Analytical Mechanics

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Analy tical Mechanics

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ANALY TICAL MECHANICS

Solutions to Problems in Classical Physics   5     5R5    5   

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2015 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20140612 International Standard Book Number-13: 978-1-4822-3940-9 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

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We can’t solve problems by using the same kind of thinking we used when we created them. Albert Einstein

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Free ebooks ==> www.Ebook777.com PREFACE

As the story goes, not everything new is also useful, and not everything old is also obsolete. In theoretical physics, one of the most convincing examples in this respect is offered by Analytical Mechanics. Created by Jean Bernoulli (1654-1705), Pierre Louis Moreau de Maupertuis (1698-1759), Leonhard Euler (1707-1783), Jean le Rond D’Alembert (1717-1783), Joseph Louis Lagrange (1736-1813), Karl Gustav Jacobi (1804-1851), William Rowan Hamilton (1805-1865), Jules Henri Poincar´e (1854-1912), and other prominent minds, Analytical Mechanics proved to be a very useful tool of investigation not only in Newtonian Mechanics, but also in almost all classical and modern branches of physics: Electrodynamics, Quantum Field Theory, Theory of Relativity, etc. It can be stated, without exaggeration, that Analytical Mechanics is essential in understanding Theoretical Physics, being a sine qua non condition of a profound training of a physicist. Due to its large field of applications, we dare to say that the term ”Analytical Mechanics” is somewhat overtaken (out of date). One of the essential properties of Analytical Mechanics is that it uses abstract, mathematical techniques as methods of investigation. By its object, Analytical Mechanics is a physical discipline, while its methods belong to various branches of mathematics: Analytical and Differential Geometry, Analysis, Differential Equations, Tensor Calculus, Calculus of Variations, Algebra, etc. That is why a physicist who studies analytical formalism must have an appropriate mathematical training. It is widely spread the idea that it is more important to learn and understand the practical applications of physics, than the theories. In our opinion, connection between Analytical Mechanics and the important chapter devoted to its applications is similar to that between physical discoveries and engineering: if the discovery of electricity, electromagnetic waves, nuclear power, etc., haven’t been put into practice, they would have remained within the laboratory frame. The purpose of this collection of solved problems is intended to give the students possibility of applying the theory (Lagrangian and v

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Free ebooks ==> www.Ebook777.com Hamiltonian formalisms for both discrete and continuous systems, Hamilton-Jacobi method, variational calculus, theory of stability, etc.) to problems concerning several chapters of Classical Physics. Some problems are difficult to solve, while others are easy. One chapter (the third), as a whole, is dedicated to the gravitational plane pendulum, the problem being solved by all possible analytical formalisms, including, obviously, the Newtonian approach. This way, one can easily observe similarities and differences between various analytical approaches, and their specific efficiency as well. When needed, some theoretical subjects are developed up to some extent, in order to offer the student possibility to follow solutions to the problems without appealing to other reference sources. This has been done for both discrete and continuous physical systems, or, in analytical terms, systems with finite and infinite degrees of freedom. A special attention is paid to basics of vector algebra and vector analysis, in Appendix B. Notions like: gradient, divergence, curl, tensor, together with their physical applications, are thoroughly developed and discussed. This collection of solved problems is a result of many years of teaching Analytical Mechanics, as the first course of theoretical physics, to the students of the Faculty of Physics. There are many excellent textbooks dedicated to applied Analytical Mechanics for both students and their instructors, but we modestly pretend to offer an original view on distribution of the subjects, the thorough analysis of solutions to the problems, and an appropriate choice of applications in various branches of Physics: Mechanics of discrete and continuous systems, Electrodynamics, Classical Field Theory, Equilibrium and small oscillations, etc. IASI, February 2014 The authors

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Free ebooks ==> www.Ebook777.com CONTENTS

CHAPTER I. FUNDAMENTALS OF ANALYTICAL MECHANICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 I.1. Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 I.1.1. Classification criteria for constraints . . . . . . . . . . . . . . . . . . 2 I.1.2. The fundamental dynamical problem for a constrained particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 I.1.3. System of particles subject to constraints . . . . . . . . . . . . . 9 I.1.4. Lagrange equations of the first kind . . . . . . . . . . . . . . . . . . 11 I.2. Elementary displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 I.2.1. Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 I.2.2. Real, possible and virtual displacements . . . . . . . . . . . . . 13 I.3. Virtual work and connected principles . . . . . . . . . . . . . . . . . . . 19 I.3.1. Principle of virtual work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 I.3.2. Principle of virtual velocities . . . . . . . . . . . . . . . . . . . . . . . . 22 I.3.3. Torricelli’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 CHAPTER II. PRINCIPLES OF ANALYTICAL MECHANICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 II.1. D’Alembert’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 II.1.1. Configuration space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 II.1.2. Generalized forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 II.2. Hamilton’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 CHAPTER III. THE SIMPLE PENDULUM PROBLEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 III.1. Classical (Newtonian) formalism . . . . . . . . . . . . . . . . . . . . . . . 47 III.2. Lagrange equations of the first kind approach . . . . . . . . . . 68 III.3. Lagrange equations of the second kind approach. . . . . . . .72 III.4. Hamilton’s canonical equations approach . . . . . . . . . . . . . . . 78 III.5. Hamilton-Jacobi method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 III.6. Action-angle variables formalism . . . . . . . . . . . . . . . . . . . . . . . 85 CHAPTER IV. PROBLEMS SOLVED BY MEANS OF THE PRINCIPLE OF VIRTUAL WORK . . . . . . . . . 92 vii

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Free ebooks ==> www.Ebook777.com CHAPTER V. PROBLEMS OF VARIATIONAL CALCULUS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 V.1. Elements of variational calculus . . . . . . . . . . . . . . . . . . . . . . . . 110 V.1.1. Functionals. Functional derivative . . . . . . . . . . . . . . . . . 110 V.1.2. Extrema of functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 V.2. Problems whose solutions demand elements of variational calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 1. Brachistochrone problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 2. Catenary problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 3. Isoperimetric problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 4. Surface of revolution of minimum area . . . . . . . . . . . . . . . 136 5. Geodesics of a Riemannian manifold . . . . . . . . . . . . . . . . . 139 CHAPTER VI. PROBLEMS SOLVED BY MEANS OF THE LAGRANGIAN FORMALISM . . . . . . . . . . . . . 156 1. Atwood machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 2. Double Atwood machine. . . . . . . . . . . . . . . . . . . . . . . . . . . . .158 3. Pendulum with horizontally oscillating point of suspension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 4. Problem of two identical coupled pendulums . . . . . . . . . 172 5. Problem of two different coupled pendulums . . . . . . . . . 178 6. Problem of three identical coupled pendulums . . . . . . . 203 7. Problem of double gravitational pendulum . . . . . . . . . . . 210 CHAPTER VII. PROBLEMS OF EQUILIBRIUM AND SMALL OSCILLATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . 227 CHAPTER VIII. PROBLEMS SOLVED BY MEANS OF THE HAMILTONIAN FORMALISM . . . . . . . . 265 CHAPTER IX. PROBLEMS OF CONTINUOUS SYSTEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 A. Problems of Classical Electrodynamics . . . . . . . . . . . . . . 313 B. Problems of Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . 332 C. Problems of Magnetofluid Dynamics and Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 APPENDICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439

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CHAPTER I FUNDAMENTALS OF ANALYTICAL MECHANICS

I.1. Constraints The concept of constraint plays an essential role in analytical mechanics. Both Newtonian and analytical mechanics work with notions like: material point (particle), velocity, acceleration, mass, force, kinetic energy, mechanical work, etc., but the notion of constraint is specific to analytical mechanics only. The difference comes from the acceptance of the concept of ”freedom”. In Newtonian mechanics, a body is free if no force acts on it. For instance, consider a body under the influence ~ = m~g . From the Newtonian mechanics point of gravitational force, G of view, the gravity acts permanently on the body, meaning that the body cannot be considered as being free. But, in view of the analytical mechanics formalism, this body is considered free, in the sense that no restriction limits its motion. In other words, within the analytical approach, a body that can move freely along any direction in space, and rotate about any axis, is said to be free. Definition. A constraint is a geometric or kinematic condition that restricts the motion of a body. The constraints are usually given as equalities and/or inequalities, either in explicit, implicit, or parametric forms. As an example, here is a constraint under the implicit form f (~r, ~v , t) = 0, (1.1) 1 where f is a function of class C on its domain of definition (as required by the Lagrange equations of the first kind formalism). Here are a few examples of geometric constraints for a particle (material point): the body is constrained to move on a certain curve (a teleferic moving along its cable of suspension), on a certain surface (a car climbing an inclined street), or inside a certain volume (a little piece of stone traveling inside a soccer ball). A constraint can be assimilated with a constraint force. The forces 1

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Free ebooks ==> www.Ebook777.com of constraint determine the body to move on a certain curve, a certain surface, or in a certain volume. Unlike Newtonian mechanics, the analytical formalism makes a clear distinction between applied and constraint forces. While Newtonian mechanics demands for knowledge of all kinds of forces, so as to let us to write the fundamental equation of dynamics, the analytical mechanics replaces the a priori knowledge of constraint forces by the cognition of the analytical expressions of the constraints (the case of Lagrange equations of the first kind), or, even, such a replacement is not necessary (Lagrange equations of the second kind, Hamilton’s canonical equations, and the Hamilton-Jacobi formalism). According to the basic concepts of analytical mechanics, the constraint forces are determined a posteriori, that is after the law of motion of the mechanical system has been determined. If fact, the analytical formalism allows one to solve even most complicated problems, where the constraint forces are not known from the beginning. I.1.1. Classification criteria for constraints There are at least three criteria for classification of the constraints, in terms of the following reasons: 1. Constraints can be expressed either by equalities, or by inequalities. 2. The time t explicitly interferes in the expression of the constraint, or it doesn’t. 3. The constraint explicitly depends on the velocity ~v , or it doesn’t. 1. Constraints expressed by equalities are called bilateral, while those given by inequalities are named unilateral. Here are a few examples of bilateral constraints: i) A heavy particle of mass m moving on the fixed sphere of radius R (spherical pendulum - see Fig.I.1): x21 + x22 + x23 = R2 , or f (x1 , x2 , x3 ) = x21 + x22 + x23 − R2 = 0.

(1.2)

ii) A particle moving on the fixed circular cone, of radius R and apex angle 2θ (see Fig.I.2) x21 + x22 = (R − x3 tan θ)2 , or f (x1 , x2 , x3 ) = x21 + x22 − x33 tan2 θ + 2Rx3 tan θ − R2 = 0. 2

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(1.3)

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Fig.I.1

Fig.I.2 iii) A particle moving on a sphere of fixed radius R whose centre moves uniformly in a straight line with velocity ~vC = (~a, ~b, ~c) (x − at)2 + (y − bt)2 + (z − ct)2 = R2 , 3

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Free ebooks ==> www.Ebook777.com or f (x1 , x2 , x3 , t) ≡ f (~r, t) = (x−at)2 +(y−bt)2 +(z−ct)2 −R2 = 0. (1.4) iv) A body (conceived as a heavy particle of mass m) moving in the gravitational field, suspended at a fixed point O by an inextensible rod of length l (rod pendulum - see Fig.I.3) x21 + x22 = l2 , or f (x1 , x2 ) = x21 + x22 − l2 = 0.

Fig.I.3 Next, we shall give some examples of unilateral constraints: i) A heavy particle of mass m suspended by means of an inextensible but flexible wire of length l (wire pendulum - see Fig.I.4) x21 + x22 ≤ l2 , or f (x1 , x2 ) = x21 + x22 − l2 ≤ 0.

(1.5)

ii) A heavy particle of mass m moving inside a rugby ball at rest. If the ball has the shape of an ellipsoid of rotation of semiaxes a, b, c, the constraint writes x21 x22 x23 + + ≤ 1, a2 b2 c2 4

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Fig.I.4 or

x21 x22 x23 + + − 1 ≤ 0. (1.6) a2 b2 c2 iii) A particle of mass m moving inside a soccer ball, conceived as a sphere of radius R, whose center moves uniformly in a straight line with the velocity ~vC = (~a, ~b, ~c) f (x1 , x2 , x3 ) =

(x − at)2 + (y − bt)2 + (z − ct)2 ≤ R2 , or f (x1 , x2 , x3 , t) ≡ f (~r, t) = (x−at)2 +(y−bt)2 +(z−ct)2 −R2 ≤ 0. (1.7) 2. If the constraint equation does not contain the time variable t explicitly [like (1.2), (1.3), (1.5), and (1.6)], it is called scleronomous or stationary, while a constraint which is time-dependent [like (1.1), (1.4), and (1.7)] is named rheonomous or non-stationary. 3. A constraint is called geometric or finite if the components of the velocity do not appear in the constraint equation [e.g. (1.2)-(1.7)]. If, on the contrary, the constraint is velocity-dependent, like (1.1), it is named kinematic or differential. As an example of kinematic constraint, consider a body whose velocity must be permanently tangent to the curve (x1 − ϕ)2 x22 − = 1, a2 b2 which is a hyperbola of semi-axes a and b, and focal length ϕ. The time derivative then yields f (x1 , x2 , x˙ 1 , x˙ 2 ) = b2 (x1 − ϕ)x˙ 1 − a2 x2 x˙ 2 = 0. 5

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(1.8)

Free ebooks ==> www.Ebook777.com In this context, let us take the time derivative of the geometric constraint f (x1 , x2 , x2 , t) = 0. The result can be written as ∂f ∂f x˙ i + =0 ∂xi ∂t

(i = 1, 3),

showing that any geometric constraint can be written as a linear kinematic constraint. (The reciprocal is not true!). Those differential constraints which can be written in a finite form are called integrable. The geometric and integrable constraints are called holonomic, while the unilateral and non-integrable constraints are called non-holonomic. It is worthwhile to mention that there are no general methods to solve problems implying non-holonomic constrains. Each case is investigated separately, by specific methods, and solution depends on how skillful the researcher is. If such a problem does not admit an analytical solution, then one must appeal to numerical methods. Usually, a constraint is investigated from all points of view at the same time. In our previous examples, the constraint (1.3) is bilateral, scleronomous and finite, (1.7) is unilateral, rheonomous and geometric, while (1.8) is bilateral, scleronomous and differential. There exists a very strong connection between constraints and the number of degree of freedom of a mechanical system. By definition, the number of real, independent parameters that uniquely determine spatial position of a body, is called the number of degree of freedom of that body. This notion can be generalized to any system of particles or rigid bodies. (N.B. A continuous, deformable medium is considered to have an infinite number of degrees of freedom). Here are some examples. A free particle (material point) in the Euclidean apace E3 has three degrees of freedom; if the particle is forced to move on a surface or a curve, its number of degrees of freedom reduces by one - or by two, respectively. In its turn, a free rigid body has six degrees of freedom: three associated with translation along some axis, and three corresponding to rotation about the axis. This number diminishes if the body is submitted to one (or more) constraints. In general, each geometric bilateral constraint reduces by one the number of degrees of freedom of a mechanical system. The number of constraints must be smaller than the number of degrees of freedom; otherwise, the mechanical problem would become senseless.

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Free ebooks ==> www.Ebook777.com I.1.2. The fundamental problem of Dynamics for a constrained particle Task. Given the mass m of a particle moving on a fixed curve (Γ), the resultant F~ of applied forces, and the initial conditions compatible with the constraints, ~r0 = ~r(t0 ), ~r˙ 0 = ~r˙ (t0 ), determine the law of ~ of the motion of the particle ~r = ~r(t), as well as the resultant L constraint forces. Solution. Let us consider the following two possibilities: i) The curve is given parametrically by xi = xi (q) (i = 1, 3), where q is a real, time-dependent parameter. Most generally, the resultant F~ of the active forces is given as F~ = F~ (~r, ~r˙ , t), so that we have the following parametric dependence: xi =xi (q) F~ (~r, ~r˙ , t) −→ F~ (q, q, ˙ t).

To solve the problem, we appeal to the fundamental equation of dynamics ~ m ~¨r = F~ + L, (1.9) ~ are unknown quantities. Projecting (1.9) on the axes where ~r and L of a three-orthogonal reference frame Oxyz, we also have m¨ xi = Fi + Li

(i = 1, 3).

(1.10)

This is a system of three ordinary differential equations with four unknowns: Lx , Ly , Lz , and q [because xi = xi (q)]. Therefore, we need ~ in one extra equation. To this end, one decomposes the vector L ~ ~ ~ ~ ~ n, two mutually orthogonal vector components Lt and Ln , L = Lt + L ~ t is tangent to the curve (Γ) at the current point P , while L ~ n is where L situated in a plane normal to the curve at the same point (see Fig.1.5). ~ t is called force of friction, and the component L ~n The component L ~ t = 0, the motion is named frictionless and the normal reaction. If L ~ n = 0 the curve (Γ) is perfectly curve is ideal or perfectly smooth. If L ~ is tangent to the curve. rough, and the force L We shall further suppose a heavy particle of mass m moving on ~ t = 0). Recalling that the instantaneous velocity the ideal curve (Γ) (L ˙ ~v = ~r is always tangent to the trajectory, which in our case is the curve (Γ), we can write ~ · ~v = xL L ˙ x + yL ˙ y + zL ˙ z = 0. 7

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(1.11)

Free ebooks ==> www.Ebook777.com This way, the problem is virtually solved: we are left with a system of four differential equations  m¨ xi = Fi + Li , (1.12) (i = 1, 3) x˙ i Li = 0, with four unknowns, q and Li (i = 1, 3).

Fig.I.5 ii) The curve is given by the intersection of two implicit surfaces  f1 (x, y, z) = 0, f2 (x, y, z) = 0. There are two ways to obtain the solution in this case: a) Solve a system of six differential equations  m¨ xi = Fi + Li ;   x˙ i Li = 0; (i = 1, 3)   f1 (x, y, z) = 0; f2 (x, y, z) = 0,

(1.13)

with six unknowns x, y, x, Lx , Ly , Lz . ~ along two directions, given b) Decompose the constraint force L by ∇f1 and ∇f2 , which are normal to the two spatial surfaces whose intersection gives rise to curve (Γ) (see Fig.1.6). Supposing, again, that the curve is ideal, we can write ~ =L ~1 + L ~ 2 = µ∇f1 + ν∇f2 , L 8

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Free ebooks ==> www.Ebook777.com where µ and ν are two scalar multipliers. The scalar components of the differential equation of motion then write m¨ xi = Fi + µ

∂f1 ∂f2 +ν ∂xi ∂xi

Therefore, we are left with five equations  ∂f1 ∂f2   m¨ xi = Fi + µ +ν , ∂xi ∂xi   f1 (x, y, z) = 0, f2 (x, y, z) = 0,

(i = 1, 3).

(i = 1, 3)

(1.14)

(1.15)

with five unknown quantities x, y, z, µ, and ν. This procedure allows one to determine both the law of motion ~r = ~r(t), and the constraint ~ force L.

Fig.I.6 Observation. We have restricted our investigation to ideal constraints, but in reality the force of friction cannot be neglected. Nevertheless, our formalism remains valid if the tangential component ~ t ) of the constraint force is known. So, instead of F~ , as resulF~f (≡ L tant of the applied forces is considered F~ ′ = F~ + F~f , and the rest of procedure remains unchanged. I.1.3. System of particles subject to constraints Consider a system of N ≥ 2 material points (particles). If at any moment t of the motion the position radius-vectors ~ri (i = 1, N ) of the 9

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Free ebooks ==> www.Ebook777.com particles and their velocities ~r˙ i (i = 1, N ) can take arbitrary values, we say that the system is free. If not, the system is subject to constraints. The most general expression of a (bilateral) constraint for a system of N particles writes f (~r1 , ~r2 , ..., ~rn , ~r˙ 1 , ~r˙ 2 , ..., ~r˙ n , t) = 0,

(1.16)

and involves both geometric and kinematic conditions obeyed by the positions and velocities of the particles. The classification criteria for a system of particles is similar to that for a single material point. For example, relations fj (~r1 , ~r2 , ..., ~rn , ~r˙ 1 , ~r˙ 2 , ..., ~r˙ n , t) = 0

(j = 1, l, l ≤ 3N )

(1.17)

constitute l bilateral, rheonomous, and differential constraints, while fj (~r1 , ~r2 , ..., ~rn , t) = 0

(j = 1, l, l ≤ 3N )

(1.18)

stand for l bilateral, rheonomous, and geometric constraints. The number l of constraints which restricts the possibilities of motion of a system of N particles cannot be bigger that 3N . If l = 3N , the status of mechanical system would be completely determined by the constraints, so that integration of the differential equations of motion would become senseless. As in the case of one particle, any geometric constraint can be written in a differential form by taking the total derivative with respect to time. For example, the constraints (1.17) can also be written as N X i=1

 ∂fj =0 gradi fj · ~r˙ i + ∂t

(j = 1, l),

which are l kinematic constraints linear in velocities, of the type N X i=1

~gij (~r1 , ~r2 , ..., ~rN , t) · ~r˙ i + g0j (~r1 , ~r2 , ..., ~rN , t) = 0

(j = 1, l).

The kinematic constraints can be integrable (holonomic) or nonintegrable (non-holonomic, or Pfaffian). All definitions met in the case of one-particle system remain valid.

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Free ebooks ==> www.Ebook777.com I.1.4. Lagrange equations of the first kind As we have seen, the fundamental equation of dynamics for the ith-particle of mass mi , which belongs to a system of N particles, writes ~ i (i = 1, N ), (1.19) mi~¨ri = F~i + L ~ i are the resultants of the applied and constraint forces, where F~i and L respectively. We also recall that, if a particle moves on the ideal surface f (x, y, z) = 0,

(1.20)

the constraint force writes ~ ≡L ~ n = µ∇f, L while in case of the motion on an ideal curve given by  f1 (x, y, z) = 0, f2 (x, y, z) = 0,

(1.21)

the constraint force is ~ = µ∇f1 + ν∇f2 . L These results can be generalized for a system of N particles, subject to l ≤ 3N ideal constraints. Thus, the constraint force acting on the ith-particle writes ~i = L

l X j=1

 µj gradi fj ,

(1.22)

and (1.19) become mi~¨ri = F~i +

l X

µj gradi fj

j=1



(i = 1, N ).

(1.23)

Equations (1.23) are called the Lagrange equations of the first kind. The differential equations (1.23), together with equations (1.18) of the constraints, namely  l X     mi~¨ri = F~i + µj gradi fj ;  i = 1, N , j = 1, l , (1.24) j=1    fj (~r1 , ~r2 , ..., ~rn , t) = 0; 11

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Free ebooks ==> www.Ebook777.com form a system of 3N + l scalar equations with 3N + l unknowns: xi , yi , zi (i = 1, N ) and µj (j = 1, l). This formalism is one of the main procedures used by analytical mechanics, and we shall apply it in the study of gravitational pendulum (see Chap.III). I.2. Elementary displacements I.2.1. Generalities Beside the concept of constraint, analytical mechanics deals with the notion of elementary displacement. By means of this new concept, we are able to eliminate even the necessity of knowledge of the analytical expressions of the constraints. This way, the ”power” of the analytical formalism sensibly increases and the problem can be solved even in those cases when identification of the constraints is difficult. Schematically, the situation can be presented as follows: Classical (Newtonian) formalism Requirement of knowledge of all forces from the beginning (or, at least, their direction and sense) ↓ Lagrange equations of the first kind formalism Knowledge of the constraint forces is not necessary, but the analytical expressions of the constraints is demanded ↓ Lagrange equations of the second kind approach Knowledge of both the constraint forces, and the analytical expressions of the constraints is not necessary Requirement of identification of the number of constraints only The concept of elementary displacement has a pragmatic reason. Obviously, solving a statics problem is much easier than investigating an application on kinematics or dynamics. Indeed, in the first case the problem reduces to a system of algebraic equations, while in the second solution is obtained by solving a system of differential equations. In 12

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Free ebooks ==> www.Ebook777.com other words, the main idea is to transform a dynamical problem into a problem of statics: determine the equilibrium of a mechanical system, instead of integrating a set of differential equations to find its law of motion. To this end, it is necessary to introduce the concepts of real, possible, and virtual elementary displacements. I.2.2. Real, possible and virtual displacements Consider a system of N ≥ 2 particles, subject to l < 3N holonomic constraints. During the time interval dt, under the action of applied force F~i , the ith-particle of the system suffers the elementary displacement d~ri , subject to both the constraints and the initial conditions compatible with the constraints. This displacement is unique and takes place effectively. It is called real elementary displacement. The difference between possible and virtual displacements can only be clearly understood if the essential role of the time variable is considered. So, let us suppose that our system is subject to l holonomic, rheonomous constraints of the form fj (~r1 , ~r2 , ..., ~rN , t) = 0

(j = 1, l).

(1.25)

If we only set the position of the ith-particle at time t, there can exist an infinite number of velocities of the particle consistent with the constraints. The displacements performed under these conditions are called possible. Among all possible displacements only one is real, namely that satisfying both the equation of motion and the initial conditions. Consider, finally, a class of elementary displacements only consistent with the constraints. These displacements are not real, but virtual. They are purely geometric and are synchronic (do not depend on the time t). Therefore, if the elementary possible displacements are considered as vectors, then all vectors representing these displacements (whose number is infinite) have the same origin, situated in some point of the variety which represents the constraint at the moment t, and their terminal points lying at any point of this variety, but considered at the moment t + dt. In contrast, the virtual displacements are always tangent to the variety representing the constraint and, being elementary, they belong in fact to this variety. In other words, the essential difference between possible and virtual displacements lies in the fact that the first ”emerge” from the variety, while the last belong to the variety. This difference can accurately be understood only if one considers the most general case of rheonomous constraints. While the 13

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Free ebooks ==> www.Ebook777.com possible elementary displacements perform in time, the virtual ones are ”instantaneous”, and can be intuitively considered as being taken on a ”frozen” constraint (or, similarly, at a ”frozen” moment of time).

Fig.I.7 To illustrate these concepts, consider a material point (particle) constrained to remain on an inclined plane which moves along a straight line with a constant velocity ~vplane (see Fig.I.7). Here are given two successive positions of the plane, separated by the infinitesimal distance d~rplane , corresponding to the infinitesimal time interval dt. At the moments t and t + dt the side AB of the plane is situated at the points Q(t) and Q′ (t + dt), while P (t) and P ′ (t + dt) denote the positions of the particle on the plane at time t and t + dt, respectively. According to our definition, the infinitesimal vectors d~r, d~r1 , d~r2 , d~r3 , etc., with their origin at P (t) and arrow-heads at P ′ , P1 , P2 , etc. represent possible elementary displacements of the particle P . Among all −−→ these displacements, only one is real, namely d~r = P P ′ . It satisfies not only the constraint, but also the equations of motion and the initial conditions. Let us now ”freeze” the plane at the moment t′ = t + dt, when the side AB is at the point Q′ (t′ ) = Q′ (t + dt). This can be imagined as taking a picture of the plane at the moment t′ . Since we are now ”out of time”, the particle can ”move” only on the plane, which becomes - for an instant - a scleronomous constraint. The elementary displacements of the particle satisfy, in this case, only the equation of the constraint. They are called virtual displacements and represent a very useful tool in analytical mechanics. Some examples of virtual displacements are shown in Fig.I.8. Here the real displacement is denoted by d~r, the possible displacements by d~r1 , d~r2 , etc., while the 14

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Free ebooks ==> www.Ebook777.com virtual displacements are marked by the Greek letter δ: δ~r1 , δ~r2 , etc., or δ~r12 , δ~r34 , etc.

Fig.I.8 Let us show that any elementary virtual displacement can be written as the difference of two possible displacements. To this end, consider a system of particles (material points) subject to l holonomic, rheonomous constraints of the type (1.25): fj (~r1 , ~r2 , ..., ~rN , t) = 0

(j = 1, l),

where ~ri = ~ri (t) (i = 1, N ). Differentiating these relations, we obtain the conditions that must be satisfied by the possible displacements d~ri (i = 1, N ) N X i=1

 ∂f gradi fj · d~ri + dt = 0 ∂t

(j = 1, l).

(1.26)

Recalling that dt ≡ δt = 0 expresses the condition for the elementary displacements δ~ri to be virtual (or, equivalently, atemporal), equation (1.26) leads to N X i=1

 gradi fj · δ~ri = 0

(j = 1, l).

(1.27)

Since the vectors gradi fj are orthogonal to the constraint surfaces fj (~r1 , ~r2 , ..., ~rN , t) = 0 (j = 1, l), relations (1.27) show that the virtual displacements are always tangent to the constraints. Obviously, this statement is not true for the real and possible displacements. For 15

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Free ebooks ==> www.Ebook777.com example, given l scleronomous constraints fj (~r1 , ~r2 , ..., ~rN ) = 0, the possible elementary displacements satisfy the relations N X i=1

 gradi fj · d~ri = 0

(j = 1, l),

(1.28)

while in case of rheonomous constraints we have N X i=1

 ∂fj gradi fj · d~ri = − dt 6= 0 ∂t

(j = 1, l).

Let us now write the relations (1.26) for two distinct sets of possible elementary displacements, d~ri ′ and d~ri ′′ : N X

 ∂fj dt = 0 gradi fj · d~ri ′ + ∂t

N X

 ∂fj gradi fj · d~ri ′′ + dt = 0 ∂t

i=1

i=1

(j = 1, l);

(j = 1, l),

and subtract them from each other. The result is N X i=1

  gradi fj · d~ri ′ − d~ri ′′ = 0

(j = 1, l).

Denoting δ~ri = d~ri ′ − d~ri ′′ we still have

N X i=1

(i = 1, N ),

 gradi fj · δ~ri = 0

(1.29)

(j = 1, l),

which are precisely relations (1.27) satisfied by the virtual displacements δ~ri . Consequently, the above statement is proved: any virtual displacement can be conceived as a difference between two possible displacements. This also explains our choice to denote the virtual displacements by two indices in Fig.I.8, such as δ~r12 = d~r2 − d~r1 , etc. , 16

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Free ebooks ==> www.Ebook777.com or, in general δ~rmn = d~rn − d~rm

 ∀ m, n = 1, N .

If one of the two possible displacements is the real one, we have δ~ri = ±d~r ∓ d~ri

(i = 1, N ).

For example, according to Fig.I.8, the elementary displacement δ~r1 writes δ~r1 = −d~r + d~r1 , while δ~r2 is given by δ~r2 = d~r − d~r2 .

Fig.I.9 As another example of a rheonomous constraint, consider a spherical balloon whose centre is fixed, being in process of inflation, as illustrated in Fig.I.9. Suppose that at two successive moments of time t and t + dt the radii of the balloon are R and R + dR, respectively. Let us presume that an ant, considered as a particle (material point) P , moves on the surface of the balloon. If at the moment t0 = 0 the radius of the balloon is zero, and the radius is increasing by a constant radial velocity vR , then the radius is R = R(t) = vR t, and the analytic expression of the constraint writes 2 2 f (x, y, z, t) = x2 + y 2 + z 2 − vR t = 0.

17

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Free ebooks ==> www.Ebook777.com If at the moment t the ant is located at the point P , then at the next moment t′ = t + dt the ant can be situated at any point (P1 , P2 , P3 , etc.) on the sphere of radius R′ = R(t′ ) = vR t′ = R + dR. In this case, the possible elementary displacements of the ant are represented by the vectors d~r1 , d~r2 , d~r3 , etc., whose number is practically infinite. Among all these possible displacements, there is one which satisfies not only the equation of the constraint, but also the differential equation of motion and the initial conditions as well: the elementary real displacement d~r. If, at the moment t′ = t + dt, the inflation of the balloon would be stopped (or ”frozen” - to use a suggestive image), the ant could move only on the sphere of radius R′ = R+dR. These imaginary, time-independent displacements are purely geometric, and have nothing to do with the real motion of the ant. They are called virtual elementary displacements. Two such displacements are indicated in Fig.I.9: δ~r23 = d~r3 − d~r2 , and δ~r4 = d~r − d~r4 . As a last example, let us suppose we are watching a movie. The motion picture is composed by a succession of images, whose frequency is settled at about 16-20 images per second, in accordance with the biological possibility of the human eye to distinguish two successive pictures. If, by any chance, the movie suddenly stops, only one static picture remains on the screen. This picture is very similar to the previous and following ones, but nevertheless distinct, being separated by infinitesimal time intervals. Any imaginary displacement, ”performed” at an instant t (δt = 0) on the static picture, can be considered as a virtual displacement. Therefore, if we denote −MRED the set of real elementary displacements ; −MP ED the set of possible elementary displacements ; −MV ED the set of virtual elementary displacements , in case of the rheonomous constraints we then have MRED ⊂ MP ED ,

MRED ∩ MV ED = ⊘,

MP ED ∩ MV ED = ⊘,

while the scleronomous constraints obey the rule MRED ⊂ MP ED ≡ MV ED . We nevertheless note that, generally speaking, in the last case the concept of possible elementary displacement cannot be defined. 18

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Free ebooks ==> www.Ebook777.com The table given below displays a synthetic image on the three types of elementary displacements, in case of rheonomous constraints.

Elementary displacements table

I.3. Virtual work and connected principles I.3.1. Principle of virtual work As we have seen in the previous paragraph, conversion of a dynamic problem into a static one brings the remarkable advantage of replacing the difficulties of solving a system of differential equations by the easier task of finding solution to an algebraic system of equations. In its turn, any problem of statics implies determination of the equilibrium conditions/positions of the system of particles under discussion. This can be done by means of the principle of virtual work. By definition, the elementary mechanical work of a force F~ by an infinitesimal displacement d~r of its point of application is given by the dot product dW = F~ · d~r. By analogy, the elementary virtual work corresponding to a virtual displacement δ~r can be defined as δW = F~ · δ~r. To understand how this notion can be applied, consider a system of N ≥ 2 particles (material points) and let δ~ri be a virtual elementary displacement of the particle Pi acted upon by the force F~i . Then, following the above definition, the infinitesimal virtual work ”performed” by F~i is δW = F~i · δ~ri . (1.30) 19

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Free ebooks ==> www.Ebook777.com Since the material point is at static equilibrium, ~¨ri = 0, and the fundamental equation of dynamics written for this point ~i mi~¨ri = F~i + L yields

~ i = 0. F~i + L

Performing the scalar product of this equation with δ~ri and summing over index i, we obtain N X i=1

F~i · δ~ri +

N X i=1

~ i · δ~ri = 0. L

(1.31)

If the constraints are ideal, meaning that the forces of constraint have ~ i ≡ (L ~ i )normal = (L ~ i )⊥ , while the vironly normal components, L tual elementary displacements are always tangent to the constraint, δ~ri ≡ (δ~ri )k , it then follows that the virtual mechanical work of the constraint forces is null: N X i=1

~ i · δ~ri = L

N X i=1

~ i )⊥ · (δ~ri )k = 0, (L

in which case (1.31) becomes N X i=1

F~i · δ~ri = 0.

(1.32)

Relation (1.32) expresses the principle of virtual work: The necessary and sufficient condition for static equilibrium of a scleronomous system subject to ideal constraints is that the virtual work of the applied forces, for virtual displacements consistent with the constraints, be null. Let our mechanical system of N particles be subject to l ideal constraints of the form fj (~r1 , ~r2 , ..., ~rN , t) = 0

(j = 1, l).

As we have seen, the virtual elementary displacements satisfy the relations N X (gradi fj ) · δ~ri = 0 (j = 1, l), (1.33) i=1

20

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Free ebooks ==> www.Ebook777.com or

N  X ∂fj i=1

∂fj ∂fj δxi + δyi + δzi ∂xi ∂yi ∂zi



=0

(j = 1, l).

(1.34)

On the other hand, the principle of virtual work (1.32) can be written as N X  Xi δxi + Yi δyi + Zi δzi = 0, (1.35) i=1

where F~i = (Xi , Yi , Zi ) (i = 1, N ). If the elementary displacements δxi , δyi , δzi would be linearly independent, then (1.35) would lead to the following system of algebraic equations (

Xi = 0, Yi = 0, Zi = 0,

(i = 1, N )

(1.36)

whose solution would furnish the equilibrium positions xi , yi , zi (i = 1, N ) for the mechanical system. In fact, the virtual displacements δxi , δyi , δzi are not independent, but have to satisfy the set of l equations (1.34). Under these circumstances, to determine the position of equilibrium of the system we shall use the method of Lagrangian multipliers. Multiplying (1.34) by the non-zero scalars λj (j = 1, l), then performing summation over j and adding to (1.35), one obtains N X i=1

"

X i +

l X j=1



λj





l X



∂fj  ∂fj  δxi + Yi + λj δyi ∂xi ∂yi j=1 

# ∂f j δzi = 0. + Z i + λj ∂z i j=1 l X

(1.37)

This relation has to be satisfied by any δxi , δyi , δzi (i = 1, N ). Since the virtual displacements δxi , δyi , δzi are submitted to the l constraints (1.34), only 3N − l virtual displacements are linearly independent. To solve the problem, one equalizes to zero l round parentheses out of the total of 3N interfering in (1.37), which produces l algebraic equations for the unknowns λj (j = 1, l). This way, in (1.37) remain only 3N − l terms (containing the already determined λj ), corresponding to the 3N − l linearly independent virtual displacements. The coefficients of these virtual displacements, namely the 3N − l round parentheses 21

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Free ebooks ==> www.Ebook777.com remaining in (1.37) have to cancel, because (1.37) now stands for a linear combination of linearly independent ”vectors”. Consequently, all the 3N round parentheses in (1.37) must cancel, which leads to a system of 3N algebraic equations with 3N + l unknowns [3N equilibrium coordinates xi , yi , zi (i = 1, N ) and l Lagrangian multipliers λj (j = 1, l)]. To have a unique solution, the system of 3N algebraic equations has to be completed with l analytical relations of the constraints fj (~r1 , ~r2 , ..., ~rN , t) = 0 (j = 1, l), that is  l X  ∂fj   = 0, X + λ  i j  ∂x  i  j=1    l  X  ∂fj  Yi + = 0, λj (i = 1, N ; j = 1, l) (1.38) ∂y i j=1    l  X  ∂fj   λj Z + = 0,  i   ∂z i  j=1   fj (~r1 , ~r2 , ..., ~rN , t) = 0. This is a system of 3N + l equations with 3N + l unknown quantities. Therefore, the problem is solved. I.3.2. Principle of virtual velocities A variant of the principle of virtual work is the so-called principle of virtual velocities. By analogy with the usual definition d~r = ~r˙ , ~v = dt one introduces the notion of virtual velocity δ~r ~v = , (1.39) δt where δt would formally represent the time interval corresponding to the virtual elementary displacement δ~r. (In fact, a virtual displacement is a purely geometric concept, associated with the time interval δt = 0). Keeping in mind this observation, the principle of virtual velocities proves to be very useful in determination of the conditions1 of kinematic equilibrium2 for a system of material points. 1

In the case of static equilibrium we are talking about ”equilibrium positions”, while the kinematics operates with ”equilibrium conditions”. 2 A system of particles is at kinematic equilibrium if it performs a uniform motion (the linear acceleration for rectilinear motions is zero, and so is the angular acceleration for circular motions). 22

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Free ebooks ==> www.Ebook777.com All previous considerations and definitions remain valid, since in both cases of static and kinematic equilibria we have ~¨ri = 0 in case i of rectilinear motions, and εi = dω ¨ i = 0 for circular motions. dt = α ¨ Indeed, the condition ~ri = 0 (or/and εi = 0) stands not only for static equilibrium, but also for kinematic equilibrium, expressed by −−−→ ~vi = ~r˙ i = const.

(ωi = const.).

(1.40)

In other words, the principle of virtual velocities can be considered as an extension of the principle of virtual work for the case when the constant appearing in (1.40) is not necessarily zero. The mathematical expression for the principle of virtual velocities is easily obtained by using the principle of virtual work δW =

N X i=1

F~i · δ~ri = 0

and dividing it by the infinitesimal ”virtual” time interval (formally, δt 6= 0) N

N

X X δW δ~ri F~i · F~i · ~vi = 0. = = δt δt i=1 i=1

(1.41)

Applying the same procedure for the circular motion, we have

δW = δt

N P

Mi δαi

i=1

δt

N X

N

X δαi Mi Mi ωi = 0. = = δt i=1 i=1

(1.42)

Since in (1.41) and (1.42) the quantity δW δt has units of power, this principle is also called principle of virtual power. I.3.3. Torricelli’s principle If, in particular, only the gravity acts on the system of material points, the principle of virtual work can also be expressed in a form given by Torricelli: The necessary and sufficient condition for a system of particles subject to ideal constraints and acted only by their own forces of gravity to be in equilibrium, is that the virtual variation of the quota of the center of gravity of the system be zero: δzG = 0. To solve a problem of static equilibrium by means of this principle, one can use the following algorithm: 23

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Free ebooks ==> www.Ebook777.com 1) An orthogonal reference frame, preferably Cartesian Oxyz, with z-axis oriented vertically, is attached to the system of N particles. 2) One identifies the number of constraints, l, settle the number of degrees of freedom of the system, n = 3N − l, and associate with each degree of freedom an independent variable quantity characteristic for the system, ξ1 , ξ2 , ..., ξn . 3) One determines the quota of the centre of gravity1 of the system by means of one of the following definitions:

i) zG =

N P

mi zi

i=1 N P

= mi

i=1

R

(D)

ii) zG = R

N 1 X mi zi , if the system is discrete, or M i=1

zρ(x, y, z) dx dy dz

1 = M ρ(x, y, z) dx dy dz

Z

zρ(x, y, z) dx dy dz, if

(D)

(D)

the system is continuously distributed in the spatial domain (D) . 4) The determined quota of the centre of gravity is expressed in terms of the n linearly independent parameters ξ1 , ξ2 , ..., ξn : zG = zG (ξ1 , ξ2 , ..., ξn ). We then calculate the differential dzG dzG = X1 (ξ1 , ξ2 , ..., ξn )dξ1 + X2 (ξ1 , ξ2 , ..., ξn )dξ2 +... + Xn (ξ1 , ξ2 , ..., ξn )dξn , identify dzG with δzG δzG = X1 (ξ1 , ξ2 , ..., ξn )δξ1 + X2 (ξ1 , ξ2 , ..., ξn )δξ2 +... + Xn (ξ1 , ξ2 , ..., ξn )δξn ,

(1.43)

and finally solve the equation δzG = 0. Since the parameters ξ1 , ξ2 , ..., 1

If the spatial extent of the system of particles is not very large, the centre of mass of the system can be identified with its centre of gravity. The following relations are written within this approximation. 24

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Free ebooks ==> www.Ebook777.com ξn are linearly independent, the virtual variations δξ1 , δξ2 , ..., δξn are also linearly independent, and (1.43) yields the following system of n algebraic equations in n variables ξ1 , ξ2 , ..., ξn :  X1 (ξ1 , ξ2 , ..., ξn ) = 0,    X (ξ , ξ , ..., ξn ) = 0,    2 1 2 . .      . Xn (ξ1 , ξ2 , ..., ξn ) = 0.

The real solutions of this system provide the equilibrium position of the system of particles.

25

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CHAPTER II PRINCIPLES OF ANALYTICAL MECHANICS

II.1. D’Alembert’s principle As shown in the last paragraph, the principle of virtual work allows one to determine either positions of static equilibrium or conditions of kinematic equilibrium for a system of N material points (particles). So, this principle offers possibility of solving a statics problem emerging from the original dynamical one (at least, principially) by means of another very important principle - the D’Alembert principle. As we shall see, this can be done by the help of force of inertia concept. Therefore, a dynamical problem can be replaced by a statics one on the real space of motion only [the space of position vectors ~ri (i = 1, N )], and working only in a non-inertial frame. Besides, this method proves to be useful only as a first step, since - as we shall see - the result is finally obtained by solving a dynamical problem, but this time in the so-called ”configuration space”. In this space, solution to the problem is obtained in an easier way. It is to be mentioned that the problem is considerably simplified by the existence of constraints (the number of second order differential equations is diminished by 2l, where l is the number of constraints). Recalling the Lagrange equations of the first kind formalism, it is not difficult to realize that the bigger the numbers N of particles and l of constraints, the more difficult the task is. Under these circumstances, the system of 3N + l equations could become unsolvable. Observing that the number of degrees of freedom n = 3N − l of the system diminishes with the increasing of the number l of constraints, Lagrange had a brilliant idea of replacing the big number of ~ri (i = 1, N ) and parameters λj (j = 1, l) by a smaller number of unknowns qk (k = 1, n). Namely, to each degree of freedom of the mechanical system Lagrange associates a parameter q. These quantities are called Lagrange variables, or generalized coordinates, and define an abstract space, named configuration space. 26

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Free ebooks ==> www.Ebook777.com The Lagrangian formalism is useful not only in Mechanics, but also in various fields of science. The choice of the generalized coordinates is not unique, and depends to some extent on the ability of the researcher, but there are certain requirements which have to be obeyed: 1) Compulsory requirements: i) to allow the position vectors of the material points to be expressed in terms of qk (and, eventually, of time): ~ri = ~ri (q1 , q2 , ..., qn , t) ≡ ~ri (q, t)

(i = 1, N );

(2.1)

ii) to satisfy the equations of constraints, that is fj (q1 , q2 , ..., qn , t)  ≡ fj ~r1 (q, t), ~r2 (q, t), ..., ~rN (q, t), t = 0

(j = 1, l);

iii) to exist the inverse transformation of (2.1), expressed as qk = qk (~r1 , ~r2 , ..., ~rN , t)

(k = 1, n).

(2.2)

2) Optional requirements: i) to take into account the symmetry proprieties of the physical system; ii) to lead to the solution in the new variables as easy as possible; iii) to allow construction of the Lagrangian function (for natural systems - see further) or the kinetic energy (for the systems which do not admit a simple or generalized potential) as simple as possible. Like the real coordinates ~ri (i = 1, N ), the generalized coordinates must be continuous functions of time, and at least twice differentiable [to assure the existence of q¨k (k = 1, n), which are the analogous to the accelerations ~ai = ~¨ri in the real space]. But unlike the real coordinates, which have the units of distance, as generalized coordinates one can take: arc elements, surface elements, angles, entropy, angular velocity, scalar or vector potential, etc. More than that, the choice of the generalized coordinates is not unique: there is usually possible to define another set of generalized coordinates q1′ , q2′ , ..., qn′ , so that qk −→ qk′ = qk′ (q1 , q2 , ..., qn , t)

(k = 1, n).

If the system of particles is not subject to constraints, then as generalized coordinates can be taken the Cartesian coordinates xi , yi , zi 27

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Free ebooks ==> www.Ebook777.com (i = 1, N ), or spherical coordinates r, θ, ϕ, or cylindrical coordinates ρ, ϕ, z, etc. For example, the system shown in Fig.I.2 is subject to the constraint f (x1 , x2 , x3 ) = x21 + x22 − x23 tan2 θ + 2Rx3 tan θ − R2 = 0, meaning that the particle has n = 3 − 1 = 2 degrees of freedom. To these degrees of freedom one associate either the pair (q1 , q2 ) = (x1 , x3 ), or the pair (q1 , q2 ) = (ϕ, x3 ), where ϕ is the angle between Ox1 and the projection on x1 Ox2 -plane of the radius vector of the particle, etc. (There exists at least two more possibilities, and it is up to the reader to identify them). In case of the first choice, relations (2.1) write x = x ,  1 q1  2 x2 = (R/ tan θ) − x3 − x21 ,  x3 = x3 , while the second choice yields (

x1 = (R − x3 tan θ) cos ϕ, x2 = (R − x3 tan θ) sin ϕ, x3 = x3 .

II.1.1. Configuration space As we have mentioned in the previous paragraph, the generalized coordinates qk (k = 1, n) define a space called configuration space. This is an abstract space, usually denoted by Rn , and has a physical (real) meaning only in some particular cases. A point P in configuration space is defined by the set of generalized coordinates q1 , q2 , ..., qn . The name ”configurations” comes from the fact that the points of this space stand for ”configurations” of the system of particles in the real, physical space. Indeed, the system of position vectors ~r1 , ~r2 , ..., ~rN , at a certain time t, define the so-called configuration of the system of particles, at time t. According to relations (2.1), the knowledge of the set of vectors ~ri (i = 1, N ) is equivalent to the cognition of the set of generalized coordinates qk (k = 1, n). This way, to know the positions of the N particles in the real space means to know the position of a single point, called representative point, in the configuration space Rn . More than that, to know the N real trajectories of all particles of the system, expressed parametrically by ~ri = ~ri (t) (i = 1, N ), is equivalent to know the trajectory of a single point (the representative point) in 28

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Free ebooks ==> www.Ebook777.com the configuration space, describing a single trajectory qk = qk (t) (k = 1, n), called generalized trajectory. Observations 1. The generalized trajectory does not represent any of the N real trajectories described by the system of particles, but ”integrates” all the real trajectories of the system. 2. The generalized trajectory can be conceived as a succession of representative points, each of them representing the configuration of the system at the considered moment of time. II.1.2. Generalized forces By differentiating (2.1) with respect to time, we have n X ∂~ri ∂~ri dqk + dt d~ri = ∂qk ∂t

(i = 1, N ),

(2.3)

k=1

which might express: i) the real elementary displacement of the particle Pi during the time interval dt, if ~ri = ~ri (q, t) satisfy the differential equations of motion, the equations of constraints, and the initial conditions compatible with the constraints, or ii) a possible elementary displacement of the same particle subject to the rheonomous constraints  fj ~r1 (q, t), ~r2 (q, t), ..., ~rN (q, t), t = 0 (j = 1, l), (2.4)

if ~ri = ~ri (q, t) satisfy only these constraints. If we set dt → δt = 0 in (2.3), then d~ri → δ~ri , dqk → δqk , and we obtain n X ∂~ri δqk (i = 1, N ), (2.5) δ~ri = ∂qk k=1

which is an infinitesimal virtual displacement of the particle Pi , subject to the rheonomous constraints (2.4). The significance of the quantities dqk and δqk appearing in (2.3) and (2.5), respectively, is similar to the displacements d~ri and δ~ri in the real space. Therefore, dqk signifies i) the real elementary displacement of the representative point in configuration space, if qk = qk (t) (k = 1, n) satisfy the equations of motion in Rn (Lagrange equations - see further), the equations of constraints (2.4), and the initial conditions compatible with the constraints qk0 = qk (t0 ), q˙k0 = q˙k (t0 ) (k = 1, n); 29

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Free ebooks ==> www.Ebook777.com ii) a possible elementary displacement of the representative point, if only the rheonomous constraint equations (2.4) are satisfied, while δqk designates a virtual elementary displacement of the representative point in the configuration space. If q1 , q2 , ..., qn are linearly independent, then δq1 , δq2 , ..., δqn have the same property. Let us now return to the concept of virtual work. In the real space, the virtual mechanical work of the forces F~i (i = 1, N ) applied to N ≥ 2 particles writes δW =

N X i=1

F~i · δ~ri ,

or, by means of (2.5), δW =

N X i=1

=

N n X X i=1

k=1

n X ∂~ri δqk ∂qk

F~i ·

k=1

∂~ri F~i · ∂qk

!

δqk =

The quantities Qk defined by Qk =

n X

! Qk δqk .

(2.6)

k=1

N X

∂~ri F~i · ∂qk i=1

(2.7)

are called generalized forces. Therefore, in the configuration space the principle of virtual work writes δW =

n X

Qk δqk = 0.

(2.8)

k=1

Since the applied forces F~i have the functional dependency of the form F~i = F~i (~r1 , ~r2 , ..., ~rN , ~r˙ 1 , ~r˙ 2 , ..., ~r˙ N , t), in view of (2.1) we conclude that the generalized forces Qk (k = 1, n) obey the following functional dependency Qk = Qk (q1 , q2 , ..., qn , q˙1 , q˙2 , ..., q˙n , t)

(k = 1, n),

(2.9)

where the quantities q˙k =

dqk dt

(k = 1, n) 30

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(2.10)

Free ebooks ==> www.Ebook777.com are called generalized velocities. Dividing (2.3) by dt, we obtain the connection between real and generalized velocities: ~r˙ i =

n X ∂~ri ∂~ri q˙k + ∂qk ∂t

(i = 1, N ).

(2.11)

k=1

Observation. As observed, the generalized forces are not forces in the Newtonian sense of the term (except for the case when the generalized coordinates have unit of length), and, consequently, their units are given by the associated generalized coordinates. Whatever the choice of the generalized coordinates, the product between a generalized coordinate and its associated generalized force has units of mechanical work [see (2.8)]. For example, if q stands for an angular variable, then the associated generalized force must have the dimension of a force moment (N · m). This observation is also valid for the principle of virtual velocities. To write this principle in the configuration space Rn , let us divide δW by δt:

δW = δt

N P F~i · δ~ri

i=1

δt

N X

N

X δ~ri F~i · F~i · = = δt i=1 i=1

n n N X X X ∂~ri δqk ∂~ri = = F~i · F~i · ∂qk δt ∂qk i=1 i=1 N X

k=1

k=1

!

n P

k=1

∂~ ri ∂qk δqk

δt

n

X δqk = Qk wk , (2.12) δt k=1

k where by wk = δq δt (k = 1, n) we denoted the virtual generalized velocities. Resuming the above observation, it is useful to realize that if wk is an ”ordinary” velocity, then the associated generalized force Qk is a Newtonian force, while if wk is an angular velocity, then Qk stands for a force moment. Consequently, in the configuration space the principle of virtual velocities writes

n

X δW = Qk wk = 0. δt

(2.13)

k=1

Let us now consider the D’Alembert’s principle. First of all, we have to mention that it belongs to the differential principles of analytical mechanics, being expressed in terms of the elementary variations of the generalized coordinates and velocities. It is the dynamic analogue to the principle of virtual work for applied forces in a static system. 31

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Free ebooks ==> www.Ebook777.com Suppose, as usual, a system of N ≥ 2 particles P1 , P2 , ..., PN , subject to the applied forces F~i (i = 1, N ) and to constrained forces Li (i = 1, N ). (The constraints are supposed to be holonomic). The fundamental equation of Newtonian mechanics for the particle Pi (lex secunda), in an inertial reference frame, then writes ~ i. mi~¨ri = F~i + L

(2.14)

By definition, the quantity (−mi~¨ri ) (no summation) is denoted by J~i = −mi~¨ri

(2.15)

and is called force of inertia. According to (2.15), this force is oriented along the acceleration vector ~¨ri , but its sense is opposite. By means of this notation, the fundamental equation of dynamics (2.14) becomes ~ i + J~i = 0 F~i + L

(i = 1, N )

(2.16)

and expresses D’Alembert’s principle: At any moment, there is an equilibrium between the applied, the constraint, and the inertia forces acting on a particle. This principle is valid in a non-inertial frame only. Indeed, equation (2.16) expresses the equilibrium condition for each particle of the system, which can also be written as ~ i + J~i = 0 mi~ai = F~i + L

(i = 1, N ).

(2.17)

This way, the dynamical problem has been converted to a kinematic −−−−−→ one (if ~r˙ i = ~vi = (const.)i ), or even a static one (if ~r˙ i = ~vi = 0), because in this frame the acceleration vector ~ai (i = 1, N ) of each particle is zero.

Fig.II.1 32

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Free ebooks ==> www.Ebook777.com To illustrate this situation, here is a simple example. Consider a parallelepipedic body of mass m performing a uniformly accelerated −−−→ linear motion (~a = const. > 0) on a horizontal plane, subject to the active force F~ > F~f (see Fig.II.1). ~ |, it follows that the accelerSince N = G = mg, and |F~f | = µ|N ation of the body is

a = |~a| =

|F~ | − µmg F = − µg > 0. m m

(2.18)

In the inertial frame (the so-called ”laboratory reference frame”) the body of mass m moves to the right with a uniform acceleration given by (2.18) (see Fig.II.2). Consider now a non-inertial frame S ′ , invariably connected to the body. This means that the body is at rest with respect to S ′ (in fact, the body itself is the reference frame S ′ ). An observer attached to the body (that is to S ′ ) moves to the right with body’s acceleration a > 0, with respect to the laboratory frame (see Fig.II.3).

Fig.II.2 ~ and N ~ acting on the body in the frame S will Forces F~ , F~f , G, also act in the frame S ′ , but the fact that the body is at rest in S ′ can only be explained if we accept that in S ′ also acts the force1 F~in = −(F~ + F~f ) = −m~a. 1

~ +N ~ =0 Here we have to take into account that G 33

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(2.19)

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Fig.II.3 This is the force of inertia. It is a real, applied force in S ′ , and by means of F~in a problem of dynamics has been turned into a statics problem. Indeed, in the frame S ′ we have ~ +N ~ + F~in = 0, m~ares = F~ + F~f + G

(2.20)

therefore the resultant acceleration is ~ares = ~¨r = 0. Unlike the forces ~ N ~ , acting in both frames S and S ′ , the force F~in acts only in F~ , F~f , G, the non-inertial frame S ′ . In other words, in the inertial frame S the force of inertia does not exist. This conversion of a dynamical problem into a statics one does not entirely solve the problem. Indeed, in order to write the principle of virtual work we should know all forces acting upon the body in the frame S ′ . But, unfortunately, the force of inertia1 F~in = −m~¨r is not known (otherwise, the dynamic problem would be solved). It seems to be a vicious circle: to solve the problem, one must know the solution! Consequently, the answer has to be searched elsewhere. The one who solved the problem was Joseph-Louis Lagrange. In this respect, let us perform the dot product of relation (2.16) expressing D’Alembert’s principle by the virtual displacement δ~ri (i = 1, N ). The result is N X i=1

F~i · δ~ri + J~i · δ~ri =

N X i=1

(F~i + J~i ) · δ~ri = 0,

(2.21)

where we took into account the fact that, in case of ideal constraints, N X i=1

1

~ i · δ~ri = L

N X i=1

~i L



· δ~ri normal



tangential

Here index ”in” comes from ”inertia”. 34

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= 0.

Free ebooks ==> www.Ebook777.com Relation (2.21) can also be written as N X i=1

 F~i − mi~¨ri · δ~ri = 0

(2.22)

and expresses D’Alembert’s principle in the form given by Lagrange: The sum of the elementary virtual works of applied and inertial forces acting on a system subject to ideal constraints is zero. This form of D’Alembert’s principle is more useful, because it does not contain the constraint forces. It can be successfully used to obtain the second-order differential equations of motion of the system in configuration space - the celebrated Lagrange equations of the second kind, or simply, Lagrange equations. These equations write: d dt



∂T ∂ q˙k





∂T = Qk ∂qk

(k = 1, n)

(2.23)

for systems which do not admit a potential, and d dt



∂L ∂ q˙k





∂L =0 ∂qk

(k = 1, n)

(2.24)

for natural systems (i.e. systems possessing a Lagrangian). Here T = T (q, q, ˙ t) is the kinetic energy of the system, Qk - the generalized forces, and L(q, q, ˙ t) - the Lagrangian of the problem. We shall derive these equations in the next paragraph, by means of an integral principle. As one can see, solving the dynamical problem in the real space cannot be totally avoided. Nevertheless, transition through a problem of statics (formulated in a non-inertial frame of the real space) to a problem of dynamics (in the configuration space) is convenient, because a problem of dynamics is easier to be solved in configuration space, than in the real space. Among other reasons, we mention the fact that the number of the second order differential equations is smaller by 2l, where l stands for the number of constraints. II.2. Hamilton’s principle This principle is of fundamental importance in many areas of physics, not only in analytical mechanics. It is an integral, variational principle. Unlike the differential principles, the integral principles consider evolution of the physical system in a finite time interval. These 35

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Free ebooks ==> www.Ebook777.com principles operate with global variations, either in real or in configuration spaces. It is worth mentioning the strong connection existing between differential and integral principles of analytical mechanics. In order to facilitate the presentation, we shall begin our investigation in the real space R3 . Consider a system of N ≥ 2 particles, subject to l holonomic constraints fk (~r1 , ~r2 , ..., ~rN , t) = 0

(k = 1, l)

(2.25)

and suppose that the law of motion of the system ~ri = ~ri (t)

(i = 1, N )

(2.26)

is known, within the time interval t1 ≤ t ≤ t2 . Consider, also, another law of motion of the system ~ri∗ = ~ri∗ (t)

(i = 1, N )

(2.27)

which satisfies only1 the equations of constraints (2.25). Making allowance for the terminology used in case of elementary displacements, we call this motion a possible motion. Following the general presentation of the variational principles (see paragraph V.1), and adapting the corresponding quantities, we demand that both trajectories given by (2.26) and (2.27) pass through the same two points P1 and P2 , corresponding to the time moments t1 and t2 (see Fig.II.4), that is ~ri (t1 ) = ~ri∗ (t1 ), and ~ri (t2 ) = ~ri∗ (t2 )

(i = 1, N ).

(2.28)

Let us define the virtual elementary displacement δ~ri as being the variation of the position vector of the particle Pi from one point of the real trajectory (Creal ) to an infinitely closed point of the possible trajectory (Cpossible ) ≡ (C ∗ ) as (see Fig.II.4) δ~ri = ~ri (t) − ~ri∗ (t)

(i = 1, N ).

(2.29)

In this case, the conditions (2.28) write δ~ri (t1 ) = 0,

δ~ri (t2 ) = 0

(i = 1, N ).

1

(2.30)

It satisfies neither the differential equations of motion, nor the initial conditions compatible with the constraints. 36

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Fig.II.4 To arrive at Hamilton’s principle, we shall use D’Alembert’s principle in the form given by Lagrange: N X i=1

or

N X i=1

(F~i − mi~¨ri ) · δ~ri = 0,

mi~¨ri · δ~ri =

N X i=1

F~i · δ~ri = δW.

Using some simple mathematical manipulations, we still have: ! N N N X X X d d mi~¨ri · δ~ri = mi~r˙ i · (δ~ri ) mi~r˙ i · δ~ri − dt i=1 dt i=1 i=1 d = dt

N X i=1

mi~r˙ i · δ~ri

!



N X i=1

mi~r˙ i · δ~r˙ i =

As can be observed, the term

N P

i=1

i=1

F~i · δ~ri = δW.

(2.31)

mi~r˙ i · δ~r˙ i comes from the vir-

tual variation of the kinetic energy T =

1 2

particles. Indeed, N

N X

N P

i=1

mi |~r˙ i |2 of the system of

N

1X 1X T = mi |~r˙ i∗ |2 = mi |~r˙ i − δ~r˙ i |2 2 i=1 2 i=1 ∗

37

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Free ebooks ==> www.Ebook777.com   1X   1X = mi |~r˙ i |2 − 2~r˙ i · δ~r˙ i + |δ~r˙ i |2 ≈ mi |~r˙ i |2 − 2~r˙ i · δ~r˙ i 2 i=1 2 i=1 N

N

N N N X X 1X 2 ˙ ˙ ˙ = mi |~ri | − mi~ri · δ~ri = T − mi~r˙ i · δ~r˙ i , 2 i=1 i=1 i=1

which yields N X i=1

mi~r˙ i · δ~r˙ i = T − T ∗ = δT.

Introducing this result into (2.31), we can write N X

d dt

i=1

N X

d = dt

i=1

mi~r˙ i · δ~ri

mi~r˙ i · δ~ri

!

!



N X i=1

− δT =

mi~r˙ i · δ~r˙ i

N X i=1

F~i · δ~ri = δW,

and, therefore d dt

N X i=1

mi~r˙ i · δ~ri

!

= δT + δW = δ(T + W ).

(2.32)

Integrating this equation between the time moments t1 and t2 , and taking into account (2.30), we still have Z

t2

t1

d dt

N X i=1

mi~r˙ i · δ~ri

or N X i=1

=

N X i=1

that is

mi~r˙ i · δ~ri (t2 ) −

N X i=1

Z

!

dt =

Z

t2

δ(T + W ) dt, t1

! t2 ˙ mi~ri · δ~ri t1

mi~r˙ i · δ~ri (t1 ) = 0 =

Z

t2

δ(T + W ) dt, t1

t2

δ(T + W ) dt = 0. t1

38

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Free ebooks ==> www.Ebook777.com R But the operators · dt and δ· are independent, therefore they commute with each other, so that the previous relation also writes δ

Z

t2

(T + W ) dt = 0,

(2.33)

t1

and represents the mathematical expression for the generalized Hamilton’s principle. If the applied force is generated by a potential function V (~r, t), then the potential V (~r ∗ , t) corresponding to a possible motion writes V (~r ∗ , t) ≡ V (~r ∗1 , ~r ∗2 , ..., ~r ∗N , t) = V (~r1 − δ~r1 , ~r2 − δ~r2 , ..., ~rN − δ~rN , t) = V (~r1 , ~r2 , ..., ~rN , t) −

N X i=1

(gradi V ) · δ~ri + O |δ~ri |2

≈ V (~r1 , ~r2 , ..., ~rN , t) − = V (~r1 , ~r2 , ..., ~rN , t) +

N X i=1

or

N X (gradi V ) · δ~ri



i=1

F~i · δ~ri = V (~r1 , ~r2 , ..., ~rN , t) + δW

V (~r ∗ , t) ≡ V ∗ = V + δW, which yields δW = −(V − V ∗ ) = −δV, and (2.33) leads to δ where

Z

t2 t1

(T − V ) dt = δ

Z

t2

L dt = 0,

(2.34)

t1

L(~r, ~r˙ , t) = T (~r, ~r˙ , t) − V (~r, t)

is the Lagrangian function expressed in the real space. Relation (2.34) expresses Hamilton’s principle in real space. Since the quantities ~ri and ~r˙ i are not independent, but must satisfy the same constraints (2.25), we have to resume this reasoning in configuration space Rn , where both generalized coordinates qj and generalized velocities q˙j are independent. 39

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Fig.II.5 Following a similar procedure, but this time in the configuration space, one observes that the analogues of (2.26) and (2.27) are qk = qk (t)

(k = 1, n)

(2.26′ )

qk∗ = qk∗ (t)

(k = 1, n),

(2.27′ )

and respectively, conditions (2.30) correspond to δqk (t1 ) = 0,

δqk (t2 ) = 0

(k = 1, n),

(2.30′ )

while the relations similar to (2.29) are δqk = qk (t) − qk∗ (t)

(k = 1, n).

(2.29′ )

Since the generalized coordinates qk are independent, except for conditions (2.29’), the virtual variations δqk (k = 1, n) are also independent. Furthermore, according to the definition, these variations are instantaneous. A formal graphical representation of the set of qk coordinates as functions of time is given in Fig.II.5. Using these observations, relation (2.34) shall be written in configuration space as δ

Z

t2

L dt = δ t1

Z

t2

L(q, q, ˙ t) dt = 0. t1

40

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(2.35)

Free ebooks ==> www.Ebook777.com Let us denote S ≡ S(q) =

Z

t2

L(q, q, ˙ t) dt

(2.36)

t1

and call this integral action. With this notation, (2.35) finally writes δS = 0.

(2.37)

This simple relation comprises the essence of analytical mechanics. It expresses Hamilton’s principle for natural systems subject to holonomic constraints: Among all possible generalized trajectories passing through two fixed points in configuration space, corresponding to two moments of time t1 and t2 , the trajectory associated with the real motion corresponds to a stationary action. Since in most cases the extremum of the action integral is a minimum, Hamilton’s principle is also called principle of minimum action. It was published in 1834 by Willian Rowan Hamilton. Let us now prove that all the important results of analytical mechanics can be obtained by means of Hamilton’s principle. 1. Lagrange equations of the second kind, written for both cases of simple and generalized potentials, on the one hand, and non-potential forces, on the other. i) To deduce Lagrange equations of the second kind for natural systems that admit a simple potential function V = V (q, t) we use Hamilton’s principle (2.37) δS = δ

Z

t2

L(q, q, ˙ t) dt = 0, t1

written for virtual and independent1 variations δq of the generalized coordinates. We therefore have Z t2 Z t2 δS = δ L(q, q, ˙ t) dt = δL(q, q, ˙ t) dt t1

= =

Z

t2 t1

∂L δqk dt + ∂qk

Z

t1

t2 t1

Z

t2 t1

 ∂L ∂L δqk + δ q˙k dt ∂qk ∂ q˙k     Z t2 d ∂L d ∂L δqk dt − δqk dt dt ∂ q˙k ∂ q˙k t1 dt



1

except for the moments of time t1 and t2 , when δqk (t1 ) = 0 and δqk (t2 ) = 0. 41

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Free ebooks ==> www.Ebook777.com =

Z

t2 t1

  t2 Z t2  d ∂L δqk dt − ∂ q˙k t1 dt t1   Z t2 Z t2 ∂L d ∂L = δqk dt − δqk dt ∂ q˙k t1 ∂qk t1 dt   Z t2   ∂L d ∂L − δqk dt = 0, =− dt ∂ q˙k ∂qk t1

∂L δqk dt + ∂qk



∂L δqk ∂ q˙k

where conditions (2.30’) have been considered. Since the virtual variations δqk are arbitrary, the last relation yields the Lagrange equations   ∂L d ∂L (2.38) − (k = 1, n). dt ∂ q˙k ∂qk ii) If our natural system admits a generalized potential V = V (q, q, ˙ t), we have to show that Hamilton’s principle (2.33) is equivalent to (2.37), on condition that the generalized forces Qk are written as   d ∂V ∂V Qk = (k = 1, n). − dt ∂ q˙k ∂qk

Indeed, we have: Z t2 Z 0=δ (T + W ) dt = t1

t2

(δT + δW ) dt = t1

t2

δT dt + t1

Z

t2

Qk δqk dt t1

  ∂V ∂V − δqk dt = δT dt + ∂ q˙k ∂qk t1 t1   Z t2 Z t2 Z t2 Z t2 d ∂V ∂V ∂V = δT dt + δqk dt − δ q˙k dt − δqk dt ∂ q˙k t1 t1 dt t1 ∂ q˙k t1 ∂qk   t2 Z t2   Z t2 ∂V ∂V ∂V = δT dt + δqk − δ q˙k + δqk dt ∂ q˙k ∂ q˙k ∂qk t1 t1 Z

=

Z

t2

t1

δT dt −

t2

Z

Z

t2

δV dt =

t1

t2

Z



d dt



Z

t1

t2

t1

δ(T − V ) dt = δ

Z

t2

L dt = δS = 0,

t1

which completes the proof. iii) To deduce Lagrange equations of the second kind for mechanical systems acted on by non-potential forces, we start with the form (2.33) of Hamilton’s principle: Z t2 Z t2 Z t2 0=δ (T + W ) dt = δT dt + δW dt t1

t1

t1

42

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Free ebooks ==> www.Ebook777.com  Z t2 ∂T ∂T = δqk + δ q˙k dt + Qk δqk dt ∂qk ∂ q˙k t1 t1   Z t2 Z t2 ∂T d ∂T = δqk dt + δqk dt ∂ q˙k t1 ∂qk t1 dt   Z t2 Z t2 d ∂T − δqk dt + Qk δqk dt ∂ q˙k t1 dt t1    t2 Z t2  Z t2 ∂T ∂T d ∂T δqk dt+ δqk − δqk dt+ Qk δqk dt ∂qk ∂ q˙k ∂ q˙k t1 dt t1 Z

=

Z

t2 t1

t2



t1

=−

Z

t2

t1



d dt



∂T ∂ q˙k



 ∂T − − Qk δqk dt ∂qk

(k = 1, n),

where conditions (2.30’) have been taken into account. Since the virtual variations δqk are arbitrary [except for (2.30’)], we are left with d dt



∂T ∂ q˙k





∂T = Qk ∂qk

(k = 1, n).

(2.39)

As one can see, by means of relation (2.33) both Lagrange equations for natural systems subject to generalized-potential and nonpotential forces can be obtained. For this reason, (2.33) expresses - as we previously asserted - the generalized Hamilton’s principle. 2. Let us define the Hamiltonian function, usually known as Hamiltonian, H(q, p, t), as H(q, p, t) = pk q˙k − L(q, q, ˙ t). Using Hamilton’s principle (2.37) and the Hamiltonian, one can obtain the system of Hamilton’s canonical equations, which are the ”heart” of the Hamiltonian formalism. We have: 0 = δS = δ

= =

Z

Z

t2

L dt = δ t1

Z

t2 t1

(pk q˙k − H) dt

t2 t1

t2

pk δ q˙k dt + t1

Z

(pk δ q˙k + q˙k δpk ) dt − Z

t2 t1

q˙k δpk dt −

Z

t2 t1

Z



t2

δH dt t1

∂H ∂H δqk + δpk ∂qk ∂pk

43

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dt

Free ebooks ==> www.Ebook777.com Z

Z t2 Z t2 d = (pk δqk ) dt − p˙ k δqk dt + q˙k δpk dt t1 dt t1 t1  Z t2  ∂H ∂H δqk + δpk dt − ∂qk ∂pk t1 Z t2 t2 Z t2 = (pk δqk ) − p˙k δqk dt + q˙k δpk dt t2

t1

t2 

t1

t1

 ∂H ∂H − δqk + δpk dt ∂qk ∂pk t1     Z t2   ∂H ∂H = δqk + q˙k − δpk dt. − p˙ k + ∂qk ∂pk t1 Z

Since the virtual variations δqk and δpk are arbitrary [except for (2.30’)], the last relation yields Hamilton’s canonical equations  ∂H    q˙k = ∂p ; k (2.40) (k = 1, n)  ∂H   p˙ k = − . ∂qk

Observations 1. Hamilton’s principle is a variational principle. Indeed, comparing our investigation with the results displayed in §1, Chap.V, concerning functionals and their extrema, the following correspondence can be observed: Independent variable x

←→

Independent variable t

Dependent variable (variational parameter) y = y(x)

←→

Dependent variable (variational parameter) q = q(t)

Function of class C 2 f = f (x, y, y ′ )

←→

Lagrange function L = L(t, q, q) ˙

RFunctional x J = x12 f (x, y, y ′ ) dx

Euler-Lagrange equations   ∂f ∂f d dx ∂y ′ − ∂yi = 0 i

(i = 1, n)

←→

←→

Action R t2functional S(q) = t1 L(t, q, q) ˙ dt

Lagrange equations of  the2nd kind d ∂L ∂L dt ∂ q˙k − ∂qk = 0 (k = 1, n)

44

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Free ebooks ==> www.Ebook777.com As we have seen in the paragraph dedicated to variational calculus, the solutions of the Euler-Lagrange equations are those ”points” in which the basic functional has an extremum (minimum or maximum). These points are called extremals. In this regard, solutions of the Lagrange equations of the 2nd kind are extremals of the acRt tion functional S = t12 L dt. In the light of the above considerations, Hamilton’s principle can be enounced as follows: The real motion of a physical system is described by the extremal of the action integral. In other words, among all functions yi = yi (x) of class C 2 defined on the interval x ∈ [x1 , x2 ], and satisfying conditions yi (x1 ) = a, yi (x2 ) = b (i = 1, 2, ...), where a and b are two real constants, the real motion of the system corresponds to that function which extremizes the action integral. 2. Since the quantities used in this paragraph can be defined in any reference frame, it follows that Hamilton’s principle does not depend on the choice of coordinates. 3. A single scalar function, the Lagrangian L, contains the whole information about the studied physical system. Once we know the Lagrangian, we can easily determine the differential equations of motion, and the associated conservation laws as well. 4. Hamilton’s principle can be also used to describe in a unitary manner some other systems, such as physical fields. This property is due to the fact that the terms contained by the Lagrangian have the units of energy. This quantity can be defined in any physical system, while not all motions/interactions can be described by means of the concept of force. This way are obtained, for example, the fundamental equations of: electrodynamics (Maxwell’s equations), theory of elasticity (Lam´e’s equations), non-relativistic quantum mechanics (Schr¨odinger’s equation), etc. 5. Hamilton’s variational principle of stationary action is widely applied in various fields of science, being one of the most general principles of nature. In order to construct the Lagrangian of a certain physical system, one must consider the following criteria (principles): - Superposition principle. If the system is composed by at least two particles, then the Lagrangian must contain three groups of terms: (i)terms describing each particle of the system, supposed to be alone; (ii)terms describing interaction between any particle and the rest of particles of the system; (iii)terms describing interaction between each particle with the external force fields (if they exist). - Invariance principle. The action has to be invariant with respect to the symmetry group characterizing the mechanical system: the Galilei-Newton group in case of Newtonian mechanics, the Lorentz45

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Free ebooks ==> www.Ebook777.com Einstein group in case of relativistic mechanics, etc. - Correspondence principle. All results obtained for a particular domain included in a more general one should be achieved by means of Hamilton’s principle applied to the more general domain. - Symmetry principle. The Lagrangian has to be not only as simple as possible, but also constructed in a way leading to differential equations of motion displaying the symmetry properties of the physical system. This can be done by a suitable choice of the generalized coordinates.

46

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CHAPTER III THE SIMPLE PENDULUM PROBLEM

The aim of this section is to solve the fundamental problem of mechanics for a constrained system, namely the simple pendulum, by means of six different methods connected to: (i) classical (Newtonian) approach; (ii) Lagrange equations of the first kind; (iii) Lagrange equations of the second kind; (iv) Hamilton’s canonical equations; (v) the Hamilton-Jacobi formalism, and (vi) the action-angle formalism. This way, we shall put into evidence the resemblances and differences between these formalisms, on the one side, and show the generality and potency of the analytical formalism, as compared to the classical one, on the other. The physical system we are going to study is represented by a material point (particle), suspended by a massless rod, constrained to move without friction on a circle in a vertical plane, under the influence of a uniform and homogeneous gravitational field. It is our purpose to study the motion of such a system only in case of free, harmonic, and non-amortized oscillations, performing motions of an arbitrary amplitude. The study of the most general case (non-linear damped and - eventually - forced oscillations) implies knowledge of notions like: bifurcation points, strange attractors, Lyapunov coefficients, etc., which overpass our approach. Our discussion stands for an application of the various methods offered by analytical mechanics which can be used for solving a problem. III.1. Classical (Newtonian) formalism According to this approach, the known elements are: the mass of the body, the acting forces (including the constraint forces), and the initial conditions compatible with the constraints. The reader is asked to find the equation of motion and the elements/characteristics of the 47

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Free ebooks ==> www.Ebook777.com motion: trajectory, period, frequency, etc. As well-known, determination of the solution is based on the second Newtonian low (lex secunda), which furnishes the differential equation of motion. As initial conditions are usually given the position and velocity of the body at some initial moment. Solution of the differential equation of motion is the low of motion, commonly written as ~r = ~r(t). One of the most important differences between the classical (Newtonian) and analytical formalisms is connected to the constraints. The classical approach demands knowledge of the constrained forces (at least their number and orientation), while the analytical procedure allows one to determine these forces at the end of calculation. ~ the force of gravity, i.e. the applied force, and Let us denote by G ~ by T the constrained force. It is convenient to choose the reference system with its origin O at the point of suspension, while the Ox and Oy axes are oriented as shown in Fig.III.1.

Fig.III.1 Suppose that, at the moment t = 0, the body is at the point P0 (x0 , y0 ) at rest (v0x = 0, v0y = 0). At the moment t, the angle between the rod and Ox is θ(t). Projecting the equation of motion m~¨r = F~ on axes, one obtains  or

Ox : Oy : 

m¨ x = mg − T cos θ ; m¨ y = −T sin θ, T x ¨=g− m cos θ ; T y¨ = − m sin θ.

48

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(3.1)

Free ebooks ==> www.Ebook777.com This is a system of two differential equations with four unknowns: x, y, θ, T . To solve it, we use the parametric equations of trajectory of the body of mass m : x = R cos θ; y = R sin θ, straightforwardly leading to  x ¨ = −R θ¨ sin θ − R θ˙2 cos θ ; y¨ = R θ¨ cos θ − R θ˙2 sin θ, and equations (3.1) become  T −R θ¨ tan θ − R θ˙2 = cosg θ − m , T 2 ¨ ˙ R θ − R θ tan θ = − m tan θ.

(3.2)

This way, we are left with two equations with two unknown quantities. By eliminating T , we have g θ¨ + sin θ = 0, R

(3.3)

which is the differential equation of motion. If θ0 (θt0 =0 ) < 4o , one can approximate sin θ ≈ θ and equation (3.3) goes to the well-known linear harmonic oscillator equation g θ¨ + θ = 0. R

(3.4)

Our result shows that, in this particular case, the body of mass m performs free, non-amortized, isochronous oscillations. Solution of the equation (3.4) can be written at least in four ways: θ = A1 eiω0 t + A2 e−iω0 t ;

(3.5 a)

θ = B1 sin ω0 t + B2 cos ω0 t ;

(3.5 b)

θ = C1 sin(ω0 t + C2 ) ;

(3.5 c)

θ = D1 cos(ω0 t + D2 ) ,

(3.5 d)

where ω02 = g/R, while the arbitrary constants A1 , A2 , B1 , B2 , C1 , C2 , D1 , D2 , are determined by means of initial conditions. For example, solution (3.5 c) leads to: θ0 = θt=0 = C1 sin C2 , θ˙0 = ω0 C1 cos C2 = 0. Since C1 cannot vanish, this gives C2 = π/2, C1 = θ0 , and the final solution reads  π θ(t) = θ0 sin ω0 t + . 2 The period of the harmonic pendulum is therefore s 2π R τ0 = = 2π , (3.6) ω0 g 49

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Free ebooks ==> www.Ebook777.com and depends only on the length R of the rod. In short, in the case of small oscillations (θ0 < 4o ), the period τ0 does not depend on the amplitude. Such motions are called isochronous or tautochronous. (In Greek, tauto means the same, and chronos means time). Let us now consider an arbitrary value of the angle θ0 . At the initial moment t0 = 0, the angle between the rod and the vertical ˙ is θ0 , while the point is at rest (θ˙0 = θ(0) = 0) at the point P0 (t0 ) (see Fig.III.1). As a result, from the mathematical point of view, we have to solve the Cauchy’s problem for equation (3.4), with the initial conditions  θt=0 = θ0 = const. (3.7) θ˙t=0 = θ˙0 = 0. If θ0 has an arbitrary value θ0 ∈ (0, π/2), integration of the differential equation of motion is more complicated. In addition, we have to be sure that the motion remains periodical. To this end, we shall go from the second derivative with respect to time to the first derivative with respect to θ, that is d  dθ  d  dθ  dθ dθ˙ ˙ 1 dθ˙2 d2 θ θ= = = . θ¨ ≡ 2 = dt dt dt dθ dt dt dθ 2 dθ Introducing now θ¨ into (3.3), separating variables and integrating, we can write Z Z θ(t) ˙ 2g θ(t) 2 ˙ sin θ dθ, dθ = − R θ(0)=θ0 ˙ θ(0)=0 or, by integration 2g θ˙2 = (cos θ − cos θ0 ), R that is θ˙ = ±

r

2g (cos θ − cos θ0 ). R

(3.8)

(3.9)

Since the angular velocity θ˙ has to be real, we must have (cos θ − cos θ0 ) ≥ 0. which means |θ| ≤ θ0 . Consequently, there are two turning points θ1 = − θ0 , θ 2 = + θ0 . As a result, the ”plus” sign in (3.9) corresponds to the intervals where the angular velocity θ˙ is positive: E → D → C [or, equivalently, θ ∈ (−θ0 , 0)], and C → B → A [or, θ ∈ (0, θ0 )] (see Fig.III.2, Fig.III.3, and Table III.1), while the ”minus” sign is associated with the intervals 50

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Free ebooks ==> www.Ebook777.com with negative angular velocity: A → B → C [or, equivalently, θ ∈ (θ0 , 0)], and C → D → E [or, θ ∈ (0, −θ0 )].

Next, let us analyze the manner of variation of the quantities ˙ θ = θ(t) and θ˙ = θ(t). To this end, we shall consider the vertical axis Ox as being a zero-axis for the values of θ: the angles determined on the right hand side of the circle are positive, while the left-oriented angles are negative. We shall start our analysis from the initial point, characterized by θi = θ0 , θ˙i = θ˙0 = 0. As one can see, it is useful to denote the starting point (position) by A, the point corresponding to θ = 0 by C, and so on (see Fig.III.2).

Fig.III.2

Fig.III.3 Inside the interval A → B → C the angle θ decreases by positive 51

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Free ebooks ==> www.Ebook777.com values until zero, while the angular velocity θ˙ decreases from zero (at A) to its minimum value θ˙min , by negative values. This behavior can be easily explained by making use of the definition of the derivative at a point, and Fig.III.3 as well. Indeed, dθ ∆θ θ2 − θ1 θ˙ = = lim = lim < 0. t2 →t1 t2 − t1 ∆t→0 ∆t dt Within the interval C → D → E the angle θ decreases by negative values up to its minimum value −θ0 . The body cannot pass over this point, going to greater negative values of θ, because θ = −θ0 is a turning point, while θ˙ increases, also by negative values, up to value zero. On the interval E → D → C angle θ increases by negative values up to value zero, while θ˙ increases, this time by positive values, up to its maximum value, corresponding to point C. Within the next interval C → B → A angle θ increases by positive values up to its maximum (initial) value θ0 and cannot pass over this point because θ = θ0 is a turning point. Angular velocity θ˙ decreases in this interval from its maximum value up to zero, by positive values. From this moment on, the motion resumes, meaning that it remains periodical. Table III.1

This analysis can be completed by the energetic approach. As reference level for the potential gravitational energy can be considered the plane orthogonal to Ox axis, passing through point C. Since the constraint is ideal (there are no energy losses), the maximum potential energy at point A is ”spent” on the account of mechanical work performed by the gravitational field. This leads to an increase of the kinetic energy of the body of mass m (which is zero at point A) up to its maximum value in the point C, where the potential energy is zero. This maximum value is numerically equal to the mechanical 52

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Free ebooks ==> www.Ebook777.com work performed by the gravitational field, and it is also equal to the initial potential energy at point A. Since the gravitational field is conservative, the total energy conserves. There exists a continuous ”transformation” of energy, the kinetic and potential energies successively passing through their maximum and minimum (zero) values, alternatively attained at points A, C, E. A synthesis of this analysis is shown in Table III.1. Let us now turn back to equation (3.9). Here the variables can easily be separated r dθ 2g dθ θ˙ = =± (cos θ − cos θ0 ) ⇒ dt = ± q , dt R 2g (cos θ − cos θ ) 0 R which gives

Z

t

dt = ±

t0

Z

θ θ0

q

dθ 2g R (cos θ

.

(3.10)

− cos θ0 )

It is more convenient to replace θ by a new variable ψ, defined as sin

θ θ0 = sin sin ψ ≡ k sin ψ, 2 2

meaning dθ =

2k cos ψ dψ . cos(θ/2)

The quantity (cos θ − cos θ0 ) appearing in (3.10) then writes (cos θ − cos θ0 ) = −(1 − cos θ) + (1 − cos θ0 ) = −2 sin2 = −2k 2 sin2 ψ + 2k 2 = 2k 2 cos2 ψ. Taking as initial moment t0 = 0, we then have: s Z R θ dθ p t=− 2g θ0 (cos θ − cos θ0 ) =−

s

=−

R 2g

s

Z

R g

arcsin

π/2

Z



arcsin π/2

1 k



sin

1 k

θ 2

sin



θ 2

2k cos ψ dψ √ cos θ2 2k cos ψ  dψ q 1 − sin2

θ 2

53

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θ θ0 + 2 sin2 2 2

Free ebooks ==> www.Ebook777.com =−

=

s

Rh g

Z

s

R g

Z

arcsin π/2



1 k

sin

Z

θ 2



dψ p 1 − k 2 sin2 ψ  

i dψ p p − 0 0 1 − k 2 sin2 ψ 1 − k 2 sin2 ψ s  1 Rh π  θ  i = F ,k , k − F arcsin sin g 2 k 2 s  1 Rh θ  i K(k) − F arcsin ,k . (3.11) = sin g k 2 π/2



arcsin

1 k

sin

θ 2

Here we took into account that, since k = sin θ20 > 0, for θ ∈ (−θ0 , θ0 ), we have |k cos ψ| = k cos ψ. We also considered that at the initial moment the pendulum departs from A, and chose the minus sign in (3.10). In (3.11) have been used the following notations: F (ψ0 , k) =

Z

ψ0 0

dψ p 1 − k 2 sin2 ψ

(3.12)

for the elliptic integral of first species, of amplitude ψ0 and modulus k, and  π  Z π/2 dψ p K(k) = F ,k = (3.13) 2 0 1 − k 2 sin2 ψ for the complete elliptic integral of the first species. In general, elliptic integrals cannot be expressed in terms of elementary functions, and are exposed in special tables. Denote g/R ≡ ω02 and observe that for a given θ0 we have K(k) ≡ the initial phase ϕ0 = const. Formula (3.11) then shows that the ”phase” of the periodical motion of pendulum is 

1

θ  ϕ(t) = ω0 t − K(k) = −F arcsin sin ,k , k 2

(3.14)

which is the implicit form of the equation of motion of pendulum. This equation can be solved in terms of θ by the help of special functions JA(u, m) and JSN (u, m), called amplitude of the Jacobi elliptic functions, and Jacoby elliptic function ”sn”, respectively, defined as: (1) Amplitude of the Jacobi elliptic functions is ”inversion” of the first kind elliptic integral, that is, if u = F (φ, m), then φ = am(u, m). 54

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Free ebooks ==> www.Ebook777.com In other words, the special function ”amplitude of the Jacobi elliptic functions” ”extracts” the amplitude φ of a first kind elliptic integral. (2) The Jacobi elliptic function ”sn” expresses the sinus of the amplitude of a first species elliptic integral: if φ is the amplitude, then sn(u, m) = sin φ. As known, the rational functions involving square roots of some quadratic forms can be integrated by the help of inverse trigonometric functions. As a result, the trigonometric functions can be defined as being inversions of the functions given by these integrals. By analogy, the elliptic functions are defined as being the inverted functions of those given by the elliptic integrals. An example has already been given above at point (1), by introducing the special function JA(u, m), called amplitude of the Jacobi elliptic functions. Here the argument m is often omitted, so that am(u, m) gets a simpler form: am(u). The Jacobi elliptic functions JSN (u, m) and JCN (u, m) are given by the relations sn(u) = sin(φ) and cn(u) = cos(φ), respectively, where φ = am(u, m). In its turn, the function JDN (u, m) is defined p as dn(u) = 1 − m sin2 φ = ∆(φ). The total number of existing Jacobi elliptic functions is twelve: JP Q(u, m), where P and Q belong to the set of four letters S, C, D, N . Each Jacobi elliptic function JP Q(u, m) satisfies the relation pq(u) = pn(u) qn(u) , with nn(u) = 1. There are several relations between the Jacobi elliptic functions, similar to those between the ordinary trigonometric functions. As a matter of fact, in the limit, the Jacobi elliptic functions turn into the usual trigonometric functions. Here are some examples: sn(u, 0) = sin(u), sn(u, 1)=tanh(u), cn(u, 0) = cos(u), cn(u, 1) = sech(u), dn(u, 0) = 1, dn(u, 1)= sech(u), etc. One of the most important properties of the elliptic functions is the fact that they are double-periodic in their complex arguments. The ordinary trigonometric functions are simple-periodic, that is they satisfy the relation f (z +sω) = f (z), for any integer s. Double periodicity is expressed by f (z + rω + sω ′ ) = f (z), for any integers r and s. The Jacobi elliptic functions sn(u, m), cn(u, m), etc. are double-periodic in the complex plane of variable u. Their periods are determined by ω = 4K(m) and ω ′ = 4iK(1 − m), where the complete elliptic integral of the first species K is given by (3.13). Taking into account these definitions, equation (3.14) can be written as  1 θ  F arcsin sin , k = K(k) − ω0 t, (3.15) k 2 55

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Free ebooks ==> www.Ebook777.com which yields   1 θ JA K(k) − ω0 t, k = arcsin sin k 2

as well as

  h 1 θ i 1 θ JSN K(k) − ω0 t, k = sin arcsin = sin , sin k 2 k 2

therefore

sin and, finally,

  θ = k JSN K(k) − ω0 t, k , 2

h  i θ ≡ θ(t) = 2 arcsin k JSN K(k) − ω0 t, k ,

(3.16)

which is the explicit form of the equation of motion of the simple pendulum. Let us now show that, if we choose the ”minus” sign in (3.10), meaning that the starting point is E (see Fig.III.2), the result is the same. Indeed, we can write: s Z dθ R −θ p t= 2g −θ0 (cos θ − cos θ0 ) s

=

=

=

=

s

Rh g

Z

=

s

s

0 −π/2

s

R 2g

p

Z

R g

R g

Z

− arcsin −π/2

Z



− arcsin −π/2

− arcsin −π/2

dψ 1 − k 2 sin2 ψ



1 k

sin



1 k

1 k

sin

+

Z

θ 2

sin

θ 2



θ 2



2k cos ψ dψ √ cos θ2 2k cos ψ  dψ q 1 − sin2

dψ p 1 − k 2 sin2 ψ  

− arcsin 0

θ 2

1 k

sin

θ 2

p



1 − k 2 sin2 ψ

 π   1 Rh θ  i − F − , k + F − arcsin sin ,k g 2 k 2 56

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i

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and, finally

s

 1 Rh θ  i K(k) − F arcsin ,k , sin g k 2 

1

θ  ω0 t = K(k) − F arcsin ,k , sin k 2  1 θ  F arcsin , k = K(k) − ω0 t, sin k 2

(3.17)

which completes the proof. As we have discussed at the beginning of this paragraph, if the angular amplitude θ is small (θ ≤ θ0 = 4o ), one can approximate sin θ ≈ θ, and equation (3.3) reduces to (3.4). Denoting ω02 = g/R, this equation can also be written as θ¨ + ω02 θ = 0,

(3.18)

which is the well known differential equation of the linear harmonic oscillator. This is an ordinary differential equation, homogeneous, with constant coefficients. The general solution is obtained by solving the attached characteristic equation r2 + ω02 = 0, with the roots r1,2 = ±iω0 . Solution is therefore [see (3.5a)] θ ≡ θ(t) = A1 er1 t + A2 er2 t = A1 eiω0 t + A2 e−iω0 t , where the arbitrary constants A1 and A2 are determined by using the initial conditions (3.7). As previously shown, solution can also be written in three equivalent forms (3.5b), (3.5c) and (3.5d). We leave to the reader to verify the equivalence of all these possible solutions. Taking (3.5c) as the solution, the initial conditions (3.7) write θ0 = θt=0 = C1 sin C2 ; θ˙0 = ω0 C1 cos C2 = 0, and the solution is  π θ(t) = θ0 sin ω0 t + = θ0 cos(ω0 t). 2 57

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(3.19)

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Fig.III.4 Using the Mathematica software, specialized in both analytical and numerical calculations, one can give graphical representations for θ = θ(t) for arbitrary oscillations [ i.e. any angular amplitude θ0 ∈ (0, π/2)], described by (3.16), as well as for small oscillations (θ0 < 4o ), associated with equation (3.19) (see Figs.III.4, III.5). To make the difference between the two types of oscillations more obvious, their superposition have also been represented graphically (see Fig.III.6). For θ0 and ω we chose the following values: θ0 = π/4 rad, ω = 3π/4 rad/s. As can be seen, at the beginning the two oscillations are

Fig.III.5 58

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Fig.III.6 similar (they are ”in-phase”). After some time, e.g. at t ≃ 11π s, the two graphic representations are in antiphase. The explanation is given by the fact that the oscillations of pendulum remain periodical, but they are not harmonic anymore. As we have discussed,pif θ0 ≤ 4o , the period of motion of pendulum is τ0 = 2π/ω0 = 2π R/g. We call these oscillation free, nonamortized, isochronous, with amplitude and initial phase depending on initial conditions. If θ > 4o , we have to appeal to (3.11). Since the motion of pendulum is identical on the four intervals (ABC), (CDE), (EDC), (CBA), the period of motion is τ = 4(∆t)(ABC) . Since (∆t)(ABC) is precisely the value of t given by (3.11), corresponding to the point C(θ = 0), we have: s  1 Rh θ i (∆t)(ABC) = tθ=0 = K(k) − F arcsin sin , k g k 2 θ=0 =

s

R K(k), g

and therefore τ =4

s

R 4 K(k) = g ω0

Z

59

π/2 0

dψ p . 1 − k 2 sin2 ψ

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(3.20)

Free ebooks ==> www.Ebook777.com We shall use this formula to express the period τ of the motion of pendulum for arbitrary amplitudes, that is θ0 ∈ (0, π/2), in terms of a series of even powers of θ0 . This procedure is very useful in applications, because allows one to determine the period of the unharmonic pendulum with any precision (Mathematica can handle approximate real numbers with any number of digits). To do this, we first observe that, ignoring the singular value θ0 = π, we have |k| < 1. A series expansion of the integrand of (3.20) gives: 1 1 2 2 1·3 4 4 p = 1 + k sin ψ + k sin ψ + ... 2 2·4 1 − k 2 sin2 ψ ∞ X 1 · 3 · 5 · · · (2n − 1) 2n 2n k sin ψ = 2 · 4 · 6 · · · 2n n=1

(3.21)

∞ ∞ X X (2n − 1)!! 2n 2n (2n − 1)!! 2n 2n k sin ψ = 1 + k sin ψ. =1+ (2n)!! 2n n! n=1 n=1

Since |k| < 1, the series (3.21) is uniformly, absolutely convergent on the interval (0, π). This fact allows us to integrate it term by term (operations of integration and summation commute, that is the series of integrals equals the integral of series). To this end, we shall first deduce some relation known as Wallis’ formula. R π/2 Let us consider the integrals of the form I2n = 0 sin2n ψ dψ, where n is an integer. We have: I2n =

=

Z Z

π/2

sin

2n

ψ dψ =

0 π/2

sin

2n−2

0

Z

π/2 0

ψ dψ −

= I2n−2 −

Z

π/2

Z

sin2n−2 ψ(1 − cos2 ψ) dψ

π/2

sin2n−2 ψ cos2 ψ dψ

0

sin2n−2 ψ cos2 ψ dψ.

0

Integrating by parts the last term, we still have Z

π/2

sin 0

2n−2

2

ψ cos ψ dψ =

Z

π/2 0

 1 2n−1 cos ψ d sin ψ 2n − 1 

iπ/2 1 h 2n−1 sin ψ cos ψ = 2n − 1 0 60

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Z

 1 sin2n−1 ψ (− sin ψ) dψ 2n − 1 Z π/2 1 = sin2n ψ dψ, 2n − 1 0

π/2 0



therefore I2n

1 = I2n−2 − 2n − 1

Z

π/2 0

sin2n ψ dψ = I2n−2 −

1 I2n , 2n − 1

which yields the following recurrence relation 2nI2n = (2n − 1)I2n−2 . Taking n = 1, 2, 3, etc., we arrive at: n=1 n=2 n=3 . . n=n−1

2I2 = I0 4I4 = 3I2 = 1·3 2 I0 6I6 = 5I4 = 1·3·5 2·4 I0 . . 2(n − 1)I2n−2 = (2n − 3)I2n−4 1·3·5···(2n−3) = 2·4·6···(2n−4) I0

n=n

2nI2n = (2n − 1)I2n−2 =

1·3·5···(2n−1) 2·4·6···(2n−2) I0 ,

leading to I2n

1 · 3 · 5 · · · (2n − 1) 1 · 3 · 5 · · · (2n − 1) = I0 = 2 · 4 · 6 · · · 2n 2 · 4 · 6 · · · 2n =

Z

π/2

dψ 0

1 · 3 · 5 · · · (2n − 1) π · , 2 · 4 · 6 · · · 2n 2

which is Wallis’ formula. We are now able to calculate K(k). Indeed, K(k) =

=

Z

π/2 0

h

Z

π/2 0

p

dψ 1 − k 2 sin2 ψ

∞ X 1 · 3 · 5 · · · (2n − 1) 2n 2n i 1+ k sin ψ dψ 2 · 4 · 6 · · · 2n n=1

61

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Free ebooks ==> www.Ebook777.com π X h 1 · 3 · 5 · · · (2n − 1) 2n = + k 2 n=1 2 · 4 · 6 · · · 2n ∞

π X h (2n − 1)!! 2n = + k 2 n=1 2n!! ∞

π X h (2n − 1)!! 2n = + k 2 n=1 2n n! ∞

Z

Z

Z

π/2

π/2

sin2n ψ dψ

0

sin2n ψ dψ

0 π/2

sin2n ψ dψ

0

i

i

i

∞ h ∞ X (2n − 1)!! 2 2n io π X h (2n − 1)!! 2 2n i π πn k · = 1+ k . = + 2 n=1 2n n! 2 2 2n n! n=1

The period of swing of a simple gravity pendulum, for an arbitrary amplitude, is then: s s ∞ h X R R πn (2n − 1)!! 2 2n io τ =4 K(k) = 4 1+ k n n! g g 2 2 n=1 = 2π

s

∞ h X Rn (2n − 1)!! 2 2n io 1+ k g 2n n! n=1

(3.22)

∞ h n X (2n − 1)!! 2 2n io = τ0 1 + k n n! 2 n=1

n

= τ0 1 +

∞ h X (2n − 1)!! 2

n=1

2n n!

sin2n

θ0 io . 2

This result is exact, meaning that we did not perform any approximation so far. If θ0 < π/2, then sin(θ0 /2) can also be developed in series according to the well-known formula ∞

X (−1)l 1 1 sin x = x − x3 + x5 − ... = x2l+1 , 3! 5! (2l + 1)! l=0

therefore X (−1)l  θ0 2l+1 θ0 θ0 θ3 θ5 sin = = 1 − 3 0 + 5 0 − ... 2 (2l + 1)! 2 2 · 1! 2 · 3! 2 · 5! ∞

l=0

Sometimes, in practical applications, only the first 4-5 terms of this series expansion are chosen. Let us write τ = τ0 (c1 + c2 θ02 + c3 θ04 + c4 θ06 + ...). 62

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Free ebooks ==> www.Ebook777.com According to (3.22), c1 = 1. To write the rest of the coefficients c2 , c3 , c4 , etc. one must observe that, for example, c4 involves the terms corresponding to n = 1, n = 2, and n = 3. We then have   1 9 225 2 θ0 4 θ0 6 θ0 τ = τ0 1 + sin + sin + sin + ... . 4 2 64 2 2304 2

Since

(3.23)

θ0 θ0 θ03 θ05 θ07 sin = 1 − + − + ... 2 2 · 1! 23 · 3! 25 · 5! 27 · 7! =

θ0 θ2 θ5 θ07 − 0 + 0 − + ..., 2 48 3840 645120

(3.24)

we can write  θ 2 θ03 θ05 θ07 θ02 0 2 θ0 = 1 − 3 + 5 − 7 + ... = + O(θ04 ). sin 2 2 · 1! 2 · 3! 2 · 5! 2 · 7! 4 (3.25) 1 1 1 2 Therefore, in view of (3.23), the coefficient of θ0 is: c2 = 4 · 4 = 16 . Next, let us determine c3 . In view of (3.24), we have:  θ 4 θ0 θ93 θ05 θ07 0 sin = 1 − + − + ... 2 2 · 1! 23 · 3! 25 · 5! 27 · 7! 2  θ θ3 θ5 θ7 0 − 3 0 + 5 0 − 7 0 + ... = 1 2 · 1! 2 · 3! 2 · 5! 2 · 7!  θ 2 θ3 θ5 θ7 0 × 1 − 3 0 + 5 0 − 7 0 + ... 2 · 1! 2 · 3! 2 · 5! 2 · 7! 2  θ 2 θ 3 θ0 θ05 θ03 θ05 0 0 7 7 = − + + O(θ0 ) − + + O(θ0 ) 2 48 3840 2 48 3840   θ2 θ6 θ0 θ03 θ0 θ5 + 2 · 0 + O(θ08 ) = 0 + 0 −2 4 2304 2 48 2 3840  θ2  θ06 θ0 θ03 θ0 θ05 0 8 × + −2 +2 · + O(θ0 ) 4 2304 2 48 2 3840  θ2   θ2 θ4 θ6 θ4 θ6 0 = 0 − 0 + 0 + O(θ08 ) − 0 + 0 + O(θ08 ) 4 48 1440 4 48 1440 4

=

θ04 θ6 − 0 + O(θo8 ). 16 96

(3.26)

To put into evidence the coefficients of θ04 and θ06 , we write (3.25) as sin2

θ 2 θ0 θ03 θ05 θ07 0 = − + − + ... 2 2 3! · 23 5! · 25 7! · 27 63

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Free ebooks ==> www.Ebook777.com = =



0

2



2 θ03 θ5 + 0 − O(θ07 ) 48 3840

θ02 θ6 θ0 θ 3 θ0 θ5 + 0 − 2 · 0 + 2 · o + O(θ08 ) 4 2304 2 48 2 3840 =

θ02 θ4 θ6 − 0 + 0 + O(θ08 ). 4 48 1440

Since, according to (3.23), the coefficients of sin2 9 , respectively, the coefficient c3 writes are 14 and 64 c3 =

(3.27) θ0 2

and sin4

θ0 2

9 1 1  1 11 + · − · = . 4 48 64 16 3072

Finally, in order to determine c4 one must consider the contribution of sin6 θ20 to the term in θ06 , that is: sin6

 θ2 3 θ0 θ4 θ6 = 0 − 0 + 0 + O(θ08 ) 2 4 48 1440

2 θ04 θ06 8 − + + O(θ0 ) = 4 48 1440  θ2  θ04 θ06 0 8 × − + + O(θ0 ) 4 48 1440  θ4  θ2  θ6 θ4 θ6 0 = 0 − 0 + O(θ08 ) − 0 + 0 + O(θ08 ) 16 96 4 48 1440  θ2 0

=

(3.28)

θ06 + O(θ08 ). 64

By means of (3.23) and (3.26)-(3.28), we then have c4 =

1 1 9  173 1 225 1 · + · − + · = . 4 1440 64 96 2304 64 737280

Consequently, for an arbitrary angle θ0 < π2 , the period increases gradually with angular amplitude, according to the formula τ = 2π

s

 R 1 11 4 173 6 1 + θ02 + θ0 + θ0 + ... g 16 3072 737280

h i 1 11 4 173 6 = τ0 1 + θ02 + θ0 + θ0 + O(θ08 ) . 16 3072 737280 64

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(3.29)

Free ebooks ==> www.Ebook777.com If the angular amplitude of oscillation is small (θ0 < 4o ), the terms involving θ02 , θ04 , etc. can be neglected as compared to unity, and we fall back on the formula corresponding to isochronous oscillations s R τ |θ0 www.Ebook777.com Presence of the term proportional to sin 3ω t is suggested by the trigonometric identity sin3 x = (3/4) sin x−(1/4) sin 3x. This way, the term involving θ3 in (3.32) will produce a term proportional to sin 3ω t, which must be compensated by the term ε sin 3ω t. The third power of this new term shall generate a term proportional to ε3 sin 9ω t, and so on. If ε ≪ 1, a rapid convergence of the series is expected. Solution (3.33) is, therefore, an approximate solution. A better approximation can be obtained if we also consider the term proportional to θ5 in (3.31). In this case, using the trigonometric identity sin5 u = (5/8) sin u − (5/16) sin 3u + (1/16) sin 5u, we search for a solution in the form θ(t) = θ0 sin ωt + εθ0 sin 3ωt + λθ0 sin 5ωt. As a result, the term proportional to θ5 generates a new series, which is more rapidly convergent than the first one, and so on. Let us now impose the solution (3.33) to verify equation (3.32). We have: 2 ¨ = d (θ0 sin ωt + εθ0 sin 3ωt) θ(t) dt2 = −ω 2 θ0 sin ωt − 9ω 2 εθ0 sin 3ωt, as well as [θ(t)]3 = θ03 (sin ωt + ε sin 3ωt)3   3 2 3 2 = θ0 sin ωt + 3ε sin ωt sin 3ωt + O(ε ) ≃ θ03

3 4

sin ωt −

 1 sin 3ωt + 3ε sin2 ωt sin 3ωt , 4

where the identity sin3 x = (3/4) sin x − (1/4) sin 3x has been used. Introducing the last two relations together with (3.33) into (3.32), one obtains −ω 2 θ0 sin ωt − 9ω 2 εθ0 sin 3ωt + ω02 (θ0 sin ωt + εθ0 sin 3ωt) 3  1 1 − ω02 θ03 sin ωt − sin 3ωt + 3ε sin2 ωt sin 3ωt = 0. 6 4 4

In order that (3.33) is an approximate solution at any time t, the sums of the coefficients of sin ωt and sin 3ωt - separately - must be zero, that is 3ω 2 −ω 2 + ω02 − 0 θ02 = 0, (3.35) 24 66

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Free ebooks ==> www.Ebook777.com and −9ω 2 εθ0 + ω02 εθ0 +

ω02 3 ω02 3 θ − θ ε sin2 ωt = 0. 24 0 2 0

(3.36)

As one observes, relation (3.34) follows immediately from (3.35), while (3.36) yields (ω 2 /24)θ02 θ02 ε = 02 ≃ , 9ω − ω02 192 where the last term (which is small as compared to the other three) has been neglected, and the approximation ω 2 ≃ ω02 has been used. Parameter ε gives the relative contribution of the term containing sin 3ωt in a solution ”dominated” by sin ωt. For example, if θ0 = 0.3 rad, (ω−ω0 ) the relative modification of frequency is ∆ω ≈ −10−2 , and ω = ω ε ≈ 10−3 . It can be easily verified that using an approximate solution of the form θ(t) = θ0 sin ωt + λθ0 sin 2ωt, and performing the required calculations, one obtains λ = 0. This means that the pendulum generates mostly the third harmonic (that is terms in sin 3ωt), while the second harmonic (corresponding to terms in sin 2ωt), as well as any other even harmonic, are absent. This result comes from the fact that the equation of motion contains a term involving θ3 (but no term in θ2 ). Here ”mostly” means that some other odd harmonics are also generated. These results can be interpreted as follows. In case of big amplitudes, does not exist a unique frequency of the periodical motion of pendulum. The most ”powerful” term is expressed as a function of sin ωt, and we call ω the fundamental frequency of the pendulum. In the first order of approximation, the fundamental frequency is given by (3.34), while (3.29) offers better approximations. The exact form of ω is given by (3.20) or (3.22). The term involving sin 3ωt is called the third harmonic of the fundamental frequency. Obviously, the real motion presents an infinite number of harmonics: the higher is their order, the smaller is their contribution (as an example, the amplitude of the third harmonic is about one thousand times smaller than the fundamental one). As we have seen at the beginning of this paragraph, to solve the problem within this formalism one must know the constraint force T~ (tension in the rod). The geometric properties of T~ (direction and sense) are shown in Fig.III.1, while its magnitude can be determined 67

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Free ebooks ==> www.Ebook777.com by means of (3.1), (3.3), and (3.8). Indeed, |T~ | = T = −



n  m R −

g R

m¨ y m(Rθ¨ cos θ − Rθ˙2 sin θ) =− sin θ sin θ 

sin θ cos θ − R

h

2g R (cos θ

i

− cos θ0 ) sin θ

sin θ

= mg(3 cos θ − 2 cos θ0 ).

o (3.37)

This way, all about the problem of simple pendulum, studied in terms of Newtonian mechanics, has been said. Next, we shall discuss the same application, by means of the analytical formalism. III.2. Lagrange equations of the first kind approach Within this formalism, the problem is presented as follows: given the mass of the body, the applied forces, the analytical equations of constraints and the initial conditions compatible with the constraints, find the low of motion of the body, the constraint forces, and, eventually, the elements of the motion like: trajectory, period of the motion, etc. As one can see, this approach (as all the other analytical formalisms) does not demand the a priori knowledge of the constrained forces; they are determined a posteriori. As a matter of fact, this is the main advantage of the analytical formalism. The Lagrangian equations of the first kind method allows to replace the knowledge of constraint forces by the analytical expressions of the constraints (given in either explicit, or implicit forms). In most cases, the constraint forces are not known from the beginning and the Newtonian approach is not able to solve the problem. Here is the algorithm for solving a problem within the Lagrangian equations of the first kind method: i) Identification of the applied forces F~i (i = 1, N ), where N is the number of the bodies (particles) composing the system, and their reference to a conveniently chosen coordinate system. ii) Fixing of the initial conditions, compatible with the constraints: ~ri0 = ~ri (t0 ), and ~r˙ i0 = ~r˙ i (t0 ). iii) Identification of the constraints, as well as their analytical expressions: fk (~r1 , ~r2 , ..., ~rN ; t) = 0 (k = 1, s), where s is the number of constraints. 68

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Free ebooks ==> www.Ebook777.com iv) Establishment and finding the solutions of the system of 3N +s equations in 3N + s unknowns: N Lagrange equations of the first kind mi~¨ri = F~i +

s X

λk gradi fk

(i = 1, N ),

k=1

together with expressions of s constraints, while the unknowns are: 3N coordinates xi = xi (t), yi = yi (t), zi = zi (t) and s non-zero constants λk (k = 1, s). ~ i (i = 1, N ), using v) Determination of the constraint forces L P s ~i = their definition L k=1 λk gradi fk . Before applying this algorithm in our case, we point out that the Lagrange equations of the first kind formalism is valid for ideal constraints only. This is a very restrictive condition, and does not correspond to reality, but, as far as the friction forces are known1 , these can be considered as applied forces and added to them. Using the same reference frame as in the previous paragraph, let us choose xOy as the plane of motion. Then, the only applied force (force of gravity) ~ = (mg, 0, 0) G is accompanied by the initial conditions ~r0 = (x0 , y0 , 0) = (R cos θ0 , R sin θ0 , 0), and

~r˙ 0 = 0.

1

The friction forces are usually determined by empirical methods. For example, in case of the relative motion of two bodies, one determines experimentally that the force of friction is proportional to the ~ , where µ is the coefficient of friction. If normal reaction, F~f = µN the motion takes place in a viscous fluid, the experiment shows that for small velocities (laminar flow) the force of friction is proportional to the relative velocity between body and fluid, F~f = −k~vr , where the coefficient k depends on viscosity of the fluid and and the form of the transverse section of the body. At high velocities (turbulent flow), the same force is proportional to the squared relative velocity, F~f = −k|~vr |2 ~ν , where ~ν is the unit vector of direction of motion of the body, and k - a ”constant” which depends on the medium density, the area of the cross section of the body, and the so-called form coefficient, tightly connected to the aerodynamic characteristics of the body. 69

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Free ebooks ==> www.Ebook777.com The body is submitted to two constraints: f1 (x, y) = x2 + y 2 − R2 = 0,

(3.38 a)

f2 (z) = z = 0.

(3.38 b)

and Since our body is a particle (point mass), the index i can be dropped. The system of equations leading to the law of motion of the body of mass m then is  ~ + P2 λk grad fk ,  m~¨r = G k=1 (3.39) f1 (x, y) = x2 + y 2 − R2 = 0,  f2 (z) = z = 0. Since the constraints are given by

grad f1 = 2x~i + 2y~j,

(3.40 a)

grad f2 = 0,

(3.40 b)

and the system (3.39) becomes  m¨ x = mg + 2λx,   m¨ y = 2λy, 2 2 2   x + y − R = 0, z = 0,

(3.41)

which is a system of four equations with four unknowns: x, y, z, λ. Here λ stands for λ1 . Relation (3.41 c) allows one to parametrize the constraint f1 (x, y) as  x = R cos θ, y = R sin θ, and the first two equations of the system (3.41) write  m(−Rθ¨ sin θ − Rθ˙2 cos θ) = mg + 2λR cos θ, m(Rθ¨ cos θ − Rθ˙2 sin θ) = 2λR sin θ.

(3.42)

This way, the problem reduces to two differential equations with two unknowns: θ(t) and the scalar λ. A rearrangement of terms, followed by division of the equations, yield: cos θ −Rθ¨ sin θ − Rθ˙2 cos θ − g = , sin θ Rθ¨ cos θ − Rθ˙2 sin θ 70

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Free ebooks ==> www.Ebook777.com leading to the differential equation for θ(t) g θ¨ + sin θ = 0, R

(3.3′ )

which is precisely equation (3.3) obtained within the Newtonian approach. From now on, the problem is solved following classical formalism. The only difference consists in determination of the force of ~ which in Lagrangian method is not previously known constraint L, (neither its direction, nor its sense or magnitude). This can be found by means of its definition ~ = L

2 X

λk grad fk = λ1 grad f1 + λ2 grad f2 = λgrad f1

k=1

= 2λx~i + 2λy~j = 2λR(~i cos θ + ~j sin θ). ~ is The magnitude of L ~ = |L|

q p L2x + L2y ≡ L = 4λ2 R2 cos2 θ + 4λ2 R2 sin2 θ = 2λR. (3.43)

The parameter λ can be determined by means of (3.42 b). Using (3.3’) and (3.7), we have m(Rθ¨ cos θ − Rθ˙2 sin θ) 2R sin θ n   h i o 2g g m R − R sin θ cos θ − R R (cos θ − cos θ0 ) sin θ λ=

=

2R sin θ

mg(2 cos θ0 − 3 cos θ) . 2R The magnitude of the constraint force therefore is =

L = 2Rλ = mg(2 cos θ0 − 3 cos θ).

(3.44)

Except for a sign difference (which shall be explained in the next paragraph), this formula coincides with (3.37) expressing |T~ | = T . This fact can also be proved if we write (3.43) as ~ ≡L= |L|

q p L2x + L2y = 4λ2 R2 cos2 θ + 4λ2 R2 sin2 θ = |2λR|, 71

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Free ebooks ==> www.Ebook777.com which obviously leads to |L| = mg|2 cos θ0 − 3 cos θ| = |3 cos θ − 2 cos θ0 | = |T |. ~ and tension in the rod T~ To conclude, the constraint force L have the same magnitude. The difference in sign appears because the two vectors have opposite senses: the gradient is oriented along the external normal to the constraint (trajectory), while the vector T~ is, by definition, oriented along the internal normal (see Fig.III.1). III.3. Lagrange equations of the second kind approach The problem to solve is the same as in the previous approach. The difference appears in the procedure of finding the result. Namely, the use of Lagrange equations of the second kind demand neither knowledge of the constraint forces, nor the analytical expressions of the constraints (essential for the first kind of Lagrange equations method), but only the number of constraints. The algorithm for solving a problem within the Lagrangian equations of the second kind method demands the following steps: i) Identification of applied forces F~i , (i = 1, N ), where N stands for the number of the bodies (particles) of the system. ii) Establishment of the initial conditions (compatible with the constraints) in real space: ~ri0 = ~ri (t0 ), and ~r˙ i0 = ~r˙ i (t0 ). iii) Identification of the number s of constraints, the number of degrees of freedom of the system, n = 3N − s, and the convenient choice of generalized coordinates qj = qj (t) (j = 1, n). iv) Determination of the relations connecting the real and configuration spaces ~ri = ~ri (q1 , q2 , ..., qn , t) ≡ ~ri (q, t) (i = 1, N ), together with their inversions qj = qj (~r1 , ~r2 , ..., ~rN , t) (j = 1, n), as well as the initial conditions in configuration space, qj0 = qj (t0 ), q˙j0 = q˙j (t0 ). v) Determination of the kinetic energy T = T (q, q, ˙ t) and the generalized forces Qj = Qj (q, t) for non-potential systems, or the Lagrangian function L(q, q, ˙ t) = T (q, q, ˙ t)−V (q, q, ˙ t) in the case of natural systems. (A system that admits a simple or generalized potential is called natural).  vi)Writing and solving the Lagrange equations of the second kind d ∂T ∂T dt ∂ q˙j − ∂qj = Qj (j = 1, n), if the applied forces are not potential,   d ∂L ∂L and dt ∂ q˙j − ∂qj = 0 (j = 1, n), in case of natural systems. Using the initial conditions, one determines the law of motion in configuration space qj = qj (t) (j = 1, n), which also represents the parametric equations of generalized trajectory in this space. 72

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Free ebooks ==> www.Ebook777.com vii) Establishment of the law of motion in the real space ~ri =  ~ri qj (t); t ≡ ~ri (t) (i = 1, N ; j = 1, n). viii) Finding the constraint forces, by means of the fundamental equation of dynamics, written for systems subject to constraints: ~ i = mi~¨ri − F~i L

(i = 1, N ).

We shall use this algorithm in order to study the problem of mathematical pendulum, performing free, non-amortized oscillations, for an arbitrary value of the angular amplitude. The only applied force is the ~ = (mg, 0, 0), its components being expressed with force of gravity G regard to the already chosen reference frame xOy (see Fig.III.7).

Fig.III.7 The initial conditions in the real space remain, obviously, the same: ~r0 = (R cos θ0 , R sin θ0 , 0), and ~r˙ 0 = 0. Liberty of the motion is limited by two constraints (already discussed in the previous paragraph), so that the system possesses a single degree of freedom: n = 3N − s = 3 · 1 − 2 = 1. The choice of the generalized coordinate associated to this degree of freedom is not unique. In this respect, one can choose the coordinate x, or coordinate y, or an arc length determined relative to an arbitrary origin, or, more convenient, the angle θ between Ox-axis and the rod. Generally speaking, the appropriate choice of the generalized coordinate(s) depends on the experience and skillfulness of the person engaged in solving a problem. In other words, if the initial choice is not accurate (convenient), meaning that solving the Lagrange equations is difficult (or 73

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Free ebooks ==> www.Ebook777.com even impossible), there always exists a possibility of choosing another set of generalized coordinates according to the ”point” transformation in configuration space: qj → qj′ = qj′ (q1 , q2 , ..., qn ; t)

(j = 1, n).

So, we choose the angular coordinate θ as generalized coordinate associated to the single degree of freedom of the simple pendulum. The relations connecting x, y, on the one hand, and θ on the other, are x = R cos θ, y = R sin θ, while the initial conditions write θt0 =0 = θ0 ,

θ˙t0 =0 = θ˙0 = 0.

In general, Lagrange equations have the form d  ∂T  ∂T − = Qj (j = 1, n). dt ∂ q˙j ∂qj

(3.45)

PN ∂~ ri are the generalized forces, and T (q, q, ˙ t) Here Qj (q, q, ˙ t) = i=1 F~i · ∂q j the kinetic energy. Our system is a natural system, that is one can define a Lagrangian function L = T − V , where the potential V can be either simple V = V (~r, t), or generalized V = V (~r, ~r˙ , t). In our case, V can be deduced as follows: dV (f ormally) . F~ = −gradV (~r, t) = − d~r Then

dV = −F~ · d~r = −mg dx,

and, by integration V = −mgx + V0 . The arbitrary constant V0 is determined by a convenient choice of the reference level of the potential energy. Since grad(V + V0 ) = gradV , where V0 is a constant, the potential energy is not uniquely determined, so it is up to us to choose a useful value for V0 . Let us take Vx=0 = 0, in which case we have V = V (x) = −mgx, 74

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(3.46)

Free ebooks ==> www.Ebook777.com or, since x = R cos θ, V (θ) = −mgR cos θ. The kinetic energy T is N

1X 1 1 T = mi |~r˙ i |2 = m(x˙ 2 + y˙ 2 ) = mR2 θ˙2 , 2 i=1 2 2

(3.47)

so that the Lagrangian can be written as L=

1 mR2 θ˙2 + mgR cos θ. 2

(3.48)

The Lagrange equation of the second kind then writes d  ∂L  ∂L = 0. − dt ∂ θ˙ ∂θ ˙ Since ∂L = mR2 θ, ∂ θ˙ equation

∂L ∂θ

(3.49)

= −mgR sin θ, we arrive at the already known

g θ¨ + sin θ = 0. (3.50) R From now on, the problem is solved as shown in the previous paragraphs. The only difference appears in determination of the constraint force, which, in view of the fundamental equation of Mechanics for a system subject to constraints, can be written as ~ i = mi~¨ri − F~i L

(i = 1, N ).

(3.51)

Before going further, it is useful to mention that θ˙ can be determined in a less complicated way than it was within the Newtonian approach. This can be done by using a first integral suggested by the problem. Since the time t does not explicitly appear in the Lagrangian (3.48), equation (3.49) admits the first integral ∂L − L = const. θ˙ ∂ θ˙ We have: ∂L 1 θ˙ − L = mR2 θ˙2 − mR2 θ˙2 − mgR cos θ 2 ∂ θ˙ 75

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Free ebooks ==> www.Ebook777.com 1 mR2 θ˙2 − mgR cos θ = const. = E, 2 where E is the total energy of the system. It can be determined in view of the initial conditions θt0 =0 = θ0 , θ˙t0 =0 = θ˙0 = 0. It then follows =

E = const = −mgR cos θ0 and the first integral yields θ˙ = ±

r

2g (cos θ − cos θ0 ), R

as expected. In general, one can also use another type of first integrals of the Lagrange equations of the second kind, namely those associated with cyclic variables (if there are any): if the Lagrangian does not explicitly depend on a certain variable (called cyclic), then the generalized momentum associated (or conjugated) with this variable is conserved. Indeed, if qα is a cyclic generalized coordinate, then ∂L/∂qα = 0, and, following the Lagrange equation for this coordinate, one obtains ∂L = pα = const. ∂ q˙α Let us now turn back to the determination of the constraint force. Since in our case N = 1, the index i in (3.51) can be dropped, so that ~ = m~¨r − F~ = m~¨r − G. ~ L

(3.52)

Unlike the Lagrange equations of the first kind formalism, where the projections on axes of the constraint force were first determined, this time we take as a reference frame a system of two orthogonal axes, connected to the body trajectory: one axis tangent to the trajectory and pointing the direction of increasing θ, of versor ~τ , and the other axis normal to the trajectory, of versor ~ν , oriented along the external normal (see Fig.III.7). The components of the constraint force on these axes then are ~ · ~τ , Lτ = L (3.53) and

~ · ~ν . Lν = L

(3.54)

In order to determine Lτ and Lν , we shall first write ~τ and ~ν in terms of ~i and ~j. With the help of Fig.III.7, we have ~τ = ~τx + ~τy = −~i sin θ + ~j cos θ, 76

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Free ebooks ==> www.Ebook777.com and

~ν = ~νx + ~νy = ~i cos θ + ~j sin θ. ~ are According to (3.52), the components of L ~ = (m¨ L x − mg, m¨ y , 0),

which allows us to write ~ · ~τ = [(m¨ Lτ = L x − mg)~i + m¨ y~j] · (−~i sin θ + ~j cos θ) = m(g − x ¨) sin θ + m¨ y cos θ, and

~ · ~ν = [(m¨ Lν = L x − mg)~i + m¨ y~j] · (~i cos θ + ~j sin θ) = m(¨ x − g) cos θ + m¨ y sin θ. Since

x ¨ = −Rθ¨ sin θ − Rθ˙2 cos θ; y¨ = Rθ¨ cos θ − Rθ˙2 sin θ,

where

2g (cos θ − cos θ0 ); θ˙2 = R

g θ¨ = − sin θ, R

we have: Lτ = m(g + Rθ¨ sin θ + Rθ˙2 cos θ) sin θ + mRθ¨ cos2 θ − mRθ˙2 sin θ cos θ   g (3.55) = mg sin θ + mRθ¨ = mR θ¨ + sin θ = 0. R This result has been expected, because the constraint is ideal (the force of friction, given by the tangent component of the constraint force, must be zero). Also, Lν = −m(g + Rθ¨ sin θ + Rθ˙2 cos θ) cos θ + mRθ¨ sin θ cos θ − mRθ˙2 sin2 θ = −mg cos θ − 2mg(cos θ − cos θ0 ) = mg(2 cos θ0 − 3 cos θ).

(3.56)

The magnitude of the constraint force is therefore L = |Lν | = mg|2 cos θ0 − 3 cos θ|.

(3.57)

As one can see, this result is identical to that obtained in the previous paragraph, but differs in sign from that found by the Newtonian approach. The explanation is that the tension T~ in the rod (see 77

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Free ebooks ==> www.Ebook777.com Fig.III.1), and the external normal ~ν (see Figg.III.7) have opposite senses. III.4. Hamilton’s canonical equations approach Within this formalism, the enunciation of the problem is the same as in the previous two paragraphs. To solve a problem by method offered by Hamilton’s canonical equations, one uses the following algorithm: i) Identification of applied forces F~i (i = 1, N ), where N is the number of bodies (particles) of the system. ii) Setting the initial conditions, compatible with the constraints, in real space: ~ri0 = ~ri (t0 ), and ~r˙ i0 = ~r˙ i (t0 ). iii) Identification of the number of degrees of freedom of the studied system n = 3N − s and a convenient choice of the generalized coordinates qj = qj (t) (j = 1, n). iv) Establishment of connection relations between the real and configuration spaces ~ri = ~ri (q1 , q2 , ..., qn , t) ≡ ~ri (q, t) (i = 1, N ). v) Determination of the Lagrange function L = L(q, q, ˙ t) and of the generalized momenta pj = ∂L/∂ q˙j (j = 1, N ) as well. vi) Writing the inverse of relation ~ri = ~ri (q, t), that is qj = qj (~r1 , ~r2 , ..., ~rN , t) ≡ qj (~r, t), and setting the initial conditions in phase space qj0 = qj (t0 ), pj0 = pj (t0 ) (j = 1, n). Pn vii) Writing the Hamiltonian by means of the definition H = j=1 pj q˙j − L = H(p, q, t). viii) Writing and solving Hamilton’s canonical equations q˙j = +

∂H ; ∂pj

p˙ j = −

∂H , ∂qj

(3.58)

which furnishes the law of motion of the system in phase space qj = qj (t), pj = pj (t) (j = 1, n). These relations also represent the parametric equations of the generalized trajectory of the representative point in phase space. ix) Determination of the law of motion in the real space: ~ri = ~ri (t). ~ i = mi~¨ri − F~i (i = x) Determination of the constraint forces: L 1, N ). In the light of this algorithm, we shall resume the problem of simple gravitational pendulum performing free, harmonic, non-amortized oscillations, for an arbitrary value θ0 of the oscillation angular amplitude. 78

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Free ebooks ==> www.Ebook777.com ~ = (mg, 0, 0), whose The only applied force is the force of gravity G components are expressed with respect to the reference system xOy (see Fig.III.7). The initial conditions are, obviously, the same: ~r0 = (x0 , y0 , 0) = (R cos θ0 , R sin θ0 , 0), and

~r˙ 0 = 0,

As we previously did, since the system possesses a single degree of freedom (n = 3N − s = 3 · 1 − 2 = 1), as generalized coordinate one can choose the angular coordinate θ. The Lagrangian being L=

1 mR2 θ˙2 + mgR cos θ, 2

the associated generalized momentum is pθ =

∂L ˙ = mR2 θ. ˙ ∂θ

(3.59)

Consequently, the phase space has two dimensions. In other words, a point in this space has two coordinates (θ, pθ ). The initial conditions write θt0 =0 = θ0 ; pθ (t0 =0) = (pθ )0 = mR2 θ˙t0 =0 = 0, while the Hamiltonian is H = pθ θ˙ − L = pθ =

2 1 pθ 2 pθ − mR − mgR cos θ mR2 2 m2 R4

1 p2θ − mgR cos θ = H(θ, pθ ). 2 mR2

(3.60)

By virtue of (3.58), we have ∂H θ˙ = ; ∂pθ

p˙θ = −

∂H , ∂θ

(3.61)

or, in view of (3.60), pθ θ˙ = ; mR2

p˙θ = −mgR sin θ. 79

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(3.62)

Free ebooks ==> www.Ebook777.com This is a coupled systems of two linear differential equations for variables θ and pθ . To obtain equation for θ, one takes the derivative with respect to time of the first equation of (3.62), then introduces the result into the second equation. The result is g θ¨ + sin θ = 0, R

(3.63)

which is precisely the equation we were looking for. From now on, the procedure follows the way already shown in the previous paragraph (Lagrange equations of the second kind), concerning determination of both the elements of motion1 , and the constraint ~ = m~¨r − G. ~ force L III.5. Hamilton-Jacobi method To solve a problem by means of the Hamilton-Jacobi formalism, one must follow the steps of this algorithm: i) Identification of applied forces F~i (i = 1, N ), where N is the number of bodies (particles) composing the system. ii) Establishment of the initial conditions (compatible with the constraints) in real space ~ri0 = ~ri (t0 ); ~r˙ i0 = ~r˙ i (t0 ). iii) Identification of the degrees of freedom of the physical system n = 3N − s and a convenient choice of the generalized coordinates qj = qj (t) (j = 1, n). iv) Establishment of relations between the real and configuration spaces ~ri = ~ri (q1 , q2 , ..., qn , t) ≡ ~ri (q, t) (i = 1, N ), their inverses qj = qj (~r1 , ~r2 , ..., ~rN , t) ≡ qj (~r, t), as well as the initial conditions in configuration space qj0 = qj (t0 ), q˙j0 = q˙j (t0 ) (j = 1, n). v) Writing the Lagrangian of the system L(q, q, ˙ t), the generalized momenta associated with the generalized coordinates pj = ∂∂L q˙j (j = 1, n), and establishment of the 2n initial conditions in configuration space qj0 = qj (t0 ), pj0 = pj (t0 ) (j = 1, n). vi) Determination of the Hamiltonian of the system H=

n X j=1

pj q˙j − L = H(q, p, t).

1

We note that even if Hamilton’s canonical equations and second kind Lagrange equations are equivalent, in this paragraph we cannot take advantage of the two first integrals established within Lagrangian approach. As a result, θ˙ is determined by means of the Newtonian formalism. 80

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Free ebooks ==> www.Ebook777.com vii) Writing and solving the Hamilton-Jacobi equation  ∂S ∂S ∂S ∂S  + H q1 , q2 , ..., qn , , , ..., , t = 0, ∂t ∂q1 ∂q2 ∂qn

(3.64)

where S(q, t) is Hamilton’s principal function. viii) Determination of the law of motion in configuration space qj = qj (t) (j = 1, n), by means of Jacobi’s theorem. ix) Setting the law of motion in real space ~ri = ~ri (t) (i = 1, N ), in view of relations established at the point (iv) of this algorithm. x) Finding the constraint forces, by means of the fundamental equation of Mechanics for systems subject to constraints ~ i = mi~¨ri − F~i L

(i = 1, N ).

We shall use this algorithm to solve the problem of plane (simple) pendulum, performing free, non-amortized oscillations, of an arbitrary amplitude θ0 . First of all, we observe that the only applied (active) ~ = (mg, 0, 0), whose components are force is the force of gravity G expressed in the xOy coordinate system (see Fig.III.7). The initial conditions are, obviously, the same as in the preceding approaches: ~r0 = (x0 , y0 , 0) = (R cos θ0 , R sin θ0 , 0); ~r˙ 0 = 0. Since the motion is subject to two constraints (see the previous methods), there is a single degree of freedom: n = 3N −s = 3·1−2 = 1. As generalized coordinate, associated with this degree of freedom, one chooses the angle θ. The coordinates of the real space x, y and the generalized coordinate θ are related by x = R cos θ;

y = R sin θ,

while the initial conditions in configuration space write θt=0 = θ0 ,

θ˙t=0 = θ˙0 = 0.

The Lagrangian and the Hamiltonian have been already deduced in paragraphs III.3 [see (3.48)] and III.4 [see (3.60)]: L=

1 mR2 θ˙2 + mgR cos θ, 2 81

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Free ebooks ==> www.Ebook777.com 1 p2θ − mgR cos θ. (3.65) 2 mR2 According to Jacobi’s theorem (see further), one can write H=

pj =

∂S ∂qj

(j = 1, n),

where S(q, t) is Hamilton’s principal function. In our case, since our system has only one degree of freedom, the generalized momentum is pθ = ∂S ∂θ , and the Hamiltonian becomes 1  ∂S 2 H= − mgR cos θ. 2mR2 ∂θ The Hamilton-Jacobi equation then writes 1  ∂S 2 ∂S + − mgR cos θ = 0. (3.66) ∂t 2mR2 ∂θ There are no general methods for solving Hamilton-Jacobi equation (3.64). Nevertheless, in some particular cases one can use the method of separation of variables. In this respect, a great help comes from Jacobi’s theorem. We shall not prove it, but apply it in our case. To do this, one must recall several important definitions. A complete integral (or complete solution) of a first-order partial differential equation is a solution of this equation which contains as many arbitrary, independent constants as there are independent variables in the equation. In Hamilton-Jacobi equation interfere (n + 1) independent variables: q1 , q2 , ..., qn , t, but, since S appears only as partial derivatives, one of the arbitrary constants is purely additive. Therefore, a complete integral of the Hamilton-Jacobi equation writes S(q1 , q2 , ..., qn , a1 , a2 , ..., an , an+1 , t) = S(q1 , q2 , ..., qn , a1 , a2 , ..., an , t) + const., where as an additive constant has been taken, for example, an+1 . Because in Hamilton-Jacobi equation appear only partial derivatives of S, this constant can be dropped. To conclude, a complete integral of the Hamilton-Jacobi equation writes S(q1 , q2 , ..., qn , a1 , a2 , ..., an , t), the constants a1 , ..., an being called essential. According to Jacobi’s theorem, if S(q1 , q2 , ..., qn , a1 , a2 , ..., an , t) is a complete integral (complete solution) of the Hamilton-Jacobi equation (3.64), then the general solution of Hamilton’s canonical equations (3.58) is given by pj =

∂S ; ∂qj

bj =

∂S ∂aj

(j = 1, n),

82

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(3.67)

Free ebooks ==> www.Ebook777.com where bj (j = 1, n) are new constants. It is not our purpose to prove this theorem, but we shall use it. Introducing S(q, a, t) into (3.67b), we obtain the law of motion in the configuration space: qj = qj (a1 , a2 , ..., an , b1 , b2 , ..., bn , t) (j = 1, n). The constants a1 , ..., an , b1 , ..., bn are determined by means of the initial conditions. This way, the law of motion in real space ~ri = ~ri (q(t), t) = ~ri (t) (i = 1, N ) can be found. In our particular case, we have to determine a complete integral S = S(θ, E, t) of equation (3.66). It depends on a single essential constant E, which has been chosen to be the total energy of the system. Since there are two independent variables θ and t, and the Hamiltonian does not explicitly depend on time, this suggests a solution of the form S(θ, a, t) = W (θ, a) + S1 (a, t),

(3.68)

where a is an essential arbitrary constant, while W and S1 are two functions, unknown for the moment. Introducing (3.68) into the Hamilton-Jacobi equation  ∂S  ∂S + H θ, = 0, ∂t ∂θ we arrive at

 ∂W (θ, a)  ∂S1 (a, t) + H θ, = 0. ∂t ∂θ

(3.69)

 ∂W  = E, H θ, ∂θ

(3.70)

Since the first term depends only on t, and the second only on θ, equation (3.69) is satisfied only if each of the two terms equals the same constant, say E. Consequently, we have:

called restricted or abbreviated Hamilton-Jacobi equation, and ∂S1 = −E. ∂t

(3.71)

Since a complete solution of (3.69) must depend on one essential constant only, this means that a must be precisely E. The constraint is scleronomous, and the Hamiltonian does not explicitly depend on time. This means that the total energy E of the system is conserved. 83

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Free ebooks ==> www.Ebook777.com It is now clear why we denoted by E the constant in Eqs. (3.70) and (3.71). Equation (3.71) yields S1 (a, t) ≡ S1 (E, t) = −Et, so that (3.68) becomes S(θ, a, t) ≡ S(θ, E, t) = −Et + W (θ, E).

(3.72)

To determine W (θ, E) one must solve the abbreviated HamiltonJacobi equation (3.70). As we shall see, in our case is not necessary to know the explicit form of this function. Introducing pθ =

∂W ∂S = ∂θ ∂θ

into Hamilton’s function, and the result into (3.70), we have 1  ∂W 2 − mgR cos θ = E, 2mR2 ∂θ and, by integration Z p 2mR2 (E + mgR cos θ) dθ. W (θ, E) = ±

 ∂S According to Hamilton-Jacobi theorem, ∂E = b [see (3.67-2)], and we have ∂W = b. (3.73) −t + ∂E Since b must have units of time, we denote it by t1 , and (3.73) becomes ∂W = t + t1 , ∂E that is Z  p ∂ 2 ± 2mR (E + mgR cos θ)dθ = t + t1 , ∂E or, recalling that E is an essential constant, and θ an independent variable, Z i ∂ hp ± 2mR2 (E + mgR cos θ) dθ = t + t1 . ∂E 84

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Free ebooks ==> www.Ebook777.com Performing the derivative, we still have ±

Z

q

dθ 2E mR2

+

2g R

= t + t1 . cos θ

The last equation says that t + t1 is the primitive of 

2g 2E + ± cos θ 2 mR R which means

or

d (t + t1 ) = ± q dθ dθ =± θ˙ = dt

r

−1/2

,

1 2E mR2

+

2g R

, cos θ

2E 2g + cos θ. 2 mR R

Let us introduce a new constant E . mgR

(3.74)

2g (cos θ − cos θ0 ). R

(3.75)

const. = cos θ0 = − This finally leads to θ˙ = ±

r

This equation has been already obtained within the frame of the Newtonian and Lagrange equations of the second kind approaches. From now on, the problem can be solved as in paragraph III.3. Observation. The way we have introduced the constant cos θ0 is in agreement with the expression of the total energy of the system, E = −mgR cos θ0 , deduced by the Lagrange equations of the second kind formalism. III.6. Action-angle variables formalism This approach is successfully used in case of the systems performing periodic motions. It emerges from the Hamilton-Jacobi method and consists in defining a new set of variables: Jj - action variables, and wj - angle variables, instead of the canonical parameters qj , pj (j = 1, n), where n is the number of degrees of freedom of the system. 85

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Free ebooks ==> www.Ebook777.com Let us briefly explain the essence of this formalism. Consider a conservative system with one degree of freedom. Denoting by E the total energy of the system, we can write H(p, q) = E.

(3.76)

This is the implicit form of a curve in the phase plane (p, q) and represents the generalized trajectory of the conservative system characterized by the constant energy E. There are two possible periodical motions: 1. If the generalized trajectory is a closed curve, the motion is called vibration. In this case, the generalized coordinate q oscillates between two constant values, both p and q being functions of time with the same period (see Fig.III.8).

Fig.III.8 2. If the solution p = p(q, E) of equation (3.76) can be written as a periodical function of q with period q0 , that is p(q + kq0 , E) = p(q, E) (k ∈ Z ), the periodical motion is called rotation, or revolution. In this case, the coordinate q can take any value (see Fig.III.9). There are systems capable to perform both vibration and rotation motions. As we shall see, such a system is precisely our case of the simple pendulum. Let us define the action variable as J=

I

p dq ,

(3.77)

where the integral is taken over a complete cycle of variation of q. It stands for both the closed area shown in Fig.III.8, and the shaded area drawn in Fig.III.9, corresponding to a period of motion of rotation. 86

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Fig.III.9 R t2 According to the definition, the quantity J and the action S = L dt have the same dimension. Taking into account the relations t1 p = p(q, E) and (3.77), we can write J = J(E) or, conversely, E = E(J). The complete integral corresponding to Hamiltonian (3.76) is S = S(q, E), with a single essential constant E. Since E = E(J), this yields S = S(q, J). (3.78) The canonical variable w associated with J is defined by w=

∂S , ∂J

(3.79)

being called angle variable. The name comes from the fact that w is associated with an angular momentum, so that its dimension is that of an angle. In view of (3.76) and E = E(J), the new Hamiltonian is H = H(J).

(3.80)

With this choice, canonical equations lead to 

J˙ = − ∂H ∂w = 0, ∂H w˙ = ∂J = ν(J).

(3.81)

It follows from (3.81)1 that J = const. (which we already know). As a result, ν is also a constant, depending on J. Integrating (3.81)2 , we obtain w = νt + α, (3.82) 87

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Free ebooks ==> www.Ebook777.com where α is an arbitrary constant of integration. To find the significance of ν, we shall determine the way of variation of w for a complete cycle of variation of q, for both vibration and rotation motions. Thus, ∆w =

I

∂w dq = ∂q

I

∂2S dq = ∂q∂J

I

∂p d dq = ∂J dJ

I

p dq = 1.

Denoting by τ the period corresponding to a complete cycle, the last two relations yield ∆w = ν ∆t = ντ = 1, which means ν=

1 . τ

(3.83)

This result shows that ν is the frequency of the periodical variation of q. This displays the fact that the period of any motion can be determined without solving the equation of motion, if one knows the dependence on time of the Hamiltonian. Inverting (3.78) and using (3.82), one can also determine the time dependence of coordinate q. Let us now apply these considerations to the ”vibration” motion of a simple pendulum. As seen in paragraph III.3, the Lagrangian of the system is 1 L = mR2 θ˙2 + mgR cos θ. 2 In our case q = θ, while the associated generalized momentum is ˙ The Hamiltonian therefore is p = ∂L = mR2 θ. ∂ θ˙ 1 p2 H = pθ˙ − L = mR2 θ˙2 − mgR cos θ = − mgR cos θ = E. 2 2mR2 This allows us to write the equation of generalized trajectory in the phase plane (θ, p) as p=±

p 2mR2 (E + mgR cos θ),

as well as the action variable J J = −2 =4

Z

Z

−θ0

[2mR2 (E + mgR cos θ)]1/2 dθ

θ0

θ0

[2mR2 (E + mgR cos θ)]1/2 dθ.

0

88

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(3.84)

Free ebooks ==> www.Ebook777.com Equation (3.81)2 can be written as ∂H 1 ν = ν(J) = = ∂J = ∂J ∂H



∂J ∂E

−1

.

According to (3.83), we then have Z

∂J d 1 =4 τ= = ν ∂E dE

=4

s

R 2g

Z

θ0

p 2mR2 (E + mgR cos θ) dθ

θ0

dθ √ . cos θ − cos θ0

0

0

! (3.85)

This result has been already obtained by means of Newtonian approach, and the remaining part of investigation can be found within this formalism. There is, nevertheless, one problem: the angle-action method is not able to determine the constraint forces (in our case, tension in the rod). Due to this inconvenience, the angle-action procedure is used only for systems performing periodical motions, in which case the main problem is determination of the period of motion. In the case of small oscillations (θ0 < 4o ), one can approximate cos θ = 1 −

θ2 θ2 + O(θ4 ) ≃ 1 − , 2 2

and the Hamiltonian writes p2 p2 − mgR cos θ ≃ + mgR H= 2mR2 2mR2



θ2 2



− mgR,

or, by omitting the constant term (−mgR), H= so that p=±

p2 mgRθ2 + = E, 2mR2 2

s

2mR2



2



 mgRθ2 E− , 2

leading to J = −2

Z

−θ0 θ0



2mR

mgRθ2 E− 2

1/2

89

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Free ebooks ==> www.Ebook777.com √ Z = 2

θ0 −θ0

= 4E

2

2

3

4mER − 2m gR θ s

R g

Z

β1 +π

 2 1/2

dθ = 4E s

R g

Z

s

R g

Z

ξ1 −ξ1

(1 − ξ 2 )1/2 dξ

β1 +π

1 + cos 2β dβ 2 β1 β1 s s " β +π #  R π sin 2β 1 R = 2πE + , = 4E g 2 4 g β1 2

cos β dβ = 4E

where the following changes of variable have been used r r mgR mgR θ = ξ = sin β, θ0 = ξ1 , β1 = arcsin(−ξ1 ). 2E 2E Therefore, J = 2πE

s

R , g

(3.86)

leading to the well-known formula of small, isochronous oscillations of a simple pendulum s R dJ = 2π . (3.87) τ= dE g Let us now turn back to the generalized trajectory given by (3.84). The value p = 0 corresponds to cos θ0 = −

E . mgR

(3.88)

If E < mgR, the pendulum performs a periodical motion called vibration, with angle θ varying between −θ0 and θ0 , with θ0 given by (3.88). If E > mgR, there are no restrictions on θ, that is the motion is the same for any θ + 2kπ (k ∈ Z ). A special situation is given by the limit case E = mgR, corresponding to θ = ±π [or, in general, θ = (2k + 1)π]. These values of θ, together with p = 0, designate the position of unstable equilibrium of the system. ∗





To conclude this chapter, let us calculate the deviation from the real time of an astronomical pendulum (θ0 = 1o 30′ ) in one year, assuming that the oscillations are isochronous, with period s R Tiso = 2π . g 90

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Free ebooks ==> www.Ebook777.com The ”exact” period, which takes into account the amplitude θ0 of the oscillations, is given by (3.9) s Z R π/2 dψ p Tnon−iso = 4 g 0 1 − k 2 sin2 ψ with k = sin θ20 . The time difference between the two values, during one period, is s " Z # π/2 dψ R p 2 − π > 0. ∆T = Tnon−iso − Tiso = 2 g 0 1 − k 2 sin2 ψ

In other words, the ”ideal” clock (which performs isochronous oscillations) advances by ∆T seconds for each period. This clock performs 1 νnon−iso = Tnon−iso complete oscillations per unit time. During one year, the number of complete oscillations is 365 × 24 × 3600 Nexact = 365 × 24 × 3600 × νnon−iso = . Tnon−iso The time difference between the two clocks, in one year’s time, is Tnon−iso − Tiso ∆t1 year = Nexact ∆T = 365 × 24 × 3600 × Tnon−iso   π = 365 × 24 × 3600 × 1 − 2K(k)   −1  Z π/2 dψ    q = 365 × 24 × 3600 × 1 − π 2 . 2 2 θ 0 0 1 − sin 2 sin ψ Performing numerical calculations, one obtains

∆t1 year = 1350 . 897 s = 22 . 515 min. The time difference in one day is therefore ∆t1 day = 3 . 701 s. The difference is too big to consider ”good” a clock performing isochronous oscillations. In case of a second pendulum (a pendulum which beats once every second, i.e. for which T = 2s), characterized by R = πg2 , the difference Tnon−iso − Tiso is ∆T(2s) = Tnon−iso − Tiso = Tnon−iso − 2 = 2 . 000157465 − 2 = 157.465 µs. 91

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CHAPTER IV PROBLEMS SOLVED BY MEANS OF THE PRINCIPLE OF VIRTUAL WORK

Problem 1 Using the virtual work principle, determine the equilibrium position of the system shown in Fig.IV.1. We are given the following data: masses m1 and m2 of the two bodies, the angle α of the inclined plane, and the distance between the top of the inclined plane and the pulley axis. The friction phenomenon and the pulley radius are neglected, while the wire is supposed to be inextensible. It is presumed that during the motion the body of mass m1 remains on the inclined plane, and the body of mass m2 only moves vertically.

Fig.IV.1 Solution Let us first identify the constraints affecting the degrees of freedom of the system. To this end, we attach to each body a reference 92

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Free ebooks ==> www.Ebook777.com frame, as shown in Fig.IV.1. The equations of constraints then are: i) f1 (z1 ) = z1 = C1 ; ii) f2 (y1 ) = y1 = 0 (the body of mass m1 does not leave the inclined plane); iii) f3 (y2 ) = y2 = C2 ; iv) f4 (z2 ) = z2 = pC3 ; v) f5 (x1 , x2 ) = x21 + 2ax1 sin α + a2 + x2 + a − L = 0. The constants C1 , C2 , C3 , without loss of generality, can be considered equal to zero. We denoted by L the constant length of the wire connecting the two bodies, and f5 was written by means of generalized Pythagoras theorem in triangle CED. We have also supposed that the two reference frames have the same origin at C, the highest point of the inclined plane. The system has, therefore, 3 · 2 − 5 = 1 degree of freedom. According to the principle of virtual work N X i=1

F~i · δ~ri = 0,

(4.1)

where N is number of bodies of the system, and F~i (i = 1, N ) the ~ 1 and G ~ 2 , so that applied forces. In our case, the applied forces are G (4.1) writes ~ 1 · δ~r1 + G ~ 2 · δ~r2 = 0. G (4.2) The vector quantities appearing in (4.2) have the following components: ~ 1 = (m1 g sin α, −m1 g cos α, 0), G G2 = (m2 g, 0, 0), δ~r1 = (−δx1 , 0, 0), δ~r2 = (δx2 , 0, 0), where we took advantage of the choice of reference frames as regarded the motion of the bodies (the body of mass m1 climbs up the inclined plane, moving in the negative sense of Ox1 axis). Using these observations, (4.2) becomes: −m1 g sin α δx1 + m2 g δx2 = 0.

(4.3)

This is a null linear combination of the quantities δx1 and δx2 , with constant coefficients m1 g sin α and m2 g. Obviously, δx1 and δx2 cannot be linearly independent: otherwise we would have m1 g sin α = 93

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Free ebooks ==> www.Ebook777.com 0 and m2 g = 0, which would be absurd. This is also shown by the fact that system has only one degree of freedom. Therefore, since the two virtual variations δx1 and δx2 are linearly dependent, one of them must be eliminated. To this end, we observe (see Fig.IV.1) that an infinitesimal variation of the coordinate x1 of the body with mass m1 identifies with an elementary variation of s = |CE|, that is δx1 ≡ δs.

(4.4)

We also observe that a small variation of the coordinate x2 of the body with mass m2 equals a small variation of l = |DE| (the wire is inextensible), so that δx2 ≡ δl. (4.5) Consequently, a relation between δx1 and δx2 , on the one hand, and a formula connecting δs and δl, on the other, are equivalent. Using the generalized Pythagoras theorem in triangle CED, we have l2 = a2 + s2 − 2as cos(π − β) = a2 + s2 + 2as cos β = a2 + s2 + 2as cos and, by differentiation,

π 2

 − α = a2 + s2 + 2as sin α,

l dl = (s + a sin α) ds. This means that we also have l δl = (s + a sin α) δs, or, in view of (4.4) and (4.5), l δx2 = (s + a sin α) δx1 . Eliminating δx2 between this relation and (4.3), we are left with [−m1 gl sin α + m2 g(s + a sin α)] δx1 = 0.

(4.6)

Since this relation is valid for any virtual infinitesimal variation δx1 , the square bracket vanishes −m1 gl sin α + m2 g(s + a sin α) = 0, 94

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Free ebooks ==> www.Ebook777.com which is the equilibrium condition. Dividing this relation by g 6= 0, we still have p m1 a2 + s2 + 2as sin α sin α = m2 (s + a sin α).

Squaring this relation and rearranging the terms, we are left with the following quadratic equation in s:

s2 (m22 −m21 sin2 α)−2sa sin α(m21 sin2 α−m22 )+a2 sin2 α(m22 −m21 ) = 0, with the solutions s1,2 =

a sin α(m21 sin2 α − m22 ) ± m22 − m21 sin2 α

q a2 sin2 α(m21 sin2 α − m22 )2 − a2 sin2 α(m22 − m21 )(m22 − m21 sin2 α) m22 − m21 sin2 α

or

,

s1,2 = a sin α 

× −1 ±

q

(m21 sin2 α − m22 )2 − (m22 − m21 )(m22 − m21 sin2 α) m22 − m21 sin2 α

The only physically acceptable solution is



.

s ≡ s1 = a sin α

 q 2 2 2 2 2 2 2 2 2 (m1 sin α − m2 ) − (m2 − m1 )(m2 − m1 sin α) − 1 , × m22 − m21 sin2 α

which can also be written as 

s ≡ s1 = a sin α 



q m1 cos α m22 − m21 sin2 α

= a sin α  q

m22 − m21 sin2 α

m1 cos α m22



m21

2

sin α





− 1

− 1 .

95

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(4.7)

Free ebooks ==> www.Ebook777.com As observed, in order that the system has an equilibrium position, the following two conditions m22 − m21 sin2 α > 0; m1 cos α q ≥1 m22 − m21 sin2 α

have simultaneously to be fulfilled. In other words, the system ( 2 2 m2 − m21 sinq α > 0; m1 cos α ≥

m22 − m21 sin2 α

(4.8)

must be compatible. The system (4.8) can also be written as (

m2 , m1 m1 ≥ m2 . | sin α| <

 We easily realize that the system is compatible. Since α ∈ 0, π2 , it also follows a condition regarding angle α: α < arcsin

m2 , m1

with m1 > m2 .

To conclude, the system is at equilibrium for   m1 cos α − 1 . s = a sin α  q 2 2 2 m2 − m1 sin α

(4.9)

Problem 2 Solve Problem 1 by means of Torricelli’s principle. Solution Since the only force acting on the mechanical system is the force of gravity, our problem can also be solved by using Torricelli’s principle. In fact, Torricelli’s principle is a particular form of the principle of virtual work, suitable to be applied to our problem. According to this principle, we have δzG = 0, (4.10) 96

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Free ebooks ==> www.Ebook777.com where zG is the vertical height of the centre of gravity of the system. Following the already known algorithm, it is convenient to attach a unique three-orthogonal reference frame Oxyz to the system of particles (bodies, in our case), with z-axis oriented vertically (see Fig.IV.2).

Fig.IV.2 If the xOy-plane is chosen as reference level for the potential energy of the system, then the height zG of the centre of gravity is1

zG =

N P

zi mi

i=1 N P

mi

i=1

N 1 X = zi mi , M i=1

(4.11)

where N is the number of the material points of the system, and M=

N X

mi

i=1

is the mass of the system. In our case N = 2, so that zG =

m1 z1 + m2 z2 . m1 + m2

1

(4.12)

It is supposed that the force of gravity does not vary within the domain occupied by the system. 97

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Free ebooks ==> www.Ebook777.com This way, our problem turns into an application of plane geometry. To ease the reasoning we shall simplify the figure, keeping only the elements necessary to determine z1 and z2 (see Fig.IV.3).

Fig.IV.3 We then have: a = |CD|,

l = |DE|,

s = |CE|,

z1 = |EN | = |AH|,

z2 = |AF |,

also |F C| + a + l = L, |AC| − z1 = s sin α, |AC| = z2 + |F C|. Therefore, |F C| = L − a − l, z1 = |AC| − s sin α, z2 = |AC| − |F C| = |AC| − L + a + l, and formula (4.12) becomes zG =

m1 z1 + m2 z2 m1 + m2

h i 1 = m1 (|AC| − s sin α) + m2 (|AC| − L + a + l) m1 + m2 =

|AC|(m1 + m2 ) − m2 (L − a) m2 l − m1 s sin α + . m1 + m2 m1 + m2 98

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Free ebooks ==> www.Ebook777.com Since |AC| = const., we can write zG = C 0 +

m2 l − m1 s sin α , m1 + m2

(4.13)

where by C0 we denoted the constant quantity C0 =

|AC|(m1 + m2 ) − m2 (L − a) . m1 + m2

Let us now differentiate (4.13), then replace differentials by virtual infinitesimal variations. The result is δzG =

m2 δl − m1 δs sin α . m1 + m2

Using the relation between δl and δs achieved in the previous problem l δl = (s + a sin α) δs, we still have δzG =

  1 m2 (s + a sin α) − m1 l sin α δs. l(m1 + m2 )

Since δs is arbitrary, the last relation yields

m2 (s + a sin α) − m1 l sin α = 0, or m2 (s + a sin α) = m1

p

s2 + a2 + 2as sin α sin α.

Squaring this equation and rearranging the terms, we finally have s2 (m22 − m21 sin2 α) − 2sa sin α(m21 sin2 α − m22 ) +a2 sin2 α(m22 − m21 ) = 0, which has been already obtained in the preceding problem. From now on, the way to solution is known. Problem 3 Using Torricelli’s principle, determine the equilibrium position of the homogeneous, rigid rod shown in Fig.IV.4. The length of the rod is 2l, the radius of the fixed semi-cylinder is R, while friction is neglected. 99

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Free ebooks ==> www.Ebook777.com Solution As observed, the equilibrium position of the rod is univoquely determined by the angle α ∈ 0, π2 only. One can intuitively anticipate that the rod can be at equilibrium only if its centre of gravity falls inside the semi-cylinder. Otherwise, due to the force of gravity, the rod would fall down outside the semi-cylinder.

Fig.IV.4 ~ = m~g , Since the only force applied to the rod is its own gravity G where m is the mass of the homogeneous rod, the problem can be solved by means of Torricelli’s principle. Choosing a Cartesian reference frame with axes oriented as shown in Fig.IV.5, and the xOy-plane passing through DN as the reference level for the potential energy, our task is to determine the segment |M C| = zG . This way, up to a certain point, our physical problem becomes a problem of plane geometry. We have: |M C| = |N C| sin α = (|AN | − |AC|) sin α = (2|AP | − l) sin α = (2R cos α − l) sin α. The vertical height of the centre of gravity of the rod then is zG = (2R cos α − l) sin α.

(4.14)

According to Torricelli’s principle, it is required to have δzG = 0. Differentiating (4.14), then passing to virtual elementary variation, we easily obtain h i δzG = − 2R sin2 α + (2R cos α − l) cos α δα = 0. 100

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Fig.IV.5 Since δα is arbitrary, the last relation implies −2R sin2 α + (2R cos α − l) cos α = 4R cos2 α − l cos α − 2R = 0, which is an algebraic equation of the second degree in cos α, with the roots √ l ± l2 + 32R2 . (cos α)1,2 = 8R   π Due to the fact that α ∈ 0, 2 , the only acceptable solution is cos α =

l+



l2 + 32R2 . 8R

(4.15)

Therefore, the equilibrium position of the rod, inside the semicylinder, is given by the angle α = arccos

l+



l2 + 32R2 8R

!

.

Since ∀α, −1 ≤ cos α ≤ 1, we must have cos α =

l+



l2 + 32R2 ≤ 1, 8R

which is equivalent to l2 + 32R2 ≤ 64R2 + l2 − 16Rl, 101

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(4.16)

Free ebooks ==> www.Ebook777.com or l ≤ 2R,

(4.17)

in agreement with our intuitive presumption. Problem 4 Two material points (particles) P1 and P2 , of masses m1 and m2 , are connected by a massless, inextensible, perfectly malleable wire of length l. Neglecting friction and supposing that the two material points can move on the semicircle x2 + y 2 = R2 , z = 0, y > 0 (see Fig.IV.6), determine the equilibrium position of the system by means of the principle of virtual work, if 2l > πR.

Fig.IV.6 Solution The system is subject to the following constraints (see Fig.IV.7): i) z1 = 0; ii) z2 = 0; iii) x21 + y12 − R2 = 0; (4.18) iv) x22 + y22 − R2 = 0; 2 v) (R2 − x21 )(R2 − x22 ) − R2 cos Rl − x1 x2 = 0. According to analytical formalism, the system possesses 3N − s = 3 · 2 − 5 = 1 degree of freedom. An intuitive analysis shows that angle α cannot be bigger than π/2; otherwise, the system could not be in equilibrium whatever α is (both points would be situated on the same side of the semicircle, with respect to y-axis). In addition, even if 0 < α < π2 , to avoid the above mentioned situation, we must have π 2 < α + β < π. In other words, to be in equilibrium, the two points 102

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Free ebooks ==> www.Ebook777.com must be on both sides with respect to y-axis. Let us prove that our intuitive observations are consistent with the results of calculation. The equilibrium position of the system can be indicated by any one of the following parameters: angle α, x1 , x2 , y1 , y2 , s, etc., where s is the arc BC of the circle (see Fig.IV.7).

Fig.IV.7 The applied forces are: ~ 1 = (0, −m1 g, 0), G

~ 2 = (0, −m2 g, 0), G

(4.19)

while the virtual displacements of the two points are δ~r1 = (δx1 , δy1 , 0),

δ~r2 = (δx2 , δy2 , 0).

(4.20)

The mathematical expression of the principle of virtual work N X i=1

F~i · δ~ri = 0

(4.21)

then yields −m1 gδy1 − m2 gδy2 = 0.

(4.22)

The virtual displacements δy1 and δy2 are not linearly independent (there is one degree of freedom). To determine the relationship between δy1 and δy2 , we observe that 

y1 = R sin[π − (α + β)] = R sin(α + β), y2 = R sin α, 103

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(4.23)

Free ebooks ==> www.Ebook777.com where β is given (fixed) due to the length l of the wire. Differentiating (4.23) and replacing dα by δα, we have 

δy1 = R cos(α + β) δα, δy2 = R cos αδα.

(4.24)

Replacing now δy1 , δy2 in (4.22) with their values given by (4.24), we have [m1 cos(α + β) + m2 cos α]δα = 0. (4.25) Since δα is arbitrary, we have m1 cos(α + β) + m2 cos α = 0, or, if cos(α + β) is developed, tan α =

m1 cos β + m2 . m1 sin β

(4.26)

Therefore, the angle α corresponding to the equilibrium of the system is   m1 cos β + m2 α = arctan , (4.27) m1 sin β or, in terms of R and l,  m1 cos(l/R) + m2 . α = arctan m1 sin(l/R) 

(4.28)

Let us now go back to equation (4.25). It can be written as 1 m1 = . m2 tan α sin β − cos β Since the ratio have

m1 m2

must be positive and different from zero, we must tan α sin β − cos β > 0,

or, equivalently, − cos(α + β) > cos α.

(4.29)

Taking into account the initial ”constraint” 0 < α < π2 , the inequality (4.29) can be satisfied only if π2 < α+β < π, which is in full agreement with our initial intuitive consideration. To conclude, the system can be at equilibrium only if the two bodies (conceived as material points) 104

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Free ebooks ==> www.Ebook777.com are on the one side and the other with respect to vertical Oy, that is, if α + β > π2 . Problem 5 Solve Problem 4 by means of Torricelli’s principle. Solution Since the only applied forces to the bodies are the forces of their own gravity, the problem can also be solved by means of Torricelli’s principle. As we know, this principle requires that the virtual variation of the height of the centre of gravity of the system is zero: δzG = 0.

(4.30)

To this end, let us first rename the coordinate axes, as shown in Fig.IV.8. With these new notations, the centre of mass of the system is given by m1 z1 + m2 z2 . zG = m1 + m2

Fig.IV.8 As easily seen,  z1 = R sin[π − (α + β)] = R sin(α + β), z2 = R sin α, so that zG =

m1 R sin(α + β) + m2 R sin α . m1 + m2 105

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Free ebooks ==> www.Ebook777.com Differentiating this relation, and then replacing the differential symbol with the symbol of first variation, we have δzG =

R [m1 cos(α + β) + m2 cos α]δα. m1 + m2

The Torricelli’s principle requires R [m1 cos(α + β) + m2 cos α]δα = 0, m1 + m2

(4.31)

which is equivalent to (4.25). From now on, the calculation follows the same way as in problem 4. Problem 6 Two material points (particles) of masses m1 and m2 , connected by a spring of length l0 (at rest) and elastic constant k, can move without friction on the walls of a gutter with the opening angle α (see Fig.IV.9). Supposing that the motion of the particles can be performed only in xy-plane, being permanently in contact with the walls of the gutter, determine the equilibrium position of the system. It is assumed that the spring is oriented along a straight line. Solution The analytical expressions of the constraints are (see Fig.IV.10): i) y1 = −x1 cot α2 ; ii) y2 = x2 cot α2 ; iii) z1 = 0; iv) z2 = 0. The system possesses 3N − s = 3 · 2 − 4 = 2 degrees of freedom. The principle of virtual work writes 2 X i=1

where

F~i · δ~ri = 0,

 ~ 1 + F~e1 , F~1 = G    ~ ~  F2 = G2 + F~e2 ,    ~ 1 = (0, −m1 g, 0),  G   ~ G2 = (0, −m2 g, 0),  F~e1 = (−Fe cos β, Fe sin β, 0),     F~e2 = −F~e1 = (Fe cos β, −Fe sin β, 0),       δ~r1 = (δx1 , δy1 , 0), δ~r2 = (δx2 , δy2 , 0), 106

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(4.32)

(4.33)

Free ebooks ==> www.Ebook777.com with |F~e | = |F~e1 | = |F~e2 | = k(lo − l).

(4.34)

Fig.IV.9

Fig.IV.10 Since x1 < 0 and x2 > 0, the equilibrium length of the deformed spring is x2 − x1 l= , (4.35) cos β so that the modulus of the elastic force acting on the two particles is   x − x 2 1 Fe = |F~e | = k l0 − , cos β

(4.36)

and the principle of virtual work (4.32) writes −m1 gδy1 − Fe cos βδx1 + Fe sin βδy1 −m2 gδy2 + Fe cos βδx2 − Fe sin βδy2 = 0. 107

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(4.37)

Free ebooks ==> www.Ebook777.com Since the system possesses two degrees of freedom, only two variables out of five x1 , x2 , y1 , y2 , and β which interfere in (4.37) are linearly independent. It is convenient to choose as linearly independent parameters the Cartesian coordinates x1 and x2 . Observing that  α α  y1 = −x1 cot ⇒ δy1 = −δx1 cot ,   2 2   α α   ⇒ δy2 = δx2 cot , y2 = x2 cot   2 2    (x1 + x2 ) cot α2    sin β = − q , 2 α 2 2 (x2 − x1 ) + (x1 + x2 ) cot 2 (4.38)    x2 − x1   cos β = q ,    2 + (x + x )2 cot2 α  (x − x ) 2 1 1 2  2    α x + x 1 2   tan β = − cot , x2 − x1 2 and using (4.36), equation (4.37) writes: " α kl0 (x2 − x1 ) δx1 m1 g cot − q 2 (x2 − x1 )2 + (x1 + x2 )2 cot2 kl0 (x1 + x2 ) cot2

α 2

α 2

#

α +k(x2 − x1 ) + q − k(x1 + x2 ) cot2 2 (x2 − x1 )2 + (x1 + x2 )2 cot2 α2 " α kl0 (x2 − x1 ) − k(x2 − x1 ) +δx2 −m2 g cot + q 2 (x2 − x1 )2 + (x1 + x2 )2 cot2 α2 # kl0 (x1 + x2 ) cot2 α2 α +q − k(x1 + x2 ) cot2 = 0. (4.39) 2 α 2 2 2 (x − x ) + (x + x ) cot 2

1

1

2

2

Since the virtual displacements δx1 and δx2 are linearly independent, relation (4.39) lead to the following system of two algebraic equations for the unknowns x1 and x2 :    m1 g cot α − kl0 (x2 − x1 ) + k(x2 − x1 )    2 R  2 α   kl (x + x ) cot α 0 1 2  2 + − k(x1 + x2 ) cot2 = 0; R 2 (4.40) α kl (x − x )  0 2 1  −m2 g cot + − k(x2 − x1 )    2 R  α  kl (x + x ) cot2  α 2  2 + 0 1 − k(x1 + x2 ) cot2 = 0, R 2 108

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Free ebooks ==> www.Ebook777.com where we denoted r α R = (x2 − x1 )2 + (x1 + x2 )2 cot2 . 2 Adding and subtracting the two equations (4.40), they can be written in a simpler form, as  α 2kl0   (m1 + m2 )g cot + 2k(x2 − x1 ) = (x2 − x1 ); 2 R   (m1 − m2 )g cot α + 2kl0 (x2 + x1 ) cot2 α = 2k(x1 + x2 ) cot2 α . 2 R 2 2 (4.41) As a result of some not difficult but long calculations, the solution which is acceptable from the physical point of view (out of two possible solutions) is found to be   m1 + m2 cos α     x1 = 2k sin α  m2 + m1 cos α     x1 = 2k sin α

g− p

2klo sin α2 m21 + m22 + 2m1 m2 cos α 2klo sin α2

!

−g + p 2 m1 + m22 + 2m1 m2 cos α

, !

(4.42) .

These relations give the equilibrium position of the system. We leave up to the reader to analyze relations between the characteristic quantities m1 , m2 , α, k, and l0 , so that x2 > 0 and x1 < 0. If, in particular, m1 = m2 = m and α = π2 , the solution given by (4.42) becomes  mg l0   x1 = − , 2k 2 mg l 0   x2 = − = −x1 , 2 2k

(4.43)

which is obvious. In this particular case the equilibrium conditions x1 < 0 and x2 > 0 coincide and write kl0 > mg.

(4.44)

The last relation shows that the equilibrium position of the system (different from the trivial one, when the bodies ”fall” in the edge of the gutter) is attained only if the elastic constant k of the spring is big enough (k > lG0 ) to exceed the weight of the two bodies.

109

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CHAPTER V PROBLEMS OF VARIATIONAL CALCULUS

V.1. Elements of variational calculus V.1.1. Functionals. Functional derivative Definition. A functional is an application defined on a linear space X, or on a part of X, with values in the field of scalars on which the linear space X is defined. (N.B. The French term for a field is corps; in English, a field is a corps commutatif). In other words, a functional can be considered as a ”function” whose domain consists of a set of functions, and whose codomain is a set of scalars (numbers, figures, etc.). This definition shows that, in fact, a functional is an operator. Let J(y) be a functional defined on the domain D of a Banach space B, with values on the real straight line R . Definition. A Banach space is a normed linear space that is a complete metric space with respect to the metric derived from its norm. Definition. A complete space (or Cauchy space) is a linear space in which any Cauchy sequence of points converges to an element of the space. Definition.  A Cauchy sequence in the metric space (X, d) is a sequence Xn n∈N of elements of X that satisfy the property: ∀ ε > 0, ∃ nε ∈ N , so that d(xn , xm ) < ε, ∀ n, m ≥ nε . Definition. A distance on a set X is any application d defined on X × X with positive real values, d : X × X → R + , that satisfies the following properties: D1 : d(x, y) = 0 if and only if x = y, (x, y ∈ X); D2 : d(x, y) = d(y, x), ∀x, y ∈ X; D3 : d(x, y) ≤ d(x, z) + d(z, y), ∀x, y, z ∈ X. Definition. A metric space is any pair (X, d), where X is a set, and d a distance defined on X. Definition. A normed space is a linear space possessing a norm, p. 110

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Free ebooks ==> www.Ebook777.com Definition. A norm is any seminorm that satisfies the property p(x) = 0 ⇒ x = 0. Definition. A seminorm is a function with real values, p : X → R , that satisfies the following properties: SN1 : p(x + y) ≤ p(x) + p(y), ∀x, y ∈ X; SN2 : p(λx) = |λ| p(x), ∀x, y ∈ X, and λ a non-zero scalar. A norm is usually denoted by: p(x) ≡ kxk. Let J(y) be a functional defined as previously shown. Let also y ∈ D, h ∈ B, so that y + h ∈ D Definition (differentiability 1): The functional J(y) is differentiable at the ”point” y if the difference J(y + h) − J(y) can be written as J(y + h) − J(y) = δ(y, h, J) + r(y, h, J), where δ(y, h, J) is a functional linear in h δ(y, αh1 + βh2 , J) = α δ(y, h1 , J) + β δ(y, h2 , J), where α and β are two scalars, while r(y, h, J) is a functional that satisfies the condition |r(y, h, J)| = 0. khk khk→0 lim

(The ”point” y is, in general, a function). Definition (differentiability 2): The functional J(y) is differentiable at the ”point” y if there exists a functional linear in h, δ(y, h, J), so that lim [J(y + h) − J(y) − δ(y, h, J)] = 0. khk→0

Definition. If the conditions of the differentiability of a functional at a ”point” are fulfilled, then the functional δ(y, h, J) is called differential of the functional J(y). Theorem: If δ(y, h, J) exists, then it is unique. (The demonstration is based on reductio ad absurdum. We skip this proof). Definition: If the functional r(y, h, J) can be written as r(y, h, J) =

1 δ2 (y, h, J) + r2 (y, h, J), 2

where δ2 (y, th, J) = t2 δ2 (y, h, J) 111

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Free ebooks ==> www.Ebook777.com and

|r2 (y, h, J)| = 0, khk2 khk→0 lim

then the functional δ2 (y, h, J) is called the second-order differential of functional J(y). For a better understanding of these definitions, next we shall present the analogy between functionals and the usual functions. Let f : I ⊂ R → R be a function defined on interval I of the real axis, with values in the set of real numbers, differentiable at the point x0 ∈ I. Then, for any x 6= x0 , we can write f (x) − f (x0 ) = f ′ (x0 )(x − x0 ) + α(x)(x − x0 ).

(∗)

Since f (x) is differentiable at the point x0 ∈ I, we have lim

x→x0

f (x) − f (x0 ) = f ′ (x0 ) + lim α(x) = f ′ (x0 ), x→x0 x − x0

which necessarily yields lim α(x) = 0.

x→x0

Denoting x − x0 = h, then x = x0 + h and we have lim α(x0 +h)=0



f (x0 + h) − f (x0 ) = hf (x0 ) + hα(x0 + h)

h→0

=

hf ′ (x0 ).

The linear function g = hf ′ (x0 ) is called the differential of f (x) at the point x0 , and it is usually denoted by g = df (x0 ). The linearity in h then writes: (αh1 + βh2 )f ′ (x0 ) ≡ df (x0 , αh1 + βh2 ) = αh1 f ′ (x0 ) + βh2 f ′ (x0 ) = αdf (x0 , h1 ) + βdf (x0 , h2 ) ≡ αh1 f ′ (x0 ) + βh2 f ′ (x0 ). We then have the following obvious analogy: J y h



f



x0



h

112

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Free ebooks ==> www.Ebook777.com δ(y, h, J)



d(x0 , h, f )

r(y, h, J)



r(x0 , h, f )

where the following notations have been used: f ′ (x0 )(x − x0 ) = hf ′ (x0 ) = df (x0 , h) ≡ d(x0 , h, f ); α(x)(x − x0 ) = hα(x0 + h) ≡ r(x0 , h, f ). To the limit condition obeyed by the functional r(y, h, J), |r(y, h, J)| = 0, khk khk→0 lim

shall therefore correspond the following condition on α(x) h α(x0 + h) = lim α(x) ≡ lim α(x) = 0. x→x0 h→0 h→0 h lim

One usually prefers the notation h = dx, so that the differential of f at the point x0 writes df (x0 ) = f ′ (x0 ) dx. To realize the analogy regarding the second differential of a functional, let us expand function f (x) in Taylor series about the point x0 : 1 f (x) = f (x0 ) + f ′ (x0 )(x − x0 ) + f ′′ (x0 )(x − x0 )2 + O(x − x0 )3 2 1 = f (x0 ) + f ′ (x0 )(x − x0 ) + f ′′ (x0 )(x − x0 )2 + β(x)(x − x0 )2 , (∗∗ ) 2 where we have used the notation β(x) (x − x0 )2 =

1 1 ′′′ f (x0 )(x − x0 )3 + f (iv) (x0 )(x − x0 )4 + ... 3! 4!

with obvious condition lim β(x) = 0.

x→x0

In view of (∗ ) and (∗∗ ), we still have α(x)(x − x0 ) =

1 ′′ f (x0 )(x − x0 )2 + β(x)(x − x0 )2 . 2 113

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Free ebooks ==> www.Ebook777.com Using notation h = x − x0 , we also have α(x)(x − x0 ) = hα(x0 + h) ≡ r(x0 , h, f ) =

1 2 ′′ h f (x0 ) + h2 β(x0 + h), 2

or, if we define the second order variation (also called ”the second order differential”) of f (x) at the point x0 as δ2 f = f ′′ (x0 )(x − x0 )2 = h2 f ′′ (x0 ) ≡ δ2 (x0 , h, f ), we can write r(x0 , h, f ) =

1 δ2 (x0 , h, f ) + r2 (x0 , h, f ), 2

where we denoted r2 (x0 , h, f ) = h2 β(x0 + h). Obviously, δ2 (x0 , th, f ) = (th)2 f ′′ (x0 ) = t2 h2 f ′′ (x0 ) = t2 δ2 (x0 , h, f ), as well as

h2 β(x0 + h) r2 (x0 , h, f ) = lim h→0 h→0 h2 h2 lim

= lim [β(x0 + h)] = lim β(x) = 0. x→x0

h→0

As one can see, the analogy completes if on the table of correspondences we add the following two relations δ2 (y, h, J)



δ2 (x0 , h, f ),

r2 (y, h, J)



r2 (x0 , h, f ).

A functional commonly used in Analytical Mechanics is J(y) = f (x, y, y ′ ) dx, where function f (x) has to satisfy certain conditions a of continuity. Let then f (x, y, z) be a function of class C 2 on the domain D : a ≤ x ≤ b, −∞ < y, z < +∞, with uniformly continuous partial derivatives with respect to y, z.  Definition (continuity at a point in the language of ε, η(ε) ): let f : I ⊂ R → R be a function, and x0 ∈ I a point within its domain of definition. We then say that f (x) is continuous at the point x0 if ∀ ε > 0, ∃ η(ε) > 0, so that |f (x) − f (x0 )| < ε, ∀x ∈ I, as soon as |x − x0 | < η(ε). Rb

114

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Free ebooks ==> www.Ebook777.com Observation. η depends only on ε if we consider continuity at a single point. But, if we refer to the continuity on the whole interval I ⊂ R, η depends on x0 as well. Definition (continuity at a point in the language of vicinities). Consider f : I ⊂ R → R , and x0 ∈ I. Function f (x) is said to be continuous at the pointx0 if ∀ U f (x0 ) , ∃V (x0 ), so that ∀ x ∈ I ∩ V (x0 ), f (x) ∈ U f (x0 ) . Definition (continuity at a point in the language of sequences): Consider f : I ⊂ R → R , and x0 ∈ I. Function f (x) is said to be continuous at the point x0 if ∀{xn }n∈N a sequence convergent to x0 , the sequence f (xn )n∈N is convergent to f (x0 ). Definition. Consider f : I ⊂ R → R . Function f (x) is said to be uniformly continuous on I ⊂ R if ∀ε > 0, ∃ η(ε) > 0, so that |f (x′′ ) − f (x′ )| < ε, ∀ x, x′ ∈ I as soon as |x − x′ | < η(ε). Observation: η depends only on ε, and is independent of x′ and x′′ which satisfy relation |x′ − x′′ | < η(ε). Let f (x) be only continuous on I ⊂ R and x′ ∈ I. If x′ is maintained fix, to a given ε corresponds an η that changes together with ε, but depends on x′ as well, η = η(ε, x′ ). If x′ covers the multitude I, ε being fixed, then the multitude of values of η(ε, x′ ) has an inferior margin, η0 (ε) =

inf ′

∀x ∈I ε f ixed



η(ε, x′ ) .

If η0 (ε) > 0, then function f (x) is said to be uniformly continuous on I. Theorem. A continuous function on a compact (i.e. closed and bounded) interval is uniformly continuous on that interval. Example: f (x) = x3 , x ∈ [1, 3]. Theorem of finite increments (Lagrange): Consider the function f : I ⊂ R → R , and a, b ∈ I any two points in I. If L1 : function f is continuous on [a, b], and L2 : function f is differentiable on (a, b), then1 ∃∗ c ∈ (a, b), so that f (b) − f (a) = (b − a)f ′ (c). Observation. There are some equivalent formulations for the relation f (b) − f (a) = (b − a)f ′ (c): 1. f (x) − f (a) = (x − a)f ′ (ξ), a < ξ < x; 1

The symbol ∃∗ stands for ”exists at least” 115

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Free ebooks ==> www.Ebook777.com 2. f (a + h) − f (a) = hf ′ (a + θh), 0 < θ < 1. Definition. It is said that the function f : I ⊂ R → R is derivable at (x0 ) the point x0 ∈ I, if the ratio f (x)−f has a finite limit at the point x−x0 x0 . This limit is called the derivative of f (x) at x0 being denoted as f ′ (x0 ): f (x) − f (x0 ) lim = f ′ (x0 ). x→x0 x − x0 Consider now the functional Z b J(y) = f (x, y, y ′ ) dx a

defined on the set of functions y = y(x) ∈ C 2 [a, b], which satisfies the conditions y(a) = 0, y(b) = 0. If we define the quantity ( ) kyk = max

sup |y(x)|, sup |y ′ (x)| ,

x∈[a,b]

x∈[a,b]

as norm on this space, then the set of these functions form a Banach space1 B. Let h(x) be a function of the Banach space B. Obviously, y(x) + h(x) ∈ B and the functional J(y + h) =

Z

b

f (x, y + h, y ′ + h′ ) dx a

is well-defined, and we can calculate the difference J(y + h) − J(y) = =

Z

b a



Z

b ′

a



f (x, y + h, y + h ) dx −

Z

b

f (x, y, y ′ ) dx a

 f (x, y + h, y ′ + h′ ) − f (x, y, y ′ ) dx.

Applying the theorem of finite increments, we still have J(y + h) − J(y) =

Z

b a

h ∂f h (x, y + θh, y ′ + θh′ ) ∂y

1

We have considered the definition of a functional on a Banach space, because on this mathematical structure all the quantities useful in analytical mechanics (convergence, continuity, differentiability and integrability) are defined. 116

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Free ebooks ==> www.Ebook777.com i ∂f ′ ′ +h (x, y + θh, y + θh ) dx, ∂y ′ ′

with 0 < θ < 1. Denoting  ∗ ∂f ∂f = (x, y + θh, y ′ + θh′ ), ∂y ∂y and



we still have

∂f ∂y ′

∗

=

J(y + h) − J(y) =

∂f (x, y + θh, y ′ + θh′ ), ∂y ′

Z

b a

(5.1)

0 0, so that  ∗ ∂f ∂f = |f0 | < ε − ∂y ∂y and

∂f ∂y

and

∂f ∂y ′ ,

it then

  ∂f ∗ ∂f − ′ = |f1 | < ε, ∂y ′ ∂y

as soon as

|(y + θh) − y| = |θh| < |h| < η(ε) and, respectively |(y ′ + θh′ ) − y ′ | = |θh′ | < |h′ | < η(ε) or - if one takes into account the definition of norm - as soon as khk < η(ε). By means of (5.2) and (5.3), we still have  Z b Z b ∂f ′ ∂f J(y + h) − J(y) = h +h dx + (hf0 + h′ f1 ) dx. (5.4) ′ ∂y ∂y a a 117

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Free ebooks ==> www.Ebook777.com Using the properties of integrals, one can easily verify that the integral  Z b ∂f ′ ∂f L(h) = h dx +h ∂y ∂y ′ a is linear in h. Indeed, L(αh1 + βh2 ) =



Z

b a

Z

b a



∂f ∂f (αh1 + βh2 ) + (αh′1 + βh′2 ) ′ ∂y ∂y



dx

   Z b ∂f ∂f ′ ∂f ′ ∂f h1 h2 + h1 ′ dx + β + h2 ′ dx ∂y ∂y ∂y ∂y a = αL(h1 ) + βL(h2 ).

We also have Z Z b  b  ′ ′ |f0 | + |f1 | dx, (hf0 + h f1 ) dx ≤ max |h|, |h | a a

that is

Z Z b  1 b ′ |f0 | + |f1 | dx ≤ 2ε(b − a). (hf0 + h f1 ) dx ≤ khk a a

(5.5)

Therefore, (5.4) gives a decomposition of J(y + h) − J(y) in a functional linear in terms of h(x), and another one that satisfies the condition (5.5). This means that the differentiability conditions are satisfied, and the differential of J(y) is δ(y, h, J) =

Z

b a



∂f ∂f h + h′ ′ ∂y ∂y



dx.

(5.6)

Since the function f (x, y, y ′ ) is of class C 2 on its domain of definition, f ∈ C 2 (D), where D : a ≤ x ≤ b, ∞ < y, y ′ < +∞, we can integrate by parts the second term in (5.6) and obtain Z

b

∂f h dx = ∂y ′ ′

a



∂f h ′ ∂y

b

a



Z

b a

d h dx



∂f ∂y ′



dx.

But 

∂f h ′ ∂y

b

a

= h(b)

  ∂f ∂f ′ ′ b, y(b), y (b) − h(a) a, y(a), y (a) , ∂y ′ ∂y ′ 118

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Free ebooks ==> www.Ebook777.com and, since h(x) ∈ B, we have h(a) = 0 and h(b) = 0, which yields δ(y, h, J) =

Z

b

a



∂f d h(x) − ∂y dx



∂f ∂y ′



dx.

(5.7)

This result can be generalized to functionals of the type Z

b a

 f x, y, y ′ , y ′′ , ..., y (n) dx,

 where f x, y, y ′ , y ′′ , ..., y (n) is a function of (n+2) variables of class C n on its domain of definition, D : a ≤ x ≤ b, −∞ < y, y ′ , ..., y (n) < +∞. A similar reasoning leads to the differential of this functional, which is

δ(y, h, J) =

Z

b a

"

∂f d h(x) − ∂y dx

dn +... + (−1)n n dx





∂f ∂y ′

∂f ∂y (n)



#

d2 + 2 dx



∂f ∂y ′′

dx.

 (5.8)

Here are two other examples of most frequently met functionals, together with their R t differentials: i) J(x) = t12 f t, x(t), x(t) ˙ dt, where x = x(t) is a vector function  with n components, x(t) = x1 (t), x2 (t), ..., xn (t) , and x(t) ˙ = dx dt . Following a similar reasoning, the differential of the functional J(y) is found as δ(x, h, J) =

Z

t2 t1

n X



∂f d hi (t) − ∂xi dt i=1



∂f ∂ x˙ i



dt.

(5.9)

  ∂z ∂z ii) J(z) = Ω f x, y, z(x, y), ∂x (x, y), ∂y (x, y) dx dy. In this case one obtains      Z ∂f ∂ ∂f ∂ ∂f − − dx dy, (5.10) δ(z, h, J) = h ∂z ∂x ∂p ∂y ∂q Ω R

where p =

∂z ∂x ,

q=

∂z ∂y .

119

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Free ebooks ==> www.Ebook777.com V.1.2. Extrema of functionals A) Necessary conditions for extremum Let J(y) be a differentiable functional, defined on a Banach space B. Definition. A value y0 is called ”point” of relative maximum of J(y) if ∃ V (y0 ) a vicinity of y0 , so that ∀ y ∈ (y0 ), J(y) ≤ J(y0 ). Definition. A value y0 is called ”point” of relative minimum of J(y) if ∃ V (y0 ) a vicinity of y0 , so that ∀ y ∈ (y0 ), J(y) ≥ J(y0 ). Theorem: If J(y) is a differentiable functional, and y0 a ”point” of extremum for J(y), then, ∀h ∈ B, δ(yo , h, J) = 0.

(5.11)

Therefore, a necessary condition for y0 to be a point of extremum for J(y) is that δ(y, h, J) equals zero at that point. Definition. The ”points” y0 where condition (5.11) is satisfied are called extremals. In other words. the extremals of a functional are those functions y(x0 ) which cancel the first differential of the functional. B) Sufficient conditions for extremum Theorem: If the functional J(y) admits the first differential δ(y, h, J) and the second differential δ2 (y, h, J), then the necessary and sufficient condition for J(y) to achieve its maximum at the ”point” y0 , ∀h ∈ B, is to have δ(y0 , h, J) = 0, r(y0 , h, J) ≤ 0, while the condition for J(y) to have a minimum at the ”point” y0 writes δ(y0 , h, J) = 0, r(y0 , h, J) ≥ 0. Indeed, since J(y) admits the first differential, we have J(y0 + h) − J(y0 ) = δ(y0 , h, J) + r(y0 , h, J). But δ(y0 , h, J) = 0 is the necessary condition for y0 to be a point of extremum for J(y), so that we are left with J(y0 + h) − J(y0 ) = r(y0 , h, J). 120

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Free ebooks ==> www.Ebook777.com Consequently, in order for y0 to be - a point of relative maximum, i.e. J(y0 + h) − J(y0 ) ≤ 0, it is necessary and sufficient that r(y0 , h, J) ≤ 0; - a point of relative minimum, i.e. J(y0 + h) − J(y0 ) ≥ 0, it is necessary and sufficient that r(y0 , h, J) ≥ 0. It can be shown that these conditions are fulfilled as follows: 1) For r(y0 , h, J) ≤ 0,  − the necessary condition is : δ2 (y0 , h, J) ≤ 0, − the sufficient condition is : δ2 (y0 , h, J) ≤ −Ckhk2 ; 2) For r(y0 , h, J) ≥ 0,  − the necessary condition is : δ2 (y0 , h, J) ≥ 0, − the sufficient condition is : δ2 (y0 , h, J) ≥ Ckhk2 , where C > 0 is a constant. Theorem. Let Ω ⊂ R n be a domain of the abstract n-dimensional space R n , x = (x1 , x2 , ..., xn ) ∈ Ω a point in this domain, and M the set of functions h(x) = h(x1 , x2 , ..., xn ), which are continuous on Ω and cancel on the frontier ∂Ω of Ω. Under these conditions, if A(x) is a continuous function on Ω and Z h(x) A(x) dx = 0, ∀ h(x) ∈ M, Ω

then A(x) = 0, ∀ x ∈ Ω. Rb Consider the functional J(y) = a f (x, y, y ′ ) dx. The necessary condition for an extremum of this functional is that its differential, given by (5.7), must be zero δ(y, h, J) =

Z

b a



∂f d h(x) − ∂y dx



∂f ∂y ′



dx ≡ G(y) = 0,

Since • h ∈ C 2 [a, b]; • h(a) = 0, h(b) = 0;

∀h(x).

  ∂f d − is • f ∈ C 2 (a ≤ x ≤ b, −∞ < y, y ′ < ∞), that is ∂f ∂y dx ∂y ′ continuous, the conditions imposed by the last theorem are fulfilled. Consequently, in order that Z

b a



∂f d h(x) − ∂y dx



∂f ∂y ′



dx = 0, ∀ h(x),

121

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Free ebooks ==> www.Ebook777.com it is necessary and sufficient to have   ∂f ∂f d = 0. − ∂y dx ∂y ′

(5.12)

This is the Euler (sometimes called Euler-Lagrange) equation of extremals. This equation can also be written as ∂ 2 f ′′ ∂2f ′ ∂2f ∂f y + y + − = 0. ′2 ′ ′ ∂y ∂y∂y ∂x∂y ∂y

(5.13)

Equations of extremals for the functional J(x) =

Z

t2 t1

 f t, x(t), x(t) ˙ dt,

 where x(t) ≡ x1 (t), x2 (t), ..., xn (t) , are written as ∂f d − ∂xi dt



∂f ∂ x˙ i



= 0,

(i = 1, n).

These equations were first obtained by Euler in 1744, and then generalized by Lagrange in Mechanics. They are known as the EulerLagrange equations. The necessary condition for an extremum of the functional Z

b a

 f x, y, y ′ , y ′′ , ..., y (n) dx

is given by Euler’s equation     d ∂f ∂f ∂f d2 − + 2 + ... ∂y dx ∂y ′ dx ∂y ′′   n ∂f n d +(−1) = 0. dxn ∂y (n) Let

 F x, y, y ′ , ..., y (n) = 0

be the implicit form of an n-th order differential equation. Definition. A differential equation of order n − k that contains k ≥ 1 arbitrary constants  ψ x, y, y ′ , ..., y (n−k) , C1 , C2 , ..., Ck = 0, 122

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Free ebooks ==> www.Ebook777.com being verified by the general integral   ϕ x, y, C1 , ..., Cn = 0

of the given differential equation, is called intermediate integral of order k or, simply, integral of order k of this equation. For k = 1, the differential equation of order n − 1  χ x, y, y ′ , ..., y (n−1) , C = 0,

that depends on a single constant of integration C, is called a first integral of the differential equation F x, y, y ′ , ..., y (n) = 0. First integrals of the Euler-Lagrange equations ∂f d − ∂yi dx



∂f ∂ y˙ i



= 0,

(i = 1, n).

1. Suppose that the function f does not explicitly depend on yi . ∂f Then ∂y = 0, and the Euler-Lagrange equations yield i ∂f = const. ∂ y˙ i

(5.14)

2. Suppose that the function f does not explicitly depend on the independent variable x. In this case, y˙ i

∂f − f = const. ∂ y˙ i

(5.15)

is a first integral of the Euler-Lagrange equations. Indeed, taking the first derivative of (5.15) with respect to x, we have d dx or

  ∂f d y˙ i − f = 0, ∂ y˙ i dx

∂f d y¨i + y˙ i ∂ y˙ i dx

that is −y˙ i





∂f ∂ y˙ i





∂f d − ∂yi dx

∂f ∂f y˙ i − y¨i = 0, ∂yi ∂ y˙ i



∂f ∂ y˙ i



= 0,

which completes the proof. 123

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Free ebooks ==> www.Ebook777.com Observation. To determine the differential of the functional Z b J(y) = f (x, y, y ′ ) dx, a

defined on the set of functions y = y(x) ∈ C 2 [a, b], we have considered y(a) = 0, y(b) = 0. Our theory remains also valid for less restrictive conditions, such as y(a) = a1 6= 0, and y(b) = b1 6= 0. It is essential, nevertheless, to have h(a) = 0, h(b) = 0 or, in other words, if the function y0 = ϕ(x) passes through the points P (a, a1 ) and Q(b, b1 ), then the curve y(x) = ϕ(x) + h(x) = y0 (x) + h(x) must also pass through the same points (see Fig.V.1).

Fig.V.1 Conditional extrema of functionals Rb Consider the functional J(y) = a f (x, y, y ′ ) dx, where the function y = y(x) is subject to a restrictive condition (constraint) defined as Z b

G(y) =

a

g(x, y, y ′ ) dx = const. ≡ C.

As can be seen, G(y) is a functional of the same type. To determine the conditional extrema of J(y) we shall appeal to the following 124

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Free ebooks ==> www.Ebook777.com Theorem. In order that y0 is a ”point” of extremum of the functional J(y), where y(x) satisfies the supplementary condition G(y) = C, it is necessary that the differential of the functional W (y) = J(y) − λG(y) cancels for y = y0 . Here λ is some non-zero scalar. Therefore, to determine the conditional extrema of the functional J(y), one search for the ordinary extrema of a new functional W (y) defined as W (y) = J(y) − λG(y) = =

Z

Z

b a



 f (x, y, y ′ ) − λg(x, y, y ′ ) dx

b

w(x, y, y ′ ) dx, a

where w = f − λg. Solution to such a problem implies determination of both the conditional extrema of J(x) and the scalar λ 6= 0. To this end, one uses the above theorem, which furnishes the following system of ”equations”  δ(y, h, W ) = 0; G(y) = C, with two unknowns: y0 and λ. V.2. Problems whose solutions demand elements of variational calculus 1. Brachistochrone problem Chronologically, this is the first problem of variational calculus. The word brachistochrone comes from Greek: brachistos (the shortest) and chronos (time). It was formulated by Jean Bernoulli in 1696 as follows: among all curves situated in a vertical plane and passing through two fixed points, determine the curve on which a heavy particle moves without friction in shortest time. Solution Choosing a reference frame as shown in Fig.V.2, where P0 (0, 0) and P1 (x1 , y1 ) are any two arbitrary points, we can write ds = v dt,

(5.16)

where v is the magnitude of the velocity of particle along the curve, and dt the infinitesimal time interval for the particle to cover the infinitesimal distance ds. 125

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Fig.V.2 According to our choice, P0 is higher than P1 . The time interval required for the particle to move from P0 to P1 then is Z t Z P1 ds t= dt = . (5.17) v 0 P0 Velocity v can be obtained by means of either the energy conservation law, or the kinetic energy theorem. Since the particle starts from rest, both ways yield 1 mv 2 = mgx, 2 that is v= On the other hand,

(5.18)

p 2gx.

(5.19)

p

(5.20)

2 (ds)2 = (dx)2 + (dy)2 = (dx)2 1 + y ′ ,

where y ′ = dy/dx. Therefore,

ds = dx

1 + y′ 2 .

Using (5.17), (5.19), and (5.20), we have Z x1 r Z x1 1 1 + y′ 2 t= √ dx = f (x, y ′ ) dx. x 2g 0 0

(5.21)

As one can see, the time t has been obtained as a functional whose ”characteristic” function f (x, y ′ ) does not depend on y = y(x). This means that the Euler equation   d ∂f ∂f − =0 ′ dx ∂y ∂y 126

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Free ebooks ==> www.Ebook777.com admits the first integral

∂f = const, ∂y ′

with ′

f (x, y ) =

r

(5.22)

1 + y′ 2 . x

(5.23)

Take now the value √12a for the constant appearing in (5.22), where a is a new constant. In view of (5.23), we then have y′ 1 q =√ ,  2a x 1 + y′ 2

and, by separation of variables, Z xr y= 0

x dx. 2a − x

(5.24)

To solve this integral, we shall make the following change of variables r x = u. (5.25) 2a − x We then have 1 √ 2 x



2a − x +



1 x √2a−x

2a − x

which yields dx =

dx = du,

√ 1 (2a − x)3/2 x du. a

(5.26)

2au2 , 1 + u2

(5.27)

4au du. (1 + u2 )2

(5.28)

Using (5.25), we have x= so that (5.26) leads to dx =

Introducing (5.25) and (5.28) into (5.24), we can write   Z u Z u 1 4au2 y= du = −2a ud 2 2 1 + u2 0 0 (1 + u ) 127

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Free ebooks ==> www.Ebook777.com #  u Z u du = −2a − 2 0 1+u 0      u u u = −2a − arctan u = −2a − arctan u . 1 + u2 1 + u2 0 "

u 1 + u2

A new change of variables

u = tan

θ 2

yields y = −2a

tan θ2 1 + tan2 = −2a



θ 2

θ − 2

!

sin θ θ − 2 2



θ θ θ = −2a sin cos − 2 2 2





= a(θ − sin θ),

so that x=

tan2 θ2 2au2 = 2a 1 + u2 1 + tan2

θ 2

= 2a sin2

θ = a(1 − cos θ). 2

Summarizing our results, we have obtained 

x = a(1 − cos θ), y = a(θ − sin θ),

(5.29)

which are the parametric equations of a cycloid, with y-axis as basis and concavity oriented upwards (see Fig.V.3).

Fig.V.3 128

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Free ebooks ==> www.Ebook777.com The constant a is not arbitrary anymore, but it represents the radius of the circle that generates the cycloid, by rolling without friction on y-axis. In our picture, the cycloid is generated by the point P1 , which is fixed with respect to the circle of radius a. According to Fig.V.3, we can write  π x = kOAk = kDCk + kEP1 k = a + kCP1 k sin θ − = a(1 − cos θ), 2 and ⌢



y = kOBk = kODk − kBDk =DP1 −kBDk =DP1 −kECk  π = aθ − a cos θ − = a(θ − sin θ). 2

To conclude, the brachistochrone curve is a cycloid (arc of a cycloid). If the body is given an initial velocity at P0 , or if friction is taken into account, then the curve that minimizes time will differ from the one described above. 2. Catenary problem Determine the curve formed by a rope or chain of uniform density and perfect flexibility, hanging freely between two points of suspension, not in the same vertical line. The rope has a fixed length (it is not extensible).

Fig.V.4 129

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Free ebooks ==> www.Ebook777.com Solution. Let l be the length of the chain, m its mass, and P1 (x1 , y1 ), P2 (x2 , y2 ) the points of suspension (see Fig.V.4). The equilibrium condition of the chain demands an extremum for the potential energy (minimum, in our case). To evaluate the potential energy, we split the chain in infinitesimal elements so that one may neglect the variation of the vertical coordinate y. If the plane xOz is taken as the reference level for the potential energy, then gravitational potential energy of the length element dl represented in Fig.V.4 is dEp = dm g y = dl λ g y,

(5.30)

where λ = l/m is the mass per unit length of the chain. Since (dl)2 = (dx)2 + (dy)2 , we then have dl =

p 1 + y ′ 2 dx;

y′ =

dy , dx

(5.31)

where only the positive solution has been considered. The potential energy of the whole chain then is Z P2 Z P2 Z x2 p Ep = dEp = λ g y dl = λg y 1 + y ′ 2 dx. (5.32) P1

P1

x1

Leaving aside the multiplicative factor λg, which in no way interferes in the extrema of the potential energy functional, one can consider the functional Z x2 p (5.33) E(y) = y 1 + y ′ 2 dx. x1

Following the variational procedure, we have to demand that the functional (5.33) has an extremum. Since the chain has a constant length Z P2 Z x2 p l= dl = 1 + y ′ 2 dx, (5.34) P1

x1

this condition tells us that we have to look for a conditional extremum. As easily seen, our constraint can be expressed through the functional Z x2 p L(y) = 1 + y ′ 2 dx = const. (5.35) x1

Therefore, to solve the problem means to find the extrema of the new functional F (y) = E(y) − µL(y), (5.36) 130

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Free ebooks ==> www.Ebook777.com where µ is an arbitrary (but non-zero) scalar. Using (5.33) and (5.35), we can write Z x2 h Z x2 i p ′2 F (y) = (y − µ) 1 + y dx = f (y, y ′ ) dx, x1

x1

where f (y, y ′ ) = (y − µ)

p

1 + y′ 2 .

(5.36′ )

The extremals of the functional (5.36) are found by solving the Euler-Lagrange equation d dx



∂f ∂y ′





∂f = 0, ∂y

(5.37)

where f (y, y ′ ) is given by (5.36’). Since f does not explicitly depend on the independent variable x, we have the first integral y′

∂f − f = const. ≡ C, ∂y ′

which yields y − µ = −C

p

(5.38)

1 + y′ 2 ,

or, by squaring and choosing the positive solution y′ =

dy 1p = (y − µ)2 − C 2 . dx C

Separation of variables leads to the following differential equation dy 1 p = dx. 2 2 C (y − µ) − C

To integrate this equation, it is convenient to make the change of variable y − µ = C cosh u, (5.39) and the solution is u=

1 x + C1 , C

so that solution (5.39) finally writes y = µ + C cosh

x

C

 + C1 .

131

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(5.40)

Free ebooks ==> www.Ebook777.com If fact, we deal with an infinite number of functions, obtained as solutions of (5.40), but only one curve passes through the points P1 and P2 and, in addition, obeys the constraint (5.34). Imposing these conditions, we are left with the following system of three algebraic equations for the unknown constants µ, C, and C1 :  x  1  + C y = µ + C cosh 1 , 1   C     x2 y2 = µ + C cosh + C1 , C Z x2 r  x     l = 1 + sinh2 + C1 dx, C x1

or

 x  1  + C1 , y1 = µ + C cosh      xC 2 + C1 , y2 = µ + C cosh  C       l = C sinh x2 + C − C sinh x1 + C . 1 1 C C

We leave the task of solving this system up to the reader. The curve given by equation (5.40) is called catenary (from the Latin catena, meaning chain). 3. Isoperimetric problem Determine the form of a plane curve of a given length l which encloses a surface of maximum area. Solution. According to Fig.V.5, the area A of the surface S bounded by the closed curve (Γ) is given by A=

Z

b a

ϕ2 (x) dx − =− 

 = −

Z

Z

Z

b

ϕ1 (x) dx = −

a

"Z

b

ϕ1 (x) dx + a

Z

Z

⌢ ϕ1 (x) dx + ⌢ AN B BM A

a b

ϕ2 (x) dx −

a

ϕ2 (x) dx b

#



 ϕ2 (x) dx = −

Z

I

b

ϕ1 (x) dx a

y dx, (Γ)

where A has been written as a difference of two areas: one bounded by the graphic of y = ϕ2 (x) between points A and B, the x-axis, and 132

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Free ebooks ==> www.Ebook777.com the straight lines x = a, x = b (the line-hatched area), and the other one bounded by the graphic of y = ϕ1 (x), between the same points A and B, and the straight lines x = a, x = b (the cross-hatched area).

Fig.V.5 To express in a more symmetric form the above relation, we shall add a ”zero” written as I I I 1 1 1 d(xy) = x dy + y dx. 0= 2 (Γ) 2 (Γ) 2 (Γ) The result is I

1 A=− y dx + 2 (Γ) 1 = 2

I

(Γ)

I

1 x dy + 2 (Γ)

I

y dx (Γ)

(x dy − y dx).

(5.41)

Suppose that the curve is defined by the parametric equations  x = x(t); y = y(t), t ∈ [a, b], with a = x(t1 ) and b = x(t2 ). Then area A writes 1 A= 2

I

(Γ)

(xy˙ − y x) ˙ dt,

133

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(5.42)

Free ebooks ==> www.Ebook777.com dy with x˙ = dx dt , y˙ = dt . The problem reduces, therefore, to the determination of extremals of the functional I 1 (xy˙ − y x) ˙ dt, (5.43) F (x, y) = 2 (Γ)

subject to the condition that the length l of (Γ) is given (and fixed): I I p I p l= dl = (dx)2 + (dy)2 = x˙ 2 + y˙ 2 dt. (Γ)

(Γ)

(Γ)

The functional expressing the constraint can be written as I p x˙ 2 + y˙ 2 dt = const. ≡ l. L(x, y) =

(5.44)

(Γ)

Toward the general theory, we construct a new functional  I  p 1 G(x, y) = F (x, y) − λL(x, y) = (xy˙ − y x) ˙ − λ x˙ 2 + y˙ 2 dt (Γ) 2 =

I

(Γ)

 g x(t), y(t), x(t), ˙ y(t) ˙ dt,

where λ is an arbitrary (but non-zero) scalar. To determine the extremals of the functional G(x, y), we have to solve the corresponding Euler-Lagrange equations    d ∂g ∂g    dt ∂ x˙ − ∂x = 0;   (5.45)  ∂g d ∂g   − = 0, dt ∂ y˙ ∂y with

g(x, y, x, ˙ y) ˙ =

p 1 (xy˙ − y x) ˙ − λ x˙ 2 + y˙ 2 . 2

Performing the derivatives in (5.45), we are left with   d     dt  d     dt

1 x˙ − y − λp 2 2 x˙ + y˙ 2

1 y˙ + x − λp 2 x˙ 2 + y˙ 2

!

!

1 − y˙ = 0; 2 1 + x˙ = 0, 2

134

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Free ebooks ==> www.Ebook777.com or

p  ˙ x+y˙ y¨ x ¨ x˙ 2 + y˙ 2 − x˙ √x¨   x˙ 2 +y˙ 2   = 0;  −y˙ − λ x˙ 2 + y˙ 2 p ˙ x+y˙ y¨  y¨ x˙ 2 + y˙ 2 − y˙ √x¨   x˙ 2 +y˙ 2   +x˙ − λ = 0, 2 2 x˙ + y˙

and, still

   x ¨y˙ − y¨x˙    y˙ −1 − λ (x˙ 2 + y˙ 2 )3/2 = 0;    x ¨y˙ − y¨x˙   x˙ +1 + λ = 0. (x˙ 2 + y˙ 2 )3/2

(5.46)

Since the solutions x˙ = 0, y˙ = 0 are not acceptable, the only remaining possibility is cancellation of the square brackets, that is x ¨y˙ − y¨x˙ 1 =− . 2 2 3/2 λ (x˙ + y˙ )

(5.47)

The nature of parameter t has not been specified so far. Let us assume that t is precisely the arc length l of the curve (Γ), determined with respect to a fixed point on (Γ), so that 2

2

x˙ + y˙ =



dx dt

2

+



dy dt

2

=

(dx)2 + (dy)2 = 1. (dl)2

(5.48)

With this choice, equation (5.47) yields 1 y˙ x ¨ − x¨ ˙y = − . λ To solve the problem, we need one more equation in x(t) and y(t). This is done by taking the derivative of (5.48) with respect to t: x¨ ˙ x + y˙ y¨ = 0. Thus, we have obtained the following system of two second-order differential equations in two variables x(t) and y(t):   y˙ x ¨ − x¨ ˙ y = − λ1 ; 

x¨ ˙ x + y˙ y¨ = 0. 135

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(5.49)

Free ebooks ==> www.Ebook777.com To facilitate the integration of (5.49), let us extract x ¨ from the second equation, and introduce the result into the first one. We then have 1 y¨ = x. ˙ (5.50) λ Repeating the procedure for y¨, one obtains 1 x ¨ = − y. ˙ λ

(5.51)

This way, we are left with a simpler system of second order differential equations, namely 

x ¨ = −λ−1 y, ˙ −1 y¨ = λ x. ˙

(5.52)

Integrating once gives 

x˙ = −λ−1 y + C1 , y˙ = λ−1 x + C2 .

With these values of x˙ and y, ˙ (5.48) becomes

or

 x 2 y 2  + C2 + , 1 = x˙ 2 + y˙ 2 = C1 − λ λ x + λC2

2

+ y − λC1

2

= λ2 .

(5.53)

This is the equation of a circle of radius R = λ, located in the xy-plane, with centre at the point xC = −λC2 , yC = λC1 . So, our problem is solved: for any constants λ, C1 , C2 (which remain undetermined), the plane curve of a given length enclosing a maximum area is a circle. 4. Surface of revolution of minimum area Let P1 (x1 , y1 ) and P2 (x2 , y2 ) be two given (and fixed) points situated in the xy-plane of a three-orthogonal trieder Oxyz. Determine the curve y = y(x) passing through P1 and P2 and generating by revolution about an axis (say, x), a surface of minimum area. Solution. The surface produced as a result of revolution about Ox is formed by three parts: areas of the two circles of the ”truncated cone”, of radii 136

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Free ebooks ==> www.Ebook777.com y1 and y2 , and the lateral area of the ”truncated cone” (see Fig.V.6). Since y1 and y2 are fixed, our problem is to determine the plane curve y = y(x) which, by its revolution about the x-axis, generates a lateral surface of minimum area.

Fig.V.6 As shown in Fig.V.6, the areas of the two circular ”bases” are S1 = πy12 = const. and S2 = πy22 = const., while the area of the elementary surface of width1 ds, and length 2πy (see Fig.V.7) is

Fig.V.7 1

This width is small enough so that one can suppose that all points on ds have the same coordinate y 137

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Free ebooks ==> www.Ebook777.com dS = 2π y ds, where ds =

(5.54)

p p (dx)2 + (dy)2 = 1 + y ′ 2 dx,

with y ′ = dy/dx. Then the lateral surface area of the ”truncated cone” is Z x2 p S = 2π y 1 + y ′ 2 dx. (5.55) x1

Ignoring the multiplicative constant 2π, which in no way affects our result, the area (5.55) is expressed as a functional, namely Z x2 p Z x2 ′2 S(y) = y 1 + y dx = f (y, y ′ ) dx. (5.56) x1

x1

To determine extrema of the functional (5.56), we have to solve Euler’s equation   ∂f ∂f d − = 0. (5.57) ′ dx ∂y ∂y

Since f (y, y ′ ) does not explicitly depend on x, there exists the first integral of (5.57) ∂f y ′ ′ − f = const. ≡ C. (5.58) ∂y We have:

so that

∂f yy ′ p = , ∂y ′ 1 + y′ 2

p yy ′ p − y 1 + y ′ 2 = C, ′2 1+y

or, after simple algebraic manipulations y′ =

dy 1p 2 = y − C 2, dx C

Considering only positive solution and integrating, we still have Z dy x p = + C1 . (5.59) C y2 − C 2

The integral on the l.h.s. can be easily solved by substitution y = C cosh u. 138

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(5.60)

Free ebooks ==> www.Ebook777.com Indeed, Z

dy

p = y2 − C 2

Z

C sinh u q du = u. 2 2 C (cosh u − 1)

In view of (5.59), we can write u=

x + C1 , C

and (5.60) finally leads to y = C cosh

x

C

 + C1 .

(5.60′ )

To give the answer to our problem, one must determine the constants C 6= 0 and C1 . This is done by imposing condition that the curve (5.60’) passes through P1 (x1 , y1 ) and P2 (x2 , y2 ), which means   y1 = C cosh 

y2 = C cosh

x1 C x2 C

 + C1 ;  + C1 .

To conclude, the curve passing through two given points, that generates by its revolution about x-axis a surface of minimum area, is a catenary. 5. Geodesics of a Riemannian manifold Determine the geodesics of a n-dimensional Riemannian manifold. Solution. This problem concerns the general theory of relativity, but we have considered it here because it is an important application of the variational calculus. Before going further, let us define the terms geodesic and Riemannian manifold (or space). In geometric approach, a geodesic is a generalization of the notion of a ”straight line” to ”curved spaces”, being (locally) the shortest path between two points in the space. From the dynamical point of view, a geodesic is the trajectory described by a free particle (material point) in space. Obviously, the shape of the geodesic depends on the structure of the space on which it is defined. For instance, in an Euclidean threedimensional space E3 , the shortest distance between two points is determined by a straight line passing through the two points. From the 139

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Free ebooks ==> www.Ebook777.com dynamical point of view, a free particle moves uniformly in a straight line, relative to an inertial frame. All these definitions can be obtained by a variational procedure as follows. In the Euclidean space E3 , the infinitesimal distance between two points is given by p p ds = |d~r| = (dx)2 + (dy)2 + (dz)2 = dxi dxi , (i = 1.3), (5.61)

where x1 = x, x2 = y, x3 = z, and Einstein’s summation convention has been used. The squared ds, that is ds2 = |d~r|2 = d~r · d~r = (dx)2 + (dy)2 + (dz)2 = dxi dxi

(i = 1.3) (5.62) is called metric of the three-dimensional Euclidean space E3 . Let us parametrize some curve in E3 by xi = xi (t)

(i = 1, 3),

where t is the parameter. Then the distance between any two points P ′ (x′ , y ′ , z ′ ) and P ′′ (x′′ , y ′′ , z ′′ ) on this curve is given by ∆s =

Z

P ′′

ds = P′

Z

P ′′ P′

Z p dxi (t)dxi (t) =

t2 t1

p

x˙ i x˙ i dt,

(5.63)

where

dxi , x′i = xi (t1 ), x′′i = xi (t2 ). dt As one observes, the distance ∆s is written as a functional of the type x˙ i =

J(x) =

Z

t2

f (t, x, x) ˙ dt, t1

where f (x) ˙ =

p x˙ i x˙ i .

Since f does not explicitly depend on x(t), the Euler equation of extremals   d ∂f ∂f − = 0 (j = 1, 3) dt ∂ x˙ j ∂xj admits the first integral ∂f = (const.)j ≡ Cj ∂ x˙ j

(j = 1, 3)

140

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(5.64)

Free ebooks ==> www.Ebook777.com or

 ∂ p x˙ j = Cj x˙ i x˙ i = √ ∂ x˙ j x˙ i x˙ i

(i, j = 1, 3).

(5.65)

Let us consider the following two cases: (a) If the arc length s is taken as the parameter t, then r √ p dxi dxi dxi dxi x˙ i x˙ i = = 1. = p ds ds (ds)2 With this result, (5.65) leads to

x˙ j = Cj

(j = 1, 3),

(5.66)

or xj = Cj s + Cj′ ,

(5.67)

where Cj′ (j = 1, 3) are three new constants of integration. To determine the six constants of integration Cj and Cj′ , one imposes the condition that P ′ and P ′′ are on the curve. Relations (5.67) can be written explicitly as x = C1 s + C1′ , y = C2 s + C2′ , z = C3 s + C3′ . Eliminating the parameter s, we obtain   C2 C2 C1′ x − C1′ ′ ′ + C2 = x+ − + C2 = k1 x + k2 ; (5.68) y = C2 C1 C1 C1   y − C2′ C3 C3 C2′ ′ ′ z = C3 + C3 = x+ − + C3 = k3 x + k4 , (5.69) C2 C2 C2 where the new constants ki (i = 1, 4) are expressed in terms of Cj and Cj′ . The relations  y = k1 x + k2 ; (5.70) z = k3 x + k4 , represent a straight line in E3 . In other words, the geodesics of the Euclidean three-dimensional space E3 are straight lines. (b) Let us now consider parameter t as being the time. Then (5.65) writes v p j = Cj (j = 1, 3), |~v |2 141

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Free ebooks ==> www.Ebook777.com or, by multiplying both numerator and denominator of the fraction by the mass m 6= 0 of the body p p j = Cj |~ p|2

(j = 1, 3).

(5.71)

If the resultant force acting on the body is zero (the body moves freely), then according to the fundamental principle of dynamics d~ p = 0, F~ = dt meaning that p~ = const., and (5.71) yields pj = Kj

(j = 1, 3),

where Kj (j = 1, 3) are constants. Therefore, (j = 1, 3).

mx˙ j = Kj

(5.72)

On the other hand, the kinetic energy theorem says that the kinetic energy of a free body is a constant. In this case, since m|~v |2 = C ′ and |~ p|2 = m2 |~v |2 = C ′′ , we have m = C ′′ /C ′ = const., where C ′ and C ′′ are also constants. Then (5.72) can be written as x˙ j = Kj′ = (const.)j , or xj = Kj′ t + x0j

(j = 1, 3),

(5.73)

where the arbitrary constants of integration x0j (j = 1, 3) can be determined by the initial conditions. In vector notation, equation (5.73) writes ~ ′ t + ~r0 . ~r = K (5.73′ ) Therefore, in the Euclidean three-dimensional space E3 , a free body moves uniformly, in a straight line. ∗





To define a Riemannian manifold one must introduce the notions of metric of a space, metric tensor, as well as covariance and contravariance of tensors. For a better understanding of the theory, we shall first consider a three-dimensional space. Here, in addition to the 142

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Free ebooks ==> www.Ebook777.com simplest coordinate system (the Cartesian frame), there are two more (but not most!) general reference frames: (i) orthogonal curvilinear coordinates (a system of curvilinear coordinates in which each family of surfaces intersects the others at right angles); (ii) non-orthogonal rectilinear coordinates (the coordinate axes are straight lines, but the angles of intersection are different from 90o ). As the most general case one can consider a combination of i) and ii): the coordinate axes are curvilinear, while the angles of intersection are different from 90o (see Fig.V.8).

Fig.V.8 Since the orthogonal system of coordinates is thoroughly discussed in textbooks, we shall analyze the case (ii), showing how nonorthogonality of axes requires introduction of the concept of variance. To facilitate the graphic representations, let us first take into account a two-dimensional space (generalization to three dimensions is trivial). Consider a vector ~a and write it in component forms as reported to two different reference systems: one orthogonal, and the other nonorthogonal (see Fig.V.9). In the first case (see Fig.V.9a) we can write  a1 = ~a · ~u1 , (5.74) a2 = ~a · ~u2 . 143

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Free ebooks ==> www.Ebook777.com Using the notations ~a1 = a1 ~u1 and ~a2 = a2 ~u2 , we also have ~a = ~a1 + ~a2 ,

(5.75)

where ~u1 and ~u2 are the unit vectors of the two axes, and a1 , a2 the components of the vector ~a (the orthogonal projections of ~a on the coordinate axes).

Fig.V.9 In the second case (see Fig.V.9b), there are two possibilities to define the components of the vector ~a: either by tracing straight lines perpendicular to the axes (components a1 and a2 ), or by drawing straight lines parallel to the axes (components a′1 and a′2 ). The first possibility allows us to write  a1 = ~a · ~u1 , (5.76) a2 = ~a · ~u2 , but, if we denote ~a1 = a1 ~u1 and ~a2 = a2 ~u2 , the relation (5.75) is not valid anymore, since now ~a 6= ~a1 + ~a2 . Using the other procedure, if we denote  ′ ~a1 = a′1 ~u1 , ′ ~a2 = a′2 ~u2 ,

(5.77)

(5.78)

then relation (5.75) remains valid, that is ′



~a = ~a1 + ~a2 , 144

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(5.79)

Free ebooks ==> www.Ebook777.com but this time



a′1 = 6 ~a · ~u1 , ′ a2 6= ~a · ~u2 ,

(5.80)

so that (5.74) loses its validity. In other words, the components a′1 and a′2 are not obtained as scalar products of the vector ~a, on the one hand, and the unit vectors ~u1 and ~u2 , on the other. To ”conciliate” these two possible choices for both cases (i.e. for the two possible choices/possibilities), one introduces the so-called dual basis. Consider, in this respect, a system of three non-coplanar, linearly independent vectors ~u, ~v , and w, ~ which form a basis in the Euclidean three-dimensional space E3 . Then any vector ~a ∈ E3 can be written as ~a = λ~u + µ~v + ν w, ~ (5.81) where the three scalars (not all zero) are called the components of the vector ~a in basis {~u, ~v , w}. ~ Let us now introduce another set of three vectors ~u ∗ , ~v ∗ , and w ~ ∗, defined as (

~u ∗ · ~u = 1, ~u ∗ · ~v = 0, ~u ∗ · w ~ = 0,

(

~v ∗ · ~u = 0, ~v ∗ · ~v = 1, ~v ∗ · w ~ = 0,

(

w ~ ∗ · ~u = 0, w ~ ∗ · ~v = 0, w ~∗·w ~ = 1.

(5.82)

It can be easily verified that the triplet of vectors ~u ∗ , ~v ∗ , and w ~∗ ~u ∗ =

~v × w ~ ; (~u, ~v , w) ~

~v ∗ =

w ~ × ~u ; (~u, ~v , w) ~

w ~∗=

~u × ~v , (~u, ~v , w) ~

(5.83)

where (~u, ~v , w) ~ = ~u · (~v × w) ~ is the mixed product of the three vectors, satisfying the definition conditions (5.82). According to their definition, the vectors ~u ∗ , ~v ∗ , and w ~ ∗ are, in their turn, linearly independent, which means that they also form a basis in R3 , called dual basis of the direct basis {~u, ~v , w}. ~ It can also be shown that the dual of dual of a vector yields the original vector, that is ∗ ∗ ∗ ~u ∗ = ~u, ~v ∗ = ~v , w ~ ∗ = w. ~ (5.84) The components of any vector ~a ∈ E3 with respect to the direct basis (e.g. the scalars λ, µ and ν that appear in (5.81)), can be expressed as scalar products of ~a and the corresponding vectors of dual basis. Indeed, by means of (5.81) and (5.82), it is easy to show that λ = ~a · ~u ∗ ,

µ = ~a · ~v ∗ ,

ν = ~a · w ~ ∗.

145

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(5.85)

Free ebooks ==> www.Ebook777.com Coming now back to our problem, in case of the frames with rectilinear, non-orthogonal coordinates we can write both a relation of type (5.75) for the unprimed components ~a = ~a1 + ~a2 , but where ~a1 = a1 ~u1∗ ,

~a2 = a2 ~u2∗ ,

(5.86)

and a relation of type (5.74) for the primed components a′1 = ~a · ~u1 , a′2 = ~a · ~u2 , but where ~u1 and ~u2 must be replaced by ~u1∗ and ~u2∗ , that is a′1 = ~a · ~u1∗ , a′2 = ~a · ~u2∗ . (5.87) Indeed, in the first case we have a1 = ~a · ~u1 , a2 = ~a · ~u2 , and ~a 6= ~a1 + ~a2 . But, if ~a1 and ~a2 are given by (5.86), even now we can write a relation of type (5.75) ~a = a1 ~u1∗ + a2 ~u2∗ = ~a1 + ~a2 , because, in view of (5.85) and (5.84) we have ∗  a1 = ~a · ~u1∗ = ~a · ~u1 , ∗ a2 = ~a · ~u2∗ = ~a · ~u2 ,

which are precisely relations (5.76). This way, the two types of relations are in ”agreement” with each other, in the sense that they are simultaneously valid. ′ ′ ′ In the second case, we have ~a = ~a1 + ~a2 , but ~a1 6= ~a · ~u1 , and ′ ~a2 6= ~a · ~u2 . But, if we replace ~u1 and ~u2 by ~u1∗ and ~u2∗ , then the two relations become equalities (a′1 = ~a · ~u1∗ , and a′2 = ~a · ~u2∗ ). This means that, in this second case, a relation of type (5.76) is also valid. Indeed, according to (5.81) and (5.85), we have ′



~a = a′1 ~u1 + a′2 ~u2 = ~a1 + ~a2 , in agreement with (5.79) and (5.79). Therefore, the problem is solved. In practice, one usually uses a different set of notations, namely:  ′  ∗ a1 = a1 , ~u1 = ~u1 , and (5.88) a′2 = a2 , ~u2∗ = ~u2 . We conclude that, in a non-orthogonal frame, any vector ~a has two sets of components: one with lower indices  a1 = ~a · ~u1 , (5.89) a2 = ~a · ~u2 , 146

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Free ebooks ==> www.Ebook777.com and the other one with upper indices  a1 = ~a · ~u1 , a2 = ~a · ~u2 .

(5.90)

Components with lower indices are called covariant, while those with upper indices are called contravariant. The direct basis is formed by covariant vectors, and the dual basis by contravariant vectors1 . In addition, as shown in Fig.V.9b, if the angle between axes becomes 90o , the two types of components are coincident (the projections orthogonal and parallel to axes become identical). In other words, in case of orthogonal coordinates, there is no difference between covariant and contravariant components of a vector. To summarize, in a non-orthogonal vector space any vector has two sets of components:  covariant, if the vector is expressed in the dual (contravariant) basis ~u1 , ~u2 , ~u3 , and contravariant,  if the vector is expressed in the direct (covariant) basis ~u1 , ~u2 , ~u3 . For example, the radius-vector ~r = x~i + y~j + z~k, in a non-orthogonal space, can be written either in terms of contravariant (dual) basis ~r = x1 ~u1 + x2 ~u2 + x3 ~u3 , or in terms of covariant (direct) basis ~r = x1 ~u1 + x2 ~u2 + x3 ~u3 . In light of the above definitions, assuming that the unit vectors of both bases are constant, while the axes of the (non-orthogonal) frame are rectilinear (so that the unit vectors of the axes are the same at any point of the space), the differential d~r of a radius-vector ~r can also be written in two ways: d~r = dx1 ~u1 + dx2 ~u2 + dx3 ~u3 , and d~r = dx1 ~u1 + dx2 ~u2 + dx3 ~u3 , in which case the metric of the space is ds2 = |d~r|2 = d~r · d~r 1

By virtue of (5.84), the terms ”direct” and ”dual” used to  desig nate the bases are relative. In fact, the two bases ~u1 , ~u2 and ~u1 , ~u2 are dual to each other. 147

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=

ր → ց

 dx1 ~u1 + dx2 ~u2 + dx3 ~u3 · dx1 ~u1 + dx2 ~u2 + dx3 ~u3 ),

  dx1 ~u1 + dx2 ~u2 + dx3 ~u3 · dx1 ~u1 + dx2 ~u2 + dx3 ~u3 ,

  dx1 ~u1 + dx2 ~u2 + dx3 ~u3 · dx1 ~u1 + dx2 ~u2 + dx3 ~u3 .

Using the new notations, we can write (5.82) in the condensed form  1, i = j, j ~ui · ~u = (5.91) 0, i 6= j, so that ds2 = |d~r|2

=

ր → ց

   dxi ~ui · dxj ~uj = ~ui · ~uj dxi dxj ,

   dxi ~ui · dxj ~uj = ~ui · ~uj dxi dxj ,

Let us denote

(5.92)

   dxi ~ui · dxj ~uj = ~ui · ~uj dxi dxj . 

~ui · ~uj = g ij , ~ui · ~uj = gij .

(5.93)

In view of (5.91), the metric (5.92) can be expressed in one of the following three forms: ds2 = g ij dxi dxj = gij dxi dxj = dxi dxi .

(5.94)

The quantities g ij and gij are components of a contravariant and a covariant tensor, respectively, called contravariant (covariant) metric tensor. As seen, relations (5.93) show that the contravariant (covariant) components of the metric tensor are given by the scalar products of the vectors of dual (direct) bases. Relation (5.94) yields, on the one hand dxi = gij dxj ,

(5.95)

dxi = g ij dxj ,

(5.96)

and

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Free ebooks ==> www.Ebook777.com on the other. This shows that, by means of a suitable choice of the metric tensor, an index can be lowered or raised, respectively. Let us now consider an Euclidean m-dimensional space Em , and let y1 , y2 , ..., ym be the Cartesian coordinates of some point in this space. The metric ds2 in Em is ds2 = dyj dyj

(j = 1, m).

(5.97)

Consider now in Em a n-dimensional variety (subspace) Rn (n < m), and let x1 , x2 , ..., xn be the coordinates of a point in Rn . Since yj = yj (x1 , x2 , ..., xn ) (j = 1, m), we can write ds2 =

∂yj ∂yj i k dx dx = gik dxi dxk i k ∂x ∂x

(j = 1, m; i, k = 1, n),

(5.98)

where by gik (x1 , x2 , ..., xn ) = gki =

∂yj ∂yj ∂xi ∂xk

(j = 1, m; i, k = 1, n)

have been denoted the covariant components of the metric tensor. This is a symmetric, second rank tensor. If gik = δik , that is, if the manifold Rn is Euclidean, we get back to the metric (5.97). If the metric (5.98) is invariant with respect to the general coordinate transformation ′



x i = x i (x1 , x2 , ..., xn )

(i = 1, n),

then the manifold Rn is called Riemannian. It is now our purpose to determine the differential equations of geodesics of a Riemannian manifold Rn . Let xi (i = 1, n) be the coordinates of a particle moving in Rn , and xi = xi (s)

(i = 1, n)

(5.99)

the parametric equations of a curve passing through two given points P1 and P2 . The arc length of the curve between the two points is Z P2 Z P2 p Z P2 p gik dxi dxk = gik x˙ i x˙ k ds, (5.100) L(x) = ds = P1

P1

P1

where x˙ i = dxi /ds. The curve given by (5.99) is a geodesic, if the functional (5.100) has an extremum (which usually is a minimum). To this end, the function p f (x, x, ˙ s) = gik x˙ i x˙ k (i, k = 1, n) (5.101) 149

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Free ebooks ==> www.Ebook777.com must satisfy the Euler-Lagrange equations d ds



∂f ∂ x˙ j





∂f =0 ∂xj

(j = 1, n).

(5.102)

Since p gik x˙ i x˙ k =

r

gik

we have:

1p ds dxi dxk = gik dxi dxk = = 1, ds ds ds ds

(5.103)

1 1 ∂f i k p p = g δ x ˙ = gjk x˙ k = gjk x˙ k ; ik j i k i k ∂ x˙ j gik x˙ x˙ gik x˙ x˙ d ds



∂f ∂ x˙ j



=

 d gjk x˙ k ds

∂gjk i k ∂gjk dxi k x˙ + gjk x ¨k = x˙ x˙ + gjk x ¨k ; i ∂x ds ∂xi   ∂f 1 ∂ p ∂ i k ix k = p = g x ˙ ˙ g x ˙ x ˙ ik ik ∂xj ∂xj 2 gik x˙ i x˙ k ∂xj =

∂gik i k 1 ∂gik i k 1 x ˙ x ˙ = x˙ x˙ . = p 2 ∂xj 2 gik x˙ i x˙ k ∂xj

With these results, equations (5.102) become gjk x ¨k + or

∂gjk i k 1 ∂gik i k x˙ x˙ − x˙ x˙ = 0, ∂xi 2 ∂xj

 1 ∂gik i k ∂gjk i k ∂gjk i k x˙ x˙ + x˙ x˙ − x˙ x˙ i i ∂x ∂x 2 ∂xj   1 ∂gji k i ∂gjk i k 1 ∂gik i k k = gjk x ¨ + x˙ x˙ + x˙ x˙ − x˙ x˙ k i 2 ∂x ∂x 2 ∂xj   1 ∂gjk ∂gij ∂gik k = gjk x ¨ + + − x˙ i x˙ k = 0. 2 ∂xi ∂xk ∂xj 1 gjk x ¨ + 2 k



If we denote Γik,j

1 = 2



∂gjk ∂gij ∂gik + − i k ∂x ∂x ∂xj



,

150

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(5.104)

Free ebooks ==> www.Ebook777.com the last equation writes gjk x ¨k + Γik,j x˙ i x˙ k = 0.

(5.105)

The quantities (5.104) are called Christoffel symbols of the first kind. Multiplying (5.104) by g jl and performing summation over j, we finally obtain the differential equation of geodesic lines in Rn x ¨l + Γlik x˙ i x˙ k = 0,

(5.106)

g jl Γik,j = Γlik

(5.107)

where are the Christoffel symbols of the second kind. It can be shown that, except for the linear transformations of coordinates, the Christoffel symbols of both kinds are not tensors. The differential equations of geodesics of a n-dimensional Riemannian manifold Rn are, at the same time, the equations of motion of a free particle in the gravitational field. In fact, the quantities al = x ¨l + Γlik x˙ i x˙ k stand for the components of the n-dimensional acceleration vector in Rn . It can be easily proved that the quantities x ¨l (l = 1, n) are not vectors. Observation. If we denote by Φ=

1 gik x˙ i x˙ k , 2

then Euler-Lagrange equations d ds



∂Φ ∂ x˙ j





∂Φ =0 ∂xj

(j = 1, n)

lead to the same result. Consequently, the variational principles δ

Z

and δ

P2 P1

Z

p gik x˙ i x˙ k ds = 0,

P2

gik x˙ i x˙ k ds = 0

P1

are equivalent. Application. Determine the geodesics of a sphere of radius one. 151

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Free ebooks ==> www.Ebook777.com In spherical coordinates, the element of the arc length writes ds2 = dr2 + r2 dθ2 + r2 sin2 θ dϕ2 ,

(5.108)

and, if r = R = 1 (R is the radius of the sphere) ds2 = dθ2 + sin2 θ dϕ2 .

(5.109)

The variational principle then writes δ

Z

ds = δ

Z

=δ ˙= with θ˙ = dθ ds , ϕ the functional

dϕ ds .

ds2 ds = δ ds2 Z

Z

dθ2 + sin2 θdϕ2 ds ds2

(θ˙2 + ϕ˙ 2 sin2 θ) ds = 0,

Therefore, we have to look for the extremals of

F (θ, ϕ) =

Z

P2

˙ ϕ) f (θ, θ, ˙ ds,

(5.110)

P1

where P1 and P2 are two given points on the surface of the sphere, and ˙ ϕ) f (θ, θ, ˙ = θ˙2 + ϕ˙ 2 sin2 θ = 1. (5.111) The Euler-Lagrange equation for the variable ϕ is   d ∂f ∂f = 0. − ds ∂ ϕ˙ ∂ϕ

(5.112)

Calculating the derivatives ∂f ∂f = 0; = 2ϕ˙ sin2 θ; ∂ϕ ∂ ϕ˙   d ∂f = 4θ˙ ϕ˙ sin θ cos θ + 2ϕ¨ sin2 θ ds ∂ ϕ˙ we are left with

ϕ¨ + 2 θ˙ ϕ˙ cot θ = 0.

(5.113)

To find the explicit equation of the geodesic, ϕ = ϕ(θ), we have to eliminate the parameter s from equations (5.111) and (5.113). To this end, we observe that (5.113) can also be written as dϕ˙ + 2ϕ˙ cot θ dθ = 0, 152

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Free ebooks ==> www.Ebook777.com or, if the variables are separated dϕ˙ cos θ dθ d(sin θ) = −2 cot θ dθ = −2 = −2 , ϕ˙ sin θ sin θ and the integration is carried out ln ϕ˙ = −2 ln(sin θ) + ln C = ln



C sin2 θ



,

which yields

C , sin2 θ where C is an arbitrary constant of integration. Observing that ϕ˙ =

dθ dθ dϕ dθ θ˙ = = = ϕ, ˙ ds dϕ ds dϕ equation (5.111) yields C2 θ˙2 + ϕ˙ 2 sin2 θ = sin4 θ

"

dθ dϕ

2

#

+ sin2 θ = 1,

or, if the variables are separated dϕ = so that C

Z

C dθ p , sin θ sin2 θ − C 2

dθ p = ϕ − ϕ0 , sin θ sin2 θ − C 2

(5.114)

(5.115)

where ϕ0 is an arbitrary constant of integration. To integrate this equation, it is convenient to make the following change of variable sin2 θ − C 2 = ζ 2 . We then have C

Z

dζ p = ϕ − ϕ0 . (ζ 2 + C 2 ) 1 − C 2 − ζ 2

Here we have an integral of the type Z dx √ , 2 2 (x + a ) b2 − x2 153

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(5.116)

Free ebooks ==> www.Ebook777.com with a = C and b = of integrals Z



1 − C 2 . The result can be found in the tables

dx 1 √ = √ arctan (x2 + a2 ) b2 − x2 a a 2 + b2

x a

r

b2 + a 2 b2 − x 2

!

,

and (5.116) writes C

Z

dζ p = arctan 2 2 (ζ + C ) 1 − C 2 − ζ 2 = arctan

so that

C

p

ζ 1 − C2 − ζ2

!

! p sin2 θ − C 2 = ϕ − ϕ0 C cos θ

p sin2 θ − C 2 = tan(ϕ − ϕ0 ), C cos θ

and, finally, 1 C cos(ϕ − ϕ0 ) = p cot θ. =√ 2 1 − C2 1 + tan (ϕ − ϕ0 )

(5.117)

This is the equation of a plane passing through the origin of the coordinate system, placed at the centre of the sphere. Therefore, the geodesics we are looking for are great circles of the sphere, obtained as a result of intersection between the plane (5.117) and the sphere (see Fig.V.10). To put into evidence the fact that (5.117) represents, indeed, the equation of a plane passing through the origin of the coordinate system, let us rewrite this equation in Cartesian coordinates. To this end, we shall use the well-known transformation relations ( x = sin θ cos ϕ, y = sin θ sin ϕ, z = cos θ, where we have considered r = 1. We have: cos(ϕ − ϕ0 ) = cos ϕ cos ϕ0 + sin ϕ sin ϕ0 = =√

x y cos ϕ0 + sin ϕ0 sin θ sin θ

C C cos θ C z cot θ = √ =√ . 1 − C2 1 − C 2 sin θ 1 − C 2 sin θ 154

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Fig.V.10 Denoting α= and β=



1 − C 2 cos ϕ0 , C



1 − C 2 sin ϕ0 , C

we finally obtain z = αx + βy,

(5.118)

which is the equation of a plane passing through tho origin of the coordinate system. Here α and β are two constants depending on C and ϕ0 , that can be determined by imposing the condition that the plane passes through the two fixed points P1 and P2 (and, obviously, through the origin O).

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CHAPTER VI PROBLEMS SOLVED BY MEANS OF THE LAGRANGIAN FORMALISM

1. Atwood machine The Atwood’s machine was invented in 1784 by Rev. George Atwood in order to verify the mechanical laws of motion with constant acceleration. It essentially consists of two bodies of masses m1 and m2 , connected by an inextensible massless string of length l over an ideal massless pulley of radius r and moment of inertia I (see Fig.VI.1). Neglecting friction between the string and the pulley, find the differential equation of motion of the system. Solution Since the only applied force is the force of gravitation, which is conservative, our system is a natural system. Supposing the pulley as

Fig.VI.1 156

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Free ebooks ==> www.Ebook777.com a third body of the system, and taking z = 0 as the plane of motion, the system is submitted to the following constraints: i) f1 (z1 ) = z1 = 0; ii) f2 (z2 ) = z2 = 0; iii) f3 (z3 ) = z3 = 0 (the pulley of mass m3 does not have a motion of translation along z-axis); iv) f4 (y1 ) = y1 = C1 (const.) (the body of mass m1 moves only along the x-axis); v) f5 (y2 ) = y2 = C2 (const.) (the body of mass m2 moves only along the x-axis); vi) f6 (y3 ) = y3 = C3 (const.) (the pulley of mass m3 does not have a motion of translation along y-axis); vii) f7 (x1 , x2 ) = x1 + x2 + (π − 2)r − l = 0 (the string is inextensible); viii) f8 (x3 ) = x3 = 0 (the pulley of mass m3 does not have a motion of translation along x-axis. Here indices 1, 2, and 3 are attached to the bodies of masses m1 , m2 , and m3 , respectively. As one can see, the system has 3 · 3 − 8 = 1 degree of freedom. Let x be the associated generalized coordinate. To write the Lagrangian of the system it is necessary to determine its kinetic and potential energies. Since the dimensions of the pulley are finite, its moment of inertia cannot be neglected. In this respect, the kinetic energy of rotation of the pulley about its own axis has also to be considered. Therefore, we have: T = T1 + T2 + T3 =

1 1 1 m1 |~v1 |2 + m2 |~v2 |2 + Iω 2 , 2 2 2

where ω is angular frequency of rotation of the pulley about its own axis. Since the string does not slide on the pulley, the linear velocity of an arbitrary point of the discus periphery, and the translation velocity of the bodies of masses m1 and m2 are the same: |~v1 | = |~v2 | = x. ˙ Then T =

1 1 x˙ 2 1 m1 x˙ 21 + m2 x˙ 22 + I 2 . 2 2 2 r

The potential energy of our conservative system is V = Vb + Vp , where Vb is the potential energy of the two bodies, and Vp = const. the potential energy of the pulley. We can write ~ 1 · d~r1 − G ~ 2 · d~r2 = −g(m1 dx1 + m2 dx2 ), dVb = −dA = −G 157

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Free ebooks ==> www.Ebook777.com so that Vb = −m1 gx1 − m2 gx2 + Vb0 . A convenient choice of the reference system for the potential energy makes it possible to consider the integration constant Vb0 as being zero. Indeed, taking Vb (x1 = 0, x2 = 0) = 0, one obtains Vb0 = 0. Therefore V = Vb + Vp = −m1 gx1 − m2 gx2 + Vp = −(m1 − m2 )gx + Vc , where Vc = −m2 gl + m2 g(π − 2)r + Vp = const. can be dropped according to the definition/property of equivalent Lagrangians. The Lagrangian of the system then writes I  2 1 m1 + m2 + 2 x˙ + (m1 − m2 )gx. L=T −V = 2 r The Lagrange equation of the second kind for the generalized coordinate x is d  ∂L  ∂L = 0. (6.1) − dt ∂ x˙ ∂x

Performing the derivatives and introducing the results into (6.1), one obtains the differential equation of motion of the system  I  m1 + m2 + 2 x ¨ − (m1 − m2 )g = 0, r

(6.2)

which yields the constant acceleration of the system a=x ¨=

(m1 − m2 )g = const. m1 + m2 + rI2

(6.3)

2. Double Atwood machine The system schematically presented in Fig.VI.2 is called double Atwood machine. Like in the previous paragraph, the masses of the pulleys is negligible, and the friction is neglected. Using the Lagrangian formalism, write the differential equations of motion of the system, and determine the accelerations of the three bodies.

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Free ebooks ==> www.Ebook777.com Solution Here we have, again, a natural system (the only active force is the force of gravity). Neglecting the radii of pulleys as compared to the lengths of the strings, the equations of constraints are: i) f1 (z1 ) = z1 = 0 (without loss of generality, one can assume that the motion takes place in the plane z = 0); ii) f2 (z2 ) = z2 = 0 (idem); iii) f3 (z3 ) = z3 = 0 (idem); iv) f4 (y1 ) = y1 = C1 (const.) (the body of mass m1 moves only along the x-axis); v) f5 (y2 ) = y2 = C2 (const.) (the body of mass m2 moves only along the x-axis); vi) f6 (y3 ) = y3 = C3 (const.) the body of mass m3 moves only along the x-axis); vii) f7 (x1 , x2 , x3 ) = 2x1 + x2 + x3 − 2l1 − l2 = 0 (the wires are inextensible1 ). This shows that the system has 3·3−7 = 2 degrees of freedom. We should mention that, unlike the ”simple” Atwood machine, when the pulley was considered one of the three material points of the system, this time the degrees of freedom involved by the existence of the two pulleys is not considered (otherwise, we would have had 13 equations to define the constraints). Let the associated generalized coordinates be ξ1 (= x1 ) and ξ2 (see Fig.VI.2). To write the Lagrangian, we have to know the kinetic and potential energies. These quantities are going to be determined by means of some simplifying conditions: the radii of the pulleys are negligible as compared to lengths of the wires, while their moments of inertia are ignored. The kinetic energy of the system is then given by the sum of the kinetic energy of translation of the three bodies along the x-axis, that is 1 1 1 T = T1 + T2 + T3 = m1 |~v1 |2 + m2 |~v2 |2 + m3 |~v3 |2 2 2 2 =

1 1 1 1 1 1 m1 x˙ 21 + m2 x˙ 22 + m3 x˙ 23 = m1 ξ˙12 + m2 x˙ 22 + m3 x˙ 23 . 2 2 2 2 2 2

1

When writing the equation of this constraint, we neglected the radii of the two pulleys. As in case of the simple Atwood machine, this approximation acts only on the potential energy of the system, which differs from the real one by a constant quantity that can be ignored in the Lagrangian. If, nevertheless, the finite radii of the pulleys are considered, then the constraint f7 writes: f7 (x1 , x2 , x3 ) = 2x1 + x2 + x3 − 2l1 − l2 + 2(π − 2)r1 + πr2 = 0. 159

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Free ebooks ==> www.Ebook777.com Here we have to express x˙ 22 and x˙ 23 in terms of ξ˙1 and ξ˙2 . Neglecting the radii of the pulleys1 , one can write x 2 = l1 + ξ 2 − x 1 = l1 + ξ 2 − ξ 1 , so that

x˙ 22 = (ξ˙2 − ξ˙1 )2 .

Fig.VI.2 Within the same approximation2 , Fig.VI.2 shows that x3 = l2 + x2 − 2ξ2 , which yields x˙ 23 = (x˙ 2 − 2ξ˙2 )2 = (ξ˙2 − ξ˙1 − 2ξ˙2 )2 = (−ξ˙1 − ξ˙2 )2 = (ξ˙1 + ξ˙2 )2 , 1

The same expression for the total kinetic energy of the system is also obtained if one considers the exact expression for x2 , that is: x2 = l1 + ξ2 − ξ1 − (π − 2)r1 − r2 , because the constant quantities disappear when taking the derivatives. 2 Writing the exact expression for x3 , that is: x3 = x2 + l2 − 2ξ2 − (π−2)r2 , leads to the same total kinetic energy, for the reason specified in the previous footnote. 160

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Free ebooks ==> www.Ebook777.com and the kinetic energy writes T =

1 1 1 m1 ξ˙12 + m2 (ξ˙2 − ξ˙1 )2 + m3 (ξ˙2 + ξ˙1 )2 . 2 2 2

(6.4)

To determine the potential energy of the system, we make allowance for the usual procedure ~ 1 · d~r1 − G ~ 2 · d~r2 − G ~ 3 · d~r3 dV = −dA = −G = −m1 gdx1 − m2 gdx2 − m3 gdx3 , so that V = −m1 gx1 − m2 gx2 − m3 gx3 + V0 . A convenient choice of the reference level for V [V (x1 = 0, x2 = 0, x3 = 0) = 0] yields V0 = 0, and, using the same approximation (the radii of pulleys are negligible as compared to the lengths of the two inextensible wires)1 , the potential energy writes V = −m1 gξ1 − m2 g(l1 + ξ2 − ξ1 ) − m3 g(l2 + x2 − 2ξ2 ) = −m1 gξ1 − m2 g(l1 + ξ2 − ξ1 ) − m3 g(l2 + l1 + ξ2 − ξ1 − 2ξ2 ) (6.5) = −g(m1 − m2 − m3 )ξ1 − g(m2 − m3 )ξ2 − gl1 (m2 + m3 ) − m3 gl2 . The Lagrangian of the system therefore is L=T −V =

1 1 1 m1 ξ˙12 + m2 (ξ˙2 − ξ˙1 )2 + m3 (ξ˙1 + ξ˙2 )2 2 2 2

+g(m1 − m2 − m3 )ξ1 + g(m2 − m3 )ξ2 ,

(6.6)

where the constant term [gl1 (m2 + m3 ) + m3 gl2 ] has been dropped. There are two Lagrange equations associated with the two generalized coordinates: d  ∂L  ∂L − = 0, (6.7) dt ∂ ξ˙1 ∂ξ1 and

d  ∂L  ∂L − = 0. dt ∂ ξ˙2 ∂ξ2

1

(6.8)

An exact calculation would demand to add the term g(π−2)(m2 r1 +m3 r2 ) + gm2 r2 to the already written expression for the potential energy, appearing with changed sign in the Lagrangian. According to definition of equivalent Lagrangians, this constant term can be omitted. 161

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Free ebooks ==> www.Ebook777.com Performing the derivatives, one easily obtains ξ¨1 (m1 + m2 + m3 ) + ξ¨2 (m3 − m2 ) − g(m1 − m2 − m3 ) = 0,

(6.9)

ξ¨1 (m3 − m2 ) + ξ¨2 (m2 + m3 ) − g(m2 − m3 ) = 0.

(6.10)

We are left with an algebraic system of two equations in two unknowns ξ¨1 and ξ¨2 . Solving the second equation for ξ¨2 in terms of ξ¨1 , and substituting the obtained value into the first equation, one gets  (m2 − m3 )2  (m2 − m3 )2 ¨ ξ1 m1 +m2 +m3 − −g −g(m1 −m2 −m3 ) = 0, m2 + m3 m2 + m3 leading to ξ¨1 =

g

h

(m2 −m3 )2 m2 +m3

+ (m1 − m2 − m3 )

m1 + m2 + m3 −

(m2 −m3 )2 m2 +m3

i

=g

m1 (m2 + m3 ) − 4m2 m3 . m1 (m2 + m3 ) + 4m2 m3 (6.11)

Then, m2 − m3 m2 − m3  m1 (m2 + m3 ) − 4m2 m3  ξ¨2 = (g + ξ¨1 ) = g 1+ m2 + m3 m2 + m3 m1 (m2 + m3 ) + 4m2 m3 =g

2m1 (m2 − m3 ) . m1 (m2 + m3 ) + 4m2 m3

(6.11′ )

Therefore, the accelerations of the three bodies are: m1 (m2 + m3 ) − 4m2 m3 ; a1 = ξ¨1 = g m1 (m2 + m3 ) + 4m2 m3 a2 = ξ¨2 = g

2m1 (m2 − m3 ) ; m1 (m2 + m3 ) + 4m2 m3

a3 = −a2 = g

(6.12)

2m1 (m3 − m2 ) . m1 (m2 + m3 ) + 4m2 m3

At the end of this investigation we mention that a1 = ξ¨1 is determined with respect to the inertial frame xOy, while a2 = −a3 = ξ¨2 is calculated relative to the pulley of radius r2 , which is a non-inertial frame.

162

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Free ebooks ==> www.Ebook777.com 3. Pendulum with horizontally oscillating point of suspension The point of suspension of a gravitational pendulum of length l and mass m performs a horizontal motion, oscillating according to the law X = a cos ω0 t (see Fig.VI.3). Determine the equations of motion of the system, in both an inertial (IRF) and a non-inertial (NIRF) reference frames. Solution I. Inertial reference frame (IRF). Let us denote by θ the angle between the (ideal) rod and the vertical line, and choose the axes of the inertial frame (fixed with respect to the laboratory frame) as shown in Fig.VI.3. The oscillations of the point O obey the law xO = X = a cos ω0 t.

Fig.VI.3 There are two constraints, given by the equations i) f1 (x, y) = (x − X)2 + y 2 − l2 = 0, ii) f2 (z) = z = 0. The system has 3 − 2 = 1 degree of freedom. Let θ be the generalized coordinate. Then, we have  x = X + l sin θ = a cos ω0 t + l sin θ; y = l cos θ, and



x˙ = −aω0 sin ω0 t + lθ˙ cos θ; y˙ = −lθ˙ sin θ,

The kinetic energy then writes T =

1 1 m(x˙ 2 + y˙ 2 ) = m(a2 ω02 sin2 ω0 t − 2alω0 θ˙ sin ω0 t cos θ + l2 θ˙2 ). 2 2 163

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Free ebooks ==> www.Ebook777.com In its turn, the potential (gravitational) energy is obtained by means of the standard procedure ~ · d~r = −mg dy. dV = −dA = −G If we choose V (y = 0) = 0, the last relation yields V = −mgy = −mgl cos θ. The Lagrangian therefore is 1 m(a2 ω02 sin2 ω0 t−2alω0 θ˙ sin ω0 t cos θ+l2 θ˙2 )+mgl cos θ. 2 (6.13) 2 1 2 2 Since the term 2 ma ω0 sin ω0 t can be written as

L = T −V =

1 d t sin 2ω0 t  1 ma2 ω02 sin2 ω0 t = ma2 ω02 − 2 2 dt 2 4ω0 i dF (t) d h1 2 = ma ω0 (2ω0 t − sin 2ω0 t) = , dt 8 dt

where F (t) = 18 ma2 ω0 (2ω0 t − sin 2ω0 t) is a function of time only, this term can be omitted and the equivalent Lagrangian writes L=

1 m(l2 θ˙2 − 2alω0 θ˙ sin ω0 t cos θ) + mgl cos θ. 2

Performing the calculations demanded by the Lagrange equation d  ∂L  ∂L = 0, − dt ∂ θ˙ ∂θ

(6.14)

we finally arrive at the desired equation lθ¨ − aω02 cos ω0 t cos θ + g sin θ = 0.

(6.15)

II. Non-inertial reference frame (NIRF). First of all, we shall show that in a NIRF the general form of the Lagrangian writes L=

1 1 m|~vr |2 + m|~ ω ×~r ′ |2 + m~vr · (~ ω ×~r ′ ) − m~a0 ·~r ′ − V (~r ′ ), (6.16) 2 2

where the significance of the quantities ~vr , ω ~ , ~r ′ , ~a0 and V (~r ′ ) shall be explained later on in this application. 164

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Free ebooks ==> www.Ebook777.com Let us consider a particle of mass m and report its motion relative to two frames S(Oxyz) and S ′ (O′ x′ y ′ z ′ ), the first being inertial (e.g. fixed with respect to the Earth), and the second non-inertial (engaged in an accelerated motion with respect to S). Any motion in the Universe can be considered as a ”composite” motion. The motion of the particle with respect to S is called absolute, and the motion of the same particle with respect to S ′ is called relative. If the particle is fixed with respect to S ′ , then the motion of S ′ relative to S is named transport motion. As an intuitive example one can consider the motion of a car with respect to the Earth (NIRF), the last one being engaged in its motion around the Sun (IRF). To find the Lagrangian (6.16) we shall study the motion of a particle P of mass m relative to both frames S and S ′ (see Fig.VI.4).

Fig.VI.4 Since ~r = ~r0 + ~r ′ , the time derivative of this relation (in Newtonian mechanics, time intervals are the same in any reference frame) yields ~v = ~r˙ = ~r˙ 0 + ~r˙ ′ .

(6.17)

Here ~v = ~r˙ is called absolute velocity. Denoting ~i ′ = ~u ′ 1 , ~j ′ = ~u ′ 2 , ~k ′ = ~u ′ 3 , we can write ~r ′ = x ′k ~u ′k , 165

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(6.18)

Free ebooks ==> www.Ebook777.com where Einstein’s summation convention has been used. Therefore ~r˙ ′ = x˙ ′k ~u ′k + x′k ~u˙ ′k = ~vr + x′k ~u˙ ′k .

(6.19)

Here ~vr = x˙ ′k ~u ′k is called relative velocity. On the other hand, let ωk′ be the components of ~u˙ ′k in the orthonormal basis ~u ′k , i.e. ′ ~u˙ ′k = ωks ~u ′s .

(6.20)

The orthogonality condition ~u ′k · ~u ′s = δks then yields d ′ ′ ′ (~u · ~u ′s ) = ~u˙ ′k · ~u ′s + ~u ′k · ~u˙ ′s = ωks + ωsk = 0, dt k ′ (k, s = 1, 3) are the components of an antisymmetric showing that ωks ′ , second rank tensor. If ω ~ (ωk′ ) is the axial vector associated with ωsk then we can write ′ = εksi ωi′ ωks

(i, k, s = 1, 3),

(6.21)

where εksi is the Levi-Civita permutation symbol. Then ′ ~ × ~u ′k . ~u ′s = εksi ωi′ ~u ′s = ωi′ ~u ′i × ~u ′k = ω ~u˙ ′k = ωks

(6.22)

By means of (6.19) and (6.22), the absolute velocity given by (6.17) writes ~ × ~u ′k ~v = ~v0 + ~vr + x′k ~u˙ ′k = ~v0 + ~vr + x′k ω = ~v0 + ~vr + ω ~ × ~r ′ ,

(6.23)

where ~v0 = ~r˙ 0 is the velocity of O′ relative to O. The absolute acceleration is obtained by taking the time derivative of (6.23). Recalling that ~vr = x˙ ′k ~u ′k , we have: ~a =

d~v = ~v˙ 0 + x ¨′k ~u ′k + x˙ ′k ~u˙ ′k + ω ~˙ × ~r ′ + ω ~ × ~r˙ ′ , dt

or, in view of (6.19) and (6.22), ~a = ~a0 + ~ar + ω ~˙ × ~r ′ + [x˙ ′k ω ~ × ~u ′k + ω ~ × (~vr + x′k ~u˙ ′k )] = ~a0 + ~ar + ω ~˙ × ~r ′ + 2~ ω × ~vr + ω ~ × (~ ω × ~r ′ ), 166

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(6.24)

Free ebooks ==> www.Ebook777.com where ~v˙ 0 = ~a0 is the acceleration of O′ with respect to O, and x ¨′ ~u ′k = ~ar the acceleration of the particle with respect to O′ , called relative acceleration. The term ω ~ × (~ ω × ~r ′ ) = ~acp is named centripetal acceleration, while 2~ ω × ~vr = ~ac is the Coriolis acceleration. If the particle is invariably attached to S ′ (in other words, the particle itself is a non-inertial frame), then ~vr = 0, ~ar = 0, and the last two relations yield ~v = ~vtr = ~v0 + ω ~ × ~r ′ ;

(6.25)

~a = ~atr = ~a0 + ω ~˙ × ~r ′ + ω ~ × (~ ω × ~r ′ ).

(6.26)

Here ~vtr and ~atr are called transport velocity and transport acceleration, respectively. If O ≡ O′ , then ~r0 = 0, ~v0 = 0, ~a0 = 0, ~r ′ = ~r, and we have ~v = ω ~ × ~r, (6.27) ~a = ω ~˙ × ~r + ω ~ × (~ ω × ~r).

(6.28)

With these notations, the absolute velocity and absolute acceleration can also be written as

Fig.VI.5 167

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Free ebooks ==> www.Ebook777.com ~v = ~vr + ~vtr ,

(6.29)

~a = ~ar + ~ac + ~atr .

(6.30)

To determine the physical significance of the pseudovector ω ~ , let us consider a special case: O ≡ O′ , and Oz ≡ Oz ′ as a fixed axis ˙ of rotation. Then (6.22) yields ~k = ω ~ × ~k = 0, meaning that ω ~ and ~ the axis of rotation are collinear: ω ~ = ω k. On the other hand, (6.29) shows that the velocity ~v of the particle is orthogonal to the plane defined by ω ~ and ~r (see Fig.VI.5), its modulus being v = |~v | = |~ ω × ~r| = ωr sin α = ωR. As well-known, the velocity of a point engaged in a uniform circular motion is v = ϕR ˙ = ωR, where ϕ˙ = ω is the angular velocity. The last two relations show that ω ~ =ω ~ (t) is a pseudovector oriented along the axis of rotation, its magnitude being equal to the angular velocity ϕ. ˙ It is called instantaneous vector of rotation. Let us now turn back to our problem. In the IRF denoted by S, the fundamental equation of motion writes F~ = m~a,

(6.31)

where F~ = −∇V (~r) is the conservative force acting upon the particle. The Lagrangian in S then is L=T −V =

1 m|~v |2 − V (~r). 2

(6.32)

To write the equation of motion of the particle in the non-inertial frame S ′ , we have to express the Lagrangian L in terms of x′i and x˙ ′i (i = 1, 3). In view of (6.23), we have |~v |2 = (~v0 + ~vr + ω ~ × ~r ′ ) · (~v0 + ~vr + ω ~ × ~r ′ ) = |~v0 |2 + |~vr |2 + |~ ω × ~r ′ |2

+2~v0 · ~vr + 2~v0 · (~ ω × ~r ′ ) + 2~vr · (~ ω × ~r ′ ).

In the non-inertial frame S ′ our Lagrangian L then writes 1 h L = m |~v0 |2 + |~vr |2 + |~ ω × ~r ′ |2 2 168

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Free ebooks ==> www.Ebook777.com i +2~v0 · ~vr + 2~v0 · (~ ω × ~r ′ ) + 2~vr · (~ ω × ~r ′ ) − V (~r ′ ).

The terms containing ~v0 can be transformed as follows:

|~v0 |2 + 2~v0 · ~vr + 2~v0 · (~ ω × ~r ′ ) = |~v0 |2 + 2~v0 · (~vr + ω ~ × ~r ′ ) = |~v0 |2 + 2~v0 · (~v − ~v0 ) = ~v0 · (2~v − ~v0 ) = ~v0 ·

d (2~r − ~r0 ) dt

d d [2(~r0 + ~r ′ ) − ~r0 ] = ~v0 · (~r0 + 2~r ′ ) dt dt h i d = ~v0 · (~r0 + 2~r ′ ) − ~a0 · (~r0 + 2~r ′ ) dt i dh = ~v0 · (~r0 + 2~r ′ ) − ~a0 · ~r0 − 2~a0 · ~r ′ . dt

= ~v0 ·

Since the expression ~v0 · (~r0 + 2~r ′ ) depends on the coordinates and/or the time only (it cannot depend on velocity), while the quantity ~a0 · ~r0 is also a function which depends only on the time (it can be written as a total derivative with respect to time of some function of time), h i d ′ the terms dt ~v0 · (~r0 + 2~r ) and (−~a0 · ~r0 ) can be omitted in L and we are left with L=

1 1 m|~vr |2 + m|~ ω ×~r ′ |2 + m~vr · (~ ω ×~r ′ ) − m~a0 ·~r ′ − V (~r ′ ), (6.33) 2 2

which is precisely (6.16). Therefore, the proof is complete. Let us now go further and use the Lagrangian formalism in order to write the equation of motion of the particle in a NIRF. To this end, it is more convenient to use vectors components in the Lagrangian (6.33), that is L=

1 1 mx˙ ′i x˙ ′i + mεijk εinl ωj′ ωn′ x′k x′l + mεijk x˙ ′i ωj′ x′k − ma′0i x′i − V (x′k ) 2 2 =

1 1 mx˙ ′i x˙ ′i + m(δjn δkl 2 2

−δjl δkn )ωj′ ωn′ x′k x′l + mεijk x˙ ′i ωj′ x′k − ma′0i x′i − V (x′k ) =

1 1 mx˙ ′i x˙ ′i + mωj′ ωj′ x′k x′k 2 2

1 − mωj′ ωk′ x′k x′j + mεijk x˙ ′i ωj′ x′k − ma′0i x′i − V (x′k ) 2 169

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Free ebooks ==> www.Ebook777.com =

1 1 mx˙ ′i x˙ ′i + mωk′ ωk′ x′i x′i 2 2

1 − m(x′i ωi′ )(x′k ωk′ ) + mεijk x˙ ′i ωj′ x′k − ma′0i x′i − V (x′k ). 2

(6.34)

We have: ∂L = mx˙ ′i δis + mεsjk ωj′ x′k = mx˙ ′s + mεsjk ωj′ x′k ; ∂ x˙ ′s d  ∂L  = m¨ x′s + mεsjk ω˙ j′ x′k + mεsjk ωj′ x˙ ′k ; dt ∂ x˙ ′s ∂L 1 = mωk′ ωk′ x′i δis − mωi′ (x′k ωk′ )δis ′ ∂xs 2

∂V 1 − mωk′ (x′i ωi′ )δks + mεijs x˙ ′i ωj′ − ma′0s − 2 ∂x′s = mωk′ ωk′ x′s − m(x′i ωi′ )ωs′ + mεijs x˙ ′i ωj′ − ma′0s −

∂V . ∂x′s

With these results, Lagrange equations of the second kind ∂L d  ∂L  − = 0 (s = 1, 3) ′ dt ∂ x˙ s ∂x′s

(6.35)

become m¨ x′s + mεsjk ω˙ j′ x′k + mεsjk ωj′ x˙ ′k − mωk′ ωk′ x′s +m(x′i ωi′ )ωs′ − mεsij x˙ ′i ωj′ + ma′0s +

∂V ∂x′s

∂V ~˙ ×~r ′ )s +2m(~ ω ×~vr )s −mωk′ ωk′ x′s +m(x′i ωi′ )ωs′ +ma′0s + ′ = m¨ x′s +m(ω ∂xs ∂V = m¨ x′s +m(ω ~˙ ×~r ′ )s +2m(~ ω ×~vr )s +m[~ ω ×(~ ω ×~r ′ )]s +ma′0s + ′ = 0. ∂xs (6.36) ′ Equation (6.36) is the xs -component of the vector equation m~ar + mω ~˙ × ~r ′ + 2m~ ω × ~vr + m~ ω × (~ ω × ~r ′ ) + m~a0 − F~ = 0, or, in a more eloquent form m~ar = F~ − m~a0 − mω ~˙ × ~r ′ − m~ ω × (~ ω × ~r ′ ) − 2m~ ω × ~vr . 170

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(6.37)

Free ebooks ==> www.Ebook777.com Taking into account (6.26), we still have m~ar = F~ − m~atr − m~ac .

(6.38)

Introducing the notations F~tr = −m~atr ;

F~c = −m~ac ,

(6.39)

we finally obtain the equation of motion of the particle with respect to the non-inertial frame S ′ m~ar = F~ + F~tr + F~c .

(6.40)

As it can be observed, the fundamental equation of motion of the particle does not keep its form when passing from the inertial frame S to the non-inertial frame S ′ . Together with the Newtonian force F~ appear two more forces F~tr and F~c , called inertial or apparent forces. The inertial forces emerge as a result of the motion of S ′ . Indeed, if ~a0 = 0, ω ~ = 0 (that is, if the frame S ′ becomes inertial), then F~tr = 0, F~c = 0, and the equation of motion gets its ”inertial” form m~ar = m~a = F~ . We should mention that, even if the inertial forces do not exist for an observer connected to S, they play the role of real forces for an observer fixed relative to S ′ . In such a frame (non-inertial) the inertial forces can be considered as being produced by some force fields which can be called (by analogy with the gravitational field) inertial force fields. As an example, the gravitational force in an inertial frame appears as an ”effect” of the gravitational field, which is a potential (or, even, conservative) field. This idea lead Einstein to elaborating his general theory of relativity, by postulating the (local) equivalence between the gravitational field and the field of inertial forces. Let us now come back to our problem and choose as the noninertial frame S ′ a coordinate system connected to the point of suspension of the rod of pendulum, O′ (see Fig.VI.6). This point oscillates with respect to O according to the law X = a cos ω0 t, with acceleration ¨ = −aω 2 cos ω0 t. As one can see, in our case ω a0 = |~a0 | = X ~ = 0, ~v0 = 0 2 ′ (−aω0 sin ω0 t, 0, 0), ~a0 = (−aω0 cos ω0 t, 0, 0), ~r = (l sin θ, l cos θ, 0), and the Lagrangian (6.33) writes LN IRF ≡ L′ = =

1 m|~vr |2 − m~a0 · ~r ′ − V (~r ′ ) 2

1 m|~v − ~v0 |2 − m~a0 · ~r ′ − V (~r ′ ). 2 171

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(6.41)

Free ebooks ==> www.Ebook777.com Since

~v − ~v0 = (x˙ − v0x , y, ˙ 0) = (lθ˙ cos θ, −lθ˙ sin θ, 0)

and V (~r ′ ) = V (y ′ ) = V (y) = −mgl cos θ, we still have LN IRF ≡ L′ =

1 2 ˙2 ml θ + malω02 cos ω0 t sin θ + mgl cos θ. 2

(6.42)

Fig.VI.6 The Lagrange equation of the second kind, written for L′ ,

gives

d  ∂L′  ∂L′ − = 0, dt ∂ θ˙ ∂θ

(6.43)

lθ¨ − aω02 cos ω0 t cos θ + g sin θ = 0,

(6.44)

which is precisely equation (6.15) obtained while working in an inertial reference frame. 4. Problem of two identical coupled pendulums Consider two identical simple pendulums, each of them with mass m and length l, connected by a spring of negligible mass. The spring is not tense when the pendulums are in equilibrium (x1 = x2 = 0 − see Fig.VI.7). The reader is asked to determine: a) Proper frequencies of the system; b) Solutions of the equations of motion, if at the initial moment t0 = 0, x1 (t0 ) = a, x2 (t0 ) = 0, and x˙ 1 (t0 ) = 0, x˙ 2 (t0 ) = 0. 172

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Free ebooks ==> www.Ebook777.com Solution a) Recalling that both gravitational and elastic forces are potential forces, we deal with a natural system. According to the notations used in Fig.VI.7, the equations of constraints are: i) f1 (z1 ) = z1 = 0; ii) f2 (x1 , y1 ) = x21 + y12 − l2 = 0; iii) f3 (z2 ) = z2 = 0; iv) f4 (x2 , y2 ) = x22 + y22 − l2 = 0, where each pendulum is reported to its own reference system. The x-axis is common for both systems, while the origins are taken in the two points of suspension. There is no restriction if we choose z = 0 as the plane of motion. For sufficiently small oscillations, the variation of coordinate y of the particles can be neglected, so that the motion can be considered as a unidimensional motion along the x-axis.

Fig.VI.7 Let x1 and x2 be the generalized coordinates associated with the 3 · 2 − 4 = 2 degrees of freedom of the system. The kinetic energy then writes T = T1 + T2 =

1 1 1 1 m|~v1 |2 + m|~v2 |2 = mx˙ 21 + mx˙ 22 . 2 2 2 2

In its turn, the potential energy has two ”components”, one of gravitational nature Vg , and the other Ve due to elastic properties of the spring: V = Vg + Ve . By means of the usual procedure, we have ~ 1 · d~r1 − G ~ 2 · d~r2 = −mgdy1 − mgdy2 , dVg = −dAg = −G or Vg = −mgy1 − mgy2 + Vg0 . 173

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Free ebooks ==> www.Ebook777.com If we choose the plane xOz as a ”reference level” for the gravitational potential energy, then Vg (y1 = 0, y2 = 0) = 0, and we may take the integration constant as Vg0 = 0. This way, q q  2 Vg = −mgl cos θ1 − mgl cos θ2 = −mgl 1 − sin θ1 + 1 − sin2 θ2 = −mgl

r

1−

 x 2 1

l

+

r

1−

 x 2 2

l

!

≃ −2mgl + mg

x21 x2 + mg 2 . 2l 2l

Similarly, dVe = −dAe = −F~e · d~r = −(−k~r · d~r) = kx dx, so that

1 2 kx + Ve0 . 2 Taking the initial state (when the deformation of the spring is zero) as the reference level for Ve , one may choose Ve0 = 0. Using again Fig.VI.7, we then have Ve =

Ve =

1 1 1 2 kx ≡ k(∆x)2 = k(x2 − x1 )2 , 2 2 2

where ∆x is the deformation of the spring. The total potential energy therefore is x2 x2 1 V = mg 1 + mg 2 + k(x2 − x1 )2 , 2l 2l 2 where the constant −2mgl has been dropped. The Lagrangian then writes L=T −V =

1 x2 x2 1 1 mx˙ 21 + mx˙ 22 − mg 1 − mg 2 − k(x2 − x1 )2 . 2 2 2l 2l 2

and Lagrange equations for x1 and x2 d  ∂L  ∂L =0 − dt ∂ x˙ j ∂xj

(j = 1, 2)

yield the system m¨ x1 = −mg

x1 + k(x2 − x1 ), l

(6.45)

m¨ x2 = −mg

x2 − k(x2 − x1 ). l

(6.46)

174

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Free ebooks ==> www.Ebook777.com Assuming that the system performs only small oscillations, we search for solutions of the form x1 = A1 sin ωt + B1 cos ωt,

(6.47)

x2 = A2 sin ωt + B2 cos ωt.

(6.48)

Introducing (6.47) and (6.48) into (6.45) and (6.46), one obtains −mω 2 A1 sin ωt − mω 2 B1 cos ωt = −mg

A1 B1 sin ωt − mg cos ωt l l

+kA2 sin ωt + kB2 cos ωt − kA1 sin ωt − kB1 cos ωt, and −mω 2 A2 sin ωt − mω 2 B2 cos ωt = −mg

B2 A2 sin ωt − mg cos ωt l l

−kA2 sin ωt − kB2 cos ωt + kA1 sin ωt + kB1 cos ωt. Identifying the coefficients of sin ωt and cos ωt of the two members of the above equations, we obtain

and

−mω 2 A1 = −mg

A1 + kA2 − kA1 , l

(6.49)

−mω 2 B1 = −mg

B1 + kB2 − kB1 , l

(6.50)

A2 − kA2 + kA1 , l B2 −mω 2 B2 = −mg − kB2 + kB1 . l −mω 2 A2 = −mg

(6.51) (6.52)

In order that the system (6.49) and (6.51) has at least one nontrivial solution, we must have −mω 2 + mg + k −k l = 0, mg 2 −k −mω + l + k which yields

or



2 mg + k = k2 , − mω + l 2

−mω 2 +

mg + k = ±k. l 175

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Free ebooks ==> www.Ebook777.com We therefore have the following two solutions for ω 2 : 2 ω1,2 =

g k k + ± , l m m

(6.53)

k g +2 , l m

(6.54)

g . l

(6.55)

with 2 ω12 ≡ ωM =

and 2 ω22 ≡ ωm =

The reader can easily prove himself that the use of (6.50) and (6.52) leads to the same result. 2 Substituting ωM in (6.49) by its value given by (6.54), we have 2 −mωM A1 = −mg

A1 + kA2 − kA1 , l

that is A2 = −A1 .

(6.56)

Using the same procedure in equation (6.50), one obtains g g −m B1 − 2kB1 = −m B1 + kB2 − kB1 , l l which means B2 = −B1 .

(6.57)

2 Taking now ωm given by (6.55) and following the same simple calculations, we get A2 = A1 , (6.58)

and B2 = B1 .

(6.59)

Since the system has two frequencies of oscillation, the solutions of the differential equations of motion (6.47) and (6.48) of the two coupled pendulums are written as a superposition of the two modes of oscillation, as x1 = C1 sin ωM t + C2 cos ωM t + C3 sin ωm t + C4 cos ωm t,

(6.60)

and x2 = −C1 sin ωM t − C2 cos ωM t + C3 sin ωm t + C4 cos ωm t, 176

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(6.61)

Free ebooks ==> www.Ebook777.com where the constants of integration Ci (i = 1, 4) are determined by means of initial conditions. b) According to the statement of the problem, x1 = a, x2 = 0 at t = 0. This implies x˙ 1 (t = 0) = a˙ = 0, x˙ 2 (t = 0) = 0. Equations (6.60) and (6.61) yield the following expressions for velocities x˙ 1 = ωM C1 cos ωM t − ωM C2 sin ωM t + ωm C3 cos ωm t − ωm C4 sin ωm t, and x˙ 2 = −ωM C1 cos ωM t+ωM C2 sin ωM t+ωm C3 cos ωm t−ωm C4 sin ωm t. Imposing the initial conditions, we have x1 (t = 0) = a = C2 + C4 ,

(6.62)

x2 (t = 0) = 0 = −C2 + C4 ,

(6.63)

x˙ 1 (t = 0) = 0 = ωM C1 + ωm C3 ,

(6.64)

x˙ 2 (t = 0) = 0 = −ωM C1 + ωm C3 .

(6.65)

It then follows C2 = C4 = a/2, and C1 = C3 = 0, and the solutions become a (6.66) x1 = (cos ωM t + cos ωm t), 2 a x2 = (cos ωm t − cos ωM t), (6.67) 2 or ω − ω  ω + ω  M m M m t cos t x1 = a cos 2 2 ω + ω  M m = Amod (t) cos t, (6.68) 2 ω + ω  ω − ω  M m M m x2 = a sin t sin t 2 2 ω + ω  M m = Bmod (t) sin t, (6.69) 2 where Amod and Bmod denote the amplitude of the two modes of oscillation. As one can see, there appears a phenomenon of beating between the two pendulums. The beating frequency is ωbeat = 2ωmod = ωM − ωm . 177

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(6.70)

Free ebooks ==> www.Ebook777.com 5. Problem of two different coupled pendulums Two simple pendulums of masses m1 , m2 and lengths l1 , l2 are coupled by a spring with spring constant k, mounted at the distance h with respect to their suspension points O1 and O2 (see Fig.VI.8). Supposing that the spring is not tensioned when pendulums are in equilibrium (θ1 = θ2 = 0), write the equations of motion of the system and determine their solutions in case of small oscillations. The motion takes place in a vertical plane.

Fig.VI.8 Solution Since both the gravitational and elastic forces are potential force fields, we have to do with a natural system. The constraints are given by i) f1 (z1 ) = z1 = 0; ii) f2 (x1 , y1 ) = x21 + y12 − l12 = 0; iii) f3 (z2 ) = z2 = 0; iv) f4 (x2 , y2 ) = x22 + y22 − l22 = 0, where with each pendulum is associated a reference system, with origins at the suspension points O1 and O2 , the axis Ox being a common axis. The system has 3 · 2 − 4 = 2 degrees of freedom. Let θ1 and θ2 be the associated generalized coordinates. The two rods are supposed to be ideal (rigid, and massless). Since x1 = l1 sin θ1 ;

y1 = l1 cos θ1 ,

x2 = l2 sin θ2 ;

y2 = l2 cos θ2 ,

178

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Free ebooks ==> www.Ebook777.com the kinetic energy of the system is T = T1 + T2 =

1 1 1 m1 |~v1 |2 + m2 |~v2 |2 = (m1 l12 θ˙12 + m2 l22 θ˙22 ), 2 2 2

while the potential energy writes V = Vg + Ve , where Vg and Ve are the gravitational and elastic potential energies, respectively. To find Vg , we apply the usual method (see previous problems): ~ 1 · d~r1 − G ~ 2 · d~r2 = −m1 gdy1 − m2 gdy2 , dVg = −dAg = −G so that Vg = −m1 gy1 − m2 gy2 + Vg0 . If we choose the plane xOz as a reference level for the gravitational potential energy, then Vg (y1 = 0, y2 = 0) = 0, and we may take the integration constant as Vg0 = 0. Therefore, Vg = −m1 gl1 cos θ1 − m2 gl2 cos θ2 . We still have dVe = −dAe = −F~e · d~r = −(−k~x · d~x) = kx dx so that

1 2 kx + Ve0 . 2 The integration constant Ve0 may be taken as zero, if we conveniently choose the initial state of the spring (x = 0) as the ”reference level” for Ve . Thus, 1 1 Ve = kx2 = (∆x)2 , 2 2 where ∆x is, obviously, the deformation of the spring (see Fig.VI.8). As one observes, Ve =

|∆x| = |x2 − x1 | = h| sin θ2 − sin θ1 |, and we still have Ve =

θ − θ   θ + θ i 1 2h 2 1 2 1 kh 4 sin2 cos2 . 2 2 2 179

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Free ebooks ==> www.Ebook777.com Suppose we restrict the pendulum’s oscillations to small angles (< 4 ). Then we can approximate o

θ − θ  θ − θ  θ − θ 3 θ2 − θ1 2 1 2 1 2 1 sin = +O ≃ ; 2 2 2 2

θ + θ   θ + θ 4 1  θ1 + θ2 2 1 2 1 2 =1− cos +O ≃ 1, 2 2 2 2 and Ve becomes 1 Ve = kh2 (θ2 − θ1 )2 . 2 Using the same approximation for cos θ1 and cos θ2 in Vg , we have  h  θ2 i θ2  Vg = −g m1 l1 1 − 1 + m2 l2 1 − 2 . 2 2 The Lagrangian of the system then writes L=T −V =

 1 m1 l12 θ˙12 + m2 l22 θ˙22 2

 1 1 − g m1 l1 θ12 + m2 l2 θ22 − kh2 θ2 − θ1 )2 , 2 2

(6.71)

where the constant term g(m1 l1 + m2 l2 ) has been dropped. Once the Lagrangian is known, the Lagrange equations of the second kind d  ∂L  ∂L − = 0 (i = 1, 2) dt ∂ θ˙i ∂θi yield

m1 l12 θ¨1 + gm1 l1 θ1 − kh2 (θ2 − θ1 ) = 0;

m2 l22 θ¨2 + gm2 l2 θ2 + kh2 (θ2 − θ1 ) = 0.

(6.72) (6.73)

Let us now turn back to the Lagrangian written for small oscilla√ tions (6.71) and introduce the following substitutions: ξ1 = m1 l1 θ1 , √ √ and ξ2 = m2 l2 θ2 . Obviously, we also have ξ˙1 = m1 l1 θ˙1 , and √ ξ˙2 = m2 l2 θ˙2 . With these notations, the Lagrangian (6.71) writes L=

 ξ 1 ˙2 ˙2  1  g 2 g  1 ξ1 2 2 ξ1 + ξ2 − ξ1 + ξ22 − kh2 − √ √ 2 2 l1 l2 2 l2 m2 l1 m1 =

1 ˙2 ˙2  1 h g kh2  2  g kh2  2 i ξ1 + ξ2 − + ξ + + ξ 2 2 l1 m1 l12 1 l2 m2 l22 2 180

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Free ebooks ==> www.Ebook777.com +

 1  kh2 1 ξ1 ξ2 = ξ˙12 + ξ˙22 − ω12 ξ12 + ω22 ξ22 + κξ1 ξ2 , (6.74) √ l1 l2 m1 m2 2 2

where the following notations have been used: 2 ω1,2 =

κ=

g l1,2

+

kh2 2 ; m1,2 l1,2

kh2 . √ l1 l2 m1 m2

(6.75)

(6.76)

Consequently, the problem reduces to the study of small oscillations of a system described by the Lagrangian (6.74). Let us write this Lagrangian in normal coordinates 1 . To this end, it is convenient to make a change of generalized coordinates, such as (ξ1 , ξ2 ) → (η1 , η2 ),

(6.77)

so that the new Lagrangian remains quadratic in the new generalized velocities, and, in addition, becomes quadratic in the new generalized coordinates (in other words, the coefficient of the mixed term vanishes). This can be accomplished by taking (6.77) as a plane rotation:  ξ1 = η1 cos ϕ − η2 sin ϕ; (6.78) ξ2 = η1 sin ϕ + η2 cos ϕ, where ϕ is a constant. As one can see ξ˙12 + ξ˙22 = η˙ 12 + η˙ 22 , which means that the Lagrangian (6.74) remains quadratic in the new generalized velocities η˙ 1 and η˙ 2 . The other two terms yield: κξ1 ξ2 = κ(η1 cos ϕ − η2 sin ϕ)(η1 sin ϕ + η2 cos ϕ) = κ sin ϕ cos ϕ(η12 − η22 ) + κη1 η2 (cos2 ϕ − sin2 ϕ), and

1

1 − (ω12 ξ12 + ω22 ξ22 ) 2 h 2 2 i 1 = − ω12 η1 cos ϕ − η2 sin ϕ + ω22 η1 sin ϕ + η2 cos ϕ 2

A set of coordinates for a coupled system such that each equation of motion involve only one of these coordinates. 181

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Free ebooks ==> www.Ebook777.com =−

  1h 2 2 η1 ω1 cos2 ϕ + ω22 sin2 ϕ + η22 ω12 sin2 ϕ + ω22 cos2 ϕ 2 i +2η1 η2 sin ϕ cos ϕ ω22 − ω12 .

In order that the Lagrangian becomes quadratic in the new variables, the mixed term must be zero, that is h  i η1 η2 κ cos2 ϕ − sin2 ϕ − sin ϕ cos ϕ ω22 − ω12 = 0,

or

cot 2ϕ =

ω22 − ω12 . 2κ

(6.79)

Therefore, if we choose ϕ according to (6.79), the new generalized coordinates η1 and η2 are normal coordinates, and the Lagrangian (6.74) writes L=

  1h 1 2 η˙ 1 + η˙ 22 − η12 ω12 cos2 ϕ + ω22 sin2 ϕ − 2κ sin ϕ cos ϕ 2 2 i +η22 ω12 sin2 ϕ + ω22 cos2 ϕ + 2κ sin ϕ cos ϕ .

Let us denote

ω 21 = ω12 cos2 ϕ + ω22 sin2 ϕ − 2κ sin ϕ cos ϕ; ω 22 = ω12 sin2 ϕ + ω22 cos2 ϕ + 2κ sin ϕ cos ϕ. Then  1 ω 21 = ω12 + ω22 − 2  1 ω 22 = ω12 + ω22 + 2

s s

 2 − ω2 ω 1 κ2 + 2 ; 4

(6.80)

 2 − ω2 ω 2 1 κ2 + . 4

(6.81)

To obtain (6.80) and (6.81) we used (6.79). Denoting cot 2ϕ =

ω22 − ω12 = a, 2κ

(6.82)

we have 1 − 2 sin2 ϕ = a ⇔ 1 + 4 sin4 ϕ − 4 sin2 ϕ = 4a2 sin2 ϕ(1 − sin2 ϕ), 2 sin ϕ cos ϕ 182

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Free ebooks ==> www.Ebook777.com with the physically acceptable solution sin2 ϕ =

 a 1 1− √ . 2 1 + a2

(6.83)

Fig.VI.9 In this case −2κ sin ϕ cos ϕ = −κ sin 2ϕ = −κ

cos 2ϕ cot 2ϕ

cos2 ϕ − sin2 ϕ κ . = −κ = −√ a 1 + a2 The rest of calculations are very simple and they remain up to the reader. So, the new Lagrangian finally writes L=

 1  1 2 η˙ 1 + η˙ 22 − ω 21 η12 + ω 22 η22 . 2 2

(6.84)

Let us now investigate the dependence of ϕ on the frequencies ω1 and ω2 . Fig.VI.9 shows this dependence for ω1 , and ω2 fixed. As one can see, ϕ varies between 0 and π/2, the interval ∆ω1 corresponding to transition from ϕ ≃ 0 to ϕ ≃ π/2 being of order κ/ω2 . Therefore, the smaller is κ, the narrower is this interval. Fig.VI.9 shows this dependence for three different values of κ, namely κ1 = 0.2, κ2 = 183

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Free ebooks ==> www.Ebook777.com 1.2, κ3 = 3.2. This has been done by means of the Mathematica software package, as follows:

The interval of variation of ϕ should be between 0 and π/2, but practically these values can only be approximated. As an example, take ω2 = 4 rad/s and κ = 0.01 (rad/s)2 and calculate v  u u u1   u  lim ϕ = lim arcsin u  1− r ω1 →0 ω1 →0  t2  

v u u1 = arcsin t 2

1− 184

 ω22 −ω12  2κ   2 2 2   ω −ω 1 + 22κ 1

ω2 p 2 4κ2 +

ω24

!

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Free ebooks ==> www.Ebook777.com s   16 1 1− √ = 0.000625, = arcsin 2 4.10−4 + 44 which can be approximated by zero. In general, no matter what the 2 individual values of ω2 and κ are, but satisfy the condition 4κ → 0, ω24 we have v   u u1 2 ω u  lim ϕ = arcsin t 1 − q 2 ω1 →0 4κ2 2 2 ω2 1 + ω 4 2 2

4κ ω4 2

→0

s   1 ω22 1 − 2 = arcsin 0 = 0. ≃ arcsin 2 ω2 In the same way one can calculate the limit lim ϕ.

ω1 →∞

Simple calculations show that, for the same values of ω2 and κ, this limit is π2 = 1.5708. It can be easily verified that, for ω1 = 87, the exv   u u u  ω 2 −ω 2  1 2 u1  2κ  already has the value π/2. r pression arcsin u 2 1 −   2 t ω 2 −ω 2 1+

2 1 2κ

This value maintains, no matter how much the increase of ω1 is. This result can also be easily verified by means of a straight calculation. In case of a weak coupling (κ ≪ |ω22 −ω12 |), the normal oscillations are ”localized”, that is, for ω1 < ω2 , the relation (6.83) yields

ϕ ω22 −ω12 κ

≫1

v   u u ω22 −ω12 u 1   u  2κ  = arcsin u 1 − r    2  t 2 ω22 −ω12 1+ 2κ ω 2 −ω 2 2

κ

1

v   u u ω22 −ω12 u 1   u  2κ  r ≃ arcsin u 1 −  = arcsin 0 = 0,   2 t 2  ω 2 −ω 2 2

1



185

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≫1

Free ebooks ==> www.Ebook777.com and (6.78) leads to ξ1 ≃ η 1 ;

ξ2 ≃ η 2 .

If ω1 > ω2 , we have: v   u u ω22 −ω12 u 1   u  2κ  r ϕ = arcsin u 1 −    2 t  2  ω22 −ω12 1+ 2κ ≃ arcsin



1=

(6.85)

|ω22 − ω12 | ≫ k ω1 > ω 2

π , 2

and (6.78) yield ξ1 ≃ −η2 ,

ξ2 ≃ η 1 .

(6.86)

In case of a strong coupling, that is |ω22 − ω12 | ≪ κ, the oscillations are not localized. Indeed, v   u u ω22 −ω12 u 1   u  2κ  = arcsin u 1 − r ϕ |ω22 −ω12 |    2  t 2 ≪1 κ ω22 −ω12 1+ 2κ |ω 2 −ω 2 | 2

v   u u ω22 −ω12 u 1   u  2κ  r = arcsin u 1 −    2  t 2 ω22 −ω12 1+ 2κ |ω 2 −ω 2 | 2

κ

1

κ

1

≪1

→0

1 π = arcsin √ = , 4 2

and (6.78) lead to 1 ξ1 ≃ √ (η1 − η2 ); 2

1 ξ2 ≃ √ (η1 + η2 ). 2

(6.87)

Relations (6.80) and (6.81) show the dependence of the normal frequencies on the parameters ω1 , ω2 , and κ. This dependence is graphically illustrated in Fig.VI.10. Here ω2 is fixed (ω2 = 4 rad/s), and the coupling is relatively weak (κ = 8(rad/s)2 ). The graphic representation has been performed using the same software package, by means of the following set of instructions: 186

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Fig.VI.10 As one observes, ω 1 < min(ω1 , ω2 ), and ω 2 > max(ω1 , ω2 ). If 187

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Free ebooks ==> www.Ebook777.com the coupling is weak (κ is small enough), except for the domain of degeneracy |ω22 − ω12 | ≃ κ, the normal frequencies practically coincide with ω1 and ω2 , respectively. Indeed, if κ can be neglected as compared ω 2 −ω 2 to 2 2 1 , then r

κ2 +

(ω22 − ω12 )2 ω 2 − ω12 ≃ 2 , 4 2

and (6.80), (6.81) yield ω 21,2

ω2 ω2 = 1 + 2 ∓ 2 2



ω22 ω2 − 1 2 2



,

namely ω 1 = ω1 , and ω 2 = ω2 . Finally, for frequencies ω1 very small (see further), one of the normal frequencies (namely ω 1 ) becomes imaginary, which means that the system is not stable anymore. Indeed, in order to be imaginary, that is r 2 2 ω ω (ω 2 − ω12 )2 < 0, ω 21 = 1 + 2 − κ2 + 2 2 2 4 we must have

or



ω12 + ω22 2

2

< κ2 +

ω1 <

(ω22 − ω12 )2 , 4

κ , ω2

which explains what we mean by ”very small p frequencies”. This can be easily seen in Fig.VI.11, where ω2 and |ω 21 | = ω 1 are represented in terms of ω1 . As observed, ω 1 presents two branches: the shortest branch, situated on the left from ω1 = ωκ2 appears as an image in a mirror situated on Oω1 axis and face downwards ofh the graphic of i κ 2 2 ω 1 = ω 1 (ω1 ) ”affected by the radical” on the interval 0, ω2 . Let us explain what ”affected by the radical” means. Function ω 1 = ω 1 (ω1 ) cannot have a graphic representation on thepinterval [0, κ/ω2 ] as a real function, because within this interval ω 1 = ω 21 does not exist. In the above defined interval, only ω 21 can be graphically represented. But the mirror image (which exists as an effect of the modulus operation) of the function ω 21 = ω 21 (ω1 ) is not exactly p the branch situated on the left of the value κ/ω2 of the graph ω 1 = |ω 21 |. The difference is due to the radical function. These observations are emphasized by Fig.VI.12, where this difference is even more obvious. 188

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Free ebooks ==> www.Ebook777.com Indeed, the two branches p on the lef the value κ/ω2 of the functions 2 2 ω 1 = ω 1 (ω1 ) and ω 1 = |ω 21 | are not as an object and its image in a horizontal mirror, situated on Oω1 axis.

Fig.VI.11

Fig.VI.12

189

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Free ebooks ==> www.Ebook777.com The law of motion is given by  η1 = A1 cos(ω 1 t + ψ1 ), η2 = A2 cos(ω 2 t + ψ2 ).

(6.88)

Here the arbitrary constants A1 , A2 , ψ1 , ψ2 are determined by means of the initial conditions. In the plane of normal coordinates, which is rotated by the angle ϕ with respect to the plane of the initial coordinates, the trajectory covers densely the rectangle delimited by −A1 ≤ η1 ≤ A1 and −A2 ≤ η2 ≤ A2 (see Fig.VI.13).

Fig.VI.13 If ω 1 and ω 2 are not commensurable, the trajectory is not closed, meaning that the motion is not periodical. Nevertheless, the projection of the representative point on both axes is periodical. But, if the normal frequencies are commensurable, ω1 n1 = ω2 n2

(n1 , n2 ∈ N ),

the trajectories become closed and appear the so-called Lissajous figures. The motion is now periodical, with the period n1 n2 τ = 2π = 2π . (6.89) ω1 ω2 Under these conditions, ξ1 (t) and ξ2 (t) write:  ξ1 (t) = A1 cos(ω 1 t + ψ1 ) cos ϕ − A2 cos(ω 2 t + ψ2 ) sin ϕ, ξ2 (t) = A1 cos(ω 1 t + ψ1 ) sin ϕ − A2 cos(ω 2 t + ψ2 ) cos ϕ. 190

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(6.90)

Free ebooks ==> www.Ebook777.com The trajectory in the plane ξ1 Oξ2 is the same, but the axis (−A1 , A1 ) of the rectangle are rotated by the angle ϕ with respect to Oξ1 axis (see Fig.VI.14). Let us now turn back to our problem and first discuss the case (a) l1 = l2 = l. The relations (6.75) and (6.76) then become: 2 ω1,2 =

κ=

g  h 2 k + ; l l m1,2

 h 2 l



k . m1 m2

(6.75′ ) (6.76′ )

Fig.VI.14 Denoting ω02 =

g l

2 = and ωe1,2

2 ω1,2 = ω02 +

κ=

k m1,2 ,

we still have:

 h 2 l

2 ωe1,2 ,

 h 2

ωe1 ωe2 . l With these notations, the normal frequencies are: ω 2 + ω22 ω 21 = 1 − 2

r

κ2 +

(ω22 − ω12 )2 4

191

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(6.75′′ ) (6.76′′ )

Free ebooks ==> www.Ebook777.com =

2ω02

+

 2 h l

+

2 ωe2 )

2

ω 22 =

and (6.79) yields

cot 2ϕ =

2 (ωe1

ω22

− 2κ



r

 2 h l

2 2 (ωe1 + ωe2 )

2

= ω02 ,

ω12 + ω22 (ω 2 − ω12 )2 + κ2 + 2 2 4  h 2 2 2 = ω02 + (ωe1 + ωe2 ). l

ω12

=

 2

  2 2 (ωe2 − ωe1 ) ωe1 1 ωe2 − . =  2 2 ωe1 ωe2 2 hl ωe1 ωe2

h l

In this case, the condition for weak coupling κ ≪ |ω22 − ω12 | becomes    2 h 2 h 2 2 ωe1 ωe2 ≪ (ωe2 − ωe1 ) , l l that is

2 2 ωe1 ωe2 ≪ |ωe2 − ωe1 |.

(6.91)

As we have seen while discussing the general case, there are two possible situations for the weak coupling: i) ω1 < ω2 which, in our case, means ωe1 < ωe2 , that is m1 > m2 , and (6.91) becomes √

k k k ≪ − , m1 m2 m2 m1

or m1 ≫ m2



m2 3− m1



,

in agreement with our hypothesis m1 > m2 . Consequently, if the coupling between the two pendulums is weak and m1 ≫ m2 , then according to the general analysis ϕ ≃ 0 and oscillations are localized. ii) ω1 > ω2 , meaning ωe1 > ωe2 , that is m1 < m2 . Then (6.91) leads to k k k ≪ − , √ m1 m2 m1 m2 or   m1 m2 ≫ m1 3 − , m2 192

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Free ebooks ==> www.Ebook777.com which emphasizes our condition m1 < m2 . Therefore, if the coupling is weak and m1 ≪ m2 , the general theory says that ϕ ≃ π2 , and oscillations are also localized. In both cases (ϕ ≃ 0 and ϕ ≃ π/2), the pendulum of mass m1 oscillates with normal frequency ω 1 = ω0 , while the other one, as the case may be, with normal frequency

ω2 = ր ց ω2 =

s

s  2  2 h h 2 2 2 2 ω0 + (ωe1 + ωe2 ) ≃ ω0 + ωe2 , (m1 ≫ m2 ), l l

s

s  2  2 h h 2 2 2 2 (ωe1 + ωe2 ) ≃ ω0 + ωe1 , (m1 ≪ m2 ). ω0 + l l

According to our general analysis, we then have: θ1 ≃ θ1 = A1 cos(ω 1 t + ψ1 ) = A1 cos(ω0 t + ψ1 ),

(6.92)

and θ2 ≃ θ2 = A2 cos(ω 2 t + ψ2 ) s

=

ր ց

A2 cos 

ω02 +



h ωe2 l

2



t + ψ2  ,

(m1 ≫ m2 ), (6.93)

s

A2 cos 

ω02 +



h ωe1 l

2



t + ψ2  ,

(m1 ≪ m2 ).

If m1 = m2 = m, then ωe1 = ωe2 = ωe , and   ω22 − ω12 1 ωe2 ωe1 cot 2ϕ = = − = 0, 2κ 2 ωe1 ωe2 that is ϕ = π/4. This means that oscillations are not localized and we have  θ1 = A1 cos(ω 1 t + ψ1 ) − A2 cos(ω 2 t + ψ2 ), (6.94) θ2 = A1 cos(ω 1 t + ψ1 ) + A2 cos(ω 2 t + ψ2 ), 193

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Free ebooks ==> www.Ebook777.com q

2 h ω , e l

q

k with ωe = m . Indeed, where ω 1 = ω0 and ω 2 = +2 the two normal modes of oscillation are associated with the normal coordinates  θ1 = A1 cos(ω 1 t + ψ1 ), θ2 = A2 cos(ω 2 t + ψ2 ),

ω02

while θ1 and θ2 are given by [see (6.78)]  θ1 = θ1 cos ϕ − θ2 sin ϕ, θ2 = θ1 sin ϕ + θ2 cos ϕ, with ϕ = π/4. This investigation leads to (6.94), where A1 =

A1 √ , 2

and

A2 A2 = √ . The constants A1 , A2 , ψ1 and ψ2 are determined by means 2 of the initial conditions. Let us now investigate three situations that are particularly important in this case, as follows: 1) The pendulums are initially displaced on the same side of the vertical axis, θ1 (0) = θ2 (0) = θ0 , and let them oscillate freely, θ˙1 (0) = 0, θ˙2 (0) = 0. Since  θ˙1 = −ω 1 A1 sin(ω 1 t + ψ1 ) + ω 2 A2 sin(ω 2 t + ψ2 ), θ˙2 = −ω 1 A1 sin(ω 1 t + ψ1 ) − ω 2 A2 sin(ω 2 t + ψ2 ),

we have

 θ = A1 cos ψ1 − A2 cos ψ2 ,   0 θ0 = A1 cos ψ1 + A2 cos ψ2 ,   0 = −ω 1 A1 sin ψ1 + ω 2 A2 sin ψ2 , 0 = −ω 1 A1 sin ψ1 − ω 2 A2 sin ψ2 .

Some very simple algebraic manipulation lead to: ψ1 = 0, ψ2 = 0, A1 = θ0 , A2 = 0. With these values of the constants, (6.94) becomes  θ1 = θ0 cos(ω 1 t) = θ0 cos(ω0 t), (6.95) θ2 = θ0 cos(ω 1 t) = θ0 cos(ω0 t), meaning that the two pendulums oscillate in phase, with frequency ω0 . 2) The two pendulums are initially displaced on one side and the other of the vertical axis, θ1 (0) = −θ2 (0) = θ0 , and let them oscillate freely, θ˙1 (0) = 0, θ˙2 (0) = 0. We then have:  +θ = A1 cos ψ1 − A2 cos ψ2 ,   0 −θ0 = A1 cos ψ1 + A2 cos ψ2 ,   0 = −ω 1 A1 sin ψ1 + ω 2 A2 sin ψ2 , 0 = −ω 1 A1 sin ψ1 − ω 2 A2 sin ψ2 , 194

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Free ebooks ==> www.Ebook777.com which yield ψ1 = 0, ψ2 = 0, A1 = 0, A2 = −θ0 , and (6.94) take the form  q  2  h 2  θ = θ0 cos(ω 2 t) = θ0 cos ω0 + 2 l ωe t ,    1 θ2 = −θ0 cos(ω q 2 t) = θ0 cos(ω 2 t + π)     2  h  ω02 + 2 l ωe t + π .  = θ0 cos

(6.96)

This result shows oscillate with the same freq that the pendulums q  2 h 2 quency, ω 2 = ω02 + 2 hl ωe = gl + 2k , but in opposition of m l phase. 3) Only one pendulum is displaced (say, the one denoted by number 1), the other one keeping its equilibrium position, θ1 (0) = θ0 , θ2 (0) = 0, and let them oscillate without initial velocity, θ˙1 (0) = 0, θ˙2 (0) = 0. Then, we have:  θ = A1 cos ψ1 − A2 cos ψ2 ,   0 0 = A1 cos ψ1 + A2 cos ψ2 ,   0 = −ω 1 A1 sin ψ1 + ω 2 A2 sin ψ2 , 0 = −ω 1 A1 sin ψ1 − ω 2 A2 sin ψ2 .

One easily finds ψ1 = 0, ψ2 = 0, A1 = −A2 = θo /2, and (6.94) yield  t)] θ1 = θ20 [cos(ω 1 t) + cos(ω  2q     θ  ω02 + 2  = 0 cos(ω0 t) + cos 2

 θ2 = θ20 [cos(ω 1 t) − cos(ω t)]  2q      = θ0 cos(ω t) − cos ω2 + 2 2

0

0

2 h l ωe t 2 h l ωe t



,



.

By means of the well-known trigonometric relations cos x − cos y = −2 sin

cos x + cos y = 2 cos

x−y x+y sin , 2 2

x−y x+y cos , 2 2

195

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(6.97)

Free ebooks ==> www.Ebook777.com we still have !  r  2   t  θ1 (t) = θ0 cos ωo − ω02 + 2 ωe hl  2    !  r   2    h t 2   2,  × cos ωo + ω0 + 2 ωe l ! r  2   t   θ2 (t) = −θ0 sin ωo − ω02 + 2 ωe hl  2    !  r  2     t h 2   × sin ωo + ω0 + 2 ωe l 2.

(6.98)

Since h < l, we still have r

r  h 2  ω 2  h 2 e ω02 + 2 ωe = ω0 1 + 2 l ω0 l



= ω0 1 +

 ω 2  h 2 e

ω0

l

  ω h 4   ω 2  h 2  e e +O ≃ ω0 1 + . ω0 l ω l 0 ωe www.Ebook777.com

Fig.VI.15 b) l1 6= l2 (e.g. l1 > l2 ). It is convenient for our investigation to find the characteristic equation. To this end, we use the equations (6.72) and (6.73), written as θ¨1 +



g kh2 + l1 m1 l12



θ1 −

kh2 θ2 = 0, m1 l12

(6.72′ )

θ¨2 +



g kh2 + l2 m2 l22



θ2 −

kh2 θ1 = 0. m2 l22

(6.73′ )

We search for solutions of the form θi = Ai sin ωt + Bi cos ωt (i = 1, 2), in order to determine the frequencies ω of the two normal modes of oscillations of the system. Since θ¨i = −ω 2 Ai sin ωt − ω 2 Bi cos ωt

(i = 1, 2),

equations (6.72’) and (6.73’) write 2

2

−ω A1 sin ωt − ω B1 cos ωt +



g kh2 + l1 m1 l12



(A1 sin ωt + B1 cos ωt)

kh2 (A2 sin ωt + B2 cos ωt) = 0, m1 l12   g kh2 2 2 −ω A2 sin ωt − ω B2 cos ωt + + (A2 sin ωt + B2 cos ωt) l2 m2 l22 −

197

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Free ebooks ==> www.Ebook777.com −

kh2 (A1 sin ωt + B1 cos ωt) = 0. m2 l22

Identifying the coefficients of sin ωt and cos ωt, we have 

 and

−ω +



kh2 g + l1 m1 l12



A1 −

kh2 A2 = 0, m1 l12

2



g kh2 + l1 m1 l12



B1 −

kh2 B2 = 0, m1 l12

2



g kh2 + l2 m2 l22



A2 −

kh2 A1 = 0, m2 l22

B2 −

kh2 B1 = 0. m2 l22

−ω +





2

−ω + 2

−ω +



g kh2 + l2 m2 l22



The system of two algebraic equations in A1 and A2 (or B1 and B2 ) has non-trivial solution if the characteristic determinant is zero   −ω 2 + g + kh2 kh2 − m1 l 2 2 l1 m l 1 1 1   = 0, 2 2 − mkh2 l2 −ω 2 + lg2 + mkh2 l2 2

2

which yields the characteristic equation 

2

−ω +



g kh2 + l1 m1 l12

 

2

−ω +



g kh2 + l2 m2 l22



k 2 h4 = 0. − m1 m2 l12 l22 (6.100)

Fig.VI.16 198

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Free ebooks ==> www.Ebook777.com In the plane (k, ω 2 ) this represents a hyperbola whose branches corresponding to the values with physical signification (we must not forget that k ∈ [0, ∞)) are represented in Fig.VI.16. Here the following numerical values have been chosen: l1 = 0.5, l2 = 0.3, m1 = 0.2, m2 = 0.3, h = 0.1, g = 10. By means of notations (6.74) and (6.75), Eq. (6.100) can be written as (ω 2 − ω12 )(ω 2 − ω22 ) − κ2 = 0, with its solutions (already determined by a different procedure, see (6.80) and (6.81)):

ω 21 and

 1 = ω12 + ω22 − 2  2

1 ω 22 = ω12 + ω2 + 2

s s

ω 2 − ω12 + 2 4

2

,

ω 2 − ω12 κ2 + 2 4

2

.

κ2

For k very small (k → 0; the spring is very weak), the normal 2 = frequencies ω 21 and ω 22 tend to the frequencies of free pendulums, ω01 2 g/l1 , and ω02 = g/l2 . Indeed, for k → 0, equation (6.100) becomes 2 2 2 (ω 2 − ω01 )(ω 2 − ω02 ) = 0, with obvious solutions ω 21 = ω01 and ω 22 = 2 ω02 . For k very big (k → ∞; the spring is very strong) the normal frequency ω 22 tends to infinity as  2  2 h h 2 2 ωe1 + ωe2 l1 l2

(6.101)

(asymptote (a) of Fig.VI.16). Indeed, let us take 1 (ω 2 + ω22 ) + ω2 α ≡ lim 2 = lim 2 1 k→∞ k k→∞

= lim

2 ω01 +

k→∞

+ lim

k→∞

s 

h2 l1 l2

2

k2 m1 m2

+

 2 h l1

1 4



k m1

2 ω02

p κ2 + (ω22 − ω12 )2 /4 k

2 + ω02 +

2k +

 2 h l2

 2

k m2

h l2



k m2

2 ω01



k

199

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 2 h l1

k m1

2

Free ebooks ==> www.Ebook777.com  2  2 h h 1 1 = + l1 2m1 l2 2m2

v "  #2 u  2 2 u h2 2 1 h h 1 1 +t + − l1 l2 m1 m2 l2 2m2 l1 2m1

v" #2 u  2  2  2  2 u h h h h 1 1 1 1 + +t + = l1 2m1 l2 2m2 l2 2m2 l1 2m1  2  2 h 1 h 1 = + , l1 m1 l2 m2

and we finally have ω 2 2(k→∞)

 2  2  2  2 h h h k k h 2 2 = kα = + = ωe1 + ωe2 , l1 m1 l2 m2 l1 l2

which is precisely the quantity (6.101). Let us next calculate ω 21 , subject to the same condition k → ∞. Dividing (6.100) by k, then taking the limit, we successively have: 1 k

("

g ω − − l1 2

#" #  2  2 h h k g k 2 ω − − l1 m1 l2 l2 m2

− or



h2 l1 l2

2

) k2 = 0, m1 m2

( # "  2  2 g h h 1 k k g + + + ω4 − ω2 k l1 l2 l1 m1 l2 m2 " # " #  2  2 h k g g h k g g + + + + l1 l2 l2 m2 l2 l1 l1 m1 )  2   2  h k2 h k2 + − = 0, l1 l2 m1 m2 l1 l2 m1 m2

or, still, ω4 ω2 − k k



g g + l1 l2



"  #  2 2 h 1 h 1 − ω2 + l1 m1 l2 m2 200

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Free ebooks ==> www.Ebook777.com  2  2 h g h 2 g2 1 1 + + = 0. l2 m2 l2 l1 m1 k l1 l2

g + l1

For k → ∞, this becomes "  #  2  2  2 2 h h g h g h 1 1 1 1 2 −ω ∞ + + + = 0. l1 m1 l2 m2 l1 l2 m2 l2 l1 m1 (6.102) which yields

ω 2∞

=

g l1

 2 h l2

1 m2

h l1

1 m1

 2

+ +

g l2



 2

h l2

h l1 2

1 m1

=g

1 m2

m1 l1 + m2 l2 2 ≡ ω∞ . m1 l12 + m2 l22

(6.103)

2 Therefore, ω 21 tends to ω∞ given by (6.103) (asymptote (b) shown in Fig.VI.16), when k tends to infinity. This is the frequency of a pendulum with two masses m1 and m2 , situated on the same ideal rod, at the distances l1 and l2 < l1 with respect to the point of suspension O (see Fig.VI.17). Let us now prove, by means of the analytical formalism, that formula (6.103) expresses the frequency of the motion of pendulum represented in Fig.VI.17. Since both bodies are on the same rod, the system has one degree of freedom, and let θ be the associated generalized coordinate. Since

xi = li sin θ;

yi = li cos θ (i = 1, 2),

and x˙ i = li θ˙ cos θ;

y˙ i = −li θ˙ sin θ (i = 1, 2),

the kinetic energy writes T = T1 + T2 =

1 1 1 m1 |~v1 |2 + m2 |~v2 |2 = θ˙2 (m1 l12 + m2 l22 ), (6.104) 2 2 2

while the potential energy is obtained by the usual procedure ~ 1 · d~r1 − G ~ 2 · d~r2 = −m1 g dy1 − m2 g dy2 . dV = −dA = −G 201

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Fig.VI.17 Choosing as the reference level the plane xOz, meaning y1 = 0, y2 = 0, the constant of integration can be assumed to be zero. Then, V = −m1 gy1 − m2 gy2 = −g cos θ(m1 l1 + m2 l2 ),

(6.105)

and the Lagrangian writes L=T −V =

1 ˙2 θ (m1 l12 + m2 l22 ) + g(m1 l1 + m2 l2 ) cos θ. 2

The Lagrange equation of the second kind   d ∂L ∂L − = 0, ˙ dt ∂ θ ∂θ

(6.106)

(6.107)

then easily yields the differential equation of motion ¨ 1 l2 + m2 l2 ) + g(m1 l1 + m2 l2 ) sin θ = 0. θ(m 1 2

(6.108)

In the limit of small oscillations (sin θ ≃ θ), if one denotes Ω=

s

g

m1 l1 + m2 l2 (≡ ω∞ ), m1 l12 + m2 l22

(6.109)

then equation (6.108) takes the expected form θ¨ + Ω2 θ = 0. 202

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(6.110)

Free ebooks ==> www.Ebook777.com Consequently, our system behaves like a simple pendulum, the oscillation frequency being given by (6.109). 6. Problem of three identical coupled pendulums Consider three identical simple pendulums, of mass m and length l, connected by two springs of negligible mass, situated at distance h with respect to the point of suspension (see Fig.VI.18). The springs are not tensioned at the state of equilibrium (θ1 = θ2 = θ3 = 0). Write the equations of motion of the system and find the solutions for the case of small oscillations.

Fig.VI.18 Solution Since both gravitational and elastic force fields are conservative, here we have again a natural system. Supposing that the motion takes place in the xOy plane, the system is subject to the following constraints: i) f1 (z1 ) = z1 = 0, ii) f2 (x1 , y1 ) = x21 + y12 − l2 = 0, iii) f3 (z2 ) = z2 = 0, iv) f4 (x2 , y2 ) = x22 + y22 − l2 = 0, v) f5 (z3 ) = z3 = 0, vi) f6 (x3 , y3 ) = x23 + y32 − l2 = 0, where each pendulum is reported to its own reference frame, with the common x-axis. The system has 3 · 3 − 6 = 3 degrees of freedom, and let θ1 , θ2 , θ3 (see Fig.VI.18) be the associated generalized coordinates. 203

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Free ebooks ==> www.Ebook777.com The kinetic energy is T = T1 + T2 + T3 = =

1 1 1 m|~v1 |2 + m|~v2 |2 + m|~v3 |2 2 2 2

1 1 1 m(x˙ 21 + y˙ 12 ) + m(x˙ 22 + y˙ 22 ) + m(x˙ 23 + y˙ 32 ). 2 2 2

Since xi = l sin θi , yi = l cos θi , (i = 1, 3), we still have T =

1 2 ˙2 ˙2 ˙2 ml (θ1 + θ2 + θ3 ). 2

In its turn, the potential energy has two components, of gravitational and elastic nature: V = Vg + Ve . The gravitational energy is obtained by the usual procedure ~ 1 · d~r1 − G ~ 2 · d~r2 − G ~ 3 · d~r3 = −mg(dy1 + dy2 + dy3 ), dVg = −dAg = −G and, by integration Vg = −mg(y1 + y2 + y3 ) + Vg0 . A convenient choice of the reference frame Vg (y1 = 0, y2 = 0, y3 = 0) = 0, allows us to take Vg0 = 0. Then, V = −mgl(cos θ1 + cos θ2 + cos θ3 ). In its turn, the elementary elastic energy dVe = −dAe = −F~e · d~r = kx dx, leads to

1 2 kx + Ve0 . 2 Taking as the ”reference level” for Ve the initial state when the deformation of the two strings is zero, Ve (x = 0) = 0, the integration constant Ve0 can also be taken to be zero. Under these assumptions, the variables x1 and x2 stand for the deformations ∆x1 and ∆x2 of the springs, and we are left with Ve =

Ve = Ve1 + Ve2 =

1 2 1 2 1 1 kx1 + kx2 = k(∆x1 )2 + k(∆x2 )2 . 2 2 2 2 204

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Free ebooks ==> www.Ebook777.com According to Fig.VI.18, we can write ∆x1 = x2 − x1 = h(sin θ2 − sin θ1 ); ∆x2 = x3 − x2 = h(sin θ3 − sin θ2 ). Since sin a − sin b = 2 sin



a−b 2



cos



a+b 2



,

we still have      θ2 − θ1 1 2 θ1 + θ2 2 2 Ve = kh 4 sin cos 2 2 2      1 2 θ3 − θ2 θ2 + θ3 2 2 + kh 4 sin cos . 2 2 2

Using the hypothesis of small oscillations, we can write 

sin

cos



θ2 − θ1 2

θ1 + θ2 2





θ2 − θ1 +O = 2

1 =1− 2





θ1 + θ2 2

θ2 − θ1 2

2

+O

3





θ2 − θ1 , 2

θ1 + θ2 2

4

≃ 1,

and, similarly, sin



θ3 − θ2 2



θ3 − θ2 ; ≃ 2

cos



θ2 + θ3 2



≃ 1.

Under these assumptions, Ve becomes Ve =

i 1 2h kh (θ2 − θ1 )2 + (θ3 − θ2 )2 , 2

and the total potential energy writes

V = Vg + Ve = −mgl(cos θ1 + cos θ2 + cos θ3 )

h i 1 + kh2 (θ2 − θ1 )2 + (θ3 − θ2 )2 . 2 The Lagrangian of the system then is L=T −V =

1 2 ˙2 ˙2 ˙2 ml (θ1 + θ2 + θ3 ) + mgl(cos θ1 + cos θ2 + cos θ3 ) 2 205

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Free ebooks ==> www.Ebook777.com h i 1 − kh2 (θ2 − θ1 )2 + (θ3 − θ2 )2 , 2 and Lagrange equations of the second kind   d ∂L ∂L − = 0 (i = 1, 3) ˙ dt ∂ θi ∂θi yield

ml2 θ¨1 + mgl sin θ1 − kh2 (θ2 − θ1 ) = 0,

ml2 θ¨2 + mgl sin θ2 + 2kh2 θ2 − kh2 (θ1 + θ3 ) = 0, ml2 θ¨3 + mgl sin θ3 + kh2 (θ3 − θ2 ) = 0,

or, if the oscillations are presumably small (sin θi ≃ θi , i = 1, 3), θ¨1 +

g kh2 + l ml2



θ1 −

kh2 θ2 = 0, ml2

 g 2kh2 kh2 + (θ1 + θ3 ) = 0, θ − 2 l ml2 ml2   2 g kh kh2 ¨ θ3 + + θ2 = 0. θ3 − l ml2 ml2

θ¨2 +





(6.111)

(6.112) (6.113)

As usual in the case of small oscillations, we are looking for solutions of the form θi = Ai sin ωt + Bi cos ωt (i = 1, 3).

(6.114)

Introducing these solutions into (6.111)-(6.113), then identifying the coefficients of sin ωt and cos ωt in each equation, one obtains the following algebraic system for the coefficients Ai , Bi (i = 1, 3) : 



g kh2 −ω + + l ml2 2



A1 −

kh2 A2 = 0, ml2

 g kh2 kh2 B − B2 = 0, −ω + + 1 l ml2 ml2   kh2 g 2kh2 kh2 2 − 2 A1 + −ω + + A − A3 = 0, 2 ml l ml2 ml2   kh2 g 2kh2 kh2 2 − 2 B1 + −ω + + B − B3 = 0, 2 ml l ml2 ml2 2

206

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(6.115)

(6.116) (6.117) (6.118)

Free ebooks ==> www.Ebook777.com  kh2 − 2 A2 + −ω 2 + ml  kh2 − 2 B2 + −ω 2 + ml

g kh2 + l ml2 g kh2 + l ml2

 

A3 = 0,

(6.119)

B3 = 0.

(6.120)

To have at least one non-trivial solution, the system of equations (6.110), (6.117), and (6.119) must obey the condition kh2 −ω 2 + g + kh2 − 0 l ml2 ml2 g kh2 2kh2 kh2 2 = 0, − ml2 −ω + l + ml2 − ml2 2 2 kh kh 0 − ml −ω 2 + gl + ml 2 2 or



kh2 g −ω 2 + + l ml2

 "

g 2kh2 −ω + + l ml2 2

−2



kh2 ml2

2 #

  g kh2 2 −ω + + l ml2

= 0.

This equation is satisfied if −ω 2 + with the solution ωI2

kh2 g + =0 l ml2

g kh2 = + , l ml2

(6.121)

or if 

g 2kh2 −ω + + l ml2 2

   2 2 g kh2 kh 2 , −ω + + = 2 l ml2 ml2

which can also be written as 4

ω − 2ω +



g kh2 + l ml2



2



g 3kh2 + l 2ml2







g 2kh2 + l ml2

with the solutions 2 ωII,III =

−2

kh2 ml2

2

= 0,

g 3kh2 + l 2ml2

207

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Free ebooks ==> www.Ebook777.com ±

s

g 3kh2 + l 2ml2

2

− =





g kh2 + l ml2

g 2kh2 + l ml2



+2



kh2 ml2

2

g 3kh2 3kh2 ± , + l 2ml2 2ml2

that is 2 2 ωII ≡ ωM =

g 3kh2 , + l ml2

(6.122)

and

g . (6.123) l The reader is advised to prove by himself that the remaining equations (6.116), (6,118), and (6.120) in the unknowns B1 , B2 , and B3 lead to the same frequencies. Changing the solution notations, the three normal frequencies of the system of three coupled pendulums are r r r g g g 3kh2 kh2 , ω = . (6.124) ω1 = , ω2 = + + 3 l l ml2 l ml2 2 2 ωIII ≡ ωm =

Introducing ω12 into (6.115), we obtain A2 = A1 .

(6.125)

Using the same procedure in (6.116), we arrive at B2 = B1 .

(6.126)

By means of (6.125) and (6.117), with ω12 given by (6.124), we arrive at A3 = A2 = A1 . (6.127) In the same way, making use of (6.118), (6.124), and (6.126), we have B3 = B2 = B1 .

(6.128)

Resuming the same operation, but this time with ω22 given by (6.124), and introducing it into (6.115), we obtain A2 = 0.

(6.129)

Next, introduce ω22 into (6.116) and get B2 = 0, 208

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(6.130)

Free ebooks ==> www.Ebook777.com while (6.117), in view of (6.129), gives A3 = −A1 ,

(6.131)

and (6.118), by means of (6.130), yields B3 = −B1 .

(6.132)

Finally, using ω32 displayed by (6.124), and the relations (6.115), (6.116), and (6.117) as well, we successively obtain A2 = −2A1 ;

(6.133)

B2 = −2B1 ;

(6.134)

A3 = A1 .

(6.135)

Using (6.134) and (6.118), one also obtains B3 = B1 .

(6.136)

The remaining relations (6.119) and (6.120) are identically satisfied by all three solutions, and we leave this up to the reader. Since the system has three oscillation frequencies, the solutions of the differential equations of motion of the three coupled pendulums are written as a superposition of the three modes of oscillations. In view if (6.114) and (6.125)-(6.136), we then have:

or

 θ1 = A sin ω1 t + B cos ω1 t + C sin ω2 t + D cos ω2 t     +E sin ω3 t + F cos ω3 t, θ2 = A sin ω1 t + B cos ω1 t − 2E sin ω3 t − 2F cos ω3 t,   θ   3 = θ1 = A sin ω1 t + B cos ω1 t − C sin ω2 t − D cos ω2 t +E sin ω3 t + F cos ω3 t,  θ1 = K1 sin(ω1 t + α1 ) + K2 sin(ω2 t + α2 )    +K3 sin(ω3 t + α3 ),  θ2 = K1 sin(ω1 t + α1 ) − 2K3 sin(ω3 t + α3 ),   θ   3 = K1 sin(ω1 t + α1 ) − K2 sin(ω2 t + α2 ) +K3 sin(ω3 t + α3 ),

or, still, in a matrix form     r  θ1 +1 g  θ2  = K1  +1  sin t + α1 l θ3 +1 209

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(6.137)

(6.138)

Free ebooks ==> www.Ebook777.com 

 ! r +1 2 g kh +K2  0  sin + t + α2 l ml2 −1   ! r +1 2 g 3kh +K3  −2  sin t + α3 , + l ml2 +1

(6.139)

where the integration constants A, B, C, D, E, F as well as Ki , αi , (i = 1, 3) are determined by means of the initial conditions: θi (t = 0) = θ0i ,

θ˙i (t = 0) = θ˙0i

(i = 1, 3).

(6.140)

7. Problem of double gravitational pendulum Study the motion of a double gravitational pendulum composed by two bodies (particles) of masses m1 , m2 , and lengths of the rods l1 , l2 (Fig.VI.19). The rods are supposed to be massless and the connections frictionless. Solution Let us first identify the constraints acting on the system. Since the motion takes place in a plane, say xOy, we can write: i) f1 (z1 ) = z1 = 0, ii) f2 (z2 ) = z2 = 0, iii) f3 (x1 , y1 ) = x21 + y12 − l12 = 0, iv) f4 (x1 , x2 , y1 , y2 ) = (x2 − x1 )2 + (y2 − y1 )2 − l22 = 0. Therefore, the system is submitted to four holonomic (bilateral, scleronomous, and finite) constraints, and has 3 · 2 − 4 = 2 degrees of freedom. Suppose that θ1 and θ2 are the generalized coordinates associated with the two degrees of freedom (see Fig.VI.19). Since the only applied forces are the gravitational forces, this is a natural system. To write the Lagrangian, we need to determine its potential and kinetic energies. Using the same procedure as in the case of the simple pendulum, we can write ~ 1 · d~r1 − G ~ 2 · d~r2 = −m1 g dx1 − m2 g dx2 , dV = −dA = −G so that V = −g(m1 x1 + m2 x2 ) + V0 . 210

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Free ebooks ==> www.Ebook777.com The integration constant V0 can be conveniently chosen by defining the reference level for the potential energy. In our case, if this level corresponds to the plane yOz, we may take V0 = 0. Indeed, V (x1 = 0, x2 = 0) = 0 ⇒ V0 = 0. Since x1 = l1 cos θ1 , x2 = l1 cos θ1 + l2 cos θ2 , we have V = −g(m1 x1 +m2 x2 ) = −gl1 (m1 +m2 ) cos θ1 −m2 gl2 cos θ2 . (6.141)

Fig.VI.19 We also observe that x1 = l1 cos θ1 , y1 = l1 sin θ1 , x2 = l1 cos θ1 + l2 cos θ2 , y2 = l1 sin θ1 + l2 sin θ2 , which means

x˙ 1 = −l1 θ˙1 sin θ1 , y˙ 1 = l1 θ˙1 cos θ1 ,

x˙ 2 = −l1 θ˙1 sin θ1 − l2 θ˙2 sin θ2 , y˙ 2 = l1 θ˙1 cos θ1 + l2 θ˙2 cos θ2 , and the kinetic energy writes T = T1 + T2 =

1 1 m1 |~v1 |2 + m2 |~v2 |2 2 2

 1 m1 (x˙ 21 + y˙ 12 ) + m2 (x˙ 22 + y˙ 22 ) 2 1 1 = l12 θ˙12 (m1 + m2 ) + m2 l22 θ˙22 + m2 l1 l2 θ˙1 θ˙2 cos(θ2 − θ1 ). 2 2 =

211

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(6.142)

Free ebooks ==> www.Ebook777.com The Lagrangian therefore is: L=T −V =

1 2 ˙2 1 l1 θ1 (m1 + m2 ) + m2 l22 θ˙22 + m2 l1 l2 θ˙1 θ˙2 cos(θ2 − θ1 ) 2 2 +gl1 (m1 + m2 ) cos θ1 + m2 gl2 cos θ2 .

(6.143)

As one can see, the Lagrangian does not explicitly depend on time. This means that the Lagrange equations of the second kind admit the energy first integral ∂L ∂L Etot = Ecin + Epot = θ˙1 + θ˙2 − L = const. ∂ θ˙1 ∂ θ˙2

(6.144)

The Lagrange equations of the second kind, associated with the generalized coordinates θ1 and θ2 , are   ∂L d ∂L − = 0 (i = 1, 2). (6.145) ˙ dt ∂ θi ∂θi Since

∂L = l12 θ˙1 (m1 + m2 ) + m2 l1 l2 θ˙2 cos(θ2 − θ1 ); ˙ ∂ θ1 ∂L = m2 l22 θ˙2 + m2 l1 l2 θ˙1 cos(θ2 − θ1 ); ˙ ∂ θ2   d ∂L = l12 θ¨1 (m1 + m2 ) + m2 l1 l2 θ¨2 cos(θ2 − θ1 ) dt ∂ θ˙1 −m2 l1 l2 θ˙22 sin(θ2 − θ1 ) + m2 l1 l2 θ˙1 θ˙2 sin(θ2 − θ1 );   d ∂L = m2 l22 θ¨2 + m2 l1 l2 θ¨1 cos(θ2 − θ1 ) dt ∂ θ˙2 +m2 l1 l2 θ˙12 sin(θ2 − θ1 ) − m2 l1 l2 θ˙1 θ˙2 sin(θ2 − θ1 );

∂L = m2 l1 l2 θ˙1 θ˙2 sin(θ2 − θ1 ) − (m1 + m2 )gl1 sin θ1 ; ∂θ1 ∂L = −m2 l1 l2 θ˙1 θ˙2 sin(θ2 − θ1 ) − m2 gl2 sin θ2 , ∂θ2 the Lagrange equations (6.145) write l12 θ¨1 (m1 + m2 ) + m2 l1 l2 θ¨2 cos(θ2 − θ1 ) − m2 l1 l2 θ˙22 sin(θ2 − θ1 ) +(m1 + m2 )gl1 sin θ1 = 0, 212

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(6.146)

Free ebooks ==> www.Ebook777.com and

m2 l22 θ¨2 + m2 l1 l2 θ¨1 cos(θ2 − θ1 ) +m2 l1 l2 θ˙12 sin(θ2 − θ1 ) + m2 gl2 sin θ2 = 0.

(6.146′ )

This way, we are left with the differential equations of motion of the system. Since their general analytical solution is not possible, one usually apply to the most important limit cases. Most frequently is studied the case of small oscillations, cos(θ2 − θ1 ) ≃ 1 and cos θi ≃ θ2 1 − 2i , (i = 1, 2). In this case, the Lagrangian (6.143) becomes L=

1 2 ˙2 1 l1 θ1 (m1 + m2 ) + m2 l22 θ˙22 + m2 l1 l2 θ˙1 θ˙2 2 2 −

1 1 gl1 (m1 + m2 )θ12 − m2 gl2 θ22 , 2 2

(6.147)

where the constant terms gl1 (m1 + m2 ) and gm2 l2 , in agreement with the general analytical formalism, have been omitted. In the following approach we shall restrict out investigation to the particular case l1 = l2 = l. Let us define as coordinates of small √ √ oscillations the quantities ξ1 = m1 lθ1 and ξ2 = m2 l(θ1 + θ2 ). The Lagrangian then writes i 1 g h 1 m2  2 L = (ξ˙12 + ξ˙22 ) − 1+2 ξ1 + ξ22 + 2 2l m1

r

m2 g ξ1 ξ2 . m1 l

(6.148)

This is precisely the Lagrangian (6.74) used in Problem No.5, with ω12 and

g = l

  m2 1+2 , m1 g κ= l

r

ω22 =

g , l

m2 . m1

(6.149)

(6.150)

The frequencies of the normal modes are obtained by means of the formulas (6.80) and (6.81) obtained in the same Problem No.5. The results are: r q h  1 2 g m m1 i 2 2 2 2 2 ω 1 = (ω1 + ω2 ) − κ2 + (ω2 − ω1 )/4 = 1+ 1− 1 + ; 2 l m1 m2 (6.151) 213

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Free ebooks ==> www.Ebook777.com ω 22

1 = (ω12 + ω22 ) + 2

r q m2  gh m1 i 2 2 2 1+ 1+ 1 + . κ + (ω2 − ω1 )/4 = l m1 m2

(6.152)

To the general analysis with respect to ω1 and κ (with ω2 fixed) developed in Problem No.5, now corresponds the behavior of the ratio m2 m . Here we shall consider only the most interesting limit cases: (1) m2 m21 by ”weak m1 ≪ 1, (2) m1 ≫ 1, and (3) m1 = m2 . What we meant 2 coupling” (localized oscillations) in Problem No.5, κ ≪ ω2 − ω12 , is now represented by r r g m2 m2 m2 g m2 ≪ 2 ≫ 1 ⇐⇒ ≫ 1, ⇐⇒ l m1 l m1 m1 m1

while the condition of ”strong coupling” (non-localized oscillations), κ ≫ ω22 − ω12 , now becomes g l

r

r g m2 m2 ⇐⇒ m2 ≪ 1 ⇐⇒ m2 ≪ 1. ≫ 2 m1 l m1 m1 m1

Let us now analyze the above mentioned three cases. 2 (1) If m m1 ≪ 1, oscillations are not localized, and according to the general analysis developed in Problem No.5, we have ξ1 ≃

η1 − η2 √ , 2

ξ2 ≃

η1 + η2 √ , 2

(6.153)

where ηi (i = 1, 2) are normal coordinates, so that we may write ηi = Ai cos(ω i t + ψi )

(i = 1, 2),

(6.154)

where the constants Ai , ψi (i = 1, 2) are determined by means of initial ω 2 −ω 2 conditions. Indeed, for cot 2ϕ = 22κ 1 we have 2 −2 gl m m1 q cot 2ϕ = =− 2 2 gl m m1

r

m2 (→ 0 through negative values) m1

meaning that ϕ → π4 through values greater than (6.78) with actual notations 

π 4,

and the relations

ξ1 = η1 cos ϕ − η2 sin ϕ, ξ2 = η1 sin ϕ + η2 cos ϕ, 214

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Free ebooks ==> www.Ebook777.com p easily lead to (6.153). Up to terms of the first order in m2 /m1 , the quantities ω i (i = 1, 2) interfering in (6.154) are given by ω 1,2 = ω0 with γ = ω0 12 the frequency

q



1 1∓ 2

r

m2 m1



= ω0 ∓ γ,

g m2 2 m1 ≪ ω0 , and ω0 = l . Indeed, ω 21 defined by (6.151) writes

v u u ω 1 = ω0 t1 + z 2 Using approximation v u u t1 + z 2 ≃

s

1+



so that ω 1 ≃ ω0

denoting z =

q

m2 m1 ,

! 1 1+ 2 . z

1 ± ζ ≃ 1 ± ζ2 , we still have

1−

z2

1−

r

(6.155)

r

1 1+ 2 z 

!

z2 −z 1+ 2 

q p = 1 + z2 − z z2 + 1 





z 1−z ≃1− , 2

  r z 1 m2 1− = ω0 1 − . 2 2 m1

Proceeding in the same manner, we also obtain   r  z 1 m2 ω 2 ≃ ω0 1 + = ω0 1 + . 2 2 m1

(6.156)

(6.157)

Suppose that initially the pendulums are at rest, θ˙1 (t0 = 0) = ˙ 0, θ2 (t0 = 0) = 0, and only the pendulum of mass m2 is displaced from equilibrium, θ1 (t0 = 0) = 0, θ2 (t0 = 0) = θ0 . Then ξ1 η1 − η2 ≃ √ θ1 = √ l m1 l 2m1  1  A1 cos(ω 1 t + ψ1 ) − A2 cos(ω 2 t + ψ2 ) , = √ l 2m1 ξ2 ξ1 η1 + η2 η1 − η2 θ2 = √ − √ − √ ≃ √ l m2 l m1 l 2m2 l 2m1 215

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(6.158)

Free ebooks ==> www.Ebook777.com 

 1 1 = √ − √ A1 cos(ω 1 t + ψ1 ) l 2m2 l 2m1   1 1 + √ A2 cos(ω 2 t + ψ2 ), + √ l 2m2 l 2m1

(6.159)

where the arbitrary constants A1 , A2 , ψ1 , and ψ2 are determined by means of the initial conditions. Since  1  θ˙1 = √ ω 2 A2 sin(ω 2 t + ψ2 ) − ω 1 A1 sin(ω 1 t + ψ1 ) , l 2m1   ω ω 1 1 θ˙2 = √ − √ A1 sin(ω 1 t + ψ1 ) l 2m1 l 2m2   ω2 ω2 − √ + √ A2 sin(ω 2 t + ψ2 ), l 2m1 l 2m2 the use of initial conditions leads to the following system of four equations in the above mentioned four constants:  1 (A1 cos ψ1 − A2 cos ψ2 ), 0 = l√2m   1        1 1  θ0 = √ 1 − √ 1 √ √ A cos ψ + + A2 cos ψ2 , 1 1 l 2m l 2m l 2m l 2m 2

1

2

1 (ω 2 A2 sin ψ2 − ω 1 A1 sin ψ1 ), 0 = l√2m   1        0 = √ω1 − √ω1 A1 sin ψ1 − √ω2 l 2m1

l 2m2

l 2m2

+

1



√ω 2 l 2m1

A2 sin ψ2 .

The third equation yields ω 1 A1 sin ψ1 = ω 2 A2 sin ψ2 , and, consequently, the fourth equation leads to ψ1 = 0, ψ2 = 0, so that the first gives A1 = A2 . Finally, the second equation says that A1 = A2 = √ θ0 l m 2 √ . Then we are left with 2 ξ1 η1 − η2 θ0 θ1 = √ ≃ √ = l m1 2 l 2m1 θ2 =



θ0 θ0 − 2 2

r

m2 m1



r

cos ω 1 t +

θ0 θ0 = (cos ω 1 t + cos ω 2 t) + 2 2

m2 (cos ω 1 t − cos ω 2 t); m1 

r

θ0 θ0 + 2 2

r

m2 m1



cos ω 2 t

m2 (cos ω 2 t − cos ω 1 t). m1

In view of the trigonometric identities x+y cos x + cos y = 2 cos x−y 2 cos 2 , x+y cos x − cos y = −2 sin x−y 2 cos 2 ,

216

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(6.160)

Free ebooks ==> www.Ebook777.com the angular coordinates θ1 and θ2 become: θ1 = θ0

= θ0

r

m2 sin m1

r



m2 (ω 2 − ω 1 )t (ω 2 + ω 1 )t sin sin m1 2 2

ω0 t 2

r

m2 m1



sin ω0 t = θ0

θ2 = θ0 cos ω0 t cos γt + θ0 

= θ0 1 +

r

m2 m1



r

r

m2 sin ω0 t sin γt, m1

m2 cos ω0 t cos γt m1

cos ω0 t cos γt ≃ θ0 cos ω0 t cos γt.

To conclude, the obtained solution (

θ1 = θ0

q

m2 m1

sin ω0 t sin γt,

θ2 = θ0 cos ω0 t cos γt,

(6.161)

shows that the pendulums oscillate ”alternatively”: each of them reaches its amplitude, when the other one is at equilibrium. At the same time, p one can see that the amplitude of the pendulum of mass m1 is m1 /m2 times smaller than the amplitude of the pendulum of mass m2 . This result can also be interpreted as being of the ”beatings” type: q we have oscillations with frequency ω0 and amplitude mod 2 θ1 (t) = θ0 m m1 sin γt, on the one hand, and oscillations with ampli-

tude θ2mod (t) = θ0 cos γt, on the other. As seen, the change with time of the first amplitude is much slower than of the second one. 2 (2) If m m1 ≫ 1, the oscillations are localized and, since ω1 > ω2 , according to the general analysis developed in Problem No.5, we have ϕ ≃ π2 , so that ξ1 ≃ −η2 , ξ2 ≃ η1 . Indeed, we have: 2 −2 gl m ω22 − ω12 m1 q cot 2ϕ = = =− 2κ 2 2 gl m m1

which means that ϕ ≃

π 2,



r

m2 (→ −∞), m1

and

ξ1 = η1 cos ϕ − η2 sin ϕ, ξ2 = η1 sin ϕ + η2 cos ϕ, 217

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Free ebooks ==> www.Ebook777.com lead to ξ1 ≃ −η2 = −A2 cos(ω 2 t+ψ2 ), and ξ2 ≃ η1 = A1 cos(ω 1 t+ψ1 ), which completes the proof. q 1 Working within the same first-order approximation for m m2 ≪ 1, we obtain

ω0 ω1 ≃ √ , 2

ω 2 ≃ ω0

r

2m2 m1



ω02

g . = l

These results can be verified as follows. Denoting z = r s   r g m2 m1 1− 1+ ω1 = 1+ l m1 m2 = ω0

r

as well as

p  1 1 + 2 1 − 1 + z 2 ≃ ω0 z

s

1 1+ 2 z

q

m1 m2 ,

we have

  z2 ω0 1−1− =√ , 2 2

r

p  1 2 1 + z 1 + z2 s √ r   2 2m2 1 z2 ≃ ω0 1 + 2 1 + 1 + = ω0 . = ω0 z 2 z m1 ω 2 = ω0

1+

Suppose, this time, that initially the pendulums are at rest, θ˙1 (t0 = 0) = 0, θ˙2 (t0 = 0) = 0, and only the pendulum of mass m1 is displaced from its equilibrium position, θ1 (t0 = 0) = θ0 , θ2 (t0 = 0) = 0. Then, we have: ξ1 η2 A2 θ1 = √ ≃− √ =− √ cos(ω 2 t + ψ2 ), l m1 l m1 l m1 ξ1 η1 η2 ξ2 θ2 = √ − √ ≃ √ + √ l m2 l m1 l m2 l m1 A1 A2 = √ cos(ω 1 t + ψ1 ) + √ cos(ω 2 t + ψ2 ). l m2 l m1 We still have ω 2 A2 θ˙1 = √ sin(ω 2 t + ψ2 ), l m1 218

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Free ebooks ==> www.Ebook777.com ω 1 A1 ω 2 A2 θ˙2 = − √ sin(ω 1 t + ψ1 ) − √ sin(ω 2 t + ψ2 ). l m2 l m1 Imposing the initial conditions, we are left with the following system of four algebraic equations in four unknowns A1 , A2 , ψ1 and ψ2 :  2 θ0 = − l√Am cos ψ2 ,   1   A 1  0 = √ cos ψ1 + √A2 cos ψ2 , l m2 l m1 A2  sin ψ2 , 0 = lω√2m  1   ω A  0 = − √1 1 sin ψ − 1 l m2

ω√2 A2 l m1

sin ψ2 .

The third equation yields ψ2 = 0, and, as a result, the last equation gives ψ1 = 0. Then the first two equations result in A2 = √ √ −lθ0 m1 , A1 = lθ0 m2 . Therefore,  r  2m2 θ1 = θ0 cos ω 2 t = θ0 cos ω0 t ; (6.162) m1  r  2m2 ω0 t θ2 = θ0 cos ω 1 t − θ0 cos ω 2 t = θ0 cos √ t − θ0 cos ω0 m1 2 q   q 2m2 2m2 √1 √1 ω0 − + ω 0 m1 m1 2 2 = 2θ0 sin t sin t (6.163) 2 2    r   r m 2m2 2 2 ≃ 2θ0 sin ω0 t = θ0 1 − cos ω0 t . 2m1 m1 ω 2 −ω12

(3) If m1 = m2 = m, by means of cot 2ϕ = 22κ where a is a notation, we have √   1+ 2 1 a 2 √ = sin ϕ = 1− √ 2 2 2 1 + a2 s √ 3π 1+ 2 √ =⇒ sin ϕ = =⇒ ϕ = , 8 2 2

= a = −1,

and     √ a 1 1 2−1 √ 1+ √ = 1− √ = 2 2 2 2 2 1+a s√ 2−1 3π √ =⇒ cos ϕ = =⇒ ϕ = . 8 2 2

1 cos2 ϕ = 2

219

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Free ebooks ==> www.Ebook777.com Then ξ1 , ξ2 write ξ1 = η1 cos ϕ − η2 sin ϕ = ξ2 = η1 sin ϕ + η2 cos ϕ =

s√

s√

s√

s√

2−1 √ η1 − 2 2 2+1 √ η1 + 2 2

2+1 √ η2 , 2 2 2−1 √ η2 . 2 2

We also have: s

ξ1 1 θ1 = √ = √  l m l m

s

1 = √  l m



2−1 √ η1 − 2 2



2−1 √ A1 cos(ω 1 t + ψ1 ) − 2 2

s√

s √

1 − √  l m

s √

1 = √  l m

2+1 √ − 2 2

1 θ˙1 = √ l m s√

s√

"s √

s√





(6.164) 

2−1  √ η2 2 2

2+1  √ η2 2 2

2 − 1 √ A1 cos(ω 1 t + ψ1 ) 2 2

2−1 √ + 2 2

as well as

s√



s√

s √

1 + √  l m

2+1 √ η1 + 2 2

2−1 √ η1 − 2 2

2+1  √ η2 2 2

2+1 √ A2 cos(ω 2 t + ψ2 ) ; 2 2

s √

1 1 θ2 = √ (ξ2 − ξ1 ) = √  l m l m



s√

(6.165)



2 + 1 √ A2 cos(ω 2 t + ψ2 ), 2 2

2+1 √ A2 ω 2 sin(ω 2 t + ψ2 ) 2 2 #

2−1 √ A1 ω 1 sin(ω 1 t + ψ1 ) , 2 2 s  s√ √ 1 2−1 2 + 1 √ − √ θ˙2 = √  A1 ω 1 sin(ω 1 t + ψ1 ) l m 2 2 2 2 −

220

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Free ebooks ==> www.Ebook777.com s √

1 − √  l m

2−1 √ + 2 2



s√

2 + 1 √ A2 ω 2 sin(ω 2 t + ψ2 ). 2 2

The initial conditions θ1 (t0 = 0) = 0, θ2 (t0 = 0) = θ0 , θ˙1 (t0 = 0) = 0, and θ˙2 (t0 = 0) = 0 then allow one to write the following four algebraic equations for the unknowns A1 , A2 , ψ1 , and ψ2 :

0=

s√

2−1 √ A1 cos ψ1 − 2 2 s √

√ lθ0 m =  s

+

0=

2+1 √ − 2 2



2−1 √ + 2 2

s√

2+1 √ A2 cos ψ2 , 2 2 

s√

2 − 1 √ A1 cos ψ1 2 2 

s√

2 + 1 √ A2 cos ψ2 , 2 2 s√

s√

2+1 2−1 √ A2 ω 2 sin ψ2 − √ A1 ω 1 sin ψ1 , 2 2 2 2 s  s√ √ 2−1 2 + 1 √ − √ 0= A1 ω 1 sin ψ1 2 2 2 2 s

−



2−1 √ + 2 2



s√

2 + 1 √ A2 ω 2 sin ψ2 . 2 2

The last two equations give ψ1 = 0 and ψ2 = 0. With these values, the first equation leads to

A2 =

s√ √

2−1 A1 , 2+1

and, finally, the second relation shows that A1 = q √

√ lθ0 m

2+1 √ 2 2



2−1 √ 4+2 2

=

+√

√ lθ0 m √

1



4−2 2



2−1 √ 4+2 2

+√

221

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,

Free ebooks ==> www.Ebook777.com so that A2 =



√ q √2−1 lθ0 m √2+1 1



4−2 2

√ 2−1 √ 4+2 2

+√

=

√ lθ0 m





2+1 √ 4−2 2

+√

1

. √

4+2 2

With these results, θ1 given by (6.164) writes   "s √ √ 1 2−1 lθ0 m  √ √  θ1 = √  cos ω 1 t 1 2−1 l m 2 2 √ √ +√ √ 4−2 2





s√

4+2 2



2+1 lθ0 m √  √ 2 2 √ 2+1√ + √ 4−2 2

1

√ 4+2 2



  cos ω 2 t

#

θ0 = √ (cos ω 1 t − cosω 2 t), 2 2

while (6.165) gives s

1 θ2 = √  l m



s √

1 + √  l m

2+1 √ − 2 2

2−1 √ + 2 2

 2 − 1  √  2 2 √

s√



lθ0 m 1

√ 4−2 2

 √ 2 + 1  lθ0 m √  √2+1 2 2 √ √ +√

s√

4−2 2

=

+

√ √ 2−1√ 4+2 2

1

√ 4+2 2

θ0 (cos ω 1 t + cos ω 2 t). 2



  cos ω 1 t



  cos ω 2 t

Recalling our initial assumptions (m1 = m2 = m, l1 = l2 = l), we also have √ √ g g ω 21 = (2 − 2); ω 22 = (2 + 2), l l and the two angular variables finally write  q  q  √ √ θ0 θ1 (t) = √ cos 2 − 2 ω0 t − cos 2 + 2 ω0 t , (6.166) 2 2  q  q  √ √ θ0 θ2 (t) = cos 2 − 2 ω0 t + cos 2 + 2 ω0 t . (6.167) 2 222

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Free ebooks ==> www.Ebook777.com Observation. This last case (m1 = m2 = m, l1 = l2 = l) can also be approached independently from the general considerations exposed in Problem No.5. With these assumptions, equations (6.146) and (6.146’) become g 2θ¨1 + θ¨2 cos(θ2 − θ1 ) − θ˙22 sin(θ2 − θ1 ) + 2 sin θ1 = 0, l g θ¨2 + θ¨1 cos(θ2 − θ1 ) + θ˙12 sin(θ2 − θ1 ) + sin θ2 = 0. l Furthermore, if the oscillations are small, cos(θ2 − θ1 ) ≃ 1, θ1 ) ≃ 0, sin θ1 ≃ θ1 , sin θ2 ≃ θ2 , and the last two equations simpler form g 2θ¨1 + θ¨2 + 2 θ1 = 0; l g θ¨2 + θ¨1 + θ2 = 0. l

(6.168) (6.168′ ) sin(θ2 − receive a (6.169) (6.169′ )

We are looking for solutions of the form θ1 = A1 eiωt , θ2 = A2 eiωt , leading to the following system   g 2 l − ω 2 A1 − ω2 A2 = 0, −ω 2 A1 + gl − ω 2 A2 = 0. This system admits non-trivial solution for the amplitudes A1 and A2 only if  g 2 2 2 − ω −ω l  = 0, (6.170) g 2 2 −ω l −ω that is

ω4 − 2

Denoting ω 2 = r, second degree in r

g l

g l

− ω2

2

= 0.

= ω02 , one obtains the following equation of the r2 − 4ω02 r + 2ω04 = 0,

with the solutions r1,2 = ω02 (2 ± or, if we return to the old variable q √ ω1 = ω0 2 − 2,



2),

ω 2 = ω0

q

2+



2.

(6.171)

These are the two normal frequencies of oscillation. It then follows that the system possesses two normal modes of oscillation, corresponding to its two degrees of freedom. 223

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Free ebooks ==> www.Ebook777.com As known, the choice of the generalized coordinates is - up to some extent - arbitrary. This fact allows us to represent the system as an assembly of linear harmonic oscillators, with one degree of freedom for each of them. In other words, to each normal frequency one can associate a periodically time-varying generalized coordinate. These coordinates are called normal. Suppose that our system oscillates with one of the normal frequencies, say ω1 . Then, since θ1 = A1 eiω1 t , θ2 = A2 eiω1 t , equation (6.169) becomes (−2ω12 + 2ω02 )θ1 = ω12 θ2 . √ Recalling that ω12 = (2 − 2)ω02 , we still have √ √ (−2 + 2 2)θ1 = (2 − 2)θ2 ,

or

√ θ2 = 2. θ1

(6.172)

Calculations for the √ second normal frequency are similar, and the result is θ2 /θ1 = − 2. Therefore, for the two normal modes of oscillation, the relations between the two generalized coordinates are √ θ2 = θ1 2 √ θ2 = −θ1 2

( for ω = ω1 ), ( for ω = ω2 ).

In the first case the modes are called symmetric, and in the second antisymmetric. √ If we choose as√ normal coordinates the quantities ηs = θ2 + 2θ1 , ηas = θ2 − 2θ1 , where the indices ”s” and ”as” stand for ”symmetric” and ”antisymmetric”, and place our discussion within the same particular case (m1 = m2 = m, l1 = l2 = l, cos(θ2 − θ1 ) ≃ θ2 1, cos θi ≃ 1 − 2i (i = 1, 2)), the Lagrangian (6.143) writes 1 1 L = ml2 θ˙12 + ml2 θ˙22 + ml2 θ˙1 θ˙2 − mglθ12 − mglθ22 2 2 2  2 η˙ s − η˙ as 1 2 η˙ s + η˙ as √ = ml + ml 2 2 2 2    η˙ s − η˙ as η˙ s + η˙ as 2 √ +ml 2 2 2 2



224

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Free ebooks ==> www.Ebook777.com 2 2  1 ηs − ηas ηs + ηas √ − mgl −mgl 2 2 2 2 h √ i mgl √ 1 2 2 = √ ml2 η˙ s2 ( 2 + 1) + η˙ as (ηs2 + ηas ). ( 2 − 1) − 4 4 2 By choosing as normal coordinates s√ 2+1 2 √ ml ; η1 = ηs 2 2 s√ 2−1 2 √ ml , η2 = ηas 2 2 

the Lagrangian writes

  g η12 1 2 η22 2 √ + √ L = (η˙ 1 + η˙ 2 ) − 2 l 2+ 2 2− 2 √ 2 √ 2i 1 2h 1 2 2 = (η˙ 1 + η˙ 2 ) − ω0 (2 − 2)η1 + (2 + 2)η2 2 2 1 2 1 = (η˙ 1 + η˙ 22 ) − (ω12 η12 + ω22 η22 ) 2 2 1 1 ≡ (η˙ 12 + η˙ 22 ) − (ω 21 η12 + ω 22 η22 ), 2 2 and we meet again the Lagrangian (6.147) for the third case (m1 = m2 = m; l1 = l2 = l). To find θi = θi (t) (i = 1, 2) we use the relations s√ s√ √ √ √ 2+1 2+1 √ √ (θ2 + 2θ1 ) η1 = lηs m =l m 2 2 2 2 = A1 cos(ω1 t + ψ1 ); (6.173) s√ s√ √ √ √ 2−1 2−1 √ √ (θ2 − 2θ1 ) η2 = lηas m =l m 2 2 2 2 = A2 cos(ω2 t + ψ2 ).

(6.174)

Using the same initial conditions, θ1 (t0 = 0) = 0, θ2 (t0 = 0) = ˙ θ0 , θ1 (t0 = 0) = 0, θ˙2 (t0 = 0) = 0, the constants A1 , A2 , ψ1 , ψ2 are found as solutions of the following algebraic system  √ q √2+1  √  lθ0 m = A1 cos ψ1 ,    √ q √2 2 √ lθ0 m 22−1 = A2 cos ψ2 , 2    0 = −ω1 A1 sin ψ1 ,   0 = −ω2 A2 sin ψ2 . 225

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Free ebooks ==> www.Ebook777.com The last two equations give ψ1 = ψ2 = 0, while the other two equations yield s√ s√ √ √ 2+1 2−1 √ ; A2 = lθ0 m √ A1 = lθ0 m 2 2 2 2 and we finally obtain  q  q  √ √ θ0 θ1 (t) = √ cos 2 − 2 ω0 t − cos 2 + 2 ω0 t ; 2 2  q  q  √ √ θ0 θ2 (t) = cos 2 − 2 ω0 t + cos 2 + 2 ω0 t , 2

which are precisely solutions (6.166) and (6.167) obtained by the first method.

226

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CHAPTER VII PROBLEMS OF EQUILIBRIUM AND SMALL OSCILLATIONS

Problem 1. A particle (material point) moves without friction inside a pipe of elliptic shape and constant cross section, rotating about its major axis with constant angular velocity ω (see Fig.VII.1). The mass m of the particle, the gravitational acceleration g, and the semi-axes a > b of the ellipse are given. Determine: a) The equilibrium positions of the particle; b) Stability of the equilibrium positions; c) Period of small oscillations of the particle about the stable positions of equilibrium.

Fig.VII.1 227

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Free ebooks ==> www.Ebook777.com Solution Obviously, if the ellipse does not rotate, position given by θ = 0 shall be a position of stable equilibrium, while θ = π would be a position of unstable equilibrium. If the ellipse rotates, then depending on the value of ω, position θ = 0 could become a position of unstable equilibrium, since upon the particle act both the centrifugal and gravitational forces, the first tending to move away the particle from the position of stable equilibrium, and the second tending to bring the particle back to this position. If ω exceeds a certain critical value ωcr , then the position θ = 0 becomes a position of unstable equilibrium, while the situation ω < ωcr corresponds to a stable equilibrium. In its turn, position given by θ = π is always a position of unstable equilibrium, no matter how big or small ω is taken. It is also possible to appear one or more intermediate states of equilibrium for θ ∈ [0, π]. For ω → ∞, the centrifugal force is far superior to the force of gravity, and θ = π/2 corresponds to a position of stable equilibrium. Using Lagrange equations of the second kind formalism, let us prove all these intuitive observations. Since this application implies a centrifugal force, which is a force of inertia, we shall investigate the problem from two different points of view, associated with inertial and non-inertial approaches. (A) Inertial reference frame (IRF) a) As we know, there is no inertial force (such as the centrifugal force) in an IRF. From the point of view of the IRF, the only applied force ~ Since the gravitational force derives from a is the force of gravity G. potential, our system is a natural one. The gravitational field being conservative, the time-independent potential is precisely the potential energy of the system. The equilibrium positions of the particle are given by the extremum (extrema) of the potential energy. According to analytical approach, dV = −dA, where

~ · d~r dA = G

is the elementary mechanical work done by the gravitational force field. The choice of the coordinate system (see Fig.VII.1) allows us to write dV = −mg dx so that V = −mgx + V0 . 228

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Free ebooks ==> www.Ebook777.com The most convenient choice for V0 is V0 = 0. This condition is fulfilled by taking the yOz-plane as the reference level for the potential energy V (x = 0) = 0 ⇒ V0 = 0. The standard form equation of the ellipse x2 y2 + =1 a2 b2 allows one the parametrization  x = a cos θ; y = b sin θ, and the potential energy of the particle becomes V = −mga cos θ.

(7.1)

Equating to zero the first derivative of V = V (θ), one obtains sin θ = 0 with ”physical” solutions θ1 = 0,

θ2 = π.

(7.2)

The stability of these equilibrium positions is determined by performing the second derivative d2 V = mga cos θ, dθ2 2

which means that θ1 = 0 is a position of stable equilibrium ( ddθV2 |θ=0 > 0), while θ2 = π corresponds to a position of unstable equilibrium 2 ( ddθV2 |θ=π < 0). These results can be intuitively anticipated if the pipe is at rest. But, if the pipe rotates, this fact has to be considered under our assumption that the frame is inertial. To solve the problem, we have to draw our attention to the ”true” value of the potential energy of the particle, due to the action of both the gravity and the motion of rotation. This is called effective potential energy. To determine the effective potential energy, let us first write the Lagrangian of the problem. To the single degree of freedom of the 229

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Free ebooks ==> www.Ebook777.com particle we shall associate the generalized coordinate q = θ, so that L = T − V (θ). Since the particle is subjected to a composite motion (a motion of translation along the ellipse, in the xOy plane, and a motion of rotation with constant angular velocity ω about the x-axis), the kinetic energy writes T = Ttr + Trot =

1 1 1 1 2 2 m~vtr + m~vrot = m(x˙ 2 + y˙ 2 ) + mω 2 y 2 2 2 2 2

1 ˙2 2 2 1 mθ (a sin θ + b2 cos2 θ) + mω 2 b2 sin2 θ. 2 2 The Lagrangian then becomes =

L=

1 ˙2 2 2 1 mθ (a sin θ + b2 cos2 θ) + mω 2 b2 sin2 θ + mga cos θ. (7.3) 2 2

As one can see, the Lagrangian does not explicitly depend on time. Consequently, Lagrange equation of the second kind admits the first integral ∂L θ˙ − L = const., ∂ θ˙ expressing conservation of the total energy of the system const. = Etot = Ecin + Epot = θ˙

∂L −L ∂ θ˙

 1 1 ˙2 2 2 mθ a sin θ + b2 cos2 θ − mω 2 b2 sin2 θ − mga cos θ. 2 2 Therefore, the total energy consists of two groups of terms, one involving the squared generalized velocity, which is the kinetic energy =

Ecin =

 1 ˙2 2 2 mθ a sin θ + b2 cos2 θ , 2

(7.4)

and the other written in terms of ω and θ, namely

1 Epot = − mω 2 b2 sin2 θ − mga cos θ. 2

(7.4′ )

We have to emphasize that, within the Lagrangian formalism, the kinetic energy is expressed in terms of the generalized velocity (in ˙ Therefore, we may call the quantity given by (7.4) the our case θ). ”true/genuine” kinetic energy. On the other hand, as seen from (7.4’), Epot contains the supplementary term − 12 mω 2 b2 sin2 θ, in addition 230

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Free ebooks ==> www.Ebook777.com to the term V (θ) = −mga cos θ, meaning that (7.4’) is the effective potential energy. We want to draw attention to the reader that the term containing ω 2 is not connected to the kinetic energy, even if ω is, in its turn, the time derivative of an angular coordinate (which is not the generalized coordinate of the problem, but it belongs to the uniform circular motion of the material point about the x-axis, i.e., it is an angular coordinate in a plane orthogonal to the x-axis). Equating to zero the first derivative of Epot , one obtains dEpot = −mω 2 b2 sin θ cos θ + mga cos θ = 0, dθ giving the equilibrium positions of the particle  sin θ = 0, ga − ω 2 b2 cos θ = 0. The first equation yields the already known positions of equilibrium θ1 = 0,

θ2 = π,

while the second equation leads to a new position of equilibrium, given by ga  (7.5) θ3 = arccos 2 2 . ω b Using notation √ ga ωcr = , (7.6) b we still have  ω 2 cr θ3 = arccos . (7.6′ ) ω b) To investigate the stability of the determined equilibrium positions, one must study the sign of the second derivative of the effective potential energy at these points. We have: d2 Epot = −mω 2 b2 cos2 θ + mω 2 b2 sin2 θ + mga cos θ dθ2 = mω 2 b2 (1 − 2 cos2 θ) + mga cos θ, leading to i) d2 Epot dθ2

θ=0



ω 2 b2 = −mω b + mga = mga 1 − ga 2 2

231

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Free ebooks ==> www.Ebook777.com "



= mga 1 −

ω ωcr

2 #

ր > 0, ω < ωcr ,

( stable equilibrium );

ց < 0, ω > ωcr ,

( unstable equilibrium );

ii) d2 Epot dθ2

θ=π

= −mω 2 b2 − mga < 0,

∀ω

( unstable equilibrium );

iii)

d2 Epot dθ2 2 2

= mω b



2 2

= mω b θ=arccos(

1−

ωcr ω

)

 ω 4  cr

ω

2



 ag 2  ga 1−2 + mga ω 2 b2 ω 2 b2

ր > 0, ω > ωcr ,

( stable equilibrium );

ց < 0, ω < ωcr , ( unstable equilibrium ).

Therefore, the equilibrium position θ1 = 0 can be of both stable equilibrium (for ω < ωcr ) and unstable equilibrium (for ω > ωcr ), position θ2 = π is always a position of unstable equilibrium, while the 2 intermediate position θ3 = arccos ωωcr can be, in its turn, a position of both stable equilibrium (for ω > ωcr ), and unstable equilibrium (for ω < ωcr ). If ω = ωcr , that is ω 2 b2 = ga, then θ3 = arccos 1 = 0, and "  4 # ω d2 Epot cr = mω 2 b2 1 − = 0.  dθ2 ω(= ωcr ) 2 ωcr θ=arccos

ω(=ωcr )

To determine the stability of equilibrium in this case (θ3 = 0, for ω = ωcr ), we have to go ”further” with derivatives. Since d d3 Epot = [mω 2 b2 (1 − 2 cos2 θ) + mga cos θ] dθ3 dθ " #  2  2 2 4ω b 2ω = mga sin θ cos θ − 1 = mga sin θ cos θ − 1 , ga ωcr we have d3 Epot dθ3

θ=arccos

ωcr ω(=ωcr )

2

=

(

mga sin θ

"

2ω ωcr

2

#)

cos θ − 1

232

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ω=ωcr

Free ebooks ==> www.Ebook777.com = mga

s

1−



ωcr ω(= ωcr

4 " 

ω(= ωcr ) 2 ωcr

2 

ωcr ω(= ωcr )

2

#

− 1 = 0.

This result shows that we have to make one step further and calculate the fourth derivative. Since ( " #) 2 d4 Epot d 2ω = mga sin θ cos θ − 1 dθ4 dθ ωcr = mga we are left with



2ω ωcr

2

(2 cos2 θ − 1) − mga cos θ,

d Epot dθ4 4

"

= mga



2ω ωcr

2

θ=arccos

2

ωcr ω(=ωcr )

2

(2 cos θ − 1) − mga cos θ

#

= 3mga > 0.

ω=ωcr

Since in this special case (ω = ωcr ) the fourth derivative at the point θ3 = 0 is strictly positive, the equilibrium position θ3 = 0 is also a position of stable equilibrium, but the small oscillations about this position are not harmonic anymore. c) Obviously, small oscillations can take place only about a position of stable equilibrium. According to our investigation, there exist two positions with this property in our case: θ1 = 0 (if ω < ωcr ), and 2 θ3 = arccos ωωcr (if ω > ωcr ). The most important question arising in a problem of small oscillations is to determine the periods of the normal modes of oscillation. Since our application involves a single degree of freedom, this part of investigation is not very difficult. There are several procedures used to determine the period(s) of oscillation of a physical system performing small oscillations. If the system possesses more than one degree of freedom, the Newtonian formalism shows to be difficult, or even useless, while the Lagrangian approach offers a simple and elegant way, by solving the characteristic equation of the system. (This is an equation of order n in ω 2 , with real and positive solutions, n being the number of degrees of freedom). Let us first briefly discuss the solution to the problem for systems with one degree of freedom, from the Newtonian point of view. This approach is based on the analogy with the simplest mechanical system performing linear (monodimensional) harmonic oscillations, which is 233

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Free ebooks ==> www.Ebook777.com a body (particle) of mass m , connected to a spring of elastic constant k. As known, the period of small oscillations of such a system is τ = 2π

r

m . k

(7.7)

In fact, this is one of the simplest mechanical systems, acted by an elastic-type force F~ = −k~x. Since the field of the elastic forces is a potential field, we have dV = −dA = −F~ · d~x = k~x · d~x = k x dx, which yields V =k

Z

x dx =

1 2 kx + V0 . 2

If we choose x = 0 as the ”reference level” (corresponding to the undeformed spring) for the potential energy, then we can write V =

kx2 . 2

(7.8)

Here are two simple, Newtonian procedures to determine the period of linear harmonic small oscillations. The first one - which we may call ”dynamical” - consists in determination of both the resulting force acting on the system, and the resulting acceleration of the system. If ~ where ξ~ is the resulting force is elastic, being of the form F~ = −K ξ, elongation of the motion, then it allows one to determine the ”elastic constant” K (analogous to k interfering in F~ = −k~x). In its turn, the resultant acceleration - by means of the fundamental equation of ¨ dynamics F~ = M~a = M ξ~ - enables determination of ”the mass” M , which is analogous to mass m of the body connected to the spring with elastic constant k. Then, by analogy with (7.7), the period of the small oscillations of the system writes τ = 2π

r

M . K

(7.9)

Another simple Newtonian method - let us call it ”energetic” implies determination of both kinetic and potential energies of the system performing small oscillations. If these quantities are expressed as 1 1 Ec = M ξ˙2 , Ep = Kξ 2 , 2 2 234

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Free ebooks ==> www.Ebook777.com where ξ has the same significance as above, then by analogy with (7.7), the period of small oscillations is also given by (7.9). Based on the Lagrangian formalism, we shall now develop a simple and efficient method of determination the oscillation period of any physical system with one degree of freedom, performing small, linear, harmonic oscillations. Working again on the above example, one can remember the potential given by (7.8), while the kinetic energy associated with the single degree of freedom is T =

1 mx˙ 2 . 2

The Lagrangian of the system therefore is L=T −V =

1 1 mx˙ 2 − kx2 . 2 2

The corresponding Lagrange equation of the second kind   d ∂L ∂L =0 − dt ∂ x˙ ∂x then leads to the well-known equation of the harmonic oscillator m¨ x + kx = 0, or x ¨ + ω02 x = 0, where ω02 = k/m. As we know (see Chap.III), the solution of this equation could have four versions, with the period given by r 2π m τ= = 2π . ω0 k Let us consider a physical system with a single degree of freedom, whose motion is unknown so far. Denote by ξ the associated generalized coordinate, and take the Lagrangian of the system of the form L = Aξ˙2 − Bξ 2 , (7.10) where A and B are two real, non-zero constants, both having the same sign. The Lagrange equation   d ∂L ∂L − =0 dt ∂ ξ˙ ∂ξ 235

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Free ebooks ==> www.Ebook777.com then yields or

Aξ¨ + Bξ = 0, ξ¨ + Ω20 ξ = 0,

with Ω20 = B/A. As one can see, we have obtained the same equation of the linear harmonic oscillator, the oscillation period being r 2π A τ= = 2π . (7.11) Ω0 B Consequently, if a physical system with a single degree of freedom has a Lagrangian of the form (7.10), then the system performs an oscillatory harmonic motion, with the period given by (7.11). Such a Lagrangian is called Lagrangian of small oscillations associated with a physical system with a single degree of freedom. Returning now to our problem, in order to determine the period of small oscillations of the particle moving without friction inside the rotating pipe, we first have to write the suitable Lagrangian. To this end, we shall use the Lagrangian given by (7.3): L=

1 ˙2 2 2 1 mθ (a sin θ + b2 cos2 θ) + mω 2 b2 sin2 θ + mga cos θ. 2 2

Fig.VII.2 236

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Free ebooks ==> www.Ebook777.com To go further with our investigation, let us suppose that the particle of mass m performs small oscillations about a position of stable equilibrium denoted by θse = const. (see Fig.7.2). The elongation ξ of the oscillating motion then is ξ = θ − θse .

(7.12)

Taking into account the expression of the Lagrangian of small oscillations (7.10), we have to consider the quadratic terms in the small quantity ξ, meaning that the elongation of small oscillations satisfies the relation O(ξ 3 ) = 0. The law of the oscillating motion can be written as ξ = ξmax sin(Ω0 t + ϕ0 ), where the constants ξmax and ϕ0 can be determined by means of initial ˙ = 0). conditions: ξ0 = ξ(t = 0), and ξ˙0 = ξ(t There are two ways of writing the Lagrangian of small oscillations in our particular case: 1. Using (7.12), the angle θ is expressed in terms of the new variable ξ θ = θse + ξ, this value is introduced into the Lagrangian (7.3), then expand L in Maclaurin series in terms of ξ about the point ξ = 0, keeping the terms up to the second power. 2. Expand the Lagrangian (7.3) in Taylor series in terms of θ about the value θ = θse and keep the terms up to the second power of (θ − θse ) = ξ. We shall use the second procedure in our investigation. Observing that L can be written as L = Ecin − Epot , the expansion in Taylor series of L implies the series expansion of Ecin and Epot in terms of θ, about the value θ = θse . Therefore, Epot (θ) = Epot (θse ) +

1 dEpot (θ − θse ) 1! dθ θ=θse

  1 d2 Epot 2 3 + (θ − θ ) + O (θ − θ ) se se 2! dθ2 θ=θse 237

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Free ebooks ==> www.Ebook777.com ≃ Epot (θse ) +

1 d2 Epot ξ 2 ≡ Epot (ξ), 2! dθ2 θ=θse

where we have used the property of the equilibrium positions dEpot = 0, dθ θ=θse

for any kind of equilibrium (stable or unstable). We also have ˙ = 1 mθ˙2 (a2 sin2 θ + b2 cos2 θ) = 1 mξ˙2 (a2 sin2 θ + b2 cos2 θ) Ecin (θ, θ) 2 2 1 dEcin ˙ ˙ = Ecin (θ, ξ) = Ecin (θse , ξ) + (θ − θse ) 1! dθ θ=θse   1 d2 Ecin ˙ (θ − θse )2 + O (θ − θse )3 ≃ Ecin (θse , ξ). + 2 2! dθ θ=θse

Here we used the fact that the term 1 dEcin (θ − θse ) 1! dθ θ=θse

 1 d 1 ˙2 2 2 = ξ (a sin θ + b2 cos2 θ) 1! dθ 2 

ξ θ=θse

contains the factor ξ˙2 ξ, which is of the third power in ξ. Therefore ˙ ≃ Ecin (θse , ξ) ˙ = 1 mξ˙2 (a2 sin2 θse + b2 cos2 θse ) ≡ Ecin (ξ). ˙ Ecin (θ, θ) 2 The term Epot (θse ) is a constant and, according to analytical formalism, can be omitted in the Lagrangian of the small oscillations Lso . Consequently, ˙ − Epot (θ) = Ecin (ξ) ˙ − Epot (ξ) Lso = Ecin (θ, θ) 1 ˙2 2 2 mξ (a sin θse + b2 cos2 θse ) 2 1 d2 Epot ˙ = Aξ˙2 − Bξ 2 , − ξ 2 ≡ L(ξ, ξ) 2! dθ2 θ=θse =

with

A=

1 m(a2 sin2 θse + b2 cos2 θse ) = const. > 0, 2 238

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Free ebooks ==> www.Ebook777.com and B=

1 d2 Epot = const. > 0. 2! dθ2 θ=θse

In view of (7.11), the period of small oscillations of the system is v u m(a2 sin2 θ + b2 cos2 θ ) u se se τ = 2π t 2 d Epot dθ 2 θ=θse

v u u = 2π t 

m(a2 sin2 θse + b2 cos2 θse )  mω 2 b2 (1 − 2 cos2 θ) + mga cos θ

v u u = 2π t

1 a2 ω 2 b2

sin2 θse +

1 − 2 cos2 θse +

θ=θse

1 2 ω 2 cos θse  ωcr 2 cos θse ω

v  u 2 a2 1 − cos2 θse + ab2 2π u 2 b t = 2 ω 1 − 2 cos2 θse + ωcr cos θse ω

=

2π √ ω 1 − e2

s

where c e= = a

1 − e2 cos2 θse , 2 1 − 2 cos2 θse + ωωcr cos θse



a 2 − b2 = a

(7.13)

s

 2 b 1− a

is the ellipse eccentricity. We are now able to calculate the oscillation period for both posi2 tions of stable equilibrium θ1 = 0 (for ω < ωcr ) and θ3 = arccos ωωcr (for ω > ωcr ), as follows: τ1 = τ |θse =0

and

2π = √ ω 1 − e2

2π =p 2 ωcr − ω 2 τ2 = τ | θ

s

1 − e2  ωcr 2 −1 ω

(ω < ωcr ),

se =arccos

( ωωcr )

2

239

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(7.14)

Free ebooks ==> www.Ebook777.com 2π = √ ω 1 − e2

s

1 − e2 cos2 θse 2 1 − 2 cos2 θse + ωωcr cos θse

v  u u 1 − e2 ωcr 4 2π ω t = √ 4 ω 1 − e2 1 − ωωcr

(ω > ωcr ).

 ω 2 

= arccos 0 =

(7.15)

2 If ω → ∞, then θ3 = arccos ωωcr , as a position of stable equilibrium for ω > ωcr , gives rise to a new position of stable equilibrium, namely: θ4 = lim θ3 = lim ω→∞

ω→∞



arccos

cr

ω

π , 2

(7.16)

in complete agreement with our initial qualitative analysis. The period of small oscillations about this position of equilibrium is τ3 = τ |θse = π2 =

1 2π 1 √ = τ0 √ , 2 ω 1−e 1 − e2

(7.17)

where notation τ0 = 2π/ω is obvious. If the ellipse degenerates into a circle (e = 0), we are left with τ3 = τ |θse = π2 = τ0 .

(7.18)

(B) Non-inertial reference frame (NIRF) a) In order to determine the equilibrium positions of the particle of mass m we shall use the same requirement of cancellation of the first derivative of potential energy. As a non-inertial reference system, one can consider a frame invariably connected to the moving particle. Unlike the inertial frame, for an observer situated in the non-inertial system not only the gravitational force, but also the centrifugal force F~cf has to be considered as an applied force. Therefore, the potential of the resultant of applied forces satisfies the relation ~ · d~r + F~cf · d~r), dV = −dA = −(G

(7.19)

where F~cf is orthogonal to the rotation axis and points radially outwards (see Fig.VII.3). 240

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Fig.VII.3 To facilitate the calculation of the scalar products in (7.19), let ~ F~cf , and d~r in terms of their components. us write the vectors G, According to Fig.VII.3, we have: ~ = (mg, 0, 0), G

F~cf = (0, mω 2 y, 0),

d~r = (dx, dy, dz),

and (7.19) becomes dV = −mg dx − mω 2 y dy, or, by integration, 1 V = −mgx − mω 2 y 2 + V0 . 2

(7.20)

Since the potential does not explicitly depend on time, it coincides with the potential energy of the system. The most convenient choice of the arbitrary constant of integration V0 is V (x = 0, y = 0) = V0 = 0, and the potential energy of the body (particle) becomes 1 1 V = V (x, y) = −mgx − mω 2 y 2 = −mga cos θ − mω 2 b2 sin2 θ 2 2 241

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Free ebooks ==> www.Ebook777.com = V (θ) ≡ Epot (θ). This result has been already obtained in our approach (A) to the problem (study in an inertial frame), but this time (working in NIRF) the formula of the potential energy was found in a more simple way. This observation is valid in general, for more complicated applications, but everything depends on the ability of the researcher. From now on, the reasoning and calculations shall follow the way developed within the inertial frame. Problem 2. A particle P of mass m and electric charge q moves without friction inside a pipe of elliptic shape and constant cross section, rotating in vacuum about its major axis with constant angular velocity ω. A point charge q ′ is fixed at the lowest point of the ellipse (see Fig.VII.4). Determine: a) The equilibrium positions of the particle with charge q; b) Stability of these positions of equilibrium; c) Period of small oscillations of the particle about the stable positions of equilibrium.

Fig.VII.4

242

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Free ebooks ==> www.Ebook777.com Solution. a) This problem is similar to the previous one, so that we shall assume the results of our preceding investigation. This time, in addition to the gravitational force, we have to consider the electrostatic force of interaction between charges q and q ′ . Before going further, we have to mention that the magnetic effects due to the motion of particle with charge q are neglected. We also specify that our reference frame is the laboratory frame, which is an inertial reference system. Since the electrostatic and gravitational fields are conservative, our system (i.e. the particle of mass m and charge q) is a natural system. The kinetic energy is the same as in Problem 1 T =

1 1 ˙2 2 2 mθ (a sin θ + b2 cos2 θ) + mω 2 b2 sin2 θ, 2 2

but to the gravitational potential energy Vg = −mga cos θ a supplementary electrostatic term has to be added Vel =

1 1 qq ′ qq ′ p q = 4πε0 (a − x)2 + y 2 4πε0 4a2 sin4 θ + b2 sin2 θ 2

qq ′ (4πε0 )−1 =q 4a2 sin4 θ2 + 4b2 sin2 =

θ 2

cos2

θ 2

=

qq ′ (4πε0 )−1 q 2| sin θ2 | (a2 − b2 ) sin2

qq ′ 1 q 8πε0 sin θ c2 sin2 2

θ 2

θ 2

+ b2

, +

b2

where we used the fact that sin θ2 > 0 on the interval θ ∈ (0, π]. The Lagrangian of our problem therefore is L=

1 ˙2 2 2 1 mθ (a sin θ + b2 cos2 θ) + mω 2 b2 sin2 θ 2 2

+mga cos θ −

qq ′ 1 q 8πε0 sin θ c2 sin2 2

θ 2

. + b2

Since L does not explicitly depend on time, the Lagrange equations of the second kind admit the first integral ∂L θ˙ − L = const., ∂ θ˙ 243

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Free ebooks ==> www.Ebook777.com where the constant is the total energy of the particle Etot ∂L 1 const. = Etot = Ecin + Epot = θ˙ − L = mθ˙2 (a2 sin2 θ + b2 cos2 θ) 2 ∂ θ˙ 1 qq ′ 1 2 2 2 q − mω b sin θ − mga cos θ + 2 8πε0 sin θ c2 sin2 2

θ 2

. + b2

By identification, we then have Ecin =

1 ˙2 2 2 mθ (a sin θ + b2 cos2 θ), 2

(7.21)

and 1 qq ′ 1 q Epot = − mω 2 b2 sin2 θ − mga cos θ + 2 8πε0 sin θ c2 sin2 2

. + b2 (7.21′ ) It is worthwhile to mention, as we previously did, that within the analytical formalism, the kinetic energy can be expressed only in terms of the squared generalized velocities. To find the equilibrium positions, one takes the first derivative of the effective potential energy Epot (θ): θ 2

dEpot = −mω 2 b2 sin θ cos θ + mga sin θ dθ q c2 sin2 θ cos θ θ 1 cos c2 sin2 θ2 + b2 + √ 2 22 θ 2 2 ′ 2 2 qq 2 c sin 2 +b − 2 2 θ θ 2 8πε0 sin 2 (c sin 2 + b2 ) = −mω 2 b2 sin θ cos θ + mga sin θ

qq ′ 2c2 sin2 θ2 cos θ2 + b2 cos θ2 − . 8πε0 2 sin2 θ2 (c2 sin2 θ2 + b2 )3/2 The equilibrium positions of the system are obtained as solutions dE of the equation dθpot = 0, that is " # θ θ 2 2 ′ 2c sin + b csc qq 2 2 sin θ mω 2 b2 cos θ − mga + = 0, 8πε0 4 sin2 θ2 (c2 sin2 θ2 + b2 )3/2 or, equivalently ( sin θ = 0; mω 2 b2 cos θ − mga +

2c2 sin θ2 +b2 csc θ2 qq ′ 8πε0 4 sin2 θ2 (c2 sin2 θ2 +b2 )3/2

244

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= 0.

Free ebooks ==> www.Ebook777.com The first equation gives two positions of equilibrium (1)

(1)

θ2 = π,

θ1 = 0,

but only the second is acceptable from the physical point of view. Denoting k = 4πεo , the second equation can also be written as 

3/2  θ 2 c sin +b ω 2 b2 cos θ − ga 2   2 θ 2 ′ 2 +qq 2c sin + b = 0. 2

θ 8k m sin 2 3

2

2

Using formula cos θ = 1 − 2 sin2 still have

θ 2

(7.22)

and squaring the last equation, we

 2 3  θ 2 2 θ 2 θ 2 2 2 +b 64k m sin c sin ga − ω b 1 − 2 sin 2 2 2 2

2

6

′ 2

−(qq )



θ + b2 2c sin 2 2

2

2

= 0.

(7.23)

This is a trigonometric equation for the unknown variable 0 < θ ≤ π. (2) Its real solutions θi (i = 1, 2, ...) give new equilibrium positions. As one can see, it is not possible to solve this equation analytically, so that we have to follow a numerical procedure. Even so, due to the multitude of values which can be taken by the constant quantities a, b, q, q ′ , m, and ω, it is a very difficult task to solve the equation. We first observe that the solutions of equation (7.23) is not affected by the sign of the electric charges. Since discussion of the general case is very complicated, we shall focus our attention upon analysis in terms of m, q, and ω. To this purpose, we shall use the software Mathematica, specialized in analytical and numerical calculations. Mathematica furnishes a simple and convenient way to solve this problem, using the graphic representation of ω, considered as an independent variable (or, better, as a parameter) in terms of θ (taken as an dependent variable). This dependence is implicitly given by means of equation (7.23). For the calculations to follow, we chose the following values for the interfering constants: k = 4πε0 = 10−9 F/m, a = 0.2 m, b = 0.15 m, q = 10−3 C, q ′ = 5.10−3 C. 1) m = 1 µg = 10−6 kg The command lines allowing the graphic representation of interdependence between ω and θ are: 245

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The dependence θ = θ(ω) is graphically represented in Fig.VII.5. As can be observed, there is a minimum value of the angular velocity ω = ωmin , below which equation has no real solutions. This value can be determined observing that θ(2) → π for ω → ωmin . Solving equation (7.23) for θ = π, one finds ωmin = 1340475.662 rad · s−1 .

Fig.VII.5 246

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Free ebooks ==> www.Ebook777.com One also observes that until a value of ω denoted ω1→3 , the variation of θ in terms of ω is univocal, but beginning with ω1→3 the graph begins to split in three branches. This shows that within interval ω ∈ (ωmin , ω1→3 ) equation (7.23) has a single root and, consequently, there exists a single equilibrium position, while for ω ∈ (ω1→3 , ∞) equation (7.23) has three solutions, corresponding to three equilibrium positions of the system, determined by three values of angle θ, (2) (2) (2) denoted in Fig.VII.5 by θ1 , θ2 and θ3 . In addition, as shown in Fig.VII.6, in the (theoretical) limit ω → (2) (2) ∞, the equilibrium value θ1 of θ tends to a minimum value θ1min , determined by the equilibrium between gravitational and electrostatic forces (if the problem is analyzed in an inertial frame), and between gravitational, electrostatic and centrifugal forces (if the problem is approached in a noninertial frame, invariably connected to the charged body, rotating about x-axis with angular velocity ω). In the same limit (2) (2) ω → ∞, the other two equilibrium values θ2 and θ3 tend to the same (2) (2) limit lim θ2 = lim θ3 = π2 . ω→∞

ω→∞

Fig.VII.6 Observation. The vertical line connecting the branches of the graph θ = θ(ω) near the value ω1→3 in Fig.VII.5 (also appearing in some forthcoming drawings, like Fig.VII.10 and/or Fig.VII.14) are due to certain specific errors in graphic representations by means of instruction ”Plot” of Mathematica. This kind of error is generated when 247

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Free ebooks ==> www.Ebook777.com the first argument in ”Plot” does not evaluate to a machine-size real number at the indicated point1 . The cause of this error can often be determined by temporarily assigning the indicated value to the plot variable, evaluating the first argument of the plotting function, and observing whether or not the result is a real number. Separation of the two distinct domains, corresponding to one and three solutions for equation (7.23), respectively, is displayed even more clearly in Fig.VII.7. As can be observed, for ω < ω1→3 equation (2) (7.23) has a single solution, θ1 , while for ω > ω1→3 there exist three (2) (2) (2) solutions θ1 , θ2 and θ3 . Not to mention the equilibrium position (1) corresponding to the analytical solution θ2 = π of equation (7.23), in addition to the equilibrium positions determined by the numerical solutions of (7.23).

Fig.VII.7 2) m = 1 g = 10−3 kg In the same way, for ωmin one obtains the value ωmin = 42389, 561 rad · s−1 ; therefore, if ω < ωmin , the system has no extra equilibrium (1) position (in addition to the analytic solution θ2 = π). This fact has a graphic representation in Fig.VII.8, where dependence θ = θ(ω) is 1

The ”indicated point” is given in every line error that appears in Mathematica during the calculations (i.e., during the time interval between the instant of launching the ”Plot” command and the instant when the graphical representation is depicted). 248

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Free ebooks ==> www.Ebook777.com given for a large scale variation of ω. As one can see, there are no major qualitative differences between the last two cases. The analysis for the case m = 10−3 kg is performed as for the previous case m = 10−6 kg. The only clear-cut difference refers to the order of magnitude of ω corresponding to the number of equilibrium positions of the system. This can be easily observed by comparing the graphic representations of the two cases (Fig.VII.6 with Fig.VII.8, or Fig.VII.7 with Fig.VII.9 and Fig.VII.10).

Fig.VII.8 We realize that an increase of three times of the order of magnitude of mass produces a two-order of magnitude reduction of the angular velocity corresponding to separation of the two domains, associated with one or three equilibrium positions of the system, respectively. Fig.VII.9 shows a graphic representation of the dependence θ = θ(ω) for a narrow interval of variation of angular velocity ω, situated to the left of ω1→3 , that is within the domain in which the system (2) possess a single equilibrium position, θ1 . The same dependence is displayed in Fig.VII.10 and Fig.VII.11, but for an interval of variation of ω situated to the right of ω1→3 , that is within the domain where the body of mass m and charge q has three equilibrium positions. This has been done to give even a more clear explanation regarding displacement of the three equilibrium positions to the limit values discussed at point 1). Indeed, comparison between Fig.VII.10 and Fig.VII.11 shows that, together with increase of ω, 249

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Free ebooks ==> www.Ebook777.com (2)

the ”separation/distance” between the equilibrium positions θ2 and (2) θ3 (both tending to the same limit for ω → ∞) decreases, while the (2) (2) (1) ”distance” between the ”pair” (θ2 , θ3 ) and θ1 increases.

Fig.VII.9

Fig.VII.10

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Fig.VII.11

Fig.VII.12 3) m = 100 g = 10−1 kg In this case, the graphic representation of θ = θ(ω) is given in Fig. VII.12. It can be seen that the qualitative dependence of θ on ω keeps the same properties as in the previous two cases. The minimum value of ω now is ωmin = 4238, 946 rad · s−1 . As in the preceding case, 251

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Free ebooks ==> www.Ebook777.com the main difference as regarded the first case consists in diminution of the order of magnitude of angular velocity ω, with respect to the mass increase. So, to an increase of five orders of magnitude (as compared to the first case), and two orders of magnitude (as compared to the second case) of the mass, corresponds a decrease of three orders of magnitude of angular velocity (with respect to the first case), and one order of magnitude (as compared to the second case), respectively. b) To determine stability of the above discussed equilibrium positions we must investigate the sign of the second derivative of the effective potential energy in these points. We have: qq ′ d2 Epot 2 2 2 2 2 2 = −mω b cos θ + mω b sin θ + mga cos θ − dθ2 8πε0  3/2 − 21 b2 sin θ2 sin2 θ2 c2 sin2 θ2 + b2 3 c2 sin2 θ2 + b2 3/2 i h 2 θ θ θ 2 θ 2 2 + b cos 2 sin 2 cos 2 c sin 2 + b 3 4 sin4 θ2 c2 sin2 θ2 + b2

2 2c2 sin θ2 cos2

θ 2

− c2 sin3

2 2 qq ′ 2 2c sin + 8πε0

θ 2

cos θ2

×

θ 2 4 θ 4 sin 2

+

2 2c2 sin2 ×

θ 2

cos θ2 + b2 cos θ2 4 sin4

θ 2

qq ′ 8πε0

h 3

3 2 2 c sin

c2 sin2

θ 2

cos θ2 c2 sin2 3 + b2 θ 2

θ 2

+ b2

1/2 i

sin−3 θ2 qq ′ 4πε0 8 c2 sin2 θ + b2 5/2 2 h   i 4 θ 2 θ 2 θ 2 θ 2 θ 4 2 2 4 × 2c sin 3 − 2 sin + c b sin (5 − 2 sin + b (2 − sin . 2 2 2 2 2 Denoting sin−3 θ2 f (θ) = 5/2 8 c2 sin2 θ2 + b2 h θ θ θ θ θ i + c2 b2 sin2 (5 − 2 sin2 + b4 (2 − sin2 , × 2c4 sin4 3 − 2 sin2 2 2 2 2 2 the second derivative with respect to θ of the potential energy becomes: = −mω 2 b2 cos2 θ +mω 2 b2 sin2 θ +mga cos θ +

d2 Epot qq ′ 2 2 2 2 2 2 = −mω b cos θ + mω b sin θ + mga cos θ + f (θ) dθ2 4πε0 252

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Free ebooks ==> www.Ebook777.com = −mω 2 b2 (1 − 2 sin2 θ) + mga cos θ + k −1 qq ′ f (θ).

(7.24)

To investigate the sign of this expression, we shall appeal to soft Mathematica. As can be observed, the sign of this expression significally depends on both mass and angular velocity, for given values of the other quantities. So, for m = 1 g, if ωmin < ω < ωe∗ = 585310, 223 rad · s−1 , equation d2 Epot = −mω 2 b2 (1 − 2 sin2 θ) + mga cos θ + k −1 qq ′ f (θ) = 0 (7.25) dθ2

has a single positive real solution, therefore  ithe function F (θ) =

d2 Epot dθ 2

changes its sigh once in interval θ ∈ 0, π , its graph being shown in

Fig.VII.13. The positive, real solution of equation (7.25), corresponding to ω = 200000 rad · s−1 (this value of ω is arbitrarily chosen, being contained in the above-mentioned interval of variation) can be determined by the command line

The physically acceptable solution of the equation in x is x0 = 0, 9299, corresponding to an angle of value θ0 = 2, 388 rad ≃ 0, 76 π rad. Next command lines allow one to give a graphic representation of F (θ) in the interval θ ∈ (0, π], for the arbitrary values m = 1 g and ω = 2 × 105 rad · s−1 .

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Fig.VII.13

Fig.VII.14 According to Fig.VII.13, the equilibrium positions (physically acceptable) obtained as solutions of equation (7.23) situated within the interval θ ∈ (0, θ0 ) are positions of stable equilibrium, while those belonging to the interval θ ∈ (θ0 , π] are positions of unstable equilibrium. 254

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Free ebooks ==> www.Ebook777.com For ω > ω ∗ , equations (7.25) admits three real, positive, physically acceptable solutions. In other words, in the interval θ ∈ (θ0 , π] function F (θ) changes its sign three times (see Fig.VII.14, which gives a graphic representation of F (θ) for a conveniently chosen ω = 6 × 106 rad · s−1 ). Consequently, the equilibrium positions (obtained as solutions S of equation (7.23)) satisfying relation θ ∈ (0, θ1 ) (θ2 , θ3 ) are positions of stable equilibrium (the second derivative of the effective potential energy S is positive), while those situated within the interval θ ∈ (θ1 , θ2 ) (θ1 , π] are positions of unstable equilibrium. The numerical analysis shows that the smallest solution of (7.25) diminishes with the increase of ω, tending to zero when ω → ∞ (see Fig.VII.17). Therefore, at high angular velocities the stable equilibrium positions are contained only within the interval θ ∈ (θ2 , θ3 ) ≈ π4 , 3π 4 . c) Small oscillations can occur, obviously, only about positions of stable equilibrium. Since we have already obtained the kinetic energy in the frame of Problem 1, the same formula

τ = 2π

s

m(a2 sin2 θse + b2 cos2 θse ) , (d2 Epot /dθ2 )|θ=θse

Fig.VII.15 can be used to find the oscillation period of the system about the position of stable equilibrium, where the values of θse are replaced with the expressions determined at the points a) and b) of this problem. 255

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Free ebooks ==> www.Ebook777.com Particular case: a = b = R (the ellipse degenerates into a circle of radius R). In this case 1 qq ′ Vel = , 4πε0 2R sin θ2 and the Lagrangian becomes L=

1 1 qq ′ 1 mR2 θ˙2 + mω 2 R2 sin2 θ + mgR cos θ − . 2 2 4πε0 2R sin θ2

This formula shows that the effective potential energy writes [see (7.21)]: 1 1 qq ′ Epot = − mω 2 R2 sin2 θ − mgR cos θ + . 2 4πε0 2R sin θ2 The equilibrium positions of the system are then obtained as # " ′ d 1 qq dEpot 1 = − mω 2 R2 sin2 θ − mgR cos θ + dθ dθ 2 4πε0 2R sin θ2 1 qq ′ cos θ2 = −mω R sin θ cos θ + mgR sin θ − = 0, 4πε0 4R sin2 θ2 2

or

2

cos θ2 16πε0 R sin2 θ2

h

− 32πε0 mω 2 R3 sin3

θ 2

i θ θ + 32πε0 mgR2 sin3 − qq ′ = 0, 2 2 which is equivalent to the following system of two equations  θ   cos = 0; 2 (7.26)   32πmε0 R2 sin3 θ (g − ω 2 R cos θ) − qq ′ = 0. 2 +64πε0 mω 2 R3 sin5

The physically accepted solution of the first equation is θ1 = π, while the second equation has no analytical solution (which is exact), while a numerical solution is almost always approximate. To solve it, we shall use again the soft Mathematica, but this time we shall focus our attention upon only one case, namely: m = 1 g, R = 0.2 m, q = 10−3 C, q ′ = 5×10−3 C. The following command lines allow one to give 256

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Free ebooks ==> www.Ebook777.com the graphic representation of θ = θ(ω), which is implicitly expressed by the equation (7.26)2 , as shown in Fig.VII.16.

Fig.VII.16 The minimum value of angular velocity ωmin = 26516, 503 rad · s−1 (i.e., under which equation (7.26)2 has no real solution) can be determined by means of the command lines: 257

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As can be observed in Fig.VII.16, for any value ω > ωmin of angular velocity, equation (7.26)2 has a solution which is physically acceptable, namely π/2 < θ2 < π. Consequently, putting our results all together, the system has two equilibrium positions, given by the analytical solution θ1 = π, and, respectively, by the numeric solution π/2 < θ2 < π. The stability of these positions can be studied by means of the second derivative with respect to θ of the potential energy: 2

"

qq ′ cos θ2

1 d Epot d −mω 2 R2 sin θ cos θ + mgR sin θ − = 2 dθ dθ 4πε0 4R sin2

= mω 2 R2 (1 − 2 cos2 θ) + mgR cos θ +

θ 2

#

qq ′ 1 + cos2 θ2 . 4πε0 8R sin3 θ2

In order to investigate the sign of this expression within the interval 0 < θ ≤ π, we consider the graphic representation of function D(θ) ≡

qq ′ 1 + cos2 θ2 d2 Epot 2 2 2 = mω R (1 − 2 cos θ) + mgR cos θ + . dθ2 4πε0 8R sin3 θ2

In this case also exists a certain value of the angular velocity ω, namely ωc∗ = 379162, 149 rad · s−1 below which equation D(θ) = 0 has a single physically accepted solution; in other words, for ω < ωc∗ , the function D(θ) changes only once its sign in the interval 0 < θ ≤ π (i.e., the graph of D(θ) passes only once through zero). For example, taking ω = 105 rad · s−1 , equation D(θ) = 0 has only one physically accepted solution (see Fig.VII.17), which is θ0 = 2, 405 rad ≃ 0, 766 π rad. Any equilibrium position 0 < θ ≤ π, physically accepted, of equation (7.26)2 is a position of stable equilibrium if θ ∈ (0, θ0 ), and, respectively, of unstable equilibrium, if θ ∈ (θ0 , π]. 258

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Fig.VII.17 d2 E

If ω > ωc∗ , then equation D(θ) = dθpot = 0 has three real solu2 tions, all of them bearing physical significance for our problem. For example, if we set ω = 5 × 106 rad · s−1 , then the graphical representation of dependence D = D(θ) looks like in Fig.VII.18, which puts into evidence two subintervals for each positive - and, respectively, negative - values of angle θ, as solution S of the equation D(θ) = 0. As observed in Fig.VII.18, if θ ∈ (0, θ1 ) (θ2 , θ3 ), then the considered equilibrium S position is a position of stable equilibrium, while if θ ∈ (θ1 , θ2 ) (θ3 , π], we have a position of unstable equilibrium.

Fig.VII.18 259

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Free ebooks ==> www.Ebook777.com The numerical analysis shows that in the limit ω → ∞, θ1 → 0 (see Fig.VII.19). Therefore, for very large angular velocities, as much as for the ellipse case, the positions of stable  equilibrium are comprised only in the interval θ ∈ (θ2 , θ3 ) ≃ π4 , 3π 4 . To determine the period of small oscillations, one uses the formula (previously obtained for ellipse, by setting a = b = R) v u m τ = 2πRu , t d2 Epot 2 dθ θ=θse

where the values of θse corresponding to the stable equilibrium states are replaced in terms of the new configuration (circle of radius R).

Fig.VII.19 Observation. If, in addition to the above conditions, the circle is at rest, our problem becomes considerably simpler. Setting ω = 0 and resuming the previous steps, we have: L=

1 1 qq ′ mR2 θ˙2 + mgR cos θ − , 2 4πε0 2R sin θ2

it which case the effective potential energy is Epot = −mgR cos θ +

qq ′ 1 . 4πε0 2R sin θ2

The equilibrium positions of the system are found by means of equation dEpot 1 qq ′ cos θ2 = mgR sin θ − =0 dθ 4πε0 4R sin2 θ2 260

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cos θ2 16πε0 R sin2

θ 2

32πε0 mgR2 sin3

 θ − qq ′ = 0, 2

which is equivalent to the system of two equations  θ   cos = 0; 2   32πε0 mgR2 sin3 θ − qq ′ = 0. 2

(7.27)

The only physically acceptable solution of the first equation is θ1 = π. Equation (7.27)2 furnishes a second equilibrium position, given by θ2 = 2 arcsin



qq ′ 32πε0 mgR2

1/3

,

provided that the charges have the same sign, and qq ′ < 32πε0 mgR2 . The type of equilibrium (stable or unstable) is shown by the sign of the second-order derivative of Epot at the points θ1 and θ2 : ”plus” for stable equilibrium, and ”minus” for unstable equilibrium. We have: # " d2 Epot d 1 qq ′ cos θ2 = mgR sin θ − dθ2 dθ 4πε0 4R sin2 θ2 qq ′ sin2 θ2 + 2 cos2 = mgR cos θ + 32πε0 R sin3 θ2

θ 2

θ qq ′ 2 − sin2 θ2 = mgR(1 − 2 sin2 ) + , 2 32πε0 R sin3 θ2 leading to

d2 Epot dθ2

and d2 Epot dθ2 "

θ=θ2

= mgR 1 − 2



θ=π

= −mgR +

qq ′ , 32πε0 R

(7.28)

  qq ′ 2 − sin2 θ22 2 θ2 = mgR 1 − 2 sin + 2 32πε0 R sin3 θ22 ′

qq 32πε0 mgR2

2/3 #

+



qq 32πε0 R

2−



qq ′ 32πε0 mgR2

qq ′ 32πε0 mgR2

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2/3

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" 2/3 #  2/3 # qq ′ qq ′ = mgR 1 − 2 + mgR 2 − 32πε0 mgR2 32πε0 mgR2 2/3  θ2 qq ′ = 3mgR cos2 = 3mgR − 3mgR > 0. (7.29) 2 32πε0 mgR 2 

According to (7.28), if charges have different signs, or have the same sign and qq ′ < 32πε0 mgR2 , then θ1 = π is a position of unstable equilibrium, while if charges have the same sign and qq ′ > 32πε0 mgR2 , then θ1 = π is a position of stable equilibrium. Relation (7.29) shows that the equilibrium position given by 1/3  qq ′ , θ = θ2 = 2 arcsin 32πε0 mgR2

if exists, is a position of stable equilibrium. The period of small oscillations about the equilibrium positions is v u m τ = 2πRu . t d2 Epot 2 dθ θ=θse

According to our previous results, if charges have the same sign and qq ′ > 32πε0 mgR2 , then about position of stable equilibrium θ1 = π occur small oscillations with period [see (7.28)] v s u m m u = 2πR τ = 2πRt 2 qq ′ d Epot −mgR + 32πε 0R dθ 2 θ=π

= 2πR

where

s

s

−1/2  32πε0 mR R qq ′ = 2π −1 qq ′ − 32πε0 mgR2 g 32πε0 mgR2  −1/2 qq ′ = τ0 −1 , 32πε0 mgR2 τ0 = 2π

s

R . g

If charges have the same sign and qq ′ < 32πε0 mgR2 , then about the 1/3 position of stable equilibrium θ2 = 2 arcsin qq ′ /32πε0 mgR2 take place small oscillations with period [see (7.29)] v v u u m 1 u τ = 2πRu = 2πR  2/3 t t d2 Epot qq ′ 2 3gR − 3gR 32πε0 mgR2 dθ θ=θ 2

262

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Free ebooks ==> www.Ebook777.com = 2π

s

= τ0

R g

( "

3 1−

( "

3 1−







qq 32πε0 mgR2

qq ′ 32πε0 mgR2

2/3 #)−1/2

2/3 #)−1/2

.

Observation. If, in particular, qq ′ = 32πε0 mgR2 , the equilibrium position given by θ2 = 2 arcsin(qq ′ /32πε0 mgR2 )1/3 shall coincide with that given by θ1 = π. In addition, we have   d2 Epot 2 θ2 = 3mgR cos = 0, dθ2 θ=θ2 =π 2 θ2 =π

therefore, to determine the stability of the equilibrium position θ2 = π in this special case (qq ′ = 32πε0 mgR2 ) we must calculate the derivatives of higher order of the effective potential energy, more precisely, up to a non-zero derivative for θ = π. We have:   d3 Epot d d2 Epot = dθ3 dθ dθ2 # " qq ′ 2 − sin2 θ2 d 2 θ mgR(1 − 2 sin ) + = dθ 2 32πε0 R sin3 θ2 = −2mgR sin qq ′ − sin4 + 32πε0 R

θ 2

cos θ2 −

3 2

θ θ cos 2 2

sin2

θ 2 6 θ sin 2

cos θ2 2 − sin2

θ 2

 6 − sin2 θ2 cos θ2 qq ′ , = −mgR sin θ − 64πε0 R sin4 θ2 which yields

d3 Epot dθ3

= 0. θ=π

This means that we also have to calculate   d4 Epot d d3 Epot = dθ4 dθ dθ3 " #  6 − sin2 θ2 cos θ2 d qq ′ =− mgR sin θ + dθ 64πε0 R sin4 θ2 263

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Free ebooks ==> www.Ebook777.com − qq ′ = −mgR cos θ − 64πε0 R

1 2

sin θ2 (6 − sin2 θ2 ) − sin θ2 cos2 sin8

qq ′ 2 sin3 + 64πε0 R = −mgR cos θ + +

θ 2

θ 2

θ 2



sin4

θ 2

cos2 θ2 (6 − sin2 θ2 ) sin8

θ 2

qq ′ 6 − sin2 θ2 + 2 cos2 64πε0 R 2 sin3 θ2

θ 2

qq ′ 2 cos2 θ2 (6 − sin2 θ2 ) . 64πε0 R sin5 θ2

Setting now θ = π and qq ′ = 32πε0 mgR2 , we still have d4 Epot dθ4 θ=π qq ′ =32πε0 mgR2

=

qq ′ 6 − sin2 θ2 + 2 cos2 64πε0 R 2 sin3 θ2 ! 2 cos2 θ2 (6 − sin2 θ2 )

−mgR cos θ +

qq ′ + 64πε0 R 5 = mgR + 4



sin5

qq ′ 32πε0 R



θ 2

θ 2

θ=π qq ′ =32πε0 mgR2

= qq ′ =32πε0 mgR2

9 mgR > 0, 4

Consequently, in this special case (qq ′ = 32πε0 mgR2 ) the equilibrium position for θ = π is a position of stable equilibrium, but oscillations about it are no longer harmonic.

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CHAPTER VIII PROBLEMS SOLVED BY MEANS OF THE HAMILTONIAN FORMALISM

Problem 1 An annular pendulum is a physical pendulum consisting of a homogeneous circular crown of radii r and R. Using the Hamiltonian technique, determine the equation of motion of such a pendulum, and the period of small oscillations about a fixed point O situated on the interior circle (see Fig.VIII.1).

Fig.VIII.1 Solution Let d be the thickness of the circular crown, and µ its mass density. Then the mass of the crown is m = πµd(R2 − r2 ). (8.1) 265

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Free ebooks ==> www.Ebook777.com Obviously, the system has only one degree of freedom. Denoting by θ the associated generalized coordinate, we are now prepared to write Hamilton’s function. Since the system is conservative, the Hamiltonian is expressed in terms of the kinetic and potential energies. The kinetic energy of the pendulum is T =

1 2 1 Iω = I θ˙2 , 2 2

(8.2)

where I is the moment of inertia of the circular crown with respect to z-axis. To determine this quantity, we appeal to Steiner’s theorem I ≡ IO = IA + mr2 ,

(8.3)

where IA is the moment of inertia of the circular crown with respect to the axis passing through its centre of symmetry A, being parallel to Oz. According to the definition, Z IA = (x′2 + y ′2 ) dm, (8.4) (D)

where x′ and y ′ are considered with respect to a reference frame with its origin at A, and (D) is the domain represented by all points of the circular crown. Due to cylindrical symmetry of the crown, the integral (8.4) can be easier calculated in cylindrical coordinates. Thus, we have:

IA =

Z

D



2

ρ dm =

Z

2

ρ µ ρ δρ dz dϕ = µ

ZR

3

ρ dρ

r

D

Zd

dz

0

R4 − r 4 1 m d 2π = π d µ(R4 − r4 ) = (R2 + r2 ). 4 2 2

Z2π



0

(8.5)

In this case, equation (8.3) leads to I = IA + mr2 =

m 2 (R + 3r2 ), 2

(8.6)

and the kinetic energy of the annular pendulum writes T =

1 ˙2 m I θ = (R2 + 3r2 )θ˙2 . 2 4 266

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(8.7)

Free ebooks ==> www.Ebook777.com ~ = (mg, 0, 0), Since the only applied force is the force of gravity G which is a conservative force, and taking into account the symmetry of the body, the potential energy of the system can be easily determined by means of ~ = −gradV, G that is

~ · d~r = −mg dx, dV = −G

so that V = −mgx + V0 . A convenient choice for the reference level of the potential energy, V (x = 0) = 0, yields V = −mgx = −mgr cos θ.

(8.8)

The Lagrangian of the system then is L=T −V =

m 2 (R + 3r2 )θ˙2 + mgr cos θ, 4

(8.9)

allowing to calculate the generalized momentum associated with θ pθ =

∂L m ˙ = (R2 + 3r2 )θ, ˙ 2 ∂θ

(8.10)

and the Hamiltonian writes m H = pθ θ˙ − L = (R2 + 3r2 )θ˙2 − mgr cos θ. 4 Since θ˙ =

(8.11)

2pθ , m(R2 + 3r2 )

we still have p2θ H = H(θ, pθ ) = − mgr cos θ. m(R2 + 3r2 )

(8.12)

Hamilton’s canonical equations then yield:  ∂H 2pθ   θ˙ = = ; ∂pθ m(R2 + 3r2 )   p˙ θ = − ∂H = −mgr sin θ. ∂θ 267

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(8.13)

Free ebooks ==> www.Ebook777.com Taking the time derivative of (8.13)1 , and using (8.13)2 , we have θ¨ +

R2

2gr sin θ = 0, + 3r2

(8.14)

which is the differential equation of motion. For small oscillations (sin θ ≈ θ), equation (8.14) becomes θ¨ +

R2

2gr θ = 0. + 3r2

(8.15)

Denoting ω02 =

R2

2gr , + 3r2

(8.16)

we are left with θ¨ + ω02 θ = 0,

(8.17)

which is the well-known differential equation of the linear harmonic oscillator, whose period is 2π T0 = = 2π ω0

s

R2 + 3r2 . 2gr

(8.18)

The last formula can be used to determine the gravitational acceleration g at the place of the Earth where the experiment is performed: g = 2π 2

R2 + 3r2 1 . r T02

(8.19)

Suppose, in this respect, that the radii r and R, and the period of small oscillations T0 are determined with high precision, while the conditions of isochronocity are fulfilled as well as possible. In this case, the relative error in determination - with the help of annular pendulum - of the gravitational acceleration g, can be estimated by means of the logarithmic finite difference method. So, we have: ln g = ln(2π 2 ) + ln(R2 + 3r2 ) − ln r − ln T02 . Differentiating this relation, then going to finite variations, we have: ∆g 2R2 ∆R R2 − 3r2 ∆r ∆T0 = 2 − 2 −2 . 2 2 g R + 3r R R + 3r r T0 268

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(8.20)

Free ebooks ==> www.Ebook777.com Denoting by δg = ∆g g the relative error on g, and by δr, δR, δT0 the relative errors committed in determination of the corresponding quantities, we finally obtain δg =

2R2 |R2 − 3r2 | δR + δr + 2δT0 , R2 + 3r2 R2 + 3r2

(8.21)

where, in order to consider the maximum relative error affecting the determination of g, all terms in (8.20) have been taken with plus sign. To conclude, the determination of g has to be written as   2 2 1 2 R + 3r ± ∆g m · s−2 , (8.22) g = 2π 2 r T0 where ∆g = g δg, or   2 2 1 2 R + 3r g = 2π m · s−2 ± δg · 100%, r T02

(8.23)

where δg is given by (8.21). Here δR, δr and δT0 are usually given by the characteristics of the measuring instruments. For example, if the radii of the circular crown are determined by the slide rule, then the absolute error affecting the determination of r and R is about 0.1mm, while if the time interval is measured by a digital chronometer, the absolute error on time could be of 0.01 s, or even smaller. In general, the errors on distances and on time are dictated by both the precision of the employed instruments, and the method of measurement. If, for instance, r = 10 cm, R = 41 cm, and T0 = 2s, in which case we have to do with a second pendulum 1 , then, according to (8.19), we have g = 2π 2

R2 + 3r2 1 = 9.8 m · s−2 . r T02

(8.24)

To conclude, in agreement with (8.22), the result regarding determination of g must be written as g = (9.8 ± 0.1)m · s−2 ,

(8.25)

g = 9.8 m · s−2 ± 1.11%,

(8.26)

or, equivalently, 1

A second pendulum is a pendulum whose period is precisely two seconds: one second for a swing in one direction, and one second for the return swing. 269

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Free ebooks ==> www.Ebook777.com where we used (8.21) to write δg = 1.11 × 10−2 .

(8.27)

The relative error on determination of g is dictated by the relative error on T0 , which is five times bigger than that on r, and more than twenty times bigger than that on R: δT0 = 5 · 10−3 ,

δr ≃ 10−3 ,

δR ≃ 2.44 · 10−4 .

Problem 2 Using the Hamiltonian formalism, write the differential equation of motion of a homogeneous circular cylinder of radius a and mass m, rolling without sliding inside a fixed cylinder of radius R, as shown in Fig.VIII.2. Determine the period of small oscillations of the cylinder about its position of stable equilibrium.

Fig.VIII.2 Solution The motion of rolling without sliding of the cylinder or radius a on the inner surface of the fixed cylinder of radius R is a motion about the instantaneous axis of rotation, which coincides with the contact line of the two cylinders. This motion can be decomposed in two motions, as follows: a motion of pure rotation of the cylinder of radius a about its own axis of rotation (passing through O′ and parallel to both generatrix line and instantaneous axis of rotation see Fig.VIII.2), and a motion of ”translation” of the symmetry centre O′ along a circle arc with all points at the same distance relative to 270

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Free ebooks ==> www.Ebook777.com the inner surface of the fixed cylinder. In fact, the second motion is also a motion of rotation, but of the cylinder of radius a ”as a whole” about z-axis. The kinetic energy associated to this motion can be determined considering that the cylinder mass is concentrated at point O′ , which rotates about z-axis on a circle of radius R − a. The motion of the moving cylinder is subject to two constraints: i) zO′ = const., 2 2 ii) x2O′ + yO ′ − (R − a) = 0, meaning that the system has one degree of freedom. Denoting by θ the angle between vertical and the straight line connecting the centers of cylinders, the linear velocity of the cylinder of radius a moving ”as ˙ − a), and the associated kinetic a whole” about z-axis is |~vO′ | = θ(R energy writes TO ′ =

1 1 m|~vO′ |2 = m(R − a)2 θ˙2 . 2 2

(8.28)

The total kinetic energy of the moving cylinder then writes as a sum of the kinetic energy of the body ”as a whole”, given by (8.28), and the kinetic energy of the moving cylinder about its symmetry axis: T ≡ TO = TO ′ + T ′ =

1 1 m|~vO′ |2 + Iz′ z′ ω 2 , 2 2

(8.29)

where Iz′ z′ is the moment of inertia of the moving cylinder with respect to its symmetry axis Iz ′ z ′ =

Z

D

2

r dm =

Za

r2 ρ h 2πr dr =

1 ma2 . 2

(8.30)

0

Here D is the set of all material points that belong to the moving cylinder, h is its length, and ρ its mass density. Relation (8.29) represents nothing else but K¨onig’s second theorem for the compound motion of the moving cylinder. The relation between ω and θ can be found at least in two different ways: (i) by means of velocity analysis, and (ii) by means of estimation of angles swept by various axes of the moving cylinder in different reference frames. The simplest way takes into account the fact that the centre of inertia of the moving cylinder is situated on its symmetry axis (passing through O′ ), while its linear velocity is ˙ v = |~vO′ | = (R − a)θ. 271

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Free ebooks ==> www.Ebook777.com The angular velocity of the motion of rotation about the instantaneous axis of rotation, which coincides with the contact lines of cylinders (passing through A and being parallel to z-axis) then is Ω=

v R−a ˙ = θ. a a

Since ω = −Ω (rotation of O′ with respect to A, with angular velocity ~ = Ω = ω is equivalent to rotation of A with respect to O′ with |Ω| ~ = Ω = ω), it follows that angular velocity |~ ω | = | − Ω| ω = |~ ω| = so that T′ =

R−a ˙ θ, a

1 1 Iz′ z′ ω 2 = m(R − a)2 θ˙2 2 4

and, therefore, T ≡ TO = TO ′ + T ′ =

3 m(R − a)2 θ˙2 . 4

(8.31)

The Lagrangian then writes L=T −V =

3 m(R − a)2 θ˙2 − V. 4

(8.32)

Since the only force applied to the moving cylinder is the force of ~ = m~g , and due to cylindrical symmetry of the body, we gravity G have ~ · d~r = −mg dx dV = −G or V (x) = −mgx + V0 .

(8.33)

A convenient choice for the reference level of the potential energy, that is V (x = 0) = 0, leads to V (x) = −mgx, or V (θ) = −mg(R − a) cos θ,

(8.34)

and the Lagrangian (8.32) takes the form L=

3 m(R − a)2 θ˙2 + mg(R − a) cos θ. 4 272

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(8.35)

Free ebooks ==> www.Ebook777.com The generalized momentum pθ , canonically conjugated with θ, is ∂L 3 ˙ = m(R − a)2 θ, ˙ 2 ∂θ

pθ =

(8.36)

and the Hamiltonian writes ˙ = θp ˙ θ −L= H(θ, θ)

3 m(R − a)2 θ˙2 − mg(R − a) cos θ. 4

(8.37)

To express H in terms of θ and pθ , one uses (8.36). The result is: H(θ, pθ ) =

p2θ − mg(R − a) cos θ. 3m(R − a)2

Hamilton’s canonical equations then yield  ∂H 2pθ   θ˙ = = ; ∂pθ 3m(R − a)2   p˙ θ = − ∂H = −mg(R − a) sin θ. ∂θ

(8.38)

(8.39)

Taking the time derivative of (8.39)1 and using (8.39)2 , we obtain θ¨ +

2g sin θ = 0. 3(R − a)

Denoting ω02 = we finally have

(8.40)

2g , 3(R − a)

θ¨ + ω02 sin θ = 0.

(8.41)

This is the differential equation of motion of the cylinder of mass m and radius a ”as a whole” relative to the reference frame Oxyz. For small angular amplitudes (sin θ ≃ θ), equation (8.41) turns into a wellknown equation of the linear harmonic oscillator. The period of small oscillations about the position of stable equilibrium is s 2π 3(R − a) T0 = = 2π . (8.42) ω0 2g This formula can be used to determine the gravitational acceleration g: 6π 2 (R − a) g= . (8.43) T02 273

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Free ebooks ==> www.Ebook777.com Problem 3 Write the Hamiltonian of a material point (particle) of mass m, situated in a conservative force field, in Cartesian, cylindrical, planepolar, spherical, parabolic, and elliptic coordinates. Solution i) Cartesian coordinates x, y, z Let us suppose that the particle moves freely in the conservative field (there are no constraints) characterized by the potential energy V (x, y, z). The Lagrangian then writes L=T −V =

1 m(vx2 + vy2 + vz2 ) − V (x, y, z) 2

1 m(x˙ 2 + y˙ 2 + z˙ 2 ) − V (x, y, z). (8.44) 2 The generalized momenta, canonically conjugated to the generalized coordinates x, y, z are:  ∂L   px = = mx˙ ;   ∂ x˙   ∂L = my˙ ; py = (8.45)  ∂ y˙      pz = ∂L = mz˙ . ∂ z˙ =

The Hamiltonian then writes H = pi q˙i − L =

1 m(x˙ 2 + y˙ 2 + z˙ 2 ) + V (x, y, z). 2

Expressing x, ˙ y, ˙ z˙ in terms of px , py , pz ,  px  x ˙ = ,   m  py y˙ = , m     z˙ = pz , m

(8.46)

(8.47)

and introducing these quantities into (8.46), we finally have H=

1 2 (p + p2y + p2z ) + V (x, y, z). 2m x

ii) Cylindrical coordinates ρ, ϕ, z 274

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(8.48)

Free ebooks ==> www.Ebook777.com In this case, the coordinate transformations are: ( x = ρ cos ϕ; y = ρ sin ϕ; z = z.

Then

( x˙ = ρ˙ cos ϕ − ρϕ˙ sin ϕ; y˙ = ρ˙ sin ϕ + ρϕ˙ cos ϕ; z˙ = z, ˙

(8.49)

(8.50)

and the squared velocity is v 2 = x˙ 2 + y˙ 2 + z˙ 2 = ρ˙ 2 + ρ2 ϕ˙ 2 + z˙ 2 , which allows us to write the kinetic energy T (ρ, ρ, ˙ ϕ, ˙ z) ˙ =

1 m(ρ˙ 2 + ρ2 ϕ˙ 2 + z˙ 2 ), 2

and the Lagrangian is L=T −V =

1 m(ρ˙ 2 + ρ2 ϕ˙ 2 + z˙ 2 ) − V (ρ, ϕ, z). 2

The associated conjugate momenta are:  ∂L   pρ = = mρ; ˙   ∂ ρ˙   ∂L = mρ2 ϕ; ˙ pϕ =  ∂ ϕ˙      pz = ∂L = mz. ˙ ∂ z˙

(8.51)

(8.52)

According to definition, the Hamiltonian is H=

3 X i=1

pi q˙i − L = pρ ρ˙ + pϕ ϕ˙ + pz z˙ − L

1 m(ρ˙ 2 + ρ2 ϕ˙ 2 + z˙ 2 ) + V (ρ, ϕ, z). 2 Expressing ρ, ˙ ϕ, ˙ z˙ in terms of generalized momenta  1  ρ˙ = pρ ;   m   1 ϕ˙ = pϕ ;  mρ2     z˙ = 1 p , z m =

275

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(8.53)

(8.54)

Free ebooks ==> www.Ebook777.com and introducing these quantities into (8.53), we finally obtain 1 H= 2m

  1 2 2 2 pρ + 2 pϕ + pz + V (ρ, ϕ, z). ρ

(8.55)

iii) Plane-polar coordinates ρ, θ The plane-polar coordinates are obtained from cylindrical coordinates by setting z = 0 and ϕ → θ. The result is: 1 H= 2m



p2ρ

1 + 2 p2θ ρ



+ V (ρ, θ).

(8.56)

iv) Spherical coordinates r, θ, ϕ Using relations between Cartesian and spherical coordinates (

x = r sin θ cos ϕ; y = r sin θ sin ϕ; z = r cos θ,

we have   x˙ = r˙ sin θ cos ϕ + rθ˙ cos θ cos ϕ − rϕ˙ sin θ sin ϕ; ˙  y˙ = r˙ sin θ sin ϕ˙ + rθ cos θ sin ϕ + rϕ˙ sin θ sin ϕ; z˙ = r˙ cos θ − rθ sin θ.

(8.57)

(8.58)

The kinetic energy writes

T (r, θ, ϕ) =

1 m(r˙ 2 + r2 θ˙2 + ϕ˙ 2 r2 sin2 θ), 2

and the Lagrangian is L=T −V =

1 m(r˙ 2 + r2 θ˙2 + ϕ˙ 2 r2 sin2 θ) − V (r, θ, ϕ). 2

(8.59)

The generalized momenta are  ∂L   = mr; ˙ pr =   ∂ r˙   ∂L ˙ pθ = = mr2 θ; ˙  ∂θ   ∂L    pϕ = = mr2 ϕ˙ sin2 θ. ∂ ϕ˙ 276

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(8.60)

Free ebooks ==> www.Ebook777.com The Hamiltonian then is H=

3 X i=1

=

pi q˙i − L = pr r˙ + pθ θ˙ + pϕ ϕ˙ − L

1 m(r˙ 2 + r2 θ˙2 + ϕ˙ 2 r2 sin2 θ) + V (r, θ, ϕ), 2

(8.61)

or, in terms of generalized coordinates and generalized momenta 1 H= 2m



p2r

1 1 + 2 p2θ + 2 2 p2ϕ r r sin θ



+ V (r, θ, ϕ).

(8.62)

v) Parabolic coordinates ξ, η, ϕ The parabolic coordinates ξ, η, ϕ can be defined starting from cylindrical coordinates ρ, ϕ, z, as follows:  

1 (ξ − η); 2 p  ρ = ξη. z=

(8.63)

The coordinate ϕ is the same in both coordinate systems, cylindrical and parabolic. Coordinates ξ and η can take any value between 0 and ∞. Eliminating η from (8.63), we have ξ 2 − 2ξz − ρ2 = 0, with the solutions ξ1,2 = z ± Setting ξ = const., this means z± or

p

(8.64)

z 2 + ρ2 .

p z 2 + ρ2 = const. = C, z=

1 (C 2 − ρ2 ), 2C

(8.65)

which is a family of paraboloids of revolution, with z-axis as axis of symmetry. Proceeding in the same manner and eliminating ξ from (8.63), we obtain p η1,2 = −z ± z 2 + ρ2 , 277

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Free ebooks ==> www.Ebook777.com so that by taking η = const. = C1 , equation z=

1 (ρ2 − C12 ) 2C1

(8.66)

represents, in its turn, a family of paraboloids of revolution, with the same axis of symmetry. Observation. Performing the change of variable r=

p z 2 + ρ2 ,

(8.67)

we can write relations (8.63) in a more convenient form. Observing that r = 21 (ξ + η) and using z = 12 (ξ − η), we obtain 

ξ = r + z; η = r − z.

(8.68)

To write the Lagrangian of a heavy particle in parabolic coordinates ξ, η, ϕ, we use the Lagrangian expressed in cylindrical coordinates (8.51) L=

1 m(ρ˙ 2 + ρ2 ϕ˙ 2 + z˙ 2 ) − V (ρ, ϕ, z). 2

Since

p  ρ = ξη;   ϕ = ϕ;   z = 1 (ξ − η), 2

we have:

Then we can write

 1  ˙  ˙   ρ˙ = 2√ξη (ξη + ξ η); ϕ˙ = ϕ; ˙   1 ˙   z˙ = (ξ − η). ˙ 2

ρ˙ 2 + ρ2 ϕ˙ 2 + z˙ 2 =

(8.69)

1 1 ˙ (ξ + ξ η) ˙ 2 + ξη ϕ˙ 2 + (ξ˙ − η) ˙ 2 4ξη 4

1 = (ξ + η) 4

ξ˙2 η˙ 2 + ξ η

!

+ ξη ϕ˙ 2 .

278

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(8.70)

Free ebooks ==> www.Ebook777.com Making allowance for (8.70), the Lagrangian (8.51) writes ! m m ξ˙2 η˙ 2 + ξη ϕ˙ 2 − V (ξ, η, ϕ). L = (ξ + η) + (8.71) 8 ξ η 2 The associated momenta then are  ∂L m ˙  pξ = = (ξ + η)ξ;   ˙  4ξ ∂ ξ   ∂L m pη = = (ξ + η)η; ˙  ∂ η˙ 4η      pϕ = ∂L = mξη ϕ. ˙ ∂ ϕ˙

(8.72)

The Hamiltonian is H=

3 X i=1

pi q˙i − L = pξ ξ˙ + pη η˙ + pϕ ϕ˙ − L

m m m (ξ + η)ξ˙2 + (ξ + η)η˙ 2 + mξη ϕ˙ 2 − (ξ + η)ξ˙2 4ξ 4η 8ξ m m − (ξ + η)η˙ 2 − ξη ϕ˙ 2 + V (ξ, η, ϕ) 8η 2 ! m ξ˙2 η˙ 2 m = (ξ + η) + + ξη ϕ˙ 2 + V (ξ, η, ϕ). 8 ξ η 2

=

(8.73)

In view of (8.72), we have  4ξpξ   ξ˙ = ;   m(ξ + η)   4ηpη η˙ = ;  m(ξ + η)    p   ϕ˙ = ϕ , mξη

(8.74)

and, introducing these expressions into (8.73), we are left with " # 16ξp2ξ 16ηp2η m H = (ξ + η) + 2 8 m2 (ξ + η)2 m (ξ + η)2 +

p2ϕ p2ϕ 2 ξp2ξ + ηp2η + V (ξ, η, ϕ) = + + V (ξ, η, ϕ). 2mξη m ξ+η 2mξη 279

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(8.75)

Free ebooks ==> www.Ebook777.com vi) Elliptic coordinates λ, µ, ϕ Transition from cylindrical to elliptic coordinates λ, µ, ϕ is performed by means of relations  p  ρ = σ (λ2 − 1)(1 − µ2 ); (8.76) ϕ = ϕ;  z = σλµ, where σ is a constant called transformation parameter. The elliptic coordinate λ takes all possible values from 1 to ∞, while µ can take all values from −1 to +1. Before going further, we observe that the relations (8.76) can be written in a more explicit geometric form. To this end, let us denote by d1 and d2 the distances, determined on z-axis, between the origin of the coordinate system and the points P1 and P2 , defined by z1 = σ and z2 = −σ, respectively. In this case, the lines λ = const. display the family of ellipsoids ρ2 z2 + =1 σ 2 λ2 σ 2 (λ2 − 1)

with foci at P1 and P2 , while the lines µ = const. represent the family of hyperboloids z2 ρ2 − = 1, σ 2 µ2 σ 2 (1 − µ2 ) with the same foci. The two segments d1 and d2 are given by p  d1 = p(z − σ)2 + ρ2 ; (8.77) d2 = (z + σ)2 + ρ2 . Using (8.76), we still have

which yields



d1 = σ(λ − µ); d2 = σ(λ + µ),

(8.78)

 d + d2  λ = 1 ; 2σ (8.79)   µ = d2 − d1 . 2σ Let us now write the Lagrangian of a heavy particle moving in a conservative force field in elliptic coordinates. In this respect, we have to introduce (8.76) into (8.51). Observing that h i  σ ˙ − µ2 ) − 2µµ(λ  2λλ(1 ˙ 2 − 1) ;  ρ˙ = p 2 2 (λ − 1)(1 − µ2 ) (8.80) ˙   ϕ˙ = ϕ; ˙ + λµ), z˙ = σ(λµ ˙ 280

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Free ebooks ==> www.Ebook777.com and, also, h i2 σ2 2 2 ˙ ρ˙ + ρ ϕ˙ + z˙ = 2 λλ(1 − µ ) − µµ(λ ˙ − 1) (λ − 1)(1 − µ2 ) 2

2

2

2

˙ + λµ) +σ 2 (λ2 − 1)(1 − µ2 )ϕ˙ 2 + σ 2 (λµ ˙ ! λ˙ 2 µ˙ 2 + σ 2 (λ2 − 1)(1 − µ2 )ϕ˙ 2 , = σ 2 (λ2 − µ2 ) − λ 2 − 1 1 − µ2

the Lagrangian becomes mσ 2 2 L= (λ − µ2 ) 2

λ˙ 2 µ˙ 2 − λ 2 − 1 1 − µ2

!

mσ 2 2 (λ − 1)(1 − µ2 )ϕ˙ 2 − V (λ, µ, ϕ). (8.81) 2 The generalized momenta, canonically conjugated to the generalized coordinates λ, µ, ϕ, are:   ∂L λ˙  2 2 2  p = = mσ (λ − µ ) ;  λ  λ2 − 1  ∂ λ˙  ∂L µ˙ (8.82) pµ = = mσ 2 (λ2 − µ2 ) 2 ;  ∂ µ˙ µ −1    ∂L   = mσ 2 (λ2 − 1)(1 − µ2 )ϕ. ˙  pϕ = ∂ ϕ˙ +

The Hamiltonian, according to its definition, then is: H=

3 X i=1

˙ λ + µp pi q˙i − L = λp ˙ µ + ϕp ˙ ϕ−L

mσ 2 2 = (λ − µ2 ) 2

λ˙ 2 µ˙ 2 − λ2 − 1 1 − µ2

!

mσ 2 2 (λ − 1)(1 − µ2 )ϕ˙ 2 + V (λ, µ, ϕ). (8.83) 2 The last step is to write the Hamiltonian in terms of generalized coordinates λ, µ, ϕ and generalized momenta pλ , pµ , pϕ . Since  2  ˙ = (λ − 1)pλ ;  λ   mσ 2 (λ2 − µ2 )   (1 − µ2 )pµ (8.84) µ ˙ = ; 2 (λ2 − µ2 )  mσ    pϕ   ϕ˙ = , 2 2 mσ (λ − 1)(1 − µ2 ) +

281

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Free ebooks ==> www.Ebook777.com the Hamiltonian finally writes: " 1 H= (λ2 − 1)p2λ + (1 − µ2 )p2µ 2mσ 2 (λ2 − µ2 ) +



1 1 + 2 λ − 1 1 − µ2



#

p2ϕ + V (λ, µ, ϕ).

(8.85)

Problem 4 Let f (q, p, t) and g(q, p, t) be two arbitrary functions of canonical variables q1 , q2 , ..., qn and p1 , p2 , ..., pn . The expression {f, g} =

∂f ∂g ∂g ∂f − ∂qi ∂pi ∂qi ∂pi

(8.86)

is called the Poisson bracket of f and g. If ~r = xi ~ui , p~ = m~r˙ , and ~l = ~r × p~ are the radius vector, the linear and the angular momenta, respectively, of a particle of mass m with respect to the Cartesian frame Oxyz of versors ~ui (i = 1, 3), prove the following relations: (i) (ii) (iii) (iv) (v)

{li , xj } = εijk xk ; {li , pj } = εijk pk ; {li , lj } = εijk lk ; {~l, ~r2 } = 0; {~l, p~2 } = 0.

Solution In order to solve the problem, we remind the reader the main properties of the Poisson brackets, together with some useful relations involving the completely antisymmetric, third rank pseudotensor εijk (the Levi-Civita symbol). So, one can easily prove that 1o . {C, f } = 0 C = const.; 2o . {Cf, g} = C{f, g}; 3o . {f, g} = −{g, f }; 4o . {−f, g} = −{f, g} = {g, f }; 5o . {f1 + f2 , g} = {f1 , g} + {f2 , g}; 6o . {f1 f2 , g} = f1 {f2 , g} + f2 {f1 , g}; 7o . D{f, g} = {Df, g} + {f, Dg}, where D is any scalar or vector ∂ ∂ differential operator, like ∂t , ∂x , ∇, etc.; o 8 . {qi , qk } = {pi , pk } = 0 , ∀ i 6= k; 282

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Free ebooks ==> www.Ebook777.com 9o . {qi , pk } = δik ; 10o . {f, {g, h}} + {h, {f, g}} {g, {h, f }} = 0 (Jacobi’s identity); P+ n ∂g o 11 . {f, g(y1 , y2 , ..., yn )} = k=1 ∂y {f, yk }, k as well as δil δim δin 12o . εijk εlmn = δjl δjm δjn ; δkl δkm δkn o 13 . εijk εimn = δjm δkn − δjn δkm ; 14o . εijk εijl = 2δkl ; 15o . εijk εijk = 3!; 16o . ~ui × ~uj = εijk ~uk (i, j, k = 1, 3); 17o . ~ui = 21 εijk ~uj × ~uk (i, j, k = 1, 3); 18o . ~ui · (~uj × ~uk ) = εijk ; 19o . εijk δij = εijk δik = εijk δjk = 0. Let us now proceed to solve our exercises (i)-(v): (i) Since ~l = ~r × p~ = εijk xj pk ~ui , we have li = εijk xj pk , therefore {li , xj } = {εilk xl pk , xj } = εilk {xl pk , xj } = εilk xl {pk , xj } + εilk pk {xl , xj } = −εilk xl δjk = εijl xl = εijk xk . (ii) {li , pj } = {εilk xl pk , pj } = εilk {xl pk , pj } = εilk xl {pk , pj } + εilk pk {xl , pj } = εilk pk {xl , pj } = εilk pk δlj = εijk pk . (iii) {li , lj } = {εilk xl pk , εjmn xm pn } = εilk εjmn {xl pk , xm pn }  = εilk εjmn xl {pk , xm pn } + pk {xl , xm pn } h   = εilk εjmn xl xm {pk , pn } + pn {pk , xm }  i +pk xm {xl , pn } + pn {xl , xm } = εilk εjmn (−xl pn δkm + pk xm δln ) = −εilk εjkn xl pn + εilk εjml pk xm = εkil εkjn xl pn − εlik εljm pk xm = (δij δln − δin δlj )xl pn − (δij δkm − δim δkj )xm pk 283

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Free ebooks ==> www.Ebook777.com not.

= δij xl pl − xj pi − δij xk pk + xi pj = xi pj − xj pi = Aij = εijk lk , where the pseudovector ~l is connected with the antisymmetric tensor Aik by relation (see Appendix B) lk =

1 1 1 εkij Aij = εkij (xi pj − xj pi ) = 2εkij xi pj = (~r × p~)k , 2 2 2

which means that the pseudovector ~l is precisely the angular momentum. (iv) {~l, ~r2 } = ~i{lx , ~r2 } + ~j{ly , ~r2 } + ~k{lz , ~r2 }. The first Poisson bracket on the r.h.s. yields {lx , ~r2 } = {ε1jk xj pk , xl xl } = ε1jk {xj pk , xl xl }

 = ε1jk xj {pk , xl xl } + pk {xj , xl xl } = ε1jk xj {pk , xl xl }

= 2ε1jk xj xl {pk , xl } = −2ε1jk xj xl δkl = −2ε1jk xj xk = 0, because xj xk = xk xj is a symmetric tensor in the pair of indices j and k, and ε1jk is antisymmetric in the same indices. Indeed, if Sij and Aij are symmetric and, respectively, antisymmetric tensors, we have: Aij Sij =

1 1 (Aij Sij + Aij Sij ) = (Aij Sij + Aji Sji ) 2 2

1 (Aij Sij − Aij Sij ) = 0. 2 Similarly, it can be shown that the other two Poisson brackets {ly , ~r2 } and {lz , ~r2 } are also zero. Consequently, =

{~l, ~r2 } = 0. (v) {~l, p~2 } = ~i{lx , p~2 } + ~j{ly , p~2 } + ~k{lz , p~2 }. Using the previous procedure, let us calculate the first Poisson bracket on the r.h.s. We have:  {lx , p~2 } = {ε1jk xj pk , pl pl } = ε1jk xj {pk , pl pl } + pk {xj , pl pl } = ε1jk pk {xj , pl pl } = 2ε1jk pk pj = 0, 284

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Free ebooks ==> www.Ebook777.com for the same reason as that displayed in the exercise (iv). In an analogous way, it can be proved that {ly , p~2 } = 0, and {lz , p~2 } = 0, which finally yields {~l, p~2 } = 0. Problem 5 Using the results obtained in Problem 4 above, calculate the Poisson brackets among the Hamiltonian of a free particle moving in a potential force field F~ = −grad V (~r) and the following functions, generically denoted by f (q, p, t): (i) f (q, p, t) = qj = xj (j = 1, 2, 3); (ii) f (q, p, t) = pj (j = 1, 2, 3); (iii) f (q, p, t) = lj (j = 1, 2, 3); (iv) f (q, p, t) = ~r = xj ~uj (j = 1, 2, 3); (v) f (q, p, t) = p~ = pj ~uj (j = 1, 2, 3); (vi) f (q, p, t) = ~r · p~ = xj pj (j = 1, 2, 3); (vii) f (q, p, t) = ~r × p~ = εijk xj pk ~ui ; (viii) f (q, p, t) = ~r2 = xj xj (j = 1, 2, 3); (ix) f (q, p, t) = p~2 = pj pj (j = 1, 2, 3); (x) f (q, p, t) = ~l 2 = lj lj (j = 1, 2, 3); Solution As well-known, the Hamiltonian of a free particle moving in a potential force field with the potential V = V (~r) equals the total mechanical energy of the particle (being a conserved quantity): H ≡E =T +V =

p~2 + V (~r). 2m

Taking into account the properties of the Poisson brackets, we then have: (i) n p~2 o {H, xj } = + V (~r), xj 2m 1 1 = {~ p2 , xj } + {V (~r), xj } = {pk pk , xj } 2m 2m  pk 1 = pk {pk , xj } + {pk , xj }pk = {pk , xj } 2m m pk pj = − δkj = − = −vj = −x˙ j , m m 285

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Free ebooks ==> www.Ebook777.com where we used the fact that V (~r) does not depend on velocity, while the variables xk and pk are independent. Therefore, x˙ j = {xj , H}. (ii) {H, pj } =

n p~2 o 1 + V (~r), pj = {~ p2 , pj } + {V (~r), pj } 2m 2m =

=

∂V ∂pj ∂V ∂pj ∂V − = δij ∂xi ∂pi ∂pi ∂xi ∂xi

∂V = [gradV (~r)]j = −Fj = −p˙j , ∂xj

so that p˙j = {pj , H}. (iii) {H, lj } = {H, εjsm xs pm } = εjsm {H, xs pm } = εjsm xs {H, pm } + εjsm pm {H, xs } = −εjsm xs p˙ m − εjsm x˙ s pm = − which can be written as

d d (εjsm xs pm ) = − lj = −l˙j dt dt

l˙j = {lj , H}.

Observation. According to the angular momentum theorem for the particle of mass m d~l ~ = ~r × F~ , =M dt our result can also be expressed as {lj , H} = (~r × F~ )j = Mj . Indeed: n p~2 o 1 {lj , H} = lj , + V (~r) = {lj , pk pk } + {lj , V (~r)} 2m 2m =

1 ∂lj ∂V ∂lj ∂V pk {lj , pk } + − m ∂xi ∂pi ∂pi ∂xi 286

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Free ebooks ==> www.Ebook777.com =

1 ∂ ∂V εjks pk ps − (εjks xk ps ) m ∂pi ∂xi

= −εjki xk

∂V = (~r × F~ )j = Mj . ∂xi

If the force is of central-type, F~ = f (r)~r, then {lj ,H} = 0. This ∂l shows that, if lj does not explicitly depend on time ∂tj = 0 , then lj = const. is a first integral of the Hamilton’s canonical equations. Let us show that this result is also valid for a particle of mass m and charge e, moving in the external static electromagnetic field ~ = −gradΦ(~r), B ~ = curlA(~ ~ r). In this case the Hamiltonian writes E H=

1 (pk − eAk )(pk − eAk ) + eΦ, 2m

while the projection on xj -axis of the angular momentum is lj = εjsn xs (pn − eAn ), so that o n 1 (pk − eAk )(pk − eAk ) + eΦ {lj , H} = εjsn xs (pn − eAn ), 2m n o 1 = εjsn xs (pn − eAn ), (pk − eAk )(pk − eAk ) 2m n o +εjsn xs (pn − eAn ), eΦ

=

1 εjsn (pk − eAk ){xs (pn − eAn ), pk − eAk } + eεjsn xs {pn − eAn , Φ} m +eεjsn (pn − eAn ){xs , Φ} =

+

1 εjsn (pk − eAk )(pn − eAn ){xs , pk − eAk } m

1 εjsn xs (pk − eAk ){pn − eAn , pk − eAk } + eεjsn xs {pn − eAn , Φ} m 1 e ∂An = εjsn (ps − eAs )(pn − eAn ) − εjsn xs (pk − eAk ) m m ∂xk e ∂Ak ∂Φ εjsn xs (pk − eAk ) − eεjsn xs m ∂xn ∂xn    ∂Φ pk − eAk ∂Ak ∂An = εjsn xs −e +e − ∂xn m ∂xn ∂xk +

287

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Free ebooks ==> www.Ebook777.com 

  ∂Φ ∂Ak ∂An = εjsn xs −e + evk − ∂xn ∂xn ∂xk h i ~ n = εjsn xs Fn = (~r × F~ )j , = εjsn xs eEn + e(~v × B)

which completes the proof. Here we used the formula   ∂An ∂Ak − ≡ vk Tnk (= −vk Tkn ) = vk εnks Bs vk ∂xn ∂xk ~ n, = εnks vk Bs = (~v × B)

where, according to (B.91), we wrote   ∂Ak ∂An ∂Ak 1 ~ s. = εsnk − = (curlA) Bs = εsnk 2 ∂xn ∂xk ∂xn (iv) {H, ~r} = {H, xj ~uj } = {H, xj }~uj + {H, ~uj }xj = {H, xj }~uj = −x˙ j ~uj = −~r˙ ,

where we used the fact that the versors of the Cartesian coordinate axes are constant quantities, as well as the result obtained at the point (i). Consequently, ~r˙ = {~r, H}. (v) {H, p~} = {H, pj ~uj } = {H, pj }~uj + {H, ~uj }pj = {H, pj }~uj = −p˙ j ~uj = −p~˙ j ,

or

p~˙ = {~ p, H},

where we used the result obtained at point (ii), and the fact that ~uj for Cartesian coordinate systems are constant quantities, as well. (vi) {H, ~r · p~} = {H, xj pj } = {H, xj }pj + {H, pj }xj = −x˙ j pj − p˙ j xj = − =− or

d (xj pj ) dt

d (~r · p~), dt

d (~r · p~) = {~r · p~, H}, dt 288

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Free ebooks ==> www.Ebook777.com where (i) and (ii) have been used. (vii) ˙ {H, ~l} = {H, lj ~uj } = {H, lj }~uj + {H, ~uj }lj = −l˙j ~uj = −~l, or

~l˙ = {~l, H},

where the point (iii) has been taken into account. (viii) {H, ~r2 } = {H, xj xj } = 2xj {H, xj } = −2xj x˙ j = −2~r · ~r˙ , or

d 2 (~r ) = {~r2 , H}, dt

where the point (i) has been used. (ix) {H, p~2 } = {H, pj pj } = 2pj {H, pj } = −2pj p˙j , or

d 2 (~ p ) = {~ p2 , H}, dt

where the result (ii) has been considered. (x) d ˙ {H, ~l 2 } = {H, lj lj } = 2lj {H, lj } = −2lj l˙j = −2~l · ~l = − (~l 2 ), dt or

d ~2 (l ) = {~l 2 , H}, dt

where the point (iii) has been used. Observation. All considered cases lead to results of the form df = {f, H}, dt where, seriatim, as function f have been chosen: xj , pj , lj , ~r, p~, ~r · p~, ~l, ~r2 , p~2 , and ~l 2 . In all these cases, f does not explicitly depend on time. In general, for f = f (q, p, t), we have df ∂f ∂f ∂f = + q˙j + p˙j dt ∂t ∂qj ∂pj 289

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Free ebooks ==> www.Ebook777.com =

∂f ∂H ∂f ∂H ∂f ∂f + − = + {f, H}. ∂t ∂pj ∂qj ∂qj ∂pj ∂t

Problem 6. Show that the temporal average values of the total kinetic and potential energies of a stable and spatially bounded system are given by the relations n Ec = H, n+2 and, respectively, V =

2 H, n+2

where H is the Hamiltonian of the system, while n is Euler’s degree (order) of homogeneity of the potential energy. Solution. Consider a stable and spatially limited (bounded) physical system composed by N particles, each particle Pk having mass mk (k = 1, N ), acted by the conservative potential forces F~kj = −gradkj Vkj

(no summation;

k, j = 1, N ),

(8.87)

where Vkj is the potential energy of interaction between particles Pk and Pj , while gradkj stands for the partial derivative with respect to the components of the vector ~rkj = ~rk − ~rj (see Fig.VIII.3). By definition, the quantity G=

N X

k=1

~rk · p~k ,

(8.88)

where p~k = mk~vk = mk ~x˙ k is the linear momentum of the particle Pk , is called the scalar virial of the system. Let us calculate the total derivative with respect to time of the virial G. We have: N

N

k=1

k=1

X d dG d X = ~rk · p~k = (~rk · p~k ) dt dt dt =

N X d~rk

k=1

dt

· p~k +

N X

k=1

~rk ·

d~ pk dt

290

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Free ebooks ==> www.Ebook777.com N N N X X X 1 ~ = p~k · p~k + ~rk · Fk = 2Ec + ~rk · F~k , mk k=1

where Ec =

N P

k=1

particles.

k=1

1 2 ~k mk p

k=1

is the total kinetic energy of the system of

Fig.VIII.3 Since N

X 1 ∂Ec pj = pk δkj = ∂pj mk mj

(no summation),

k=1

we can write N N N X X X 1 1 ∂Ec 2Ec = p~k · p~k = pk pk = pk , mk mk ∂pk k=1

k=1

(8.89)

k=1

expressing the fact that the total kinetic energy of the system of particles is a Eulerian homogeneous function of the second degree. Here we remind the reader that the function f (x1 , x2 , ..., xn ) is called homogeneous of degree s with respect to the independent variables x1 , x2 , ..., xn , if f (tx1 , tx2 , ..., txn ) = ts f (x1 , x2 , ..., xn )

(t ∈ R∗ )

291

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Free ebooks ==> www.Ebook777.com in which case is valid Euler’s theorem regarding homogeneous functions of degree s: n X ∂f = sf. (8.90) xk ∂xk k=1

We have therefore shown that N

X dG = 2Ec + ~rk · F~k , dt

(8.91)

k=1

where F~k is the total force acting on the particle Pk , F~k =

N X

F~kj

(k = 1, 2, ..., N ).

(8.92)

j=1 j6=k

Since forces F~kj are conservative, deriving from the potential Vkj (|~rkj |), that is F~kj = −∇~rkj Vkj (|~rkj |), (8.93) we can calculate the second term of the r.h.s. of Eq. (8.91). Indeed, N X

k=1

~rk · F~k =

=

N X

k=1

N X N X

~rk · F~kj +

N X N X

~rk · F~kj −

k=1 j=1 j www.Ebook777.com where σ is the transformation parameter, and denoting by ~r1 and ~r2 the radius vectors of the points A1 (0, 0, σ) and A2 (0, 0, −σ) (see Fig.VIII.6), we have p  r1 = p(z − σ)2 + ρ2 , (8.138) r2 = (z + σ)2 + ρ2 , or, in view of (8.137),

 r + r1 λ = 2 , 2σ  µ = r2 − r1 . 2σ

Denoting

 ee1  α1 = , 4πε0  α2 = ee2 , 4πε0

relation (8.134) writes

V =

α1 α2 + . r1 r2

(8.139)

(8.140)

(8.141)

Fig.VIII.6 As one can see, α1 α2 a(λ) + b(µ) + ≡ . r1 r2 λ 2 − µ2 Indeed, on the one side α1 α2 α1 r2 + α2 r1 + = , r1 r2 r1 r2 305

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(8.142)

Free ebooks ==> www.Ebook777.com and on the other, in view of (8.139), a(λ) + b(µ) a+b σ 2 (a + b) ≡ = . λ 2 − µ2 (λ + µ)(λ − µ) r1 r2 As a final step, one must show that α1 r2 + α2 r1 = σ 2 (a + b).

(8.143)

2 2 Since a = a(λ) = α1 +α λ and b = b(µ) = α1 −α µ relation (8.143) σ σ follows immediately. Since our system is conservative, we are now able to write the abbreviated Hamilton-Jacobi equation (8.132) in elliptic coordinates. Using (8.133) and (8.134), we have "  2  2 ∂W ∂W 1 2 2 (λ − 1) + (1 − µ ) 2mσ 2 (λ2 − µ2 ) ∂λ ∂µ

+



λ2

1 1 + − 1 1 − µ2



∂W ∂ϕ

2 #

+

a(λ) + b(µ) = E, λ 2 − µ2

(8.144)

where the expressions for the generalized momenta pλ =

∂W , ∂λ

pµ =

∂W , ∂µ

pϕ =

∂W ∂ϕ

(8.145)

have been considered. Recalling that [see(8.131)] S(q, t) = W (q) − Et,

(8.146)

our main purpose is to determine the reduced action W (q) by solving the abbreviated Hamilton-Jacobi equation. This can be done by method of reparation of variables. Since ϕ is also a cyclic coordinate, we search for a solution of the form W (q) = W (λ, µ, ϕ) = pϕ ϕ + W1 (λ) + W2 (µ).

(8.147)

Here pϕ is nothing else but the constant value of the generalized momentum pϕ = ∂S ∂ϕ , associated with cyclic coordinate ϕ, while W1 (λ) and W2 (µ) are solutions of the equations 2

(λ −1)



dW1 dλ

2

+

p2ϕ +2mσ 2 a(λ)−2mσ 2 (λ2 −1)E = β, (8.148) λ2 − 1 306

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Free ebooks ==> www.Ebook777.com and 2

(1−µ )



dW2 dµ

2

+

p2ϕ +2mσ 2 b(µ)+2mσ 2 (1−µ2 )E = −β, (8.149) 1 − µ2

respectively. Here β is an arbitrary constant. The last two equations have been obtained by introducing (8.147) into (8.144), then multiplying by 2mσ 2 (λ2 − µ2 ). After some arrangements of terms, this yields "

(λ2 − 1)



dW1 dλ

2

p2ϕ + 2 + 2mσ 2 a(λ) − 2mσ 2 λ2 E + 2mσ 2 E λ −1

#

# 2 p ϕ + + 2mσ 2 b(µ) + 2mσ 2 µ2 E − 2mσ 2 E = 0. + (1 − µ2 ) 1 − µ2 (8.150) Since the first square bracket depends only on λ, and the second only on µ, one can write [..λ..] = −[..µ..] = β, and equations (8.148) and (8.149) follow immediately. Integrating (8.148) and (8.149), one can easily obtain W1 (λ) and W2 (µ). Introducing these results into (8.147), and then into (8.146), we are left with the following complete integral of the Hamilton-Jacobi equation: "



dW2 dµ

2

Z s p2ϕ β − 2mσ 2 a(λ) S = −Et + pϕ ϕ + 2mσ 2 E + − dλ λ2 − 1 (λ2 − 1)2 Z s p2ϕ β + 2mσ 2 b(µ) + 2mσ 2 E − − dµ, 1 − µ2 (1 − µ2 )2

(8.151)

or, by means of (8.136) and (8.140), S = −Et + pϕ ϕ +

Z s

2mσ 2 E +

p2ϕ β − 2mσ(α1 + α2 )λ − dλ λ2 − 1 (λ2 − 1)2

Z s p2ϕ β + 2mσ(α1 − α2 )µ + 2mσ 2 E − − dµ. 1 − µ2 (1 − µ2 )2

(8.152)

The complete integral (8.152) depends on the arbitrary constants pϕ , β, and E. Taking derivatives with respect to these quantities, and 307

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Free ebooks ==> www.Ebook777.com equating the results to some new constants, one can find the general solution of the equations of motion. The reader is invited to prove that the constant β is connected to the conservation of the quantity β = σ 2 p2ρ − M 2 + 2mσ(α1 cos θ1 + α2 cos θ2 ), ~ is the total angular momentum of the particle, and θ1 , θ2 where M are the angles shown in Fig.VIII.6. Problem 10 Show that the transformation    sin p  ; Q = ln q  P = q cot p

(8.153)

is canonical and find all possible types of generating function. Solution To show that the transformation (q, p) → (Q, P ) is canonical, it is sufficient to prove that the absolute value of the Jacobian J of the transformation equals 1. Indeed,   ∂Q ∂Q q − sin p q cos p ∂q ∂p sin p q2 sin p q   J = ∂P ∂P = cot p q − sin12 p ∂q ∂p 1 − = q cot p

As can be observed,

J=

cot p 1 − cot2 p = 1. = q 2 − sin2 p sin p

∂Q ∂P ∂Q ∂P − = {Q, P } = 1, ∂q ∂p ∂p ∂q

where {Q, P } is the Poisson bracket of quantities Q and P . Since our problem concerns one generalized coordinate and one generalized momentum, the maximum number of generating functions is four: F1 (q, Q), F2 (q, P ), F3 (p, Q), F4 (p, P ). The canonicity condition for the transformation (q, p) → (Q, P ) then furnishes relation p dq − P dQ = dF1 (q, Q), (8.154) 308

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Free ebooks ==> www.Ebook777.com or, in view of (8.153) (p + cot p)dq − q cot2 p dp = dF1 .

(8.155)

In order to be a total differential, function F1 must satisfy the relation ∂ ∂ (−q cot2 p) = (p + cot p), ∂q ∂p which proves to be true. By virtue of this property, we can write dF1 =

∂F1 ∂F1 dq + dp. ∂q ∂p

(8.156)

Comparing (8.156) and (8.155), one obtains  ∂F 1  = p + cot p,  ∂q   ∂F1 = −q cot2 p. ∂p

(8.157)

Equation (8.157)1 then yields F1 =

Z

(p + cot p)dq + f (Q) = pq + q cot p + f (Q),

(8.158)

where f (Q) is an arbitrary function of Q. According to (8.157)2 we then have Z F1 = − q cot2 p dp = −q(−p − cot p) + g(q), (8.159) with g(q) another arbitrary function of q. We therefore found 

F1 (q, Q) = qp + q cot p + f (Q), F1 (q, Q) = qp + q cot p + g(q),

which means that F1 (q, Q) = qp + q cot p = q arcsin qe

 Q

+

p

1 − q 2 e2Q , eQ

(8.160)

is the generating function of type F1 (q, Q) of the canonical transformation (8.153). 309

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Free ebooks ==> www.Ebook777.com The other three types of generating functions are obtained by means of the appropriate Legendre transformations. Let us start with " # p 2 e2Q  1 − q dF1 = (p + cot p)dq − q cot2 p dp = arcsin qeQ + dq qeQ 1 − q 2 e2Q − qe2Q or

p

eQ 1 − q 2 e2Q

dq + p

qeQ 1 − q 2 e2Q

!

dQ ,

p

1 − q 2 e2Q dQ. (8.161) eQ In order to determine F2 (q, P ) let us perform Legendre transformation dF1 + d(P Q) = d(F1 + P Q) ≡ dF2 ! p 2 e2Q 1 − q − P dQ + QdP. (8.162) = arcsin(qeQ )dq − eQ Q

dF1 = arcsin(qe )dq −

To be independent of Q, the coefficient of dQ in (8.162) must be null, that is p 1 − q 2 e2Q , (8.163) P = eQ which yields p Q = − ln q 2 + P 2 , (8.164) eQ = (q 2 + P 2 )−1/2 ,

(8.165)

e2Q = (q 2 + P 2 )−1 ,

(8.166)

and (8.162) becomes h i p dF2 (q, P ) = arcsin q(q 2 + P 2 )−1/2 dq − ln q 2 + P 2 dP.

(8.167)

One can easily be proved that dF2 is an exact differential. Indeed,  h io p ∂  ∂ n − ln q 2 + P 2 = arcsin q(q 2 + P 2 )−1/2 . ∂q ∂P The third generating function F3 (p, Q) can be determined by means of Legendre transformation p 1 − q 2 e2Q dF1 − d(pq) = arcsin(qeQ )dq − dQ − q dp − p dq eQ 310

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Free ebooks ==> www.Ebook777.com 

 = arcsin(qe ) − p dq − Q

p 1 − q 2 e2Q dQ − q dp. eQ

To depend only on coordinates (p, Q), one must have p = arcsin(qeQ ), that is sin p = qeQ . Written in coordinates (p, Q), the generating function F3 (p, Q) = F1 − qp is given by F3 = e−Q sin p(p + cot p),

(8.168)

with exact differential dF3 = −e−Q sin p dp −

cos p dQ, eQ

(8.169)

which can be easily verified, since  ∂  cos p  ∂ − Q = −e−Q sin p . ∂p e ∂Q

To determine the fourth (and final!) type of generating function F4 = F4 (p, P ) of the canonical transformation (8.153) one performs the Legendre transformation p  1 − q 2 e2Q Q dQ dF1 + d(P Q − pq) = arcsin qe dq − eQ

+P dQ + Q dP − p dq − q dp ! p 2 e2Q    1 − q = arcsin qeQ − p dq − − P dQ + Q dP − q dp. eQ (8.170) In order that the generating function F4 = F1 + P Q − pq = F2 − pq = F3 + P Q depends only on coordinates p and P , the parentheses in (8.170) must be zero, that is      arcsin qeQ = p, p 2 2Q   1 − q e = P, eQ which coincide with (8.153)1 and (8.163). The generating function F4 (p, P ) therefore is:

F4 = F1 + P Q − pq = q cot p + P Q = P + P ln 311

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sin p P cot p

!

Free ebooks ==> www.Ebook777.com = P + P ln while its differential is

 cos p  , P

dF4 = −P tan p dp + ln

 cos p  P

(8.171)

dP.

(8.172)

This is an exact differential, because ∂ h  cos p i ∂ ln = (−P tan p) . ∂p P ∂P Summarizing, the four generating functions of the canonical transformation (8.153) are     F1      F2       F   3 F4

= F1 (q, Q) = q arcsin qe

 Q

+

p

1 − q 2 e2Q , !eQ

q p q2 + P 2 = F3 (p, Q) = e−Q (p + cot p)sin p, p = F4 (p, P ) = P + P ln cos , P = F2 (q, P ) = q arcsin

+P −

P ln(q 2 + P 2 ), 2

while their exact differentials write p  1 − q 2 e2Q  Q   dF1 = arcsin(qe )dq − dQ,   eiQ h    dF = arcsin q(q 2 + P 2 )−1/2 dq − ln pq 2 + P 2 dP, 2 cos p   dF3 = −e−Q sin p dp − Q dQ,    e     dF4 = −P tan p dp + ln cos p dP. P

312

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CHAPTER IX PROBLEMS OF CONTINUOUS SYSTEMS

A. Problems of Classical Electrodynamics Problem 1 An electrized particle of charge q and mass m moves with velocity ~ B), ~ described by the electro~v in the variable electromagnetic field (E, ~ y, z, t). Neglecting quantum magnetic potentials V (x, y, z, t) and A(x, and relativistic effects, show that the equation of motion of the particle can be written in a Lagrangian form, and determine the Lagrangian function L. Solution Since

~ ~ = −grad V − ∂ A ; E ∂t ~ = curl A, ~ B

(9.1) (9.2)

the equation of motion of the particle

also writes

~ + ~v × B) ~ m~¨r = q(E

(9.3)

! ~ ∂ A ~ . + ~v × curl A m~¨r = q −grad V − ∂t

(9.4)

The last term of (9.4) can be transformed as follows: ~ = εijk vj (∇ × A) ~ k ~ui = εijk vj εklm (∂l Am )~ui ~v × curlA = εkij εklm vj (∂l Am )~ui = (δil δjm − δim δjl )vj (∂l Am )~ui

~ ui − (~v · ∇)A ~ = ∂i (vj Aj )~ui − (vj ∂j )(Ai ~ui ) = ∂i (~v · A)~ 313

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Free ebooks ==> www.Ebook777.com ~ − (~v · ∇)A, ~ = ∇(~v · A) where we made allowance for the fact that, within the Lagrangian approach, the generalized coordinates [in our case xi (i = 1, 3)], and generalized velocities [here vi (i = 1, 3)], are independent quantities. Thus vi , together with ~ui = const. have been introduced under the derivative sign ∂i = ∂/∂xi . Equation (9.4) then becomes " # ~ ∂ A ~ − (~v · ∇)A ~ m~¨r = q −grad V − + ∇(~v · A) ∂t !# ~ ∂ A ~ ~ − + (~v · ∇)A = q −∇(V − ~v · A) ∂t " # ~ d A ~ + . = −q ∇(V − ~v · A) dt "

But ~¨r =

d~ r˙ dt

=

d~ v dt ,

(9.5)

and (9.5) yields

n h io  d ~ − ∇ q −V + ~v · A ~ = 0. m~v + q A dt

Identifying (9.6) with Lagrange equations of the 2nd kind   d ∂L ∂L = 0, − dt ∂~v ∂~r

(9.6)

(9.7)

we obtain

∂L ~ = m~v + q A; ∂~v n h io io ∂L ∂ n h ~ ~ = ∇ q −V + ~v · A = q −V + ~v · A , ∂~r ∂~r

(9.8) (9.9)

where L is the Lagrangian function, while ∂L ∂~ r and  

∂L ∂~ v stand  for a formal ∂L ∂L ∂vy , ∂vz , respectively.

∂L ∂L ∂L ∂L ∂L writing of ∂L ∂~ r = ∂x , ∂y , ∂z , and ∂~ v = ∂vx , According to Eq.(9.9), it then follows that

~ + f (~v ), L(~r, ~v , t) = −qV + q(~v · A) where f (~v ) is an arbitrary function of class C 0 at least. It can be determined by means of (9.8): df 1 = m~v ⇒ f (~v ) = m~v 2 , d~v 2 314

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Free ebooks ==> www.Ebook777.com and the Lagrangian becomes ~ + 1 m~v 2 . L(~r, ~v , t) = −qV + q(~v · A) 2

(9.10)

We conclude that the equation of motion (9.3) can be written in the Lagrangian form (9.7), with L given by (9.10). It is important to mention that, in our case, L(~r, ~v , t) = T (~r, ~v , t) − U (~r, ~v , t),

(9.11)

where T (~r, ~v , t) is the kinetic energy, and ~ U (~r, ~v , t) = qV − q(~v · A)

(9.12)

the generalized (velocity dependent) potential. As well-known, the electromagnetic potentials are defined only up to a gauge transformation V′ =V −

∂φ , ∂t

~′ = A ~ + ∇φ, A

(9.13)

where φ(~r, t) is a function of class C 1 in R3 . Let us investigate the ~ → L(V ′ , A ~ ′ ). We have: transformation L(V, A)   1 ∂φ ′ ′ ′ 2 ~ ~ + ∇φ) + 1 mv 2 L = −qV + q(~v · A ) + mv = −q V − + q~v · (A 2 ∂t 2   ∂φ d ∂φ + q~v · ∇φ = L + q + (~v · ∇)φ = L + (qφ). (9.14) =L+q ∂t ∂t dt

The last relation shows that, if two Lagrangians L and L′ differ by a total derivative with respect to time of a scalar function F = F (~r, t), they are equivalent.

Problem 2 Using the Lagrangian technique, determine the equation of motion of a particle of charge q and mass m, moving in the electromagnetic ~ B). ~ field (E, Solution According to Lagrangian formalism, the differential equations of motion of the particle write   ∂T d ∂T − = Qk (k = 1, n), (9.15) dt ∂ q˙k ∂qk 315

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Free ebooks ==> www.Ebook777.com where T is the kinetic energy, Qk are the generalized forces, and n is the number of degrees of freedom of the system. If the applied force F~ can be derived from a generalized potential energy function (or, in short, a velocity dependent potential) U (q, q, ˙ t) = Cj q˙j + U0 = U1 + U0

(j = 1, n),

(9.16)

where Cj = Cj (q, t), then we can add the quantity ∂U d − ∂qk dt



∂U ∂ q˙k



to both sides of (9.15), and obtain     d ∂(T − U ) ∂(T − U ) d ∂U ∂U − = Qk − + . dt ∂ q˙k ∂qk dt ∂ q˙k ∂qk If the generalized forces Qk are given by   d ∂U ∂U , Qk = − dt ∂ q˙k ∂qk

(9.17)

(9.18)

and denote L(q, q, ˙ t) = T (q, q, ˙ t) − U (q, q, ˙ t),

(9.19)

then we are left with the well-known Lagrange equations for natural systems   d ∂L ∂L = 0. (9.20) − dt ∂ q˙k ∂qk Here L(q, q, ˙ t) ≡ L(q1 , q2 , ..., qn , q˙1 , q˙2 , ..., q˙n , t), etc. To establish the differential equation of motion of the particle of ~ B), ~ one must charge q and mass m in the electromagnetic field (E, first construct the Lagrangian function. Since no constraints act on the particle, as generalized coordinates and generalized velocities one can choose the Cartesian coordinates qi ≡ xi (i = 1, 3) and the velocity components q˙i ≡ x˙ i (i = 1, 3), respectively. If only the electric field is present, the Lagrangian writes L = T −U = T −qV , where V (~r, t) is the electric potential. If the magnetic field is also present, we have to consider a new term in the Lagrangian, ~ r, t). To this end, we expressed in terms of the vector potential A(~ take advantage of the fact that the Lagrangian is a scalar (invariant) ~ can only appear as one of the dot products: A ~· function, so that A ~ A· ~ ~r˙ , and A· ~ ~¨r. Since B 2 does not appear in the equation of motion, A, 316

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Free ebooks ==> www.Ebook777.com the first possibility is excluded. On the other hand, the usual equations of motion met in physics are second-order differential equations, which means that the third possibility is also unacceptable. Therefore, the Lagrangian writes L=

1 ~ mv 2 − qV + q~v · A. 2

(9.21)

We then successively have: ∂L ∂L ≡ = mvi + qAi ; ∂ q˙i ∂vi       d ∂L d ∂L ∂Ai ∂Ai ≡ = m¨ xi + q + x˙ k ; dt ∂ q˙i dt ∂ x˙ i ∂t ∂xk ∂L ∂V ∂Ak ∂L ≡ = −q + qvk ∂qi ∂xi ∂xi ∂xi

(i, k = 1, 2, 3),

and Lagrange’s equations (9.20) become     ∂Ai ∂Ak ∂Ai ∂V − − m¨ xi = q − + qvk , ∂xi ∂t ∂xi ∂xk or m¨ xi = qEi + qvk Tik ,

(9.22)

where by Tik has been denoted the antisymmetric tensor Tik =

∂Ak ∂Ai − . ∂xi ∂xk

(9.23)

Let us show that Bj are the components of the axial vector (pseudovector) associated with the antisymmetric tensor (9.23). Indeed, Bj = =

1 1 ∂Ak 1 ∂Ai εjik Tik = εjik − εjik 2 2 ∂xi 2 ∂xk

1 ∂Ak 1 ∂Ak 1 ∂Ak ~ j. εjik − εjki = 2 · εjik = (curlA) 2 ∂xi 2 ∂xi 2 ∂xi

~=B ~ right from the beginThis result explains why we denoted curlA ning. The equation of motion (9.22) then finally writes ~ i, m¨ xi = qEi + qvk (εikl Bl ) = qEi + q(~v × B) 317

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Free ebooks ==> www.Ebook777.com or, in vectorial form  ~ + ~v × B ~ . m~¨r = q E

(9.24)

Problem 3 Solve Problem 2, using the Hamiltonian formalism. In addition, investigate the following two particular cases: (1) The particle is an electron of charge −e, the electromagnetic ~ B, ~ and the initial velocity ~v0 of the field is static and uniform, with Ek particle is orthogonal to the field. (2) The particle is an electron of charge −e, the electromagnetic ~ and B ~ have different direcfield is static and uniform, but this time E tions, while the initial velocity ~v0 is orthogonal to the plane determined ~ and B. ~ by E Solution Let us write the Lagrangian (9.21) in the form L=

1 mvk vk − qφ + qvk Ak 2

(k = 1, 2, 3),

(9.25)

where Einstein’s summation convention has been used. According to their definition, the generalized momenta pi (i = 1, 2, 3) are pi ≡

∂L ∂L = = mvk δik + qAk δik = mvi + qAi ∂ q˙i ∂vi

(i = 1, 2, 3). (9.26)

The Hamilton therefore is H(q, p, t) ≡

n X

k=1

Since vk =

pk q˙k − L(q, q, ˙ t) =

1 (pk − qAk ) m

1 mvk vk + qφ. 2

(k = 1, 2, 3),

we still have H=

1 (pk − qAk )(pk − qAk ) + qφ. 2m

Hamilton’s canonical equations  ∂H  ;  q˙i ≡ x˙ i = ∂pi   p˙i = − ∂H ≡ − ∂H , ∂qi ∂xi

(i = 1, 2, 3)

318

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(9.27)

(9.28)

Free ebooks ==> www.Ebook777.com then yield x˙ i =

∂H 1 1 = (pk − qAk )δik = (pi − qAi ), ∂pi m m

p˙ i = −

∂H q ∂Ak ∂φ = (pk − qAk ) −q , ∂xi m ∂xi ∂xi

or, in a synthetic form  1   x˙ i = (pi − qAi ); m q ∂A ∂φ   p˙ i = (pk − qAk ) k − q . m ∂xi ∂xi

(9.29)

Let as now take the time derivative of (9.29.)1 . In view of (9.29)2 , we then have: m¨ xi = p˙ i − q A˙ i   q ∂Ak ∂φ ∂Ai ∂Ai = (pk − qAk ) −q −q + x˙ k , m ∂xi ∂xi ∂t ∂xk

or

∂φ ∂Ai m¨ xi = −q −q + q x˙ k ∂xi ∂t



∂Ak ∂Ai − ∂xi ∂xk



,

(9.30)

~ = A(~ ~ r, t) has been taken into where the functional dependence A account. Using the relations between field and potentials ~ ~ = −∇φ − ∂ A , E ∂t

~ = ∇ × A, ~ B

we still have m¨ xi = qEi + qvk Tik , which is precisely equation (9.22), with Tik given by (9.23). From now on, calculations are identical to those developed in Problem 2. Let us now consider the particular cases (1) and (2): Case 1 The equation of motion of the electron in the static and uniform ~ B ~ is [see (9.24)] electromagnetic field E,  ~ + ~v × B ~ , m~¨r = −e E 319

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(9.31)

Free ebooks ==> www.Ebook777.com where m is the electron mass, and −e its electric charge. Without loss ~ B ~ are oriented along the z-axis, of generality, let us suppose that E, and ~v0 along the x-axis. Since ~i ~j ~ = x˙ y˙ ~v × B 0 0

~k ˙ ~i − xB ˙ ~j, z˙ = yB B

the components of vector equation (9.31) along the coordinate axes are ( Ox : m¨ x = −eB y, ˙ (9.32) Oy : m¨ y = eB x, ˙ Oz : m¨ z = −eE. Recalling that E = const., the last equation of (9.32) yields z(t) = −

C2 eE 2 C1 t + t+ , 2m m m

where C1 and C2 are two arbitrary constants of integration. To find them, one takes z(0) = 0, and z(0) ˙ = 0 at the initial moment of time t = 0. Then we obtain C1 = 0, C2 = 0, so that z(t) = −

eE 2 t . 2m

(9.33)

To integrate the system of coupled ordinary differential equations 

m¨ x = −eB y, ˙ m¨ y = eB x, ˙

we shall use the√complex variable ξ = x + iy. Multiplying the last equation by i = −1 and adding the result to the first, we have eB ˙ ξ¨ − i ξ=0 m

(9.34)

ξ = aeiωt + b,

(9.35)

with the solution where ω = eB m , while a = ax + iay , and b = bx + iby are two complex constants of integration. Returning to the old variables, we can write 

x(t) = ax cos ωt − ay sin ωt + bx , y(t) = ax sin ωt + ay cos ωt + by . 320

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(9.36)

Free ebooks ==> www.Ebook777.com The four arbitrary, real constants of integration ax , ay , bx , by can be univocally determined by means of the following four initial conditions: 

x(0) = 0, x(0) ˙ = 0,

y(0) = 0, y(0) ˙ = 0.

(9.37)

Some simple calculations then yield v0 sin ωt, ω

(9.38)

v0 (1 − cos ωt). ω

(9.39)

x(t) = and y(t) =

Using the last two relations, one easily obtains h v2 v 0 i2 = 02 . x2 (t) + y(t) − ω ω

(9.40)

This means that the projection of the trajectory of electron on xyplane, at any time, is a circle of constant radius R = v0 /ω and centre  at the point C(x0 , y0 ) = C 0, vω0 . On the other hand, x˙ 2 (t) + y˙ 2 (t) = v02 , which shows that the motion of rotation about the field is uniform. The angular velocity is ω=

v0 eB = . R m

The spatial (three-dimensional) motion of the electron is obtained by composing the uniform rotation given by (9.38) and (9.39), with the accelerated motion given by (9.33). The trajectory is, therefore, a helix of variable pitch, wrapped around the circular cylinder of radius R. Case 2 The motion of the electron in the static, uniform electromagnetic ~ B ~ is governed by equation (9.31) (we rewrite it, for convefield E, nience)  ~ + ~v × B ~ . m~¨r = −e E (9.41) 321

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Free ebooks ==> www.Ebook777.com ~ the yzSuppose, this time, that the z-axis is oriented along B, ~ ~ plane is determined by E and B, while the initial conditions demand ~v0 = (v0 , 0, 0). Since ~i ~j ~k ~ = x˙ y˙ z˙ = yB ~v × B ˙ ~i − xB ˙ ~j, 0 0 B the components of vector equation (9.31) along the coordinate axes are ( Ox : m¨ x = −eB y, ˙ (9.42) Oy : m¨ y = −eEy + eB x, ˙ Oz : m¨ z = −eEz .

Using the initial conditions ~r0 = (0, 0, 0) and ~v0 = (v0 , 0, 0), the last equation (9.42) yields eEz 2 z(t) = − t . (9.43) 2m Introducing the complex variable ξ = x + iy, the first two equations (9.42) can be compressed to give one equation eB ˙ e ξ¨ − i ξ = −i Ey . m m

(9.44)

This is a second order linear, non-homogeneous, ordinary differential equation with constant coefficients. Its solution is obtained by adding a particular solution of the non-homogeneous equation to the general solution of the homogeneous equation. Solution of the homogeneous equation has been already obtained [see (9.35)]: ξ0 (t) = aeiωt + b, where a and b are two complex constants. Since the non-homogeneous equation contains only derivatives of the unknown variable, we cannot use the term that causes inhomogeneity. So, we shall appeal to the variation of constants method. Following this procedure, since the fundamental set of solutions of the homogeneous equation is  ξ1 (t) = eiωt , ξ2 (t) = 1, the general solution to the problem is given by Z Z iωt ′ ξ(t) = e a (t) dt + b′ (t) dt, 322

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Free ebooks ==> www.Ebook777.com where a′ (t) and b′ (t) are the solutions of the first-order, differential equations system ( ξ1 (t)a′ (t) + ξ2 (t)b′ (t) = 0, e ξ1′ (t)a′ (t) + ξ2′ (t)b′ (t) = −i Ey , m or

(

eiωt a′ (t) + b′ (t) = 0, e iω eiωt a′ (t) = −i Ey . m The second equation yields a(t) = −

ieEy −iωt e + c, mω 2

while the first equations gives eEy t + d, mω

b(t) =

where c and d are two arbitrary, complex constants of integration. The general solution of equation (9.44) therefore is ξ(t) = ceiωt −

eEy Ey ieEy + t + d = ceiωt + t + g, 2 mω mω B

(9.45)

where g is a new complex constant of integration, whose value can be easily found. Separating the real and imaginary parts in (9.45), we find ( Ey t + gx , x(t) = cx cos ωt − cy sin ωt + (9.46) B y(t) = cx sin ωt + cy cos ωt + gy . The initial conditions  x(0) = 0; x(0) ˙ = v0 ; lead to

so that

( (

cx + gx = 0; Ey −cy ω + = v0 ; B

cx = 0; Ey v0 cy = − ; ωB ω

y(0) = 0; y(0) ˙ = 0, cy + gy = 0; cx ω = 0, gx = 0; v0 Ey gy = − . ω ωB

323

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(9.47)

Free ebooks ==> www.Ebook777.com Denoting α =

v0 ω



Ey ωB ,

the relations (9.46) write

(

Ey t, B y(t) = α(1 − cos ωt).

x(t) = α sin ωt +

Introducing a new notation α =

E α0 ωBy ,

with α0 =

(9.48)

B Ey

 v0 −

Ey B



in (9.48) and recalling (9.43), we finally have the parametric equations of the trajectory  Ey  x(t) = (ωt + α0 sin ωt),    ωB  Ey (1 − cos ωt), y(t) = α0  ωB     z(t) = − eEz t2 . 2m

(9.49)

Fig.IX.1 shows an example for the trajectory described by (9.49), for the specified values of the physical quantities. All quantities are expressed in SI unit system. The graphic representations has been performed by means of software Mathematica, according to the following command lines:

324

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Fig.IX.1

Problem 4 A magnetic dipole of moment µ is placed in the magnetic field ~ B. Using Hamiltonian approach, determine the interaction energy ~ between the dipole and the field, Wm = −~ µ · B. Solution ~ be the homogeneous and constant magnetic induction field Let B in which is placed the point charge q, and let µ ~ be the magnetic moment produced by a uniform circular motion of q (see Fig.IX.2). Such a magnetic field possesses the vector potential ~ × ~r, ~ = 1B A 2

(9.50)

where ~r is the position vector (radius vector) of some point P (~r) of the field with respect to an arbitrary chosen reference frame Oxyz (see Fig.IX.3). 325

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Fig.IX.2 Indeed, ~ =∇× ∇×A =



1~ B × ~r 2



1 ~ ~ + (~r · ∇)B ~ − (B ~ · ∇)~r] [B ∇ · ~r − ~r ∇ · B 2 =

1 ~ ~ = B, ~ (3B − B) 2

~ is source-free (∇ · B ~ = 0) where we took into account that the field B ~ = 0], while and constant [(~r · ∇)B ~ · ∇)~r = Bi (B

∂ ~ (xk ~uk ) = Bi ~uk δik = Bi ~ui = B. ∂xi

Fig.IX.3 326

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Free ebooks ==> www.Ebook777.com According to (9.27) (see Problem 3), the Hamiltonian of our physical system is H=

1 (pk − qAk )(pk − qAk ) + qφ, 2m

or, if we use vector notation H=

1 q2 ~ 2 q ~ + qφ. |~ p|2 + |A| − p~ · A 2m 2m m

By means of formula (9.50), we still have H=

2 1 q ~ × ~r) + q |B ~ × ~r|2 . |~ p|2 + qφ − ~v · (B 2m 2 8m

(9.51)

Fig.IX.4 Usually, the last term is much smaller than the rest of the terms, so that it can be neglected. Then we are left with H = H0 + Hint , where H0 =

1 |~ p|2 + qφ 2m

(9.52)

(9.53)

~ is absent, and is the dipole energy when the external field B q ~ × ~r) Hint = − ~v · (B 2 327

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(9.54)

Free ebooks ==> www.Ebook777.com is the energy of interaction between dipole and the external field. All we have to do now is to show that the two formulas (9.54) and ~ express the same thing. To this end, we shall use the Wm = −~ µ·B geometric interpretation of the cross product. As far as one knows, the modulus of a cross product of two vectors equals the area of a parallelogram with the vectors for sides (see Fig.IX.4). In the limit ∆t → 0, the area ∆S swept by the radius vector ~r of the particle is given by 1 lim ∆S = lim |~r × ∆~r|, (9.55) ∆t→0 ∆t→0 2 where ∆~r is the variation of ~r during the time interval ∆t. Dividing (9.55) by ∆t, one obtains the areolar velocity of the particle ~ ~ dS 1 ∆S = lim = ~r × ~v , ∆t→0 ∆t dt 2 and, therefore ~ q ~ × ~r) = − q B ~ · (~r × ~v ) = −q B ~ · dS . Hint = − ~v · (B 2 2 dt

(9.56)

Let us now write equation (9.56) for a period T of one complete uniform revolution of the particle, that is ~ = −B ~ · I∆S ~ = −~ ~ ~ · q ∆S µ · B. Hint = −B T

(9.57)

~ = πr2 I ~n is the dipole moment of the magnetic sheet, where µ ~ = I∆S produced by the circular current of radius r. Here ~n is the unit vector of the external normal to the circuit surface.

Fig.IX.5 328

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Free ebooks ==> www.Ebook777.com ∆S Observation. Equality dS dt = T which we have previously used is sustained by the uniform feature of the circular motion of the particle. Indeed, in case of uniform, circular motion we have (see Fig.IX.5)

dS =

dθ ω dt dt ∆S = ∆S = ∆S . 2π 2π T

Problem 5 A simple pendulum of length l, made out of a conductor material, moves in such a way that its inferior end slides without friction on a conductor support of circular form. The pendulum arm is a rigid of mass m, moving in the external static and homogeneous magnetic ~ orthogonal at any point to the pendulum rod. One end field |B|, of the metallic support and the point of support of pendulum are connected by an ideal condenser of capacity C. The electric circuit closes through the pendulum arm, as shown in Fig.IX.6. Neglecting the electric resistance and inductance of the circuit, determine the period T of the pendulum. Solution During the infinitesimal time interval dt, the pendulum arm of length l describes on the circular support the distance ds = l dα and sweeps the elementary surface of area dS = l ds/2 = l2 dα/2 (see ~ is orthogonal to this surface, Fig.IX.7). Since, at any point, the field B ~ · dS ~ = B dS = the elementary magnetic flux through dS is dΦ = B 2 Bl dα/2. Since dΦ is variable, it induces in the pendulum conductor arm the electromotive tension (voltage) e=−

Bl2 dα Bl2 dΦ =− =− α. ˙ dt 2 dt 2

Fig.IX.6 329

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(9.58)

Free ebooks ==> www.Ebook777.com This tension generates in the circuit shown in Fig.IX.7 an electric current, so that on the condenser plates shall appear the time variable electric charge CBl2 q = C|e| = α. ˙ (9.59) 2

Fig.IX.7 Choosing as generalized coordinate the angle α between the pendulum arm and vertical, the Lagrangian writes L = T − V.

(9.60)

Here

1 2 1 Iω = ml2 α˙ 2 , (9.61) 2 2 where I = ml2 is the moment of inertia of the pendulum rod, with respect to point axis of rotation, and T =

V = Vg + Vem

(9.62)

is the total potential energy of the system, where Vg and Vem stand for contributions of the gravitational and electromagnetic fields, respectively. If the horizontal plane containing O is chosen as the reference level for the gravitational potential energy (see §3, Chap.III), then Vg = −mgl cos α,

(9.63)

while the electromagnetic potential energy is [see (9.58) and (9.59)] Vem =

1 1 qe = − CB 2 l4 α˙ 2 . 2 8

(9.64)

In view of (9.60)-(9.64), the Lagrangian of the system is L=

4ml2 + CB 2 l4 2 α˙ + mgl cos α. 8 330

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(9.65)

Free ebooks ==> www.Ebook777.com The canonical momentum pα conjugate to the generalized coordinate α is   ∂L CB 2 l4 2 α, ˙ (9.66) pα = = ml + ∂ α˙ 4 and the Hamiltonian writes   2 ml CB 2 l4 α˙ 2 − mgl cos α, H = pα α˙ − L = + 2 8 or, in terms of α and pα , p2α 2 4  − mgl cos α. 2 ml2 + CB4 l

H=

The system of Hamilton’s canonical equations then is  ∂H pα   =  α˙ = 2 4 , ∂pα ml2 + CB4 l  ∂H   p˙α = − = −mgl sin α. ∂α

(9.67)

(9.68)

The differential equation describing the motion of the pendulum is obtained taking the time derivative of the first equation (9.68), then substituting p˙α given by the second equation (9.68). The result is: α ¨+ or



mgl sin α 2 4 = 0, ml2 + CB4 l

CB 2 l4 ml + 4 2



α ¨ + mgl sin α = 0.

(9.69)

If the angle α is small, one can approximate sin α ≃ α, and the differential equation (9.69) becomes   CB 2 l4 2 ml + α ¨ + mglα = 0. 4 Denoting ω2 =

ml2

mgl 2 4 = + CB4 l

equation (9.70) writes

l g

1 2 l2  , 1 + CB 4m

α ¨ + ω 2 α = 0. 331

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(9.70)

(9.71)

(9.72)

Free ebooks ==> www.Ebook777.com This is the well-known second order, linear, homogeneous, differential equation with constant coefficients known as the harmonic oscillator equation. The oscillation period T is s   2π l CB 2 l2 T = = 2π 1+ , (9.73) ω g 4m where the term CB 2 l2 /4m represents the ”contribution” of the electromagnetic effects, in addition to the gravitational influence. Indeed, as we have seen in Chap.III, the period of gravitational pendulum, for small oscillations, is given by s l T0 = 2π . g

B. Problems of Fluid Mechanics Euler-Lagrange Equations for Continuous Systems The analytical formalism can be successfully applied in the study of continuous systems (i.e. systems with an infinite number of degrees of freedom): fields, fluids, magnetofluids, etc. Let us first deduce the fundamental equations of analytical technique, applied to continuous systems. Consider the functional Z h ∂ϕ i ∂ϕ , .., J(ϕ) = L x1 , .., xn ; ϕ(x1 , .., xn ), dx1 ...dxn , (9.74) ∂x1 ∂xn Dn

defined on the bounded domain Dn of a n-dimensional space Rn , where L is a continuous and differentiable function, admitting as many partial derivatives as necessary, and ϕ a function of class C 2 in Dn . Assuming that the values of ϕ on the closed hypersurface Sn−1 which bounds the domain Dn are given, let us determine the function ϕ which makes J(ϕ) an extremum. Suppose that ϕ(x1 , .., xn ) performs the stationary value of J(ϕ). In this case, for any infinitesimal variation ϕ → ϕ + δϕ, where δϕ = ǫη(x1 , .., xn ), with η(x1 , .., xn )|Sn−1 = 0, (9.75) 332

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Free ebooks ==> www.Ebook777.com the first variation δJ(ϕ) of integral (9.74) must be zero. Since n   X ∂ϕ ∂ϕ  ∂L ∂η  ∂η ∂L L x, ϕ + ǫη, = L x, ϕ, + ǫη +ǫ +ǫ + ..., ∂x ∂x ∂x ∂ϕ ∂xi ∂ϕ,i i=1

where ϕ,i = ∂ϕ/∂xi , the first variation of J(ϕ) is Z  n ∂L X ∂η ∂L  δJ(ϕ) = ǫ η + dΩ, ∂ϕ i=1 ∂xi ∂ϕ,i

(9.76)

Dn

with dΩ = dx1 dx2 ...dxn . Integrating by parts the second term in (9.76), we have: Z X n ∂η ∂L dΩ ∂x ∂ϕ i ,i i=1 Dn

Z X Z X n n ∂  ∂L  ∂  ∂L  η dΩ − η dΩ. = ∂xi ∂ϕ,i ∂xi ∂ϕ,i i=1 i=1

(9.77)

Dn

Dn

But dΩ = dx1 ...dxk ...dxn = dxk dSk

(no summation),

(9.78)

where dSk = dx1 ...dxk−1 dxk+1 ...dxn is the element of hypersurface orthogonal to dxk . Using Green-Gauss theorem and the boundary condition (9.75), we see that the first integral on the r.h.s. of (9.77) vanishes: Z X Z X n n ∂  ∂L  ∂L η dΩ = η dSi = 0. ∂xi ∂ϕ,i ∂ϕ,i i=1 i=1 Dn

Sn−1

Thus, the first variation of J(ϕ) is δJ(ϕ) = ǫ

Z

Dn

n X ∂  ∂L i η − dΩ. ∂ϕ i=1 ∂xi ∂ϕ,i

h ∂L

The necessary and sufficient condition for a stationary value of J(ϕ) requires that, for any function η (except for condition (9.75)), we have ∂L X ∂  ∂L  − = 0. ∂ϕ i=1 ∂xi ∂ϕ,i n

333

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(9.79)

Free ebooks ==> www.Ebook777.com Assuming that L is a function of h variables ϕ(1) (x), ..., ϕ(h) (x), let us consider the functional Z (s) J[ϕ ] = L[x, ϕ(s) (x), ϕ(s) (s = 1, h). (9.80) ,x ]dx1 ...dxn Dn

Following a similar procedure as for a single variable ϕ, the stationary condition of functional J[ϕ(s) ] yields the following system of secondorder, partial differential equations: X ∂  ∂L  ∂L − =0 ∂ϕ(s) i=1 ∂xi ∂ϕ(s) n

(s = 1, h),

(9.81)

,i

called the Euler-Lagrange equations of our variational problem. In order to use these equations in CDM (Continuous Deformable Media) mechanics, we choose x1 = x, x2 = y, x3 = z, x4 = t.

(9.82)

With this choice, the functional (9.80) becomes

J[ϕ

(s)

]=

Zt2 Z

t1 V

(s)

(s) (s) L[x, y, z, t; ϕ(s) (x, y, z, t); ϕ(s) ,x , ϕ,y , ϕ,z , ϕ,t ]

×dx dy dz dt

(s = 1, h),

(9.83)

and Euler-Lagrange equations read: ∂  ∂L  ∂L ∂  ∂L  − − =0 ∂xi ∂ϕ(s) ∂t ∂ϕ(s) ∂ϕ(s) ,t

(s = 1, h),

(9.84)

,i

where the summation convention has been used for index i = 1, 2, 3. Comparing (9.83) with the action integral (2.36), we realize that there can be settled an equivalence between them if we choose L=

Z

L dx dy dz =

V

Z

L dτ.

(9.85)

V

Therefore function L stands for the Lagrangian per unit volume. It is called the Lagrangian density. 334

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Free ebooks ==> www.Ebook777.com With this notation, Hamilton’s principle (2.35) reads

δ

Zt2Z

t1

V

 Ldτ dt = 0,

(9.86)

and can be used as a fundamental postulate in the study of holonomic CDM, while Euler-Lagrange equations (9.84) are the equations of motion of these systems. Comparing equations (9.81) with Lagrange’s equations for systems with a finite number of degrees of freedom (2.24), we realize that they are different in certain respects. In case of continuous systems the role of generalized coordinates is played by the functions ϕ(s) , called dependent variables or variational parameters, while x1 , .., xn play the role of independent variables. Our choice (9.82) shows that both the space coordinates x, y, z and the time t are now taken as independent parameters, while ϕ(s) are selected from the physical variables which characterize a given system. In view of these considerations, equations (9.81) can be regarded as an infinite chain of Lagrange’s-type differential equations, each of them being obtained by a successive fixation of space variables x, y, z. Since L is a Lagrangian density, all quantities appearing in it must be represented by their densities, such as: mass density ρ, entropy density s, current density ~j, etc. Equations (9.81) are particularly useful in analytical formalism of CDM, because they can be applied not only in the study of condensed media (solid, fluid), but also in the derivation of fundamental equations governing the fields. Observation. Euler-Lagrange’s equations (9.81) do not change their form if instead of L we choose ′

L (x, ϕ

(s)

(s) , ϕ,i )

= L(x, ϕ

(s)

(s) , ϕ,i )

n X ∂ + Fk (x, ϕ(s) ), ∂xk

(9.87)

k=1

provided the integration domain Dn remains unchanged, and the field variables ϕ(s) take fixed values on the boundary Sn−1 of Dn . To prove this, we integrate the last relation on Dn and, using the generalized Green-Gauss theorem, we obtain: Z

Dn



L dΩ =

Z

Dn

Z X Z Z n ∂Fk LdΩ + dΩ = LdΩ + ∂xk Dn

k=1

Dn

Sn−1

335

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n X

k=1

Fk dSk .

Free ebooks ==> www.Ebook777.com Applying now the operator δ to this relation, we still have: δ

Z

Dn



L dΩ = δ

Z

LdΩ +

Dn

Z

n X n h X ∂Fk

Sn−1 i=1 k=1

∂xi

δxi +

i ∂Fk (s) δϕ dSk . ∂ϕ(s)

But, by hypothesis, on the boundary Sn−1 we have δxi = 0, δϕ(s) = 0, and therefore Z Z L′ dΩ = δ

δ

LdΩ

Dn

Dn

which means that the condition of stationary value of J[ϕ(s) ] δJ[ϕ

(s)

]=δ

Z

LdΩ = 0

Dn

does not change at the transformation (9.87). As a result, EulerLagrange’s equations (9.81) do not change their form. In other words, two Lagrangian densities which differ from one another by a divergence term are equivalent. Problem 6 Find the equation of motion of an ideal, compressible fluid, performing isentropic motion in an external potential field1 (Euler’s equation). Solution Let us denote by V ∗ (~r, t), s(~r, t), and ε(~r, t) the potential of the exterior field (e.g. the gravitational field), entropy and internal energy, taken per unit mass, respectively. Then the kinetic energy density is 1 v |2 , where ~v (~r, t) is the velocity field, while the potential energy is 2 ρ|~ composed by two terms: ρε and ρV ∗ , corresponding to internal and external forces, respectively. Nevertheless, the expression Lo =

1 2 ρ~v − ρ(ε + V ∗ ) 2

(9.88)

cannot be used as a Lagrangian density, because it contains only some of the physical variables which define the system. In turn, this is due 1

J.W.Herivel, Proc. Camb. Phil. Soc., 51, 1955, p.344. 336

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Free ebooks ==> www.Ebook777.com to the fact that we did not take into consideration the constraints acting on the fluid, which in our case are the equation of continuity ∂ρ + ∇ · (ρ~v ) = 0, ∂t

(9.89)

and the equation of conservation of entropy ∂s ds = + (~v · ∇)s = 0. dt ∂t

(9.90)

A suitable Lagrangian density is constructed by using the method of Lagrangian multipliers. In this respect we amplify equations (9.89) and (9.90) by the multipliers α(~r, t) and β(~r, t), respectively, and add the result to (9.88). This yields L=

h ∂ρ i  ∂s  1 2 ρ~v − ρ(ε + V ∗ ) − α + ∇ · (ρ~v ) − βρ + ~v · ∇s . (9.91) 2 ∂t ∂t

It is more convenient to our purpose to use the Lagrangian density in a slightly modified form. Taking advantage of the property (9.87), we shall add to (9.91) the divergence ∂ (αρvj ) ∂xj

(j = 1, 4),

where we choose x1 = x, x2 = y, x3 = z, x4 = t, v1 = vx , v2 = vy , v3 = vz , v4 = 1. Since  ∂ρ



 ∂α  ∂ −α + ~v · ∇ρ + ρ∇ · ~v + ∇ · (αρ~v ) + (αρ) = ρ + ~v · ∇α , ∂t ∂t ∂t we finally obtain L=

 ∂α   ∂s  1 2 ρ~v − ρ(ε + V ∗ ) + ρ + ~v · ∇α − βρ + ~v · ∇s . (9.92) 2 ∂t ∂t

Choosing s, ρ, vx , vy , vz as variational parameters ϕ( i), i = 1, 5 in (9.84), we then have: i) ϕ(1) = s. The corresponding Euler-Lagrange equation is ∂L ∂  ∂L  ∂  ∂L  − − = 0. ∂s ∂xi ∂s,i ∂t ∂s,t 337

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(9.93)

Free ebooks ==> www.Ebook777.com Using the thermodynamic fundamental equation of equilibrium processes 1 T ds = dε(ρ, s) + p d , ρ we find

∂L ∂L ∂ε = = −ρT ; ∂s ∂ε ∂s

∂L = −βρvi ; ∂s,i

∂L = −βρ. ∂s,t

Introducing these results in (9.93) and using the equation of continuity, after simplifying by ρ 6= 0 we arrive at ∂β + ~v · ∇β = T. ∂t

(9.94)

ii) ϕ(2) = ρ. We have ∂  ∂L  ∂  ∂L  ∂L − − = 0. ∂ρ ∂xi ∂ρ,i ∂t ∂ρ,t Performing the derivatives 1 p ∂α ∂L = ~v 2 − (ε + V ∗ ) − + + ~v · ∇α ; ∂ρ 2 ρ ∂t ∂L =0; ∂ρ,i we obtain

∂L = 0, ∂ρ,t

1 2 p ∂α ~v − (ε + V ∗ ) − + + ~v · ∇α = 0, 2 ρ ∂t

(9.95)

which is a Bernoulli-type equation. iii) ϕ(3,4,5) = vk (k = 1, 2, 3). In this case we have three scalar equations ∂L ∂  ∂L  ∂  ∂L  − − = 0. ∂vk ∂xi ∂vk,i ∂t ∂vk,t Since

∂α ∂s ∂L = ρvk + ρ − βρ ; ∂vk ∂xk ∂xk ∂L =0; ∂vk,i

∂L = 0, ∂vk,t

338

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Free ebooks ==> www.Ebook777.com we arrive at ~v = −∇α + β∇s,

(9.96)

which is a Clebsch transformation. Therefore, the functions α(~r, t), β(~r, t), and s play the role of Clebsch potentials. The last step is now to eliminate the multipliers α and β from the equations (9.94) - (9.96). To do this, we shall first replace ∇α = −~v + β∇s into (9.95): 1 p ∂α − |~v |2 − (ε + V ∗ ) − + + β~v · ∇s = 0. 2 ρ ∂t Applying to this equation the operator gradient we still have: −~v × curl~v − (~v · ∇)~v −

p 1 p ∗ ∇ρ − T ∇s − ∇V − ∇p + ∇ρ ρ2 ρ ρ2

∂ (β∇s − ~v ) + β∇(~v · ∇s) + (~v · ∇s)∇β = 0. ∂t But, by virtue of (9.90) and (9.94): +

−~v × curl~v − T ∇s +

∂s  ∂β ∇s + β∇ + β∇(~v · ∇s) + (~v · ∇s)∇β ∂t ∂t

= −~v × curl(β∇s) − T ∇s − (~v · ∇β)∇s + T ∇s + (~v · ∇s)∇β = −~v × (∇β × ∇s) − (~v · ∇β)∇s + (~v · ∇s)∇β = 0, and therefore we are left with Euler’s equation ∂~v 1 1 + (~v · ∇)~v = −∇V ∗ − ∇p = F~ − ∇p. ∂t ρ ρ

(9.97)

Observation. Euler’s equation can also be written as ∂~v 1 1 + ∇(~v · ~v ) − ~v × (∇ × ~v ) = F~ − ∇p. ∂t 2 ρ v An irrotational [∇×~v = 0 ⇒ ~v = ∇ϕ(~r, t)], stationary ( ∂~ ∂t = 0) flow of a non-turbulent, perfect, incompressible (ρ = const.), and barotropic [ρ = ρ(p)] fluid is then governed by

1 ρ|~v |2 + ρg z + p = const., 2 called Bernoulli’s equation. 339

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(9.97′ )

Free ebooks ==> www.Ebook777.com Problem 7 An ideal incompressible liquid of density ρ is situated in a vertical cylindrical container, rotating about its axis with constant angular velocity ω. At rest, the height of the liquid column is h. Supposing that the fluid and the container rotate together, as a solid body, determine: a) The shape of the surface of the rotating liquid; b) The pressure at every point inside the liquid; c) The pressure exerted at every point on the bottom of the container; d) The total force and the total pressure exerted by the fluid on the bottom of the container. Solution a) Let us attach a reference frame to the rotating liquid, with z-axis along the ascendant vertical, as shown in Fig.IX.8. The components of velocity of some particle of the liquid, with respect to this frame, are: vx = −ωy, vy = ωx, vz = 0, because ~i ~j ~v = ω ~ × ~r = 0 0 x y

~k ω = ωx~j − ωy~i = vx~i + vy~j + vz~k. z

As we already know, the equation of motion of an ideal, incompressible fluid is Euler’s equation (9.97) 1 ∂~v + (~v · ∇)~v = F~ − ∇p, ∂t ρ

(9.97)

where F~ is the mass-specific force ~ ∆G ~g ∆m F~ = = = ~g = −g~k. ∆m ∆m v Because there is no time variation of ~v , ∂~ ∂t = 0, and the projections on axes of (9.97) write  1 ∂p  Ox : (~v · ∇)vx = − ,    ρ ∂x   1 ∂p Oy : (~v · ∇)vy = − , (9.98)  ρ ∂y    1 ∂p   Oz : (~v · ∇)vz = −g − . ρ ∂z

340

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Free ebooks ==> www.Ebook777.com

Fig.IX.8 Since   ∂ ∂ (~v · ∇)vx = −ωy + ωx (−ωy) = −ω 2 x, ∂x ∂y   ∂ ∂ + ωx (~v · ∇)vy = −ωy (ωx) = −ω 2 y, ∂x ∂y and vz = 0, we are left with the following system  1 ∂p  ,  ω2 x =   ρ ∂x   1 ∂p ω2 y = ,  ρ ∂y      1 ∂p + g = 0. ρ ∂z

Observing that, by integration, equations (9.99) give  p 1 2 2   = ω x + f1 (y, z),   2  ρ p 1 = ω 2 y 2 + f2 (z, x),  ρ 2   p    = −gz + f3 (x, y), ρ 341

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(9.99)

Free ebooks ==> www.Ebook777.com where fi (i = 1, 2, 3) are arbitrary functions of their arguments, the general solution of the system (9.99) is p 1 = ω 2 (x2 + y 2 ) − gz + C, ρ 2

(9.100)

where C is a constant of integration. Formula (9.100) can also be obtained by working in a non-inertial frame. Consider a frame rotating together with the liquid about z-axis with constant angular velocity ω. Since the fluid acceleration relative v to this frame ~a = ∂~ v · ∇)~v equals zero, Euler’s equation writes ∂t + (~ 1 1 0 = F~tot − ∇p = F~g + F~cf − ∇p, ρ ρ

(9.101)

where F~g = −g~k is the mass-specific gravitational force, and F~cf = ω 2~r1 = ω 2 (x~i + y~j) the mass-specific centrifugal force (N.B. in the non-inertial frame, F~cf is considered as an applied force). Performing the dot product between (9.101) and d~r = dx~i + dy ~j + dz ~k, we obtain   1 1 ~ ~ 0 = Fg + Fcf − ∇p · d~r = −g dz + ω 2 (xdx + ydy) − dp ρ ρ     1 2 2 1 = d(−gz) + d ω r1 + d − p , 2 ρ where we took into account that ∇p · d~r =

∂p dxi = dp, ∂xi

and r12 = x2 + y 2 . Integrating, we have p 1 = ω 2 (x2 + y 2 ) − gz + C, ρ 2 where C is an arbitrary constant of integration. This is precisely the equation (9.100), as expected. Since the pressure p of the liquid at the free surface is zero, formula (9.100) yields  ω2 2 0= x + y 2 − gz + C, (9.102) 2 or  ω2 2 z= x + y2 + C ′ , (9.102′ ) 2g 342

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Free ebooks ==> www.Ebook777.com with C ′ = C/g. We therefore conclude that the shape of the surface of the liquid, rotating about the symmetry z-axis of the cylinder, is a paraboloid of revolution of equation (9.102). b) The constant of integration C appearing in equation (9.102) can be determined by equalizing the volumes of liquid at rest, V = πR2 h, and in its motion. The last one writes

V =

Z

r1 dr1 dϕdz =

=

r1 dr1

0

r1 dr1

0

(V )

ZR

ZR

Z2π

ω2 2g



2

C

Zr1 + g

=

ω2 2g

(x2 +y 2 )+ C g

Z



0

dz = 2π

ZR

dz

0

r1 dr1

0

0

0

Z2π



ω2 2 C r + 2g 1 g



πω 2 R4 πCR2 + . 4g g

Equalizing the two formulas for the fluid volume, we find ω 2 R2 . 4

C = gh −

Introducing this result into (9.100), we obtain the pressure at any point inside the rotating liquid   ρω 2 R2 2 2 p = ρg(h − z) + x +y − . (9.103) 2 2 c) The pressure of liquid at any point of the bottom of the container is immediately obtained by making z = 0 in (9.103):   ρω 2 R2 2 2 p = p(x, y) = ρgh + x +y − . (9.104) 2 2 d) The total force exerted by the fluid on the bottom of the container is Z Z Z F z=0, S=πR2 = dF = p(x, y) dS = p(x, y) dxdy (S)

=

Z

(S)

(S)

p(r1 )r1 dr1 dϕ =

(S)

ZR 0

p(r1 )r1 dr1

Z2π



0

343

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Free ebooks ==> www.Ebook777.com = 2π

ZR 

ρω 2 ρgh + 2

0



r12

R2 − 2



r1 dr1

 R  ZR 2 Z ZR 2 2 ω 3 ω R = 2πρ  ghr1 dr1 + r1 dr1 − r1 dr1  2 4 0

0

0

= πR2 ρgh = M g,

(9.105)

where M = ρV = πR2 ρh is the total mass of the liquid. This result shows that the total force equals the weight of the liquid enclosed in the container. The total pressure exerted on bottom of the container is F z=0,S=πR2 = ρgh. (9.106) P = p z=0,S=πR2 = S Problem 8 A layer of viscous liquid of thickness h is upperly limited by a free surface, and by a fixed plane on its lower side. The angle of inclination of the plane with respect to horizontal is α, as shown in Fig.IX.9. Find: 1) Velocity of the liquid under the action of its own weight; 2) Mass flow through a cross section of the liquid layer.

Fig.IX.9 Solution 1) The liquid under investigation is homogeneous, viscous, and incompressible, its motion is plane-parallel and unidimensional, while 344

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Free ebooks ==> www.Ebook777.com the flow is stationary. The main equations describing such a fluid are: the equation of continuity ∂ρ + div(ρ~v ) = 0, ∂t

(9.107)

and Stokes-Navier equation 

 ∂~v ρ + (~v · ∇)~v = ρF~ − ∇p + (λ + µ)∇θ + µ∆~v , ∂t

(9.108)

where ~v = ~v (x, y, z, t) is the velocity field, F~ is the mass-specific force, λ and µ are dynamic coefficients of viscosity, and θ = div~v . In case of the homogeneous and incompressible fluids (i.e. liquids) ρ = ρ(~r, t) = const. and (9.107) yields θ = div~v = 0, while Stokes-Navier equation becomes 1 ∂~v + (~v · ∇)~v = F~ − ∇p + ν∆~v , ∂t ρ

(9.108′ )

where ν = µ/ρ is the kinematic coefficient of viscosity of the fluid. Since the fluid flow is stationary and performs in x-direction (see Fig.IX.9), the only non-zero component of the velocity is ~v (x, y, z, t) ≡ ~ivx = ~iu(x, y). Then equation div~v = 0 gives div~v =

∂vx (x, y) ∂u = = 0, ∂x ∂x

that is u = u(y) and, consequently, (~v · ∇)~v = vx

 ∂~v ∂ ~ = vx iu(y) = 0. ∂x ∂x

Taking into account that the motion is stationary Stokes-Navier (9.108) becomes

∂~ v ∂t

ρF~ − ∇p + µ∆~v = 0. 345

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 = 0 , equation (9.109)

Free ebooks ==> www.Ebook777.com Projecting this vector equation on the two axes of the xOy frame attached to the liquid (see Fig.IX.9), we have: ρg sin α −

∂p d2 u + µ 2 = 0, ∂x dy

and −ρg cos α −

∂p = 0. ∂y

(9.110)

(9.111)

This way, our problems turns into a problem of mathematical physics equations. Namely,  we have to solve the system (9.110)-(9.111) within the domain D = x, y| − ∞ < x < +∞, 0 < y < h , with the following boundary conditions: (i) u(y) = 0, for y = 0; (ii)

du dy

= 0,

for y = h;

(iii) p(x, y) = p0 = const.,

for y = h.

Integrating (9.111), we obtain p = p(x, y) = −ρgy cos α + p1 (x). The arbitrary function p1 (x) can be determined by using the boundary condition (iii), which gives p0 = −ρgh cos α + p1 (x), so that p = p(x, y) = p0 + ρg(h − y) cos α.

(9.112)

Introducing this result into (9.110), we have d2 u ρg sin α =− , 2 dy µ and, by integration, u=−

ρg sin α 2 y + C1 y + C2 . 2µ

The integration constants C1 and C2 are determined by means of conditions (i) and (ii), which lead to C2 = 0;

C1 =

ρg sin α h, µ

346

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Free ebooks ==> www.Ebook777.com so that u = u(y) =

ρg (2h − y)y sin α. 2µ

(9.113)

2) To calculate the mass flow through a cross section of the liquid layer (i.e. the mass of liquid going through a unit surface orthogonal to x-axis), we use the definition of this quantity for a stationary flow of a homogeneous and incompressible fluid, which is Qm

qm 1 = = Lz h Lz h

Zh

dqm ,

0

where dqm =

dm ρdV ρ u(y)∆t Lz dy = = = ρ u(y)Lz dy ∆t ∆t ∆t

is the elementary mass crossing the elementary surface of any width Lz (along z-axis) and thickness dy (along y-axis), in unit time. In view of (9.113), we have:

Qm

1 = Lz h

Zh

1 ρu(y)Lz dy = h

0

Zh 0

=

ρ2 g (2h − y)y sin αdy 2µ

ρ2 gh2 sin α . 3µ

(9.114)

Problem 9 Investigate the motion of a viscous, heavy liquid, flowing under the action of its own weight, in a rectangular pipe of sides 2a (width) and h (height). The angle of inclination of the pipe with respect to horizontal is α (Fig.IX.10). At the surface of liquid the velocity varies 2 according to the law u(y, h) = um 1 − ay2 , where um is the maximum velocity at surface, in the middle of the pipe. Solution To solve the problem, we appeal to the equation of continuity (9.107), and Stokes-Navier equation (9.108). The components of the mass-specific force F~ and velocity ~v are Fx = g sin α,

Fy = 0,

Fz = −g cos α,

347

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Free ebooks ==> www.Ebook777.com and vx = u,

vy = vz = 0,

respectively. Since the fluid is homogeneous and incompressible (liquid), ρ = ρ(~r, t) = const., and the equation of continuity (9.107) yields div ~v = ∂u ∂x = 0 ⇒ u = u(y, z), −∞ < x < +∞. On the other v hand, taking into account that the motion is stationary ∂~ ∂t = 0 and   ∂ ∂ ~ i u(y, z) = 0, the components of vector (~v · ∇)~v = vx ∂x ~v = u ∂x equation (9.108) along the coordinate axes are   2  ∂p ∂ u ∂2u   + 2 , 0 = ρg sin α − +µ  2  ∂x ∂y ∂z   ∂p (9.115) 0=− ,  ∂y     ∂p  0 = −ρg cos α − . ∂z

Fig.IX.10 Equation (9.115)2 yields p = p(x, z), while (9.115)3 gives p = −ρgz cos α + p1 . Since p = p0 for z = h, we have p1 (x) = p0 + ∂p ρgh cos α = const. In this case ∂x = 0 and equation (9.115)1 becomes ∂2u ∂2u + 2 = k, ∂y 2 ∂z

−a < y < a, 0 < z < h,

(9.116)

α where we denoted k = − ρg sin . µ Equation (9.116) is a Poisson-type equation, with the boundary conditions:

348

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Free ebooks ==> www.Ebook777.com (i) u(y, 0) = 0,

−a ≤ y ≤ a;

(ii) u(−a, z) = 0, u(a, z) = 0, 0 ≤ z ≤ h;   2 (iii) u(y, h) = um 1 − ay2 , −a ≤ y ≤ a. Since the motion of the liquid is symmetrical with respect to xzplane, it is sufficient to be studied only in the domain 0 < y < a of the yz-plane. To this end, it is convenient to introduce new, dimensionless coordinates, defined as u = um U,

y = aY,

z = aZ,

h = l, a

k=

um K. a2

This way, equation (9.116) writes ∂2U ∂2U + = K, ∂Y 2 ∂Z 2

0 < Y < 1, 0 < Z < l.

(9.117)

This is a Poisson-type equation for U = U (Y, Z), with the boundary conditions (i′ ) U (Y, 0) = 0, 0 ≤ Y ≤ 1; (ii′ ) U (1, Z) = 0, 0 ≤ Z ≤ l; (iii′ ) U (Y, l) = 1 − Y 2 , 0 ≤ Y ≤ 1. The boundary conditions are of the first kind, so that we have to find the solution of the Dirichlet problem for Poisson’s equation. We look for a solution of (9.117) of the form U (Y, Z) =

∞ X

An cos

n=1



(2n − 1)πY 2



ζn (Z).

(9.118)

Since  2   ∞ X (2n − 1)π (2n − 1)πY ∂2U =− An cos ζn (Z), ∂Y 2 2 2 n=1   ∞ X ∂2U (2n − 1)πY d2 ζn (Z) = An cos , 2 ∂Z 2 2 dZ n=1

equation (9.117) gives ∞ X

n=1

An cos



(2n − 1)πY 2

"

d2 ζn (Z) − dZ 2



(2n − 1)π 2

2

349

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#

ζn (Z) = K. (9.119)

Free ebooks ==> www.Ebook777.com Let us now prove that the r.h.s. of (9.119) can be written as   ∞ X 4(−1)n−1 (2n − 1)πY K=K , cos (2n − 1)π 2 n=1

(9.120)

which is possible because   ∞  1, X cos (2n − 1)ξ 4 (−1)n−1 =  0, π n=1 2n − 1

π for 0 ≤ ξ < , 2 π for ξ = . 2

(9.121)

To prove (9.121), we shall use the following uniformly, absolutely convergent (for |z| < 1), series 1 = 1 − iz − z 2 + iz 3 + ..., 1 + iz

1 1 + z2

1 = 1 + iz − z 2 − iz 3 + ..., 1 − iz ∞ X 2 4 6 = 1 − z + z − z + ... = (−1)n−1 z 2n−2 , n=1

where z = ξ + iη is a complex number. Integrating these relations, we have: Z Z   1 dz = −i ln(1 + iz) = (1 − iz − z 2 + iz 3 + z 4 + ...) dz 1 + iz Z

z3 z4 z5 z2 − +i + + ..., 2 3 4 5 Z   dz = i ln(1 − iz) = (1 + iz − z 2 − iz 3 + z 4 + ...) dz =z−i

1 1 − iz

z2 z3 z4 z5 − −i + + ..., 2 3 4 5 Z Z   1 dz = arctan z = (1 − z 2 + z 4 − z 6 + ...) dz 1 + z2 =z+i

∞ X z3 z5 z7 z 2n−1 =z− + − + ... = (−1)n−1 . 3 5 7 2n − 1 n=1

Adding the first two relations, we obtain Z Z 1 1 1 + iz dz + dz = −i ln(1 + iz) + i ln(1 − iz) = −i ln 1 + iz 1 − iz 1 − iz 350

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   z2 z3 z4 z5 z2 z3 z4 z5 z−i − +i + + .. + z + i − −i + + .. 2 3 4 5 2 3 4 5   z5 z3 + − ... , =2 z− 3 5

so that ∞ X  −i 1 + iz z3 z5 z 2n−1 = arctan z . ln =z− + − ... = (−1)n−1 2 1 − iz 3 5 2n − 1 n=1 (9.122) Choosing z = eiξ , we can write:

1 + iz 1 + i(cos ξ + i sin ξ) = 1 − iz 1 − i(cos ξ + i sin ξ) =

(1 − sin ξ + i cos ξ)(1 + sin ξ + i cos ξ) (1 + sin ξ − i cos ξ)(1 + sin ξ + i cos ξ) =

cos ξ i π i cos ξ = e 2. 1 + sin ξ 1 + sin ξ

Denoting f (z) = and observing that

1 + ieiξ cos ξ i π = e 2, iξ 1 − ie 1 + sin ξ

ln f (z) = ln |f (z)| + i arg f (z), where

Im[f (z)] π = , Re[f (z)] 2

arg f (z) = arctan the relation (9.122) becomes −i −i ln f (z) = ln 2 2 i = − ln 2



cos ξ 1 + sin ξ





cos ξ i π e 2 1 + sin ξ

π i i − ln ei 2 = − ln 2 4





cos ξ 1 + sin ξ

2

Then we can write  2 ∞ X i cos ξ π eiξ(2n−1) − ln ± = (−1)n−1 4 1 + sin ξ 4 2n − 1 n=1 351

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+

π . 4

Free ebooks ==> www.Ebook777.com  cos(2n − 1)ξ sin(2n − 1)ξ = , (−1) +i 2n − 1 2n − 1 n=1   S 3π  where the ”plus” sign corresponds to the case ξ ∈ 0, π2 2 , 2π , and the ”minus” sign to the interval ξ ∈ π2 , 3π 2 . Identifying the real and imaginary parts in the last relation, one obtains: ∞ X

n−1



π π  , for 0 ≤ ξ < , cos(2n − 1)ξ 4 2 = (−1)n−1  0, for ξ = ± π , 2n − 1 n=1 2 ∞ X

which is relation (9.121), and ∞ X

(−1)

n=1

1 − 1)ξ = ln 2n − 1 4

n sin(2n



cos ξ 1 + sin ξ

Formula (9.121) then yields (9.120) by setting ξ = (9.120), equation (9.119) then writes ∞ X

An cos

n=1



(2n − 1)πY 2

"

d2 ζn (Z) − dZ 2



2

π 2Y

(2n − 1)π 2

.

. By means of 2

ζn (Z)

#

  ∞ X (2n − 1)πY 4(−1)n−1 cos =K . (2n − 1)π 2 n=1   of both sides, we are left with Equalizing coefficients of cos (2n−1)πY 2 the following ordinary differential equations: ζn′′ (Z) −

(2n − 1)2 π 2 (−1)n−1 4K ζn (Z) = 4 (2n − 1)πAn

with boundary conditions ζn (0) = 0, and

∞ P

(n ∈ N∗ ),

An cos

n=1



(2n−1)πY 2

(9.123) 

ζn (l)

= 1 − Y 2. For any value of n ∈ N∗ , the general solution of the corresponding non-homogeneous equation (9.123) writes as a sum of the general (o) solution ζn of the homogeneous equation ζn′′ (Z) −

(2n − 1)2 π 2 ζn (Z) = 0 4

(n ∈ N∗ )

352

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(9.124)

Free ebooks ==> www.Ebook777.com (p)

and a particular solution ζn of the non-homogeneous equation, that is ζn (Z) = ζn(o) + ζn(p) , n = 1, 2, 3, ... The homogeneous equations (9.124) have solutions of the form ζn(o) = Bn sinh

(2n − 1)πZ (2n − 1)πZ + Cn cosh , 2 2

while a particular solution of the corresponding non-homogeneous equation must be chosen as a constant (the term producing inhomogeneity is a constant!), for each n ∈ N∗ : ζn(p) = Pn = (const.)n .

(9.125)

The constants Pn must verify the inhomogeneous equations (9.123) (2n − 1)2 π 2 (−1)n−1 4K − Pn = , 4 (2n − 1)πAn which gives Pn =

(−1)n 16K . (2n − 1)3 π 3 An

Therefore, the general solutions of (9.123) are (2n − 1)πZ (2n − 1)πZ (−1)n 16K + Cn cosh + . 2 2 (2n − 1)3 π 3 An (9.126) The arbitrary integration constants Bn and Cn are determined from the boundary conditions: ζn (0) = 0, ζn (Z) = Bn sinh

and

∞ X

An cos

n=1



(2n − 1)πY 2



ζn (l) = 1 − Y 2 .

The first boundary condition yields ζn (0) = Cn +

(−1)n 16K = 0, (2n − 1)3 π 3 An

that is Cn = −

(−1)n−1 16K (−1)n 16K = , (2n − 1)3 π 3 An (2n − 1)3 π 3 An 353

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(9.127)

Free ebooks ==> www.Ebook777.com and ζn (Z) becomes (2n − 1)πZ (−1)n−1 16K (2n − 1)πZ ζn (Z) = Bn sinh + cosh 2 (2n − 1)3 π 3 An 2 +

(−1)n 16K (2n − 1)3 π 3 An

  (−1)n 16K (2n − 1)πZ (2n − 1)πZ . + 1 − cosh = Bn sinh 2 (2n − 1)3 π 3 An 2 (9.128) The second boundary condition then gives ∞ X

n=1

An cos



(2n − 1)πY 2



ζn (l) =

∞ X

n=1

An cos



(2n − 1)πY 2



   (2n − 1)πl (−1)n 16K (2n − 1)πl × Bn sinh + 1 − cosh 2 (2n − 1)3 π 3 An 2   ∞  X (−1)n 16K (2n − 1)πl (2n − 1)πl + 1 − cosh = An Bn sinh 3 π3 2 (2n − 1) 2 n=1   (2n − 1)πY × cos = 1 − Y 2. (9.129) 2 Next, we shall express 1 − Y 2 as a series of the same type as that on the l.h.s. To this end, we first observe that, since the series [see (9.121)] ∞ X π cos(2n − 1)ξ = (−1)n−1 2n − 1 4 n=1

is absolutely and uniformly convergent, it can be integrated term by term. Therefore: # Z "X ∞ cos(2n − 1)ξ (−1)n−1 dξ 2n − 1 n=1 =

∞ Z  X

(−1)

  − 1)ξ dξ 2n − 1

n−1 cos(2n

n=1

=

∞ X

n=1

(−1)n−1

sin(2n − 1)ξ π = ξ + K1 . 2 (2n − 1) 4 354

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Free ebooks ==> www.Ebook777.com To determine the integration constant K1 we impose condition for the last relation to be valid for any ξ, including ξ = 0. This gives K1 = 0, and the last relation becomes ∞ X

(−1)n−1

n=1

A new integration yields Z "X ∞

(−1)

sin(2n − 1)ξ π = ξ. 2 (2n − 1) 4 # − 1)ξ dξ (2n − 1)2

n−1 sin(2n

n=1

∞ Z  X

=

(−1)

  − 1)ξ dξ (2n − 1)2

n−1 sin(2n

n=1

=− or

∞ X

(−1)n−1

n=1

∞ X

(−1)n−1

n=1

cos(2n − 1)ξ π = ξ 2 + K2 , 3 (2n − 1) 8

π 2 cos(2n − 1)ξ = − ξ + K3 , (2n − 1)3 8

where K3 = −K2 is a new constant of integration. Using the already known procedure, we set ξ = 0 and obtain ∞ X (−1)n−1 π3 K3 = = , (2n − 1)3 32 n=1

so that ∞ X

(−1)

π − 1)ξ = 3 (2n − 1) 8

n−1 cos(2n

n=1

Taking now ξ of the form ξ =

π 2Y



 π2 2 −ξ , 4

0≤ξ≤

π . 2

, we still have

  ∞ X (−1)n−1 32 (2n − 1)πY 1−Y = cos . 3 π3 (2n − 1) 2 n=1 2

With these results, (9.129) becomes ∞  X

n=1

  (2n − 1)πl (−1)n 16K (2n − 1)πl An Bn sinh + 1 − cosh 2 (2n − 1)3 π 3 2 355

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  ∞ X (2n − 1)πY (−1)n−1 32 × cos = cos . 3 π3 (2n − 1) 2 n=1   Identifying the coefficients of cos (2n−1)πY of both sides, we still 2 have   (2n − 1)πl (2n − 1)πl (−1)n 16K An Bn sinh 1 − cosh + 2 (2n − 1)3 π 3 2 (2n − 1)πY 2



= which gives ( Bn =

(−1)n−1 32 , (2n − 1)3 π 3

 ) (−1)n−1 32 (−1)n−1 16K (2n − 1)πl + 1 − cosh (2n − 1)3 π 3 (2n − 1)3 π 3 2 ,

An sinh

(2n − 1)πl 2

   , (−1)n−1 16 (2n − 1)πl (2n − 1)πl = 2 + K 1 − cosh An sinh . 3 3 (2n − 1) π 2 2 Introducing Bn into (9.128), we have ζn (Z) =

(

   (2n − 1)πl (−1)n−1 16 2 + K 1 − cosh (2n − 1)3 π 3 2

,

) (2n − 1)πl (2n − 1)πZ An sinh sinh 2 2   (2n − 1)πZ (−1)n 16K 1 − cosh + (2n − 1)3 π 3 An 2 (2n−1)πZ

1 sinh 2 = An sinh (2n−1)πl 2    n−1 (−1) 16 (2n − 1)πl × 2 + K 1 − cosh (2n − 1)3 π 3 2   1 (−1)n 16K (2n − 1)πZ + 1 − cosh . An (2n − 1)3 π 3 2 356

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Free ebooks ==> www.Ebook777.com To simplify this formula, let us denote β(Z) = 1 − cosh and γ(Z) = sinh

(2n − 1)πZ 2

(2n − 1)πZ . 2

We then have: 1 (−1)n−1 16 ζn (Z) = An (2n − 1)3 π 3

 

  γ(Z) 2 + Kβ(l) − Kβ(Z) . γ(l)

(9.130)

In view of (9.130), the solution (9.118) writes   ∞ X  γ(Z) (−1)n−1 16  U (Y, Z) = 2 + Kβ(l) − Kβ(Z) (2n − 1)3 π 3 γ(l) n=1 × cos



(2n − 1)πY 2

A new notation δ(Y ) = cos





.

(2n − 1)πY 2



finally yields   ∞ X  γ(Z) (−1)n−1 16 δ(Y )  U (Y, Z) = − Kβ(Z) . 2 + Kβ(l) 3 π3 (2n − 1) γ(l) n=1 (9.131) Our final purpose is to determine u = u(y, z). Let us then go back to the dimensional quantities u = um U, y = aY, z = aZ, ha = l, and k = uam2 K. Since   ∞ X (−1)n−1 16 (2n − 1)πY U (Y, Z) = cos 3 π3 (2n − 1) 2 n=1 ×

(



(2n − 1)πl 2 + K 1 − cosh 2



sinh (2n−1)πZ 2

sinh (2n−1)πl 2  ) (2n − 1)πZ −K 1 − cosh , 2 357

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Free ebooks ==> www.Ebook777.com we have   ∞ X (2n − 1)πy (−1)n−1 16 u(y, z) = um cos (2n − 1)3 π 3 2a n=1

×

(

ka2 2+ um



(2n − 1)πh 1 − cosh 2a

ka2 − um





(2n − 1)πz 1 − cosh 2a

sinh (2n−1)πz 2a sinh (2n−1)πh 2a

)

,

or, still,   ∞ X (2n − 1)πy (−1)n−1 16 u(y, z) = um cos 3 π3 (2n − 1) 2a n=1

×

(

ρg a2 sin α 2− µum



(2n − 1)πh 1 − cosh 2a

ρg a2 sin α + µum





(2n − 1)πz 1 − cosh 2a

sinh (2n−1)πz 2a sinh (2n−1)πh 2a

)

.

(9.132)

Taking a = h = 1 m, µ = ρ sin α, um = 1 m · s−1 and g ≈ 10 m · s−2 , we finally obtain   ∞ X (2n − 1)πy (−1)n−1 32 u(y, z) = cos 3 π3 (2n − 1) 2 n=1

×

(



(2n − 1)π 1 − 5 1 − cosh 2 



(2n − 1)πz +5 1 − cosh 2

sinh (2n−1)πz 2 sinh (2n−1)π 2 )

.

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(9.133)

Free ebooks ==> www.Ebook777.com Fig.IX.11 shows the curves of equal velocity, obtained by means of software Mathematica, if the following command lines are used:

As one can see, the flow velocity of the viscous fluid, under the action of its own gravity, is maximum at the depth of about 0.35 m. 359

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Fig.IX.11

C. Problems of Magnetofluid Dynamics and Quantum Mechanics Hamilton’s Canonical Equations for Continuous Systems The purpose of this chapter is to write Hamilton’s canonical equations for continuous systems (fields, fluids, plasma) and show how this formalism can be applied in Magnetohydrodynamics. Recalling definition of the Hamiltonian for discrete systems H(q, p, t) =

n X i=1

pi q˙i − L(q, q, ˙ t),

where qi (i = 1, n) are the generalized coordinates, q˙i (i = 1, n) the generalized velocities, and pi (i = 1, n) the generalized momenta, let us define the Hamiltonian density, that is the Hamiltonian per unit volume, as H=

h X s=1

(s)

π(s) ϕ,t − L, 360

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(9.134)

Free ebooks ==> www.Ebook777.com (s)

where L is the Lagrangian density, ϕ,t are the partial derivatives with respect to time of the variational parameters ϕ(s) , and π(s) stand for the momentum densities associated with ϕ(s) π(s) =

∂L (s)

∂ϕ,t

.

(9.135)

If we choose x1 = x, x2 = y, x3 = z, x4 = t, the functional dependence of the Lagrangian density L is  (s)  (s) L = L xi , t, ϕ(s) (xi , t), ϕ,i , ϕ,t

(i = 1, 2, 3; s = 1, h),

(9.136)

(s)

where ϕ,i = ∂ϕ(s) /∂xi . In this case, we have   (s) H = H xi , t, ϕ(s) (xi , t), ϕ,i , π(s)

(i = 1, 2, 3; s = 1, h).

The Hamiltonian of a continuous system then writes Z   (s) H = H xi , t, ϕ(s) , ϕ,i , π(s) dτ.

(9.137)

(9.138)

V

Recalling that ϕ(s) are continuous and derivable functions of the independent variables x, y, z, t, let us perform an arbitrary variation δH of H for some fixed values of x, y, z (δxi = 0). In view of (9.137), and using the summation convention for index i = 1, 2, 3, we have δH =

Z

V

Z h h X ∂H ∂H δϕ(s) δt + δH dτ = (s) ∂t ∂ϕ s=1 V

h X ∂H

(s) + δϕ,i (s) s=1 ∂ϕ,i

h i X ∂H + δπ(s) dτ. ∂π(s) s=1

(9.139)

The variation δH can be written in an alternative form. Using (9.134) and (9.136), we find: δH =

Z

V

δ

h X s=1

(s) π(s) ϕ,t



− L dτ =

Z hX h 

V

(s)

(s)

π(s) δϕ,t + ϕ,t δπ(s)

s=1



h  i X ∂L ∂L ∂L ∂L (s) (s) (s) − δt − δϕ + δϕ + δϕ dτ. (9.140) ,t ,i (s) (s) ∂t ∂ϕ(s) ∂ϕ,t ∂ϕ,i s=1

361

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Free ebooks ==> www.Ebook777.com But

Z X h ∂L

(s) δϕ,i (s) s=1 ∂ϕ,i

V

dτ =

Z

V

 ∂ X ∂ (s) δϕ dτ ∂xi s=1 ∂ϕ(s) h

,i

Z X h ∂  ∂L  (s) − δϕ dτ. (s) ∂x i ∂ϕ s=1

(9.141)

,i

V

The first integral on the r.h.s can be transformed into a surface integral over the boundary S of the domain of volume V . Since ϕ(s) have fixed values on S, we obtain Z

 ∂ X ∂L (s) δϕ δτ = ∂xi s=1 ∂ϕ(s) h

,i

V

I X h ∂L

(s) ∂ϕ,i S s=1

δϕ(s) dSi = 0.

(9.142)

Introducing these results into (9.140) and performing some reduction of terms, by means of Euler-Lagrange’s equations (9.81) we arrive at δH =

Z nX h

(s) ϕ,t δπ(s)

s=1

V

=

,i

Z hX h 

V

h h X ∂  ∂L i (s) o ∂L ∂L − δt − δϕ dτ − (s) (s) ∂t ∂x ∂ϕ i ∂ϕ s=1

s=1

 ∂L i (s) δt dτ. ϕ,t δπ(s) − π(s),t δϕ(s) − ∂t

(9.143)

A similar integration by parts can be done in (9.139), which yields Z nX h ∂H ∂H δH = δt δπ(s) + ∂t V s=1 ∂π(s) +

h h X ∂H ∂  ∂H i (s) o − δϕ dτ. (s) (s) ∂x ∂ϕ i ∂ϕ s=1

(9.144)

,i

Equalizing the coefficients of the same arbitrary variations δϕ(s) , δπ(s) , and δt in (9.143) and (9.144), we obtain the following system of equations (s)

ϕ,t =

∂H ; ∂π(s)

π(s),t = −

∂H ∂  ∂H  + ∂ϕ(s) ∂xi ∂ϕ(s)

(i = 1, 2, 3; s = 1, h),

,i

362

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(9.145)

Free ebooks ==> www.Ebook777.com as well as the identity

∂H ∂L =− . (9.146) ∂t ∂t The system of 2h partial differential equations (9.145) is analogous to the Hamilton’s system of canonical equations (2.40), while the identity (9.146) is also valid for physical systems with a finite number of degrees of freedom. Equations (9.145) can be written in a symmetric form by making allowance for the notion of functional derivative. To do this, we R shall calculate the partial derivative of L = V L dτ with respect to ϕ(s) (xj ) (j = 1, 4) for x1 = x, x2 = y, x3 = z fixed. Let us consider the family of functions ϕ(s) (xj , ε), where ε is a parameter chosen in such a way that ϕ(s) (xj , ε)|S = ϕ(s) (xj , 0) ≡ ϕ(s) (xj ),

(9.147)

where S is the closed surface bounding the domain of volume V . Consider the derivative dL = dε

Z X (s) h h ∂L ∂ϕ,i i ∂L ∂ϕ(s) + dτ. (s) ∂ϕ(s) ∂ε ∂ϕ,i ∂ε s=1

V

But

Z X Z X (s) h h ∂L ∂ϕ,i ∂L ∂  ∂ϕ(s)  dτ = dτ (s) ∂ε (s) ∂x ∂ε i ∂ϕ ∂ϕ s=1 s=1 ,i ,i

V

V

Z X Z X h h ∂  ∂L ∂ϕ(s)  ∂  ∂L  ∂ϕ(s) = dτ − dτ ∂xi ∂ϕ(s) ∂ε ∂xi ∂ϕ(s) ∂ε s=1 s=1 ,i

V

,i

V

I X Z X h h ∂L ∂ϕ(s) ∂  ∂L  ∂ϕ(s) dS − dτ. = i (s) ∂ε (s) ∂x ∂ε i ∂ϕ ∂ϕ s=1 s=1 ,i

S

,i

V

According to condition (9.147), the surface integral vanishes, and hence dL = dε

Z X h h ∂  ∂L i ∂ϕ(s) ∂L − dτ. (s) (s) ∂x ∂ε ∂ϕ i ∂ϕ s=1

(9.148)

,i

V

The choice of ϕ(s) (xj , ε) tells us that they are equal to ϕ(s) (xj ) everywhere, except for a vicinity Q of the fixed point x, y, z. The 363

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Free ebooks ==> www.Ebook777.com same property holds valid for ∂ϕ(s) /∂ε. At the limit Q → 0, we can write dL/dε δL ∂L ∂  ∂L  (i = 1, 4). (9.149) = = − Q→0 ∂ϕ(s) /∂ε ∂xi ∂ϕ(s) δϕ(s) ∂ϕ(s) lim

,i

This expression is called the functional derivative or variational derivative of the Lagrangian L with respect to the field variables ϕ(s) . Using this definition, Euler-Lagrange’s equations (9.145) become δL = 0. δϕ(s)

(9.150) (s)

Observing that H does not depend on ϕ,t and ∂π(s) /∂xi , the functional derivatives of the Hamiltonian H with respect to π(s) and ϕ(s) are ∂H ∂  ∂H  δH = − (j = 1, 2, 3), ∂xj ∂ϕ(s) δϕ(s) ∂ϕ(s) ,j (9.151) which allows us to write equations (9.145) in a symmetric form δH ∂H = ; δπ(s) ∂π(s)

(s)

ϕ,t =

δH ; δπ(s)

π(s),t = −

δH . δϕ(s)

(9.152)

Problem 10 Using the Hamiltonian approach, find the equation of motion of an ideal magnetofluid, performing isentropic motion in an external electromagnetic field. It is assumed that the gravitational field is also taken into account. Solution The first step is to write the Lagrangian density of our problem. This should be composed by three terms: one term Lfo corresponding to the fluid, one term Lem corresponding to the electromagnetic field, o and one term Lint expressing the interaction between the two fields: L = Lfo + Lem o + Lint .

(9.153)

~ φ [see In this formulation, the usual electromagnetic potentials A, (9.1) and (9.2)] can be chosen as variational parameters. 364

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Free ebooks ==> www.Ebook777.com But this choice is not unique. As an alternative approach to the variational formalism of our problem, we shall use a different representation of the electromagnetic field, which makes it possible to simplify the Lagrangian density and, consequently, the solution of this application. Since our model implies the existence of conduction, convection, and displacement current densities, the usual Maxwell’s source equations e ~ = µ0~j; ∇ · E ~ =ρ ∇×B (9.154) ε0 should be written as ~ 1 ~ = ~j + ρe~v + ǫo ∂ E ; ∇×B µo ∂t

~ = ρe . ǫo div E

(9.155)

These equations can be expressed in a symmetric form, similar to that of the source-free Maxwell’s equations ~ ~ = − ∂B ; ∇×E ∂t

~ =0 ∇·B

(9.156)

by using the Lagrangian density L′ =

 1 ~ 2 1 ~ 2 ~ ~ ~ − 1 ~j ǫo |E| − |B| + P · E + ~v × B 2 2µo λ  ~ ~ · curl E ~ + ∂ B − ψ div B, ~ −M ∂t

(9.157)

~ + ~v × B ~ = 1 ~j E λ

(9.158)

where the source-free Maxwell’s equations (9.156), and Ohm’s law

~ (~r, t), and have been used as equations of constraint, while P~ (~r, t), M ψ(~r, t) are Lagrangian multipliers. If we choose as variational parameters Ei , Bi (i = 1, 2, 3), the Euler - Lagrange equations (9.81) yield: ~ = 1 (∇ × M ~ − P~ ), E ǫo

 ~  ∂M ~ ~ B = µo ∇ψ + P × ~v + . ∂t

(9.159)

~ B ~ in terms of the These relations define the electromagnetic field E, ~ generalized antipotentials M , ψ. As one can see, the appearance of the terms P~ and P~ × ~v generalizes the usual antipotentials, defined in 365

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Free ebooks ==> www.Ebook777.com the case ~j = 0, ρe = 0. Taking the divergence of (9.159)1 and the curl of (9.159)2 , we find the source Maxwell’s equation in the symmetric form 1  ~ − P~ × ~v = ∂ (ǫo E ~ + P~ ), ∇× B µo ∂t ~ + P~ ) = 0. ∇ · (ǫo E

(9.160)

These equations have been first obtained by Calkin 1 , who called vector field P~ the ’polarization’. Comparing (9.160) with (9.157), we can write ~ ~j = ~v ∇ · P~ + ∇ × (P~ × ~v ) + ∂ P , ∂t

ρe = −∇ · P~ ,

(9.161)

which satisfy identically the equation of continuity ∂ρe + ∇ · (~j + ρe~v ) = 0. ∂t

(9.162)

~, ψ Like the usual electromagnetic potentials, the antipotentials M can be related by a Lorentz-type condition. Introducing (9.159) into Maxwell’s source-free equations (9.156), we have  ~ ∂ψ  ∂2M ~ ~ ∆ M − ǫ o µo 2 = ∇ ∇ · M + ǫ o µo ∂t ∂t +ǫo µo

∂ ~ (P × ~v ) − ∇ × P~ . ∂t

~ satisfies the homogeneous D’Alembert’s wave equaIn order that M tion, the following two conditions must be fulfilled: ~ + ǫo µo ∂ψ = 0; div M ∂t

(9.163)

∂ ~ (P × ~v ) = curl P~ . (9.164) ∂t Relation (9.163) is the Lorentz condition for our antipotentials, and we shall use it as a constraint in the Lagrangian density. The advantage of this representation is that the Lagrangian density of the electromagnetic field ǫ o µo

Lo =

1 ~ 2 1 ~ 2 ǫo |E| − |B| , 2 2µo

1

(9.165)

M.G.Calkin, An Action Principle for Magnetohydrodynamics, Can. J. Phys., 41, 1963, p. 2241. 366

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Free ebooks ==> www.Ebook777.com ~ B ~ are given in terms of M ~ , ψ, includes the interaction in which E, between field and particles. This is possible because the interaction between the electromagnetic field and the point sources has been replaced by an interaction between the electromagnetic and the ’polarization’ fields. Using Herivel’s Lagrangian density (9.92), we postulate the Lagrangian density for our model to be 1 ~ 2 1 ~ 2 1  ∂ψ 2 ~ L= |B| − ǫo |E| − ∇ · M + ǫ o µo 2µo 2 2ǫo ∂t

 ∂α   ∂s  1 + ρ|~v |2 − ρ(ε + V ∗ ) + ρ + ~v · ∇α − βρ + ~v · ∇s . (9.166) 2 ∂t ∂t Before going further, we need to define the explicit relation between velocity and the ’polarization’ fields. Using Euler-Lagrange equations (9.84) with ϕ(s) ≡ vi (i = 1, 2, 3) as variational parameters, we have: ∂L =0; ∂vi,k

∂L =0; ∂vi,t

Since

∂L ∂L ∂Bk = + ρvi + ρα,i − βρs,i = 0. ∂vi ∂Bk ∂vi ∂L 1 = Bk ; ∂Bk µo

∂Bk ∂ = [µo (ψ,k + ǫkjm Pj vm + Mk,t )] = µo ǫkjm δim Pj = µo ǫkji Pj , ∂vi ∂vi we finally find

or, equivalently

1 vj = −α,j + βs,j − ǫjkm Bk Pm , ρ

(9.167)

1~ ~v = −∇α + β∇s − B × P~ , ρ

(9.168)

which is a generalized Clebsch transformation 1 . According to (9.135), the momentum densities πMj , πPj , πψ , πα , πs , associated with the field variables Mj , Pj , ψ, α, s, are: πMj =

∂L = Bj ; ∂Mj,t

πPj =

∂L =0; ∂Pj,t

1

Merches, I., Variational principle in magnetohydrodynamics, Phys. Fluids 12 (10), 2225, (1969). 367

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Free ebooks ==> www.Ebook777.com πψ =

∂L = −µo (Mi,i + ǫo µo ψ,t ) = 0 ; ∂ψ,t

πα =

∂L =ρ; ∂α,t

πs =

(9.169)

∂L = −βρ. ∂s,t

Using (9.134), we now can write the Hamiltonian density H: H=

6 X s=1

(s)

π(s) ϕ,t − L = Bj Mj,t − µo (Mj,j + ǫo µo ψ,t )ψ,t +

1 1 1 1 (Mj,j + ǫo µo ψ,t )2 + ρα,t − βρs,t − Bj Bj + ǫo Ej Ej − ρvj vj 2ǫo 2µo 2 2 +ρ(ε + V ∗ ) − ρ(α,t + vj α,j ) + βρ(s,t + vj s,j ). The form of the Hamiltonian density can be simplified if we observe that, by virtue of (9.167), we can write Bj Mj,t = Bj =

i h1 Bj − (P~ × ~v )j − ψ,j µo

1 Bj Bj + vj (ρvj + ρα,j − βρs,j ) − Bj ψ,j . µo

Introducing this result into H, we obtain the Hamiltonian density in terms of the field variables, and their partial derivatives with respect to coordinates and time: H= +

1 1 Bj Bj + ǫo Ej Ej − Bj ψ,j 2µo 2

1 1 [Mj,j Mk,k − ǫ2o µ2o (ψ,t )2 ] + ρvj vj + ρ(ε + V ∗ ). 2ǫo 2

(9.170)

In order to apply the Hamiltonian technique, it is necessary to express H in terms of the field variables Mj , Pj , ψ, α, s, and their conjugate momentum densities πMj , πPj , πψ , πα , πs . Using (9.159)1 , and (9.169), we get the Hamiltonian density in the final form: H= −

1 1 πMj πMj + (ǫjkm Mm,k − Pj )(ǫjli Mi,l − Pj ) − πMj ψ,j 2µo 2ǫo

 1 1  1 1 πψ (Mj,j − ǫo µo ψ,t ) + πα − α,j − πs s,j − ǫjkm πMk Pm 2ǫo µo 2 πα πα 368

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Free ebooks ==> www.Ebook777.com   1 1 × − α,j − πs s,j − ǫjli πMl Pi + πα ε + πα V ∗ , πα πα

(9.171)

where ε = ε(πα , s). Utilizing the calculation following formula (9.96) and applying canonical equations (9.145), we obtain the following system: πMj ,t = −ǫjkm Em,k ; Mj,t =

1 πMj − ψ,j − ǫjkm Pk vm ; µo

πPj ,t = 0 = Ej + ǫjkm vk πMm ; Pj,t = 0 ; πψ,t = −πMj ,t ; ψ,t = −

1 (Mj,j − ǫo µo ψ,t ) ; 2ǫo µo

πα,t = −(πα vj ),j ; 1 1 p; α,t = − vj vj − vj α,j + ε + V ∗ + 2 πα πs,t = −T πα − (πs vj ),t ; s,t = −vj s,j . Rearranging these equations and using the vector notation, we have: ~ ∂B ~ ; ∇·B ~ = 0, = −∇ × E ∂t ~ + ~v × B ~ = 0, E

 ~  ~ = µo ∇ ψ + P~ × ~v + ∂ M , B ∂t ~ + ǫo µo ∂ψ = 0, ∇·M ∂t ∂s + ~v · ∇s = 0, ∂t ∂ρ + ∇ · (ρ~v ) = 0, ∂t ∂ P~ = 0, ∂t 369

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Free ebooks ==> www.Ebook777.com ∂β + ~v · ∇β = T, ∂t p ∂α 1 2 |~v | − V ∗ − ε − + + ~v · ∇α = 0. 2 ρ ∂t We therefore found the source-free Maxwell’s equations (9.156), Ohm’s ~ in terms of the antipotentials law for infinite conductivity, the field B ~ , ψ, the Lorentz condition (9.163), the equation of conservation M of entropy (9.90), the equation of continuity (9.89), as well as the equations (9.94) and (9.95), which we also obtained in the case of uncharged fluids. Recalling that our final purpose is to obtain the equation of motion, the next step consists in eliminating the Lagrangian multipliers from the equations 1~ × P~ = 0, ~v + ∇α − β∇s + B ρ 1 2 p ∂α |~v | − ǫ − V ∗ − + + ~v · ∇α = 0, 2 ρ ∂t

(9.172)

∂β + ~v · ∇β = T. ∂t Extracting ∇α from (9.172)1 and introducing this expression into (9.172)2 , then taking the gradient of the result, we have   1 p 1 ∂ 2 ∗ ~ ~ ∇(β~v · ∇s) + (∇α) − ∇|~v | = ∇ ε + V + + ~v · B × P , ∂t 2 ρ ρ or, by using again (9.172)1  1 ∂~v ∂ 1 ~ ∂ ~ (β∇s) − − B × P − ∇|~v |2 + ∇(β~v · ∇s) ∂t ∂t ∂t ρ 2   p 1 ∗ ~ ~ = ∇ ε + V + + ~v · B × P . ρ ρ

Using the vector formula

~ · B) ~ =A ~ × (∇ × B) ~ +B ~ × (∇ × A) ~ ∇(A ~ · ∇)B ~ + (B ~ · ∇)A, ~ +(A 370

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Free ebooks ==> www.Ebook777.com and the relation (9.163), we still have  1 ∂ 1 ~ ∂~v + (~v · ∇)~v = −∇V ∗ − ∇p + B × P~ ∂t ρ ∂t ρ

h 1 i   ~ × P~ + ∇ 1 ~v · B ~ × P~ . +~v × ∇ × B ρ ρ

(9.173)

If ~a, ~b, ~c are any three vector fields, it is not difficult to prove the following vector identity: ~a × [∇ × (~b × ~c)] + ~b × [∇ × (~c × ~a)] + ~c × [∇ × (~a × ~b)] = (~b × ~c)∇ · ~a + (~c × ~a)∇ · ~b + (~a × ~b)∇ · ~c + ∇(~c · ~a × ~b). Multiplying this relation by lations, we arrive at ∇

=

1

1 ρ

(9.174)

and performing some simple calcu-

 1 i 1 h ~c · ~a × ~b − (~c · ~a × ~b)∇ + ∇ × ~a × ~b × ~c ρ ρ ρ h 1 i 1 − ∇ × (~a × ~b) × ~c + (~a × ~b) ∇ · ~c ρ ρ

1 1 [∇ × (~a × ~c) + ~c ∇ · ~a] × ~b + ~a × [∇ × (~b × ~c) + ~c ∇ · ~b]. ρ ρ

~ ~c = ~v and replace the term If in this expression we put ~a = P~ , ~b = B, 1 v by means of the equation of continuity, we still have ρ∇ · ~  h 1 i  ∂ 1 ~ ~ ~ ~ ~ ~ B×P + P ×B ∇ ~v · P × B + ~v × ∇ × ρ ρ ∂t ρ 1

h  i h   i ~ × ∇ 1 − ∇ 1 × (P~ × B) ~ × ~v +~v × (P~ × B) ρ ρ

=

1 ∂ ~ ~ 1 ~ 1 P~ ×[∇×(B×~ ~ v )+~v ∇·B], ~ (P × B)+ [∇×(P~ ×~v )+~v ∇·P~ ]× B+ ρ ∂t ρ ρ

or, after some reduction of terms ∇

 h  i   ~ + ~v × ∇ × 1 B ~ × P~ + ∂ 1 P~ × B ~ ~v · P~ × B ρ ρ ∂t ρ

1

=

i 1 h ∂ P~ ~ + ∇ × (P~ × ~v ) + ~v ∇ · P~ × B ρ ∂t 371

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Free ebooks ==> www.Ebook777.com h ∂B i ~ 1 ~ × ~v ) + ~v ∇ · B ~ . + P~ × + ∇ × (B ρ ∂t

In view of Maxwell’s equation (9.156)2 , the induction equation for infinite conductivity (9.156)1 , and the relation (9.161)1 , the r.h.s. of the ~ Introducing this result into (9.173), last equation reduces to ρ1 ~j × B. we finally arrive at ρ

h ∂~v ∂t

i ~ − ∇p + ρF~ ′ , + (~v · ∇) ~v = ~j × B

which is the expected equation of motion. Here F~ ′ = −gradV ∗ is the external (e.g. gravitational) mass-specific force (see Problem 6). Problem 11 Using the analytical formalism, derive time-dependent Schr¨ odinger equation for a microparticle of mass m, described by the wave function ψ(~r, t), using the following two procedures: a) The Lagrangian approach for continuous systems, by means of the Lagrangian density L=

 ¯ ∗ h ¯2 h ∗ ∇ψ · ∇ψ ∗ + V ψψ ∗ + ψ ψ,t − ψψ,t ; 2m 2i

(9.175)

b) The Hamiltonian method for continuous systems. Solution a) In this case, Schr¨ odinger equation for the microparticle of mass m, described by the wave function ψ, follows from the Euler-Lagrange equation !   ∂ ∂L ∂ ∂L ∂L − − = 0 (i = 1, 2, 3), ∗ ∂ψ ∗ ∂xi ∂ψ,i∗ ∂t ∂ψ,t or, using the formal vector formulation for spacial derivatives     ∂L ∂L ∂ ∂L −∇ − = 0. (9.176) ∗ ∂ψ ∗ ∂(∇ψ ∗ ) ∂t ∂ψ,t Using (9.175), we have: ∂L ¯ h = V ψ + ψ,t ; ∗ ∂ψ 2i 372

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Free ebooks ==> www.Ebook777.com ∂L h2 ¯ = ∇ψ; ∂(∇ψ ∗ ) 2m   ∂L h2 ¯ ¯2 h ∇ = ∇ · ∇ψ = ∆ψ; ∂(∇ψ ∗ ) 2m 2m ∂L h ¯ = − ψ; ∗ ∂ψ,t 2i   ∂L ∂ h ¯ = − ψ,t . ∗ ∂t ∂ψ,t 2i Introducing these results into (9.176), we obtain Vψ+

h2 ¯ ¯ ∂ψ h ¯ ∂ψ h − ∆ψ + = 0, 2i ∂t 2m 2i ∂t

or i¯ h

∂ψ h2 ¯ =− ∆ψ + V ψ, ∂t 2m

(9.177)

which is the equation we were looking for. In operatorial form, equation (9.177) can be written as

where

ˆ = Hψ, ˆ Eψ

(9.178)

∂ ˆ = i¯ E h ∂t

(9.179)

is the energy operator, and 2

h ˆ =−¯ H ∆+V 2m

(9.180)

is the (Hermitian) Hamiltonian operator for our physical system: a particle of mass m moving in the field with potential energy V . As one can see, Schr¨ odinger equation in its general form i¯ h

∂ ˆ r, t) ψ(~r, t) = Hψ(~ ∂t

is closely related to the Hamilton-Jacobi equation   ∂ ∂S S(qi , t) = H qi , ,t , ∂t ∂qi 373

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Free ebooks ==> www.Ebook777.com where S is the action, and H the Hamilton function. Here the generalized coordinates qi (i = 1, 2, 3), used in the Hamilton-Jacobi formalism, can be set to the p position in Cartesian coordinates ~r = ~r(x, y, z). Substituting ψ = ρ(~r, t) exp iS(~r, t)/¯ h, where ρ is the probability density, into Schr¨odinger equation, and then taking the limit h ¯ → 0, we get the Hamilton-Jacobi equation. We therefore can conclude that the motion of the particle of mass m, described by Schr¨ odinger equation, in the limit h ¯ → 0 is governed by Hamilton-Jacobi equation. Using a similar procedure, we find Schr¨ odinger equation for the ∗ wave function ψ , which is ∂ψ ∗ h2 ¯ −i¯h =− ∆ψ ∗ + V ψ ∗ , ∂t 2m

(9.181)

ˆ ∗ ψ∗ = H ˆ ∗ ψ ∗ = Hψ ˆ ∗, E

(9.182)

or

ˆ ∗ = H. ˆ ˆ ∗ = −i¯h ∂ , and H where E ∂t b) The Hamiltonian density for a continuous system is defined by [see (9.134)]: h X (s) H= π(s) ϕ,t − L, (9.183) s=1

where

π(s) =

∂L (s)

∂ϕ,t

are the generalized momentum densities. In this case, Hamilton’s canonical equations [see (9.152)] write (s)

ϕ,t =

δH ; δπ(s)

δH , δϕ(s)

(9.184)

(j = 1, 2, 3),

(9.185)

π(s),t = −

where δ ∂ ∂  ∂  = − ∂xj ∂ϕ(s) δϕ(s) ∂ϕ(s) ,j

is the functional (or variational) derivative with respect to the field variables ϕ(s) . By choosing ψ ∗ as field variable, we have: πψ∗ =

∂L h ¯ = − ψ, ∗ ∂ψ,t 2i 374

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(9.186)

Free ebooks ==> www.Ebook777.com and the Hamiltonian density writes ∗ H = πψ∗ ψ,t −L=−

=−

¯ h ¯2 h ∇ψ · ∇ψ ∗ − V ψψ ∗ − ψ ∗ ψ,t 2m 2i

2i ¯2 h h ¯ ∇ψ · ∇ψ ∗ + ψ ∗ πψ∗ − ψ ∗ ψ,t , 2m h ¯ 2i

(9.187)

where L is given by (9.175). Using (9.187), we then have: 2i ∂H h ¯ = V πψ∗ − ψ,t , ∗ ∂ψ ¯ h 2i ∂H h2 ¯ =− ∇ψ, ∂(∇ψ ∗ ) 2m and therefore ∇



∂H ∂(∇ψ ∗ )



=−

h2 ¯ ¯2 h ∇ · ∇ψ = − ∆ψ. 2m 2m

The functional derivative of H with respect to the field variable ψ is   ∂H δH ∂H = −∇ δψ ∗ ∂ψ ∗ ∂(∇ψ ∗ ) ∗

h ¯ ¯2 h 2i ∗ ∆ψ, = V πψ − ψ,t + h ¯ 2i 2m so that (πψ∗ ),t = −

δH 2i h ¯ ¯2 h ∗ = − V π + ψ − ∆ψ, ψ ,t δψ ∗ h ¯ 2i 2m

or, in view of (9.186), ¯ h 2i − ψ,t = − V 2i h ¯ that is i¯h



 ¯ h h ¯ ¯2 h − ψ + ψ,t − ∆ψ, 2i 2i 2m

∂ψ h2 ¯ =− ∆ψ + V ψ, ∂t 2m

which is precisely Schr¨odinger equation. 375

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Free ebooks ==> www.Ebook777.com A similar calculation can be performed if the wave function ψ is taken as field variable. Indeed, we have: ∂L h ¯ πψ = = ψ∗ , ∂ψ,t 2i so that the Hamiltonian density writes ¯2 h h ¯ ∗ H = πψ ψ,t − L = − ∇ψ · ∇ψ ∗ − V ψψ ∗ + ψψ,t 2m 2i 2 ¯h 2i h ¯ ∗ =− ∇ψ · ∇ψ ∗ − V ψπψ + ψψ,t . 2m h ¯ 2i Thus,

∂H 2i ¯ ∗ h = − V πψ + ψ,t ; ∂ψ h ¯ 2i

h2 ¯ ∂H =− ∇ψ ∗ ; ∂(∇ψ) 2m   ∂H h2 ¯ ∇ ∆ψ ∗ . =− ∂(∇ψ) 2m The functional derivative of H with respect to ψ then is   ∂H δH ∂H = −∇ δψ ∂ψ ∂(∇ψ) =−

¯2 h 2i ¯ ∗ h + V πψ + ψ,t ∆ψ ∗ , h ¯ 2i 2m

so that

δH 2i ¯ ∗ h ¯2 h = V πψ − ψ,t − ∆ψ ∗ , δψ h ¯ 2i 2m or, in terms of ψ ∗ and its derivatives, (πψ ),t = −

h2 ¯ ∂ψ ∗ =− ∆ψ ∗ + V ψ ∗ , −i¯h ∂t 2m which is Schr¨odinger equation for ψ. In operatorial form, this equation writes ˆ ∗ ψ∗ = H ˆ ∗ ψ ∗ = Hψ ˆ ∗, E with ˆ∗ = E and



∂ i¯ h ∂t

∗

= −i¯ h

∂ ∂t

 2 2 ˆ~2 ¯ h 1 h ¯ p ˆ =− H =H ∆+V = ∇ +V = + V, 2m 2m i 2m ˆ~ = h¯ ∇ is the momentum operator. where p i ˆ∗

376

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Appendix A

GREEK ALPHABET

No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Capital letter A B Γ ∆ E Z H Θ I K Λ M N Ξ O Π P Σ T Υ Φ X Ψ Ω

Small letter α β γ δ ǫ (variant : ε) ζ η θ (variant : O) ι κ λ µ ν ξ o π (variant : ̟) ρ (variant : ̺) σ (variant : ς) τ υ φ (variant : ϕ) χ ψ ω

377

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Name alfa beta gamma delta epsilon dzeta eta theta iota kappa lambda miu niu csi omicron pi rho sigma tau upsilon fi hi psi omega

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Appendix B ELEMENTS OF TENSOR CALCULUS B.1. Scalar and vector quantities The quantities usually met in physics can be classified in terms of their tensor nature: scalar = 0th rank (order), vector = 1st rank, 2nd rank, etc. tensor quantities1 . Note the difference between mathematical definition of a tensor (scalar, vector, etc.), and tensor physical quantities. It is also essential to mention the nature of the space on which a tensor is defined. Our present discussion is carried on the Euclidean three-dimensional space E3 . Any physical quantity which is completely determined by knowing a (real or complex) number, associated with it and expressed in physical units, is called scalar physical quantity. The fundamental property of scalar quantities is that they do not depend on any coordinate system. In this category fall: mass, mechanical work, temperature, entropy, electric charge, etc. These quantities are usually called, simply, scalars. Any quantity which, in addition to its numerical value (magnitude, or modulus), is characterized by a spatial orientation (direction and sense), is called vector quantity or, simply, vector. For example, the mechanical force acting on a rigid body is determined by its magnitude, direction and sense. Quantities like: linear velocity, acceleration, momentum, electric field intensity, etc., are also vectors. The physical reality displays quantities whose determination requires more than three associated real numbers. These quantities are called tensor quantities and can be found, for example, in the study of rigid body and continuous deformable media mechanics. Here are some examples: the inertia tensor Iαβ , (α, β = 1, 2, 3), Cauchy’s stress tensor Tik (i, k = 1, 2, 3), the momentum density flow tensor Πik = ρvi vk + pδik (i, k = 1, 2, 3), etc. 1

In various branches of physics we meet even tensors of higher rank. For instance, the elasticity tensor (also called stiffness tensor) Cijkl (i, j, k, l = 1, 2, 3) is a fourth rank tensor. 378

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Free ebooks ==> www.Ebook777.com In the Euclidean three-dimensional space E3 , a vector is a mathematical entity represented by an oriented segment of straight line, the algebraic components on the axes being the vector components. A vector is usually denoted by an arrow over a letter (e.g. F~ - the force −−→ vector) or a group of two letters (e.g. AB), etc. In the last example, A is the origin (or starting point, or tail) of the vector, B its terminal point (also called head or tip), the sense goes from A to B, while the vector modulus is determined by the length of the segment kABk. ~ is denoted by |A|, ~ or, simpler, by Usually, the modulus of a vector A A. A vector with magnitude one (in a certain unit system) is called unit vector. In geometry and physics, the versor of an axis ∆ or of a ~ is a unit vector indicating its direction and sense. vector A In terms of its origin, a vector can be: 1) free, if its origin (point of application) is arbitrary (in the laboratory reference frame). Examples: the gravitational acceleration, the magnetic induction of the Earth, etc. 2) bound, if its tail is fixed. For example, a force applied to a certain point of an elastic body, or the electric field intensity attached to an electric point-charge, are bound vectors. 3) sliding, if its origin can be moved anywhere on its direction. A classical example in this respect is the force acting on a rigid body. The angular velocity ω ~ describing rotation of a rigid body about a fixes axis is also an example of sliding vector. There are various methods used to describe a vector quantity: descriptive, geometric, numerical, etc. For instance, a free vector is fully characterized by its three projections on axes of a coordinate frame in E3 or, equivalently, its magnitude and the two angles formed by the vector with anyone of the coordinate axes (the third angle can be determined by means of the other two angles, on condition that the property of acute or obtuse angle is known). To define a sliding vector one must know its magnitude, the two angles between the vector direction and any two axes, as well as two coordinates of the intersection point of the vector direction with any of coordinates planes. In case of a bound vector, one must know six quantities: three coordinates of the origin, and three of the head, or, equivalently, three coordinates of the origin and three projections of the vector on axes. B.2. Elements of vector algebra Position of an arbitrary point P in space can be univoquely defined by a vector with its point of application at the origin of a conveniently chosen coordinate system, and head at point P . It is usually 379

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Free ebooks ==> www.Ebook777.com called position vector or radius vector, being denoted by ~r. An axis is a straight line on which a sense has been chosen, denoted by an arrow. The opposite sense is called negative. Consider an axis (∆) and an arbitrary vector ~a, with origin at A and head at B. The projection of ~a on ∆ is the vector defined by the orthogonal projections A′ and B ′ of A and B on axis ∆ (see Fig.B.1). −−−→ Let a∆ be the algebraic magnitude of vector projection A′ B ′ . By definition, the sign of the magnitude is ” + ” if the axis and the projection have the same sense, and ” − ” if the senses are opposite. As observed, we can write a∆ = a cos α, which mean that projection a∆ is positive, negative, or zero, in terms of α < π2 , > π2 , and π2 , respectively. We shall use notation a∆ = pr∆~a.

(B.1)

Fig.B.1 If ~a equals the unit vector ~i, then (B.1) writes pr∆~a = cos α.

(B.2)

Let Oxyz be a right-handed orthogonal frame, and ~a an arbitrary vector with origin at P1 and its terminal point at P2 (see Fig.B.2). The magnitudes of the vector projections ~ax , ~ay , ~az on axes are called components of the vector ~a. According to Fig.B.2, we have ax = x2 − x1 ,

a y = y2 − y1 ,

az = z 2 − z 1 ,

(B.3)

where x1 , y1 , z1 and x2 , y2 , z2 are the coordinates of the terminal points of ~a. 380

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Fig.B.2 Two (or more) vectors with common magnitudes and orientations, but parallel directions, are called equipolent (or equivalent). The equipolent vectors are determined by the same relations (B.3). The vector projections ~ax , ~ay , ~az are the vector components of ~a. A relation of equality between two vectors can be established only for vectors of the same type. For instance, two sliding vectors ~a and ~b are equal if they have the same direction, sense, and magnitude. This equality is expressed by the vector relation ~a = ~b, or, equivalently, by the scalar relations a x = bx ,

a y = by ,

a z = bz ,

obtained by projecting the vector relation ~a = ~b on coordinate axes. The equality relation has the following properties: 1. Reflexivity: ~a = ~a 2. Symmetry: if ~a = ~b, then ~b = ~a, and reciprocally. 3. Transitivity: if ~a = ~b and ~b = ~c, then ~a = ~c. Vectors with the same direction are called collinear. If two collinear vectors have opposite senses, they are called opposite. B.2.1. Vector operations The most common operations with free vectors are: - addition (subtraction); - multiplication by a scalar; 381

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Free ebooks ==> www.Ebook777.com - scalar (dot) product; - vector (cross) product; - mixed product. By means of operations of scalar and vector products can also be defined some other vector operations: dot product of a vector product, double vector product, etc. Vector addition Consider two free vectors ~a and ~b. To draw their sum, we take two vectors equipolent to ~a and ~b, with their common origin at an arbitrary point O (see Fig.B.3). By definition, the diagonal of the parallelogram constructed on the two vectors is called resultant vector or vector sum of ~a and ~b, being denoted by ~c = ~a + ~b.

Fig.B.3 The resultant of two vectors can also be drawn by using the polygon rule. At the terminal point of one vector one constructs a vector equipollent to the second one: the resultant is then a vector with origin at the origin of the first vector, and head at the head of the second vector. The parallelogram rule can be generalized for n vectors n P ~ai ~a1 , ~a2 , ..., ~an . In this case, the vector sum ~a = ~a1 +~a2 + ... +~an = i=1

is traced by applying n − 1 times the parallelogram rule, as shown in Fig.B.4. As observed, the vectors ~a1 , ~a2 , ..., ~an are placed on sides of a polygonal contour. If the contour is closed, then the resultant is zero. The following properties of the vector sum are obvious: - commutativity: ~a + ~b = ~b + ~a; 382

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Free ebooks ==> www.Ebook777.com - associativity: (~a + ~b) + ~c = ~a + (~b + ~c). The above considerations show that a vector can be fully defined by means of its three projections on axes of a Cartesian coordinate system Oxyz. According to Fig.B.2, we have ~a = ~ax + ~ay + ~az .

(B.4)

If ~i, ~j, ~k are the versors of the axes Ox, Oy, and Oz, respectively, then we can write ~ax = ax~i,

~ay = ay~j,

~az = az~k,

and ~a becomes ~a = ax~i + ay~j + az~k,

(B.5)

which is the analytical expression of vector ~a. Using notations ax = a1 , ~i = ~u1 ,

ay = a 2 , ~j = ~u2 ,

az = a3 , ~k = ~u3 ,

we can also write (B.5) as ~a = a1 ~u1 + a2 ~u2 + a3 ~u3 =

3 X

ai ~ui .

i=1

Fig.B.4 383

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(B.6)

Free ebooks ==> www.Ebook777.com Here and hereafter we shall use Einstein’s summation convention: when an index is repeated in a term that implies a sum over all possible values for that index, then the summation is implicitly understood and the summation symbol is omitted. This way, relation (B.6) becomes ~a = ai ~ui

(i = 1, 3).

(B.7)

The position vector of any point P relative to a Cartesian frame Oxyz can be written according to (B.7). Position of the point P is uniquely determined either by coordinates x, y, z, or by its position vector ~r, with origin at O and head at P . In fact, to any point corresponds a position vector ~r with respect to coordinate frame Oxyz. Observing that x, y, z are the components of ~r on axes, and making allowance for (B.6), the analytical expression of ~r writes ~r = x~i + y~j + z~k = xi ~ui

(i = 1, 3).

(B.8)

Let us consider three vectors ~a, ~b, ~c with the components ~a = ~a(ax , ay , az ),

~b = ~b(bx , by , bz ),

~c = ~c(cx , cy , cz ).

(B.9)

Using (B.7), we realize that the vector relation ~c = ~a + ~b is equivalent to the following three scalar relations: c i = a i + bi

(i = 1, 3).

(B.10)

Vector subtraction If ~a and ~b are two free vectors, then relation d~ = ~a − ~b define the operation of subtraction. In fact, this operation is a consequence of the vector addition, since the difference ~a − ~b can be written as ~a +(−~b), where −~b is the opposite of ~b. The resultant is obtained by means of the parallelogram rule: the vector difference d~ has its origin at the head of the subtrahend vector ~b, and its terminal point at the head of the minuend vector ~a. Projecting on axes the vector relation d~ = ~a − ~b, we obtain the following three equivalent scalar relations: d i = a i − bi

(i = 1, 3).

384

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(B.11)

Free ebooks ==> www.Ebook777.com Multiplication of a vector by a scalar Let ~a be a free vector, and λ a real scalar. Their product, denoted by λ~a, is a vector collinear to ~a, of magnitude |λ~a| = |λ||~a|, having the same sense as ~a if λ > 0, and opposite sense if λ < 0. If λ = 0, or ~a = 0 (null vector), we have λ~a = 0. The fact that two vectors ~a and ~b are collinear can be expressed as ~b = λ~a, (B.12)1 or

~a = µ~b.

(B.12)2

Multiplication of a vector by a scalar has the following properties: i) (λ + µ)~a = λ~a + µ~a; ii) λ(~a + ~b) = λ~a + λ~b; iii) λ(µ~a) = µ(λ~a) = λµ~a; iv) 1 · ~a = ~a · 1 = ~a. Consider n vectors ~a1 , ~a2 , ..., ~an . If there exist n scalars λ1 , λ2 , ..., λn , not all zero, such that λ1~a1 + λ2~a2 + ... + λn~an = 0,

(B.13)

then vectors ~a1 , ~a2 , ..., ~an are said to be linearly dependent. If relation (B.13) holds only for λ1 = λ2 = ... = λn = 0, then ~a1 , ~a2 , ..., ~an are linearly independent. As an example, consider the linear combination of two non-collinear vectors ~a1 and ~a2 : ~a3 = λ1~a1 + λ2~a2 ,

(B.14)

where λ1 and λ2 are not simultaneously zero. In this case ~a3 is a vector situated in the plane determined by directions of ~a1 and ~a2 , which means that ~a1 , ~a2 , and ~a3 are coplanar. The coplanarity condition (B.14) can also be written as λ1~a1 + λ2~a2 + λ3~a3 = 0,

(B.15)

which means that the three vectors are linearly dependent. Reciprocally, if three vectors are linearly dependent, then they are coplanar. The vector equation (B.13) shows that vectors ~a1 , ~a2 , ..., ~an are linearly independent, (λ1 = λ2 = ... = λn = 0), if the rank of the matrix coefficients equals the number of unknowns λ1 , λ2 , ..., λn . Therefore, in the Euclidean space E3 exist at most three linearly independent vectors, which form a basis of this space. 385

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Free ebooks ==> www.Ebook777.com Consider the system of versors ~i, ~j, ~k of axes of the Cartesian system Oxyz. These vectors are linearly independent, meaning that they form a basis in E3 , and, consequently, any vector ~a can be written in terms of its components ax , ay , az and the basis elements ~i, ~j, ~k according to (B.5): ~a = ax~i + ay~j + az~k. Scalar (dot) product The scalar (or dot, or inner) product of two vectors ~a and ~b, denoted ~a · ~b, is defined as   c ~a ·~b = |~a||~b| cos ~a, ~b = ax bx + ay by + az bz = ai bi

(i = 1, 3). (B.16)

According to definition (B.16), the dot product is a scalar. Properties: i) the dot product is commutative: ~a · ~b = ~b · ~a; ii) the dot product is distributive with respect to vector addition: ~a · (~b + ~c) = ~a · ~b + ~a · ~c; iii) if ~a = ~b, then ~a · ~a = |~a|2 = a2 ; iv) for any scalar λ, we have: ~a · (λ~b) = ~b · (λ~a) = λ(~a · ~b); v) if ~a · ~b = 0, then either the vectors are orthogonal, or at list one of them is null. vi) if ~u1 , ~u2 , ~u3 are the versors of an orthogonal reference frame, we have: ~u1 · ~u2 = ~u2 · ~u3 = ~u3 · ~u1 = 0, and ~u21 = ~u22 = ~u23 = 1. The last two relations can be written in a condensed form ~ui · ~uk = δik , where δik =



1, 0,

i = k, i 6= k

is the Kronecker symbol. 386

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(B.17)

(B.18)

Free ebooks ==> www.Ebook777.com Using (B.5) and (B.18), one can express the analytic form of the dot product of vectors ~a(ax , ay , az ) and ~b(bx , by , bz ) as ~a · ~b = (ai ~ui ) · (bj ~uj ) = ai bj δij = ai bi

(i = 1, 3).

(B.19)

In particular, if ~a = ~b, ~a · ~a = |~a|2 = a2 = ai ai = a21 + a22 + a23 .

(B.20)

The magnitude of projection of the vector ~a on axis Oxi (i = 1, 2, 3) of a Cartesian frame Oxyz is given by the scalar product (~a)i = ~a · ~ui = ak ~uk · ~ui = ak δik = ai ,

(B.21)

where ~ui is the versor of the xi -axis. In physics, the dot product is used, for example, to express the elementary work done by a force F~ which, acting on a body, displaces its tail by quantity d~r: dL = F~ · d~r, or the infinitesimal flow of the ~ through surface dS: ~ dΦ = B ~ · dS. ~ magnetic induction B Vector (cross) product Let ~a and ~b be two free vectors. Their vector or cross product is a vector denoted ~c = ~a × ~b, with the following properties: i) its direction is normal to the plane determined by ~a and ~b; ii) its sense is given by the right-hand screw rule ( or right-hand rule); iii) its magnitude equals the area of the parallelogram that the vectors span (vectors ~a and ~b have the same origin at O - see Fig.B.5)

Fig.B.5 387

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Free ebooks ==> www.Ebook777.com By definition, the cross product of vectors ~a and ~b is given by   c ~c = ~a × ~b = |~a||~b| sin ~a, ~b ~uc = εijk aj bk ~ui ,

(B.22)

where ~uc is the versor of vector ~c, and εijk (i, j, k = 1, 2, 3) is the completely antisymmetric unit third rank pseudotensor (the Levi-Civita symbol). The cross product of two vectors has the following properties: i) it is anticommutative: ~a × ~b = −~b × ~a; ii) if λ is a scalar, then: (λ~a) × ~b = ~a × (λ~b); iii) it is distributive with respect to vector addition: ~a × (~b + ~c) = ~a × ~b + ~a × ~c; iv) if one of the vectors is null, or the two vectors are collinear, the cross product is also null. In the last case we have: ~a × ~b = ~a × (λ~a) = λ(~a × ~a) = 0; v) the versors ~u1 , ~u2 , and ~u3 of an orthogonal reference frame obey the following rules: 

~u1 × ~u1 = 0, ~u1 × ~u2 = ~u3 ,

~u2 × ~u2 = 0, ~u2 × ~u3 = ~u1 ,

~u3 × ~u3 = 0; ~u3 × ~u1 = ~u2 .

(B.23)

These relations can be written in a condensed form by means of the Levi-Civita symbol, defined as εijk =

(

+1, if i, j and k are an even permutation of 1,2, and 3; −1, if i, j and k are an odd permutation of 1,2, and 3; 0, if any of the indices are equal,

e.g. ε123 = +1, ε312 = −1, ε233 = 0, etc. With this notation, relations (B.23) write (B.24) ~ui × ~uj = εijk ~uk (i, j, k = 1, 3). One can easily verify that the Levi-Civita symbol satisfies the following relation:

εijk εpqs

δip = δjp δkp

388

δiq δjq δkq

δis δjs . δks

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(B.25)

Free ebooks ==> www.Ebook777.com Setting i = p and summing over this index, we have: εijk εiqs = δjq δks − δjs δkq , where all indices go from 1 to 3. Making j = q in the last relation and summing again, we obtain εijk εijs = δjj δks − δjs δkj = 3δks − δks = 2δks . By virtue of this property, let us multiply (B.24) by εijs and perform summation over indices i and j. The result is εijs ~ui × ~uj = εijs εijk ~uk = 2δsk ~uk = 2~us , which yields ~us =

1 εsij ~ui × ~uj 2

(i, j, s = 1, 3).

The analytic form of the cross product, in view of (B.24), therefore is: ~a × ~b = (ai ~ui ) × (bj ~uj ) = εijk ai bj ~uk

(i, j, s = 1, 3).

(B.26)

Another modality to express the cross product of vectors ~a and ~b is to write (B.26) as follows: ~a × ~b = εijk ai bj ~uk = (ay bz − az by )~i + (az bx − ax bz )~j + (ax by − ay bx )~k ~i ~j ~k = ax ay az , bx by bz

(B.27)

which means that the cross product can be expressed as a third order symbolic determinant. Projecting the cross product ~a ×~b onto direction ~up , i.e. performing the dot product ~a × ~b · ~up , we have   ~a × ~b = ~a × ~b · ~up = εijk ai bj ~uk · ~up p

= εijk ai bj δkp = εijp ai bj = εpij ai bj . 389

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Free ebooks ==> www.Ebook777.com If the vectors ~a and ~b are collinear, the determinant (B.27) is null, because the second and third rows of the determinant have proportional elements. In physics, the cross product is used to express, for example, the following quantities: ~ = ~r × F~ , - Moment of a force F~ with respect to a pole O: M where ~r is ”connecting” the pole O with the point of application of F~ ; ~ = 1 ∇ × ~v = 1 curl ~v in fluid mechanics, where - Vortex vector1 Ω 2 2 ~v is the velocity field of the fluid; H Id~l×~r ~ of a linear current, B ~ = µ - Magnetic induction B 4π r3 . (Γ)

where µ is the magnetic permeability of the medium in which the electric circuit is displaced, d~l is an arc/line element whose sense is given by the electric current of intensity I of the linear electric circuit Γ, and ~r is the vector ”connecting” the origin of d~l with the point ~ is determined; where the induction B - Energy flux density of an electromagnetic field, i.e. Poynting’s ~ = E ~ × H, ~ where E ~ and H ~ are the electric and magnetic vector S components of the electromagnetic field, respectively. Mixed product of three vectors Let ~a, ~b and ~c be three free vectors, with their origin at O. The quantity ~a · (~b × ~c) = ai εijk bj ck = (ai ~ui ) · (bj ~uj × ck ~uk )

where2

= ai bj ck ~ui · (~uj × ~uk ) ax ay az = bx by bz , cx cy cz

(B.28)

~ui · (~uj × ~uk ) = εijk , is called the mixed product of the vectors ~a, ~b, and ~c. ~ is not a polar vector, but a pseudovector. In fact, Ω This way, we can give a different definition to the Levi-Civita symbol, as being the mixed product of versors of an orthogonal frame on the Euclidean space E3 : εijk = ~ui · (~uj × ~uk ), as the Kronecker symbol δij is given by the dot product δij = ~ui · ~uj . 1 2

390

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Free ebooks ==> www.Ebook777.com According to (B.28), the properties of the mixed product are closely connected to the properties of determinants: the mixed product is null if two rows of the determinant are proportional, and changes its sign if two rows are inverted. Also, a cyclic permutation of vectors (rows in the determinant) leaves the mixed product unchanged: ~a · (~b × ~c) = ~b · (~c × ~a) = ~c · (~a × ~b). In physics, the mixed product of three vectors is used - for example - to express the moment of a force F~ about an axis ∆: ~ O, M∆ = ~u∆ · (~r × F~ ) = ~u∆ · M ~ O the moment of F~ about an where ~u∆ is the versor of axis ∆ and M arbitrary point (pole) O on the axis. It can be shown that the choice of the point O is arbitrary. Mixed product of four vectors (Dot product of cross products) The mixed product of four vectors can be obtained from the mixed product of three vectors, by replacing the vector under dot product ~ Using analytical method, with a cross product, such as: (~a ×~b)·(~c × d). we have: ~ = (~a × ~b)i (~c × d) ~ i = εijk aj bk εipq cp dq (~a × ~b) · (~c × d) = (δjp δkq − δjq δkp )aj bk cp dq = aj cj bk dk − aj dj bk ck ~ − (~a · d)( ~ ~b · ~c). = (~a · ~c)(~b · d)

(B.29)

Double cross product Consider three free vectors ~a, ~b and ~c, with their common origin at some point O. The double cross product of these vectors is defined as d~ = ~a × (~b × ~c). (B.30) According to our previous considerations one observes that, from the geometric point of view, the vector d~ is orthogonal to both vectors ~a and (~b × ~c), therefore is coplanar with ~b and ~c. Using analytical approach, we have: ~a × (~b × ~c) = εijk aj (~b × ~c)k ~ui = εijk εklm aj bl cm ~ui = (δil δjm − δim δjl )aj bl cm ~ui = aj cj (bi ~ui ) − aj bj (ci ~ui ) 391

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Free ebooks ==> www.Ebook777.com = (~a · ~c)~b − (~a · ~b)~c.

(B.31)

In physics, the double cross product is used, for example, in order to express the macroscopic force of interaction between two linear electric circuits: µ F~12 = −F~21 = 4π

I I

Γ1 Γ2

d~l2 × (d~l1 × ~r12 ) , 3 r12

where µ is the magnetic permeability of the medium where the electric circuits are placed, I1 and I2 are the current intensities, d~l1 and d~l2 are two vector arc elements of circuits Γ1 and Γ2 , respectively, while ~r12 = −~r21 is connecting the tails of d~l1 and d~l2 . Lagrange’s identity Squaring the cross product of two free vectors ~a and ~b, we have: (~a × ~b)2 = (~a × ~b) · (~a × ~b) = (~a × ~b)i (~a × ~b)i = εijk aj bk εimn am bn = (δjm δkn − δjn δkm )aj am bk bn = aj aj bk bk − aj ak bk bj = (aj aj )(bk bk ) − (aj bj )(ak bk ) = ~a2~b2 − (~a · ~b)2 ,

(B.32)

called Lagrange’s identity. This is a particular case of Lagrange’s identity written for any two sets {a1 , a2 , ..., an } and {b1 , b2 , ..., bn } of real or complex numbers: n X

k=1

=

n−1 X

n X

i=1 j=i+1

a2k

!

n X

b2k

k=1

(ai bj − aj bi )2 =

!



n X

a k bk

k=1

!2

n n 1X X (ai bj − aj bi )2 . 2 i=1 j=1,j6=i

B.3. Tensors and pseudotensors Since teaching experience of the authors showed that some students meet difficulties in understanding the notion of pseudotensor, in this paragraph we shall focus our attention on this subject. For a physicist, the concepts of pseudoscalar and pseudovector, as particular cases of a pseudotensor (a pseudoscalar is a pseudotensor of rank zero, while a pseudovector is a pseudotensor of rank one), are of significant importance. If we agree to call scalar of the first type a true scalar, 392

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Free ebooks ==> www.Ebook777.com then a pseudoscalar can be named scalar of the second type. In the same way, a vector can be polar (vector of the first type) or axial (vector of the second type). From a strictly mathematical viewpoint, these definitions are rather simple, but in physics all these concepts have to be connected to our intuition. Let us begin our investigation with the notion of tensor. There are various possibilities to define the notion of tensor, but in our case the most appropriate is to start from the general coordinate transformation concept. Without going too deep in the subject, we shall underline the most important definitions and their physical significance. Position in space of an arbitrary point (particle) P can be determined by the system of three Cartesian coordinates x1 , x2 , x3 with respect to the three-orthogonal reference frame Oxyz (see Fig.B.6). Since both position of O and direction of coordinate axes are arbitrary (due to homogeneity and isotropy of vacuum space), it is necessary to define the law of transformation of the coordinates of point P , when passing from the initial coordinate system S(Ox1 x2 x3 ) to another system S ′ (Ox′1 x′2 x′3 ). Obviously, the ”new” coordinates x′i (i = 1, 3) are functions of the ”old” coordinates xi (i = 1, 3): x′i = fi (x1 , x2 , x3 )

(i = 1, 3).

Fig.B.6 The relations x′i = const. (i = 1, 3) represent equations of three planes 393

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Free ebooks ==> www.Ebook777.com with respect to S only if the above transformation is linear, that is x′i = Rij xj + Xi

(i, j = 1, 3),

where Xi are coordinates of the origin O′ of the frame S ′ , with respect to the frame S. The nine quantities Rij can be considered as a matrix, called matrix of coordinate transformation. In order to determine the same distance between any two points, no matter which coordinate system is considered, the quantities Rij have to obey the orthogonality condition Rij Rik = δjk (i, j, k = 1, 3). This condition diminishes up to three the number of independent parameters Rij . The linear transformation x′i = Rij xj + Xi , where the quantities Rij obey the orthogonality condition, is called non-homogeneous coordinate transformation. If Xi = 0, the transformation is called homogeneous.

Fig.B.7 It can be shown that ˆ ≡ (det Rij ) = ±1. det R

ˆ = +1 are called proper transTransformations characterized by det R ˆ = −1 are named improper transforformations, while those with det R mations. For example, translations x′i = Xi (i = 1, 3) and/or rotations x′i = Rij xj (i, j = 1, 3) are proper transformations, while inversions (reflections in a mirror of the coordinate axes) are improper transformations. Fig B.7 shows a graphic representation of an improper transformation, whose matrix is   −1 0 0 ˆ =  0 1 0. R 0 0 1 394

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Free ebooks ==> www.Ebook777.com It can be proved that the set of non-homogeneous orthogonal transformations form a group structure with respect to the operation of composition of these transformations, called non-homogeneous orthogonal coordinate transformation group. The properties of this group, as well as of its main subgroups, can be found in the recommended literature. For a better understanding of the concepts of pseudoscalar and pseudovector, as particular cases of a pseudotensor, it is necessary to define the notion of frame orientation. A frame S is called righthanded if rotation of a right screw (or drill) oriented along Ox3 by an angle of 90o produces superposition of axis Ox1 over axis Ox2 (see Fig.B.8). In other words, if ~u1 × ~u2 = ~u3 , where ~u1 , ~u2 and ~u3 are versors of the axes Ox1 , Ox2 and Ox3 , respectively, then Oxyz is a right-handed reference frame. This way, connection between the cross product operation and the frame orientation becomes more obvious. A left-handed frame can be obtained by an odd number of inversions of coordinate axes. Within the framework of orthogonal coordinate transformations, the above considerations can be expressed in a concise form as follows. Consider a right-handed frame S. Then any frame S ′ obtained by an orthogonal coordinate transformation with ˆ = +1 is a right-handed frame; → det R ˆ = −1 is a left-handed frame. → det R

Fig.B.8 According to these observations, one can define the main algebraic 395

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Free ebooks ==> www.Ebook777.com quantities as follows: Definition. A scalar is a quantity characterized by a number which remains unchanged under any coordinate transformation. A scalar physical quantity is defined in the same way, but the unit system has to be mentioned. Definition. A scalar with the same algebraic form in any coordinate system is called invariant. Definition. A vector ~v is a quantity characterized by an ordered system of three numbers {v1 , v2 , v3 }, called vector components, which under an orthogonal coordinate transformation change according to the law vi′ = Rij vj (i, j = 1, 3), where vi′ are the vector components in S ′ , and Rij are the elements of the transformation matrix. Definition. A nth rank tensor in the Euclidean space Em is a quantity characterized by an ordered set of mn numbers ti1 i2 ...in (i1 , i2 , ..., in = 1, m), called tensor components, which under an orthogonal coordinate transformation change according to the law t′i1 i2 ...in = Ri1 j1 Ri2 j2 ...Rin jn tj1 j2 ...jn (i1 , i2 , ..., in , j1 , j2 , ..., jn = 1, m). Definition. A nth rank pseudotensor in the Euclidean space Em is a quantity characterized by an ordered set of mn numbers t∗i1 i2 ...in (i1 , i2 , ..., in = 1, m), called pseudotensor components, which under an orthogonal coordinate transformation change according to the law ′ ˆ i j Ri j ...Ri j t∗ t∗i1 i2 ...in = (det R)R 1 1 2 2 n n j1 j2 ...jn

(i1 , i2 , ..., in , j1 , j2 , ..., jn = 1, m). ˆ= Consequently, under a proper orthogonal transformation (det R +1) pseudotensors transform in the same way as ordinary tensors, ˆ = −1) appears a while under an improper transformation (det R change of sign. A pseudovector is also called an axial vector, while an ordinary vector is called a polar vector. For example, under an inversion of axes (improper orthogonal transformation) xi → x′i = −xi the components of a vector transform according to vi′ = −vi (i = 1, 3), ′ while a pseudovector obeys the rule vi∗ = vi∗ (i = 1, 3). Keeping in mind the connection between frame orientation and the proper/improper character of coordinate transformation, we can say that tensors change their sign when orientation changes, while pseudotensors don’t. 396

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Fig.B.9 To illustrate these theoretical considerations, let us take into account a physical application, namely the formula expressing the axial vector of magnetic induction produced in vacuum by a continuous electric current I, situated in the plane x1 Ox2 (see Fig.B.9). As wellknown, the appropriate formula is the Biot-Savart law ~ = dB

~ µo I d~l × (~r − ρ ~) µo I d~l × R = , 3 3 4π R 4π |~r − ρ ~|

~ has its tip at the point where the field is determined, where vector R and its tail at the origin of the current element I d~l. Since the electric current lies in x1 Ox2 plane, we have ~u1 ~u2 ~u3 ~ = Idl1 Idl2 0 I d~l × R X1 X2 X3 = IX3 dl2 ~u1 + IX2 dl1 ~u3 − IX1 dl2 ~u3 − IX3 dl1 ~u2 ,

leading to the following components of the elementary magnetic field ~ at the point P (~r) dB  µo I X3 dl2  dB1 = + ;    4π R3  µo I X3 dl1 dB2 = − ; 3  4π R     dB = + µo I (X2 dl1 − X1 dl2 ) . 3 4π R3 397

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Free ebooks ==> www.Ebook777.com Let us now perform the improper orthogonal transformation xi → x′i = −xi

(i = 1, 3).

~ ′ | = |R|, ~ I′ = As a result, the scalar quantities remain unchanged (|R ~′ I, µ′o = µo ), but the vector quantities change their sign (d~l′ = −d~l, R ~ so that = −R), ~′ = dB that is

~′  µo I µ′o I ′ d~l′ × R µo I  ~ ~ ~ = dB, ~ = (−d l)×(− R) = (d~l× R) 4π 4π R3 4π R3 R′3  ′  dB1 = dB1 , dB ′ = dB2 ,  2′ dB3 = dB3 ,

which is in complete agreement with the above considerations: under an improper orthogonal transformation (or, equivalently, under the change of the frame orientation) the pseudotensor components remain unchanged. Let us now get back to the special connection between the cross product operation and the pseudo/non-pseudo character of tensors. As we have seen, a mirror-image of a reference frame means the change of orientation of axes. But this change determines the change of tensors sign and maintains the sign of pseudotensors. Since the effect of inversion can algebraically be expressed by means of the cross product, the connection between the operation of cross product and pseudo/nonpseudo character of tensors becomes obvious. As we have seen, a right-handed reference frame is characterized by ~u1 × ~u2 = ~u3 . Performing one (or an odd number of) inversion(s) of coordinate axes (e.g. x′3 = −x3 ), the frame orientation changes, becoming a left-handed frame: ~u1 × ~u2 = −~u3 . Due to the noncommutative property of the cross product, nothing changes if the inversion affects any other coordinate axis. Indeed, if we perform inversion x′2 = −x2 , then (see Fig.B.10) ~u3 = ~u2 × ~u1 = −~u1 × ~u2 , or ~u1 × ~u2 = −~u3 , which is precisely the same relation as in case of inversion of Ox1 . 398

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Fig.B.10 If, on the contrary, the number of inversions is even, then neither the frame orientation, nor the pseudo/non-pseudo character of tensors is affected. Algebraically, this property can also be expressed by means of the cross product. Indeed, if two inversions (say, of axes Ox1 and Ox3 ) take place, then according to Fig.B.11 we can write ~u1 ×~u2 = ~u3 , meaning that the new frame is also right-handed, as it was before inversions.

Fig.B.11 The above discussion suggests a simple and useful method of identification of the character of tensors (pseudo, or non-pseudo). To this end, we must observe that the change of the frame orientation (followed by the change of sign of tensor components, on the one hand, and by maintenance of the sign of pseudotensor components, on the other) depends on the number (even or odd) of inversions of coor399

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Free ebooks ==> www.Ebook777.com dinate axes. This way, if we conventionally associate minus sign to pseudotensor quantities, and plus sign to tensor quantities, and observe that, as a result of this association, the cross product has the character of ”pseudo-”, due to its anticommutativity (associated with (−1) sign), then the problem is solved: the character of tensor quantities (of any kind) is given by the ”resultant” sign associated with expressions defining these quantities [(+1) or (−1)]. Here are a few examples. 1. As we know, the position vector of the tail of a force F~ , with respect to a point (pol) is a polar vector (as well as the force F~ ). In view of the above defined convention, to both vectors one associates the sign (+1). Since the operation of cross product has a character of pseudovector, it is associated with sign (−1). Consequently, the ~ of the force F~ relative to a pol is a pseudovector. Indeed, moment M ~ M |{z} (−1)

=

~ Ω |{z} (−1)

=

~r |{z} (+1)

× |{z} (−1)

F~ |{z} . (+1)

 ∂ ∂ ∂ ~ ~ ~ 2. The operator nabla ∇ = i ∂x + j ∂y + k ∂z is a polar vector, and so is the linear velocity of a moving particle of fluid, therefore the quantity called vorticity 

1 2

∇ |{z} (+1)

× |{z} (−1)

~v |{z} (+1)

is a pseudovector (or axial vector). Applying the same procedure, one can show that the curl of any polar vector is a pseudovector. 3. The force of interaction between two circuits Γ1 and Γ2 traveled by the electric currents I1 and I2 is a polar vector. Indeed, since the current elements I1 d~l1 and I2 d~l2 are polar vectors, and so is the position vector of I2 d~l2 with respect to the origin of I1 d~l1 , ~r12 = −~r21 , we have F~12 = −F~ 21 |{z} | {z } (+1) (+1) I I d~l2 × (d~l1 × ~r12 ) µ 1 |{z} |{z} |{z} |{z} |{z} = I1 I2 . 3 4π r12 (+1) (−1) (+1) (−1) (+1) (Γ1 ) (Γ2 )

4. The moment of a force F~ with respect to an axis ∆ is: ~ O, M∆ = ~u∆ · (~r × F~ ) = ~u∆ · M 400

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Free ebooks ==> www.Ebook777.com ~ O is the moment of F~ with respect where ~u∆ is the versor of ∆, while M to an arbitrary point (pol) of the axis. This quantity is a pseudoscalar. Indeed, F~ ) M∆ = ~u∆ · (|{z} ~r × |{z} |{z} . |{z} |{z} (−1) (+1) (+1) (−1) (+1) 5. If ~a, ~b and d~ are axial vectors, and ~c a polar vector, then ~ is a pseudoscalar. Indeed, (~a × ~b) · (~c × d) p |{z} (−1)

=

(|{z} ~a (−1)

× |{z} (−1)

~b ) · |{z} (−1)

(|{z} ~c (+1)

× |{z} (−1)

d~ ) |{z} . (−1)

Next we shall give some examples of physical quantities, of 0th and 1st rank tensor, together with their units in SI. i) Examples of scalar physical quantities: mass, m[kg]; temperature, T [K]; pressure, p[N · m−2 ]; mechanical work, L[J]; concentration of a chemical substance in a solution, c[mol · m−3 ]; power, P [W ]; density of a substance, ρ[kg · m−3 ]; magnetic permeability of vacuum, µ0 [H · m−1 ]; electric potential in a point, V [V ]; electric flux, Φe [V · m], etc. ii) Examples of pseudoscalar physical quantities: moment of a force with respect to an axis, M∆ [N · m]; magnetic flux, Φm [T · m2 ], etc. iii) Examples of vector physical quantities: linear velocity, ~v [m · −1 s ]; Newtonian force, F~ [N ]; linear acceleration, ~a[m · s−2 ]; electric ~ · m−1 ]; electromagnetic flux density (Poynting vecfield intensity, E[V −2 ~ [J ·m−2 ·s−1 ]; electric field induction, D[C·m ~ tor), Y = N ·m−1 ·V −1 ]; electromagnetic momentum density, ~g [kg · m−2 · s−1 ], etc. iv) Examples of pseudovector physical quantities: magnetic induc~ ]; angular velocity, ω tion, B[T ~ [rad · s−1 ]; moment of a force relative to ~ [n · m]; fluid vorticity, Ω[s−1 ], etc. a pole, M B.4. Elements of vector analysis B.4.1. Vector functions Consider a vector ~a depending on a real, variable parameter t. If to each value of t corresponds a vector ~a, we say that ~a is a vector function of t: ~a = ~a(t). If ~r = ~r(t) is the position vector of some point P with respect to the Cartesian coordinate system Oxyz, then the curve described by P is given by the parametric equations x = x(t), y = y(t), z = z(t). Let P be a material point (particle) whose position is defined by vector ~a(t) with respect to origin O of the frame Oxyz. A variation of 401

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Free ebooks ==> www.Ebook777.com the parameter t produces a variation of ~a(t), so that its head describes a curve C. If radius vector is proportional to the velocity of the moving particle, then the curve ~a = ~a(t) is called hodograph or velocity diagram (see Fig.B.12). B.4.2. Derivative and differential of a vector function Let ~a = ~a(t) be a continuous vector function of parameter t. Then its derivative with respect to t is defined as ∆~a d~a = lim , ∆t→0 ∆t dt where ∆~a = ~a(t + ∆t) − ~a(t) is the variation ∆~a of the function ~a(t), corresponding to a variation ∆t of the parameter t (which determines displacement of the particle a from position P1 to position P2 ), while ∆~ a. ∆t is a vector collinear to ∆~

Fig.B.12 a Quantity ∆~ ∆t is a vector parallel to tangent to the curve C at point P , its sense being given by the increasing variation of parameter a t. The magnitude (modulus) of derivative d~ dt is

s 2  2  2 d~a da da da x y z = + + . dt dt dt dt

Differential of a vector function ~a = ~a(t) can be defined by relation a d~a = d~ dt dt, where dt is an elementary, arbitrary variation of parameter 402

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Free ebooks ==> www.Ebook777.com t. If vector ~a is a function of several variables x1 , x2 , ..., xn , the total differential of ~a(x1 , x2 , ..., xn ) is defined as ∂~a dxi ∂xi

d~a =

(i = 1, n),

∂ is the operator of partial derivative with respect to xi . where ∂x i Keeping in mind the properties of vector operations and of the differentiation of scalar functions, we have:

d d~a d~b (~a ± ~b) = ± ; dt dt dt d d~a (λ~a) = λ dt dt

(λ = const.);

d~b ~ d~a d (~a · ~b) = ~a · +b· ; dt dt dt d~b d~a ~ d (~a × ~b) = ~a × + × b; dt dt dt   d~a  d ~ d · ~b × ~c + ~a · b × ~c ; ~a · ~b × ~c = dt dt dt  d~a   d d ~ b × ~c . ~a × ~b × ~c = × ~b × ~c + ~a × dt dt dt

B.4.3. Integral of a vector function The (definite) integral of a continuous vector function ~a(t) can be defined in much the same way as of scalar functions, except that the integral is a vector. In Euclidean space E3 the following three types of integrals are defined: P R2 (a) Curvilinear (line) integral: ~a · d~r, along a curve C, between P1

points P1 and P2 , where vector ~a has its origin on the curve, and d~r is a vector element of C. This integral is called circulation of vector ~a along curve C, betweenHpoints P1 and P2 . If C is a closed curve, circulation is denoted by ~a · d~r. (C)

(b) Double integral:

R

S

~ where ~a is a vector with its origin ~a · dS,

~ an oriented surface element. The integral is called flux at S, and dS of vectorH~a through surface S. If S is a closed surface, the integral is ~ denoted ~a · dS. S

403

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Free ebooks ==> www.Ebook777.com (c) Triple integral:

R

~a dτ , where D is a domain of volume V , dτ

D

a volume element, and ~a has its origin at an arbitrary point of D. B.4.4. Scalar and vector fields If a physical quantity has a well-determined value in any point of a spatial domain D, then the multitude of these values define the field of the physical quantity. For example, a heated body produces a temperature field. In each point of the body, the temperature has a certain value. Or, as another example, in each point of a flowing liquid can be determined the velocity of some fluid particle at that point. To define a field, one must establish a correspondence between each point of the domain D and the value of physical quantity at that point. This correspondence defines a point function. Since each point is determined by its coordinates x, y, z (or, equivalently, by its position vector ~r) relative to the origin of a coordinate reference system Oxyz, the fields of various physical quantities are determined by knowledge of the specified (scalar, vector, or tensor) functions depending on the variables x, y, z. If to each point P ∈ D of Euclidean space E3 one can associate a scalar quantity ϕ(P ), then we say that in domain D has been defined a scalar field ϕ(x, y, z). Such a scalar function is, for example, temperature: at each point of some domain D the temperature has a certain value T (P ). Correspondence between the points of the considered domain and the values of temperature at those points defines a scalar function T (P ) = T (x, y, z). Another example is offered by pressure p of the air: at each point of the surrounding atmosphere one can settle a scalar function p(P ) = p(x, y, z). If to each point P ∈ D one can associate a vector quantity ~a(P ) = ~a(x.y.z), we say that in D has been defined a vector field. Such a field, exists for example, in the region surrounding an electrized body: at each point of this domain one can define a vector field called electric ~ Velocities associated to each molecule of a fluid in field intensity E. motion is another example of vector field. ~ field is called non-stationary if ϕ (reA scalar ϕ (or vector A) ~ depend explicitly on time. If not, the field is called spectively, A) ~ together with their partial derivatives, stationary. Functions ϕ and A, are supposed to be continuous.

404

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Free ebooks ==> www.Ebook777.com B.5. First-order vector differential operators 1 B.5.1. Gradient Let ϕ(x, y, z) be a function of class C 1 (the first derivatives exist and are continuous), defined on a certain domain D ⊂ E3 . The above investigation shows that ϕ(x, y, z) defines in D a scalar field. The differential of ϕ is dϕ =

∂ϕ ∂ϕ ∂ϕ ∂ϕ dx + dy + dz = dxi ∂x ∂y ∂z ∂xi

(i = 1, 3).

(B.33)

In view of (B.16), this relation can be written as dϕ =

=





  ∂ϕ~ ∂ϕ~ ∂ϕ ~ k · dx~i + dy ~j + dz ~k i+ j+ ∂x ∂y ∂z

∂ϕ ∂ϕ ∂ϕ ~u1 + ~u2 + ~u3 ∂x1 ∂x2 ∂x3



 · dx1 ~u1 + dx2 ~u2 + dx3 ~u3 ,

therefore it can be considered as the dot product of vectors d~r = dx~i + dy ~j + dz ~k and

~ = ∂ϕ~i + ∂ϕ~j + ∂ϕ ~k = ~ui ∂ϕ A ∂x ∂y ∂z ∂xi

(i = 1, 3).

(B.34)

~ whose Cartesian components are ∂ϕ (i = 1, 3), The vector field A, ∂xi is called gradient of scalar field ϕ and usually is denoted by gradϕ. With this notation, relation (B.33) writes dϕ = gradϕ · d~r.

(B.35)

Vector gradϕ determines variation of function ϕ about a point, ∂ϕ , represents the rate of while its component on xi -axis, which is ∂x i variation of ϕ on xi -direction. The direction of gradϕ at some point is direction of the most rapid variation of scalar function ϕ, starting from that point, while the sense of gradϕ is given by the increase of 1

In vector calculus, ”grad”, ”div”, and ”curl” are usually called operators. If fact, the only true operator is nabla, and ∇ϕ = gradϕ, ~ = divA, ~ and ∇ × B ~ = curlB ~ are expressions obtained as a ∇·A result of different ways of application of nabla upon scalar or vector quantities. Note that the Laplacian ∆ is, also, a true operator. 405

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Free ebooks ==> www.Ebook777.com ϕ. Relation (B.33) shows that the vector gradϕ can be obtained by applying the operator ∂ ∂ ∂ ∂ ∇ = ~i + ~j + ~k = ~ui ∂x ∂y ∂z ∂xi

(i = 1, 3)

(B.36)

to the scalar function ϕ(xi ). This operator, denoted by symbol ∇, is called nabla operator or Hamilton’s operator. By means of nabla, the gradient writes gradϕ = ~ui

∂ϕ = ∇ϕ ∂xi

(i = 1, 3).

(B.37)

~ defined by (B.37) is called potential field, and The vector field A ϕ is the potential of this field. For example, using relation between ~ and the electrostatic potential V , the electrostatic field intensity E ~ E = −gradV , we can determine one quantity, if the other is known. (The sign ”minus” is introduced by convention, to show that the field ~ go out of ”plus” point charges and enter the ”minus” point lines of E charges). Definition (B.36) shows that ∇ is a linear, vector differential operator. Projecting ∇ϕ(xi ) on xi -direction, we have:   ∂ϕ ∂ϕ ∂ϕ ∂ϕ ~uk · ~ui = δik = , (gradϕ)i = (∇ϕ) · ~ui = ~uk · ~ui = ∂xk ∂xk ∂xk ∂xi or, by means of the commonly used notations in vector analysis gradi ϕ = ∇i ϕ =

∂ϕ = ∂i ϕ = ϕ,i . ∂xi

(B.38)

Equipotential surfaces Suppose that function ϕ(P ) is uniform and continuous in D, together with its partial derivatives up to the second order, and consider all points where this function has the same value. These points form a surface in D. Indeed, if x, y, z are the coordinates of P , relation ϕ(P ) = K(const.) can also be written as ϕ(x, y, z) = K,

(B.39)

which is the implicit form of a surface in D ⊂ E3 . In view of (B.35) and (B.39), we then have dϕ = 0 = gradϕ · d~r. 406

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(B.40)

Free ebooks ==> www.Ebook777.com This relation is valid at any point of surface (B.39). Since equality (B.40) has to be true for any vector of the dot product, and because d~r lies in the plane tangent to surface (B.39), it follows that at any point of this surface the vector ∇ϕ is oriented along the normal to surface. Giving values to the constant K in (B.39) one obtains a family of surfaces satisfying condition (B.40), called equipotential surfaces or level surfaces. Equation (B.40) is called equation of equipotential surfaces. Field lines ~ and an arbitrary curve given Consider a stationary vector field A ~ is tangent to by its parametric equations xi = xi (s) (i = 1, 3). If A the curve at any of its points, then the curve is called field line or ~ The differential equations of field lines are line of force of the field A. ~ × d~s = 0. Projecting this vector obtained from the obvious relation A relation on the axes of a Cartesian orthogonal frame, we have dx dy dz = = . Ax Ay Az

(B.41)

~ = gradϕ is orthogonal to the According to (B.40), the vector A arc element d~r which, in turn, is tangent to equipotential surfaces. ~ = gradϕ and d~r are collinear, it follows that Since the field lines of A they are also orthogonal to the equipotential surfaces ϕ = const. Directional derivative Let (∆) be an axis whose direction and sense are given by the r versor ~u∆ = d~ r| is the magnitude of the line element ds , where ds = |d~ of the axis (∆). The directional derivative of a scalar function ϕ = ϕ(~r) along the given axis (∆) is the projection of gradϕ on the axis: gradϕ · ~u∆ =

dϕ dϕ d~r · = . d~r ds ds

Here, a formal writing for the gradient, gradϕ =

dϕ d~ r,

has been used.

B.5.2. Divergence of a vector field Since ∇ is a vector operator, one can perform the dot product ~ where A ~ = A(A ~ x , Ay , Az ) is an arbitrary vector field. This ∇ · A, ~ and is denoted by divA. ~ Note that, operation is called divergence of A ~ ~ Indeed, even if the scalar product is commutative, A · ∇ 6= ∇ · A. ~ is a well-defined quantity (scalar or pseudoscalar, the divergence of A ~ depending on the tensor nature of A)   ∂ ∂Ak ∂Ai ~ ∇ · A = ~ui · (Ak ~uk ) = δik = ∂xi ∂xi ∂xi 407

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Free ebooks ==> www.Ebook777.com =

∂Ax ∂Ay ∂Az + + , ∂x ∂y ∂z

(B.42)

while the quantity    ∂ ∂ ∂ ~ ~ ~ ~ ~ ~ ~ A · ∇ = Ax i + Ay j + Az k · i +j +k ∂x ∂y ∂z = Ax

∂ ∂ ∂ + Ay + Az ∂x ∂y ∂z

is, in its turn, an operator. Using notations (B.38), we can also write ~ = ∇i Ai = divA

∂Ai = ∂i Ai = Ai,i ∂xi

(i = 1, 3).

(B.43)

~ with zero divergence, divA ~ = 0, is called diverA vector field A gence-free (or source-free or solenoidal). The lines of such a field are ~ 6= 0, we have a source-field, with the following closed curves. If divA two possibilities: ~ > 0, ր divA ~ divA 6= 0 ~ < 0. ց divA

Fig.B.13a ~ > 0) the sources of the field are positive In the first case (divA (springs, or wellheads), while in the second sources are negative (or sinks). For a source-field, there always exists a field line which does not close, but goes to infinity. It is conventionally accepted that the lines of a source field emerge from positive sources, and enter the negative sources. Obviously, the field lines of a solenoidal field are ~ of closed curves. Such a field is, for example, the magnetic field B 408

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Free ebooks ==> www.Ebook777.com a solenoid (Fig.B.13a), similar to the field of a permanent magnet, shown in (Fig.B.13b).

Fig.B.13b No matter how small is a magnet, it always has two poles, north and south. P.A.M.Dirac predicted the existence of the magnetic monopole, but its real existence has not been detected so far. The almost perfect analogy between electrostatic end magnetostatic fields, for the simplest case of two electric/magnetic charges, is graphically sketched in Fig.B.13a and Fig.B.13b. The main property of the electrostatic field is expressed by Gauss’s ~ = 1 ρ 6= 0, where ρ = ρ(~r) = lim ∆q = dq is the electric law divE ε ∆τ dτ ∆τ →0

charge density, while Gauss’s law for the magnetostatic field writes ~ = 0. In those regions of space where no electric charges are divB ~ = 0, present (ρ = 0), the two equations have the same form: divE ~ and divB = 0. In other words, in such regions of space the two fields behave identically, which allows one to speak not about electric and magnetic field separately, but about an electromagnetic field. The mathematical ”equivalence” of the two fields allowed their unification under the same theory. This was done by J.C. Maxwell, who wrote his celebrated system of equations1 . 1

Theory developed by J.C.Maxwell was the first successful classical unified field theory. Later, in the Standard Model, the weak and the electromagnetic interactions have been combined into a unified electroweak theory. In its turn, this theory was unified with theory of strong interactions (quantum chromodynamics), leading to a new theory capable to give a unitary description of electromagnetic and nuclear (weak and strong) interactions. Gravity has yet to be successfully included in a theory of everything. Simply trying to combine the graviton with the strong and electroweak interactions runs into fun409

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Free ebooks ==> www.Ebook777.com B.5.3. Curl of a vector field ~ x , Ay , Az ) and let us perform the cross Consider a vector field A(A ~ The result of this operation is a vector field B, ~ B ~ = product ∇ × A. ~ called curl of the vector field A. ~ If A ~ is a polar vector, then B ~ ∇ × A, ~ = ∇×A ~ is obtained is an axial vector. The analytical expression of B by means of (B.27): ~ =∇×A ~ = εijk ∂j Ak ~ui B       ∂Ax ∂Ay ∂Az ∂Ay ~ ∂Az ~ ∂Ax ~ i+ j+ k = − − − ∂y ∂z ∂z ∂x ∂x ∂y ~i ~k ~j ∂ ∂ ∂ . = ∂x (B.44) ∂y ∂z Ax Ay Az

Performing dot product of (B.44) and versor ~us , we obtain the ~ = curlA ~ on the direction defined by ~us : projection of B ~ s = (εijk ∂j Ak ~ui ) · ~us = εijk ∂j Ak δsi = εsjk ∂j Ak . (curlA)

(B.45)

~ is: For example, the component along x-axis of curlA ~ 1 = (curlA) ~ x = ε1jk ∂j Ak = ∂2 A3 − ∂3 A2 = ∂Az − ∂Ay . (curlA) ∂y ∂z ~ having the property curlA ~ = 0 is called irrotaA vector field A tional or curl-free. Such a field is, for example, the electrostatic field ~ intensity E. damental difficulties since the resulting theory is not renormalizable. Theoretical physicists have not yet formulated a widely accepted, consistent theory that combines general relativity and quantum mechanics. The incompatibility of the two theories remains an outstanding problem in the field of physics. Some theoretical physicists currently believe that a quantum theory of general relativity may require frameworks other than field theory itself, such as string theory or loop quantum gravity. A Grand Unified Theory (GUT) is a model in particle physics in which at high energy, the three gauge interactions of the Standard Model which define the electromagnetic, weak, and strong interactions, together with gravitational interactions, are merged into one single theory. At the present there is no accepted unified field theory, and thus it remains an open line of research. 410

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Free ebooks ==> www.Ebook777.com B.6. Vector identities Here are the most frequently used identities of vector analysis, together with their justification. 1. grad(ϕψ) = ϕ gradψ + ψ gradϕ. (B.46) Indeed, grad(ϕψ) = ~ui 2.

~ = ϕdivA ~+A ~ · gradϕ. div(ϕA) Indeed, ~ = div(ϕA)

3.

∂(ϕψ) ∂ψ ∂ϕ = ~ui ϕ + ~ui ψ = ϕ gradψ + ψ gradϕ. ∂xi ∂xi ∂xi (B.47)

∂(ϕAi ) ∂Ai ∂ϕ ~+A ~ · gradϕ. =ϕ + Ai = ϕdivA ∂xi ∂xi ∂xi

~ = ϕ curlA ~ + (gradϕ) × A. ~ curl(ϕA) Indeed,

(B.48)

~ = εijk ∂j (ϕAk )~ui = ϕ(εijk ∂j Ak ~ui ) + εijk (∂j ϕ)Ak ~ui curl(ϕA) ~ + (gradϕ) × A. ~ = ϕ curlA

4.

~ × B) ~ =B ~ · curlA ~−A ~ · curlB. ~ div(A Indeed,

(B.49)

~ × B) ~ = ∂i ( A ~ × B) ~ i = ∂i (εijk Aj Bk ) div(A = εijk Aj ∂i Bk + εijk Bk ∂i Aj = −Aj εjik ∂i Bk + Bk εkij ∂i Aj ~ j + Bk (curlA) ~ k = −A ~ · curlB ~ +B ~ · curlA. ~ = −Aj (curlB) 5.

~ · B) ~ =A ~ × (curlB) ~ +B ~ × (curlA) ~ grad(A ~ · ∇)B ~ + (B ~ · ∇)A. ~ +(A We first observe that a component of the gradient reads

(B.50)

~ · B) ~ = ∂j (Ak Bk ) = Ak ∂j Bk + Bk ∂j Ak . ∂j ( A ~ s = ǫslm ∂l Bm by ǫsjk and perNext, let us multiply relation (curl B) form summation over indices l and m: ~ s = (δlj δmk − δlk δmj )∂l Bm = ∂j Bk − ∂k Bj . ǫsjk (curl B) Using this result, we still have: ~ · B) ~ = ǫjks [Ak (curl B) ~ s + Bk (curl A) ~ s ] + (Ak ∂k )Bj + (Bk ∂k )Aj , ∂j ( A 411

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Free ebooks ==> www.Ebook777.com or, in a vector form ~ · B) ~ =A ~ × curl B ~ +B ~ × curl A ~ + (A ~ · ∇)B ~ + (B ~ · ∇)A. ~ grad(A ~ × B) ~ = Adiv ~ B ~ − Bdiv ~ ~ + (B ~ · ∇)A ~ − (A ~ · ∇)B. ~ curl(A A Indeed,

6.

(B.51)

~ × B) ~ = ~ui εijk ∂j (A ~ × B) ~ k = ~ui εijk εklm ∂j (Al Bm ) curl(A = ~ui εijk εklm Al ∂j Bm + ~ui εijk εklm Bm ∂j Al = ~ui (δil δjm − δim δjl )Al ∂j Bm + ~ui (δil δjm − δim δjl )Bm ∂j Al = ~ui Ai ∂j Bj − ~ui Aj ∂j Bi + ~ui Bj ∂j Ai − ~ui Bi ∂j Aj = (Ai ~ui )(∂j Bj ) − (Bi ~ui )(∂j Aj ) + (Bj ∂j )(Ai ~ui ) − (Aj ∂j )(Bi ~ui ) ~∇·B ~ −B ~ ∇·A ~ + (B ~ · ∇)A ~ − (A ~ · ∇)B. ~ =A 7.

~ = 0. div(curlA) Indeed,

(B.52)

~ = ∂i (curlA) ~ i = εijk ∂i (∂j Ak ) = div(curlA)

 1 εijk ∂i ∂j Ak +εijk ∂i ∂j Ak 2

 1  1 εijk ∂i ∂j Ak − εijk ∂j ∂i Ak = εijk ∂i ∂j Ak − εijk ∂i ∂j Ak = 0, 2 2 ~ y, z) satisfy the conditions dewhere we have supposed that A(x, manded by Schwartz’ theorem, which ”assure” the equality of mixed second-order derivatives of Ak , that is =

∂ 2 Ak ∂ 2 Ak = ⇐⇒ ∂i ∂j Ak = ∂j ∂i Ak . ∂xi ∂xj ∂xj ∂xi Therefore, since the pseudotensor εijk is antisymmetric, and ∂i ∂j Ak symmetric in the same pair of indices (ij), we have εijk ∂i ∂j Ak = εijk 8.

∂ 2 Ak = 0. ∂xi ∂xj

curl(gradϕ) = 0. Indeed,

(B.53)

curl(gradϕ) = ~ui εijk ∂j (gradϕ)k = ~ui εijk ∂j ∂k ϕ ≡ ~ui εijk 412

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∂2ϕ = 0, ∂xj ∂xk

Free ebooks ==> www.Ebook777.com 2

ϕ where the pair of summation indices in the product εijk ∂x∂j ∂x is (jk). k

9. div(gradϕ) = ∇ · (∇ϕ) = (∇ · ∇)ϕ = ∇2 ϕ = ∆ϕ, where the operator (denoted as ∆) ∇2 = ∆ =

(B.54)

∂2 ∂2 ∂2 ∂2 + + = ≡ ∂i ∂i ∂x2 ∂y 2 ∂z 2 ∂xi ∂xi

is called the Laplace operator or Laplacian (in Cartesian coordinates). Equation ∆ϕ = 0 is called Laplace’s equation. Solutions of Laplace’s equation are called harmonic functions. This operator can be generalized in four dimensions, by choosing three spacial coordinates and an additional fourth coordinate for the time (temporal coordinate). It is called D’Alembert’s operator or Dalembertian, being denoted by ⊓ ⊔. Taking x1 = x, x2 = y, x3 = z, and x4 = ivt, where v is a constant velocity, we have ⊔= ⊓

4 X

∂2 ∂2 ∂2 ∂2 1 ∂2 ∂2 = + + + = ∆ − . ∂xµ ∂xµ ∂x21 ∂x22 ∂x23 ∂x24 v 2 ∂t2 µ=1

Choosing v = c, where c is velocity of light in vacuum, we meet the Minkowski space-time. ~ = grad(divA) ~ − ∆A. ~ 10. curl(curlA) (B.55) Indeed, ~ = ~ui εijk ∂j (curlA) ~ k = ~ui εijk εklm ∂j ∂l Am curl(curlA) = ~ui (δil δjm − δim δjl )∂j ∂l Am = ~ui ∂m ∂i Am − ~ui ∂j ∂j Ai = ~ui ∂i ∂m Am − ∂j ∂j Ai ~ui

~ − ∂j ∂j A ~ = ∇(∇ · A) ~ − ∆A. ~ = ~ui ∂i (divA)

To conclude, here are some final remarks. Nabla (or del) is a vector differential operator. When applied to a scalar field, nabla denotes the gradient. When applied to a vector field, this can be done by means of dot product (divergence), or cross product (curl). The dot product of nabla with itself is an operator called Laplacian. In its turn, the Laplacian can be applied to both scalar and vector1 fields. The following Table synthesizes these final remarks. 1

As and example, here is the equation satisfied by the vector ~ in an isotropic and homogeneous medium : ∆A ~ = −µ~j, potential A where ~j is the current density vector. 413

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Table B.1. Various ways of acting of the nabla operator. B.7. Orthogonal curvilinear coordinates. First-order differential operators in orthogonal curvilinear coordinates Let ~r be the radius-vector of a point P (~r) in tridimensional Euclidean space E3 and x1 , x2 , x3 its Cartesian coordinates with respect to a three-orthogonal Cartesian frame Ox1 x2 x3 . Consider, also, three independent real parameters q1 , q2 , q3 , so that coordinates xi (i = 1, 3) are functions of class C 1 of these parameters: xi = xi (q1 , q2 , q3 )

(i = 1, 3).

(B.56)

Fig.B.14 Suppose that between the set of coordinates xi on the one hand, and of the parameters qj , on the other, exists a one-to-one correspondence (this is an alternate name for a bijection), that is to a set of 414

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Free ebooks ==> www.Ebook777.com coordinates x corresponds a single set of parameters q, and vice-versa. Under these circumstances, transformation (B.56) is locally reversible if the functional determinant (Jacobian) is different from zero: J=

∂(x1 , x2 , x3 ) 6= 0. ∂(q1 , q2 , q3 )

(B.57)

If two out of the three parameters qj (j = 1, 3), say q2 and q3 , have fixed values, then one obtains the line (curve) of coordinate q1 . This means that through any point in space pass three lines of coordinates (see Fig.B.14). Parameters q1 , q2 and q3 are called curvilinear coordinates, or general coordinates of the point P . If ~ei (i = 1, 3) are vectors tangent to the curve of coordinates qi (i = 1, 3), then according to (B.57) are linearly independent, therefore they form a basis in E3 . Indeed,   ∂~r ∂~r ∂~r ~e1 · (~e2 × ~e3 ) = · × ∂q1 ∂q2 ∂q3 ∂x1 ∂x2 ∂x3 ∂q1 ∂q1 ∂q1 ∂x1 ∂x2 ∂x3 ∂(x1 , x2 , x3 ) = J 6= 0. = ∂q2 ∂q2 ∂q2 = ∂(q1 , q2 , q3 ) ∂x1 ∂x2 ∂x3 ∂q3

∂q3

∂q3

Fig.B.15 If at any point of the domain D(q) , defined by the set of all possible values of parameters (q), the vectors ~ei form a right-handed orthogonal trieder, then the three curvilinear coordinates define an orthogonal 415

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Free ebooks ==> www.Ebook777.com curvilinear coordinate system. If ~ei are linearly independent, but are not orthogonal to each other, then qi (i = 1, 3) define a general curvilinear coordinate system. Two of the most commonly used orthogonal curvilinear coordinate systems are cylindrical and spherical coordinates. The cylindrical coordinates of a point P are ρ, ϕ and z (see Fig.B.15). The range of variation of these parameters is: n o D(cyl) = (ρ, ϕ, z) ρ ≥ 0, 0 ≤ ϕ < 2π, −∞ < z < +∞ . In this case, relations (B.56) write ( x ≡ x = ρ cos ϕ, 1

x2 ≡ y = ρ sin ϕ, x3 ≡ z = z,

(B.58)

while the inverse relations qj = qj (x1 , x2 , x3 ) (j = 1, 3) are p   q1 ≡ ρ = x2 + y 2 , y  q2 ≡ ϕ = arctan x , q3 ≡ z = z.

Fig.B.16 The spherical coordinates which define position of some point P are r, θ, ϕ (see Fig.B.16). The range of variation of these parameters is: n o D(spher) = (r, θ, ϕ) r ≥ 0, 0 ≤ θ ≤ π, 0 ≤ ϕ < 2π . 416

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Free ebooks ==> www.Ebook777.com Relations (B.56) then write: (

x1 ≡ x = r sin θ cos ϕ, x2 ≡ y = r sin θ sin ϕ, x3 ≡ z = r cos θ,

(B.59)

and the inverse relations qj = qj (x1 , x2 , x3 ) (j = 1, 3) are  p  q1 ≡ r = x2 + y 2 + z 2 ,   z  q2 ≡ θ = arccos p , x2 + y 2 + z 2     q2 ≡ ϕ = arctan y . x

In order to define a new coordinate system, parameters (q1 , q2 , q3 ) have to vary between certain limits, so that the set of points of space and the set of curvilinear coordinates qj = qj (x1 , x2 , x3 ) be in univocal correspondence. Let eˆ1 , eˆ2 and eˆ3 be the versors of the three vectors ~ei (i = 1, 3). In the new coordinate system, the radius vector ~r of some point P (~r) is given by ~r = ~r(q1 , q2 , q3 ).

(B.60)

∂~ r (i = 1, 3) are vectors tangent to the coordinate lines Since ∂q i qi (i = 1, 3), we can write

∂~r = ~ei = hi eˆi ∂qi

(i = 1, 3),

(B.61)

where we denoted s 2  2  2 ∂~r ∂x1 ∂x2 ∂x3 hi = + + = ∂qi ∂qi ∂qi ∂qi

(i = 1, 3). (B.62)

These quantities are called Lam´e coefficients. They stand for the moduli of the vectors ~ei , tangent to the coordinate lines qi . Let us also consider the versors ~n1 , ~n2 , and ~n3 oriented along normals to the surfaces q1 = q1o = const., q2 = q2o = const., and q3 = q3o = const., respectively. Since vectors grad qi are also normal to surfaces qi = qio = const., denoting by gi the moduli of vectors gradqi , we can write grad qi = gi~ni

(no summation;

i = 1, 3).

417

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(B.63)

Free ebooks ==> www.Ebook777.com Since grad qi =

∂qi ∂qi ∂qi ∂qi~ ∂qi ~ ∂qi ~ eˆ1 + eˆ2 + eˆ3 ≡ i+ j+ k, ∂x1 ∂x2 ∂x3 ∂x ∂y ∂z

we have:



2  2  2 ∂qi ∂qi ∂qi = + + ∂x1 ∂x2 ∂x3 2  2  2  ∂qi ∂qi ∂qi + + . = ∂x ∂y ∂z

gi2

(B.64)

The quantities gi (i = 1, 3) are called differential parameters of the first order. Definition. We say that vectors ~a1 , ~a2 and ~a3 , on the one side, and vectors ~b1 , ~b2 and ~b3 , on the other, form a system of reciprocal vectors with respect to each other, if the following relations are fulfilled: ~ai · ~bi = 1

(no summation;

~ai · ~bj = 0

i = 1, 3),

(i = 1, 3; i 6= j).

∂~ r Let us show that gradqi (i = 1, 3) and ∂q (i = 1, 3) form a system of i reciprocal vectors. To this end, we must prove that

gradqi ·

∂~r =1 ∂qi

and gradqi ·

(no summation;

∂~r =0 ∂qj

i = 1, 3),

(i = 1, 3; i 6= j).

Taking the dot product between vector relation d~r =

∂~r ∂~r ∂~r dq1 + dq2 + dq3 ∂q1 ∂q2 ∂q3

and gradqi , we have

=



gradqi ·

∂~r ∂q1



dqi = gradqi · d~r     ∂~r ∂~r dq1 + gradqi · dq2 + gradqi · dq3 ∂q2 ∂q3

and, since dq1 , dq2 and dq3 are independent, it follows that only the coefficient of dqi (i = 1, 3) equals 1, the other two coefficients being zero, which completes the proof. 418

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Free ebooks ==> www.Ebook777.com In case of a Cartesian system of coordinates, versors eˆ1 ≡ ~i, eˆ2 ≡ ~j, and eˆ3 ≡ ~k are orthogonal two-by-two. If the origin O of the frame ”travels” through the whole space, each of them form a system of equipolent vectors. Unlike Cartesian coordinates, in case of curvilinear orthogonal coordinates the versors eˆ1 , eˆ2 , and eˆ3 form a right-handed orthogonal system at any point P in space, but the versors associated to some other point P ′ 6= P are not equipolent to those with origin at P . In addition, vectors ~e1 , ~e2 and ~e3 do not keep their magnitude when the point P ”travels” throughout the space. Finally, in case of any other curvilinear coordinates both vector systems ~e1 , ~e2 , ~e3 , and ~n1 , ~n2 , ~n3 have any orientation (but remain, nevertheless, linearly independent). Coordinate lines of the orthogonal curvilinear coordinates are orthogonal two-by-two at any point P of space. It then follows that, in case of orthogonal curvilinear coordinates, the versors ~n1 , ~n2 and ~n3 coincide with eˆ1 , eˆ2 , and eˆ3 , respectively, being orthogonal two-by-two. Therefore, we have: ~ni = eˆi ;

eˆi · eˆj = 0;

~ni · ~nj = 0

(i 6= j).

Following these relations, in order to be orthogonal, it is necessary and sufficient for a system of curvilinear coordinates to satisfy ∂~r ∂~r ∂x ∂x ∂y ∂y ∂z ∂z · = · + · + · =0 ∂qi ∂qj ∂qi ∂qj ∂qi ∂qj ∂qi ∂qj

(i 6= j), (B.65)

or, gradqi ·gradqj =

∂x ∂x ∂y ∂y ∂z ∂z · + · + · =0 ∂qi ∂qj ∂qi ∂qj ∂qi ∂qj

(i 6= j). (B.65′ )

If the curvilinear coordinates are orthogonal, one can find a relationship between Lam´e’s coefficients hi , and the first-order differential parameters gi . Using relations grad qi = gi eˆi (no summation), and ∂~ r ˆi (no summation), and reciprocal relations as well, we ob∂qi = hi e tain: ∂~r grad qi · = gi eˆi · hi eˆi = gi hi = 1, ∂qi so that gi =

1 hi

(i = 1, 2, 3),

and, in particular, grad qi =

eˆi hi

(no summation; i = 1, 3). 419

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(B.66)

Free ebooks ==> www.Ebook777.com Line (arc) element in orthogonal curvilinear coordinates In a system of orthogonal curvilinear coordinates we can write d~r =

∂~r ∂~r ∂~r dq1 + dq2 + dq3 ∂q1 ∂q2 ∂q3

= h1 dq1 eˆ1 + h2 dq2 eˆ2 + h3 dq3 eˆ3 .

(B.67)

The components |d~r1 | ≡ |d~s1 | = |h1 dq1 eˆ1 | = ds1 = h1 dq1 , ds2 = h2 dq2 , and ds3 = h3 dq3 of the vector d~r along directions defined by eˆ1 , eˆ2 and eˆ3 are called curvilinear components of d~r. Squaring (B.67), we obtain d~r2 = d~r · d~r ≡ ds2 = h21 (dq1 )2 + h22 (dq2 )2 + h23 (dq3 )2 , or, by convention, ds2 = h21 dq12 + h22 dq22 + h23 dq32 .

(B.68)

This is the squared line element (called metric) in orthogonal curvilinear coordinates. Surface element in orthogonal curvilinear coordinates The surface element in curvilinear coordinates can be obtained by using the definition of the surface element in Cartesian coordinates. As we know, from geometric point of view, the cross product of two vectors ~a and ~b is a (pseudo)vector whose direction is given by the right-hand screw rule, and magnitude equal to the area of the parallelogram that the vectors span (see Fig.B.17).

Fig.B.17 420

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Free ebooks ==> www.Ebook777.com In view of these remarks, a vector surface element in orthogonal curvilinear coordinates is given by the cross product of two line elements, also considered in curvilinear coordinates. For a better understanding, let us first consider a Cartesian coordinate system. In such a ~x = dydz~i, dS ~y = system, there exists three vector surface elements: dS ~z = dxdy~k, given by cross product of three line elements: dzdx~j, and dS d~rx = dx~i, d~ry = dy ~j, and d~rz = dz ~k. More precisely, the vector sur~x = d~ry × d~rz = dy dz(~j × ~k) = dy dz ~i is an elementary face element dS vector whose magnitude equals the area of the rectangle determined by d~ry and d~rz , its direction being orthogonal to the area dy dz, and sense given by the right-hand screw rule. The other two vector surface elements are obtained in a similar manner (see Fig.B.18): ~y = d~rz × d~rx = dz dx(~k × ~i) = dzdx~j, dS and

~z = d~rx × d~ry = dx dy(~i × ~j) = dxdy~k. dS

Fig.B.18 The vector surface element therefore is ~ = dS ~ x + dS ~ y + dS ~z = dy dz ~i + dz dx ~j + dx dy ~k. dS 421

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Free ebooks ==> www.Ebook777.com By analogy with Cartesian coordinates, the surface element constructed on line elements d~ri = hi dqi eˆi and d~rj = hj dqj eˆj (no summation) is dSeij = d~ri × d~rj = hi hj dqi dqj (ˆ ei × eˆj ),

where we have used symbol ∼ over S to display the tensor character of cross product of the two line elements. The vector associated with the antisymmetric (pseudo)tensor dSeij is [see (B.91)]: ~i = 1 εijk dSejk = 1 εijk d~rj × d~rk ≡ 1 εijk d~sj × d~sk . dS 2 2 2 For example, taking i = 1, we have: ~ 1 ≡ dS ~q = 1 ε1jk dSejk dS 1 2  1 = ε123 d~r2 × d~r3 + ε132 d~r3 × d~r2 2  1 = h2 h3 dq2 dq3 eˆ2 × eˆ3 − h2 h3 dq2 dq3 eˆ3 × eˆ2 2 = h2 h3 dq2 dq3 eˆ2 × eˆ3 = h2 h3 dq2 dq3 eˆ1 . (B.69)

Proceeding in the same manner, the other two vector surface elements ~2 = h3 h1 dq3 dq1 eˆ2 and dS ~3 = h1 h2 dq1 dq2 eˆ3 are obtained. The vector dS surface element in orthogonal curvilinear coordinates then writes ~ = h2 h3 dq2 dq3 eˆ1 + h3 h1 dq3 dq1 eˆ2 + h1 h2 dq1 dq2 eˆ3 . dS (B.70) In the above relation hi hj dqi dqk is the vector surface element normal to the ”plane” qi P qj (i, j = 1, 3), its orientation being given by versor eˆk (k = 1, 3). Volume element in orthogonal curvilinear coordinates The volume element in orthogonal curvilinear coordinates can be obtained by generalizing the volume element written in Cartesian coordinates,   dV = d~r1 · (d~r2 × d~r3 ) = dxdydz ~i · ~j × ~k = dxdydz, taking into account the geometric interpretation of the mixed product of three vectors. Consider two infinitely near points M (q1 , q2 , q3 ) and N (q1 +dq1 , q2 +dq2 , q3 + dq3 ). Tracing the three coordinate surfaces passing through M , and the three passing through N , one obtains an elementary curvi∂~ r ∂~ r ∂~ r linear parallelepiped, with edges ∂q dq1 , ∂q dq2 , and ∂q dq3 . The vol1 2 3 ume of this elementary parallelepiped is given by the mixed product   dV = d~s1 · (d~s2 × d~s3 ) = h1 h2 h3 dq1 dq2 dq3 eˆ1 · eˆ2 × eˆ3 = h1 h2 h3 dq1 dq2 dq3 . 422

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(B.71)

Free ebooks ==> www.Ebook777.com Examples of orthogonal curvilinear coordinates 1. Cylindrical coordinates Let us apply the previous general considerations to cylindrical coordinates, and calculate Lam´e’s coefficients, the arc element, the metric, the surface and volume elements. Since  x = ρ cos ϕ, y = ρ sin ϕ, z = z, q1 = ρ, q2 = ϕ, q3 = z, the radius vector ~r of some point P of the space writes ~r = x~i + y~j + z~k = ρ cos ϕ~i + ρ sin ϕ ~j + z~k, so that

∂~r = cos ϕ~i + sin ϕ ~j, ∂ρ ∂~r = −ρ sin ϕ~i + ρ cos ϕ ~j, ∂ϕ ∂~r ~ = k. ∂z

One first observes that ∂~r ∂~r · = 0; ∂ρ ∂ϕ

∂~r ∂~r · = 0; ∂z ∂ϕ

∂~r ∂~r · = 0, ∂ρ ∂z

which, in agreement with (B.65), prove that our frame is an orthogonal coordinate system. Next, one can calculate Lam´e’s coefficients h1 = 1,

h2 = ρ,

h3 = 1,

the arc element d~r = dρˆ e1 + ρdϕˆ e2 + dzˆ e3 = dρ~uρ + ρdϕ~uϕ + dz~k, the metric ds2 = dρ2 + ρ2 dϕ2 + dz 2 , the vector surface element ~ = ρdϕdzˆ dS e1 + dρdzˆ e2 + ρdρdϕˆ e3 = ρdϕdz~uρ + dρdz~uϕ + ρdρdϕ~uz , 423

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Free ebooks ==> www.Ebook777.com and the volume element dV = ρdρdϕdz. Spherical coordinates Following the same steps as shown above, we have: x = r sin θ cos ϕ, q1 = r,

y = r sin θ sin ϕ, q2 = θ,

z = r cos θ,

q3 = ϕ,

~r = r sin θ cos ϕ~i + r sin θ sin ϕ~j + r cos θ~k, ∂~r = sin θ cos ϕ~i + sin θ sin ϕ~j + cos θ~k, ∂r ∂~r = r cos θ cos ϕ~i + r cos θ sin ϕ~j − r sin θ~k, ∂θ ∂~r = −r sin θ sin ϕ~i + r sin θ cos ϕ~j, ∂ϕ One observes that ∂~r ∂~r · = 0, ∂r ∂θ

∂~r ∂~r · = 0, ∂ϕ ∂θ

∂~r ∂~r · = 0, ∂r ∂ϕ

which, in agreement with (B.65), show that this is also a system of orthogonal curvilinear coordinates. Next, let us calculate: Lam´e’s coefficients h1 = 1, h2 = r, h3 = r sin θ, the arc element d~r ≡ d~s = drˆ e1 + rdθˆ e2 + r sin θdϕˆ e3 = dr~ur + rdθ~uθ + r sin θdϕ~uϕ , the metric ds2 = dr2 + r2 dθ2 + r2 sin2 θdϕ2 , the vector surface element ~ = r2 sin θdθdϕˆ dS e1 + r sin θdrdϕˆ e2 + rdrdθˆ e3 = r2 sin θdθdϕ~ur + r sin θdrdϕ~uθ + rdrdθ~uϕ , 424

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Free ebooks ==> www.Ebook777.com and the volume element dV = r2 sin θdrdθdϕ. Gradient, curl, divergence and Laplacian in orthogonal curvilinear coordinates (OCC) Gradient in OCC As well known, finding solution to a physical problem is considerably facilitated if one takes into account its symmetry. Since usually we have to do with applications involving spherical, cylindrical, polar plane, etc. symmetry, it is necessary to know the expressions of the differential operators (gradient, curl, divergence, Laplacian) in these coordinates. Let ϕ = ϕ(x, y, z) ≡ ϕ(~r) be a scalar field of class C 1 on a domain D ⊂ E3 . Then, if q1 , q2 , q3 are the system of OCC, we have: gradϕ(q1 , q2 , q3 ) =

∂ϕ ∂ϕ ∂ϕ gradq1 + gradq2 + gradq3 . ∂q1 ∂q2 ∂q3

Since gradqi =

eˆi hi

(no summation),

we can write grad ϕ(q1 , q2 , q3 ) =

1 ∂ϕ 1 ∂ϕ 1 ∂ϕ eˆ1 + eˆ2 + eˆ3 . h1 ∂q1 h2 ∂q2 h3 ∂q3

(B.72)

For example, the gradient in cylindrical coordinates writes grad φ(ρ, ϕ, z) =

∂φ 1 ∂φ ∂φ ~ ~uρ + ~uϕ + k, ∂ρ ρ ∂ϕ ∂z

(B.73)

while in spherical coordinates is grad φ(r, θ, ϕ) =

∂φ 1 ∂φ 1 ∂φ ~ur + ~uθ + ~uϕ . ∂r r ∂θ r sin θ ∂ϕ

(B.74)

Curl in OCC ~ = Vi eˆi = V1 eˆ1 + V2 eˆ2 + V3 eˆ3 of class Consider the vector field V 1 ~ . We have: C on the domain D ⊂ E3 , and let us calculate curl V ~ = curl(V1 eˆ1 + V2 eˆ2 + V3 eˆ3 ) = V1 curl eˆ1 + (gradV1 ) × eˆ1 curl V 425

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Free ebooks ==> www.Ebook777.com +V2 curl eˆ2 + (gradV2 ) × eˆ2 + V3 curl eˆ3 + (gradV3 ) × eˆ3 = Vi curl eˆi + (gradVi ) × eˆi .

(B.75)

To express curl eˆi we use the already known relation grad qi =

1 eˆi hi

(no summation;

i = 1, 3).

According to vector identity (B.53) curl(grad qi ) = 0

∀qi ,

and therefore curl(grad qi ) = curl 1 = curl eˆi + grad hi



1 hi





× eˆi = 0

1 eˆi hi



(no summation),

so that 1 curl eˆi = −grad hi But grad



1 hi





1 hi

=−



× eˆi

1 gradhi h2i

(no summation).

(B.76)

(no summation),

and (B.76) becomes 1 1 curl eˆi = 2 gradhi × eˆi hi hi

(no summation;

i = 1, 2, 3).

Introducing this relation into (B.75), we can write ~ = Vi curl eˆi + (gradVi ) × eˆi curlV Vi gradhi × eˆi + (gradVi ) × eˆi hi   Vi 1 = gradhi + gradVi × eˆi = grad(Vi hi ) × eˆi hi hi =

=

1 1 1 grad(V1 h1 ) × eˆ1 + grad(V2 h2 ) × eˆ2 + grad(V3 h3 ) × eˆ3 . h1 h2 h3 426

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Free ebooks ==> www.Ebook777.com Consequently, ~ = curlV For example:

1 grad(Vi hi ) × eˆi hi

(i = 1, 3).

~ )1 = curlV ~ · eˆ1 (curlV

(B.77)

 1 1 1 grad(V1 h1 ) × eˆ1 + grad(V2 h2 ) × eˆ2 + grad(V3 h3 ) × eˆ3 ·ˆ e1 = h1 h2 h3     1 1 grad(V2 h2 ) × eˆ2 · eˆ1 + grad(V3 h3 ) × eˆ3 · eˆ1 = h2 h3   V2 = gradV2 × eˆ2 + gradh2 × eˆ2 · eˆ1 h2   V3 + gradV3 × eˆ3 + gradh3 × eˆ3 · eˆ1 h3    1 ∂V2 1 ∂V2 1 ∂V2 = eˆ1 + eˆ2 + eˆ3 × eˆ2 · eˆ1 h1 ∂q1 h2 ∂q2 h3 ∂q3    V2 1 ∂h2 1 ∂h2 1 ∂h2 + eˆ1 + eˆ2 + eˆ3 × eˆ2 · eˆ1 h2 h1 ∂q1 h2 ∂q2 h3 ∂q3    1 ∂V3 1 ∂V3 1 ∂V3 + eˆ1 + eˆ2 + eˆ3 × eˆ3 · eˆ1 h1 ∂q1 h2 ∂q2 h3 ∂q3    V3 1 ∂h3 1 ∂h3 1 ∂h3 + eˆ1 + eˆ2 + eˆ3 × eˆ3 · eˆ1 h3 h1 ∂q1 h2 ∂q2 h3 ∂q3   V2 ∂h2 1 ∂V2 + (ˆ e3 × eˆ2 ) · eˆ1 = h3 ∂q3 h2 h3 ∂q3   1 ∂V3 V3 ∂h3 + + (ˆ e2 × eˆ3 ) · eˆ1 h2 ∂q2 h2 h3 ∂q2 

=−

1 ∂V2 V2 ∂h2 1 ∂V3 V3 ∂h3 − + + h3 ∂q3 h2 h3 ∂q3 h2 ∂q2 h2 h3 ∂q2   1 ∂ ∂ = (V3 h3 ) − (V2 h2 ) . h2 h3 ∂q2 ∂q3

The other two components are obtained in an analogous way:   1 ∂ ∂ ~ (curlV )2 = (V1 h1 ) − (V3 h3 ) , h3 h1 ∂q3 ∂q1 427

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Free ebooks ==> www.Ebook777.com and ~ )3 = (curlV

  1 ∂ ∂ (V2 h2 ) − (V1 h1 ) . h1 h2 ∂q1 ∂q2

Therefore, ~ = curlV

  ∂ ∂ 1 (V3 h3 ) − (V2 h2 ) eˆ1 h2 h3 ∂q2 ∂q3

  1 ∂ ∂ + (V1 h1 ) − (V3 h3 ) eˆ2 h3 h1 ∂q3 ∂q1   ∂ ∂ 1 (V2 h2 ) − (V1 h1 ) eˆ3 . + h1 h2 ∂q1 ∂q2 ~ writes For example, in cylindrical coordinates curlV ~ = curlV



∂Vϕ 1 ∂Vz − ρ ∂ϕ ∂z



~uρ +



∂Vz ∂Vρ − ∂z ∂ρ

  1 ∂(ρVϕ ) ∂Vρ ~ − k, + ρ ∂ρ ∂ϕ



~uϕ

while in spherical coordinates writes   ∂ ∂V 1 θ ~ = (Vϕ sin θ) − curlV ~ur r sin θ ∂θ ∂ϕ

   1 ∂Vr 1 ∂ 1 ∂ ∂Vr − (rVϕ ) ~uθ + (rVθ ) − + ~uϕ . r sin θ ∂ϕ r ∂r r ∂r ∂θ 

Divergence in OCC ~ = Vi eˆi = V1 eˆ1 + V2 eˆ2 + V3 eˆ3 of class Consider the vector field V 1 ~ is C on the domain D ⊂ E3 . The divergence of V ~ = div(Vi eˆi ) = div(V1 eˆ1 + V2 eˆ2 + V3 eˆ3 ) divV = V1 divˆ e1 +(gradV1 )·ˆ e1 +V2 divˆ e2 +(gradV2 )·ˆ e2 +V3 divˆ e3 +(gradV3 )·ˆ e3 = Vi divˆ ei + (gradVi ) · eˆi .

(B.78)

To calculate divˆ ei (i = 1, 3) we take into account that: eˆ1 = eˆ2 × eˆ3 , eˆ2 = eˆ3 × eˆ1 , eˆ3 = eˆ1 × eˆ2 . For i = 1, we have: div eˆ1 = div(ˆ e2 × eˆ3 ) = eˆ3 · curlˆ e2 − eˆ2 · curlˆ e3 428

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Free ebooks ==> www.Ebook777.com 1 1 gradh2 × eˆ2 − eˆ2 · gradh3 × eˆ3 h2 h3     gradh3 grad(h2 h3 ) gradh2 + = eˆ1 · . = eˆ2 × eˆ3 · h2 h3 h2 h3 = eˆ3 ·

By circular permutations, we also have: divˆ e2 = eˆ2 ·



grad(h3 h1 ) h3 h1



,

divˆ e3 = eˆ3 ·



grad(h1 h2 ) h1 h2



.

and

Introducing these results into (B.78), we obtain: ~ = V1 eˆ1 · divV

grad(h3 h1 ) grad(h2 h3 ) + gradV1 · eˆ1 + V2 eˆ2 · h2 h3 h3 h1

grad(h1 h2 ) + gradV3 · eˆ3 h1 h2     grad(h3 h1 ) grad(h2 h3 ) + gradV1 + eˆ2 · V2 + gradV2 = eˆ1 · V1 h2 h3 h3 h1   grad(h1 h2 ) +ˆ e 3 · V3 + gradV3 h1 h2 +gradV2 · eˆ2 + V3 eˆ3 ·

grad(V1 h2 h3 ) grad(V2 h3 h1 ) grad(V3 h1 h2 ) + eˆ2 · + eˆ3 · h2 h3 h3 h1 h1 h2 " 1 = grad(V1 h2 h3 ) · h1 eˆ1 + grad(V2 h3 h1 ) · h2 eˆ2 h1 h2 h3 #

= eˆ1 ·

+grad(V3 h1 h2 ) · h3 eˆ3 .

(B.79)

Since gradϕ · h1 eˆ1 = h1 eˆ1 · = h1 eˆ1 ·





∂ϕ ∂ϕ ∂ϕ gradq1 + gradq2 + gradq3 ∂q1 ∂q2 ∂q3

∂ϕ eˆ1 ∂ϕ eˆ2 ∂ϕ eˆ3 + + ∂q1 h1 ∂q2 h2 ∂q3 h3



=

∂ϕ , ∂q1

429

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Free ebooks ==> www.Ebook777.com and, analogously,

∂ϕ , ∂q2 ∂ϕ , gradϕ · h3 eˆ3 = ∂q3 it follows from (B.79) that   1 ∂ ∂ ∂ ~ divV = (V1 h2 h3 ) + (V2 h3 h1 ) + (V3 h1 h2 ) . h1 h2 h3 ∂q1 ∂q2 ∂q3 (B.80) For example, the divergence in cylindrical coordinates is ~ = 1 ∂ (ρVρ ) + 1 ∂Vϕ + ∂Vz , divV ρ ∂ρ ρ ∂ϕ ∂z whereas in spherical coordinates is written as   ∂ 2 ∂ ∂ 1 ~ (r sin θ Vr ) + (r sin θ Vθ ) + (r Vϕ ) . divV = 2 r sin θ ∂r ∂θ ∂ϕ gradϕ · h2 eˆ2 =

Laplacian in OCC Following the definition, we can write: ∆Φ = div(gradΦ).

(B.81)

Using (B.72), (B.80), and (B.81), we have: "     1 ∂ h2 h3 ∂Φ ∂ h3 h1 ∂Φ ∆Φ = + h1 h2 h3 ∂q1 h1 ∂q1 ∂q2 h2 ∂q2 #   h1 h2 ∂Φ ∂ + . (B.82) ∂q3 h3 ∂q3 For example, in cylindrical coordinates the Laplacian writes        1 ∂Φ ∂Φ ∂ ∂ ∂Φ 1 ∂ ρ + + ρ ∆Φ = ρ ∂ρ ∂ρ ∂ϕ ρ ∂ϕ ∂z ∂z   2 2 1 ∂ ∂Φ 1 ∂ Φ ∂ Φ = + , ρ + 2 ρ ∂ρ ∂ρ ρ ∂ϕ2 ∂z 2 while in spherical coordinates is expressed as "     1 ∂ ∂Φ ∂ ∂Φ 2 ∆Φ = 2 r sin θ + sin θ r sin θ ∂r ∂r ∂θ ∂θ #   ∂ 1 ∂Φ + ∂ϕ sin θ ∂ϕ     1 ∂ 1 ∂ ∂Φ 1 ∂2Φ 2 ∂Φ = 2 r + 2 sin θ + 2 2 . r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂ϕ2 430

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Free ebooks ==> www.Ebook777.com B.8. Some applications of vector analysis Exercise 1 Let ~r(x, y, z) be the radius vector of the point P , with respect to the origin of the curl ~r, div ~r,  Cartesian frame Oxyz. Calculate:  grad r, grad 1r r6=0 , ∆ 1r r6=0 , and ∆ 1r .

Solution

curl ~r = ~ui εijk ∂j xk = ~ui εijk δjk = 0; ∂xi = 3; ∂xi ∂r √ ∂r = ~ui xi xi grad r = ~ui ∂xi ∂xi ∂xk xk δki xi ~ui ~r 1 2xk = ~ui = = = ~ur ; = ~ui √ 2 xk xk ∂xi r r r     1 ∂ 1 1 ∂r grad = ~ui = −~ui 2 r r6=0 ∂xi r r ∂xi div~r =

1 ∂ √ 1 xk δki ~r xi ~ui x x = −~ u = − ; = − k k i r2 ∂xi r2 r r3 r3        1 1 ~r ∆ = div grad = −div r r6=0 r r3     1 1 1 ∂r3 3 = − 3 div~r − ~r · ∇ ~ui = − 3 − ~r · − 6 r r3 r r ∂xi

= −~ui

3 ~r · ~ui ∂ 3 ~r · ~ui 3 ∂xk + 6 (xk xk )3/2 = − 3 + 6 (xk xk )1/2 2xk 3 r r ∂xi r r 2 ∂xi 3 3 = − 3 + 5 ~r · ~r = 0. r r  1 To determine ∆ r for any value of r, that is including value r = 0, let us appeal to some simple knowledge of electrostatics. As well-known, Poisson’s equation for an electric charge of volume density ρ(~r), situated in vacuum, writes =−

∆V (~r) = −

1 ρ(~r). ε0

(B.83)

In case of a point charge distribution Q, whose position is defined by the radius vector ~r0 , we can write ρ(~r) = Q δ(~r − ~r0 ), 431

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(B.84)

Free ebooks ==> www.Ebook777.com where δ(~r − ~r0 ) = δ(x − x0 )δ(y − y0 )δ(z − z0 ) is the delta Dirac distribution. Since in this case V (~r) =

Q 1 , 4πε0 |~r − ~r0 |

(B.85)

by substituting (B.84) and (B.85) into (B.83), we have: ∆



Q 1 4πε0 |~r − ~r0 |

or ∆





1 |~r − ~r0 |

=− 

ρ(~r) Q = − δ(~r − ~r0 ), ε0 ε0

= −4π δ(~r − ~r0 ).

If the electric point charge is situated at the origin of the coordinate system (~r0 = 0), the desired equation follows immediately:   1 ∆ = −4πδ(r). r Exercise 2 ~ is a constant vector, calculate: div(A ~ × ~r), curl(A ~ × ~r), and If A ~ · ~r). grad(A Solution According to the vector identity (C.49), we have

~ × ~r) = ~r · curlA ~−A ~ · curl~r = ~r · 0 − A ~ · 0 = 0. div(A Using (B.51), we can write ~ × ~r) = A ~ div ~r − ~r divA ~ + (~r · ∇)A ~ − (A ~ · ∇)~r curl(A ~ div~r − (A ~ · ∇)~r = 3A ~−A ~ = 2A. ~ =A Finally, by means of (B.52), we have ~ r) = A×(∇×~ ~ ~ ~ ~ = (A·∇)~ ~ ~ grad(A·~ r)+~r ×(∇×A)+( A·∇)~ r +(~r ·∇)A r = A. Exercise 3 If ~a is a constant vector, and ~r the radius vector of some point in the Cartesian frame Oxyz, calculate: ~ = ~a × (~r × ~a); (a) divergence and curl of the vector V 432

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Free ebooks ==> www.Ebook777.com ~ (b) divergence and curl of the vector V ~ (c) divergence and curl of the vector V ~ (d) divergence and curl of the vector V ~ (e) divergence and curl of the vector V

= ~r × (~a × ~r); = (~a × ~r) + (~a · ~r)~a; r = ~ar×~ 2 ; = grad r.

Solution (a) We have:   ~ = div ~a × (~r × ~a) = (~r × ~a) · curl~a − ~a · curl(~r × ~a) divV

and

 = −~a · curl(~r × ~a) = −~a · ~r div~a − ~a div ~r  +(~a · ∇)~r − (~r · ∇)~a = −~a · (−3~a + ~a) = 2a2 ,

  ~ = curl ~a × (~r × ~a) curl V   = ~a div(~r × ~a) − (~r × ~a) div ~a + (~r × ~a) · ∇ ~a − (~a · ∇)(~r × ~a) = −(~a · ∇)(~r × ~a) = −al ∂l (εijk xj ak )~ui = −εijk al ak δlj ~ui  1 εijk aj ak + εijk aj ak ~ui 2  1 = − εijk aj ak + εikj ak aj ~ui 2  1 = − εijk aj ak − εijk aj ak ~ui = 0. 2

= −εijk aj ak ~ui = −

(b) In an analogous way, we can write:

~ = div[~r × (~a × ~r)] = (~a × ~r) · curl ~r − ~r · curl(~a × ~r) div V = −~r · (2~a) = −2~r · ~a, and

  ~ = curl ~r × (~a × ~r) = ~r div(~a × ~r) curl V   −(~a × ~r) div~r + (~a × ~r) · ∇ ~r − (~r · ∇)(~a × ~r)

= −3(~a × ~r) + (~a × ~r)i ∂i (xk ~uk ) − xj ∂j (~ui εilm al xm ) = −3(~a × ~r) + (~a × ~r)i ~uk δik − ~ui εilm xj al δjm = −3(~a × ~r) + (~a × ~r) − (~a × ~r) = −3(~a × ~r). 433

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Free ebooks ==> www.Ebook777.com ~ = (~a × ~r) + (~a · ~r)~a, we have: (c) For V

  ~ = div (~a × ~r) + div (~a · ~r)~a div V   = div (~a · ~r)~a

= (~a · ~r)div ~a + ~a · grad(~a · ~r) = ~a · ~a = ~a2 , and

  ~ = curl (~a × ~r) + curl (~a · ~r)~a curl V

= 2~a + (~a · ~r)curl ~a + grad(~a · ~r) × ~a = 2~a. r ~ = ~a×~ (d) Let us now calculate divergence and curl of V r 2 , with ~a = const. We have:     1 ~a × ~r 1 ~ div V = div = 2 div(~a × ~r) + (~a × ~r) · grad 2 r r r2



   1 ∂ = (~a × ~r) · grad = (~a × ~r) · ~ui ∂xi r2   2 1 ∂xk 2xk = −(~a × ~r) · ~ui 3 r 2r ∂xi

=− and

1 r2



  2 2 (δ x ~ u ) · (~ a × ~ r ) = − ~ r · (~ a × ~ r ) = 0, ik k i r4 r4

    ~a × ~r 1 1 ~ curl V = curl = 2 curl(~a × ~r) + grad × (~a × ~r) 2 r r r2 =

 2~a   2 2 2 2 2~a − ~ r × (~ a × ~ r ) = − r ~ a − (~ a · ~ r )~ r = (~ a · ~ r )~ r . r2 r4 r2 r4 r4

~ = grad r, we have: (e) Finally, for V

    ~r 1 1 ~ divV = div(gradr) = div = div ~r + ~r · grad r r r   3 ~r 3 1 2 = + ~r · − 3 = − = , r r r r r

and

~ = curl(gradr) = 0. curlV 434

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Free ebooks ==> www.Ebook777.com Exercise 4 Given the vector field w ~ with properties div w ~ = 0, and curl w ~ = 0, ~ show that E = grad(~r · w) ~ + curl(~r × w) ~ +w ~ = 0. Solution Using vector identities (B.50) and (B.51), we have: ~ = grad(~r · w) E ~ + curl(~r × w) ~ +w ~ = ~r × curl w ~ +w ~ × curl ~r + (~r · ∇)w ~ + (w ~ · ∇)~r +~r div w ~ −w ~ div ~r + (w ~ · ∇)~r − (~r · ∇)w ~ +w ~ = 2(w ~ · ∇)~r − 2w ~ = 2wi ∂i (xk ~uk ) − 2w ~ = 2w ~ − 2w ~ = 0. Exercise 5 Given the vector field ~v with properties div ~v = ~a2 , and curl ~v = 2~a, where ~a is a constant vector, show that L = div(~v × curl ~v ) − 4div ~v = 0. Solution In view of vector identity (B.49), we have: L = div(~v × curl~v ) − 4div ~v = curl~v · curl ~v − ~v · curl(curl ~v ) − 4div ~v = 4~a · ~a − 2~v · curl ~a − 4~a2 = 0. B.9. Isomorphism between the set of second rank antisymmetric tensors and the set of vectors, defined on the same space E3 Consider a second rank covariant tensor Aij (i, j = 1, 3), defined on the three-dimensional Euclidean space E3 . In general, a tensor has nm components. Here n represents the number of dimensions of the space, and m = p + q indicates the tensor rank, where p stands for the number of covariant indices, and q for the number of contravariant indices. In our case, q = 0, m = p = 2, and n = 3, so that we have 32 = 9 components. In matrix form, the tensor Aij writes   A A A 11 12 13  Aij =  A21 A22 A23  . (B.86) A31 A32 A33 If the tensor Aij is antisymmetric Aij = −Aji

∀i, j = 1, 3,

435

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(B.87)

Free ebooks ==> www.Ebook777.com the elements disposed on the principal diagonal are zero, and we are left with   0 A12 A13  Aij =  A21 0 A23  . (B.88) A31 A32 0

In view of (B.87), this relation can be written in eight equivalent ways, as follows:

Aij



1



0  = −A12 −A13 

A12 0 −A23

 A13 A23  ; 0

 0 −A A 21 13  Aij 2 =  A21 0 A23  ; −A13 −A23 0   0 A12 −A31  Aij 3 =  −A12 0 A23  ; A31 −A23 0   0 A A 12 13  Aij 4 =  −A12 0 −A32  ; −A13 A32 0   0 −A −A 21 31  Aij 5 =  A21 0 A23  ; A31 −A23 0   0 −A21 A13  Aij 6 =  A21 0 −A32  ; −A13 A32 0   0 A −A 12 31  Aij 7 =  −A12 0 −A32  ; A31 A32 0   0 −A −A 21 31  Aij 8 =  A21 0 −A32  . A31 A32 0

These relations show that a second-rank antisymmetric tensor, defined on a three dimensional space, has only three essentially distinct components. It can be shown that, in general, a p-times covariant and 436

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Free ebooks ==> www.Ebook777.com q-times contravariant tensor, defined on a n-dimensional space, antisymmetric in a indices, has Cna np+q−a essentially distinct components. In our case, C32 32+0−2 = C32 = 3.2 2 = 3. A similar relation can be found for a tensor symmetric in s indices, in which case the number of essentially distinct components is s C n np+q−s , where s

Cn =

n(n + 1)(n + 2)...(n + s − 1) s!

is the number of combinations with repetition of n elements taken s times. Coming now back, out of eight possibilities of choosing Aij we shall consider the third, that is

Aij



3



0  = −A12 A31

A12 0 −A23

 −A31 A23  . 0

(B.89)

This choice was made on purpose, because according to (B.89) this is the only possibility of having as distinct components A12 , A23 , and A31 . Since in a three-dimensional space vectors have three components, one can consider the three essentially distinct components of a second rank antisymmetric tensor as the components of a vector in this space, which are: A12 → v3 , v1 = A23 , A23 → v1 , ⇒ v2 = A31 , (B.90) A31 → v2 , v3 = A12 . This way, the choice of indices agrees with the cyclic permutation of indices 1, 2 and 3. In one relation, correspondences shown by (B.90) can be written as 1 (B.91) vi = εijk Ajk , 2 where εijk (i, j, k = 1, 3) is the Levi-Civita symbol. This relation can be easily verified for any group of indices. Conversely, we can ask whether the reciprocal correspondence is valid, that is if the three distinct components vi (i = 1, 3) of a vector in a three-dimensional space could stand for the three essentially distinct components of a second rank antisymmetric tensor on the same space. The answer is affirmative and the correspondence writes 437

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Free ebooks ==> www.Ebook777.com v1 → A23 , v2 → A31 , v3 → A12 ,



A12 = v3 , A23 = v1 , A31 = v2 .

(B.92)

These correspondences can also be written in a single relation as Aij = εijk vk .

(B.93)

The two relations (B.91) and (B.93) express the isomorphism between the set of second rank antisymmetric tensors and the set of vectors on E3 . In fact, for the sake of rigurosity, to a second rank antisymmetric tensor, defined on E3 , one can associate a pseudovector (axial vector) and conversely, to an ordinary vector one can associate a second rank pseudotensor.

438

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Free ebooks ==> www.Ebook777.com REFERENCES

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Free ebooks ==> www.Ebook777.com 16. Cronin, J.A., Greenberg, D.F., Telegdi, V.L.: University of Chicago Graduate Problems in Physics with Solutions, AddisonWesley, Reading, Mass. (1967) 17. de Lange, O., Pierrus, J.: Solved Problems in Classical Mechanics: Analytical and Numerical Solutions with Comments, Oxford University Press, USA (2010) 18. Fassano, A., Marmi, S.: Analytical Mechanics: An Introduction, Oxford University Press (2006) 19. Forray, M.J.: Variational Calculus in Science and Engineering, McGraw-Hill, New-York (1968) 20. Fox, W.R.: Introduction to Fluid Mechanics, Wiley, NewYork (1973) 21. Fox, R.W., McDonald, A.T., Pritchard, P.J.: Introduction to Fluid Mechanics, Wiley, New-York (2009) 22. Gantmacher, F.: Lectures in Analytical Mechanics, Mir, Moskow (1975) 23. Gignoux, C., Silvestre-Brac, B.: Solved Problems in Lagrangian and Hamiltonian Mechanics, Grenoble Sciences, Springer (2009) 24. Goldstein, H.: Classical Mechanics, Addison-Wesley Publishing Co. Mass., 2nd edn. (1980) 25. Grechko, L.G., Sugakov, V.I., Tomasevich, O.F., Fedorchenko, A.M.: Problems in Theoretical Physics, Mir Publishers, Moscow (1977) 26. Greenwood, D.T.: Principles of Dynamics, 2nd edn., Englewood Cliffs (1988), Dover, New-York (1997) 27. Hand, L.N., Finch, J.D.: Analytical Mechanics, Cambridge University Press (1998) 28. Ionescu-Pallas, N.: Introduction to Modern Theoretical Mechanics, Academy Printing House, Bucharest (1969) 29. Irodov, I., Saveliev, I., Zamcha, O.: Recueil de probl`emes de ´ physique g´en´erale, Editions Mir, Moscou (1976) 30. Jeffery, D.J.: Classical Mechanics Problems, Portpentagram Publishing (self-published), New Mexico (2001) 31. Johns, O.D.: Analytical Mechanics for Relativity and Quantum Mechanics, Oxford University Press (2005) 32. Kibble, T.W.: Classical Mechanics, McGraw-Hill, London (1973), Addison-Wesley Longman (1986) 33. Kittel, C., Kight, W.D., Ruderman, M.A.: Mechanics, Berkeley Physics Course, vol. I, McGraw-Hill, New-York (1973) 34. Kotkin, L.G., Serbo, V.G.: Collection of Problems in Classical Mechanics, Pergamon Press, Oxford, 1st edn. (1971) 440

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Free ebooks ==> www.Ebook777.com 35. Lagrange, J.L.: Analytical Mechanics, Kluwer Academic Publishers (2010) 36. Landau, L.D., Lifshitz, E.M.: Fluid Mechanics, 2nd edn. Pergamon (1987) 37. Landau, L.D., Lifshitz, E.M.: The Classical Theory of Fields, Pergamon (1987) 38. Landau, L.D., Lifshitz, E.M.: Mechanics, 3rd edn., Pergamon Press (1976) 39. Lurie, A.I.: Analytical Mechanics, Springer-Verlag, Berlin Heidelberg (2002) 40. Marion, J.: Classical Dynamics, Academic Press Inc., NewYork (1965) 41. Merches, I., Burlacu, L.: Applied Analytical Mechanics, The Voice of Bucovina Press, Iasi (1995) 42. Meshcherskii, I.V.: A Collection of Problems of Mechanics, Pergamon Press, Oxford (1965) 43. Moore, E.L.: Theoretical Mechanics, Wiley (1983) 44. Morin, D.: Introduction to Classical Mechanics with Problems and Solutions, Cambridge University Press (2008) 45. Papastavridis, J.G.: Analytical Mechanics: A Comprehensive Treatise on the Dynamics of Constrained Systems for Engineers, Physicists and Mathematicians, Oxford University Press (2002) 46. Pellegrini, C., Cooper, R.K.: Modern Analytical Mechanics, Springer-Verlag, Berlin Heidelberg (1999) 47. Pollard, H.: Mathematical Introduction to Celestial Mechanics, Englewood Cliffs, N.J.: PrenticeHall, Inc. (1966) 48. Rossberg, K.: A First Course in Analytical Mechanics, Wiley, New-York (1983) 49. Saletan, E.J., Cromer, A.H.: Theoretical Mechanics, John Wiley, New-York (1971) 50. Serrin, J.: Mathematical Principles of Classical Fluid Mechanics, Handbuch der Physik, Springer-Verlag, Berlin, vol. VIII/1, Fluid Dynamics I (1959) 51. Spiegel, M.R.: Theory and Problems of Theoretical Mechanics, McGraw-Hill, New-York (1967) 52. Ter Haar, D.: Elements of Hamiltonian Mechanics, NorthHolland Publishing Co., Amsterdam (1964) 53. Tomasevich, O.F.: A Collection of Problems in Theoretical Physics, Kiev University Press, Kiev (1958) 54. Torok, J.S.: Analytical Mechanics: With an Introduction to Dynamical Systems, Wiley-Interscience (1999) 441

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Free ebooks ==> www.Ebook777.com 55. Vladimirov, V., Mikhailov, V., Chabounine, M., Karimova, Kh., Sidorov, Y., Vacharine, A.: Recueil de probl`emes d’´equations de ´ physique math´ematique, Editions Mir, Moscou (1976) 56. Wess, J.: Theoretische Mechanik, Springer-Verlag, Berlin Heildelberg (2007) 57. Ziwet, A., Field, P.: Introduction to Analytical Mechanics, Scholarly Publishing Office, University of Michigan (2005)

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