ϕ2
0
16
+ . . .
]
.
Thus, the period increases with increasing amplitude.
In the case ϕ0 = 1
2π the period diverges: The pendulum remains at rest
in the indifferent equilibrium.
2.2 Motion in a Plane
In this paragraph we continue with the treatment of a single particle, but
we leave the one-dimensional motion and consider the motion in a plane.
Examples are the plane pendulum or, as one will see in Chap. 5, the motion
of a particle in a central force field. In many cases there is a force center, and
the motion in the plane suggests the use of plane polar coordinates10 (with
the origin in the force center).
2.2.1 Fixed and Moving Basis
The transformation between plane polar coordinates and cartesian coordi-
nates is
x = r cosϕ
y = r sinϕ.
10 For the extension to spherical polar coordinates and to cylindrical coordinates in
the three-dimensional case see Appendix C.
42 2 Newtonian Mechanics: First Applications
ey
r
er
ex
eϕ
ϕ
eϕ
ϕ
ey
er
ex
r
y
X
Fig. 2.6. The fixed (ex, ey) and moving (er, eϕ) basis. The trajectory r(t) is
indicated by the broken line
In addition, it is practical to introduce a fixed (cartesian) and a moving basis
(with radial and tangential unit vectors er and eϕ, respectively). (The basis
vectors point into the direction of increasing coordinates!) The position vector
can be written in the alternative forms
r = xex + y ey
r = r er
as shown in the left part of Fig. 2.6. The relation between the basis vectors is
(
er
eϕ
)
=
(
cosϕ sinϕ
− sinϕ cosϕ
)(
ex
ey
)
, (2.15)
see the left part of Fig. 2.6, with the inverse relation
(
ex
ey
)
=
(
cosϕ sinϕ
− sinϕ cosϕ
)−1(
er
eϕ
)
=
(
cosϕ − sinϕ
sinϕ cosϕ
)(
er
eϕ
)
.
The matrix of the inverse transformation is thus the transpose of the original
transformation matrix, since the transformation is a unitary transformation
(since the determinant of the transformation matrix is equal to unity).
Representation of a Vector in the Fixed and Moving Basis
Analogously to the vector r of the position other vectors can be represented
in the fixed and in the moving basis. An example of the vector F of the force
is given in Fig. 2.7.
2.2.2 Time Dependence of the Moving Basis
The basis vectors of the moving basis are varying with time (except for a
purely radial motion); for the derivatives with respect to time11 (Fig. 2.8) one
obtains
11 For the space derivatives see Appendix D.
2.2 Motion in a Plane 43
Fig. 2.7. The decomposition of the force F (r) in the (left) fixed and (right) moving
basis
ėr = ϕ̇eϕ
ėϕ = −ϕ̇er.
(2.16)
Proof:
(1) Intuitive derivation: The derivative of the radial unit vector with respect
to time is
ėr = lim
Δt→ 0
Δer
Δt
, Δer = er(t+ Δt) − er(t).
The arc length is
|Δer| = Δϕ |er| = Δϕ,
and the direction for |Δer| � 1 (Δϕ � π) is
Δer ‖ eϕ,
thus
ėr = lim
Δt→ 0
Δer
Δt
=
der
dt
=
eϕ dϕ
dt
= ϕ̇eϕ.
Analogously one finds ėϕ.
er(t)
Dϕ
er (t+Dt) = er¢
Dϕ
Der¢
¢
eϕ
er
eϕ
er
D e ϕ
Fig. 2.8. The time dependence of the moving basis: In the left panel the trajectory
of the particle is indicated by the broken line; also shown are the basis vectors at two
different times. In the right panel the same basis vectors are shown on an expanded
scale
44 2 Newtonian Mechanics: First Applications
ϕ
vr
Û
vÛ
vϕ
Û
rÛ
Fig. 2.9. The decomposition of the velocity v in its radial and angular part. The
trajectory r(t) is indicated by the solid broken line
(2) Formal way: From
(
er
eϕ
)
=
(
cosϕ sinϕ
− sinϕ cosϕ
)(
ex
ey
)
one finds
(
ėr
ėϕ
)
= ϕ̇
(
− sinϕ cosϕ
− cosϕ − sinϕ
)(
ex
ey
)= ϕ̇
(
− sinϕ cosϕ
− cosϕ − sinϕ
)(
cosϕ − sinϕ
sinϕ cosϕ
)(
er
eϕ
)
= ϕ̇
(
0 1
−1 0
)(
er
eϕ
)
= ϕ̇
(
eϕ
−er
)
.
�	
2.2.3 Velocity, Acceleration, etc. in Plane Polar Coordinates
Position, Velocity, Acceleration
In addition to the position, also the velocity, acceleration, etc. can now be
decomposed into radial and tangential components. The position vector has,
of course, only a radial component,
r = r er.
For the velocity one obtains
ṙ = ṙ er + r ėr
(2.16)
= ṙ er + r ϕ̇eϕ
= vr er + vϕ eϕ
(2.17)
2.2 Motion in a Plane 45
(the first term of the result is the component of the velocity in the direction
of the position vector, and the second term is the component perpendicular
to it), see Fig. 2.9. From this one obtains with er · eϕ = 0 the kinetic energy
T = 1
2mṙ2 = 1
2m
(
v2
r + v2
ϕ
)
= 1
2m
(
ṙ2 + r2ϕ̇2
)
. (2.18)
For the acceleration one obtains analogously with (2.16) and (2.17)
r̈ = (r̈ er + ṙ ėr) + (ṙ ϕ̇eϕ + r ϕ̈eϕ + r ϕ̇ ėϕ)
= r̈ er + ṙ ϕ̇ eϕ + ṙ ϕ̇ eϕ + r ϕ̈ eϕ − r ϕ̇ ϕ̇er
=
(
r̈ − r ϕ̇2
)
er + (r ϕ̈+ 2 ṙ ϕ̇) eϕ (2.19)
= br er + bϕ eϕ.
The four terms in (2.19) are the radial, centripetal, angular and (negative)
Coriolis acceleration (of the particle).
Inertial Forces
The centrifugal force mrϕ̇2 and the Coriolis force −2mṙϕ̇ are also called
inertial forces,12 which appear in a moving (here rotating) system, even if the
particle is at rest in this system and even if there are no external forces acting.
Momentum and Angular Momentum
From the velocity one obtains the momentum
p = m ṙ = m (ṙ er + r ϕ̇eϕ)
and the angular momentum (with respect to the coordinate origin)
l = r × p = m (r × v) = mr er × (ṙ er + rϕ̇ eϕ)
= m
(
rṙ er × er + r2ϕ̇ er × eϕ
)
= 0 +mr2ϕ̇eω = mr2 ω = θω
with the angular velocity ϕ̇ = ω and the vector ω = ω eω of the angular
velocity13 as well as the moment of inertia θ = mr2 of a point mass m at a
distance r from the axis of rotation.14
12 More on inertial forces further below in Sect. 7 on moving coordinate systems.
13 Compare the infinitesimal vector in (1.36).
14 More on moments of inertia and time dependent axes of rotation in Chap. 8 on
rigid bodies.
46 2 Newtonian Mechanics: First Applications
Radial and Circular Motion
Special Case: Radial motion: ϕ = const, ϕ̇ = 0. One obtains
r = r er
v = ṙ = ṙ er
a = r̈ = r̈ er
p = m ṙ = mṙ er
l = r × p = 0
T = 1
2mṙ2 = 1
2mṙ
2.
Special Case: Circular motion r = const, ṙ = 0. One obtains
r = r er
v = ṙ = rϕ̇ eϕ
a = r̈ = −rϕ̇2 er + rϕ̈ eϕ
p = mṙ = mrϕ̇ eϕ = mvϕ eϕ
l = r × p = mr2ϕ̇eω = mr2ω
T = 1
2mṙ2 = 1
2mr
2ϕ̇2.
2.2.4 Example: Plane Pendulum
Again this example:15 In the homogeneous gravitational field the pendulum
mass is subject to the gravitational force mg and to the tension S by the
pendulum rod (assumed as massless) (Fig. 2.10),
F = mg + S.
The equation of motion
F = m r̈
can be decomposed into the radial and tangential components,
Fr = F · er = mg cosϕ− S = m
(
r̈ − r ϕ̇2
)
= m
(
0 − lϕ̇2
)
= −ml ϕ̇2
Fϕ = F · eϕ = −mg sinϕ = m (rϕ̈+ 2ṙϕ̇)
= m (lϕ̈+ 0) = mlϕ̈
with the radial and tangential component of the acceleration from (2.19).
(Because of the appearance of the tangential component Fϕ the force is not
a central force.) The second equation is a differential equation for ϕ,
15 See also Sect. 2.1.5.
2.2 Motion in a Plane 47
ϕ l
S
m
mg
mg cos ϕ
mg sin ϕ
Fig. 2.10. The plane pendulum, cf. Fig. 2.3
ϕ̈ = −g
l
sinϕ, (2.20)
cf. Sect. 2.1.5. The first equation is an algebraic equation for the stress S,
S = m
(
g cosϕ+ l ϕ̇2
)
, (2.21)
which is compensated by the sum of the radial component of the fugal force
and of the gravitational force (both oriented outward). In this form one obtains
the stress as a function of ϕ and ϕ̇.
One can also express ϕ̇ by other quantities, e.g., by the energy or by the
amplitude, see problem 2.8.
Small Displacements
For small displacements,
sinϕ ≈ ϕ,
the differential (2.20) for ϕ takes the form
ϕ̈+
g
l
ϕ = 0.
The eigen solutions are
ϕ±(t) ∝ e±iωt
with
ω2 =
g
l
.
The general solution is the superposition of the eigen solutions,
ϕ(t) = a+eiωt + a+e−iωt
48 2 Newtonian Mechanics: First Applications
With the (initial) conditions (by choice)
ϕ(t = 0) = 0, ϕmax = ϕ0
the solution is
ϕ = ϕ0 sin(ωt).
The stress S as a function of time t can now be obtained from the known
solution for ϕ(t),
S
(2.21)
= m
(
g cosϕ+ lϕ̇2
)
≈ m
[
g
(
1 − 1
2
ϕ2
)
+ lϕ̇2
]
= mg
[
1 − 1
2
ϕ2
0 sin2(ωt) +
l
g
ω2ϕ2
0 cos2(ωt)
]
= mg
[
1 + ϕ2
0 −
3
2
ϕ2
0 sin2(ωt)
]
with ω2 = g/l and cos2(ωt) = 1 − sin2(ωt).
2.3 Two-Particle Systems
Now we leave the one-particle systems. For the following it is essential that
the interaction energy between the two particles is assumed to depend only
upon the relative coordinate,
F (r1, r2) = F (r1 − r2),
(and not in a more complex way upon the individual coordinates of the two
particles). Because of the homogeneity of space this condition is practically
always fulfilled, and with additional isotropy one even has
F (r1, r2) = F (r1 − r2) = F (|r1 − r2|)
as in the case of the gravitational or Coulomb interaction. Then the two-
particle problem can be reduced to two one-particle problems.
2.3.1 Center-of-Mass and Relative Coordinates
Definition 16. (see Fig. 2.11)
Center-of-mass coordinate R =
m1r1 +m2r2
m1 +m2
Relative coordinate r = r1 − r2
Total mass M = m1 +m2
Reduced mass μ
1
μ
=
1
m1
+
1
m2
.
2.3 Two-Particle Systems 49
Rr 2
r1
r
Fig. 2.11. The center-of-mass and relative coordinates R and r, respectively
The inverse relations are
r1 = R +
μ
m1
r
r2 = R − μ
m2
r.
2.3.2 Equations of Motion
The equations of motion, expressed in terms of the coordinates of the two
particles, are
m1 r̈1 = F 1(r1) + F 12(r1 − r2) (2.22)
m2 r̈2 = F 2(r2) + F 21(r1 − r2) (2.23)
with the external forces F i and the internal forces F ij . If one introduces
center-of-mass and relative coordinates, one obtains the equations of motion,16
M R̈ = F 1(r,R) + F 2(r,R) (2.24)
μ r̈ = F 12(r) +
(
μ
m1
F 1(r,R) − μ
m2
F 2(r,R)
)
. (2.25)
Proof: For (2.24): Adding the two equations (2.22) and (2.23) yields
(2.22) + (2.23)
= m1 r̈1 +m2 r̈2 =
d2
dt2
(m1r1 +m2r2) =
d2
dt2
M R = M R̈
= F 1 + F 2 + F 12 + F 21 = F 1 + F 2.
For (2.25): The (weighted) difference of the two equations (2.22) and (2.23)
yields
μ
m1
(2.22) − μ
m2
(2.23) = μ (r̈1 − r̈2) = μr̈
=
μ
m1
F 1 −
μ
m2
F 2 +
μ
m1
F 12 −
μ
m2
F 21
=
μ
m1
F 1 −
μ
m2
F 2 + μ
(
1
m1
+
1
m2
)
F 12
=
μ
m1
F 1 −
μ
m2
F 2 + F 12 . �	
16 Equation (2.24) is identical to (1.21).
50 2 Newtonian Mechanics: First Applications
Comment: The equations of motion (2.22) and (2.23) for r1 and r2 are
coupled through the inner forces F ij , and the equations of motion (2.24) and
(2.25) for r and R are coupled only through the external forces F i; in a closed
system (i.e., without external forces) or in a system with constant external
forces the latter equations are thus decoupled; this is the reason for the trans-
formation from single-particle to center-of-mass and relative coordinates.
2.3.3 Closed Systems
In a closed system (F 1 = 0, F 2 = 0) (2.24) and (2.25) are simplified to the
form
M R̈ = 0
μ r̈ = F 12(r).
The center-of-mass and relative motions are decoupled: The center-of-mass
motion is force-free; the relative motion is determined by the relative force
F 12. The relative motion is the same as that of one particle in the field of the
other (assumed as fixed); only the mass (m1 or m2) is to be substituted by the
reduced mass (μ).
Comments:
• Except for the influence of the mutual gravitational force the motion of
a planet around the sun is also subject to the forces of moons and other
planets in this example, but these latter forces are comparatively weak,
such that the planet and the sun can be considered as a closed system in
good approximation. On the other hand, the system of moon and earth
is not a closed system; because of nearly the same distance from the sunone has F 1/m1 ≈ F 2/m2, and even though the external forces in (2.25)
for the relative motion approximately vanish, the center-of-mass motion,
however, is subject to the influence of the force F , which the sun exerts
on the total mass of earth and moon.
• For the center-of-mass motion in the closed system, of course, the center-
of-mass energy, momentum, and angular momentum are conserved.17 Of
interest is thus the relative motion.
• In the limit m1 � m2 one obtains (in the case of vanishing external forces,
i.e., in a closed system)
R = r2
M = m2
μ = m1
17 See Sect. 1.6.
Summary: Newtonian Mechanics – First Applications 51
and
m2 R̈ = 0
m1 r̈1 = F 12.
The light particle moves in a field of the practically resting (or uniformly
moving) heavy mass. If one chooses an inertial system, in which one has
R = r2 = 0, then the second equation
m1r̈1 = F 12
has the form like the equation of motion (of particle 1) in an open system.
This is the basics of the dynamics of one-particle-systems in external fields,
like they are caused by sufficiently heavy systems, on which the dynamics
of the light mass practically does not act back on the heavy mass.
Summary: Newtonian Mechanics – First Applications
One-Dimensional Motion:
mẍ
(2.1)
= F (x, ẋ, t)
F (x, ẋ, t) = F (x) is always conservative:
V (x)
(2.2)
= −
∫
F (x) dx
with E = T + V .
Motion in a Plane (central force F (r) = F (r) er):
(
er
eϕ
)
(2.15)
=
(
cosϕ sinϕ
− sinϕ cosϕ
)(
ex
ey
)
.
d
dt
(
er
eϕ
)
(2.16)
= ϕ̇
(
eϕ
−er
)
r̈
(2.19)
=
(
r̈ − rϕ̇2
)
er + (rϕ̈ + 2ṙϕ̇)eϕ
= − 1
m
∇V = − 1
m
(
∂V
∂r
er +
1
r
∂V
∂ϕ
eϕ
)
52 2 Newtonian Mechanics: First Applications
Two-Body Problem:
m1 r̈1
(2.22)
= F 1(r1) + F 12(r1, r2)
m2 r̈2
(2.23)
= F 2(r2) + F 21(r1, r2)
M R̈
(2.24)
= F 1(r,R) + F 2(r,R)
μ r̈
(2.25)
= F 12(r) +
(
μ
m1
F 1(r,R) − μ
m2
F 2(r,R)
)
F i(ri) = 0 ⇒ free center-of-mass motion
⇒ relative motion in plane: μ r̈ = F 12(r).
Problems
2.1. Free Particle. Solving the equation of motion describe the motion of a
point mass without forces. Write down the solution for the initial conditions
r (t) = r0 and v (t) = v0 for t = 0, and illustrate the trajectory of the point
mass with a sketch. Determine the angular momentum. Write down the con-
served quantities.
Hint: Avoid a special choice of the reference point (namely that on the tra-
jectory).
2.2. Motion in a Homogeneous Gravitational Field. Determine the so-
lution of the equation of motion
mr̈ = mg
with the initial conditions r(0) = r0 and ṙ(0) = v0. Employ two different
methods:
(1) Decompose the equation of motion in cartesian coordinates.
(2) Use a coordinate-free procedure.
2.3. Freely Falling Body from Large Height. A body falls from large
height h perpendicularly to the earth’s surface (radius R of the earth) with
the force
F = −γ mM
r2
.
Determine the velocity upon hitting the earth. Compare the result with that
one which results for small height h � R.
Up to which height is the difference between these two calculations less
than 1%?
2.4. Vertical Throw, Gravitational Field. Discuss the vertical throw of
a mass m in the gravitational field of the earth
Problems 53
(a) Let the initial velocity in the throw of the mass from the earth’s surface
be v0. Determine the velocity v of mass as a function of the distance z
from the center of the earth.
(b) What is the minimum value of v0 such that the mass leaves the gravi-
tational range of the earth? (G = 6.67 × 10−11 Nm2 kg−2, M = 5.98 ×
1024 kg, R = 6.37 × 106 m)
2.5. The Inclined Throw. For an inclined throw, determine duration,
height, and range of the throw. Let the initial velocity be v0; the initial angle
be α.
2.6. Freely Falling Body in a Homogeneous Gravitational Field with
Friction. Investigate the motion of a body initially at rest under the influence
of the homogeneous gravitational field and of the Stokes and alternatively of
the Newton friction, and sketch the results. Discuss the result for large times,
and determine the correction to v(t) by the friction by the air if this is small.
2.7. Motion with Friction on an Inclined Plane. Investigate the mo-
tion of an initially resting body on an inclined plane as a function of the
inclination angle α (against the horizontal direction) under the influence of
the homogeneous gravitational fields and of the Coulomb friction (Fig. 2.12).
What changes, if one introduces an additional force on the body, acting in the
direction parallel or perpendicular to the inclined plane?
α
mg Fig. 2.12. Motion on an inclined plane
2.8. The Plane Pendulum of Sect. 2.2.4. Prove with the help of the energy
conservation theorem that the energy can be expressed by the amplitude ϕ0
of the vibration. Prove then that the stress in the hinge is given by
S = mg (3 cosϕ− 2 cosϕ0) .
2.9. Motion in a Central Force Field. A point mass moves in a central
force field
F = −m a
r3
er
with a > 0. Investigate the constants of the motion as well as the form of the
motion for large and small values of a.
54 2 Newtonian Mechanics: First Applications
2.10. Circular Trajectory and Potential. Consider a circular orbit r =
const. of a particle. Under which condition for the r-dependence of the poten-
tial energy V (r, ϕ) is this orbit a solution of the equation of motion?
2.11. Rocket in the Homogeneous Gravitational Field. A rocket starts
from ground vertically. The initial mass of the rocket is m0, the rate of mass
change is constant in time (ṁ = −α), and the exhaust velocity u0 of the
material from the engine relative to the rocket is kept constant in time.
(a) At which time tE does the engine stop burning fuel, if the final mass of
the rocket is mE?
(b) Which velocity relative to the earth does the exhaust matter have at
leaving the rocket, if the rocket moves with velocity v relative to the
earth?
(c) Set up the equation of motion of the rocket in an appropriate coordinate
system.
Hint: Formulation of the momentum conservation for the rocket and the
ejected mass.
(d) Which condition has to hold for the rocket to lift off at t = 0?
(e) Write down the velocity v(t) as well as its maximum value vmax, and sketch
v(t) in a graph for t ≤ tE .
(f) Write down the distance s(t) traveled by the rocket, and sketch its time
development for t ≤ tE .
2.12. Dynamics of the Universe. From the red shift of the light from
far-away galaxies one knows the expansion of the universe (Hubble law: the
change of the distance between two galaxies with time is proportional to this
distance) which – if it always existed – which has started with the Big Bang.
The expansion of the universe counteracts the gravitation.
(a) Give an expression for the equation of motion for a galaxy of mass m at
the distance R(t) from the origin. Rationalize that only the mass of the
galaxies inside of the sphere of radius R(t) act on m. Assume an average
mass density ρ(t). Write down the equation of motion for R(t) with the
assumption of the mass conservation in the universe.
(b) Reformulate the equation of motion such that one can notice a constant of
the motion. What is the meaning of this quantity, and by which quantity
is it determined?
(c) Solve the differential equation for R(t), where the integration constant
from b) occurs as a parameter (distinguish different cases; use an integral
table), and discuss the different cases of the solution by means of the
graphs R(t).
3
Lagrangian Mechanics
In this chapter an alternative formulation of the mechanics of point masses will
be treated, which will turn out to be of particular advantage, if the motion
is restricted by constraints. A first example for a motion with constraints
has been encountered in the context of the plane pendulum in Sect. 2.1.5:
The constraints restrict the configuration space. (In the example of the plane
pendulum the configuration space is no longer the R
3 but a circle, an R
1; the
configuration space of the spherical pendulum is the surface of a sphere, an
R
2.) Another example is the force which keepsa train on the rails.
In other cases it may be more difficult to determine the forces, which re-
strict the motion of one or several particles to the reduced configuration space.
In this chapter we will thus present the Lagrangian formulation of the equa-
tions of motion with implicit1 and explicit2 consideration of the constraints.
We will first consider the various forms of the constraint3 and then the forces
of constraint,4 which would be needed for the Newtonian dynamics. But we
will see that the need of these forces, which may be very difficult to obtain,5
can be circumvented.6
Like many other phenomena of physics the Lagrangian equations of motion
can be derived from an extremum principle.7
In the following we will continue to assume inertial reference frames as
postulated by Newton’s first law.
1 See Sect. 3.5.
2 See Sect. 3.9.
3 See Sect. 3.3.
4 See Sect. 3.4.
5 The stress in the pendulum rod has been determined only after the equation of
motion has been solved, see Sect. 2.2.4.
6 See Sect. 3.5.
7 See Sect. 3.10
56 3 Lagrangian Mechanics
3.1 Motion with Constraints
According to the second Newtonian axiom the motion of point masses is sub-
ject to the influence of forces,
mr̈i = F ext
i +
∑
j
F ij + Zi = F i + Zi. (3.1)
Even though in many cases the (internal and external) forces F i are known,
the so-called
forces of constraint Zi,
which force (“constrain”) the motion of the ith particle to a subspace of the
R
3, are not known a priory.
On the other hand, the constraints themselves are in most cases (as in the
example of the pendulum) very apparent and easy to formulate. In addition,
it may be of an advantage to employ coordinates other than cartesian coordi-
nates, e.g., polar coordinates as known from the treatment of the pendulum.
Often it is also of an advantage to use the (still to be developed) Lagrangian
equations of motion instead of the Newtonian equations of motion, and there
are very general prescriptions for setting up these Lagrangian equations of
motion; this makes the Lagrangian method often very much easier to handle
than the Newtonian method.
Three Good Reasons
Thus the motivation for the Lagrangian formulation will be firstly the easy
consideration of the constraints. Secondly, it will be of an advantage that
one will have to deal with scalars, while the Newtonian equations of motion
are vectorial relations. Finally, the form invariance8 of the Lagrangian equa-
tions of motion under coordinate transformation, i.e., the independence of
the equations of motion in various (also curvilinear) coordinate systems, is an
advantage.
Comment: The Newtonian equation of motion
mr̈ = F
is not form-invariant under coordinate transformation: In cartesian represen-
tation one has for the motion in a plane
mẍ = Fx
mÿ = Fy
8 See also Sect. 3.8.2 further below.
3.2 Constraints and Generalized Coordinates 57
and in plane polar coordinates
m
(
r̈ − rϕ̇2
)
= Fr
m (rϕ̈+ 2ṙϕ̇) = Fϕ
with the notation as in Fig. 2.7. Even though the Newtonian equations of
motion in abstract representation are form-invariant under Galilean transfor-
mations (from one inertial system to another), they have a different form in
the different representations. This will be not so in the Lagrangian equations
of motion.
3.2 Constraints and Generalized Coordinates
The example of the plane pendulum has shown that it may be more practical
to work with coordinates other than cartesian coordinates.
3.2.1 Examples of Constraints
If the motion of a particle is restricted to a sheet in R
3, then the coordinates
of the particle must obey the equation
G(r) = 0
describing this sheet.
As an example, the spherical pendulum moves on the surface of a sphere;
G(r) = x2 + y2 + z2 − l2 = 0 (cartesian coordinates)
G(r) = r2 − l2 (spherical coordinates).
This sheet may also be varying in time (e.g., a sheet fixed in the laboratory
but moving along with the laboratory on the rotating earth),
G(r, t) = 0.
The motion of a particle may also be restricted to a curve in space; the curve
is an intersection of two sheets, and in this case one has two constraints,
G1 = 0
G2 = 0,
e.g., for a plane pendulum (in the (x, z)-plane)
0 = G1 = y
0 = G2 = x2 + y2 + z2 − l2 = x2 + z2 − l2.
58 3 Lagrangian Mechanics
Other constraints cannot be expressed in the form of equations. If in the
case of the spherical pendulum the mass is not fixed to a (rigid) rod, but to
a (flexible) string, then the mass can reach every point in the interior of a
sphere,
x2 + y2 + z2 − l2 ≤ 0. (3.2)
Analogous relations hold for more particles, e.g., one has for a rigid dumb-bell
with two masses at a distance l
G(r1, r2) = (r1 − r2)
2 − l2 = (x1 − x2)
2 + (y1 − y2)
2 + (z1 − z2)
2 − l2 = 0.
3.2.2 Classification of Constraints
Consider a system of N particles with 3N coordinates and k independent
constraints of the form
3N∑
i=1
Gjidxi +Gj0 dt = 0 (j = 1, . . . , k) (3.3)
then the jth constraint is called
(a) Rheonomous (“running,” “flowing”), if it is varying in time,
Gj0 �= 0
(b) Scleronomous (“rigid,” “solid”), if it is independent of time,
Gj0 = 0
(c) Holonomic (“total”), if it can be written in integrated (not only in differ-
ential) form
Gj = cj = const,
i.e., if one has
Gji =
∂Gj
∂xi
, Gj0 =
∂Gj
∂t
.
The integrability condition (x0 ≡ t)
∂2Gj
∂xi∂xl
=
∂2Gj
∂xl∂xi
(i, l = 0, 1, . . . , 3N and j = 1, . . . , k)
thus reads
∂Gjl
∂xi
=
∂Gji
∂xl
.
(d) Nonholonomic, if the constraint cannot be written in the form as in case
(c). Inequalities like, e.g., (3.2) are nonholonomic constraints.
3.2 Constraints and Generalized Coordinates 59
Fig. 3.1. The plane pendulum
Fig. 3.2. The bead on the rotating rod
Examples:
For (b) and (c): Plane pendulum, see Fig. 3.1. The constraint is scleronomous
and holonomic, in cartesian coordinates
G = x2 + z2 − l2 = 0
⇒ 2xdx+ 2z dz = 0
or, in plane polar coordinates,
r − l = 0 ⇒ dr = 0.
For (a) and (c): Bead on a rod rotating uniformly (with the angular velocity
ω), see Fig. 3.2. In plane polar coordinates the angular dependence is given by
ϕ = ωt ⇒ dϕ− ω dt = 0
(in plane polar coordinates). The constraint is rheonomous and holonomic,
60 3 Lagrangian Mechanics
y
x
qj
Fig. 3.3. For the constraints of a skate
y − x tan(ωt) = 0
⇒ dy − tan(ωt) dx− xω
cos2(ωt)
dt = 0
(in cartesian coordinates).
For (d): Skate (ideally touching the ice at one point), see Fig. 3.3.9 The system
is not a point particle and has three degrees of freedom, the position (x, y)
and the orientation θ. The constraint is10
tan θ =
dy
dx
or
tan θ dx− dy = 0
or
0 = tan θ dx− 1d y + 0 dθ
= Gx dx+Gy dy +Gθ dθ.
This is not of the form
dG(x, y, θ) = 0
and, therefore, cannot be integrated to yield a form like G(x, y, θ) = const.
Namely, one has
9 A similar example is the wheel, see Sect. 3.9.3 below.
10 A rigid body has six degrees of freedom, three of the position and three of the
orientation; the trivial constraints, namely the restriction to the positions in the
plane and the restriction to one degree of freedom of the orientation are not
mentioned here explicitly.
3.2 Constraints and Generalized Coordinates 61
∂Gx
∂θ
=
∂2G
∂θ ∂x
=
d
dθ
tan θ �= 0
∂Gθ
∂x
=
∂2G
∂x∂θ
=
∂
∂x
0 = 0
and thus
∂Gx
∂θ
=
∂2G
∂θ ∂x
�= ∂2G
∂x∂θ
=
∂Gθ
∂x
.
The constraint is nonholonomic.
3.2.3 Degrees of Freedom
A system of N (point-like) particles without constraints has 3N degrees of
freedom, and the motion is described in a 3N -dimensional configuration space
and can thus be represented by 3N independent coordinates. Through k inde-
pendent constraints of the form (3.3) given above the number of the indepen-
dent degrees of freedom (the dimension of the configuration space) is reduced
by k and only is
f = 3N − k.
Example: Plane pendulum. A particle has 3 coordinates; the plane pendulum
has 2 constraints (motion in a plane, motion on a sphere) and, therefore, has
3 − 2 = 1 degree of freedom.
Comments:
• The nonholonomic constraint (3.2),
x2 + y2 + z2a string does not reduce the number of the degrees
of freedom in comparison to that of a free particle (but this condition
restricts the domain of definition of the coordinates).
• Of the 1
2N(N − 1) (pairwise) constraints
|ri − rj | = cij
for all interparticle distances of the N particles of a rigid body only 3N−6
are independent of each other: Given three distances, the position of a
particle in a rigid body is (nearly unambiguously) fixed. The number n =
3N − 1
2N(N − 1) would be negative for sufficiently large N :
N 2 3 4 5 6 7 8
n 5 6 6 5 3 0 −4
Therefore one has indeed f = 6 degrees of freedom, three each for the
positions and for the orientation.
62 3 Lagrangian Mechanics
r
x
y
ϕ
Fig. 3.4. Two different sets of coordinates, (x, y) and (r,ϕ), respectively, for the
description of the motion in a plane
3.2.4 Generalized Coordinates
In principle, instead of the 3N cartesian coordinates x1, . . . , x3N one could
employ any other coordinates q1, . . . , q3N for the description of the particle
positions
xi(t) = xi(q1, . . . , q3N , t) (i = 1, . . . , 3N),
in the example of the motion in the (x, y)-plane, e.g., the plane polar coordi-
nates (r, ϕ) instead of the cartesian coordinates (x, y), see Fig. 3.4.
Comment: For practical reasons one chooses (if possible at all) such coor-
dinates, in which the constraints
Gj(r1, . . . , rN , t) = Gj(r1(q1, . . . , q3N , t), . . . , t) = G̃j(q1, . . . , q3N , t) = 0
each describe a whole coordinate sheet in R
3N (i.e., a sheet qi = const.). Then
it is often very easy to find the 3N − k = f independent coordinates.
Example: The spherical pendulum. Transition from cartesian coordinates x,
y, and z to spherical polar coordinates r, ϑ, and ϕ leads to the constraint
0 = G(x, y, z) = x2 + y2 + z2 −R2
= r2 −R2 = G̃(r, ϑ, ϕ).
The mentioned coordinate sheet is the surface of the sphere r = R = const.
The independent coordinates are (q1, q2) = (ϑ, ϕ).
Definition 13. (Generalized coordinates): The (arbitrarily chosen) indepen-
dent coordinates q1, . . . , qf are denoted as the generalized11 coordinates.
3.2.5 Example: Plane Double Pendulum
The motion shall take place in the (y, z)-plane. The constraints are (in carte-
sian coordinates)
11 Very rarely the q1, . . . , q3N are denoted as generalized coordinates.
3.3 Forces of Constraint 63
z
ϑ1
ϑ1ϑ2
y
l1
l2
m1
m2
Fig. 3.5. The notations for the plane double pendulum
G1 = x1 = 0
G2 = x2 = 0
G3 = y2
1 + z2
1 = l21
G4 = (y2 − y1)
2 + (z2 − z1)
2 = l22.
The first two conditions describe the motion in the plane x = 0, and the last
two describe the fixed pendulum lengths. For practical reasons one chooses
ϑ1, ϑ2, as the two remaining independent variables (the generalized coordi-
nates), see Fig. 3.5.
Introduction of the generalized coordinates:
x1 = 0 x2 = 0
y1 = l1 sinϑ1 y2 = y1 + l2 sinϑ2 = l1 sinϑ1 + l2 sinϑ2
z1 = −l1 cosϑ1 z2 = z1 − l2 cosϑ2 = −l1 cosϑ1 − l2 cosϑ2
The equations of motion and their solution are treated in Sect. 3.5.7.
3.3 Forces of Constraint
If a particle moves on a predetermined trajectory or sheet, then except for
the external and internal forces also further forces, the so-called forces of
constraint, must act on the particle, which force (“constrain”) it on this curve
or sheet. These forces of constraint themselves do not accelerate the particle(s)
along a curve or sheet. Also, the forces of constraint do not have a component
parallel to the respective curves or sheets and thus can act only perpendicular
to the respective curves or sheets. The constraint G = 0 describes a sheet in
coordination space, and the force of constraint has thus the direction of
∇G(r, t)
64 3 Lagrangian Mechanics
(for a single force of constraint). The force of constraint for a motion on a
sheet can then be written as
Z = λ∇G
with a still to be determined function λ, the so-called Lagrangian multiplier,
or, for a motion along the intersecting line of two sheets with constraints
G1 = 0, G2 = 0, as
Z = λ1∇G1 + λ2∇G2.
In general one has for several particles
Gj(r1, . . . , rN , t) = 0 (j = 1, . . . , k),
and the force of constraint on the ith particle is
Zi =
k∑
j=1
λj∇iGj =
k∑
j=1
λjGji (i = 1, . . . , 3N) (3.4)
with Gij from (3.3).
Comment: Frictional forces may be connected with the motion along a
sheet. Frictional forces are not forces of constraint. Like all other forces they
influence the trajectory of the particle (or of the particles), but they do not
force the particle(s) onto a fixed sheet or trajectory.
3.4 The Principle of d’Alembert
Even though the forces of constraint are not known because of the unknown
Lagrangian parameters, in this section we will derive some properties of the
forces of constraint which are essential for what follows. With the principle
of d’Alembert considered in this section or with the equivalent principle of
virtual work statements are made concerning the forces of constraint.
Comment: Later in this chapter we will derive the equations of motion from
the rather general Hamilton principle.12 In contrast to this latter principle,
the former principles are tailored for applications in mechanics.
3.4.1 Real and Virtual Displacements
Consider the motion of particles with constraints. In a time interval dt the
particle (index i) moves by the real “displacements” dri.
Definition 18. (Virtual Displacement): A virtual displacement δri is a dis-
placement with fixed time and consistent with the constraints.
12 See Sect. 3.10.
3.4 The Principle of d’Alembert 65
t = const
t1
t2
δr dr
Fig. 3.6. Real trajectory (thick line) and varied trajectories (thin line) in an R
2
with a real displacement dr and a virtual displacement δr
q(t)+δq(t)
tt2t1
q
q(t)δq
Fig. 3.7. Time dependence of a “real” coordinate q and of the virtually displaced
coordinate q + δq
Comments:
• The virtual displacement δri with fixed time is in contrast to the real
displacement dri along the real trajectory.
• The virtual displacements are thus tangential vectors in configuration
space. The vectors δri thus point at positions on different, geometrically
possible trajectories of the ith particle at a given time. For example, a
given trajectory can be realized by a given initial conditions, but δri may
also point at other imaginative trajectories, see Figs. 3.6 and 3.7.
Example: Bead on a Rotating Rod
At every instance, the virtual displacements δr are all oriented along the rod,
which has a given orientation at a given time. In general, the real displace-
ments dr are not oriented along the rod, since the rod rotates during the time
interval dt, see Fig. 3.8.
If one considers a particle and its different trajectories r(t, ε), “varied” by
a (small) parameter ε, with r(t, 0) = r(t), then
δr(t) = r(t, ε) − r(t, 0) = ε
dr
dε
∣∣∣
ε=0
+O(ε2)
is the virtual displacement, and
dr(t) = r(t+ dt) − r(t) = dt
dr
dt
+O(dt2)
is the real displacement.
66 3 Lagrangian Mechanics
dr
dr
Fig. 3.8. The real and the virtual displacements of a bead on the rotating rod with
the axis perpendicular to the paper plane
If one expresses the displacements by generalized coordinates ql, then one
has
ri = ri(q1, . . . , qf , t) with f = 3N − k,
and the infinitesimally small real and virtual displacements are
dri =
f∑
l=1
∂ri
∂ql
dql +
∂ri
∂t
dt
δri =
f∑
l=1
∂ri
∂ql
δql + 0 δt ≡ 0 (!).
3.4.2 The Principle of Virtual Work
Consider the motion of a particle with a constraint G = 0. The virtual dis-
placement δr is oriented tangential to the sheet described by the constraint,
the force of constraint Z ∝ ∇G is oriented perpendicular to the sheet G = 0;
thus one has
Z ⊥ δr ⇔ Z · δr = 0.
(In Fig. 3.8 the vector of the force of constraint is oriented in the plane of the
figure and perpendicular to the rod.) Thus the so-called virtual work of the
force of constraint vanishes.
In generalization of the preceding argument to N particles, the principle
of virtual work states that the forces of constraint do not perform any virtual
work:
N∑
i=1
Zi · δri = 0 (3.5)with the forces of constraint Zj of Sect. 3.3.
3.5 Lagrangian Equations of the Second Kind 67
Comments:
• Virtual work and real work:∑
iZi · δri is the virtual work of the forces of constraint∑
i F i · δri is the virtual work of the internal and external forces
−
∑
i ṗi · δri is the virtual work of the inertial forces∑
iZi · dri is the real work of the forces of constraint
• If the constraints are rheonomous, Zi = Zi(t), the forces of constraint can
perform real work,
∑
i
Zi · dri �= 0;
for example, the bead sliding on the rotating rod of Fig. 3.8 is accelerated
in the direction of the rod.
• For scleronomous constraints (for one particle) one has
dr ⊥ Z,
because not only δr but also dr is oriented tangentially to the sheet G = 0
fixed in time.
3.4.3 The Principle of d’Alembert
If one combines the Newtonian law
ṗi
(3.1)
= F i + Zi
with the principle of the virtual work
N∑
i=1
Zi · δri
(3.5)
= 0,
one obtains the equivalent principle of d’Alembert
N∑
i=1
(F i − ṗi) · δri = 0. (3.6)
3.5 Lagrangian Equations of the Second Kind
In this section finally the equations of motion with implicit13 consideration of
the constraints are set up.
13 Explicit consideration will be dealt with in Sect. 3.9.
68 3 Lagrangian Mechanics
3.5.1 External and Internal Forces
Starting from the principle of d’Alembert one can write the first term in
(3.6) as
N∑
i=1
F i · δri =
f∑
l=1
Ql δql
with the number f = 3N −k of the (independent) degrees of freedom and the
generalized forces (type I)14
Ql =
N∑
i=1
F i ·
∂ri
∂ql
(l = 1, . . . , f). (3.7)
Proof: via
δri =
f∑
l=1
∂ri
∂ql
δql. (3.8)
3.5.2 Conservative Forces
For conservative forces one has
F i = −∇iV,
and one obtains the generalized forces again as the (negative) derivative of
the potential, here, however, with respect to generalized coordinate,
Ql = −∂V
∂ql
. (3.9)
Proof:
Ql =
N∑
i=1
F i ·
∂ri
∂ql
= −
N∑
i=1
∇iV · ∂ri
∂ql
= −∂V
∂ql
.
�	
14 There is no standard definition of the generalized force; further customary def-
initions are given in Sects. 3.6 and 3.7 below. To differentiate between the dif-
ferently defined generalized forces they are characterized by an addition in this
book. While reading other text books one should look up the definitions.
3.5 Lagrangian Equations of the Second Kind 69
3.5.3 Inertial Forces
The second part in (3.6) of d’Alembert’s principle can be written as
∑
i
ṗi · δri =
∑
l
[
d
dt
(
∂T
∂q̇l
)
− ∂T
∂ql
]
δql (3.10)
with the kinetic energy15
T =
1
2
∑
i
miv
2
i with vi =
d
dt
ri(q1, . . . , qf , t).
Proof: Let ri = ri({ql}, t). Then one has
ṙi =
d
dt
ri =
∑
l
∂ri
∂ql
q̇l +
∂ri
∂t
(3.11)
and from this
∂ṙi
∂q̇l
=
∂ri
∂ql
. (3.12)
From
T =
∑
i
1
2m (ṙi)
2
follows
∑
l
[
d
dt
∂T
∂q̇l
]
δql =
∑
l
[
d
dt
(
∂
∂q̇l
∑
i
mi
2
ṙ2
i
)]
δql
=
∑
l
∑
i
mi
[
d
dt
(
ṙi ·
∂ṙi
∂q̇l
)]
δql
(3.12)
=
∑
i
∑
l
mi
[
d
dt
(
ṙi ·
∂ri
∂ql
)]
δql
=
∑
i
∑
l
mi
(
r̈i ·
∂ri
∂ql
+ ṙi ·
∂ṙi
∂ql
)
δql
(3.8)
=
∑
i
mi (r̈i · δri + ṙi · δṙi)
=
∑
i
(ṗi · δri + pi · δṙi)
15 T = 1
2
∑
i miv
2
i only holds in non-relativistic mechanics. In Special Relativity one
has
T =
m0c
2
√
1 − v2/c2
−m0c
2.
See the Course on Special Relativity theory.
70 3 Lagrangian Mechanics
and
∑
l
∂T
∂ql
δql =
∑
l
[
∂
∂ql
(∑
i
mi
2
ṙ2
i
)]
δql
=
∑
l
∑
i
miṙi ·
∂ṙi
∂ql
δql =
∑
i
pi · δṙi
as well as finally
∑
l
[
d
dt
(
∂T
∂q̇l
)
− ∂T
∂ql
]
δql =
∑
i
ṗi · δri.
�	
Comments:
• The two terms on the right side of (3.11) lead to the consequence that the
kinetic energy T can be expressed by the generalized coordinates and their
derivatives with respect to time,
T =
1
2
∑
ll′
μll′ q̇lq̇l′ +
∑
l
λlq̇l + κ (3.13)
with
μll′(q1, . . . , qf , t) =
3N∑
i=1
mi
∂xi
∂ql
∂xi
∂ql′
(3.14)
λl(q1, . . . , qf , t) =
∑
i
mi
∂xi
∂ql
∂xi
∂t
(3.15)
κ(q1, . . . , qf , t) =
1
2
∑
i
mi
(
∂xi
∂t
)2
. (3.16)
The kinetic energy T is thus of quadratic form in the velocities (in the q̇i),
and since the xi may depend nonlinearly upon the ql, the coefficients λ, μ
and κ in general depend upon the coordinates ql. For example the kinetic
energy in plane polar coordinates reads T = 1
2m(ṙ2 + r2ϕ̇2).
• One has λl = 0 and κ = 0 for scleronomous constraints.
• In contrast, for the bead on the uniformly rotating rod with rheonomous
constraint one has ϕ̇ = ω = const. and T = 1
2m(ṙ2 +r2ω2) (i.e., μ(r) = m,
λ(r) = 0, κ(r) = 1
2mr
2ω2).
3.5.4 Lagrangian Equations of Motion
Substituting in the principle of d’Alembert, one finds
3.5 Lagrangian Equations of the Second Kind 71
0
(3.6)
=
N∑
i=1
(ṗi − F i) · δri
(3.10)
=
f∑
l=1
[
d
dt
∂T
∂q̇l
− ∂T
∂ql
−Ql
]
δql (3.17)
with f = 3N−k. While in the case of existing constraints the ri and thus the
δri are not linearly independent, the δql are so (if the constraints are holo-
nomic). From the vanishing of expression (3.17) for arbitrary δql one concludes
that the expression in the square brackets must vanish,
d
dt
(
∂T
∂q̇l
)
− ∂T
∂ql
= Ql l = 1, . . . , f. (3.18)
This equation describes the equality of the inertial forces to the internal and
external forces.
Newtonian Equation of Motion as a Special Case
Only if constraints are missing, Zi = 0, all the δri are linearly independent.
Then the Newtonian equation of motion ṗi = F i of axiom II follows from the
left side of (3.17). (Alternatively one can use for example cartesian coordinates
xi, yi, and zi as generalized coordinates and combine the three components
of the coordinates of a particle to form the vector ri.)
Conservative Force as a Frequent Special Case
• For a conservative force there is a potential energy V (q1, . . . , qf ) with
Ql
(3.9)
= −∂V
∂ql
,
∂V
∂q̇l
= 0.
With the Lagrangian function L
L = T − V , T = T ({ql}, {q̇l}, t)
V = V ({ql})
(3.19)
one finds from (3.9) and (3.18)
d
dt
∂L
∂q̇l
− ∂L
∂ql
= 0. (3.20)
The equations (3.20), occasionally also (3.18), are denoted as the
Lagrangian equations of motion of the second kind
and sometimes as the
Euler–Lagrangian equations.16
16 In general, the Euler–Lagrangian equations are the result of a variational method,
see, e.g., Hamilton’s principle treated further below in Sect. 3.10.
72 3 Lagrangian Mechanics
• If one can separate the forces into conservative forces Ql or F i (which
can be derived from a potential V ) and non-conservative forces QRl or FR
i
(e.g., frictional forces), then one has
d
dt
∂T
∂q̇l
− ∂T
∂ql
= Ql +QRl
d
dt
∂L
∂q̇l
− ∂L
∂ql
= QRl
(3.21)
with
L = T − V
Ql = −∂V
∂ql
QRl =
∑
i
FR
i · ∂ri
∂ql
.
(3.22)
Form Invariance
The form of the Lagrangian equation of motion (3.18) or (3.20) is independent
of the special choice of the coordinate transformation xi(q1, . . . qf ), and thus
the form of (3.18) is independent of the choice of the generalized coordinates,
in contrast to the Newtonian equation of motion.17
3.5.5 Example: Plane Motion in a Central Potential
For practical reasons one chooses plane polar coordinates (in the plane z = 0)
as generalized coordinates. Then the kinetic and potential energies as well as
the Lagrangian function are
T = 1
2mṙ2 = 1
2m
(
ṙ2 + r2ϕ̇2
)
V = V (r)
L = T − V = 1
2m
(
ṙ2 + r2ϕ̇2
)
− V (r).
For the equations of motion one needs
∂L
∂r
= mrϕ̇2 − ∂V
∂r
∂L
∂ṙ
= mṙ
⇒ 0 =
d
dt
∂L
∂ṙ
− ∂L
∂r
= m
(
r̈ − rϕ̇2
)
+
∂V
∂r
17 See also the comment at the end of Sect. 3.1 and in particular Sect. 3.8.2.
3.5 Lagrangian Equations of the Second Kind 73
Fig. 3.9. The bead sliding on a uniformly rotating rod, cf. Fig. 3.2
∂L
∂ϕ
= 0
∂L
∂ϕ̇
= mr2ϕ̇
⇒ 0 =
d
dt
∂L
∂ϕ̇
− ∂L
∂ϕ
=
d
dt
(
mr2ϕ̇
)
+ 0 =
d
dt
l
with the angular momentum l as in Sect. 2.2.3.
3.5.6 Example: Bead Sliding on a Uniformly Rotating Rod
We repeat the example of Sect. 3.2.2 (Fig. 3.9). One has to deal with the
motion in a plane (z = 0). The (additional) constraint can be given most
easily in plane polar coordinates,
ϕ = ωt.
The (only) generalized coordinate isr. With this one obtains
V = 0
L = T = 1
2mṙ2 = 1
2m
(
ṙ2 + r2ϕ̇2
)
= 1
2m
(
ṙ2 + r2ω2
)
and
∂L
∂r
= mrω2
∂L
∂ṙ
= mṙ
⇒ 0 =
d
dt
∂L
∂ṙ
− ∂L
∂r
= m
(
r̈ − rω2
)
.
The equation of motion is thus
r̈ = ω2r.
For the solution of this equation see problem 3.10.
74 3 Lagrangian Mechanics
3.5.7 Example: Plane Double Pendulum
Here we repeat the double pendulum of Sect. 3.2.5. With the coordinates
x1 = 0 x2 = 0
y1 = l1 sinϑ1 y2 = y1 + l2 sinϑ2 = l1 sinϑ1 + l2 sinϑ2
z1 = −l1 cosϑ1 z2 = z1 − l2 cosϑ2 = −l1 cosϑ1 − l2 cosϑ2
one obtains the velocities (which one needs for the kinetic energy)
ẏ1 = l1ϑ̇1 cosϑ1
ż1 = l1ϑ̇1 sinϑ1
ẏ2 = l1ϑ̇1 cosϑ1 + l2ϑ̇2 cosϑ2
ż2 = l1ϑ̇1 sinϑ1 + l2ϑ̇2 sinϑ2
and thus
ṙ2
1 = l21ϑ̇
2
1
ṙ2
2 = l21ϑ̇
2
1 + l22ϑ̇
2
2 + 2l1l2ϑ̇1ϑ̇2 (cosϑ1 cosϑ2 + sinϑ1 sinϑ2)
= l21ϑ̇
2
1 + l22ϑ̇
2
2 + 2l1l2ϑ̇1ϑ̇2 cos(ϑ1 − ϑ2).
The energies are
V = g {m1l1(1 − cosϑ1) +m2[l1(1 − cosϑ1) + l2(1 − cosϑ2)]}
T = 1
2mṙ2
1 + 1
2mṙ2
2
= 1
2m1l
2
1ϑ̇
2
1 + 1
2m2[l21ϑ̇
2
1 + l22ϑ̇
2
2 + 2l1l2 cos(ϑ1 − ϑ2)ϑ̇1ϑ̇2].
The Lagrangian equations are obtained with
∂L
∂ϑ̇1
=
∂T
∂ϑ̇1
= m1l
2
1ϑ̇1 +m2[l21ϑ̇1 + l1l2 cos(ϑ1 − ϑ2)ϑ̇2]
∂L
∂ϑ̇2
= m2[l22ϑ̇2 + l1l2 cos(ϑ1 − ϑ2)ϑ̇1]
∂L
∂ϑ1
= −m2l1l2 sin(ϑ1 − ϑ2)ϑ̇1ϑ̇2 − g[m1l1 sinϑ1 +m2l1 sinϑ1]
∂L
∂ϑ2
= +m2l1l2 sin(ϑ1 − ϑ2)ϑ̇1ϑ̇2 − gm2l2 sinϑ2
d
dt
∂L
∂ϑ̇1
− ∂L
∂ϑ1
= m1l
2
1ϑ̈1 +m2[l21ϑ̈1 − l1l2 sin(ϑ1 − ϑ2)(ϑ̇1 − ϑ̇2)ϑ̇2
+l1l2 cos(ϑ1 − ϑ2)ϑ̈2]
+m2l1l2 sin(ϑ1 − ϑ2)ϑ̇1ϑ̇2
+g[m1l1 sinϑ1 +m2l1 sinϑ1]
3.5 Lagrangian Equations of the Second Kind 75
= (m1 +m2)l21ϑ̈1 +m2l1l2 cos(ϑ1 − ϑ2)ϑ̈2 +m2l1l2 sin(ϑ1 − ϑ2)ϑ̇2
2
+g(m1 +m2)l1 sinϑ1
d
dt
∂L
∂ϑ̇2
− ∂L
∂ϑ2
= m2[l22ϑ̈2 − l1l2 sin(ϑ1 − ϑ2)(ϑ̇1 − ϑ̇2)ϑ̇1 + l1l2 cos(ϑ1 − ϑ2)ϑ̈1]
−m2l1l2 sin(ϑ1 − ϑ2)ϑ̇1ϑ̇2 + gm2l2 sinϑ2
= m2l
2
2ϑ̈2 −m2l1l2 sin(ϑ1 − ϑ2)ϑ̇2
1 +m2l1l2 cos(ϑ1 − ϑ2)ϑ̈1
+gm2l2 sinϑ2.
Case of Small Displacements
For small displacements one obtains
0 = (m1 +m2)l21ϑ̈1 +m2l1l2ϑ̈2 + g(m1 +m2)l1ϑ1
0 = m2l
2
2ϑ̈2 +m2l1l2ϑ̈1 + gm2l2ϑ2.
One divides the first equation by (m1 +m2)l21 and the second by m2l1l2, and
with the abbreviations
μ =
m2
m1 +m2
and λ =
l2
l1
one can write the system of equations in the form
(
1 μλ
1 λ
)(
ϑ̈1
ϑ̈2
)
+
g
l1
(
1 0
0 1
)(
ϑ1
ϑ2
)
= 0.
With the ansatz
(
ϑ1
ϑ2
)
∝ e−iωt
one obtains
( g
l1
− ω2 −μλω2
−ω2 g
l1
− λω2
)(
ϑ1
ϑ2
)
= 0.
The solubility condition for this homogeneous system of equations is the van-
ishing of the secular determinant of the coefficient matrix,
(
g
l1
− ω2
)(
g
l1
− λω2
)
− ω4μλ = 0
⇒ ω4(λ− μλ) − ω2(λ + 1)
g
l1
+
(
g
l1
)2
= 0
⇒ ω4 − ω2λ+ 1
λ
1
1 − μ
g
l1
+
(
g
l1
)2 1
λ(1 − μ)
= 0
76 3 Lagrangian Mechanics
with the solution
ω2
± =
1
2
λ+ 1
λ
1
1 − μ
g
l1
(
1 ±
√
1 − 4(1 − μ)
λ
(λ + 1)2
)
.
Case of Equal Pendulum Lengths (and Small Displacements)
For the case of equal pendulum lengths, l1 = l2 = l,
λ = 1,
one obtains
(
g
l
− ω2
)2
= ω4μ ⇒ ω2 − g
l
= ±ω2√μ ⇒ ω2
± =
g
l
1
1 ∓√
μ
.
With the knowledge of the eigen values the eigen vectors can be determined:
one inserts the eigen values into the eigen value equation,
(
g
l
− ω2
±
)
ϑ1,± = μω2
±ϑ2,±
= ∓ω2
±
√
μϑ1,± = μω2
±ϑ2,±
ϑ1,±
ϑ2,±
= ∓√
μ.
The general solution is the superposition of the eigen solutions,
ϑi(t) = a+ϑi,+e−iω+t + a−ϑi,−e−iω−t + c.c.
Comments:
• Vibrations are treated in more detail in Chap. 4.
• At this point it shall be stressed that in an eigen vibration the two oscil-
lators vibrate with a fixed phase.
• The term abbreviated as “c.c.” denotes the complex-conjugate of the pre-
ceding expression(s).
3.5.8 Separable Systems
In general, the Lagrangian equations of motion are coupled differential equa-
tions for the coordinates q1, . . . qf . However, if there are two (or more) mu-
tually independent systems as sketched in Fig. 3.10, i.e., if the Lagrangian
function decomposes into two (or more) parts such that the coordinates and
their corresponding velocities are contained in separate parts,
L(q1, q̇1, . . . , qf , q̇f , t) = L1(q1, q̇1, . . . , qμ, q̇μ, t)
+ L2(qμ+1, q̇μ+1, . . . , qf , q̇f , t),
3.5 Lagrangian Equations of the Second Kind 77
Fig. 3.10. The system of the particles i = 1, . . . , μ is separated from the system of
the particles i = μ+ 1, . . . , N
i.e., if one can separate the coordinates (here the coordinates q1, . . . , qμ from
the coordinates qμ+1, . . . , qf ), then the time dependence of the coordinates
q1, . . . , qμ is independent of that of the coordinates qμ+1, . . . , qf . Then there
are thus two (or more) decoupled sets of (coupled) differential equations. This
follows immediately from
0 =
d
dt
∂L
∂q̇l
− ∂L
∂ql
=
d
dt
∂L1
∂q̇l
− ∂L1
∂ql
(l = 1, . . . , μ)
and correspondingly for the other coordinates qj (with j = μ+1, . . . , f). Thus,
the equation of motion for q1 contains also the coordinates q2, . . . , qμ, but not
the coordinates qμ+1, . . . , qf .
Example: Relative and Center-of-Mass Coordinates
For two particles without external forces the relative and center-of-mass mo-
tion can be separated from each other: While one has, e.g., in the picture of
the individual particles the (nonseparable) Lagrangian function
L(r1, r2) =
ṙ2
1
2m1
+
ṙ2
2
2m2
− V (r1 − r2), (3.23)
one has in the picture of the relative and center-of-mass motion the (separable)
Lagrangian function
L(R, r) =
Ṙ
2
2M
+
ṙ2
2μ
− V (r) = L1(R) + L2(r). (3.24)
While the equations of motion for the coordinates r1 and r2 following from the
Lagrangian function (3.23) are coupled, the equations of motion for R and r
following from the Lagrangian function (3.24) are decoupled (separated) from
each other.
78 3 Lagrangian Mechanics
3.6 Generalized Momentum, Force, etc.
The conjugate coordinates ql and momenta pl introduced in the following
play a central role in the further development of theoretical physics, in partic-
ular in the Hamiltonian mechanics and in the commutation rules of quantum
mechanics.
3.6.1 Generalized Force and Generalized Momentum
So far, the generalized quantities
ql generalized coordinates
q̇l generalized velocities
have been introduced. In the following, additional generalized quantities will
be defined.
Definition 19. (Canonical momentum): The generalized (also: canonical or
conjugate18) momentum is
pl =
∂L
∂q̇l
. (3.25)
Definition 20. (Generalized force II): The generalized force (type II)19 is
Fl =
∂L
∂ql
.
Comments:
• The generalized force Fl of type II is distinguished from the force Ql
(3.9)
=
−∂V/∂ql of type I by
Fl = Ql +
∂T
∂ql
.
• With these generalized quantities one can write the Lagrangian equations
d
dt
∂L
∂q̇l
=
∂L
∂ql
+QRl (3.26)
in a form similar to the Newtonian equations as
ṗl = Fl +QRl . (3.27)
18 Conjugate to the generalized coordinate ql
19 See footnote 14 in Sect. 3.5.1: This notation is not customary in general.
3.6 Generalized Momentum, Force, etc. 79
• For velocity independent potentials20 one has
∂V
∂q̇l
= 0 ⇔ ∂L
∂q̇l
=
∂T
∂q̇l
,
and in systems without constraints one can use the cartesian coordinates as
the generalized coordinates, ql = xl, and from (3.26) or (3.27) one imme-
diately obtains the Newtonian law of motion in cartesian representation.
• Of the inertial forces21 appearing in rotating coordinate systems the cen-
trifugal force is contained in the generalized force (type II), the Coriolis
force in ṗϕ.
Example: Plane Motion in a Central Potential
The potential V (r) depends only upon the coordinate r, but not upon the
velocities ṙ, ϕ̇ (and also not upon ϕ). The generalized momenta are
pr =
∂L
∂ṙ
=
∂T
∂ṙ
= mṙ
pϕ =
∂L
∂ϕ̇
=
∂T
∂ϕ̇
= mr2ϕ̇
and represent the component of the momentum in the radial direction and
the angular momentum, respectively. The generalized forces of type I are
Qr = −∂V
∂r
= −V ′
Qϕ = −∂V
∂ϕ
= 0.
With the kinetic energy
T = 1
2m
(
ṙ2 + r2ϕ̇2
)
the difference between the generalized forces of type I and type II is
Fr −Qr =
∂T
∂r
= mrϕ̇2
Fϕ −Qϕ =
∂T
∂ϕ
= 0.
20 For velocity dependent potentials seespecial case, however, which can be treated in the context of this Course,
are extended rigid or (slightly) deformable solid bodies.
Regensburg, Dieter Strauch
January 2009
5 Electromagnetism is likewise a field theory, see the Course on Electromagnetism.
6 See the Course on Thermodynamics and Statistics.
Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
1 The Newtonian Mechanics of Point-Mass Systems:
General Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Point Masses, Open and Closed Systems . . . . . . . . . . . . . . . . . . . . 1
1.1.1 The Point Mass as an Idealization . . . . . . . . . . . . . . . . . . . 1
1.1.2 Open and Closed Systems, Internal and External Forces 2
1.2 Newton’s Axioms (1687): A New Era . . . . . . . . . . . . . . . . . . . . . . 4
1.3 Einstein’s Equivalence Principle (1916): Another New Era . . . . 8
1.3.1 Mass, Weight, Force, and the Like . . . . . . . . . . . . . . . . . . . 8
1.3.2 Einstein’s Equivalence Principle . . . . . . . . . . . . . . . . . . . . . 9
1.4 Types of Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4.1 Central Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4.2 Conservative Forces (I) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4.3 Central Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.4.4 Potential and Potential Energy . . . . . . . . . . . . . . . . . . . . . . 12
1.4.5 The Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.4.6 Frictional Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.5 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5.2 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.6 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.6.1 Conservation of Momentum. . . . . . . . . . . . . . . . . . . . . . . . . 17
1.6.2 Angular-Momentum Conservation and Central Forces . . 18
1.7 Energy-Conservation Theorem and Conservative Forces . . . . . . 18
1.7.1 Conservative Forces (II) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.7.2 Potential Energy, Work, and Power . . . . . . . . . . . . . . . . . . 19
1.7.3 Energy Conservation Theorem . . . . . . . . . . . . . . . . . . . . . . 20
1.8 Invariances and Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . 20
1.8.1 Translations and Rotations . . . . . . . . . . . . . . . . . . . . . . . . . 20
x Contents
1.8.2 Invariances and Conservation Laws:
Noether’s Theorem (I) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.8.3 The Usefulness of the Conservation Laws . . . . . . . . . . . . . 23
1.8.4 The Usefulness of Symmetry Analysis . . . . . . . . . . . . . . . . 23
1.9 * Virial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.10 The Basic Problem of the Mechanics of a Point Mass . . . . . . . . 25
Summary: Newtonian Mechanics – General Properties . . . . . . . . . . . . 25
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2 Newtonian Mechanics: First Applications . . . . . . . . . . . . . . . . . . 29
2.1 One-Dimensional Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.1.1 Constant Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.1.2 Time-Dependent Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.1.3 Velocity-Dependent Force . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.1.4 Coordinate-Dependent Force . . . . . . . . . . . . . . . . . . . . . . . . 32
2.1.5 Example: Plane Pendulum (with Small Amplitude) . . . . 34
2.1.6 * Example: Plane Pendulum with Large Amplitude . . . . 38
2.2 Motion in a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
2.2.1 Fixed and Moving Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
2.2.2 Time Dependence of the Moving Basis . . . . . . . . . . . . . . . 42
2.2.3 Velocity, Acceleration, etc. in Plane Polar Coordinates . 44
2.2.4 Example: Plane Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . 46
2.3 Two-Particle Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
2.3.1 Center-of-Mass and Relative Coordinates . . . . . . . . . . . . . 48
2.3.2 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
2.3.3 Closed Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Summary: Newtonian Mechanics – First Applications . . . . . . . . . . . . . 51
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3 Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
3.1 Motion with Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.2 Constraints and Generalized Coordinates . . . . . . . . . . . . . . . . . . . 57
3.2.1 Examples of Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
3.2.2 Classification of Constraints . . . . . . . . . . . . . . . . . . . . . . . . 58
3.2.3 Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
3.2.4 Generalized Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
3.2.5 Example: Plane Double Pendulum . . . . . . . . . . . . . . . . . . . 62
3.3 Forces of Constraint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
3.4 The Principle of d’Alembert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.4.1 Real and Virtual Displacements . . . . . . . . . . . . . . . . . . . . . 64
3.4.2 The Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . 66
3.4.3 The Principle of d’Alembert . . . . . . . . . . . . . . . . . . . . . . . . 67
3.5 Lagrangian Equations of the Second Kind . . . . . . . . . . . . . . . . . . 67
3.5.1 External and Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . 68
3.5.2 Conservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Contents xi
3.5.3 Inertial Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
3.5.4 Lagrangian Equations of Motion . . . . . . . . . . . . . . . . . . . . 70
3.5.5 Example: Plane Motion in a Central Potential . . . . . . . . 72
3.5.6 Example: Bead Sliding on a Uniformly
Rotating Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
3.5.7 Example: Plane Double Pendulum . . . . . . . . . . . . . . . . . . . 74
3.5.8 Separable Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.6 Generalized Momentum, Force, etc. . . . . . . . . . . . . . . . . . . . . . . . . 78
3.6.1 Generalized Force and Generalized Momentum . . . . . . . . 78
3.7 Velocity-Dependent Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
3.7.1 Generalized Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
3.7.2 Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
3.8 Gauge Invariance and Form Invariance . . . . . . . . . . . . . . . . . . . . . 81
3.8.1 (Un)Ambiguity of the Lagrangian Function:
Gauge Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
3.8.2 * Form Invariance of the Lagrangian Equations
under Point Transformations . . . . . . . . . . . . . . . . . . . . . . . . 83
3.9 * Lagrangian Equations of the First Kind . . . . . . . . . . . . . . . . . . 84
3.9.1 Lagrange Multipliers . . . . . .Sect. 3.7 further below.
21 For the inertial forces and the Coriolis force see Sect. 7.2.4 further below.
80 3 Lagrangian Mechanics
3.7 Velocity-Dependent Forces
3.7.1 Generalized Potential
The equation of motion
QRl =
d
dt
∂L
∂q̇l
− ∂L
∂ql
with L = T − V ({ql})
for a particle under the influence of conservative and other forces, −∂V/∂ql
and QRl , respectively, suggest the question whether some of these other forces
QRl can be described by a generalized potential
V R({ql}, {q̇l}, t);
the generalized forces (type III)22 are then
QRl =
d
dt
∂V R
∂q̇l
− ∂V R
∂ql
. (3.28)
Comment: For a velocity independent potential one obtains from (3.28) the
generalized force of type I.
3.7.2 Lorentz Force
The definition (3.28) would be superfluous, if there were no application,
namely, e.g., that to the Lorentz force: A particle with the mass m and the
charge23 q in an electromagnetic field is subject to the (velocity dependent)
Lorentz force24
F = q
(
E +
[c]
c
v × B
)
.
The Lorentz force is obtained as a generalized force from the generalized
potential25
V = qφ− q
[c]
c
A · v (3.29)
with the scalar potential φ(r, t) and the vector potential A(r, t) and the
fields26
22 See footnote 14 in Sect. 3.5.1.
23 Not to be confused with the generalized coordinate ql.
24 [x] = x in SI-units, [x] = 1 in Gauß units, cf. footnote 1.18.
25 The potential energy of a charge in the electromagnetic field is Epot = qφ+q [c]
c
A ·
v. Notice the different sign of the term with the vector potential A in the energy
and in (3.29).
26 See the Course on Electrodynamics.
3.8 Gauge Invariance and Form Invariance 81
E = −∇φ− [c]
c
Ȧ
B = ∇× A.
Proof: The force (type III) is given by (v = ṙ)
F
(3.28)
=
d
dt
∇ṙV −∇rV
=
d
dt
(
−q [c]
c
A
)
− q∇
(
φ− [c]
c
(A · v)
)
= −q [c]
c
(
∂A
∂t
+ ṙ · ∇A
)
− q∇φ+ q
[c]
c
[
v × (∇× A) + v · ∇A
]
= qE + q
[c]
c
v × B
with ∇(A · v)
(D.26)
= v × (∇× A) + v · ∇A. �	
For the particle (without constraints) in an electromagnetic field one obtains
the canonical (generalized) momentum p,
p =
∂L
∂v
= mv + q
[c]
c
A. (3.30)
Comments:
• For Lorentz (and other velocity dependent) forces one has thus to distin-
guish between
– The canonical momentum p = ∂L/∂v and
– The dynamical (kinetic) momentum π = mv = mṙ = p − q [c]
c A.
• The Lagrangian function
L =
m
2
v2 + q
[c]
c
A · v − qφ (3.31)
here has not the simple form L = Ekin − Epot.
• The Lagrangian function is not unique but subject to gauge transforma-
tions, see the following section. In the form given here the form is known
as the “minimal coupling.”
3.8 Gauge Invariance and Form Invariance
3.8.1 (Un)Ambiguity of the Lagrangian Function:
Gauge Transformation
As known from the Newtonian formalism the potential energy is determined
except for an additive constant and is thus not unique; likewise the Lagrangian
function L is not unique:
82 3 Lagrangian Mechanics
Proposition: Gauge invariance of the Lagrangian equations of mo-
tion (of the second kind):
Given two Lagrangian functions L and L′, which are connected to each other
by a so-called gauge transformation
L′ = L+
dF
dt
with
∂F
∂q̇l
= 0 (3.32)
with the gauge function of the form F = F ({ql}, t) (notice ∂F/∂q̇l = 0), thus
with
dF
dt
=
∂F
∂t
+
∑
l
∂F
∂ql
q̇l. (3.33)
Then, the functions L and L′ of (3.32) lead to the same equations of motion,
d
dt
(
∂L′
∂q̇l
)
− ∂L′
∂ql
=
d
dt
(
∂L
∂q̇l
)
− ∂L
∂ql
= 0
Proof: (for one coordinate q):
d
dt
(
∂L′
∂q̇
)
− ∂L′
∂q
=
d
dt
(
∂L
∂q̇
)
− ∂L
∂q
+X
with the additional term
X =
d
dt
(
∂
∂q̇
dF
dt
)
− ∂
∂q
dF
dt
,
which is to be shown to vanish. With
dF
dt
=
∂F
∂q
q̇ +
∂F
∂t
one obtains (∂F/∂q̇ = 0)
X =
d
dt
∂
∂q̇
dF
dt
− ∂
∂q
dF
dt
=
d
dt
∂
∂q̇
[
∂F
∂t
+
∂F
∂q
q̇
]
− ∂
∂q
dF
dt
=
d
dt
[
0 +
∂F
∂q
]
− ∂
∂q
dF
dt
=
d
dt
∂F
∂q
− ∂
∂q
dF
dt
= 0.
�	
Comments:
• With the gauge transformation (3.32) the canonical momentum transforms
according to
p′l = pl +
∂F
∂ql
, (3.34)
see problem 3.23.
3.8 Gauge Invariance and Form Invariance 83
3.8.2 * Form Invariance of the Lagrangian Equations
under Point Transformations
Let
qi = qi(q′j , t)
be a point transformation. In the form given here transformations to moving
coordinates are also included. With the equation of motion
d
dt
∂L
∂q̇i
− ∂L
∂qi
= 0
one has
d
dt
∂L′
∂q̇′j
− ∂L′
∂q′j
= 0,
where
L′(q′j , q̇
′
j , t) = L(qi(q′j , t), q̇i(q
′
j , q̇
′
j , t), t)
arises simply by inserting.27 Notice that with
q̇i =
dqi
dt
=
∂qi
∂q′j
q̇′j +
∂qi
∂t
the (generalized) velocity q̇i depends upon q′β, q̇
′
β , and t.
The Lagrangian equations of motion are thus form-invariant under point
transformations. In contrast, the Newtonian equations are not form-invariant
under transformations to curvilinear coordinates (and to noninertial reference
systems): Form invariance of the Newtonian equations would mean that
mẍi = Fi(xi) in the system S
and
mẍ′i = F ′
i (x
′
i) in the system S′
would be true; but, in fact, this is true only for a Galilean transformation of one
inertial system S to another inertial system S′ in the standard configuration.28
In addition, for the motion in a plane29 in a central potential V = α/r with
F = F (r)er, F (r) = −α/r2, one has
mẍ = α
x
(x2 + y2)
3
2
mÿ = α
y
(x2 + y2)
3
2
,
27 Without proof.
28 For details on coordinate transformations see Chap. 7.
29 See, for example, Sects. 2.2.4 or 5.1.
84 3 Lagrangian Mechanics
but
mr̈ =
α
r2
+mrϕ̇2
mϕ̈ = −2m
ṙϕ̇
r
.
3.9 * Lagrangian Equations of the First Kind
In the foregoing, the dependent coordinates (because of the constraints) and
along with it the forces of constraint have been eliminated from the equations
of motion. As a result, the Lagrangian equations of the second kind have been
obtained. Now these considerations shall be extended to the explicit treatment
of the dependent coordinates and of the forces of constraint.
One starts from
(1) The Newtonian equations of motion,
miẍα
(3.1)
= Fα + Zα α = 1, . . . , 3N. (3.35)
(2) The constraints (Gjα = ∂Gj/∂xα for holonomic constraints),
dGj
(3.3)
=
3N∑
α=1
Gjα dxα +Gj0 dt = 0 (3.36)
δGj =
3N∑
α=1
Gjα δxα = 0 j = 1, . . . , k. (3.37)
(3) The principle of the virtual work,
3N∑
α=1
Zα δxα
(3.5)
= 0. (3.38)
3.9.1 Lagrange Multipliers
Because of the k constraints only f = 3N − k of the coordinates xα of the
N particles are independent, and thus only f of the virtual displacements
δxα are linearly independent. The dependent displacements δxα can now be
eliminated. To this end one takes the difference of the equations (3.37) and
(3.38),
(3.38) −
k∑
j=1
λj(3.37) =
3N∑
α=1
(
Zα −
k∑
j=1
λjGjα
)
δxα = 0. (3.39)
3.9 * Lagrangian Equations of the First Kind 85
Comment: The λj are called Lagrange multipliers. They depend upon the
time, λj = λj(t), but are independent of the coordinates, ∂λj/∂xα = 0.
One divides the sum over α in (3.39) into a sum over the independent
coordinates (α = 1, . . . , f with f = 3N − k) and a second sum over the k
dependent coordinates (α = f + 1, . . . , 3N),
3N∑
α=1
(
Zα −
k∑
j=1
λjGjα
)
δxα =
f∑
α=1
(
Zα −
k∑
j=1
λjGjα
)
δxα
+
3N∑
α=f+1
(
Zα −
k∑
j=1
λjGjα
)
δxα. (3.40)
One chooses the Lagrange multipliers λj (j = 1, . . . , k) such that in the second
sum in (3.40) the summands (Zα−
∑k
j=1 λjGjα) vanish. These are k linear al-
gebraic equations for the k different Lagrange multipliers λj . In the remaining
first sum in (3.40) the virtual displacements δxα for α = 1, . . . , f are linearly
independent. From the linear independence of these δxα it follows then that
all summands of the first sum vanish. Thus one has for all α
Zα =
k∑
j=1
λjGjα α = 1, . . . , 3N, (3.41)
one part because of the choice of the λj , the other part because of the linear
independence of the virtual displacements δxα.
3.9.2 Lagrangian Equations of the First Kind
Instead of (3.35) and (3.36) one now has because of (3.41) the equations
∑
α
Gjα dxα +Gj0 dt = 0 j = 1, . . . , k (3.42)mαẍα = Fα +
k∑
j=1
λjGjα α = 1, . . . , 3N. (3.43)
These are 3N + k (differential) equations for the 3N + k unknown quantities
xα and λj .
If one rewrites (3.42) and (3.43) in terms of arbitrary, not necessarily inde-
pendent and not necessarily cartesian coordinates,30 one obtains, analogously
to the results of Sect. 3.5.4,
∑
l
G̃jl dql + G̃j0 dt = 0 j = 1, . . . , k (3.44)
30 See, e.g., Goldstein [1] Sects. 2–4.
86 3 Lagrangian Mechanics
d
dt
∂L
∂q̇l
− ∂L
∂ql
= QRl +
∑
j
λjG̃Jl l = 1, . . . , 3N (3.45)
with
xα = xα({ql}, t)
dxα =
∑
l
∂xα
∂ql
dql +
∂xα
∂t
dt
G̃jl =
∑
α
Gjα
∂xα
∂ql
G̃j0 = Gj0 +
∑
α
Gjα
∂xα
∂t
.
Comments:
• For scleronomous constraints one has
Gj0 = 0,
∂xα
∂t
= 0.
• For holonomic constraints one has
Gj({xα}, t) = 0 ⇔ dGj =
∑
α
∂Gj
∂xα
dxα +
∂Gj
∂t
dt, (3.46)
and (3.42) can be written in the form dGj = 0; thus one has in (3.43)
Gjα = ∂Gj/∂xα and Gj0 = ∂Gj/∂t, thus
mαẍα = Fα +
k∑
j=1
λj
∂Gj
∂xα
α = 1, . . . , 3N (3.47)
or, instead of (3.45),
d
dt
∂L
∂q̇l
− ∂L
∂ql
=
∑
j
λj
∂G̃j
∂ql
+QRl l = 1, . . . , 3N
with G̃j({ql}, t) = Gj({xα({ql}, t)}, t).
3.9.3 Example: Atwood Machine
Two masses m1 and m2 are connected to each other by a string of length l
strung over a (massless) pulley and can move along the vertical (z) direction,
see Fig. 3.11. The (holonomic) constraint reads
G = z1 + z2 − l = 0. (3.48)
3.9 * Lagrangian Equations of the First Kind 87
m1
m2
Fig. 3.11. The Atwood machine
The equations of motion (of the first kind) read
mαẍα = Fα + λGα,
or, with ∂G/∂zi = 1,
m1z̈1 = −m1g + λ
∂G
∂z1
= −m1g + λ (3.49)
m2z̈2 = −m2g + λ
∂G
∂z2
= −m2g + λ. (3.50)
In (3.49) λ is the force of constraint on m1, in (3.50) it is the force of constraint
on m2, i.e., in both cases the tension in the string.
For the solution of (3.48)–(3.50) one eliminates λ,
(3.49)− (3.50) = m1z̈1 −m2z̈2 = − (m1 −m2) g
m1
d2
dt2
(3.48) = m1z̈1 +m1z̈2 = 0.
Subtraction of the first from the second equation leads to
(m1 +m2) z̈2 = (m1 −m2) g
or
z̈2 =
m1 −m2
m1 +m2
g.
This can be integrated trivially; with z2(t0) = z2,0 and ż2(t0) = v2,0 one
obtains
z2(t) =
1
2
m1 −m2
m1 +m2
g (t− t0)
2 + v2,0 (t− t0) + z2,0.
The solution for z1 is
z1 = l− z2.
88 3 Lagrangian Mechanics
z
z–h
h x
x
α
Fig. 3.12. For the moving plane the angle α is fixed, and the height h(t) (at x = 0)
is a function of time
For the force of constraint one obtains from the equations (3.49) and (3.50),
respectively,
λ = m1 (z̈1 + g) = m2 (z̈2 + g)
= m2
[
m1 −m2
m1 +m2
+ 1
]
g
= 2
m1m2
m1 +m2
g.
The force of constraint has a positive z (upward) component.
Example: Point Mass on a Moving Inclined Plane
With the choice of the coordinate system as in Fig. 3.12 the constraint is
G = z − h(t) − x tanα = 0. (3.51)
The constraint is holonomic and rheonomous.31 With the Lagrangian function
L = 1
2m
(
ẋ2 + ẏ2 + ż2
)
−mgz
the Lagrangian equations of the first kind are
mẍ− 0 = λ
∂G
∂x
= −λ tanα (3.52)
mÿ − 0 = λ
∂G
∂y
= 0
mz̈ +mg = λ
∂G
∂z
= λ (3.53)
or
mr̈ = F + Z
31 For the treatment with the Lagrangian equations of the second kind see
problem 3.15.
3.9 * Lagrangian Equations of the First Kind 89
with the components
m
⎛
⎝
ẍ
ÿ
z̈
⎞
⎠ =
⎛
⎝
0
0
−mg
⎞
⎠+
⎛
⎝
−λ tanα
0
λ
⎞
⎠ .
The vector of the force of constraint is thus
Z = λ
⎛
⎝
− tanα
0
1
⎞
⎠ .
The equation ÿ = 0 can be integrated trivially and leads to uniform motion
in the y direction. There remain the three coupled equations for x, z, and λ
to be solved.
For the elimination of λ from the equations of motion one has
(3.52) ⇒ λ = −mẍ cotα
(3.53) ⇒ λ = m (z̈ + g)
or
ẍ cotα+ z̈ + g = 0
and from (3.51)
−G̈ = ẍ tanα− z̈ + ḧ = 0
These are two equations for x and z. Elimination of z̈ by adding these two
equations leads to
ẍ (cotα+ tanα) + g + ḧ = 0
or
ẍ = − g + ḧ
cotα+ tanα
= −
(
g + ḧ
)
sinα cosα.
Now this can be integrated for a given time dependence of h(t). Now one can
insert successively. One obtains z(t) from the constraint,
z = h+ x tanα,
and finally
λ = m (z̈ + g) = −mẍ cotα = +m
(
g + ḧ
)
cos2 α.
See also problem 3.15.
90 3 Lagrangian Mechanics
z
y
x
v
ϑ
ϕ
Fig. 3.13. The definition of the angles ϕ and ϑ for the wheel
Example: The Vertical Wheel
In this section the motion of a wheel shall be investigated. The wheel (with
the radius R) shall roll on a plane (the (x, y)-plane); an inclination of the
wheel shall be neglected. As in the example of the skate of Sect. 3.2.2 one has
to deal not with a system of point masses, but with an extended body, whose
properties will be treated in a more general context of rigid bodies further
below.32 With the notation as in Fig. 3.13 the generalized coordinates are x,
y, ϑ, and ϕ.
The constraint (the condition for rolling) is
ds = R dϕ
or, if one decomposes the path into its components,
0 = G1xdx+G1ϕdϕ = dx−R cosϑ dϕ (3.54)
0 = G2ydy +G2ϕdϕ = dy −R sinϑ dϕ. (3.55)
These are nonholonomic constraints as in the example of the skate in
Sect. 3.2.2. The Lagrangian function is
L = T = 1
2m
(
ẋ2 + ẏ2
)
+ 1
2Iϕϕ̇
2 + 1
2Iϑϑ̇
2
with the moments of inertia Iϕ and Iϑ = Iz. The Lagrangian equations of the
first kind are
0 =
d
dt
∂L
∂q̇l
− ∂L
∂ql
−
∑
j
λjGjl
q1 = x : 0 = mẍ− 0 − λ1
q2 = y : 0 = mÿ − 0 − λ2
q3 = ϕ : 0 = Iϕϕ̈− λ1 (−R cosϑ) − λ2 (−R sinϑ)
q4 = ϑ : 0 = Iϑϑ̈− 0 − 0.
32 See Chap. 8 further below.
3.9 * Lagrangian Equations of the First Kind 91
From the last equation (for q4 = ϑ) one obtains
ϑ̈ = 0
⇒ ϑ̇ = ϑ̇0 = Ω = const.
⇒ ϑ = Ωt+ ϑ0
with ϑ(t) = ϑ0 for t = 0.
The forces of constraint can be expressed by ϑ and ϕ: With the time
derivative of the constraints (3.54) and (3.55)
ẋ = Rϕ̇ cosϑ
ẏ = Rϕ̇ sinϑ ⇒ ẍ = Rϕ̈ cosϑ−Rϕ̇ϑ̇ sinϑ
ÿ = Rϕ̈ sinϑ+Rϕ̇ϑ̇ cosϑ
one obtains
λ1 = mẍ = mR
(
ϕ̈ cosϑ− ϕ̇ϑ̇ sinϑ
)
λ2 = mÿ = mR
(
ϕ̈ sinϑ+ ϕ̇ϑ̇ cosϑ
)
.
(3.56)
This one can insert into the equation for q3 = ϕ and obtains
0 = Iϕϕ̈+R (λ1 cosϑ+ λ2 sinϑ)
=
(
Iϕ +mR2
)
ϕ̈.
Because of Iϕ +mR2 �= 0 one obtains
ϕ̈ = 0 ⇒ ϕ̇ = ϕ̇0 = ω = const.
and with this from (3.56)
λ1 = −mRϕ̇0ϑ̇0 sinϑ
λ2 = mRϕ̇0ϑ̇0 cosϑ.
If one inserts this into (3.54) and (3.55) for the constraints, one finally obtains
ẋ = Rϕ̇0 cosϑ
ẏ = Rϕ̇0 sinϑ ⇒ v =
(
ẋ
ẏ
)
= Rϕ̇0
(
cosϑ
sinϑ
)
with
ϑ(t) = Ωt+ ϑ0 , ϕ̇0 = ω
and
Z =
(
Zx
Zy
)
=
(
λ1
λ2
)
= mRϕ̇0ϑ̇0
(
− sinϑ
cosϑ
)
.
92 3 Lagrangian Mechanics
z
ϑ0 > 0
Fig. 3.14. The direction of the force of constraint for the wheel
The direction of the force of constraint is perpendicular to the velocity, see
also Fig. 3.14. The force of constraint points into the direction of the wheel
axis, i.e., the torque for the rotation around the wheel axis vanishes. (⇒ ϕ̇ =
const.!) The motion of the wheel is such that the contact point runs around a
circular orbit. However, for this motion a torque arises, which would lead to
an inclination of the wheel. Thus, in this example also a torque of constraint
must act in addition, which compensates this inclination.
3.10 Hamilton’s Principle
In this paragraph it shall be shown that the Lagrangian equations of motion,
which had been derived above with physical arguments from d’Alembert’s
principle, also can be derived as the Euler equations of an extremum principle,
in this case of Hamilton’s principle.
Comment: In an axiomatic approach to classical mechanics, Hamilton’s
principle could be a starting point, alternative to the Newtonian axioms.
3.10.1 The Action
Definition 20. (Action): The action W is33
W =
∫ t2
t1
L dt (3.57)
with the Lagrangian function
L = L({ql(t)}, {q̇l(t)}, t) .
33 The action is often denoted by S.
3.10 Hamilton’s Principle 93
q
t
t1 t2
Fig. 3.15. Two varied trajectories with the same
initial and final points
Comments:
• The notation “action” is an unfortunate choice. The physical action has
nothing to do with performance, but rather with effectiveness.
• The action W is a functional34 of the trajectories ql(t),
W = W [ql(t)] .
3.10.2 Hamilton’s Principle
The actionfor the real trajectories is “stationary,” i.e., extremal,
δW = δ
∫ t2
t1
L dt = 0, (3.58)
with fixed initial and final points,
δql(t1) = 0 = δql(t2).
Comments:
• There are thus several varied trajectories ql(t) + δql(t), which all have the
same initial and final point, compared with the real trajectory ql(t) of a
single particle, see Fig. 3.15.
• It may look as if, according to Hamilton’s principle, for the prediction of
the trajectory the total trajectory should be known (because otherwise one
could not evaluate the integral), and as if this principle would contradict
the principle of the causality. However, one could argue against this that
only a piece of the trajectory (namely that between the arbitrarily chosen
times t1 and t2) must be known. Thus, the Newtonian laws enable the
prediction of the trajectory, starting from any given time,35 i.e., if one
knows the coordinates ql as well as, e.g., the velocities q̇l at a given time.
34 See Appendix E.1.
35 See, for example, the discussion in Sommerfeld [6].
94 3 Lagrangian Mechanics
• The result of the variational method36 (with an integral over time) is a set
of Euler–Lagrangian equations (of the second kind) (3.20) (at an arbitrary
time), see the following.
• One finds that the inanimate nature apparently behaves such that some
quantities become extremal. For example, a ray of light, emitted from a
point, does not “know”, of course, where it will eventually arrive. In fact,
the “trajectory” of a ray of light is such that the arrival at this final point is
in the shortest time possible. (Additional examples for extreme quantities:
minimum energy, maximum entropy, etc.)
• Hamilton’s principle has applications not only in mechanics, but also in
other fields of physics and technology.
The variational problem (3.58) implies37 the Lagrangian function (3.20) of
the second kind,38
δW = 0 ⇔ d
dt
∂L
∂q̇l
− ∂L
∂ql
= 0. (3.59)
Proof:
δW = δ
∫ t2
t1
L dt =
∫ t2
t1
⎛
⎝
∑
j
∂L
∂ql
δql +
∑
j
∂L
∂q̇l
δq̇l +
∂L
∂t
δt
⎞
⎠ dt. (3.60)
The third term vanishes because of δt = 0; partial integration of the second
term leads to ∫ t2
t1
∂L
∂q̇l
δq̇l dt = −
∫ t2
t1
δql
d
dt
∂L
∂q̇l
dt, (3.61)
where the integrated term vanishes because of vanishing variations δq(t1) =
0 = δq(t2) at the limits, and one obtains
0 = δW =
∫ t2
t1
∑
j
(
∂L
∂ql
− d
dt
∂L
∂q̇l
)
δqldt. (3.62)
Since the limits of integration are arbitrary, the integrand (the sum) vanishes,
and since the δql are linearly independent, the summands (in the brackets)
vanish. �	
Comment: The gauge transformation (3.32)
L′ = L+
dF
dt
36 See Appendix E.
37 See Appendix E.2.
38 For the derivation of the Lagrangian equations of the first kind from Hamilton’s
principle see, e.g., Goldstein [1], Sects. 2–4.
3.11 Symmetries and Conservation Laws 95
leaves the variation of the action unchanged,
δW ′ = δW.
Proof:
δW ′ = δ
∫ t2
t1
(
L+
dF
dt
)
dt
= δW + δ
∫ t2
t1
dF
dt
dt = δW + δ
∫ t2
t1
dF
= δW + δF
∣∣∣
t2
t1
= δW + [F (q + δq) − F (q)]
∣∣∣
t2
t1
= δW + 0
because of δq(t1) = 0 = δq(t2). �	
3.11 Symmetries and Conservation Laws39
Further above,40 the conservation laws have been inferred from the form of the
Newtonian forces. In this paragraph we shall analyze the Lagrangian function
for conservation laws.
3.11.1 Noether’s Theorem (II)
Noether’s theorem: [15]
For each invariance
of the Lagrangian function (or of the action integral)
under a continuous transformation
there is a conservation theorem.
In the following we will start from the Lagrangian equations of the second
kind,
d
dt
∂L
∂q̇l
− ∂L
∂ql
= 0,
here for reasons of simplicity without constraints,
d
dt
∂L
∂ṙi
− ∂L
∂ri
= 0.
39 According to Hill [13]
40 See Sect. 1.6.
96 3 Lagrangian Mechanics
Examples:
invariance of the system under . . . conservation theorem for . . .
. . . (cyclic) variable, in particular : . . . canonical momentum
. . . time translation (homogeneity of time) . . . Hamiltonian function
. . . space translation (homogeneity of space) . . . momentum
. . . space rotation (isotropy of space) . . . angular momentum
. . . space inversion . . . parity
. . . Galilean transformation . . . center of mass
Comments:
• It is important to observe the fact that the transformation is continuous.
For example, the symmetry operation, which maps a crystal onto itself (by
a shift of a lattice vector, by a mirror operation, etc.), is not a continuous
transformation, since it cannot be described by a continuous parameter
(an arbitrary angle of rotation, etc.).
• Additional invariances are connected with additional conserved quantities
like the charge, the number of baryons, etc.; the invariance under Lorentz
transformations of Special Relativity theory41 is connected with a maxi-
mum velocity (the velocity of light).
• The Lagrangian densities (for the continuous quantum-mechanical fields
of the elementary particles instead of the discrete coordinates of the point
mechanics) play a role not only in classical field theories like the continuum
theory of deformable bodies42 and in classical electrodynamics43 but in
particular also in the modern physics of the elementary particles with
correspondingly conserved quantities.
3.11.2 Cyclic Coordinate and Conservation
of the Conjugate Momentum
Definition 21. (Cyclic Coordinate): A coordinate ql is called cyclic, if the
Lagrangian function L is independent of this coordinate,
∂L
∂ql
= 0 ⇔ ql is cyclic. (3.63)
Proposition: The conjugate momentum pl corresponding to a cyclic coordi-
nate ql is a conserved quantity (if the system can be described by a Lagrangian
function, i.e., if all forces are conservative),
∂L
∂ql
= 0 ⇔ pl = const. (3.64)
41 See the Course on Special Relativity.
42 See Chap. 10.
43 See the Course on Classical Electrodynamics.
3.11 Symmetries and Conservation Laws 97
Proof: For the derivative of the generalized momentum with respect to time
one obtains with the use of the Lagrangian equation of motion
0
(3.63)
=
∂L
∂ql
(3.20)
=
d
dt
∂L
∂q̇l
(3.25)
= ṗl. (3.65)
�	
Comments:
• The existence of conserved quantities very obviously facilitates the inte-
gration of the equations of motion. Thus the aim is to make a clever choice
of the coordinate system (i.e., the system of the generalized coordinates
ql) such that as many coordinates as possible are cyclic.
Example: Motion in a Central Potential
In cartesian coordinates the Lagrangian function is
L = L(x, y, z, ẋ, ẏ, ż) = 1
2m
(
ẋ2 + ẏ2 + ż2
)
− V (
√
x2 + y2 + z2),
and none of these coordinates is cyclic. In spherical polar coordinates it is
L = 1
2m
(
ṙ2 + r2ϕ̇2 sin2 ϑ+ r2ϑ̇2
)
− V (r),
and ϕ is a cyclic coordinate,
∂L
∂ϕ
= 0 ⇒ pϕ =
∂L
∂ϕ̇
= mr2ϕ̇ sin2 ϑ = const.
In this example, the z component of the angular momentum is thus conserved.
3.11.3 Homogeneity in Time and Energy Conservation
The invariance of the Lagrangian function L of a system under a (continuous)
time translation τ is equivalent to writing
L(ql, q̇l, t) = L(ql, q̇l, t+ τ)
for arbitrary τ , or equivalent to the fact that the Lagrangian function L has
no explicit time dependence,
∂L
∂t
= 0,
but in general an implicit dependence, dL/dt �= 0, because of the time depen-
dence of the coordinates ql(t). One obtains from the homogeneity of time the
so-called Hamiltonian function
98 3 Lagrangian Mechanics
H =
f∑
l=1
q̇lpl − L (3.66)
as a conserved quantity,
∂L
∂t
= 0 ⇒ dH
dt
= 0. (3.67)
Proof: One obtains
dL
dt
=
∑
l
(
∂L
∂q̇l
q̈l +
∂L
∂ql
q̇l
)
+
∂L
∂t
(3.20)
=
∑
l
(
∂L
∂q̇l
q̈l + q̇l
d
dt
∂L
∂q̇l
)
+ 0 =
∑
l
d
dt
(
∂L
∂q̇l
q̇l
)
⇒ 0 =
d
dt
(∑
l
∂L
∂q̇l
q̇l − L
)
=
d
dt
(∑
l
plq̇l − L
)
=
dH
dt
.
�	
Comments:
• Equation (3.66) is a so-called Legendre transformation,44 by which the
function L(q, q̇) is transformed into a function H(q, p),
L(q, q̇) → H(q, p).
• Legendre transformations play a vital role in thermodynamics in the con-
text of the different thermodynamic potentials.
With the additional assumptionsT =
1
2
∑
ll′
μll′({ql})q̇lq̇l′ (3.68)
V = V ({ql}),
∂V
∂q̇l
= 0 (for all l) (3.69)
then follows
H = T + V = E = const.
If the system can be described by a Lagrangian function L = T − V with the
given restrictions (i.e., if there are no velocity dependent forces), the energy
E is thus a conserved quantity (a “constant of the motion”).
44 See the Appendix J.
3.11 Symmetries and Conservation Laws 99
Fig. 3.16. The bead sliding on a rotating rod, cf. Fig. 3.2
Proof: With
∑
l
q̇lpl =
∑
l
q̇l
∂L
∂q̇l
=
∑
l
q̇l
∂T
∂q̇l
= 2T
one obtains
0 =
dH
dt
=
d
dt
(∑
l
q̇lpl − L
)
=
d
dt
(2T − (T − V )) =
d
dt
(T + V )
=
d
dt
E.
�	
Example: Bead Sliding on the Rotating Rod
For the example of the bead sliding on the rod rotating (uniformly) with the
angular velocity ω (Fig. 3.16) from Sect. 3.5.6, the Lagrangian function
L = T = 1
2m
(
ṙ2 + r2ω2
)
is time independent. With the canonical momentum
p =
∂L
∂ṙ
= mṙ
the Hamiltonian function
H = ṙp− L = 1
2m
(
ṙ2 − ω2r2
)
=
p2
2m
− m
2
ω2r2 = H(r, p)
is a conserved quantity. Since here the kinetic energy is not of the form (3.68)
(because of the term 1
2mr
2ω2 originating from the constraint), the energy
E = H −mω2r2
is not conserved. In fact, here the force of constraint does not do virtual, but
real work.
100 3 Lagrangian Mechanics
3.11.4 Space Homogeneity and Conservation of Momentum
Space homogeneity of the system is equivalent to the invariance of the system
under continuous45 translations,
L(ri, ṙi, t) = L(ri + a, ṙi, t)
with an arbitrary vector a or
∑
i
∂L
∂ri
≡
∑
i
∇iL = 0.
This implies the conservation of the (canonical) total momentum P =
∑
i pi,
∑
i
∇iL = 0 ⇒ dP
dt
= 0. (3.70)
Proof: 46
(i) From
0 = L(ri + a, ṙi, t) − L(ri, ṙi, t) =
∑
i
a · ∇iL+ O(a2)
and for an arbitrary translation a one has
∑
i
∇iL = 0.
(ii) From the Lagrangian equation(3.20) with ∂L/∂ṙi = ∂
∑
j
1
2mjṙ
2
j/∂ṙi =
miṙi = pi one has
0 =
∑
i
∂L
∂ri
(3.20)
=
d
dt
∑
i
∂L
∂ṙi
=
d
dt
∑
i
pi =
d
dt
P = 0.
�	
3.11.5 Isotropy and Angular-Momentum Conservation
The isotropy of a system is equivalent to the invariance under continuous
rotations. Rotations are described by
ri → r′
i = ri + Δϕ × ri
45 The importance of the continuity of an invariance has already pointed out in
Sect. 1.8.
46 The proof is exactly along the lines of that in Sect. 1.8.
Summary: Lagrangian Mechanics 101
with the rotation vector47 Δϕ, thus
L(ri, ṙi, t) = L(ri + Δϕ × ri, ṙi + Δϕ × ṙi, t).
This implies the conservation of the angular momentum,
d
dt
L =
d
dt
∑
i
ri × pi = 0. (3.71)
Proof: Analogously to the proof of (3.70) one obtains
0 = L(ri + Δϕ × ri, ṙi + Δϕ × ṙi, t) − L(ri, ṙi, t)
=
∑
i
[
(Δϕ × ri) ·
∂L
∂ri
+ (Δϕ × ṙi) ·
∂L
∂ṙi
]
+ O(Δϕ2)
(3.20)
= Δϕ ·
∑
i
[
ri ×
d
dt
∂L
∂ṙi
+ ṙi ×
∂L
∂ṙi
]
+ O(Δϕ2)
= Δϕ · d
dt
∑
i
ri ×
∂L
∂ṙi
+ O(Δϕ2).
Since Δϕ is arbitrary, one concludes
0 =
d
dt
∑
i
ri ×
∂L
∂ṙi
=
d
dt
∑
i
ri × pi =
d
dt
∑
i
li =
d
dt
L.
�	
Comment: Here, the canonical momentum corresponding to the “coordi-
nate” ϕ is the total angular momentum L.
Summary: Lagrangian Mechanics
Alternative Principles:
Principle of virtual work Zi · δri
(3.5)
= 0
Principle of d’Alembert (ṗi − F i) · δri
(3.6)
= 0
Principle of Hamilton δW = δ
∫
L dt
(3.58)
= 0
Euler–Lagrangian Equations (Lagrangian equations of the second kind):
d
dt
∂L
∂q̇l
− ∂L
∂ql
(3.21)
= QRl
d
dt
∂T
∂q̇l
− ∂T
∂ql
(3.21)
= Ql +QRl
(l = 1, . . . , f ; f = 3N − k)
47 See also Sect. 1.8.1.
102 3 Lagrangian Mechanics
Gauge Transformation leaves equation of motion invariant:
L′ (3.32)
= L+
dF
dt
with
∂F
∂q̇l
= 0
Lagrangian Equations of the first kind:
d
dt
∂L
∂q̇l
− ∂L
∂ql
(3.45)
= QRl +
k∑
j=1
λjG̃jl
(l = 1, . . . , 3N)
and G̃jl dql + G̃j0 dt
(3.44)
= 0 (j = 1, . . . , k)
In most cases:
L = T ({ql}, {q̇l}, t) − V ({ql}, t) ≡ L(q, q̇, t)
Canonical Momentum:
pl
(3.25)
=
∂L
∂q̇l
Hamiltonian Function: (Legendre transformation):
H
(3.66)
=
∑
l
plq̇l − L = H({ql}, {pl}, t)
Conservation Laws due to symmetries/invariances (Noether’s theorem):
ql cyclic ⇔ ∂L
∂ql
= 0
(3.64)⇔ pl = const.
∂L
∂t
= 0
(3.67)⇔ H = const.
Charge q in an Electromagnetic Field:
L(r, ṙ, t)
(3.31)
= 1
2mṙ2 − qφ(r, t) + q
[c]
c
ṙ · A(r, t)
H(r,p, t)
(3.72)
=
1
2m
(
p − q
[c]
c
A(r, t)
)2
+ qφ(r, t)
Problems
3.1. Brachistochrone in the Homogeneous Gravitational Field I. Two
points P1(x1, y1) and P2(x2, y2) shall be connected by a curve y(x), on which
a point mass sliding without friction moves from P1 to P2 under the influence
of the homogeneous gravitational field (in the y-direction) in minimal time.
The velocity of the point mass is zero at the point P1.
Problems 103
(a) Determine the time T which the point mass needs to get from P1 to P2, as
a function of the velocity and of the path traveled. Formulate the problem
in a variational form
δT = δ
∫ 2
1
f(y, y′, x) dx = 0.
(b) Using the Euler equation for f(y, y′, x) write down the differential equation
for y(x), and solve the differential equation.
Comment: With the substitution y′ = u(y) the resulting differential equa-
tion 2yy′′ + y′2 + 1 = 0 can be reduced to a differential equation of first
order for u(y). Then the differential equation y′ = u(y) can be solved with
the substitution y = a sin2 ϕ. With x1 = y1 = 0 the solution then has the
form of a cycloid,
x =
a
2
(2ϕ− sin 2ϕ), y =
a
2
(1 − cos 2ϕ).
(c) Which parameter regime ϕ of the cycloid is traced by the point mass for
P1 = (0, 0) and P2 = (πa4 − a
2 ,
a
2 ), P2 = (3πa
4 + a
2 ,
a
2 ), or P2 = (πa, 0)?
3.2. Brachistochrone in the Homogeneous Gravitational Field II. A
particle of mass m may move without friction under the influence of gravity
from a point (0, y0) to (x0, 0) with the initial velocity v = 0. What is the form
of the trajectory of the particle, if the point (x0, 0) shall be reached in the
shortest possible time?
(a) With the help of the energy conservation determine the total time T as a
functional of the cartesian position coordinate.
(b) What is the variational problem (boundary conditions)?
(c) What is the Euler equation? To this end show that one has
∫ x0
0
∂F
∂y′
δy′dx = −
∫ x0
0
d
dx
(
∂F
∂y′
δy
)
dx.
(d) The functional F does not depend explicitly upon x. Show that
F − y′
∂F
∂y′
= const.
is a first integral of the Euler–Lagrangian equation.
(e) With the help of the transformation ∂y/∂x = − cot(θ/2) determine the
parameter representation of the trajectory B = B(x(θ), y(θ)).
3.3. Catenary Curve. A string of length l is suspended in the homogeneous
gravitational field between two points the distance of which is smaller than
the string length.
(a) Give an expression for the extremal condition.
(b) Write down the constraint and the boundary conditions.
104 3 Lagrangian Mechanics
(c) Integrate the resulting differential equation
k dy =
√
(y + μ)2 − k2 dx
(Lagrange multiplier μ; k = const.). With the transformation
y + μ = k cosh z
a direct integration is possible.
(d) Determine the constants with the help of the constraint and of the bound-
ary conditions.
3.4. Atwood Machine. Two masses m1 and m2 are connected by a string
of length l strung without friction across a massless wheel (see Fig. 3.17).
(a) Give an expression for the equation of motion for the total mass m1 +m2.
(b) What is the constraint, and of which kind is it?
(c) Give an expression for the d’Alembert principle, and derive from this the
equations of motion.
(d) Introduce the coordinate of a mass as the generalized coordinate q, and
express the coordinate of the other mass by q, by the radius of the wheel,
and by the string length �.
(e) Write down the Lagrangian function, and set up the Lagrangian equation
of the second kind.
(f) Solve the equation of motion. How does the acceleration change in the
equation of motion, if the mass of the wheel has to be taken into account
(qualitative arguments)?
R
m1
m2
α1 α2 x
y
Fig. 3.17. Left: Atwood machine (for problem 3.4). Right : A modified Atwood
machine (for problem 3.5)
Problems 105
3.5. A Modified Atwood Machine. Two masses m1 and m2 are positioned
on inclined planes with the inclination angles α1 and α2, respectively. They
are connected by a massless rope of constant length, see Fig. 3.17.
(a) How many constraints apply to the two masses? Write the constraints in
the form
fi (x1, y1, x2, y2) = 0.
(b) With the help of Lagrange multipliers set up the equations of motion for
the two masses in cartesian coordinates.
(c) Determine the acceleration ÿ1 and the modulus of bi = (ẍi, ÿi) of each
particle.
Hint: First, eliminate the Lagrange multipliers and then, with the help of
the constraints, all coordinates except for y1.
(d) Determine the constraining forces for the case α1 = α2 = 90◦.
3.6. The Wheel and Axis. The sketched wheel and axis of mass m2 is
fixed by a rope to the ceiling (Fig. 3.18); the rope is tied around the smaller
disk (radius r). Tied around the larger disk (radius R > r) runs a rope at
which the mass m1 pulls under the influence of gravity.
r R
m2
m1
Fig. 3.18. The wheel and axis (for problem
3.6)
(a) Was happens without the mass m1?
(b) Give an expression (now with m1) for the equilibrium condition in the
static case. What are the constraints? How is the mass m1 to be chosen
for static equilibrium with r, R, and m2 given.
(c) Now consider the wheel and axis with the attached mass m1 as a dy-
namical system. Write down the kinetic and the potential energies, and
formulate the Lagrangian function in an appropriately chosen generalized
coordinate.
(d) What is the Lagrangian equation of the second kind for the system? In-
terpret this equation, in that you identify cause and effect of the motion.
Set up the relation to the static case.
106 3 Lagrangian Mechanics
(e) Now write down the Hamiltonian function and the canonical equations.
3.7. Motion on a Spiral. A particle moves under the influence of the grav-
itational force without friction on a spiral of radius R and of threading 2π/α
which rotates with the angular velocity ω. The equation for the geometrical
position of all points on the spiral thus is
x = R cos (αz − ωt) ; y = R sin (αz − ωt) .
(a) Set up the equations of motion for the particle’s coordinates.
(b) Solve the equation for the motion in the z-direction.
(c) Determine the constraining forces acting on the particle, and discuss their
origin.
3.8. Particle on Logarithmic Spiral. Let the trajectory of a point mass
be r(φ) = a ebφ with a, b > 0 (logarithmic spiral).
Which central force acts on the point mass?
Hint: φ̇ can be eliminated using conserved quantities.
3.9. Motion Along a Parabola. A point mass moves in the gravitational
field g = (0,−g) in the (x, y)-plane on the parabola y = 1
2a x
2.
(a) From the information given above, are there conserved quantities? If so,
which? Of which kind is the constraint?
(b) Write down the kinetic and the potential energies for the point mass. What
is the Lagrangian function, if one chooses x as the generalized coordinate?
(c) Give an expression for the Lagrangian equation of the second kind for
x(t), and write down the canonical momentum p conjugate to x.
(d) Determine the Hamiltonian function, and approximate this with x � a
such that one obtains a quadratic form in x and p.
(e) Write down the canonical equations for this approximate Hamiltonian
function, and solve this with the initial conditions x(0) = 0, p(0) = p0.
3.10. Bead Sliding on a Uniformly Rotating Rod. A bead slides without
friction on a horizontal wire rotating about a vertical axis with the angular
velocity ω (Fig. 3.19).
Fig. 3.19. The bead sliding on a rotating rod (like
Fig. 3.9) (for problems 3.10 and 3.11)
Problems 107
(a) What are the constraints, and of which kind are they?
(b) The equation of motion is
r̈ = ω2r.
Determine the eigensolutions and the general solution.
(c) Determine the solution for the initial conditions r(0) = r0 and ṙ(0) = v0.
Explain the possible forms of motion.
(d) What are the conserved quantities of the system? Compare in particular
the Hamiltonian function with the total energy.
3.11. Motion of a Bead on a Rotating Rod. A bead of mass m moves
in the gravitational field on a rod which rotates horizontally with constant
angular velocity ω. The frictional force on the bead is of the form F r =
−γAv/|v|, where v is the velocity along the rod; A is the force pressing the
bead to the rod.
(a) Set up the Lagrangian equations of the first kind. Which physical quantity
is hidden in the Lagrange multiplier? Write down the constraining force,
and determine the force by which the mass is pressed against the rod.
(b) Solve the equations of motion (neglecting gravity) with the initial condi-
tions r(t) = r0 and v(t) = 0 for t = 0.
3.12. A Bead on a Rotating Wire. A bead of mass m slides along a
straight wire which is fixed to a vertical axis at an angle α; the vertical axis
rotates with the angular velocity ω.
Solve the equation of motion for the initial conditions r(0) = 0, ṙ(0) = v.
3.13. The Rotating Plane Pendulum. A plane pendulum with its hinge in
the coordinate origin swings in the (x, z) plane (Fig. 3.20). This plane rotates
with constant angular velocity ω around the z-axis (antiparallel to the ho-
mogeneous gravitational field). Employ spherical polar coordinates (r, ϑ, ϕ).
(a) Set up the constraints.
(b) Determine the Lagrangian function.
(c) Determine the equation of motion.
(d) A stationary solution is given by a vanishing acceleration (ϑ̈ = 0).
Show that there are one or more stationary solutions, and determine the
corresponding equilibrium positions ϑ0 as solutions of the stationary equa-
tion.
(e) Investigate the (time dependent) motions in the neighborhood of the sta-
tionary solutions. To this end, make the ansatz ϑ = ϑ0 + δ, and expand
the equation of motion to lowest order in δ; this will result in a linear
differential equation. Determine the solutions δ(t).
An equilibrium position ϑ0 is called stable, if the corresponding solution
δ is restricted to a (finite) neighborhood of ϑ0; otherwise the equilibrium
position is called unstable. Which equilibrium positions are thus stable or
unstable?
108 3 Lagrangian Mechanics
Fig. 3.20. Rotating plane pendulum (for
problem 3.13)
3.14. Point Mass on Inclined Plane. A point mass m slides on an inclined
plane with inclination angle α.
What is a useful generalized coordinate to describe the motion of the point
mass? Set up the Lagrangian function, and from this derive the equation of
motion.
3.15. Point Mass on a Moving Inclined Plane. A point mass moves
without friction under the influence of the homogeneous gravitational field
(in negative z-direction) on an inclined straight line (Fig. 3.21). The straight
line moves in the x direction, such that the height h(t) (along the z axis) is a
function of time t.
z
z–h
h x
x
α
Fig. 3.21. Point mass on a moving inclined plane
(for problem 3.15)
(a) Set up the constraint. How many degrees of freedom does the system have?
(b) Determine the Lagrangian function of the mass.
(c) Determine the equation of motion of the mass.
(d) Determine x(t) and z(t) for the cases h(t) = const., h(t) = vt, and
h(t) = 1
2at
2.
(e) Determine the force of constraint.
Problems 109
3.16. Particle Sliding on a Cone. A point mass m moves without friction
under the influence of gravity on the surface of a circular cone.
(a) Write down the kinetic and the potential energies in cartesian coordinates.
Give an expression for the constraint.
(b) Introduce the coordinates x = r cosϕ and y = r sinϕ, and employ the
relation z = r cotα from the constraint.
(c) What are the Lagrangian equations of the second kind for the generalized
coordinates r and ϕ? Which conserved quantity can be derived directly
from these equations? What is the physical meaning of this quantity?
(d) Employ this conserved quantity to eliminate one of the variables. Interpret
the (one-dimensional) equation of motionfor the radial motion.
(e) Write down the condition for the point mass not to accelerate radially.
3.17. Crank Mechanism. Consider a piston engine (e.g., a steam engine, see
Fig. 3.22), which runs slowly enough such that inertial forces can be neglected.
Via the crank mechanism the piston forceD is transformed into the crank force
K which is tangential to the radius r.
r Dα
β
K
x
Fig. 3.22. Crank mechanism (for problem 3.17)
(a) Give an expression for the condition for static equilibrium (principle of
virtual work). Write down the constraint relating the virtual displacements
δx of the piston to the rotation angle δα.
Hint: First employ the angle β in addition to x and α, and eliminate β by
geometrical considerations.
(b) How can K be represented as a function of D and α?
3.18. Kinetic Energy as a Homogeneous Function. Show that for a
mechanical system with holonomous, skleronomous constraints the kinetic
energy is a homogeneous function of the generalized velocities of order n = 2,
i.e., T (αq̇1, . . . , αq̇m) = α2T (q̇1, . . . , q̇m).
3.19. A Particle in an Electric and Magnetic Field. The Lagrangian
function for a particle of mass m and of charge q, which moves in an electrical
field E and a magnetic field B, has the following form:
110 3 Lagrangian Mechanics
L =
1
2
mṙ2 + q · Aṙ − qφ.
The fields E and B are related to the scalar potential φ(r) and the vector
potential A(r) via the relations B = ∇ × A and E = −∇φ− ∂A/∂t.
(a) Using the Lagrangian equations of the second kind show that
mr̈ = q(E + ṙ × B).
(b) Determine the canonical momentum conjugate to r, and determine the
Hamiltonian function.
(c) Is the total energy of the system conserved? Try a physical interpretation
of this matter.
(d) At time t = 0 the particle is placed into an electrical field E = (E, 0, 0)
and a magnetic field B = (0, 0, B) at the point r0 = 0, where it has the
initial velocity v0.
• Solve the equations of motion for the particle under the given initial
conditions, and describe the trajectory of the particle.
Hint: To solve for x(t) and y(t) introduce the complex quantity u(t) =
x(t) + iy(t).
• Discuss the trajectory for the case E = 0.
• How does the motion of the particle change, if the electrical field also
has a z-component?
3.20. A Point Mass on a Sphere. A point mass may move under the
influence of gravity on the surface of a sphere (Fig. 3.23).
φ
mg
m
Fig. 3.23. A point mass on the surface of a sphere (for
problem 3.20)
(a) Write down the constraints and the constraining force in polar coordinates.
(b) Give an expression for the equation of motion, and decompose this into
the radial and azimuth components.
(c) Integrate the azimuth equation of motion to obtain the energy conserva-
tion for the initial conditions φ(t = 0) = 0, φ̇(t = 0) = φ̇0.
(d) Via the radial equation of motion determine the Lagrange multiplier and
the constraining force.
Problems 111
(e) At which angle does the point mass detach from the surface of the sphere?
3.21. Particle on a Rotation Paraboloid. A particle of mass m moves
under the influence of gravity on the internal area of a rotation paraboloid
x2 + y2 = az
without friction.
(a) Set up the equation of motion for the particle in Cartesian coordinates.
Employ the Lagrangian equations of the first kind.
(b) How large is the angular velocity ω of the particle, if it moves on the
circle which results from the cut of the rotation paraboloid with the plane
z = h? How large is the radius ρ0 of the circle?
3.22. Dissipation Function and Free Case with Friction. The friction
as a force depending upon the velocity can be accounted for via the dissipation
function
D =
1
2
n∑
l,m=1
βlm q̇l q̇m
in the Lagrangian formalism by a generalized force Q(R)
j = −∂D/∂q̇j. The
Lagrangian equation of the second kind then reads
d
dt
∂L
∂q̇j
− ∂L
∂qj
+
∂D
∂q̇j
= 0.
(a) Give an expression for the differential energy change connected with the
frictional forces Q(R)
j , and express the time rate of change of this energy
by the dissipation function D.
(b) A point mass falls under the influence of gravity and of friction, which
is described by the dissipation function D = α
2 v
2. Starting from the La-
grangian function and the dissipation function give an expression for the
equation of motion. Determine the velocity for the point mass to be at
rest at t = 0 (as the initial condition). Discuss the result with the help of
the graph for v(t).
3.23. Gauge Transformation of the Canonical Momentum. Given be
the gauge transformation
L′ (3.32)
= L+
dF
dt
with
∂F
∂q̇l
= 0.
Show that the canonical momenta transform according to
p′l = pl +
∂F
∂ql
.
112 3 Lagrangian Mechanics
3.24. Hamiltonian Function of a Charged Particle. Given a particle
with charge q with the Lagrangian function
L(r, ṙ, t)
(3.31)
= 1
2mṙ2 − qφ(r, t) + q
[c]
c
ṙ · A(r, t).
Prove that the Hamiltonian function is
H(r,p, t) =
1
2m
(
p − q
[c]
c
A(r, t)
)2
+ qφ(r, t). (3.72)
4
Harmonic Vibrations
In this chapter vibrations shall be investigated, and in particular an oscillator
• Without and with damping
• Without and with coupling to other oscillators, as well as
• Without and with external force
In the context of the external force we shall encounter a very simple case of a
Green function, which plays an important role in statistical mechanics.
In this context we shall study the tools to treat
• (Un)coupled, (in)homogeneous, linear differential equations
• Eigen values and eigen vectors of matrices, and finally
• The Fourier transformation
The prototype of the equation of motion is the equation of motion of a
simple harmonic oscillator, for which the restoring force is proportional to the
displacement (Hooke’s law, ω2
0), damped by Stokes friction1 (γ), and forced
by an external field (f(t)),
ẍ+ 2γẋ+ ω2
0x = f(t). (4.1)
In the case of the oscillator without damping and without external force one
obtains the simpler equation
ẍ+ ω2
0x = 0 (ω2
0 = k/m).
The restoring force −kx = −mω2
0x of a spring with the spring constant k can
be derived from the potential (Fig. 4.1)
V (x) = 1
2kx
2.
The potential is symmetrical with respect to the origin, V (x) = V (−x);
thus the turning points are located symmetrically. For a given energy E (as a
1 See Sect. 1.4.
114 4 Harmonic Vibrations
0
1
2
3
4
5
V
/f
−3 −2 −1 0 1 2 3
x
−3
−2
−1
0
1
2
3
K
/f
Fig. 4.1. The potential V (top) and the force F (bottom) of the simple harmonic
oscillator in units of the force constant k as a function of the displacement x
constant of the motion!) the form of V (x) is arbitrary for all x with V (x) >
E. (This is wrong in the quantum-mechanical case because of the tunneling
effect.)
The harmonic oscillator is a model for many systems:
Spring
For the spring one has a force proportional to the change from the equilibrium
length l0 of the spring, and for a displacement in the direction of the axis of
a spiral spring (chosen as the z direction) one has
mz̈ = −k (z − z0) . (4.2)
With the substitution
ζ = z − z0
one obtains
mζ̈ = −kζ.
Comments:
• For a real spring Hooke’s law is only approximately valid and only for
small elongations. For large elongations one observes the phenomenon of
flowage (see Fig. 4.2), for which the spring does not return to its original
form after the tension is released.
• The restoring force for a small displacement perpendicular to the axis of an
unextended spring is proportional to the third power of the displacement.2
2 See problem 4.3.
4 Harmonic Vibrations 115
x
V
Fig. 4.2. The phenomenon of flowing of a material at larger stretching
Three-Dimensional Oscillator
In the case of the three-dimensional oscillator one has in generalization of
(4.2)
mr̈ = −k (r − r0) .
This equation can be decomposed into the three orthogonal cartesian com-
ponents, and for each single component one has a one-dimensional equation
(4.2) as in the case of the pendulum,
mẍi = −k (xi − xi,0) .
Molecule
The (potential) energy of the diatomic moleculeas a function of the distance
of the cores has a minimum and is likewise approximately harmonic near the
minimum (the broken curve in Fig. 4.3).
Crystals
In crystals the potential between two atoms as a function of their distance
qualitatively has the form as in the case of the molecule. For small displace-
ments out of the equilibrium positions the restoring forces are again harmonic.
v
r
Fig. 4.3. The potential energy V as a function of the interatomic distance r for the
diatomic molecule (schematic)
116 4 Harmonic Vibrations
RLC Circuit
The equation of motion for the electrical charge Q (with the current I = Q̇)
as a function of the voltage U
LQ̈+RQ̇+
1
C
Q = U
of the RLC circuit, see Fig. 4.4, is the equation of a damped forced oscillator.
R
L
C
U
Fig. 4.4. The RLC circuit
Mathematical Pendulum
The potential energy is a periodic function of the angle, see Fig. 4.5,
V (ϕ) = mgl (1 − cosϕ)
(with V (0) = 0). Differently from the other mentioned examples, here there
are bound and free motions,3 depending on the energy (0 ≤ E mgl, respectively). For small displacements from the minimum position
the potential can be expanded into a Taylor series,
V (ϕ) = 1
2mglϕ
2 (ϕ � π).
The potential is thus approximately harmonic in the neighborhood of the
minimum (and only there) and can be approximated by a parabola (broken
curve in Fig. 4.5). In this approximation one obtains the equation of motion
ml2ϕ̈ = −mglϕ (ϕ� π).
3 See the reasoning in the context of Fig. 2.1 on p. 33.
4.1 The Simple Oscillator 117
j2p
V
0
Fig. 4.5. The potential energy V as a function of the displacement angle ϕ for the
mathematical pendulum
Molecular Crystals
The rotational potential of a molecule in a lattice (assumed as rigid) of a
molecular crystal is likewise periodic in most cases, even though not of the
simple form as in the case of the mathematical pendulum.
4.1 The Simple Oscillator
Here the “simple” oscillator shall be investigated, by which we mean the one-
dimensional, free (i.e., without external force), harmonic oscillator without
damping.
4.1.1 Eigen Solutions of the Homogeneous Differential Equation
The equation of motion of the simple oscillator reads
mẍ+ kx
(4.1)
= 0 (4.3)
or
ẍ+ ω2
0x = 0 (4.4)
with ω2
0 = k/m.
Eigen Solutions
For the solution of the linear differential (4.4) one makes an exponential ansatz
eλt or, since one expects oscillations, the ansatz4
x(t) = x0 e−iωt.
4 The sign of the phase is chosen here as in nonrelativistic quantum mechanics. In
classical mechanics one could also choose the opposite sign.
118 4 Harmonic Vibrations
Inserting this into the differential (4.4) one obtains
(
−ω2 + ω2
0
)
x0e−iωt = 0.
With this equation fulfilled for all times t, the factor in front of the exponential
function must vanish, and one obtains an algebraic equation
(
−ω2 + ω2
0
)
x0 = 0
for the amplitude x0 instead of the differential (4.4) for the displacement x(t).
For nontrivial solutions x0 �= 0 the secular equation
−ω2 + ω2
0 = 0
must be fulfilled with the solutions
ω± = ±ω0 = ω1,2. (4.5)
Thus, one obtains two eigen solutions (for the differential equation of second
order)
x1(t) = e−iω1t and x2(t) = e−iω2t.
General Solution
The general solution is the superposition of the eigen solutions,
x(t) = a1x1(t) + a2x2(t) = a1e−iω1t + a2e−iω2t (4.6)
with two (superposition) constants a1 and a2.
Comment: Instead of the solutions e±ω0t one also could have obtained the
eigen solutions
sin(ω0t) and cos(ω0t),
(as a combination of the exponential functions) and the general solution
would be
x(t) = c sin(ω0t) + d cos(ω0t).
Reality of the Solution
The factors e±iωt in (4.6) are complex,
e±iz = cos z ± i sin z.
But since the displacement x(t) is real, the superposition coefficients ai must
be complex. For example, one obtains from (4.6) the general solution
x(t) = a cos(ω0t) + b sin(ω0t)
4.1 The Simple Oscillator 119
with real coefficients a and b,
a = a1 + a2, ib = a1 − a2.
Proof: From the reality of the solutions follows
x(t) = a1 e−iω0t + a2 e
+iω0t
= x∗(t) = a∗1 e+iω0t + a∗2 e−iω0t.
This must hold for all times t, and from the linear independence of the func-
tions e±iω0t one has
a1 = a∗2 (a∗1 = a2).
If one sets
a1 = 1
2 (a+ ib) ,
then one gets
a2 = 1
2 (a− ib) .
With this one obtains for the solution
x(t) = 1
2 (a+ ib) [cos(ω0t) − i sin(ω0t)] + 1
2 (a− ib) [cos(ω0t) + i sin(ω0t)]
= a cos(ω0t) + b sin(ω0t).
�	
Comments:
• Of the four constants, which determine the real and imaginary part of
a1 and a2 of the general solution (4.6), only two are independent of each
other.
• Thus one can write the solutions in the alternative forms
x(t) = a cos(ω0t) + b sin(ω0t) (4.7)
= a1 e−iω0t + a∗1 eiω0t (4.8)
= a1 e−iω0t + c.c. (4.9)
with
a1 = 1
2 (a+ ib)
and where “c.c.” stands for the complex conjugate of the preceding
expression.
120 4 Harmonic Vibrations
Comment: (Concerning the use of complex quantities) Because of the conve-
nience offered by the exponential function one mostly works with the complex
quantities e±iωt, without saying explicitly that one means the real part of the
corresponding expression. This works fine for all linear operations (addition,
differentiation, etc.), e.g.,
d
dt
(
Re (α e−iωt) + Re (β eiωt)
)
=
d
dt
Re
(
α e−iωt + β eiωt
)
= Re
d
dt
(
α e−iωt + β eiωt
)
.
Taking the square and other nonlinear operations one has first to take the real
part and then to square because of
Re (z1z2) �= Re z1 · Re z2.
4.2 The Free Oscillator with Damping
Here now the one-dimensional, free (i.e., without external force), harmonic
oscillator with damping shall be investigated. The development follows very
closely that one of the previous section.
4.2.1 Eigen Solutions of the Homogeneous Differential Equation
For Stokes friction5 the homogeneous equation of motion of a linear oscillator
reads
mẍ+ �ẋ+ kx
(4.1)
= 0 (4.10)
or
ẍ+ 2 γ ẋ+ ω2
0x = 0 (4.11)
with 2γ = �/m and ω2
0 = k/m.
Eigen Solutions
For the solution of the linear differential (4.11) one makes an exponential
ansatz eλt or, since one expects oscillations, the ansatz6
x(t) = x0 e−iωt.
5 See Sect. 1.4.
6 The sign of the phase is chosen here as in nonrelativistic quantum mechanics. In
classical mechanics one could also choose the opposite sign.
4.2 The Free Oscillator with Damping 121
If one inserts this into the differential (4.11), one obtains
(
−ω2 − 2iωγ + ω2
0
)
x0 e−iωt = 0.
With this equation fulfilled for all times t, the factor in front of the exponential
function must vanish, and one obtains an algebraic equation
(
−ω2 − 2iωγ + ω2
0
)
x0 = 0
for the “amplitude” x0 instead of the differential (4.11) for the displacement
x(t). For nontrivial solutions x0 �= 0 the secular equation
−ω2 − 2iωγ + ω2
0 = 0
must be fulfilled with the solutions
ω± = ω1,2 =
{
−iγ ±
√
ω2
0 − γ2 = −iγ ±Ω for γ2 ω2
0
(4.12)
with
Ω =
√
ω2
0 − γ2 and μ =
√
γ2 − ω2
0 .
Thus, one obtains two eigen solutions (for the differential equation of second
order)
x1(t) = e−iω1t and x2(t) = e−iω2t.
General Solution
The general solution is the superposition of the eigen solutions,
x(t) = a1x1(t) + a2x2(t) = a1 e−iω1t + a2 e−iω2t + c.c. (4.13)
with two (superposition) constants a1 and a2.
The Case γ = ω0
In the case γ = ω0 (i.e., ω1 = ω2 = −iγ) it looks as if the two eigen solutions
would be the same. However, for a twofold zero of the secular equation one
has the eigen solutions
e−γt and t e−γt
and the general solution
x(t) = (c1 + c2t) e−γt (4.14)
with two constants c1 and c2.
122 4 Harmonic Vibrations
4.2.2 Reality of the Solution
In the case γ2 �= ω2
0 the two eigen solutions (4.12) for the frequency ω are
complex; also e−iωt is complex,
ez = ex+iy = [coshx+ sinhx] [cos y + i sin y]
with x and y real. But since the displacement x(t) is real, the superposition
coefficients ai must be complex in addition to the exponential functions.
Case γin the case ω2
0 > γ2, one obtains from (4.12) the general solution
x(t) = e−γt [a cos(Ωt) + b sin(Ωt)] , Ω =
√
ω2
0 − γ2 real
with real coefficients a and b,
a = a1 + a2, ib = a1 − a2.
Proof: From the reality of the solutions follows
x(t) = e−γt
(
a1 e−iΩt + a2 e
+iΩt
)
= x∗(t) = e−γt
(
a∗1 e+iΩt + a∗2 e−iΩt
)
.
Since this holds for all times t, from the linear independence of the functions
e±iΩt one has
a1 = a∗2 (a∗1 = a2).
If one sets
a1 = 1
2 (a+ ib) ,
then one gets
a2 = 1
2 (a− ib) .
With this one obtains for the solution
x(t) = e−γt 12 {(a+ ib) [cos(Ωt) − i sin(Ωt)]
+ (a− ib) [cos(Ωt) + i sin(Ωt)]}
= e−γt [a cos(Ωt) + b sin(Ωt)] .
�	
4.2 The Free Oscillator with Damping 123
Comment: Of the four constants, which determine the real and imaginary
part of a1 and a2 of the general solution (4.13), only two are independent of
each other.
Thus one can write the solutions in the alternative forms
x(t) = e−γt [a cos(Ωt) + b sin(Ωt)]
= e−γt
(
a1 e−iΩt + a∗1 eiΩt
)
= e−γta1 e−iΩt + c.c.
(ω2
0 > γ2) (4.15)
with
a1 = 1
2 (a+ ib)
Ω =
√
ω2
0 − γ2.
Case γ > ω0 (Overdamped Case)
In the case ω2
0 ω0 (Overdamped Case)
With
−iω1,2 = −γ ±
√
γ2 − ω2
0 = −γ ± μ
one obtains
x(t) =
1
−2μ
{
[x0 (−γμ) − v0] e−γt+μt − [x0 (−γ + μ) − v0] e−γt−μt
}
= e−γt
[
x0γ + v0
μ
sinh(μt) + x0 cosh(μt)
]
.
7 The multiplication of matrices with each other and with vectors is denoted here
and in the following without a dot (product sign).
4.3 The Forced Harmonic Oscillator 125
0 10 20 30
ω0t
−1
−0,5
0
0,5
1
(ω
0/
v 0)
 x
γ/ω
0
 = 0.0
γ/ω
0
 = 0.2
γ/ω
0
 = 1.0
γ/ω
0
 = 2.0
Fig. 4.6. The displacement x of the oscillator hit by an impulsive force at time
t = 0 for various damping constants
Case γ = ω0 (Critically Damped Case)
From
γ → ω0, Ω → 0
one obtains
x(t) = lim
Ω→0
e−γt
[
(x0γ + v0)
sin(Ωt)
Ω
+ x0 cos(Ωt)
]
= e−γt [x0 + (x0γ + v0) t] .
Comment: The time dependence of x(t) for the case x0 = 0 is shown in
Fig. 4.6 and is the same as for the Green function g(t) considered further
below in Sect. 4.3.4 and represented in Fig. 4.12.
4.3 The Forced Harmonic Oscillator
4.3.1 Equation of Motion
In the following a simple oscillator under the influence of a driving force f(t)
shall be considered. The equation of motion is
mDtx ≡ mẍ+ �ẋ+ kx
(4.1)
= f(t) (4.18)
126 4 Harmonic Vibrations
with the differential operator Dt given by
mDt = m
d2
dt2
+ �
d
dt
+ k.
This is an inhomogeneous linear differential equation; the inhomogeneity here
is the driving force f(t).
The general solution x of the inhomogeneous linear differential (4.18)
mDtx(t) = f(t)
is a superposition of the general solution xhom of the homogeneous equation,8
mDtxhom(t) = 0,
with any particular solution xpart of the inhomogeneous equation (4.18),
x(t) = xhom(t) + xpart(t).
In the case of the damped oscillator the solution (4.13) or (4.14) of the ho-
mogeneous equation (4.10) decays in time and after a sufficiently long time is
small in comparison to the particular solution. This section thus deals with
the construction of a particular solution.
4.3.2 Solution of the Inhomogeneous Linear Differential Equation
by Fourier Transformation
For linear differential equations there is, in principle, a general procedure,
namely that of the Fourier transformation.9 One performs a Fourier decom-
position of the driving force,
f(t) =
∫ ∞
−∞
dω
2π
e−iωtF (ω) (4.19)
with
F (ω) =
∫ ∞
−∞
dt eiωtf(t).
For the particular solution one makes thus the ansatz analogous to (4.19)
x(t) =
∫
dω
2π
e−iωtX(ω) (4.20)
8 See Sect. 4.2 above.
9 See Appendix I.
4.3 The Forced Harmonic Oscillator 127
(under omission of the suffix “part”), and by inserting the ansatz into the
equation of motion
Dtx = ẍ+ 2γẋ+ ω2
0x− f(t)
m
= 0
one obtains
∫
dω
2π
e−iωt
[(
−ω2 − 2iωγ + ω2
0
)
X(ω) − 1
m
F (ω)
]
= 0.
Since this must hold for all times, it follows that the contents of the square
brackets must vanish,
D(ω)X(ω) :=
(
−ω2 − 2iωγ + ω2
0
)
X(ω) =
1
m
F (ω).
Now one can solve for X(ω) and obtains for the particular solution
X(ω) =
1
m
1
ω2
0 − 2iωγ − ω2
F (ω) =
1
mD(ω)
F (ω) = G(ω)F (ω) (4.21)
with the Green function (response function), see Fig. 4.7,
G(ω) =
1
mD(ω)
=
1
m
1
ω2
0 − 2iωγ − ω2
. (4.22)
Comments:
• One obtains x(t) from (4.20) with X(ω) from (4.21).
• By the use of the Fourier transformation
f(t) =
∫ ∞
−∞
dω
2π
e−iωtF (ω)
−3 −2 −1 0 1 2 3
ω/ω0
−1
0
1
R
e 
G
(ω
) 
m
ω
02
γ/ω
0
 = 0.2
γ/ω
0 
= 1.0
γ/ω
0
 = 2.0
−3 −2 −1 0 1 2 3
ω/ω0
−2
−1
0
1
2
Im
 G
(ω
) 
m
ω
02
γ/ω0 = 0.2
γ/ω0 = 1.0
γ/ω0 = 2.0
Fig. 4.7. Real part (left panel) and imaginary part (right panel) of the Green
function (in units of 1/mω2
0) for three different damping constants
128 4 Harmonic Vibrations
– because of the property of the exponential function – the Fourier trans-
form of the time derivatives is particularly simple; with h(t) = df/dt one
obtains
h(t) =
∫
dω
2π
e−iωtH(ω) =
d
dt
f(t) =
d
dt
∫
dω
2π
e−iωtF (ω)
=
∫
dω
2π
e−iωt(−iω)F (ω)
and thus
h(t) = df/dt ⇔ H(ω) = −iωF (ω). (4.23)
This is one of the reasons why one often prefers to work in frequency space
instead of working in time space.
• In particular, by the Fourier transformation the differential operator Dt
has turned into a multiplicative factor D(ω).
• The property
G(ω) = G∗(−ω) (4.24)
is a very general property of any response function (Green function), which
describes the reaction (here the displacement) of a system (here of the
oscillator) to an external perturbation (here to the driving force).
• The Green function is determined only by the properties of the system
(here of the oscillator) and is independent of the externally applied field
(here of the force). With the introduction of the Green function the
problem has not become more simple, but more general, since with the
knowledge of the Green function one can find the particular solution to a
differential equation with an arbitrary inhomogeneity. The larger difficulty
generally lies in the determination of the Green function and less in the
subsequent application.
Periodic External Force
For a periodic external force with the period T one can expand the force (like
any other periodic function) in terms of a Fourier series,
f(t) =
∞∑
n=−∞
Fn e−iωnt
with
ωn = n
2π
T
and with the expansion coefficients10
10 See Appendix I.3.
4.3 The Forced Harmonic Oscillator 129
Fig. 4.8. The vectors of the force F (ω) and of the displacement X(ω) (in arbitrary
units) in the complex plane. The arrows indicate the evolution with time in clockwise
direction
Fn =
1
T
∫ T
0
dt f(t) eiωnt.
For the corresponding expansion of the displacement of the oscillator then one
obtains analogously
Xn = G(ωn)Fn.
HarmonicExternal Force
See problem 4.13.
4.3.3 Green Function of the Damped Oscillator
in Frequency Space
Since the force f(t) is real, the Fourier transform F (ω) is complex; likewise
X(ω) is complex. The Green function (4.22) in frequency space is
G(ω) =
1
m
1
ω2
0 − 2iωγ − ω2
= |G(ω)| eiϕ(ω).
With X(ω) = G(ω)F (ω) the displacement X(ω) has a phase shift ϕ(ω) rela-
tive to the driving force F (ω), see Figs. 4.8 and 4.9.
The Green function can be written in the form
G(ω) = − 1
m
1
(ω − ω1)(ω − ω2)
= − 1
m
1
ω1 − ω2
(
1
ω − ω1
− 1
ω − ω2
)
with
ω1,2
(4.12)
= −iγ ±
√
ω2
0 − γ2 =
{
−iγ ±Ω (γ ω0).
130 4 Harmonic Vibrations
−3 −2 −1 0 1 2 3
ω/ω0
0
1
2
3
|G
(ω
)|
 m
ω
02
γ/ω0 = 0.2
γ/ω0 = 1.0
γ/ω0 = 2.0
−3 −2 −1 0 1 2 3
ω/ω 0
−0,5
0
0,5
φ/
π
γ/ω0 = 0.2
γ/ω0 = 1.0
γ/ω0 = 2.0
Fig. 4.9. The modulus (left panel) and the phase ϕ ≡ φ (right panel) of the Green
function for various strengths of the damping constant
Fig. 4.10. The position of the poles of the Green function in the complex frequency
plane for increasing strength of the damping (from left to right, see also Fig. 4.11)
0 0, 5 1 1, 5 2
γ/ω0
−4
−3
−2
−1
0
1
Ω
/ω
0,
 −
(γ
 +
 μ
)/ω
0
Ω/ω0
−(γ+μ)/ω0
Fig. 4.11. Position of the real and imaginary part of the poles of the Green function
as a function of the damping strength in the complex frequency plane, see also
Fig. 4.10
Comments:
• The poles of the Green function lie at ω = ω1,2 in the lower complex
frequency half-plane and are given by the eigen values ω1,2 of the secular
equation, see Figs. 4.9–4.11.
4.3 The Forced Harmonic Oscillator 131
• In the underdamped case (ω0 > γ), with
Ω =
√
ω2
0 − γ2,
one can write the Green function in the form
G(ω) =
1
m
1
Ω2 − (ω + iγ)2
=
1
2Ωm
[
1
Ω − (ω + iγ)
+
1
Ω + (ω + iγ)
]
.
The Green function has poles at the positions
ω = ±Ω − iγ.
In particular, the real part of the pole is equal to the eigen frequency, and
the imaginary part of the pole is equal to the damping constant.
• In the overdamped case one has to set μ = iΩ as in Sect. 4.2, and the poles
are located at ω = i(−γ ± μ) on the negative imaginary frequency axis.
4.3.4 Time-Dependent Green Function
From X(ω) = G(ω)F (ω) (in frequency space) from (4.21) one obtains accord-
ing to the convolution theorem11
x(t) =
∞∫
−∞
dt′g(t− t′) f(t′)
(in time space) with
g(t) =
∫
dω
2π
G(ω) e−iωt =
1
m
i
ω1 − ω2
θ(t)
(
e−iω1t − e−iω2t
)
(4.25)
= θ(t)
e−γt
m
t×
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
sin(Ωt)
Ωt
withΩ =
√
ω2
0 − γ2 for ω0 > γ
sinh(μt)
μt
with μ =
√
γ2 − ω2
0 for ω0time dependence or, for a force with a periodic time
dependence,
Pn =
∑
m
VmFn−m,
respectively; the mean power is (in the periodic case)
〈p〉t =
1
T
∫ T
0
p(t) dt = P0 =
∑
n
VnF−n =
∑
n
VnF
∗
n .
With
Vn = −iωnXn = −iωnG(ωn)Fn
one obtains
ReP0 = Re
∑
n
−iωnG(ωn) |Fn|2 =
∑
n
ωnImG(ωn) |Fn|2 .
16 See Appendix I.1.2.
136 4 Harmonic Vibrations
−3 −2 −1 0 1 2 3
ω/ω0
0
0,5
1
1,5
2
2,5
ω
ω
0 
Im
 G
(ω
)
γ/ω0 = 0.2
γ/ω0 = 1.0
Fig. 4.13. Absorbed power of the damped oscillator as a function of frequency
Harmonic External Force
For a purely harmonic external force
f(t) = F cos(ωt) =
F
2
(
e−iωt + eiωt
)
with the period T = 2π/ω the coefficients of the Fourier series are
Fn =
{
F/2 n = −1, 1
0 otherwise
with F real. The mean supplied (respectively dissipated) power is then
ReP0 =
∑
n=±1
ω ImG(ω) |Fn|2 = 1
2ω ImG(ω)F 2.
Absorption
The imaginary part of the Green function thus is connected to the energy
dissipation. This is a general result for generalized susceptibilities (as relations
between the reactions of systems and the forces causing these reactions). For
example, the absorption of light is described by the imaginary part of the
dielectric susceptibility, which in the case of the absorption of infra-red light
by lattice vibrations has a form which is very similar to the form of the Green
function of the simple oscillator.
With the Green function of the simple oscillator of Sect. 4.3.3 one obtains
the absorption (except for constants thus the absorbed power) as a function
of the frequency given by
a(ω) = mω ImG(ω)
(4.22)
= ω Im
1
ω2
0 − ω2 − 2iωγ
=
2ω2γ
(ω2
0 − ω2)2 + (2ωγ)2
.
4.3 The Forced Harmonic Oscillator 137
This function in the form of a Lorentz curve is shown in Fig. 4.13. The maxi-
mum of the absorption lies at the (undamped) eigen frequency,
ωmax = ±ω0
with
a(ωmax) =
1
2γ
with a half width (FWHM)17 of
Δω ≈ 2γ (γ � ω0),
defined via
a(ωmax ± 1
2Δω) = 1
2a(ωmax).
Proof:
(i) One obtains the maximum from
0 =
da(ω)
dω
= 2ω
da(ω)
dω2
= 2ω
d
dω2
2ω2γ
(ω2
0 − ω2)2 + (2ωγ)2
= 4ωγ
⎛
⎜⎝
1
(ω2
0 − ω2)2 + (2ωγ)2
− ω2 2
(
ω2 − ω2
0
)
+ (2γ)2
[
(ω2
0 − ω2)2 + (2ωγ)2
]2
⎞
⎟⎠ .
If one introduces the common denominator, from the vanishing of the
numerator one has
0 =
(
ω2
0 − ω2
)2
+ (2ωγ)2 + 2ω2
(
ω2
0 − ω2
)
− (2ωγ)2
=
(
ω2
0 − ω2
) (
ω2
0 + ω2
)
.
For real frequencies one obtains thus the maximum at
ω = ωmax = ±ω0.
(ii) If one inserts this into the expression for the absorption, one obtain
amax = a(ωmax) =
2ω2
maxγ
0 + (2ωmaxγ)2
=
1
2γ
.
(iii) The half width (FWHM) Δω = 2γ one obtains from
a(ω) = 1
2amax
17 Full Width at Half Maximum; some authors denote the width by γ.
138 4 Harmonic Vibrations
or
2ω2γ
(ω2
0 − ω2)2 + (2ωγ)2
=
1
4γ
or
(2ωγ)2 = 1
2
[(
ω2
0 − ω2
)2
+ (2ωγ)2
]
or
(2ωγ)2 = (ω2
0 − ω2)2
with the solution
ω2 = ω2
0 + 2γ2 ± 2γ
√
ω2
0 + γ2 ≈ ω2
0 ± 2γω0 (γ � ω0)
and
ω ≈ ω0 ± γ.
�	
4.4 Coupled Oscillators
We shall start with a simple example in this section and then continue with
the general case in the following section.
4.4.1 Introductory Example:
Stretching Vibrations in the CO2 Molecule
In this section we will treat the linear molecule. We will disregard the bending
vibrations and concentrate on the stretching vibrations only; we will also
neglect the effect of damping. Assuming a Hooke-like behavior the restoring
forces are proportional to the change of the binding length. For a stretching
vibration the change in length is equal to the difference of the displacements.
Either from the consideration, which displacements lead to which forces and
along with it to which accelerations, or (more easily) from the Lagrangian
function
L = 1
2
[
M
(
u̇2
1 + u̇2
3
)
+mu̇2
2
]
− 1
2k
[
(u1 − u2)
2 + (u3 − u2)
2
]
(for the numbering see Fig. 4.14) one obtains the equations of motion
Mü1 = k (u2 − u1)
mü2 = k [(u1 − u2) + (u3 − u2)]
Mü3 = k (u2 − u3)
4.4 Coupled Oscillators 139
f
O
1 2 3
C O
f
Fig. 4.14. The numbering of the atoms of the CO2 molecule
or, in matrix notation,
⎛
⎝
M
m
M
⎞
⎠
⎛
⎝
ü1
ü2
ü3
⎞
⎠ =
⎛
⎝
−k k 0
k −2k k
0 k −k
⎞
⎠
⎛
⎝
u1
u2
u3
⎞
⎠ ,
in short
Mü = Ku, (4.28)
This is a system of three coupled linear homogeneous differential equations
of second order (with constant coefficients), which can be solved with an
exponential ansatz.
There are different procedures for the solution of a system of linear homo-
geneous differential equations:
(I) Direct diagonalization
(II) Decoupling by clever linear combinations
(III) Transformation
each of which will be considered in the following.
(I) Direct Diagonalization
With the ansatz,
⎛
⎝
u1(t)
u2(t)
u3(t)
⎞
⎠ =
⎛
⎝
w1
w2
w3
⎞
⎠ e−iωt , in short u(t) = w e−iωt (4.29)
with the phase being the same for all “particles” in the differential (4.28) one
obtains
−ω2
⎛
⎝
M
m
M
⎞
⎠
⎛
⎝
w1
w2
w3
⎞
⎠ e−iωt =
⎛
⎝
−k k 0
k −2k k
0 k −k
⎞
⎠
⎛
⎝
w1
w2
w3
⎞
⎠ e−iωt.
Because of the still to be determined values of ω this system represents an
eigen value problem. Since this must hold for all times t,
140 4 Harmonic Vibrations
⎛
⎝
k − ω2M −k 0
−k 2k − ω2m −k
0 −k k − ω2M
⎞
⎠
⎛
⎝
w1
w2
w3
⎞
⎠ = 0,
in short (
K − ω2M
)
w = 0 (4.30)
has to hold. By the ansatz (4.29) for the displacements ui(t) the system of lin-
ear differential equations has turned into a system of linear algebraic equations
for the amplitudes wi. These equations are homogeneous (like the differential
equations). The condition for the solubility of any system of linear homoge-
neous equations is the vanishing of the secular determinant,
0 = det
(
ω2M − K
)
= det
⎛
⎝
ω2M − k k 0
k ω2m− 2k k
0 k ω2M − k
⎞
⎠
=
(
ω2M − k
)2 (
ω2m− 2k
)
− 2k2
(
ω2M − k
)
=
(
ω2M − k
) [(
ω2M − k
) (
ω2m− 2k
)
− 2k2
]
=
(
ω2M − k
) [
ω4Mm− ω2k (2M +m)
]
=
(
ω2M − k
)
ω2Mm
(
ω2 − k
2M +m
Mm
)
with the three solutions
ω2
1 = 0
ω2
2 = k
1
M
ω2
3 = k
(
2
m
+
1
M
)
.
The first part of the eigensolutions thus consists of the eigen values ±ωj
(j = 1, 2, 3).
Comments:
• The eigen values ω2
j are the eigen values of the differential equation (4.28)
or of the algebraic equation (4.30). One obtains the eigen value problem of
the theory of matrices in the standard form, if one multiplies (4.30) with
M−1/2 (which exists, since M is diagonal),
(
M−1/2KM−1/2 − ω21
)(
M1/2w
)
≡
(
K̃ − ω21
)
w̃ = 0.
• The coefficient matrix K or K̃ is symmetrical because of Newton’s reaction
law and thus has real eigen values ω2.18
18 See Appendix G.3.
4.4 Coupled Oscillators 141
• The stability of the system under displacements from the equilibrium con-
figuration (like for the minimum of the potential of the single oscillator)
has the consequence that the eigen values ω2
j not only are real but that
also ω2
j ≥ 0 is true.
With the knowledge of the eigen frequencies one can determine the eigen
vectors as the second part of the eigen solution. For the calculation of the
eigen vector belonging to an eigen value one substitutes the eigen value in the
eigen value equation (4.30).
Case ω2 = ω2
1 = 0: The eigen value (4.30) takes the form
⎛
⎝
−k k 0
k −2k k
0 k −k
⎞
⎠
⎛
⎜⎝
w
(1)
1
w
(1)
2
w
(1)
3
⎞
⎟⎠ = 0.
If one writes explicitly for the first row, one obtains
−k
(
w
(1)
1 − w
(1)
2
)
= 0,
from which one obtains
w
(1)
1 = w
(1)
2
and analogously from the third row
w
(1)
3 = w
(1)
2 .
Of course, the eigen vector19 is determined up to a normalization factor N ,
w(1) = N (1)
⎛
⎝
1
1
1
⎞
⎠ .
Case ω2 = ω2
2 = k/M : The eigen value equation (4.30) takes the form
⎛
⎝
0 k 0
k k(mM − 2) k
0 k 0
⎞
⎠
⎛
⎜⎝
w
(2)
1
w
(2)
2
w
(2)
3
⎞
⎟⎠ = 0.
From the first row one obtains
kw
(2)
2 = 0 ⇒ w
(2)
2 = 0.
19 The three components of the eigen vector are the displacements of the three atoms
in the binding direction.
142 4 Harmonic Vibrations
j = 1: Translation of the molecule as a
rigid unit; all eigen vector components
are equal; there is no restoring force,
for which. . . . . . . . . . . . . . . . . . . . . . . . . 84
3.9.2 Lagrangian Equations of the First Kind . . . . . . . . . . . . . . 85
3.9.3 Example: Atwood Machine . . . . . . . . . . . . . . . . . . . . . . . . . 86
3.10 Hamilton’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
3.10.1 The Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
3.10.2 Hamilton’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
3.11 Symmetries and Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . 95
3.11.1 Noether’s Theorem (II) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
3.11.2 Cyclic Coordinate and Conservation
of the Conjugate Momentum. . . . . . . . . . . . . . . . . . . . . . . . 96
3.11.3 Homogeneity in Time and Energy Conservation . . . . . . . 97
3.11.4 Space Homogeneity and Conservation of Momentum . . . 100
3.11.5 Isotropy and Angular-Momentum Conservation . . . . . . . 100
Summary: Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4 Harmonic Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
4.1 The Simple Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
4.1.1 Eigen Solutions of the Homogeneous Differential
Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
4.2 The Free Oscillator with Damping . . . . . . . . . . . . . . . . . . . . . . . . . 120
4.2.1 Eigen Solutions of the Homogeneous Differential
Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
4.2.2 Reality of the Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
4.2.3 Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
xii Contents
4.3 The Forced Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
4.3.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
4.3.2 Solution of the Inhomogeneous Linear Differential
Equation by Fourier Transformation . . . . . . . . . . . . . . . . . 126
4.3.3 Green Function of the Damped Oscillator
in Frequency Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
4.3.4 Time-Dependent Green Function . . . . . . . . . . . . . . . . . . . . 131
4.3.5 Energy Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
4.4 Coupled Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
4.4.1 Introductory Example: Stretching Vibrations
in the CO2 Molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
4.4.2 General Coupled Vibrations . . . . . . . . . . . . . . . . . . . . . . . . 145
4.4.3 Normal Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
Summary: Harmonic Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
5 Central Potentials and the Kepler Problem . . . . . . . . . . . . . . . . 157
5.1 Central Force and Motion in a Plane . . . . . . . . . . . . . . . . . . . . . . . 157
5.1.1 Central Potential, Central Force, and
Angular-Momentum Conservation . . . . . . . . . . . . . . . . . . . 157
5.1.2 Central Potential and Effective Radial Potential . . . . . . . 159
5.1.3 Central Potential and Trajectory . . . . . . . . . . . . . . . . . . . . 160
5.2 Kinematics of the Kepler Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 163
5.2.1 Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
5.2.2 Polar Representation of the Conics . . . . . . . . . . . . . . . . . . 165
5.2.3 Determination of the Force and Potential
from the Trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
5.2.4 Determination of the Trajectory from the Potential . . . . 168
5.2.5 Trajectory and Rotation Periods . . . . . . . . . . . . . . . . . . . . 171
5.2.6 The Laplace–Runge–Lenz Vector . . . . . . . . . . . . . . . . . . . . 173
5.2.7 Perihelion Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
Summary: Central Potentials and the Kepler Problem . . . . . . . . . . . . 177
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
6 Collision and Scattering Problems . . . . . . . . . . . . . . . . . . . . . . . . . 183
6.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
6.1.1 Scattering and Collision . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
6.1.2 Momentum Change and Impulsive Force . . . . . . . . . . . . . 184
6.1.3 Laboratory System and Center-of-Mass System . . . . . . . 185
6.1.4 Consequences of the Conservation of Momentum . . . . . . 186
6.1.5 Elastic and Inelastic Scattering . . . . . . . . . . . . . . . . . . . . . . 187
6.1.6 Consequences of the Angular-Momentum
Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
Contents xiii
6.2 Collision of Hard Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
6.2.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
6.2.2 Elastic Collision of Smooth Spheres
(Laboratory System) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
6.3 Scattering by a Central Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 194
6.3.1 Central Potential and Scattering Angle . . . . . . . . . . . . . . . 194
6.3.2 Scattering by the Gravitational Potential . . . . . . . . . . . . . 195
6.4 * The Cross-Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
6.4.1 The Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
6.4.2 Scattering of Many Probe Particles
by Many Target Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
6.4.3 Scattering from a Particle at Rest . . . . . . . . . . . . . . . . . . . 197
6.4.4 The Differential Cross-Section . . . . . . . . . . . . . . . . . . . . . . . 198
6.4.5 The Total Cross-Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
6.4.6 The Rutherford Cross-Section . . . . . . . . . . . . . . . . . . . . . . . 200
6.4.7 Scattering Cross-Section for a Collision
of Hard Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
Summary: Collision and Scattering Problems . . . . . . . . . . . . . . . . . . . . 202
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
7 Moving Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
7.1 Translations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
7.1.1 The Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
7.1.2 Inertial Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
7.2 Rotation Around a Fixed Point . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
7.2.1 Active and Passive Rotation . . . . . . . . . . . . . . . . . . . . . . . . 209
7.2.2 Infinitesimal Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
7.2.3 Representation in Different Coordinate Systems . . . . . . . 211
7.2.4 Uniformly Rotating System: Centrifugal
and Coriolis Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
7.2.5 The Foucault Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
7.3 Galilean and Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . 217
7.3.1 The Relativity Principlereason one has ω = 0
j = 2: Vibration of even parity; upon
inversion the vibrational pattern re-
mains unchanged
j = 3: Vibration of odd parity; upon
inversion all displacements change
their sign
Fig. 4.15. The eigen vectors (these are the displacement amplitudes of the atoms)
of the three stretching eigen vibrations with j = 1, 2, 3 (from top to bottom) of the
CO2 molecule of Fig. 4.14
The same is obtained from the third row. The second row is, with w
(2)
2 = 0,
k
(
w
(2)
1 + w
(2)
3
)
= 0 ⇒ w
(2)
1 = −w(2)
3 .
The eigen vector belonging to the second eigen value ω2
2 is (up to a normal-
ization factor)
w(2) = N (2)
⎛
⎝
1
0
−1
⎞
⎠ .
Case ω2
3 = k( 2
m + 1
M ): With the analogous procedure one obtains
k
⎛
⎝
2Mm 1 0
1 m
M 1
0 1 2Mm
⎞
⎠
⎛
⎜⎝
w
(3)
1
w
(3)
2
w
(3)
3
⎞
⎟⎠ = 0
and from the first and from the third row
w
(3)
2 = −2
M
m
w
(3)
1 = −2
M
m
w
(3)
3 .
The eigen vector is (up to a normalization factor)
w(3) = N (3)
⎛
⎝
m
−2M
m
⎞
⎠ .
Comments:
• One obtains here three so-called normal vibrations (eigen vibrations),
which are defined as those vibrations in which all particles vibrate with the
4.4 Coupled Oscillators 143
same frequency and fixed phase relation. In systems with n degrees of free-
dom one obtains correspondingly n normal vibrations. The corresponding
frequencies are the eigen frequencies.
• The interpretation of the normal vibrations (eigen vibrations) with the
frequencies ωj and eigen vectors w(j) can be taken from Fig. 4.15.
• The “vibration” with j = 1 describes the motion of the center of mass.
For the vibrations with j = 2 and j = 3 the center of mass is at rest. (The
eigenvectors are orthogonal to each other.20)
The eigen solutions are thus
u1,+(t) = w(1) , u1,−(t) = w(1)t,
u2,±(t) = w(2)e±iω2t,
u3,±(t) = w(3)e±iω3t
(ω1 = 0 is a twofold zero!21). These are six eigen solutions, from the three
equations of second order. The general solution is a superposition of the eigen
solutions,
u(t) = w(1) (a+ bt) +
3∑
j=2
(
ajw
(j)e−iωjt + c.c.
)
.
where c.c. stands for the complex-conjugate expression.
One can compare the nonzero frequencies by eliminating the force
constant k. The theoretical value for the frequency ratio from this simple
model is
ω3
ω2
=
√
2
m + 1
M
1
M
=
√
2
M
m
+ 1 =
√
2
16
12
+ 1 = 1.915.
From the experimentally determined frequencies22 of the stretching vibrations
CO2 one finds23 1/λ3 = 2, 396 cm−1 and 1/λ2 = 1, 351cm−1 and thus
ω3
ω2
=
1/λ3
1/λ2
=
2396
1351
= 1.77.
The ratio of the frequencies is thus obtained in the correct order of magnitude
with an error of approximately 10%. Actually, there is also a coupling between
the two oxygen atoms, which has been neglected in the model and which is
responsible for the deviation of the theoretical from the experimental value.
The force constant for the stretch vibration can thus be determined with an
error of approximately 20%.
20 See Appendix G.4.
21 See (F.5).
22 The energy unit customary in infra-red spectroscopy is the wave number λ =
1/λ = ν/c = ω/(2πc).
23 From a very old compilation of spectroscopic data of molecules [12].
144 4 Harmonic Vibrations
(II) “Clever” Linear Combination
In systems with inversion symmetry, i.e., in systems, which are transformed
onto themselves under space inversion, one can classify the vibrations as those
with either even or odd parity, i.e., displacements, which under space inversion
are transformed either into themselves or into their negative. If the positions
of the particles with the indices i and −i are related to each other by inversion
symmetry, one can form the combinations
ui ± u−i.
One obtains for the CO2 molecule
⎛
⎝
M
M
m
⎞
⎠
⎛
⎝
ü1 − ü3
ü1 + ü3
ü2
⎞
⎠ =
⎛
⎝
−k 0 0
0 −k 2k
0 k −2k
⎞
⎠
⎛
⎝
u1 − u3
u1 + u3
u2
⎞
⎠ .
From here on the exponential ansatz must be made again for the solution.
From the block-diagonal form of the force-constant matrix one can see imme-
diately that there is one vibrations of even and two vibrations of odd parity,
and the solution ω2
+ = k/M and w
(+)
1 = −w(+)
3 for the vibration with of even
parity (with j = 2 from above, see also Fig. 4.15) can be read off immediately
from the block-diagonal form of the force-constant matrix.
For the subsequent evaluation the exponential ansatz can be used again.
Comment on Practical Implications:
Usefulness of the Symmetry Analysis
At this place the symmetry of the system is employed in order to reduce the
expense. In the example of the CO2 molecule one has, besides the identity
only one further symmetry operation, namely the inversion (identical with a
mirror operation). In the case of other molecules as well as of solids often
there are essentially more symmetry operations, and the consequence for the
symmetry properties of the eigensolutions is the topic of the theory of groups
and representations.
In most cases, the form of the eigen vectors is not apparent, except possi-
bly for the parity in systems with inversion symmetry. However, for a closed
system there are always the rigid translations in the three orthogonal direc-
tions with eigen frequencies ω = 0 because of the conservation of the center
of mass; in addition the other eigen vectors have to be orthogonal to these
eigen vectors of the rigid translation, i.e., keeping the center of mass at rest.
(III) Transformation
The decomposition in vibrations of even and of odd parity one can also per-
form by the (orthogonal) transformation24
24 The transformation matrix T is orthogonal, T−1 = TT , see Appendix G.2.
4.4 Coupled Oscillators 145
T =
⎛
⎝
b 0 −b
b 0 b
0 1 0
⎞
⎠ with b =
1√
2
One obtains
T
⎛
⎝
M
m
M
⎞
⎠
⎛
⎝
ü1
ü2
ü3
⎞
⎠
= T
⎛
⎝
M
m
M
⎞
⎠T−1T
⎛
⎝
ü1
ü2
ü3
⎞
⎠ =
⎛
⎝
M
M
m
⎞
⎠
⎛
⎝
b (ü1 − ü3)
b (ü1 + ü3)
ü2
⎞
⎠
= T
⎛
⎝
−k k 0
k −2k k
0 k −k
⎞
⎠T−1T
⎛
⎝
u1
u2
u3
⎞
⎠
=
⎛
⎝
−k 0 0
0 −k 2bk
0 2bk −2k
⎞
⎠
⎛
⎝
b (u1 − u3)
b (u1 + u3)
u2
⎞
⎠ .
For the consecutive evaluation the exponential ansatz can be used again.
4.4.2 General Coupled Vibrations
The equation of motion for a system of N coupled oscillators can be written
in the form
⎛
⎝
m1 0 . . .
0 m2 . . .
...
...
. . .
⎞
⎠
⎛
⎝
ü1
ü2
...
⎞
⎠ = −
⎛
⎝
a11 a12 . . .
a21 a22 . . .
...
...
. . .
⎞
⎠
⎛
⎝
u1
u2
...
⎞
⎠ ,
in short
Mü = −Ku. (4.31)
The displacement vector u contains the 3N components of the N oscillators,
and the matrices K and M are (3N × 3N)-matrices. With the ansatz
u(t) = w e−iωt
one obtains from the system of (homogeneous linear) differential equation
(4.31)
−ω2Mw e−iωt = −Kw e−iωt
a system of (homogeneous linear) algebraic equations
(
K − Mω2
)
w = 0. (4.32)
146 4 Harmonic Vibrations
The solubility condition for the linear homogeneous system of equations (4.32)
is the vanishing of the secular determinant,
det
(
K − Mω2
)
= 0,
from which one obtains the eigen values ω2
j and with these the eigen vectors25
wj ; eigen values and eigen vectors fulfill the equation
(
K − Mω2
j
)
wj = 0. (4.33)
and in combination form the eigen solutions,26
wj e±iωjt (j = 1, . . . , 3N).
The eigen vectors fulfill the orthogonality relation27
w†
i · M · wj = δi,j (4.34)
and the completeness relation
∑
j
wjw
†
j · M =
∑
j
M · wjw†
j =
∑
j
M1/2 · wjw†
j · M1/2 = 1, (4.35)
i.e., the vectors M1/2wj are orthonormal. Here, in the completeness relation
the tensor product28 wjw
†
j of the eigen vectors is to be taken. The completeness
relation says that each displacement u can be described as a superposition of
eigen displacements wj ,
u = 1 · u =
∑
j
wj
(
w†
j · M · u
)
.
Proof:
(i) Orthogonality: Since K and M are symmetrical matrices,
K† = K ⇔ aij = aji,
one has
w†
j · K = (K · wj)† =
(
ω2
jM · wj
)†
= ω2
jw
†
j · M,
25 For simplicity we use a lower index here.
26 For multiple zeroes see (F.5).
27 The notation w† stands for the complex-conjugate and transpose of the quan-
tity w; the transpose of a row vector is a column vector and vice versa, see
AppendixG.2.
28 The tensor product ab of two vectors a and b (with the elements ai and bj ,
respectively) is a tensor with the elements (ab)ij = aibj . Sometimes the tensor
product is also written as a⊗b. To distinguish the tensor product from the scalar
product it is thus important that the scalar product (dot product, inner product)
a · b is written with a multiplication dot.
4.4 Coupled Oscillators 147
and thus one obtains from the eigen value equation
0 = w†
j · (K − ω2
iM) · w(i) = w†
j · M · wi(ω2
j − ω2
i ).
This expression vanishes for different frequencies only if the orthogonality
relation is fulfilled. For equal eigen values the eigen vectors can be chosen
orthogonal. The normalization factor in front of the Kronecker-symbol in
(4.34) is chosen to be unity.
(ii) Completeness: See problem 4.4.
�	
Comment: For the case of the CO2 molecule see problem 4.4. The general
solution is a superposition of the eigen solutions29
u(t) =
∑
j
ajwje−iωjt + c.c.
Here, u(t) and wj are vectors, and aj are superposition coefficients.
4.4.3 Normal Coordinates
In Sect. 4.4.1 use has been made of the symmetry of the CO2 molecule in order
to bring the matrix of the force constant to a block-diagonal form; even though
these blocks describe the vibrations of even parity decoupled from those of odd
parity, the oscillators within a given parity are still coupled to each other. The
diagonalization of the force-constant matrix described by (4.33) has effected
a transformation from coupled harmonic oscillators to completely decoupled
so-called normal vibrations.
One writes the displacements u(t) (the “normal” coordinates with 3N
components for N particles) as a (time dependent) superposition of the eigen
displacements wj ,
u(t) =
∑
j
wjQj(t). (j = 1, . . . , 3N) (4.36)
The superposition coefficients Qj(t) are the so-called normal coordinates. The
reverse of (4.36) is
Qj = w†
j · M · u. (4.37)
Proof:
w†
j · M · u (4.36)
= w†
j · M ·
∑
j′
w†
j′Qj′
(4.34)
=
∑
j′
δj,j′Qj′ = Qj .
�	
29 For multiple zeroes see (F.5).
148 4 Harmonic Vibrations
The equation of motion for a normal coordinate is
Q̈j = −ω2
jQj. (4.38)
Proof: With (4.37) one obtains
Q̈j = w†
j · M · ü = −w†
j · K · u
= −
∑
j′
w†
j · K · wj′Qj′ = −
∑
j′
w†
j · M · wj′ω2
j′Qj′
= −
∑
j′
δj,j′ω
2
j′Qj′ = −ω2
jQj .
�	
Comments:
• The eigen solutions Qj(t) ∝ e±iωjt of (4.38) are obvious.
• The normal coordinates are the starting point for the quantization of vi-
brations in quantum mechanics.
Summary: Harmonic Vibrations
Equation of Motion of Harmonically Coupled Damped Vibrations:
(A, B, . . . are (N ×N)-matrices, u, w are (N × 1) vectors):
Dtü ≡ Aü + Bu̇ + Cu ≡
(
Ad2
t + Bdt + C
)
u = f(t) (with dt ≡ d/dt)
with
f(t) =
∫
dω
2π
e−iωtFω.
The General Solution is the superposition of the general solution of the
homogeneous equation and of a particular solution of the inhomogeneous
equation,
u(t) = uhom(t) + upart(t)
with
Dtu
hom(t) = 0
With the ansatz
uhom(t) = w e−iωt
Problems 149
one obtains
D(ω)w ≡
(
−ω2A − iωB + C
)
w
with the solution condition for the N values of ω2 (secular equation)
det
(
−ω2A − iωB + C
)
= 0
A Particular Solution is
upart(t) =
∫
dω
2π
e−iωtUpart
ω
with
Upart
ω = [D(ω)]−1
Fω .
Problems
4.1. Normal Coordinates. Prove that the energy of a system of coupled
oscillators, expressed by normal coordinates Qj , can be written in the form
E = 1
2
∑
j
(
Q̇jQ̇
†
j + ω2
jQjQ
†
j
)
.
4.2. The Harmonic Oscillator. For the harmonic oscillator the force F
acting on a particle is proportional to its displacement r from the equilibrium
position and oriented oppositely,
F = −kr.
(a) The One-dimensional Harmonic Oscillator:
Set up the equation of motion for a one-dimensional harmonic oscillator,
and solve this.
From the integration of the energy conservation theorem one obtains
t− t0 =
∫ x
x0
dx′√
2
m [E − U(x′)]
with x0 = x(t0). From this derive x(t).
(b) The n-Dimensional Harmonic Oscillator: Given be an n-dimensional har-
monic oscillator.
By integration of the energy conservation theorem derive an equation
equivalent to the case (a). Rationalize your procedure.
150 4 Harmonic Vibrations
4.3. Transverse Elongation of a Spring. Consider the sketched system of
a mass m and two equal springs (spring constant k, equilibrium length l0)
shown in Fig. 4.16. One end of each spring is attached to the mass, and the
other end is fixed. The distance of the two fixed ends is 2l. The potential
energy of a spring of the length l obeying Hooke’s law is
V = 1
2k (l − l0)
2
.
Calculate the restoring forces for a (small) displacement z of the mass in the
direction perpendicular to the springs for the cases l = l0 and l �= l0 to leading
order of z.
Fig. 4.16. Transverse displacements of springs (for problem 4.3)
4.4. Eigenvectors of Coupled Oscillators. Determine the normalization
constants N (j) (1 ≤ j ≤ 3) of the eigen vectors w(j) of the CO2 molecule of
Sect. 4.4.1, and prove the orthogonality and completeness of the eigen vectors.
4.5. Coupled Pendulums. Two equal pendulums (small amplitudes) are
coupled to each other by a harmonic spring.
(a) Write down the normal vibrations of the system.
(b) Solve the eigen value problem, if at time t = 0 these two pendulums are
in the equilibrium positions and the left pendulum has the velocity v.
4.6. Diatomic Molecule. Two atoms of masses m1 and m2, respectively,
(for example Na and Cl or C and O,but also H and H) shall be coupled to each
other by a harmonic potential (spring constant f) see Fig. 4.17. Determine the
eigen vibrations in the direction of the connecting line (valence vibrations) by
the solution of the coupled equations of motion. Identify the eigen vectors.
M1 M2
f Fig. 4.17. The model of the diatomic molecule
(for problem 4.6)
Problems 151
4.7. Eigen Modes of a Linear Array of Atoms. Consider a linear chain
of three equal masses. The masses are coupled to each other and to the walls
by springs of equal strength. The springs shall be stretched even in the equi-
librium state of the system by the force F . Let the equilibrium distance of the
masses be l0 (see Fig. 4.18).
(a) Set up the system of the equations of motion for transverse vibrations
with small displacements. Here, consider only the interaction of a mass
with its nearest neighbors.
(b) Solve the equations of motion, and determine the eigen frequencies and
eigen vectors for transverse vibrations. Represent the calculated eigen vec-
tors in a graph.
1 2 3
l0
l
Fig. 4.18. The chain of three particles (for problem 4.7)
Hint: It is useful to interpret the system of equations to be solved as an
eigen value problem.
4.8. Valence vibrations of the CH4 Molecule. In the equilibrium state
of the CH4 molecule the H atoms are positioned at the corners of a symmet-
rical tetrahedron around the C atom, see Fig. 4.19. To describe the valance
vibrations, during which the H atoms move along the lines connecting the
edges of the tetrahedron to the central C atom, assume that forces act only
between C and H which are described by a potential of the form
V (xH
i ) =
1
2
f
(
xH
i − xC
)2
(i = 1, . . . , 4). Here, xH
i and xC denote the positions of the H atoms and of
the C atom, respectively.
(a) Write down the position vector for the center of mass of the molecule,
and set up the equations of motion of the single atoms with respect to an
arbitrary coordinate origin. Do the equations of motion change upon the
transition to the center-of-mass system?
(b) Which simplifications of the equations of motion are possible, if only the
valence vibrations are to be determined? Give an expression for the equa-
tions of motion for the valence vibrations.
152 4 Harmonic Vibrations
C
H1
H2
H3
H4
x C
x1
→
→
H
Fig. 4.19. The CH4 molecule (for prob-
lem 4.8)
(c) Solve the coupled equations of motion, and write down the eigen frequen-
cies and eigen vectors. What influence does the special configuration of
the H atoms have on the eigen vibrations? Illustrate the eigenvectors by
sketches.
4.9. A Pendulum with Mobile Suspension. A mass m2 bound to a string
of length � oscillates in a plane under the influence of gravity. The suspension
is formed by a mass m1 which is freely mobile along a line in horizontal
direction (like a trolley on a (fixed) crane’s horizontal extension arm and a
bucket of concrete hanging from it), see Fig. 4.20.
Fig. 4.20. The pendulum with mobile suspension (for prob-
lem 4.9)
(a) Set up the Lagrangian function in appropriate generalized coordinates, and
from this derive the equations of motion. Which quantities are conserved?
(b) Solve the equations of motion for small angular displacements of the mass
m2, and determine the vibrational frequency.
Hint: It is useful first to approximate the equations of motion for small
angles and then to separate the independent variables.
(c) Discuss the solution, in particular in the limits m1 � m2 and m2 � m1.
What kind of motion is performed by the mass m1?
4.10. The Cycloid Pendulum After Huygens (Patent 1657). Two
blocks of cycloidic form constrain the space of motion of a pendulum. Let A
be the end point of the line touching the blocks, see the left part of Fig. 4.21.
Problems 153
The parameter representation of a cycloid results from the motion of a point
on a circle, if the circle rolls along a straight line, ξ = R (ϕ − sin ϕ), ζ =
R (1 − cos ϕ).
(a) Introduce the angle ϕ as the generalized coordinate, and determine the
coordinates of the point P .
(b) Choose the length � of the pendulum such that a most simple coordinate
representation results (trigonometry!)
(c) Write down the kinetic and potential energy of the point mass as a function
of ϕ. Choose the potential energy to vanish at z = −�.
(d) Introduce a new generalized coordinate q which replaces the old coordinate
ϕ, and determine the Lagrangian function.
(e) Write down the equation of motion for q. Is there something special with
the vibrational period?
A
P
x
z
0
z
j
x
Fig. 4.21. Left : The cycloid pendulum. Right : Geometrical construction of the
cycloid (for problem 4.10)
4.11. Diatomic Linear Chain. Consider a diatomic linear chain of atoms,
sketched in Fig. 4.22, with masses m and M ; a is the lattice constant (i.e., the
periodicity length). Between neighboring atoms there is a harmonic force with
the spring constant k. In equilibrium the distance of the equilibrium positions
x0
n and y0
n, respectively, of neighboring atoms is equal to 1
2a, see Fig. 4.22.
f m M
Fig. 4.22. The linear chain (for problem 4.11)
(a) Set up the Lagrangian function for the system. Use the displacements
un = xn−x0
n and vn = yn− y0
n out of the equilibrium position x0
n and y0
n,
154 4 Harmonic Vibrations
respectively, as coordinates. Then derive the equations of motion, and
compare with the Newtonian equations of motion.
(b) Solve the system of differential equations with the ansatz (“Bloch ansatz”)
un = Aei(qna−ωt) and vn = Bei(qna−ωt), (4.39)
respectively, for the two different kinds of atoms, i.e., determine the eigen
frequencies and eigen vectors as well as the general solution.
(c) The system has infinitely many degrees of freedom, and therefore one
obtains a continuum of possible eigen frequencies. Derive the relation be-
tween the frequency ω and the wave number q, the so-called “dispersion
relation.”
(d) Assume at first M > m. Sketch the dispersion relation, and comment on
the periodicity of ω(q). What are the frequencies of the two dispersion
branches at the points q = 0 and q = π/a? Calculate and sketch the
displacement pattern of the atoms at the special points q = 0 and q =
π/a. The two branches of the dispersion relation are called “optical” and
“acoustic” branch, respectively, why?
(e) Calculate the “sound velocity” in the chain.
(f) The frequencies of the two branches fill the so-called allowed bands. Be-
tween the bands there is a band gap. What kind of wave vector belongs
to a (so-called forbidden) frequency outside of the allowed bands? What
is the consequence for the position dependence of the wave, if one forces
the system with a frequency outside of the allowed bands? Sketch the
dispersion relation including complex wave vectors.
(g) Now set M = m. How does the dispersion relation change? What would
one have obtained, if one had started the calculation with a monatomic
chain?
4.12. Monatomic Linear Chain – Transverse Vibrations. Consider a
monatomic linear chain at rest at the ends of which a force F acts along
the chain outwards (mass m, lattice constant a of the stretched chain, spring
constant k).
(a) The masses of the chain are now displaced transversely (perpendicular to
the chain). Show that in linear approximation the forces along the chain
are independent of the displacement.
(b) Set up the equations of motion for the transverse displacements, and solve
them with the ansatz
un(q, t) = ei(qan−ωt).
How does ω depend upon q?
(c) Perform the limit to the string (a → 0). How does one have to scale the
masses and the force constant k with a?
Problems 155
4.13. Harmonically Driven Oscillator with Friction. The external
forces acting on a vibrating body are the frictional force and a periodic force
F0 cos(ωt), such that one has the following equation of motion for the body,
mẍ+ kẋ+Dx = F0 sin(ωt).
(a) Show that the displacement due to this force is x(t) = x0 sin(ωt+ ϕ).
(b) Sketch the vibrational amplitude x0 as a function of ω, and discuss the
limiting case of vanishing damping (k → 0).
(c) Sketch and discuss the phase shift ϕ between the periodic force and the
vibrational displacement as a function of ω.
(d) How does the result x(t) = x0 sin(ωt + ϕ) correlate to the expression
x(t) = x1 cos(ωt) + x2 sin(ωt)?
(e) Determine the most general solution of the equation of motion.
5
Central Potentials and the Kepler Problem
In this chapter the investigations of the Sects. 2.3 and 2.2 are applied to the
bound1 motion of a particle (of mass m) in a central potential
V (r) = V (|r|) = V (r). (5.1)
Included in this chapter is the case of the relative motion of two particles of a
closed system (i.e., without external forces), the interaction energy of which
is described by a central potential; this case can be reduced to the case of a
single particle, if only the mass m of the single particle is substituted by the
reduced mass μ of the pair.2 The specialization to the gravitational potential
(i.e., to the Kepler problem) will be made in Sect. 5.2.
5.1 Central Force and Motion in a Plane
5.1.1 Central Potential, Central Force, and Angular-Momentum
Conservation
Central Force
The isotropy of the potential (5.1) leads to a conservative central force
F (r) = −∇V (r) = −dV (r)
dr
er = F (r)er (5.2)
and thus the angular momentum3 l (with respect to the center) and the
energy4 E are conserved quantities.
1 The free (unbound) motion is treated in the following Chap. 6 on the scattering
of particles.
2 See Sect. 2.3.
3 See Sect. 1.6.2.
4 See Sect. 1.7.
158 5 Central Potentials and the Kepler Problem
Alternatively: The system of a particle in a time independent potential is
invariant under translations in time, and with this (because of the absence
of any, in particular rheonomous, constraints) the energy is a constant of the
motion.5 In addition, the system of a particle in a central potential V (r) is
invariant under (space) rotations, and with this the angular momentum is a
constant of the motion.6
Plane Motion
Because of
l = r × p = const. ⇒ r(t) ⊥ l and p(t) ⊥ l
the position r and the momentum p always lie in a plane perpendicular to the
angular momentum l, i.e., the motion takes place in a plane perpendicular
to l. Well-known examples are the motion of planets in the gravitational field
of the sun (neglecting the influence of other planets) or that of a satellite in the
field of the earth (neglecting the influence of sun, moon, and other planets).
The investigation of the generally three-dimensional motions under the
action of central forces can thus be reduced to twodimensions.7 (Here the
use of plane polar coordinates is then of an advantage, and one obtains, e.g.,
l = mr2ϕ̇ for the modulus of the angular momentum.)
Conservation of Areal Velocity
In the context of the planetary motion in former years one has talked of
the conservation of the areal velocity Ḟ , which states nothing else than the
conservation of the angular momentum l = |l|,
l = 2mḞ = const. (5.3)
The conservation of areal velocity Ḟ says that the area F covered by the radius
per unit time is a constant. Even though mentioned in general in the context
of the gravitational potential, one has the conservation of areal velocity for
any central potentials.
Proof: The intuitive derivation of the conservation of areal velocity goes
as follows: In a small time interval dt the position vector changes from r to
r + dr, see Fig. 5.1. The area covered during this time interval is
dF = 1
2 |r × dr|
5 See Sect. 3.11.3.
6 See Sect. 3.11.5.
7 The vanishing of the coordinate perpendicular to the plane, in which the motion
takes place, is then not a constraint but a consequence of the form of the potential.
5.1 Central Force and Motion in a Plane 159
dF
r
dr
Fig. 5.1. Notations in the context of the conservation
of areal velocity
and
dF
dt
= 1
2 |r × ṙ|
(2.17)
= 1
2 |r er × (ṙ er + rϕ̇eϕ)|
= 1
2 |r ṙ er × er + r2ϕ̇er × eϕ|
= 1
2 |0 + r2ϕ̇ez| = 1
2 r
2 ϕ̇ = const.
Except for the factor 2m this is just the angular momentum. �	
5.1.2 Central Potential and Effective Radial Potential
As obtained in Sect. 5.1.1 the energy and angular momentum are conserved
quantities for a central potential,
E = T + V =
m
2
ṙ2 + V (r) =
m
2
(
ṙ2 + r2ϕ̇2
)
+ V (r) = const. (5.4)
l = mr2 ϕ̇ = const. (5.5)
If one expresses the angular velocity ϕ̇ in (5.4) by the angular momentum l
of (5.5), one obtains
E =
m
2
ṙ2 +
l2
2mr2
+ V (r). (5.6)
The second term of the right-hand side is the so-called centrifugal potential. If
one combines this centrifugal potential with the external potential V (r), one
obtains an effective potential
Veff(r) = V (r) +
l2
2mr2
, (5.7)
valid for the radial motion. Instead of (5.6) the energy can thus be written
also in the form
E =
m
2
ṙ2 + Veff(r). (5.8)
An example is plotted in Fig. 5.2.
160 5 Central Potentials and the Kepler Problem
10
x
−5
0
5
−1/x
0.12/x2
−1/x+ 0.12/x2
Fig. 5.2. The form of the effective potential (full line)
− 1
x
+
l2
x2
as the superposition of the true potential (dash-dotted line; here the gravitational
(or Coulomb) potential) −1/x and the centrifugal potential (dashed line) l2/x2 for
the case l = 0.1. See also Fig. 5.4 below
Comments:
• Introducing the effective potential one has to remember that it is for just
the radial motion. The angular motion has been eliminated by the use of
the conservation theorem for the angular momentum.
• For the (effectively) one-dimensional radial motion one can use all results of
Sect. 2.1 for the one-dimensional motion with the replacement x(t) → r(t).
5.1.3 Central Potential and Trajectory
Dynamics of a Particle in a Central Potential
Equation (5.8) for the energy can be integrated as in the case of the one-
dimensional motion in Sect. 2.1. From (5.8) one obtains
1
ṙ
=
dt
dr
= ± 1√
2
m [E − Veff(r)]
(5.9)
dt = ± dr√
2
m [E − Veff(r)]
(5.10)
5.1 Central Force and Motion in a Plane 161
and from there8
t(r) − t0 = ±
∫ r
r0
dr′√
2
m (E − Veff(r′))
(5.11)
with r0 = r(t0). From this one obtains the time t(r) as a function of the
distance r and from this the distance r(t) as a function of time t, if the
mapping r → t is injective.
Now (5.5) for the angular momentum – in principle – can also be inte-
grated,
ϕ(t) − ϕ0 =
l
m
∫ t
t0
dt′
r2(t′)
(5.12)
with r(t′) from the inversion of (5.11).
Proof: With the angular momentum l = mr2 ϕ̇ one obtains
dϕ
dt
=
l
mr2(t)
⇒ ϕ(t) − ϕ0 =
l
m
∫ t
t0
dt′
r2(t′)
with ϕ0 = ϕ(t0). �	
Kinematics of a Particle in a Central Potential
The functions r(t) and ϕ(t) together serve as a parameter description of the
trajectory (with the time t as the parameter). Eliminating t one obtains the
trajectory in the form r(ϕ).
The form r(ϕ) of the trajectory one can also be deduced directly from the
conservation laws,
ϕ(r) − ϕ0 =
∫ r
r0
dr
l/mr2√
2
m [E − Veff(r)]
. (5.13)
8 According to Goldstein [1], Sect. 3.5, the integral
dt = ± dr√
2
m
[E − V (r)]
with V (r) = a rn
can be expressed in term of
trigonometric functions for n = −2,−1, 2 ,
elliptical functions for n = −6,−4,−3, 1, 4, 6 .
162 5 Central Potentials and the Kepler Problem
From this one obtains the angle ϕ(r) as a function of the distance r and from
this the distance r(ϕ) as a function of the angle, if the mapping r → ϕ is
injective.
Proof: Eliminating dt from the energy and angular-momentum conservation
theorems one obtains
dt
(5.10)
=
dr√
2
m [E − Veff(r)]
⇒ dϕ
(5.5)
=
l
mr2
dt
(5.10)
=
l
mr2
dr√
2
m [E − Veff(r)]
(5.14)
⇒ ϕ− ϕ0 =
√
l2
2m
∫ r
r0
dr
r2
√
E − Veff(r)
.
�	
Determination of the Potential from the Trajectory
Reversely, one can determine the potential from the form r(ϕ) of the trajectory
with the help of (5.6),
V (r) = E − l2
2mr2
− l2
2mr4
(
dr
dϕ
)2
= E − l2
2m
[(
d1
r
dϕ
)2
+
1
r2
]
.
(5.15)
For the force one obtains
F = −∇V (r) = −dV
dr
er = −F (r)er
F (r) = −dV
dr
=
l2
mr4
[
d2r
dϕ2
− 2
r
(
dr
dϕ
)2
− r
]
= − l2
m
1
r2
[
d2 1
r
dϕ2
+
1
r
]
.
(5.16)
Proof: With
d
dt
=
dϕ
dt
d
dϕ
(5.5)
=
l
mr2
d
dϕ
(from l = mr2ϕ̇) one obtains for the radial component of the force
5.2 Kinematics of the Kepler Motion 163
Fr
(2.19)
= mr̈ −mrϕ̇2 = m
[(
l
mr2
d
dϕ
)2
r − r
(
l
mr2
)2
]
= m
(
l
mr2
)2
[
d2r
dϕ2
− 2
r
(
dr
dϕ
)2
− r
]
= −m
(
l
mr
)2 [d2 1
r
dϕ2
+
1
r
]
.
�	
Comment: The final forms in (5.15) and (5.16) will turn out to be practical
in the context of the trajectories (conics), along which a particle moves under
the influence of the gravitational potential, see the following section.
5.2 Kinematics of the Kepler Motion
In this section the investigations of Sect. 5.1 on the motion in a central force
field will be specialized to the gravitational field. Again, in this section only
the relative motion in the gravitational potential will be investigated. In this
section the reduced mass will be denoted again by m. Since, in the case of the
motion of the earth around the sun, the mass of the sun (m = 1.99× 1030 kg)
is very much larger than the mass of the earth (m = 5.98 × 1024 kg), thus
the reduced mass is nearly equal to the mass of the earth. One has similar
conclusions for the motion of the moon (m = 7.34×1022 kg) around the earth.
One could start from the known gravitational potential
V (r) = −G m1m2
|r1 − r2|
and deduce the planetary motion. However, in the following we want to start
from Kepler’s laws and from those draw conclusions about the existence and
the form of the potential.
5.2.1 Kepler’s Laws
Kepler’s First Law (1609)
The planet trajectories are ellipses, the focus of which is the sun.
Comment: In fact, there is an additional effect, the perihelion rotation, which
is caused by the influence of the other planets and by relativistic corrections.9
From the fact of the periodic motion one concludes that one has energy
conservation for the motion, properly speaking actually only on the time av-
erage; but if one assumes in the following that the energy is conserved also
locally in time (which would have to be proven), then the force must also be
conservative, and thus a potential energy V must exist.
9 See Sect. 5.2.7 further below.
164 5 Central Potentials and the Kepler Problem
Kepler’s Second Law (1609)
Conservation of areal velocity: The radius vector (from the sun to the planets)
traces equal areas Δa in equal times Δt,
ȧ = 0,
compare Fig. 5.1. Together with the plane of the motion claimed in the first law
this theorem is identical to the angular-momentum conservation theorem,10
l
(5.3)
= 2mȧ.
The force must thus be a central force.Together with the first Kepler law the
potential energy thus must depend upon the distance only,
V (r) = V (r) (5.17)
Proof: See (1.13) in Sect. 1.4. In plane polar coordinates one has11
F = −∇V (r, ϕ) = −er
∂V
∂r
− eϕ
1
r
∂V
∂ϕ
.
From F ‖ er follows
∂V
∂ϕ
= 0.
�	
Comments:
• From the elliptic form of the trajectory one obtains
V (r) = −k
r
with k =
l2
mp
. (5.18)
For the proof of this relation and for the determination of the constant k
see (5.24) of Sect. 5.2.3 further below.
• The first two laws already suffice to fix the distance dependence of the
gravitational potential.
Kepler’s Third Law (1619)
The cubes of the large half-axes of the different planets are proportional to
the squares of the rotation periods,
a3
T 2
=
(
l
2πm
)2 1
p
=
k
4π2m
= x = const. (5.19)
For the determination of the proportionality constant see Sect. 5.2.5 further
below.
10 See Sect. 5.1.1.
11 See the representation of the nabla operator in cylindrical coordinates in Ap-
pendix D.
5.2 Kinematics of the Kepler Motion 165
Comments:
• Actually this law is true only in the limit of the mass of the planet being
small compared to the mass of the sun, i.e., of the reduced mass thus
being approximately equal to the mass of the planet, see Sect. 5.2.5. This
is equivalent to the approximate assumption that the sun is fixed (in the
galaxy or in the universe).
• For the constant x in (5.19) to be independent of the mass m of the
planets one has to conclude that the constant k in the potential (5.18) is
proportional to the mass m of the planet, and because of the equality of
action and reaction (actio and reactio) thus proportional to the mass M
of the central star,
V (r) = −GmM
r
(5.20)
with the gravitational constant12
G = 6.67 × 10−11 m3 kg−1 s−2. (5.21)
• With Kepler’s third law then the gravitational potential is fixed (except
for the constant G). In fact, Kepler’s laws have been the experimental
basis for the derivation of the gravitational potential.13 (Kepler himself
had assumed a 1/r force rather than the actual 1/r2 force.)
5.2.2 Polar Representation of the Conics
As is well known the trajectories of the particles in the gravitational potential
are conics, the focus of which coincides with the center of the force. If one
chooses the focus as the origin of a system of plane polar coordinates, then
the conics are given by
r =
p
1 + e cosϕ
, (5.22)
Here e is the so-called eccentricity. The form of the conics is (Fig. 5.3)
e = 0 circle
0 1 hyperbola.
Comments:
• If one replaces e by −e, the conics are rotated by π.
12 A more precise value can be found in Appendix A.
13 See Sect. 1.4
166 5 Central Potentials and the Kepler Problem
b
a ea p
r
ϕ
b
a
j
p
r
Fig. 5.3. The polar representation of the ellipse (left) and of the hyperbola (right)
• For the bound motion,14 the points of minimum and maximum distance
from the force center (perihelion and aphelion, respectively) is given by
ϕ = 0 and ϕ = π, respectively,
rmin =
p
1 + e
and rmax =
p
1 − e
(|e| ≤ 1),
and in the perpendicularly oriented direction (ϕ = π/2) the distance of the
ellipse from the focus is equal to the half parameter p (the latus rectum).
The large and small half-axes of the ellipse are
a =
p
1 − e2
, b =
p√
1 − e2
, (5.23)
respectively.
5.2.3 Determination of the Force and Potential
from the Trajectory
Determination of the Potential from the Trajectory
One obtains
V (r) = −k
r
with k =
l2
mp
(5.24)
as stated in (5.18).
Proof: With
1
r
=
1
p
(1 + e cosϕ)
one obtains
V (r)
(5.15)
= E − l2
2mr2
− l2
2m
(
d1
r
dϕ
)2
(5.22)
= E − l2
2mp2
(1 + e cosϕ)2 − l2
2mp2
(−e sinϕ)2
= E − l2
2mp2
(
1 + e2 + 2e cosϕ
)
,
14 For the unbound motion see Sect. 6.3.2.
5.2 Kinematics of the Kepler Motion 167
and with
2e cosϕ
(5.22)
=
2p
r
− 2
one obtains from this
V (r) = E +
l2
2mp2
(
1 − e2
)
− l2
mp
1
r
.
With V (r) → 0 for r → ∞ one obtains not only
V (r) = − l2
mp
1
r
but also
E = − l2
2mp2
(
1 − e2
)
. (5.25)
�	
Comments:
• The potential V (r) is determined solely by the parameter p of the trajec-
tory. (The parameter e describes only the particular type and form of the
conic.)
• Kepler’s third law is not needed here.
Determination of the Force from the Trajectory
From the form of the trajectory one can infer the form of the (gravitational)
force: According to Kepler’s second law the force is a central force, F =
Fr(r)er, and as a conservative force it must then have the form F = Fr(r)er.
The radial component of the equation of motion is
Fr = m
(
r̈ − rϕ̇2
)
.
Looking for the r-dependence of Fr one obtains
Fr = − k
r2
with k =
l2
mp
. (5.26)
Proof: One obtains for the force
F
(5.16c)
= −m
(
l
mr
)2 [d2 1
r
dϕ2
+
1
r
]
.
With the polar representation
1
r
(5.16c)
=
1 + e cosϕ
p
168 5 Central Potentials and the Kepler Problem
of the trajectory’s ellipse one obtains
d2 1
r
dϕ2
= −1
p
e cosϕ =
1
p
− 1
r
and thus
F (r) = − l2
mp
1
r2
= − k
r2
.
Alternatively the force can be obtained from F = −∇V with V from (5.24).
�	
Comment: Analogous to the potential V (r) thus the force F (r) is determined
solely by the parameter p of the trajectory. (The parameter e describes only
the particular form of the trajectory.)
Parameters of the Trajectory and Conserved Quantities
In reverse of the foregoing one can relate the parameter of the polar represen-
tation of the ellipse to the constants of the motion, i.e., to the energy E and
to the angular momentum l,
r =
p
1 + e cosϕ
(5.27)
with
p =
l2
mk
and e =
√
1 +
2El2
mk2
. (5.28)
Proof: One determines p from (5.24) or (5.26) as well as e from p and (5.25).
�	
5.2.4 Determination of the Trajectory from the Potential
Kepler’s laws (reversely) can be deduced from the potential V (r) = −k/r:
The trajectories are conic curves (5.22) with the parameters from equation
(5.28).
Proof: With (5.13) for the trajectory of arbitrary central potentials and with
V (r) = −k/r one obtains
ϕ(r) =
l√
2m
∫
dr
r2
√
E + k
r − l2
2mr2
.
5.2 Kinematics of the Kepler Motion 169
Here it is practical to substitute u = 1/r (and du = −dr/r2),
ϕ(u) − ϕ0 = −
∫
du√
2mE
l2 + 2mk
l2 u− u2
= −
∫
du√
a+ bu+ cu2
GR 2.261=
1√
−c
arcsin
2cu+ b√
b2 − 4ac
(c 1 ⇔ E > 0 hyperbola
10 x
−5
0
−1
/x
 +
 l
2 /x
2
l = 0.0
l = 0.2
l = 0.3
l = 0.4
Fig. 5.4. The form of the effective potential in the case of the gravitational (or
Coulomb) potential for different angular momenta.
If one considers the effective potential
Veff(r)
(5.7)
= −k
r
+
l2
2mr2
for the radial motion (plotted in Fig. 5.4) one finds,16 that there is a minimum
distance for l �= 0. In the case E 0 there is no maximum
distance; the particle arrives from infinitely large distances and then vanishes
there; one denotes this by a scattering or unbound state. For a bound state
the ratio of the half-axes is
b
a
=
√
1 − e2 =
√
2|E|l2
mk2
.
Comments:
• Withincreasing angular momentum (at fixed energy) or with decreasing
energy (increasing modulus of the energy) (at fixed angular momentum)
15 See also p. 165.
16 See Sect. 2.1.4.
5.2 Kinematics of the Kepler Motion 171
the ratio b/a approaches unity. In contrast, in quantum mechanics one has
spherically-symmetrical situations for vanishing angular momentum.
• In the general case V (r) = krn−1 (except for n = 0 and n = 3) the
trajectories of the bound motion are not closed, but form rosette-like curves
with minimum and maximum distances. The point of nearest approach
to the center of the force is different from one revolution to the next
(perihelion rotation). One would obtain closed trajectories only if the angle
for a full revolution (from one maximum value of the distance r to the next)
would be a rational multiple of 2π. It seems to be an exceptional accident
that the time variation of the angle and of the distances in the Kepler
problem have the same period.17
5.2.5 Trajectory and Rotation Periods
First Alternative: Conservation of Areal Velocity
Kepler’s third law can be verified with the quantities e and p of the parameter
description (5.22). From the conservation of areal velocity one obtains for the
rotation period T of the periodic motion
l
2m
(5.3)
= Ḟ =
F
T
=
πab
T
and from this with the parameters a
(5.23)
= p/(1− e2) and b
(5.23)
= p/
√
1 − e2 of
the ellipse
a3
T 2
= a3
(
l
2πabm
)2
=
(
l
2πm
)2
p
1 − e2
1 − e2
p2
=
(
l
2πm
)2 1
p
.
One can now relate the trajectory parameter p to the constant k of the po-
tential, see (5.24) or (5.26)
a3
T 2
=
(
l
2πm
)2 1
p
=
(
l
2πm
)2
mk
l2
=
k
4π2m
.
This result no longer depends upon the parameters p and e of the trajectory,
thus not upon the energy E or the angular momentum l of the planet, but
upon its mass m. If one substitutes the reduced mass, which was named m so
far, by the actual expression mM/(m +M) of the two masses involved, and
the constant k by GmM , then
a3
T 2
=
G(m+M)
4π2
≈ GM
4π2
(5.29)
17 Actually the perihelion rotation does not vanish because of the influence of the
other planets and because of effects of General Relativity theory. For the perihe-
lion rotation by a perturbation of the (1/r)-potential see Sect. 5.2.7.
172 5 Central Potentials and the Kepler Problem
is approximately independent of the mass of the planets. Thus one finds that
Kepler’s third law is valid only approximately (for a comparably large mass
M of the central star).
Second Alternative: Integration of the Motion
Along the Trajectory
Kepler’s third law makes two statements, namely firstly the proportionality
between the square of the period and third power of the half-axis and secondly
the independence of the proportionality factor from the mass of the planets.
The first statement is a consequence of the first and of the second law: From
the angular momentum conservation theorem (conservation of areal velocity)
l = mr2ϕ̇ ⇔ dt =
mr2
l
dϕ (5.30)
one has for a rotation period
T
(5.30)
=
m
l
∫ 2π
0
r2dϕ
(5.22)
=
mp2
l
∫ 2π
0
dϕ
(1 + e cosϕ)2
=
2mp2
l
∫ π
0
dϕ
(1 + e cosϕ)2
GR 2.554.3= −2mp2
l
1
1 − e2
[
e sinϕ
1 + e cosϕ
∣∣∣∣
π
0
−
∫ π
0
dϕ
1 + e cosϕ
]
GR 2.553.3=
2mp2
l
1
1 − e2
[
0 +
2√
1 − e2
arctan
√
1 − e2 tan(ϕ/2)
1 + e
∣∣∣∣∣
π
0
]
=
2mp2
l
1
1 − e2
2√
1 − e2
arctan∞ =
4mp2
l
1
(1 − e2)3/2
π
2
.
If one now inserts also 1 − e2
(5.23)
= p/a and p
(5.24)
= l2/mk, one obtains
T 2 =
(
2πmp2
l
)2(
a
p
)3
=
(
2πm
l
)2
l2
mk
a3 = 4π2m
k
a3
with k ∝ m.
Third Alternative: Mechanical Similarity
Reversely, the power of the central potential can be obtained from Kepler’s
third law as follows:18 One starts from a potential V (r) = krn−1, and the
Newtonian equation of motion is
18 Compare Landau and Lifshitz [3] Sect. 10.
5.2 Kinematics of the Kepler Motion 173
m
d2r
dt2
= −∇V (r) = ∇krn−1 = (n− 1) krn−2er.
If one scales the time with α and the length with β,
t = αt′ r = βr′,
one obtains
β
α2
m
d2r′
dt′2
= βn−2∇′kr′n−1
or
m
d2r′
dt′2
= α2βn−3∇′kr′n−1.
One obtains for the scaled and for the unscaled quantities identical equations
of motion, if one chooses
α2 = β3−n.
One obtains thus
(T/T ′)2
(a/a′)3
=
α2
β3
= β−n.
Thus, Kepler’s third law
(T/T ′)2
(a/a′)3
= 1
leads again to n = 0.
5.2.6 The Laplace–Runge–Lenz Vector
In the case of the motion in a potential V (r) = −k/r there is a further
conserved quantity besides that of the energy and the angular momentum,
namely the Runge–Lenz vector19
Λ =
p × l
mk
− r
r
= e eperihelion Λ̇ = 0 |Λ| = e. (5.31)
(p is the momentum, l is the angular momentum, eperihelion is the vector from
the focus to the perihelion, and e is the eccentricity of the trajectory.) The
Runge–Lenz vector thus stands for the constancy of the perihelion direction.
(In the more general case V (r) = f(r) with f(r) �= r−1 the direction of the
perihelion is time dependent.)
19 Often kΛ and sometimes mkΛ (in Goldstein [1], for example) is denoted as the
Runge–Lenz vector.
174 5 Central Potentials and the Kepler Problem
p
l
r
Λ
Λ p x l /mk
−er
−er
Fig. 5.5. On the Runge–Lenz vector
Proof: For the notation see Fig. 5.5.
(i) For the derivative with respect to time one obtains
Λ̇ =
1
mk
(
ṗ × l + p × l̇
)
− 1
r
ṙ + r
1
r3
r · ṙ.
With constant angular momentum
l̇ = 0
and with the equation of motion
ṗ = − k
r3
r
one obtains
Λ̇ = − 1
mr3
r × l + 0 − 1
r
ṙ + r
1
r3
r · ṙ
= − 1
r3
r × (r × ṙ) + 0 − 1
r
ṙ + r
1
r3
r · ṙ (B.3)
= 0.
(ii) Since the Runge–Lenz vector is a constant, it can be evaluated at an
arbitrary point of the trajectory, in particular at the perihelion (index 0,
r = r0, etc.). At the perihelion one has ϕ0 = 0 and ṙ0 = 0 and thus
v0 = r0ϕ̇0e
0
ϕ.
With
l = r × p = mr2ϕ̇ez
one obtains at the perihelion
p × l = mr0ϕ̇0e
0
ϕ ×mr20ϕ̇0ez = m2r30ϕ̇
2
0e
0
r
5.2 Kinematics of the Kepler Motion 175
and
Λ =
(
mr30ϕ̇
2
0
k
− 1
)
e0
r .
The energy is an additional constant of the motion and can be evaluated at
the perihelion, where the radial part of the kinetic energy vanishes. With
E =
l2
2mr20
− k
r0
l = mr20ϕ̇0
and
2El2
mk2
(5.28)
= e2 − 1
one obtains
r0 =
k
2E
(e− 1)
ϕ̇0 =
2E
l
e+ 1
e− 1
.
If one inserts for r0 and ϕ̇0 in Λ, one obtains the results. �	
5.2.7 Perihelion Rotation
While the nonrelativistic two-body theory yields a fixed perihelion (constant
Runge–Lenz vector), a rotation of the perihelion direction of the planet tra-
jectories is actually observed. The origins of it are firstly the influence of the
other planets, which have been neglected so far, and secondly the effects of
the General Relativity theory. For the planet Merkurium the first effect leads
to a rotation by 531′′ per century and the second by 43′′. The qualitative form
of the trajectory is shown in Fig. 5.6 in a much exaggerated form.
For a small perturbation δV of the gravitational potential one obtains
an open trajectory; the position of the perihelion changes from rotation to
rotation. One obtains the rotation angle δϕ of the perihelion caused by the
perturbation δV
δϕ =
d
dl
2m
l
∫ π
0
dϕr2(ϕ) δV (r(ϕ)) + O(δV 2).
Proof: From
dϕ
dr
(5.14)
=
l
r2
√
2m
1√
E − V (r) − l2
2mr2
176 5 Central Potentials and the Kepler Problem
one obtains the polar angle ϕ as a function of the distance r replacing V (r) →
V (r) + δV (r),
ϕ(r) − ϕ0 =
l√
2m
∫ r
r0
dr
r2
√
E − V (r) − δV (r) − l2
2mr2
= −
√
2m
d
dl
∫ r
r0
dr
√
E − V (r) − δV (r) − l2
2mr2
.
Fig. 5.6. A rosette-type trajectory; the two arrows mark the positions of the peri-
helion at two consecutive revolutions
In particular one obtains half a rotation (from perihelion to aphelion), if one
integrates from rmin to rmax,
ϕ = −2
√
2m
d
dl
∫ rmax
rmin
dr
√
E − V (r) − δV (r) − l2
2mr2
= −2
√
2m
d
dl
∫ rmax
rmin
dr
⎡
⎣
√
E − V (r) − l2
2mr2
−
1
2δV (r)
√
E − V (r) − l2
2mr2
+ O(δV 2)
⎤
⎦ .
The first term yields just the full rotation angle of 2π for the unperturbed
gravitationalpotential. The second term yields the perihelion rotation angle
δϕ. If one substitutes in this term the square root by dϕ/dr, see above, one
obtains
δϕ = 2
√
2m
d
dl
∫ rmax
rmin
dr 1
2δV (r)
1√
E − V (r) − l2
2mr2
Summary: Central Potentials and the Kepler Problem 177
(5.14)
=
d
dl
2m
l
∫ rmax
rmin
dr
dϕ
dr
r2δV (r)
=
d
dl
2m
l
∫ π
0
dϕr2(ϕ) δV (r(ϕ)).
�	
In particular one obtains for a perturbation potential of the form
δV (r) =
w
r2
a perihelion rotation of
δϕ =
d
dl
2m
l
∫ π
0
dϕr2(ϕ) δV (r(ϕ)) =
d
dl
2m
l
∫ π
0
dϕw = −2πmw
l2
.
Summary: Central Potentials and the Kepler Problem
V (r) = V (r) ⇒ F (r) = F (r)er
As conserved quantities one has the angular momentum
l
(5.4)
= μr2ϕ̇
and the energy
E =
μ
2
ṙ2 + V (r) =
μ
2
ṙ2 +
l2
2μr2
+ V (r)
(5.7)
= 1
2μṙ
2 + Veff(r)
From the energy conservation one finds
ṙ =
dr
dt
⇒ dt
(5.10)
=
dr√
(2/μ)[E − Veff(r)]
(5.11)⇒ t(r) ⇒ r(t)
From the angular-momentum conservation one finds
dϕ
(5.10)
=
l
μr2(t)
dt ⇒ ϕ(t)
(5.14)
=
dr
r2
√
2μ[E − Veff(r)]
(5.13)⇒ ϕ(r) ⇒ r(ϕ)
In particular for the gravitational potential
V (r)
(5.18)
= −k
r
(5.20)
= −GMm
r
178 5 Central Potentials and the Kepler Problem
one obtains Kepler’s laws
conics r
(5.22)
=
p
1 + e cosϕ
conservation of areal velocity l
(5.3)
= const.
rotation periods/axes
a2
T 3
(5.29)
= const.
with
p
(5.28)
=
l2
kμ
e
(5.28)
=
√
1 +
2El2
μk2
as well as the conserved Runge–Lenz vector
Λ
(5.31)
=
p × l
μk
− r
r
= e eperihelion
Problems
5.1. Kepler’s Laws. Why does a month last for about four weeks? Deter-
mine the length of the month from the following quantities: G = 6.68 ×
10−11m3
moon kg−1 s−2, mmoon = 7.34 × 1022 kg, mearth = 5.98 × 1024 kg and
earth–moon distance R = 3.84 × 108 m.
Which of these quantities is not needed (rationalize)?
5.2. Kepler’s Laws and Circular Orbits.
(a) Assume that Kepler’s first law describes the planetary orbits as circles
with the sun in the center. Which consequences result for Kepler’s second
and third laws?
(b)Using Kepler’s modified laws from (a) determine the force underlying the
motion of the planets.
5.3. Central Potential and Circular Orbit. Under the influence of a cen-
tral potential a mass m moves on a circular orbit which passes through the
potential center.
Determine the potential as a function of the distance from the potential center.
5.4. Trajectories for Motion with Central Force. The energy-conser-
vation theorem (as a first integral of the equation of motion) reads
E =
μ
2
ṙ2 +
L2
2μr2
+ U(r).
To obtain the expression for the trajectory r(ϕ) (in plane polar coordinates)
perform the second integration.
Problems 179
5.5. Motion in the Central Field: Determination of the potential
from the knowledge of the motion.
(a) Determine the radial and azimuthal components of the position vector r,
velocity vector ṙ, and acceleration vector r̈ of a moving particle.
Hint: Use plane polar coordinates: Let
er = (cosϕ, sinϕ) and eϕ = (− sinϕ, cosϕ) be the unit vectors in radial
and azimuthal direction, respectively.
Show that ėr = ϕ̇eϕ and ėϕ = −ϕ̇er.
(b)Which central force has to act on a point mass for the trajectory to be
(i) a logarithmic spiral r = eaϕ,
(ii) a reciprocal Archimedes spiral r−1 = bϕ
(a > 0, b > 0)? Sketch!
5.6. Double Stars. Two bodies (masses M1 and M2) rotate around each
other at constant distance.
(a) Determine the position of the center of mass for a given distance R .
What is the position of the center of mass of the moon–earth system, what
is it in the earth–sun system?
Mmoon = 7.34 × 1022 kg,Mearth = 5.98 × 1024 kg,Msun = 1.99 × 1030 kg
Rmoon−earth = 3.84 × 108 m, Rsun−earth = 1.496 × 1011 m.
For comparison: radius of the earth rearth = 0.637 × 107 m , radius of the
sun rsun = 0.696 × 109 m)
(b)Which forces occur in a system with the origin in the center of mass and
with one of the coordinate axes along the connecting line between the two
bodies? How can one conclude the period of the double-star system from
the constancy of the distance? Determine the period for the two systems
treated in (a).
(c) Which period would an artificial satellite have which is in an orbit close
to the earth?
(d)Can one imagine a double-star system with relativistic velocities?
5.7. Applications in Astronomy.
(a) The parallaxe of the sun yields the average distance between sun and earth
as aE = 1.49×108 km. The period of Jupiter around the sun is 11.86 years.
How large is the average distance between sun and Jupiter?
(b)When sun and Jupiter are in opposition the trajectory radius of Jupiter’s
satellite Io appears under an angle of 2.3′. How large is the radius of the
trajectory of Io?
(c) The period of Io’s orbit around Jupiter is 1.77 days. What is the mass of
Jupiter?
(d) In the telescope Jupiter is resolved as a disk with an apparent equatorial
radius of 38′′, if the sun and Jupiter are in quadrature to each other. How
180 5 Central Potentials and the Kepler Problem
large is the radius RJ of Jupiter (oblateness neglected), and thus how large
is its average density? How large is the gravitational acceleration at the
surface of Jupiter?
Astronomical notations:
Opposition: A constellation in which, observed from the earth, the elongation
(the length difference) between sun and star is 180◦.
Quadrature: A constellation in which, observed from the earth, the elongation
(the length difference) between sun and star is 90◦.
5.8. A particle in the Central Force Field. Determine the orbit of a point
mass m in the central force field V (r) = −k/r2, (k > 0).
(a) First, investigate qualitatively the possible forms of the trajectories for the
radial motion with the help of an effective potential
Veff(r) = V (r) +
l2
2mr2
.
Discriminate the cases l2/2m > k, l2/2m 0, and E 0) in the form
dϕ
dr
= ± L
mr2
1√
2
m
(
E + λ
r − L2
2mr2
) (5.32)
(b)Determine the aphelion and perihelion from (a) as an expression in terms
of the conserved quantities. Write down corresponding formulations for
the parameters e and p of the conic.
(c) Using r+ for aphelion and r− for perihelion show that ϕ(r) can be ex-
pressed as
dϕ
dr
= ±1
r
√
r+r−
(r − r−) (r+ − r)
(5.33)
(d) For a (1/r) potential aphelion, perihelion, and center of force always lie
on a straight line. Prove this claim by integrating the differential equation
(5.33).
Hint: Use ϕ(r = r−) = 0 for the integration.
(e) Now determine ϕ(r) for the case
V (r ) = −λ
r
+
B
r2
,
and discussthe angular difference of successive perihelion positions and
the angular velocity of the perihelion rotation as a function of B.
(f) What may be the origin of such an additional term in the potential?
5.12. Ballistic Trajectory. Neglecting the friction by the air and the accel-
eration phases, the trajectory of a ballistic rocket describes a section of an
ellipse limited by the earth’s surface (Radius RE), see the sketch in Fig. 5.7.
182 5 Central Potentials and the Kepler Problem
θ
β
α
vo
→
RE
h
Fig. 5.7. The ballistic trajectory
(a) Using the Runge-Lenz vector show that the range βRE of the rocket is
determined by
tan(β/2) =
sinα cosα
gRE
v2
0
− sin2 α
,
where α denotes the initial angle from the vertical direction and v0 the
initial velocity. (g is the gravitational acceleration.)
(b)Determine the initial angle which leads to the maximum range for a given
initial velocity v0. (Here, 0 R1 +R2.
6.1.2 Momentum Change and Impulsive Force
During a scattering process the momentum p of the particle is changed due
to the force F acting on it,
ṗ = F .
The equation can be integrated over the (short) time interval t1 ≤ t ≤ t2 of
the interaction with the result
p(t2) − p(t1) =
∫ t2
t1
dtF . (6.1)
The integral on the right side is called the impulsive force.
1 For the corresponding treatment of the scattering process see the Rutherford
scattering in the Kepler problem, Sect. 6.4.6.
6.1 Kinematics 185
E2,p2 ¢ ¢E2,p2
E1,p1 ¢ ¢E1,p1
Fig. 6.2. The notation in the collision or
scattering process. (A more precise sketch
one finds in Fig. 6.3)
Comments:
• In the example of the collision of hard spheres the time interval, dur-
ing which the force acts, is infinitesimally small. In practically all cases,
however, the integral (6.1) is not evaluated; rather, for the analysis of a col-
lision of macroscopic spheres only the conservation of energy, momentum,
and angular momentum is employed.2
• In the case of scattering by a potential, the form of the potential will enter
the scattering angle3 and the cross-section.4 In reverse, the analysis of the
latter can be used to determine the potential.
6.1.3 Laboratory System and Center-of-Mass System
In the following the quantities long before and long after the collision or
scattering process shall be characterized by unprimed and primed symbols,
respectively, see Fig. 6.2.
In principle, the kinematics of the scattering or collision process can be
described in any inertial system. Particularly useful, however, are either the
laboratory system, in which one of the scattering partners (the so-called target)
is at rest before the scattering process, or the center-of-mass system, in which
the center of mass is at rest.
In the laboratory system the velocities v1 and v2 of the two collision
or scattering partners and the velocity V of the center of mass before the
scattering process are
vL
1 �= 0, vL
2 = 0, V L =
m1v
L
1
m1 +m2
;
in the center-of-mass system one has before the scattering process
vCM
1 �= 0, vCM
2 �= 0, V CM =
m1v
CM
1 +m2v
CM
2
m1 +m2
= 0.
2 See Sect. 6.2 further below.
3 See Sect. 6.3.1 further below.
4 See Sect. 6.4.4.
186 6 Collision and Scattering Problems
b
1
2
1θ1
θ2
θs
2
Fig. 6.3. The motion of two particles with repulsive interaction in the laboratory
system (left panel) and in the center-of-mass system (right panel) (schematic)
The relation between the quantities in both systems is
vCM
1 = vL
1 − V L =
m2
m1 +m2
vL
1
vCM
2 = vL
2 − V L = − m1
m1 +m2
vL
1
or
μvL
1 = m1v
CM
1 = −m2v
CM
2 (6.2)
with the reduced mass μ.
In the following the velocities after the collision or scattering process are
to be determined.
6.1.4 Consequences of the Conservation of Momentum
Since there are no external forces assumed to act, the total momentum is a
conserved quantity,
P = p1 + p2 = p′
1 + p′
2.
In the center-of-mass system the center-of-mass momentum vanishes, P =
0, and one has before the scattering process
p1 = −p2 (6.3)
and after the scattering process
p′
1 = −p′
2. (6.4)
In the center-of-mass system, the two scattering partners have thus oppositely
equal momenta before as well as after the scattering process.
6.1 Kinematics 187
In the laboratory system one has
P = p1 + 0 = p′
1 + p′
2. (6.5)
For the decomposition of the momenta (after the scattering process) into
components parallel and perpendicular to the direction of the incoming mo-
mentum one obtains
p′1‖ + p′2‖ = p1
p′1⊥ + p′2⊥ = 0.
The two angles (θ1 and θ2 in the left part of Fig. 6.3) denote the directions
in which the particles travel after scattering process and which are unknown
so far.
The momentum conservation theorem yields three conditions for the orig-
inally six unknown quantities (the three components of the two momenta p′
1
and p′
2). An additional condition is obtained from the energy conservation
theorem.5 Central potentials lead to a motion in a plane and with this to an
additional condition.6 Then one last quantity remains unknown, which can be
determined from the special form of the potential.7
6.1.5 Elastic and Inelastic Scattering
Definition 22. (Elastic and inelastic scattering):An elastic collision or scat-
tering process is a process, in which the (mechanical) energy (of the two col-
lision or scattering partners) is a conserved quantity; in an inelastic collision
or scattering process a part of the mechanical energy is converted into other
forms of the energy, e.g., into deformation energy, excitation energy of the
atomic nucleus or of the electron cloud, or any other type of energy.
Comments:
• In a completely inelastic collision or scattering process the two scattering
partners remain attached to each other after the collision or scattering
process.
• An example for a completely inelastic collision or scattering process is
the capture reaction of a particle hitting an atomic nucleus, where the
projectile remains stuck in the nucleus; the reverse process is that of the
decay (into two particles).8
If one denotes the kinetic energies (long before and long after the scatter-
ing) by
Ei =
p2
i
2mi
and E′
i =
p′2
i
2mi
,
5 See the following Sect. 6.1.5.
6 See the Sect. 6.1.6.
7 See the Sect. 6.3.1.
8 Here, one must take account of effects of Special Relativity.
188 6 Collision and Scattering Problems
respectively, then the energy conservation theorem (because of negligible po-
tential energies) reads
E1 + E2 = E′
1 + E′
2 +Q,
where Q is the energy converted into deformation, heat, or other excitation
energy. One obtains
Q = E1 − E′
1 + E2 − E′
2 =
1
2m1
(p2
1 − p′2
1 ) +
1
2m2
(p2
2 − p′2
2 ). (6.6)
Scattering in the Center-of-Mass System
In the center-of-mass system9 one has p1
(6.3)
= −p2, thus p1 = p2 and analo-
gously p′1 = p′2. With this one obtains from (6.6)
Q =
1
2
(
1
m1
+
1
m2
)(
p2
1 − p′21
)
=
1
2μ
(
p2
1 − p′21
)
and thus
{
p1 = p′1 for Q = 0 (elastic scattering process)
p1 �= p′1 for Q �= 0 (inelastic scattering process). (6.7)
Comments:
• The maximum energy, which can be converted, is
Qmax =
p2
1
2μ
for p′21 = 0 (completely inelastic scattering process); this is of interest for
nuclear reactions, which one induces by collisions (in a linear accelerator
or in a storage ring). Notice that this maximum energy is given by the
momentum in the center-of-mass system; since there the total momentum
vanishes, the total energy can be converted into excitation energy; this is
not so in the laboratory system, in which the total momentum does not
vanish.
Elastic Process in the Laboratory System
In the laboratory system10 the target particle is at rest initially, p2 = 0. The
energy conservation theorem reads
p2
1
2m1
=
p′21
2m1
+
p′22
2m2
. (6.8)
9 Compare Fig. 6.3 (right).
10 Compare Fig. 6.3 (left).
6.1 Kinematics 189
Fig. 6.4. The three cases of scattering: central scattering, no scattering, and oblique
scattering (from left to right)
From the momentum conservation theorem
p1 = p′
1 + p′
2
one obtains taking squares
1
2m1
p2
1 =
1
2m1
(
p′21 + p′22 + 2p′
1 · p′
2
)
. (6.9)
If one subtracts the two equations (6.8) and (6.9), one obtains
1
2m1
(
p′22 + 2p′
1 · p′
2
)
=
1
2m2
p′22 ⇒ p′
1 · p2 =
m1 −m2
m2
p′22 .
Case m1 = m2: In the case of equal masses of the two scattering partners one
obtains
p′
1 · p′
2 = 0
and from this, cf. Fig. 6.4,
either p′
2 = 0 ⇒ p′
1 = p1 (no scattering/collision)
or p′
1 = 0 ⇒ p′
2 = p1 (central collision with billiards)
or p′
1 ⊥ p′
2 (oblique collision with billiards).
Comment: For the oblique collision of equal masses the scattering partners
in the laboratory system thus travel at right angles after the collision.11
6.1.6 Consequences of the Angular-Momentum Conservation
Central potentials lead to central forces and thus to the conservation of angular
momentum; thus the scattering process takes place in a plane.
Definition 23. (Impact parameter): The distance of one of the scattering
partners from the asymptote of the trajectory of the other scattering partner
is the impact parameter.12
11 This is modified, if one takes into account the effects of Special Relativity.
12 This is denoted by b in Fig. 6.3.
190 6 Collision and Scattering Problems
The angular momentum is related to the collision parameter. In an elastic
scattering process this distance (i.e., the impact parameter) is the same before
and after the scattering process because of the angular-momentum conserva-
tion; this one can see most easily in the center-of-mass system, see Fig. 6.3;
the angular momentum of the system in an elastic scattering process is
l = bpCM
i = b′p′CM
i (center-of-mass system), (6.10)
and since the moduli of the momenta are conserved because of the conservation
of the kinetic energy (in an elastic scattering process in the center-of-mass
system), one has thus
b = b′.
6.2 Collision of Hard Spheres
6.2.1 Notations
In an collision process the contact plane is the instantaneous common tan-
gential plane. The collision normal is the normal of the tangential plane at
the point of contact. For a collinear collision the velocity vectors v1 and v2
of the two collision partners lie parallel to each other; otherwise the collision
is called oblique. For a central collision the centers of the masses of the two
collision partners lie on the collision normal before the collision takes place;
otherwise the collision is eccentric. Ideally, for a smooth collision the impul-
sive force has the direction of the collision normal; otherwise it is rough. in
the smooth collision of spheres no angular momentum is exchanged.
Comment: In a rough collision of macroscopic spheres thus a component of
the impulsive force is acting in the tangential plane, and angular momentum
is exchanged; in the (ideally) rough collision and at the moment of contact
the two collision partners roll against each other without sliding (like tooth
wheels).
6.2.2 Elastic Collision of Smooth Spheres (Laboratory System)
In this case the scattering angle is fixed by the scattering geometry (and is
not calculated from the hard-core potential): In a collision of smooth spheres
the impulsive force acts only along the collision normal, and the angular mo-
menta of both spheres are unchanged. Since the impulsive force has only a
component along the collision normal, only the momentum component of the
two particles along the collision normal is changed, while the perpendicular
components remain unchanged. The two particles may have the masses m1
and m2, respectively; their radii shall be R1 and R2, respectively.
In the laboratory system the particle at rest before the collision thus does
not have any momentum component perpendicular to the collision normal
6.2 Collision of Hard Spheres 191
b
θ1
V’1❘❘
V’1 ❘
θ2
V’
➛
2
V1
V1❘❘
❘V1
V’1
Fig. 6.5. The collision of hard spheres
after the collision; after the collision it thus travels in the direction of the
collision normal. Let the angles of the directions after the collision against the
incoming direction be θ1 and θ2 (here for simplicity without a prime; the two
angles are oriented oppositely, but are both taken as positive), see Fig. 6.5.
The impact parameter is b. The momentum and velocity components in the
direction of the collision normal (index ‖) and perpendicular (index ⊥) to
it are
v1‖ = v1 cos θ2
v1⊥ = v1 sin θ2
v′2‖ = 0
v′2⊥ = 0
v′1‖ = v′1 cos(θ1 + θ2)
v′1⊥ = v′1 sin(θ1 + θ2)
v′2‖ = v2
v′2⊥ = 0
The angle θ2 is given by the collision parameter b,
sin θ2 =
b
R1 +R2
(b ≤ R1 +R2). (6.11)
The conservation of momentum,
p1 + p2 = p′
1 + p′
2,
leads to
m1v1 cos θ2 + 0 = m1v
′
1 cos(θ1 + θ2) +m2v
′
2
for the components parallel to the collision normal and to
m1v1 sin θ2 + 0 = m1v
′
1 sin(θ1 + θ2) + 0
for the components perpendicular to the collision normal. This is supple-
mented by the energy conservation
m1
2
v2
1 =
m1
2
v′21 +
m2
2
v′22 .
192 6 Collision and Scattering Problems
cotθ
cotθ2
cotθ2
0
θ
ππ
2
θ2
m1− m2
m1+ m2
θ1+ θ2
Fig. 6.6. For the solution of (6.12): Determination of θ1 for a given θ2. θ1 + θ2 is
indicated for the case m1 > m2
These are three equations for v′1, v′2, and θ1 depending upon θ2, b, and v1.
Oneobtains
v′2 = v1
2m1
m1 +m2
cos θ2
v′1 = v1
√(
m1 −m2
m1 +m2
)2
+
4m1m2
(m1 +m2)2
sin2 θ2
cot(θ1 + θ2) =
m1 −m2
m1 +m2
cot θ2. (6.12)
Since one has |(m1 −m2)/(m1 +m2)| 0), and the mass moves along a hyperbola. For very large distances
(r → ∞) the trajectory approaches the asymptote of this hyperbola, see
Fig. 6.7. The corresponding polar angle ψ is given by
cosψ
(5.22)
= −1
e
(r → ∞). (6.17)
The energy and the angular momentum are conserved quantities, and if
one evaluates the two quantities for r → ∞, one obtains the relations14
l = msv∞
E =
m
2
v2
∞ =
l2
2ms2
. (6.18)
13 Properly speaking one has to take the kinetic energy of the relative motion with
a reduced mass μ.
14 Differently from Sect. 6.1.6, the impact parameter is denoted here by s (and not
by b) in order to avoid a confusion with the notation for the small half-axis.
196 6 Collision and Scattering Problems
The scattering angle is the angle θ between the asymptotes. The impact
parameter is the distance s of the focus from the asymptote.
With
sin(1
2θ)
(6.14)
= sin(ψ − 1
2π) = − cosψ
(6.17)
=
1
e
one obtains
tan(1
2θ) =
sin(1
2θ)
cos(1
2θ)
=
1/e√
1 − (1/e)2
=
1√
e2 − 1
(5.28)
=
√
mk2
2El2
(6.18)
=
k
2Es
.
�	
Fig. 6.8. Scattering of a projectile (probe S) from the attractive potential (V 0, right panel) (schematic)
6.4 * The Cross-Section
6.4.1 The Problem
The motion of the planets or comets (i.e., of macroscopic particles) along their
trajectories can be observed, and from this one can infer the form of the po-
tential.15 In contrast, one can neither observe the trajectories of microscopic
projectiles, nor can one fix (or even observe) a certain impact parameter ex-
perimentally.
The problem thus lies in the fact that in general one does not know the
impact parameter. Rather, in general a beam of probe particles (projectiles)
(with a given distribution of impact parameters) hits on a more or less ex-
tended distribution of target particles. The distribution of impact parameters
leads to a distribution of scattering angles. One is thus interested in the an-
swer to the question of how many particles are deflected by an angle θ (and
15 See (5.15).
6.4 * The Cross-Section 197
arrive on a detector with the opening angle dΩ). The distribution of the scat-
tered particles over different scattering angles is described by the (differential)
cross-section.16
A beam of (“probe”) particles strikes on a substance (“target”). The beam
consists of many particles (e.g., of electrons, ions, photons, etc.), which are
scattered by the many particles of the target (electrons, nuclei, etc.). The
scattering is caused by the interaction between the probe and target particles,
which is described by an interaction potential V .
If one picks one probe and one target particle, then the probe particle is
deflected towards the target for an attractive potential V and away from the
target for a repulsive potential, see Fig. 6.8. The deflection is the larger the
stronger the potential is; as a rule (for a potential decreasing with increasing
distance), the probe particle is thus the less affected the larger the impact
parameter is.
6.4.2 Scattering of Many Probe Particles
by Many Target Particles
If radiation penetrates matter a given projectile can undergo multiple scat-
tering processes. For the following a number of assumptions shall be made:
1. The beam of the incoming projectiles consists of particles of a single kind.
2. The distribution of the target atoms is sufficiently dilute and/or the inter-
action potential is of sufficiently short range, such that a single scattering
process is a good approximation (i.e., a particle scattered once is not scat-
tered again).17
3. The incoming beam is collimated, i.e., the velocities of the incoming par-
ticles are parallel.
4. The incoming beam is monochromatic, i.e., the incoming particles all have
the same energy (and thus the same velocity).
5. The density of the incoming probe particles is homogeneous. (The leads
to an axially symmetric distribution of scattering angles.)
6.4.3 Scattering from a Particle at Rest
In the following we shall start in particular. . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
7.3.2 General and Special Transformation . . . . . . . . . . . . . . . . . 218
7.3.3 The Galilean Transformation . . . . . . . . . . . . . . . . . . . . . . . 219
7.3.4 Galilean Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
7.3.5 The Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . . . 221
Summary: Moving Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
8 Dynamics of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
8.1 The Rigid Body as a System of N Point Masses . . . . . . . . . . . . . 227
8.2 Translational and Rotational Energy: Inertia Tensor . . . . . . . . . 228
8.2.1 The Center of Mass as the Specific Point . . . . . . . . . . . . . 228
8.2.2 The Instantaneously Fixed Point as the Specific Point . . 231
xiv Contents
8.3 Transition to the Continuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
8.3.1 Example: Center of Mass of a Homogeneous
Hemisphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
8.3.2 Examples of Inertia Tensors . . . . . . . . . . . . . . . . . . . . . . . . 234
8.4 Change of the Reference System: Steiner’s Theorem. . . . . . . . . . 238
8.4.1 Steiner’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
8.4.2 Angular Momentum and Torque . . . . . . . . . . . . . . . . . . . . 239
8.4.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
8.5 Principal Moments of Inertia and Principal Axes . . . . . . . . . . . . 242
8.5.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
8.5.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
8.6 Rotation Around a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
8.6.1 Inertia Tensor and Moment of Inertia . . . . . . . . . . . . . . . . 248
8.6.2 The Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
8.6.3 Example: Motion in the Homogeneous Gravitational
Field: The Physical Pendulum . . . . . . . . . . . . . . . . . . . . . . 249
8.6.4 Example: A Cylinder Rolling on an Inclined Plane . . . . . 251
8.7 Rotation Around a Fixed Point: The Top . . . . . . . . . . . . . . . . . . . 253
8.7.1 Space-Fixed (Inertial) and Body-Fixed Coordinate
System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
8.7.2 * Euler Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
8.7.3 * The Euler Equations of Motion . . . . . . . . . . . . . . . . . . . . 257
8.8 The Force-Free Symmetrical Top . . . . . . . . . . . . . . . . . . . . . . . . . . 258
8.8.1 The Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
8.8.2 * Stability of the Rotational Motion of the Top . . . . . . . 260
Summary: Dynamics of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
9 Hamiltonian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
9.1 Hamiltonian Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 269
9.1.1 The Hamiltonian Function . . . . . . . . . . . . . . . . . . . . . . . . . . 269
9.1.2 Canonical (Hamiltonian) Equations of Motion . . . . . . . . . 271
9.1.3 * (Un)Ambiguity of the Hamiltonian Function:
Gauge Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274
9.2 * Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275
9.2.1 Definition 29. (Poisson Brackets) . . . . . . . . . . . . . . . . . . . . 275
9.2.2 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276
9.2.3 Fundamental Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . 277
9.2.4 Properties of the Poisson Brackets . . . . . . . . . . . . . . . . . . . 277
9.2.5 Example: Harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . 277
9.2.6 Transition to Quantum Mechanics . . . . . . . . . . . . . . . . . . . 278
9.3 Configuration Space and Phase Space . . . . . . . . . . . . . . . . . . . . . . 278
9.3.1 Configuration Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
9.3.2 Phase Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279
Contents xv
9.4 * The Modified Hamilton Principle . . . . . . . . . . . . . . . . . . . . . . . . 281
9.5 * Canonical Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282
9.5.1 Point Transformation in Configuration Space . . . . . . . . . 282
9.5.2 Point Transformation in Phase Space . . . . . . . . . . . . . . . . 283
9.5.3 Canonical Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 284
9.5.4 Time Development as a Canonical Transformation . . . . . 287
9.5.5 Canonical Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
9.5.6 Generating Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
9.5.7 The Theorem of Liouville . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
9.6 * Hamilton–Jacobi Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
9.6.1 Action Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
9.6.2 The Characteristic Function . . . . . . . . . . . . . . . . . . . . . . . . 298
9.6.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
Summary: Hamiltonian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304
10 * Introduction to the Mechanics of Continua . . . . . . . . . . . . . . 309
10.1 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
10.2 Lagrangian Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
10.2.1 Kinetic Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
10.2.2 Potential Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . . . 312
10.2.3 Lagrangian Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314
10.3 Hamilton’s Principle and Lagrangian Equations . . . . . . . . . . . . . 314
10.4 The Energy–Momentum Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
10.4.1 The Conservation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 317
10.4.2 Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318
10.4.3 Energy–Current Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
10.4.4 Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320
10.4.5 The Strain–Momentum Density . . . . . . . . . . . . . . . . . . . . . 320
10.4.6 The Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
10.4.7 Conservation of the Strain Momentum . . . . . . . . . . . . . . . 323
Summary: Mechanics of Continua . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
A Physical Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
B Scalars, Vectors, Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
B.1 Definitions and Simple Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
B.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
B.1.2 Behavior Under Inversion. . . . . . . . . . . . . . . . . . . . . . . . . . . 330
B.2 Vectors . . . . .from assumption (2), i.e., from the
assumption of a single scattering process. For further simplification we shall
make the following assumptions in addition:
6. We will use the frame of the target particles initially at rest.
7. The target particle is so heavy or fixed to its surrounding that it is subject
to a negligibly small recoil. (Then the laboratory and the center-of-mass
systems are identical, and one deals with the motion of a projectile around
a fixed force center.)
16 See Sect. 6.4.4 further below.
17 In quantum mechanics this approximation is known as the Born approximation.
198 6 Collision and Scattering Problems
8. For sufficiently large distances of the probe particle from the target the
interaction energy is so weak that it is a good approximation to consider
the motion of the probe particle as free long before the scattering process
(t → −∞) and long after the scattering process (t→ +∞).
9. The target has no internal degrees of freedom, to which energy of the probe
particle can be transferred. This is thus an elastic scattering process with
ECM(t → −∞) = ECM(t→ +∞)
and thus
vi = vs.
10. A central potential
V (r) = V (|r|) = V (r)
shall be assumed. (The above-mentioned hard-core potential is an exam-
ple.) Then the angular momentum conservation theorem applies: The tra-
jectory of the probe particle lies in a plane (the so-called scattering plane),
zq
f
y
x
y
Fig. 6.9. Axial symmetry and notation of the angles in the scattering process
and with the angular-momentum conservation the distance of the target
from the asymptote of the trajectory before the scattering process (i.e.,
the impact parameter) is equal to the one after the scattering process.
Namely, the angular momentum (perpendicular to the plane of motion) is
bvi = bvs. The problem has axial symmetry in addition, i.e., for a given
impact parameter b the scattering angle θ is independent of the azimuth
angle φ, see Fig. 6.9.
11. The interaction potential V (r) is a monotonic18 function of the distance r
(like the gravitational potential).
6.4.4 The Differential Cross-Section
Particles with different impact parameters b are scattered with different scat-
tering angles θ. A measure for the number of the particles, which are scattered
into a certain space angle element dΩ, is the differential cross-section
18 For nonmonotonic interaction potentials see Goldstein [1] Sects. 3–10.
6.4 * The Cross-Section 199
dσ
dΩ
=
d2N/(dt dΩ)
d2N/(dt df)
=
df
dΩ
. (6.19)
d2N/(dt dΩ) is the number of particles, which are scattered into the space
angle element dΩ in the time unit dt, and d2N/(dt df) is the particle flux
density of the incoming particles (i.e., the number of the particles, which pass
through an areal element df per time unit dt.)
All particles which pass through an areal element df = b db dϕ per unit
time are scattered into a space angle element dΩ = sin θ dθ dϕ. This can
be inverted, since θ is monotonic in the impact parameter b because of the
assumption (11): All particles, which are scattered into a space angle element,
have passed through an areal element df .19 For the notation see Fig. 6.10.
b
db
b
Θ
dΘ
Fig. 6.10. Notation for the differential cross-section with one trajectory drawn
In order to arrive at an unambiguous20 relation, one makes the additional
assumption:
12. The maximum scattering angle is π.
Now one can write the differential scattering cross-section (6.19) as
dσ
dΩ
=
∣∣∣∣
df
dΩ
∣∣∣∣ =
∣∣∣∣
b db dϕ
sin θ dθ dϕ
∣∣∣∣ =
b
sin θ
∣∣∣∣
db
dθ
∣∣∣∣ (6.20)
with dθ/db from (6.14) and (6.15). (Typically the impact parameter b de-
creases with increasing impact parameter b; for the case db/dθProblems 203
scattering angle (V (r) = −k/r)
tan
θ
2
(6.16)
=
k
2Es
differential cross-section
dσ
dΩ
(6.19)
=
d2N/dt dΩ
d2N/dt df
=
df
dΩ
(6.20)
=
b
sin θ
∣∣∣∣
db
dθ
∣∣∣∣
Rutherford scattering cross-section (V (r) = −k/r)
dσ
dΩ
(6.21)
=
(
k
4E
)2 1
sin4 1
2θ
Problems
6.1. Energy Transfer in the Laboratory System. A target of mass m2
at rest is hit by a particle of mass m1 and kinetic energy E. Which fraction
of this energy is at disposal for the physically interesting interaction energy?
Discuss the limits m1 � m2 and m1 ≈ m2.
6.2. Scattering of an α-Particle under Nuclear Excitation. An
α-particle of mass mα and velocity vα collides with a nucleus of mass mK
at rest; it excites the nucleus and is deflected by an angle β from its original
direction. The excitation energy is Q. (Given are mα, vα, mK , β, and Q.)
How large is the velocity v′
α of the α-particle after the collision in the labo-
ratory system?
6.3. Particle Decay. A particle with the energy E (internal energy) de-
cays spontaneously (i.e., without external influence) into two fragments “1”
and “2.”
(a) Give an expression for the energy conservation for the decay in the center-
of-mass system. Write down the “decay energy.” This is that energy which
is converted into kinetic energy in the decay process.
(b) Let v be the velocity of the decaying particle in the laboratory system,
u1 the velocity of the fragment “1” in the center-of-mass system, and v1
that in the laboratory system. Discuss v1 as a function of v and u1. First,
assume v > u1, and then investigate the case v 0 and q1q2 a
with V0 > 0.
(a) Discuss the scattering process qualitatively, and make a sketch for the
trajectory of the particle. Include the scattering angle θ and the impact
parameter s in the sketch.
(b)Where on the trajectory does a force act on the particle? Determine the
relation between the directions of the momentum for r > a and rpositions
r − ri = r′ − r′
i
are invariant under the translation (7.1), and thus the force F is invariant
under a translation,
F (r) = F ′(r′). (7.2)
• Successively performed translations commute,
R = R1 + R2 = R2 + R1,
i.e., the succession of the translations is arbitrary.
7.1.2 Inertial Forces
The equation of motion in the system S is
mr̈ = F
and in the system S′ (with r̈ = r̈′ + R̈)
mr̈′ = F −mR̈. (7.3)
Comments:
• An acceleration of the system S′ leads thus (in S′) to an additional force,
oriented oppositely to the relative acceleration (−R̈); this force does not
appear in the inertial system S. Such additional forces in noninertial sys-
tems are called inertial forces.5 (The force, with which one is pressed
against the back in an accelerated vehicle, is such an inertial force.)
• Even though the force is invariant under a translation according to (7.2),
the Newtonian equations of motion are in general not invariant under a
translation, as (7.3) shows.
• The special case of R̈ = 0 (Galilean transformation) will be taken up in
Sect. 7.3.3.
7.2 Rotation Around a Fixed Point
7.2.1 Active and Passive Rotation
The active rotation is the rotation of a vector in a fixed coordinate system. The
passive rotation is the rotation of a coordinate system, i.e., the representation
of the same vector in different coordinate systems. Here we will treat the
passive rotation.
5 Inertial forces appear not only in some translated systems but in particular also
in all rotating systems, see Sect. 7.2.5.
210 7 Moving Reference Frames
ey
ex
ex
y′
x′
x
ey
ry
Fig. 7.3. The representation of a vec-
tor r in two different coordinate systems
(in R
2) rotated against each other by an
angle ϕ
The transformation of the coordinates of a point from one coordinate sys-
tem to that of a rotated coordinate system is described by an orthogonal
transformation U,6
r = U · r′. (7.4)
Comments:
• If the rotation angles are time independent and if the vectors r0 and v
in the translation vector R = r0 + vt of Sect. 7.1.1 are time independent,
then one speaks of a
general Galilean transformation,
r = U0 · r′ + r0 + vt with U̇0 = 0 and ṙ0 = 0. (7.5)
• The specialty of the special Galilean transformation is that the coordinate
systems are not rotated against each other, U0 = 1 (as well as r0 = 0).
• For a rotation in R
2 one has
(
x′
y′
)
=
(
cosϕ sinϕ
− sinϕ cosϕ
)(
x
y
)
as in Fig. 7.3, and thus the transformation matrix for this rotation is
U =
(
cosϕ sinϕ
− sinϕ cosϕ
)
.
7.2.2 Infinitesimal Rotations
Consider two systems S′ and S. The system S′ shall have the same origin as the
inertial system S, but it rotates with the (generally time dependent) angular
velocity ω = ϕ̇ relative to the system S. Consider a fixed position vector r′ in
the rotating system S′. In an infinitesimally small time interval dt the system
S′ rotates by an angle dϕ, and the vector r in the system S changes by
6 More details are given in Sect. 8.7.2.
7.2 Rotation Around a Fixed Point 211
Fig. 7.4. On the coordinate transformation for
infinitesimal rotations; see also Fig. 1.11
dr
(1.35)
= dϕ × r′ (7.6)
(infinitesimal rotation), see Fig. 7.4.
Comments:
• From (7.6) one has
ṙ = ϕ̇ × r′.
• Infinitesimal rotations commute.
• Finite rotations generally do not commute.
Proof: Consider two consecutively performed infinitesimal rotations dϕ1 and
dϕ2, which lead to changes
dr1 = dϕ1 × r1
dr2 = dϕ2 × r2 with r2 = r1 + dr1.
The total change of r is
dr = dr1 + dr2
= dϕ1 × r1 + dϕ2 × r2
= dϕ1 × r1 + dϕ2 × (r1 + dϕ1 × r1)
= (dϕ1 + dϕ2) × r1 + dϕ2 × (dϕ1 × r1).
In the first term the order of the rotations is arbitrary. The last term shows
that the order of the rotations is not arbitrary, but for infinitesimally small
rotations is small of second order and in this case can be neglected. �	
7.2.3 Representation in Different Coordinate Systems
The quantities position, momentum, etc. are represented differently in different
systems. Consider two coordinate systems S and S′ with the same origin. The
inertial system S′ with the axes e′
i rotates with the angular velocity ω relative
212 7 Moving Reference Frames
to the system S with the axes ei. With (7.6) one has thus for the axes in the
coordinate system S
ė
(S)
i = 0
ė′
i
(S) = ω × e′
i
(S′)
ė′
i
(S′) = 0.
The same position vector r is represented differently in the two systems,
depicted in Fig. 7.3,
r =
∑
i
xiei = r(S)
=
∑
i
x′ie
′
i = r(S′).
Comments:
• Notice that the vectors r(S) and r(S′) with different upper indices S and
S′, respectively, are the same quantity and that the index only denotes
the different representation of the same quantity in different coordinate
systems, see also Fig. 7.2, where r′ ≡ r(S′).
• The vector ω describes an instantaneous axis of rotation.
For the velocity one obtains with (7.6)
v = ṙ(S)
= ṙ(S′) + ω × r(S′). (7.7)
Proof: In the two representations with ėi = 0 one obtains
v = ṙ(S) =
d
dt
∑
i
xiei =
∑
i
ẋiei
and, with ė′
i = ω × e′
i,
v =
d
dt
∑
i
x′ie
′
i =
∑
i
(
ẋ′ie
′
i + x′iė
′
i
)
=
∑
i
(ẋ′ie
′
i + x′iω × e′
i)
= ṙ(S′) + ω × r(S′).
�	
Comments:
• The second term in (7.7) is the velocity (in system S) of a point resting in
the rotating system S′, see (7.6); the first term is the additional velocity,
with which a point moves in the system S′.
7.2 Rotation Around a Fixed Point 213
• One can abbreviate the relation between the time derivatives in the two
systems by the operator identity
(
d
dt
. . .
)(S)
=
(
d
dt
. . .+ ω × . . .
)(S′)
. (7.8)
For the acceleration one obtains analogously
r̈(S) = r̈(S′) + ω × (ω × r(S′)) + 2ω × ṙ(S′) + ω̇ × r(S′) (7.9)
or alternatively
[(
d
dt
)(S)
]2
. . . =
[(
d
dt
+ ω×
)(S′)
]2
. . . .
Proof: With the result for the velocity one obtains (with r′ = r(S′))
r̈ =
d
dt
[
ṙ′ + ω × r′]
=
d
dt
[
∑
i
ẋ′ie
′
i + ω ×
∑
i
x′ie
′
i
]
=
∑
i
[(
ẍ′ie
′
i + ẋ′iė
′
i
)
+ ω ×
(
ẋ′ie
′
i + x′iė
′
i
)
+ ω̇ × x′ie
′
i
]
=
∑
i
[(ẍ′ie
′
i + ẋ′iω × e′
i) + ω × (ẋ′ie
′
i + x′iω × e′
i) + ω̇ × x′ie
′
i]
=
(
r̈′ + ω × ṙ′)+ ω ×
(
ṙ′ + ω × r′)+ ω̇ × r′
= r̈′ + ω × (ω × r′) + 2ω × ṙ′ + ω̇ × r′.
Alternatively
r̈ =
(
d
dt
+ ω×
)(S′) [
ṙ′ + ω × r′]
=
[
r̈′ + ω̇ × r′ + ω × ṙ′]+ ω ×
[
ṙ′ + ω × r′]
= r̈′ + ω × (ω × r′) + 2ω × ṙ′ + ω̇ × r′.
�	
Comment: Notice ω̇ �= 0.
214 7 Moving Reference Frames
7.2.4 Uniformly Rotating System: Centrifugal and Coriolis Force
If one uses (7.9) in the Newtonian equation of motion, one obtains
r
ω x r
ω
–ω ´ ( ω ´ r )
Fig. 7.5. On the centrifugal force
mr̈(S) = F
mr̈(S′) = F −mω × (ω × r(S′))
− 2mω × ṙ(S′) (7.10)
with ω̇ = 0. This shows that in the rotating system S′ inertial forces appear,
see Fig. 7.5, namely
the centrifugal force,
−mω × (ω × r(S′)) = mω2r(S′) −mω(ω · r(S′)) (7.11)
and the Coriolis force
−2mω × ṙ(S′)
as well as an additional (unnamed) force −mω̇ × r′, which vanishes in a
uniformly rotating system.
Example: On the northern hemisphere of the earth the Coriolis force leads
to a clockwise deflection, cf. Fig. 7.6.
7.2.5 The Foucault Pendulum
Without consideration of the rotation of the earth one deals with a spheri-
cal pendulum or, with appropriate initial conditions, a plane pendulum. For
small displacements from the equilibrium position the vertical displacements
of the pendulum are small in comparison with the horizontal displacements
(Δz′ � Δx′max and Δz′ � Δy′max), such that the problem can be reduced
approximately to a two-dimensional (horizontal) problem. For small displace-
ments one obtains the equation of motion
r̈′ = −ω2
0r
′
7.2 Rotation Around a Fixed Point 215
r
v
–ω ´ v
ω
r
v
–ω ´ v
ω
Fig. 7.6. The Coriolis force for a motion along a latitude (left) and along a longitude
(right) on the earth
R
x′
y′
β
Ω
Fig. 7.7. For the Foucault pendulum
with ω2
0 = g/l as well as r′ = (x′, y′) and with the origin on the surface on
the earth, see Fig. 7.7.
The equation of motion with consideration of the rotationof the earth
(with constant angular velocity Ω) reads
r̈′ (7.10)
= −ω2
0r
′ − Ω × [Ω × (R + r′)] − 2Ω × ṙ′
with
Ω = Ω (0, 0, 1)
Ω′ = (Ωx′ , Ωy′ , Ωz′) = Ω (0, cosβ, sinβ)
216 7 Moving Reference Frames
in coordinate systems which are fixed in space and fixed on the earth with
unprimed and primed, respectively, notation (at R as in Fig. 7.7), where β is
the geographic latitude. If one neglects the centrifugal term which is quadratic
in Ω (i.e., small), the equation of motion is
r̈′ + ω2
0r
′ + 2Ω × ṙ′ = 0
or, in components,
ẍ′ + ω2
0x
′ − 2Ωz′ ẏ′ = 0
ÿ′ + ω2
0y
′ + 2Ωz′ ẋ′ = 0.
Here the motion in the vertical (z′-)direction is neglected (ż′ � ẋ, ẏ) again.
Thus, two coupled linear differential equations evolve.
For the solution of this system of equations one can look at the components
x′ and y′ as components of a complex quantity
ξ = x′ + iy′ (7.12)
and obtains by multiplication of the second equation with i and subsequent
addition
ξ̈ + ω2
0ξ + 2iΩz′ ξ̇ = 0.
This is a linear differential equation. With the exponential ansatz
ξ ∝ eiαt
one obtains (
−α2 + ω2
0 − 2Ωz′α
)
ξ = 0.
The vanishing of the secular equation yields the solutions for α,
α = −Ωz′ ±
√
Ω2
z′ + ω2
0
≈ −Ωz′ ± ω0
because of Ωz′ � ω0.
Without consideration of the rotation of the earth (Ω = 0, α = ±ω0) the
solution, known from Sect. 4.2, is reproduced,
ξ0(t) = A eiω0t +B e−iω0t.
With consideration of the rotation of the earth one can write the solution in
the form
ξ = ξ0(t) e−iΩz′ t. (7.13)
On the northern hemisphere one has
Ωz′ = Ω sinβ > 0
7.3 Galilean and Lorentz Transformation 217
because of β > 0. The initial conditions shall be chosen such that, neglecting
the rotation of the earth, the pendulum would move in a plane. Now con-
sider the rotation of the earth with the same initial conditions. Because of
the sign in the exponent in (7.13) this pendulum plane rotates clockwise (in
mathematically negative sense) with the frequency Ωz′ .
Comments:
• If one chooses ξ0(t) real one has
x = Re ξ = ξ0 cos(Ωt)
y = Im ξ = −ξ0 sin(Ωt).
• For Vancouver one has β ≈ 49◦, sinβ ≈ 0.75, and the period is
T =
2π
Ω sinβ
=
2π
Ω
1
sinβ
≈ 24 h
1
0.75
= 32 h.
• A second solution would be obtained by
ξ̃ = x′ − iy′.
and
α̃ = Ωz′ ±
√
Ω2
z′ + ω2
0
≈ Ωz′ ± ω0
If one chooses ξ0 as real for simplicity one has
(
x′(t), y′(t)
)
=
(
Re ξ(t), Im ξ(t)
)
=
(
Re ξ̃(t),−Im ξ̃(t)
)
= ξ0(t)
(
cos(Ωz′t),− sin(Ωz′t)
)
,
thus in both cases a clockwise rotation.
• The two solutions one could have obtained in a standard way with the
ansatz (
x′
y′
)
∝
(
x0
y0
)
eiαt
much more laboriously. The comfortable way for problems involving rota-
tions (e.g., motion of a charged particle in a magnetic field) thus is often
a complex superposition of the coordinates like in (7.12).
7.3 Galilean and Lorentz Transformation
The Special Relativity theory is treated in detail in a separate Course.7 Here
the (nonrelativistic) Galilean transformation is contrasted to the (relativistic)
Lorentz transformation, the latter being valid in the relativistic and nonrela-
tivistic regime.
7 See the course Special Relativity.
218 7 Moving Reference Frames
7.3.1 The Relativity Principle
In the theory of Special Relativity considerations are made concerning the laws
of nature in inertial systems moving relative to each other. Inertial systems
are those systems, which are defined in the first Newtonian axiom8 Since a
force-free particle moves at constant speed in any inertial system, different
inertial systems must move with constant velocity relative to each other. In
principle it is not possible to ascertain which of the different inertial systems
is at rest. One can only determine the relative velocity between two (or more)
different inertial systems.
The relativity principle reads:
Relativity principle:
The laws of nature have the same form
in all inertial systems.
(7.14)
Comments:
• In the case of the Newtonian relativity the transformation from one inertial
system to another is performed by the Galilean transformation and in the
case of the Einstein relativity by the Lorentz transformation.
• It turns out that the Newtonian relativity is a limit of the Einstein rel-
ativity (i.e., if the relative velocity is small compared to the velocity of
light) and that the laws of mechanics and of electrodynamics (the Newto-
nian and the Maxwellian theory) have a common transformation behavior
under the Lorentz transformation.
• The influence of forces due to external masses and thus of gravitational
potentials is described by the General Relativity theory; in Special Rel-
ativity one assumes that those masses, if there are any, are infinitely far
away (and thus without any effect).
7.3.2 General and Special Transformation
General Transformation
Consider an inertial system S′, which moves with constant velocity v relative
to an inertial system S. In the general transformation the two coordinate
systems can be rotated against each other by a fixed angle (matrix U) from
(7.4) and translated by a fixed vector (translation vector r0), see Fig. 7.8,
r′ (7.5)
= U · r − vt+ r0, U = U−1
0
but U and r0 shall not be of interest in the following.
8 See Sect. 1.2.
7.3 Galilean and Lorentz Transformation 219
S
S’
v
r’
r
r0
vt0
Fig. 7.8. The (general) transformation
between different inertial systems
Special Transformation
Choose the x-directions of the coordinate systems S′ and S along v. At time
t = 0 the two inertial systems shall coincide. This is the special transformation,
depicted in Fig. 7.9. We will also refer to these systems as them being in the
standard configuration.
Fig. 7.9. Inertial systems moving relative
to each other in the standard configuration
Agreement 9 In the following always the standard configuration shall be
chosen, if not stated otherwise.
7.3.3 The Galilean Transformation
In the standard configuration the (special) Galilean transformation is
x′ = x− vt
y′ = y
z′ = z
t′ = t.
(7.15)
Comments:
• The last of these equations seems obvious, but will be modified in Special
Relativity. In the light of the latter an absolute time is (wrongly) assumed
and can be looked at as a principle of Newtonian mechanics.
220 7 Moving Reference Frames
• The Galilean transformation turns out to be the nonrelativistic limit of
the Lorentz transformation (in which the considered velocities v are small
compared to the velocity of light).9
• To facilitate the comparison with the Lorentz transformation below one
can write the coordinate transformation in the form10
⎛
⎜⎝
t′
x′
y′
z′
⎞
⎟⎠ =
⎛
⎜⎝
1
−v 1
1
1
⎞
⎟⎠
⎛
⎜⎝
t
x
y
z
⎞
⎟⎠ . (7.16)
7.3.4 Galilean Invariance
Galilean Invariants
Definition 25. (Galilean invariants): Galilean invariants are quantities,
which are invariant under Galilean transformations.
The relative positions are invariant under (special) Galilean transformations,
r1 − r2 = r′
1 − r′
2,
and in particular the distances are invariant under (general) Galilean trans-
formations.
Only because one assumes in Newtonian mechanics that the internal forces
typically depend upon the instantaneous11 relative positions of two particles,
these forces are then invariant under Galilean transformations.
Invariant under Galilean transformations are many scalars, besides the
distance among others the mass m.
Form Invariance Under Galilean Transformation
Claim:
The Newtonian equations of motion
are invariant under a Galilean transformation,
mr̈ = F (r) ⇔ mr̈′ = F ′(r′).
(7.17)
Proof: According to (7.2) the force F is invariant under a translation,
F = F ′, and thus in particular under a Galilean transformation. In addition,
the acceleration is invariant under a Galilean transformation, r̈ = r̈′. Finally,
the mass is an invariant scalar, m = m′. �	
9 See the Course on Special Relativity theory.
10 Missing elements in the transformation matrix stand for a zero.
11 Actually, this is an approximation.
7.3 Galilean and Lorentz Transformation 221
Fig. 7.10. Inertial systemsmoving relative
to each other in the standard configuration,
cf. Fig. 7.9
Comments:
• Equation (7.17) exhibits the equations of motion of two different iner-
tial systems in the same form; this is the form invariance under Galilean
transformation.
• The Newtonian equations of motion are form-invariant only in their
coordinate-free notation.
7.3.5 The Lorentz Transformation
In addition to the relativity principle (7.14) the Special Relativity theory is
subject to a second principle:
The maximum velocity is the velocity of light. (7.18)
The corresponding (special) Lorentz transformation can be derived under
the following assumptions:
1. The transformation of space and time is linear.
2. The relativity principle holds.
3. The maximum velocity is the velocity of light, c.
Under these assumptions the Lorentz transformation in the standard con-
figuration (Fig. 7.10) turns out to be
⎛
⎜⎝
ct′
x′
y′
z′
⎞
⎟⎠ =
⎛
⎜⎝
γ −βγ
−βγ γ
1
1
⎞
⎟⎠
⎛
⎜⎝
ct
x
y
z
⎞
⎟⎠ (7.19)
with
β =
v
c
, γ =
1√
1 − β2
.
Comments:
• With the first two assumptions one finds that there is a maximum veloc-
ity,12 which then is postulated as being the velocity of light.
12 See the Course on Special Relativity.
222 7 Moving Reference Frames
• The inverse transformation is13
⎛
⎜⎝
ct
x
y
ct
⎞
⎟⎠ =
⎛
⎜⎝
γ βγ
βγ γ
1
1
⎞
⎟⎠
⎛
⎜⎝
ct′
x′
y′
z′
⎞
⎟⎠ (7.20)
• In the nonrelativistic case v � c one obtains the Galilean transformation
(7.16) from the Lorentz transformation (7.19).14
• The concept (or postulate) of an absolute time must be given up in Special
Relativity, where one finds the effect of the time dilation.
• The distance is a Galilean invariant,15 but not a Lorentz invariant; instead,
one finds the effect of the length contraction.
• With the validity of the Lorentz transformation (rather than the Galilean
transformation) the mechanics (Einstein) and the electrodynamics
(Maxwell) have a unified transformation behavior. Until the beginning of
the twentieth century the different behavior of the two theories (Newto-
nian and Maxwellian, respectively) presented a serious puzzle, since both
theories were interconnected by the Lorentz force.
• The invariance of the Maxwell equations in electrodynamics against
Lorentz transformations was originally considered merely as a form in-
variance, and as such one could have transposed the Lorentz transforma-
tion to mechanics. Einstein’s prime achievement was the interpretation
of the Lorentz transformation as the transformation between inertial
systems.
For more details see the first chapters in the Course on Special Relativity,
where the effects of length contraction and time dilation are discussed and
where subsequently also the transformation of velocity, momentum, energy,
force, etc. are treated.
Summary: Moving Reference Frames
General transformation (rotation and translation)
r′(t) = U(t) · r(t) − R(t)
rotating systems (S′ rotates with ω relative to S)
(
d
dt
)(S)
. . .
(7.8)
=
(
d
dt
+ ω×
)(S′)
. . .
13 For the proof see problem 7.5.
14 For the proof see problem 7.6.
15 See Sect. 7.3.4.
Problems 223
external, centripetal, Coriolis, additional force:
mr̈(S) = F
mr̈(S′) (7.10)
= F −mω × (ω × r(S′)) − 2mω × ṙ(S′) −mω̇ × r(S′)
General Galilean transformation (between inertial systems)
r′ (7.5)
= U0 · r − r0 − vt with U̇0 = 0 and ṙ0 = 0
Special Galilean transformation (r0 = 0, U0 = 1, v = vex)
x′
(7.15)
= x− vt
y′ = y
z′ = z
t′ = t
⇔
⎛
⎜⎝
ct′
x′
y′
z′
⎞
⎟⎠ =
⎛
⎜⎝
1
−β 1
1
1
⎞
⎟⎠
⎛
⎜⎝
ct
x
y
z
⎞
⎟⎠ .
Relativity Principle:
The laws of nature are the same in all inertial systems.
This is a result of the Galilean transformation in the context of classical
mechanics, and this is a first postulate of Special Relativity.
Second postulate of Special Relativity:
The maximum velocity is the speed of light.
Special Lorentz transformation
⎛
⎜⎝
ct′
x′
y′
z′
⎞
⎟⎠
(7.19)
=
⎛
⎜⎝
γ −βγ
−βγ γ
1
1
⎞
⎟⎠
⎛
⎜⎝
ct
x
y
z
⎞
⎟⎠
with
β =
v
c
, γ =
1√
1 − β2
.
Problems
7.1. Earth as an Accelerated Reference System. Even though the equa-
tions of motion are more simple in inertial systems, one usually describes a
motion on the earth in a reference system (laboratory system) rotating along
with the earth. Strictly speaking, this laboratory system is not an inertial
system because of the rotation of the earth. At a point on the earth with
the geographic latitude ϕ a Cartesian coordinate system S′ is defined, see
Fig. 7.11:
x̄3-axis vertically upwards
x̄2-axis towards north
x̄1-axis towards east.
224 7 Moving Reference Frames
For the angular velocity of the earth one has
|ω| =
2π
24
h−1 = 7.27 × 10−5 s−1.
(a) What is the equation of motion of a point mass in this coordinate system
close to the earth’s surface (neglect the terms in ω2)?
(b)How does the Coriolis force depend upon the geographic latitude?
(c) Orient the coordinate system S′ such that the x̄3-axis is perpendicular to
the earth’s surface. Which equations of motion are then to be solved for a
point mass close to the earth’s surface? (The Coriolis force can be taken
from (b).)
(d)A body initially at rest is dropped from the height H . Solve the equation
of motion in (c) under the supposition that ˙̄x1 and ˙̄x2 remain negligibly
small during the time of the free fall. Determine the deflection towards
east originating from the earths’s rotation.
Determine the value of the deflection towards east for a height H = 100 m
at ϕ = 45◦.
R
x3
j
j
w w
r0
r0
x2
Fig. 7.11. For the deflection towards
east in the free fall (for problem 7.1)
7.2. Coriolis Force. In which direction does the Coriolis force point (on
earth), if the velocity has radial direction?
7.3. Centrifugal and Coriolis Force on the Southern Hemisphere.
Compare the direction of the centrifugal and of the Coriolis force on the
southern and northern hemispheres of the earth.
7.4. Coriolis Force and Meteorology.
(a) Explain into which directions the winds from north, east, south, and west
are deflected on the northern and southern hemisphere (see left part of
Fig. 7.12). Neglect the centrifugal acceleration, and assume that the air
moves at constant height.
Problems 225
Z
Q P
R
ρ
E2E1
e1
E3
Y
X
z
y
x
r
0
e2
ω1ω
ω2e3
λ
60�
0�
T
H
Fig. 7.12. Left : For the Coriolis-forces on the earth. Right : High- and low-pressure
regions on the meteorological chart (for problem 7.4)
(b)Explain the origin of cyclones (see the right part of Fig. 7.12). Which sense
of rotation do they have on the northern and southern hemisphere?
7.5. Inverse Lorentz Transformation. Prove that the transformation ma-
trix of the inverse Lorentz transformation (in the standard configuration) is
given by
⎛
⎜⎝
γ βγ
βγ γ
1
1
⎞
⎟⎠ .
7.6. The Galilean Transformation as the Nonrelativistic Limit of
the Lorentz Transformation. Prove that the Galilean transformation is
the nonrelativistic limit of the Lorentz transformation.
7.7. Lorentz Transformation of the Velocity. Prove that the Lorentz
transformation of the velocity (in the standard configuration) is given by
⎛
⎝
u′x
u′y
u′z
⎞
⎠ =
1
1 − vux/c2
⎛
⎝
ux − v
uy/γ
ux/γ
⎞
⎠ .
Hint: Write ux etc. as dx/dt.
8
Dynamics of a Rigid Body
In the previous chapters the motion of one or two particles has been inves-
tigated, whose point-like extension is a meaningful ansatz, if one deals, e.g.,
with the motion of the center of mass or if the extension of the two particles
under consideration is small compared to their distance.
In condensed matter the latter approximation is not valid: The extension
of the atoms is comparable to the distance between the neighboring atoms.
In this chapter extended particles shall thus be investigated. We have in
mind, however, the ideal case of a rigid body, and therefore we exclude elastic
matter1 as well as fluids in particular.
8.1 The Rigid Body as a System of N Point Masses
One can think of an extended body as to be composed of point masses (atoms).
If the body is rigid, the positionsof the various point masses are connected to
each other by constraints such that the distances of pairs (i, k) of atoms are
fixed (time independent). These constraints are
G(i k) = r2i k − c2i k = 0 (i k) = 1, . . . , N (N−1)
2 ;
they are not independent of each other, because for a sufficiently large N one
would have more constraints (≈ 1
2N
2) as coordinates (3N).2
The forces of constraint are
Zi k = λ(i k) ∇iG(i k) = λ(i k)ri k
Zk i = λ(i k) ∇kG(i k) = −λ(i k)ri k = −Zi k
(from the independent constraints). The forces of constraint cancel pairwise.
1 For elastic materials, see Chap. 10.
2 This was noted already in Sect. 3.2.3.
228 8 Dynamics of a Rigid Body
Comments:
• One is not interested in the trajectory of every single particle at all. How
ever will one survey 1023 to 1024 trajectories (per mole), and what could
one do with them (even if one had them)?
• The large number of constraints makes their explicit consideration unprac-
tical.
• In addition, the pairwise oppositely oriented forces of constraint do not
enter the conservation of total momentum, angular momentum, and
energy:
Ṗ = F ext F ext =
∑
i
F ext
i
L̇ = N ext N ext =
∑
i
ri × F ext
i
In the following thus the translation and rotation (three degrees of freedom
each, possibly reduced due to additional constraints) of the rigid body will be
investigated. While a point mass has only translational energy, an extended
body has thus rotational energy in addition. The latter is the primary object
of this chapter.
8.2 Translational and Rotational Energy: Inertia Tensor
For an extended rotating body there are two specific points: the center of
mass3 and the point at instantaneous rest.4 The instantaneously fixed point
may lie in the interior of the body (as in the case of the wheel rotating around
a fixed axis) or outside (as in the case of the pendulum) or also on the surface
(as in the case of the rolling body).
8.2.1 The Center of Mass as the Specific Point
If one chooses the center of mass as the specific point, one can divide the ki-
netic energy into translational and rotational energy. To this end one decom-
poses the coordinates ri of the point masses into the center-of-mass coordinate
R and the coordinates r′
i relative to the center of mass, see Fig. 8.1,
ri = R + r′
i
with
R =
1
M
∑
i
miri ,
∑
i
mir
′
i = 0.
3 See Sect. 8.2.1.
4 See Sect. 8.2.2.
8.2 Translational and Rotational Energy: Inertia Tensor 229
0
S
R
ri
′ri
Fig. 8.1. The notation of the coordinates with re-
spect to an arbitrary reference point O and with
respect to the center of mass S
Kinetic Energy and Inertia Tensor
One obtains for the kinetic energy
T = 1
2
∑
i
miṙ
2
i = 1
2MṘ
2
+ 1
2ω · Θ(S) · ω. (8.1)
Here the inertia tensor (relative to the center of mass S) is
Θ(S) =
∑
i
mi
(
r′i
2
1 − r′
i ⊗ r′
i
)
(8.2)
with
1 =
⎛
⎝
1 0 0
0 1 0
0 0 1
⎞
⎠ and r ⊗ r =
⎛
⎝
xx xy xz
yx yy yz
zx zy zz
⎞
⎠
and with the components
Θ
(S)
ij =
∑
i
mi
(
r′i
2
δi,j − x′i,ix
′
i,j
)
. (8.3)
Proof: (i) Inserting one obtains
T = 1
2
∑
i
miṙ
2
i = 1
2
∑
i
mi
(
Ṙ + ṙ′
i
)2
= 1
2
∑
i
mi
(
Ṙ
2
+ 2ṙ′
i · Ṙ + ṙ′
i
2
)
= 1
2MṘ
2
+ Ṙ ·
∑
i
miṙ
′
i + 1
2
∑
i
miṙ
′
i
2.
The term in the middle vanishes, since
∑
imir
′
i gives the (fixed) position of
the center of mass in the center-of-mass system, and this is the origin for the
primed coordinates.
230 8 Dynamics of a Rigid Body
(ii) For the rotation of a rigid body around the center of mass with the (in-
stantaneous) angular velocity ω the velocity of a point mass i is
ṙ′
i
(7.7)
= ω × r′
i.
With this one obtains for the last term (with ω2 = ω · 1 · ω)
1
2
∑
i
miṙ
′
i
2 = 1
2
∑
i
mi (ω × r′
i)
2 (B.4)
= 1
2
∑
i
mi
[
r′i
2
ω2 − (ω · r′
i)
2
]
= 1
2ω ·
∑
i
mi
[
r′i
2
1 − r′
i ⊗ r′
i
]
· ω = 1
2ω · Θ(S) · ω.
�	
Comments:
• According to (8.1) the kinetic energy can thus be decomposed into a part,
which contains the translational energy of the total mass, thought of as
concentrated at the center of mass, and into a part, which contains the
rotational energy relative to the center of mass.
• Compare this result for the center of mass with the result for the instan-
taneously fixed point in Sect. 8.2.2 further below.
Angular Momentum and Inertia Tensor
In a similar way one obtains for the angular momentum of a body rotating
with the angular velocity ω
L = MR × Ṙ + L(S) (8.4)
and for the angular momentum with respect to the center of mass
L(S) = Θ(S) · ω (8.5)
with the components
L
(S)
i =
3∑
j=1
Θ
(S)
ij ωj.
Proof:
L(S) =
∑
i
mir
′
i × ṙ′
i =
∑
i
mir
′
i × (ω × r′
i)
(B.3)
=
∑
i
mi[r′i
2
ω − r′
i(r
′
i · ω)] =
∑
i
mi[r′i
2
1 − r′
i ⊗ r′
i] · ω
= Θ(S) · ω.
�	
8.2 Translational and Rotational Energy: Inertia Tensor 231
0
P
ρ
ri
′′ri
Fig. 8.2. The notation of the coordinates
relative to an (instantaneous) point P
Comment: The angular momentum L(S) and the rotation vector ω are not
parallel in general; they are parallel only if the inertia tensor is proportional
to the unit matrix, i.e., if the rotation vector is parallel to one of the three
principal axes.5
8.2.2 The Instantaneously Fixed Point as the Specific Point
The instantaneously fixed point with the position vector ρ shall be denoted
by P. One introduces the coordinates r′′
i of the point masses relative to the
point P, see Fig. 8.2,
ri = ρ + r′′
i .
As an alternative to (8.1) the kinetic energy can be written as
T = 1
2
∑
i
miṙ
2
i = 1
2ω · Θ(P) · ω (8.6)
with the inertia tensor Θ(P) with respect to the instantaneously fixed point P
Θ(P) =
∑
i
mi
(
r′′i
2
1 − r′′
i ⊗ r′′
i
)
.
Proof: As above with R → ρ, but with ρ̇ = 0.
Comment: Θ(P) and Θ(S) are different from each other in general. The
relation between the inertia tensors with respect to the center of mass and
with respect to other points, respectively, is given by Steiner’s theorem.6
Summarizing one can treat the motion of a rigid body either as a pure
rotation around an (instantaneously) fixed point or as a translation of the
center of mass and a rotation around the center of mass.
5 See Sect. 8.5 further below.
6 See Sect. 8.4 further below.
232 8 Dynamics of a Rigid Body
8.3 Transition to the Continuum
Definition 26. (Mass density): The mass density is
ρ(r) =
dm(r)
dV (r)
= lim
ΔV (r)→0
Δm(r)
ΔV (r)
= lim
ΔV (r)→0
1
ΔV (r)
ri∈ΔV (r)∑
i
mi. (8.7)
Comments:
• The limit (8.7) to the continuum has its limitations because of the cor-
puscular nature of matter. Thus, the limits are always to be understood
in the way that the considered volumes ΔV are small in comparison to
macroscopic dimensions but always still large in comparison to atomic di-
mensions (such that the average is taken over a sufficiently large number
of atoms to create a continuous function of the position). See Fig. 8.3.
Fig. 8.3. Macroscopic and micro-
scopic view of solid matter
• Exactly the same consideration is made in electrodynamics with the defi-
nition, e.g., of the electrical charge density.
• In fact, the Heisenberg uncertainty relation of quantum mechanics tells
that one cannot locate the position of a mass or of a charge (simultaneously
with the momentum), but that there is a probability distribution for the
position (and momentum).
If one considers the point masses in the limit of infinitesimally small (and
correspondingly many) masses and the following sum as a Riemann sum, one
obtains for any function f(r)
∑
i
mif(ri) →
∫
dm(r)f(r) =
∫
d3r ρ(r) f(r). (8.8)
These and the following relations one formally obtains with the density
8.3 Transition to the Continuum 233
ρ(r) =
∑
i
miδ(r − ri) (8.9)
of a point-mass distribution. In particular one has
the total mass
M =
∑
i
mi =
∫
d3rρ(r); (8.10)
the position vector of the center of mass
R =
1
M
∑
i
miri =
1
M
∫
d3rρ(r) r; (8.11)
the total momentum
P =
∑
i
miṙi =
∫
d3rρ(r) ṙ; (8.12)
the total angular momentum
L =
∑
i
miri × ṙi =
∫
d3rρ(r) r × ṙ; (8.13)and the components of the inertia tensor
Θij
(8.3)
=
∑
i
mi
(
r2i δi,j − xi,ixi,j
)
=
∫
d3rρ(r)
(
r2δi,j − xixj
)
. (8.14)
8.3.1 Example: Center of Mass of a Homogeneous Hemisphere
Consider a hemisphere with the mass M and the radius R. For reasons of
symmetry the center of mass lies on the rotational axis, which is chosen as
the z axis. Then the z component of the center of mass is
Z
(8.11)
=
1
M
∫
d3r ρ(r) z
in the coordinate system of Fig. 8.4.
With spherical coordinates as the coordinates adapted to the problem one
obtains
z = r cosϑ
d3r = r2dr sinϑ dϑ dϕ
and the mass density
ρ(r) =
M
V
θ(R − r) θ(1
2π − ϑ) with V =
1
2
4π
3
R3
234 8 Dynamics of a Rigid Body
z
x
y
Fig. 8.4. The coordinate system for the hemisphere
as well as
Z =
1
2π
3 R3
∫ R
0
r2 dr
∫ π/2
0
sinϑ dϑ
∫ 2π
0
dϕr cosϑ.
With (t = cosϑ)
∫ R
0
r3dr =
R4
4
∫ π/2
0
cosϑ sinϑ dϑ =
∫ 1
0
cosϑ d(cosϑ) =
∫ 1
0
t dt =
1
2
∫ 2π
0
dϕ = 2π
one obtains
Z =
1
2π
3 R3
· R
4
4
· 1
2
· 2π =
3
8
R.
8.3.2 Examples of Inertia Tensors
(i) Inertia Tensor of a Homogeneous Sphere
with Respect to the Center of Mass
With the mass M and the radius R one has the (homogeneous) mass density
ρ(r) =
M
V
θ(R − r), V =
4π
3
R3
and the moment of inertia
Θij =
2
5
MR2δi,j .
8.3 Transition to the Continuum 235
Fig. 8.5. Coordinate system for the sphere
Proof: Using the coordinates of Fig. 8.5 the moment of inertia is
Θij
(8.14)
=
∫
d3r ρ(r)
(
r2δi,j − rirj
)
.
In cartesian coordinates one obtains
Θzz =
∫
d3r ρ(r)
(
r2 − z2
)
= Θxx = Θyy
Θxy =
∫
d3r ρ(r) (0 − xy) = Θyx = Θxz = . . . .
As in Sect. 8.3.1 the integrations is most comfortably performed with spherical
coordinates,
x = r sinϑ cosϕ
y = r sinϑ sinϕ
z = r cosϑ
d3r = r2 dr sinϑ dϑ dϕ.
With this one obtains for the zz-element of the inertia tensor
Θzz =
M
V
∫ R
0
r2 dr
∫ π
0
sinϑ dϑ
∫ 2π
0
dϕ
(
r2 − r2 cos2 ϑ
)
=
M
V
∫ R
0
r4 dr
∫ π
0
(
1 − cos2 ϑ
)
sinϑ dϑ
∫ 2π
0
dϕ.
With the substitution t = cosϑ one obtains
∫ π
0
(
1 − cos2 ϑ
)
sinϑ dϑ =
∫ 1
−1
(
1 − t2
)
dt = 2 − 2
3
=
4
3
and thus
Θzz =
M
4π
3 R
3
R5
5
· 4
3
· 2π =
2
5
MR2.
236 8 Dynamics of a Rigid Body
z
y
x Fig. 8.6. The coordinate system for the cuboid
Alternatively one employs symmetry and obtains
Θzz =
1
3
(Θxx + Θyy +Θzz)
=
1
3
∫
2r2ρ(r) d3r =
M
R3
∫ R
0
2r4dr =
2
5
MR2.
Furthermore,
Θxy = −M
V
∫ R
0
r2 dr
∫ π
0
sinϑ dϑ
∫ 2π
0
dϕ (r sinϑ cosϕ) (r sinϑ sinϕ)
= −M
V
∫ R
0
r4 dr
∫ π
0
sin3 ϑ dϑ
∫ 2π
0
dϕ cosϕ sinϕ = 0 ,
since the integral over ϕ vanishes. (Or: The integrand in
∫
ρ(r)xixjd3r is odd
in xi and/or xj with respect to the integration regime). �	
(ii) Inertia Tensor of a Homogeneous Cuboid
with Respect to the Center of Mass
Cartesian coordinates are adapted to the problem, see Fig. 8.6. One chooses
the coordinate system with the origin in the center of mass and the axes
parallel to the edges (with the edge lengths Li) of the cuboid.
With the mass density
ρ =
M
V
θ(1
2Lx − |x|) θ(1
2Ly − |y|) θ(1
2Lz − |z|)
with V = LxLyLz one obtains the inertia tensor
Θ
(S)
ij = δi,j
M
12
(L2
x + L2
y + L2
z − L2
i ).
8.3 Transition to the Continuum 237
Proof: The moment of inertia is
Θij =
∫
ρ(r)
(
r2δi,j − xixj
)
d3r .
With
r2 = x2 + y2 + z2
d3r = dxdy dz
one obtains
Θxx =
M
V
∫ Lx/2
−Lx/2
dx
∫ Ly/2
−Ly/2
dy
∫ Lz/2
−Lz/2
dz (y2 + z2)
=
M
V
[∫ Lx/2
−Lx/2
dx
∫ Ly/2
−Ly/2
y2dy
∫ Lz/2
−Lz/2
dz + (y ↔ z)
]
=
M
LxLyLz
[
Lx
2
3
(
Ly
2
)3
Lz + (y ↔ z)
]
=
M
12
(L2
y + L2
z)
Θxy = −M
V
∫
dx
∫
dy
∫
dz xy
= −M
V
∫ Lx/2
−Lx/2
xdx
∫
y dy
∫
dz = 0,
since the integrand in the integral over x (and in that over y) is odd in the
(symmetrical) integration regime, and the integral(s) thus vanish(es). �	
(iii) Inertia Tensor of a Homogeneous Cuboid
with Respect to a Corner
One chooses the coordinate system as in the previous example, but with the
origin in a corner, see Fig. 8.7. Then the mass density is
ρ(x, y, z) =
M
V
θ(Lx − x) θ(Ly − y) θ(Lz − z)
×θ(x) θ(y) θ(z)
with V = LxLyLz.
The moment of inertia is
Θ
(0)
ij =
{ 1
3 M [(L2
x + L2
y + L2
z) − L2
i ] i = j
− 1
4 MLiLj i �= j.
238 8 Dynamics of a Rigid Body
z
x
y
Ly
Lz
Lx
Fig. 8.7. The coordinate system for the
cuboid
Proof: The moment of inertia is
Θij =
∫
ρ(r)
(
r2δij − xixj
)
d3r
Θxx =
M
V
∫ Lx
0
dx
∫ Ly
0
dy
∫ Lz
0
dz
(
y2 + z2
)
=
M
V
[∫ Lx
0
dx
∫ Ly
0
y2dy
∫ Lz
0
dz +
∫ Lx
0
dx
∫ Ly
0
dy
∫ Lz
0
dz z2
]
=
M
V
[
Lx
L3
y
3
Lz + LxLy
L3
z
3
]
=
M
3
(
L2
y + L2
z
)
Θxy =
M
V
∫ Lx
0
dx
∫ Ly
0
dy
∫ Lz
0
dz (0 − xy)
= −M
V
∫ Lx
0
dxx
∫ Ly
0
dy y
∫ Lz
0
dz
= −M
V
L2
x
2
L2
y
2
Lz = −M
4
LxLy.
(The integrand is an odd function of x and y, but not with respect to the
integration regime; thus the integral does not vanish.) �	
8.4 Change of the Reference System: Steiner’s Theorem
8.4.1 Steiner’s Theorem
Steiner’s theorem gives a relation between the inertia tensor relative to the
center of mass (S) and that with respect to an arbitrary point (Q). Let R be
the vector,7 which points from the point Q to the center of mass S (or the
7 Equation (8.15) is invariant under the replacement R → −R.
8.4 Change of the Reference System: Steiner’s Theorem 239
Fig. 8.8. Center of mass S and an arbitrary point Q,
which may also be outside of the body (The vector R
may equivalently point from Q to S)
reverse), see Fig. 8.8 analogous to Fig. 8.2 (if one substitutes there the point
P by the point Q). Then one has
Θ(Q) = Θ(S) +M
(
R21 − R ⊗ R
)
(8.15)
with the total mass
M
(8.10)
=
∫
ρ(r) d3r.
Proof: One chooses Q as the coordinate origin and r = R + r′. Then the
inertia tensor with respect to the point Q is
Θ(Q) =
∫
ρ(r)
(
r21 − r ⊗ r
)
d3r
=
∫
ρ(r′)
[
(R + r′)2 1 − (R + r′) ⊗ (R + r′)
]
d3r′
=
∫
ρ(r′)
[(
R2 + 2r′ · R + r′2
)
1
− (R ⊗ R + r′ ⊗ R + R ⊗ r′ + r′ ⊗ r′)
]
d3r′.
Now one has ∫
ρ(r′) r′ d3r′ = 0,
since the vectors r′ have their origin at the center of mass. Of the seven terms
thus the second, fifth, and sixth term vanish, and one obtains
Θ(Q) =
∫
ρ(r′)
(
r′21 − r′ ⊗ r′) d3r′ +
(
R21 − R ⊗ R
) ∫
ρ(r′) d3r′
= Θ(S) +M
(
R21 − R ⊗ R
)
.
�	
8.4.2 Angular Momentum and Torque
In a way analogous to the inertia tensor one finds a decomposition of the other
quantities connected with the rotation, of the angular momentum, and of the
torque
L(Q) = L(S) + R × P (8.16)
240 8 Dynamics of a Rigid Body
and
N (Q) = N (S) + R × F . (8.17)
Proof: (i) The angular momentum with respect to the point Q is
L(Q) =
∫
ρ(r) (r × ṙ) d3r
=
∫
d3r′ρ(r′) (R + r′) × (Ṙ + ṙ′)
=
∫
d3r′ρ(r′) [R × Ṙ + r′ × Ṙ + R × ṙ′ + r′ × ṙ′).
Since the primed coordinate system has its origin at the center of mass,
∫
ρ(r′) r′ d3r′ = 0,
the second term vanishes; with
ṙ′ = ω × r′
thus the third vanishes, and one obtains
L(Q) =
∫
d3r′ρ(r′) r′ × ṙ′ +
∫
d3r′ρ(r′)R × Ṙ
= L(S) + R × P
with
P = M · Ṙ, M =
∫
ρ(r′) d3r′.
Alternatively (and much simpler):
L(Q) = Θ(Q) · ω = Θ(S) · ω +M
(
R2ω − RR · ω
)
= L(S) +MR × (ω × R) = L(S) + R × P .
(ii) The proof for the torque is quite analogous:
N (Q) = L̇
(Q)
= L̇
(S)
+ Ṙ × P + R × Ṗ = N (S) + 0 + R × F .
�	
Comments:
• A quantity connected with the rotation (torque, angular momentum) with
respect to an arbitrary point Q is thus equal to that with respect to the
center of mass plus a term which describes the rotation of the total mass
thought of as assembled at the center of mass. In particular, the total
angular momentum is equal to the sum of orbital angular momentum
R × P and eigen angular momentum L(S) (spin, as called in quantum
mechanics).
8.4 Change of the Reference System: Steiner’s Theorem 241
Fig. 8.9. Center of mass S and a point Q at a corner
of a cuboid
8.4.3 Examples
(i) Homogeneous Cuboid
The homogeneous cuboid has been treated further above in the examples (ii)
and (iii) of Sect. 8.3.2. The result was
Θ
(S)
ij =
{ 1
12M (L2
x + L2
y + L2
z − L2
i ) i= j
0 i �= j
Θ
(Q)
ij =
{
1
3M (L2
x + L2
y + L2
z − L2
i ) i = j
− 1
4M LiLj i �= j
with the notation as in Fig. 8.9.
Here one can verify the Steiner theorem: On the one hand, one obtains
Θ
(Q)
ij −Θ
(S)
ij =
{
1
4M (L2
x + L2
y + L2
z − L2
i ) i = j
− 1
4M LiLj i �= j.
With the center-of mass-coordinate (with respect to the selected corner)
R =
1
2
(Lx, Ly, Lz)
one obtains on the other hand
Θ
(Q)
ij − Θ
(S)
ij
= M(R21 − R ⊗ R)
=
M
4
⎡
⎣(L2
x + L2
y + L2
z
)
⎛
⎝
1 0 0
0 1 0
0 0 1
⎞
⎠−
⎛
⎝
L2
x LxLy LxLz
LyLx L2
y LyLz
LzLx LzLy L2
z
⎞
⎠
⎤
⎦
=
M
4
⎛
⎝
L2
y + L2
z −LxLy −LxLz
−LyLx L2
z + L2
x −LyLz
−LzLx −LzLy L2
x + L2
y
⎞
⎠ .
242 8 Dynamics of a Rigid Body
(ii) Inertia Tensor of a Homogeneous Full Sphere
with Respect to a Point on the Surface
The coordinate system shall be chosen such that the specific point lies in
the origin; in addition the z axis shall coincide with the symmetry axis as in
Fig. 8.10. Then the vector to the center of mass is
R = (0, 0, R)
and the inertia tensor
Θ(0) = Θ(S) +M (R21 − R ⊗ R)
with
Θ(S) =
2
5
MR21
of example (i) of Sect. 8.3.2. With this one obtains
Θ(0) = M R2
⎡
⎣
(
2
5
+ 1
)⎛
⎝
1 0 0
0 1 0
0 0 1
⎞
⎠−
⎛
⎝
0 0 0
0 0 0
0 0 1
⎞
⎠
⎤
⎦ ,
thus
Θ(0)
xx = Θ(0)
yy =
7
5
MR2
Θ(0)
zz =
2
5
MR2
Θ
(0)
ij = 0 (i �= j).
Here one deals with a symmetrical top with the principal axes8 ez and, e.g.,
ex and ey.
8.5 Principal Moments of Inertia and Principal Axes
8.5.1 Definitions
Since the inertia tensor is not a scalar, the vectors L of the angular momentum
and ω of the angular velocity are not parallel in general because of
L
(8.5)
= Θ · ω.
But one can ask whether there are directions in which L and ω are parallel,
L = Θ · ω = I ω.
8 See the following Sect. 8.5.
8.5 Principal Moments of Inertia and Principal Axes 243
S
0
x
y
z
Fig. 8.10. The coordinate system for the sphere
This poses an eigen value problem.9 Since Θ is a real and symmetrical tensor,
one finds indeed always (three) real eigen values I with orthogonal eigen
vectors e,
Θ · ei = Ii ei i = 1, 2, 3. (8.18)
Notations:
• The eigen values Ii are called principal moments of inertia, and the eigen
vectors ei are called principal axes.
• If two of the principal moments of inertia are equal, e.g., I1 = I2 �= I3, one
speaks of a symmetrical body.
• If all three principal moments of inertia are equal, I1 = I2 = I3, one speaks
of a spherical body.
• If one point is fixed,10 one speaks of a top (instead of a body).
For the case in which the vector of the angular velocity is oriented along
one of the principal axes of inertia, then also the angular momentum is ori-
ented along this direction. For a symmetrical body with I1 = I2 one has this
parallelity for all vectors which lie in the plane spanned by e1 and e2: Let
ω = a e1 + b e2.
Then one has
L = Θ · ω = Θ · (ae1 + be2) = aI1e1 + bI2e2 = I1 (ae1 + be2) .
For the spherical body finally one has the parallelity for the vectors in
arbitrary direction. In the general case one can use the property (9) of
9 See Appendix G.
10 See Sect. 8.7 further below.
244 8 Dynamics of a Rigid Body
z
y
x Fig. 8.11. The coordinate system for the cube
Appendix G.4 and obtains the angular-momentum components in the basis
of the principal axes (normalized to unity),11
Θ =
∑
i
eiIie
T
i with |ei| = 1 ⇒ L = Θ · ω =
∑
i
eiIie
T
i · ω =
∑
i
eiLi.
Comment: The reference point P is arbitrary; however, the inertia tensor
Θ(P) and the moments of inertia I(P) depend upon the reference system.
8.5.2 Examples
(i) Inertia Tensor of a Homogeneous Cube
with Respect to the Center of Mass
Let the edge length be L (Fig. 8.11). The inertia tensor is according to example
(ii) of Sect. 8.3.2
Θij = δi,j
1
6
ML2.
The inertia tensor is already diagonal. The principal moments of inertia are
all equal,
Ii =
1
6
ML2 i = 1, 2, 3.
this is a spherical top. The principal axes of inertia, are for example,
e1 = ex, e2 = ey, e3 = ez.
11 If e is a (column) vector, then eT denotes the transposed (row) vector.
8.5 Principal Moments of Inertia and Principal Axes 245
(ii) Inertia Tensor of a Homogeneous Cube
with Respect to a Corner
According to example (iii) of Sect. 8.3.2 the inertia tensor is, in the notation
given in Fig. 8.12,
z
x
y
Ly
Lz
Lx
Fig. 8.12. The coordinate system for the cube
(Lx = Ly = Lz = L)
Θij =
{
2
3ML2 i = j
− 1
4ML2 i �= j
Θ =
ML2
12
⎛
⎝
+8 −3 −3
−3 +8 −3
−3 −3 +8
⎞
⎠ .
The principal moments of inertia are
Ii = ML2γi
with
γ1 =
1
6
, γ2 = γ3 =
11
12
with the principal axes of inertia (normalized to 1)
e1 =
1√
3
⎛
⎝
1
1
1
⎞
⎠ , e2 =
1√
2
⎛
⎝
0
1
−1
⎞
⎠ , e3 =
1√
6
⎛
⎝
−2
1
1
⎞
⎠ .
Proof: The eigen values I = ML2γ are obtained from the secular equation
det
⎡
⎢⎣ML2
⎛
⎜⎝
2
3 − γ − 1
4 − 1
4
− 1
4
2
3 − γ − 1
4
− 1
4 − 1
4
2
3 − γ
⎞
⎟⎠
⎤
⎥⎦ = 0.
246 8 Dynamics of a Rigid Body
This is an equation of third order for γ,
0 =
(
2
3
− γ
)3
− 2
(
1
4
)3
− 3
(
2
3
− γ
)(
1
4
)2
=
[(
2
3
− γ
)
− 2
4
] [(
2
3
− γ
)2
+
2
4
(
2
3
− γ
)
+
(
1
4
)2
]
=
[(
2
3
− γ
)
− 2
4
] [(
2
3
− γ
)
+
1
4
]2
.
From this one can determine the eigen values easily,
γ1 =
2
3
− 2
4
= 2
4 − 3
12
=
1
6
γ2 = γ3 =
2
3
+
1
4
=
8 + 3
12
=
11
12
.
For the eigen vectors one obtains by inserting the eigen values into the eigen
value equation (as solution of the homogeneous equation)
e1 =
1√
3
⎛
⎝
1
1
1
⎞
⎠ , e2 =
1√
2
⎛
⎝
0
1
−1
⎞
⎠ , e3 =
1√
6
⎛
⎝
−2
1
1
⎞
⎠ .
�	
Comment: If one can guess the eigen vectors, the eigen values are easy to
find. By comparison of the right and left side of
M e = λe with M =
⎛
⎝
a b b
b a b
b b a
⎞
⎠
one obtains
e1 =
⎛
⎝
1
1
1
⎞
⎠ , M e1 = (a+ 2b)e1 λ1 = a+ 2 b =
2
3
+ 2
(
−1
4
)
=
1
6
e2 =
⎛
⎝
0
1
−1
⎞
⎠ , M e2 = (a − b)e2 λ2 = a− b =
2
3
+
1
4
=
11
12
e3 =
⎛
⎝
2
−1
−1
⎞
⎠ , M e3 = (a − b)e3 λ3 = (a− b) =
11
12
The third of the eigen vectors is less obvious than the other two. However, if
one has determined two eigen vectors e1 and e2, one obtains the third from
e3 = ±e1 × e2
8.6 Rotation Around a Fixed Axis 247
(for right- or left-handed systems) because of the orthogonality of the eigen
vectors. The eigen vectors are fixed except for a factor (the so-called normal-
ization factor), which in general is determined such that the eigen vectors are
unit vectors.
8.6 Rotation Around a Fixed Axis
The motion of the rigid body around a fixed axis is a pure rotation around
this axis. The motion has one degree of freedom: the rotation angle ϕ. If one
denotes the unit vector in the direction of the (fixed) axis of rotation by eω,
then one can assign a rotation vector ϕ to the rotation,
ϕ
(7.6)
= ϕeω
as in Fig. 8.13.
Fig. 8.13. On the infinitesimal rotations (pre-
viously shown in Figs. 1.11 and 7.4)
Let a point be described by the position vector r′
i in the body-fixed system
(with the origin on any point on the axis of rotation). Then the change dri
of a position ri in the space fixed system is described by
dri
(1.35)
= dϕ × r′
i.
The velocity of a point with the position vector r′
i is then
ṙi = ω × r′
i (8.19)
with
ω = ωeω = ϕ̇eω = ϕ̇ and ėω = 0.
Comment: In contrast to the single degree of freedom here, the motion of
a top (i.e., the motion around an instantaneous axis with one fixed point),
considered in the following paragraph 8.7, has three degrees of freedom: the
rotational angle (one degree of freedom) and the orientation of ω (two further
degrees of freedom). This motion is likewise described by (8.19), but with
ėω �= 0.
In this paragraph the more simple case of the fixed axis shall be treated.
248 8 Dynamics of a Rigid Body
8.6.1 Inertia Tensor and Moment of Inertia
The part of the energy connected with the rotation around the fixed axis is
Trot = 1
2ω · Θ · ω = 1
2ω eω · Θ · eω ω = 1
2I ω
2
with the moment of inertia (rotational inertia)
I = eω · Θ · eω. (8.20)
For the representation of the body by point masses one obtainsI =
∑
i
r2i⊥mi (8.21)
and for the representation by a mass distribution
I =
∫
r2⊥ dm =
∫
ρ(r) r2⊥ d3r, (8.22)
where r⊥ is the distance of the point with the position vector r from the axis
of rotation, cf. Fig. 8.14.
Fig. 8.14. The definition of r⊥
Proof:
I = eω · Θ · eω =
∑
i
mi eω · (r2i 1 − ri ⊗ ri) · eω
=
∑
i
mi
(
r2i − (eω · ri)2
)
=
∑
i
mi (eω × ri)2 =
∑
i
mir
2
i⊥.
�	
In cartesian (x, y, z) or sometimes better in cylindrical coordinates
(r⊥, ϕ, z) with r2⊥ = x2 + y2 one chooses customarily
ez = eω .
8.6 Rotation Around a Fixed Axis 249
Then one has
I = Θzz =
∑
i
mi(x2
i + y2
i )
and
I =
∫
dx
∫
dy
∫
dz ρ(x, y, z)(x2 + y2).
Also here Steiner’s theorem (8.15) can be of use, if one knows the moment
of inertia I(S) = Θ
(S)
zz with respect to the center of mass or if one can calculate
it easily, then
I = Θzz = Θ(S)
zz +MR2
⊥,
where R⊥ is the distance of the center of mass from the axis of rotation.
8.6.2 The Equation of Motion
The equation of motion reads in general
N = L̇, L = Θ · ω,
and in general12 one has Θ̇ �= 0, since the moment of inertia changes with
the change of the orientation with time. In this paragraph, however, one deals
with the motion around a fixed axis of rotation,
ω = ω eω
N = N eω
I = eω · Θ · eω,
and then one has
ėω = 0 ⇒ İ = 0,
thus
N = I ω̇ (fixed axis of rotation). (8.23)
Comment: With eω = ez fixed also Θzz is fixed, but for example Θxx (in
the space coordinate system) changes with time. However, the change of Θxx
and Θyy and possibly Θxy is of no interest for the motion with the fixed axis
ez, unless one wants to calculate the torque of constraint.
8.6.3 Example: Motion in the Homogeneous Gravitational Field:
The Physical Pendulum
The plane physical pendulum, shown in Fig. 8.15, is an example for the above-
treated rotation around a fixed axis of rotation.
12 See Sect. 8.7 further below.
250 8 Dynamics of a Rigid Body
0
S
ϕ
eω
→
R
→
K = Mg
→ →
Fig. 8.15. The notations for the physical pendulum
In the homogeneous gravitational field the torque is
N = R × F = MR × g,
where R is the vector from a point on the axis of rotation to the center of
mass and F = Mg the force on the total mass M thought of as assembled in
the center of mass.
Proof: For a system of point masses the torque is
N =
∑
i
ri × F i =
∑
i
ri ×mi g
=
∑
i
mi ri × g = M R × g = R × F .
�	
The component in direction of the (horizontal) axis of rotation eω is
N = eω · N = eω · R × F = eω · (−RMg sinϕeω) = −RMg sinϕ,
where ϕ is the angle between R and F . The angular momentum is
Lz = Θzzϕ̇ = Iϕ̇.
For a pure rotational motion the equation of motion is thus
Iϕ̈ = −RMg sinϕ.
The equation of motion is the pendulum equation
ϕ̈ = −RMg
I
sinϕ.
For comparison: The equation of motion of the mathematical pendulum is
ϕ̈ = −g
l
sinϕ.
8.6 Rotation Around a Fixed Axis 251
In order to make the similarity more apparent, one introduces an effective
length
l∗ = I/RM
and obtains
ϕ̈ = − g
l∗
sinϕ.
For small amplitudes one has then the frequency
ω =
√
g
l∗
=
√
RMg
I
.
8.6.4 Example: A Cylinder Rolling on an Inclined Plane
In this example the direction of the axis is fixed; however, the body can execute
a translation in addition. In the (x, y)-system of Fig. 8.16 the forces on the
body are (1) the gravitational force
Mg = Mg (sinα,− cosα);
it causes an acceleration in the x direction and a torque around the point of
contact A, but not, however, around the center of mass; (2) the force N 0 of
the support (as a force of constraint); it does not cause an acceleration or a
torque, neither around the point of contact nor around the center of mass;
(3) the frictional force R0 (as a force of constraint); it prevents sliding and
can be substituted by the condition of rolling; it causes an acceleration in
negative x direction and a torque with respect to the center of mass (but
not with respect to the point of contact). The rolling condition is (with the
angle ϕ as in Fig. 8.16, which is not the customary angle ϕ of the plane polar
coordinates)
Fig. 8.16. The coordinate system for the body rolling on the inclined plane
252 8 Dynamics of a Rigid Body
dx = r dϕ ⇔ x = rϕ+ x0,
and as an additional constraint one can use y = const = 0.
As treated in Sect. 8.2 one can choose the position of the axis of rotation.
Either the axis through the center of mass or the one through the (instanta-
neously fixed) point of contact are obvious.
Method I: Axis of Rotation Through the Center of Mass
If one chooses a coordinate system as in Fig. 8.16, the equations of motion are
Mẍ = Mg sinα−R0
Mÿ = −Mg cosα+N0 = 0
I(S)ϕ̈ = r R0.
The gravitational force Mg here does not cause any torque (since the lever
vanishes). With ϕ̈ = ẍ/r one obtains on the one hand
R0 = I(S) ϕ̈
r
= I(S) ẍ
r2
from the angular momentum conservation and on the other hand
R0 = Mg sinα−Mẍ
from the momentum conservation and from this
ẍ
(
I(S)
r2
+M
)
= Mg sinα
or
ẍ = g
sinα
1 + I(S)
M r2
,
i.e., a constant acceleration.
Alternatively with the Lagrangian instead of the Newtonian equations of mo-
tion: The Lagrangian function is
L = T − V =
1
2
Mẋ2 +
1
2
I(S)
(
ẋ
r
)2
+Mgx sinα.
The equation of motion is
0 =
d
dt
∂L
∂ẋ
− ∂L
∂x
=
(
M +
I(S)
r2
)
ẍ−Mg sinα
with the same result
ẍ =
g sinα
1 + I(S)/Mr2
.
8.7 Rotation Around a Fixed Point: The Top 253
Method II: Instantaneous Axis of Rotation
The torque is now caused by the gravitational force (and not by the frictional
force),
N (A) = r ×M g = rMg sin(π − α)e = rMg sinα e
= L̇
(A)
= I(A)ω̇
with e = −ez and with the moment of inertia (with respect to the point of
contact, according to the theorem of Steiner)
I(A) = I(S) +M r2.
With
ω̇ = ω̇ e, ω̇ = ϕ̈ = ẍ/r
one obtains
rMg sinα = I(A)ϕ̈ =
(
I(S) +M r2
)
ẍ/r
or
ẍ =
r2Mg sinα
I(S) +M r2
=
g sinα
1 + I(S)
M r2
as above.
8.7 Rotation Around a Fixed Point: The Top
8.7.1 Space-Fixed (Inertial) and Body-Fixed Coordinate System
In the previous Sect. 8.6 the motion of the rigid body around a fixed axis has
been investigated. In this paragraph the motion of the top shall be treated,
i.e., the motion around an instantaneous axis with one fixed point.13
The velocity of a point mass is
ṙi
(8.19)
= ω × r′
i.
In contrast to the previous Sect. 8.6 here the direction of the rotational vector
ω of the angular velocity is not fixed. This motion has three degrees of freedom:
the rotational angle (one degree of freedom) and the orientation of ω (two
additional degrees of freedom).
The inertia tensor Θ of a body is a fixed quantity in the body system;
however, in the space coordinate system it depends upon the instantaneous
orientation of the body. To investigate the free14 motion of a body it would
13 The fixed point can be the (possibly force-free) center of mass.
14 For the motion in an external torque field see the heavy top in Sect. 8.7.3 or
Goldstein [1], Sects. 5–7.
254 8 Dynamics of a Rigid Body
thus be very disadvantageous, to work in an inertial system (S) (in which the
inertia tensor Θ is time dependent); rather it is advantageous to work in the
body system (S′) (in which Θ is independent of time).
For the construction of a Lagrangian function one needs, among others,
the kinetic energy of the rotation
Trot = 1
2ω · Θ · ω,
and it would thus be advantageous to express the angular velocity ω (and the
inertia tensor Θ) in the body system (S′).
z�
x�
z
ψ
ϕ ϕ
.
.
Fig. 8.17. Exemplary description of the rotation in the body system
As a special example15 for a body-fixed system a body is considered,
which rotates with the angular velocity ϕ̇ around its body x′ axis as shown
in Fig. 8.17. The x′ axis of the body rotates around the (perpendicularly ori-
ented) space z axis with the angular velocity ψ̇. Then the vector of the angular
velocity in the space system is given by
ω(S) = ϕ̇(cosψ, sinψ, 0) + ψ̇(0, 0, 1)
and in the body system by
ω(S′) = (ϕ̇, 0, 0) + ψ̇(0, sinϕ, cosϕ).
8.7.2 * Euler Angles
The TransformationThe transformation of the coordinates of a point from one coordinate system
to a second coordinate system rotated relative to the first is described by an
orthogonal transformation U,
r = U · r′.
15 The general motion is described by the Euler angles and their derivatives with
respect to time. See Sect. 8.7.2.
8.7 Rotation Around a Fixed Point: The Top 255
Fig. 8.18. The Euler angles
A general rotation is given by three angles16 and can be described by three
consecutive rotations.
Definition 27. (Euler angles): The orientation of a body can be described by
the Euler angles ϕ, θ, and ψ, which are defined as follows:17 The primed and
unprimed coordinates are fixed in the body and space system, respectively. The
first rotation is around the z axis with the angle ϕ, compare Fig. 8.18,
⎛
⎝
x1
y1
z1
⎞
⎠ =
⎛
⎝
cosϕ sinϕ 0
− sinϕ cosϕ 0
0 0 1
⎞
⎠
⎛
⎝
x
y
z
⎞
⎠ ⇔ r1 = U1 · r.
The second rotation is around the x1 axis with the angle θ,
⎛
⎝
x2
y2
z2
⎞
⎠ =
⎛
⎝
1 0 0
0 cos θ sin θ
0 − sin θ cos θ
⎞
⎠
⎛
⎝
x1
y1
z1
⎞
⎠ ⇔ r2 = U2 · r1.
The third rotation is around the z2 axis with the angle ψ,
⎛
⎝
x′
y′
z′
⎞
⎠ =
⎛
⎝
cosψ sinψ 0
− sinψ cosψ 0
0 0 1
⎞
⎠
⎛
⎝
x2
y2
z2
⎞
⎠ ⇔ r′ = U3 · r2.
16 For example the orientation of a point of the body is determined by two angles;
the third angle describes then the rotation of the body around the axis, which
goes through this point and the origin.
17 See Goldstein [1]; Sommerfeld [6] has interchanged the notation of ϕ and ψ.
256 8 Dynamics of a Rigid Body
With
r1 = U1 · r, r2 = U2 · r1, r′ = U3 · r2
one obtains
r′ = U · r with U = U3 · U2 · U1. (8.24)
Comments:
• In contrast to translations consecutively performed rotations do not com-
mute,
U1 · U2 �= U2 · U1,
i.e., the succession of the rotations is not arbitrary.
• Of this one can convince oneself easily with an example. To this end one
considers, e.g., the rotations Ry and Rz of a position vector r = (x, y, z)
around the y and around the z axis by 90◦ each with
Ry
⎛
⎝
x
y
z
⎞
⎠ =
⎛
⎝
0 0 1
0 1 0
−1 0 0
⎞
⎠
⎛
⎝
x
y
z
⎞
⎠ =
⎛
⎝
z
y
−x
⎞
⎠
Rz
⎛
⎝
x
y
z
⎞
⎠ =
⎛
⎝
0 −1 0
1 0 0
0 0 1
⎞
⎠
⎛
⎝
x
y
z
⎞
⎠ =
⎛
⎝
−y
x
z
⎞
⎠ .
If one interchanges the order of the two rotations, one obtains a different
result,
RzRy =
⎛
⎝
0 −1 0
1 0 0
0 0 1
⎞
⎠
⎛
⎝
0 0 1
0 1 0
−1 0 0
⎞
⎠ =
⎛
⎝
0 −1 0
0 0 1
−1 0 0
⎞
⎠
RyRz =
⎛
⎝
0 0 1
0 1 0
−1 0 0
⎞
⎠
⎛
⎝
0 −1 0
1 0 0
0 0 1
⎞
⎠ =
⎛
⎝
0 0 1
1 0 0
0 1 0
⎞
⎠ .
• The angles θ and ϕ describe the orientation of the body axis; the angle ψ
describes the (eigen) rotation of the body around the body axis.
• The angular velocity of the precession is ϕ̇, the one of the nutation is θ̇,
and the one of the eigen angular momentum (spin) is ψ̇.
The Angular Velocity
If one has the kinetic energy T = 1
2ω ·Θ ·ω in the space system, then not only
the angular velocity ω but also the inertia tensor Θ is time dependent; thus
one wants to know these quantities in the body system, in which the inertia
tensor is constant.
The angular velocity one be described by the Euler angles and their time
dependence. In the space system, the angular velocity ωϕ of the rotation with
ϕ̇ around the body axis (z3 axis) is
8.7 Rotation Around a Fixed Point: The Top 257
ωϕ = U3 · U2 · U1 · ezϕ̇ = ϕ̇
⎛
⎝
sinψ sin θ
cosψ sin θ
cos θ
⎞
⎠
The angular velocity ωθ of the rotation with θ̇ around the x1 axis is
ωθ = U3 · U2 · exθ̇ = θ̇
⎛
⎝
cosψ
− sinψ
0
⎞
⎠
The angular velocity ωψ of the rotation with ψ̇ around the z axis is
ωψ = ezψ̇ = ψ̇
⎛
⎝
0
0
1
⎞
⎠
The total angular velocity is
ω = ωϕ + ωθ + ωψ.
8.7.3 * The Euler Equations of Motion
In the space (inertial) system S the equations of motion for the top (pure
rotations, no translations) is
L̇
(S)
= N (S)
with
L(S) = Θ(S) · ω(S).
The angular velocity ω as well as the inertia tensor Θ in the space system (S)
are varying in time. The transformation from the space to the body system
(S′) is
N (S) = L̇
(S) (7.8)
= L̇
(S′)
+ ω × L(S′).
One has thus
L̇
(S′)
= N (S) − ω × L(S′), (8.25)
where N (S) and ω are in the space system S.
Agreement: For the rest of this paragraph the angular momentum L and the
angular velocity ω shall be considered as given in the body-fixed system, and
the indices S′ shall be omitted.
With L = Θ · ω, Θ̇ ≡ 0 (in the system S′) and Θij = Iiδi,j in the
body principal-axis system one then finds the so-called Euler equations as
the component-wise representation of (8.25)
I1ω̇1 = N1 + (I2 − I3)ω2 ω3
I2ω̇2 = N2 + (I3 − I1)ω3 ω1
I3ω̇3 = N3 + (I1 − I2)ω1 ω2.
(8.26)
258 8 Dynamics of a Rigid Body
Proof: In the principal-axis body system one has on the one hand
Li = ωiIi ⇒ L̇i = ω̇iIi;
on the other hand, one has
L̇
(8.25)
= N − ω × L, ⇒ L̇i = Ni −
∑
jk
εijkωjLk
and from these two relations
ω̇iIi = Ni −
∑
jk
εijkωjωkIk
(and cyclic) with the Levi-Civita tensor with the elements
εijk =
{
+1 if (ijk) is an even permutation of (123)
−1 if (ijk) is an odd permutation of (123)
0 otherwise.
�	
Comment: Notice that the Euler equations (8.26) are nonlinear coupled
differential equations (of first order) (for ω1, ω2 and ω3). Since there is thus
in general18 no “harmonic” approximation, the treatment of the dynamics of
the angular momenta is much more difficult than that of oscillations.
8.8 The Force-Free Symmetrical Top
8.8.1 The Equations of Motion
In the case of the symmetrical top the nonlinear system of (8.26) can be
treated analytically. Let the principal moments of inertia be
I1 = I2 �= I3.
The force-free top is (by definition) without external torque,
N = 0,
thus, either the external field vanishes, or the top is placed in the homogeneous
gravitational field and is suspended at the center of mass. Then the Euler
equations of motion (8.8) reduce to
I1 ω̇1 = −(I3 − I1)ω3 ω2
I1 ω̇2 = (I3 − I1)ω3 ω1
I3 ω̇3 = 0.
18 See but the following Sect. 8.8.
8.8 The Force-Free Symmetrical Top 259
From this one finds immediately a first solution
ω3 = const. (8.27)
With this a system of two coupled linear equations for ω1 and ω2 is obtained,
which one can solve with an exponential ansatz
ωi ∝ eλt (i = 1, 2).
One obtains (
I1 λ (I3 − I1)ω3
−(I3 − I1)ω3 I1 λ
) (
ω1
ω2
)
= 0.
The solubility condition is the vanishing of the secular determinant,
(I1 λ)2 + [(I3 − I1)ω3]2 = 0.
One obtains the eigen values
λ = ±iω3
I3 − I1
I1
= ±iΩ (8.28)
and the eigen vectors (
1
∓i
)
(except for a phase and except for a normalization factor).
The general solution is
(
ω1
ω2
)
= a
(
1
−i
)
eiΩt + b
(
1
i
)
e−iΩt
Ωt+ϕ
ω
e2
e3
e1
Fig. 8.19. The motion of the force-free sym-
metrical top (I3 > I1) in the body system
260 8 Dynamics of a Rigid Body
or
ω1 = A cos(Ωt+ ϕ)
ω2 = A sin(Ωt+ ϕ)
ω3 = C
with
Ω
(8.28)
= ω3
I3 − I1
I1
.
The body does not rotate about a fixed symmetry axis. Rather, in the body
system one has thus the following picture: The vector ω of the angular velocity
of the rotating body precesses with constant angular velocity Ω (for I3 > I1 in
the same direction as ω3) around the body symmetry axis e3 with ω3 = const,
see Fig. 8.19.
Example: The motion of the poles of the earth:19 The earth is flattened at
the poles, such that one finds20 for the moments of inertia
I3 − I1
I1
≈ 1
300
.
From this one obtains
Ω ≈ ω3
300
, T3 =
2π
ω3
= 1 d ⇒ T =
2π
Ω
≈ 300 d,
i.e., the instantaneous axis of rotation of the earth should precess with the
so-called Euler period of about 300 days. However, the yearly period (of 365
days) causes atmospheric redistributions of the mass, which changes the Euler
period of 300 days. In addition, the earth is not rigid but elastic and primarily
liquid in the interior; this reduces the effect; the resulting so-called Chandler
period is T = 415 to 433 days [14, 16]. The radius of the pole motion is
A ≈ 10 m, but varying in time/“Chandler wobble”).
The rotation of the inclination with time against the ecliptic is another
effect, which originates from the forces of the. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
B.2.1 Rules for Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
B.2.2 Dyadic Product (Tensor Product) of Vectors . . . . . . . . . . 331
B.2.3 Vector Product (Cross Product) of Vectors . . . . . . . . . . . 331
B.2.4 Triple Scalar Product of Vectors . . . . . . . . . . . . . . . . . . . . . 332
B.2.5 Multiple Products of Vectors . . . . . . . . . . . . . . . . . . . . . . . . 332
xvi Contents
C Rectangular Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 333
C.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
C.2 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
C.3 Spherical Polar Coordinates (Spherical Coordinates) . . . . . . . . . 334
C.4 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335
C.5 Plane Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335
C.6 Inverse Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336
D Nabla (Del) Operator and Laplace Operator . . . . . . . . . . . . . . . 339
D.1 Representations of the Nabla and Laplace Operators . . . . . . . . . 339
D.2 Standard Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340
D.3 Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
D.4 Integral Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342
D.4.1 The Theorem of Gauß . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342
D.4.2 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
D.4.3 The Theorem of Green . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
E Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
E.1 Functions and Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347
E.2 Variational Problem and Euler Equation . . . . . . . . . . . . . . . . . . . 348
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350
F Linear Differential Equations with Constant Coefficients . . . 351
F.1 Homogeneous Linear Differential Equations . . . . . . . . . . . . . . . . . 351
F.2 Inhomogeneous Linear Differential Equations . . . . . . . . . . . . . . . 352
F.3 Stability of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353
G Quadratic Matrices and Their Eigen Solutions . . . . . . . . . . . . . 355
G.1 The Eigen Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355
G.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355
G.3 Properties of the Eigen Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
G.4 Properties of the Eigen Vectors of Hermitian Matrices . . . . . . . . 358
H Dirac δ-Function and Heaviside Step Function . . . . . . . . . . . . . 361
H.1 Properties of the Dirac δ-Function and of the Heaviside
Step Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
H.2 Representation of the δ-Function by Functional Sequences . . . . 362
H.3 Integral Representation of the δ-Function . . . . . . . . . . . . . . . . . . . 363
H.4 Periodic δ-Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
H.5 The δ-Function in R
3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
H.6 The δ-Function as an Inhomogeneity of the Poisson Equation . 364
Contents xvii
I Fourier Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365
I.1 The Transformation: Fourier Integral . . . . . . . . . . . . . . . . . . . . . . 365
I.1.1 Examples and Applications . . . . . . . . . . . . . . . . . . . . . . . . . 366
I.1.2 Convolution Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
I.1.3 Parseval’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
I.1.4 Uncertainty Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372
I.2 Fourier Transformation in R
4: Plane Waves . . . . . . . . . . . . . . . . . 374
I.2.1 The Whole R
3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374
I.2.2 Normalization Volume V . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
I.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376
I.3.1 The Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376
I.3.2 Examples and Applications . . . . . . . . . . . . . . . . . . . . . . . . . 378
I.3.3 Convolution Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
I.3.4 Parseval’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382
I.3.5 Fourier Series in R
3: Lattices . . . . . . . . . . . . . . . . . . . . . . . 382
I.3.6 Functions with Lattice Periodicity . . . . . . . . . . . . . . . . . . . 383
Summary: Fourier Integral and Fourier Series . . . . . . . . . . . . . . . . . . . . 384
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385
J Change of Variables: Legendre Transformation . . . . . . . . . . . . . 387
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391
Symbols
a Acceleration, general vector
da Surface (“area”) element (oriented outwards)
A Work
A Electromagnetic vector-potential field
b Impact parameter
b General vector
B Magnetic-induction field
c General vector
C Curve
d Differential
d General vector
D General matrix
e eccentricity
e Unit vector, eigenvector
e Exponential function
E Electrical field
f Force-density field
f Force (density)
F Generating function, area, gauge function (Sect. 3.8)
g Green matrix
g Gravitational acceleration, general vector field
G Reciprocal-lattice vector
h Lattice vector
H Hamiltonian function
H Hamiltonian density
i Imaginary unit
I Moment of inertia
j Energy current density
k Force constant, wave vector in R
1
k Wave vector
K Hamiltonian-type function
xx Symbols
l Angular momentum
L Lagrangian function, canoid lengths
L Lagrangian density
L Total angular momentum
m (Point) Mass
M Mass matrix, general matrix
n Vector normal to a surface element
N Total number of particles
N Torque
p Canonical momentum
p Momentum, electrical dipole
P Power, transformed canonical momentum
P Total momentum
q Canonical coordinate, electrical charge
q Vector within Brillouin zone
Q Generalized force, transformed canonical coordinate, electrical charge
r Position vector
R Radius of sphere
R Center of mass, translation vector
s Impact parameter
s Small deviation from ω (Sect. 8.8.2)
ds Line element
S Tension
t Time
T Kinetic energy
T Energy-momentum tensor
T Stress tensor
u Displacement field, general vector
U Transformation matrix, unitary matrix
v Velocity, general vector
V Potential energy
V Center-of-mass velocity
Z Force of constraint
α Stokes frictional force constant (1.18), angle, cartesian index
β Velocity relative to velocity of light (Sect. 7.3), cartesian index
γ Damping constant, constant in SR
δ Dirac delta function
Δ Difference
ε Levi-Civita tensor, small quantity
ε Reduced energy
η Viscosity coefficient
θ Heaviside theta function,sun and the moon; the corre-
sponding period is T ≈ 26,000 years.
The geometrical north pole is the cut of the body symmetry axis through
the surface of the earth; the kinematical north pole is that of ω.
8.8.2 * Stability of the Rotational Motion of the Top
Consider a free (asymmetrical) top with three different principal axes Ii. The
uniform rotations
19 The center of mass of the earth is considered as a fixed point; the rotation around
the sun is thus neglected.
20 (I3 − I1)/I1 = 1/306 according to Goldstein [1], Sect. 5.6.
8.8 The Force-Free Symmetrical Top 261
ωi = ωiei, ωj = 0 (j �= i) (8.29)
around each of the three principal axes of inertia ei are (stationary) solutions
of the force-free equations of motion (8.26) (with N = 0)
I1ω̇1 − (I2 − I3)ω2ω3 = 0
I2ω̇2 − (I3 − I1)ω3ω1 = 0
I3ω̇3 − (I1 − I2)ω1ω2 = 0.
(8.30)
Not all stationary solutions (8.29) are stable:21 Now let ω = (0, 0, ω3) be
the solution (8.27), the stability of which is to be determined. To this end one
investigates the behavior of a neighboring solution ω + s
(0, 0, ω3) + (s1, s2, s3) with si � ω3.
If one inserts this into the equation of motion (8.30) and, because of the
smallness of the deviations si, takes into account only the terms linear in si
and neglects the terms of higher order, one obtains a system of coupled linear
equations,
I1ṡ1 − (I2 − I3)ω3s2 = 0
I2ṡ2 − (I3 − I1)ω3s1 = 0
I3ṡ3 = 0.
(8.31)
One finds from the last of the three equations s3 = const. For the two
other components one makes the ansatz
si ∝ eλt i = 1, 2 (8.32)
and obtains (
λI1 −(I2 − I3)ω3
−(I3 − I1)ω3 λI2
)(
s1
s2
)
= 0.
The vanishing of the secular determinant yields
λ2 =
(I2 − I3)(I3 − I1)
I1I2
ω2
3 . (8.33)
Case λ2 0: If I3 is between I1 and I2, the solution (8.32) has one part
decreasing and one part increasing exponentially with time. Because of the
increasing term the solution for s(t) moves away from the stationary solution
ω = (0, 0, ω3), which is thus unstable.
21 For the stability see Appendix F.3.
262 8 Dynamics of a Rigid Body
Case I1 = I2 (λ2 a (Fig. 8.21). First, determine this
from the rotational motion of the body around the momentary touching line
of these two cylinders. Then compare with the result of the determination of
T from the translation of the center of mass of the rolling cylinder and of the
rotation around its symmetry axis.
R
a
Fig. 8.21. The cylinder of radius a rolling inside of a
hollow cylinder of radius R (for problem 8.8)
8.9. Cylinder Rolling on an Inclined Plane. Consider a cylinder (radius
R) which moves on an inclined plane (inclination α) under the influence of
the gravity see Fig. 8.22.
α
R
Fig. 8.22. Cylinder on an inclined plane
(for problem 8.9)
(a) Give an expression for the Lagrangian function(Euler) angle, scattering angle
Symbols xxi
Θ Moment of inertia
Θ Moment-of-inertia tensor
κ Kinetic-energy expansion coefficient, counter for function zeroes
λ Lagrangian multiplier, kinetic-energy expansion coefficient,
wavelength, eigenvalue
λ Runke-Lenz vector
Λ Lagrangian-type function
μ Sliding-friction coefficient, reduced mass,
kinetic-energy expansion coefficient, four-index in SR
ν Frequency, four-index in SR
ξ Complex coordinate, curvilinear coordinate
π Number pi, canonical momentum density
Π Transformation of canonical coordinates
ρ Mass density, charge density, particle density
� Damping constant, cylindrical radial coordinate
σ Cross section
τ Time increment
φ Electromagnetic scalar potential
ϕ Rotation angle
ϕ Angular vector of rotation
χ Electromagnetic gauge function
ψ Euler angle
ω Vector of angular velocity
ω Angular velocity
dΩ Space angle element
1
The Newtonian Mechanics of Point-Mass
Systems: General Properties
Starting from the Newtonian axioms general conclusions concerning the mo-
tion of masses shall be drawn in this chapter, and we will concentrate on
theorems which one obtains without explicit calculation, i.e., without explicit
solutions of the equations of motion. The integration of the equations of mo-
tion will then be carried out for various examples in Chap. 2.
The primary quantity of this chapter is the force, while in Chap. 3 the
primary quantity is the (kinetic and potential) energy. Even though the force
on the one hand and the energy on the other as the cause of the motion should
be known from the elementary lecture on phenomenological (experimental)
mechanics, the force as agens movendi shall be put in the foreground in this
introduction, since the Newtonian mechanics is more intuitive than the more
abstract Lagrangian theory, even though in many cases the latter method is
more useful and comfortable in setting up the equations of motion.
1.1 Point Masses, Open and Closed Systems
1.1.1 The Point Mass as an Idealization
Of course, all bodies are extended and not point-like, but in most of those
cases, which shall be investigated here, the extension of the various bodies
is small compared to their mutual distances, and in this case it is a good
approximation to consider these bodies as point-like.1 Then one talks of point
masses.
Examples: The extension of moon and earth is small compared to their
mutual distance. – The extension of the earth–moon system is small compared
to the distance to the sun. – The extent of the system of the sun with its
planets is small compared to the distance to the nearest star. – The extension
of the molecules of a dilute gas is small compared to the average distance
1 See also the comment to the center of mass at the end of Sect. 1.5.
2 1 Newtonian Mechanics: General Properties
Fig. 1.1. The internal and external forces in a system of particles
between the molecules; but for dense gases at high pressures one may approach
the limit of the validity of the approximation by a point-like “extent”.2 –
Idealizations similar to the massive particles one has likewise for the extended
electrically charged particles.
1.1.2 Open and Closed Systems, Internal and External Forces
Definition 1. (Internal and External Forces): If one combines a chosen num-
ber of particles to a system then the forces between these selected particles of
the system are inner forces, and the (“externally” acting) forces of the other
(external) particles on these selected (“inner”) particles are external forces.
The force F i(ri) on a particle (index i) at position ri thus is composed of
the external force F ext
i (ri) and of the internal forces F ij , which are caused
by the particles j at the positions rj ,3
F i(ri) =
∑
j
F ij(r1, r2, . . .) + F ext
i (ri), (1.1)
see Fig. 1.1.
Definition 2. (Closed System): A closed system is a system without external
forces.
Definition 3. (Open System): An open system is a system with external
forces.
2 Extended solid bodies are treated in Chap. 8 and extended elastic bodies in
Chap. 10.
3 The other way around, Scheck denotes F ij as the force of the particle i on the
particle j.
1.1 Point Masses, Open and Closed Systems 3
Fig. 1.2. Left: Earth and moon with negligible further external forces as a closed
system. Right: Earth and moon as an open system with external forces exerted by
the sun
Example: One can consider the motion of the moon around the earth and
the motion these two around the sun (neglecting the influence of the other
planets) in (at least) two ways: Firstly, one can combine moon and earth to
an open system; then the force between moon and earth is an internal force,
and the forces by the sun on the system of earth and moon are external forces.
Alternatively, one can combine all three bodies to a (closed) system; then all
forces between these three bodies are internal forces. Finally, if one would take
into account the forces of the other planets and stars acting on the three-body
system of sun, moon, and earth, then those forces would be external forces,
and the three-body system would be an open system.
Further Examples:
• Planets in the gravitational field of the sun (open)
• Planets and sun (closed), see Fig. 1.2
• Charge in a condenser (open)
• Charge and condenser (and possibly the source of the current) (closed),
see Fig. 1.3
Comment: Whether one considers an open or a closed system, often is a
question of expediency. For example, in an approximate way one can often
investigate the motion of the planets in the field of a central star assumed as
fixed; the only change for the planetary motion taking account of the moving
Fig. 1.3. A charge in a condenser as an (left) open system and (right) closed system
4 1 Newtonian Mechanics: General Properties
Fig. 1.4. The internal forces Fij and external forces F ext
i acting on particle i
central star would be the replacement of the mass of the planet by the reduced
mass (of planet and central star).4 In the (academic) limit of an infinitely
heavy central star the two systems are identical.
1.2 Newton’s Axioms (1687): A New Era
The original formulation of the Newtonian axioms have been different, of
course. In modern formulation they read:
(I) Law of Inertia: There are reference systems, namely so-called inertial sys-
tems, in which each particle (index i) remains in rectilinear uniform motion
(in particular at rest), if there is no resulting force acting on it,
F i = 0 ⇔ ṗi = 0 ⇔ pi = const. (1.2)
(II) Law of Motion: In inertial systems the time change of the momentum is
equal (in modulus and direction) to the acting force,
F i = ṗi. (1.3)
(III) Reaction Law: The force F ji of a particle i acting on a particle j is
oppositely equal to the force F ij of particle j acting on particle i,
F ji = −F ij . (1.4)
(IV) Superposition Principle: The resulting force F i of all forces F ij acting
on a particle i are superposed according to the rule of vector addition,
F i
(1.1)
=
∑
j
F ij + F ext
i . (1.5)
Cf. Fig. 1.4.
4 See the two-body problem in Sect. 2.3.
1.2 Newton’s Axioms (1687): A New Era 5
Agreement 1 (time derivatives): A time derivative of a quantity will be de-
noted by a dot over the symbol.
Comments:
• These laws presume that space and time exists, that one can thus measure
distances |Δr| and time differences Δt, and that the length and time scales
are constant in space and time. The time is homogeneous, and the space
is homogeneous and isotropic. Newton assumed an absolute space and an
absolute time which is not quite correct, as one knows since Einstein’s for-
mulation of the Special Relativity theory at the beginning of the twentieth
century.5
• Starting from these laws, the motion of particles under the influence of
forces shall be described in this chapter. (In the following we will investi-
gate primarily point masses.) Thus the main object is to describe the form
of the trajectories of all particles, for example the Kepler ellipses in the
case of the planetary motion. This is the topic ofthe kinematics. Or, and
this is the more frequent case, one is rather interested in how, i.e., with
which time dependence, the ith particle moves along the trajectory. Then
one is interested in the time dependence of the coordinate ri(t). This is
the topic of the dynamics. From the latter the form of the trajectory is
given in one of many other possible forms of parametric representations,
namely that with the time t as parameter.
• Furthermore, the differentiability etc. is assumed, from which we will al-
ways start in the following, if not stated otherwise, such that the following
quantities (of a particle) are defined:
velocity v(t) = ṙ(t) ≡ dr
dt = limΔt→ 0
r(t+Δt)−r(t)
Δt
acceleration a = v̇ = r̈.
In classical mechanics one also uses, e.g., the mass density dm/dV , which
presents difficulties in the microscopic regime because of the corpuscular
nature of matter.
• The systems postulated in the first Newtonian law are the so-called inertial
systems. In contrast, an isolated particle in all accelerated systems is thus
subject to forces, even if there are no external forces acting.6 Since the
forces decrease at large distances between the particles7 and since forces
originate only from other particles, one can define an inertial system equiv-
alently by the requirement that such a system is sufficiently far away from
5 Even though one can measure distances Δl = |r1 − r2| and time differences
Δt = t1 − t2 (Euklidian structure of the Rt × R
3), these may depend upon the
reference system: There are the effects of length contraction and time dilation,
see Sect. 7.3.
6 These are so-called inertial forces, see, e.g., Sects. 7.1.1 and 7.2.4.
7 This holds for the gravitational force and for the Coulomb force and, as far as
one knows, also for all other forces, like e.g., nuclear forces – with the exception
of the confinement forces between the quarks.
6 1 Newtonian Mechanics: General Properties
all other particles and does not rotate against the system of the far-away
particles or is in any other way accelerated.8
• Newton’s first law may seem as a special case of the second law. This is not
the case, since it postulates the existence of the inertial reference systems
and defines them.
• Useful (approximate) inertial systems are
– for many problems: earth
– for Coriolis effects on the earth: planetary system, galaxy
• With his formulation of the second axiom in the form of F = ṗ (and not
F = mv̇) Newton was (unknowingly) ahead of his time.9
• In general one has
ṗ = m v̇ + ṁv,
and the equation F = mv̇ is a special case. – The case ṁ �= 0 appears,
e.g., in the treatment of the rocket, the mass of which decreases with time
due to burning of the fuel.
• Implicitly two-particle interactions are assumed in the third law.10 This
applies to the gravitational interaction of macroscopic physics as well as to
the Coulomb interaction of macroscopic and microscopic physics. It does
not apply to the phenomenological frictional force,11 which one would have
to derive from microscopic interactions, as well as to the phenomenological
angular forces in molecules and solids.
• The sum of the internal forces vanishes because of Newton’s third axiom,
∑
ij
F ij = 0, (1.6)
even if they are not central forces.12
8 In accelerated reference systems one has the appearance of the so-called inertial
forces, see Sect. 7.2.4.
9 In the Special Relativity theory one has
p =
m0v(t)√
1 − (v(t)/c)2
with the rest mass m0. This is sometimes written in the form
p = mv with m =
m0√
1 − (v(t)/c)2
.
But properly speaking, in Special Relativity it is not the mass, which appears as
velocity dependent, but it actually is the derivative with respect to time which is
modified, e.g., in v = ṙ. See the Course on Special Relativity.
10 Cf. Fig. 1.5.
11 See Sect. 1.4.6.
12 See Fig. 1.5 and Sect. 1.4.
1.2 Newton’s Axioms (1687): A New Era 7
Fig. 1.5. The oppositely equal forces; left : central forces; right : noncentral forces
• In inertial systems the Newtonian laws are form-invariant under Galilean
transformations,13 i.e., the Newtonian laws have the same mathematical
form, independent of the special inertial system.
• Newton’s first and second laws are fundamental insofar as they
united the celestial and earthly mechanics. Before Newton the idea
was that processes on earth depended on action by man or nature and were
unforeseeable. In contrast, the motion of the stars and planets have been
extremely steady, thus they were not only considered as extraterrestrial
but as supernatural; in fact, in some ages before that a “soul” has been
assigned to them because of their seemingly fore-free motion. Actually,
Galileo Galilei has had similar ideas.
• Newton’s first and second laws have been also a remarkable rev-
olution of physical thought, insofar as from empirical experience
it needed a permanent force to keep a body in steady motion.
• The case of vanishing friction was not realized before Newton’s time, except
possibly by Galileo Galilei. He was among the first who, among many
other accomplishments, started from notions, assumptions, and models to
draw conclusions in a way which we consider today as scientific. However,
instead of the rectilinear motion he assumed the circular motion to be the
force-free motion like that of the celestial bodies.
Agreement 2 (constant mass): If not stated otherwise, the mass m of a
particle will always be assumed as constant in the following
ṁ = 0 .
Agreement 3 (reference system): If not stated otherwise, an inertial system
will always be assumed in the following.
Definition 4. (Configuration space): The trajectory14 r(t) : Rt → R
3 of a
single particle is a mapping of the one-dimensional space Rt of the time onto
13 See Sect. 7.3.3.
14 In Special Relativity theory this is the so-called world line, see Sect. 7.3.
8 1 Newtonian Mechanics: General Properties
the R
3, the so-called configuration space. If one has N particles with their
trajectories ri(t) (with i = 1, . . . , N) the configuration space is analogously
the N -fold product R
3 × R
3 × · · · × R
3 = R
3N .
1.3 Einstein’s Equivalence Principle (1916):
Another New Era
An implicit additional principle is Einstein’s equivalence principle of the equal-
ity of gravitational and inertial mass:
1.3.1 Mass, Weight, Force, and the Like
With length and time as measurable quantities one can easily define velocity,
acceleration, volume, etc. But there remains the problem of measuring mass,
weight, force, momentum, energy or the like without a new unit.
Even if well known to the physicist from the introductory lectures on
phenomenological mechanics one should be aware that the latter of the above-
mentioned quantities have been difficult to define some ages ago. While a
spring can be stretched by a muscle independent of whether the spring is
aligned horizontally or vertically, the extension of the spring by a mass in
the gravitational field depends on the orientation of the spring relative to the
vertical direction; there seems to be a difference between “mass” and “weight”
(of course, as we know today). Anyway, another unit had to be introduced.
In Newton’s second law either the mass or alternatively the force or the
momentum is defined. With force (and momentum) given, the relation
p = miv
defines the inertial mass mi.15 The gravitational mass mg is defined by the
gravitational force
F = mgg
in a (homogeneous) gravitational field. By equating these forces,
F = mgg and F = miv̇, (1.7)
one obtains
v̇ =
mg
mi
g
for all bodies, thus a proportionality of inertial and gravitational mass.
15 This is the easy route: Of course, if the mass were given, the force (and momen-
tum) would be derived quantities.
1.4 Types of Forces 9
1.3.2 Einstein’s Equivalence Principle
The equality
mg = mi (1.8)
is known as the Einstein equivalence principle.
Comments: Even though the mass equivalence is usually referred to as being
stated in the 1905 paper of Einstein, this latter paper contains the energy-
mass equivalenceof Special Relativity, while the mass equivalence is the basic
assumption of the General Relativity theory of the 1916 paper.
Originally postulated, the equality has been found from experiments to
hold to within 10−12. In fact, already in 1916 the mass equality was experi-
mentally confirmed to within some parts of 10−6 by Eötvös (1828–1919).
Agreement 4 (mass): Henceforth we will not distinguish between inertial
and gravitational mass.
1.4 Types of Forces
1.4.1 Central Forces
Definition 5. (Fields): A scalar field or a vector field shall be understood as
a scalar or vectorial quantity, respectively, depending on a coordinate, i.e., a
mapping of configuration space to a scalar or vectorial space.
Comment: The notion of a field may be more familiar from electromag-
netism: The connection of the electrical field E and of the force F on a charge
q at the position r is F (r) = qE(r). A scalar field is thus a mapping of the
R
3 onto an R
1, and a vector field is a mapping of the R
3 onto an R
3.
Definition 6. (Central force field): A central force field is a radially oriented
vector field (for F 0 oriented to or from the center, respectively)
F (r) = F (r)er, er =
r
r
. (1.9)
er is the unit vector (with modulus 1) pointing in the radial direction, see the
left part of Fig. 1.5.
Agreement 5 (unit vectors): In general we will denote unit vectors by e.
10 1 Newtonian Mechanics: General Properties
Basic experience for masses (gravitational force): The gravitational
force between two (point) masses m1 and m2 at the positions r1 respectively
r2 is oriented radially and proportional to the squared inverse of their dis-
tance,16
F Grav
ij = −G mimj
|ri − rj |2
eij (1.10)
with the unit vector
eij =
ri − rj
|ri − rj |
. (1.11)
Comments:
• Like the axioms the basic experiences cannot be proven on mathematical
grounds. They can only be verified by experiment to within a certain
approximation; presently the error is less than 10−10.
• With the mass and the force defined in (1.7) and with the symmetry of the
interaction, the force in (1.10) must necessarily contain a proportionality
constant, the gravitational constant G.
• Kepler’s laws17 laid the basis for the gravitational potential. They could
be taken as a set of alternative basic experiences. However, from the one
gravitational potential all three of Kepler’s laws can be concluded.
Further Examples:
The Coulomb force18 between two point charges q1 and q2 at the positions
r1 respectively r2 is qualitatively similar to the gravitational force,
F Coul
ij =
1
4πε0
qiqj
|ri − rj |2
eij , (1.12)
see Fig. 1.6. An example for a non-central force is the force, which an electrical
point dipole p at r = 0 exerts on a charge Q at r (or also reversely),
F (r) = − 1
4πε0
p − 3 er p · er
r3
,
16 The gravitational constant is G = 6.67× 10−11 m3 kg−1 s−2; a more precise value
can be found in Appendix A.
17 See Sect. 5.2.1.
18 Unfortunately still two unit systems are customary in electromagnetism; both
systems can be used here simultaneously if one defines
[x] =
{
x in the SI system
1 in the cgs system.
In particular one has
1
4πε0
= 4πc2 × 10−7
with the velocity of light c = 2.997 924 58 × 108 m s−1.
1.4 Types of Forces 11
Fig. 1.6. The force F on an electrically (left) positive and (right) negative charge
q in an electrical field E
Fig. 1.7. The forces of an electrical dipole p on
positive charges at equal distances but different posi-
tions, namely parallel, respectively, perpendicular to the
dipole orientation
see Fig. 1.7 (The force field is, however, a superposition of central (Coulomb)
forces originating from the two charges forming the dipole, see the Course on
Electricity.)
Additional examples for noncentral forces one finds for the interactions
between elementary particles.
1.4.2 Conservative Forces (I)
Definition 7. (Conservative force field): A force field F (r) is called conser-
vative,19,20 if it can be written as a gradient of a scalar field V (r),
F (r) = −∇V (r). (1.13)
Notation: This scalar field V (r) is called potential energy.
Agreement 6 (del operator): Instead of the nabla operator (del operator) ∇
we will often formally write ∂/∂r,
F (r) = −∇V (r) ≡ −∂V (r)
∂r
,
well knowing that one cannot divide by a vector. However, this will turn out
to be useful in the context of the chain rule of differentiation.
Agreement 7 (conservative forces): If not stated otherwise, we will treat the
inner forces as conservative forces. In particular for the gravitational and for
the Coulomb force one has
19 Alternative definitions will be found in Sect. 1.7.
20 The denotation stems from the fact that the energy is conserved in a system with
conservative forces; see Sect. 1.7.
12 1 Newtonian Mechanics: General Properties
V Grav
ij = −G mimj
|ri − rj |
(1.14)
V Coul
ij =
1
4πε0
qiqj
|ri − rj |
. (1.15)
with
∇i
1
|ri − rj |
=
d 1
|ri−rj |
d|ri − rj |
∇i|ri − rj | = − 1
|ri − rj |2
eij
as in (1.10) and (1.12).
1.4.3 Central Potentials
Definition 8. (Central potential): A central potential is radially symmetric
and has the form
V (r) = V (|r|) = V (r). (1.16)
Comments A central potential leads to a conservative and central force, see
Problem 1.4.
1.4.4 Potential and Potential Energy
In mechanics the notions potential and potential energy are mostly used with
same meaning. Properly speaking, e.g., the gravitational potential U(r) at the
position r, caused by a mass M at the position rM , is given by
U(r) = −G M
|r − rM | ,
and the potential energy of a mass m at the position rm in this potential is
V (rm) = mU(rm) = −G mM
|rm − rM | .
Comment: For reasons of lingual simplification the notion “potential” is
often incorrectly used in the sense of “potential energy.” In electromagnetism
one is more careful in differentiating between these two notions.
1.4.5 The Lorentz Force
Basic experience for charges in electromagnetic fields: The force of
electromagnetic fields on a point charge q does not only depend upon its
coordinate r but also upon its velocity and is21
21 See footnote 1.18.
1.4 Types of Forces 13
Fig. 1.8. The vectors of the magnetic induction B, of
the velocity v, and of the Lorentz force F = qv × B
F (r,v, t) = q
[
E(r, t) +
[c]
c
v(r, t) × B(r, t)
]
, (1.17)
see Fig. 1.8. Here the first term is the Coulomb force, and the second term is
the Lorentz force, the latter depicted in Fig. 1.8.
In general, an external force on a particle can thus depend upon its coor-
dinate r as well as upon its velocity ṙ and explicitly upon the time t,
F = F (r, ṙ, t).
1.4.6 Frictional Forces
Frictional forces are primarily velocity dependent.
Examples for velocity-dependent forces are
• Stokes frictional force (fast ballistics)
F = −αvev. (1.18)
(For a sphere with radius R, which moves in a medium with the viscosity
coefficient η, one has α = 6πRη.)
• Newtonian frictional force (slow ballistics)
F = −βv2ev. (1.19)
• Coulomb friction (sliding and static friction)
F =
{
−μGF⊥ev for v �= 0
−F ‖ for v = 0 (1.20)
with F‖consists of only a single particle.
1.5.1 Definitions
Definition 9. (for a single particle, characterized by the index i):
mi the mass of the particle;
ri the coordinate (the position) of the particle;
pi = miṙi the momentum of the particle;
li = ri × pi the angular momentum of the particle with respect
to the origin (at r = 0)
F i the force acting on the particle;
Ti = 1
2miṙ
2
i kinetic (translational22) energy23 of the particle.
Definition 10. (for a system of N particles):
The sums are understood as
∑
≡
∑N
i=1
total mass M =
∑
mi
center of mass R = 1
M
∑
mi ri
total momentum P =
∑
pi =
∑
miṙi = MṘ
total angular momentum L =
∑
li =
∑
ri × pi
internal total energy E =
∑ 1
2mi
p2
i + 1
2
∑
ij Vij
(conservative systems)
22 Extended particles have also rotational energy in addition to the translational
energy, see Chap. 8.
23 This is generally not equal to 1
2
mṙ2, see, e.g., the motion in a plane, Sect. 2.2.
1.5 Equations of Motion 15
Fig. 1.10. The vectors of the position r, of the
momentum p, and of the angular momentum l; l
points into the plane of the drawing
as well as
(external) total force F ext =
∑
F ext(ri)
(external) total torque N ext =
∑
r × F ext
i (ri)
work of the external field Aext =
∑∫
F ext(ri) · dri
power of the external field P ext = Ȧext .
Definition 11. (Vectors and Pseudovectors):
1. Vectors, which transform into their negative under space inversion, are
also called polar vectors.
2. Vectors, which are invariant under space inversion, are also called axial
vectors or pseudo vectors.
Examples for (polar) vectors: The position vector r and its derivatives
ṙ, r̈ with respect to time, the momentum vector p, the force F , etc.
Examples for pseudovectors: The angular momentum vector l = r × p,
see Fig. 1.10, or all cross products of polar vectors.
1.5.2 Equations of Motion
The equations of motion for the total system are:
(i) Equation of motion for the (total) momentum: The time change of the
momentum of the total system is equal to the external force (axiom II),
Ṗ = F ext. (1.21)
(ii) Equation of motion for the (total) angular momentum: The time change
of the angular momentum of the total system is equal to the external
torque,
L̇ = N ext, (1.22)
where the internal forces are assumed as central forces.
16 1 Newtonian Mechanics: General Properties
(iii) Equation of motion for the (total) energy: The time change of the energy
of the system is equal to the external power,
Ė = P ext. (1.23)
Proof:
(i) According to the second axiom one has
Ṗ
(1.3)
=
∑
i
ṗi =
∑
i
⎡
⎣
∑
j
F ij + F ext
i
⎤
⎦ (1.6)
= 0 + F ext.
(ii)
L̇ =
d
dt
∑
i
ri × pi =
∑
i
ṙi × pi +
∑
i
ri × ṗi
=
∑
i
mi ṙi × ṙi +
∑
i
ri × F i (pi = miṙi , ṗi = F i)
= 0 +
∑
i
ri × F i =
∑
i
ri ×
⎛
⎝
∑
j
F ij + F ext
i
⎞
⎠ .
The first term in the brackets can be rewritten after interchange of the
indices and with the third axiom,
∑
ij
ri × F ij = 1
2
∑
ij
ri × F ij + 1
2
∑
ij
rj × F ji
= 1
2
∑
ij
(ri − rj) × F ij .
According to the assumption the internal forces are central forces, such
that one has
F ij ‖ ri − rj
and thus
L̇ = 0 +
∑
i
ri × F ext
i = N ext .
(iii) With the chain rule one has
Ė =
d
dt
⎡
⎣
∑
i
1
2mṙi · ṙi + 1
2
∑
ij
Vij(ri, rj)
⎤
⎦
=
∑
i
mr̈i · ṙi + 1
2
∑
ij
ṙi · ∇iVij + 1
2
∑
ij
ṙj · ∇jVij
1.6 Conservation Laws 17
(1.13)
=
∑
i
mr̈i · ṙi − 1
2
∑
ij
ṙi · F ij − 1
2
∑
ij
ṙj · F ji
(1.4)
=
∑
i
mr̈i · ṙi − 1
2
∑
ij
ṙi · F ij − 1
2
∑
ij
ṙi · F ij
=
∑
i
⎛
⎝F i −
∑
j
F ij
⎞
⎠ · ṙi
(1.5)
=
∑
i
F ext
i · ṙi
with the axioms II and IV. �	
Comments:
• The preceding equations of motion are valid in particular for a single par-
ticle.
• Only the external, but not the internal forces contribute to the equations
of motion of the total quantities P , L, and E.
• The equation of motion for the total momentum implies that the center
of mass moves in such a way as if the total external force acts on a point
mass, consisting of the total mass and located at the center of mass. This is
the ground on which for the idealization by point masses in the mechanics
of massive bodies rests.
1.6 Conservation Laws
The next to last comment also implies that in a closed system (i.e., with
F ext
i = 0) the total of the momentum, angular momentum and energy are
conserved quantities.
Denotation: The conserved quantities are also called constants of the motion
or integrals of the motion.
Comment: The last name originates from the fact that the conserved con-
stant can be given as a result of a (first) integration of the equations of motion;
then the (first) integrals of the motion contain only the first derivative of the
coordinate with respect to time, while the equations of motion themselves
contain the second derivative. For example, the (conserved) total angular mo-
mentum L =
∑
imiri×ṙi only contains the first derivative, while the equation
of motion (1.22) contains the second derivative.
1.6.1 Conservation of Momentum
Conservation of Momentum: The momentum is a conserved quantity, if there
are no external forces,
F ext = 0 ⇔ Ṗ = 0 ⇔ P = const. (1.24)
Of course, this is clear from the second axiom, and it also follows from (1.21).
18 1 Newtonian Mechanics: General Properties
1.6.2 Angular-Momentum Conservation and Central Forces
According to (1.22) the angular momentum is a conserved quantity, if the
external torque vanishes. This is trivially fulfilled for vanishing external forces.
But this is also fulfilled for central forces:
Angular-momentum Conservation Theorem: If the force is a central force,
then there is no torque, and the angular momentum is conserved,
F i = Fi(r)er,i ⇒ N = 0 ⇔ L̇ = 0 ⇔ L = const.
(1.25)
Proof:
N = r × F
(1.9)
= r × Fer =
F
r
r × r = 0 .
�	
1.7 Energy-Conservation Theorem
and Conservative Forces
1.7.1 Conservative Forces (II)
Definition 12. Alternative to Definition 7 (conservative force): A force F is
called conservative, if alternatively one has:
(a) The conservative force field F (r) can be written as the gradient of a
potential energy V (r),
F (r)
(1.13)
= −∇V (r). (1.26)
(b) The conservative force field F is rotation-free,
∇× F (r) = 0. (1.27)
(c) The line integral over the conservative force F is independent of the path
C (dependent only upon the initial and final point of the path C),
∫
C
F (r) · dr independent of C. (1.28)
(d) The closed line integral over the conservative force F vanishes,
∮
F (r) · dr = 0 for all closed paths. (1.29)
1.7 Energy-Conservation Theorem and Conservative Forces 19
Proofs (extracts):
(a) ⇒ (b)
∇× F
(1.26)
= −∇×∇V (D.31)
= 0
(b) ⇒ (d)
∮
∂F
F · dr
(D.36)
=
∫
F
∇× F · da
(1.27)
= 0 for area F arbitrary
(a) ⇒ (c)
∫
C
F · dr
(1.26)
= −
∫
∇V · dr = −
∫
dV
with the last integral depending only upon boundary points. �	
Comment: Possibly, the condition (b) is not defined everywhere; for exam-
ple, the force field of a point mass or of a point charge at their respective
positions is not defined. (This is one of the difficulties with the idealizations;
in reality the masses or charges are extended, and then the force field at least
does not diverge, even though the direction still is not defined everywhere.)
1.7.2 Potential Energy, Work, and Power
The reverse of (1.13)=(1.26) is: The potential energy V (r) belonging to a
conservative force F (r) is the indefinite line integral
V (r) = −
∫
F (r) · dr + V0 (1.30)
and is defined up to a constant V0. The definite integral is
V (r) − V (r1) = −
∫ r
r1
F (r) · dr.
Definition 13. (Work): The work of the external force fields on a particle
(not necessarily conservative forces) is
A =
∫
dA = +
∫
F · dr. (1.31)
Definition 14. (Power): The power of the external force fields on a particle
(not of the particle on the field) is
P = Ȧ = F (r) · ṙ (1.32)
(from dA = F · dr ⇒ dA
dt = F · dr
dt ).
Comment: The line integral over a force exists for non-“pathological” force
fields, even if this force field