Cornerstones
Series Editors
Charles L. Epstein, University of Pennsylvania, Philadelphia
Steven G. Krantz, Washington University, St. Louis
Advisory Board
Anthony W. Knapp, State University of New York at Stony Brook, Emeritus
Emmanuele DiBenedetto
Classical Mechanics
Theory and Mathematical Modeling
Emmanuele DiBenedetto
Department of Mathematics
Vanderbilt University
Nashville, TN 37240
USA
em.diben@vanderbilt.edu
ISBN 978-0-8176-4526-7
e-ISBN 978-0-8176-4648-6
DOI 10.1007/978-0-8176-4648-6
Springer New York Dordrecht Heidelberg London
Library of Congress Control Number: 2010936443
Mathematics Subject Classification (2010): 34B25, 34C75, 37J05, 37J10, 37K05, 37K10, 37N05, 37N10,
70A05, 70B05, 70B10, 70E05, 70E15, 70E17, 70E18, 70H03, 70H05, 70H14, 70H15, 70H25, 70J10,
70J25, 70J30, 70J35, 76A02, 36D05, 70G75
c Springer Science+Business Media, LLC 2011
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Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .xvii
1
GEOMETRY OF MOTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Trajectories in R3 and Intrinsic Triads . . . . . . . . . . . . . . . . . . . . .
2
Areolar Velocity and Central Motions . . . . . . . . . . . . . . . . . . . . . .
3
Geometry of Rigid Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
The Euler Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Rotation Matrices and Angular Velocity . . . . . . . . . . . . . . . . . . . .
6
Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1
Translations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2
Precessions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3
Rototranslations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
The Axis of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
Relative Rigid Motions and Coriolis’s Theorem . . . . . . . . . . . . . .
9
Composing Rigid Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1
Connecting the Euler Angles and ω . . . . . . . . . . . . . . . . . .
10 Fixed and Moving Axodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.1 Precessions: Fixed and Moving Cones . . . . . . . . . . . . . . . .
11 Plane Rigid Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.1 Center of Instantaneous Rotation . . . . . . . . . . . . . . . . . . . .
11.2 Centrodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
3
5
6
7
10
10
10
11
11
12
13
15
16
17
18
19
20
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2c Areolar Velocity and Central Motions . . . . . . . . . . . . . . . . . . . . . .
2.1c Cycloidal Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2c The Brachistochrone and Tautochrone . . . . . . . . . . . . . . .
2.3c Hypocycloidal and Epicycloidal Trajectories . . . . . . . . . .
2.4c Kepler’s Gravitational Laws . . . . . . . . . . . . . . . . . . . . . . . .
2.5c Apsidal Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6c Elliptic Trajectories of Some Central Motions . . . . . . . . .
20
20
20
20
21
22
22
23
v
vi
2
Contents
8c
9c
Relative Rigid Motions and Coriolis’s Theorem . . . . . . . . . . . . . .
Composing Rigid Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1c Connecting the Euler Angles and ω . . . . . . . . . . . . . . . . . .
10c Fixed and Moving Axodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.1c Cone with Fixed Vertex and Rolling without Slipping . .
10.2c Cylinder Rolling without Slipping . . . . . . . . . . . . . . . . . . .
11c Plane Rigid Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.1c Parametric Equations of Fixed and Moving Centrodes .
11.2c Centrodes for Hypocycloidal Motions . . . . . . . . . . . . . . . .
11.3c The Cardano Device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.4c More on the Cardano Device . . . . . . . . . . . . . . . . . . . . . . . .
24
25
25
25
25
26
27
27
29
29
30
CONSTRAINTS AND LAGRANGIAN COORDINATES .
1
Constrained Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Constrained Mechanical Systems . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1
Actual and Virtual Displacements . . . . . . . . . . . . . . . . . . .
2.2
Holonomic Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3
Unilateral Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Intrinsic Metrics and First Fundamental Form . . . . . . . . . . . . . .
4
Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Examples of Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1
Geodesics in a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2
Geodesics on a Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3
Geodesics on Surfaces of Revolution . . . . . . . . . . . . . . . . .
33
33
35
36
37
37
38
39
41
41
41
42
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1c Constrained Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1c Elliptic Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2c Parabolic Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3c Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4c Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2c Constrained Mechanical Systems . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1c Holonomic and Nonholonomic Constraints . . . . . . . . . . . .
2.2c Disk Rolling without Slipping on a Line . . . . . . . . . . . . . .
2.3c Sphere Rolling without Slipping in a Plane . . . . . . . . . . .
2.4c Rigid Rod with Constrained Extremities . . . . . . . . . . . . .
3c Intrinsic Metrics and First Fundamental Form . . . . . . . . . . . . . .
3.1c A Parameterization of the Torus . . . . . . . . . . . . . . . . . . . .
4c Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5c Examples of Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1c The Clairaut Theorem [29] . . . . . . . . . . . . . . . . . . . . . . . . .
43
43
43
44
46
46
46
46
47
48
49
51
51
52
52
52
Contents
3
vii
DYNAMICS OF A POINT MASS . . . . . . . . . . . . . . . . . . . . . . . .
1
Newton’s Laws and Inertial Systems . . . . . . . . . . . . . . . . . . . . . . .
2
Mathematical Formulations of (1.1)–(1.2) . . . . . . . . . . . . . . . . . . .
3
General Theorems of Point-Mass Dynamics . . . . . . . . . . . . . . . . .
3.1
Positional and Central Forces . . . . . . . . . . . . . . . . . . . . . . .
3.2
Conservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
The Two-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1
Gravitational Trajectories . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Newton’s and Kepler’s Laws and Inertial Systems . . . . . . . . . . .
6
Dynamics of a Point Mass Subject to Gravity [133] . . . . . . . . . .
6.1
On the Notion of Weight, Vertical Axis, and Gravity . . .
6.2
Gravitational Motion near the Surface of Earth . . . . . . .
7
Motion of a Constrained Point Mass . . . . . . . . . . . . . . . . . . . . . . .
7.1
Smooth Constraints and Relative Energy . . . . . . . . . . . . .
7.2
Rough Constraints and Relative Energy . . . . . . . . . . . . . .
7.3
Remarks on Fixed Constraints . . . . . . . . . . . . . . . . . . . . . .
7.4
Remarks on the Case of F Conservative . . . . . . . . . . . . . .
8
The Mathematical Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
55
57
57
58
59
60
61
62
64
65
66
67
68
70
71
72
72
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3c General Theorems of Point-Mass Dynamics . . . . . . . . . . . . . . . . .
3.1c Elastic Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2c Point Mass Moving in a Fluid . . . . . . . . . . . . . . . . . . . . . . .
3.3c Elastic Motions in Viscous Media . . . . . . . . . . . . . . . . . . . .
3.4c Forced Oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5c Phase Delay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6c Amplitude of Forced Oscillations . . . . . . . . . . . . . . . . . . . .
4c The Two-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1c Resolving Kepler’s Motions . . . . . . . . . . . . . . . . . . . . . . . . .
4.2c Stable Circular Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3c Radial Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4c Closed Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5c The n-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6c Dynamics of a Point Mass Subject to Gravity [133] . . . . . . . . . .
6.1c Motion of {P ; m} in the Air: Viscous Regime . . . . . . . . .
6.2c Motion of {P ; m} in the Air: Hydraulic Regime . . . . . . .
6.3c Upward Initial Velocity (vo > 0) . . . . . . . . . . . . . . . . . . . . .
6.4c Downward Initial Velocity (vo < 0) . . . . . . . . . . . . . . . . . .
7c The Mathematical Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.1c The String Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2c The Cycloidal Pendulum (Huygens [81]) . . . . . . . . . . . . . .
7.3c The Spherical Pendulum (Lagrange [101]) . . . . . . . . . . . .
7.4c Small Oscillations of a Spherical Pendulum . . . . . . . . . . .
7.5c The Foucault Pendulum [11, 57] . . . . . . . . . . . . . . . . . . . . .
75
75
75
75
76
77
77
78
78
78
79
80
82
83
84
85
85
85
86
86
86
87
88
89
90
viii
Contents
7.6c
7.7c
4
Point Mass Sliding on a Circle . . . . . . . . . . . . . . . . . . . . . . 91
Point Mass Sliding on a Sphere . . . . . . . . . . . . . . . . . . . . . 92
GEOMETRY OF MASSES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
1
Material Systems and Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
1.1
Continuous Distribution of Masses . . . . . . . . . . . . . . . . . . . 93
1.2
Discrete Distribution of Masses . . . . . . . . . . . . . . . . . . . . . 94
2
Center of Mass and First-Order Moments . . . . . . . . . . . . . . . . . . . 94
3
Second-Order Moments and Huygens’s Theorem . . . . . . . . . . . . . 94
4
Ellipsoid of Inertia and Principal Axes . . . . . . . . . . . . . . . . . . . . . 96
5
Miscellaneous Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
5.1
Center of Mass and Principal Axes of Inertia . . . . . . . . . . 99
5.2
Planar Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
6
Computing Some Moments of Inertia . . . . . . . . . . . . . . . . . . . . . . 100
6.1
Homogeneous Material Segment . . . . . . . . . . . . . . . . . . . . . 100
6.2
Homogeneous Material Rectangle . . . . . . . . . . . . . . . . . . . . 101
6.3
Homogeneous Material Parallelepiped . . . . . . . . . . . . . . . . 101
6.4
Homogeneous Material Circle and Disk . . . . . . . . . . . . . . . 102
6.5
Homogeneous Material Right Circular Cylinder . . . . . . . 102
6.6
Homogeneous Material Sphere and Ball . . . . . . . . . . . . . . 103
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
2c Center of Mass and First-Order Moments . . . . . . . . . . . . . . . . . . . 103
3c Second-Order Moments and Huygens’s Theorem . . . . . . . . . . . . . 104
3.1c Inertia Tensor for a Homogeneous Ellipsoid . . . . . . . . . . . 104
5c Miscellaneous Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
5.1c Finding the Principal Axes of a Planar System . . . . . . . . 105
5.2c Planar Systems and Mohr’s Circle . . . . . . . . . . . . . . . . . . . 106
6c Computing Some Moments of Inertia . . . . . . . . . . . . . . . . . . . . . . 107
6.1c Right Circular Cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
6.2c Homogeneous Material Triangle . . . . . . . . . . . . . . . . . . . . . 108
6.3c Homogeneous Material Right Triangle . . . . . . . . . . . . . . . . 110
6.4c Spiral Ramp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
5
SYSTEMS DYNAMICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
1
Flow Map and Derivatives of Integrals . . . . . . . . . . . . . . . . . . . . . 111
2
General Theorems of System Dynamics . . . . . . . . . . . . . . . . . . . . 112
2.1
D’Alembert’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
2.2
Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
2.3
Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2.4
Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
3
Cardinal Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
3.1
Cardinal Equations in Noninertial Systems . . . . . . . . . . . 116
3.2
Motion Relative to the Center of Mass . . . . . . . . . . . . . . . 116
3.3
König’s Theorem [24, 96] . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
Contents
4
5
6
ix
Mechanical Quantities of Rigid Systems . . . . . . . . . . . . . . . . . . . . 117
4.1
Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
Workless Constraints: Discrete Systems . . . . . . . . . . . . . . . . . . . . 120
5.1
Proof of Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 121
The Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
6.1
The Principle of Virtual Work for Rigid Systems . . . . . . 122
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
3c Cardinal Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
3.1c Disk Rolling on a Slanted Guide as in Figure 3.1c . . . . . 123
3.2c Rod Moving in a Plane with a Fixed Extreme . . . . . . . . . 124
3.3c Ring Sliding on a Material Spinning Segment . . . . . . . . . 125
3.4c A Material Segment with Extremities on Rectilinear
Guides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
3.5c Vibrations of a Flywheel . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
4c Mechanical Quantities of Rigid Systems . . . . . . . . . . . . . . . . . . . . 131
4.1c Disk Rolling without Slipping on a Line . . . . . . . . . . . . . . 131
4.2c Disk Rolling in a Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
4.3c Rigid Rod with Constrained Extremities . . . . . . . . . . . . . 132
4.4c Rigid Rod Moving in a Plane . . . . . . . . . . . . . . . . . . . . . . . 132
4.5c The Double Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
4.6c Cone of §10.1c of Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . 134
4.7c Square Plate with a Vertex on a Rectilinear Guide . . . . . 135
4.8c Two Hinged Rods with Constrained Extremities . . . . . . . 137
4.9c Triangle Rotating about an Axis . . . . . . . . . . . . . . . . . . . . 140
6
THE LAGRANGE EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . 141
1
Kinetic Energy in Terms of Lagrangian Coordinates . . . . . . . . . 141
1.1
Kinetic Energy for Rigid-Body Motion . . . . . . . . . . . . . . . 143
2
The Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
3
The Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
3.1
Fixed Constraints and the Energy Integral . . . . . . . . . . . . 146
3.2
Motion along Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
4
Lagrangian Function for Conservative Fields . . . . . . . . . . . . . . . . 147
4.1
Further Integrals of Motion and Kinetic Momenta . . . . . 148
5
Lagrangian and Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
5.1
Canonical Form of the Lagrange Equations . . . . . . . . . . . 149
5.2
Partial and Total Time Derivative of the Hamiltonian . . 150
6
On Motion in Phase Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
6.1
Integrals of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
7
Equilibrium Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
7.1
Equilibrium for Rigid Systems . . . . . . . . . . . . . . . . . . . . . . 152
8
Canonical Form of the n-Body Problem . . . . . . . . . . . . . . . . . . . . 153
9
Lagrangian Coordinates Relative to {Po ; mo } . . . . . . . . . . . . . . . 154
10 Inertial Systems and Further Integrals of Motion . . . . . . . . . . . . 155
x
Contents
11
12
13
The Planar 3-Body Problem (Lagrange [99]) . . . . . . . . . . . . . . . . 156
Configuration of the Three Bodies . . . . . . . . . . . . . . . . . . . . . . . . . 157
12.1 Configuration of an Equilateral Triangle . . . . . . . . . . . . . . 159
12.2 Configuration of a Segment . . . . . . . . . . . . . . . . . . . . . . . . . 159
Collapse of the n Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
13.1 The Lagrange–Jacobi Identity . . . . . . . . . . . . . . . . . . . . . . . 161
13.2 The Sundman Inequality [142] . . . . . . . . . . . . . . . . . . . . . . 161
13.3 A Collapse Would Occur in Only Finite Time . . . . . . . . . 161
13.4 A Total Collapse Would Occur Only If M = 0 . . . . . . . . 162
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
1c Kinetic Energy in Lagrangian Coordinates . . . . . . . . . . . . . . . . . . 163
1.1c Homogeneous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
2c The Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
2.1c An Example of Nonconservative Lagrangian
Components of Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
4c Lagrangian Function for Conservative Fields . . . . . . . . . . . . . . . . 164
4.1c Rigid Rod with Extremities Sliding on a Circle . . . . . . . . 164
4.2c Points Sliding on Rectilinear Guides . . . . . . . . . . . . . . . . . 165
4.3c Pulleys and Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
4.4c Homogeneous Material Frame . . . . . . . . . . . . . . . . . . . . . . . 168
4.5c Disk Rolling on a Slanted Guide . . . . . . . . . . . . . . . . . . . . . 169
5c Lagrangian and Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
5.1c The Legendre Transform of the Lagrangian . . . . . . . . . . . 170
7c Equilibrium Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
7
PRECESSIONS AND GYROSCOPES . . . . . . . . . . . . . . . . . . . . 173
1
The Euler Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
2
Precessions by Inertia or Free Rotators . . . . . . . . . . . . . . . . . . . . . 174
3
Moving and Fixed Polhodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
3.1
Equations of the Moving Polhode . . . . . . . . . . . . . . . . . . . . 176
3.2
Geometry of Moving Polhodes . . . . . . . . . . . . . . . . . . . . . . 177
3.3
Ellipsoids of Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
3.4
Principal Axes of Inertia as Axes
of Permanent Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
4
Integrating Euler’s Equations of Free Rotators (Jacobi [88]) . . . 179
4.1
Integral of (4.3) when I = Ii , i = 1, 2, 3 . . . . . . . . . . . . . . . 180
4.2
Integral of (4.3) when I = Ii for Some i = 1, 2, 3 . . . . . . . 180
5
Rotations about a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
5.1
Rotations about a Principal Axis of Inertia . . . . . . . . . . . 181
6
Gyroscopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
7
Spinning Top Subject to Gravity (Lagrange) . . . . . . . . . . . . . . . . 184
7.1
Choosing the Initial Data: Sleeping Top . . . . . . . . . . . . . . 184
7.2
First Integrals and Constant Solutions . . . . . . . . . . . . . . . 186
8
Precession with Zero Initial Velocity . . . . . . . . . . . . . . . . . . . . . . . 186
Contents
9
10
11
12
xi
8.1
Oscillatory Nutation and Its Period . . . . . . . . . . . . . . . . . . 187
8.2
Precession about the Fixed Vertical Axis e3 . . . . . . . . . . 188
8.3
Visualizing the Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
Precession with Arbitrary Initial Velocity . . . . . . . . . . . . . . . . . . . 190
Precessions of Constant Direction (g(θo ) ≥ 0) . . . . . . . . . . . . . . . 191
Precessions That Invert Their Direction (g(θo ) < 0) . . . . . . . . . . 192
Spinning Top Subject to Friction . . . . . . . . . . . . . . . . . . . . . . . . . . 193
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
1c The Euler Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
1.1c Precession Relative to the Center of Mass . . . . . . . . . . . . 195
1.2c Precession of Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
1.3c Poinsot and Astronomical Precessions of Earth . . . . . . . . 196
1.4c Poinsot Precession of Earth . . . . . . . . . . . . . . . . . . . . . . . . . 196
1.5c Astronomical Precession of Earth . . . . . . . . . . . . . . . . . . . . 197
5c Rotations about a Fixed Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
5.1c The Compound Pendulum (Huygens [81]) . . . . . . . . . . . . 197
6c Gyroscopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
7c Spinning Top Subject to Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . 200
7.1c Intrinsic Equation of a Gyroscope . . . . . . . . . . . . . . . . . . . 200
7.2c The Gyroscopic Compass or Gyrocompass . . . . . . . . . . . . 201
7.3c Integrating the Intrinsic Equations . . . . . . . . . . . . . . . . . . 203
10c Precessions of Constant Direction (g(θo ) ≥ 0) . . . . . . . . . . . . . . . 204
10.1c Asymptotic Amplitude of the Nutation . . . . . . . . . . . . . . . 204
10.2c Asymptotic Period of the Nutation . . . . . . . . . . . . . . . . . . 205
8
STABILITY AND SMALL OSCILLATIONS . . . . . . . . . . . . . . 207
1
Notion of Stability in Phase Space . . . . . . . . . . . . . . . . . . . . . . . . . 207
1.1
Equilibrium Configurations . . . . . . . . . . . . . . . . . . . . . . . . . 208
2
Lyapunov Stability Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
3
Criteria of Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
3.1
Some Generalizations and the Četaev Theorem . . . . . . . . 210
4
Dirichlet Stability Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
5
Stability and Instability of the Poinsot Precessions . . . . . . . . . . . 212
6
Linearized Motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
6.1
Stability and Instability of Linearized Motions . . . . . . . . 215
7
Small Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
8
Vibrations of Masses Subject to Elastic Forces . . . . . . . . . . . . . . 217
9
Normal Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
10 Degenerate Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
11 Fundamental Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
2c Lyapunov Stability Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
2.1c Harmonic Oscillator and Exponential Decay . . . . . . . . . . 222
2.2c Damped Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
xii
Contents
4c
5c
6c
7c
9
Dirichlet Stability Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
4.1c Point Mass Constrained on a Curve . . . . . . . . . . . . . . . . . . 223
4.2c On the Dirichlet Instability Criterion . . . . . . . . . . . . . . . . 224
4.3c Rigid Rod with Extremities on a Parabola . . . . . . . . . . . . 224
4.4c Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
4.5c Material Rod and Point Mass {P ; m} Connected
by a Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
Stability and Instability of Poinsot Precessions . . . . . . . . . . . . . . 226
5.1c Instability of Rotations about the Intermediate Axis . . . 226
5.2c Ellipsoids of Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
Small Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
6.1c Small Oscillations of the System in §4.5c . . . . . . . . . . . . . 227
6.2c Computing the Normal Frequencies . . . . . . . . . . . . . . . . . . 228
6.3c Double Pendulum in §4.5c of the Complements
of Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
Degenerate Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
VARIATIONAL PRINCIPLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
1
Maxima and Minima of Functionals . . . . . . . . . . . . . . . . . . . . . . . . 231
1.1
Stationary Points of Functionals . . . . . . . . . . . . . . . . . . . . . 232
2
The Least Action Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
2.1
Reduced Action and Maupertuis Principle . . . . . . . . . . . . 234
3
The Hamilton–Jacobi Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 236
4
The Functional of Geometrical Optics . . . . . . . . . . . . . . . . . . . . . . 237
5
Huygens Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
6
Canonical Form of Huygens Systems . . . . . . . . . . . . . . . . . . . . . . . 240
6.1
Contact Virtual Differential Forms . . . . . . . . . . . . . . . . . . . 241
7
Wave Fronts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
7.1
First Definition of Wave Front . . . . . . . . . . . . . . . . . . . . . . 242
7.2
Second Definition of Wave Front . . . . . . . . . . . . . . . . . . . . 243
7.3
Equivalence of the Two Definitions . . . . . . . . . . . . . . . . . . 243
7.4
Velocity of the Light Ray and of the Wave Front . . . . . . 244
7.5
Normal Velocity and Normal “Slowness” of the Wave
Front . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
8
The Huygens Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
8.1
Optical Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
1c Maxima and Minima of Functionals . . . . . . . . . . . . . . . . . . . . . . . . 246
1.1c A Functional with Stationary Points and No Extrema . . 246
1.2c A Functional with a Minimum That Is Not Achieved . . 247
1.3c The Brachistochrone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247
1.4c Motion along Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
1.5c An Isoperimetric Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 250
Contents
2c
4c
xiii
The Least Action Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
2.1c On the Two-Point Boundary Value Problem (2.1) . . . . . 252
2.2c Stationary Points of the Action Cannot Be Maxima . . . 252
2.3c Critical Points of the Action Are Local Minima . . . . . . . 253
The Functional of Geometrical Optics . . . . . . . . . . . . . . . . . . . . . . 255
10 CANONICAL TRANSFORMATIONS . . . . . . . . . . . . . . . . . . . . 257
1
Changing the Variables of Motion . . . . . . . . . . . . . . . . . . . . . . . . . 257
1.1
Pointwise Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 258
2
Variational and Canonical Transformations . . . . . . . . . . . . . . . . . 258
2.1
Canonical Transformations . . . . . . . . . . . . . . . . . . . . . . . . . 259
3
Constructing Classes of Canonical Transformations . . . . . . . . . . 260
4
Constructing Canonical Transformations by Other Pairs
of Independent Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
5
Examples of Canonical Transformations . . . . . . . . . . . . . . . . . . . . 262
5.1
The Flow Map Is a Canonical Transformation . . . . . . . . . 262
¯ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
5.2
On the Primitives of dω
5.3
Jacobi Integration of Hamilton Equations [87] . . . . . . . . . 263
6
Symplectic Product in Phase Space and Symplectic Matrices . . 264
6.1
Symplectic Linear Transformations . . . . . . . . . . . . . . . . . . 264
7
Characterizing Canonical Transformations by Symplectic
Jacobians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
8
Poisson Brackets [131] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
8.1
Invariance of the Poisson Brackets by Canonical
Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
9
The Jacobi Identity [89] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
10 Generating First Integrals of Motion by the Poisson Brackets . 268
11 Infinitesimal Canonical Transformations . . . . . . . . . . . . . . . . . . . . 269
11.1 Variations for Infinitesimal Canonical Transformations . 270
12 Variational and Canonical Transformations . . . . . . . . . . . . . . . . . 271
12.1 Proof of Lemma 12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
4c Constructing Canonical Transformations by Other Pairs
of Independent Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
5c Examples of Canonical Transformations . . . . . . . . . . . . . . . . . . . . 274
5.8c Linear Canonical Transformations . . . . . . . . . . . . . . . . . . . 276
6c Symplectic Product in Phase Space and Symplectic Matrices . . 276
7c Characterizing Canonical Transformations by Symplectic
Jacobians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
7.1c Linear Canonical Transformations by Symplectic
Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
7.2c The Poincaré Recurrence Theorem . . . . . . . . . . . . . . . . . . 277
7.3c A Note on Liouville’s Theorem . . . . . . . . . . . . . . . . . . . . . . 278
7.4c Integral Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279
xiv
Contents
8c
Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
8.1c Lagrange Brackets [103] . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
8.2c Invariance of Lagrange Brackets by Canonical
Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
11 INTEGRATING HAMILTON–JACOBI EQUATIONS
AND CANONICAL SYSTEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
1
Complete Integrals of Hamilton–Jacobi Equations . . . . . . . . . . . 283
1.1
Examples of Complete Integrals of (1.1) . . . . . . . . . . . . . . 283
1.2
Hamiltonian Independent of t . . . . . . . . . . . . . . . . . . . . . . . 284
2
Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
2.1
An Example of Complete Separability . . . . . . . . . . . . . . . . 286
2.2
Cyclic Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
3
Reducing the Rank of a Canonical System . . . . . . . . . . . . . . . . . . 286
3.1
Canonical Systems with the Energy Integral . . . . . . . . . . 287
3.2
Cyclic Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
4
The Liouville Theorem of Quadratures . . . . . . . . . . . . . . . . . . . . . 288
4.1
Poisson Brackets in 2(N + 1) Variables . . . . . . . . . . . . . . . 289
4.2
Proof of Liouville’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 289
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291
1c Complete Integrals of Hamilton–Jacobi Equations . . . . . . . . . . . 291
1.1c Envelopes of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291
2c Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292
2.1c Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292
2.2c Parabolic Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
2.3c Elliptic Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295
2.4c Force-Free {P ; m} in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297
2.5c The Cycloidal Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . 297
12 INTRODUCTION TO FLUID DYNAMICS . . . . . . . . . . . . . . . 299
1
Geometry of Deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299
1.1
Incompressible Deformations . . . . . . . . . . . . . . . . . . . . . . . . 300
1.2
The Equation of Continuity . . . . . . . . . . . . . . . . . . . . . . . . . 300
2
Cardinal Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
3
The Stress Tensor and Cauchy’s Theorem . . . . . . . . . . . . . . . . . . 302
3.1
Symmetry of the Stress Tensor . . . . . . . . . . . . . . . . . . . . . . 304
3.2
Miscellaneous Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304
4
Perfect Fluids and Cardinal Equations . . . . . . . . . . . . . . . . . . . . . 305
4.1
Barotropic Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305
4.2
Fluid in Uniform Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . 306
5
Rotations and Deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
6
Vortex Filaments and Vortex Sheets . . . . . . . . . . . . . . . . . . . . . . . 308
6.1
Circulation and Helmholtz Theorem . . . . . . . . . . . . . . . . . 309
7
Equations of Motion of Ideal Fluids . . . . . . . . . . . . . . . . . . . . . . . . 310
Contents
8
9
10
11
12
13
14
xv
7.1
Flow past an Obstacle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
Barotropic Flows with Conservative Forces . . . . . . . . . . . . . . . . . 311
8.1
Incompressible Barotropic Fluids Subject to Gravity . . . 313
Material Lines and Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
Transport of the Vorticity in Barotropic Fluids Subject to
Conservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314
Barotropic Potential Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316
11.1 Barotropic, Potential, Stationary Flows . . . . . . . . . . . . . . 318
11.2 Stationary, Potential, Incompressible Flows . . . . . . . . . . . 318
Stationary, Incompressible Potential Flow past an Obstacle . . . 319
12.1 The Paradox of D’Alembert . . . . . . . . . . . . . . . . . . . . . . . . 319
Friction Tensor for Newtonian Viscous Fluids . . . . . . . . . . . . . . . 320
The Navier–Stokes Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321
14.1 Conservation and Dissipation of Energy . . . . . . . . . . . . . . 322
14.2 Dimensionless Formulation, Reynolds Number,
and Similarities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
Problems and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
7c Equation of Motion of Ideal Fluids . . . . . . . . . . . . . . . . . . . . . . . . 324
7.1c Transmission of Sound Waves . . . . . . . . . . . . . . . . . . . . . . . 324
7.2c Continuity Equation in Cylindrical Coordinates . . . . . . . 325
8c Barotropic Flows with Conservative Forces . . . . . . . . . . . . . . . . . 325
8.1c Torricelli’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
11c Barotropic Potential Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326
11.1c Bidimentional Incompressible Flows . . . . . . . . . . . . . . . . . 326
12c Stationary, Incompressible Potential Flow past an Obstacle . . . 327
12.1c Proof of Proposition 12.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 327
13c Friction Tensor for Newtonian Viscous Fluids . . . . . . . . . . . . . . . 329
13.1c Isotropic Tensors of the Fourth Order . . . . . . . . . . . . . . . . 329
14c The Navier–Stokes Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332
14.1c A Paradox of Ideal Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . 332
14.2c Viscous Fluids: Solving the Paradox . . . . . . . . . . . . . . . . . 333
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
Preface
Millennia of astronomical observations were fully understood only when the
seminal ideas of Galileo, Kepler, and Newton were impacted by mathematics.
The subsequent theoretical elaborations of the laws of motion by d’Alembert,
Lagrange, Hamilton, Liouville, and others are at the basis of all mechanical devices that affect modern life in essentially all its practical aspects. Lagrange, in
the preface to his Traité de Mécanique Analytique [101], sets his vision of mechanics as a branch of mathematics, building on physical principles.1 Classical
mechanics stands as perhaps the most successful example of what contemporary scientists call “interdisciplinary” research. The complexity of astronomy
in Newton’s day is countered today by the complexity of disciplines such as
chemistry and biology. Classical mechanics is a chief example of the scientific
method of organizing a “complex” collection of information into theoretically
rigorous unifying principles. In this sense it represents one of the highest forms
of modeling. The elegance and depth of the theoretical thinking coupled with
its ubiquitous applications make it comparable, in applied sciences, to Euclid’s
geometry. It also has a foundational value comparable to calculus, both as a
fundamental language of applied sciences and as a catalyst of new concepts
and discoveries.
This book collects my lectures in rational mechanics delivered at the School
of Engineering of the University of Rome, Tor Vergata, from 1986 to 1998. The
main vision was mathematical modeling, and the layout is theoretical. The
required background includes a working knowledge of linear algebra (vector
calculus, algebra of matrices), multivariate calculus, basic theory of ordinary
1
From the Preface of [101]: “. . . On ne trouvera point de Figures dans cet Ouvrage.
Les méthodes que j’y expose ne demandent ni constructions, ni raisonnements
géométriques ou mécaniques, mais seulement des opérations algébriques, assujetties á une marche régulière et uniforme. Ceux qui aiment l’Analyse, verront avec
plaisir la Mécanique en devenir une nouvelle branche, . . .
Tel est le plan que j’avais tâché de remplir . . .
On a conservé la notation ordinaire du Calcul différentiel, parce qu’elle répond au
système des infiniment petits . . . ”
xvii
xviii
Preface
differential equations, and elementary physics. While the Lebesgue integral is
used, a working knowledge of it is not required. Its use is mainly intended as
a unifying feature, bridging from discrete systems whose material properties
are described by a series to weighted Dirac masses, to continuum systems. In
practice, in problems and examples only the Riemann integral is used.
The geometry of rigid motions is presented along with some of its implications to mechanical devices and the theory of Poinsot precessions. The latter is
remarkable, since it explains the phenomenon of equinoctial precessions only
by the geometrical rolling of the Poinsot cones. While Lagrange’s equations are
often assumed as a principle, here they are derived and put on a mathematical footing. Conversely, the cardinal equations, which are often passed over in
favor of the Lagrangian and Hamiltonian formalism, are shown to be the basis
of such a formalism. Classical topics, such as gyroscopes, precessions, spinning
tops, effects of rotation of Earth on gravitational motions, variational principles, n-body problem, and celestial mechanics, are revisited to underscore the
role of mathematics, without which they can be only perceived but not fully
explained. Attention is paid to the theory of small oscillations and Lyapunov’s
stability, including stability and instability of Poinsot precessions and celestial
motions. The connection between mechanics and geometrical optics is traced
to their common variational principles. The former leads to a maximum-rank
Hamiltonian system, and the latter yields a degenerate Huygens canonical
system. The degeneracy is overcome by Euler’s theorem of homogeneous functions. The Hamilton variational formalism naturally leads to the symplectic
formalism and canonical transformations. These identify, among the transformations that preserve the variational and canonical nature of the Hamilton
equations, those that preserve the hidden geometry of the motion (Lie), such
as Poisson and Lagrange brackets and volumes (Liouville’s theorem). They
also transform paths in phase space into “immobile points” in the transformed phase space, thereby providing an integration method of Hamiltonian
systems (Jacobi). The basic techniques of integration of the Hamilton–Jacobi
equations (complete integrals, envelopes, separability, etc), are presented in
their natural symplectic formalism. Some classical facts have been given new
proofs based on more modern mathematical language. These include the local
minimality of the stationary points of the action, the notion of envelopes, and
the Hamiltonian form of Huygens systems in geometrical optics. In the last
chapter we give a very brief introduction to continuum and fluid mechanics,
mainly to underscore the mathematical and modeling ideas needed in transitioning from finite degrees of freedom to nonrigid continuum systems.
Classical mechanics has evolved from these seminal principles into
numerous fields, including statistical mechanics, relativity, ergodic theory,
symplectic geometry, and continuum mechanics including elasticity and fluid
dynamics. The methods of investigation are diverse, ranging from probability and measure theory and classical analysis to algebra, topology, and
Riemannian geometry. We have refrained from going into any of them both
because of the specialized and massive nature of each of them, and because
Preface
xix
of our goal of giving essentially a “calculus-type” introduction to the applied
sciences. As such, all topics are mutually interlaced, each providing the background for the indicated several directions that mechanics takes on. However,
some more mathematical parts might be omitted at a first reading. They
include the parameteric equations of fixed and moving centrodes (§11c of
Chapter 1); the theory of radial potential (§4.3c), which, while used for some
atomic potentials, in this context has mostly a theoretical value; finding the
principal axes of inertia of a planar rigid system (§5c of Chapter 4), and some
mathematical remarks on the stationary points of a functional (§§1.1c–1.2c
and §1.5c of Chapter 9).
I learned classical mechanics from Giorgio Sestini (1908–1991) at the University of Florence, Italy, back in the mid 1970s. Sestini’s teachings impressed
upon me the artful balance between physics and mathematics, neither being
permitted to be self-absorbing. These notes reflect those lectures, including
some exercises and problems taken from my old class notes.
I would like to extend special thanks and express my deep appreciation to
my close collaborators Fabrizio Davı́, Alessandro Tiero, Luciano Teresi, Luigi
Chierchia and Vieri Mastropietro. These fine researchers helped me, over the
years and at various stages, in teaching Classical Mechanics at the Univ. of
Rome Tor Vergata. They conducted recitation sessions for the students and
assisted them in practice and problem solving sessions. They suggested a
number of ways of improving my draft notes into a usable tool and suggested
topics and problems aimed at expanding the scope and clarity of the course.
I am very much indebted to them.
I would like to thank the numerous students who, over the years, have
pointed out misprints and imprecise statements, among them Giovanni
Caruso, Roberto Cespa, Ernesto Mottola, Daniele Ludovisi, Giuseppe Vairo,
Michael O’Leary, Luca Andreassi, Alessio Lupoi, Vito Introna, Luciano
Cantone, Francesco Vivio, Sonia Marfia, and Riccardo Baudille. The help
of Daniele Andreucci needs to be singled out. He read a large part of the
manuscript, picking up imprecise statements and misprints and pointing out
unclear passages; I am much indebted to Daniele.
I am also grateful to a number of people for having examined the
manuscript at various stages and for providing suggestions and feedback on the
material and its presentation, among them Bob Glassey, Steve Davis, Franco
Maceri, Giovanni Galdi, Tinsley Oden, Morton Gurtin, David Kinderlehrer,
Ralph Showalter, Avner Friedman, Don Saari, John Ockendon, David Hoff.
I would like to extend many thanks to Robert Sekerka for his critical reading of the chapter on canonical transformation and pointing unclear points.
His remarks led to a rewriting of that chapter.
Special thanks go to James Serrin for having examined in detail the final
version of the manuscript; he suggested changes in some terminology and
presentation, made me aware of further references, pointed out a more general
version of Stokes’s theorem on the friction stress tensor in fluids [139], and
xx
Preface
made precise some thermodynamical facts in the chapter on fluid dynamics.
I am very much grateful to James.
My special thanks go also to Franco Maceri, former dean of the School
of Engineering of the University of Rome Tor Vergata, for his support and
encouragement in writing these notes and for affording me the freedom of
experimenting with less-traditional ways of teaching classical mechanics.
Partial support of the NSF is gratefully acknowledged.
Emmanuele DiBenedetto
1
GEOMETRY OF MOTION
1 Trajectories in R3 and Intrinsic Triads
A triple of vectors {e1 , e2 , e3 } is positively oriented if ei ∧ ej = ek for every
cyclic permutation {i, j, k} of the indices. A positive triad is a Cartesian reference system {O; e1 , e2 , e3 } in R3 with origin in O and positively oriented
unit vectors {e1 , e2 , e3 }. A vector-valued function
(a, b) ∋ t → v(t) = xj (t)ej =
3
xj (t)ej
j=1
from an interval (a, b) ⊂ R into R3 is continuous or differentiable at a point
of (a, b) or in the whole (a, b) if so are the scalar functions xj (·) for j = 1, 2, 3.
The summation notation will be used throughout, that is, a monomial
expression with repeated indices is intended to be summed over those indices.
Calculus operations are effected in terms of components. If v and w are
differentiable vector-valued functions, then
(v ∧ w)′ = v′ ∧ w + v ∧ w′ ,
(v · w)′ = v′ · w + v · w′ .
The latter equality implies that (v2 )′ = 2v · v′ . Therefore if v has constant length, then v and v′ are orthogonal. Let t → P (t) be a continuously
differentiable vector-valued function defined in (a, b) and such that
Ṗ = ẋ21 + ẋ22 + ẋ23 > 0 in (a, b).
The set of points {P (t)}t∈(a,b) is a curve γ in R3 and the vector P ′ is tangent
to γ in P . The intrinsic parameter of γ is its arc length
t
s=
P ′ (τ )dτ,
t ∈ (a, b).
a
With improper but suggestive symbolism, we denote by s → P (s) the function
s → P(s) = P (t(s)), to stress that corresponding values of the parameters t
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 1,
c Springer Science+Business Media, LLC 2011
1
2
1 GEOMETRY OF MOTION
and s identify the same geometric point P on γ. In terms of s the unit tangent
t(s) and unit normal n(s) to γ in P (s) are
t(s) =
d
Ṗ (t)
P (s) =
,
ds
Ṗ (t)
n(s) =
t′ (s)
,
t′ (s)
provided t′ (s) = 0.
If t′ = 0 the normal n is not defined. Since t has constant length, t · t′ = 0.
Therefore n is normal to t and in this sense is the unit normal to γ. Set
t′ = κn,
κ = t′ ,
b = t ∧ n.
The quantity κ(s) is the curvature of γ at P (s) and it measures by how
much γ deviates from t(s) by an infinitesimal variation of s. This formula also
determines the orientation of n and hence the orientation of b. The unit vector
b is the binormal to γ at P . The triple of unit vectors {t, n, b} is positive and
the triad {P ; t, n, b} is called the intrinsic triad to γ at P . Taking derivatives
of b2 ≡ 1 and b · t ≡ 0 with respect to the parameter s gives
b · b′ = 0,
t · b′ + b · t′ = 0.
These imply that db/ds is orthogonal to both b and t. Therefore b′ is parallel
to n. Thus there exists a scalar function s → λ(s) such that
b′ = λn,
|λ| = b′ .
The quantity λ(s) is the torsion of γ at P (s) and it measures by how much
the curve γ deviates from the plane through P (s) and normal t(s) ∧ n(s) by
an infinitesimal variation of the parameter s. These relations between tangent
t, normal n, and binormal b of a curve γ are called Frenet formulas [58]. If
the parameter t is time, the map t → P (t) is interpreted as a point moving
along its trajectory γ. Its velocity v = Ṗ and its acceleration a = P̈ may
be expressed in terms of a Cartesian triad {O; e1 , e2 , e3 } or in terms of its
intrinsic triad as
t
Ṗ = v = (ẋ1 , ẋ2 , ẋ3 ) = tṡ,
Ṗ = v = ẋ21 + ẋ22 + ẋ23 = |ṡ|,
t
P̈ = a = (ẍ1 , ẍ2 , ẍ3 ) = s̈t + κṡ2 n,
a = ẍ21 + ẍ22 + ẍ23 = s̈2 + κ2 ṡ4 .
Note that the acceleration a has zero component along the binormal b. If
s̈ ≡ 0, the motion is uniform. If κ ≡ 0, it is a straight-line motion. The
uniform straight-line motions are those for which κ ≡ 0 and s̈ ≡ 0.
Throughout we will avoid specifying the range of variation of t, and for
all times will mean for all times within the range of variation of t.
2 Areolar Velocity and Central Motions
3
2 Areolar Velocity and Central Motions
A motion is planar if its trajectory lies in a plane π. Select a Cartesian system
{O; e1 , e2 , e3 } with origin O ∈ π and e3 ⊥ π and set
P − O = ρu,
ρ = P − O,
u = cos ϕe1 + sin ϕe2 ,
ϕ = e
1 u,
where ϕ from e1 to P −O is spanned counterclockwise. By time differentiation,
u̇ = ϕ̇u⊥ ,
ü = −ϕ̇2 u + ϕ̈u⊥ .
One verifies that u⊥ = e3 ∧ u and therefore the triple {u, u⊥ , e3 } is positive.
From this one computes the expression of velocity and acceleration in terms
of its radial (i.e., along u) and transversal (i.e., along u⊥ ) components as
Ṗ = ρ̇u + ρϕ̇u⊥ ,
(2.1)
P̈ = ρ̈ − ρϕ̇2 u + (2ρ̇ϕ̇ + ρϕ̈) u⊥ .
For a planar motion t → P (t) and a fixed point O, denote by 2A the moment
of the velocity Ṗ with respect to the pole O, i.e.,
2A = (P − O) ∧ Ṗ = ρ2 ϕ̇u ∧ (e3 ∧ u) = ρ2 ϕ̇e3 .
(2.2)
e2
P (t + dt)
½ +d½
P (t )
' +d'
½
'
e1
O
Fig. 2.1.
The vector A is normal to the plane of the motion, and it is called areolar
velocity. Consider the area swept out by the vector radius P −O in an elemental
time dt during which the angle ϕ undergoes a variation dϕ. Such an area is
given by 21 ρ2 dϕ up to terms of higher order in dt. Formally dividing by dt gives
A = 12 ρ2 |ϕ̇|. This justifies the name areolar velocity given to A. A motion
t → P (t) is central if there exists a point O, called center of motion, such that
(P − O) ∧ P̈ = 0
for all times.
4
1 GEOMETRY OF MOTION
Equivalently, the motion is central if the vector radius P − O and the
acceleration P̈ are parallel at all times. If t → P (t) is central, then
2Ȧ =
d
(P − O) ∧ Ṗ = 0,
dt
which implies
A = const.
Therefore, apart from the case A = 0, the motion takes place in the plane
π through O and normal to A. This implies that central motions are planar.
The vector radius P − O sweeps out equal areas in equal times and it keeps its
rotation orientation at all times. Also, there exists a constant ao > 0, called
the area constant, such that
A = 21 ao ,
ϕ̇ =
ao
.
ρ2
(2.3)
The second equality implies the formal operation
d
d
ao d
= ϕ̇
= 2
.
dt
dϕ
ρ dϕ
(2.4)
In particular, the radial component of the velocity is
ρ̇ =
d 1
ao d
ρ = −ao
.
ρ2 dϕ
dϕ ρ
From the second of (2.3), by taking the time derivatives we obtain ρ (2ρ̇ϕ̇+
ρϕ̈) = 0. Therefore in view of (2.1), the transversal component of the acceleration is zero. To compute the radial component, observe that
ρ̈ = −ao
d d 1
a2 d2 1
= − 2o 2 ,
dt dϕ ρ
ρ dϕ ρ
ρϕ̇2 =
a2o
.
ρ3
Combining these remarks with (2.1) yields the expressions of velocity and
acceleration of a central motion in the form
v = −ao
d 1
ao
u + u⊥ ,
dϕ ρ
ρ
a=−
a2o
ρ2
d2 1
1
+
ρ dϕ2 ρ
u.
(2.5)
These are known as the Binet formulas. Their interest is in that the geometric
trajectory alone determines velocity and acceleration. A motion is circular if
its trajectory is a circle. In such a case the Binet formulas take the form
v=
ao ⊥
u ,
ρ
where ρ is the radius of the trajectory.
a=−
a2o 1
u,
ρ2 ρ
(2.6)
3 Geometry of Rigid Motions
5
3 Geometry of Rigid Motions
A set E ⊂ R3 is in rigid motion if the mutual distance of any two of its points
is constant in time, that is, if
for all P, Q ∈ E
t → P (t) − O(t) = (const)O,P
for all t.
The rigid motion of E is determined by the motion of any triple {P1 , P2 , P3 }
of noncollinear points of E. Any other point Q ∈ E is uniquely determined,
along its motion, from the three independent relations Q − Pi = (const)i ,
i = 1, 2, 3. Having fixed a triple of noncollinear points {P1 , P2 , P3 } in E, the
three equations
Pi − Pj = (const)ij ,
i, j = 1, 2, 3,
i = j
are linearly independent. Therefore of the nine coordinates
x1 (Pi ), x2 (Pi ), x3 (Pi ) , i = 1, 2, 3,
only six are linearly independent. The rigid motion of E is uniquely determined
by any six, linearly independent, of these functions of t, and therefore a rigid
motion has at most six degrees of freedom.
The rigid motion of E is described with respect to a triad Σ =
{Ω; e1 , e2 , e3 } with origin at Ω and positively oriented unit vectors
{e1 , e2 , e2 }. Such a Cartesian reference system will be called fixed. Introduce now a triad S = {O; u1 , u2 , u3 }, clamped to E, with origin at O and
positively oriented unit vectors {u1 , u2 , u3 }. These are uniquely determined
by O and three points Qj , j = 1, 2, 3, by the formula
ui = Qi − O,
(Qi − O) · (Qj − O) = δij ,
i, j = 1, 2, 3,
where δij is the Kronecker delta. The uj are equivalence classes, and this
formula identifies them through representatives. The points O and Qi might
or not belong to E. The triad {O; u1 , u2 , u3 } is clamped to E, in the sense
that every P ∈ E is required to satisfy
P − O = (const)o
P − Qi = (const)i ,
i = 1, 2, 3.
Therefore the rigid motion of the system E coincides with the rigid motion
of the system E ′ = E ∪ {O; Q1 , Q2 , Q3 }. Thus the rigid motion of E may be
regarded as the motion of the triad S with respect to the fixed triad Σ. In this
sense, the notion of rigid motion is independent of the presence of a possible
material rigid body E and is identified with the geometric positions of the
triad S with respect to the fixed triad Σ, following a parameter t. For this
reason S is called the moving triad. For a point P denote by
P = P − O = (x1 , x2 , x3 ) = xi ui ,
P = P − Ω = (y1 , y2 , y3 ) = yi ei ,
its representations with respect to the moving triad S and the fixed triad Σ.
6
1 GEOMETRY OF MOTION
The first three linearly independent functions, determining the rigid
motion of S, will be chosen as the functions t → yo,i (t), i = 1, 2, 3, coordinates of O − Ω with respect to Σ. The remaining three could be, for example,
the Euler angles of S with respect to Σ of the three components of the
instantaneous angular velocity vector ω. In the next sections we introduce
these choices and trace their connection.
4 The Euler Angles
Modulo a translation, assume that Ω = O. If e3 and u3 are not parallel,
the fixed plane {e1 ; e2 } and the moving plane {u1 ; u2 } intersect along a line
called line of the nodes. Denote by n the unit vector along the line of the
nodes oriented so that the triple {n, e3 , u3 } is positively oriented. The Euler
angles are defined as follows [54], [46, pp. 99–125].
Angle ϕ (of Precession): The angle ϕ ∈ [0, 2π) between e1 and n counted
from e1 , counterclockwise with respect to e3 .
Angle ψ (of Proper Rotation): The angle ψ ∈ [0, 2π) between n and u1 ,
counted from n counterclockwise with respect to u3 .
Angle θ (of Nutation): The angle θ ∈ [0, π] between e3 and u3 counted
from e3 counterclockwise with respect to n.
e3
u2
u3
µ
e2
'
e1
Ã
u1
n
Fig. 4.1.
By keeping Ω = O fixed, the triad Σ can be moved to coincide with S by
effecting sequentially the following three rotations:
5 Rotation Matrices and Angular Velocity
7
1. Rotation of Σ about e3 of an angle ϕ to bring e1 to coincide with n.
The new position of Σ is the positive triad Σ1 = {O; n, e′2 , e3 }, and it is
realized by the rotation matrix
⎞
⎛
cos ϕ sin ϕ 0
A1 = ⎝ − sin ϕ cos ϕ 0 ⎠ .
0
0 1
2. Rotation of Σ1 about n of an angle θ to bring e3 to coincide with u3 .
The new position of Σ1 is the positive triad Σ2 = {O; n, e′′2 , u3 }, and it is
realized by the rotation matrix
⎞
⎛
1 0
0
A2 = ⎝ 0 cos θ sin θ ⎠ .
0 − sin θ cos θ
3. Rotation of Σ2 about u3 of an angle ψ to bring n to coincide with u1 . This
last rotation takes Σ2 into S, and it is realized by the rotation matrix
⎞
⎛
cos ψ sin ψ 0
A3 = ⎝ − sin ψ cos ψ 0 ⎠ .
0
0 1
The composition of these matrices, in the indicated order, is A = A3 A2 A1 ,
A=
cos ψ cos ϕ − cos θ sin ϕ sin ψ
cos ψ sin ϕ + cos θ cos ϕ sin ψ sin ψ sin θ
− sin ψ cos ϕ − cos θ sin ϕ cos ψ − sin ψ sin ϕ + cos θ cos ϕ cos ψ cos ψ sin θ
sin θ sin ϕ
− sin θ cos ϕ
cos θ
.
(4.1)
The matrix A carries points y = (y1 , y2 , y3 ) in Σ into points x = (x1 , x2 , x3 )
in S by the formula x = Ay. One verifies that A is unitary and that A−1 = At .
The composition A3 A2 A1 is not commutative, since interchanging the order
of any two of these matrices does not carry Σ into S.
5 Rotation Matrices and Angular Velocity
For a vector w denote by wS and wΣ its representations in S = {O; u1 , u2 , u3 }
and Σ = {O; e1 , e2 , e3 }, i.e.,
wS = x1 u1 + x2 u2 + x3 u3 ,
wΣ = y 1 e 1 + y 2 e 2 + y 3 e 3 ,
where
where
xi = w · ui ,
yj = w · ej .
The transformation from wS to wΣ is realized by the matrix (αij ) = (ei · uj ).
This is unitary and (αij )−1 = (αji ). Therefore
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
y1
y1
x1
x1
⎝ y2 ⎠ = αij ⎝ x2 ⎠ and ⎝ x2 ⎠ = (αji ) ⎝ y2 ⎠ .
(5.1)
y3
x3
x3
y3
8
1 GEOMETRY OF MOTION
If w is fixed in S and moves with it, the components xj of wS are constant in
time, whereas the components yi of wΣ are, in general, nonconstant functions
of the parameter t. The velocity u̇i of the unit vector ui is a free vector in
space, representable in S and Σ. Representing it in S,
u̇i,S = (u̇i · uℓ ) uℓ .
The coefficient (u̇i · uℓ ) is zero for ℓ = i. Let j, k be the values of the index ℓ
for which ℓ = i. If the permutation {i, j, k} is even, then
u̇i,S = (u̇i · uj )uj + (u̇i · uk )uk
= (u̇i · uj )uk ∧ ui − (u̇i · uk )uj ∧ ui
= [−(u̇i · uk )uj + (u̇i · uj )uk ] ∧ ui .
If the permutation {i, j, k} is odd, the same conclusion holds up to a sign
change on the right-hand side. Let ω be the free vector whose components in
S are given by
ωk,S = (u̇i · uj )ǫijk ,
where
ǫijk = (ui ∧ uj ) · uk .
(5.2)
The symbol ǫijk , called Ricci alternator. It is zero if any two of the indices
i, j, k are equal and equals ±1 according to the parity of the permutation
{i, j, k} [134, 135]. With this notation,
u̇i,S = ω ∧ ui
for
i = 1, 2, 3.
If w = xi ui is fixed with S, then
ẇS = xi u̇i,S = ω ∧ xi ui = ω ∧ wS .
Since the exterior and scalar product of two vectors are independent of their
representation,
ẇ = ω ∧ w
for every vector w fixed with S.
(5.3)
Although defined in terms of its components in S, the vector ω is intrinsic and
it can be equivalently represented in the coordinates of Σ. The vector ω is an
intrinsic characteristic of the rigid motion of S with respect to Σ and, as such,
is independent of the representation of S. Indeed, let S ′ = {O′ ; u′1 , u′2 , u′3 } be
a new triad, clamped to S and following the same rigid motion of S. Denoting
by ω ′ the vector defined as in (5.2) with ui replaced with u′i ,
ω′ ∧ u′i = u̇′i = (u′i · uj )u̇j = (u′i · uj )ω ∧ uj = ω ∧ (u′i · uj )uj = ω ∧ u′i .
From this, (ω − ω ′ ) ∧ u′i = 0 for i = 1, 2, 3. Therefore ω − ω ′ is parallel to
three nondegenerate mutually orthonormal vectors, and hence it must be zero.
The vector ω is the angular velocity of the rigid motion of S, or the vector of
instantaneous rotation.
5 Rotation Matrices and Angular Velocity
9
The same vector calculus can be effected starting from the transformation
formulas (5.1). Given a vector w, identify wS and respectively wΣ with the
column vector of its components in S and Σ. Let w be fixed with S. Then by
taking the time derivative of the second equality of (5.1),
(αji )ẇΣ + (α̇ji ) wΣ = 0.
(5.4)
Using the first equality of (5.1),
ẇS = − [(α̇ji )(αij )] (αji )wΣ = − [(α̇ji )(αij )] wS .
The element of position ij of the product matrix (α̇ji )(αij ) is computed from
[(α̇ji )(αij )]ij = α̇ℓi αℓj = u̇i · uj ,
ui = (α1,i , α2,i , α3,i )t .
From this,
⎛
⎞
0
u̇1 · u2 u̇1 · u3
0
u̇2 · u3 ⎠ wS
ẇS = − ⎝ u̇2 · u1
u̇3 · u1 u̇3 · u2
0
⎛
⎞
0
ω3,S −ω2,S
ω1,S ⎠ wS = (ω ∧ w)S .
= − ⎝ −ω3,S 0
ω2,S −ω1,S 0
One arrives at the same formula in terms of the components in Σ. From (5.4),
ẇΣ = −(αij )(α̇ji )wΣ .
The element of position ij in the product matrix is
[(αij )(α̇ji )]ij = αiℓ α̇jℓ = (ei · uℓ )(ej · u̇ℓ )
= (ei · uℓ )[ej · (u̇ℓ · uh )uh ] = (u̇ℓ · uh )(ei · uℓ )(ej · uh )
=
(u̇ℓ · uh ) [(ei · uℓ )(ej · uh ) − (ei · uh )(ej · uℓ )]
ℓ<h
= ωn,S (ei ∧ ej ) · un = αkn ωn,S ǫijk = ωk,Σ ǫijk .
Therefore
⎛
⎞
0
ω3,Σ −ω2,Σ
0
ω1,Σ ⎠ wΣ = (ω ∧ w)Σ .
ẇΣ = − ⎝ −ω3,Σ
ω2,Σ −ω1,Σ 0
While written in terms of the coordinates of Σ or S, these formulas are intrinsic and are equivalent to (5.3).
10
1 GEOMETRY OF MOTION
6 Velocity and Acceleration
For a moving point P denote by v(P ) = Ṗ its velocity and by a(P ) = P̈ its
acceleration. If P − O is fixed to S, its derivatives are computed from (5.3) as
v(P ) = v(O) + ω ∧ (P − O),
(6.1)
a(P ) = a(O) + ω̇ ∧ (P − O) + ω ∧ (P − O)′
(6.2)
= a(O) + ω̇ ∧ (P − O) + ω ∧ [ω ∧ (P − O)].
These are the Poisson formulas of the rigid motion of S [132]. Velocities
and accelerations are those recorded by an observer on the fixed triad Σ.
However, these vectors are independent of reference frames and as such, could
be represented equivalently in S or Σ. All kinematic information on the motion
of any point P fixed with S are included in the two vector-valued functions
of time v(O) and ω, which are called the characteristics of the rigid motion.
Examples of rigid motions are obtained by specifying the form of these two
functions.
6.1 Translations
A rigid motion is a translation if ω ≡ 0. The degrees of freedom reduce to
three. From (6.1)–(6.2), it follows that all points in S have the same velocity and acceleration of O, and the system moves with the axes of the triad
{O; u1 , u2 , u3 } remaining parallel. If v(O) = const, the translation occurs
along a straight line.
6.2 Precessions
A rigid motion is a precession if S has a fixed point O, called a pole of the
precession. The degrees of freedom are three, and (6.1) takes the form
v(P ) = ω ∧ (P − O)
for all P ∈ S.
Therefore at every instant t, all the points of the straight line
ℓ(t) = {P − O = λω(t),
λ real parameter}
have zero velocity. Such a straight line is called the axis of instantaneous
rotation, and it varies with t. If the direction of ω is constant, the line ℓ is
constant in time, and all points of S rotate about ℓ with the same angular
velocity. If the direction of ω is constant, the precession is a rotation. If ω is
constant, the precession is uniform. The name precession given to these rigid
motions will be justified in §1.5c of Chapter 7.
7 The Axis of Motion
11
6.3 Rototranslations
A rigid motion is a rototranslation if the directions of v(O) and ω are constant
in time. The degrees of freedom are two. If v(O)∧ω ≡ 0, the system translates
along the axis through O and parallel to ω with velocity v(O) and rotates
about the same axis with angular velocity ω. If v(O) · ω ≡ 0, the system
translates along the constant direction of v(O) and rotates about the variable
axis through O and directed as ω. Instantaneously, i.e., during an infinitesimal variation dt of the parameter t, any rigid motion can be regarded as an
instantaneous translation followed by an instantaneous rotation.
7 The Axis of Motion
For a point P fixed with S, we wish to compute the component of its velocity
v(P ) along ω. Assuming ω = 0 and using the Poisson formula (6.1),
v(P ) = (v(P ) · ω)
ω
ω def
= (v(O) · ω)
= I(ω),
ω2
ω2
(7.1)
where O is any point fixed in S. Therefore v(P ) = I(ω) for all P ∈ S.
Equivalently, the velocity of every point P ∈ S has the same component along
ω. Since I(ω) is independent of P ∈ S, it is called the vectorial invariant of
the rigid motion. We decompose the velocity of a point P ∈ S as
v(P ) = v(P ) + v(P )⊥ = I(ω) + [v(P ) − I(ω)]
= I(ω) + v(O)⊥ − (P − O) ∧ ω ,
where we have used the Poisson formula (6.1). The first vector is parallel to
ω and is independent of P . The vector in brackets is perpendicular to ω and
depends on P . Since these two vectors are orthogonal,
v(P )2 = I(ω)2 + v(O)⊥ − (P − O) ∧ ω2 .
For a fixed t, we seek the set of all points P ∈ S for which
v(O)⊥ − (P − O) ∧ ω = 0.
(7.2)
Equivalently, we look for the geometric locus ℓ(t) of those points P ∈ S whose
velocity, at time t, is parallel to ω(t) and has least modulus. Indeed, a point
P fixed with S satisfies (7.2) if and only if
v(P ) ∧ ω = 0
and
v(P ) = inf v(Q).
Q∈S
(7.3)
Let P and P ′ be any two distinct points of ℓ(t). Writing (7.2) for each of
them and subtracting gives (P − P ′ ) ∧ ω = 0. Therefore the locus ℓ(t) is a line
parallel to ω(t). Such a line is denoted by μ(ω) and is called instantaneous
12
1 GEOMETRY OF MOTION
axis of motion, or axis of Mozzi [122]. If Po is a point in ℓ(t), the parametric
equation of μ(ω) is P = Po +λω for λ ∈ R. Therefore determining the equation
μ(ω) reduces to finding one of its points Po . For example, Po might be the
projection on μ(ω) of a point O ∈ S. To identify such a projection, write (7.2)
for such a Po and form the exterior product by ω to obtain
(Po − O)ω2 = ω ∧ v(O) =⇒ Po = O +
ω ∧ v(O)
,
ω2
ω = 0.
Therefore the parametric equation of μ(ω) is [22, pages 92–120],
P −O =
ω ∧ v(O)
+ λω,
ω2
ω = 0,
λ ∈ R.
(7.4)
The axis of motion μ(ω) is independent of the point O. That is, if (7.4) were
written with O replaced by any other point Q ∈ S, it would describe the same
set of points ℓ(t). Moreover, (7.4) is written in intrinsic vectorial form and as
such is independent of any reference system. Having identified it, one could
write it alternatively in the coordinates of Σ or S.
The velocity of every point P ∈ μ(ω) is parallel to ω. If P ∈ S has zero
velocity, then P ∈ μ(ω). The vectorial invariant I(ω) vanishes if and only if
all points of the axis of motion μ(ω) have instantaneous zero velocity. In a
precession of pole O, the axis of motion is a line through O and direction ω;
moreover, I(ω) = 0. In a translation, ω ≡ 0 and the axis of motion is not
defined. More generally, μ(ω) is not defined for those values of the parameter
t for which ω(t) = 0.
8 Relative Rigid Motions and Coriolis’s Theorem
Let S be in rigid motion with respect to Σ with characteristics v(O) and ω.
If P is a point moving with respect to S,
P (t) − O = xi (t)ui ,
(8.1)
the velocity and acceleration of P with respect to S are vS (P ) = ẋi ui and
aS (P ) = ẍi ui . Regard now P and O as a pair of points moving with respect
to Σ. Taking the time derivative of (8.1) and using the differentiation formula
(5.3) for vectors fixed with S gives
Ṗ = Ȯ + ẋi ui + xi u̇i = vS (P ) + [v(O) + ω ∧ (P − O)].
(8.2)
By the Poisson formula (6.1), the vector in brackets is the velocity of P as if
it were fixed with S and moving following the same rigid motion of S; it is
called the transport velocity of P and is denoted by
vT (P ) = v(O) + ω ∧ (P − O).
(8.3)
9 Composing Rigid Motions
13
Combining these remarks, one obtains the expression of the velocity vΣ (P )
with respect to Σ in the form
vΣ (P ) = vS (P ) + vT (P ).
(8.4)
From (8.1) by double differentiation,
P̈ − Ö = ẍi ui + 2ẋi u̇i + xi üi .
By the differentiation formula (5.3),
xi üi = xi (ω ∧ ui )′ = xi ω̇ ∧ ui + xi ω ∧ (ω ∧ ui )
= ω̇ ∧ (P − O) + ω ∧ [ω ∧ (P − O)].
Therefore the acceleration aΣ (P ) of P with respect to Σ is given by
aΣ (P ) =a(O) + ω̇ ∧ (P − O) + ω ∧ [ω ∧ (P − O]
+ aS (P ) + 2ω ∧ vS (P ).
By the second Poisson formula (6.2) the sum of the first three terms on the
right-hand side is the acceleration of P as if it were fixed with S and moving
following the same rigid motion of S; it is called the transport acceleration of
P , and it is denoted by
aT (P ) = a(O) + ω̇ ∧ (P − O) + ω ∧ [ω ∧ (P − O)].
(8.5)
The last term is the deflection, or Coriolis acceleration, denoted by
aC (P ) = 2ω ∧ vS (P ).
(8.6)
Theorem 8.1 (Coriolis [34]). The acceleration aΣ (P ) of P with respect to
Σ is the sum of the relative acceleration aS (P ) of P with respect to S, the
transport acceleration aT (P ) of P , regarded as instantaneously fixed with S,
and the deflection acceleration aC (P ), i.e.,
aΣ (P ) = aS (P ) + aT (P ) + aC (P ).
(8.7)
9 Composing Rigid Motions
Let Σ, S, and S ′ be triads in mutual rigid motion, i.e.,
Σ = {Ω; e1 , e2 , e3 },
S = {O; u1 , u2 , u3 },
S ′ = {O′ ; u′1 , u′2 , u′3 },
where S is in rigid motion with respect to Σ with characteristics vΣ (O) and
ω, and S ′ is in rigid motion with respect to S with characteristics vS (O′ )
14
1 GEOMETRY OF MOTION
and ω ′ . By (8.4), the velocity vΣ (P ) with respect to Σ of a point P ∈ S ′ is
the sum of vT (P ) given by (8.3) and vS (P ) given by
vS (P ) = vS (O′ ) + ω′ ∧ (P − O′ ).
Therefore
vΣ (P ) = vS (O′ ) + vΣ (O) + ω ∧ (O′ − O) + (ω + ω′ ) ∧ (P − O′ ).
The first term vS (O′ ) is the velocity of O′ relative to S, whereas the vector
in brackets is the transport velocity of O′ , regarded as instantaneously fixed
with S. By (8.2) the sum of these first two vectors on the right-hand side is
the velocity of O′ with respect to Σ. Combining these remarks yields
vΣ (P ) = vΣ (O′ ) + (ω + ω ′ ) ∧ (P − O′ )
(9.1)
for all pairs of points P, O′ ∈ S ′ . The motion of S ′ with respect to Σ results
from the composition of the motion of S with respect to Σ and the motion of
S ′ with respect to S.
Proposition 9.1 The triad S ′ moves with rigid motion with respect to Σ
with characteristics vΣ (O′ ) and ω + ω ′ .
Proof. Write (9.1) in the form
Ṗ − Ȯ′ = (ω + ω′ ) ∧ (P − O′ ).
Taking the scalar product of this by P − O′ gives
P − O′ = (const)(P,O′ )
for every pair of points P, O′ ∈ S ′ .
Therefore the motion of S ′ with respect to Σ is rigid and has characteristics
vΣ (O′ ) and ω∗ given by (5.2) with S replaced by S ′ and uj replaced by u′j .
Since P − O′ is fixed with S ′ , by the differentiation formula (5.3),
[ω∗ − (ω + ω ′ )] ∧ (P − O′ ) = 0
for all pairs of points P, O′ ∈ S ′ . Thus ω∗ = ω + ω ′ .
Proposition 9.2 Let So = Σ be a fixed triad and let Si , i = 1, . . . , n, be
triads such that Si is in rigid motion with respect to Si−1 with characteristics
vSi−1 (Oi ) and ω i . Then the motion
n of Sn with respect to Σ is rigid with
characteristics vΣ (On ) and ω = i=1 ω i .
Therefore the composition of finitely many rigid motions is a rigid motion.
Conversely, any rigid motion can be decomposed into one or several rigid
motions. The decomposition, however, in general is not unique.
The composition of n translations is a translation. The composition of
n precessions with the same pole is a precession with the same pole. If the
9 Composing Rigid Motions
15
precessions are uniform, the composite precession is uniform. The composition
of two precessions of poles O1 and O2 and angular characteristics ω1 and ω 2
is a rigid motion of characteristics
vΣ (O2 ) = ω 1 ∧ (O2 − O1 ),
ω = ω1 + ω 2 .
This is not a precession unless vΣ (O2 ) ≡ 0.
As an example consider a horizontal platform rotating about a fixed axis
with angular velocity ω = ωu3 for a scalar ω. A cylinder is installed on the
platform with axis directed as u2 , and at a distance d > 0 from the axis of
motion of the platform. The cylinder rotates about its axis with angular velocity ω ′ = ω ′ u2 . Determine the characteristics of the composite rigid motion,
identifying the triads Σ, S, and S ′ . Write down the expression of the velocity
of the generic point P of the cylinder with respect to the fixed triad Σ.
9.1 Connecting the Euler Angles and ω
Let Σ and S be two triads with the same origin Ω = O and assume that S
moves by a precession of pole O with respect to Σ, with angular characteristic ω. Let ϕ, ψ, and θ be the instantaneous Euler angles relative to the two
triads. The nodal line has direction n and is the intersection of the planes
through the pole O and normals e3 and u3 respectively. The precession of S
can be regarded as the composition of three precessions with the same pole
O as follows:
(a) Precession about e3 with angular velocity ω 1 = ϕ̇e3 ;
(b) Precession about n with angular velocity ω 2 = θ̇n;
(c) Precession about u3 with angular velocity ω3 = ψ̇u3 .
The rigid motion of S with respect to Σ is the precession of pole O and
angular characteristic
ω = ϕ̇e3 + θ̇n + ψ̇u3 .
(9.2)
From the definition of the line of the nodes we obtain
n = cos ψu1 − sin ψu2 ,
n = cos ϕe1 + sin ϕe2 .
(9.3)
Moreover, from the transformation matrix from S to Σ, introduced in (4.1),
e3 = sin ψ sin θu1 + cos ψ sin θu2 + cos θu3 .
Putting this in (9.2) gives the expression of ω in the coordinates of the moving
triad S in terms of the Euler angles
ω = (ϕ̇ sin ψ sin θ + θ̇ cos ψ)u1
+ (ϕ̇ cos ψ sin θ − θ̇ sin ψ)u2 + (ϕ̇ cos θ + ψ̇)u3 .
(9.4)
The expression of ω in the coordinates of the fixed triad Σ is
ω = (ψ̇ sin θ sin ϕ + θ̇ cos ϕ)e1
− (ψ̇ sin θ cos ϕ − θ̇ sin ϕ)e2 + (ψ̇ cos θ + ϕ̇)e3 .
(9.5)
16
1 GEOMETRY OF MOTION
10 Fixed and Moving Axodes
Equation (7.4) of the axis of motion is defined for a fixed value of the
parameter t. As t ranges over its domain of definition, (7.4) may be regarded
as a map
ω(t) ∧ v(O(t))
(t, λ) → P (t; λ) = O(t) +
+ λω(t)
(10.1)
ω(t)2
of the pair of parameters (t, λ). With this interpretation, (10.1) defines a
surface, which we denote by G. Since for every t fixed (10.1) represents a line,
such a surface is a ruled surface, called an axode. The equation (10.1) of the
axode is intrinsic, i.e., independent of any reference system. When written in
coordinates of Σ it represents a surface fixed with Σ called a fixed axode and
denoted by GΣ . When it is written in the coordinates of S, it represents a
!
Fig. 10.1.
surface fixed with S and moving with it, called a moving axode and denoted
by GS . The surfaces GΣ and GS are geometric representations of the positions
of the axis of motion μ(ω) along its motion relative to the triads Σ and S.
Therefore μ(ω) = GΣ ∩ GS . A geometric point P ∈ μ(ω) can be regarded as
the instantaneous position of a point moving on GΣ . As such, it has velocity
time derivative of the right-hand side
vΣ (P ) =
of (10.1) written in the coordinates of Σ.
Similarly, the same point P ∈ μ(ω) can also be regarded as the instantaneous
position of a point moving in GS , and as such it has velocity
time derivative of the right-hand side
vS (P ) =
of (10.1) written in the coordinates of S.
Finally, the very same P ∈ μ(ω), regarded as instantaneously fixed with S,
and thus instantaneously following the rigid motion of S, has a velocity given
10 Fixed and Moving Axodes
17
by the vectorial invariant I(ω). Thus the same geometric point P ∈ μ(ω) can
be interpreted in three different ways as the instantaneous position of
t → PΣ (λ; t),
t → PS (λ; t),
t → P (t; λ),
motion of P (λ; ·) on the surface GΣ ;
motion of P (λ; ·) on the surface GS ;
motion of P (λ; ·) transported by S.
Theorem 10.1. The two surfaces GΣ and GS are mutually tangent at points
P ∈ μ(ω) such that vΣ (P ) ∧ ω = 0.
Proof. By definition of the axis of motion, the transport velocity of a point
P ∈ μ(ω) is the vectorial invariant I(ω). From (8.4),
vΣ (P ) − vS (P ) = vT (P ) = I(ω).
Since vΣ (P ) is tangent to GΣ at P and μ(ω) is entirely contained in σΣ , the
tangent plane to GΣ at a point P ∈ μ(ω) has normal vΣ (P ) ∧ ω. Similarly,
vS (P ) is tangent to GS at P , and the tangent plane to GS at such a point
has normal vS (P ) ∧ ω. Taking the exterior product of the previous relation
by ω yields
vΣ (P ) ∧ ω = vS (P ) ∧ ω.
Therefore at every point of the axis of motion for which vS (P ) ∧ ω = 0, the
two surfaces GΣ and GS have the same tangent plane.
The surface GS is attached to S, and as such it shares the same rigid
motion of S. Moreover, for each value of the parameter t it must be tangent
to GΣ along the axis of rotation. Therefore GS rolls over GΣ and slides along
μ(ω). The sliding velocity is I(ω). If the vectorial invariant is zero, then GS
rolls without sliding over GΣ . The relevance of these ruled surfaces is in that a
rigid motion can be realized as the motion of rolling and sliding of a ruled surface GS over another ruled surface GΣ along their common generators μ(ω).
This correspondence between rigid motions and mutual rolling and sliding of
two ruled surfaces with the same generators is at the foundation of the theory
of mechanical gears. If ω, although variable, has constant direction, then the
axodes GS and GΣ are both cylindrical surfaces.
Using (10.1), we write down the equations of these cylindrical surfaces and
examine the case that there is a point O ∈ S such that v(O) ∧ ω ≡ 0.
10.1 Precessions: Fixed and Moving Cones
In a precession of pole O and characteristic ω the axodes are cones with vertex
at O, called Poinsot cones [129]. If ω, although variable, forms a constant angle
θ ∈ (0, π2 ) with e3 , then GΣ is a right circular cone. In general, however, GS ,
while a cone, need not be circular. If ω has constant direction, one might take
e3 = ω/ω. Then the fixed axode GΣ degenerates into an axis through the
18
1 GEOMETRY OF MOTION
e3
e3
u3
u3
O
O
Fig. 10.2.
pole of the precession and directed like ω. In this case the precession is a
rotation. The moving axode GS is a right circular cone, possibly degenerate.
For this it suffices to show that the angle θ between e3 and u3 is constant.
Since u3 is fixed with S, by the differentiation formula (5.3),
d
(e3 · u3 ) = −θ̇ sin θ = e3 · (ω ∧ u3 ) = 0.
dt
Let Σ, S1 , S2 be triads with the same origin O. Assume that S2 precesses with
respect to S1 with characteristic ω2 u3 . Assume, moreover, that S1 precesses
with respect to Σ with characteristic ω1 e3 . The composition is a precession
of pole O and characteristic ω = ω2 u3 + ω1 e3 . If ω1 and ω2 are constant,
then the ruled surfaces GΣ and GS2 relative to the resulting precession are
both right circular cones. Indeed, the parallelogram of the two vectors ω2 u3
and ω1 e3 keeps a constant configuration along the rigid motion. Therefore its
diagonal that is directed as ω forms a constant angle with the axes e3 and u3 .
11 Plane Rigid Motions
Let Σ = {Ω; e1 , e2 , e3 } and S = {O; u1 , u2 , u3 } be two triads and denote by
π the coordinate plane y3 = 0 in Σ and by p the coordinate plane x3 = 0 in S.
If S moves with rigid motion with respect to Σ in such a way that π = p at all
times, the rigid motion of S is said to be planar. In such a case e3 ≡ u3 and the
trajectory of O is in π, i.e., (Ω − O) · u3 ≡ 0. Since u̇3 = ω ∧ u3 = ė3 ≡ 0,
the vector ω is always parallel to u3 and the plane p slides over π in the sense
that the trajectory of every point in p is in π. Let P ∈ S and denote by Pp
its projection on p. By the Poisson formula (6.1),
v(P ) = v(O) + ω ∧ (P − O) = v(O) + ω ∧ (Pp − O) = v(Pp ).
Therefore the velocity of any point P ∈ S is uniquely determined by its
projection on p. In particular, v(P ) is normal to ω. In this sense the plane p
is called the representative plane of the rigid motion.
11 Plane Rigid Motions
19
11.1 Center of Instantaneous Rotation
The axis of motion is parallel to u3 at all times. Its trace C on the
representative plane p is called the center of instantaneous rotation. By
equation (7.4) of the axis of motion such a trace may be realized, with no loss
of generality, for λ = 0. Therefore
C =O+
ω ∧ v(O)
,
ω2
provided ω = 0.
(11.1)
Since the vectorial invariant I(ω) is zero, the velocity v(C) of the center of
instantaneous rotation is identically zero. This permits an entirely geometrical
determination of C.
!
°p
°¼
C
¼=p
Fig. 11.1.
Theorem 11.1 (Chasles [27]). If ω = 0, the center C is on the normal line
to the trajectory of any point P of the representative plane p, drawn through P .
Proof. Write the Poisson formula (6.1) for O = C, and equation (11.1) with
O replaced by the generic point P . This gives
v(P ) = ω ∧ (P − C),
C=P+
ω ∧ v(P )
.
ω2
Chasles’s theorem implies that C is geometrically determined by the trajectories of any two points P1 and P2 of the representative plane p. The normal
lines to these trajectories drawn through P1 (t) and P2 (t) intersect at C(t).
If ω(t) = 0, the center of instantaneous rotation C(t) is not defined. If ω is
identically zero, the motion is a translation and the trajectories of any two
points are parallel lines. In such a case C might be defined as the point at
infinity of the normal bundle of such parallel lines.
20
1 GEOMETRY OF MOTION
P2
v(P 2 )
C
v(P 1 )
P1
Fig. 11.2.
11.2 Centrodes
The traces of GΣ and GS on the fixed plane π and the representative plane p
are two curves Γπ and Γp , called centrodes. Since I(ω) ≡ 0, the moving centrode Γp rolls without slipping over the fixed centrode Γπ , and their tangency
point is the center of instantaneous rotation.
Problems and Complements
2c Areolar Velocity and Central Motions
2.1c Cycloidal Trajectories
A circle of center O and radius R rolls without slipping on a line, as in
Figure 2.1c. A point P of the rolling circle traces a curve called a cycloid.
Assume that P starts from the origin Ω. Denote by C the contact point of
the rolling circle with the line and by ϕ the angle formed by OC and OP ,
⌢
measured counterclockwise from OC. With this notation, P C= ΩC = Rϕ,
and the parametric equations of the cycloid are
x1 = R(ϕ − sin ϕ),
x2 = R(1 − cos ϕ),
and the corresponding Cartesian form is
R − x
2
− 2Rx2 − x22 .
x1 = R arccos
R
Compute t(ϕ), n(ϕ), and the curvature κ(ϕ).
2.2c The Brachistochrone and Tautochrone
Consider an arc of a cycloid, inverted and translated as in Figure 2.2c.
A point “falls” from A along a curve γAB of extremities A and B, subject
2c Areolar Velocity and Central Motions
21
only to gravity. The time it takes the point to reach B depends on the curve
γAB . The problem of the brachistochrone is that of finding the curve γAB
that minimizes this time. The arc of the cycloid through A and B, as in Figure 2.2c, is the curve of least time (§1.3c of the Complements of Chapter 9).
Points starting to “fall” from A, M1 , M2 , . . . = B at the same instant all reach
B simultaneously. In this sense the brachistochrone is also the tautochrone
through A and B.
x2
O
R
Ω
x1
2¼R
C
Fig. 2.1c.
x2
yo
A
M1
B
M2
xo
xo+2¼R
x1
Fig. 2.2c.
2.3c Hypocycloidal and Epicycloidal Trajectories
A circle of center O and radius ρ rolls without slipping in the interior of a fixed
circle of center Ω and radius R > ρ (left of Figure 2.3c). A point P of the
rolling circle traces a curve called a hypocycloid. If P starts from Po = (R, 0)
and ϕ = P
o ΩO, the parametric equations of the hypocycloid are
R−ρ
ϕ,
ρ
R−ρ
ϕ.
x2 = (R − ρ) sin ϕ − ρ sin
ρ
x1 = (R − ρ) cos ϕ + ρ cos
For R = 4ρ, the Cartesian form of the hypocycloid is the astroid
2/3
x1
2/3
+ x2
= R2/3 .
22
1 GEOMETRY OF MOTION
A circle of center O and radius ρ rolls without slipping on the exterior of a
fixed circle of center Ω and radius R (right of Figure 2.3c). A point P of the
rolling circle traces a curve called an epicycloid. If P starts from Po = (R, 0)
and ϕ = P
o ΩO, the parametric equations of the epicycloid are
R+ρ
ϕ,
ρ
R+ρ
ϕ.
x2 = (R + ρ) sin ϕ − ρ sin
ρ
x1 = (R + ρ) cos ϕ − ρ cos
x2
x2
O
P
P
O
x1
'
Ω
R
Po
x1
'
Ω
R
Po
Fig. 2.3c.
2.4c Kepler’s Gravitational Laws
Planets move about the Sun by Kepler’s laws ([92, 93]):
1. The orbits of the planets are ellipses and the Sun occupies one of its foci.
2. The vector radius from the Sun to one of the planets sweeps equal areas
in equal intervals of time (law of areas).
3. The square of the period of revolution of a planet is proportional to the
cube of the semimajor axis of the planet’s orbit (harmonic law).
It follows that the motion of the planets is central. Assume that a planet
revolves along an ellipse of semiaxes a > b. Denote by S the focus on the
semiaxis (0, a) and by ϕ the angle between the generic position P − S of
the planet and the positive direction of the major axis of the ellipse. Using
the second of the Binet formulas (2.5), prove that a = −a a2o /b2 ρ2 u. By the
harmonic law, the proportionality factor a a2o /b2 is the same for all planets
(§4.1 and §5 of Chapter 3).
2.5c Apsidal Points
Given a central motion t → P (t) of center O, decompose Ṗ and P̈ in their
radial and transversal components as in (2.1). A point Po , on the trajectory
of P is an apsis if the radial velocity at Po is zero. If ρ̈(Po ) = 0, an apsis
2c Areolar Velocity and Central Motions
23
is an extremal point for the function t → ρ(t). A maximum is an apocenter,
whereas a minimum is a pericenter. If Po is an apsis, the line trough the center
of motion O and Po is an apsidal axis and ρo = Po − O is an apsidal radius.
In a circular motion all points of the trajectory are apsidal, and there is only
one apsidal radius. Assume now that the acceleration a(ρ) is a known function
Qo
Po
O
Po
Po
Fig. 2.4c.
of ρ, and is explicitly independent of t. An example is the gravitational motion
of a planet about the Sun. In such a case the apocenter and pericenter are
called aphelion and perihelion respectively. Then the motion is governed by
the system of differential equations
ρ̈ − ρϕ̇2 = a(ρ) · u,
2ρ̇ϕ̇ + ρϕ̈ = a(ρ) · u⊥ ,
(2.1c)
complemented by some initial data that we choose at an apsis Po . Denoting
by ao the area constant and using (2.3), such initial data take the form
ρ(0) = ρo = Po − O,
ρ̇(0) = 0,
ϕ̇(0) =
ao
.
ρ2o
(2.2c)
The system (2.1c) does not change by changing t into −t. Therefore the solution of (2.1c)–(2.2c) is symmetric with respect to the apsidal axis through Po .
Proposition 2.1c The trajectory of a central motion whose acceleration depends only on ρ is symmetric with respect to any of its apsidal axes. Moreover,
there exist at most two apsidal radii.
Proof. Let Po be an apsis. If there are no other apsides, the statement is trivial.
This occurs, for example, if the trajectory is a parabola (§5 of Chapter 3).
If Po′ is the next apsis, its symmetric Po′′ with respect the apsidal axis
through Po is also an apsis. By symmetry, Po′ − O = Po′′ − O. The same
argument, starting from Po′ , shows that the apsis next to Po′ must have a
distance from O equal to Po − O. Iteration of the same argument shows
that Po − O and Po′ − O are the only possible apsidal radii.
2.6c Elliptic Trajectories of Some Central Motions
If a point P moves with acceleration −k(P − O), for a given positive constant
k and a fixed point O, then its trajectory is an ellipse, possibly degenerate.
24
1 GEOMETRY OF MOTION
The motion is central and thus planar. In a suitable coordinate system ẍ =
−kx and ÿ = −ky, from which
x = a cos(ωt + ϕ)
y = b sin(ωt + ψ),
where ω 2 = k
for nonnegative constants a, b and real constants ϕ, ψ. Therefore
x2
y2
xy
+ 2 − 2 sin(ψ − ϕ) = cos2 (ψ − ϕ).
2
a
b
ab
8c Relative Rigid Motions and Coriolis’s Theorem
A point P moves with constant velocity v along a straight line ℓ, which in
turn spins about one of its points O, on a horizontal plane, with constant
angular velocity ωe3 . Take a fixed system with origin at O and coordinate
plane y3 = 0 coincident with the horizontal plane where ℓ spins. Choose a
moving triad S = {O; u1 , u2 , u3 } with u3 = e3 and u1 as the unit direction of
ℓ. Then S precesses with respect to Σ, with characteristic ω = ωe3 . Assume
that initially u1 = e1 and set v = v. Then
(P − O)S = vtu1 ,
(P − O)Σ = vt(cos ωte1 + sin ωte2 ).
From these and the formulas (8.5)–(8.7) of relative kinematics one computes
vS (P ) = v(cos ωte1 + sin ωte2 )
vT (P ) = ωvt e3 ∧ (cos ωte1 + sin ωte2 )
= ωvt(− sin ωte1 + cos ωte2 ),
vΣ (P ) = vS (P ) + vT (P ) = v(cos ωte1 + sin ωte2 )
+ ωvt(− sin ωte1 + cos ωte2 ) = (P − O)′Σ ,
aT (P ) = ω̇ ∧ (P − O) + ω ∧ [ω ∧ (P − O)]
= ω 2 vte3 ∧ (e3 ∧ u1 ) = −ω 2 vtu1
= − ω 2 vt(cos ωte1 + sin ωte2 ),
aC (P ) = 2ω ∧ vS (P ) = 2ωve3 ∧ u1 = 2vωu2
= 2ωv(− sin ωte1 + cos ωte2 ),
aΣ (P ) = (−ω 2 vt cos ωt − 2ωv sin ωt)e1
+ (−ω 2 vt sin ωt + 2ωv cos ωt)e2 = (P − O)′′Σ .
Making use of the expression of (P − O)Σ , the parametric equations of the
trajectory of P with respect to Σ are
y1 = vt cos ωt
y12 + y22 = ρ = vt.
=⇒
y2 = vt sin ωt
Setting θ = ωt, the polar form of the trajectory is ρ = (v/ω) θ, which
is the Archimedean spiral.
10c Fixed and Moving Axodes
25
9c Composing Rigid Motions
9.1c Connecting the Euler Angles and ω
A moving point in R3 is represented by its Cartesian coordinates, as functions
of time, and the Cartesian components of its velocity are the derivatives of
such coordinates. The Cartesian representation (9.5) might suggest that ω
might be represented in a similar way. The issue is then whether there exist
scalar functions fi (θ, ϕ, ψ), i = 1, 2, 3, of the Euler angles such that setting
w = f1 (θ, ϕ, ψ)e1 + f2 (θ, ϕ, ψ)e2 + f3 (θ, ϕ, ψ)e3 ,
one might compute ω = ẇ, i.e.,
∂fi
∂fi
∂fi
θ̇ +
ψ̇,
ϕ̇ +
ωi = f˙i (θ, ϕ, ψ) =
∂θ
∂ϕ
∂ψ
i = 1, 2, 3.
From the first component of ω given by (9.5), it follows that
∂f1
= sin θ sin ϕ,
∂ψ
∂f1
= cos ϕ.
∂θ
Take the θ-derivative of the first and the ψ-derivative of the second to get
∂ 2 f1
= cos θ sin ϕ,
∂ψ∂θ
∂ 2 f1
= 0.
∂ψ∂θ
Therefore no such functions fi exist. The vector ω is said to be a nonintegrable
combination of the rotation parameters, and the components of ω in (9.5) are
called pseudocoordinates.
10c Fixed and Moving Axodes
10.1c Cone with Fixed Vertex and Rolling without Slipping
The vertex O of a right circular cone of height
√ h, opening 2α, and radius R is
fixed on a vertical axis at distance 0 ≤ d ≤ R2 + h2 from a horizontal plane
π, whereas the base of the cone rolls without slipping on π as in Figure 10.1c,
with instantaneous contact point C. Let Ω be the projection of O on π, and
denote by ϕ the angle between a fixed direction on π and C − Ω.
Identify the axodes and write down the expression of ω in terms of ϕ̇.
Choose a fixed triad S with origin at the vertex of the cone and u3 directed as Q − O, and denote by θ the angle between (C − O) and −e3 . The
instantaneous velocity of C as transported by the rigid motion of S is zero.
Therefore C is on the axis of instantaneous rotation. Since O is fixed, it also
is on such an axis. Thus the axodes are cones with vertex at O generated by
26
1 GEOMETRY OF MOTION
e3
!
O
d
Ω
e2
'
Q
e1
C
Fig. 10.1c.
the lines through O and C. For every value of d, the moving axode GS is the
cone with vertex at O and aperture 2α. The fixed axode GΣ is the cone with
vertex at O and aperture 2θ. If d √
= 0, the axode GΣ degenerates into the
plane π through Ω = O, and if d = R2 + h2 , it degenerates into the vertical
axis through O.
The vector ω is directed as O − C, and by (6.1), Q̇ = ω ∧ (Q − O).
The trajectory of Q is the circle of radius h sin (α + θ) on the fixed plane
y3 = d − h cos(α + θ), and its angular velocity is ϕ̇e3 . Therefore
Q̇ = ϕ̇h sin (α + θ)(− sin ϕe1 + cos ϕe2 )
= ωsign{ϕ̇}h sin α(− sin ϕe1 + cos ϕe2 ),
ω = |ϕ̇|(cos θ + sin θ cot α).
From these,
ω = −ω sign{ϕ̇} [sin θ(cos ϕe1 + sin ϕe2 ) − cos θe3 ]
= −ϕ̇(sin θ + cos θ cot α) [sin θ(cos ϕe1 + sin ϕe2 ) − cos θe3 ] .
C−Ω
. If d =
If d = 0, then θ = 12 π and ω = −ϕ̇ |C−Ω|
ω = ϕ̇ cot αe3 .
√
R2 + h2 , then θ = 0 and
10.2c Cylinder Rolling without Slipping
Consider the rigid motion of characteristics v(O) = ve1 and ω = (v/ρ)e2 ,
where v and ρ are given positive constants. If initially O = (0, 0, ρ), one has
O(t) = (vt, 0, ρ). From (7.4), or equivalently (10.1), written in Σ, a point P
on the axis of motion has coordinates
y1 (t; λ) = vt,
v
y2 (t; λ) = λ ,
ρ
y3 (t; λ) = 0.
11c Plane Rigid Motions
27
e3
!
e2
O=Ω
'
Q
e1
C
Fig. 10.2c.
Therefore μ(ω) is represented in Σ as the line through (vt, 0, 0), in the plane
y3 = 0, and parallel to the coordinate axis y2 . The fixed axode GΣ is the
plane y3 = 0 regarded as the union of lines parallel to e2 . The moving triad
is taken with origin at O and
u2 = e2 ,
u3 = sin ωte1 + cos ωte3 ,
u1 = u2 ∧ u3 ,
ω=
v
.
ρ
Expressing (7.4), or equivalently (10.1), in S gives
P (t; λ) = ρ(e2 ∧ e1 ) + λωe2 = ρ sin ωtu1 + λωu2 − ρ cos ωtu3 .
The axis μ(ω) is parallel to u2 and goes through ρ(sin ωt, 0, − cos ωt). Therefore the moving axode GS is the cylinder x21 + x23 = ρ2 . For P ∈ μ(ω),
vS (P ) = v(cos ωtu1 + sin ωtu3 ) = ve1 .
Therefore the vectorial invariant I(ω) is zeroand the cylinder x21 + x23 = ρ2
rolls without sliding on the plane y3 = 0.
11c Plane Rigid Motions
11.1c Parametric Equations of Fixed and Moving Centrodes
The triad S = {O; u1 , u2 , u3 } moves with plane rigid motion with respect to
Σ = {Ω; e1 , e2 , e3 } with e3 = u3 . Denote by θ the angle between e1 and u1
and set ω = θ̇e3 . Then
u1 = cos θe1 + sin θe2 ,
u2 = − sin θe1 + cos θe2 ,
e1 = cos θu1 − sin θu2 ,
e2 = sin θu1 + cos θu2 .
28
1 GEOMETRY OF MOTION
y3
x3
x1
y2
Ω
!
O
x2
¹ (!)
y1
Fig. 10.3c.
Denote by
O = yo,1 e1 + yo,2 e2 ,
v(O) = ẏo,1 e1 + ẏo,2 e2 ,
the positions and velocity of O in Σ. Writing (11.1) in Σ yields
1
(11.1c)
C = (yo,1 , yo,2 ) + (−ẏo,2 , ẏo,1 ).
θ̇
These are the time-parametric equations of the
fixed centrode Γπ . If the trajectory of O is known in terms of θ, i.e., θ → ηo,1 (θ), ηo,2 (θ) , then
1 d
d
yo,1 (t), yo,2 (t) =
ηo,1 (θ), ηo,2 (θ) .
dθ
θ̇ dt
Substituting these in (11.1c) gives the equations of Γπ in terms of θ:
′
η1 (θ) = ηo,1 (θ) − ηo,2
(θ),
′
η2 (θ) = ηo,2 (θ) + ηo,1
(θ).
(11.2c)
For the equation of the moving centrode Γp compute
Ȯ = ẏo,1 (cos θu1 − sin θu2 ) + ẏo,2 (sin θu1 + cos θu2 ).
Writing (11.1) in S and taking into account that O ≡ 0 in S, one gets
1
1
(11.3c)
(ẏo,1 sin θ − ẏo,2 cos θ)u1 + (ẏo,1 cos θ + ẏo,2 sin θ)u2 .
θ̇
θ̇
These are the time-parametric equations of the moving centrode. If the trajectory of the origin of S is known as a function of the parameter θ, one obtains
from (11.3c) the parametric equations of Γp in terms of θ only:
C=
′
′
ξ1 (θ) = sin θ ηo,1
(θ),
(θ) − cos θ ηo,2
′
′
ξ2 (θ) = cos θ ηo,1
(θ).
(θ) + sin θ ηo,2
(11.4c)
Thus if the trajectory of O is known in Σ as a function of θ, then fixed and
moving centrodes can be regarded as geometric curves independent of motion.
11c Plane Rigid Motions
29
11.2c Centrodes for Hypocycloidal Motions
A right circular cylinder of center O and radius ρ rolls without slipping in the
cavity of a right circular cylinder of center Ω and radius R > 2ρ. A normal
cross section is as in Figure 11.1c. Denote by P a point fixed on the moving
circle, by Po a point fixed on the fixed circle, and by C the contact point
between the two circles. Set also ϕ = P
o ΩC and θ = COP . Find a relation
between ϕ̇ and θ̇ and write down the parametric equations of fixed and moving
centrodes in terms of t and in terms of θ. Compute the acceleration of P .
For a hypocycloidal motion, compute geometrically and analytically fixed
and moving centrodes.
Ω
'
½
O
µ
C
P
Po
Fig. 11.1c.
11.3c The Cardano Device
A rigid rod of length 2ρ moves with its extremities A and B constrained on
the axes of a Cartesian system with origin in Ω, as in Figure 11.2c. Compute
fixed and moving centrodes, geometrically and analytically.
The trajectories of A and B are the coordinate axes y2 = 0 and y1 = 0.
Therefore by Chasles’s theorem, the center C is at the intersections of the
normals to the coordinate axes through A and B. The angle
ACB is a right
angle and the triangle ABC can be inscribed in a semicircumference of radius
ρ. It follows that the moving centrode is the circle of diameter AB. Moreover,
the distance ΩC equals 2ρ for all positions of C. Thus the fixed centrode is
the circle centered at Ω and radius 2ρ.
Introduce fixed and moving triads Σ and S as in Figure 11.2c. Then
ηo,1 = 2ρ sin θ,
ηo,2 ≡ 0.
Therefore from (11.2c), the parametric equations of Γπ in terms of θ are
η1 = 2ρ sin θ,
η2 = 2ρ cos θ.
30
1 GEOMETRY OF MOTION
From (11.4c) one obtains the parametric equations of the moving centrode Γp
in terms of θ,
ξ1 = 2ρ sin θ cos θ,
ξ2 = 2ρ cos2 θ,
whose Cartesian form is
ξ12 + (ξ2 − ρ)2 = ρ2 .
By construction, the circle of radius ρ, fixed with S, rolls without slipping in
the interior of the fixed circle centered at Ω and of radius 2ρ. Therefore the
Cardano device generates and is generated by a hypocycloidal motion.
y2
x2
x1
C
B
P
µ
Ω
A=O
y1
Fig. 11.2c.
Theorem 11.1c (Cardano [18, 19]). Every point of the moving centrode
traces a diameter of the fixed centrode.
Proof. The velocity of a point P on the moving centrode is by Chasles’s
theorem normal to P − C. Since the triangle ΩP C is inscribed in a semicircle, the direction of the velocity of P goes through Ω at all times. Thus
the trajectory of P is a curve whose tangent at each of its points goes
through Ω.
11.4c More on the Cardano Device
Every hypocycloidal motion can be realized as the motion of a rigid rod with
extremities constrained on two intersecting, not necessarily orthogonal, guides.
As an example consider a rigid rod of length 2ρ moving with its extremities A
and B constrained on two axes forming an angle α ∈ (0, 12 π]. Prove, geometrically and analytically, that (a) the moving centrode is the circle through Ω,
11c Plane Rigid Motions
31
A, and B, and radius ℓ/2 sin α; (b) the fixed centrode is the circle of center Ω
and radius ℓ/ sin α; (c) every point of the moving centrode traces a diameter
of the fixed centrode (see also §3.4.1c of the Complements of Chapter 5).
For the Cardano device (α = 21 π) prove that every point of the rod traces
an ellipse, possibly degenerate. Prove that the same conclusion holds for α ∈
(0, 12 π).
2
CONSTRAINTS AND LAGRANGIAN
COORDINATES
1 Constrained Trajectories
Constraints are limitations imposed on the geometrical or kinematic
configuration of a mechanical system. For example, in a rigid motion any
two points are required to be at constant mutual distance. This is a rigidity
constraint. A system with one of its points constrained on a surface is an
example of a constrained mechanical system. Assume that a point P moves,
being constrained to a surface S ⊂ R3 . Such a surface can be represented,
at least locally, as the level set of a regular function f defined in a domain
G ⊂ R3 , i.e.,
S = {P [f (P ) = 0] and ∇f (P ) > 0}, P ∈ G.
(1.1)
By the implicit function theorem, one of the coordinates, say for example x3 ,
P (t)
[f = 0 ]
Fig. 1.1.
may be represented explicitly in terms of the remaining two. This provides a
local parameterization of S in terms of (x1 , x2 ). Such a parameterization is
not unique. Indeed, choosing any pair of parameters q = (q1 , q2 ), the surface
S can be represented, at least locally, by
⎧
⎨ x1 = x1 (q1 , q2 ),
∂(x1 , x2 )
S = x2 = x2 (q1 , q2 ),
= 0.
provided
det
⎩
∂(q1 , q2 )
x3 = x3 x1 (q), x2 (q) ,
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 2,
c Springer Science+Business Media, LLC 2011
33
34
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
If P moves on S, the Cartesian representation
of its motion is determined by
the function of time t → q(t) = q1 (t), q2 (t) , through the composition
P (q) = x1 (q), x2 (q), x3 (q) .
Conversely, the Cartesian representation t → P (t) of the motion of P permits
one, by inversion, to determine t → q(t). The velocity of P is given by
Ṗ =
dP
q̇h = ∇q P · q̇.
∂qh
The parameters q = (q1 , q2 ) are the Lagrangian coordinates, whereas q̇ =
(q̇1 , q̇2 ) are the Lagrangian velocities of P [101]. The system has two degrees
of freedom. Assume now that P is constrained on a regular curve γ ⊂ R3 .
[f2 = 0 ]
P (t)
[f1 = 0 ]
Fig. 1.2.
Such a curve can be represented, at least locally, as the intersection of the
level sets of two smooth functions f1 and f2 defined in a domain G ⊂ R3 , i.e.,
P [f1 (P ) = 0] ∩ [f2 (P ) = 0]
,
P ∈ G.
(1.2)
γ=
(∇f1 , ∇f2 ) of rank 2
Then, at least locally, two of the coordinate variables, say for example x2
and x3 , can be represented explicitly in terms of x1 . This provides a local
parametric representation of γ in terms of the parameter x1 . At times it might
be more convenient to introduce directly a parameter q and parameterize γ
as q → P (q). Such a parameterization can be recast, at least locally, in terms
of x1 , or any other parameter, provided P ′ (q) > 0, which we assume. The
geometric trajectory of P is γ, and the motion is determined by the function
of time t → q(t). The velocity is Ṗ = (dP/dq)q̇. The parameter q is the
2 Constrained Mechanical Systems
35
Lagrangian coordinate of P , whereas q̇ is its Lagrangian velocity. A moving
point P constrained on γ has one degree of freedom.
The choice of the Lagrangian coordinates is not unique. In the applications it often occurs that one may introduce them directly, as suggested by
the mechanical problem at hand, with no reference to the Cartesian representations of the constraints. The constraints in (1.1)–(1.2) are independent
of time and are called fixed or workless. It is conceivable that P might move
on a surface, or a curve, itself depending on time. As an example consider a
point P moving in a horizontal plane x3 = 0 and constrained by
x1 sin ωt − x2 cos ωt = 0,
ω ∈ R.
This is a time-dependent restriction on the configurations of P . For a fixed t,
the constraint is a straight line. As t changes, the constraint contributes to the
determination of the geometric trajectory and the time-law of motion. Taking
q = P − O as Lagrangian coordinate, we have
P (q; t) = (q cos ωt, q sin ωt).
Therefore the position of P depends on q, and explicitly also on time. By
differentiation,
dP
∂P
Ṗ =
q̇ +
.
dq
∂t
The first of these vectors is the velocity of P on the constraint as if the
constraint were independent of time. The second is the transport velocity due
to the movement of the constraint. Constraints of this kind are moving.
2 Constrained Mechanical Systems
Consider n points Pℓ = (xℓ,1 , xℓ,2 , xℓ,3 ), ℓ = 1, . . . , n, subject to m constraints
fj (P1 , . . . , Pn ; t) = 0,
j = 1, . . . , m,
(2.1)
where fj are m smooth functions of their arguments. They are defined in
G × I, where G is an open subset of R3n and I is an interval of R. We assume
that
m
S(t) =
[fj (·; t) = 0] = ∅ for all t ∈ I
j=1
and that for all P ∈ S(t), the Jacobian matrix
∂fj (P ; t) ∂fj (P ; t) ∂fj (P ; t)
∂xℓ,1
∂xℓ,2
∂xℓ,3
(2.2)
has maximum rank for all t ∈ I. For example, if m < 3n, such a matrix has
rank m and the system has 3n − m degrees of freedom. This defines, at least
36
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
locally, a (3n − m)-dimensional moving manifold S(t). Such a manifold can
be parameterized, for all t ∈ I, in terms of 3n − m Lagrangian coordinates
q ∈ R3n−m , i.e.,
∂Pℓ
= 3n − m ∀t ∈ I.
S(t) ∋ Pℓ = Pℓ (q; t) rank of
∂q
If the points Pℓ move on their constraints, their motion is determined by the
3n − m functions t → q(t). Their velocity is
Ṗℓ =
∂Pℓ
∂Pℓ
.
q̇h +
∂qh
∂t
The first is the instantaneous velocity of Pℓ as moving on S(t), as if this
surface were instantaneously fixed. The second is the velocity of transport of
the constraint S(t). The constraints in (2.1) are in general moving constraints.
If they do not depend on time, they are fixed, or workless. In such a case S
and its parameterization, in terms of Lagrangian coordinates, are independent
of t. The points Pℓ = Pℓ (q) are represented only in terms of q and have no
explicit dependence on time. Therefore
∂Pℓ
=0
∂t
and
Ṗℓ =
∂Pℓ
q̇h
∂qh
(fixed constraints).
2.1 Actual and Virtual Displacements
An elemental displacement of the n points Pℓ ,
P1 , . . . , Pn ; t −→ P1 + dP1 , . . . , Pn + dPn ; t + dt ,
is said to be actual or admissible if it is compatible with the constraints in
(2.1) along their time evolutions, i.e.,
fj (P1 , . . . , Pn ; t) = 0,
fj (P1 + dP1 , . . . , Pn + dPn ; t + dt = 0,
j = 1, . . . , m.
From these we obtain
dfj =
∂fj
∂fj
dPℓ +
dt = 0,
∂Pℓ
∂t
j = 1, . . . , m.
(2.3)
An elemental displacement of the n points Pℓ of the form
(P1 , . . . , Pn ; t) −→ (P1 + δP1 , . . . , Pn + δPn ; t)
is said to be virtual if it is compatible with the constraints (2.1) regarded as
fixed at time t, i.e.,
fj (P1 , . . . , Pn ; t) = 0,
fj (P1 + δP1 , . . . , Pn + δPn ; t) = 0,
j = 1, . . . , m.
2 Constrained Mechanical Systems
37
These imply
∂fj
· δPℓ = ∇Pℓ fj · δPℓ = 0,
∂Pℓ
j = 1, . . . , m,
(2.4)
where the symbol δ denotes an elemental virtual differential. If the constraints
in (2.1) are fixed, then virtual and actual displacements coincide.
2.2 Holonomic Constraints
A constraint, fixed or moving, is holonomic if it imposes restrictions only on
the geometrical configuration of the points Pℓ , and imposes no restriction
on their time variations Ṗℓ , P̈ℓ , etc. The constraints in (2.1) are holonomic.
Consider two configurations E = (P1 , . . . , Pn ; t) and E ′ = (P1′ , . . . , Pn′ ; t′ ) of
the n points Pℓ . These are compatible with the constraints (2.1) if they both
satisfy the equations of the constraints. However, no restriction is placed on
the displacements of the system needed to move E into E ′ . For this reason,
constraints of the type of (2.1), moving or fixed, are called also configurational constraints. A constraint that would impose restrictions on how E has
to move into E ′ is not holonomic. For example, a constraint that would impose restrictions of the curvature of the trajectories of the points Pℓ is not
holonomic.
2.3 Unilateral Constraints
A point P = (x1 , x2 , x3 ) subject to the limitation x3 ≥ x21 + x22 is constrained
to move within a paraboloid, possibly up to its boundary. Similarly, the constraint P ≤ 1 forces P to move within the unit ball about the origin of R3 ,
possibly up to its boundary.
Let f ∈ C 1 (G) be such that ∇f > 0 in G. A point P is said to be
subject to a unilateral constraint if it is required to satisfy f (P ) ≤ 0. If P
is in the open set [f < 0], its elemental displacements δP , starting at P ,
are unrestricted. Suppose now that P ∈ [f = 0] and undergoes an elemental
displacement δP , starting from this configuration. Since ∇f (P ) points outside
the set [f ≤ 0], the displacement δP will be compatible with the constraint
only if the angle between ∇f and δP is right or obtuse. Therefore elemental
displacements δP from boundary configurations P ∈ [f = 0] are admissible
only if ∇f (P ) · δP ≤ 0. It follows that elemental displacements of a point P
subject to a unilateral constraint are, in general, not reversible.
A system of n points Pℓ is subject to a unilateral constraint if
m
{P1 , . . . , Pn } ∈
fj ≤ 0 ,
fj ∈ C 1 (G × I).
j=1
It is assumed that such an intersection is not empty and that (2.2) is in force.
If a point Pℓ is in the interior of its constraint then its elemental displacements
δPℓ are unrestricted. If Pℓ belongs to one of the surfaces [fj = 0], its virtual
displacements δPℓ must satisfy ∇fj · δPℓ ≤ 0.
38
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
3 Intrinsic Metrics and First Fundamental Form
A surface S ⊂ R3 is the image of a smooth vector-valued function
G ∋ (u, v) −→ P (u, v) = x1 (u, v), x2 (u, v), x3 (u, v) ,
defined in a connected open set G ⊂ R2 , such that the matrix
⎞ ⎛
⎞
⎛
∂x1 ∂x2 ∂x3
∂P
⎜ ∂u ⎟ ⎜ ∂u ∂u ∂u ⎟
⎟ ⎜
⎟
⎜
⎟=⎜
⎟
⎜
⎝ ∂P ⎠ ⎝ ∂x ∂x ∂x ⎠
1
2
3
∂v
∂v ∂v ∂v
has maximum rank. Set
∂P
A=
∂u
2
3
∂xi
∂P ∂P
·
=
=
∂u ∂u
∂u
i=1
3 ∂x ∂x
∂P ∂P
i
i
·
=
,
B=
∂u ∂v
i=1 ∂u ∂v
2
3
∂P
∂xi
∂P ∂P
C=
·
=
=
∂v
∂v ∂v
∂v
i=1
2
,
2
,
and consider the quadratic form
AB
ξ
= Aξ 2 + 2Bξη + Cη 2
ξη
η
BC
1
(Aξ + Bη)2 + (AC − B 2 )η 2 ,
=
A
(3.1)
where (ξ, η) ∈ R2 is arbitrary. By the Cauchy inequality, AC − B 2 ≥ 0.
Therefore the quadratic form in (3.1) is positive definite. It is called the first
fundamental form of the surface S. An elemental variation (du, dv) of the
parameters (u, v) induces an infinitesimal variation dP on S, whose modulus is
(ds)2 = dP · dP =
2
∂P
∂P
du +
dv
∂u
∂v
2
(3.2)
2
= A(du) + 2Bdudv + C(dv) .
This is the intrinsic metric on S. It is intrinsic, since it depends only on
the geometry of S, and it is independent of its parameterization (see §3c
of the Complements). To a regular curve γ ⊂ G, parameterized by t, there
corresponds a curve Γγ ⊂ S by the correspondence
γ = t → u(t), v(t)
⇐⇒ Γγ = t → P u(t), v(t) .
(3.3)
4 Geodesics
39
If the elemental variation (du, dv) occurs along γ, (3.2) gives an elemental arc
length on the corresponding Γγ . If γ is parameterized by v or respectively by
u, the elemental arc length of Γγ is computed from (3.2) as
Γ°
P (u; v)
v
x3
°
x2
u
x1
Fig. 3.1.
A(du)2 + 2Bdudv + C(dv)2
2
du
du
+ C dv
= A
+ 2B
dv
dv
2
dv
dv
+C
= A + 2B
du.
du
du
ds =
(3.4)
4 Geodesics
Consider now those curves in G for which one of the two parameters is fixed:
γu = {u → (u, vo ), vo = const}
γv = {v → (uo , v), uo = const}
⇐⇒
⇐⇒
Γu = {u → P (u, vo )} ,
Γv = {v → P (uo , v)} .
The vector ∂P/∂u is tangent to Γu and ∂P/∂v is tangent to Γv . These two
vectors are linearly independent, since the matrix (∂P/∂u, ∂P/∂v) has maximum rank. They permit one to compute the elemental area dσ on S and its
normal unit vector ν by the formulas
∂P ∂P
du dv,
dσ =
∧
∂u
∂v
∂P ∂P
∧
∂u
∂v
ν=
∂P ∂P .
∂u ∧ ∂v
A curve Γγ ⊂ S is a geodesic if its normal n is parallel to ν at any of its
points, i.e., if n ∧ ν = 0. This occurs if
n·
∂P
=0
∂u
and
n·
∂P
= 0.
∂v
40
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
If Γγ is parameterized as in (3.3), set
∆ = Au̇2 + 2B u̇v̇ + C v̇ 2 = ṡ2 .
Then the unit tangent t and the unit normal n to Γγ are
1
t= √
∆
∂P
∂P
u̇ +
v̇ ,
∂u
∂v
κn =
d
∂P
dt
1
∂P
√
u̇ +
v̇
.
dt
∂v
ds
∆ ∂u
Imposing now that n be normal to ∂P/∂u and discarding the factor dt/ds
gives
∂P d
∂P
1
∂P
√
u̇ +
v̇
∂u dt
∂v
∆ ∂u
2
1
1
∂P
d ∂P
∂P ∂P
∂P
∂P
d
√
v̇ − √
u̇ +
v̇
u̇ +
=
dt
∂u
∂u
∂v
∂u
∂v
dt
∂u
∆
∆
d
1
1
∂P ∂ 2 P 2 ∂P ∂ 2 P
√ (Au̇ + B v̇) − √
=
u̇ +
u̇v̇
dt
∂v ∂u2
∆
∆ ∂u ∂u2
∂P ∂ 2 P 2
∂P ∂ 2 P
1
u̇v̇ +
v̇
−√
∂v ∂u∂v
∆ ∂u ∂u∂v
∂C 2
1
∂B
∂A 2
d ∂ √
u̇ + 2
u̇v̇ +
v̇
∆− √
=
dt ∂ u̇
∂u
∂u
∂u
2 ∆
d ∂ √
∂ √
=
∆−
∆.
dt ∂ u̇
∂u
0=
Proposition 4.1 A curve Γγ ⊂ S parameterized as in (3.3) is a geodesic
if and only if the functions t → u(t), v(t) are solutions of the system of secondorder differential equations
d ∂ √
∂ √
∆−
∆ = 0,
dt ∂ u̇
∂u
d ∂ √
∂ √
∆−
∆ = 0.
dt ∂ v̇
∂v
(4.1)
These may be rewritten in terms of u or v taken as local parameters. For
example, taking u as local parameter, one may express locally v = v(u) as a
function of u. Then the second equality of (4.1) takes the form
d ∂ √
∂ √
∆−
∆ = 0.
du ∂v ′
∂v
(4.2)
Remark 4.1 The same equations arise by regarding the geodesics as the
curves of least path on S between any two of its points (§1.4c of Chapter 9).
5 Examples of Geodesics
41
5 Examples of Geodesics
5.1 Geodesics in a Plane
Let S be the plane of equation ai xi = b. Assuming a3 = 0, we may take
u = x1 and v = x2 and compute
A=
a21 + a23
,
a23
B=
a 1 a2
,
a23
C=
a22 + a23
.
a23
In view of (4.2), the problem reduces to solving
d
Cv ′ + B
√
= 0,
du A + 2Bv ′ + Cv ′2
which implies v ′′ = 0. Therefore the geodesics in a plane are line segments.
5.2 Geodesics on a Sphere
From the parameterization of the sphere of radius R in terms of polar coordinates, we have
∆ = R2 sin2 v + v ′2 .
A = R2 sin2 v, B = 0, C = R2 ;
Therefore, by (4.2), the geodesics v = v(u) on a sphere are solutions of
v′
d
sin v cos v
−
= 0.
du sin2 v + v ′2
sin2 v + v ′2
Multiplying by v ′ and performing elementary manipulations yields
!
"
d
v ′2
2
− sin v + v ′2 = 0.
du
sin2 v + v ′2
It follows that for a constant c ∈ (0, 1),
#
#
1
dv
dv
$
=
u(v) = c
2
4
2
2
sin v
1−c
sin v − c2 sin v
− cot2 v
c2
c
= − arcsin √
cot v + c̄.
1 − c2
This finally implies
c
R cos v.
1 − c2
Therefore the geodesics are the intersection of the sphere with the planes
through the origin,
c
(sin c̄)x1 − (cos c̄)x2 = √
x3 .
1 − c2
The meridians on a sphere are geodesics, whereas none of the parallels, except
the equator, is a geodesic.
(sin c̄)R sin v cos u − (cos c̄)R sin v sin u = √
42
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
5.3 Geodesics on Surfaces of Revolution
If S is a surface of revolution, after a possible rotation and relabeling of the
coordinate axes, it can be parameterized as
x1 = u cos v,
x2 = u sin v,
x3 = f (u).
(5.1)
Here f is a smooth function of the variable u, and S is interpreted as a surface
obtained by rotating the graph of x3 = f (x1 ) about the x3 axis. From such a
x3
f(u)
x2
v
u
x1
Fig. 5.1.
parametric representation, we compute
A = 1 + f ′2 (u),
B = 0,
C = u2 ;
∆ = 1 + f ′2 (u) + u2 v ′2 .
If Γγ ⊂ S is a geodesic parameterized by u, then by (4.2),
d ∂
u2 v ′
d
′2 (u) + u2 v ′2 =
= 0.
1
+
f
du ∂v ′
du 1 + f ′2 (u) + u2 v ′2
Therefore, for a constant c ∈ R and u > c,
u v = c 1 + f ′2 (u) + u2 v ′2 =⇒ v ′ = ±c
4 ′2
2
1 + f ′2 (u)
√
.
u u 2 − c2
From these one finds the implicit equation of the geodesics in the form
# u
1 + f ′2 (η)
v − vo = ±c
dη.
uo η η 2 − c2
Recalling the geometric meaning of ∆, we obtain
1
du
=
,
ds
1 + f ′2 (u) + u2 v ′2
1c Constrained Trajectories
43
where ds is the elemental arc length on the geodesic. With this symbolism,
the previous differential equations of a geodesic can be rewritten in the form
1 + f ′2 (u) du
d
dv
2 dv
u
= 0,
= ±c √
.
(5.2)
ds
ds
ds
u u2 − c2 ds
If c = 0, then dv/ds = 0 and therefore the curves v = const are geodesics.
These are the meridians traced on S. The parallels, i.e., the curves u = const
on S, in general are not geodesics. From (5.2) it follows that for a parallel to
be a geodesic, we must have dv/ds = const = 0, and moreover,
√
du
u u2 − c2 dv
c
= ±
,
i.e.,
f ′2 (u) = ∞.
ds
1 + f ′2 (u) ds
Geometrically, a parallel P ⊂ S is a geodesic if S is tangent along P to a right
circular cylinder with vertical axis. Thus for a sphere, only the equatorial
parallel is a geodesic. For a cylinder the geodesics are curves normal to a
generator at each of their points.
Problems and Complements
1c Constrained Trajectories
A class of Lagrangian coordinates arises by a change of variables in R3 as
indicated by the following examples.
1.1c Elliptic Coordinates
Let ℓ be a fixed positive parameter and define
⎧
⎨ x1 = ℓ sinh u cos ϕ sin θ, u ∈ R+ ,
x2 = ℓ sinh u sin ϕ sin θ, ϕ ∈ [0, 2π),
⎩
x3 = ℓ cosh u cos θ,
θ ∈ [0, π].
(1.1c)
From these we obtain
x23
x21 + x22
+
= 1,
2
ℓ2 sinh u ℓ2 cosh2 u
ℓ2
x2 + x2
x23
− 21 2 2 = 1.
2
cos θ ℓ sin θ
Therefore the surfaces u = const > 0 are ellipsoids of revolution about the
x2 -axis. The semiaxes are a1 = a2 = ℓ sinh u and a3 = ℓ cosh u. The surfaces
θ = const are hyperboloids of two sheets. They are of revolution about the
x3 -axis. Determine the “coordinate planes” ϕ = const.
44
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
The velocity of a moving point P expressed in elliptic coordinates is
⎛
⎞
u̇ cosh u cos ϕ sin θ − ϕ̇ sinh u sin ϕ sin θ + θ̇ sinh u cos ϕ cos θ
Ṗ = ℓ ⎝ u̇ cosh u sin ϕ sin θ + ϕ̇ sinh u cos ϕ sin θ + θ̇ sinh u sin ϕ cos θ ⎠ ,
u̇ sinh u cos θ − θ̇ cosh u sin θ
and its modulus squared is
Ṗ 2 = ℓ2 u̇2 + θ̇2 sinh2 u + sin2 θ + 2ℓ2 ϕ̇2 sinh2 u sin2 θ.
If u = const > 0, one obtains from these the expressions of the velocity and
its modulus for a point constrained on an ellipsoid of revolution about the
x3 -axis. Set
A(u, θ) = sinh2 u + sin2 θ
and compute from (1.1c)
u x1 =
cosh u cos ϕ sin θ
,
ℓ A(u, θ)
θx 1 =
sinh u cos ϕ cos θ
,
ℓ A(u, θ)
u x2 =
cosh u sin ϕ sin θ
,
ℓ A(u, θ)
θx 2 =
sinh u sin ϕ cos θ
,
ℓ A(u, θ)
u x3 =
sinh u cos θ
,
ℓ A(u, θ)
θx 3 =
− cosh u sin θ
,
ℓ A(u, θ)
ϕx1 =
− sin ϕ
,
ℓ sinh u sin θ
ϕx2 =
cos ϕ
,
ℓ sinh u sin θ
ϕx3 = 0.
From these, the Jacobian of the transformation from Cartesian coordinates
into elliptic coordinates is
JCart→ell = ℓ3 A(u, θ) sinh u sin θ.
Let x → f (x) be a smooth function in a domain G ⊂ R3 . Set
F (u, ϕ, θ) = f ℓ sinh u cos ϕ sin θ, ℓ sinh u sin ϕ sin θ, ℓ cosh u cos θ
and verify that
∇x f 2 =
Fu2
2
ℓ A(u, θ)
+
Fθ2
2
ℓ A(u, θ)
+
Fϕ2
ℓ2 sinh2 u sin2 θ
.
1.2c Parabolic Coordinates
Set
√
⎧
⎪
⎨ x1 = √uv cos ϕ,
x2 = uv sin ϕ,
⎪
⎩ x3 = u − v .
2
u, v ≥ 0,
ϕ ∈ [0, 2π),
(1.2c)
1c Constrained Trajectories
45
For u > 0 fixed, compute v from the first two equalities and put it in the last
to get
$
u2 − r 2
,
r = x21 + x22 .
x3 =
2u
For u = const > 0 these are paraboloids with vertex at (0, 0, 12 u). Analogously,
keeping v constant, a similar calculation gives the paraboloids
r2 − v 2
x3 =
,
v > 0.
2v
Therefore the generalized coordinate surfaces u = const or v = const are
paraboloids. For this reason the variables (u, v, ϕ) are called parabolic coordinates. Describe the surfaces ϕ = const.
The velocity of a point P in terms of parabolic coordinates is
&
⎛ &
⎞
√
1
v
u
1
⎜ 2 u̇ u cos ϕ + 2 v̇ v cos ϕ − ϕ̇ uv sin ϕ ⎟
⎜
⎟
⎜
⎟
&
&
⎜
⎟
⎜1
⎟
√
1
v
u
Ṗ = ⎜ u̇
,
sin ϕ + v̇
sin ϕ + ϕ̇ uv cos ϕ ⎟
⎜2
⎟
u
2
v
⎜
⎟
⎜
⎟
⎝
⎠
1
(u̇ − v̇)
2
and its modulus squared is
u + v u̇2
v̇ 2
2
Ṗ =
+ ϕ̇2 uv.
+
4
u
v
From the first two equalities of (1.2c) we have uv = r2 . From this and the
third equality of (1.2c),
u xi =
2xi
,
u+v
vxi =
2xi
,
u+v
ux3 =
2u
,
u+v
vx3 =
−2v
,
u+v
ϕx2 =
1
cos2 ϕ.
x1
ϕx1 = −
x2
cos2 ϕ,
x21
i = 1, 2,
From these the Jacobian of the transformation from Cartesian coordinates to
parabolic coordinates is
JCart→parab = 14 (u + v).
Let x → f (x) be a smooth function in a domain G ⊂ R3 . Set
√
√
F (u, v, ϕ) = f uv cos ϕ, uv sin ϕ, 12 uv
and verify that
∇x f 2 =
1 2
4 2
Fu u + Fv2 v +
F .
u+v
uv ϕ
46
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
1.3c Spherical Coordinates
Compute the velocity of a point P ∈ R3 in
⎧
⎨ x1 = ρ cos ϕ sin θ,
x2 = ρ sin ϕ sin θ,
⎩
x3 = ρ cos θ,
terms of spherical coordinates
ρ ≥ 0,
ϕ ∈ [0, 2π),
θ ∈ [0, π].
(1.3c)
Verify that
⎛
⎞
ρ̇ cos ϕ sin θ − ϕ̇ρ sin ϕ sin θ + θ̇ρ cos ϕ cos θ
Ṗ = ⎝ ρ̇ sin ϕ sin θ + ϕ̇ρ cos ϕ sin θ + θ̇ρ sin ϕ cos θ ⎠
ρ̇ cos θ − θ̇ρ sin θ
and
Ṗ 2 = ρ̇2 + ϕ̇2 ρ2 sin2 θ + θ̇2 ρ2 .
Compute the expressions of the velocity of a point constrained to move in the
cavity of a sphere (spherical pendulum).
1.4c Cylindrical Coordinates
Compute the velocity of a moving point
dinates
⎧
⎨ x1 = r cos ϕ,
x2 = r sin ϕ,
⎩
x3 = x3 ,
P ∈ R3 in terms of cylindrical coorr ≥ 0,
ϕ ∈ [0, 2π),
y3 ∈ R.
(1.4c)
Verify that
⎞
ṙ cos ϕ − ϕ̇r sin ϕ
Ṗ = ⎝ ṙ sin ϕ + ϕ̇r cos ϕ ⎠ ,
ẋ3
⎛
Ṗ 2 = ṙ2 + ϕ̇2 r2 + ẋ23 .
Compute the expression of the velocity of a point moving on a right circular
cylinder with vertical axis.
2c Constrained Mechanical Systems
2.1c Holonomic and Nonholonomic Constraints
Let a mechanical system be described by N independent Lagrangian parameters q = (q1 , . . . , qN ). A holonomic constraint on the system is of the form
f (q; t) = const, or by taking derivatives in t,
fqh q̇h + ft = 0
along the motion.
2c Constrained Mechanical Systems
47
On the other hand, a constraint of the type
Ah (q; t)q̇h + Ao (q; t) = 0
is in general not holonomic, since it imposes limitations on the Lagrangian
configurations q and the Lagrangian velocities q̇. However, if there exists a
smooth function f (q; t) such that
Ah (q; t) = fqh (q; t),
i = 1, . . . , N,
Ao (q; t) = ft (q; t),
then such a constraint can be rewritten as f˙(q; t) = 0 or f (q; t) = const and is
therefore holonomic. As an example consider a point P constrained by Ṗ = u,
where u is a fixed vector. Such a constraint is holonomic, since it requires only
that the trajectory be a straight line.
The constraint Ṗ = c, where c is a given positive constant, restricts the
modulus of the velocity, and it cannot be reduced to a holonomic constraint by
integration. Notice that, in contrast to the previous example, such a constraint
does not restrict the configurations of P . In particular, P might go from P1
to P2 along an arbitrary path, provided the motion occurs at constant speed.
e2
O
'
R
Ω
e1
C
y1
Fig. 2.1c.
2.2c Disk Rolling without Slipping on a Line
A disk of center O and radius R is constrained to move on a linear horizontal
guide while remaining in a fixed vertical plane, as in Figure 2.1c. The system
has two degrees of freedom, and we may choose as Lagrangian coordinates the
angle ϕ between C −O and a fixed radius. Requiring that the disk roll without
slipping means to impose on the contact point C, regarded as part of the rigid
motion of the disk, to have zero velocity,
Ċ = Ȯ − ϕ̇e3 ∧ (C − O) = 0.
This can be written in the form ẏ1 − Rϕ̇ = 0, which is equivalent to
y1 − Rϕ = const.
Therefore for a disk on a guide, the constraint of “rolling without slipping” is
holonomic. Assume that the disk moves on a parabola, an ellipse, or a cycloid,
48
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
remaining on a fixed vertical plane. Write down the analytical expression of
the constraint “rolling without slipping” and conclude that in all cases, the
constraint is holonomic.
2.3c Sphere Rolling without Slipping in a Plane
A sphere of center O and radius R is required to roll without slipping in
a horizontal plane, as in Figure 2.2c. As Lagrangian coordinates take the
Cartesian coordinates y1 , y2 of the center O and the Euler angles ϕ, ψ, θ,
formed by a moving triad S with origin O and fixed with the sphere, with a
fixed triad Σ. The constraint of “rolling without slipping” translates into
Ċ = ẏ1 e1 + ẏ2 e2 + ω ∧ (C − O) = 0.
Using the expression of the vector ω in terms of the Euler angles (formula
y3
y2
R
y1
O
C
Fig. 2.2c.
(9.5) of Chapter 1), this can be rewritten as
ẏ1 + R(ψ̇ sin θ cos ϕ − θ̇ sin ϕ) = 0,
ẏ2 + R(ψ̇ sin θ sin ϕ + θ̇ cos ϕ) = 0.
Such a constraint cannot be integrated, that is, cannot be expressed as
f y1 , y2 , θ, ϕ, ψ; t = const
for some smooth function f . If such an f were to exist, it would have to satisfy
fy1 ẏ1 + fy2 ẏ2 + fθ θ̇ + fϕ ϕ̇ + fψ ψ̇ + ft = 0.
Put into this the previous expressions of ẏ1 and ẏ2 to get
−fy1 R(ψ̇ sin θ cos ϕ − θ̇ sin ϕ) − fy2 R(ψ̇ sin θ sin ϕ + θ̇ cos ϕ)
+ fθ θ̇ + fϕ ϕ̇ + fψ ψ̇ + ft = 0.
(∗)
2c Constrained Mechanical Systems
49
Taking the derivative with respect to ϕ̇ gives ∂f /∂ϕ = 0. Therefore f is
independent of ϕ. Taking now the derivative with respect to ϕ, and keeping
in mind that f is independent of ϕ, yields
sin θ(fy1 sin ϕ − fy2 cos ϕ)ψ̇ + (fy1 cos ϕ + fy2 sin ϕ)θ̇ = 0.
Since the displacements dθ and dϕ are arbitrary, this generates the algebraic
homogeneous linear system
fy1 sin ϕ − fy2 cos ϕ = 0,
fy1 cos ϕ + fy2 sin ϕ = 0,
in the unknowns fyi . The system admits only the trivial solution fyi = 0,
i = 1, 2. Therefore f is independent of y1 and y2 . The independence of ϕ, y1 , y2
permits one to rewrite (∗) as
fθ θ̇ + fψ ψ̇ + ft = 0.
Taking now the derivative with respect to θ̇ gives fθ = 0, and analogously
we also have fψ = 0. Therefore f is independent of θ and ψ. Finally, it
is also independent of t. The contradiction implies that no such f exists.
Therefore for a sphere moving in a plane, the constraint of “rolling without
slipping” is not holonomic. The nonexistence of f means that the Lagrangian
parameters (y1 , y2 , θ, ϕ, ψ) are not restricted, i.e., the sphere might take any
configuration in the plane. Thus the constraints must act by limiting the
Lagrangian velocities.
2.4c Rigid Rod with Constrained Extremities
The extremities A and B of a rigid rod of length h are constrained to move in
two orthogonal planes π1 and π2 as in Figure 2.3c. One of the extremities,
say B, is connected to a point C ∈ π2 through a rigid rod BC of length ℓ. The
other extremity, A, is connected to a point O ∈ π1 ∩ π2 by a rigid rod OA, of
length ℓ. The point C is at distance ℓ from π1 ∩ π2 . Take a Cartesian system
with origin in O, and x-axis as π1 ∩ π2 , oriented so that C = (−ℓ, 0, ℓ).
(a). Determine the number of degrees of freedom of the system. Write down
the equations of the constraints and form their Jacobian matrix.
(b). Compute the determinant of all minors of maximum rank and find conditions on h and ℓ for the Jacobian matrix to have maximum rank.
The system has one degree of freedom and the constraints are
zA = 0,
yB = 0,
2
x2A + yA
= ℓ2 ,
(xB + ℓ)2 + (zB − ℓ)2 = ℓ2 ,
50
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
¼2
z
C
B
A
y
O
Fig. 2.3c.
and in addition, |B − A| = h. Using the third and fourth equations of the
constraints, this can be rewritten as
2xA xB + 2ℓ (xB − zB ) + h2 = 0.
From these one computes the Jacobian matrix
⎛
⎞
0 0 1/2 0
0
0
⎜ 0 0 0
0 1/2 0 ⎟
⎜
⎟
x
y
0
0
0
0 ⎟
J =⎜
⎜ A A
⎟.
⎝ 0 0 0 xB + ℓ 0 zB − ℓ ⎠
xB 0 0 xA + ℓ 0
−ℓ
The minors of order 5 with nonzero determinant must contain the third and
fifth columns. Therefore the problem reduces to extracting the nontrivial non
minors of order three out of the last three rows. These are
⎛
⎛
⎞
⎞
xA yA 0
xA yA
0
D1 = ⎝ 0 0 xA + ℓ ⎠ ,
D2 = ⎝ 0 0 z B − ℓ ⎠ ,
xB 0 −ℓ
xB 0 xA + ℓ
⎛
⎛
⎞
⎞
yA
xA
0
0
0
0
D4 = ⎝ 0 xB + ℓ zB − ℓ ⎠ .
D3 = ⎝ 0 xB + ℓ zB − ℓ ⎠ ,
0 xA + ℓ −ℓ
xB xA + ℓ −ℓ
By direct calculation,
= yA xB (xB + ℓ),
= yA xB (zB − ℓ),
= −xA [(xB + ℓ)ℓ + (zB − ℓ)(xA + ℓ)] ,
= −yA [(xB + ℓ)ℓ + (zB − ℓ)(xA + ℓ)] .
For J to have maximum rank we must have 4i=1 det2 Di > 0. Using the last
two equations of the constraints, we obtain
det D1
det D2
det D3
det D4
4
i=1
2
2 2 2
det2 Di = yA
xB ℓ + [(xB + ℓ)ℓ + (zB − ℓ)(xA + ℓ)] ℓ2 .
3c Intrinsic Metrics and First Fundamental Form
51
Therefore J is not of maximum rank if
yA xB = 0
and
(xB + ℓ)ℓ = −(zB − ℓ)(xA + ℓ).
(∗)
If xB = 0, then zB = ℓ = 0 and h2 = 2ℓ2 . If xB = 0 and yA = 0, then xA = ±ℓ.
If xA = −ℓ, then also xB = −ℓ and h2 = 2ℓ2 . Examine the remaining cases.
3c Intrinsic Metrics and First Fundamental Form
A new parameterization of S is a smooth invertible transformation
G ∋ (u, v)
u = u(u′ , v ′ )
u′ = u′ (u, v)
v = v(u′ , v ′ )
v ′ = v ′ (u, v)
,
(u′ , v ′ ) ∈ G′ ,
from G into a domain G′ ⊂ R2 . The matrix is invertible if the Jacobian
determinant is nonzero, i.e., if
⎛
⎞
∂u ∂u
⎜ ∂u′ ∂v ′ ⎟
⎜
⎟
det J = 0.
J =⎜
⎟,
⎝ ∂v ∂v ⎠
∂u′ ∂v ′
The surface may be then parameterized by
G′ ∋ (u′ , v ′ ) −→ Q(u′ , v ′ ) = P u(u′ , v ′ ), v(u′ , v ′ ) ,
and one computes (du, dv) = (du′ , dv ′ )J t . From these,
du
AB
AB
= (du′ , dv ′ )J t
ds2 = (du, dv)
dv
BC
BC
′ ′ ′
du
A B
= ds′2 ,
= (du′ , dv ′ )
B′ C ′
dv ′
J
du′
dv ′
where A′ , B ′ , C ′ are the coefficients of the first fundamental form, relative to
the new parameterization of S.
3.1c A Parameterization of the Torus
A torus is the surface obtained by a rigid revolution about the x3 -axis of a
circumference of center (xo , 0, 0) and radius R ∈ (0, xo ). A parameterization
of the torus is
x1 (u, v) = (xo + R cos u) cos v, u ∈ [0, 2π],
x2 (u, v) = (xo + R cos u) sin v, v ∈ [0, 2π],
x3 (u, v) = R sin u.
Prove that the surface is nondegenerate, i.e., its first fundamental form is
positive definite at each of its points.
52
2 CONSTRAINTS AND LAGRANGIAN COORDINATES
4c Geodesics
Let Γγ ⊂ S be as in (3.3). The condition for Γγ to be a geodesic may be
expressed using the intrinsic parameterization in terms of the arc length.
Denoting by κ the curvature of Γγ , we have
∂P du ∂P dv
dP
=
+
,
ds
∂u ds
∂v ds
2
d2 P
∂ 2 P dv
∂ 2 P du
∂ 2 P du dv
κn =
+
=
+2
ds2
∂u2 ds
∂u∂v ds ds
∂v 2 ds
2
2
∂P d u ∂P d v
+
+
.
∂u ds2
∂v ds2
t=
2
Imposing the condition that Γγ be a geodesic yields the differential system
2
2
1 ∂A du
d2 v
∂A du dv ∂P ∂ 2 P dv
d2 u
+
+
B
= 0,
+
+
A
2 ∂u ds
∂v ds ds
∂u ∂v 2 ds
ds2
ds2
∂P ∂ 2 P
∂v ∂u2
du
ds
2
+
1 ∂C
∂A du dv
+
∂u ds ds 2 ∂v
dv
ds
2
+B
d2 v
d2 u
+
C
= 0.
ds2
ds2
Observing that
∂P ∂ 2 P
1 ∂C
∂B
−
,
=
2
∂u ∂v
∂v
2 ∂u
∂P ∂ 2 P
1 ∂A
∂B
−
,
=
2
∂v ∂u
∂u
2 ∂v
this system can be rewritten as
Au′′ + Bv ′′ = − 21 [Au u′2 + 2Av u′ v ′ + (2Bv − Cu )v ′2 ],
Bu′′ + Cv ′′ = − 21 [(2Bu − Av )u′2 + 2Au u′ v ′ + Cv v ′2 ].
Solving it, we arrive at the system in normal form:
u′′ = − c111 u′2 + 2c112 u′ v ′ + c122 v ′2 ,
v ′′ = − c211 u′2 + 2c212 u′ v ′ + c222 v ′2 .
(4.1c)
(4.2c)
The coefficients ckij , i, j, k = 1, 2, are called the Christoffel symbols, and can
be computed explicitly from (4.1c).
Prove that (4.1c) is equivalent to (4.1). Compute the Christoffel symbols
in the case that S is a plane, a sphere, or a surface of revolution.
5c Examples of Geodesics
5.1c The Clairaut Theorem [29]
Assume that the constant c in (5.2) is not zero, thereby excluding that the
geodesic is a meridian. From the parametric representation (5.1) it follows
5c Examples of Geodesics
53
that the unit vector tangent to a parallel (u = const) is u = (− sin v, cos v, 0).
Given now a geodesic that is not a meridian, compute its unit tangent at the
generic point of curvilinear coordinate s. From (5.1), written in terms of the
parameter s,
circular section
of radius co
meridian
geodesic
µ
x2
x1
Fig. 5.1c.
t=
cos v
dv
du
dv
du
du
− u sin v , sin v
+ u cos v , f ′ (u)
.
ds
ds
ds
ds
ds
Let θ(s) be the angle formed by the geodesic at s, with the meridian passing
through the same point. From the expression of u and t,
u · t = sin θ = u
dv
.
ds
Combining this with the first equality of (5.2) gives the Clairaut theorem [29]
u sin θ = const,
along geodesics that are not meridians. Give a geometric interpretation of this
fact in the particular case when f (u) ր ∞ as u ց 0.
3
DYNAMICS OF A POINT MASS
1 Newton’s Laws and Inertial Systems
A point mass {P ; m} is in a uniform mechanical state if its velocity is constant.
Departures from a uniform state occur only by variations of velocity caused by
solicitations external to {P ; m} and acting on it. Such external solicitations
are called forces. The vector equation
F = ma,
m ∈ R+ ,
(1.1)
encompasses the first and second Newton’s laws and it describes how an external force causes its variation from a uniform mechanical state [123]. The
first law asserts that in absence of external solicitations (F = 0), a uniform
mechanical state remains uniform (a = 0).1 The second law asserts that variations from a uniform mechanical state (a ≡ 0) are proportional to the acting
solicitation (F = 0).2 The proportionality factor m in (1.1) is the inertial
1
Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter
in directum nisi quatenus illud a viribus impressis cogitur statum suum mutare, [123,
§13, page 54]. The first law was perceived by Leonardo da Vinci although in a nonmathematical formalism: “...ogni moto attende al suo mantenimento, ovvero ogni
corpo mosso sempre si muove, in mentre che la potenzia del motore in lui si rinserra, . . . ogni corpo seguirà tanto la via del suo corso per linea retta quanto durerà
in esso la natura della violenza fatta. . . ” Codex Atlanticus (1478–1518). A physical
notion of the first law appears in G. Galilei [61] ... Una nave che vadı̀a movendosi
per la bonaccia del mare . . . è disposta, quando le fusser rimossi tutti gli ostacoli accidentarii ed esterni, a muoversi, con l’inpulso concepito una volta, incessabilmente
e uniformemente... .
2
Lex II: Mutationem motus proportionalem esse vi motrici impressae et fieri secundum lineam rectam qua vis illa imprimitur..., [123, §13, page 54]. ... Vis impressa
est actio in corpus exercita, ad mutandum ejus statum vel quiescendi vel movendi
uniformiter in directum..., [123, §2, page 40]; this is one of the Definitiones preceding the Axiomata, (Def. III). The vis impress is also called by Newton vis motrix in
Def. VIII, book I of [123], page 44.
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 3,
c Springer Science+Business Media, LLC 2011
55
56
3 DYNAMICS OF A POINT MASS
mass of P , and it is positive, since it expresses that accelerations resulting
from external forces are directed as such acting forces.3 These laws and their
mathematical formulation in (1.1) are formulated with respect to a reference
system, termed inertial, whose existence is postulated.4 We define as inertial
any system Σ within which (1.1) holds.
Let S be a triad in rigid motion with respect to Σ, with characteristics
vΣ (O) and ω. If S translates with respect to Σ with vΣ (O) = const (e.g.,
ω = 0 and v̇Σ (O) = 0), by the Coriolis theorem, an observer in S detects
the same acceleration as an observer in Σ, and (1.1) continues to hold in
S. Therefore, if Σ is inertial, along with Σ are inertial those and only those
systems in uniform, straight-line translation with respect to Σ. More generally,
multiplying (8.7) of Chapter 1 by m and using (1.1) gives
maS (P ) = F + FT + FC ,
(1.2)
where
FT = −maT (P )
and
FC = −maC (P )
are the forces due to transport and Coriolis acceleration respectively. Thus
the inertial law (1.1) continues to hold in S, provided that in the account of
the external forces one includes the forces due to transport of S and the ones
due to the Coriolis acceleration.
The third law asserts that if a point mass {P1 ; m1 } exerts a force F on
another point mass {P2 ; m2 }, then the latter exerts a force −F on the former,
so that the applied system of vectors {(F; P1 ), (−F; P2 )} forms a couple of
zero moment.5
For the third law to hold it is not required that the two point mass be in
contact. Actions and reactions are postulated to occur simultaneously even
at distance. This is equivalent to postulating that mechanical effects between
material points propagate at infinite speed.
A further assumption in these laws is the existence of an absolute time,
independent of any reference system.6 The infinite speed of propagation of
mechanical effects is a consequence of the postulate of an absolute time.
3
Materiae vis insita et potentia resistendi qua corporis unumquodque, quantum
in se est, perserverat in statu suo vel quiescendi vel movendi uniformiter in directum.
Hanc autem quantitatem (materiae) sub nomine corporis vel massae in sequentibus
passim intelligo... [123, §1, page 39].
4
Newton set such an inertial system in the fixed stars, e.g., those stars whose
relative position and configuration had not significantly changed up to the eighteenth
century since the astronomical observations of Ptolemy, about 130 CE.
5
Lex III: Actioni contrariam semper et aequalem esse reactionem: sive corporum
duorum actiones in se mutuo semper esse aequales et in partes contrarias dirigi,
[123, §14, page 55].
6
Tempus absolutum, verum, et mathematicum, in se et natura sua, sine relatione
ad esternum quodvis aequabiliter fluit, alioque nomine dicitur duratio. . . , [123, §11,
page 52].
3 General Theorems of Point-Mass Dynamics
57
2 Mathematical Formulations of (1.1)–(1.2)
A force F acting on {P ; m} is given through a smooth vector-valued function
(P, Ṗ ; t) −→ F(P, Ṗ ; t) = (F1 (x, ẋ; t), F2 (x, ẋ; t), F3 (x, ẋ; t))
defined in a region of R7 with values in R3 . With this symbolism, (1.1) is a
vector differential equation of the second order, or equivalently a system of
three scalar differential equations, e.g.,
m P̈ = F(P, Ṗ ; t)
or
m ẍj = Fj (x, ẋ; t),
j = 1, 2, 3.
(2.1)
Either one of these describes the evolution of t → P (t) starting from some
position Po and velocity Ṗo at some prescribed time to . The typical problem
of the dynamics of a point mass {P ; m} consists in integrating the system
(2.1) starting from such “initial data.” The system (2.1) may be rewritten as
a system of six differential equations of the first order:
mQ̇ = F (P, Q; t),
Ṗ = Q,
or
mẏj = Fj (x, y; t),
ẋj = yj , j = 1, 2, 3.
If P moves on a constraint one has to add the equation of the constraint,
yielding a problem in the dynamics of a constrained point mass. If the trajectory of P is known, one might write (2.1) in terms of its intrinsic triad, e.g.,
Ft (s, ṡ; t) = ms̈,
Fn (s, ṡ; t) = mκṡ2 ,
Fb (s, ṡ; t) = 0,
(2.2)
where t(s), n(s), and b(s) are respectively the tangent, normal, and binormal
unit vectors to the trajectory for the value s of the parameter. The third
of these further signifies the parallelism of external solicitation and resulting
acceleration. In particular, the acting force is always on the osculating plane
to the trajectory.
3 General Theorems of Point-Mass Dynamics
The elemental work done by F for an elemental displacement dP is dL =
F · dP . The work done by F in displacing {P ; m} along a smooth curve
γ = {t → P (t), t ∈ [to , t1 ]},
P (to ) = Po ,
is given by
L=
γ
F · dP =
t1
to
F · Ṗ dt.
P (t1 ) = P1 ,
(3.1)
58
3 DYNAMICS OF A POINT MASS
One defines momentum Q and kinetic energy T of {P ; m} as7
Q = mṖ ,
T =
1
2
m Ṗ 2 .
From these and (1.1), by taking derivatives, we obtain Q̇ = F and Ṫ = F · Ṗ .
Therefore
T (t1 ) − T (to ) = L,
or in differential form,
dT = dL.
(3.2)
Theorem 3.1. (i) The time derivative of the momentum of {P ; m} equals
the external force acting on it.
(ii) In the absence of external solicitations, the momentum remains constant.8
(iii) The variation of kinetic energy in some time interval equals the work
done by the external forces in the same time interval.
3.1 Positional and Central Forces
Forces P → F(P ) dependent only on P and independent of Ṗ and t, are
positional. If F(·) is defined in a region G ⊂ R3 , the pair {G; F} defines a
positional field. A field is uniform if F is constant; an example is the gravitational field F(P ) = −mge3 .9 A positional field is central if there exists a
point O, called the center of the field, such that
(P − O) ∧ F(P ) = 0
∀P ∈ G.
An example is the gravitational field generated by a point mass {O; mo }. Any
other material point {P ; m} in the field is subject to the force10
F(P ) = −γ
P −O
.
P −O
(3.3)
where k > 0 is Hooke’s constant.
(3.4)
mmo
P −O
2
Elastic forces provide a further example of central fields. A spring of fixed
endpoint O and mobile endpoint P is extended from its position of rest P = O.
Then P is acted on by a force given by Hooke’s law,
F(P ) = −k(P − O),
A more general example of a central field is
F(P ) = f (P )(P − O),
O ∈ R3 fixed,
(3.5)
where f is a smooth function defined in G.
7
The original terminology for momentum was quantitas motus, literally quantity
of motion [123, Liber I, Def. II, page 40]. The kinetic energy was initially called vis
viva, i.e., “living force,” by G. W. Leibniz, in his Theoria Motus Abstracti. Leibniz
conceived an elemental motion as an instantaneous elemental insurgence of the vis
mortua into vis viva.
8
Observed first by Newton in [123, Corollarium III, § II, page 59].
9
At sea level g = 9.8066 m/s2 . It is, however, a function of altitude [76, F–158].
10
Here γ = 6.7 · 10−11 m3 Kg/s2 is the gravitational constant [76, F–87].
3 General Theorems of Point-Mass Dynamics
59
3.2 Conservative Forces
A positional force F(P ) defined in G is conservative if there exists a function
U ∈ C 1 (G), called a potential, uniquely determined up to a constant, such
that
F(P ) = ∇U (P ) for all P ∈ G.
Denote by dP an elemental displacement of P along a curve γ as in (3.1) and
contained in G. If F is conservative, then
By integration,
γ
dU = ∇U · dP = F · dP = dL.
F · dP =
dL =
γ
P1
Po
dU = U (P1 ) − U (Po )
for every curve γ ⊂ G with endpoints Po and P1 . Therefore the work done by
F in displacing {P ; m} from Po to P1 , within G, is independent of γ. By (3.2),
d(T − U ) = mP̈ · dP − ∇U · dP = F · dP − dU = 0,
(3.6)
or in integral form,
T (t) − U P (t) = T (to ) − U (Po ).
(3.7)
The quantity −U (P ) is the potential energy of {P ; m}, whereas E = T (Ṗ ) −
U (P ), sum of the kinetic and potential energies, is the energy of {P ; m}. The
previous relations assert that E is constant along the motion. The conservation
of energy expressed formally by (3.6) or equivalently by (3.7) is called the
energy integral of the motion.11
For k ∈ R consider the level sets
[U = k] = {P ∈ G U (P ) = k, ∇U (P ) > 0}.
By the implicit function theorem these sets, if nonempty, are at least locally
smooth surfaces; they are called equipotential surfaces and have unit normal
n=
∇U
∇U
pointing in the direction of increasing U .
If Po and P are in [U = k], the work done by F in displacing {Po ; m} into
{P ; m}, along any path γ lying or not on such an equipotential surface, is zero.
From this, T (Po ) = T (P1 ). Therefore equipotential surfaces are also surfaces
of constant kinetic energy. Uniform fields F = u are conservative, and their
potential is
U (P ) = (P − O) · u + const
in R3 .
11
A formal notion of integral of motion, is in §6.1 of Chapter 6.
60
3 DYNAMICS OF A POINT MASS
The gravitational force in (3.3) is conservative, and its potential is
U (P ) = γ
mmo
+ const,
P −O
in R3 − {O}.
The elastic force in (3.4) is conservative, and its potential is
U (P ) = − 21 k P − O
2
in R3 .
+ const,
The central force in (3.5) is conservative if f is radial. Indeed,
P −O
f (P ) = f ( P − O ) =⇒ F(P ) = ∇
rf (r)dr.
The next proposition characterizes all fields of the form (3.5) that are also
conservative.
Proposition 3.1 The field (3.5) is conservative if and only if f is radial.
Proof. If F is conservative with potential U , then
∇U = ρf (P )∇ρ
where
ρ= P −O .
4 The Two-Body Problem
Assimilate the Sun to a point mass {O; mo } and Earth to a point mass {P ; m}.
With respect to an inertial system Σ, Earth is acted upon by the gravitational
force F given by (3.3), which imparts to it an acceleration aΣ (P ). By Newton’s
third law, the Sun is acted upon by the force −F, which imparts to it an
acceleration aΣ (O). By Newton’s first law,
aΣ (P ) =
F
m
and
aΣ (O) = −
F
.
mo
Choose a triad S with origin in the Sun and axes kept at all times parallel to
those of the inertial triad Σ so that S translates with respect to Σ with some
velocity v(O). Since aΣ (O) = 0, the translation velocity v(O) is not constant,
e.g., the motion of S with respect to Σ is not a uniform straight-line motion.
Therefore S is not inertial.
The two-body problem consists in describing the motion of Earth with
respect to the Sun, i.e., the motion of {P ; m} with respect to the triad S. By
Coriolis’s theorem, since ω = 0,
aS (P ) = aΣ (P ) − aΣ (O) =
m + mo
F.
m mo
Therefore
mS aS (P ) = F,
where
mS =
m mo
m + mo
is the reduced mass.
(4.1)
4 The Two-Body Problem
61
Proposition 4.1 Earth moves with respect to the Sun as a point mass with
reduced mass mS , acted upon by the gravitational force F in (3.3), as if S
were inertial.
The same arguments continue to hold for the motion of any planet about
the Sun, regarded as an isolated system. More generally, one might consider
two material bodies, assimilated to point masses {O; mo } and {P ; m} and ask
to describe the motion of one relative to the other. The terminology two-body
problem originates from such a more general setting.
4.1 Gravitational Trajectories
The first equality of (4.1) with the gravitational force F given by (3.3) takes
the form
m(m + mo ) P − O
maS (P ) = −γ
.
(4.2)
P −O 2 P −O
It follows that the motion is central and thus planar. Setting P − O = ρ,
the second of the Binet formulas in (2.5) of Chapter 1 implies
d2 1 1
γ(m + mo )
+ =
,
dϕ2 ρ ρ
a2o
where ao is the area constant and the angle ϕ is measured from a fixed direction in S to P − O. The general integral of this differential equation with
respect to the variable 1/ρ is
1
m + mo
= A sin (ϕ + α) + γ
,
ρ
a2o
where A and α are arbitrary constants. Choosing α = 21 π and setting
ρo =
a2o
,
γ(m + mo )
e = Aρo ,
(4.3)
gives the polar equation of the trajectory in the form
ρ=
ρo
.
1 + e cos ϕ
(4.4)
This is the polar equation of a conic with one of its focii in O, parameter ρo ,
and eccentricity e. If e < 1, the conic is an ellipse; if e = 1, it is a parabola;
and if e > 1, it is a hyperbola.12
12
On a plane π fix a line ℓ and a point O not in ℓ. A conic is the geometric locus
of all points in π such that the ratio of their distance to O and to ℓ is constant. The
constant value of such ratios is denoted by e and is called the eccentricity of the
conic. The point O is a focus and the line ℓ is the directrix. The elliptic, parabolic,
or hyperbolic nature of these orbits was observed by Newton in [123, Liber I, De
Motu Corporum §66, page 134].
62
3 DYNAMICS OF A POINT MASS
The force on the right-hand side of (4.2) is conservative and has potential
U =γ
m(m + mo )
.
ρ
Therefore by the energy integral,
1
m(m + mo )
mvS2 − γ
= E,
2
ρ
(4.5)
where E is the total energy of the system, which remains constant along
an arbitrary but fixed trajectory. From the first of Binet’s formulas (2.5) of
Chapter 1, we compute
2
d 1
a2
2
2
vS = ao
+ 2o .
dϕ ρ
ρ
Combining this with (4.3)–(4.5) gives
E = a2o m
e2 − 1
.
2ρ2o
It follows that if the total energy of the system is negative, then the trajectory
is an ellipse. In the limiting case e = 0, the trajectory is a circle of radius ρo ,
as expressed by (4.4). From the second equality of (4.3) it follows that this
occurs only if A = 0. In such a case,
E=−
a2o m
1
= − mvS2 ,
2 ρ2o
2
so that the total energy equals the kinetic energy with opposite sign.
The cases E ≥ 0 are characterized similarly.
5 Newton’s and Kepler’s Laws and Inertial Systems
The discussion of the previous section assumes that the gravitational force F
has the form (3.3) stipulated by Newton. The conclusion is that the motion
of the planets about the Sun is planar and their trajectories are ellipses.13
Moreover, the areolar velocity is constant and the vector radius P − O sweeps
equal areas in equal times. These conclusions are precisely the first two laws
of Kepler. Let now a and b be the major and minor semiaxes of a planetary
orbit, and let T be the corresponding period of revolution. From the definition
of the area constant and the geometric properties of ellipses,
ao = 2π
13
ab
;
T
ρo =
b2
.
a
Theoretically they could be ellipses, parabolas, or hyperbolas. Astronomical
observations confirm that they are ellipses. Parabolic and hyperbolic orbits are observed in the motion of comets [49, 28–104], [50, 105–251], [52].
5 Newton’s and Kepler’s Laws and Inertial Systems
63
Therefore
a3
γ(m + mo )
=
.
T2
4π 2
The same formula holds for any other planet of mass m′ orbiting along an
ellipse of major semiaxis a′ with period T ′ , e.g.,
a′3
γ(m′ + mo )
m′ + mo γ(m + mo )
m′ − m a3
=
=
= 1+
.
T ′2
4π 2
m + mo
4π 2
m + mo T 2
Since the mass of a planet is negligible with respect to the mass of the Sun,14
m′ − m
a′3
a3
1+
≈1
=⇒
≈ 2.
′2
m + mo
T
T
This is Kepler’s third law. Thus Newton’s gravitational law validates, although
approximately, Kepler’s third law. Conversely, Kepler’s laws validate Newton’s
gravitational law.
Having then accepted these laws one as a mutual validation of the other,
it follows that the plane of the motion of a planet is fixed with respect to
the inertial system Σ, whose existence has been postulated. In this sense the
orbital planes of the planets are inertial.
It must be stressed, however, that these conclusions follow from having
assumed the system Sun–planet to be isolated. In reality, the gravitational
contribution of the other celestial bodies is nonzero, and as a consequence,
the orbital planes are only approximately inertial.15
In what follows, in describing mechanical phenomena on Earth, we will
assume that the orbital plane of Earth is inertial, within the indicated approximations. To be specific, we will take to coincide with the coordinate
plane of {e1 , e2 } of the inertial system Σ = {Ω; e1 , e2 , e3 }.
Since the mass of Earth is negligible with respect to the mass of the Sun,
from the definition of reduced mass in (4.1)16 we obtain
m
mo
= m 1−
≈ m.
mS = m
m + mo
m + mo
14
Let mE be the mass of Earth. The mass mJ of Jupiter, the planet of largest
mass in the solar system, is mJ = 318 mE . The mass mM of Mercury, the planet
of smallest mass in the solar system, is mM = .05 mE . The mass of the Sun is
mo = 331950 mE . For these values,
mJ − mM
≈ 0.958 · 10−3 .
mM + mo
The numerical data are taken from [76, F–145 and F–165].
15
See for example the formulation of the n-body problem in §4.5c of the Complements. Newton was well aware of such a mutual gravitational interaction . . . Coelos
nostros infra coelos fixarum in orbem revolvi volunt, et planetas secum deferre; singuale coelorum partes et planetae qui relative quidem in coelis suis proximis quiescunt, moventur vere. . . ; Newton [123], §11, page 52.
16
From the previous numerical data, mS = m 0.99999699.
64
3 DYNAMICS OF A POINT MASS
This, along with Proposition 4.1, is a further validation that the system centered in the Sun and congruent to Σ = {Ω; e1 , e2 , e3 } may be assumed to be
inertial with respect to gravitational phenomena occurring on Earth.17
6 Dynamics of a Point Mass Subject to Gravity [133]
Let Σ = {Ω; e1 , e2 , e3 } be an inertial triad with origin in the Sun and such
that the plane of {e1 , e2 } coincides with the orbital plane of Earth. We also
assume, still approximately, that the axis of rotation of Earth is normal to
its orbital plane and therefore is directed as e3 . Let also S = {O; u1 , u2 , u3 }
be a triad fixed with Earth, with origin on its center O, and with u3 = e3
oriented from south to north. The triad S is in rigid rototranslation with
respect to Σ with characteristics ω = ωu3 and vΣ (O), which we assume
given.18 Assume also that the acceleration aΣ (O) of the center of Earth
with respect to Σ is negligible.19 Therefore the transport acceleration aT (P )
of a point P transported by the rigid motion of S is (formula (8.5) of
Chapter 1)
aT (P ) = aΣ (O) + ω̇ ∧ (P − O) + ω ∧ (ω ∧ (P − O))
= ω ∧ [ω ∧ (P − O)] = −ω 2 (P − Q),
17
Newton elaborates on the approximate nature of such an inertial system and
on the conceptual difficulty of identifying an inertial system other than as mathematical postulate: . . . Motus quidem veros corporum singulorum cognoscere et ab
apparentibus actu discriminare difficillium est; propterea quod partes spatii illius immobilis in quo corpora vere moventur, non incurrunt in sensus, [123, §11, page 52].
Despite Newton’s attempt to ground mechanics on a purely rationalistic basis, mechanical phenomena are based on observations that are true only within some order
of approximation. This interplay between rational mechanics and experimental mechanics was clear in Galileo’s thinking: . . . Prendiamo per ora questo come postulato,
la verità assoluta del quale ci verrà poi stabilita dal vedere altre conclusioni, fabbricate sopra questa ipotesi, rispondere e puntualmente confrontarsi con l’esperienza. . .
[61].
18
The average speed of the center of Earth about the Sun is 29.8 km/s, or
2.98 · 106 cm/s [76, F–145]. Therefore vΣ (O) ≈ 3 · 106 cm/s. The direction
of vΣ (O) is determined by the trajectory of O according to Kepler’s first law.
Since Earth completes a self-revolution about u3 in one sidereal day, e.g., 86,164 s
[76, F–103–105; F–146], one computes ω = 2π/86,164 s−1 = 7.292 · 10−5 s−1 .
19
This is the centripetal acceleration of O directed as Ω − O. Let R denote the
average distance from Earth to the Sun. Its numerical value is R = 149, 5 · 106 km, or
1,495 · 1013 cm [76, F–145]. Therefore, assuming that the trajectory is approximately
circular, by (2.6) of Chapter 1, aΣ (O) = vΣ (O)2 /R = 0.6 cm/s2 . This value is
less than one-thousandth of the mean acceleration of gravity g = 9.83225 m/s2 [76,
F–147–148].
6 Dynamics of a Point Mass Subject to Gravity [133]
65
North
u3
P
centrifugal force
force of gravity weight
O
Ã
¸
u2
u1
South
Fig. 6.1.
where Q is the projection of P on the coordinate axis of u3 . It follows, by
Coriolis’s theorem, that the acceleration of P relative to S is
aS (P ) = aΣ (P ) + ω 2 (P − Q) − 2ω ∧ vS (P ).
If {P ; m} is a moving point mass, multiplying this expression by m and taking
into account (1.2) gives
maS (P ) = FΣ + mω 2 (P − Q) − 2mω ∧ vS (P ).
(6.1)
6.1 On the Notion of Weight, Vertical Axis, and Gravity
The point mass {P ; m} moves by gravity if the only force FΣ detected by
the inertial system Σ is the gravitational force (3.3), where mo is the mass of
Earth regarded as concentrated in its center O. Therefore
maS (P ) = −
γmmo
P −O
2
P −O
+ mω 2 (P − Q) − 2mω ∧ vS (P ).
P −O
(6.2)
Weight is measured at rest, e.g., with the point mass {P ; m} on the surface
of Earth and vS (P ) = 0. On {P ; m} act the gravitational force (3.3) and the
centrifugal force mω 2 (P − Q). What one measures as weight is
−mg k = −γ
mmo
P −O
2
P −O
+ mω 2 (P − Q).
P −O
(6.3)
The unit vector k, referred to as vertical, does not coincide with the unit
vector normal to the surface of Earth, unless P is at one of the poles or at the
equator. Assume, without loss of generality, that P is in the plane {u2 , u3 } as
66
3 DYNAMICS OF A POINT MASS
in Figure 6.1. The angle ψ formed by P − O with the horizontal unit vector
u2 is called the geocentric latitude, whereas the angle λ formed by the vertical
unit vector k with u2 is the astronomical latitude.
If one were to take into account the transport acceleration aΣ (O) of the
center of Earth with respect to the Sun, this would have to be added to the
right-hand side of (6.3). In such a case the unit vector k would not lie in
the plane {u2 , u3 }. We estimate the deflection of k from such a plane and
conclude that it is negligible.
6.2 Gravitational Motion near the Surface of Earth
To describe the motion of {P ; m} with respect to an observer on the surface
of Earth, choose a triad So = {Po ; i, j, k}, with origin at the initial position
of P , with k vertical as in (6.3), the unit vector j from west to east and i
chosen so that So is positive. Measures of weight leading to (6.3) are carried
k
axis of gravity
axis of weight
north
¸
west
j
east
i
south
Fig. 6.2.
on a fixed position Po on the surface of Earth. If P departs from Po , vertically
or longitudinally, both the constant g and the vertical unit vector k change.
We will assume that the motion of {P ; m} takes place in a sufficiently small
region about Po where both g and k might be taken as constants. This occurs,
for example, if along the motion, P − Po is negligible with respect to the
radius of Earth.20 From (6.1)–(6.3),
aS (P ) = −gk − 2ωu3 ∧ vS (P ),
and by integration,
vS (P ) = −gtk − 2ωu3 ∧ (P − Po ) + vo ,
where vo is the initial velocity of P . Denote by (x, y, z) the coordinates of
P in So . Moreover, u3 = − cos λi + sin λk. Therefore the previous expression
of vS (P ) written in the coordinates of So generates the system of first-order
differential equations
20
For a range of values of g from the equator to the poles in terms of latitude, as
well as in terms of altitude from sea level, see [76, F–133, F–151, F–158].
7 Motion of a Constrained Point Mass
ẋ = 2ωy sin λ + ẋo ,
ẏ = −2ω (z cos λ + x sin λ) + ẏo ,
67
(6.4)
ż = −gt + 2ωy cos λ + żo ,
with initial conditions P (0) = Po . Taking the derivative of the second equation
and putting, in the expression so obtained, ẋ and ż given by the first and third
equations, we obtain
A = 2ωg cos λ,
ÿ = −4ω 2 y + At − B, where
B = 2ω (żo cos λ + ẋo sin λ) ,
whose general integral is
y = C sin (2ωt + α) +
At − B
.
4ω 2
(6.5)
The first term is the general integral of the associated homogeneous equation,
and it depends on two arbitrary parameters C and α. In the case of freefall
from rest, vo = 0 and B = α = 0 and C = −A/8ω 3 . Therefore
y=
g cos λ
[2ωt − sin (2ωt)].
4ω 2
(6.6)
Put now y(t) in the first and third equations of (6.4) and integrate. This gives
x=
g sin 2λ 2 2
ω t − sin2 ωt ,
4ω 2
g cos2 λ 2 2
1
ω t − sin2 ωt .
z = − gt2 +
2
2ω 2
(6.7)
(6.8)
If in (6.6)–(6.8) we let ω → 0, we recover the classical laws of freefall of
a material body. If ω = 0, then (6.6) detects an eastward deflection that
is zero at the poles (λ = ±π/2) and largest at the equator. The x(t) in
(6.7) is a deflection along a meridian. Such a deflection is toward the south
in the northern hemisphere (λ > 0) and toward the north in the southern
hemisphere. An estimate of such deflections is in §6c of the Complements.
7 Motion of a Constrained Point Mass
If {P ; m} is subject to a single constraint [f (P ; t) = 0], then (2.1) are augmented by the equation of the constraint and become
⎧
⎨ mP̈ = F(P, Ṗ ; t) + R(P, Ṗ ; t),
f (P ; t) = 0,
(7.1)
⎩
P (to ) = Po , Ṗ (to ) = Ṗo ,
68
3 DYNAMICS OF A POINT MASS
where Po and Ṗo are given vectors. The force R is the reaction due to the
constraint. The level sets [f (·; t) = 0] are smooth surfaces in R3 , which might
be regarded as moving following the parameter t. At each instant t the point
P lies on one of them. If f is independent of time, the constraint is fixed, and
P moves on the fixed surface [f = 0].
If P is subject to a double constraint, the equations of motions are
⎧
⎨ mP̈ = F(P, Ṗ ; t) + R(P, Ṗ ; t),
fj (P ; t) = 0, j = 1, 2,
(7.2)
⎩
P (to ) = Po , Ṗ (to ) = Ṗo .
The point P (t) lies at the intersection of the level sets [fj (·; t) = 0], j = 1, 2.
If both functions fj are independent of t, e.g., if the constraints are fixed,
their intersection γ is the trajectory of P . The reaction R, while not a priori
Rn
R
P
Rt
[f = 0 ]
Fig. 7.1.
known, must be included in the account of all external forces, since it arises
from enforcing the constraints, which are external to {P ; m}. The system (7.1)
consists of four scalar equations, whereas the system (7.2) consists of five scalar
equations. In either case the unknowns are the three scalar functions making
up t → P (t) and the three components of R. Therefore the problem of motion
of a constrained point mass is underdetermined, and its solvability hinges upon
the availability of further information on the nature of the constraints.
7.1 Smooth Constraints and Relative Energy
Assume first that {P ; m} is subject to a single constraint [f (·; t) = 0]. Such a
constraint is smooth or frictionless if at each time t, it only generates reactions
normal to the surface [f (·; t) = 0], equivalently if there exists a function
t → λ(t) ∈ R such that
R(P, Ṗ ; t) = λ(t)∇f (P ; t).
In such a case (7.1) becomes
⎧
⎨ mP̈ = F(P, Ṗ ; t) + λ∇f (P ; t),
f (P ; t) = 0,
⎩
P (to ) = Po , Ṗ (to ) = Ṗo .
(7.3)
7 Motion of a Constrained Point Mass
69
This is a system of four equations in the four scalar unknown functions of time
{P, λ} and therefore, at least in principle, well posed. For fixed constraint, the
fixed surface [f = 0] is smooth or frictionless if it does not oppose the sliding
of P on it, and offers a reaction only to motions that would let P leave the
surface. Let δP be an elemental virtual displacement and denote by δLR the
elemental virtual work done by R for such a virtual displacement. Then
δLR = R · δP = λ(t)∇f (P ; t) · δP = 0.
Therefore the virtual work done by the reaction offered by the frictionless
constraint [f (·; t) = 0] is zero. For an actual displacement dP , one has df = 0
and ∇f · dP + ft dt = 0. Therefore the reaction due to the constraint does the
actual elemental work
dLR = R · dP = −λ(t)ft (P ; t) · dt.
Multiplying the first equation of (7.3) by Ṗ gives
Ṫ (t) = F · Ṗ − λ(t)ft ,
or in differential form,
d(T − LF ) = −λft dt,
where dLF is the elemental work done by the external forces F applied to
{P ; m}. This relation shows how variations of kinetic energy are affected by
the motion of the constraint. For a fixed constraint one has d (T − LF ) = 0.
A double constraint [fj (·; t) = 0], j = 1, 2, is smooth or frictionless if there
[f2 = 0 ]
¸1 Ñ f1
¸2 Ñ f2
P
[f1 = 0 ]
Fig. 7.2.
exist two scalar functions t → λj (t) for j = 1, 2 such that
R(P, Ṗ ; t) = λj (t)∇fj (P ; t).
70
3 DYNAMICS OF A POINT MASS
In such a case (7.2) takes the form
⎧
⎨ mP̈ = F(P, Ṗ ; t) + λj ∇fj (P ; t),
fj (P ; t) = 0, j = 1, 2,
⎩
P (to ) = Po , Ṗ (to ) = Ṗo .
(7.4)
This is a system of five scalar equations in the five scalar unknown functions
{P, λ1 , λ2 }, which, at least in principle, is well posed. For fixed constraints, the
trajectory of P is determined by the constraints. Such a trajectory is smooth
or frictionless if it does not oppose the sliding of P on it and resists only
motions that would let P abandon its trajectory. By energetic considerations
analogous to that for single constraints,
δLR = R · δP = 0,
dLR = R · dP = −λj fj,t dt,
d (T − LF ) = −λj fj,t dt.
For fixed constraints, the last of these reduces to (iii) of Theorem 3.1.
F
Rn
b = arctan
a b
Rt
Rn
Rt
P
[f = 0 ]
cone of friction
Fig. 7.3.
7.2 Rough Constraints and Relative Energy
A single constraint [f = 0] acting on {P ; m} generates a reaction
R = R t + Rn ,
t=
Ṗ
,
Ṗ
n=
∇f
,
∇f
where Rt and Rn denote the components of R along t and n. The constraint
is rough if there exists a positive constant γ > 0 such that
Rt = γ Rn .
(7.5)
7 Motion of a Constrained Point Mass
71
The constant γ, called the dynamic friction coefficient, depends on the nature of the contact and is determined experimentally. In the case of smooth
constraints information was provided on the components of the reaction, normal, and tangent to the constraint. Formula (7.5), called Coulomb’s law, has
the same role for rough constraints. Similar considerations hold in the case
of rough double constraints, modulo the obvious changes in the meaning of
t and n. The mechanical problems (7.1)–(7.2) augmented by (7.5) are well
posed.
The component Rt opposes the motion of P , so that R · dP ≤ 0. This in
turn implies d(T − LF ) ≤ 0. Therefore rough constraints dissipate energy.
b = arctan
F
Rn
a b
Rt
Rn
Rt
P
[f1 = 0 ] ∩ [f2 = 0 ]
cone of friction
Fig. 7.4.
7.3 Remarks on Fixed Constraints
Assume that {P ; m} is subject to a single fixed constraint [f = 0], smooth or
rough, and rewrite (7.1) in the components
man = Fn (P, Ṗ ; t) + Rn ,
mat = Ft (P, Ṗ ; t) + Rt .
(7.6)
By the constraint, f (P ) = 0 along the motion, so that by differentiation,
Ṗ · ∇f = 0. Differentiating this once again gives
a · ∇f = −Ṗ t fxi ,xj Ṗ ,
a = P̈ .
Therefore if ∇f > 0, as we have assumed, then
Ṗ t fxi ,xj Ṗ
∇f
=−
.
an = a ·
∇f
∇f
Thus an can be expressed as an explicit function of P and Ṗ . But then, by
virtue of (7.6), also Rn can be expressed as an explicit function of P and Ṗ .
If a relation is known between Rt and Rn , such as for example (7.5), also Rt
can be expressed as an explicit function of P and Ṗ . This implies that the
right-hand side of the second equation of (7.6) is a known explicit function of
P and Ṗ , and therefore the integration can be effected.
72
3 DYNAMICS OF A POINT MASS
In the case of a double fixed constraint, the geometric trajectory γ is
known, but not its temporal parameterization. Parameterizing it by its arc
length s and introducing its intrinsic triad {t, n, b}, the system (7.2) can be
rewritten as
ms̈ = F · t + Rt ,
mκṡ2 = F · n + Rn ,
0 = F · b + Rb ,
(7.7)
with given s(to ) = so and ṡ(to ) = vo . If the constraint is smooth, then Rt =
Rb = 0 and the first equation of (7.7) can be integrated to resolve the motion
t → s(t). The remaining two equations permit one to compute the reaction
λj ∇fj in terms of the two functions t → λj (t), j = 1, 2.
If the constraints are rough, the second equation of (7.7) implies that Rn
is an explicit function of (s, ṡ; t), and in view of (7.5), the same is true of
Rt . Therefore the right-hand side of the first equation of (7.7) is an explicit
function of (s, ṡ; t). Thus the integration can be effected to resolve the motion
t → s(t).
7.4 Remarks on the Case of F Conservative
For smooth fixed constraints, δP = dP and the reactions are workless. An elemental variation of kinetic energy is balanced only by the elemental work
dLF = F · dP done by the the forces F applied to {P ; m}. If F is conservative,
then by the integral of the energy,
T − U (P ) = To − U (Po ),
∇U = F.
As a particular case consider the motion of {P ; m} subject to gravity −mgu3 ,
and constrained to move on a smooth and fixed surface or curve. Irrespective
of the constraint, the integral of the energy can be given the form
Ṗ
2
= 2g(a − x3 ),
(7.8)
where a is the level at which P has zero velocity. The constraint might keep
P from reaching such a level; however, (7.8) continues to hold.
8 The Mathematical Pendulum
A point mass {P ; m} is constrained to move on a vertical, smooth, fixed
circumference C = {x2 + z 2 = ℓ2 }, subject to gravity −mgk. The point P
is kept on C by a rigid, weightless rod of length ℓ, called the length of the
pendulum, with one of its extremities hinged on the center of C. The system
has one degree of freedom, and as Lagrangian coordinate one might take the
8 The Mathematical Pendulum
73
k
Q
Q¢
O
i
'
n
t
P
A
-mgk
Fig. 8.1.
angle ϕ between A − O and P − O spanned counterclockwise starting from
A − O. The trajectory is known and has intrinsic tangent and normal
t = cos ϕi + sin ϕk,
n = − sin ϕi + cos ϕk.
Therefore
−mgk = −mg(cos ϕ n + sin ϕ t),
and (7.2) written with respect to the intrinsic triad as in (7.7) becomes
ms̈ = −mg sin ϕ,
m
ṡ2
= −mg cos ϕ + R,
ℓ
(8.1)
where R is the reaction due to the constraint acting only along n, since C is
smooth. By the energy integral (7.8),
2mg
(a − z)
= −mg cos ϕ + R.
ℓ
From this one computes the reaction R in terms of the level z of P and the
level a where Ṗ = 0, e.g.,21
R = mg
(2a − 3z)
ℓ
(since z = −ℓ cos ϕ).
(8.2)
Putting this in the second equation of (8.1) gives ṡ2 = 2g(a − z). Therefore
if a > ℓ, the point P never stops and revolves indefinitely on C. If a < ℓ, the
point P comes to a stop at some Q ∈ C, at level zQ < ℓ, and then swings
to Q′ ∈ C, which is symmetric to Q with respect to the coordinate vertical
21
Such a level might or might not be attainable. See (7.8) and the related remarks.
74
3 DYNAMICS OF A POINT MASS
axis. It then continues to perform periodic oscillations between Q and Q′ with
some period T . Assume a < ℓ, set s = ℓϕ and a = −ℓ cos α, and rewrite the
energy integral in the form
ϕ̇2 =
2g
4g 2 1
sin 2 α − sin2 21 ϕ .
(cos ϕ − cos α) =
ℓ
ℓ
Separating the variables yields
ϕ
dθ
g
t=
2
.
ℓ
2
1
0
sin 2 α − sin2 21 θ
This is an implicit resolution of the motion t → ϕ(t) in terms of the Lagrangian
parameter ϕ. Choosing the initial datum ϕ(0) = 0, the point P will reach
Q, for ϕ = α, after a time equal to one-fourth of the period of oscillations.
Therefore
dθ
ℓ α
T =2
.
g 0
sin2 1 α − sin2 1 θ
2
2
To compute this integral, introduce the change of variables
η=
sin 21 θ
,
sin 21 α
k = sin 12 α,
sin2 21 α − sin2 21 θ = k 2 (1 − η 2 ),
kdη = 12 cos 12 θdθ = 21 1 − k 2 η 2 dθ.
Therefore the period T is computed from the elliptic integral
ℓ 1
dη
T =4
.
2
g 0
1 − η 1 − k2 η2
Expand the integrand in a Taylor series with respect to the parameter k and
integrate term by term to get
2
ℓ
1 2
13
4
1+ k +
T = 2π
k + ··· ,
k = sin2 21 α.
g
4
24
For small
oscillations, e.g., α ≈ 0, one recovers Galileo’s approximate formula
T = 2π ℓ/g. Such a formula can be arrived at directly by setting sin ϕ ≈ ϕ
in (8.1). The approximate equation of motion would then be
g
g
ϕ̈ + ϕ = 0,
which implies
ϕ = A cos
t+B
ℓ
ℓ
for two
real constants A and B. This describes an oscillatory motion of period
T = 2π ℓ/g.
3c General Theorems of Point-Mass Dynamics
75
Problems and Complements
3c General Theorems of Point-Mass Dynamics
3.1c Elastic Forces
Let {P ; m} be subject to (3.4) starting from the initial position Po ≡ O with
initial velocity Ṗo , which will be assumed to be parallel to Po − O. The motion
is central, and it takes place in the plane through O and normal (Po − O) ∧ Ṗo .
On such a plane introduce Cartesian axes x1 , x2 , with origin at O and the
x1 -axis oriented as Po − O. Then (3.4) yields the system
k
(3.1c)
ẍj + ω 2 xj = 0, j = 1, 2; ω 2 = ,
m
whose general integral is
xj = Aj sin (ωt + αj ),
Aj , αj ∈ R,
j = 1, 2.
Compute the constants Aj and αj in terms of the initial data. The system admits the energy integral; moreover, the areolar velocity is constant. Therefore
Ṗ
2
+ ω2 P − O
2
= const1 ,
Ṗ ∧ (P − O) = const2 .
(3.2c)
Compute these constants in terms of the initial data and show that (3.1c) and
(3.2c) are equivalent.
3.2c Point Mass Moving in a Fluid
The fluid opposes the motion of {P ; m} with a resistance R = −f ( Ṗ )Ṗ ,
where f is a smooth, nonnegative function whose form is determined from
experiments. For sufficiently slow motions, f ( Ṗ ) = const (in the air
Ṗ ≤ 2 m/s). In such a case the motion is said to be in viscous regime.
If {P ; m} is assimilated to a ball of sufficiently small radius ρ, then
f ( Ṗ ) = 6πµρ
for
Ṗ ≪ 1 (viscous regime),
where µ is the kinematic viscosity of the fluid.22 For larger speeds, f ( Ṗ ) is
proportional to Ṗ and the motion is said to be in hydraulic regime (in the air
22
The dynamic viscosity is a measure of a resistance offered by a fluid when forced
to change its shape. It is a sort of internal friction measured as the resistance elicited
by two ideal parallel planes immersed in the fluid when forced into a mutual sliding
motion. The unit of measure is the poise, after J.L.M. Poiseuille. It is measured
in dyne/s per cm2 and is the force distributed tangentially on a planar surface of
1 cm2 , needed to cause a variation of velocity of 1 cm/s between two ideal parallel
planes immersed in the fluid and separated by a distance of 1 cm. For water at 20◦ C,
the dynamic viscosity is 0.01002 poise. The kinematic viscosity is the ratio of the
dynamic viscosity to the density of the fluid. The c.g.s. unit of kinematic viscosity
is the stoke, after G.G. Stokes. Numerical values of dynamic and kinematic viscosity
for several fluids are in [76, F–36–45.].
76
3 DYNAMICS OF A POINT MASS
2 m/s< Ṗ ≤ 200 m/s). For a point mass assimilated to a ball of sufficiently
small radius ρ,
f ( Ṗ ) = 5πµρ2 Ṗ
(hydraulic regime).
3.3c Elastic Motions in Viscous Media
A point mass {P ; m}, assimilated to a ball of radius ρ ≪ 1, is attracted to a
fixed point O by a spring of elastic constant k. Assume that Po − O is parallel
to Ṗo , so that the motion takes place along the line through O and direction
Ṗo . Assume, moreover, that Ṗ ≪ 1, so that the motion is in viscous regime.
Denoting by x the coordinate variable along the trajectory of P , the only
nontrivial equation of motion is
ẍ + 2εẋ + ω 2 x = 0,
ω2 =
k
,
m
whose general integral is
x = e−εt C1 eβt + C2 e−βt ,
2ε = 6π
β=
µρ
,
m
(3.3c)
ε2 − ω 2 ,
where Cj , j = 1, 2, are real or complex arbitrary constants. If ε2 > ω 2 , the
general integral is
x = e−εt (A sinh βt + B cosh βt) ,
where A and B are real constants to be determined from the initial data. If,
for example, x(0) = 0 and ẋ(0) = 1, one computes B = 0 and A = 1/β, and
x=
1 −εt
e
sinh βt,
β
xmax =
2
ω
ω
ε+β
ε/β
,
lim x = 0.
t→∞
If ε2 = ω 2 , the general integral is
x = (At + B) e−εt .
If x(0) = 0 and ẋ(0) = 1, the solution increases from zero to its maximum
value 1/(eε), and goes to zero as t → ∞. If ε2 < ω 2 , the general integral is
β∗ = ω 2 − ε2 .
x = Ae−εt sin (β∗ t + α),
√
These are damped oscillations (ε > 0), of period T∗ = 2π/ ω 2 − ε2 . Moreover,
x (t + nT∗ ) = e−nεT∗ x(t),
∀ n ∈ N,
so that the amplitude of the damped oscillations decays exponentially to zero.
If ε = 0, the oscillations are free and have period T = 2π/ω. Thus the presence
of the fluid (ε > 0) lengthens the period of the free oscillations.
3c General Theorems of Point-Mass Dynamics
T
77
T¢
t
Fig. 3.1c.
3.4c Forced Oscillation
On the previous system impose an external forcing term, periodic with period
2π/θ, acting on the same line of the trajectory. Then (3.3c) is modified into
ẍ + 2εẋ + ω 2 x = A sin θt,
(3.4c)
where A and θ are given constants. The general integral of (3.4c) is given by
the general integral xh (·) of the associated homogeneous equation (3.3c), e.g.,
one of the previous cases, augmented by a particular integral xp (·), which will
be sought of the form
xp = B sin (θt − γ) = B(sin θt cos γ − cos θt sin γ).
Employing to this to solve (3.4c) yields
2εθ
,
ω 2 − θ2
From these we compute
B(ω 2 − θ2 ) cos γ + 2Bεθ sin γ = A.
tan γ =
φ2
,
φ
=
(ω 2 − θ2 )2 + 4ε2 θ2 ,
(ω 2 − θ2 )2
ω 2 − θ2
A
2εθ
, cos γ =
, B= .
sin γ =
φ
φ
φ
1 + tan2 γ =
(3.5c)
The general integral of (3.4c) is then x = xh + xp , which is interpreted as the
superposition of the damped oscillations xh and the forced harmonic oscillation xp .
3.5c Phase Delay
The phase of the forced oscillation is θt − γ, whereas the phase of the external
forcing term is θt. Therefore xp exhibits a phase delay of γ with respect to
the external forcing term, given by
2εθ
−1
γ(θ) = tan
.
ω 2 − θ2
78
3 DYNAMICS OF A POINT MASS
° (µ )
O
¼
2
!
µ
¼
2
Fig. 3.2c.
If θ increases from zero to ω, the phase delay γ increases from zero to π/2.
If θ increases from ω to +∞, then γ increases from π/2 to π. If the forcing
term A sin θt has the same frequency ω/2π of the free oscillations, then the
phase delay is always π/2, e.g., the forced vibrations are in quadrature of
phase with respect to the external forcing term. If θ = 0, the oscillations are
in concurrence of phase with the forcing term.
3.6c Amplitude of Forced Oscillations
The amplitude B of the forced oscillations is given by the third equation of
(3.5c). Its behavior with respect to θ hinges on the behavior of the function
θ → φ(θ). By taking the derivative of φ with respect to θ, we get
2φφ′ = 4θ θ2 − ω 2 − 2ε2 .
√
Assume first ω 2 − 2ε2 > 0. If θ increases from zero to ω 2 − 2ε2 , then φ
decreases, and the amplitude of the forced oscillations increases from A/ω 2 to
its maximum
A
Bmax = √
.
2ε ω 2 − ε2
When such a maximum is reached, the external forcing term √
and the resulting
forced oscillations are in resonance. When θ increases from ω 2 − 2ε2 to ∞,
then φ increases and the amplitude B(θ) decreases to zero. If ω 2 − 2ε2 ≤ 0,
then B(θ) decreases for all values of θ, and tends to zero as θ → ∞.
Thus in an oscillating mechanical system, forced by an external vibration
of frequency θ/2π,√for small forcing frequencies the forced vibrations reach a
maximum for θ = ω 2 − 2ε2 and tend to disappear as θ → ∞.
4c The Two-Body Problem
4.1c Resolving Kepler’s Motions
Write the energy integral (4.5) in polar coordinates ρ and ϕ:
1 2
m ρ̇ + ρ2 ϕ̇2 − U = E,
2
U =γ
m(m + mo )
.
ρ
4c The Two-Body Problem
79
B(µ )
"=0
1
µ
!
Fig. 3.3c.
Taking into account that ρ2 ϕ̇ = ao , these can be rewritten as
1
mρ̇2 − Ueff = E,
2
1 a2
Ueff = U − m 2o .
2 ρ
(4.1c)
The function Ueff is called effective potential. This relation provides a time
resolution of Kepler’s motion in the implicit form
ρ
m
dr
t=±
.
(4.2c)
2 ρo E + Ueff (r)
4.2c Stable Circular Orbits
These relations may be interpreted as the energy integral of a mechanical
system with one degree of freedom and subject to the potential Ueff . The only
equation of motion of such a system is
mρ̈ =
d
Ueff
dρ
(4.3c)
with some given initial data. Formula (4.2c) is the implicit time resolution of
the motion generated by (4.3c). The energy E in (4.1c) depends on the initial
data, e.g., Eo = E(ρo , ρ̇o ). In what follows, rather than prescribing ρo and ρ̇o ,
we will prescribe equivalently ρo and Eo . By (4.1c)–(4.3c) the orbit of P is
circular if and only if ρo and Eo are solutions of
d
γm(m + mo ) ma2o
Ueff
= −
+
= 0,
dρ
ρ2
ρ3 ρ=ρo
ρ=ρo
γm(m + mo ) 1 a2o
− m 2
Eo = −
ρ
2 ρ
ρ=ρo
80
3 DYNAMICS OF A POINT MASS
e.g.,
ρo =
a2o
,
γ(m + mo )
Eo =
1 a2o
m .
2 ρ2o
(4.4c)
In the trivial case that the area constant ao is zero, the circular orbit degenerates to a point. If ao > 0, then
d2
γ
Ueff
= − 3 (m + mo ) < 0,
2
dρ
ρ
ρ=ρo
so that Ueff has a maximum at ρo and the corresponding potential energy
Veff has a minimum at ρo . Physically, a point mass {P ; m} tends to take a
configuration that minimizes its potential energy, e.g., small variations from
that position tend, roughly speaking, to be damped so that P can resume the
position of minimum potential energy. In this sense ρ = ρo is a configuration
of stable equilibrium.23
In the context of the systems Earth–Sun, if Earth were in the position
ρo with energy Eo given by (4.4c), it would move along a stable, circular
trajectory about the Sun. If ao = 0, equation (4.3c), irrespective of the initial
data, forces P to be attracted by O. Thus if Earth did not rotate with respect
to the Sun (i.e., ao = 0), then irrespective of its initial position ρo and its
initial speed ρ̇o , it would ultimately fall into the Sun.
4.3c Radial Potentials
The analysis of the two-body problem is independent of the particular form
of the gravitational potential. An energy integral similar to (4.5) would continue to hold starting from a potential that would ensure that the motion is
central and thus planar. Potentials that generate central motions are radial
(Proposition 3.1). Therefore (4.1c)–(4.2c) continue to hold for any smooth
radial function ρ → U (ρ) defined in (0, ∞). As an example consider the case
U = γρα
and
1 a2
Ueff = γρα − m 2o ,
2 ρ
where γ and α are given real constants. For γ > 0 and α = −1 this is the
gravitational potential; for γ > 0 and α = 2 this is the elastic potential. Other
values of γ and α occur in atomic potentials [14, Chap. IX, §67].
Denote by {Po , mo } a fixed point and by {P ; m} a point mass in relative
motion with respect to {Po ; mo }. One might then ask whether the motion
generated by (4.3c), with these potentials, admits circular orbits and whether
such orbits, if any, are stable. One might also investigate whether {P ; m} will
23
We are referring here to an intuitive notion of stability. A mathematical notion
is in §1.1 of Chapter 8. By this notion, maxima for the potential correspond to
configurations of stable equilibrium (Dirichlet stability criterion, §4 of Chapter 8).
4c The Two-Body Problem
81
ultimately fall onto {Po ; mo }, or whether the two points move apart indefinitely. Circular orbits ρ = ρo are possible if and only if the initial data ρo and
Eo satisfy
d
1
α+2
2
Ueff ρ=ρo = 3 γαρ
+ mao
= 0,
dρ
ρ
ρ=ρo
1 a2o
α
Eo = − γρ − m 2
.
2 ρ
ρ=ρo
These orbits, if they exist, are stable if
d2
1
Ueff
= 4 γα(α − 1)ρα+2 − 3ma2o
2
dρ
ρ
ρ=ρo
< 0.
ρ=ρo
These orbits are unstable if this condition is violated. That is, the sole absence of maximality for the potential suffices for one to conclude that the
configuration is unstable (Corollary 6.1 of Chapter 8).
4.3.1c Circular Orbits
Prove that circular orbits are possible only if γα ≤ 0. Moreover, if γα > 0,
then regardless of the initial data, P − Po → ∞ as t → ∞. Finally, if γα < 0
and the area constant ao is zero, then P falls onto Po .
Assume next that γα < 0 and that ao > 0. Prove that if α = −2 then
circular orbits are admissible and
ρα+2
=
o
−ma2o
,
γα
Eo =
ma2o α + 2
.
ρ2o
2α
(4.5c)
Moreover, for these values,
d2
ma2o
U
(ρ)
(α + 2).
=
−
eff
dρ2
ρ4o
ρ=ρo
(4.6c)
Therefore, these possible circular trajectories are stable for α > −2 and unstable for α < −2.
4.3.2c The Case −2 < α < 0
This includes the gravitational potentials for γ > 0 and α = −1. Prove that
regardless of initial data (ρ∗ , E∗ ), P never falls onto Po , e.g., there exists some
Ro > 0, determined by (ρ∗ , E∗ ), such that ρ(t) ≥ Ro for all t ≥ 0. Moreover,
if E∗ > 0, then P − Po → ∞ as t → ∞, whatever the initial radius ρ∗ .
82
3 DYNAMICS OF A POINT MASS
4.3.3c The Case α > 0
This includes the elastic potentials for γ < 0 and α = 2. Prove that regardless
of initial data (ρ∗ , E∗ ) however fixed, P never falls onto Po , nor will it be arbitrarily far from it, e.g., there exist two positive numbers Ro < R1 , determined
in terms of (ρ∗ , E∗ ), such that Ro ≤ ρ(t) ≤ R1 , for all times.
4.3.4c The Case α < −2
Let (ρo , Eo ) be the initial data given by (4.5c)–(4.6c) characterizing possible
circular orbits. Prove that if E∗ > Eo , then regardless of the initial datum ρ∗ ,
the point P will ultimately fall onto Po . Moreover,
E∗ < Eo
and
ρ ∗ > ρo
=⇒
E∗ < Eo
and
ρ ∗ < ρo
=⇒
lim
t→∞
lim
t→∞
P (t) − Po = ∞,
P (t) − Po = 0.
4.3.5c The Case α = −2
Prove that if 2γ = ma2o , circular orbits are not admissible. Moreover, if 2γ >
ma2o , then P ultimately falls onto Po , and if 2γ < ma2o , then P − Po → ∞ as
t → ∞. Finally, if 2γ = ma2o , the only orbits that are solutions of (4.1c)–(4.2c)
are circular.
4.4c Closed Orbits
Return now to (4.1c), where U (·) is any any smooth radial function defined
in (0, ∞). Regarding ρ as a function of ϕ and using the formal differentiation
formula (2.4) of Chapter 1, (4.1c) can be rewritten as
1 a2o
m
2 ρ4
dρ
dϕ
2
= E + Ueff (ρ).
Integrating this by separation of variables gives the polar equation ρ = ρ(ϕ)
of the trajectory, in the implicit form
ρ
dr
m
ϕ − ϕ∗ = ±ao
.
2 ρ∗ r2 E + Ueff (r)
Assume now that the trajectory is confined between two limiting circles centered at Po with radii 0 < Ro < R1 , for example as in the case α > 0. Choosing
ϕ∗ = 0 as the angle for which ρ(ϕ∗ ) = Ro , the previous formula takes the form
ρ
m
dr
.
,
co = a o
ϕ = co
2
2
E + Ueff (r)
Ro r
4c The Two-Body Problem
83
An orbit generated by (4.1c) for such a choice of initial data is closed if after
ϕ having spanned the unit circle an integer number of times, the point P
returns to its initial position. One then asks whether (4.1c) generates orbits
that although not necessarily circular or elliptic, are closed.
Prove that closed orbits are possible if and only if there exist positive
integers m and n such that
2π m = n co
R1
Ro
r2
dr
.
E + Ueff (r)
Such an occurrence is rather special and, in general, will depend on the nature
of the potential U (ρ) or the initial data (ρ∗ , E∗ ), or both. The following theorem underscores the importance of the gravitational and elastic potentials.
Theorem 4.1c (Bertrand [9]). The gravitational and elastic potentials are
the only ones for which (4.1c) generates closed orbits for any choice of initial
data (ρ∗ , E∗ ).
4.4.1c More on the Polar Equation of the Trajectory
The polar equation ρ = ρ(ϕ) of the trajectory could be derived from (4.3c)
by the formal differentiation formula (2.4) of Chapter 1. Thus
d
a o d ao d
ρ
=
Ueff (ρ).
m 2
ρ dϕ ρ2 dϕ
dρ
For gravitational potentials this formula was derived by Clairaut in [30].
4.5c The n-Body Problem
Given n material points {Pi ; mi }, i = 0, 1, . . . , n − 1, subject to their mutual
gravitational attraction, one would like to describe their motion with respect
to one of them, say for example {Po ; mo }. Let Σ be an inertial system and
let S be a triad centered at Po whose axes, along the motion of {Po ; mo },
are parallel to those of Σ. Thus S is in rigid motion with respect to Σ, with
characteristics vΣ (Po ) and ω = 0. It follows from Coriolis’s theorem that
aS (Pi ) = aΣ (Pi ) − aΣ (Po ),
i = 0, 1, . . . , n − 1.
Moreover, by (1.1), for all i = 0, 1, . . . , n − 1,
mi aΣ (Pi ) =
n−1
j=0
j=i
γ
mi m j
Pj − Pi
2
Pj − Pi
.
Pj − Pi
(4.7c)
84
3 DYNAMICS OF A POINT MASS
For i = 0 this gives
mo aΣ (Po ) =
n−1
j=1
γ
mo m j
Pj − Po
2
Pj − Po
.
Pj − Po
Multiplying the ith equation (4.7c) by mi gives
mi aS (Pi ) =
n−1
γ
j=0
j=i
mi m j
Pj − Pi
2
n−1
Pj − Pi
mi mj
−
γ
Pj − Pi
Pj − Po
j=1
2
Pj − Po
Pj − Po
mi (mi + mo ) Pi − Po
Pi − Po 2 Pi − Po
n−1
Pj − Pi
Pj − Po
.
γmi mj
−
+
Pj − Pi 3
Pj − Po 3
j=1
= −γ
j=i
For n = 1 this reduces to the two-body problem. While the two-body problem
is solvable, as indicated in §4, the n-body problem is, in general, still open
(see [121]). A further discussion on the n-body problem is in §§8–12 of Chapter 6. The three-body problem, with the further assumption that the motion
is planar, was solved by Lagrange and Euler (§§11–12 of Chapter 6).
6c Dynamics of a Point Mass Subject to Gravity [133]
Expand (6.6)–(6.8) in a Maclaurin series in t and discard the terms of order
higher than four to obtain
x=
y=
z=
2
4
1
12 gω sin 2λ t + · · · ,
3
1
3 gω cos λ t + · · · ,
− 21 gt2 + 61 gω 2 cos2 λ t4
+ ··· .
For t ≪ 1, discarding the terms of order higher than three, this exhibits only
an eastward deflection along j, and the trajectory is
9gy 2 + 8ω 2 cos2 λ z 3 = 0.
In the second series discard the terms of order higher than three and take λ ≈
0.287 π (astronomical latitude of Rome, Italy). Then y(t) = 0.0172t3 cm/s3 .
After a fall of 4 s, one computes an eastward deflection of ≈10 mm. Assume
next that in (6.4), vo = 0 and prove that the vertical and meridian deflections
are of the order of ω, and therefore not negligible with respect to the eastward
deflection.
6c Dynamics of a Point Mass Subject to Gravity [133]
85
6.1c Motion of {P ; m} in the Air: Viscous Regime
To extract the effect of air resistance, discard the Coriolis force and assume
that initial velocity is vertical, so that the motion takes place along the vertical
axis, of unit ascending vector k. The point mass is assimilated to a ball of
radius ρ ≪ 1; it is also assumed that its speed is so low that the motion occurs
in viscous regime, as defined in §3c of the Complements. This yields the single
equation of motion and initial conditions
µρ
z̈ + β ż + g = 0, β = 6π ;
z(0) = 0, ż(0) = vo ,
m
whose integral is
g
g
1
vo +
1 − e−βt − t.
z=
β
β
β
Prove that as β → 0 one recovers the classical Torricelli’s laws for the free
gravitational fall of a point mass. If vo = −g/β, then {P ; m} falls in uniform
straight-line motion. Moreover, regardless of vo ,
g
(limiting velocity).
lim ż = żlim = −
t→∞
β
This limiting velocity is reached with good approximation in finite time. From
the values of g and żlim one computes the kinematic viscosity µ.
6.2c Motion of {P ; m} in the Air: Hydraulic Regime
With the same framework as the viscous regime, the motion is driven by the
initial value problem
mz̈ + λ|ż|ż + mg = 0,
λ = 5πµρ2 ,
z(0) = 0,
ż(0) = vo .
This mathematical problem is well posed, at least for small times. However,
physically it is meaningful only for speeds that are sufficiently large to justify
the hydraulic regime. Thus we will assume that |vo | is large enough so that
at least for small times, the hydraulic regime is in force.
6.3c Upward Initial Velocity (vo > 0)
For sufficiently small times that guarantee ż(t) ≥ 0,
mg
ż 2
.
a2 =
z̈ = −g 1 + 2 ,
a
λ
From this we obtain
v
v − a tan (gt/a)
a
o
o
= (0, t̄ ).
, t ∈ 0, arctan
ż = a
a − vo tan (gt/a)
g
a
Since z̈ < 0, there exists a time t′ ∈ (0, t̄) such that ż(t′ ) is so small as to
evoke the viscous regime. From such a time forward such a regime will be in
force, and for t → ∞ the point mass will have the same limiting velocity żlim
as before.
86
3 DYNAMICS OF A POINT MASS
6.4c Downward Initial Velocity (vo < 0)
The equation of motion becomes
ż 2
z̈ = −g 1 − 2 ,
a
ż(0) = vo < 0,
(6.1c)
so long as ż ≤ 0. If vo = −a, the only solution is ż = vo and {P ; m} falls in
uniform straight-line motion along the vertical. If vo = a, then (6.1c) admits
the implicit integral
a
a + ż a − vo
a + ż a − vo
t = − ln
> 0.
(6.2c)
, provided
2g
a + vo a − ż
a + vo a − ż
Lemma 6.1c ż ≤ 0 at all times.
Proof. If vo < −a, then z̈ > 0 in some time interval t ∈ (0, t∗ ) and ż increases
from vo < 0 to −a. If t∗ < ∞, then starting from t∗ ,
ż 2
z̈ = −g 1 − 2 ,
ż(t∗ ) = −a,
t ≥ t∗ .
a
This has the only solution ż = −a. If vo > −a, then z̈ < 0 in some time
interval t ∈ (0, t∗ ) and ż decreases from vo < 0 to −a. Starting from t∗ ,
ż 2
ż(t∗ ) = −a,
t ≥ t∗ .
z̈ = −g 1 − 2 ,
a
It follows from the lemma that (a − vo )/(a − ż) > 0 at all times. Therefore
the inequality in (6.2c) is verified if (a + ż)/(a + vo ) > 0. If vo < −a, then
(ż(t) + a) < 0 for t ∈ (0, t∗ ). In such a time interval the inequality of (6.2c) is
verified, and the implicit integral can be rewritten in the explicit form
a + ż
a + vo −2gt/a
=
e
,
a − ż
a − vo
vo < −a,
t ∈ (0, t∗ ).
This shows that the value ż = −a is approached asymptotically, and therefore
t∗ = ∞. In a similar manner one proves that if vo > −a, then t∗ = ∞.
7c The Mathematical Pendulum
7.1c The String Pendulum
The point mass {P ; m} is fixed at one of the extremities of an ideal, inextensible, massless string of length ℓ, whose other extremum is fixed at O.
This is an example of unilateral constraint, since P is allowed to abandon the
7c The Mathematical Pendulum
87
circumference C. The string exerts a traction Rn directed toward O. Therefore
for P not to abandon C we must have
Rn ≥ 0,
i.e., from (8.2),
2a ≥ 3z.
If this holds, the pendulum never abandons C, and it behaves as if it were
held by a rigid rod. For this to hold, the level z of P , the level a, real or
ideal, and the length ℓ must satisfy a > −ℓ; otherwise, the pendulum would
not move; z < a by the energy integral; −ℓ ≤ z ≤ ℓ by the equation of the
constraint. Given such restrictions, the condition on the traction is verified in
the following two cases:
−ℓ < a ≤ 0 oscillatory case,
2a ≥ 3ℓ revolving case.
Except for these two cases, there exists on C a point at level 0 < 3z ≤ 2a < 3ℓ
where Rn = 0. Starting from this point, {P ; m} abandons C and it describes
a parabola. The equation of this parabola is found by observing at such an
instant, taken as initial, one has
z(0) = 32 a, v(0) = 23 ag = 0 along the tangent to C.
Write down the equation of such a parabola and show that it is tangent to C
at the point of level 3z(0) = 2a. Show that the reaction R(ϕ) at such a point.
Compute the instant t1 when the parabola intercepts C and show that the
function t → P (t) is continuous in a neighborhood of t1 .
3
ℓ
2
2a
3
O
C
A
Fig. 7.1c.
7.2c The Cycloidal Pendulum (Huygens [81])
Let {P ; m} be constrained on a smooth, fixed cycloid
x = R(ϕ + sin ϕ),
z = R(1 − cos ϕ),
ϕ ∈ (−π, π).
88
3 DYNAMICS OF A POINT MASS
The arc length is s = 4R sin 12 ϕ, which implies s2 = 8Rz. Therefore the
z-component of the unit tangent to the cycloid is
k·t =
dz
s
=
.
ds
4R
By the energy integral, Ṗ 2 = 2g(a − z). To resolve the motion in terms of
the parameter t write (1.1) in the triad intrinsic to the cycloid as in (2.2). The
equation along t is
ms̈ = F · t = −mgk · t = −
mg
s.
4R
This describes an oscillatory motion of period T = 2π 4R/g. Thus the approximate law of Galileo is exact for oscillations of a point mass along an arc
of cycloid [81], [80, Vol. XVIII].
7.3c The Spherical Pendulum (Lagrange [101])
A point mass {P ; m} moves in the cavity of a smooth, fixed sphere of center
O and radius ℓ, subject to gravity. Introduce a triad S = {O; i, j, k} with k
along the ascending vertical and set
x = ℓ sin θ cos ϕ,
y = ℓ sin θ sin ϕ,
z = ℓ cos θ.
By the energy integral,
2 2
1
2 mℓ (θ̇
+ ϕ̇2 sin2 θ) − mgℓ cos θ = Eo .
Since the constraint is smooth, the reaction is directed as P − O, and its
projection on the plane z = 0 is on a line through O. It follows that the
motion of the projection P∗ of P in the plane z = 0 is central, and its areolar
velocity is constant, i.e.,
(P∗ − O) ∧ Ṗ∗ = ℓ2 ϕ̇ sin2 θ = ao .
If ao = 0, the motion takes place in a plane and {P ; m} behaves like a mathematical pendulum. Assuming ao > 0, the energy integral and the area integral
imply
sin2 θ = Ao + A1 cos θ + A2 cos2 θ + A3 cos3 θ,
m
Ao = E − 2 a2o , A1 = mgℓ, A2 = −E, A3 = −mgℓ.
2ℓ
2 2
1
2 mℓ θ̇
The right-hand side is nonnegative for θ in some interval [θ1 , θ2 ]. Therefore the motion takes place in the annular, spherical sector θ ∈ (θ1 , θ2 ).
7c The Mathematical Pendulum
89
Choosing the initial time so that θ(0) = θ1 , the implicit time resolution of the
motion is
dη
1 2 cos θ
t = ± mℓ
.
2
Ao + A1 η + A2 η 2 + A3 η 3
cos θ1
To compute the reaction, observe that along the motion, P − O
this by double differentiation,
Ṗ 2 + (P − O) · P̈ = 0,
and moreover,
2
= ℓ2 . From
mP̈ = −mgk + R.
Therefore
mṖ 2 − mgk · (P − O) + R · (P − O) = 0,
from which, by the energy integral,
2
1
2 mṖ
+ mgz − 23 mgz = 12 Rn ℓ
=⇒
Rn =
2Eo − 3mgz
.
ℓ
7.4c Small Oscillations of a Spherical Pendulum
Assume θ ≈ π, ż, z̈ ≈ 0 and z ≈ −ℓ. Since the sphere is smooth, R =
λ(P − O)/ℓ. If z ≈ −ℓ, one also has R ≈ mgk, so that λ ≈ −mg, and the
approximate equations of motion for θ ≈ 0 take the form
ẍ + (g/ℓ)x = 0,
ÿ + (g/ℓ)y = 0.
(7.1c)
Each of these describes a harmonic oscillation of period 2π ℓ/g. The motion
takes place in the plane z = −ℓ and the trajectory is an ellipse, possibly
degenerate.
Fig. 7.2c.
90
3 DYNAMICS OF A POINT MASS
7.5c The Foucault Pendulum [11, 57]
This is a spherical pendulum, suspended at one of the extremities of a long,
ideal, inextensible, massless string of length ℓ, whose second extreme is fixed
at O. Its relevance is that it elucidates the influence of the rotation of Earth
on the motion of a spherical pendulum. Let Σ be an inertial system and fix a
triad S = {O; i, j, k} with origin at the center of the spherical pendulum, and
with unit vectors chosen as in §6.2. By Coriolis’s theorem,
aS (P ) = aΣ (P ) − 2ω ∧ vS (P )
= aΣ (P ) + 2ω (cos λi − sin λk) ∧ (ẋ, ẏ, ż),
where ω is the angular speed of the rotation of Earth about its axis, and λ
is the astronomical latitude of the point of coordinates (0, 0, −ℓ). Assuming
the oscillations are small, the equations of motion is S are
g
ẍ + x − 2ωo ẏ = 0,
ℓ
g
ÿ + y + 2ωo ẋ = 0,
ℓ
ωo = ω sin λ.
Multiply the second by the imaginary unit i and add it to the first to get
z̈ + 2iωo ż + (g/ℓ)z = 0,
where
z = x + iy.
This has the general integral
√
√
2
2
z = e−iωo t Aei (g/ℓ)+ωo t + Be−i (g/ℓ)+ωo t ,
where A and B are constants to be determined in terms of the initial conditions. To interpret the motion of z, assume that ωo2 is negligible with respect
to (g/ℓ). In such a case,
√
√
√
√
2
2
Aei (g/ℓ)+ωo t + Be−i (g/ℓ)+ωo t ≈ Aei g/ℓ t + Be−i g/ℓ t
= (x∗ + iy∗ ) = z∗ ,
where the components x∗ and y∗ are solutions of (7.1c). Therefore t → z∗ (t)
describes an ellipse in the complex plane, and it represents the motion of small
oscillations of a spherical pendulum when the effects of the rotation of Earth
are neglected. With these approximations, the previous general integral can
be rewritten in the approximate form
z = e−iωo t z∗ ,
and it is interpreted as a point that describes ellipses on planes that are ideally
superposed to the plane z = 0 and that are themselves rotating about k with
angular velocity ωo k. This was indeed the outcome of Foucault’s experiments
[57, 1851], mathematically formalized by Binet [11].
7c The Mathematical Pendulum
91
k
Q
ℓ
t
n
P
R
a
i
O
Fig. 7.3c.
7.6c Point Mass Sliding on a Circle
A point mass {P ; m} is constrained on a smooth, fixed circle of center O and
radius R, and set in a vertical plane. The point P is subject to gravity and
is acted upon by an elastic force F = k(Q − P ) as in Figure 7.3c. Compute
the reaction due to the constraint in terms of the position of P . Let α be the
angle between P − O and Q − O. Prove that if initially α ∈ [0, 12 π), then α
remains in such an interval for all times, provided k(R + ℓ) > mg. In such
a case prove that the motion is oscillatory, and compute the period of small
oscillations (sin α ≈ α).
Since the constraint is smooth, R = λn for λ ∈ R, whereas the external
forces are given by
F(e) = k(Q − P ) − mg cos αn − mg sin αt
= [k(R + ℓ) cos α − kR − mg cos α] n
+ [k(R + ℓ) sin α − mg sin α] t.
Therefore, in terms of the intrinsic triad to the trajectory,
2
mRα̈ − [k(R + ℓ) − mg] sin α = 0,
mRα̇ − [k(R + ℓ) cos α − kR − mg cos α] − λ = 0.
The first of these determines the motion, in terms of the Lagrangian coordinate
α, starting from some initial conditions. Therefore t → α(t) is known, and the
second determines λ, and thus the reaction due to the constraints. Starting
from α(0) ∈ (0, 21 π), the first equation keeps α(t) in the same interval only if
92
3 DYNAMICS OF A POINT MASS
k(R + ℓ) > mg. In such a case the motion is oscillatory and the period of the
small oscillations is
mR
.
T = 2π
k(R + ℓ) − mg
7.7c Point Mass Sliding on a Sphere
A point mass {P ; m}, subject to gravity, starts moving from its rest position
at the top of a sphere having been activated by an elemental disturbance.
Find the point and the time when it leaves the sphere.
4
GEOMETRY OF MASSES
1 Material Systems and Measures
A distribution of masses within a bounded set E ⊂ R3 is described by a
measure µ. The symbol dµ(P ) is the elemental mass about P as measured by
µ. The measure µ is required to be finite, that is,
{total mass of the system} =
dµ < ∞,
E
and supported in E, that is,
dµ = 0 for every compact
K ⊂ R3 − Ē.
K
The mechanical quantities of a material system are described by scalar- or
vector-valued functions
Q −→ Φ(Q) = f (P − Q)dµ(P ),
where f (·−Q) is µ-integrable for all Q. The pair {M; dµ} will mean a material
system M whose masses are measured by a given such a measure µ. In what
follows we will refer to either continuous or discrete systems.
1.1 Continuous Distribution of Masses
Let dx denote the Lebesgue measure in E. A nonnegative, Lebesgue measurable function ρ(·) defined in E may be regarded as the density function of a
continuous distribution of masses in Ω. In such a case dµ = ρ dx. The measure
dµ is absolutely continuous with respect to the Lebesgue measure and ρ is its
Radon–Nikodym derivative. The total mass contained in E is
m=
ρ dx = dµ,
provided ρ ∈ L1 (E).
E
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 4,
c Springer Science+Business Media, LLC 2011
93
94
4 GEOMETRY OF MASSES
In the last integral the domain of integration has been omitted, since µ is
supported in Ē. If {M; dµ} is distributed over a line or a plane, then µ is
the Lebesgue measure in R1 or R2 respectively, and ρ is the linear or planar
density of the system.
1.2 Discrete Distribution of Masses
The measure describing a system of n material points {Pj ; mj } is
dµ =
n
mj δ(Pj ),
j=1
where δ(Pj ) is the Dirac mass concentrated in Pj . The total mass is
n
n
mj δ(Pj ) =
mj .
m = dµ =
j=1
j=1
2 Center of Mass and First-Order Moments
Given a system {M; dµ} and O ∈ R3 , the equation
m = dµ,
m(Po − O) = (P − O)dµ(P ),
(2.1)
defines a point Po , independent of O, called center of mass of {M; dµ}. For
continuous or discrete systems, (2.1) takes the form
n
(Pj − O)mj .
m(Po − O) =
m(Po − O) = (P − O)ρ dx,
j=1
Let π{Q; e} denote the plane through Q and normal to the unit vector e. The
moment of {M; dµ} with respect to π{Q; e} is
(P − Q) · edµ(P ) = m(Po − Q) · e.
(2.2)
Thus the moment of {M; dµ} with respect to π{Q; e} equals the moment of
the point mass {Po ; m} with respect to the same plane. The integrals in (2.1)
and (2.2) are moments of the first order, or of degree one.
3 Second-Order Moments and Huygens’s Theorem
For a material system {M; dµ} define [81]:
(i) Polar Moment of Inertia of {M; dµ} with respect to a point O,
Io = P − O2 dµ(P ).
3 Second-Order Moments and Huygens’s Theorem
95
(ii) Planar Moment of Inertia of {M; dµ} with respect to a plane
π{Q; e},
Iπ{Q;e} = |(P − Q) · e|2 dµ(P ).
(iii) Axial Moment of Inertia of {M; dµ} with respect to an axis ℓ{Q; e}
through Q and directed as the unit vector e,
Iℓ{Q;e} = (P − Q) ∧ e2 dµ(P ).
(iv) Deflection Moment of Inertia1 of {M; dµ} with respect to a pair of
nonparallel planes π{Q; e} and π{Q; e′ },
Iπ{Q;e};π{Q;e′ } =
(P − Q) · e (P − Q) · e′ dµ(P ).
The moments of inertia are nonnegative, whereas the deflection moments of
inertia are of variable sign. Introduce a triad S with origin in O and set
3
2
Io =
xi dµ,
Ii = Iplane{xi =0} = x2i dµ,
i=1
2
Iii = Iaxis xi =
xj dµ.
j =i
Then
Iii =
j =i
Ij ,
3
i=1
Iii = 2
3
Ij = 2Io .
(3.1)
j=1
The distances P − O are intrinsic to {M; dµ} and are invariant under a
rotation of S. Thus Io is also an intrinsic quantity of {M; dµ}. The deflection
moments of inertia with respect to pairs of coordinate planes are denoted by
−Iij = I{xi =0};{xj =0} = xi xj dµ,
i, j = 1, 2, 3, i = j.
These inertia quantities, with the indicated notation, are then organized into
the symmetric matrix
⎛
⎞
I11 I12 I13
σ = ⎝ I21 I22 I23 ⎠,
I31 I32 I33
called an inertia tensor. From the previous remarks, the trace of σ is 2Io ,
which is invariant under rotations of S.
1
The terminology deflection moments will be justified in §5 of Chapter 7 and in
particular, Remark 5.1.
96
4 GEOMETRY OF MASSES
Theorem 3.1 (Huygens [82]). For every O ∈ R3 and every plane π{Q; e},
Io = IPo + mPo − O2 ,
(3.2)
2
Iπ{Q;e} = Iπ{Po ;e} + m|(Po − Q) · e| ,
2
Iℓ{Q;e} = Iℓ{Po ;e} + m(Po − Q) ∧ e ,
Iπ{Q;e};π{Q;e′ } = Iπ{Po ;e};π{Po ;e′ } ,
+ m (Po − Q) · e (Po − Q) · e′ .
(3.3)
(3.4)
(3.5)
Proof. To prove (3.2) compute
Io = P − O2 dµ(P ) = (P − Po ) + (Po − O)2 dµ(P )
2
2
dµ(P ) + 2(Po − O) · (P − Po )dµ(P )
= P − Po dµ(P ) + Po − O
= IPo + mPo − O2 ,
since the last term vanishes by definition of center of mass. As for (3.4),
compute
Iℓ{Q;e} = (P − Q) ∧ e2 dµ(P )
= (P − Po ) ∧ e + (Po − Q) ∧ e2 dµ(P )
= Iℓ{Po ;e} + m(Po − Q) ∧ e2
+ 2 (Po − Q) ∧ e ·
(P − Po )dµ(P ) ∧ e .
From this (3.4) follows, since the last term vanishes. The remaining formulas
are proved analogously.
Remark 3.1 The center of mass Po of a system {M; dµ} is the point for
which the polar moment is minimal.
Remark 3.2 Given a bundle of parallel lines (planes), the one through Po is
the axis (plane) whose axial (planar) moment of inertia is minimal.
4 Ellipsoid of Inertia and Principal Axes
Proposition 4.1 For any two unit vectors a and b,
at σb = Io a · b − Iπ{O;a};π{O;b} ,
at σa = Iℓ{O;a} .
4 Ellipsoid of Inertia and Principal Axes
97
Proof. From the definition of deflection moment of inertia with respect to a
pair of planes,
(P − O) · a (P − O) · b dµ(P )
Iπ{O;a};π{O;b} =
2
xi xj ai bj dµ
= xi ai xj bj dµ = xi ai bi dµ +
i=j
= (Io − Iii )ai bi −
= Io a · b − at σb.
Iij ai bj
i=j
This establishes the first statement of the proposition. Putting now a = b
and using the definition of planar moment of inertia yields
at σa = Io a2 − Iπ{O;a}
= P − O2 dµ(P ) − |(P − O) · a|2 dµ(P )
2
P − O2 − |(P − O) · a| dµ(P )
=
= (P − O) ∧ a2 dµ(P ) = Iℓ{O;a} .
Corollary 4.1 The matrix σ is positive semidefinite. Moreover,
xt σx = x2 Ix
for all x ∈ R3 \{0},
where Ix is the axial moment of inertia of {M; dµ} with respect to the axis
through the origin and direction x.
Proof. By Proposition 4.1, xt σx ≥ 0 for all x ∈ R3 . The corollary follows by
putting a = x/x into the second equation of Proposition 4.1.2
This formula gives a practical way of computing the axial moments of inertia
of a system {M; dµ} with respect to axes through the origin whenever the
inertia tensor σ is known. Assume now that σ is not degenerate. Having fixed
xo ∈ R3 \{O}, compute Ixo and set
λ2 = xo 2 Ixo .
The points x ∈ R3 satisfying the equation xt σx = λ2 define an ellipsoid, denoted by Eλ , and called an inertia ellipsoid or Cauchy ellipsoid [23]. It follows
from the definition that
Ix =
2
λ2
,
x2
for every x ∈ Eλ .
In general σ is not positive definite. For an axial distribution of masses σ is
degenerate.
98
4 GEOMETRY OF MASSES
Thus, knowing the axial moment of inertia with respect to some point xo ∈ Eλ
permits one to compute the axial moment with respect to any other point of
the inertia ellipsoid Eλ . These remarks suggest putting σ into diagonal form,
or equivalently, writing Eλ in its canonical form. Since σ is symmetric and
positive definite, its eigenvalues Ii are nonnegative, and there exists a unitary
matrix U such that
⎛
⎞
I1 0 0
UσU t = ⎝ 0 I2 0 ⎠ = (Ii δij ) .
0 0 I3
The matrix U rotates S into a new system S ′ , whose coordinates we denote
by X = (X1 , X2 , X3 ). If x ∈ Eλ , then
λ2 = xt σx = (Ux)t UσU t (Ux) = X t (UσU t )X = X t (Ii δij )X.
Therefore, in the rotated system S ′ , the ellipsoid Eλ takes its canonical form
Ii Xi2 = λ2 .
√
The semiaxes ai = λ/ Ii of Eλ are inversely proportional to the eigenvalues
of σ. In particular, the largest semiaxis corresponds to the least eigenvalue.
The next proposition asserts that the transformed matrix UσU t keeps the
same material-geometric significance as σ. Thus σ identifies a tensor.
Proposition 4.2 Let σ ′ be the inertia tensor of {M; dµ} with respect to S ′ .
Then σ ′ = UσU t .
Proof. The entries Iij of σ can be written concisely as
2
Iij = δij P − O dµ(P ) − xi xj dµ.
Since U is unitary, the entries Iij of the transformed matrix UσU t are
Iij = αiℓ Iℓk αjk = αiℓ Io αjℓ − αiℓ xℓ xk αjk dµ
= Io δij − Xi Xj dµ.
Thus Iij are the entries of the inertia tensor of {M; dµ} in S ′ .
Corollary 4.2 The eigenvalues Ii of σ, for i = 1, 2, 3, are the axial moments
of inertia of {M; dµ} with respect to the coordinate axes of S ′ .
The new coordinate axes are called principal axes of inertia and are directed as
the eigenvectors of σ. In S ′ , the inertia tensor σ ′ is diagonal, and the elements
of the diagonal are the axial moments of inertia of {M; dµ} with respect to
the principal axes of inertia.
5 Miscellaneous Remarks
99
5 Miscellaneous Remarks
Assume that the entries of the ith row and ith column of σ are all zero except
Iii . For example, assume that σ is of the form
⎞
⎛
I11 I12 0
(5.1)
σ = ⎝ I21 I22 0 ⎠.
0 0 I33
Since I33 is an eigenvalue of σ, the x3 -axis is a principal axis of inertia. The
ellipsoids Eλ are of the form
I11 x21 + 2I12 x1 x2 + I22 x22 + I33 x23 = λ2 .
From the definitions and the form of σ it follows that the x3 -axis is a principal
axis of inertia if and only if I13 and I23 are zero, equivalently, if and only if
Eλ is symmetric with respect to the plane {x3 = 0}.
5.1 Center of Mass and Principal Axes of Inertia
The inertia tensor σ and the principal axes of inertia depend on the choice of
O and are called principal axes of inertia with respect to O. If the principal
triad of inertia has origin at the center of mass Po of the material system
{M; dµ}, it is called a central principal triad of inertia, and the coordinate
axes are central principal axes of inertia.
Proposition 5.1 If an axis is principal with respect to O and contains Po ,
then it is central principal.
Proof. If Po = O, the conclusion is trivial. If Po = O, let ℓ be the axis through
O and Po and let SO be the principal triad of inertia with origin at O. By
assumption, ℓ is a principal axis of inertia, and without loss of generality we
label it as the x1 -axis of SO , so that
−I13 = x1 x3 dµ = 0.
−I12 = x1 x2 dµ = 0,
Moreover, since Po = (a, 0, 0) for some a = 0,
x3 dµ = x2 dµ = 0.
From these one computes the deflection moment of inertia:
i = 2, 3.
I{x1 =a};{xi =0} = (x1 − a)xi dµ = 0,
Thus ℓ is a central principal axis of inertia.
100
4 GEOMETRY OF MASSES
Corollary 5.1 The principal axes of inertia with respect to a point of a central principal axis are parallel to the central principal axes of inertia.
Corollary 5.2 If a principal axis of inertia with respect to a point O contains
Po , then the principal axes with respect to O are parallel to the central principal
axes of inertia.
5.2 Planar Systems
Let {M; dµ} be a system of masses distributed in a plane π. Since π can be
regarded as a plane of material symmetry for {M; dµ}, any axis normal to π
is a principal axis of inertia. Let S be a triad with origin O ∈ π and with u3
normal to π. With respect to S, the tensor of inertia σ has the form (5.1).
Moreover,
2
I11 = x2 dµ,
I22 = x21 dµ
I33 = I11 + I22 .
(5.2)
6 Computing Some Moments of Inertia
6.1 Homogeneous Material Segment
A segment of length a > 0 is identified with the interval (− 21 a, 21 a) of the x1 axis. By material symmetry the coordinate axes are principal axes of inertia.
We compute I11 = 0 and Iij = 0, i = j. Moreover,
IO = I22 = I33 = ρ
a/2
−a/2
x21 dx1 =
2
1
12 ma ,
m = ρa.
(6.1)
The axial moment with respect to any axis ℓ normal to the segment through
one of the extremities is computed by Huygens’s theorem as
Iℓ =
a
2
x2
2
1
12 ma
+ 14 ma2 = 13 ma2 .
b
2
x2
a
2
x1
a
2
a
2
b
2
Fig. 6.1.
x1
6 Computing Some Moments of Inertia
101
6.2 Homogeneous Material Rectangle
The triad S as in the left-hand picture Figure 6.1 is a central principal axis
of inertia. One might regard the rectangle as a material segment of constant
density bρ, obtained by condensing on the segment M N those points at equal
distance from the x2 -axis. Then by (6.1),
2
1
12 ma ,
I22 =
and by symmetry
I11 =
2
1
12 mb .
Moreover, by (5.2),
I33 = I11 + I22 =
2
1
12 m(a
+ b2 ).
By Huygens’s theorem,
I
with resp. to axis
{x3 =0}∩{x1 =−a/2}
I
with resp. to axis
{x3 =0}∩{x2 =b/2}
=
2
1
12 ma
+ 14 ma2 = 31 ma2 ,
=
2
1
12 mb
+ 41 mb2 = 31 mb2 .
For a material homogeneous square of edge a,
I11 = I22 =
2
1
12 ma ,
I33 = 16 ma2 .
x3
c
2
x1
b
2
a
2
x2
Fig. 6.2.
6.3 Homogeneous Material Parallelepiped
Referring to Figure 6.2, regard the parallelepiped as a material rectangle of
sides a and b and of constant surface density ρc. The mass of such a rectangle
is m = ρabc. By the previous calculations,
I11 =
2
1
12 m(b
+ c2 ),
I22 =
2
1
12 m(a
+ c2 ),
I33 =
2
1
12 m(a
+ b2 ).
Let ℓ be the axis parallel to u3 and through the point 21 (a, b), whose distance
√
from the origin is 12 a2 + b2 . By Huygens’s theorem,
Iℓ = I33 + 41 m(a2 + b2 ) = 13 m(a2 + b2 ).
102
4 GEOMETRY OF MASSES
Consider now the axis ℓ′ through the origin and 12 (b, a, c) and set
⎞
⎛ 2
0
0
b + c2
m⎝
(a, b, c)t
0
a 2 + c2
0 ⎠,
.
σ=
v= √
12
a 2 + b 2 + c2
0
0
a2 + b 2
From Proposition 4.1,
Iℓ′ = vt σv =
1 a2 (a2 + c2 ) + b2 (b2 + c2 ) + c2 (a2 + b2 )
m
.
12
a 2 + b 2 + c2
6.4 Homogeneous Material Circle and Disk
The x3 -axis through Po and normal to the plane of the circle or disk is a
central principal axis of inertia. By symmetry, any pair of axes orthogonal in
Po and normal to the x3 -axis is central principal and
I11 = I22 ,
I33 = 2I11 = 2I22 .
Therefore it will suffice to compute I33 . For the circle,
I33 = mR2 ,
I11 = I22 = 21 mR2 .
For the disk, I11 = I22 = 14 mR2 , and
I33 = ρ
(x21 + x22 )dx1 dx2 = 21 mR2 .
x21 +x22 <R2
x3
h
2
R
x2
R
h
2
x1
Fig. 6.3.
6.5 Homogeneous Material Right Circular Cylinder
The cylinder has height h and section a disk of radius R. The axis of the
cylinder is central principal. The remaining central principal axes are any
two orthogonal lines through Po and normal to the axis of the cylinder. The
density is constant and the mass of the cylinder is m = ρπR2 h. To compute
I33 one regards the mass of the cylinder concentrated into a homogeneous
material disk of center Po . Therefore I33 = 12 mR2 . To compute I11 = I22 ,
2c Center of Mass and First-Order Moments
103
regard the cylinder as a stack of homogeneous material disks of radius R and
infinitesimal thickness dx3 , and each of mass πρR2 dx3 . By Huygens’s theorem,
the moment of the generic of these disks whose center has distance x3 from
Po is given by
πρR2 14 R2 + x23 dx3 .
By integration in dx3 over (− 21 h, 12 h),
I11 = I22 = 14 m R2 + 31 h2 .
x3
a3
R
a1
a2
x2
Fig. 6.4.
6.6 Homogeneous Material Sphere and Ball
By symmetry, any triad with origin in Po is central, principal of inertia, and
I11 = I22 = I33 . Moreover, from the definition of polar moment with respect
to Po ,
I11 + I22 + I33 = 3I11 = 2Io .
For the sphere of radius R,
Io = mR2 ,
Iii = 32 mR2 ,
i = 1, 2, 3.
For the ball of radius R,
Io = 53 mR2 ,
Iii = 52 mR2 .
Problems and Complements
2c Center of Mass and First-Order Moments
2.1. Prove that Po is independent of the choice of O. Moreover, Po is in the
convex envelope of the support of dµ(P ).
104
4 GEOMETRY OF MASSES
2.2. In a material homogeneous triangle the center of mass is at the intersection of the medians, one-third the distance from the intersection of any
one of them with the side of the triangle they bisect.
P0
P0
R
® ®
O
Fig. 2.1c.
2.3. The center of mass of a material homogeneous arc of circumference of
center O, radius R, and aperture 2α ∈ (0, π) is Po = (0, R sin α/α).
2.4. For a material homogeneous circular sector of center O, radius R, and
aperture 2α, compute Po = (0, 2R sin α/3α).
2
2.5. For the material homogeneous
parabolic segment x2 ≤ 2px1 , x1 ∈ (0, a),
3
a, p > 0, compute Po = 5 a, 0 .
x2
O
Po
a
x1
Po
R
Fig. 2.2c.
2.6. The center of mass of a material homogeneous hemisphere of radius R
is at distance 83 R from the equatorial plane.
3c Second-Order Moments and Huygens’s Theorem
Let σ be the tensor of inertia of a material system {M; dµ} with respect to a
triad S. Prove that tr(σ) and det(σ) are invariant by orthogonal transformations of S. Give a geometrical interpretation of det(σ).
3.1c Inertia Tensor for a Homogeneous Ellipsoid
Let E be a material homogeneous ellipsoid of density ρ, mass m, and semiaxes
a1 , a2 , a3 , referred to a triad S as in Figure 6.4. By symmetry σ = Iii δij .
5c Miscellaneous Remarks
105
The change of variables Xi = xi /ai , i = 1, 2, 3, whose Jacobian is (a1 a2 a3 ),
maps E into the unit ball B1 centered at the origin. Therefore
2
4π
1
I33 =
x1 + x22 dx
3m
a1 a2 a3 E
2
2
2
X1 dX + a2
X22 dX = F1 + F2 .
= a1
B1
In polar coordinates
2
F1 = a1
π/2
−π/2
Analogously F2 =
4 2
15 a2 π.
0
2π
B1
1
2
4
15 πa1 .
r4 cos3 θ cos2 ϕ dρ dθ dϕ =
0
Therefore, I33 = 51 m(a21 + a22 ).
5c Miscellaneous Remarks
5.1c Finding the Principal Axes of a Planar System
Let {M; dµ} be a planar system and let S be a triad with the plane {x3 = 0}
on the plane of the system. Let now Sϕ be a triad obtained by rotating
S through an angle ϕ, counterclockwise about u3 . It is required to find the
angle ϕ for which Sϕ is a principal triad of inertia. The corresponding rotation
matrix is
⎞
⎛
cos ϕ sin ϕ 0
R = ⎝ − sin ϕ cos ϕ 0 ⎠.
0
01
Therefore the tensor of inertia σϕ with respect to Sϕ is
⎛
⎛
⎞
⎞
I11 (ϕ) I12 (ϕ) 0
I11 I12 0
σϕ = R ⎝ I21 I22 0 ⎠ Rt = ⎝ I21 (ϕ) I22 (ϕ) 0 ⎠,
0
0 I33
0 0 I33
where
I11 (ϕ) = I11 cos2 ϕ + 2I12 sin ϕ cos ϕ + I22 sin2 ϕ,
2I12 (ϕ) = −I11 sin 2ϕ + 2I12 cos 2ϕ + I22 sin 2ϕ,
2
(5.1c)
2
I22 (ϕ) = I11 sin ϕ − 2I12 sin ϕ cos ϕ + I22 cos ϕ.
Therefore Sϕ is principal of inertia if the angle ϕ is chosen so that
I12 (ϕ) = 0, e.g.,
(I11 − I22 ) sin 2ϕ = 2I12 cos 2ϕ.
106
4 GEOMETRY OF MASSES
If I11 = I22 one chooses
2ϕ = tan−1
2I12
I11 − I22
∩ 0,
π
.
2
If I11 = I22 then I12 cos 2ϕ = 0. If I12 = 0 then ϕ = π/4, whereas if I12 = 0,
the original triad S was already principal of inertia.
Remark 5.1c Let ℓ(ϕ) be the axis through the origin and of unit vector
v = (cos ϕ, − sin ϕ, 0)t . By Proposition 4.1 and (5.1c),
Iℓ(ϕ) = vt σv = I11 (ϕ)
and
2I12 =
d
Iℓ(ϕ) .
dϕ
Therefore the angle ϕ that identifies the principal axes of inertia is a stationary point for the function ϕ → Iℓ(ϕ) . Prove that such a point is indeed an
extremum.
5.2c Planar Systems and Mohr’s Circle
Assume that the triad S is principal of inertia, so that I12 + 0. To examine
the functions ϕ → I(ϕ) starting from S, rewrite (5.1c) in the form
I11 − I22
I11 + I22
+
cos 2ϕ,
2
2
I22 − I11
sin 2ϕ,
=
2
I22 − I11
I11 + I22
+
cos 2ϕ.
=
2
2
I11 =
I12
I22
From the first two equations,
2
2
(I11 (ϕ) − a) + I12
= b2 ,
where
I11 − I22
I11 + I22
,
b=
.
2
2
This implies that the pair (I11 (ϕ), I12 (ϕ)) represents a point in the circle
of center (a, 0) and radius |b|, called Mohr’s circle. It permits a graphical
determination of the axial and deflection moments of inertia in a planar system
as functions of ϕ. Having determined I11 and I22 , one constructs such a circle
as in the right-hand picture Figure 5.1c. Then, for a given ϕ, one traces
through (a, 0) and slope tan 2ϕ. The coordinates of M are I11 and I12 . Prove
that I22 (ϕ) is the abscissa of M ′ .
a=
6c Computing Some Moments of Inertia
x3 = x⬘3
I22
x⬘2
M ≡ (I11, I22)
2'
'
x1
(a,0)
O
x2
107
I11
M⬘
x⬘1
Fig. 5.1c.
6c Computing Some Moments of Inertia
6.1c Right Circular Cone
Compute the inertia tensor σ of a material homogeneous right circular cone of
height h and radius R with respect to a triad centered at its vertex and with
u3 directed as the axis of the cone as in the left-hand picture Figure 6.1c.
The cross section normal to the axis of the cone and at level z is a disk of
radius r = Rz/h whose moments of inertia with respect to its central principal
axes are
I11 (z) = I22 (z) = 14 dm(z)r2 = 41 πρr4 dz,
I33 = 12 dm(z)r2 = 21 πρr4 dz,
where ρ is the density of the cone. Therefore by Huygens’s theorem,
h
2
2
R
πρ R
3
I11 = I22 =
m R2 + 4h2 .
z 4 dz = 20
+4
4
h
h
0
Moreover,
I33 = 12 πρ
h
r4 dz =
0
2
3
10 mR .
By symmetry, Iij = 0 for i = j. Therefore the tensor of inertia is
⎛ 2
⎞
R + 4h2
0
0
3 ⎝
0
R2 + 4h2 0 ⎠.
m
σo =
20
0
0
2R2
6.1.1c Right Circular Cone—I
Verify that the center of mass is on the axis of the cone at distance 34 h from
its vertex. Use this and Huygens’s theorem to compute the tensor of inertia
with respect to the central principal triad of the cone. Such a tensor is
σPo
⎛ 2 1 2
⎞
R + 4h
0
0
3 ⎝
0
R2 + 14 h2 0 ⎠.
m
=
20
0
0
2R2
108
4 GEOMETRY OF MASSES
x3
R
x3
h
h
z
z
O
x2
x2
x1
x1
Fig. 6.1c.
6.1.2c Right Circular Cone—II
Compute the tensor of inertia of the previous cone with respect to a triad
with origin at the center of its base and u3 on the axis of the cone as in
the right-hand picture Figure 6.1c. The cross section normal to the axis
of the cone and at distance z from the plane {x3 = 0} is a disk of radius
r = R(h − z)/h whose moments of inertia with respect to its central principal
axes are computed as before. Then by Huygens’s theorem,
h
1 4
2 2
3
I11 = I22 = πρ
dz = 10
m 12 R2 + 13 h2 .
4r + r z
0
Moreover,
I33 =
1
2 πρ
h
r4 dz =
0
2
3
10 mR .
By symmetry, Iij = 0 for i = j. Therefore
σo =
⎛1
3
⎝
10 m
2R
2
⎞
+ 31 h2
0
0
1 2
1 2
0
0 ⎠.
2R + 3h
0
0
R2
Deduce σo from σPo and Huygens’s theorem.
6.2c Homogeneous Material Triangle
Let ∆(ABC) be a homogeneous material triangle of density ρ, mass m, vertices A, B, C, and sides a = BC, b = AC, c = AB. Assume that BC is
the largest side, denote by O the projection of A on BC, and set h = OA.
Introduce a fixed triad S centered at O, with u3 directed as (C − O) and u1
directed as (A − O). Compute the tensor of inertia σS of the triangle with
respect to S.
6c Computing Some Moments of Inertia
109
u3 = e3
C
C
b
u2
a
e1
W=0
0
h
A
B
u1
c
h
γ
A
B
u1
Fig. 6.2c.
In computing I33 , the triangle is regarded as the infinite union of
contiguous rectangles of infinitesimal width dx1 whose base is parallel to
BC. Such a rectangle is at distance x1 from BC and has infinitesimal mass
ρa(h − x1 )dx1 /h. Therefore
ρa
h
I33 =
h
0
x21 (h − x1 )dx1 = 16 mh2 .
In computing I11 , the triangle is regarded as the union of the triangles
∆1 (OAB) and ∆2 (AOC). Therefore using the previous calculation,
I11 = ρ
x23 dx1 dx3
x23 dx1 dx3 + ρ
∆1
∆2
2
= 16 m1 a − b2 − h2 + 16 m2 (b2 − h2 )
= 61 m a2 − 3 b2 − h2 a − b2 − h2 .
As for I13 , we compute
I13 = −ρ
∆1
x1 x3 dx1 dx3 − ρ
x1 x3 dx1 dx3
∆2
2
a − b2 − h2 − (b2 − h2 )
= 61 mh 12 a − b2 − h2 .
=
2
1
24 ρh
Finally, since the system is planar, I12 = 0 and I22 = I11 + I33 .
110
4 GEOMETRY OF MASSES
6.3c Homogeneous Material Right Triangle
If ∆(ABC) is a right triangle with hypotenuse b and legs h and a =
then B = O and
⎛ 2
⎞
a
0
− 12 ah
σS = 16 m ⎝ 0 a2 + h2 0 ⎠.
− 21 ah
0
h2
√
b 2 − h2 ,
6.4c Spiral Ramp
Compute the moment of inertia with respect to the x3 -axis of the homogeneous
material spiral ramp given in cylindrical coordinates by x3 = θ, ρ ∈ [0, 1],
θ ∈ [0, π].
5
SYSTEMS DYNAMICS
1 Flow Map and Derivatives of Integrals
Let G(t) be the configuration at time t of a material system {M; dµ} in motion
from its initial configuration Go . Every point Po ∈ Go follows its trajectory
to arrive at the position P (t) ∈ G(t) at time t; vice versa, a point P ∈ G(t)
may be regarded as originating from the motion of some Po ∈ Go . The flow
map is the transformation
P = Φ(Po ; t) : Go −→ G(t);
Φ ∈ C ∞ (Go × I),
where I is the interval of time on which where the motion is defined. It is
assumed that for each fixed t ∈ I, the flow map is a smooth bijection between
Go and G(t), with smooth inverse. From the definition,
Ṗ = Φt (Po ; t),
P̈ = Φtt (Po ; t),
...
for all t ∈ I.
Kinematic and dynamic information on the motion are expressed by scalar- or
vector-valued smooth functions (P, Ṗ ; t) → F (P, Ṗ ; t). These can be regarded
as defined in Go × I by composing them with the flow map
F (P, Ṗ ; t) = F (Φ(Po ; t), Φt (Po ; t); t) .
Consider material systems for which the elemental mass dµ(P ) about P is the
same as the elemental mass dµ(Po ) about the initial position Po , e.g.,
dµ(P ) = dµ Φ(Po ; t) = dµ(Po ).
This occurs, for example, if {M; dµ} is discrete or a nondeformable continuum.1 In such a case
1
If it is a continuum, then dµ(P ) = ρ(P )dV (P ), where ρ(·) is the density and
dV (P ) is the Lebesgue measure of an elemental volume about P . If it is nondeformable, the configuration G(t) is obtained from Go by a rigid motion, so that the
Jacobian of the transformation is one.
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 5,
c Springer Science+Business Media, LLC 2011
111
112
5 SYSTEMS DYNAMICS
G(t)
F (P, Ṗ ; t)dµ(P ) =
Go
F (Φ, Φt ; t) dµ(Po ).
Proposition 1.1 Let {M; dµ} be nondeformable or discrete. Then
d
d
F (P, Ṗ ; t)dµ(P ).
F (P, Ṗ ; t)dµ(P ) =
dt G(t)
G(t) dt
The proof is a direct consequence of the definition and the assumed smoothness of the flow map and F . In what follows we consider only discrete systems
or nondeformable continua, so that the proposition is in force.
2 General Theorems of System Dynamics
2.1 D’Alembert’s Principle
A material system {M; dµ} is regarded as the union of point masses
{P ; dµ(P )} of elemental mass dµ(P ). If {M; dµ} is in motion, each of its
points is acted upon by forces
f (P, Ṗ ; t)dµ(P ) = f (i) (P, Ṗ ; t) + f (e) (P, Ṗ ; t) dµ(P ).
The function f is smooth and it has the physical dimensions of a force per
unit of mass. The force f dµ(P ) is the resultant of forces f (i) dµ(P ) internal to
the system and forces f (e) dµ(P ) external to it. The former may be regarded
as solicitations exerted on {P ; dµ(P )} by the remaining points of the system,
whereas the latter are generated by external causes. If {M; dµ} is subject to
constraints, these will exert on each {P ; dµ(P )} a reaction r(P, Ṗ ; t)dµ(P ),
which in general is unknown. Since these are due to causes external to the
system, they are compounded in f (e) dµ(P ). The total force acting on {M; dµ}
is F = F(i) + F(e) , where
F(e,i) = f (e,i) (P, Ṗ ; t)dµ(P ).
The moment of f dµ(P ) with respect to a pole O is
(P − O) ∧ f (i) + f (e) dµ(P ).
The total resulting moment is M = M(i) + M(e) , where
M(e,i) = (P − O) ∧ f (e,i) dµ(P ).
By Newton’s third law, the system of the internal forces consists of couples of
zero wrench. Therefore F(i) = M(i) = 0.
2 General Theorems of System Dynamics
113
The d’Alembert principle states that every {P ; dµ(P )} making up {M; dµ}
moves according to Newton’s first law, as if it were removed from {M; dµ},
provided the internal forces are included in the compound of forces acting on,
it, e.g., [36],
P̈ dµ(P ) = f (i) + f (e) dµ(P ).
(2.1)
Its relevance is in that every dynamic problem can be regarded as a problem of pointwise equilibrium of all forces acting on every point of the system,
provided one includes the internal and external forces, including possible reactions due to constraints, and inertial forces. From this, integrating in dµ(P ),
(e)
F = P̈ dµ(P )
(since F(i) = 0).
(2.2)
2.2 Momentum
The momentum of the elemental point mass is Ṗ dµ(P ). Integrating gives the
momentum of the system
Q = Ṗ dµ(P ).
Theorem 2.1 (of the Momentum). The derivative of the momentum of the
system {M; dµ} equals the resultant F(e) of all external forces acting on it.
Proof. Take the derivative in the expression of Q and use Proposition 1.1 and
(2.2) to get
Q̇ =
P̈ dµ(P ) = F(e) .
Theorem 2.2 (Center of Mass).
(i) The momentum of the system equals the momentum of the center of mass
Po where ideally the entire mass m of {M; dµ} is concentrated, e.g., Q =
mṖo .
(ii) The center of mass Po moves as a point mass {Po ; m} solicited by the
resultant F(e) of the external forces, e.g., mP̈o = F(e) .
Proof. From the equation of the center of mass, by taking the time derivative,
mṖo = Ṗ dµ(P ).
This implies (i). Differentiating a second time and taking into account the
theorem of the momentum proves (ii).
114
5 SYSTEMS DYNAMICS
2.3 Angular Momentum
The angular momentum of the elemental point mass {P ; dµ(P )} with respect
to a pole O is
dK = (P − O) ∧ Ṗ dµ(P ).
Integrating in dµ(P ) gives the resultant angular momentum
K = (P − O) ∧ Ṗ dµ(P ).
The point O might be fixed or mobile and it might or might not belong to
the system. By taking the derivative of K, we have
K̇ = (P − O) ∧ f (i) + f (e) dµ(P ) − mȮ ∧ Ṗo
= M(e) + mṖo ∧ Ȯ.
The last term vanishes if O = Po or if O is fixed.
Theorem 2.3 (of the Angular Momentum). The time derivative of the
resultant angular momentum of a system {M; dµ}, taken with respect to the
center of mass Po of a fixed point O, equals the resultant total moment M(e)
of the external forces, taken with respect to the same point.
Remark 2.1 If {M; dµ} is isolated then M(e) = 0, and if moments are taken
with respect to the center of mass Po , then K = const. Therefore if K = 0, the
plane through Po and normal K is constant. The solar system is approximately
isolated and its center of mass is approximately in the Sun. The plane through
the Sun and normal K was called by Laplace the invariant plane of the solar
system ([109]; see also §§4–5 of Chapter 3).
2.4 Energy
The kinetic energy of an elemental point mass is
dµ(P ) gives the energy of the system
1
Ṗ 2 dµ(P ).
T =
2
1 2
2 Ṗ dµ(P ).
Integrating in
Taking the differential in dt yields
(i)
f (P, Ṗ ; t) + f (e) (P, Ṗ ; t) · Ṗ dµ(P )
dT = Ṗ · P̈ dµ(P ) =
= dL(i) + dL(e) ,
3 Cardinal Equations
115
where the differential
dL
(e,i)
=
f (e,i) · dP dµ(P )
is the elemental work done by all the internal and external forces respectively.
Integrating over the time interval (to , t) gives T (t) − T (to ) = L, where L is
the work done by the internal and external forces in the same time interval.
Notice that the internal forces being in equilibrium (F(i) = M(i) = 0) does
not imply that the system of internal forces is workless.
Theorem 2.4 (Kinetic Energy). The variation in kinetic energy in a time
interval (to , t) equals the total work done in the same time interval by the
internal and external forces acting on {M; dµ}.
If the force densities f (i,e) are conservative, then f (e,i)i = ∇U (e,i) for two
smooth functions P → U (e,i) (P ). Setting U = U (i) + U (e) , the differential
form of the kinetic energy implies
d(T − U ) = 0,
e.g.,
t → [T (t) − U (t)] = const.
The function t → −U (t) is the potential energy of the system. Thus the sum of
the kinetic and potential energies of the system is conserved along the motion.
3 Cardinal Equations
A consequence of the previous remarks is that the motion of a material system
{M; dµ} must satisfy the equations
mP̈o = F(e) = Q̇,
K̇ = M(e) + mṖo ∧ Ȯ = M(e) + Q ∧ Ȯ.
(3.1)
(3.2)
If the pole O is either fixed or chosen as the center of mass Po , then K̇ = M(e) .
The resultant F(e) is inclusive of possible reactions due to constraints;
similarly, M(e) includes the moments of such possible reactions.
The general problem of system dynamics is to determine, starting from
some initial mechanical configuration of a system {M; dµ}, its evolution and
the reactions that constrain its evolving configurations.
Equations (3.1)–(3.2) are necessary for any system {M; dµ} and for this
reason are called the cardinal equations of dynamics. However, they are not
sufficient for the actual determination of the motion. Indeed, they are global
in nature (e.g., integral) and provide no information on the motion of the
single points of the system. They consist of six scalar equations and thus
sufficient only for a mechanical system with at most six degrees of freedom.
For example, they are necessary and sufficient to determine the motion of an
116
5 SYSTEMS DYNAMICS
unconstrained rigid system. The cardinal equations involve only free vectors
in space, and as such they are independent of a reference system. However,
having been derived from Newton’s laws, they have mechanical meaning only
when referred to an inertial system Σ = {Ω; e1 , e2 , e3 }, which we assume has
been selected and fixed.
3.1 Cardinal Equations in Noninertial Systems
Let S = {O; u1 , u2 , u3 } be a triad in rigid motion with respect to Σ with given
characteristics v(O) and ω. Given a vector-valued smooth function t → v(t),
denote by xj (t) its components in S, so that v = xj uj . By differentiation,
dv
+ ω ∧ v,
v̇ = ẋj uj + ω ∧ v =
dt S
where (dv/dt)S denotes the derivative of v relative to S. Applying such a
differentiation rule to the vectors Q and K in (3.1)–(3.2) gives
dQ
dK
+ ω ∧ Q = F(e) ,
+ ω ∧ K = M(e) .
(3.3)
dt S
dt S
3.2 Motion Relative to the Center of Mass
Let S be a triad with origin at the center of mass Po and translating with
respect to Σ with velocity Ṗo . The motion of {M; dµ} described with respect
to S is called relative to the center of mass [51,65]. By the formula for relative
velocity,
Ṗ = (Ṗ )S + Ṗo ,
e.g.,
vΣ (P ) = vS (P ) + vΣ (Po ).
Therefore the momentum of the system is
Q = (Ṗ )S dµ(P ) + mṖo .
By the theorem of the center of mass, mṖo = Q. Therefore
QS = (Ṗ )S dµ(P ) = 0.
(3.4)
The elemental angular momentum, taken with respect to a fixed point O ∈ Σ,
is given by
(P − O) ∧ (Ṗ )S + Ṗo dµ(P ).
By integration in dµ(P ),
K = (P − O) ∧ (Ṗ )S dµ(P ) + (P − O) ∧ Ṗo dµ(P )
= KS + (Po − O) ∧ mṖo
= KS + (Po − O) ∧ Q.
4 Mechanical Quantities of Rigid Systems
117
3.3 König’s Theorem [24, 96]
The kinetic energy of {M; dµ}, expressed in terms of the motion relative to
the center of mass is
2
(Ṗ )S + Ṗo dµ(P )
2T =
2
2
dµ(P
)
+
Ṗ
dµ(P
)
+
2
(Ṗ )S Ṗo dµ(P ).
= (Ṗ )S
o
From (3.4) it follows that the last term is zero. Therefore
1
1
2
(Ṗ )2S dµ(P ).
where TS =
T = mṖo + TS ,
2
2
(3.5)
The first term on the right-hand side is the kinetic energy of the center of
mass, where one ideally concentrates the entire mass of the system and TS is
the kinetic energy relative to S. Formula (3.5) is König’s theorem.
4 Mechanical Quantities of Rigid Systems
If {M; dµ} is rigid, choose an inertial triad Σ and a triad S = {Ω; e1 , e2 , e3 }
fixed with the system, in rigid motion with respect to Σ with characteristics
v(O) and ω. The momentum of the elemental point mass {P ; dµ(P )} is
[v(O) + ω ∧ (P − O)] dµ(P ).
From this by integration,
Q = m [v(O) + ω ∧ (Po − O)] = mṖo .
The angular momentum of {P ; dµ(P )} with respect to a pole O is
(P − O) ∧ [v(O) + ω ∧ (P − O)] dµ(P ),
and by integration,
K = m(Po − O) ∧ v(O) +
(P − O) ∧ ω ∧ (P − O) dµ(P ).
Write (P − O) = xi ui and ω = ωi ui in the coordinates of S, and compute
(P − O) ∧ ω ∧ (P − O) = x22 + x23 ω1 − x1 x2 ω2 − x1 x3 ω3 u1
+ −x1 x2 ω1 + x21 + x23 ω2 − x2 x3 ω3 u2
+ −x1 x3 ω1 − x2 x3 ω2 + x21 + x22 ω3 u3 .
118
5 SYSTEMS DYNAMICS
Therefore
(P − O) ∧ ω ∧ (P − O) dµ(P ) = (I11 ω1 + I12 ω2 + I13 ω3 ) u1
+ (I21 ω1 + I22 ω2 + I23 ω3 ) u2
+ (I31 ω1 + I32 ω2 + I33 ω3 ) u3 ,
where (−1)1+δij Iij are the moments of inertia of {M; dµ}, axial (i = j), or
deflection (i = j), computed in S. Thus
K = m(Po − O) ∧ v(O) + σω,
(4.1)
where σ = (Iij ) is the inertia tensor of {M; dµ} with respect to the triad S
fixed with {M; dµ}. If the system has a fixed point, choosing the origin of S
to coincide with such a fixed point, the first term in the right-hand side of
(4.1) is zero. Such a term also vanishes it the origin of S is at the center of
mass of {M; dµ}. In either of these cases (4.1) becomes
(4.1)′
K = σo ω,
where σo is the inertia tensor of {M; dµ} with respect to the fixed triad S with
origin either in a fixed point O or in the center of mass Po . If S is principal
of inertia relative to O, then
(4.1)′′
K = I1 ω1 u1 + I2 ω2 u2 + I3 ω3 u3 ,
where Ij are the axial moments of inertia with respect to the coordinate
principal axes through O. These formulas continue to hold instantaneously,
e.g., for elemental motions for which v(O) = 0. This occurs, for example, if O
is a point of the instantaneous axis of rotation and in addition v(O) · ω = 0.
4.1 Kinetic Energy
The kinetic energy of the elemental point mass {P ; dµ(P )} is
dT = 12 [v(O) + ω ∧ (P − O)]2 dµ(P ),
and by integration,
2T = mv2 (O) + 2mv(O) · ω ∧ (Po − O) +
2
(P − O) ∧ ω dµ(P ). (4.2)
Let ℓ(O; ω) be the axis through O with direction ω and let Iℓ(O;ω ) denote
the moment of inertia of {M; dµ} with respect to such an axis. From the
definition of axial moment of inertia,
2
(P − O) ∧ ω dµ(P ) = Iℓ(O;ω ) ω2 .
4 Mechanical Quantities of Rigid Systems
119
With this notation,
2T = mv2 (O) + 2mv(O) · ω ∧ (Po − O) + Iℓ(O;ω ) ω2 .
(4.2)′
Corollary 4.1 If {M; dµ} has a fixed point O, then
2T = Iℓ(O;ω ) ω2 = ω t σo ω = K · ω,
(4.3)
where σo is the inertia tensor with respect to a triad S fixed with {M; dµ} and
with origin in O.
Another special case occurs if the origin of S coincides with the center of mass
of the system.
Corollary 4.2 Let SPo be a triad fixed with {M; dµ} and with origin at the
center of mass Po of the system. Then
2T = mṖo2 + Iℓ(Po ;ω) ω2
(4.4)
= mṖo2 + ω t σPo ω = mṖo2 + K · ω,
where σPo is the tensor of inertia of {M; dµ} with respect to SPo and Iℓ(Po ;ω)
is the axial moment of inertia with respect to the axis ℓ(Po ; ω), through Po
and directed as ω.
Remark 4.1 If O is a point of the instantaneous axis of rotation, then v(O)
is parallel to ω and (4.2)′ takes the form
2T = mv2 (O) + Iℓ(O;ω ) ω2 ,
O ∈ {axis of motion}.
(4.2)′′
Remark 4.2 By Huygens’s theorem,
Iℓ(O;ω ) = Iℓ(Po ;ω) +
m(Po − O) ∧ ω2
ω2
ω = 0 .
Therefore (4.2) can be given the form
2T = mv2 (O) + 2mv(O) · ω ∧ (Po − O)
+ Iℓ(Po ;ω) ω2 + m(Po − O) ∧ ω2 .
(4.2)′′′
If O is a point of the instantaneous axis of rotation, then
2T = mv2 (O) + Iℓ(Po ;ω) ω2 + m(Po − O) ∧ ω2 .
(4.2)iv
120
5 SYSTEMS DYNAMICS
5 Workless Constraints: Discrete Systems
A discrete material system {M; dµ} consisting of n point masses {Pℓ ; mℓ } is
subject to m independent constraints [fj = 0]. These exert reactions Rℓ on
the points {Pℓ ; mℓ }, which are in general unknown. Assuming m < 3n, the
equations of motion are
mℓ P̈ℓ = Fℓ + Rℓ ,
ℓ = 1, 2, . . . , n,
fj (P1 , . . . , Pn ; t) = 0, j = 1, 2, . . . , m < 3n,
(5.1)
where Fℓ includes the internal and external forces acting on Pℓ , but not the
reactions due to the constraints. This system consists of (3n + m) scalar equations with the 6n scalar unknowns Pℓ (t) and Rℓ (t) and therefore is not sufficient to describe the motion of {M; dµ}, and further information is needed
on the nature of the constraints. The elemental virtual work done by Rℓ
during an elemental virtual displacement δPℓ compatible with the constraints
is Rℓ · δPℓ . The elemental virtual work done by the system of the reactions is
n
δΛ =
ℓ=1
Rℓ · δPℓ .
The constraints [fj = 0] are smooth or workless if2
for all virtual displacements
compatible with the constraints.
δΛ = 0
(5.2)
Theorem 5.1 (Lagrange [101] Chapter XV). Let the constraints [fj = 0]
be workless. Then there exist m constants λj such that
m
Rℓ =
j=1
λj ∇Pℓ fj ,
ℓ = 1, 2, . . . , n.
(5.3)
Corollary 5.1 The equations of motion of a discrete system subject to workless constraints are
n
mℓ P̈ℓ = Fℓ +
j=1
λj ∇Pℓ fj ,
fj (P1 , P2 , . . . , Pn ; t) = 0,
ℓ = 1, 2, . . . , n,
(5.4)
j = 1, 2, . . . , m < 3n.
These are (3n + m) scalar equations in the (3n + m) unknowns t → Pℓ (t)
for ℓ = 1, . . . , n and t → λj (t) for j = 1, . . . , m. Therefore the motion of the
system and its reactions are determined from its initial conditions.
2
A more general notion of workless that would include unilateral constraints
would be that δΛ ≥ 0 for every elemental virtual displacement δP compatible with
the constraints.
5 Workless Constraints: Discrete Systems
121
5.1 Proof of Lagrange’s Theorem
Relabel the coordinates of the points Pℓ by setting
y2 = x1,2 , y3 = x1,3 ,
y1 = x1,1 ,
y4 = x2,1 ,
y5 = x2,2 , y6 = x2,3 ,
...
...
...
y3n−2 = xn,1 , y3n−1 = xn,2 , y3n = xn,3 .
Likewise relabel the components of reactions Rℓ as
R2 = R1,2 , R3 = R1,3 ,
R1 = R1,1 ,
R4 = R2,1 ,
R5 = R2,2 , R6 = R2,3 ,
...
...
...
R3n−2 = Rn,1 , R3n−1 = Rn,2 , R3n = Rn,3 .
With this notation, an elemental virtual displacement is denoted by
δP = (δP1 , . . . , δPn ) = (δy1 , . . . , δy3n ) = δy.
Compatibility of such a displacement with the constraints and the assumption
that the constraints are workless are written in the form
3n
∂fj
∂fj
δyi = −
δys ,
s=m+1 ∂ys
i=1 ∂yi
m
(5.5)
3n
m
i=1
j = 1, . . . , m,
Ri δyi = −
Rs δys .
(5.6)
s=m+1
Since the constraints are independent, the matrix (∂fj /∂yi ) has rank m. Without loss of generality assume that the minor of the first m rows and m columns
has nonzero determinant. The first m equations in (5.5) can be regarded as a
linear algebraic system in the m unknowns δyi , i = 1, . . . , m. These unknowns
are uniquely determined by an arbitrary choice of the (3n − m)-tuple δys
for s = m + 1, . . . , 3n. The assumption states that every solution δy of (5.5)
is also a solution of (5.6). This implies that the last equation of the system
(5.5)–(5.6) must be a linear combination of the preceeding ones. Therefore
there exist m real parameters λj such that
m
λj
Ri =
j=1
and
∂fj
,
∂yi
3n
i = 1, . . . , m,
3n
m
λj
Rs δys =
s=(m+1)
j=1
∂fj
δys .
s=m+1 ∂ys
Rewrite the last relation as
3n
s=m+1
m
Rs −
λj
j=1
∂fj
δys = 0
∂ys
122
5 SYSTEMS DYNAMICS
and observe that the (3n − m)-tuple δys is arbitrary to deduce
m
λj
Rs =
j=1
∂fj
,
∂ys
s = m + 1, . . . , 3n.
This implies the theorem by recalling the relabeling of the coordinates of Pℓ
and of the components of Rℓ .
6 The Principle of Virtual Work
A constraint [f = 0] imposed on a point mass {P ; m} is smooth if it generates reactions normal to the surface [f = 0]. This in turn implies that the
work done by the reaction is zero for every elemental virtual displacement
δP compatible with the constraint. Conversely, if the virtual work of the
reaction is zero for every elemental virtual displacement δP compatible with
the constraint, then the reaction is normal to the surface [f = 0]. The theorem of Lagrange asserts that for a discrete system of n points subject to
m constraints, the constraints are workless if and only if the reactions Rℓ
due to the constraints are normal to the intersection of all surfaces [fj = 0].
Equivalently, the constraints are smooth if and only if the total work done
by the reactions Rℓ is zero for every virtual displacement δPℓ of the points
Pℓ compatible with the constraints. This notion of smooth constraints will
be adopted for all mechanical systems, discrete or not. We will say that a
set of constraints imposed on the motion of a material system {M; dµ} is
smooth or workless if the total work done by the reactions is zero for every
virtual displacement of the system compatible with the constraints. Such an
assumption placed on the constraints is the principle of virtual work.
6.1 The Principle of Virtual Work for Rigid Systems
The principle of virtual work as expressed by (5.2) is satisfied by the rigidity
constraints of a rigid system {M; dµ}. Let S be a triad fixed with {M; dµ} and
in rigid motion with respect to an inertial triad Σ. The rigidity constraints
require that the mutual distance of any two points P, Q ∈ S be constant
in time. The constraints are time independent, so that virtual and actual
displacements coincide. These constraints generate on P a reaction
P −Q
R(P, Q) = γ
for some γ ∈ R.
P − Q
By Newton’s third law the constraints generate on Q a reaction −R. Therefore
the work done by the reactions acting on the pair of points P, Q ∈ S is
P −Q
P −Q
δP − γ
δQ
P − Q
P − Q
(P − Q) · δ(P − Q)
=γ
= γδP − Q = 0.
P − Q
δΛ(P, Q) = γ
3c Cardinal Equations
123
Problems and Complements
3c Cardinal Equations
3.1c Disk Rolling on a Slanted Guide as in Figure 3.1c
e3
R
R
Po
e1
C
weight
®
Fig. 3.1c.
It is a rigid motion of characteristics vPo and ω. The center of mass Po
moves on a line parallel to the guide, with acceleration g sin αe1 . Therefore v(Po ) = gt sin αe1 . The angular momentum with respect to Po is
K = 12 M R2 ω. If the constraint is smooth, the reaction R is on the normal to the guide through Po and the weight is applied in Po . Therefore K̇ = 0
and ω = const. The velocity of C is computed from Poisson’s formula, as
v(C) = (g sin αt − Rω)e1 , where ω = ω. Thus if the guide is smooth,
the disk never rolls without slipping. Assume now that the guide is rough
and exerts a reaction R so that the disk rolls without slipping. Decomposing
R = R + R⊥ into its components, parallel and normal to the guide, impose
that R ≤ γR⊥ , where γ is the friction coefficient. Let x be the abscissa
of Po on the guide. By the theorems of the momentum and of the center of
mass,
M ẍ = M g sin α − R ,
2
1
2 M R ω̇
= RR .
Rolling without slipping implies Ṗo = ẋ = ωR. Therefore
R = 31 M g sin α ≤ γR⊥ = γM g cos α
=⇒
tan α ≤ 3γ.
Therefore the constraints resulting from the guide and the “rolling without
slipping” impose a limitation on α and the friction coefficient γ. Assuming
the latter being satisfied, v(C) = 0 and R · δC = 0. Therefore the resulting
constraints are workless. The potential U and the kinetic energy T of the
disk are
U = −M gx sin α + const,
T = 12 M ẋ2 + 41 M R2 ω 2 = 34 M ẋ2 .
In computing T we have used König’s theorem. From the above, taking into
account the expression of R ,
124
5 SYSTEMS DYNAMICS
Ėdisk = M ( 32 ẍ − g sin α)ẋ = 0.
Thus the energy is conserved.
3.2c Rod Moving in a Plane with a Fixed Extreme
A rigid material rod of length ℓ is constrained to move, subject to its weight,
in a vertical plane π with one of its extremities O kept fixed by a workless
cylindrical hinge. The second extreme Q is connected to a spring, of elasticity
constant k, whose fixed center C is on the horizontal through O and at distance
ℓ from O. Assume that the density function on the rod is ρ(P ) = P − O,
and that initially Q(0) = C and Q̇(0) = 0.
Compute the momentum of the system and the angular momentum with
respect to O, and write down the cardinal equations. Compute the period of
the small oscillations and the reactions due to the constraints.
Let ϕ be the angle between (C − O) and (Q − O) and, for the generic point
on the rod, set x = P − O. Every point P on the rod moves over a circle
centered at O of radius x and with of unit tangent t(ϕ) and unit normal n(ϕ).
Mass and position of the center of mass are computed from
ℓ
ℓ
1 2
m=
xdx = 12 ℓ2 ,
x2 dx = 13 ℓ3 ,
ℓ
x
=
Po = 32 ℓn.
o
2
0
0
The expression of the elastic force in terms of t and n is
F = −k(Q − C) = −kℓ [sin ϕt + (1 − cos ϕ)n] .
Let dw, dQ, and dK, denote the elemental weight, momentum, and angular
momentum with respect to O of the element of rod contiguous to P . In terms
of t and n,
dw = xg(cos ϕt + sin ϕn)dx,
dQ = ϕ̇x2 tdx,
dK = ϕ̇x3 n ∧ tdx.
By integration over the rod,
w = 21 gℓ2 (cos ϕt + sin ϕn),
Q = 31 ℓ3 ϕ̇t,
K = 14 ℓ4 ϕ̇e,
where e = n ∧ t. The moment of dw with respect to O is x2 g cos ϕedx, and
by integration the moment of the weight is 31 ℓ3 g cos ϕe. The moment of the
elastic force with respect to O is −kℓ2 sin ϕe. Next compute Ṗo = − 23 ℓϕ̇t and
P̈o = 32 ℓ ϕ̈t − ϕ̇2 n . Therefore the first cardinal equation is
1 3
3 ℓ (ϕ̈t
− ϕ̇2 n) = (−kℓ sin ϕ + 21 gℓ2 cos ϕ + Rt )t
+ (−kℓ(1 − cos ϕ) + 12 gℓ2 sin ϕ + Rn )n,
where Rt and Rn are the components of the reactions due to the constraints
along t and n respectively. In components,
3c Cardinal Equations
125
3k
3g
3
cos ϕ − 2 sin ϕ + 3 Rt ,
2ℓ
ℓ
ℓ
3k
3g
3
ϕ̇2 = − sin ϕ + 2 (1 − cos ϕ) − 3 Rn .
2ℓ
ℓ
ℓ
ϕ̈ =
(∗)
By the theorem of angular momentum,
4k
1
4g
1 4
ℓ ϕ̈ = ℓ3 g cos ϕ − kℓ2 sin ϕ =⇒ ϕ̈ =
cos ϕ − 2 sin ϕ.
4
3
3ℓ
ℓ
(∗∗)
Therefore
3
k
g
cos ϕ.
Rt (ϕ) = 2 sin ϕ −
ℓ3
ℓ
6ℓ
Multiply the second equation of (∗∗) by ϕ̇ and integrate in dt. Using the initial
conditions ϕ̇(0) = 0 gives
ϕ̇2 =
8k
8g
sin ϕ − 2 (1 − cos ϕ).
3ℓ
ℓ
From this and the second equation of (∗) one computes
11k
25g
3
sin ϕ + 2 (1 − cos ϕ).
Rn = −
3
ℓ
6ℓ
ℓ
3.3c Ring Sliding on a Material Spinning Segment
A homogeneous rod of length ℓ and mass M spins on a horizontal plane about
its midpoint Po with angular speed ω, starting from some angular speed ωo .
A point mass {P ; m} slides with no friction on the rod starting from Po .
Determine the velocity of P with respect to the rod in terms of its distance x
from Po .
The weight on P and the reactions are workless, so that the kinetic energy
is conserved and equals
T = 12 Iωo2 ,
where
I=
2
1
12 M L .
Therefore
(I + mx2 )ω 2 + mẋ2 = Iω 2 ,
i.e.,
ẋ2 = Iωo2 − (I + mx2 )ω 2 .
Since the angular momentum is conserved, (I + mx2 )ω = Iωo . Therefore
√
x
|ẋ| = ωo Im √
.
I + mx2
126
5 SYSTEMS DYNAMICS
3.4c A Material Segment with Extremities on Rectilinear Guides
A material homogeneous rigid rod of mass m and length ℓ moves with its
extremities A and B constrained on two smooth rectilinear guides intersecting
at O and forming an angle α =
AOB ∈ [ 21 π, π). The points A and B are
attracted to O by two elastic forces of elasticity constants kA and kB .
(i) Find the instantaneous center of motion C of the corresponding plane
rigid motion and determine the fixed and moving centrodes.
(ii) As the sole Lagrangian coordinate take the angle θ between (B − A) and
(O − A) and write down the cardinal equations taking C as the pole.
Justify such a choice of pole.
(iii) Assuming α = 21 π and kA = 2kB , determine the configurations of equilibrium and the corresponding reactions due to the constraints at the
points A and B. (See also §4.4c of the Complements of Chapter 8.)
(iv) Assuming further kA = kB = 0 and θ ≪ 1, prove that the system effects
small oscillations about the equilibrium configuration θ = 0, with period
T = 2π 2g/3ℓ.
RB
B
O
`B
®
RA
µ
Po
A
weight
`A
C
Fig. 3.2c.
3.4.1c Geometry of the Rigid Motion
The trajectories of A and B are the lines ℓA and ℓB respectively. By Chasles’s
theorem the center of instantaneous rotation C is at the intersection of the
= OAC
= 1 π, the quadrilatnormals to ℓA and ℓB respectively. Since OBC
2
eral of vertices O, A, C, B can be inscribed in the circle through O, A, B and
3c Cardinal Equations
127
diameter OC. The segment AB is a chord of such a circle, seen by O under
a constant angle α. Therefore the moving centrode is such a circle. Since
OC = ℓ/ sin α, the distance from C to O remains constant along the motion. Therefore the fixed centrode is the circle of center O and radius ℓ/ sin α
(§§11.3c–11.4c of Chapter 1.)
3.4.2c Second Cardinal Equation
First compute
OB = ℓ cos π − (α + θ) − OA cos(π − α),
OA = ℓ cos θ − OB cos(π − α),
OB =
ℓ
ℓ
sin θ; OA =
sin(α + θ),
sin α
sin α
BC =
ℓ
ℓ
cos θ; AC = −
cos(α + θ).
sin α
sin α
Introduce a fixed triad with origin in O, with e1 the unit vector of (B − O),
with e2 ascending vertically and e3 such that the triad Σ = {O; e1 , e2 , e3 } is
positive. With respect to Σ,
ℓ
(sin θe1 − cos θ e2 ) ,
sin α
ℓ
ℓ
ℓ
Po − O =
sin θ − cos π − (α + θ) e1 − sin π − (α + θ) e2 ,
sin α
2
2
ℓ
Po − C = cos(α + θ)e1 + (Po − C) · e2 e2
2
ℓ
ℓ
ℓ
cos θ − sin(α + θ) e2 ,
= cos(α + θ)e1 +
2
sin α
2
C −O =
where Po is the center of mass of the rod. The moments of the elastic forces
FA = −kA (A − O) and FB = −kB (B − O), with respect to C, are
ℓ 2
sin(α + θ) cos(α + θ)e3 ,
sin α
ℓ 2
sin θ cos θe3 .
(B − C) ∧ FB = kB
sin α
(A − C) ∧ FA = kA
The moment of the weight, still with respect to C, is
−mg(Po − C) ∧ e2 = − 21 ℓmg cos(α + θ)e3 .
128
5 SYSTEMS DYNAMICS
The moment of the reactions RA and RB with respect to C are zero. Therefore
the resultant moment of the external forces is
ℓ 2
M(e) =
kA sin(α + θ) cos(α + θ) + kB sin θ cos θ e3
sin α
− 12 ℓmg cos(α + θ)e3 .
The angular momentum with respect to C is computed from
K=
(P − C) ∧ Ṗo + ω ∧ (P − Po ) dµ(P )
rod
= m(Po − C) ∧ Ṗo +
(P − C) ∧ [ω ∧ (P − C)] dµ(P )
rod
+
(P − C) ∧ [ω ∧ (C − Po )] dµ(P )
rod
= m(Po − C) ∧ Ṗo + ω ∧ (C − Po )
(P − C) ∧ [ω ∧ (P − C)] dµ(P ).
+
rod
The velocity of C as part of the rigid motion of the rigid material rod is zero,
and it is given by Poisson’s formula
v(C) = Ṗo + ω ∧ (C − Po ) = 0.
This is the reason for having chosen C as the pole with respect to which
moments are computed. From these remarks, since also ω = −θ̇e3 ,
(P − C) ∧ [ω ∧ (P − C)] dµ(P ) = −θ̇Iℓ(C;e3 ) e3 .
K=
rod
By Huygens’s theorem,
Iℓ(C;e3 ) = Iℓ(Po ;e3 ) + mC − Po 2
1
ℓ2 ℓ 2
ℓ2
mℓ2 + m
+
cos θ sin(α + θ)
=
cos2 θ −
12
4
sin α
sin α
ℓ 2
1 2
= mℓ + m
cos α cos θ cos(α + θ).
3
sin α
Therefore the second cardinal equation for this system is
ℓ 2
1 2
d
θ̇ cos θ cos(α + θ)
mℓ θ̈ + m
cos α
3
sin α
dt
ℓ 2
=−
kA sin(α + θ) cos(α + θ)
sin α
1
+ kB sin θ cos θ + ℓmg cos(α + θ).
2
3c Cardinal Equations
129
3.4.3c First Cardinal Equation
The resultant of the external forces is
ℓ
kA sin(α + θ) cos α + kB sin θ − RA · e1 e1
sin α
ℓ
kA sin(α + θ) sin α + mg − RA · e2 − RB · e2 e2 .
−
sin α
F(e) = −
Moreover,
Po − O =
ℓ
ℓ
ℓ
sin θ + cos(α + θ) e1 − sin(α + θ)e2 ,
sin α
2
2
from which
d
(Po − O) − θ̇2 (Po − O).
dθ
Therefore the first cardinal equation takes the form
P̈o = θ̈
m θ̈
d
(Po − O) − θ̇2 (Po − O) = F(e) .
dθ
3.4.4c The Case α =
1
π
2
and kA = 2kB
From the second cardinal equation,
2
1
3 mℓ θ̈
= 21 ℓ2 kB sin 2θ − 21 ℓmg sin θ.
From this one finds the equilibrium configurations θ = 0 and cos θ = mg/2ℓkB .
For the configuration θ = 0, the reactions are computed from the first cardinal
equation RA = 0 and RB = mg+ℓkA . The second equilibrium configuration
holds only if kB ≥ mg/2ℓ. In such a case,
mg 2
,
RB = 2mg.
RA = ℓkB 1 −
2kB ℓ
3.4.5c The Case α =
1
π
2
and kA = kB = 0
This case yields θ̈ = −(g/L) sin θ, where L = 2ℓ/3. For θ ≪ 1 the rod effects
oscillations about its equilibrium configuration, of period T = 2π g/L.
3.5c Vibrations of a Flywheel
A homogeneous flywheel of mass M is mounted on springs of elasticity constant k and is equipped with a damping device as in Figure 3.3c. The flywheel spins about its axis, so that its center O can move only on the vertical
y-axis starting from its rest position Ω. The flywheel has a material impurity
130
5 SYSTEMS DYNAMICS
modeled by a small mass m placed at the point P∗ , at distance ρ from the
center O. The system has two degrees of freedom, and as Lagrangian coordinates we take the ordinate y of the center of the flywheel with respect to its
rest position y = 0, and the angle θ as in Figure 3.3c. The generic point P
of the flywheel has coordinates
P = P − O cos θu1 + (y + P − O sin θ) u2 ,
and it has velocity
Ṗ = −P − Oθ̇ sin θe1 + ẏ + P − Oθ̇ cos θ e2 .
The kinetic energy and the momentum of the system are
2T = (I + mρ2 )θ̇2 + (M + m)ẏ 2 + 2mρθ̇ẏ cos θ,
Q = mρθ̇ sin θe1 + (M + m)ẏ + mρθ̇ cos θ e2 .
P
µ
y
O
Po
P
viscous
fluid
Fig. 3.3c.
‘
The damping device, realized by a grid in a viscous fluid, opposes the motion with a force −cẏe2 for a given positive constant c depending on the fluid.
The spring opposes the motion with an elastic force −kye2 , and the force of
gravity of the flywheel is −(M + m)ge2 . An external motor keeps the flywheel
spinning at constant angular speed θ̇ = ω. At such constant angular speed,
the first cardinal equation, along the e2 -axis, takes the form
ÿ +
k
m
c
ẏ +
y=−
ρω 2 cos ωt − g.
M +m
M +m
M +m
Prove that as ω → ∞, the y-component of the velocity of the center of mass
tends to zero. Analyze the reactions due to the constraints, as functions of ω.
4c Mechanical Quantities of Rigid Systems
131
4c Mechanical Quantities of Rigid Systems
4.1c Disk Rolling without Slipping on a Line
A homogeneous disk of radius R and mass m rolls without slipping on a
horizontal rectilinear guide, as in §2.2c of the Complements of Chapter 2.
Compute the angular momentum KΩ and the kinetic energy T .
The constraint of rolling without slipping implies ω = −(ẏ1 /R)e3 . Moreover, by the theorem of the momentum, Q = mẏ1 e1 . Let σPo be the inertia
tensor with respect to a central principal triad with the axis ℓ(Po , e2 ) normal
to the disk. From §6.4 of Chapter 4,
⎞
⎛
1 0 0
1
σPo = mR2 ⎝ 0 1 0 ⎠.
4
0 0 2
Next,
KΩ =
(P − Po ) ∧ Ṗ dµ(P ) +
(P − Ω) ∧ Ṗ dµ(P )
= KPo + (Po − Ω) ∧ Q
= σPo (0, 0, −ẏ1/R)t + (y1 e1 + Re2 ) ∧ mẏ1 e1
= − 23 mRẏ1 e3 .
The kinetic energy is
T = 12 mṖo2 + 21 I33 ω2 = 43 mẏ12 .
4.2c Disk Rolling in a Plane
Assume now that the disk is constrained to move only in the vertical plane
y3 = 0. The system has three degrees of freedom, and as Lagrangian coordinates one may choose the coordinates (y1 , y2 ) of Po and the angle ϕ formed
by e1 with a fixed radius of the disk. With these choices,
Q = m (ẏ1 e1 + ẏ2 e2 ) ,
t
KPo = σPo (0, 0, ϕ̇) = 21 mR2 ϕ̇e3 .
For a fixed pole Ω,
KΩ = KPo + (Po − Ω) ∧ Q,
2T = mṖo2 + I33 ω2 .
(4.1c)
Therefore
KΩ = m
1
2R
2
ϕ̇ + ẏ1 y2 − y1 ẏ2 e3 ,
T = 21 m(ẏ12 + ẏ22 ) + 41 mR2 ϕ̇2 .
132
5 SYSTEMS DYNAMICS
4.3c Rigid Rod with Constrained Extremities
A material homogeneous rigid rod of length ℓ and mass m moves with its
extremities A and B constrained on two rectilinear orthogonal guides intersecting at Ω, as in the Cardano device (§11.3c of Chapter 1). Compute KΩ
and T . Choose Σ with origin in Ω and e1 and e2 along the guides. Choose
also S = {Po ; u1 , u2 , u3 } centered at Po , with u2 along (B − A) and u3 = e3 .
The system has one degree of freedom, and as Lagrangian coordinate choose
the angle ϕ = e
1 u1 . For these choices,
2Po = ℓ(cos ϕe1 + sin ϕe2 ),
2Q = mℓϕ̇ (− sin ϕe1 + cos ϕe2 ) .
Using the calculations of §6.1 of Chapter 4, compute the inertia tensor σPo of
the rod with respect to S. Then, using (4.1c) one computes 3KΩ = mℓ2 ϕ̇e3
and 6T = mℓ2 ϕ̇2 .
4.4c Rigid Rod Moving in a Plane
If the rod is constrained to the plane y3 = 0, the system has three degrees
of freedom, which are chosen as the coordinates (y1 , y2 ) of Po and the angle
ϕ = e
1 u1 . For such choices,
KΩ =
T =
2
1
12 mℓ ϕ̇e3 + m(y1 e1 + y2 e2 )
2
2
2 2
1
1
24 mℓ ϕ̇ + 2 m ẏ1 + ẏ2 .
∧ (ẏ1 e1 + ẏ2 e2 ),
Compute KΩ and T for a rod with one of the extremes constrained on a guide
and the other free in the plane y3 = 0.
4.5c The Double Pendulum
A homogeneous material rod of length ℓ and mass m is hinged at one of
its extremities to a fixed point Ω. The second extremity, denoted by P∗ , is
hinged to one extremity of a second homogeneous material rod, of length L
and mass M . The system is constrained to move in the plane y3 = 0. Compute
KΩ and T .
The system has two degrees of freedom for the choice of the angles ϕ and
θ as in Figure 4.1c. For these choices,
2Po = ℓ (sin ϕe1 − cos ϕe2 ) ,
P∗ = ℓ (sin ϕ e1 − cos ϕ e2 ) ,
2Qo = (2ℓ sin ϕ + L sin θ) e1 − (2ℓ cos ϕ + L cos θ) e2 .
Therefore
2Q = ℓϕ̇(m + 2M ) cos ϕ + LM θ̇ cos θ e1
+ ℓϕ̇(m + 2M ) sin ϕ + LM θ̇ sin θ e2 .
4c Mechanical Quantities of Rigid Systems
133
e2
e1
Ω
'
Po
µ
Qo
Fig. 4.1c.
The first rod is in precession with ω = ϕ̇e3 . Choose S fixed with the rod
with origin at Ω, and u1 as (Po − Ω). The second rod is in rigid motion with
characteristics Ṗ∗ and ω ′ = θ̇e3 . Choose a moving triad S ′ , fixed with the
second rod, with origin in P∗ and u′1 as (Qo − P∗ ).
The inertia tensors σΩ and σP∗ with respect to S and S ′ are
⎞
⎞
⎛
⎛
000
000
1 2⎝
1
0 1 0 ⎠,
σP′ ∗ = M L2 ⎝ 0 1 0 ⎠.
σΩ = mℓ
3
3
001
001
The angular momentum is computed from
(P − Ω) ∧ Ṗ dµ(P ) +
(P − Ω) ∧ Ṗ dµ(P )
KΩ =
first rod
second rod
= σΩ (0, 0, ϕ̇)t +
(P − P∗ ) ∧ Ṗ dµ(P ) + (P∗ − Ω) ∧ M Q̇o
second rod
= σΩ (0, 0, ϕ̇)t + σP∗ (0, 0, θ̇)t + M (Qo − P∗ ) ∧ Ṗ∗ + (P∗ − Ω) ∧ M Q̇o .
From the expression of Qo compute Q̇o and substitute it above to conclude
that
KΩ = 31 mℓ2 ϕ̇ + M L2 θ̇ e3 − 12 M ℓ 2ℓϕ̇ + L(ϕ̇ − θ̇) cos (θ − ϕ) e3 .
To compute T , regard the second rod as in rigid motion, with characteristics
Q̇o and θ̇e3 . Then
2
2T =
Ṗ dµ(P ) +
Ṗ 2 dµ(P )
first rod
second rod
2
Q̇o − θ̇e3 ∧ (P − Qo ) dµ(P )
= 13 mℓ2 ϕ̇2 +
= 31 mℓ2 ϕ̇2 +
second rod
2 2
1
12 M L θ̇ +
M Q̇2o .
134
5 SYSTEMS DYNAMICS
4.6c Cone of §10.1c of Chapter 1
Assume that the cone is homogeneous with mass m and that the plane π
is smooth. Assume that θ ∈ (0, π2 ) and compute the kinetic energy and the
angular momentum with respect to O. Moreover, assuming ω = const,
compute the reactions offered by the plane π and the vertical axis through O.
4.6.1c Mechanical Quantities of the Cone
In the elemental time dt the contact point C covers a circular arc of elemental
length ϕ̇dt R sin θ/ sin α in the fixed plane y3 = 0. In the same time interval,
the extreme C of the vector (C − Q) covers a circular arc of elemental length
ψ̇dt R in the moving plane x3 = h containing the base of the cone. Since the
cone rolls without slipping,
sin αψ̇ = ϕ̇ sin θ
(which defines ψ).
The vector ω in the coordinates of the moving triad S is
ω = −ωsign{ψ̇} [(cos ψu1 − sin ψu2 ) sin α + cos αu3 ] .
The inertia tensor σo with respect to S has been computed in §6.1c of the
Complements of Chapter 4 (see also the left Figure 6.1c). Therefore the
angular momentum with respect to O is given, in the coordinates of S, by
⎛
⎞
(R2 + 4h2 ) cos ψ sin α
3
K = σo ω = − sign{ψ̇}mω ⎝ −(R2 + 4h2 ) sin ψ sin α ⎠.
20
2R2 cos α
For the kinetic energy,
2T = ω t σo ω =
2
3
20 mω
2R2 + 4h2 − R2 sin2 α .
The trajectory of the center of mass Po is the circle of radius 34 h sin (α + θ) in
the fixed plane y3 = d − 34 h cos (α + θ). Since its angular velocity is ϕ̇e3 , one
computes
Ṗo = 34 ϕ̇h sin (α + θ) (− sin ϕe1 + cos ϕe2 ) .
Verify that the same expression for T could be derived using Corollary 4.2
and the previous expression for Ṗo . Use also the inertia tensor σPo computed
in §6.1.1c of the Complements of Chapter 4.
4.6.2c Cardinal Equations and Reactions
The component of the reaction in O normal to the vertical axis is
R⊥e3 = − 43 mϕ̇2 h sin (α + θ) (cos ϕ e1 + sin ϕ e2 ).
4c Mechanical Quantities of Rigid Systems
135
The moment of the reaction in C with respect to O has modulus RC h sin θ,
lies on the moving plane x3 = 0, and is normal to the projection of ω on such
a plane. Moreover, it forms a positive triad with such a projection and the
moving axis of u3 .
Denote by ψ the angle formed by u1 with the projection of ω on the
moving plane x3 = 0. Then the expression of the moment of RC with respect
to O, written in the coordinates of S, takes the form
−RC h sin θu,
where u = (sin ψu1 + cos ψu2 + 0u3 ) .
By analogous considerations, the moment of the weight with respect to O, in
the coordinates of S, is 34 hmg sin (α + θ)u. Therefore the resultant moment
with respect to O of the external forces in the coordinates of S is
M(e) = 43 hmg sin (α + θ) − RC h sin θ u.
From the expression of K in S,
K̇ =
dK
dt
3
+ ω ∧ K = 20
mω|ψ̇|(R2 + 4h2 ) sin αu
S
⎞
⎞ ⎛
⎛
cos ψ sin α
(R2 + 4h2 ) cos ψ sin α
3
mω2 ⎝ −(R2 + 4h2 ) sin ψ sin α ⎠ ∧ ⎝ − sin ψ sin α ⎠.
+ 20
cos α
2R2 cos α
The exterior product of the last two vectors equals (R2 − 4h2 ) sin α cos αu.
Therefore the second cardinal equation, in the coordinates of S, takes the
form
3
K̇ = 20
mω (R2 + 4h2 )|ψ̇| sin α + ω(R2 − 4h2 ) sin α cos α u
= 34 hmg sin (α + θ) − RC h sin θ u = M(e) .
In the particular case d = R, e.g., (α + θ) = 12 π,
ω =
|ψ̇|
|ϕ̇|
=
sin α
cos α
and
2 2
3
20 mω R
sin 2α =
3
4 hmg
− RC h cos α .
4.7c Square Plate with a Vertex on a Rectilinear Guide
A homogeneous material square plate of edge ℓ and mass m moves on a
horizontal plane x3 = 0, with one of the extremities A constrained on a
guide as in Figure 4.2c. The point A is attracted to a point O, fixed on the
horizontal guide, by a spring of elastic constant k. The plane x3 = 0 rotates
about the e3 -axis with constant angular velocity ωe3 . Resolve the motion.
136
5 SYSTEMS DYNAMICS
x2
Po
µ
O
x1
A
x3
Fig. 4.2c.
Introduce a fixed inertial triad Σ = {O; e1 , e2 , e3 } with origin at O and
a moving triad S = {O; u1 , u2 , u3 } with u3 = e3 , as in Figure 4.2c. The
system has two degrees of freedom, and as Lagrangian coordinates choose the
abscissa x of A on the guide and the angle θ. With these choices the center of
mass Po of the plate in the coordinates of S is
π
π
ℓ
ℓ
u1 + √ sin θ +
u2 .
Po = (x + √ cos θ +
4
4
2
2
The velocity vS (Po ) and the acceleration aS (Po ) of Po with respect to S are
π
π
ℓ
ℓ
u1 + √ θ̇ cos θ +
u2 ,
vS (Po ) = ẋ − √ θ̇ sin θ +
4
4
2
2
ℓ
π
π
ℓ
− √ θ̇2 cos θ +
u1
aS (Po ) = ẍ − √ θ̈ sin θ +
4
4
2
2
π
π
ℓ
− θ̇2 sin θ +
u2 .
+ √ θ̈ cos θ +
4
4
2
The plate moves with rigid motion with respect to S with characteristics
vS (Po ) and θ̇e3 . Therefore the velocity of a point P of the plate with respect
to S is
vS (P ) = vS (Po ) + θ̇e3 ∧ (P − Po ).
The deflection velocity of P as part of the rigid motion of S with respect to
Σ is ωe3 ∧ (P − O). Therefore the velocity of the generic point P of the plate
with respect to the fixed triad Σ is
vΣ (P ) = vS (Po ) + θ̇e3 ∧ (P − Po ) + ω e3 ∧ (P − O)
= vS (Po ) + ω + θ̇ e3 ∧ (P − Po ) + ωe3 ∧ (Po − O).
4c Mechanical Quantities of Rigid Systems
137
Denote by aS (P ), aT (P ), and aC (P ) respectively the acceleration of P with
respect to S, the transport acceleration of P as part of the rigid motion of S,
and the Coriolis acceleration
aS (P ) = aS (Po ) + θ̈e3 ∧ (P − Po ) + θ̇2 (P − Po ),
aC (P ) = 2ωe3 ∧ vS (Po ) + 2ω θ̇(P − Po ),
aT (P ) = −ω 2 (P − O).
Momentum and angular momentum, both computed with respect to Po , are
QPo = m ]vS (Po ) + ωe3 ∧ (Po − O)] ,
KPo = J (ω + θ̇)e3 ,
where J is the axial moment of the inertia of the plate with respect to the
axis ℓ (Po ; e3 ). Verify that the cardinal equations reduce to identities and do
not provide information on the motion. The kinetic energy and the potential
U are given by
2T = mvS2 (Po ) + J (ω + θ̇)2 + mω 2 Po2 + 2mωvS (Po ) · e3 ∧ (Po − O),
2U = −kx2 + const.
Verify that
∂T
= mvΣ (Po ) · u1 ,
∂ ẋ
∂T
∂ θ̇
= mvΣ (Po ) · e3 ∧ (Po − A) + J (ω + θ̇),
∂T
∂T
= mωvΣ (Po ) · u2 ,
= −m(ω + θ̇)vΣ (Po ) · (Po − A).
∂x
∂θ
4.8c Two Hinged Rods with Constrained Extremities
Two rigid homogeneous equal rods ΩM and M N , each of mass m and length
ℓ, are hinged at their common extremity M as in Figure 4.3c, and are constrained in a vertical plane. The first extremity Ω of the first rod is kept fixed
at Ω, and the second extremity N of the second rod is constrained on the line
segment line (C − Ω) of length 2ℓ and forming an angle π4 with respect to the
horizontal e1 . The point N is attracted by C by an elastic force of Hooke’s
constant k. The hinges in Ω and M and the constraint on N are workless.
The system has one degree of freedom, and as Lagrangian coordinate we take
the angle ϕ formed by (M − Ω) and e1 . Let also ψ = ϕ + π4 denote the angle
between (M − Ω) and (C − Ω). We have
P1 − Ω = 21 ℓ (cos ϕe1 − sin ϕ e3 ) ,
P2 − Ω = 12 ℓ (2 cos ϕ − sin ϕ)e1 + (cos ϕ − 2 sin ϕ)e3 ,
√
√
N − Ω = 2ℓ cos ψ e1 + e3
C − Ω = 2ℓ e1 + e3 .
138
5 SYSTEMS DYNAMICS
e3
C
N
Ω
'
e1
P2
P1
-mge3
M
Fig. 4.3c.
4.8.1c The Energy
The sum of the weights and the elastic force is a conservative field with
potential
U = −mg{(P1 − Ω) + (P2 − Ω)} · e3 − 21 kC − N 2 + const
= 12 mgℓ(3 sin ϕ − cos ϕ) − 2kℓ2 (1 − cos ψ)2 + const.
The kinetic energy TΩM of the rod ΩM is TΩM = 61 mℓ2 ϕ̇2 . The kinetic energy
TMN of the rod M N is computed by König’s theorem (3.5):
TMN =
=
2 2
1
24 mℓ ϕ̇
2 2
1
24 mℓ ϕ̇
+ 12 mṖ22
+ 12 mℓ2 ϕ̇2
5
4
− sin 2ϕ .
Combining these calculations gives the kinetic energy of the system in the
form
T = 12 mℓ2 ϕ̇2 53 − sin 2ϕ .
The total energy E = T − U is then
E = 12 mℓ2 ϕ̇2 53 − sin 2ϕ − 12 mgℓ(3 sin ϕ − cos ϕ)
+ 2kℓ2 (1 − cos ψ)2 + const.
4.8.2c Momentum and Angular Momentum
The center of mass Po has coordinates
Po − Ω = 41 ℓ [(3 cos ϕ − sin ϕ)e1 + (cos ϕ − 3 sin ϕ)e3 ] .
4c Mechanical Quantities of Rigid Systems
139
By Theorem 2.2 on the center of mass, Q = 2mṖo . Therefore
Q = − 21 mℓϕ̇ [(3 sin ϕ + cos ϕ)e1 + (3 cos ϕ + sin ϕ)e3 ] .
The angular momentum of the rod ΩM with respect to Ω is KΩM (Ω) =
1
2
3 mℓ ϕ̇e2 . The rod M N rotates about it center of mass P2 with angular velocity −ϕ̇e2 . Therefore the angular momentum of M N with respect to P2 is
1
KMN (P2 ) = − 12
mℓ2 ϕ̇e2 . The angular momentum of M N with respect to Ω
is computed from
KMN (Ω) = KMN (P2 ) + (P2 − Ω) ∧ QMN ,
where QMN is the momentum of the rod M N . Expanding these calculations,
one computes
K(Ω) = KΩM (Ω) + KMN (Ω) = mℓ2 ϕ̇e2 .
4.8.3c Cardinal Equations
The first cardinal equation is
1
2 mℓ ϕ̈ [(3 sin ϕ + cos ϕ)e1 + (3 cos ϕ + sin ϕ)e3 ]
+ ϕ̇2 [(3 cos ϕ − sin ϕ)e1 + (cos ϕ − 3 sin ϕ)e3 ]
= −2mge3 − k(N − C) + RΩ + RN ,
where the last two terms are the reactions exerted by the constraints at Ω
and N . The second cardinal equation with respect to the pole Ω takes the
form
mℓ2 ϕ̈e2 = 21 mgℓ(3 cos ϕ − sin ϕ)e2 + (N − Ω) ∧ RN .
If the motion were resolved, e.g., if one knew the Lagrangian function t → ϕ(t),
these equations would provide the reactions due to the constraints.
4.8.4c Resolving the Motion
Since the constraints are workless, the energy E is conserved. Taking the time
derivative of E gives
1 d
E =mℓ2 ϕ̈ 53 − sin 2ϕ + mℓ2 ϕ̇2 cos 2ϕ
ϕ̇ dt
− 12 mgℓ(3 cos ϕ + sin ϕ) + 4kℓ2 (1 − cos ψ) sin ψ = 0.
This differential equation in the unknown t → ϕ(t) is independent of the
reactions due to the constraints. Solving it, starting from some prescribed
initial configuration, for example N (0) = C, resolves the motion.
140
5 SYSTEMS DYNAMICS
4.9c Triangle Rotating about an Axis
Let ∆(ABC) be the triangle of §6.2c of the Complements of Chapter 4. The
triangle rotates about the axis u3 , while O remains fixed. The system has one
degree of freedom and as Lagrangian coordinate take the angle ϕ = u
1 e1 .
Momentum, kinetic energy, and angular momentum with respect to the pole
O in terms of ϕ are given by
1
Q = − 31 mhϕ̇ sin ϕe1 + cos ϕe2 ,
T = 12
mh2 ϕ̇2 ,
KO = σS ω = ϕ̇ (I13 e1 + I33 e3 ).
If ∆(ABC) is a right triangle with the right-angle vertex at B = O, then
KO =
1
12 mhϕ̇ (ae1
+ 2he3 ).
6
THE LAGRANGE EQUATIONS
1 Kinetic Energy in Terms of Lagrangian Coordinates
Let {M; dμ} be a material system whose mechanical state is described by N
Lagrangian coordinates q = (q1 , . . . , qN ). Since every point P ∈ {M; dμ} is
identified along its motion by the map (q, t) → P (q, t), the configuration of
the system is determined, instant by instant, by the map t → q(t) : R → RN .
The latter can be regarded as the motion of some abstract point in some
N -dimensional space, called configuration space. Since N is the least number
of parameters needed to identify uniquely the position of each point P of the
system, each of the maps {M; dμ} ∋ P → ∂P/∂qh, h = 1, . . . , N , is not
identically zero. Equivalently, we have the following lemma.
Lemma 1.1 Let ξ ∈ RN be fixed. Then ∇q P · ξ = 0 for all P ∈ {M; dμ} if
and only if ξ = 0.
The actual and virtual displacements dP and δP of P ∈ {M; dμ} are
dP =
∂P
∂P
dt,
dqh +
∂qh
∂t
δP =
∂P
δqh ,
∂qh
where dq and δq represent the actual and virtual displacement of the point q
in the configuration space. The velocity Ṗ of a point P of the system and the
“velocity” q̇ of its Lagrangian representation, are related by
Ṗ =
∂P
∂P
,
q̇h +
∂qh
∂t
from which
∂ Ṗ
∂P
=
.
∂ q̇h
∂qh
(1.1)
Using the expression for Ṗ in terms of q and q̇, compute
2
1
∂P
∂P
1
T =
Ṗ 2 dμ(P ) =
q̇h +
dμ(P )
2
2
∂qh
∂t
2
2
1
∂P
∂P
∂P
∂P
+
dμ(P ).
=
q̇h + 2
q̇h ·
2
∂qh
∂qh
∂t
∂t
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 6,
c Springer Science+Business Media, LLC 2011
141
142
6 THE LAGRANGE EQUATIONS
Set
Ahk (q; t) =
∂P ∂P
dμ(P ),
∂qh ∂qk
∂P ∂P
dμ(P ),
∂qh ∂t
∂P
q̇k dμ(P )
∂qk
Bh (q; t) =
(1.2)
and compute
∂P
q̇h
∂qh
2
dμ(P ) =
=
N
∂P
q̇h
∂qh
Ahk (q; t)q̇h q̇k .
h,k=1
Then the expression of the kinetic energy in terms of the Lagrangian velocity
q̇ can be given the form
T = T0 + T1 + T2 ,
(1.3)
where
T0 (q; t) =
1
2
∂P
∂t
2
dμ(P ),
and
T2 (q, q̇; t) =
T1 (q, q̇; t) = Bh (q; t)q̇h ,
N
1
Ahk (q; t)q̇h q̇k .
2 h,k=1
(1.4)
(1.5)
The term T2 is a homogeneous function of degree two in the variables q̇,
whereas T1 is a homogeneous function of degree one in q̇, and T0 is a homogeneous function of degree zero in q̇. For time-independent constraints,
T0 = T1 = 0 and the kinetic energy reduces to T2 .
Proposition 1.1 T2 (q, ξ; t) is a positive definite quadratic form in the variables (ξ1 , . . . , ξN ).
Proof. From the definition of T2 and Ahk compute
∂P
∂P
2
T2 (q, ξ; t) =
ξh
ξk dμ(P ) = |∇q P · ξ| dμ(P ).
∂qh ∂qk
Therefore T2 (q, ξ; t) ≥ 0 for all ξ ∈ RN . Moreover, for a fixed ξ T2 (q, ξ; t)
vanishes if and only if ∇q P · ξ = 0 for all P ∈ {M; dμ}. By Lemma 1.1 this is
impossible unless ξ = 0.
Corollary 1.1 det (Ahk ) = 0.
Corollary 1.2 The kinetic energy of a mechanical system subject only to
holonomic and fixed constraints is a positive definite quadratic form of the
Lagrangian velocities.
2 The Principle of Virtual Work
143
1.1 Kinetic Energy for Rigid-Body Motion
Choose a fixed triad Σ and a moving triad S = {O; e1 , e2 , e3 } whose axes
are principal axes of inertia for the material system {M; dμ}. As Lagrangian
coordinates take the coordinates of the center of mass Po in Σ and the Euler’s
angles ϕ, ψ, θ of S with respect to Σ. Then1
2T = mṖo2 + ωt σω,
m = dμ(P ),
(1.6)
where σ is the inertia tensor of the system with respect to S and ω is the
angular velocity of the rigid motion written in the coordinates of S. The first
term is a quadratic form of the coordinates of Po , whereas the second term is
a quadratic form of the Lagrangian velocities ϕ̇, ψ̇, θ̇. Indeed,
ω t σω = I1 (ϕ̇ sin ψ sin θ + θ̇ cos ψ)2
+ I2 (ϕ̇ cos ψ sin θ − θ̇ sin ψ)2
(1.7)
+ I3 (ϕ̇ cos θ + ψ̇)2 ,
where Ii are the axial moments of inertia of the system with respect to the
principal axes of S. If two of the axial moments of inertia coincide, say, for
example, I1 = I2 = I, then the kinetic energy takes the form
2T = I(ϕ̇2 sin2 θ + θ̇2 ) + I3 (ϕ̇ cos θ + ψ̇)2 .
(1.8)
A rigid body with such a property is a gyroscope, and the axis of u3 is the
gyroscopic axis.
2 The Principle of Virtual Work
The system {M; dμ} is subject to constraints and is acted upon by a distribution of forces
(P, Ṗ ; t) −→ [f (P, Ṗ ; t) + r(P, Ṗ ; t)]dμ(P ).
The function f is the pointwise distribution of internal and external forces per
unit mass, whose functional form is assumed to be known. The function r is the
pointwise distribution, per unit mass, of the reactions due to the constraints,
1
Corollary 4.2 of Chapter 5. The components of ω in S, in terms of ϕ̇, ψ̇, and θ̇,
are computed in (9.4) of Chapter 1.
144
6 THE LAGRANGE EQUATIONS
and its functional form is in general unknown. The elementary work done by
the forces f and the reactions r for an elemental virtual displacement δq is
f (P, Ṗ ; t) + r(P, Ṗ ; t) · δP dμ(P )
δL =
∂P
δqh dμ(P )
f (P, Ṗ ; t) + r(P, Ṗ ; t) ·
=
∂qh
∂P
∂P
dμ(P ) + δqh r ·
dμ(P ).
= δqh f ·
∂qh
∂qh
The function
(q, q̇; t) → Φh (q, q̇; t) =
f·
∂P
dμ(P )
∂qh
(2.1)
is the hth component, in configuration space, of the Lagrangian resultant of
the distribution of forces f (P, Ṗ ; t) acting on the system. The quantity
∂P
δΛ = δqh r ·
dμ(P )
∂qh
is the virtual work due to the reactions r(P, Ṗ ; t). With this notation
δL = Φh δqh + δΛ.
(2.2)
If f (P, Ṗ ; t) is conservative, there exists a function P → U (P ) such that
f = ∇U . In such a case Φh takes the form
∂
∂P
dμ(P ) =
V,
(2.1)V
Φh = ∇U ·
∂qh
∂qh
where the function
(q; t) −→ V (q; t) =
U dμ(P )
is the potential of the system. Therefore for conservative distributions of forces
f (P, Ṗ ; t), the Lagrangian components Φh of the force are the derivatives of
the potential with respect to qh . Equivalently,
Φ = (Φ1 , . . . , ΦN ) = ∇q V
and
Φh δqh = ∇q V · δq = δV.
With this notation, (2.2) can be written as
δ(L − V ) = δΛ.
(2.2)V
The principle of virtual work stipulates that instant by instant, the work
δΛ done globally by all the reactions due to the constraints is zero for every
virtual displacement δq compatible with the constraints [101].2
2
More generally, to include the case of unilateral constraints, one might require
that δΛ have a sign.
3 The Lagrange Equations
145
Remark 2.1 The principle of virtual work is a condition on the nature of
the constraints and not on the particular motion of the system. Indeed, it is
required that δΛ be zero for every virtual displacement δq, not being restricted
to a possible Lagrangian trajectory of the motion. Constraints satisfying such
a requirement are smooth.
Consider now d’Alembert’s principle in the form [36]
P̈ − f (P, Ṗ ; t) − r(P, Ṗ ; t) dμ(P ) = 0
∀P ∈ {M; dμ}.
Multiplying this by δP and integrating in dμ(P ) gives
∂P
∂P
δqh (P̈ − f ) ·
dμ(P ) = δqh r
dμ(P ).
∂qh
∂qh
The last integral is the elemental work δΛ done by the reactions. If the constraints are smooth, the right-hand side is zero for all virtual displacement δq.
Thus
∂P
δqh (P̈ − f ) ·
dμ(P ) = 0.
∂qh
Since δq is arbitrary, this gives the N equations
∂P
(P̈ − f ) ·
dμ(P ) = 0,
h = 1, . . . , N.
(2.3)
∂qh
These are necessary conditions to be satisfied by the motion of any system
subject to holonomic constraint that obey the principle of virtual work.
3 The Lagrange Equations
For such a system, start from the kinetic energy
1
Ṗ 2 (q, q̇; t)dμ(P )
T =
2
and take the time derivative with respect to q̇h . Taking into account the second
equation of (1.1),
∂T
∂ Ṗ
∂P
= Ṗ ·
dμ(P ) = Ṗ ·
dμ(P ).
∂ q̇h
∂ q̇h
∂qh
Next take the time derivative with respect to time and apply the principle of
virtual work in the form (2.3) to obtain
d ∂T
∂P
∂ Ṗ
= P̈ ·
dμ(P ) + Ṗ ·
dμ(P )
dt ∂ q̇h
∂qh
∂qh
∂ 1
∂P
Ṗ 2 dμ(P )
dμ(P ) +
= f·
∂qh
∂qh 2
∂
T,
= Φh (q, q̇; t) +
∂qh
146
6 THE LAGRANGE EQUATIONS
where Φh is the hth component in configuration space of the Lagrangian
resultant of all the forces, internal and external, acting on the system. These
N equations, rewritten in the form [101]
d ∂T
∂T
−
= Φh ,
dt ∂ q̇h
∂qh
h = 1, . . . , N,
(3.1)
are the equations of Lagrange. They form a system of N ordinary differential equations of the second order, in the unknown Lagrangian coordinates
(q1 , . . . , qN ). The system is of rank N in the variables q̈h . Indeed, using in (3.1),
the expression (1.3)–(1.5) of the kinetic energy gives, for all h = 1, . . . , N ,
Ahk q̈k = −Ȧhk q̇k − Ḃh +
∂T
def
+ Φh = gh (q, q̇; t),
∂qh
where gh are known functions of (q, q̇; t) and are independent of q̈. Since
det(Ahk ) = 0 and Ahk (q; t) is independent of q̇ and q̈, this in turn implies
q̈h = fh (q, q̇; t),
h = 1, . . . , N,
(3.1)′
for some known functions fh of the arguments (q, q̇; t) only.
Since the rank of the system (3.1) equals the number of degrees of freedom,
the Lagrange equations determine unambiguously the motion of the system,
starting from some prescribed initial data.
3.1 Fixed Constraints and the Energy Integral
If the constraints are time-independent, the kinetic energy does not have an
explicit dependence on time and reduces to T2 (q; q̇). By taking the derivative
with respect to time, we obtain
d
∂T
∂T
T =
q̇h +
q̈h .
dt
∂qh
∂ q̇h
From (1.5),
2T = Ahk (q)q̇h q̇k =
∂T
q̇h .
∂ q̇h
(3.2)
Taking the time derivative yields
2
∂T
d ∂T
d
T = q̇h
+ q̈h
.
dt
dt ∂ q̇h
∂ q̇h
From this last formula subtract the previously obtained expression for
Taking into account the Lagrange equations (3.1) gives us
d
dqh
d ∂T
∂T dqh
T =
= Φh
.
−
dt
dt ∂ q̇h
∂qh dt
dt
d
dt T .
The differential form of this equality is dT = dL, where dL = Φh dqh is the
elemental work done by the Lagrangian resultant Φ of the external and internal
forces during a displacement dq in configuration space. In integral form, one
obtains the energy integral T − L = const.
4 Lagrangian Function for Conservative Fields
147
3.2 Motion along Geodesics
A point P not subject to any external forces moves on a smooth, workless
surface S of parametric equations (u, v) → P (u, v). Choosing q = (u, v) as
Lagrangian coordinates, one computes
Ṗ 2 = q̇ t
AB
BC
q̇ = ∆ (q, q̇) ,
where the functions q → A(q), B(q), C(q) are the coefficients of the first fundamental form of the surface S (§§3, 3c, of Chapter 2). Since there are no
external forces, the energy coincides with the kinetic energy, up to a constant,
and the Lagrange equations (3.1) take the form
d ∂
∂
∆−
∆ = 0,
dt ∂ q̇i
∂qi
i = 1, 2.
(3.3)
√
Dividing these by 2 ∆ gives
d ∂ √
∂
d
∂ √
1
∆ ∆,
∆−
∆ = − 3/2
dt ∂ q̇i
∂qi
4∆ ∂ q̇i dt
i = 1, 2.
However, by the energy integral, ∆(q, q̇) is constant along the motion. Therefore q is a solution of (3.3) if and only if it is a solution of the system
d ∂ √
∂ √
∆−
∆ = 0,
dt ∂ q̇i
∂qi
i = 1, 2.
(3.4)
These are the parametric equations of the geodesics on the surface S. Thus a
force-free point on a surface S moves along a geodesic (§4 of Chapter 2 and
§1.4c of Chapter 9).
4 Lagrangian Function for Conservative Fields
If the distribution of forces f (P, Ṗ ; t) is conservative, the components Φh are
given by (2.1) and (2.1)V . Since V is independent of q̇, the Lagrange equations
(3.1) may be rewritten in the form [101]
d ∂(T + V ) ∂(T + V )
−
= 0,
dt
∂ q̇h
∂qh
h = 1, . . . , N.
The function
def
(q, q̇; t) −→ L(q, q̇; t) = T (q, q̇; t) + V (q; t)
(3.1)V
148
6 THE LAGRANGE EQUATIONS
is the Lagrangian function of the mechanical system. The differential system
(3.1)V is of rank N , since
2
2
∂ T
∂ L
= det
= det(Ahk ) = 0.
det
∂ q̇h q̇k
∂ q̇h ∂ q̇k
By taking the derivative of L with respect to time and taking into account
(3.1)V , we obtain
d
∂L
∂L
∂L
L = q̇h
,
+ q̈h
+
dt
∂qh
∂ q̇h
∂t
Add these two relations to obtain
∂L
d
∂L
d
q̇h
L=
,
+
dt
dt
∂ q̇h
∂t
0 = q̇h
i.e.,
∂L
d ∂L
− q̇h
.
dt ∂ q̇h
∂qh
d
∂L
H=−
,
dt
∂t
(4.1)
where we have set
∂L
− L.
(4.2)
∂ q̇h
If the constraints are time independent, both L and H are explicitly independent of t. In such a case (4.1) implies
H(q, q̇; t) = q̇h
H(q, q̇) = const
along the motion (fixed constraints).
(4.3)
Thus H(q, q̇) is an integral of motion. Such an integral, however, does not give
new information on the motion of the system. Indeed from (3.2),
H=
∂T
q̇h − (T + V ) = 2T − T − V = T − V.
∂ q̇h
Thus (4.3) coincides precisely with the energy integral.
4.1 Further Integrals of Motion and Kinetic Momenta
A Lagrangian coordinate qh is called cyclic or ignorable if the Lagrangian L
is independent of qh . Then Lqh = 0, and (3.1)V gives
∂L
(q, q̇; t) = const
∂ q̇h
along the motion.
This then is an integral of motion. The quantities
ph =
∂L
∂T
=
= Ahk (q; t)q̇k + Bh (q; t),
∂ q̇h
∂ q̇h
h = 1, . . . , N,
(4.4)
constants or not, are called kinetic momenta and have the dimensions of
a momentum. In the case of a single unconstrained point {P ; m}, the kinetic momentum is precisely the momentum of {P ; m}. Indeed, by taking the
Lagrangian coordinates as the Cartesian coordinates, one has Ahk = δhk and
ph = mẋh . In terms of kinetic momenta, the hth Lagrange equation in (3.1)V
can be written as ṗh = Lqh . Thus if qh is cyclic or ignorable, then ph = const
is an integral of the motion.
5 Lagrangian and Hamiltonian
149
5 Lagrangian and Hamiltonian
The function H introduced in (4.2) is Hamilton’s function or the Hamiltonian
of the system. Even though it has been derived from the Lagrangian L and
the Lagrange equation of motion, its relation to the Lagrangian is of a mathematical nature and is independent of the motion. First regard the Lagrangian
as a function of two N -tuples of independent variables
(q, ξ) −→ L(q, ξ; t),
q, ξ ∈ RN ,
t fixed.
Then introduce two new N -tuples of independent variables (q, η), by setting
ηh =
∂L(q, ξ; t)
= Ahk (q; t)ξk + Bh (q; t),
∂ξh
h = 1, . . . , N.
(5.1)
The variables η are uniquely determined by the two N -tuples (q, ξ). Conversely, since det(Ahk ) = 0, the equations (5.1) permit one to compute the
variables ξ in terms of the two N -tuples (q, η). In this way (5.1) can be regarded as a mutual transformation between two pairs of N -tuples of independent variables (q, ξ) and (q, η), each of which can be taken as a set of 2N
independent variables. Choose (q, η) as independent variables and regard ξ
as functions of these. The Hamiltonian is defined, independently of motion,
as [72]
R2N ∋ (q, η) −→ H(q, η; t) = η · ξ − L(q, ξ; t),
where the variables ξ are meant to be computed in terms of (q, η) by the transformations (5.1). From this definition, taking into account the independence
of η and q, compute
∂L(q, ξ; t) ∂L(q, ξ; t) ∂ξk
∂H(q, η; t)
∂ξk
= ηk
−
−
,
∂qh
∂qh
∂qh
∂ξk
∂qh
∂ξk
∂H(q, η; t)
∂ηk
∂L(q, ξ; t) ∂ξk
=
ξk + ηk
−
.
∂ηh
∂ηh
∂ηh
∂ξk
∂ηh
5.1 Canonical Form of the Lagrange Equations
In the framework of mechanics the variables ξ are identified with the Lagrangian velocities q̇, and the variables η are the kinetic momenta p. Along
the motion, the Hamiltonian H(q, p; t) is a function of the Lagrangian coordinates q and the momenta p. Since q satisfy the Lagrange equations (3.1)V ,
the previous differentiation formulas can be rewritten as
∂H
∂ q̇k
∂ q̇k
∂ q̇k
∂L
∂L ∂ q̇k
= pk
−
−
= pk
− ṗh − pk
= −ṗh ,
∂qh
∂qh
∂qh
∂ q̇k ∂qh
∂qh
∂qh
∂H
∂ q̇k
∂ q̇k
∂pk
∂L ∂ q̇k
∂L ∂ q̇k
=
q̇k + pk
−
= q̇h + ṗk
−
= q̇h .
∂ph
∂ph
∂ph
∂ q̇k ∂ph
∂ph
∂ q̇k ∂ph
150
6 THE LAGRANGE EQUATIONS
Therefore the Lagrange equations (3.1)V in terms of the variables (q, p) take
the equivalent form
q̇h =
∂H(q, p; t)
,
∂ph
ṗh = −
∂H(q, p; t)
,
∂qh
h = 1, . . . , N.
(5.2)
This is the canonical form of the Lagrangian system (3.1)V . The variables p
and q are called conjugate or canonical variables [74].
5.2 Partial and Total Time Derivative of the Hamiltonian
Multiply the first equation of (5.2) by ṗh and the second by q̇h and subtract
the expression so obtained. This gives
0=
∂H
∂H
dH ∂H
−
.
q̇k +
ṗk =
∂qk
∂pk
dt
∂t
Therefore, along the motion, the total derivative of H q(t), p(t); t and the
partial derivative of H(q, p; t) with respect to time are the same. For timeindependent constraints, the Lagrangian and the Hamiltonian are both explicitly independent of time. In such a case,
d
∂
H= H=0
dt
∂t
=⇒
t → H q(t), p(t) = const.
This gives back the energy integral, since for fixed constraints, the Hamiltonian
coincides with the total energy of the system.
6 On Motion in Phase Space
As a summary, consider the motion of a system subject only to holonomic
constraints, satisfying the principle of virtual work. Having introduced N
Lagrangian parameters q, the movement is determined by the system (3.1)
starting from a set of initial data. If in particular the external forces are conservative, denoting by L(q, q̇; t) the Lagrangian of the system, the movement
is determined by the N differential equations of the second order
d ∂L(q, q̇; t) ∂L(q, q̇; t)
−
= 0,
dt
∂ q̇h
∂qh
h = 1, . . . , N,
(6.1)
starting from some given initial conditions. In the Hamiltonian formalism,
one regards the kinetic momenta ph as N new independent variables. Having established that the two sets of variables (q, q̇) and (q, p) are mutually
equivalent, the Hamiltonian
H(q, p; t) = ph q̇h − L(q, q̇; t),
ph =
∂L
,
∂ q̇h
(6.2)
7 Equilibrium Configurations
151
satisfies the canonical system (5.2). The system of N ordinary differential
equations of the second order (6.1) is equivalent to the system of 2N ordinary
differential equations of the first order (5.2). The space R2N , where the first
N -tuple of variables represent the Lagrangian configurations and where the
second N -tuple of variables are the kinetic momenta, is called phase space,
and the solutions of (5.2) can be regarded as curves t → [q(t), p(t)] in phase
space. From a mathematical point of view, determining the motion of the
system reduces to the integration of (6.1) or equivalently (5.2), starting from
some given initial data.
6.1 Integrals of Motion
The general integral of (5.2) is a set of 2N functions t → q(t; χ), p(t; π) depending upon 2N parameters (χ, π) and satisfying (5.2) identically as the
parameters range within their domain of definition. An integral of the system
(5.2) is a nontrivial relation f q(t), p(t); t = const, satisfied identically by
all solutions of (5.2). Equivalently, an integral is a smooth function f (q, p; t),
that remains constant along any orbits of a solution of the system (5.2). The
constant depends, in general, on the particular orbit along which f is being
computed. Such a relation is nontrivial in the sense that ∇q,p f (q, p; t) > 0
within its domain of definition.
For fixed constraints, the energy is an integral of motion, since it remains constant along solutions of (5.2). However, the value of such a constant depends on the particular orbit, or equivalently it depends on the initial
conditions.
If the Lagrangian L does not depend on one of the coordinates qh , then
also the corresponding Hamiltonian is independent of qh . If so, the variable qh
is called cyclic or ignorable. If qh is cyclic, (5.2) implies that the corresponding
kinetic momentum is constant along the motion. Thus ph = const is an integral
of motion.
7 Equilibrium Configurations
The problem of equilibrium for a mechanical system consists in finding those
configurations qo for which the system (3.1) with initial conditions q(0) = qo
and q̇(0) = 0 admits the only solution q = qo .
Proposition 7.1 Assume that the constraints are fixed and satisfy the principle of virtual work. Then (qo ; to ) is an equilibrium configuration for {M; dμ}
if and only if Φh (qo ; t) = 0, for all h = 1, . . . , N .
Proof (Sufficiency). Since the constraints are time-independent, the equilibrium condition for (3.1) takes the form
Ahk q̈k = Φh (q; t) + 12 Aℓk;qh q̇k q̇ℓ − Ahk;qℓ q̇k q̇ℓ ,
q(to ) = qo , q̇(to ) = 0.
152
6 THE LAGRANGE EQUATIONS
This has a unique solution, and since q = qo solves the system, it must be the
only solution.
Proof (Necessity). If the system is in equilibrium in the position qo , then Ṗ
and P̈ are identically zero for all P ∈ {M; dμ}. If δP is a virtual displacement,
from (2.2)V and the principle of virtual work in the form (2.3),
f · δP dμ(P ) = −δΛ = 0,
(7.1)
where δΛ is the infinitesimal virtual work done by the reactions due to the constraints. In terms of the correspondent Lagrangian coordinates, this becomes
∂P
f (P ; t) · δP dμ(P ) = δqh f (P ; t) ·
dμ(P )
∂qh
= δqh Φh (q; t) = 0
for all arbitrary virtual δq.
Remark 7.1 The interest of (7.1) is that it can be seperated from the reactions r, e.g., the equilibrium configurations can be determined without actually knowing the forces reacting to the constraints.
Remark 7.2 If the distribution f of the forces applied to the system is conservative with potential U (q), then the equilibrium positions, if any, are determined by ∇q U = 0.
Remark 7.3 For a single point P acted upon by a force F and otherwise
unconstrained, the equilibrium is realized if F · δP = 0 for all δP . Therefore
P is in equilibrium if and only if F = 0. If P is constrained on a smooth fixed
surface, the equilibrium is realized if F · δP = 0 for all δP compatible with
the constraint. Therefore P is in equilibrium if and only if F(P ) is normal to
the surface constraining P .
7.1 Equilibrium for Rigid Systems
Let {M; dμ} be rigid and let S be a triad fixed with the rigid body and moving
with it. A virtual displacement of a point P ∈ {M; dμ} is of the form
δP = δO + δω ∧ (P − O),
where O is an arbitrary point of S. Putting this in (7.1) gives
δO · f (P, t)dμ(P ) = −δω · (P − O) ∧ f (P, t)dμ(P ).
8 Canonical Form of the n-Body Problem
153
If the rigid motion is unconstrained, there are no reactions and the virtual
displacements δO and δω are arbitrary. Therefore
f (P, t)dμ(P ) = 0,
(P − O) ∧ f (P, t)dμ(P ) = 0.
(7.2)
These characterize the equilibrium configurations of a rigid unconstrained
body.
8 Canonical Form of the n-Body Problem
Consider a system of n point masses {Pi ; mi }, i = 0, 1, . . . , (n − 1), subject
to their mutual gravitational forces and otherwise unconstrained. The system
has 3n degrees of freedom and the Hamiltonian H depends on some choice
of 3n independent Lagrangian coordinates. There is no natural or preferred
choice of such coordinates, each exhibiting advantages and shortcomings. In
this and the next two sections we make three different choices of Lagrangian
coordinates, and in terms of them write the Hamiltonian and the corresponding canonical system.
First, as Lagrangian coordinates choose the Cartesian coordinates of each
Pi , with respect to a fixed inertial triad Σ = {Ω; e1 , e2 , e3 }, e.g., qj = Pj − Ω,
for j = 0, 1, . . . , n − 1. The kinetic energy and the potential V in terms of
these Lagrangian parameters are given by
2T =
n−1
j=0
mj q̇2j ,
2V =
n−1
i,j=0
j=i
γ
mi m j
.
qj − qi
(8.1)
Having introduced the Lagrangian L = T + V , the equations (3.1)V yield
mj q̈j = −
n−1
i=0
i=j
γ
qj − qi
mi m j
,
qj − qi 2 qj − qi
j = 0, 1, . . . , n − 1.
(8.2)
The kinetic momenta p are the components of the vectors pj = ∂L/∂ q̇j , and
in terms of these, the Hamiltonian is computed from
H=
n−1
j=0
pj · q̇j − L = T − V =
1 2 1 n−1
1 n−1
mi m j
.
pj −
γ
2 j=0 mj
2 i,j=0 qj − qi
j=i
Thus the canonical form of (8.2) is
1
pj ,
mj
n−1
qj − qi
mi m j
,
ṗj =
γ
qj − qi 2 qj − qi
i=0
q̇j =
i=j
j = 0, 1, . . . , n − 1.
154
6 THE LAGRANGE EQUATIONS
Summing the second of these over the subscript j gives the integral of motion
d n−1
ṗj = 0
dt j=0
=⇒
n−1
mj Ṗj = Q = const.
j=0
This is precisely the conservation of momentum. Next we interpret such an integral of motion in terms of cyclic variables, for a suitable choice of Lagrangian
coordinates.
9 Lagrangian Coordinates Relative to {Po; mo}
As Lagrangian parameters now choose the coordinates of {Po ; mo } with respect to the inertial triad Σ and the coordinates of the remaining points Pi
relative to Po , i.e.,
qo = Po − Ω,
qj = Pj − Po ,
j = 1, . . . , n − 1.
(9.1)
In terms of these, the kinetic energy and the potential are given by
2T =
n−1
j=0
= q̇2o
2V =
mj (Ṗj − Ṗo )2 + mṖo2 + 2mj (Ṗj − Ṗo ) · Ṗo
n−1
j=1
n−1
γ
i,j=0
j=i
=
n−1
j=1
γ
mj +
n−1
j=1
mj q2j + 2q̇o
n−1
mj q̇j ,
j=1
m i mj
(Pj − Po ) − (Pi − Po )
n−1
mi m j
mo m j
+
.
γ
qj
qj − qi
i,j=1
j=i
It follows that L is independent of the first three coordinates of qo . Therefore
these are cyclic variables and give rise to the integrals of motion ∇q̇o L = 0.
To interpret such integrals of motion, compute the kinetic momenta relative
to the Lagrangian coordinates in (9.1). This gives
∇q̇j L = pj = mj Ṗj
for j = 1, . . . , n − 1,
n−1
mi Ṗi for j = 0.
∇q̇o L = po =
(9.2)
i=0
Thus for j = 1, . . . , n − 1, the components of the kinetic momenta relative to
the Lagrangian variables Pj −Po coincide with the kinetic momenta one would
obtain by taking as Lagrangian coordinates the absolute coordinates of the
points Pj with respect to the inertial triad Σ. However, for j = 0, the components of the kinetic momenta relative to the variables Po are the components
of the momentum of the system. Thus the integral of motion ∇q̇o L = 0 is
10 Inertial Systems and Further Integrals of Motion
155
precisely the conservation of momentum. To write the Hamiltonian in terms
of the variables q and p, first observe that
n−1
n−1
n−1
mo q̇o =
pj .
mj Ṗj = po −
mj Ṗj −
j=1
j=1
j=0
Then compute
2 n−1
n−1
1 2
1
H=
po −
pj
pj +
2mo
j=1 2mj
j=1
−
n
1 n−1
mo m j
1
mi mj
−
.
γ
γ
2 j=1 qj
2 i,j=0 qj − qi
(9.3)
j=i
From this, one verifies that H is independent of qo , and writes down the
corresponding Hamiltonian system
n
1
po −
q̇o =
pj ,
mo
j=1
n
1
1
po −
pi ,
pj +
q̇i = −
mo
mi
(9.4)
j=1
ṗo = 0,
n
γmo mi qi
qj − qi
γmi mj
,
ṗi =
−
2
2
2
qi qi
j=0 qj − qi qj − qi
j=i
where in the second and last equations, the index i ranges from 1 to n − 1.
The choice (9.1) of Lagrangian coordinates is due to Poincaré, and (9.4) is
called the Poincaré canonical form of the n-body problem [127].3
10 Inertial Systems and Further Integrals of Motion
Having fixed the inertial triad Σ, choose a new triad Σo whose axes are
parallel to those of Σ and with origin at the center of mass Ωo of the n
bodies. By conservation of momentum, the center of mass translates with
constant velocity with respect to any inertial reference system. It follows that
Σo is inertial and the center of mass is at rest with respect to Σo . This, in
turn, implies that since there are no external forces acting on the system,
the angular momentum K, taken in Σo and with respect to any point, fixed
or moving, is conserved (§2 of Chapter 5). Computing it with respect to the
central body Po gives
K=
n−1
j=0
3
(Pj − Po ) ∧ mj Ṗj = const.
The choice (9.1) is motivated by a class of linear canonical transformations. See
§5.8.1c of the Complements of Chapter 10.
156
6 THE LAGRANGE EQUATIONS
The system has also the energy integral H = const. Denote by q the
Lagrangian coordinates introduced in (9.1) and by p the corresponding kinetic
momenta, and identify (q, p) as a point in phase space R2N , where N = 3n.
With this symbolism, we rewrite the energy integral and the integral of angular momentum as
H(q, p) = (const)o ,
Kj (q, p) = (const)j ,
j = 1, 2, 3,
where Kj are the components of K in Σo . These identify locally and implicitly
four surfaces in phase space, and the motion takes place on the intersection
of these surfaces. Along the motion,
∇H · q̇, ṗ = 0,
∇Kj · q̇, ṗ = 0,
j = 1, 2, 3.
Thus at all times, the vectors ∇H and ∇Kj are coplanar and lie on the hyperplane normal to the trajectory of (q, p) in phase space. It follows that there
exist real-valued parameters λj such that
∇H = λj ∇Kj
along the motion.
(10.1)
This is the method of Lagrange multipliers, where the parameters λj are, in
general, functions of time. Summarizing, with respect to the inertial system
Σo , the n-body problem has the following integrals of motion:
Q = 0,
K̇ = 0,
Ω̇o = 0,
Ḣ = 0,
conservation of momentum,
conservation of angular momentum,
the center of mass is at rest in Σo ,
conservation of energy.
The first three equations represent nine scalar integrals of motion. Complemented with the last one, the n-body problem has ten integrals of motion.
It turns out that these are the only integrals possible, e.g., any integral of
motion of the n-body problem is a linear combination of these [16, 121, 127].
11 The Planar 3-Body Problem (Lagrange [99])
Assume that n = 2 and that the three points Po , P1 , P2 move in the inertial plane x3 = 0. The inertial reference system Σo being fixed, we choose
Lagrangian coordinates as in (9.1). Let (xo , yo ) denote the coordinates of Po
with respect to Σ, and let (Pi − Po ) = (xi , yi ), i = 1, 2, be the coordinates of
Pi relative to the central body {Po ; mo }. The Lagrangian coordinates q and
the corresponding kinetic momenta p are
q = (xo , yo , x1 , y1 , x2 , y2 ),
p = (mj Ṗj , m1 Ṗ1 , m2 Ṗ2 ).
12 Configuration of the Three Bodies
157
The conservation of angular momentum in this context takes the form
(11.1)
xi (mi Ṗi ) · e2 − yi (mi Ṗi ) · e1 = const.
i=1,2
This and (10.1) imply that for ω = λ3 ,
∂H
= −ωyi ,
∂[(mi Ṗi ) · e1 ]
∂H
= ωxi ,
∂[(mi Ṗi ) · e2 ]
i = 1, 2.
Therefore the differential equations of the canonical form (9.4), relative to the
variables (Pi − Po ), are
ẋi + ωyi = 0,
ẏi − ωxi = 0,
i = 1, 2.
(11.2)
Thus P1 and P2 , spin about Po with constant angular velocity ω. Subtracting
yields
(x1 − x2 )′ + ω(y1 − y2 ) = 0,
(11.3)
(y1 − y2 )′ − ω(x1 − x2 ) = 0.
Multiply the first equation of (11.2) by xi and the second by yi and add the
resulting equations. Also multiply the first equation of (11.3) by (x1 − x2 ) and
the second by (y1 − y2 ) and add the resulting equations. These operations give
d
Pi − Po 2 = 0,
dt
i = 1, 2,
and
d
P1 − P2 2 = 0.
dt
Thus, along the motion, the mutual distance of the three bodies remains
constant. As a consequence, their mutual geometric configuration remains
constant.
12 Configuration of the Three Bodies
From (11.1) also using (10.1), compute
∂H
= ω(mi Ṗi ) · e2 ,
∂xi
∂H
= −ω(mi Ṗi ) · e1 ,
∂yi
i = 1, 2.
(12.1)
Moreover, having chosen Σo with the origin at the center of mass of the
system, the momentum is identically zero. Therefore
mo Ṗo = −m1 Ṗ1 − m2 Ṗ2
= −m1 (Ṗ1 − Ṗo ) − m2 (Ṗ2 − Ṗo ) − (m1 + m2 )Ṗo .
From this,
Ṗo = −
m1
m2
(Ṗ1 − Ṗo ) −
(Ṗ2 − Ṗo ),
m
m
m = (mo + m1 + m2 ).
158
6 THE LAGRANGE EQUATIONS
Return to the first of (12.1) and write it for the index i = 1:
∂H
= ω[m1 (Ṗ1 − Ṗo ) + m1 Ṗo ] · e2
∂x1
m
m2
1
(Ṗ1 − Ṗo ) +
(Ṗ2 − Ṗo ) · e2
= ωm1 ẏ1 − ωm1
m
m
ωm1
(m1 ẏ1 + m2 ẏ2 ) .
= ωm1 ẏ1 −
m
Compute the last term by putting ω ẏi = −ẍi , which follows from (11.2). Using
also the conservation of momentum,
ωm1
m1
(m1 ẏ1 + m2 ẏ2 ) = −
(m1 ẍ1 + m2 ẍ2 )
m
m
m1 d
m1 (Ṗ1 − Ṗo ) + m2 (Ṗ2 − Ṗo ) · e1
=−
m dt
2
m1 d
mi Ṗi − mṖo · e1 = m1 ẍo .
=−
m dt j=0
Moreover, from (9.3),
∂H
mo m1 x1
m1 m2 (x1 − x2 )
=γ
+γ
.
∂x1
P1 − Po 3
P1 − P2 3
Therefore, from this and the second equation of (11.2),
γ
m2 (x1 − x2 )
mo x1
+γ
= ω 2 x1 − ẍo .
3
P1 − Po
P1 − P2 3
Finally, from (8.2) for the index j = 0,
ẍo = γ
m2 x2
m1 x1
+γ
.
P1 − Po 3
P2 − Po 3
Putting this equation in the preceding one and factoring out x1 and x2 gives
A1 x1 + A2 x2 = 0,
(12.2)
where
γ(mo + m1 )
γm2
+
− ω2,
P1 − Po 3
P1 − P2 3
γm2
γm2
A2 =
−
.
P2 − Po 3
P1 − P2 3
A1 =
In a similar way compute
B1 x1 + B2 x2 = 0,
(12.3)
12 Configuration of the Three Bodies
159
where
γm1
γm1
−
,
3
P1 − Po
P1 − P2 3
γm1
γ(mo + m2 )
B2 =
+
− ω2.
3
P2 − Po
P1 − P2 3
B1 =
By the symmetry of the variables x and y, or by direct calculation we also
have
A1 y1 + A2 y2 = 0,
B1 y1 + B2 y2 = 0,
(12.4)
with the same coefficients Ai and Bi .
12.1 Configuration of an Equilateral Triangle
Let (x1 , x2 ) be a solution of the algebraic system (12.2)–(12.3), and similarly,
let (y1 , y2 ) be a solution of the system (12.4). If these solutions are linearly
independent, then (x1 y2 −x2 y1 ) = 0. This implies that the linear homogeneous
system
A1 x1 + A2 x2 = 0,
A1 y1 + A2 y2 = 0,
regarded in the unknowns Ai , has only the trivial solution A1 = A2 = 0.
Similarly, we must also have B1 = B2 = 0. It follows from the definitions of
the parameters Aj and Bj that the three points Po , P1 , P2 are the vertices
of an equilateral triangle. The motion of the three bodies then is a uniform
rotation about the center of mass Ωo with angular velocity
ω =
γm
.
P1 − P2 3
The possible motions are then ∞2 , corresponding the choice of ω and the
orientation of ω, with respect to the inertial plane x3 = 0.
12.2 Configuration of a Segment
If the solutions (x1 , x2 ) of (12.2)–(12.3) and (y1 , y2 ) of (12.4) are linearly
dependent, the three bodies Po , P1 , P2 remain collinear during the motion, and
at constant mutual distance. Apart from a possible reordering, we may assume
that Po is between P1 and P2 . In the plane x3 = 0 introduce a Cartesian system
with the x-axis on the line of the collinear points Po , P1 , P2 with origin at Po
and positive orientation from Po to P1 . With this notation,
P1 − Po = x1 ,
P2 − Po = −x2 ,
y1 = y2 = 0.
160
6 THE LAGRANGE EQUATIONS
The the system (12.2)–(12.3) takes the form
γ(mo + m1 ) γm2
γm2
,
− 2 +
x21
x2
(x1 − x2 )2
γm1
γ(mo + m2 )
γm1
ω 2 x2 = 2 −
−
.
x1
(x1 − x2 )2
x22
ω 2 x1 =
(12.5)
Eliminating ω 2 and setting X = −x2 /x1 gives
(mo + m1 )X 5 + (2mo + 3m1 )X 4 + (mo + 3m1 )X 3
− (mo + 3m2 )X 2 − (2mo + 3m2 )X − (mo + m2 ) = 0.
This algebraic equation has a single positive root.4 Such a root permits one to
compute the ratio X = −x2 /x1 . Then, having fixed an arbitrary x1 > 0, either
equation of (12.5) permits us to compute ω. Also in such a case the possible
motions are ∞2 , corresponding to the choice of x1 and the orientation of ω.
13 Collapse of the n Bodies
The n-body problem originates from the motion of planets. It is then natural
to ask whether at some future time, the n bodies might collapse to a point.
Choose an inertial triad Σo {Ωo ; e1 , e2 , e3 } with origin at the center of mass
Ωo as indicated in §10, and as Lagrangian coordinates take the coordinates of
the points Pj , as indicated in §8. A total collapse will occur if there exists a
time, finite or infinite, such that
n−1
i,j=0
mi mj Pi − Pj 2 =
n−1
i,j=0
mi mj qi − qj 2 = 0.
The polar moment of inertia of the system with respect to Ωo is
J=
n−1
j=0
mj Pj − Ωo 2 =
n−1
j=0
mj q2j .
The notion of total collapse can be expressed in terms of J by computing
n−1
i,j=0
mi mj qi − qj 2 =
=
n−1
mi
j=0
i=0
n−1
mi
i=0
=2
n−1
n−1
i=0
n−1
j=0
mj qi − qj 2
mj qi 2 − 2qi · qj + qj 2
mi qi 2
n−1
mj = 2mJ,
j=0
where m is the total mass of the system. Thus a total collapse occurs at some
time t if and only if at that time J = 0.
4
The existence of a unique positive root follows from Lagrange’s method of alternation of the signs of the coefficients [102].
13 Collapse of the n Bodies
161
13.1 The Lagrange–Jacobi Identity
Computing the second time derivative of J gives
J¨ = 2
n−1
j=0
mj q̇2j + 2
= 4T + 2
n−1
i,j=0
i=j
γ
n−1
j=0
mj qj · q̈j
mi m j
qi · qj − q2i ,
qj − qi 3
where we have used the expression (8.1) of the kinetic energy and (8.2). From
the identity
2(qi · qj − q2i ) = (q2j − q2i ) − qj − qi 2
compute
2
n−1
i,j=0
i=j
γ
n−1
mi mj
mi m j
= −V.
(qi · qj − q2i ) = −
γ
3
qj − qi
qj − qi
i,j=0
i=j
Combining these remarks proves the Lagrange–Jacobi identity
1 ¨
2J =
2T − V = T + E.
(13.1)
13.2 The Sundman Inequality [142]
Let M be the angular momentum of the system with respect to Ωo , e.g.,
M=
n−1
j=0
mj (Pj − Ωo ) ∧ (Ṗj − Ω̇o ) =
By the Cauchy–Schwarz inequality,
M2 ≤
n−1
j=0
mj qj 2
n−1
j=0
n−1
j=0
mj qj ∧ q̇j .
mj q̇j 2 = 2JT.
Putting into this the expression of the kinetic energy T computed from the
Lagrange–Jacobi identity proves the Sundman inequality [142]
M2 ≤ J(J¨ − 2E).
13.3 A Collapse Would Occur in Only Finite Time
Suppose that a collapse occurs as t → ∞, e.g.,
n−1
1
lim
mi mj qi − qj 2 = 0.
t→∞ 2m t→∞ i,j=0
lim J = lim
t→∞
(13.2)
162
6 THE LAGRANGE EQUATIONS
From the expression of V given by the second equation of (8.1),
1 n−1
mi mj
= ∞.
γ
t→∞ 2 i,j=0 qi − qj
lim V = lim
t→∞
i=j
Since the energy T − V is conserved, T → ∞, and by the Lagrange–Jacobi
identity also J¨ → ∞ as t → ∞. Therefore J¨ ≥ 1 for t sufficiently large, and
J≥
1 2
t + at + b
2
for t sufficiently large
for some real constants a and b. As a consequence, J → ∞, contradicting the
collapse.
13.4 A Total Collapse Would Occur Only If M = 0
If there is a total collapse at some finite time t∗ , then
lim J = lim
t→t∗
t→t∗
1 n−1
mi mj qi − qj 2 = 0,
2m i,j=0
lim J¨ = lim T = lim V = ∞.
t→t∗
t→t∗
t→t∗
Therefore there exists to ∈ [0, t∗ ) such that J¨ > 0 for all t ∈ [to , t∗ ]. From this,
t∗
˙ ∗ ) ≥ J(t)J(t)
˙ +
0 = J(t∗ )J(t
J˙2 dτ ≥ J(t)J˙(t)
t
for all t ∈ [to , t∗ ]. Therefore if a collapse occurs at some finite time t∗ , there
exists an interval [to , t∗ ] such that J˙ ≤ 0 for all t ∈ [to , t∗ ]. Multiply the
˙
Sundman inequality by J/J
for t ∈ [to , t∗ ]. Since J˙ ≤ 0,
J˙
1
˙ .
M2 ≥ J˙J¨ − 2JE
J
2
Integrate this inequality over the interval [to , t] ⊂ [to , t∗ ) to obtain
M2 ln
1
J(t)
˙ o )2 − 2E J(t) − J(to ) ≥ − 1 J˙2 (to ).
≥ J˙2 (t) − J(t
J(to )
2
2
From this,
J˙2 (to )
M2 ≤
2 ln J(to ) − ln J(t)
Thus M → 0 as t → t∗ .
for all t ∈ [to , t).
1c Kinetic Energy in Lagrangian Coordinates
163
Problems and Complements
1c Kinetic Energy in Lagrangian Coordinates
1.1c Homogeneous Functions
A function η → F (η) continuous in RN − {0} is homogeneous of degree λ if
F (τ η) = τ λ F (η)
for all η ∈ RN and for all τ > 0.
As an example, let aij ∈ R for i, j = 1, . . . , N . Then
F (η) = aij ηi ηj
F (η) =
is homogeneous of degree 2,
|aij ηi ηj | is homogeneous of degree 1.
More generally, if F is homogeneous of degree 2, then |F | is homogeneous
of degree 1. Every function of the ratio η/η is homogeneous of degree
zero.
A remarkable fact observed by Euler is that the mere homogeneity of F
implies its differentiability along rays through the origin. For a fixed nonzero
vector η ∈ RN , denote by Fη the derivative of F along η, whenever it
exists.
Theorem 1.1c (Euler). Let F be homogeneous of degree λ. Then for every
nonzero η ∈ RN , the directional derivative Fη exists and ηFη (η) = λF (η).
Moreover, setting u = η/η,
λF (η) = η · ∇u F (η).
(1.1c)
If F is differentiable, then this relation characterizes homogeneous functions
in the following sense.
Proposition 1.1c A function F ∈ C 1 (RN ) is homogeneous of degree λ if
and only if
λF (η) = η · ∇F,
∀η ∈ RN .
(1.2c)
If F is homogeneous and regular, then its directional derivatives are also
homogeneous. This is made precise in the following proposition.
Proposition 1.2c Let F ∈ C 1 (RN ) be homogeneous of degree λ. Then the
partial derivatives Fηi are homogeneous of degree (λ − 1).
164
6 THE LAGRANGE EQUATIONS
2c The Principle of Virtual Work
2.1c An Example of Nonconservative Lagrangian Components
of Forces
Two point masses {Pi ; mi }, i = 1, 2, move in a horizontal plane, constrained
to be at constant mutual distance ℓ and subject to the nonconservative field
f (P ) =
u ∧ (P − O)
,
u ∧ (P − O)2
where u is the unit vector perpendicular to the plane of motion. Having introduced a reference system with origin at O and e3 = u, we describe P1 by
its polar coordinates (ρ, ϕ). We describe P2 by the angle ψ between (P2 − P1 )
and (P1 − O). Therefore the Lagrangian coordinates are (ρ, ϕ, ψ) and
P1 − O = ρ cos ϕe1 + ρ sin ϕe2 ,
P2 − O = {ρ cos ϕ + ℓ cos(ϕ − ψ)}e1
+ [ρ sin ϕ + ℓ sin(ϕ − ψ)]e2 .
The measure dμ(P ) in (2.1) is a combination of Dirac masses acting on the
points Pi . Therefore the Lagrangian components of the resultant of forces
along the “directions” (ρ, ϕ, ψ) are
∂P
∂P1
∂P2
dμ(P ) = f (P1 )
+ f (P2 )
;
∂ρ
∂ρ
∂ρ
∂P1
∂P2
∂P
Φϕ = f (P )
dμ(P ) = f (P1 )
+ f (P2 )
;
∂ϕ
∂ϕ
∂ϕ
∂P
∂P1
∂P2
dμ(P ) = f (P1 )
+ f (P2 )
.
Φψ = f (P )
∂ψ
∂ψ
∂ψ
Φρ =
f (P )
These permit one to compute such components explicitly.
4c Lagrangian Function for Conservative Fields
4.1c Rigid Rod with Extremities Sliding on a Circle
The extremities A and B of a rigid rod of length R and mass m slide without
friction on a circle of mass M and radius R. The extremities are attracted
to
√
a point P fixed on the circle by two springs of elasticity constant k/ 3 > 0.
The circle, in turn, is free to rotate about its center O, remaining in a fixed
horizontal plane (Figure 4.1c). The system has two degrees of freedom. In
the plane of the system fix a line through O and let Po = (R, 0). Then as
4c Lagrangian Function for Conservative Fields
165
O
R
α
β
B
R
P
A
P0
Fig. 4.1c.
Lagrangian parameters choose the angles α = P
OPo and β = P
o OA. Then
AOB = π/3 and
√
P − A = 2R 1 − cos (α + β),
√
P − B = 2R 1 − cos (α + β + π/3).
The kinetic energy T and the potential V are
T = 21 R2 (M α̇2 + 56 mβ̇ 2 ),
V = kR2 cos (α + β + 61 π) + const.
The Lagrangian is L = T + V , and the Lagrange equations (3.1)V are
α̈ +
k
sin (α + β + 61 π) = 0,
M
β̈ +
6k
sin (α + β + 61 π) = 0.
5m
These imply that α = (5m/6M )β and reduce to the equation of a pendulum,
π
k(5m + 6M )
(5m + 6M )
α + , ν2 =
.
5m
6
5mM
Resolve the motion starting from the rest position A = Po = P , i.e.,
θ̈ + ν 2 sin θ = 0,
θ=
α(0) = β(0) = α̇(0) = β̇(0) = 0
or
θ(0) = 61 π,
θ̇(0) = 0.
Compute the reactions due to the constraints at A, B, and O.
4.2c Points Sliding on Rectilinear Guides
Two points P1 and P2 , of equal mass m, slide on frictionless rectilinear horizontal guides ℓi , i = 1, 2, at mutual distance d. The two points Pi are attracted
166
6 THE LAGRANGE EQUATIONS
by points Oi ∈ ℓi , fixed at the same ordinate y = 0, by springs of elasticity
constant k. They are also mutually attracted by an elastic force of constant k.
The points have coordinates P1 = (0, y1 ) and P2 = (d, y2 ), and the variables
yi can be taken as Lagrangian coordinates. The Lagrangian is
2L = m(ẏ12 + ẏ22 ) − k y12 + y22 + (y2 − y1 )2 + const.
The two Lagrange equations are
mÿ1 + k(2y1 − y2 ) = 0,
mÿ2 + k(2y2 − y1 ) = 0.
If the initial conditions are homogeneous, the trivial solution is the only one.
Assume then
ẏ1 (0) = vo = 0,
ẏ2 (0) = y1 (0) = y2 (0) = 0.
Introducing the new variables u = y1 + y2 and v = y1 − y2 , and the constant
ν 2 = k/m, the system is transformed into
ü + ν 2 u = 0, u̇(0) = vo , u(0) = 0,
v̈ + 3ν 2 v = 0, v̇(0) = vo , v(0) = 0.
This has solutions
u=
vo
sin νt,
ν
√
vo
v = √ sin 3νt.
3ν
Assume now that the guides ℓi are parallel and vertical so that the Pi
are subject also to their weight. Write down the Lagrangian and solve the
resulting Lagrangian equations.
4.3c Pulleys and Weights
Two equal pulleys of radius R and mass M are connected as in Figure 4.2c
by perfectly flexible, not extensible, weightless ropes, not allowed to slide in
the grooves. The two material points {Pi ; mi }, i = 1, 2, are subject to their
weight, and P1 is also acted upon by the elastic force F = −k(P1 − A), for a
given k > 0.
Denoting by zP the vertical coordinate of a point P :
2
T = 12 m1 żP2 1 + 21 m2 żP2 2 + 14 M R2 (ϕ̇2 + ψ̇ 2 ) + 21 M żO
.
1
Moreover,
żP1 = −żO1 , żP2 = 2żO1 ,
żP1 = −Rϕ̇, żO1 = Rψ̇,
which implies
ϕ̇ = ψ̇.
4c Lagrangian Function for Conservative Fields
167
O
ϕ
P1
Q
O1
Q1
ψ
P2
e3
e1
Fig. 4.2c.
Choosing z = zP1 as the only Lagrangian coordinate, we obtain
T = 12 (m1 + 4m2 + 2M )ż 2 .
The total of gravitational and elastic potentials is
V = −m1 gzP1 − m2 gzP2 − M gzO1 − 12 kzP2 1 + const.
Moreover,
zP1 + zO1 = const,
2zO1 − zP2 = const,
which implies
2zP1 + zP2 = const.
Therefore
V = (−m1 + 2m2 + M )gz − 12 kz 2 + const.
The energy integral is then
(m1 + 4m2 + 2M )ż 2 + 2(m1 − 2m2 − M )gz + kz 2 = const.
Take the time derivative of this and divide by ż, to get
(m1 + 4m2 + 2M )z̈ + kz = (2m2 − m1 + M )g.
Therefore the system exhibits harmonic oscillations of period T about its
equilibrium configuration zo , where
g
m1 + 4m2 + M
.
zo = (2m2 − m1 + M ),
T = 2π
k
k
168
6 THE LAGRANGE EQUATIONS
Write down the Lagrangian L and verify that the corresponding Lagrangian
equation is precisely the previous one. The equilibrium configuration can also
be found by setting Vz = 0 (see also Remark 7.2).
4.4c Homogeneous Material Frame
A homogeneous square material frame of edge 2ℓ and mass m constrained in
a horizontal plane π can rotate about the midpoint O of one of its edges by a
fixed cylindrical, workless hinge. On the opposite side, a point P of mass M
can slide with no friction, and it is attracted by an elastic force of constant k
by the center Q of that side (Figure 4.3c). Write down the integrals of the
energy and angular momentum. Compute the Lagrangian, write down the
Lagrangian system, and put it in canonical form. The system has two degrees
e
P
α
O
Q
Fig. 4.3c.
of freedom, and as Lagrangian coordinates take the angle ϕ formed by (Q−O)
with a fixed direction of the plane π and the distance x from P to Q. Denote
by e the ascending unit normal to π. The kinetic energy is
T = 21 I ϕ̇2 + M Ṗ 2 ,
I = 37 mℓ2 ,
Ṗ 2 = (2ℓϕ̇ − ẋ)2 + x2 ϕ̇2 .
Here I is the moment of inertia of the frame with respect to the vertical axis
through O, and it can be computed by Huygens’s theorem, whereas Ṗ can be
computed by the formula of relative velocities. Therefore
T = 67 mℓ2 ϕ̇2 + 21 M (2ℓϕ̇ − ẋ)2 + x2 ϕ̇2 .
The potential of the elastic force is 2V = −kx2 + const. Therefore the energy
integral takes the form
7mℓ2 ϕ̇2 + 3M (2ℓϕ̇ − ẋ)2 + x2 ϕ̇2 + 3kx2 = const.
4c Lagrangian Function for Conservative Fields
169
The angular momentum with respect to O is the sum of the angular
momentum K1 of the frame and the angular momentum K2 = M (P − O) ∧ Ṗ
of the point P . One computes
K1 = Iω = − 37 mℓ2 ϕ̇e,
K2 = M 2ℓẋ − (4ℓ2 + x2 )ϕ̇ e.
Therefore the integral of the angular momentum takes the form
( 37 mℓ2 + 4M ℓ2 + M x2 )ϕ̇ + 2M ℓẋ = const.
The Lagrangian of the system is
L = 76 mℓ2 ϕ̇2 + 12 M (2ℓϕ̇ − ẋ)2 + x2 ϕ̇2 − 21 kx2 + const.
From this we obtain
∂L
∂ ẋ
d ∂L
dt ∂ ẋ
∂L
∂ ϕ̇
d ∂L
dt ∂ ϕ̇
= −M (2ℓϕ̇ − ẋ),
= −M (2ℓϕ̈ − ẍ),
∂L
= M xϕ̇2 − kx,
∂x
∂L
= 0,
∂ϕ
= 37 mℓ2 ϕ̇ + M 2ℓ(2ℓϕ̇ − ẋ) + x2 ϕ̇ ,
= 37 mℓ2 ϕ̈ + M 2ℓ(2ℓϕ̈ − ẍ) + 2xẋϕ̇ + x2 ϕ̈ .
Thus the Lagrange equations are
2
7
3 mℓ ϕ̈
M (2ℓϕ̈ − ẍ) + M xϕ̇2 − kx = 0,
+ M 2ℓ(2ℓϕ̈ − ẍ) + 2xẋϕ̇ + x2 ϕ̈ = 0.
For the Hamilton canonical form, introduce the variables
q1 = x,
and compute
q2 = ϕ,
p1 =
∂L
,
∂ ẋ
p2 =
∂L
,
∂ ϕ̇
2
2H =
3 (p2 + 2ℓp1 )
p21
+
+ kq12 .
(7mℓ2 + 3M q12 ) M
4.5c Disk Rolling on a Slanted Guide
Consider the Problem 3.1c of Chapter 5, and continue to assume that the disk
rolls without slipping on the guide. A point mass {P ; m} is placed on a smooth
circular guide centered at Po with radius 0 < ρ < R, grooved on the disk.
Compute the total kinetic and potential energy of the system. Write down the
Lagrangian and the corresponding Lagrangian equations. Determine the equilibrium configurations, if any, in terms of α ∈ [0, π/2). Determine those initial
170
6 THE LAGRANGE EQUATIONS
configurations, if any, starting from which the angle between (P − Po ) and e1
remains constant. If such configurations exist, determine the corresponding
reaction force exerted by the constraint on P .
Let ϕ be the angle formed by e1 and (P − Po ). Then
Ṗ =
d
(P − Po ) + Ṗo = (ẋ − ρϕ̇ sin ϕ) e1 + ρϕ̇ cos ϕe3 .
dt
From this, the kinetic energy TP of the point P is
2TP = m ẋ2 + ρ2 ϕ̇2 − 2ρẋϕ̇ sin ϕ .
Therefore the total kinetic energy of the system is given by
T =
3
4M
+ 21 m ẋ2 + 21 mρ2 ϕ̇2 − mρẋϕ̇ sin ϕ.
The total potential is
V = −(M + m)gx sin α − mgρ sin(ϕ − α) + const.
From the Lagrangian L = T + V , one find the equations of motion
3
2M
+ m ẍ − mρϕ̈ sin ϕ − mρϕ̇2 cos ϕ + (M + m)g sin α = 0,
mρ2 ϕ̈ − mρẍ sin ϕ + mρẋϕ̇ cos ϕ + mgρ cos(ϕ − α) = 0.
Equilibrium occurs for ẋ = ẍ = ϕ̇ = ϕ̈ = 0. From the Lagrange equations,
this occurs only for sin α = 0 and cos(ϕ − α) = 0, i.e., only if α = 0 and
ϕ = ±π/2.
If ϕ remains constant, then ϕ̇ = 0 and the Lagrange equations give
ẍ = kg sin α,
tan α tan ϕ =
−1
;
k+1
k=
2(M + m)
.
3M + 2m
Thus such a motion is possible if initially, and hence for all later times, ϕ is
given to satisfy such a relation. The reaction is then computed from
mP̈ = −mgu + R,
u = − sin αe1 + cos αe3 .
5c Lagrangian and Hamiltonian
5.1c The Legendre Transform of the Lagrangian
For q and t fixed the function ξ → L(q, ξ; t) is convex and
1
L(q, ξ; t)
V (q; t)
ξh ξk
= lim
Ahk (q; t)
+
= ∞.
ξ→∞
ξ
ξ→∞ 2
ξ
ξ
lim
(∗)
7c Equilibrium Configurations
171
The Legendre transform of L with respect to ξ is
L∗ (q, η; t) = sup η · ξ − L(q, ξ; t) .
ξ∈RN
By the structure of L as a function of ξ, and the growth condition (∗), the
supremum is achieved at a unique ξ, at finite distance from the origin, and
satisfying
∂L(q, ξ; t)
ηh =
,
h = 1, 2, . . . , N.
∂ξh
This permits expressing ξ = ξ(q, η; t). For such a ξ,
L∗ (q, η; t) = η · ξ − L(q, ξ; t) = H(q, η; t).
Therefore the Hamiltonian η → H(q, η; t) is the Legendre transform of the
Lagrangian ξ → L(q, ξ; t). Prove that the Hamiltonian is a convex function
of η, and it satisfies a growth condition similar to (∗) as η → ∞. Thus one
might take the Legendre transform ξ → H∗ (q, ξ; t) of η → H(q, η; t). Prove
that H∗ = L. Therefore L∗∗ = L and H∗∗ = H. It is said that the Lagrangian
and the Hamiltonian are in involution with respect to the Legendre transform.
7c Equilibrium Configurations
A material smooth semicircle of mass M , radius R, and center C is constrained
to move in a vertical plane π as in Figure 7.1c. One of the extremes O is fixed
by a cylindrical hinge, and the second extreme Q is attracted to a point P on
the horizontal through O and at a distance 2R from it by a spring of elasticity
constant k. Along the arc a ring A of mass m can slide without friction.
Determine the equilibrium configurations of the system. Denote by e1 the unit
vector along (P − O) and by e2 the unit ascending vertical. As Lagrangian
coordinates take the angle ϕ between (Q − O) and e1 and the angle ψ between
P
O
Q
Fig. 7.1c.
172
6 THE LAGRANGE EQUATIONS
(A − C) and e1 . Denoting by Po the center of mass of the semicircle, we have
Po − C = 2R/π. Then in terms of ϕ and ψ,
2
2
Po = R cos ϕ − sin ϕ e1 − R sin ϕ + cos ϕ e2 ,
π
π
Q = 2R cos ϕ e1 − 2R sin ϕ e2 ,
A = R (cos ϕ + cos ψ) e1 − R (sin ϕ + sin ψ) e2 ,
(P − Q) = 2R [(1 − cos ϕ) e1 + sin ϕe2 ] .
Taking virtual variations yields
2
2
cos ϕ e1 + cos ϕ − sin ϕ e2 δϕ,
π
π
δQ = −2R(sin ϕe1 + cos ϕe2 )δϕ,
δA = −R[(sin ϕδϕ + sin ψδψ)e1 + (cos ϕδϕ + cos ψδψ)e2 ].
δPo = −R
sin ϕ −
The external forces are
−M ge2 on Po ,
−mge2 on A,
−k(Q − P ) on Q.
Therefore the elemental work is computed as
δL = −M ge2 · δPo − mge2 · δA − k(Q − P ) · δQ
= R [gπ(M + m) cos ϕ − (2M g + 4kπR) sin ϕ] δϕ/π
+ mgR cos ψδψ.
For the equilibrium, δL = 0 for all virtual variations. This is realized for
ψ=
π
,
2
tan ϕ =
πg(m + M )
.
2gM + 4kπR
7
PRECESSIONS AND GYROSCOPES
1 The Euler Equations
Let {M; dµ} be a rigid system in precession about a pole O. Introduce a fixed
inertial triad Σ and a moving triad S, both with origin at O, so that S is in
rigid motion with respect to Σ with angular characteristic ω. The latter is
the unknown of the motion. The system is acted upon by external forces that
generate a resultant moment M(e) with respect to the pole O. The constraint
that keeps O fixed and other possible constraints give rise to reactions of
resultant moment M with respect to O. It is assumed that the moment M(e)
and M are known functions of ω, or equivalently of the Euler angles. For such
a system the cardinal equations (3.1)–(3.2) of Chapter 5 are necessary and
sufficient to resolve the motion.
The moving triad S will be chosen so that its coordinate axes are principal
axes of inertia for the system. For such a choice, the inertia tensor σ with
respect to such a triad takes the form
⎛
⎞
I1 0 0
σ = ⎝ 0 I2 0 ⎠ ,
0 0 I3
where Ii are the moments of inertia of {M; dµ} with respect to the principal
coordinate axes. From the formula of time differentiation relative to S,
K̇ =
dσω
dt
S
+ ω ∧ σω = σ(ω̇)S + ω ∧ σω.
Therefore the second cardinal equation (3.2), written in the coordinates of S,
takes the form
σ ω̇ + ω ∧ σω = M(e) + M.
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 7,
c Springer Science+Business Media, LLC 2011
(1.1)
173
174
7 PRECESSIONS AND GYROSCOPES
Equivalently, in term of the components in S,
(e)
I1 ω̇1 = (I2 − I3 )ω2 ω3 + M1 + M1 ,
(e)
I2 ω̇2 = (I3 − I1 )ω1 ω3 + M2 + M2 ,
(1.2)
(e)
I3 ω̇3 = (I1 − I2 )ω1 ω2 + M3 + M3 .
These are called the Euler equations of the precession.
2 Precessions by Inertia or Free Rotators
A special case is M(e) = M = 0. This would occur, for example, for a rigid
body constrained to move about its center of mass, by a workless constraint.
Another example is the rotation of Earth about its longitudinal axis. In such
a case, the system (1.2) becomes
I1 ω̇1 = (I2 − I3 )ω2 ω3 ,
I2 ω̇2 = (I3 − I1 )ω1 ω3 ,
I3 ω̇3 = (I1 − I2 )ω1 ω2 ,
(2.1)
and the motion is called either precession by inertia or precession of a free
rotator or Poinsot precession. The first integral of motion follows from the
second cardinal equation and gives K = Ko for a given vector Ko constant in
Σ. The second integral is that of the energy, and it is derived from (2.1) by
multiplying the ith equation by ωi and summing over i = 1, 2, 3. This gives
0=
1 d
1 d t
1 d
dT
Ii ωi2 =
ω σω =
Iω ω2 =
.
2 dt
2 dt
2 dt
dt
Therefore the kinetic energy remains constant along the motion. The two
integrals
K = Ko
constant vector in Σ,
(2.2)
T = To
given nonnegative constant,
permit a remarkable geometrical description of the Poinsot precessions. If T
and K are zero, the system (2.1) admits only the trivial solution. Assume
henceforth that T > 0 and K > 0, set
√
λ = 2T ,
(2.3)
and construct the corresponding inertia ellipsoid
Eλ = x ∈ R3 xt σx = λ2 .
From the origin O draw the semiaxis directed as ω, and denote by P its
intersection with the inertia ellipsoid Eλ . Consider now the plane π tangent
to Eλ at P . The next proposition asserts that such a plane remains constant,
and therefore it is a geometric integral of the motion.
2 Precessions by Inertia or Free Rotators
175
Proposition 2.1 (Poinsot [129]) Along the motion, K remains normal
to π. Moreover, π = πo for a given constant plane πo .
Proof. The gradient of the function f (x) = xt σx is 2σx, and it is normal to
the tangent plane to the surface xt σx = λ2 at x. Therefore σω = K is normal
to π and points outside the ellipsoid. Since K = Ko , the plane π has, along
the motion, a constant normal vector. To establish that indeed π is a constant
plane, it will suffice to show that the distance from the center O of Eλ to π
is constant. Denote by H the intersection of the normal from O to π and
continue to denote by P the point at which π is tangent to the ellipsoid Eλ .
The angle α formed by P − O and H − O is the same as the angle between ω
and K regarded as vectors starting at O. The angle α is acute, since
K · ω = ω t σω = 2T > 0.
(2.4)
Moreover,
H − O = P − O cos α,
cos α =
ω·K
.
ω K
Since P ∈ Eλ ,
λ
P − O2 Iω = (P − O)t σ(P − O) = λ2 , i.e., P − O = √ .
Iω
Therefore
λ Iω ω2
ω t σω
λ
= √
H − O = √
Iω ω K
Iω ω K
√
2T
λ
λ 2T
=
.
=
Iω ω2 =
K
K
K
(2.5)
Remark 2.1 By (2.4), the components of ω, regarded as a vector starting at
O, satisfy the equation of Eλ . Therefore ω ∈ Eλ and ω = P − O.
Remark 2.2 The axis through O and parallel to ω is the axis of motion of
the rigid motion of S with respect to Σ. Therefore its intersection P with
Eλ , as part of the rigid motion of S, has instantaneous zero velocity (§7 of
Chapter 1).
Since the inertial ellipsoid Eλ is fixed with S, one might think of it as a
rigid material ellipsoid and identify its rigid motion, as the rigid motion of S.
Along the motion, Eλ remains tangent to a fixed plane π normal to K. Since
the point of tangency P has instantaneous zero velocity, the ellipsoid rolls
without slipping on π.
176
7 PRECESSIONS AND GYROSCOPES
O
p =p0
H
K = K0
Fig. 2.1.
3 Moving and Fixed Polhodes
The geometrical locus of the contact points of Eλ and π can be described
either in S or in Σ. When interpreted in S it is a curve traced on Eλ called
a moving polhode. When regarded in Σ it is a curve traced in the fixed plane
π, called a fixed polhode. This double description is closely related to that of
moving and fixed axodes GS and GΣ (§10 of Chapter 1). For a precession, the
Poinsot cones are generated by the axis of motion of the intrinsic parametric
equation
(λ, t) −→ P (λ; t) = O + λω.
When regarded in Σ this generates a fixed cone, and when interpreted in S it
generates a moving cone. For λ > 0, this equation gives the positive Poinsot
semicone generated by the half-lines from O and directed as ω. The moving
and fixed polhodes are the intersections of the positive Poinsot semicones
with the inertia ellipsoid Eλ (moving polhode) and the fixed plane π (fixed
polhode). Conversely, given the moving and fixed polhodes and projecting on
them the origin O generates the moving and fixed positive Poinsot semicones.
3.1 Equations of the Moving Polhode
Let x be a point of the moving polhode. The tangent plane to Eλ at x is
(y − x) · K = (y − x)t σω = Ii xi (yi − xi ) = 0,
and the distance h from such a plane to O is
h=
Ii x2i
Ii2 x2i
=
λ2
Ii2 x2i
.
(3.1)
3 Moving and Fixed Polhodes
177
The equations of the moving polhode are then
I1 x21 + I2 x22 + I3 x23 = λ2 ,
I1 (I1 − I)x21 + I2 (I2 − I)x22 + I3 (I3 − I)x23 = 0,
(3.2)
where we have set
λ 2
.
(3.3)
h
The first equation of (3.2) expresses that x ∈ Eλ . The second is derived by
rewriting (3.1) in the form
I=
Ii2 x2i =
Ii x2i
h
2
= I Ii x2i .
The number I defined by (3.3) has the dimensions of a moment of inertia. For
a fixed λ, the largest (smallest) value of I along the motion occurs when h
equals the length of the smallest (largest) of the semiaxes of the ellipsoid Eλ .
3.2 Geometry of Moving Polhodes
Assume first that the axial moments of inertia Ii are all distinct and have, for
example, the ordering
I1 < I2 < I3 .
(3.4)
Then the parameters h and I range over the intervals
λ
λ
,
I ∈ [I1 , I3 ] .
h∈ √ ,√
I3
I1
The parameters λ, h, and I are mutually and uniquely determined by the
initial data To and Ko , through (2.2), (2.5), and (3.3). Thus the possible
geometrical configurations of the moving polhodes, for different values of these
parameters, corresponds to the possible motions originating from different
initial data.
3.2.1 Either I1 < I < I2 < I3 or I1 < I2 < I < I3
If the first of these occurs, the second of (3.2) implies
I1 (I − I1 )x21 = I2 (I2 − I)x22 + I3 (I3 − I)x23 .
This is the equation of a cone whose axis is along the major semiaxis of Eλ .
Therefore the moving polhodes are closed curves traced on Eλ and surrounding
its major semiaxis. Such an axis bears the least moment of inertia I1 . In the
second occurrence the moving polhodes are closed curves traced on Eλ and
surrounding its minor semiaxis. Such an axis bears the largest moment of
inertia I3 .
178
7 PRECESSIONS AND GYROSCOPES
3.2.2 The Degenerate Cases I1 = I and I = I3
√
If I = I1 , then the only solutions of (3.2) are the two points (±λ/ I1 , 0, 0),
and the moving polhodes degenerate in either of them. In such a case h equals
the length of the major axis of Eλ . Such an axis remain fixed along the motion
and Eλ rotates about it. A similar degeneracy occurs if I = I3 .
3.2.3 The Degenerate Case I2 = I
The parameter h equals the length of the intermediate axis of Eλ , and the
second equation of (3.2) degenerates into the two planes
I1 (I2 − I1 )x1 ±
I3 (I3 − I2 )x3 = 0.
Each of these traces an ellipse on the ellipsoid Eλ . Each√of these ellipses is, in
turn, divided into two arcs by the two points (0, ±λ/ I2 , 0). Each of these
four arcs
√ is a degenerate limiting moving polhode. Moreover, the two points
(0, ±λ/ I2 , 0) are two distinct, independent, degenerate polhodes. Indeed, by
the remarks of §3.4, rotations about the principal axes are permanent.
e3
e2
e1
Fig. 3.1.
3.3 Ellipsoids of Rotation
If two of the moments of inertia Ii coincide, Eλ is an ellipsoid of rotation. The
geometric axis of rotation is the gyroscopic axis, and the material system to
which Eλ corresponds is a gyroscope.
Proposition 3.1 Assume, for example, that Eλ is an ellipsoid of rotation
about u3 and that I1 = I2 = I. Then ω is constant, and the Poinsot cones
4 Integrating Euler’s Equations of Free Rotators (Jacobi [88])
179
are right circular cones, possibly degenerate. The moving polhode is a circle,
possibly degenerate, traced on Eλ , and the fixed polhode is a circle, possibly
degenerate, traced in the fixed plane π.
Proof. By (2.1) the third component ω3 of ω is constant. From the assumptions,
K = σω = I(ω1 u1 + ω2 u2 ) + I3 ω3 u3 = Iω + (I3 − I)ω3 u3 .
Taking the scalar product by ω gives
ω2 =
I3 2
I3 2
K·ω
2To
+ 1−
+ 1−
ω3 =
ω3 = const.
I
I
I
I
3.4 Principal Axes of Inertia as Axes of Permanent Rotation
Let ω be a nontrivial solution of (2.1), and recall that such a system corresponds to (1.1) with the right-hand side identically zero. Assume that at
some instant, say for example t = 0, the vector ω(0) = ω o is directed as one
of the coordinate axes of S. Since the triad S is principal of inertia, one has
σω o = ηω o for some η ∈ R and therefore ωo ∧ σω o = 0. It follows that the
only solution of
σ ω̇ + ω ∧ σω = 0
with ω(0) = ωo
is ω = ωo identically. Thus if at some instant, ω is directed as one of the
principal axes of inertia, it remains constant in that configuration and the
rotation about that axis is permanent. For this reason the principal axes of
inertia are also referred to as the axes of permanent rotation. By analogous
considerations, if ω is constant, it must be directed as one of the principal
axes of inertia.
4 Integrating Euler’s Equations of Free Rotators
(Jacobi [88])
The initial data in (2.1) are given in terms of the first integrals
Ii ωi2 = 2T,
Ii2 ωi2 = K2 ,
K2 = 2T I.
(4.1)
The first is the energy integral; the last one follows from (2.5), by choosing the
parameter λ from (2.3), and by using the definition (3.3) of the “moment of
inertia” I. The second follows from (3.1) with x replaced by ω, by Remark 2.1.
The function ω2 , the second component of ω, will be computed by assuming
that the moments of inertia Ii are ordered as in (3.4). Multiply the first
180
7 PRECESSIONS AND GYROSCOPES
equation of (4.1) by I1 and by I3 , subtract the resulting expressions, and
take into account the last equation of (4.1) to obtain
I1 (I1 − I3 )ω12 = 2To (I − I3 ) − I2 (I2 − I3 )ω22 ,
I3 (I3 − I1 )ω32 = 2To (I − I1 ) − I2 (I2 − I1 )ω22 .
These are rewritten in the form
ω12 = A21 ν12 − ω22 ,
where
I2 (I2 − I3 )
,
I1 (I1 − I3 )
2To I − I3
ν12 =
,
I2 I2 − I3
A21 =
ω32 = A23 ν32 − ω22 ,
I2 (I2 − I1 )
,
I3 (I3 − I1 )
2To I − I1
ν32 =
.
I2 I2 − I1
A23 =
Putting (4.2) into the second equation of (2.1) gives
ω̇2 = ±A ν12 − ω22 ν32 − ω22 ,
where
A=
(I3 − I2 )(I2 − I1 )
.
I1 I3
(4.2)
(4.2)′
(4.3)
(4.3)′
4.1 Integral of (4.3) when I = Ii , i = 1, 2, 3
Assume, for example, that I1 < I < I2 and set
ν3
ω2
,
k= .
η=
ν3
ν1
The definitions of νi , i = 1, 3, imply |η| < 1 and k ∈ (0, 1). Then (4.3) can be
rewritten in terms of η as
η̇ = ±Aν1
1 − η2
1 − k2 η2 .
This is the equation of a mathematical pendulum of period (§8 of Chapter 3)
1
ds
4
√
√
.
Aν1 0
1 − s2 1 − k 2 s 2
4.2 Integral of (4.3) when I = Ii for Some i = 1, 2, 3
If I = I1 , then ν3 = 0. The polhodes are degenerate, and the system spins
about the axis u1 with constant angular velocity ω. Indeed, (2.1) and (4.2)–
(4.2)′ imply ω1 = const and ω2 = ω3 = 0. The case I = I3 is analogous. If
I = I2 , then
2To
ν12 = ν32 =
= ν2,
k = 1,
I2
5 Rotations about a Fixed Axis
181
and (4.3) takes the form
ω̇2 = ±A ν 2 − ω22 ,
ω2 (0) = ω2,o .
(4.4)
If ω2,o = ±ν, one has the only solution (0, ±ν, 0) resulting in a constant
rotation about the intermediate axis u2 . If |ω2,o | < ν, then (4.4) can be
integrated explicitly as
ν + ω2
ν + ω2,o ±2Aν t
=
e
.
ν − ω2
ν − ω2,o
These solutions tend to ±ν as t → ±∞.
5 Rotations about a Fixed Axis
The rigid system {M; dµ} is constrained to spin, without slipping, about a
fixed axis ℓ of unit direction u. Such a constraint can be realized by a spherical
hinge placed on a point O ∈ ℓ and a cylindrical hinge placed on another point
Q ∈ ℓ. The constraints generate reactions RO and RQ as vectors applied in
O and Q respectively. By the nature of the constraint, RQ is normal to ℓ,
whereas RO could take any direction. Fixed and moving triads are taken as
S = {O; u1 , u2 , u3 = u},
Σ = {O; e1 , e2 , e3 = u}.
For such a choice, the axis ℓ need not be principal of inertia and the corresponding inertia matrix σ = (Iij ) need not be diagonal. As the only Lat
grangian coordinate one may choose the angle ϕ = e
1 u1 , so that ω = (0, 0, ϕ̇) .
With these choices, (1.1), written in components, takes the form
I13 ϕ̈ − I23 ϕ̇2 = M(e) · u1 − Q − O RQ · u2 ,
I23 ϕ̈ + I13 ϕ̇2 = M(e) · u2 + Q − O RQ · u1 ,
(5.1)
I33 ϕ̈ = M(e) · u3 + µ,
where µ is the moment along ℓ generated by the possible presence of friction.
Here M(e) and µ are known functions of (ϕ, ϕ̇). The last of these permits the
determination of ϕ starting from some given initial data, whence ϕ is known
as a function of time; it can be put in the first two equations to determine the
reactions due to the constraints.
5.1 Rotations about a Principal Axis of Inertia
If ℓ is a principal axis of inertia, then Ii3 = 0, i = 1, 2. and (5.1) become
M(e) · u1 = Q − O RQ · u2 ,
M(e) · u2 = −Q − O RQ · u1 ,
I33 ϕ̈ = M
(e)
· u3 + µ.
(5.2)
182
7 PRECESSIONS AND GYROSCOPES
Thus the role of the reactions RQ of the constraint is to generate a moment
with respect to O that will balance the component normal to ℓ of the resultant moment M(e) of the external forces acting on the system. This occurrence characterizes the principal axes of inertia as formalized by the following
theorem.
Proposition 5.1 Let {M; dµ} be in rigid rotation about a fixed axis ℓ, and
let M⊥ℓ denote the component, normal to ℓ, of the resultant moment of the
external forces and the reaction due to constraints acting on {M; dµ}. Then
{ℓ principal of inertia}
⇐⇒
{M⊥ℓ = 0} .
Proof. It suffices to prove the implication ⇐=. By the assumption the first two
equations of (5.1) must hold with the right-hand sides identically zero. Multiplying the first by I23 and the second by I13 and subtracting the expressions
so obtained gives
2
2
2
ϕ̇ = 0.
I23 + I13
Remark 5.1 Suppose ℓ is not a principal axis of inertia, and that the external
forces generate a moment M(e) whose component normal to ℓ is zero. Then the
material system {M; dµ} spontaneously generates reaction moments normal
to ℓ. For this reason the moments Iij , i = j, are called deflecting moments.
6 Gyroscopes
For a material system {M; dµ} let S = {Po ; u1 , u2 , u3 } be its central principal
triad of inertia (§5.1 of Chapter 4). The system is a gyroscope if the ellipsoid
of inertia Eλ with respect to S is of rotation, say for example if I1 = I2 = I.
In such a case the axis u3 is called a gyroscopic axis. Assume that the center
of mass Po is fixed and that the gyroscope is in precession about Po . One
might perturb the motion of the system by applying a force F = F u1 , for
some F ∈ R, to a point of the gyroscopic axis other than Po . Such a force
generates a moment M = M u2 , for some M ∈ R. In this setting, the Euler
equations (1.2) take the form
ω̇1 = −νω2 ,
ω̇2 = νω1 +
ω3 = ωo,3 .
M
,
I
ν=
(I3 − I)
ωo,3 ,
I
By taking the time derivative we obtain
ω̈1 + ν 2 ω1 = −ν
M
,
I
ω̈2 + ν 2 ω2 = 0.
(6.1)
6 Gyroscopes
183
This can be integrated explicitly starting from given initial data
ω1 (0) = ωo,1 , ω̇1 (0) = −νωo,2 ,
ω2 (0) = ωo,2 , ω̇2 (0) = νωo,1 +
The explicit solution is
ω1 +
where
A=
M
= A cos (νt + α),
Iν
ωo,1 +
M
Iν
2
2 ,
+ ωo,2
M
.
I
ω2 = A sin (νt + α),
(6.2)
Iνωo,2
.
Iνωo,1 + M
(6.2)′
tan α =
Assume that the rotation about u3 is large with respect to the remaining
parameters of the motion, in the sense that
M
≪ 1,
Iνωo,3
ωi
≪ 1,
ωo,3
i = 1, 2.
(6.3)
It follows from (6.2)–(6.2)′ that if (6.3) holds at some time, say for example
for t = 0, then it continues to hold at all times. In particular, if ωo,i , i = 1, 2,
are negligible with respect to the rotation |ωo,3 | about the gyroscopic axis,
then they continue to be negligible at all times.
The velocity of the point u3 = (0, 0, 1) as part of the rigid motion of S is
u̇3 = ω ∧ u3 = ω2 u1 − ω1 u2
and u̇3 = ω12 + ω22 .
Thus if (6.3) holds at some time, then at all times the motion of the gyroscopic
axis, as detected by u3 , remains negligible with respect to ωo,3 . This effect is
called tenacity of the gyroscopic axis, in the sense that it resists perturbations.
The functions ωi for i = 1, 2 are periodic with period 2π/|ν| and frequency |ν|
proportional to |ωo,3 |. Thus u3 exhibits periodic oscillations with frequency
proportional to |ωo,3 |. This is the quivering of the gyroscopic axis. From (6.2),
u̇3 = A sin (νt + α)u1 − A cos (νt + α)u2 +
Integrating over a period yields
1 T
M
u2 .
u̇3 dt =
T 0
Iν
M
u2 .
Iν
(6.4)
Therefore the average of the velocity of u3 over a period exhibits a deflection
along u2 . Recall that the perturbation F = F u1 has been applied along u1
and the resultant moment M = M u2 acts along u2 . Thus it might appear
that the gyroscope would deflect toward the moments rather than toward the
perturbing force. This is called the paradox of parallelism (see also Remark 8.1
and §6c of the Complements).
184
7 PRECESSIONS AND GYROSCOPES
7 Spinning Top Subject to Gravity (Lagrange)
Let {M; dµ} be a gyroscope subject to gravity and in precession about a pole
O on the gyroscopic axis at distance d > 0 from its center of mass Po , as in
Figure 7.1. The fixed triad Σ = {O; e1 , e2 , e3 } is chosen with e3 ascending
vertically. The moving triad S = {O; u1 , u2 , u3 } is chosen with u3 along the
gyroscopic axis. Therefore S is principal of inertia, although not central, and
the inertia tensor σ with respect to such a triad takes the form
⎛
⎞
I 0 0
⎝ 0 I 0 ⎠.
0 0 I3
The Lagrangian coordinates are chosen as the Euler angles. The kinetic energy
in terms of these angles is (§1.1 of Chapter 6)
2T = I(ϕ̇2 sin2 θ + θ̇2 ) + I3 (ϕ̇ cos θ + ψ̇)2 .
Denoting by m the mass of {M; dµ}, the potential of the gravitational force is
V (ϕ, ψ, θ) = −mgd cos θ + const.
Therefore the Lagrangian is given, up to a constant, by
2L(ϕ, ψ, θ) = I(ϕ̇2 sin2 θ + θ̇2 ) + I3 (ϕ̇ cos θ + ψ̇)2 − 2mgd cos θ.
(7.1)
Since the constraints are fixed, the energy E = T − V is conserved and gives
the first integral
2E = I(ϕ̇2 sin2 θ + θ̇2 ) + I3 (ϕ̇ cos θ + ψ̇)2 + 2mgd cos θ
= I(ϕ̇2o sin2 θo + θ̇o2 ) + I3 (ϕ̇o cos θo + ψ̇o )2 + 2mgd cos θo ,
(7.2)
where for a generic function h of time, we have set h(0) = ho . Since the
Lagrangian is independent of ϕ and ψ, these are cyclic or ignorable Lagrangian
coordinates, and give rise to the two first integrals (§4.1 of Chapter 6)
∂L
= I ϕ̇ sin2 θ + I3 (ϕ̇ cos θ + ψ̇) cos θ
∂ ϕ̇
(7.3)
= I ϕ̇o sin2 θo + I3 (ϕ̇o cos θo + ψ̇o ) cos θo ,
∂L
= I3 (ϕ̇ cos θ + ψ̇) = I3 (ϕ̇o cos θo + ψ̇o ).
∂ ψ̇
(7.4)
7.1 Choosing the Initial Data: Sleeping Top
Assume that initially the gyroscopic axis u3 has no nutation with respect to
the fixed vertical axis e3 , that is, θ̇o = 0. Gravity tends to pull down the
7 Spinning Top Subject to Gravity (Lagrange)
e3
185
u3
Θ
mg
O
ψ
e2
ϕ
u1
u2
n
e1
Fig. 7.1.
system by generating a nutation θ̇ = 0. The initial angles ϕo and ψo in (7.2)–
(7.4) can be chosen arbitrarily. Choosing them both zero amounts to having
the moving axis u1 and the fixed axis e1 both coincide, initially, with the
nodal axis. The initial vector ωo is computed from (9.4) of Chapter 1 with
θ̇o = ϕo = ψo = 0,
ωo = ϕ̇o sin θo u2 + (ϕ̇o cos θo + ψ̇o )u3 .
Putting the integral (7.4) into (7.3) and (7.2), for such a choice of initial data,
gives
I ϕ̇ sin2 θ = I ϕ̇o sin2 θo + I3 ωo,3 (cos θo − cos θ),
I θ̇2 = 2mgd(cos θo − cos θ) + I(ϕ̇2o sin2 θo − ϕ̇2 sin2 θ),
(7.5)
ωo,3 = (ϕ̇ cos θ + ψ̇) = (ϕ̇o cos θo + ψ̇o ).
If θo = 0, the second of these equations has the only solution θ = 0, so that the
gyroscopic axis remains vertical. In such a case the nodal axis and the angles ϕ
and ψ are not defined. The first equation of (7.5) holds identically, regardless
of the meaning of these angles. The last one can be interpreted as ω = ωo,3 e3 .
Since along the motion the gyroscopic axis remains vertical, gravity does not
generate a moment with respect to the pole O, and the motion reduces to
the Poinsot precession of a free rotator. For such a motion the remarks of
§3.4 are in force, and the precession reduces to a constant rotation about the
gyroscopic axis. For this reason, the case of the initial datum θo = 0 is referred
to as the case of the sleeping top. A similar analysis holds if θo = π.
186
7 PRECESSIONS AND GYROSCOPES
7.2 First Integrals and Constant Solutions
Starting from the expression (7.1) of the Lagrangian, write the Lagrange equation for the variable θ, and use (7.3)–(7.4). This gives
θ̈ = sin θ
θ(0) = θo ,
mgd
I3
+ ϕ̇2 cos θ − ϕ̇ ωo,3 ,
I
I
θ̇(0) = θ̇o = 0.
(7.6)
For the sleeping top, θo = 0, and (7.5) has the only solution θ = 0, which is
compatible with (7.6). For θo ∈ (0, π), the system (7.5) admits the constant
solution θ = θo , ϕ̇ = ϕ̇o , ψ̇ = ψ̇o . Such a solution is compatible with (7.6)
only if
I3
mgd
ϕ̇2o cos θo − ϕ̇o ωo,3 +
= 0.
(7.7)
I
I
In particular, the constant solution θ = θo ∈ (0, π) is not possible if ϕ̇o = 0.
8 Precession with Zero Initial Velocity
Gravitational effects are felt only if θo ∈ (0, π). To single them out, assume
that the initial angular velocity of nutation and precession is zero, i.e., θ̇o =
ϕ̇o = 0. The motion then is described by (7.5) starting from the initial data
ϕ(0) = ψ(0) = θ̇o = ϕ̇o = 0,
θ(0) = θo ∈ (0, π),
ψ̇(0) = ωo,3 .
The equations of the resulting precession are ([97, Vol. XII]; also in [132])
I3 (cos θo − cos θ)
,
I
sin2 θ
I3 (cos θo − cos θ)
cos θ,
ψ̇ = ωo,3 − ωo,3
I
sin2 θ
2
2mgd
2 I3 cos θo − cos θ
(cos θo − cos θ) − ωo,3
θ̇2 =
2
I
I
sin θ
ϕ̇ = ωo,3
(8.1)
(8.2)
2
(8.3)
def
= f (θ).
We will seek nonconstant solutions of (8.3), since the constant solution θ = θo
is not compatible with (7.6). Equation (8.3) is well defined only for f (θ) ≥ 0,
i.e., for θ ≥ θo . The function f (·) vanishes for θ = θo , and one computes
f ′ (θ)
θ=θo
=
2mgd
sin θo > 0.
I
Therefore f (θ) is positive in a right neighborhood of θo . Moreover, f (θ) → −∞
as θ → π. Therefore there exists some angle θ1 ∈ (θo , π) such that f (θ) > 0
8 Precession with Zero Initial Velocity
187
for θ ∈ (θo , θ1 ) and f (θo ) = f (θ1 ) = 0. In such an interval, by separation of
variables,
θ
θ
dτ
dτ
t=
(8.4)
or t =
for θ ∈ [θo , θ1 ] .
f (τ )
f (τ )
θo
θ1
These permit one to compute the integral of (8.3) in implicit form. The angle
θ effects oscillations between θo and θ1 and remains confined in this interval.
8.1 Oscillatory Nutation and Its Period
To establish such an oscillatory behavior and to compute its period, write
2
ωo,3
I32
2mgd (cos θo − cos θ) 2
(cos θo − cos θ)
sin θ −
2
I
2mgdI
sin θ
2mgd (cos θo − cos θ)
=
g(θ),
I
sin2 θ
f (θ) =
(8.5)
where we have set
g(θ) = sin2 θ − a(cos θo − cos θ),
a=
2
ωo,3
I32
.
2mgdI
The angle θ1 is the first zero, following θo , of the function g(·), and for such
an angle,1
2 cos θ1 = a − a2 − 4a cos θo + 4.
Regarding θ1 as a function of the parameter a, one computes
lim θ1 (a) = θo ,
a→∞
lim θ1 (a) = π.
a→0
(8.6)
The parameter a involves the geometric-material parameters h, m, I, I3 ,
and the kinematic parameter ωo,3 . Assume first that the geometric-material
parameters are fixed and examine the behavior of the precession as ωo,3 → ∞.
From g(θ1 ) = 0,
2mgdI
cos θo − cos θ1 = 2 2 sin2 θ1 .
(8.6)′
ωo,3 I3
Therefore, as ωo,3 increases, the amplitude of the nutation decreases up to
vanishing as ωo,3 → ∞. For ωo,3 ≫ 1, the period T of the nutation can be
computed from the first equation of (8.4), and we have
θ1
sin θ dθ
I
T =2
.
(8.7)
2mgd θo √cos θ − cos θ sin2 θ − a(cos θ − cos θ)
o
o
1
Prove that this is the only admissible solution of g(θ) = 0 for all a ≥ 0. Prove
also that f (θ) < 0 for θ > θ1 .
188
7 PRECESSIONS AND GYROSCOPES
The last integral is computed by the change of variables
a(cos θo − cos θ) = sin2 θ sin2 s,
s ∈ (0, π/2).
Taking differentials, we obtain
a sin θ dθ = 2 sin θ cos θ sin2 s dθ + 2 sin2 θ sin s cos s ds,
sin θ dθ =
2 sin2 θ sin s cos s ds
2 sin2 θ sin s cos s ds
,
=
a (1 − η)
a − 2 cos θ sin2 s
where we have set
2 cos θ sin2 s
.
a
Since a ≫ 1, one also has η ≪ 1. In terms of the new variables,
π/2
ds
I
4
T =√
a 2mgd 0
1−η
π/2
∞
I
4
=√
1+
η i ds
a 2mgd 0
i=1
η(θ, s, a) =
=
2π I
1
.
+o
ωo,3 I3
ωo,3
Thus for ωo,3 ≫ 1 the gyroscopic axis effects oscillatory nutations of frequency
−2
of the order of ωo,3 and amplitude of the order of ωo,3
. As ωo,3 → ∞ the
vibrations are of high frequency and low amplitude, thereby exhibiting another
occurrence of the quivering of the gyroscopic axis.
8.2 Precession about the Fixed Vertical Axis e3
Whence the nutation t → θ(t) is known, the remaining aspects of the precession are determined by (8.1). The gyroscopic axis starts from rest and begins
to precess about the fixed axis e3 with speed ϕ̇. Denoting by Tϕ the period
of such a precession, one computes from (8.1)
Tϕ
I3 Tϕ cos θo − cos θ
2π =
dt.
ϕ̇dt = ωo,3
I 0
sin2 θ
0
For ωo,3 ≫ 1, by virtue of (8.6)′ , the last integrand is of the order of a−1 .
Therefore
2π
1
2mgd
.
(8.8)
=
+o
Tϕ
ωo,3 I3
ωo,3
The left-hand side is the average speed, over a period, of the precession of the
top about the fixed vertical axis e3 . Thus the faster the top spins about its
gyroscopic axis (ωo,3 ≫ 1), the slower, on average over a period, it precesses
about e3 .
8 Precession with Zero Initial Velocity
189
−2
,
Remark 8.1 As ωo,3 → ∞ the amplitude of nutation is of the order of ωo,3
′
as indicated by (8.6) . On the other hand, the average of the angular velocity ϕ̇,
−1
over a period, is of the order of ωo,3
. It might appear, then, that the gyroscope
would deflect first in the direction of the moment of its weight, rather than
along it. This is a further occurrence of the paradox of parallelism. Formula
(8.3) resolves the paradox by revealing an incipient motion in the direction
of gravity. If ωo,3 = 0, the system exhibits harmonic oscillations about the
equilibrium position θ = π, and it behaves like a compound pendulum (§5.1c
of the Complements). Indeed, taking the time derivative of (8.3) with ωo,3 = 0
and setting α = π − θ gives
α̈ +
g
sin α = 0,
L
L=
I
.
md
Remark 8.2 For ωo,3 fixed, one might trace the amplitude of the resulting
nutation from the geometric-material parameters h, m, I, I3 of the system.
For example, the amplitude of the nutation tends to zero as d → 0, or as
I3 → ∞. In the first case the center of mass tends to coincide with the pole
O, and the precession tends to a Poinsot precession. In the second case the
gyroscope tends to be a very flat top. If, on the other hand, I3 → 0, the
top tends to resemble, so to speak, a material stick, of cross section a small
radius. In such a case the second equation of (8.6) implies that θ1 → π, and
the “stick” tends to “fall” on its vertical position. If d → 0, the pole of the
precession tends to coincide with the center of mass, and the motion tends to
a Poinsot precession.
e3
u3
q0
q1
O
Fig. 8.1.
8.3 Visualizing the Motion
A visualization can be realized by the geometric configurations of the point
(ϕ, θ) → P (ϕ, θ), traced by the positive gyroscopic semiaxis, on the unit sphere
190
7 PRECESSIONS AND GYROSCOPES
of Σ. For θ = θo , nutation and precession both have zero velocity. From (8.1)
it follows that ϕ̇ > 0 for all θ ∈ (θo , θ1 ). From (8.1) and (8.3), regarding ϕ as
a function of θ, one also computes
lim
θ→θo
∂ϕ
ϕ̇
= lim
= 0,
θ→θo θ̇
∂θ
lim
θ→θ1
∂ϕ
ϕ̇
= lim
= ∞.
θ→θ1 θ̇
∂θ
Therefore as θ travels from θo to θ1 and then back to θo , the angle of precession
ϕ increases, undergoes an instantaneous stop at θ = θo , and has infinite speed
at θ = θ1 . The corresponding P (ϕ, θ) always rotates counterclockwise about
the fixed vertical axis e3 ; it stops whenever θ = θo and has infinite speed
whenever θ = θ1 . Its configurations are traced in Figure 8.1. Because of such
a visual rendering of the motion, the gyroscopic axis is also called the figure
axis.
9 Precession with Arbitrary Initial Velocity
We will continue to assume that of θ̇o = 0, so that the equations of (7.5) are in
force. If no further assumptions are made on the initial velocities, they yield
I3 (cos θo − cos θ)
sin θo 2
+ ϕ̇o
,
2
I
sin θ
sin θ
I3 (cos θo − cos θ)
sin θo
cos θ − ϕ̇o
ψ̇ = ωo,3 − ωo,3
I
sin θ
sin2 θ
2
(cos
θ
−
cos
θ)2
I
2mgd
o
3
2
(cos θo − cos θ) − ωo,3
θ̇2 =
I
I2
sin2 θ
2
sin θo
+ ϕ̇2o
cos2 θo − cos2 θ
sin θ
2ωo,3ϕ̇o I3 sin θo 2
def
−
(cos θo − cos θ) = f (θ).
I
sin θ
ϕ̇ = ωo,3
(9.1)
2
cos θ,
(9.2)
(9.3)
The last is well defined for f (θ) ≥ 0. Write
f (θ) =
where
g(θ) =
cos θo − cos θ
g(θ),
sin2 θ
(9.4)
2
2mgd
2 I3
sin2 θ − ωo,3
(cos θo − cos θ)
2
I
I
(9.5)
I3
sin2 θo .
I
In the case ϕ̇o = 0, it follows from (8.1) that for θ > θo , both ωo,3 and ϕ̇ have
the same sign. In particular, ϕ̇ has constant sign and the precession occurs
always in the same direction. By analogy, it is assumed that
+ ϕ̇2o sin2 θo (cos θo + cos θ) − 2ϕ̇o ωo,3
ϕ̇o
and
ωo,3
are both positive.
(9.6)
10 Precessions of Constant Direction (g(θo ) ≥ 0)
191
This, however, does not imply that the precession has constant direction.
It will be shown in the next sections that if g(θo ) ≥ 0, the precession has
constant direction, whereas if g(θo ) < 0, the precession might invert its direction. The sign of g(θo ) determines also the direction of the incipient nutation.
Indeed, computing (7.6) for t = 0 gives
g(θo )
.
2 sin θo
θ̈(to ) =
Therefore, since θ̇o = 0, if g(θo ) > 0, the gyroscopic axis begins to move away
from the vertical axis, whereas if g(θo ) < 0, it begins to approach it.
10 Precessions of Constant Direction (g(θo) ≥ 0)
Assume first g(θo ) > 0. Then θ → f (θ) vanishes for θ = θo , and one computes
f ′ (θ)
θ=θo
=
g(θo )
> 0.
sin θo
Therefore f (θ) is positive in a right neighborhood of θo . Moreover, f (θ) → −∞
as θ → π. Therefore there exists a first angle θo < θ1 < π such that f (θ)
remains positive in the interval (θo , θ1 ), it vanishes at θ1 , and it is negative
in a right neighborhood of θ1 . In such an interval θ is computed explicitly as
in (8.4) starting from (9.3), with the proper new meaning of the function f .
Since f (θ) would change sign across θo and θ1 , the nutation angle θ remains
confined in such an interval and effects oscillations of period2
T =2
θ1
θo
dτ
.
f (τ )
(10.1)
It follows from (9.1) and the assumption (9.6) that the angle of precession ϕ
is nondecreasing and the precession has constant direction. In particular, the
figure axis effects a counterclockwise rotation, about the fixed vertical axis e3 .
Regarding ϕ as a function of θ, we compute, starting from (9.1) and (9.2),
lim
θ→θo
∂ϕ
ϕ̇
= lim
= ∞,
θ→θo θ̇
∂θ
lim
θ→θ1
∂ϕ
ϕ̇
= lim
= ∞.
θ→θ1 θ̇
∂θ
Following the trace θ → P (θ) of the figure axis on the unit sphere of Σ gives
a visual rendering of the motion as in Figure 10.2.
The limiting case g(θo ) = 0 implies
I3
mgd
= ϕ̇o ωo,3
− ϕ̇2o cos θo .
I
I
2
For an asymptotic analysis as ωo,3 → ∞ see §10c of the Complements.
(10.2)
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7 PRECESSIONS AND GYROSCOPES
This coincides precisely with the compatibility condition (7.7) for solutions of
the system (7.5). From that discussion it follows that the system (9.1)–(9.2)
admits the unique solution
θ = θo ,
ϕ̇ = ϕ̇o ,
ψ̇ = ωo,3 − ϕ̇o cos θo .
The nutation is zero and the motion reduces to a precession of constant
velocity ϕ̇ = ϕ̇o , of period Tϕ = 2π/ϕ̇o .
e3
u3
q2
q1
O
Fig. 10.2.
11 Precessions That Invert Their Direction (g(θo) < 0)
The condition g(θo ) < 0 imposes on ωo,3 the restriction
ωo,3
I3
mgd
> ϕ̇o cos θo +
.
I
I ϕ̇o
Rewrite g(θ) in the form
g(θ) =
2
2mgd
2 I3
sin2 θ − ωo,3
+ ϕ̇2o sin2 θo (1 − cos θ)
I
I2
sin2 θo
I3
− ϕ̇o (1 + cos θo )
ωo,3
+
(1 + cos θo )
I
2
(11.1)
.
The nature of the precession depends on various combinations of the initial
data ωo,3 and ϕ̇o with the geometric material parameters of the gyroscope.
To single out the inversion of the direction of the precession we assume
ωo,3
I3
2mgd
>
≥ 2ϕ̇o .
I
I ϕ̇o
(11.2)
12 Spinning Top Subject to Friction
193
For such a choice, the last term of (11.1) is a positive constant. The function
f (·) vanishes for θ = θo and
f ′ (θ)
θ=θo
=
g(θo )
< 0.
sin θo
Therefore f (θ) is positive in a left neighborhood of θo . Moreover, by (11.2),
f (θ) → −∞ as θ → 0. Thus there exists an angle 0 < θ1 < θo such that
f (θ1 ) = 0. Such an angle is a zero of g(·). Now g is a quadratic polynomial in
cos θ, and g(θo ) < 0 and g(0) > 0. It follows that g(θ) = 0 has the only root
θ = θ1 . We conclude that there here exists θ1 ∈ (0, θo ) such that f (θ) > 0 for
θ ∈ (θ1 , θo ), f (θ1 ) = f (θo ) = 0, and f (·) is negative in (0, θ1 ).
The gyroscopic axis, starting from θo , with zero velocity of nutation, moves
up, so to speak, to θ1 and effects oscillations within [θ1 , θo ]. Amplitude and
period of such oscillations, as well as their asymptotic behavior as ωo,3 → ∞,
are computed as in §8.1 (see also §10c of the Complements).
The function θ → sin2 θϕ̇(θ) is increasing in (0, θo ), it is positive for θ = θo ,
and
I3
lim ϕ̇ sin2 θ = −ωo,3 (1 − cos θo ) + ϕ̇o sin2 θo
θ→0
I
sin2 θ
I3
o
− ϕ̇o (1 + cos θo )
< 0.
= − ωo,3
I
1 + cos θo
Therefore there exists a unique angle θ∗ ∈ (0, θo ) such that ϕ̇(θ∗ ) = 0. Such
an angle is computed from
I ϕ̇o sin2 θo = ωo,3 I3 (cos θ∗ − cos θo ) .
The angle θ∗ is in the interval of positivity of f (·). To prove this we refer to
the second equation of (7.5) with ϕ̇(θ∗ ) = 0, and use (11.2). This gives
I θ̇2
θ=θ∗
= (ωo,3 I3 ϕ̇o − 2mgd) (cos θ∗ − cos θo ) > 0.
When θ spans the interval [θo , θ∗ ), the velocity of precession ϕ̇ is positive, the
angle of precession ϕ increases, and the figure axis rotates counterclockwise
about the fixed axis e3 . When θ spans (θ∗ , θ1 ], the velocity of precession ϕ̇
is negative, ϕ decreases, and the figure axis regresses, effecting a clockwise
rotation about e3 . To gain a qualitative description of the motion, compute
∂ϕ
ϕ̇
ϕ̇
∂ϕ
= lim
= lim
= −∞,
lim
= ∞.
θ→θo ∂θ
θ→θ1 θ̇
θ→θo θ̇
∂θ
The trace of the figure axis on the fixed unit sphere is visualized in
Figure 11.1.
lim
θ→θ1
12 Spinning Top Subject to Friction
Let {M; dµ} be a gyroscope in precession about a pole O of the gyroscopic
axis. The moving triad S = {O; u1 , u2 , u3 } is chosen with u3 along the gyroscopic axis. It is assumed that the possible external forces generate a zero
194
7 PRECESSIONS AND GYROSCOPES
e3
u3
q0
q1
O
Fig. 11.1.
resultant moment and that friction opposes the motion by generating a moment directed as ω,
M = −λω = −λ ωi ui
for some λ > 0.
Therefore M(e) = 0 and (1.2) take the form
I ω̇1 = (I − I3 )ω2 ω3 − λω1 ,
I ω̇2 = (I3 − I)ω1 ω3 − λω2 ,
(12.1)
I3 ω̇3 = −λω3 .
The third equation is integrated explicitly as
ω3 = ωo,3 e−(λ/I3 )t .
(12.2)
Put this in the first two equations and multiply the resulting equations by
exp {(λ/I)t}. This transforms the system into
d
(I − I3 )
ω
1 =
ωo,3 e−(λ/I3 ) t ω
2 ,
dt
I
d
(I3 − I)
ω
2 =
ωo,3 e−(λ/I3 )t ω
1 ,
dt
I
where
ω
i = e(λ/I)t ωi .
(12.3)
Multiplying the first equation by ω
1 and the second by ω
2 and adding gives
Therefore the solutions are
2
2
ω
12 + ω
22 = ωo,1
+ ωo,2
.
2 + ω 2 e−(λ/I)t cos α,
ωo,1
o,2
2 + ω 2 e−(λ/I)t sin α,
ω2 = ωo,1
o,2
ω1 =
(12.4)
1c The Euler Equations
where
α=
I3 (I3 − I)
ωo,2
.
ωo,3 1 − e−(λ/I3 )t + tan−1
λI
ωo,1
195
(12.5)
This expression for α is derived by putting (12.4) into (12.3). This gives a
differential equation for α̇, which is integrated by elementary quadratures.
From (12.2) and (12.4) one estimates
ω2 ≤ ωo 2 e−λt/ max{I;I3 }
for t ≥ 0.
From this, the kinetic energy is estimated as
T = ωt σω ≤ max{I; I3 }ωo 2 e−λt/ max{I;I3 } .
Therefore T → 0 exponentially as t → ∞.
If λ → 0, the motion tends to a Poinsot precession.
If the ellipsoid Eλ is a sphere, then I3 = I and (12.2)–(12.4) reduce to
ωi = ωo,i e−(λ/I)t ,
i = 1, 2, 3.
Problems and Complements
1c The Euler Equations
1.1c Precession Relative to the Center of Mass
Let {M; dµ} be in rigid motion with characteristics Ṗo and ω. For such a
system the cardinal equations (3.1)–(3.2) of Chapter 5 are in force, where
F(e) and M(e) are known functions of Po , Ṗo , and ω. If F(e) were independent
of ω, the first cardinal equation would provide the kinematics of the center of
mass Po ; this in turn could be put in the second to determine ω. In general,
however, F(e) depends on ω, and the first cardinal equation is not directly
integrable. One might then think of determining first ω, which put in the first
cardinal equation would determine the motion of Po . The key idea is that
the characteristic ω does not change by referring the rigid motion to a triad
centered at Po and translating with velocity Ṗo . Thus the determination of ω
reduces to integrating the second cardinal equation, written for Po = Ṗo = O.
This is the motion of {M; dµ} relative to its center of mass (§3.2 of Chapter 5).
1.2c Precession of Earth
The Sun’s planar apparent path over a year is called the ecliptic. The center of
mass Po of Earth moves in the plane of the ecliptic. Let Σ = {Po ; e1 , e2 , e3 }
be a triad with e3 normal to the plane of the ecliptic. With respect to Σ,
196
7 PRECESSIONS AND GYROSCOPES
Earth is in precession of pole Po and is acted upon by moments due to the
presence of the Sun, Moon, and other celestial bodies. Even assuming that
the Sun and the Moon are point masses, these moments are not zero, since
Earth is not perfectly spherical, being slightly flat at its poles. The average
equatorial radius is 6,378 km, whereas the difference between the equatorial
and polar radii is 21.5 km [76, F–145].
1.2.1. Assuming the Sun and the Moon to be point masses, prove that if
Earth were perfectly spherical, the resultant gravitational moment would be
zero. Moreover, if the gyroscopic axis of Earth, from south to north, were
normal to the plane of the ecliptic, the resultant gravitational moment due to
the Sun would be zero.
1.2.2. The maximum inclination of the gyroscopic axis in the plane of the
ecliptic is about 23◦ 29′′ [76, F–144]. Using this value estimate the size of the
component along the gyroscopic axis of Earth, of the resultant gravitational
moment due to the Sun and the Moon, and conclude that they are both
negligible.
1.3c Poinsot and Astronomical Precessions of Earth
Let S = {Po ; u1 , u2 , u3 } be the moving triad, clamped to Earth, oriented so
that the angle u
3 e3 is acute. Independently of the motion of Po , the triad
S is in precession with respect to Σ, with characteristic ω. The resultant
gravitational moment M is a function of ω only, and it is rewritten as
M = Mu3 + M⊥u3 ,
Mu3 = M · u3 ,
M⊥u3 = M − Mu3 .
The precession of Earth can be regarded, with good approximation, as the
composition of two precessions with the same pole Po (§9 of Chapter 1). The
first is acted upon by Mu3 only. Using 1.2.2 above, Mu3 ≈ 0 and the resulting
precession is a Poinsot precession. The second is acted upon by the moment
M⊥u3 and it is called astronomical precession.
1.4c Poinsot Precession of Earth
While not perfectly spherical, Earth is a gyroscope. Setting Mu3 = 0, the
is given by
equations (6.1) with M = 0 are in force, and their explicit integral
2 + ω 2 = 0,
(6.2). If the equatorial component of ω is zero, that is, if ωo,1
o,2
then the precession is a constant rotation about the gyroscopic axis of Earth.
However, astronomical observations reveal that while small, such a term is
not zero. This implies that the axis of rotation does not coincide with the
gyroscopic axis. These axes form an angle, called the constant of nutation,
or solar parallax, estimated to be about 0.0087′′ sexagesimal seconds [76, F–
144]. The rotation is from west to east (counterclockwise) and ω is oriented
accordingly. Then the solar parallax is counted from the axis of ω to the
gyroscopic axis, oriented from south to north.
5c Rotations about a Fixed Axis
197
1.4.1. Using this value
of the solar parallax and the average radius of Earth,
2 + ω 2 . The points where the positive gyroscopic
give an estimate of ωo,1
o,2
semiaxis and the positive semiaxis of rotation meet the surface of Earth are
the geographic and boreal poles. Verify that the distance between these poles
does not exceed 50 cm.
1.5c Astronomical Precession of Earth
This is the slow counterclockwise rotation of gyroscopic axis of Earth about
the fixed axis e3 normal to the plane of the ecliptic, whose period is estimated
to be about 26,000 years. It might be described by the angle of precession ϕ
between the fixed axis e1 and the nodal axis determined by the configurations
of Σ and S. The intersection of the nodal positive semiaxis with the ecliptic
is the vernal equinox, and it occurs about March 21. The intersection of the
nodal negative semiaxis with the ecliptic is the autumnal equinox, and it occurs
about September 23. For this reason the nodal axis is also called axis of the
equinoxes.3
1.5.1. Prove that because of the astronomical precession, the alternation
of the equinoxes does not occur at equal intervals of time. Indeed, every year
the Sun reaches an equinox slightly earlier than the preceding year. Because
of this phenomenon of preceding, these motions are called precessions.4
5c Rotations about a Fixed Axis
5.1c The Compound Pendulum (Huygens [81])
Consider a rigid body, subject to its weight, rotating without friction about
a fixed horizontal axis ℓ of unit direction u, through a fixed point O, but not
containing the center of mass Po , as in Figure 5.1c. As the sole Lagrangian
3
From Latin equi noctis, since when the Sun is in that position, day and night
have the same duration.
4
Observed first by Hipparchus of Nicaea, (circa 190–126 BCE). Hipparchus had
compiled a map of the sky including about 800 stars along with their coordinates and
their relative brightness. Comparing his map with the one compiled by Timocharis
of Alexandria about 50 years earlier, he noticed a difference of about 2◦ in the
position of the same stars. He explained the difference by the precession of equinoxes,
whose advance he estimated to be about 36 seconds. Modern calculations have it
at about 50 seconds. Hipparchus’s map remained essentially unchanged up to the
fifteenth century, except for the addition of new stars by Ptolemy (second century
CE). Generations of astronomers based their investigations on that map, including
Copernicus (1473–1543) and Galileo (1564–1642).
198
7 PRECESSIONS AND GYROSCOPES
parameter take the angle between Po − O and the descending vertical. From
the third equation of (5.1) with µ = 0,
I33 ϕ̈ = −mgPo − O sin ϕ,
m = dµ(P ),
which is rewritten as
ϕ̈ +
g
sin ϕ = 0,
L
L=
I33
.
m Po − O
(5.1c)
Thus the motion reduces to that of a pendulum of length L, and for this
reason the system it is called a compound pendulum. Set I33 = IO and denote
O
ϕ
P0
O′
Fig. 5.1c.
by IPo the moment of inertia with respect to the axis through Po and parallel
to ℓ. By Huygens’s theorem,
IO = IPo + mPo − O2 .
Therefore
L = Po − O +
IPo
> Po − O.
m Po − O
Starting from O and moving a distance L along Po − O, determine a point O′
and denote by ℓ′ an axis through O′ and parallel to ℓ. The point O′ and the
axis ℓ′ have a remarkable property established by Huygens.
Proposition 5.1c (Huygens [81]) The pairs {O; ℓ} and {O′ ; ℓ′ } are interchangeable in the sense that gravity-driven, frictionless precessions about ℓ
with pole O and about ℓ′ with pole O′ have the same period.
6c Gyroscopes
199
Proof. Let IO′ be the axial moment of inertia with respect to ℓ′ and let
L′ =
IO′
m Po − O′
be the length of the new compound pendulum. It will suffice to show that
L = L′ . By Huygens’s theorem,
L′ = Po − O′ +
IPo
.
m Po − O′
From the expression of L and the definition of O′ ,
Po − O′ = L − Po − O =
IPo
.
m Po − O
Putting this in the previous expression of L′ gives
L′ = Po − O + Po − O′ = L.
The Lagrangian and Hamiltonian are
L = 21 I33 ϕ̇2 + mgPo − O cos ϕ + const,
H=
1
2
p2
− mgPo − O cos ϕ + const.
I33
The sole Lagrange equation coincides with (5.1c). The Hamilton canonical
equations are
p
ϕ̇ =
,
ṗ = −mgPo − O sin ϕ.
I33
6c Gyroscopes
6.1.
Write down (6.2) with initial data ωo,i = 0, i = 1, 2. Compute
ü3 = ω̇ ∧ u3 + ω ∧ (ω ∧ u3 )
and show that the incipient acceleration of (0, 0, 1) is parallel to F = F u1 .
Show that ü3 is parallel to the resultant force, at all times, provided in compounding the forces one includes those due the transport and Coriolis accelerations (§8 of Chapter 1). Show that the same occurs if ωo,i are not zero.
6.2. A material homogeneous rigid ellipsoid E of mass m and semiaxes ai ,
i = 1, 2, 3, is in precession about its center Po with characteristic ω. The
central principal triad S is taken with the unit vectors ui clamped along the
homologous axes. The ellipsoid rotates about its a3 axis with velocity ϕ̇u3 ,
which, in turn, rotates about a fixed unit vector e with velocity θ̇e.
Express the kinetic energy in terms of ϕ, θ, and their time derivatives.
Write down the Euler equation of the corresponding
Poinsot precession and
√
find an explicit integral in the case a1 = a2 = 3a3 .
200
7 PRECESSIONS AND GYROSCOPES
The inertia matrix is σ = Ii δij , where the Ii are computed from §3.1c of
the Complements of Chapter 4. Denoting by ϕ the angle between u1 and e,
ω = θ̇ cos ϕ u1 − θ̇ sin ϕ u2 + ϕ̇ u3 .
The kinetic energy is computed from
2T = ω t σω = I1 θ̇2 cos2 ϕ + I2 θ̇2 sin2 ϕ + I3 ϕ̇2 .
There are no external forces, and the moments with respect to Po of the
reactions due to the constraints are zero. Therefore (1.1) yields
(a22 + a23 )(θ̈ cos ϕ − θ̇ϕ̇ sin ϕ) = (a22 − a23 )θ̇ ϕ̇ sin ϕ,
(a21 + a23 )(θ̈ sin ϕ + θ̇ϕ̇ cos ϕ) = (a23 − a21 )θ̇ ϕ̇ cos ϕ,
(a21 + a22 )ϕ̈ = (a21 − a22 )θ̇2 sin ϕ cos ϕ.
Multiplying the first equation by θ̇ cos ϕ, the√second by θ̇ sin ϕ, and the third
by ϕ̇ and adding gives Ṫ = 0. If a1 = a2 = 3a3 , the third equation implies
ϕ̈ = 0, and the first two equations became
2θ̈ cos ϕ − 3θ̇ϕ̇ sin ϕ = 0,
2θ̈ sin ϕ + 3θ̇ϕ̇ cos ϕ = 0.
Multiplying the first by cos ϕ and the second by sin ϕ and adding gives θ̈ = 0.
7c Spinning Top Subject to Gravity
7.1c Intrinsic Equation of a Gyroscope
The trace P of the positive gyroscopic semiaxis on the unit sphere of Σ is a
curve γ that can be parameterized by the arc length s. As the positive intrinsic
triad of γ take {t, v, u3 }, where
t=
du3
,
ds
v = u3 ∧ t,
and also
t = v ∧ u3 .
(7.1c)
Taking the s-derivative of the second equation and using the Frenet formulas
(§1 of Chapter 1) gives
dv
= u3 ∧ κn = κu3 ∧ (cos βu3 − sin βv) = κ sin βt,
ds
(7.2c)
3 . The curvature κ can be expressed in terms of β. Indeed,
where β = nu
taking the s-derivative of the third equation of (7.1c) and using the Frenet
formulas yields
κn = v ∧ t + (u3 ∧ κn) ∧ u3 = −u3 − κ(n · u3 )u3 + κn.
7c Spinning Top Subject to Gravity
201
From this we obtain
κ(n · u3 )u3 = −u3 ,
or
κ=−
1
.
cos β
Putting this in (7.2c) gives the differentiation formula
dv
= − tan β t
ds
and also
v̇ = −ṡ tan β t.
(7.3c)
Next decompose ω along the unit vectors of the intrinsic triad {t, v, u3 }.
Applying Poisson’s formula u̇3 = ω ∧ u3 , one gets
ṡ t = u̇3 = [(ω · t)t + (ω · v)v + (ω · u3 )u3 ] ∧ u3
= −(ω · t)v + (ω · v)t.
Taking now the exterior product of both sides by u3 and using (7.1c) gives
ω = ṡ v + ω3 u3 .
(7.4c)
In particular, the component of ω along t is zero. Since u3 is directed along
the gyroscopic axis, the inertia tensor σ does not change if it is computed with
respect to the triad S = {O; t, v, u3 } centered at the pole O of the precession
and with axes directed along the intrinsic triad to γ. Then
⎛
⎞⎛ ⎞
I 0 0
0
K = σω = ⎝ 0 I 0 ⎠ ⎝ ṡ ⎠ = I ṡ v + I3 ω3 u3 .
0 0 I3
ω3
Taking the time derivative and using the differentiation formulas (7.3c) gives
K̇ = (−I ṡ tan β + I3 ω3 )ṡ t + I s̈ v + I3 ω̇3 u3 .
From this expression for K̇, the second cardinal equation takes the form
[−I ṡ tan β + I3 ω3 ]ṡ = Mt ,
I s̈ = Mv ,
I3 ω̇3 = Mu3 ,
(7.5c)
where Mt , Mv , and Mu3 are the components of the resultant moment of the
external forces along the indicated vectors. These are the intrinsic equations
of the gyroscope [105, art. 55], [124].
7.2c The Gyroscopic Compass or Gyrocompass
This device is a top in precession about its center of mass Po with the figure
axis constrained to remain in the fixed plane π through Po and tangent to the
surface of Earth. Such a constraint can be realized by a Cardan joint. Earth
is regarded as a homogeneous sphere rotating with constant angular velocity
ωT N, where N is the unit vector of the south pole→north pole axis. Denote
202
7 PRECESSIONS AND GYROSCOPES
by e3 the ascending unit normal to π at Po and let α = e
3 N. We assume that
Po is not at either pole, so that α ∈ (0, π). The projection of N on π defines
a unit direction e1 normal to e3 . This identifies a triad ST = {Po ; e1 , e2 , e3 }
clamped to Earth. Let S = {Po ; u1 , u2 , u3 } be the moving triad clamped to
the top, with the unit vector u3 chosen along the figure axis and regarded as
applied in Po . Thus u3 remains on π, and its second extreme moves along a
unit circle centered in Po and lying on π. Therefore v = e3 and β = 0. The
triad S is in precession with respect to ST , with ω given by (7.4c), and the
intrinsic equations (7.5c) become
I3 ω3 ṡ = Mt ,
I s̈ = Me3 ,
I3 ω̇3 = Mu3 .
(7.6c)
The constraints that keep the figure axis on π exert on that axis reactions
normal to it, so that the corresponding resultant moment, which is in general
unknown, is directed as t. Thus Me3 and Mu3 do not involve moments of the
reactions. If Me3 and Mu3 are known as functions of (s, ω3 ), then the last two
equations of (7.6c) can be integrated and provide s and ω3 as functions of
time. Putting these in the first equation gives the component along t of the
resultant moment. The remaining active forces are those due to gravity and
those due to the rotation of Earth. The sum of the forces of gravity and the
forces due to the centrifugal acceleration of Earth is by definition the weight
of the top (§6 of Chapter 3), and as such, it does not generate a moment with
respect to Po . Therefore the functional form of the components Me3 and Mu3
is determined only from the Coriolis forces. These generate the moment
M = −2 (P − Po ) ∧ [ωT N ∧ vST (P )]dµ(P ),
where P is the generic point of the top, contiguous to the elemental mass
dµ(P ), and vST denotes its velocity with respect to the triad ST . Expressing
vST by the Poisson formula and using the expression (7.4c) of ω, compute
M = −2 (P − Po ) ∧ [ωT N ∧ vST (P )] dµ(P )
= −2ωT (P − Po ) ∧ {N ∧ [ω ∧ (P − Po )]} dµ(P )
= 2ωT (P − Po ) ∧ {[ω · N](P − Po ) − [N · (P − Po )]ω} dµ(P )
= 2ωT [N · (P − Po )] [ω ∧ (P − Po )] dµ(P ).
Because of the gyroscopic structure of the top, the triad S = {Po ; t, v, u3 } is
principal central of inertia. Writing
N = Nt t + Nv v + Nu3 u3 ,
(P − Po ) = xt + yv + zu3 ,
7c Spinning Top Subject to Gravity
203
the last integrand can be rewritten as
(Nt x + Nv y+Nu3 z)[(ṡz − ω3 y)t + ω3 xv − ṡxu3 ]
=(ṡNu3 z 2 − ω3 Nv y 2 )t + ω3 Nt x2 v − ṡNt x2 u3
vectorial terms of coefficients
+
.
of mixed-type xy, xz, yz
The integral of these mixed terms gives the deflection moments of the top
These are zero, since S
with respect to the coordinate planes of the triad S.
is central principal of inertia. The remaining terms are computed using the
gyroscopic structure of the top, and give
2
2
2 x dµ(P ) = 2 y dµ(P ) = I3 ,
z 2 dµ(P ) = I.
Therefore the moment generated by the Coriolis forces is
M
= [2ṡNu3 I − ω3 Nv I3 ]t + ω3 Nt I3 v − ṡNt I3 u3 .
ωT
Putting this in the last two equations of (7.6c) yields the differential system
I s̈ = I3 ω3 ωT Nt ,
I3 ω̇3 = −I3 ṡωT Nt .
(7.7c)
7.3c Integrating the Intrinsic Equations
A first integral of (7.7c) is found by multiplying the first equation by ṡ and
the second by ω3 and adding, which gives the energy integral in the form
I ṡ2 + I3 ω32 = const.
More generally, (7.7c) can be integrated starting from some given initial data.
Let θ = u
3 e1 in the plane π. Since the second extreme of u3 moves over the
unit circle about Po on π, one has s = θ. Moreover, Nt = − sin α sin θ. With
these remarks, the system (7.7c) takes the form
θ̈ = −aω3 sin θ,
ω̇3 = bθ̇ sin θ,
θ(0) = θo ,
ω3 (0) = ωo,3 ,
I3 ωT
sin α,
I
b = ωT sin α.
a=
(7.8c)
The second equation can be integrated explicitly, and we obtain
ω3 = ωo,3 + b(cos θo − cos θ).
Putting this in the first equation yields
b
(cos θo − cos θ) sin θ = 0.
θ̈ + a ωo,3 1 +
ωo,3
(7.9c)
204
7 PRECESSIONS AND GYROSCOPES
If θo = 0, i.e., if initially u3 points to the north pole, then the only solution
of (7.9c) is θ = 0, that is, the figure axis continues to point toward the north
pole at all times. Assume now that θo = 0, and that the gyroscope is set,
initially, in rapid rotation about its figure axis, with ωo,3 ≫ 1 so large that
b/ωo,3 is negligible. Then an approximate integral of (7.9c) is
θ̈ + a ωo,3 sin θ = 0.
This is the equation of a pendulum of length ℓ = g/aωo,3 and frequency
√
ν = aωo,3 . Therefore the gyroscopic axis exhibits approximate periodic oscillations about e1 , which itself is directed toward the north pole. In this sense
this device is a compass.
10c Precessions of Constant Direction (g(θo) ≥ 0)
Rewrite g(θ) in the form
2mgd
(b + sin2 θ) − a(cos θo − cos θ) ,
I
2
I
2 I3
ωo,3
+ ϕ̇2o sin2 θo ,
a=
2mgd
I2
I3
I
2ϕ̇2o cos θo − ϕ̇o ωo,3
sin2 θo .
b=
2mgd
I
g(θ) =
(10.1c)
The condition g(θo ) > 0 implies
ωo,3
I3
2mgd
< 2ϕ̇o cos θo +
.
I
I ϕ̇o
Therefore if ωo,3 → ∞, the condition holds if either ϕ̇o → ∞ or if ϕ̇o → 0. In
either case a → ∞.
10.1c Asymptotic Amplitude of the Nutation
Since θ1 is the first zero of g(·),
cos θo − cos θ1 =
(b + sin2 θ1 )
.
a
Therefore if ϕ̇o → 0 and ωo,3 → ∞, then θ1 → θo , in agreement with (8.6)–
(8.6)′ . If ϕ̇o → ∞ while ωo,3 = O(1), then
cos θo − cos θ1 = 2 cos θo ,
i.e.,
θ1 = π − θ o .
Since θo < θ1 < π, this is possible only if θo ∈ (0, π/2). Therefore the nutation
angle θ exhibits periodic oscillations of amplitude π − 2θo , symmetric about
θ = π/2. In the limiting case θ = π/2 the nutation is zero, and the motion
reduces to a precession of infinite speed.
10c Precessions of Constant Direction (g(θo ) ≥ 0)
205
10.2c Asymptotic Period of the Nutation
The expression (8.7) of the period T can be rewritten as
θ1
sin θ dθ
I
T =2
.
2mgd θo √
cos θo − cos θ (b + sin2 θ) − a(cos θo − cos θ)
Introduce the change of variables
a(cos θo − cos θ) = (b + sin2 θ) sin2 s,
s ∈ (0, π/2),
and compute the differentials
sin θ dθ =
2(b + sin2 θ) sin s cos s ds
,
a(1 − η)
η=
2 cos θ sin2 s
.
a
For a ≫ 1 one also has η ≪ 1. Therefore proceeding as in the case of (8.7),
one gets an asymptotic estimate of T as either ωo,3 → ∞ or ϕ̇ → ∞ in the
form
I
1
.
T = 2π
+o
ωo,3 + ϕ̇o
2 I 2 + ϕ̇2 I 2 sin2 θ
ωo,3
o
o
3
10.2.1. Compute the period of the precession starting from (8.1) and using
an argument similar to the one leading to (8.7). Investigate the behavior as
a → ∞.
10.2.2. If ωo,3 → 0, the motion tends to a spherical pendulum (§7.3c of the
Complements of Chapter 3). This is called a spherical compound pendulum.
Compute the length of such a compound pendulum.
8
STABILITY AND SMALL OSCILLATIONS
1 Notion of Stability in Phase Space
Let {M; dµ} be a mechanical system with N degrees of freedom, by the
Lagrangian parameters (q1 , . . . , qN ). We will assume that {M; dµ} is subject to fixed holonomic constraints satisfying the principle of virtual work and
is acted upon by conservative forces. The motion is then determined either by
the Lagrange equations (3.1)V of Chapter 6 or the Hamilton equations (5.2)
of the same chapter, or by some form of the cardinal equations (3.1)–(3.2) of
Chapter 4, such as, for example, the system of Poinsot precessions (2.1) of
Chapter 7. Since the constraints are fixed, the structural functions of these
equations, such as the Lagrangian or Hamiltonian, are explicitly independent of time. These dynamical systems can be given a unifying mathematical
formalism. Let E be an open set in Rn , for some n ∈ N, whose points are denoted by x = (x1 , . . . , xn ). Given a locally Lipschitz-continuous vector-valued
function
E ∋ x −→ F(x) = F1 (x), . . . , Fn (x) ,
the first-order system of ordinary differential equations
ẋ = F(x),
x(to ) = xo ∈ E,
(1.1)
admits a unique solution for t about to . We assume that proper conditions
are placed on F and xo so that (1.1) is solvable for all t ∈ R. Up to a possible
time shift, we will assume that to = 0 and will avoid specifying the domain
of definition E of F. The Hamiltonian system (5.2) of Chapter 6 is of this
form for n = 2N , x = (p, q), and F = (∇p H, −∇q H). The system of Poinsot
precessions (2.1) of Chapter 7 has the form (1.1) for n = 3 and x = ω.
Referring back to (3.1)′ of Chapter 6, also the Lagrange equations can be
recast in the form (1.1), for n = 2N and
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 8,
c Springer Science+Business Media, LLC 2011
207
208
8 STABILITY AND SMALL OSCILLATIONS
xj =
qj for j = 1, . . . , N,
q̇j for j = N + 1, . . . , 2N,
Fj =
xj+N for j = 1, 2, . . . , N,
fj
for j = N + 1, . . . , 2N.
This last example indicates that the equation of motion can be recast in a
form like (1.1) even for moving constraints and forces not necessarily conservative. If the constraints are moving, F bears an explicit dependence on time.
Dynamical systems of the form (1.1) with F explicitly independent of t are
called autonomous.
1.1 Equilibrium Configurations
A point xo ∈ E is an equilibrium configuration for (1.1) if for the initial datum
xo = xo , the system admits the unique solution t → x(t) = xo .
An equilibrium configuration xo is stable if for every ε > 0 there exists
δ > 0 such that for all initial data xo − xo < δ, the corresponding solutions
x(·; xo ) remain in the ball centered at xo with radius ε, i.e., if x(t; xo )−xo <
ε for all t ∈ R. An equilibrium configuration xo is asymptotically stable if there
exists ε > 0 such that for all initial data xo − xo < ε, the corresponding
solutions x(·; xo ) tend to xo as t → ∞, i.e., if x(t; xo ) − xo → 0 as t → ∞.
An equilibrium configuration xo is unstable if for every pair of positive
numbers ε > 0 and δ, there exists xo − xo < δ such that the corresponding
solution x(·; xo ) does not remain in the ball centered at xo with radius ε, i.e.,
if for all t′ > 0 there exists t > t′ such that x(t; xo ) − xo > ε.
Remark 1.1 Because of the physical significance of these notions, their definition pertains only to positive times. However, solutions of (1.1) might exist
for all t ∈ R, and the corresponding notions would have to be specified for
t → ±∞. For example, the scalar equation ẋ = λx for λ < 0 has solutions
x = xo eλt . For it, the origin is a configuration of stable equilibrium for t → ∞
and unstable equilibrium for t → −∞.
2 Lyapunov Stability Criteria
Let xo be an equilibrium configuration for the dynamical system (1.1). A
smooth function F : E → R, is a Lyapunov function for the pair {F; xo } if
F (x) > 0
F (xo ) = 0,
F(x) · ∇F(x) ≤ 0
for all x ∈ E − xo ,
(2.1)
for all x ∈ E.
The first two conditions imply that xo is an isolated minimum for F . The last
one implies that F is nonincreasing along solutions of (1.1). This is verified
2 Lyapunov Stability Criteria
209
if F = const is a first integral of (1.1). Thus first integrals are candidates to
be Lyapunov functions. For example, in the case of the Hamiltonian system
(5.2) of Chapter 6, with Hamiltonian explicitly independent of t, the energy
is conserved along trajectories of motion in phase space. Therefore the energy
(q, p) → E(p, q) is a candidate Lyapunov function for such a system.1
Theorem 2.1 (Lyapunov [118]). If {F; xo } admits a Lyapunov function,
then xo is a configuration of stable equilibrium for (1.1).
Proof. Denote by Bρ (xo ) the open ball in Rn , centered at xo with radius ρ,
and for σ > 0 set
[F < σ] = {x ∈ E F (x) < σ}.
Since xo is an isolated minimum for F , for every ε > 0 there exists σ > 0 such
that [F < σ] ⊂ Bε (xo ). The set [F < σ] is open, and it contains a ball
Bδ (xo ) ⊂ [F < σ] ⊂ Bε (xo ).
For xo ∈ Bδ (xo ) the corresponding trajectories t → x(t; xo ) remain in [F < σ],
and thus in Bε (xo ), since F (xo ) < σ and F is nonincreasing along them.
Assume now that {F; xo } admits a Lyapunov function satisfying the more
stringent condition
for all x ∈ E − xo .
F(x) · ∇F(x) < 0
(2.2)
This would force F to be strictly decreasing along any solution of (1.1).
Theorem 2.2. If {F; xo } admits a Lyapunov function satisfying (2.2), then
xo is a configuration of asymptotically stable equilibrium for (1.1).
Proof. By the previous theorem, xo is a configuration of stable equilibrium.
Having fixed ε > 0, let Bδ (xo ) be such that all orbits t → x(t; xo ) originating from an initial datum xo ∈ Bδ (xo ) remain in Bε (xo ). If xo were not
asymptotically stable, there would exist xo ∈ Bδ (xo ) such that
lim x(t; xo ) − xo > 0.
t→∞
In particular,
lim F x(t; xo ) = σ
for some σ > 0.
Since F is strictly decreasing along trajectories, F x(t; xo ) > σ for all t > 0.
On the other hand, by the continuity of F ,
t→∞
[F < σ] ⊂ Bε′ (xo )
1
for some 0 < ε′ < ε.
A Lyapunov function need not be differentiable. It is required only to be continuous and to admit a continuous right derivative along solutions of (1.1) [110].
210
8 STABILITY AND SMALL OSCILLATIONS
Therefore x(·; xo ) remains confined in the closed annulus ε′ ≤ x − xo ≤ ε.
On such a compact set there exists λ > 0 such that
max
ε′ ≤x−xo ≤ε
F(x) · ∇F(x) = −λ < 0.
Therefore F x(t; xo ) → −∞, contradicting the continuity of F .
3 Criteria of Instability
Let xo be an equilibrium configuration and assume that there exists a smooth
function F : E → R such that
F (xo ) = 0,
∃{xn } → xo such that F (xn ) > 0,
F(x) · ∇F(x) > 0 ∀x ∈ E − xo .
(3.1)
The last of these implies that F strictly increases along solutions of (1.1).
Theorem 3.1. If {F; xo } admits a function F satisfying (3.1), then xo is a
configuration of unstable equilibrium.
Proof. For any balls Bδ (xo ) ⊂ Bε (xo ) however small, there exists xo ∈ Bδ (xo )
such that F (xo ) > 0. For such a fixed positive number, there exists 0 < ε′ < ε
such that F (x) < F (xo ) for all x ∈ Bε′ (xo ). Since F increases along solutions
of (1.1), the orbit t → x(t : xo ) never penetrates Bε′ (xo ). However, such an
orbit must exit Bε (xo ). Indeed, if not, it would remain confined to the closed
spherical annulus ε′ ≤ x − xo ≤ ε. On such a compact set,
min
ε′ ≤x−xo ≤ε
F(x) · ∇F(x) ≥ λ
for some λ > 0.
Therefore F x(t; xo ) → ∞ as t → ∞.
Remark 3.1 The proof requires only that for any fixed ε > 0, the function
F be increasing along solutions of (1.1) that remain in Bε (xo ). Therefore the
last inequality of (3.1) could be replaced by
Ḟ x(t; xo ) > 0 along orbits of (1.1) that remain in Bε (xo ).
(3.1)′
3.1 Some Generalizations and the Četaev Theorem
The proof of Theorem 3.1 uses only the last inequality of (3.1) for orbits
originating from points xo close to the points of {xn }. Since F is continuous,
for all n ∈ N there exists an open set On containing xn and such that
F (x) > 0
for all x ∈ O(xn ).
4 Dirichlet Stability Criteria
Setting O =
211
O(xn ), the requirement (3.1) implies
∃ an open set O ⊂ E such that
xo ∈ ∂O; F > 0 in O; F = 0 on ∂O,
F(x) · ∇F(x) > 0
(3.2)
∀x ∈ O.
Theorem 3.2 (Četaev [26]). If {F; xo } admits a function F satisfying
(3.2), then xo is a configuration of unstable equilibrium.
Remark 3.2 The proof is analogous to the previous arguments. It shows that
given ε > 0, the last inequity of (3.2) needs to be verified only along solutions
of (1.1) whose orbits remain in O ∩ Bε (xo ).
4 Dirichlet Stability Criteria
Let {M; dµ} be a mechanical system subject to fixed holonomic constraints,
satisfying the principle of virtual work, and acted upon by conservative forces
of potential V . Its dynamics are governed by the Hamilton equations for the
variables (p, q). By the definition of kinetic momenta p,
2T (q, q̇) = Ahk (q)q̇h q̇k ⇐⇒ 2T (q, p) = ahk (q)ph pk ,
where ahk (q) is the inverse matrix of Ahk (q) . The Hamiltonian is explicitly
independent of t, and it takes the form
H(p, q) = 12 ahk (q)ph pk − V (q).
Theorem 4.1 (Dirichlet [43]). If q o is an isolated maximum of V , then
(0, q o ) is a configuration of stable equilibrium.
Proof. Up to a change of variables, we may assume that q o = 0 and V (0) = 0.
Since the constraints
are fixed, H(q, p) = E(q, p) is conserved along the motion. Since ahk (q) is positive definite, if q o is an isolated maximum for V ,
the point (q o , 0) is an isolated minimum for E. Therefore (q, p) → E(q, p) is a
Lyapunov function, relative to such a point, for the corresponding Hamiltonian
system. The conclusion then follows from Theorem 2.1.
Corollary 4.1 If q o is an isolated minimum for the potential energy, then
(0, q o ) is a configuration of stable equilibrium.
Theorem 4.2. If q o is an isolated minimum of V , then (0, q o ) is a configuration of unstable equilibrium.
212
8 STABILITY AND SMALL OSCILLATIONS
Proof. We may assume that q o = 0 and V (0) = 0. The function F (p, q) = p · q
satisfies the first two requirements of (3.1), relative to (0, 0) and the Hamiltonian system. The conclusion will follow from Theorem 3.1 upon verifying
that F satisfies (3.1)′ . Along solutions of the Hamiltonian system (see (6.2)
of Chapter 6),
q̇h ph = H(p, q) + L(q, q̇) = 2T (p, q).
Therefore, along such orbits,
Ḟ (p, q) = q̇h ph + qh ṗh = 2T − ∇q H · q = 2T − ∇q (T − V ) · q
= ahk (q)ph pk − 21 ahk,qℓ (q)qℓ ph pk + ∇V · q
≥ ao p2 − 12 ahk,qℓ (q)qℓ ph pk + ∇V · q
for a positive number ao . Since the functions ahk (·) are regular, there exists
εo > 0 such that
|ahk,qℓ (q)qℓ ph pk | ≤ ao p2
for q < εo .
Therefore, along solutions of the Hamiltonian system confined in q < εo ,
Ḟ (p, q) ≥ 21 ao p2 + ∇V · q.
If V has an isolated minimum at the origin, the number εo can be chosen so
that ∇V (q) · q > 0 for all 0 < q < εo .
5 Stability and Instability of the Poinsot Precessions
Consider the precession by inertia, or Poinsot precession of a system {M; dµ}
about a pole O. Introduce a moving triad S, clamped with {M; dµ}, with
origin at O, and principal of inertia. Denoting by Ii the moments of inertia
of {M; dµ} with respect to the coordinate axes, the dynamics of the system
are determined by (§2 of Chapter 7)
I1 ω̇1 = (I2 − I3 )ω2 ω3 ,
I2 ω̇2 = (I3 − I1 )ω1 ω3 ,
(5.1)
I3 ω̇3 = (I1 − I2 )ω1 ω2 .
The system has integrals of energy and angular momentum, i.e.,
T = To
K2 = Ko 2
2
⇐⇒ Ii ωi2 = Ii ωo,i
,
2 2
2 2
⇐⇒ Ii ωi = Ii ωo,i ,
for a given initial configuration ωo = (ωo,1 , ωo,2 , ωo,3 ). By the remarks of §3.4
of Chapter 7, a rotation about a principal axis of inertia is permanent. Thus in
particular, the configurations ωo = (ωo,1 , 0, 0) are of equilibrium. The stable
5 Stability and Instability of the Poinsot Precessions
213
or unstable nature of such equilibrium configurations depends uniquely on
the material geometry of {M; dµ}, or more precisely, on the structure of the
ellipsoid of inertia Eλ . Introduce the functions
2
F1 (ω) = Ii ωi2 − I1 ωo,1
,
2
F2 (ω) = Ii2 ωi2 − I12 ωo,1
,
F (ω) = F12 (ω) + F22 (ω).
The first two of these are first integrals of (5.1) and vanish for ω = ωo . The
last one is a nonnegative first integral of (5.1) vanishing for ω = ω o .
Proposition 5.1 The equilibrium configuration ωo = (ωo,1 , 0, 0) is stable in
any one of the following cases:
I1 < min{I2 ; I3 },
I1 > max{I2 ; I3 },
I1 = I2 = I3 .
Proof. In the last case the ellipsoid of inertia is a sphere, and every axis
through the pole of the precession is a principal axis of inertia. Therefore
the assertion follows from the remarks of §3.4 of Chapter 7. To establish the
assertion for the remaining two cases we will show that F is a Lyapunov
function for the system (5.1) and the configuration ω o . First, F is a non
negative first integral of (5.1) vanishing for ω = ω o . Thus it suffices to show
that F (ω) > 0 for all ω = ωo . For one such ω, either F1 (ω) = 0 or F1 (ω) = 0.
In the former case there is nothing to prove. In the latter case,
2
I1 ω12 = I1 ωo,1
− I2 ω22 − I3 ω32 .
Since ω = ωo , this implies that at least one of the components ω2 , ω3 is not
zero. Putting this in the expression of F2 gives
2
F22 (ω) = I2 (I2 − I1 )ω22 + I3 (I3 − I1 )ω32 > 0.
Proposition 5.2 Assume that the moments of inertia Ii are not all equal.
The equilibrium configuration ω o is unstable in either of the following two
cases:
I2 ≤ I1 ≤ I3 ,
I3 ≤ I1 ≤ I2 .
Proof. Assume that the first of these holds and that ωo,1 > 0. The function
F (ω) = ω2 ω3 is regular and the set O = [F > 0] is open and nonempty.
Moreover, ω o ∈ ∂O. Thus F satisfies the first two conditions of (3.2). The
last one will be verified in the weaker form indicated by Remark 3.2. Along
solutions of (5.1),
Ḟ (ω) = ω1
I1 − I2 2 I3 − I1 2
ω2 +
ω3 .
I3
I2
214
8 STABILITY AND SMALL OSCILLATIONS
For ω ∈ O the quantity in parentheses is positive. Since ωo,1 > 0, there exists
ε > 0 sufficiently small that ω1 > 0 for ω − ω o < ε. Therefore
Ḟ (ω) > 0
along solutions of (5.1) in O ∩ Bε (ω o ).
The conclusion now follows from the Četaev instability theorem.
Theorem 5.1. Let Eλ be the ellipsoid of inertia of a free rotator. If Eλ is a
sphere, every rotation is stable. Otherwise, rotations about the extreme axes
are stable, whereas rotations about the intermediate axes are unstable.
6 Linearized Motions
Let {M; dµ} be subject to fixed, smooth, holonomic constraints, and acted
upon by forces of potential V , and let
L(q, q̇) = 12 Ahk (q)q̇h q̇k + V (q)
be its Lagrangian, uniquely determined by stipulating that V (0) = 0. Assume
that q = 0 is a stationary point of V , so that ∇V (0) = 0. By Taylor’s formula,
L(q, q̇) = 12 Ahk (0)q̇h q̇k + 12 Vqh qk (0)qh qk + O(|(q, q̇)|3 ).
The linearized motion is that for which the terms of order higher than two
are neglected, and is thereby described by the linearized Lagrangian
2Lo = Ahk (0)q̇h q̇k + Vqh qk (0)qh qk = q̇ t Ahk (0) q̇ + q t Vqh qk (0) q.
The matrix Ahk (0) is symmetric and positive definite. Therefore there exists
a unitary matrix A that diagonalizes it, i.e.,
⎛
⎞
λ1 0 0 · · · 0
⎜ 0 λ2 0 · · · 0 ⎟
⎜
⎟ def
A Ahk (0) A−1 = ⎜
. ⎟ = Λ,
⎝· · · · · · · · · . . . .. ⎠
0 0 · · · 0 λN
where λj are the eigenvalues of Ahk (0) . These eigenvalues are all posi1
1
tive with their multiplicity, and therefore the matrices Λ 2 and Λ− 2 are well
defined. With this formalism, the approximate Lagrangian Lo can then be
rewritten as
2Lo = (Aq̇)t A Ahk (0) A−1 (Aq̇) + (Aq)t A Vqh qk (0) A−1 (Aq)
= (Aq̇)t Λ(Aq̇) + (Aq)t A Vqh qk (0) A−1 (Aq)
1
1
1
1
1
1
= (Λ 2 Aq̇)t I(Λ 2 Aq̇) + (Λ 2 Aq)t (Λ− 2 A) Vqh qk (0) (Λ− 2 A)t (Λ 2 Aq)
= Q̇t IQ̇ + Qt (chk )Q,
6 Linearized Motions
215
where we have set
λh (Aq)h
1
Q = Λ 2 (Aq),
1
− 12
(chk ) = (Λ A) Vqh qk (0) (Λ− 2 A)t .
Qh =
i.e.,
The matrix (chk ) is symmetric, its eigenvalues µh are all real, and there exists
a unitary matrix B such that
⎞
⎛
µ1 0 0 . . . . . . 0
⎜ 0 µ2 0 . . . . . . 0 ⎟
⎜
⎟
B(chk )B −1 = ⎜
. ⎟.
⎝. . . . . . . . . . . . . . . .. ⎠
0 0 0 . . . 0 µN
Introducing new Lagrangian coordinates
1
θh = (BQ)h = (BΛ 2 Aq)h ,
h = 1, . . . , N,
(6.1)
the linearized Lagrangian Lo and the linearized potential Vo take the form
Lo =
N
1
|θ̇h |2 + V (θ),
2 h=1
Vo (θ) =
N
1
µh θh2 ,
2 h=1
(6.2)
and the corresponding Lagrange equations are
θ̈h − µh θh = 0,
h = 1, . . . , N.
(6.3)
These are the equations of motion, linearized about the origin of the phase
space R2N for the variables (q̇, q). They approximate the actual motion if by
prescribing initial data (θ̇o , θo ) close to the origin, the resulting solutions (θ̇, θ)
remain near the origin at all times. This occurs, for example, if the origin is
a configuration of stable equilibrium for the system.
In these arguments, the origin of the configuration space RN for the variables q could be replaced by any fixed configuration q o . While q o is a stationary
point of the potential V , no restrictions are placed on the Hessian Vqh qk (q o ) .
Since
the eigenvalues
µh of (chk ) coincide with the eigenvalues of the Hessian
Vqh qk (q o ) , information on the nature of the stationary point q o can be given
in terms of these eigenvalues.
6.1 Stability and Instability of Linearized Motions
If the eigenvalues µh are all negative, the linearized potential Vo has an
isolated minimum at the origin. Therefore by Dirichlet’s stability criterion
of Theorem 4.1, the origin is a configuration of stable equilibrium for the
linearized motion (6.3).
Proposition 6.1 If µh > 0 for some h = 1, . . . , N , then the origin is a
configuration of unstable equilibrium for the linearized motion (6.3).
216
8 STABILITY AND SMALL OSCILLATIONS
Proof. Assume µ1 > 0. Then for the initial data θo = (θo,1 , . . . , 0) and θ̇o = 0,
the system (6.3) has the unique solution
√
θ1 = θo,1 cosh µ1 t,
and
θh = 0
for h = 2, . . . , N.
Thus for every ε > 0 there exist initial configurations in Bε (0) whose orbits
depart from the origin for t → ∞.
If q o is an isolated maximum for the original potential V , then (0, q o ) is
a configuration of stable equilibrium for the original motion. On the other
hand, this occurs if and only if the eigenvalues of (chk ) are all negative. Thus
configurations of stable equilibrium of the original motion remain of stable
equilibrium for the linearized motion.
It is natural to ask whether instability configurations for the linearized
system (6.3) are also instability configurations for the original system.
Proposition 6.2 (Lyapunov) If all the eigenvalues of (chk ) are not zero,
and if one of them is positive, then the origin is a configuration of unstable
equilibrium for the original motion.
Since the eigenvalues of (chk ) coincide with the eigenvalues of the Hessian of V
at the origin, the assumptions of the
imply that the origin is not a
proposition
maximum point for V . Thus if det Vqh qk (0) = 0, the mere lack of maximality
of V implies that the origin is a configuration of unstable equilibrium.
o
Corollary
6.1
point for the potential V such that
Let (0, q ) be a stationary
o
det Vqh qk (q ) = 0. Then if (0, q o ) is not a maximum, it is a configuration of
unstable equilibrium for the system.
7 Small Oscillations
If the eigenvalues of (chk ) are all negative, set µh = −ωh2 and rewrite the
linearized system (6.3) in the form
θ̈h + ωh2 θh = 0,
h = 1, . . . , N.
(7.1)
These represent N independent mathematical pendulums, or equivalently N
uncoupled harmonic oscillators, each with frequency ωh . These are called normal modes or principal frequencies. The coordinates θh introduced in (6.1) are
the normal coordinates.
Proposition 7.1 Let {M; dµ} be a mechanical system subject to fixed,
smooth, holonomic constraints, and acted upon by potential forces of potential V . For every configuration (0, q o ) of stable equilibrium there exists a local
system of normal coordinates such that the corresponding linearized motion
about (0, q o ) can be separated into N independent harmonic oscillators.
8 Vibrations of Masses Subject to Elastic Forces
217
The solutions of (7.1) are of the form
θh = Ah cos(ωh t + ϕh ),
h = 1, . . . , N.
It follows from (6.1) that the original Lagrangian coordinates qh are a linear
combination of the normal coordinates [8]. Therefore
qh =
N
Bhk cos (ωk t + ϕk ) ,
h = 1, . . . , N,
(7.2)
k=1
where Bhk and ϕk are real constants to be determined from the initial data.
Each of the N independent solutions of (7.1) is a fundamental vibration and
the motion. In phase space (p, q), the motion results from the composition of
these fundamental vibrations. While each of the normal coordinates describes
a periodic motion, the composite motion t → q(t) is not periodic in general.
If a period T exists such that q(t + T ) = q(t) for all t, then (7.2) implies
cos ωh (t + T ) + ϕh = cos(ωh t + ϕh ) ∀t ∈ R, h = 1, 2, . . . , N.
Since t is arbitrary, this is possible if and only if there exist positive integers
nh such that ωh T = 2πnh for all h = 1, . . . N . This in turn implies
ωh
ωk
=
,
nh
nk
h, k = 1, 2, . . . , N.
Thus the principal frequencies ωh must be commensurable. Summarizing, the
composite motion t → q(t) is periodic if and only if the normal modes are commensurable. A similar treatment occurs in acoustical phenomena, when sound
is decomposed into its principal frequencies. In that case the sound is clear, and
an instrument is tuned, if the principal frequencies are commensurable. Ordering such principal frequencies in increasing order ω1 < ω2 < · · · < ωN , the
first is the first harmonic and corresponds to the bass, whereas the following
ones are the second harmonic, etc., up to the highest frequency corresponding
to the treble.
8 Vibrations of Masses Subject to Elastic Forces
Two points {P1 ; m} and {P2 ; m} of equal mass m are attracted by a third
point mass {P3 ; M } by elastic forces of equal constant k. Assume that the
system moves in R3 and it is in a configuration of stable equilibrium when
the three masses are aligned and P3 is equidistant from P1 and P2 . Choose a
Cartesian system with the horizontal axis as the axis of equilibrium and such
that the configuration of stable equilibrium is
P1,o = (ℓ, 0, 0),
P3,o = (0, 0, 0),
P2,o = (−ℓ, 0, 0),
218
8 STABILITY AND SMALL OSCILLATIONS
for a given positive constant ℓ. Translate the coordinates of the points about
their equilibrium configuration, by setting
ξ = P1 − P1,o ,
η = P2 − P2,o ,
ζ = P3 .
It is assumed that the system exhibits small oscillations about its equilibrium
configuration and that in addition, the horizontal components of these oscillations are negligible with respect to those normal to the equilibrium axis, that
is, (Pi − Pi,o ) · Pi,o ≈ 0. This system is taken as a model for the vibrations
of a 3-atom molecule, such as for example CO2 (carbon dioxide), when the
atomic potentials are approximated by elastic potentials [14, Chap. IX, §67].
With this symbolism and stipulations, the kinetic energy T and the elastic
potential V are
2T = m(Ṗ12 + Ṗ22 ) + M Ṗ32 = m(ξ˙2 + η̇ 2 ) + M ζ̇ 2 ,
2V = −k(P1 − P1,o ) − P3 2 − k(P2 − P2,o ) − P3 2
= −k(ξ 2 + η 2 + 2ζ 2 − 2ξ · ζ − 2η · ζ),
or in matrix form,
⎞⎛ ⎞
⎛ ⎞t ⎛
ξ̇
mI O O
ξ˙
2T = ⎝η̇ ⎠ ⎝ O mI O ⎠ ⎝η̇ ⎠ ,
O O MI
ζ̇
ζ̇
⎞⎛ ⎞
⎛ ⎞t ⎛
ξ
ξ
I O −I
2
− V = ⎝η ⎠ ⎝ O I −I⎠ ⎝η ⎠ ,
k
ζ
ζ
−I −I 2I
where I is the 3 × 3 identity matrix and O is the 3 × 3 zero matrix. Set
µ2 = m/M and rewrite T in the form
⎞⎛
⎞
⎛ ⎞t ⎛
⎞⎛ ⎞ ⎛
⎞t ⎛
ξ˙
ξ˙
I O O
ξ˙
ξ˙
I OO
2
T = ⎝η̇ ⎠ ⎝O I O ⎠ ⎝η̇ ⎠ = ⎝ η̇ ⎠ ⎝O I O⎠ ⎝ η̇ ⎠ .
m
O O I/µ2
OO I
ζ̇/µ
ζ̇
ζ̇
ζ̇/µ
Similarly, rewrite the potential V as
⎞t ⎛
⎞
⎛
⎞⎛
ξ
ξ
I O −µI
2
− V = ⎝ η ⎠ ⎝ O I −µI ⎠ ⎝ η ⎠ .
k
ζ/µ
ζ/µ
−µI −µI 2µ2 I
With this notation, the system of the Lagrange equation can be written as
⎞⎛
⎞
⎞
⎛
⎛
⎞⎛
ξ¨
I OO
I O −µI
ξ
⎝O I O⎠ ⎝ η̈ ⎠ + ν 2 ⎝ O I −µI ⎠ ⎝ η ⎠ = 0,
(8.1)
OO I
−µI −µI 2µ2 I
ζ/µ
ζ̈/µ
where we have set ν 2 = k/m.
9 Normal Coordinates
219
9 Normal Coordinates
The normal frequencies are found by diagonalizing the second matrix, whose
eigenvalues are the roots of the algebraic equation
⎛
⎞
(1 − λ)I
O
−µI
(1 − λ)I
−µI ⎠ = 0.
det ⎝ O
−µI
−µI (2µ2 − λ)I
Subtract the second row from the first and then add the first column to the
second. This reduces the calculation of the determinant to
⎛
⎞
(1 − λ)I
O
O
(1 − λ)I
−µI ⎠ .
det ⎝ O
−µI
−2µI (2µ2 − λ)I
Now multiply the second row by 2µ and add it to the last one. Then multiply
the last column by 2µ and subtract it from the second. This reduces the
calculation to
⎛
⎞
(1 − λ)I
O
O
(1 + 2µ2 − λ)I −µI⎠ = −λ3 (1 − λ)3 (1 + 2µ2 − λ)3 .
det ⎝ O
−I
O
−λI
Therefore the eigenvalues of the second matrix in (8.1) are λ = 0, λ = 1, and
λ = (1 + 2µ2 ), each with multiplicity 3. The corresponding eigenvectors are
for λ = 0
for λ = 1
for λ = 1 + 2µ2
⎧
(µ, 0, 0, µ, 0, 0, 1, 0, 0)
⎪
u1 =
⎪
,
⎪
⎪
1 + 2µ2
⎪
⎪
⎨
(0, µ, 0, 0, µ, 0, 0, 1, 0)
,
u2 =
⎪
1 + 2µ2
⎪
⎪
⎪
(0, 0, µ, 0, 0, µ, 0, 0, 1)
⎪
⎪
⎩ u3 =
,
1 + 2µ2
⎧
(1, 0, 0, −1, 0, 0, 0, 0, 0)
⎪
√
v1 =
,
⎪
⎪
⎪
2
⎪
⎨
(0, 1, 0, 0, −1, 0, 0, 0, 0)
√
,
v2 =
⎪
2
⎪
⎪
⎪
−1, 0, 0, 0)
⎪
⎩ v3 = (0, 0, 1, 0, 0,
√
,
2
⎧
(1, 0, 0, 1, 0, 0, −2µ, 0, 0)
⎪
w1 =
,
⎪
⎪
⎪
2(1 + 2µ2 )
⎪
⎪
⎨
(0, 1, 0, 0, 1, 0, 0, −2µ, 0)
,
w2 =
⎪
2(1 + 2µ2 )
⎪
⎪
⎪
(0, 0, 1, 0, 0, 1, 0, 0, −2µ)
⎪
⎪
⎩ w3 =
.
2(1 + 2µ2 )
220
8 STABILITY AND SMALL OSCILLATIONS
Therefore the unitary matrix that diagonalizes the second matrix in (8.1) is
given by
⎞
⎛
µI
µI
I
⎜
1 + 2µ2
1 + 2µ2
1 + 2µ2 ⎟
⎟
⎜
⎟
⎜
−I
I
⎟.
√
√
O
B=⎜
⎟
⎜
2
2
⎟
⎜
⎠
⎝
I
−2µI
I
2
2
2
2(1 + 2µ )
2(1 + 2µ )
2(1 + 2µ )
By means of B the Lagrange equations (8.1) can be rewritten as
⎛
⎞
⎞
⎞ ⎛
⎛
ξ¨
ξ
OO
O
⎠ B ⎝ η ⎠ = 0.
O
B ⎝ η̈ ⎠ + ν 2 ⎝O I
2
ζ/µ
O O (1 + 2µ )I
ζ̈/µ
(9.1)
10 Degenerate Vibrations
The first three components of the normal coordinates written in vector
form are
1
µ
θ1 =
ξ +η + 2ζ .
2
µ
1 + 2µ
Recalling the definition of ξ, η, ζ and that µ2 = m/M , compute
1
µ
[m(P1 + P2 ) + M P3 ]
θ1 =
2
1 + 2µ m
µ
2m + M m(P1 + P2 ) + M P3
=
2
m
2m + M
1 + 2µ
1 + 2µ2
Po ,
=
µ
where Po is the center of mass of the system. Therefore the first three equations
of (9.1), written in vector form, are
P̈o = 0,
or
Ṗo = const.
(10.1)
This integral was already known expressing the conservation of momentum.
Thus the effect of the first three normal coordinates is to permit vibrations
to occur in a reference system that translates rectilinear uniform motion with
respect to the center of mass. The Lagrange equations (9.1) are in intrinsic
vector form and independent of a reference frame. Having assumed that a
configuration of the system is in stable equilibrium does not preclude such a
configuration being realized in some other inertial system. This is referred to
as degenerate fundamental vibration.
11 Fundamental Vibrations
221
Fix now an inertial Cartesian system Σ centered at Po and in (10.1) choose
Ṗo = 0. This implies θ1 = 0 identically and
Po = O
and
P1 + P2 = −
P3
.
µ2
(10.2)
11 Fundamental Vibrations
The remaining six normal coordinates are, in vector form,
(P1 − P1,o ) − (P2 − P2,o )
√
,
2
(P1 − P3 ) + (P2 − P3 )
θ3 =
.
2(1 + 2µ2 )
θ2 =
The corresponding Lagrange equations are, still in vector form,
k
2
,
θ̈2 + ω2 θ2 = 0, ω2 =
m
(11.1)
k(1 + 2µ2 )
2
θ̈3 + ω3 θ3 = 0, ω3 =
.
m
Since these are independent oscillators, the physical significance of each of
them is drawn by setting the remaining ones equal to zero. Setting θ3 = 0
and taking into account the last of (10.2) implies that P3 is fixed at the origin
and consequently P1 = −P2 . Then the first equation of (11.1) can be written
equivalently in terms of P1 − P1,o or P2 − P2,o as
(Pi − Pi,o )′′ + ω22 (Pi − Pi,o ) = 0,
i = 1, 2.
These have solutions
Pi − Pi,o = Ai cos(ω2 t + ϕi ),
i = 1, 2,
where Ai are constant vectors and ϕi are real constants. Thus the fundamental
vibration θ2 describes the oscillation of Pi , i = 1, 2, about their equilibrium
configurations Pi,o . Since P1 + P2 = 0, the two oscillations have the same
amplitudes and are in opposition of phase, i.e., ϕ1 = ϕ2 + π.
For the fundamental vibration corresponding to ω3 set θ2 = 0. Then taking
into account (10.2),
(P1 − P1,o ) − (P2 − P2,o ) = 0,
P̈3 + ω32 P3 = 0.
Taking into account that P1,o + P2,o = 0, compute
−P3 = µ2 (P1 + P2 )
= µ2 [(P1 − P1,o ) + (P2 − P2,o )]
= B cos(ω3 t + ϕ),
222
8 STABILITY AND SMALL OSCILLATIONS
where B is a constant vector and ϕ is a real constant. From these, we obtain
Pi − Pi,o =
B
cos(ω3 t + ϕ),
2µ2
P3 = B cos(ω3 t + ϕ + π).
Therefore the points Pi exhibit oscillations of frequency ω3 about their equilibrium configurations Pi,o , of equal amplitude and in concurrence of phase.
The point P3 exhibits oscillations about the origin of amplitude B, frequency ω3 , and in opposition of phase with respect the two “exterior” points
P1 and P2 .
Problems and Complements
2c Lyapunov Stability Criteria
2.1c Harmonic Oscillator and Exponential Decay
For fixed ω ∈ R consider the two dynamical systems
q̇ = p,
q̈ ± ω 2 q = 0, or equivalently
ṗ = ∓ω 2 q.
(2.1c)
Verify that F (q, p) = 12 (ω 2 q 2 + p2 ) is a Lyapunov function for (2.1c)+ and the
origin, and conclude that such a point is a configuration of stable equilibrium.
Verify that F (q, p) = qp is a function relative to (2.1c)− that satisfies (3.2)
and identify the configurations of unstable equilibrium.
2.2c Damped Oscillator
Consider the damped oscillator (3.3c) of Chapter 3,
q̇ = p,
2
q̈ + 2εq̇ + ω q = 0,
or equivalently
ṗ = −(2εp + ω 2 q).
(2.2c)
A Lyapunov function relative to the origin is
2F (q, p) = ω 2 + 2ε2 q 2 + p2 + 2εqp.
This is positive in a neighborhood of the origin of R2 , except at the origin.
Prove that along solutions of (2.2c),
Ḟ (q, p) = −ε ω 2 q 2 + p2 ,
and conclude that the origin is a configuration of asymptotically stable equilibrium. Observe that the energy is not conserved and that asymptotic stability
is possible only because of dissipations, i.e., only if ε > 0.
4c Dirichlet Stability Criteria
223
4c Dirichlet Stability Criteria
4.1c Point Mass Constrained on a Curve
Let a curve on the vertical plane of (x, y) be represented as the graph of a
smooth function f defined in R and such that f (0) = f ′ (0) = 0. A point mass
{P ; m} slides on it, with no friction, and subject to its weight. The system
has one degree of freedom, and taking q = x as Lagrangian coordinate, one
computes
V = −mgf (q),
T = 21 mq̇ 2 1 + f ′2 (q) ,
E(q, q̇) = 21 mq̇ 2 1 + f ′2 (x) + mgf (q).
Since the energy is conserved, E = Eo and one computes
q̇ 2 =
2 Eo − mgf (q)
,
m 1 + f ′2 (q)
provided
f (q) ≤
Eo
.
mg
(4.1c)
4.1.1c Isolated Maxima of the Potential V
If the origin is an isolated minimum for f , then Eo > 0, and the equation
f (q) = Eo /g has two distinct roots
q1 < 0 < q2
such that
f (q) <
Eo
g
in the interval
(q1 , q2 ).
Therefore any initial position qo ∈ (q1 , q2 ) and initial velocity q̇o for which
E(qo , q̇o ) = Eo generate a motion whose Lagrangian trajectories remain in
the interval (q1 , q2 ). Quantify the configuration of stable equilibrium of the
origin in terms of the two variables (q, q̇).
The stable nature of the origin as an equilibrium configuration can be
established indirectly by tracing the trajectories (q, q̇) in phase space. Using
the first equation of (4.1c), prove that such trajectories are symmetric with
respect to the q-axis are closed curves surrounding the origin and represent
periodic motions.
4.1.2c Isolated Minima of the Potential V
If the origin is an isolated maximum for f , then Eo is of variable sign. If
Eo > 0 the first equation of (4.1c) implies that |q̇| > 0. Therefore the abscissa
q starting from qo is always increasing or decreasing, and {P ; m} departs
indefinitely from its equilibrium configuration. Assuming Eo > 0, trace the
curves (q, q̇) in phase space and show that while these might temporarily
approach the origin, they will eventually depart from it indefinitely.
If Eo < 0, the motion is possible only outside the interval (q1 , q2 ). Prove
that there exist trajectories that depart indefinitely from such an interval.
Prove that a trajectory originating from a point qo outside such an interval
never approaches either q1 or q2 . Discuss the case Eo = 0.
224
8 STABILITY AND SMALL OSCILLATIONS
4.1.3c Nonisolated Minima of the Potential V
Assume that there is an interval (−δ, δ) where f = 0. Prove that the origin
is a configuration of unstable equilibrium. Examine the cases in which the
points of (−δ, δ) are maxima or minima or neither for f .
4.2c On the Dirichlet Instability Criterion
For systems subject only to their weight, the Dirichlet criterion can be given
the following form.
Proposition 4.1c Let {M; dµ} be a mechanical system subject only to its
weight and constrained by smooth, fixed, holonomic constraints. Then the configurations of stable equilibrium are the isolated minima of the quote of the
center of mass. Stationary points of the quote of the center of mass that are
not minima are configurations of unstable equilibrium.
4.3c Rigid Rod with Extremities on a Parabola
The extremities A and B of a material homogeneous rod are constrained to
slide on the workless parabola 2hy = x2 , for some given h ∈ R.
The system has one degree of freedom and as Lagrangian parameter choose
the abscissa q of the center of mass Po of the rod. The coordinates (xA , yA )
and (xB , yB ) of the extremities of the rod satisfy the equation of the parabola,
and the coordinates of the center of mass satisfy
2(xPo , yPo ) = [(xA + xB ), (yA + yB )] = 2(q, y(q)).
From these, we have
2hy(q) = q 2 +
h2 ℓ 2
.
4(h2 + q 2 )
Assuming first that h > 0, prove that:
(i) If h < ℓ, there are three equilibrium configurations. Of these, two are
stable, whereas the one corresponding to q = 0 is unstable.
(ii) If h ≥ ℓ, then q = 0 is the only equilibrium configuration, and it is stable.
Assuming now that h < 0 prove that:
(i) If −h < ℓ, there are three equilibrium configurations, of which only q = 0
is stable.
(ii) If −h ≥ ℓ, then q = 0 is the only equilibrium configuration, and it is
unstable.
4c Dirichlet Stability Criteria
225
4.4c Miscellaneous Problems
Complete part (c) of the problem of §3.4c of the Complements of Chapter 5,
by determining the nature of the equilibrium configurations.
Complete the problems of §3.4.4c and §3.4.5c of the Complements of
Chapter 5, by determining the nature of the equilibrium configurations.
Complete the problem of §7c of the Complements of Chapter 6, by determining the nature of the equilibrium configurations.
Assume that the double pendulum of §4.5c of the Complements of
Chapter 5 is subject only to its weight. Find all the equilibrium configurations, and prove that ϕ = θ = 0 is the only configuration of stable
equilibrium.
4.5c Material Rod and Point Mass {P ; m} Connected by a Spring
A material homogeneous rod of extremities O and B, mass M , and length 2ℓ
moves with its extremity O constrained by a fixed cylindrical hinge. A point
mass {P ; m} slides on the horizontal workless guide taken as the axis of abscissas. The second extremity B of the rod is connected to P by a spring
of elasticity constant k. Determine the equilibrium configurations and their
nature.
The system has two degrees of freedom, and as Lagrangian coordinates,
formed by the rod with the
take the abscissa q of P and the angle ϕ = BOP
positively oriented horizontal axis. The potential of the external forces is
V = ℓM g sin ϕ − 12 k q 2 − 4ℓq cos ϕ + const,
and the equilibrium configurations are determined by setting
∇V = −kq + 2ℓk cos ϕ,
ℓM g cos ϕ − 2kℓq sin ϕ = 0.
Therefore the equilibrium configurations are the solutions of the system
q = 2ℓ cos ϕ,
cos ϕ(M g − 4kℓq sin ϕ) = 0.
Apart from the trivial solution (0, ±π/2), the systems admits the configurations
1
Mg
2
2
,
(4kℓ) − (M g) , arcsin
4kℓ
2k
(4.2c)
−1
Mg
2
2
,
(4kℓ) − (M g) , π − arcsin
2k
4kℓ
provided (M g/4kℓ) ≤ 1. The Hessian of V is
−k
−2kℓ sin ϕ
Hess(V ) = det
−2kℓ sin ϕ −ℓM g sin ϕ − 2kℓq cos ϕ
= kℓ (M g sin ϕ + 2kq cos ϕ) − 4k 2 ℓ2 sin2 ϕ.
226
8 STABILITY AND SMALL OSCILLATIONS
From this one computes
Hess(V )0,±π/2 = ±kℓ (M g ∓ 4kℓ) ,
Vqq = −k < 0.
Therefore (0, −π/2) is a configuration of unstable equilibrium, whereas
(0, π/2) is stable if M g > 4kℓ and unstable if M g < 4kℓ. In this latter
case also the two expressions of (4.2c) are equilibrium configurations. Computing the Hessian of V in each of them gives the same value
4k 2 ℓ2 − 41 M 2 g 2 > 0.
Thus if M g < 4kℓ, the configurations (4.2c) are of stable equilibrium. Finally,
is M g = 4kℓ, one recovers the equilibrium configurations ϕ = ±π/2.
5c Stability and Instability of Poinsot Precessions
5.1c Instability of Rotations about the Intermediate Axis
Let the moments of inertia Ii and the semiaxes ai of the ellipsoid of inertia
be ordered as
I1 < I2 < I3 ,
a1 > a2 > a3 ,
λ
ai = √ .
Ii
Consider a Poinsot precession with initial datum ω o close to the intermediate
axis, say, for example,
2
3
ωo,1
+ (a2 − ωo,2 )2 + ωo,3
≤ ε2 ,
0 < ε ≪ a2 .
Compute I from (3.3) and (4.1) of Chapter 7,
I=
2
2
Ii2 ωo,i
I 2ω2
λ
= i 2i =
2 ,
h
Ii ωi
Ii ωo,i
and observe that I < I3 . There are three possible cases:
I1 < I < I2 < I3 ,
I1 < I2 < I < I3 ,
I1 < I2 = I < I3 .
In the first case, the polhode, i.e., the trace of ω on Eλ , is a curve surrounding the major semiaxis a1 . In the second case, the polhode is a closed curve
traced on Eλ and surrounding the minor semiaxis. This follows from the classification of the polhodes in §3.2.1 of Chapter 7. In either case ω departs
from its initial configuration ωo , along its polhode, thereby generating instability. Using the integration procedure of §4 of Chapter 7, prove that ω covers
the entire polhode, with a period that can be computed only in terms of Ii
and I. In the third case, the polhode is the intersection of Eλ with the degenerate Poinsot cones. These intersections consist of two ellipses each through
6c Small Oscillations
227
the points (0, ±a2 , 0). This follows from §3.2.3 of Chapter 7. The vector ω
departs from its initial configuration ω o along one of these ellipses.
Using the integration procedure of §4 of Chapter 7, prove that ω covers the
entire polhode in an infinite time. Thus ω moves very slowly along its polhode.
For this reason the instability of such a motion is essentially undetectable, and
for small times the motion appears to be stable.
5.2c Ellipsoids of Rotation
Let I1 = I2 = I and take an initial datum close to any one of the equatorial
axes, say, for example,
λ
a= √ .
I
Computing I, one has I1 = I2 < I < I3 . By Proposition 3.1 of Chapter 7,
the polhodes are circles traced on Eλ , parallel to the equatorial circle and
surrounding the gyroscopic axis. Thus the configuration of ω on an equatorial
axis is unstable. Using the integration procedure of §4 of Chapter 7, prove
that ω covers the entire polhode, with a period that is inversely proportional
to the number ε. Thus ω moves very slowly along its polhode, and the motion
appears to be stable. For ε = 0, the initial datum ω o lies on an equatorial
axis; such an axis, remains fixed, and the resulting rotation is stable. Thus,
roughly speaking, as ε → 0, the slow instability tends to stability.
Finally, if Eλ is a sphere, all rotations are permanent, and every configuration is of stable equilibrium.
2
(ωo,1 − a cos θ)2 + (ωo,2 − a sin θ)2 + ωo,3
≤ ε2 ,
6c Small Oscillations
6.1c Small Oscillations of the System in §4.5c
Assume M g > 4kℓ and study the small oscillations about the equilibrium
configuration (0, π/2). One computes
Vqq Vqϕ
k 2kℓ
.
=−
2kℓ ℓM g
Vqϕ Vϕϕ
(q,ϕ)=(0,π/2)
Therefore for (q, ϕ) near (0, π/2),
q
k 2kℓ
3
+ O |(q, ϕ), (q̇, ϕ̇)| .
V (q, ϕ) = −(q, ϕ)
ϕ
2kℓ ℓM g
One also computes
2
2T = mq̇ +
2 2
4
3 M gℓ ϕ̇
q̇
m
0
.
= (q̇, ϕ̇)
ϕ̇
0 43 M gℓ2
Determine the normal coordinates and the normal modes about the configuration of stable equilibrium (0, π/2).
228
8 STABILITY AND SMALL OSCILLATIONS
6.2c Computing the Normal Frequencies
The following method provides an efficient procedure to compute the normal
modes ωh independently of the normal coordinates. Linearize the Lagrangian
and write down the linearized equations
Ahk q̈k + Bhk qk = 0,
Akh = Ahk (0),
Bhk = Vqh qk (0).
Motivated by (7.2), we look for solutions of the type
qh = Ch cos(ωt + ϕh ),
h = 1, 2, . . . , N,
where Ch and ϕh are constants to be determined. Putting this in the linearized
Lagrange equations gives
Ahk Ck ω 2 − Bhk Ck = 0,
h = 1, 2, . . . , N.
This is a homogeneous linear algebraic system in the unknowns Ck , which
has a nontrivial solution C = (C1 , C2 , . . . , CN )t if the determinant of the
coefficient is zero, i.e., if
det Ahk ω 2 − Bhk = 0.
The latter in an algebraic equation of degree N that admits N solutions, real
or complex, counted with their multiplicities.
Prove that if the origin is a configuration of stable equilibrium, these roots
are precisely the normal modes of the system.
6.3c Double Pendulum in §4.5c of the Complements of Chapter 5
Kinetic energy T and potential V are computed as
2T = ℓ2 (M + 31 m)ϕ̇2 + 13 M L2 θ̇2 + M ℓLϕ̇θ̇ cos(θ − ϕ),
V = 12 mgℓ cos ϕ + (ℓ cos ϕ + 21 L cos θ)M g + const.
Linearizing these about the configuration of stable equilibrium ϕ = θ = 0
gives
1
⎞
⎛ 2
1
ℓ M + 3 m 2 M ℓL
⎠ ϕ̇ ,
2To = ϕ̇ θ̇ ⎝
θ̇
1
1
2
2 M ℓL
3ML
⎛
2Vo = −g ϕ θ ⎝
ℓ(M + 12 m)
0
0
1
2 LM
⎞
⎠ ϕ .
θ
7c Degenerate Vibrations
229
Therefore the linearized Lagrange equations are
⎞
⎛ 2
⎞
⎛
ℓ (M + 31 m) 21 M ℓL
0
ℓ M + 21 m
ϕ̈
⎝
⎠
⎠ ϕ = 0.
+g⎝
θ
θ̈
1
1
1
2
0
2 M ℓL
3ML
2 LM
Find the normal coordinates and the normal modes in the case that L = ℓ
and hence M = m.
7c Degenerate Vibrations
We have assumed that the configuration {m; P1,o }, {m; P2,o }, {M ; 0} is of stable equilibrium. We will derive a condition on the masses m and M that would
guarantee such a stability independently of the elastic constant k. Eliminating
the variable ζ from (10.2), one computes the kinetic energy T and the potential
V only in terms of ξ and η:
2
T = (1 + µ2 )(ξ˙2 + η̇ 2 ) + 2µ2 ξ˙ · η̇,
m
2
V = −(1 + 2µ2 )(ξ 2 + η 2 ) − 4µ2 (1 + µ2 )ξ · η.
k
In matrix form,
2
T =
m
t
ξ˙
ξ˙
,
(Ahk )
η̇
η̇
2
− V =
k
t
ξ
ξ
,
(Bhk )
η
η
where the matrices (Ahk ) and (Bhk ) are given by
µ2 I
(1 + µ2 )I
,
(Ahk ) =
µ2 I
(1 + µ2 )I
(1 + 2µ4 )I 2µ2 (1 + µ2 )I
(Bhk ) =
.
2µ2 (1 + µ2 )I (1 + 2µ4 )I
The indicated equilibrium configuration is stable if (Bhk ) is positive definite,
i.e., if for all (ξ, η)t ∈ R6 − {0},
t
ξ
ξ
= (1 + 2µ4 )(ξ 2 + η 2 ) + 4µ2 (1 + µ2 )ξ · η
(Bhk )
η
η
= 2µ2 (1 + µ2 )(ξ + η)2 + (1 − 2µ2 )(ξ 2 + η 2 ) > 0.
The condition is then 2µ2 < 1, i.e., M > 2m.
9
VARIATIONAL PRINCIPLES
1 Maxima and Minima of Functionals
Given two points qo and q1 in RN and an interval [to , t1 ] ⊂ R, consider the
convex set K of all smooth curves parameterized with t ∈ [to , t1 ], and of
extremities qo and q1 , i.e.,
K = q ∈ C 1 [to , t1 ] q(to ) = qo , q(t1 ) = q1 .
Given q ∈ K and a vector-valued function ϕ ∈ Co∞ (to , t1 ), the curve {q + λϕ}
is still in K for all λ ∈ R. More generally, any modified path q + δq of a curve
q ∈ K remains in K if δq is smooth and δq(to ) = δq(t1 ) = 0. In such a case δq
is called a synchronous variation of the curve q ∈ K.
If q ∈ K is the orbit of a mechanical system in configuration space, a
synchronous variation δq might be regarded as a virtual variation from the
Lagrangian path q into the virtual Lagrangian path q + δqthat keeps
its extremities fixed. Given a function (ξ, η; t) → F (ξ, η; t) ∈ C 2 R2N +1 , consider
the functional
t1
K ∋ q −→ J(q) =
F q(t), q̇(t); t dt.
to
For each q ∈ K, the functional returns a real number. We ask whether J(q)
has a greatest lower bound as q ranges over K, i.e., if
t1
inf
F q(t), q̇(t); t dt > −∞.
q∈K
to
The infimum might not exist, or if it does, it might not be achieved by any
q ∈ K (see Problems 1.1c and 1.2c of the Complements). If the infimum exists
and it is achieved by some q ∈ K, such an orbit is called a minimum point
for J. Similarly one might ask whether J has a least upper bound. If such a
supremum exists and it is achieved for some q ∈ K, the orbit q is a maximum
point for the functional J. If q is a minimum point for J, for any ϕ ∈ Co∞ (to , t1 )
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 9,
c Springer Science+Business Media, LLC 2011
231
232
9 VARIATIONAL PRINCIPLES
the function of one real variable λ → J(q + λϕ) has a minimum for λ = 0.
Therefore
d
J(q + λϕ) λ=0 = 0
∀ϕ ∈ Co∞ (to , t1 ).
(1.1)
dλ
The same holds if q is a maximum point for J. However, if (1.1) holds for
some q ∈ K, such an orbit need not be a maximum or a minimum point (see
Problem 1.1c of the Complements).
1.1 Stationary Points of Functionals
If q ∈ K undergoes a synchronous variation δq, its derivative q̇ undergoes the
variation δ q̇. If δq and δ q̇ are both infinitesimal, we compute the corresponding
variation of F (q, q̇; t) as
δF q, q̇; t = F q + δq, q̇ + δ q̇; t − F q, q̇; t
=
∂F (q, q̇; t)
∂F (q, q̇; t)
δqh +
δ q̇h
∂qh
∂ q̇h
=−
∂F
d ∂F
−
dt ∂ q̇h
∂qh
δqh +
∂F
δqh .
∂ q̇h
d
dt
The corresponding variation of the functional J is
t1
δJ(q) =
F q + δq, q̇ + δ q̇; t − F q, q̇; t dt
to
=
t1
to
δF q, q̇; t dt
t1
∂F
d ∂F
∂F
δqh dt
δqh dt −
−
=
∂
q̇
dt
∂
q̇
∂q
h
h
h
to
to
t1 t1
∂F
d ∂F
∂F
δqh dt.
=
δqh −
−
∂ q̇h
dt
∂
q̇
∂q
h
h
to
to
t1
d
dt
(1.2)
Since δq(to ) = δq(t1 ) = 0, we finally compute
δJ(q) = −
t1
d ∂F
∂F
−
dt ∂ q̇h
∂qh
to
δqh dt.
(1.3)
Whenever the synchronous variation δq is of the type δq = ϕdλ, for some
ϕ ∈ Co∞ (to , t1 ), the variation of J in (1.3) takes the form
δJ(q) = −dλ
t1
to
d ∂F
∂F
−
dt ∂ q̇h
∂qh
ϕh dt.
(1.3)′
2 The Least Action Principle
233
Remark 1.1 The differential in (1.3) is, in general, topologically different
from the differential in (1.3)′ . The latter is called a Gâteaux differential. Having fixed q ∈ K and ϕ ∈ Co∞ (to , t1 ), such a differential is computed along
the orbits λ → q + λϕ. These can be regarded as “lines” in K through q
and with fixed “slope” ϕ. The differential appearing in (1.3) is the Fréchet
differential. For q ∈ K fixed, such a differential is computed along any “path”
q + δq ∈ K. In what follows we will not distinguish them further, since for
smooth F (ξ, η; t), these two notions of differential coincide [4].
Having (1.1)–(1.3)′ as a guideline, we say that q ∈ K is a stationary point
for the functional J if δJ(q) = 0. Extremum points, if any, are stationary
points. Indeed, if q ∈ K is a minimum point, (1.2) implies that δJ(q) ≥ 0 for
all δ q̂. Writing (1.2) for δq and −δq gives ±δJ(q) ≥ 0, and thus δJ(q) = 0.
The next proposition provides a characteristic condition for q ∈ K to be a
stationary point for J.
Proposition 1.1 A curve q ∈ K is a stationary point for the functional J if
and only if it is a solution of the system of differential equations
d ∂F
∂F
−
= 0,
dt ∂ q̇h
∂qh
h = 1, 2, . . . , N.
(1.4)
Proof. If q ∈ K solves (1.4), it follows from (1.3)–(1.3)′ that it is a stationary
point for J. Conversely, if q ∈ K is a stationary point for J, then (1.3)′ gives
t1
to
d ∂F
∂F
−
dt ∂ q̇h
∂qh
ϕh dt = 0
∀ϕ ∈ Co∞ (to , t1 ).
By virtue of (1.4), stationary points for J are only those that satisfy a
system of differential equations similar to the Lagrange equations. The characteristic condition (1.4) does not contain sufficient information to determine
the nature of a stationary point. Conditions that would identify a stationary
point as an extremum involve the second variation of J and thus the structure of the function F [144], [12, Chapter II, 37–64]. Likewise, (1.4) does not
contain sufficient information for one to infer uniqueness of stationary points.
2 The Least Action Principle
The motion t → q(t) of a mechanical system subject to smooth, holonomic
constraints satisfying the principle of virtual work and acted upon only by
conservative forces is determined by the Lagrange equations through the
Lagrangian L, starting from some given initial conditions. Fix a time interval
(to , t1 ) along the time evolution of the motion and set q(to ) = qo and
q(t1 ) = q1 . Then for t ∈ (to , t1 ), the orbit t → q(t) might also be regarded
234
9 VARIATIONAL PRINCIPLES
as a solution of the two-point boundary value problem (see §2.1c of the
Complements)
d ∂L
∂L
−
= 0, h = 1, . . . , N ,
dt ∂ q̇h
∂qh
(2.1)
q(to ) = qo , q(t1 ) = q1 .
Having determined the Lagrangian L of a mechanical system, the functional
t1
L(q, q̇; t)dt
(2.2)
K ∋ q → S(q) =
to
is the action or the Hamiltonian action of the system. The variational principle
of Proposition 1.1 in this context takes the following form.
Theorem 2.1 (Hamilton Principle of Stationary Action). The motion
of a mechanical system subject to smooth, holonomic constraints satisfying the
principle of virtual work and acted upon only by conservative forces evolves
along a Lagrangian trajectory q ∈ K in configuration space, which is a stationary point for the Hamiltonian action.
Remark 2.1 It can be shown that stationary points of the Hamiltonian action cannot be maxima (§2.2c of the Complements). Moreover, while solutions
of (2.1) need not be a minimum for the action q → S(q), they are always minima in some “local sense”(§2.3c of the Complements). For this reason the
Hamilton principle of stationary action is commonly referred to as the principle of least action.1
2.1 Reduced Action and Maupertuis Principle
If the constraints are fixed, the Lagrangian, and hence the Hamiltonian, are
independent of time, and H(p, q) = γ, for some constant γ, is an integral of
motion. Rewrite the action as
t1
S(q) =
ph q̇h − H(q, p) dτ
to
q1
=
qo
1
ph dqh −
t1
to
H(q, p)dτ.
A first form of such a principle was introduced by Hamilton in [69]. While motivated by the principle of path of least time of geometrical optics (see §4), Hamilton
perceived its general breadth. From his introduction: . . . A certain quantity which
in one physical theory is the “action” and in another “time,” expended by light in
going from any first to any second point, is found to be less than if the light had
gone in any other than its actual path . . . The mathematical novelty of my method
consists in considering this quantity as a function. . . and in reducing all researches
respecting optical systems of rays to the study of this single function: a reduction
which presents mathematical optics under an entirely novel view, and one analogous
(as it appears to me) to the aspect under which Descartes presented the application
of Algebra to Geometry. . . . Applications to mechanics are in [72, 73].
2 The Least Action Principle
235
In these integrals the parameter t need not be time, and it could be replaced
by any other parameter s ∈ [so , s1 ], provided q(so ) = qo and q(s1 ) = q1 .
Extrema of S(q) could then be found for curves s → q(s) in configuration
space, expressed in terms of such a new parameter. The energy integral H = γ
suggests looking for the stationary points of the action among those curves in
phase space, parameterized to enforce the energy integral. Set
[so , s1 ] ∋ s → q(s), p(s) ∈ C 1 [so , s1 ] such that
Kγ =
.
q(so ) = qo , q(s1 ) = q1 , and H(q, p) = γ
For such curves in phase space,
S(q) =
q1
qo
ph dqh − γ(t1 − to ).
The integral on the right-hand side is the reduced action and is denoted by
So . Since stationary points of S(q) yield a curve in Kγ , they can be found as
the stationary points of the reduced action in Kγ . Thus, for fixed constraints,
the motion can be determined by
q1
δ γ So = δγ
ph dqh = 0,
(2.3)
qo
where δγ denotes the variation effected in Kγ . This is a special case of the
Hamilton principle of stationary action, and is referred to as the Maupertuis
principle. It was visualized by Maupertuis in 1744 [119, Vol. II, no 328] and
mathematically formalized by Euler [53] and Lagrange [98].
The variational principle in (2.3) is of purely geometrical nature, since it involves curves in Kγ irrespective of their parameterization, provided H = const.
As such it determines the geometrical form of the trajectories irrespective of
the time-law on them, as illustrated by the following example.
2.1.1 Motion along Geodesics
A point moves on a fixed smooth surface S and is unsolicited and unconstrained otherwise. Then denoting by ds its elemental arc length, we obtain2
1 ds
2 dt
2
=T =L=H
and
ph q̇h =
∂T
q̇h = 2T.
∂ q̇h
The reduced action takes the form
t1 √
t1
q1
ds
dt = 2γ
ph q̇h dt =
ds,
2T
So =
dt
to
to
qo
whose stationary points are the geodesics on S.
2
Combine the remarks of §1.4c with those in §§3.2 and 4 of Chapter 6.
236
9 VARIATIONAL PRINCIPLES
3 The Hamilton–Jacobi Equation
If in (2.2) the second limit of integration (q1 ; t1 ) is regarded as variable (q; t),
the Hamiltonian action can be regarded as a function of (q; t):
V (q; t) =
t
to
L(q, q̇; τ )dτ =
t
to
[ph q̇h − H(q, p; τ )] dτ.
For a fixed t consider a variation q + δq from the path q that keeps the
Lagrangian configuration q(to ) and velocity q̇(to ) at time to and is otherwise
unrestricted. Thus in particular, δq(to ) = δ q̇(to ) = 0, and no further restrictions are placed at time t. Computing the corresponding variation of V (q; t)
gives
t
δV (q; t) = δ
L(q, q̇; τ )dτ
to
t
∂L(q, q̇; τ )
∂L(q, q̇; τ )
δqh +
δ q̇h dτ
∂qh
∂ q̇h
to
t
t
d ∂L(q, q̇; τ ) ∂L(q, q̇; τ )
∂L(q, q̇; τ )
δqh −
−
=
∂ q̇h
dt
∂ q̇h
∂qh
to
to
=
δqh dτ.
Therefore if q is a solution of (2.1),
δV (q; t) = ph δqh .
Since the virtual variation δq is arbitrary,
∂V (q; t)
= ph ,
∂qh
h = 1, . . . , N.
(3.1)
Starting from (3.1), we now compute the total derivative of t → V q(t); t .
This gives
∂V (q; t) ∂V (q; t)
d
V (q; t) =
+
q̇h = L(q, q̇; t).
dt
∂t
∂qh
From this and (3.1),
∂V (q; t)
= L(q, q̇; t) − ph q̇h = −H(q, p; t).
∂t
Therefore the action V (q; t) is a solution of the Hamilton–Jacobi equation
[73, 90]
∂V
+ H (q, ∇q V ; t) = 0.
(3.2)
∂t
4 The Functional of Geometrical Optics
237
4 The Functional of Geometrical Optics
The Fermat principle states that a light ray travels between any two points
qo and q1 of a medium along the path of least time [56, Vol. I, 170–172].
Parameterizing the trajectory q of the ray with either time t or arc length s,
the instantaneous unit tangent along the direction of propagation is3
τ (s) = t(t) =
d
q̇
=
q(s).
q̇
ds
The speed v (q, t) of propagation of a light ray is, in general, a function of
the position q of the ray and its direction t. For example, in crystals the speed
of a light ray depends on its direction of propagation. The refraction index
ν(q, t) is defined as
1
def
ν(q, t) =
.
v (q, t)
A medium is optically isotropic if its refraction index is independent of t;
it is optically homogeneous if the refraction index is constant. The refraction
index is a property of the medium; it gives the speed of a light ray when it
goes instantaneously through the position q with direction t.
With this symbolism, the infinitesimal time dt it takes the light ray to
cover the infinitesimal distance ds is given by
dt = ν(q, τ )ds = ν(q, t) q̇ dt.
If the trajectory t → q(t) of a light ray from qo to q1 is known, the time of
travel is
t1
q1
T =
ν(q, τ )ds.
ν(q, t) q̇ dt =
to
qo
In the latter of these equalities the time dependence is eliminated and the
integral depends only on the geometrical properties of the trajectory, through
the refraction index. This suggests that we seek the trajectory of a light ray
by expressing Fermat’s principle of least time in the form
t1
q1
T = inf
ν(q, t) q̇ dt =
inf
ν(q, τ )ds,
(4.1)
q∈K
to
q∈K(qo ,q1 )
qo
where K(qo , q1 ) is the set of all Smooth curves of endpoints qo and q1 parameterized with their intrinsic arc length. For optically homogeneous media the
latter equality implies that the curve of least time is the line segment joining
qo to q1 . Since qo and q1 are arbitrary, this implies that in media with constant
refraction index, light rays travel along straight lines. By Proposition 1.1 the
3
With perhaps improper symbolism we have used the same symbol to denote
q(t) and q(s), the instantaneous position of the ray, with different parametric
representations.
238
9 VARIATIONAL PRINCIPLES
trajectory of least time must be a solution of the system of the Euler–Lagrange
equations4
d ∂ν(q, t) q̇
∂ν(q, t) q̇
−
= 0,
dt
∂ q̇h
∂qh
h = 1, 2, . . . , N.
(4.2)
Moreover, recalling that the refraction index is the reciprocal of the speed of
propagation of a light ray,
ν(q, t) q̇ = 1
along trajectories solutions (4.2).
(4.3)
Equation (4.2)–(4.3) are (N +1) relations to be satisfied by the N components
of the trajectory. However, of these only N are independent, since
q̇ → ν(q, t) q̇
is homogeneous of degree 1 in the variable q̇.
(4.4)
In the next sections we investigate the structure of these systems, and we will
put them in their canonical form.
5 Huygens Systems
Let (ξ, η) → F (ξ, η) ∈ C 2 (R2N ) be homogeneous of degree 1 in the variables
η, for all fixed ξ. By Euler’s theorem on homogeneous functions (§1.1c of the
Complements of Chapter 6),
F (ξ, η) =
∂F (ξ, η)
ηh .
∂ηh
Taking the derivative with respect to the generic variable ηk gives
∂ 2 F (ξ, η)
ηh = 0,
∂ηh ∂ηk
k = 1, . . . , N.
This can be regarded as an algebraic homogeneous linear system in the unknowns ηh . Since η is arbitrary, such a system admits nontrivial solutions.
Therefore
∂ 2 F (ξ, η)
det
= 0.
∂ηh ∂ηk
In particular, the system (4.2) is not normal. Assume momentarily that the
refraction index is independent of t. Then one verifies that
rank
∂ 2 ν(q) q̇
∂ q̇h ∂ q̇k
= N − 1.
Although optical trajectories are curves in R3 , we will regard them as in RN ,
to stress that the principles of geometrical optics are dimension-independent.
4
5 Huygens Systems
239
Therefore for isotropic media, (4.3) is the extra equation that makes the system normal. One also verifies that q̇ → [ν(q) q̇ ]2 is homogeneous of degree
2 and
∂ 2 (ν(q) q̇ )2
rank
= N.
∂ q̇h ∂ q̇k
In analogy with light propagation in isotropic media, consider systems of
the type
d ∂F (q, q̇) ∂F (q, q̇)
−
= 0,
h = 1, . . . , N ,
dt ∂ q̇h
∂ q̇h
(5.1)
F (q(t), q̇(t)) = 1,
where F (ξ, η) is homogeneous of degree 1 in η and
rank
∂ 2 F (q, q̇)
∂ q̇h ∂ q̇k
= N − 1,
∂ 2 F 2 (q, q̇)
∂ q̇h ∂ q̇k
rank
= N.
(5.2)
Systems of the type (5.1) in which F satisfies (5.2) are called Huygens systems
or systems of geometrical optics.5
Proposition 5.1 A smooth trajectory t → q(t) is a solution of the Huygens
system (5.1)–(5.2) if and only if is a solution of the normal system
d ∂ 21 F 2 (q, q̇) ∂ 12 F 2 (q, q̇)
−
= 0,
dt
∂ q̇h
∂qh
h = 1, . . . , N.
(5.3)
Proof. Let q be a solution of (5.1). Multiplying the first equation by F (q, q̇)
gives
d ∂ 12 F 2 (q, q̇) ∂ 21 F 2 (q, q̇)
∂F (q, q̇) d
F (q, q̇).
−
=
dt
∂ q̇h
∂qh
∂ q̇h dt
Conversely, if q solves (5.3), then
F (q, q̇)
d ∂F (q, q̇) ∂F (q, q̇)
−
dt ∂ q̇h
∂ q̇h
=
∂F (q, q̇) d
F (q, q̇).
∂ q̇h dt
Therefore the converse will follow if F (q, q̇) = 1 is an integral of (5.3). This, in
turn, is a consequence of the canonical form of (5.3). Such a system is normal
and has the structure of the Lagrange equations for the “Lagrangian”
q̇) = 1 F 2 (q, q̇)
L(q,
2
5
The property of η → F (η) being homogeneous of degree 1 does not imply any
restriction on the rank of the Hessian of F . Give examples of homogeneous functions
of degree 1 whose Hessian has rank 1, 2, . . . , (N − 1).
240
9 VARIATIONAL PRINCIPLES
whose Hessian with respect to the variables q̇ has maximum rank. Then,
introducing the “kinetic momenta”
ph =
∂ 21 F 2 (q, q̇)
,
∂ q̇h
h = 1, . . . , N,
the pair of variables (q, p) and the Hamiltonian
p) = ph q̇h − 1 F 2 (q, q̇)
H(q,
2
(5.4)
satisfy the canonical system
q̇h =
p)
∂ H(q,
,
∂ph
ṗh = −
p)
∂ H(q,
,
∂qh
h = 1, . . . , N.
(5.5)
Since q̇ → F 2 (q, q̇) is homogeneous of degree 2, by Euler’s theorem we have
ph q̇h =
∂ 12 F 2 (q, q̇)
q̇h = F 2 (q, q̇).
∂ q̇h
From this and (5.4),
p) = 1 F 2 (q, q̇) = 1 ph q̇h .
H(q,
2
2
(5.4)′
p) coincides with the Lagrangian 1 F 2 (q, q̇),
Therefore the Hamiltonian H(q,
2
whenever in the latter q̇ is expressed in terms of the new independent variables
independent of time, have the
(q, p). Now, systems of the type (5.5), with H
“energy integral” (§5.2 of Chapter 6)
p) = 1 F 2 (q, q̇) = const.
H(q,
2
6 Canonical Form of Huygens Systems
Even though the system (5.1) is not normal, the previous remarks permit one
to put it in canonical form. From (5.4) and (5.4)′ ,
p) =
2H(q,
p)
∂ H(q,
ph .
∂ph
p) is, for all fixed
Therefore by Euler’s theorem and its converse, H(q,
N
q ∈ R , homogeneous of degree 2 in the variables p. Write now the canonical
Hamiltonian system (5.5) and the Hamiltonian (5.4), computed along the
curve t → q(t), a solution of the Lagrangian system (5.1). Setting
p) = F (q, q̇)
H(q, p) = 2H(q,
6 Canonical Form of Huygens Systems
241
and taking into account (5.4)′ gives
H(q, p) =
and
q̇h =
∂H(q, p)
,
∂ph
p) = 1
2H(q,
ṗh = −
∂H(q, p)
,
∂qh
(6.1)
h = 1, . . . , N.
(6.2)
These is the canonical form of (5.1) with Hamiltonian H(q, p) = F (q, q̇),
where q̇h is regarded as expressed in terms of (q, p). Since the Hamiltonian H
is homogeneous of degree 2 in p, the new Hamiltonian H is homogeneous of
degree 1 in p, and by Euler’s theorem,
H(q, p) =
∂H(q, p)
ph ,
∂ph
∀q ∈ RN .
(6.3)
Computing this along solutions of (6.1)–(6.2), we obtain
H(q, p) = ph q̇h ≡ 1.
(6.4)
Proposition 6.1 Light rays travel along trajectories (q, p) in phase space,
such that ph dqh = dt.
6.1 Contact Virtual Differential Forms
The differential form ph δqh , where δ denotes a virtual variation, is called of
contact. A remarkable property of these forms is that they are invariant along
solutions of the canonical system (6.1)–(6.2).
Proposition 6.2 (Lie [114]) Let (q, p) be a solution of (6.1)–(6.2). Then,
as a function of time, pδq = const.
Proof. Since the time derivative
d
dt
and the virtual variation δ commute,
d
(ph δqh ) = ṗh δqh + ph δ q̇h
dt
∂H(q, p)
∂H(q, p)
δqh + ph δ
=−
∂qh
∂ph
∂H
∂H
∂H(q, p)
=−
δqh −
δph + δ
ph
∂qh
∂ph
∂ph
∂H(q, p)
ph .
= −δH + δ
∂ph
The assertion follows, since H(q, p) is homogeneous of the degree 1 in p and
(6.3) holds.
242
9 VARIATIONAL PRINCIPLES
7 Wave Fronts
For each pair (q, p) consider the elemental portion of the plane through q and
normal p, i.e.,
dπ(q, p) = ξ ∈ RN p · (ξ − q) = 0; ξ − q < δq ,
where δq is an infinitesimal variation of q. The portion dπ(q, p) is an infinitesimal front normal to p. Consider now the canonical system (6.1)–(6.2) with
initial conditions
q(to ) = qo ∈ RN fixed,
p(to ) = po ∈ RN arbitrary.
(7.1)
Thus the initial origin qo of the light ray is specified, but its initial direction of
propagation po is not restricted and it may vary in RN . Following the previous
geometric interpretation of a pair (p, q) ∈ R2N , these initial data represent a
bundle of ∞(N −1) elemental fronts, all through qo . Since light propagates in
all directions (in general with variable speed), studying the canonical system
(6.1)–(6.2) with the initial conditions (7.1) amounts
to investigating light
propagation from a point source. Denote by t → q(t; po ), p(t; po ) the solution
of (6.1)–(6.2) originating from the initial datum (qo , po ). The first quantity
is the trajectory of the ray, and the second describes the evolution of an
elemental front, normal to p(t; po ) at time t, originating from an elemental
front, normal to po at time to .
7.1 First Definition of Wave Front
The wave front at time t of light originating from a point source in qo is the
geometric set
Φ(qo ; t) =
q(t; po ).
(7.2)
po ∈ RN
Proposition 7.1 The wave front Φ(qo ; t) is a smooth (N − 1)-dimensional
surface, at least for t sufficiently close to to .
Proof. Since H(q, p) is homogeneous of degree 1 in p, the first equation of
(6.2) can be rewritten in the form (Proposition 1.1c of the Complements of
Chapter 6)
q̇h =
∂H(q, p/ p )
,
∂ph / p
h = 1, . . . , N.
Set
n(t) =
p(t; po )
,
p(t; po )
no =
po
.
po
(6.2)′
7 Wave Fronts
243
As po ranges over RN , the unit vectors no range over the whole unit sphere
SN −1 , and n(t) ranges over a subset of SN −1 . For t close to to , from (6.2)′ we
have
qh (t; n) = qo,h +
∂H(qo , n)
(t − to ) + oh (|t − to |; n).
∂nh
This is a smooth map from SN −1 into RN , and it represents a smooth manifold
whose dimension is the rank of the matrix
∂qh
∂nk
=
∂oh (|t − to |; n)
∂ 2 H(qo , n)
(t − to ) +
∂nh ∂nk
∂nk
.
Since H(q, p) = F (q, q̇) and the Hessian of F with respect to q̇ has rank
N − 1, also the Hessian of H with respect to n has rank N − 1. Therefore, for
t sufficiently close to to ,
rank
∂qh
∂nk
= rank
∂ 2 H(qo , n)
∂nh ∂nk
= N − 1.
7.2 Second Definition of Wave Front
Each of the ∞N −1 elemental portions of planes dπ(qo , po) is transformed,
at time t = to , into the infinitesimal portion of plane dπ q(t; po ), p(t; po ) .
The envelope of these infinitesimal wave fronts, as po ranges over the unit
sphere of RN , is the wave front of light generated at time t by a point source
in qo .
Such a front can be visualized as the “assembled motion” of the infinitesimal portions of planes of the bundle centered at qo . They move from their
initial position at time to and become reassembled, at each further time t, as
infinitesimal tiles of a surface.
7.3 Equivalence of the Two Definitions
The surface Φ(qo ; t) might be regarded as a moving constraint for its own
points. Therefore, the virtual displacements of a point q ∈ Φ(qo ; t) are those
tangent to Φ(qo ; t). Since Φ(qo ; to ) = qo , the only virtual displacements of qo
are zero. It follows from Proposition 6.2 that
p(t; po ) δq(t; po ) = po δqo = 0
for every virtual displacement δq(t; po ). Thus, for all po , the vector p(t; po ) is
normal to the tangent plane to the front Φ(qo ; t) at q(t, po ).
244
9 VARIATIONAL PRINCIPLES
7.4 Velocity of the Light Ray and of the Wave Front
Let the function F (q, q̇) appearing in the Huygens system (5.1)–(5.2), be given
by ν(q; t) q̇ . From this one computes the “kinetic momenta”
ph =
∂ 12 F 2 (q, q̇)
∂ν
= q̇h ν 2 + q̇ ν
∂ q̇h
∂tk
δhk −
q̇h q̇k
q̇
.
(7.3)
In isotropic media, ν depends only on the position q, and in such a case, (7.3)
becomes
ν 2 (q) q̇h = ph ,
h = 1, . . . , N.
(7.3)′
Therefore in isotropic media, the Lagrangian velocity q̇ of a light ray, along
its trajectory q, is normal to the wave front to which q belongs.
In an isotropic media this is no longer the case, and the direction of propagation of the front Φ(qo ; t), differs from the direction of a light ray that
penetrates it.
In isotropic and optically homogeneous media, the refraction index is a
constant νo . In such a case the second equation of (6.2) implies that all components of the initial unit sphere po = 1 remain constant along the motion.
Therefore by (7.3), every light ray travels along half-lines originating at qo at
constant speed νo−1 . Thus the wave front Φ(qo ; t) is a sphere centered at qo
and radius t/νo .
7.5 Normal Velocity and Normal “Slowness” of the Wave Front
The normal velocity of a wave front Φ(qo ; t) at one of its points q is the
projection of the velocity of q along the normal to the front, i.e., q̇h ph / p .
Since this is computed along optical trajectories,
q̇h
ph
1 ∂H(q, p)
1
H(q, p)
=
=
.
ph =
p
p
∂ph
p
p
(7.4)
Here we have taken into account that H(q, p) ≡ 1 along optical trajectories.
The same expression could be computed starting from (7.3), or alternatively
from Proposition 6.2. It follows from (7.4) that the larger the value of p ,
the slower the front moves. For this reason, Hamilton called p the vector of
normal slowness [71].
8 The Huygens Principle
Let Φ(qo ; t) be the wave front at time t, originating from a point source qo .
Every q ∈ Φ(qo ; t) might be considered itself as a point source, which generates
its own front Φ(q; τ ) at time τ > 0. The Huygens principle connects the front
Φ(qo ; t + τ ) originating at qo with the fronts Φ(q; τ ) as q ranges over Φ(qo ; t).
8 The Huygens Principle
245
Theorem 8.1 (Huygens [83, 84]). The front Φ(qo ; t + τ ) is the envelope of
the fronts Φ(q; τ ), as q ranges over Φ(qo ; t).
Proof. It will suffice to establish that for all ξ ∈ Φ(qo ; t + τ ), there exists some
q ∈ Φ(qo ; t) such that the front Φ(q; τ ) is tangent to Φ(qo ; t + τ ) at ξ. Fix
ξ ∈ Φ(qo ; t + τ ), and let q be the intersection of its trajectory of least time,
with Φ(qo ; t). Such a trajectory, when restricted to its portion between q and
ξ, is the trajectory of least time connecting these two points. This implies
that ξ ∈ Φ(q; τ ). Moreover, the same trajectory defines in ξ the same “kinetic
momentum” p. Therefore the two surfaces Φ(qo ; t + τ ) and Φ(q; τ ) have the
same tangent plane at ξ.
F (q0 ; t + r)
q
q0
F (q0 ; t)
Fig. 8.1.
8.1 Optical Length
A similar argument, still based on the local nature of the Fermat principle,
permits one to characterize further the wave fronts. For q ∈ RN let T (qo ; q)
be the time it takes a light ray originating at qo to reach q. The function
q → T (qo ; q) is called optical length of the path from qo to q. Now, all points q ∈
Φ(qo ; t) reach the wave front, starting from qo , at the same time t. Therefore
the wave front Φ(qo ; t) is a level set for the optical length, i.e.,
Φ(qo ; t) = q ∈ RN T (qo ; q) = t .
This implies that the gradient of q → T (qo ; q) is parallel to the “kinetic
momentum” p. The next proposition asserts that indeed these two vectors
coincide.
246
9 VARIATIONAL PRINCIPLES
Proposition 8.1 ∇q T (qo ; q) = p.
Proof. Having fixed q ∈ Φ(qo ; t), let ∆q be a variation of q and denote by ∆t
the corresponding variation of t, linked by
q + ∆q ∈ Φ(qo ; t + ∆t),
∆t = O(|∆q|).
By the Fermat principle, there exists a ray originating at q ∈ Φ(qo ; t) that
reaches Φ(qo ; t + ∆t) in the least time ∆t.6 By the local nature of such a
principle, this ray must be the extension to the time interval (t, t + ∆t) of the
ray that in the time interval (to , t) goes from qo to q. For such a ray, by (6.4)
the trajectory of such a ray, there must hold, by
ν(q, t) q̇ = F (q, q̇) = H (q, p) = ph q̇h ,
t=
q̇
.
q̇
Therefore
T (qo ; q + ∆q) − T (qo ; q) =
=
t+∆t
ν(q, t) q̇ dτ
t
t+∆t
ph q̇h dτ
t
= ph ∆qh + o(|∆q|).
It follows from (7.4) that those surfaces defined intrinsically by
∇q T (qo ; q) = const
(8.1)
are those wave fronts of constant normal velocity. Equation (8.1), introduced
by Hamilton in [71], is called the eikonal equation of geometrical optics. Special
solutions are functions whose level sets are spheres or hyperplanes.
Problems and Complements
1c Maxima and Minima of Functionals
1.1c A Functional with Stationary Points and No Extrema
Let N = 1 and consider the functional
1
J(q) =
q 3 (t)dt,
K = {q ∈ C 1 [0, 1] q(0) = q(1) = 0}.
0
Verify that q ≡ 0 is a stationary point for J and prove that J admits no
maxima or minima in K.
6
It is not asserted that the front Φ(qo ; t + ∆t) is reached at q + ∆q, only that
such a front is reached in the least time ∆t.
1c Maxima and Minima of Functionals
247
1.2c A Functional with a Minimum That Is Not Achieved
Let N = 1 and consider the functional
1
q̇ 2 (t)
dt,
K = {q ∈ C 1 [0, 1] q(0) = 1, q(1) = 0}.
J(q) =
2 (t)
1
+
q̇
0
Verify that 0 < J(q) ≤ 1 for all q ∈ K and inf q∈K J(q) = 0. However, such an
infimum is not achieved in K. One such q would have to satisfy
d
q̇
= 0.
dt 1 + q̇ 2
This implies q̇ = const, and therefore q(t) = 1 − t. However, J(1 − t) = 21 .
Consider now the sequence of curves
1 − nt t ∈ 0, n1 ,
t → qn (t) =
0
t ∈ n1 , 1 .
Verify that J(qn ) → 0 as n → ∞. The functions qn are not in C 1 [0, 1]. Modify
their construction to get the same conclusions for functions qn ∈ K.
1.3c The Brachistochrone
In a vertical plane fix two points Po = (xo , yo ) and P1 = (x1 , y1 ) such that
xo < x1 and yo > y1 . Denote by K the set of all curves joining Po and P1 that
can be parameterized with x ∈ (xo , x1 ), i.e.,
K = y ∈ C 1 [xo , x1 ], y(xo ) = yo , y(x1 ) = y1 .
A point mass “falls” by gravity from Po to P1 along one of these curves.
Denoting by s the arc length along one such curve,
ds = ṡ dt = ẋ2 + ẏ 2 dt = 1 + y ′2 (x) dx,
and the speed of the point is v = ṡ. Therefore the time it takes from the
“falling” point to reach P1 is
x1
1 + y ′2 (x)
dx.
T =
v
xo
If vo is the initial speed, by the energy integral we have
v 2 − vo2 = −2g(y − yo ),
which implies
v=
2g(y∗ − y),
2gy∗ = vo2 + 2gyo.
248
9 VARIATIONAL PRINCIPLES
Therefore,
1
T = √
2g
x1
xo
1 + y ′2 (x)
dx.
y∗ − y(x)
(1.1c)
The problem of the brachistrochrone is to find the curve in K for which the
time T is least. If such a curve exists, by Proposition 1.1 it must be a solution of
1 + y ′2 (x)
1 + y ′2 (x)
d ∂
∂
−
= 0,
′
dx ∂y
∂y
y∗ − y(x)
y∗ − y(x)
y(xo ) = yo ,
(1.2c)
v(xo ) = vo .
The problem of the brachistochrone, introduced formally by J. Bernoulli in
1696, is regarded as the beginning of the calculus of variations.
1.3.1c Finding the Brachistochrone (Abel [1, 2])
Assume first vo = 0 and consequently y∗ = yo . Prove that (1.2c) implies
2R − (y∗ − y)
′
√
y (x) = −
y∗ − y
for a given positive constant R. From this,
√
yo − y
dy.
x(y) = −
2R − (yo − y)
By the substitution (yo − y) = 2R sin2 (ϕ/2), this integral gives
x = xo + R(ϕ − sin ϕ),
y = yo − 2R(1 − cos ϕ).
This is a cycloid through (xo , yo ) for ϕ = 0. Choose R so that it also goes
through (x1 , y1 ) for ϕ = π/2. Investigate the case that the point has positive
initial kinetic energy and find the corresponding cycloid.
1.3.2c On the Minimality of the Brachistochrone
The previous calculations show only that cycloids are stationary points of the
functional in (1.1c). Having determined the curve y = y(x), let ϕ ∈ Co∞ (xo , x1 )
and set
x1
1 + (y + λϕ)′2
T (ϕ; λ) =
T (ϕ; λ)dx,
,
B(ϕ; λ) =
y∗ − (y + λϕ)
xo
1c Maxima and Minima of Functionals
249
where λ is a real parameter. The function λ → B(ϕ; λ) has a critical point for
λ = 0, which is a minimum if
∂2
B(ϕ; λ)
> 0.
∂λ2
λ=0
To verify such a condition, compute
′
d
(y + λϕ) ϕ′
T (ϕ; λ) =
dλ
′2
y∗ − (y + λϕ) 1 + (y + λϕ)
′2
ϕ 1 + (y + λϕ)
+
,
2 [y∗ − (y + λϕ)]3/2
ϕ′2
3 ϕ2 1 + y ′2
d2
T (ϕ; λ) λ=0 = √
+
dλ2
4 (y∗ − y)5/2
y∗ − y (1 + y ′2 )3/2
′ ′
yϕϕ
+
.
3/2
(y∗ − y)
1 + y ′2
From this, by Cauchy’s inequality, for all ε > 0,
d2
(1 − ε)ϕ′2
T
(ϕ;
λ)
≥
+
√
λ=0
3/2
dλ2
y∗ − y (1 + y ′2 )
1
3
−
4 4ε
ϕ2 1 + y ′2
.
(y∗ − y)5/2
Choose ε = 13 .
1.4c Motion along Geodesics
Let S be a smooth surface in R3 of parametric equations P = P (u, v) as in
§3 of Chapter 2. Let also γ be a curve lying on S parameterized with t and
parametric equations
S ∋ P (t) = P u(t), v(t) .
The infinitesimal arc length on γ is
√
ds = ∆ dt,
where
∆ = Au̇2 + 2B u̇v̇ + C v̇ 2 ,
where A, B, C are the elements of the first fundamental form of S. The length
of the portion of γ between any two of its points Po = P (to ) and P1 (t1 ) is
L(γ) =
t1
√
∆ dt.
to
The curve γ is a geodesic on S if for any two of its points Po and P1 the length
L(γ) is the least among all curves lying on S with extremities Po and P1 .
250
9 VARIATIONAL PRINCIPLES
By Proposition 1.1 the parametric representation t → P (t) of one such
curve, if it does exist, must satisfy the system (4.1) of Chapter 2. From the
remarks of §3.2 of Chapter 6 it follows that the motion of a point mass, constrained on a surface S and unsolicited otherwise, takes place along geodesics
on S. Equivalently, a point constrained on a surface S to which no other forces
are applied moves spontaneously along trajectories of least length on S.
1.5c An Isoperimetric Problem
Among all smooth closed curves in a plane of fixed length ℓ, one seeks the one
that encloses the largest area. Denote by (x, y) the coordinates of the plane
and regard these curves as parameterized by
[0, 2π] ∋ t → P (t) = x(t) e1 + y(t) e2 ,
P (0) = P (2π).
The area enclosed by such a curve and its length are
2π
1 2π
A=
(xẏ − ẋ y) dt,
ℓ=
ẋ2 + ẏ 2 dt.
2 0
0
(1.3c)
The problem reduces to maximizing the first integral subject to the constraint
represented by the second. Denoting by λ a Lagrange multiplier, introduce the
functional
2π
2π
def
J=
(xẏ − ẋy) + 2λ ẋ2 + ẏ 2 dt =
F (x, y; ẋ, ẏ)dt.
0
0
Compute
2λ
∂F
= −y +
ẋ,
∂ ẋ
ẋ2 + ẏ 2
∂F
2λ
= x+
ẏ,
∂ ẏ
ẋ2 + ẏ 2
∂F
= ẏ,
∂x
∂F
= −ẋ.
∂y
Therefore, by Proposition 1.1, the curve that encloses the largest area, for a
given fixed perimeter ℓ, satisfies the system
d
λẋ
y−
dt
ẋ2 + ẏ 2
= 0,
By integration,
λẏ
,
x − xo = −
ẋ2 + ẏ 2
λẏ
d
x+
dt
ẋ2 + ẏ 2
= 0.
λẋ
y − yo =
,
ẋ2 + ẏ 2
where xo and yo are generic integration constants. Squaring and adding these
gives
P − Po = |λ|,
Po ≡ (xo , yo ).
1c Maxima and Minima of Functionals
251
Thus these curves are circles of radius |λ| centered anywhere in the plane.
The parameter λ is computed by imposing the constraint 2π|λ| = ℓ. Choosing
Po = O and λ > 0, the parametric equations of such a curve are
[0, 2π) ∋ t → P (t) = λ cos te1 + sin te2 .
This circle is a stationary point for the functional J. To prove that it actually
is a maximum, we will show that there exists no closed curve P + δP , the
perturbation of such a circle, of the same circumference and of larger area.
Such a perturbation is represented as δP = (δx, δy), where t → δx(t), δy(t) are
smooth functions in [0, 2π] and δP (0) = δP (2π). The perturbation is small in
the sense (δP, δ Ṗ ) < ε, for some 0 < ε ≪ 1. The length of such a perturbed
curve must be the same, i.e.,
2π
2π
2
2
(ẋ + δ ẋ) + (ẏ + δ ẏ) dt =
ẋ2 + ẏ 2 dt,
(1.4c)
0
0
and its area must be larger, i.e.,
2π
(x + δx)(ẏ + δ ẏ) − (y + δy)(ẋ + δ ẋ) dt >
0
2π
0
(xẏ − y ẋ) dt.
(1.5c)
By a Taylor expansion of the integrand on the left-hand side of (1.4c),
2π
δx cos t + δy sin t dt
0
(1.4c)′
2π
2
3
1
δ ẋ cos t + δ ẏ sin t dt + O ε = 0.
+
4λ 0
A similar Taylor expansion in (1.5c) gives
2π
δx cos t + δy sin t dt +
0
2π
δx δ ẏ dt > 0.
(1.5c)′
0
Without loss of generality, the perturbation δP might be taken of the form
δP = εϕ(t) {cos te1 + sin te2 } ,
where ε is positive and ϕ is a smooth periodic function of t with period 2π.
Putting this expression of δP in (1.4c)′ –(1.5c)′ yields
2π
2π
ε
ϕdt +
ϕ̇2 dt = O(ε2 ),
4λ 0
0
(1.6c)
2π
ε 2π 2
ϕ dt > 0.
ϕdt +
2 0
0
From these we obtain
2π
1 2π
1
1 2π 2
ϕ dt <
ϕdt = −
ϕ̇2 dt + O(ε).
−
2 0
ε 0
4λ 0
252
9 VARIATIONAL PRINCIPLES
Assume first that ϕ does not have zero average in [0, 2π). Then this is impossible for ε sufficiently small. If, on the other hand, ϕ has zero average in
[0, 2π), then the first equation of 1.6c implies
2π
ϕ̇2 dt = O(ε).
0
This, in turn, is impossible for ε sufficiently small unless ϕ̇ ≡ 0. This would
imply ϕ = ϕo and the perturbed curve P + δP would be
[0, 2π) ∋ t → (P + δP ) (t) = (λ + εϕo ) cos te1 + sin te2 .
However, this is a circle whose length is not ℓ = 2πλ unless εϕo = 0.
A similar property holds in RN , i.e., among all “smooth, oriented, closed”
hypersurfaces of equal given area, the sphere is the one that encloses the
largest volume [38].
2c The Least Action Principle
2.1c On the Two-Point Boundary Value Problem (2.1)
The existence of a unique solution to (2.1) is implied by the a priori knowledge
of the motion t → q(t). Indeed, the extreme points qo and q1 have been chosen
accordingly. However, if one were to prescribe arbitrary points qo and q1 , the
system (2.1) in general does not have a solution, nor is such a solution, if any,
unique. As an example, consider
q̈ + q = 0,
t ∈ [0, τ ],
q(0) = 0, q(τ ) = δ.
Prove that the parameters δ and τ can be chosen so that such a problem has
no solution. If (2.1) is solvable, the solution in general is not unique, as shown
by the following problem:
q̈ + q = 0,
q(0) = q(π) = 0.
This has the solutions q ≡ 0 and q(t) = sin t.
2.2c Stationary Points of the Action Cannot Be Maxima
Assume first that the constraints are workless. By Proposition 1.1 and
Corollary 1.2 of Chapter 6, the kinetic energy T is a positive definite quadratic
form of q̇. Therefore
t1
t1
sup
|q̇|2 dt − const.
L(q, q̇; t)dt ≥
sup
co
q∈K
q∈K;q≤1
to
Choose a sequence of curves qn ∈ K such that
qn ≤ 1
∀n ∈ N
and
sup
qn
Modify the argument for moving constraints.
to
t1
to
|q̇n |2 dt = ∞.
2c The Least Action Principle
253
2.3c Critical Points of the Action Are Local Minima
An orbit q ∈ K, a solution of (2.1), is a local minimum for the Hamiltonian
action S(·) if there exists ε ∈ (0, 1) such that
S(q) < S(q + δq) ∀ synchronous δq such that
sup δq < ε.
[to ,t1 ]
Proposition 2.1c Let q ∈ K be a solution of (2.1). There exist two numbers
ε, σ ∈ (0, 1), which can be determined a priori only in terms of L and q, such
that if |t1 − to | ≤ σ, then q is a local minimum for the Hamiltonian action.
Proof. Assume first that the constraints are workless and compute the variation δL of the Lagrangian, corresponding to a synchronous small variation δq
of the orbit q:
δL = L(q + δq, q̇ + δ q̇) − L(q, q̇)
= 21 Ahk (q + δq)[(q̇h + δ q̇h )(q̇k + δ q̇k ) − q̇h q̇k ]
+ 12 [Ahk (q + δq) − Ahk (q)]q̇h q̇k + U (q + δq) − U (q)
= 21 Ahk (q + δq)δ q̇h δ q̇k + Ahk (q + δq)q̇h δ q̇k
+ 21 [Ahk (q + δq) − Ahk (q)]q̇h q̇k + U (q + δq) − U (q)
= 21 Ahk (q + δq)δ q̇h δ q̇k + [Ahk (q + δq) − Ahk (q)]q̇h δ q̇k
∂
Ahk (q)δqk q̇h q̇k
+ 21 Ahk (q + δq) − Ahk (q) −
∂qk
∂
+ U (q + δq) − U (q) −
U(q)δqk
∂qk
∂
∂
+
L(q, q̇)δ q̇k +
L(q, q̇)δqk .
∂ q̇k
∂qk
(2.1c)
Set also
A(q) =
∂A (q) ∂ 2 A (q) ∂ 2 U(q)
hk
hk
+
+
.
∂qi
∂qi ∂qj
∂qi ∂qj
h,k,i,j=1
N
Since Ahk (·) and U(·) are smooth, one may choose ε sufficiently small, possibly
dependent on L and q, such that
sup A q(t) = C and
sup A q(t) + δq(t) ≤ 2C.
[to ,t1 ]
[to ,t1 ]
Since ξ → Ahk ξh ξk is a positive definite quadratic form, there exists a positive
constant co , independent of δq such that
1
2 Ahk (q
+ δq) δ q̇h δ q̇k ≥ co δ q̇ 2 .
254
9 VARIATIONAL PRINCIPLES
Moreover,
|Ahk (q + δq) − Ahk (q)||q̇h ||δ q̇k | ≤ 2C q̇
δq
δ q̇
2C 2
≤ 12 co δ q̇ 2 +
q̇ 2 δq
co
≤ 12 co δ q̇ 2 + C1 δq 2 ,
2
where we have used Cauchy’s inequality and have set
2C 2
sup q̇
co [to ,t1 ]
C1 =
2
.
Since q ∈ K is known as a solution of (2.1), the quantity C1 can be regarded
as a known constant. The remaining terms in (2.1c), except the last, are
estimated as
1
∂Ahk (q)
δqk |q̇h ||q̇k |
Ahk (q + δq) − Ahk (q) −
2
∂qk
∂
+ U (q + δq) − U (q) −
U (q) δqk
∂qk
≤ 2C(1 + q̇(t) 2 ) δq
2
≤ C2 δq 2 .
The last is transformed by using that t → q(t) is a solution of (2.1),
∂L
∂L
δ q̇k +
δqk = −
∂ q̇k
∂qk
=
d ∂L
∂L
−
dt ∂ q̇k
∂qk
δqk +
d
dt
∂L
δqk
∂ q̇k
∂L
δqk .
∂ q̇k
d
dt
Combining these estimates in (2.1c) gives
δL ≥
1
co δ q̇
2
2
− C3 δq
2
+
d
dt
∂L
δqk ,
∂ q̇k
where the constant C3 = (2C + C1 + C2 ) can be regarded as known a priori.
Integrate this dt over (to , t1 ) and recall that δq(to ) = δq(t1 ) = 0 to get
S(q + δq) − S(q) ≥
1
co
2
t1
to
δ q̇ 2 dt − C3 (t1 − to ) sup δq 2 .
[to ,t1 ]
Using that the virtual variations δq vanish at to and t1 , estimate
t1
δ q̇ 2 dt.
sup δq 2 ≤ (t1 − to )
[to ,t1 ]
to
4c The Functional of Geometrical Optics
255
Combining these estimates gives
S(q + δq) − S(q) ≥
1
2
2 co − C3 (t1 − to )
t1
δ q̇ 2 dt.
to
It follows that if the time interval (to , t1 ) is small enough to satisfy
1
2 co
− C3 (t1 − to )2 ≥ 41 co > 0,
then S(q) < S(q + δq), and the stationary orbit q ∈ K, a solution of (2.1), is
a local minimum for the Hamiltonian action.
Modify the proof to include the case of moving constraints. The proposition
was first proved, by a different entirely geometrical argument, by Darboux
[37, Tome II, no 246, 545, 568].
4c The Functional of Geometrical Optics
Assume that the entire space R3 is divided into two connected optically homogeneous media Ωo and Ω1 , separated by a smooth surface Σ, and with
refraction indices νo and ν1 respectively. In each of them, light rays propagate along straight lines, with speeds νo−1 and ν1−1 respectively. Therefore
the minimum problem (4.1) reduces to finding the point q ∈ Σ for which
the time
t = νo q − qo + ν1 q1 − q
is the least. Computing its variation, we have
q − qo
q1 − q
− ν1
· δq = 0.
δt = νo
q − qo
q1 − q
Since q varies on Σ, there exists a parameter λ ∈ R such that
νo
q1 − q
q − qo
− ν1
= λn(q),
q − qo
q1 − q
(4.1c)
where n(q) is the unit normal to Σ in q. This implies that the three vectors
q − qo
,
q − qo
q1 − q
,
q1 − q
n(q),
are coplanar. Let then π be the plane that contains them and let t be
the tangent to Σ in q lying in π. Taking the scalar product of (4.1c) by t
gives
νo sin i − ν1 sin r = 0,
(4.2c)
256
9 VARIATIONAL PRINCIPLES
where
i ≡ angle between q − qo
q − qo
r ≡ angle between
and n(q) is the angle of incidence,
q1 − q
q1 − q
and n(q) is the angle of refraction.
Consider the case of an optical medium consisting (at least locally) of (n + 1)
layers of optically homogeneous media Ωh with refraction indices νh , separated
by n portions of smooth surfaces Σh .
Formulas (4.1c)–(4.2c) are Snell’s laws (1621) reported by Huygens in [84]
and Descartes in [39].
n
q0
W0
i
r
W1
q1
Fig. 4.1c.
10
CANONICAL TRANSFORMATIONS
1 Changing the Variables of Motion
Let H(p, q; t) be the Hamiltonian of a mechanical system with N degrees of
freedom. By the least action principle, an orbit t → (p(t), q(t)) in phase space
R2N is a solution of the Hamilton canonical equations if and only if it is a
stationary point of the action, i.e.,
t1
ṗ = −∇q H,
δ
[pq̇ − H(p, q; t)]dt = 0 ⇐⇒
(1.1)
q̇
= ∇p H,
to
where (to , t1 ) is any subinterval of the time of motion, and the virtual variation δ is synchronous in such an interval. Consider now smooth, invertible
transformations
⎛
⎞
∂P ∂P
P = P (p, q; t),
Q = Q(p, q; t),
p = p(P, Q; t),
q = q(P, Q; t),
⎜ ∂p
∂q ⎟
⎟
J =⎜
⎝ ∂Q ∂Q ⎠ ,
∂p
(1.2)
∂q
with det J = 0. Among these look for those that preserve the variational structure of (1.1). Precisely those for which, for any given Hamiltonian H(p, q; t),
there exists a new one K(P, Q; t) such that the orbits satisfying (1.1), when
transformed by (1.2), are those and only those for which
t1
Ṗ = −∇Q K,
δ
[P Q̇ − K(P, Q; t)]dt = 0 ⇐⇒
(1.3)
Q̇ = ∇P K,
to
where the virtual variation δ is synchronous in (to , t1 ). We call these variational transformations, since they mutually keep the form of the Hamilton
equations and their corresponding least action principle. These transformations, whenever they exist, are local, i.e., they are well defined, in general
within a sufficiently small neighborhood of a given point (po , qo ; to ), taken as
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 10,
c Springer Science+Business Media, LLC 2011
257
258
10 CANONICAL TRANSFORMATIONS
initial data in (1.1). In what follows we will avoid specifying the local nature
of (1.2) and their respective domains of definition.
Remark 1.1 It is not required that the new Hamiltonian K be obtained
from H by substituting (p, q) in terms of (P, Q; t), nor that Q and P have
the physical significance of Lagrangian configurations and kinetic momenta.
It is required only that the variational structures of (1.1) and (1.3) be mutually preserved, for any Hamiltonian H, and a corresponding new one K. For
this reason, the variational transformations are independent of H, as further
clarified by the next example.
1.1 Pointwise Transformations
The choice of the Lagrangian coordinates is not unique, and from a given
choice q = (q1 , . . . , qN ), one can construct another one, by prescribing a
smooth, locally invertible transformation Q = Q(q; t). These are called pointwise transformations of the Lagrangian coordinates. Describing the motion in
terms of these, one constructs a new Lagrangian L′ (Q, Q̇; t), and defines new
kinetic momenta P = ∇Q̇ L′ . Then, along the motion, the new Hamiltonian
K(P, Q; t) = Ph Q̇h − L′ (P, Q; t)
satisfies the Hamilton equations and its corresponding least action principle,
as indicated in (1.3). Thus pointwise transformations are variational transformations, and because of their construction, they are independent of the
original Hamiltonian H(p, q; t). The new variables (Q, P ) keep their physical
meaning of Lagrangian configurations and kinetic momenta. However, neither
the new Lagrangian L′ nor the new Hamiltonian H′ is obtained from L or H
by substitution of variables.
2 Variational and Canonical Transformations
The requirement that the transformation (1.2) be variational amounts to imposing conditions on the integrals in (1.1) and (1.3), so that the vanishing of
anyone of them, for all the proper virtual synchronous variations, implies the
vanishing of the other for all corresponding virtual synchronous variations.
One such condition is the existence of a smooth function
(p, q, P, Q; t) → F (p, q, P, Q; t)
(2.1)
pdq − Hdt = P dQ − Kdt + dF.
(2.2)
such that
Indeed, when computed along any curve t → (p(t), q(t)) in phase space and the
corresponding transformed curve t → (P (t), Q(t)), the integrands in (1.1) and
2 Variational and Canonical Transformations
259
(1.3) differ by the virtual synchronous variation of the total time derivative
of F . The differentials dq and dQ and dF are meant with respect to the set
of variables {p, q, P, Q; t}. Given an invertible transformation as in (1.2), only
2N of {p, q, P, Q} are independent. To avoid specifying the set of independent
variables, we consider them as a full set of variables, possibly dependent or
independent. Introduce the differential d¯ at frozen time, acting on the variables
{p, q, P, Q} only [111]. Thus for a smooth function f depending on the full set
of variables {p, q, P, Q; t},
¯ = df − ft dt = ∇p f dp
¯ + ∇q f dq
¯ + ∇P f dP
¯ + ∇Q f dQ,
¯
df
(2.3)
¯ dq,
¯ dP
¯ , dQ
¯ are expressed in terms of the 2N variables chosen as
where dp,
independent in (1.2). Notice, however, that if ξ is an independent variable,
¯ = dξ. Thus if, for example, Q is chosen as independent,
then ξt = 0 and dξ
¯ = dQ. With this notation rewrite (2.2) in the equivalent
then Qt = 0 and dQ
form
¯ − P dQ
¯ − dF
¯ = [(H − pqt ) − (K − P Qt ) + Ft ]dt.
pdq
(2.4)
This relation, roughly speaking, separates the differentials with respect to the
space variables {p, q, P, Q} from the differential with respect to time. The
latter is satisfied if
¯ def
¯ − P dQ
¯ = dF
¯
dω
= pdq
(2.5)
and
def
K = H − (pqt − P Qt ) + Ft .
(2.6)
Given an invertible transformation as in (1.2), if a function F exists satisfying
¯ in the space variables {p, q, P, Q} only
(2.5), then the differential form dω
is exact. Then with such a function at hand, (2.6) gives the form of the new
¯ is exact, there exists
Hamiltonian K in terms of the old one H. Conversely, if dω
a function F as in (2.1) such that, defining K by (2.6), the sufficient condition
(2.2) is satisfied. This, in turn, implies that the invertible transformation (1.2)
preserves the variational structure of (1.1) and (1.3).
2.1 Canonical Transformations
A locally invertible transformation as in (1.2) is canonical if the differential
¯ in the variables {p, q, P, Q} is exact, equivalently if there exists a
form dω
smooth function F as in (2.1) satisfying (2.5). The function F is the primitive
¯ The requirement that dω
¯ be exact is a condition on the transformation
of dω.
(1.2) alone and not on the Hamiltonian H. Thus, once a transformation of
¯ is exact, any Hamiltonian
the type of (1.2) has been identified for which dω
system as in (1.1), for any given Hamiltonian H, is transformed into a new
Hamiltonian system of the form (1.3) for a new Hamiltonian K defined by
(2.6). A canonical transformation is completely canonical if it is explicitly
independent of time. In such a case d¯ = d and dω = dF and K = H. Canonical
260
10 CANONICAL TRANSFORMATIONS
transformations form a strict subclass of the variational transformations. The
transformation
Q = q,
P = λp,
λ = 1,
N = 1,
K(·, ·; t) = H(λ·, ·; t),
(2.7)
preserves the variational form (1.1) for the new Hamiltonian K; however,
there is no F such that dω = dF . We postpone to §7 the characterization of
all variational transformations, and focus next on canonical transformations,
in view of their remarkable invariance properties (§§7–8).
3 Constructing Classes of Canonical Transformations
The function F could in principle be dependent on all four N -tuples of variables p, q, P, Q. However, by the mutual invertibility requirement in (1.2), only
2N of these are independent. Assume momentarily that the transformation in
(1.2) is canonical and that the variables (q, Q) can be chosen as independent,
so that the remaining variables can be expressed in terms of these. Then,
¯ set
having identified the function F, a primitive of dω,
(q, Q; t) → F1 (q, Q; t) = F p(q, Q; t), q, P (q, Q; t), Q; t .
(3.1)
¯ being exact implies
Then the condition of dω
p = ∇q F1 (q, Q; t),
P = −∇Q F1 (q, Q; t).
(3.2)
This suggests constructing canonical transformations for which (q, Q) are independent as follows [120]. Fix a smooth function F1 (q, Q; t) satisfying
det
∂ 2 F (q, Q; t)
1
= 0
∂qh ∂Qk
(3.3)
in its domain of definition. Then define p and P from (3.2). The first of these
permits expressing Q in terms of (p, q; t). Putting this in the second expresses
P in terms of (p, q; t). This identifies the first two equations of (1.2). The form
of the second two is determined by inversion, which is guaranteed by (3.3).
The form of the new Hamiltonian is
K(P, Q; t) = H(p, q; t) + F1,t (q, Q; t).
(3.4)
The function F1 (q, Q; t) is called the generator of the transformation. The
construction procedure is independent of the Hamiltonian H and identifies
the subclass of canonical transformations for which (q, Q) can be taken as
independent variables. Such a class would not include the pointwise transformations, since q = q(Q; t) and in general the kinetic momenta (p, P ) cannot
be expressed in terms of (q, Q; t) alone. A trivial example of this is the identity
transformation q = Q and p = P .
4 Constructing Canonical Transformations by Other Pairs of IV
261
4 Constructing Canonical Transformations by Other
Pairs of Independent Variables
The requirement that the differential form dω be exact can be written in any
of the equivalent forms
¯ − P dQ
¯ = dF,
¯
pdq
¯ + QdP
¯ = dF
¯ 2,
pdq
¯ + P dQ
¯ = dF
¯ 3,
q dp
¯ + QdP
¯ = dF
¯ 4,
q dp
(4.1)
where
F2 = F + P Q,
(4.2)
where
F3 = F − pq,
(4.3)
where
F4 = F + P Q − pq.
(4.4)
The form of (4.1) suggests constructing the class of canonical transformations
for which (q, P ) are independent by a similar procedure. First, if (q, P ) are
independent, rewrite (2.4) in the form
¯ + QdP
¯ − dF
¯ 2 = (H − K + F2,t )dt.
pdq
(4.5)
Next, fix a function F2 = F2 (q, P ; t) satisfying
det
∂ 2 F (q, P ; t)
2
= 0
∂qh ∂Pk
(4.6)
in its domain of definition, and set
p = ∇q F2 (q, P ; t),
Q = ∇P F2 (q, P ; t).
(4.7)
Similarly, the class of canonical transformations for which (p, Q) can be taken
as independent is constructed by taking F3 = F3 (p, Q; t) satisfying the analogue of (3.3) and (4.6) and the analogue of the differential equalities (2.4)
and (4.5). Then one sets
q = ∇p F3 (p, Q; t),
P = −∇Q F3 (p, Q; t).
(4.8)
Finally, the class of canonical transformations for which (p, P ) can be taken as
independent is constructed by taking F4 = F4 (p, P ; t) satisfying the analogue
of (3.3) and (4.7) and the analogue of the differential equalities (2.4) and (4.5),
and setting
q = −∇p F4 (p, P ; t),
Q = ∇P F4 (p, P ; t).
(4.9)
In all cases, the new Hamiltonian is given by
K(P, Q; t) = H(p, q; t) + Fj,t ,
j = 1, 2, 3, 4.
(4.10)
262
10 CANONICAL TRANSFORMATIONS
5 Examples of Canonical Transformations
The function F2 (q, P ) = qP generates through (4.7) the identity transformation. The function F1 (q, Q) = qQ generates through (3.2) the transformation
Q = p and P = −q. The original kinetic momenta p become the Lagrangian
coordinates in the transformed system, whereas the original Lagrangian coordinates q become the opposite of the “new” kinetic momenta P . This example
indicates that in general, canonical transformations do not preserve the physical meaning of the original variables. The pointwise transformations of §1.1
are generated by F2 (q, P ; t) = Q(q; t)P . Given a smooth positive function
t → f (t), the transformation
P =
1
p,
f (t)
Q = f (t)q is canonical with primitive F = P Q.
(5.1)
5.1 The Flow Map Is a Canonical Transformation
The flow map transforms (P, Q) into variables
p = p(P, Q; t),
q = q(P, Q; t),
ṗ = −∇q H,
p(0) = P,
with
q̇ = ∇p H,
q(0) = Q.
solutions of
(5.2)
For |t| sufficiently small, the transformation is locally smooth and invertible.
It is also canonical with primitive F given by
t
F (P, Q; t) =
[pqt − H(p, q; s)] ds,
(5.3)
0
where in the integrand p and q are expressed in terms of {P, Q; t}, as indicated
in (5.2). By direct calculation,
∇Q F =
=
t
(qt ∇Q p + p∇Q qt − ∇p H∇Q p − ∇q H∇Q q) ds
0
t
(qt ∇Q p + p∇Q qt − qt ∇Q p + pt ∇Q q) ds
0
=
0
t
(p∇Q q)t ds = p∇Q q − P.
Similarly, one computes ∇P F = p∇P q. Therefore
¯ = p∇Q qdQ + p∇P qdP − P dQ = dω.
¯
dF
Finally, since Ft = pqt − H, it follows from (2.6) that K = 0, since Q is an
independent variable.
5 Examples of Canonical Transformations
263
¯
5.2 On the Primitives of dω
Given a canonical transformation (1.2), the form of the primitive F depends
on a choice of 2N independent variables. In (5.1) and in the flow map, the
independent variables have been taken as (P, Q). However, in these examples,
the pair (p, Q) or (q, P ) could also be taken as independent, and the primitive
could be rewritten in terms of these.
The 2N independent variables are selected N out of (p, q) and N out of
the (P, Q). Thus in principle, there could be 2(2n!)/(n!) forms of a primitive
F, each corresponding to a pair of N -tuples of independent variables. If any
one of the pairs of N variables (q, Q), (q, P ), (p, Q), (p, P ) can be taken as
independent, the generators Fj can be found as the primitives of the respective differential forms in (4.1)–(4.4). If none of these pairs can be taken as
independent variables, then the primitive F differs from the Fj . However, it
can be written as one of these, modulo the interchange of some of the kinetic
momenta p, or Lagrangian coordinates q, into the Lagrangian coordinates Q
and the kinetic momenta P respectively. Indeed, these transformations are
canonical. In this sense the Fj are essentially all the primitives of a canonical
transformation.
5.3 Jacobi Integration of Hamilton Equations [87]
For a given Hamiltonian H(·, ·; t), choose a smooth function F1 (q, Q; t), a
solution of the Hamilton–Jacobi equation (§3 of Chapter 9)
∂F1 (q, Q; t)
+ H (∇q F1 (q, Q; t), q; t) = 0.
(5.4)
∂t
The variables Q are regarded as parameters, and F1 (·, Q; t) is regarded as a
family of solutions of (5.4) parameterized with Q. Assuming that (5.4) admits
a family of smooth solutions satisfying (3.3), one might use such F1 (q, Q; t) as
the generator of a canonical transformation as in (3.2). These transformations
resolve the motion in the following sense. First, by (3.4) the transformed
Hamiltonian K(P, Q; t) is identically zero. Therefore, if t → (p(t), q(t)) is a
solution of the canonical Hamiltonian system (1.1), the transformed system
is Ṗ = 0 and Q̇ = 0 identically. Thus P and Q are constants, and their value
can be computed from the resulting transformations (1.2) at some fixed time
to , that is,
P = Q(po , qo ; to ),
Q = P (po , qo ; to ),
by using the initial data (qo , po ) associated to the original canonical system.
Putting these constants into the second equation of (1.2) determines the motion in phase space
t → p(t) = p(P, Q; t),
t → q(t) = q(P, Q; t).
(5.5)
While the method is simple and elegant, it rests upon solving the Hamilton–
Jacobi equation (5.4) depending on N scalar parameters Q. Solution
264
10 CANONICAL TRANSFORMATIONS
techniques of (5.4) will be given in the next chapter. One could start from
any one of the generators of the form Fj and arrive at the same solution
methods. The choice of Fj in this method is not unique, and it depends on
the possibility of solving the associated Hamilton–Jacobi equation.
6 Symplectic Product in Phase Space
and Symplectic Matrices
The variables p and q in the Hamilton equations play an antisymmetric role
in the sense
∇p H
ṗ
O −I
=E
.
,
where E =
q̇
I O
∇q H
The matrix E is the antisymmetric or symplectic identity, and it satisfies
Et = E−1 = −E
and
E2 = −I.
Given any two vectors u = (p, q)t and v = (P, Q)t in R2N , the matrix E
defines a symplectic scalar product in R2N , by the formula
[u, v] = Eu · v = −Ev · u,
where “·” is the usual operation of scalar product in R2N . From the definition
we have
[u, v] = −[v, u],
[u, u] = 0,
[u, v] = ph Qh − qh Ph .
If N = 1, then u and v are vectors in R2 , and if they are nontrivial, the
number
[u, v] = p1 Q1 − q1 P1
is the only nontrivial component of the exterior product v ∧ u. Moreover,
p1 Q1 − q1 P1 is the measure of the “oriented area” of the parallelogram of u
and v. If N > 1, the vectors u and v form a parallelogram in R2N , whose
projections on the planes of the coordinate axes (qh , ph ) are parallelograms of
oriented area ph Qh − qh Ph . Therefore [u, v] may be interpreted as the sum of
the oriented areas of such projections.
6.1 Symplectic Linear Transformations
Let T be a linear transformation of R2N into itself, which we identify with its
representative matrix. The transformation T is symplectic if it preserves the
symplectic product, that is, if
[T u, T v] = [u, v]
for all u and v in R2N .
7 Characterizing Canonical Transformations by Symplectic Jacobians
265
The antisymmetric identity E preserves the symplectic product, since
[Eu, Ev] = E2 u · Ev = −u · Ev = Eu · v = [u, v].
If T is symplectic, then
Eu · v = [u, v] = [T u, T v] = ET u · T v = (T t ET )u · v.
Since this has to hold for all vectors u and v in R2N , a linear transformations T : R2N → R2N is symplectic if and only if either one of the following
equivalent conditions holds:
T t ET = E,
T −1 = −ET t E,
(T −1 )t E(T −1 ) = E.
(6.1)
Therefore the transpose and inverse of a symplectic matrix are symplectic
matrices. If T1 and T2 are linear symplectic transformations of R2N into itself,
one verifies that
(T1 T2 )t E(T1 T2 ) = E.
Therefore the product of two symplectic matrices is symplectic. Thus the
collection of all linear symplectic transformations from R2N into itself form
a group under composition, called the symplectic group of order N . By (6.1),
det(T ) = ±1, and it can be shown that indeed det(T ) = 1.
Proposition 6.1 A linear, symplectic transformation T : R2N → R2N , preserves orientations and volumes.
7 Characterizing Canonical Transformations
by Symplectic Jacobians
Proposition 7.1 The transformation (1.2) is canonical if and only if the
Jacobian matrix J is symplectic.
Proof. For a smooth curve s → [q(s), p(s)] and its transform by (1.2), denote
by δ a synchronous virtual variation of {p, q, P, Q}. These variations are effected at time t frozen, and are synchronous with respect to the parameter s.
¯ be the differential form defined in (2.5) and compute
Let dω
¯ = d(pδq
¯
¯
δ dω
− P δQ) + δ∗ dω,
where we have set
¯ = (δpdq
¯ − δq dp)
¯ − (δP dQ
¯ − δQdP
¯ ).
δ∗ dω
¯ is exact if and
By the Pfaff–Jacobi theorem [86, 126], the differential form dω
¯ = 0 for all virtual synchronous variations of curves in (p, q)-space
only if δ∗ dω
and the corresponding variations in (P, Q)-space. Compute
¯ = (δp, δq)E − (δP, δQ)EJ (dp,
¯ dq)
¯ t
−δ∗ dω
¯ dq)
¯ t.
= (δp, δq) E − J t EJ (dp,
266
10 CANONICAL TRANSFORMATIONS
If (1.2) is canonical, then J is symplectic and det J = 1. Therefore we have
the following corollary.
Corollary 7.1 Canonical transformations are volume-preserving. In particular, the flow map is volume-preserving.
The last statement is known as Liouville’s theorem.
8 Poisson Brackets [131]
For any two scalar-valued smooth functions (p, q; t) → F (p, q; t), G(p, q; t), the
Poisson brackets of F and G are defined by
∂F ∂G
∂F ∂G
−
∂ph ∂qh
∂qh ∂ph
= E∇p,q F · ∇p,q G = [∇p,q F, ∇p,q G].
{F, G} =
(8.1)
One verifies the calculus properties
{F, G} = −{G, F },
{F, C} = 0 for all constants C,
{F1 + F2 , G} = {F1 , G} + {F2 , G},
{F1 F2 , G} = F1 {F2 , G} + F2 {F1 , G},
{qh , qk } = {ph , pk } = 0,
{F, qh } = Fph ,
(8.2)
{ph , qk } = δhk ,
{F, ph } = −Fqh .
Moreover, if x is any scalar variable taken out of (p, q; t), then
{F, G}x = {Fx , G} + {F, Gx }.
(8.3)
By these properties, the Hamiltonian system (1.1) can be written in the form
ṗh = {H, ph },
q̇h = {H, qh }.
(8.4)
8.1 Invariance of the Poisson Brackets by Canonical
Transformations
The last term in (8.1) is the symplectic product of the two vectors ∇q,p F and
∇q,p G. Performing on F and G the change of variables (1.2) yields
{F, G}p,q = [∇p,q F, ∇p,q G] = [J∇P,Q F, J∇P,Q G].
Therefore if J preserves the symplectic product, then {F, G}p,q = {F, G}P,Q ,
that is, J preserves the Poisson brackets. In particular, if F = Qh and G =
Pk , then
{Ph , Pk }q,p = {Ph , Pk }P,Q = {Qh , Qk }p,q = {Qh , Qk }P,Q = 0,
{Ph , Qk }p,q = {Ph , Qk }P,Q = δhk .
(8.5)
9 The Jacobi Identity [89]
267
For any transformation as in (1.2) with Jacobian J, symplectic or not, one
computes
{Ph , Pk } {Ph , Qk }
t
.
JEJ =
{Qh , Pk } {Qh , Qk }
Therefore J is symplectic if and only if equations (8.5) hold. In particular,
(8.5) characterizes canonical transformations.
Proposition 8.1 The transformation (1.2) is canonical if and only if
{ph , pk }P,Q = {qh , qk }P,Q = 0,
{Ph , Pk }p,q = {Qh , Qk }p,q = 0,
{ph , qk }P,Q = δhk ,
{Ph , Qk }p,q = −δhk .
(8.6)
Equivalently, the transformation (1.2) is canonical if and only if it preserves
the Poisson brackets.
9 The Jacobi Identity [89]
Proposition 9.1 For every triple F , G, H of smooth functions of (p, q; t),
{F, {G, H}} + {G, {H, F }} + {H, {F, G}} = 0.
(9.1)
Proof. Regard (9.1) as a homogeneous linear form of the second derivatives
of F . These occur only in the last two terms
{G, {H, F }} − {H, {G, F }}.
(9.2)
We show first that such an expression does not contain second derivatives of
F . The action of G through the Poisson brackets is formally given by
DG =
∂G ∂
∂G ∂
−
.
∂ph ∂qh
∂qh ∂ph
Introducing the notation
y = (y1 , y2 , . . . , yN , yN +1 , . . . , y2N ) = (q, p),
the operation DG can be expressed as DG = ah ∂/∂yh , where the coefficients
ah depend only on the first derivatives of G. Similarly, the action of H through
the Poisson brackets is given by DH = bk ∂/∂yk . With this notation,
{G, {H, F }} − {H, {G, F }} = DG DH F − DH DG F
∂ ∂F
∂ ∂F
bk
− bh
ak
.
= ah
∂yh
∂yk
∂yh
∂yk
268
10 CANONICAL TRANSFORMATIONS
In this expression the terms containing the second derivatives of F are
ah b k
∂2F
∂2F
− ak b h
= 0.
∂yh ∂yk
∂yh ∂yk
It follows that the expression in (9.2) involves only first derivatives of F , and
therefore it takes the form
{G, {H, F }} − {H, {G, F }} = Ah
∂F
∂F
+ Bh
,
∂ph
∂qh
(9.3)
where the coefficients Ah and Bh depend on the first and second derivatives
of G and H, and are independent of F . Such independence permits one to
compute them by choosing F = ph and F = qh . This gives
{G, {H, ph }} − {H, {G, ph }} = Ah ,
{G, {H, qh }} − {H, {G, qh }} = Bh .
Therefore
Ah = {Hqh , G} + {H, Gqh } = −{G, H}qh ,
Bh = {Gph , H} + {G, Hph } = {G, H}ph .
These in (9.3) give
{G, {H, F }} + {H, {F, G}} = {G, H}ph Fqh − {G, H}qh Fph
= −{F, {G, H}}.
10 Generating First Integrals of Motion
by the Poisson Brackets
The total time derivative of a smooth function F of (q, p; t) along a solution
of the canonical system (1.1) is computed as
∂F
∂F
d
∂F
F =
+
q̇h +
ṗh
dt
∂t
∂qh
∂ph
∂H ∂F
∂H ∂F
∂F
∂F
+
+ {H, F }.
−
=
=
∂t
∂ph ∂qh
∂qh ∂ph
∂t
Proposition 10.1 A smooth function (q, p; t) → F (q, p; t) is an integral of
motion if and only if Ft + {H, F } = 0. If F is explicitly independent of time,
it is an integral of motion if and only if {H, F } = 0.
Theorem 10.1 (Poisson [131]). If F and G are two integrals of motion,
then {F, G} is also an integral of motion.
11 Infinitesimal Canonical Transformations
269
Proof. Compute the derivative of {F, G} along any orbit solution of (1.1).
Using the Jacobi identity (9.1) and the calculus properties (8.2) of the Poisson
brackets, we obtain
d
{F, G} = {F, G}t + {F, G}ph ṗh + {F, G}qh q̇h
dt
= {F, G}t − {F, G}ph Hqh + {F, G}qh Hph
= {F, G}t + {H, {F, G}}
= {F, G}t − {F, {G, H}} − {G, {H, F }}
= {Ft , G} + {F, Gt } + {F, {H, G}} + {{H, F }, G}
= {Ft + {H, F }, G} + {F, Gt + {H, G}} = 0.
Remark 10.1 While the theorem guarantees that given any two integrals
of motion F and G, one can generate another one by {F, G}, it does not
assert that the latter is independent of F and G. If this were the case, by
iteration one could generate infinitely many independent integrals of motion,
contradicting that the mechanical system at hand has finitely many degrees
of freedom.
11 Infinitesimal Canonical Transformations
A family of canonical transformations parameterized by a small parameter ε
is close to the identity if it is generated by (§4)
F2 (q, P ; t, ε) = qP + εG(q, P ; t, ε),
|ε| ≪ 1,
(11.1)
for a smooth function G of its arguments. The transformation is close to the
identity, since it appears, at least formally, as a small perturbation of qP ,
which is the generator of the identity (§5). From (11.1) one computes the new
variables
∇q F2 = p = P + ε∇q G(q, P ; t, ε),
∇P F2 = Q = q + ε∇P G(q, P ; t, ε),
and the increments
δp = P − p = −ε∇q G(q, P ; ε),
δq = Q − q = ε∇P G(q, P ; ε).
These transformations are canonical for all ε, so small as to guarantee that
(4.6) is satisfied. Therefore, for all such ε, the variables (p, q) and (P, Q) satisfy
the characteristic condition (8.6) for the transformation to be canonical. Although the generator is F2 , we say that G is the generator of the infinitesimal
canonical transformation, close to the identity.
270
10 CANONICAL TRANSFORMATIONS
Even though G depends on (q, P ; ε), it can be regarded as a function of
(p, q; ε), up to an infinitesimal of higher order in ε. Indeed, by Taylor’s formula,
G(q, P ; t, ε) = G(q, p; t, ε) + O(ε). With this notation,
δp = −ε∇q G(p, q; t, ε) + o(ε),
δq = ε∇p G(p, q; t, ε) + o(ε).
(11.2)
Remark 11.1 Transformations of the type of (11.2), with generic terms o(ε),
are close to the identity, but not necessarily canonical. They are canonical if
they are generated by a family of functions G(q, P ; t, ε) as in (11.1).
As an example consider the flow map from (P, Q) into (p, q) as defined in
(5.2). For an infinitesimal dt, such a map is close to the identity, since
p = P + dtṗ + o(dt)
= p − dt∇q H + o(dt),
q = Q + dtq̇ + o(dt)
= q + dt∇p H + o(dt).
It is also canonical since the flow map is a canonical transformation. Therefore
εG = (dt)H, where H is computed in terms of (q, P ), and where dt plays the
role of the small parameter ε appearing in (11.1). The motion can be interpreted as a continuous progression of infinitesimal canonical transformations.
Given a smooth function G of the arguments (p, q; t, ε), for a small parameter ε, consider transformations (p, q) → (P, Q) defined by (11.2) without
higher-order corrections, that is,
P − p = δp = −ε∇q G(p, q; t, ε),
Q − q = δq = ε∇p G(p, q; t, ε).
(11.3)
These in general are not canonical, since the characteristic conditions (8.6)
are satisfied only up to terms o(ε) of higher order in ε. While not necessarily
canonical, they are called infinitesimal canonical transformations close to the
identity.
11.1 Variations for Infinitesimal Canonical Transformations
The variation of a smooth function F (p, q; t) induced by (11.3) is computed as
δF = ∇p F δp + ∇q F δq = −ε
If F = H, then
∂F ∂G
∂F ∂G
+ε
= ε{G, F }.
∂ph ∂qh
∂qh ∂ph
δH(p, q; t) = ε{G, H}.
If G(p, q) is explicitly independent of t and ε, and is an integral of motion, then
{G, H} = 0 by Proposition 10.1, and consequently δH = 0. Thus the corresponding infinitesimal canonical transformation generated by (11.3) for such a
G preserves the Hamiltonian. Conversely, if (11.3) preserves the Hamiltonian,
and G is explicitly independent of t, then G is an integral of motion.
12 Variational and Canonical Transformations
271
Proposition 11.1 The integrals of motion explicitly independent on t are
those and only those G that generate, by (11.3), an infinitesimal canonical
transformation that preserves the Hamiltonian H.
Remark 11.2 The proposition permits one to seek integrals of motion among
those transformations that preserve the Hamiltonian. For example, if H has a
symmetry with respect to some of its variables, those rotations that keep the
symmetry of H are first integrals of motion.
If H has a cyclic variable, say for example qh , then H remains unchanged
for transformations of the type δqk = εδhk and δph = 0. The function G that
generates such a transformation through (11.3) is G = ph + const. Such a
G remains constant along the motion, according to Proposition 11.1 and the
remarks of §6.1 of Chapter 6.
12 Variational and Canonical Transformations
Let (1.2) be variational, so the it preserves the variational structure (1.1).
To every smooth Hamiltonian (p, q; t) → H(p, q; t) there corresponds a new
smooth Hamiltonian (P, Q; t) → KH (P, Q; t), dependent upon H. Denote by
Ko the function KH corresponding to the choice H = 0. Moreover, for every
Hamiltonian H, denote by H its pointwise transform through (1.2), that is,
(P, Q; t) → H(P, Q; t) = H(q(P, Q; t), p(P, Q; t); t).
Theorem 12.1 (Lie [113]). Let (1.2) be variational. There exists t → f (t)
such that for every Hamiltonian H,
(P, Q; t) → KH (P, Q; t) = f (t)H(P, Q; t) + Ko (P, Q; t).
Moreover, the transformation (1.2) is canonical if and only if f (t) = 1.
Proof. If the transformation preserves the variational structure of the Hamiltonian system, then
Ṗ
∇P KH
=E
∇Q KH
Q̇
and also
∂
Ṗ
ṗ
+
=J
q̇
Q̇
∂t
Therefore
JEJ
t
∇P H
∇Q H
P
Q
= JEJ
∂
+
∂t
P
Q
t
∇P H
∇Q H
∂
+
∂t
∇P KH
=E
,
∇Q KH
P
.
Q
272
10 CANONICAL TRANSFORMATIONS
for all smooth functions (P, Q; t) → H(P, Q; t). Taking H = 0, we obtain
∂ P
∇P Ko
=E
.
∇Q Ko
∂t Q
Putting this in the previous expression gives the identity
∇P (KH − Ko )
t ∇P H
.
=E
JEJ
∇Q (KH − Ko )
∇Q H
(12.1)
Lemma 12.1 Let (1.2) be variational, so that (12.1) holds. There exists a
smooth function t → f (t) determined only in terms of (1.2) such that JEJ t =
f (t)E.
Assuming the lemma for the moment, (12.1) provides the form of the new
Hamiltonian, as claimed by Proposition 12.1. The transformation (1.2) is
canonical if and only if J is symplectic. By the structure (6.1) of the symplectic
matrices, (1.2) is canonical if and only if f (·) = 1.
12.1 Proof of Lemma 12.1
Write the matrix JEJ t in the form
F1,1 F1,2
t
, where Fi,j are N × N matrices.
JEJ =
F2,1 F2,2 )
With this symbolism rewrite (12.1) as
F1,1 ∇P H + F1,2 ∇Q H = −∇Q (KH − Ko ),
F2,1 ∇P H + F2,2 ∇Q H = ∇P (KH − Ko ).
(12.2)
Take the divergence of the first equation with respect to P and of the second
with respect to Q. Adding the resulting expressions gives
divP (F1,1 ∇P H + F1,2 ∇Q H) + divQ (F2,1 ∇P H + F2,2 ∇Q H) = 0,
(12.3)
valid for all smooth functions (Q, P ; t) → H(Q, P ; t). Fix two indices k and
ℓ, and in (12.3) choose first H = Pk and then H = Pℓ Pk . This gives the 2N
relations
∂
∂
F1,1;hk +
F2,1;hk = 0,
∂Ph
∂Qh
∂
∂
(F1,1;hk Pℓ ) +
(F2,1;hk Pℓ ) = 0.
∂Ph
∂Qh
These imply F1,1 = O. Similarly, choosing first H = Qk and then H = Qℓ Qk
gives F2,2 = O. This permits one to rewrite (12.3) as
divP (F1,2 ∇Q H) + divQ (F2,1 ∇P H) = 0.
4c Constructing Canonical Transformations by Other Pairs of IV
273
Choosing H = Qk or H = Pk implies that F1,2;hk are independent of Ph , and
F2,1;hk are independent of Qh . Choosing H = Qℓ Pk gives
F1,2;hk δℓk + F2,1;hk δhℓ = 0,
from which
F1,2;hk = −F2,1;hk = 0
for h = k
and
F1,2;hh = −F2,1;hh .
Moreover, these functions are independent of Ph and Qh respectively. These
remarks applied to the first equation of (12.2) give
F1,2;kk
∂
∂H
=−
(KH − Ko ).
∂Qk
∂Qk
Write this for two indices h and k. Then take the derivative of the one corresponding to the index h with respect to Qk , and take the derivative of the one
corresponding to the index k with respect to Qh . Subtracting the resulting
expressions gives the identities
∂
∂H
∂H
∂
F1,2;kk
=
F1,2;hh
∂Qh
∂Qk
∂Qk
∂Qh
for h, k = 1, . . . , N,
(12.4)
valid for every smooth function (Q, P ; t) → H(Q, P ; t). Therefore F1,2;hh =
F1,2;kk for all h, k = 1, . . . , N , and (F1,2 ) = f (t)I.
Problems and Complements
4c Constructing Canonical Transformations by Other
Pairs of Independent Variables
Write (4.1)–(4.4) with F = F1 , where F1 is defined in (3.1). By (4.1), the
functions Q → F1 (q, Q; t) and P → F2 (q, P ; t), for fixed q ∈ RN and t ∈ R,
are affine and therefore convex. Prove that
F2 (q, P ; t) = sup [Ph ξh + F1 (q, ξ; t)].
ξ∈RN
That is, P → F2 (q, P ; t) is the Legendre transform of Q → −F1 (q, Q; t) (see
§5c of the Complements of Chapter 6). The latter is not coercive nor strictly
convex. Prove that nevertheless, the supremum is achieved for some Q at finite
distance from the origin. Prove that for such a Q, one has P = −∇Q F1 (q, Q; t),
which is precisely the second equation of (3.2). This connection between
F1 (q, Q; t) and F2 (q, P ; t) does not imply that if one of them generates a
274
10 CANONICAL TRANSFORMATIONS
canonical transformation the other does. Give an example of a canonical
transformation for which the variables (q, P ) can be taken as independent
and (q, Q) cannot. In such a case F2 generates a canonical transformation and
F1 does not. Neverthless, the relation (4.2) between them continues to hold.
Prove that the generators Fj can be constructed as the Legendre transform
of some Fi , for i = j, with respect to suitable variables.
4.1. Find the canonical transformation generated by F2 (p, Q) = e−2Q − p2
(N = 1).
4.2. Find the canonical transformation generated by F3 (p, Q) = p(1 − eQ ),
for N = 1. Answer: P = p(1 + q) and Q = ln(1 + q).
4.3. Let F2 (p, Q; ε) = qP + εq 2 (1 + cos P ) for N = 1. Find the values of
ε for which F2 generates a canonical transformation. Verify that such
transformations are
1
1 − 1 − 4εQ sin P .
p = P + 2εq(1 + cos P ),
q=
2ε sin P
4.4. Verify that the transformation (N = 1)
√
√
√
Q = ln(1 + p sin q), P = 2(1 + p sin q) p cos q
satisfies 2(P dQ − pdq) = d(p sin 2q). Therefore it is completely canonical
with generator F3 (p, Q) = −(eQ − 1)2 cot q.
4.5. Verify that the transformation (N = 1)
p = 2et P Q ln P,
q = e−t P Q
¯ + QdP
¯ is exact.
is canonical, by verifying that the differential form q dq
Compute the generator F2 as the primitive of such a form.
5c Examples of Canonical Transformations
5.1. Prove that any t-dependent canonical transformation is the flow map
(5.2) for some H.
5.2. Verify that the transformation P = p and Q = q − pt is canonical and
the Hamiltonian K associated to any Hamiltonian H is
K(P, Q; t) = H(P, Q + P t; t) − 21 P 2 .
5.3. Let N = 2, assume that in (1.2) the variables {p2 , q1 , P1 , Q2 } can be
taken as independent, and seek a primitive F̄ dependent on these vari¯ rewrite (4.1) for N = 2 as
ables and t. If F is a primitive of dω,
¯ 2 = d¯F̄ ,
¯ 1 − P2 dQ
¯ 2 + Q1 dP
¯ 1 − q2 dp
p1 dq
(5.1c)
5c Examples of Canonical Transformations
275
where F̄ = F + (P1 Q1 − p2 q2 ) and where {p2 , q2 , P1 , Q1 } are expressed
in terms of the independent variables {p2 , q1 , P1 , Q2 }. Then set
p1 = F̄q1 ,
q2 = −F̄p2 ,
P2 = −F̄Q2 ,
Q1 = F̄P1 .
In practice, having identified {p2 , q1 , P1 , Q2 } as independent, one writes
(5.1c) and finds a primitive F̄ of the differential form on the left-hand
side.
5.4. Verify that the transformation (N = 2)
P1 P2 − Q1 Q2
,
P12 + Q22
P2 Q2 − P1 Q1
q2 =
,
P12 + Q22
q1 =
p1 = −P1 Q1 ,
1
p2 = (P12 − Q22 ),
2
¯ is exact.
is canonical, by verifying that the differential form dω
¯ is exact and that the
5.5. For the indicated transformation, verify that dω
functions F (p, q; t) and F1 (q, Q; t) are primitives:
q
P = − arctan ,
p
Q=
1 2
(p + q 2 ),
2
q
1 2
1
(p + q 2 ) arctan + pq,
2
p
2
1
q
+ q 2Q − q 2 .
F1 (q, Q) = Q arcsin √
2
2Q
F (p, q) =
¯ is exact and that the
5.6. For the indicated transformation, verify that dω
functions F (p, q; t) and F1 (q, Q; t) are primitives:
Q = 2q e−t cos p.
P = 2q et sin p,
F (p, q) = pq − q sin p cos p,
1
rQe−t
− Qe−t 2q − Q2 e−2t .
F1 (q, Q) = Q arccos √
2
2q
¯ is exact and that the
5.7. For the indicated transformation, verify that dω
functions Fj below, for j = 1, 2, 3, 4, are all generators:
P = q cot p,
Q = ln
sin p
,
q
1 − q 2 e2Q + e−Q 1 − q 2 e2Q ,
q
F2 (q, P ) = q arctan + P 1 − ln q 2 + P 2 ,
P
cos p
−Q
F3 (p, Q) = e cos p, F4 (p, P ) = P + P ln
.
P
F1 (q, Q) = q arccos
276
10 CANONICAL TRANSFORMATIONS
5.8c Linear Canonical Transformations
Given constant invertible N × N matrices (αhk ) and (βhk ), consider the linear
transformation on R2N
Ph = αhk pk ,
Qh = βhk qk .
(5.2c)
Prove that (5.2c) is canonical if and only if it preserves the bilinear form pq,
that is, if
ph qh = Ph Qh =⇒ ph qh = αhk βhℓ pk qℓ .
Therefore (αhk ) must be the transpose of the inverse of (βhk ).
5.8.1c An Example Related to the n-Body Problem
Transform the Lagrangian coordinates q into
Q1 = q1 ,
Qh = qh − q1 ,
h = 2, . . . , N.
(5.3c)
Transform now p so that the resulting transformation is linear and completely
canonic. This occurs for
⎞
⎛
⎞
⎛
1 0 0 ··· 0
1 1 ··· 1 1
⎜−1 1 0 · · · 0⎟
⎟
⎜
⎜ 0 1 0 · · · 1⎟
⎟
⎟
⎜
⎜
.
(βhk ) = ⎜−1 0 1 · · · 0⎟ , (αhk ) = ⎜
. . .. ⎟
⎟
⎜
⎝
.
· · · · · · · · · . .⎠
⎝ · · · · · · · · · . . . .. ⎠
0 0 0 ··· 1
−1 0 0 · · · 1
The corresponding p transformations are (see §9 of Chapter 6, and in particular (9.1))
P1 = p1 + p2 + · · · + pN ,
Ph = ph ,
h = 2, . . . , N.
(5.4c)
6c Symplectic Product in Phase Space
and Symplectic Matrices
Let T be a linear transformations of R2N into itself, identified with its representative matrix, and up to an isomorphism write
T11 T12
,
(6.1c)
T =
T21 T22
where (Tij ) are N × N matrices.
6.1. Prove that T is symplectic if and only if
t
t
t
t
T21
T11 − T11
T21 = O, T21
T12 − T11
T22 = −I,
t
t
t
t
T22 T11 − T12 T21 = I, T22 T12 − T12 T22 = O.
7c Characterizing Canonical Transformations by Symplectic Jacobians
277
6.2. Prove that if T is symplectic, then
T −1 =
t
t
−T12
T22
t
t
−T21 T11
.
6.3. The symplectic group of order 2 consists of 2 × 2 matrices whose determinant is 1. Rotation matrices of R2 are symplectic transformations.
7c Characterizing Canonical Transformations
by Symplectic Jacobians
7.1c Linear Canonical Transformations by Symplectic Matrices
Consider the linear transformation with constant coefficients
Ph = αp,hk pk + αq,hk qk ,
Qh = βp,hk pk + βq,hk qk ,
of the Jacobian
J=
(αp,hk ) (αq,hk )
.
(βp,hk ) (βq,hk )
Therefore this transformation is canonical if and only if
(βp,hk )t (αp,hk ) − (αp,hk )t (βp,hk ) = O,
(βp,hk )t (αq,hk ) − (αp,hk )t (βq,hk ) = −I,
(βq,hk )t (αp,hk ) − (αq,hk )t (βp,hk ) = I,
(βq,hk )t (αq,hk ) − (αq,hk )t (βq,hk ) = O.
Verify that if (αq,hk ) = (βp,hk ) = O, one finds the same condition of canonical
transformation as that in §5.8c. Set
(αp,hk ) = (βq,hk ) = O;
(αq,hk ) = (αhk ),
(βp,hk ) = (βhk ).
Find a condition on these coefficients for the resulting transformation to be
canonical. In particular, for N = 1 justify why the transformation in (2.7) is
not canonical unless λ = 1.
7.2c The Poincaré Recurrence Theorem
Assume that the Hamiltonian system (1.1) is autonomous, and that it admits
an invariant region E. Specifically, H is independent of time and there exists
a region E ⊂ R2N such that for all (po , qo ) ∈ E, the unique solution
t → S t [(po , qo )] = (p(t), q(t))
of (1.1) originating at (po , qo ) remains in E for all times. Here we have denoted
by S t the solution operator at time t. In particular, S o = I.
278
10 CANONICAL TRANSFORMATIONS
A set B ⊂ E is mapped at time t into
S t [B] =
S t [(po , qo )].
(po ,qo )∈B
If B is measurable, also S t [B] is measurable, and by Liouville’s theorem,
|B| = |S t [B]|.
Theorem 7.1c (Poincaré [127 Vol. 3 Chap. 26]). Assume that E is of
finite measure. For every measurable B ⊂ E of positive measure and every
to ∈ R,
(a) There exists t > to such that S t [B] ∩ B = ∅.
(b) There exists a set of measure zero Bo ⊂ B such that all orbits originating
at B − Bo return to intersect B − Bo infinitely many times.
Remark 7.1c The first part of the theorem asserts that among all orbits
originating in B, there are some that return to B after a sufficiently long
time. The theorem does not assert that all orbits originating in B return to
B. However, the second part of the theorem asserts that this occurs for all
orbits originating at all points of B, except possibly for a set Bo of measure
zero.
Remark 7.2c The set B could be a ball of arbitrarily small radius. In such
a case all orbits originating at almost all points in B return, to be arbitrarily
close to their initial position, after a sufficiently long time. For this reason this
is referred to as the recurrence theorem.
Proof (of (a)). Pick to > 0, set Bo = B, and for j ∈ N, set Bj = S jto [B].
There exist two indices jo ≥ 0 and jo + i for some i > 0 such that Bjo ∩
Bjo +i = ∅. Indeed, otherwise the sets Bj would be disjoint and contained in
E, contradicting that |E| < ∞. If jo = 0, the assertion is proven. If jo > 0,
then Bjo −1 ∩ Bjo −1+i = ∅. If jo = 1, this proves the assertion; otherwise,
Bjo −2 ∩ Bjo −2+i = ∅. Repeating this argument jo times proves (a).
Proof (of (b)). Denote by Bo the subset of B such that every orbit originating
at Bo never intersects B. If Bo had positive measure, by the same argument
of (a), this would contradict that |E| < ∞. The second part of the assertion
is proved by iteration.
7.3c A Note on Liouville’s Theorem
Liouville’s theorem could be proved more directly from the definition of flow
map. Assume that a bounded open set E(t) ⊂ RN with smooth boundary
∂E(t) is mapped by a smooth velocity field t → v(·; t) into E(t + dt) in an
7c Characterizing Canonical Transformations by Symplectic Jacobians
279
infinitesimal time interval dt. Then by elementary calculus considerations and
the Gauss–Green theorem,
d
|E(t)| =
v · ndσ =
div vdv,
dt
∂E(t)
E(t)
where n is the unit normal to ∂E(t) pointing outside E(t) and dσ and dv are
respectively the surface measure on ∂E(t) and the volume measure on E(t).
For the flow map in phase space,
v = (ṗ, q̇) = (−∇q H, ∇p H)
and
div v = 0.
More generally, the flow map generated by a smooth irrotational vector field
t → v(·; t) is volume-preserving.
7.4c Integral Invariants
Let γ be a smooth closed curve in R2N of parametric equations
p = p(τ ), q = q(τ ) for τ ∈ [τ1 , τ2 ]
.
γ=
with p(τ1 ) = p(τ2 ), q(τ1 ) = q(τ2 )
Denote by Γ the image of γ by the transformation (1.2) of the parametric
equations
P = P (τ ) = P (p(τ ), q(τ ); t) with P (τ1 ) = P (τ2 )
Γ =
.
Q = Q(τ ) = Q(p(τ ), q(τ ); t) with Q(τ1 ) = Q(τ2 )
¯ = dF
¯ . Therefore
If (1.2) is canonical with generator F , then dω
τ2
d
¯ =
¯ +
F (q(τ ), Q(τ ); t)dτ.
pdq
P dQ
τ1 dτ
γ
Γ
The last integral vanishes, since γ and Γ are closed, and we have the following
result.
Proposition 7.1c The integral of the differential form pdq in R2N along a
smooth closed curve γ ⊂ R2N is invariant by canonical transformations.
7.4.1c Poincaré Surface Invariants [20, 127]
Denote by pi and qi the coordinate unit vectors along the axes of the pi
and qi respectively. To the differential form pdq there is associated the vector
F = pi qi + 0i pi . If γ ⊂ R2N is a smooth closed curve as above, denote by σ
a smooth surface whose boundary is γ and denote by σh the projection of σ
on the hyperplane {ph , qh }, whose unit normal is ph ∧ qh . Let also n be the
280
10 CANONICAL TRANSFORMATIONS
unit normal to σ chosen so that the orientation on γ is counterclockwise with
respect to the orientation of n. By Stokes’s theorem,
∇p,q ∧ F · ndσ
curl F · ndσ =
pdq =
σ
σ
γ
=
σ
N
∂
pk ph ∧ qk · ndσ =
ph
h=1
dph dqh .
σh
The right-hand side is the sum of the oriented areas of the projections of a
smooth surface σ ⊂ R2N on the coordinate planes {ph , qh } and is called the
Poincaré surface invariant. Let Γ and Σ be the transforms of γ and σ by (1.2)
and denote by Σh the projection of Σ on the coordinate plane {Ph , Qh }. If
(1.2) are canonical, then by Proposition 7.1c,
N
h=1
dph dqh =
σh
N
h=1
dPh dQh .
Σh
8c Poisson Brackets
8.1c Lagrange Brackets [103]
Let (u, v) → F(u, v), G(u, v) be smooth vector-valued functions of two
N -tuples of variables (u, v). Pick a scalar variable u out of u and a scalar
variable v out of v. The Lagrange brackets of (u, v) with respect to F and G
are defined by
[[u, v]]F,G =
∂Fi ∂Gi
∂Fi ∂Gi
−
.
∂u ∂v
∂v ∂u
One verifies that
[[u, v]]F,G = −[[v, u]]F,G = [[v, u]]G,F .
Let z be a smooth, locally invertible, 2N -valued function of (p, q) so that the
transformations
(p, q) → z(p, q) = z1 (p, q), . . . , z2N (p, q) ,
z → (p(z), q(z)),
are well defined. There exists a remarkable relation between the Lagrange
brackets of any pair of variables (zh , zℓ ) with respect to the pair of vector
functions q and p and the Poisson brackets of any pair of functions (zℓ , zk )
with respect to the variables (p, q).
Proposition 8.1c [[zh , zℓ ]]q,p {zℓ , zk }p,q = δhk .
8c Poisson Brackets
281
Proof. from the definitions,
∂q ∂p
∂qi ∂pi ∂zℓ ∂zk
∂zℓ ∂zk
i
i
−
−
∂zh ∂zℓ ∂zℓ ∂zh ∂pj ∂qj
∂qj ∂pj
∂pi ∂zℓ ∂zk ∂qi
∂qi ∂zℓ ∂zk ∂pi
=
+
∂zℓ ∂pj ∂qj ∂zh
∂zℓ ∂qj ∂pj ∂zh
∂pi ∂zℓ ∂qi ∂zk
∂qi ∂zℓ ∂pi ∂zk
−
−
.
∂zℓ ∂qj ∂zh ∂pj
∂zℓ ∂pj ∂zh ∂qj
[[zh , zℓ ]]q,p {zℓ , zk }p,q =
Since p and q are independent, the last two terms are zero. For the first two,
∂pi ∂zℓ
∂pi
=
= δij ,
∂zℓ ∂pj
∂pj
∂qi ∂zℓ
∂qi
=
= δij .
∂zℓ ∂qj
∂qj
Therefore
[[zh , zℓ ]]q,p {zℓ , zk }p,q =
∂zk ∂qi
∂zk ∂pi
∂zk
δij +
δij =
= δhk .
∂qj ∂zh
∂pj ∂zh
∂zh
Remark 8.1c The proposition asserts that the two matrices ([[zh , zℓ ]]q,p ) and
({zℓ , zk }p,q ) are inverses of each other. This is based only on the definitions of
Lagrange and Poisson brackets and on the mutual invertibility of the variables
z and (p, q). In particular, it is not required that these transformations be
canonical.
8.2c Invariance of Lagrange Brackets by Canonical
Transformations
Given a smooth invertible transformation as in (1.2), construct z(p, q; t) by
setting
Pi (q, p; t), i = 1, 2, . . . , N,
zi (q, p; t) =
Qi (q, p; t), i = N + 1, N + 2, . . . , 2N.
With this notation,
([[Ph , Pk ]]q,p ) ([[Ph , Qk ]]q,p )
,
([[zh , zℓ ]]q,p ) =
([[Qh , Pk ]]q,p ) ([[Qh , Qk ]]q,p )
({Ph , Pk }p,q ) ({Ph , Qk }p,q )
({zh , zℓ }p,q ) =
.
({Qh , Pk }p,q ) ({Qh , Qk }p,q )
By Proposition 8.1, the transformation (1.2) is canonical if and only if
({Qh , Qk }q,p ) ({Qh , Pk }q,p )
O I
.
=
−I O
({Ph , Qk }q,p ) ({Ph , Pk }q,p )
It follows from Proposition 8.1c that (1.2) is canonical if and only if
([[Qh , Qk ]]q,p ) ([[Qh , Pk ]]q,p )
O −I
.
=
I O
([[Ph , Qk ]]q,p ) ([[Ph , Pk ]]q,p )
282
10 CANONICAL TRANSFORMATIONS
Proposition 8.2c The transformations (1.2) are canonical if and only if
[[ph , pk ]]P,Q = [[qh , qk ]]P,Q = 0;
[[ph , qk ]]P,Q = δhk ,
(8.1c)
[[Ph , Pk ]]p,q = [[Qh , Qk ]]p,q = 0;
[[Ph , Qk ]]p,q = δhk .
Corollary 8.1c The transformation (1.2) is canonical if and only if it preserves the Lagrange brackets.
11
INTEGRATING HAMILTON–JACOBI
EQUATIONS AND CANONICAL SYSTEMS
1 Complete Integrals of Hamilton–Jacobi Equations
A complete integral of the Hamilton–Jacobi equation
Ft + H(q, ∇q F ; t) = 0
(1.1)
is a smooth function F of q and t that depends on N arbitrary parameters
Q = (Q1 , , . . . , QN ) and that satisfies (1.1) as q and Q range within the domain
of definition of F . The dependence on the parameters Q is required to be
“essential” in the sense that
2
∂ F (q, Q; t)
det
= 0
(1.2)
∂qh ∂Qk
within its domain of definition. Finding a complete integral of (1.1) resolves
the motion, as indicated in §5.3 of Chapter 10.
1.1 Examples of Complete Integrals of (1.1)
Since F appears in (1.1) only through its derivatives, from a complete integral
F (q, Q; t) one can construct a family of integrals by adding a constant. If H
depends only on p, then (1.1) has the complete integral
F (q, Q; t) = q · Q − tH(Q) + C.
(1.3)
A complete integral of the eikonal equation ∇q F = 1 is
F (q, Q) = q · Q + C
for C ∈ R
and Q = 1.
A complete integral of Clairaut’s equation
q · ∇q F + f (∇q F ) = F
(1.4)
for a smooth f : RN → R is
F (q, Q) = q · Q + f (Q).
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 11,
c Springer Science+Business Media, LLC 2011
(1.5)
283
284
11 INTEGRATING HAMILTON–JACOBI EQUATIONS
1.2 Hamiltonian Independent of t
If H is independent of t, we seek a complete integral of (1.1) of the form
F (q, Q; t) = Fo (q, Q) − E(Q)t
(1.6)
for a smooth E : RN → R, where Fo (q, Q) is a complete integral of the
stationary Hamilton–Jacobi equation
H(q, ∇q Fo ) = E(Q).
(1.7)
If such a complete integral can be determined, the corresponding canonical
transformations take the form
p = ∇q Fo (q, Q),
P + ωt = ∇Q Fo (q, Q),
ω = ∇Q E(Q).
(1.8)
Since H is independent of time, it coincides with the energy of the system,
which is conserved. On the other hand, (1.7), written along the motion, gives
t → H(q(t), p(t)) = E(Q).
Therefore the constant E(Q) is the energy of the system determined through
the N parameters Q, which remain constant along the motion.
2 Separation of Variables
Denote by q ′ = (q2 , . . . , qN ) the last N − 1 of the variables q. Assume that the
Hamiltonian is of the form
H(q, ∇q F ; t) = H q ′ , ∇q′ F, f1 (q1 , Fq1 ); t ,
(2.1)
where f1 (·, ·) is a smooth function of two scalar variables only and is independent of t. Thus H depends on q1 and Fq1 only through a function f1 of these,
independent of time. Examples of such Hamiltonians are
H(q, ∇q F ; t) =
N
j=1
fj (qj , Fqj ),
H(q, ∇q F ; t) =
N
j=1
fj (qj , Fqj ),
where fj (·, ·) are smooth functions each of two scalar variables and independent of t. More generally, the Hamiltonian could be of the form
H(q, ∇q V ) = H (f1 (q1 , Fq1 ; f2 (q2 , Fq2 ; f3 (· · · ; fN (qN , VqN ; · · · )))) .
Hamiltonians that do not exhibit the structure (2.1) include
H(q, ∇q F ) =
divq F
,
1 + q
H(q, ∇q F ; t) =
Fq1
.
q12 + t2
(2.2)
2 Separation of Variables
285
Assume that H has the structure (1.1) and seek a solution of (1.1) of the form
F (q; t) = FN −1 (q ′ ; t) + h1 (q1 )
for smooth functions FN −1 : RN → R and h1 : R → R. If a solution of this
form exists, then we must have
∂
FN −1 + H (q ′ , ∇q′ FN −1 , f1 (q1 , h1,q1 ); t) = 0
∂t
(2.3)
identically as q1 ranges within its domain of definition. Therefore f1 (·, ·) must
be constant, that is,
(q1 , h1,q1 ) → f1 (q1 , h1,q1 ) = Q1 ,
(2.4)
for some constant Q1 . The latter is a first-order partial differential equation,
whose complete integral F1 (q1 ; Q1 ) is computed by quadratures. Putting now
(2.4) into (2.3) gives
∂
FN −1 + H (q ′ , ∇q′ FN −1 , Q1 ; t) = 0.
∂t
(2.5)
This is a Hamilton–Jacobi equation involving only N − 1 independent variables. Thus if the Hamiltonian has the structure (2.1), then the variable q1
can be separated from the remaining variables q ′ .
Denote by q ′′ = (q3 , . . . , qN ) the last N − 2 of the variables q and assume
that the Hamiltonian in (2.5) is of the form
H(q ′ , ∇q′ FN −1 , Q1 ; t) = H (q ′′ , ∇q′′ FN −1 , f2 (q2 , FN −1,q2 ; Q1 )) ,
where f2 (·, ·; Q1 ) is a smooth function of two scalar variables and is independent of t. Then we seek a solution of (2.4) of the form
FN −1 (q ′ ; t) = FN −2 (q ′′ ; t) + h2 (q2 ; Q1 ).
If a solution of this form exists, then
(q2 , h2,q2 ) → f2 (q2 , h2,q2 ; Q1 ) = Q2 ,
for some constant Q2 . The latter can be integrated by quadratures to give the
complete integral F2 (q2 ; Q1 , Q2 ). Therefore an integral of (1.1) has the form
F (q; t) = FN −2 (q ′′ ; t) + F1 (q1 ; Q1 ) + F2 (q2 ; Q1 , Q2 ),
where FN −2 (q ′′ ; t) is a solution of the Hamilton–Jacobi equation in N − 2
independent variables
∂
FN −2 + H (q ′′ , ∇q′′ FN −2 , Q1 , Q2 ; t) = 0.
∂t
If this process can be repeated N times, all the variables q are separated, and
the Hamilton–Jacobi equation (1.1) is called completely separable.
286
11 INTEGRATING HAMILTON–JACOBI EQUATIONS
2.1 An Example of Complete Separability
Assume that the Hamiltonian H in (1.1) is independent of t and has the
general structure (2.2). Then set
Fo (q, Q) = F1 (q1 , Q1 ) + F2 (q2 ; Q1 , Q2 ) + · · · + FN (qN ; Q1 , . . . , QN ),
where for j = 1, . . . , N , the function Fj is the complete integral of the firstorder equation
fj (qj , Fj,qj ; Q1 , . . . , Qj−1 ) = Qj
in the single variable qj . Regarding now t as the (N + 1)st variable, we seek
a complete integral of (1.1) of the form
(q, Q; t) → F (q, Q; t) = Fo (q, Q) + FN +1 (t, Q).
Putting this into (1.1) and taking into account the construction of the fj and
the structure (2.2) of the Hamiltonian gives
∂
FN +1 (t, Q) = −H(q, ∇q Fo ) = −E(Q).
∂t
Thus FN +1 = −tE(Q), and the complete integral of (1.1) is
(q, Q) → F (q; Q) = Fo (q, Q) − tE(Q),
where (q, Q) → Fo (q, Q) is the complete integral of the corresponding stationary Hamilton–Jacobi equation.
2.2 Cyclic Variables
If H has a cyclic variable, say for example q1 , then it is of the form (2.1)
with f1 (q1 , Fq1 ) = Fq1 . Now Fq1 is the kinetic moment p1 corresponding to
the Lagrangian coordinate q1 (see (3.2) of Chapter 10 and §5.3 of the same
chapter). Since q1 is cyclic, Fq1 = Q1 for a constant Q1 . Therefore a complete
integral of (1.1) can be sought of the form
F (q, Q; t) = F ′ (q ′ ; t) + Q1 q1 ,
where F ′ is a complete integral of the Hamilton–Jacobi equation (2.5) in N −1
independent variables. Further cyclic variables permit one to reduce further
the number of independent variables.
3 Reducing the Rank of a Canonical System
The rank of the canonical system
ṗ = −∇q H(q, p; t),
q̇ = ∇p H(q, p; t),
(3.1)
3 Reducing the Rank of a Canonical System
287
is defined as the nonnegative integer
rank of the matrix
∂ 2 H(p, q; t)
∂ph ∂qk
.
The general theory of systems of first-order differential systems ensures that
the knowledge of a first integral of (3.1) lowers the rank of the system from
2N to (2N − 1). However, for canonical systems as in (3.1), the knowledge of
a first integral lowers the rank from 2N to 2(N − 1).
In the next sections we will verify this property for specific physical integrals, such as the energy integral and the integrals of the kinetic momenta.
3.1 Canonical Systems with the Energy Integral
If H is independent of t, the mechanical system has the energy integral
t → H p(t), q(t) = E = const
for t about some initial to . Without loss of generality we may assume that for
at least one of the variables q, say for example qN ,
|q̇N | > 0
and
|HpN (p, q)| > 0
(3.2)
for all t about to . Then by the implicit function theorem t can be locally resolved in terms of qN , which in turn may be taken as an independent variable.
To stress such a choice of independent variables, set
qN = η,
t = t(η)
in a neighborhood of to .
For such a choice, along the motion t → p(t), q(t) , a solution of (3.1), set
η → p̄(η) = (p1 (t(η)), . . . , pN −1 (t(η))) ,
η → q̄(η) = (q1 (t(η)), . . . , qN −1 (t(η))) .
The second inequality of (3.2) and the implicit function theorem permit one
also to compute
η → pN (η) = ϕ(p̄(η), q̄(η); η)
for a smooth (p̄, q̄; η) → ϕ(p̄, q̄; η). Next for h = 1, . . . , N − 1, compute
∂H def
d
∂H
ṗh
p̄h =
/
=−
= Gh (q̄, p̄, η, ϕ(q̄, p̄, η)) ,
dη
q̇N
∂qh
∂pN
q̇h
d
∂H def
∂H
q̄h =
/
=
= Fh (q̄, p̄, η, ϕ(q̄, p̄, η)) .
dη
q̇N
∂ph
∂pN
This is a system whose rank does not exceed 2(N − 1).
288
11 INTEGRATING HAMILTON–JACOBI EQUATIONS
3.2 Cyclic Variables
Assume that H in (3.1) is independent of ℓ components of the Lagrangian
configuration q, say, for example after a possible reordering, (q1 , . . . , qℓ ). In
such a case the first ℓ kinetic momenta are constant along the motion, say for
example t → pj (t) = po,j for j = 1, . . . , ℓ. Therefore H might be considered as
independent of the first ℓ Lagrangian positions qj as well as the corresponding
kinetic momenta pj . Setting
p̃ = (pℓ+1 , . . . , pN ),
q̃ = (qℓ+1 , . . . , qN ),
H̃(p̃, q̃; t) = H(p̃, q̃, po,1 , . . . , po,ℓ ; t),
the canonical system (3.1) is transformed into the system
ṗh = −
∂ H̃(q̃, p̃; t)
,
∂qh
q̇h =
∂ H̃(q̃, p̃; t)
∂ph
for h = ℓ + 1, . . . , N,
of rank not exceeding 2(N − ℓ). If H is independent of all the N components
of the Lagrangian positions q, then all N kinetic momenta p are constant.
These are N integrals of motion, and the resulting system is of rank zero.
Equivalently, if all the Lagrangian coordinates are cyclic, the N integrals of
the corresponding kinetic momenta resolve the motion. This is a special case
of the next Liouville theorem.
4 The Liouville Theorem of Quadratures
Assume that the canonical system (3.1) has N independent first integrals
(p, q; t) → Fi (p, q; t) = Qi
i = 1, . . . , N,
(4.1)
where Q = (Q1 , . . . , QN ) are constants. These are independent in the sense
that the N × 2N matrix
∂Fi (p, q; t) ∂Fℓ (p, q; t)
(4.2)
∂pj
∂qk
i,j,ℓ,k=1,...,N
has maximum rank in a neighborhood of some fixed initial configuration.
Assume, moreover, that these first integrals are in involution with respect to
the Poisson brackets, that is,
N
∂Fi ∂Fj
∂Fi ∂Fj
{Fi , Fj } =
= 0, i, j = 1, . . . , N. (4.3)
−
∂qh ∂ph
h=1 ∂ph ∂qh
Theorem 4.1 (Liouville [116, 117]). N independent first integrals of (3.1),
in involution, resolve the motion.
4 The Liouville Theorem of Quadratures
289
Remark 4.1 This is a special case of a more general theorem of Lie ([115];
also in [64, Chap.VIII]), which asserts that ℓ independent first integrals in
involution reduce (3.1) into another one, still canonical and of rank 2(N − ℓ).
Liouville’s theorem follows from this for ℓ = N .
The integration of (3.1) hinges upon finding a complete integral of the
Hamilton–Jacobi equation (1.1). The proof of Liouville’s theorem consists in
constructing such a complete integral from the independent first integrals Fj
in involution.
4.1 Poisson Brackets in 2(N + 1) Variables
Denote the time t by qo and introduce the two (N + 1)-tuples of variables
(po , p) = (po , p1 , . . . , pN ),
(qo , q) = (qo , q1 , . . . , qN ),
where po is a scalar auxiliary variable. Set also
(po , p, qo , q) → Fo (po , p, qo , q) = po + H(p, q; qo )
and regard the first integrals Fj as functions of the 2(N + 1) variables
(po , p, qo , q), independent of po .
Lemma 4.1 The (N + 1) functions {Fo , F1 , . . . , FN } are in involution with
respect to the two (N + 1)-tuples of variables (po , p) and (qo , q), i.e., denoting
by {{·, ·}} the Poisson brackets with respect to these two (N + 1)-tuples of
variables
∂Fi ∂Fj
∂Fi ∂Fj
{{Fi , Fj }} =
−
+ {Fi , Fj } = 0
∂po ∂qo
∂qo ∂po
for all i, j = 0, 1, . . . , N .
Proof. If i, j = 0, the assertion follows since Fj are independent of po and are
in involution with respect to the two N -tuples of variables p and q. If i = 0
and j = 0, then
{{Fo , Fj }} = Fj,t + {H, Fj } = 0
by Proposition 10.1 of Chapter 10.
4.2 Proof of Liouville’s Theorem
Without loss of generality, we assume that the independence of the first integrals {F1 , . . . , FN } as expressed in (4.2) is realized with respect to the kinetic
momenta pj , that is,
∂Fi (q, p; t)
= 0
(4.4)
det
∂pj
i,j=1,...,N
290
11 INTEGRATING HAMILTON–JACOBI EQUATIONS
in a neighborhood of some fixed initial configuration. Then by the implicit
function theorem, each of the pj for j = 1, . . . , N can be expressed in terms
N
of (q, t; Q) on the set σ = i=1 [Fi = Qi ]. Such a set is nonempty, since the
Fj are first integrals of (3.1). Now on σ, also po can be expressed in terms of
(q, Q; t) by setting Fo = 0. Thus on σ,
pi = pi (q, Q; t)
for i = 1, . . . , N
and
po = −H q, p1 (q, Q; t), . . . , pN (q, Q; t); t .
Lemma 4.2 The differential form
pdq = po (q, Q; qo )dqo +
N
i=1
is exact on σ.
pi (q, Q; qo )dqi = −Hdt +
Proof. Consider the (N + 1) × (N + 1) matrices
∂Fi
∂Fi
,
B=
,
A=−
∂qj
∂pj
C=
N
pi dqi
i=1
∂pi
∂qj
.
It suffices to show that C is symmetric in σ. Indeed, in such a case
∂pi
∂pj
=
∂qj
∂qi
for all i, j = 0, 1, . . . , N,
which is the characteristic condition for pdq to be exact. On σ,
0=
dFi
∂Fi
∂Fi ∂pℓ
=
+
.
dqj
∂qj
∂pℓ ∂qj
Therefore A = BC on σ, and since B is nondegenerate, C = B −1 A. Thus C
is symmetric if
B −1 A = At (B t )−1 ⇐⇒
B t A = BAt ⇐⇒ B t A − BAt = O.
The latter, written in components, is precisely the condition {{Fi , Fj }} = 0
for all i, j ∈ {0, 1, . . . , N }, ensured by Lemma 4.1.
Let (q, Q; t) → F (q, Q; t) be a primitive of pdq whose existence is ensured by
the previous lemma. Then
pi (q, Q; t) =
∂F (q, Q; t)
∂qi
for i = 0, 1, . . . , N.
1c Complete Integrals of Hamilton–Jacobi Equations
291
For i = 0 this gives p = ∇q F , and for i = 0,
∂
F (q, Q; t) = −H(q, ∇q F (q, Q; t); t).
∂t
Therefore F is a family of solutions of the Hamilton–Jacobi equation (1.1),
depending on the N -dimensional parameter Q. To show that it is a complete
integral, it remains to show that it satisfies (1.2). Since Fh = Qh on σ,
∂ 2 F (q, Q; t)
∂Fh (q; t)
=
= δhk .
∂qh ∂Qk
∂Qk
Problems and Complements
1c Complete Integrals of Hamilton–Jacobi Equations
1.1c Envelopes of Solutions
Let (q, Q; t) → F (q, Q; t) be a complete integral of (1.1) and seek a smooth
function (q; t) → U (q; t) whose graph, for t fixed, is tangent at each of its points
(η, U (η; t)) to the graph of (q, Q(η); t) → F (q, Q(η; t); t) for some η within the
range of the parameters Q. Such a function, if it exists, is the envelope of the
family of solutions F (·, Q; t) of (1.1) parameterized by Q. From the definition,
U (q; t) = F (q, Q(q; t); t)
and
∇U (q) = ∇q F (q, Q; t)
Q=Q(q;t)
.
Therefore finding the envelope of a complete integral F (q; Q; t) reduces to
finding Q = Q(q; t) and putting this in the expression for F .
From the first of these, for fixed t,
∇U = ∇q F = ∇q F + ∇Q F · ∇q Q.
Therefore
FQi (q, Q(q; t); t)Qi,qj = 0,
j = 1, . . . , N.
(1.1c)
This is a linear homogeneous system in the unknowns FQi (q, Q(q; t); t) that
admits only the zero solution if det(Qi,qj ) = 0. In such a case the functions
Q = Q(q; t) can be computed from
FQi (q, Q(q; t); t) = 0,
i = 1, . . . , N.
292
11 INTEGRATING HAMILTON–JACOBI EQUATIONS
1.1.1c Examples of Envelopes
(a) In the Clairaut equation (1.4) take f (∇q F ) = − 21 ∇q F 2 , and show that
the corresponding envelope is U (q) = 21 q2 .
(b) For the eikonal equation show that the envelope of its complete integral
is U (q) = q. This is a case when the determinant of the coefficients in
(1.1c) vanishes. Discuss how the calculation Q = Q(q) is actually done.
(c) If H = p2 , the envelope of the complete integral of corresponding
Hamilton–Jacobi equation is U (q) = q2 /4t.
2c Separation of Variables
Consider a point mass {P ; m} acted upon by a potential x → V (x, x) with
nontrivial dependence on x. The Lagrangian is
L(x, ẋ) = 12 mẋ2 + V (x, x).
(2.1c)
The kinetic momenta are pi = mẋi and the Hamiltonian is
H(x, p) =
1
p2 − V (x, x).
2m
(2.2c)
The corresponding Hamilton–Jacobi equation is
1
∇x F 2 − V (x, x) = 0.
(2.3c)
2m
Since V has a nontrivial dependence on x, the Hamiltonian does not have
the structure (2.1) with respect to any of the pairs of variables (xi , Fxi ). As
an example, consider the case of a gravitational potential V (x) = x−1 .
However, having fixed V , at times it might be possible to choose Lagrangian
coordinates for which H has the structure (2.1). In the next sections we will
indicate some of these choices.
Ft +
2.1c Spherical Coordinates
For these coordinates the Lagrangian has the form (§1.3c of the Complements
of Chapter 2)
L = 12 m ρ̇2 + ϕ̇2 ρ2 sin2 θ + θ̇2 ρ2 + V (ρ, ϕ, θ).
(2.4c)
From this one computes the kinetic momenta and the Hamiltonian,
pρ = mρ̇,
H=
pϕ = mϕ̇ρ2 sin2 θ,
pθ = mθ̇ρ2 ,
p2ϕ
1
p2
p2ρ + 2 2 + θ2 − V (ρ, ϕ, θ).
2m
ρ sin θ ρ
(2.5c)
2c Separation of Variables
293
Assume that the potential V is of the form
−2mV (ρ, ϕ, θ) = g(ρ) +
h(θ)
ρ2
for smooth functions g(·) and h(·). Then
H=
p2ϕ
p2 + h(θ)
1
p2ρ + g(ρ) + θ 2
+ 2 2 ,
2m
ρ
ρ sin θ
and the corresponding Hamilton–Jacobi equation is
Ft +
1
1
1
Fρ2 + g(ρ) + 2 Fθ2 + h(θ) +
Fϕ2
2m
ρ
sin2 θ
= 0.
(2.6c)
Since H is cyclic with respect to the variable ϕ, we seek a solution of the form
F (ρ, ϕ, θ; t) = F ′ (ρ, θ, Qϕ ; t) + Qϕ ϕ,
where Qϕ is a scalar constant and F ′ is a solution of the Hamilton–Jacobi
equation
Ft′ +
Q2ϕ
1
1
Fρ′2 + g(ρ) + 2 Fθ′2 + h(θ) +
2m
ρ
sin2 θ
=0
in the sole variables ρ and θ. Since this has the structure (2.1) with respect
to θ, we seek a solution of the form
F ′ (ρ, θ, Qϕ ; t) = F ′′ (ρ, Qϕ , Qθ ; t) + W (θ, Qϕ , Qθ ),
where Qθ is a constant and W is a complete integral of the first-order partial
differential equation
2
Q
Q2ϕ
ϕ
=
Q
dθ.
Wθ2 + h(θ) +
=⇒
W
=
Q
−
h(θ)
−
θ
θ
sin2 θ
sin2 θ
The function ρ → F ′′ (ρ, Qϕ , Qθ ; t) is a solution of the Hamilton–Jacobi
equation
1
Qθ
Ft′′ +
Fρ′′2 + g(ρ) + 2 = 0
2m
ρ
in the sole variable ρ. This last equation is separable with respect to ρ and t
and gives an integral of the form
F ′′ (ρ, Qϕ , Qθ ; t) = W ′ (ρ, Q) − tE(Q)t,
Q = (Qρ , Qϕ , Qθ ),
294
11 INTEGRATING HAMILTON–JACOBI EQUATIONS
where Q → E(Q) is a smooth function of Q, and W ′ is a complete integral of
the stationary Hamilton–Jacobi equation
Q
Qθ
θ
Wρ′2 + g(ρ) + 2 = 2mE(Q) =⇒ W ′ =
2mE(Q) − g(ρ) − 2 dρ.
ρ
ρ
Combining these remarks, the complete integral of (2.6c) is given by
F (ρ, ϕ, θ, Qρ , Qϕ , Qθ ; t) = −tE(Qρ , Qϕ , Qθ )t + Qϕ ϕ
Q2ϕ
Qθ
dθ +
Qθ − h(θ) −
2mE(Q) − g(ρ) − 2 dρ.
+
ρ
sin2 θ
Using this complete integral resolves the motion.
2.2c Parabolic Coordinates
Lagrangian, kinetic momenta, and Hamiltonian are (§1.2c of the Complements
of Chapter 2)
L=
1
u + v u̇2
v̇ 2
m
+
+ ϕ̇2 uv + V (u, v, ϕ),
2
4
u
v
pu = m
u+v
u̇,
4u
pv = m
u+v
v̇,
4v
pϕ = muv ϕ̇,
1 p2ϕ
2 up2u + vp2v
+
− V (u, v, ϕ).
m u+v
2m uv
Assume that V is of the form
H=
V (u, v, ϕ) = −
g(u) + h(v)
u+v
for smooth functions g(·) and h(·). Then
H=
1
(u + v) 2
p + m (g(u) + h(v)) ,
2(up2u + vp2v ) +
m(u + v)
2uv ϕ
and the corresponding Hamilton–Jacobi equation is
Ft +
(u + v) 2
1
Fϕ + m (g(u) + h(v)) = 0.
2uFu2 + 2vFv2 +
m(u + v)
2uv
Since H is cyclic in the variables ϕ and t, we seek a complete integral of the
form
F (ρ, ϕ, θ, Q, E) = F ′ (u, v, Q) + Qϕ ϕ − tE(Q),
2c Separation of Variables
295
where Q = (Qu , Qv , Qϕ ) and E(Q) are constants and F ′ is a complete integral
of the stationary Hamilton–Jacobi equation
1
(u + v) 2
Qϕ + m (g(u) + h(v)) = E(Q)
2uFu′2 + 2vFv′2 +
m(u + v)
2uv
in the sole variables u and v. Multiplying by (u+v) and collecting homologous
terms gives the completely separable equation
2uFu′2 + mg(u) +
Q2ϕ
Q2ϕ
− mE(Q)u + 2vFv′2 + mh(v) +
− mE(Q)v = 0.
2u
2v
The complete integral of this can be sought in the separated form
(u, v, Q) → F ′ (u, v, Q) = F1′ (u, Q) + F2′ (v, Q),
1
2mu [uE(Q) − g(u)] − Q2ϕ + 2uQv du,
2u
1
2mv [vE(Q) − h(v)] − Q2ϕ − 2vQu dv.
F2′ (v, Q) =
2v
F1′ (u, Q)
=
Write down the complete integral and resolve the motion.
2.3c Elliptic Coordinates
Lagrangian, kinetic momenta, and Hamiltonian are (§1.1c of the Complements
of Chapter 2)
L=
1
m ℓ2 (u̇2 + θ̇2 )(sinh u2 + sin2 θ) + 2ℓ2 ϕ̇2 sinh2 u sin2 θ + V (u, ϕ, θ),
2
pu = mℓ2 (sinh2 u + sin2 θ)u̇,
2
pϕ = 2mℓ2 (sinh u sin θ) ϕ̇,
pθ = mℓ2 (sinh2 u + sin2 θ)θ̇,
H=
p2ϕ
1
p2u + p2θ
− V (u, ϕ, θ).
+
2mℓ2 sinh2 u + sin2 θ 2(sinh u sin θ)2
Assume that V is the gravitational potential generated by two masses situated
at two points P1 and P2 . Take first a Cartesian coordinate system such that
the two points are on the x3 -axis and symmetric with respect to the origin,
say for example P1 = (0, 0, ℓ) and P2 = (0, 0, −ℓ) for some ℓ > 0. Such a
potential has the form
P → V (P ) = −
γ1
γ2
−
,
P − P1 P − P2
296
11 INTEGRATING HAMILTON–JACOBI EQUATIONS
where γi for i = 1, 2 are given positive constants. Expressing P first in Cartesian coordinates and then in terms of elliptic coordinates gives
γ1
γ2
−V (P ) = 2
+ 2
2
2
2
(x1 + x2 ) + (x3 − ℓ)
(x1 + x2 ) + (x3 + ℓ)2
ℓγ2
ℓγ1
+
.
=
cosh u − cos θ cosh u + cos θ
From this we obtain
H=
p2ϕ
1
p2u + p2θ
+
2mℓ2 sinh2 u + sin2 θ 2(sinh u sin θ)2
ℓγ2
ℓγ1
+
,
+
cosh u − cos θ
cosh u + cos θ
and the associated Hamilton–Jacobi equation is
Ft +
Fϕ2
Fu2 + Fθ2
1
+
2mℓ2 sinh2 u + sin2 θ 2(sinh u sin θ)2
ℓγ2
ℓγ1
+
= 0.
+
cosh u − cos θ cosh u + cos θ
Since ϕ and t are cyclic, we seek an integral of the form
F (u, ϕ, θ) = F ′ (u, θ, Q) + Qϕ ϕ − tE(Q),
where Q = (Qu , Qv , Qϕ ) and E(Q) are constants and F ′ is a complete integral
of the stationary Hamilton–Jacobi equation
Q2ϕ
1
Fu′2 + Fθ′2
+
2
2
2mℓ2 sinh u + sin θ 2(sinh u sin θ)2
ℓγ2
ℓγ1
+
= E(Q)
+
cosh u − cos θ cosh u + cos θ
in the sole variables u and θ. Multiply both sides by
sinh2 u + sin2 θ = cosh2 u − cos2 θ = (cosh u − cos θ)(cosh u + cos θ)
and collect homologous terms to get
Q2ϕ
− 2mℓ2 E(Q) cosh2 u + 2mℓ2 (γ1 + γ2 ) cosh u
2 sinh2 u
Q2ϕ
+ 2mℓ2 E(Q) cos2 θ + 2mℓ2 (γ1 − γ2 ) cos θ = 0.
+ Fθ′2 +
2 sin2 θ
Fu′2 +
This is a completely separable stationary Hamilton–Jacobi equation. Compute
a complete integral and resolve the motion.
2c Separation of Variables
297
2.4c Force-Free {P ; m} in R3
If V = 0, then (2.3c) has the complete integral
F (x, Q; t) = Q · x −
1
Q2 t,
2m
Q = (Q1 , Q2 , Q3 ).
From this and the integration method of §5.3 of Chapter 10, we have
mẋi = Qi ,
Pi + ωi t = −xi ,
ωi =
Qi
,
m
i = 1, 2, 3,
where Qi and Pi are determined by the initial data.
2.5c The Cycloidal Pendulum
The Lagrangian of a harmonic oscillator, its kinetic moment, and its Hamiltonian are (§7.2c of Chapter 3; here q is the arc length on the cycloid and
ω 2 = g/4R)
L=
m 2
(q̇ + ω 2 q 2 ),
2
p = mq̇,
H=
1 2
(p + m2 ω 2 q 2 ),
2m
The corresponding Hamilton–Jacobi equation is
Ft +
1
(F 2 + m2 ω 2 q 2 ) = 0,
2m q
whose complete integral is
F (q, Q; t) =
2mE(Q) − m2 ω 2 q 2 dq − tE(Q).
Resolve the motion starting from this integral.
12
INTRODUCTION TO FLUID DYNAMICS
1 Geometry of Deformations
A bounded open connected set in Eo ⊂ RN deforms in time to E in the
sense that points y ∈ Eo are in one-to-one correspondence with points x ∈ E
through smooth, nonintersecting trajectories t → x(t) such that x(0) = y and
x(t) = x. This defines a flow map and a velocity field
x = Φ(y, t),
v(x, t) = Φt (y, t).
(1.1)
The functions Φ(·, t) may be regarded as a family of transformations defined
in Eo and parameterized with t. These transformation will be assumed to be
smooth and invertible independent of t. In the Lagrangian formalism, kinematic information on x(t) ∈ E is provided by the trajectories t → x(t),
independently of their membership to E, as an open connected subset of RN ,
this bearing a role only in the determination of such paths [100]. In the Eulerian formalism, kinematic information on points x ∈ E is provided by the
flow map Φ(·, t), which bears the “globality” of Eo and E [47, 48]. In both
formalisms these quantities must coincide. Therefore ẋ = v(x, t) and
ẍ =
d
∂
ẋ = v(x, t) + ẋ · ∇x v = Dt v,
dt
∂t
(1.2)
where the operator Dt formally defined by
Dt =
∂
+ v∇x
∂t
(1.3)
is the total or material derivative along Lagrangian paths. For t fixed, the
Jacobian of the transformation Φ(·, t) is
∂Φ (y, t)
∂Φk (y, t)
i
J(x, t) = J[Φ(y, t), t)] = det
= Aij
= Jδik ,
∂yj
∂yj
where Aij is the determinant of the algebraic complement of the (ij)th entry
of the Jacobian matrix ∇Φ.
E. DiBenedetto, Classical Mechanics: Theory and Mathematical Modeling,
Cornerstones, DOI 10.1007/978-0-8176-4648-6 12,
c Springer Science+Business Media, LLC 2011
299
300
12 INTRODUCTION TO FLUID DYNAMICS
Proposition 1.1 (Euler [47]) Dt J = J div v.
Proof. From the previous expression of J, we have1
Dt J(x, t) =
∂Φ (y, t)
∂ ∂Φi (y, t)
∂
i
det
= Aij
∂t
∂yj
∂yj
∂t
= Aij
∂vi
∂vi ∂Φk (y, t)
∂vi
= Aij
=
J.
∂yj
∂xk ∂yj
∂xi
1.1 Incompressible Deformations
If an infinitesimal portion about any y ∈ Eo moves by possibly changing
its shape and/or configuration, but keeping fixed its infinitesimal volume,
then Dt J = 0 and consequently div v = 0, and the deformation is called
incompressible. Conversely, a deformation is incompressible if and only if
div v = 0.
1.2 The Equation of Continuity
Let G ⊂ RN be open and let E(t) ⊂ G be a deforming subdomain of G with
smooth boundary ∂E(t). For a smooth function (x, t) → ρ(x, t) defined in a
neighborhood of E(t), by the previous proposition,
d
d
ρ(Φ(y, t), t)Jdy
ρ(x, t)dx =
dt E(t)
dt Eo
d
ρ[Φ(y, t), t)J]dy
=
dt
Eo
=
[(Φt · ∇x ρ + ρt )J + ρJt ]dy
Eo
=
[ρt + div(ρv)]Jdy
Eo
=
[ρt + div(ρv)Jdx.
E(t)
If ρ(x, t) is the material density of a body occupying the domain G, then for
every deforming subset E(t) ⊂ G,
ρ(x, t)dx = mass of the body in E(t).
E(t)
1
The derivative of the determinant of n × n matrix is the sum of n determinants
obtained from the original matrix upon substitution of each row (column) by the
row (column) of the corresponding derivatives.
2 Cardinal Equations
301
If elements of G evolve conserving their mass, then
d
ρ(x, t)dx = 0 for all deforming subdomains E(t) ⊂ G.
dt E(t)
Since E(t) ⊂ G is arbitrary, local deformations of G preserve the mass if and
only if
ρt + div(ρv) = 0
pointwise in G.
(1.4)
This is the continuity equation, and it expresses conservation of mass.
2 Cardinal Equations
Along the motion, points x ∈ E ⊂ G are acted upon by a material distributions of forces f (x, ẋ; t)ρ(x, t)dx, and by reactions acting on ∂E due to the
remaining portion G − E that opposes the possible deformation of E. These
are a priori unknown, depend on the material structure of G, and should not
depend on the particular subdomain E ⊂ G. In the Cauchy formalism they
are represented by a smooth vector-valued function
G × S1 × R ∋ (x, n, t) → T(x, n, t) ∈ R3 ,
where S1 is the unit sphere in R3 . Then, assuming that ∂E is smooth, reaction
forces of G − E acting on ∂E are described by
{reactions opposing deformations of E} =
T(x, n, t)dσ,
∂E
where dσ is the surface measure on ∂E and n is the outward unit normal
to ∂E at x ∈ ∂E. The component (T · n)n of T along n is the traction or
compression force, whereas the component T − (T · n)n tangent to ∂E at x is
the shear force. By d’Alembert principle, the motion of any subdomain E ⊂ G
is a sequence of instantaneous equilibrium states, parameterized with time, of
all forces acting on that portion, including the reactions to deformation. Thus
T(x, n, t)dσ,
(2.1)
[ẍ − f (x, ẋ, t)]ρdx =
E
x ∧ [ẍ − f (x, ẋ, t)]ρdx =
E
∂E
x ∧ T(x, n, t)dσ,
(2.2)
∂E
for all subdomains E ⊂ G.
Lemma 2.1 T(·, n, t) = −T(·, −n, t).
Proof. Fix P ∈ G and n ∈ S1 . For 0 < ε, δ ≪ 1 consider the disk Dε (P )
centered at P with radius ε, normal to n, and the right cylinder Cδ (P ) of
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12 INTRODUCTION TO FLUID DYNAMICS
base Dε (P ) and height δ. Write (2.1) over Cδ (P ) and let δ → 0 by keeping
ε > 0 fixed, to obtain
T(x, −n, t)dσ.
T(x, n, t)dσ = −
Dε (P )
Dε (P )
Divide both sides by |Dε (P )| and let ε → 0.
3 The Stress Tensor and Cauchy’s Theorem
Having fixed a triad Σ = {O; e1 , e2 , e3 }, represent n ∈ S1 by its director
cosines n = (α1 , α2 , α3 ) with respect to the coordinate axes of Σ.
Theorem 3.1 (Cauchy). For all n = (α1 , α2 , α3 ) ∈ S1 ,
T(·, n, t) = αi T(·, ei , t).
Proof. Fix P ∈ G and n ∈ S1 and write down (2.1), where E is the tetrahedron
with vertex in P , height 0 < ε ≪ 1, base ∆ABC normal to n, and faces
∆AP B, ∆BP C, ∆CP A, parallel to the coordinate planes. By setting ∆σ =
|∆ABC|, one has |E| = 13 ε∆σ, and
|∆AP B| = α3 ∆σ,
|∆BP C| = α1 ∆σ,
|∆AP C| = α2 ∆σ.
For these choices, (2.1) takes the form
T(x, −e1 , t)dσ
T(x, n, t)dσ +
[ẍ − f (x, ẋ; t)]ρdx =
∆BP C
∆ABC
E
T(x, −e3 , t)dσ.
T(x, −e2 , t)dσ +
+
∆AP C
∆AP B
Dividing both sides by ∆σ gives
ε
1
[ẍ − f (x, ẋ; t)]ρdx =
T(x, n, t)dσ
3|E(t)| E
|∆ABC| ∆ABC
α1
+
T(x, −e1 , t)dσ
|∆BP C| ∆BP C
α2
+
T(Q, −e2 , t)dσ
|∆AP C| ∆AP C
α3
+
T(x, −e3 , t)dσ.
|∆AP B| ∆AP B
Let ε → 0 by keeping the vertex P of the tetrahedron fixed and the base
∆ABC normal to n.
3 The Stress Tensor and Cauchy’s Theorem
303
While computed at ei , the vectors T(·, ei , t), need not be directed along
the homologous coordinate axes. The components τij (·, t) = T(·, ej , t) · ei
of T(·, ej , t) along ei define a matrix
⎛
⎞
τ11 τ12 τ13
T = (τij ) = ⎝τ21 τ22 τ23 ⎠ ,
τ31 τ32 τ33
called a stress tensor. The entries τii are traction or compression stresses, and
τij , for i = j, are shear stresses. The shear force acting on an infinitesimal
plane surface normal to e1 is τ21 e2 + τ31 e3 . In general,
{shear force relative to ei } =
τij ej .
j =i
Corollary 3.1 T(·, n, t) = T · n = (τij )n.
Proof. From the definitions and Theorem 3.1,
T(·, n, t) = αj T(·, ej , t)=αj [T(·, ej ; t) · ei ]ei
⎞⎛ ⎞
⎛ ⎞ ⎛
⎛ ⎞
⎛ ⎞
α1
τ11 τ12 τ13
τ13
τ12
τ11
= α1 ⎝τ21 ⎠ +α2 ⎝τ22 ⎠ +α3 ⎝τ23 ⎠ = ⎝τ21 τ22 τ23 ⎠ ⎝α2 ⎠ = T · n.
α3
τ31 τ32 τ33
τ33
τ32
τ31
Corollary 3.2 Let G ⊂ R3 be an open set identified with a material system
of density ρ(·, t) whose points x ∈ G are in motion under the external force
density f (x, ẋ, t) and the internal stress tensor T. Then
[ẍ − f (x, ẋ, t)]ρ = div T
in G.
(3.1)
Proof. Let E be any portion of G with smooth boundary ∂E. By the Gauss–
Green theorem and Corollary 3.1,
div Tdx.
T · ndσ =
T(x, n, t)dσ =
E
∂E
∂E
Therefore (2.1) takes the form
[ẍ − f (x, ẋ, t)]ρdx =
div Tdx
E
for all subdomains E ⊂ G.
E
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12 INTRODUCTION TO FLUID DYNAMICS
3.1 Symmetry of the Stress Tensor
Proposition 3.1 (τij ) = (τji ).
Proof. By the Gauss–Green theorem,
xh τij eh ∧ ei αj dσ
x ∧ T(x, n, t)dσ =
∂E
∂E
∂
=
(τij xh )eh ∧ ei dx
∂x
j
E
∂τij
=
xh eh ∧ ei dx +
τij δhj eh ∧ ei dx
∂xj
E
E
=
x ∧ div Tdx −
τij ei ∧ ej dx.
E
E
Put this in the second cardinal equation (2.2) and take into account (3.1) to
obtain
τij ei ∧ ej dx =
[(τ23 − τ32 )e1 + (τ31 − τ13 )e2 + (τ12 − τ21 )e3 ]dx = 0
E
E
for all subdomains E ⊂ G.
3.2 Miscellaneous Remarks
The matrix T is intrinsic to the system and independent of its representations
in the following sense. Let Σ ′ = {O; e′1 , e′2 , e′3 } be a new triad obtained by Σ
by a rotation of the coordinate axes realized by a unitary matrix U , so that
in particular e′j = U ej . By Corollary 3.1,
τij′ = T(·, e′j , t) · e′i = (τhk )U ej · U ei = eti [U t (τhk )U ]ej = [U t (τhk )U ]ij .
The tensor T is a linear map in R3 whose matrix (τij ) is a representative. We
will call both T and its matrix representations stress tensors.
The unknowns of the motion are the trajectories t → x(t) of the points of
G, the density function ρ(·, t), and the nine components τij of T. The second
cardinal equation (2.2), which amounts to three scalar equations, has been
used to establish the symmetry of T and thus reduce by three the unknowns of
the motions. The remaining first cardinal equation, in the pointwise form (3.1),
amounts to three scalar equations, which alone are insufficient to resolve the
motion. One needs to provide additional information on the material structure
and on the tensorial state of the system both in the interior of G and on its
boundary ∂G. For example, for rigid systems ρ = const, and the T is the
rigidity constraint.
4 Perfect Fluids and Cardinal Equations
305
4 Perfect Fluids and Cardinal Equations
A fluid is a continuum material system whose equilibrium configurations are
possible if and only if the stress tensor T is proportional to the identity I, that
is, if
T(·, n) = T · n = −p(·)n
in steady state,
where p(·) is a smooth function defined in G, called pressure. This formula
is a mathematical rendering of Pascal’s principle, by which the pressure in
any point of the fluids exerts equal force by unit surface in all directions
[125]. This mathematical definition of fluid reflects the intuitive idea of a
material continuum system that does not oppose the mutual sliding of its
ideal internal layers. If in the fluid at rest the shear components of its stress
tensor were not zero, these would generate an incipient shearing of internal
layers, since the system does not have a mechanism to oppose it. Likewise, an
ideal material surface traced in the fluid at rest remains in equilibrium only
if acted upon by forces normal to it. In a real fluid in motion, the kinematic
viscosity generates shear stresses that oppose layer sliding (see §3.2c of the
Complements of Chapter 3). Then real fluids are classified as more viscous
(oil, paraffin, etc.) or less viscous (alcohol, ether, gas, etc.) according to the
size of these shear stresses. A real fluid is ideal of perfect if the shear stresses
are negligible even in dynamic regime, that is, if
T(·, n, t) = T · n = −p(·, t)n
in G and for all times.
(4.1)
In such a case div T = −∇p(·, t), and (3.1) takes the form
ρ[ẍ − f (x, ẋ, t)] + ∇p = 0
in G for all t.
(4.2)
Equation (4.1) is the constitutive law of ideal fluids, and (4.2) is the cardinal
or momentum equation of an ideal fluid.
4.1 Barotropic Fluids
These are fluids for which a link is known between pressure and density, say,
for example,
ρ = ρ(p) ∈ C 1 (R+ ),
ρ(·) ≥ ρo for some ρo > 0.
(4.3)
Relations of this kind, called equations of state, are experimental and include,
for example, homogeneous fluids for which ρ(·, t) = ρo . In view of the continuity equation (1.4) these are also incompressible. Adiabatic fluids are those for
which heat transfer is far slower than variations of pressure. Thermodynamic
considerations lead to the equation of state [138, §16, page 150, and §30, pages
172–176]
p = cρ1+α for given positive constants c and α.
(4.4)
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12 INTRODUCTION TO FLUID DYNAMICS
The constant 1 + α is the ratio between the heat capacity at constant volume
and heat capacity at constant pressure. If α = 0, these heat capacities are the
same, and both pressure and density are essentially constant. For barotropic
fluids, (4.2) takes the form
ẍ = f (x, ẋ, t) − ∇
p(x,t)
ds
.
ρ(s)
(4.5)
If the fluid is in equilibrium, ẋ = ẍ = 0 identically and
f (x) = ∇
p(x)
ds
.
ρ(s)
If f is conservative with potential V , then
p
ds
∇ V −
= 0,
which implies
ρ(s)
V −
p
ds
= const.
ρ(s)
Summarizing, we may state the following.
Proposition 4.1 A barotropic fluid admits an equilibrium configuration if
and only if f is conservative. Moreover, the equipotential surfaces are also
curves of constant pressure (isobaric) and surfaces of constant pressure.
As an example consider a barotropic fluid subject only to gravity. Then f (x) =
−ge3 , where e3 is the unit ascending vertical, and V (x) = −gx3 + const. If
the fluid is adiabatic and satisfies (4.4) with α = 0 (that is, both pressure and
density are essentially constant), then
−gx3 = c ln
p
,
po
where po is the pressure at level x3 = 0.
From this follows the barometric formula
p(x) = po e−(g/c)x3 .
This implies that the equipotential and isobaric surfaces are horizontal planes.
If the fluid is separated from the air by a surface, its pressure along such a
surface must equal that of the air, which is constant. Therefore the separation surface is isobaric and therefore is a horizontal plane. This is known as
Stevin’s law.
4.2 Fluid in Uniform Rotation
A homogeneous fluid confined in a right circular cylinder is in uniform rotation
about the axis of the cylinder with angular speed ω and is subject to its weight.
Introduce a triad Σ = {O; u1 , u2 , u3 }, fixed with the fluid, with origin at the
5 Rotations and Deformations
307
axis of the cylinder and u3 ascending vertically. The generic point x of the
fluid is acted upon by gravity and the Coriolis forces, so that
f (x) = ρ(ω 2 x1 , ω 2 x2 , −g).
Therefore equilibrium is possible only if ∇p = ρω 2 (x1 , x2 ), −ρg . From this
by integration,
p(x) = po + ρ[ 21 ω 2 (x21 + x22 ) − g x3 ].
Therefore the isobaric surfaces are the paraboloids
x3 =
1 2 2
ω x1 + x22 + const.
2g
In particular, the free surface in contact with air is a paraboloid.
5 Rotations and Deformations
Let v(·, t) be the velocity field generated by the flow map in (1.1) and assume
that the fluid at time t undergoes an elemental rigid motion of characteristics
blv(xo , t) and ω, where xo is an arbitrary but fixed point in the instantaneously
rigid fluid. By the Poisson formula (6.1) of Chapter 1,
v(x, t) = v(xo .t) + ω ∧ (x − xo ).
Since the motion is instantaneously rigid, ω does not depend on the variables
x of the generic point in the fluid. Taking the curl of both sides gives
curl v = (v3,x2 − v2,x3 )e1 + (v1,x3 − v3,x1 )e2 + (v1,x2 − v1,x2 )e3 = 2ω. (5.1)
Therefore curl v(x, t) gives, apart the factor 2, the angular velocity of the
infinitesimal element of fluid about x, regarded as instantaneously rigid. For
this reason curl v(·, t) is called a vorticity field. If curl v(·, t) = 0, the field is
irrotational. If G is simply connected, an irrotational field is also potential,
that is, there exists a function ϕ(·, t) ∈ C 1 (G), called a kinetic potential, such
that v(·, t) = ∇ϕ(·, t). The flow is called potential and the velocity field v(·, t)
is normal to the instantaneous equipotential surfaces [ϕ(·, t) = const(t)]. If
the velocity field is stationary, the kinetic potential is independent of t and
the trajectories of the fluid particles are normal to the equipotential surfaces.
Next expand v(·, t) in a Taylor series about a point xo in the fluid, to
obtain
v(x, t) = v(xo , t) + [∇v(xo , t)] · (x − xo ) + O(|x − xo |2 ).
Therefore up to terms of higher order,
vi (x, t) = vi (xo , t) + vi,xj (xj − xo,j ),
i = 1, . . . , N.
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12 INTRODUCTION TO FLUID DYNAMICS
For fixed indices i, j,
vi,xj =
1 ∂vi
∂vj 1 ∂vi
∂vj
+
= Dij + Rij .
+
−
2 ∂xj
∂xi
2 ∂xj
∂xi
(5.2)
The entries Dij and Rij define two tensors D and R. The first is symmetric
and is called a deformation tensor. The second is skew-symmetric and is called
a rotation tensor. With this notation, the previous Taylor expansion takes the
approximate form
v(x, t) = v(xo , t) + D · (x − xo ) + R · (x − xo ).
(5.3)
Consider an infinitesimal
arc d(x − xo ) within the fluid and along its motion,
of length dℓ = d(x − xo )2 . Then
d 2
dℓ = 2d(x − xo ) · d(ẋ − x˙o )
dt
= d(xi − xo,i )vi,xj d(xj − xo,j )
= 2d(x − xo )P t · D · d(x − xo ).
Therefore D tracks the deformations of infinitesimal lengths along the motion.
In a rigid motion lengths are preserved and D = 0. If D = λI, then the
deformation occurs uniformly along the coordinate axes and the fluid expands
if λ > 0 and contracts if λ < 0. From the definition of R,
R · (x − xo ) =
1
2
curl v ∧ (x − xo ) = ω ∧ (x − xo ).
Therefore R gives the angular velocity of the system as if it were in instantaneous rigid motion. These remarks and (5.3) suggest we regard the
infinitesimal motion of a fluid as the sum of (i) an infinitesimal translation
along v(xo , t), (ii) an infinitesimal deformation along the coordinate axes, and
(iii) an infinitesimal rigid rotation about the axis through xo and directed as
curl v(xo , t). This is known as Cauchy’s theorem.
6 Vortex Filaments and Vortex Sheets
A vortex filament is a curve in G tangent, for fixed t, to the vortex field
curl v(·, t). A point y where curl v(y, t) = 0 is a vortex point. Through a
vortex point y passes a unique vortex filament τ → x(τ ), a solution of
curl v(·, t) ∧
x(0) = y,
dx(τ )
= 0,
dτ
τ ∈ R a parameter,
(t fixed).
6 Vortex Filaments and Vortex Sheets
309
If curl v(y, t) = 0, the vortex filament through y is not defined. Let γ = {s →
x(s)} be a smooth curve in G whose points are vortex points. From each of
the points of γ trace the corresponding vortex filament
curl v[x(s), t] ∧
dx(τ ; s)
= 0,
dτ
(t fixed)
(6.1)
x(0; s) = x(s),
to obtain, locally in τ , a surface called a vortex sheet. If γ is closed, then
(6.1) generates tubelike surfaces, called vortex tubes. Let T be the portion of
a vortex tube cut by two smooth nonintersecting sections Σ1 and Σ2 . The
oriented boundary ∂T consists of Σ1 and Σ1 , with unit normal ni oriented
outward T and the portion Σo , of ∂T lying on the vortex tube, with outward
unit normal n. By the Gauss–Green theorem,
curl v · ndσ
div curl vdx =
Σo
T
+
curl v · n1 dσ
Σ1
+
curl v · n2 dσ.
Σ2
Since Σo is made out of portions of vortex filaments, curl v · n = 0 on Σo .
Moreover, div curl v = 0. Therefore
−
curl v · n2 dσ.
(6.2)
curl v · n1 dσ =
Σ1
Σ2
Since the sections Σ1 and Σ2 are arbitrary, the flow of vortex entering a vortex
tube across a section Σ equals the flow of vortex exiting the tube through any
other section.
6.1 Circulation and Helmholtz Theorem
Let γ ⊂ G be a smooth closed curve, and let Σ be any smooth oriented surface
with unit normal n whose boundary is γ. By Stokes’s theorem,
curl v(x, t) · n(x)dσ =
v(x, t) · dx = Cγ (t).
(6.3)
Σ
γ
The quantity Cγ (t) is the circulation of the vector field v(·, t) along γ. If G
is simply connected, (6.3) implies that the motion is potential if and only
if Cγ (t) = 0 for any smooth closed curve γ ⊂ G. Likewise, if G is simply
connected, the motion is irrotational if and only if is potential. A smooth
closed curve traced on a vortex tube surrounds it if it cannot be deformed
with continuity while remaining on the tube to be shrunk to a point. On a
310
12 INTRODUCTION TO FLUID DYNAMICS
vortex sheet one can draw curves homotopic to a point. If Σ is a portion of
the vortex sheet included by γ, then by (6.3), Cγ (t) = 0, since on a vortex
sheet curl v · n = 0.
Theorem 6.1 (Helmholtz). Let γ1 and γ2 any two smooth closed curves
traced on the same vortex tube and surrounding it. Then
Cγ1 (t) =
v(x, t) · dx = Cγ2 (t).
v(x, t) · dx =
γ1
γ2
Proof. Combine (6.2) and (6.3).
The circulation along a smooth closed curve γ surrounding a vortex tube
is called the intensity of the vortex tube. The intensity of a vortex tube is
independent of the curve γ lying on the tube surrounding it.
7 Equations of Motion of Ideal Fluids
The motion is driven by the momentum equation (4.2) and by the assumption
that mass is conserved. Then recalling the form (1.1) of ẍ in terms of v, the
equations of motion of an ideal fluid are
ρ[vt + (v · ∇)v − f ] = −∇p,
momentum equation,
ρt + div(ρv) = 0,
some equation of state
conservation of mass,
or a derivative of it.
(7.1)
The equation of state could be, for example, (4.4). Another example is that
of a link of the type p = F (E)ρ between pressure p and density ρ. here
E = E(x, t) is the entropy of the elemental mass of fluid about x at time
t, and F (·) is a smooth experimentally given function of the entropy. For
adiabatic transformations the entropy is constant along the Lagrangian path
of t → x(t), or equivalently along the flow map [146]. Therefore
p
Dt
=0
in G.
(7.2)
ρ
This might take the place of the last of equation (7.1). The system (7.1)–(7.2)
is called the system of Euler equations for ideal isentropic flows.
These systems, irrespective of the form of the last equation of (7.1), consist of five scalar equations from which one seeks to resolve the motion by
determining the five scalar unknowns v, ρ, p. The domain G might or might
not be bounded, and solvability requires some information on the behavior of
the fluid at the boundary of G. Even so, these equations, while not amenable
to resolving the motion, may be used to derive some specific dynamic issues,
as indicated by the next example.
8 Barotropic Flows with Conservative Forces
311
7.1 Flow past an Obstacle
A rigid impermeable body occupies a bounded, simply connected domain
C ⊂ R3 and is immersed in a fluid occupying R3 − C. One seeks to compute
the force exerted by the fluid on C. Combining the momentum equation and
the continuity equation in (7.1) gives
ρf − ∇p = ρvt + ρ(v · ∇)v = (ρv)t + (ρvj v)xj .
Integrate this over BR − C, where BR is the ball of radius R centered at the
origin and R is so large that C̄ ⊂ BR . Using the Gauss–Green theorem and
taking into account that C is impermeable and therefore v · n = 0 on ∂C, this
gives
x
x
−
dσ +
p + ρv v ·
f dx,
(ρv)t dx −
pndσ =
R
R
BR −C
BR −C
|x|=R
∂C
where n is the unit normal internal to C. On the other hand, by the constitutive law (4.1) of ideal fluids, the force acting on ∂C is
pndσ.
T(x, n, t)dσ = −
F=
∂C
∂C
Putting this in the previous relation and letting R → ∞ gives
x
x
dσ
p + ρv v ·
F = lim
R→∞ |x|=R
R
R
+ lim
(ρv)t dx − lim
f dx,
R→∞
BR −C
R→∞
BR −C
provided the limits exist. If the motion of the fluid is stationary and there are
not external forces, then
x
x
dσ.
(7.3)
F = lim
p + ρv v ·
R→∞ |x|=R
R
R
8 Barotropic Flows with Conservative Forces
The acceleration of a point x in motion inside the fluid is computed from
(1.1). Also by standard vector calculus,
(v · ∇)v = 21 ∇v2 − v ∧ curl v.
Therefore the acceleration of a generic x ∈ G can be written in any of the
equivalent forms
vt + (v · ∇)v,
ẍ =
(8.1)
vt − v ∧ curl v + 12 ∇v2 .
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12 INTRODUCTION TO FLUID DYNAMICS
Using the second of these in the momentum equation (7.1), the latter takes
the form
p
ds
vt − v ∧ curl v = −∇B,
− V.
(8.2)
B = 12 v2 +
ρ(s)
The first term in the expression of B is the specific (i.e., per unit mass) kinetic
energy of the particle about x. The last term is the specific potential energy of
such a particle. The middle term is the specific internal energy due to pressure.
Therefore B is the total specific energy of the particle about x. The quantity
B is the Bernoulli trinomial [7].
Computing the material derivative of B along Lagrangian paths yields
Dt B(P, t) = Bt + v · ∇B
1 ∂
1
p
= v · vt + (v · ∇)v + ∇p − ∇V +
ρ
ρ ∂t
1 1 ∂
= v · ẍ − f + ∇p +
p.
ρ
ρ ∂t
Enforcing now the continuity equation (7.1) gives
Dt B =
1 ∂
p.
ρ ∂t
(8.3)
Therefore in general, the specific energy is not conserved along material
paths, due to the variation of its internal energy. If the motion is stationary,
then pt = 0 and every particle conserves its specific energy along its material
trajectory. Moreover,
∇B = v ∧ curl v.
(8.4)
Therefore, for stationary motions, the energy is conserved along streamlines
of the velocity field v. Since the motion is stationary, the streamlines of v
are actually material paths, since they are trajectories of elemental particles
along their stationary motion. By (8.4), the energy is also conserved along
streamlines of the vorticity field curl v. Such an energy, however, might change
from point to point. If the motion is stationary and irrotational, then the
energy is constant for all fluid particles.2
2
More generally, the energy is constant for motions that while not irrotational,
satisfy v∧curl v = 0. In such a case the particles move along v and effect infinitesimal
rotations about their material path. They are called helicoidal or Beltrami flows.
9 Material Lines and Surfaces
313
8.1 Incompressible Barotropic Fluids Subject to Gravity
If the external forces are due only to the weight, then V (x) = −gx3 + const.
Therefore if the fluid is incompressible (ρ = const), the Bernoulli trinomial
becomes
1
v2
p
B=
+ x3 +
g
2g
ρg
in G.
The first term is called the kinetic level, since it gives the height from which a
point mass has to fall for an impact speed of |v| (see §6c of the Complements
of Chapter 3). The second term is the actual level of a point x. The third is
the piezometric level or head, since it gives the height of a column of fluid that
exerts pressure p on x. By (8.3), variations of the total energy of a fluid particle
along its Lagrangian path are due only to variations of the piezometric head.
If the motion is stationary, the piezometric head is constant and the energy
is preserved along material paths of the fluid and streamlines of the vorticity
field curl v. This occurrence is referred to as the theorem of the three levels.
9 Material Lines and Surfaces
Continue to assume that the fluid in G is barotropic and acted upon only by
conservative forces, so that the momentum equation (4.5) has the form
p ds
+V .
(9.1)
ẍ = −∇
ρ(s)
Let γ(·, to ) be a smooth curve in G at some fixed instant to . Identify the
points of γ(·, to ) with the actual material particles lying on it, and follow
their motion following the parameter t. The flow map transforms γ(·, to ) into
a smooth curve γ(·, t) ⊂ G.3
The points of γ(·, t) are the same material points of γ(·, to ), and for this
reason γ(·, t) are called material lines. Similarly, one defines smooth material
surfaces Σ(t) ⊂ G as images by the flow map of smooth surfaces Σ(to ) ⊂ G,
so that Σ(t) contains the same material particles as Σ(to ).
Theorem 9.1 (Kelvin). For a barotropic fluid subject only to conservative
forces, the circulation of the velocity field v(·, t) along closed material lines
γ(·, t) is constant in t. Equivalently,
v(x, t1 ) · dx
v(x, to ) · dx =
γ(·,to )
γ(·,t1 )
for any two smooth closed material lines γ(·, to ) mutually connected by the
flow map.
3
While this occurs for |t − to | < ε for sufficiently small ε, we will avoid specifying
such a locality.
314
12 INTRODUCTION TO FLUID DYNAMICS
Proof. Let γ(·, t) be smooth closed material lines, parameterized by τ ∈ [−δ, δ]
for some δ > 0, for all |t − to | < ε for sufficiently small ε. Then by the
momentum equation (9.1),
d
dx(τ, t)
d δ
dτ
v x(τ, t), t ·
v(x, t) · dx =
dt γ(·,t)
dt −δ
dτ
δ
1 δ d 2
dx(τ, t)
dτ +
ẋ (τ, t)dτ
ẍ(τ, t) ·
=
dτ
2 −δ dτ
−δ
δ p
dx(τ, t)
ds
+V ·
dτ
=−
∇
ρ(s)
dτ
−δ
p ds
+ V · dx = 0.
∇
=−
ρ(s)
γ(·,t)
Proposition 9.1 For a barotropic fluid subject only to conservative forces:
i. Vortex sheets are material surfaces. Equivalently, if Σ(to ) is a vortex sheet,
then all material surfaces Σ(t) generated from Σ(to ) by the flow map are
vortex sheets.
ii. Vortex filaments are material lines. Equivalently, if γ(·, to ) is a vortex
filament, then all material lines γ(·, t) generated from γ(·, to ) by the flow
map are vortex filaments.
iii. The intensity of a vortex tube is constant in time.
iv. If the motion is irrotational at some instant to , it remains irrotational for
all times.
Proof. Let Σ(t) be a material surface generated by a vortex sheet Σ(to ). If
Σ(t) were not a vortex sheet, it would contain a smooth closed curve γ(·, t) ⊂
Σ(t), where the circulation of v(·, t) is not zero. By definition of material
surface, γ(·, t) is generated by a smooth closed curve γ(·, to ) ⊂ Σ(to ). Since
Σ(to ) is a vortex sheet, the circulation of v(·, to ) about γ(·, to ) is zero, and
it remains zero in time by Kelvins’s theorem. The contradiction proves (i).
The statement (ii) for vortex filaments follows from (i), since vortex filaments
lie on vortex sheets. Statement (iii) follows from (ii) when applied to vortex
filaments on vortex tubes. To establish (iv) it suffices to observe that if the
motion were not irrotational at some instant t, there would exist a curve γ(·, t)
traced on a material surface Σ(t) about which the circulation of v(·, t) would
not be zero.
10 Transport of the Vorticity in Barotropic Fluids
Subject to Conservative Forces
Take the curl of both sides of (9.1) to get
wt − curl(v ∧ w) = 0,
where
w = curl v.
10 Transport of the Vorticity in Barotropic Fluids
315
The first component of curl(v ∧ w) is
[curl(v ∧ w)]1 = (v1 w2 − v2 w1 )x2 − (v3 w1 − v1 w3 )x3
= (w · ∇)v1 − (div v)w1 − (v · ∇)w1 .
Computing the remaining equations similarly, we arrive at the vector equation
wt − (w · ∇)v + (div v)w + (v · ∇)w = 0.
Transform the various terms by the continuity equation and by introducing
the vorticity
ω=
curl v
w
=
ρ
ρ
(angular velocity per unit mass).
With this notation compute
w = ρωt + ωρt ,
(w · ∇) · v = ρ(ω · ∇)v,
(div v)w = ρ(div v)ω,
(v · ∇)w = ρ(v · ∇)ω + ω(v · ∇)ρ.
Putting these in the previous vector equation gives
ρ[ω t + (v · ∇)ω − (ω · ∇)v] = −ω[ρt + (v · ∇)ρ + ρ(div v)].
The term on the right-hand side is zero by the continuity equation, whereas the
first two terms in brackets on the left-hand side represent the total derivative
of the vorticity. Therefore
Dt ω = (ω · ∇)v.
(10.1)
This is the law by which the vorticity is transported along the material trajectories, or equivalently by the flow map.
Proposition 10.1 Let x = Φ(y, t) be the flow map defined, say, for |t| < δ
for some δ > 0. Then
ω Φ(y, t), t = ∇Φ(y, t) · ω(y, 0).
Proof. Consider the two vector-valued functions
A(y, t) = ω Φ(y, t), t ,
+
G × R ∋ (y, t) −→
B(y, t) = ∇Φ(y, t) · ω(y, 0).
They coincide for t = 0, since
B(y, 0)∇Φ(y, 0) · ω(y, 0) = Iω(Φ(y, 0), 0) = A(y, 0).
(10.2)
316
12 INTRODUCTION TO FLUID DYNAMICS
The transport law (10.1), applied to A and written in components is
∂
∂
Ai = Aj
vi (Φ(y, t), t).
∂t
∂Φj
Also one verifies that
∂
∂
Φi,t ωh (y, 0)
Bi =
∂t
∂yh
∂
vi Φ(y, t), t ωh (y, 0)
=
∂yh
∂
∂
vi Φ(y, t), t
Φj (y, t)ωh (y, 0)
=
∂Φj
∂yh
∂
= Bj
vi Φ(y, t), t .
∂Φj
Therefore A and B satisfy the same differential first-order equation. Since
they are the same at t = 0, they coincide for all times.
Corollary 10.1 (Lagrange–Cauchy)4 If the motion of a barotropic fluid
subject only to conservative forces is irrotational at some time, it remains
irrotational at all times.
11 Barotropic Potential Flows
A potential barotropic fluid with equation of state (4.4) moves in a domain
G ⊂ R3 , with the velocity field v(·, t) generated by a kinetic potential u(·, t),
so that v = ∇u. If there are no exterior forces acting on the fluid (f = 0), the
momentum equation (7.1) takes the form
∂
px
u xi + u xj u xi xj = − i ,
∂t
ρ
More concisely,
i = 1, 2, 3.
p
ds
1
2
∇ ut + |∇u| +
= 0.
2
0 ρ(s)
Using the equation of state, compute
p
1+α p
ds
=
.
α
ρ
0 ρ(s)
4
[21, Vol. 1, 5–318]; [97, Vol. 4, 695–748].
(11.1)
(11.1)′
11 Barotropic Potential Flows
317
Combining these remarks gives the Bernoulli law for barotropic potential
fluids
1
1+α p
2
ut + |∇u| +
= h,
(11.2)
2
α
ρ
where h(·) is a function only of t. The positive quantity
dp
= (1 + α)
G × R ∋ (x, t) → c(x, t) =
dρ
+
2
p
ρ
has the dimensions of a square velocity, and c(x, t) is the speed of sound in
the fluid at x at time t. Multiply the ith equation in (11.1) by uxi and sum
over i = 1, 2, 3, to get
1 ∂
1
1
|∇u|2 + ∇u · ∇|∇u|2 = − ∇p · ∇u.
2 ∂t
2
ρ
(11.3)
Using the continuity equation yields
1
p 1
p
− ∇p · ∇u = −∇
∇ρ · ∇u
∇u −
ρ
ρ
ρ ρ
p 1
p
p
ρt +
· ∇u +
∆u.
= −∇
ρ
ρ ρ
ρ
From the equation of state (4.4) compute
p
1
p
1
p
+ Dt α = 0.
Dt 1+α = α Dt
ρ
ρ
ρ
ρ ρ
Expand the total derivative and recall that v = ∇u to obtain
∂ p
p
p
+ ∇u · ∇
− α 2 [ρt + ∇u · ∇ρ] = 0.
∂t ρ
ρ
ρ
By the continuity equation,
ρt + ∇u · ∇ρ = ρ∆u.
Therefore
∂ p
p
p
· ∇u =
+α
∆u.
−∇
ρ
∂t ρ
ρ
Putting this in (11.3) gives
1
1 ∂
1 ∂
c2 ∆u − ∇u · ∇|∇u|2 =
|∇u|2 −
p.
2
2 ∂t
ρ ∂t
(11.4)
318
12 INTRODUCTION TO FLUID DYNAMICS
11.1 Barotropic, Potential, Stationary Flows
If in addition the flow is stationary, rewrite (11.4) in the form
Aij uxi xj = 0,
1
(Aij ) = δij − 2 uxi uxj .
c
where
(11.5)
the matrix (Aij ) admits the two eigenvalues
λ1 = 1 −
|∇u| 2
c
,
λ2 = 1.
The ratio M = |∇u|/c is called the Mach number. This equation can be
regarded also in terms of relative motion. For example, one might think of
a rigid body in stationary motion in a potential barotropic fluid with no
further solicitations, say for example an airplane or a bullet. With respect to
an observer fixed with that rigid body, the body is at rest and the fluid moves
around it. If M < 1, the flow is subsonic, the matrix (Aij ) has two positive
eigenvalues, and the partial differential equation in (11.5) is called elliptic. If
M > 1, then the speed of the fluid exceeds the speed of sound and the flow
is supersonic; the matrix (Aij ) has one positive and one negative eigenvalue
and the corresponding equation is hyperbolic. If M = 1, the flow is sonic,
the matrix (Aij ) has one zero eigenvalue, and the equation is parabolic [41,
Preliminaries]. Mach 1 is the speed of sound, Mach 2 is twice the speed of
sound, and so on.
11.2 Stationary, Potential, Incompressible Flows
The velocity field v(·, t) in G is given by a kinetic potential u(·, t) defined in
G. Since the flow is stationary, both are independent of t and v = ∇u. Since
the fluid is incompressible, div v = 0 (§1.1), and therefore div ∇u = 0 in G.
Some information is given on the behavior of the fluid near the boundary
∂G, which itself is assumed to be smooth. For example, denoting by n(x) the
outward unit normal to ∂G and x ∈ ∂G, one may assume that the quantity
of fluid crossing ∂G at x per unit surface is known, that is,
v · n = ∇u · n = h
for some given h ∈ C(∂G).
Resolving the motion hinges on finding the kinetic potential u. From the
previous remarks, such a potential is a solution of the Neumann problem
u ∈ C 2 (G) ∩ C 1 (Ḡ),
∆u = 0 in G,
∇u · n = h on ∂G.
(11.6)
Necessary and sufficient conditions of solvability, solution techniques, and a
discussion on uniqueness are in [40, Chapters III and IV]. Here we report a
physical characterization of such solutions observed by Kelvin.
12 Stationary, Incompressible Potential Flow past an Obstacle
319
Theorem 11.1 (Kelvin [143]). The motion corresponding to (11.1) is the
one that minimizes the kinetic energy among all irrotational vector fields v
with the same flux h on ∂G. Equivalently, if u is a solution of (11.1), then
2
2
T∇u =
|v|2 dx = Tv
|∇u|2 dx ≤
ρ
ρ
G
G
for all divergence-free vector fields defined in Ḡ and such that v ·n = h on ∂G.
Proof. The difference between the kinetic energy due to ∇u and that due
to v is
2
|∇u|2 − |v|2 dx
(T∇u − Tv ) =
ρ
G
2
|∇u − v| dx + (∇u − v) · ∇udx
=−
G
G
(∇u − v) · ∇udx = 0.
≤
G
12 Stationary, Incompressible Potential Flow
past an Obstacle
Let C be a bounded simply connected domain in R3 with no cavities, which
will be identified with a material rigid impermeable body immersed in a fluid
in motion and occupying R3 − C̄. By a change of reference system we may
assume that C is still and the fluid moves around it. Assuming that the fluid
is stationary, potential, and incompressible, the potential u satisfies
∆u = 0 in R3 − C̄,
∇u · n = 0 on ∂C,
lim ∇u = v∞ ,
|x|→∞
(12.1)
where v∞ is a given vector and n is the unit normal to ∂C, interior to C.
The next proposition gives precise information on the behavior of u and ∇u
as |x| → ∞.
Proposition 12.1 For any arbitrary constant γ, for |x| ≫ 1,
u(x) = γ + x · v∞ + O(|x|−2 ),
∇u(x) = v∞ + O(|x|−3 ).
(12.2)
The proof of this proposition is given in §12c of the Complements.
12.1 The Paradox of D’Alembert
The formulation (12.1) requires only that the fluid be potential and incompressible. Assume in addition that the fluid if barotropic and there are no
320
12 INTRODUCTION TO FLUID DYNAMICS
forces, external or internal. Since the flow is stationary ut = 0 and incompressible (ρ = const), and there are no external forces, the Bernoulli law
(11.2) gives
p = γo + γ1 v2
for given positive constants γo and γ1 .
The asymptotic behavior of Proposition 12.1 then gives for |x| ≫ 1,
p(x) = γo + γ1 v∞ + O(|x|−3 )
2
= γ∗ + O(|x|−3 ),
where γ∗ = γo + γ1 v2 . Compute now the force exerted by the fluid on C. By
(7.3) and the indicated asymptotic behavior of p one computes
x
x
dσ + ρv∞
F = lim γ∗
v∞ · dσ + O(R−3 )4πR2 = 0.
R→∞
R
|x|=R
|x|=R R
Therefore, having assumed that the fluid is perfect, incompressible, potential,
and that there are no forces internal and external yields the paradox that the
fluid exerts no force on C. This suggests that we formulate more physically
reasonable assumptions on the fluid, such as, for example, that the fluid is
viscous, and as a consequence the presence of internal friction.
13 Friction Tensor for Newtonian Viscous Fluids
In real fluids the friction generated by the mutual sliding of infinitesimal layers
generates shear forces that oppose the motion. The stress tensor T takes the
more general form
τij = −pδij + σij ,
(13.1)
where σij are due to friction. Two infinitesimal layers slide over one another
if their velocities are different. Therefore the σij = σij (∇v) depend on the
gradient of the velocity. Moreover, σij = 0 if ∇v = 0. Assuming that the
σij (·) are smooth functions of their arguments, they can be expanded in Taylor
series about the origin of their arguments to give
∂σij
σij (∇v) = γijhk vh,xk + Oij (∇v2 ),
where γijhk =
∂vh,xk ∇v=0
for i, j = 1, 2, 3, where Oij (·) are infinitesimal of higher order in |∇v|.
A fluid is Newtonian if (σij ) depends linearly on ∇v, so that the higher-order
terms in the previous Taylor expansions are negligible. Water and alcohol are
Newtonian, whereas paints and gels are not.
The numbers γijhk as the indices i, j, h, k run over 1, 2, 3, represent a
fourth-order tensor that quantifies the stresses due to the presence of internal friction in a fluid. By its physical nature such a tensor must be isoptropic, that is, must be independent of rotations of the Cartesian system of
its representation.
14 The Navier–Stokes Equations
321
Lemma 13.1 Let (γijhk ) for i, j, h, k = 1, 2, 3 be a representation of an
isotropic tensor σ. Then there exist numbers λ and µ1 , µ2 such that
γijhk = λδij δhk + µ1 δih δjk + µ2 δik δjh .
The lemma is established in §13.1c of the Complements.5 Assuming it for the
moment, it implies that σij must be of the form
σij = λδij vh,xh + µ1 vi,xj + µ2 vj,xi .
Since (σij ) must also be symmetric (Proposition 3.1),
σij = λδij vh,xh + µ1 vj,xi + µ2 vi,xj .
Adding these two expressions for σij gives
σij = 12 λ div vδij + 12 µ̄(vi,xj + vj,xi ) = 21 λ div vδij + 12 µ̄Dij ,
where µ̄ = µ1 + µ2 and (Dij ) is the deformation tensor introduced in (5.2).
This representation of the friction stress tensor in Newtonian fluids is due to
Stokes [141]. If the fluids are also incompressible, then
σij = 12 µ̄Dij .
(13.2)
The constant 21 µ̄ is called the kinematic viscosity, and is determined experimentally (§3.2c). By thermodynamic considerations, µ > 0 [138, page 213]
and [108, Chapter V].
14 The Navier–Stokes Equations
A Newtonian, viscous, incompressible fluid moves in a domain G ⊂ R3 . The
momentum equations for such a fluid are those in (3.1). Taking into account
the form (8.1) of the acceleration ẍ and the form (13.1)–(13.2) of the stress
tensor T, these equations take the form
[vt + (v · ∇)v − f ]ρ − ∇p = 12 µ̄ div(Dij ) = µ̄∆v + ∇ div v.
Therefore, since the fluid is incompressible,
1
vt − µ∆v + (v · ∇)v + ∇p = f,
ρ
div v = 0,
5
in G × R+ ,
A more general stress–deformation relation is due to Serrin [139].
(14.1)
322
12 INTRODUCTION TO FLUID DYNAMICS
where µ = µ̄/2ρ. The constant µ is the kinematic viscosity, and its physical
dimensions are length squared over time (§3.2c of the Complements of
Chapter 3).
14.1 Conservation and Dissipation of Energy
Assume that there are no external forces, so that f = 0. Multiply (14.1) by
v and perform standard vector calculus operations. Recalling the definition
(8.2) of the Bernoulli trinomial, we get
Dt B =
3
pt
+ µ∆ 12 v2 − µ
|∇vi |2 ,
ρ
i=1
p
where B = 21 v2 + .
ρ
(14.2)
For µ = 0 this coincides with Bernoulli’s equation (8.3) for f = 0 and ρ =
const. As indicated in §8, the term B is the specific energy of a material particle
about x. Therefore the left-hand side of (14.2) is the material derivative of such
a specific energy. The first term on the right-hand side is the time variation of
the internal energy about x. The second term can be regarded as a dissipation
of kinetic energy due to viscosity. The last term is the energy dissipation due
to the rough mutual sliding of infinitesimal layers over one another. Thus the
variation of energy along Lagrangian paths is balanced by the time variation
of the internal energy and the dissipation of energy due to viscosity.
14.2 Dimensionless Formulation, Reynolds Number,
and Similarities
The Navier–Stokes equations (14.1) are written in their physical dimensions.
To render them dimensionless, select length and time units ℓ and τ and introduce dimensionless variables and quantities6
x′ =
x
,
ℓ
t′ =
t
,
τ
v′ =
τ
v,
ℓ
p′ =
τ2 p
,
ℓ2 ρ
f′ =
τ2
f.
ℓ
Then (14.1) becomes
vt′ ′ −
1 ′ ′
∆ v + (v′ · ∇′ )v′ + ∇′ p′ = f ′ ,
R
div′ v′ = 0,
where
R=
6
in G′ × R′+ ,
ρ ℓ2
1 ℓ2
=
.
µτ
µτ
Recall that f is a specific force, that is, force per unit mass.
(14.3)
14 The Navier–Stokes Equations
323
Here ∆′ , ∇′ , and div′ denote the analogous differential operations with respect
to the variables x′ , and G′ is the dimensionless description of G. The number
R is called the Reynolds number. From the dimensions of µ it follows that R
is dimensionless.
Two motions are similar if they take place in homothetic domains with the
same Reynolds number. Roughly speaking, the two domains have the same
geometry and are mutually rescaled by a given length scale. The length scale
being fixed, one then rescales the time to obtain the same Reynolds number.
For example, in building a vessel one is interested in investigating a priori
how the shape of the hollow impacts the motion of the surrounding fluid. One
builds a model vessel, to be used in a limited laboratory environment, of the
same shape but of reduced size, by rescaling the geometry by a fixed length.
Experiments are performed with such a model in the same fluid where the
vessel is intended to operate, so that the two fluids have the same viscosity.
Finally, having fixed the length scale, one introduces a new time scale so that
the Reynolds number remains the same. The two motions are then similar,
and experimental laboratory operations correspond to those of the real fluid
up to inverse length and time scales.
These remarks imply that the mathematical investigation of motions modeled by the Navier–Stokes equations reduces to an investigation of (14.3) with
R = 1, since space and time scales can always be chosen so that R = 1. Denoting again by v, p, and f the indicated dimensionless quantities, and by ∆,
∇, div the homologous operations with respect to the indicated rescaled, dimensionless variables, the mathematical problem consists in finding a velocity
field v defined in G × R+ such that
vt − ∆v + (v · ∇)v + ∇p = f ,
div v = 0,
v(·, 0) = vo ,
in G × R+ ,
(14.4)
in G for t = 0.
Here vo is the initial velocity field defined in G and assumed to be known.
The determination of v hinges on further information on its behavior on the
boundary of G. For example, since the fluid is viscous, it adheres to the boundary of its container G, so that v = 0 on ∂G. This is a Dirichlet datum of v
on ∂G. On the other hand, the container might be impermeable, so that no
fluid flows out of it at ∂G, that is, v · n = 0 on ∂G for all times. This is a
Neumann datum of v on ∂G. This boundary information need not be homogeneous or could be intertwined, so that for example, a Dirichlet datum is
given on a portion ∂1 G of ∂G and a Neumann datum is prescribed on the
remaining portion ∂2 G = ∂G − ∂1 G. While the physical formulation is simple,
the corresponding mathematical problems are still not well understood and
are the object of current investigations [59, 78].
324
12 INTRODUCTION TO FLUID DYNAMICS
Problems and Complements
7c Equation of Motion of Ideal Fluids
7.1c Transmission of Sound Waves
An ideal compressible fluid moves in a region G ⊂ R3 satisfying the momentum equation and the conservation of mass (7.1). Assume the following
physical modeling assumptions:
(a) The fluid moves with small relative velocity and small time variations
of density. Therefore second-order terms of the type vi vj,xh and ρt vi are
negligible with respect to first-order terms.
(b) Heat transfer is slower than pressure drops, i.e., the process is adiabatic
and ρ = h(p) for some h ∈ C 2 (R).
Expanding h(·) about the equilibrium pressure po , renormalized to be zero,
gives
ρ = a o p + a1 p 2 + · · · .
Assume further that the pressure is close to the equilibrium pressure, so that
all terms of order higher than one are negligible when compared to ao p. These
assumptions in the momentum equation yield
(ρv)t = −∇p + f
in G × R.
Now take the divergence of both sides to obtain
∂
div(ρv) = −∆p + div f
∂t
in G × R.
From the continuity equation,
div(ρv) = −ρt = −ao pt .
Combining these remarks gives the equation of the pressure in the propagation
of sound waves in a fluid, in the form [130]
∂2p
− c2 ∆p = f
∂t2
in G × R,
(7.1c)
where c2 = 1/ao and f = − div f /ao . This is the wave equation in three space
dimensions [41, Chapter VI].
8c Barotropic Flows with Conservative Forces
325
7.2c Continuity Equation in Cylindrical Coordinates
Refer back to the cylindrical coordinates of §1.4c of the Complements of
Chapter 2 and deduce the formulas of formal differentiation
sin ϕ ∂
∂
∂
−
,
= cos ϕ
∂x1
∂r
r ∂ϕ
∂
cos ϕ ∂
∂
+
.
= sin ϕ
∂x2
∂r
r ∂ϕ
(7.2c)
From these compute
∂
∂
∂
ρ(ṙ cos ϕ − rϕ̇ sin ϕ) +
ρ(ṙ sin ϕ + rϕ̇ cos ϕ) +
(ρż)
∂x
∂y
∂x3
sin ϕ ∂
∂
ρ(ṙ cos ϕ − rϕ̇ sin ϕ)
= cos ϕ ρ(ṙ cos ϕ − rϕ̇ sin ϕ) −
∂r
r ∂ϕ
cos ϕ ∂
∂
∂
ρ(ṙ sin ϕ + rϕ̇ cos ϕ) +
+ sin ϕ ρ(ṙ sin ϕ + rϕ̇ cos ϕ) +
(ρż)
∂r
r ∂ϕ
∂x3
∂
∂
1 ∂
(ρrṙ) +
(ρrϕ̇) +
(ρrż) .
=
r ∂r
∂ϕ
∂x3
div(ρṖ ) =
Therefore the continuity equation takes the form
∂ρ 1 ∂
∂
∂
+
ρrvr +
ρrvϕ +
ρrvz = 0,
∂t
r ∂r
∂ϕ
∂z
(7.3c)
where (vr , vϕ , vx3 ) are the Lagrangian components of the velocity of the system
in cylindrical coordinates. The derivative along v is computed starting from
(7.2c) as
∂
∂
∂
+ vx2
+ vx3
∂x1
∂x2
∂x3
sin ϕ ∂
∂
−
= (ṙ cos ϕ − rϕ̇ sin ϕ) cos ϕ
∂r
r ∂ϕ
∂
cos ϕ ∂
∂
+
+ ẋ3
+ (ṙ sin ϕ + rϕ̇ cos ϕ) sin ϕ
∂r
r ∂ϕ
∂x3
∂
∂
∂
+ ϕ̇
+ ẋ3
= ṙ
.
∂r
∂ϕ
∂x3
v · ∇ = vx1
Therefore the operation of total derivative in terms of these coordinates is
Dt =
∂
+ (vr , vϕ , vx3 ) · ∇(r,ϕ,x3 ) .
∂t
8c Barotropic Flows with Conservative Forces
8.1c Torricelli’s Theorem
Consider the outflow of an ideal barotropic fluid subject to gravity from a
small orifice on the bottom of a container of large horizontal cross section.
326
12 INTRODUCTION TO FLUID DYNAMICS
The free surface exposed to air is taken as a reference level x3 = 0. The outflow
is so small that the flow is taken as stationary with good approximation, and
as a consequence the height h from the free surface x3 = 0 to the orifice is
constant. Also, the level of the free surface is lowered so slowly that the velocity
of the points on the plane x3 = 0 can be taken as zero. It is assumed also that
the horizontal planes within the container are isobaric surfaces (see the final
example of §4.1). Let po be the atmospheric pressure at the free surface x3 = 0.
Since the orifice is in contact with air, the atmospheric pressure there must
also be po and such a value of the pressure is the same on the horizontal plane
through the orifice and within the container. Since the Bernoulli trinomial is
constant B(0) = B(−h), we have
po
v2
po
=
−h+ ,
ρg
2g
ρg
which implies
|v| =
2gh.
Therefore the outflow speed depends only on the height of the fluid above the
orifice. In particular, it is independent of the orientation of the nozzle.
11c Barotropic Potential Flows
11.1c Bidimentional Incompressible Flows
The fluid moves in a simply connected domain G ⊂ R2 with smooth boundary
∂G and it does not flow out of G, so that v · n = 0 on ∂G, where n is the
unit outward normal at ∂G. Assume also that ρ = 1, so that ω = curl v.
Incompressibility div v = 0 implies v1,x1 = −v2,x2 . Therefore, for fixed t, the
differential form v2 (x, t)dx1 − v1 (x, t)dx2 is exact and there exists a smooth
function ϕ defined in G such that
v = (ϕx2 , −ϕx1 ) = ∇∗ ϕ,
which implies
ω = −∆ϕ,
where ω is the only nonzero element of ω. If n = (n1 , n2 ) is the unit normal
exterior to ∂G, the vector τ = (−n2 , n1 ) is the unit vector tangent to ∂G
oriented counterclockwise. Using these remarks, we compute on ∂G
0 = v · n = ∇ϕ · τ , which implies ϕ ∂G = c(t),
where c(t) is constant in the space variables x ∈ ∂G and depends at most
on time. Notice also that the right-hand side of (10.1) is zero, since the only
nonzero component of ω is along the x3 -axis, and v is independent of x3 . The
flow is then determined by the system
∆ϕ = −ω,
Dt ω = 0,
ϕ ∂G = const(t)
v = ∇∗ ϕ
in G × R+ ,
on ∂G × R+ .
(11.1c)
(11.2c)
12c Stationary, Incompressible Potential Flow past an Obstacle
327
From the second of these one computes
−ωt = det(∇ϕ, ∇ω)t .
Therefore the motion is stationary, that is, ωt = 0, if and only if ω and ϕ are
functionally independent. Prove that if ωt = 0 at some instant, then ωt ≡ 0.
12c Stationary, Incompressible Potential
Flow past an Obstacle
12.1c Proof of Proposition 12.1
Look for solutions of (12.1) of the form
u(y) = y · v∞ + w(y),
where w satisfies
∆w = 0 in R3 − C,
∇w · n = −v∞ · n on ∂C,
lim ∇w(y) = 0,
|y|→∞
lim w(y) = 0.
(12.1c)
|y|→∞
These follow from (12.2), and the last equation corresponds to taking γ = 0
in (12.2). Let BR denote the ball of radius R about the origin, and let R > 0
be so large that C ⊂ BR . Introduce the cutoff function
⎧
for 0 ≤ r < R,
⎪
⎨1
π r − R
2
r → ζ(r) = cos
for R ≤ r < 2R,
⎪
2 R
⎩
0
for r > 2R.
By definition ζ ∈ C 2 (R+ ), and
⎧
for 0 < r ≤ R,
⎪
⎨0
r − R
1
′
ζ (r) =
sin π
for R < r ≤ 2R,
⎪
R
⎩R
0
for r > 2R,
⎧
for 0 < r ≤ R,
⎪
⎨0
r − R
π
′′
ζ (r) =
for R < r ≤ 2R,
cos π
⎪
R
⎩ R2
0
for r > 2R.
Fix x ∈ BR − C, multiply the first equation of (12.1c) by
y→
ζ(|y|)
,
|y − x|
y = x,
328
12 INTRODUCTION TO FLUID DYNAMICS
and integrate by parts over Gε = B2R − C ∪ Bε (x) , where 0 < ε ≪ 1 is
so small that Bε (x) ⊂ B2R − C. The boundary of Gε consists of ∂C, with
exterior normal n, and the two spheres |y − x| = ε and |y| = 2R. The unit
normals exterior to G on ∂C and |x − y| = ε are interior to C and Bε (x)
respectively. By the Gauss–Green theorem,
v∞ · n
ζ(|y|)
y−x
−
dy =
dσ +
dσ.
∇w∇ ·
∇w ·
|y
−
x|
|y
−
x|
|y
− x|
Gε
|y−x|=ε
∂C
The integral on the left-hand side is further integrated by parts to give
y−x
ζ(|y|)
1
·
dσ = −
dy
w∆
w∇
|y − x| |y − x|
|y − x|
Gε
|y−x|=ε
v∞ · n
y−x
1
dσ +
dσ +
· ndσ.
=
∇w ·
w∇
|y − x|
|y − x|
|y−x|=ε
∂C |y − x|
∂C
(12.2c)
The various integrals on the right-hand side are transformed and are estimated
as follows. First one verifies that ∆|y − x|−1 = 0 for y = x; then
∆
1
1
ζ(|y|)
= 2∇
· ∇ζ(|y|) +
∆ζ (|y|)
|y − x|
|y − x|
|y − x|
1 ′′
2
y−x y
′
= 2ζ ′ (|y|)
+
ζ
.
(|y|)
+
ζ
(|y|)
|y − x|3 |y| |y − x|
|y|
By the structure of the cutoff function ζ(·), the right-hand side is nonzero
only on the spherical annulus R < |y| < 2R. For fixed x, we may take R so
large that x ∈ B 21 R , so that |y − x| ≥ 21 R. For these choices,
ζ(|y|) const
≤
∆
|y − x|
R3
for R ≤ |y| ≤ 2R.
Therefore the first integral on the right-hand side of (12.2c) is estimated as
ζ(|y|)
dy ≤ const sup |w| → 0 as |y| → ∞.
w∆
|y − x|
R≤|y|≤2R
Gε
The second integral on the right-hand side is estimated as
y−x
dσ ≤ 4πε sup |∇w| → 0 as ε → 0.
∇w ·
|y
−
x|
|y|≤2|x|
|y−x|=ε
Computing the derivative in the integral on the left-hand side gives
y−x
∂ 1
1
1
·
=
= − 2.
∇
|y − x| |y − x|
∂r r r=ε
ε
13c Friction Tensor for Newtonian Viscous Fluids
329
Putting these calculations in (12.2c) and taking the limit first as R → ∞ for
ε fixed, and then as ε → 0, yields the implicit representation formula
v∞ · n
1
1
1
dσ −
· ndσ.
(12.3c)
w(x) = −
w∇
4π ∂C |y − x|
4π ∂C
|y − x|
Proposition 12.1c For |x| ≫ 1 we have the asymptotic behavior
1
1
w(x) = O
,
∇w(x) = O
.
|x|2
|x|3
Proof. Such a behavior is obvious for the second integral on the right-hand
side of (12.3c). To prove it for the first integral, observe that for y ∈ ∂C and
|x| ≫ 1,
1 1
1
1 1
|y|
≤
, where t =
.
|x| 1 + t |y − x|
|x| 1 − t
|x|
Therefore by Taylor’s formula,
1
|y|
1
=
1+O
.
|y − x|
|x|
|x|
Thus
∂C
1
v∞ · n
dσ =
|y − x|
|x|
v∞ · ndσ +
∂C
O
∂C
|y|
v∞ · ndσ .
|x|
Since v∞ is a constant vector, the first integral vanishes, thereby proving the
assertion.
13c Friction Tensor for Newtonian Viscous Fluids
13.1c Isotropic Tensors of the Fourth Order
Given a continuously differentiable velocity field v defined in R3 , consider the
expression
Tij = γijhk vhk ,
where vhk =
∂vh
,
∂xk
i, j, h, k = 1, 2, 3.
(13.1c)
The nine numbers Tij are the representative entries of a tensor T of order 2,
with respect to a Cartesian triad Σ. Similarly, (γijhk ) is the Σ-representative
of a fourth-order tensor Γ . Let now Σ ′ be a new Cartesian triad obtained
from Σ by a rotation, realized by a unitary matrix A : Σ → Σ ′ . The vector
field x → v(x) is transformed into
Σ ′ ∋ y −→ v′ (y) = Av(A−1 y),
(13.2c)
330
12 INTRODUCTION TO FLUID DYNAMICS
and the representation of T in Σ ′ is
′
′
Tℓm
= γℓmrs vrs
,
′
where vrs
=
∂vs′
,
∂ξs
ℓ, m, r, s = 1, 2, 3.
Using (13.2c), compute
′
vrs
= Arh Ask
∂vh
= Arh Ask vhk .
∂xk
Therefore
′
Tℓm
= Arh Ask γℓmrs vhk .
The tensor T is isotropic if its action on vectors is independent of the reference
Cartesian triad, that is, if for all w ∈ Σ,
(Tij )w = A−1 (Tij′ )Aw,
∀w ∈ Σ.
Since w ∈ Σ is arbitrary,
′
Tij = Aℓi Tℓm
Amj ,
i, j = 1, 2, 3.
Using these representations, it follows that T is isotropic if
γijhk vhk = Aℓi Amj Arh Ask γℓmrs vhk ,
i, j = 1, 2, 3,
for all unitary matrices A. This in turn implies
γijhk = Aℓi Amj Arh Ask γℓmrs ,
i, j, h, k = 1, 2, 3.
(13.3c)
This is the condition for a fourth-order tensor Γ to be isotropic.
Proposition 13.1c Let Γ be a fourth-order isotropic tensor. Then its representation with respect to a Cartesian triad Σ is
γijhk = λδij δhk + µ1 δih δjk + µ2 δik δjh ,
(13.4c)
where the constants λ, µ1 , and µ2 are independent of Σ.
Proof. In (13.3c) take the rotation matrix
⎞
⎛
−1 0 0
A = ⎝ 0 −1 0⎠ .
0 0 1
For such a choice, the entries in (13.3c) are nonzero only if the quadruple ijhk
coincides with ℓmrs, and in such a case
γijhk = Aii Ajj Ahh Akk γijhk .
13c Friction Tensor for Newtonian Viscous Fluids
331
From the structure of the matrix A above, one verifies that if in the quadruple ijhk, the index 3 occurs an odd number of times, then γijhk = −γijhk .
Therefore
if in the quadruple ijhk the index 3
γijhk = 0
occurs an odd number of times.
Repeating the same arguments for the choices of rotation matrices
⎞
⎛
⎞
⎛
1 0 0
−1 0 0
A = ⎝0 −1 0 ⎠ ,
A = ⎝ 0 1 0 ⎠,
0 0 −1
0 0 −1
one concludes that
γijhk = 0
if in the quadruple ijhk any one of the indices
1, 2, 3 occurs an odd number of times.
Therefore the only nonzero elements are of the form
γiihh ,
γihih ,
γihhi ,
where repeated indices are not meant to be added. From (13.3c) compute
γ1133 = Aℓ1 Am1 Ar3 As3 γℓmrs = δℓm δrs γℓmrs ,
γ2233 = Aℓ2 Am2 Ar3 As3 γℓmrs = δℓm δrs γℓmrs .
Therefore γ1133 = γ2233 , and by symmetry,
γ1122 = γ1133 = γ2233 = γ3311 = γ2211 = λ.
If, on the other hand, all indices are equal, (13.3c) gives the identity
γiiii = Aℓi Ami Ari Asi γℓmrs = δℓmrs γℓmrs = γℓℓℓℓ ,
i, ℓ = 1, 2, 3.
Analogous considerations for the remaining terms imply that there exist constants λ, µ1 , µ2 , θ such that
γiihh = λ,
γihih = µ1 ,
i=h
Putting this in (13.3c) gives
γijhk =
γihhi = µ2 ,
γiiii = θ,
Aℓi Amj Arh Ask γℓmrs
indices of the form
iihh, i=h
+
Aℓi Amj Arh Ask γℓmrs
indices of the form
ihih, i=h
for all i, h = 1, 2, 3.
332
12 INTRODUCTION TO FLUID DYNAMICS
+
Aℓi Amj Arh Ask γℓmrs
indices of the form
ihhi, i=h
+
Aℓi Amj Arh Ask γℓmrs
indices of the form
iiii
= λδij δhk + µ1 δih δjk + µ2 δik δjh +θδij δik
i=h
= λδij δhk + µδih δjk + σδik δjh + [θ − (λ + µ1 + µ2 )]δijhk .
To conclude the proof it will be shown that this form of the tensor (γijhk )
satisfies (13.3c) for every unitary matrix A if and only if θ = λ + µ1 + µ2 .
Indeed, from (13.3c),
λδij δhk + µ1 δih δjk + µ2 δik δjh + [θ − (λ + µ1 + µ2 )]δijhk
= Aℓi Amj Arh Ask λδℓm δrs + µ1 δℓr δms + µ2 δℓs δmr + [θ − (λ + µ1 + µ2 )]δijhk
= λδij δhk + µ1 δih δjk + µ2 δik δjh + Aℓi Amj Arh Ask [θ − (λ + µ1 + µ2 )]δℓmrs .
!
Therefore the tensor on the left-hand side satisfies (13.3c) for all unitary
matrices A if
[θ − (λ + µ1 + µ2 )](δijhk − Aℓi Amj Arh Ask δℓmrs ) = 0
for all unitary matrices A. This is possible only if the coefficient independent
of the indices is zero.
14c The Navier–Stokes Equations
14.1c A Paradox of Ideal Fluids
Denote by (x, y, z) the coordinates of R3 . An incompressible fluid of constant
density ρo fills the slab 0 < y < 1 and moves so that the velocity field has
constant direction, say for example along the x-axis, and is driven by a constant pressure difference between two sections normal to the direction of the
velocity. Thus with obvious symbolism,
t
v(x, y, z, t) = v(x, y, t), 0, 0 ,
p(x1 ) = p1 ,
p(x2 ) = p2 ,
p(x, y, z) = p(x),
∆p = p2 − p1 ,
ℓ = x2 − x1 > 0.
Assume that there are no external forces. Incompressibility div v = 0 implies
vx = 0 and v(x, y, t) = v(y, t). If the fluid is ideal, it must satisfy the Euler
equations (7.1), that is,
ρo vt = −px .
14c The Navier–Stokes Equations
333
Taking the x-derivative of both sides gives
pxx = 0,
from which
p(x) = p1 +
∆p
(x − x1 ).
ℓ
(14.1c)
Putting this in the previous Euler equation yields
vt = −
∆p
,
ρo ℓ
from which
v(y, t) = −
∆p
t + f (y),
ρo ℓ
where f (·) is a smooth arbitrary function of y, subject only to the conditions
f ′ (0) = f ′ (1) = 0, that is, that the fluid does not flow out of the slab. This
solution of the Euler equations implies that the velocity grows to infinity as
t → ∞, even though the pressure difference between the sections x = x1 and
x = x2 is finite and constant.
14.2c Viscous Fluids: Solving the Paradox
If the fluid is viscous, velocity and pressure satisfy the Navier–Stokes equations
with f = 0. Recalling that only the first component of v is nonzero, and that
it depends only on y, the Navier–Stokes equations for such a fluid give
vt −
1
1
vyy + px = 0
R
ρo
in [0 < y < 1] × R+ ,
(14.2c)
where R is the Reynolds number of the system. Since the fluid is viscous, it
adheres to the walls y = 0 and y = 1 of the slab, so that v(0, t) = v(1, t) = 0
for all times. Taking the x-derivative of (14.2c), one computes the pressure as
in (14.1c), which substituted into (14.2c) gives
vt −
∆p
1
vyy +
= 0,
R
ρo ℓ
whose solution is independent of t and is given by
v(y) =
R ∆p
y(1 − y).
2ρℓ
Thus the fluid is in stationary regime with zero velocity at the walls and
distributed along a parabolic profile for y ∈ (0, 1).
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Index
n-body problem, 83, 153, 160, 276
canonical form, 153
integral(s) of motions, 156
Poincaré canonical form, 155
2-body problem, 60, 78, 80, 83
reduced mass of, 60
3-body problem, 156, 159
planar figuration(s), 157
Bernoulli trinomial, 312, 322
Bernoulli, Johann, 248
Bertrand, Joseph Louis François, 83
Binet formulas, 4, 22, 61
borealis pole, 197
brachistochrone, 20, 247, 248
minimality of, 248
Bruns, H., 156
absolute time, postulate of, 56
acceleration, 2, 10, 13, 123
in a central motion, 4
centrifugal, 202
Coriolis, 13, 56, 137, 199
transport, 13, 64, 137
angular momentum, 114, 116–119, 123,
124, 128, 131, 133, 134, 137, 155,
156, 161, 169
conservation of, 155–157, 161, 212
theorem of, 114
aphelion, 23
apocenter, 23
apsis, 22, 23
area constant, integral, 4, 23, 61, 62, 80,
88
areolar velocity, 3, 20, 75, 88
axis of inertia
principal, 100, 103, 105, 106, 108,
118, 143, 173, 179, 181, 182, 213
central, 99, 100, 103, 105, 106, 108
axode, 16–18, 25–27, 127, 176
canonical
transformation(s), 258–260
canonical form of
Hamiltonian systems, 240, 283
Huygens systems, 240
Lagrange equations, 149, 150, 153,
239
canonical transformation(s), 257, 266,
271, 277, 282, 284
completely canonical, 259, 276
generator of, 259, 260, 262, 273
in terms of Lagrange brackets, 281
in terms of Poisson brackets, 266
induced by the flow map, 262
induced by the Hamilton–Jacobi
equation, 263
infinitesimal, 269–271
integral invariant(s) by, 279
invariance of volumes by, 266, 278
linear, 276, 277
Liouville theorem for, 266, 278
pointwise, 258
preserve Lagrange brackets, 281, 282
preserve Poisson brackets, 266, 267
preserve symplectic matrices, 265
barometric formula, 306
Bernoulli law, 317, 320
343
344
Index
Cardano, Girolamo, 29, 30
cardinal equation(s), 115, 116, 123, 124,
126–130, 134, 135, 137, 139, 155,
173, 174, 195, 201, 301, 304
Cauchy, Augustin, 97
centrode, 20, 27–30, 126
Chasles, Michel, 19
Christoffel symbols, 52
circulation, 309, 310
Clairaut equation, 283, 292
Clairaut, Alexis-Claude, 83
collapse of the n bodies, 160
configuration space, 141, 144, 146, 231,
234
constitutive law
of ideal fluids, 305, 311
constraint, 33, 49, 50, 145, 201
fixed, 35, 36, 69–72, 142, 146, 151,
184, 207, 211, 214, 217, 224
holonomic, 37, 46, 47, 142, 145, 150,
207, 211, 214, 217, 224, 233, 234
moving, 35, 36, 208, 243, 252
of rigidity, 33, 122
of rolling without slipping, 47–49,
123, 131
rough, 70–72, 123
smooth, 68, 69, 88, 91, 122, 123, 145,
147, 214, 217, 224
transport due to, 35, 36
unilateral, 37, 120, 144
workless, 35, 36, 72, 120–122, 137,
139, 147, 174, 252, 253
continuity equation, 300, 305, 310, 324
in cylindrical coordinates, 325
Copernicus, 197
Coriolis
acceleration, 13, 56, 137, 199
force, 85, 199, 203, 307
theorem, 24
Coriolis, Gustave Gaspard, 12
cycloid, 20, 87, 248
d’Alembert paradox, 319
d’Alembert principle, 112, 113, 145, 301
d’Alembert, Jean Le Ronde, 112
Darboux, Gaston, 255
deflection(ing)
acceleration, 13
along a meridian, 67, 84
eastward, 67, 84
velocity, 136, 183
deformation tensor, 308, 321
deformations, 301, 307
geometry of, 299
incompressible, 300
reactions due to, 301
DeGiorgi, Ennio, 252
Descartes law, 256
differential form(s)
exact, 265, 290
of contact, 241
virtual, 241
Dirac mass, 94, 164
directrix, 61
Dirichlet criterion of stability, 80, 211,
215, 223, 224
Dirichlet, Gustav Peter Lejeune, 211
dissipation of energy, 322
due to viscosity, 322
dynamical system, 207, 208, 222
autonomous, 208
Earth, 201
center, speed of, 64
gyroscopic axis of, 196
mass of, 63, 65
precession of the, 195
radius of, 197
eccentricity, 61
ecliptic, plane of the, 195–197
effective potential, 79, 80
eikonal equation, 246, 283, 292
ellipsoid of inertia, 96–100, 103, 105,
106, 108, 174–176, 182, 213, 226
elliptic coordinates, 43, 44, 295
elliptic integral, 74
energy integral, 59, 62, 71–73, 75, 78,
80, 88, 146–148, 150, 151, 156,
167, 168, 174, 179, 184, 203, 212,
240, 247, 287, 312
entropy, 310
envelope
of solutions, 291, 292
of wave fronts, 243
epicycloid, 21, 22, 29
equation of continuity, 300, 305
equation of state, 305
Index
equilibrium configuration(s), 126, 129,
151, 152, 167, 168, 171, 208, 210,
305
asymptotically stable, 208, 209, 222
for a rigid system, 152
of barotropic fluids, 306, 307
stable, 80, 208, 209, 211, 213,
215–218, 220, 222–226, 228, 229
unstable, 208, 210, 211, 213, 216, 222,
224, 226
equinoctial precession, 175, 197
equinox(es), axis of, 197
Euler equations, 173, 174, 179, 195, 199
of ideal fluids, 310
Euler’s angles, 6, 15, 25, 48, 143, 173,
184
nutation, 6, 191, 196, 204
precession, 6, 190, 191, 193, 197
proper rotation, 6
Euler, Leonhard, 6, 95
Euler–Lagrange equations, 238
Eulerian formalism, 299
Fermat principle, 237, 245, 246
Fermat, Pierre de, 237
figure axis, 190, 191, 193, 201, 202, 204
flow map, 111, 112, 299
canonical transformation(s) of, 262
generated by irrotational fields, 279
infinitesimal transformations induced
by, 270
volume preserving, 266, 278
fluid(s)
adiabatic, 305, 324
barotropic, 305, 306, 311, 313, 314,
316, 318
homogeneous, 305
incompressible, 305, 313, 318, 321
isentropic, 310
Newtonian, 320, 321, 329, 333
perfect, ideal, 305, 320, 332
equations of, 310
perfect,ideal
equations of, 310
potential, 316, 318
viscous, 320, 321, 329, 333
force(s)
central, 59, 75
centrifugal, 65
345
compression, 301
conservative, 59, 62, 71, 72, 115, 144,
147, 150, 152, 207, 211, 214, 233,
234
Coriolis, 56, 85, 199, 202, 203, 307
external,internal, 69, 91, 112–115,
120, 128, 129, 135, 143, 146, 147,
150, 155, 172, 173, 194, 200, 201
gravitational, 56, 60–62, 65, 130, 153,
202
inertial, 113, 115
of transport, 56, 199
positional, 58, 75
shear, 301, 303
traction, 301
workless, 115
Foucault pendulum, 90
Foucault, Jean Bernard, Léon, 90
Frechét differential, 233
free rotator(s), 174, 179, 185, 214
Frenet formulas, 2
Frenet, Jean Frédéric, 2
frequency(ies)
bass, 217
commensurable, 217
normal, 216, 219, 227–229
of nutation, 188
of oscillations, 78, 183, 216, 217
principal, 216, 219, 227–229
treble, 217
friction tensor, 320
functional, 231
extrema of, 231, 232, 234, 246
of geometrical optics, 237, 255
stationary point(s) of, 231–234, 246,
248, 252, 253
Gâteaux differential, 233
Galileo, Galilei, 55, 64, 88, 197
geodesic, 39–41, 43, 52, 53, 147, 235,
249, 250
as curve of least path, 40
equation of, 42, 43
on a cylinder, 43
geographic pole, 197
geometrical optics
functional of, 237, 255
gyrocompass, 201, 204
346
Index
gyroscope(s), 143, 178, 182–184, 189,
192, 194, 196, 199
intrinsic equations of, 200–203
gyroscopic axis, 143, 178, 182–185, 188,
190, 191, 193, 194, 196, 197, 201,
204, 227
of Earth, 196
quivering of, 183, 188
tenacity of, 183
gyroscopic compass, 201, 204
Hamilton, 244
canonical equation(s), 199, 207, 211
canonical system, 149, 207, 209, 211,
212, 240, 241, 283
rank of, 286
normal slowness, 244
principle of stationary action, 234
Hamilton, Sir William Rowan, 149
Hamilton–Jacobi equations, 236, 263,
283, 293, 294, 296, 297
complete integrals of, 283, 284, 286,
297
completely separable, 285, 296
separation of variables, 284, 285, 292
stationary, 284, 286, 294–296
Hamiltonian, 149–151, 153, 155, 170,
199, 207, 211, 240, 257–263, 270,
271, 284, 292, 294, 295, 297
Hamiltonian action, 234, 236, 257
local minima of, 234, 253, 255
reduced, 234
stationary point(s) of, 234, 252, 253,
257, 258
Helmholtz theorem, 309, 310
Hipparchus of Nicaea, 197
homogeneous function, 142, 163,
238–242
Huygens, 168, 198, 244
principle, 244
systems, 238, 239, 244
in canonical form, 240
theorem, 119, 168
Huygens, Christiaan, 88, 95
hypocycloid, 21, 30
incidence angle, 256
incompressible
deformation, 300
index of refraction, 237, 238, 244, 255
inertia tensor, 95, 97–99, 104, 105, 107,
108, 119, 131–134, 143, 173, 181,
184, 200, 201
inertial mass, 56
inertial system, triad, 55, 56, 60, 63,
64, 83, 90, 116, 122, 136, 153–156,
160, 173, 220
instability
Četaev criteria, theorem, 210, 214
of linearized motion(s), 215
of Poinsot precessions, 212, 226
integral invariant(s), 279
integral of angular momentum, 156,
169, 212
integral(s) of motion, 59, 148, 151, 154,
156, 174, 179, 184, 186, 270, 271,
287
by Poisson brackets, 268, 269
in involution, 288, 289
of the n-body problem, 156
irrotational, 309, 312, 314, 318
field, 307
permanence of, 316
isoperimetric, 250, 252
isotropic tensor, 320, 329, 330
Jacobi identity, 267, 269
Jacobi, Carl Gustav Jacob, 236, 265
Jupiter, 63
König, Samuel, 117
Kelvin theorem
of minimum energy, 318
of the circulation, 313
Kepler’s laws, 22, 62–64
Kepler, Johannes, 22, 62
kinetic energy, 58, 59, 62, 69, 71, 72,
114, 115, 118, 119, 123, 130, 131,
134, 137, 138, 141, 142, 145–147,
153, 154, 163, 165, 168, 174, 184,
199, 200, 218, 248, 252
of center of mass, 117
theorem of, 115
kinetic level, 313
kinetic momentum, 148–151, 153, 154,
156, 211, 240, 244, 245, 258, 262,
292, 294, 295, 297
Index
Lagrange, 156, 160
brackets, 280
equation(s), 141, 145–150, 163, 165,
166, 169, 186, 199, 207, 211, 215,
218, 220, 221, 233, 239, 241
canonical form of, 149, 150, 153
linearized, 228, 229
multiplier(s), 156, 250
Lagrange brackets, 280
invariant by canonical transformations, 282
Lagrange, Joseph Louis, 34, 120
Lagrange–Jacobi identity, 161, 162
Lagrangian, 148–151, 164, 170, 186,
199, 207, 211, 214, 233, 240, 294,
295, 297
components of forces, 144
configuration(s), 47, 236, 258
formalism, 299
linearized, 214, 215
path, trajectory, 111, 145, 234
virtual, 231
resultant of forces, 144, 146
velocity, 34, 35, 47, 49, 142, 143, 149,
244
Lagrangian coordinate, 34–36, 43,
46–49, 72, 74, 92, 126, 130–132,
137, 141, 143, 146, 147, 149, 150,
152–154, 156, 160, 163–165, 168,
181, 184, 207, 215, 217, 276
cyclic, 148, 151, 154, 184, 271, 286,
288
ignorable, 148, 151, 154, 184, 271
Laplace, Pierre Simon, 114
latitude, astronomical, geocentric, 66,
90
of Rome, Italy, 84
least action principle, 233, 234, 252,
257, 258
Legendre transform, 170, 171, 273
of the Lagrangian, 170
Leibniz, Gottfried Wilhelm, 58
Leonardo da Vinci, 55
Lie, Marius Sophus, 241, 271, 289
light ray(s), 237, 242, 245
in crystal(s), 237
propagation of, 237, 255
from a point source, 242, 243
speed of, 237, 238
347
trajectory of, 237, 238, 242, 244, 246
velocity of, 244
linearized
Lagrange equations, 228, 229
Lagrangian, 214, 215
motion, 214, 217
stability, instability of, 215, 216
potential, 215, 216
system, 216
Liouville, Joseph, 266, 288
Lyapunov, 208, 216
function, 208, 209, 211, 213, 222
isolated minima of, 208, 209
stability criteria, 208, 222
Lyapunov, Aleksandr Mikhailovich, 209
Mach number, 318
material
derivative, 299
lines, surfaces, 313
path, 312
Maupertuis principle, 234, 235
Maupertuis, Pierre Louis Moreau
de, 234
Mercury, 63
Mohr’s circle, 106
molecule 3-atom, 218
moment, 183, 196, 203
of a couple, 56
of a force, 112, 114, 127, 182
of reaction, 115, 128, 135, 182, 202
of the velocity, 3
resultant, 173, 182, 183, 202
moment of inertia, 95, 118, 173,
177–179, 198, 212, 213, 226
axial, 95–100, 103, 105, 106, 108, 118,
119, 137, 143, 168, 177, 199
deflection, 95–99, 106, 108, 118, 119,
182, 203
of first order, 94
of second order, 95
planar, 95–100, 103, 105, 106, 108
polar, 95–100, 103, 105, 106, 108, 160
momentum, 58, 113, 116, 117, 124, 130
angular, 114, 116–119, 123, 124, 128,
131, 133, 134, 137, 138, 155–157,
161, 169
conservation of, 58, 154–156, 220
equation, 305, 310, 312, 313, 324
of the center of mass, 113
348
Index
Navier–Stokes equations, 321, 332
in dimensionless form, 322, 323
Neumann problem, 318
Newton’s laws, 55, 60, 62, 63, 112, 113,
116, 122
Newton, Isaac, 55, 58, 61–63
Newtonian fluid, 320, 321, 329, 333
nodal line, 6, 15, 185, 197
normal
coordinates, 216, 217, 219–221, 227
frequency, 216, 219, 227–229
commensurable, 217
modes, 216, 217, 219, 227–229
slowness, 244
nutation, 6, 184, 185, 187, 190–192, 196,
204
amplitude of, 187–189, 204
asymptotic, 193, 204
angle of, 191
frequency of, 188
period of, 187, 188, 193
asymptotic, 193, 205
optical length, 245
oscillations, 187, 204, 221
amplitude of, 78, 204, 221, 222
concurrence of phase, 77, 222
damped, 76, 77, 222
degenerate, 220, 229
delay of phase, 77
forced, 76–78
resonance, 78
free, 76, 77
frequency of, 78, 183, 216, 217, 222
fundamental, 217, 220, 221
composition of, 217
harmonic, 76, 77, 89, 167, 189, 216,
217, 222, 297
independent, 216, 221
uncoupled, 216, 221
opposition of phase, 221
period of, 74, 76, 77, 89, 92, 126, 129,
167, 191, 193, 216, 217
quadrature of phase, 77
small, 74, 89, 90, 92, 126, 129, 216,
227
parabolic coordinates, 44, 45, 294
paradox of d’Alembert, 319
paradox of parallelism, 183, 189
parallax solar, 196
Pascal’s principle, 305
path of least time, 238
pendulum, 165, 204
as unilateral constraint, 86
compound, 189, 197–199, 204, 205
cycloidal, 87, 297
double, 132, 225, 228
Foucault, 90
mathematical, 72, 86, 88, 180, 216
period of, 180
spherical, 46, 88, 90, 205
string, 86
pericenter, 23
perihelion, 23
Pfaff, Johann Friedrich, 265
phase space, 150, 151, 156, 207, 217,
223, 241
trajectory in, 209
piezometric head, 313
planar moment of inertia, 106
Poincaré surface invariants, 279, 280
Poincaré, Jules-Henri, 155
Poinsot cone, 17, 176, 179
degenerate, 226
fixed, moving, 17, 176, 179
Poinsot precession, 174, 185, 189, 195,
199, 207
of Earth, 196
stability, instability of, 212, 226
Poinsot, Louis, 17
Poisson brackets, 266–269, 280, 281
in 2(N + 1) variables, 289
invariant by canonical transformations, 266
Poisson formulas (see rigid motions), 10
Poisson Siméon Denis, 10
polhode, 176, 177, 180, 226, 227
degenerate, 178–180
fixed, moving, 176–179
postulate of absolute time, 56
potential, 144, 153, 154, 211, 214
atomic, 218
effective, 79, 80
energy, 59, 115, 123
gravitational, 295
isolated extrema of, 211, 215, 216,
223
Index
kinetic, for a fluid, 307, 316, 318
linearized, 215
stationary point of, 215, 216
precession, 10, 15, 17, 174, 182, 186,
190, 196, 202
astronomical of Earth, 196, 197
of constant direction, 190, 191, 204
of equinoxes, 175, 197
period of, 188, 192, 197, 198
Poinsot (see Poinsot prec.), 174
pole of a, 10, 173, 194, 196
relative to the center of mass, 195
that invert direction, 191, 192
pressure, 305
principle
of Maupertuis, 235
principle of
d’Alembert, 112, 113, 145, 301
Fermat, 237, 245, 246
geometrical optics, 238
Huygens, 244
least action, 233, 234, 238, 252, 257,
258
least time, 238
Maupertuis, 234
Pascal, 305
stationary action, 234
variational, 234
virtual work, 122, 143–145, 150–152,
164, 207, 211, 233, 234
for rigid systems, 122
pseudocoordinates, 25
reaction, 89, 182, 202
due to constraints, 68–73, 91, 92, 112,
113, 115, 120, 122–124, 126, 129,
130, 134, 139, 144, 152, 165, 173,
181, 182, 200, 202
due to deformations, 301
due to rigidity constraints, 122
work of, 144, 145
workless, 71, 72
reduced action, 234
refraction
angle, 256
index, 237, 238, 244, 255
Reynolds number, 322, 323
Ricci symbols, 8
Ricci-Curbastro Gregorio, 8
349
rigid motion(s), 33, 307
axis of motion or rotation, 10–12, 16,
17, 25
in a precession, 12
axis of Mozzi, 12
axode, 17, 18
characteristics of, 8, 10, 12–15, 17,
18, 24, 56, 64, 83, 116, 117, 133,
173, 195, 196, 199
composition of, 13, 14, 25
precessions, 14, 18, 196
Coriolis theorem, 12
flow map for, 111
geometry of, 5, 126
instantaneous, 308
kinetic energy for, 143
plane, 18
Poisson formulas, 10, 123, 128, 201,
202
precession, 10
relative, 12, 24
rolling and slipping, 17
rolling without slipping, 17, 20–22,
27, 29, 47–49, 123, 131, 134, 169,
175
rotation, center of, 10, 18–20
triad, fixed, moving, 5, 15, 24, 27, 29,
48, 56, 64, 118, 119, 127, 133, 134,
136, 143, 152–156, 160, 173, 181,
184, 196, 202, 212
vectorial invariant, 11, 16, 17, 19, 27
rigidity constraints, 33, 122
rotation tensor, 308
sidereal day, 64
Snell’s law, 256
Snellius van Royen Willebrord, 256
solar parallax, 196
solar system, invariant plane of, 114
speed of sound, 317, 318
spherical pendulum, 46
compound, 205
stability, 80, 207
asymptotic, 208, 209, 222
Dirichlet criterion of, 80, 211, 215,
223, 224
Lyapunov criteria of, 208, 222
of Poinsot precessions, 212, 226
stationary action principle, 234
350
Index
Stevin’s law, 306
streamlines, 312
stress
compression,traction, 303
due to friction, 320
shear, 303, 305
stress tensor, 302, 303, 321
as a linear map, 304
representative of, 304
symmetry of, 304
subsonic flow, 318
Sundman inequality, 161
symplectic, 264
group of order N , 265
identity, 264
matrix, 264, 265, 277
canonical transformations, 265
preserve volume, 265, 266
product, 264, 266
transformation, 264, 265,
276, 277
preserve volume, 266
tautochrone, 21
tensor
isotropic, 320, 329, 330
of deformation, 308, 321
of friction, 320
of rotation, 308
of stress, 302, 321
symmetry of, 304
tensor of inertia, 95, 97–99, 104–108,
119, 131–134, 143, 173, 181, 184,
200, 201
theorem of
Četaev, of instability, 210, 214
Cauchy, 302, 304
Chasles, 19, 126
Clairaut, 52, 53
Coriolis, 13, 56, 60, 65, 83, 90
Euler, of homogeneous functions, 163,
238, 240, 241
Helmholtz, 309, 310
Huygens, 95, 100, 101, 103–108, 128,
198, 199
König, 117, 123
Kelvin
of minimum energy, 318
of the circulation, 313
Lagrange, 121, 122
Lie, 289
Liouville
of quadratures, 288, 289
of volumes, 266, 278
Lyapunov, 209
Poincaré of recurrence, 277, 278
the angular momentum, 114
the center of mass, 113, 116, 117, 123
the kinetic energy, 115
the momentum, 113, 123
the three levels, 313
Torricelli, 325
Timocharis of Alexandria, 197
top, 184, 188, 201
sleeping, 184–186
spinning, 184, 200
subject to friction, 193
Torricelli theorem, 325
total derivative, 299
trajectory
in phase space, 156, 209
of least length, 250
optical, 238, 244
transformation
canonical, 258–260, 271, 284
symplectic, 264, 276
variational, 257, 258, 271
triad, 5, 6, 16, 18, 61, 95
inertial, 122, 136
intrinsic, 1, 57, 88, 91, 202
principal of inertia, 99, 100, 103, 105,
106, 108, 179, 184, 212
central, 99, 100, 103, 105, 106, 108,
131, 182, 199, 202, 203
variation
of a functional, 232, 236
synchronous, 231, 232, 253, 257–259,
265
virtual, 172, 231, 236, 241, 254,
257–259, 265
velocity
angular, 6–8, 24, 26, 143, 157, 159,
180, 186, 189, 201, 308
of nutation, 186
of precession, 186, 193
per unit mass, 315
areolar, 3, 88
Index
potential (of a fluid), 307
transport, 12, 14, 17, 25, 35, 36
virtual differential, 37
virtual displacement(s), 36, 37, 120–122,
141, 144–146, 152, 243
virtual work, 69, 120, 122, 152
due to reaction, 144, 152
principle of, 122, 144, 145, 150–152,
164, 207, 211, 233, 234
viscosity, dynamic, kinematic, 75, 85,
305, 321, 322
vortex
filament, 308, 309
as material lines, 314
sheet, 309, 310
as material surfaces, 314
tube, 309
intensity of, 310, 314
vorticity, 314, 315
field, 307, 308
transport of, by flow map, 315
wave equation, 324
wave front, 242–245
envelope of, 243
from a point source, 244
infinitesimal, 242, 243
of constant normal velocity, 246
velocity of, 243
work, 57, 59, 115, 144
done by reaction, 122
elemental, 144, 145, 172
351