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A Student’s Guide to Lagrangians and Hamiltonians
A concise but rigorous treatment of variational techniques, focusing primarily on Lagrangian and Hamiltonian systems, this book is ideal for physics, engineering and mathematics students. The book begins by applying Lagrange’s equations to a number of mechanical systems. It introduces the concepts of generalized coordinates and generalized momentum. Following this, the book turns to the calculus of variations to derive the Euler–Lagrange equations. It introduces Hamilton’s principle and uses this throughout the book to derive further results. The Hamiltonian, Hamilton’s equations, canonical transformations, Poisson brackets and Hamilton–Jacobi theory are considered next. The book concludes by discussing continuous Lagrangians and Hamiltonians and how they are related to field theory. Written in clear, simple language, and featuring numerous worked examples and exercises to help students master the material, this book is a valuable supplement to courses in mechanics.
patrick hamill is Professor Emeritus of Physics at San Jose State University. He has taught physics for over 30 years, and his research interests are in celestial mechanics and atmospheric physics.




A Student’s Guide to Lagrangians
and Hamiltonians
PATRICK HAMILL San Jose State University


University Printing House, Cambridge CB2 8BS, United Kingdom
Published in the United States of America by Cambridge University Press, New York
Cambridge University Press is part of the University of Cambridge.
It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence.
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© P. Hamill 2014
This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.
First published 2014
Printed and bound in the United Kingdom by TJ International Ltd. Padstow Cornwall
A catalog record for this publication is available from the British Library
Library of Congress Cataloging in Publication data Hamill, Patrick. A student’s guide to Lagrangians and Hamiltonians / Patrick Hamill. pages cm Includes bibliographical references. ISBN 978-1-107-04288-9 (Hardback) – ISBN 978-1-107-61752-0 (Paperback) 1. Mechanics, Analytic–Textbooks. 2. Lagrangian functions–Textbooks. 3. Hamiltonian systems–Textbooks. I. Title. QA805.H24 2013 515′.39–dc23 2013027058
ISBN 978-1-107-04288-9 Hardback ISBN 978-1-107-61752-0 Paperback
Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.


Contents
Introduction page ix Acknowledgements x
I LAGRANGIAN MECHANICS 1
1 Fundamental concepts 3 1.1 Kinematics 3 1.2 Generalized coordinates 5 1.3 Generalized velocity 7 1.4 Constraints 9 1.5 Virtual displacements 11 1.6 Virtual work and generalized force 12 1.7 Configuration space 13 1.8 Phase space 15 1.9 Dynamics 15 1.9.1 Newton’s laws of motion 15 1.9.2 The equation of motion 16 1.9.3 Newton and Leibniz 16 1.10 Obtaining the equation of motion 18 1.10.1 The equation of motion in Newtonian mechanics 19 1.10.2 The equation of motion in Lagrangian mechanics 19 1.11 Conservation laws and symmetry principles 25 1.11.1 Generalized momentum and cyclic coordinates 27 1.11.2 The conservation of linear momentum 30 1.11.3 The conservation of angular momentum 33 1.11.4 The conservation of energy and the work function 36 1.12 Problems 41
v


vi Contents
2 The calculus of variations 44 2.1 Introduction 44 2.2 Derivation of the Euler–Lagrange equation 45 2.2.1 The difference between δ and d 52 2.2.2 Alternate forms of the Euler–Lagrange equation 55 2.3 Generalization to several dependent variables 58 2.4 Constraints 60 2.4.1 Holonomic constraints 60 2.4.2 Non-holonomic constraints 64 2.5 Problems 67
3 Lagrangian dynamics 70 3.1 The principle of d’Alembert. A derivation of Lagrange’s equations 70 3.2 Hamilton’s principle 73 3.3 Derivation of Lagrange’s equations 75 3.4 Generalization to many coordinates 75 3.5 Constraints and Lagrange’s λ-method 77 3.6 Non-holonomic constraints 81 3.7 Virtual work 83 3.7.1 Physical interpretation of the Lagrange multipliers 84 3.8 The invariance of the Lagrange equations 86 3.9 Problems 88
II HAMILTONIAN MECHANICS 91
4 Hamilton’s equations 93 4.1 The Legendre transformation 93 4.1.1 Application to thermodynamics 95 4.2 Application to the Lagrangian. The Hamiltonian 97 4.3 Hamilton’s canonical equations 98 4.4 Derivation of Hamilton’s equations from Hamilton’s principle 100 4.5 Phase space and the phase fluid 101 4.6 Cyclic coordinates and the Routhian procedure 104 4.7 Symplectic notation 106 4.8 Problems 107
5 Canonical transformations; Poisson brackets 109 5.1 Integrating the equations of motion 109 5.2 Canonical transformations 110


Contents vii
5.3 Poisson brackets 117 5.4 The equations of motion in terms of Poisson brackets 119 5.4.1 Infinitesimal canonical transformations 120 5.4.2 Canonical invariants 124 5.4.3 Liouville’s theorem 127 5.4.4 Angular momentum 128 5.5 Angular momentum in Poisson brackets 129 5.6 Problems 132
6 Hamilton–Jacobi theory 134 6.1 The Hamilton–Jacobi equation 135 6.2 The harmonic oscillator – an example 137 6.3 Interpretation of Hamilton’s principal function 139 6.4 Relationship to Schrödinger’s equation 140 6.5 Problems 142
7 Continuous systems 144 7.1 A string 144 7.2 Generalization to three dimensions 150 7.3 The Hamiltonian density 151 7.4 Another one-dimensional system 154 7.4.1 The limit of a continuous rod 156 7.4.2 The continuous Hamiltonian and the canonical field equations 160 7.5 The electromagnetic field 162 7.6 Conclusion 166 7.7 Problems 166
Further reading 168 Answers to selected problems 169 Index 171




Introduction
The purpose of this book is to give the student of physics a basic overview of Lagrangians and Hamiltonians. We will focus on what are called variational techniques in mechanics. The material discussed here includes only topics directly related to the Lagrangian and Hamiltonian techniques. It is not a traditional graduate mechanics text and does not include many topics covered in texts such as those by Goldstein, Fetter and Walecka, or Landau and Lifshitz. To help you to understand the material, I have included a large number of easy exercises and a smaller number of difficult problems. Some of the exercises border on the trivial, and are included only to help you to focus on an equation or a concept. If you work through the exercises, you will better prepared to solve the difficult problems. I have also included a number of worked examples. You may find it helpful to go through them carefully, step by step.
ix


Acknowledgements
I would like to acknowledge the students in my graduate mechanics classes whose interest in analytical mechanics were the inspiration for writing this book. I also wish to acknowledge my colleagues in the Department of Physics and Astronomy at San Jose State University, especially Dr. Alejandro Garcia and Dr. Michael Kaufman, from whom I have learned so much. Finally, I acknowledge the helpful and knowledgeable editors and staff at Cambridge University Press for their support and encouragement.
x


PART I
Lagrangian mechanics




1
Fundamental concepts
This book is about Lagrangians and Hamiltonians. To state it more formally, this book is about the variational approach to analytical mechanics. You may not have been exposed to the calculus of variations, or may have forgotten what you once knew about it, so I am not assuming that you know what I mean by, “the variational approach to analytical mechanics.” But I think that by the time you have worked through the first two chapters, you will have a good grasp of the concept. We being with a review of introductory concepts and an overview of background material. Some of the concepts presented in this chapter will be familiar from your introductory and intermediate mechanics courses. However, you will also encounter several new concepts that will be useful in developing an understanding of advanced analytical mechanics.
1.1 Kinematics
A particle is a material body having mass but no spatial extent. Geometrically, it is a point. The position of a particle is usually specified by the vector r from the origin of a coordinate system to the particle. We can assume the coordinate system is inertial and for the sake of familiarity you may suppose the coordinate system is Cartesian. See Figure 1.1. The velocity of a particle is defined as the time rate of change of its position and the acceleration of a particle is defined as the time rate of change of its velocity. That is,
v =dr
dt =  ̇r, (1.1)
3


4 1 Fundamental concepts
x
y
z
r
m
Figure 1.1 The position of a particle is specified by the vector r.
and
a =dv
dt = r ̈. (1.2)
The equation for a as a function of r, r ̇, and t is called the equation of motion. The equation of motion is a second-order differential equation whose solution gives the position as a function of time, r = r(t). The equation of motion can be solved numerically for any reasonable expression for the acceleration and can be solved analytically for a few expressions for the acceleration. You may be surprised to hear that the only general technique for solving the equation of motion is the procedure embodied in the Hamilton–Jacobi equation that we will consider in Chapter 6. All of the solutions you have been exposed to previously are special cases involving very simple accelerations. Typical problems, familiar to you from your introductory physics course, involve falling bodies or the motion of a projectile. You will recall that the projectile problem is two-dimensional because the motion takes place in a plane. It is usually described in Cartesian coordinates. Another important example of two-dimensional motion is that of a particle moving in a circular or elliptic path, such as a planet orbiting the Sun. Since the motion is planar, the position of the body can be specified by two coordinates. These are frequently the plane polar coordinates (r, θ) whose relation to Cartesian coordinates is given by the transformation equations
x = r cos θ,
y = r sin θ.
This is an example of a point transformation in which a point in the xy plane is mapped to a point in the rθ plane.


1.2 Generalized coordinates 5
You will recall that in polar coordinates the acceleration vector can be resolved into a radial component and an azimuthal component
a =  ̈r = ar ˆr+aθ θˆ . (1.3)
Exercise 1.1 A particle is given an impulse which imparts to it a velocity v0. It then undergoes an acceleration given by a = −bv, where b is a constant and v is the velocity. Obtain expressions for v = v(t) and x = x(t). (This is a one-dimensional problem.) Answer: x(t) =
x0 + (v0/b) (1 − e−bt ) .
Exercise 1.2 Assume the acceleration of a body of mass m is given by a = −(k/m)x. (a) Write the equation of motion. (b) Solve the equation of motion. (c) Determine the values of the arbitrary constants (or constants of integration) if the object is released from rest at x = A. (The motion
is called “simple harmonic.”) Answer (c): x = A sin (√k/mt + π /2) =
A cos √k/mt.
Exercise 1.3 In plane polar coordinates, the position is given by r = r ˆr. Obtain expressions for the velocity and acceleration in terms of r, θ , rˆ, θˆ .
(Hint: Express rˆ and θˆ in terms of iˆ and jˆ.) Answer: a = (r ̈ −rθ ̇ 2)ˆr+(rθ ̈ + 2r ̇θ ̇ )θˆ .
1.2 Generalized coordinates
In the preceding section we stated that the position of the particle was given by the vector r. We assumed an inertial Cartesian coordinate system in which the components of r were (x, y, z). Of course, there are many other ways we could have specified the position of the particle. Some ways that immediately come to mind are to give the components of the vector r in cylindrical (ρ, φ, z) or in spherical coordinates (r, θ , φ). Obviously, a particular problem can be formulated in terms of many different sets of coordinates. In three-dimensional space we need three coordinates to specify the position of a single particle. For a system consisting of two particles, we need six coordinates. A system of N particles requires 3N coordinates. In Cartesian coordinates, the positions of two particles might be described by the set of numbers (x1, y1, z1, x2, y2, z2). For N particles the positions of all the particles are
(x1, y1, z1, x2, y2, z2, . . . , xN , yN , zN ).


6 1 Fundamental concepts
As you know, some problems are more easily solved in one coordinate system and some are more easily solved in another. To avoid being specific about the coordinate system we are using, we shall denote the coordinates by qi and the corresponding velocities by q ̇i. We call qi the “generalized coordinates.” Of course, for any particular problem you will chose an appropriate set of coordinates; in one problem you might use spherical coordinates in which case as i ranges from 1 to 3, the coordinates qi take on the values r, θ , φ, whereas in another problem you might use cylindrical coordinates and qi = ρ, φ, z. To convert from Cartesian coordinates to generalized coordinates, you need to know the transformation equations, that is, you need to know the relations
q1 = q1(x1, y1, z1, x2, y2, . . . , zN , t) (1.4)
q2 = q2(x1, y1, z1, x2, y2, . . . , zN , t)
...
q3N = q3N (x1, y1, z1, x2, y2, . . . , zN , t).
The inverse relations are also called transformation equations. For a system of N particles we have
x1 = x1(q1, q2, . . . , q3N , t) (1.5)
y1 = y2(q1, q2, . . . , q3N , t)
...
zN = zN (q1, q2, . . . , q3N , t).
It is often convenient to denote all the Cartesian coordinates by the letter x. Thus, for a single particle, (x, y, z) is written (x1, x2, x3) and, for N particles, (x1, y1, . . . , zN ) is expressed as (x1, . . . , xn), where n = 3N.
We usually assume that given the transformation equations from the qs to the xs, we can carry out the inverse transformation from the xs to the qs, but this is not always possible. The inverse transformation is possible if the Jacobian determinant of Equations (1.4) is not zero. That is,
∂(q1, q2, . . . , qn)
∂ (x1, x2, . . . , xn) =
∣∣∣∣∣∣∣
∂q1/∂x1 ∂q1/∂x2 · · · ∂q1/∂xn
...
∂qn/∂x1 ∂qn/∂x2 · · · ∂qn/∂xn
∣∣∣∣∣∣∣
= 0.
The generalized coordinates are usually assumed to be linearly independent, and thus form a minimal set of coordinates to describe a problem. For example, the position of a particle on the surface of a sphere of radius a can be described in terms of two angles (such as the longitude and latitude). These two angles


1.3 Generalized velocity 7
form a minimal set of linearly independent coordinates. However, we could also describe the position of the particle in terms of the three Cartesian coordinates, x, y, z. Clearly, this is not a minimal set. The reason is that the Cartesian coordinates are not all independent, being related by x2 + y2 + z2 = a2. Note that given x and y, the coordinate z is determined. Such a relationship is called a “constraint.” We find that each equation of constraint reduces by one the number of independent coordinates. Although the Cartesian coordinates have the property of being components of a vector, this is not necessarily true of the generalized coordinates. Thus, in our example of a particle on a sphere, two angles were an appropriate set of generalized coordinates, but they are not components of a vector. In fact, generalized coordinates need not even be coordinates in the usual sense of the word. We shall see later that in some cases the generalized coordinates can be components of the momentum or even quantities that have no physical interpretation. When you are actually solving a problem, you will use Cartesian coordinates or cylindrical coordinates, or whatever coordinate system is most convenient for the particular problem. But for theoretical work one nearly always expresses the problem in terms of the generalized coordinates (q1, . . . , qn). As you will see, the concept of generalized coordinates is much more than a notational device.
Exercise 1.4 Obtain the transformation equations for Cartesian coordinates to spherical coordinates. Evaluate the Jacobian determinant for this transformation. Show how the volume elements in the two coordinate systems are related to the Jacobian determinant. Answer: dτ = r2 sin θ drdθ dφ.
1.3 Generalized velocity
As mentioned, the position of a particle at a particular point in space can be specified either in terms of Cartesian coordinates (xi) or in terms of generalized coordinates (qi). These are related to one another through a transformation equation:
xi = xi (q1, q2, q3, t).
Note that xi, which is one of the three Cartesian coordinates of a specific particle, depends (in general) on all the generalized coordinates.


8 1 Fundamental concepts
We now determine the components of velocity in terms of generalized coordinates. The velocity of a particle in the xi direction is vi and by definition
vi ≡ dxi
dt .
But xi is a function of the qs, so using the chain rule of differentiation we have
vi =
∑3
k=1
∂ xi
∂ qk
d qk
dt + ∂xi
∂t = ∑
k
∂ xi
∂ qk
q ̇k + ∂xi
∂t . (1.6)
We now obtain a very useful relation involving the partial derivative of vi with respect to q ̇j . Since a mixed second-order partial derivative does not depend on the order in which the derivatives are taken, we can write
∂
∂ q ̇j
∂ xi
∂t = ∂
∂t
∂ xi
∂ q ̇j
.
But ∂xi
∂q ̇j = 0 because x does not depend on the generalized velocity q ̇. Let us
take the partial derivative of vi with respect to q ̇j using Equation (1.6). This yields
∂ vi
∂ q ̇j
=∂
∂ q ̇j
∑
k
∂ xi
∂ qk
q ̇k + ∂
∂ q ̇j
∂ xi
∂t
=∂
∂ q ̇j
∑
k
∂ xi
∂ qk
q ̇k + 0
=∑
k
(∂
∂ q ̇j
∂ xi
∂ qk
)
q ̇k + ∑
k
∂ xi
∂ qk
∂ q ̇k
∂ q ̇j
=0+∑
k
∂ xi
∂ qk
δij
= ∂xi
∂ qj
.
Thus, we conclude that
∂ vi
∂ q ̇j
= ∂xi
∂ qj
. (1.7)
This simple relationship comes in very handy in many derivations concerning generalized coordinates. Fortunately, it is easy to remember because it states that the Cartesian velocity is related to the generalized velocity in the same way as the Cartesian coordinate is related to the generalized coordinate.


1.4 Constraints 9
1.4 Constraints
Every physical system has a particular number of degrees of freedom. The number of degrees of freedom is the number of independent coordinates needed to completely specify the position of every part of the system. To describe the position of a free particle one must specify the values of three coordinates (say, x, y and z). Thus a free particle has three degrees of freedom. For a system of two free particles you need to specify the positions of both particles. Each particle has three degrees of freedom, so the system as a whole has six degrees of freedom. In general a mechanical system consisting of N free particles will have 3N degrees of freedom.
When a system is acted upon by forces of constraint, it is often possible to reduce the number of coordinates required to describe the motion. Thus, for example, the motion of a particle on a table can be described in terms of x and y, the constraint being that the z coordinate (defined to be perpendicular to the table) is given by z = constant. The motion of a particle on the surface of a sphere can be described in terms of two angles, and the constraint itself can be expressed by r = constant. Each constraint reduces by one the number of degrees of freedom.
In many problems the system is constrained in some way. Examples are a marble rolling on the surface of a table or a bead sliding along a wire. In these problems the particle is not completely free. There are forces acting on it which restrict its motion. If a hockey puck is sliding on smooth ice, the gravitational force is acting downward and the normal force is acting upward. If the puck leaves the surface of the ice, the normal force ceases to act and the gravitational force quickly brings it back down to the surface. At the surface, the normal force prevents the puck from continuing to move downward in the vertical direction. The force exerted by the surface on the particle is called a force of constraint.
If a system of N particles is acted upon by k constraints, then the number of generalized coordinates needed to describe the motion is 3N − k. (The number of Cartesian coordinates is always 3N, but the number of (independent) generalized coordinates is 3N − k.)
A constraint is a relationship between the coordinates. For example, if a particle is constrained to the surface of the paraboloid formed by rotating a parabola about the z axis, then the coordinates of the particle are related by z − x2/a − y2/b = 0. Similarly, a particle constrained to the surface of a sphere has coordinates that are related by x2 + y2 + z2 − a2 = 0.


10 1 Fundamental concepts
If the equation of constraint (the relationship between the coordinates) can be expressed in the form
f (q1, q2, . . . , qn, t) = 0, (1.8)
then the constraint is called holonomic.1 The key elements in the definition of a holonomic constraint are: (1) the equals sign and (2) it is a relation involving the coordinates. For example, a possible constraint is that a particle is always outside of a sphere of radius a. This constraint could be expressed as r ≥ a. This is not a holonomic constraint. Sometimes a constraint involves not just coordinates but also velocities or differentials of the coordinates. Such constraints are also not holonomic. As an example of a non-holonomic constraint consider a marble rolling on a perfectly rough table. The marble requires five coordinates to completely describe its position and orientation, two linear coordinates to give its position on the table top and three angle coordinates to describe its orientation. If the table top is perfectly smooth, the marble can slip and there is no relationship between the linear and angular coordinates. If, however, the table top is rough there are relations (constraints) between the angle coordinates and the linear coordinates. These constraints will have the general form dr = adθ , which is a relation between differentials. If such differential expressions can be integrated, then the constraint becomes a relation between coordinates, and it is holonomic. But, in general, rolling on the surface of a plane does not lead to integrable relations. That is, in general, the rolling constraint is not holonomic because the rolling constraint is a relationship between differentials. The equation of constraint does not involve only coordinates. (But rolling in a straight line is integrable and hence holonomic.) A holonomic constraint is an equation of the form of (1.8) relating the generalized coordinates, and it can be used to express one coordinate in terms of the others, thus reducing the number of coordinates required to describe the motion.
Exercise 1.5 A bead slides on a wire that is moving through space in a complicated way. Is the constraint holonomic? Is it scleronomous?
1 A constraint that does not contain the time explicitly is called scleronomous. Thus x2 + y2 + z2 − a2 = 0 is both holonomic and scleronomous.


1.5 Virtual displacements 11
Exercise 1.6 Draw a picture showing that a marble can roll on a tabletop and return to its initial position with a different orientation, but that there is a one-to-one relation between angle and position for rolling in a straight line.
Exercise 1.7 Express the constraint for a particle moving on the surface of an ellipsoid.
Exercise 1.8 Consider a diatomic molecule. Assume the atoms are point masses. The molecule can rotate and it can vibrate along the line joining the atoms. How many rotation axes does it have? (We do not include operations in which the final state cannot be distinguished from the initial state.) How many degrees of freedom does the molecule have? (Answer: 6.)
Exercise 1.9 At room temperature the vibrational degrees of freedom of O2 are “frozen out.” How many degrees of freedom does the oxygen molecule have at room temperature? (Answer: 5.)
1.5 Virtual displacements
A virtual displacement, δxi, is defined as an infinitesimal, instantaneous displacement of the coordinate xi, consistent with any constraints acting on the system. To appreciate the difference between an ordinary displacement dxi and a virtual displacement δxi, consider the transformation equations
xi = xi(q1, q2, . . . , qn, t) i = 1, 2, . . . , 3N.
Taking the differential of the transformation equations we obtain
dxi = ∂xi
∂ q1
dq1 + ∂xi
∂ q2
dq2 + · · · + ∂xi
∂ qn
dqn + ∂xi
∂t dt
=
∑n
α=1
∂ xi
∂ qα
dqα + ∂xi
∂t dt.
But for a virtual displacement δxi, time is frozen, so
δxi =
n ∑
α=1
∂ xi
∂ qα
dqα. (1.9)


12 1 Fundamental concepts
1.6 Virtual work and generalized force
Suppose a system is subjected to a number of applied forces. Let Fi be the force component along xi. (Thus, for a two-particle system, F5 would be the force in the y direction acting on particle 2.) Allowing all the Cartesian coordinates to undergo virtual displacements δxi, the virtual work performed by the applied forces will be
δW =
3∑N
i=1
Fi δxi .
Substituting for δxi from Equation (1.9)
δW =
3∑N
i=1
Fi
( 3∑N
α=1
∂ xi
∂ qα
δqα
)
.
Interchanging the order of the summations
δW =
3∑N
α=1
( 3∑N
i=1
Fi
∂ xi
∂ qα
)
δqα .
Harkening back to the elementary concept of “work equals force times distance” we define the generalized force as
Qα =
3∑N
i=1
Fi
∂ xi
∂ qα
. (1.10)
Then the virtual work can be expressed as
δW =
3∑N
α=1
Qα δqα .
Equation (1.10) is the definition of generalized force.
Example 1.1 Show that the principle of virtual work (δW = 0 in equilibrium) implies that the sum of the torques acting on the body must be zero.
Solution 1.1 Select an axis of rotation in an arbitrary direction . Let the body rotate about  through an infinitesimal angle . The virtual displacement of a point Pi located at Ri relative to the axis is given by
δRi =  × Ri .
The work done by a force Fi acting on Pi is
δWi = Fi · δRi = Fi · (  × Ri ) = · (Ri × Fi ) =  · Ni ,


1.7 Configuration space 13
where Ni is the torque about the axis acting on Pi. The total virtual work is
δW = ∑
i
 · Ni = · ∑
i
Ni =  · Ntot.
For a non-rotating body, the direction of  is arbitrary, so Ntot = 0.
Exercise 1.10 Consider a system made up of four particles described by Cartesian coordinates. What is F10?
Exercise 1.11 A particle is acted upon by a force with components Fx and Fy. Determine the generalized forces in polar coordinates. Answer: Qr = Fx cos θ + Fy sin θ and Qθ = −Fxr sin θ + Fyr cos θ .
Exercise 1.12 Consider the arrangement in Figure 1.2. The string is inextensible and there is no friction. Show by evaluating the virtual work that equilibrium requires that M = m/ sin θ . (This can be done trivially using elementary methods, but you are asked to solve the problem using virtual work.)
1.7 Configuration space
The specification of the positions of all of the particles in a system is called the configuration of the system. In general, the configuration of a system consisting of N free particles is given by
(q1, q2, . . . , qn), where n = 3N. (1.11)
Consider a single particle. Its position in Cartesian coordinates is given by x, y, z, and in a rectilinear three-dimensional reference frame the position of the particle is represented by a point. For a system consisting of two particles, you might represent the configuration of the system by two points. However, you could (at least conceptually) represent the configuration by a single point in a six-dimensional reference frame. In this 6-D space there would be six mutually perpendicular axes labeled, for example, x1, y1, z1, x2, y2, z2.
Similarly, the configuration of a system composed of N particles is represented by a single point in a 3N-dimensional reference frame.
m
M
q
Figure 1.2 Two blocks in equilibrium.


14 1 Fundamental concepts
The axes of such a multi-dimensional reference frame are not necessarily Cartesian. The positions of the various particles could be represented in spherical coordinates, or cylindrical coordinates, or the generalized coordinates, q1, q2, . . . , qn. Thus, the configuration of a system consisting of two particles is expressed as (q1, q2, . . . , q6) where q1 = x1, q2 = y1, . . . , q6 = z2 in Cartesian coordinates. In spherical coordinates, q1 = r1, q2 = θ 1, . . . , q6 = φ2, and similarly for any other coordinate system. Consider the three-dimensional reference frame of a single particle. As time goes on the values of the coordinates change. This change is continuous and the point that represents the position of the particle will trace out a smooth curve. Similarly, in the n-dimensional configuration space with axes representing the generalized coordinates, the point representing the configuration of the system will trace out a smooth curve which represents the time development of the entire system. See Figure 1.3. It might be noted that when we transform from one set of coordinates to a different set of coordinates, the character of the configuration space may change dramatically. Straight lines may become curves and angles and distances may change. Nevertheless, some geometrical properties remain unaltered. A point is still a point and a curve is still a curve. Adjacent curves remain adjacent and the neighborhood of a point remains the neighborhood of the point (although the shape of the neighborhood will probably be different).
Exercise 1.13 A particle moves along a line of constant r from θ 1 to θ 2. Show that this is a straight line in r–θ space but not in x–y space.
Exercise 1.14 The equation y = 3x + 2 describes a straight line in 2-D Cartesian space. Determine the shape of this curve in r–θ space.
q1
q2
q3
Figure 1.3 A path in configuration space. Note that time is a parameter giving the position of the configuration point along the curve.


1.9 Dynamics 15
1.8 Phase space
It is often convenient to describe a system of particles in terms of their positions and their momenta. The “state” of a system is described by giving the values of the positions and momenta of all the particles at some instant of time. For example we could represent the position of a particle by the Cartesian coordinates xi, i = 1, 2, 3. Similarly, the momentum of the particle can be represented by pi, i = 1, 2, 3. Imagine drawing a 2n-dimensional coordinate system in which the axes are labeled thus: x1, . . . , xn; p1, . . . , pn. The space described by this set of axes is called phase space. The positions of all the particles as well as the momenta of all the particles are represented by a single point in phase space. As time progresses the position and the momentum of each particle changes. These change continuously so the point in phase space moves continuously in the 2n coordinate system. That is, the time development of the system can be represented by a trajectory in phase space. This trajectory is called the “phase path” or, in some contexts, the “world line.”2 We will shortly define “generalized momentum” and then phase space will be described in terms of the generalized coordinates (q1, . . . , qn) and the generalized momenta (p1, . . . , pn).
1.9 Dynamics
1.9.1 Newton’s laws of motion
Dynamics is the study of the laws that determine the motion of material bodies. Your first introduction to dynamics was almost certainly a study of Isaac Newton’s three laws.
(1) The first law or the law of inertia: a body not subjected to any external force will move in a straight line at constant speed.
(2) The second law or the equation of motion: the rate of change of momentum of a body is equal to the net external force applied to it.
(3) The third law or action equals reaction: if a body exerts a force on a second body, the second body exerts and equal and opposite force of the same kind on the first one.
The first law tells us that a free particle will move at constant velocity.
2 The concept of phase space is very important in the study of chaotic systems where the intersection of the trajectory of the system with a particular plane in phase space (the so-called “surface of section” ) can be analyzed to determine if chaotic motion is taking place. Also in statistical mechanics, a basic theorem due to Liouville states that, if one plots the phase space trajectories of an ensemble of systems, the density of phase space points in the vicinity of a given system will remain constant in time.


16 1 Fundamental concepts
The second law is usually expressed by the vector relation
F = dp
dt . (1.12)
If the mass is constant this relationship reduces to the well known equation F = ma.
The third law can be expressed in either in the strong form or in the weak form. The strong form states that the forces are equal and opposite and directed along the line joining the particles. The weak form of the third law only states that the forces are equal and opposite. These laws (especially the second law) are extremely useful for solving practical problems.
1.9.2 The equation of motion
The second law is often used to determine the acceleration of a body subjected to a variety of unbalanced forces. The equation for the acceleration in terms of positions and velocities is called the “equation of motion.” (In introductory physics courses the determination of the acceleration involved isolating the system, drawing a free body diagram, and applying the second law, usually in the form F = ma.) Since forces can be expressed in terms of positions, velocities and time, the second law yields an expression for the acceleration having a form such as
x ̈ = x ̈(x, x ̇, t).
This is the equation of motion. The position of a particle as a function of time can then be determined by integrating the equation of motion twice, generating
x = x(t).
The procedure is called determining the motion. Newton’s laws are simple and intuitive and form the basis for most introductory courses in mechanics. It will be convenient to occasionally refer to them, but our study is not based on Newton’s laws.
1.9.3 Newton and Leibniz
It is well known that Isaac Newton and Gottfried Leibniz both invented the calculus independently. It is less well known that they had different notions concerning the time development of a system of particles. Newton’s second law gives us a vector relationship between the force on a particle and its


1.9 Dynamics 17
acceleration. (For a system of N particles, we have N vector relations corresponding to 3N scalar second-order differential equations. These equations are often coupled in the sense that the positions of all N particles may be present in each of the 3N equations of motion. To determine the motion we must solve all these coupled equations simultaneously.) On the other hand, Leibniz believed that the motions of particles could be better analyzed by considering their vis-viva or (as we would call it today) their kinetic energy. Essentially, Newton considered that the quantity mivi was conserved during the motion, whereas Leibniz believed that miv2
i was constant. In modern terminology, Newton believed in the conservation of momentum whereas Leibniz believed in the conservation of kinetic energy. It soon became clear that kinetic energy (T ) is not conserved during the motion of a system, but when the concept of potential energy (V ) was introduced, Leibniz’s theory was extended to state that the total energy (E = T + V ) is constant, which is, of course, one of the basic conservation laws of mechanics. Thus, you might say, both Newton and Leibniz were correct. In your introductory physics course you often switched between using Newton’s laws and using the conservation of energy. For example, the problem of a falling object can be solved using F = ma or by applying the conservation of energy. Later, Euler and Lagrange elaborated on Leibniz’s idea and showed that the motion of a system can be predicted from a single unifying principle, which we now call Hamilton’s principle. Interestingly, this approach does not involve vectors, or even forces, although forces can be obtained from it. The approach of Euler and Lagrange is referred to as “analytical mechanics” and is the primary focus of our study. (Since Newton’s approach is very familiar to you, we shall occasionally use it when considering some aspect of a physical problem.) Analytical mechanics does not require Newton’s three laws and, in particular, does away with the third law. Some physicists have suggested that Newton introduced the third law as a way of dealing with constraints. (A bouncing ball is “constrained” to remain inside the room. In the Newtonian picture, the ball bounces off a wall because the wall exerts a reaction force on it, as given by the third law.) The Lagrangian method incorporates constraints into the framework of the analysis, and neither the second nor the third law is necessary. However, the forces of constraint as well as the equation of motion can be determined from the method. Furthermore, all of the equations in analytical mechanics are scalar equations, so we do not need to use any concepts from vector analysis. However, sometimes vectors are convenient and we shall use them occasionally.


18 1 Fundamental concepts
This book is concerned with developing the concepts and techniques of analytical mechanics using variational principles.3 I am sure you will find this to be a very beautiful and very powerful theory. In the introduction to a well known book on the subject,4 the author, speaking about himself, says, “Again and again he experienced the extraordinary elation of mind which accompanies a preoccupation with the basic principles and methods of analytical mechanics.” (You may not feel an “elation of mind”, but I think you will appreciate what the author was expressing.)
1.10 Obtaining the equation of motion
In this section we review the methods for obtaining and solving the equation of motion. As you might expect, solving the equation of motion yields an expression for the motion, that is, for the position as a function of time. It happens that there are several different ways of expressing the equation of motion, but for now let us think of it as an expression for the acceleration of a particle in terms of the positions and velocities of all the other particles comprising the system. As pointed out by Landau and Lifshitz,5 “If all the coordinates and velocities are simultaneously specified, it is known from experience that the state of the system is completely determined and that its subsequent motion can, in principle, be calculated. Mathematically this means that if all the coordinates
q and velocities q ̇ are given at some instant, the accelerations .q. at that instant are uniquely defined.” In other words, the equations of motion can be expressed as relations of the type
.q.i = .q.i (q1, q2, . . . , qn; q. 1, q. 2, . . . , q. n; t).
If we know the accelerations (.q.i) of the particles, then we can (in principle) determine the positions and velocities at a subsequent time. Thus, a knowledge of the equations of motion allows us to predict the time development of a system.
3 Be aware that the fields of analytical mechanics and the calculus of variations are vast and this book is limited to presenting some fundamental concepts. 4 Cornelius Lanczos, The Variational Principles of Mechanics, The University of Toronto Press, 1970. Reprinted by Dover Press, New York, 1986.
5 L. D. Landau and E. M. Lifshitz, Mechanics, Vol 1 of A Course of Theoretical Physics, Pergamon Press, Oxford, 1976, p.1.


1.10 Obtaining the equation of motion 19
1.10.1 The equation of motion in Newtonian mechanics
If the masses are constant, an elementary way to obtain the equation of motion is to use Newton’s second law in the form
r ̈ = F/m. (1.13)
This is a second-order differential equation for r that can, in principle, be solved to yield r = r(t). Of course, this requires having an expression for the force as a function of r, r ̇, and t. As a simple one-dimensional example, consider a mass m connected to a spring of constant k. The force exerted by the spring on the mass is F = −kx. Therefore, Newton’s second law yields the following equation of motion
mx ̈ + kx = 0. (1.14)
It is important to note that the second law is applicable only in inertial coordinate systems. Newton was aware of this problem and he stated that the second law is a relationship that holds in a coordinate system at rest with respect to the fixed stars.
1.10.2 The equation of motion in Lagrangian mechanics
Another way to obtain the equation of motion is to use the Lagrangian technique. When dealing with particles or with rigid bodies that can be treated as particles, the Lagrangian can be defined to be the difference between the kinetic energy and the potential energy.6 That is,
L = T − V . (1.15)
For example, if a mass m is connected to a spring of constant k, the potential energy is V = 1
2 kx2 and the kinetic energy is T = 1
2 mv2 = 1
2 mx ̇2. Therefore the Lagrangian is
L=T −V = 1
2 mx ̇2 − 1
2 kx2.
It is usually easy to express the potential energy in whatever set of coordinates being used, but the expression for the kinetic energy may be somewhat difficult to determine, so whenever possible, one should start with Cartesian
6 Although Equation (1.15) is pefectly correct for a system of particles, we shall obtain a somewhat different expression when considering continuous systems in Chapter 7. In general, the Lagrangian is defined to be a function that generates the equations of motion.


20 1 Fundamental concepts
coordinates. In the Cartesian coordinate system the kinetic energy takes on the particularly simple form of a sum of the velocities squared. That is,
T =1
2 m(x ̇2 + y ̇2 + z ̇2).
To express the kinetic energy in terms of some other coordinate system requires a set of transformation equations. For example, for a pendulum of length l, the potential energy is V = −mgl cos θ and the kinetic energy is T = 1
2 mv2 = 1
2 m(lθ ̇ )2. (Here θ is the angle between the string and the perpendicular.) Therefore the Lagrangian is
L=T −V = 1
2 ml2θ ̇ 2 + mgl cos θ .
The kinetic energy is a function of the velocity. (You are familiar with
expressions such as T = 1
2 mx ̇2 and T = 1
2 ml2θ ̇ 2.) The velocity is the time
derivative of a position coordinate (x ̇ and lθ ̇ ). The potential energy is usually a function only of the position. In generalized coordinates, the position is expressed as q and the velocity as q ̇. The Lagrangian is, therefore, a function of q and q ̇. That is, L = L(q, q ̇), or more generally, L = L(q, q ̇, t). If we are concerned with a single particle free to move in three dimensions, there are three qs and three q ̇s. Then L = L(q1, q2, q3, q ̇1, q ̇2, q ̇3, t). For a system of N particles, if n = 3N we write
L = L(qi, q ̇i, t); i = 1, . . . , n.
I assume you were introduced to the Lagrangian and Lagrange’s equations in your course on intermediate mechanics. In the next two chapters you will find derivations of Lagrange’s equations from first principles. However, for the moment, I will simply express the equation without proof and show you how to use it as a tool for obtaining the equation of motion. For a single coordinate q, Lagrange’s equation is
d
dt
∂L
∂q ̇ − ∂L
∂q = 0. (1.16)
If there are n coordinates, there are n Lagrange equations, namely,
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
= 0, i = 1, . . . , n. (1.17)
It is important to realize that Lagrange’s equations are the equations of motion of a system.


1.10 Obtaining the equation of motion 21
For example, for a mass on a spring the Lagrangian is L = T − V =
1
2 mx ̇2 − 1
2 kx2. Plugging this into the Lagrange equation we obtain
d
dt
(∂L
∂ x ̇
)
− ∂L
∂x = 0,
d
dt
∂
∂ x ̇
(1
2 mx ̇2 − 1
2 kx2
)
−∂
∂x
(1
2 mx ̇2 − 1
2 kx2
)
= 0,
d
dt (mx ̇) − kx = 0,
so
mx ̈ + kx = 0,
as expected. (Compare with Equation 1.14.) To illustrate the use of Lagrange’s equations to obtain the equations of motion, we consider several simple mechanical systems.
Example 1.2 Atwood’s machine: Figure 1.4 is a sketch of Atwood’s machine. It consists of masses m1 and m2 suspended by a massless inextensible string over a frictionless, massless pulley. Evaluate the Lagrangian and obtain the equation of motion.
Solution 1.2 The kinetic energy of the masses is
T=1
2 m1x ̇2
1+1
2 m2x ̇2
2,
and the potential energy is
V = −m1gx1 − m2gx2,
where we selected V = 0 at the center of the pulley. The system is subjected to the constraint x1 + x2 = l = constant. The Lagrangian is,
V=0
m2
m1
x1
x2 = l - x1
Figure 1.4 Atwood’s machine.


22 1 Fundamental concepts
L=T −V = 1
2 m1x ̇2
1+1
2 m2x ̇2
2 + m1gx1 + m2gx2.
But using x2 = l − x1 we can rewrite the Lagrangian in terms of a single variable,
L= 1
2 m1x ̇2
1+1
2 m2x ̇2
1 + m1gx1 + m2g(l − x1),
=1
2 (m1 + m2)x ̇2
1 + (m1 − m2)gx1 + m2gl.
The equation of motion is
d
dt
∂L
∂ x ̇1
− ∂L
∂ x1
= 0,
d
dt (m1 + m2)x ̇1 − (m1 − m2)g = 0,
x ̈1 = m1 − m2
m1 + m2
g.
Example 1.3 A cylinder of radius a and mass m rolls without slipping on a fixed cylinder of radius b. Evaluate the Lagrangian and obtain the equation of motion for the short period of time before the cylinders separate. See Figure 1.5
Solution 1.3 The kinetic energy of the smaller cylinder is the kinetic energy of translation of its center of mass plus the kinetic energy of rotation about its center of mass. That is,
T=1
2m
(
r ̇2 + r2θ ̇ 2)
+1
2 I φ ̇ 2,
where r is the distance between the centers of the cylinders and I = 1
2 ma2. The potential energy is V = mgr cos θ . Therefore
qb
a
f
V=0
Figure 1.5 A cylinder rolling on another cylinder.


1.10 Obtaining the equation of motion 23
L= 1
2 m(r ̇2 + r2θ ̇ 2) + 1
2 I φ ̇ 2 − mgr cos θ .
But there are two constraints, namely, r = a + b and aφ = bθ . Therefore r ̇ = 0 and φ ̇ = (b/a)θ ̇ . So,
L= 1
2 m(a + b)2θ ̇ 2 + 1
2
(1
2 ma2
)(b
a θ ̇
)2
− mg(a + b) cos θ
=1
2m
(
a2 + 2ab + 3
2 b2
)
θ ̇ 2 − mg(a + b) cos θ
and the equation of motion is
d
dt
∂L
∂q ̇ − ∂L
∂q = 0
or
m
(
a2 + 2ab + 3
2 b2
)
θ ̈ − mg(a + b) sin θ = 0.
Example 1.4 A bead of mass m slides on a massive frictionless hoop of radius a. The hoop is rotating at a constant angular speed ω about a vertical axis. See Figure 1.6. Determine the Lagrangian and the equation of motion.
Solution 1.4 If the hoop is massive enough, the motion of the bead will not affect the rotation rate of the hoop. Thus we ignore the energy of the hoop. The kinetic energy is the energy of the bead due to the rotation of the hoop as well as kinetic energy of the bead due to its motion on the hoop which is associated with changes in the angle θ . Therefore,
L= 1
2 ma2θ ̇ 2 + 1
2 ma2ω2 sin2 θ + mga cos θ .
w
qa
mg
m
Figure 1.6 A bead slides on a rotating hoop.


24 1 Fundamental concepts
V=0
f
q
x
y
z
m
Figure 1.7 A spherical pendulum.
The equation of motion is
ma2θ ̈ − ma2ω2 sin θ cos θ + mga sin θ = 0.
Example 1.5 A spherical pendulum consists of a bob of mass m suspended from an inextensible string of length l. Determine the Lagrangian and the equations of motion. See Figure 1.7. Note that θ, the polar angle, is measured from the positive z axis, and φ, the azimuthal angle, is measured from the x axis in the x–y plane.
Solution 1.5 The Lagrangian is
L= 1
2 m(x ̇2 + y ̇2 + z ̇2) − mgz.
But
x = l sin θ cos φ
y = l sin θ sin φ
z = l cos θ
so
L= 1
2 ml2(θ ̇ 2 + sin2 θ φ ̇ 2) − mgl cos θ .
The equations of motion are
d
dt
(
ml2θ ̇
)
− ml2φ ̇ 2 sin θ cos θ − mgl sin θ = 0,
d
dt (ml2 sin2 θ φ ̇ ) = 0.


1.11 Conservation laws and symmetry principles 25
l2
l1
m1
m2
Figure 1.8 A double planar pendulum.
Exercise 1.15 Obtain the Lagrangian and the equation of motion for an Atwood’s machine if the pulley has a moment of inertia I. Answer: a = g(m2 − m1)/(m1 + m2 + I /R2).
Exercise 1.16 Obtain the Lagrangian for a simple pendulum and for a double pendulum. (The double pendulum is illustrated in Figure 1.8. Assume the motion is planar.) Answer: For the double pendulum
L= 1
2 (m1 + m2)l2
1 θ ̇ 2
1+ 1
2 m2
(
l2
2 θ ̇ 2
2 + 2l1l2θ ̇ 1θ ̇ 2 cos(θ 1 − θ 2)
)
+ m1gl1 cos θ 1 + m2g(l1 cos θ 1 + l2 cos θ 2).
Exercise 1.17 A flyball governor is a simple mechanical device to control the speed of a motor. As the device spins faster and faster, the mass m2 rises. The rising mass controls the fuel supply to the motor. See Figure 1.9. The angle θ is formed by one of the rods and the central axis, and the angle φ is the rotation angle about the axis. The angular speed at any moment is ω = φ ̇ . Assume the system is in a uniform gravitational field. Show that the Lagrangian is given by
L = m1a2 (θ ̇ 2 + ω2 sin2 θ
)
+ 2m2a2θ ̇ 2 sin2 θ + 2 (m1 + m2) ga cos θ .
1.11 Conservation laws and symmetry principles
The three most important conservation laws in classical mechanics are the conservation of linear momentum, the conservation of angular momentum and the conservation of energy. Although you are very familiar with these laws,


26 1 Fundamental concepts
w
V=0
q
aa
aa
m1 m1
m2
Figure 1.9 A flyball governor.
it is interesting to consider them in terms of generalized coordinates and the Lagrangian. We will show that these three basic conservation laws are consequences of the homogeneity and isotropy of space and the homogeneity of time. When we say a region of space is homogeneous we mean that it is the same in one location as in another. Consequently, a system will be unaffected by a displacement from one point to another. As a simple example, I claim that my grandfather clock behaved the same on one side of the room as on the other (but it would behave differently on the Moon). Similarly, if space is isotropic, it looks the same in all directions and a physical system is unchanged under a rotation. My grandfather clock behaved the same after I rotated it through 90◦ about a vertical axis. However, the space in my room is not isotropic for a rotation about a horizontal axis. (When I turned my grandfather clock upside-down it stopped working.) Finally, if time is homogeneous, a physical system will behave the same at two different times. My clock behaved the same today as it did last week. Time is also isotropic in the sense that physical systems behave the same whether time is running forward or backward. When we say that a system has a particular symmetry, we mean that the system is invariant with respect to a change in some particular coordinate. For example, a sphere is an object that appears the same no matter how it is rotated. Spherical symmetry implies an invariance with respect to rotations, i.e., with respect to a change in the angular orientation of the object. Similarly, if a system is unchanged when it undergoes a displacement in some particular


1.11 Conservation laws and symmetry principles 27
direction, we say that it has translational symmetry in that direction. When a system is invariant with respect to a change in time, we say it has temporal symmetry. As an example of translational symmetry consider an object sitting on an infinite horizontal surface in a uniform, vertical gravitational field. If the object is moved in the horizontal plane, the system is unchanged. If, however, the object is moved vertically (if it is raised), the potential energy changes. Thus this system has translational symmetry in the horizontal plane but it does not have translational symmetry for vertical displacements. In a region where no forces are acting, space is homogeneous in all three directions. A mechanical system is completely described by its Lagrangian. If the Lagrangian does not depend explicitly on some particular coordinate qi, then a change in qi does not affect the system. The system is said to be symmetric with respect to changes in qi.
1.11.1 Generalized momentum and cyclic coordinates
Generalized momentum A free particle is described by the Lagrangian
L= 1
2 m(x ̇2 + y ̇2 + z ̇2).
Taking the partial derivative with respect to x ̇ we obtain
∂L
∂x ̇ = mx ̇.
But mx ̇ is just the x component of the linear momentum. That is,
px = ∂L
∂x ̇ . (1.18)
Similarly,
py = ∂L
∂y ̇ pz = ∂L
∂z ̇ .
The Lagrangian for a freely rotating wheel with moment of inertia I is
L= 1
2 I θ ̇ 2
and
∂L
∂θ ̇ = I θ ̇.
But I θ ̇ is the angular momentum.


28 1 Fundamental concepts
In both of these examples, the momentum was the derivative of the Lagrangian with respect to a velocity. We carry this idea to its logical conclusion and define the generalized momentum by
pi = ∂L
∂ q ̇i
.
The generalized momentum pi is associated with the generalized coordinate qi and is sometimes referred to as the conjugate momentum. Thus, we have seen that the linear momentum px is conjugate to the linear coordinate x and the angular momentum I θ ̇ is conjugate to the angular coordinate θ .
Cyclic coordinates
If a particular coordinate does not appear in the Lagrangian, it is called “cyclic” or “ignorable.” For example, the Lagrangian for a mass point in a gravitational field is
L= 1
2 m(x ̇2 + y ̇2 + z ̇2) − mgz.
Since neither x nor y appear in the Lagrangian, they are cyclic. (Note, however, that the time derivatives of the coordinates, x ̇ and y ̇, do appear in the Lagrangian, so there is an implicit dependence on x and y.) Lagrange’s equation tells us that
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
= 0.
But if qi is cyclic, the second term is zero and
d
dt
∂L
∂ q ̇i
= 0.
Since
pi = ∂L
∂ q ̇i
we have
d
dt pi = 0
or
pi = constant.
That is, the generalized momentum conjugate to a cyclic coordinate is a constant.


1.11 Conservation laws and symmetry principles 29
Exercise 1.18 Identify any conserved momenta for the spherical pendulum.
Exercise 1.19 Write the Lagrangian for a planet orbiting the Sun. Determine any cyclic coordinates and identify the conserved conjugate
momenta. Answer: L = (1/2)(mr ̇2 + mr2θ ̇ 2) + GmM/r2.
Exercise 1.20 Show that as long as the kinetic energy depends only on velocities and not on coordinates, then the generalized force is given by Qi = ∂L
∂qi .
The relation between symmetries and conserved quantities
We have determined that the generalized momentum conjugate to an ignorable (cyclic) coordinate is conserved. If the coordinate qi is cyclic, it does not appear in the Lagrangian and the system does not depend on the value of qi. Changing qi has no effect on the system; it is the same after qi is changed as it was before. If qi = θ, then a rotation through θ changes nothing. But that is, essentially, the definition of symmetry. Consequently, if a coordinate is cyclic, the system is symmetric under changes in that coordinate. Furthermore, each symmetry in a coordinate gives rise to a conserved conjugate momentum. For example, for an object sitting on a horizontal surface, the horizontal components of linear momentum are constant. If the x and y axes are in the horizontal plane, then there is no force in either the x or y directions. If there were a force in, say, the x direction, then the potential energy would depend on x and consequently the Lagrangian would contain x and it would not be an ignorable coordinate. The idea that conserved quantities are related to symmetries is highly exploited in studies of the elementary particles. For example, particle physicists frequently utilize the law of conservation of parity and the law of conservation of “strangeness.” These conservation laws are associated with observed symmetries in the behavior of elementary particles. Parity conservation is an expression of the symmetry of right-handedness and left-handedness in reactions involving elementary particles. (As you probably know, there are some reactions in which parity is not conserved.) As an example from the field of elementary particle physics, it is observed that some reactions never occur. There is no apparent reason why a reaction such as π − + p → π o +  does not take place. It must violate some conservation law. However, it does not violate any of the everyday conservation laws such as charge, mass-energy, parity, etc. But since the reaction does not


30 1 Fundamental concepts
occur, using the principle that “what is not forbidden is required,” physicists concluded that there must be a conservation principle at work. They called it conservation of strangeness. Studying reactions that do occur allowed them to assign to each particle a “strangeness quantum number.” For example, the strangeness of a π particle is 0, that of a proton is also 0 and the strangeness of a  particle is −1. The total strangeness on the right-hand side of the reaction is not equal to the total strangeness on the left-hand side. Therefore, in this reaction, strangeness is not conserved and the reaction does not take place. This is not any weirder than requiring that linear momentum be conserved during a collision. However, we feel at home with momentum conservation because we have an analytical expression for momentum and we known that conservation of momentum implies that no net external forces are acting on the system. We do not have an analytical expression for strangeness nor do we know what symmetry is implied by strangeness conservation. We have no idea what the associated cyclic coordinate might be. None the less, the basic idea is the same. In conclusion, we can state that for every symmetry, there is a corresponding constant of the motion. This is, essentially, Noether’s theorem.7
Exercise 1.21 Find the ignorable coordinates for the spherical pendulum and determine the associated symmetry.
1.11.2 The conservation of linear momentum
In this section we will study the theoretical basis for the law of conservation of linear momentum. However, before we begin, let us note that there are several different ways to arrive at this conservation law. For example, in your introductory physics course, you learned the law of conservation of linear momentum by considering Newton’s second law in the form
F = dP
dt .
If there is no net external force acting on the system, F = 0, and hence the time derivative of the total momentum P is zero, that is, the total linear momentum of the system is constant. In your intermediate mechanics course, you probably considered the conservation of linear momentum from a more sophisticated point of view. Perhaps
7 Emmy Noether (1882–1932), a mathematical physicist, proved the theorem that is named after her.


1.11 Conservation laws and symmetry principles 31
you considered a situation in which the potential energy did not depend on some particular coordinate, say x, and the kinetic energy also did not contain x. When you wrote the Lagrangian, L = T − V , you obtained an expression that did not contain x and consequently x was a cyclic coordinate. That is,
∂L
∂x = 0.
Now the Lagrange equation for x is
d
dt
∂L
∂x ̇ − ∂L
∂x = 0,
which reduces to
d
dt
∂L
∂x ̇ = 0.
But ∂L
∂x ̇ = px , so
d
dt px = 0.
Again we see that if the Lagrangian does not depend on x, then px is constant. We now show that the conservation of linear momentum is a consequence of the homogeneity of space, that is, we shall show that in a region of space whose properties are the same everywhere, the total linear momentum of a mechanical system will be constant. Our argument not only guarantees the conservation of linear momentum, but also leads to Newton’s second and third laws, as we now demonstrate. Consider a system composed of N particles situated at rα (α = 1, . . . , N ). (For the sake of variety, the argument here is formulated in terms of vectors.) If every particle is displaced by the same infinitesimal distance , the positions of all particles will change according to rα → rα + . We assume that  is a virtual displacement in which the all the particles are displaced infinitesimally but the velocities are unchanged and time is frozen. Recall that if f = f (qi, q ̇i, t), then the differential of f, according to the rules of calculus, is
df =
n∑
i=1
∂f
∂ qi
dqi +
∑n
i=1
∂f
∂ q ̇i
dq ̇i + ∂f
∂t dt.
But the change in f due to an infinitesimal virtual displacement is
δf = ∂f
∂ q1
δq1 + ∂f
∂ q2
δq2 + · · ·
=
N ∑
α=1
∂f
∂ ra
· δrα,


32 1 Fundamental concepts
where we have used the fact that time is frozen and that a displacement of the entire system will have no effect on velocities.8 Consequently, the change in a Lagrangian due to a virtual displacement  is
δL = ∑
α
∂L
∂ rα
· δrα = ∑
α
∂L
∂ rα
·  = · ∑
α
∂L
∂ rα
,
where we used that fact that δrα = , that is, all particles are displaced the same amount. Note that we are summing over particles (α = 1, . . . , N ), not components (i = 1, . . . , 3N ). If the space is homogeneous, the Lagrangian does not change and δL = 0. Therefore,
∑
α
∂L
∂ rα
= 0.
But by Lagrange’s equations,
∂L
∂ rα
=d
dt
∂L
∂ vα
,
so
d
dt
∑
α
∂L
∂ vα
=d
dt
∑
α
pα = d
dt Ptot = 0.
This means that
Ptot = constant,
as expected. We are not using any particular set of coordinates, as we have been formulating the problem in terms of vectors. The vector definition of kinetic energy is T = (1/2)mv · v. As long as we express our relations in terms of vectors
8 We are using the notation of Landau and Lifshitz involving the derivative of a scalar with respect to a vector. For example, if there is only one particle (α = 1), we define
∂F
∂r ≡ ˆi ∂F
∂x + jˆ ∂F
∂y + kˆ ∂F
∂z .
Thus, the derivative of a scalar with respect to a vector is defined as the vector whose components are the derivatives of the scalar with respect to the components of the vector. Note that
∂F
∂r = ∇F.


1.11 Conservation laws and symmetry principles 33
the kinetic energy depends only on the square of the velocity and not on the coordinates9 and we can write
∂L
∂ rα
= − ∂V
∂ rα
= Fα.
But if
∑
α
∂L
∂ rα
= 0,
then
∑
α
Fα = 0.
That is, the sum of all the forces acting on all the particles in the system is zero. If the system consists of just two particles, this tells us that F1 + F2 = 0, which is Newton’s third law. Additionally,
∂L
∂ rα
=d
dt
∂L
∂ vα
=d
dt pα = Fα,
and we have also obtained Newton’s second law.
1.11.3 The conservation of angular momentum
Our next task is to show that the conservation of angular momentum is a consequence of the isotropy of space. But before we tackle that problem, let us consider the conservation law first from an elementary and then from an intermediate point of view. The elementary level consideration of the law of conservation of angular momentum usually starts with the definition of the angular momentum of a particle:
l = r × p,
where r is the position of the particle and p =mv is its linear momentum. Differentiating with respect to time yields
dl
dt = d
dt (r × p) = dr
dt × p + r× dp
dt .
9 This is also true in Cartesian coordinates in which T = (1/2)m
(
x ̇2 + y ̇2 + z ̇2)
. But in most
other coordinate systems the kinetic energy will depend on coordinates as well as velocities. For example, in spherical coordinates
T = (1/2)m(r ̇2 + r2θ ̇ 2 + r2 sin2 θ φ ̇ 2).


34 1 Fundamental concepts
Since dr
dt = v, the first term is m(v × v) = 0. The second term contains dp
dt
which, by Newton’s second law, is the net force acting on the particle. Therefore the second term is r × F = N = torque. That is
dl
dt = N.
One can then show that the law holds for a system of particles (which requires invoking Newton’s third law), and one concludes that the angular momentum of a system of particles is constant if there is no net external external torque acting on it. A familiar example is the constancy of angular momentum of a particle in a central force field (such as a planet under the influence of the Sun). Since a central force has the form F = f (r)ˆr, the torque acting on the particle is
N = r × F =rf (r)(ˆr × ˆr) = 0,
and the conjecture is proved. An intermediate level discussion of the conservation of angular momentum might begin by considering a particle moving in a plane in a region of space where the potential energy is V = V (r). (For example, V = −GMm/r.) The transformation equations from Cartesian to polar coordinates are
x = r cos θ
y = r sin θ.
Therefore,
x ̇ = r ̇ cos θ − rθ ̇ sin θ
y ̇ = r ̇ sin θ + rθ ̇ cos θ ,
and
T =1
2m
(
x ̇2 + y ̇2)
=1
2m
(
r ̇2 + r2θ ̇ 2)
so
L=T −V = 1
2m
(
r ̇2 + r2θ ̇ 2)
− V (r).
The momentum conjugate to θ is
pθ = ∂L
∂θ ̇ = ∂
∂ θ ̇
[1
2m
(
r ̇2 + r2θ ̇ 2)
− V (r)
]
= mr2θ ̇ ,
which we recognize as the angular momentum of the particle. Observe that θ is ignorable, so the angular momentum, mr2θ ̇ is constant.


1.11 Conservation laws and symmetry principles 35
q
df
df
ra sin qdf = |dra|
ra
ra (after rotation)
From above
df
va (after)
va (before)
dra = df × ra
ra sin q
Figure 1.10 The rotation of the system changes the position of each particle by δrα and the velocity of each particle by δvα.
We now address this question in a more advanced manner and show that the conservation of angular momentum is a consequence of the isotropy of space. Consider a mechanical system that is rotated through a virtual angle δφ around some axis. Let δφ be a vector directed along the axis of rotation and having magnitude δφ. Place the origin of coordinates on the axis of rotation. Owing to the rotation, every particle in the system is displaced by some distance δrα and if a particle has velocity vα, the direction of the velocity vector will change by δvα. Figure 1.10 shows the effect of the rotation on the position vector rα. If the space is isotropic, the rotation will have no effect on the Lagrangian so δL = 0. That is,
0 = δL = ∑
α
( ∂L
∂ rα
· δrα + ∂L
∂ vα
· δvα
) .
(Note that in this situation the velocities of the particles are not constant.) Again, we use Lagrange’s equation to replace ∂L
∂rα with d
dt
∂L
∂vα = d
dt pα =  ̇pα and write
∑
α
(p ̇ α · δrα + pα · δvα
) = 0.
From the figure, |δr| = r sin θ δφ. But since δr is perpendicular to both r and δφ, we can write δr = δφ × r. Similarly, δv = δφ × v. Therefore, ∑
α
(p ̇ α · δφ × rα + pα · δφ × vα
) = 0.
Interchanging the dot and the cross we obtain
∑
α
δφ·(rα ×  ̇pα +  ̇rα × pα
) = δφ· ∑
α
d
dt
(rα × pα
) = 0.


36 1 Fundamental concepts
Consequently, since δφ =0
∑
α
d
dt (rα × pα) = d
dt
∑
α
rα × pα = dL
dt = 0,
where L is the total angular momentum. That is, the isotropy of space leads to the conservation of angular momentum. (Be careful not to confuse the total vector angular momentum, L, with the Lagrangian L.)
Exercise 1.22 Using the elementary mechanics approach, show that the angular momentum of a system of particles is constant if no net external torque acts on the system. Note that you must invoke Newton’s third law in the strong form.
1.11.4 The conservation of energy and the work function
Consider the kinetic energy of a single particle. In introductory physics we learned the work–energy theorem, which states that the net work done on the particle by external forces will result in an equal increase in the kinetic energy of the particle. By definition, work is
W=
∫
F · ds.
Consequently, the increase in kinetic energy of a particle acted upon by a force F as it goes from point 1 to point 2 is
T2 − T1 =
∫2
1
F · ds.
If the force is conservative, the quantity F · ds is the differential of a scalar quantity denoted by U and called the “work function.” The concept of work function has been replaced in modern terminology by potential energy (V ) defined as the negative of the work function.
V = −U.
This means that if the force is conservative it can be expressed in terms of the gradient of a scalar function, thus
F = −∇V .
Then the work energy theorem yields
T2 − T1 =
∫2
1
−∇V ·ds.


1.11 Conservation laws and symmetry principles 37
But ∇V · ds = dV , so
T2 − T1 = −(V2 − V1),
or
T2 + V2 = T1 + V1.
This equation expresses the conservation of mechanical energy; it indicates that there is a quantity that remains constant as the system is taken from one configuration to another under the action of conservative forces. This quantity is, of course, the total mechanical energy E = T + V . The condition for a force to be conservative is that it is equal to the negative gradient of a scalar function. This is equivalent to requiring that the curl of the force be zero, i.e. ∇ × F = 0. The preceding discussion of the conservation of energy was fairly elementary. We now consider energy from an advanced point of view and show that the conservation of energy follows from the homogeneity of time. In general, the derivative of the Lagrangian L = L(q, q ̇, t) with respect to time is
dL
dt = ∑
i
∂L
∂ qi
d qi
dt + ∑
i
∂L
∂ q ̇i
d q ̇i
dt + ∂L
∂t .
By Lagrange’s equation, ∂L
∂qi = d
dt
∂L
∂q ̇i , so
dL
dt = ∑
i
[ q ̇i
d
dt
( ∂L
∂ q ̇i
)
+ ∂L
∂ q ̇i
d q ̇i
dt
]
+ ∂L
∂t .
But note that
d
dt
∑
i
q ̇i
∂L
∂ q ̇i
=∑
i
d
dt
( q ̇i
∂L
∂ q ̇i
)
=∑
i
[ q ̇i
d
dt
( ∂L
∂ q ̇i
)
+ dq ̇i
dt
∂L
∂ q ̇i
] ,
so the summation in the preceding equation can be replaced by d
dt
∑
i
q ̇i ∂L
∂q ̇i ,
and
dL
dt = d
dt
∑
i
q ̇i
∂L
∂ q ̇i
+ ∂L
∂t ,
or
∂L
∂t = d
dt
(
L−∑
i
q ̇i
∂L
∂ q ̇i
) .


38 1 Fundamental concepts
We now introduce a new quantity called the energy function h that is defined by
h = h(q, q ̇, t) = ∑
i
q ̇i
∂L
∂ q ̇i
− L,
so that
∂L
∂t = −dh
dt .
Note that h depends on the generalized positions and the generalized velocities. Let us assume temporal homogeneity so that time does not appear explicitly in the Lagrangian. Then ∂L
∂t = 0 and consequently
d
dt
(
L−∑
i
q ̇i
∂L
∂ q ̇i
)
= −dh
dt = 0.
That is, the energy function is constant. But does this imply that the energy (E = T + V ) is constant? To answer this question we note that the kinetic energy has the general form
T = T0 + T1 + T2,
where T0 does not depend on velocities, T1 is a homogeneous function of the first degree in velocities and T2 is a homogeneous function of second degree in velocities.10 (In Cartesian coordinates, T0 and T1 are both zero, and T2 is not only of second degree in velocities, but is actually a quadratic function of velocities, i.e. it contains x ̇2 but not x ̇y ̇.) Recall that a function f (x1, . . . , xn) is homogeneous of degree k if
f (αx1, αx2, . . . , αxn) = αkf (x1, . . . , xn).
Euler’s theorem on homogeneous functions states that, if f (xi) is homogenous of degree k, then
∑
i
xi
∂f
∂ xi
= kf.
Since the kinetic energy has the general form T0 + T1 + T2, the Lagrangian also has this form: L = L0 + L1 + L2. Then by Euler’s theorem,
10 This can be proved by starting with the kinetic energy in Cartesian coordinates and then using the transformation equations xi = xi (q1, q2, . . . , qn, t), and the fact that
x ̇i = ∑
j
∂∂qxji q ̇j + ∂∂xti . See Problem 1.12 at the end of the chapter.


1.11 Conservation laws and symmetry principles 39
∑
i
q ̇i
∂ L0
∂ q ̇i
=0
∑
i
q ̇i
∂ L1
∂ q ̇i
= L1
∑
i
q ̇i
∂ L2
∂ q ̇i
= 2L2.
Therefore,
∑
i
q ̇i
∂L
∂ q ̇i
=∑
i
q ̇i
∂(L0 + L1 + L2)
∂ q ̇i
= L1 + 2L2,
and
∑
i
q ̇i
∂L
∂ q ̇i
− L = (L1 + 2L2) − (L0 + L1 + L2) = L2 − L0.
If the transformation equations do not involve t explicitly, T = T2, that is, T is a homogeneous function of second degree in the velocities. And if V does not depend on velocities, L0 = −V . Then
h = T + V = E.
As long as these conditions are satisfied, the energy function is equal to the total energy and the total energy is conserved. If we replace ∂L
∂q ̇i by pi, the energy function is transformed into the Hamiltonian, which is a function of the generalized coordinates and the generalized momenta:
H = H (q, p, t) = ∑
i
q ̇i pi − L.
The energy function and the Hamiltonian are essentially the same thing, but they are expressed in terms of different variables, that is, h = h(q, q ̇, t) and H = H (q, p, t). (Whereas h depends on velocity, H depends on momentum.)
Exercise 1.23 (a) Show that F = −yiˆ+xjˆ is not conservative. (b) Show that F = yiˆ+xjˆ is conservative.
Exercise 1.24 Show that F = 3x2yiˆ + (x3 + 1)jˆ + 9z2 ˆk is conservative.
Exercise 1.25 Prove that the curl of a conservative force is zero.
Exercise 1.26 Prove Euler’s theorem. (Hint: Differentiate f (tx, ty, tz) = tkf (x, y, z) with respect to t, then let t = 1.)


40 1 Fundamental concepts
Exercise 1.27 Show that z2 ln(x/y) is homogeneous of degree 2.
Exercise 1.28 Show that ∑
i
q ̇i ∂L2
∂q ̇i = 2L2.
The virial theorem
The virial theorem states that for a bounded conservative system (such as a planet going around a star) the average value of the kinetic energy is proportional to the average value of the potential energy. (For the planet/star we find 〈T 〉 = −(1/2) 〈V 〉 .) We now prove the theorem. In Cartesian coordinates the kinetic energy is a homogenous quadratic function depending only on the velocities, so Euler’s theorem yields
∑
α
vα · ∂T
∂ vα
= 2T .
As long as the potential energy does not depend on velocities, the momentum can be expressed as
pα = ∂T
∂ vα
,
and hence
2T = ∑
α
pα · vα = ∑
α
pα · drα
dt .
But
d
dt
∑
α
pα · rα = ∑
α
pα · drα
dt + ∑
α
d pα
dt · rα.
Therefore
2T = d
dt
∑
α
pα · rα − ∑
α
d pα
dt · rα.
Now by Newton’s second law, for a conservative system,
d pα
dt = Fα = − ∂V
∂ rα
,
so
2T = d
dt
∑
α
pα · rα + ∑
α
∂V
∂ rα
· rα.
Let us evaluate the average value of all the terms in this equation:
〈2T 〉 =
〈d
dt
∑
α
pα · rα
〉 +
〈∑
α
rα · ∂V
∂ rα
〉 .


1.12 Problems 41
Recall that the time average of a function f (t) is
〈f (t)〉 = τli→m∞
1
τ
∫τ
0
f (t)dt,
s〈o d
dt
∑
α
pα · rα
〉
= τli→m∞
1
τ
∫τ
0
d
dt
∑
α
pα · rαdt = τli→m∞
1
τ
∫τ
0
d
(∑
α
pα · rα
)
= τli→m∞
1
τ
[∑
α
pα · rα
]τ
0
= τli→m∞
∑
α
pα · rα
∣∣∣∣τ
−∑
α
pα · rα
∣∣∣∣0
τ = 0,
where we assumed that ∑
α
pα · rα remains finite at all times. In other words,
we are assuming the motion is bounded, so the numerator is finite but the denominator goes to infinity. We are left with
2 〈T 〉 =
〈∑
α
rα · ∂V
∂ rα
〉 .
If the potential energy is a homogeneous function of position of degree k, then by Euler’s theorem the right-hand side is kV , and
2 〈T 〉 = k 〈V 〉 .
Thus, for example, for a particle of mass m in a gravitational field, V = − GMm
r so k = −1 and
2 〈T 〉 = − 〈V 〉 .
Since E = T + V , we see that 〈E〉 = − 〈T 〉, indicating that the total energy is negative, and that the magnitude of the average kinetic energy is one half the magnitude of the average potential energy. As another example, consider a mass on a spring. The potential energy is V = (1/2)Kx2 so k = 2 and 2 〈T 〉 = 2 〈V 〉, that is, 〈T 〉 = 〈V 〉 .
1.12 Problems
1.1 There is a story about a physicist who got a traffic ticket for running a red light. Being a clever person the physicist proved to the judge that it was neither possible to stop nor to continue without either running the light or breaking the speed limit. In this problem we determine if the story is credible or is just a figment of the imagination of an overworked graduate


42 1 Fundamental concepts
student. Assume the physicist is driving a car at a constant speed v0, equal to the speed limit. The car is a distance d from an intersection when the light changes from green to yellow. The physicist has a reaction time τ and the car brakes with a constant deceleration ac. The light remains yellow for a time ty. Determine the conditions such that the car can neither stop nor continue at v0 without running the red light. Also, using realistic values for τ , ty, ac and v0 determine whether or not this situation could actually arise. (Solve using simple kinematic relations.) 1.2 A cylinder of mass M and radius R is set on end on a table at a distance L from the edge, as shown in Figure 1.11. A string is wound tightly around the cylinder. The free end of the string passes over a frictionless pulley and hangs off the edge of the table. A weight of mass m is attached to the free end of the string. Determine the time required for the spool to reach the edge of the table.
m
L
M
Figure 1.11 A cylinder on a frictionless table.
1.3 A string passes over a massless pulley. Each end is wound around a vertical hoop, as shown in Figure 1.12. The hoops tend to descend, unwinding the string, but if one hoop is much more massive than the other, it can cause the lighter hoop to rise. The hoops have masses M1 and M2 and radii R1 and R2. Show that the tension in the string is τ = gM1M2 (M1 + M2)−1 .
M1
M2
Figure 1.12 Two hoops hang from a pulley. The string unwinds as the hoops descend.


1.12 Problems 43
1.4 The transformation equations for the bipolar coordinates η and ζ are
x = a sinh η
cosh η − cos ζ , y = a sin ζ
cosh η − cos ζ .
(a) Show that the inverse transformation exists. (b) Obtain expressions for η and ζ in terms of x and y. 1.5 A disk of mass m and radius a rolls down a perfectly rough inclined plane of angle α. Determine the equations of motion and the constraints acting on the disk. 1.6 A pendulum of length l and mass m is mounted on a cart of mass M that is free to roll along a track. There is no friction. Write the Lagrangian for the pendulum/cart system. 1.7 Starting with the definition
T =1
2
N ∑
i=1
m
(
x ̇ 2
i + y ̇2
i + z ̇2
i
)
and the transformation equations, show that
T=
3∑N
j =1
3∑N
k=1
1
2 Ajkq ̇j q ̇k +
3∑N
j =1
Bj q ̇j + T0,
where
Ajk =
N ∑
i=1
mi
( ∂xi
∂ qj
∂ xi
∂ qk
+ ∂yi
∂ qj
∂yi
∂ qk
+ ∂zi
∂ qj
∂ zi
∂ qk
) ,
Bj =
N ∑
i=1
mi
( ∂xi
∂ qj
∂ xi
∂t + ∂yi
∂ qj
∂yi
∂t + ∂zi
∂ qj
∂ zi
∂t
) ,
T0 =
N ∑
i=1
1
2 mi
[( ∂xi
∂t
)2
+
( ∂yi
∂t
)2
+
( ∂zi
∂t
)2] .


2
The calculus of variations
2.1 Introduction
The calculus of variations is a branch of mathematics which considers extremal problems; it yields techniques for determining when a particular definite integral will be a maximum or a minimum (or, more generally, the conditions for the integral to be “stationary”). The calculus of variations answers questions such as the following.
• What is the path that gives the shortest distance between two points in a plane? (A straight line.) • What is the path that gives the shortest distance between two points on a sphere? (A geodesic or “great circle.”) • What is the shape of the curve of given length that encloses the greatest area? (A circle.) • What is the shape of the region of space that encloses the greatest volume for a given surface area? (A sphere.)
The technique of the calculus of variations is to formulate the problem in terms of a definite integral, then to determine the conditions under which the integral will be maximized (or minimized). For example, consider two points (P1 and P2) in the x–y plane. These can be connected by an infinite number of paths, each described by a function of the form y = y(x). Suppose we wanted to determine the equation y = y(x) for the curve giving the shortest path between P1 and P2. To do so we note that the distance between the two end points is given by the definite integral
I=
∫ P2
P1
ds,
where ds is an infinitesimal distance along the curve. Our task is to determine the function y = y(x) such that I is minimized.
44


2.2 Derivation of the Euler–Lagrange equation 45
As an analogy, let us recollect the introductory calculus method of finding the maximum or minimum (that is, a “stationary point”) of some function. For example, consider an ordinary function, say h(x, y), which we will assume represents a topological map with h being the altitude. Imagine a mountain whose height is greater than the height of any other point on the map. That is, at some point (say x0, y0) the height h is an absolute maximum. How can we determine x0 and y0? As you know from elementary calculus, the technique is to find the point at which the partial derivatives of h with respect to x and y are zero. (Clearly, as we go through a maximum, the slope of a curve changes from positive to negative, so it must be zero at the maximum itself.) But if the derivative of h with respect to x is zero, this means that points that are infinitesimally distant from the peak along the x direction will have the same value of h as the peak itself! That is, if ∂h/∂x = 0, then there is no change in h for a small change in x. To resolve this conundrum, we note that near the peak, the function h(x, y) can be expanded in a Taylor’s series, thus,
h(x + dx, y + dy) = h(x0, y0) + dx ∂h
∂x
∣∣∣∣x0,y0
+ dy ∂h
∂y
∣∣∣∣x0,y0
+ second-order terms.
The derivatives ∂h/∂x and ∂h/∂x on the right-hand side are zero so in the infinitesimal neighborhood of a stationary point the value of the function is, to first order, equal to its value at the stationary point. To find a change in h we need to go to the second-order terms. These terms will also tell us the nature of the stationary point: if they are positive the stationary point is a minimum, if they are negative it is a maximum and if they are positive in one direction and negative in another, the stationary point is a saddle point. Of course, we are not dealing with the problem of finding a stationary point for a function, but rather for an integral. Furthermore, we are not varying the coordinates, but rather the path along which the integral is evaluated. Nevertheless, to find the stationary point for our integral we will use the concept that to first order the integral has the same value at all points in an infinitesimal neighborhood of the stationary point.
2.2 Derivation of the Euler–Lagrange equation
We now derive the basic equation of the calculus of variations. It is called the Euler–Lagrange equation, or simply the Euler equation. As we derive the


46 2 The calculus of variations
relation, we shall try to make things more concrete by considering a simple example, namely the problem of finding the length of the shortest curve between two points in a plane. The curve is described by the function y = y(x). An infinitesimal element of the curve has length ds, where
ds =
√
dx2 + dy2.
The length of the curve is, then,
I=
∫f
i
ds =
∫f
i
√
dx2 + dy2 =
∫f
i
√
1+
( dy
dx
)2 dx
=
∫f
i
√
1 + y′2dx,
where the integral is taken between the end points (initial and final) of the curve and the symbol y′ is defined to be y′ = dy
dx . (For convenience we denote differentiation with respect to x by a prime.) See Figure 2.1. Now suppose you want to find the curve that corresponds to the shortest path length between the two end points. That means you want to find a function y = y(x) such that the integral I is a minimum. The integral can be written in the following way
I=
∫f
i
(y) dx, (2.1)
where  is a function of y. But y itself is a function (it is a function of x). So  is a function of a function. We call it a functional.
i
f
dx
dy
ds
y
x
Figure 2.1 A curve y = y(x) between points i and f .


2.2 Derivation of the Euler–Lagrange equation 47
The problem is to find the function y(x) that minimizes the integral of the functional (y). For the situation we are discussing, the functional is obviously
(y) =
√
1 + y′2.
For different problems, as you can imagine, there are different functionals. Here is a different problem. Determine the shape z = z(x) of a frictionless wire such that a bead subjected to the gravitational force will slide from an initial point (xi, zi) to a final point (xf , zf ) in the minimum amount of time. (This is a famous problem that was solved by Isaac Newton in a few hours; it is called the Brachistochrone problem.) In this problem we want to minimize the time. The time it takes an object to go from one point to another is given by the distance divided by the velocity. For a particle sliding down a wire the velocity can be determined from conservation of energy in the form 1
2 mv2 = mgh, yielding
v = √2gh,
where h is the distance below the initial point. If we measure z downwards
from zi we can write v = √2gz. The time to slide a distance ds along the curve is, then,
dt = ds
√2gz =
√
dx2 + dz2
2gz =
√
1 + z′2
2gz dx.
The quantity to be minimized in this problem is
I=
∫f
i
√
1 + z′2
2gz dx,
and the functional is
(z, z′) =
√
1 + z′2
2gz . (2.2)
Example 2.1 A geodesic is the shortest distance between two points on a spherical surface. This is a segment of a great circle. Eventually we will want to determine the equation for a great circle, but here we will simply determine the appropriate functional.


48 2 The calculus of variations
a
adq
a sin q
q
f
a sin q d f
ds
Figure 2.2 An element of path, ds, on the surface of a sphere.
Solution 2.1 An element of path on a sphere of radius a is denoted ds in Figure 2.2. From the geometry of the sphere it is clear that
ds2 = a2dθ 2 + a2 sin2 θ dφ2.
The length of a curve on the sphere is, therefore
I=
∫
ds =
∫
adθ
√
1 + sin2 θ φ′2,
where φ′ = dφ/dθ . The functional is
=
(
1 + sin2 θ φ′2)1/2
.
The functionals obtained in these examples are representative of those found in calculus of variations problems. In general, the functionals we will be using depend on two functions, such as y and y′, as well as the parameter x, thus:
 = (x, y, y′). (2.3)
Note that y = y(x) and y′ = y′(x) = dy(x)
dx . That is, x is the independent
parameter and y and y′ are functions of x. (Later we shall formulate problems in which the independent parameter is the time, t.) What we have done so far is to describe how the functional is obtained for problems such as the shortest distance, the Brachistochrone and the geodesic. We have not, however, shown how to determine the function that minimizes the integral.


2.2 Derivation of the Euler–Lagrange equation 49
y
x
i
f
e=0
Figure 2.3 A family of curves Y = y(x) + η(x). The shortest distance curve is indicated by = 0. Note that all the curves meet at the end points, so η(xi ) = η(xf ) = 0.
Let us now derive the condition for the integral to be stationary. I will use the shortest path between two points as an example, but the derivation is general. Imagine drawing a straight line between the points. As you know, this is the path that gives the shortest distance. Imagine drawing nearby paths between the same two points (i.e. paths that differ infinitesimally from the shortest path). See Figure 2.3. The nearby paths will differ slightly from the “true” or minimal path. If y = y(x) is the equation of the minimal path, the equation of one of the nearby paths can be expressed as
Y (x, ) = y(x) + η(x), (2.4)
where is a small quantity and η(x) is an arbitrary function of x but which has an important condition on it, namely that η(x) must be zero at the end points because at those points the paths all meet. For a given function η(x), different values of will yield different paths, all belonging to same family of curves. (Selecting a different function η(x) will generate another family of curves, each characterized by the value of .) Assuming a given η(x), the path Y (x) is a function of as well as x. That is why we wrote Y = Y (x, ). The length of any one of these curves will, of course, depend on the value of and can be expressed as
I( ) =
∫ xf
xi
(x, Y, Y ′)dx,
where  has the form
=
√
1 + Y ′2.
We wrote I as a function only of because the dependence on x has been integrated out.


50 2 The calculus of variations
We want to determine the function y(x) that makes the integral I stationary. In ordinary calculus we would set the differential to zero (dI = 0). But now we are asking which of the functions y(x) makes I stationary, so we set the variation to zero (δI = 0). This is often called the “first variation” because δI is defined in terms of the Maclaurin expansion
I ( ) = I (0) +
[dI
d
]
=0
+ O( 2).
Keeping only the first term in we define
δI ( ) = I ( ) − I (0) =
[dI
d
]
=0
.
Consequently, δI = 0 is equivalent to [ dI
d
]
=0 = 0. Inserting the integral expression for I , gives us
[d
d
∫ xf
xi
(x, Y, Y ′)dx
]
=0
= 0. (2.5)
Y and Y ′ are both functions of . Taking the derivative under the integral we have
∫ xf
xi
[∂
∂Y
∂Y
∂ + ∂
∂Y′
∂Y′
∂
]
=0
dx = 0. (2.6)
The second term can be integrated by parts as follows:
∫ xf
xi
[ ∂
∂Y′
∂Y′
∂
]
dx =
∫ xf
xi
∂
∂Y′
∂
∂
( dY
dx
)
dx =
∫ xf
xi
∂
∂Y′
d
dx
(∂Y
∂
)
dx
=
∫ xf
xi
∂
∂Y′ d
(∂Y
∂
)
= ∂
∂Y′
∂Y
∂
∣∣∣∣
xf
xi
−
∫ xf
xi
d
dx
( ∂
∂Y′
) ∂Y
∂ dx.
But Y = y(x) + η(x) so ∂Y
∂ = η(x) and at the end points η(x) = 0 (because the curves meet there). Hence
∂
∂Y′
∂Y
∂
∣∣∣∣
xf
xi
= ∂
∂Y′
[η(xf ) − η(xi)] = 0.
Consequently, Equation (2.6) is
∫ xf
xi
[( ∂
∂Y − d
dx
[ ∂
∂Y′
]) ( ∂Y
∂
)]
=0
dx = 0. (2.7)


2.2 Derivation of the Euler–Lagrange equation 51
The integrand is the product of two functions. Since ∂Y
∂ = η(x), Equation (2.7) has the form
∫b
a
f (x)η(x)dx = 0, (2.8)
where
f (x) = ∂
∂Y − d
dx
( ∂
∂Y′
) .
The function η(x) is arbitrary (except at the end points) so Equation (2.8) can be true if and only if f (x) = 0 for all values of x. You might think that requiring f (x) = 0 for all x is too restrictive, but it is easy to show that it is not. You can appreciate this by assuming η(x) is zero everywhere except at some point between xi and xf , say at x = ξ . Next consider the integral
∫ xf =ξ +ε
xi =ξ −ε
f (x)η(x)dx = 0,
where ε is a very small quantity. Since the integral is zero everywhere, it is zero in the range of integration. Furthermore, f (x) is essentially constant over the small region of integration. Denote it by f (ξ ) and take it outside the integral sign,
f (ξ )
∫ xf =ξ +ε
xi =ξ −ε
η(x)dx = 0.
The integral is not zero, so f (ξ ) must be zero. But the point x = ξ could be anywhere in the interval xi to xf , so f (x) is zero at all points along the path.
Another argument that ∫ b
a f (x)η(x)dx = 0 implies f (x) = 0 is to assume that f (x) is not zero and use the fact that η(x) is arbitrary. For example, we could require that η(x) be negative everywhere that f (x) is negative and that η(x) be positive wherever f (x) is positive. But then the product f (x)η(x) would be positive everywhere and the condition (2.8) would not be met. Once again, we conclude that Equation (2.8) is true if and only if f (x) = 0. Consequently, Equation (2.7) reduces to
[∂
∂Y − d
dx
( ∂
∂Y′
)]
=0
= 0,
which is equivalent to requiring that
∂
∂y − d
dx
(∂
∂y′
)
= 0. (2.9)


52 2 The calculus of variations
This is called the Euler–Lagrange equation.1 It gives a condition that must be met if y = y(x) is the path that minimizes the integral
∫ xf
xi
(x, y, y′)dx.
For example, in determining the equation for the minimal distance between two points in a plane, the functional was
(x, y, y′) =
√
1 + y′2.
Plugging this into the Euler–Lagrange equation we obtain
−d
dx
∂
∂y′
(
1 + y′2) 1
2 = 0,
and this leads, after a bit of algebra, to
y = mx + b,
where m and b are constants. But y = mx + b is the equation of a straight line!
Exercise 2.1 Show that the functional  = √1 + y′2 leads to y = mx+b.
Exercise 2.2 Find the equation for the shortest distance between two points in a plane using polar coordinates. Answer: r cos(θ + α) = C, where α and C are constants.
Exercise 2.3 Determine and identify the curve y = y(x) such that
∫ x2
x1 [x(1 + y′2)]1/2dx is stationary. Answer: A parabola.
Exercise 2.4 Determine and identify the curve y = y(x) such that ∫ x2
x1
ds x
is stationary. Answer: A circle.
2.2.1 The difference between δ and d
The difference between δ and d is more than notational. When applied to a variable, δ represents a virtual displacement, which we usually think of as a change in a coordinate carried out with time frozen. More generally, it is an infinitesimal change in some variable that is virtual; that
1 You have met a variant of the Euler–Lagrange equation in Chapter 1 (Equation (1.16)).


2.2 Derivation of the Euler–Lagrange equation 53
is, carried out in an arbitrary but kinematically allowable manner while keeping the independent parameter constant. In contrast, we think of d as representing an actual change in a variable which occurs during some finite period of time. When applied to a function, the symbol δ, which was introduced by Lagrange, represents the variation of the function, whereas d is the differential of the function. As a simple example, consider the function F (x, x ̇, t), that is a function that depends on position, velocity and time. But position and velocity both depend on time. That is, time is the independent parameter, and we can write
F = F (x(t), x ̇(t), t).
According to the rules of calculus, the differential of F is
dF = ∂F
∂x dx + ∂F
∂x ̇ dx ̇ + ∂F
∂t dt.
On the other hand, the variation of F is
δF = ∂F
∂x δx + ∂F
∂x ̇ δx ̇.
The most significant difference between the differential and the variation is that the differential takes us to a different point of the same function whereas the variation leads to a different function. This is shown schematically in Figure 2.4, where we consider two functions of the independent variable x. These are y = f (x), the “true” function (i.e. the function that minimizes the definite
dx
dy f (x)
g(x)
Initial point
Final point
y
x
dy
Figure 2.4 An illustration of the difference between δ and d.


54 2 The calculus of variations
integral) and y = g(x) = f (x) + η(x), which is a nearby function with the same end points, i.e. the “varied” function. The variation δy is given by
δy = g(x) − f (x).
That is, δy gives the difference between the true function and the varied function at the same value of x. On the other hand, if y = f (x), then dy is the change in f (x) due to a change in x. That is
dy = f (x + dx) − f (x).
The difference between δ and d becomes particularly important when applied to definite integrals because both kinds of changes are often acting simultaneously. In the derivation of the Euler–Lagrange equation we used as a parameter that indicated whether or not we were on the “true” path (for which = 0) or on one of the varied paths ( = 0). It is clear that the integral I giving the length of a path depends on . That is why we wrote I = I ( ). Our next step was to take the derivative of I with respect to and set it equal to zero. (The different values of take us to different curves, i.e. to different functions y = y(x).)
We could also have expressed our problem in terms of variations, such as
δy = η(x)d ,
and
δy′ = η′(x)d .
These relations agree with the expression Y (x) = y(x) + η(x). If we recall that Y (x) represents a family of curves y(x, ), then replacing Y (x) with y(x, ) we have
δy = δ[y(x, )] =
[dy(x, )
d
]
=0
δ=
[d
d (y(x) + η(x))
]
=0
δ = η(x)δ .
Since I = I (y, y′, x) we have
δI = ∂I
∂y δy + ∂I
∂y′ δy′.
(Note that x is frozen.) But
δI = δ
∫ x2
x1
(y, y′, x)dx
=
∫ x2
x1
δdx


2.2 Derivation of the Euler–Lagrange equation 55
=
∫ x2
x1
(∂
∂y δy + ∂
∂y′ δy′
)
dx
=
∫ x2
x1
[∂
∂y η(x)d + ∂
∂y′ η′(x)d
]
dx.
Thus we could express the condition on I in either one of two ways, namely by setting
[dI
d
]
=0
=0
as we did in Equation (2.7), or by stating that the variation δI is zero. These two statements are equivalent. A final comment on notation. The “functional derivative” of  = (y, y′, x) is defined by
δ
δy = ∂
∂y − d
dx
∂
∂y′ . (2.10)
This is sometimes called the “variational derivative.”
Exercise 2.5 Show that δ and d commute.
Exercise 2.6 Given
δ = ∂
∂y δy + ∂
∂y′ δy′,
demonstrate that δI = 0 means the same thing as [ dI
d
]
=0 = 0.
2.2.2 Alternate forms of the Euler–Lagrange equation
We derived the Euler–Lagrange equation in the form
∂
∂y − d
dx
(∂
∂y′
)
= 0.
There are two variants of this expression. One of them is useful for problems in which  does not depend on y and the other is useful for problems in which  does not depend on x. If  does not depend on y, the Euler–Lagrange equation reduces to
d
dx
(∂
∂y′
)
= 0,


56 2 The calculus of variations
from which we conclude that
∂
∂y′ = constant.
(This was the situation when we solved the “shortest distance in a plane” problem.) Next assume that  does not depend on x. Then, since  = (y, y′), the partial derivative of  with respect to x is zero and the total derivative of  with respect to x will be
d
dx = ∂
∂y
dy
dx + ∂
∂y′
dy′
dx .
But dy′
dx = d
dx
dy
dx = y′′, so
d
dx = ∂
∂y
dy
dx + y′′ ∂
∂y′ = y′ ∂
∂y + y′′ ∂
∂y′ . (2.11)
If we multiply the Euler–Lagrange equation by y′ we obtain
y′ ∂
∂y − y′ d
dx
(∂
∂y′
)
= 0.
Add y′′ ∂
∂y′ to both sides:
y′ ∂
∂y − y′ d
dx
(∂
∂y′
)
+ y′′ ∂
∂y′ = y′′ ∂
∂y′
y′ ∂
∂y + y′′ ∂
∂y′ = y′′ ∂
∂y′ + y′ d
dx
(∂
∂y′
) .
The left-hand side is just d
dx (see Equation (2.11)) and the right-hand side is
d dx
(
y′ ∂
∂y′
)
, so
d
dx = d
dx
(
y′ ∂
∂y′
) ,
or
d
dx
(
 − y′ ∂
∂y′
)
= 0.
Therefore,
 − y′ ∂
∂y′ = constant. (2.12)


2.2 Derivation of the Euler–Lagrange equation 57
Example 2.2 The Brachistochrone problem: The problem is to determine the shape of a wire such that a bead on the wire will slide from some initial point to some lower final point in the least amount of time. Note that the two end points (x1, z1) and (x2, z2) are fixed. In Section 2.2 we obtained the functional. We now solve for the curve z = z(x).
Solution 2.2 The functional was given by Equation (2.2) as
(z, z′) =
√
1 + z′2
2gz .
Note that  does not depend on x. Therefore,
 − z′ ∂
∂z′ = constant.
The factor 2g can be discarded. Carrying out the mathematics, we obtain √
1 + z′2
z − z′2
√z√1 + z′2 = constant = 1/C. (2.13)
This can be expressed as
z(1 + z′2) = C. (2.14)
The details are left as an exercise. We can obtain a solution to this differential equation by setting
z′ = cot θ .
Then 1 + z′2 = 1/ sin2 θ , and our differential equation can be written
z(1 + cot2 θ ) = z/ sin2 θ = C,
or
z = C sin2 θ .
Next note that
dx
dθ = dx
dz
dz
dθ = 1
z′
dz
dθ = tan θ dz
dθ = tan θ d
dθ (C2 sin2 θ ) = C(1 − cos 2θ ).
Therefore
x=
∫
C(1 − cos 2θ )dθ = C
[
θ − sin 2θ
2
] .


58 2 The calculus of variations
Letting C = 2A and 2θ = φ we can write the following parametric equations for x and z:
x = A(φ − sin φ)
z = A(1 − cos φ).
These are the parametric equations for a cycloid. Consequently, the bead slides in minimum time from (x1, z1) to (x2, z2) if the wire is bent into a cycloid.
Exercise 2.7 Show the equivalence of the following two forms of the Euler–Lagrange equation
∂
∂y − d
dx
∂
∂y′ = 0
and
∂
∂x − d
dx
(
 − y′ ∂
∂y′
)
= 0,
assuming y′ = 0.
Exercise 2.8 Obtain Equation (2.14) starting with Equation (2.13).
Exercise 2.9 Determine the minimum time for a bead to slide down a wire
from the point (−π , 2) to (0, 0) (meters). Answer: π /√g.
2.3 Generalization to several dependent variables
We have been considering problems whose functionals have the form  = (x, y, y′). In this section we generalize to functionals that depend on n dependent variables, y1, y2, . . . , yn and their derivatives, y′
1, y′
2, . . . , y′n. We
assume the variables are independent.2 The functional for the system is now (y1, y2, . . . , yn, y′
1, y′
2, . . . , y′n, x). We will show that this leads to n Euler–Lagrange equations of the form:
d
dx
(∂
∂yi′
)
− ∂
∂yi
= 0, i = 1, . . . , n.
2 This section is included for the sake of completeness. The procedure is a generalization of the the simpler situation described in the the previous section (2.2). You may skip this section without loss of continuity, but note that some of the results obtained here are used in Section 3.4.


2.3 Generalization to several dependent variables 59
The derivation is similar to that presented in the preceding section, but is more general. Recall that our intention is to determine the functions y1(x), y2(x), . . . which will minimize (or maximize) the definite integral
I=
∫ xf
xi
(y1, . . . , yn, y′
1, . . . , y′n, x)dx.
That is, we want to obtain the functions yi(x) such that
δI = 0.
As before, there are an infinite number of paths between the end points. The paths that are infinitesimally near the minimal or true path can be described by
Yi(x) = yi(x) + ηi(x).
The integral I depends on which path is chosen, so it can be considered a function of . The variation of I is
δI = δ
∫ xf
xi
dx =
∫ xf
xi
δdx =
∫ xf
xi
∑
i
( ∂
∂ Yi
δYi + ∂
∂ Yi′
δYi′
)
dx.
Considering Yi and Yi′ as functions of , we write
δI =
(∂I
∂
)
=0
d=
[∫ xf
xi
∑
i
( ∂
∂ Yi
∂ Yi
∂ + ∂
∂ Yi′
∂ Yi′
∂
)
dx
]
=0
d.
The second term can be integrated by parts. (Note that the mathematical procedure here is nearly identical to that from Equation (2.6) to Equation (2.7).) We have
∫ xf
xi
∂
∂ Yi′
∂ Yi′
∂ d dx =
∫ xf
xi
∂
∂ Yi′
∂ 2 Yi
∂ ∂x dxd =
∫ xf
xi
∂
∂ Yi′
∂
∂
( ∂Yi
∂x dx
) d
= ∂
∂ Yi′
∂ Yi
∂
∣∣∣∣
x2
x1
−
∫ xf
xi
∂ Yi
∂
d
dx
∂
∂ Yi′
dxd
=−
∫ xf
xi
∂ Yi
∂
d
dx
∂
∂ Yi′
dxd ,
where we have used the fact that ∂Yi
∂
∣∣∣x2
x1
= 0 because all the curves pass through
the end points.


60 2 The calculus of variations
Consequently,
δI =
∫ xf
xi
∑
i
( ∂
∂ Yi
∂ Yi
∂ − ∂Yi
∂
d
dx
∂
∂ Yi′
)
d dx
=
∫ xf
xi
∑
i
( ∂
∂ Yi
−d
dx
∂
∂ Yi′
) ∂Yi
∂ d dx
=
∫ xf
xi
∑
i
( ∂
∂ Yi
−d
dx
∂
∂ Yi′
)
δYi dx.
When = 0, δI = 0 because I is minimized along the path yi = yi(x). So
[δI ] =0 = 0 =
∫ xf
xi
∑
i
(∂
∂yi
−d
dx
∂
∂yi′
)
δyidx. (2.15)
Since the δyis are independent, the only way this expression can be zero is for all the terms in the parenthesis to be zero. Hence,
∂
∂yi
−d
dx
(∂
∂yi′
)
= 0, i = 1, . . . , n, (2.16)
as expected.
2.4 Constraints
Many problems in the calculus of variations involve constraints or, as they are sometimes called, “auxiliary conditions.” For example, a particle might be constrained to the z = 0 plane or to move along a specific curve.
2.4.1 Holonomic constraints
Let us begin by considering holonomic constraints. Recall from Section 1.4 that a holonomic constraint is a relationship between the dependent variables that can be expressed as a function that is equal to zero. In the presence of a constraint, the dependent variables are not independent of one another, being related by the auxiliary conditions. For example, let us consider a problem described by  = (y1, . . . , yn; y′
1, . . . , y′n; x) where the n dependent variables are related by m constraints of the form
f1(y1, y2, . . . , yn, x) = 0,
...
fm(y1, y2, . . . , yn, x) = 0.


2.4 Constraints 61
Each constraint reduces by one the number of degrees of freedom. If there are n dependent coordinates and m constraints we can use the constraints to reduce the number of degrees of freedom to n − m and then apply n − m Euler–Lagrange equations to solve the problem. This is conceptually simple, but may be fairly complicated in practice. A better approach is to keep all n variables and use the method of Lagrange multipliers. We now consider this method (often referred to as Lagrange’s λ-method). We begin by taking the variation of the constraint equations, thus:
δf1 = ∂f1
∂y1
δy1 + ∂f1
∂y2
δy2 + · · · + ∂f1
∂yn
δyn = 0,
...
δfm = ∂fm
∂y1
δy1 + ∂fm
∂y2
δy2 + · · · + ∂fm
∂yn
δyn = 0. (2.17)
Next, we multiply each of these relations by an “undetermined multiplier” λi and sum them to obtain
λ1δf1 + λ2δf2 + · · · + λmδfm = 0. (2.18)
The sum is zero because each δf is zero. We also know that the variation of the functional  vanishes at the stationary point. That is,
δ = ∂
∂y1
δy1 + ∂
∂y2
δy2 + · · · + ∂
∂yn
δyn = 0. (2.19)
The sum of Equations (2.18) and (2.19) must also be zero, so we have
0 = ∂
∂y1
δy1 + ∂
∂y2
δy2 + · · · + ∂
∂yn
δyn
+ λ1
( ∂f1
∂y1
δy1 + ∂f1
∂y2
δy2 + · · · + ∂f1
∂yn
δyn
)
+ · · · + λm
( ∂fm
∂y1
δy1 + ∂fm
∂y2
δy2 + · · · + ∂fm
∂yn
δyn
)
. (2.20)
The expression is complicated so, to illustrate the procedure, I will assume that there is a single constraint. (The generalization to m constraints is left as a problem.) If there is only one constraint, Equation (2.18) can be written as
λδf = λ
( ∂f
∂y1
δy1 + ∂f
∂y2
δy2 + · · · + ∂f
∂yn
δyn
)
= 0, (2.21)


62 2 The calculus of variations
and Equation (2.20) reduces to
∂
∂y1
δy1 + ∂
∂y2
δy2 + · · · + ∂
∂yn
δyn
+λ
( ∂f
∂y1
δy1 + ∂f
∂y2
δy2 + · · · + ∂f
∂yn
δyn
)
= 0.
Re-arranging yields
∑n
i=1
(∂
∂yi
+ λ ∂f
∂yi
)
δyi = 0.
Since we introduced λ arbitrarily, we are free to give it any value we wish. Let us select λ such that the nth term of the sum is zero. That is, the value of λ will be determined by requiring that
∂
∂yn
+ λ ∂f
∂yn
= 0. (2.22)
Then the nth term of the summation is eliminated, leaving us with
n−1
∑
i=1
(∂
∂yi
+ λ ∂f
∂yi
)
δyi = 0.
Now all the δys are independent and the summation is zero if and only if each coefficient of a variation δyi (i = 1, . . . , n − 1) is zero. That is,
∂
∂yi
+ λ ∂f
∂yi
= 0 i = 1, 2, . . . , n − 1. (2.23)
Note that we have not eliminated any variables. We can sum Equations (2.21), (2.22) and (2.23) to obtain
δ + λδf = 0.
This can be written as
δ( + λf ) = 0, (2.24)
because δ(λf ) = f δλ + λδf = λδf since f = 0.
Equation (2.24) states that the variation of +λf is zero for arbitrary variations in the generalized coordinates. Thus, we have reformulated the problem. Originally we had δ = 0 subject to the constraint f (y1 . . . yn) = 0, but now we have δ( + λf ) = 0 with no constraints on the coordinates. The original problem had n−1 degrees of freedom whereas the reformulated problem treats all the coordinates as unconstrained and introduces another variable λ giving n + 1 degrees of freedom. Note further that δ( + λf ) = 0 yields n equations, and f = 0 is one more equation. Therefore we have n + 1 equations in n + 1 unknowns.


2.4 Constraints 63
x
y
Φ
Circle in xy plane
(x − 2)2 + (y − 2)2 = 1
Plane: Φ = x + y
Figure 2.5 The circle in the xy plane is projected upwards onto the shaded surface.
Example 2.3 As a very simple application of the foregoing, consider the plane (x, y) = x + y. A circle in the xy plane is projected onto the  plane, as shown in Figure 2.5. Determine the location of the maximum and minimum points of the projected circle, given the constraint (x − 2)2 + (y − 2)2 = 1.
Solution 2.3 Given
(x, y) = x + y,
f (x, y) = (x − 2)2 + (y − 2)2 − 1 = 0,
the condition
δ( + λf ) = 0,
when inserted into the Euler–Lagrange equation, leads to the following three equations
∂
∂x ( + λf ) = 0 ⇒ 1 + λ(2(x − 2)) = 0,
∂
∂y ( + λf ) = 0 ⇒ 1 + λ(2(y − 2)) = 0,
∂
∂λ ( + λf ) = 0 ⇒ (x − 2)2 + (y − 2)2 − 1 = 0.


64 2 The calculus of variations
The third equation is simply the constraint. The first and second equations can be solved for λ. Equating the two expressions for λ we obtain
1
2x − 4 = 1
2y − 4
which yields x = y. Inserting this into the third equation we obtain
2(x − 2)2 = 1
from which
x = 2 ± 1/√2
and consequently, the extrema are at
(1.29, 1.29) and (2.71, 2.71).
2.4.2 Non-holonomic constraints
Differential constraints
Suppose the constraint is not an algebraic relation between the variables, but a differential relation. For example, the “rolling without slipping” constraint is expressed as a relation between the linear displacement and the angular displacement in the form
ds = rdθ .
Dividing by dt you note that this is a relationship between velocities rather than coordinates. Now obviously if you can integrate the relation between differentials, you will get a holonomic constraint, and you can proceed as before. If the constraint is truly non-holonomic, all is not lost because for some nonholonomic constraints one can apply the Lagrangian λ-method described above. Assume the constraint is expressed in the form
A1δy1 + A2δy2 + · · · + Anδyn = 0.
There are two things to be noted here. First of all, the partial derivatives that appeared in Equation (2.17) are replaced by the coefficients Ai which are functions of the dependent variables but they are not derivatives of some function. Secondly, we cannot eliminate a variable because we do not have the equations between them that we need to eliminate one variable in terms of the others. Nevertheless, the Lagrangian λ-method can be applied. Define the quantity δf :
δf = A1δy1 + A2δy2 + · · · + Anδyn = 0.


2.4 Constraints 65
Multiply δf by λ and add it to the variation of the functional :
δ + λδf = 0.
Now, once again, you can treat all the vis as independent. If the auxiliary conditions are time dependent (rheonomous) the procedure is a bit more complicated and we will not consider it. The interested student may read Section 2.13 of Lanczos3 for a brief discussion.
Isoperimetric constraints
Some constraints are expressed as integrals. Such a constraint is called an “isoperimetric condition” because the most famous problem of this type is the “Queen Dido” problem of finding the curve of constant perimeter that encloses the greatest area.4 In general the constraint is given as a definite integral which has a fixed given value, thus
∫ x2
x1
f (x, y, y′)dx = C, (2.25)
where C is known. Such a problem is solvable using Lagrange’s λ-method. Assume the integral to be maximized is
I=
∫ x2
x1
(x, y, y′)dx. (2.26)
For simplicity let us consider a system subjected to a single constraint. (The generalization to several conditions of the form (2.25) is straightforward.) The variation of the condition (2.25) is
δ
∫ x2
x1
f (x, y, y′)dx =
∫ x2
x1
( ∂f
∂y δy + ∂f
∂y′ δy′
)
dx = 0. (2.27)
Multiply Equation (2.27) by the undetermined constant λ and add it to δI to obtain
δ
∫ x2
x1
( + λf ) dx = 0.
Thus the Lagrangian λ-method implies that the integral
∫ x2
x1
( + λf ) dx
3 Lanczos, op. cit. 4 The legend tells us that Queen Dido, escaping from her brother, went to Carthage. She offered to buy land, but the ruler of the town haughtily told her she could only have as much land as she could enclose in the hide of an ox. Dido cut the hide of the largest ox she could find into the thinnest strip possible. The problem is to show that, given the length of the strip, the largest area is enclosed by a circle.


66 2 The calculus of variations
is stationary and hence the integrand satisfies the Euler–Lagrange equation. Note that the constraint and the integral of the functional have the same limits of integration.
Example 2.4 Determine the shape of the curve of length L that encloses the greatest area.
Solution 2.4 Recall from the calculus that the area enclosed by a curve C is given by
Area =  = 1
2
∮
C
(xdy − ydx).
Expressing x and y parametrically as x(t) and y(t) we write
= 1
2
∮
C
(
x dy
dt − y dx
dt
)
dt = 1
2
∮
C
(xy′ − yx′)dt.
(Here t is just a parameter – it is not the time.) The constraint that the curve has length L is
f =∮
C
√
x′2 + y′2dt − L = 0.
Note that  = (x, y, x′, y′, t) with t as the independent parameter. Since  + λf is stationary, it obeys the following Euler–Lagrange equations:
∂( + λf )
∂x − d
dt
[ ∂ ( + λf )
∂x′
]
= 0,
∂( + λf )
∂y − d
dt
[ ∂ ( + λf )
∂y′
]
= 0,
yielding
1
2y′ − d
dt
[
−1
2 y + λx′
√x′2 + y′2
]
= 0,
−1
2x′ − d
dt
[ 1
2 x + λy′
√x′2 + y′2
]
= 0.
Integrating, we obtain
1
2y + 1
2 y − λx′
√x′2 + y′2 = C2,
−1
2x − 1
2 x − λy′
√x′2 + y′2 = −C1,


2.5 Problems 67
where C1 and C2 are constants of integration. Rearranging we write
y − C2 = λx′
√x′2 + y′2 ,
x − C1 = λy′
√x′2 + y′2 .
Consequently, squaring and adding yields
(x − C1)2 + (y − C2)2 = λ2
x′2 + y′2 (x′2 + y′2) = λ2,
which is the equation of a circle of radius λ with center at (C1, C2).
2.5 Problems
2.1 Generalize the Lagrange λ-method to a system with n coordinates and m constraints (m < n) and show that
δ( + λ1f1 + · · · + λmfm) = 0,
where the λs are obtained from
∂
∂ qi
+ λ1
∂f1
∂ qi
+ · · · + λm
∂fm
∂ qi
= 0, i = n − m + 1, . . . , n).
2.2 Determine the equation of the curve giving the shortest distance between two points on the surface of a cone. Let r2 = x2 + y2 and z = r cot α. 2.3 Determine the equation for shortest curve between two points on the surface of cylinder (r = 1 + cos θ ). 2.4 Consider a solid of revolution of a given height. Determine the shape of the solid if it has the minimum moment of inertia about its axis. 2.5 Consider the variational problem for variable end points. (a) Let
I=
∫b
a
(x, y, y′)dx,
where y(b) is arbitrary. Show that
δI = η ∂
∂y′
∣∣∣∣
b
a
+
∫b
a
(∂
∂y − d
dx
∂
∂y′
)
ηdx,
with the condition
∂
∂y′
∣∣∣∣x=b
= 0.


68 2 The calculus of variations
r
a
b
Figure 2.6 A frictionless tunnel for rapid transits between two points on Earth.
(b) Assume y is fixed at x = a but the other end point can lie anywhere on a curve g(x, y) = 0. Show that in addition to the Euler–Lagrange condition we have (
 − y′ ∂
∂y′
) ∂g
∂y − ∂
∂y′
∂g
∂x = 0.
2.6 A frictionless tunnel is dug through the Earth from point a to point b. An object dropped in the opening will slide through the tunnel under the action of gravity and emerge at the other end with zero velocity. The gravitational potential at a point inside the Earth is φ = −Gms/r where ms is the mass of the material enclosed by a sphere of radius r. (We are assuming a constant density Earth.) Use polar coordinates and
find the equation of the path that minimizes the time ∫ dt. Determine the transit time.5 See Figure 2.6. 2.7 Two rings of radius a are placed a distance 2b apart. (The centers of the rings lie along the same line and the planes of the rings are perpendicular to this line.) Find the shape of a soap film formed between the rings, noting that the film will have minimal surface area. 2.8 A particle of mass m is in a two-dimensional force field given by
F = −G Mm
r2 ˆr.
The curve requiring the minimum time for the particle to fall from one point to another is the solution of the differential equation
dr
dθ = f (r).
Determine f (r). 2.9 Consider the function
f (x, y, z) = x2 + 2y2 + 3z2 + 2xy + 2xz,
5 See P. W. Cooper, Through the Earth in forty minutes, Am. J. Phys., 34, 68 (1966) and G. Venezian, Terrestrial brachistochrone, Am. J. Phys., 34, 701 (1966).


2.5 Problems 69
subject to the condition
x2 + y2 + z2 = 1.
What is the minimum value of f (x, y, z)? 2.10 The speed of light in a medium of index of refraction n is v = c/n = ds/dt. The time for light to travel from point A to point B is ∫B
A
ds
v.
Obtain the law of reflection and the law of refraction (Snell’s law) by using Fermat’s principle of least time. Using polar coordinates, show that if n is proportional to 1/r2, that the path followed by a ray of light is given by sin(θ + c) = kr, where c and k are constants. 2.11 (This is known as “Newton’s problem.”) Consider a solid of revolution generated by a curve from the origin to a point B in the first quadrant by rotating the curve about the x axis. It is desired to determine the shape of this curve if the air resistance on the solid is a minimum. (The solid is moving towards the left.) Isaac Newton assumed the resistance was given by the expression
2π ρV 2
∫
y sin2 dy,
where  = y′ = the slope of the curve, ρ is the air density and V 2 is the velocity. (This is not a very good expression for air resistance, but let us assume it is correct.) Obtain an expression for the integral to be minimized. 2.12 Determine the curve of length l that joins points x1 and x2 on the axis, such that the area enclosed by the curve and the x axis is maximized. 2.13 Determine the shape of a cylinder (that is, the ratio of radius to height) that will minimize the surface area for a given volume.


3
Lagrangian dynamics
This chapter is an introduction to Lagrangian dynamics. We begin by considering d’Alembert’s principle and derive Lagrange’s equations of motion from it. We discuss virtual work in detail. Then we present Hamilton’s principle and use it to carry out a second derivation of Lagrange’s equations. (This is similar to the derivation of the Euler–Lagrange equation presented in Chapter 2.) It should be noted that Hamilton’s principle is the fundamental principle of analytical mechanics. The fact that it is deceptively simple should not obscure the fact that it is very profound. We then consider how to determine the forces of constraint, and finish up by discussing the invariance of Lagrange’s equations.
3.1 The principle of d’Alembert. A derivation of Lagrange’s equations
Recall that the Lagrange equations of motion are
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
= 0, i = 1, 2, . . . , n. (3.1)
In this section we present a simple derivation of Lagrange’s equations based on d’Alembert’s principle. This principle is, in a sense, a re-statement of Newton’s second law, but it is expressed in a way that makes it a very useful concept in advanced mechanics. For a system of N particles, d’Alembert’s principle is
N ∑
α=1
(Fext
α − p ̇ α
) · δrα = 0, (3.2)
where Feαxt is the external force acting on particle α and  ̇pα is the change in momentum of that particle with respect to time. By Newton’s second law it is clear that the term in parenthesis is zero. Therefore, multiplying this term
70


3.1 The principle of d’Alembert 71
by δrα has no effect. Furthermore, summing over all particles is just a sum of terms all equal to zero. Thus the fact that the expression is equal to zero is obviously true. What is not so obvious (yet) is why this particular formulation is of any value. We will begin by expressing d’Alembert’s principle in Cartesian coordinates and then transforming to generalized coordinates. For N particles there are n = 3N Cartesian coordinates, and we can express d’Alembert’s principle in scalar form as
∑n
i=1
(F ext
i − p ̇i
) δxi = 0. (3.3)
If there are n Cartesian coordinates and k constraints, we can (in principle) express the problem in terms of n − k generalized coordinates. The transformation equations are
x1 = x1(q1, q2, . . . , qn−k, t)
x2 = x2(q1, q2, . . . , qn−k, t)
...
xn = xn(q1, q2, . . . , qn−k, t).
Note that although there are n Cartesian coordinates xi there are only n − k generalized coordinates because each constraint reduces by one the number of degrees of freedom and hence the number of generalized coordinates. The quantity δri in (3.2) or δxi in (3.3) is a virtual displacement. According to Equation (1.9),
δxi =
n−k
∑
j =1
∂ xi
∂ qj
δqj .
Therefore, the first term in d’Alembert’s principle can be written
∑
i
F ext
i δxi = ∑
i
F ext
i
⎛
⎝∑
j
∂ xi
∂ qj
δqj
⎞
⎠=∑
i,j
(
F ext
i
∂ xi
∂ qj
)
δqj = ∑
j
Qj δqj .
The last quantity is the “virtual work.” (It is the work done during the virtual displacement. See Section 1.6.) Next consider the other term in d’Alembert’s principle, namely p ̇iδxi:
∑n
i
p ̇i δxi =
∑n
i
p ̇i
n−k
∑
j
∂ xi
∂ qj
δqj =
∑n
i
mi x ̈i
n−k
∑
j
∂ xi
∂ qj
δqj ,


72 3 Lagrangian dynamics
where we assumed the masses of the particles are constant. But,
d
dt
(
mi x ̇i
∂ xi
∂ qj
)
= mi x ̈i
∂ xi
∂ qj
+ mi x ̇i
d
dt
∂ xi
∂ qj
.
Recall Equation (1.7):
∂ xi
∂ qj
= ∂x ̇i
∂ q ̇j
,
and note further that the last term of the previous equation involves the expression
d
dt
∂ xi
∂ qj
= ∂vi
∂ qj
= ∂x ̇i
∂ qj
.
Consequently,
∑
i
p ̇i δxi = ∑
i,j
[d
dt
(
mi x ̇i
∂ xi
∂ qj
)
− mi x ̇i
∂ x ̇i
∂ qj
]
δqj ,
=∑
i,j
[d
dt
(
mi x ̇i
∂ x ̇i
∂ q ̇j
)
− mi x ̇i
∂ x ̇i
∂ qj
]
δqj .
But
∂T
∂ qj
=∂
∂ qj
∑
i
(1
2 mi x ̇2
i
)
=∑
i
mi
( x ̇i
∂ x ̇i
∂ qj
) ,
and
∂T
∂ q ̇j
=∂
∂ q ̇j
∑
i
(1
2 mi x ̇2
i
)
=∑
i
mi
( x ̇i
∂ x ̇i
∂ q ̇j
) ,
so finally,
∑
i
p ̇i δxi = ∑
j
[d
dt
∂T
∂ q ̇j
− ∂T
∂ qj
]
δqj ,
and d’Alembert’s principle reads
∑n
i=1
(F ext
i − p ̇i
) δxi = ∑
j
Qj δqj − ∑
j
[d
dt
∂T
∂ q ̇j
− ∂T
∂ qj
]
δqj = 0,
0=∑
j
(
Qj − d
dt
∂T
∂ q ̇j
+ ∂T
∂ qj
)
δqj .


3.2 Hamilton’s principle 73
Since the δqj are independent, each term in the series must individually be equal to zero, so
d
dt
∂T
∂ q ̇j
− ∂T
∂ qj
= Qj . (3.4)
This is called the Nielsen form of Lagrange’s equations. Thus we have derived Lagrange’s equations in terms of generalized forces and kinetic energy. If the generalized forces are derivable from a potential that is independent of the generalized velocities, we can write Qj = −∂V /∂qj , and consequently,
d
dt
∂T
∂ q ̇j
− ∂T
∂ qj
= − ∂V
∂ qj
.
That is,
d
dt
∂(T − V )
∂ q ̇j
− ∂(T − V )
∂ qj
= 0.
This yields the familiar form of Lagrange’s equations:
d
dt
∂L
∂ q ̇j
− ∂L
∂ qj
= 0. (3.5)
Equations (3.4) and (3.5) are equivalent expressions for Lagrange’s equations.
Exercise 3.1 Using the Nielsen form, determine the equation of motion for a mass m connected to a spring of constant k.
Exercise 3.2 Using the Nielsen form, determine the equations of motion
for a planet in orbit around the Sun. (Answer: mr ̈ − mrθ ̇ 2 = − GMm
r2 and
mrθ ̈ + 2mr ̇θ ̇ = 0.)
3.2 Hamilton’s principle
A mechanical system composed of N particles can be described by n = 3N Cartesian coordinates. All of these coordinates may not be independent. If there are k equations of constraint, the number of independent coordinates is n−k. The set of values for all the coordinates at a given instant of time is called the configuration of the system at that time. That is, the configuration is given by {q1, q2, . . . , qn−k}. We shall assume that all of the generalized coordinates can be varied independently. (When we consider how to determine the forces


74 3 Lagrangian dynamics
of constraint, we will formulate the problem in terms of generalized coordinates that are not all independent. But, for now, keep in mind that all the qis are independent.) Recall that the system can be represented as a point in the (n − k) dimensional space called configuration space. As time goes on, the coordinates of the particles will change in a continuous manner and the point describing the system will move in configuration space. As the system goes from an initial configuration to a final configuration this point traces out a path or trajectory in configuration space. Obviously there are infinitely many paths between the initial point and the final point. We know, however, that a mechanical system always takes a particular path between these two points. What is special about the path actually taken? Hamilton’s principle answers this question. It states that the path taken by a system is the path that minimizes the “action.” The action of a mechanical system is defined as the line integral
I=
∫ t2
t1
Ldt, (3.6)
where
L = L(q1, q2, . . . , qn−k; q ̇1, q ̇2, . . . , q ̇n−k; t),
is the Lagrangian of the system and the limits on the integral, t1 and t2, are the initial and final times for the process. Hamilton’s principle tells us that the system behaves in such a way as to minimize this integral. Therefore, the variation of the integral is zero. In equation form, Hamilton’s principle is
δ
∫ t2
t1
Ldt = 0. (3.7)
To interpret this relation, imagine a configuration space and an initial and final point for some process as indicated in Figure 3.1. The figure shows several
Initial point
Final point
i
f
Figure 3.1 Three paths in configuration space.


3.4 Generalization to many coordinates 75
paths between the end points i and f. Assume that one of these paths is the
one that minimizes the action. (It is the “true” path.) Then, along that path,
δ ∫ t2
t1 Ldt = 0. The integral is a minimum for the path the system actually follows, and along any other path the value of the action will be greater.
3.3 Derivation of Lagrange’s equations
We can now derive Lagrange’s equations from Hamilton’s principle, recognizing that a simple change of variables from (x, y, y′) to L(t, q, q ̇) leads to the nearly trivial reformulation of Hamilton’s principle into the language of the calculus of variations. Hamilton’s principle tells us that a mechanical system will follow a path in configuration space given by q = q(t) such that the integral
∫ tf
ti
L(t, q, q ̇)dt
is minimized. (Note that here the Lagrangian is to be considered a functional.) Now if q = q(t) is the minimizing path, then L must satisfy the EulerLagrange equation
d
dt
(∂L
∂ q ̇
)
− ∂L
∂q = 0. (3.8)
It is customary to call this equation simply “Lagrange’s” equation. It can easily be generalized to systems described by an arbitrary number of generalized coordinates.
Exercise 3.3 Show that Equation (3.7) leads to Equation (3.8). Use the argument given in Section 2.2 as a guide.
3.4 Generalization to many coordinates
Consider a system described by the generalized coordinates q1, q2, . . . , qn. We assume the coordinates are all independent. The Lagrangian for this system is L(q1, q2, . . . , qn, q ̇1, q ̇2, . . . , q ̇n, t). Our plan is to show that Lagrange’s equations in the form
d
dt
( ∂L
∂ q ̇i
)
− ∂L
∂ qi
= 0, i = 1, . . . , n


76 3 Lagrangian dynamics
follow from Hamilton’s principle. The derivation is similar to that presented in Section 2.3. Recall the definition of the variation of a function. If
f = f (qi, q ̇i, t),
then
δf = ∑
i
[( ∂f
∂ qi
)
δqi +
( ∂f
∂ q ̇i
)
δq ̇i
] .
Note that in this expression, the time is assumed constant. Consider again Hamilton’s principle in the form of Equation (3.7), which is
δ
∫ t2
t1
Ldt = δ
∫ t2
t1
L(q1, q2, . . . , qn, q ̇1, q ̇2, . . . , q ̇n, t)dt = 0,
or
δI = 0,
where the action is represented by I, and the integral is taken over the “path” actually followed by the physical system. Note that for this path the variation in the action is zero. However, as before, there are an infinite number of possible paths between the end points, and we express them in the form
Qi(t) = qi (t) + ηi(t).
In this case, the action can be considered a function of . The Lagrangian L is a function of Qi, Q ̇ i and (indirectly) of . Therefore, the variation of the action is
δI =
∫ t2
t1
δLdt =
∫ t2
t1
∑
i
( ∂L
∂ Qi
δQi + ∂L
∂Q ̇ i
δQ ̇ i
)
dt.
Noting that Qi and Q ̇ i are functions of , we write
δI = ∂I
∂d =
∫ t2
t1
∑
i
( ∂L
∂ Qi
∂ Qi
∂ d + ∂L
∂Q ̇ i
∂Q ̇ i
∂d
)
dt.
The second term can be integrated by parts. (Note that the mathematical procedure here is nearly identical to that from Equation (2.6) to Equation (2.7).)


3.5 Constraints and Lagrange’s λ-method 77
We have
∫ t2
t1
∂L
∂Q ̇ i
∂Q ̇ i
∂ d dt =
∫ t2
t1
∂L
∂Q ̇ i
∂ 2 Qi
∂ ∂t dtd =
∫ t2
t1
∂L
∂Q ̇ i
∂
∂
( ∂Qi
∂t dt
) d
= ∂L
∂Q ̇ i
∂ Qi
∂
∣∣∣∣
t2
t1
−
∫ t2
t1
∂ Qi
∂
d
dt
∂L
∂Q ̇ i
dtd
=−
∫ t2
t1
∂ Qi
∂
d
dt
∂L
∂Q ̇ i
dtd ,
where we have used the fact that ∂Qi
∂
∣∣∣t2
t1
= 0 because all the curves pass
through the end points. Consequently,
δI =
∫ t2
t1
∑
i
( ∂L
∂ Qi
∂ Qi
∂ − ∂Qi
∂
d
dt
∂L
∂Q ̇ i
)
d dt
=
∫ t2
t1
∑
i
( ∂L
∂ Qi
−d
dt
∂L
∂Q ̇ i
) ∂Qi
∂ d dt
=
∫ t2
t1
∑
i
( ∂L
∂ Qi
−d
dt
∂L
∂Q ̇ i
)
δQi d t .
When = 0, δI = 0 because I is minimized along the path qi = qi(t). So
[δI ] =0 = 0 =
∫ t2
t1
∑
i
( ∂L
∂ qi
−d
dt
∂L
∂ q ̇i
)
δqidt. (3.9)
Since the δqis are independent, the only way this expression can be zero is for all the terms in the parenthesis to be zero. Hence, Lagrange’s equations are proved.
3.5 Constraints and Lagrange’s λ-method
We now apply Lagrange’s λ-method of Section 2.4.1 to determine the forces of constraint acting on a system. Let a system be described by n generalized coordinates q1, q2, . . . , qn. Assume, further, that the system is subjected to k holonomic constraints, so there are k equations relating the coordinates. Obviously, not all of the generalized coordinates are independent, so we could use the k equations of constraint to reduce the number of generalized coordinates to n − k. However, we prefer, for the moment, to use all n of the coordinates.


78 3 Lagrangian dynamics
The behavior of the system is described by Hamilton’s principle,
δI = δ
∫ t2
t1
Ldt = 0,
and the k equations of constraint
fj (q1, q2, . . . , qn) = 0, j = 1, . . . , k. (3.10)
If all the coordinates were independent, Hamilton’s principle would lead to n equations of the form
d
dt
( ∂L
∂ q ̇i
)
− ∂L
∂ qi
= 0, i = 1, . . . , n. (3.11)
But we cannot write this (yet) because as noted following Equation (3.9), this result depends on all the δqs being independent. Equations (3.10) represent holonomic constraints. Consequently
δfj = ∂fj
∂ q1
δq1 + ∂fj
∂ q2
δq2 + · · · + ∂fj
∂ qn
δqn = 0.
That is,
∑n
i=1
∂fj
∂ qi
δqi = 0 j = 1, . . . , k. (3.12)
Here the coefficients ∂fj /∂qi are functions of the qi. If Equation (3.12) holds, then multiplying it by some quantity, call it λj , will have no effect, so we can write it as
λj
∑n
i=1
∂fj
∂ qi
δqi = 0.
The λs are undetermined multipliers. There are k such equations and adding them together still yields zero, so
k ∑
j =1
λj
∑n
i=1
∂fj
∂ qi
δqi = 0.
Further, we can integrate from ti to tf and still not change the fact that the value is zero. This yields the useful relation
∫ t2
t1
dt
⎛
⎝
k ∑
j =1
λj
∑n
i=1
∂fj
∂ qi
δqi
⎞
⎠ = 0. (3.13)


3.5 Constraints and Lagrange’s λ-method 79
Equation (3.9) can be written as
∫ t2
t1
dt
∑
i
( ∂L
∂ qi
−d
dt
∂L
∂ q ̇i
)
δqi = 0. (3.14)
Adding Equations (3.13) and (3.14) we obtain
∫ t2
t1
dt
∑
i
⎛
⎝ ∂L
∂ qi
−d
dt
∂L
∂ q ̇i
+
k ∑
j =1
λj
∂fj
∂ qi
⎞
⎠ δqi = 0. (3.15)
The δqs are not all independent; they are related by the k equations of constraint. However, n − k of them are independent and for these coordinates the term in parenthesis in Equation (3.15) must be zero. For the remaining k equations we can select the undetermined multipliers λj such that
∂L
∂ qi
−d
dt
∂L
∂ q ̇i
+
k ∑
j =1
λj
∂fj
∂ qi
= 0.
Consequently, we have, for all qis, the following relations
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
=
k ∑
j =1
λj
∂fj
∂ qi
, i = 1, . . . , n. (3.16)
Equations (3.16) are n equations and Equations (3.10) give us k equations so we have n + k equations which can be solved to obtain expressions for the qs and for the λs. We now show how the λs are related to the generalized forces of constraint. Consider Lagrange’s equations written in the Nielsen form (Equation (3.4)). Let us write the equation in the following way, in which we separate the conservative forces Qc
i derivable from a potential function (V ) from the forces
of constraint Qnc
i (which in general cannot be derived from a potential that depends only on the coordinates):
d
dt
∂T
∂ q ̇i
− ∂T
∂ qi
= Qc
i + Qnc
i.
The conservative force (Qc) can be expressed as − ∂V
∂qi . Assuming that
∂V /∂q ̇i = 0, we can write this equation as
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
= Qnc
i . (3.17)


80 3 Lagrangian dynamics
a
qR
s
Figure 3.2 A disk rolling without slipping on an inclined plane.
But Equations (3.16) and (3.17) must be identical. Hence we have the following expression
Qnc
i=
k ∑
j =1
λj
∂fj
∂ qi
,
and we have obtained the desired expression for the forces of constraint.
Example 3.1 Consider a disk of radius R rolling down an inclined plane of length l and angle α. Find the equations of motion, the angular acceleration, and the force of constraint. See Figure 3.2.
Solution 3.1 The moment of inertia of a disk is I = 1
2 MR2, and the kinetic
energy is T = 1
2 Ms ̇2 + 1
2 I θ ̇ 2. The potential energy is V = Mg(l − s) sin α. Therefore,
L= 1
2 Ms ̇2 + 1
4 MR2θ ̇ 2 + Mg(s − l) sin α.
The disk is constrained to roll without slipping. Therefore,
f (s, θ ) = s − Rθ = 0.
This is a holonomic constraint. Consequently,
d
dt
∂L
∂s ̇ − ∂L
∂s = λ ∂f
∂s where ∂f
∂s = 1,
d
dt
∂L
∂θ ̇ − ∂L
∂θ = λ ∂f
∂θ where ∂f
∂θ = −R.


3.6 Non-holonomic constraints 81
Carrying out the indicated operations we obtain the two equations of motion
d
dt (Ms ̇) − Mg sin α = λ,
d
dt
(1
2 MR2θ ̇ 2
)
= −Rλ.
Furthermore, from the constraint equation we obtain one more equation, namely,
θ ̈ = s ̈/R.
You can easily show that
θ ̈ = 2
3
g sin α
R,
and
s ̈ = 2
3 g sin α.
It follows that
λ = Ms ̈ − Mg sin α = − 1
3 Mg sin α.
Now the forces of constraint are given by
Qs = λ ∂f
∂s = −1
3 Mg sin α = normal force,
and
Qθ = λ ∂f
∂θ = 1
3 MgR sin α = torque.
Qs and Qθ are the generalized forces required to keep the disk rolling down the plane without slipping.
Exercise 3.4 Fill in the missing steps in the example above to show that θ ̈ = (2/3)g sin α/R and s ̈ = (2/3)g sin α.
3.6 Non-holonomic constraints
Certain non-holonomic constraints can also be treated by Lagrange’s λ-method, as described in Section 2.4.2. Suppose the non-holonomic constraints are


82 3 Lagrangian dynamics
expressed as relations among the differentials of the coordinates. For example, if there were only one such constraint, it would be given by
A1dq1 + A2dq2 + · · · + Andqn = 0,
where the As are known functions of the qs. If this relation holds for the dqs, it will also hold for the δqs, so we can write
A1δq1 + A2δq2 + · · · + Anδqn = 0. (3.18)
But this last expression has the same form as the expression for δf for a holonomic constraint, which we wrote as
δf = ∂f
∂ q1
δq1 + ∂f
∂ q2
δq2 + · · · + ∂f
∂ qn
δqn = 0.
Therefore, we can use exactly the same reasoning as before, simply replacing the ∂f
∂q s with As. (Note, however, that although the As are known quantities, they are not partial derivatives of some known function.) Consequently, if our single non-holonomic constraint can be expressed as in Equation (3.18), the Lagrange equations are
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
= λAi, i = 1, 2, . . . , n.
If there is more than one non-holonomic constraint, we need to generalize this relation. For example, if there are m non-holonomic constraints we will have i equations of the form
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
=
∑m
k=1
λkAki i = 1, 2, . . . , n.
We can even use this technique for time dependent (“rheonomic”) constraints having the form
n∑
i=1
Aki dqi + Bkt dt = 0, k = 1, 2, . . . , m.
Here the coefficients (As and Bs) are, in general, functions of both the coordinates and time. As before, we can replace d with δ and write these constraints in the form
n∑
i=1
Aki δqi + Bkt δt = 0, k = 1, 2, . . . , m.


3.7 Virtual work 83
The coefficients of δt (the Bs) do not enter into the equation of motion because time is frozen during the virtual displacements, δqi. Thus the equations of motion take on the familiar form
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
=
∑m
k=1
λkAki, i = 1, 2, . . . , n.
Note, however, that the Bs enter into the relations between the velocities, thus
Ai1q ̇1 + Ai2q ̇2 + · · · + Ainq ̇n + Bi = 0, i = 1, 2, . . . , n.
3.7 Virtual work
We now consider the concept of virtual work in more detail than was done in Section 1.6. A mechanical system will be in equilibrium if and only if the total virtual work of all the “impressed” forces is zero. (Impressed forces are those applied to the system and do not include forces of constraint.1) The principle of virtual work for a system in equilibrium is expressed mathematically as
δW = ∑
j
Qj δqj = 0.
Here the virtual displacements δqi satisfy all the constraints. Note that the principle of virtual work is a variational principle. It is interesting to contrast the principle of virtual work with the Newtonian concept of equilibrium. In Newtonian mechanics when a system is in equilibrium, the vector sum of all of the forces acting on it is zero. That is, Newtonian mechanics replaces the constraints by forces. Analytical mechanics, on the other hand, does away with the reaction forces and only considers the impressed forces; however, it requires that any virtual displacements satisfy the constraints on the system. (As we shall see, considering virtual displacements that violate the constraints allows us to determine the forces of constraint.) In vector notation the generalized forces are just the components of the actual forces and the principle of virtual work can be expressed as
F1 · δr1 + F2 · δr2 + · · · + FN · δrN = 0.
In terms of Cartesian coordinates, this is
F1x δx1 + F1y δy1 + · · · + FNzδzN = 0.
1 The forces of constraint are often referred to as “reaction” forces.


84 3 Lagrangian dynamics
When written in this form, we appreciate that the scalar products of the forces and the virtual displacements are all zero. This means that the force Fi is perpendicular to any allowed displacement of particle i. For a free particle, all displacements are allowed, so the net force must be zero. For a particle restricted to a table top, the force must be perpendicular to the table top. If an equilibrium problem involves constraints, then the equilibrium condition can be determined using the Lagrange λ-method. We have justified the principle of virtual work, but we have not proved it. Let us leave it as a postulate. (In fact, it is the only postulate of analytical mechanics and is applicable to dynamical as well as equilibrium situations.2) Postulate The virtual work for a system in equilibrium is zero,
δW = 0.
Since Newtonian dynamics states that the total force on a system in equilibrium is zero, we conclude that the reaction forces are equal and opposite to the impressed forces. Thus, the postulate can be stated in terms of reaction forces as follows. Postulate The virtual work of any force of reaction is zero for any virtual displacement that satisfies the constraints on the system.
Exercise 3.5 Assume the impressed force is derivable from a scalar function. Show that the equilibrium state is given by the stationary value of the potential energy, δV = 0.
3.7.1 Physical interpretation of the Lagrange multipliers
In equilibrium, the virtual work of the impressed forces is zero:
δW = 0.
If the impressed forces can be derived from a potential energy (Fi = −∂V /∂qi) then the virtual work is equal to the negative of the variation of the potential energy because
δW = i Fiδqi = − i (∂V /∂qi )δqi = −δV .
If the physical system is subjected to a holonomic constraint given by
f (q1, . . . , qn) = 0,
2 For a discussion of the postulate, see Lanczos, op. cit. page 76.


3.7 Virtual work 85
then the Lagrange λ-method requires that
δ(L + λf ) = 0.
(See Equation (2.24).) Since λ is undetermined, we can just as well use −λ and define a modified Lagrangian by
L = L − λf.
Since L = T − V , this is equivalent to defining a modified potential energy V + λf. Therefore, δW = 0 is equivalent to
δ(V + λf ) = 0.
If we permit arbitrary variations of the generalized coordinates (not just those that satisfy the constraint), then reaction forces will act on the system. Consequently, the term λf can be interpreted as the potential energy related to the force of constraint. Thus, the xi component of the reaction force is
Fi = − ∂(λf )
∂ xi
= −λ ∂f
∂ xi
− ∂λ
∂ xi
f.
But f = 0, so
Fi = −λ ∂f
∂ xi
.
We conclude that a holonomic constraint is maintained by forces that can be derived from a scalar function. As we have seen, some non-holonomic constraints can also be dealt with using the Lagrangian λ-method; however, these forces (friction is an example) cannot be derived from a scalar function.
Example 3.2 You have probably proved by using calculus (perhaps in a previous mechanics course) that the curve formed by a hanging chain or heavy rope is a catenary. That problem can also be solved using variational techniques. The quantity to be minimized is the potential energy, V = V (y). The problem is to determine y = y(x) subject to the constraint that the length of the chain is a given constant. In variational terms, we want to determine y = y(x) given
that δV = 0 and subject to the constraint ∫ x2
x1 ds = constant = length = l.
Solution 3.2 The constraint can be expressed as
l=
∫ x2
x1
√
1 + y′2dx.


86 3 Lagrangian dynamics
The potential energy of a length of chain ds is dV = (dm)gh = ρgyds. Dropping unnecessary constants,
V=
∫ x2
x1
yds =
∫ x2
x1
y
√
1 + y′2dx.
The Lagrange λ-method leads to
δ
∫ x2
x1
(y + λ)
√
1 + y′2dx = 0.
Determining the equation of the curve is left as an exercises.
Exercise 3.6 The functional in the preceding example does not depend on x. Reformulate the problem as discussed in Section 2.2.2 and show that the curve is a catenary.
Exercise 3.7 Consider a simple pendulum composed of a mass m constrained by a wire of length l to swing in an arc. Assume r and θ are both variables. Obtain the Lagrange equations and using the λ-method, determine the tension in the wire. Check by using elementary methods.
3.8 The invariance of the Lagrange equations
The Lagrange equations are invariant under a point transformation. A point transformation is a transformation from a set of coordinates q1, q2, . . . , qn to a set s1, s2, . . . , sn such that for every point Pq in “q-space” there corresponds a point Ps in “s-space.” The transformation equations have the form
q1 = q1(s1, s2, . . . , sn, t),
...
qn = qn(s1, s2, . . . , sn, t).
It is not difficult to show that the Lagrange equations are invariant under a point transformation. (That is, they retain the same form. See Problem 3.3 at the end of the chapter.) However, it may be somewhat more difficult to appreciate the significance of this invariance, so let us consider it in more detail. Figure 3.3 shows an n-dimensional configuration space. At a given instant, the system is represented by a single point. As time goes on, this point moves


3.8 The invariance of the Lagrange equations 87
q1
q2
t
s1
s2
t
A
B A
B
C
C
~
~
~
Figure 3.3 “True” path in two different coordinate systems.
along a curve.3 The system moves from end point A to end point B along the curve C that minimizes the action. By considering varied paths between the same end points and requiring that the action integral be stationary for the “true” path, we obtain the conditions
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
= 0, i = 1, . . . , n.
Let the coordinates qi be transformed to si by a point transformation. The end points are transformed to A ̃, B ̃ and the curve C is transformed to C ̃, as shown in the right-hand panel of Figure 3.3. The action integral is minimized by the “true” path between the end points in any coordinate system, consequently the Lagrange equations are valid in the new coordinate system. Note, however, that although the value of L is the same in both coordinate system, its form may be quite different. The invariance of the Lagrange equations allows us to select whatever set of coordinates makes the equations easiest to solve. (The invariance principle becomes a postulate in the theory of special relativity in which Einstein postulated that the laws of physics are invariant under transformations between inertial reference frames.) The Lagrangian equation of motion for a particular i will not, in general, be the same in the two systems, but the complete sets of equations are equivalent. For this reason, one should actually refer to the equations as co-variant rather than in-variant. For example, if
L= 1
2 x ̇2 + 1
2 y ̇2 + 1
(x2 + y2)1/2 ,
3 We can include the time as another coordinate, and represent the system as a point in an (n + 1)-dimensional configuration space. In this space, the system traces out a curve or “world line” that gives the history of the system.


88 3 Lagrangian dynamics
a point transformation, x = r cos θ , y = r sin θ , yields the Lagrangian
L= 1
2 r ̇2 + 1
2 r2θ ̇ 2 + 1
r.
The equations of motion in terms of x and y are
x ̈ = − x
(x2 + y2)3 y ̈ = − y
(x2 + y2)3 ,
whereas the equations of motion in terms of r and θ are
r ̈ − rθ ̇ 2 + 1
r =0 d
dt (r2θ ̇ ) = 0.
Although these equations of motion are quite different in form, they yield the same result and for a particular set of the position and velocity parameters they yield the same numerical value for the Lagrangian.
3.9 Problems
3.1 Assume the generalized force is derivable from a potential independent of the generalized velocities. Show that Equations (3.1) follow from Equations (3.4). 3.2 Use d’Alembert’s principle to determine the equilibrium condition for the system illustrated in Figure 3.4. 3.3 If one expresses the Lagrangian in terms of some set of generalized coordinates, qi, the Lagrange equations of motion have the form
d
dt
∂L
∂ q ̇i
− ∂L
∂ qi
= 0.
Assume that we transform from the qi to a new set of coordinates si by a so-called “point transformation”
qi = qi (s1, s2, . . . , sn, t).
mM
qf
Figure 3.4 A system in equilibrium.