The Fundamentals of Newtonian Mechanics: For an Introductory Approach to Modern Physics [2192-4805, 1 ed.] 9783031472886, 9783031472893

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Undergraduate Lecture Notes in Physics

Maurizio Spurio

The Fundamentals of Newtonian Mechanics For an Introductory Approach to Modern Physics

Undergraduate Lecture Notes in Physics Series Editors Neil Ashby, University of Colorado, Boulder, CO, USA William Brantley, Department of Physics, Furman University, Greenville, SC, USA Matthew Deady, Physics Program, Bard College, Annandale-on-Hudson, NY, USA Michael Fowler, Department of Physics, University of Virginia, Charlottesville, VA, USA Morten Hjorth-Jensen, Department of Physics, University of Oslo, Oslo, Norway Michael Inglis, Department of Physical Sciences, SUNY Suffolk County Community College, Selden, NY, USA Barry Luokkala , Department of Physics, Carnegie Mellon University, Pittsburgh, PA, USA

Undergraduate Lecture Notes in Physics (ULNP) publishes authoritative texts covering topics throughout pure and applied physics. Each title in the series is suitable as a basis for undergraduate instruction, typically containing practice problems, worked examples, chapter summaries, and suggestions for further reading. ULNP titles must provide at least one of the following: • An exceptionally clear and concise treatment of a standard undergraduate subject. • A solid undergraduate-level introduction to a graduate, advanced, or non-standard subject. • A novel perspective or an unusual approach to teaching a subject. ULNP especially encourages new, original, and idiosyncratic approaches to physics teaching at the undergraduate level. The purpose of ULNP is to provide intriguing, absorbing books that will continue to be the reader’s preferred reference throughout their academic career.

Maurizio Spurio

The Fundamentals of Newtonian Mechanics For an Introductory Approach to Modern Physics

Maurizio Spurio Department of Physics and Astronomy University of Bologna Bologna, Italy

ISSN 2192-4791 ISSN 2192-4805 (electronic) Undergraduate Lecture Notes in Physics ISBN 978-3-031-47288-6 ISBN 978-3-031-47289-3 (eBook) https://doi.org/10.1007/978-3-031-47289-3 Translation from the Italian language edition: “Meccanica Newtoniana per un approccio propedeutico alla fisica moderna” by Maurizio Spurio, © 2023. Published by Edises edizioni. All Rights Reserved. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland Paper in this product is recyclable.

Non ho delle pretese, il merito l’è tutto della scuola bolognese! (G. Puccini: Gianni Schicchi, libretto di G. Forzano)

As a thank to the greatest teacher I have met, Prof. Attilio Forino.

Preface

This book introduces the concepts and applications of classical mechanics into Newtonian formalism by thinking of students of scientific disciplines who will continue their studies by encountering modern physics. Classical mechanics is the basis for any university-level study of technicalscientific disciplines. One of the problems I encountered when I started teaching Mechanics for students of the degree program in physics, was the selection of a suitable textbook. When I look at most of the existing manuals, I feel that the emphasis is on idealized applications on the scale of human dimensions while the basic aspects are not always sufficiently highlighted. A significant fraction of the texts is occupied by examples based on winches, rods, inclined planes,…, and students are often disoriented because the fundamental concepts are drowned in a very large number of artificial examples. This has the consequence that university students often expect from the class of Mechanics a repetition of what faced in high school, perhaps with a more advanced mathematical formalism, and not the presentation of the theory that is also at the foundation of modern physics. Newtonian dynamics is not only an efficient tool for solving applied physics and engineering problems but is a fascinating theory anchored in the basic questions posed since the times of the Greek philosophers. Think about the concepts of space, the flow of time, what are the measurable physical quantities, how to compare observations with models and build theories, which are the physical principles and the mathematical structure needed to describe the Universe. My aim was to encourage students to consider Newtonian mechanics as a starting point for these fundamental aspects and to introduce how, in modern physics, they will be addressed. Examples, applications, and how to get a result from a problem are important aspects, both for students of the basic sciences and for those oriented toward more technical disciplines. However, I believe that students of undergraduate programs in disciplines who will deal with modern physics will benefit more from an approach in which the aspects of classical mechanics are preliminarily translated in terms of current knowledge. For example, although at the time of Newton nothing was known about the atomic structure of matter, it is evident that this has a decisive role on the macroscopic aspects covered by classical mechanics. ix

x

Preface

The book does not have the same degree of difficulty from beginning to end. As the Harry Potter saga begins with the style of a children’s book when the magician starts school magic and ends with the dark tones of the civil war of the last volume, here the first chapters assume that the student has the basic knowledge of high schools, while the following incorporate the current growth and gradually use what is acquired in the parallel courses of the semester. This also applies to questions and exercises: in the last chapters, compound exercises, which requires what is introduced in the previous ones, are more frequent. Classical mechanics is not the ultimate theory, either in the Newtonian version shown here or in the Lagrangian and Hamiltonian versions that students will encounter as an intermediate step before modern physics. The theory has had enormous success (not only to send man to the moon) but also deep crises that have emerged since the beginning of the last century. I have tried to highlight these aspects of success and the limits that will lead to developments in relativity and quantum mechanics. This is not a book of history of physics, and it does not follow a strict timeline of how things have followed from Galilei onwards. I do not enter into the diatribes and disputes (even very lively) about the priorities of the discoveries. I have sometimes reported brief biographical notes of the main protagonists: they are only useful for framing the historical periods in which the studies were carried out and to highlight the acceleration that occurred in the scientific discoveries. Some sections are marked with the symbol (*). This means that the material or the mathematics involved in the paragraph is on average more complex and the subject can be addressed in the second reading. A problem I felt was the choice of notation: with which letter to indicate each of the physical quantities introduced. In the end, I decided to align the text with the ISO 80000 standard on quantities and units of measurement https://en.wikipe dia.org/wiki/ISO/IEC_80000. ISO is the acronym of International Organization for Standardization, the world organization for the definition of technical standards. The numerical solution of queries and exercises requires mental mechanisms other than those engaged in understanding a text. The solution of the exercises requires a necessary training with a much more active and autonomous role on the part of the student. Although there are on the market excellent manuals with proposed and solved exercises, my suggestion is that the student should work on exercises without looking at the solutions. In this book, traditional exercises coexist with more innovative ones and are inserted at the end of each chapter with the numerical solutions reported in an Appendix. Even failed attempts give important returns, because they develop processing skills. In particular, as cooperation is a key ingredient in modern physics, the students are encouraged to interact with other students (in small or large groups) to solve problems. The interaction with others is an added value that you never find in any book. The text is based on the lessons held in the first year of the degree program in physics (9 ECTS) at the University of Bologna, Italy, and drafts of the text in the Italian version have been reviewed by many students. However, inaccuracies, typos, or more serious errors can always remain and I am grateful to anyone who wants to report them to me. For the exercises, I want to thank my colleagues during the

Preface

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years who contributed with me to elaborate, modify, and resolve those proposed here: Prof. L. Guiducci, Dr. Nicolò Masi, Dr. Giulia Illuminati, and Dr. Filippo Sala. The translation in English from the original Italian version (who benefit from the feedback of many colleagues) was made by the author with the help of Dr. Eleonora Spurio. Bologna, Italy

Maurizio Spurio [email protected]

Contents

1

Physical Quantities and Units of Measurement . . . . . . . . . . . . . . . . . . . 1.1 Physics and the Scientific Method . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Measurable Quantities and the International System . . . . . . . . . . 1.2.1 Distances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.2 Times . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.3 Masses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Measuring Physical Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Direct and Indirect Measurements . . . . . . . . . . . . . . . . . . 1.3.2 The Distance Ladder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Mass Measurements Out of Range for Scales . . . . . . . . 1.3.4 From Pico to Tera . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 The Concept of Space and Time . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 The Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Newtonian Space and Time . . . . . . . . . . . . . . . . . . . . . . . 1.4.3 Four-Dimensional Spacetime (*) . . . . . . . . . . . . . . . . . . . 1.4.4 Cosmological Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Time Synchronization and Dissemination . . . . . . . . . . . . . . . . . . . 1.6 Time Measurements Without Periodic Phenomena (*) . . . . . . . . 1.6.1 The Law of Radioactive Decay . . . . . . . . . . . . . . . . . . . . 1.6.2 Radiocarbon Dating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Questions and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 5 6 6 8 8 9 9 10 11 11 12 13 14 15 17 18 20 22 24

2

Vectors and Operations with Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction: Scalars, Vectors and Tensors . . . . . . . . . . . . . . . . . . 2.2 Cartesian Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Right-Handed Coordinate System and Right-Hand Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . .

27 27 28 29

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Contents

2.3

Representation of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Vectors in Intrinsic Representation . . . . . . . . . . . . . . . . . 2.3.2 Vectors in Cartesian Representation . . . . . . . . . . . . . . . . 2.3.3 Unit Vectors in Cartesian Representation . . . . . . . . . . . . Product of a Scalar and a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . Sum and Difference of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Scalar Product in Intrinsic Representation . . . . . . . . . . . 2.6.2 Scalar Product in Cartesian Representation . . . . . . . . . . Vector Product Between Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 Vector Product in Intrinsic Representation . . . . . . . . . . . 2.7.2 Vector Product in Cartesian Representation . . . . . . . . . . Area and Volume in Vector Spaces (*) . . . . . . . . . . . . . . . . . . . . . . The Equals Are Not All Equal . . . . . . . . . . . . . . . . . . . . . . . . . . . . Laws and Principles, Physics and Mathematics . . . . . . . . . . . . . . Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30 30 31 32 34 34 36 37 37 38 38 40 40 43 44 46

3

Kinematics of the Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Uniform and Uniformly Accelerated Motion . . . . . . . . . . . . . . . . 3.3 Equation of Motion, Velocity and Acceleration . . . . . . . . . . . . . . 3.3.1 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Velocity, Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.3 Acceleration, Definition . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 The Direct Kinematics Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Infinitesimal Displacement and Infinitesimal Path . . . . 3.4.2 Composition of Motions and Trajectory . . . . . . . . . . . . . 3.5 The Inverse Kinematics Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Uniform and Non-uniform Circular Motion . . . . . . . . . . . . . . . . . 3.6.1 Uniform Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 The Cartesian Representation . . . . . . . . . . . . . . . . . . . . . . 3.6.3 Non-uniform Circular Motion . . . . . . . . . . . . . . . . . . . . . 3.7 Polar Plane Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.1 Circular Motion with Co-moving Unit Vectors . . . . . . . 3.7.2 Definition of Polar and Cylindrical Coordinates . . . . . . 3.8 The Intrinsic Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Poisson’s Rules for Moving Unit Vectors . . . . . . . . . . . . . . . . . . . 3.10 Motion on any Trajectory (*) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11 Questions and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47 47 49 56 57 59 60 60 61 61 62 64 65 65 67 68 68 69 70 72 73 74

4

Forces and the Dynamics of the Particle . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Measuring Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Weight and Anthropomorphic “effort” . . . . . . . . . . . . . . 4.2.2 The Dynamometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3 The Weight is a Force . . . . . . . . . . . . . . . . . . . . . . . . . . . .

79 79 81 81 82 84

2.4 2.5 2.6

2.7

2.8 2.9 2.10 2.11

Contents

4.3 4.4 4.5 4.6 4.7 4.8

xv

The Vector Nature of Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Constraint Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Contact Friction Between Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . Newton’s First Law of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . Newton’s Second Law of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . Kinematic Effects of Some Forces . . . . . . . . . . . . . . . . . . . . . . . . . 4.8.1 Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8.2 Fall with the Presence of Viscous Friction . . . . . . . . . . . Harmonic Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9.1 The Simple Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9.2 The Elastic Force and the Harmonic Oscillator . . . . . . . What Do We Know Today About Forces . . . . . . . . . . . . . . . . . . . . 4.10.1 Fundamental Interactions . . . . . . . . . . . . . . . . . . . . . . . . . 4.10.2 Force Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.10.3 The Arrow of Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Questions and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

86 88 89 92 93 94 94 95 98 98 103 104 104 106 106 108

5

Frames of Reference in Relative Motion . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Galilean Relativity Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Inertial Frames of Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Non-inertial Frames of Reference . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Pseudo-forces in Non-inertial Frames . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Dynamics in an Accelerated Vehicle . . . . . . . . . . . . . . . . 5.4.2 Dynamics in a Rotating Frame . . . . . . . . . . . . . . . . . . . . . 5.4.3 Earth’s Drag Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Coriolis Force in Guglielmini’s Experiment (*) . . . . . . . . . . . . . . 5.6 The Coriolis Force on the Motion of Terrestrial Fluids . . . . . . . . 5.7 Generalized Principles of Relativity . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Questions and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

113 113 116 119 123 123 125 126 128 132 133 134

6

Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Definition of Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Work in Vectors’ Intrinsic Representation . . . . . . . . . . . 6.2.2 Work in Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . 6.2.3 Measurement Units of Work . . . . . . . . . . . . . . . . . . . . . . . 6.3 The Work-Energy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Examples of Calculating the Work of a Force . . . . . . . . . . . . . . . . 6.4.1 Work of the Weight Force . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.2 Work of the Elastic Force . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.3 Work of the Dynamic Friction . . . . . . . . . . . . . . . . . . . . . 6.4.4 Anthropomorphic Work (“work” in Common Parlance) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Conservative and Non-conservative Forces . . . . . . . . . . . . . . . . . .

137 137 138 138 139 140 140 142 142 143 143

4.9

4.10

4.11

145 146

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Contents

6.6 6.7

Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Mechanical Energy and Its Conservation . . . . . . . . . . . . . . . . . . . 6.7.1 Definition of Mechanical Energy . . . . . . . . . . . . . . . . . . . 6.7.2 Mechanical Energy with Non-conservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exact Differentials and Potential Energy . . . . . . . . . . . . . . . . . . . . 6.8.1 Force is the Negative Gradient of Potential Energy . . . Differential Operators: Divergence, Curl and Laplacian (*) . . . . 6.9.1 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9.2 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9.3 Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9.4 Schwarz’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Conservative Forces: Null Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . Central Forces are Conservative . . . . . . . . . . . . . . . . . . . . . . . . . . . Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Towards the Principle of Energy Conservation . . . . . . . . . . . . . . . Questions and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

148 149 149

Dynamics of Mechanical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Mass Distribution: Discrete and Continuous Systems . . . . . . . . . 7.1.1 Systems of Material Points . . . . . . . . . . . . . . . . . . . . . . . . 7.1.2 Continuous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 The Center of Mass of a System of Points . . . . . . . . . . . 7.3.2 The Center of Mass of a Continuous Body . . . . . . . . . . 7.4 Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Linear Momentum and Angular Momentum . . . . . . . . . . . . . . . . 7.5.1 Linear Momentum of a System . . . . . . . . . . . . . . . . . . . . 7.5.2 Angular Momentum of a System . . . . . . . . . . . . . . . . . . . 7.6 Conservation of Linear Momentum . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Euler’s First Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 Conservation of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . 7.7.1 Euler’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8 Newton’s Third Law of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.1 Comment on Newton’s Laws . . . . . . . . . . . . . . . . . . . . . . 7.9 Properties of the Center of Mass Frame . . . . . . . . . . . . . . . . . . . . . 7.9.1 Linear Momentum P’ in the C.M. Frame . . . . . . . . . . . . 7.9.2 Intrinsic Angular Momentum (Spin) L’ in the C.M. Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.9.3 Euler’s Second Law in the C.M. Frame . . . . . . . . . . . . . 7.9.4 Kinetic Energy T’ in the C.M. Frame . . . . . . . . . . . . . . 7.10 Equilibrium of a Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.11 Questions and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

165 165 165 166 168 170 170 172 173 177 177 178 179 180 182 183 183 185 186 187

6.8 6.9

6.10 6.11 6.12 6.13 6.14 7

150 151 151 152 152 153 153 153 154 155 157 158 159

187 188 189 190 191

Contents

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Collisions and Decays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Impulsive Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Elastic Collision Between Two Bodies . . . . . . . . . . . . . . . . . . . . . 8.3.1 The One-Dimensional Case . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Two-Dimensional Case and Constraining Forces . . . . . 8.3.3 Let Us Reflect on the Nature of the Constraining Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Collisions in the Center-of-Mass Frame . . . . . . . . . . . . . . . . . . . . 8.4.1 One-Dimensional Case in the Center-of-Mass Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Motion of a Rocket . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Energy-Mass Conservation in Collisions and Decays . . . . . . . . . 8.7 Conservation Laws in Two or Three Body Decays . . . . . . . . . . . 8.7.1 Alpha Decay (Two-Body Process) (*) . . . . . . . . . . . . . . 8.7.2 Beta Decay (Three-Body Process) (*) . . . . . . . . . . . . . . 8.8 Partially Elastic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9 Questions and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

195 195 197 199 200 202

Considerations on Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Isotropy and Homogeneity of the Universe . . . . . . . . . . . . . . . . . . 9.2 Translation of Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Rotation of Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Reflexion of Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Not all Triplets are Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Polar and Axial Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6.1 How Can Axial Vectors be Distinguished . . . . . . . . . . . 9.6.2 Scalar and Pseudoscalar Quantities . . . . . . . . . . . . . . . . . 9.6.3 Pseudoscalar Quantities in Nuclear Phenomena . . . . . .

225 225 227 229 232 234 235 236 238 239

10 Newton’s Law of Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Pre-Galilean Astronomical Measurements . . . . . . . . . . . . . . . . . . 10.2.1 Earth’s Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 Earth-Moon Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.3 Earth-Sun Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 The Apple and the Moon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Law of Universal Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.1 The Inverse of the Square of the Distance . . . . . . . . . . . 10.5 Inertial Mass and Gravitational Mass . . . . . . . . . . . . . . . . . . . . . . . 10.6 Gravitational Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.1 Potential Energy of the Weight Force . . . . . . . . . . . . . . . 10.6.2 Classical Limits of Relativity and Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 Escape Speed From a Celestial Body of Mass M . . . . . . . . . . . . . 10.7.1 Event Horizon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

241 241 242 242 244 244 245 247 248 250 252 253

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203 204 205 207 210 213 215 216 219 221

254 255 257

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Contents

10.8 10.9

Measurement of G: The Torsion Balance . . . . . . . . . . . . . . . . . . . Spherical Coordinates (*) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.9.1 Line, Surface and Volume Elements . . . . . . . . . . . . . . . . 10.9.2 Gravitational Potential Energy: Second Method . . . . . . 10.10 Mass in the Center of the Sphere (*) . . . . . . . . . . . . . . . . . . . . . . . 10.11 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

259 262 263 266 266 270

11 Motions in Gravitational Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Historical Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Kepler’s Empirical Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 The Two-Body System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Angular Momentum, Kepler’s 1st and 2nd Laws . . . . . . . . . . . . 11.4.1 Kepler’s First Law, Plane Orbits . . . . . . . . . . . . . . . . . . . 11.4.2 Kepler’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Kepler’s Third Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Mechanical Energy of the Two-Body System . . . . . . . . . . . . . . . . 11.6.1 Qualitative Solutions for the Effective Potential . . . . . . 11.7 Kepler’s First Law, Elliptic Orbits (*) . . . . . . . . . . . . . . . . . . . . . . 11.7.1 Conic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7.2 First Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7.3 Eccentricity Versus Energy and Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7.4 Elliptic Solutions: Semi-axes as a Function of E, L . . . 11.7.5 Degeneracy in Classical and Quantum Physics . . . . . . . 11.8 Kepler’s Third Law, Revised . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.8.1 The Black Hole in the Center of the Galaxy . . . . . . . . . 11.9 Two Neutron Stars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.10 Triumphs and Falls of Newtonian Theory . . . . . . . . . . . . . . . . . . . 11.11 Gravitational Indications for Dark Matter . . . . . . . . . . . . . . . . . . . 11.12 Questions and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

273 273 276 278 280 281 282 283 284 285 288 289 290

12 Dynamics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Angular Momentum and Angular Velocity . . . . . . . . . . . . . . . . . 12.3 Moment of Inertia of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.1 Moment of Inertia of a Cylinder and Hollow Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3.2 Moment of Inertia of a Rod . . . . . . . . . . . . . . . . . . . . . . . 12.3.3 Moment of Inertia of a Rod on a Disk . . . . . . . . . . . . . . . 12.4 Conservation of Angular Momentum and Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.5 An Application of the II Euler’s Law . . . . . . . . . . . . . . . . . . . . . . 12.6 Huygens-Steiner Theorem for Moments of Inertia . . . . . . . . . . . . 12.7 Center of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.7.1 Physical Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.7.2 Torsion Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

311 311 312 315

292 294 295 296 297 299 302 304 307

315 316 317 318 319 321 323 325 326

Contents

12.8 12.9

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Moment of Inertia Tensor (*) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Rotational Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.9.1 Body Rolling Without Sliding . . . . . . . . . . . . . . . . . . . . . 12.9.2 The Motion of the Wheel . . . . . . . . . . . . . . . . . . . . . . . . . 12.10 The Motion of the Spinning Top (*) . . . . . . . . . . . . . . . . . . . . . . . 12.11 Questions and Summary Exercise . . . . . . . . . . . . . . . . . . . . . . . . .

327 330 330 332 333 335

13 Considerations on Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Work Needed to Build a System . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Potential Energy of a Bound Spherical System . . . . . . . . . . . . . . 13.2.1 Age of the Sun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.2 Energy Conservation in Stellar Gravitational Collapse (*) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Gravitational Field and Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 First Integral from Energy Conservation . . . . . . . . . . . . . . . . . . . . 13.4.1 Application to a Falling Particle . . . . . . . . . . . . . . . . . . . 13.5 Motion in a Potential Energy Field . . . . . . . . . . . . . . . . . . . . . . . . . 13.5.1 Stable and Unstable Equilibrium . . . . . . . . . . . . . . . . . . 13.5.2 Permitted and Forbidden Regions . . . . . . . . . . . . . . . . . . 13.6 More on Harmonic Motion (*) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6.1 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6.2 Harmonic Oscillator with Complex Numbers . . . . . . . . 13.6.3 Mechanical Energy of the Harmonic Oscillator . . . . . . . 13.7 Damped Harmonic Oscillator (*) . . . . . . . . . . . . . . . . . . . . . . . . . . 13.7.1 Overdamped Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.7.2 Discussion on Mechanical Energy and Developments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.8 Developments and Problems of Classical Mechanics . . . . . . . . . 13.8.1 Lagrangian and Hamiltonian Formalisms . . . . . . . . . . . . 13.8.2 Determinism in Newtonian Mechanics . . . . . . . . . . . . . . 13.9 Epilogue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.10 Questions and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

345 345 348 350 352 355 357 358 359 359 361 362 363 364 366 366 368 371 372 372 373 374 375

Appendix: Numerical Solution of the Exercises . . . . . . . . . . . . . . . . . . . . . . . 379 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383

Chapter 1

Physical Quantities and Units of Measurement

Abstract Physics is a discipline based on observable and measurable quantities and founded on the Galilean scientific method. Seven fundamental basic units of measurement are necessary to define all quantities used and all other units of measurement are derived. For the study of mechanics, three of the seven basic units are necessary: distances, time, and masses. In this chapter, in addition to the definition of physical dimension and of the unit of measurement, we deal about direct and indirect measures. Direct measures refer usually to distances, time intervals and masses at the human scale, but physics deals with quantities from the sub-microscopic to the cosmological worlds that needs to be measured. The chapter introduces the concepts of space, time and mass and their connection with modern physics and how physics solve the problem to measure microscopic and macroscopic distances, times, and masses.

1.1 Physics and the Scientific Method The word physics comes from the ancient Greek and it literally means nature. Initially, by the time of Aristotle, physics meant the study of all natural laws. The use of the term, has become more and more narrower over time. Still at the time of Newton and Galileo Galilei, the term merged all the empirical disciplines (sometimes with confusion in the context of what today we would call pseudo-sciences, such as alchemy). During the 1700s, the word . physics began to delimit the study of inorganic nature, from which chemistry subsequently separated, acquiring specific skills. The turning point that proved to be decisive for separating science from pseudoscientific disciplines is the procedure (developed by Galilei and other subsequent authors: Descartes, Newton, …) which led to the definition of the scientific method. This represents the effective tool for selecting hypotheses or laws: the method requires submit models to the judgment of the facts, the experiments, eliminating the ideas that do not work.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Spurio, The Fundamentals of Newtonian Mechanics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-47289-3_1

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There is no single protocol that characterizes the scientific method. Typically, the process is based on a series of steps. The first aspect is trying to characterize the subject of investigation as best as possible. It is advisable to try to remove all those aspects that may present complications (for example, friction), which can then be introduced later. Then the subsequent step is gathering information with observations. In many cases, these are pre-existing, public observations of verified reliability.1 Based on these observations, we can try to formulate an explanatory hypothesis: a simple model, or a more structured mathematical theory. In the case of several possible alternative models, we can choose the one that appears simpler to us, with fewer ad hoc hypotheses to explain the phenomenon. Subsequently, we must try to deduce the consequences of the model we have built, possibly developing experiments that can verify or disprove the formulated hypothesis. In particular, we need to predict which observations and experiments could prove it false (process of falsifiability). I will present a particularly instructive example in Sect. 9.6 in the case of mirror symmetry properties. The subsequent phase of experiments execution and data collection in a reproducible manner can take place on even long time scales, and with experimental groups working in parallel, in a cooperative or competitive way. For example, starting from (about) 1980 various experimental groups in the world began to think of building an instrument for the detection of gravitational waves. For two decades, the groups were in competition with each other: each of them managed to significantly improve some technical aspects, but never being able to accomplish the feat. At one point, they decided to pool their knowledge, and the collaboration between several US and European groups led to the detection of the first gravitational wave on September 14th 2015 and the opening of a new branch of astrophysics. The data analysis phase, in particular the information on how the acquired data is processed, must however be public. In 2011, the claim that an experiment at the Gran Sasso Laboratories (in Italy) on the speed of neutrinos (elementary particles) had shown that they travel faster than light caused a great sensation. The open discussion in the community led in a few months to highlight a hardware error in a optic fiber connection of the experiment. Once corrected the error, the neutrinos returned to travel at a speed than is not exceeding the speed of light. The next step of interpreting data and drawing up a conclusion can serve to confirm (or refute) the starting model, or to better adapt it on the basis of the observations. In the various stages, the results and data/experiment comparisons are always public, published in journals that adopt the peer-review procedure. Now, we are going to describe how a peer-review procedure works. After an author (or group) submits the manuscript of the article to a journal, an editor competent in the subject, sends the article to 2 or 3 reviewers (referees), renowned experts of the

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Today, for example, we cannot rely on material simply found on the web. The data could be genuine, but could have been acquired with some form of bias which could affect the correctness of the conclusions. Normally, only information that has been peer-reviewed (see below) should be used.

1.2 Measurable Quantities and the International System

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same sector, which will always remain anonymous and unknown to the author (or for a certain period, usually at least 50 years). Referees can advise or advise against the publication of the article, or advise publication only after a review, or invite the authors to respond to some objections. Based on the opinion of the reviewers, the editor decides whether or not to publish the article or whether to wait for the next revision. This protocol is useful as a possible filter against method errors and other intentional or inadvertent flaw, but also to avoid pollution by unscientific or pseudo-scientific works. For example, I receive between 10 and 20 requests for article evaluations per year. I don’t feel sufficiently expert on all topics, and I refuse to evaluate some of them. Anything that does not adopt this method is to be classified as pseudo-scientific. Care must be taken: pseudo-sciences are methods or practices that claim to be scientific (or that want to appear scientific) but which do not respect the scientific method described above in some fundamental aspects, mainly by contravening the requirements of verifiability. While in many cases some pseudo-sciences are futile but harmless (astrology), others are potentially very dangerous as they replace, for example, certified medical practices with water (homeopathy), or they claim that vaccinations are harmful, denying their role in safeguarding one’s own life and the lives of others.

1.2 Measurable Quantities and the International System Physics investigates in the field of observable and measurable quantities. For example, even if we all know what it is, we cannot quantify a quantity like how much we love Mom. The amount of love towards mother, not being a measurable quantity, does not fall within the scope of physics. Since ancient times, it was realized that in order to exchange information, some method was needed to make quantities such as the amount of matter (the.mass),.times and measurements of.distances objective and quantifiable. Linked to distances, there are derived quantities such as area and volume measurements. Until the time of the French Revolution and the advent of Napoleon, each nation, city-state, duchy, fraction of commune did as it pleased in terms of measurements of lengths and masses. For example, in Bologna the units of length to be used for trade are located in the square of the former market (Fig. 1.1). The Bolognese foot, basic length of the local metric system, corresponding to about 38 cm; the perch, equivalent to 10 feet; the arm, about 64 cm long; finally the double arm. Alongside these there are also models of a brick and a tile of standard dimensions. If you visit the close city of Ferrara (less than 50 km far), at the entrance to the Estense Castle there are similar samples, obviously with different lengths. Now, put yourself in the shoes of the poor merchants of the time when they had to go or trade “abroad” and think how hard it must have been.

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Fig. 1.1 Near Piazza Maggiore in Bologna (middle Italy), on the eastern side of Palazzo Comunale and close to the Piazza Maggiore, the samples of the units of measurement in use in the city during the Middle Ages have been located since 1547 Table 1.1 The seven fundamental quantities: name, dimensional symbol, unit of measurement used in the SI and abbreviation. All the other quantities that will be introduced will be derived from these seven, with appropriate dimensions. For example, speed has dimensions of [L T.−1 ] Name Dimensional symbol Unit of measurement Abbreviation Length Time Mass Electric current Temperature Luminous intensity Amount of substance

[L] [T] [M] [I] [.Θ] [J] [N]

Meter Second Kilogram Ampere kelvin Candela Mole

m s kg A K cd mol

Napoleon decided that French standards suited everyone.2 The current International System of Units (in force since 1961), abbreviated to SI,3 is the modern version of the metric system developed by a commission of French revolutionaries chaired by Lagrange from 1791. Over time, in addition to the meter, other basic units of measurement were added, up to seven quantities (in 1971). Table 1.1 lists them. All other units of measurement are derived from these and are therefore called derived units. To each fundamental quantity is assigned a physical dimension, which is indicated in square brackets: for lengths, the symbol is [L], for example. The physical dimensions are given in the second column. The name of the unit of measurement, and the symbol adopted in the SI context, are instead shown in the third and fourth column.

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Obviously, the British disagreed. SI stands for Système International, which is an abbreviated form of its full French name Système international d’unités. 3

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Within the framework of classical mechanics, only the first three quantities are necessary: distances, time and masses. The other quantities that we will encounter (velocity, acceleration, force, energy,…) are based on the three fundamentals which have dimensions [L], [T] and [M], respectively. The other four fundamental quantities will be introduced in later courses. All the quantities that are necessary for physics studies are deduced from the seven fundamental ones listed in the table. There is a branch of physics that deals precisely with metrology, i.e. the issues related to the measurement of physical quantities, to dimensional analysis and calculation, to the choice of systems of units of measurement. In fact, over the years there have been various ways in which the fundamental quantities have been defined. In all cases, the requirements that the measurement standards must fulfill are those of precision, accessibility, reproducibility and inalterability. From May 20th, 2019 the definition of the units of measurement in the SI associated with the seven fundamental physical quantities is obtained starting from the set numerical value of 7 physical constants. More information can be found on the Bureau International des Poids et Mesures website https://www.bipm.org/en/home. For a complete understanding of this “final version”, however, you must first study modern physics to understand what quantities the seven chosen constants correspond to. On the other hand, I believe that a brief (partial) summary of the historical path that leads to the definition of the samples of the fundamental units of measurement that we will use in the course is useful: meter, second, kilogram.

1.2.1 Distances The first attempt to unify the measurement of distance was made in 1795 by the French Academy of Sciences, which defined the meter as the fraction .1/(10 × 106 ) of the part of the terrestrial meridian (between the Pole North and the Equator) which passes through Paris. In 1799 a prototype was built using a platinum ruler. A second more accurate sample was built in 1889, with a platinum-iridium X-section rod kept at a temperature of 0 .◦ C (to avoid thermal expansion), which allowed an accuracy of 0.2 .µm. Only 30 samples were built, then distributed in various national metrology institutes around the world. After the advent of techniques using atomic physics, in 1960 the meter was redefined as equal to 1650763.73 times the vacuum wavelength of red-orange light emitted by the .86 Kr, a definition which assumed a precision of up to 0.01 .µm. Finally, before the latest redefinition in 2019, in 1983 it was decided to define the meter as the distance traveled by light in vacuum in 1/299792458 of a second.

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1.2.2 Times Since ancient times, it has been thought to measure the passing of time in terms of a periodic phenomenon, i.e. a phenomenon that returns to itself after a defined time interval. In one year the Earth completes a full revolution around the Sun: it was the most used periodic phenomenon to measure the passage of time. The .second was defined in 1832 as the fraction 1/86400 of the mean solar day, averaged over a year. After 1950, with the advent of the atomic clock technique (and therefore, using periodicity of the atomic type), however, discrepancies were observed with respect to the terrestrial periodicity. In 1967 the second was redefined as a multiple of the oscillation period (9192631770 oscillations) of a particular frequency of electromagnetic radiation emitted by .133 Cs atoms. I will come back a little later, Sect. 1.5 on some fundamental aspects related to synchronization operations and on the possibility of measuring time using nonperiodic phenomena, Sect. 1.4.4.

1.2.3 Masses The mass of an object can be established by referring to masses of objects taken as a sample. In 1795, the unit mass was defined as the amount of water at melting point (0.◦ C) contained in one cubic centimeter. This corresponded to one gram. Since commercial exchanges often concern much more massive objects, and being the sample made of water inconvenient and unstable, in 1799 it was chosen to refer to the volume of a liter of water in its most stable state of density, i.e. at a temperature of 4 .◦ C, at which the density of water is maximum. This corresponded to the object with mass equal to 1 kg. However, the need arose to create a sample based on a metallic artifact for reasons of reproducibility and inalterability, and a first platinum sample was created with a mass equivalent to one kilogram as defined above, which remained in use for ninety years. Since 1889 the kilogram has been defined as the mass of a new artifact made of platinum-iridium in the shape of a right circular cylinder, with a height equal to a diameter of 39.17 mm. The prototype, which is kept at the Bureau International des Poids et Mésures in Sèvres, near Paris, served as a standard until the latest revision in 2019. Until very recently, the kilogram was the only unit of measurement referring to a material object.

“Weight” Problems To measure the mass of an object, an instrument called a balance is generally used (there are different types of it). In the Laboratory course, you will learn several procedures for determining masses using balances. Daily empirical experience shows

1.2 Measurable Quantities and the International System

7

us how the mass of a compound object is the sum of the masses of the constituents. However, if we look at the structure of matter, we know that it is made up of a finite number of fundamental objects: atoms. In turn, atoms4 are not indivisible. They are made up of nuclei and electrons. In turn, the nuclei are composed of protons and neutrons. Which, in turn, are made up of quarks. Which, in turn …we don’t know! From the atomic level to the size of the smallest objects, however, we know that the mass of a compound is not the sum of the masses of the constituents. The mass of a nucleus is not the sum of its components, protons and neutrons. The most common (for example) carbon nucleus is made of . Z = 6 protons and . N = 6 neutrons. The sum . A = Z + N is called mass number; the electric charge is given by the atomic number. Z . Each nucleus X (the initial of the element name, for example C for carbon) is symbolically indicated with . ZA X. The mass of the nucleus of .126 C is about 0.6% smaller than the sum of the masses of 6 protons and 6 neutrons. In the formation of the nucleus, what is lost in mass is gained in energy, as we will discover in Chap. 8. Also in the atomic case we have an analogous situation; but the fractional mass loss is less than .10−8 and it is hardly measurable. It therefore seems easy: at the subatomic level, when objects are formed, there is a loss of mass and a gain of energy.5 However, if we consider the proton, we know that it is made up of other particles called quarks. In the simple quark structure model of the proton, it is composed of three quarks each of mass6 about 2 MeV/c.2 . However, the proton does not have three times the mass of a quark, but almost 500 times more! There is another particular feature of the proton. We don’t know the dimensions of elementary particles, like the electron and the quark: we only know that their mass (whatever that means) is confined to a region of unknown dimensions and less than −16 .10 cm in length. The proton mass of 938 MeV/c.2 is instead within a known region of about .10−13 cm in radius; in here, besides the three constituents of mass 2 MeV/c.2 there must be something else that supplies the mass to the proton (and to the neutron, for which the same situation holds). So, what is mass? It is a good question, which we have only been trying to answer for a few years. An important step took place in 2012, with the discovery of the Higgs boson (see note of Sect. 8.1 for the definition of boson) at CERN. The discovery confirmed a theory that is the first step towards understanding what mass 4

The word atom comes from Greek atomos that means ‘indivisible’ , based on a- ‘not’ + temnein ‘to cut’. 5 On the other hand, . E = mc2 is the only universally known formula of physics, even if many people do not grasp its meaning. 6 Practical units. Although we will always use the SI units in this book, it is important to know that there exist, and they are universally used, other units of measurement which are called . practical For example, an astrophysicist struggles to measure distances in m; he prefers to use the parsec (pc), which corresponds to .∼ 3 × 1016 m, or 3 light-years (the distance traveled by light in 1 year). Similarly, particle physicists measure masses not in kilograms, but in units of MeV/c.2 . If it existed, the mass of a 1 MeV/c.2 particle = .1.79 × 10−30 kg. The electron has a mass of 0.511 MeV/c.2 . The proton of 938 MeV/c.2 , the neutron 0.14% more than the proton.

8

1 Physical Quantities and Units of Measurement

of matter is, an adventure that in the future will surely involve someone who is now reading these lines. For now, the important thing is to be able to say that we know how to measure masses.

1.3 Measuring Physical Quantities 1.3.1 Direct and Indirect Measurements The measurement of a physical quantity is called direct when the measured quantity is directly compared with the unit sample of measurement or its multiples or sub-multiples. For example, the determination of mass with a equal-arm balance (with calibration weights) or the length of a table with a graduated ruler are direct measurements. A measurement performed using pre-calibrated instruments is also a direct measurement, such as a calibrates scale for the mass, or the measurement of temperature using a thermometer, or that of a force using a dynamometer (Chap. 4). On the other hand, indirect measures are those in which we do not measure the quantity of interest, but others which are linked to it by some functional relationship. The speed of a car can be determined directly by measuring how long it takes to travel a stretch of road of known length, or by using a speed camera. This system is based on the fact that by bouncing light of known wavelength fired from a laser source onto a car, the light returning to the detector slightly changed its wavelength, in a way that depends on the speed of the car. The speed of the car is therefore based on the physical law that describes what is called the Doppler effect, which you will study in the electromagnetism course. The dividing line between direct and indirect measurements is labile: how could we imagine an instrument for the direct measurement of the solar mass? Or how is it possible to directly measure the speed with which the Sun rotates with respect to the center of the Galaxy? In fact, most of the observations outside the human size ranges rely on indirect methods. This is particularly true in the field of the microcosm (that is, at the atomic level and its sub-constituents) and in that of the macrocosm (of astrophysics and cosmology). How can we rely that the result of an indirect measurement is compatible with a hypothetical direct measurement of the quantity involved? One of the strongest arguments supporting the fact that we can get reliable information from indirect measurements is that there are several methods to measure the same quantity. If the different methods give a consistent result within the errors, we are confident in the goodness of the result and in the fact that the laws we apply are “right”. Conversely, if we have faith in a physical law, and something appears to be wrong, usually something interesting is hidden, as I will show in Sect. 11.10. In that case, the confidence in Newton’s laws and in the gravitational law made it possible to discover an unknown planet, Neptune.

1.3 Measuring Physical Quantities

9

This book is also meant to show that we can have confidence in the indirect observations, starting with those that are based on the laws of mechanics.

1.3.2 The Distance Ladder Modern physics deals with scales of distances ranging from the size of the femtometer (1 fm .= 10−15 m) to gigaparsecs (1 Gpc .= 3 × 1025 m). This means that it is not possible to have a single procedure for measuring a distance, by comparing it with a sample ruler. It is necessary to have some kind of distance ladder. Going to dimensions much larger than the terrestrial ones, the scale of cosmic distances is the set of methods that astronomers use to determine the distances of celestial bodies. The ladder consists of successive steps, and each step is used to determine distances in the next. Some objects have distances that can be measured with different techniques, allowing relative inter-calibration. However, the further you go up the ladder, the more the uncertainties add up and the distances can be affected by errors. At the base of the ladder are the radar observations of some nearby planets, which allow to determine the distances within the solar system from which it is possible to trace the parameters of the Earth’s orbit. Subsequently, the exploited techniques use the Earth’s orbit as a basis (the stellar parallax). The next steps involve the use of standard candles, i.e. objects whose brightness are thought to be reliably known. By comparing the object’s own luminosity with the apparent one, the distance is obtained. This technique is based on the dependence of the flux on the inverse square of the distance, as described in Sect. 10.4. Sequentially ordered, these standard candles are: the Cepheid variables, the Tully-Fisher relation for galaxies and type Ia supernovae. Finally, there is the measure of the red-shift of extremely distant objects. From the point of view of the microcosm, distances cannot be measured by comparing rulers at sub-micrometer levels. Atomic distances (for example in crystals) are measured through interference and diffraction phenomena using X-rays. Below the atomic dimensions, five orders of magnitude smaller, we find the nuclear dimensions at the femtometre level. In this case, through deflection measurements of high energy particles (which replace X-rays) a parameter is measured, called cross section, which represents a sort of geometric area of the target. Similarly, the dimensions of the elementary particles are estimated (when possible) by measuring the cross section.

1.3.3 Mass Measurements Out of Range for Scales The mass of objects larger than the ones a truck can usually carry cannot be measured with scales. In this case, the physical laws that we will gradually find must be used to determine the masses. It is obviously not possible to put the Earth on a scale. The mass of very large objects (a mountain) can be determined by trying to calculate

10

1 Physical Quantities and Units of Measurement

their volume and estimating their density. It was finally possible to determine the mass of Earth when it was known how the masses interact gravitationally, and after having measured with a torsion pendulum the value of the universal constant .G which enters the interaction formula (Chap. 10). The value of the mass of the Sun can be determined through Kepler’s III law (Sect. 11.5), as well as the same equation allows to determine the mass of black holes around which bright stellar objects rotate, Sect. 11.8.1. From the sub-microscopic point of view, the concept of mass is fundamental for understanding particles. Their masses are determined through a relativistic relationship that links the particle’s energy with its momentum. The two quantities will be introduced in the following chapters, and both correspond to quantities that can be determined by particle detectors.

1.3.4 From Pico to Tera Often, the quantities (fundamental or derivative) used in the SI are too small or too large to express a result. Although powers of 10 can easily be used, the result are conventionally expressed in terms of multiples and submultiples of the magnitude of one thousand. The prototype is the kg, which means 1000 g. Instead of writing 1000, a “k” is added to the unit of measure. The scale factors used are indicated in Table 1.2. They are also used for quantities that do not concern physics. Note that the multiplicative factor symbols are all capitalized (except the kilo). A Tbyte therefore represents a device that has a memory capacity for your computer of 10.12 bytes. Factor symbols that represent submultiples are shown in lowercase letters. One millionth of a second is .μs.

Table 1.2 Multiple and submultiples factors of the units of measurement of physical quantities Prefix Symbol Exponent Prefix Symbol Exponent .10

3

.10

6

.10

9

12 .10 .10

15

18 .10

kilo Mega Giga Tera Peta Exa

k M G T P E

.10

−3

.10

−6

.10

−9

−12 .10 .10

−15

−18 .10

milli micro nano pico femto atto

m .μ

n p f a

1.4 The Concept of Space and Time

11

1.4 The Concept of Space and Time 1.4.1 The Euclidean Space The Euclidean space is a space in which the axioms and postulates of Euclidean geometry hold. In particular: • • • •

space has three “dimensions”; two straight lines having a common point define a plane in the space; the sum of the internal angles of a triangle is exactly .π rad (or 180.◦ ); the Pythagorean theorem holds.

In the second half of the 1800s, non-Euclidean geometries begin to develop, built by denying or not accepting some Euclidean postulates. In some of these geometries, for example, the sum of the internal angles of a triangle is .> 180◦ , or .< 180◦ , see Fig. 1.2. All mathematical geometries are internally consistent, but obviously only one of the infinite possible solutions is the one adopted by Nature. If Nature has chosen the Euclidean plane geometry, for example, the sum of the internal angles of a triangle must be .= 180◦ . From a physical point of view, the problem is of experimental nature, and can be translated into the question: Is the structure of the Universe in which we live that of a Euclidean space or a non-Euclidean space? Obviously, in the second case, you will also want to know the curvature parameter, a value linked to the sum of the internal angles of a triangle. When you will study Einstein’s general relativity, you will find that the geometry of space really depends on a parameter of curvature. The curvature parameter .k allows us to determine the type of universe and its geometry: for .k = 0, we obtain a flat universe (Euclidean geometry). The determination of this parameter is experimental in nature. In the last 20–30 years, experiments have been

Fig. 1.2 Relationship between the sum of the internal angles of a triangle in the case of Euclidean and non-Euclidean plane geometry. The deformation of the triangles is caused by the fact that the space is bent

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1 Physical Quantities and Units of Measurement

carried out to determine the curvature of the Universe, which you will study with observational cosmology. The current experimental answer is that the visible Universe can be described by the parameters of Euclidean geometry not only on a local scale (Solar System, nearby stars, Galaxy) but also on a cosmological scale.

1.4.2 Newtonian Space and Time The fact that space is Euclidean allows the introduction and use of a fundamental mathematical quantity in physics: the vector. This will be the topic of the next chapter. A vector is the most appropriate tool for describing the position of an object in a three-dimensional Euclidean space. As we will see starting from Chap. 4, Newton’s 1687 Principia book represents the cornerstone of classical mechanics. Newton must use the notions of space and time to describe his new concepts. In his book, he explains his vision of absolute time, space, and motion, where absolute means without relation to anything external. I reproduce, translating into English from Latin, what Newton wrote about time. I. The absolute, true, mathematical time, by itself and by its nature without relation to anything external, flows uniformly, and it is also named duration; the relative, apparent and vulgar, is a sensible and external measure (accurate or approximate) of duration by means of motion, which is commonly employed in place of true time: such are the hour, the day, the month, the year . Similarly, there is the definition of space. II. Absolute space, by its nature without relation to anything external, always remains the same and motionless; relative space is a movable dimension or measure of the absolute space, which our senses define in relation to its position relative to bodies. Newtonian space and time are, therefore, absolute. The universe is a substantial space, endowed with reality, an empty container, indifferent to the matter contained in it and to the observer who analyzes the movements of matter in it. Similarly, absolute time indicates an eternal flow, free from space and existing independently of its vulgar measure in hours, days and years. Modern physics, first with special relativity and then with general relativity, will completely overturn this vision. However, in the physics within the solar system that we will address, the Newtonian view represents a working hypothesis that we can use, as long as the speeds of objects remain far from the speed of light. Furthermore, we could assume that in this huge “box” representing Newtonian space, each point (or at least, a sufficiently dense grid of points) is synchronized. Today this is operationally possible via GPS or similar devices, Sect. 1.5.

1.4 The Concept of Space and Time

13

1.4.3 Four-Dimensional Spacetime (*) As we have seen, in the Newtonian view the three-dimensional space is quite distinct from time and both are considered absolute. This is no longer the case in modern physics, and it is worthwhile here to briefly anticipate some of the facts presented in the following chapters and courses, starting with that of electromagnetism. By the late 1800s, in fact, the predictions of J.C. Maxwell’s theory of electromagnetism were being confirmed with an experimental observation of paramount importance: the speed of propagation of electromagnetic waves (the speed of light, c) was the same in all reference frames, regardless of the speed with which the reference frame moves. This observation falsifies what are called Galilei’s transformations (which we shall meet in Sect. 5.1). In the space-time of Galilei and Newton, the distance between two objects in space and between two events in time is an absolute quantity, which does not depend on the inertial reference frame in which the observer is placed. This is demonstrably not true when the distance between two objects (or the simultaneity of two events) is measured by the use of laser, sonar or other electromagnetic radiation. And as we have seen, outside the scales of “human” distances (say, beyond a few kilometers) these methods of measurement are unavoidable. The conclusion (both theoretical and experimental) is that observers in relative motion find different distances between two objects as observed in their respective reference frames, and two simultaneous events in one reference frame are not in the other. In the early 1900s, it became inevitable to consider that there is an indissoluble link between space and time and that it was necessary to abandon the concept of their absolute character. Such considerations were formulated independently by works of J.H. Poincaré, H.A. Lorentz, in the special relativity of A. Einstein (1905), and in the work of mathematician H. Minkowski. The mathematical structure needed to describe space-time as tightly connected in a four-dimensional structure (with the fourth dimension given by ct, that is, time multiplied by the speed of light, which dimensionally is a length) is the one initiated by Minkowski himself, after whom it is conventionally named. The novelty lies in the “mixing” between the three spatial and the temporal dimensions, whose “separation” varies depending on the reference frame in which the observer is located. The mathematical formalism due to Minkowski provides a simple “local” model, most useful for considering those physical phenomena involving a finite and constant speed for light, such as those described by special relativity. Galilei’s transformations for going from one inertial reference frame to another must be replaced by Lorentz transformations, as discussed in Sect. 5.7. Within a few years, however, Einstein in his theory of general relativity (1915) realized that Minkowski’s space-time was not usable to describe the universe as a whole. In general relativity, Minkowski’s space-time is only a model that locally approximates space-time, which is actually “distorted” by masses. A second fundamental empirical observation comes into play here: the equivalence of inertial mass and gravitational mass, which we will discuss in Sect. 10.5.

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1 Physical Quantities and Units of Measurement

General relativity describes gravitational interaction no longer as the action at a distance between massive bodies of Newtonian theory (as described in Chap. 10), but as the effect of a perturbation of the geometry of space-time (more specifically, the curvature) due to the distribution and flux of mass, energy and momentum. Inertial reference frames (Chap. 5) are identified with the coordinates relating to free-falling observers, who move in the geometry of space-time along geodesic trajectories, that is, along the shortest curve joining two points in space. The weight force results in this way as an apparent force (Chap. 5) observed in non-inertial reference frames. In the view of modern physics, therefore, the distinction being made between kinematics (Chap. 3) and dynamics (Chap. 4) makes less sense than in the classical physics discussed later in the book. However, the modern view is impossible to understand without the solid foundation provided by Newtonian concepts.

1.4.4 Cosmological Time Many physicists (starting from Newton, as we have seen) have attempted to explain what time is. Some have written books on time so confused but so well packaged, flavored with the scent of oriental philosophy and without the help of mathematical formulas, which have had enormous success with the general public. Measuring a time interval is a procedure that is clear to everyone. It has a start, a stop and a tool that measures the passage of time. It is what Newton calls the duration of an event, Sect. 1.4. Whether the Newtonian equivalent of absolute time exists and what it is represents a much more difficult problem. Although space and time are inextricably linked, and for example two observers in relative motion can measure different durations of an observable event, we live in a single Universe and cosmic history can be uniquely defined. The theory of general relativity makes it possible to define a cosmological time, time measured by an observer integral to the reference frame of comoving coordinates. Adequate treatment cannot do without knowledge of relativistic physics, which you will acquire in more advanced courses. I will try to explain it in the simplest way: the laws of physics can be written and are valid in any coordinate system. This is valid both in classical physics (Chap. 9) and in relativistic physics. However, there are coordinate systems where the involved math is simpler. Observational cosmology has developed a paradigm, the Big Bang theory supported by an enormous amount of experimental data, Fig. 1.3. To describe the evolution of the universe, the most natural choice is that of comoving coordinates: they are the coordinates of an observer who “gets carried away” by the expansion of the universe, a co-moving observer. Such an observer is characterized by perceiving the universe (and in particular its cosmic background radiation) as isotropic. The fossil background radiation, first observed in 1965 by Penzias and Wilson, is an observable microwave radiation containing information about the structure of the universe about 300,000 years after the Big Bang. The background radiation measured by any other observer, not at rest with respect to co-moving coordinates,

1.5 Time Synchronization and Dissemination

15

Fig. 1.3 Sketch of the expansion of the universe, which began 13.8 billion years ago with the Big Bang, as the cosmological time varies. The change in curvature represents the acceleration of the expansion, which began in the middle of the expansion and is still ongoing. Credits: NASA/WMAP Science Team

will be shifted to blue in some regions of the sky and to red in others, in relation to the direction and intensity of the observer’s velocity. The time measured by a co-moving observer is called co-moving time or cosmological time; instant zero is obviously that of the Big Bang and the value of cosmological time coincides with the age of the universe. The events that have occurred in the universe can be uniquely placed at a certain cosmological time. Therefore, it is possible to measure this time in some way. Based on the most precise estimates obtained by the European Space Agency’s Planck Surveyor space probe-telescope, the universe has evolved since the Big Bang and has a calculated age of .13.798 ± 0.037 billion years.

1.5 Time Synchronization and Dissemination A fundamental aspect in physics is the synchronization of clocks and the dissemination of times. Time synchronization is the ability to share the same time (within a predefined error) from distributed systems. The operation that brings the distributed signal to all stations is called signal dissemination. Until twenty years ago, this required very complicated operations (laying hundreds or thousands of

16

1 Physical Quantities and Units of Measurement

km of cables!). Today it is possible to synchronize computer clocks with an atomic clock using the Internet network. The typical difference (that is, the time difference recorded by two different clocks) is a few tens of milliseconds. Atomic clocks play a key role in technology today: their time signals are those that are used by GPS receivers to track location. The Global Positioning System (GPS) is, as you surely know, the US-operated satellite navigation system, freely accessible by anyone equipped with a GPS receiver. Through a dedicated network of artificial satellites in orbit, GPS provides any mobile terminal, or GPS receiver, with information about its geographical coordinates and its time in all weather conditions, anywhere on the surface of the Earth where there is an obstacle-free contact with at least four satellites in the system. The location is achieved through the transmission of a radio signal by each satellite and the processing of the received signals by the receiver. Its current degree of accuracy is on the order of a few meters,7 depending (among other factors) on weather conditions, availability and position of satellites relative to the receiver, quality and type of receiver. The GPS system is the most effective proof of the theory of general relativity that you will study in the following years: only with the corrections related to the gravitational effects the GPS manages to get to the mentioned precision.8 In general, a GPS receiver always has a precise time indication of the order of a few ns, which is independent of the position and usable while on the move. An example of application in physics of the time synchronization of different parts of an apparatus is the Pierre Auger Observatory (PAO), Fig. 1.4. It is a huge distributed detector system located in Argentina consisting of 1600 stations spread over approximately 3000 km.2 . Each station is about 1.5 km from the others. It studies the arrival on Earth of particle cascade induced by the so-called cosmic rays, mainly protons and heavier nuclei accelerated by cosmic high-energy mechanisms. The physical signal is characterized by the almost simultaneous switching on of about ten stations. Simultaneity requires each station to have the possibility of registering the arrival of a small fraction of the whole cascade of particles with a clock that is synchronized (within a few ns) with that of all the other stations in order to be able to reconstruct the signal. As can be seen from the photo on the right, each of the 1600 stations is equipped with a GPS antenna, as well as solar panels and batteries for powering the electronics necessary for the detector to operate. To reliably achieve sub-nanosecond time synchronization in distributed detectors, configurations using patch cables must be used. In recent years, open-source solutions have been developed such as the protocol called White Rabbit, used by a large community, such as the one of the experiments at CERN. This method is also used to synchronize the thousands of modules that are part of the KM3NeT neutrino For easily understandable military reasons, the spatial accuracy .ΔL of the GPS is limited to about 3–5 m. Since the light travels at c = 30 cm/ns, the corresponding temporal resolution is .Δt = ΔL/c ∼ 10 ns. The GPS is actually intrinsically more accurate, up to 1–3 ns. 8 An alternative positioning system to GPS, called Galileo, has been developed in Europe, and is it coming into operation. It should perform better than GPS. 7

1.6 Time Measurements Without Periodic Phenomena (*)

17

Fig. 1.4 The Pierre Auger Observatory (PAO), an experiment for the measurement of cosmic rays of the highest energies, created by an international collaboration in Argentina. In the drawing to the left, each dot indicates one of 1600 surface detectors that require time synchronization. The four dots with names in blue refer to other detectors always used for the detection of cosmic rays and synchronized with the others. The distance scale indicates that the extent is approximately 70 km. On the right, one of these 1600 surface detectors. Note the solar panels for power, and the GPS antenna for receiving time signals and transmitting data to the central station (Courtesy of Pierre Auger Collaboration)

telescope, under construction in the Mediterranean Sea, Fig. 1.5. Underwater, GPS does not reach and necessarily wired systems are used, which are also needed to transmit power and receive the acquired data.

1.6 Time Measurements Without Periodic Phenomena (*) One of the techniques for dating events in the cosmological time defined in Sect. 1.4.4 uses radioactive decays. Radioactivity is a natural phenomenon involving nuclei and particles. About nuclei, I will talk about it in a more complete way in Chap. 8. I suggest reading this section (especially the part where I introduce the mathematics) a second time while reading that chapter. When there is a situation in the cosmological history of the universe which, some time after the Big Bang, produces an event, this event can cause as a side effect the production of unstable nuclei (or particles), which may transform into something else in a later epoch. This transformation is what we call radioactive decay. How much later the decay of a particular nucleus will occur, after its production, nobody knows or can know: it is a phenomenon that occurs on a statistical basis. This is exactly what happens when a puppy born: at its birth, we know that it will die. But no one can know when the process will take place. We know that if the newborn is a human,9 or an elephant, his life expectancy (otherwise known as mean life) it will be about 75 years. If it is a puppy dog, the life expectancy span will be about 12 years. 9

Provided he was born in a nation of the Western world, in an averagely wealthy family.

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1 Physical Quantities and Units of Measurement

Fig. 1.5 Illustrative drawing of the KM3NeT detector under construction in the Mediterranean Sea. When complete, this telescope for observing neutrino interactions will have 230 strings about 1 km high anchored at a depth of 3500 m. Each string consists of 18 spheres of 43 cm in diameter (one is visible in the foreground), spaced 90 m from the others. Each sphere contains 31 photomultipliers, devices capable of detecting the light produced by the particles originating from the interaction of neutrinos. Each device must be synchronized with the others with a precision of 1 ns to accurately reconstruct the neutrino trajectory (Courtesy of the KM3NeT Collaboration)

1.6.1 The Law of Radioactive Decay It is possible to deduce a mathematical law that describes the situation. I consider a sample with a certain number of elements, . N0 , of any radioactive substance (for example, radium nuclei) at a certain initial time, .t = 0. If I surround the sample under observation with suitable measuring instruments, I can observe that a certain number of electrons are emitted in a certain time interval .Δt. I know that every emission of an electron involves the decay of a nucleus. Based on the observations, I find empirically a relationship between the number of nuclei . N (t) present at a certain instant of time .t and those present in a subsequent instant . N (t + Δt): .

N (t + Δt) = N (t) − λN (t)Δt .

(1.1)

The relation expresses this observational fact: the number of nuclei has decreased (the .− sign) by an amount that depends on the observation time .Δt, on the number of nuclei . N (t) that are present before the elapsed time interval and a constant, .λ, which depends only on the type of nuclei used for the measurement. Wherever we repeat this observation, in any circumstance (different place, on a plain, on the mountains or underground, at different pressure, temperature, applied external forces,…) we

1.6 Time Measurements Without Periodic Phenomena (*)

19

will always find the same relation (1.1) with the same value of .λ for each radioactive element considered.10 The (1.1) can also be written as: ΔN ≡ [N (t + Δt) − N (t)] = −λN (t)Δt

(1.2)

ΔN = −λN (t) . Δt

(1.3)

.

that is, also as: .

The relation just found is valid even if we make the time interval tend to zero, i.e. if we write: dN dN ΔN ≡ lim −→ = −λN (t) . . (1.4) Δt→0 Δt dt dt The equation we just wrote is called differential equation, and it expresses the following problem: what is the function . N (t) whose derivative (left-hand side of the equation) corresponds to the function itself, multiplied by a negative constant (righthand side)? Using a table of the derivatives of notable functions, it can be immediately seen that the exponential function with negative exponent has this property, i.e. the function: −λt . N (t) = N0 e (1.5) is exactly the function that satisfies (1.4). In Sect. 4.8.2 we will see a more elaborate mathematical way to solve the same problem, called variable separation. For now, if you don’t know the derivatives of simple functions, consult https://en.wikipedia. org/wiki/Differentiation_rules and do the check, or trust me. Actually, any other multiplicative constant in place of . N0 would satisfy the equation. However, only the value . N0 guarantees that the number of nuclei at the initial time .t = 0 is exactly what we had at the beginning. As we will see in the next section, the constant .λ can only be a quantity that has dimensions [T.−1 ]. Statistical methods (which you will address in the Physics Laboratory course) will guarantee that the quantity .τ with the dimensions of a time equal to 1 .τ = (1.6) λ is precisely what we call the mean lifetime. The mean lifetimes of the nuclei can go from .μs (there are also much smaller ones, but we can’t even measure them) to billions of years. The most common isotope of uranium (.238 92 U) has an mean lifetime 9 .τ = 6.4 × 10 years, comparable with the age of the universe. If the value of .τ (or .λ) is found, and we are able to obtain information on the value at .t = 0 of the number . N0 , it is possible to determine how much time has elapsed since the instant .t = 0 and date the event in time by measuring how many nuclei ∗ ∗ . N (t ) remained at a certain time .t > 0. This is also possible if we know the initial 10

Those who study physics of nuclei and the Fermi theory of weak interactions which explains radioactive decay will see how to correlate the value of .λ with the properties of nuclei.

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1 Physical Quantities and Units of Measurement

ratio between the . N0 values of two different elements, with different mean lifetimes. To understand how, follow the example below.

1.6.2 Radiocarbon Dating Carbon 12 (that is, .126 C) is the most abundant carbon isotope in nature and it is stable, i.e. it never decays. However, another isotope is present in the Earth’s atmosphere, carbon 14, .146 C. This isotope differs from the previous one (as you can see from the mass number) because it has two more neutrons in the nucleus. For chemical reactions, which only care about the number of electrons in the atom, the two carbons are indistinguishable. They are also indistinguishable for the chemical reactions involving the growth and metabolism of plants that grow, of sheep that eat grass, of carnivores that eat sheep, or that use their wool. The .146 C has an mean lifetime .τ = 1/λ = 8270 years. It is formed in the atmosphere by the bombardment of cosmic rays (those studied by the Pierre Auger Observatory). Neutrons are produced by the interaction of cosmic rays in the atmosphere,11 and those newly-produced neutrons can be absorbed by stable nuclei and form .146 C. The ratio between the quantities of the two isotopes present in the atmosphere and on the ground is stable and it is equal to: 14 6C = 1.17 × 10−12 . (1.7) .R = 12 C 6 This same relation has remained unchanged even in distant times of the past (at least tens of millions of years). Let’s try to practice: if we extract 1 gram of natural carbon obtained from a newly occurred combustion of vegetables, how many radioactive decays per minute can I count? Using the knowledge of chemistry, the number of atoms (and therefore of nuclei) present in a gram of matter is equal to . N A /A where . N A = 6.023 × 1023 is the Avogadro’s number and . A = 12 is the mass number. Thus, 1 g of carbon consists of .

NA = 5.0 × 1022 nuclei , A

(1.8)

of which: .

11

N 14 = R ×

NA = 1.17 10−12 × 5.0 1022 = 5.8 × 1010 nuclei A

(1.9)

Free neutrons are not normally available, as they are not stable particles. Because the neutron has greater mass than the proton, an isolated neutron tends to decay into a proton and other particles, as we shall see in Sect. 8.7.2, with an average lifetime of about 15 min. Neutrons bound in nuclei have a different effective mass and do not have sufficient energy to decay.

1.6 Time Measurements Without Periodic Phenomena (*)

21

belong to the isotope 14. The number of decays per time interval is given by the (1.3), i.e.: .

|ΔN 14 | 1 = λN 14 = 5.8 × 1010 = 14 decay/min . (1.10) Δt 8270 × 365 × 24 × 60

In the relation, .365 × 24 × 60 represents the number of minutes in a year. This quantity (which measures the decays per time interval) is called activity. Thus, a perfectly efficient counter for measuring electrons from radioactive decay surrounding 1 g of carbon freshly extracted from burning vegetables would measure about 14 decays per minute due to the very small fraction of a radioactive isotope. Now, we want to establish the era of a piece of coal extracted from an ancient Etruscan furnace, showing an activity of 10.0 counts per minute. What happened to produce this activity variation? For a living organism, with a metabolism interacting with the external environment, the carbon it ingest (by growing, if it is a vegetable; by eating, if it is a herbivore or carnivore) is maintained exactly in the ratio given by (1.7). As soon as the exchange with the outside ceases (the vegetable is cut, the animal dies), there is no more exchange. The nuclei of .126 C remain unchanged, while 14 12 . 6 C begins to decay, transforming into another element (nitrogen, in this case). Thus, the . R ratio of (1.7) decreases: the numerator decreases over time with the law (1.5), while the denominator remains unchanged. With this information, the physics problem is finished and only the mathematical step needed to determine how old the piece of coal is needed. You will obtain that the studied sample was produced in the Etruscan furnace 2350 years ago (see Exercise 1.1). Dating with .146 C can determine the age of an object, i.e. the period in which the object was formed. However, since its lifetime .τ is just over 8000 years, it is not possible to date objects older than about 100000 years. In general, it is difficult to date objects older than 10.τ , as the number of decaying nuclei has decreased by a factor .e−10 = 5 × 10−5 . It is necessary to change the radioactive element, choosing one with a longer mean lifetime. This has been done to date the Earth and the Solar System with techniques similar to those mentioned. The Earth has an age of 4.54 billion years (Gy), with an accuracy of about 0.05 Gy. The Sun has an age of about 4.6 Gy. Overall, the age of the Solar System is approximately 1/3 of the age of the Universe. To obtain this result, the ages of various bodies present within the Solar System were studied, and particular minerals called zircons were analyzed. These minerals are based on silica, oxygen and zirconium and can be mainly observed in igneous rocks. Under certain circumstances, they can be found in stable geological contexts and remain virtually unchanged for billions of years. Zirconia often contains small amounts of lead. The lead nucleus represents the stable element at the end of a long chain of decay processes that start from uranium .238 92 U, which has an mean lifetime of over 5 Gy. The ratio between the relative

12

The decay process of this isotope is illustrated in Sect. 8.7.2.

22

1 Physical Quantities and Units of Measurement

quantities of all the elements present in the chain (from uranium that has Z = 92, to Pb with Z = 82) allows us to accurately date the mineral, and therefore the period of its formation corresponding to that of the planet (the Earth) that contains it.

1.7 Dimensional Analysis Dimensional analysis is the decomposition of physical quantities into fundamental quantities in the context of the formulas which establish their relations. Dimensional analysis is a procedure used in physics to verify the correct use of units of measurement in formulas and equations, and it is used to check the consistency of the result with respect to the quantities used in the calculations. In my opinion, the ability to carry out dimensional analysis, to produce estimates by orders of magnitude and the ability to interpret graphs are the three main characteristics that distinguish physicists from any other professional figure.13 Dimensional analysis can be used as a consistency check in the various steps leading to a formula or in checking the final expression, using the fact that physical dimensions (those defined in the second column of Table 1.1 within the symbol “[ ]”) can be treated as algebraic quantities. These quantities can be summed or subtracted from each other only if they have the same dimensions. In addition, the terms of each member of an equation must have the same dimensions. Dimensional analysis can be used as a valuable aid in judging a priori an expression, since the necessary (but not sufficient) condition for an equality relation to be correct is that the dimensions of the physical quantities in both members of the equation are the same. It is not easy to show the power of dimensional analysis in the first chapter of a physics book, when few formulas has been introduced! For example, dimensional analysis is used to understand the units of an introduced quantity. In the first equation presented in this book, (1.1), I introduced three quantities: . N is dimensionless (it is the number of some kind of objects), .Δt is a time interval. Therefore, necessarily for the equation to be correct, the unknown quantity .λ must have dimensions of an inverse of time, i.e. .[T −1 ]. Generally, in the case of not too complicated formulas, the dimensional analysis can be performed mentally. For example, I recall that the period of oscillation of a pendulum (Chap. 4) depends on the ratio between the gravitational acceleration .g, which is an acceleration (Chap. 3) and therefore has dimensions .[L T −2 ], and the length .l of the wire, which has dimensions .[L]. I remember the ratio is within the square root, but I never remember if .g or .l is in the numerator. Doing dimensional analysis

13

During the lessons while writing on the blackboard, dimensional analysis is frequently used. In a written text this can be avoided, but I always suggest that the student practice dimensional analysis of more complex equations.

1.7 Dimensional Analysis

/ .

T ∝

23

l g

−→

[T ] = [L]1/2 [L T −2 ]−1/2 = [L 0 T 1 ]

(1.11)

I immediately verify that the correct relation has .l in the numerator. Of course, I can’t check which constant of proportionality is missing (.2π ). Normally, the verification that I formally wrote in (1.11) is done quickly mentally or on a small piece of paper. Among the conditions imposed by the dimensional analysis, there is the one for which the argument of every transcendent function must be dimensionless. The physical reason is that if I change the units of measurement in a transcendental function in which a dimensional quantity appears, the result depends on the chosen units of measurement. For example, if you wrote the law of radioactive decay as: −t . N (t) = No e the result would depend on which units of measurement I choose for the time: I would get a different number if I use for the calculation .t = 1 (min), or .t = 60 (s), even if it is the same quantity of time. There is no ambiguity if the law is written as in (1.5): . N (t) = No e−t/τ , being .τ a constant with precise units of measure (seconds, minutes or years). The mathematical reason for the statement above is linked to the fact that every function can be represented, near a point, as an expansion of polynomials. This mathematical technique is called the expansion in Taylor series. The Taylor series of a function at a point is the representation of the function as a polynomial with coefficients calculated starting from the derivatives of the function itself at that point. Normally this topic is addressed at the end of the math analysis course, but for now you can limit yourself to verifying easily, using a pocket calculator, that if .x « 1 then the following notable developments hold: .

sin x ≃ x −

x3 + o(x 3 ) 6

(1.12)

.

cos x ≃ 1 −

x2 + o(x 2 ) 2

(1.13)

x3 x2 + + o(x 3 ) 4 6

(1.14)

ex ≃ 1 + x +

.

.

ln(1 + x) ≃ x −

x2 + o(x 2 ) 2

(1 + x)a ≃ 1 + ax + o(x)

.

(1.15) (1.16)

The term .o(x) means that the following terms are all of higher powers than those expressed inside the symbol. Since .x « 1, each successive term to a higher power is much smaller than the previous one. This expansion in terms of powers of .x means that each polynomial term would have different physical dimensions if .x were not dimensionless!

24

1 Physical Quantities and Units of Measurement

Fig. 1.6 Definition of the angle measurement in radians

A particularly relevant case is that of the amplitude of the angles. In common language degrees are used. Degrees contain an unnatural convention, namely that one round angle equals .360◦ : it could have been any other number. Instead the radian is a dimensionless unit of measurement, and for this reason it is adopted in the SI. In fact, an angle measured in radians, as can be seen from Fig. 1.6, is equal to the ratio between the arc of circumference .s to its radius. The round angle corresponds to an angle .2πr/r = 2π radians.

1.8 Questions and Exercises Questions 1. The mole is the unit for measuring the amount of substance, and is defined as the amount of substance that contains .6.0225 × 1023 elementary entities. This corresponds to the numerical value of Avogadro’s constant,. N A . The molecular mass of a substance is the ratio of the mass of a given amount of that substance to the number of moles of the same amount of that substance. It is commonly expressed in units of atomic mass (u). Molecular mass can be calculated as the sum of the atomic masses of all the constituent elements of the molecule. For example, the molecular mass of water (H.2 O) is the sum of two atomic masses of hydrogen (1.00794 u) and one of oxygen (15.9994 u). How many water molecules are there in one gram? [A: .3.34 × 1022 ] 2. With the data from the previous question, how many water molecules are there in one cm.3 (under standard pressure and temperature conditions)? [A: .3.34 × 1022 ] 3. Determine the number .n of atoms per present in gold (. Z = 79, A = 197, density 19.3 g/cm.3 ) and estimate the distance between the centers of two atoms, assuming the atoms are arranged on a cubic lattice. cm.3

[A: .n = 6 × 1022 cm.−3 ; distance=2.6 .10−8 cm.] 4. In an experiment in which helium nuclei are sent against a target consisting of 1 mm of gold, it is observed that 99.9% reaches a detector that is on the beam line. Assuming that all the atomic mass is concentrated in the spherical-shaped nucleus, and neglecting the size of the

1.8 Questions and Exercises

25

helium nucleus, determine the radius of the gold nucleus. Such an experiment was used by Rutherford to determine the size of nuclei. Make use of the dimensional analysis and the number .n determined in the question 3. [A: .∼ 2 × 10−13 cm.] 5. The mass of an atom is concentrated in the nucleus. The nuclear radius is expressible in terms of the mass number . A by the relation . R = 1.2 A1/3 fm, where 1 fm=.10−15 m. The quantity . A represents the number of nucleons, that is, the sum of protons and neutrons. Estimate the nuclear density in terms of the density of water. [A: 2.3 .1014 density of water.] 6. A neutron star is an object with a mass of about 1.4. Mʘ , where the solar mass. Mʘ = 2 × 1030 kg. The neutron star is aggregated to have density equal to that of the nuclear matter determined in the question 5. Determine the radius of a neutron star. [A: .∼ 15 km] 7. The density of the interstellar medium in our Galaxy is measured to be .10−21 kg m.−3 and consists mainly of hydrogen. Estimate how many hydrogen atoms per cm.3 are present in the Galaxy. [A: 0.6 atoms/cm.3 ] 8. The Galaxy can be schematized as a disk of radius 15 kpc and thickness 200 pc. Since we know that 1 pc=.3.08 × 1016 m, determine the volume of the Galaxy and the total mass of gas in the interstellar medium. Use the data from the question 7. [A: .V ∼ 4 1060 m.3 ; . M ∼ 4 1039 kg.3 ] 9. There are about 10.11 stars in the Galaxy with mass comparable to the Sun. Determine total mass . Ms in the stars and the fraction of the Galaxy mass in the form of interstellar medium (ism). Use the value determined in the question 8 for the mass of the interstellar medium. [A: . Ms ∼ 2 1041 kg; . f ism ∼ 2%] 10. A glass filled with water has radius 3 cm; left open, in 4 h the level has dropped 1 mm. Determine, in grams/hour, the rate at which water evaporates. Also determine how many water molecules evaporate in one second per cm.2 of surface area. [A: .vevp ∼ 0.7 g/h; .∼ 5 1017 molecules cm.−2 s.−1 ] 11. A beam sending neutrinos from CERN, Geneva–Switzerland, to the Gran Sasso Laboratories in Italy, near L’Aquila, was operational between 2006 and 2012. The distance as the crow flies (i.e., along the Earth’s maximum circle passing between the two sites) is 730 km. The midpoint is approximately at the height of Pontremoli city (Tuscany region). Determine the straight-line path of the neutrinos, and how many meters below Pontremoli the beam passed. The Earth’s radius is about 6370 km.14 [A: L = 729 km; the beam passes 10.4 km below Pontremoli] 12. Looking at the relationship (1.12), determine the percent error made using the approximation ◦ ◦ ◦ .sin θ = θ for angles equal to 10. , 5. and 1. . [A: 0.5%; 0.1%; 0.01% ] 14

Note to Italian education ministers: no tunnel is needed between CERN and Gran Sasso for neutrino propagation.

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1 Physical Quantities and Units of Measurement

13. Three fundamental magnitudes connecting quantities defined in mechanics are: .(i) the speed of light in vacuum,.c = 299792458 m s.−1 ;.(ii) Planck’s reduced constant,.h = 1.05459 10−34 J s. It defines the “quantum” of angular momentum, a quantity introduced in Chap. 7; .(iii) the universal gravitation constant .G = 6.67259 10−11 N m.2 kg.−2 , introduced in Chap. 10. In the standard model describing the macrocosm (Big Bang theory) the combination of these fundamental quantities assumes an extremely important role. Using dimensional analysis, obtain by combining .h, c and .G a quantity with the dimension of a mass. This quantity is denoted by .m P and is called Planck mass. Determine its value the kg of .m p . [Note: 1 N (newton) = 1 kg m s.−2 ; 1 J (joule) = 1 kg m.2 s.−2 ]. [A: .m P = 2.17 10−8 kg ] 14. Still making use of the three quantities introduced in the question 13, using dimensional analysis get the quantity with the dimensions of a time, .t P , called Planck time. Express .t P in s. [A: .t P = 5.4 10−44 s ] 15. Still making use of the three quantities introduced in the question 13, using dimensional analysis get the quantity with the dimensions of a length, .l P , called Planck’s length. Express .l P in m. [A: .l P = 1.16 10−35 m ] Exercise 1.1 The law of radioactive decay is given by the relation . N (t) = No e−λt , where . No is the initial number of nuclei of a certain substance, . N (t) those remaining after a certain time .t and .λ a constant characteristic of each radioactive substance. The half-life, .t1/2 , is the time required to observe the halving of the initial number of elements. The activity . A(t) is the number of decays per unit time (units: 1 disintegration/s = 1 Bq). The .14 6 C isotope of carbon has half-life .t1/2 = 5730 years and is naturally occurring at a concentration, relative to the stable isotope .12 6 C, equal 12 −12 . The .12 C has concentration in nature equal to 99% of the total. to .r = 14 6 C/. 6 C = 1.17 × 10 6 Avogadro’s constant is equal to 6.02 10.23 mol.−1 . Determine the functional relation between .t1/2 and .λ and the value of .λ for .14 6 C. Show that the functional relation between the activity . A(t) and .λ is given by . A(t) = λN (t). Determine the activity of 1 g of natural carbon freshly obtained by combustion. The measured activity of a piece of coal of mass m = 25 g from an Etruscan furnace is 250 decays/minute. Determine the age of the sample. 5. The dating limit using the .14 6 C technique is about 50000 years. In a sample with such an age, what is the percentage of .14 6 C radioisotopes remaining? 1. 2. 3. 4.

Chapter 2

Vectors and Operations with Vectors

Abstract A scalar is a quantity uniquely described by a real number, as mass, temperature, volume, pressure,... Other quantities, such as the position or the displacement, need additional information. These quantities are called vectors, and each is characterized by an intensity and a direction. In this chapter, after introducing the most widely used coordinate system (the right-handed Cartesian orthogonal one) I introduce the vectors, their representation, the operations (sum, difference, scalar and vector products), and the reason why they are so important in the study of kinematics and dynamics.

2.1 Introduction: Scalars, Vectors and Tensors In physics we will find several classes of quantities. The first, most common, are scalar quantities. A scalar quantity is uniquely described by a real number, called .scalar , associated if necessary with a unit of measurement. Examples of scalar quantities are mass, temperature, volume, pressure,…Other quantities, such as position, need additional information. These quantities are called vectors and each is characterized by an intensity and a direction. The prototype of vectors is the displacement: not only do we need to know how much to move, but also where to go, that is, in which direction. In mathematics a vector is an element of a particular algebraic structure, called vector space. Vector quantities are, in addition to position and displacement, velocity, acceleration, force, angular momentum,…Finally, we will encounter auxiliary quantities called tensors. These quantities make it possible to link scalar quantities or vector quantities together. For example, in Chap. 12 we will see that the most general relationship between angular momentum and angular velocity (both vectors) passes through a tensor quantity.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Spurio, The Fundamentals of Newtonian Mechanics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-47289-3_2

27

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2 Vectors and Operations with Vectors

In this chapter, after introducing the most widely used coordinate system (the Cartesian orthogonal one) I will introduce the vector quantities so that they can be used for kinematics and dynamics studies. We will return to vectors and their symmetry properties in the Chapter 9, going into more detail about what, from the point of view of physics, are the characteristics that make them so important.

2.2 Cartesian Coordinate Systems The task of the first part of our studies is to locate the position of objects and their variation over time. For this reason, it is necessary to define a coordinate system. This is a system against which a certain physical phenomenon or physical object is observed and measured. One thing we will have to take into account (and we will do so over several chapters) is that motion is always relative to a reference frame with respect to which we observe it. We will see also, at the beginning of the next chapter, the connection between reference frames and coordinate systems. Experience has taught us that the physical space in which we live is a threedimensional space of Euclidean nature. It takes three pieces of information to locate a point in this space. In many situations, we do not need all three coordinates: when we observe a car (or a cyclist, more ecological) moving on a defined circuit, the only information needed is where it is along the circuit, how many kilometers it has traveled since the start. Implicitly, we are assuming that the size of the car or cyclist is small and negligible compared to the size of the circuit. In order to study an event that occurs on this pre-determined path, only one coordinate is needed. In other circumstances, motion occurs on a plane. In this case, there is no need to mention the quota (which remains constant, or pre-defined). It only takes two pieces of spatial information to completely localize the object. One possibility for doing this is to use a two-dimensional orthogonal Cartesian coordinate system, Fig. 2.1. In this system, the position of point P is uniquely defined by two numbers: the distance of the projection of P on the x-axis, .x P , and the corresponding distance on the y-axis, . y P . The two axes of this system (as the name implies) are orthogonal to each other. Later, we will see other possible coordinate systems, starting with the one with plane polar coordinates, Sect. 3.7. The most general case is when the body can move freely. Its position P also needs the third information, Fig. 2.2, which is the distance of the projection of P along the vertical axis, .z P . The .z-axis is mutually perpendicular to the .x and . y axes of the two-dimensional system. In this way, the coordinate is uniquely defined by the triplet: .(x P , y P , z P ) (2.1)

2.2 Cartesian Coordinate Systems

29

Fig. 2.1 Cartesian orthogonal coordinate system in two dimensions. The position of point P is uniquely defined by two numbers .(x P , y P )

Fig. 2.2 Right-handed Cartesian orthogonal coordinate system in three dimensions. The position of point P is uniquely defined by the tern of numbers .(x P , y P , z P )

2.2.1 Right-Handed Coordinate System and Right-Hand Rule In the previous discussion, there is an important aspect to specify: how the direction of the .z axis should be relative to the other two. I use Fig. 2.2 to illustrate the situation. In the figure, the .x axis is pointing toward the reader, and the . y axis lies on the horizontal plane of the paper. In this configuration, should the .z axis go upward (as it is drawn) or downward? There is no way to make the problem disambiguous, without fixing a convention. A convention is a rule tacitly accepted by all, so as

30

2 Vectors and Operations with Vectors

Fig. 2.3 Illustration of how to apply the right-hand rule to a triplet of Cartesian axes: the palm of the hand along the x-axis, the rotation of the fingers toward the y-axis. The orientation of .z is given by the thumb. The same operation will apply when we define the vector product between vectors

to break the ambiguity. The rule defined in this case (and in any other situation in the future where similar ambiguity arises again) is called right-hand rule. There are many ways of defining and applying it. My preferred method is the one depicted in Fig. 2.3. The right-hand rule: Using your right hand, imagine to accommodate in the palm the first of the vectors that appears in a triad (as the three axes.(x, y, z) of a Cartesian system), or in an equation. Now imagine rotating your palm as if you wanted to bring the first vector onto the second, with a rotation of the smaller angle. The orientation of the thumb indicates the positive direction of the resulting vector. In the case of the coordinate system of Fig. 2.2 there is a triplet of objects, the three axes .x, . y and .z. According to the rule just defined, we imagine rotating the first one (.x axis) about the second one (. y axis) in accordance with the indicated minor angle (the opposite direction would imply a rotation of 270.◦ ). The thumb results facing up, so the positive direction of .z is exactly as indicated in the drawing. If you try to do the same operation with your left hand, you will find the direction of .z reversed! This is why everyone must agree on one convention (even if you are left-handed). Similarly, the direction of .z is reversed if you swap the .x axis with the . y axis.

2.3 Representation of Vectors 2.3.1 Vectors in Intrinsic Representation The historically useful way of representing the position of an object is through the use of the concept of vector. If you want to know the position of a point, with respect to a reference point O, just “shoot an arrow” (figuratively speaking) from point O to point P, as shown in the left-hand side of Fig. 2.4. To get to the point, just follow the arrow, without needing any predefined coordinate system. It may be not practical,

2.3 Representation of Vectors

31

Fig. 2.4 Left: a vector in intrinsic representation. Only indicated is the start (O) and the arrival (P) connected with an arrow. The same vector is represented within a Cartesian coordinate system, in ˆ are indicated which Cartesian unit vectors .(ˆi, ˆj, k)

but it is a perfectly legitimate way of indicating the position of a point-like object, called intrinsic representation. Mathematically vectors are treated as oriented segments. Two oriented segments are said to be equipollent if they have the same length (magnitude) and direction. The operations between vectors (necessary for calculating segment lengths, included angles, areas, volumes, etc.) are the translation into algebraic form of possible operations between space-oriented segments, in accordance with the laws of Euclidean geometry. A vector is referred to in a printed text as a quantity written in bold type, .a. On the blackboard, or when writing by hand, the notation of placing an arrow above the letter denoting the vector, .a, is used.

2.3.2 Vectors in Cartesian Representation If the universe were made up of only one observer, the use of intrinsic vectors would suffice. However, it is clear that any sort of communication between the observer in O and others becomes extremely uncomfortable if a coordinate system is not predefined. Again, we can make use of a Cartesian coordinate system1 like the one shown on the right-side of Fig. 2.4. The same vector, denoted.a, can now be measured by the length of its three components: a = (ax , a y , az ) .

.

1

(2.2)

From now on, when not necessary, we no longer mention that the axes must be orthogonal and the system necessarily right-handed.

32

2 Vectors and Operations with Vectors

The three quantities to the right of the equality are not in bold: each value indicates a scalar distance with respect to a defined axis (as if operating on the predefined trajectory represented by each axis). From (2.2) it is clear that you need three quantities to define a vector. In a Cartesian representation, the three quantities are precisely the distances of the projections along the three axes. If we wanted to simply express the length of the vector, its magnitude, without caring about its orientation, it is clear from Fig. 2.4 that we can use the Pythagorean theorem. The magnitude of a vector is a completely different quantity from the vector2 and it has a different notation. It can often be denoted by the vector symbol within bars, or by simply transcribing it not in bold or without the arrow above it. In the case of the vector .a defined above, its magnitude is: . ax2 + a 2y + az2 .

|a| = a =

.

(2.3)

In most cases when the vector represents a generic physical quantity, its magnitude (2.3) is also called the intensity of the vector. Notice also that a representation of a vector in Cartesian coordinates as given by (2.2) is equivalent to defining a vector in intensity and direction as mentioned in the introductory section. In fact, the intensity corresponds to the length (magnitude) of the vector. The direction is given by the orientated line containing the segment defining the vector .a, and that can be established through geometric methods using the components.

2.3.3 Unit Vectors in Cartesian Representation Still referring to Fig. 2.4 there is an alternative way of representing a vector, and that is by using unit vectors. A unit vector is, by definition, a vector divided by its magnitude. Thus, if we locate the point P through the vector called .r with Cartesian coordinates: . (2.4) .r = (x, y, z) with magnitude r = x 2 + y 2 + z 2 its unit vector is defined as: .

rˆ ≡

r r

(2.5)

Note that as it is defined, a unit vector is always a dimensionless quantity. Unit vectors are always defined with the symbol of a French circumflex accent,3 as it appears in (2.5). 2

One must be very strict and not confuse, when speaking and writing, a vector with its magnitude. In many textbooks, unit vectors are denoted by the symbol .uˆ other where “other” is the symbol I use to denote the unit vector. For example, the unit vector along the .x axis in this notation becomes: .uˆi . I prefer the streamlined version. 3

2.3 Representation of Vectors

33

The concept of a vector is so important in physics that we will discover many quantities that have vector behavior. These vector quantities may have physical dimensions other than .[L], which is the dimension of the position vector. The magnitude (intensity) of these quantities by definition will have the same physical dimensions as the vector, and thus the same units in the SI. A unit vector, because it is defined as the ratio of any vector to its magnitude, has no physical dimensions whatever are the units of the considered vector. It only indicates the direction of the arrow that represents direction. The particularly interesting case is when the vector is along the three Cartesian axes. The quantities are defined as Cartesian unit vectors: ˆi ≡ (r, 0, 0) ; ˆj ≡ (0, r, 0) ; kˆ ≡ (0, 0, r ) r r r

.

(2.6)

The three unit vectors are shown in Fig. 2.4. Thus, any vector quantity can be written as the sum of three vectors along the three components. For example, the position is indicated as: .r = x ˆ i + y ˆj + z kˆ (2.7) and the vector a in (2.2) as: a = ax ˆi + a y ˆj + az kˆ .

.

(2.8)

In the rest of the book, we will indifferently use (2.2) or (2.8) notation to represent a vector in Cartesian coordinates. A third notation for vectors will introduced in Chap. 9. Kinematics (Chap. 3) deals with studying how the vector r varies over time, describing the position of a moving object, i.e. r = r(t). Using the notation just introduced, the unit vector co-moving with the position becomes: rˆ ≡

.

x ˆi + y ˆj + z kˆ r =. . r x 2 + y2 + z2

(2.9)

In this notation, Cartesian coordinates are understood to be functions of time, [x(t), y(t), z(t)]. Other unit vectors will be defined later, along with other coordinate systems that are not Cartesian orthogonal, and in different frames of references. Note ˆ which are fixed, anchored the difference between the Cartesian unit vectors, (.ˆi, ˆj, k), to the reference frame, and the unit vector .rˆ , which is free to vary if the point P moves; this is why it has been defined as co-moving.

.

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2 Vectors and Operations with Vectors

2.4 Product of a Scalar and a Vector The position of a point as defined by (2.2) or (2.7) represents the prototype of a vector, an object composed of three components. The properties of this quantity are such that they are independent of the choice of the frame of reference. On this, we will return in detail in Chap. 9. Now we want to define possible operations on these quantities. The first simple relationship is the following: what happens if I multiply the vector by a scalar quantity? Intuitively, if I have an arrow that is “2” units long and multiply it by 3, I will get an arrow that is “6” units long and always oriented like the previous one. This is easy to see by operating, for example, on (2.2): if .α is any real number: αa = α(ax , a y , az ) = (αax , αa y , αaz ) .

.

(2.10)

What happens if the scalar is negative? In the case of.α = −1, the coordinates of (2.2) would all change sign. Looking at the drawing in Fig. 2.4 you would find that the arrow points to the quadrant where all the coordinates are negative, that is, the arrow would point in exactly the opposite direction of that represented. Multiplication by a negative number reverses the direction of the vector. What happens if the scalar is a dimensional quantity? In that case, the new quantity is always obtainable by the relation (2.10), but its physical dimensions are different and it therefore corresponds to a different physical quantity. For example, when we multiply the vector acceleration, .a, by the scalar mass, .m, the resulting quantity is a force, a quantity that have different physical dimensions with respect to acceleration.

2.5 Sum and Difference of Vectors If we take as the prototype of a vector the displacement of a particle (the object whose position is indicated by the vector), we can consider two successive displacements of the same particle. For example, consider the box on the left side of Fig. 2.5. We

Fig. 2.5 Left: sum between two vectors of vectors using the triangle rule. Center: sum between two vectors using the parallelogram rule. Right: difference between two vectors

2.5 Sum and Difference of Vectors

35

move first by a vector .a and subsequently by a vector .b. The result is that we have arrived at the position given by the tip of vector .b. Each vector .a and .b is an oriented segment of a line in space. The two vectors, and thus the two lines, have one point in common, the point where the first vector ends and the second begins. According to Euclidean geometry, two lines in space that have a point in common define a plane. Always remember this, namely that we are operating within the framework of Euclidean geometry. On this plane (defined by the vectors .a and .b) we can then draw another oriented segment starting at the beginning of .a and ending at the final point of arrow of .b. We can define this new vector as .a + b, drawn with green color in the figure. This new vector is the displacement corresponding to the sum of the individual vectors in a plane: this graphical property of summing vectors is called triangle rule. Note (middle frame of Fig. 2.5) that we could take an equipollent vector .b (i.e., a vector with the same magnitude and direction but with a different point of application). The sum of the two vectors is now represented by the diagonal of the parallelogram: this graphical property of summing vectors is called parallelogram rule. In Cartesian coordinates, if the vector .a is given by (2.2) and .b = (bx , b y , bz ), then the sum of the two vectors is the vector c = a + b = (ax + bx , a y + b y , az + bz ) .

.

(2.11)

or, using the notation with the unit vectors introduced in (2.8): c = (ax + bx )ˆi + (a y + b y )ˆj + (az + bz )kˆ .

.

(2.12)

The demonstration is intuitive, if you project the lengths of the vectors in Fig. 2.5 along the Cartesian axes. Important: Note that from the point of view of physics we are constrained by the fact that the dimensions of the two vectors that sum must be identical. We can sum accelerations with accelerations, for example, and not accelerations with other quantities with different dimensions. Equations in physics must be homogeneous, so we cannot sum non-homogeneous quantities. Homogeneous means that the quantities must necessarily have the same physical dimensions. Two quantities that can be added can also be subtracted. Given two vectors, .a and .b, the difference between .a and .b is denoted by .a − b. It corresponds to that vector which when added to .b results in the vector .a. It is immediate to verify this property in the right-hand box of Fig. 2.5. The graphical method for the difference between two vectors, if they have the starting point in common as in the figure, is to draw an arrow from the tip of the second vector to the tip of the first. Such a rule is sometimes called the parallelogram rule for the difference. It is left to the Algebra course to demonstrate (admittedly, rather simply) that vectors form a vector space with algebraic properties such as: • there exists a neutral element (the zero) such that: .a + 0 = a; • there exists the inverse element (opposite vector) such that .a + (−a) = 0;

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2 Vectors and Operations with Vectors

Fig. 2.6 Graphical representation of a vector .r12 that, instead of originating in the origin of the Cartesian coordinate system, starts at a point 1 and ends at a point 2

• the sum between vectors has the commutative property: .a + b = b + a; • the sum has the associative property: .(a + b) + c = a + (b + c); The difference between two vectors will play a significant role in the definition of velocity in the next chapter. The difference between vectors is important in defining a vector with origin different from the origin of the frame of the coordinate system. Referring to Fig. 2.6 we may be interested in describing a vector that starts at position 1: .r1 ≡ (x1 , y1 , z 1 ) and ends at position 2, given by .r2 ≡ (x2 , y2 , z 2 ). In this way, the projections along the three axes of the Cartesian coordinate system correspond to .(x2 − x1 , y2 − y1 , z 2 − z 1 ); thus, the difference between the two vectors can be written as: r − r1 ≡ r12 = (x2 − x1 , y2 − y1 , z 2 − z 1 ) .

. 2

(2.13)

2.6 Scalar Product Two vectors can be multiplied by each other. There are two ways to multiply two vectors: one operation that provides a scalar quantity, and a second operation that provides a vector quantity. Here, I present the first case in two different representations.

2.6 Scalar Product

37

Fig. 2.7 Definition in intrinsic representation of the scalar product between two vectors

2.6.1 Scalar Product in Intrinsic Representation In the intrinsic representation (i.e., freed from any coordinate system) two vectors with a common point are coplanar, Fig. 2.7. The scalar product between two vectors is denoted by .a · b (read: a dot b), and produces the scalar result: c = a · b ≡ ab cos θ

.

(2.14)

where .cos θ is the angle between the two vectors. This operation is also called dot product between vectors, because of the symbol (.·) used to define it. Note that the quantity .b cos θ can be interpreted as the length of the projection of the vector .b onto .a as shown in Fig. 2.7 (or, quite equivalently .a cos θ is the projection of the first vector onto the second). Again, demonstrations are left to the Algebra course that the scalar product between vectors has the: • commutative property: .a · b = b · a; • associative property with respect to a scalar: .(αa) · b = a · (αb) = α(a · b); • distributive property: .(a + b) · c = a · c + b · c; Note that the square of a vector corresponds to the scalar product of a vector by itself: a2 = a · a

.

(2.15)

and thus, the scalar product between two different vectors represents a kind of extension to the concept of the square of a vector when using two different vectors.

2.6.2 Scalar Product in Cartesian Representation We have shown that vectors have a Cartesian representation in which, for example, a = (ax , a y , az ) and .b = (bx , b y , bz ). In that representation, their scalar product is:

.

a · b = a x b x + a y b y + az bz

.

(2.16)

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Let us show that (2.16) is equivalent to the intrinsic definition (2.14). We use the fact that any frame of reference (translated or rotated) can be used to represent vectors. The formal demonstration of this is deferred to Chap. 9. We then choose the coplanar Cartesian system with the vectors .a and .b, such that .az = bz = 0. Further, we choose the .x axis such that it is coincident with the vector .a, so that its Cartesian coordinates are .a = (a, 0, 0). In this way (see Fig. 2.7) the vector .b is rotated with respect to the first and has components: .b = (b cos θ, b sin θ, 0). Thus, by applying the (2.16) we obtain: .a · b = a x b x + a y b y + az bz = ab cos θ + 0 + 0 (2.17) that corresponds exactly to (2.14). There is an alternative way of looking at it. We have defined the three unit vectors along the three Cartesian axes in (2.6). By definition, each of them is orthogonal to the others, so the scalar product of two different unit vectors is null (they have .cos(π/2) = 0). Conversely, the scalar product of the unit vector with itself gives 1 (the magnitude is unit, and .cos 0 = 1). Ultimately, we have the following relations: ˆi · ˆi = 1 ; ˆj · ˆj = 1 ; kˆ · kˆ = 1 .

ˆi · ˆj = 0 ; ˆi · kˆ = 0 ; ˆj · kˆ = 0

(2.18)

Therefore, if I represent the two vectors in terms of unit vectors a = ax ˆi + a y ˆj + az kˆ .

b = bx ˆi + b y ˆj + bz kˆ

(2.19)

then their scalar product is exactly that given by (2.17), based on the distributive property of the scalar product and the (2.18) rules on unit vectors.

2.7 Vector Product Between Vectors 2.7.1 Vector Product in Intrinsic Representation The vector product between two vectors .a and .b is defined as c =a×b,

.

(2.20)

(read: a cross b) the vector that has magnitude equal to the area of the parallelogram formed by the two vectors .a and .b (see Fig. 2.8), i.e. |a × b| = ab sin θ .

.

(2.21)

2.7 Vector Product Between Vectors

39

Fig. 2.8 Definition in intrinsic coordinates of vector product (or cross product) between two vectors. The magnitude of the cross product .a × b corresponds to the area of the parallelogram that the two vectors span

The vector .c is perpendicular to both vectors, and thus normal to the plane containing a and b. The direction is determined by the right-hand rule, as expressed in Sect. 2.2.1. The operation defined above is also called cross product between vectors, because of the symbol (.×) used to define it. Also in this case, demonstrations of the following properties of the vector product are left to the Algebra course: • • • •

anti-commutative property .a × b = −b × a associative property for a scalar: .(αa) × b = a × (αb) = α(a × b); distributive property: .(a + b) × c = a × c + b × c; property of the BAC-CAB: .(a × b) × c = b(a · c) − c(a · b) (Question 13).

In particular, we will make use of the triple vector product property called (to remember it mnemonically) of the BAC-CAB. Rules similar to those of the scalar product for unit vectors are extremely useful. We have that the vector product between the same unit vector is null (the sine of the angle is null), while the vector product between two different unit vectors gives the third unit vector (with the + or .− sign depending on whether cyclic or anti-cyclic order is followed): ˆi × ˆi = 0 ; ˆj × ˆj = 0 ; kˆ × kˆ = 0 .

ˆi × ˆj = kˆ ; ˆj × kˆ = ˆi ; kˆ × ˆi = ˆj ˆj × ˆi = −kˆ ; kˆ × ˆj = −ˆi ; ˆi × kˆ = −ˆj

note that the last line is obtainable from the previous one.

(2.22)

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2.7.2 Vector Product in Cartesian Representation Using the definition of the vectors .a and .b as given in (2.19), the vector product between two vectors is written: c = a × b = (a y bz − az b y )ˆi + (az bx − ax bz )ˆj + (ax b y − a y bx )kˆ

.

(2.23)

You can easily develop the .3 × 3 terms of the product and use the relations between unit vectors shown in (2.22) with three of the nine terms being null as unit vectors multiplied by themselves. Mnemonically, it is very useful to represent the vector product in terms of the development of the determinant.4 In this way you can write the (2.23) as the development of a .3 × 3 determinant with on the first line the three unit vectors, and on the second and third lines the Cartesian components of the two vectors in the exact order in which they are written: ˆi jˆ kˆ c = a × b = a x a y az b x b y bz

.

(2.24)

We use the two vectors as written in (2.17) and shown in a three-dimensional frame of reference in Fig. 2.8 a = aˆi ; b = b cos θˆi + b sin θˆj

(2.25)

c = a × b = ab cos θ(ˆi × ˆi) + ab sin θ(ˆi × ˆj) = ab sin θkˆ

(2.26)

.

then

.

i.e., the same as the definition given in the intrinsic representation, (2.21).

2.8 Area and Volume in Vector Spaces (*) The fact that the magnitude of the vector .c given by (2.20) corresponds to the plane area of the parallelogram formed by the two vectors .a and .b, suggests that we can associate the “surface” with a vector quantity. In fact, if we have two oriented segments .a and .b lying in a plane with one end in common (exactly as in Fig. 2.8), we can define the area A (vector) of the parallelogram formed by them as A≡a×b,

.

4

(2.27)

When you have done this the Algebra and Geometry. If this is not the case, you can go to the next section for the time being.

2.8 Area and Volume in Vector Spaces (*)

41

Fig. 2.9 Left: Projection of a flat surface onto the Cartesian orthogonal planes. Right: A closed surface is represented by a null surface!

which corresponds exactly to the definition of cross product (2.20). The direction of A is perpendicular to the plane containing the two oriented segments. Like all definitions, this one is useful when it can be made more general and, especially, if it has applications in physics. In the case of generalization: what happens if the surface is not that of a parallelogram? What if the surface is not flat? What if the surface is closed (like that of a rugby ball, for example)? Suppose first that the surface is flat, but has any shape such as in Fig. 2.9 on the left. We can define the area vector A with the magnitude that has its surface area (measured in any way) and direction perpendicular to the surface. How to identify the orientation? That is, which positive direction do we choose? In the case of the two oriented segments in Fig. 2.8 the ambiguity (upward, not downward direction) is broken by the right-hand rule. But for any surface, this is not possible. However, we can solve it this way: the surface has an edge (its circuital perimeter), and the positive direction of A corresponds to that indicated by the thumb of the right hand, when the palm that encircles the circuit moves in a counter-clockwise direction. What is the utility of this? The components of A have a simple geometric meaning. Let us take a plane surface that forms an angle .θ with the horizontal plane .x y (Fig. 2.9 on the left). The geometric projection of the area considered on the .x y plane is . A cos θ. But the normal to the plane of which the surface A is a part forms an angle .θ with the .z axis, and this corresponds to the direction of that vector. Thus, the . z component of the vector A drawn in the figure is precisely . A z = A cos θ. In the situation of the figure, you can easily verify that . A x = 0 and . A y = A sin θ. In the case where the area A was instead oriented completely randomly in the considered Cartesian coordinate system, you would find that . A x correspond to the geometric projection of the surface on the . yz plane and . A y to that on the .x z plane. Thus, the

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three components of the vector A correspond to the areas of the projections onto the three orthogonal planes defined by the Cartesian axes. What happens if the surface is not flat? It can always be divided into a large number, .i = 1, 2, . . . , N , of very small flat areas and each represented by the vector .Ai . Each very small area is like a small tile that must cover a dome, or a curved surface. The vector representing the curved surface is then A ≡ A1 + A2 + · · · + A N =

N .

.

Ai .

(2.28)

1

Note that in this case the magnitude of A is not equal to the area . of the curved Ai = N A 1 surface, which corresponds to the sum of the areas of the tiles, i.e.,. A = (assuming in the last equality that all the tiles have the same area). However, even in this case the values of the three components of A are equal to the areas of the projections on the three planes .x y, x z, yz of a Cartesian frame of reference. If you were to buy hilly land to cultivate it you would be interested in area . A, but if you plan to build a house you would be interested only in . A z . Finally, considering a closed surface, as in Fig. 2.9 on the right, we obtain that the value of its area vector is zero. In the case of closed surfaces, the area vector of each surface element is conventionally directed outward from the surface. Thus, we can always associate small surfaces in pairs such that their combined projection on a plane is zero. In the figure the two areas .A1 and .A2 (with . A1 /= A2 ) have the same projection on the .x y plane, but with opposite signs, i.e. in the Cartesian system . A 1z = a and . A 2z = −a. By similar reasoning, imagining to decompose the closed . surface into many symmetrically arranged surfaces, we have that .A = i Ai = 0. The vector representing a closed surface is zero. Finally, is it possible to define the volume using this vector representation? Solid geometry provides the solution. If you have a parallelepiped with a certain base area, its volume is given by the base area times the projection of the third segment along the direction at the normal to the base, see Fig. 2.10. Let us call c the third vector

Fig. 2.10 Three vectors define a parallelepiped. The volume is given by the product between the area of the base parallelogram and the height

2.9 The Equals Are Not All Equal

43

and let .α be the angle it forms with the direction perpendicular to the base. The projection of c along that direction is .h = c cos α, while the magnitude of the base area is . A z = |a × b| = ab sin θ. Thus, the volume of the parallelepiped corresponds to .V = abc cos α sin θ = |a × b|c cos α. You can then easily convince yourself that this corresponds to the relation for volume: .

V = (a × b) · c .

(2.29)

Note that volume is always a scalar quantity. For those interested in algebra, they can verify that the volume of the parallelepiped corresponds to the development of the determinant given by:

.

c x c y cz V = (a × b) · c = ax a y az b x b y bz

(2.30)

The concept of area as a vector quantity, of sum (or integration) over surfaces of any shape and the closed circuits that represent the edge of those surfaces, and the implications between circuits, areas and volumes will be aspects of fundamental importance when you study the laws of electromagnetism. We will introduce the concept of infinitesimal area and infinitesimal volume in Sect. 10.9.

2.9 The Equals Are Not All Equal You may have already noticed that in previous pages two symbols (.≡ and .=) have been used to represent an equality in an equation. First, I clarify that I use the symbol “.≡” when I intend to define, baptize, a certain quantity. For example, I just defined in the previous section the vector area, and I will define linear momentum as .p ≡ mv and angular momentum .L ≡ r × p. The definitions are harmless; I could define as many things as I like.5 However, in physics only those definitions that will produce new quantities that, in addition to having mathematical consistency, can be measured will be of interest. I can define mathematical entities that have no interest in the real world (they do not refer to quantities that I can measure). These are perhaps elegant objects, but certainly totally useless. Instead, I use the symbol “.=” when the right-hand side of the equation represents a mathematical consequence of the left-hand side. For example, in the case I write .(x 2 − 1) = (x + 1)(x − 1). This is the most common case. If I want to express that the equality is only approximated, I use the symbol “..” . For example, when I sum two quantities, one of which is very small compared to the other. In the case of an object falling a few meters from the ground, the initial 5

For example, in Chap. 9 I will try to define a quantity which, unfortunately, will not be successful.

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distance of the object from the Earth center is . RT + h . RT , if . RT = 6300 km represents the Earth’s radius. Or, when making use of the approximations of Taylor’s theorem, as described in Sect. 1.7. In case the equality expresses only a physical approximation by orders of magnitude or by dimensional analysis, I use “.∼”. For example, from the dimensional analysis the period .T of a pendulum depends on the length .l and the acceleration of gravity .g via the relation .T ∼ (l/g)1/2 . Missing, in my opinion, is the symbol to represent the most important of the equalities in physics: the equality that expresses correspondence between quantities based on empirical, experimental observations. We have already encountered one such equality in (1.1). We will find many others, such as when we write .F = ma. This relation expresses the second law of dynamics, we are NOT introducing a definition: it is not at all a definition. We are not even doing algebrical steps that require equality. We are not writing an approximate expression. We are writing a relationship that arises from empirical observation. Maybe I could invent a symbol like: N atural Law .

. .. . F = ma

N

or

.... F = ma

(2.31)

but maybe it would be excessive. The important thing to understand is that these laws are not mathematically deductible (needless to look up a mathematical proof ), but are provable or refutable only in the laboratory, with measuring instruments and observations. This is the goal of your future work, if you are interested in becoming a physicist. During your career as a student you will encounter many equalities similar to (2.31) that are not expressed in this way, but which you must recognize as such.

2.10 Laws and Principles, Physics and Mathematics The working scheme of mathematics takes Euclidean geometry as its prototype. Euclid’s geometry consists of the assumption of five simple and intuitive concepts, called axioms or postulates, and, derived from these postulates, of the theorems, which are propositions without contradiction with the axioms. The other disciplines of mathematics that are based on postulates do not need to verify that these postulates do not contradict reality of the world in which we are living. Mathematics is not an experimental discipline. In physics we cannot rely on postulates, as in mathematics. Our work is based on N

.... the derivation of empirical laws, the ones we should denote by “. = ”. The power of physics is that empirical laws can be independently verified by different experimental groups (sometimes in cooperation, sometimes in competition with each other), at different times, with different tools. We know very well that some laws are not “absolute” and have validity only for certain values of the parameters. All linear mechanical systems are based on an equation of the type . F = −kx, where . F is

2.10 Laws and Principles, Physics and Mathematics

45

a force and .x is the displacement with respect to an equilibrium position. No one dreams that the law holds for “every possible” value of .x. The law of universal gravitation, which we will find and discuss in Chap. 10 is much more general and holds for an enormously wider range of distances .r . However, no one opposes those who try to test whether the law of gravitation must be modified for very large (or very small) values of .r . Thus, even though Newton’s laws of dynamics were (and are!) enormously successful because they are predictive and allowed to achieve unhoped-for goals6 no one scrambled when Einstein found a more accurate formulation of the dynamics. This new dynamics (the one called relativistic) allows for more accurate predictions when the velocities of bodies are close to the speed of light but, in the case of velocities .v . c, the usual classical Newtonian formulas (limit of low velocities) return. However, even in physics we have principles. A principle consists of an assumption on which several theories are based and is assumed to be true on the basis of a huge number of independent experimental observations, which for this reason leave to assume a general validity. There are some principles that will be stated later in this text. For example, the principle of: • • • • • • •

inertia; relativity of frames of reference; conservation of linear momentum; conservation of angular momentum; conservation of energy; equivalence between inertial and gravitational mass; homogeneity and isotropy of the universe.

In the continuation of your studies you will find more: the principles of thermodynamics; the principle of conservation of electric charge; Heisenberg’s uncertainty principle; Pauli’s exclusion principle; … Please, use the word principle only in cases where the statement is really a physical principle in the modern sense (the one above). Often the word principle is used totally out of place: when we speak of “principles of Dynamics” or even worse “Newton’s principle of action and reaction”. These are not principles at all: the Newtonian laws of dynamics are a special (low-speed) case of the equations of relativistic dynamics; the law of action and reaction is valid under some simple circumstances (I describe it in detail in Sect. 7.8).

6

Don’t just think about the fact that man went to the Moon. Think about how difficult it is to make a photocopy or a printout, guiding the microscopic ink droplets precisely to form the letters you want! The motion of these droplets is dictated by Newtonian laws.

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2.11 Questions 1. What relation must be valid between the vectors a and b, which are different from each other and nonzero, so that the relation: .(a + b) × (a − b) = 0 is verified? [A: angle .θ = 0 + nπ] 2. Show that if the magnitudes of the sum and difference between two vectors are equal, then the vectors are perpendicular to each other. 3. In a Cartesian reference system two vectors are defined as .a = 2cˆi − ˆj + 4kˆ and .b = 4ˆi + ˆ For which values of .c are the two vectors orthogonal? (c − 1)ˆj + 5k. [A: .c = −3] 4. Show that if the sum and difference between two vectors are perpendicular, then the magnitudes of the two vectors are equal. 5. Two vectors a and b are equal in magnitude. Their sum has magnitude 4 and their vector product magnitude 16. Determine the magnitude of the two vectors. √ [A: . 20] 6. Two vectors a and b comply with the following conditions:.(i).a · b = 20;.(ii).(a + b) · a = 36; .(iii) .(a + b) · b = 45. Determine the magnitude of the two vectors and the angle .α between them. [A: .a = 4, b = 5, α = 0◦ ] 7. A displacement s of magnitude 3 m is made in a Cartesian system at an angle .θ = 30◦ with the .x-axis. Express the vector in Cartesian coordinates. ˆ m] [A: .s = (2.60ˆi + 1.5ˆj + 0k) ˆ and .b = (2ˆi − ˆj + k) ˆ gives 8. Find the vector .c which added to the vectors .a = (4ˆi + 3ˆj − 2k) ˆ as a resultant .r = (4ˆi + 3ˆj + 1k) [A: .c = (−2ˆi + ˆj)] 9. Given the two vectors a and b of the Question 8 express their sum, difference, scalar product and vector product in Cartesian coordinates. What is the angle .α between them? [A: .α = 77◦ ] 10. Given two vectors a and b, show that in intrinsic representation their vector product.A = a × b corresponds to the oriented area of the parallelogram defined by the two vectors. Calculate the magnitude of the area in the case of the two vectors defined in the Question 8. [A: . A = 12.9.] 11. Given three vectors a, b and c show that in intrinsic representation the magnitude: .V = (a × b) · c corresponds to the volume of the parallelepiped defined by the three vectors. Calculate the volume in the case of the vectors a, b defined in the Question 8, and .c = (−2ˆi + ˆj). 12. Given three vectors a, b and b show, making use of the representation with the determinant that: .(a × b) · c = a · (b × c) = b · (a × c). If not, the definition of parallelepiped volume would have been problematic. 13. Demonstrate the property of BAC-CAB: .(a × b) × c = b(a · c) − c(a · b), verifying it by direct development after arranging the vector .c along the .x axis and the vector .b in the plane .(x, y).

Chapter 3

Kinematics of the Particle

Abstract Kinematics is the branch of classical mechanics aiming at a quantitatively description of the motion of particles and extended bodies, resorting exclusively to the notions of space and time, independently of the causes (forces) of the motion itself. The main kinematics derived quantities (velocity and acceleration) are defined, and the concept of direct and indirect kinematics problems. After presenting the uniform and uniformly accelerated motion in one dimension, I discuss the composition of motions. Great emphasis is given on the uniform and non-uniform circular motion either in intrinsic, Cartesian, and polar representations. The deep comprehension of the property of the circular motion is the first necessary step to enter in the world of waves and quantum mechanics.

3.1 Introduction and Definitions Kinematics is the branch of classical mechanics aiming at a quantitatively description of the motion of particles and extended bodies, resorting exclusively to the notions of space and time, independently of the causes (forces) of the motion itself. Kinematics originated with the studies of Galileo Galilei, but its modern structuring, using the principles of infinitesimal calculus, starts with Newton himself and several other scholars in the early 1700s. The material point, the particle To avoid superfluous complications, we begin to treat the motion of a body as if it were a simple geometric point: in practice, we consider a body of negligible size with respect to the trajectory in which it moves. In kinematics such a point is also called material point or simply particle (I adopt mostly this second naming). The frame of reference The position of such a particle is associated with coordinates in a frame of reference, in many cases a Cartesian one. In physics, a frame of reference (or reference frame) corresponds to a coordinate system whose origin, orientation, and scale are specified by a set of reference points whose position is identified on the axes both © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Spurio, The Fundamentals of Newtonian Mechanics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-47289-3_3

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mathematically (with numerical coordinate values) and sometimes physically (signaled by conventional markers). In our Euclidean 3-dimensional space, three scales are needed to fully define a reference frame. In the Cartesian coordinate system, a reference frame may be defined with a reference point at the origin and a reference point at one unit distance along each of the three coordinate axes. Having defined a reference system, the particle can be identified by the position vector from the origin of the reference frame to the point. Since the particle is moving, the time coordinate must also be specified. The complete kinematic information is thus defined by four quantities: three spatial coordinates and one temporal coordinate. Because the particle kinematics is the study of its motion in space versus time, the fact that the particle has a mass .m is irrelevant. This will be crucial in the context of the dynamics of the particle, discussed in the following chapter. The equation of motion The purpose of kinematics is therefore to determine the equation of motion, that is, the function: .r(t) = x(t)ˆ i + y(t)ˆj + z(t)kˆ (3.1) which describes the position of the particle as a function of time. Equation 3.1 is expressed in Cartesian coordinates; however other representations can be used. In any representation, however, three independent variables as a function of time are needed to describe the free motion of the particle in space. Knowledge of the equation of motion makes it possible to determine how an object’s position varies with time, i.e., its velocity, and how its velocity varies with time, i.e., its acceleration. Velocity and acceleration are both vector quantities, as the position r. The trajectory The set of positions in space that the particle takes as time changes is called the trajectory. The trajectory of a particle thus corresponds to a path in space of pure geometric character whose knowledge needs three spatial coordinates. As we shall see, on the Earth’s surface (neglecting friction with air) any moving particle follows a parabolic trajectory, which in a frame of reference generally integral with the thrower, takes on the equation: z = ax 2 + bx + c (.z is the vertical coordinate, .x a coordinate in the plane of the Earth’s surface), which corresponds to a parabolic motion in the.(x, z) plane. Apparently, the third coordinate is missing. In fact, it is present but takes the value . y = 0 for each time. The quantities .a, b, c are constants that can be determined. Obviously, the particle is in motion and will assume different.(x, z) coordinates at different times. However, for the definition of trajectory, the instant of time at which a certain position is occupied is irrelevant. Knowing the equation of motion not only means knowing the trajectory of the particle. It also means knowing at what time the particle will be at a certain point

3.2 Uniform and Uniformly Accelerated Motion

49

in the trajectory. Thus, the equation of motion makes it possible to locate moving bodies not only in space, but also in time and for its complete determination four variables are necessary (three spatial and the time).

3.2 Uniform and Uniformly Accelerated Motion Everyday experience tells us that any body in motion in a plane stops after a while. This consideration had prompted the ancients to deduce that a continuous “stress” was required to keep a body in motion. In a different way, along the vertical direction a body in free fall travels paths that increase as time increases, but differently if to fall is (for example) a feather or a stone. Pre-Galileian philosophy explained that the stone had greater weight, and therefore greater “affinity” to rejoin the ground, which is massive. The feather, having less “affinity,” proceeds more slowly. Galilei’s first extraordinary discoveries refer precisely to these two motions occurring near the Earth’s surface. About motion on a horizontal plane, Galilei deduced that if all forms of friction are eliminated a body persists in its state of stillness or motion covering equal spaces in equal times. I will discuss what friction is in Sect. 4.5. Galilei then guessed that a motion with constant speed (i.e., motion covering equal spaces in equal times) is a kind of natural state, for the maintenance of which no external solicitation is required.1 The situation is completely different in the case of the vertical direction. In the case of motion along the vertical direction, Galilei deduced from observations that, if all forms of friction can be eliminated, every object that falls from rest from the same height increases the distance traveled during the same time intervals but arrives at the ground taking the same time. With the technology of the Galilei’s time, both observations turn out to be of extraordinary relevance. Today, experimental verification in educational laboratories is very simple, as you might have already experienced: see below for further details. The property of maintaining the state of motion, if undisturbed, also occurs in outer space. Outside the Earth’s atmosphere, where it is indeed possible to significantly remove all forms of friction, extraordinary experiments have been carried out, faithfully reported by high-quality footage.

1

Galilei, however, mistakenly believed that this natural state was not on a straight direction, but on a circumference.

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3 Kinematics of the Particle

Equations of motion of the particle on Earth In the laboratory, one can measure how the position of a moving particle varies with time in a plane parallel to the ground, along (for example) the .x axis. A frictionreducing device is needed, such as dry ice,2 a calibrated axis to locate the position of the particle during motion, and a video camera to take pictures. These normally have internally a device for measuring time. The particle starts at .x0 : with a small “thrust” it is set in motion at the moment when the time measurement begins (.t = 0).3 The video camera (or cameras, if the path is very long) allow you to record the position of the particle (coordinate .x) as time changes. Normally, pictures can be taken at regular time intervals, in which the particle, its position and time at which the picture was taken are framed. The equation of motion describes the .x coordinate as a function of time .t and is very simple: .

x(t) = At + x0

(3.2)

where . A is a constant characteristic of that particular experiment, specially of the way the particle was initially set in motion. A similar law would be found if one were to move the object on the other axis parallel to the ground, the . y axis. A different situation arises if I study the fall of an object, that is, motion along the . z axis. Again, Galilei makes a remarkable extrapolation of his observations. If you drop a stone and a feather, the stone reaches the ground first. This is an effect due to the misleading role of friction with air molecules. Today we can perform the vertical drop experiment in a vacuum tube, such as the one shown in Fig. 3.1, where a significant fraction of atmospheric gas can be removed and thus almost completely suppress the friction exerted by the gas molecules. A really professional version from BBC with a stone and a feather falling in a vacuum chamber is available at: https://www.youtube.com/watch?v=E43-CfukEgs. Performing the test, we would find that different objects arrive at the ground at the same time. The equation of motion we can derive is extremely interesting. The device drops the object from a standstill, that is, without an initial thrust, and at the same initial height .z 0 . In that case, the equation of motion obtained with the rock and the feather (or whatever you like to test) is still the same. We then obtain a first empirical universal law, one that I would like to denote by the notation introduced with (2.31). This empirical law, valid for any object falling on the Earth’s surface without an initial thrust, corresponds to the equation of motion: .

1 z(t) = − gt 2 + z 0 . 2

(3.3)

You can practice, using the data given in the right-hand side of Fig. 3.1, to show that z quota decreases (hence the reason for the—sign) with a quadratic progression with

.

2

Dry ice is nothing more than frozen carbon dioxide, which passes from the solid to the gaseous state by sublimation, that is, without passing through the liquid state. 3 The “thrust” to provide the motion corresponds to an impulse, a physical quantity that we will discuss in Sect. 8.2.

3.2 Uniform and Uniformly Accelerated Motion

51

Fig. 3.1 Vacuum tube for demonstration of the Galilean theory of falling bodies, made in collaboration with the Italian Institute of Nuclear Physics (INFN). The table on the right contains the data for an experiment performed. The camera made 30 frames per second; the altitude is measured in meters, starting from the top. The initial velocity was zero

respect to time. Note the difference with (3.2), which instead depends linearly on the parameter .t. Moreover, unlike constant . A in (3.2) the parameter .g/2 that appears4 is a universal parameter, which does not change if you change the conditions. In fact, I can test it by changing devices, perhaps using one with a vacuum tube in which I can vary .z 0 and, most importantly, impart a small initial thrust (upward or downward) to the object when it is dropped. By repeating the test, I would find the following equation of motion: .

1 z(t) = − gt 2 + A' t + z 0 . 2

(3.4)

The parameter multiplying .t 2 has not changed; it is always the same. Instead, a term ' . A appears, which, as in the case of motion in the plane parallel to the ground, depends on the initial conditions. The parameters denoted by . A and . A' in the (3.2) and (3.4) 4

The reason for the apparently annoying factor 2 will become clear in a moment.

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3 Kinematics of the Particle

have dimensions .[L T −1 ]: in the International System they are measured by m s.−1 . In contrast, the parameter .g that appears in (3.4) is a universal parameter (close to the Earth’s surface: I could use the same device on different continents and verify that the (3.4) does not change, at least as long as I do not use very refined measuring instruments) that has dimensions .[L T −2 ]: in the International System units, its value is about 9.8 m s.−2 . (Spoiler alert: of course, we could deduce the (3.4)! But we need the ingredient we don’t have yet: the force that makes objects to fall). Average and instantaneous speed The main difference between the two motions described by (3.2) and (3.4) is that the distance traveled for equal time intervals is different. While in the former case equal spaces are traversed in equal time intervals, in the latter case for equal time intervals (.Δt = 0.033 s, in the example) the space traversed increases progressively, see upper part of Fig. 3.2.

Fig. 3.2 Upper: graphical representation of the data present in tabular form in Fig. 3.1. Superimposed on the data, a dashed curve fits the points with a second-order polynomial. The parameters of the equation are overlaid on the graph. Bottom: the ratio .Δz/Δt is reported calculated for each line on the sheet, always as time changes. For example, the height corresponding to the third shot is .z 3 = 2.249 m, while .z 2 = 2.266 m. Thus .Δz = −0.016 m, .Δt = 0.033 s and .Δz/Δt = −0.49 m/s. It corresponds to the second point from the left. In this case, the equation interpolating the data corresponds to a straight line

3.2 Uniform and Uniformly Accelerated Motion

53

An interesting quantity is how the coordinate varies, i.e. .Δx in (3.2), .Δz in (3.4), as .Δt varies. The quantity representing the ratio between variations is called average speed. I begin by studying motion along the .z axis, of which I have given in the bottom panel of Fig. 3.2, for each frame of video, the value of the ratio that corresponds to its average speed: Δz [z(t + Δt) − z(t)] = . .vz (t) ≡ (3.5) Δt Δt In the figure caption I also exemplified how the values were calculated using the spreadsheet. As you can verify, the value of average speed (in absolute value) increases as time changes. Our photographic images refer only to discrete instants of time (.Δt = 0.033 s), however, we are confident that if we had taken the photos at smaller time intervals, the only change in the graphical representation of motion would have been a larger number of dots. The additional points would always be on the dashed line interpolating the data. I can imagine, in essence, that there is a continuous function that describes the situation even when I perform the mathematical passage to the limit for .Δt → 0. Thus, I can define: dz Δz ≡ . Δt→0 Δt dt

v (t) ≡ lim

. z

(3.6)

The quantity in (3.6) is called the speed of the object that is moving along the .z axis. The speed has dimensions of .[L T −1 ] and is therefore measured in m/s in the SI. In physics, it is common to denote the derivative of a function with the notation used in (3.6). The average speed of the object that is moving on the .x axis is similarly v =

. x

Δx [x(t + Δt) − x(t)] = Δt Δt

(3.7)

and, again I can perform the passage to the limit and define the speed along the direction of the .x axis as: v (t) ≡ lim

. x

Δt→0

dx Δx ≡ . Δt dt

(3.8)

In the case of our two experiments, we already deduced from the data the analytical function describing the equation of motion both along the .x axis, (3.2), and on the vertical .z axis, (3.4). We can immediately obtain the speed for the two motions. Along the .x axis: d(At + x0 ) dx = = A. .v x (t) ≡ (3.9) dt dt It follows that the coefficient I had called . A is nothing more than the (constant) speed with which the particle is moving along the .x axis.

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3 Kinematics of the Particle

For the falling motion along .z direction, however: v (t) ≡

. z

dz d(−1/2gt 2 + A' t + z 0 ) = = −gt + A' . dt dt

(3.10)

As we expected from the bottom panel in Fig. 3.2, the equation describing the speed is that of a straight line (with negative slope), with equation .−gt. In the case of the figure, . A' = 0. In general at .t = 0 is .vz (0) = A' ≡ v0,z , that is, the constant corresponds to the speed it has at the initial time. The (3.4) can then be rewritten as: .

1 z(t) = − gt 2 + v0,z t + z 0 . 2

(3.11)

Without informing you, I did above the first computation of derivatives, a common practice for a physicist. The fundamental properties and notable derivatives of the most commonly used functions are covered in the analysis course. However, you can find the remarkable derivatives and the differentiation rules at https://en.wikipedia. org/wiki/Differentiation_rules. On the web, there are different apps available that compute derivatives, just entering the analytical expression of the function. Scalar acceleration Speed can also vary; you all know that in a car you start moving at low speeds, then you can increase it- within the allowed limits, of course. We can define the average acceleration along the .z axis the quantity az (t) ≡

Δvz . Δt

The corresponding scalar acceleration is the quantity obtainable by making the transition to the limit of infinitesimal time intervals of .az (t), that is: Δvz dvz ≡ Δt→0 Δt dt

a (t) ≡ lim

. z

(3.12)

and, of course, similarly for the particle moving along the .x axis. The scalar acceleration has dimensions .[L T −2 ] and thus is measured in m/s.2 in the SI. In the case of motion in the horizontal plane, (3.9), it is immediate to verify that a (t) ≡

. x

dvx =0 dt

(3.13)

because the derivative of a constant is always zero. In the case of free-fall motion, however: dvz d(−gt) (3.14) .az (t) ≡ = = −g . dt dt

3.2 Uniform and Uniformly Accelerated Motion

55

We have the following remarkable result: near the Earth’s surface, when all forms of friction are carefully removed, all objects fall downward with scalar acceleration equal to g, which is worth about 9.8 m s.−2 . Now you understand why we had introduced a factor of .1/2 into the equation of motion. Summarizing the characteristics of the two motions discussed. Along the .x axis we have a motion that is called uniform rectilinear motion with the following features: • the particle travels through equal spaces in equal times; • its speed remains constant over time; • its scalar acceleration is zero. Along the .z axis we have a uniformly accelerated motion with the following characteristics: • the particle travels through spaces that increase with the square of the time; • its speed (in absolute value) increases linearly with time; • its scalar acceleration is constant, directed downward, and equal in magnitude to the parameter .g; • The value of .g is identical for all falling bodies near the Earth’s surface. Galileo Galilei (1564–1642) is considered the father of modern science for explicitly introducing the scientific method (also called the Galilean method or experimental method). As professor at the University of Padua, on August 25, 1609, Galilei (Fig. 3.3) presented the telescope of his own construction (made on the inspiration of Dutch instruments) to the government of Venice, which doubled his salary and offered him a life-long teaching contract. He used the instrument not for commercial purposes, but to observe the sky and came to the conclusion that in addition to the stars known and visible to the naked eye, there are countless others never observed before. The Universe, therefore, is larger than then known. Again thanks to the telescope, he concludes that the Moon has structures similar to those on the surface of the Earth, so there is no difference in nature between the Earth and the Moon, contrary to the Aristotelian-Ptolemaic view of the universe. On January 7, 1610, he first observed four satellites of Jupiter, which Galilei named Medicean stars in honor of Cosimo II de’ Medici. It is an observation of extraordinary importance, so much so that Galilei himself will immediately report it (with the observation of lunar roughness) by publishing on March 12, 1610 the book (Fig. 3.4) Sidereus Nuncius: exactly three months only after the first observation! If I have to find the birth date of modern Science, for me it is indeed January 7, 1610: Galilei inaugurated the scientific method by making observations with an instrument, deduced conclusions from his study, and published the results. In 1624 Galilei began his new work, the Dialogo sopra i due massimi sistemi del mondo (Dialogue Concerning the Two Chief World Systems) in Italian. I will mention again this book in Chap. 5. Using the expedient of a dialogue between three imaginary interlocutors (Sagredo, Salviati and Simplicio), he exposes the theories of cosmology of that time, and thus also the Copernican one, but without showing any personal commitment to any of them, although everyone understands for whom he holds. In fact, the rhetorical expedient was not enough: suspected of heresy and accused of wanting to subvert Aristotelian natural philosophy and the Holy Scriptures, Galilei was subjected to a long and exhausting investigation for him by the Holy Office. The process ended on June 22, 1633, after he was forced to abjure the astronomical conceptions reported in his books and confined to his own villa in Arcetri, near Florence.

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3 Kinematics of the Particle

Fig. 3.3 Portrait of Galileo Galilei, by Justus Sustermans (1635). Image in the public domain thanks to the Uffizi Gallery, Florence (Italy)

After the process, Galilei dictated (he had been forbidden to write!) and published in 1638 in the Netherlands the great treatise Discorsi e dimostrazioni matematiche intorno a due nuove scienze (Discourses and mathematical demonstrations relating to two new sciences). It is the book from which the kinematic considerations given above are derived. This second text is also organized as a dialogue that takes place over four days between the same three protagonists as the previous Dialogue. After 359 years, in 1992, the plenary session of the Pontifical Academy of Sciences acknowledged the errors made in the Holy Office’s investigative process, rehabilitating Galilei.

3.3 Equation of Motion, Velocity and Acceleration Let us generalize the quantities defined in the previous section, the scalar velocity and acceleration, to the more general case. We use a Cartesian reference frame (later, the quantities can also be defined in other coordinate systems).

3.3 Equation of Motion, Velocity and Acceleration

57

Fig. 3.4 The discovery of the moons of Jupiter. Bottom, copy of part of the sheet containing hand notes taken during Galilei’s observation of the nights from January 7, 1610 and after. Above, I have reproduced part of what is inferred, with the relative position of Jupiter and observable satellites. You can find the text of Sidereus Nuncius at https://la.wikisource.org/wiki/Sidereus_nuncius

3.3.1 Equation of Motion Suppose we have a Cartesian reference frame with origin at the point O. With respect to this frame, the position of the particle to be studied at a generic time .t is given by the relation: .r(t) = x(t)ˆ i + y(t)ˆj + z(t)kˆ . (3.15) However, imagine that we want to study the particle with respect to a different point Ω of the reference frame with coordinates .rΩ ≡ (xΩ , yΩ , z Ω ). For some reasons, may be of experimental nature, referring the study with respect to the point .ω might be more practical, and I want to express the equation of motion with respect to this point. At a time .t1 the vector joining .Ω with the particle is .s(t1 ), which, as shown in Fig. 3.5, is given by the sum of two vectors and corresponds to

.

r(t1 ) = s(t1 ) + rΩ

.

−→

s(t1 ) = r(t1 ) − rΩ

(3.16)

58

3 Kinematics of the Particle

Fig. 3.5 Left: Definition of the infinitesimal displacement vector in Cartesian coordinates between the position occupied by the particle in two generic times .t and .t + Δt. The direction of .Δs tends to become tangent to the path when .Δt tends to zero. Right: the resultant velocity vector is always tangent to the trajectory

Using Cartesian coordinates, without making the vector .rΩ explicit, we have: s(t1 ) = x(t1 )ˆi + y(t1 )ˆj + z(t1 )kˆ − rΩ

.

(3.17)

At a later time .t1 + Δt the particle has moved to the point s(t1 + Δt) = x(t1 + Δt)ˆi + y(t1 + Δt)ˆj + z(t1 + Δt)kˆ − rΩ

.

(3.18)

having traveled a path: Δs ≡ (r(t1 + Δt) − rΩ ) − (r(t1 ) − rΩ ) .

= [x(t1 + Δt) − x(t1 )]ˆi + [y(t1 + Δt) − y(t1 )]ˆj + [z(t1 + Δt) − z(t1 )]kˆ = Δx ˆi + Δy ˆj + Δz kˆ

(3.19) Thus, the equation of motion (3.15) is equivalent to the use of the variable .s(t), and .Δr = Δs. I try to explain the reason for this apparent complication of having introduced the vector s. Notice first that the vector s(t) and the vector r(t) coincide if .rΩ = 0, that is, if the point .Ω corresponds to the origin O of the reference frame. The first reason, as mentioned, is that we need not be bound to the origin of the reference frame. Thus, we have verified that the equation of motion is independent of the choice of the point in which the observer is located. The second is that the definition of the displacement .Δs avoids the use of the notation .Δr found in some textbooks, because this second notation causes extreme confusion when introducing spherical coordinates. In spherical coordinates, we have a radial variable called .r and its differential quantity .dr is different from the vector quantity .dr obtained if we confuse position and displacement. The scalar magnitude

3.3 Equation of Motion, Velocity and Acceleration

59

dr is fundamental to the study of gravitation (Chap. 11) and when you study atomic physics. Let us now analyze the geometrical meaning of the magnitude .Δs, with the help of Fig. 3.5 on the left. The magnitude corresponds to the oriented segment representing a chord to the trajectory followed by the particle. The smaller the time interval .Δt considered, the more the segment tends to coincide with the trajectory itself and then become tangent to it.

.

3.3.2 Velocity, Definition Suppose now that we know the equation of motion, (3.16), that is, we know .r(t), or equivalently .s(t) (the two quantities could differ by one additive constant). This corresponds to know the infinitesimal displacement .ds instant by instant. Similarly to the one-dimensional motion, we can determine how quickly the position changes, i.e., calculate the incremental ratio between .Δs and .Δt, letting the time interval tend to zero: Δs ds v(t) ≡ lim = Δt→0 Δt dt dy d x . (3.20) ˆi + ˆj + dz kˆ = dt dt dt = vx (t)ˆi + v y (t)ˆj + vz (t)kˆ this quantity is defined as the velocity of the particle at the considered point of the trajectory. Referring to the right panel of Fig. 3.5, the velocity vector by construction is tangent point by point to the trajectory. Velocity and speed Some scientific terms in English have a well defined meaning. This is the case of velocity and speed. The velocity is a vector quantity. The magnitude of the velocity is the speed, which is a scalar quantity. According to the definition (3.20) above, the particle speed is / v(t) ≡

vx2 (t) + v 2y (t) + vz2 (t) .

Only for one-dimensional motion (when the direction is fixed) we can indifferently use the two terms. I will try to be rigorous on this, never using the term speed when I talk about the vector quantity, and viceversa.

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3 Kinematics of the Particle

3.3.3 Acceleration, Definition Empirically, we observe that the velocity of a particle can also vary over time. We can therefore study how the velocity varies over a time interval .Δt, then let the time interval tend to zero. This quantity is called acceleration of the particle: dv Δv = Δt dt dvx ˆ dv y ˆ dvz ˆ i+ j+ k = dt dt dt = ax (t)ˆi + a y (t)ˆj + az (t)kˆ .

a(t) ≡ lim

Δt→0

.

(3.21)

Acceleration is the first derivative of the velocity function, which in turn is the first derivative of the equation of motion. For this reason, acceleration can also be expressed as the second derivative of the equation of motion: d ds d 2s = 2 dt dt dt d2x ˆ d2 y ˆ d2z ˆ = 2i+ 2j+ 2k dt dt dt

a(t) = .

(3.22)

3.4 The Direct Kinematics Problem One aspect now becomes clear: given the equation of motion (3.16), velocity and acceleration can be determined by the mathematical differentiation rules. There is no technical difficulty: the derivatives of the main functions can be found in tabulated form, or they can be calculated by computer programs. Given the equation of motion, the determination of velocity and acceleration of the particle is only a technical, math, problem that completely solves the kinematic problem. In essence: .

if r(t) [s(t)] is known, then: v(t) ≡

dv ds and a(t) ≡ are calculable. dt dt

(3.23)

This means, for example, that we will be able to locate the future position of a particle known the current coordinates. One may ask whether knowing the derivative of acceleration (i.e., the third derivative of the equation of motion) gives any additional information. The answer is no: from the point of view of the characterization of the motion of particles, it is enough to know up to the second derivative of the equation of motion.

3.4 The Direct Kinematics Problem

61

3.4.1 Infinitesimal Displacement and Infinitesimal Path In (3.19) we determined a finite displacement. When tending to the limit for infinitesimal time intervals, this quantity correspond to a differential quantity. In general, in mathematics the differential of a function quantifies the infinitesimal variation of the function with respect to an independent variable.5 The definition of velocity in (3.20) allows us to write the differential form ds = v(t)dt

.

(3.24)

where this quantity correspond to a differential of the equation of motion, i.e.: ds = s(t + dt) − s(t) = d x ˆi + dy ˆj + dz kˆ

.

(3.25)

We will make frequent use of this differential quantity, called infinitesimal displacement. Note that it is a vector quantity whose direction is always tangent, point by point, to the trajectory followed by the particle during its motion. Sometimes it is useful to define the scalar quantity that corresponds to the magnitude of the infinitesimal displacement; this scalar quantity is defined as infinitesimal path and, in Cartesian coordinates, using the Pythagorean theorem, is equal to ds =

.

√ d x 2 + dy 2 + dz 2 .

(3.26)

3.4.2 Composition of Motions and Trajectory I have already defined the trajectory as the geometric path in space through which the particle passes. The equation of the trajectory in general can be derived from the parametric form of the equation of motion (3.16), where the parameter is the time. An example can clarify the situation. The two motions studied separately by Galilei (uniform rectilinear motion in the plane and uniformly accelerated motion in the vertical direction) can occur simultaneously. This is the case, for example, with 5

Using differential calculus, it is possible to relate the infinitely small changes of various variables to each other using derivatives. If f is a function of x, then the differential df is related to dx by the formula df dx df = dx where . dd xf denotes the derivative of f with respect to x. This formula summarizes the intuitive idea that the derivative of f with respect to x is the limit of the ratio of differences.Δ f /Δx as.Δx becomes infinitesimal. For instance, in the case when f corresponds to a spatial coordinate (e.g., .z) and .x to the time, the relation above corresponds to dz = vz dt where .vz is the .z component of the velocity.

62

3 Kinematics of the Particle

a cannon that fires a projectile with speed .v0 at an angle .θ from the horizontal. In this way, the initial components of the velocity vector are: v = (v0,x , v0,y , v0,z ) = (v0 cos θ, 0 , v0 sin θ )

. 0

(3.27)

With this initial component of velocity, the equation of motion in the vertical direction coincides with the (3.11). The motion on the .x axis remains analogous to (3.2), with obviously . A = v0,x . Along the . y component, nothing varies and so we can assume that for any time . y(t) = 0. In order to set the initial coordinates, is convenient to choice the position of the particle at .t = 0 coincident with the origin of the reference frame, thus we have .x0 = 0, y0 = 0, z 0 = 0. With this simplification, the equation of motion takes the form: .

1 x(t) = v0 t cos θ ; z(t) = − gt 2 + v0 t sin θ . 2

(3.28)

The trajectory is a curve in a.(x, z) plane that can be described by a function of the type z = z(x). To do this, we remove the time that serves as a parameter. I made explicit it from the first equation (.t = x/v0 cos θ ) and put it into the second, obtaining:

.

.

1 x2 z=− g + x tan θ . 2 (v0 cos θ )2

(3.29)

This is the searched equation, the trajectory, which corresponds to a parabolic path with the concavity pointing downward. In the section on questions and exercises you will manipulate this equation to determine the range, which corresponds to the distance .x G traveled horizontally by the body before it touches the ground: x =

. G

2v02 sin θ cos θ v 2 sin 2θ = 0 . g g

(3.30)

It is also possible to derive the angle which maximize the range, which corresponds to the angle .θ = 45◦ . See also questions (4), (5) and (6).

3.5 The Inverse Kinematics Problem With the present technology, it is possible to have instruments that directly measure velocity or acceleration. Therefore, the question is relevant: is it possible, knowing the acceleration or the velocity (remember, both vector quantities), to determine the equation of motion of the particle? This will become the key issue when we study dynamics: knowing the forces on the particle will allow us to obtain the acceleration. From this, we will want to know how the particle moves. Let us assume that we know the function .a(t) in (3.21). Since acceleration is the first derivative of velocity, component by component I have:

3.5 The Inverse Kinematics Problem

a (t) =

. x

dvx dt

63

; a y (t) =

dv y dvz ; az (t) = . dt dt

(3.31)

To get the velocity (e.g., component .x) I can write the differential form dvx = ax (t)dt

.

(3.32)

and integrate both the left and right sides: {

vx

.

vx (t0 )

{

t

dvx =

ax (t)dt .

(3.33)

t0

On the right-hand side I have to integrate the function .ax (t) between the initial time and a generic time .t; on the left-hand side, correspondingly, I have to integrate from the initial velocity (component .x) equal to .vx (t0 ) to a generic final velocity .vx (t). The left-hand integration is immediate: {

t

v (t) − vx (t0 ) =

. x

ax (t)dt

(3.34)

t0

while the integration of the right-hand side is possible, because we have assumed to know the function .ax (t). Normally, if this is an analytic function, integration is possible by the methods of mathematical analysis. In many real situations, the integration may not be simple. However, we may make use of numerical techniques to solve it. The same thing we can repeat for the other two components, and at the end we can write (3.34) together with the other components in vector form as: {

t

v(t) = v(t0 ) +

.

a(t)dt .

(3.35)

t0

The problem is completely solved as long as the initial .v(t0 ) velocity is known. Integration can always be done (by hand or with the computer) but the problem needs knowledge of what the initial conditions are in order to be solved. In the absence of the knowledge of the initial conditions, the problem is indeterminate, i.e.: there are infinite similar solutions that differ by an initial constant. In no way can this initial constant be deduced mathematically: the only possible solution is observation. Knowing the velocity vector from the (3.35), let us see how to obtain the equation of motion. Let us consider as usual the component along the .x axis: .

dx = vx (t) dt

−→

d x = vx (t)dt

(3.36)

from which, integrating the differential form under the same conditions as the .x component of the velocity:

64

3 Kinematics of the Particle

{

x

.

x(t0 )

{ dx =

t

{ vx (t)dt

−→

t

x(t) − x(t0 ) =

t0

vx (t)dt .

(3.37)

t0

Exactly as in the case of velocity, we can collect the three scalar equations into a vector equation and have: {

t

r(t) = r(t0 ) +

.

v(t)dt .

(3.38)

t0

Again, the problem is fully determined as long as the position at the initial time is known. In the absence of knowledge of the initial conditions, the problem is indeterminate (i.e., there are infinitely many similar solutions that differ by an initial constant). The inverse kinematics problem makes it possible, by integration, to trace the equation of motion when the acceleration of the body is known. For the problem to be totally determined, however, it is necessary to know two initial constants. Since the system is described by vector equations, the two constants to be known (e.g., velocity and initial position) are two vectors. The problem of initial constants (or boundary conditions, in the case of integrals in spatial coordinates) is an absolutely typical situation in physics whenever you have to trace a quantity by the operation of integration. No matter how powerful your mathematical training may be, the problem remains indeterminate unless you have data that constrain the studied system at the initial time (or in the boundary region). When in the next chapter I have introduced the concept of force, and the relationship between force and acceleration (Sect. 4.7), the importance of the inverse kinematics problem for determining the equations of motion will become clear.

3.6 Uniform and Non-uniform Circular Motion Circular motion represents a fundamental type of motion for the understanding of several aspects, as will be seen in the following chapters. We will study circular motion sequentially on three possible representations: the Cartesian one, that of a polar coordinate system (which we will introduce for the occasion) and in intrinsic vector representation. As the word itself indicates, circular motion occurs on a predetermined trajectory, which is a circumference whose equation in the Cartesian plane can be written as: .

where .r represents its radius.

x 2 + y2 = r 2

(3.39)

3.6 Uniform and Non-uniform Circular Motion

65

3.6.1 Uniform Circular Motion We start with uniform motion: consider the point P in motion on the circumference (Fig. 3.6), at a certain angle .ϕ with respect to the .x axis. Uniform means that the radius .r sweeps equal .ϕ angles in equal time intervals .t. If we call .T the period, that is, the time the point P takes to travel the entire circumference, we will have the following similarity: 2π 2π ϕ = −→ ϕ = t . (3.40) t T T from which I can derive: ϕ = ωt

where:

.

ω=

2π . T

(3.41)

The (3.40) expresses the concept of sweeps equal angles at equal times. We need to start thinking in differential terms: the relation (3.40) is equivalent to writing that .

dϕ ≡ ϕ˙ = ω . dt

(3.42)

Two things are to be noticed: For the first time, I use a simplified notation of derivative function, represented by the function symbol with a dot on it. It is used only when it corresponds to the derivative with respect to time, and when the notation is simplified and unambiguous (as it will be in our case). The second thing to note is that I have equated this constant derivative with a quantity called .ω, with dimensions −1 .[T ]. Its units are therefore s.−1 or rad/s (radians are dimensionless). This scalar quantity is called angular speed or angular frequency.6 Next, we will drop the condition of uniform motion, so the angular speed may vary over time, .ω = ω(t).

3.6.2 The Cartesian Representation Let us return to the point P on the circumference of Fig. 3.6. To fix ideas, at the initial time .t = 0 the particle is at the position .r(t = 0) = (r, 0) along the .x axis. In this study, it is convenient to get the equation of motion with respect to the origin of the coordinate system. The coordinates of point P as.t changes in Cartesian representation correspond to the equation of motion: { r(t) ≡

.

6

x(t) = r cos ϕ(t) . y(t) = r sin ϕ(t)

(3.43)

The term angular speed is usually used when dealing with circular motion; very often the equations we will find for circular motion also apply to other physical phenomena (e.g., waves or vibrations), and in that case the term angular frequency is more appropriate.

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3 Kinematics of the Particle

Fig. 3.6 Particle in motion on a circumference. The point P can be represented either in Cartesian coordinates or with unit vectors co-moving with the point itself

From the equation of motion I can easily derive the velocity of the point, remembering how composite functions are derived, that is: d[cos ϕ] dϕ dϕ(t) d[cos ϕ(t)] = = − sin ϕ(t) = − sin ϕ(t)ϕ˙ dt dϕ dt dt whereby, using the convention adopted in (3.42), i.e. .ϕ˙ = ω: { v(t) ≡

.

vx (t) = v y (t) =

dx dt dy dt

= −r ω sin ϕ(t) . = r ω cos ϕ(t)

(3.44)

How is the vector .v(t) placed with respect to the vector .r pointing from O to P? Let’s consider what happens at .t = 0: .vx (0) = 0, while .v y (0) = r ω: at the initial time the velocity is directed along the . y axis (as in Fig. 3.6). Since at .t = 0 the radial vector r is along .x, it turns out that velocity and position vector are perpendicular to each other. This situation holds for all subsequent times. One way to prove that two vectors are perpendicular is to verify that their scalar product, (2.16), is null. We use (3.43) and (3.44) r · v = x(t)vx (t) + y(t)v y (t) = r ω(− cos ϕ sin ϕ + sin ϕ cos ϕ) = 0 .

.

(3.45)

Therefore, position and velocity vectors are orthogonal to each other not only at t = 0, but also at each subsequent time. Since the trajectory is a circumference, it follows that the velocity is always tangent to the circumference. This is what we

.

3.6 Uniform and Non-uniform Circular Motion

67

expected for the direction of the velocity based on the considerations discussed in Sect. 3.3.2. Let us now determine the acceleration, deriving the velocity (3.44), and remembering that in this case .ω˙ = 0 because the motion is uniform. We will have: { a≡

.

ax (t) = a y (t) =

dvx dt dv y dt

= −r ω2 cos ϕ(t) = −ω2 x(t) . = −r ω2 sin ϕ(t) = −ω2 y(t)

(3.46)

Here we used the (3.43) that had defined Cartesian coordinates. Note the following remarkable result: the components of the acceleration are equal to those of the position vector, less than a .ω2 factor, which is necessary to maintain the dimensions of .[L T −2 ], and a sign reversal. The acceleration is directed along the radius, but pointing toward the center. In uniform circular motion, therefore, the acceleration of the particle is centripetal, that is, instant by instant points in the direction of the center of the circumference.

3.6.3 Non-uniform Circular Motion Let us look the changes if we drop the condition that the motion is uniform. This corresponds to the fact that in (3.42) we no longer require that the derivative be constant: now .ω = ω(t). Taking this into account, nothing formally changes in the equation of motion, (3.43). Nothing also changes in the equation expressing velocity, (3.44). 7 Only to notice is that now the term .ω is time-varying: we will have to take this into account when we derive the acceleration by doing the derivative of the velocity. In particular, nothing changes about the direction of .v: it is still a vector perpendicular to the radius and tangent to the trajectory. There is, however, a change in the expression of the acceleration. The derivative of the velocity vector now becomes, taking into account that we must also derive the .ω term of the (3.44): { a(t) =

.

ax (t) = a y (t) =

dvx dt dv y dt

= −r ω˙ sin ϕ(t) − r ω2 cos ϕ(t) . = +r ω˙ cos ϕ(t) − r ω2 sin ϕ(t)

(3.47)

That is, the acceleration is no longer just centripetal, but has an additional term (the one with .ω) ˙ that, we can infer, has a tangential direction. To find out whether this is correct, there is a more convenient coordinate system to use.

I wrote the (3.44) that way, without the usual .sin ωt and .cos ωt just so that I wouldn’t have to change it in the case of nonuniform motion

7

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3 Kinematics of the Particle

3.7 Polar Plane Coordinates 3.7.1 Circular Motion with Co-moving Unit Vectors In this section, I repeat what I did above but express the position, velocity and acceleration in terms of moving unit vectors, co-moving with the point P. The first unit vector I construct is the one expressing the position (3.43), rewritten in terms of the Cartesian unit vectors as: r(t) = r [cos ϕ(t)ˆi + sin ϕ(t)ˆj] .

.

(3.48)

The unit vector co-moving with position, which originates from the center O and points toward the moving point P, is defined as: rˆ (t) ≡

.

r(t) = cos ϕ(t)ˆi + sin ϕ(t)ˆj r

(3.49)

Similarly, the velocity (3.44) can be expressed as: v(t) = r ω[− sin ϕ(t)ˆi + cos ϕ(t)ˆj]

.

(3.50)

and again it is straightforward to verify that .r(t) · v(t) = 0. Therefore, I can use the (3.50) to define a second unit vector co-moving with the point, always perpendicular to the radial one and tangent to the trajectory. I define this second unit vector as: ˆ ϕ(t) ≡

.

v(t) = − sin ϕ(t) ˆi + cos ϕ(t) ˆj rω

(3.51)

The third unit vector can be defined using the vector product between these two unit vectors. Since it will be perpendicular to the plane, it is coincident with the Cartesian unit vector of the .z axis, and I use the same symbol: kˆ ≡ rˆ × ϕˆ

.

(3.52)

Let us rewrite the acceleration (3.47) in the non-uniform case in terms of these unit vectors: a(t) = [−r ω˙ sin ϕ(t) − r ω2 cos ϕ(t)] ˆi + [r ω˙ cos ϕ(t) − r ω2 sin ϕ(t)] ˆj . = r ω[− ˙ sin ϕ(t) ˆi + cos ϕ(t)ˆj] − r ω2 [cos ϕ(t) ˆi + sin ϕ(t) ˆj] = r ω˙ ϕˆ − ω2 r rˆ

(3.53)

In the case of uniform motion, when .ω˙ = 0, the acceleration is only centripetal, that is, along the .rˆ direction. In the case of non-uniform motion, there is also a tangential ˆ which is parallel to the velocity and component, i.e. directed as the unit vector .ϕ, which provides for increasing (or decreasing) the speed along the trajectory.

3.7 Polar Plane Coordinates

69

3.7.2 Definition of Polar and Cylindrical Coordinates By studying circular motion, we almost necessarily introduced a second and widely used coordinate system. This is the polar coordinate system that is a twodimensional coordinate system in which each point in the plane is identified by an angle and a distance from a fixed point as in Fig. 3.7. The angle is conventionally denoted with the Greek letter.ϕ and is measured from the.x axis in a counter-clockwise direction. The cylindrical coordinate system extends the two-dimensional polar system by adding a third coordinate, which measures the height of the point from the base plane. This introduces the third dimension, which corresponds to the height .z in the Cartesian plane. The three cylindrical coordinates can be converted to Cartesian coordinates with the relations ⎧ ⎨ x = r cos ϕ y = r sin ϕ . (3.54) ⎩ z=z. This coordinate system has three unit vectors. The substantial difference from the unit vectors of a Cartesian coordinate system (which do not depend on the position P of the object being studied) is that these unit vectors change direction as time changes, always remaining “hooked” to the moving point. The first of the three unit vectors, .rˆ , is in fact along the direction of the radius .r joining the origin with the point P. The second, .ϕˆ is always in the .x y plane, and always perpendicular to the first unit vector (with the property .rˆ · ϕˆ = 0). Finally, the third is the unit vector .kˆ analogous to that of Cartesian coordinates and defined by the condition (3.52). This last unit vector is not considered in plane polar coordinates. The relationship between the two polar unit vectors and the Cartesian ones, as we have shown above and as shown in the figure, is as follows:

Fig. 3.7 Transition from an orthogonal Cartesian coordinate system to a polar plane coordinate system

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3 Kinematics of the Particle

{ .

rˆ = cos ϕ ˆi + sin ϕ ˆj ϕˆ = − sin ϕ ˆi + cos ϕ ˆj .

(3.55)

Note that the following properties apply: { .

d rˆ · dϕ = (− sin ϕˆi + cos ϕ ˆj) · dϕ = ϕ˙ ϕˆ r˙ˆ = ddtrˆ = dϕ dt dt ˙ϕˆ = d ϕˆ = d ϕˆ · dϕ = (− cos ϕˆi − sin ϕ ˆj) · dϕ = −ϕ˙ rˆ dt dϕ dt dt

(3.56)

where we used the convention of calling .ϕ˙ the derivative with respect to time of the angular variable. Having determined the change in time of the unit vectors, we can determine the velocity expressed in polar coordinates. Here, the vector .r = r rˆ is composed of two terms (the magnitude of .r and its unit vector); both can vary in time independently. Thus: d rˆ dr d(r rˆ ) dr rˆ + r (3.57) . = = dt dt dt dt which, using the first of the (3.56) becomes v=

.

dr = r˙ rˆ + r ϕ˙ ϕˆ . dt

(3.58)

In general, the velocity has a component perpendicular to the radius and a radial component. As for acceleration, we have already shown in (3.53) for a non-uniform circular motion that it has radial and tangential components: a = −ω2 r rˆ + r ω˙ ϕˆ .

.

(3.59)

/ Note that, according to (3.50), the speed is .v = vx2 + v 2y = ωr , the relation (3.59) can also be written with the centripetal acceleration in terms of the speed: a=−

.

v2 rˆ + r ω˙ ϕˆ . r

(3.60)

These notations will be used to study the equation of motion of a particle on any trajectory.

3.8 The Intrinsic Coordinates A third way to study circular motion is to use vectors in intrinsic representation. We start by defining the angular velocity as a vector with symbol .ω having as magnitude the angular speed. For an object rotating in a plane, we can establish (conventionally) that .ω has a direction perpendicular to the plane in which it rotates. Thus, from

3.8 The Intrinsic Coordinates

71

Fig. 3.7, if the plane of rotation is that of the Cartesian plane containing the .x, y axes, the direction of .ω is along the .z axis. How to choose the positive direction? Again, we can make use of the right-hand rule convention, Sect. 2.2.1. From Fig. 3.7, the positive rotation (.x toward . y) corresponds to a counter-clockwise rotation. Conventionally, we take the positive orientation of.ω coincident with the positive . z axis for a particle rotating in the counter-clockwise direction (as in our example). According to our convention we define the angular velocity the vector ω ≡ ωkˆ

.

(3.61)

which is positive when the rotation of the particle is counter-clockwise. In this way, it is immediate to verify that the velocity (3.50) can be written also as .

v=ω×r .

(3.62)

Let’s verify this by making use (for practice) of the rule established by (2.24): ˆi jˆ kˆ v = 0 0 ω = ω(−yˆi + x ˆj) = r ω[− sin ϕ(t) ˆi + cos ϕ(t) ˆj] x y z

.

(3.63)

(alternatively, you can directly use the (2.23)) having used the coordinate values in (3.54). The expression obtained for the velocity is exactly the same as in (3.50). Similarly, it is extremely instructive to obtain the acceleration by directly deriving the (3.62) dω dr dv = ×r+ω× = ω˙ × r + ω × v . (3.64) .a = dt dt dt thus: .

a = ω˙ × r + ω × v = ω˙ × r + ω × (ω × r) .

(3.65)

As usual, the acceleration will consist of a centripetal term (the last one) and a tangential term. The tangential one is present only when the magnitude of the angular velocity changes with time (.ω˙ /= 0). You can check the centripetal term using the rules of the vector product between unit vectors,8 noting that: v = ω × r = ωkˆ × r rˆ = ωr (kˆ × rˆ ) = ωr ϕˆ

.

(3.66)

and then the last term in (3.65) corresponds to ˆ = −ω2 r rˆ ω × (ω × r) = ω2 r (kˆ × ϕ)

.

These equations in intrinsic representation will be widely used later. 8

ˆ in (2.22) is now being replaced by .(ˆr, ϕ, ˆ ˆ k) The tern .(ˆi, ˆj, k)

(3.67)

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3 Kinematics of the Particle

3.9 Poisson’s Rules for Moving Unit Vectors Polar coordinates introduced a relevant novelty: unit vectors whose direction varies with time. Since the unit vectors are hinged with a reference frame, the variation in time of the unit vectors is due to rotations of the reference frame with respect to another non-rotating reference frame.9 Rotations are always characterized by the angular velocity vector. I have already shown what the time derivatives of the unit vectors used in polar coordinate in (3.56), reported here: { d rˆ .

dt d ϕˆ dt

= ϕ˙ ϕˆ = −ϕ˙ rˆ .

(3.68)

Poisson’s rule states that the time derivative of a unit vector is equal to the vector product of the angular velocity of the system times the unit vector itself. In the case of our two unit vectors: ⎧ d rˆ ⎨ dt = ω × rˆ (3.69) . ⎩ d ϕˆ ˆ = ω × ϕ . dt Let us verify that the relations in (3.69) coincide with those in (3.68). In polar coordinates, rotation occurs about the .z axis and the angular velocity is .ω = ωkˆ = ˆ I can write the (3.69) as: ϕ˙ k. { .

d rˆ dt d ϕˆ dt

= ω × rˆ = ωkˆ × rˆ = ϕ( ˙ kˆ × rˆ ) = ϕ˙ ϕˆ ˆ ˆ = −ϕ˙ rˆ = ω × ϕˆ = ω(k × ϕ)

(3.70)

which is exactly what was intended to be proved. This can be extended also to Cartesian systems. In general, if we have a Cartesian frame of reference S’ that is rotating with respect to another Cartesian frame of reference S, the rotating system S’ is characterized by three moving unit vectors .(ˆ i' , ˆj' , kˆ ' ). These unit vectors have coordinates that vary with time with respect to the unit vectors defined in S, and the system S’ rotates with angular velocity with three nonzero components, .ω = (ωx , ω y , ωz ). If the rotation occurs with respect ˆ and the two Cartesian rotating verses to the .z-axis, we have .ω = (0, 0, ωz ) = ωk, ˆ Again, the time can coincide with those of the polar coordinates, .ˆi' = rˆ e .ˆj' = ϕ. derivatives of the moving unit vectors follow the Poisson rule, and are given by: ⎧ ˆ' di ' ⎪ ⎨ dt = ω × ˆi ˆj' d . = ω × ˆj' ⎪ ⎩ ddtkˆ ' = ω × kˆ '

(3.71)

dt

9

Further on, in Chap. 5 we will define two classes of frames of reference: inertial ones, which have no rotations, and non-inertial frames of reference in which rotations can occur. The moving coordinate system we are considering will be precisely a non-inertial reference frame.

3.10 Motion on any Trajectory (*)

73

The use of this relationship will be important when we define the properties of noninertial frames of reference in Chap. 5. You can easily verify that (3.71) coincides ˆ with (3.69) when .ω = ωk.

3.10 Motion on any Trajectory (*) The motion on a generic trajectory involves knowledge of the equation of the path (the curve denoted .γ in Fig. 3.8). For this reason, a general study is rather difficult. The position vector is written as a function of the intrinsic coordinate, .s which is obtained by integrating the infinitesimal path on the trajectory: { s=

.

γ

ds =

{ √ d x 2 + dy 2 + dz 2 .

(3.72)

γ

The quantity .s represents the distance in meters from the origin. The description of the motion of a particle can be reduced to a scalar function that associates with each instant of time a certain numerical value of the coordinate, .s = s(t). We can get more information about what generally happens by observing point P and two nearby points (P’ to its left and P” to its right) on the curve .γ . In Euclidean geometry, three points define a plane. If we consider the circumference passing through P, P’ and P”, we can see that the circumference tends to a limit position when

Fig. 3.8 Representation of what happens on a predetermined trajectory,.γ , in the vicinity of a point P. For Euclidean geometry, three points (in the drawing P, P’ and P”) define a plane (called the osculating plane). In addition, for a sufficiently small neighboring region of P passes an osculating circle of radius .ρ that locally better approximates the curve .γ . The velocity of the point is always along the direction.tˆ tangent to the trajectory at the point. By convention, the unit vector.ρˆ is directed toward the center

74

3 Kinematics of the Particle

P’, P” tend towards P, always remaining on the curve .γ . This limiting circumference is called osculating circle. Its radius .ρ is called curvature radius of the curve .γ at the point P. Thus, locally, the motion on any trajectory can be approximated by that on a circumference of radius .ρ (the value of which will change as the point moves), and the conclusions we drew for circular motion apply: • the velocity is always tangent point by point to the trajectory. We can locally define the tangent vector, .tˆ, whereby v = v tˆ =

.

ds tˆ . dt

(3.73)

• In the plane, together with the unit vector .tˆ we can define the unit vector perpenˆ with orientation pointing to the center of dicular and directed along the radius, .ρ, the osculating circumference. A third unit vector is perpendicular to the plane (it is called bi-normal) and is defined as: bˆ ≡ ρˆ × tˆ . • acceleration always lies in the osculating plane, with a radial component (directed toward the center of the osculating circle) and a tangential component. Using the notation used in (3.60), taking into account that in this case the unit vector .ρˆ is directed toward the center, we have: a = v˙ tˆ +

.

v2 ρˆ . ρ

(3.74)

Interestingly, the infinitesimal displacement vector .ds is defined in intrinsic coordinates as the vector that has magnitude equal to the infinitesimal path and direction always tangential. For this reason, it can be written as: ds = ds tˆ.

.

(3.75)

3.11 Questions and Exercises Questions 1. An automobile A travels on a straight road at the constant speed .v A = 75 km/h. A second car B arrives at the speed .v B = 130 km/h and starts braking inducing a constant deceleration 2 .a B = −9.0 m/s. when it is at the distance .d from A. Determine the minimum distance .d such that the collision between the cars is avoided. [A: .d > 16.7 m]

3.11 Questions and Exercises

75

2. A particle moves in uniformly accelerated motion on a straight line. After.t1 = 4 s it has traveled 60 m and has a velocity .v1 = 33 m/s. Determine the acceleration and initial velocity. [A: .v0 = 3 m/s; .a = 9 m/s.2 ] 3. The equation of motion of a material point is expressed by the relation .x(t) = αt 3 − βt 2 − γ , with the constants .α, β, γ real positive. Determine velocity and acceleration. What type of motion is involved? [A: Variable acceleration] 4. Verify that the range .x G of a projectile is given by the (3.30). 5. Two cannons are placed in the same position at different altitudes, .h 1 and .h 2 . Two projectiles are fired simultaneously and horizontally. Calculate what ratio the two initial velocities .v1 and .v2 must be for the two projectiles to have the same range. √ [A: .v1 /v2 = h 2 / h 1 ] 6. Verify that the launch angle .θ that produces the maximum range corresponds to .45◦ . Verify further that in the absence of friction, although the trajectory changes, the range remains the same for angles that differ by the same amount (positive or negative) from the angle.45◦ . Verify this numerically for a projectile fired at an angle.θ = 30◦ or.θ = 60◦ . In the case of the presence of friction with the atmosphere, for which angle between .30◦ , .45◦ and .60◦ would the range be higher? 7. A particle moves on a straight line with acceleration .a(t) = αt + β, with .α = 18 m/s.3 and 2 .β = −8 m/s. . Calculate its velocity at time .t1 = 3 s knowing that the velocity at the initial time is .v0 = 2 m/s. Determine the space traveled between the initial time and .t1 . [A: .v1 = 61 m/s; .s1 = 51 m] 8. One wants to determine the depth .h of a well experimentally. For this purpose, you throw a stone into it, and you hear the thud at the bottom after a time .τ = 3.0 s. The speed of sound is worth .vs = 340 m/s. What is the value .h? [A: .h = 41 m] 9. In the case of the previous question, if the effect of the finite speed of sound were neglected, would .h be larger or smaller? What percentage error would be made? [A: error of 9%.] 10. Determine the angular speed of the hour and minute hands of a clock (of course, with hands and not digital!). If at 3:00 (a.m. or p.m.) the angle between the hour hand and minute hand form an angle of .90◦ , calculate after how long the hands are exactly overlapping. [A: After 16 min 22 s.] 11. An observer is stationary on a merry-go-round at a distance . R = 3 m from the axis of rotation. The merry-go-round, starting from a standstill, begins to rotate with constant angular acceleration .ω˙ = 0.1 rad/s.2 . Determine the time .t1 at which the merry-go-round reaches the angular velocity of .0.1 revolutions per second and the magnitude of the observer’s felt acceleration at time .t1 . [A: .t1 = 6.28 s; .a1 = 1.22 m/s.2 ] 12. The position vector along a trajectory expressed in terms of the scalar distance .s from the origin is given by the relation .r = as 2 + bs + c, with the condition that .a = 0 Determine the dimensions of the vectors .a, b, c and show that the trajectory is that of a branch of a parabola passing through c. Draw the trajectory.

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3 Kinematics of the Particle

13. A particle is constrained to move on a circular guideway of radius . R = 3.00 m, on which it can slide without friction, according to the motion equation law .s(t) = kt 3 , with.k = 2.0 m/s.3 . Calculate the tangential component .at and the normal component .ac of the acceleration at time .t1 =1.0 s. [A: .at = 12.0 m/s.2 ; .ac = 27.0 m/s.2 ] 14. A particle moves on a predetermined trajectory with the equation of motion .s(t) = kt 2 , with .k constant and with magnitude of the acceleration equal to .a = 2k. Show by using (3.74) that 2 the radius of curvature is given by the relation .ρ = √ v2 2 . Show that in the present case the a −¨s motion is rectilinear. 15. In the case considered in the question/(14), show what the trajectory corresponds to in case the magnitude of acceleration is .a = 2k 1 +

t T

, where .T .= cost.

[A: A circumference] 16. The equation of motion of a particle is given in Cartesian coordinates by the relation 2 2 .r(t) = αt ˆi + βt ˆj. Determine .(i) the trajectory of the motion and .(ii) the scalar function .s(t) indicating the distance traveled as time varies. The initial condition at .t = 0 is .s(0) = 0. Verify the dimensional correctness of the equations found. √ [A: .(i) Straight line with slope .b/a. .(ii) .s(t) = α 2 + β 2 t 2 ] 17. A projectile is launched from the Earth’s surface with velocity .v0 = 50.0 m/s, at an angle ◦ .θ = 60 to the vertical. Determine the radius of curvature of the projectile’s trajectory at the time immediately following the launch. [A: .ρ = 294 m.]

Exercise 3.1 The position of a particle is defined by the position vector r(t) = at 3 ˆi + bt 2 ˆj + ct kˆ where numerically .a = 0.33, b = 0.71, c = 1.0. Determine: 1. 2. 3. 4. 5.

the dimensions of .a, b, c; the function describing the velocity and acceleration vectors; the average speed .vm in the time interval between .t = 0 s and .t = 2 s; the normal and tangent components of acceleration; the radius of curvature .ρ at time .t = 1 s.

Exercise 3.2 As an exercise in ballistics, we use the example of a soccer ball approximated as a point object (i.e., not subject to rotations, and the “effects” of the ball related to it). We also neglect all dissipative sources (friction, deformation, etc.). A player kicks a free kick from a distance. L = 19 m from the opponent’s goal (whose regulation height is .h = 2.44 m). The speed, estimated with the Video Assistant Referee (VAR) systems with which the ball is kicked is .v0 = 18 m/s, at an angle ◦ .α = 27 with the plane of the playing field. Determine: 1. the expression of the maximum height .z max reached by the balloon as a function of .v0 and .α. Determine the numerical value in the case presented. 2. The maximum range .xmax (again as a function of .v0 and .α), that is, how many meters away the ball would touch the plane of the soccer field again (neglecting the presence of any stands). 3. Determine whether the ball arrives at the goal position above or below the crossbar. 4. At what velocity .v1 would the ball have to be kicked for it to arrive under the crossbar, keeping the angle .α unchanged. 5. In that case, would it pass the opponents’ barrier (approximately 2 m high) placed 9.15 m away?

3.11 Questions and Exercises

77

Fig. 3.9 Figure for exercise (3.3)

Exercise 3.3 A particle P is moving on a circumference of radius . R = 3.50 m according to an angular speed given by .ω(t) = kt 2 , with .k = 13.18 degrees/s.3 and starting from rest. Consider a system of Cartesian axes placed on the plane of the circle, with origin corresponding to the center of the circle and oriented as in the Fig. 3.9 left. The point . A = (R, 0) shown in the figure represents the position of particle P at the initial time. Determine: 1. 2. 3. 4. 5.

the equation of the angular displacement .θ(t); the Cartesian coordinates of the point . B where the particle is located at the time .t B = 3.00 s; the time .tC to needed arrive at point .C, i.e., the point of Cartesian coordinates .(0, −R). the average speed .vm of the particle in the motion between point A and point C; the acceleration vector at the time .tC .

Chapter 4

Forces and the Dynamics of the Particle

Abstract The dynamics is the study of how a system might move or develop over time and study the causes of those changes. The great merit of Isaac Newton was to clearly define the concept of force, its vector nature, and the effect of forces on free objects. This chapter introduces the force, the instrument to measure it, and few simple examples of forces observable on human scales. In the second part of the chapter, after the introduction of the second law of dynamics, I study the kinematic effects of the forces introduced in the first part when applied on a massive point-like object. The chapter ends with what we know today about fundamental forces of the nature.

4.1 Introduction After introducing kinematic quantities in the previous chapter, here we start to consider aspects related to dynamics, that is, the study of how a system might move or develop over time and study the causes of those changes. Few premises are fundamental: • This is not a historical or historiographical text. I do not intend to report the exact sequence in which concepts were formulated, to attribute them to a particular author, to enter into the controversies about the primogeniture of discoveries. Even the formulas or laws that I report are in the modern formulation, the one most useful to the student of physics. For example, the function expressing the law of universal gravitation contained in Chap. 10 is not the Newton’s formulation, but that of later authors. In his book Principia (see Fig. 4.1) Newton illustrates the gravitational law with a series of theorems relating to the motion of planets and geometrical propositions. • The physical quantities that I introduce starting from this chapter (force, linear momentum, impulse, momentum of forces, angular momentum, work, kinetic energy, potential energy,…) began to be fully understood by the scientific community gradually, starting from the time of Galilei until the first half of the 1800s. Again, in introducing these quantities I will not use a historical sequence, but the one most convenient from the point of view of scientific exposure. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Spurio, The Fundamentals of Newtonian Mechanics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-47289-3_4

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Fig. 4.1 Isaac Newton, in a portrait by Sir Godfrey Kneller, 1702, oil on canvas. Left: title page of the first edition of the Principia. https://archive.org/details/philosophiaenat00newt/page/n5/mode/ 2up?ref=ol&view=theater

• I use the Galilean technique of removing accidents and initially reducing the exposition to the essential concepts. In particular, in exposing the concept of force I imagine that the body on which it acts is a point-like particle and not an extended body, and that the frame of reference in which I am operating does not affect the observation of phenomena. The topics of extended bodies and inertial and non-inertial frames of reference will be introduced in later chapters. • I assume that I know how to operationally measure the mass of the particle. Mass at the moment is assumed as an intrinsic property of matter, which can be represented by a scalar value and is conserved in time and space, remaining constant in every isolated system. Isaac Newton (1642–1727) is one of the greatest scientists of all time. In his lifetime, he also held the positions of president of the Royal Society, director of the English Mint and was a Member of Parliament. His monumental work Philosophiae Naturalis Principia Mathematica (shortened as Principia) appeared in 1687 in its first edition, then revised and expanded in two more subsequent editions. The Principia contains the fundamental rules of classical mechanics through its laws of motion, the formulation of gravitation in terms of a universal law, and an explanation of Kepler’s phenomenological laws: Newton thus credited the heliocentric theory definitively. A Latin-to-English translation of the book is available at the link: https://en.wikisource.org/wiki/The_Mathematical_ Principles_of_Natural_Philosophy_(1729). With the aim of describing the laws of nature in mathematical language, Newton developed the infinitesimal calculus beginning in 1666, anticipating the work of Gottfried Wilhelm von Leibniz

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by about ten years. A long diatribe arose between the two over the primogeniture of the infinitesimal calculus. Newton did not publish the mathematical aspects until 1704, in De Quadratura Curvarum. Newton also dealt with the nature of light, and first understood that white light is composed of the sum of all the colors of the spectrum. In his 1704 book Opticks he advanced the hypothesis that light was composed of particles. This corpuscular theory was in contrast to the wave theory of light, supported by contemporary authors such as Robert Hooke and (mainly) Christiaan Huygens. Of the Principia it particularly struck the contemporaries the interaction at a distance introduced by the law of universal gravitation. On the causes of this distance attraction Newton (at least in his writings) did not pronounce himself. He declared hypotheses non fingo, intending to limit himself to giving a quantitative description of the phenomenon, without formulating hypotheses about its causes. This is the basis of the scientific method of rejecting any explanation of nature without solid experimental verifications, and not imagining hypotheses that have not been induced by a rigid concatenation of experiments and reasoning based on the relationship of cause and effect. In terms of social and personal life, Newton had to be an extremely closed and irascible person. There are no known affections, mottoes of spirit (it is said that he never smiled), and many of his letters are of controversy and contrast with colleagues of the time. In his library of personal writings, in addition to papers on physics, astronomy and mathematics, documents on topics of theology and alchemy have been found. At a time when the principles of chemistry were not yet clear, alchemy investigated the nature of substances by referring to hermetic traditions through tests and experiments aimed at verifying hypotheses later revealed to have no scientific basis. Newton published nothing on these topics.

4.2 Measuring Forces Newton’s starting point begins with Galilei’s two observations reported in Sect. 3.2. The property that a body has of not varying its velocity if left undisturbed was called by Newton principle of inertia. It corresponds to the first law of dynamics, which I will discuss in detail in Chap. 5, considering what will be called inertial frames of reference. Galilei had tried to speculate about his observation concerning the fall along the vertical direction, but he failed to find an adequate explanation. Newton solved the problem by introducing the concept of force.

4.2.1 Weight and Anthropomorphic “effort” I try to introduce the physical concept of force from the anthropomorphic stress, called effort 1 in the following. Since Galilei, we have understood that all objects on the surface of the Earth fall with the same acceleration, the big stone and the goose feather. I know that if I operate in the absence of friction, on a perfectly smooth horizontal plane, I feel no effort to move an object. The problem is the lifting, the displacement along the vertical direction, where we know from Sect. 3.2, that there is present a downward 1

Effort is not a physical quantity. We will forget about this after this section.

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directed acceleration of magnitude .g. Experience tells me that my body is sensitive to some quantity other than the simple acceleration of gravity, which is the same for all bodies. The effort I exert is in fact proportional to the mass .m of the object I have to lift. My body is well aware that I must make a greater effort to lift the stone instead of the feather. I can, empirically, define the following quantity .

P ≡ mg

(4.1)

which I call the weight of the object. Weight is obviously a measurable quantity that has physical dimensions equal to [M L T.−2 ], i.e., different than those of a mass. In the SI it is measured in kg m s.−2 . Intuitively, I can imagine that the physiological effort is proportional to the weight of the object. Effort, however, is a subjective quantity: during the last move, I saw lifting objects that I never imagined could be moved. A subjective quantity is not, as we saw in Chap. 1, a physical quantity.

4.2.2 The Dynamometer Some devices have the characteristic of being able to undergo elongation or shortening and then return (if the deformation is not too great) to the initial rest configuration. Altogether, these devices are called springs, Fig. 4.2. The physiological effort that must be exerted to stretch the spring depends (fixed the spring) on how large the elongation is. Called with .k the characteristic property of springs (which depends on the material, the thickness of the coils, the number of coils per unit length,…), and by .l its elongation with respect to the rest position, I find that: physiological effort ∝ kl .

.

(4.2)

However, the spring can be used to operationally define the weight of objects. We construct identical objects (e.g., small cylinders) of the same volume and homogeneous material, which therefore have equal mass. By the definition (4.1), the weight of all the small cylinders is identical. I can, for example, construct small cylinders that have a unit weight, equal to 1 kg m s.−2 . If I arrange the spring along the vertical direction (as in Fig. 4.2), I can annotate the rest position and see the elongation .z when I hang one of the small cylinders. In case I hang two identical small cylinders, I notice that the elongation is doubled, that is, .2z. I can therefore arrange to construct a graduated scale, corresponding to the elongation of the spring for gradually increasing numbers of small cylinders. What I am constructing is a calibrated instrument for measuring the weight of objects. The total weight is given by the sum of the weights of the constituents: if the weight of a small cylinder is 1 kg m s.−2 , the weight of 5 small cylinders corresponds to 5 kg m s.−2 . However, I must verify that each time I add a new small cylinder, the

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Fig. 4.2 A springs and the consequences of Hooke’s law. On the left, the spring is at rest. Two identical objects, hooked at the end, produce elongations .z from the equilibrium position proportional to the weights (Credits: Wikipedia CC BY-SA 3.0)

spring returns to the equilibrium position when I remove everything. Normally, the calibrated scale results linear for a certain maximum number of small cylinders. A linear scale means that the displacement .z is directly proportional to the number of hanging cylinders. I must be careful of hysteresis phenomena: if I hang up too much weight, the spring will deform irreparably and can no longer be used as a measuring instrument. Such a calibrated spring constitutes an instrument for measuring the weight of unknown objects, and is called a dynamometer. Now, each unit mark drawn near the spring corresponds to the weight of 1 kg m s.−2 . The value of the constant .k is determinable. In fact, if a unit weight . P1 produces a unit displacement .z 1 , then: .k = P1 /z 1 . (4.3) A small needle or arrow at the end of the dynamometer allows me to interpolate the position in the case of objects having unknown weights intermediate between those of two small cylinders. I can then define the phenomenological law .

P = kz

(4.4)

which allows me to determine the weight . P of an unknown object by measuring with the graduated scale the elongation .z of the spring. The quality of the process is ensured by the fact that one can construct different dynamometers (differences on sizes, materials, construction methods), but still obtain for the measurement of the same object the same weight (within measurement errors, of course).

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Fig. 4.3 The weight of a small cylinder, directed downward, is balanced by the restoring force of the spring

4.2.3 The Weight is a Force Newton had the merit of recognizing that weight is a physical quantity that falls into the class of quantities that he will call by the name of forces. From now on, I call this quantity by the term weight force, leaving the term weight to denote its magnitude (4.1). We will have to verify that the weight force, acting along the vertical direction, is a quantity of a vector nature. In that case, in a Cartesian reference frame, the ˆ direction corresponds to that of the unit vector .k: F P = −mg kˆ .

.

(4.5)

Similarly, the pull-back property of the spring that opposes the weight force is itself a force. In fact, if you hang a weight on a spring, it reaches an equilibrium position: only a homogeneous physical quantity can countereact a force (Fig. 4.3). This restoring force of the spring always acts along the direction of the axis of the spring. For example, if I arrange the spring in the horizontal plane along the .x axis, I can write .F H = −kx ˆ i (4.6) where the quantity .x represents the displacement from the rest position .x0 = 0 of the spring.2 If I had arranged the spring along the vertical direction, I would have an equation analogous to (4.6), with only the replacement of the .z variable instead of

2

Obviously, I could put the origin of the .x axis at a position other than the rest position of the spring. In this case, the formula (4.6) becomes unnecessarily complicated in .F H = −k(x − x0 )ˆi.

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85

Fig. 4.4 A dynamometer with maximum capacity of 20 N, error on measurement . R) from the center of the sphere of mass . M and radius . R, feels a gravitational potential equal to .

VG = −

G Mm o , z

(10.56)

that is, as if all the mass . M were concentrated in its center. Again referring to Fig. 10.9, we can consider a mass element of the sphere .dm: dm = ρd V = ρr 2 d(cos θ)dϕdr

.

(10.57)

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which, being placed at a distance . D from .m o , contributes an infinitesimal potential energy on .m o equal to G(dm)m o ρr 2 d(cos θ)dr Gm o ρr 2 d(cos θ)dϕdr =− = (−2πGm o ) . D D D (10.58) In the last step, I integrated over .dϕ (that variable never enters any other quantity), obtaining the factor .2π. The overall effect of all elements of mass .dm is obtained by integrating the (10.58) over the volume. The distance D is also expressed in the vector relation: d VG = −

.

D=z−r

(10.59)

.

as can be seen from the vectors drawn in Fig. 10.9, i.e. .

D 2 = (z − r)2 = (z 2 + r 2 − 2z · r) = (z 2 + r 2 − 2zr cos θ) .

(10.60)

The potential energy on the .m o particle is then given by the solution of the double integral: {

{

{

cos θ=1

d(cos θ) . − 2zr cos θ)1/2 0 cos θ=−1 (10.61) Let us address the double integral: we first integrate on the variable .

VG =

d VG = (−2πGm o )

R

ρ(r )r 2 dr

.

(z 2

+ r2

cos θ ≡ χ

and then we integrate the result obtained with respect to the variable .r . I do (for convenience) the following variable changes: .

A ≡ z2 + r 2 ;

B ≡ 2zr

that will prove useful because .

A + B = (z + r )2 ;

A − B = (z − r )2 .

(10.62)

Now I take the external integration of (10.61) and make the proposed notation changes { 1 { 1 d(cos θ) dχ . = . (10.63) 2 + r 2 − 2zr cos θ)1/2 (z (A − Bχ)1/2 −1 −1

10.10

Mass in the Center of the Sphere (*)

269

This integration has as a primitive function: [ [ ] ] dχ 2( A − Bχ)1/2 1 2(A − Bχ)1/2 −1 = = 1/2 −B 2zr −1 ( A − Bχ) −1 1 [ [ ] 1/2 1/2 ] (A + B) − (A − B) (z + r ) − (z − r ) . = = zr zr [ ] 2r 2 = = zr z (10.64) where I made use of the relations (10.62) defined above. The result is remarkable: the complicated integration on the polar angle removes the dependence on .r , leaving us with .1/z as the dimensional factor and a numerical factor of 2. We now complete the work, substituting the (10.64) into the (10.61) {

1

.

VG =

−4πGm o z

{

R

ρ(r )r 2 dr =

0

−Gm o z

{

R

ρ(r )4πr 2 dr .

(10.65)

0

In the case when .ρ = cost we have {

R

ρ

.

4πr 2 dr = 4πρ

0

R3 = ρV = M 3

(10.66)

but also in the case where .ρ = ρ(r ) the magnitude .4πr 2 dr corresponds to the spherical volume element .d V for which {

R

.

4πρ(r )r 2 dr = M .

(10.67)

0

Ultimately, the (10.65) becomes: .

VG =

−Gm o G Mm o M =− z z

(10.68)

That is: the object of mass .m o at distance .z from the center of a sphere of radius R and mass . M feels a gravitational potential as if all the mass of the sphere were concentrated in the center of the sphere. You may have understood now why Newton needed to invent integral calculus to prove all this. Something that seems trivial and immediate to us to prove is not at all.

.

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10.11 Questions 1. Determine the intensity of the gravitational force exerted by a sphere of radius . R = 15 cm and . M = 158 kg and one of radius .r = 3 cm with mass .m = 0.73 kg whose surfaces are 5 cm apart [A: . F = 1.45 10−7 N] 2. Determine the intensities of the gravitational force of attraction between Earth and the Moon and between Earth and the Sun, and calculate their ratio . R. [A: . R = 6 10−3 ] 3. A quantity of mass . M in space splits into two fractions, of mass .αM and .(1 − α)M. Assuming fixed separation . D between the two fragments, for what value of .α is the gravitational interaction maximum? [A: .α = 1/2] 4. Determine to what altitude you need to rise above the Earth’s surface for the acceleration of gravity to change by 1%. Repeat the operation, imagining that you can descend by digging into the Earth’s interior. [A: ascend 32 km; descend 64 km.] 5. The writer has a mass (measured by the scale) of 73 kg. Assuming the Earth’s radius doubles, determine .(i) my weight if . MT remained constant, and .(ii) if the Earth’s density remained constant, [A: .i) 180 N; .ii) 1430 N.] 6. Determine the value of the Earth’s mass using the value of .g = 9.81 m/s.2 and that of .G given in (10.30). Earth’s radius is . RT = 6370 km. [A: . MT = 5.96 1024 kg] 7. If two planets have the same mass, but . A has twice the radius of . B, determine what the ratio of the accelerations of gravity is worth .g A /g B . [A: .g A /g B = 1/4] 8. If two planets have the same density, but . A has twice the radius of . B, determine how much the ratio of the accelerations of gravity is worth .g A /g B . [A: .g A /g B = 2] 9. The mass of the proton is .m p = 1.67 10−27 kg, that of the electron 1840 less than .m p . The Coulomb potential energy for the electromagnetic interaction between electron and proton in the hydrogen atom (fundamental state) is of the order of 10 eV, where 1 eV=.1.6 10−19 J. Estimate the ratio . R between gravitational potential energy and electromagnetic potential energy. [A: . R ∼ 10−38 ] 10. Determine the height .h relative to the Earth’s surface of a geostationary satellite traveling in a circular orbit in the equatorial plane. Also determine its speed .v. [A: .h = 35790 km; .v = 3.0 km/s.] 11. Neglecting friction with the atmosphere, determine how fast a projectile must be launched in order to travel in a circular orbit around the Earth so that it can return to its starting point. [A: .7.9 km/s]

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271

12. The ISS (International Space Station) is on an orbit 420 km above the Earth’s surface, and is moving with a speed of about .7.66 km/s. What is the value of the acceleration of gravity .g I SS at that height? What is the acceleration .a I SS worth on a body inside the station? Why do we read in the newspapers that experiments are carried out inside the station in the so-called “microgravity” regime? Is this statement justified? [A: .g I SS = 8.63 m s.−2 ; .a I SS ∼ 0] 13. Determine the work required to transport 10 kg of material from Earth to the ISS, the International Space Station, in orbit 420 km above the Earth’s surface. [A: .3.2 × 108 J] 14. An artificial satellite of mass .m = 3.4 ton is in a circular orbit at the altitude .h 1 = 5000 km. Due to various causes, it gradually loses energy and reaches an altitude .h 2 = 600 km. By how much is its kinetic energy varied? [A: .ΔT = 3.7 1010 J] 15. A meteorite is at a distance from the center of the Earth equal to 5 times the Earth’s radius . RT with negligible speed and direction of impact toward the ground. Determine the speed with which it reaches the Earth’s surface. [A: .v = 10.0 km/s] 16. An artificial satellite of mass .m = 2.5 ton is in a circular orbit around the Earth at a distance of 1600 km from the surface. Determine the magnitude of the satellite’s angular momentum . L, its kinetic energy . T , potential energy . V and total energy. [A: . L = 1.4 1014 kg m.2 /s; .T = 0.63 1011 J; .V = −2T ; . E = −T .] 17. We normally think that dissipative forces tend to decrease the velocity of an object. This is correct for isolated systems. Consider the case of an artificial satellite of mass .m in a circular orbit of radius .r around the Earth. Because of friction with the Earth’s atmosphere, the radius of the orbit decreases very slowly by a quantity.Δr as it approaches the surface. Determine:.(i) the change in velocity .Δv due to the change in the radius of the orbit; .(ii) the corresponding change in kinetic energy.ΔT and potential energy.ΔV ;.(iii) the work.ΔW done by the frictional forces. Recall that .Δr < 0. / MT [A: .Δv = (Δr/2) GrM3 T ; .ΔT = −(Δr ) Gm ; .ΔV = |ΔT |; .ΔW = −|ΔT |] 2r 2 18. Show that the force field expressed in spherical coordinates.F = f (r, θ)r is conservative if and only if the function . f is independent of .θ.

Chapter 11

Motions in Gravitational Fields

Abstract The chapter deals with the grandiose applications of the law of universal gravitation combined with the Newtonian laws of dynamics. For almost all known history, the universe for mankind was roughly limited to our solar system. It was only after 1920 that we comprehend that there were hundreds of billions of galaxies like our Galaxy, where the Sun is nothing more than one among about 100 billion stars. After introducing the Kepler’s empirical laws, I will show how Newton’s theory succeeds in explaining the three phenomenological relations derived by Kepler. After detailing the equations of motion of a two-body system, I use them not only in the case of planets of the solar system, but also of stars around the supermassive black hole in the Galactic center, in the motion of a two-neutron star system, and in the derivation of the gravitational indications for dark matter. At the end of the chapter, a discussion about triumphs and falls of Newtonian theory.

11.1 Historical Introduction The first grandiose application of the law of universal gravitation combined with the Newtonian laws of dynamics is related to the motions of the planets of the solar system. Allow me a brief historical introduction necessary to understand the impact of Newton’s laws not on the history of physics, but on that of humanity. For almost all of known history, the universe for mankind was roughly limited to our solar system. The Aristotelian-Ptolemaic system, the conception of the geocentric universe, was for many centuries the cosmological system of reference, accepted for nearly two millennia until the Renaissance. It seems that this system can be traced back to the Greek astronomer Eudoxus of Knidos (408–355 B.C.) but it is likely that earlier cultures had a similar vision. One is impressed to realize that what we know today has been acquired in a very recent period of human history. It was only after 1920 that we comprehend that there were hundreds of billions of galaxies like our Galaxy,1 where the Sun is nothing more than one among about 100 billion stars. In the last 20 years, thanks to observational 1

When we refer to our Galaxy, sometimes called the Milky Way, we use the capital letter.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Spurio, The Fundamentals of Newtonian Mechanics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-47289-3_11

273

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cosmology we have a knowledge of the history and evolution of the Universe at such a level that was previously totally unimaginable. The first to realize (too early) that the universe might be enormously larger than was believed was Giordano Bruno (1548—Rome, February 17, 1600). His burning at Campo dei Fiori by the Roman church was a warning and heavy conditioning for contemporary and later thinkers. Aristotle (384–322 B.C.) is the philosopher who translated in an organic way all the previous knowledge on the conformation of the universe in two of his works, the Physics and the De coelo. Aristotle proposed a geocentric model, with the Earth at the center of the universe. The structure was extremely complicated: according to Aristotle, the heavenly bodies (in order: the Moon, Mercury, Venus, the Sun, Mars, Jupiter, Saturn, and the sky of fixed stars) move on 55 concentric spheres. Claudius Ptolemy (c. 100–168) worked in Alexandria, Egypt, and authored the monumental astronomical treatise known as Almagestus in which he systematized the Aristotelian model in an organic way. The name of the book means in Arabic “the greatest”. After the darkness that followed the fall of the Roman Empire, the initial rediscovery in Europe of the scientific and philosophical culture of classical Greece occurred mainly through Arab manuscripts. Ptolemy formulated a geocentric model, made up of epicycles and deferents: the epicycle is a circumference whose center is placed on the circumference of a circle of greater radius called deferent, Fig. 11.1. The Ptolemaic structure is of remarkable beauty and complexity, and it was also predictive: the positions in the future of celestial objects could be calculated with some accuracy. In Europe, the scientific method began to flourish again around the year 1000 thanks to the advent of the line of thought called scholasticism. Scholasticism dealt with the philosophy of nature based both on observations of events and on the auctoritas of some authors, including Aristotle in the metaphysical field and Ptolemy in the astronomical one. But come the time of the reform of Luther (1483–1546)

Fig. 11.1 Ptolemaic diagram of a hypothetical planet orbiting Earth. The larger (dashed) orbit is the deferent and “x” represents its center; the smaller orbit is the epicycle. The Earth’s position is indicated

11.1 Historical Introduction

275

Fig. 11.2 Reproduction of the diagram of orbits in De revolutionibus orbium coelestium, Nicholas Copernicus (ed. 1566) at the INAF-Brera (Milan, Italy) Astronomical Observatory Library. Book available at https://archive.org/details/nicolaicopernici00cope_1/mode/2up

and any substantial change in thought could be seen as subversive work, both in the Catholic and Protestant camps. A decisive step toward the modern vision was taken by Nicholas Copernicus (1473–1543), a contemporary of Luther. Copernicus graduated in canon law in Ferrara (Italy), and also studied astronomy in Bologna. Copernicus’ theory placed the Sun, not the Earth, at the center of the orbits of the planets. Such a system for describing the observed universe is called heliocentric. The Copernican theory contains, for today’s eye, severe flaws: the assumption that the orbits of the planets are circular, and the maintenance of eccentrics and epicycles. They were in the heliocentric theory required a number of “circles” comparable to those used by Ptolemy. Conscious of the revolutionary message of his work De revolutionibus orbium coelestium (On the revolutions of the celestial spheres), Fig. 11.2, Copernicus published it in the very year of his death.2 Tycho Brahe (1546–1601) was a Danish nobleman with a passion for astronomy, and he built a dedicated observatory for his studies. Brahe realized that progress in astronomical science could only be achieved through systematic and rigorous obser2 I hope you noticed that I used the word “revolution” with the meaning of “extraordinary upheaval, radical change”: this meaning takes its very name from the effect of Copernicus’ book with that title.

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11 Motions in Gravitational Fields

vation, night after night, and through the use of instruments that were as accurate as possible. In 1572 he had the opportunity to observe the penultimate visible supernova (SN 1572) exploded in our Galaxy: his book De Nova Stella (1573) coined the term “nova” to indicate the advent of a “new star” (from which modern astronomers have derived .nova and supernova). The purpose of his methodical and systematic observations, with enormous amounts of accumulated data, was to refute the Copernican heliocentric theory. The great authority acquired by Brahe delayed the establishment of the Copernican system, but his data favored the complete abandonment of the Ptolemaic system. John Kepler (1571–1630) was called in late 1599 by Tycho Brahe as his assistant. In 1601, on Brahe’s death, he became his successor in the post of mathematician, astronomer and imperial astrologer in Prague. In 1604 he observed a supernova (the last one so far observed in the Galaxy), SN 1604.3 Kepler, precisely using the data accumulated by Tycho, became a staunch supporter of the Copernican system. His first two laws were enunciated in 1609, when he published his masterpiece Astronomia nova. His third law was made known in 1619 in the work Harmonices Mundi. In 1620 Kepler’s mother was arrested after being accused of witchcraft by the Protestant church: the trial lasted six years, and Kepler assumed her defense.

11.2 Kepler’s Empirical Laws Kepler’s laws are phenomenological laws: they define relationships between quantities that are based on observed data. One intuits that there are fundamental rules that determine the relations observed, but one is unable to formulate them. To find those fundamental relations, it will necessary to wait 80 years the Newtonian Principles. Below, Kepler’s three laws are stated; later in the chapter I will show how Newton’s theory succeeds in explaining the phenomenological relations derived from Kepler. The statements of Kepler’s three laws are as follows: 1. (Law of Elliptical Orbits, 1609) The orbit described by each planet occurs on a plane and is an ellipse, with the Sun at one of the two foci. 2. (Law of Areas, 1609) The line segment joining the Sun to a planet sweeps out equal areas during equal intervals of time (Fig. 11.3). 3. (Law of Periods, 1619) The ratio of the square of the time that each planet takes to travel through the orbit (period .T ) to the cube of length of the semi-major axis of the orbit, .a, is the same for all planets, i.e., .T 2 /a 3 = cost. See Table 11.1.

3

Prior to 1572, the last reported supernova had appeared in 1064. During the critical phase of the discussion of the Aristotelian immutability of the sky, two supernovae in 30 years are a happy coincidence, a divine sign in favor of scientists. Since then, no other supernova has been visible to the naked eye.

11.2 Kepler’s Empirical Laws

277

Fig. 11.3 Schematic illustration of Kepler’s law of areas. The position when the planet is closest to the Sun is called the perihelion (or, periastron if it is any other celestial object). The farthest point in the orbit is called aphelion (or, apoastron if it is not the Sun) Table 11.1 Today’s data related to Kepler’s third law for the planets of the Solar System. Mass is expressed in terms of the Earth’s mass, (. MT = 5.97 × 1024 kg). The Astronomical Unit (A.U.) corresponds to .1.5 108 km. The distance from the Sun is indicative of the semi-major axis of the ellipse, .a. The ratio .T 2 /a 3 is in the units used for .T and .a. The eccentricity of the elliptical orbit is defined in Fig. 11.10. All orbits are nearly circular except for Mercury and Pluto. The inclination of the orbit shows that the planes in which the orbits occur are practically parallel, except for Pluto. Also for these reasons, Pluto in 2006 was downgraded by the International Astronomical Union to a dwarf planet 2 3 . M/M T Sun Period . T /a Eccentricity Inclination Planet ◦ distance (years) .( ) (A.U.) Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

0.055 0.815 1.00 0.107 318 95.2 14.5 17.1 0.002

0.387 0.723 1.000 1.520 5.200 9.540 19.20 30.100 39.440

0.241 0.615 1.000 1.880 11.9 29.4 84.0 165.0 248.0

1.002 1.001 1.000 1.006 1.007 0.996 0.997 0.998 1.003

0.206 0.007 0.017 0.093 0.048 0.056 0.047 0.009 0.249

7.01 3.39 0 1.85 1.31 2.49 0.77 1.77 17.2

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11 Motions in Gravitational Fields

11.3 The Two-Body System The study of the Solar System (consisting of the Sun and planets, and the satellites of the planets) is that of a complicated system of many bodies. The solution adopted historically has been to study two-by-two interactions (e.g., the Sun and the Earth; the Sun and Venus), taking the planet of which you are interested and the object of higher mass, or the closer, whose gravitational interaction is dominant. Once the analytical solution (the orbit that the planet will follow, with the possibility of locating it in space once the initial conditions are known) is found, the perturbative theory is adopted. That is, one goes to see how the orbit of Earth (for example) can be affected by the orbit of Venus or the Moon (or the other nearest planets). Because of this, the resolution of the two-body system is very important. Consider observing, in an inertial reference frame, two masses, .m 1 and .m 2 (Fig. 11.4) gravitationally interacting through (10.10). The two-mass system can be considered isolated; we neglect all other possible external forces. The center of mass of the system is therefore at rest (or moving in uniform rectilinear motion) with respect to the inertial frame. The body of mass .m 1 will be described by the vector .r1 , and the body of mass .m 2 will be described by the vector .r2 . The force of attraction between points 1 and 2 passes along the line joining them, defined by the relation: r

. 12

= r1 − r2

(11.1)

which, as the drawing shows and by the parallelogram rule, is a vector starting at .m 2 and arriving at .m 1 . By construction, the unit vector rˆ

. 12

starts at 2 and points at 1.

Fig. 11.4 Two-body system observed from an inertial frame of reference. The vectors involved in the problem are defined

=

r1 − r2 r12

(11.2)

11.3 The Two-Body System

279

The equations of dynamics describing the problem are then ⎧ 2 2 ⎪ m 1 ddtr21 = F1 = − Gmr 21 m 2 rˆ 12 = − Gmr 21 m 2 r1r−r ⎨ 12 12 12 .

⎪ ⎩m

(11.3) d 2 r2 2 dt 2

= F2 =

+ Gmr 21 m 2 rˆ 12 12

=

2 + Gmr 21 m 2 r1r−r 12 12

notice that the distance between the two points is .r12 and that in the second equation the sign .+ appears to indicate that the force .F2 is just as direct as the unit vector ˆ 12 defined in (11.2). As they are written, the (11.3) represents a system of coupled .r differential equations: the first equation describes the second derivative of the first vector, but this quantity is also contained in the second equation. And vice versa. The resolution is not trivial. The system (11.3) can be simplified in the following way: I first carry the mass term of both equations to the right-end side, without simplifying:

.

⎧ d2r 1 1 ⎪ ⎨ dt 2 = − m 1

Gm 1 m 2 r1 −r2 2 r12 r12

⎪ ⎩ d 2 r2 = + 1 dt 2 m2

Gm 1 m 2 r1 −r2 2 r12 r12

(11.4)

then, I subtract the second term from the first: ( 0 d 2 (r1 − r2 ) Gm 1 m 2 r1 − r2 1 1 . =− + . 2 dt 2 r12 m1 m2 r12 I define .

1 ≡ μ

(

1 1 + m1 m2

0 −→

μ≡

m1m2 . m1 + m2

(11.5)

(11.6)

The quantity .μ, with the dimension of a mass, is called the reduced mass of the two-body system. Using this definition, the (11.5) becomes: μ

.

Gm 1 m 2 d 2 r12 rˆ 12 . =− 2 2 dt r12

(11.7)

We traced the two-body problem back to a single-body problem. The (11.7) says that the two-body system can be reduced to the study of the motion of a body with mass equal to the reduced mass of the system, moving at the distance .r12 and directed opposite to the unit vector .rˆ 12 . Let us now look at the two limit cases, to understand the physical meaning of the reduced mass: • A dominant mass, .m 2 >> m 1 . In the limit where .m 1 /m 2 → 0, we have from the (11.6) that .μ = m 1 , that is, the reduced mass coincides with the smaller mass, which feels a force directed toward .m 2 . In practice, we go back to what we studied

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11 Motions in Gravitational Fields

in the previous chapter, when we assumed .m 1 = m > m.

11.6 Mechanical Energy of the Two-Body System

287

dynamics of system allows all values of .r until you reach .r = 0 (=collision). In fact, as .m approaches the massive object . M the potential energy decreases, and then (to conserve energy) the kinetic energy of.m increases: it is falling toward the massive object. This is the mechanism everyone imagines for objects that are in motion near a black hole (or a star). On the contrary, most objects do NOT precipitate on the black hole (or the star), just as the Earth is not swallowed up by falling into the Sun. The reason is that situations where objects have . L > 0 are much more likely. I now consider the situation where . L /= 0: with reference to Fig. 11.7, this means that the two vectors, .v and .r, are not parallel. The object of mass .m is approaching . M but not along the radial direction. In this case |L| = |r × μv| = μvb

.

(11.33)

where .v is the initial speed of the body of mass .m. The quantity .b = r sin ϕ (shown in the drawing) is called impact parameter, and it corresponds to the minimum distance the body in motion would have with respect to object . M in case there was no interaction. You can easily understand that for randomly moving objects, it is much more likely to have nonzero relative angular momentum rather than zero angular momentum. From (11.33) you can also see that solutions with .b = 0 (i.e., collision) are not allowed for objects in motion if . L > 0, unless some internal or external mechanism induces energy dissipation. This is way celestial bodies do not normally fall on our heads. Situations such as those scripted in the movie Don’t Look Up5 are extremely unlikely due to the presence of angular momentum. Although rare and unlikely these conditions are, it is still necessary to keep the situation under control of telescopes. If. L > 0, the term in the effective potential .+L 2 /2μr 2 acts as a kind of centrifugal potential, which prevents the celestial bodies from collapsing on one another. Let us look, with reference to Fig. 11.6 what the possible solutions are, based on the total mechanical energy of the system .

E = Trad + Ve f f

(11.34)

where the radial kinetic energy term is the first in (11.30). The following situations are possible: • . E = E hyp > 0. In this regime, the radial kinetic energy is very large, allowing the moving object to get closer and closer, until a minimum distance is reached (the one obtained by projecting the contact point of . E = E hyp with .Ve f f on the .x axis). After the moment the minimum distance is reached, the body will begin to move away. As we shall see in the accurate geometrical description of the next section, the motion will assume a hyperbolic shape. The following situations are possible:

5

Movie written and directed by Adam McKay (2021). Also worth watching for the presence of Jennifer Lawrence and Meryl Streep.

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11 Motions in Gravitational Fields

• . E = E par = 0. The situation is similar to the previous case: there is always a point of minimum distance, and the trajectory takes the form of a parabola. • . E = E ell < 0. In such a regime, the radial energy is not very high, and the only allowed regions are those between the points intercepted by the line . E = E ell with the curve .Ve f f , that is, the region between .r1 and .r2 . Thus, the object is confined to an elliptical orbit. • . E = E cir c < 0. It corresponds to the fact that the radial kinetic energy is zero. In that case (never varying the value of the radius .r0 ) the orbit is circular. The discussion so far has necessarily been qualitative. To be more quantitative, and to derive the parameters of the elliptical orbits observed by Kepler, we need to go into some mathematical aspects presented in the following sections. It is interesting to note that in all two-body systems where there is a gravitationally bound system (i.e., with elliptical or circular orbits), the total mechanical energy is always negative. It must be so: gravitational potential energy is negative, kinetic potential energy is positive. If the latter prevails, it means that objects can move away from each other at infinite distances (remember the concept of escape speed). Conversely, if mechanical energy is negative, it means that the “bond” caused by it prevents the bodies from moving away (questions 3 and 4). This is also the case when studying N-particle systems. In a stable, isolated system of N interacting particles through a potential .V (r ), it can be shown that there is a relationship between the average kinetic energy . of the system and the potential energy. In the case of a potential of type.V (r ) ∝ r −1 such as the gravitational potential or the Coulomb potential of electrostatic interactions, we have that 2 = −

.

(11.35)

where. represents the average (negative) potential energy of the system itself. The relation (11.35) (the demonstration of which is postponed to the statistical mechanics course) is fundamental in gas-related studies, from the chemistry up to macrosystems that form stars and galaxies, and is called the virial theorem.

11.7 Kepler’s First Law, Elliptic Orbits (*) Ellipses, parabolas and hyperbolas are the plane geometry solutions of the gravitational two-body system. Such functions, in analytic geometry and projective geometry, correspond to curves obtained from a cone’s surface intersecting a plane; for this reason, they are often called conics and conic functions their analytical representation. Some mathematical reminders are necessary to arrive at the elliptical orbits of planets.

11.7 Kepler’s First Law, Elliptic Orbits (*)

289

11.7.1 Conic Functions Conic sections can be defined based exclusively on concepts of plane geometry. Referring to Fig. 11.8, give a straight line called the directrix (the vertical one in the drawing) and a point F outside the directrix called the focus. A conic section consists of the locus of points whose distance . P F from the focus F and distance . P Q from the directrix are in constant ratio .e to each other, that is: PF .

PQ

=e.

(11.36)

The quantity .e is defined positive, and is called (for the geometric meaning we shall see) eccentricity. Calling now the distance between the focus and the directrix . F O = β, the distance . P F = r , from the geometry of the drawing you can see that PF .

PQ

=

r =e. β − r cos ϕ

(11.37)

Solving the equation with respect to the variable .r (which is the one we are interested in, because it describes the locus of the points defining the searched curve) we have r=

.

Fig. 11.8 Geometric parameters needed to define functions called .conics, i.e., parabolas, hyperbolas, and ellipses

eβ . 1 + e cos ϕ

(11.38)

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11 Motions in Gravitational Fields

You can appreciate that this function describes a curve in the plane in polar coordinates, .r = r (ϕ), with origin in the focus F. Recalling that .cos ϕ varies between -1 and 1, it becomes clear that three possible cases can occur: • .e < 1. In this case, the denominator is always defined and the possible values of .r eβ eβ are between .r ∈ [ 1+e ; 1−e ]. The geometric locus of the points never goes too far from the focus and directrix. The drawn curve is like the one in blue in the figure, that is, a ellipse. • .e = 1. In this case, the denominator of (11.38) can cancel. The locus is unbounded ; ∞]. The geometric locus is the one in purple in the figure, i.e., a and .r ∈ [ eβ 2 parabola. • .e > 1. In this case, the denominator of (11.38) can cancel, but the locus can get eβ ; ∞]. The geometric locus is the one even closer to the directrix because .r ∈ [ 1+e in green in the figure, which is a hyperbola.

11.7.2 First Integral Let us start again from the equation of motion (11.9) for the two-body system and see how, in polar coordinates, the solutions admitted for the orbits are functions described by (11.38), that is, conics. In the equation of motion μ

.

Gm 1 m 2 dv =− rˆ dt r2

(11.39)

I make use of the relations on the derivatives of the unit vectors given in (3.56), i.e.: .

r˙ˆ = ϕ˙ ϕ ˆ

and

˙ˆ = −ϕˆ ϕ ˙r

(11.40)

and from the second of which I obtain .

rˆ = −

ˆ 1 dϕ , ϕ˙ dt

(11.41)

Substituting this relationship into (11.39), and using .α = Gm 1 m 2 : μ

.

α 1 dϕ ˆ dv = 2 dt r ϕ˙ dt

−→

dv α dϕ ˆ − =0. 2 dt μϕr ˙ dt

(11.42)

We recognize in the denominator the quantity that in (11.19) represents the magnitude of angular momentum, which is a constant quantity. Therefore, by replacing . L = μϕr ˙ 2 I can rewrite (11.42) as | | α d v− ϕ . ˆ =0 dt L

−→

| | d L v−ϕ ˆ =0. dt α

(11.43)

11.7 Kepler’s First Law, Elliptic Orbits (*)

291

Fig. 11.9 The velocity assumed by the planet at three different positions along the orbit. Only in aphelion and perihelion v is perpendicular to the radial vector r

I have thus reduced the differential equation of motion (which is of second order) to a first integral of motion: a function that remains constant over time when evaluated along the possible trajectories assumed by the system. In particular, from (11.43) I obtain | | L (11.44) . v−ϕ ˆ = cost α As with all problems arising from a second-order differential equation, there will be two initial conditions to know, two constants to impose. Note that, as it is written, (11.44) is dimensionless: all unit vectors are dimensionless. To determine the constant of motion appearing in (11.44), I consider a particular position, namely that corresponding to perihelion, see Fig. 11.9. At this point, the velocity vector v is perpendicular to the radial vector .r = r rˆ . It follows that v will be ˆ which in perihelion is parallel to the direction indicated by the moving unit vector .ϕ, along the . y axis of a Cartesian reference frame centered on the Sun, as in Fig. 11.9. In this way, I can write the constant of (11.44) at perihelion as: | .

| L v−ϕ ˆ = eˆj α

(11.45)

where .e is a positive scalar constant. The constant, as written, must remain the same throughout the planet’s trajectory. A constant, after all, is a constant! I now consider the left and right-hand members of (11.45) and perform the scalar product of both ˆ members with the unit vector .ϕ: | L v−ϕ ˆ = eˆj · ϕ ˆ . .ϕ ˆ · α |

(11.46)

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11 Motions in Gravitational Fields

Recalling from (3.55) the relationship between the moving unit vectors in polar coordinates and the unit vectors in Cartesian coordinates, specifically: ϕ ˆ = − sin ϕˆi + cos ϕˆj

(11.47)

ˆj · ϕ ˆ = cos ϕ .

(11.48)

.

I obtain

.

By virtue of this simple relationship, the (11.46) becomes | .

| L ϕ ˆ · v − 1 = e cos ϕ . α

(11.49)

Using the velocity written in polar coordinates, (3.58), I have ϕ ˆ ·v =ϕ ˆ · (˙r rˆ + r ϕ˙ ϕ) ˆ = r ϕ˙ .

.

Then, (11.49) becomes:

| .

| L r ϕ˙ − 1 = e cos ϕ . α

(11.50)

(11.51)

˙ 2 , the Because in polar coordinates the magnitude of angular momentum is . L = μϕr above relationship corresponds to: .

L2 − 1 = e cos ϕ μαr

−→

1 μα = (1 + e cos ϕ) 2 r L

(11.52)

or, equivalently: .

r=

L2 1 . μα (1 + e cos ϕ)

(11.53)

This equation has the same structure as (11.38) describing conic functions. As a result, the solutions for the motion can be ellipses, parabolas or hyperbolas depending on the initial conditions, that is, on the value of the arbitrary parameter .e introduced and of the magnitude of the angular momentum. The next step is then to find the relationship between the geometric parameter .e and the total mechanical energy . E of the system.

11.7.3 Eccentricity Versus Energy and Angular Momentum We found, by solving the first integral of motion, that the possible orbits are conic. To determine which, we need to tie the free parameter .e (eccentricity) of (11.53) with the energy and angular momentum of the particular two-body system considered.

11.7 Kepler’s First Law, Elliptic Orbits (*)

293

In the case of elliptical solutions it will be necessary to determine the values of the semi-major (.a) and semi-minor (.b) axes, that are linked together by the parameter .e. The values of .a and .b (and thus the eccentricity) will be determined by knowing the total mechanical energy and the magnitude of the angular momentum. Why these two quantities? Because we know they are constant of the motion. And they are two physical quantities, while .a and .b are geometric parameters. We are physicists interested in the real motions of individual objects, not only in the pure abstract forms of geometry. The total mechanical energy of the system given by (11.32) is conserved over time, and thus along the trajectory of motion. I usually consider a convenient ˆ point, perihelion. At perihelion, .r = r P and the velocity (3.58), i.e., .v = r˙ rˆ + r ϕ˙ ϕ, is exactly perpendicular to the radius, so the .r˙ component is zero. Therefore, the total energy in perihelion is. | | 2 L α . .E = − (11.54) rP 2μr P2 I will now consider the relation that provides the equation of the orbit (11.53) found above at the perihelion position (which corresponds, by definition, to the value .cos ϕ = +1): L2 (11.55) .r P = μα(1 + e) I insert the expression of .r P already determined into (11.54): | L 2 [μα(1 + e)]2 α[μα(1 + e)] − 2μL 4 L2 | | 2 μα2 μα (1 + e)2 μα2 (1 + e) = = − [(1 + e)2 − 2(1 + e)] 2 L2 L2 2 L2 μα2 2 (e − 1) . = 2 L2 |

E= .

α L2 − rP 2μr P2

|

|

=

(11.56)

Now I can obtain explicitly the parameter .e from the (11.56): / e=

.

1+

2E L 2 . μα2

(11.57)

Note that .e depends (given the masses of the two objects) on the mechanical energy and angular momentum (squared) of the system. Remembering the geometric meaning of the parameter .e as mentioned in Sect. 11.7.1, we will have that: • .e < 1, solutions that are compatible with elliptical trajectories, are possible only in the case of negative mechanical energy, . E < 0; • .e = 1, parabolic trajectory, possible only in the case of zero mechanical energy, . E = 0;

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11 Motions in Gravitational Fields

• .e > 1, solutions that are compatible with hyperbolic trajectories, only possible in the case of positive mechanical energy, . E > 0. This is consistent with what was determined in the previous qualitative discussion of Fig. 11.6.

11.7.4 Elliptic Solutions: Semi-axes as a Function of E, L I consider the only solutions involving elliptical trajectories, those with negative energy. For clarity, I denote them with. E = −|E|. First, I can find the relation between the semi-major axis of the ellipse and the total energy starting from (11.56). With the change of notation . E = −|E|, and reversing numerator and denominator, I rewrite it as: 2 L2 1 = . (11.58) . |E| μα2 (1 − e2 ) Making use of the relation (11.53) I can determine the distance of the focus from perihelion .(cos ϕ = +1) and aphelion .(cos ϕ = −1), i.e.: r =

. P

L2 L2 ; rA = μα(1 + e) μα(1 − e)

(11.59)

in such a way that the semi-major axis can be written as (see Fig. 11.10): a=

.

Fig. 11.10 Geometric parameters of an ellipse of semi-major axis .a and semi-minor axis .b. The major axis is .2a = r p + r A . The area of the ellipse is given by the relation . A = πab. The eccentricity is / defined as .e =

1−

b2 a2

L2 rP + rA = 2 μα(1 − e2 )

(11.60)

11.7 Kepler’s First Law, Elliptic Orbits (*)

295

and from comparing (11.60) with (11.58) I can recognize that

.

a=

Gm M α = . 2|E| 2|E|

(11.61)

We obtained the remarkable result that the semi-major axis of the elliptical orbit depends only on the total mechanical energy . E of the system. Finally, we need the relation that determines the value of the semi-minor axis, to uniquely define the orbit. I use the relation (11.57), which, with the convention used for negative energy, can be written as / e=

.

2|E|L 2 1− μα2

/ ←→

e=

1−

b2 a2

(11.62)

which I compared, in the same line, with the geometric relation connecting eccentricity with the major .a, and minor .b, semi-axis of the ellipse. Equalizing the two negative terms: .

2|E|L 2 b2 = μα2 a2

−→

b2 = a 2

2|E|L 2 α2 2|E|L 2 = μα2 4|E|2 μα2

(11.63)

having in the last step used (11.61). Simplifying, I then obtain the relation:

.

L b= √ 2μ|E|

(11.64)

Thus: the semi-minor axis depends on both the total mechanical energy and the magnitude of angular momentum.

11.7.5 Degeneracy in Classical and Quantum Physics The situation illustrated in Fig. 11.11 is what in physics is called degeneracy. It will be one of the fundamental aspects of studying quantum mechanics but, as you see, it also exists in classical mechanics. In atomic and subatomic physics, degeneracy occurs when several distinct quantum states have the same energy level. In the classical case seen, degeneracy is broken by the value of angular momentum: knowledge of. L determines the actual shape of the ellipse traveled by the planet. Also in the atomic, molecular or nuclear case a certain value of energy may correspond to several distinct states characterized by different values of angular momentum. In the quantum case, degeneracy is broken by knowing . L 2 (the square of the angular momentum) and one of its components (conventionally, the one called . L z ).

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11 Motions in Gravitational Fields

Fig. 11.11 These ellipses have the same value of .a but different values of .b. A particle having a certain value of mechanical energy can orbit indifferently on one of these ellipses. There is thus a .degeneracy with respect to the energy parameter (i.e., different admissible solutions). Removal of the degeneracy is done by considering another physical quantity, in this case the magnitude of angular momentum . L that uniquely fixes (known . E) the length of the semi-minor axis

11.8 Kepler’s Third Law, Revised Knowing now the orbital parameters of the ellipse from Kepler’s first law, we can easily explain its III law without making the approximation of circular orbits. From the relation (11.22), which expresses the infinitesimal area as a function of angular momentum, I can derive: L dt .d A = 2μ

{ −→

T

A= 0

L L dt = T 2μ 2μ

(11.65)

having integrated over a period (i.e., over one revolution of the planet around the orbit, sweeping the complete area of the ellipse). From the geometry (Fig. 11.10) we have . A = πab. Therefore, if I square the period obtained in (11.65): .

T2 =

4μ2 4μ2 2 A = 2 π 2 a 2 b2 2 L L

(11.66)

and replacing the value of parameter .b obtained in (11.64), I get: .

T2 =

4μπ 2 a 2 α 4μπ 2 a 3 4μ2 2 2 L 2 = = . π a L2 2μ|E| α 2|E| α

(11.67)

In the last step, I recognized the relation (11.61). Now entering explicitly the value of .α and remembering the definition of reduced mass .μ, (11.67) becomes:

.

T2 =

4π 2 a 3 G(m + M)

−→

G(m + M) a3 = 2 T 4π 2

(11.68)

11.8 Kepler’s Third Law, Revised

297

which is analogous to the relation we had obtained in (11.27) with the assumption of circular orbits.

11.8.1 The Black Hole in the Center of the Galaxy Kepler’s third law is a very powerful relation: it allows us to know physical quantities that are difficult to determine (for example, the mass of a central body as the Sun) by measuring orbital parameters of objects in motion around the central body, such as the semi-major axis of ellipses and revolution periods. The latest, extraordinary example of the use of Kepler’s third law is the Nobel Prize-winning case in 2020. Two different groups (one in Europe and one in the U.S.) have measured the mass of the supermassive black hole (SMBH) existing in the center of our Galaxy. Thanks to the data published and collected in Fig. 11.12, the determination of the mass of the SMBH is also possible from our side. It takes time (expressed in years) to measure periods of revolution of astrophysical objects: the data shown cover about 26 years. The observations were made using telescopes sensible to infrared radiation, a band that makes it possible to see stars in the central part of our Galaxy. Visible light is in fact absorbed by galactic material that is along

Fig. 11.12 This figure is taken from a recent article that contains the observations that led to the Nobel Prize in Physics in 2020 https://doi.org/10.1051/0004-6361/201833718. See the text for further information

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11 Motions in Gravitational Fields

the line of sight. In the following, I will impose the discussion while the reader can perform the calculations. The figure shows observations related to the motion of the star (which emits electromagnetic radiation) called S2. The symbols are the experimental data, taken at various epochs, and the curves the analytical fits; the colors indicate the year, as from scale on the right. The motion appears to occur around a fixed point (the focus) [ does not appear denoted as SgrA*. In frame A) on the left this point denoted by .× to be in the focus of an ellipse, because the image taken from Earth is projected onto a plane that is not perpendicular to the observer. In particular, the major visible axis of the ellipse is rotated by an angle equal to .θ = 43◦ with respect to the view from Earth, so the true value is obtained by correcting by a factor of .1/ cos θ what appears in the figure. The distance of the Galactic center (and thus of the S2-SgrA* system) from Earth is D .= 25640 light-years. From Box (A), you can estimate the value of the semi-major axis, .a, of the ellipse by making the suggested correction. Use the scale in the center of the figure, or the angular distance given along the . y axis, parallel to the major axis. Recall that 1 AU corresponds to .1.49 × 1011 m while arcsec is an angular unit that corresponds to 1/3600 of a degree, i.e., 1 arcsec .= .4.85 × 10−6 rad. The value of the revolution period of S2 can be determined from picture (B), which shows its radial speed. Recall that the velocity is maximum in periastrum, so the value of .T corresponds to how much time passes between two successive maxima. The scale along the x-axis is in years. After calculating the values of .a and .T you can determine the mass . M of the object placed in the focus of the ellipse, and corresponding to the position denoted as SgrA*. Make use of Kepler’s III law, thus assuming negligible mass .m of S2. To have a more intelligible number, express the mass . M in units of solar masses, rather than in kg, remembering that . Mo = 2 × 1030 kg. In the focus of the ellipse, where the . M mass should be, there is nothing to be [ It is therefore what we seen! No radiation of any kind is coming from the region .×. call a black hole. There is something there, because it produces a gravitational effect, but nothing can be seen. Try to determine the value of the event horizon, using the relation (10.32) or (10.33). Express this value in Astronomical Units, to compare it with the size of our Solar System. You can verify, by inspecting Box (C), that the star S2 always orbits at distances much greater than the event horizon and does not fall inside the black hole. Finally, you can estimate the speed of S2 at the periastrum, in terms of the speed of light. If you didn’t want to do the math, here are the values: the length of the semimajor axis of the ellipse is .a = 1.5 × 1014 m and the period about 16 years, or .T = 5 × 108 s. The mass of the central object is therefore . M = 4 106 × Mo (correct: four .million solar masses!). It can only be a SMBH with event horizon equal to ∗ . R = 0.08 AU (small: it would be within the orbit of Mercury, which is about 0.4 AU). The star S2 is always far from the SMBH and when in perihelion its speed is 0.02 c.

11.9 Two Neutron Stars

299

11.9 Two Neutron Stars Perfectly isolated two-body systems are hard to find on Earth. In astrophysical environments, binary systems (i.e., composed of two interacting stars) are very common: probably a significant fraction of stars make up binary systems. A binary star system is little affected by other objects much farther apart, and thus constitutes an excellent approximation of an ideal isolated system. A particularly interesting case is that of binary systems composed of two neutron stars.6 The prediction for the mass of a neutron star is in a rather small range near 1.5 . Mo . This means that all neutron stars have approximately the same mass. They are also extremely compact because they have a radius of the order of 15 km. Objects that could be rapidly rotating neutron stars were first observed in 1967 and classified as pulsars. Today, we know that pulsars are precisely young neutron stars (there is a catalog containing a few thousand pulsars) that have a very high angular velocity, and emit radiation along an axis that can intercept Earth’s line of sight. See also Fig. 12.4 of the next chapter. The incoming pulsed signal has a well-defined periodicity and represents a nearly perfect clock, Sect. 12.4. A binary system composed of two neutron stars was first observed by Russell A. Hulse and Joseph H. Taylor in 1974 using the Arecibo radio telescope located in Puerto Rico. One of the two neutron stars was very young, and emitted pulsed radiation. This binary system was a practically ideal case of an isolated system composed of two objects of equal mass, Fig. 11.13. In the framework of Newtonian mechanics, the mechanical energy of an isolated system of two neutron stars (given by the sum of potential and kinetic energy) remains constant. The distance .r between the two stars varies (as seen in Fig. 11.13) from a minimum value when both are in periastron to a maximum value when they are in apastron (the distance .r3 in the figure). Consequently, the velocity v also depends on the position of the stars on the orbit, and the total energy is given by .

E=

1 2 m2 μv − G . 2 r

(11.69)

Since the orbit is closed, from Sect. 11.6.1 we know that . E < 0. For a perfectly isolated system, classical mechanics predicts eternally equal orbital motion, with . E = cost.

6

A neutron star is an object formed as a result of the gravitational collapse of a star with a mass greater than 8 . Mo . Based on observations and theory of formation and evolution, stars emit energy for the process of nuclear fusion, initially fusing hydrogen nuclei to form helium nuclei, and gradually forming heavier nuclei until they reach nuclei with the mass number of iron. Once the iron is synthesized in its core, a star can no longer obtain energy from fusion and becomes unstable due to gravitational pressure, which, as we know, pulls the mass of the star toward its center. The tremendous gravitational compression in the stellar core leads to the formation of a degenerate object, having a density equal to that of atomic nuclei, predicted by quantum mechanics, composed only of neutrons (particles with no electric charge) and called a “neutron star” for this reason.

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Fig. 11.13 Orbits of two neutron stars as seen from the laboratory system (Earth). The cross indicates the position of the center of mass. The relative position of the two stars in the orbit is shown for three instants of time .t1 , t2 , t3 , with their relative distance .r . If we place ourselves in the reference frame of one of the two objects, the motion of the other corresponds to that of an object of mass equal to the reduced mass .μ = m/2 moving on a closed orbit (elliptical, in this case)

General relativity, on the other hand, predicts that a binary system moving on the orbits of Fig. 11.13 can emit gravitational waves. These are energy-carrying perturbations of space-time that can (if sufficiently intense) be detected by instruments called gravitational interferometers. The emission of gravitational waves (which propagate through space) causes the total mechanical energy . E of the system not to remain constant, but to decrease over time, with emitted power equal to: .

dE = −PGW dt

(11.70)

where . PGW represents the amount of energy per unit time radiated in the form of gravitational waves. This quantity depends substantially only on the masses .m of the two objects and their reciprocal distance .r , and in the case as simple as that shown in the figure it is analytically determinable by general relativity. This was what occurred in the case of the binary system discovered by Hulse and Taylor, assuming that the masses of the two objects corresponded to that of neutron stars, .m = 1.5Mo , since the relative distance was determinable from the measurement of the orbital period using Kepler’s III law. The emission of energy in the form of gravitational waves causes the system to slow down, resulting in a delay in reaching the periastrum compared to the Newtonian prediction. In fact, the total mechanical energy becomes even more negative, decreasing the kinetic energy, varying the minimum and maximum distance in a way that decreases .Ve f f and converging toward the orbital radius .r0 , Fig. 11.6. See questions 3, 4. The variation of orbital parameters can thus be determined analytically, and compared with observations as time evolves. Figure 11.14 shows the slowdown predicted by general relativity superimposed on data related to system observations.

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Fig. 11.14 Change in orbital period of the binary system discovered by Hulse and Taylor. The red dots are experimental data indicating the observed change at periastrum (when the two objects are closer together) as a function of the year. The blue line illustrates the theoretically predicted change based on general relativity. Data from J. M. Weisberg and J. H. Taylor, Relativistic Binary Pulsar B1913+16: Thirty Years of Observations and Analysis, https://arxiv.org/abs/astroph/0407149

Data points and prediction match perfectly! Therefore, the hypothesis that the masses of the two objects correspond to those of two neutron stars is correct. The discovery of the system and its analysis earned Hulse and Taylor the 1993 Nobel Prize in Physics for “the discovery of a new type of pulsar, a discovery that has opened up new possibilities for the study of gravitation.” This is an indirect observation of gravitational waves: one sees the effect on orbits by admitting that they exist. Based on predictions, it can be estimated (problem 11.4) that the two stars will enter coalescence in time scale on the order of a few hundred million years. This means that as they continue to lose energy, the two-body system finally collides and merges into a single object. What happens when the two neutron stars approach coalescence? The emission of energy in the form of gravitational waves becomes so intense that it can be detected with the generation of detectors developed from 2015. The signal contained in the gravitational waves and recorded by the instruments contains the information to derive the mass of the two objects, their mutual distance, and the distance of their center of mass from Earth. Again, the simple Newtonian dynamics of the two-body system helps predict the signal to be searched (technically: it provides the template). The coalescence event of two neutron stars recorded on September 17, 2017, was of fundamental importance for many aspects of astrophysics. One of the major finding is that it is the processes of melting neutron stars that originate the heavier elements of the periodic table, including gold and platinum.

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11.10 Triumphs and Falls of Newtonian Theory If every planet in the Solar System was affected by the gravitational interaction with the Sun alone, each of them would move on trajectories described exactly by perfect ellipses. These correspond to the negative-energy solutions of the general two-body problem. The trajectories predicted by the two-body problem are called unperturbed motion solutions. However, each body in the Solar System is affected not only from the Sun’s attraction, but also from that of all other nearby bodies orbiting around. Distant objects have little influence, because the intensity of the interaction decreases as the square of the distance. In fact, no member of the Solar System moves on a strictly hyperbolic, parabolic, elliptical or even circular trajectory. Deviations of the actual trajectories of bodies from conical orbits are called perturbations, and the motion is called perturbed. In order of distance from the Sun, the eight planets in the solar system are: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune. As you can see in Table 11.1 (which also contains the former planet Pluto), the eccentricities of the orbits are very small (except for the first and last). It is interesting to note that the plane of the orbit (“inclination” column) is very close to the Earth’s orbital plane. Not only the orbits are in a plane, but the orbital planes are almost parallel to each other (except for Pluto, and this is one of the main reasons for the “radiation” from the club of planets). In short: when you study how planetary systems are formed, you will have to find some mechanism that forces objects to stay almost on the same plane; or, in other words, to have all angular momenta parallel in direction. This will also help to understand why spiral galaxies like the Milky Way are flat. As mentioned in the introduction, until Newton’s time, planets up to Saturn were known. Uranus was identified as something other than a star only on March 13, 1781 by William Herschel. His discovery was completely unexpected: the planets visible to the naked eye (up to Saturn) had been known for millennia, and no one suspected the existence of other planets until Herschel’s discovery. This extraordinary event multiplied observational efforts (including telescope construction) to study of our solar system. An important activity, with the development of observational technology, was the monitoring of deviations to the perfect elliptical orbit of Jupiter and Saturn. They are the two planets with the highest mass, and their mutual perturbation can be calculated numerically. The perturbative effects are such that the orbit of each of the two is not a perfect ellipse, but there is an “oscillation” with respect to the trajectory. With wonder, despite difficult calculations, perfect agreement was found between observations and predictions after perturbative corrections. Soon after the discovery of Uranus, its perturbations induced by the proximity of the two massive planets were also studied. However, calculations for Uranus presented difficulties, and the interactions with Jupiter and Saturn could not explain the observed motion. A possible solution (immediately recognized by many authors) was the existence of another planet not yet observed that could further perturb the orbit of Uranus.

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The British astronomer John C. Adams became convinced of this hypothesis and thought of solving the problem (between 1843 and 1845) with what is now called the inverse problem, which is an attempt to deduce the parameters of a mathematical model from observed data. Although the problem is simple for modern mathematics after the advent of electronic calculators, at the time it required long and laborious manual calculations. Adams calculated the path of Uranus using the assumed position of a perturbing body, derived the difference between the path he calculated and the observations, and with the latter tried to determine the parameters (trajectory) of the perturbing body. In September 1845, Adams reported the result of his work to the director of the Cambridge Astronomical Observatory, with a request to observe the possible presence of the new planet based on his data. At the same time, Urbain Le Verrier (unaware of Adams’ work), in late 1845 presented a study to the French Academy of Sciences in Paris showing that preexisting theories had failed to explain the motion of Uranus. He in turn began a study similar to the Adams’ one and in June 1846, in a second memoir submitted to the same Academy, he communicated the position of the proposed perturbing body, without giving any indication of mass or orbit. This led to an interesting international intrigue between France, England and Germany. British astronomers (who had treated Adams’ request as outlandish) realized that perhaps they had wasted their time. Le Verrier had, in addition to his colleagues in Paris, also provided the information to the Berlin Observatory. In Berlin they received Le Verrier’s letter on September 23, 1846, and the observatory immediately scheduled observations in the region suggested by Le Verrier. They were also lucky: the new object (later named Neptune) was discovered shortly after midnight, after less than an hour of searching and less than one degree from Le Verrier’s predicted position. For the entire scientific community, it was a triumph of the predictivity and power of Newton’s theory. And on the strength of this success, a second study began, concerning the first planet, Mercury (the one closest to the Sun). What was known is that Mercury’s orbital ellipse has peculiarities related to its proximity to the Sun; in particular, it is not fixed in space. If we consider perihelion as a specific point, this has a precession motion. Perihelion precession of Mercury’s orbit means the rotation of the perihelion position of the planet’s orbit as shown in Fig. 11.15. The effect is explained by the perturbation on Mercury of the other planets. The calculations (known at the time of Le Verrier) were that the estimated total effect was a rotation of about 5557 arcsec/century. As mentioned, 1 arcsec is equal to 1/3600 of a degree. Thus, the expected deviation is about 1.5.◦ per century. Le Verrier, first, discovered that this planet is advancing faster than the theory predicts: observations showed that perihelion is ahead by 5600 arcsec per century, about 43 arcsec more than predicted. Le Verrier, using the same method that had enabled the sensational discovery of Neptune, proposed in 1859 the existence of a hypothetical planet Vulcan, whose orbit would be internal to that of Mercury. On the strength of credibility, now everyone

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Fig. 11.15 Diagram of Mercury’s perihelion precession (greatly exaggerated in order to visualize it!)

pointed telescopes, but nothing was found. This remained a fundamental problem in classical physics for about 60 years. In 1916 Albert Einstein enunciated his theory of gravitation based on general relativity. It also allowed the calculation of the perihelion precession of the planets. In 1919 Einstein was able to report that the amount of precession calculated for Mercury corresponded to the observed deviation. The fact that such an established theory as the Newtonian one failed (even for a tiny effect!) allowed the development of a more general and precise theory, that of general relativity, which you will study in the following years.

11.11 Gravitational Indications for Dark Matter There is a modern version, conceptually very similar to the old question of invisible planets, which is giving rise to the so-called dark matter problem. We observe in astrophysical systems, ranging from the size of galaxies, galaxy clusters, up to super-clusters, certain anomalies in motion that seem to be explained only by assuming the existence of a large amount of “dark” matter. This matter is called dark matter because it does not emit electromagnetic radiation of any kind, and has so far manifested itself only through its gravitational interaction with ordinary matter. Even the simple considerations of the Sun’s motion in the Galaxy (questions 7, 8 and 9) indicate that the mass required to account for the Sun’s observed speed is about an order of magnitude greater than the mass emitting electromagnetic radiation of any kind (the “luminous” mass). Today we can detect radiation from the lowest to the highest energies, from radio (with radio telescopes) to X-rays and gamma-rays (through satellite experiments).

11.11 Gravitational Indications for Dark Matter

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The first indications of the possibility of dark matter (in the modern sense) were inferred in 1933 by the astronomer Fritz Zwicky through measurements of the velocities of some of the .∼1,000 galaxies that make up the cluster called Coma. To explain the observed velocities and the dynamics of the system by Newton’s laws, it is necessary for the mass present to be as much as 400 times the luminous mass. However, it was not until the 1970s that observations of discrepancies between the ratio of “dynamical mass” (detectable through its gravitational interactions) and “luminous mass” (detectable because it emits electromagnetic radiation) began to be carried out in a systematic way, using different techniques and on objects whose size scales varied considerably. The most solid evidence for dark matter emerges from analysis of the rotation curves of spiral galaxies. Spiral galaxies are objects that, like our Galaxy, contain about 100 billion of stars. The number is not a guess: it is determined by the observed number density of stars multiplied by the visible volume of the galaxy. The stars are located largely in a central core (called the bulge) and to a small extent in a flattened disk in which they are placed on nearly circular orbits around the galactic center. As with the orbits of planets, according to Newtonian dynamics, stars that are farther from the center should have lower orbital velocities, as represented by the dashed line in Fig. 11.16. A simple reasoning allows us to determine the trend of the speed of stars as a function of their distance .r from the galactic center. Spiral galaxies have a radius of about 15 kpc and thickness between 200 and 300 pc. The main simplification is related to the fact that observations indicate that about 90% of existing stars are

Fig. 11.16 Rotation curve of stars in a spiral galaxy. The black dots indicate the measured rotational speed of stars in the spiral galaxy NGC3198, as a function of their distance.r from the galactic center (1 kpc corresponds to.∼3 1019 m). The dashed line represents the expected contribution determined on the basis of the luminous matter in the galactic core and galactic disk. The red dash-dot line includes the contribution of a halo of non-luminous matter, but that must be present to explain the observed data, so as to form the black solid line superimposed on the experimental data that represents the effect of what I call “dynamical mass”

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enclosed in the bulge of radius . R B =2 kpc. In practice, we can consider (unless than a small correction) all the mass . M of a galaxy in its central region of radius . R B . In such a case, considering a star of mass .m at distance .r > R B in a circular orbit, the force with which it is attracted by the galactic mass . M is exactly offset by its centripetal force, with magnitude: Gm M v2 . = m from which v = r2 r

/

GM r

(11.71)

Even when correcting for the fact that what is measured is the radial component √ of the velocity, and not its magnitude .v, the expected trend is of the type .1/ r , as indicated by the dashed line in Fig. 11.16. However, measurements of orbital velocities, measured by exploiting the Doppler effect,7 of stars in the peripheral regions of a large number of spiral galaxies agree: in no case is the Newtonian prediction respected. On the contrary, instead of decreasing at great distances from the center of the respective galaxies, the orbital velocities remain to a good approximation constant, as seen in Fig. 11.16 for the data relating to one of the galaxies studied. One possible explanation for the anomaly is that the mass present, but not luminous, increases with increasing radius, as the dash-dot line in Figure. In particular, the orbital velocity of the most peripheral stars is such that, if the galaxy were composed only of the luminous matter, these stars would leave in a short time, escaping into intergalactic space. This would happen because the kinetic energy possessed by the star is greater than the gravitational potential energy that holds it bound to the galaxy. Since no galaxies are known from which stars disperse in this way, it is believed that there must be additional mass in all galaxies that does not emit electromagnetic radiation, hence “dark”. The existence of dark matter, of which we have so far only indirect evidence, is also believed to be valid in the current standard model of cosmology. Without the presence of dark matter from the earliest stages of the Universe, the formation of galaxies and clusters of galaxies from about a billion years after the Big Bang could not be explained. Accurate computer simulations of the evolutionary dynamics of the Universe starting from the epoch when atoms formed (.∼300000 years after the Big 7 According to quantum mechanics, each perturbed atom or molecule emits radiation with a specific combination of wavelengths. Thus, each element has a specific “signature”, that is, a combination of radiation at given wavelengths that, when detected, allows the identification of the emitting elements. However, if the source and observer are relatively approaching or moving away, the emitted radiation is observed with a slightly shifted wavelength value. The observed shift depends on the radial component of the velocity, through a relationship known as Doppler effect. For example, hydrogen atoms emit radiation with a specific combination of wavelengths, which can be identified with special instruments. If all the lines are shifted to longer wavelengths (toward the red region, in visible light), the Doppler effect makes it possible to estimate the radial velocity with which observer and source are moving apart. Conversely, if the shift of all lines is toward shorter wavelengths (toward blue, in the visible region) observer and source are approaching. In practice, by observing the radiation emitted by a source, the relative radial velocity between source and observer can be estimated.

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Bang) indicate that it would have taken a much longer time to form galaxies without dark matter; in some cases, the formation of the structures observed today cannot even be achieved. Even after forming, assuming gravitational force as the only force that shapes the large-scale Universe, it would not explain how galaxies could be kept intact, since luminous matter is unable to develop sufficient gravitational attraction. But what is dark matter? It is a good problem whose solution we still do not know. To answer this question, we need the contribution of new generations of physicists (you!).

11.12 Questions and Exercises Questions 1. Calculate the radius. R S of the orbit of a geostationary satellite using the lunar period,.T = 27.3 d, and the Earth-Moon distance, . DT L = 3.8 105 km, as data. [A: . R S = 4.2 104 km ] 2. A dwarf planet moves around the Sun on an elliptical orbit with semi-major axis.a, semi-minor axis .b and period of revolution .T . Determine the speed when its direction is perpendicular to the semi-minor axis of the ellipse. [A: .v = 2πa/T ] 3. Show that in the case of a binary system in which the orbit is circular of radius .r , the kinetic energy corresponds to half the absolute value of the potential energy. In other words, the total mechanical energy corresponds to . E = − 21 Gmr1 m 2 . 4. In a gravitationally bound binary system on a circular orbit of radius .r , there is energy dissipation by some mechanism (radiation emission, gravitational wave emission). Determine whether, as a result of the emission, the radius .r of the orbit increases or decreases with time. [A: .dr/dt < 0, then decreases. ] 5. Discuss what changes, if any, should be made in the question 3 in the case of elliptical, rather than circular, orbits. 6. In the case of a gravitationally bound two-body system on an elliptical orbit in which there is energy dissipation as in the case of the question 4, explain why the system first tends to have a circular orbit and later to reduce the value of the radius. Hint: look carefully at Fig. 11.6. 7. As the Earth revolves around the Sun, the Sun revolves around the center of our Galaxy (where the black hole described in Fig. 11.12 resides). The Sun’s distance from the Galactic center is 16 m), and the speed of solar revolution is .v = 220 km/s. Using . D = 8.5 kpc (1pc= .3.0 10 o these data, estimate the total mass . Ms of the Galaxy matter enclosed in the sphere of radius 30 kg, estimate the . D. Assuming that the stars all have the same mass as the Sun, . Mo = 2 10 number . Ns of stars in the Galaxy if all the mass determined above were concentrated in the stars. The estimated number of stars in our Galaxy by electromagnetic radiation is between 100 and 400 billion stars. Is the result obtained compatible with this estimate? [A: . Ms = 7 1042 kg; . Ns = 3.5 1012 . ] 8. Using the data and the number of stars in the Galaxy estimated in the question 7, determine the average distance .d (in pc) between star and star assuming for the Galaxy a spherical volume.

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11 Motions in Gravitational Fields [A: .d = 0.9 pc. ]

9. Using the data and the number of stars in the Galaxy estimated in the question 7, determine the average distance .d (in pc) between star to star assuming for the Galaxy a flat disk of height 300 pc. The average observed distance between stars in our Galaxy exceeds pc. Is the result obtained consistent with the observations? [A: .d = 0.3 pc. ] 10. Using the expression of the orbit of an ellipse of small eccentricity .e in polar coordinates, determine the ratio between the velocities of a planet at apogee, .v A , and perigee, .v P , as a function of eccentricity. [A: .v A /v P = (1 − 2e) ] 11. Using the expression of the orbit of an ellipse of small eccentricity .e in polar coordinates, show that the variation of the planet’s angular velocity as a function of eccentricity is equal to Δω . ω = 4e. 12. Halley’s comet is visible when it is close to perihelion, which is 0.586 AU from the Sun. The last perihelion passage was in 1986 and the next one will be in 2061. Determine the distance to aphelion in AU. [A: .r A = 35.6 AU ] 13. Neutron stars have masses equal to.1.5Mo . Knowing that quantum mechanics sets their volume density equal to .∼1.4 × 1014 g/cm.3 , estimate the radius. [A: . R∼15 km ]

Exercise 11.1 Ganymede is a satellite of the planet Jupiter, like the Moon for Earth. Using the data below, determine: 1. the acceleration of gravity .gG on Jupiter (for comparison, the acceleration of gravity on Earth is 9.81 m/s.2 ); 2. the mass density of Jupiter; 3. the period of revolution of Ganymede around Jupiter, expressed in Earth days; 4. the escape velocity of an object from Jupiter’s surface (for comparison, the escape velocity from Earth is 11.2 km/s). Numerical data. Mass of Jupiter;.1.90 × 1027 kg; Mass of Ganymede;.1.48 × 1023 kg; Radius of the orbit of Ganymede; .1.07 × 106 km; Radius of Jupiter: .69.9 × 103 km; .G: .6.67 × 10−11 N m.2 /kg.2 . Exercise 11.2 A body of mass .m gravitationally attracted to a much larger body of mass . M describes an elliptical trajectory if the total mechanical energy . E T is strictly negative, while it has a trajectory corresponding to a hyperbola branch if . E T > 0. An artificial satellite of mass 3 24 kg) on an .m = 2.00 × 10 kg, considered point-like, moves around the Earth (. M T = 5.97 × 10 elliptical orbit, with perigee value .d p = 1.10 × 106 m and apogee value .da = 4.10 × 106 m (both calculated with respect to the Earth’s surface, whose radius is . RT = 6370 km). The speed the satellite has at perigee is .v p = 7.90 km/s. Recall that .G = 6.67 × 10−11 N m.2 /kg.2 . 1. Verify that the data given actually contemplate an elliptical orbit. 2. Determine the velocity of the satellite at apogee. 3. Knowing that the eccentricity of the orbit is .e = 1/6, determine the equation of the satellite’s trajectory with respect to a Cartesian reference system that has the axis of symmetry along the line joining perigee with apogee and origin in the center of symmetry of the trajectory. 4. Prove, with the position and velocity at perigee given in the text, that the eccentricity of the orbit is 1/6.

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Fig. 11.17 Figure for exercise 11.3

Exercise 11.3 An artificial satellite is initially located at a point A at a distance.r0 = 4.0 × 104 km from the center of the Earth, as in Fig. 11.17. Neglecting any friction, calculate (.G = 6.67 × 10−11 N m.2 /kg.2 , . MT = 5.97 × 1024 kg; . RT = 6.37 × 103 km): 1. The speed with which it would reach the Earth’s surface if it were left free at rest in a reference frame integral with the Earth; 2. the speed .v0 that the satellite would have to have to travel a circular orbit of radius .r0 around the Earth; 3. briefly discuss what would happen to the satellite if it were to increase the speed .v0 at point A. 4. If, on the other hand, at A the satellite had velocity .v A = v0 /2 it would travel in an elliptical orbit, as shown in the figure. Determine the distance .r1 of the second vertex P of the orbit from the center of the Earth. 5. Determine the velocity at that point P. Exercise 11.4 A neutron star (NS) is composed of neutrons (particles with no electric charge, mass .m N = 1.67 × 10−27 kg). The distance between neutrons is .d = 2 × 10−15 m, as is the case between protons and neutrons within atomic nuclei. Theory and observations confirm that NSs have a mass about 1.5 times that of the Sun, . Mo = 1.99 × 1030 kg. Recalling that .G = 6.67 × 10−11 N m.2 /kg.2 ): 1. estimate the density (in km/m.3 ) of matter in a NS, using .d as the distance between neutrons. Determine the radius . R N S of the NS assuming it has mass . M = 1.5Mo . NSs are observed when they rotate rapidly on themselves, emitting electromagnetic radiation (and are called pulsars) or in binary systems. The most famous of the latter is the Hulse-Taylor system, discovered in 1973, consisting of two NSs that lose energy by emitting gravitational waves. 2. Determine the value of the reduced mass of a binary system of two NSs. 3. The Hulse-Taylor system has measured period T = 7.75 h. Determine the distance R between the two NSs, and compare it with the solar radius (. Ro = 0.7 × 109 m). 4. Determine the kinetic energy possessed by the Hulse-Taylor binary system.

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5. Knowing that the energy loss (assumed constant) by gravitational wave emission corresponds to .d E/dt = −7.35 × 1024 J/s, estimate how long all the kinetic energy will be dissipated. 6. Considering the total mechanical energy of the system, show whether the two NSs approach or recede from each other by gravitational wave emission, .d E/dt. Exercise 11.5 The Sun has mass . Mo = 1.99 1030 kg, radius . Ro = 7.0 108 m and rotates on itself (approximately) in 30 days. Using the data in Table 11.1, determine: 1. the acceleration of gravity on the surface of the Sun, .g S ; 2. the fraction of the total mass of the solar system that is concentrated in the Sun. 3. the total angular momentum . L o of the Sun. Note: Use the relation: . L o = J ω, where . J = 2 is the momentum of inertia of a sphere, see Chap. 12. 2/5Mo Ro 4. the total angular momentum . L SS of the Solar System, using the data in the table. Jupiter and Saturn contribute 85% of the angular momentum.

Chapter 12

Dynamics of Rigid Bodies

Abstract In Chap. 7 on mechanical systems, we determined that an ideal rigid body has six degrees of freedom. Three degrees of freedom compete with the motion of the center of mass. I return here on the problem to describe the remaining three degrees of freedom with the Euler’s second law. This law allows us to study the variation of angular momentum in the center-of-mass frame using the moments of the external forces present in this frame, even if it is non-inertial. I start deriving the relation between angular momentum and angular velocity in the case of a fixed rotation axis, afterward with the definition of the moment of inertia of rigid bodies and their determination in few simple cases, and finally with examples of application of the II Euler’s law. Then, I describe the situation in the most general case of rotation with the moment of inertia tensor. The chapter ends with some examples, as the spinning top.

12.1 Introduction After traveling through space, this chapter returns to Earth. In particular, we will deal with the study of ideal rigid bodies, for which the rigidity relation (7.4) d = |ri − r j | =

. ij

/ (xi − x j )2 + (yi − y j )2 + (z i − z j )2 = cost .

(12.1)

holds, where .i and . j are two volume elements localized by the vectors .ri and .r j in an inertial frame of reference. Also in Chap. 7 on mechanical systems, we determined that an ideal rigid body has six degrees of freedom. Three degrees of freedom compete with the motion of the center of mass (C.M.), a point that moves as if all the mass of the system were concentrated there. The motion of the C.M. is determined by the Euler’s first law, (7.36): dP (12.2) . = Fext = MaC M dt

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Spurio, The Fundamentals of Newtonian Mechanics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-47289-3_12

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which allows us to determine the translation of the system as a whole. To describe the remaining three degrees of freedom, we have the Euler’s second law, (7.50)1 : .

dL = τ ext . dt

(12.3)

This relationship allows us to study the variation of angular momentum in the centerof-mass frame using the moments of the external forces present in this frame, even if it is non-inertial. The moments of the forces (torques) induce rotations of the system. In general, if we study both translations with (12.2) and rotations with (12.3), we are studying the rototraslations of the mechanical system. This chapter places the observer astride the C.M. position, and studies the system (mostly) by looking at it from that point. In many practical situations, the C.M’s frame may coincide with with that of the laboratory.

12.2 Angular Momentum and Angular Velocity The purpose of this section is to make (12.3) less abstract: it is necessary to relate angular momentum to the geometric and physical parameters (mass, size, rotation) of the mechanical system, the rigid body, that we are considering. This situation will also be addressed in two stages: let us first study the simplest case, the one in which the rigid body is rotating with respect to a single axis of rotation. This is the situation depicted in Fig. 12.1. To fix the situation, the axis of rotation is the .z-axis passing through the center of mass. Later, in Sect. 12.8, we will address the more general case. The pivot point for calculating the angular momentum .L is along the axis of rotation: in the following, for simplicity, I assume the pivot point coincident with the C.M., although this is not strictly necessary. In the configuration of Fig. 12.1, the angular velocity is given by: ω = ω kˆ .

.

(12.4)

The total angular momentum of the system is defined by (7.23) L≡

∑

.

ri × m i vi .

(12.5)

i

Rigid bodies have an extended volume: I have already defined what volume elements are (Sect. 10.9), and how to treat them in Cartesian or spherical coordinates. Let us then imagine to divide the volume of the object in Fig. 12.1 into a discrete number of very small volumes .Vi , each of mass .m i = ρVi . The mass element under consideration lies on a plane at height .z i from the pivot point, in the plane perpendicular to In this chapter, it is not necessary to keep the superscript “.' ” because the observer’s frame will coincide with that of the center of mass.

1

12.2 Angular Momentum and Angular Velocity

313

Fig. 12.1 Rigid body: left, side view. The object is rotating with angular velocity directed along The mass element considered is in a plane at height .z from the pivot point and distance .d and direction .ρˆ from the rotation axis. Right, top view of the circular trajectory carried out by the mass element for rotation. Its velocity is perpendicular to the two unit vectors shown before, along the direction .tˆ

ˆ .k.

ˆ and at a radial distance .di from the axis of rotation. the direction of the unit vector .k, ˆ In this plane, I define two unit vectors that move with the point: the unit vector .ρ, directed from the rotation axis to the point, and the unit vector tangent to the circular trajectory, .tˆ, parallel to the velocity. The three unit vectors are then such that: .

ρˆ × tˆ = kˆ

(12.6)

Each mass element .m i is located relative to the pivot point by the vector: r = z i kˆ + di ρˆ .

. i

(12.7)

The velocity in circular motion is given by (3.62). The velocity is contained in the plane, always in a direction perpendicular to the radial vector, that is: v = ω × ri = ω kˆ × di ρˆ = ωdi tˆ .

. i

Thus, the total angular momentum (12.5) of the system becomes:

(12.8)

314

12 Dynamics of Rigid Bodies

L=

∑

∑ (z i kˆ + di ρ) ˆ × (m i ωdi tˆ)

ri × (m i ωdi tˆ) =

i

=

∑ i

.

= −ω

i

z i m i ωdi (kˆ × tˆ) + ∑

z i m i di ρˆ +

∑

i

= −ω

∑

∑

m i ωdi2 (ρˆ × tˆ) =

i

m i di2 ω kˆ =

(12.9)

i

∑

z i m i di ρˆ + (

i

m i di2 )ω .

i

Note that I have made use of unit vector relationships as given in (12.6). Two terms appear in (12.9): ∑ • .( i m i di2 )ω contains a definite positive quantity that multiplies the angular velocity. This quantity: ∑ .J ≡ m i di2 (12.10) i

is defined∑ moment of inertia of the body. Its units are kg m.2 . ˆ is more complicated. Numerically it is smaller than the • .L⊥ ≡ ω( i z i m i di ρ) previous one, because it sums terms that can be positive and negative. In particular, if the object has rotational symmetry, for each mass element at a certain distance ˆ there will be one at the same distance in the negative in the positive direction .ρ, direction (as in Fig. 7.7), and the sum is exactly zero. In practice, in most simple problems this term cancels out. Some engineering technical aspects are aimed at making this term as small as possible. Ultimately, in the case of a body with a single axis of rotation, the relationship between angular momentum and angular velocity becomes: .

L = J ω + L⊥ .

(12.11)

For a massive extended object with rotational symmetry .L⊥ = 0. The scalar quantity J is determined by summation (12.10) or, in the case of a continuous distribution of matter (passing from sum to integral) as:

.

{ .

J≡

ρ(r )r 2 d V

(12.12)

V

where .r represents the distance to the rotation axis of the .d V infinitesimal volume element. Equation 12.11 expresses an important relationship: the angular momentum of a body is a quantity composed of two terms, one of which (the first) is directed along the rotation axis and is usually the dominant component. The second term (null in

12.3 Moment of Inertia of Rigid Bodies

315

the case of objects with rotational symmetry) lies in the plane perpendicular to the rotation axis. In case we can neglect the second term, (12.11) is simplified to a scalar equation: ˆ = J ω kˆ −→ L = J ω .L = L k (12.13) since the only not-null component is directed along the .z-axis. The reason for this is that the only active component of the momentum of forces, .τz is along the .z axis. The second Euler’s law becomes the scalar equation:

.

J

dω = τz . dt

(12.14)

The (12.14) is the one that is used to solve 95% of the exercises on rotations of rigid bodies.

12.3 Moment of Inertia of Rigid Bodies 12.3.1 Moment of Inertia of a Cylinder and Hollow Cylinder Let us see how the moment of inertia of rigid bodies with a regular shapes can be derived from the definition (12.12). I consider a cylinder of uniform density .ρo , radius . R and height . L as in Fig. 12.2. A disk can be considered as a cylinder with . L « R. Let us consider a shell of radial thickness .dr = r e − ri . Such a shell will have infinitesimal area .d A = 2πr dr . It is not necessary to specify whether .r = ri or .r = r e , because the two values differ precisely by an infinitesimal quantity. Similarly, the infinitesimal volume will be: .d V = d A · L = 2πr Ldr . Let us calculate the moment of inertia of the cylinder with respect to its vertical axis. Each mass element in the shell.d A along the cylinder will be at the same distance from the axis. Then, in the integration of (12.12) we have { .

R

J=

{

R

ρo r (2πr Ldr ) = 2πρo L 2

0

] 1 R4 = M R2 r dr = 2πρo L 4 2 [

3

0

(12.15)

given that . M = ρo (π R 2 L) is the mass of the cylinder. In case we had a hollow cylinder (i.e., only filled between .r = ri and .r = re ) we would have to integrate between these extremes and thus: { re . J = 2πρo L r 3 dr (12.16) ri

316

12 Dynamics of Rigid Bodies

Fig. 12.2 Geometric parameters for the calculation of the moment of inertia of a cylinder or cylindrical ring with respect to its central axis

This is equivalent to saying that, if you have considered a full object, a matter subtraction (for a hole, a milling, etc.) can be considered in the calculation of the moment of inertia as the sum of an object with negative mass. Note that for physical dimensions, the moment of inertia can only depend on the mass . M of the system and the characteristic length (or lengths) of the system. A cylinder has two parameters that have dimensions of a length: the radius and the height. In the case of the cylinder rotating with respect to the axis, the parameter that enters the mass distribution is the radius. In this particular case, the height of the cylinder (or disk) is irrelevant to the result.

12.3.2 Moment of Inertia of a Rod Let us now consider another simple case, a homogeneous rod of length . L like the one in Fig. 7.6. Let us imagine that the rod can be constrained at one end: it can rotate by completing a maximum circumference of radius . L. We can calculate the moment of inertia immediately, since we know that the volume element is .d V = Ad x, so: { .

J=

L

{

L

ρo x (Ad x) = ρo A 2

0

0

x 2 d x = ρo A

1 L3 = M L2 . 3 3

(12.17)

Again, as expected, enter in . J the mass of and the length (squared) of the rod.

12.3 Moment of Inertia of Rigid Bodies

317

A rod can also rotate constrained at another point: for example, its center. In this case, the center of the rod also corresponds to its center of mass. We must expect the moment of inertia to change, because the distribution of the distances of the individual masses changes. By placing the axis of rotation in the center of the rod, the calculation of the moment of inertia (12.17) is modified, changing the extremes of integration. The left end will be in the position .x = −L/2 while the right end will be in the position .x = +L/2 and thus: { .

J = ρo A

L/2 −L/2

[ x 2 d x = ρo A

x3 3

] L/2 −L/2

=

[ ] L3 1 ρo A L 3 + = M L 2 . (12.18) 3 8 8 12

The result, as expected, is different. In the case of rotation with respect an axis passing to the center of mass, we have a numerically smaller moment of inertia . J . The values of the moments of inertia, calculated with respect to the axis of symmetry passing through the center of mass of the most common solids (assuming constant density) can be found tabulated in several books and sites, for example: https://en.wikipedia.org/wiki/List_of_moments_of_inertia. Once . J is known for an axis passing through the center of mass, it can be obtained for any other axis of rotation, as explained in the next section.

12.3.3 Moment of Inertia of a Rod on a Disk Having known the moment of inertia of a simple object (disk, rod, sphere,…) we can calculate the moment of inertia of compound solids. The concept is that the moment of inertia with respect to an axis corresponds to an integral over a volume, (12.12). Integrals represent sums: if I add another element of which I know the moment of inertia calculated with respect to the same axis, the two moments of inertia add up: { .

{

J=

ρ(r )r d V +

ρ(r )r 2 d V

2

V1

(12.19)

V2

See the example in the figure (Fig. 12.3): I glue a rod of negligible thickness and length . R over a disk of the same radius. The integral over the volume of the disk of mass . M gives . J D = 1/2M R 2 ; the integral over the volume of the rod gives . J A = 1/3m R 2 ; these correspond to the integrals over the volumes .V1 and .V2 of (12.19), so . J = J D + J A . If, on the other hand, I remove from the disk a region of mass .m and radius . R of similar shape to the rod, the result would be . J = J D − J A .

318

12 Dynamics of Rigid Bodies

Fig. 12.3 Moment of inertia of two objects, calculated with respect to the same axis. The result is the sum of the moments of inertia

12.4 Conservation of Angular Momentum and Angular Velocity For an isolated system, without moments or forces applied, the conservation of angular momentum allows us to write (12.14) as: .

J

dω =0 dt

−→

J ω = cost

(12.20)

that is: the product of moment of inertia and angular speed must remain constant. The typical example is that of the dancer (or the skater); if the dancer does a twirl starting from outstretched arms and suddenly narrows them, his/her angular speed increases because decreasing the radius (shortening the arms) decreases the moment of inertia. The same situation occurs in particularly interesting astrophysical systems. I have already mentioned in Sect. 11.9 that a supernova explosion of a star of mass .>8Mʘ can produce a neutron star by pressing part of its mass into a spherical region of 10–15 km radius. The mass of the formed system, the neutron star, is determined by quantum-mechanical and nuclear properties and is .∼1.5Mʘ . The remaining of the mass of the initial star is ejected and manifests as the supernova remnant. Before undergoing collapse, the star is rotating. The rotation of stars has a characteristic period on the order of a month. The central mass . M of the star (which will later become the neutron star) has an initial radius, . Ri , similar to that of our Sun, . Ri = 0.7 × 106 km and is obviously rotating with the same angular speed as the star, −6 −1 .ωi = 2 × 10 s . After the sudden collapse (in a few seconds), this part of the star is compressed into a region of radius . R f = 10 km. We apply (12.20) to figure out by how much the angular speed of the compact object increases. The moment of inertia of a sphere is .cost · M R 2 . You need not be concerned about the constant: you will find it tabulated (it is 2/5) but it is not necessary. Sphere was before, sphere remains, the constant will simplify. For conservation of angular momentum, you will have cost (M Ri2 )ωi = cost (M R 2f )ω f

.

(12.21)

12.5 An Application of the II Euler’s Law

319

Fig. 12.4 Schematic diagram of a . pulsar , that is, a neutron star of radius about 10 km rotating rapidly on itself. The rotation axis (and the direction of angular momentum) is along the vertical; the axis representing the magnetic moment (in green) is tilted. Electromagnetic radiation is emitted along this magnetic axis, which can point toward the Earth; in that case, we receive a very regular pulsed signal. Image from Chap. 6 https://link.springer.com/ book/10.1007/978-3-31996854-4

from which, in terms of final angular speed and entering the numerical values we obtain: ( )2 Ri ≃ 2 × 10−6 (0.5 × 1010 ) = 104 s−1 (12.22) .ω f = ωi Rf That is, the newly formed neutron star rotates on itself up to about 1000 times per second! Also because of rotation, the object emits electromagnetic radiation directed mostly along one direction (Fig. 12.4), and slows down its angular speed. When, due to rotation, the emission axis points toward the Earth, we receive an electromagnetic pulse. Compact sources that emit with periods from milliseconds to a few seconds are known and are called . pulsar s.

12.5 An Application of the II Euler’s Law A school case to illustrate problems related to the rotation of rigid bodies is that of a pulley. A homogeneous disk of radius . R and mass . M is constrained to rotate by a completely frictionless mechanism (pulley). The edge of the disk is wrapped by an inextensible wire of negligible mass, which can be unwind by the presence of an object of mass .m tied to one end, as in Fig. 12.5. The problem may ask at what angular speed the pulley rotates, or at what speed (or acceleration) the weight drops.

320

12 Dynamics of Rigid Bodies

Fig. 12.5 A dynamical system consisting of a pulley totally free of friction, to which a rigid and homogeneous disk of radius . R and mass . M is bound via a central hub. The disk can rotate because an inextensible wire is wrapped around its edge, and it can unfold by following the fall of an attached weight

The main point of these problems is to understand the number of unknowns. In this case, as the figure shows, there are two unknowns: the tension of the wire, and the acceleration with which the object of mass .m falls. You may think about others unknown (for instance: the magnitude of force with which the object of mass .m falls), but the linearly independent are always two. Note that the body is not in free fall (there is tension .T of the wire) and the acceleration of fall .az will have to be less than .g. Once .az is known, the speed .vz can be derived. So, we will need to find two equations for the unknowns .az and .T . The fall causes the disk to rotate, with a certain speed and angular acceleration. The tension is the same along the entire length of the wire, and particularly at point A where the tension transmits motion to the disk in the pulley. This rotates with angular speed .ω and, since the rope is at distance . R from the pivot, the velocity with which the wire unwinds must be . Rω = vz . The equation of motion (along the .z axis, downward) for the mass .m is: maz = mg − T .

.

(12.23)

12.6 Huygens-Steiner Theorem for Moments of Inertia

321

A second equation is required, which is given by the rotation motion of the disk with moment of inertia . J , that is, by (12.14): τ = J ω˙ .

(12.24)

.

The torque .τ that makes the disk rotate holds: τ = RT sin 90◦ = RT

(12.25)

.

as in point A arms and tension are perpendicular vectors. In Sect. 12.3 I have shown how the moment of inertia . J can be calculated: the moment of inertia of a disk for an axis passing its center is . J = 21 M R 2 . Plugging this value into (12.24) we have: τ = RT = J ω˙

.

−→

RT =

1 M R 2 ω˙ 2

(12.26)

We managed to have a system with two (12.23 and 12.26) and two unknowns: { .

T = 21 M(R ω) ˙ = 21 Maz maz = mg − T

(12.27)

z having used the relation .az = dv = R ω˙ between the acceleration of the disk and the dt acceleration of fall. The system thus becomes

{ .

T = 21 Maz maz = mg − 21 Maz

(12.28)

from which we can finally derive the two unknowns: { .

T = 21 Maz m g. az = m+M/2

(12.29)

As expected, the acceleration of fall is less than .g and is the smaller the bigger the mass . M of the disk. The fact that the spinning object is a disk (and has no other shape) enters the numerical factor .2 in the denominator of the mass.

12.6 Huygens-Steiner Theorem for Moments of Inertia The Huygens-Steiner theorem (also known as the theorem of parallel axes), allows calculation of the moment of inertia of a solid with respect to an axis parallel to that passing through the center of mass. In many situations this avoids laborious direct calculations.

322

12 Dynamics of Rigid Bodies

Fig. 12.6 Geometry for the proof of the Huygens-Steiner theorem. The coordinates of each volume element (the black dot in the drawing) in the C.M. frame are .(xi' , yi' ), while in the frame with origin in O (contained in the rotation axis) they are .(x i , yi )

The statement of the Huygens-Steiner theorem is as follows: The moment of inertia with respect to one axis, parallel to another passing through the center of mass, is obtained by adding to the moment of inertia calculated with respect to the C.M., . JC M , the product of the mass of the body itself and the square of the distance between the axes: 2 . J = JC M + Md (12.30) Let us verify this by considering an object (Fig. 12.6) whose moment of inertia with respect to the center of mass is known. We now want to determine the moment of inertia of the object with respect to an axis passing through the point O, at which we fix the origin of the frame S. In the frame S, the center of mass has coordinates .(x C M , yC M ) and is .d meters away from the origin O. If the coordinates of the mass element at the point P are .(xi' , yi' ) in the C.M. frame, the coordinates of the same point in the S frame are: { xi = xC M + xi' . (12.31) yi = yC M + yi' . So, from the definition of moment of inertia: J=

∑

m i (xi2 + yi2 ) =

∑

i

.

=

∑

m i [(xC M + xi' )2 + (yC M + yi' )2 ] =

i

2 m i [x ' i

+

2 y'i ]

i

+

∑ i

2m i [(xi' xC M ) + (yi' yC M )] +

∑

m i [xC2 M + yC2 M ]

i

= JC M + Md 2 . (12.32)

12.7 Center of Gravity

323

Here the central summation because we have made use of the fact that, in the ∑ is zero∑ center-of-mass frame, . i m i xi' = i m i yi' = 0. Using the Huygens-Steiner theorem, it is immediate to derive that the moment of inertia of the rod with respect to one end is given by: .

J = JC M + M

L2 L2 L2 L2 =M +M =M 4 12 4 3

having taken into account the value of. J calculated in relation to the center as obtained in (12.17).

12.7 Center of Gravity In practical problems of mechanical engineering (particularly, in those related to statics) it is important to know where the weight force is applied. Obviously, the weight force acts on each element of mass. However, the problem can be simplified using the concept of center of gravity. Imagine having an object composed of many mass elements, .m i on each of which a weight .Fi = m i g acts (the vector .g points down). The equivalent system is the system on which there is only one force, which is applied on only one point (the center of gravity). What are the conditions for the two systems to behave identically? For the two dynamical systems to be considered equivalent, not only must they have the same resultant force: ∑ .F B = Fi (12.33) i

but also that the resulting moment of force is equal to that of the summation of the moments of the individual forces: ∑ .τ B = ri × Fi = r B × F B (12.34) i

The point B that satisfies the condition (12.34) is called the center of gravity of the system. The condition defining the center of gravity is normally applied for the local gravitational field, but it can be extended to other uniform force fields. In some situations, the center of gravity may coincide with the center of mass of a body, but the two terms are not always interchangeable. For the center of gravity to coincide with the center of mass of a body, it must have uniform density (or the density of the body must have some properties of symmetry). In addition, the gravitational field must be considered uniform, i.e. the object must not be too large. On a large scale, the direction of the weight force applied on a particle changes. I show that, under these conditions, the center of gravity and center of mass of the object coincide, using the simple example in Fig. 12.7. Consider two masses .m 1 and

324

12 Dynamics of Rigid Bodies

Fig. 12.7 Definition of center of gravity: two forces are applied on two masses. The center of gravity is the position where the resultant torque is equal to the sum of the moments of the forces

m 2 subject to their weights and spaced .d apart. The equivalent system is the one with point B with both forces applied, for which (12.33) and (12.34) are valid. I calculate the moment of the forces with respect to a generic point O:

.

τ = x1 F1 + x2 F2 = x1 F1 + (x1 + d)F2

.

(12.35)

If there is a point B at a distance .x1 + b from the pivot point where the resultant forces . FB = F1 + F2 are applied, with respect to point B the moment of the forces will be .τ B = (x 1 + b)(F1 + F2 ) . (12.36) Equalizing the two relations x F1 + (x1 + d)F2 = (x1 + b)(F1 + F2 )

. 1

(12.37)

I derive the coordinate .b: b=

.

F2 m2g m2 d= d= d. F1 + F2 (m 1 + m 2 )g (m 1 + m 2 )

(12.38)

In this case, as can be seen by comparing with (7.8), the position of the center of gravity coincides with that of the center of mass. The preceding discussion can be made more general by imagining having any body (e.g., the sort of pumpkin of Fig. 12.6) divided into many elements of mass .m i with resultant force ∑ ∑ ∑ .F B = Fi = mi g = g m i = Mg . (12.39) i

i

i

The torque is: τ =

∑

.

i

ri × Fi =

∑ i

∑

ri × (m i g) =

(

i

m i ri ) × Mg ≡ r B × F B . M

(12.40)

12.7 Center of Gravity

325

Again, I showed that the point B where we can imagine applying all the forces exists. Note that in both (12.39) and (12.40) I have taken the vector quantity .g out of the summation. This is of course possible if we assume that the vector maintains the same direction for all masses considered. This turns out to be correct if the all points are not too far apart.

12.7.1 Physical Pendulum As an application of the above, I study the motion of a rigid body of mass .m constrained (without friction) at point O, as shown in Fig. 12.8. In the equilibrium configuration, the vertical line passing through the constraint O also passes through the center of gravity B of the system (which coincides with the C.M.), at a distance .d. In the case where the center of gravity is below O, the equilibrium is called stable. If it is higher, the equilibrium is unstable. I will clarify in Sect. 13.5.1 the meaning of these two terms. Suppose we displace the system by an initial angle .θo relative to the stable equilibrium position. The total weight of the body can be applied at the center of gravity. We also know that the moment of the equivalent forces, (12.40), is given by the relation, with .|r B | = d: .τ = r B × F B −→ τ = −mgd sin θ (12.41)

Fig. 12.8 Physical pendulum: a rigid body is displaced by an angle .θ from the equilibrium position. The body is constrained at the point O. The center of gravity (point B) is below O and along the axis of symmetry. The weight of the whole body can be imagined to be applied at B

326

12 Dynamics of Rigid Bodies

reducing in this case to a scalar equation (only the component perpendicular to the plane of the sheet is nonzero). The minus sign is related to the fact that bringing the vector .r B to .F B requires a clockwise rotation. The equation of motion for rotations (the II Euler’s equation), (12.14), in this case becomes: .

J ω˙ = τ = −mgd sin θ .

(12.42)

˙ the differential equation describing the motion becomes: Since .ω = θ, .

J θ¨ = −mgd sin θ ≃ −mgd θ

(12.43)

which is similar to that found for the simple pendulum, Sect. 4.9.1, and which can be solved only by approximating (to the first order of a series development, (1.12)) .sin θ ≃ θ. The equation: ¨ + mgd θ = 0 .θ (12.44) J has, similarly to the simple pendulum, solution given by a periodic variation of the angle from the initial position .θo : / θ(t) = θo cos(ωt)

.

with ω =

mgd . J

(12.45)

The corresponding oscillation period is / J 2π = 2π . .T = ω mgd

(12.46)

Such a system can be used to measure in a operative way the unknown moment of inertia . J with respect to an axis by knowing the position of the center of gravity and measuring the period of oscillation of the system. This is one of the didactic experiments normally present in the Physics Laboratory. Finally, note that a simple pendulum consists of a mass .m tied to an inextensible wire of negligible mass at a distance .d from the constraint. His moment of = md 2 , which when inserted into the (12.46) gives a period inertia is/simply . J / .

T = 2π

md 2 mgd

= 2π

d , g

as we had already obtained.

12.7.2 Torsion Pendulum The case of the torsion pendulum, defined in describing Cavendish’s experiment in Sect. 10.8 is quite analogous. Here, the aim is to find a way to measure the value of the constant of the wire .κ, which produces a torque

12.8 Moment of Inertia Tensor (*)

327

τ = −κθ

.

(12.47)

in response to a .θ displacement of the rod from the equilibrium position. The differential equation of motion .

κθ =0 J θ¨ = −κθ ; θ¨ + J

(12.48)

is similar to (12.44); analogous therefore is the solution and the period of oscillation, in this case, is given by: / J . T = 2π (12.49) . κ Since .κ is to be derived, the moment of inertia of the system must be known. For a homogeneous rod of length .d and mass . Ma which can rotate with respect to its center, the moment of inertia corresponds (12.18) to . Ja = 1/12Ma d 2 . There are two masses .m hanging from the ends at distance .d/2 from the center, so in total: .

J = Ja + 2

Ma d 2 md 2 md 2 = + . 4 12 2

(12.50)

When measured in your laboratory, you will find that the period .T is quite long (tens of minutes). You will also find that by observing some complete oscillations, friction cannot be completely neglected but produces a damping of oscillation amplitude, as we will derive in Sect. 13.7.

12.8 Moment of Inertia Tensor (*) Let us turn to analyze the case in which the rotation of the rigid object is as general as possible, and there is no single default axis of rotation (as we considered in Sect. 12.2). In that case, the angular velocity vector in Cartesian coordinates is written as: ω = (ωx , ω y , ωz ) .

.

(12.51)

In navigation or flight, the effects of the three possible rotations are well known (see Fig. 12.9). The angular momentum vector (12.5), in this situation, corresponds to: L≡

∑

.

i

ri × m i vi =

∑

ri × m i (ω × ri ) .

(12.52)

i

Here I make use of on the triple vector product property called “BAC-CAB” (Sect. 2.7.1): .a × (b × c) = b(a · c) − c(a · b) (12.53)

328

12 Dynamics of Rigid Bodies

Fig. 12.9 The three angular degrees of freedom, applied to the case of an airplane. The three angles of rotation translate into roll, pitch, and yaw. The names are not relevant: they serve as a reminder that there are three independent modes of rotation

to rewrite the (12.52) in terms of scalar products: L=

∑

.

m i [ω(ri · ri ) − ri (ri · ω)] = ω

∑

i

m i ri2 −

∑

i

m i ri (ωx xi + ω y yi + ωz z i ) .

i

(12.54) The individual components, developing the vector .ri : ∑

L x = ωx

m i (xi2 + yi2 + z i2 ) −

i

m i (xi2

+

yi2

+

z i2 )

−

i

L z = ωz

∑

m i (ωx xi2 + ω y xi yi + ωz xi z i )

i

∑

L y = ωy

.

∑ ∑

m i (ωx xi yi + ω y yi2 + ωz yi z i )

(12.55)

i

m i (xi2 + yi2 + z i2 ) −

∑

i

m i (ωx xi z i + ω y z i yi + ωz z i2 )

i

or, collecting on the components of the angular velocity: L x = ωx

∑ i

.

L y = − ωx

∑ ∑

m i (xi yi ) − ωz

∑

i

m i (xi yi ) + ω y

i

L z = − ωx

∑

m i (yi2 + z i2 ) − ω y

∑

i

m i (xi2 + z i2 ) − ωz

i

m i (xi z i ) − ω y

∑

i

m i (z i yi ) + ωz

∑

i

m i (xi z i )

∑

m i (yi z i )

(12.56)

i

m i (xi2 + yi2 )

i

I can compact the notation by defining: Jx x ≡

∑

m i (yi2 + z i2 ) ;

i .

Jx y ≡ −

∑ i

Jyy ≡

∑ i

m i (xi yi ) ;

Jzy ≡ −

m i (xi2 + z i2 ) ; ∑ i

Jzz ≡

∑ i

m i (z i yi ) ;

Jx z ≡ −

m i (xi2 + z i2 )

∑

m i (xi z i ) .

i

(12.57)

12.8 Moment of Inertia Tensor (*)

329

The three equations of (12.52) can then be written as in the form:

.

L x = Jx x ωx + Jx y ω y + Jx z ωz L y = Jyx ωx + Jyy ω y + Jyz ωz L z = Jzx ωx + Jzy ω y + Jzz ωz

(12.58)

For a body of any shape and mass distribution, the relationship between angular momentum and angular velocity does not go through a simple scalar, but through a .(3 × 3) matrix called the inertia tensor. Thus .L does not in general have the same direction as .ω. This is the cause of the complicated behavior that, in general, rotating bodies can exhibit. Imagining as in Chap. 9 of writing the vectors as .(3 × 1) column vectors. Then, the relation (12.58) can be written as L ≡ [J ] ω

.

(12.59)

where .[J ] is the .(3 × 3) matrix representing the inertia tensor. The matrix is real, has positive elements on the diagonal, and is symmetric (it is immediate to verify that ∑ ∑ . J yx = − i m i (yi x i ) = − i m i (x i yi ) = Jx y , and similarly for other non-diagonal elements). As you study in the course of Algebra, a real and symmetric matrix is a diagonalizable matrix. It means that you can find a linear transformation that makes .[J ] with nonzero elements only on the diagonal. Translated into physics language, a linear transformation means that you can find a new reference frame by appropriate rotations (representing linear transformations) with respect to which the inertia tensor consists of only the diagonal elements. These diagonal elements are normally denoted by . J1 , J2 , J3 . In practice, we have defined for the body under study a reference frame with a set of mutually perpendicular principal axes, with respect to which the inertia tensor has three principal moments of inertia. The order with which they are usually defined is such that . J1 ≥ J2 ≥ J3 . The angular momentum in such a case is written i + J2 ωx ˆj + J3 ωz kˆ . .L = J1 ω x ˆ (12.60) For a body with symmetries, the principal axes are immediately identifiable (for example, the axis of a cylinder definitely represents a principal axis). Sometimes it is possible for two principal moments to be identical; in the case of the homogeneous sphere we have . J1 = J2 = J3 . The determination of the principal axes of inertia and its eigenvalues-so in algebra it is also called the search for diagonal elements-is an important aspect in many branches of engineering. Operations in which you will have to diagonalize a matrix or search for its eigenvalues will become a frequent problem in quantum mechanics.

330

12 Dynamics of Rigid Bodies

12.9 Rotational Kinetic Energy In Sect. 7.9.4 I showed that kinetic energy can be seen as the sum of a term related to the motion of the center of mass (thus, to its translation) and a term related to what happens with respect to the center of mass frame. We now have the tools to derive that this second term depends only on rotations, and that it can be related to the moment of inertia of the system and its angular velocity. As usual, let us begin with the most common case, in which rotation occurs with respect to a single axis, as in the example in Fig. 12.1. Placing ourselves in the C.M., the object does not translate and all its volume elements away from the axis rotate with angular speed .ω. The rotational kinetic energy .T R (I replace this symbol for the ' . T used in Sect. 7.9.4 for clarity) will be: 1∑ 1∑ 1∑ m i vi2 = m i vi · vi = m i (ω × ri ) · (ω × ri ) 2 i 2 i 2 i 1∑ 1∑ = m i (ωdi tˆ) · (ωdi tˆ) = m i ω 2 di2 2 i 2 i 1 1 ∑ m i di2 )ω 2 = J ω 2 = ( 2 i 2

TR =

.

(12.61)

having used in the second line, for a rotation about the .z-axis, the relation for velocity as in (12.8). In the more general case of three rotations, the mathematical treatment of the scalar product between vector products is algebraically not very simple. However, it is easy to see that, in the case we have chosen as the reference system that of the principal axes, and the inertia tensor is diagonal, the kinetic energy can be written as: 1 R .T = (J1 ωx2 + J2 ω 2y + J3 ωz2 ) . (12.62) 2

12.9.1 Body Rolling Without Sliding As an application, we can study the pure rolling motion of a rigid body, such as a cylinder or a sphere. It is called pure rolling motion when a body rolls without sliding, that is, the velocity of the connection point between the object and the plane of contact is zero: this is, for example, the case of the wheel. Normally, it is the force of static friction that ensures the immobility of the contact point: after an infinitesimal time, the contact point has moved to a new neighboring point. Consider the situation in Fig. 12.10: a disk of radius . R and mass . M starts from a height .h at rest. At what speed does it reach the ground? When we speak of speed in this case we mean the speed of the center of mass. It is clear that each point moves

12.9 Rotational Kinetic Energy

331

Fig. 12.10 How to deal with the case of a disc/sphere falling from an inclined plane rolling without sliding

with this drag speed, plus a rotational velocity that depends on how far it is from the center of the disk. The most immediate method is to solve the problem using conservation of mechanical energy. Taking the lowest point of the disk as the reference point for potential energy, its potential energy at the beginning will be .V i = Mgh, while its kinetic energy will be zero. During the rolling motion, the angular speed of the disk will be related instant by instant to the dragging speed by the relation: .

.

Rω = vC M .

(12.63)

At ground, the potential energy of the disc will be zero, while the kinetic energy T f = TC M + T R , i.e., 1 1 MvC2 M + J ω 2 2 2 1 1 1 1 1 = MvC2 M + ( M R 2 )ω 2 = MvC2 M + MvC2 M 2 2 2 2 4 3 2 = MvC M 4

Tf = .

(12.64)

where I made use of the (12.63) and the value of the moment of inertia of the disk.2 Therefore, applying energy conservation, we have 3 2 . MvC M = Mgh 4

/ −→

vC M =

4 gh 3

(12.65)

√ value less than . 2gh obtained for a point-like object. The reason is clear: in this case, in addition to translation, some of the kinetic energy is used for rotations.

2

If it had been a sphere, the numerical coefficient would have been 2/5.

332

12 Dynamics of Rigid Bodies

12.9.2 The Motion of the Wheel The wheel has been of fundamental importance in the development of society, and its motion is well described (in a first approximation) by a pure rolling motion. Rotation occurs around the instantaneous point of contact that has zero velocity. In Fig. 12.11, the instantaneous contact point is marked by B. At this point, the static frictional force provides grip on the ground and allows the wheel to move. After an infinitesimal time, the contact point changes, becoming an infinitesimal neighboring point of the wheel and so on. Under the condition of pure rolling motion, the static friction force does not perform any work. The reason is that, instant by instant, the wheel at point B has zero resultant velocity. This can be derived very easily by remembering that for a body rolling without sliding, (12.63) holds. We then observe the motion of the wheel by placing ourselves in the system of the center of mass, Fig. 12.11 on the left. Each point on the surface of the wheel (in the drawing were indicated 8 positions) has a radial velocity vector of magnitude .ω R and oriented as in the figure (in the case of counter-clockwise rotation). Placing ourselves in the laboratory reference frame (figure at right), that is, of an outside observer who sees the wheel moving, we see the translational motion of the frame with velocity (in magnitude) .vcm directed to the left. This corresponds to the fact that the translation velocity vector (drawn in red) applies to every point on the wheel. At each point, for the outside observer, the velocity of the wheel is given by the vector sum of the two vectors (red-black). Specifically, at point A (top), the velocity of the point on the wheel surface is twice as high as .vcm . At the contact point B, on the other hand, the resultant velocity is zero.

Fig. 12.11 Left: The motion of a wheel (e.g., of a bicycle) seen in the the center-of-mass frame. Each point rotates with angular speed .ω and linear velocity .ω R equal to the velocity of the center of mass, if it rotates without sliding. Right: the motion as seen from the lab frame, that is, from an observer watching the bike move. At each point you have to sum the two vectors using vector rules. In A and B the resulting velocities are reported. The wheel is in ideal rolling condition, see text

12.10 The Motion of the Spinning Top (*)

333

You need not worry about the wheel moving up or down: notice that for every component of the black upward vector, there is in a symmetrical position a component pointing downward. The resultant of the velocity component orthogonal to the plane is zero. Therefore, under ideal conditions the static frictional force does not do any work, because at point B the velocity .v B = 0 and therefore the power dissipated .F · v = 0. The above is true, as specified, only under ideal conditions. Due to the deformation of the point of contact, in practice, there is a small dissipation of energy: this effect is quantified by the quantity called rolling friction. In any case, this form of energy dissipation is generally much less than what there would be if the contact point .str etched. In this case, point B would have a nonzero velocity, and dynamic friction would come into play. Consequently, much less energy is required to maintain pure rolling motion than is required to make objects crawl.

12.10 The Motion of the Spinning Top (*) In conclusion of this chapter, I consider a not trivial problem that can be solved using approximation methods. It involves studying the motion of a spinning top, which exhibits two characteristic rotations (Fig. 12.12). The first, most evident, is the rotation of the spinning top about itself, about its axis of symmetry. It is the rotation denoted by .ωs in the figure. The second, slower rotation is the so-called precession: the change in the direction of the axis of rotation of a rotating body. This precession is such that the tip of the spinning top makes a second circular motion around the .z-axis, indicated in the figure by the angular velocity .Ω p . How can we find the parameters related to the two rotations? We must, as usual, start from the second Euler equation, (12.3), imagining that all the weight of the system is applied in the center of gravity . B placed along the axis of the spinning top. In this way dL (12.66) . = τ B = r B × Fg with Fg = −Mg kˆ dt The axis of rotation coincides with the axis of symmetry passing through the center of gravity. We denote this moving axis in the considered reference frame by the unit vector .rˆ B , which exits the origin of the frame and points toward the center of gravity. The dominant angular velocity can then be written as ω s = ωs rˆ B

.

(12.67)

The angular momentum of the system will be a complicated function of the two rotations, passing through the inertia tensor. However, under the assumption that the axis of rotation definitely has the largest principal moment of inertia, and that

334

12 Dynamics of Rigid Bodies

Fig. 12.12 Rotational and precessional motion of a spinning top. The spinning top rotates rapidly with respect to its axis of symmetry, with angular speed .ωs . However, it is noted that the outer tip of the spinning top rotates in turn, with a motion called precession and angular speed .Ω P « ωs . The center of gravity is along the direction of the axis of symmetry

ωs ≫ Ω p , we can initially neglect the contribution due to precession motion and assume with good approximation that

.

L ≃ J ωs rˆ B .

.

(12.68)

Then, as shown in the drawing, the angular momentum vector has the same direction as the axis passing through the center of gravity. Note that, in case you have thrown ˆ then the two vectors .r B and .Fg the spinning top perfectly vertical for which .rˆ B = k, in the (12.66) are parallel, and .L = cost: the spinning top would continue to rotate around itself without any variation (this, of course, neglecting any friction). It is now necessary to take into account how the unit vector .rˆ B varies with time. Since the unit vector is rotating, I will use Poisson’s rules of derivation, Sect. 3.9 to obtain: d rˆ B = Ω p × rˆ B . (12.69) . dt Note that the rotation of unit vector .rˆ B is around the .z axis, thus with angular speed Ω p (see figure). We now use the previous considerations by fitting the quantities found in the (12.66) dL = Jω ˙ = r B × Fg (12.70) . dt

.

with ω ˙ =

.

d rˆ B d(ωs rˆ B ) = ωs = ωs (Ω p × rˆ B ) dt dt

(12.71)

12.11 Questions and Summary Exercise

335

Equation 12.70 becomes: .

Jω ˙ = J ωs (Ω p × rˆ B ) = r B × Fg = −Fg × r B .

(12.72)

As you can see from the last equality, the vector .Ω p must have direction along the z-axis (as the weight force vector). We can now calculate the magnitude of (12.72), using the geometric relations shown in the figure:

.

(J ωs )(Ω p sin θ) = Mg sin θr B

.

(12.73)

simplifying and making explicit .Ω p : Ωp =

.

Mgr B Mgr B .60 = J ωs L

(12.74)

The spinning top precedes with respect to the .z-axis with an angular speed that is smaller the larger .ωs is (other conditions being equal), but which is still constant because the angular momentum (12.68) of the isolated system is constant. Such situations are extremely important, for example, in the study of the Earth frame, with the phenomenon of Earth precession. Indeed, the Earth possesses precessional motion: its axis of rotation rotates slowly (with a cycle of 25,800 years) around the perpendicular to the plane of its orbit, with respect to which it is inclined by about 23.5.◦ .

12.11 Questions and Summary Exercise Note: This section collects general and summary exercises involving rigid bodies and physical quantities also presented in other chapters, e.g. forces, constraints and friction (Chap. 4), work, energy, and energy conservation (Chap. 6), moments and conservation laws (Chap. 7), elastic and inelastic shocks (Chap. 8).

Questions 1. A homogeneous rigid rod AB, of mass m = 4 kg and length . L = 500 cm rotates about an axis passing through endpoint A and forming an angle .α = 45◦ with the rod itself. Calculate the moment of inertia of the rod with respect to this axis. [A: . J = 0.125 kg m.2 ] 2. On a thin bar of negligible mass 1 m long are arranged 5 bodies each of mass 0.10 kg at distances of 25 cm from each other starting from one end. Determine the moment of inertia of the whole system with respect to an axis perpendicular to the bar and passing .(i) through one end; .(ii) through the center of mass. Repeat the study in the case of a bar of mass M=1 kg. [A: .(i) J = 0.188 kg m.2 ; .(ii) J = 0.063 kg m.2 ]

336

12 Dynamics of Rigid Bodies

3. Determine the moment of inertia of a homogeneous circular corona, of surface density.σ = 1.25 kg m.−2 , with inner radius .r1 = 0.30 m and outer radius .r2 = 0.50 m. [A: . J = 0.107 kg m.2 ] m.−2

4. A rectangular plate of density .σ = 1.25 kg is arranged with the longer side .l = 75 cm along the .x axis of a Cartesian reference frame, and with the shorter side .h = 30 cm along the . y axis. Calculate the moment of inertia . Jx with respect to rotations about the .x-axis and the moment of inertia . Jy with respect to rotations about the . y-axis. 102

[A: . Jx = 0.84 kg m.2 ; . Jy = 5.3 kg m.2 ] 5. A rectangular slab of the same size and arranged in the same way as that in the question 4 has a surface density that varies with position according to the function .σ(x, y) = a0 + a1 x y where .a0 = 0.15 103 kg m.−2 and .a1 = 0.60 103 kg m.−4 . Determine the moment of inertia with respect to the . y-axis. 6. A thin, homogeneous disk of radius R = 50 cm and mass m = 200 g lies in a plane. Calculate the moment of inertia with respect to any of its diameters. [A: . J =

1 2 4MR

= 0.0125 kg m.2 ]

7. A yo-yo consists of a homogeneous slotted cylinder, radius R = 7.0 cm and mass m = 100 g. The inner slot is of negligible width, and does not significantly affect the value of the moment of inertia . J . In the inner slot, of radius .r = 3.0 cm is wrapped a string, attached at the other end to a fixed point (a hook, for example). Calculate the acceleration .acm of the yo-yo when it is allowed to fall freely. [A: .acm =

g 1+J/(mr 2 )

= 2.64 m s.−2 ]

8. The homogeneous circular corona of the question 3 rotates about its axis of symmetry by making one revolution every 90 s around the axis of symmetry. Determine magnitude of the angular momentum. [A: . L = 7.5 10−3 kg m.2 s.−1 ] 9. Show that the moment of inertia for an axis perpendicular to the conjunction of a system of two bodies .m 1 , m 2 spaced .r apart and passing through C.M. is . J = μr 2 , where .μ is the reduced mass of the system. Apply this to the CO system (data given in question 3). [A: . J = 6.8 10−46 kg m.2 s.−1 ] 10. Find the moment of inertia of the CO.2 molecule (data given in the question 3) with respect to an axis perpendicular to the molecule and passing through the C.M. 11. Find the moment of inertia of the H.2 O molecule (data given in the question 3) with respect to the axis perpendicular to the molecule and passing through the C.M. 12. Find the moment of inertia of the NH.3 molecule (data given in the question 3) with respect to the axis perpendicular to the plane containing the three H atoms and passing through the nitrogen atom N. [A: . J = 4.4 10−47 kg m.2 s.−1 ] 13. Using a spherical coordinate system, show that the moment of inertia of a solid sphere of radius 2 2 . R and mass . M with respect to any diameter passing through the center is . M R . 5 14. In geometry, an ellipsoid constitutes the three-dimensional analogue of the ellipse in the plane. The equation of an ellipsoid in a Cartesian coordinate system is given by .

y2 z2 x2 + 2 + 2 =1, a2 b c

where a, b and c are fixed real numbers such that .a ≥ b ≥ c > 0 represent the semi-axes of the ellipsoid. In the case of different a, b, c, we have a scalene ellipsoid; if .a > b = c, we

12.11 Questions and Summary Exercise

337

have a prolate ellipsoid; in the case .a = b > c, we have a oblate ellipsoid. Without making use of integrals, but relying on the result of the question 13 show that the moment of inertia of a homogeneous ellipsoid of mass . M with respect to the axis containing .a is given by the 2 2 relation . Ja = M(b 5+c ) and that similar relations hold for the moments of inertia with respect to the other axes .b (. Jb =

M(a 2 +c2 ) ) 5

and .c (. Jc =

M(a 2 +b2 ) ). 5

Exercise 12.1 A solid spherical ball of radius . R = 5.0 cm and mass . M = 100.0 g descends along an inclined plane of length . L = 1.5 m and with an angle of inclination with respect to the ground of .α = 30◦ . The ball starts at the top of the inclined plane from a stationary position and rolls without crawling all the way down. At the end of the inclined plane, the ball falls toward the smooth (frictionless) ground from a height .h 0 = 1.0 m. After drawing the system, calculate (neglecting air friction, rolling friction due to non-perfectly elastic bodies, and the effect of the “edge” at the end of the inclined plane): 1. the angular speed of the ball as it leaves the inclined plane and begins to fall toward the ground, and 2. its total kinetic energy at that instant; 3. the position where the ball hits the ground; and 4. the total kinetic energy at that instant; 5. assuming nearly instantaneous ground impact, the mechanical energy lost in the impact if the bouncing ball rises to a maximum height .h 1 = 0.80 m above the ground. The moment of inertia of a sphere of radius . R and mass . M is: . J = 2/5M R 2 . Exercise 12.2 A disk of mass . M = 470 g and radius . R = 15 cm rolls without crawling on a horizontal plane. A point mass .m = 240 g impacts on it and remains attached, Fig. 12.13 (left). At the instant immediately preceding the impact, the velocity of the mass .m is vertical and .v0 = 25 m/s in magnitude. The projection of the impact point on the horizontal support plane is at a distance .b = 5 cm from the origin O. 1. After a certain time, the mass .m comes into contact with the support plane at a point O’. Determine the distance between O’ and O. 2. Determine the moment of inertia of the system with respect to the origin O immediately after the collision. 3. Determine the angular speed .ω of the disk+mass system immediately after the collision. 4. Determine the angular speed .ω ' of the system when the mass .m comes into contact with the supporting plane. Exercise 12.3 A homogeneous sheet of mass. M = 8.0 kg, side. D = 50 cm and negligible thickness can rotate without friction about the horizontal axis passing through point . A and coincident with the side, perpendicular to the plane of the sheet (see Fig. 12.13 right). Gravity acts along the .z direction, downward. The sheet, initially at rest in its equilibrium position, is struck in the center by a projectile of mass .m = 40 g moving with velocity .v1 = 300 m/s perpendicular to the sheet. After piercing the slab, the projectile continues, on the same trajectory, with speed .v2 ; the slab goes into rotation with angular speed .ω = 3.0 rad/s. 1. Show that the moment of inertia of the slab with respect to the horizontal axis passing through 2 . A is . M D /3, regardless of the value of side . H . 2. Determine the value of the speed .v2 . 3. Determine the maximum inclination .θmax of the plate with respect to the equilibrium position, .θ = 0, shown in the figure. 4. Calculate the mechanical energy lost in the collision.

338

12 Dynamics of Rigid Bodies

Fig. 12.13 Left: Figure for Exercise 12.2. Right: Figure for Exercise 12.3

5. Is linear momentum conserved? Discuss what phenomenon, if any, must be considered for the total linear momentum to be conserved. Exercise 12.4 A homogeneous disk of mass . M = 300 g and radius . R = 23.0 cm rotates with constant angular speed .ω0 = 25.0 rad/s around a fixed vertical axis, coincident with its geometric axis, Fig. 12.14 left. The disk is fixed to the axis and is constrained to rotate in a horizontal plane. By symmetry, under these conditions, the axis is not stressed by any radial force. At a certain instant, a projectile of mass .m = 28.0 g is fired at the disk, sticking to it (completely inelastic impact). The projectile (considered point-like) has velocity .v = 40 km/h parallel to the axis of rotation of the disk and distance .d = 19.0 cm from it (see figure). 1. Calculate the value of the magnitude of the angular momentum of the rotating disk before the collision; 2. determine the angular speed of the disk after the impact; 3. determine the amount of energy lost in the impact process; 4. determine the radial force with which the axis of rotation is stressed in the final situation, with the projectile embedded in the disc. Justify why this radial force must exist. Exercise 12.5 As shown in Fig. 12.14 right, an object that can be considered point-like with mass = 0.50 kg is resting on a rough horizontal surface with dynamic friction coefficient .μ=0.30. At the initial time, .t = 0 s, it is observed that the object is located at a distance .d = 30 cm from the . B end of a rod and is moving toward that point with speed 4.0 m/s. The rod is homogeneous, of length . L, mass . M = m and negligible transverse dimensions; it is free to rotate around the endpoint O. The rod has length . L = 2.0 m. Neglect any friction on the axis of rotation of the rod. At a certain instant, the object collides with the rod at point B, and the collision is such that immediately after the collision the two bodies have a kinetic energy equal to 4/5 of the kinetic energy possessed by the object .m immediately before the collision. Determine: .m

1. the speed with which the object arrives at point B; 2. the speed of the object after the collision and the angular speed of the rod: discuss which of the two permissible solutions is physically possible.

12.11 Questions and Summary Exercise

339

Fig. 12.14 Left: Figure for Exercise 12.4. Right: Figure for Exercise 12.5

Fig. 12.15 Figure for Exercise 12.6

3. Determine whether the collision between the two bodies is elastic, partially inelastic or completely inelastic. 4. Determine at what maximum angle .θmax the rod reaches as a result of the collision. Exercise 12.6 A cylinder of mass . M = 2.0 kg and radius . R = 10 cm is resting on a horizontal plane, and its center O is connected to a point P on the plane by a spring (with negligible length at rest) of spring constant .k = 150 N/m, see Fig. 12.15. The cylinder is initially stationary, its axis is perpendicular to the sheet, and the resting point of the cylinder is to the left of P, at a distance .d = 40 cm from it. In the region to the left of point P there is no friction; to the right of point P the plane is rough and the cylinder is constrained to roll without crawling. Determine: 1. The velocity of the center of mass of the cylinder when the point of contact arrives at P (immediately before). 2. The moment of inertia of the cylinder with respect to an axis perpendicular to the sheet passes through the point P, when the point of contact is in P. 3. the angular speed of the cylinder immediately after the point of contact arrives at P; 4. the maximum distance, .d A , traveled by the contact point of the cylinder in the region with friction immediately to the right of the point P. (Note: The cylinder always moves while keeping its axis parallel to itself.) Exercise 12.7 Consider two rods of equal length . L = 17.2 cm and masses . M and .3M (with = 153 g) glued to each other as in Fig. 12.16, left. The system of the two rods is constrained to rotate (with negligible friction) about an axis passing through an extreme . O. Initially, the two-rod

.M

340

12 Dynamics of Rigid Bodies

Fig. 12.16 Left: Figure for Exercise 12.7. Right: Figure for Exercise 12.8

system is stationary in the stable equilibrium position. A mass .m = M/2 impacts the system at point . D at a distance .h = 7/4L from the axis of rotation. The velocity of .m before the collision has magnitude .v0 = 3.70 m/s and forms an angle .θ = 45◦ with the perpendicular to the rod. After the impact, the mass remains embedded in the rod. Calculate: 1. 2. 3. 4.

the moment of inertia . J of the system with respect to the axis of rotation . O; the distance .d of the center of mass of the system from the axis of rotation; the angular speed .ω of the system after impact; the horizontal and vertical components of the impulse due to the constraining force that the axis of rotation applies to the bar, which is necessary to conserve linear momentum.

Exercise 12.8 The roller in the Fig. 12.16 (right) consists of two mutually integral (i.e., welded) disks of the same material and density. The roller can rotate without friction around a fixed pivot coincident with the axis, arranged horizontally. Two bodies .m 1 = 1.00 kg and .m 2 hang from wires (inextensible and of negligible mass) wound in opposite directions respectively on a disk of radius .r 1 = 20.0 cm and on a smaller disk of radius .r 2 = 10.0 cm, the latter of mass . M2 = 589 g. At the initial time, the system is left free to set itself in motion. It is also known that the mass .m 2 is such that, during motion, the tension .T acting on the two wires are equal. It is asked to calculate: 1. The moment of inertia of the roller composed of the two overlapping disks with respect to the central pivot; 2. the angular acceleration with which the roller rotates; 3. the accelerations, .a1 and .a2 , of the two bodies (specify which of the two bodies drops); 4. the tension .T of the wires and the value of the unknown mass .m 2 . Exercise 12.9 The rigid system in Fig. 12.17 (left) consists of a homogeneous bent bar (thick line), of negligible cross section, total mass.m = 5.00 kg and total length. L = 1.10 m. The two horizontal sections are both .d = 30.0 cm long. The system is rotating without friction and steadily, making 115 revolutions/minute with respect to the vertical axis of rotation . AB. 1. Determine the moment of inertia of the system with respect to the axis of rotation . AB. 2. Calculate the kinetic energy of the system. 3. Calculate the total force acting on the system, applied at the center of gravity.

12.11 Questions and Summary Exercise

341

Fig. 12.17 Left: Figure for Exercise 12.9. Right: Figure for Exercise 12.10

4. Calculate the axial component (i.e., parallel to the axis of rotation. AB) of the angular momentum of the system, taking the center of mass of the bar as pivot point O. 5. Calculate the transverse component of the angular momentum of the system, again with respect to the same pivot O. Exercise 12.10 Consider the system in Fig. 12.17 right. A device with cylindrical symmetry, which rolls without crawling on a horizontal plane, has mass . M = 10.0 kg and radius . R = 8.0 cm. The device is rotated by an inextensible wire with negligible mass connected to a suspended object of mass.m = 5.7 kg. The pulley and all the joints (not shown in the figure) that are used to transmit the motion have negligible mass and produce no friction. Initially, the system is stationary with mass m positioned at a height .h = 50 cm above the floor. Calculate: 1. the speed of .m when it comes to impact with the floor and the angular speed of the cylinder at the same instant; 2. the acceleration with which the mass .m falls; and 3. the frictional force between the floor and the cylinder during the fall. 4. The work done by the frictional force between the plane and the cylinder during the fall. Exercise 12.11 A block of mass .m = 100 g is constrained to move in a smooth radial groove of a uniform disk of mass . M = 1.00 kg and radius . R = 10.0 cm. The block is connected to a rope that runs in the groove and passes through a hole in the center of the disk. An external tension .T can be applied to the rope, as in Fig. 12.18 on the left. Initially, the disk+block system is in frictionless rotation about a vertical axis passing through the center of the disk and perpendicular to the disk with angular speed .ω = 10 rad/s. The block initially lies on the edge of the disk. 1. Calculate the tension .T of the string required to keep the block balanced at the edge of the disk. 2. A tension greater than the equilibrium tension is applied to the string and the block approaches the center of the disk. Justify whether, relative to the block+disc system, the following quantities are conserved in the motion: (2.a) linear momentum, (2.b) energy, and (2.c) angular momentum with respect to the axis of rotation. 3. Calculate the angular speed of the disk+block system at the instant when the block is located, stationary in the rotating reference frame with the disk, at distance . R1 = 1/2R from the center of the disk.

342

12 Dynamics of Rigid Bodies

Fig. 12.18 Left: Figure for Exercise 12.11. Right: Figure for Exercise 12.12

4. Calculate the work done by the string tension in moving the block from the initial position to the position . R1 . Exercise 12.12 A homogeneous sphere of mass . M and radius . R initially stationary on top of the inclined plane in A with its center at an altitude .h A = 5.0 m rolls along the plane AB inclined by an angle .α = 60o , Fig. 12.18 left. Past the horizontal section BC the sphere goes up the plane CD inclined by an angle .β = 70o . Assuming that all along the way the sphere rolls without crawling (i.e., with pure rolling motion) 1. 2. 3. 4.

Determine the speed of the center of the sphere when it arrives at D, at elevation .h D = 2.0 m. Draw, in a clear manner, the forces acting on the sphere at a generic point in the section AB. Determine, again in the section AB, the acceleration of the center of mass of the sphere and the minimum value of the dynamic friction coefficient .μd for the motion to be pure rolling.

The moment of inertia of a sphere of radius. R and mass. M is. J = 2/5M R 2 . Neglect possible effects in the variations of slopes at points B and C. Exercise 12.13 A projectile of mass.m = 2.5 kg is shot tangentially (see Fig. 12.19 left) at the edge of a ring having radius . R = 50 cm and whose mass is equal (within measurement errors) to that of the projectile. The ring, free to move on a frictionless horizontal plane, is initially stationary. Before the collision the projectile’s speed is .v0 = 3.0 m/s; after the collision the ring and the projectile remain together. Calculate, immediately after the collision: 1. the position of point D representing the center of mass of the projectile+ring system and the distance .d from the origin C of the Cartesian reference frame; 2. the velocity .v D of the center of mass of the system; 3. the angular speed .ω of the system; 4. the fraction of mechanical energy dissipated in the collision. Exercise 12.14 Consider the homogeneous bar, of negligible cross section, mass . M = 1.333 kg and length . L = 60.0 cm in Fig. 12.19 right. It is constrained to rotate in a vertical plane about its center of mass O, which for simplicity coincides with the origin of a reference system .(x, z). For times .t < 0, the bar is stationary vertically, as in the figure. At time .t = 0, a projectile of mass .m = 63.0 g impacts the bar completely inelastically. The velocity of the projectile .v0 = 1.88 m/s is horizontal and the impact occurs at a distance .d = 22.0 cm from the center O. Calculate: 1. the angular speed .ω0 of the bar after the impact; 2. the impulse .Δp of the forces applied by the constraint at point O to the bar during the impact (specify magnitude and direction).

12.11 Questions and Summary Exercise

343

Fig. 12.19 Left: Figure for Exercise 12.13. Right: Figure for Exercise 12.14 Fig. 12.20 Figure for Exercise 12.15

3. Following the collision, the bar rotates without friction around O. Calculate the angular speed ◦ .ω1 of the bar when it is parallel to the horizontal axis (i.e., it has made a rotation of .90 ). 4. Show that the potential energy of the bar does not change if it is in a vertical position (as in the figure) or if it is parallel to the horizontal axis. Exercise 12.15 A homogeneous disk, having mass . M = 2.50 kg and radius . R = 15.0 cm, can rotate, without friction, about a fixed horizontal axis, passing through the point O of its circumference and, thanks to the stretched wire AB, is in the equilibrium condition shown in Fig. 12.20. Under these conditions determine: 1. the tension .T to which the wire AB (of negligible mass) is subjected; 2. the force .Qo due to the constraining reaction present at the point O. At a certain instant, the wire is cut and the disk starts rotating about the axis passing through O. Calculate, at the time when point A of the disk comes to rest on the vertical line passing through O: 3. the angular speed .ω with which the disk rotates; 4. the magnitude .Q’o of the force exerted by the constraint at O.

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12 Dynamics of Rigid Bodies

Fig. 12.21 Figure for Exercise 12.16

Exercise 12.16 The system in Fig. 12.21 consists of a homogeneous disk of mass . M = 300 g and radius . R = 40.0 cm. At the disk, a slit has been produced along the entire length . R of the disk, obtained by removing a bar of mass equal to 1/20 of . M from the disk. Into the radial slit is inserted a cube of mass .m = 40.0 g. The cube is initially at distance . R/3 from center O and is constrained to slide without friction along the slit. The disk, in the horizontal plane, can rotate without friction about a fixed vertical axis passing through O. With respect to this axis, calculate: 1. the moment of inertia of the system, neglecting effects due to the transverse dimensions of the slit and considering the cube to be point-like. 2. The initially stationary system is then struck by a projectile of mass .m P = 20.0 g, which sticks into the cube; the cube protrudes from the surface of the disk as shown in the figure. Before the impact, the projectile has speed .v P = 40.0 m/s in the plane of the disk, directed orthogonally to the slit as in the figure. Determine the angular speed .ω1 of the disk immediately after the impact. 3. Next, the block with the projectile in it slides (without friction) along the slit until it is ejected. Calculate, for the instant the cube reaches the edge, the angular speed .ω2 of the disc and 4. the radial speed of the cube.

Chapter 13

Considerations on Energy

Abstract This final chapter generalizes some concepts introduced in Chap. 6: how is it possible to define the work done by force fields on a system of points, or on a continuous system, and (if the forces are conservative) define the potential energy of such a system. For a continuous system, I obtain the potential energy of a bound spherical object. This quantity allows to estimate the age of the Sun and discuss the energy conservation in stellar gravitational collapses. The study of possible motions in a potential energy field is used to introduce the concept of stable, unstable, and indifferent equilibrium positions in one-dimensional problems. Always in one dimension, I return on the elastic force and harmonic motion, also in presence of dumping. I conclude with some considerations on developments and problems of classical mechanics, and on the determinism in Newtonian mechanics.

13.1 Work Needed to Build a System In Chap. 6 I introduced the concept of work done by a force. The definition was rather limiting, because it only considered the work done on a point-like particle. Similarly, the definition of the potential energy associated with a conservative force. How is it possible to broaden the concept of work done by a field of forces on a system of points, or on a continuous system, and (if the forces are conservative) define the potential energy of such a system? I start considering the case of a central force F=−

.

α21 rˆ . r2

(13.1)

where the constant .α21 = Gm 1 m 2 in the gravitational case of two point-like masses, (10.10), or .α21 = −kq1 q2 in the case of the Coulomb interaction of two electrically charged particles. Let us imagine that, initially, the two particles are at a very large distance, the math enthusiast would say infinite distance, .∞. Starting from this initial point, they approach along the line joining them up to a distance .r21 as in Fig. 13.1. The calculation of the work done by the force (13.1) using the definition (6.1) is very

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Spurio, The Fundamentals of Newtonian Mechanics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-47289-3_13

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13 Considerations on Energy

Fig. 13.1 Left: two particles 1 and 2 are pushed to a distance.r12 from an enormously large starting distance. Right: A third particle approaches from an infinite distance to the system of the two particles locked in the position show on left-hand side. The trajectory followed by particle 3 is any, and the involved forces are central

simple. It doesn’t matter whether particle 1, particle 2, or both, move. Let’s put the origin of the reference frame centered (for example) on particle 2, and see particle 1 approaching. The force on 1 due to 2 (as in Fig. 11.4) does a work equal to { .

W21 =

r21

∞

−

α21 α21 dr = . 2 r r21

(13.2)

Work is positive, as we know, in the case of gravitational interactions. In the electrostatic case, the work done by the force is positive when the two particles have opposite sign electric charges. Conversely, two charges with the same sign repel each other, and the work done by the Coulomb force is negative, that is, some external force must bring the two electric charges approaching together, it is not a spontaneous process.1 You can easily verify that you would obtain numerically the same result by placing the reference frame on particle 1 and determining the work done on particle 2 seen as in motion. Making use of path integration in spherical coordinates as introduced in Sect. 10.9, it is also possible to verify that the same result is obtained if the two particles approach following a generic trajectory, and not along a straight line. What happens if, keeping the positions of particles 1 and 2 unchanged, I transport from an infinite distance up to the coordinate point . P3 a third particle 3 that feels the 1 For example, protons are pressed together in the nucleus of stars because of the pressure due to the gravity force exerted by external layers of matter.

13.1 Work Needed to Build a System

347

interaction (gravitational or electrostatic) of the first two particles? I can calculate the work done on particle 3 by imagining that we are witnessing the process in the reference frame where 1 and 2 are at rest. Then, { .

W3 =

P3 ∞

{ F3 · ds =

P3

∞

(F31 + F32 ) · ds

(13.3)

having used here the principle of superposition of forces. The force is always given by (13.1), and indexes such as 32 simply indicate the particles involved. Recalling that the only relevant path element for calculating the work of a central force in spherical coordinates is that of the radial direction (d.r ), we obtain: .

W3 =

α31 α32 + . r32 r31

(13.4)

The total work done to create the final configuration on the right side of the Fig. 13.1 figure is then: α21 α32 α31 . W = W21 + W3 = + + . (13.5) r21 r32 r31 Note that (13.5) is totally symmetrical with respect to the masses .m 1 , m 2 , m 3 (or, with the electric charges .q1 , q2 , q3 ) and their positions . P1 , P2 , P3 , even if particle 3 was introduced last. The same result would have been achieved even if we had (for example) created system 23 first and then brought particle 1 closer. Furthermore, the involved forces are conservative: therefore the quantity .W is representative of the potential energy function of the system, which depends on the configuration assumed by the system. It doesn’t matter the order or the time it takes to arrive at the final configuration: only the mutual position of the three objects matters. In practice, we could call the quantity . V = −W the configuration energy of the system; the change of sign is to adopt the convention used in Chap. 6. Unfortunately, this term is not frequently used in physics, and thus we call it also the potential energy of the system. The method used to calculate the work done to form the system in the previous example can be extended to any discrete and continuous systems. In the next section, we will consider the case of a continuous system. In the case of discrete systems, we need a summation that spans over . N objects. If the position of each object is defined (for example, the positions of the positive and negative ions of a crystal lattice with defined geometry), the configuration energy of a system of . N objects is determined by summing over the pairs of objects: 1 ∑ ∑ α jk . 2 j=1 k/= j r jk N

.

V =−

(13.6)

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13 Considerations on Energy

Double summation notation means: take . j = 1 and add the values of the quantity corresponding to the values .k = 2, 3, 4, . . . N ; then, take . j = 2 and add the values corresponding to .k = 1, 3, 4, . . . N ; continue like this until . j = N . Clearly, each pair is counted twice, which explains the 1/2 factor. The above expression can be implemented in calculator programs to determine the binding energy of systems such as clusters of stars, crystal lattices or grains of salt. Negative (configuration) potential energy means that work must be done by external forces to dissociate the system.

13.2 Potential Energy of a Bound Spherical System Many objects are spherical in shape: not just marbles and raindrops, but planets and stars as well. If the various parts interact with each other by attraction, positive work is done by the forces involved during the formation of the system. This means that if we want to decompose the system (evaporate a drop, explode a star) work must be done on the system itself. Consider a sphere whose parts feel the gravitational interaction. A sphere has only two relevant parameters: its mass . M and its radius . R. From the point of view of dimensional analysis, it is easy to obtain a quantity that has the dimensions of an energy, i.e. [M L.2 T.−2 ]. When we use these parameters and the fact that the constant . G must necessarily be introduced when the gravitational interaction is involved, we obtain: G M2 . Vspher e ∝ (13.7) R (verify!). The numerical factor that depends on the specific geometry of the system is of course not deducible from dimensional analysis. In this case with spherical symmetry, we can obtain the numerical factor using the method defined in Sect. 10.9, and assuming that the density .ρ of the forming object is constant in the volume. Let us first consider (Fig. 13.2) the work done by the force of gravity exerted between a solid spherical nucleus, of radius .r , volume .(4/3π)r 3 , mass .m = (4/3π)ρr 3 , and the mass of a spherical shell that is approaching from a very large distance. A spherical shell corresponds to a spherical surface of area . A = 4πr '2 (where .r ' is the distance from the center) and infinitesimal thickness .dr , so that its infinitesimal volume is '2 .d V = 4πr dr . As we have seen in Sect. 10.9, the mass .m of the spherical nucleus is as if it were concentrated in its center, which we take as the origin of the reference frame. The mass corresponding to each element of the spherical shell is always at the distance ' .r . The spherical shell can approach up to the distance .r , and then settle on top of the pre-existing spherical nucleus. At this distance from the center, the density always assumes the constant value .ρ, and therefore the total mass of the spherical shell

13.2 Potential Energy of a Bound Spherical System

349

Fig. 13.2 Geometry related to the determination of the work done by gravity for the creation of a sphere of radius . R by overlapping spherical shells of thickness d.r . See text

corresponds to .dm = ρd V = 4πρr 2 dr . As in (13.2), the (infinitesimal) work done by gravity to bring the (infinitesimal) mass .dm closer to .m up to the distance .r is dW = G

.

( 4 πρr 3 )(4πρr 2 dr ) m · dm 1 =G 3 = G(4πρ)2 r 4 dr r r 3

(13.8)

Now, we can think of a sphere of radius . R as the superposition of many spherical shells, starting from the radius .r = 0 up to .r = R. This implies the integration of (13.8) with these integration extremes: {

.

R

{

R

1 G(4πρ)2 r 4 dr 3 0 0 5 1 3 4π 1 2R = G( ρR 3 )2 = G(4πρ) 3 5 5 3 R 3 G M2 = 5 R

W =

dW =

(13.9)

ρR 3 is the total mass of the system. The quantity .W represents the where . M = 4π 3 work done by the gravitational force to form the sphere of radius. R and gravitationally bound mass . M. Its potential (or, as it would be better to say configuration) energy corresponds to the work changed in sign, i.e.: .

Vspher e = −

3 G M2 . 5 R

(13.10)

350

13 Considerations on Energy

We can assume that at the beginning, the total mechanic energy of the system is null. In fact, the potential energy is null at very large distances, as the kinetic energy of individual mass shells. Thus, if we assume energy conservation, the energy (13.10) can be “stored” in the system as if compressed springs are in actions between mass elements of the sphere. Alternatively, energy is released in some form during the formation of the system to the external world. In both cases, positive work must be provided to decompose the system, releasing the stored energy or by external forces, numerically equal to (13.10) in absolute value.

13.2.1 Age of the Sun An example of the use of the above equation concerns the possibility of estimating (incorrectly!) the age of the Sun. The Sun has a mass . MΘ , radius . RΘ equal to .

MΘ = 2 × 1030 kg ;

RΘ = 7 × 108 m

(13.11)

and emits a power in the form of radiation equal to .

PΘ = 3.8 × 1026 W .

(13.12)

For a time during the 1800s, in the absence of other physical hypotheses, it was believed that the Sun gave off energy by combustion processes, as if it were composed of a burning hydrocarbon. Based on this model, the Sun could function for about 6,000 years.2 To determine this value, it is necessary to know how much energy the combustion of a hydrocarbon provides. To get this number, go to the pantry and look at the olive oil label (or butter, or any other edible oil or fat, if you are not from the Mediterranean area). What I have, it says that about 3500 kJ (or 830 kcal) are supplied per 100 g of product, i.e., about 35 MJ/kg. Hydrocarbons produce approximately (per gram) the same energy that oil, butter or other fatty matter supplies to the body: the elementary physical process of chemical conversion releasing energy produces (as a first approximation) the same quantity of energy per gram of matter. Therefore, 1 kg of “highly flammable” matter produces, for processes of a chemical nature (that is, for processes in which electromagnetic interactions are involved at the atomic level) about .

2

Q chem ≃ 40 MJ/kg (calorific value of hydrocarbons).

(13.13)

The value is very close to creationist estimates of the date of appearance of Adam and Eve on Earth. We would be very close to the end, as the Egyptian pyramids are about 4500 years old, and older historical artifacts are known.

13.2 Potential Energy of a Bound Spherical System

351

You can check the values of different materials on the page https://en.wikipedia.org/ wiki/Heat_of_combustion. On the basis of these data, the duration .τchem of the Sun due to chemical combustion can be determined as: τ

. chem

=

MΘ · Q chem = 2 × 1011 s ≃ 6000 years . PΘ

(13.14)

Such a small value is totally inconsistent with any kind of observation. In the mid-1800s one of the most authoritative physicists of the time, now known as Lord Kelvin3 imagined that the power emitted by the Sun could correspond to the slow release of the energy stored for its gravitational formation. Based on this hypothesis, the solar potential energy stored in the system is given by (13.10), which corresponds with parameters (13.11), to .VS = −2.3 × 1041 J. If this stored energy were released with the power . PΘ , the Solar lifetime, .τgrav , determined assuming conversion of gravitational energy would correspond to: τ

. grav

=

|VS | = 6 × 1014 s ≃ 2 × 107 years . PΘ

(13.15)

This estimate had a very strong impact on another great scientist, Charles Darwin4 . In the first edition of the Origin book Darwin had estimated the minimum age for the formation of a particular region in the south of England in about three hundred million years. This was the order of magnitude of the elapsed time needed for the evolution of species, within which he had conceived his theory of evolutionary change. However, this estimate was inconsistent with the duration of the Sun activity determined by Lord Kelvin. Conditioned on the authority of the physicist, in later editions of the book Darwin removed the paragraph: the (13.15) estimate of 20 million years for the age of the Sun (and therefore of the solar system) was not absolutely compatible with the slow evolutionary Darwinian model. The question was resolved in the 1900s with the development of knowledge in nuclear physics. In particular, in 1939 Hans Bethe published the article Energy production in stars,5 where he speculates that the process capable of generating the energy emitted by the Sun and other stars is nuclear fusion. This process, for each

3

Lord William Thomson (1824–1907), 1st Baron Kelvin, commonly known as Lord Kelvin. He is known, in addition to thermodynamic studies, for having developed the Kelvin scale, which measures absolute temperature. The title of baron was conferred upon him on the merits of his discoveries. 4 Charles Robert Darwin (1809–1882) was a British biologist, naturalist and geologist, who contributed decisively to the history of science with the theory of evolution. The publication of the first edition of the work On the origin of species in 1859 represents a fundamental event. In the book, the evolution of plant and animal species is described as a result of natural selection acting on the variability of hereditary characters, and their diversification and multiplication by descent from a common ancestor. His theory is based on the many data and observations obtained during a voyage around the world on the ship HMS Beagle, and in particular during his stop in the Galápagos islands. 5 https://journals.aps.org/pr/abstract/10.1103/PhysRev.55.434.

352

13 Considerations on Energy

atomic nucleus involved, generates an amount of energy about 6 orders of magnitude greater than the chemical reactions (question 2). Therefore, assuming . Q nucl ∼ 106 Q chem , the estimate of the lifetime of the Sun based on the nuclear reactions is of the order τ

. nucl

=

MΘ · Q nucl ≃ 6 × 109 years . PΘ

(13.16)

The detailed model predicts that the Sun will last about 10 billion years, 5 of which have already lived. Is it possible to somehow verify the above model? The answer is yes. The Sun (like other stars) mainly produces energy in its central nucleus (called core), by fusing 4 protons into a helium (He) nucleus and releasing 27 MeV of energy under the form of highly energetic photons (also called .γ-rays). The protons, which repel each other by Coulomb interaction, in the .cor e are forced to approach each other due to the enormous pressure of the force of gravity exerted by the external layers. Once the protons are close enough to make contact (the proton has a radius of about 10.−15 m), nuclear forces (vastly stronger than the Coulomb repulsion) set off the fusion reactions. The process .4 p →42 He does not balance the electric charge (the helium nucleus is composed of 2 protons and 2 neutrons). In fact, the process foresees (in addition to the emission of Q = 27 MeV of energy) also the emission of 2 positrons (the antiparticle of the electron, with a positive electric charge and indicated with .e+ ) and two electron neutrinos (.νe ). Photons take about 100,000 years to reach the surface of the Sun, meanwhile exerting the pressure that opposes gravity and keeps the star in balance. Due to the fact that they are extremely weakly interacting, neutrinos immediately flee from the core without exchanging energy with the Sun matter and after about 8 min they can reach the Earth. Since about 1980, the neutrino flux from the Sun has been measured by several underground experiments. The progenitor is an experiment in a former mine in the United States (Homestake, in South Dakota), which earned its creator R. Davis Jr. the Nobel Prize in 2002. These experiments (SuperKamiokande in Japan, SNO in Canada, Gallex/GNO and Borexino at Gran Sasso) have accurately measured the flux of neutrinos from the Sun originating from the fusion reactions of light nuclei and thus verifying the correctness of the estimate reported in (13.16).

13.2.2 Energy Conservation in Stellar Gravitational Collapse (*) Nucleosynthesis in stars, i.e. the fusion of light nuclei to obtain heavier nuclei as in the case of reaction.4 p →42 He + 2.e+ + 2.νe +. Q, is exothermic (energy is releases,. Q > 0, in the form of photons) until the iron nucleus is produced. When the production of photons, which keep the star in equilibrium against gravity, due to fusion reactions stops, the star undergoes a gravitational collapse. Stellar gravitational collapse is one

13.2 Potential Energy of a Bound Spherical System

353

Fig. 13.3 Image of the remnant of SN1572 exploded in the Cassiopeia constellation (inside our Galaxy) at the time of Tycho Brahe. The image is formed by observations made in the visible light by the Hubble Space Telescope (yellow), in X-rays by the Chandra satellite (blue and green) and in the infrared by the Spitzer Space Telescope (red). The material that releases the image corresponds to a shock wave that moves away with speed.v/c ∼ 10−2 from the center of what used to be the star. The velocity can also be deduced from the size of the cloud (a few pc) and from the.∼ 500 years since the explosion. Credits: X-ray: NASA/CXC/SAO; Optical: NASA/STScI; Infrared: NASA/JPL-Caltech

of the astrophysical processes that can be observed and called by astronomers supernova (SN). The appearance of a supernova is a spectacular and very rare phenomenon to see with the naked eye, but quite frequent when using powerful telescopes. The last SNe visible without the aid of instruments occurred on November 11, 1572 (reported by Tycho Brahe), and in October 1604, observed by Kepler. Figure 13.3 shows what Tycho’s SN looks like today. Over the next 5 billion years, the Sun will fuse H nuclei (i.e., protons) in He and subsequently He in C at its center. Stars more massive than the Sun, by virtue of their greater gravitational pressure, are able to implement fusion cycles which from C lead to the production of increasingly heavier elements. They therefore evolve in an “onion” structure, with a sequence of concentric layers within which nuclear reactions of different nature take place. The outermost layer is made up of hydrogen; proceeding towards the center of the star, the layers of helium, carbon, oxygen are found in sequence, and with the central core composed of iron nuclei. Only stars with mass . M > 8MΘ manage to have a pressure in the core so high as to synthesize

354

13 Considerations on Energy

iron nuclei. Nuclei heavier than iron cannot be formed without expending energy. Therefore, when iron is synthesized inside a star, the production of energy ceases and the star undergoes gravitational collapse. When iron has formed in the core of the star, the external states continue to “precipitate” towards the center. At this point, the compressed core becomes so dense that each proton in the iron nucleus can capture an electron, transforming into a neutron and emitting a neutrino. Thus begins the formation of what will become a neutron star, which I mentioned in Sect. 11.9. In the compression process, the force of gravity does work, which is transformed into energy in the form of creation of particles (mainly neutrinos) and kinetic energy of the particles. Through (13.10) we can determine the variation of potential energy in the transition from a spherical object of mass . M and radius . R ≃ 109 m as in (13.11) to a radius . R N S ≃ 104 m and . M N S = 1.5MΘ , as in a neutron star. This change in energy is equal to ΔV = |

.

3G M N2 S 3G M N2 S 3G M 2 |≃ ≃ 3 × 1046 J , − 5R 5R N S 5R N S

(13.17)

relation practically independent from the initial values of . M, R. When the star disintegrates as a supernova, the work done by gravity must correspond to the available energy given above. However, the observations show that the electromagnetic radiation emitted contributes to 0.01% of .ΔV , a very small fraction. Only 1% of .ΔV competes to the kinetic energy of all the material emitted that propagates throughout the galaxy, as seen by telescopes (see Fig. 13.3 and question 8). For some time this non-conservation of the energy observed in gravitational collapses with respect to the expected value was a fascinating enigma. Finally, starting from the end of the ’60s, it began to hypothesize that the missing energy ended up in the form of neutrinos, and that these also played a fundamental role in the stellar gravitational collapse. These models, which predict that .∼99% of the available energy is transformed into neutrinos, were supported starting in the 1980s by computer simulations. Even if the interaction probability of neutrinos is very small (remember that they flee from the Sun instantaneously, while photons are trapped for a long time), the density of matter in a collapsing star is so high that, even if neutrinos multiply in number, they remain trapped. Eventually, the neutrino density becomes so high that (like when a balloon is overinflated) the system explodes. The matter is expelled outwards (the supernova explosion takes place) and the neutrinos are free to propagate in the universe. The whole process I described takes a few seconds. This model was experimentally verified on February 23, 1987 when a supernova observed with a small telescope in a satellite galaxy of the Milky Way (the Large Magellanic Cloud) produced events due to neutrino interactions in three different detectors (Kamiokande in Japan, IMB in the United States and Baksan in Russia). Taking into account the distance of the SN, the probability of interaction of neutrinos with matter, and the mass of the detectors, the number of neutrino interactions observed (about 25 in total) was perfectly in agreement with the supernova explosion model and with the conservation of energy. Masatoshi Koshiba in 2002 was awarded

13.3 Gravitational Field and Potential

355

the Nobel Prize for contributions to neutrino astrophysics through the observation of the supernova with the Kamiokande experiment.

13.3 Gravitational Field and Potential In the relation (10.10) which describes the gravitational interaction between two masses .m and . M, and in the (10.19) which describes the gravitational potential energy, the role of two masses is equal and symmetrical. However in many situations, as we implicitly considered in Chap. 10, the situation is not symmetrical at all. One of the two masses, for example the one indicated with the capital letter . M, is dominant because it is numerically very large compared to .m. This is certainly the case of weight, where the mass of the Earth dominates over the mass of the pencil that is falling off my desk. Even though the force that the Earth feels and the force that the pencil feels are equal in magnitude, the effects are different because the accelerations are different. And accelerations are, in magnitude, given by the ratio between force and mass. In many situations we have a dominant object of mass . M, or a system of points of total mass . M and well-defined relative interactions, and an object of mass .m entering the system. An example would be the solar system being approached by a comet from outside the system. What we can imagine, therefore, is that the dominant body (or the system) of mass . M creates in Newtonian space a vector quantity given by the ratio between the force and the small mass .m, which we could call test mass. From this definition, in the case of a single massive object of mass . M, the gravitational vector field is given by: F M (13.18) .F ≡ = −G 2 rˆ . m r In fact, you will not hear much about gravitational vector field: this quantity has physical dimensions of an acceleration. In the case of the Earth, the magnitude of the quantity on the right in the (13.18), is equivalent to the acceleration .g that a mass .m feels in the presence of the mass . M of the Earth. Therefore it is unnecessary and uneconomical to call the same thing by two different names! The introduction of an auxiliary vector field becomes much more important and full of consequences in the case of electrical forces that you will study in the near future. In that case, the force F exerted between two electric charges . Q and .q is proportional to their product . Qq (and inversely proportional to the square of their distance). Also in this case there could be an asymmetry situation, in which . Q ≫ q, and it could be useful to define a new vector field as the ratio between the force F and the test charge, q. The new quantity E≡

.

F q

(13.19)

356

13 Considerations on Energy

is called electric field and it is a completely new physical quantity compared to the acceleration felt by the test charge .q of mass .m q , given by .a = F/m q . Given the electric field defined by (13.19), the force that a test electric charge feels is equal to its charge times the field. The vector .E is a different quantity from acceleration, and it assumes such an important physical meaning that it deserves the wide space it receives in the course of electromagnetism. An important property of fields, inherited from the properties of forces, is the superposition principle, introduced in Sect. 4.3 and which we have also seen in action in Sect. 13.1. If the effects of multiple sources (whether they are masses . M1 , M2 , . . . in the case of a gravitational field, or electric charges . Q 1 , Q 2 , . . . in the case of electric field) are applied to a test particle, the resulting force is given by the sum of the forces (gravitational or electric) acting on the test particle. Thus, the resulting force field on the test particle is given by the superposition of the individual effects. In this way, we can have a mathematical representation of the resulting field at any point of a reference frame, field that will act on any test particle that we want to consider.6 Similarly to how we defined a vector field for the gravitational interaction, we can define a scalar field starting from (10.19). We define the gravitational potential produced by a massive object . M as: VG ≡

.

M VG = −G . m r

(13.20)

This quantity does not have an equivalent that we have already defined with another name, so it is extremely interesting. In many situations it is convenient to study the system using the concept of gravitational potential .VG . See for example the questions 6 and 7. Once the potential acting on a test particle is known, the potential energy is simply given by the product of the test mass .m by .VG . Symmetrically, in the course of electromagnetism you will define an electric potential from the interaction between electric charges. The electric potential will have a fundamental function in the description of atomic and subatomic phenomena in quantum mechanics.

Alert spoiler. An interesting effect occurs when the masses. M1 , M2 , . . . (or the electric charges Q 2 , . . . ) that create the resulting field are not at rest. In this situation, the relative distances between the test mass (charge) and any object creating the field change over time, and the resulting gravitational (or electric) field is non-static, changing strength and direction as a function of time. In the case of moving electric charges, you will see, this gives rise to a second vector field called magnetic field. Both the electric and magnetic fields at any fixed location are a function of time, and can be defined at any point of a reference frame even if no electric charge is present at that point. It’s how an electromagnetic wave is formed. General relativity predicts that in particular circumstances, the movement of masses induces a gravitational wave. 6

.Q1,

13.4 First Integral from Energy Conservation

357

13.4 First Integral from Energy Conservation Based on what we have seen so far, there are two ways to determine the equation of motion (and the trajectory) of a particle in the presence of a known potential: using Newton’s second law directly or relying on the law of conservation of energy. I stated that the two approaches are equivalent from the mathematical point of view, but up to now I have used the energetic approach only applied to the complicated case of the motion of the Sun-planet system (Sect. 11.7). In that situation, I defined the first integral of the motion and noticed that the problem is uniquely solved knowing the total mechanical energy and the magnitude of angular momentum. To better understand the importance of the concept of potential energy (or simply of the “potential”), limiting the complications connected with the use of vectors, in the following I will consider examples in one-dimensional situations. In this context, the potential energy of the system is described as a function of a single coordinate, 2 . V (x). The kinetic energy of a particle of mass .m is given by . T = mv /2 and, in the absence of dissipative forces, the total mechanical energy is constant. By applying (6.31) to study the motion of the particle we can use the relation .

mv 2 + V (x) = E = cost . 2

(13.21)

For a given potential .V (x), depending on the initial conditions the particle can have very different trajectories. Initial conditions define the particle’s position and velocity at the initial time .t0 in Newton’s II law .

x¨ =

1 dV F(x) =− . m m dx

(13.22)

When the initial constants are known, the double integration of (13.22) allows us to determine unequivocally the equation of motion .x(t). On the other hand, if we have .x(t0 ) = x0 and .v(t0 ) = v0 , the mechanical energy is uniquely defined as . E = mv02 + V (x0 ). Then, at the cost of a single integration, (13.21) allows to obtain the 2 equation of the motion, as it represents a first integral of the motion (valid in case the forces are conservative) written as: .

m 2 x˙ + V (x) = E 2

(13.23)

To show that (13.23) corresponds to the integration of (13.22), I derive it with respect to time, remembering the rules of derivation of a composite function: m x˙ x¨ +

.

d V (x) d x =0 d x dt

−→

m x¨ = −

d V (x) , dx

(13.24)

having simplified .x˙ in the two members. After verified this, I rewrite (13.23) as:

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13 Considerations on Energy

/ dx 2 . =± [E − V (x)] . dt m

(13.25)

The presence of the double sign does not scare you: it simply indicates that there are two possible directions of motion of the particle for a given position .x. Mathematically, (13.25) is a separable differential equation and its solution can be immediately obtained in the form of the integral / .

±

m 2

{

x

√ x0

dx = E − V (x)

{

t

dt = t − t0 .

(13.26)

t0

If the analytical expression of the potential .V (x) is known, the (13.26) can be integrated. Only one integration is required to arrive at the equation of motion, and not two as using the (13.22): in some situations, this leads to a simplification of the computational effort. As expected, however, in many problems the integral can be obtained only with approximate methods, or numerically. Note that we always need to know two constants: in this case, the total mechanical energy . E and the initial position .x0 . In this respect, there is no “discount” in switching from Newton’s II law to the use of a prime integral.

13.4.1 Application to a Falling Particle I will now use the procedure defined by (13.26) to obtain the equation of motion in the case of a falling particle. When we encountered this problem in Chap. 3, we agreed on the convention that this one-dimensional motion is along the .z axis: I use this variable instead of the .x from the previous section. The potential energy is given by (6.27), i.e. .V (z) = mgz. We assume the object of mass .m is initially at rest, .v0 = 0 and then .T (0) = 0, with initial position . z(t = 0) = 0. By this choice, . V (0) = 0 and therefore the total mechanical energy is . E = 0. The body, as in the device of Sect. 3.2, is falling in the .z < 0 region and consequently .z˙ < 0. This last consideration implies that between the two signs .± of the (13.26) we will have to choose the negative one. Therefore, in this situation I can write the (13.26) as7 : / .

−

m 2

{

z z0

/ { z dz ' m dz ' =t. = − √ √ 2 0 −mgz ' 0 − (mgz ' )

(13.27)

Don’t be shocked by the “–” sign under the root, because .z ' < 0. But it is convenient anyway to change the integration variable as . y = −mgz ' and .dy = −mgdz ' : in this way, . y assumes positive values. I write now the quantities within the integral as: 7 .z

To avoid confusion, I use the mathematical convention by calling .z ' the variable to integrate and the integration limit.

13.5 Motion in a Potential Energy Field

359

/ { z { z m 1 d(−mgz ' ) dy 1 .t = − =√ √ √ 2 0 (−mg) −mgz ' y 2mg 0

(13.28)

√ This integration is simple, with primitive given by .[2 y]. Returning to the original variable, the solution to the problem becomes: [ ]z / √ 2√ 2 −mgz ' = .t = √ −z . g 2mg 0 1

(13.29)

Again, the negative sign under the root remember us that .z is negative. We can compare this solution with what we obtained in Chap. 4, by squaring (13.29): t2 =

.

2 (−z) g

−→

1 z = − gt 2 2

(13.30)

which is exactly what is obtained in (4.17) with the same initial conditions. Thus the two methods of deriving the equation of motion (the use of Newton’s II law or its first integral given by the law of conservation of energy) produce absolutely identical results, as must be. The method of integrating the first integral of motion can also be used to determine the period of oscillation for a harmonic oscillator, .V (x) = 21 kx 2 (6.28). However, I believe that the mathematical effort to solve the already known problem can be spared, and the search for the solution can be left to students who are passionate about more formal problems.

13.5 Motion in a Potential Energy Field The conservation of mechanical energy allows a general classification of the types of permitted motions, including movements limited or not to a region of space. A useful approach is to consider motions that involve only small deviations from an equilibrium position. In this case, the solutions can be simplified to a special universal form, which we already know. To do this, I must first define what a equilibrium situation is.

13.5.1 Stable and Unstable Equilibrium A dynamic system is in equilibrium (Sect. 7.10) when the resultant of the forces and that of the moments of the forces are both zero. Any dynamic system in equilibrium can be perturbed by small external perturbations. These can lead to displacements from the equilibrium position, Fig. 13.4 (see question 9). The considered system is

360

13 Considerations on Energy

Fig. 13.4 The allowed positions for a particle moving in a region of space under the effect of the potential V(x) depend on the value of its total mechanical energy E. Also indicated are the three positions of stable equilibrium (A), unstable (B), and indifferent (C), discussed in the text

in stable equilibrium if, after the small displacement from the equilibrium position, it tends to return to it. As we will see, the small perturbation causes an elastic-type recall of the system towards the equilibrium point. Instead, the system is in unstable equilibrium when, slightly moved from its equilibrium position, it tends to move further away. A very small perturbation is therefore sufficient for the system to move away from the initial position, possibly in search of a new equilibrium condition. Finally, the system is in indifferent equilibrium when, due to any perturbation of reasonable intensity, it remains stably in the new position, without returning to the initial one and without moving further away. Illustratively, the structure of solids is such that every element of mass is in an equilibrium position. For example, the table top is balanced before anything is placed on it. When I put my laptop on it, a small perturbation is applied to the system, which corresponds to a stable equilibrium situation. In fact, once the stress has been removed, the table top is not modified. But how does the table respond during the solicitation? It responds with a force that opposes the perturbation (the weight of the object) a force that, as a first approximation, can be considered elastic in nature. Suppose there are internal forces described by a potential energy function, .V . Under certain conditions the system is in equilibrium. This could mean, for example, that all molecules are .ro apart, or something similar. The potential energy function depends on the coordinates: let’s simplify, imagining it is a function of a single variable, .r , which is needed to quantify the displacement of the system from the equilibrium position. By definition, the force is zero at equilibrium: .

( ) dV F =− =0 dr ro

(13.31)

13.5 Motion in a Potential Energy Field

361

where the convention of writing the subscript .ro indicates that the derivative is calculated right at the point .r = ro . The quantity under parenthesis is therefore not a function, but the value assumed by the function at a specified value of the independent variable. The function.V (r ) is unknown, but if the equilibrium is stable we can imagine that, in a sufficient small region, it has a shape similar to what is shown in Fig. 13.4 around position A. Moving by a small amount .(r − ro ) from equilibrium, the function .V can be approximated with a Taylor series expansion (Sect. 1.7): ( .

V (r − ro ) = V (ro ) +

dV dr

)

( (r − ro ) +

ro

d2V dr 2

) (r − ro )2 + . . .

(13.32)

ro

Since the second term is zero at equilibrium, the equation can be written as: .

V (r − ro ) = α + β(r − ro )2

(13.33)

where .α, β are constants corresponding to the function and its second derivative calculated in .ro . Interestingly, as a first approximation, potential energy is quadratic and takes in a plot the form of a parabola. This parabola (if .β > 0) has the concavity towards the top, as in the case near the position indicated by A in the figure. If (13.33) approximates the unknown potential energy, the corresponding force is: dV = −(2β)(r − ro ) = −k(r − ro ) . F(r − r o ) = − (13.34) dr with .k taking on a positive value if .β > 0, i.e., if the configuration was in stable equilibrium. What we have obtained is that, following small external stresses, the “internal” potential energy has such an effect that the dynamic response of the system, as a first approximation, is equivalent to the return force of a spring. This is why we often call the behavior of solids elastic in many situations. Conversely, if the equilibrium had been unstable (point B in the figure) the value of the parameter would have been .β < 0. In this case, the sign of the (13.34) force would have been positive, and it would have been a repulsive force, which pushes the system further and further away from equilibrium. In the case C (sometimes called metastable), the second derivative would also be zero and there would be no force, at least to the first order of (.r − ro ).

13.5.2 Permitted and Forbidden Regions We can now generalize the above discussion to classify the possible motions of a particle in a given potential energy V(r). The idea behind this classification is quite simple, and can be summed up in the fact that the kinetic energy .T cannot be negative. As we did in the case of the effective gravitational potential, Sect. 11.6.1,

362

13 Considerations on Energy

if we put the curves V(r) and a horizontal line E =constant on the same graph, the intervals of .r that are allowed for the motion are only those which satisfy the relation . T = E − V (r ) ≥ 0. For positions .r such that . E − V (r ) < 0, the presence of the particle (or system) is prohibited. Typically, the higher the energy . E, the wider for the particle are the allowable ranges of .r . Referring to the Fig. 13.4 and to the 4 defined energy values, we have: • for . E = E 0 , the kinetic energy is zero. The particle is constrained to remain at the point . A, where the potential energy has a minimum, and where the particle (or system) is in stable equilibrium. • For . E = E 1 , the movement is restricted to the region .r3 ≤ r ≤ r4 , and the system can also be classified as bounded. The particle is as if trapped in a well of potential. The velocity vanishes at the points.r3 , r4 : normally it is a point in which the velocity function changes its sign (the motion reverses, as in the pendulum which reaches the extreme point). The velocity is maximum when .V is minimum, i.e., coinciding with the passage through point A. • For . E = E 2 , the values .r2 ≤ r ≤ r5 , or the values .r ≥ r6 are allowed, depending on the initial condition at the starting point. If at the instant .t = 0 the particle is in the region to the right of .r6 , it will have to remain forever in the region with .r ≥ r 6 , and can never be in the other possible region .r 2 ≤ r ≤ r 5 . There are two regions allowed, but motion can occur exclusively in either the first or the second. • For . E = E 3 , any position .r ≥ r1 is possible. Note that Newtonian mechanics is strict on the aspect of the forbidden regions, as seen in the cases . E = E 1 and . E = E 2 : if the energy conditions do not allow to enter, the forbidden regions cannot be occupied by the particle or by the considered system. This is another aspect that will need to be reviewed when you study quantum mechanics. The state of the system will be described by a function, called wave function which describes the probability of occupying a certain region of space. Even though there are forbidden regions in quantum mechanics, the transition probability in a forbidden region is not completely zero. This probability is not arbitrary, but can be calculated for example with what is called the tunneling effect. Even if quantum mechanics involves processes that are totally forbidden in classical mechanics, the probability of occurrence of these processes is calculable.

13.6 More on Harmonic Motion (*) Let us now examine the situation in which we slightly perturb an object from the stable equilibrium position .ro . As we have seen in (13.34), the perturbed object feels a linear force .

F(r − ro ) = −k(r − ro )

−→

r¨ = −

k (r − ro ) m

(13.35)

13.6 More on Harmonic Motion (*)

363

where I made use of Newton’s second law. How is it possible to verify that (13.35) has the same solutions found in Sect. 4.9? I can rewrite (13.35), using the fact that we can add or subtract any constant quantity in a differential, as: .

d 2 (r − ro ) k = − (r − ro ) 2 dt m

(13.36)

and finally making the change of variable .x ≡ (r − ro ), write: .

d2x k =− x 2 dt m

(13.37)

which corresponds exactly to (4.42) or to (4.36) (with different variable name). Therefore the problem has already been solved: but it is useful to deal with it using a new point of view and making use of something that I hope you have started studying in your math class.

13.6.1 Complex Numbers You surely know that an equation like .

x 2 = −1

(13.38)

has no solutions in the real field. To have solutions, it is necessary to expand the numbers we have available (after natural, integer, fractional, real): for this reason, complex numbers are introduced. The specific solution of the (13.38) is the number . x = i; the complex number .i (initial of imaginary) is by definition such that i 2 ≡ −1 .

.

(13.39)

In general, a complex number .a is a number such that it can always be written in the form: .a = p + iq (13.40) where . p, q are two real numbers. With the introduction of complex numbers, any algebraic equation can be solved. Let mathematics teachers demonstrate that the algebraic rules are satisfied by the definition given above (distributive property, commutative, existence of the neutral element,…). In (13.40), the term . p is called real part and the number .q imaginary part of the complex number. Sometimes the real part of the complex number .a is indicated with .Ra and its imaginary part .Ja; both the real and imaginary parts are real numbers.

364

13 Considerations on Energy

The definition of .i as the number whose square is –1 is ambiguous, because if you chose .−i its square would also be .−1. Therefore the conjugate of a complex number is defined as that number which has the same real part of the complex number but opposite imaginary part. The conjugate complex number is conventionally represented with an asterisk. Hence if a = p + iq is a complex number, a ∗ = p − iq is its conjugate.

.

It is easy to verify that the product of a complex number and its conjugate is a real number: ∗ 2 2 .aa = p + q Complex numbers can also be raised to a power. One of the most remarkable properties is that the complex number .eit can be written as: eit = cos t + i sin t

.

(13.41)

relation which is also called Euler’s formula (or identity). Euler’s formula is remarkable, because it represents a sort of unification of algebra and geometry in a plane. Let’s imagine we have two real numbers, where .r represents a radial distance and .θ an angle (as in polar plane coordinates), a complex number can be written in a general way as .a = r eiθ , i.e.: r eiθ = r (cos θ + i sin θ) = (x + i y)

.

(13.42)

as if we were projecting the radius .r of a circle onto the two orthogonal axes of a Cartesian reference frame. In this frame, the real part represents√the abscissas and the imaginary part the ordinates. You can easily verify that .r = x 2 + y 2 and .tan θ = y/x. The derivation properties remain unchanged when applied to numbers and complex functions. For example, if .a = r eiθ then: .

deiθ da =r = ir eiθ = ia . dθ dθ

(13.43)

13.6.2 Harmonic Oscillator with Complex Numbers After having briefly recalled the basic properties of complex numbers (without the ambition to have been exhaustive), I return to the problem offered by the differential (13.37), verifying that one of its solutions is given by the complex function: .

x(t) = C1 e−iωo t

(13.44)

13.6 More on Harmonic Motion (*)

365

where .C1 is also a complex, time-independent number. It can be verified on the basis of the (13.43) property that .

x¨ = (−iωo )2 C1 e−iωo t = −ωo2 x

as long as .ωo is fixed to the real number: 2 .ωo

k = m

/ −→

ωo =

k . m

The solution is not unique: the function .

x(t) = C2 eiωo t

(13.45)

is also an equally good solution of (13.36), and the most general solution must be of the form: 1 (C1 e−iωo t + C2 eiωo t ) . x(t) = (13.46) 2 where the need for the .1/2 factor will become clear shortly. Interpreting (13.46) as a solution to the physical problem of the harmonic oscillator puts us in difficulty for two reasons:.(i) a complex number cannot represent a physical displacement, which must necessarily be a real number;.(ii) the solution of a second order differential equation shall contain only two integration constants, defined by the initial conditions about position and velocity. The complex function (13.46) contains four indeterminate constants, corresponding to the real and imaginary parts of .C1 and .C2 . These two conditions place restrictions on the possible values of .C1 and .C2 . Let us verify that the situation in which the two constants one complex conjugate of the other allows to have real solutions, that is with two real constants to be determined. I write the two quantities using two real numbers . A, B: C1 = A + i B ; C2 = C1∗ = A − i B .

.

In this way, using Euler’s formula, (13.46) can be written as: 1 [(A + i B)(cos ωo t − i sin ωo t) + (A − i B)(cos ωo t + i sin ωo t)] 2 . = A cos ωo t + B sin ωo t . (13.47) This explains the inclusion of the factor 1/2 in (13.46). Taking into account the cosine addition properties, (13.47) can be written as: x(t) =

.

x(t) = C cos(ωo t + φ) with C =

√

A2 + B 2 ; cos φ = √

A A2

+ B2

(13.48)

366

13 Considerations on Energy

In this way, we have obtained exactly what is reported in Chap. 4, with the (13.47) equivalent to the (4.38) and the (13.48) to (4.39). See questions 11 and 12. In (13.48) the amplitude .C is by definition positive, and typically the phase is chosen so that .0 ≤ φ ≤ 2π. Classically, in a harmonic oscillator, the frequency of the oscillations is independent of the amplitude.

13.6.3 Mechanical Energy of the Harmonic Oscillator Now, I determine the mechanical energy of the oscillator, verifying that it remain constant. Using the above results in (13.48), remembering that .mωo2 = k and the derivation properties, I obtain: mC 2 ωo2 m x˙ 2 kC 2 = sin2 (ωo t + φ) = sin2 (ωo t + φ) 2 2 2 kx 2 kC 2 . V = = cos2 (ωo t + φ) 2 2 E = T + V = mC 2 ωo2 = cost . T =

(13.49)

As expected, the total mechanical energy of the harmonic oscillator is constant, Fig. 13.5, but it is variably divided between kinetic energy and potential energy as a function of time. As usual, the period of the system is given by the relation .T = 2π/ωo . After a period .T , the situation repeats the same way. The total mechanical energy is thus proportional to the square of the amplitude .C and of the system’s own angular frequency, .ωo , and is independent of the phase .φ.

13.7 Damped Harmonic Oscillator (*) The ideal harmonic oscillator described by (13.37) is practically non-existent. There is no existing device without any dissipative effect which keeps the mechanical energy strictly constant forever. Dissipative effects must therefore be parametrized in some way, and the equation of motion modified accordingly. One way to do this is, as we have already seen in Sect. 4.8.2, to insert a force proportional to the speed, . FV = −ηv. The differential equation that includes this effect in (13.37) is: .

η dx d2x k . =− x− dt 2 m m dt

(13.50)

This formulation of the friction force is not the most general: however, it is extremely useful to deal with and in many situations it very well approximates experimental observations. In a more compact form, (13.50) can be written as

13.7 Damped Harmonic Oscillator (*)

367

Fig. 13.5 Total mechanical energy distribution . E versus time during a period .T . In the case of the figure, at the initial time the energy is purely potential, as in the case of a tense and firm spring, or a pendulum moved from its equilibrium position. The sum of kinetic and potential energy, in the absence of dissipative effects, remains constant

.

x¨ + 2γ x˙ + ωo2 x = 0

(13.51)

where .ωo2 = k/m as usual, and .γ ≡ η/(2m) is called the damping coefficient. To solve (13.51), I use the same approach that was applied in the previous section. A possible solution is a function in the complex field: .

x(t) = X o eαt

(13.52)

where at the moment the meaning of . X o is irrelevant, it is just a complex constant. Recalling the derivation properties of the exponential function (.x˙ = αx , x¨ = α2 x), inserting this test function in (13.51) I obtain, associated with the differential equation, the following characteristic algebraic equation: α2 + 2γα + ωo2 = 0 .

.

(13.53)

This quadratic equation always has solutions in the complex field given by: α1,2 = −γ ±

.

/

/ γ 2 − ωo2 = −γ ± i ωo2 − γ 2 .

Three different cases can be distinguished.

(13.54)

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13 Considerations on Energy

13.7.1 Overdamped Motion The first situation considered is that in which the damping due to viscosity or friction is very large: numerically, we have .γ 2 − ωo2 > 0. In this case, the two solutions of (13.54) are both in the real field and α1,2 = (−γ ±

.

/

γ 2 − ωo2 ) < 0

(13.55)

that is, they are two real and negative numbers. The general solution of the differential (13.51) can be written as .

x(t) = Aeα1 t + Beα2 t .

(13.56)

with . A, B real constants. This result indicates that there are actually no oscillations at all! After a more or less long time, depending on the values of the (negative) constants .α1 and .α2 , the system exponentially approaches the equilibrium position, as can be seen from the three examples in Fig. 13.6. This behavior is typical of a system with high strength and relatively low elasticity.

Fig. 13.6 Oscillation amplitude (arbitrary units, a.u.) over time in a overdamped oscillatory motion. In this example, I have considered a system with proper period .T = 2 s, i.e. .ωo = 3.14 s.−1 . The amplitudes . A = B = 0.5 of (13.56) are arbitrarily taken equal in order to give initial amplitude . x(t = 0) = 1.0. The damping coefficient is .γ > ωo to have solutions with real .α1,2 . Three values have been assumed: .γ = 8.0 s.−1 (green line); .γ = 6.0 s.−1 (blue line); .γ = 4.0 s.−1 (red line)

13.7 Damped Harmonic Oscillator (*)

13.7.1.1

369

Critical Damping

The case .γ = ωo is special because the two solutions .α1,2 in (13.54) are identical. Without going into the solution details, you can easily verify by inspection that a solution of (13.54) is given by the function: .

x(t) = e−γt (At + B) .

(13.57)

always with . A, B real constants. Qualitatively, there is not a large difference with the previous case: after a much longer time than .γ −1 (quantity having the dimensions of a time), the system approaches the equilibrium position exponentially.

13.7.1.2

Damped Oscillatory Motion

The case with small damping is particularly interesting in many situations. This is what happens when, for example, you excite the string of a musical instrument, such as the piano. Hear the characteristic sound, with a dominant frequency of the note being played, which fades more or less quickly as if lost in the distance. You can do this by using pedals, which can increase the damping of the sound. A real damped oscillatory motion occurs under the condition .γ 2 < ωo2 . In this case, I can write / / .

γ 2 − ωo2 = iω with ω ≡

ωo2 − γ 2 ,

(13.58)

corresponding to the fact that .ω is a real number. Consequently, the two solutions of the characteristic algebraic equation are given by α1,2 = −γ ± iω

.

(13.59)

The general solution of the dynamic equation has the following solution in the complex field: −γt . x(t) = e [C1 eiωt + C2 e−iωt ] . (13.60) which can be rewritten, using Euler’s formula, as: .

x(t) = e−γt [(C1 + C2 ) cos ωt + i(C1 − C2 ) sin ωt] .

(13.61)

The requirement that the solution be real imposes, as usual, that the two complex numbers .(C1 , C2 ) are conjugates of each other. So, if we write: C1 = a + ib ; C2 = a − ib

.

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13 Considerations on Energy

we get the relations: (C1 + C2 ) = 2a ; (C1 − C2 ) = 2ib

.

and the solution (13.61) can be rewritten as: .

x(t) = e−γt [2a cos ωt − 2b sin ωt]

(13.62)

that is, making use of the addition properties of trigonometric functions, as .

x(t) = X 0 e−γt sin (ωt + φ) .

(13.63)

The constants relating to amplitude. X 0 and phase.φ are connected to the real numbers a, b defined above by the relations:

.

/ .

X 0 = −2b 1 +

a2 a ; tan φ = − . 2 b b

(13.64)

Obviously, the constants . X 0 , φ are fixed by knowing the initial conditions on the ˙ 0), in a similar way to the case of the undamped position .x(t = 0) and speed .x(t = harmonic oscillator. In Fig. 13.7 you can see the amplitude of the oscillation as a

Fig. 13.7 Damped oscillatory motion: amplitude (a.u.) of the oscillation over time. In the example, I have considered a system with proper period .T = 2 s, i.e. .ωo = 3.14 s.−1 . The amplitudes .a = −b = 0.5 of (13.64) are arbitrarily taken equal in order√to give initial amplitude .x(t = 0) = 1.0. With these initial constants, the amplitude value . X o = 2 and the phase .φ = π/4. The damping coefficient is .γ < ωo to have complex solutions for .α1,2 . Three values have been assumed: .γ = 0.6 s.−1 (yellow line); .γ = 0.2 s.−1 (red line); .γ = 0.1 s.−1 (purple line)

13.7 Damped Harmonic Oscillator (*)

371

function of time in three situations of weakly damped motion. The proper period considered is always .T = 2.0 s, but it can be appreciated with a ruler that there is a small dependence that depends on the damping term .γ, since the effective angular frequency of the system is given by (13.58). For damping values.γ « ωo the result (13.63) can be seen as describing a harmonic oscillation with amplitude decreasing slowly with time, .

X (t) = X 0 e−γt .

(13.65)

The inverse of the constant .γ is a parameter with the dimensions of a time, and therefore it is a characteristic time of the system. Normally, .τ ≡ 1/γ can be written and this quantity is called damping time constant. After a time interval equal to.t = τ , the amplitude decreases to a value .1/e ≃ 0.37 of the initial one.

13.7.2 Discussion on Mechanical Energy and Developments We can calculate the variation of the total mechanical energy of the system as a function of time. By virtue of (13.63), it is evident that mechanical energy is not conserved due to damping, which is responsible for energy dissipation. The analytical determination of the energy decrease in the general case is mathematically tedious, even if not difficult. For this reason, I only consider the particular case of a very weakly damped motion in which the amplitude varies as (13.65). This case has technically simple solutions, because all terms proportional to .γ are small and can be neglected. In this way, the time dependence of potential and kinetic energy can be written as: kx 2 k X 02 −2γt 2 V (t) = sin (ωt + φ) = e 2 2 . (13.66) m x˙ 2 m X 02 ω 2 −2γt 2 T (t) = cos (ωt + φ) . ≃ e 2 2 Note that in general .ω 2 /= k/m. However, in the chosen approximation we can actually assume .ω ≃ ωo = k/m. In this way we have for the total mechanical energy: .

E(t) = V (t) + T (t) =

m X 02 2 −2γt = E 0 e−2γt ω e 2

(13.67)

where . E 0 is the total mechanical energy at the initial time. Thus, it can be seen that the time dependence of mechanical energy has an exponential decrease, i.e., it is proportional to the square of the amplitude, . X 2 (t). In many interesting examples of macroscopic harmonic oscillations, the energy dependence has a behavior that is approximately described by (13.67). The missing mechanical energy is normally dissipated in other forms: vibrations of the system (phonons), heat, mechanical waves (sound), …according to the considered system.

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At this point we could ask ourselves: is it possible to restore the energy that is dissipated by the system? Certainly yes, through the external solicitation of a force. But here we would enter a field of extraordinary importance and complexity. In fact we should start by choosing a solicitation with a particular angular frequency .ω, so that we can solve the problem. We could justify this particular choice with what is called Fourier analysis, which shows how each periodic signal can be seen as the superposition of a series of signals with a fixed angular frequency. After this, we would notice that the solicitation can have a very large effect in the vicinity of angular frequency close to the value of the proper one, an effect which is called resonance. Mathematically, resonance has a specific description of how the amplitude of the signal varies as a function of the angular frequency of the external force. Basically, to do this and discuss consequences I would need almost another semester of teaching!

13.8 Developments and Problems of Classical Mechanics 13.8.1 Lagrangian and Hamiltonian Formalisms In the early 1800s, two alternative formulations of classical mechanics to the Newtonian one were developed. They are named after their inventors, the French mathematician and astronomer J.L. Lagrange (born in Turin in 1736-died in Paris in 1813) and the Irish mathematician Sir W. R. Hamilton (1805–1865). The Lagrangian and Hamiltonian formulations of mechanics are equivalent to the Newtonian one from the point of view of the purposes, that is to determine the equation of motion and the trajectory of the object under study given the dynamics of the system. However, in some complex situations the two formulations make it easier to reach the solution from the mathematical point of view. The complete development of the Lagrangian and Hamiltonian mechanics is the task of the courses normally called analytical mechanics or rational mechanics of the following semester. The Lagrangian and Hamiltonian formulations are preparatory to the formulation of quantum mechanics and to the field theory of modern physics, where the fundamental concepts of Newtonian mechanics are maintained: think of the principles of conservation of energy, momentum and angular momentum, experimentally valid in any physical process. For a preview, I refer to Chap. 19 of the second volume of Feynman’s Physics, https://www.feynmanlectures.caltech.edu/II_19.html. Feynman implemented the use of concepts described in classical mechanics in his formalism of quantum mechanics. Even the revisions to the dynamic definitions required when the velocities of the objects are not negligible with respect to the speed of light .c are easily implemented. As you are introduced to special relativity and quantum mechanics, you will return to what aspects of classical mechanics need to be reviewed.

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13.8.2 Determinism in Newtonian Mechanics There is another aspect of Newtonian mechanics which is worth mentioning. Arriving in the second half of the 19th century, many believed that, with the setting of classical dynamics and the new discoveries of electromagnetic phenomena recently formalized in Maxwell’s equations, the picture of physics was complete. Not only that, all physical processes were believed to be deterministic. In this context, Newtonian determinism means that, once the initial conditions are set at .t0 , at any later time .t > t0 you will only get to a determined configuration. That is, reciprocally, given the current configuration at the instant .t, by traversing the time coordinate backwards, the initial condition at the instant .t0 can be uniquely determined. Even before the crisis triggered by the theory of relativity and the advent of quantum mechanics, Newtonian determinism as defined above presented problems. For example, H. Poincaré (1854–1912) from 1899 began to study the stability of the motion of three gravitationally interacting bodies. He concluded that the issue of stability is of a fundamental nature, particularly by observing how small differences in initial conditions can produce very different effects in phenomena. This is one of the essential characteristics of what is now called chaotic behavior, which also appears using deterministic equations such as the Newton ones. The study of the chaotic effects started in 1963 by E. N. Lorenz (1917–2008). He observed that a simplified atmospheric model shows very different numerical solutions for extremely small changes in the initial conditions. The simplified atmospheric model, derived from hydrodynamic equations based on the Newtonian formalism, consisted of three nonlinear ordinary differential equations of the first order. This is what in scientific literature is called the butterfly effect. This term includes the concept of a sensitive dependence of the final state on the initial conditions, which is the founding basis of the mathematical theory of chaos. I have repeatedly recalled that the two initial conditions, on position and velocity, are fundamental to uniquely determine the equation of motion. The solution (i.e., the path of the motion) is unique when the two initial constants are fixed: in three dimensions we need the initial position .r0 and initial velocity .v0 , to be determined by observations. However, to any observation needed for the determination of .r0 and .v0 is intrinsic an indeterminacy, what we normally call as measurement error. So we don’t actually know .r0 , but .(r0 ± Δr0 ), where the second term represents the uncertainty on the value of the initial position. And the same for the initial velocity. This implies that not a single position .r(t) is allowed after a certain period of time t based on the dynamics of the system, but a range of possible values, defined by the propagation of the initial uncertainty. Due to the nature of the interactions present in the case of systems with many bodies, even small differences in the initial conditions of a system described by coupled differential equations can produce significant differences in the final state.

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13 Considerations on Energy

13.9 Epilogue It may seem strange to have a section in a physics textbook called Epilogue. The epilogue is the concluding part of a story, which may seem fitting at the end of a novel. In this context, the epilogue is perhaps similar to the one corresponding to the conclusion of the first season of television series. Here, you have learned about the context, the characters, the places, the stories, the relationships and the mutual contrasts. At the end of the first season, there is a temporary conclusion, a situation of (partial) balance achieved between the various events and characters of the story. However, it is clear that beyond that point you have reached, there is something else that will be introduced to you in the following seasons. Here, I have presented the dynamics of the material point and of rigid bodies, the extended solid objects whose volume elements are bound by the condition (7.4). In the next “season”, you will extend your knowledge to the other two ordinary states of matter: fluids and gases. Fluids are substances, or mixtures of several substances, that have no form of their own and cannot sustain a tangential force for any appreciable time. Fluid mechanics deals with the study of fluids under different conditions in relation to its characteristics. The discipline is distinguished into fluidostatics, if the fluid is studied in a reference frame in which it appears stationary, and fluid dynamics, if the fluid is in motion. Newton began covering these topics in the second part of his Principia “The Motion of Bodies (in resisting mediums)”. The subject has evolved so much since then, including interests on applied physics, that it needs a devoted course. Gases are fluids that have no volume of their own and are composed of ideally non-interacting particles that tend to occupy all the available volume. The study of gases, on the other hand, involves the study of systems with so many degrees of freedom that it requires a different paradigm and a statistical approach. That is what you will do in the thermodynamics course. But beyond that there is much more. For example, in the penultimate paragraph I introduced complex numbers and began to describe their use in the classical treatment of elastic systems. Continuing on that subject would be of extreme interest. But you will be able to discover in the second season that electromagnetic phenomena produce electromagnetic waves, described by the mathematics we have just introduced. In the following seasons you will learn that quantum mechanics is often also called wave mechanics: the behavior of microscopic systems is described by wave functions, treated from a mathematical point of view in a similar way to what is introduced in this chapter. And towards the end of the series, you’ll discover why a class of elementary particles are called resonances: their behavior during production is formalized in a way analogous to a mechanical system forced to oscillate with a particular frequency. When you have developed the Lagrangian formulation, you will see that for each symmetry of the quantity called Lagrangian (that is, for each continuous transformation of the generalized coordinates defined in this formulation, which leaves it unchanged) a conserved quantity corresponds. And this sheds new light on the conservation laws of energy, momentum and angular momentum that we have defined. Later you will see that these quantities will become operators,

13.10 Questions and Exercises

375

i.e., mathematical objects that apply on the wave functions that describe the state of a microscopic system, which admit eigenvalues corresponding to the measurable quantities. You won’t have to wait long to have the following.seasons available, where you’ll experience new problems (relativistic speeds, divergence of physical quantities, disagreement between theory and data,…) and new characters (operators, quanta,…). Don’t think that this analogy I’m using is too trivial: when I attended the first year of Physics, I awaited the lectures of Physics I class with the same trepidation with which my children awaited the release of the final episodes of Game of Thrones. My justification is that we didn’t have television platforms at the time, but above all that the professor’s lectures were at the level of a theatrical monologue, such was the care with which he prepared and presented them. Continuing my career, however, I understood that the first season forms the structure and the architrave of the whole series, and that I was truly fortunate to have followed it with extreme attention and passion. Woe to forget the main protagonists presented here, on pain of total misunderstanding of the events that you will meet in the following seasons.

13.10 Questions and Exercises Questions 1. Determine the energy flux on Earth from the Sun, which is .150 × 106 km away, using (13.12). Taking into account that a photovoltaic panel can have an efficiency of 20%, estimate the electrical power produced by a solar panel placed in a direction perpendicular to the Sun’s rays and in the absence of clouds. [R: 1366 W/m.2 ; 270 W/m.2 , but in really optimal conditions.] 2. Using the energy scale used in the microscopic world, chemical processes release about 10 eV per elementary reaction, while a nuclear reaction releases about 10 MeV. Multiply the number of molecules per mole to get the equivalent in the nuclear case of the value given in (13.13). 3. Show that if you have a mass. M distributed of on a spherical shell, that is, on a sphere of radius . R hollow inside, and a point mass .m at a distance .h from the sphere, the gravitational potential mM . This again is equivalent to the fact that all the mass of the spherical energy is .V = −G (R+h) shell can be considered to be concentrated in the center of the sphere. 4. Show that if you have a mass . M distributed of on a spherical shell, and a point mass .m lies inside from the sphere, the gravitational potential energy is .V = −G mRM , that is, constant and not dependent on the position of .m. 5. Show that in the case described by the question 4 the point mass .m inside the sphere feels no gravitational force from the spherical shell. 6. Determine the gravitational potential .VG in the case of a homogeneous solid sphere of radius . R as a function of .r > R. 7. Making use of the result obtained in the question 3, determine the gravitational potential .VG in the case of a homogeneous solid sphere of radius . R as a function of .r ≤ R.

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13 Considerations on Energy

8. In the stellar gravitational collapse of a star of mass 12 . MΘ all mass (minus that which remains concentrated in the neutron star) is ejected from the explosion with velocity.v ∼ 10−2 c (Fig. 13.3). Determine the total kinetic energy that competes with the matter in the system after the explosion, and compare the value with (13.17). [R: .T ≃ 1044 J; .< 10−2 ΔV ] 9. Show that a necessary condition for a system to be in equilibrium under the conditions of the potential shown in Fig. 13.4, it must be at the position corresponding to the coordinates of point A, or B, or C. Are there other equilibrium points in the region of .r shown in Fig? 10. Verify that if .a = pa + iqa and .b = pb + iqb are two complex numbers, then the product of the two is the complex number. Get real and imaginary part. [R: .c = ab = ( pa pb − qa qb ) + i( pa qb + pb qa ) ] 11. Determine the values of amplitude.C and phase.φ in the relation / (13.48) of harmonic oscillator given by (13.36). The proper frequency of the system is .ωo = .v(0) = vo e . x(0) = x o .

k m

and the initial conditions

/

[R: .C =

xo2 +

vo2 ωo2

; .φ = arctan

vo ωo x o ]

12. Determine the values of amplitude.C and phase.φ in the relation / (13.48) of harmonic oscillator given by (13.36). The proper frequency of the system is.ωo = . E and the initial speed .v(0) = vo < 0.

[R: .C =

k m , the total mechanical energy

/

2E mωo2

; .φ = − arcsin vo

/

m 2E ]

13. Show that the elastic force exhibited by a bouncing ball of radius . R, assumed perfectly elastic, can be written as .F = −k(r − R)b fˆ r and that it admits potential energy .V = 21 k(r − R)2 . Make use of Cartesian coordinates. Exercise 13.1 A rod of length . D = 150 cm, mass . M = 150 g is constrained to rotate freely in the plane about one end, Fig. 13.8 left. At a distance of . D/5 from the constraint, the rod is attached to a spring of elastic constant .k = 64 N/m, fixed in the plane at the other end; the spring is at rest when the rod is in the initial position, shown in the figure (top view). A point filler ball is thrown orthogonally to the rod and at time .t = 0 hits it exactly in the center, sticking to it. The mass of the ball is .m = 0.025 kg and its speed is .v = 4 m/s. Determine:

.x y

1. the moment of inertia of the rod with respect to the constraint; 2. the angular speed .ω of the rod immediately after the impact of the ball. As a result of the impact and the presence of the spring, the rod begins to oscillate. 3. The period .T of the small oscillations of the rod; 4. the maximum angle reached by the rod; 5. Draw the graph of the coordinate.x of the ball as a function of time.t in the time interval between .t = −2T and .t = 2T, where . T is the period of the oscillations. 6. Note that following the collision, it appears that the linear momentum of the system is not conserved. This is related to the fact that the constraint exerts an impulsive force on the system. Determine what is the impulse provided by the constraint in the impact. Exercise 13.2 The CO (carbon monoxide) molecule is composed of one carbon atom (A = 12) and one oxygen atom (A = 16). The mass number A represents the sum of the number of protons and neutrons in the nucleus. Proton and neutron have mass that we assume to be identical, equal to .m p = 1.67 × 10−27 kg. The electron mass is negligible (.m e = 1/1840 m p ). In equilibrium, the distance .r between the two C-O nuclei takes the value .r0 = 1.1 × 10−10 m. The nuclei can be regarded as point-like, since their size is negligible with respect to .r0 .

13.10 Questions and Exercises

377

Fig. 13.8 Figure for Exercise 13.1

1. Determine the position of the center of mass of the two-body CO system when in equilibrium. Use a drawing to define the reference frame. When stimulated, the molecule can vibrate, varying the distance .r between the two nuclei. The molecule opposes the solicitation with a force that can approximate as an elastic force (as with a “spring”), with an elastic constant equal to .k, and with mass subject to acceleration equal to the reduced mass of the CO molecule (see question 7. below). It is observed that the molecule vibrates with angular frequency .ω = 0.6 × 1015 rad/s (measured as radiation in microwaves). 2. Determine the value of the reduced mass of the system. 3. Determine the value of the constant .k. It is known that the dissociation energy of CO gas amounts to 1062 kJ/mole, where 1 mol is equivalent to .6.02 × 1023 molecules of CO. Assuming that there is a single “spring” that binds C-O: 4. Determine the energy to dissociate a single CO molecule. In our mechanical equivalent, this means stretching the “spring” too far and breaking it. 5. By how much does the “spring” have to be stretched percentage-wise relative to the distance at equilibrium .r0 to break? 6. Prove the above, namely, that if the force opposing the change in the CO system grows linearly with .r , then the equation describing the system can be approximated precisely by the equation: d2r .μ 2 = −k(r − r 0 ). dt Exercise 13.3 In the device shown in the Fig. 8.15 the bumper of mass . M = 26.2 g is subject, in addition to the action of the spring of spring constant .k = 3.40 N/m, also to a braking frictional force with magnitude. FA = −bV , which is proportionally opposed to the speed of. M with constant −3 kg/s. If the initial speed of the bumper is . V = 1.73 m/s: .b = 6.0 × 10 0 1. show that the amplitude. A of the oscillations decreases with time as with exponential attenuation . A(t) = A0 exp(−t/τ ) 2. determine .τ and draw the function . A(t); 3. determine after how much time the amplitude decreases by 90% from the maximum.

Appendix: Numerical Solution of the Exercises

Answer of Exercise 1.1 (1) .λ = 3.84 × 10−12 s.−1 ; (3) . A .= 0.22 Bq (4) 2350 y (5) 0.024% Answer of Exercise 3.1 (3) .vm = 2.33 m/s; (5) .ρ = 2.83 m Answer of Exercise 3.2 (2) . R = 26.8 m; (3) Unfortunately for the player, no (height 2.8 m) (4) .v1 = 17.5 m/s; (5) Yes: the ball would pass 2.9 meters high, going over the barrier. Answer of Exercise 3.3 (2).(−1.68, 3.07) m; (3).tC .= 3.95 s; (4).vm .= 4.18 m/s; (5).ac = (6.36ˆi, 45.1ˆj) m/s.2 . Answer of Exercise 4.1 (1) .T = 1.8 N; (2) . R = 0.34 m; (3) .α = 58.5◦ ; (4) .ωmax = 6.12 rad/s Answer of Exercise 4.2 (2) .θs .= 0.681 rad; . F .= 238 N; (3) . F1 .= 180 N; (4) Increased (156 N rather than 139 N) Answer of Exercise 4.3 (2) 3.16 m; (4) Magnitude of frictional force = 8 N. Answer of Exercise 4.4 (1) Tension= 4.80 N; . FA .= 1.75 N; (2) .μs .= 0.30; (3) .v = 0.53 m/s Answer of Exercise 4.5 (1) .a = 1.21 m/s.2 ; (2) .v = 4.0 m/s; (3) .μ∗s .= 0.123; (4) .s = 3.18 m. Answer of Exercise 4.6 (1) . Mmin .= 0.033 kg; . Mmax .= 0.47 kg; (2) .T .= 6.04 N; (3) 2.75 m/s; (4) .t .= 0.73 s. Answer of Exercise 4.7 (1) .T1 = 8.8 N; (2) . L eq 0.28 m; (4) . L max .= 0.46 m; .T1 =19.6 N; (5) .vmax = 1.98 m/s.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Spurio, The Fundamentals of Newtonian Mechanics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-47289-3

379

380

Appendix: Numerical Solution of the Exercises

Answer of Exercise 4.8 (1) 0.28 m; (2) .vxA = 0.80 m/s; .v yA .= –3.43 m/s; (3) 0.041 N; (4) 10.7 m/s.2 ; (5) –0.55 from the surface of the water. It does not touch the bottom. Answer of Exercise 5.1 (1) 2.07 m/s = 7.46 km/h; (3) v=0 when it is in contact. Answer of Exercise 5.2 (1) 3.0 m/s; (2) 4.24 m/s; (3) 0.82 m Answer of Exercise 6.1 (1) .μd .= 0.66; (2) .V P .= 4.87 m/s; (3) .V P (3.0; –3.84) m/s; (4) . O P = 1.17 m. Answer of Exercise 6.2 (1) .r2 = 0.317 m; (2) .v B = 3.05 m/s; (3) .v E = 2.49 m/s; (4) 0.162 N. Answer of Exercise 6.3 (2) . D = 7.37 m; (4) .t ∗ .= 2.90 s Answer of Exercise 6.4 (2) 0.35 m; (4) 0.175 s; (5) . L 1 = 2L; .t1∗ = t ∗ (same time). Answer of Exercise 6.5 (1) .T .= 1.42 s; (2) .v0 = 0.23 m/s; .t = T /4 .= 0.35 s (3) .v0 = 4.5 m/s; (4) 2 2 .ac = 6.93 m/s. ; .a T = 6.93 m/s. . Answer of Exercise 6.6 (1) .v0 = 2.6 × 106 m/s, less than 1% c; . F = 1.6 × 10−16 N; (2) 78 ns; (3) W=0; (4) .v A = v0 . Answer of Exercise 7.1 (1) 3000 kg/m.3 ; (2) . F = 0.42 N; (3) .τ = 0; (4) .θ = 23.6o = 0.41 rad. Answer of Exercise 7.2 (1) .v1 = 10.5 m/s; (2) .l = 3.50 m; (4) .T .= 779 N. Answer of Exercise 7.3 (2) .τz = 0; (4) .vmin .= 1.9 m/s Answer of Exercise 7.4 (2) .ω0 .= 7.0 rad/s; (3) W=19.8 J; (5) .t ∗ .= 0.2 s Answer of Exercise 7.5 (1) .v0 .= 70 m/s (2) .v1,x .= –35.0 m/s; .v2,x .= 11.6 m/s; other components are null; (3) 6640 J; (4) 333 m. Answer of Exercise 8.1 (1) .θ2 = 36.6◦ ; (2) 6.82 kg m/s; (3) .k = 1.45 105 N/m; (4) Dissipated . E = 9.37 J Answer of Exercise 8.2 (1) .Vsled = 2.31 m/s, .Vm .= –7.69 m/s; (2) Velocities change sign relative to point 1); (3) Objects return to quiet. Answer of Exercise 8.3 (1) 3.13 m/s outward, along x-axis, –1.40 m/s return, same axis; (2) . M .= 26.2 g; (3) .V .= 1.73 m/s; (4) . Ao .= 15.2 cm. Answer of Exercise 8.4 (1) .v f .= –10.8 m/s; (2) .Vcm .= 3.6 m/s; (3) . K .= 1.04 J; (4) . K ’ = 0.52 J; (5) 5.8 cm; (6) T = 0.051 s;

Appendix: Numerical Solution of the Exercises

381

Answer of Exercise 8.5 (1) .ΔE .= 12.0 J; (2) .vcm .= 1.20 m/s; (3) .Δx .= 1.55 cm; (4) .ω = 129 s.−1 . Answer of Exercise 8.6 (1) .v0 = 8.21 m/s; (2) . J = −1.194 N s; (3) . K s = 6.17 J; (4) .ΔE = 0.575 J. Answer of Exercise 11.1 (1) .gG = 25.94 m/s.2 ; (2) 1330 kg/m.3 ; (3) .TG .= 7.15 days; (4) .v f .= 60.2 km/s Answer of Exercise 11.2 (1) The total energy . E T = −4.46 × 1010 J is negative; (2) .va = 5.63 m/s; (3) y2 x2 + 78.22 = 1012 . Equation of the ellipse (.x, y expressed in m): . 80.46 Answer of Exercise 11.3 (1) 10.4 km/s; (2) .v0 = 3.15 km/s; (4) .r1 = r0 /7; (5) 11.0 km/s Answer of Exercise 11.4 (1) .2.3 × 1017 kg/m.3 ; . R N S = 13 km; (2) .μ = 1.5 × 1030 kg; (3) .2.5RΘ ; (4) 4 6 .1.5 × 10 4 J; (5) .650 10 years Answer of Exercise 11.5 (1) .g S = 260 m/s.2 ; (2) . MΘ /M SS .= 99.84% (3) . L Θ = 9.4 1041 kg m.2 s.−1 (4) 43 . L SS = 3.2 10 kg m.2 s.−1 Answer of Exercise 12.1 (1) 64.8 rad/s; (2) 0.735 J; (3) 0.89 m (along x); (4) 1.72 J; (5) 0.33 J. Answer of Exercise 12.2 (1) 0.42 m; (2) J = 0.037 kg m.2 ; (3) .ω 8.2 rad/s; (4) .ω' .= 14.1 rad/s; Answer of Exercise 12.3 (2) .v2 .= 100 m/s; (3) .θmax = 0.56 rad; (4) .ΔE .= 1600 J Answer of Exercise 12.4 (1) 0.198 kg m.2 /s; (2) 22.2 rad/s; (3) –2.01 J; (4) 2.62 N . Answer of Exercise 12.5 (1) 3.77 m/s; (2) V = 2.41 m/s, .ω .= 2.05 rad/s or V = 3.25 m/s, .ω .= 0.78 rad/s. Only the first solution is physically possible. (4) .θmax .= 0.77 rad Answer of Exercise 12.6 (1) .vcm = 3.46 m/s; (2) . J P = 3.0 × 10−2 kg m.2 ; (3) .ω = 23.1 rad/s; (4) .d A .= 0.33 m Answer of Exercise 12.7 (1) 0.0332 kg m.2 ; (2) 21.5 cm; (3) 1.50 rad/s; (4) on .x: –0.032 N s; on . y: –0.200 N s. Answer of Exercise 12.8 (1) 0.050 kg m.2 ; (2) 14 rad/s.2 ; (3) .a2 .= –1.4 m/s.2 (rises); (4) .m 2 .= 0.63 kg. Answer of Exercise 12.9 (1) . J = 8.16 × 10−2 kg m.2 ; (2) . K = 5.87 J; (3) sum forces is zero (4) . L || = 0.98 kg m/s; . L ⊥ = 1.22 kg m/s. Answer of Exercise 12.10 (1) .v .= 1.64 m/s, .ω .= 20.5 rad/s; (2) 2.7 m/s.2 ; (3) 13.5 N; (4) null work

382

Appendix: Numerical Solution of the Exercises

Answer of Exercise 12.11 (1) .T .= 1.00 N; (3) 11.4 rad/s; (4) 0.041 J. Answer of Exercise 12.12 (1) .v D .= 6.5 m/s; (3) .a = 6.1 m/s.2 ; (4) .μd = 0.78 . Answer of Exercise 12.13 (1) D=.(0, −R/2), .d = 25 cm; (2) .v D .= 1.5 m/s; (3) .ω = 2.0 rad/s; (4) 1/3 . Answer of Exercise 12.14 (1) .ω0 .= 0.60 rad/s; (2) . J .= –0.11 N s; (3) .ω1 = 2.60 rad/s. Answer of Exercise 12.15 (1) .T = 12.3 N direction opposite to weight; (2) . Q o = T ; (3) .ω .= 9.33 rad/s; (4) . Q 'o .= 57.2 N. Answer of Exercise 12.16 (1) 0.0239 kg m.2 ; (2) .ω1 .= 4.41 rad/s; (3) .ω2 =3.25 rad/s; (4) .vr .= 1.44 m/s. Answer of Exercise 13.1 (1) 0.113 kg m.2 ; (2) .ω .= 0.593 rad/s; (3) .T .= 0.93 s; (4) .θmax .= 0.088 rad; (6) –0.022 kg m/s Answer of Exercise 13.2 (1) along the line that connects CO a .d = 0.57r0 = 6.3 × 10−11 m from C; (2) −26 .μ = 1.15 × 10 kg; (3).k = 4.0 × 103 N/m; (4). E = 1.77 × 10−18 J; (5) 30%.

Index

A Absolute space, 12 Absolute time, 12, 14 Acceleration, 60, 93 average, 54 scalar, 54 tangential, 68 Accessibility, 5 Action-reaction law, 185 Active Galactic Nucleus (AGN), 259 Activity, 21 Adams J.C., 303 Age of the Sun, 350 Alchemy, 1 Alpha decay, 210, 211, 215–217 Alpha-spectroscopy, 215 Analytical mechanics, 372 Angular frequency, 65, 102 Angular momentum, 178, 182, 197, 280, 302, 312 Angular speed, 65 Angular velocity, 70, 71, 236, 312, 328 Anthropomorphic work, 145 Aphelion, 291 Apoastron, 286 Apparent force, 14 Approximate solution, 100 Area, 40 Arecibo radio telescope, 299 Aristarchus, 244 Aristotelian system, 273 Aristotle, 241 De coelo, 274 Physics, 274 Arrow of time, 107

2001: A Space Odyssey, 125 Atom, 7 Atomic number, 7 Avogadro’s constant, 24 Axial vector, 235 Axiom, 44 B BAC-CAB, 39, 327 Ballistic pendulum, 220 Becquerel H., 210 Beta decay, 158, 210, 216, 217 Bethe H., 351 Big Bang theory, 14, 306 Binding forces, 105 Black hole, 258 Bohr N., 218, 255 Bose S.N., 196 Boson, 196 Boundary conditions, 64 Bounded regions, 362 Bound spherical system, 348 Brahe T., 275, 353 De Nova Stella, 276 Bruno G., 274 Bulge, 305 Butterfly effect, 373 C Calorific value of hydrocarbons, 350 Cardinal equations, 177 Cartesian coordinate system, 28, 48, 168 Cavendish experiment, 326 Cavendish H., 260

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 M. Spurio, The Fundamentals of Newtonian Mechanics, Undergraduate Lecture Notes in Physics, https://doi.org/10.1007/978-3-031-47289-3

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384 Center of gravity, 323 Center of mass, 169, 170, 311, 323 frame, 186, 204 IInd theorem, 181 Ist theorem, 181 Central force, 139, 155, 247 Centrifugal force, 99, 123, 126 Centrifugal potential, 285 Centripetal acceleration, 67 CERN, 197 Chadwick J., 201 Chandra satellite, 353 Characteristic algebraic equation, 367 Chemical elements, 217 Chemistry, 1 Circular motion non-uniform, 67 uniform, 65 Circular orbit, 288 Coalescence, 301 COBE experiment, 118 Coefficient of restitution, 220 Cohesion, 105 Collisions, 195, 210 elastic, 196 Color charge, 106 Column vector, 229 Coma cluster, 305 Comoving coordinates, 14 Comoving observer, 14 Comoving unit vector, 313 Complex conjugate, 364 Complex numbers, 363 Composition of velocities, 116 Configuration energy, 347 Conic functions, 289 Conservation of mechanical energy, 149 Conservation laws, 180, 210 Conservative forces, 146, 157 Constant speed, 49 Constraining forces, 202 Constraint reaction, 88 Constraints, 168 Contact forces, 85, 104, 106 Continuous body, 172 Continuous systems, 166, 168 Coordinate system, 28, 47, 67 Copernicus N., 275 De revolutionibus, 275 Coriolis force, 123, 128, 129 Correspondence principle, 255 Cosmic background radiation, 14, 118

Index Cosmic rays, 16 Cosmological principle, 226 Cosmological time, 14 Coulomb C.A. de, 105 Coulomb force, 204, 249 Coulomb potential, 288 Covariance for rotation, 226 Covariant transformation, 231 Creation of particles, 354 Cross product, 39 Cross section, 9, 196 Curl, 153 Curvature parameter, 11 Curvature radius, 74 Cylindrical coordinate system, 69 D Damped harmonic oscillator, 366 Damping, 103 Damping time constant, 371 Dark matter, 249, 304 Darwin C.R., 351 Decay processes, 210 Deferent, 274 Degeneracy, 295 Degrees of freedom, 168 Density, 166 Descartes R., 1 Diagonalizable matrix, 329 Differential equation, 19 Differential operators, 152 Differentiation rules, 60 Dimensional analysis, 22 Dipole anisotropy, 118 Direct kinematics problem, 60 Directrix, 289 Displacement, 27, 34, 225 differential, 61 Dissemination, 15 Dissipated energy, 219 Dissipative force, 150, 158, 195 Distance, 5 Distance ladder, 9 Divergence, 152 Doppler effect, 8, 306 Dot product, 37 Drag acceleration, 124 Dynamic sliding friction, 91 Dynamics, 14, 79 in a rotating frame, 125 Dynamometer, 82, 85, 225, 260

Index E Earth density, 167 meridian, 127 parallel, 127 precessional motion, 335 reference frame, 128 Earth-Moon distance, 244 Earth’s drag acceleration, 126 Earth’s radius, 242, 243 Earth-Sun distance, 244 Eccentricity, 289, 292, 295 Editor, 2 Einstein A., 13, 104, 134, 196, 212, 304 Elastic collision, 196, 199, 203, 220 Elastic system, 361 Electric charge, 106 Electric field, 105, 106, 356 Electric potential, 356 Electromagnetic interaction, 104 Electromagnetic radiation, 319 Electromagnetic wave, 356 Electromagnetism, 104, 148, 152 Electroweak interaction, 105 Elementary particle, 7, 196, 219 Ellipse, 290 Elliptical orbit, 286, 288, 294 Empirical law, 50 Energy forms, 158 Energy of the system configuration, 347 Energy spectrum, 218 Eötvös L., 251 Epicycle, 274 Equation of motion, 48, 48, 53, 57 Equatorial plane, 127 Equilibrium, 359 indifferent, 360 of a rigid body, 190 stable, 360 unstable, 360 Eratosthenes, 242 Error propagation, 373 Escape speed, 255, 288 Euclidean geometry, 11, 31 Euclidean space, 11, 28 Eudoxus of Knidos, 273 Euler first law, 180, 181 identity, 364 matrix for rotations, 230 second law, 183, 315 Event duration, 14 Event horizon, 258

385 Event Horizon Telescope (EHT), 259 Exact differential, 146, 151 Exact solution, 100 Experimental method, 55 External forces, 179 External torque, 183

F Fermi E., 219 Feynman R., 92, 152, 372 diagrams, 200 Lectures on Physics, 92 Field theory, 137 First integral, 290, 357 Fluid dynamics, 168 Force, 80, 81, 93, 137 as gradient of potential energy, 151 central, 155 centrifugal, 123 constraint, 88 Coriolis, 123 friction, 89 impulsive, 197 vector nature, 86 Force acting at a distance, 85 Force fields, 113, 138 classical, 106 Forms of energy, 158 Foucault’s pendulum, 128 Fourier analysis, 372 Frame of reference, 47, 116 Frequency, 102 Friction, 55, 89 Frictional force, 333 Fundamental interactions, 104, 137

G Galactic black hole, 298 Galilean method, 55 Galilean relativity, 116 Galilean relativity principle, 113 Galilean transformations, 115 Galilei G., 1, 13, 47, 49, 55, 79, 226, 250 Dialogo, 55, 113 Discorsi, 56 Sidereus Nuncius, 55 Generalized principles of relativity, 133 General principle of relativity, 134 General relativity, 16, 104, 137, 249 Geocentric model, 274 Geodesic trajectory, 14

386 Global Positioning System (GPS), 16 Gradient, 152, 252 Gradient of potential energy, 151 Gravitational collapse, 352 Gravitational effective potential, 285 Gravitational field, 249 Gravitational interaction, 104, 179 Gravitational mass, 250 Gravitational potential, 356 Gravitational potential energy, 252, 266 Gravitational waves, 2, 159, 197, 300, 356 Gravitation constant G, 248 Guglielmini’s experiment, 128

H Half-life, 215 Harmonic motion, 362 Harmonic oscillator, 103 dumping, 366 Heat, 196 Helicity, 239 Heliocentric system, 275 Hellenistic physics, 242 Herschel W., 302 Higgs boson, 7, 131, 196 Homogeneity, 226 Hooke R., 85, 241 Hooke’s law, 84 Hubble Space Telescope, 353 Hulse-Taylor binary system, 301 Huygens-Steiner theorem, 321 Hydrocarbons energy, 350 Hyperbola, 290 Hyperbolic orbit, 287 Hysteresis, 83

I Imaginary part, 363 Impact parameter, 287 Impulse, 50, 199 Impulsive force, 197, 200 Inalterability, 5 Indirect observation, 301 Inelastic collisions, 196 Inertial frame of reference, 117 Inertial mass, 250 Inertial reference frame, 186 Inertia tensor, 329 Inextensible rope, 86 Infinitesimal area, 263

Index path, 263 volume, 263 Infinitesimal displacement, 58, 139, 263 Infinitesimal path, 61 Infinitesimal triangle, 282 Infinitesimal volume, 166, 314 Information trasmission, 185 Initial conditions, 116 Initial constants, 63, 64 Internal energy, 205 Internal forces, 179, 185 International System of units, 4 Intrinsic angular momentum, 188 Invariance for translation, 226 Inverse kinematics problem, 62 Inverse problem, 303 Irrotational vector field, 154 Isolated system, 179, 183, 184 Isotropy, 226

J James Webb telescope, 257 Joule J.P., 140 Joule unit, 140

K Kepler, 276 1st law, 281 2nd law, 282 3rd law, 283, 296 Astronomia nova, 276 Harmonices Mundi, 276 laws, 276 Kilogram, 6 Kinematics, 14, 33, 47 Kinetic energy, 141, 189, 201 in C-M. system, 190 KM3NeT detector, 18 König theorem for kinetic energy, 190

L Laboratory reference frame, 205 Laplacian operator, 153 Large Hadron Collider (LHC), 197 Laser, 213 Latitude, 262 Law of elastic force, 85 Lee T.D., 239 Leibniz G.W. von, 242 Le Verrier U., 303 Limit angle, 90

Index Linear density, 167 Linear momentum, 93, 177, 199, 206 Linear transformation, 329 Line integrals, 138 Live forces theorem, 141 Lorentz H.A., 13 transformations, 134 Luther M., 274 M Magnetic field, 105, 106 Many bodies system, 186 Mass, 3, 6, 80, 158, 204 additive property, 165 creation, 213 gravitational, 250 inertial, 93, 250 variable, 208 Mass-energy, 212 Mass number, 7 Material point, 47 Matrix of rotation, 230 Maxwell equations, 212 Maxwell J.C., 13, 212 Mean lifetime, 19 Measurement direct, 8 error, 373 indirect, 8 Mechanical energy, 149, 288 conservation, 149 Mechanical equivalent of heat, 140 Medicean stars, 55 Mercury orbit, 303 Meridian, 263 Messier catalog, 259 Meter, 5, 5 Metrology, 5 Minkowski H., 13 Mirror reflection, 237 Mirror symmetry, 233 MOdified Newtonian Dynamics, MOND, 249 Mole, 24 Moment of force, 175 Moment of inertia, 314, 315 Moment of inertia tensor, 327 Monochromatic, 213 Monoenergetic process, 213 Moon centripetal acceleration, 246 Mutationem motus, 93

387 N Nabla operator, 152 Nature, 1 Neutrino, 18, 105, 159, 219, 354 Neutron, 105, 201 Neutron star, 258, 280, 299, 318, 354 Newton, I., 1, 13, 80, 159, 180, 226, 242 De Quadratura, 81 determinism, 373 Opticks, 81 optics, 248 Principia, 12, 79, 80, 92, 183, 241, 250, 374 Noether, 180 theorem, 183 Non-conservative force, 147 Non-Euclidean geometries, 11 Non-inertial accelerations, 122 Non-inertial forces, 123 Non-inertial frames of reference, 119 Nuclear fusion, 299, 351 Nuclear radioactivity, 210 Nuclei, 217 chart, 217 O Observational cosmology, 12, 226, 274 Osculating circle, 73, 74 Osculating plane, 73 P Parabola, 290 Parallel, 263 Parallelogram rule, 35 Parity, 236 Partial derivatives, 138, 151, 153 Partially elastic collisions, 219 Particle, 47 Particle accelerators, 213 Particle physics, 196 Path in space, 48 Pauli W., 219 Peer-review, 2 Periastron, 286 Perihelion, 291 Period, 102 Periodic phenomenon, 6 Perturbation, 302 Perturbative theory, 278 Phase factor, 101 Photomultiplier (PMT), 18 Physical dimension, 4

388 Physical pendulum, 325 Physical principle, 180, 183 Physics, 1 Pierre Auger Observatory (PAO), 17 Pitch, 328 Pivot point, 176, 178, 183, 188, 221 Planck constant, 26, 178, 196, 255 length, 26 mass, 26 time, 26 Planck satellite, 118 Plane orbits, 281 Plane polar coordinates, 282 Plumb line, 129 Poincaré H., 373 Poincaré J.H., 13 Poisson’s rules, 72, 122 Polar angle, 262 Polar coordinates, 168 Polar coordinate system, 69 Polar plane coordinates, 68 Polar vector, 235 Polarity, 236 Position, 33 Positron, 217 Postulate, 44 Potential energy, 148, 345 gradient, 151 Potential energy of spherical system, 348 Potential well, 362 Power, 157 Practical units, 7 Precession, 303, 333 Precision, 5 Preliminary reference Earth model (PREM), 167 Primitive function, 146 Principal axes, 329 Principal moments of inertia, 329 Principle angular momentum conservation, 45,183 energy conservation, 45 momentum conservation, 45, 180 of conservation of energy, 158 of equivalence, 45, 252 of general relativity, 134 of homogeneity of the universe, 45 of inertia, 45, 81, 116, 119 of isotropy of the universe, 45 of relativity, 45, 133 of special relativity, 133 Principle in physics, 45

Index Proton, 105 Pseudo-accelerations, 122 Pseudo-forces, 123 Pseudoscalar, 238 Pseudo-sciences, 1 Pseudo-vectors, 235 Ptolemaic system, 273 Ptolemy Almagestus, 274 Pulsar, 299 Pythagorean theorem, 11

Q Quantum ChromoDynamics (QCD), 105 Quantum ElectroDynamics (QED), 105 Quantum mechanics, 137, 186, 215, 329 Quantum numbers, 255 Quark, 7, 105, 106, 195

R Radial kinetic energy, 284 Radioactive decay, 17 Radioactive nuclei, 217 Range, 62 Rational mechanics, 372 Real part, 363 Referee, 2 Reference frame, 28, 47 inertial, 117 Reference points, 47 Reflection, 232 Relativistic dynamics, 45 Relativistic physics, 186 Reproducibility, 5 Residual forces, 105 Resonance, 372 Restoring force, 84 Restricted principle of relativity, 133 Reversibility, 107 Right-handed coordinate system, 29, 227 Right-hand rule, 30, 227 Rigid bodiy, 168 Rigidity constraint, 168 Rigidity relation, 311 Rocket motion, 207 Rolling friction, 92, 333 Roll, 328 Roll without sliding, 330 Rotating reference frame, 121 Rotation, 189, 225, 226 Rotational kinetic energy, 330

Index Rotation curve of stars, 305 Rotations, 190, 312 Rototraslations, 312 Ruler, 9 Rutherford scattering, 195, 204 Rutherford E., 210 S Scalar, 238 Scalar product, 37, 139, 235 Scalar quantity, 27 Scattering, 195 Scholasticism, 274 Schwarz’s theorem, 153, 157 Scientific method, 1 Second, 6, 6 Separation of variables, 97 Simple pendulum, 98 Sliding friction, 90 Solar system, 179, 277 Spacetime, 13, 104 Speed, 53, 59 average, 53 Speed of light, 13 Spherical coordinates, 167, 168, 262, 312, 347 Spin, 188, 196 Spiral galaxies, 302 Spitzer Space Telescope, 353 Spring, 82 Standard Model (SM), 105 Static, 90 Static friction, 330 Static frictional force, 333 Stellar black holes, 258 Stellar parallax, 9 Strong interaction, 105 Subnuclear physics, 195 Sun age, 350 Super-massive black holes (SMBH), 258 Supernova, 276, 353 Superposition principle, 87, 356 Surface density, 167 Symmetry, 180 Synchronization, 15 Système International, 4 Systems of material points, 165 T Taylor development, 254 Taylor expansion, 23

389 Template, 301 Tension, 86 Tensor, 27, 327 Theorem of parallel axes, 321 Thermodynamics, 107, 158, 168 Thomson W. (lord Kelvin), 351 Time, 6 Time measurement, 17 Time reversal, 107, 210 Torque, 170, 175, 178, 312 external, 182 internal, 182 Torsion balance, 260 Torsion pendulum, 251, 326 Total differential, 146 Total linear momentum, 177 Totally inelastic collision, 196 Total mechanical energy, 149 Trade winds, 132 Trajectory, 48, 61, 139 Translation, 189, 225–227, 312 Triangle rule, 34, 35 Tribology, 89 Tunneling effect, 362 Two-body decay, 211, 215 Two-body system, 278, 284

U Uniform accelerated motion, 50, 55 Uniform rectilinear motion, 50, 55 Unit vector, 32 Cartesian, 33 co-moving, 33, 68 Unperturbed motion, 302 Upper limits, 251 Uranus discovery, 302

V Vacuum, 207 Vacuum chamber, 50 Van der Waals forces, 105 Variable separation, 19 Vector, 27 area, 40 axial, 235 Cartesian representation, 32 difference, 35 direction, 32 equipollent, 31, 35 homogeneous, 35 intensity, 32

390

Index

intrinsic representation, 30 magnitude, 32 polar, 235 sum, 35 Vector field, 106 irrotational, 154 Vector product, 38, 39 Vector space, 27 Velocity, 59, 59 Velocity of light, 212 Virial theorem, 288 Viscous friction, 92 Volume, 40 Volume density, 166

Wheel motion, 332 White Rabbit, 16 Wind tunnels, 96 WMAP satellite, 118 Work, 138, 345 anthropomorphic, 145 Cartesian coordinates, 139 of dynamic friction, 143, 150 of elastic force, 143 of weight force, 142 on a closed path, 148 Work-energy theorem, 141 Workforce, 145 Wu C.S., 240

W Watts, J., 157 units, 157 Wave function, 362 Weak interaction, 105 Weight, 82 Weight force, 84 Well of potential, 362

Y Yang C.N., 239 Yaw, 328

Z Zenith, 243 Zwicky F., 305