Classical newtonian gravity 978-3-030-25846-7

Table of contents :
Preface......Page 7
Introduction......Page 9
Contents......Page 10
Symbols......Page 13
1.1 Vector Functions of Real Variables......Page 15
1.2 Limits of Vector Functions......Page 16
1.3 Derivatives of Vector Functions......Page 17
1.3.1 Geometric Interpretation......Page 18
1.4 Integration of Vector Functions......Page 21
1.5.1 Gradient and in Spherical Polar Coordinates......Page 23
1.5.2 Gradient and in Cylindrical Coordinates......Page 26
1.6 The Divergence Operator......Page 28
1.7 The Curl......Page 32
1.8.1 Spherical Polar Coordinates......Page 36
1.8.2 Cylindrical Coordinates......Page 37
1.9 Vector Fields......Page 38
1.9.1 Lines of Force of a Vector Field......Page 39
1.10 The Divergence Theorem......Page 42
1.10.1 The Stokes' Theorem......Page 46
1.11 Velocity Fields......Page 54
1.12 Meaning of the Divergence of a Vector Field......Page 57
1.13 Link Between Divergence and Volume Variation......Page 58
1.14.1 Flux Tubes......Page 60
1.15 Further Readings......Page 61
2.1 Single Particle Gravitational Potential......Page 62
2.2 Motion of a Particle Gravitating in a Resisting Medium......Page 63
2.3 The Gravitational N-Body Case......Page 65
2.3.1 Potential of N Gravitating Bodies......Page 67
2.4 The Scalar Virial Theorem......Page 68
2.4.1 Consequences of the Virial Theorem......Page 70
2.5 Continuous Distributions of Matter......Page 72
2.5.1 Poisson's and Laplace's Equations......Page 73
2.6 Gauss' Theorem......Page 75
2.7 Gravitational Potential Energy......Page 77
2.8 Newton's Theorems......Page 80
2.9 Further Readings......Page 83
3.1 The Potential and Force Generated by a Spherical Matter Distribution......Page 84
3.1.1 Calculating Spherical Potentials via Poisson's Equation......Page 86
3.2 Motion in a Spherical Potential......Page 88
3.2.1 Circular Trajectories......Page 89
3.3 Potential Generated by a Homogeneous Sphere......Page 90
3.3.1 Trajectories in a Homogeneous Sphere......Page 91
3.3.2 Radial Motion in a Homogeneous Sphere......Page 93
3.4.1 The Plummer Sphere......Page 94
3.4.2 The Isochrone Potential......Page 97
3.5 Quality of Motion......Page 98
3.5.1 The Keplerian Case......Page 99
3.5.2 The Homogeneous Sphere Case......Page 100
3.6 Periods of Oscillations......Page 104
3.6.1 Radial Period in the Keplerian Potential......Page 108
3.6.2 Radial Period in the Homogeneous Sphere Potential......Page 110
3.6.3 Radial Period in the Plummer Potential......Page 112
3.6.4 Radial Period in the Isochrone Potential......Page 114
3.7 Azimuthal Period......Page 115
3.7.1 Fully Periodic Motion......Page 116
3.8 The Inverse Problem in a Central Force Field......Page 117
3.8.1 From Elliptic Trajectories to their Central Force Field......Page 118
3.9 Further Readings......Page 120
4.1 Fundamental Solution of Laplace's Equation......Page 121
4.2 Harmonic Functions......Page 123
4.3 Legendre's Polynomials......Page 131
4.3.1 Some Properties of Legendre's Polynomials......Page 133
4.4.2 Second Recurrence Relation......Page 134
4.5 The Legendre's Differential Equation......Page 136
4.6 Orthogonality of Legendre's Polynomials Pm(u)......Page 137
4.7 Development of a Function in Series of Legendre's Polynomials......Page 139
4.8 Rodrigues' Formula......Page 141
4.9 Harmonic and Homogeneous Polynomials......Page 144
4.10 Surface Spherical Harmonics......Page 148
4.11 Solution of the Differential Equation for Sm(θ,)......Page 149
4.12 The Solution in : Pm(p)(cos)......Page 151
4.13 Note on the Associated Legendre's Equation......Page 153
4.14 Zonal, Tesseral and Sectorial Spherical Harmonics......Page 154
4.15 Orthogonality of Surface Spherical Harmonics......Page 159
4.16 Developments in Terms of Spherical Harmonics......Page 161
4.17 Development of the Potential Generated by a Regular Mass Distribution......Page 162
4.17.1 External Potential Series Development......Page 164
4.18 Further Readings......Page 175
5.1 Approximate Axisymmetry of the Earth Potential......Page 176
5.2 Satellite Motion......Page 178
5.3 A Short Note on the Perturbation Theory: Osculating Elements......Page 179
5.4 Search for the Time-Dependent Elliptic Elements......Page 180
5.5 Further Readings......Page 181
A Two- and N-Body Adimensional Motion Equations......Page 182
B Two Body Regularization......Page 184
BookmarkTitle:......Page 186

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UNITEXT for Physics

Roberto A. Capuzzo Dolcetta

Classical Newtonian Gravity A Comprehensive Introduction, with Examples and Exercises

UNITEXT for Physics Series Editors Michele Cini, University of Rome Tor Vergata, Roma, Italy Attilio Ferrari, University of Turin, Turin, Italy Stefano Forte, University of Milan, Milan, Italy Guido Montagna, University of Pavia, Pavia, Italy Oreste Nicrosini, University of Pavia, Pavia, Italy Luca Peliti, University of Napoli, Naples, Italy Alberto Rotondi, Pavia, Italy Paolo Biscari, Politecnico di Milano, Milan, Italy Nicola Manini, University of Milan, Milan, Italy Morten Hjorth-Jensen, University of Oslo, Oslo, Norway

UNITEXT for Physics series, formerly UNITEXT Collana di Fisica e Astronomia, publishes textbooks and monographs in Physics and Astronomy, mainly in English language, characterized of a didactic style and comprehensiveness. The books published in UNITEXT for Physics series are addressed to graduate and advanced graduate students, but also to scientists and researchers as important resources for their education, knowledge and teaching.

More information about this series at http://www.springer.com/series/13351

Roberto A. Capuzzo Dolcetta

Classical Newtonian Gravity A Comprehensive Introduction, with Examples and Exercises

123

Roberto A. Capuzzo Dolcetta Department of Physics Sapienza University of Rome Rome, Italy

ISSN 2198-7882 ISSN 2198-7890 (electronic) UNITEXT for Physics ISBN 978-3-030-25845-0 ISBN 978-3-030-25846-7 (eBook) https://doi.org/10.1007/978-3-030-25846-7 © Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, 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

This book is dedicated to my wife Daniela and my son Flavio, both essential ingredients of my life.

Preface

The scope of this book is to put on paper many things which, although basic, are not easily found in a single text, and are not always trivial to understand and manage. Actually, this book intends giving an easy, but complete, introduction to classical Newtonian theory of gravitation, as a basic piece of Physics of particular relevance for Astrophysics. It is known, indeed, that among the four fundamental forces of Physics, gravity has the unique aspect of being an unscreened force, which permeates the whole universe at all scales. On the other side, in addition to the above property (which is often referred to as infrared divergence), Newtonian interaction among individual point masses is characterized, also, by the ultraviolet divergence: the force exerted by one point mass on another one is unlimited when the two bodies approach too closely each other. These two characteristics of Newtonian gravitation are not real issues for limited, in size, and more regular (extended objects instead of point masses) distributions of matter. Another relevant characteristic, and conceptual, limitation of the purely Newtonian gravity interaction is the intrinsic assumption that it generates a force field such that in every point in the physical space the intensity and direction of the force instantly feel of the variation of the location of the set of massive bodies. This corresponds to the unlikely, and strongly criticized by Einstein, infinite interaction speed. In spite of these practical and conceptual limitations, Newtonian gravity is an exceptional tool for the understanding of the mechanics on the Earth and in the sky over an enormous range of space- and time-scales and over a wide range of velocities. In other words, whenever both the approximations of weak gravity field and that of low velocities (respect to the speed of light in vacuum, c) of the objects under study are valid, General Relativity (or similar theories) and Special Relativity are irrelevant to a precise determination of the objects motion. For example, on the Earth, the consideration of General Relativity effects has a role just when an extremely high accuracy of position determination is required (GPS systems), and Special Relativity has no relevance for mechanics out of the submillimeter scales.

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Preface

It is useless to remind the exceptional successes of classical Newtonian gravitation theory as a tool to launch and control with enormous precision interplanetary probes, at least when the Sun is not too closely approached (remember that even for the Mercury’s orbit, the General Relativistic deviation from the classical understanding was just of *43″ per tropical century against the measured 574″ which is an effect less than 8%). So the classical celestial mechanics (few-body problem) up to stellar dynamics (large N-body problem) can be treated with high precision in the frame of classical Newtonian dynamics, at least when the distances between the bodies are large respect to the Schwarzschild radii of the moving objects. With this book, I hope I was able to give a mathematically reliable, although simple, treatment of the main issues in classical Newtonian gravitational theory for particle systems and regular distributions of matter. I added, at that purpose, a series of exercises with a full solution for most of them. The book is mainly thought for an audience of undergraduate and graduate students in Physics, Mathematics and Astronomy-Astrophysics. Anyway, it should be appreciated by any reader with a knowledge of mathematics and physics at the college level. Rome, Italy June 2019

Roberto A. Capuzzo Dolcetta

Acknowledgements I wrote some parts of this book while hosted at the Astronomical Rechen Institut in Heidelberg (Germany), at the Aspen Center for Physics (Aspen, CO, USA) and at the Leiden Observatory (Leiden, The Netherlands), institutions which I warmly acknowledge.

Introduction

This book gives an introduction to classical Newtonian gravitation and potential theories, as pieces of Physics essential for understanding classical mechanics and particularly relevant for Astrophysics. Among the four fundamental forces of Physics, gravity has the unique aspect of being an unscreened force which permeates the whole universe. Moreover, although Einstein’s general relativity provides a more extended framework for gravity, for most of the practical purposes, both in the field of pure scientific investigation and in the applicative one, Newtonian gravity provides much simpler and sufficiently approximated results, whenever applied in the regime of weak field. To reach the aim, the book is structured as follows. In the first chapter, some essential elements of vectorial calculus are recalled, especially to provide the formalism used in the following chapters. In the second chapter, classical Newtonian gravity theory for one and a generic number N of point masses is presented and discussed. The theory for point masses is naturally extended to the continuous case in the same chapter. In the third chapter, the paradigmatic case of spherical symmetry in the mass density distribution (central force) is dealt with the introduction of the useful tool of qualitative treatment of motion. In chapter four, the general case of nonsymmetric mass density distribution is discussed. In this chapter, classical potential theory is presented, with elements of harmonic theory, which is essential to understand the series development of the potential discussed in the second part of the chapter. The short, final, chapter five deals with the specific case of the motion of a satellite around the Earth. Examples and exercises are presented throughout the book to clarify aspects of the theory.

ix

Contents

1 Elements of Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Vector Functions of Real Variables . . . . . . . . . . . . . . . 1.2 Limits of Vector Functions . . . . . . . . . . . . . . . . . . . . . 1.3 Derivatives of Vector Functions . . . . . . . . . . . . . . . . . . 1.3.1 Geometric Interpretation . . . . . . . . . . . . . . . . . 1.4 Integration of Vector Functions . . . . . . . . . . . . . . . . . . 1.5 The Formal Vector Operator r . . . . . . . . . . . . . . . . . . 1.5.1 Gradient and r in Spherical Polar Coordinates 1.5.2 Gradient and r in Cylindrical Coordinates . . . 1.6 The Divergence Operator . . . . . . . . . . . . . . . . . . . . . . 1.7 The Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Formal Calculation of Divergence and Laplacian by the Operator r . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8.1 Spherical Polar Coordinates . . . . . . . . . . . . . . 1.8.2 Cylindrical Coordinates . . . . . . . . . . . . . . . . . 1.9 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9.1 Lines of Force of a Vector Field . . . . . . . . . . . 1.10 The Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . 1.10.1 The Stokes’ Theorem . . . . . . . . . . . . . . . . . . . 1.11 Velocity Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.12 Meaning of the Divergence of a Vector Field . . . . . . . . 1.13 Link Between Divergence and Volume Variation . . . . . 1.14 Solenoidal Vector Fields . . . . . . . . . . . . . . . . . . . . . . . 1.14.1 Flux Tubes . . . . . . . . . . . . . . . . . . . . . . . . . . 1.15 Further Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2 Newtonian Gravitational Interaction . . . . . . . . . . . . . . . . . . 2.1 Single Particle Gravitational Potential . . . . . . . . . . . . . 2.2 Motion of a Particle Gravitating in a Resisting Medium 2.3 The Gravitational N-Body Case . . . . . . . . . . . . . . . . . .

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Contents

2.4 2.5 2.6 2.7 2.8 2.9

2.3.1 Potential of N Gravitating Bodies . . . 2.3.2 Mechanical Energy of the N Bodies . The Scalar Virial Theorem . . . . . . . . . . . . . . 2.4.1 Consequences of the Virial Theorem . Continuous Distributions of Matter . . . . . . . . 2.5.1 Poisson’s and Laplace’s Equations . . Gauss’ Theorem . . . . . . . . . . . . . . . . . . . . . . Gravitational Potential Energy . . . . . . . . . . . . Newton’s Theorems . . . . . . . . . . . . . . . . . . . Further Readings . . . . . . . . . . . . . . . . . . . . .

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3 Central Force Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 The Potential and Force Generated by a Spherical Matter Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Calculating Spherical Potentials via Poisson’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Motion in a Spherical Potential . . . . . . . . . . . . . . . . . . . 3.2.1 Circular Trajectories . . . . . . . . . . . . . . . . . . . . . 3.3 Potential Generated by a Homogeneous Sphere . . . . . . . 3.3.1 Trajectories in a Homogeneous Sphere . . . . . . . 3.3.2 Radial Motion in a Homogeneous Sphere . . . . . 3.4 Some Relevant Spherical Models . . . . . . . . . . . . . . . . . . 3.4.1 The Plummer Sphere . . . . . . . . . . . . . . . . . . . . 3.4.2 The Isochrone Potential . . . . . . . . . . . . . . . . . . 3.5 Quality of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 The Keplerian Case . . . . . . . . . . . . . . . . . . . . . 3.5.2 The Homogeneous Sphere Case . . . . . . . . . . . . 3.6 Periods of Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Radial Period in the Keplerian Potential . . . . . . 3.6.2 Radial Period in the Homogeneous Sphere Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.3 Radial Period in the Plummer Potential . . . . . . . 3.6.4 Radial Period in the Isochrone Potential . . . . . . 3.7 Azimuthal Period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.1 Fully Periodic Motion . . . . . . . . . . . . . . . . . . . 3.8 The Inverse Problem in a Central Force Field . . . . . . . . 3.8.1 From Elliptic Trajectories to their Central Force Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Further Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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54 55 55 57 59 60 62 64 67 70

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4 Potential Series Developments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 4.1 Fundamental Solution of Laplace’s Equation . . . . . . . . . . . . . . . 109 4.2 Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

Contents

4.3 4.4

4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17

4.18

xiii

Legendre’s Polynomials . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Some Properties of Legendre’s Polynomials . . Recurrence Relations . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 First Recurrence Relation . . . . . . . . . . . . . . . 4.4.2 Second Recurrence Relation . . . . . . . . . . . . . The Legendre’s Differential Equation . . . . . . . . . . . . . Orthogonality of Legendre’s Polynomials Pm ðuÞ . . . . . Development of a Function in Series of Legendre’s Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Rodrigues’ Formula . . . . . . . . . . . . . . . . . . . . . . . . . Harmonic and Homogeneous Polynomials . . . . . . . . . Surface Spherical Harmonics . . . . . . . . . . . . . . . . . . . Solution of the Differential Equation for Sm ðh; uÞ . . . . ðpÞ The Solution in u: Pm ðcos uÞ . . . . . . . . . . . . . . . . . . Note on the Associated Legendre’s Equation . . . . . . . Zonal, Tesseral and Sectorial Spherical Harmonics . . . Orthogonality of Surface Spherical Harmonics . . . . . . Developments in Terms of Spherical Harmonics . . . . . Development of the Potential Generated by a Regular Mass Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.17.1 External Potential Series Development . . . . . . Further Readings . . . . . . . . . . . . . . . . . . . . . . . . . . .

5 The Earth Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Approximate Axisymmetry of the Earth Potential . . . 5.2 Satellite Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 A Short Note on the Perturbation Theory: Osculating Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Search for the Time-Dependent Elliptic Elements . . . 5.5 Further Readings . . . . . . . . . . . . . . . . . . . . . . . . . .

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Appendix A: Two- and N-Body Adimensional Motion Equations . . . . . . 171 Appendix B: Two Body Regularization . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

Symbols

List of symbols used in this book. M R G AU jvj, jjvjj

Value of the Sun mass (1.989  1030 in kilograms). Value of the Sun radius (6.955  105 in kilometers). Newtonian constant of gravitation (6.67408  10−11 in MKS units). Astronomical Unit, i.e. the average Sun-Earth distance ’1.5  108 km. With the symbols jvj or jjvjj we intend the Euclidean norm of a vector v,  n 1=2 P 2 i.e., in Rn , jvj  jjvjj ¼ vi : Alternatively, we use also v to i¼1

ei ev r R, Z, Z n ∇ , ^ n det(A) r, h, u R, h, z SR ðrÞ

indicate the norm of v. This symbol refers to the unit vector in the ith coordinate direction in Rn . Unit vector in the direction of v, ev  v=v. n P Represents the position vector in Rn , i.e. r ¼ xi ei . i¼1

With these symbols we refer, respectively, to the set of real, integer, and positive (+) or negative (−) integer numbers. This symbol indicates the outward unit vector in a point of a surface. With this symbol we intend, in Rn , the formal vector operator n P r ei @x@ i . i¼1

These symbols represent, respectively, scalar () and vector (^) product between vectors. Symbol for difference between sets. Determinant of matrix A. These symbols represent spherical polar coordinates in R3 . These symbols represent cylindrical coordinates in R3 . With this symbol we indicate, in Rn , the spherical hypersurface of radius R and center in r.

xv

xvi

E, Q, L xi vi   n k log Ck L2 1k ¼) (¼ ()

Symbols

These symbols represent, respectively, the mechanical energy, momentum and angular momentum per unit mass. We adopt the Einstein’s summation convention that implies summation n P over repeated indexes, i.e., in Rn , xi vi  xi vi . i¼1

For integer n and k, 0  k  n, it gives the usual binomial coefficient n! k!ðnkÞ!. Represents the natural logarithm (in base e). Represents the space of functions continuous up to the kth derivative. Represents the space of square-integrable functions. Indicates that there are k independent arbitrary choices for a parameter. Logical implication: a ¼) b means a implies b, i.e. that if proposition a is true then proposition b is also true. Inverse logical implication: a (¼ b means b implies a, i.e. that if proposition b is true then proposition a is also true. Double implication, that means identity.

Chapter 1

Elements of Vector Calculus

1.1 Vector Functions of Real Variables Let v = v(t) a real function of one real variable. If v ∈ R3 , we can write v(t) = vx (t)i + v y (t)j + vz (t)k,

(1.1)

where i, j, k are the unit vectors in the directions of the axes of an orthogonal coordinate system, such that i · i = j · j = k · k = 1, i · j = i · k = j · k = 0.

(1.2) (1.3)

Example 1.1.1 The circumference of radius R is represented with the function r(θ ) = R cos θ i + R sin θ j, with 0 ≤ θ < 2π . It corresponds to the vector function r : [0, 2π ) −→ [−R, R] × [−R, R]. Example 1.1.2 Let us give the Cartesian and parametric representations of a parabola as  x = t, 2 y = ax , y = at 2 .

© Springer Nature Switzerland AG 2019 R. A. Capuzzo Dolcetta, Classical Newtonian Gravity, UNITEXT for Physics, https://doi.org/10.1007/978-3-030-25846-7_1

1

2

1 Elements of Vector Calculus

Fig. 1.1 A double-valued function plot

In vector formalism, it corresponds to r : R −→ R+ ,

r = ti + at 2 j.

Note that hereafter we will deal only with single-valued vector functions (monodrome functions), excluding multivalued function, which do not have an inverse function (Fig. 1.1).

1.2 Limits of Vector Functions Given the vector function v = v(t), if ∀ > 0, ∃δ > 0 such that for |t − t0 | < δ , then |v(t) − V| < , we say that lim v(t) = V.

t→t0

Note that, if v(t) = vx (t) i + v y (t) j + vz (t) k.

(1.4)

1.2 Limits of Vector Functions

3

then ⎧ lim vx = Vx , ⎪ ⎪ t→t 0 ⎨ lim v y = Vy , lim v = V ⇐⇒ t→t 0 t→t0 ⎪ ⎪ ⎩ lim vz = Vz .

(1.5)

t→t0

A vector function v = v(t) is said to be continuous in t0 if lim v(t) = v(t0 ).

t→t0

(1.6)

Clearly, the function v(t) is continuous if and only if all its components, vx (t), v y (t), vz (t), are continuous. The usual properties of scalar continuous functions still hold for vector functions: • • • •

If v is continuous then αv, where α ∈ R, is also continuous. If v and w are continuous then v · w is also continuous. If v and w are continuous then v ∧ w is also continuous. If v and w are continuous then v ± w is also continuous. Moreover, some additional properties for the vector functions limits hold. Let v(t) e w(t) two vector functions such that lim v(t) = V,

lim w(t) = W,

t→t0

t→t0

(1.7)

then • lim αv = αV (α ∈ R), t→t0

• lim |v| = |V|, t→t0

• lim v · w = V · W, t→t0

• lim v ∧ w = V ∧ W. t→t0

1.3 Derivatives of Vector Functions Let v = v(t) be a vector function of a real variable t defined in an interval I . For any, arbitrary, point t = t0 in I let us consider the difference quotient v(t) − v(t0 ) . t − t0 If the difference quotient (1.8) has a finite limit

(1.8)

4

1 Elements of Vector Calculus



dv dt

 ≡ v˙ 0 = lim

t→t0

t0

v(t) − v(t0 ) . t − t0

(1.9)

(the dot over the vector is hereafter indicating derivative with respect to t) we say that the function v = v(t) is differentiable in the point t0 of the domain I and such a limit value is called derivative of v(t) in t0 . The following relations hold (i) (ii) (iii) (iv)

d d (αv) = α (v) ∀α ∈ R, dt dt d ˙ (v + w) = v˙ + w, dt d ˙ (v · w) = v˙ · w + v · w, dt d ˙ (v ∧ w) = v˙ ∧ w + v ∧ w. dt

Theorem 1.1 If the vector function v(t) is differentiable in t0 , it results to be continuous in t0 . Proof Let v(t) differentiable in the point t = t0 . It implies that the limit lim

t→t0

v(t) − v(t0 ) = v˙ (t0 ), t − t0

(1.10)

exists and is finite, thus

lim (v(t) − v(t0 )) = lim

t→t0

t→t0

v(t) − v(t0 ) (t − t0 ) = v˙0 lim (t − t0 ) = 0, t→t0 t − t0

(1.11)

which, indeed, corresponds to the continuity of v(t) in t0 . Note While all differentiable functions are continuous, the viceversa is not true, i.e. there are functions which are continuous but not differentiable. An example is f (x) = |x|, which is continuous everywhere except in x = 0.

1.3.1 Geometric Interpretation Let v(t) a continuous function and define δv = v(t + δt) − v(t) which is a vector in the direction pointing from the point P to Q (Fig. 1.2).

1.3 Derivatives of Vector Functions

5 P

Fig. 1.2 Geometric view of vector function derivative

δv

v(t)

O

Q

v(t+dt)

−v(t)

If the limit dv δv ≡ δt→0 δt dt lim

(1.12)

dv is finite, we easily understand (see Fig. 1.2) that the vector derivative has the dt direction of the straight line tangent to the curve in P. Exercise 1.1 Assuming v(t) to be a vector function of constant modulus (|v| ≡ v = const.) show that it results v · v˙ = 0 (i.e. v˙ ⊥ v). Solution Actually, being |v| constant, we have d d (v · v) = 2v · v˙ = |v|2 = 0. dt dt Exercise 1.2 Given the vector functions a(t), b(t) and c(t), calculate the derivative d (a ∧ b · c). dt Solution d d dc (a ∧ b · c) = c · (a ∧ b) + (a ∧ b) · = dt dt   dt da db dc =c· ∧b+a∧ + (a ∧ b) · . dt dt dt Exercise 1.3 Calculate the derivative Solution Left to the reader.

d (a ∧ b ∧ c). dt

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1 Elements of Vector Calculus

Exercise 1.4 Show that er ∧ der =

r ∧ dr r2

r where er ≡ is the radial versor. r Solution der = d

r

r

=

1 r dr − 2 dr, r r

hence we have  er ∧ der = er ∧

r 1 dr − 2 dr r r

 =

1 r ∧ dr. r2

er e˙ r = Moreover, because er is perpendicular to e˙ r (i.e. 

π ), it results 2

|er ∧ e˙ r | = |˙er |. Exercise 1.5 Show that v ∧ v˙ = 0 if and only if v has invariable direction (ev = const.) Solution Geometrically, the solution is intuitive because the fixed direction of v (although v may vary) implies Δv ∝ v, i.e. Δv  v and so v ∧ v˙ = 0; viceversa, if v ∧ v˙ = 0 then Δv  v. In a more rigorous way. v Let ev = the unit vector in the (fixed) direction of v, then we can write v = vev , v hence v˙ = ve ˙ v + v˙ev , so that ˙ v + v˙ev ) = v 2 ev ∧ e˙ v , v ∧ v˙ = vev ∧ (ve which implies v ∧ v˙ = 0 ⇐⇒ e˙ v = 0 ⇐⇒ ev = const.

1.3 Derivatives of Vector Functions

7

Exercise 1.6 Calculate d (r · r˙ ) dt d (b) (r ∧ r˙ ) dt (a)

Solution d (r · r˙ ) = r˙ · r˙ + r · r¨ = |˙r|2 + r · r¨ dt d (b) (r ∧ r˙ ) = r˙ ∧ r˙ + r ∧ r¨ = r ∧ r¨ dt (a)

1.4 Integration of Vector Functions Definition 1.1 Let f = f(t) a real vector function of the real variable t in the domain [a,b]. This function is said integrable over [a,b] if exists a continuous function F(t) dF = f, that formally translates into such that dt  F(t) = f(t) dt. (1.13) Upon this definition, F(b) − F(a) =

b

f(t) dt 1 Of course, the above vector rela-

a

tion corresponds (in R3 ) to the following scalar equalities dF y = fy , dt

dFx = fx , dt

dFz = fz . dt

(1.14)

Exercise 1.7 Calculate the integral  r ∧ r¨ dt. Solution An integration by parts leads to 

 r ∧ r¨ dt =

1 Here

r∧

d r˙ dt = dt



 r ∧ d r˙ = r ∧ r˙ −

r˙ ∧ dr = r ∧ r˙ + c,

we define [F(x)]ab ≡ F(b) − F(a), to use throughout the book.

8

1 Elements of Vector Calculus

given that r˙ ∧ r˙ = 0 and with c a constant vector. In the case of a particle of mass F m subjected to a force F, we have r ∧ r¨ = r ∧ , which means that for a central m force field (r ∧ F = 0) it results r ∧ r˙ ≡ mL = c , where c is a constant vector which expresses the integral of angular momentum.2 Exercise 1.8 Solve, with respect to r, the equation a ∧ r¨ = b, with a and b constant vectors. Solution A double integration in time leads to

a ∧ r˙ = b t + c,

a∧r=

1 2 bt + ct + d, 2

then we have for r this implicit expression a ∧ (a ∧ r) = (a · r)a − (a · a)r =

1 (a ∧ b)t 2 + (a ∧ c)t + (a ∧ d) 2

Exercise 1.9 Verify that v · v˙ = 0 implies |v| ≡ v = const.. Note that such result together with that of Exercise 1.1 constitutes a necessary and sufficient condition. Solution Integrating both sides of the relation v · v˙ = 0 gives   1 v · v˙ dt = v · dv = v · v = c, 2 √ where c is a constant, and so |v| = 2c = const. Equation 1.9 represents the kinetic energy theorem: when the acceleration a ≡ v˙ is orthogonal to v then the unit mass particle kinetic energy T = 21 v · v is constant. Note that the above relation is the inverse of that proved in Exercise 1.1, so that v · v = 0 and |v| = const. are equivalent.

2 An integral

of motion is a function (either scalar or vector) which keeps constant along all possible solutions of the motion equations and which is independent of t.

1.5 The Formal Vector Operator ∇

9

1.5 The Formal Vector Operator ∇ The operator nabla, ∇, is defined as a (formal) differential vector ∇≡i

∂ ∂ ∂ +j +k . ∂x ∂y ∂z

(1.15)

As well known, the gradient of a scalar function f (r) of class C 1 (Ω), Ω ⊆ Rn , is a vector field whose components are the partial derivatives of the function. This formally translates into

∇ f : Ω −→ Rn ,

∇ f ≡ grad f = ei

∂f . ∂ xi

(1.16)

By its definition, the vector operator ∇ is an application which projects a scalar field f of class C 1 (Ω) in a continuous vector field ∇ f . Exercise 1.10 Show that ∇ f (x, y, z) is a vector normal at the point (x, y, z) to the generic level surface of f (x, y, z), f (x, y, z) = c, where c is an arbitrary real constant. Solution dr is a vector tangent to a curve represented by r = r(s), where As known, et = ds s is a curvilinear abscissa. To solve the exercise it is sufficient to show that the orthogonality condition dr ∇f · =0 ds holds for an arbitrary curve belonging to the surface f (x, y, z) = c, passing through the point (x, y, z). This is easily seen because, indeed, on the level surface, where f is constant, results ∇f ·

∂ f dx ∂ f dy ∂ f dz df dr = + + = = 0. ds ∂ x ds ∂ y ds ∂z ds ds

1.5.1 Gradient and ∇ in Spherical Polar Coordinates The transformation from spherical polar to Cartesian coordinates is ⎧ ⎨ x = r sin ϕ cos θ, y = r sin ϕ sin θ, ⎩ z = r cos ϕ,

with r ≥ 0, 0 ≤ ϕ ≤ π and 0 ≤ θ < 2π,

(1.17)

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1 Elements of Vector Calculus

which have the inverse transformation r=

x2

+

y2

+

z2,

θ = 2 arctan

y x 2 + y2 + x

, ϕ = arctan

x 2 + y2 . z (1.18)

It results that the spherical polar unit vectors are: r er = = sin ϕ cos θ i + sin ϕ sin θ j + cos ϕ k, r ∂r 1 eθ = = − sin θ i + cos θ j, ds ∂θ dθ eϕ =

(1.19)

∂r 1 = cos θ cos ϕ i + cos ϕ sin θ j − sin ϕ k, ds ∂ϕ dϕ

where we have made use of the relation ds 2 = d x 2 + dy 2 + dz 2 , so that ds = dθ



dx dθ

2

 +

dy dθ

2

 +

dz dθ

2 ,

ds . dϕ The gradient operator of a function can be written formally as

and analogous for

∇ f = (∇ f · er ) er + (∇ f · eθ ) eθ + (∇ f · eϕ ) eϕ ,

(1.20)

so as its explicit expression requires writing the polar components ∇ f · er , ∇ f · eθ and ∇ f · eϕ in terms of spherical polar coordinates. This is done taking into account that ⎧ ∂f ∂f ∂f ⎪ ⎪ ∇ f · er = sin ϕ cos θ + sin ϕ sin θ + cos ϕ, ⎪ ⎪ ∂x ∂y ∂z ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂f ∂f ∇ f · eθ = − sin θ + cos θ, (1.21) ⎪ ∂ x ∂y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂f ∂f ∂f ⎪ ⎩ ∇ f · eϕ = cos θ cos ϕ + cos ϕ sin θ − sin ϕ, ∂x ∂y ∂z

1.5 The Formal Vector Operator ∇

and that ⎧ ∂f ⎪ ⎪ = ⎪ ⎪ ∂r ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂f = ⎪ ∂θ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂f ⎪ ⎩ = ∂ϕ

11

∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f ∂f ∂f + + = sin ϕ cos θ + sin ϕ sin θ + cos ϕ, ∂ x ∂r ∂ y ∂r ∂z ∂r ∂x ∂y ∂z ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f ∂f + + = − r sin ϕ sin θ + r sin ϕ cos θ, ∂ x ∂θ ∂ y ∂θ ∂z ∂θ ∂x ∂y ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f ∂f ∂f + + = r cos θ cos ϕ + r cos ϕ sin θ − r sin ϕ. ∂ x ∂ϕ ∂ y ∂ϕ ∂z ∂ϕ ∂x ∂y ∂z (1.22)

To obtain the complete expression of the spherical polar components of the gradient of the function f , it is necessary to solve for ∂∂ xf , ∂∂ yf , ∂∂zf the system of equations (1.22). The system (1.22) is of the type ⎛ ∂f ⎞ ⎜ ∂r ⎜ ∂f ⎜ ⎜ ⎜ ∂θ ⎝ ∂f ∂ϕ

⎛

∂f ⎜ ∂x ⎟ ⎜ ⎟ ⎟ T ⎜ ∂f ⎟=J ⎜ ⎜ ∂y ⎟ ⎝ ∂f ⎠ ∂z

⎞ ⎟ ⎟ ⎟ ⎟, ⎟ ⎠

(1.23)

where J T is the transposed of the Jacobian matrix of the Cartesian-spherical polar coordinate transformation ⎛ ∂x ∂x ⎜ ∂r ∂θ ∂(x, y, z) ⎜ ⎜ ∂y ∂y J= =⎜ ∂(r, θ, ϕ) ⎜ ∂r ∂θ ⎝ ∂z ∂z ∂r ∂θ

∂x ∂ϕ ∂y ∂ϕ ∂z ∂ϕ

⎞ ⎟ ⎟ ⎟ ⎟. ⎟ ⎠

(1.24)

The formal solution of the problem is thus given by expressions Eq. 1.21, where ∂f ∂f ∂f the partial derivatives , , are functions of r, θ, ϕ through the solution of ∂ x ∂ y ∂z the system (1.22), which, formally, is ⎛

∂f ⎜ ∂x ⎜ ∂f ⎜ ⎜ ⎜ ∂y ⎝ ∂f ∂z

⎞

⎛ ∂f ⎞

⎜ ∂r ⎟ ⎜ ⎟ ⎟ T −1 ⎜ ∂ f ⎟ = (J ) ⎜ ⎜ ∂θ ⎟ ⎝ ∂f ⎠

⎟ ⎟ ⎟ ⎟. ⎟ ⎠

(1.25)

∂ϕ

The general relation (JT )−1 = (J−1 )T allows to obtain the solution of the above system by the knowledge of J−1 . In this case, JT is orthogonal, so that (JT )−1 = J. The

12

1 Elements of Vector Calculus

gradient polar components can even more easily obtained by a straight comparison of 1.21 and 1.22, leading straightforwardly to ∂f , ∂r 1 ∂f ∇ f · eθ = , r sin ϕ ∂θ 1 ∂f . ∇ f · eϕ = r ∂ϕ

∇ f · er =

(1.26)

Finally, the gradient in spherical polars is ∇f =

∂f 1 ∂f 1 ∂f er + eθ + eϕ . ∂r r sin ϕ ∂θ r ∂ϕ

(1.27)

That means that in spherical polar coordinates the formal vector ∇ is written as ∇ = er

∂ ∂ 1 ∂ 1 + eθ + eϕ . ∂r r ∂θ r sin ϕ ∂ϕ

(1.28)

Note In some textbooks the variables θ and ϕ are exchanged respect to our definition, so that 1 ∂f 1 ∂f ∂f er + eθ + eϕ . (1.29) ∇f = ∂r r ∂θ r sin θ ∂ϕ

1.5.2 Gradient and ∇ in Cylindrical Coordinates Let us now express the gradient of a function in cylindrical coordinates. Let (ρ, θ , z) be the cylindrical coordinates, graphically identified by Fig. 1.3, whose relation with Cartesian coordinates is ⎧ ⎨ x = ρ cos θ, y = ρ sin θ, with 0 ≤ θ < 2π (1.30) ⎩ z = z. The system basis versors are ⎧ eρ = cos θ i + sin θ j, ⎪ ⎪ ⎪ ∂r 1 ⎨ eθ = = − sin θ i + cos θ j, ∂θ ds ⎪ ⎪ ⎪ dθ ⎩ ez = k.

(1.31)

1.5 The Formal Vector Operator ∇

13

Fig. 1.3 Spherical polar (r, θ, ϕ) and cylindrical (ρ, θ, z) coordinates

z z

r y θ

ρ

x

In such a basis, the gradient of a function f (r) is written as ∇ f = (∇ f · eρ ) eρ + (∇ f · eθ ) eθ + (∇ f · ez ) ez ,

(1.32)

⎧ ∂f ∂f ⎪ ⎪ ∇ f · eρ = cos θ + sin θ, ⎪ ⎪ ∂ x ∂y ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂f ∂f sin θ + cos θ, ∇ f · eθ = − ⎪ ∂x ∂y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∂f ⎪ ⎩ ∇ f · ez = . ∂z

(1.33)

where

The partial derivatives of f with respect to ρ, θ , z are written as ⎧ ∂f ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f ∂f ⎪ ⎪ = + + = cos θ + sin θ, ⎪ ⎪ ∂ x ∂ρ ∂ y ∂ρ ∂z ∂ρ ∂x ∂y ⎪ ⎨ ∂∂ρf ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f ∂f = + + =− sin θ + cos θ, ⎪ ∂θ ∂ x ∂θ ∂ y ∂θ ∂z ∂θ ∂x ∂y ⎪ ⎪ ⎪ ∂f ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f ⎪ ⎩ = + + = , ∂z ∂ x ∂z ∂ y ∂z ∂z ∂z ∂z

(1.34)

and the general procedure to obtain the ρ, θ, z components of the gradient vector is the same as in the previous subsection. Also in this case, the comparison of Eqs. 1.33 and 1.34 allows their straightforward determination as ∇f =

∂f 1 ∂f ∂f eρ + eθ + ez . ∂ρ ρ ∂θ ∂z

(1.35)

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1 Elements of Vector Calculus

Finally, in cylindrical coordinates the formal vector ∇ is written as ∇ = eρ

∂ 1 ∂ ∂ + eθ + ez . ∂ρ ρ ∂θ ∂z

(1.36)

1.6 The Divergence Operator Definition 1.2 Given a differentiable vector field v = v(r) in a domain C ∈ Rn , we call divergence of the vector v the sum of the partial derivatives of the vector components respect to homonymous coordinates

div v : C −→ R

div v ≡ ∇ · v =

∂vi , ∂ xi

(1.37)

adopting, hereafter, Einstein’s summation convention. Clearly, the divergence operator transforms a vector field into a scalar one. Formally, div v is expressible as the scalar product of ∇ with v, ∇ · v. A vector field with null divergence, ∇ · v = 0, is said solenoidal. As we will see in the following, the gravitational field generated by a regular mass distribution is solenoidal in the (empty) space external to the mass distribution. Exercise 1.11 Letting r ∈ Rn , show that ∇ · r = n.

(1.38)

Solution We have ∇ ·r=

∂ xi = 1 + 1 + · · · + 1 = n. ∂ xi n

times

Exercise 1.12 For r ∈ Rn , show that ∇ · er =

n−1 . r

(1.39)

Solution ∂ xi ∇ · er = = ∂ xi r

nr −

n 

(xi2 /r )

i=1 r2

=

nr − r n−1 = . 2 r r

The above result implies that, in R3 , ∇ · er = 2/r .

1.6 The Divergence Operator

15

Exercise 1.13 Let f : Rn −→ R represent a scalar field and v : Rn −→ Rn a vector field. Show that the following relation holds. ∇ · f v = f ∇ · v + v · ∇ f.

(1.40)

Solution ∇ · fv =

∂f ∂vi ∂ ( f vi ) = f + vi . ∂ xi ∂ xi ∂ xi

By mean of (1.40), the result (1.39) of Exercise 1.12 is obtained. Actually, we have

∇ · er = ∇ ·

1 1 n r 1 n−1 r n = ∇ ·r+r·∇ = −r· 2 = − = . r r r r r r r r

(1.41)

Exercise 1.14 Calculate in R3 ∇ · (r ∧ a), with a = const. Solution ∇ · (r ∧ a) =

∂ ∂ ∂ (yaz − za y ) − (xaz − zax ) + (xa y − yax ) = 0. ∂x ∂y ∂z

Such result is obtained also by applying the rules of the scalar triple product to the formal vector ∇, r and a.3 Exercise 1.15 Show that in Rn ∇ · v = ∇vi · ei . Solution ∇vi · ei = ek

∂vi ∂vi · ei = = ∇ · v. ∂ xk ∂ xi

(1.42)

Exercise 1.16 In Rn , given the scalar function ϕ(r), calculate in a direct way the divergence of the gradient of ϕ, i.e. ∇ · ∇ϕ ≡ (∇ · ∇)ϕ = ∇ 2 ϕ, where ∇ 2 ≡

n  ∂2 is the Laplacian formal operator, often written also as . ∂ xi2 i=1

3 The scalar triple product of 3 vectors satisfies various identities, among which a · (b ∧ c) = det(a, b, c).

16

1 Elements of Vector Calculus

Solution ∇ · ∇ϕ = ∇ 2 ϕ = ei

∂ ∂ϕ ∂ 2ϕ · ei = ≡ ϕ. ∂ xi ∂ xi ∂ xi2

Exercise 1.17 Compute the gradient of the divergence of er , ∇ ∇ · er . Solution ∇ ∇ · er = ∇

1 1−n n−1 1−n r = (n − 1)∇ = ∇r = er = (1 − n) 2 . 2 2 r r r r r

Exercise 1.18 Given two vector fields a and b, show that ∇ · (a∇b − b∇a) = a∇ 2 b − b∇ 2 a. Solution We leave the proof to the reader. Exercise 1.19 Given r ∈ R3 and a differentiable function f (r ), show that ∇ · f er =

1 d  2  r f . r 2 dr

(1.43)

Solution By mean of (1.40) it results ∇ · f er = f ∇ · er + er · ∇ f = f

 n−1 1  + er · f  (r )er = 2 (n − 1) f r + r 2 f  , r r

where the apex,  , indicates derivative with respect to r . For n = 3 the expression above reduces to ∇ · f er =

 1 d  2  1  r f . 2 f r + r2 f  = 2 2 r r dr

Let us note that f (r ) er could represent a central force field, attractive in the region where f (r ) < 0 and repulsive where f (r ) > 0. Exercise 1.20 In Rn , calculate explicitly 1 (i) ∇ · ∇ , r 1 (ii) ∇ · ∇ log . r

1.6 The Divergence Operator

17

Solution With regard to (i)

∇ ·∇

    1 1 1 r 1 1 = ∇ · − 2 ∇r = ∇ · − 2 = − 3∇ · r − r · ∇ 3 = r r r r r r 1 3−n 3 n 3 r = − 3 n + r · 4 ∇r = − 3 + 4 r · = , r r r r r r3

(1.44)

so that ∇ ·∇

1 1 3−n =0 = ∇2 = r r r3

⇐⇒

n = 3.

(1.45)

1 This means that in R3 the function is harmonic, because its Laplacian is zero. r With regard to (ii)   1r 1 1 1 1 ∇ · ∇ log = ∇ · r − 2 ∇r = −∇ · = − 2∇ ·r −r ·∇ 2 = r r rr r r n r 2−n = − 2 + 2 3 · ∇r = , r r r2 so that ∇ · ∇ log

1 2−n = =0 r r2

⇐⇒

n = 2.

1 As a consequence, the function log (and log r ) is harmonic in R2 (it is the so r called logarithmic potential). Exercise 1.21 Verify that in Rn ∇ ·∇

1 = 0 ∀ integer n ≥ 0. r n−2

(1.46)

Solution Actually, it results that ∇ ·∇

1 r n−2

r 1−n ∇ · r + r · ∇(2 − n)r −n = r + (2 − n)r · (−n)r −n−1 ∇r = r·r + (2 − n)(−n)r −n−1 = r + (2 − n)(−n)r −n = 0.

= ∇ · (2 − n) r 1−n ∇r = (2 − n) = (2 − n)nr −n = (2 − n)nr −n = (2 − n)nr −n

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1 Elements of Vector Calculus

1 In particular, if n = 2 then n−2 = 1, which is a constant and so all its derivatives r 1 are zero. If n = 1, then n−2 = r = |x| and, clearly, every 1st degree polynomial in r the single variable x, p1 (x) = ax + b, is such to give ∇ · ∇ p1 (x) = 0.

1.7 The Curl The curl of a vector field v of class C 1 (Ω), where Ω ∈ R3 , is also a vector field, written as  curl v =

curl v : Ω −→ R3 ,      ∂v y ∂v y ∂vz ∂vx ∂vx ∂vz − i− − j+ − k. ∂y ∂z ∂x ∂z ∂x ∂y

(1.47)

As for the divergence, which can be written as the formal scalar product of ∇ and v, the curl of v can be formally written as the vector product between ∇ and v    i j k     ∂ ∂ ∂  . (1.48) curl v = ∇ ∧ v =    ∂ x ∂ y ∂z   v x v y vz  Contrary to the divergence, the curl transforms a vector field in another vector field only in R2 and R3 . Definition 1.3 A vector field v ∈ C 1 (Ω) is called irrotational if ∇ ∧ v = 0.

(1.49)

Definition 1.4 A vector field v ∈ C 0 is called conservative if it can be written as gradient of a scalar function, i.e. if exists a scalar field f (r) of class C 1 (the potential of v) such that v = ∇ f . Lemma 1.1 Let v be a conservative vector field of class C 1 , whose potential is a scalar field f of class C 2 . It results curl v = ∇ ∧ ∇ f = 0.

(1.50)

The easy demonstration is left to the reader. Corollary 1.2 A conservative vector field of class C 1 (Ω) is irrotational. The viceversa, in general, is not true, unless Ω is a simply connected domain.

1.7 The Curl

19

Exercise 1.22 Show that ∇ ∧ r = 0.

(1.51)

Solution   i   ∂ curl r = ∇ ∧ r =   ∂x  x because every partial derivative

∂ xi ∂x j

j ∂ ∂y y

k ∂ ∂z z

     = 0,   

is identically zero.

Exercise 1.23 Given the vector function F : R3 −→ R3 of class C 1 , show that ∇ ∧F=0

⇐⇒

F = ∇ϕ,

(1.52)

where ϕ(r) is a scalar function of class C 2 . Solution (⇐=) If F = ∇ϕ, then the curl of F is   i  ∂   ∇ ∧ F =  ∂x  ∂ϕ   ∂x

j ∂ ∂y ∂ϕ ∂y

k ∂ ∂z ∂ϕ ∂z

      = (ϕzy − ϕ yz ) i − (ϕx z − ϕzx ) j + (ϕ yx − ϕx y ) k,   

where ϕx y ≡ ∂∂x ∂ϕ and similarly for the other derivatives. The above curl vanishes ∂y because the function ϕ is of class C 2 and so, for Schwarz’s theorem, the order of partial derivation can be exchanged, i.e. ϕx y = ϕ yx , etc. (=⇒) On the other side, if ∇ ∧ F = 0 then  ∇ ∧F=

∂ Fy ∂ Fz − ∂z ∂y



 i−

∂ Fz ∂ Fx − ∂x ∂z



 j+

∂ Fy ∂ Fx − ∂x ∂y

 k = 0,

that implies ∂ Fy ∂ Fz = , ∂z ∂y

∂ Fz ∂ Fx = , ∂x ∂z

∂ Fy ∂ Fx = . ∂x ∂y

The expressions above represent the conditions for the differential form F · dr = Fx d x + Fy dy + Fz dz to be an exact differential. This means that it does exist a continuous and differentiable function ϕ(r) whose differential reproduces the given

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1 Elements of Vector Calculus

form, dϕ = to

∂ϕ ∂ϕ ∂ϕ dx + dy + dz = Fx d x + Fy dy + Fz dz, which is equivalent ∂x ∂y ∂z F = ∇ϕ.

In conclusion, all irrotational fields are conservative, i.e. they can be derived from scalar (potential) functions. Exercise 1.24 Show that if F admits a potential then it results ∇ · F ∧ r = 0. Solution Since F ∧ r = (Fy z − Fz y)i − (Fx z − Fz x)j + (Fx y − Fy x, )k, it is ∂ ∂ ∂ (Fy z − Fz y) − (Fx z − Fz x) + (Fx y − Fy x) = ∂x ∂y ∂z ∂ Fy ∂ Fy ∂ Fz ∂ Fx ∂ Fz ∂ Fx = z− y− z+ x+ y− x= ∂x ∂x ∂y ∂y ∂z ∂z = r · ∇ ∧ F.

∇ ·F∧r=

In this hypothesis of the exercise ∇ ∧ F = 0, whence ∇ · F ∧ r = 0. Exercise 1.25 Given a scalar field f and a vector field v, show that ∇ ∧ f v = f ∇ ∧ v − v ∧ ∇ f = f ∇ ∧ v + ∇ f ∧ v.

(1.53)

Solution The solution is left to the reader. Note that the expression Eq. 1.53 is similar to that of the scalar product of ∇ with f v (Eq. 1.40), paying attention to the sign. Exercise 1.26 Determine the real values of a, b and c for which the field F = (x + 2y + az)i + (bx − 3y − z)j + (4x + cy − 2z)k is conservative, and find the potential function.

1.7 The Curl

21

Solution a = 4, b = 2, c = −1.   3 2 x2 + 2x(y + 2z) − y y − z − z 2 + const. U (x, y, z) = 2 2 3 Exercise 1.27 Show that ∇ ∧ (∇ ∧ v) = ∇(∇ · v) − ∇ 2 v, where ∇ 2 v is the vector whose components are ∇ 2 vx , ∇ 2 v y , ∇ 2 vz . Solution By a straight formal application of the triple product rule it results ∇ ∧ ∇ ∧ v = (v · ∇)∇ − (∇ · ∇)v, whence, substituting (v · ∇)∇ with ∇(∇ · v) we have the seeked relation. A longer way to the same result is via a thorough development of the calculations !   ∂v y ∂2 ∂vz ∂2 ∂2 ∂ ∂vx + + i+ ∇ ∧∇ ∧v=− + + i + v j + v k) + (v x y z ∂x ∂x ∂y ∂z ∂x2 ∂ y2 ∂z 2     ∂v y ∂v y ∂ ∂vx ∂vz ∂ ∂vx ∂vz + + j+ + + k = −∇ 2 v + ∇∇ · v. ∂y ∂x ∂y ∂z ∂z ∂ x ∂y ∂z

Exercise 1.28 Show that a central force field F(r) = f (r)er is conservative if, and only if, it is spherically symmetric, i.e. f (r) = f (r ). Solution Given a central force field F(r) = f (r)er , it is ∇ ∧ f er = f ∇ ∧ er − er ∧ ∇ f, and, being ∇ ∧ er = 0, it results. ∇ ∧ f er = −er ∧ ∇ f. Now (i) If the field is conservative, i.e. ∇ ∧ f er = 0, the above equation gives ∇ f parallel to er i.e. f is a function of r ; (ii) if the field is spherically symmetric ( f (r) = f (r )) ∇ f is (see Exercise 1.10) orthogonal to any spherical surface centered at the origin, which are its level

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1 Elements of Vector Calculus

surfaces, and thus ∇ f is radially oriented and ∇ ∧ f er = 0. This result is contained in the formula ∇ f (r ) = f  (r )er

=⇒

er ∧ ∇ f = 0.

1.8 Formal Calculation of Divergence and Laplacian by the Operator ∇ 1.8.1 Spherical Polar Coordinates It is easy to verify that in spherical polar coordinates the formal vector ∇ is written as (Eq. 1.27) ∇ = er

∂ ∂ 1 1 ∂ + eθ + eϕ . ∂r r sin ϕ ∂θ r ∂ϕ

(1.54)

Given a generic vector field F in spherical polar coordinates F = Fr er + Fr eθ + Fϕ eϕ , its divergence is, so, formally expressed as   1 1 ∂ ∂ ∂ + eθ + eϕ ·F ∇ · F = er ∂r r sin ϕ ∂θ r ∂ϕ 1 ∂F ∂F 1 ∂F + eθ · + eϕ · . = er · ∂r r sin ϕ ∂θ r ∂ϕ

(1.55)

(1.56)

Keeping into account the orthonormality of the polar basis and the expressions (1.19), it results that ∂eθ ∂er ∂eϕ ∂er = = = = 0. ∂r ∂r ∂ϕ ∂r

(1.57)

∂er = − sin ϕ sin θ i + sin ϕ cos θ j = sin ϕ eθ ∂θ ∂er = cos ϕ cos θ i + cos ϕ sin θ j − sin ϕk = eϕ ∂ϕ ∂eϕ = − sin θ cos ϕ i + cos ϕ cos θ j = cos ϕ eθ , ∂θ

(1.58)

Moreover

1.8 Formal Calculation of Divergence and Laplacian by the Operator ∇

23

which imply for the divergence of F the expression     1 1 ∂ Fθ ∂ Fϕ ∂ Fr + Fr sin ϕ + + cos ϕ Fϕ + Fr + = ∇ ·F= ∂r r sin ϕ ∂θ r ∂ϕ   ∂ Fr 1 ∂ Fθ ∂ Fϕ 2 1 cos ϕ Fϕ + sin ϕ = = + Fr + + ∂r r r sin ϕ ∂θ r sin ϕ ∂ϕ  1 ∂  1 ∂  2  1 ∂ Fθ r Fr + + sin ϕ Fϕ . = 2 r ∂r r sin ϕ ∂θ r sin ϕ ∂ϕ (1.59) The expression of the Laplacian of a scalar function U (r) is, consequently, obtained simply by letting F = ∇U in the above expression     ∂ 2U ∂ 1 1 1 ∂U 2 ∂U + r + 2 sin ϕ = ∂r r sin ϕ ∂θ 2 r sin ϕ ∂ϕ r ∂ϕ   ∂ 2 ∂U 1 ∂ 2U 1 ∂U ∂ 2U + + 2 + sin ϕ . = 2 2 2 ∂r 2 r ∂r ∂θ r sin ϕ ∂ϕ ∂ϕ r sin ϕ

1 ∂U ∇ U ≡ ∇ · ∇U = 2 r ∂r 2

(1.60)

1.8.2 Cylindrical Coordinates In such a coordinate system the formal vector ∇ is written as (Eq. 1.36) ∇ = eρ

∂ 1 ∂ ∂ + eθ + ez , ∂ρ ρ ∂θ ∂z

(1.61)

and the vector F on the cylindrical basis is F = Fρ eρ + Fθ eθ + Fz ez .

(1.62)

  ∂ 1 ∂ ∂ + eθ + ez · F. ∇ · F = eρ ∂ρ ρ ∂θ ∂z

(1.63)

Consequently,

Now, given the orthonormality of the cylindrical vector basis, and given the expressions (1.31), we have ∂eρ ∂eθ ∂eθ ∂ez ∂ez ∂ez ∂eρ = = = = = = = 0, ∂ρ ∂z ∂ρ ∂z ∂ρ ∂θ ∂z and

(1.64)

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1 Elements of Vector Calculus

∂eρ = − sin θ i + cos θ j, ∂θ ∂eθ = − cos θ i − sin θ j, ∂θ

(1.65)

 1 ∂ Fθ 1 ∂  ∂ Fz ρ Fρ + ++ , ρ ∂ρ ρ ∂θ ∂z

(1.66)

so that Eq. 1.63 becomes ∇ ·F=

and, so, the Laplacian of a scalar function U (r) is   ∂U 1 ∂ 2U 1 ∂ ∂ 2U 2 ρ + 2 + . ∇ U ≡ ∇ · ∇U = ρ ∂ρ ∂ρ ρ ∂U 2 ∂z 2

(1.67)

1.9 Vector Fields In Rn a generic vector field is a function of n real variables (x1 , x2 , ..., xn ) which can be written on a basis {ei } as F(r) = Fi ei . For n = 3, it is F(x, y, z) = Fx i + Fy j + Fz k.

(1.68)

A typical example is that of the force field generated by a charged particle (Coulombian “charges”). In this case the force is radially directed toward the particle (source of the field) if the test particle in the generic position r has a charge of opposite sign or in the opposite side for same sign. The gravitational force is always attractive. In the case of a single particle which exerts a force, we speak of central force field. This particular central force field has, obviously, a spherically symmetric intensity, i.e. F = f (r )er . Of course, the sign of the function f (r ) discriminates between the attractive and repulsive case which scales, in the gravitational and Columbian cases, as ⎧ a2 ⎪ ⎪ ⎨ f (r ) = − 2 < 0, actractive force, r (1.69) ⎪ a2 ⎪ ⎩ f (r ) = > 0, repulsive force, r2 where the constant a 2 accounts for the coupling physical constant. In the Newtonian case of a point of mass m, a 2 = Gm, where G is the universal gravitational constant and m the mass sited in the origin and generating the field.

1.9 Vector Fields

25

1.9.1 Lines of Force of a Vector Field Definition 1.5 The lines of force of a vector field F are curves in Rn whose tangent at any point is in the direction of the field. As we will see, an information on the intensity is possible for solenoidal fields, and it is given by the density of the lines of force (see Sect. 1.14.1). Mathematically, the lines of force are represented by the curves r(s) (s is the curvilinear abscissa along the curve) such that dr = cF(x, y, z), ds where, for any arbitrary value of the non-zero constant c, the curve. In R3 the above relation gives

(1.70) dr ds

is a vector tangent to

dx dy dz dr = i+ j + k = cFx i + cFy j + cFz k, ds ds ds ds

(1.71)

dx dy dz = = Fx Fy Fz

(1.72)

so that

represent the system of 3 differential equations whose solution gives the lines of force of F. Note From its definition, the tangent to a line of force in any point P of the space gives the direction of the vector field evaluated in that point; it is so possible to assign an orientation to the line as that of the vector field in that point. Exercise 1.29 Determine the lines of force of a generic central field, expressed, in R3 , by F = f (x, y, z)er =

f (x, y, z) (xi + yj + zk) . r

Solution In this case, the conditions (1.72) are written as dx dy dz dy dz dx = x = = y = = z, Fx F F y z f f f r r r

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1 Elements of Vector Calculus

equivalent to dy dz dx = = . x y z Integrating, we have log |y| = log |x| + c1 , log |z| = log |x| + c2 , log |z| = log |y| + c3 , that is |y| = ec1 |x|, |z| = ec2 |x|, |z| = ec3 |y|.

(1.73)

The first of Eq. 1.73 corresponds, in the x, y plane, to y = ec1 x (1st and 4th quadrant) and y = ec1 (−x) (2nd and 3rd quadrant), which are straight lines passing through the origin. The same holds for the lines of force in the (x, z) and (y, z) planes. Exercise 1.30 Determine the lines of force of the gravitational field generated by two points of masses m 1 and m 2 sited in P1 = (−a, 0, 0) and P2 = (a, 0, 0), where a > 0. Discuss the existence of an equilibrium in the case m 1 = m 2 . Solution Let r1 = (−a, 0, 0) and r2 = (a, 0, 0) the 2 particles position vectors; on the generic point P with radius vector r= and unitary mass, the combined force exerted by the 2 particles is (Fig. 1.4) F(r) = G

m2 m1 (r1 − r) + G (r2 − r), 3 |r1 − r| |r2 − r|3

Fig. 1.4 Positions of the two particles of masses m 1 e m 2 placed on the x coordinate axis

z

m m1

r r1 r2 m2 x

y

1.9 Vector Fields

27

whose components are m1 m2 (−a − x) + G (a − x), 3 |r1 − r| |r2 − r|3 m m1 m2 (−y) + G (−y), Fy = G 3 |r1 − r| |r2 − r|3 m1 m2 (−z) + G (−z). Fz = G 3 |r1 − r| |r2 − r|3

Fx = G

The conditions to determine the lines of force dy dz dx = = , Fx Fy Fz lead to dx dy = = m1 m2 m1 m2 G (−a − x) + G (a − x) G (−y) + G (−y) |r1 − r|3 |r2 − r|3 |r1 − r|3 |r2 − r|3 dz = . m1 m2 G (−z) + G (−z) 3 3 |r1 − r| |r2 − r|

The rightmost equality above reduces to dz dy = , y z or, equivalently, log |y| = log |z| + c1

⇐⇒

|y| = ec1 |z|.

This means that the lines of force lie on planes through the x axis, which means an invariance by rotation around that axis and so the lines of force are contained in dz dy = and surfaces of revolution around the x axis. This is made clear by taking y z by multiplying and dividing by y e z, so to have dy dz ydy zdz 1 dy 2 1 dz 2 1 d(y 2 + z 2 ) dx = = 2 = 2 = = = = . 2 2 2 2 y z y z 2 y 2 z 2 y +z Fx The last equality is a differential relation linking x and y 2 + z 2 , thus the solutions are ∞1 surfaces of revolution around the x axis. The conclusion of the exercise is left to the reader.

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1 Elements of Vector Calculus

1.10 The Divergence Theorem Definition 1.6 Let x(t), y(t) and z(t) be single value functions continuous in [t1 , t2 ] and such that x(t1 ) = x(t2 ),

y(t1 ) = y(t2 ),

z(t1 ) = z(t2 ).

(1.74)

Upon these conditions the curve written by the parametric expressions x = x(t), y = y(t) and z = z(t) is a closed curve. Note The further condition that x(u) = x(v), y(u) = y(v) and z(u) = z(v) occurs only for u = v corresponds to no intersections in the curve, which is so called closed and simple. Let us suppose that the functions x(t), y(t) and z(t) are piecewise differentiable in the interval [t1 , t2 ] and give the following Definition 1.7 A domain N of R3 is said normal if (i) N is a convex polyhedron; (ii) N is bounded by a surface S composed by a finite number of plane portions and by a curved surface F, and is such that for some orientation of a Cartesian reference frame O(x, y, z) results: (a) the projection F of F onto the plane (x,y) is limited by a simple and closed curve consisting of a finite number of arcs, each with continuous tangent; the projection of every side of S onto (x,y) divides this plane into a finite number of regions each limited by a simple and closed curve; (b) any straight line parallel to the z axis intersects N in a single (connected) segment and F is represented by a Cartesian functional relation z = f (x, y), with f (x, y) a monodrome function continuous with its first order partial derivatives in the region F. (c) same conditions hold for (x, z) and (y, z). Note A sphere is not a normal domain because condition (ii) (b) does not hold, but it is union of a finite number of normal domains. Note A sub region OABC of a sphere (see Fig. 1.5) is normal as far as the angles  AOC,  AO B and  B OC are small enough.

1.10 The Divergence Theorem

29

Fig. 1.5 The region OABC is a normal domain only when the angles  AOC, AO B and  B OC are small enough

A B O C

Theorem 1.3 (Divergence theorem) Let A(x,y,z) a vector field continuous with its first order partial derivatives. If N is a normal domain, then the divergence theorem (referred also to as Ostrogradskij theorem or, also, Green’s theorem in space) holds. The thesis of the theorem is:   ∇ · A dV = A · n dσ, (1.75) N

Σ

where Σ ≡ ∂ N is the surface contouring N and n is its outward unit vector (Fig. 1.6). Proof Let S the surface contouring N , while σ is the projection onto the plane (x, y) of one, σ , of the sides of Σ, and v is the spatial region given by the intersection of N with the cylindrical surface along the sides of σ (Fig. 1.6). The region v is thus characterized by two surfaces z = f 1 (x, y) e z = f 2 (x, y) such that f 1 (x, y) ≤ f 2 (x, y),

Fig. 1.6 Schematic representation of the normal domain N

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1 Elements of Vector Calculus

where one of the 2 surfaces is planar. Let us proof the theorem in the region v. Consider, first, only the vertical component, A z , of A.  v

∂ Az dV = ∂z =

 v

 ( σ

∂ Az d xd ydz = ∂z

 σ

⎡ ⎢ ⎣

z=f 2 (x,y)

z= f 1 (x,y)

⎤ ∂ Az ⎥ dz ⎦ d xd y = ∂z

) A z (x, y, f 2 (x, y)) − A z (x, y, f 1 (x, y)) d xd y.

Indicating with Δσ2 an element of the upper, curved, portion F of the surface Σ, which is represented by z = f 2 (x, y), and with Δσ2 the projection of Δσ2 onto the plane (x, y), it results  Δσ2 =

 Δσ2

dσ =

 Δσ2

dσ cos α =

Δσ2

k · n2 dσ2 ,

where α is the angle between the outward normal, n2 , to Δσ2 and the positive z axis. On another side, on the inferior surface (that represented by z = f 1 (x, y)) we have  Δσ1 =

Δσ1

 dσ = −

 Δσ1

dσ cos β =

Δσ1

−k · n1 dσ1 ;

with β the angle between the outward normal, n1 , to Δσ1 and the positive z axis; consequently  ( ) A z (x, y, f 2 ) − A z (x, y, f 1 ) d xd y =  σ  = A z (x, y, f 2 ) d xd y − A z (x, y, f 1 ) d xd y = σ σ  2  1 = A z (x, y, z)k · n dσ2 + A z (x, y, z)k · n dσ1 = σ2 σ1   = A z k · n dσ = A z k · n dσ, σ1 ∪σ2 =σz

∂v

taking into account that on the lateral sides of ∂v it is, obviously, k · n = 0. Thanks to the property (ii) (c) of the normal domains, the same result above holds, also, for A x and A y

1.10 The Divergence Theorem

31

 ∂ Ax d xd ydz = A x i · n dσ, ∂x   v  ∂v ∂ Ay A y j · n dσ. d xd ydz = v ∂y ∂v



Now, due to property (ii) (a), the projection of all sides of Σ onto a coordinate plane is a finite number, n, of regions contoured by simple and closed curves, and so 

n   ∂ Ax ∂ Ax d xd ydz = d xd ydz = ∂ x N vk ∂ x k=1  n   A x i · n dσ = A x i · n. k=1

σk,x

Σ

This occurs, in perfect analogy, also for A y e A z , and so by summation of the integrals and by noting that A x i · n + A y j · n + A z k · n = A · n, the thesis of the theorem is obtained:   ∇ · A dV = A · n dσ. N

(1.76)

Σ

Note The divergence theorem can be extended to domains for which property (ii) (b) of Definition 1.7 does not hold; it suffices to divide the domain in sub-regions for any of which (b) actually holds. Obviously, the outward normal to the surface of separation bounding two sub-regions has to be considered, when performing the surface integral, once with a direction and once with its opposite so that the whole contribution of such separation surface to the surface integral cancels out leaving only the contribution of the overall bounding surface. As example, think to a spherical domain and intersect with a diametral plane. The planar cross section is the separation surface mentioned above.

Theorem 1.4 (Divergence theorem in the plane) Let A be a vector field, continuous with its first order partial derivatives, and S a planar normal domain, such that S¯ = S ∪ ∂ S and C = ∂ S where C is a simple, closed planar curve. Show that in R2 the divergence theorem assumes the form (Green theorem)  * *     ∂ Ay ∂ Ax + dσ = A · n ds = (1.77) A x dy − A y d x . ∂ x ∂ y S C C

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1 Elements of Vector Calculus

Fig. 1.7 Graphical sketch for the divergence theorem in

y

R2

t ds

S α n

x

Proof Indicating by ds the line element vector counter-clockwise oriented along the contour of S, and by α the angle it forms with the direction parallel to the x axis, from Fig. 1.7 it is clear that d x = ds cos α = ds sin(π/2 − α) = −ds n · j = −n y ds, dy = ds sin α = ds cos(π/2 − α) = −ds n · i = n x ds, by which 

 A · n ds =

C







A x n x + A y n y ds =

C



 A x dy − A y d x .

(1.78)

C

1.10.1 The Stokes’ Theorem Let us introduce now the Stokes’ theorem. Definition 1.8 A surface S is said open, two sided (or orientable) when the outward normal vector is uniquely defined at every point of the surface. Note From the above definition, a torus is an open, two sided (orientable), surface while the Möbius band (see Fig. 1.8) and the Klein bottle are not. Theorem 1.5 (Stokes’ Theorem) Let S be any open, two-sided, surface bounded by a simple closed curve C, and A a vector field continuous with its first order partial derivatives in a region containing S. Under these hypotheses, it results  * A · dr = ∇ ∧ A · n dσ. (1.79) C

S

1.10 The Divergence Theorem

33

Fig. 1.8 A Möbius strip

In words, the line integral of the tangential component of A along C equals the surface integral of the normal component of ∇ ∧ A on every open, two-sided, surface having C as contour. Proof Let us proof the theorem in the hypothesis that S satisfies the same conditions of the curved surface F in the divergence theorem, i.e. it satisfies the conditions (ii) (a), (b) and (c) of the Definition 1.7 of a normal domain, and that its projections onto the coordinate planes are regions where the Green’s theorem (divergence theorem in R2 ) holds. We have to show that 



  ∇ ∧ A x i + A y j + A z k · n dσ =

(∇ ∧ A) · n. dσ = S

S

* A · dr. (1.80) C

Since ∇ ∧ Ax i = we can write 

  (∇ ∧ A x i) · n. dσ =

S

∂ Ax ∂ Ax j− k, ∂z ∂y

S

∂ Ax ∂ Ax j·n− k·n ∂z ∂y

(1.81)

 dσ.

(1.82)

If the Cartesian expression for S is z = f (x, y), then A x over S is, actually, A x (x, y, f (x, y)), and so

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1 Elements of Vector Calculus

A x (x, y, f (x, y)) ≡ F(x, y), ∂F ∂ Ax ∂ Ax ∂ f = + , ∂y ∂y ∂z ∂ y which, by substitution of Eq. 1.82 as

(1.83) (1.84)

∂ Ax through the above relation, allows writing the rhs of ∂z

  + S

∂F ∂ Ax − ∂y ∂y



j·n ∂f ∂y

∂ Ax k·n − ∂y

, dσ.

(1.85)

The position vector of a generic point on S is r = xi + yj + f (x, y)k, so that ∂∂ry =

j+

∂f k ∂y

is a vector lying on the plane tangent, in this point, to S. As a consequence n·

∂f ∂f n·j ∂r =0=n·j+ n · k ⇐⇒ =− ; ∂y ∂y ∂y n·k

which reduces the integral (1.85) to    ∂F ∂F ¯ ∂F − n · k dσ = − dS = − d xd y, S ∂y S¯ ∂ y S¯ ∂ y

(1.86)

(1.87)

where S¯ is the projection of S onto the plane (x, y) (Fig. 1.9). For the divergence theorem in R2 , it is possible to write *  ∂F d xd y = F d x, (1.88) − ∂ S¯ S¯ ∂ y ¯ which is a simple closed curve, Γ . Finally, we have where ∂ S¯ is the contour of S,  * F d x. (1.89) (∇ ∧ A x i) · n. dσ = S

Γ

Noting that F(x, y) = A x (x, y, f (x, y)), we have that F over Γ is equal to A x over C and, due to that d x is the same on C and on Γ , it results *  A x d x. (1.90) (∇ ∧ A x i) · n. dσ = S

C

Similarly, on the planes (x, z) and (y, z) it is *    ∇ ∧ A y j · n. dσ = A y dy, *C S A z dz. (∇ ∧ A z k) · n. dσ = S

C

(1.91)

1.10 The Divergence Theorem

35

Fig. 1.9 Schematic representation of the surface S and of its projection S¯ onto the (x, y) plane

The thesis of the theorem is so obtained by straight summation of the lhs and rhs of (1.90) and (1.91): *  A · dr. (1.92) (∇ ∧ A) · n = S

C

Corollary 1.3 A direct consequence of the Stokes Theorem is * ∇ ∧ A = 0 ⇐⇒ A · dr = 0,

(1.93)

C

where C is an arbitrary simple closed curve. Proof The relation (1.79) implies * (i)

∇ ∧A=0

=⇒

A · dr = 0, C

and, by the continuity of both A and its first order partial derivatives, it results also * A · dr = 0

(ii)

=⇒

∇ ∧ A = 0.

C

Relation (ii) is obtained by that, if it would exist a point P where ∇ ∧ A = 0 then, by continuity, it should exist a whole neighbourhood V of P where ∇ ∧ A = 0 and so, taking a surface S ⊂ V and its outward normal versor n, it would be ∇ ∧ A · n = 0 on the surface and thus  * A · dr = ∇ ∧ A · n dσ = 0 CS

S

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1 Elements of Vector Calculus

where C is the contour of S, against the hypothesis. Exercise 1.31 Given, in R2 , the vector field F=

y x i − 2 j, 2 r r

show that for it the Stokes theorem does not hold over the all x, y plane. Solution The curl of F in two dimensions is 

∂ Fy ∂ Fx − ∇ ∧F=k ∂y ∂x

 ,

which vanishes everywhere (but the origin where it is not defined) because y ∂ Fx r 2 − 2yr r x 2 − y2 = = , 4 ∂y r r4 ∂ Fy −r 2 + 2xr rx x 2 − y2 = = . ∂x r4 r4

If ∇ ∧ F = 0 everywhere but the origin, obviously  (∇ ∧ F) · n dσ = 0, S

for any planar region S not containing the origin. * Actually, if we perform the line integral

F · dr along a circumference C cen-

tered at the coordinate origin, using at this purpose polar coordinates (r, θ ), we can write 2π

* F · dr = C

0

1 (−r 2 cos2 θ − r 2 sin2 θ ) dθ = − r2

2π dθ = −2π. 0

The residual, −2π , derives by the integration around the pole (which is the origin). The singularity is easily eliminated by integrating along the path Γ = Γ1 ∪ Γ2 as described in Fig. 1.10. In R2 \ (0, 0) the potential of F is   x + c, U (x, y) = arctan y as it is easily verified.

1.10 The Divergence Theorem

37

Fig. 1.10 Path apt to exclude the singularity residual

y Γ1

Γ2 x

Exercise 1.32 If S is a closed surface union of two surfaces for which the Stokes’ theorem holds, show that  ∇ ∧ A · n dσ = 0. S

Solution Left to the reader. Exercise 1.33 Let S R (0) a sphere of radius R centered at the coordinate origin, and let A = a(x, y, z) er a continuous vector field; calculate  S R (0)

∇ · A d V.

Solution By the divergence theorem we have   ∇ · A dV = S R (0)

∂ S R (0)

A · n dσ,

that, in spherical polar coordinates (dσ = R 2 sin ϕ dθ dϕ) writes as   A · n dσ = R 2 a(x, y, z) sin ϕ dθ dϕ. ∂ S R (0)

∂ S R (0)

In the simplifying assumption that a(x, y, z) = a(r ), it results  R2 a(r ) sin ϕ dθ dϕ = 4π R 2 a(R). ∂ S R (0)

38

1 Elements of Vector Calculus

Another way to get the result is through the following development 

 S R (0)

∇ · A dV =

 R  2 da a + 4πr 2 dr = (a∇ · er + er · ∇a) d V = r dr S R (0) 0

R =

R

1 d 2 (r a) 4πr 2 dr = 4π r 2 dr

d(r 2 a) = 4π R 2 a(R),

0

0

which vanishes only if a(R) = 0. Exercise 1.34 Same of the previous exercise, in the case a(R, θ, ϕ) = −a(R, θ, π − ϕ), i.e. for a vector field which is antisymmetric respect to the x y plane. Solution π2π

 N

∇ · A d V = R2

a(R, θ, ϕ) sin ϕ dθ dϕ = 0 0

⎡ π ⎤ 2π 2 π ⎢ ⎥ ⎥ = R2 ⎢ ⎣ a(R, θ, ϕ) sin ϕ dϕ + a(R, θ, ϕ) sin ϕ dϕ ⎦ dθ = 0

0

π 2

0

0

π 2

⎡ π ⎤ 2π 2 π ⎢ ⎥ ⎥ = R2 ⎢ ⎣ a(R, θ, ϕ) sin ϕ dϕ − a(R, θ, π − ϕ) sin ϕ dϕ ⎦ dθ = ⎤ ⎡ π 2π 2 ⎥ ⎢ = 2R 2 ⎣ a(R, θ, ϕ) sin ϕ dϕ ⎦ dθ. 0

0

Exercise 1.35 Show that the radius, R, of a sphere centered in the origin can be expressed as  R=

1 4π

 S R (0)

r · n dσ

 13

,

where ∂ S R (0) is the sphere surface. Solution By the Divergence theorem, it is   r · n dσ = ∂ S R (0)

S R (0)

∇ · r d V = 3VR = 4π R 3 .

1.10 The Divergence Theorem

39

Exercise 1.36 Prove that the vector field F = (2x y + 3) i + (x 2 − 4z) j − 4y k is conservative and find its potential function. Solution The existence of the potential is guaranteed by that      ∂ Fy ∂ Fy ∂ Fz ∂ Fx ∂ Fx ∂ Fz − i+ − j+ − k= ∇ ∧F= ∂y ∂z ∂x ∂z ∂x ∂y = (−4 + 4) i + 0 · j + (2x − 2x)k = 0. 

We are looking for a function U (x, y, z) such that ∂U = Fx , ∂x

∂U = Fz . ∂z

∂U = Fy, ∂y

Consequently, given the above expression of F, a first integration over x gives  U (x, y, z) = Fx d x + f (y, z) = x 2 y + 3x + f (y, z), where f (y, z) is an arbitrary function of y and z only, to be determined by derivation of the above function U with respect to y ∂U ∂f = x2 + = Fy = x 2 − 4z, ∂y ∂y which, integrating with respect to y, gives  f (y, z) = − 4z dy = −4zy + g(z). Substituting in Eq. 1.36 we have U (x, y, z) = x 2 y + 3x − 4zy + g(z) whose derivation with respect to z leads to dg ∂U = −4y + = Fz = −4y ∂z dz

=⇒

dg =0 dz

=⇒

g(z) = c,

40

1 Elements of Vector Calculus

which, eventually, leads to U (x, y, z) = x 2 y + 3x − 4zy + c.

1.11 Velocity Fields If the fluid is considered as a set of N ‘particles’, in line of principle the perfect description of the fluid motion is through the knowledge at every time t of all the i = 1, 2, .., N position vectors ri (t), that could be determined by the knowledge of the total forces acting on every particle followed by the integration of their Newtonian equations of motion. This would involve an overwhelming evaluation of O(N 2 ) pair interactions. Let us think about a one-liter bottle of water as constituted by N H2 O molecules considered as fluid elementary particles. The number of water molecules in a liter of water is N  3.35 × 1025 so that the number of force interactions to compute is N (N − 1)/2 ∼ 5 × 1050 . This means that a perfectly deterministic particle-by-particle motion calculation is absolutely unviable. For this reason, some approximation is needed. There are two ways to approach the study of the motion of a fluid: Eulerian or Lagrangian. The Eulerian frame corresponds to a field view, and the relevant field is the velocity one. On the other side, the Lagrangian specification of the fluid flow is a way of looking at fluid motion where the ‘observer’ follows an individual fluid parcel (a representative, and small, piece of fluid) as it moves through space and time. Plotting the position of an individual parcel through time gives the pathline of the parcel. Mathematically speaking, supposedly known the position of a given ith fluid particle at a given time t0 , the Lagrangian description corresponds to the knowledge, at any time t, of the 3 functions ⎧ ⎪ ⎨xi = xi (xi0 , yi0 , z i0 ; t), (1.94) yi = yi (xi0 , yi0 , z i0 ; t), ⎪ ⎩ z i = z i (xi0 , yi0 , z i0 ; t), supposed continuous for every t (if t → t0 , then x → x0 , y → y0 , z → z 0 ). The time derivatives of the functions (1.94) ⎧ ⎪ ⎨x˙i = x˙i (xi0 , yi0 , z i0 ; t), (1.95) y˙i = y˙i (xi0 , yi0 , z i0 ; t), ⎪ ⎩ z˙ i = z˙ i (xi0 , yi0 , z i0 ; t), give the velocity of every particle at any time as a function of its initial position (that at time t0 ). The mathematical procedure of elimination of x0 , y0 , z 0 in the

1.11 Velocity Fields

41

rhs of Eq. 1.94, which is done (if possible) by its inversion, allows passing from Lagrangian to Eulerian specification, which corresponds to the identification of the velocity field (fluid flow) v = vx i + v y j + vz k at any time t and in any point x, y, z of the 3-dimensional space x˙ = x(x, ˙ y, z; t) = vx ,

y˙ = y˙ (x, y, z; t) = v y ,

z˙ = z˙ (x, y, z; t) = vz . (1.96)

The velocity flow may depend explicitly on t. In that case, the flow lines also depend on time and constitute, so, a set of ∞1 curves defined by the usual (see 1.72) differential conditions dy dz dx = = . vx (x, y, z; t) v y (x, y, z; t) vz (x, y, z; t)

(1.97)

If the velocity field is independent of t (stationary flux), the curves satisfying Eq. 1.97 are the actual trajectories of fluid motion. Exercise 1.37 Given a Lagrangian fluid view  r = x0 et i + y0 e−t j + z 0 et k, r(0) = r0 , or, in scalar form y = y0 e−t ,

x = x 0 et ,

z = z0 ,

(1.98)

z˙ = 0,

(1.99)

it is asked to obtain its Eulerian description. Solution By time differentiation of Eq. 1.98: x˙ = x0 et ,

y˙ = −y0 e−t ,

thus the Eulerian form, that is obtained by elimination of x0 and y0 in the above relations, is x˙ = x = vx ,

y˙ = −y = v y ,

z˙ = 0 = vz .

This is a planar stationary flow, whose flux lines (which are trajectories of motion) are dy dx = x −y

=⇒

log |x| = − log |y| + c,

42

1 Elements of Vector Calculus

Fig. 1.11 The cylindrical domain C0 of Exercise 1.40

z z=1

C0

z=0 x y

or, equivalently |x||y| = ec = C hyperbola. The above Cartesian expression of trajectories obviously coincides with that obtained by a straight elimination of the t parameter in Eq. 1.99, leading to x = x0 et and y = y0 e−t , so that x y = x0 y0 et e−t = x0 y0 . Exercise 1.38 Given the (Eulerian) velocity field v = − cos t i + sin t j, determine fluid trajectories and stream lines.4 Solution Left to the reader. Exercise 1.39 Study the properties of the fluid motion represented by the Lagrangian expression  r=

4A

x0 + y0 t x0 − y0 −t e + e 2 2



 i+

x0 + y0 t x0 − y0 −t e − e 2 2

 j + z 0 k.

stream line has in any its point the direction of the velocity field. When the flow is stationary, stream lines coincide with fluid trajectories.

1.12 Meaning of the Divergence of a Vector Field

43

1.12 Meaning of the Divergence of a Vector Field A fluid in motion can expand or contract: a fluid element that at t0 covers a portion of space C0 at time t will extend, generally, over a portion of space C such that, in general, vol(C0 ) = vol(C). In such case we speak of a compressible fluid. Given the mass continuity equation which, in Lagrangian form, writes as Dρ = −ρ∇ · v, Dt

(1.100)

where D/Dt represents the Lagrangian derivative (that evaluated along the flow), and defining the specific volume V ∗ = 1/ρ as that occupied by the unitary fluid mass, we clearly have that in the incompressible case V ∗ = const. and thus DV ∗ 1 Dρ =− 2 = 0. Dt ρ Dt So, from Eq. 1.100 an incompressible flow is solenoidal, ∇ · v = 0. Exercise 1.40 Let us consider the Lagrangian flow defined by Eq. 1.37 r = x0 et i + y0 e−t j + k.

(1.101)

If at t = t0 the domain C0 is given by the cylinder (Fig. 1.11) 0 ≤ z ≤ 1, x02 + y02 = a 2 , show that at a generic t the domain C results defined by  x = x0 et , x expansion 0 ≤ z ≤ 1, y = y0 e−t , y compression. Solution From Eq. 1.101 we get x2 y2 + = x02 + y02 = a 2 , e2t e−2t that is x 2 y 2 + = 1, aet ae−t which is the equation of the ellipse of semiaxes |a|et (x expansion) and |a|e−t (y contraction). Note the invariance of the domain volume in time vol(C) = πaet ae−t = πa 2 = vol(C0 ).

44

1 Elements of Vector Calculus

This agrees with the fluid incompressibility, guaranteed by that ∇ · v = 0 (indeed, v = x i − y j). Exercise 1.41 Suggest examples of flows for which vol(C0 ) = vol(C).

1.13 Link Between Divergence and Volume Variation If C is the space domain occupied, at time t, by a fluid which at time t0 occupied C0 , in the Lagrangian framework r = r(r0 ; t) the volume of C is expressed as   V (t) = d 3r = J (r0 ; t) d 3 r0 , (1.102) C

C0

where J (r; t) is the Jacobian determinant of the coordinate transformation from r to r0    ∂x ∂x ∂x      ∂ x0 ∂ y0 ∂z 0   ∂r  ∂y ∂y ∂y  = J (r0 ; t) ≡ det .  ∂ x0 ∂ y0 ∂z 0  ∂r0  ∂z ∂z ∂z      ∂ x0 ∂ y0 ∂z 0 If J (r0 ; t) is time-differentiable with continuity, it is   d J (r; t) d 3 r0 = J˙(r0 ; t) d 3 r0 . V˙ (t) = dt C0 C0 The integrand in the rightmost integral above is       J˙(r0 ; t) =     

  ∂2x ∂2 y ∂2z     ∂t∂ x0 ∂t∂ x0 ∂t∂ x0   ∂x ∂y ∂z   + ∂ y0 ∂ y0 ∂ y0   ∂x ∂y ∂z     ∂z 0 ∂z 0 ∂z 0

∂x ∂y ∂z ∂ x0 ∂ x0 ∂ x0 ∂2x ∂2 y ∂2z ∂t∂ y0 ∂t∂ y0 ∂t∂ y0 ∂x ∂y ∂z ∂z 0 ∂z 0 ∂z 0

            +        

∂x ∂y ∂z ∂ x0 ∂ x0 ∂ x0 ∂x ∂y ∂z ∂ y0 ∂ y0 ∂ y0 ∂2x ∂2 y ∂2z ∂t∂z 0 ∂t∂z 0 ∂t∂z 0

      .    

In the hypothesis that the first and second partial derivatives are continuous, then

1.13 Link Between Divergence and Volume Variation

    ∂ ∂x   ∂ x0 ∂t J˙(r0 , t) =  0   0   1   0   +  ∂x  ∂  ∂z ∂t 0

45

     1  0  0          ∂ ∂ ∂ ∂x ∂y ∂z  + +   ∂y 1 0 ∂t ∂ y ∂t ∂ y ∂t  0 0 0      0 1 0 0 1   0 0  1  0     , ∂ ∂ ∂y ∂z  ∂z 0 ∂t ∂z 0 ∂t  ∂ ∂ x0



∂y ∂t



∂ ∂ x0



∂z ∂t

more compactly written as  ∂v  x   J˙(r0 , t) =  ∂ x0  0  0  ∂vx = + ∂ x0

    ∂v y ∂vz   1 0 0   1 0 0    ∂v ∂v ∂v    y z   0 1 0  ∂ x0 ∂ x0  +  x + =   1 0   ∂ y0 ∂ y0 ∂ y0   ∂vx ∂v y ∂vz      0 0 1 0 1  ∂z 0 ∂z 0 ∂z 0  ∂v y ∂vz + = (∇ · v)t=t0 . ∂ y0 ∂z 0 t=t0

Given the arbitrariness of t0 , we thus get   ∇ · v d 3 r = V˙ (t). V˙ (t0 ) = (∇ · v)t=t0 d 3 r0 = C0

C

As a consequence, if ∇ · v = 0 everywhere in space, then V˙ (t) = 0

⇐⇒ V (t) = constant.

Viceversa, if V˙ (t) = 0, ∀t, then the continuity of v and of its first derivatives imply ∇ · v = 0: the fluid is said to be incompressible. Applying the mean theorem for integrals, we have V˙ = ∇ · v, V

lim

C→(x,y,z)

V˙ = ∇ · v, V

where the brackets represent an average over the domain. Clearly, ∇ · v is the fractional rate of expansion (∇ · v > 0) or compression (∇ · v < 0) of the fluid around the generic point (x, y, z). Exercise 1.42 Show that the volume of a fluid domain C is expressible by the formula  1 r · n dσ. V (C) = 3 ∂C Solution Left to the reader.

46

1 Elements of Vector Calculus

1.14 Solenoidal Vector Fields 1.14.1 Flux Tubes A flux tube is a particular type of vector field (see Sect. 1.9), defined as that surface whose lateral surface has tangent in every point which is parallel to the field lines, assuming that an outward normal can be defined in every point of such a surface. The simplest example is that of a fluid which moves into a physical tube without crossing its lateral sides. We can prove that in the case of an incompressible flow, the fluid speed increases (decreases) when the flux tube cross section shrinks (widens). Let us consider a region of space T limited by a flux tube and by two orthogonal cross sections S1 and S2 to the tube, as in Fig. 1.12. If the generic vector field F is solenoidal (like the velocity field of an incompressible fluid) then     ∇ · F dV = F · n dσ = F · n dσ + F · n dσ = 0, ∂T

T

S1 ∪S2

L

(1.103)

where ∂ T is the union of S1 , S2 and of the lateral surface, L, of the tube. Clearly, the surface integral over the lateral surface is zero because there F · n = 0, so that Eq. 1.103 reduces to  F · n dσ = 0, S1 ∪S2

which implies 

 F · n dσ = − S1

F · n dσ. S2

F n n

F

S1 Fig. 1.12 A flux tube

S2

1.14 Solenoidal Vector Fields

47

A straight application of the integral mean theorem leads to F1 ar ea(S1 ) = F2 ar ea(S2 ), where F1 and F2 are the norms of the F vector field outward normal components over the two cross sectional surfaces. This, obviously, leads to the scaling ar ea(S2 ) F1  = , F2  ar ea(S1 ) that is: the intensity of a solenoidal vector field increases (the field line density increases) where the flux tube shrinks while its intensity decreases (as the field lines density) across larger cross sections of the flux tube. The ratio is inversely proportional to the area of the orthogonal cross sections.

1.15 Further Readings This chapter contains all the notions that are needed to understand the remainder of the book. Readers wishing to deepen their knowledge about theory of functions of many variables and vector analysis can refer to many excellent introductory textbooks, like, e.g., [23, 25]. Aspects of fluid-dynamics are covered by [15].

Chapter 2

Newtonian Gravitational Interaction

Discrete Mass Distributions 2.1 Single Particle Gravitational Potential Let P a point of mass m p located at position r in a Cartesian reference frame whose origin is occupied by another point of mass m. The Newtonian gravity force exerted by the particle of mass m on the particle of mass m P is F(r) = −G

m mp m mp er = −G 3 r, r2 r

then, if m p = 1 F(r) = −G Assuming m P = 1 and letting U (r ) = G ∇U = ∇G

m r. r3

(2.1)

m , it results r

m 1 m m = Gm ∇ = −G 2 ∇r = −G 2 er = F, r r r r

(2.2)

so that U (r ) is the single particle Newtonian potential (Fig. 2.1). r 3−n Because, in Rn \ {0}, ∇ · Gm 3 = G m 3 , we have r r ∇ · ∇U ≡ ∇ 2 U = 0.

© Springer Nature Switzerland AG 2019 R. A. Capuzzo Dolcetta, Classical Newtonian Gravity, UNITEXT for Physics, https://doi.org/10.1007/978-3-030-25846-7_2

49

50

2 Newtonian Gravitational Interaction z

Fig. 2.1 A Cartesian reference frame centered on the particle of mass m. The particle in P(x,y,z) feels the potential generated by m

r

P(x,y,z)

m O

y

x

This means that (i) F is a solenoidal vector field, and (ii) U (r) is a harmonic function1 in the empty space around the particle m placed in the origin. The vector equation of motion of the P particle with given initial conditions is ⎧ m ⎪ ⎨ r¨ = −G 3 r r r(0) = r0 ⎪ ⎩ r˙ (0) = r˙ 0 .

(2.3)

The Eq. 2.3 constitute a system of 3 scalar, 2nd order, ordinary differential equations. The motion of the P particle in the space around the particle of mass m sitting at the coordinate frame and giving the potential is the typical case of one-body problem. Of course, due to that by Newton’s third principle the particle of mass m is also feeling the acceleration induced by the particle of mass m P , the coordinate frame anchored in the m particle is not inertial. It becomes inertial in the limit m  m P (as, for instance, in the case of the Sun-Earth system). Note that the 2nd order system in Eq. 2.3 can be easily reduced to a system of 6, 1st order, differential equations, by the introduction of the auxiliary vector v = r˙ . This is not relevant for the one-body (and two-body) point like Newtonian cases because the explicit solution is known, while it is needed in the case of more general interaction potentials where the explicit solution does not exist and should be approximated by numerical methods.

2.2 Motion of a Particle Gravitating in a Resisting Medium The equation of motion of a particle of mass m P submitted to both a conservative force Fc = m P ∇U and a dissipative force in the regime of linear dependence on the velocity, Fd = −λ˙r (λ > 0), is 1A

function is called harmonic if its Laplacian is zero.

2.2 Motion of a Particle Gravitating in a Resisting Medium

r¨ =

51

Fc + Fd λ r˙ . = ∇U − mP mP

(2.4)

A scalar multiplication by r˙ of both sides of Eq. 2.4 and an integration over time leads to t t t λ r˙ · r¨ dt = ∇U · r˙ dt − r˙ · r˙ dt, (2.5) mP 0

0

0

whence 1 λ 1 r˙ · r˙ − U (r) = r˙ 0 · r˙ 0 − U (r0 ) − 2 2 mP

t r˙ · r˙ dt.

(2.6)

0

Defining the mechanical energy per unit mass, E, as E = 21 r˙ · r˙ − U (r), we get t λ E(t) = E 0 − r˙ · r˙ dt. mP

(2.7)

0

Let us note that λ Ld = − mP

t

r(t) r˙ · r˙ dt = −

0

λ r˙ · dr mP

(2.8)

r(0)

represents the (always negative) work (per unit mass) done by the dissipative force Fd in the time interval [0, t]. Consequently the expression λ r˙ · r˙ = Fd · v ≤ 0 E˙ = − mP gives the power being dissipated at any time t. Note Let us recall a concept of differential calculus. Definition 2.1 Let A an open set in R3 and X d x + Y dy + Z dz a differential form, where X, Y, Z are continuous and differentiable functions of x, y, z in A. The form is called exact if exists a function f (x, y, z) such that d f = X d x + Y dy + Z dz, where d f is the total differential of f . Given that, we have this theorem:

(2.9)

52

2 Newtonian Gravitational Interaction

Theorem 2.1 The differential form X d x + Y dy + Z dz, where X, Y, Z are continuous and differentiable functions of x, y, z in a simply connected set A, is an exact differential if and only if the following equalities hold ∂X ∂Y = , ∂y ∂x

∂X ∂Z = , ∂z ∂x

∂Y ∂Z = . ∂z ∂y

(2.10)

Going back to what said above on the motion of a particle in presence of a dissipative force, to recover a constant of motion in the case of dissipative force of the type in Eq. 2.4, it is needed that ∇ ∧ r˙ = 0, which corresponds to a series of equalities similar to those in Eq. 2.10 ∂ x˙ ∂ y˙ = , ∂y ∂x

∂ x˙ ∂ z˙ = , ∂z ∂x

∂ y˙ ∂ z˙ = . ∂z ∂y

(2.11)

These relations would imply (by the previous theorem) the existence of a scalar function Φ(r) whose total differential is the differential form xd ˙ x+ y˙ dy + z˙ dz and thus r˙ = ∇Φ(r), (2.12) which would imply λ − mP

r(t) λ ∇Φ · dr = − [Φ(r) − Φ(r0 )] , mP

(2.13)

r(0)

so to have a new, allegedly generalized, “energy” constant of motion E+

λ λ Φ(r) = E 0 + Φ(r0 ) = const. mP mP

(2.14)

Anyway, the questions are: • does this make sense? • what is its physical meaning?

2.3 The Gravitational N-Body Case Given a set of isolated N particles subjected to their mutual, pairwise, interaction force, the force acting on the ith particle, Fi (i = 1, 2, ..., N ), is given by the sum over j = 1, 2, ..., N (excluding j = i) of the forces contributed by every other jth particle, Fi j . In the case of purely Newtonian interaction, if we indicate with ri the position vector of the ith particle and with m i its mass, the force exerted by the jth particle of mass m j on the ith is, ∀i = j,

2.3 The Gravitational N -Body Case

Fi j = Fi j so that Fi =

53

r j − ri m j mi with Fi j = G , |r j − ri | |r j − ri |2

N 

Fi j =

N 

j=1 i= j

G

j=1 i= j

m j mi (r j − ri ), |r j − ri |3

(2.15)

(2.16)

is the total force exerted by the other N − 1 bodies of the system on the ith. Clearly, Fi j = −F ji ,

Fi j = F ji .

(2.17)

The total momentum of the forces acting on the N -body system is the sum of the momentum of the above internal forces and that of external forces (F(e) ) (i)

M=M

(e)

+M

=

N 

ri ∧ Fi +

N 

i=1

=

N 

ri ∧

i=1

N 

Fi j +

i=1 i= j

ri ∧ Fi(e) =

i=1 N 

(2.18) ri ∧ Fi(e) .

i=1

If we consider the system as composed by the whole set of pairs (i, j), the momentum of internal forces can  be written as the sum of the contribution of all these pairs, N N (N − 1) , so that which are, in number, = 2 2 M(i) =

N  i=1

ri ∧

N  j=1 j=i

Fi j =

N N   i=1 j=1 j=i

ri ∧ Fi j =



(ri ∧ Fi j + r j ∧ F ji ) =

pairs

 mi m j (ri − r j ) ∧ G (r j − ri ) = 0, = |r − r j |3 i pairs

(2.19) ∀i, j.

Due to that M(i) = 0, the second cardinal equation of dynamics writes as N d  L˙ = ri ∧ m i r˙ i = M(e) . dt i=1

Consequently, for a system which is either isolated or submitted to a spherically symmetric force field,it is M(e) = 0 and so the angular momentum is an integral of motion, L = L0 = const.

54

2 Newtonian Gravitational Interaction

2.3.1 Potential of N Gravitating Bodies Given N points whose masses and positions are m k and rk , k = 1, 2..., N , if they interact via Newtonian gravity only, the function defined as N N N N −1  N −1    mi m j G mi m j G  mi m j U (r1 , . . . r N ) = =G = , 2 i, j=1 |ri − r j | r 2 ri j ij i=1 j=i+1 j=1 i= j+1 i= j

(2.20) (where ri j ≡ |ri − r j | = r ji ), is the potential2 of the Newtonian force because  Fk = ∇k U,

∇k ≡

with

∂ ∂ ∂ , , ∂xk ∂xk ∂xk

.

This is easy to verify; indeed3 N N N ∂U 1  m i m j ∂ri j 1  m i m k ∂rik 1  m k m j ∂rk j =− G 2 =− G − G = ∂xk 2 i, j=1 2 i=1 rik ∂xk 2 j=1 rk j ∂xk ri j ∂xk j=k

i= j

=−

N  l=1 l=k

G

j=k

m k ml (xk − xl ) = Fkx , rkl3

and similarly for the yk and z k coordinates. Given this potential, the equations of motion of the generic ith body of the N -body set are synthetically written as ⎧ ∇U ⎪ ⎨ r¨ i = mi r (0) = ri0 ⎪ i ⎩ r˙ i (0) = r˙ i0 ,

(2.21)

where the initial conditions for both position and velocity have been specified. The particular case of a test particle of mass m, identified by its position vector r and moving (without influencing the motion of the other bodies) in the force field generated by other N bodies yields to its equation of motion

2 The

potential in Eq. 2.20 is NOT per unit mass. the antisymmetry of the action of ∇k respect to the pair distance; actually

3 Note

∇k (|rk − r j |) =

rk − r j = −∇ j (|rk − r j |). |rk − r j |

2.3 The Gravitational N -Body Case

r¨ =

55 N 

G

j=1

mj (r j − r), |r j − r|3

(2.22)

which corresponds to the force deriving from this potential U=

N 

G

j=1

mj m . |r j − r|

(2.23)

2.3.2 Mechanical Energy of the N Bodies The total mechanical energy of the N body system is E=

N 1 1  mi m j m i vi2 − G = T − U, 2 i=1 2 i, j=1 ri j

(2.24)

i= j

where T is the total kinetic energy. E is an integral of motion (i.e. it is constant along any possible solution of the motion equations and does not depend explicitly on t), how can be easily seen by taking the time derivative of E along the motion (we leave this to the reader as exercise). It is relevant noting that only the total energy E is conserved while the individual particle energy E i 4 is not. This means that, even if during the system evolution a particle assumes a positive energy, E i > 0, which corresponds to a speed larger than the local escape velocity, it is not guaranteed it will be “expelled” from the N -body system because it could interact with other field objects and reduce E i to a negative value again. This is of course unlikely whenever the speed over the local escape velocity is kept until reaching a very low dense environment.

2.4 The Scalar Virial Theorem A relevant relation involving the location in the phase space5 of a system of N interacting objects takes the name of virial theorem. Given a set of N point masses which in a Cartesian reference frame are identified by position vectors ri , i = 1, 2, ..., N , their polar moment of inertia, I , is 4E

i

= 21 m i vi2 −

j=1 i= j

G

mi m j ri j

.

5 The phase space is a 6-dimensional space where every point (x,

and velocity of a particle.

y, z, vx , v y , vz ) represents position

56

2 Newtonian Gravitational Interaction

I =

N 

m i ri · ri .

(2.25)

i=1

Taking its second derivative with respect to time we have I¨ = 2

N 

m i (˙ri · r˙ i + ri · r¨i ),

(2.26)

i=1

which may be written as  1¨ ri · ∇i U. I = 2T + 2 i=1

(2.27)

N   ∂U ∂U ∂U xi , ri · ∇i U = + yi + zi ∂xi ∂ yi ∂z i i=1 i=1

(2.28)

N

The quantity N 

is called Clausius’ virial after Rudolf Clausius who introduced the so called virial theorem in 1870. Let us recall an important mathematical aspect. Definition 2.2 Let f (r) be a real function of m real variables; this function is called homogeneous of degree n if for any α = 0 the relation f (αr) = αn f (r)

∀r ∈ Rm ,

(2.29)

holds. When Eq. 2.29 holds for α > 0 or α < 0 we speak of f (r) as positively or negatively homogeneous, respectively. Lemma 2.1 If the interaction potential U (r1 , . . . , r N ) is a homogeneous function of degree n, it results N  ri · ∇i U = nU. (2.30) i=1

Proof If U (r1 , . . . , r N ) is homogeneous of degree n it satisfies the relation Eq. 2.29, and so by differentiation with respect to α of both sides of that equation, we have ∂  n ∂ U (αr1 , . . . , αr N ) = α U (r1 , . . . , r N ) , ∂α ∂α

(2.31)

that develops into N   ∂U ∂αxi ∂U ∂αyi ∂U ∂αz i + + = nαn−1 U, ∂αx ∂α ∂αy ∂α ∂αz ∂α i i i i=1

(2.32)

2.4 The Scalar Virial Theorem

57

equivalent to N  (∇αri U ) · ri = nαn−1 U.

(2.33)

i=1

Given the arbitrariness of α we can set α = 1 in the above equation and the relation 2.30 is obtained. Thanks to lemma (2.1) we can write Eq. (2.27) as 1¨ I = 2T + nU, 2

(2.34)

which is the actual thesis of the virial scalar theorem. Due to that the potential of a set of N particles interacting via Newtonian classic gravity is given by 1  mi m j G , (2.35) U= 2 i, j=1 ri j i= j

which is (as easily seen) a positively homogeneous function of degree n = −1, the virial theorem thesis assumes the form 1¨ I = 2T − U = 2T + Ω = E + T 2

(2.36)

where Ω ≡ −U is the potential energy of the system. Note that if I¨ = 0 then E = −T , that means d E/dT = −1: the thermal capacity of a self-gravitating system is negative!

2.4.1 Consequences of the Virial Theorem By dividing the 2 sides of the virial relation (Eq. 2.34), where we set n = −1) by |Ω| = −Ω we have 2T 1 I¨ = − 1, (2.37) 2 |Ω| |Ω| which, upon definition of the virial ratio as Q ≡ 2T /|Ω| ≥ 0, translates into Q=

1 I¨ + 1. 2 |Ω|

(2.38)

A system is said viralized if it is globally steady, i.e. if I¨ = 0 and so Q = 1, ∀t. It is subviral when Q < 1, which corresponds to a negative acceleration of the

58

2 Newtonian Gravitational Interaction

polar inertia moment, I¨ < 0 (initial contraction) and supervirial when Q > 1, which corresponds to a positive acceleration of the polar inertia moment, I¨ > 0 (initial expansion). Let us examine some particular cases concerning I¨ = 0. If the system motion is periodic with period P, performing an integral average over the period of left side and second rhs of Eq. 2.36 we get 11 1 ¨ I P = 2 2P

P

1 1 ˙ I (P) − I˙(0) = 0 = 2T P + Ω P . I¨ dt = 2P

(2.39)

0

In the above equation, we use that, if I (t) is periodic, also I˙(t) is periodic. This is a general property. Indeed, to say that I (t) is periodic means that I (t + P) = I (t) for every t. Consequently I (t + P + Δt) − I (t + P) I (t + Δt) − I (t) = lim = I˙(t) I˙(t + P) = lim Δt→0 Δt→0 Δt Δt Equation 2.39 can be, equivalently, expressed as < Q > P = 1. In general, performing an average over a generic time interval τ , the result is 11 1 ¨  I τ = 2 2τ

τ 0

11 ˙ I (τ ) − I˙(0) = 2T τ + Ω τ . I¨ dt = 2τ

(2.40)

Thus, if τ is properly chosen of the order of the oscillation time of the system, it is I˙(τ )  I˙(0) or, equivalently, < Q >τ  1, i.e. the system is globally virialized over the time scale τ . In the case of a motion which is non-periodic but limited in the phase space, i.e. ∃L , M > 0 :

ri (t) ≤ L , |˙ri | ≤ M ∀t,

(2.41)

it follows that I˙ is limited, i.e. ∃H > 0 such that | I˙| ≤ H, ∀t and so 1 ¨ 1 ˙ 1  I ∞ = lim I (τ ) − I˙(0) ≤ 2 2 τ →∞ τ 1 1 1 ˙ 2H | I (τ )| + | I˙(0)| ≤ lim = 0. ≤ lim 2 τ →∞ τ 2 τ →∞ τ A last, but relevant, result is that if the system energy is positive, E > 0, the system is unstable (unbound in physical space). Actually 1¨ I = E + T ≥ E > 0, 2

2.4 The Scalar Virial Theorem

59

and, by two integrations over time from 0 to t I (t) ≥ Et 2 + I˙(0)t + I (0).

(2.42)

Given that E > 0, the latter formula implies, lim I (t) = +∞,

(2.43)

t→∞

that means the system is indeed unstable, in what exists at least an escaping particle, for which ri (t) → ∞.

2.5 Continuous Distributions of Matter Let F(r) be the gravity force exerted on the unitary mass in the point r (see Fig. 2.2) by a regular matter distribution with volume density ρ(r) defined in a bound domain C. The generalization to a continuous distribution of the expression by sum of the total force exerted by a set of N point masses on a unit test mass sited in r yields straightforwardly to the integral  F(r) = G C

r − r ρ(r )d 3 r . |r − r|3

(2.44)

Clearly, the potential of this force is  U(r) = G C

ρ(r ) 3  d r, |r − r|

(2.45)

given that it results ∇U = F(r).

(2.46) z

Fig. 2.2 Radius vector, r, of the particle feeling the infinitesimal force generated by particle of mass dm  in r

r−r’ r

dm’=dm(r’) r’

y x

60

2 Newtonian Gravitational Interaction

2.5.1 Poisson’s and Laplace’s Equations Let us see some characteristics of the force field Eq. 2.44 and its potential Eq. 2.45. The divergence of the force field is the divergence of the integral in Eq. 2.44. Now, we distinguish the internal case, r ∈ C, from the external, r ∈ / C. We have  r − r ∇ · F(r) = G ∇·  ρ(r ) d 3 r = |r − r|3 C   1 1   ρ(r )d 3 r . =G ∇ · (r − r) + (r − r) · ∇   3 |r − r|3 C |r − r| (2.47) For every r and r in R3 , ∇ · (r − r) = −3, and ∇

1 r − r 3 1  , = − ∇|r − r | = 3 |r − r |3 |r − r |4 |r − r |4 |r − r|

(2.48)

consequently  ∇ · F(r) = G C

 3 3(r − r) · (r − r) −  ρ(r )d 3 r . + |r − r|3 |r − r|5

(2.49)

Now, if we evaluate the divergence in an external point, r ∈ / C, it is r = r , and the integral in Eq. 2.49 is a proper integral whose integrand is zero. Consequently, ∇ · F = 0,

(2.50)

in the empty space around a distribution of matter. This means that the force field is solenoidal and the Laplace’s equation ∇ · ∇U ≡ ∇ 2 U = 0

(2.51)

holds. In other words, the field equation for the classical gravitational force in the empty space is just the Laplace’s equation. On the other hand, the evaluation of the force on a point within the matter distribution implies to that the above integral is improper (the integrand is singular in r = r. We can evaluate this improper integral as  ∇ · F = G lim 

r →r

ρ(r )∇ · C

r − r 3  d r = G lim h→0 |r − r|3

 Sh (r)

ρ(r )∇ ·

r − r 3  d r, |r − r|3

2.5 Continuous Distributions of Matter

61

  where Sh (r) = r : |r − r| ≤ h is the sphere of radius h centered in r. The integral is evaluated by applying the mean theorem for integrals, first, and, after, the divergence theorem. The mean theorem allows taking out of the integral the density as evaluated in a (unknown) point rh within Sh (r), while the divergence theorem can be used after we swapped ∇r with −∇r ,6 thanks also to the continuity of ρ(r ). Finally  r − r 3  ∇r ·  d r = ∇ · F = −G lim ρ(rh ) h→0 |r − r|3 S (r) (2.52)  h r − r · n dσ, = −G lim ρ(rh )  3 h→0 ∂ Sh (r) |r − r| where ∂ Sh (r) is the sphere surface (r : |r − r| = h) and n is its normal unit vector. Obviously, r − r , (2.53) n = er −r =  |r − r| so that  ∇ · F = −G lim ρ(rh ) h→0

∂ Sh (r)

1 ρ(rh ) dσ  = −G lim h→0 h 2 |r − r|2

 ∂ Sh (r)

dσ  =

ρ(rh ) 4πh 2 = −4πGρ(r), = −G lim h→0 h 2 because of the continuity of the density ρ(r). The final result corresponds to the so called Poisson’s equation ∇ · F = ∇ · ∇U = −4πGρ(r),

(2.54)

which is the gravitational field equation in the matter domain. To resume, the field equations linking a regular matter density distribution ρ(r) and the Newtonian potential it generates are 

if r ∈ / C, ∇ · F = ∇ · ∇U (r) = ∇ 2 U = 0, ∇ · F = ∇ · ∇U (r) = ∇ 2 U = −4πGρ(r), if r ∈ C.

(2.55)

Note The definition of the force per unit mass by the integral Eq. 2.44 is wherever valid. Actually, the integral is improper just for r = r , but it always converges. Indeed, it results

6 The exchange is not a general property; in this case it is allowed because applied to (r

− r)/|r − r|.

62

2 Newtonian Gravitational Interaction

   0 ≤  lim G

  ρ(r ) d 3 r  3  ≤ G lim (r − r) d r ρ(r )  h  3  2 h→0 h→0 Sh (r) |r − r| Sh (r) |r − r|  h   2 |r − r|  = G lim ρ(rh ) 4π   d |r − r| = 4πGρ(rh ) lim h = 0 (2.56) h→0 h→0 − r|2 |r   0

2.6 Gauss’ Theorem Theorem 2.2 (Gauss’ Theorem) The integral of the outward normal component of the force F(r) to any closed surface containing a quantity M of matter is equal to −4πG M. In other words, the outward force flux through the surface is −4πG times the mass enclosed (in the electrostatic case the mass is substituted by the electric charge). Proof By integration of both sides of the Poisson’s equation over a generic subdomain N of the domain where the density is non zero we have, for the divergence theorem,    ∇ · ∇U d 3 r = ∇U · n dσ = −4πG ρ(r) d 3 r = −4πG M. ∂N

N

N

Exercise 2.1 Show whether the following force fields are solenoidal in R3 , and discuss possible singular points.  x y 1 1 (a) F = 2 j+  i + arctan k 2 y + z2 z z y + z2 x y (b) F = − 2 i + 2 j r r (c) F = er r Solution To check whether the fields are solenoidal it is necessary to verify where ∇ · F = 0. Exercise (a)   x y 1 1 j+  i + arctan k = ∇ ·F=∇ · y2 + z2 z z y2 + z2      ∂ 1 x y ∂ 1 ∂  + arctan + = = ∂x y 2 + z 2 ∂y z z ∂z y2 + z2  z 1 −z 2 y2 + z2 − z 1 + + = 2 3 =  3 . y + z2 z y2 + z2 y2 + z2 2 y2 + z2 2 

2.6 Gauss’ Theorem

63

It results that F is not solenoidal. Exercise (b)  Given that r = x 2 + y 2 , F(r) becomes F=−

x2

y x i+ 2 j 2 +y x + y2

 ∂ x = ∂ y x 2 + y2 2x y 2x y = 2 −  2 = 0. x 2 + y2 x 2 + y2

and

∇ ·F=

∂ ∂x

 −

y 2 x + y2



+

Consequently ∇ · F = 0,

∀(x, y) ∈ 2 \ {0, 0}.

Exercise (c) ∇ · F = er ∇ · r + r · ∇er = 3er + er r · ∇r = er (3 + r ) = 0,

∀r,

(2.57)

and the field is not solenoidal. Anyway, it is possible to scale it with a scalar function α(r ) to make it solenoidal. Actually, ∇ · αF = α∇ · F + F · ∇α = α∇ · F + so to have

dα F · er = 0, dr

1 dα ∇ ·F 3+r er (3 + r ) =− =− , =− α dr F · er er r r

which implies α=

c , r 3 er

where c = 0 is an arbitrary constant. The field αF =

c r is thus solenoidal for r = 0. r3

Exercise 2.2 Compute the flux through the unitary radius sphere centered at the coordinate origin, S1 (0), of the force fields: (a) F = r (b) F = x i

64

2 Newtonian Gravitational Interaction

Solution Exercise (a)  φ=

 ∂ S1 (0)

F · n dσ =

π 2π

∂ S1 (0)

r dσ =

sin ϕ dθdϕ = 4π 0

0

Exercise (b)  φ=

∂ S1 (0)

 F · n dσ =

∂ S1 (0)

x2 dσ = r

sin2 ϕ cos2 θ sin ϕ dθ dϕ = 0 0

2π =

π2π

π cos θ dθ

sin3 ϕ dϕ,

2

0

0

but π

−1 1 4 2 2 sin ϕ dϕ = − sin ϕ dcos ϕ = (1 − cos2 ϕ) dcos ϕ = − + 2 = , 3 3 3

0

−1

1

so to have  2π   2π 1 + cos 2θ 1 4 2π 4 dθ = θ + sin θ cos θ 0 = π. φ= cos2 θ dθ = 3 0 2 2 3 0 Clearly, the same result above could have been obtained much faster by application of the divergence theorem  φ=

S1 (0)

∇ · F dV = V =

4 π. 3

2.7 Gravitational Potential Energy As we said in Sect. 2.5, the Newtonian gravitational force is a conservative field. Indeed  r − r ∇ ∧ F(r) = G ∇∧  ρ(r ) d 3 r = 0, (2.58) |r − r|3 C as it is easily seen because, for a point external to the mass domain (r = r )

2.7 Gravitational Potential Energy

∇∧

65

r − r 1 1 =  ∇ ∧ (r − r) − (r − r) ∧ ∇  = |r − r|2 |r − r|3 |r − r|3  r − r −3 = 0. = −(r − r) ∧ |r − r|4 |r − r|

If the point where we calculate the curl of F is in C, we can see, in the same way followed to obtain Poisson’s equation, that  ∇ ∧ F(r) = lim G h→0

Sh (r)

∇∧

r − r ρ(r ) d 3 r = 0. |r − r|3

(2.59)

Given its conservativity, it is possible to obtain useful information on F(r) from its potential function, U (r) (see Eq. 2.45), and potential energy. We give here the definition of gravitational potential energy. Definition 2.3 Let C a material domain characterized by a regular density distribution ρ(r) vanishing out of C. We define as gravitational energy, Ω, of the system C the work done against gravitational forces to bring all the matter elements from infinite to their actual configuration in C. Let us see two ways to obtain the expression for Ω. The work done in transferring a quantity of matter δm from infinity to position r is  δΩ = −

r ∞

 d L = −δm

r

∞

 F · dr = −δm U (r) − lim U (r) = −δm U (r), r→∞

consequently a local increment in density δρ(r) corresponds to an increment of potential energy  δρ(r)U (r) d 3 r. (2.60) δΩ = − C

According to Poisson’s equation we have ∇ 2 (U + δU ) = −4πG(ρ + δρ), i.e. ∇ 2 δU = −4πGδρ, whence δρ(r) = −

1 ∇ 2 δU. 4πG

(2.61)

Consequently, the expression Eq. 2.60 becomes 1 δΩ = 4πG



1 U ∇ δU d r ≡ 4πG C 2

 U ∇ · ∇δU d 3 r.

3

C

(2.62)

66

2 Newtonian Gravitational Interaction

Let us recall the Theorem 2.3 (Integration by parts) Let f (x, y, z) be a scalar function and A = A(x, y, z) a vector function, both defined over a 3D domain C and such that f , A, and C satisfy the hypotheses of the divergence theorem. The formula of integration by parts in 3D holds 

 ∇ · ( f A) d r =

 (f∇ ·A+A·∇ f) d r =

3

3

C

C

∂C

f A · n dσ. (2.63)

Note that the above formula derives by a straight application of the divergence theorem to the function f A. If we apply the above formula to the integral in Eq. 2.62, we get δΩ =

1 4πG

 



∂R3

U ∇δU · n dσ −

R3

∇δU · ∇U d 3 r .

(2.64)

The first integral in the above expression vanishes because for large r ⎧ 1 ⎪ ⎪ ⎪ ⎨ U (r ) ∝ r , 1 |∇δU (r )| ∝ 2 , ⎪ ⎪ r ⎪ ⎩ dσ ∝ r dr, 

so that it results

∂R3

U ∇δU · n dσ ∝ lim

r →∞



−

1 = 0. r

(2.65)

Consequently, Eq. 2.64 reduces to 1 δΩ = − 4πG



1 1 δ ∇δU · ∇U d r = − 3 4πG 2 R



3

R3

∇U · ∇U d 3 r. (2.66)

If we now sum up all the contributions δΩ we, finally, have the expression for the potential energy Ω  1 |∇U |2 d 3 r. (2.67) Ω=− 8πG R3 An alternative form for Ω is obtained by setting A ≡ ∇U and f ≡ U at the r.h.s. of Eq. 2.67 in the above expression and using, again, the formula for integration by parts, which leads to Ω=−

1 8πG



  1 U ∇ 2U d 3 r = − U (r)ρ(r) d 3 r, 2 R3 R3

 ∂R3

U ∇U · n dσ −

because the first integral in Eq. 2.68 vanishes and ∇ 2 U = −4πGρ(r).

(2.68)

2.7 Gravitational Potential Energy

67

A much faster way to get the above expression of the potential energy of a distribution of matter contained in a domain C is by using its definition as generalization to a double integral of the double sum of Eq. 2.20, that is 1 Ω=− G 2

  C C

1 dm dm  =− |r − r| 2

 C



1 dm  =− dm G  |r − r| 2 C

 U (r)ρ(r) d 3 r. C

(2.69)

Exercise 2.3 Calculate the expression for the potential energy of a homogeneous sphere of density ρ0 and radius R. Solution 1 Ω=− 2

R 

 S R (0)

U (r ) dm = −πGρ0

= −4π

Gρ20

ρ0 4πr 2 dr =

0

 2

r2 R − 3 2

R5 R5 − 3 15

=−

16π 2 3 G M2 Gρ0 R 5 = − . 15 5 R

Let us note that for a generic density distribution in spherical symmetry cut at a given radius R the potential energy assumes always a form Ω = −A2 (G M 2 )/R where A2  1/2 is a coefficient which depends on the “shape” of the distribution, being larger for more compact density distributions.

2.8 Newton’s Theorems Let us briefly deal with the classical Newton’s theorems about gravitational potential and force. Theorem 2.4 (Newton’s First Theorem) The gravitational force internal to a spherically symmetric, homogeneous, material shell is null. The theorem is proven by (see Fig. 2.3) noting that on any internal point the diametral opposite matter shells, A1 and A2 , intersected by the same solid angle, α, Fig. 2.3 In the internal point P the contribution of the 2 given spherical shells to the force cancels out

68

2 Newtonian Gravitational Interaction

have a mass which scales with the same square power of the distance to the point of the (inverse) square scaling of their (opposite in direction) gravitational force on the point (ΔFi ). That is Δm 1 = ασr12 and Δm 2 = ασr22 , where σ is the (constant) surface mass density, so that ΔF1 Δm 1 r22 = 2 = 1. ΔF2 r1 Δm 2 The proof of the Newton’s first theorem results, also, as a direct corollary of Newton’s second theorem: Theorem 2.5 (Newton’s Second Theorem) The gravitational force external to a spherical and homogeneous material shell is the same exerted by all of its mass as concentrated at a point at its centre. Proof Let P a point of unitary mass out of the material shell of radius R and constant density σ, identified by its position vector r. Let us compute explicitly the force vector F(r). If the (constant) surface density of the shell is σ, the mass element in a generic point on the shell surface is dm  = σ R sin ϕ dθ Rdϕ (see Fig. 2.4), so we have π2π F(r) = G 0 0

r − r σ sin ϕ dθ dϕ |r − r|3 π

= 2πG R σ(−er ) 2

3

0

Fig. 2.4 Schematic representation of the infinitesimal element of mass on the spherical surface, its radius vector r , |r | = R, and that, r, of the external point P

(r − R cos ϕ ) sin ϕ dϕ (R 2 + r 2 − 2r R cos ϕ ) 2

,

2.8 Newton’s Theorems

69

where the last equality was obtained by a little manipulation (multiplying and dividing by |r − r|) and noting that (see Fig. 2.4)  = r − r  cos ϕ = r − R cos ϕ . |r − r| cos P Letting D ≡ |r − r|, we have |r − r|2 = R 2 + r 2 − 2r R cos ϕ = D 2 ,

(2.70)

whence R cos ϕ =

R2 + r 2 − D2 2r

=⇒ r − R cos ϕ =

r 2 + D2 − R2 . 2r

The integration variable ϕ can be substituted by D, taking into account that d D 2 = 2Dd D = 2r R sin ϕ dϕ =⇒ sin ϕ dϕ =

D d D, rR

so that the force may be written as R F(r) = Gπ 2 σ(−er ) r

r +R 

r −R

r 2 − R2 1+ D2

dD =

r +R

R (r 2 − R 2 ) 4πGσ R 2 σ(−e ) D − = (−er ) = r r2 D r2 r −R GM = − 2 er r

= Gπ

(2.71)

where M = 4πσ R 2 is the shell mass. Although, historically, Newton gave the demonstration of what is called Newton’s first theorem by a geometrical arguments before introducing his second theorem, its thesis is easily derived as a corollary of his second theorem, as we now see. Corollary 2.1 (Demonstration of Newton’s First Theorem) Proof Using expression Eq. 2.71 and taking into account that D(ϕ = 0) = R − r when r ≤ R, we have R F(r) = Gπ 2 σ(−er ) r

r +R  r 2 − R2 1+ d D = 0. D2

(2.72)

R−r

Another way to get the theses of both the Newton’s theorems is by mean of the potential computation, as follows.

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2 Newtonian Gravitational Interaction

The gravitational potential, U (r), of a spherically symmetric material shell is given by the integral π 2π 

Us (r) = G 0

0 2 2πσ

σ R sin ϕ dθ R dϕ R 2 + r 2 − 2r R cos ϕ 2r R 

=

dy

= + r2 − y  2r R R = −G 2πσ R2 + r 2 − y = −2r R r

= GR

2r R

−2r R

R2

⎧ R2 ⎪ ⎨ 4πG σ se r ≥ R   R r = −2πG σ |R − r | − (R + r ) = ⎪ r ⎩ 4πG Rσ se r < R, so that the force is ⎧ R2 ⎪ ⎨ −4πG σer , if r ≥ R r2 F(r) = ∇Us = ⎪ ⎩ 0, if r < R,

(2.73)

which is equivalent to the results of first and second Newton’s theorem. It is trivial the extensions to material sphere of finite size and spherically symmetric density. It suffices substituting σ with ρ(R) in the integral expression of the potential and integrate over R ∞ U (r ) = Us (R) d R. 0

2.9 Further Readings The classical, Newtonian, theory of gravitation is one of the most studied, and applied, fields in theoretical and experimental Physics. It is so found an enormous quantity of textbooks, useful, for the curious reader, to enlarge and deepen the notions presented in this chapter. As mere examples, I suggest [10, 13, 16]. Some deep theoretical insight to gravity is found in the 2-volume book [8], with extension to General Relativity in the fundamental texts [14, 20]. An astrophysical careful insight comes from reading [4, 18, 22].

Chapter 3

Central Force Fields

3.1 The Potential and Force Generated by a Spherical Matter Distribution The Newtonian potential generated by matter distributed over a domain C according to a density law ρ(r) is, by definition   dm  ρ(r ) 3  = G d r, U (r) = G (3.1)   C |r − r| C |r − r| so that if C is limited in space, then lim U (r) = 0.

r →∞

The Euclidean distance between the two points r and r can be written as   , |r − r| = r 2 + r 2 − 2r r cos rr

(3.2)

 is the angle between the r and r vectors; in spherical polar coordinates where rr (r, θ, ϕ), the cosine theorem gives  = cos ϕ cos ϕ  + sin ϕ sin ϕ  cos(θ − θ  ). cos rr

(3.3)

Substitution by mean of (3.2) and (3.3) into Eq. 3.1, and taking into account that in spherical polar coordinates ρ(r )d 3 r = ρ(r  , θ  , ϕ  )r  2 sin ϕ  dθ  dϕ  dr  , makes evident that the integral (3.1) for the potential is very complicated to compute. Anyway, in the case of spherical symmetry for the density, i.e. ρ(r ) = ρ(r  ), the problem is clearly invariant by rotations around the origin, and so we can choose the polar reference frame setting the axis from which evaluate the ϕ angles along the r direction so that (in the new spherical polar coordinates for r (Fig. 3.1) ) ϕ = 0, yielding © Springer Nature Switzerland AG 2019 R. A. Capuzzo Dolcetta, Classical Newtonian Gravity, UNITEXT for Physics, https://doi.org/10.1007/978-3-030-25846-7_3

71

72

3 Central Force Fields

Fig. 3.1 Coordinate frame set after rotation leading the z axis along the r direction

|r − r| =

 r  2 + r 2 − 2r r cos ϕ  .

(3.4)

In the case of a spherical C domain of radius R, we have Rπ2π U (r ) = G 0 0 0

R

r  2 sin ϕ  dθ  dϕ  dr  ρ(r  )  = r  2 + r 2 − 2r r cos ϕ 

ρ(r  )r  dr 

π

2

= 2π G 0

0

(3.5) 

sin ϕ  dϕ  . 2  2   r + r − 2r r cos ϕ

Letting t = cos ϕ  in the above integral, it becomes R

ρ(r  )r  dr 

t=−1 

2

U (r ) = 2π G 0

R = 2π G

t=1

d(r  2 + r 2 − 2r r t)  = 2r r r  2 + tr 2 − 2r r t (3.6)

2 1 ρ(r  )r  (r  + r − |r  − r |)dr  . rr 

0

Now, we distinguish between the potential (i) external to the distribution (r ≥ R) and (ii) internal (0 ≤ r ≤ R). Actually (i) when r ≥ R (and r  ≤ R ≤ r ), |r  − r | = r − r  , so that

3.1 The Potential and Force Generated by a Spherical Matter Distribution

4π G U (r ) = r

R

ρ(r  )r  dr  = G 2

73

M(R) ≡ Uint (r ); r

(3.7)

0

(ii) when 0 ≤ r ≤ R, |r  − r | = r − r  , while |r  − r | = r  − r when 0 ≤ r ≤ r  , leading to 4π G U (r ) = r

 r

2





R

r ρ(r ) dr + r r

0

=

 r  ρ(r  ) dr  =

G M(r ) + 4π G r

R

(3.8)

r  ρ(r  ) dr  ≡ Uint (r ) + Uext (r ).

r

Note that while the potential on a generic point P at distance r from the origin is contributed by both the matter within r (Uint (r ) = G M(r )/r ) and out of r (Uext (r )), the force has only the internal contribution, as already stated by Newton’s theorems. Indeed, for any choice of integrable spherically symmetric density law, the force generated by the spherical potential is: 



F = ∇U = U (r )er =

4π G − 2 r

 r

2





R







r ρ(r ) dr +

r ρ(r ) dr + r 0



r

  R 4π G 2  G M(r ) 2   2  r ρ(r ) + r ρ(r ) dr −  + r ρ(r ) er = − er .  r r2

(3.9)

r

3.1.1 Calculating Spherical Potentials via Poisson’s Equation The potential generated by a spherical matter distribution can be easily obtained by solving the Poisson’s equation in spherical polar coordinates. If ρ(r) is spherically symmetric, also U (r) is spherically symmetric, so that the Poisson’s equation reduces to

 1 d 2 dU r = −4π Gρ(r ). (3.10) r 2 dr dr

74

3 Central Force Fields

A first integration of Eq. 3.10 over r gives 

dU = −G M(r ), d r2 dr

(3.11)

dU dU − lim r 2 = −G M(r ). r →0 dr dr

(3.12)

r 0

which writes as r2

Now, it is lim r 2 dU = 0, because any regular matter distribution is broader than dr r →0

that of the point-like case (a delta function) which produces the U ∝ 1/r potential, thus Eq. 3.12 simply gives the first and second Newton’s theorem theses, resumed as dU M(r ) = −G 2 . dr r

(3.13)

A further r integration by parts leads to ∞

∞ dU = −G

r

r

  ∞ M(r ) d M(r ) M(r ) ∞ = dr = G −G r2 r r r r



M(r ) M(r ) − = G lim r →∞ r r



∞ − 4π G

(3.14)

ρ(r )r dr, r

which reduces to M(r ) + 4π G U (r ) = G r

∞ ρ(r )r dr + U∞ ,

(3.15)

r

assuming lim

r →∞

M(r ) r

= 0 and U∞ ≡ lim U (r ) = 0. Note that the condition lim r →∞

r →∞

M(r ) r

= 0 means that the asymptotic (large r ) behaviour of ρ(r ) is ρ ∝ 1/r 2+β with β > 0. This asymptotic behaviour for ρ(r ) guarantees, also, convergence of the integral in the rhs of Eq. 3.15. Exercise 3.1 Compute the flux through a spherical surface of radius R of the gravitational force exerted by a spherical matter distribution, ρ(r ).

3.1 The Potential and Force Generated by a Spherical Matter Distribution

75

Solution The flux is given by π2π

 F(r) · n dσ = − S R(0)

G M(R) 2 R sin ϕ dθ dϕ = −4π G M(R), R2

0 0

as expected by the Gauss’ theorem.

3.2 Motion in a Spherical Potential As we saw in Sect. 1.4 of Chap. 1, in the case of a central force field (per unit mass), F = f (r) er , the orbital angular momentum (per unit mass) L is conserved L˙ = r ∧ f (r) er = 0,

(3.16)

so that the motion is planar. The particle acceleration, in spherical polar coordinates defined on the motion plane, writes as a = (¨r − r θ˙ 2 ) er + (r θ¨ + 2˙r θ˙ ) eθ , and the equation of motion in scalar form reduces to r¨ − r θ˙ 2 = f (r ), r θ¨ + 2˙r θ˙ = 0,

(3.17)

(3.18)

where we have also assumed spherical symmetry for f (r), to guarantee conservativity. The second equation in the system 3.18 corresponds to d 2  r θ˙ = 0, dt

(3.19)

that is equivalent to L z = r 2 θ˙ = const., i.e. the conservation of the orbital angular momentum (which was guaranteed because the force field is central). The angular momentum conservation allows the elimination of θ˙ in the first equation in Eq. 3.18, so to have the, decoupled, equation for the radial motion r¨ −

L2 = f (r ), r3

where, for simplicity, we let hereafter L ≡ L z .

(3.20)

76

3 Central Force Fields

dU , and so dr dUe f f L2 dU ≡ , r¨ = 3 + r dr dr

Being the force conservative, f (r ) =

(3.21)

where the effective potential, Ue f f (r ; L), is defined as  2 L 1 L2 dr + U = − + U + c, Ue f f (r ; L) = r3 2 r2 with c an integration constant, which is zero under the two conditions lim U (r ) = 0 r →∞

and lim Ue f f (r ; L) = 0. Note that the term −L 2 /(2r 2 ) in the above equation is the r →∞ centrifugal potential.

3.2.1 Circular Trajectories In the above scheme, provided that the force is attractive, circular trajectories exist of any radius r , upon the proper choice of the angular momentum L. Indeed Eq. 3.21, under the assumption r = const. for every t, that implies r˙ = r¨ = 0, gives for the angular momentum of the unit mass particle on the circular obit of radius r the value  dU . (3.22) L c = −r 3 dr Note that the existence of circular trajectories of motion stands on attractivity of the force, dU < 0, otherwise L 2c < 0. dr Because L = r 2 θ˙ = r vθ , the circular speed, vc (r ), is given by  Lc dU = −r . (3.23) vc (r ) ≡ r dr In principle, circular orbits in spherical potentials may be both stable or unstable; however, if the potential has a gravitational origin (i.e. it satisfies the Poisson’s equation) the unstable case is never realized (see note in Sect. 3.6).  The simplest case, that of motion in a point mass potential, gives vc (r ) = which is the Keplerian circular velocity law.

GM , r

Exercise 3.2 Show that among all the possible orbits of given energy E in a spherical potential, the circular one is that of maximum angular momentum.

3.2 Motion in a Spherical Potential

77

Solution Given a spherical potential U (r ), the energy per unit mass is E=

 1 1 L2 1 2 r˙ + r 2 θ˙ 2 − U (r ) = r˙ 2 + − U (r ), 2 2 2 r2

so that  L = 2r 2

2

 1 2 E − r˙ + U (r ) ≤ 2r 2 [E + U (r )] = L 2c , 2

where r˙ = 0 corresponds to the circular orbit whose angular momentum is indicated by L c .

3.3 Potential Generated by a Homogeneous Sphere As we saw in Sect. 3.1, the expression of the gravitational potential of a generic, spherically symmetric, distribution of matter is (Eq. 3.8) M(r ) U (r ) = G + 4π G r

R

r  ρ(r  ) dr  ,

(3.24)

r

which, assuming ρ(r ) = ρ0 = const. for r ≤ R, and ρ(r ) = 0 for r > R, gives

U (r ) =

⎧  R   ⎪ ⎪ 4 r 2 1 2 ⎪ 2 2 ⎪ R if r ≤ R, π Gρ r r + 4π Gρ = 2π Gρ − 0 0 0 ⎪ ⎨3 2 3 r

⎪ ⎪ ⎪ 3 ⎪ ⎪ ⎩ 4 π Gρ0 R 3 r

(3.25) if r > R,

that is U (r ) =

⎧ ⎨ har monic, for r ≤ R, ⎩

K eplerian, for r > R.

78

3 Central Force Fields

3.3.1 Trajectories in a Homogeneous Sphere As we saw above, the force inside a homogeneous sphere is harmonic, F = −kr er , 4 where the elastic constant is k = Gπρ0 ≡ ω2 > 0. The vector, 2nd order, differ3 ential equation of motion is r¨ = −ω2 r er , which, once projected onto the plane of motion, gives  x¨ = −ω2 x (3.26) y¨ = −ω2 y whose general solutions are x(t) = ax cos ωt + bx sin ωt and y(t) = a y cos ωt + b y sin ωt, with ax , bx , a y , b y constants to be determined by the initial conditions ˙ = x˙0 , y(0) = y0 , y˙ (0) = y˙0 . on r and r˙ , namely x(0) = x0 , x(0) Upon the given i.e., the solutions specify as

x(t) = x0 cos ωt +

x˙0 sin ωt, ω

y(t) = y0 cos ωt +

y˙0 sin ωt, ω

(3.27)

which are the parametric expressions of the trajectory. The choice of the particular i.c. x0 > 0, x˙0 = 0, y0 = 0, y˙0 > 0 (see Fig. 3.2), which correspond to an initial velocity vector orthogonal to the initial position vector, assumed on its turn on the x axis, leads, after squaring and summing member to member in Eq. 3.27, to the canonical Cartesian equation of an ellipse x2 y2 +

 = 1, x02 y˙0 2 ω

(3.28)

where a 2 ≡ x02 and b2 ≡ ( y˙0 /ω)2 are the (squared) semi axes.

Fig. 3.2 The particle starts moving with x0 > 0, x˙0 = 0, y0 = 0, y˙0 > 0

y

. yo/w . yo xo

x

3.3 Potential Generated by a Homogeneous Sphere

79

Actually, we can see that the trajectories are conics for any choice of i.c.. Indeed, solving Eq. 3.27 for cos ωt and sin ωt, and using the trigonometric identity, it results cos2 ωt + sin2 ωt = 1 =

( yω˙0 x −

( yω˙0 x0 −

x˙0 y)2 ω x˙0 y )2 ω 0

+

(x0 y − y0 x)2 ( yω˙0 x0 −

x˙0 y )2 ω 0

,

(3.29)

and, after some manipulations,    

x˙02 y˙02 y˙0 1 2 2 2 + y0 x − 2 x˙0 + x0 y0 x y + + x0 y 2 = 2 (x0 y˙0 − y0 x˙0 )2 . 2 2 2 ω ω ω ω (3.30) The above expression is that of a conic section, whose general equation is



Ax 2 + Bx y + C y 2 = D,

(3.31)

with, in our case y˙2 A = 02 + y02 ≥ 0, ω x˙2 C = 02 + x02 ≥ 0, ω

 y˙0 B = −2 x˙0 + x0 y0 , ω2

D=

1 (x0 y˙0 − y0 x˙0 )2 ≥ 0. ω2

If A = C and B = 0 the curve is a circumference. In general, defined the discriminant of the conic as Δ ≡ B 2 − 4 AC, it is known, from analytic geometry, that ⎧ ⎨ Δ < 0 for ellipse, Δ = 0 for parabola, ⎩ Δ > 0 for hyperbola. In our case

Δ = −4

x˙0 y˙0 x0 − y0 ω ω

2 = −4D ≤ 0,

so that trajectories are either ellipses (D > 0) or parabolas (D = 0), these latter being degenerate. Actually, D = 0 ⇐⇒ (x0 y˙0 − y0 x˙0 )2 = |r0 ∧ v0 |2 = L 20 = 0, meaning that r0 v0 . Due to that L is constant, it results r(t) v(t) for every t, that means the parabola degenerates in two coinciding straight lines (purely radial motion).

80

3 Central Force Fields

Formally 

  

˙2

x0 y˙0 y˙02 2 2 2 2 + y0 x + +x y − 2 x˙0 + x0 y0 x y = 0 ω2 ω2 0 ω2

2 y˙0 x˙0 x − y + (y0 x − x0 y)2 = 0 ⇐⇒ ω ω ⎧ y˙0 x˙0 ⎪ ⎨ x− y=0 ω ω ⎪ ⎩ y0 x − x0 y = 0.

⇐⇒

The two straight lines above are coincident because y˙0 /x˙0 = y0 /x0 .

3.3.2 Radial Motion in a Homogeneous Sphere As said above, the case of i.c. such that L0 = 0 gives, as always in a central force field, a purely radial motion. Remembering the variable separation introduced in Sect. 3.2, and letting L = 0, it is r¨ =

4 dU = − π Gρ0 r = −ω2 r, dr 3

(3.32)

whose solution is a radial oscillation across the origin for negative mechanical energy, E, and an unlimited straight line motion for E ≥ 0 (see Sect.3.5.2). In the first case, the radial oscillation occurs in a time (called radial period)  π 1 2π 1 3π = Tr = = . (3.33) 2 ω 2 Gρ0 4 π Gρ0 3 Notably, the above time is independent of both r (0) ≤ R and r˙ (0). Exercise 3.3 Calculate the time needed to a person which falls into an imaginary tunnel crossing diametrically the Earth to reach the antipodes. Compare this time with that required by a normal line aircraft to reach the same point. Solution Left to the reader.

3.4 Some Relevant Spherical Models

81

3.4 Some Relevant Spherical Models In the class of spherical potentials, all admissible stationary and stable mass density distributions should be less concentrated than the, extreme, point mass case (Dirac’s dρ delta function ρ(r) = δ(r)) while keeping a radially decreasing profile < 0. So dr the density distributions, and the corresponding potentials, should lay in between the Keplerian (point mass) and harmonic (homogeneous sphere) cases, which are the only spherical potentials giving bound trajectories that are closed (fixed, nonprecessing). This statement is, basically, the thesis of the Bertrand’s theorem [3], whose demonstration is rather complicated and beyond the scopes of this book. Anyway, we will present in Chap. 4 an Exercise (4.14) whose solution constitutes a partial demonstration of the theorem. A direct consequence of the Bertrand’s theorem is that in any realistic model of stationary, stable, spherical self-gravitating system (but the 2 cases above mentioned) the bound orbits are precessing. It may be useful, so, to study a bit better some spherical distributions that are intermediate between the two extreme cases of pointmass and uniform density. Two relevant examples are the so called Plummer’s and isochrone potentials.

3.4.1 The Plummer Sphere The Plummer (or Schuster, see [4]) sphere is an unbound spherical distribution of matter whose density law is ρ0 , ρ(r ) =   r 2  25 1+ b

(3.34)

where ρ0 is the central density and b > 0 is a length scale. It is worth noting that the above density law is an exact solution of the LaneEmden equation. The Lane-Emden equation is the radially symmetric equilibrium equation of a self gravitating, polytropic, matter distribution. A polytrope is a matter distribution where density and pressure are related in a power law form, of the type p = kρ 1+1/n , where k > 0 and n > 0 are constants, the latter called polytropic index. Another relevance of the Plummer distribution is that, upon projection, it constitutes a good fit for the light distribution of globular star clusters1 up to distances not too far from the center. The mass contained in a sphere of radius r centered in the symmetry center is (letting x = r/b) clusters are star clusters satellites of our Milky Way, composed by 105 − 107 stars distributed in a nearly spherical volume of diametral size ∼100 pc.

1 Globular

82

3 Central Force Fields

r M(r ) = 4πρ0 0

r 2 dr = 4πρ0 b3   r 2  25 1+ b

r/b

x2 dx

5 . 1 + x2 2

0

(3.35)

x Adopting u = x and dv =  5 d x in the usual formula of integration by 1 + x2 2 parts, we have ⎧ ⎫ r/b r/b ⎨ 1 ⎬ x d x 1 − M(r ) = 4πρ0 b3 + . (3.36) 3 ⎩ 3 (1 + x 2 ) 23 0 3 (1 + x 2 ) 2 ⎭ 0

By addition and subtraction of x 2 in the numerator of the integral in the r.h.s. of Eq. 3.36, this integral rewrites as r/b

r/b

dx 3

(1 + x 2 ) 2

0

= 0

dx

r/b

− √ 1 + x2

x2 dx 3

0

(1 + x 2 ) 2

.

(3.37)

By integration by parts of the second integral in the r.h.s of Eq. 3.37 we have r/b 0

r/b

dx (1 +

3

x 2) 2

= 0



1 (−2) x − √ √ 2 2 1+x 1 + x2 dx

r/b

r/b +

0

0

dx = √ 1 + x2

r/b =  r 2 . 1+ b

(3.38)

Finally ⎛

⎞

⎜ 1 ⎟ r/b r/b 1 ⎜ ⎟ = M(r ) = 4πρ0 b3 ⎜−  +   r 2 ⎟  r 2  23 ⎝ 3 ⎠ 3 1+ 1+ b b 3 4 (r/b) 4 = πρ0 b3  =⇒ Mtot = lim M(r ) = πρ0 b3 . 3  r →∞   3 3 r 2 2 1+ b

(3.39)

3.4 Some Relevant Spherical Models

83

The force per unit mass felt by a particle at r is F(r ) = −G Mtot  1+

r/b e,  r 2  23 r

(3.40)

b

which, at first order in r/b, behaves as the force in a homogeneous sphere for r/b  1 and has Keplerian behaviour at large distances from the center. The potential in the Plummer model is easily obtained as U (r ) = 

G Mtot 1 + (r/b)2

.

(3.41)

Let us note that the above potential constitutes a “mollification” of the point mass potential where the mollification length scale (called also softening scale) is b. The substitution of the purely Keplerian interaction with this softened one avoids the UV divergence but of course does not preserve the overall Newtonian behaviour of the system. Some further comments at this regard are found in Appendix B. The potential energy of this distribution of matter is obtainable, for instance, by Eq. 2.67, as follows

1 Ω=− 8π G 1 =− 2



2 G Mtot

1 |∇U | d r = − 8π G R3 2

∞

b 0

∞ 

3

G M(r ) r2

2

0

r 

2 1 G Mtot (r/b) = − d    r 2 3 b 2 b [1 + b 4

4πr 2 dr = ∞ 0

(3.42)

x4

3 d x. 1 + x2

The improper integral above converges because for x  1 the integrand scales x as 1/x 2 . Also this integral is solved by parts, letting u = x 3 and dv = 3 , 1 + x2 leading to ∞ 0

x4



x3 1

3 d x = −  4 1 + x2 2 1 + x2

∞ + 0

3 4

∞

0

x2 1 + x2

2 d x =

3 4

∞

0

x2 1 + x2

2 d x.

(3.43) A further integration by parts of the integral in the rightmost hand side of Eq. 3.43 gives

84

3 Central Force Fields

∞ 0



x2

1 x

2 d x = − 2 2 1 + x2 1+x

∞ 0

1 + 2

∞ 0

dx 1

= 2 2 1+x

∞ 0

dx

 , (3.44) 1 + x2

so that, finally ∞ 0

x4

3

3 d x = 8 1 + x2

∞ 0

dx 3π 3

 = [arctan x]∞ , 0 = 2 8 16 1+x

(3.45)

which eventually yields, for the potential energy, the expression Ω=−

2 3π 3 Gρ02 b5 3π G Mtot =− . 32 b 18

(3.46)

3.4.2 The Isochrone Potential Another potential which, like the Plummer’s, satisfies the condition of being nearly harmonic in the central region and falling off as Keplerian at large distances from the center is the isochrone potential. The name derives from that it is the most general potential that makes the radial oscillation period independent of the angular momentum. Actually, the potential is U (r ) =

1 GM GM  = √

2 , 2 2 b b+ b +r 1 + 1 + br

(3.47)

where b is a length scale, one of the 2 free parameters of the model (the other is the total mass M). The corresponding force per unit mass is F=

r/b GM dU er = − 2

er ,   dr b

r 2 2 

r 2 1+ 1+ b 1+ b

which, at first order in r/b, yields 1 − GbM dU 2 (r/b)2 = 1 GM r dr − 4 b2 b

r b r b

 1,  1.

(3.48)

(3.49)

Using Poisson’s equation, it is easily seen that the density distribution which generates the potential (3.47) is ρ(r ) =

3M (t + 1)t 2 − 13 x 2 (3t + 1) , 4π b3 (t + 1)3 t 3

(3.50)

3.4 Some Relevant Spherical Models

85

√ where x ≡ r/b and t = t (x) ≡ 1 + x 2 . The above expression gives a finite central density ρ(0) = 3M/(16π b3 ). Note that the asymptotic behaviours (for small and large distances from the center) of the density is ⎧ 5 2 ⎪ ⎪ ⎨ρ(0)(1 − 12 x ), x  1, ρ(x) ≈

⎪ 3M ⎪ ⎩ 4πb 3

(3.51) 1 , (1+x 2 )5/2

x  1.

The total mass is x M = lim 4π b3

ρ(x)x 2 d x,

x→∞

(3.52)

0

where the integral convergence is guaranteed by that, for x  1, x 2 ρ(x) ∝ x −3 . The potential energy ∞ Ω = −2π G Mb

2 0

ρ(x)x 2 d x, √ 1 + 1 + x2

(3.53)

is a complicated integral, whose explicit computation is left as exercise to the reader. We limit to note that the above integral is convergent because for x  1 the integrand behaves as xρ(x) ∝ x −4 .

3.5 Quality of Motion Although the spherical symmetry represents a significant simplification, the integration of the equation of motion remains a task that cannot, in general, be faced on an analytical basis but requires a numerical approach. Anyway, when the density, and so the potential, are spherically symmetric, a qualitative approach to the problem is possible, which helps because it makes a restriction of the phase space allowed to the motion. The bases of the qualitative treatment of the motion are the constant of motions.2 The constant of motions, indeed, pose constraints to the actual motion. In the case of spherical potentials, the constants of motions are the mechanical energy, E, and the angular momentum, L, here both per unit mass (i.e. specific quantities), which are also integrals of motion.

2 A constant of motion is a function of phase-space coordinates and time that remains constant along

every trajectory of motion; an integral of motion is a particular constant of motion that depends only upon phase-space coordinates and not upon time.

86

3 Central Force Fields

The mechanical energy per unity mass of a particle of mass m in a spherical potential U (r ) ≡ −V (r ) is (in spherical polar coordinates) E=

1 1 L2 1 2 + V (r ), (˙r + r 2 θ˙ 2 ) + V (r ) = r˙ 2 + 2 2 2 r2

(3.54)

so that the for the motion to be real the requirement is 1 2 1 L2 − V (r ) = E − Ve f f (r ; L) ≥ 0, r˙ = E − 2 2 r2

(3.55)

i.e. E ≥ Ve f f (r ; L), which is a limitation for the physical space allowed to the motion of the particle of given energy and angular momentum. The function Ve f f (r ; L) is 2 the effective potential energy, Ve f f (r ; L) ≡ −Ue f f (r ; L) = 21 Lr 2 − U (r ).

3.5.1 The Keplerian Case The Keplerian, point mass, potential is the simplest to approach and so it is the archetype of the qualitative analysis of motion. Whenever the test particle mass, m, is much smaller than the other mass, M, i.e. m  M, the problem is, in practice, a one body problem, because the barycenter almost coincides with the position of the massive particle. In any case, also if m/M is not much smaller than 1, qualitative considerations hold for the relative motion, i.e. that of one of the 2 bodies around the other. The effective potential energy is 1 L2 GM , − 2 2r r

Ve f f (r ; L) =

(3.56)

so that the motion is possible whenever the test particle energy is E ≥ Ve f f,min . In the case of L 2 = 0, the motion is possible (and radial) for every energy E, because lim+ Ve f f (r ; 0) = −∞, while for L 2 > 0, Ve f f (r ; L) takes its minimum r →0

2

value, Ve f f,min = − 21 ( GLM )2 , at rmin = GLM . Clearly, the function Ve f f (r ; L) is continuous for every r but the origin, which is singular for both L 2 = 0, where Ve f f diverges to −∞, and L 2 > 0 where it diverges positively (centrifugal barrier). For the sake of clarity we can distinguish the non-radial case (L 2 > 0) from the radial one (L 2 = 0). Case L 2 > 0 Qualitatively speaking, this case corresponds to the two example curves in Fig. 3.3 where arbitrary units are chosen. The upper curve corresponds to a value of L 2 twice of the other curve. The stationary values of r (t) are solutions of the equation r˙ = 0, which writes as

3.5 Quality of Motion

87

E − Ve f f (r ; L) = E −

1 L2 GM = 0, + 2 r2 r

(3.57)

1 2 L = 0, 2

(3.58)

that is the quadratic equation Er 2 + G Mr − whose solutions r1,2 =

−G M ±



(G M)2 + 2E L 2 , 2E

(3.59)

are (i) complex conjugate when (G M)2 + 2E L 2 < 0. Actually, this corresponds to E < − 21 ( GLM )2 = Ve f f,min , and means no motion; (ii) two real and coincident when (G M)2 + 2E L 2 = 0. This means that for E = Ve f f,min , pericenter, r− , and apocenter, r+ , distances are equal, so that the motion is along a circle (circular orbit). (iii) two real and distinct when (G M)2 + 2E L 2 > 0. If E > 0, the solution with the − sign in front of the square root is negative while the one with + is positive, and corresponds to the pericenter, r− . If E < 0 the solutions are both positive and r− corresponds to taking the + sign and r+ to − sign. Case L 2 = 0 This case corresponds to the lower curve in Fig.3.3, which plots the function Ve f f (r ; 0) = − GrM , which goes to −∞ for r → 0+ and is always negative. The absence of centrifugal barrier implies that the percenter is r− = 0. The equation to determine extremes of radial motion reduces to Er + G M = 0 which has, obviously, only one solution. It is positive for E < 0, which gives the apocenter distance r+ = −(G M)/E, while for E > 0 it is negative. This means that for E ≥ 0 the motion is unbound, like in the case L 2 > 0.

3.5.2 The Homogeneous Sphere Case In the case of a homogeneous sphere of mass density ρ0 , we have (see Eq. 3.25) ⎧

 1 L2 1 2 ⎪ ⎪ 2 ⎪ r − R , if r ≤ R, + 2π Gρ0 ⎪ ⎨ 2 r2 3 (3.60) Ve f f = ⎪ 2 3 ⎪ ⎪ 1 L R 4 ⎪ ⎩ if r > R. − π Gρ0 , 2 r2 3 r We can distinguish the non-radial case (L 2 > 0) from the radial one (L 2 = 0).

88

3 Central Force Fields

Fig. 3.3 Behaviour of Ve f f (r ; L), in arbitrary units, of the Keplerian potential for L = 0 (lower curve) and two non zero values of L 2 . The dotted, short-dashed and long-dashed horizontal lines refer to examples of level of energies for which oscillations are radially limited

Case L 2 > 0 From Eq. 3.60 it is lim Ve f f (r ; L) = +∞,

r →0+

lim Ve f f (r ; L) = 0− ,

r →∞

conditions that, by continuity of Ve f f (r ; L) for r > 0, imply that Ve f f (r ; L) has either only one stationary point or it has an odd number of stationary points. In this specific case (Eq. 3.60), we have3 ⎧ L2 4 ⎪ ⎪ − ⎪ ⎨ r 3 + 3 π Gρ0 r, if r ≤ R, Vef f = (3.61) ⎪ 2 3 ⎪ L R 4 ⎪ ⎩− + π Gρ0 2 , if r > R, r3 3 r so that

Vef f = 0 ⇐⇒

3 Hereafter,

⎧  14

⎪ 3 ⎪ 2 ⎪ , ⎪ ⎨ r1 = L 4π Gρ0 ⎪ ⎪ ⎪ ⎪ ⎩ r2 = L 2

3 . 4π Gρ0 R 3

we indicate by apices,  , derivatives with respect to r .

(3.62)

3.5 Quality of Motion

89

The above zeros of the Ve f f (r ; L) radial derivative have sense if r1 ≤ R and r2 ≥ R, which reflect into conditions for L 2 : 4 π Gρ0 R 4 , 3 4 r2 ≥ R ⇐⇒ L 2 ≥ π Gρ0 R 4 . 3 r1 ≤ R ⇐⇒ L 2 ≤

(3.63) (3.64)

Obviously, a low angular momentum implies a more limited extension of the radial oscillation respect to a high angular momentum. The above threshold value for L 2 corresponds  to the square of the angular momentum of the circular orbit of radius R, L c (R) = R 2

4 π Gρ0 . 3 2

Summarizing, if L ≤ L 2c (R) then r1 is the stationary point, otherwise the stationary point is r2 . Actually, r1 and r2 are minima for Ve f f (r ; L), because Vef f (r1 ; L) = and Vef f (r2 ;

L) =

16π Gρ0 > 0, 3

R3 4π Gρ0 3/2 3 L

(3.65)

4 > 0.

(3.66)

Moreover  Ve f f (r1 ; L) = L c (R)

4π Gρ0 − 2π Gρ0 R 2 , 3

(3.67)

so that the condition L ≤ L c (R) corresponds to  4π π − 2π Gρ0 R 2 = − Gρ0 R 2 < 0. Ve f f (r1 ; L) ≤ L c (R) 3Gρ0 3 On the other hand, for r2 it is 1 Ve f f (r2 ; L) = − 2



4π R3 Gρ0 3 L

2 < 0.

In conclusion, for any value of L a radially periodic motion is possible only for E < 0. It is worth noting that for |L| < L c (R) the trajectories always penetrate the sphere (r− < R, ∀E), while for |L| > L c (R) it is r+ > R, ∀E. √ Radially limited orbits all external to the sphere require |L| > 2L c (R).

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3 Central Force Fields

Summarizing: (a) (b) (c) (d)

no motion, E < Ve f f,min : circular motion, r− = r+ = r , E = Ve f f,min : Ve f f,min < E < 0: radially periodic motion, r− ≤ r ≤ r+ ,, E ≥ 0: radially unbound trajectories, r+ = ∞.

Note that case (d) is that of a particle faster than escape velocity (unbound), because E ≥ 0 corresponds to  1 2 1 (˙r + r 2 θ˙ 2 ) = v 2 ≥ U (r ) ⇐⇒ v ≥ 2U (r ) ≡ vesc (r ). 2 2

(3.68)

The pericenter, r− (E, L), and apocenter, r+ (E, L), distances are stationary points for r (t) and, so, are the roots of the equation $ % r˙ 2 = 2 E − Ve f f (r ; L) = 0.

(3.69)

We leave to the reader the discussion of the existence of those roots and their dependence upon E and L. Case L 2 = 0 This case corresponds to r r˙ ∀t, that is a radial motion. If L 2 = 0, Ve f f (r ; L) = V (r ) and, in the specific case of a homogeneous sphere, V (0) = −2π Gρ0 R 2 < 0 and V  (r ) = 0 both for r = 0 and r → ∞, moreover ⎧ 4 ⎪ ⎪ r≤R ⎨ π Gρ0 > 0, 3 

3 V (r ) = R 8 ⎪ ⎪ − π Gρ0 < 0, r > R. ⎩ 3 r

(3.70)

Expressions above say that the minimum is attained in r = 0 which is a stable equilibrium point. The qualitative characteristics of the L 2 = 0 motion are summarized as: (a) (b) (c) (d) (e)

E < −2π Gρ0 R 2 , no motion, E = −2π Gρ0 R 2 , particle sitting in the origin −2π Gρ0 R 2 < E < 0, radial oscillation through the origin: 0 ≤ r ≤ r+ , E = 0, asymptotically unbound motion, E > 0, unbound motion.

It is worth noting that the case L 2 = 0 has as solution r (t) that of the 3D harmonic oscillator for E ≤ V (R) (the whole oscillation is within the sphere) or of a harmonic oscillation for r ≤ R matched to a Keplerian motion for r > R for E > V (R). Actually, the equation of motion reduces to

3.5 Quality of Motion

91

⎧ 4 ⎪ ⎨ − π Gρ0 r, r ≤ R, dU 3 r¨ = = 3 ⎪ dr ⎩ − 4 π Gρ0 R , r > R. 3 r2

(3.71)

Note In the homogeneous sphere potential, which is Keplerian out of the sphere boundary R, as we saw the condition√for a particle to be unbound is E ≥ 0 which is equivalent to v ≥ vesc (r ) ≡ 2U (r ). On this purely classical basis, John Michell in 1783 deduced that a sufficiently massive planet could have been completely dark because even light would have not been fast enough to escape the gravitational attraction. Under the hypothesis of uniform mass density, ρ0 , the condition for the  planet to be dark would be that the escape speed at its surface vesc (R) = R 8π Gρ0 3 is greater than the speed of light in vacuum, c. This means, for a planet with the average density of the Earth (5.51 g cm−3 ), to have a radius R ≥ 1.7 × 108 km  243R  1.13 AU! Although interesting, this implicit deduction of the existence of an event horizon is physically flawed, because it bases on the wrong assumption that the light behaves as “particles” with a non zero finite mass, so to allow writing their mechanical energy per unit mass in the usual way.

Exercise 3.4 Compute, for L 2 = 0, the pericenter and apocenter distances for oscillation within the homogeneous sphere. Solution r˙ = 0 ⇐⇒E − Ve f f (r ) = 0, i.e.

2  r 2 E − 2π Gρ0 − R = 0, 3 2 − π Gρ0 r 2 + E + 2π Gρ0 R 2 = 0 3

 =⇒ r+ =

3R 2 +

E . (2/3)π Gρ0

As expected, there is only one solution acceptable for the quadratic equation (positive determination of the square root). The pericenter r− = 0 is not a point where the time derivative of r (t) vanishes: at the time t of origin crossing, the function r (t) is indeed not differentiable.

3.6 Periods of Oscillations As seen above, in a spherical potential, once given the (constant) values of E and L the motion is allowed in the radial region where E ≤ Ve f f (r ; L). In particular, the solutions for r of the equation E = Ve f f (r ; L) give the radial stationary points

92

3 Central Force Fields

Fig. 3.4 Behaviour of Ve f f (r ; L) vs radius in the 2 regimes of low (upper panel) and high (lower panel) specific orbital angular momenta, in a homogeneous sphere. In both panels, horizontal and vertical solid lines refer to zero energy and to the sphere radius location, respectively. In the upper panel the straight lines at decreasing energies corresponds to cases of: (i) oscillation through the sphere boundary (dotted line); (ii) oscillations touching internally the sphere boundary (short dashed line); (iii) oscillations all within the sphere (long dashed line)

(where r˙ = 0) which define pericenter and apocenter distances, r− = r− (E, L) and r+ = r+ (E, L) respectively. The number of different radial intervals of oscillations, and so the number of pairs (r− , r+ ), depends upon the shape of the potential U (r ), because for any minimum of Ve f f (r ; L) there is always a local radial oscillation occurring in a finite time for energies lower than the lowest closer maximum of Ve f f (r ; L) (see Fig. 3.4). The time needed to make a complete radial oscillation (radial period, Tr ) is formally given by r+ Tr (E, L) = r−

dr + r˙

r+ 

=2 r−

r− r+

dr = r˙

r+ r−

dr  + 2(E − Ve f f )

r− r+

dr  = − 2(E − Ve f f )

dr . 2(E − Ve f f )

(3.72) This characteristic time, which involves an improper integral with singularity in both r− and r+ , can, indeed, be finite or infinite in dependence on convergence or divergence of the integral.

3.6 Periods of Oscillations

93

We distinguish two cases: (i) Vef f (r− ; L) < 0 and Vef f (r+ ; L) > 0; (ii) Vef f (r− ; L) = 0 and/or Vef f (r+ ; L) = 0. Case (i) In this case the integral at the r.h.s. of Eq. 3.72 converges. Actually, possible divergence could come just in the neighborhood of r− or r+ , so we can develop Ve f f (r ; L) around r− and r+ to get Ve f f (r ; L) = Ve f f (r± ; L) + Vef f (ξ± ; L)(r − r± ),

(3.73)

where ξ± is an unknown point in the neighbor of r− (if we study convergence around r− , so we call it ξ− ) or r+ (if we study convergence around r+ , so we call it ξ+ ). Note that by continuity Vef f (ξ− ; L) < 0 and Vef f (ξ+ ; L) > 0. This leads to 

1 1 = & &, 2(E − Ve f f ) & & 2 &Vef f (ξ± ; L)(r − r± )&

(3.74)

which gives a convergent integral r+ Tr (E, L) = 2 r−

√ r+ dr 2 dr  = =  √ |Ve f f (ξ± ; L)| |r − r± | 2(E − Ve f f ) √

2 =  |Ve f f (ξ± ; L)|

r−

x+ x−

√  dx 2 2 =  |r+ − r− |. √ |Ve f f (ξ± ; L)| |x|

(3.75)

Case (ii) Let us consider Vef f (r− ; L) = 0 (the case Vef f (r+ ; L) = 0 is equivalent). Figure 3.5 (where r− corresponds to the local maximum) displays an example of such a situation, whose gravitational significance is discussed in the following Note box. The lowest order approximation is, in this case 1 Ve f f (r ; L) = Ve f f (r− ; L) + Vef f (η− ; L)(r − r− )2 , 2 if Vef f (r− ; L) < 0, otherwise the approximation is at higher order in r − r− . Consequently 

1 1 1 1 = = , 2(E − Ve f f ) |Vef f (η− ; L)| r − r− −Vef f (η− ; L)(r − r− )2

94

3 Central Force Fields

Fig. 3.5 Sketch of an effective potential with the centrifugal barrier and having two wells. The horizontal line refers to the energy required to a particle to reach the relative maximum separating the two wells

so that Tr (E, L) = 

r+

2 |Vef f (η− ;

L)| r

−

dr , r − r−

(3.76)

which diverges logarithmically in the lower integration limit. Note In the spherical gravitational case, the possibility of more than one potential well is excluded. Actually, given a density ρ(r ), it generates a potential U (r ) and so a force whose absolute value is G M(r )/r 2 . The centrifugal equilibrium condition is r θ˙ 2 = G

M(r ) , r2

which, by elimination of θ˙ by the integral of angular momentum, becomes L 2 = G M(r )r. The above equation has a unique solution, for any chosen value of L 2 , because the function M(r )r is a continuous increasing function of r which is zero in the origin and diverges at infinite. This means that, for any given value of the angular momentum, there is only one circular orbit, which is characterized, indeed, by Vef f (r ; L) = 0.

3.6 Periods of Oscillations

95

3.6.1 Radial Period in the Keplerian Potential In this case, U (r ) = G M/r where M is the mass of the point generating the field, so that the period of radial oscillations writes as r+ Tr = 2 r−

√  = 2   r+

dr



1 L2 GM 2 E− + 2 2r r

r−

r dr Er 2 − 21 L 2 + G Mr

.

(3.77)

As usual, r± are the minimum and maximum radial distances, solutions of Er 2 + G Mr −

1 2 L = 0. 2

(3.78)

As the qualitative study of the Kepler potential taught us, the motion is bound (r+ is finite) or unbound (r+ → ∞) when Ve f f,min (L) ≤ E < 0 or E ≥ 0, respectively. When L 2 = 0 then r− = 0; we have to discuss dependence of r± on E when 2 L > 0. • Ve f f,min (L) ≤ E < 0: the Eq. 3.78 has two real (the radicand is ≥ 0) and distinct (because Ve f f,min (L) ≤ E) solutions  L2 −G M ∓ G 2 M 2 + 2E 2 m , r± = (3.79) 2E with r− < r+ < ∞, taking sign + for r− and − for r+ in Eq. 3.79. The circular orbit (r− = r+ ) corresponds to E = Ve f f,min . 2 • E = 0: Eq. 3.78 reduces to a linear equation, with r− = 21 GLM as root. Consequently, the motion is unbound, r+ → ∞. Note that lim |˙r | = 0, that means that r →∞ the particle of zero energy in the Keplerian field reaches an infinite distance from the origin in an infinite time with eventual zero speed. The integral in Equation 3.77 can be rewritten as follows, after letting a ≡ E < 0, b ≡ G M > 0 and c ≡ − 21 L 2 < 0 √  r dr , Tr = 2 √ 2 ar + br + c r+

(3.80)

r−

√ and can be solved by parts, with the choices u = r and dv = dr/ ar 2 + br + c, so to have

96

3 Central Force Fields

⎫ ⎧    ⎬



r+ √ ⎨ r 1 2ar + b 2ar + b r+ −√ dr . Tr = 2 +√ arcsin arcsin √ √ ⎭ ⎩ −a −a Δ Δ r− r−

(3.81) Now, ⎧ ⎫ √  

t+ ⎬ √ ⎨ r Δ 1 2ar + b r+ −√ Tr = 2 +√ arcsin t dt , arcsin √ ⎩ ⎭ −a −a 2a Δ r−

(3.82)

t−

where we introduced t≡

2ar + b , √ Δ

2ar− + b 2ar+ + b and t+ = √ . which implies t− = √ Δ Δ ' The integral arcsin t dt is, also, solvable by parts (u = arcsin t and dv = dt) obtaining   arcsin t dt = t arcsin t + 1 − t 2 , to give, finally  r+  √ √  √ 1 Δ Δ 1 t + Tr = 2 − √ = 1 − t2 arcsin t r − √ 2a 2a −a −a r−   



√ 1 1 b b +√ = = 2 −√ arcsin(−1) − arcsin(1) − 2a 2a −a −a √

  √  √ b π b 2 2 GM 1 − = π π = 2 2√ = . 2a 2 2 (−a) 23 2 (−E) 23 −a Note that Tr depends only on E and not on L.

(3.83)

3.6 Periods of Oscillations

97

3.6.2 Radial Period in the Homogeneous Sphere Potential In any spatially limited spherical distribution of density (ρ(r ) > 0 for r ≤ R) the possible cases, in dependence on E and L, are: (i) r− < r+ ≤ R, (ii) r− < R ≤ r+ , (iii) R ≤ r− < r+ . In the case of uniform density (ρ(r ) = ρ0 for r ≤ R), case (iii) occurs, for r+ < 8 +∞, only if L 2 ≥ π Gρ0 R 4 . 3 In the 3 cases we have: r+ 

(i) Tr = 2 r−

dr 2(E − Veint ff)

R 

(ii) Tr = 2 r−



(iii) Tr = 2 r−

r+

dr 2(E − Veint ff)

r+

,

dr 2(E − Veext ff)



+ R

dr 2(E − Veext ff)

,

(3.84)

,

ext where Veint f f (r ; L) and Ve f f (r ; L) are the expressions of the effective potential energy inside and out of the homogeneous sphere, respectively. Let us treat extensively case (i). We have

√  Tr = 2 √ r+

r−

with

r dr Ar 4

+ Br 2 + C

⎧ 2 ⎪ ⎪ ⎪ A = − π Gρ0 < 0, ⎨ 3 B = E + 2π Gρ0 R 2 ≥ 0, ⎪ ⎪ 1 ⎪ ⎩ C = − L 2 ≤ 0. 2

,

(3.85)

98

3 Central Force Fields



Note that B ≥ 0 because, to have motion, E ≥ Ve f f (r1 ; L) = L  Gρ0 > 0. and, consequently, B ≥ L 4π 3

4π Gρ0 − 2π Gρ0 R 2 , 3

The integral in Eq. 3.85 is an elliptic integral.4 If L 2 = 0 (purely radial motion) the integral simplifies, because L 2 = 0 ⇒ C = 0, and it reduces to r+ √ r+ √  dr 2 dr =√ Tr = 2 √ =  2 A Ar + B B 0 0 1 + r2 B  √ y+ √ − BA dy 2 2 y  =√ =√ [arcsin y]0+ , −A B 1 − y2

(3.86)

0

 after setting − BA r 2 = y 2 so as y = − BA r . From Exercise 3.4, it is y+ = 1, whence √ Tr = 

2

2 π Gρ0 3

[arcsin y]10 =

1 π ·  2 π Gρ0 3

(3.87)

Note The above result could have also been obtained taking into account that in a homogeneous sphere when L = 0 the motion is radial and harmonic with π = · pulsation ω2 = 43 π Gρ0 which gives the motion period T = 2π) ω π Gρ0 3 By its definition, the radial period is Tr = T /2. Let us now consider the generic L 2 > 0 case. In this case we have √ x+ √ r+2 2 dr 2 2 dx Tr = = , √ √ 2 2 Ar 4 + Br 2 + C Ax 2 + Bx + C r−2

(3.88)

x−

with the substitution x = r 2 . Given that A < 0 √ and Δ ≡ B 2 − 4 AC > 0, the integral in Eq. 3.88 can be solved by the substitution Ax 2 + Bx + C = t (x − α), where 'x √  “elliptic integral” is any function f (x) which can be expressed as f (x) = R t, P(t) dt, c

√  where c is a constant, R t, P(t) is a rational function of both arguments and P is a polynomial of degree 3 or 4.

4 An

3.6 Periods of Oscillations

99

α is one of the two (real and distinct) roots of the radicand x− and x+ (x+ > x− ), giving √  

2 1 2 Ax + B x+ Tr = − . (3.89) arcsin √ √ 2 −A Δ x− The values of x± are solutions of Ax 2 + Bx + C = 0, that are √ −B ∓ Δ , (3.90) x± = 2A √ √ so that r− = x− , r+ = x+ . Substituting in Eq. 3.89 x− and x+ as from (3.90) we get √ √ 1 2 2π π Tr (E, L) = − √ , (3.91) = π [arcsin(−1) − arcsin(1)] = √ 2 3 Gρ0 2 −A 2 −A which is the same results of the case L 2 = 0. This is relevant because it means that Tr does not depend neither on E nor on L, so it corresponds to a fully isochrone radial oscillation. Actually, the oscillation period depends only upon ρ0 , other than (as obvious) upon G. This characteristic reminds the small oscillations of a simple pendulum whose period does not depend on the initial conditions (and so on the energy and angular momentum) but just on the length of the wire and on the gravitational acceleration. Exercise 3.5 In the case of a homogeneous sphere, calculate Tr for orbits such that r− < R ≤ r+ (case (ii) in Eq. 3.84). Solution Left to the reader.

3.6.3 Radial Period in the Plummer Potential A Plummer sphere of mass M and scale radius b gives the potential (see Eq. 3.41) GM , U (r ) = √ r 2 + b2

(3.92)

and the effective potential energy is GM 1 L2 + , Ve f f (r ; L) = − √ 2 r2 r 2 + b2

(3.93)

100

3 Central Force Fields

consequently, the radial period of oscillations for given E and L is r+ √  Tr = 2  r−

√ r 4 r 2 + b2 √ 1 √ Er 2 r 2 + b2 + G Mr 2 − L 2 r 2 + b2 2

dr,

(3.94)

where, as usual, r− and r+ are the pericenter and apocenter distances. In general cases for E and L, neither the integral Eq. 3.94 nor the equation E − Ve f f (r ; L) = 0 can be explicitly solved, requiring a numerical approach. Given this, we limit here to the study of the case of small amplitude oscillations of radial orbits (L = 0) around the sphere center, i.e. the case of low E  −G M/b. For L = 0, and for  energies,  GM 2 − 1, and any E, we have r− = 0 and r+ = b bE √  Tr = 2  r+

0

dr E+

√G M r 2 +b2

,

which, by a Taylor expansion up to 4th order, gives √   2 B C Ar+ + r+3 − r+5 + O(r+7 ) , Tr = √ 3 5 E + (G M)/b

(3.95)

(3.96)

where ⎧ A = 1, ⎪ ⎨ GM B = 4b2 (Eb+G , M) ⎪ ⎩ C = 3G M(2Eb+G M) .

(3.97)

32b4 (Eb+G M)2

The radial period assumes so the expression √

Tr = √

  GM 2 2 − 3G M(2Eb + G M) r 4 + O(r 6 ) , r+ 1 + r + E + (G M)/b 12b2 (Eb + G M) + 160b4 (Eb + G M)2 +

(3.98)

which, at lowest order, approximates as   GM 2 −1 √ 2 bE Tr = √ 2b. r+ = √ E + (G M)/b E + (G M)/b √

(3.99)

3.6 Periods of Oscillations

101

3.6.4 Radial Period in the Isochrone Potential As we saw, the isochrone potential summarizes some characteristics of a homogeneous matter distribution (at small radii) and of a Keplerian potential at large radii. Actually, its potential is U (r ) =

1 GM GM  = √

2 , 2 2 b b+ b +r 1 + 1 + br

(3.100)

where b is a length scale, one of the 2 free parameters of the model (the other is the total mass M). The radial period of bound oscillations (E < 0) is r+ √  Tr = 2  r−

dr E−

1 L2 2 r2

+

GM b

 1 2 1+ 1+( br )

.

(3.101)

With the change of variables GM =1+ s= bU (r )

 1+

 r 2 b

≥ 2,

it results dr = b √

s−1 ds, s(s − 2)

and the integral in Equation 3.101 transforms into Tr =

√

s+ 

2b s−

(s − 1)ds , 

2 G M 1 L G M s−2 − Es 2 − 2E − b b 2 b2

(3.102)

so that, indicating with s− and s+ (s− < s+ ) the two roots of the denominator, it can be rewritten as 

2 Tr = b −E

s+ s−

   1 (s − 1) 2 −1 + (s− + s+ ) . ds = π b √ −E 2 (s − s− )(s+ − s) (3.103)

102

3 Central Force Fields

As it is easily seen 1 GM (s− + s+ ) = 1 − , 2 2bE and finally √ Tr =

2 GM . π 2 (−E) 32

(3.104)

Note that this result (independent of both b and L) is the same as for the Keplerian potential of a mass M.

3.7 Azimuthal Period As for the radial period, we can define an azimuthal period as the (constant) time needed to make a revolution around the origin, i.e. the time needed to increase of 2π the angular coordinate, θ , on the motion plane. Formally, Tθ is given by t (θ+2π) 

Tθ =

θ+2π 

dt = θ

t (θ)

1 dθ = L θ˙

θ+2π 

1 r dθ = L

r (θ+2π) 

2

θ

r r (θ)

2 dθ

dr

r (θ+2π) 

dr = r (θ)

dr , r˙

(3.105) where we made use of the angular momentum conservation, L = r 2 θ˙ = const. To evaluate Tθ , we can choose as initial θ the pericenter azimuth, for which r (θ ) ≡ r− . We can thus write Tθ as r+ Tθ = r−

dr + r˙

r(2π)

r+

dr , r˙

(3.106)

so that if r (2π ) < r− , Tθ < Tr ; if r (2π ) = r− , Tθ = Tr ; if r (2π ) > r− , Tθ > Tr . Clearly, indicating with Δθ the angular variation of the particle in a complete radial oscillation in a time Tr , it is given by θ(r  +)

Δθ =

θ(r  −)

dθ + θ(r− )

θ(r+ )

r+

dθ = 2 r−

L dr  , r 2 2[E − Ve f f (r ; L)]

(3.107)

3.7 Azimuthal Period

103

y

r −

y

r −

r + x

r + x

Fig. 3.6 A Keplerian ellipse (left panel) and a canonic ellipse (right panel). The arrow indicates the usual direction of increasing azimuth

so that



Δθ ≥ 2π, if Tθ ≤ Tr , Δθ ≤ 2π, if Tθ ≥ Tr .

(3.108)

For the Keplerian case and the homogeneous sphere, Tr = Tθ and Tr = Tθ /2, respectively (Fig. 3.6).

3.7.1 Fully Periodic Motion From what we said above, a planar motion in an external radial force field may be radially periodic and/or azimuthally periodic, and this happens when exists a constant time, Tr , such that r (t + Tr ) = r (t) and another constant of time, Tθ , such that θ (t + Tθ ) = θ (t), for every t. Obviously, the full (or simple) periodicity corresponds to the existence of a single constant of time T for which r(t + T ) = r(t), for every t. In a planar motion this is equivalent to r (t + T ) = r (t) and θ (t + T ) = θ (t). It is easy to see that if Tr and Tθ are commensurable, i.e. their ratio is a rational number, Tθ /Tr = l/m where l and m are positive integers, the motion is periodic and the period is T = lTr = mTθ . Let us show, indeed, that, upon the above hypotheses, r (t + T ) = r (t) and θ (t + T ) = θ (t), ∀t. Actually, if Tr is the radial period, i.e. r (t + Tr ) = r (t), then r (t + T ) = r (t + lTr ) = r (t + (l − 1)Tr + Tr ) = r (t + (l − 1)Tr ) = r (t + (l − 2)Tr + Tr ) = r (t + (l − 3)Tr ) = ... = r (t + (l − l + 1)Tr + Tr ) = r (t + Tr ) = r (t). By the same token it can be shown that θ (t + T ) = θ (t + mTθ ) = θ (t) and so the periodicity is proven.

104

3 Central Force Fields

3.8 The Inverse Problem in a Central Force Field In a central force field (per unit mass) whose equiforce surfaces are spherical, F = f (r )er , the spherical polar equations of motions are   ar = r¨ − r θ˙2 er = f (r ) er , (3.109)

 aθ = r θ¨ + 2˙r θ˙ eθ = 0. As we saw in Sect. 3.2, integration of the second of the equations above leads to L z = r 2 θ˙ = const., so that we can eliminate θ˙ in the first equation. Actually, letting again L = L z , r˙ =

dr dr L , θ˙ = dθ dθ r 2

(3.110)

and θ¨ =

d θ˙ L d = 2 dt r dθ



L r2

 = −2

L 2 dr , r 5 dθ

(3.111)

implying

 

2 2 dr d r L 2 dr 2 d 2 r 2 dr L ˙ + ¨= − 2 . θ θ˙ = θ dθ dθ 2 dθ r2 dθ 2 r 5 dθ

(3.112)

Excluding radial orbits (L = 0) which cross the origin, and letting u ≡

1 , we have r

r¨ =

d dt



1 du d 2r dr 2 =− 2 −→ = 3 dθ u dθ dθ 2 u



du dθ

2 −

1 d 2u , u 2 dθ 2

which leads to writing the first of Eqs. 3.109 in terms of u and θ   

2 2 2 2 du d u 1 du 2 4 2 L u − 2 2 − 2L u − L 2u3 = u 3 dθ u dθ dθ   2 2d u 3 = −L u + u = f (1/u). dθ 2 2

The final form of the equation for the trajectory is so

2  d u 1 u2 + u = − 2 f (1/u). dθ 2 L

(3.113)

(3.114)

(3.115)

(3.116)

3.8 The Inverse Problem in a Central Force Field

105

Given the force field f (1/u) the solution of the above non-linear differential equation in the unknown u(θ ) would give the trajectory r (θ ) = 1/u(θ ) of the body, once that ) are provided. Both these conditions are obtained from initial (in u(θ0 ) and ( du dθ 0 time) conditions r (0) = r0 and r˙ (0) = r˙0 , being (with the choice θ0 as the anomaly at time t = 0)



 1 du 1 dr r˙0 (3.117) and =− 2 =− . u(θ0 ) = , r0 dθ 0 L r0 dθ 0 The search of solution for r (θ ) once given f (r ) corresponds, indeed, to the direct problem of dynamics. On the other side, supposedly known the planar trajectory r (θ ), the search for a central force field compatible with that trajectory constitutes the inverse problem of dynamics. The existence and uniqueness of a solution for this inverse problem is not guaranteed. Under some conditions the central force field compatible with a given family of trajectories is given by

2  d u f (1/u) = −L 2 u 2 + u . (3.118) dθ 2 Of course, this explicit solution is found whenever the trajectory u(θ ) is (i) twice differentiable and (ii) the right hand side of Eq. 3.118 can be expressed as explicit function of u (which is not always the case).

3.8.1 From Elliptic Trajectories to their Central Force Field A Keplerian ellipse is an elliptical trajectory one focus of which is in the coordinate origin (where the massive point is located), so that its polar expression is r (θ ) =

p 1 = , 1 + e cos(θ − θ0 ) u

(3.119)

where e < 1 is the eccentricity, p the ellipse’s parameter and θ0 the angular position of the pericenter (indeed, r− = r (θ0 ) = p/(1 + e), while r+ = r (θ0 + π ) = p/(1 − e)). Commonly, the choice θ0 = 0 is done, which means counting θ angles starting from the pericenter position. For the given elliptic trajectory, it is du e sin(θ − θ0 ) =− dθ p so that Eq. 3.118 gives

=⇒

d 2u e cos(θ − θ0 ) , =− dθ 2 p

(3.120)

106

3 Central Force Fields

 e cos(θ − θ0 ) 1 + e cos(θ − θ0 ) 1 + = −L 2 u 2 , (3.121) f (1/u) = −L 2 u 2 − p p p which in terms of r gives the Keplerian force field f (r ) = −

L2 1 . p r2

(3.122)

On another hand, ellipses centered in the origin have the so called “canonical” (Cartesian) form x2 y2 + = 1. a2 b2 In planar polar coordinates, Eq. 3.123 writes as x = a cos θ, y = b sin θ,

(3.123)

(3.124)

Thanks to (3.124) and letting u ≡ 1/r , Eq. 3.123 transforms into u2 =

a2

cos2

1 . θ + b2 sin2 θ

(3.125)

Differentiating the above expression we get du 2 du = 2u = 2(a 2 − b2 ) sin θ cos θ u 4 , dθ dθ

(3.126)

that leads, in sequence, to du = (a 2 − b2 ) sin θ cos θ u 3 , dθ

(3.127)

and $ % d 2u = (a 2 − b2 ) 3(a 2 − b2 )u 5 sin2 θ cos2 θ + u 3 (cos2 θ − sin2 θ ) . 2 dθ

(3.128)

Given the above expression, Eq. 3.118 writes as ( ) * + f (1/u) = −L 2 u 2 (a 2 − b2 ) 3u 5 (a 2 − b2 ) sin2 θ cos2 θ + u 3 (cos2 θ − sin2 θ) + u ,

(3.129) and so, eliminating sin2 θ and cos2 θ by mean of Eq. 3.125, we finally have

3.8 The Inverse Problem in a Central Force Field

f (1/u) = −

107

L2 L2 1 = − r. a 2 b2 u a 2 b2

(3.130)

We have thus obtained the harmonic force, f (r ) ∝ −r , as expected. Exercise 3.6 In spherical symmetry, what potential admits trajectories of type r = aebθ , with a, b ∈ R? Solution f



 

 1 1 b2 −bθ 1 1 b2 1 = −L 2 2 e = −L 2 2 + = + u r a r r r r  dU 1 . = −L 2 3 b2 + 1 = r dr

By straightforward integration U (r ) =

(b2 + 1)L 2 1 + c, 2 r2

whence, using Poisson’s equation   2  1 1 1 d 2L 2 r b +1 3 = ρ(r ) = 4π G r 2 dr m r

  1 1 1 L2 2 b + 1 2 − 2 < 0, = 4π G m r r

∀b ∈ R.

The above density is always negative, that means that the logarithmic spiral trajectories are not compatible with a dissipationless Newtonian gravitational field.

3.9 Further Readings The contents of this chapter are strictly related to the ones of the previous chapter. So the relevant additional readings partly overlap, and the suggested books are [4, 5, 8, 10, 13, 14, 16, 18, 20, 22].

Chapter 4

Potential Series Developments

4.1 Fundamental Solution of Laplace’s Equation We saw in Chap. 1 that in Rn ⎧ 1 ⎪ ⎪ ⎪ ⎨ ∇ · ∇ log r = 0, if n = 2 ⎪ ⎪ 1 ⎪ ⎩∇ · ∇ = 0, if n > 2, n−2 r

(4.1)

  n  with r ≡  xi2 . i=1

Let us consider the generic point r ∈ Rn as ‘pole’ and define r˜ = r − r . If f = n−1 , it results f (˜r ), recalling that ∇ · er = r ∇ 2 f (˜r ) = ∇ · ∇ f (˜r ) = ∇ · f  (˜r )er˜ = f  (˜r )∇ · er˜ + er˜ · ∇ f  (˜r ) = n−1  n−1  f (˜r ) + er˜ · f  (˜r )er˜ = f (˜r ) + f  (˜r ). = r˜ r˜ Consequently, the Laplace’s partial differential equation ∇ 2 f (˜r ) = 0 reduces to an ordinary differential equation n−1  f (˜r ) + f  (˜r ) = 0, r˜

(4.2)

which, in the case n = 3, writes also as 1 d 2  r˜ f (˜r ) = 0. r˜ 2 d r˜ © Springer Nature Switzerland AG 2019 R. A. Capuzzo Dolcetta, Classical Newtonian Gravity, UNITEXT for Physics, https://doi.org/10.1007/978-3-030-25846-7_4

(4.3) 109

110

4 Potential Series Developments

In general, the solution, as dependent only upon r˜ = |r − r |, of the Laplace’s equation above is given by1 ⎧ trivial case ⎪ ⎨ f (˜r ) = const., c0 f (˜r ) = n−2 + c1 , if n > 2 ⎪ r˜ ⎩  f (˜r ) = c0 log r˜ + c1 , if n = 2. Actually, the differential equation (4.2) is solved by letting g(˜r ) = f  (˜r ) such that the 2nd order equation transforms into the 1st order system ⎧  ⎨ f (˜r ) = g(˜r ) ⎩ n−1 r˜

(4.4)

g(˜r ) + g  (˜r ),

which has solution g(˜r ) = c0 r˜ 1−n , so that f (˜r ) comes by a quadrature of g(˜r ), to get ⎧ r˜ 2−n ⎪ ⎪ ⎨ f (˜r ) = c0 + c1 , if n > 2 2−n ⎪ ⎪ ⎩ f (˜r ) = c0 log r˜ + c1 , if n = 2.

(4.5)

Letting, in Eq. 4.5, c0 = −G,

c1 = 0,

for n > 2,

c0 = G,

c1 = 0,

for n = 2,

one obtains the so called fundamental solution of the Laplace’s equation (or fundamental harmonic) ⎧ 1 ⎪ ⎪ , if n > 2, ⎨ Γn (|r − r |) = −G (2 − n)|r − r |n−2 (4.6) ⎪ ⎪ ⎩ Γ (|r − r |) = G log |r − r |, if n = 2, 2

whence the definition of volume potential as a convolution product

UC (r) =

ρ(r )Γn (|r − r |) d n r ≡ ρ ∗ Γn ,

(4.7)

C

where C is a matter domain and the operator ∗ represents the convolution product. 1 In

R every linear function f (x) = ax + b is harmonic.

4.1 Fundamental Solution of Laplace’s Equation

111

In the case n = 3 we obtain the already known (see Sect. 2.5.1) expression for the potential



ρ(r ) 3  d r, (4.8) UC (r) = G  C |r − r | with ρ(r ) bound and integrable function in the domain C. If n = 2 the convolution gives

UC (r) = G

ρ(r ) log |r − r | d 2 r ,

(4.9)

C

and so ∇ · ∇UC = −2πGρ

(4.10)

which is the Poisson’s equation in 2D. In the general case of n dimensions it is ∇ · ∇UC = −nωn Gρ,

(4.11)

where nωn = σn represents the area of the unit ball whose volume measure is ωn .

4.2 Harmonic Functions Let S represent the ball centered in O and unitary radius and, on its surface, consider   (opposed to the curved sides the spherical triangle abc with angles at corners A, B, C a, b, c obtained as intersection of the spherical surface with three planes through the sphere center) (Fig. 4.1).

Fig. 4.1 The spherical triangle abc. Source https://commons.wikimedia.org/wiki/File: Spherical_trigonometry_legendre.svg

112

4 Potential Series Developments

For this kind of triangle the following theorems of spherical trigonometry hold: Theorem 4.1 (Sine theorem)   sin  B sin C sin A = = . sin a sin b sin c

(4.12)

Theorem 4.2 (Cosine theorem)  cos a = cos b cos c + sin b sin c cos A, cos b = cos a cos c + sin a sin c cos  B,  cos c = cos a cos b + sin a sin b cos C,  = − cos   + sin   cos a, cos A B cos C B sin C  cos C  + sin A  sin C  cos b, cos  B = − cos A     cos C = − cos A cos B + sin A sin  B cos c.

(4.13) (4.14) (4.15) (4.16) (4.17) (4.18)

Note that Theorems 4.1 and 4.2 are the equivalent, in spherical trigonometry, of the sine theorem and of the cosine (Carnot’s) theorem in the plane, which give, respectively   sin  B sin C sin A = = , a b c  a 2 = b2 + c2 − 2bc cos A.

(4.19) (4.20)

Now, let r and r the position vectors of two arbitrary points P and Q of unit mass. Setting G = 1, the Newtonian potential in P ≡ r as due to the presence of Q ≡ r (and viceversa), is given by 1 , Γ3 (|r − r |) = |r − r | 

with |r − r | = r 2 + r  2 − 2rr  cos γ, 

(4.21)

 . Resorting to spherical polar coordinates such that after defining the angle γ = rr    r = (r, θ, ϕ) and r = (r , θ , ϕ ), and applying the cosine Theorem 4.2 to the P  P  S spherical triangle drawn on the spherical surface of radius |r | (see Fig. 4.2), it results that  P  = cos rr 

 ≡ cos γ = cos ϕ cos ϕ + sin ϕ sin ϕ cos(θ − θ ), cos P

so to give 1 1 = = f (r, θ, ϕ, r  , θ , ϕ ) =  2 2  |r − r | r + r − 2rr  cos γ 1 1 =    . 2 r r r + 1 − 2 cos γ r r

(4.22)

4.2 Harmonic Functions

113

θ−θ

S ϕ ϕ P r

P’’

γ

P’

r’

O θ θ Fig. 4.2 The spherical triangle P  P  S

Letting in the above relation μ ≡

r and u ≡ cos γ, it writes as r

1 1 1 =  . |r − r | r μ2 + 1 − 2μu

(4.23)

Our purpose is now that of building a proper development of the above function in series of μ. Actually, the function (1 + x)n admits a development in series of x, such that (i) (ii) (iii) (iv)

it converges for |x| < 1, if n ≤ −1; it converges for −1 < x ≤ 1, if −1 < n < 0; it converges for |x| ≤ 1, if n > 0 and n ∈ / Z+ it reduces to the usual Newton’s binomial expansion, if n ∈ Z+ .

In general, the formal expression is (note the alternate change of sign) n(n − 1) · · · (n − k + 1) k n(n − 1) 2 x ± · · · + (±1)k x + 2 k! ∞   ∞   n n + ······ ≡ (±x)k = (−1)k x, k k k=0 k=0 (4.24) where in the above sum the sign + is chosen if k is even and − if k is odd, and where we have introduced a double round parenthesis symbol as a generalization of the usual binomial coefficient (which is defined when both n and k are non negative integers, while here n is not necessarily a positive integer). (1 ± x)n = 1±nx +

114

4 Potential Series Developments

Actually, for n ∈ Z+ , we have, by our definition     n n(n − 1) · · · (n − k + 1) n! n ≡ = = . k k! k!(n − k)! k Going back to Eq. 4.23, the function 

1 μ2

+ 1 − 2μu

 − 1 = 1 + (μ2 − 2uμ) 2

(4.25)

is of the type (1 + x)n , once assumed x ≡ μ(μ − 2u) and n = −1/2. Consequently, 

k f actor s

     1 3 5 1    1  − − − ··· −k n −2 2 2 2 2 = = = k k(k − 1) · · · · · · · · · 2 · 1 k       1 3 5 2k − 1 − − − ··· − 2 2 2 2 = = k(k − 1) · · · · · · · · · 2 · 1 

= (−1)k

1 1 · 3 · 5 · · · (2k − 1) = 2k k(k − 1) · · · · 2 · 1

= (−1)k

1 · 3 · 5 · · · (2k − 1) = 2k · 2(k − 1) · · · 4 · 2 k 

(2l − 1)

k l=1

= (−1)

(4.26)

k 

, 2l

l=1

where the final ratio is that between the product of the first k odd integers to that of the first k even integers. As consequence of what above, we have 1·3 2 1 μ (μ − 2u)2 + = 1 − μ(μ − 2u) + 2 2·4 + 1 − 2μu 1 · 3 · · · (2k − 1) k 1·3·5 3 μ (μ − 2u)3 + · · · + (−1)k μ (μ − 2u)k + · · · (4.27) − 2·4·6 2 · 4 · · · 2k



1

μ2

The above series is not a simple power series but, rather, a polynomial series in μ, convergent for −1 < x ≡ μ(μ − 2u) ≤ 1,

4.2 Harmonic Functions

115

which is equivalent to2 

√ |μ| ≤ 2 − 1 ⇐⇒ 0 < γ < π. −1 < u < 1

The series (4.27) can be written in the more compact form ∞  1  1 −2  μk (μ − 2u)k = = k μ2 + 1 − 2μu k=0   ∞ ∞  1  k k   1 −2 k l −2 k k−l k−l μ (−1) (2u) = μ = μk+l Vl,k , l k k k=0 l=0 k=0 l=0 (4.28) where we set   k (2u)k−l . (4.29) Vl,k = (−1)k−l l

The coefficient of the generic μm is given by the sum of the coefficients of all the terms μk+l , with k and l variables under the constraint k + l = m with l ≤ k according to Table 4.1. Accordingly, the coefficient of μm is given by       n n n V0,m + V0,m−1 + V0,m−2 + · · · · · · + = m m−1 m−2 1 · 3 · 5 · · · (2m − 1) (−1)m (2u)m + = (−1)m    2 · 4 · · · 2m V0,m

+ (−1)m−1

1 · 3 · · · (2(m − 1) − 1) m−1 (−1)m−2 (2u)m−2 + 2 · 4 · · · 2(m − 1) 1!   

(4.30)

V1,m−1

1 · 3 · · · (2(m − 2) − 1) (m − 2)! (−1)m−4 (2u)m−4 + · · · + (−1)m−2 2 · 4 · · · 2(m − 2) 2!(m − 4)!    V2,m−2

m  It represents the summation of + 13 alternate (in sign) terms and it is actually 2 a polynomial of degree m in the variable u, Pm (u), whose general expression comes from the below development4

2 See

Exercise 4.1 in Sect. 4.3. is the integer part of m/2. 4 Note that (−1)2m−3l = (−1)l . 3 [m/2]

116

4 Potential Series Developments

Table 4.1 Values of k and l allowed upon the conditions k + l = m and l ≤ k

if m is even if m is odd



1 μ2

=

+ 1 − 2μu ⎡

=

2

k=0

l

m m–1 .. .

0 1 .. .

.. .

.. .

m 

m 

2 m 

2 m 

2

∞  1  k −

k

k

l=0

+1

2

μk+l Vl,k = ⎤

  ∞ ⎢  1  ⎥ −2 k ⎢ ⎥ (2u)k−l ⎥ μm = (−1)k−l ⎢ ⎣ ⎦ l k m=0 lk    l+k=m

(4.31)

Vl,k

⎡

⎤  1    ∞ −2 m − l m−2l m−2l ⎥ m ⎢ = 2 (−1)m−2l u ⎣ ⎦μ = l m − l m=0 lk l+k=m

⎡ =

⎤

m 

⎥ ∞ ⎢ 2 ⎢ ⎥ ⎢ (−1)m−l 1 · 3 · · · (2(m − l) − 1) (−1)m−2l (m − l)! 2m−2l u m−2l ⎥ μm = ⎢ ⎥ 2 · 4 · · · 2(m − l) l!(m − 2l)! ⎦ m=0 ⎣ l=0    2m−l (m−l!!

⎡ m  ⎤ ∞ 2 1 · 3 · · · (2(m − l) − 1) 1 1 ⎢ ⎥ X  u m−2l ⎦ μm = = 2m−l (mX −X l)! l (−1)l  ⎣ X  X  m−l l!(m − 2l)! 2 X 2 (m −X l)!  X m=0 l=0 ⎡ m  ⎤ ∞ 2 1 · 3 · · · (2(m − l) − 1) ⎢ ⎥ (−1)l u m−2l ⎦ μm = = ⎣ 2l l! (m − 2l)! m=0

≡

∞

l=0

Pm (u)μm ,

m=0

(4.32)

4.2 Harmonic Functions

117

where we set Pm (u) =

[ m2 ] 1 · 3 · · · (2(m − l) − 1) l=0

2l l! (m − 2l)!

(−1)l u m−2l .

(4.33)

Pm (u) is called Legendre’s polynomial of degree m, and it has a relevant importance in mathematical physics. Hereafter we will often refer to 

1 μ2 + 1 − 2μu

as “generating” function of Legendre’s polynomials. Note Note that, because 

m − 2l, is even when m is even, m − 2l, is odd when m is odd,

and because (−u)m−2l = (−1)m−2l u m−2l , it results Pm (−u) = (−1)m Pm (u)

(4.34)

Note ⎧m  m ⎪ = , when m is even ⎪ ⎨ 2 2   ⎪ ⎪ ⎩ m = m − 1 , when m is odd 2 2 The terms in the Legendre’s polynomials are of alternating sign, and the highest power of u has always a positive sign. Pm (u) contains only even powers of u, if m is even, and only odd powers of u, if m is odd. This resumes as P2m (u) = P2m (−u), P2m+1 (u) = −P2m+1 (u),

so P2m (u) is an even function so P2m+1 (u) is an odd function

Finally, the development for the fundamental harmonic ∞ ∞ 1 1 1 1 rm m  = = P (u)μ = P (u) m m |r − r| r  μ2 + 1 − 2μu r  m=0 r m+1 m=0

is such that the following theorem holds:

(4.35)

118

4 Potential Series Developments

Theorem 4.3 The series ∞ 1 1 Pm (u)μm , = |r − r | r  m=0

(4.36)

converges for μ=

r < 1, r

u = cos γ ∈ [−1, 1] ⇐⇒ γ ∈ [0, π].

Proof Using the known Euler’s formula u = cos γ = it results 

eiγ + e−iγ , 2

1

1 = $ % $ %= μ2 + 1 − 2μu 1 − μeiγ 1 − μe−iγ $ %− 1 $ %− 1 = 1 − μeiγ 2 1 − μe−iγ 2 = =

∞

αk ekiγ μk

k=0

∞

(4.37)

αk e−kiγ μk

k=0

where we developed in series of x = μeiγ and y = μe−iγ the two expressions (1 − 1 1 x)− 2 and (1 − y)− 2 . The two series are converging ∀γ ∈ R if |μ| < 1, so that the original series with coefficients Pm (u) converges for μ < 1. The expression Eq. 4.37 allows writing Pm (u) as a trigonometric polynomial: Pm (u) = 2α0 αm cos mγ + 2α1 αm−1 cos(m − 2)γ+ + 2α2 αm−2 cos(m − 4)γ + · · · · · · .

(4.38)

Actually, Pm (u) is obtained by the sum of trigonometric terms in (4.37) with indexes that give m as sum. For example, the second addend above come this way 2α1 αm−1 cos(m − 2)γ = α1 αm−1 eiγ e−i(m−1)γ + αm−1 α1 ei(m−1)γ e−iγ = $ % = α1 αm−1 eiγ(2−m) + e−iγ(2−m) = = 2α1 αm−1 cos(2 − m)γ, and, thus, for the (k + 1)th term we have 2αk αm−k cos(2k − m)γ. Moreover, αk > 0 ∀k, because

(4.39)

4.2 Harmonic Functions

119

 1  ∞ ∞ −2 1 · 3 · 5 · · · (2k − 1) k k xk = (1 − x) = (−1) (−1)2k x . k 2 · 4 · 6 · · · 2k  k=0 k=0  n

αk >0

Being all coefficients positive, Pm (u) takes its maximum value in u = cos γ = 1; actually Pm (u) =

[ m2 ]

2αk αm−k cos(2k − m)γ ≤

k=0

[ m2 ]

2αk αm−k = Pm (1).

k=0

Finally, let us note that, if m is even, in the expression of Pm (u) the constant term α m+1 cos γ. 2α2m appears, otherwise there is the term 2α m−1 2 2 2

4.3 Legendre’s Polynomials The generic Legendre’s polynomial of degree m is written as Pm (u) =

[ m2 ]

(−1)l

l=0

1 · 3 · · · (2(m − l) − 1) m−2l u = 2l l! (m − 2l)! m−l &

m 2

=

[ ]

(−1)l

l=0

(4.40)

(2k − 1)

k=1

2l l! (m − 2l)!

u m−2l .

Plots of Legendre’s polynomials up to the 5th order are shown in Fig 4.3. Actually, Legendre’s polynomials with m = 0, . . . , 5 are explicitly given by P0 (u) = 1, P1 (u) = u, 1 P2 (u) = (3u 2 − 1), 2 1 1·3·5 3 1·3 u − u = (5u 3 − 3u), P3 (u) = 3! 2 2 1 P4 (u) = (35u 4 − 30u 2 + 3), 8 1 P5 (u) = (63u 5 − 70u 3 + 15u). 8

120

4 Potential Series Developments 2

1.5

P0

1

P1 0.5

P3 P5

P4

0

-0.5 P2 P0(x) P1(x) P2(x) P3(x) P4(x) P5(x)

-1

-1.5

-2 -1

-0.5

0

0.5

1

Fig. 4.3 Legendre’s polynomials Pm for 0 ≤ m =≤ 5

The zeros of such polynomials are P1 (u) = 0 ⇐⇒

u 1 = 0, u 1,2 = ± 13 , P2 (u) = 0 ⇐⇒ P3 (u) = 0 ⇐⇒ u 1 = 0, u 2,3 = ± 35 , √ √ P4 (u) = 0 ⇐⇒ u 1,2 = ± 15+235 30 , u 3,4 = ± 15−235 30 , √ √ P5 (u) = 0 ⇐⇒ u 1 = 0, u 1,2 = ± 35+263 70 , u 3,4 = ± 35−263 70 . Remembering that for odd values of m the polynomial Pm (u) contains the linear term in u, it results ∀k ∈ Z+ . P2k+1 (0) = 0,

4.3 Legendre’s Polynomials

121

4.3.1 Some Properties of Legendre’s Polynomials Note that

Pm (1) = 1, Pm (−1) = (−1)m ,

for every non-negative integer m, as we easily now show. Actually, letting u = 1, it results ∞ 1  = Pm (1)μm . (4.41) 2 μ + 1 − 2μ m=0 At the same time, for |μ| < 1, it is also (as geometric series in μ) 

∞

1

=

μ2 + 1 − 2μ

1 = um , 1 − μ m=0

(4.42)

so that, by comparison of Eqs. 4.41 and 4.42 it results Pm (1) = 1,

(4.43)

for any non-negative integer m. Additionally, because (see Eq. 4.34) Pm (−u) = (−1)m Pm (u),

(4.44)

letting u = 1 in the above formula we get Pm (−1) = (−1)m Pm (1) = (−1)m . Exercise 4.1 Show that the series ∞

1  = μ2 + 1 − 2μu k=0 converges for |μ| < Solution Left to the reader.

√

 1  −2 μk (μ − 2u)k k

2 − 1 and for −1 < u < 1.

(4.45)

122

4 Potential Series Developments

4.4 Recurrence Relations 4.4.1 First Recurrence Relation ' m Because the series ∞ m=0 Pm (u)μ is uniformly convergent for |μ| < 1, it is term by term differentiable with respect to μ and the obtained series converges to the 1 . derivative (with respect to μ) of  2 μ + 1 − 2μ From this, it results that ∞

∞

d 1 u−μ dμm  = Pm (u) Pm (u)mμm−1 , = 3 =  dμ μ2 + 1 − 2uμ dμ 2 2 μ + 1 − 2uμ m=0 m=0 which, after some manipulations, leads to ∞

1 u−μ  = Pm (u)mμm−1 , 2 2 μ + 1 − 2uμ μ + 1 − 2uμ m=0

(4.46)

and to (u − μ)

∞

Pm (u)μ = (μ + 1 − 2uμ) m

2

m=0

∞

Pm (u)mμm−1 ,

(4.47)

m=0

and, carrying the coefficients into the summation ∞

∞ % % $ $ Pm (u) uμm − μm+1 = Pm (u)m μm+1 + μm−1 − 2uμm .

m=0

(4.48)

m=0

Applying to this latter expression the polynomial identity principle (coefficients of same power of μ must be equal on both sides) we finally have this recurrence relation (m + 1)Pm+1 (u) − (2m + 1)u Pm (u) + m Pm−1 (u) = 0, ∀m ≥ 1.

(4.49)

4.4.2 Second Recurrence Relation The formal series development of the derivative with respect to u of the generating is function √ 2 1 μ +1−2μu

4.4 Recurrence Relations

123 ∞

1 d μ  =$ P m (u)μm . 3 = % 2 du μ + 1 − 2μu 2 2 μ + 1 − 2μu m=0

(4.50)

We now verify that P m (u) =

d Pm (u). du

(4.51)

At this purpose, we integrate with respect to u from 0 to generic u the above expression, integrating term by the term the series

0

u

$

μ

% 23 du = 

1

−

1

+ 1 − 2uμ 1 + μ2 μ2 + 1 − 2μu

u ∞ ∞ ∞ μm P m (u) du = μm Pm (u) − μm Pm (0), m=0

0

μ2

m=0

which implies



u

(4.52)

m=0

P m (u) du = Pm (u) − Pm (0),

(4.53)

0

and so P m (u) =

d Pm (u) ≡ Pm (u), du

(4.54)

which finally yields to ∞

d 1 μ  = Pm (u)μm . 3 = 2 du μ2 + 1 − 2μu 2 μ + 1 − 2μu) m=0

(4.55)

A multiplication of both the 2nd and 3rd sides of relation Eq. 4.55 by u − μ gives ∞

(u − μ)

μ$ μm Pm (u), % 3 = (u − μ) μ2 + 1 − 2uμ 2 m=0 ( ) ∞ 1 d  μm Pm (u), μ = (u − μ) dμ μ2 + 1 − 2uμ m=0 ) (∞ ∞ d m μ μ Pm (u) = (u − μ) μm Pm (u), dμ m=0 m=0 μ

∞ m=0

mμm Pm (u) =

∞ m=0

Pm (u)(uμm − μm+1 ),

124

4 Potential Series Developments

whence the second recurrence relation:  (u) ∀m ≥ 1. Pm (u)m = u Pm (u) − Pm−1

(4.56)

4.5 The Legendre’s Differential Equation The first and second recurrence relations Eqs. 4.49 and 4.56 lead to the differential equation satisfied by the generic Legendre’s polynomial Pm (u). Differentiation with respect to u of Eq. 4.49 gives   (u) − (2m + 1)Pm (u) − (2m + 1)u Pm (u) + m Pm−1 (u) = 0, (m + 1)Pm+1  (u) as obtained by Eq. 4.56, leads to which, by eliminating Pm−1

 (m + 1))Pm+1 (u) − (2m + 1)Pm (u) − (2m + 1)u Pm (u) + m(u Pm (u) − m Pm (u) = 0.

If we write the above expression reducing by 1 the indexes (m + 1 → m) we get  m Pm (u) − [2(m − 1) + 1]Pm−1 (u) − [2(m − 1) + 1]u Pm−1 (u)+  +(m − 1)[u Pm−1 (u) − (m − 1)Pm−1 (u) = 0.

After collecting coefficients one gets m2

−m         m Pm (u) + u Pm−1 (u) [m − 1 − 2(m − 1) − 1] −Pm−1 (u) [2(m − 1) + 1 + (m − 1)2 ] = 0,

which reduces to

 (u) − m Pm−1 (u) = 0, Pm (u) − u Pm−1

(4.57)

 (u) by mean of (4.56), becomes which, by elimination of Pm−1

(1 − u 2 )Pm (u) + mu Pm (u) − m Pm−1 (u) = 0,

(4.58)

that, upon differentiation with respect to u, leads to d  [(1 − u 2 )Pm (u)] + m Pm (u) + mu Pm (u) − m Pm−1 (u) = 0, du

(4.59)

 which, after a further elimination of Pm−1 (u) with (4.56), leads to the 2nd order Legendre’s differential equation

4.5 The Legendre’s Differential Equation

125

d [(1 − u 2 )Pm (u)] + m(m + 1)Pm (u) = 0. du

(4.60)

 A third recurrence relation can be obtained by relating Pm (u), Pm−2 (u), and Pm−1 (u) by mean of (4.56) and (4.57) (the reader is suggested to verify it, as exercise)  (u) − (2m − 1)Pm−1 (u) = 0. Pm (u) − Pm−2

(4.61)

4.6 Orthogonality of Legendre’s Polynomials Pm (u) Let I = [−1, 1] ⊂ R be a closed interval, and Pm (u), Pn (u) two Legendre’s polynomials of arbitrary degrees m and n. Theorem 4.4 The polynomials Pm (u) and Pn (u) are orthogonal in the interval I . More specifically, it results

1 (Pm (u), Pn (u)) ≡

Pm (u) Pn (u) du = −1



where δmn =

2 δmn , 2m + 1

0, m = n 1, m = n

(4.62)

(4.63)

is the usual Kronecker symbol. Proof Let us consider first m = n. Polynomials Pm (u) and Pn (u) satisfy the Legendre’s equation (4.60), i.e. d [(1 − u 2 )Pm (u)] + m(m + 1)Pm (u) = 0, du d [(1 − u 2 )Pn (u)] + n(n + 1)Pn (u) = 0, du

(4.64) (4.65)

so that by multiplication of both sides by Pn (u) and Pm (u), respectively, and by a subsequent subtraction of the second from the first equation we get  + d * (1 − u 2 ) Pm (u)Pn (u) − Pn (u)Pm (u) = du = [m(m + 1) − n(n + 1)] Pm (u)Pn (u).

(4.66)

126

4 Potential Series Developments

Note that the left hand side of the equation above is obtained by simple operations, as follows d d [(1 − u 2 )Pm (u)] − Pm (u) [(1 − u 2 )Pn (u)] = du du d d 2  = [(1 − u )Pm (u)Pn (u)] − [(1 − u 2 )Pn (u)Pm (u)] = du du  + d * (1 − u 2 ) Pm Pn (u) − Pn Pm (u) . = du Pn (u)

(4.67)

Integrating with respect to u between −1 and 1 both sides of Eq. 4.66 it results 

(1 − u

2

)(Pm Pn

−

1 Pn Pm ) −1

1 = [m(m + 1) − n(n + 1)]

Pm (u)Pn (u) du,

−1

(4.68) where the lhs is, clearly, zero, because Pm (u), Pn (u) ∈ C∞ ([−1, 1]). It then follows that

1 Pm (u)Pn (u) du = 0, −1

as we wanted to show. Let us now treat the case m = n. Consider the generating function of Pm (u) and square both sides of the series development, to obtain (∞ )2 ∞ ∞ 1 m m = P (u)μ = P (u)μ Pn (u)μn = m m μ2 + 1 − 2uμ m=0 m=0 n=0 =

∞ ∞

(4.69)

Pm (u)Pn (u)μm μn ,

m=0 n=0

which, upon integration in [−1, 1], yields

1 −1

⎞ ⎛ 1

∞ ∞ 1 ⎝ Pm (u)Pn (u) du ⎠ μm+n . du = μ2 + 1 − 2uμ m=0 n=0

(4.70)

−1

The integrals at rhs are all zeros for n = m, as we saw above, so we set n = m obtaining5

5 Using

the developments of log(1 + μ) and log(1 − μ), valid for |μ| ≤ 1.

4.6 Orthogonality of Legendre’s Polynomials Pm (u)

−

127

∞ ∞ (1 − μ)2 1+μ 1 μ2m+1 μ2m 1 1 log log = 2 = 2 = = 2μ (1 + μ)2 μ 1−μ μ m=0 2m + 1 2m + 1 m=0

=

∞

(4.71)

1 μ2m

m=0

Pm2 (u) du −1

that, for the identity by series principle, implies

1 Pm2 (u) du = −1

2 . 2m + 1

Finally, for every (m, n) positive integer pair, it is (Pm (u), Pn (u)) =

2 δmn . 2m + 1

(4.72)

From the above theorem it results the following

0 1 2m+1 Lemma 4.5 The set of polynomials P m (u) = Pm (u) is orthonormal in 2 the interval [−1, 1]. Moreover, because the set of Pm (u) polynomials is dense in the Hilbert space L 2 ([−1, 1]), the set P m (u) represents an orthonormal base of the Hilbert space L 2 ([−1, 1]). It is also possible to show that any other set of polynomials of alternating powers of u of degree m (including u 0 ) is orthogonal in [−1, 1] and reduces to the form c Pm (u) where c is a constant.

4.7 Development of a Function in Series of Legendre’s Polynomials Consider a function f (u) defined over the real interval [−1, 1]. It can be developed + in * P (u) = series of Legendre’s polynomials adopting the orthonormal function base m 3 2 2m + 1 Pm (u) , if f (u) satisfies the following Dirichlet conditions: 2 a. f (u) ∈ C 0 ([−1, 1]) except for a finite number of discontinuity points; b. f (u) has a finite number of minima and maxima in the [–1, 1] interval. Upon the above constraints, as further sufficient but not necessary condition, the series ∞ ( f, P m )P m (u), (4.73) m=0

128

4 Potential Series Developments

converges to: (i) f (u), if f is continuous in u; f (u + 0) + f (u − 0) , if u is one of the points in the finite set of discontinuity (ii) to 2 points for f (u).6 The series is written as f (u) =

∞

cm P m (u),

m=0

with $

%

1

cm = f, P m (u) =

f (u) P m (u) du.

(4.74)

−1

Indeed, if the series converges to f (u) and it is integrable term by term, it follows $

%

1 f (u) P n (u) du =

f, P n =

∞

1

−1

P m (u)P n (u) du = cn ,

cm

m=0

−1

due to P m (u) orthonormality. Exercise 4.2 Develop the Heaviside step function, defined in the interval [−1, 1] as  f (u) =

0, 1,

−1 ≤ u < 0, 0 ≤ u ≤ 1,

in series of Legendre’s polynomials (Fig. 4.4). Solution The development is possible because f (u) is continuous in the [−1, 1] interval, excluding the origin, so that (for any m ≥ 1)

1 cm =

1 f (u) P m (u) du =

−1

 =

P m (u) du = 0

1 2m + 1 2 2m + 1



1

    Pm+1 (u) − Pm−1 (u) du =

0

  1 2m + 1 Pm+1 (1) − Pm+1 (0) − Pm−1 (1) + Pm−1 (0) = = 2 2m + 1    1 2m + 1 1 − Pm+1 (0) −
1 + Pm−1 (0) = =
2 2m + 1 6 Here

we define f (u ± c) ≡ lim f (u ± c). c→0

4.7 Development of a Function in Series of Legendre’s Polynomials Fig. 4.4 Sketch of the Heaviside function restricted to the [−1, 1] interval

129

1

f(x+0)+f(x−0) 2

−1

 =

1

  1 2m + 1 Pm−1 (0) − Pm+1 (0) , 2 2m + 1

where we used the relation Eq. 4.61 and that Pk (1) = 1, ∀k. We have, finally

cm =

⎧ √2 ⎪ ⎪ 2 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ 0,

if m = 0

if m is even ⎪ 5 4 ⎪ ⎪ ⎪ m−1 1 · 3 · · · (m − 2) m+1 1 · 3 · · · m ⎪ 2m+1 1 ⎪ 2 ⎪ − (−1) 2 m+1 , if m is odd ⎩ m−1 m−1 2 2m+1 (−1) 2 2 ( 2 )! 2 2 ( m+1 2 )!

4.8 Rodrigues’ Formula Proposition 1 Let Pm (u) the generic Legendre’s polynomial. We show that it can be expressed as %m 1 dm $ 2 u −1 , Pm (u) = m (4.75) m 2 m! du which constitutes the Rodrigues’ formula. Proof m

Pm (u) =

2

l=0

m−l &

(2k − 1)

k=1 u m−2l = (−1)l l 2 l! (m − 2l)!

1 · 3 · · · (2m − 1) m 1 · 3 · · · (2m − 3) m−2 1 · 3 · · · (2m − 5) m−4 u − u = + − ··· = u m! 2(m − 2)! 22 2 (m − 4)!

130

4 Potential Series Developments

=

=

1 · 3 · · · (2m − 1)  m m! u − u m−2 + m! 2(m − 2)!(2m − 1)  m! u m−4 − · · · = + 2 2 2(m − 4)!(2m − 1)(2m − 3)

6 7 m 1 & m(m − 1) m−2 m(m − 1)(m − 2)(m − 3) m−4 (2k − 1) u m − + − · · · . u u m! 2(2m − 1) 22 2(2m − 1)(2m − 3) k=1

By m subsequent integrations over the limits 0 and u of Pm (u) and of the finite product here above it results



Pm (u) d m u ≡

du



du · · · du Pm (u) =   m times

=

6 m 1 & u 2m (2k − 1) m! (m + 1)(m + 2) · · · 2m k=1

− +

u 2m−2 m(m − 1) + 2(2m − 1) (m − 2 + 1)(m − 2 + 2) · · · (2m − 2)

7 u 2m−4 m(m − 1)(m − 2)(m − 3) − · · · = 22 2(2m − 1)(2m − 3) (m − 3)(m − 2) · · · (2m − 4)

6 m u 2m−2 1 & (2m)! = (2k − 1) u 2m − + (2m)! k=1 2(2m − 1) (2m − 2)!

7 u 2m−4 (2m)! − ··· = + 2 2 2(2m − 1)(2m − 3) (2m − 4)!

=

6 7 m 1 & m(m − 1) 2m−4 (2k − 1) u 2m − mu 2m−2 + − ··· = u (2m)! k=1 2

  m m 1 & l m (u 2 )m−l = = (2k − 1) (−1) l (2m)! k=1 l=0 m $ %m %m 1 & 1 $ 2 u −1 . (2k − 1) u 2 − 1 = m (2m)! k=1 2 m!

4.8 Rodrigues’ Formula

131

Now, by m differentiations of the first left side and last right side of the above series of relations, we eventually get the Rodrigues’ formula Pm (u) =

1 2m m!

%m dm $ 2 u −1 . m du

Exercise 4.3 Show that the function ym (u) = k

dm 2 (u − 1)m , du m

satisfies the Legendre’s equation, and that k =

1 if ym (1) = 1. 2m m!

Solution dx = k 2um(u 2 − 1)m−1 , and, multiplying Let x(u) = k(u 2 − 1)m ; it follows that du by (u 2 − 1) both sides, it results (u 2 − 1)

dx = k 2mu(u 2 − 1)m = 2mux. du

Making m + 1 differentiations of the lhs and rhs 6 7 d m+1 d m+1 dx 2 (u = 2m m+1 (ux). − 1) m+1 du du du Applying to the above expression the relation m+1 m + 1 d m+1 f (k) g (m+1−k) , ( f · g) = du m+1 k k=0

which is valid for any m + 1 times differentiable functions f (u) and g(u), we have d m+1 du m+1

6 (u 2 − 1)

dx du

7 = (u 2 − 1) = 2mu

d m+2 x d m+1 x m(m + 1) d m x 2 m = + (m + 1) 2u m+1 + m+2 2 du du du

dm x d m+1 x + 2m(m + 1) m , m+1 du du

which reduces to (u 2 − 1)

d m+2 x d m+1 x dm x + 2u − m(m + 1) = 0. du m+2 du m+1 du m

(4.76)

132

4 Potential Series Developments

Upon the position ym =

dm x du m

the Eq. 4.76 assumes the form (u 2 − 1)

d 2 ym dym + 2u − m(m + 1)ym = 0, du 2 du

and, collecting terms, finally 6 7 d2 dym 2 (u − 1) − m(m + 1)ym = 0. du 2 du The proof that k =

1 if ym (1) = 1 is left to the reader. m!

2m

Exercise 4.4 Using the Rodrigues’ formula, show that Pm (u) has m real and distinct zeroes in the (−1, 1) interval. Show also that, by application of Rolle’s theorem, it results that a. the zeroes of P2m (u) are symmetric with respect to the origin; b. the zeroes of P2m+1 (u) are symmetric with respect to the origin and P2m+1 (0) = 0. Solution Left to the reader.

4.9 Harmonic and Homogeneous Polynomials Let us remind the series development of the inverse of the Euclidean distance between two points r ≡ (x, y, z) and r ≡ (x  , y  , z  ) in space ∞ ∞ ∞ m 1 1 m m r = P (u)μ = P (u)μ ≡ Hm (x, y, z), m m |r − r | r  m=0 r  m+1 m=0 m=0

where Hm (x, y, z) is a homogeneous polynomial of degree m in x, y, z, when taking x  , y  , z  ) as parameters. We recall also that μ≡ where

r r · r x x  + yy  + zz  , u ≡ cos γ = = ,   r rr rr 

4.9 Harmonic and Homogeneous Polynomials

r=



133

x 2 + y2 + z2, r  =



x 2 + y2 + z2,

so that the function Hm = Pm (u)

rm r  m+1

,

depends upon r through u m r m , u m−2 r m , u m−4 r m , . . ., that is through r 0 , r 2 , r 4 , . . ., i.e. only through powers of r with exponents ≤ m. Consequently, Hm (x, y, z) is a homogeneous polynomial of mth degree7 being, in its turn, u a homogeneous function of degree 0 in x, y, z. 1 admits a power series development8 This implies that the quantity |r − r | ∞ 1 ≡ H (x, y, z) = ai jk x i y j z k . m |r − r | m=0 i, j,k

Because the series Eq. 4.77 converges to

(4.77)

1 in a neighborhood of the origin, |r − r |

1 is analytical 9 in every point r = r . |r − r | 1 is a harmonic function which has a developIn conclusion, the function |r − r | ment in terms of homogenous polynomials, equivalent to a development in power series convergent in a neighborhood of the origin. This means that the series (4.77) is term by term differentiable, and thus the function

∇2

∞ 1 = 0 = ∇ 2 Hm (x, y, z). |r − r | m=0

A power series cannot converge to zero in a domain containing the origin if its coefficients are not all zero, that is ∇ 2 Hm (x, y, z) = 0,

∀m.

The above conditions imply that all the Hm polynomials are harmonic: they are called “solid spherical harmonics”. Table 4.2 gives the first Hm (x, y, z), with m = 0, 1, 2. 7 All

(x x  + yy  + zz  )m−2k

r 2k . r  2m−2k+1 8 The development is of the type ai jk (x − x0 )i (y − y0 ) j (z − z 0 )k with x0 = y0 = z 0 = 0. 9 In what it has a power series development. the addends are of the type

'

134

4 Potential Series Developments

Table 4.2 Harmonic polynomials of the first 3 orders m 0 1 2

Hm (x, y, z) 1 r x x  + yy  + zz  r

2

1  2r

5

2

x (2x 2 − y 2 − z 2 ) + y  (2y 2 − z 2 − x 2 ) + z  (2z 2 − x 2 − y 2 )+ 2

+6y  z  yz + 6x  y  x y + 6x  z  x z

2



Table 4.3 All the possible spherical harmonics of first 3 orders m Spherical harmonics 0 1 2

Number of s. h.

1 x, y, z 2x 2 − y 2 − z 2 , 2y 2 − z 2 − x 2 , 2z 2 − x 2 − y 2 , yz, x y, x z

1 3 6

The spherical harmonics Hm depend upon x  , y  , z  as parameters, and the Hm are harmonics for any value of parameters, so that every single addend in the polynomial is also a spherical harmonic. This means that the spherical harmonics in the case 0 ≤ m ≤ 2 are those of Table 4.3. The 2nd order (m = 2) harmonics are not all independent; actually, any of the first three of them in Table 4.3 can be written as linear combination of the two remaining. For instance, the first one can be written as follows 2x 2 − y 2 − z 2 = a(2y 2 − z 2 − x 2 ) + b(2z 2 − x 2 − y 2 ),

with a = b = −1 ,

reducing to 5 the number of linearly independent spherical harmonics of 2nd order. This, together with that the number of independent spherical harmonics of order m = 0 and m = 1 is, respectively, 1 and 3, suggests that, in general, the number of independent spherical harmonics of arbitrary order m would be 2m + 1. This conjecture results to be true, as we now see. Actually, the generic homogeneous polynomial of degree m can be written as Fm =

i jk

 ai jk x i y j z k ,

where

i + j + k = m, i, j, k = 0, 1, ..., m,

(4.78)

1 where ai jk are (m + 1)(m + 2) constants, as we now show. The constraint i + 2 j + k = m implies that, choosing arbitrarily j and k among their m + 1 possible values (k + j ≤ m), i, which is always given by i = m − j − k, assumes the values reported in the following Table 4.4.

4.9 Harmonic and Homogeneous Polynomials

135

Table 4.4 For every value of j the table reports in the other columns the allowed values of k, the corresponding number of choices and the constrained values of i j k Number of choices i =m− j −k 0 1 2 .. .

0, 1, . . . , m 0, 1, . . . , m − 1 0, 1, . . . , m − 2 .. .

m+1 m m−1 .. .

m, m − 1, . . . , 0 m − 1, m − 2, . . . , 0 m − 2, m − 3, . . . , 0 .. .

m−1 m

0,1 0

2 1

1,0 0

From the table it is clear that the total number of choices is the sum of the values in the 3rd column, 1 + 2 + · · · + (m − 1) + m + (m + 1) that is the arithmetic progression a + (a + d) + (a + 2d) + · · · + (a + md), whose sum is m

1 (a + kd) = (m + 1)a + (m + 1)md. 2 k=0

In our case a = d = 1 so to give m 1 1 (1 + k) = (m + 1) + (m + 1)m = (m + 1)(2 + m). 2 2 k=0

When the polynomial Fm other than homogeneous is also harmonic, it should be such that   ∇ 2 Fm = 0 = ai jk i(i − 1)x i−2 y j z k + j ( j − 1)x i y j−2 z k + k(k − 1)x i y j z k−2 . i jk

∇ 2 Fm is, in its turn, a homogeneous polynomial of degree m − 2. It is so possible writing ∇ 2 Fm =



Ai jk x i y j z k = 0 ,

i + j + k = m − 2,

i jk

where the coefficients Ai jk are linear combinations of the ai jk and are such that all the m(m − 1) m(m − 1) coefficients Ai jk are zeros. This implies the existence of linear 2 2

136

4 Potential Series Developments

(m + 1)(m + 2) coefficients aik . As a conclusion, a homoge2 1 1 neous and harmonic polynomial of degree m has (m + 1)(m + 2) − (m − 1)m = 2 2 2m + 1 arbitrary coefficients. This statement corresponds to thesis of the following

relations among the

Theorem 4.6 The number of linearly independent homogeneous and harmonic polynomials of degree m is 2m + 1. Obviously, any of the independent homogeneous and harmonic polynomial can be obtained by making one of the 2m + 1 possible choices for the coefficients. The most general homogeneous and harmonic polynomial would be given by the expression Fm (x, y, z) =

2m+1

α(k) Fm(k) (x, y, z),

k=1

where the Fm(k) functions are the 2m + 1 independent homogenous and harmonic polynomials.

4.10 Surface Spherical Harmonics When considering spherical polar coordinates (r, θ, ϕ), r ≥ 0,

0 ≤ θ < 2π, 0 ≤ ϕ ≤ π,

the generic order m spherical harmonic Fm (x, y, z) is transformed into a function of spherical polar coordinates this way Fm (x, y, z) = Fm (r sin ϕ cos θ, r sin ϕ sin θ, r cos ϕ) = =

m

ai jk r i+ j+k (sin ϕ cos θ)i (r sin ϕ sin θ) j (r cos ϕ)k =

(i jk)=0

= rm

m

ai jk (sin ϕ cos θ)i (r sin ϕ sin θ) j (r cos ϕ)k =

(i jk)=0

= r Fm (sin ϕ cos θ, sin ϕ sin θ, cos ϕ) ≡ r m Sm (θ, ϕ), m

where Sm (θ, ϕ) ≡ Fm (sin ϕ cos θ, sin ϕ sin θ, cos ϕ) is a homogeneous and harmonic polynomial in the quantities sin ϕ cos θ, sin ϕ sin θ, cos ϕ, called surface spherical harmonic. It is defined on the unitary radius sphere and is also named Laplace’s spherical function of order m. As the solid spherical harmonics, also the linearly independent surface spherical harmonics of order m are in number of 2m + 1. In spherical polars, the Laplacian ∇ 2 Fm writes as

4.10 Surface Spherical Harmonics

137

    1 1 ∂ Fm ∂ Fm ∂ 2 Fm ∂ + r2 + 2 2 sin ϕ = ∂r ∂ϕ r sin ϕ ∂θ2 r 2 sin2 ϕ ∂ϕ   2 ∂ Sm 1 ∂ 2 m−1 1 1 m ∂ Sm m ∂ sin ϕ = = 2 (r mr Sm ) + 2 2 r + 2 2 r r ∂r ∂θ2 ∂ϕ r sin ϕ r sin ϕ ∂ϕ   r m−2 ∂ 2 Sm r m−2 ∂ ∂ Sm = m(m + 1)Sm r m−2 + + sin ϕ . sin ϕ ∂ϕ ∂ϕ sin2 ϕ ∂θ2

∇ 2 Fm =

1 ∂ r 2 ∂r

The above expression, valid for every r > 0, implies the following equation to be satisfied by the corresponding surface harmonic Sm (θ, ϕ) (m + 1)m Sm (θ, ϕ) +

  ∂ Sm 1 ∂ 2 Sm 1 ∂ sin ϕ = 0. + sin ϕ ∂ϕ ∂ϕ sin2 ϕ ∂θ2

(4.79)

Letting cos ϕ = t, the Eq. 4.79 can be written as (m + 1)m Sm (θ, ϕ) +

6 7 1 ∂ 2 Sm ∂ 2 ∂ Sm (1 − t = 0, + ) 1 − t 2 ∂θ2 ∂t ∂t

(4.80)

which reduces to the Legendre’s equation for Sm as function of t = cos ϕ alone (this rm is no surprise because Hm (x, y, z) =  m+1 Pm (u), so that Pm is surface harmonic r of order m and a proper choice of axis allows ϕ = 0, that gives u = cos ϕ. andPm does not depend on θ). The Eq. (4.80) is invariant upon change of m with −(m + 1), implying that to every solid harmonic r m Sm corresponds the harmonic r −(m+1) Sm .

4.11 Solution of the Differential Equation for Sm (θ, ϕ) We look for 2m + 1 particular solutions for the differential equation (4.80) by variable separation Sm( p) (θ, t) = f (θ)g(t),

p = 1, . . . , 2m + 1. p

A linear combination of the 2m + 1 solutions Sm θ, t) will then represent the most general solution Sm (θ, t) =

2m+1 p=1

Substituting Sm with f (θ)g(t) in Eq. 4.80

α p Sm( p) (θ, t) .

138

4 Potential Series Developments

6 7 1 ∂2 f ∂ 2 ∂g (1 − t ) = 0, m(m + 1) f g + g + f 1 − t 2 ∂θ2 ∂t ∂t

(4.81)

which can be written separating in θ and t 10 6 7 1 d2 f 1 − t2 d 2 dg (1 − t ) =− m(m + 1)(1 − t ) + g dt dt f dθ2 2

(4.82)

which is admissible only if 6 7 1 − t2 d 2 dg (1 − t ) = ν, m(m + 1)(1 − t ) + g dt dt 1 d2 f = ν, − f dθ2 2

(4.83) (4.84)

where ν is a real constant. Because the characteristic equation associated to (4.84) ( p) is ξ 2 = −ν, a non divergent solution for Sm requires ν ≥ 0.11 Additionally, to have ( p) a solution which is periodic of period to guarantee the univocity of Sm √ 2π, in order on the unitary sphere, it must be ν = q ∈ Z± ,12 i.e. ν = q 2 . Finally f (θ) = a cos qθ + b sin qθ,

q = 0, ±1, ±2, . . . . (q)

Once known a particular integral of (4.83), indicated by Pm (t), a particular solution of the Eq. 4.80 is given by Pm(q) (t)

f (θ) =

Pm(q)

 ×

a cos qθ + b sin qθ a cos(−qθ) + b sin(−qθ).

Then, any linear combination of the type Sm(q) = αPm(q) [a cos qθ + b sin qθ] + β Pm(q) [a cos(−qθ) + b sin(−qθ)] = = (α + β)Pm(q) a cos qθ + (α − β)Pm(q) sin(−qθ)

(4.85)

is still a solution of the equation of the surface spherical harmonics. A particular solution in terms of cosines only is obtained for α = β, while it 1 is in terms of sines only when α = −β. The further constraints α = β = and 2a 1 lead to the so called Laplace’s functions of order q α = −β = − 2b 1 − t2 . fg 11 In such case the solution writes as f (θ) = a cos √νθ + b sin √νθ. 12 cos √ν(θ + 2π) = cos √νθ cos √ν2π − sin √νθ sin = cos √νθ only if √ν is a positive or √ negative integer (similar reasoning for sin ν(θ + 2π). 10 It

suffices multiplying Eq. 4.81 by

4.11 Solution of the Differential Equation for Sm (θ, ϕ) (q)

Sm,1 = Pm(q) cos qθ ,

139

(q)

Sm,2 = Pm(q) sin qθ .

(4.86)

Clearly, the most general solution of the surface spherical harmonic equation is given by summation of terms like the two in the rightmost side of Eq. 4.85 varying q between 0 and m (so that q and −q assume the 2m + 1 integer values = 0, ±1, ±2, . . . , ±m), how it happens for a homogeneous and harmonic polynomial of degree m. Defining amp = (α + β)a and bmp = (α − β)b, it results Sm (θ, ϕ) =

m

Sm( p) (θ, ϕ) =

p=0

=

m $ % amp Pm( p) cos pθ + bmp Pm( p) sin pθ = p=0

am0 Pm(0)

+ 0 + am1 Pm(1) cos θ + bm1 Pm(1) sin θ + · · · +

+ amm Pm(m) cos mθ + bmm Pm(m) sin mθ

(4.87)

(2m + 1 terms) .

Obviously, also the single addends in the above expression are surface spherical harmonics, individually obtained by choices of the kind amk = 0, ∀k, and bmk = 0, ∀k = l, which lead to Sm( p) (θ, ϕ) = bmp Pm( p) sin pθ, called also elementary surface spherical harmonic.

( p)

4.12 The Solution in ϕ: Pm (cos ϕ) The Eq. 4.83 for the generic ν = p 2 can be written, after division by 1 − t 2 = 0, as 6 7 p2 d 2 dg (1 − t ) − g(t) = 0, m(m + 1)g(t) + dt dt 1 − t2

(4.88)

which for p = 0 reduces to m(m + 1)g(t) +

6 7 d dg (1 − t 2 ) = 0, dt dt

(4.89)

that is the Legendre’s differential equation, one particular solution of which we know to be13 Pm(0) (t) = Pm (t) = Pm (cos ϕ). 13 So

Pm is the Laplace function of order zero.

140

4 Potential Series Developments ( p)

If p = 0, the differential equation satisfied by Pm is called associated Legendre’s differential equation, whose solution in terms of Legendre’s polynomial is due to Poincarè p p d Pm( p) (t) = (t 2 − 1) 2 p Pm (t), p ≤ m; (4.90) dt and, using Rodrigues’ formula Pm( p) (t) =

p+m p d 1 2 2 (t − 1) (t 2 − 1)m . 2m m! dt p+m

(4.91)

( p)

The Pm (t) functions are called associated Legendre’s function of the first kind. Let us verify that the expression (4.91) actually satisfies the Eq. (4.88). By introducing the auxiliary function z(t) defined as p

g(t) = (t 2 − 1) 2 z(t) the Eq. (4.88) becomes (1 − t 2 )z  − 2t ( p + 1)z  + (m 2 + m − p 2 − p)z = 0

(4.92)

where the apex stays for derivative respect to t. On the other side, a p-times differentiation of the Legendre’s equation gives m(m + 1)

 d p+1  d p Pm (1 − t 2 )Pm = 0. + p p+1 dt dt

(4.93)

By using the following development14  p+1  d p+1 p + 1 dk d p+1−k 2  (1 − t )P = (1 − t 2 ) p+1−k Pm = m p+1 k k dt dt dt k=0

=

 p+1  p+1 k=0

k

[ p+2]

= (1 − t 2 )Pm

[ p+2]

(1 − t 2 )[k] Pm

= [ p+1]

+ ( p + 1)(−2t)Pm

+

( p + 1)! [ p] (−2)Pm , 2!( p − 1!)

and making a substitution in Eq. (4.93), we arrive to m(m + 1)Pm[ p] + (1 − t 2 )Pm[ p+2] − 2t ( p + 1)Pm[ p+1] − p( p + 1)Pm[ p] = = (1 − t 2 )Pm[ p+2] − 2t ( p + 1)Pm[ p+1] + (m 2 + m − p 2 − p)Pm[ p] = 0,

14 The

symbol [ p] stands for pth derivative with respect to t.

(4.94)

( p)

4.12 The Solution in ϕ: Pm (cos ϕ)

141 [ p]

where the last equality corresponds to satisfying Eq. 4.92 with z = Pm , so that the solution g(t) of Eq. (4.88) is indeed p

p

g(t) = Pm( p) (t) = (t 2 − 1) 2 Pm[ p] (t) ≡ (t 2 − 1) 2

dp Pm (t) . dt p

( p)

It is also possible to verify that the set of functions Pm (t) above defined is an orthogonal system in (−1, 1), giving (Pm( p) ,

Pn( p) )

1 =

Pm( p) Pn( p) dt = (−1) p

−1

(m + p)! 2 δmn , (m − p)! 2m − 1

∀ p ≥ 0, m = p, p + 1, . . . ; n = p, p + 1, . . .

(4.95)

4.13 Note on the Associated Legendre’s Equation The associated Legendre’s equation with m and p non negative integers has two families of solutions, so that its general solution is ( p) , g(t) = c1 Pm( p) + c2 Q m ( p)

where Q m are called associated Legendre’s functions of the second kind p

( p) Qm = (−1) p (t 2 − 1) 3

d p Qm . dt p

where Q m (t) are the Legendre’s function of the second kind, i.e. the other family of solutions, other than Pm (t), of the Legendre’s equation. Being t real and |t| ≤ 1, the function in Eq. 4.90 p

Pm( p) (t) = (t 2 − 1) 2

dp Pm (t), dt p

is real only for even values of p, so that it is often advantageous replace it with ( p)

p

P m (t) = (1 − t 2 ) 2

dp Pm (t), dt p

and/or

(4.96)

dp Tm( p) (t) = (−1) p (1 − t 2 ) Pm (t), dt p p 2

142

4 Potential Series Developments

these latter real functions being called Ferrer function of the first kind (we will use ( p) them in the following indicating them as Pm ). Exercise 4.5 Determine the conditions that the function f : R3 −→ R supposed dependent on x, y, z only by linear combinations of x, y, z (i.e. f (ax + by + cz)) must satisfy to be harmonic (a, b, c are real constants). Solution Let t = ax + by + cz, it is ∇2 f =

d2 f d2 f d2 f ∂2 f ∂2 f ∂2 f d2 f + + = a 2 2 + b2 2 + c2 2 = (a 2 + b2 + c2 ) 2 , 2 2 2 ∂x ∂y ∂z dt dt dt dt

d2 f which is zero either by the trivial condition a 2 + b2 + c2 = 0 or 2 = 0, this latter dt meaning that f is linear in t ≡ ax + by + cz f (x, y, z) = αt + β = α(ax + by + cz) + β . Exercise 4.6 Given a harmonic function F(x, y, z), determine under what condiF is also harmonic. tions r Solution The condition is ∇ 2 Fr = 0, where ∇2

  F 1 F F =∇ ·∇ =∇ · ∇ F − 2 ∇r = r r r r   F 1 F 1 = ∇ 2 F + ∇ F · ∇ − 2 ∇ 2 r + ∇r · ∇ − 2 = r r r r 1 F2 F = 0 − 2 ∇ F · er − 2 − er ∇ 2 = r r r  r  2 2 1 1 ∂F − 3 F − er · − 3 Fer + 2 ∇ F = =− 2 r ∂r r r r 2 ∂F . =− 2 r ∂r

4.14 Zonal, Tesseral and Sectorial Spherical Harmonics The elementary surface spherical harmonics of order m are am0 Pm (t), for p = 0 (1 harmonic), and ( p) Pm (t)(amp cos pθ + bmp sin pθ), for 1 ≤ p ≤ m (2m harmonics).

4.14 Zonal, Tesseral and Sectorial Spherical Harmonics

143

Fig. 4.5 Unitary sphere zone subdivision for P5 (ϕ). Symbol + or − identify the region of positivity or negativity of the polynomial P5

We can distinguish three cases which characterize the various locations on the unitary sphere of the zeros of these harmonics (i) p = 0: zonal harmonics, (ii) 1 ≤ p ≤ m: tesseral harmonics, (iii) p = m: sectorial harmonics. (i) case p = 0: zonal harmonics If p = 0, then the surface spherical harmonic Pm(0) (t) is the Legendre’s polynomial in the variable t = cos ϕ, which has m distinct roots, symmetric respect to π π t = cos ϕ = 0 ⇐⇒ ϕ = (if m is odd also t = 0, i.e. ϕ = is a root of the poly2 2 nomial) (Fig. 4.5). The roots lays, so, on parallel circles dividing the unitary sphere in zones, whence the name of zonal harmonics. We remind that at the poles (ϕ = 0, ϕ = π) Pm (1) = 1 and Pm (−1) = (−1)m . (ii) case 1 ≤ p ≤ m: tesseral harmonics If 1 ≤ p ≤ m, the harmonic takes the form 6 p (−1) (t 2 − 1) 2 p

7 1 d m+ p 2 m (t − 1) (amp cos pθ + bmp sin pθ), 2m m! dt m+ p

whose roots are those of 1 d m+ p 2 p (a) (t 2 − 1) 2 m (t − 1)m = 0, and/or 2 m! dt m+ p (b) amp cos pθ + bmp sin pθ = 0. Case (a) gives 1 − t 2 = 0 ⇐⇒ cos ϕ = ±1 ⇐⇒ ϕ = 0, π and/or

(north and south poles)

dp 1 d m+ p 2 m (t − 1) = Pm (t) = 0, 2m m! dt m+ p dt p

this latter equation (as it could be demonstrated) has m − p real roots in the interval −1 < t < 1 symmetrically distributed as for the zonal harmonics.

144

4 Potential Series Developments

Fig. 4.6 Sign distribution of a spherical harmonic ( p) Sm (θ, ϕ)

Case (b) gives tan pθ = −

amp bmp

that is     amp amp k 1 + kπ ⇐⇒ θk = arctan − + π. pθk = arctan − bmp p bmp p The above values of θk define on the unitary sphere p great circles spaced by Δθ = π θk − θk−1 = . The 2m − 2 zeros lie at the poles and on both parallel and meridian p circles which divide the sphere surface on a set of tesseras of alternating sign for the harmonic, whence the name of tesseral harmonics (Fig. 4.6). ( p) The sign of Sm (θ, ϕ) changes crossing a parallel and a meridian. (iii) case p = m: sectorial harmonics If p = m the generic harmonic is 6 m (−1) (t 2 − 1) 2 m

7 1 d 2m 2 m (t − 1) (amm cos mθ + bmm sin mθ). 2m m! dt 2m

It is easily seen that d 2m 2 (t − 1)m = (2m)! dt 2m and so the zeros of the Pm(m) (θ, ϕ) harmonic are obtained by m

(t 2 − 1) 2 = 0, and/or amm cos mθ + bmm sin mθ = 0,

4.14 Zonal, Tesseral and Sectorial Spherical Harmonics

145

Fig. 4.7 Sign subdivision of a generic sectorial harmonic Sm(m) (θ, ϕ)

+

that is

−

+

−

+

⎧ 2 poles ⎨ t = ±1 ⇐⇒ϕ = 0,π, k amm + π, m great circles. ⎩ θk = arctan − bmm m

The zeros location on the unitary sphere along meridians and the subsequent subdivision in spherical sectors justifies the name of sectorial harmonics (Fig. 4.7). Exercise 4.7 Write the explicit expression of P2(1) (t), P3(2) (t), P2(3) (t). Solution Being p

Pm( p) (t) = (−1) p (1 − t 2 ) 2

dp Pm (t), dt p

it results 6 7  3 2 1 d 1 d P2 (t) = −(1 − t 2 ) 2 (t − ) = −3t 1 − t 2 dt dt 2 3 7 2 2 6 1 3 d d (5t − 3t) = 15t (1 − t 2 ) P3(2) (t) = (1 − t 2 ) 2 P3 (t) = (1 − t 2 ) 2 dt dt 2 7 3 3 6 3 2 1 (3) 2 23 d 2 23 d P2 = −(1 − t ) (t − ) = 0. P2 (t) = −(1 − t ) dt 3 dt 3 2 3 P2(1) (t) = −(1 − t 2 ) 2

1

Exercise 4.8 Verify that P3(2) (t) given in the Exercise 4.7 satisfies its associated Legendre’s equation. Solution Given the Legendre’s equation m(m + 1)g +

 p2 d  (1 − t 2 )g  − g = 0, dt 1 − t2

146

4 Potential Series Developments

letting p = 2 and m = 3, it becomes 12g +

 4 d  (1 − t 2 )g  − g = 0. dt 1 − t2

Taking g(t) = P3(2) (t) and substituting into the above equation  d  4 (1 − t 2 )(15 − 45t 2 ) − 15t (1 − t 2 ) = dt 1 − t2   = 15t 12(1 − t 2 ) + 12t 2 − 12 = 0,

12 × 15t (1 − t 2 ) +

as required. Exercise 4.9 Write the associated Legendre’s equation in the form f (g, g  , g  ; t)=0. Solution The associated Legendre’s equation writes as m(m + 1)g − 2tg  + (1 − t 2 )g  −

p2 g = 0, 1 − t2

or, equivalently 7 6 p2 g = 0. (1 − t 2 )g  − 2tg  + m(m + 1) − 1 − t2    f (g,g  ,g  ;t)

p p d Clearly, if g = (t 2 − 1) 2 p Pm is solution of the above equation, also g = cg where dt c is a constant, is a solution. Consequently, to any imaginary solution of the Legendre’s equation corresponds p a real one obtained by simple multiplication by (−1) 2 .

Exercise 4.10 Show that for 0 ≤ ϕ ≤ π 6 7 1 (3) 2! (3) 4! (3) P3 (t) + 11 × P5 (t) + 15 × P7 (t) + · · · cos 3θ. − sin ϕ cos 3θ = 8 7 × 6! 8! 10!

Solution Left to the reader.

4.15 Orthogonality of Surface Spherical Harmonics

147

4.15 Orthogonality of Surface Spherical Harmonics Theorem 4.7 The set of elementary harmonics {Ymp (θ, ϕ)} = {Pm( p) (ϕ) cos pθ, Pm( p) (ϕ) sin pθ},

p = 0, 1, . . . , m,

is a system of orthogonal functions on the unitary sphere. p

p

Proof Compute (Ym , Yn ), treating the various possible cases.



(a)

( p) ( p) Pm cos pθ Pn sin pθ dσ =

S1 (0)

2π

π ( p) ( p) cos pθ sin pθ dθ Pm Pn sin ϕ dϕ = 0, 0

because

0

2π cos pθ sin pθ dθ = 0. 0



Pm( p)

(b)

cos

pθ Pn( p)

2π cos pθ dσ =

S1 (0)

π 2

cos pθ dθ 0

1 = π(1 + δ0 p )

Pm( p) Pn( p) sin ϕ dϕ =

0

Pm( p) Pn( p) dt = π(1 + δ0 p )(Pm( p) , Pn( p) ),

−1

where we made use of that ⎧

2π if p = 0, ⎨ 2π, 7 6 2 pθ θ=2π cos pθ dθ = 1 sin pθ cos pθ ⎩p + = π, if p = 0. 2 2 θ=0 0

(c)

S1 (0)

Pm( p)

sin

pθ Pn( p)

2π sin pθ dσ =

1 2

sin pθ dθ

Pm( p) Pn( p) dt =

−1

0

= π(1 − δ0 p )(Pm( p) , Pn( p) ) (for p = 0 the corresponding harmonic is identically null). It remains to compute (Pm( p) ,

Pn( p) )

1 = −1

Pm( p) (t)Pn( p) (t) dt

1 = (−1)

2p −1

(1 − t 2 ) p Pm[ p] Pn[ p] dt,

148

4 Potential Series Developments

which, integrating by parts15 gives the following (Pm( p) ,

Pn( p) )

% 1 $ = 1 − t 2 Pm[ p] Pn[ p−1] −1 −

1 −1

Pn[ p−1]

 d  (1 − t 2 ) p Pm[ p] dt. dt (4.97)

By mean of Eq. 4.92 we have, after multiplication by (1 − t 2 ) p  d  (1 − t 2 ) p+1 Pm[ p+1] = ( p 2 + p − m 2 − m)(1 − t 2 ) p Pm[ p] = dt = −(m − p)( p + m + 1)(1 − t 2 ) p Pm[ p] , that, by substitution in the above integral (provided the change p + 1 −→ p) $

% Pm( p) , Pn( p) = (m − p + 1)(m + p)

1

Pm[ p−1] Pm[ p−1] (1 − t 2 ) p−1 dt =

−1

% $ = (m − p + 1)(m + p) Pm( p−1) Pn( p−1) =

% $ = (m − p + 1)(m + p)(m − p + 1)( p − 1 + m) Pm( p−2) Pn( p−2) = (m + p)! 2 = ··· = δmn . (m − p)! 2m + 1 In conclusion, we have shown that $

% 2 (m + p)! Ym( p) , Yn( p) = (1 ± δ0 p )π δmn , (m − p)! 2m + 1

that proves the theorem.

0 1 ( p) Thanks to the above result, the set of Ym for any p = 0, 1, . . . , m results to be a system of orthogonal functions for 0 ≤ θ < 2π and 0 ≤ ϕ ≤ π, that can be used as basis functions for development in series of functions satisfying the Dirichlet’s conditions. By the result of the theorem it is easily shown (left to the reader as exercise) that, if Sm (θ, ϕ) and S n (θ, ϕ) are surface spherical harmonics, then (Sm , S n ) = 0,

if m = n, 4

5 m 2π (m + k)! (amk a mk + bmk bmk ) (Sm , S m ) = 2am0 a m0 + , 2m + 1 (m − k)! k=1 % p [ p] $ [ p] u = 1 − t 2 Pm and dv = Pm dt; remember that the symbol [ p] here means pthderivative respect to t.

15 With

4.15 Orthogonality of Surface Spherical Harmonics

where Sm (θ, ϕ) =

149

n (anp cos( pθ) + bnp sin( pθ)) Pm( p) , p=0

n S n (θ, ϕ) = (a np cos( pθ) + bnp sin( pθ)) Pn( p) . p=0

In particular, if S m = Pm (cos ϕ) (as from the choices a m0 = 1 and a mp = bmp , ∀ p > 0), it results (Sm , Pm ) =

4π 4π am0 = Sm (0, 0), 2m + 1 2m + 1

due to that Sm (0, 0) = am0 Pm (1) +

m

(4.98)

amp Pm( p) (1) = am0 ,

p=1 ( p)

being Pm (1) = 1 ∀m, and Pm (1) = 0, ∀m with p ≤ m.

4.16 Developments in Terms of Spherical Harmonics Given that the set of Ym (θ, ϕ) constitutes a basis, a sufficiently regular function f (θ, ϕ) can be developed as f (θ, ϕ) =

∞ m

(amp cos pθ + bmp sin pθ)Pm( p) (cos ϕ),

(4.99)

m=0 p=0

whose coefficients amp (for p = 0, 1, 2, . . . , m) and bmp (for p = 1, 2, . . . , m), are 88 amp = 88

S1 (0)

( p)

f Pm cos pθ dσ

( p) 2 S1 (0) (Pm cos pθ) dσ 88 ( p) S1 (0) f Pm sin pθ dσ

bmp = 88

( p) S1 (0) (Pm

sin

pθ)2

dσ

= =

2m + 1 (m − p)! 2π(1 + δ p0 ) (m + p)! 2m + 1 (m − p)! 2π (m + p)!





S1 (0)

S1 (0)

f Pm( p) cos pθ dσ,

f Pm( p) sin pθ dσ,

(4.100) ( p) ( p) as obtained by multiplication of both sides of Eq. 4.99 by Pm cos pθ and Pm sin pθ and integration over the unitary sphere, keeping in mind what obtained in the preceding Sect. 4.15. Note that the coefficient bm0 is not defined because it does not contribute to the development Eq. 4.99 (it multiplies sin 0 = 0).

150

4 Potential Series Developments

Note The unique determination of coefficients of the series means that if the above series development of f (θ, ϕ) exists it is unique.

4.17 Development of the Potential Generated by a Regular Mass Distribution Remembering that ∞ ∞ 1 rm 1 m P (cos γ)μ = P (cos γ) = m m |r − r | r  m=0 r  m+1 m=0

r < 1, it is clear that if we divide the mass domain (where r r r ρ(r) ≥ 0) in the two parts (a)  < 1, and (b)  > 1, the following two developments r r hold, respectively

is convergent for μ =

∞

(a)

1 rm r = P (cos γ) , convergent for  < 1, m m+1   |r − r | m=0 r r

(b)

1 r m r = P (cos γ) , convergent for < 1. m  m+1 |r − r | m=0 r r

∞

Example 4.17.1 Let C = {r : ρ(r) ≥ 0} be the domain of mass M and consider a generic point r ∈ C. Let also consider the sphere of radius r  centered at the origin of the reference system. In the Fig. 4.8 (a) represents the open ball Sr  (0), (b) represents C \ S r  (0), where S r  (0) indicates the closed ball of radius r  and center in the origin. The two series developments above can be considered as internal and external to S r  (0), i.e. due to the contribution to the Newtonian potential U (r) given by the matter inside and out the ball. The potential in the generic point r = (x  , y  , z  ), in its integral form is



U (r ) = G C

+

∞

6 ∞

1 dm = G |r − r | m=0

Sr (0)



m=0 3 R \Sr  (0)

Pm (cos γ)

Pm (cos γ)

rm r  m+1

ρ(r) d 3 r+

7 r 3 ρ(r) d r = Uint (r ) + Uext (r ), r m+1 m

4.17 Development of the Potential Generated by a Regular Mass Distribution

151

z

Fig. 4.8 Sketch of the C matter domain, related to the potential expansion

(b) (a)

r’

y

x

and can be obtained by considering that the integration term by term is possible because the two series are uniformly convergent being bounded by the geometric series of argument μ < 1.16 The resulting development is U (r ) =

∞

Uint,m

m=0

1 r  m+1

+

∞

Uext,m r  = m

m=0

1 1 1 = Uint,0  + Uint,1  2 + · · · + Uint,k  k+1 + · · · + Uext,0 + Uext,1 r  + · · · r r r k + Uext,k r  + · · · (4.101) where the Uint,m and Uext,m coefficients depend also on the direction of vector radius r , and are given by



Uext,m

Uint,m = G



=G

R3 \Sr (0)

Sr (0)

Pm (cos γ)r m ρ(r) d 3 r,

Pm (cos γ)

1 ρ(r) d 3 r. r m+1

(4.102) (4.103)

Obviously, to the evaluation of the potential in points out of the minimum sphere containing C, whose radius we indicate with R (see Fig. 4.9), the contribution comes only from the internal terms . This means that an evaluation of the potential in the space between the surface delimiting C and the surface of the ball of radius R requires a direct evaluation of the potential integral, because here the series development is not valid.

16 Note

that |Pm (cos γ)| ≤ 1.

152

4 Potential Series Developments

Fig. 4.9 The dashed curve represents the minimum circle containing the mass domain C

R

4.17.1 External Potential Series Development We now deal with the potential series development in the external convergence region, r  > R, that is ∞



U (r ) = Uint (r ) = G

S R (0)

m=0

=

∞





Um (θ , ϕ )

m=0

Pm (cos γ)

1 rm+1

rm r  m+1

ρ(r) d 3 r = (4.104)

,

where we shortened Uint,m in Um . The problem is, now, the evaluation of the integrals Um (θ , ϕ ) = G



Pm (cos γ)r m ρ(r) d 3 r,

(4.105)

S R(0)

which are called moments of the matter distribution (they are indeed the average of the generic mth power of r weighted by the function G Pm (cos γ) ρ(r)). In spherical polar coordinates it is Um (θ , ϕ ) = G



Pm (cos γ)r m+2 ρ(r) sin ϕ dr dθdϕ,

S R (o)

with cos γ = cos ϕ cos ϕ + sin ϕ sin ϕ cos(θ − θ ). Let’s compute the two first coefficients, U0 e U1 . • If m = 0, then

ρ(r) d 3 r = G M, P0 (cos γ) = 1 =⇒ U0 = G C

(4.106)

4.17 Development of the Potential Generated by a Regular Mass Distribution

whence

153

1 GM =  ,  r r

U0

which is the monopole term. It corresponds to the lowest order approximation of the potential as due by a point of mass M in the coordinate origin, approximation which is more and more valid at larger distances from the origin because the contribution m+1 of higher moments reduces as 1/r  at increasing distances whilst the coefficients Um are bound. • If m = 1, then

P1 (cos γ) = cos γ

=⇒ U1 = G =G

cos γ r ρ(r) d r = G 3

C 

r · r

r · r r ρ(r) d 3 r = rr 

C

r dm =

G Mr · rG , r

C

8 r dm where rG = 8C gives the center of mass position of the matter domain C. C dm  The term U1 /r in the series is the dipole term. It vanishes upon the choice of the coordinate origin in the mass distribution center of mass (i.e. letting rG = 0). With other coordinate choices, the dipole term is equal to the component of the vector G Mr along rG . The term with m = 2 is called quadrupole term, and so on. Try now to express the generic Um (θ , ϕ ). To do this we keep in mind that the “Laplace’s coefficient”, Pm (cos γ), is a surface spherical harmonic in θ and ϕ (thinking of the pair (θ,ϕ) variable upon the unitary sphere and of (θ ,ϕ ) as fixed17 ) so that it can be written on the harmonic basis Pm (cos γ) = am0 Pm (cos ϕ) +

m

  Pm( p) (cos ϕ) amp cos( pθ) + bmp sin( pθ) ,

p=1

(4.107) where coefficients amp and bmp depend upon the angular coordinates (θ , ϕ ) of r on the unitary sphere. The method to find amp and bmp is the same as for the generic function f (θ, ϕ) development (see Eq. 4.99) leading to am0 amp bmp 17 Or

2m + 1 = 4π 2m + 1 = 2π 2m + 1 = 2π

viceversa.

Pm (cos γ)Pm (cos ϕ) dσ,

(m − p)! Pm (cos γ)Pm( p) (cos ϕ) cos pθ dσ ( p = 1, 2, . . . , m), (m + p)!

(m − p)! Pm (cos γ)Pm( p) (cos ϕ) sin pθ dσ ( p = 1, 2, . . . , m). (m + p)!

154

4 Potential Series Developments

The practical computation of the above integrals takes advantage by a proper change of reference frame, centering at (θ , ϕ ), letting θ = θ − θ and ϕ = ϕ − ϕ . Upon this transformation Pm (cos γ) = Pm (cos ϕ) so that by relation Eq. 4.98 it results.

2m + 1 (X m , Ym ) = am0 = Pm (cos γ)Pm (cos ϕ) dσ = Pm (cos ϕ ), 4π (m − p)! ( p) P (cos ϕ ) cos pθ , amp = 2 (m + p)! m (m − p)! ( p) bmp = 2 P (cos ϕ ) sin pθ . (m + p)! m Due to the above result the series coefficients are simply

2m + 1 Pm (cos γ)Pm (cos ϕ) dσ = Pm (cos ϕ ), 4π (m − p)! ( p) P (cos ϕ ) cos pθ =2 ( p = 1, 2, . . . , m), (m + p)! m (m − p)! ( p) P (cos ϕ ) sin pθ =2 ( p = 1, 2, . . . , m). (m + p)! m

am0 = amp bmp

Corollary 4.8 Given the above expressions of the series coefficients, the explicit development of Pm (cos γ) is Pm (cos γ) = Pm (cos ϕ )Pm (cos ϕ)+ +2

m (m − p)! ( p) Pm (cos ϕ)Pm( p) (cos ϕ )(cos pθ cos pθ + sin pθ sin pθ ), (m + p)! p=1

which is equivalent, applying a trigonometric identity, to Pm (cos γ) = Pm (cos ϕ )Pm (cos ϕ)+ +2

m (m − p)! ( p) Pm (cos ϕ)Pm( p) (cos ϕ ) cos p(θ − θ ). (m + p)! p=1

The above is called Addition Theorem. Note Clearly, (θ, ϕ) and (θ , ϕ ) are interchangeable in Pm (cos γ). We can now calculate the generic Um (θ , ϕ ) in the potential development. Actually, it is

4.17 Development of the Potential Generated by a Regular Mass Distribution

6

Um = G +2





Pm (cos γ)r ρ(r) d r = G Pm (cos ϕ )

C m p=1

m

3

(m − p)! ( p) P (cos ϕ ) (m + p)! m

C

155

Pm (cos ϕ) r m dm+ C

7 Pm( p) (cos ϕ) cos p(θ − θ )r m dm . (4.108)

Let us consider, first, the first addend in the square bracket in Eq. 4.108 Pm (cos ϕ )



Pm (cos ϕ) r m dm = M R m Pm (cos ϕ )

C

Pm (cos ϕ) C

 r m dm ≡ R M

≡ −M R m Pm (cos ϕ )Jm , having defined the adimensional quantity

Jm ≡ −

Pm (cos ϕ) C

 r m dm , R M

which is, indeed, the opposite of the (adimensional) mth moment of r weighted with ρ(r)Pm (cos ϕ). Finally, the corresponding generic mth contribution to the external potential is so GM −  r

 m R Jm Pm (cos ϕ ). r

Let us now consider the sum over p in the square brackets of Eq. 4.108, whose generic term is18

(m − p)! ( p) Pm (cos ϕ ) Pm( p) (cos ϕ) cos p(θ − θ )r m dm = (m + p)! C 6

 r m dm (m − p)! m ( p)   Pm (cos ϕ ) cos pθ Pm( p) (cos ϕ) cos pθ + = 2M R (m + p)! R M C

 r m dm 7 + sin pθ Pm( p) (cos ϕ) sin pθ = R M C % $ = M R m Pm( p) (cos ϕ ) cos pθ Jmp + sin pθ J mp , 2

where Jmp J mp

(m − =2 (m + (m − =2 (m +

 r m dm p)! , Pm( p) (cos ϕ) cos pθ p)! C R M

 r m dm p)! , Pm( p) (cos ϕ) sin pθ p)! C R M

are adimensional quantities. 18 Using

cos(θ − θ ) = cos pθ cos pθ + sin pθ sin pθ .

(4.109)

156

4 Potential Series Developments

Finally, the generic term in the external potential series expansion is GM 1 =  Um  r m+1 r ⎡

 m R × r

× ⎣−Jm Pm (cos ϕ) +

m

⎤

(Jmp cos pθ + J mp sin pθ )Pm( p) (cos ϕ )⎦ .

p=1

The potential results to be a series where a radial, power decaying, part is modulated by a (θ , ϕ ) dependence through surface spherical harmonics with coefficients depending on the distribution of matter density. As expected, if we consider only the terms with m = 0, 1 and choosing the frame origin in the barycenter, O ≡ G, the development reduces to U (r  ) =

GM , r

because J0 = −1, J1 = J11 = J 11 = 0. Exercise 4.11 Determine the expression of the gravitational potential generated by a massive homogeneous ring  R of mass M = 2πλ0 R, where λ0 is the (constant) ring linear mass density and R its radius. Solution (i) If r < R then U (x, y, z) = (ii) if r > R then U (x, y, z) =

∞ m=0 ∞ m=0

Uext,m r m ; Uint,m

1 r m+1

.

Case (i), r < R. By virtue of the axial symmetry of the problem, once assumed ϕ = π/2 on the ring plane,

where

'

Pm (cos γ)

1



1 Pm (cos ϕ) m+1 dm = ' m+1 r r R R  m+1

GM GM dm R = − m+1 Pm J˜m , = m+1 Pm (cos ϕ) ' Pm (cos ϕ) R r M R R

Uext,m = G

dm = G Pm (cos ϕ)

4.17 Development of the Potential Generated by a Regular Mass Distribution

J˜m = −

=

'

⎧ R −1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎨

157

 m+1 dm R = −Pm (0) = Pm (cos ϕ) r M if m = 0, if m is odd, m

2 ⎪ & ⎪ ⎪ ⎪ (2k − 1) ⎪ ⎪ ⎪ ⎪ m k=1 ⎪ ⎪ % if m is even. m $ ⎩ −(−1) 2 2 2 m2 !

It turns out that ∞  r m  GM  ˜ − Jm Pm (cos ϕ) = R m=0 R ⎡ ⎤ m 2 & ⎢ ⎥ (2k − 1) ∞ ⎢  r m ⎥ m k=1 ⎢ ⎥ = 2πG ⎢1 + (−1) 2 Pm ⎥= m $m % ⎢ ⎥ 2 R 2 ! 2 m=2 ⎣ ⎦ m even

U (x, y, z) =

7 6  r 4 1·3 1  r 2 = 2πGρ0 1 − P2 + 2 P4 − ··· . 2 R 2 ·2 R (ii), r > R

4  m 5 ∞ GM R U (x, y, z) = Jm Pm (cos ϕ) ; 1− r r m=2

upon the choice of barycentral reference system (J1 = 0). The zonal coefficients are

Jm = −

=

so that

Pm (cos ϕ)

'

⎧ −1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎨

R

 r m dm = −Pm (0) = R M if m = 0, if m is odd,

m ⎪ 2 & ⎪ ⎪ ⎪ ⎪ (2k − 1) ⎪ ⎪ ⎪ m k=1 ⎪ ⎩ −(−1) 2 if m even, m m 2 2 ( 2 )!

158

4 Potential Series Developments

5 4  2  4 GM R 1·3 R 1 U (x, y, z) = + 2 P4 (cos ϕ) − ··· = 1 − P2 (cos ϕ) r 2 r 2 ·2 r 5 4  3  5 R R R 1·3 1 = 2πGρ0 + 2 P4 (cos ϕ) − ··· . − P2 (cos ϕ) r 2 r 2 ·2 r Exercise 4.12 Evaluate the gravitational force exerted by the homogeneous ring considered in the previous exercise in points at distances r > R from the ring center and determine the number of terms in the series development of the potential required to have an error in the radial component of the force not greater than 10−2 times the lowest order (monopole) contribution. Solution In axisymmetry, U is independent of θ, and the corresponding force is F = ∇U =

∂U 1 ∂U er + eϕ , ∂r r ∂ϕ

where ⎡ ⎤  m  m ∞ ∞ GM ⎢ ∂U R R ⎥ = − 2 ⎣1 − Fr = Jm Pm (cos ϕ) − Jm Pm (cos ϕ)m ⎦= ∂r r r r m=2 m even

⎡

=−

⎤

m=2 m even

 m ∞ GM ⎢ R ⎥ 1 − (m + 1)J P (cos ϕ) ⎣ ⎦. m m r2 r m=2 m even

The summation over even values of m ≥ 2 in the above equation corresponds to l 

(2k − 1)

l k=1

m = 2l, l ≥ 1 =⇒ (m + 1)Jm = −(2l + 1)(−1)

2l l!

where the fraction numerator is the product of the first l odd positive integers. It turns out that the radial force expansion is an alternating series, whose first three terms are 5 4  2  4 GM R 3 R 1 Fr = − 2 1 − P2 (cos ϕ) + P4 (cos ϕ) − ··· . r 2 r 8 r The ϕ component of the force is  m ∞ GM 1 ∂U R  Fϕ = = 2 Jm Pm (cos ϕ) sin ϕ , r ∂ϕ r m=2 r

4.17 Development of the Potential Generated by a Regular Mass Distribution

159

where the apex means derivative with respect to t = cos ϕ. The series above is, also, an alternating series GM Fϕ = 2 r

4

1  P (cos ϕ) sin ϕ 2 2

5  2  4 R 1·3  R − 2 P4 (cos ϕ) sin ϕ + ··· . r 2 ·2 r

Being the series alternated in sign, when considering up to the mth term, the error bound in the force radial component is || ≤ (m + 2)|Jm | =

GM (m + 3)|Jm+2 Pm+2 | r2

 m+2 R , r

so that the condition required by the exercise is satisfied when the error in the radial component of the force relative to the monopole is |r | ≡

|| (m + 1)(m − 1) · · · 3 · 1 ≤ (m + 3)|Jm+2 | = (m + 3) ≤ 10−2 , m GM 2 2 +1 ( m2 + 1)! r2

that corresponds to an inequality for m to be solved numerically. Exercise 4.13 Given a spherical massive shell of uniform surface mass density σ0 and generic radius R, it is asked to show that the gravitational force vanishes inside the shell. Solution The potential series development in points internal to the shell (r < R) is U (x, y, z) = where

∞  r m  GM  ˜ − Jm Pm (cos ϕ) , R m=0 R

 m+1 R σ0 r 2 sin ϕ dθdϕ Pm (cos ϕ) = r M R 

1 σ0 R 2 2π 1, if m = 0, Pm (t) dt = = 0, ∀m = 0, M

J˜m = −



!

−1

so that U (x, y, z) = U (r ) =

GM = constant R

=⇒

F = ∇U = 0, ∀r < R.

Note that this is an alternative demonstration of the 1st Newton’s theorem.

160

4 Potential Series Developments

Exercise 4.14 Show that any radial perturbation δU (r ) to a Keplerian potential leads to a precession of the motion of the unit mass particle. Find the precession angle in α the case δU (r ) = 2 , which corresponds to a dipole perturbation. r Solution Given a spherical potential U0 (r ) a perturbation leads, in general, to U (r, θ, ϕ) = U0 (r ) + δU (r, θ, ϕ) =

k + δU (r, θ, ϕ). r

where the last equality holds in the assumption of an unperturbed Keplerian potential, k U0 (r ) ≡ with k > 0. Of course, a purely radial perturbation is such that δU depends r only on r . First, note that the condition for the perturbation to be small is 9 9 9 δU 9 9 9 9 U 9  1, 0 that defines the region of space where linear considerations are valid. For example, α assuming δU (r ) = β with α a non-zero constant and β > 0, the previous inequality r |α| leads to r β−1  . In the case of a dipole perturbation, β is equal to 2 and the k |α| . perturbation is small only in regions sufficiently out of the sphere of radius R = k Let us evaluate the precession angle Δθ, defined in Chap. 3, Sect. 3.7, as

r+ Δθ = 2L r−

1 |dr |   r2 2 E +U −

L2 2r 2

.

To transform the integrand from an irrational to a rational function, a formal derivation and an indefinite integration over L is performed, to give

Δθ =

∂ ∂L



⎤ ⎡ :

r+

r+  ⎥ ⎢ ∂ 1 dr L2  ⎥ = −2 dr. 2 E + U − d L2L ⎢    ⎦ ⎣ r2 ∂L 2r 2 2 2 E +U − L2 r− r− 2r

 Letting f (U ) ≡ U0 is19

19 The

 L2  2 E + U − 2 , its development up to 1st order in U around 2r

apex,  , stands for derivative respect to U0 .

4.17 Development of the Potential Generated by a Regular Mass Distribution

 

f (U ) = f (U0 ) + f (U0 )δU =

161

 δU k L2  2 E+ − 2 + ,  r 2r L2  k 2 E+ − 2 r 2r

by which

r+

r+ ∂ δU ∂ Δθ = −2 f (U0 ) dr − 2 dr .   ∂L ∂L L2  k r− r− 2 E+ − 2    2r  r   K eplerian=2π

(4.110)

δθ

The first integral in Eq. 4.110 is simply the Keplerian θ variation, i.e. 2π (the trajectory is fixed); it remains to compute the second integral, giving the variation δθ respect to the Keplerian value, which requires, again, a Keplerian approximation for r˙ that, using the angular momentum conservation, gives (for r− ≤ r ≤ r+ ) dr dr r2 = = dt = dθ.  r˙ L k L2  2 E+ − 2 r 2r so that ∂ δθ = −2 ∂L

θ+ θ−

r2 δU dθ. L

An explicit computation of δθ would require the knowledge of both θ− ≡ θ(r− ) and θ+ ≡ θ(r+ ), thing that implicitly requires the knowledge of the solution of the motion α trajectory problem. The dipole perturbation, δU (r ) = 2 , yields to r   ∂ 1  2α  = 2 θ(r+ ) − θ(r− ) . δθ = −2α θ(r+ ) − θ(r− ) ∂L L L Assuming, again, for θ± their Keplerian values, θ− = 0 and θ+ = π, we finally have  α Δθ = 2π 1 + 2 . L Clearly, • If α > 0 the precession is forward (counterclockwise), because Δθ > 2π. • If α < 0 the precession is backward (clockwise), because Δθ < 2π.

162

4 Potential Series Developments

Fig. 4.10 Forward precession (α > 0 =⇒ Δθ > 2π)

Δθ>2π

Fig. 4.11 Backward precession (α < 0 =⇒ Δθ < 2π)

Δθ 0. = M R2 2

J2 =

The coefficient J2 is positive because z is along the Earth minor axis (which is along the spin direction) so that I3 > I2 and I3 > I1 . From satellite measurements of M, R, I1 , I2 , I3 , the value J2 = 1.0827 × 10−3 is obtained (see also Table 5.2).

5.2 Satellite Motion

167

5.2 Satellite Motion The vectorial equation of motion for a point mass satellite in the axisymmetric Earth field, neglecting atmospheric drag, is

 ∞  m  R GM r¨ = ∇U = ∇ Jm Pm (cos ϕ) . 1− r r m=2 The potential U (r) can be thought as a “perturbed” monopole potential U (x, y, z) =

GM + R, r

where, in axisymmetry, the perturbing function R is R(r, ϕ) = −

∞  GM  R m Jm Pm (cos ϕ). r m=2 r

The above function actually constitutes a small perturbation over the monopole term, because (see Table 5.2) J2 (which is by far the largest) is only ∼10−3 . Using the perturbing function, the full 1st order system of equations of motion for the satellite is ⎧ x˙ = u ⎪ ⎪ ⎪ ⎪ y˙ = v ⎪ ⎪ ⎪ ⎪ z ˙=w ⎪ ⎪ ⎪ ⎪ GM ∂R ⎪ ⎪ ⎪ ⎨ u˙ = − r 3 x + ∂ x (5.3) ⎪ ⎪ G M ∂R ⎪ ⎪ v˙ = − 3 y + ⎪ ⎪ ⎪ r ∂y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ GM ∂R ⎪ ⎪ ⎩w , ˙ =− 3 z+ r ∂z with initial conditions ⎧ ⎨ x(0) = x0 y(0) = y0 ⎩ z(0) = z 0 ,

⎧ ⎨ u(0) = x˙0 v(0) = y˙0 ⎩ w(0) = z˙ 0 .

Of course, in the limit R → 0 the motion is Keplerian. The magnitude of the relative perturbation over the monopole

168

5 The Earth Potential

   R     GM  ,  r2 

(5.4)

is of order J2 , so that what it is expected is a “secular” (i.e. occurring on a time scale significantly longer than the Keplerian orbital period) departure of the actual orbit respect to the Keplerian, unperturbed, one. The solution of the system Eq. 5.3 is modernly reachable at the highest precision with numerical methods. Anyway, it is useful a short treatment of the classical perturbation method behind this problem.

5.3 A Short Note on the Perturbation Theory: Osculating Elements The unperturbed solution of the equation of motion is by definition that obtained for R = 0. It depends on the six constants (x0 , y0 , z 0 ) and (x˙0 , y˙0 , z˙ 0 ). Usually, another set of six constants is used at their place, called elliptic elements, which are linked to the initial conditions. The adjective “elliptic” derives from that the unperturbed solution of negative energy is a Keperian ellipse. The elliptic elements are Ω, the longitude of the ascending node. It defined as the angle formed by the x axis of the plane of reference with the intersection line between the orbital plane and the reference plane (called line of nodes) in the direction of the ascending node which is the point where the satellite passes from negative to positive values of z coordinate; i, the inclination angle between the orbital plane and the reference plane (0◦ ≤ i ≤ 180◦ ); a, the trajectory semi-major axis; ω, the position angle of the perigee; e, the trajectory eccentricity; τ , the time of passage of the satellite at its perigee. The unperturbed solution is, so, fully identified by the above set of values, r = r(t; Ω, i, a, w, e, τ ). The solution of the perturbed problem consists in finding the set of elliptic element in the form of slowly varying time functions such that at any time t the values of the elliptic elements at that time define a Keplerian elliptic orbit which osculates the real satellite orbit.

5.4 Search for the Time-Dependent Elliptic Elements

169

5.4 Search for the Time-Dependent Elliptic Elements Without entering in details, we sketch here the standard procedure to solve the perturbed satellite problem. We indicate by (α1 , α2 , α3 , α4 , α5 , α6 ) = (Ω, i, a, ω, e, τ ), the set of elliptic elements, supposed to depend, slowly, on time. The scope is the determination of such functions αi (t), i = 1, 2, · · · , 6, as satisfying Eq. 5.3 in the form of i = 1, 2, · · · , 6, αi (t) = αi0 + si (t) + pi (t) where αi0 = αi (t0 ) is the osculating elliptic element evaluated at t = t0 . The function si (t) is the secular part of the solution and depends on the osculating elliptic elements and on the Jm coefficients, other than, linearly, on t; pi (t) is the periodic part of the solution. Were si (t) and pi (t) known, an observation of the satellite at an initial time t0 would give the whole set of αi0 and so the actual unknowns would be the Jm quantities. Thus, many different observations of different satellites at different times could constitute a way to evaluate some of the Jm , as well as the product G M and R. and so of the monopole component J0 . In Tables 5.1 and 5.2 we report values obtained in this way. A more refined study would keep into account the dependence, even if small, of the Earth potential on θ , that causes the presence in the perturbation function R of some tesseral harmonics, which are hard to determine. In conclusion, given the definition of geoid as the shape of the ocean surface under the Earth’s gravity, it is an equipotential surface of the effective potential, i.e. 1 Ue f f (x, y, z) = U (x, y, z) + ω2 (x 2 + y 2 ) = c, in Cartesian coordinates, 2 1 2 2 2 Ue f f = U (r, θ, ϕ) + ω r sin ϕ = c, in spherical polar coordinates 2 where c is a constant. Available data on the Earth’s potential from study of satellites motion show that the geoid deviates from the reference ellipsoid of revolution around the polar axis for no more than ±100 m. Note that Newton was the first to evaluate the value of Table 5.1 Values of G, M, and R

G M = 3.986032 × 106 cm3 sec−2 R = 6378.165 km = 6.378165 × 108 cm M = 5.97319 × 1027 g G = 6.673231 × 10−8 dyn cm2 g−2

170 Table 5.2 Some values of Jm from the motion of 9 satellites (from Kozai, Y., New Determination of Zonal Harmonic Coefficients of the Earth’s Gravitational Potential. SAO Special Report. 165. (1964))

5 The Earth Potential m

Jm /10−6

2 3 4 5 6 7 8 9 10 11 12 13 14

1082.645 −2.546 −1.649 −0.210 0.646 −0.333 −0.270 −0.053 −0.054 0.302 −0.357 −0.114 0.179

the relative deviation of the Earth’s ellipsoid from sphericity obtaining the value 1 . s = 1/230 which is quite similar to modern determinations, ∼ 298.25

5.5 Further Readings This chapter presented a short spot of classical celestial mechanics theory of perturbation as well as some geodetical information on the Earth potential. The reader who wish to go deeper in the topics of celestial mechanics may take advantage by both the classical text [21] and the more modern one [7]. For a deeper approach to the Earth gravity field see [6].

Appendix A

Two- and N-Body Adimensional Motion Equations

For practical computational convenience it is useful transforming the equations of motion of a classical, Newtonian, N -body system in an adimensional form, so that solution can be partially scaled to various sets of initial parameters. Let us use the simple 2-body case as paradygm. The 2nd order vectorial equation of the relative motion for two objects of masses m 1 and m 2 writes as μ¨r = −G

m1m2 r, r3

(A.1)

where μ = m 1 m 2 /(m 1 + m 2 ) is the reduced mass and r ≡ r2 − r1 (or, equivalently r ≡ r1 − r2 ) gives the relative position of the two objects, identified by r1 and r2 in an inertial reference frame. Equation A.1 writes also as r¨ = −G

m1 + m2 r. r3

(A.2)

A choice of proper mass and length units, M and L, which can be assumed as M = m 1 + m 2 and, for negative orbital energy (elliptic trajectory), as L = a where a is the ellipse’s semi major axis, leads to write Eq. A.2 as r¨ G M m 1 + m 2 (r/L) =− 3 . L L M (r/L)3

(A.3)

   We can define √ new, adimensional, variables r = r/L, m i = m i /M, t = t/T , where 3/2 T = L / G M. If E < 0, T /(2π) is the orbital period (note also that the adoption of M, L and T as units of measure implies G = 1). This choice of units transforms Eq. A.3 into the adimensional form

d 2 r r 3 = − . r dt  2 © Springer Nature Switzerland AG 2019 R. A. Capuzzo Dolcetta, Classical Newtonian Gravity, UNITEXT for Physics, https://doi.org/10.1007/978-3-030-25846-7

(A.4)

171

172

Appendix A: Two- and N -Body Adimensional Motion Equations

Such procedure of adimensionalization can be followed also for the generic N -body case, where the equations of motion are (see Sects. 2.3 and 2.3.1), for every 1 ≤ i ≤ N ⎧ N  r −r ⎪ ⎪ ¨ r = G m j jr 3 i , i ⎪ ⎨ ij j=1 j=i

⎪ ⎪ ⎪ ⎩ ri (0) = ri0 , r˙ i (0) = r˙ i0 ,

(A.5)

where ri j = |ri − r j | is the Euclidean distance between particle i and j. The above system of 6N second-order scalar differential equations is reduced at first order by letting vi = r˙ i (i = 1, 2, . . . , N ) to have ⎧ r˙ = vi , ⎪ ⎪ i ⎪ ⎪ ⎪ ⎪ ⎪ N  ⎨ r −r m j jr 3 i , v˙ i = G ij j=1 ⎪ ⎪ j=i ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ri (0) = ri0 , vi (0) = r˙ i0 .

(A.6)

Once written as a system of 3N , first order differential equations, its numerical solution can fruit of a plethora of available numerical algorithms. As a matter of fact, only a very careful numerical treatment allows to obtain a reliable solution, overcoming the U V and I R divergence of the interaction force and the strong non-linearity of the equations system. I limit here to say that a straightforward way to overcome the problem of the divergent acceleration when 2 of the N bodies approach at vanishing distance is both the regularization of the 2-body interaction as described in Appendix B and that of softening the purely Newtonian interaction potential. This means that the Ui j ∝ 1/ri j interaction potential is replaced by a “softened”, non singular, potential. One way to soften the interaction potential is (for instance) the√Plummer’s one described in Sect. 3.4.1 of this book, which has the form Ui j ∝ 1/ r 2 + 2 where  (the constant b of the Plummer sphere) is the softening parameter (length scale of potential smoothing). To have a simultaneous preservation of a proper global Newtonian behaviour of the N -body system and elimination of the small scale divergence, a proper choice of  must be done. It is easily seen that a √ proper choice is a fraction of the average closest neighbour distance, dcn ∼ R/ 3 N , where R is the physical size of the N -body system. At regard of the N -body dynamics topic we limit to cite the book [1] which is comprehensive of most of the aspects of the hard topic of gravitational N -body simulations.

Appendix B

Two Body Regularization

The Newtonian interaction potential between two point mass particles, Ui j ∝ 1/ri j , diverges for two particles approaching toward zero distance (called U V divergence). This causes a relevant problem in the numerical integration of the evolution of an N -body system because in order to follow with sufficient accuracy the, rare, close encounters, the time step must decrease by orders of magnitude (to the order of the flyby time of the encounter, τ ∼ 2d/vr , where d is the closest approach distance and vr the two body relative speed) so that the overall integration becomes computationally overwhelming unless particular treatments are introduced (for example, methods apt to separate the i, j binary motion from the rest of the system, whose motion is essentially frozen over τ , to be resynchronized afterwards). Another interesting, and more correct, way to deal with the U V divergence of Ui j is via its regularization, a procedure first introduced by the mathematician LeviCivita and, later, faced and refined in many different ways by other researchers, among whom Kustaneeimo, Stiefel, etc. Let us consider two bodies of masses m i and m j and position vectors ri and r j in a given reference frame. The equation of their relative motion, after setting r ≡ r j − ri , G = 1, and m i + m j = 1 is1 r¨ = −

r . r3

(B.1)

As said above, the force is singular in r = 0. By introducing the differential spacetime transformation dt = r n dτ (with given, but arbitrary, n) which, equivalently, is dτ = r −n dt, it results dr dτ dr −n r˙ = (B.2) = r , dτ dt dτ

1 These choices imply that the unit of time is such that the orbital period of the bound 2-body motion

is equal to 4π 2 . © Springer Nature Switzerland AG 2019 R. A. Capuzzo Dolcetta, Classical Newtonian Gravity, UNITEXT for Physics, https://doi.org/10.1007/978-3-030-25846-7

173

174

Appendix B: Two Body Regularization

so that the final, 2nd order, differential (with respect to τ ) equation of motion is dr dr d 2r − nr −1 = −r 2n−3 r. 2 dτ dτ dτ

(B.3)

Choosing n = 1 and indicating with an apex,  , differentiation with respect to τ , the above equation writes as 1 r (B.4) r − r  r = − . r r For didactic convenience, we limit here to develop the one-dimensional case, letting thus r = x i. The above equation reduces to a scalar differential equation x  −

x 2 = −1. x

By mean of the energy integral, E = 21 x˙ 2 − 2

x x

1 x

=

(B.5) 1 2

  2 x x

− x1 , we can substitute

= 2(E x + 1) in Eq. B.5 obtaining x  − 2E x = 1,

(B.6)

which is non singular. A further change of variable, u = 2E x + 1, transforms Eq. B.6 into a classic harmonic oscillator in the variable u u  − 2Eu = 0,

(B.7)

whose solution in terms of sine and cosine functions is well known. The seeked function x(t) is finally obtained by the formula x(τ ) = (u(τ ) − 1)/(2E), with the τ τ (t) relation coming by the inversion of the integral relation t (τ ) = x(τ )dτ . 0

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