The principles of classical physics, though superseded in specific fields by such theories as quantum mechanics and gene

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*English*
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*Table of contents : Contents......Page 7Figures......Page 13Preface......Page 18Acknowledgments......Page 201.1 Energy......Page 221.1.1 Car driving......Page 241.2 Mass......Page 251.3 Momentum......Page 282.1 Whether to Stop or Run Through?......Page 312.2 Vertical Jump......Page 342.2.1 Height equation: from conservation law......Page 352.2.2 Jumps of animals......Page 362.3 Hourglass......Page 372.4 Motion of a Chain in a Tube......Page 382.5.1 Gravitation......Page 402.5.3 Strong interaction......Page 423.1 Cartesian vs Polar Coordinates......Page 443.2 Coriolis Force......Page 483.3 Rotation Group......Page 494.1.1 Mechanics......Page 504.1.2 Electric circuits......Page 514.1.3 Optics......Page 524.2 The Principle of Least Action......Page 534.3 More Thoughts on Why “(T −V )”?......Page 555 Work and Energy......Page 565.1 At the T-junction......Page 575.2 Motion of a Heavy Particle on a Smooth Curve in a Vertical Plane......Page 605.3 Motion of a Heavy Particle, Placed on the Outside of a Smooth Circle in a Vertical Plane and Allowed to Slide Down......Page 615.4 Motion in a Vertical Plane of a Heavy Particle Attached by a Fine String to a Fixed Point......Page 625.5 Conservative Force......Page 635.5.1 Interpretation of grad V......Page 645.6.1 Galilean invariant......Page 655.6.2 Example......Page 676 Mechanics of a System of Particles......Page 696.1.1 Simple ‘usual’ pendulum......Page 706.1.2 Leaking bob......Page 716.2 Work-energy Theorem Revisited......Page 746.3 Displacement......Page 766.4 Rotation......Page 776.5 Rigid Body Motion: Basic Ideas......Page 786.6 Rotation of a Rigid Body about an Arbitrary Axis......Page 816.6.1 Special cases......Page 826.7 Moments of Inertia of Simple Bodies......Page 836.8 Principal Axes — Stationary Points of Kinetic Energy......Page 856.9 Euler’s Equations......Page 877.1 Non-conservative Forces and Energy Loss......Page 917.1.1 Energy loss......Page 927.2 Bowling — Physics of the Rolling Ball......Page 937.3 Squealing and Squeaking......Page 988.1.1 Direct impact......Page 998.1.2 Poisson’s hypothesis......Page 1008.1.3 Kinetic energy lost by impact......Page 1018.1.4 Generalization of Newton’s rule......Page 1028.2.1 Deformation energy......Page 1038.2.2 Impact force......Page 1048.3 Falling Pencil on a Table......Page 1079.1 The Two-body Problem......Page 1129.1.1 Bounded orbits......Page 1159.2.1 Case I......Page 1179.2.2 Case II......Page 1189.3 Satellite Paradox......Page 1199.3.1 Descending path on a near-circular orbit......Page 1209.4 Rotation Curves: an Anomaly......Page 1229.5 The Rosetta-Philae Comet Mission......Page 12310.1 Black Holes at LHC......Page 12510.2 Nuclear Explosion......Page 12610.3 Insect Flight......Page 12711.1 Free Oscillations......Page 12911.3 Compound Pendulum......Page 13011.4 Damped Harmonic Oscillator......Page 13111.5 Driven Damped Simple Harmonic Oscillator......Page 13211.7 Another Instance of Simple Harmonic Motion......Page 13511.8 Two Coupled Oscillators......Page 13711.9 Three Coupled Oscillators......Page 14011.10 Many Coupled Oscillators......Page 14111.11 Dissipation by a Rapidly Oscillating Potential......Page 14312.1 Transverse Modes of a String......Page 14512.2.1 Reﬂection and transmission of waves on a string......Page 14712.3 Standing Waves on Planar Membranes......Page 14912.4.1 Newton’s derivation......Page 15212.4.2 Correct derivation (Laplace)......Page 15413.1 Physics of Music......Page 15713.2 Western Classical Music......Page 16013.3 Transposition, Musical Scales, and Algebraic Groups......Page 16114 Fluid Mechanics......Page 16314.2 Euler’s Equation......Page 16514.2.1 Applications......Page 16614.4 Streamlines......Page 16714.5 Speed of Sound Inside a Fluid......Page 16814.5.1 Effect of bubbles......Page 16914.6 Sound of a Brook......Page 17114.7 Why is Water Watery?......Page 17215.1 Gravity Waves in Liquid......Page 17415.1.2 Shallow water waves (Tsunami)......Page 17615.2 Capillary Waves......Page 17716 The Kinetic Theory of Gases......Page 18016.1 Equipartition of Kinetic Energy, Ideal Gas Law......Page 18116.2 Football Game: Kinetic Theory Perspective......Page 18316.3 Adiabatic Reversible Compression......Page 18516.4 Adiabatic Reversible Compression (from Mechanics and Kinetic Theory)......Page 18617 Concepts and Laws of Thermodynamics......Page 18817.1 Adiabatic Transitions and Accessibility of States of a System - Empirical Entropy, First and Second Laws......Page 18917.2 Sears’ Illustration of Caratheodory’s Treatment......Page 19117.3.1 Reversible process......Page 19517.3.3 Irreversibility......Page 19617.4 Order or Disorder......Page 19717.5 How Does Entropy Look Like?......Page 19818.1 Thermodynamic Potentials......Page 20218.2 Van der Waals Equation for Real Gases......Page 20518.2.1 Liquefaction of gases......Page 20618.3 The Third Law of Thermodynamics......Page 20718.4.1 Diffusion......Page 20818.4.2 Law of mass action......Page 21018.5 Chemical Potential......Page 21118.6 Van’t Hoff Equation of State for Dilute Solutions......Page 21319.1 Gibbs and Boltzmann Entropies......Page 21519.2 Boltzmann Factor: Application to “Phases of Matter”......Page 21819.2.1 Gases and solids......Page 21919.2.2 Liquids......Page 22119.3 Failure of Classical Physics......Page 222Bibliography......Page 224Index......Page 230*

Mechanics, Waves and Thermodynamics Despite the fact that classical physics has been superseded in specific fields by theories such as quantum mechanics and general relativity, it still retains importance in a broad range of applications. Classical thermodynamics and Newtonian mechanics are extremely useful to describe certain physical and chemical systems: for instance in explaining why one cannot jump standing erect; why small insects jump so easily; in applying mechanics to athletics and bowling; in explaining the wobbling of the earth, ripples on a lake to tsunamis, oscillations in spring systems to sound, music scales and their mathematics, etc. This book presents the fundamental concepts of classical physics in a coherent and logical manner. Besides elaborating the basic concepts of energy, mass, momentum and entropy, it also discusses topics like mechanics of single particle, oscillations and waves to enhance learners’ understanding. A unique feature of this book lies in the discussions on concepts such as central forces in the context of a recent spacecraft landing on a comet, Caratheodory’s axioms, applications of thermodynamics, understanding states of matter with the simple usage of Boltzmann factor, which are generally not found in books on classical physics. The concepts are explained in detail through live examples and solved problems. Classical theories are taught in all universities offering courses in Physics. This book is specially designed for undergraduate students of science and engineering. Sudhir Ranjan Jain is Scientific Officer (H) at the Bhabha Atomic Research Centre, Mumbai. He is also Professor at the Homi Bhabha National Institute and adjunct Professor at the UM-DAE Centre for Excellence in Basic Sciences, Mumbai. He has been a visiting researcher at the Universit´e Libre de Bruxelles, Belgium, University of Utrecht, Netherlands, and University of Maryland, USA. Professor Jain has written many articles published in peer-reviewed journals. His research interests include nonlinear dynamics, quantum chaos, statistical physics, nuclear theory, and mathematical physics.

CAMBRIDGE–IISc SERIES

Cambridge–IISc Series aims to publish the best research and scholarly work on different areas of science and technology with emphasis on cutting-edge research. The books will be aimed at a wide audience including students, researchers, academicians and professionals and will be published under three categories: research monographs, centenary lectures and lecture notes. The editorial board has been constituted with experts from a range of disciplines in diverse fields of engineering, science and technology from the Indian Institute of Science, Bangalore. IISc Press Editorial Board: G. K. Ananthasuresh, Professor, Department of Mechanical Engineering K. Kesava Rao, Professor, Department of Chemical Engineering Gadadhar Misra, Professor, Department of Mathematics T. A. Abinandanan, Professor, Department of Materials Engineering Diptiman Sen, Professor, Centre for High Energy Physics Titles in print in this series: • Continuum Mechanics: Foundations and Applications of Mechanics by C. S. Jog • Fluid Mechanics: Foundations and Applications of Mechanics by C. S. Jog • Noncommutative Mathematics for Quantum Systems by Uwe Franz and Adam Skalski

Cambridge - IISc Series

Mechanics, Waves and Thermodynamics An Example-based Approach

Sudhir Ranjan Jain

4843/24, 2nd Floor, Ansari Road, Daryaganj, Delhi - 110002, India Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781107145191 c Sudhir Ranjan Jain 2016

This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2016 Printed in India A catalogue record for this publication is available from the British Library Library of Congress Cataloguing in Publication data Names: Jain, Sudhir Ranjan, author. Title: Mechanics, waves and thermodynamics : an example-based approach / Sudhir Ranjan Jain. Description: New York, NY : Cambridge University Press, 2016. | ?2016 | Includes bibliographical references and index. Identifiers: LCCN 2015051229| ISBN 9781107145191 (hardback) | ISBN 1107145198 (hardback) Subjects: LCSH: Mechanics. | Waves. | Thermodynamics. Classification: LCC QC125.2 .J35 2016 | DDC 531–dc23 LC record available at http://lccn.loc.gov/2015051229 ISBN 978-1-107-14519-1 Hardback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.

Dedicated to my father Suresh Kumar Jain

Contents Figures Preface Acknowledgments 1 Energy, Mass, Momentum 1.1 Energy 1.1.1 Car driving 1.2 Mass 1.3 Momentum

xiii xvii xix 1 1 3 4 7

2 Kinematics, Newton’s Laws of Motion 2.1 Whether to Stop or Run Through? 2.2 Vertical Jump 2.2.1 Height equation: from conservation law 2.2.2 Jumps of animals 2.3 Hourglass 2.4 Motion of a Chain in a Tube 2.5 Forces 2.5.1 Gravitation 2.5.2 Electro-weak interaction 2.5.3 Strong interaction

10 10 13 14 15 16 17 19 19 21 21

3 Circular Motion 3.1 Cartesian vs Polar Coordinates 3.2 Coriolis Force 3.3 Rotation Group

23 23 27 28

4 The Principle of Least Action 4.1 Action of the Principle 4.1.1 Mechanics 4.1.2 Electric circuits 4.1.3 Optics 4.2 The Principle of Least Action 4.3 More Thoughts on Why “(T −V )”?

29 29 29 30 31 32 34

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Contents

5 Work and Energy 5.1 At the T-junction 5.2 Motion of a Heavy Particle on a Smooth Curve in a Vertical Plane 5.3 Motion of a Heavy Particle, Placed on the Outside of a Smooth Circle in a Vertical Plane and Allowed to Slide Down 5.4 Motion in a Vertical Plane of a Heavy Particle Attached by a Fine String to a Fixed Point 5.5 Conservative Force 5.5.1 Interpretation of grad V 5.5.2 Relation with curl of the force 5.6 Work-energy Theorem and Galilean Invariance

36 36 39 40 41 42 43 44 44

5.6.1 Galilean invariant 5.6.2 Example

44 46

6 Mechanics of a System of Particles 6.1 Analysing the Leaky Pendulum

48 49

6.1.1 Simple ‘usual’ pendulum 6.1.2 Leaking bob

49 50

Work-energy Theorem Revisited Displacement Rotation Rigid Body Motion: Basic Ideas Rotation of a Rigid Body about an Arbitrary Axis

53 55 56 57 60

6.6.1 Special cases

61

6.2 6.3 6.4 6.5 6.6

6.7 Moments of Inertia of Simple Bodies 6.8 Principal Axes — Stationary Points of Kinetic Energy 6.9 Euler’s Equations 7 Friction 7.1 Non-conservative Forces and Energy Loss 7.1.1 Energy loss 7.2 Bowling — Physics of the Rolling Ball 7.3 Squealing and Squeaking

62 64 66 70 70 71 72 77

8 Impulse and Collisions 8.1 Impact of Smooth Spheres

78 78

8.1.1 Direct impact 8.1.2 Poisson’s hypothesis

78 79

Contents

8.1.3 Kinetic energy lost by impact 8.1.4 Generalization of Newton’s rule 8.2 Forward Karate Punch 8.2.1 Deformation energy 8.2.2 Impact force 8.3 Falling Pencil on a Table 9 Central Forces 9.1 The Two-body Problem 9.1.1 Bounded orbits 9.2 Two Bodies Under their Own Gravitational Interaction 9.2.1 Case I 9.2.2 Case II 9.2.3 Case III 9.3 Satellite Paradox 9.3.1 Descending path on a near-circular orbit 9.4 Rotation Curves: an Anomaly 9.5 The Rosetta-Philae Comet Mission

ix 80 81 82 82 83 86 91 91 94 96 96 97 98 98 99 101 102

10 Dimensional Analysis 10.1 Black Holes at LHC 10.2 Nuclear Explosion 10.3 Insect Flight

104 104 105 106

11 Oscillations 11.1 Free Oscillations 11.2 Transverse Oscillations in Mass-spring System 11.3 Compound Pendulum 11.4 Damped Harmonic Oscillator 11.5 Driven Damped Simple Harmonic Oscillator 11.6 Beats 11.7 Another Instance of Simple Harmonic Motion 11.8 Two Coupled Oscillators 11.9 Three Coupled Oscillators 11.10 Many Coupled Oscillators

108 108 109 109 110 111 114 114 116 119 120

11.10.1 Phase velocity 11.10.2 Three-dimensional long-range order 11.11 Dissipation by a Rapidly Oscillating Potential

122 122 122

x

Contents

12 Waves 12.1 Transverse Modes of a String 12.2 Standing Waves in One Dimension 12.2.1 Reflection and transmission of waves on a string 12.3 Standing Waves on Planar Membranes 12.4 Speed of Sound in Air 12.4.1 Newton’s derivation 12.4.2 Correct derivation (Laplace) 13 Sound of Music 13.1 Physics of Music 13.2 Western Classical Music 13.3 Transposition, Musical Scales, and Algebraic Groups 14 Fluid Mechanics 14.1 Equation of Continuity 14.2 Euler’s Equation 14.2.1 Applications 14.3 Bernoulli’s Equation 14.4 Streamlines 14.4.1 Applications

125 125 127 127 128 131 131 133 136 136 139 140 142 144 144 145 146 146 147

14.5 Speed of Sound Inside a Fluid

147

14.5.1 Effect of bubbles

148

14.6 Sound of a Brook 14.7 Why is Water Watery? 15 Water Waves 15.1 Gravity Waves in Liquid 15.1.1 Deep water waves 15.1.2 Shallow water waves (Tsunami) 15.2 Capillary Waves 16 The Kinetic Theory of Gases 16.1 Equipartition of Kinetic Energy, Ideal Gas Law 16.2 Football Game: Kinetic Theory Perspective 16.3 Adiabatic Reversible Compression 16.4 Adiabatic Reversible Compression (from Mechanics and Kinetic Theory) 16.5 Maxwellian Distribution of Velocities of Gas Molecules

150 151 153 153 155 155 156 159 160 162 164 165 167

Contents

xi

17 Concepts and Laws of Thermodynamics 168 17.1 Adiabatic Transitions and Accessibility of States of a System - Empirical Entropy, First and Second Laws 169 17.2 Sears’ Illustration of Caratheodory’s Treatment 170 17.2.1 Temperature as a property 17.3 Reversible and Irreversible Adiabatic Processes

174 174

17.3.1 Reversible process 17.3.2 Carnot cycle 17.3.3 Irreversibility

174 175 175

17.4 Order or Disorder 17.5 How Does Entropy Look Like?

176 177

18 Some Applications of Thermodynamics 18.1 Thermodynamic Potentials 18.2 Van der Waals Equation for Real Gases 18.2.1 Liquefaction of gases 18.3 The Third Law of Thermodynamics 18.4 Gaseous Mixture 18.4.1 Diffusion 18.4.2 Law of mass action 18.5 Chemical Potential 18.6 Van’t Hoff Equation of State for Dilute Solutions

181 181 184 185 186 187 187 189 190 192

19 Basic Ideas of Statistical Mechanics

194

19.1 Gibbs and Boltzmann Entropies

194

19.1.1 Entropy and “energy-spreading” 19.2 Boltzmann Factor: Application to “Phases of Matter” 19.2.1 Gases and solids 19.2.2 Liquids 19.3 Failure of Classical Physics Bibliography Index

197 197 198 200 201 203 209

Figures 1.1 Log of maximum power of the engine is plotted against 3 times log of maximum speed of cars from a randomly chosen selection for which data was available on the web. 2.1 A schematic distance-time graph for a vehicle approaching a stoplight. The shaded portion shows the region that will correspond to an illegal choice of speeding. 2.2 Each of the three figures show a human preparing to jump. The position of F(oot), S(hin), T(high), and B(ack) are taken from realistic data (not drawn to scale). The filled dot marks the Centre of Gravity which goes higher by the height of the jump. 2.3 Grains of sand fall freely under gravity and hit the bottom, exerting pressure derivable from Newton’s second law. 2.4 A chain of length ` inside a tube is falling under gravity. The prelation between velocity v and the length x in the column BC follows v ∼ g/`x (2.13) 2.5 Pressure, P due to a falling chain as a function of the length x reaching the table. The first derivative of pressure falls discontinuously to one-third of its value as soon as the entire length rests on the table. 2.6 To calculate the gravitational force due to a spherical body on a mass m at a distance r from the centre of the sphere, we may divide the sphere into spherical shells, shells into rings, and eventually add all the contribution by integrating. Inside the sphere, the force turns out to be zero. 3.1 In traversing around the circle half-way, the velocity reverses. This presents a simple way to see how the average acceleration comes proportional to square of speed divided by the radius. 3.2 Polar coordinates and the corresponding mesh. 4.1 Current is divided in two branches with resistances R1 and R2 . Applying the principle of least action in an appropriate form leads to Kirchhoff’s law. 4.2 As a ray of light enters a denser medium, 2 from 1, it bends towards the normal. The angles of incidence and refraction are respectively i and r. The vertical distance of the point A (B) from where a ray ensues (terminates) is y1 (y2 ) from the plane CD across which the medium changes. 5.1 A car is approaching a T-junction, avoiding to collide. Should it turn or apply brakes?

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5.2 The diagram shows that while driving, one can dodge an obstacle by a turn through 15 degrees. 5.3 Components of acceleration (left) and force (right) of a particle moving along a curved path are shown. 5.4 Motion of a heavy particle moving along a smooth circle in a vertical plane, under gravity. 5.5 A particle attached to a string of radius R, and moving along a circle in a vertical plane. 6.1 Water filled in the spherical bob can be described in terms of cylindrical coordinates most naturally. 6.2 The centre of mass (CM) of the bob is shown as a function of the height of the water level normalized to the inner radius a of the bob, ξ = h/a. The bob is made of steel of density 7.85 g/cm3 , its outer radius is 5 cm and thickness is 5 mm. Density of water is 1 g/cm3 . We see the non-monotonicity in the variation. 6.3 Applying work-energy theorem to understand uncoiling of a chain on application of force F on a frictionless floor. 6.4 Displacement of a rigid body in its plane consists in a rotation about a point. 6.5 Instantaneous centre of rotation of a sliding rod. 6.6 A rolling circular rigid lamina. 6.7 A marble of mass m moves along a straight line down an inclined plane. The path is along a groove so that it remains straight. The turntable slows down (moves faster) as the marble rolls up (down) without slipping. 6.8 The body shown is symmetric about the z0 -axis. In any plane perpendicular to z0 , there would be points rotating along a circle where there are points P and P0 diametrically opposite to each other. This axis must pass through the centre of mass, C. The expressions Eq. (6.53) must hold good. 6.9 Axial precession of the earth occurs with a period of 25,800 years, shown here schematically. 7.1 Forces on a rolling ball. 7.2 As the ball slips and rolls, the linear velocity decreases and the angular velocity increases. Finally, the ball just rolls. 7.3 The rolling ball is in contact with a surface, this presents a torque in a direction that opposes the rolling motion. 7.4 Forces when the ball is in contact with a surface. 7.5 The components of the forces are shown in detail to clearly state the equations of motion for a rolling ball with a surface-contact of the ground. 8.1 The height of the ball hA is proportional to the square of its velocity. This simple idea is used in estimating the coefficient of restitution by considering two balls made up of different materials and colliding with another common ball or surface.

38 39 40 41 52

53 54 55 55 56

58

62 69 72 73 74 75 76

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8.2 The collision of a ball is here generalized with unequal angles of incidence and reflection. The Newton’s rule can be generalized for this collision also. 8.3 A bone is supported at two fixed ends. On application of a force as shown, it bends. The inner surface compresses whereas the outer surface expands, hence, there is a neutral surface in between (shown as dotted line). 8.4 Section of the bone shown in Fig. 8.3. 8.5 Fall of a pencil presents an interesting instance of nonlinear dynamics where friction plays an essential role. 8.6 The force, F is shown here as a function of the angle for a typical pencil. F is quite small, to see the numbers in a nice way we have multiplied it by 1000. 8.7 The reaction, N is shown here as a function of the angle for a typical pencil. N is multiplied by 100 so that the numbers look reasonable. Why is this graph non-monotonic? 9.1 A satellite is at an inclination ϕ w.r.t. reference axis. A drag, D is opposing the velocity v w.r.t. the local horizontal. The role of drag is discussed in the motion of the satellite. 11.1 A mass is performing transverse oscillations, which happen to be nonlinear. This is unlike the longitudinal oscillations of a mass connected to two springs on either sides. 11.2 A horizontal plate performs harmonic motion when supported by two rotating wheels. 11.3 Two equal masses attached to identical springs, there appear two normal modes. 11.4 Dispersion relation for ω0 equal to one. 12.1 The tension on the two sides of an element of a string shows us the resultant force. The Newton’s equation for the element with this force gives us the wave equation. 12.2 The vibrations set on a rectangular membrane settle into a standing wave pattern. We have taken L = π and B = 1. The case shown is for (m, n)=(4, 5). The number of nodal domains ν4, 5 is 20. 12.3 Nodal domains for circular (left panel) and elliptical (right panel) annular membranes, the modes shown are respectively with quantum numbers (8, 9) and (8, 5). 12.4 The confocal parabolae form a domain as shown here in terms of parabolic coordinates. The membrane has the shape as shown, clamped at the boundary. The solutions of the Helmholtz equation is possible in terms of the Kummer functions. See the nodal domains and nodal lines of the solutions in Fig. 12.5. 12.5 Nodal patterns for two kinds of solutions - symmetric and antisymmetric about the central vertical line.

xv

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84 85 87 89

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109 115 116 121

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Figures

12.6 Nodal patterns for isosceles triangle and equilateral triangle [69]. Note that the nodal curves are quite complicated and the domains are irregularly shaped. Only recently have we found the counting function of nodal domains. Even so, there remain many unsolved problems. 13.1 Tanpura — tension of four strings can be adjusted by tightening the knobs. 13.2 Basic notes are shown here on a schematic keyboard of a piano. 14.1 Force across an area element. 14.2 Change of pressure with height of atmosphere. 14.3 A bubble pulsates, and in turn, sets the surrounding water into vibration. 16.1 This diagram gives the three-dimensional view of the relative velocities before and after the collision of two particles. The complexity is explained in the text. 17.1 Two cylinders of volume VA and VB are attached end-to-end with pistons PA and PB on the outer sides. The cylinders may contain different real gases, their walls and pistons providing an adiabatic envelope. The dividing wall is rigid, hence, permeable by radiant heat. We may, in some processes, remove the adiabatic boundary and allow a contact with a heat reservoir. 17.2 A typical Carnot cycle. 17.3 Entropy versus enthalpy is very instructive in understanding the role played by the concavity of this graph.

133 137 140 143 143 150

160

171 176 180

Acknowledgments The Department of Atomic Energy in collaboration with University of Mumbai established the Centre for Excellence in Basic Sciences where I taught the course entitled “Physics-I”. The chairman of the Academic Board, Shashikumar M. Chitre always encouraged and supported me. R. V. Hosur, Director of the Centre kindly invited me to teach several courses, widening my scope of understanding and bringing me in contact with bright students from the first to the tenth semesters. In each year, there were many students whose interest and enthusiasm provided me great stimulus. I would like to thank all my students over these years. I am thankful to Jayant V. Narlikar who sent me some of the older Cambridge tripos question papers which have played an important role in shaping my lectures. B. Srinivasan (BARC), M. S. Bhatia (BARC), and Ameeya Bhagwat (CBS) shared rare books and insights. I am extremely grateful to them. For the invaluable help in drawing the figures, my thanks are due to Sunder Sahayanathan (BARC). In writing Section 9.4 on the rotational curves of galaxies, I thank Sandeep Joshi for his help. For the figures of nodal patterns in Section 12.2, it gives me a great pleasure to acknowledge the help and very fruitful collaborative times with Rhine Samajdar and Naren Manjunath. I am grateful to K. L. Sebastian (IISc Bangalore) who suggested me to submit the manuscript to the Cambridge-IISc series. I am also grateful to the editorial board of this Series and the reviewers for their constructive remarks. The editorial staff of the Cambridge University Press has been very cooperative and patient; in particular, I would like to place on record my gratitude to Gauravjeet Singh Reen in this regard. My immediate and extended family in general have been very supportive. In particular, my wife, Alka, and my son, Manan have been by my side at all times. My journey in music during last twenty-five years with my wife has been vastly enriching. Chapter 13 has been written in close discussions with her. Her insight into music has been very helpful; in fact, that chapter, some day, could be expanded to stand on its own in collaboration with her. Without the very necessary, affectionate pressure created by my daughter, Swareena, this book would not have been completed in time. My brother, Shaleen has been my lifelong companion in sharing and discussing academic matters, concepts, research and didactic matters. My parents, and particularly my father, have emphasized the need of lucidity in any explanation that might be undertaken. As a humble tribute to his constructive criticism, I dedicate this book to him on his seventy-fifth birthday.

1

Chapter

Energy, Mass, Momentum The idea of energy predates, is implicit in, and outlives Newtonian mechanics. — Leon Cooper

1.1 Energy Energy - a concept which we commonly use is surprisingly abstract [2]. All we know from our experience is that the total energy is conserved. The law of conservation states that there is a quantity, called energy that doesn’t change with all the changes occurring around us. But, what is energy? We will try to become familiar with this important concept. Kelvin and Rankine defined the concept of energy in order to understand the physical principles underlying the heat engines [3]. Engines were simplified as devices that raised weights by a certain height, given some amount of coal to burn [4]. In 1840s, there were very significant discoveries and publications by Joule and Mayer where they proved that heat is simply a mode of motion. The definition by Kelvin and Rankine was inspired by this. With slightly different terminology, Helmholtz proved that the ability to perform work is conserved in a wide range of processes. When we strike a matchstick to light it, the energy provided by us and the chemical reaction produces heat and light. In a moment, several transformations of energy take place. We can also ask - is there a macroscopic quantity that has changed when motion ceases “without having caused another motion”? Answer could be that objects get warmer. In his remarkable essay, the French engineer Sadi Carnot (1824) asked: Is the motive power of heat invariable in quantity, or does it vary with the agent which one uses to obtain it? After a number of subsequent developments, Helmholtz (1870 ca) concluded that energy has many forms and that the total energy should be taken as a sum of all possible forms of energy. We can appreciate it as we realize [1] that 4.18 Joules (J) is associated with all the following - (i) work done in pushing with a force 4.18 N through a distance of 1 m; (ii) can produce 1 calorie of heat; (iii) can move 2.6 × 1019 electrons through a potential difference of 1 volt; (iv) can raise 100 g by 418 cm. From the work of Carnot and others, we were led to the concept of efficiency of engines. This led to a technological revolution where one was attempting to invent the most efficient engine. Underlying this and alongwith the conservation of energy is the following major result - there can be no perpetual motion. To understand these ideas, we illustrate by considering a simple weight-lifting machine. Here, we are going to lift a larger mass by a certain height by lowering a smaller mass by a unit distance. Let us

2

Mechanics, Waves and Thermodynamics

consider two types of such machines, a reversible machine A and an irreversible machine B. As the name suggests, if we run an irreversible machine backwards, the original condition will not be realized. The initial condition will be restored for a reversible machine. Let us denote by X (Y) the height by which a reversible (irreversible) machine lifts the larger mass upon lowering the smaller mass by a unit distance. We can show that X > Y. Let us prove this by assuming the contrary, i.e., Y > X. We first operate the machine B so that the larger mass is lifted by a height Y. Then, at no cost, we can lower the mass to X. We can now operate the machine A backwards and bring the masses back to the original configuration. The initial and final configuration being the same, we see that free power is generated. This means that there is perpetual motion. Thus, the assumption, Y > X is false. We therefore learn that reversible machines are the best. This important conclusion is due to the conservation law of energy. Rankine’s definition is [3]

The term ‘energy’ comprehends every state of a substance which constitutes the capacity for performing work. Quantities of energy are measured by the quantities of work which they constitute the means of performing. For the example of the weightlifting machine, we call the sum of weights times heights as the gravitational potential energy. The notion of ‘potential energy’, although coming from the times of Aristotle, was made sharper by Rankine. It is associated with the configuration energy and represents the potential the system possesses. There was the concept of energy associated with motion in the sense that a faster moving ball possesses more energy. Whereas this notion of kinetic energy was clearly palatable, the understanding of potential energy took some time to be cleared. It was Maxwell who, recognizing the brilliant work of Rankine and attributing to him, cleared that the energy residing in a body even while it is not in motion is the potential energy. This is available to the body for spending however, is unspent until that time. Experimentally, it was simply demonstrated by observing that a freely falling ball reaches the ground faster when it starts higher. Does the mass of a battery reduce when it is completely discharged? [5]

A battery converts chemical energy to electrical energy which is then transferred out of the battery, decreasing the total energy and hence the total mass of the battery. To estimate the mass, we need to estimate the (useful) energy stored in the battery. We estimate that the power consumption of a laptop is certainly more than a night bulb (approximately 4 Watt) and lesser than a 100 Watt bulb. Let us take the geometric mean, 20 W to be the power consumption. My laptop discharges in about 3 hours (approximately 10,000 seconds). Hence, energy E = Power × time

= (20 W) × (10, 000 s) = 2 × 105 J.

Energy, Mass, Momentum

3

Corresponding change in mass is m=

E 2 × 105 J = kg ≈ 2 ng. c2 9 × 1016

How do we measure this small mass? So, when we say that the chemical reactions conserve mass, we are not completely right as we ignore the changes of the order of a few nanograms ! Students will find it interesting to explore methods to measure small weights like nanograms. 1.1.1 Car driving Here, we want to demonstrate how an understanding of kinetic energy helps us in estimating the power required to drive a car (this discussion follows [6]). A car journey involves three stages: acceleration from rest to speed v, moving at constant velocity, slowing down to rest. Why do we need any energy at all to move our car at a constant velocity? The acceleration of the car is zero, so is the net force acting on it. The force applied by the car is certainly non-zero as it is counter-acting the air resistance. Generally, we ignore the air resistance, however, here it is important. The air around a moving car is quite complicated. We can make a simple (average) picture. Before being stuck with the car, the air is roughly at rest. Afterwards, it drags with the car at a velocity v. Car does work in increasing the kinetic energy of the air of mass Ma from zero to Ma v2 /2. The air which starts moving is related to the column that forms behind the moving car. If A is the frontal area and d is the distance the car travels, Ma = ρa Ad. Power to drive is P=

1 (ρa Ad )v2 1 Ma v2 = 2 t 2 t

(1.1)

where t is the time. Combining d and t, we have P = ρa Av3 /2. We can check this prediction. Taking data for all sorts of cars from Rs 1,50,000, 33 horsepower Tata Nano to Rs 15,00,00,000, 1200 horse power Bugatti Veyron, it is found that this dependence holds well (Fig. 1.1). We can see that the discussion above is independent of the shape of the car. Mathematically, car’s frontal area A is replaced with Cdrag A where Cdrag is the drag coefficient. Cdrag is a number between 0 and 1. There is an interesting observation that we should present now. Consider Honda Accord which has A = 3 m2 and Cd = 0.3. The power required by this car to move at a speed of 25 m/s is 21 (1.2) (3) (0.3) (25)3 W, i.e., approximately 8 kW. Let us make another calculation based on the mileage of this car, which is 13 km/L. Upon burning gasoline gives 0.34 × 108 J/L. This amounts to a power, 1L 8J 25 ms 13000 m 0.3 × 10 L which is 57700 J/s or 57.7 kW. The question obviously is why the car consumes this much power when it should only consume about 8 kW? An interesting related point was brought out in [6]. If car’s mileage was measured while it was driving closely behind a large truck, it was found to be better. In this manner, the car

4

Mechanics, Waves and Thermodynamics

was riding in the air column dragged by the truck and hence, had to expend “no power” to accelerate the air. The power consumption dropped by 25 per cent. Why doesn’t the measurement show an infinite mileage?

Fig. 1.1

Log of maximum power of the engine is plotted against 3 times log of maximum speed of cars from a randomly chosen selection for which data was available on the web. The cars that appear in this selection are Tata Nano, Maruti Suzuki 800, Hyundai Grandilo, Toyota Etios, Honda Mobilio, Renault Duster, Honda Amaze, Ford Ecosport, Toyota Corolla, Honda City, Hyundai Verna, Ford Torino, Audi Q7, BMW X6, Porsche Macan, Lamborghini Huracan, Mercedes SLS, Ferrari Enzo, F1, Bugatti Veyron. The scatter of points can be seen to lie quite close to a straight line. This means that log P ∝ log v3 , a hypothesis discussed in the text.

The answer to this question brings us to the domain of the second law of thermodynamics. It is not possible to convert thermal energy released by burning gasoline to kinetic energy with 100 per cent efficiency. For the type of engines we are discussing, the maximum efficiency is about 50 per cent. Hence, we see that even if the power required to drive at 25 m/s is about 8 kW, the fuel consumption will correspond to about 57 kW. To save fuel for future, it would be preferable to make use of (i) a bicycle which has minimal frontal area, or, (ii) a train which takes advantage of the bogies lined up close to each other, hence, well within the air column of the bogie in front.

1.2 Mass ... inspite of all the strenous efforts of physicists and philosophers, the notion of mass ... is still shrouded in mystery. — Max Jammer

Johannes Kepler (1618) first introduced the concept of inertia as a spontaneous tendency in each body to come to rest [7, 8, 9]. Newton [10] considered inertia as an innate property of

Energy, Mass, Momentum

5

matter. He adopted a two-stage strategy for discovering how weight and quantity of matter are related: (i) Using pendula, he demonstrated experimentally that all bodies fall with the acceleration due to gravity, independent of their weight or nature or texture; (ii) Then, applying his second law that the ratio of the forces of gravity acting on any two bodies, at a given location, will be equal to the compound ratio of their quantities of matter and their accelerations. However, given that the accelerations are equal, he concluded that the ratio of weights is equal to the ratios of the quantities of matter. Euler (1745) quantified inertia more explicitly than Newton. He compares the inertias of two bodies by comparing the forces required to accelerate them. He identified mass (and quantity of matter) with “quantity of inertia”? What did he mean? Quantity of inertia is a notional division of the body into particles of equal inertia. “Quantity of matter” no longer meant the volume of primary matter. Newton’s concept of “quantity of matter” had geometrized mass, distancing the concept from mechanical powers. Euler made the concept of mass explicitly dynamical and less hypothetical. Mensurationally, Mach (1868) defined mass entirely kinematically - without any need to measure forces. The ratio is a0 m = − m0 a where a0 and a are accelerations which the bodies of mass m and m0 produce in each other. Mach [11] defines mass as a special and distinct property determinative of accelerations. Mach evidently attempted to merge the roles of inertial and gravitational attraction in a single mass concept. For the first time in physics, mass was defined abstractly. Early 20th century brought two new concepts of mass: that which identified mass with ‘inertia’; and the concepts “inertial mass” and “gravitational mass”, which were given prominence by Einstein (1907) [22]. Bondi (1957) made a further classification of mass into “active gravitational mass” and “passive gravitational mass” [13]. With all the different ideas introduced above, let us now see how we might approach the concept of mass. Newton’s law of gravity for weak fields, produced by a gravitational source M1 will be expressed as GM1 = g1 , r2

(1.2)

where g1 is the gravitational acceleration experienced by a small, slowly moving body at a distance r from the source. If g1 is the free acceleration experienced by a small, quasi-static gravitating source M2 , then the latter will induce an acceleration g2 in the original source, if it is also free. This obeys g2 =

GM2 . r2

(1.3)

Also, g2 M2 = . g1 M1

(1.4)

6

Mechanics, Waves and Thermodynamics

If we now hold each of them at a fixed distance r apart - by a rod (say) - then the magnitude of the contact force required to stop the accelerations, exerted by the rod on each body is expressed by F1 = i1 g2 =

i1 M2 , r2

F2 = i2 g1 =

i2 M1 r2

(1.5)

where i j ’s are linear inertias of the bodies. Since, F1 and F2 must be equal and opposite, we have M1 i1 = , i2 M2

(1.6)

which means that inertial mass of a body is proportional to its gravitational mass. If two bodies are in equilibrium on a balance of unequal arms, then the support forces exerted by the scale pans on each body, to prevent them from accelerating, are F1 = i1 g,

F2 = i2 g.

(1.7)

However, these are equal and opposite to the contact forces, W1 and W2 immediately exerted by the bodies on the scale-pans. So, W1 = i1 g,

W2 = i2 g,

(1.8)

and i1 W1 = . i2 W2

(1.9)

This means that the ratio of dead weights of any two static bodies at a given location is equal to ratio of their linear inertias. The concept of mass is also related to the total energy of the body. According to the Thomson–Rankine definition, energy is the capacity of an object or system to perform work. This is the ‘mass energy’ or ‘body energy’, a loss of this energy causes the loss of the very substance of the body. However, to say that matter is energy is incoherent (to quote [9]) because it identifies an object with an abstract property of the body. Einstein’s equation, E = mc2 implies that, for two static bodies, E1 m1 i1 = = E2 m2 i2

(1.10)

as mass is measured by inertia. From the theory of relativity, we know that if the velocity of a body√prior to acceleration is along the z-axis, then the longitudinal component of inertia, il is i0 / 1 − v2 /c2 (:= γ 3 i0 )

Energy, Mass, Momentum

7

and transverse component is it = γi0 . Although, it is alright to talk of longitudinal and transverse inertia, it does not make sense to talk of longitudinal and transverse mass. For more than two centuries the concept of mass in physics has an increasingly dynamic interpretation. A unified definition of mass can be given by noting that all material particles must have energy, inertia, a gravitational field and a capacity for weight. A distinct concept of mass is justified through the following relation for two neighbouring bodies at rest: m1 E1 i1 M1 W1 = = = = . m2 E2 i2 M2 W2

(1.11)

1.3 Momentum The commonly used definition of momentum, mass times velocity, does not hold good for relativistic momentum, radiation momentum, or to the momentum of a photon. We discuss a brief history of the concept of momentum, eventually arriving at a definition that applies universally (following [14]). A concept akin to momentum appeared in the work of John Philoponus of Alexandria, about sixth century A.D. In the context of projectile motion, it appeared as ‘some incorporeal motive force imparted by the projector to the projectile’. In European middle ages, the internal force was called ‘impetus’. Jean Buridan (1295–1358) defined: The impetus imparted by the projectile varies as the velocity and as the quantity of matter of the body.” He meant this to be the force that initiates and maintains the motion. Accepting this, Galileo (1564–1642) called this as ‘momentum’. He introduced a unit for momentum, however, his conception was that momentum causes impact. The concept of inertia was first introduced by Kepler (1571–1630) as a spontaneous tendency of a body to come to rest. The French philosopher, Ren´e Descartes (1644), six years after Galileo published his book [15], describes impetus as ‘force of a body in motion’. Descartes discovered that when a body in motion collides with a similar body, the speed of the combined body is halved. Hence, he concluded that God conserves the total quantity of motion - product of quantity of matter and speed. This so-called Cartesian theory of impact was greatly improved during 1668–1669 by John Wallis (1616–1703), Christopher Wren (1632–1723), and Christiaan Huygens (1629–1695). They showed that this quantity is ‘directed’, and remains the same before and after the impact. Wallis, in fact, was the first one to write momentum, V , as V = (pondus (weight) P) × (celeritas (speed) C).

(1.12)

It is natural to come to the connection between the quantity of motion and force. Galileo, Huygens, and Hooke made models for understanding force by considering tension in a string, a spring attached to a weight, pressure of a fluid etc. It took an extraordinary leap of imagination for Newton to recognize that a force does not remain in a freely moving body, once its action is ceased. Stating his first law, Newton notes, ‘every body is only with difficulty put out of its state of rest or motion’. The second law is stated as ‘a change of motion is proportional to the motive force impressed’ [16]. The well-known ‘Force = mass

8

Mechanics, Waves and Thermodynamics

times acceleration’ does not belong to Newton, this came in the nineteenth century [17]. George Atwood (1745–1807) invented and analyzed the working of his (Atwood) machine using Newton’s second law, hence, providing its verification. Newton laid the foundations of the law of conservation of momentum by combining his first and the third law. He did not consider the notions like work, kinetic energy, and impulse. In 1686, Gottfried Leibnitz (1646–1716) discussed the idea of ‘living force’ in place of momentum. He proved that ‘power’ is proportional to mass as well as speed. By advocating that the ‘living force’ was always combined with ‘impetus’, he argued that momentum and kinetic energy must be related. The concept of the ‘force of moving body’ was rejected by d’Alembert (1717–1783) and Euler (1707–1783). They showed that during an impact, a contact force F is not only exerted for a distance ds however, R R also 1 2 during a time dt. Mathematically, this is equivalent to 2 mv = Fds and mv = Fdt. Hence, with respect to the idea of ‘persisting impetus’, both Leibnitz and Descartes were wrong. Around 1767, this led to disappearance of momentum (as mv) and kinetic energy (as mv2 /2) as physical concepts for about a century; they remained as mathematical concepts though. Joseph Louis Lagrange (1736–1813) argues [18]: ‘The quantity of movement of a body is the measure of that force which the body is capable of exercising against an obstacle, and which is called percussion’. He follows Newton in associating the product of mass and the velocity to what he calls as ‘finite force’ (force acting during a short time of no motion) of R a body. Jean Belanger (1790–1874) replaced percussion by impulsion, defined as 0t Fdt. Thus, came James Clerk Maxwell (1831–1879), combining Belanger and Lagrange to state in 1868: the impulse of a force is equal to the momentum produced by it. William Rankine (1820–1872) gave interpretation of the term ‘energy’ as the capacity of performing work. On the other hand, William Thomson (1824–1907) gave an interpretation to kinetic energy (in 1867, a century after Euler and d’Alembert) as the capacity of a moving body to performing work. As pointed out in a compelling manner by Roche [14], although new concepts of momentum emerged during the late nineteenth century, the great creative period of defining new dynamical concepts had passed. Radiation momentum was introduced by John Poynting (1852–1914) in 1884 [19]; J. J. Thomson (1856–1940) observed that a stationary charge experiences an impulse near a current [20]; relativistically, Max √ Planck (1858–1947) observed [21] that the momentum should be expressed as mx/ ˙ 1 − v2 /c2 (similarly, othet two components) where v is the speed of the particle; Albert Einstein (1879–1955) found that photon has a momentum hν/c [22]. Following Roche [14], we present a general definition of momentum - The momentum of a body will be defined as the capability to generate an impulse as it stops. Note that radiation impulses act during absorption. √ For a relativistic particle moving with a speed v (γ = 1/ 1 − v2 /c2 ), the impulse on the target is related to the inertial force, F (= ma) by J=

Z t 0

Fdt 0 = −

Z t 0

γ 3 madt 0

Energy, Mass, Momentum

=−

Z 0

9

γ 3 mdv

v

=−

Z 0

d (γmv) = γmv.

(1.13)

v

It is understood that the body and the target move with an infinitesimal velocity after the impact. Let us apply the definition of electromagnetic momentum: if a electromagnetic pulse is absorbed, then during a short time interval ∆t, impulse generated is Z 1 ∂ (B × E)dV ∆t F∆t = ∆J = − ∂t c2 µ0

=−

∴

J=

∂ PEM ∆t = −∆PEM . ∂t

Z t 0

∆J = −

Z 0 PEM

dPEM = PEM .

(1.14)

Total field momentum prior to absorption is impulse generated during absorption. Suppose a charge is at rest, next to the field from an electric circuit or magnet. If the current is suddenly made zero, then an impulse acts on the charge: J=

Z t

qEdt = −

0

= −q

Z t ∂A

q

0

Z 0

dA = qA

∂t

dt

(1.15)

A

where A is the vector potential. We see that the definition of momentum proposed by Roche gives all the cases satisfactorily.

2

Chapter

Kinematics, Newton’s Laws of Motion I do not know what I may appear to the world, but to myself I seem to have been only like a boy playing at the edge of the sea-shore, and diverting myself now and then finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me. — Isaac Newton

We have seen that familiar concepts like energy, mass, and momentum have witnessed interesting and rich historical development. Now, we are ready to apply these in relation with some physical situations. For motion in one or more dimensions, we know the relations between the distance travelled, velocity, acceleration, and time. We shall revise these concepts and equations as we apply them to a situation in traffic, calculating the limits of high jump of human beings and animals, etc.

2.1 Whether to Stop or Run Through? We are approaching a road intersection, the traffic light is green and we keep up the speed thinking that we will be through. Before that happens, the stop-light turns yellow. We are now forced to make a decision to either stop or quickly run through before the light turns red [23]. Using the physical laws involving the kinematics of linear acceleration, we will suggest the range of safe speeds. Let the initial vehicle speed be v0 ; the initial distance of the vehicle from the intersection be d0 ; duration of yellow light be T ; width of intersection be s; ability of the vehicle to accelerate (decelerate) be a+ (a− ). We need to integrate x¨ = a with the condition that either (a) displacement (d0 + s) is achieved in time T or (b) zero speed be achieved in less than distance, d0 . The integration of the equation of motion give us the following conditions: 1 to run through d0 ≤ v0 T + a+ T 2 − s 2 to stop d0 ≥

v20 . 2a−

(2.1)

Kinematics, Newton’s Laws of Motion

11

Let us represent the situation in terms of a distance-time graph. To do so, we need to give a range of possible values to all the relevant quantities: • Vehicle speed, v0 (20 km/h - 80 km/h) • Distance to intersection, d0 (7 m - 50 m) • Intersection width, s (7 m - 20 m) • Yellow light interval, T (2 s - 4s) • Acceleration or deceleration, a± (1 m/s2 - 3 m/s2 )

The distance-time graph (Fig. 2.1) schematically shows the possibilities of legal stop or run through (L) or an illegal path (I). The marked rectangle is the region to be avoided for legal driving.

Fig. 2.1

A schematic distance-time graph for a vehicle approaching a stoplight. The shaded portion shows the region that will correspond to an illegal choice of speeding.

To make the conditions in Eq. (2.1) universal, we can re-define the dimensionless variables as follows: (i) dimensionless speed, σ0 = v0 /(s/T ), (ii) dimensionless distance, δ0 = d0 /s, (iii) dimensionless acceleration, α ± = a± /(s/T 2 ), dimensionless time, τ = t/T . The equations now become to run through δ0 ≤ σ0 +

to stop δ0 ≥

σ02 . 2α −

α+ −1 2 (2.2)

12

Mechanics, Waves and Thermodynamics

Exercise: legally.

Draw the distance-speed graph and discuss the region which would be allowed

Problem A point moving with uniform acceleration describes distances s1 , s2 in successive intervals t1 , t2 . Find the acceleration. Distance travelled is: 1 s1 = u1t1 + at12 , 2 1 s2 = u2t2 + at22 , 2 with u2 = u1 + at1 . Substituting this in the above equation, and multiplying the equation for s1 (s2 ) by t2 (t1 ) and subtracting them, we find a=

2(s2t1 − s1t2 ) . t1t2 (t1 + t2 )

Problem A point mass moves from rest with a uniform acceleration. Show that in any time interval, the space-average of the velocity is 4/3 of its time-average. Why? Given the initial speed, u = 0. The final speed is v = at, and also we have v√2 = 2ax, x = 1 2 2 at . In an interval T , let the point mass travel a distance, L. Clearly, T = 2L/a. The time-average of the speed is 1 hvit = T

=

Z T 0

1 vdt = T

Z T

at dt 0

p aT = aL/2. 2

The space-average of v is hvis =

1 L

Z L

vdx =

0

1 L

Z L√

2ax dx 0

√ 2 2aL = . 3 We see that √ r 4 2 2a 2 hvis = hvit = hvit . 3 a 3 The reader is invited to reflect on this finding.

Kinematics, Newton’s Laws of Motion

13

Problem Prove that, if a point moves with a velocity varying as any power not less than unity of its distance from a fixed point which it is approaching, it will take an infinite time to reach that point. Let us consider a particle located at P, approaching O. The distance is x, and with µ > 0, dx/dt = −µxn . Integrating, Z

−n

x dx = −

Z

µdt,

=⇒

1 = µt + c. (n − 1)xn−1

If n > 1, then t → ∞ as x → 0. When n = 1, log x = −µt − c1 . Hence, as x → 0, log x → −∞ which implies that t → ∞. Problem F∝

Is the following a valid law of force (with symbols having usual interpretation): m1 + m2 rˆ12 ? m1 − m2

(2.3)

No. On interchanging ‘1’ and ‘2’, we do not get an equal and opposite force (violates Newton’s third law).

2.2 Vertical Jump How high can we jump? We restrict our discussion to a “standing vertical jump”. An equation for the height of jump is derived from (a) kinematic equations and Newton’s laws of motion, (b) the principle of conservation of energy applied to potential and kinetic energies at two positions of the jump. Of course, as we are interested in the jumps of living beings, aspects related to the physiology do play an important role. Could one jump higher if legs were longer, or, does the height of jump depend whether one is in Himalayas or if one were in Varanasi or Utrecht or in Accra? How do we compare jumps of a man with those of a grasshopper or a kangaroo? Height equation

With an initial velocity, v0 and final velocity, v f , a uniform acceleration, a, the distance d covered in time t, we have the following kinematic equations: v f = v0 + at, v2f = v20 + 2ad.

(2.4)

Total distance over which Centre of Gravity (C.G.) can displace can be divided into two segments (Fig. 2.2) - a stretching segment, S and a free flight path H. (Stretching segment, S extends from the beginning of the crouched position to the erect position before contact with ground is lost.) For stretching part, the acceleration, a=

average net upward force on the jumper, Fn . mass, m

14

Mechanics, Waves and Thermodynamics

As v0 = 0, the final velocity at the end of segments, v j0 is given by v2j0 =

2Fn S, m

(2.5)

where Fn = Fr − mg,

(2.6)

Fr being the average reaction force of the ground on jumper. For the free flight path, final velocity at the highest position is zero. The acceleration is −g and v0 = v j0 . Then, 0 = v2j0 − 2gH.

(2.7)

From Eqs (2.5) and (2.7), we get H=

Fn S . mg

(2.8)

2.2.1 Height equation: from conservation law The equation for height can also be calculated employing the principle of conservation of energy [38]. Take the crouched position as zero reference potential (Fig. 2.2). Work done on the jumper by the floor during the push off period is theRsum of change in potential energy, mgS, and, kinetic energy imparted at position, S (= 0S dS0 Fn ). At the top of the jump, kinetic energy is zero, and potential energy is mg(H + S). Conservation of energy gives

Fig. 2.2

Each of the three figures show a human preparing to jump. The position of F(oot), S(hin), T(high), and B(ack) are taken from realistic data (not drawn to scale). The filled dot marks the Centre of Gravity which goes higher by the height of the jump.

Kinematics, Newton’s Laws of Motion

mg(H + S) = mgS +

Z S 0

RS

hence,

H=

0

15

Fn dS0 ,

Fn dS0 . mg

(2.9)

This is valid even for a non-uniform acceleration. Let us make a typical calculation for a man of mass 70 kg. Average reaction force during take off is typically equivalent to 150 kg. Taking S as 1.4 ft or 42 cm, we get H=

(150 − 70) × 42 cm = 48 cm(1.6 ft). 70

The jump is about one-third to one-fourth of the total height. 2.2.2 Jumps of animals Why grasshopper (and other smaller insects) jumps much higher than its size? There are various biological reasons for this, let us list the major ones: (i) a grasshopper takes (1/30)th of a second to take off whereas a man takes one-half second; (ii) a grasshopper has longer hind legs; (iii) angle between S (shin) and T (thigh) is obtuse for a man whereas it is (acutely) acute for a grasshopper; (iv) a grasshopper has an extensor muscle (which straightens the leg) which is larger than its flexor muscle (which bends it). It can lift off the ground a weight ten times its own and develops a tension of about 250 times its weight. The biological constraints, developed over the varied habitat over a long evolutionary period, are such that a good athlete can jump one-third of her height, and, a 5 ft kangaroo can jump 8 ft. Let us present a scaling argument to explain why typically all the smaller insects can jump much higher than their own dimensions. Assume that the strength of an animal to exert a force F that is proportional to cross-sectional area of its muscular tissue A which, in turn, is proportional to L2 (L being typical dimension). Mass of the animal is proportional to L3 (assuming constant density). Acceleration is F/m ∝ L−1 . As S ∼ L, by Eq. (2.8), H is independent of L (however, inversely proportional to g). Athletics

The spectacular sight of a high jumper performing at the Olympic games (for instance) can be summarized into an anxious urgence of the start-up run, gracious take-off leading to the body gliding over the bar, landing on the soft mattress. Similarly, marvellous sights can be witnessed while watching the long-jumpers, triple jumpers, pole-vaulters, and other athletes. To deal with the problems precisely, we will have to invoke rigid body mechanics (although we know well enough that our bodies are not entirely rigid). For the present, we will content ourselves by reducing the motion of our body to the motion of its centre of mass (CM).

16

Mechanics, Waves and Thermodynamics

So let us take ourselves on a run over a horizontal plane where after running a certain distance, the horizontal speed is v0 and we launch ourselves at an angle θ0 w.r.t. the direction of the run. Our final position of landing is such that the CM has dropped vertically through a height H and it is at a horizontal distance D from the point of launch. Show that " # s v20 sin 2θ0 2gH 1+ 1+ 2 2 D= . (2.10) 2g v0 sin θ0 Show that for the event of long jump, the maximum range occurs for θ0 approximately 25◦ .

2.3 Hourglass Grains of sand carry time with them. No matter how the net force on a particular grain of sand varies as it slides, falls freely, and then hits the bottom, the change in momentum is zero, since it starts from rest and it comes to rest. Hence, the time average of the net force experienced by each grain must therefore be just its weight (Fig. 2.3).

Fig. 2.3

Grains of sand fall freely under gravity and hit the bottom, exerting pressure derivable from Newton’s second law.

In the steady state, the weight of the column of falling sand is about the same as if it were not falling. For the idealized case when the sand starts at rest and drops through a fixed distance, this can be shown as follows. Let n be the number of grains of sand falling per unit time, m be the mass of a grain, v be the speed acquired by falling, g be the acceleration due to gravity, T be the time required for a grain to fall, and mv be the momentum acquired by a grain. Then, nmv is the momentum lost per unit time by the sand striking the bottom which, in turn, is the average force exerted by the falling sand due to its momentum. On the other hand, number of grains falling freely at any time is nT . Weight of the column of falling sand is (nT )(mg). However, v = gT . So, nmv = nT mg. Therefore, force due to impacting sand = weight of the falling column

Kinematics, Newton’s Laws of Motion

17

The amount by which the LHS increases the weight of the hourglass, RHS decreases the weight by an equal amount. This is a beautiful illustration of Newton’s second law: Force = rate of change of momentum.

2.4 Motion of a Chain in a Tube Consider the rectilinear motion of a uniform chain [36]. Suppose, that a chain of length ` and mass density µ per unit length is placed in a tube (e.g., a gold chain in a glass tube, with no friction), ABC. The tube is bent at a right angle at B with AB (horizontal) and BC (vertically downwards) as straight parts (Fig. 2.4). At time t, a length of the chain in the column, BC is denoted by x and (` − x) is in AB. Let v be the velocity at time t, and let T denote the chain tension at B. For the horizontal part, tension will be the rate of change of linear momentum in the horizontal part: µ (` − x )

Fig. 2.4

dv dv dv = µ (` − x ) = µ (` − x )v = T . dt dx/v dx

(2.11)

A chain of length ` inside a tube is falling under gravity. p The relation between velocity v and the length x in the column BC follows v ∼ g/`x (2.13)

For the vertical part: µxv

dv = µgx − T dx

(2.12)

Adding Eqs (2.11) and (2.12), µ`v

dv = µgx, or, `v2 = gx2 + C dx

where C is given by the initial conditions.

(2.13)

18

Mechanics, Waves and Thermodynamics

Problem A uniform fine chain of length ` and linear mass density, µ is suspended with its lower end just touching a horizontal table, and allowed to fall. Find the pressure on the table when a length x has reached it. When the velocity is v, a length vδt is brought to rest in a short interval of time, δt. So, a mass µvδt is brought to rest and therefore the amount of momentum destroyed is µv2 δt in time δt. Hence, the pressure on the table due to the rate of destruction of momentum is µv2 . However, since the upper end falls freely, the velocity when a length x has reached the table is given by v2 = 2gx. Since, the weight on the table is µgx, the total pressure on the table is 3µgx. This holds good so long as x < `. As soon as x = `, the force due to impacts ceases and the pressure becomes µg` (Fig. 2.5).

Fig. 2.5

Pressure, P due to a falling chain as a function of the length x reaching the table. The first derivative of pressure falls discontinuously to one-third of its value as soon as the entire length rests on the table.

Falling of chain requires application of momentum balance to variable-mass systems. We would like to draw the attention of readers to a particularly interesting exposition of these systems, as a project for further study [37]. Problem A rope of mass M and length L is hanging at rest from a fixed end, under the influence of gravity. The linear mass density along the height (z-axis) is µ (z) = µ0 sin(πz/L) where µ0 is a constant. Determine µ0 and tension along the rope. Knowing that M=

Z L 0

µ (z)dz =

2Lµ0 , π

hence, giving the constant µ0 = πM/2L.

(2.14)

Kinematics, Newton’s Laws of Motion

19

As a function of the length, tension is given by T (z) = g

Z z 0

πz Mg 1 + cos . µ (z )dz = 2 L 0

0

(2.15)

Problem A car is naturally sliding down an inclined plane that makes an angle α with the horizontal. A ball is thrown out in a direction perpendicular to the incline from the moving car in the same vertical plane as that of car’s motion. Will the ball return to the car? Let us analyze the motion of the ball in the rest frame of the car. This frame is accelerating with respect to the inertial frame attached to the sloping surface of the incline, with acceleration g sin α in the direction of motion of the car. In this frame, the ball experiences a pseudo-force mg sin α in the direction opposite to the car’s motion. This is equivalent to the situation in which the ball is thrown vertically from the level ground, with the ball’s weight mg replaced by mg cos α. Hence, the ball will return to the car.

2.5 Forces At this moment of our understanding of nature, there are following interactions: gravitational, electro-weak, and strong. Here, we try to become slightly familiar with each of these. 2.5.1 Gravitation We have considered, for most part, point particles with some mass. Earth is not a point, and still, Moon goes around it along an orbit that is described by assuming that both are point objects. Let us calculate the force due to Earth assuming it to be a solid sphere of radius R and mass, Me . First of all, let us observe that if a mass Mm revolves around the Earth, the acceleration is independent of Mm . The force on Moon, at a distance r, due to Earth is Fm = −

GMe Mm rˆem r2

(2.16)

where G is the gravitational constant, 6.672 × 10−11 Nm2 kg−2 , and, rˆem is the unit vector in the direction from Earth to Moon. The gravitational force will accelerate the Moon by changing its state which, in turn, will depend on its mass as appearing in the Newton’s law. Let us denote this mass by Mminertial . The force in Eq. (2.16) is equated to Mminertial am . If Mm = Mminertial , then am is independent of Mminertial . The equality of mass has been found to be true experimentally upto a great accuracy. This is unexpected and surprising, as we don’t understand it completely.

20

Mechanics, Waves and Thermodynamics

The acceleration due to gravity on an object on the surface of the Earth is (radius of the Earth being Re ∼ 6 × 106 m) 6.672 × 10−11 .(6 × 1024 ) GMe g= 2 ∼ ∼ 10 m/s2 . (2.17) Re (6 × 106 )2 Problem Calculate the (a) gravitational force between two balls of mass 60 kg at a distance of 1 m from each other, (b) the distance between the Earth (Me = 6 × 1024 kg) and the Moon (Mm = 7 × 1022 kg), knowing that the force between them is 1020 N, (c) the distance between the Earth and the Sun (MS = 2 × 1030 kg), knowing that the force between them is 1022 N. Let us now calculate the force of a spherical body on another mass m. The strategy is to divide the sphere into thin spherical shells, thin spherical shells into rings, and begin with the gravitational pull of the ring.

Fig. 2.6

To calculate the gravitational force due to a spherical body on a mass m at a distance r from the centre of the sphere, we may divide the sphere into spherical shells, shells into rings, and eventually add all the contribution by integrating. Inside the sphere, the force turns out to be zero.

Let us consider a ring at an angle θ (Fig. 2.6). At the centre of the sphere, this subtends an angle dθ and has a radius R sin θ . The width of the ring will be Rdθ . Taking a small thickness t ( R), the mass of the ring is the product of the density, ρ = M/(t4πR2 )), and the volume, t2πR2 sin θ dθ . The mass m sees the ring of mass M sin θ dθ /2 at a distance away from the line joining the centre of the sphere and the mass in a symmetrical manner. Hence, all the transverse components cancel. Denoting the distance of m from the ring by s, and by α the angle between the direction of the force and the line joining the centres, the gravitational pull of the ring on m is Gm(M sin θ dθ /2) cos α/s2 . The total force is F=

Z

GmM sin θ dθ . 2s2

(2.18)

Kinematics, Newton’s Laws of Motion

Making the transformation, s = F=

21

√ R2 + r2 − 2rR cos θ , and, t = r − R cos θ

GMm I (r ). 2R

(2.19)

We now recall a standard result: I (r ) =

Z r +R

dt r−R

1 = 2 2r

t 3

(R2 − r2 + 2rt ) 2

r+R p r2 − R2 2 2 . R − r + 2rt − √ R2 − r2 + 2rt r−R

(2.20)

With this, F=

GMm , r2

= 0,

r > R, r < R.

(2.21)

The force is as if the entire mass of the sphere is concentrated at its centre. There is a gravitational field due to a mass M at a point r, (GM/r3 )r. The systematic treatment of the fields is the subject of classical and quantum field theory. 2.5.2 Electro-weak interaction Besides mass, another fundamental attribute of a particle is its charge, q. Unlike masses, charges could be positive or negative. There are two different kinds of charges, positive and negative - like charges repel, and, unlike charges attract. Two charges q1 and q2 at a distance r from each other experience an inverse-square force along the direction connecting the two charges: Fe = k

q1 q2 rˆ = q1 E r2

(2.22)

1 with ε0 is free space dielectric permittivity. where k is the constant of proportionality, 4πε 0 This is the Coulomb’s law. In the above equation, E is the electric field. It has been shown by Glashow, Salam, and Weinberg that the electromagnetic and weak interactions have the same origin. Weak interactions are important to explain the decay of neutron by what is called the β -decay.

2.5.3 Strong interaction From the alpha scattering experiment, it became known that atom has in its centre a positively charged nucleus. It was later realized by Rutherford that nucleus has protons which are about 1840 times more massive than the electrons and are positively charged. Furthermore, Chadwick discovered that there are other particles inside nuclei — called

22

Mechanics, Waves and Thermodynamics

neutrons. These are neutral, and about the same mass as protons. However, there is an important question that poses itself - how do like charged particles remain bound in a region of the size of the order of Fermi (10−15 m). The repulsive force between two protons at a distance of one Fermi is 9 × 109 × 1 × 1 ' 1040 N! 10−15 × 10−15

(2.23)

The nuclear force has to be attractive such that it can overcome this repulsion. Hence, a new force is required. This is a milestone in our understanding of nature. The range of this force is of the order of Fermis. Compare, this with gravitational and electromagnetic interactions which are both infinite ranged. The forces are mediated by a certain particle. Electromagnetic interaction is mediated by a photon (of zero mass). For instance, an electron exchanges a photon with another electron. The range of the force is inversely proportional to the mass of the exchange particle. The Japanese physicist, Hideki Yukawa proposed that the nuclear force is mediated by another particle with mass of about 140 MeV. This particle was named as meson by the Indian physicist, Homi Jehangir Bhabha.

3

Chapter

Circular Motion All you get from a circular argument is dizzy. — Darrin Bell

What is the force required to keep a body moving along a circle at a constant speed? What is the acceleration of a body moving along a circle of radius R, at a constant speed v? Clearly, one can see from the Fig. 3.1 that the velocity is not constant, it is directed upwards at 1 and downwards at 2. So, by the definition of acceleration, the body is accelerating. The average acceleration over half a circle is given by average acceleration =

change in velocity . time interval

Change in velocity = v2 − v1 = ↓ − ↑. The magnitude of v2 − v1 is 2v, and its direction is down (↓). Time taken to traverse half the circle is πR/v. Hence, the average acceleration has the magnitude, 2 v2 2v = . πR/v π R

(3.1)

This is a simple way to see how v2 /R appears in circular motion. This finds a more natural description in terms of radial and angular coordinates (r, θ ), called the polar coordinates instead of Cartesian coordinates.

3.1 Cartesian vs Polar Coordinates The position of a particle moving in the xy-plane can be written in terms of unit vectors ˆi, ˆj along x and y directions: r = r ˆi cos ωt + ˆj sin ωt with r and ω as constants. Let us find the trajectory, the velocity, and the acceleration? The trajectory r (t ) is given by |r|(t ) =

p

r2 cos2 ωt + r2 sin2 ωt = r (t ) = constant,

24

Mechanics, Waves and Thermodynamics

which is a circle (Fig. 3.2). The particle moves along anti-clockwise direction around the circle, starting from (r, 0) at time, t = 0. It traverses the circle in time T such that ωT = 2π; ω is called the angular velocity and is measured in radians per second. The time period is T (Fig. 3.1).

Fig. 3.1

In traversing around the circle half-way, the velocity reverses. This presents a simple way to see how the average acceleration comes proportional to square of speed divided by the radius.

The linear velocity is v=

dr = rω −ˆi sin ωt + ˆj cos ωt . dt

Note that r.v = 0, so r ⊥ v, or v is tangent to the trajectory. Hence, v = |v| = rω. Instantaneous acceleration is a=

dv = −ω 2 r. dt

Thus, a k r and hence it is radially directed. This is the centripetal acceleration.

Fig. 3.2

Polar coordinates and the corresponding mesh.

Circular Motion

25

For uniform or non-uniform acceleration, a(t ), we can generally integrate the equation of motion (EOM) to get the velocity at time t, starting with velocity v0 at time t = t0 : v(t ) = v0 +

Z t

a(t 0 )dt 0 .

t0

Recall that, for uniform acceleration 1 r(t ) = r0 + v0t + at 2 . 2 ˆ and with the usual notation, e.g., in gravitational field, the uniform acceleration is a = −gk, the following well-known equations are written: x = x0 + v0xt, y = y0 + v0yt, z = z0 + v0zt − 12 gt 2 . With r0 = 0 at t = 0, we can see that z is a parabolic function of x: z=

v0z g x − 2 x2 . v0x 2v0x

(3.2)

This is what we observe - a parabolic path. Polar coordinates If the point P is in the x-y plane (Fig. 3.2), p y r = x2 + y2 , θ = tan−1 . x

(3.3)

The mutually perpendicular unit vectors (ˆi, ˆj) change to mutually perpendicular (rˆ , θˆ ) by a rotation through an angle θ : rˆ = ˆi cos θ + ˆj sin θ , θˆ = −ˆi sin θ + ˆj cos θ .

(3.4)

Position vector of points along a circular path, r = xˆi + yˆj becomes r = rˆr in plane polar coordinates. Note that rˆ .θˆ = 0. Instantaneous velocity is v=

=

dr d = (rˆr) dt dt d rˆ dr rˆ + r . dt dt

(3.5)

26

Mechanics, Waves and Thermodynamics

The first term is obviously radially directed, what about the second? The second term can be shown to be angular: dθ ˆ dθ dθ ˆ d rˆ ˆ = i(− sin θ ) + j(cos θ ) = θ. dt dt dt dt

(3.6)

Hence, v=

dθ dr rˆ + r θˆ . dt dt

(3.7)

Also, d dθˆ = −ˆi sin θ + ˆj cos θ dt dt

= −ˆr

dθ . dt

(3.8)

The acceleration in polar coordinates is dv d a= rˆ ˙ r + rθ˙ θˆ = dt dt = r¨ − rθ˙ 2 rˆ + 2r˙θ˙ + rθ¨ θˆ .

(3.9)

This is an involved expression, let us interpret each term. The term, r¨ gives linear acceleration, rθ˙ 2 gives centripetal acceleration (directed towards the centre), 2r˙ θ˙ is Coriolis term (perpendicular to rˆ ), and rθ¨ is the linear acceleration in the tangential, (θˆ -) direction. That is why when you are trying to walk radially from the centre of a carousel, you tend to fall sideways. Problem If the radial velocity of a particle is proportional to its transverse velocity, then the path of a particle is an equiangular spiral. With k as a proportionality constant, r˙ = krθ˙ . That is, dr/r = kdθ . Integrating, we get r = cekθ - an equiangular spiral. Problem Suppose the particle considered in the above problem describing r = cekθ is such that its acceleration has no radial component. Prove that its angular velocity is constant and that the magnitude of both the velocity and acceleration are proportional to r. We know that r¨ − rθ˙ 2 = 0. From the equation of equiangular curve, r˙ = ceθ θ˙ = rθ˙ , and, r¨ = r˙ θ˙ + rθ¨ = rθ˙ 2 + rθ¨ . Therefore, rθ¨ = r¨ − rθ˙ 2 = 0. Hence, θ¨ = 0. This implies θ˙ =constant, K. Then, r˙ = Kr. So, v2 = r˙ 2 + (rθ˙ )2 = 2K 2 r2 . Therefore, v varies as r. As radial acceleration is zero, we have non-zero transverse acceleration = 2r˙ θ˙ + rθ¨ = 2r˙ θ˙ = 2K 2 r. Hence, acceleration varies as r.

Circular Motion

27

Problem A point is moving in a circle. Show that at any instant its angular velocity about a point on the circumference is half the angular velocity about its centre. Hint Angle subtended by an arc on the circumference is half the angle subtended at the centre. Problem On a frictionless, horizontal table, there is a hole at the centre. A body 1 of mass m1 is connected to another body 2 of mass m2 by an inextensible string of length `, passing through the hole. While the body 1 is moving along a circle of radius r0 with a constant angular speed ω0 , the body 2 is held at a vertical distance z below the surface of the table. What is the minimum angular speed so that the body 2 moves upwards when we release it. The only force acting on 1 is the tension T whereas on 2, there is gravity also. The Newton’s equations can be written: (radial motion of 1) − T = m1 r¨ − rθ˙ 2 0 = m1 rθ¨ + 2r˙ θ˙

(tangential motion of 1) (vertical motion of 2)

m2 g − T = m2 z¨.

(3.10)

As r + z = `, r¨ = −¨z. Simple manipulation yields the acceleration, z¨ =

m2 g − m1 r02 ω02 m1 + m2

Hence, if m1 r0 ω02 > m2 g, i.e., if r m2 g ω0 > , m1 r0

(3.11)

(3.12)

then the body 2 moves upward on release.

3.2 Coriolis Force A spinning dancer or an ice skater knows that the spinning speed increases as the hands are drawn in. For those who are neither dancers nor skaters (like me), we have watched this. There are real forces involved in moving the arms inward or outward, however, these are radial forces. Radial forces cannot generate a torque and hence, cannot explain change in angular velocity. The force that is acting is a tangential force which is a fictitious force called as the Coriolis force . Let us now understand this by a simple mathematical argument [26]. Consider two equal masses m1 and m2 connected by a relatively massless rigid rod. Each mass is at a distance r from the axis of rotation. The total mass is M = m1 + m2 and the angular velocity

28

Mechanics, Waves and Thermodynamics

be ω. The angular momentum is L = Iω = Mr2 ω. The angular momentum is conserved, hence, L˙ = 0. The rate of change of angular momentum is ˙ L˙ = I ω˙ + Iω

= Mr2

dω dr + 2Mrω ; dt dt

(3.13)

this pushes the masses outward. The tangential velocity and acceleration is given by vt = rω and at = rdω/dt w.r.t. the frame of reference rotating at constant ω. With Mdr/dt = pr , the radial momentum, L˙ = rMat + r2pr ω.

(3.14)

Now, conservation of angular momentum implies that Mat = −2pr ω.

(3.15)

We see that the dancer and skater experience a tangential acceleration caused by the force (−2pr ω ) - the Coriolis force.

3.3 Rotation Group The Eq. (3.4) effects a rotation and can be re-written in a matrix form as follows: " # " # ˆi rˆ = R(θ ) ˆj θˆ

(3.16)

where " R(θ ) =

cos(θ )

sin(θ )

− sin(θ ) cos(θ )

# (3.17)

is a rotation matrix. It can be easily verified that R(θ ) R(β ) = R(θ + β ). With this, it also becomes clear that (i) β = −θ will give us no rotation, so R(−θ ) is the inverse of R(θ ) with respect to matrix multiplication, if (ii) R(0) defines a unique identity. Consider all the 2 × 2 matrices with respect to matrix multiplication. Product of any two results in a matrix of the same kind. Hence, the set of matrices considered is closed under this binary operation. The matrices also obey the associative law w.r.t. to multiplication. Due to the existence of unique identity and a unique inverse for each element, this set constitutes a group. This is called the rotation group. Due to the fact that the matrices are orthogonal, i.e., transpose is equal to the inverse, the rotations make an orthogonal group O(2). Furthermore, as the determinant of each matrix is unity, the group is called the Special Orthogonal group, SO(2).

4

Chapter

The Principle of Least Action For since the fabric of the universe is most perfect and the work of a most wise Creator, nothing at all takes place in the universe in which some rule of maximum or minimum does not appear. — Leonhard Euler

Approach to physics through general principles is superior to particular formulations. Here we discuss one such principle which provides the governing equations and laws in mechanics, waves, and even in understanding electric circuits [27]. This is the principle of least action. As an application from optics, this principle was employed to understand the transmission of radio signals between two places on the Earth via ionosphere.

4.1 Action of the Principle 4.1.1 Mechanics Systems where the total energy can be written in terms of kinetic energy, T and potential energy, V , with an interchange between these two forms as the motion progresses, the motion of a particle from a position s1 at time t1 to s2 at time t2 is described in terms of the action, defined by S=

Z s2

mvds =

s1

Z t2

mv(vdt ) =

t1

Z t2

2T dt.

(4.1)

t1

Mass of the particle is m, and v is its speed. In going along this path, the energy spent is minimum. The integral above is also a minimum for the path followed, out of all possible paths one might imagine. Denoting the variation of the value of the integral by δ as the path is slightly changed (varied), we can write the Maupertius-Euler principle of least action as a statement embodying the fact that this variation is minimum (or extremum) by setting Z s2

δ s1

mvds = δ

Z t2

T dt = 0.

(4.2)

t1

During the motion, the interchange between the values of T and V occurs in a way so that the integral of the difference between the two is a minimum. The difference, (T − V )

30

Mechanics, Waves and Thermodynamics

is called the kinetic potential or the Lagrangian. For this case, we have the Hamilton’s principle of least action, Z t2

δ

(T −V )dt = 0.

(4.3)

t1

Further, if a system with total energy E loses an energy of amount L and gains an amount G, then the sum, (T + V ) equals (E − L + G). In this case, the Hamilton’s principle in a generalized form is Z t2

δ

(T −V − L + G)dt = 0.

(4.4)

t1

4.1.2 Electric circuits Consider a circuit where current I is divided to pass through parallel branches, each containing a resistor (Fig. 4.1). The current flowing through the branches, I1 and I2 through the resistors R1 and R2 respectively, sums to I: I = I1 + I2 .

Fig. 4.1

(4.5)

Current is divided in two branches with resistances R1 and R2 . Applying the principle of least action in an appropriate form leads to Kirchhoff’s law.

In each branch, heat developed is I12 R1t and I22 R2t, summing to total heat, Q = I12 R1t + I22 R2t

= I12 R1t + (I − I1 )2 R2t.

(4.6)

The Principle of Least Action

31

The current will divide (equivalent to possible “paths”) such that Q is minimized. Applying the principle of least action, 0=

∂Q ∂ R1

= 2I1 R1 − 2 (I − I1 ) R2 ; =⇒ I1 R1 = I2 R2 ,

(4.7)

the Kirchoff’s law. 4.1.3 Optics The oldest version of the principle of least action comes up when applying it to the propagation of light. Suppose E is the energy of the radiation being transported at a constant rate in an interval between t1 and t2 . Then the principle of least action may be written as 0 = δ

Z t2 t1

= Eδ

E dt = E δ

Z t2

dt t1

Z s2 ds s1

c

(4.8)

where c is the speed of light in the medium. This is precisely the Fermat’s principle of least time. Let us employ this principle to obtain the Snell’s law of refraction of light as it propagates from a point A in one medium to a point B in another (Fig. 4.2). Referring to the figure, the path lengths in the two media are denoted by s1 and s2 . The total path length is q q 2 2 (4.9) s = s1 + s2 = y1 + x + y22 + (d − x)2 . Total time taken to traverse this path is q q y21 + x2 y22 + (d − x)2 t= + v1 v2

(4.10)

where v1 , v2 are velocities of light in the two media. To find the minimum time light takes to travel as the medium changes, we need to set dt/dx to zero: d −x x 1 1 q =q , v v y21 + x2 1 y22 + (d − x)2 2

implying that the Snell’s law.

sin i sin r = , v1 v2

(4.11)

32

Mechanics, Waves and Thermodynamics

If the medium remains the same, then the angles will be equal. If we place a perfectly reflecting mirror at the horizontal plane passing through O, then light will reflect such that the angle of reflection is equal to the angle of incidence as that is the straight line connecting the reflected point in the absence of the mirror.

Fig. 4.2

As a ray of light enters a denser medium, 2 from 1, it bends towards the normal. The angles of incidence and refraction are respectively i and r. The vertical distance of the point A (B) from where a ray ensues (terminates) is y1 (y2 ) from the plane CD across which the medium changes.

4.2 The Principle of Least Action Before Newton proposed his law of motion, the second law: F = ma,

(4.12)

Fermat had proposed the principle of least time discussed above. In Eq. (4.12), the Left Hand Side (LHS) is the applied force and Right Hand Side (RHS) is the inertial term which describes the manifestation of the force on an object of mass m. We would like to employ the Hamilton’s principle to derive the possible equation of motion. Consider the motion of a mass from a point 1 to 2 under some force field which is derivable from a potential V (x). Let us further imagine that there are several independent paths that the mass could follow. One of the paths is according to the solution of Eq. (4.12). The principle of least action states that the solution of Eq. (4.12) gives a path for which the quantity, action defined as S=

Z t2 t1

[T (t ) −V (t )]dt,

(4.13)

The Principle of Least Action

33

is minimum. Hence, we want to find the curve connecting 1 to 2 for which the action is minimum. This problem belongs to the subject of calculus of variations. Here, we will try to solve this for the case of one space dimension, the generalization is immediate. Let us call the path resulting from the solution of the second law by xN (t ), and the corresponding action be SN . Further, let us denote the path by x(t ) and the corresponding action by S. We assume that the path x(t ) = xN (t ) + η (t ) where η (t ) is a small variation. The paths are from 1 to 2 - fixed points where η (t ) is zero. The action is " # Z m dx 2 S= −V (x) dt. (4.14) 2 dt Substituting x(t ) = xN (t ) + η (t ) in Eq. (4.14), # " Z t2 m dxN dη 2 + −V (xN + η ) dt S = 2 dt dt t1 "

Z t2 m dxN 2

=

2

t1

dt

dxN dη m +m + dt dt 2

dη dt

#

2

−V (xN + η ) dt.

(4.15)

The term (dη/dt )2 is much smaller than the other terms, so we ignore it. As η is small, V (x + η ) can be expanded in Taylor series to obtain dV η 2 d 2V V (xN + η ) = V (xN ) + η + + ··· dx x=xN 2 dx2 x=xN

(4.16)

Ignoring the second and higher order terms, "

S=

Z t2 m dxN 2 2

t1

dt

# dV dxN dη −V (xN ) + m −η dt dt dt dx x=xN

(4.17)

The first two terms correspond to the action for the (N)ewtonian path SN . We want to concentrate on the difference between S and SN , which we write as δS =

Z t2 t1

# dV dxN dη dt. −η m dt dt dx x=xN

"

(4.18)

Recalling the integration by parts, we can write Z

dxN dη dt = dt dt

Z

d (xN (t )η (t ))dt − dt

Z

η

d dxN dt. dt dt

(4.19)

34

Mechanics, Waves and Thermodynamics

Hence, t2 Z t Z t2 2 d dx dV dxN N η (t ) − m η (t )dt − δS = m η (t )dt. dt dt t1 dx x=x t1 dt t1 N The first term is zero as η (t ) is zero at t1 ,t2 . So the variation of action is # " Z t2 d 2 xN dV δS = η (t )dt. −m 2 − dt dx x=xN t1 For this to be zero, the integrand has to be zero. That is, d 2 xN dV m 2 =− , dt dx x=xN

(4.20)

(4.21)

(4.22)

which is exactly Newton’s second law.

4.3 More Thoughts on Why “(T – V)”? Consider the path of a particle of mass m going from a region with potential V1 to another region with energy of the particle is E. Velocities in the two regions p potential V2 . The total p are v1 = 2m(E −V1 ) and v2 = 2m(E −V2 ). Following the principle of least action, we want to find the function that will give the desired path upon minimization [29]. The path is just as in Fig. 4.2 showing a refracted path, i.e., in time t, in going from A to B, the path satisfies mv1 sin i = mv2 sin r.

(4.23)

The perpendicular distances of A and B from CD are y1 and y2 respectively, and d is the horizontal distance between them. Maupertius–Euler action as a function of x and E is q q S0 (x, E ) = mv1 (E ) y21 + x2 + mv2 (E ) y22 + (d − x)2 . (4.24) Assuming that we know dE and dx, the variation of S0 is dS0 (x, E ) = (mv1 sin i − mv2 sin r )dx +

∂ S0 dE. ∂E

Note that dv1,2 /dE = 1/(mv1,2 ). we have q q 2 + x2 0 y y22 + (d − x)2 ∂ S (x, E ) 1 = + = t1 + t2 = t. ∂E v1 v2 Eq. (4.25) doesn’t give Eq. (4.23) due to the term (∂ S0 /∂ E )dE.

(4.25)

(4.26)

The Principle of Least Action

35

If Et is subtracted from S0 , the desired quantity is obtained; let us denote this by S. S = S0 − Et = (mv1 `1 − Et1 ) + (mv2 `2 − Et2 ) = mv21 − E t1 + mv22 − E t2 = (T1 −V1 ) t1 + (T2 −V2 ) t2 .

(4.27)

This is the quantity to be minimized. The difference between T and V appears naturally. Hence, we have illustrated in this Chapter that not only Newton’s laws however, but also laws governing the propagation of light or current in circuits are based on one unifying principle.

5

Chapter

Work and Energy The animal frame, though destined to fulfill as many other ends, is as a machine more perfect than the best contrived steam-engine - that is, is capable of more work with the same expenditure of fuel. — James Prescott Joule

As we remarked right in the beginning of the book, energy is a central concept in our understanding of nature. We put any change in perspective with changes in forms of energy. With the equations describing motion and the principle of least action underlying it, we have now come to a stage that we see how we can calculate energy. Using these, we can also find out the work done in a process. And indeed, what is the relationship between work and energy. We shall illustrate all the working through examples. The reader is encouraged to think of new instances or variations and make her (his) own calculations of work and energy.

5.1 At the T-junction You are driving a car at a high speed. Suddenly, at a crossing, a large truck turns into the road in front of you. You have to decide whether to stop or to turn away. To analyze this commonplace situation, we make some simplifying assumptions [24]. Let us assume that the car does not skid. That is, the coefficient of friction is the same as in the direction of motion. Let us also assume that if the driver is not using the force to provide centripetal acceleration to turn, (s)he will use it by putting on the brakes to provide deceleration in the direction of motion. Clearly, there are three choices: (1) to steer straight and apply brakes fully; (2) to turn in a circular arc without applying brakes, using all available force to produce centripetal acceleration; (3) to combine (1) and (2). Simplifying further, let us approximate the truck by a wall and the (avoidable) event as a car heading into a wall on a T-junction as shown in the Fig. 5.1. Denote by ` the original distance of the car from the wall and by v0 the original speed of the car. First, we look at the choice (2). Force required to turn a particle of mass m with speed v0 in a circular arc of radius `, Fr =

mv20 ` .

On the other hand, force required to stop a v2

particle of mass m and speed v0 in a distance `, Fs = ma = m 2`0 . This means that Fr = 2Fs . If the car can be turned in a circular arc without hitting the wall, it can be stopped in only half the distance to the wall.

Work and Energy

37

We will now argue that the proper choice is straight stop, choice (1). Let us consider the direction of original motion of the car as the X-axis, q the car starting from the origin. It is acted upon by a force F with magnitude F = |F| = Fx2 + Fy2 at a variable angle θ which is a function of x. We want to find θ (x) such that the particle travels along a path, y = y(x) and stop (vx = 0) within a distance s, measured along X-direction. This distance s can be determined by equating the work done on the particle to the change in kinetic energy: Z s 0

1 1 Fx dx = mv2s − mv20 . 2 2

Fig. 5.1

(5.1)

A car is approaching a T-junction, avoiding to collide. Should it turn or apply brakes?

At x = s, vs = 0, and we want to find θ (x) for which s is minimum in Z s 0

1 F cos θ (x)dx = − mv20 . 2

(5.2)

We re-cast the problem to avoid dealing with a variable upper limit s. We take the equation Z ` 0

1 F cos θ (x)dx = m v2` − v20 2

(5.3)

where we find θ (x) such that (v2` − v20 ) will be a maximum. We can also ask - if v` = 0, find the path so that a car with largest initial speed can be stopped. We determine an extremum (maximum, minimum, or a stationary value) for the integral by demanding Z `

δ 0

F cos θ (x)dx = 0.

(5.4)

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Mechanics, Waves and Thermodynamics

As F is constant, Z `

δ

cos θ (x)dx = 0

0

Z `

implying 0

δ (cos θ (x))dx = 0 =

Z `

(− sin θ )δ θ (x)dx = 0.

(5.5)

0

With δ θ (x) an arbitrary function, we arrive at the condition: sin θ (x) = 0, i.e., θ (x) = 0 or π.

(5.6)

This proves that the path is a straight line parallel to the direction of initial velocity. If θ (x) = 0, force is in the same direction as the initial velocity and the car will strike the wall with a maximal velocity (not necessarily the largest velocity). If θ (x) = π, force is opposite to the motion and v` = 0, hence, −F` = 12 mv20 . This establishes the first choice. If the wall is too close for the car to be stopped in the straight path, which path will give the least component of velocity perpendicular to the wall at the time of impact? This means that we fix ` and v0 in Eq. (5.3) where 12 mv20 > F` and find θ (x) for which v` will be minimum. Solution of this problem will be the same as discussed above. Inspite of any natural impulse to turn, the proper thing to do is to apply brakes and go straight. In doing so, the component perpendicular to the X-axis is zero and hence the impact with the wall will be least severe. Related problem Suppose we are driving on a highway at a speed such that the minimum stopping distance is D. What is the width W of an obstacle that can just be dodged by turning to the side in a circular arc? We have seen that the turning radius is twice the stopping distance. Therefore, the angle made at the vertex A, α = 30 degrees (see Fig. 5.2). We see that β = 15 degrees.

Fig. 5.2

The diagram shows that while driving, one can dodge an obstacle by a turn through 15 degrees.

39

Work and Energy

Plot the path of a particle that is acted upon by a force of constant magnitude at a constant angle γ with the path, so that the tangential and normal components of the acceleration are proportional to cos γ and sin γ respectively. Show that the path is an equiangular spiral of the form ρ = ρ0 exp(Kφ ) where K = cot Ψ = 2 cot γ, and Ψ is the constant angle between radial lines from the pole on the graph and the path [35]. (The factor 2 is coming here for the same reason as discussed above.) Remarks

In turning, since the wheels are not skidding sideways, we are concerned with static friction. Force of static friction f and maximum force of static friction Fmax are different. It is only Fmax = (coefficient of friction) × (normal reaction), f < Fmax .

5.2 Motion of a Heavy Particle on a Smooth Curve in a Vertical Plane The diagrams show (Fig. 5.3) components of acceleration and force separately. If m is the mass of the particle, forces are its weight mg and the normal reaction, R. The components v2 of acceleration are v dv ds along the tangent and ρ along the inward normal. By resolving along the tangent, mv

dv dy = −mg sin ψ = −mg . ds ds

(5.7)

Integrating, mv2 = C − mgy. 2

Fig. 5.3

(5.8)

Components of acceleration (left) and force (right) of a particle moving along a curved path are shown.

40

Mechanics, Waves and Thermodynamics

With the condition that v = v0 at y = y0 , we have from energy conservation, 1 m v2 − v20 = mg(y0 − y). 2

(5.9)

Resolving along the normal, mv2 = R − mg cos ψ. ρ Substituting v from (5.9), we get m 2 R = mg cos ψ + v0 + 2g (y0 − y) . ρ

(5.10)

(5.11)

Given the form of the curve, ρ and ψ will be known and hence, R can be written down. If we equate R to zero, we shall have an equation to determine the point, if any, at which the particle leaves the curve.

5.3 Motion of a Heavy Particle, Placed on the Outside of a Smooth Circle in a Vertical Plane and Allowed to Slide Down If a particle starts at Q at an angular distance α from the highest point A (Fig. 5.4), and R is the radius of the circle and v is the velocity at P where the angular distance is θ from A, then v2 = 2gR(cos α − cos θ ).

(5.12)

Resolving along the inward normal, mv2 = mg cos θ − R, R

Fig. 5.4

(5.13)

Motion of a heavy particle moving along a smooth circle in a vertical plane, under gravity.

Work and Energy

41

where R is the outward reaction. Therefore, m R = mg cos θ − 2gR(cos α − cos θ ) R

= mg(3 cos θ − 2 cos α ).

(5.14)

The particle flies off the curve as R vanishes, hence, at θ given by cos θ =

2 cos α. 3

(5.15)

5.4 Motion in a Vertical Plane of a Heavy Particle Attached by a Fine String to a Fixed Point Suppose that a particle starts with velocity u from its lowest position B. If v is the velocity at P and θ is the angle the string makes with the vertical, then the energy is given by (Fig. 5.5) 1 m(v2 − u2 ) = −mgR(1 − cos θ ). 2

(5.16)

By resolving the inward normal, mv2 = T − mg cos θ R

Fig. 5.5

(5.17)

A particle attached to a string of radius R, and moving along a circle in a vertical plane.

42

Mechanics, Waves and Thermodynamics

where T is the tension of the string. Therefore, mu2 − 2mg(1 − cos θ ) R u2 = m 3g cos θ − 2g + . R

T = mg cos θ +

(5.18)

To find the height of ascent, we put v = 0 in Eq. (5.16) and get 2gR cos θ = 2gR − u2

(5.19)

and by putting T = 0 in Eq. (5.18) we find that the tension vanishes when 3gR cos θ = 2gR − u2 .

(5.20)

The following cases are of interest: 1. If u2 < 2gR, the string does not reach the horizontal position and the tension does not vanish; 2. If u2 = 2gR, the string just reaches the horizontal position, the tension vanishes for θ = π/2, and the particle through a quadrant on each side of the vertical; 3. If 2gR < u2 < 5gR, we find that there is a value of θ , an obtuse angle, given by Eq. (5.20) smaller than that given by Eq. (5.19), so that the string becomes slack before the velocity vanishes and the particle will fall away from the circular path and move in a parabola till the string again becomes taut; 4. If u2 = 5gR, the tension just vanishes at the highest position, however, v doesn’t vanish, so that the circular path persists; 5. If u2 > 5gR, neither v nor T vanishes.

5.5 Conservative Force Under the influence of a force F(r), the trajectory of a single particle at the position vector r is obtained on solving the equation, m

d2r = F(r). dt 2

(5.21)

The force is said to be conservative if the potential energy V (r) is such that F(r) = −∇V (r) kˆ where ∇ = ˆi ∂∂x + ˆj ∂∂y + kˆ ∂∂z . The force is written as negative of a gradient operator. Multiplying the above equation by dr/dt, we can write dr d 2 r m d dr 2 dr = .F(r). (5.22) m . 2 = dt dt 2 dt dt dt

Work and Energy

43

Hence, if a particle goes from points (r1 ,t1 ) to (r2 ,t2 ) along some path, we may integrate over time and get " # Z Z (r2 ,t2 ) (r2 ,t2 ) m dr 2 = F.dr. (5.23) d 2 dt (r1 ,t1 ) (r1 ,t1 ) Hence, we arrive at the work-energy theorem, stated as KE(r2 ,t2 ) − KE(r1 ,t1 ) =

Z (r2 ,t2 )

F.dr

(r1 ,t1 )

= Work done by the force, F.

(5.24)

Given a potential V in terms of which the force can be expressed, the RHS of Eq. (5.24), Z (r2 ,t2 )

F.dr = −

(r1 ,t1 )

Z (r2 ,t2 )

∇V .dr (r1 ,t1 )

= −[V (r2 ) −V (r1 )],

(5.25)

independent of the path connecting the initial and final points. Defining total energy as the sum of the kinetic and potential energy, we see that the total energy at the initial point is the same as at the final point. 5.5.1 Interpretation of grad V Consider the equipotential surfaces (V = constant, V0 ). In three dimensions, the surfaces will be two dimensional. Now consider two points r and r + dr of the surface, V = V0 . Then, ∇V .dr = 0. To see this, ∇V .dr = dx

∂V ∂V ∂V + dy + dz ∂x ∂y ∂z

= V (r + dr) −V (r) = V0 −V0 = 0.

(5.26)

This means that ∇V is perpendicular to dr lying on the surface, V = V0 . |∇V | is a measure of the rate at which the potential changes in space in a direction perpendicular to it, given by the unit vector n. ˆ We note an important relation: ∇V .dr = |∇V ||dr| = V (r + dr) −V (r) = dV0 , or

|∇V | =

dV0 . |dr|

(5.27)

44

Mechanics, Waves and Thermodynamics

5.5.2 Relation with curl of the force Given a force, how do we know if it is conservative? A force can be expressed as −∇V if and only if curl F is zero. Assuming that the force is gradient of a scalar, it can be seen that curl is zero by looking at the z-component of curl F is

(curlF)z =

∂ Fy ∂ Fx − ∂x ∂y

= −

∂ ∂V ∂ ∂V + = 0. ∂x ∂y ∂y ∂x

(5.28)

Proof of the converse: If, however, curl F is zero, then we use the Stokes’ theorem, Z Z

curl F.dS =

S

I C

F.dr.

(5.29)

Hence, the work done along a closed path is zero. Since the path could be any simple closed curve, it can be broken into any two paths that will combine into one path. That is 0=

I C

I

F.dr =

C1

F.dr +

I C2

F.dr.

(5.30)

So, I C1

F.dr = −

I C2

F.dr.

(5.31)

Since C1 and −C2 are arbitrary paths, we conclude that the work done along any two paths connecting the same points is identical. This concludes the proof that the force is conservative if curl F is zero.

5.6 Work-energy Theorem and Galilean Invariance It is a common experience that the outcome of experiments performed is the same while we are at rest or in a uniform motion. This invariance of experimental results is called Galilean invariance. We show that Galilean invariance implies the impulse theorem. Moreover, impulse theorem implies that there exists another Galilean invariant - the difference between the work done and change in kinetic energy [25]. 5.6.1 Galilean invariant Consider an N-particle system observed from two inertial reference frames S and S’. With respect to the origin of S, the position vector of S’ is R. Let V be the constant velocity of the frame S’ w.r.t. S. The position vectors of the ith particle w.r.t. S and S’ are related by the following equations:

Work and Energy

r0i = ri − R,

v0i = vi − V.

45

(5.32)

Denoting the total mass by m = ∑i mi , and total momentum by P = ∑i mi vi , the total kinetic energy in S’ is T0 =

1 N mi v0i .v0i 2 i∑ =1

1 = T + MV.V − P.V 2

(5.33)

1 N mi vi .vi . 2 i∑ =1 Let us now consider that this system transits from a state A to a state B. By Eq. (5.33), the changes in kinetic energies will be

where T =

0 ∆TAB = ∆TAB − V.∆PAB .

(5.34)

Let Fi be the force exerted upon the ith particle. The work done in taking the system from A to B is a function of paths, denoted by Γ and Γ0 in S and S’. Since, V is constant, we have for the difference between the work done and change in kinetic energy in S and S’: 0 0 − ∆TAB ) (WAB − ∆TAB ) − (WAB N

=

∑

Fi .dri − ∆TAB −

Z

N

i=1 Γ

N

=

!

Z

∑

i=1 Γ

Fi .vi dt − ∑

Z

i=1 Γ

Z = −V. ∆PAB −

N

∑

Z

0 i=1 Γ

! 0 Fi .dri − ∆TAB

Fi .v0i dt − V.∆PAB

tB

Fdt

(5.35)

tA

where F = ∑Ni=1 Fi . Hence, if the work-energy theorem holds, and is Galilean Rinvariant, then the LHS of Eq. (5.35) is zero. This implies the impulse theorem, ∆PAB = ttAB Fdt ). Moreover, we see that WAB − ∆TAB is a Galilean invariant. 0 = W 0 . Using the relations Galilean invariance of work-energy theorem means ∆TAB AB above, we get 0 = WAB

Z

N

∑ Fi .dr0i

Γ0AB i=1

46

Mechanics, Waves and Thermodynamics

Z tB N

∑ Fi .v0i dt

=

tA i = 1 N

=

Z

∑

i=1

ΓAB

Fi .dri −

= WAB − V.IAB IAB =

Z tB tA

Fi .Vdt

where

Z tB N

∑ Fi dt.

(5.36)

tA i = 1

It is now easily seen that the change in kinetic energy in S’ frame can be written as Z tB N Z 0 ∆TAB = ∑ Fi .dri − V. Fi dt . (5.37) i=1

ΓAB

tA

For the special case of one particle subjected to time-independent forces, Eq. (5.37) can be written as 0 ∆TAB + (tB − tA )V.F

=

Z ΓAB

F.dr

(5.38)

The LHS of this equation depends on vA , vB , tA , tB , V whereas RHS depends only on the position and velocity vectors of states A and B. As suggested by [25], it should be re-written as V.F = G

(5.39)

where F, G depend on A, B. Let us illustrate the usage of this by an example. More examples can be seen in [25]. 5.6.2

Example

Particle in a uniform gravitational field

Let m be the mass of the particle moving in a gravitational field, the acceleration due to gravity is denoted by g. Let the frame S be such that y-axis is vertical and x-axis is horizontal. Let A be the initial state where we know the position and the velocity vA . Given that the final position of the particle is (xB , yB ), we want to find the final velocity and the time taken. Let us consider a frame S’ moving with a constant velocity V. Equation (5.38) becomes v2B − v2A − 2V.(vB − vA ) = −2g(yB − yA ) + 2Vy g(tB − tA ).

(5.40)

Work and Energy

47

Arranging this equation to look like Eq. (5.39), we make the identifications: F = −2(vB − vA ), G = −2g(yB − yA ) + 2Vy g(tB − tA ) + v2A − v2B .

(5.41)

Equation (5.40) can be re-written in components: 2Vx (vBx − vAx ) + 2Vy [g(tB − tA ) + (vBy − vAy )] = 2g(yB − yA ) + v2B − v2A .

(5.42)

Due to the fact the V is arbitrary, the coefficients of its components must vanish: vBx = vAx , vBy − vAy = −g(tB − tA ), v2B − v2A = −2g(yB − yA ).

(5.43)

Apply the above to the case of a projectile fired from the origin with an initial velocity of magnitude v0 at an angle θ w.r.t. x-axis. Find the time it takes before it returns to the x-axis.

6

Chapter

Mechanics of a System of Particles Consider a collection of N particles of mass mi . The position of each particle is specified by three coordinates, xi , yi , zi . If they are free to move, then the number of coordinates needed to specify a state is 3N. In a rigid body, the relative position of two particles does not change with time. In that case, the total number of coordinates to characterize the motion is much lesser. To count, we have three coordinates for the first particle. With respect to the first particle, the position of the second particle is specified by the two angles (θ , φ ). For the third particle, we have to specify the angle with respect to the line connecting the first two particles. Hence, we need six coordinates in all. We can interpret three coordinates for specifying the centre of mass, and three others to denote the angles of rotation. Each particle, located at ri experiences a force fi due to which it accelerates with an acceleration r¨i . By Newton’s second law, we can write fi =

dpi . dt

(6.1)

ext The total force fi is a sum of internal (fint i ) and external (fi ) forces. If we consider the equation of motion of all the particles together, subjected to a total force, we will have to ext sum over i. The sum, ∑i fext i = F . The internal force between any two particles will appear in a pair, i on j being exactly opposite to j on i, by Newton’s third law. Hence, total internal force will be exactly zero. So, we have for the system of particles with a total momentum P:

Fext =

d dP = ∑ mi r˙i = ∑ mi r¨ i . dt dt i i

(6.2)

¨ where M = ∑i mi , we can write the force experienced by the collection of With P = M R, particles, as dP d d = M R˙ = ∑ mi r˙i . dt dt dt i

(6.3)

Mechanics of a System of Particles

49

This suggests that the system of particles moves with a “common” position vector, defining its centre of mass (CM) by R=

1 mi ri . M∑ i

(6.4)

To illustrate, let us consider an example where we have two bodies of masses m1 , m2 attached to each other by a rigid rod of length `. We would like to find the CM of this object and its equation of motion when it is thrown under the influence of gravity. Let the position vectors of the masses be r1 and r2 . The position vector of the CM is R=

m1 r1 + m2 r2 . m1 + m2

(6.5)

The vectors joining CM to the masses m1 , m2 are r01 = r1 − R =

m2 (r1 − r2 ), m1 + m2

r02 = r2 − R = −

m1 m1 (r1 − r2 ) = − r01 ; m1 + m2 m2

(6.6)

we see that they are proportional to each other. This proves the intuitively obvious fact that the CM lies on the line joining the two masses. The magnitude of (r1 − r2 ) is `. The total force is F = m1 g + m2 g. The equation of CM is ¨ = (m1 + m2 )g. (m1 + m2 )R

(6.7)

We therefore find that the CM follows a parabolic path under gravity. For a system made of a large number of particles placed very close to each other, we can define the CM in a continuum form: R

R= R

dmr dm

(6.8)

where r is the position vector of an infinitesimal mass element dm. Calculate the CM of a rigid rod of linear density λ (x) = λ0 (1 − (x/`)3 ) and length `.

6.1 Analysing the Leaky Pendulum In this Section, we bring out an interesting application of the idea of centre of mass by considering a pendulum with a leaky bob. We follow the analysis by Arun [39]. 6.1.1 Simple ‘usual’ pendulum Given p that the mass of the bob is m, length of the string is L, the time period is T0 = 2π L/g = 2π/ω0 . In terms of the angle, θ between the string containing the bob with

50

Mechanics, Waves and Thermodynamics

the vertical, the force is F = −mg sin θ = m

dv dt

(6.9)

with v = Ldθ /dt. Hence, d2θ = −g sin θ ' −gθ , dt 2 g implying θ¨ = − θ = −ω02 θ . L L

(6.10)

The approximation, sin θ ∼ θ is actually valid for not so small angles, close to π/6 radians, which is numerically nearly 1/2 = sin π/6. 6.1.2 Leaking bob Consider the case of a leaky water-filled bob. The mass is changing, so that under the assumption of small oscillations, we can write d d F= m (Lθ ) = −mgθ . (6.11) dt dt This simplifies to an equation for θ (t ): d2θ 1 dm dθ + + ω02 θ = 0. dt 2 m dt dt

(6.12)

The motion is akin a damped harmonic oscillator with time period decreasing from T0 = 2π/ω0 as mass changes: T0

T=q

1−

L 4g

1 dm 2 m dt

.

(6.13)

Experiments on oscillations of burette whose liquid drips demonstrate that T increases initially (as in Eq. (6.13)), and then starts decreasing (not borne out by the above equation). We need to consider the change in the position of the centre of mass also. This, in turn, will result in change in length of the string holding the bob. Hence, in Eq. (6.11), L also changes in time, leading to the following equation and the corresponding angular frequency: v " u # ug ¨ m˙ L˙ 1 2L˙ m˙ 2 L ω =t + + − + . (6.14) L L mL 4 L m The CM moves and hence, changes the time period. Hence, the length of the pendulum can be written as L = string (`)+ radius of the bob (r0 )+ position of CM w.r.t. bob’s centre (origin) (x). As water drips, CM moves down with positive dL/dt. At some instant when

Mechanics of a System of Particles

51

the water content is low, the CM moves towards the origin, dL/dt acquiring a negative sign. We now need to calculate the Centre of Mass. We illustrate the calculation in three parts - shell, water body, leaking bob. Spherical shell

Although it is obvious that the CM must be at the centre of the shell, taken as origin, let us just indicate the multiple integrals in spherical polar coordinates [40]. With ρshell denoting the density of the shell, the mass is mshell = ρshell

=

Z

dV = ρshell

Z a0

Z 2π

Z π/2

dr

r cos θ dφ

rdθ −π/2

a

4π ρshell a30 − a3 . 3

0

(6.15)

CM is located at ρshell x = mshell

Z a Z π/2 Z 2π −π/2 0

a0

r2 cos θ drdθ dφ r cos θ cos φ ˆi + r cos θ sin φ ˆj + r sin θ kˆ

ˆ = 0ˆi + 0ˆj + 0k,

(6.16)

as expected. Water body

Water body is initially spherical with radius as the inner radius of the sphere. Consider now a disc of thickness dz at a distance z from the centre. The radius of the disc is r and hence the area is A = πr2 . The volume of the disc is dV = πr2 dz. Mass of water in this disc will be dM = ρwater πr2 dz. The amount of water depends on the z and r2 = a2 − z2 (Fig. 6.1). In the case the sphere is not completely filled, and the height of the water is h, Z h

Mwater = ρwater π

dz a2 − z2

−a

= ρwater π

2a3 h3 + a2 h − 3 3

.

(6.17)

The position of the CM is xwater =

ρwater π Mwater

ρwater π = Mwater

Z h

dzz a2 − z2

−a

4 a a2 h2 h4 − + − . 4 2 4

(6.18)

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Mechanics, Waves and Thermodynamics

Fig. 6.1

Water filled in the spherical bob can be described in terms of cylindrical coordinates most naturally.

Leaking bob

The CM of the leaking bob can be found by employing the above results in the following formula: xbob =

mshell xshell + Mwater xwater . mshell + Mwater

(6.19)

Recalling that the CM of the shell will remain at the origin, and, ρwater is unity in C.G.S. units, we have 2 4 aρw ξ2 − ξ4 − 14 xbob = (6.20) . 4ρs a30 ξ3 2 − 1 + ρw ξ − 3 + 3 3 a3 where ξ = h/a. Plotting xbob vs ξ to see that the CM is not a monotonic function (Fig. 6.2). Exercise:

Plot the time-period of the bob in the above discussion as a function of ξ .

Question Petrol starts leaking from a car travelling at speed v on a frictionless track. What is the speed of the car at a later time? No force is applied by the car on the petrol. Hence, petrol applies no force on the car and can’t alter its momentum. Question Consider a composite disc, with a main section of mass density ρ and radius r, and a subsection of mass density 2ρ and radius r/2. Let the centre of the larger disc be the origin, and, the centre of the smaller disc be located at r/2 from the origin. Where is the centre of mass (CM) of the composite disc?

Mechanics of a System of Particles

Fig. 6.2

53

The centre of mass (CM) of the bob is shown as a function of the height of the water level normalized to the inner radius a of the bob, ξ = h/a. The bob is made of steel of density 7.85 g/cm3 , its outer radius is 5 cm and thickness is 5 mm. Density of water is 1 g/cm3 . We see the non-monotonicity in the variation.

Using the CM formula, superposing the centre of masses of two discs of radii r and r/2 and density ρ, one finds that the CM is at a distance r/10 from O towards the centre of the smaller disc.

6.2 Work-energy Theorem Revisited There are instances where for a system of many particles, one confuses work with what is more suitably termed as pseudo-work. For instance [42] - “when a block slides down an inclined plane, the change in kinetic energy is equal to the work done by gravity minus the work done by the frictional force. The block gets warmer. This increase in thermal energy should appear in the Work-Energy theorem.” This is an instance of a situation involving systems where the objects cannot be treated as point particles. To see the above point, let us consider a system of particles where the ith particle is subjected to an external force, Fi . Newton’s second law is

∑(Fi )ext = MaCM

(6.21)

i

where M is the total mass and aCM is the acceleration of the centre of mass (CM). Integrating w.r.t. the displacement of the CM, we have

54

Mechanics, Waves and Thermodynamics

Z

∑(Fi )ext .drCM =

Z

M

i

Z

∑

(Fi )ext .drCM

i

dvCM .drCM dt

1 2 =∆ Mv . 2 CM

(6.22)

LHS is called “pseudo-work” [43] because the forces have been multiplied by the CM displacement rather than individual displacements. RHS is also basically the kinetic energy of the CM. Equation (6.22) is “pseudowork-energy theorem” as Z

(Fi )ext .drCM 6=

Z

(Fi )ext .dri .

(6.23)

For point particles, ri = rCM so Eq. (6.22) is then the usual work-energy theorem. Problem We consider uncoiling a uniform chain of length L made of metal links. Apply a force on the bunched chain so that it uncoils on a frictionless floor. The speed of the chain after opening out is v. Referring to the Fig. 6.3, we can write the two equations as follows: Centre of mass

F (d − L/2) = mv2 /2,

First Law of Thermodynamics

Fd = mv2 /2 + ∆E.

(6.24)

Here, ∆E is the change in internal energy of the chain. This may be thought of as related to an increase in temperature after the inelastic collisions among the links have damped. Find v and ∆E.

Fig. 6.3

Applying work-energy theorem to understand uncoiling of a chain on application of force F on a frictionless floor.

Mechanics of a System of Particles

55

6.3 Displacement We want to show that the displacement of a rigid body in its plane is equivalent to a rotation about a point in the plane (Fig. 6.4).

Fig. 6.4

Displacement of a rigid body in its plane consists in a rotation about a point.

Let A and B be any two points of the rigid body (distance between any two points is fixed) which becomes displaced to A’B’. Draw the perpendicular bisectors of AA’, BB’ meeting in I. Then, IA = IA’, IB = IB’, and AB = A’B’. Therefore, 4 IAB ' 4 IA’B’. Hence, 4 IA’B’ is obtained from the first by a rotation through an ∠ AIA’ or ∠ BIB’. The point I above is called the instantaneous centre of rotation . As another example, let us consider a rod AB constrained to move along given lines OX and OY (Fig. 6.5) in a plane. The point I is the instantaneous centre of rotation.

Fig. 6.5

Instantaneous centre of rotation of a sliding rod.

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Mechanics, Waves and Thermodynamics

6.4 Rotation A circular rigid lamina rolls along in contact with a straight line. We want to find the velocity and acceleration of a point on the circumference. (Fig. 6.6).

Fig. 6.6

A rolling circular rigid lamina.

Let C be the centre of a circle of radius R and P be the point of contact with the x-axis at a time t. Let A be a point on the circumference whose velocity and acceleration are required. The angular velocity of the circle is θ˙ . Let O be the origin, so OP = arc AP, i.e., O is the position of A in the rolling motion. With coordinates of A as (x, y), we have x = OP − AC sin θ = R(θ − sin θ ), y = CP − AC cos θ = R(1 − cos θ ).

(6.25)

Hence, the components of velocity are x˙ = Rθ˙ (1 − cos θ ), y = Rθ˙ sin θ .

(6.26)

The components of acceleration are x¨ = Rθ¨ (1 − cos θ ) + Rθ˙ 2 sin θ , y¨ = Rθ¨ sin θ + Rθ˙ 2 cos θ .

(6.27)

Note that at the point of contact (θ = 0), x˙ = 0 = y. ˙ Hence, the point of contact of a rolling circle is instantaneously at rest. However, the acceleration components are x¨ = 0, y¨ = Rθ˙ 2 , showing that the point of contact has an acceleration Rθ˙ 2 towards the centre.

Mechanics of a System of Particles

57

6.5 Rigid Body Motion: Basic Ideas Rotation of a top or a planet or a pulsar - all evoke fascination. Treating these as rigid bodies, we know that as they rotate about an axis (z-axis, say), each of its constituent ˆ The velocity v of particle rotates about the axis with the same angular velocity ω = ω k. the ith particle is given in terms of the components: vi = ω × ri = (−ωyi )ˆi + (ωxi )ˆj.

(6.28)

The angular momentum of the ith particle about the axis of rotation is (i)

= −ωmi zi xi ,

(i)

= −ωmi zi yi ,

(i)

= ωmi xi2 + y2i .

Lx Ly Lz

(6.29)

Analogous to the manner in which we write the linear momentum as mass times velocity, we would like to write the angular momentum as “a mass-like object” times angular velocity. This “mass-like object” is the moment of inertia which is not a scalar. This is simply understood if one tries to recall one’s experience in trying to rotate any solid object. It is easier to rotate in certain orientations than another. A common example is a water bottle - it is easier to spin about axes that pass closer along its length. It has nine components which can be arranged in a nice symmetric manner in the form of a 3 × 3 real symmetric matrix. The matrix elements are: Ixx = ∑ mi y2i + z2i , Iyy = ∑ mi xi2 + z2i , Izz = ∑ mi xi2 + y2i , i

i

i

Ixy = − ∑ mi xi yi = Iyx , i

Iyz = − ∑ mi yi zi = Izy , i

Izx = − ∑ mi zi xi = Ixz .

(6.30)

i

Hence, angular momentum of a rigid body is given by ω L = I .ω

(6.31)

where I is the moment of inertia tensor with components as above. If the centre of mass is at rest, and there is no translational energy, the total rotational energy of the system is given by

∑ i

1 mi 2 vi = ω 2 ∑ mi xi2 + y2i . 2 2 i

(6.32)

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Mechanics, Waves and Thermodynamics

This can be written as 1 (0 2

0

ω ) . I . (0

0

ω )T .

(6.33)

This is the case when the rotation is about the z-axis. In the case when the rotation is about a general axis, nˆ where n. ˆ nˆ = 1. The unit vector can be generally written as nˆ = nx xˆ +ny yˆ + nz zˆ. The angular velocity is ω = ω n. ˆ The expression for angular momentum is L = I .ω n. ˆ

(6.34)

Similarly, rotational kinetic energy is given by 1 TR = (ω nˆ ).I .(ω nˆ )T . 2

(6.35)

Problem Consider a marble of mass m and moment of inertia Im which rolls without slipping down an inclined plane. The inclined plane is rigidly attached to a turntable. The marble is constrained to move along a straight-line path on the inclined plane (along a groove, for instance, see Fig. 6.7). The turntable is light and rotates freely. The angular velocity is not necessarily constant. The moment of inertia of the combined system of the inclined plane and the turntable is It . Assume that the marble stays on the inclined plane, find its position as a function of time.

Fig. 6.7

Solution

A marble of mass m moves along a straight line down an inclined plane. The path is along a groove so that it remains straight. The turntable slows down (moves faster) as the marble rolls up (down) without slipping.

Let the speed of the marble be v, then (Fig. 6.7)

v2 = x˙2 + φ˙ x2 cos2 θ .

(6.36)

Energy conservation implies that 1 1 1 E = M x˙2 + It φ˙ 2 + Mx2 φ˙ 2 cos2 θ + mgx sin θ 2 2 2

(6.37)

Mechanics of a System of Particles

59

where It is the moment of inertia of the Table and M is the effective mass of the marble (see the next example if you find the effective mass mysterious), M = m+

Im . r2

(6.38)

As E is independent of φ , ∂ E/∂ φ˙ is the constant of the motion. This is the angular momentum L=

∂E = It + Mx2 cos2 θ φ˙ . ∂ φ˙

(6.39)

Using L, we can rewrite the energy as E =

=

1 2 L2 M x˙ + + mgx sin θ 2 2(It + Mx2 cos2 θ ) 1 2 M x˙ + Ueff (x). 2

(6.40)

As the turntable rotates, the marble rolls up the incline (along the groove), its potential energy increases. Because it is also rotating about the centre of the turntable, its kinetic energy has also increased. To conserve the energy, the turntable will slow down. There is a speed of rotation of the turntable for which the marble will be at rest. Around this value, the marble will oscillate up and down. The reader is encouraged to calculate the frequency of this oscillation. Problem A marble rolling along a curved groove lies in the xy-plane in a uniform gravitational field. Let the mass of the marble be m, radius r, and moment of inertia, Im . It rolls without slipping, so the linear velocity v is equal to rω. Show that its effective mass is M = m+ Solution E =

Im . r2

(6.41)

Conservation of energy implies 1 2 1 mv + Im ω 2 + mgy 2 2

=

1 2 1 v 2 mv + Im + mgy 2 2 r

=

1 Im m + 2 v2 + mgy. 2 r

We see that there is an effective mass, M = m + Im /r2 .

(6.42)

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Mechanics, Waves and Thermodynamics

6.6 Rotation of a Rigid Body about an Arbitrary Axis We have seen that rotational motion is characterized by angular momentum and torque and their relation with angular velocity and angular acceleration. We want to derive these relations for a rigid body rotating about a fixed axis (which may not be passing through the Centre of Mass (CM)) [30]. Given a rigid body with position vector ri of the ith constituent particle of mass mi w.r.t. a fixed point O. The velocity and linear momentum of this particle is respectively vi and pi . The angular momentum about O is LO =

∑ mi (ri × vi ),

discrete collection of particles,

i

Z

=

dm(r × v),

continuum of particles.

(6.43)

The velocity is v = ω × r, hence, LO =

Z

dm(r × (ω × r)).

(6.44)

Let us consider body-set of axes (x0 , y0 , z0 ) with origin at O, orientation of the axes is left arbitrary. The space-set of axes (x, y, z) constitutes an inertial frame w.r.t. the rigid body. The body-set of axes is fixed and the dynamics is easier to express in its terms: LO =

=

Z

ω) dm ωr2 − r.(r.ω "

Z

dm ∑

ωi0 eˆi r2 − xi0 eˆ0i

i

=

Z

!#

∑

x0j ω 0j

j

dm ∑ δi j r2 − xi0 x0j ω 0j eˆ0i i, j

=

∑ Ii0j ω 0j eˆ0i .

(6.45)

i, j

This implies that angular momentum need not be parallel to angular velocity. Elements of the moment of inertia tensor are Ii0j

=

Z

dm[δi j r2 − xi0 x0j ].

(6.46)

Angular momentum in the two sets of coordinate systems are related by a matrix containing time-independent coefficients, LO = S L0O

(6.47)

Mechanics of a System of Particles

61

with elements of S being Si j = eˆ0i .eˆ j .

(6.48)

Establish the relation between torque and angular acceleration, as above. The torque τO = r×

dp dLO = . dt dt

(6.49)

In body-set axes, ith component of torque is

(τ 0O )i = ∑ I i j j

dω 0j dt

+ ∑ εi jk ω 0j Ikl0 ωl0

(6.50)

j,k,l

where εi jk is a completely antisymmetric tensor. 6.6.1

Special cases

Rotation about a body axis

Let us consider the case when a rigid body is rotating about a body axis at an angular speed ω. Let us orient the body coordinate system in such a way that the z0 axis identifies itself with the axis of rotation. All the points on this axis are at rest, we can choose any point on the axis as the origin of body-coordinate system. The components of angular momentum are 0 0 0 LOx = Ixz0 ω; LOy = Iyz0 ω; LOz = Izz0 ω.

(6.51)

Show that the components of torque are given by 0 τOx = Ixz0

dω − Iyz0 ω 2 , dt

0 τOy = Iyz0

dω − Ixz0 ω 2 , dt

0 τOx = Izz0

dω . dt

(6.52)

From Eq. (6.51), we see that angular momentum components are non-zero even along the direction perpendicular to the axis of rotation. The moment of inertia matrix is real-symmetric, we may diagonalize this matrix. The eigen directions along which the moment of inertia matrix is diagonal are called the principal axes. This is independent of the point chosen as O.

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Mechanics, Waves and Thermodynamics

If the body rotates about a body principal axis (say z0 ) at angular speed ω, the angular momentum will have components only along z0 and hence Ixz0 and Iyx0 are zero. Hence, we can write for angular momentum and torque: 0 0 0 LOx = LOy = 0, LOz = Izz0 ω,

0 0 0 = Izz0 = 0, τOz = τOy τOx

dω . dt

(6.53)

Rigid body possessing an axis of symmetry

For a body possessing an axis of symmetry z0 (see Fig. 6.8), we see that for any point P there is a point P’ diametrically opposite to it. This axis must pass through CM, C. We may now choose origin of the body coordinate system at C. Hence, Eq. (6.53) holds.

Fig. 6.8

The body shown is symmetric about the z0 -axis. In any plane perpendicular z0 , there would be points rotating along a circle where there are points P and diametrically opposite to each other. This axis must pass through the centre mass, C. The expressions Eq. (6.53) must hold good.

to P0 of

6.7 Moments of Inertia of Simple Bodies For carrying out the calculations, we will need the moments of inertia (MI) of the bodies. 1. Rectangle: Let us consider a rectangular plate of length 2a and breadth 2b. We would like to find the MI about a line passing through the centre, parallel to the

Mechanics of a System of Particles

63

length. We identify the x-axis with this line and y-axis parallel to the breadth, through the centre. The MI is 2

∑ my

Z bZ a

or −b −a

σ y2 dxdy

(6.54)

where σ is the surface mass density of the plate. On integration, the MI is 4σ ab3 /3 or mass of the rectangle × b2 /3. 2. Ellipse:

(6.55)

Consider a uniform elliptic plate described by the equation,

x2 y2 + = 1. a2 b2

(6.56)

We would like to find the MI about the X-axis. MI is √ Z Z a

a2 −x2

(b/a)

√

I =

−a −(b/a)

= σ (πab)

a2 −x2

σ y2 dydx

b2 . 4

(6.57)

Problem Show that the MI of a circle of radius b about a diameter is (mass of circle) × b2 /4. 3. Ellipsoid: equation,

Consider a uniform solid ellipsoidal of density ρ described by the

x2 y2 z2 + + = 1. a2 b2 c2

(6.58)

We would like to find the MI about the x-axis. MI is Z Z Z

ρ (y2 + z2 )dxdydz.

(6.59)

To find the integral, we define new variables ξ = x/a, η = y/b, ζ = z/c. The integral becomes ρab3 c

Z Z Z

η 2 dξ dηdζ + ρab3 c

Z Z Z

ζ 2 dξ dηdζ .

The integration is over the sphere, ξ 2 + η 2 + ζ 2 = 1. The required MI is I = ρabc b2 + c2

Z Z Z

= πρabc b2 + c2

Z

1

−1

ξ 2 dξ dηdζ

ξ 2 1 − ξ 2 dξ

(6.60)

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Mechanics, Waves and Thermodynamics

=

4 πρabc b2 + c2 15

= Mass of the ellipsoid ×

1 2 b + c2 . 5

(6.61)

4. Triangle: Consider now a uniform triangular plate of surface density σ and mass M and find its moment of inertia with respect to any line in its plane. Let us denote the lengths of the perpendiculars dropped from the vertices to this line by α, β , and γ. Denoting by (x, y, z) the coordinates of a point on the triangular plate, the perpendicular distance from this point to the plate is (αx + β y + γz). The moment of inertia is I=σ

Z Z

(αx + β y + γz)2 dS

(6.62)

where dS is the area element on the plate. We quote the result here [31] I =

1 σ Area α 2 + β 2 + γ 2 + αβ + β γ + γα dS 6

1 = ×M× 3

"

α +β 2

2

β +γ + 2

2

γ +α + 2

2 # .

(6.63)

Two bodies are called equimomental if their moments of inertia about all the lines are equal. It can be readily seen from Eq. (6.63) that a triangle is equimomental to three particles of mass one-third of the triangle situated respectively at the midpoints of its sides [32]. A uniform solid tetrahedron of volume V and mass M is equimomental to a system of five particles, four of which are of mass M/20 situated at the vertices, while the fifth of mass 4M/5 is at the centre of mass of the tetrahedron. More about equimomental bodies can be referred to in [33].

6.8 Principal Axes — Stationary Points of Kinetic Energy The shape of terrestrial animals (and humans) mostly determines their axis of rotation. However, they adjust their posture to turn about another axis, for example in dance or gymnastics. This motivates us to consider how the distribution of mass around the axis of rotation influences the rotational motion. We shall see that there is a way to think whereby it turns out that minimum effort leads to the idea of principal axes of rotation. We are going to minimize the kinetic energy of rotation to demonstrate this point [34]. Let us consider a simple example — an inhomogeneous and rigid thin plate of an arbitrary shape in the xy-plane.

Mechanics of a System of Particles

65

How to choose direction of axis of rotation so that the kinetic energy is minimized, i.e., work done to produce rotation is minimized. Let the body-set of axes be (x0 , y0 , z0 ). At time t = 0, the origins of space-set, (x, y, z) and body-set coincide. The position vector of the ith particle is ri = xiˆi + yi ˆj, and velocity vi = ω × ri .

(6.64)

Kinetic energy of the plate is T=

1 N 1 N 2 m v = i i ∑ mi (ω i × ri )2 . 2 i∑ 2 =1 i=1

(6.65)

With ω = ωx i0 + ωx j0 , ω i × r|2 = ωy2 x2 + ωx2 y2 − 2ωx ωy xy. |ω

(6.66)

The kinetic energy is T =

=

1 1 mi xi2 ωy2 + ∑ mi y2i ωx2 − ∑ mi xi yi ωx ωy 2∑ 2 i i i 1 1 Ixx ωx2 + Iyy ωy2 + Ixy ωx ωy , 2 2

(6.67)

where Ixx = ∑i mi y2i , Iyy = ∑i mi xi2 , and Ixy = ∑i mi xi yi . Ixy may be positive, negative, or zero. Let us calculate the extremum of Eq. (6.67) for the case when the angular velocity vector is constant, i.e., q ωx2 + ωy2 = ω 2 = constant, or ωy = ± ω 2 − ωx2 . (6.68) Equation 6.67 can be written as 1 1 T (ωx ) = Ixx ωx2 + Iyy (ω 2 − ωx2 ± Ixy ωx 2 2

q

ω 2 − ωx2 .

(6.69)

Exercise: Show that for Ixx = Iyy , Ixy = 0, T (ωx ) is constant, independent of the axis of rotation. For Ixx > Iyy , Ixy = 0, minima of kinetic energy is for ωx = 0 and ωy = ω; the axis of rotation lies along the y-axis. We shall use Ixy 6= 0. Extremal points can be obtained by dT /dωx = 0, i.e., ωx4 − ω 2 ωx2 +

Ixy ω 4 = 0. 4Ixy2 + (Ixx − Iyy )2

(6.70)

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Mechanics, Waves and Thermodynamics

Four solutions for ωx , combining this with ωx2 + ωy2 = ω 2 = constant, we get eight ordered pairs (ωx , ωy ). Exercise: List all the solutions and argue that the four set of these eight solutions which are physically distinct are given by " #1 2 Ixx − Iyy ω ω± = √ 1 ± . 4Ixy2 + (Ixx − Iyy )2 2

(6.71)

The solutions are: {(ω+ , ω− ), (−ω− , ω+ )} for Ixy > 0, {(−ω+ , ω− ), (ω− , ω+ )} for Ixy < 0.

(6.72)

For the case in Eq. (6.72), Ixy > 0, the kinetic energy is 1 T± = 2

Ixx + Iyy ±

q

4Ixy2 + (Ixx − Iyy )2 2

ω 2.

(6.73)

These correspond to the extrema of kinetic energy. Exercise: For Ixy < 0, obtain 6.73) with appropriate solutions from Eq. (6.72). From T− (Eq. 6.73) and T ≥ 0, it follows that Ixx Iyy ≥ Ixy2 . Axes fixed by Eq. (6.72) are principal axes of inertia of a thin rigid plate of arbitrary shape. These axes are mutually perpendicular. This is because for Ixy > 0, (ω− /ω+ )× (−ω+ /ω− ) = −1. The quantities in brackets in Eq. (6.73) are the principal moments of inertia. These have been found by minimizing the kinetic energy.

6.9 Euler’s Equations In considering the rotation of a rigid body, we have used two obvious sets of axes - bodyfixed and space-fixed. For a rotating body, the body-fixed frame is non-inertial whereas the space-fixed frame is inertial. It is convenient to assume that the CM is at rest, and consider the rotations of the body about it. In the body-fixed axis, we have the advantage of choosing the moment of inertia to be diagonal and constant in time, i.e., (b) (b) (b) I (b) = diag Ixx , Iyy , Izz (6.74) where the superscript (b) stands for the body-fixed axis.

Mechanics of a System of Particles

67

In the space-fixed axis, the components of moment of inertia I (s) will generally change with time, and, the off-diagonal elements will not all vanish. If the body is rotating with an angular velocity ω , then we know that the velocity of a particle w.r.t. the two axes are related by ω ×r v(b) = v(s) −ω where r(b) = r(s) = r. Hence, (b) (s) dr dr ω × r. = −ω dt dt It turns out that this is valid in general, i.e., (b) (s) dS dS ω ×S = −ω dt dt for any vector S. For instance, if we use this formula where S is L, we get (b) (s) dL dL ω × L. = −ω dt dt

(6.75)

(6.76)

(6.77)

(6.78)

Note that in the last term on the RHS, L(b) = L(s) = L because CM is at rest. If N is the (s) external torque acting on the body, then dL = N. Hence, we get the Euler’s equation dt for a rigid body: (b) dL ω × L. = N −ω (6.79) dt For N equal to zero, one gets the Euler’s equation for a torque-free motion. If the body axes are chosen such that the moment of inertia tensor is diagonal, then Lx = Ixx ωx , Ly = Iyy ωy , Lz = Izz ωz where ω = ωx xˆ(b) + ωy yˆ(b) + ωz zˆ(b) , and moment of inertia is defined w.r.t. the body-axis. Since the components of I remain constant in time, Euler’s equations become Ixx

dωx − ωy ωz (Iyy − Izz ) = 0, dt

Iyy

dωy − ωz ωx (Izz − Ixx ) = 0, dt

Izz

dωz − ωx ωy (Ixx − Iyy ) = 0. dt

(6.80)

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Mechanics, Waves and Thermodynamics

We now consider some special cases. 1. Consider a body for which Ixx = Iyy 6= Izz . Then, we see that dωz /dt = 0 and ω˙ x +

ωy ωz (Izz − Ixx ) = 0, Ixx

ω˙ y +

ωx ωz (Ixx − Izz ) = 0. Ixx

(6.81)

Differentiating the first equation w.r.t. t and substituting in the second equation, and using ω˙ z = 0, we get ω¨ x + Ω2 ωx = 0

(6.82)

where ωz Ω= (Izz − Ixx ) . Ixx

(6.83)

These equations imply that ωx and ωy vary sinusoidally as cos(Ωt + φ ) and sin(Ωt + φ ). W.r.t. the body-axis, the component of ω along zˆ remains constant however, the x−, y− components vary with time with a period 2π/Ω. This motion is called precession about the z− axis. The motion of the earth illustrates our discussion. The shape of the earth is a geoid, it is oblate. Along the equator, the radius is slightly larger than the radius along the pole. It is known that 1 Izz − Ixx ' 0.00327 ' . Ixx 305 Hence, Ω ' ωz /305. It is q also known that ωx and ωy are very small compared to ωz

for the earth, hence, ωz ' ωx2 + ωy2 + ωz2 , which is 2π radians per 24 hours. This means that Ω is 2π radians per 305 days. So the earth’s rotation axis should precess around its north-south axis with a time period of 305 days. Although Euler predicted this in 1755, it was measured by Chandler in 1891. The difficulty in measurement q 2 is due to the small value of the amplitude of precession, ωx + ωy2 /ωz is about 10−6 . This is called the Chandler wobble. The actual time period is about 433 days because earth is not a rigid body. This is the precession of the axis of rotation about the North-South axis. If one travels to the Arctic attempting to find the axis of earth’s rotation, one finds that its position changes with a period of seven years. This is due to the pressure changes due to the fluctuations of salinity, temperature etc. at the bottom of the oceans. Changes in the phase of the Chandler’s wobble are not understood. For instance, in 1920, the phase change was π radians. Why?

Mechanics of a System of Particles

69

There is the axial precession where the axis of rotation of earth precesses about an axis fixed in space. This has a larger amplitude (Fig. 6.9). The origin of this is the torque acting on earth due to the gravitational force of sun and moon, and, owing to the fact that earth is not spherically symmetric. To study this, one has to use Euler’s equations including torque. One needs to find how L changes in space-fixed axis. One finds that this precession has a much larger time period of approximately 25,800 years [41]. However, due to its larger amplitude than Chandler wobble and the fact that it is w.r.t. the space-fixed axis, it has observable consequences. For example, this leads to a change in the ‘north’ star and all the constellations in the sky which are all fixed in the space axis. 2. Suppose that we have a highly asymmetric object such that all three moments of inertia are different: Ixx < Iyy < Izz . One can show that for a torque-free motion, rotations about x- and z- axes (smallest and largest) are stable while rotations about the y-axis is unstable. By stable (unstable) point we mean where small deviation from that point does not (does) grow with time. This can be easily verified by throwing a book spinning about three different axes, one by one. It is observed that for rotations about two of the axes, the book does not wobble (stable) however, about one axis the book wobbles a lot (unstable).

Fig. 6.9

Axial precession of the earth occurs with a period of 25,800 years, shown here schematically.

This kind of motion has also some interesting manifestations in the physics of triaxial nuclei or metallic clusters. However, we shall leave the reader here to verify the above ideas with more accessible examples, and encourage her (him) to play with tops.

7

Chapter

Friction One of the dirty little secrets is that there is no generally accepted explanation of the basic laws of friction. — David A. Kessler

We have gone ahead and confidently set up the equations of motion, found a principle on which it is based. We have assumed that the surfaces are smooth, forces are conservative, and there are no losses. However, as we all know, world is not an ideal place, surfaces come in contact and produce frictional forces. The exact origin of friction remains an outstanding open problem. In this Chapter, we include frictional force in our analysis of motion.

7.1 Non-conservative Forces and Energy Loss We have discussed in Section 5.5, a conservative force can be expressed as a negative gradient of a potential. Many forces do not fall in this category. The most common example is the frictional force. We shall discuss examples for rolling and sliding friction in the following Sections. Here, we discuss an instance of drag force on an object moving through a fluid, which is velocity-dependent. Consider a raindrop falling down in the vertical direction with ˆ The drag force is velocity, v = vk. Fdrag = −γvkˆ

(7.1)

where γ is the drag coefficient. The equation of motion is m

dv = −γv dt

(7.2)

in the absence of any other force. This can be solved to give v(t ) ∼ e−γt/m . If the initial velocity is v0 , then we can write the general solution as v(t ) = v0 e−γt/m . Now including gravity, the EOM is m

dv = −γv + mg dt

(7.3)

71

Friction

In addition to the one-dimensional homogeneous solution, a particular solution of this equation is mg/γ. The general solution is v(t ) = Ce−γt/m +

mg γ

(7.4)

with C as an arbitrary constant. At t = 0, v(0) = 0 implies that C = −mg/γ. Hence, the general solution is v(t ) =

mg (1 − e−γt/m ). γ

(7.5)

As t → ∞, velocity tends to mg/γ - terminal velocity, vT . Let us find more about vT by imagining the raindrops falling through air. The mass of the drop is given by ρ 43 πr3 . For modelling γ, it is natural to assume that drag force experienced by a drop will be proportional to the area in contact with air. Let us consider γ = αr2 where α is the constant of proportionality. The terminal velocity is 3

mg ρ 4πr 3 vT = = ∝ r. γ αr2

(7.6)

This implies that the terminal velocity of larger drops is more than that of smaller drops. 7.1.1 Energy loss We would like to show that total energy decreases when a particle is moving in the presence of a frictional force. Let us write the EOM of a particle of mass m in the presence of conservative and frictional forces: m¨r = F + Ffr ,

(7.7)

where F = −nabla V . Multiplying by r˙ and integrating from (r1 ,t1 ) to (r2 ,t2 ), we get KE at(t2 ) − KE at(t1 ) = − [V (t2 ) −V (t1 )] +

Z (r2 ,t2 ) (r1 ,t1 )

dt r˙ .Ffr .

(7.8)

That is, Total energy(t2 ) − Total energy(t1 ) =

Z (r2 ,t2 ) (r1 ,t1 )

dt r˙ .Ffr .

(7.9)

Because r˙ .Ffr < 0 at all times, total energy at t2 is lesser than at t1 . If the form of the frictional force is −γ r˙ , then the energy loss is −γ

Z t2 t1

dt (r˙ )2 < 0 because γ > 0.

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The problems with variable mass also involve change in total energy, the examples are related to physics of rockets, the reader is referred to [58].

7.2 Bowling — Physics of the Rolling Ball Consider a rigid sphere, solid and uniform, of mass M and radius r which is thrown horizontally along a rigid, horizontal surface with an initial velocity v0 . To study the motion, we note the two phases - one when the ball is initially rolling and slipping, followed by just rolling and eventually slowing down [59]. Let there be a sliding frictional force, f acting on the sphere in the direction opposite to the motion of the centre of mass (CM) of the ball (Fig. 7.1). The sphere rolls and slips with equations of motion, Sum of all forces

= − f = MaCM ,

Sum of all torques = f r = ICM α

(7.10) (7.11)

where aCM is the acceleration of the CM, and, ICM is the moment of inertia of the sphere about the CM, which is simply 25 Mr2 . It is important to keep in mind that torques should be calculated about CM. Because slipping between surface and sphere occurs, with the coefficient of kinetic friction, µk , we can write f = µk Mg.

(7.12)

Equations (7.10) and (7.12) imply aCM = −µk g,

Fig. 7.1

Forces on a rolling ball.

(7.13)

Friction

α =

fr ICM

=

5µk g . 2r

73

(7.14)

Because aCM and α are constants, we obtain vCM = v0 − µk gt, ω =

5µk g t, 2r

(7.15) (7.16)

assuming that the initial angular speed is zero. Note that the linear and rotational quantities are not related by vCM = rω, aCM = rα because the sphere slips and rolls at the same time. From Eq. (7.15), we see that while the sphere is acquiring angular speed, its linear speed decreases. When vCM = rω = 52 µk gtR , the sphere just rolls, the frictional force disappears. The time tR follows from 2v0 5 µk gtR = v0 − µk gtR or tR = . 2 7µk g

(7.17)

This leads to the second phase when the ball has a uniform motion (Fig. 7.2).

Fig. 7.2

As the ball slips and rolls, the linear velocity decreases and the angular velocity increases. Finally, the ball just rolls.

Question How can frictional force disappears? Recall that frictional force appears (i) when there is a relative motion between two surfaces, or, (ii) when there is a force that attempts to cause a relative motion, though without achieving it. Question rotation?

If the frictional force disappears, what causes the torque providing the

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Mechanics, Waves and Thermodynamics

When a rolling body rolls without slipping, the point of contact of the body with the surface is always instantaneously at rest with respect to the sustaining surface. A torque acting on a body causes a change in its angular speed. Rolling friction It seems that once the sphere reaches the pure rolling state (without slipping), the motion continues indefinitely with a constant speed. However, our experience shows that the body stops. To explain this, we must analyse the problem realistically. The problem is with our assumption of both the body and the surface being perfectly rigid. So we must assume that rolling ball and/or the plane deform slightly such that there is not just a contact point between them, however, a contact area. Usually, both surfaces deform. However, we consider two cases: (I) deformable body rolling freely on a rigid horizontal surface, (II) rigid rolling body moving freely on a deformable horizontal surface. Case I This is the case of an automobile tyre when it is not under traction (with less air inside). In this case an asymmetric deformation takes place, in such a way that the normal force exerted by the plane (i.e., horizontal surface) on the tyre now acts at a point slightly displaced from the theoretical contact point (Fig. 7.3).

Fig. 7.3

The rolling ball is in contact with a surface, this presents a torque in a direction that opposes the rolling motion.

The normal force exerts a torque about the CM opposed to tyre rotation, which causes reduced angular speed. This torque is τ f = ρN

(7.18)

where ρ is the displacement of N w.r.t. the action line of W . ρ depends on (a) the elastic properties of the rolling body, (b) its radius, (c) the effective load that the tyre bears. Because the torque opposes the tyre rotation, it is referred to as rolling friction and is usually many times smaller than the sliding friction. This is why balls and rollers are used to reduce friction.

Friction

75

Case II This is the case of a billiard ball moving on a cloth-covered billiard table. Here the deformation of the ball can be neglected in relation to the deformation of the cloth. The situation we analyse is in Fig. 7.4.

Fig. 7.4

Forces when the ball is in contact with a surface.

From Fig. 7.5, we can write the equations of motion: −Rx = MaCM , Ry −W = 0, Rx h − Ry δ = ICM α,

(7.19) (7.20) (7.21)

where torques have been taken about the CM, and the moment of inertia (MI) is about the axis through the same point. If the ball rolls without slipping (aCM = rα) then, taking account the MI (ICM = 2Mr2 /5) and assuming small deformations (h ∼ r), it follows from Eq. (7.19) that Ry δ = Rx h − ICM α 2 aCM ' Rx r − Mr2 . 5 r 2 Rx 7 = Rx r. = Rx r − Mr − 5 M 5

(7.22)

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Mechanics, Waves and Thermodynamics

Fig. 7.5

The components of the forces are shown in detail to clearly state the equations of motion for a rolling ball with a surface-contact of the ground.

This shows that Ry δ > Rx r.

(7.23)

Hence, deceleration of the ball is justified by the action of Rx against the velocity of the CM of the ball, and, a resultant torque that decelerates the rotation. Hence, force R plays the role of rolling resistive force. Last of Eq. (7.22) can also be written as 5δ 5 Rx = = sin θ . Ry 7r 7

(7.24)

For small deformations, sin θ ' θ , sin β ' β , so Rx = R sin β 5 ' Rβ = Ry θ 7

=

5 Mgθ 7

(7.25)

because Ry = W = Mg. Rolling resistive force is a constant force, determined by the deformation of the surface, however, independent of the velocity of the ball. The reason that a rolling body comes to rest is due to the fact that while the point A is instantaneously at rest, A’ is not. Due to this, a relative motion between the two surfaces is set up and hence, there is a sliding friction acting at the point A’.

Friction

77

Problem I am standing near a swimming pool, not quite willing to take the plunge. While I stand contemplating - to jump or not to jump - my daughter gently pushes me . . . Model me as a uniform rigid rod of height, L = 160 cm and mass, M = 60 kg. The horizontal force, F applied on me was about 100 N, some h equal to 120 cm above the ground for about 0.32 s. The rod (i.e., me) pivots at an angular velocity ω about an axis that passes through my feet. So, F × h = (ML2 /3) × dω/dt. The CM will rotate at a speed vCM = 80 CM × ω. Taking the coefficient of friction, µ to be 0.1, we may write for the acceleration of CM : (60 kg) dvCM /dt = 100 N − (0.1 × 60 kg × 10 m/s2 ). Calculate ω and vCM . What is the combination of µ, F, and h to prohibit my fall?

7.3 Squealing and Squeaking Writing by a chalk on a blackboard produces squealing, doors squeak, and cars squeal to a sudden stop. What exactly determines this, and what determines the pitch? This is an instance of sticking and slipping [44, 45, 46]. Incorrect holding of chalk leads to sticking to the board. To keep writing, the writer bends the chalk. Chalk suddenly slips and begins vibrating and periodically striking the board. This is heard as a squeal. Let us make a slightly more expanded discussion, although very inadequate to the satisfaction of a puritan. The coefficient of friction between blackboard and tip of the chalk is denoted by µ. The force F at the tip, making an angle θ w.r.t. the direction of motion is F=

µN . 1 − µ tan θ

(7.26)

N is the normal reaction at the tip. If θ = cot−1 µ, then F is very large (chalk can break). There is no slip between the moving tip and the surface, and θ increases. As the tip moves on, the moment at the pivot (point where the chalk is held by fingers) opposing the deflection of the chalk becomes so large that “finger + chalk” becomes rigid. The force then decreases so that the elastic energy stored in the fingers tends to swing the tip away from the board. Slip occurs. The chalk is set into vibration as there is a periodically varying tangential force. These vibrations lead to emitting of loud sound, called squeal.

8

Chapter

Impulse and Collisions We are all creatures of impulse. — Mark Twain

The collision of particles can be understood by collecting the equations of motion and laws of conservation. The change in momentum is equal to impulse. We shall consider the collision of spheres to begin with. After explaining basic concepts, we calculate the change in kinetic energy in an inelastic collision. Then, we come to the main part where we apply these ideas to understand karate punch. The common occurrence of a falling pencil leads to a rather nontrivial analysis, we see how nonlinear is such a simple looking event.

8.1 Impact of Smooth Spheres Motion of smooth spheres after impact is governed by (i) principle of conservation of momentum, (ii) experimental law of Newton - “The relative velocity of the spheres along the line of centres immediately after the impact is (−e) times the relative velocity before the impact: ∆v = −e∆u.” The quantity e is called the coefficient of restitution, which depends on the substance. For example, e (steel/ivory) ∼ 1 (perfectly elastic) whereas e ∼ 0 (inelastic material). Experimentally, e can be determined by considering an experiment where two spheres collide and one measures the ratio of initial and final velocities in terms of height (Fig. 8.1). This is owing to the simple relation, v2 ≈ height, h. Then,

(vA − vB ) (initially) = −e. (vA − vB ) (finally) For e = 1, vA − vB = −(vA − vB ), hence vA = vB . For e = 0, vA = vB . 8.1.1 Direct impact Let m1 , m2 be the masses of the two spheres, u1 , u2 be their velocities before impact, and v1 , v2 be their velocities after the impact and let the motion be along the lines of the centres. The momentum along the line of motion is unaltered by the impact, so that m1 v1 + m2 v2 = m1 u1 + m2 u2 .

(8.1)

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Impulse and Collisions

Fig. 8.1

The height of the ball hA is proportional to the square of its velocity. This simple idea is used in estimating the coefficient of restitution by considering two balls made up of different materials and colliding with another common ball or surface.

By Newton’s rule, v1 − v2 = −e(u1 − u2 ).

(8.2)

Multiplying Eq. (8.2) by m, we can manipulate the equations and find v1 =

m1 u1 + m2 u2 − em2 (u1 − u2 ) , m1 + m2

v2 =

m1 u1 + m2 u2 + em1 (u1 − u2 ) . m1 + m2

(8.3)

The impulse between the spheres which reduces the velocity of the first from u1 to v1 is m1 u1 + m2 u2 − em2 (u1 − u2 ) mu1 − mv1 = m u1 − m1 + m2

=

(1 + e)m1 m2 (u1 − u2 ) . m1 + m2

(8.4)

For the second sphere, m2 (u2 − v2 ) = −

(1 + e)m1 m2 (u1 − u2 ) . m1 + m2

(8.5)

Note that m2 (v2 − u2 ) is the same as m1 (u1 − v1 ). 8.1.2 Poisson’s hypothesis “When the bodies come in contact, there is a short interval in which they undergo compression followed by another short interval in which the original shape is restored. At

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Mechanics, Waves and Thermodynamics

the instant of greatest compression the bodies have a common velocity along the lines of centres, and the impulsive pressure between the bodies during restitution is less than the impulsive pressure during compression in the ratio e:1.” If the bodies were inelastic (e = 0) they would acquire a common velocity, (m1 u1 + m2 u2 )/(m1 + m2 ) and there would be no restitution. Hypothesis is that when the bodies are elastic, the whole impulsive forces during impact is (1 + e) times the impulse during compression. And this implies the Newton’s rule. Let us show this. Let I be the impulse during compression, and U be the common velocity at the instant of greatest compression, then m1 (u1 −U ) = I, m2 (U − u2 ) = I, implies

u1 − u2 =

I I + . m1 m2

(8.6)

By Poisson hypothesis, the whole impulse = (1 + e)I. Therefore, the whole impulse is m1 (u1 − v1 ) = (1 + e)I, m2 (v2 − u2 ) = (1 + e)I; u1 − u2 − (v1 − v2 ) = (1 + e)I

1 1 + m1 m2

= (1 + e)(u1 − u2 ) implying thereby v1 − v2 = −e(u1 − u2 ).

(8.7)

8.1.3 Kinetic energy lost by impact We need to consider only direct impact, because, in oblique impact, the square of the velocity is the sum of the squares of the perpendicular components. The components at right angles to the line of centres remains the same by impact. Hence, only the components of velocity along the line of centres can effect a change in kinetic energy. Kinetic energy before the impact can be shown to be expressed in the following manner: Tb =

1 1 m1 u21 + m2 u22 2 2

1 m1 u1 + m2 u2 2 1 m1 u1 + m2 u2 2 = (m1 + m2 ) + m1 u1 − 2 m1 + m2 2 m1 + m2 1 m1 u1 + m2 u2 2 + m2 u2 − . 2 m1 + m2

(8.8)

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Impulse and Collisions

+m2 u2 Note that m1mu11 + is the velocity of the C.M., and in the last two brackets of Eq. (8.8) m2 we have the velocities of spheres relative to C.M., and in the last two brackets of Eq. (8.8) we have velocities of the spheres relative to the C.M. (So, kinetic energy of the spheres = kinetic energy of the C.M. + kinetic energy of spheres in relative motion w.r.t. the C.M.) The RHS of Eq. (8.8) can be further written by expanding the two squares: m1 (m1 u1 + m2 u2 )2 2m1 u1 (m1 u1 + m2 u2 ) (m1 u1 + m2 u2 )2 1 + m1 u21 + RHS = − 2(m1 + m2 ) 2 (m1 + m2 )2 m1 + m2

+

m2 u22 +

m2 (m1 u1 + m2 u2 )2 2m2 u2 (m1 u1 + m2 u2 ) − (m1 + m2 )2 m1 + m2

(m1 u1 + m2 u2 )2 (m1 u1 + m2 u2 )2 1 2 2 + m1 u1 + m2 u2 − = 2(m1 + m2 ) 2 m1 + m2 1 (m1 u1 + m2 u2 )2 2 2 + m1 m2 u1 + m1 m2 u2 − 2m1 m2 u1 u2 = 2(m1 + m2 ) 2(m1 + m2 )

=

(m1 u1 + m2 u2 )2 1 m1 m2 + (u1 − u2 )2 . 2(m1 + m2 ) 2 m1 + m2

(8.9)

Similarly, the kinetic energy of the spheres after the impact is Ta =

=

1 1 m1 v21 + m2 v22 2 2

(m1 v1 + m2 v2 )2 1 m1 m2 + (v1 − v2 )2 . 2(m1 + m2 ) 2 m1 + m2

(8.10)

We know that m1 u1 + m2 u2 = m1 v1 + m2 v2 , and v1 − v2 = −e(u1 − u2 ). Therefore, the loss of kinetic energy is Ta − Tb =

1 m1 m2 (1 − e2 )(u1 − u2 )2 2 m1 + m2

= (1 − e2 ) KE of the spheres w.r.t. C.M. before the impact.

(8.11)

8.1.4 Generalization of Newton’s rule Suppose a sphere moving with velocity u strikes a smooth plane in a direction making an angle θ with the normal to the plane, and that it rebounds with velocity v making an angle φ with the normal Fig. (8.2).

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Mechanics, Waves and Thermodynamics

Fig. 8.2

The collision of a ball is here generalized with unequal angles of incidence and reflection. The Newton’s rule can be generalized for this collision also.

The above rule makes v cos φ = eu cos θ

(8.12)

and because the velocity parallel to the plane is unaltered v sin φ = u sin θ .

(8.13)

Therefore, cot φ = e cot θ .

(8.14)

Angles of incidence and reflection of such a sphere impinging on the plane are governed by this law.

8.2 Forward Karate Punch One of the peculiar features of Karate is that the strikes to an opponent are designed to terminate several centimetres inside an opponent’s body. We examine the scientific basis for this [60]. Also, we examine the forces involved in a strike to a bone of the component, e.g., a forearm bone. Two important aspects are calculations/estimation of deformation energy and impact force. 8.2.1 Deformation energy This is the energy lost to deformation of the body at the area of contact. Karate strike is a collision between a body of mass M1 at rest and a mass M2 moving at a speed v. The energy lost is ∆E =

1 − e2 M1 M2 2 v 2 M1 + M2

(8.15)

Impulse and Collisions

83

where e is the coefficient of restitution. In forging, the above equation tells us that the maximum fraction of initial energy lost to deformation is by putting e = 0 in ∆E 1 2 2 M2 v

= (1 − e2 )

M1 . M1 + M2

(8.16)

So, e equal to zero refers to inelastic collision. This fraction is larger for smaller values of M2 , and hence, a light hammer is more desirable than a heavy one. In pile driving, no deformation of pile is sought, and one attempts to have elastic collisions where e = 1. The Karate strike is more complicated as e is dependent on where the opponent is hit. Even though hitting muscle and fat would involve a greater e, and hence less ∆E, than hitting a bone covered by a thin layer of skin, a strike to a high-e area may result in more pain to the opponent. We are more interested in the absolute magnitude of energy transfer unlike in forging where one is interested in an efficient transfer of energy to deformation. So, we are interested in Eq. (8.15) more than Eq. (8.16). Since, Eq. (8.15) depends on v2 , the Karate fighter attempts to maximize the speed of the strike. Karate black belts are said to complete this motion in about 0.2 s with maximum speeds of about 7 m/s occurring between 70 pre cent and 80 pre cent of the way through motion. Maximum energy transferred to deformation would occur if impact is made at 75 per cent point. Let us assume that M2 be the mass of only one striking arm (= 10 pre cent of man’s total mass, 70 kg). With e = 0, M1 = 70 kg, M2 = 7 kg, this energy is 1 70 × 7 × × 7 × 7 = 156 J, 2 77 which is the maximum energy transferred to deformation damage. Introductory Karate students often believe that the rotation of the fist in the forward punch adds to the energy delivered to the opponent. If we treat the arm as a solid, uniform cylinder of radius r = 3 cm and we use the arm mass and punching time from above, we find that the rotational energy Er is 1 1 2 Er = M2 r ω 2 2 2

= 0.4 J which is negligible compared to 156 J. 8.2.2 Impact force Forward punch with the arm as a free mass of about 7 kg, the maximum momentum is 7 kg × 7 m/s = 49 kg m/s. With time of contact of about 10 ms, average impact force is

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Mechanics, Waves and Thermodynamics

4900 N. We must calculate the stress created by impact force and then compare that value with the maximum value that can be withstood by a human bone. If a static force F is applied to a material over an area A, then the stress σ = F/A. If a straight bone is pulled at each end by a force F, then the bone will be elongated by an amount δ ` that is proportional to the force and the initial length L. The ratio δ `/L is the strain ε and is related to stress via the Young’s modulus, Y of elasticity: σ = Y ε.

(8.17)

If a static force causes bending of a bone (Fig. 8.3), then there will be compression along the inside of the curve and elongation along the outside curve. Somewhere inside the bone will be a surface called the neutral surface along which there is neither compression nor elongation.

Fig. 8.3

A bone is supported at two fixed ends. On application of a force as shown, it bends. The inner surface compresses whereas the outer surface expands, hence, there is a neutral surface in between (shown as dotted line).

The magnitude of strain anywhere in the bone is expressed as a function of the distance y away from the neutral surface by ε=

δ` yθ y ' = , L Rθ R

(8.18)

R being the radius of curvature of the bent bone. The stress is σ=

Yy . R

(8.19)

The external forces on the bone causes torques on the bone, and these torques are in equilibrium. Consider the cross-section of the bone on the near-side of the imaginary cut indicated in Fig. 8.4.

Impulse and Collisions

Fig. 8.4

85

Section of the bone shown in Fig. 8.3.

Stresses over the cross-section cause a torque on the cross-section about the axis OO’. The bone does not turn about that axis, however, because there is an equal and opposite, torque, FL 2 2 about OO’ due to the support force at the end of the bone. The torque on the cross-sectional area element dA at a distance y from the axis OO’ due to stress is dτ = yσ dA.

(8.20)

Total torque τ=

Z

(8.21)

yσ dA.

Using Eq. (8.19), τ=

Y R

Z

y2 dA =

YI R

(8.22)

where I = y2 dA is the moment of inertia. For equilibrium, we must have R

FL Y = τ = I. 22 R

(8.23)

The maximum stress occurs at the bone surfaces along the inside and outside of the curve. If the stresses there exceed the ultimate bending stress σu , which is called the modulus of rupture, then the bone ruptures. Let us approximate the distance between the neutral surface and the surface of the bone along the inside curve to the radius a of the bone. The maximum force that can be endured follows from Eqs (8.23) and (8.19) as Fmax =

4IY 4Iσu = LRmin La

(8.24)

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Mechanics, Waves and Thermodynamics

because by (8.19), σu =

Ya . Rmin

(8.25)

Ultimate bending stress for wet human long bones ∼ 2×108 N/m2 . If the bone is a uniform cylinder 1 cm in radius and 20 cm in length, then I =

Fmax =

πa4 , 4 22 1 2 × 108 × 4 = 3142 N. × 20 1 7 10 × 104 100 × 100

The centre of bone is deflected by less than 1 cm.

8.3 Falling Pencil on a Table A pencil of mass M is falling on a table, the EOM are [47] (Fig. 8.5) F = M

dvx , dt

N − Mg = M

dvy , dt

Icm

dω = NH sin θ − FH cos θ . dt

(8.26)

Icm is the moment of inertia about an axis through the C.M., ω = dθ /dt, (vx , vy ) are components of velocity. We will take the simple case where at time t = 0, ω, vx , vy are all zero. Simple observation indicates that ω increases with time t. So, from the third equation of Eq. (8.26), dω > 0, implies that NH sin θ > FH cos θ , dt N tan θ > F

(8.27)

during the fall. If the contact end slides on the table, then F = µN implies that tan θ > µ. If tan θ < µ, F will be less than µN which means that contact end grips the table instead of sliding. The pencil will then pivot about the contact end which will be presented from sliding by a relatively small static friction force. We assume that the static and sliding friction coefficients to be equal. The velocity is

Impulse and Collisions

87

v = ω ×r

= ω −kˆ × H sin θ ˆi + H cos θ ˆj

= −ωH sin θ ˆj + ωH cos θ ˆi

Fig. 8.5

(8.28)

Fall of a pencil presents an interesting instance of nonlinear dynamics where friction plays an essential role.

giving us the components, vx , vy . Hence, F = M

dvx dt

dωH cos θ dt dω − ω 2 sin θ = MH cos θ dt

= M

(8.29)

and N = Mg + M

dvy dt

d (−Hω sin θ ) dt dω 2 = Mg − MH sin θ + ω cos θ . dt

= Mg + M

(8.30)

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Mechanics, Waves and Thermodynamics

Substituting Eqs (8.29) and (8.30) in Eq. (8.26), we get MgH MH 2 dω MH 2 2 dω = sin θ − − ω (sin θ cos θ − sin θ cos θ ), dt Icm Icm dt Icm d2θ = MgH sin θ , dt 2

or

Icm + MH 2

or

MgH d2θ = sin θ 2 dt Icm + MH 2

(8.31)

where we can identify the frequency by ω02 = MgH/I0 with I0 = Icm + MH 2 being the moment of inertia of the pencil about an axis through the contact end. Equation (8.31) is valid only when the contact point is at rest. For small θ , sin θ ≡ θ . The solution of Eq. (8.31) will be θ=

θ0 ω0t (e + e−ω0t ) 2

(8.32)

which shows that the angular displacement grows exponentially with time. Multiplying Eq. (8.31) by θ˙ , we get 1 d ˙2 θ + 2ω02 cos θ = 0. 2 dt

(8.33)

That is, θ˙ 2 = ω 2 = −2ω02 cos θ + c.

(8.34)

Assuming that at t = 0, θ (t ) = θ0 , ω 2 = 2ω02 (cos θ − cos θ0 ).

(8.35)

One may also obtain this relation from the law of conservation of energy. Exercise:

Derive the following equations:

F = MHω02 sin θ (3 cos θ − 2 cos θ0 ), N = Mg − MHω02 (1 + 2 cos θ cos θ0 − 3 cos2 θ ).

(8.36)

Figures 8.6 and 8.7 shows the variation of F and N with θ . If the pencil is released from rest from a near-vertical position, then F N. The pencil will start to fall without sliding. Equation (8.36) shows that for cos θ0 close to unity, F increases with time to a maximum value of 0.3MHω02 at θ approximately equal to 26.7 degrees. Further, it decreases to zero at θ ∼ 48.2 degrees, and then reverse its sign. On the other hand, N decreases with time to a minimum value Mg − 4MHω02 when θ = cos−1 0.33 ≡ 70.5 degrees.

Impulse and Collisions

89

Fig. 8.6

The force, F is shown here as a function of the angle for a typical pencil. F is quite small, to see the numbers in a nice way we have multiplied it by 1000.

Fig. 8.7

The reaction, N is shown here as a function of the angle for a typical pencil. N is multiplied by 100 so that the numbers look reasonable. Why is this graph nonmonotonic?

If the pencil has a uniform mass distribution so that I0 = 4

MH 2 MH 2 = + MH 2 , 3 3

(8.37)

then the minimum value of N is zero. N can decrease to a value where |F| > µN in which case Eqs (8.30) and (8.36) are invalid. If F/N remains lesser than µ until F reverses its

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Mechanics, Waves and Thermodynamics

sign, the magnitude of F/N will increase to µ after F reverses its sign. Therefore, the pencil will start to slide either forward or backward after the initial grip phase, depending on µ. Equation 8.36 show that the maximum positive value of F/N is 0.37 when the pencil is released from near-vertical position. F/N is maximum when θ is about 35 degrees. Hence, the bottom end of the pencil will start to slide backward (forward) if µ < 0.37 (µ > 0.37). The analysis of this very common instance illustrates how complex the working of the world is. Also, it demonstrates the analytical power of the simple laws of mechanics - the simplicity that is not only awe-inspiring however, also a hallmark of the nature of truth.

9

Chapter

Central Forces ... the planets revolve around the Sun, and the Sun himself is a member of the galaxy or Milky Way system which revolves in a remarkable way. How did all of these rotatory motion come into being? What secures their permanence ... — Edmund Taylor Whittaker

Some of the most important force laws related to direct observations are the central forces - forces that are directed from or towards a point (centre). Well-known examples are gravitational forces and electrostatic forces.

9.1 The Two-body Problem Consider two point particles with masses m1 and m2 at positions r1 and r2 . We assume that they are interacting through a central potential energy which only depends on the magnitude r = |r2 − r1 |. That is, V = V (r ). Then, the mutual force acts along the line joining the particles: F12 = −∇r1 V =

r2 − r1 f (r ) r

= ∇r1 V = −F21 , with f (r ) = dV /dr. The unit vector rˆ =

r1 −r2 r

(9.1)

points from r2 to r1 . The total energy is

1 1 Etotal = m1 r˙ 21 + m2 r˙ 22 + V (r ). 2 2

(9.2)

Newton’s equations are m1 r¨ 1 = −ˆr f (r ), m2 r¨ 2 = rˆ f (r ).

(9.3)

We can rewrite these equations in terms of the CM coordinate R, total mass M = m1 + m2 , the relative coordinate r = r1 − r2 and reduced mass µ = m1 m2 /M. As we have seen in

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the last Chapter, this two body problem can be broken into a CM part and a part in terms of the relative coordinate. i.e., 1 1 Etotal = M R˙ 2 + µ r˙ 2 + V (r ). 2 2

(9.4)

¨ = 0 which implies that the CM acceleration is zero. Also, Newton’s equations give M R µ r¨ = −ˆr f (r ). Since, the CM moves with a constant velocity, we will henceforth concentrate only on relative motion which is like a one-particle problem. The angular momentum for the one-particle problem is l = r × µ r˙ . One can immediately prove that l is a constant of the motion (for any central force), For this, we see that the rate of change of angular momentum is l˙ = r˙ × µ r˙ + r × µ r¨ = 0

(9.5)

as r¨ k rˆ for a central force. Let us choose the z-direction in the direction of l. Hence, r and r˙ must be perpendicular to l. Hence, the motion of r lies in a plane. Let us introduce polar coordinates: x = r cos φ , y = r sin φ . The angular momentum is l = µr × rˆ ˙ r + rφ˙ φˆ ˆ = µr2 φ˙ k. Hence, r2 φ˙ = |l|/µ. The area swept out in time dt is 12 r2 dφ which is 12 r2 φ˙ dt = Thus, we get the Kepler’s second law - this holds for any central force. For the gravitational force, the potential is V (r ) = −

Gm1 m2 Gm1 m2 , and f (r ) = . r r2

(9.6) l 2µ dt.

(9.7)

The conservation of the energy in the relative coordinate gives E =

1 2 µ r˙ + V (r ) 2

=

1 µ r˙ 2 + r2 φ˙ 2 + V (r ) 2

=

1 2 l2 Gm1 m2 µ r˙ + − 2 2 2µr r

using r2 φ˙ = l/µ. As energy is a constant in time, s dr 2G(m1 + m2 ) 2E l2 =± + − 2 2. dt r µ µ r

(9.8)

(9.9)

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93

Using φ˙ = l/µr2 , we may write dr r˙ = ˙ dφ φ µr2 = ± l

s

2G(m1 + m2 ) 2E l2 + − 2 2. r µ µ r

(9.10)

Calling 1/r by α, dα = −dr/r2 . Eq. (9.10) can be rewritten as dα µ =± dφ l

s 2G(m1 + m2 )α +

2E l 2 α 2 − 2 . µ µ

(9.11)

It can be verified that the solution is 1 α = [1 + ε cos(φ − φ0 )] a

(9.12)

where (using µ = m1 m2 /(m1 + m2 )) a =

l2 l 2 (m1 + m2 ) = , Gµ 2 (m1 + m2 ) Gm21 m22

r ε =

a2 2Eµ = l2

s

2l 2 E G2 µ 3 (m1 + m2 )2

(9.13)

and φ0 is an arbitrary constant. Hence, the orbit is given by a = 1 + ε cos(φ − φ0 ). r

(9.14)

We can orient the plane to make φ0 equal to zero. If we do that, then we are left with a/r = 1 + ε cos φ , ε is called the eccentricity of the orbit. Now we need to consider different cases: Case I ε = 0: this gives r = a, a circle. Case II 0 < ε < 1: in this case, a = r + εr cos φ = r + εx.

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This implies (a − εx)2 = r2 = x2 + y2 , which can be re-arranged as 2 εa a2 2 2 (1 − ε ) x + + y = . 1 − ε2 1 − ε2 This can be re-expressed as 2 εa x + 1−ε 2 y2 + =1 a2x a2y

(9.15)

√ where ax = a/(1 − ε 2 ), ay = a/ 1 − ε 2 . Since 0 < ε < 1, ax > ay . We have an ellipse whose origin is at (x, y) = (−εa/(1 − ε 2 ), 0). The semi-major and semi-minor axes are respectively ax and ay . Further, it must be noted that to get 0 < ε < 1, we require the energy E to be negative as can be verified immediately from the expression of ε.The ensuing orbits are bound orbits. In this case, the two particles never get infinitely far from each other. The well-known examples are planets revolving around the sun. Case III ε = 1: then we have (a − x)2 = x2 + y2 . Hence, we have a parabola, a2 − 2ax = y2 .

(9.16)

Case IV ε > 1: this gives 2 εa a2 , y − (ε − 1) x + = 1 − ε2 1 − ε2 2

2

(9.17)

an equation of hyperbola. Cases III and IV do not yield bound orbits as the particles eventually get infinitely separated from each other. The expression of eccentricity shows that the parabolic orbit corresponds to E = 0 whereas the hyperbolic orbit corresponds to E > 0. 9.1.1 Bounded orbits Let us return to the bound orbits with E < 0. The area of the ellipse is πax ay . Owing to Kepler’s second law, the area swept per unit time is l/2µ. Hence, time period T of one revolution must be related to the area of the ellipse as πax ay /T = l/2µ. This implies that T=

2µ πa2 . l (1 − ε 2 )3/2

However, s l=

Gm21 m22 a . m1 + m2

(9.18)

(9.19)

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We thus have √ 2µ m1 + m2 πa3/2 T = Gm21 m22 (1 − ε 2 )3/2 2π

= p

G(m1 + m2 )

a3/2 x .

(9.20)

Now, m2 , mass of the Sun is much larger than m1 , mass of a planet. So, to a good approximation that T2 =

4π 2 3 a . GmSun x

(9.21) 2

4π This is Kepler’s second law with proportionality constant Gm which is independent of Sun any characteristics of planets. In fact, knowing T and ax for any planet gives us the mass of Sun.

Question Far away from the centre of attraction, a body (approximately looking like a point particle) is orbiting along an elliptical path with period T . Let this body be a rocket whose fuel gets over and its tangential velocity becomes small, thus it begins to “fall”. How long will it take for the rocket to “fall” on to the centre? As the velocity changes, the body will switch its orbit from one ellipse to another. The period T will change to T 0 as the semi-major axis changes from a to a0 . Kepler ’s third law implies T 02 a03 = . T2 a3

(9.22)

03 1/2 = T2 aa3 . In the limit of long time, the body comes to a complete halt, and hence, falls along a straight line, 2a0 will just become a [48]. Hence, r 1 0 1 a03 T T = T = √ . 03 2 2 8a 4 2

Time taken to fall (along the semi-major axis) is

T0 2

If the body were the Earth and the centre be the Sun, T is one year, and T 0 will be 64 days. Knowing that light takes approximately 8 minutes to reach the Earth, we can estimate the average velocity of the Earth as it falls to be 26,800 km/s. Problem Show that the orbit of Earth around the Sun never changes the direction of circulation about the Sun. The conservation of angular momentum L implies that both the magnitude and direction of L are conserved separately [49]. To see that the magnitude L is conserved, note that

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˙ L˙ = d/dt (L2 /2) = 0. This means that L does not change with time. L˙ = 0 implies that L. For direction, observe that for fixed L, L = mr2 θ˙ ≥ 0. This means that θ˙ ≥ 0. Hence, θ (t ) increases monotonically with time for non-zero L; the sign of θ does not change. Of course, this applies to all central forces. Following [50], we now discuss a set of physical systems under gravitational interaction.

9.2 Two Bodies Under their Own Gravitational Interaction 9.2.1 Case I Let us assume that a mass m1 is at rest, situated at origin, and, a mass m2 at a distance x from it. The equation of motion (EOM) is m2

d2x Gm1 m2 =− . 2 dt x2

(9.23)

We can also arrive at the EOM by conservation of energy. The gravitational potential energy of the two masses separated by a distance L is −Gm1 m2 /L. The motion of m2 takes place at the expense of the potential energy as it reduces to −Gm1 m2 /x. In turn, the kinetic energy of m2 is Gm1 m2 Gm1 m2 m2 2 x˙ = − − − . (9.24) 2 L x Hence, the EOM is r dx 2Gm1 (L − x) =− , dt Lx

(9.25)

the negative sign on the RHS is consistent with the fact that x is decreasing. Question

Show that one can simply integrate Eq. (9.23), and obtain Eq. (9.25).

Dimensionless variables

It is very instructive to work with dimensionless variables. For this, we define typical length and time scales. The typical length scale is `, so the natural definition is x0 = x/`. For the time scale, we use G, `, m and perform dimensional analysis [50]. Writing with unknown exponents and requiring the combination to yield time: we write first, `a Gb mc = [L]a+3b [M ]c−b [T ]−2b , implying a + 3b = 0, c − bp= 0, −2b = 1, yielding b = −1/2, c = −1/2, a = 3/2. This gives the time-scale, t0 = `3 /(Gm), and the dimensionless time variable is t 0 = t/t0 . Now we re-write the EOMs in the dimensionless variables. To avoid writing primes everywhere, we avoid the primes in the following. The equations are x¨ = −x−2 ,

(9.26)

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97

and r x˙ = −

2(1 − x ) . x

Equation 9.27 can be simply solved (substituting x = cos2 θ ) to yield √ q √ 2 cos−1 x + x(1 − x) . t= 2

(9.27)

(9.28)

Equation (9.26) can’t be solved so easily as it is a nonlinear equation. A linear equation is one where if x1 (t ) and x2 (t ) are two solutions of the equation, then a linear combination of these, αx1 (t ) + β x2 (t ), is also a solution for constants a, b. For a nonlinear equation, the linear superposition is not a solution. Let us write a power series solution in t. We may imagine that the particle does not start at rest and would be moving away. As a result, the series will only consist of even powers: ∞

x(t ) =

∑ a2nt 2n ,

(9.29)

n=0

with a0 = 1 as x(0) = 1. The solution is found to be x(t ) = 1 −

t 2 t 4 11t 6 73t 8 − − − −... 2 12 360 5040

(9.30)

9.2.2 Case II If the two objects have masses m1 and m2 , then the change in gravitational potential energy in bringing them from a distance L to x will be connected with the kinetic energy of the two masses: m1 x˙12 m2 x˙22 1 1 + = Gm1 m2 − . (9.31) 2 2 x L To get further, we need the conservation of linear momentum: m1 x˙1 + m2 x˙2 = 0

(9.32)

Observing that the relative velocity x˙ is x˙2 − x˙1 , and going over to dimensionless variables, the final equation turns out to be r dx0 2 ( 1 − x0 ) = − . (9.33) dt 0 x0 The solution is √ q √ 2 0 −1 0 0 0 t = x (1 − x ) + cos x . 2

(9.34)

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9.2.3 Case III Let us now endow the two objects with charges q1 and q2 in addition to their masses. The conservation of energy gives m1 x˙12 m2 x˙22 1 1 + = (Gm1 m2 − kq1 q2 ) − . (9.35) 2 2 x2 − x1 L Show that the EOM for x = x2 − x1 is kq1 q2 1 1 2 x˙ = 2G(m1 + m2 ) 1 − − . Gm1 m2 x L

Question

(9.36)

Obtain the expression for the velocity. Furthermore, prove that the expression for t

t0 = p

L3 / [ G ( m

1 + m2 )(1 − (kq1 q2 / (Gm1 m2 )))]

(9.37)

in terms of x0 = x/L is √ q √ kq1 q2 2 0 x0 (1 − x0 ) + cos−1 x0 , 0 ≤ x0 < 1, < 1, t = 2 Gm1 m2 √ q √ 2 kq1 q2 −1 0 0 0 = x (x − 1) + cosh x , x0 > 1, > 1. 2 Gm1 m2

(9.38)

Question Calculate the distance from the earth at which the gravitational pull of the sun on a particle of matter is equal to the gravitational pull of the earth on the same particle. The distance is about 2,68,333 km, which is about two-thirds as great as the average distance from the earth to the moon. Our moon is constantly under greater pull from the sun than it is from the earth [51]. In fact, the pull of the sun is more than twice as large that of the earth. At 14,88,00,000 km from the sun, the gravitational field strength due to the sun is approximately 6 × 10−3 N/kg; at 38,400 km from the earth, the field strength due to the earth is only 2.7 × 10−3 N/kg.

9.3 Satellite Paradox The linear velocity of an artificial satellite increases as it descends in a near-circular orbit. Its acceleration in the direction of motion is found to be the same as if the (minus the) air drag force were pushing the satellite [52] and we would like to understand this paradoxical situation. The mass of the satellite at a distance r from the centre of the earth is denoted by m. First of all, consider the case of zero drag and near-circular orbit where its speed is denoted by vc . In addition to the gravitational force w due to earth of radius R, there is the

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drag force D. On a stable circular orbit, centrifugal force equals the gravitational force. Mass of the earth can be re-written as GmM gR2 . = mg implying M = R2 G

(9.39)

GmM mgR2 mv2c , = = r2 r2 r

(9.40)

So, w=

giving the velocity in the circular orbit, v2c =

gR2 . r

(9.41)

9.3.1 Descending path on a near-circular orbit Consider a satellite travelling in a near-circular orbit, along a spiral with an instantaneous descent angle θ , measured from local horizontal. The direction of drag is assumed to be exactly opposite to the direction of satellite velocity. Let us equate the components of forces parallel and perpendicular to the motion. Along the direction of velocity of the satellite, w sin θ − D = ma1 = m

dv . dt

(9.42)

The perpendicular component of acceleration a2 appears in w cos θ = ma2 ,

(9.43)

and has a natural form, v2 /r. Now we assume that the orbit is nearly circular, with radius d (θ +φ ) of path curvature, rc and angular velocity, ωc = dt (Fig. 9.1): dθ dφ v a2 = v = vωc = v + . rc dt dt

(9.44)

The radial velocity is dr = −v sin θ . dt

(9.45)

The component of v perpendicular to r is r

dφ 1 v cos θ = v cos θ , from where = . dt dt rdφ

(9.46)

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Fig. 9.1

A satellite is at an inclination ϕ w.r.t. reference axis. A drag, D is opposing the velocity v w.r.t. the local horizontal. The role of drag is discussed in the motion of the satellite.

Thus, we can write dφ v cos θ = , dt r dθ dθ v cos θ = . dt dφ r

(9.47)

Substituting Eq. (9.47) in Eq. (9.44) gives dθ v2 cos θ a2 = 1+ . r dφ

(9.48)

Using this, Eqs (9.40) and (9.43) v2 dθ mgR2 m 1+ =w= 2 . r dφ r

(9.49)

Hence, −1 −1 gR2 dθ dθ 2 1+ = vc 1 + . v = r dφ dφ 2

As we are assuming that the descent angle is small, we can neglect Consequently, v2 ≡ v2c .

(9.50) dθ dφ

in comparison to 1.

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101

Now we need to express a relation combining descent angle with D. From Eq. (9.45), we use 1/dt in Eq. (9.42) gives w sin θ − D =

vdv (−m sin θ ). r

(9.51)

We know that v ≡ vc , hence v

dvc v2 dv = vc =− c. dr dr 2r

(9.52)

With this, Eq. (9.51) gives w sin θ − D =

w sin θ , 2

(9.53)

or, sin θ =

2D . w

(9.54)

Finally, we would like to use this result in determining the rate of increase of satellite velocity as a function of the drag force. We see that w sin θ − D = w

2D dv −D = m , w dt

(9.55)

and whence m

dv = D. dt

(9.56)

What happens, in fact, is [52] that gravity accelerates (naturally) rather than the drag. However, the drag force produces a descent angle in a manner that the resultant of gravitational and drag has a component in the direction of the satellite velocity.

9.4 Rotation Curves: an Anomaly Astronomers in the mid-twentieth century, while studying the motion of the galaxies, found that the galaxies in the outer parts of the galactic clusters were moving at a speed which is far too greater than that could be accounted for by the visible matter. With that speed, it seems impossible for the galaxies to remain gravitationally bound to the cluster. Similar conclusions were made on the basis of observations on stars at the edge of the galaxy. By using Newton’s laws, the orbital velocity of stars in the galaxy can be estimated. Such curves are called rotational curves of galaxies. The discrepancy between theoretical estimate and observation remains an outstanding open problem. To understand the problem, let us consider the motion of a star of mass ms under the gravitational pull of the galaxy of mass M. Let us assume the galaxy of spherical shape of

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constant density and radius r. The gravitational force will result in a circular motion of a star with orbital velocity, vs : r GMms GM ms v2s = . (9.57) , implying vs = 2 r r r For a star moving on the periphery of the galaxy we have M = Mgalaxy as the whole galaxy attracts. On the other hand, for the stars in the interior of the galaxy, M < Mgalaxy which depends on the matter distribution in the galaxy. For the assumed “spherical galaxy” of uniform density, we have ! Mgalaxy M (r ) = × 4πr3 . (9.58) 4πR3galaxy For these two cases, s GMgalaxy vs = r , Rgalaxy r

=

GMgalaxy , r

r < Rgalaxy ,

r ≥ Rgalaxy .

(9.59)

Thus, the velocity of stars within the galaxy increases linearly with distance from the centre while at the periphery, the velocity varies as the inverse square root of the distance. Clearly, real galaxies have a complex matter distribution, the general variation of the orbital velocity is similar. Since real galaxies are not exact spheres however, they have a gradient of mass distribution, the two parts of the curves smoothly join each other in the region r ∼ Rgalaxy . Over the years, observation on thousands of galaxies given rotation curves in disagreement with the predictions based on Eq. (9.59). The observational rotation curve is such that it does not decay with r as in Eq. (9.59), it just remains flat. This suggests four possibilities: (a) Eq. (9.57) is wrong; (b) the formula of gravitational force, GM m/r2 is incorrect; (c) Newton’s law, F = ma needs correction; (d) there is some hidden mass, some dark matter. At the length-scale to which these observations are referring to, it is currently believed that the resolution will come by an understanding of the physics of dark matter, or/and, modification of Newton’s law for accelerations lesser than 10−12 m/s2 .

9.5 The Rosetta-Philae Comet Mission Our understanding of central forces, and the nature of orbits comes to a real test when we plan to send a rocket out in the space. There have been many missions, very exciting. However, perhaps the most challenging until now was landing on a comet. On November 12, 2014, after a voyage of ten years and hundreds of millions of kilometres, the European

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Rosetta–Philae spacecraft caught up with comet 67P and dropped a lander, Philae, upon it. This is simply amazing, at least. It becomes a thriller if you now consider that the size of this comet is just 4 km across, with an escape velocity of 2 m/s - you could ‘walk and escape’. √ We have seen that velocity of a near-surface satellite in an orbit is gR where g is the acceleration due to gravity of the body of radius R. For the Earth, this about 8 km/s. A near-surface satellite of Earth covers a distance of approximately 40,000 km in about 90 minutes. For 67P, radius is smaller by a factor of 3200 and g by a factor of 10,000. The landing speed was planned to be 1 m/s. Like in any thriller, the little lander dropped but due to a small rocket burn and due to failure of anchoring, it bounced. It went up a hundred metres and landed again, now perfectly, a kilometre away. Calculate (a) the velocity with which it must have bounced off, and, (b) the time it took for it to land again, (c) what must be the weight of the lander of mass 200 kg on 67P?

10

Chapter

Dimensional Analysis We need units and dimensions of physical quantities as soon as we begin to describe a natural phenomenon. More often than not, the phenomenon we wish to describe is more complicated than the basic tenets known to us. To discover dependence on basic relevant quantities without the knowledge of all the details of its derivation, the powerful method of dimensional analysis is employed. This is done by constructing dimensionless quantities by knowing the physical quantities and the dimensions involved in the description. Buckingham π theorem guarantees (n − k) dimensionless quantities for n physical quantities and k dimensions [53]. There is a function of all the dimensions satisfying f (π1 , . . . , πn−k ) = 0 where πi is the i-th dimensionless quantity. For example, the oscillations of a pendulum are described in terms of the mass of the bob m, the length of the rod `, time period of the oscillations t, and the acceleration due to gravity, g - four quantities in all. The dimensions involved are mass M, length L, and time T . Hence, we are looking for (4 − 3 =) one dimensionless quantity. That is, ma `bt c gd is dimensionless, implying [M ]a [L]b+dp [T ]c−2d = constant. We see that a is zero, and with c = 1, d is 1/2 and b is -1/2, or f (t g/` p) = 0. There is just one dimensionless quantity in this case. Experiments reveal that t g/` = 2π.

10.1

Black Holes at LHC

There is a speculation that the particle physics experiment at the Large Hadron Collider (LHC) at CERN (near Geneva, Switzerland) will produce black holes. Assume that its mass m is much smaller than the mass of earth, ME with density ρ which is larger than near the centre of the earth. The black hole will be attracted towards the centre of the earth, performing oscillation. Describe the motion of the black hole, and estimate its radius and mass. The black hole will perform a simple harmonic motion, which we estimate using dimensional analysis. The relevant quantities are ME , G, RE , ρ. The time period of the black hole is given by the formula: s R3E 1 TB = or ∼ √ . (10.1) GME Gρ

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105

We now compute the force experienced by a black hole as a function of radius from the centre of the earth. Once the black hole enters the earth, the pull experienced by it is from the earth’s mass interior to its radial position: F = m¨rrˆ = − where ME (r ) = so, r¨ = −

GmME (r ) rˆ, r2

(10.2)

4πρr3 , 3

GME r. R3E

(10.3)

This gives the angular frequency and hence the time period, 2π = 2π TB = ω

s

R3E or = GME

s

3π . Gρ

(10.4)

Assume that the collision producing a black hole involves an approximate amount of energy, 5 × 1020 Nm to bring a black hole to rest. The equivalent mass is m = E/c2 ∼ 5000 kg. For the radius, dimensional analysis gives R ∼ Gm/c2 . This gives the radius, 5 × 10−24 m approximately, a billionth of the proton’s size!

10.2

Nuclear Explosion

The quantities involved are the energy yield E, radius of the fireball R at time t, density of air ρ (= 1.2 kg/m3 ). With four quantities and three dimensions, there can be only one dimensionless quantity. We can construct it as E a Rbt c ρ d =

[M ][L]2 [T ]−2 )a [L]b [T ]c ([M ][L]−3

d

= [M ]a+d [L]2a+b−3d [T ]−2a+c = dimensionless.

(10.5)

Hence, c = 2a, d = −a, b = 3d − 2a = −5a. Choosing b = 1, we get a = −1/5, c = −2/5, d = 1/5. So, we have the remarkable estimate obtained by Sir Geoffrey Taylor in 1947, R∼

Et 2 ρ

51 .

(10.6)

106

10.3

Mechanics, Waves and Thermodynamics

Insect Flight

Let us obtain a formula relating wing-beat frequency with mass and wing area for a flying insect [54]. There are five basic variables: 1. ρ, the density of air (ML−3 ), 2. m, the mass of insect (M), 3. g (LT −2 ), 4. ν, the frequency of the characteristic wing beat (T −1 ), 5. A, the effective area of the wings (L2 ). According to the Buckingham pi theorem, with five quantities and three dimensions, the number of dimensionless ratios are two. These could be ρA3/2 /m and ν 2 A1/2 /g. Hence, we can write F (ρA3/2 /m, ν 2 A1/2 /g) = 0.

(10.7)

To obtain the dimensionless quantities, let us write ρ a mb gc ν d Ae = [M ]a+b [L]−3a+c+2e [T ]−2c−d .

(10.8)

We have the following equations: a + b = 0, − 3a + c + 2e = 0, − 2c − d = 0. Case I If d = 0, then c = 0, e = 3a/2, b = −a. We get ρ a m−a A3a/2 , i.e., ρA3/2 /m. Case II The choice of a = 0 implies b = 0. This gives e = d/4, c = −d/2. We get the other quantity, ν 2 A1/2 /g. Equation (10.7) can be re-written as 3/2 ρA ν 2 A1/2 = φ , g m 1/2 −1/4

or ν = g

A

ψ

ρA3/2 m

.

(10.9)

Let us estimate for an ordinary bee (apis mellifera) ρ ∼ 1.2 kg/m3 , m = 10−6 kg, A ∼ 3/2

0.006 cm2 . With these parameters, ρAm ∼ 5 × 10−4 , a small positive number. Using this fact, we may expand ψ in a power series: ψ (x) = kxα (1 + a1 x + a2 x2 + . . .) ∼ kxα to a leading order.

(10.10)

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107

Substituting this in Eq. (10.9), we have 1

ν = g 2 ρ α km−α A

3α − 1 2 4

.

(10.11)

Define L = m/A, 1

α

1

ν = g 2 ρ α km 2 − 4 L−

3α − 1 2 4

.

Experimentally, ν ∝ L. The above equation gives ν ∝ L− frequency is 1 1 g 2 m 2 . ν =k ρ A

(10.12) 3α − 1 2 4

. Hence, α = −1/2. The

(10.13)

This formula gives for a honey bee, ν ∼ 350 s−1 whereas the experimental observation gives 200 − 250 s−1 . The discrepancy is explained in [55, 56].

11

Chapter

Oscillations Opinion is like a pendulum and obeys the same law. — Arthur Schopenhauer

In Chapter 5, we discussed the simple harmonic motion of a pendulum, and the case when bob is of variable mass. In this Chapter, we now consider this subject in more detail. This will also lay the foundation of further chapters where we consider more general situations of coupled oscillators, eventually leading us to waves and the phenomenon of sound. Oscillations and waves surround us. Clocks, string of a tanpura, tea sloshing back and forth in a cup, pulse travelling along a rope, ocean waves approaching the shore, light, sound, matter waves . . . - amazing !

11.1

Free Oscillations

The motion of a simple pendulum, a mass attached to a spring moving on a frictionless surface, oscillation of charge in an LC circuit - all correspond to free oscillations. Oscillations are due to an interplay of inertia and a return force. To a “displacement” ψ, the return force tries to bring the system to equilibrium by imparting suitable “velocity”, dψ/dt - larger the ψ, larger the return force. Inertia opposes the change dψ/dt. Hence, there is an oscillation that is set up. In the example of an LC circuit, ψ stands for the charge. Because electrons repel each other, they distribute themselves on a capacitor plate as uniformly as possible. This involves a change in charge with time, which is opposed by the inductance (providing the inertial term). The frequency of oscillation ω is given by ω2 =

return force . (unit displacement) × (unit mass)

(11.1)

For instance, for a simple pendulum, the ω 2 = Mgψ/(`ψM ) = g/`. For a one-dimensional mass-spring system where a mass M has two identical springs of spring constant K on either side, with other ends connected to fixed walls, the frequency is given by ω 2 = (2KM )/(Mψ ).

Oscillations

11.2

109

Transverse Oscillations in Mass-spring System

Let us consider the motion of a mass attached to two springs as shown in Fig. 11.1, without gravity. In the rest position, the springs will be horizontal, each of length ‘a’ along the x-axis, the oscillations take place along y-axis. With spring constant, K and unstretched length a0 , tension is T = K (` − a0 ) along the axis.

Fig. 11.1

A mass is performing transverse oscillations, which happen to be nonlinear. This is unlike the longitudinal oscillations of a mass connected to two springs on either sides.

If the mass moves along y direction, each spring contributes a return force T sin θ along y-axis. We may approximate sin θ by y/`. Newton’s second law can be written: M

y a0 d2y = F = −2T sin θ = −2K ( ` − a ) = −2Ky 1 − . y 0 dt 2 ` `

(11.2)

The length of the spring ` is a function of y. The return force is not linearly proportional to y, thus the equation is not that of a simple harmonic oscillation. Let us now assume that the spring is a slinky - a helical spring with a0 /a 1 (about 1/60). Also, because ` > a, a0 /` can be neglected. With this, Eq. (11.2), we get d2y = −ω 2 y, dt 2

ω2 =

2K 2T0 = (for a0 = 0) M Ma

(11.3)

where T0 = K (a − a0 ). The ratio of the longitudinal to transverse frequency is ωlong. 1 =√ . ωtrans. 1 − a0 /a

(11.4)

Longitudinal oscillations are more rapid than the transverse oscillations.

11.3 Compound Pendulum Consider a compound pendulum, where a rigid rod of mass M and length L hangs from one end and oscillates under gravity. If θ is the angle with respect to the vertical direction, the potential energy due to gravity is V (θ ) = −Mg L2 cos θ . The kinetic energy is the rotational

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kinetic energy, 12 I ω 2 where I is the moment of inertia of the rod about one end, ML2 /3 and ω = dθ /dt. Total energy is E =−

1 ML2 ˙ 2 M gL cos θ + θ . 2 2 3

(11.5)

Since total energy is conserved, dE/dt = 0, implying ML2 ˙ ¨ 1 MgL sin θ θ˙ + θθ = 0 2 3 3g θ¨ + sin θ = 0. 2L

(11.6)

This is a nonlinear equation. It linearizes if sin θ can be replaced by θ . This is possible for θ < π/6 to a reasonable approximation as sin(π/6) = 1/2 ≡ π/6. The usage of the Taylor expansion of sin θ = θ − θ 3 /3! + θ 5 /5! − . . . gives us, to linear term: p θ¨ + ω02 θ = 0 where ω0 = 3g/2L. (11.7) Comparing the simple harmonic oscillations of this pendulum with a compound pendulum, √ we see that the frequency here is higher by a factor 3/2 for the same length.

11.4 Damped Harmonic Oscillator Suppose that there is a damping force acting on the oscillator which is proportional to the velocity and opposes the motion. This can be due to some frictional force. For instance, if this force is Bx˙ with positive B, then the equation of motion is mx¨ + kx = −Bx. ˙

(11.8)

Calling B/m as 2γ, we rewrite the equation as x¨ + 2γ x˙ + ω02 x = 0.

(11.9)

This is the equation describing the motion of a damped harmonic oscillator. There are three different cases to consider. Case I γ < ω0 : underdamped case The equation above is an ordinary differential equation with constant coefficients. The solution will be such that its derivatives are simply proportional to it. The only function of a real variable is an exponential, exp(αt ). Substituting this function into the equation yields values q of α in terms of ω0 , γ. qThe solutions in this case are a linear combination of e−γt cos(

ω02 − γ 2t ) and e−γt sin(

ω02 − γ 2t ).

Case II γ = ω0 : critically damped case The two solutions are e−γt and te−γt .

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111

Case III γ > ω0 : overdamped √ case −γt± γ 2 −ω02 t . These are both exponentially decaying functions. The two solutions are e The equation of a damped harmonic oscillator also appears in many areas in physics. For example, consider a LCR circuit. If the two plates of the capacitor have charges +Q and −Q, we have the equation LQ¨ + RQ˙ +

Q = 0. C

(11.10)

Comparing with (11.9), 2γ = R/L and ω02 = 1/LC.

11.5 Driven Damped Simple Harmonic Oscillator Let us now consider subjecting the damped harmonic oscillator to a time-dependent force f (t ). That is, x¨ + 2γ x˙ + ω02 x =

f (t ) . m

(11.11)

In general, this is hard to solve. However, we can solve when the force is a simple harmonic function of time, let f (t )/m = α cos(ωt ) where ω need not be equal to ω0 . The general solution of (11.11) can be found on application of the theory of ordinary differential equations. The solutions of the homogeneous equation are known from the previous Section. For a general solution, we need to add a particular solution. We may try a solution like x(t ) = c1 cos ωt + c2 sin ωt.

(11.12)

Substituting this in the above equation gives − c1 ω 2 cos ωt − c2 ω 2 sin ωt + 2γ (−c1 ω sin ωt + c2 ω cos ωt )

+ ω02 (c1 cos ωt + c2 sin ωt ) − α cos ωt = 0

(11.13)

Since cos ωt and sin ωt are linearly independent functions of time, it implies that a1 cos ωt + a2 sin ωt = 0

(11.14)

for all values of t; thus a1 = 0 = a2 . Using this idea above, we obtain two equations in unknowns c1 , c2 : −c1 ω 2 + 2γc2 ω + c1 ω02 − α = 0, −c2 ω 2 − 2γc1 ω + c2 ω02 = 0.

(11.15)

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Solving these, we get α (ω02 − ω 2 ) , (ω02 − ω 2 )2 + 4γ 2 ω 2

c1 =

2γαω

c2 =

(ω02 − ω 2 )2 + 4γ 2 ω 2

.

(11.16)

Let us now define two quantities R, φ such that q α R = c21 + c22 = q (ω 2 − ω02 )2 + 4γ 2 ω 2

(11.17)

and R cos φ = c1 , R sin φ = c2 . We have, therefore, the expressions cos φ = sin φ =

ω02 − ω 2 , (ω02 − ω 2 )2 + 4γ 2 ω 2 2γω

(ω02 − ω 2 )2 + 4γ 2 ω 2

.

(11.18)

We then have x(t ) = c1 cos ωt + c2 sin ωt

= R[cos ωt cos φ + sin ωt sin φ ] = R cos(ωt − φ )

(11.19)

which has to be compared to the driving term, α cos ωt. So we see that the response differs by a phase from the driving term. The above solution, Eq. (11.19) is a particular solution of the equation: x¨ + 2γ x˙ + ω02 x = α cos ωt. To get the general solution, we need to add the solution of the damped harmonic oscillator. We found that there are two solutions, both of which decay exponentially to zero as time becomes large, for γ > 0. The decaying solutions are called transients as they are only important for a finite time. For the oscillator to settle down, we need to wait for a long time so that the transients vanish and we have x(t ) = R cos(ωt − φ ). Resonance

For instance, for an oscillator, the energy is mx˙2 /2 + mω02 x2 /2 which is proportional to R2 rather than R. We see that 2 1 R = 2 . 2 α (ω − ω0 )2 + 4γ 2 ω 2

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113

Now let us consider the case of weak damping, i.e., γ ω0 . We need to find the value of ω for which (R/α )2 is maximum. This will correspond to the minimum of the denominator above - differentiating it w.r.t. ω 2 and setting it to zero, we get ω 2 = ω02 − 2γ 2 . Hence, ω=

q

ω02 − 2γ 2 ' ω0 if γ ω0 .

(11.20)

If ω0 γ, then we can write alternatively

(ω 2 − ω02 )2 + 4γ 2 ω 2 ' (ω − ω0 )2 (ω + ω0 )2 + 4γ 2 ω 2 ' 4ω02 [(ω − ω0 )2 + γ 2 ]. Hence, 2 R 1 1 . ' 2 α 4ω0 (ω − ω0 )2 + γ 2

(11.21)

(11.22)

A plot of (R/α )2 vs ω gives the lineshape at ω0 with width 2γ. For ω γ0 , we have a high peak at ω0 . This is called resonance because driving frequency is equal to the resonant frequency. At ω − ω0 equal to ±γ, the peak falls to half. The ratio Q = ω0 /2γ is called the quality factor of the resonance. The larger the value of Q, the higher and narrower the resonance will be. Let us now discuss about the phase difference φ . For ω0 γ, we see that cos φ = p

ω0 − ω

(ω − ω0 )2 + γ 2

sin φ = p

γ

(ω − ω0 )2 + γ 2

,

.

(11.23)

For ω ω0 , we find that φ ∼ 0; for ω ω0 , φ ∼ π, and, for ω = ω0 , φ = π/2. Exercise: For a harmonic oscillator with natural frequency ω0 , forced at a frequency ω with frictional force with coefficient 2γ (> 0), the differential equation governing the displacement is x¨ + 2γ x˙ + ω02 x =

F cos ωt. m

(11.24)

Show that in the steady state, on performing an average over 2π/ω, all the energy given by the driving source is dissipated by the frictional force. Thus, E = x˙2 /2 + ω02 x2 /2 is such that dE/dt = 0.

114

11.6

Mechanics, Waves and Thermodynamics

Beats

A superposition of two harmonic oscillations of frequencies ω1 and ω2 produces beats and modulation. Let ψ1 (t ) = A cos ω1t and ψ2 (t ) = A cos ω2t. Their superposition will yield (ω1 − ω2 )t (ω1 + ω2 )t ψ (t ) = ψ1 (t ) + ψ2 (t ) = 2A cos cos . (11.25) 2 2 We can define ωav = (ω1 + ω2 )/2 and ωmod = (ω1 − ω2 )/2 so that ω1 = ωav + ωmod , ω2 = ωav − ωmod . Equation (11.25) can be re-written as ψ (t ) = Amod (t ) cos ωavt,

with

Amod (t ) = 2A cos ωmodt.

(11.26)

If ω1 ≈ ω2 , then ωmod ωav , thus during the fast oscillation cos ωavt, Amod (t ) varies only slightly. Amod (t ) has a period of 2π after which it returns to the original value; however, it becomes zero two times. As a result, during a period, while listening to the beats, we hear nothing on two occasions. In between silences, we hear an average pitch. The cosine changes sign while passing through a zero however, we just hear loud or soft sound, not the sign. The ear-brain system is thus called a “square-law detector” [74]. Amod (t )2 has two minima for each modulation cycle. Repetitive rate of large values is called the beat frequency, ωbeat = 2ωmod = ω1 − ω2 .

(11.27)

With Amod (t ) = 2A cos ωmodt, we can see the square law algebraically: A2mod (t ) = 4A2 cos2 ωmodt = 2A2 [1 + cos ωbeatt ].

(11.28)

11.7 Another Instance of Simple Harmonic Motion Let us consider an interesting problem wherein a horizontal plate of weight W = mg is resting on two rapidly rotating wheels [78]. The wheels are separated by 2a (Fig. 11.2). The coefficient of friction between the plate and wheels is µ. Let us assume that at rest the CM of the plate is at a distance ‘a’ from each of the wheels. We displace the plate by x from equilibrium, measured towards the right from the equilibrium position. Let us call the points of contact of wheels with the plate are P and Q. We must consider the conditions of static equilibrium: Equilibrium of forces: Equilibrium of moments:

FP + FQ = W , 2aFQ − (a + x)W = 0.

(11.29)

From Eq. (11.29), FP =

a−x W, 2a

FQ =

a+x W. 2a

(11.30)

Oscillations

Fig. 11.2

115

A horizontal plate performs harmonic motion when supported by two rotating wheels.

Forces of friction at P and Q are in opposite sense with magnitudes µFP and µFQ respectively. Now we consider the cases where the wheels rotate inwards or outwards. 1. Inward rotation Motion of the plate will be in the left direction relative to the point P and to the right relative to the point Q. The forces of friction at these points are therefore + µFP and −µFQ respectively. The resultant horizontal force acting on the table is F = µFP − µFQ = −

µW x. a

(11.31)

The equation of motion of the plate is mx¨ = −

µW x. a

(11.32)

This shows p that the plate performs a simple harmonic motion with angular frequency, ωi = µg/a. 2. Outward rotation The plate moves to the right relative to A and to the left relative to B. Respective forces of friction are −µFP and + µFQ , with resultant F = −µFP + µFQ = +

µW x. a

(11.33)

The equation of motion is mx¨ =

µW x. a

(11.34)

The solution in this case is non-oscillatory: x(t ) = C1 eΩt + C2 e−Ωt

(11.35)

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where C1 ,C2 are constants, and, Ω = the initial conditions.

p

µg/a. The constants C1 ,C2 are obtained by

3. Rotation in the same sense Suppose the rotation is towards right, the forces of friction will be + µFP and + µFQ . Hence, F = µFP + µFQ = µW .

(11.36)

In this case, x¨ = constant.

11.8 Two Coupled Oscillators Consider two identical springs and two identical masses arranged in a manner that a mass m is at one end and the other one is between the two springs. This system is hung vertically under gravity. The spring constant is denoted by k and the equilibrium length of a spring is `. If x1 and x2 are the vertical positions of the two masses (increasing downwards) (Fig. 11.3), the kinetic energy is 1 1 KE = mx˙12 + mx˙22 . 2 2

Fig. 11.3

(11.37)

Two equal masses attached to identical springs, there appear two normal modes.

The potential energy is 1 1 PE = K (x1 − `)2 + K (x2 − x1 − `)2 − mgx1 − mgx2 . 2 2

(11.38)

Let us first find the values of x1 and x2 where the potential V (x1 , x2 ) has the minimum value. To compute this,

Oscillations

117

∂V = K (x1 − `) − K (x2 − x1 − `) − mg, ∂ x1 ∂V = K (x2 − x1 − `) − mg. ∂ x2

(11.39)

Let us denote the equilibrium solutions to these equations by x1 = `10 and x2 = `20 , given by ∂V ∂V = = 0 at x1 = `10 , x2 = `20 . ∂ x1 ∂ x2

(11.40)

To capture the nature of the potential (“cup” - or “cap” - like), we expand V (x1 , x2 ) about the equilibrium values. This is given by the expression ∂V ∂V + (x1 − `20 ) V (x1 , x2 ) = V (`10 , `20 ) + (x1 − `10 ) ∂ x1 x1 =`10 ,x2 =`20 ∂ x2 x1 =`10 ,x2 =`20 2 2 1 1 ∂ V ∂ V + + (x2 − `20 )2 (x1 − `10 )2 2 ∂ x12 x1 =`10 ,x2 =`20 2 ∂ x22 x1 =`10 ,x2 =`20 2 ∂ V + (x1 − `10 )(x2 − `20 ) + ... (11.41) ∂ x1 ∂ x2 x1 =`10 ,x2 =`20 Further, note that ∂ 2V ∂ 2V ∂ 2V = 2K, = K, = −K. ∂ x1 ∂ x2 ∂ x12 ∂ x12

(11.42)

We can rewrite Eq. (11.41) near equilibrium points: 1 V (x1 , x2 ) = V (`10 , `20 ) + (x1 − `10 )2 2K 2 1 + (x2 − `20 )2 K − (x1 − `10 )(x2 − `20 )K. 2

(11.43)

Let (x1 − `10 ) = y1 , (x2 − `20 ) = y2 . Then V (y1 , y2 ) = V (`10 , `20 ) + Ky21 +

K 2 y − Ky1 y2 . 2 2

(11.44)

The kinetic energy is my˙21 /2 + my˙22 /2. Newton’s equations give us my¨1 = −

∂V = −2Ky1 + Ky2 , ∂ y1

my¨2 = −

∂V = Ky1 − Ky2 . ∂ y2

(11.45)

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Define ω02 = k/m. These are linear, ordinary differential equations with constant coefficients, we expect exponential solutions. Let y1 by a1 e−iωt and y2 by a2 e−iωt . The differential equations then turn algebraic in a1 and a2 : −ω 2 a1 = −2ω02 a1 + ω02 a2 , −ω 2 a2 = −ω02 a2 + ω02 a1 .

(11.46)

We have to find the frequency ω and the ratio a2 /a1 . We have the equations,

(2ω02 − ω 2 )a1 = ω02 a2 , (ω02 − ω 2 )a2 = ω02 a1 .

(11.47)

Multiplying these equations, we have ω 4 − 3ω02 ω 2 + ω04 = 0, thus, ω 2 =

√ ω02 (3 ± 5). 2

There are two solutions. The frequency r √ √ 1 1+ 5 (3 + 5) = ω0 ' 1.618ω0 . ω+ = ω0 2 2 √ The number (1 + 5)/2 is the golden ratio. The ratio √ 2ω02 − ω 2 a2+ 1− 5 = ' −0.618 < 0. = a1+ 2 ω02 The other frequency is r √ √ 1 5−1 (3 − 5) = ω0 ' 0.618ω0 ω− = ω0 2 2

(11.48)

(11.49)

(11.50)

(11.51)

and the ratio is √ 2ω02 − ω 2 a2− 1+ 5 = ' 1.618 > 0. = a1− 2 ω02

(11.52)

The general solution of the Newton’s equations are y1 (t ) = a1− cos(ω−t + φ− ) + a1+ cos(ω+t + φ+ ), y2 (t ) = a2− cos(ω−t + φ− ) + a2+ cos(ω+t + φ+ ),

(11.53)

with four unknowns a1− , φ− , a1+ , φ+ . The coefficients a2− , a2+ depend on a1− and a1+ . These unknown parameters are fixed by the initial conditions on yi ’s and y˙i ’s.

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119

The larger frequency ω+ corresponds to a mode in which a2+ /a1+ < 0. That is, two masses move out of phase w.r.t. each other. This motion stretches or compresses the second spring more and therefore gives a larger frequency. The smaller frequency ω− corresponds to a mode in which a2− /a1− > 0. That is, two masses move in phase w.r.t. each other. In this mode, the second spring is extended or compressed lesser, hence the oscillation frequency is smaller. These are called the ‘normal modes’ of this system. There are two normal modes because there are two masses and they are moving in one dimension.

11.9

Three Coupled Oscillators

Let us consider a system of three identical masses, m joined by two springs (with equilibrium length, `). Hence, one mass is in the middle and the other two are at the endpoints. We now expect that there will be three normal modes. If xi (i = 1, 2, 3) denote the positions of the three masses, total kinetic energy is ∑i mx˙i2 /2. Potential energy is K 2 2 2 [(x2 − x1 − `) + (x3 − x2 − `) ]. The Newton’s equations are mx¨1 = −K (x1 + ` − x2 ), mx¨2 = −K [(x2 − x1 − `) − (x3 − x2 − `)], mx¨3 = −K (x3 − x2 − `).

(11.54)

Defining k/m = ω02 , and y1 = x1 − `, y2 = x2 − 2`, y3 = x3 − 3`, we get y¨1 = −ω02 (y1 − y2 ), y¨2 = −ω02 (2y2 − y1 − y3 ), y¨3 = −ω02 (y3 − y2 ).

(11.55)

As above, we seek solutions with the same frequency for the three masses: yi = ai eiωt . We find the three normal modes explained below: √ 1. Highest frequency The highest frequency is ω = 3ω0 , with a2 /a1 = −2 and a3 /a1 = 1. In this mode, particles 1 and 3 move in phase whereas the particle 2 in the middle moves in an opposite phase and twice the amplitude. This leads to maximum stretching and compressing of the springs and thus the frequency of this mode is the highest. 2. Middle frequency The frequency is ω = ω0 where the particles 1 and 3 move in opposite directions and the particle 2 is at rest. That is, a2 = 0 and a3 /a1 = −1. Because the middle particle doesn’t move at all, the oscillation frequency is just equal to the frequency of one spring with a fixed end.

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3. Smallest frequency The smallest frequency is zero where all the particles move by the same amount, a1 = a2 = a3 . the springs are unstretched in this case. The CM moves in this case. Note that the CM is at rest for the other two cases. The general solution with all three modes is √ y1 = a1 cos( 3ω0t + φ1 ) + b1 cos(ω0t + φ2 ) + c1 + d1t, y2 = −2a1 cos(ω0t + φ1 ) + c1 + d1t, √ y3 = a1 cos( 3ω0t + φ1 ) − b1 cos(ω0t + φ2 ) + c1 + d1t,

(11.56)

where we have six unknown parameters a1 , φ1 , b1 , φ2 , c1 , d1 , to be decided by the initial conditions.

11.10 Many Coupled Oscillators Let us now consider the case of an infinite number of particles of same mass m, with springs attached between the masses of the kind considered above. The position of the nth particle is xn where n is an integer. As above, the kinetic and potential energies can be written as T =

m +∞ 2 x˙n , 2 n=∑ −∞

V =

K +∞ (xn+1 − xn − `)2 . 2 n=∑ −∞

(11.57)

The EOM are mx¨n = −

∂V = −K [xn − xn−1 − ` + xn + ` − xn+1 ] ∂ xn

= −K [2xn − xn+1 − xn−1 ].

(11.58)

In terms of the displacement of the particles from their equilibrium position, yn = xn − n`, we can write the EOM as follows: y¨n = −ω02 [2yn − yn+1 − yn−1 ].

(11.59)

This is a model of particles oscillating back and forth in a crystalline solid - a solid in which the particles are uniformly spaced in equilibrium, with a separation of `. The oscillation modes here are longitudinal modes (not transverse). The frequency is given by ω02 = k/m. To solve Eq. (11.59), let us try again yn = αn exp(−iωt ) where we have to

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121

imagine the nature of the dependence of αn on n. We may guess an exponential dependence, i.e., y = exp(−iωt + λn ). We would obviously prefer that the oscillations about the equilibrium positions, yn should remain finite as n tends to ±∞. Thus the real part of λ should be exactly zero, and we should have λ = ik where k is real. Since ei2πn = 1 where n is an integer, or, k and k + 2π yields the same answer. We restrict the range of k to −π < k < π. On substituting yn = exp[i(kn − ωt )] in the equation for yn , we get from Eq. (11.59) −mω 2 = −K (2 − eik − e−ik ), i.e.,

ω2 =

K k (2 − 2 cos k) = 4ω02 sin2 . m 2

(11.60)

Hence, we obtain what is known as the dispersion relation, ω = 2ω0 | sin(k/2)|,

− π < k < π.

(11.61)

Through this relation, we know how ω depends on k, we write it as ωk (Fig. 11.4). Hence, there are an infinite number of normal modes of frequencies labelled by k. For a given value of k, the solution is yn = ak cos(kn − ωk t + φk )

Fig. 11.4

Dispersion relation for ω0 equal to one.

(11.62)

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with two arbitrary constants ak and φk . As k → 0, ωk → 0; this is the mode corresponding to a uniform motion of all the masses where no spring is compressed or stretched. These normal modes are longitudinal modes of sound in a one-dimensional crystalline solid. They are called phonons in a quantum mechanical description. 11.10.1 Phase velocity Given the form yn (t ) = ak cos(kn − ωk t + φk ), the points with the same phase (defining a wavefront) satisfy kn − ωk t + φk =constant. Hence n=

φk ωk t − + constant. k k

(11.63)

The average position of the nth particle is xn = n` = (ωk `/k)t + constant. So the velocity of this point is dxn /dt = (ωk `/k). This is called the phase velocity, given by the case discussed above by 2ω0 ` k v ph = sin . (11.64) k 2 This could be positive or negative depending on whether k > 0 or k < 0. That is, the wave could travel to x → ∞ or x → −∞. Also, note that v ph depends on k in this model. 11.10.2 Three-dimensional long-range order About ten years ago, a Coulomb crystal was created in a trap with 2700 Calcium ions (40Ca+ ) [79]. Their confining radius was about 90 µm, amounting to a density of approximately 1015 m−3 . Consider three such ions subjected to a confining potential mω 2 x2 /2, with ω/2π as trap frequency, 600 kHz, the equations of motion can be immediately set up: e2 1 1 2 x¨1 − ω x1 + + , (11.65) m (x2 − x1 )2 (x3 − x1 )2 with two other equations for x2 , x3 obtaining by cyclically rotating 1, 2, 3 in the equation. Discuss the normal modes bout an equilibrium configuration when the middle ion is at rest and the other two are moving in opposite directions.

11.11 Dissipation by a Rapidly Oscillating Potential We have seen that a periodically driven system can be turned into a resonance. Here we show that a particle of mass m can be made to lose energy by subjecting it to a rapidly oscillating potential, r 1 2 k V (x,t ) = V0 (x) cos(ωt + φ ); V0 (x) = kx , ω . (11.66) 2 m

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123

Due to the sign change of cos(ωt + φ ), the potential changes its character from attractive to repulsive over a period, T = 2π/ω. The average over a period of V (x,t ) is zero. The equations of motion is mx¨ = −

∂V (x,t ) = −kx cos(ωt + φ ). ∂x

(11.67)

The motion of the particle can be thought of as composed of two parts - a faster part, x˜(t ) that follows the oscillatory force, and, slower part, X (t ). While we can average over the faster motion, the particle essentially ends up being described by X (t ) on a longer timescale, which we need to determine. Thus, k X¨ + x¨˜ = − (X + x˜) cos(ωt + φ ). m

(11.68)

Averaging over time from zero to 2π/ω, and denoting the average by angular brackets, we expect that hXi will remain unchanged; so ¨ = X¨ = − k hx˜ cos(ωt + φ )i. hXi m

(11.69)

From Eqs (11.68), (11.69), we get k k x¨˜ = − X cos(ωt + φ ) − [x˜ cos(ωt + φ ) − hx˜ cos(ωt + φ )i]. m m

(11.70)

As x˜ ∼ cos(ωt + φ ), the last term in (11.70) becomes cos2 (ωt + φ ) - hcos2 (ωt + φ )i, which is cos(2ωt + 2φ )/2. We neglect this as it is a higher harmonic. Thus, simply we have k x¨˜ = − X cos(ωt + φ ). m

(11.71)

Taking X as constant over 2π/ω, an integral gives us x˜ =

k X cos(ωt + φ ). mω 2

(11.72)

For X, we have mX¨ = −kh

k k2 2 X cos ( ωt + φ ) i = − X, mω 2 2mω 2

(11.73)

which implies that there is an effective potential k2 X 2 /4mω 2 dictating the evolution of X. Question Show that the phase jumps from 0 to π/2 at respectively t− and t+ , the total variation in the energy of slow variable is m ˙2 ∆E = X (t+ ) − X˙ 2 (t− ) . (11.74) 2

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Using X˙ (t+ ) = X˙ (t− ) + kX (t )/mω, show that m kX (t ) kX (t ) + 2X˙ (t− ) . ∆E = 2 mω mω

(11.75)

Energy associated with slow variable is 1 k2 X 2 E = mX˙ 2 + . 2 4mω 2

(11.76)

This gives r X˙ (t− ) = ±

2E k2 X 2 − 2 2. m 2m ω

Eventually, show that with each jump, the energy loss is by about 27 per cent [80].

(11.77)

12

Chapter

Waves Like as the waves make towards the pebbl’d shore, so do our minutes, hasten to their end. — William Shakespeare

From a detailed discussion of Newtonian mechanics, now we have come to a treatment of delocalized objects. The ideas of particle mechanics will be suitably generalized to treat propagation of disturbances. Perhaps the most common instance consists in making a transverse wave propagate along a string. How the waves reflect, refract, and stand in one and two-dimensional systems will also be studied.

12.1

Transverse Modes of a String

Consider an infinitely long string over a real line, with points x ∈ [−∞, ∞]. At equilibrium, let us assume the string is straight and lying at y = 0. When the string is oscillating, y coordinate of each point varies with x and t. We will assume that the amplitude of oscillations is small, i.e., |dy/dx| 1 for all x and t. The string will be assumed to have a uniform density, µ and string tension T. The concept of string tension T is as follows: at any point in the string, the two parts of the string have forces T1 and T2 acting tangentially on them as shown. By Newton’s third law, T1 = −T2 . The magnitudes of both T1 and T2 are the same. Now, we would like to derive an equation describing the transverse waves associated with the motion of the string. Consider an infinitesimal section of the string of length dx. With µ as the linear density, the mass of this section is µdx. Mass times acceleration will give µdx∂ 2 y/∂t 2 . This must equal the vertical force. If we measure the angle from the horizontal on both sides of the line element (Fig. 11.5). At point x, the vertical force is −|T2 sin ϕ2 | = −T sin ϕ2 . We can write the force by dy (12.1) −T sin θ2 ' −T θ2 ' −T tan θ2 = −T dx x which will be small due to the slope being small. The vertical force at (x + dx) is upward, |T1 sin θ1 |. The net vertical force on the infinitesimal length is 2 dy dy d y T −T =T dx. (12.2) dx x+dx dx x dx2

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Mechanics, Waves and Thermodynamics

Fig. 12.1

The tension on the two sides of an element of a string shows us the resultant force. The Newton’s equation for the element with this force gives us the wave equation.

We now have the Newton’s equation 2 d y ∂ 2y , µdx 2 = T dx ∂t dx2

(12.3)

i.e., ∂ 2y T = ∂t 2 µ

d2y . dx2

The dimensions of T /µ is square of velocity. Let us define a velocity v = called the wave equation. It can be verified that a solution is y(x,t ) = ak cos(kx − ωk t + φk )

(12.4) p T /µ. This is

(12.5)

where ωk = v|k|. The wavefront (constant phase) is given by kx − ωk t + φk = constant. This implies x = ωk t/k + constant. The phase velocity v ph is ωk /k where v ph = ±v and |v ph | is independent of the magnitude of k. Here, k can be any real number. There can be two transverse modes of oscillations for each k. These correspond to oscillations along y and z directions. Moreover, what we have discussed is true for an infinite string. The waves we have discussed are having solutions given by Eq. (12.5). The variables p y is a function of x,t, the wave carrying the disturbance is travelling at a speed T /µ whereas the phase of cosine is also moving at ωk /k. This is called, for obviously natural reasons, a travelling or a progressive wave.

Waves

12.2

127

Standing Waves in One Dimension

Consider a string of length, `. There are two important cases: 1. The string has both ends fixed, so y = 0 at x = 0, `. The displacement of the string is given by y(x,t ) = A sin(kx) sin(ωk t + φ )

(12.6)

where k = nπ/`, n = 1, 2, 3, . . ., and ωk /k = v, the phase velocity. When y is zero at x = 0 and `, the wave travels to these endpoints and reflects. The amplitude is zero at these points. 2. The string is fixed at one end (x = 0) and free at x = `. Hence, y = 0 at x = 0 and ∂y ∂ x = 0 at x = `. The displacement is given by Eq. (12.6) with k = (n + 1/2)π/`, n = 0, 1, 2, . . . with ωk = kv. 12.2.1 Reflection and transmission of waves on a string Consider a non-uniform string made of two parts - part 1 for x < 0 has mass density µ1 , part 2 for x > 0 has mass density µ2 . We would like to see how a wave travels through this string and study what happens at the junction where the density changes. The tension T must be the same everywhere otherwise the stringp will start moving p along +xˆ or −xˆ direction. The wave velocity in x < 0 (x > 0) is v1 = T /µ1 (v2 = T /µ2 ). The wave equation is ∂ 2y ∂ 2y = v21 2 2 ∂t ∂x

for x < 0,

∂ 2y ∂ x2

for x > 0.

= v22

(12.7)

Let a wave of unit amplitude be incident from x = −∞, where k1 = ω/v1 > 0. This wave has the phase velocity v ph = ω/k1 = v1 > 0. When the wave strikes the point x = 0, it can get partly reflected and partly transmitted. Let us say the reflected wave is rei(−k1 x−ωt ) where phase velocity is −ω/k1 = −v < 0 where r is the reflection amplitude. Let the transmitted amplitude τ, so τei(k2 x−ωt ) where phase velocity is ω/k2 = v2 > 0. The frequency is the same at all places however, k1 , k2 are different (i.e., wavelengths 2π/k1 and 2π/k2 for x < 0 and x > 0). Putting everything together, we have y(x,t ) = ei(k1 x−ωt ) + rei(−k1 x−ωt ) ,

= τei(k2 x−ωt ) ,

x > 0.

x < 0, (12.8)

Let us now try to express r and τ in terms of v1 and v2 . As the wave equation is of second order, it is good to assume that both y and ∂ y/∂ x are continuous at x = 0, i.e., y(x)|x→0− = y(x)|x→0+ for all t,

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∂ y ∂ y = ∂ x x→0− ∂ x x→0+

for all t.

(12.9)

These equations yield 1 + r = τ,

ik1 − ik1 r = ik2 τ.

(12.10)

As ki = ω/vi , (i = 1, 2), we get (1 − r )/v1 = τ/v2 . We finally get r=

2v2 v2 − v1 and τ = . v2 + v1 v1 + v2

(12.11)

There are some interesting special cases: 1. µ1 = µ2 implies v1 = v2 There is no difference between the parts, so r = 0 and τ = 1. 2. µ2 µ1 implies v2 v1 Right part of the string is much heavier than the left part, so r = −1 and τ = 0. For x < 0, y = ei(k1 x−ωt ) − ei(−k1 x−ωt )

= 2i sin(k1 x)e−iωt .

(12.12)

3. µ2 µ1 implies v2 v1 Right part is much lighter than the left part. Hence, we get r = 1 and τ = 2. For x = 0, we see that y = ei(k1 x−ωt ) + ei(−k1 x−ωt )

= 2 cos(k1 x)e−iωt .

12.3

(12.13)

Standing Waves on Planar Membranes

A membrane of a certain shape, clamped at the edges, can be set into vibrations by tapping. The disturbance will travel to the edges, reflect and interfere, and eventually form a standing wave. It is a pattern of undulations, depending on the shape of the membrane and the energy given to the membrane. Let us assume that the membrane is in the z = 0 plane. Hence, the amplitude of vibrations, ψ (x, y) will satisfy the two-dimensional Helmholtz equation: ∂ 2ψ ∂ 2ψ + 2 + k2 ψ = 0. ∂ x2 ∂y

(12.14)

Waves

129

The membrane is at rest on the edges, so at the boundary ψ (x, y) = 0. For different shapes, we need to solve this equation. To begin with, let us consider a rectangular shape with length L and breadth B. The reader can immediately verify by substitution into Eq. (12.14) that the solutions are: r mπx nπy 4 sin sin (12.15) ψ (x, y) = LB L B where m, n are integers. The wavenumber cannot take all possible real values, it is quantized. Simple substitution results in k2 =

mπ 2 L

+

nπ 2 B

=

π2 2 (m + α 2 n2 ), L2

(12.16)

denoting L2 /B2 by α 2 . For simplicity, we may choose L = π. The solution makes a checkerboard pattern (see e.g., Fig. 12.1). The domains corresponding to the sign of the function - up or down. They are separated by lines on which the membrane is at rest. These are nodal lines and the domains are called nodal domains.

Fig. 12.2

The vibrations set on a rectangular membrane settle into a standing wave pattern. We have taken L = π and B = 1. The case shown is for (m, n) = (4, 5). The number of nodal domains ν4,5 is 20.

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Mechanics, Waves and Thermodynamics

The nodal lines can be immediately seen to be straight lines by setting Eq. (12.15) to zero for any m, n. For the case shown in the figure, the vibration is zero when either sin 4x = 0 or sin 5y = 0. The condition on sin 4x implies that 4x = mπ or the straight lines, x = m(π/4); m can be 0, 1, 2, 3, 4. These correspond to four vertical lines. Similarly, there are five horizontal lines. The total number is twenty. In general, it will be mn. There are special shapes for which the solutions of the Helmholtz equation can always be written as a single product of some functions. Unfortunately, even for nice shapes like a circle or an ellipse or a confocal parabola, the solutions are respectively written in terms of Bessel functions, Mathieu functions, and Kummer functions [66]. However, till we learn about them, we can enjoy the patterns they make. We see the annular membranes made by circles and ellipses, set into vibration (Fig. 12.2). For the shape of the membrane as confocal parabolae (Fig. 12.3), the Eq. (12.14) is solvable and the nodal patterns made by the solutions are seen in Fig. 12.4. For all the above shapes, it is easy to find an expression for the number of nodal domains in terms of the quantum numbers [67, 68]. For triangular-shaped membranes, the analytical solutions of the Helmholtz equation are not known. For certain special cases - right isosceles, equilateral, hemi-equilateral triangles, the solutions can be written down exactly in terms of sums of trigonometric products. For example, for the right isosceles triangular membrane of side length π, the solutions are ψ (x, y) ∼ sin(mx) sin(ny) − sin(nx) sin(my).

(12.17)

For these non-separable triangular shapes, counting the number of nodal domains has been an open problem for a long time. Some recent results (see Fig. 12.5) can be seen in [69].

Fig. 12.3

Nodal domains for circular (left panel) and elliptical (right panel) annular membranes, the modes shown are respectively with quantum numbers (8, 9) and (8, 5).

Waves

Fig. 12.4

131

The confocal parabolae form a domain as shown here in terms of parabolic coordinates. The membrane has the shape as shown, clamped at the boundary. The solutions of the Helmholtz equation is possible in terms of the Kummer functions. See the nodal domains and nodal lines of the solutions in Fig. 12.5.

12.4 Speed of Sound in Air In the last Chapter, we discussed the case of many coupled oscillators. We found the dispersion relation, the slope of which gives the speed of sound. Sound has only longitudinal modes in gases and liquids. In solids, sound has both longitudinal and transverse modes. These modes consist of density oscillations. We now find an expression for the speed of sound in greater detail. In the longitudinal mode, the oscillations of the particles are in the same direction as the direction of propagation of the sound. In gas or liquid, the pressure tends to be same everywhere, thus opposing changes in density. There is a tendency to restore constant density everywhere. Hence there occur longitudinal oscillations. However, if the particles move in the transverse direction, there is no change in density anywhere. There is no restoring force, and hence no oscillation. In solids, the restoring force is not provided by pressure. Particles in solid are subjected to restoring forces which can be imagined to be mediated by springs. These springs oppose relative motions of the particles regardless of whether they are in transverse or longitudinal direction. Hence, they can cause both transverse and longitudinal oscillations of density. 12.4.1 Newton’s derivation The speed of sound was first calculated by Newton. He assumed that temperature remains constant because Newton didn’t know about atoms and molecules.

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Mechanics, Waves and Thermodynamics

Fig. 12.5

Nodal patterns for two kinds of solutions - symmetric and antisymmetric about the central vertical line.

In equilibrium (pressure P0 ), let us consider two planes of cross-sectional areas A, separated by dx, so the volume is Adx. Suppose that in a longitudinal mode travelling along x-direction, the two planes deviate from their positions by amounts y and y + dy. So the two planes are at x + y and x + dx + y + dy. Here y is a function of x, so dy = ∂∂ xy dx. The new volume between the planes is V + ∆V = A(dx + dy) ∂y = Adx 1 + , ∂x

(12.18)

implying ∆V = Adx ∂∂ xy . Due to this change in volume, the pressure between the planes changes to P = P0 + ∆P. To find the relation between ∆V and ∆P, Newton used Boyle’s law where T remains constant, then PV = constant. Hence, V ∆P + P0 ∆V = 0

(12.19)

where we are assuming that ∆P/P0 and ∆V /V are very small. Hence, ∂y

∆P = −P0

= −P0

Adx ∂ x ∆V = −P0 V Adx ∂y , or ∂x

∂y P = P0 1 − . ∂x This relation is incorrect. Let us now derive the correct relation.

(12.20)

Waves

133

12.4.2 Correct derivation (Laplace) The problem is that PV =constant is not true. The frequency of sound is such that there is not enough time for heat to be exchanged between different regions. The correct assumption is that different regions are thermally isolated from each other and we should use adiabatic expression, PV γ =constant. For air, which is mainly consisting of diatomic molecules like N2 , O2 , γ = 1.4. If a molecule has D degrees of freedom, it turns out that γ = (D + 2)/D. A gas containing monoatomic molecules has γ = 5/3 since a molecule with only one atom has D = 3, three translational degrees of freedom in three dimensions. For diatomic molecules, D should be six - three translational degrees of freedom of the CM, two rotational degrees of freedom (around the two axes which are perpendicular to the axis of the molecule), and one vibrational degree of freedom (vibrations along the axis of the molecule).

Fig. 12.6

Nodal patterns for isosceles triangle and equilateral triangle [69]. Note that the nodal curves are quite complicated and the domains are irregularly shaped. Only recently have we found the counting function of nodal domains. Even so, there remain many unsolved problems.

It turns out that at temperatures around 20 degrees Celsius, the vibrational motion at frequency ω is not possible because the corresponding energy h¯ ω kB T . For air, D = 5 (three translations and two rotations) and γ = 7/5 = 1.4. With the adiabatic equation, we can find the change in pressure as follows:

(P0 + ∆P)(V + ∆V )γ = P0V γ , ∆P ∆V γ or 1 + 1+ = 1, P0 V i.e.,

∆P ∆V +γ = 0. P0 V

(12.21)

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Mechanics, Waves and Thermodynamics

Hence, ∆P = −γP0

∂y ∆V = −γP0 . V ∂x

(12.22)

Now we apply Newton’s second law to a small volume element of air. Mass between the two planes separated by a distance dx is ρV = ρAdx, and the acceleration is ∂ 2 y/∂t 2 where y is the deviation from the equilibrium position. The net force on this volume of air is due to the pressure difference at x and x + dx, i.e., Fx = A[P(x) − P(x + dx)] ∂y ∂y = A P0 1 − γ − P0 1 − γ ∂x x ∂ x x+dx ∂y ∂y ∂ 2y = AP0 γ − = AP0 γ 2 dx. ∂ x x+dx ∂x x ∂x

(12.23)

The Newton’s law gives 2 ∂ 2 y P0 γ ∂ 2 y 2∂ y = = v ∂t 2 ρ ∂ x2 ∂ x2

where v is the speed of sound,

(12.24) p γP0 /ρ.

Problem Calculate the speed of sound for an ideal gas at STP and at some temperature, 25◦ Celsius. Solution We know an ideal gas obeys PV = nRT with R = 8.314 J/(mol K). Taking one mole and STP (T = 273K, P = 1atm = 1.01 × 105 N/m2 , V = 22.4 L = 22.4 ×10−3 m3 ), we see that PV = 2262 J. Moreover, RT = 8.314 KJ × 273K = 2270 J. The close agreement shows that air is indeed close to an ideal gas at STP. At STP, the density of air is about (29 g/mol)/(22.4 L/mol) = 1.3 kg/m3 . The mass is calculated by taking an average of 80 per cent Nitrogen and 20 per cent Oxygen. Hence, the speed of sound at 0◦ C is r 1.01 × 105 × 1.4 = 329.8 m/s. (12.25) v0◦ C = 1.3 What is the velocity of sound at 20◦ C? Going back to the ideal gas law, we need to find P/ρ. Divide PV = RT by 29 g/mol, then P

(29 × 10−3 kg)/(22.4 × 10−3 m3 ) RT P = ' 287 T. ρ 29 × 10−3 kg

=

RT 29 × 10−3 kg (12.26)

Waves

135

Hence, s v=

γP p = γ287 T. ρ

The speeds of sound at two different temperatures are related as r vT1 T1 = . vT2 T2

(12.27)

(12.28)

We then find that r v25◦ C = v0◦ C

298 = 344.6 m/s, 273

in agreement with the observations made experimentally.

(12.29)

13

Chapter

Sound of Music As the life and processes of the world go on the actions which take place are accompanied by these tremors, and we live in the world of sound. — William Henry Bragg

Hindustani classical music is based on what are called ragas. Ragas are constructed by choosing at least five notes, specifying their usage. One of the most striking differences of this music with Western classical music is that the raga is born as soon as one begins to sing or play an instrument. In Western classical music, the music is completely written and the player presents her (his) interpretation. In Hindustani classical music, given the rules, the artist begins and the cognoscenti immediately recognize the raga. The untrained people cannot recognize the raga. However, something remarkable happens. The moment an artist commits a mistake, a lay person invariably understands the discord. This has made the author feel that there is something more fundamental about music in its relationship with humans, or may be even animals and plants. Music provides a bridge between arts and sciences. In the following, we present very elementary ideas on physics of music.

13.1

Physics of Music

The ear-brain system immediately makes out the difference between music, news message, or traffic. This system understands the changes in acoustic energy in terms of loudness and pitch. To understand this, we need to understand the propagation of acoustic waves in the context of musical instruments. For a pure sinusoidal oscillation, pitch and frequency are closely related. The pitch increases with frequency, related by [65] Pitch interval = K log2

ν2 ν1

(13.1)

where ν1 , ν2 are frequencies of the two notes, K is taken as 1200. Musical instruments with vibrating strings are based on standing waves. As we have seen earlier, a standing wave is formed when the number of waves propagating down a string or a slinky is equal to that reflected back and an integral number of the wavelengths is equal to the length of the string. A standing wave transmits no energy. The strings are tied to the bridge (as seen in Fig. 13.1). This coupling damps the vibrations, eventually

Sound of Music

Fig. 13.1

137

Tanpura — tension of four strings can be adjusted by tightening the knobs.

stopping them. In guitars, there are frets which can be used to change the length of the vibrating string. This changes the pitch. The same purpose is served by the tuning pegs in a tanpura or a sitar. The larger diameter and mass per unit length of the fourth string of a tanpura gives a bass sound whereas the other strings give treble sound. The fundamental frequency is preferred because it has lowest kinetic energy - so its easiest to excite. Tanpura, Guitar and violin all sound differently. To understand, note that a vibrating string emits very small amount of acoustic energy (can hardly be heard). Most of the acoustic energy is emitted by the resonator. The geometry of the resonator gives different sounds. Nature of wood also plays an important role. Wood is an extremely anisotropic material, its composition is variable. So, desired resonant properties of a violin (say) demands conditioning the wood, and those who know, know that it can take years to make a near-perfect violin! There are more wind instruments than string instruments. There are two main reasons for this - (i) human lungs, wind tract and mouth are suited to provide air stream, (ii) wind instruments are stronger resonators than string instruments. The wind instrument is excited by blowing across an edge (of a flute, for instance). When the air is forced into restrictive dimensions, Bernoulli principle will create the pressure difference. Also, it creates eddies and vortices, creating source of noise, stimulating allowed resonances. For standing waves to be generated, there must be node and an antinode. That is why all wind instruments are like open pipes. Pitch of a pipe is determined by its length and the medium. Instead of air, if there is helium gas then the wave velocity in helium is greater than in air. The sound changes - however, if the pitch of the human voice is produced by changing the tension of the vocal chords, then why the medium will modify the resonant frequency? This is

138

Mechanics, Waves and Thermodynamics

because the throat, mouth, larynx act like resonant chambers, leading to a change in voice. Temperature and humidity also affect the velocity and hence the pitch of a wind instrument. Now, another question - why is the sound quality different for the same note played on different instruments. The quality or timbre of a note depends on the intensity of other harmonics of the note. Many factors are involved here - the speed of the air blown across the edge, material of the pipe, thickness of the pipe. Sch¨onberg’s music was written for a scale of 12 notes, African music makes use of quarter and eighth tones, Indian music has 12 tones with quarter, eighth, one-sixteenth notes, making 22 different “notes” called shruti. Indian classical music has its origins from thousands of years ago. Western classical music evolved from the Greeks who used a four note pattern (tetrachord). Two tetrachords were then joined to give various modes. One of these modes, the Lydian, corresponds to our modern major scale. Pythagoras was the first person to form a scientific theory of the structure of this scale. Using just the two intervals, the fifth and the octave, one can find all the notes in the major scale. These intervals correspond to simple fractions of the length of the vibrating string with frequency ν. Let us explain this first in the context of Indian (Hindustani) classical music. There are seven notes, called “Sa, Re, Ga, Ma, Pa, Dha, Ni”. Next octave will begin with ˙ Re, ˙ is 2ν; ˙ . . .. If the frequency of Sa is ν, then the frequency of Sa Sa again, written as Sa, Pa will be at 32 ν; Ma at 43 ν. Given that Sa occurs at 240 Hz, Pa will be at 360 Hz, Ma at 320 Hz. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 Sa Re Ga Ma Pa Dha Ni Srinivas Pandit [70] illustrated the notes by relating them to the frequency of oscillation of string tied at ends of certain length. This relation is described in the table below. Note Name of the Note Length of string (inches) Frequency (Hz) Sa Re Ga Ma Pa Dha Ni ˙ Sa

Shadj Rishabh Gandhar Madhyam Pancham Dhaiwat Nishad Shadj(2ν )

36 32 30 27 24 21 13 20 18

240 270 288 320 360 405 432 480

In Hindustani classical music, in addition to the above tabulated pure notes (called sur in Hindi), there are some notes which are either softer or sharper. There is a softer Re which corresponds to the length of string equal to 33 13 inches. There are three other softer notes which are Ga, Dha, Ni and one sharper note Ma.

Sound of Music

139

Let us observe the ratios of the frequencies of the notes shown in the Table. We begin with the frequency of the note Sa, which is 240 Hz. The note, Pa is 3/2 of 240 Hz, i.e. 360 Hz. Remarkably, the notes succeeding Pa are of frequencies, respectively 3/2 of the frequencies of the notes succeeding Sa. For instance, Dha (405 Hz) is 3/2 of Re (270 Hz), and so on. In units of 240 Hz, we see that the successively higher notes are at fractions, 9/8, 6/5, 4/3, 3/2, 27/16, 9/5, 2/1 of 240 Hz. Later, we will see these fractions in the context of Western classical music and observe the differences. In Hindustani classical music, one of the main distinctions is that twelve notes are divided into ten families. The notes consisting them are given below alongwith their names: 1. Bilawal (all pure notes) 2. Kalyan (sharp Ma) 3. Khamaj (soft Ni) 4. Bhairav (soft Re, Dha) 5. Kafi (soft Ga, Ni) 6. Marwa (soft Re, sharp Ma) 7. Asavari (soft Ga, Dha, Ni) 8. Poorvi (soft Re, Dha, and, sharp Ma) 9. Todi (soft Re, Ga, Dha, and sharp Ma) 10. Bhairavi (soft Re, Ga, Dha, Ni) Ragas made out of each family is associated with two further aspects - (i) the time of singing in the day or night, and, (ii) mood or sentiment it generates. It requires an in-depth knowledge and experience to understand these two subtleties. It is a fact that a raga like Bhairav with soft-Re, Dha brings out the nuances of early morning whereas raga Yaman of the Kalyan family refers to the first phase of the night, just after sunset.

13.2 Western Classical Music There are commonalities between Western and Indian (Hindustani) classical music but there are more differences. We discuss certain aspects of the Western music here, more details could be seen in [71]. Musical scale is similar but different. Between fundamental frequency ν and its higher harmonic note at 2ν, we have a scale - as in Hindustani music. The perfect fifth at 3ν/2 is exactly the note Pa. Consonances created by fundamental frequencies and perfect fifth are at the basis of harmony. It is interesting to note that the Indian classical accompanying instrument Tanpura (Fig. 13.1) has four strings, tuned at Pa, Sa, Sa, Sa . , is just an instance of creating a consonance. Western scale was created by Pythagoras around 500 B.C., based on a monochord (called Ektara in ancient India). This consists of a string tied at two ends, which could be divided into two by a movable bridge. The two sides produced notes when plucked.

140

Mechanics, Waves and Thermodynamics

C D E F G Fig. 13.2

A B

Basic notes are shown here on a schematic keyboard of a piano.

Tension is constant throughout, so there is no change in pitch of the two parts. The ratio 2:1 creates the fundamental frequency and its higher harmonic. The ratio 3:2 makes a perfect fifth. Using this approach, a diatonic 7-note scale can be generated which is at the foundation of Western music. One may begin with a note F (Fig. 13.2) with a frequency νF and make a perfect fifth at 3νF /2. Continuing to create perfect fifths from this note (and dividing by two whenever the frequency becomes larger than 2νF ) all the frequencies are obtained. One may base the calculation on F or C or any other note. For comparison with the Table given for Hindustani classical music, we give a Table for Western music here showing diatonic scales [71]. Note Pythagorean scale Just scale Equal temperament based on C) in C (Hz based on A = 440Hz C D E F G A B

13.3

1 9/8 81/64 4/3 3/2 27/16 243/128

1 9/8 5/4 4/3 3/2 5/3 15/8

261.9 293.7 329.6 349.2 392.0 440.0 493.9

Transposition, Musical Scales, and Algebraic Groups

Some wind musical instruments are transposing instruments - when one plays a note C on a B-flat clarinet, tone B-flat is heard. We are using the term note for what is written, and tone for what is heard, following Helmholtz [72]. Whenever the instrument is tuned to a scale other than C-major, transposition results. We tabulate below transposed notes, well-known to the musicians. Subsequently, following [73], we show that there are some interesting properties hidden here.

Sound of Music

scale → scale played ↓

C

C#

D

D#

E

F

F#

G

G#

A

A#

B

C C# D D# E F F# G G# A A# B

C C# D D# E F F# G G# A A# B

C# D D# E F F# G G# A A# B C

D D# E F F# G G# A A# B C C#

D# E F F# G G# A A# B C C# D

E F F# G G# A A# B C C# D D#

F F# G G# A A# B C C# D D# E

F# G G# A A# B C C# D D# E F

G G# A A# B C C# D D# E F F#

G# A A# B C C# D D# E F F# G

A A# B C C# D D# E F F# G G#

A# B C C# D D# E F F# G G# A

B C C# D D# E F F# G G# A A#

141

To see that the twelve elements listed in the above Table form a group, we need to first recognize a binary operation, ×. Subsequently, we will have to establish closure w.r.t. ×, find a unique identity element, and, a unique inverse for each element. Finally, we need to prove the associative w.r.t. ×. The binary operation corresponds to the multiplication of scales which corresponds to a superposition of flats and sharps. For instance, D × F = G,

D# × F = G#,

and so on. The unit element is the scale C major, as can be seen in the Table. Its existence allows us to identify the inverse elements: C−1 = C, (C#)−1 = B, (D#)−1 = A, and so on. The inverse element in fact corresponds to a scales such that the number of flats in one is equal to the number of sharps in the other. The flats are cancelled by the sharps. Moreover the group is commutative or Abelian as, e.g., A × F# = D# = F# × A. The associative law is fulfilled because the order in which flats and sharps appear in a product doesn’t matter. There is one more observation: D1 = D, D2 = E, D3 = F, . . . These elements make a cyclic group of order 12. Thus we have seen a beautiful connection of harmony in music and groups which are a mathematical abstraction of symmetries.

14

Chapter

Fluid Mechanics The very ink with which history is written is merely fluid prejudice. — Mark Twain

A fluid (liquid or gas) is something which cannot oppose a shearing stress, i.e., a force acting tangentially to any surface of the fluid. If a tangential force F is exerted, the fluid will start flowing. Different fluids respond to F differently - for instance, water flows almost instantaneously compared to honey or tar. Solids can withstand much more shearing stress - on application of shear on an eraser, it just bends. Here too, there is variation, shearing a ruler will bend it and eventually break it. When a truck-load of bricks is unloaded all at once, the bricks almost “flow”. Similar behaviour can be recalled in filling wheat or rice grains or even sugar in boxes, we sort of “pour” the grains in. These examples constitute granular matter, their behavior is quite different in many ways compared to fluids and solids. Water is seen to rest on the floor or in a vessel - thus, it does oppose a force acting on it perpendicular to the surface. This is pressure, p. If dS is the surface area element at a point on the surface of the fluid, the outward force at that point is pdS (Fig. 14.1). Pressure is a scalar. If we consider different dSi ’s then the forces are pdSi where p is the same in all the cases. However, p is a function of space and time coordinates, (x, y, z,t). If a fluid is at rest, then across any surface element dS inside it, the forces from the two sides are equal and opposite. This is because dSL = −dSR , implying pdSL + pdSR = 0. What is the pressure inside a liquid at rest at a distance h below its surface, when subjected to gravity? The surface in contact with the atmosphere is at rest, and is at the atmospheric pressure, p = patm (Fig. 14.2). At the depth h, the pressure is p = patm + ρgh = patm − ρgz where ρ is the density of the fluid. We can prove this by considering a column of liquid of height h and its weight on the bottom surface. Using this, we can prove the following. Archimedes’ principle: Consider a closed surface S which is entirely inside a liquid and ˆ encloses a volume V . The buoyant force F acting on S is ρgV k. We use the relation,

Proof: Z S

PdS =

Z

∇PdV . V

(14.1)

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The fluid exerts the force on S is (− ∇ P dS), the negative sign is because the force is pointed inwards. So, R

F = −

=

Z

Z S

PdS = −

Z

∇PdV V

ˆ ˆ ρgkdV = ρgV k.

(14.2)

Hence the proof. Any fluid element can be specified by six independent quantities - density (ρ), pressure (P), components of velocity (vx , vy , vz ), temperature (T ). For the case of constant temperature, we need five equations to determine these quantities. The equation determining p as a function of ρ and T is the equation of state. For an ideal gas, the equation is P = cρT where c is a constant. At NTP, we have seen that c=

p 1.01 × 105 Nm−2 = = 287. ρT 1.29 kg m−3 × 273 K

For liquids, the equation of state is much more complicated.

Fig. 14.1

Force across an area element.

Fig. 14.2

Change of pressure with height of atmosphere.

(14.3)

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14.1

Equation of Continuity

This basic equation is about the conservation of mass. Mass flux density is given by the product, ρv. With the surface element dS, ρv.dS gives the amount of material fluid going through dS per unit time. Now consider the integral of this quantity over some surface S enclosing a volume V which gives mass of fluid coming out of S per unit time. Conservation of mass gives d ρv.dS = − dt S By Gauss’ theorem, Z

Z S

ρv.dS =

Z

Z

ρdV = −

Z

dV

∂ρ . ∂t

(14.4)

∇.(ρv)dV

(14.5)

V

implying thereby Z ∂ρ + ∇.(ρv) dV = 0. V ∂t

(14.6)

Due to the fact that volume is arbitrary, for Eq. (14.6) to hold, the integrand must vanish thus giving us the equation of continuity. R If the integral, S ρv.dS is zero, then clearly the product of velocity and cross-sectional area is constant. For instance, as water flows down a tap under gravity, its velocity increases. Hence, cross-section area has to decrease, consistent with our everyday observation. Blood flows inside arteries blocked by cholesterol deposits much faster due to constriction. Same idea applies to traffic flow also.

14.2

Euler’s Equation

Euler’s equation is just an expression of Newton’s second law of motion. Consider a small volume element dV of the fluid, of mass ρdV . Suppose it is moving with velocity v. The inertial term of the Newton’s law is mass × acceleration = (ρdV )

dv . dt

(14.7)

The force on a surface element is a sum of the force of bulk, and, an external force (e.g., gravity). Using Gauss’s theorem, we can write force = −

Z

Pds + | {z } force of bulk on surface

= −∇PdV + (ρdV )fext where fext is force per unit mass.

Fext |{z} e.g., gravity

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To find dv/dt, we have to note that as the volume element dV moves with a velocity v, its position, r at a time t, becomes r + vdt at a time (t + dt ). Hence dv = v(r + vdt,t + dt ) − v(r,t )

= v(x + vx dt, y + vy dt, z + vz dt,t + dt ) − v(x, y, z,t ) ∂ ∂ ∂ ∂ v + vy dt v + vz dt v + dt v ∂x ∂y ∂z ∂t ∂ = dt v + (v.∇)v , ∂t

= vx dt

thus

dv ∂ = v + (v.∇)v dt ∂t

Euler noted that this not the same as ∂ v/∂t. Hence, we obtain the Euler equation of fluid mechanics: ∂ ∇P v + (v.∇)v = − + fext (14.8) ∂t ρ 14.2.1 Applications Fluid at rest

For fluid at rest, v = 0, so Hence,

∇P ρ

= fext . If the external force is due to gravity, then fext = −gˆz.

∂P ∂P ∂P = = 0 and = −ρg. ∂x ∂y ∂z

(14.9)

For a liquid whose volume is not very large (ρ and g are assumed independent of z), we get P = −ρgz + constant. Gas

For a gas, ρ is a function of z. We are familiar with this as the atmosphere gets rarer as we go higher up in altitude. Let us assume that g is constant, equal to 9.8 m/s2 . We shall also assume that temperature T is a constant. It is not a good assumption in an absolute sense, however, good in a relative sense as temperature decreases much slower than density or pressure. So, if are interested in the variation of P (or ρ) with z, we may assume that T is constant. Let us say that T = 293 K. This implies that P = 287 × 293ρ. This implies dP gP 9.8P P = − =− =− dz cT 287 × 293 8.6 km

=⇒ P(z) = P(0)e−z/8.6 km .

(14.10)

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Here, we have the expression for the pressure at a height z above the earth’s surface; we may consider P(0) to be 1 atm. on the surface of earth. Hence, the pressure decreases by a factor of e ' 2.72 for every 8.6 km we go up. Note that the height of Mount Everest is similar, 8.85 km.

14.3 Bernoulli’s Equation We start with the Euler equation. One can use the vector identity, 1 (v.∇)v = ∇(v.v) − v × (∇ × v). 2

(14.11)

Next, let fext = −∇φ ext where φ ext is the external potential per unit mass. Let us consider P as a function of ρ, so that ∇P/ρ can be written as ∇w where w is a function of ρ. For example, if ρ is constant, then w = P/ρ. If P = cρ where c is a constant, then w = c log ρ. Hence, we get ∂v + ∇(v2 /2) − v × (∇ × v) = −∇ω − ∇φ ext . ∂t

(14.12)

Now, consider a steady state such that the partial derivative w.r.t. time is zero at all points r. This implies that 2 v ext ∇ +ω +φ = v × (∇ × v). (14.13) 2

14.4 Streamlines A streamline is a line along which a particle of the fluid moves. A fluid may have many streamlines - their pattern being the same for a uniform flow however, changing for an unsteady flow. A tangent vector at any point on the streamline gives the velocity v at that point. Any two close by points, r and r + dr on a streamline, separated by dr, should satisfy dr = εv where ε is a small constant number. The change in the value of v2 /2 + ω + φ ext between these points is dr.∇(v2 /2 + ω + φ ext ) = εv.∇(v2 /2 + ω + φ ext )

= εv.v × (∇ × v) = 0.

(14.14)

Hence, the values of v2 /2 + ω + φ ext at two nearby points must be equal. Extending this argument, we see that all points on a given streamline must have the same value of this quantity - the Bernoulli’s principle. On different streamlines, the values are different.

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14.4.1 Applications 1. Torricelli’s law A very well-known problem is posed by the water coming out of a small hole in a bucket at some depth below the surface. One would like to find the water velocity v just outside the hole. Consider a streamline going through the hole. Consider two points A and B on that streamline, one inside the bucket where the velocity is zero and the other outside where the velocity is v. The value of ω for a liquid of constant density is P/ρ which, in turn, is given by P0 /ρ + gd at A and P0 /ρ at B where P0 is the atmospheric pressure. Being close to each other, the points A and B are at the same height, thus φ ext = gz is the same at A and B. Bernoulli’s equation gives P0 v2 P0 + gd = + , ρ ρ 2 p v2 = 2gd, or v = 2gd.

(14.15)

2. If you open a window of a room when a strong wind is blowing outside, you observe that small objects (pieces of paper) will fly out. This is because on a streamline going out from the room, v2 /2 is larger outside, thus P is smaller outside than inside. So, small objects will be pushed out from a region of high pressure to a region of low pressure. You have surely observed the curtain in the bathroom flying in as you open the shower. 3. A similar experience is felt by people, who try to throw candy-wrappers outside a fast-moving train - they can’t succeed as the wrapper keeps returning to the train. On the other hand, if you are standing on a platform when a fast train (Rajdhani or Shatabdi express in India or TGV in France) passes by, you are well-advised to stay away from it as you might be sucked in. The air very close to the train moves with the velocity of the train. This is called the ‘no-slip’ condition - the velocity of a fluid close to a boundary of the fluid must be zero with respect to the boundary. The air away from the train is not moving, thus there is a lower pressure closer to the train.

14.5 Speed of Sound Inside a Fluid “Inside the fluid” means far from the boundaries. Let us consider a fluid which has only longitudinal sound waves, i.e., the velocity of the fluid v must be in the same direction as ˆ A fluid does not have transverse sound modes the direction of propagation of the sound, k. ˆ there is no restoring force which can bring the fluid back to equilibrium because if v ⊥ k, - a fluid cannot oppose a shearing force. In equilibrium, we have ρ = ρ0 and P = P0 , and v = 0. So, for a sound wave, we will assume that ρ = ρ0 + δ ρ, P = P0 + δ P and v 6= 0, with δ ρ, δ P, v all small quantities. Substituting these in the equation of continuity and ignoring all second-order quantities, we get ∂δρ + ρ0 ∇.v = 0. ∂t

(14.16)

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Similarly, Euler’s equation gives (ignoring external force) 1 1 dP ∂v = − ∇δ P = − ∇δ ρ. ∂t ρ0 ρ0 dρ ρ0

(14.17)

dP The quantity ( dρ )ρ0 is to be calculated for the adiabatic case where no heat is exchanged between different parts of the fluid. Equations (14.16) and (14.17) give us

∂ 2δ ρ ∂ ∇.v =0 + ρ0 2 ∂t ∂t and

∂v 1 ∇. =− ∂t ρ0

dP dρ

∇2 δ ρ.

(14.18)

ρ0

Combining these two equation gives ∂ 2δ ρ dP = ∇2 δ ρ. ∂t 2 dρ ρ0

(14.19)

dP Calling ( dρ )ρ0 = v2 where v is the velocity of the wave, we get the wave equation. Let us now try to explore if there are solutions for progressive waves, which could be written simply as

δ ρ = a exp[i(k.r − ωt )];

(14.20)

substituting in the wave equation gives ω 2 = v2 k2 , =⇒ ω = v|k|.

(14.21)

Going back to the Eq. (14.17), we see that v=

av ˆ k exp[i(k.r − ωt )] ρ0

(14.22)

ˆ where we have used the fact that k/ω = k/v|k| = k/v. We see that v is in the direction of k. Indeed, it is a longitudinal wave. It is well-known that at temperature of about 293 K, the velocity of sound in air is dP )ρ0 is much 340 m/s whereas its velocity in water is almost 1500 m/s. This is because ( dρ larger in water, i.e., a much larger change in pressure is required to change the density by the same amount as in air. 14.5.1 Effect of bubbles Perform the following experiment. Take a tall cylindrical glass, an effervescent tablet and a spoon. Powder the tablet and drop in glass. Add water to fill the glass and immediately

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149

start tapping the bottom of the glass with the spoon. As one taps and listens to the pitch of the sound, one observes that the pitch goes down and then increases to a high tone. We would like to qualitatively discuss why the pitch changes. Because the tablet releases bubbles, it is possible to guess that change in pitch is due to bubbles. To test this, we can shine light [61] on the glass from a side and observe the scattered light while tapping. we find that when the density of bubbles is higher, the pitch of the sound is lower. Which type of oscillations produce a low pitch, and, which produce a high pitch? Low pitch is produced by the standing waves in the medium that fills the cylinder. This behaves like a tube closed at one end and opened at the other. If L is the length of the column and cm is the speed of sound in medium, then the resonant frequencies in such a tube are approximately νn = n

cm , 4L

n = 1, 3, 5, . . .

(14.23)

We know that cm is about four and a half times larger than the speed of sound in air. For the high pitch, it is produced mainly by the vibration of the glass container. This has been experimentally verified in [61]. Now, what is the mechanism that explains how bubbles in water affect the speed of sound? It is reasonable to guess that the speed of sound in a water-bubble mixture is somewhere between the speed of sound in pure water and that in pure air. The fundamental frequency of the empty glass is 610 Hz and the glass filled with water is 3200 Hz. However the lowest fundamental frequency found experimentally is 230 Hz. This is almost three times lower than the fundamental frequency of the glass filled with air. Crawford found in his work, called “Hot chocolate effect” [62], that the speed of sound in water-bubble mixture can go down by a factor eight compared to air. Why? The speed of sound in gases and liquids is given by [61] s c=

1 , ρχ

(14.24)

where χ and ρ are the compressibility and the density of the medium. The density of the water is approximately 800 times larger than the density of air. However, the compressibility of water is smaller than that of air by a factor of 15000. We have to see what is the speed of sound in water with bubbles? Approximately, although the density of water with bubbles is roughly the same as the density of water, however, it is much more compressible. The speed of sound in this case may be significantly lower than the speed of sound in air.

150

14.6

Mechanics, Waves and Thermodynamics

Sound of a Brook

We have heard drops of water falling in a bucket filled with water. In his beautiful book, “The World of Sound”, W. H. Bragg [75] quoted an explanation given by Richard Paget. Paget found that each drop makes a cavity in water, which then closes and gives a sound like a resonator with rigid walls. Minnaert [76] investigated the sounds of running water by letting air escape in bubbles from the orifice of a tube immersed in water. He found that each bubble gives a sound of definite pitch. We want to describe this argument, which turns out to be correct. Paget’s argument is invalid as the bubbles have diameter approximately 3 mm to 6 mm. If we consider a spherical resonator of this size, frequency is about 100000 Hz, which is inaudible. Following Minnaert, let us suppose that the bubbles make a sound because they pulsate in closing. A bubble contracts and expands, and consequently the surrounding water is set in vibration. The elasticity is due to air of the bubble. What is the frequency of such oscillations? The radius of the bubble can be written as (Fig. 14.3) r (t ) = r0 + a sin

2πt . T

(14.25)

Assuming that the compression of air is adiabatic, and the ratio of the specific heats at constant pressure to that at constant volume be γ, we can write the ratio of the pressures of compressed to uncompressed states as γ 3γ P V0 r = = . (14.26) P0 V r−x

Fig. 14.3

A bubble pulsates, and in turn, sets the surrounding water into vibration.

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151

Here the radius is decreased by an amount x, and volume V is 4πr3 /3. Hence, assuming that x r, employing the binomial expansion, we have P − P0 =

3γPx . r

(14.27)

Using this to calculate the potential energy at minimum volume: −

Z V V0

(P − P0 )dV =

Z a 3γPx 0

r

4πr2 dx = 6πγPra2 .

(14.28)

The kinetic energy is determined by the (radial) motion of water. The velocity of the wall of the bubble is dr (t )/dt = (2πa/T ) cos(2πt/T ). On the other hand, the velocity of a particle at a distance R is (r/R)2 (2πa/T ) cos(2πt/T ). Total kinetic energy of all the volume elements of density ρ is 1 2

Z

dr dt

2

ρ = 2 max

Z ∞ 4 r 4π 2 a2 r

R4 T 2

4πR2 dR =

8π 3 ρr3 a2 . T2

(14.29)

Equating (14.28) and (14.29), we obtain the time period. T and frequency, ν: 4π 2 ρr2 , T2 = 3γP

1 νM = 2πr

s

3γP . ρ

(14.30)

If it were a spherical resonator, the frequency would be 0.715 νsph = r

r

γP s

(14.31)

where s is the density of the gas in the bubble. Comparing these frequencies, we see that Minnaert’s frequency νM is about 80 times smaller than νsph . This reduction brings the sound in audible range, in agreement with what we hear.

14.7

Why is Water Watery?

Moving our finger slowly in water is much easier than moving it in honey. If we move the finger at a speed where a finger crosses an A4 size paper across its diagonal in one second, then the water seems to suffer no frictional resistance. There are two forces of resistance of motion in a liquid - inertial and viscous. If the linear size of the dipped finger is `, u is its speed and ρ is the density of the liquid, then the inertial force is estimated by taking the time as `/u: u Fi = Ma ≡ `3 ρ = `2 ρu2 . (14.32) `/u

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Viscous force equals the viscous momentum transfer per second from moving parts of water to the part at rest: Fv ≡ η × velocity gradient × area of moving water u = η `2 = ηu` `

(14.33)

where η is the coefficient of viscosity. Reynolds number is the ratio, R=

Fi `ρu = . Fv η

(14.34)

Because Fi ∝ u2 and Fv ∝ u, viscous force is more in relatively slowly moving objects. For the moving finger, u ∼ 1 − 100 cm/s, η ∼ 10−2 g/cm s, so R ∼ 103 to 104 . So, viscous resistance is negligible, so water feels “watery”. If we consider micro-organisms with ` ∼ 1 µm, u ∼ 10 µm/s, R ∼ 10−4 × 1 × 10−3 /10−2 ∼ 10−5 . This shows that viscous force of water is enormously large for the micro-organisms. Swimming of micro-organisms makes an interesting subject, elaborated by Purcell [77].

15

Chapter

Water Waves I swim across a sea of quotes, splashing in the words and riding the waves of wisdom. — Terri Guillemets

15.1

Gravity Waves in Liquid

In liquids, instead of variations in pressure, the waves are driven by the force of gravity. We observed that sound waves travelled deep inside the fluid. The gravity waves are strongest at the surface of the liquid. Let us consider a column of liquid of depth D with bottom specified by z = −D and surface by z = 0. Let us consider an incompressible liquid, i.e., ρ = constant. The continuity equation gives us ∇.v = 0. In the Euler equation, we assume that v is small, so that to the first order, we have 1 ∂v = − ∇P + fext . ∂t ρ

(15.1)

Let fext be −∇φ ext where φ ext is the potential energy per unit mass (gz for gravity). Noting that ∇ × ∇P = 0 and ∇ × ∇φ ext = 0, taking the curl of (15.1), we get ∂∇×v =0 ∂t

=⇒ ∇ × v is a function of x, y, z only.

(15.2)

Since, we are looking for wave solutions, quantities must depend on time also. If not, time-derivatives must be zero. We expect that v will oscillate around zero. So if ∇ × v is independent of time, it can only be zero. Hence, we have ∇ × v = 0 as well as ∇.v = 0. Hence, there exists a scalar function, the velocity potential f such that v = ∇ f . Thus, ∇.v = 0 becomes ∇2 f = 0, the Laplace’s equation. Hence, we need to solve this equation with boundary conditions at the two surfaces. At z = 0 (upper surface), we have ∂f P ext ∇ + +φ = 0, ∂t ρ

=⇒

∂f P + + φ ext = function of t alone. ∂t ρ

(15.3)

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Mechanics, Waves and Thermodynamics

We are looking for solutions that represent progressive waves. So we must have the expression above that is constant independent of x, y, z,t. Let the profile of the surface of water be h(x, y,t ) which gives the height at a point (x, y) at a time t. This profile is w.r.t. a flat equilibrium profile. At the surface, the pressure P0 is atmospheric. So we have ∂f P ∂t + ρ + gz equal to constant. This implies that ∂2 f ∂z + g = 0 at the surface. ∂t 2 ∂t

(15.4)

Now ∂ z/∂t = vz which is the vertical component of the velocity. This, in turn, is equal to ∂ f /∂ z since v = ∇ f . Hence, we have at the surface z = 0 the condition: ∂2 f ∂f = 0. +g ∂t 2 ∂z

(15.5)

Now we need a condition at z = −D. Here vz = 0. So we want ∂∂ zf = 0 at z = −D. However, vx = ∂ f /∂ x need not be zero at z = −D. Let us assume that the wave is travelling along x-direction. Then, f is a function of x, z,t and not y. Since a wave propagating along x-direction looks like cos(kx − ωt ), let us assume f (x, z,t ) = ψ (z) cos(kx−ωt ) and see if this satisfies all the equations and boundary conditions. ∇2 f = 0 implies that −k2 ψ (z) cos(kx − ωt ) +

d2ψ cos(kx − ωt ) = 0, or dz2

d2ψ = k2 ψ (z); thus ψ (z) = aekz + be−kz dz2

(15.6)

in general. Next we want vz = ∂ f /∂ z = 0 at z = −D. This gives ae−kD − bekD = 0 =⇒ b = ae−2kD . Thus, f (x, z,t ) = a ekz + e−2kD−kz cos(kx − ωt ) = ae−kD ekz+kD + e−kD−kz cos(kx − ωt )

= 2ae−kD cosh [k(z + D)] cos(kx − ωt ).

(15.7)

Let us call 2ae−kD by b. Let us now impose ∂ 2 f /∂t 2 + g∂ f /∂ z = 0 at z = 0. This gives −ω 2 b cosh(kD) cos(kx − ωt ) + gbk sinh(kD) cos(kx − ωt ) = 0.

(15.8)

This implies that ω 2 = gk tanh(kD).

(15.9)

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155

A general relation between the frequency and wavenumber is known as the dispersion relation. 15.1.1 Deep water waves If the depth of water D wavelength, λ = 2π/k, i.e., kD 2π. Then, tanh kD ' 1 and we get ω 2 ' gk. The phase and group velocities are r ω g vphase = = , k k

vgroup

dω 1 = = dk 2

r

g vphase = . k 2

(15.10)

Next, let us look at the velocity profile: vx =

∂f = −kb cosh[k(z + D)] sin(kx − ωt ), ∂x

vz =

∂f = kb sinh[k(z + D)] cos(kx − ωt ). ∂z

(15.11)

Suppose that we are close to z = 0. Then, k(z + D) ' kD 2π, which implies cosh k(z + D) ' ek(z+D) /2, sinh k(z + D) ' ek(z+D) /2 also. Hence, vx = −

vz =

=⇒ v2x + v2z =

kb k(z+D) e sin(kx − ωt ), 2

kb k(z+D) e cos(kx − ωt ), 2 k2 b2 2k(z+D) e . 4

(15.12)

That is, the velocity profile is a circle of radius (kb/2)ek(z+D) . The amplitude (kb/2)ek(z+D) decreases as we go down from the surface (z = 0) to more and more negative values of z. The water moves less and less as we go deeper. 15.1.2 Shallow water waves (Tsunami) Here, we assume that the depth D √ λ , so Dk 2π. Hence tanh(kD) ' kD and we get 2 2 ω = gDk √ , which implies that ω = kD k. We see that both phase and group velocities are v = gD. Tsunami waves are examples of shallow water waves even though they are in the middle of an ocean. This is because the wavelength of these waves is hundreds of

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kilometers which is much greater than depth of water in the sea (about 10 km). The velocity is then q p v = gD ∼ (10 m/s2 )(10 km) ∼ 300 m/s (15.13) which is about as fast as the speed of sound of air. As the tsunami wave approaches a seashore, the depth D decreases which implies that the velocity decreases. Then, the gravitational potential energy must increase in order to keep the total energy constant. So the height of tsunami increases to increase the potential energy accordingly. What about the velocity profile in this case? Since kD 2π and 0 ≤ (z + D) ≤ D at all points, we have 0 ≤ k(z + D) ≤ 2π. This implies that cosh k(z + D) ' 1, sinh k(z + D) ' k(z + D). The velocity components are vx = −kb sin(kx − ωt ), vz = k2 b(z + D) cos(kx − ωt ).

(15.14)

The amplitude of vx is kb, independent of z; the amplitude of vz is k2 b(z + D) - this varies from k2 bD to zero from the upper surface to the lower surface. Moreover, kD 2π which implies that k2 bD kb. So, vz is always much smaller than vx . In the above analysis, we assumed that the amplitude of oscillations is much smaller than both the wavelength λ and D. If the amplitude of oscillations becomes comparable to λ , we have to include the effect of surface tension.

15.2 Capillary Waves At the interface of two phases that do not mix, there is an asymmetry of the intermolecular force acting on the molecules at the interface. This leads to a surface energy density which increases with the surface area. Another way to express this is in terms of surface tension, σ . This acts in a direction tangent to the surface to minimize the area. At an interface, it is like the tension in a stretched membrane. Presence of surface waves increases the area and hence the surface energy. This, in turn, adds to the gravitational potential energy associated with the waves. We now need to account for both surface tension and gravity. Let us denote the surface by z = η (x, y,t ). If the surface tension at x is σ , at a nearby point (x + ∆x), it will be σ + (∂ σ /∂ x)∆x. Accordingly, if the slope of the surface at x is ∂ η/∂ x, then at the neighbouring point it will be ∂ η/∂ x + [(∆x)2 /2]∂ 2 η/∂ x2 . Further, on the two sides of the interface, let us denote the pressure inside by p(x,t ) and outside by p0 (atmospheric pressure). For the interface to be in a mechanical equilibrium along the vertical (z-) direction, all the forces should sum to zero: ∂ η ∂ 2η ∂η ∂σ 2 ( p − p0 )∆x + σ + ∆x + 2 (∆x) − σ = 0. (15.15) ∂x ∂x ∂x ∂x

Water Waves

157

Keeping upto the first order terms in ∆x, we have

( p − p0 ) + σ

∂ 2η ∂ σ ∂ η = 0. + ∂ x2 ∂x ∂x

(15.16)

If we assume that σ is constant, then p(x,t ) = p0 − σ

∂ 2η . ∂ x2

Bernoulli’s principle allows us to write ∂ ∂ 2η ∂3 f ∂p = −σ 2 = −σ 2 ∂t ∂t ∂x ∂x ∂z

(15.17)

(15.18)

because z-component of velocity, (∂ η/∂t ) = ∂ f /∂ z. On the surface, the boundary condition is ∂2 f σ ∂3 f ∂f ( x, z = 0,t ) − (x, 0,t ) + g (x, 0,t ) = 0. 2 2 ∂t ρ ∂x ∂z ∂z

(15.19)

The boundary condition at the bed remains the same as in the last Section. We know that f (x, y,t ) can be written as f (x, z,t ) = (const, C) cos[k(x − ct )]{tan h kD sin h kz + cos hkz}.

(15.20)

Substituting this into Eq. (15.19), and simplifying the algebra, we get the dispersion relation, c2 1 σ k2 = 1+ tan h kD. (15.21) gD kD ρg Hence, we see that a non-zero σ increases the speed of the wave. In deep liquids (D λ ), tanh kD is almost unity, thus c2 σ k2 1 = 1+ . gD kD ρg

(15.22)

If surface tension term dominates, σ k2 /ρg 1, then we get the capillary waves for which c2 kσ = . gD ρgD This immediately gives us the speed of the capillary waves to be c = σ k/ρ.

(15.23)

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Mechanics, Waves and Thermodynamics

Exercise: Plot c2 /gD versus 1/kD and identify the regions where we have capillary waves, deep and shallow water waves. Problem For the water-air system, with σ = 72 dynes/cm2 , ρ = 1 g/cm3 , calculate the wavelength at which the speed of capillary waves is minimum (on Earth, Moon).

16

Chapter

The Kinetic Theory of Gases I believe that we do not know anything for certain, but everything probably. — Christiaan Huygens

Gas is made of molecules interacting by certain short-range interactions. The range of interaction is much smaller than the mean free path of the molecules. The mean free path, in turn, depends on the density of the gas. For a dilute gas, the number of molecules in a given volume is small. Hence, the mean free path is much larger than in the case of a dense gas. One may compare one’s situation in a bus or a metro that is lesser and more crowded. The fundamental force due to which gas molecules interact is electromagnetic, which is infinite-ranged. The effective interaction is modified due to the fact that there is a collection of particles. These interactions are known by various names, such as van der Waals, Lennard-Jones interaction, etc. The particles collide with the walls and due to the resulting change in momentum, there is pressure exerted on the walls. For sake of simplicity, let the atom (molecule) be moving with a velocity v = vxˆi and it collides with a hard wall, the component of the momentum normal to the wall will change sign. The x-component of the momentum mvx will change to (−mvx ). Hence, the net change in momentum is 2mvx . Now suppose we let a cylindrical piston of cross-section A compress the gas. If N particles are occupying a volume V , the number of particles colliding during a time interval t are (N/V )Avxt. Denoting N/V by n, the force on the piston wall is F

= (change in momentum in one collision) × (number of collisions per unit time) = 2mvx .nAvx = 2nAmv2x .

Pressure exerted on the wall P = F/A = nmv2x . A factor of two is missing because the particles are incident on only one side of the piston. This implies PV = Nmv2x . The square of the velocity components must be averaged over all the particles, and, also over the three directions due to isotropy. Thus, averaged v2x , denoted by hv2x i is hv2 i/3 where v2 = v2x + v2y + v2z . The pressure P is nmhv2 i/3. We can now write PV =

2 2 m 2 N hv i = U 3 2 3

where U is the total internal energy. This is a relation of a great significance.

160

16.1

Mechanics, Waves and Thermodynamics

Equipartition of Kinetic Energy, Ideal Gas Law

Consider a mixture of two gases A and B with masses of their basic constituent respectively given by mA and mB . Assume that the range of interaction is much lesser than the inter-particle distance. Hence, the particles interact over a very short interval of time. Assume that the collisions among the particles are perfectly elastic, thus conserving linear momentum. Let us assume furthermore that the kinetic energy is only translational (ignoring for the moment the rotational energy, for instance). Denote the velocities of the particles by vA and vB before a collision, and, wA and wB after the collision. Conservation of linear momentum and kinetic energy give us the following equations: mA vA + mB vB = mA wA + mB wB ,

(16.1)

1 1 1 1 mA v2A + mB v2B = mA w2A + mB w2B . 2 2 2 2

(16.2)

As a consequence of Eq. (16.1), the absolute value of relative velocity remains the same before and after a collision. Denoting (vB − vA ) by V and (wB − wA ) by W, |V| = |W|.

(16.3)

With respect to the Centre of Mass (CM) frame, the effect of collisions is to change the direction of the relative velocity (Fig. 16.1). It is natural to expect that a large number of collisions will make the distribution of V isotropic. That is, the probability that the direction of V belongs to a elemental solid angle dΩ depends only on the size of the solid angle.

Fig. 16.1

This diagram gives the three-dimensional view of the relative velocities before and after the collision of two particles. The complexity is explained in the text.

The Kinetic Theory of Gases

161

We now show that collisions can only reduce the correlations between vCM and V. Let us assume that at a given time, there is a joint distribution f (vCM , V, θ ) where θ is the angle between vCM and V for a pair of molecules that are about to collide. Hence, one has hvCM .Vi =

Z

Z

dvCM

Z

dV

dθ dφ sin θ f (vCM , V, θ )vCMV cos θ ,

(16.4)

where z-axis is along vCM and the xz-plane is defined by vCM and V. V and W define a plane α which forms an angle φ1 with the xz-plane. The angle between V and W is θ1 , and, the angle between W and vCM is Θ. Isotropy of V implies that the distribution is invariant with respect to rotations about vCM , i.e., the distribution is independent of φ . Because vCM determines a special direction, isotropy of V does not imply hcos θ i 6= 0. Note that each collision leaves |vCM | and |V| unchanged, thus its effect is to give a direction (θ1 , φ1 ), in polar coordinates, to the relative velocity W of the outgoing particles with respect to relative velocity V of the ingoing particles. Invariance w.r.t. rotations around V intimates that the probability density PV (θ1 , φ1 ) does not depend on φ1 . On the other hand, PV (θ1 , φ1 ) is determined only by the laws of the collision, and should satisfy Galilean invariance; thus it cannot depend on vCM . We can now evaluate the average of vCM .V by integrating over relative direction (θ1 , φ1 ) of W w.r.t. V, then on the relative direction of V w.r.t. vCM , and finally over V and vCM . We can see that [81] cos Θ = cos θ cos θ1 − sin θ sin θ1 cos φ1 .

(16.5)

To realize this, one can set V along the z-axis and write W as a function of (θ1 , φ1 ). Then, apply a rotation by an angle θ around the y-axis. When we average upon φ1 , the second term vanishes. Hence, when vCM and V are fixed, we have hcos Θi = hcos θ cos θ1 i = hcos θ ihcos θ1 i

(16.6)

which implies that |hcos Θi| ≤ |hcos θ i|.

(16.7)

This will be true even if we average over vCM and V . Thus, |hvCM .Wi| ≤ |hvCM .Vi|.

(16.8)

Hence, collisions reduce correlations. Hence, we may conclude that correlations should vanish at equilibrium: hvCM .Vieq = 0.

(16.9)

Noting that vCM =

m1 v1 + m2 v2 , m1 + m2

(16.10)

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Mechanics, Waves and Thermodynamics

we have hvCM .Vieq = h

=

(m1 v1 + m2 v2 ).(v2 − v1 ) ieq m1 + m2

hm2 v22 − m1 v21 ieq − (m2 − m1 )hv1 .v2 ieq . m1 + m2

(16.11)

Assume that velocities of the colliding particles is independent, so that hv1 .v2 ieq = 0 (stosszahlansatz). Then, at equilibrium, due to Eq. (16.9), we have 1 1 m2 hv22 ieq = m1 hv21 ieq , 2 2

(16.12)

the equipartition of energy ! In one dimension, the above argument does not hold. In this case, the only recoil possibility corresponds to W = −V , and thus |hcos θ i| = 1.

16.2

Football Game: Kinetic Theory Perspective

Molecular motion is all around us however, thankfully, we don’t perceive it directly. The kinetic energy associated with this motion is expressed as heat. To visualize the molecular motion, we compare it with the motion of a football as it moves around in a game. This interesting discussion has been carried out in [82]. There are similarities and important differences, thus it is not immediately clear what would be the ultimate conclusions. Like atoms or molecules in gases, football spends a lot of time in free flight. The average distance of the free flight is the mean free path. In air at room temperature for a given pressure, P the mean free path is approximately given by λ (mm) =

6.6 . P(Pa)

(16.13)

This will be about 66 nm at normal pressure. The diameter, d of nitrogen or oxygen molecule is approximately 0.35 nm. The ratio 66/0.35 ' 200. The diameter of the football is 25 cm - a billion times a gas molecule. On scaling the dimensions, the corresponding mean free path for football ∼ (200 × 0.25 m) = 50 m. This is about half the length the ground (110 m × 75 m). Goalkeepers regularly hit the ball to this distance. The passes are shorter, nevertheless the comparison looks encouraging. Let us list the differences between the motion of molecules and football: 1. Speeds in the two cases are quite different. The average speed of the molecule is about 400 m/s (comparable to the speed of sound). Football speed is, at maximum, about 30 m/s (or 100 km/h).

The Kinetic Theory of Gases

163

2. High speed and small distances imply very frequent collisions in gases. The average time of collision is τ=

66 × 10−9 m = 1.6 × 10−10 s. 400 m/s

(16.14)

The average time of collisions in football is few seconds or even fraction of a second. 3. Force of gravity is very important in football. In gases, there is no such force. 4. The football is given energy by the player as she (he) kicks it, and, friction with air and ground, and gravity tends to stop the ball. In gases, the collisions are taken as elastic. There can be two cases: (a) isothermal conditions - here, the gas is considered to be in contact with a thermal reservoir (e.g., in a container). Energy is continuously exchanged with the reservoir, usually through the walls. The molecules interacting with a wall maintains the energy balance on an average. Over the average, there are fluctuations; (b) adiabatic conditions - the walls of the container are perfectly elastic or rigid, energy is conserved on each rebound. Total energy is conserved, there are no thermal fluctuations. However, there are local fluctuations in energy because the kinetic energy is exchanged during collisions. Mean free path would be almost the same in the two cases. In football game, the boundary conditions are open. The ball leaves the ground occasionally and is thrown back in a short time. This is closer to isothermal regime as the ball that is thrown in may have different energy from the one that went out. The open boundary condition and the energy exchange at the walls, make football closer to isothermal case of the gas. The average energy of the football is independent of the thermal reservoir and depends on the energy supplied by the players. We may consider here an analogy with temperature. For instance, if we compare a professional and an amateur match, the average energy in the former will be more, and hence we may state the same for the temperature. Now we have a question - given a picture of the trajectories of the ball, will we be able to make out the difference with a molecular simulation of the trajectories of an atom or molecule of a gas? Yes, statistics of the free paths can be different. The main difference will be in angular correlations. In CM frame, angular dispersion of hard sphere collision is uniform, i.e., any angle is uniformly probable. In a football match, both teams would like to score against each other. Hence, there will be a strong bias in the directions of the ball. Also, both teams would prefer to keep the ball inside the playfield. The probability of hitting a wall in a gas can be deducted from the number of molecules colliding wall per second is 3 × 1023 s−1 cm−2 [83] for three dimensions. The play volume of the football is 110 m × 75 m × 15 m. To make a comparison with the gas molecules, let us make a micro-field 110 nm × 75 nm × 15 nm.

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Mechanics, Waves and Thermodynamics

One can verify that the number of molecules in this micro-field at NTP will be (3 × 103 ). Number of collisions across lateral area is perimeter/height, 1.6 ×1013 s−1 . The collision time can be found to be about 0.2 ns.

16.3

Adiabatic Reversible Compression

Consider one mole of an ideal gas inside a box. Movement of a piston compresses or expands the gas. Assume that the compression is very slow and reversible. Further, assume that there is no heat exchange with the surroundings, meaning that the walls are adiabatic (δ Q = 0). Work done to compress is δW = pdV where p is the pressure and dV is the change in volume. As a result, the change in internal energy is dU = δ Q − δW = −pdV .

(16.15)

For an ideal gas, the change in internal energy is proportional to the change in temperature: dU = CV dT where CV is the isochoric specific heat. We also know that ideal gas obeys p = RT /V with R as the Gas Constant. Using Eq. (16.15), we obtain CV dT = −

that is,

RT dT R dV dV , implying =− , V T CV V

dT dV = −(γ − 1) T V

(16.16)

noting that R = CP −CV , γ = CP /CV , where CP is the isobaric specific heat. For an ideal, monoatomic gas, γ = 5/3 whereas for a diatomic gas, it is 7/5. Integrating Eq. (16.16) from (Ti ,Vi ) to (T f ,V f ), we get Tf = Ti

Vi Vf

γ−1 ,

(16.17) γ

γ

which is the same as piVi = p f V f . This is the standard result derived using thermodynamic considerations. We now present a derivation of this result using kinetic theory. Problem Airplanes fly at an altitude of about 30,000 ft where the temperature is about 233 K and pressure 0.3 atm. Nevertheless, while exchanging cold air from outside, air conditioners (and not heaters) are used, in order to maintain comfortable temperature inside at a normal pressure of 1 atm. Why? As the outside air comes inside, it is quickly compressed adiabatically. Work done on the air to compress it increases its internal energy and hence the temperature. Calling the initial and final values as i and f , we can use the constancy of TV γ−1 to write the ratio, Ti /T f = (V f /Vi )γ−1 and PV γ = constant, it can be shown that the final temperature will be close to 330 K. Hence, we need air conditioners.

The Kinetic Theory of Gases

16.4

165

Adiabatic Reversible Compression (From Mechanics and Kinetic Theory)

Let us denote the mass of a molecule by m and that of the wall of the piston by M. The area of the piston seen by the molecules is denoted by A and the distance compressed (expanded) by x. We also identify the x-axis with the direction of motion of the piston. The x component of the velocity of the molecule changes upon collision with moving piston at a speed w from ux to u0x . The speed of the wall changes to w’. The collisions of the molecules with the walls of the cylindrical container are elastic (equivalent to adiabaticity). From conservation of energy and linear momentum, m M 02 m 2 M 2 ux + w = u02 w , x + 2 2 2 2 mux + Mw = mu0x + Mw0 .

(16.18)

From these, we can express u0x , w0 in terms of ux , w, denoting the ratio m/M by ε: u0x = [−ux (1 − ε ) + 2w](1 + ε )−1 , w0 = [2ux ε + w(1 − ε )](1 + ε )−1 .

(16.19)

Since, the wall is much heavier than a molecule, ε is a small quantity and we shall drop terms that are higher than linear order in it. Also, the wall is moving much slowly than a molecule, thus w/ux is nearly zero - this can be seen as defining a quasi-static process. Employing these, we can see the ratio u0x /ux = [−(1 − ε ) + 2(w/ux )](1 − ε ) employing binomial expansion, thus u0x ≡ −ux + 2w.

(16.20)

Similarly, w0 /ux ≡ 2ε + (1 − ε )(w/ux ) ≡ w/ux ,

(16.21)

thus w0 = w. From the conservation of energy, we can write 2 −2 u02 x = [−ux (1 − ε ) + 2w] (1 + ε )

=

u2x

4w2 4w 1 − 2ε + 2 − (1 − ε ) (1 − 2ε ) ux ux

' u2x − 4wux .

(16.22)

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Mechanics, Waves and Thermodynamics

To relate ux to the root mean square speed of a typical molecule of an ideal gas at equilibrium at temperature T , we recall that the average kinetic energy of a molecule is hEi = mhvi2 /2 = 3kB T /2, by equipartition of energy. Let us now determine the number of collisions, Nc (x) that molecules undergo with the moving piston as a function of distance, x it moves. In a time interval ∆t, the number of collisions encountered with the piston is 21 ux ∆tAρ where ρ is the density of the gas and the factor one-half corresponds to the fraction approaching the piston. Since, one mole of gas contains an Avogadro number NA of molecules, the density can be written as NA /(Ax). Therefore, the number of collisions per unit time is Nc =

NA ux . 2x

(16.23)

At each collision, ux changes according to Eq. (16.22). Change in KE for the x-motion is ∆Ex = ∆(mu2x /2) = −2mux w. The rate of change of Ex is dEx NA ux 2w = Nc ∆Ex = − .2mux w = − Ex dt 2x x

(16.24)

where Ex = 21 mu2x NA . The relative change in total energy can now be related to the change in Ex by using the equipartition theorem. Employing the linear relation between energy and temperature, we have dT 2w 2 dx dE = =− dt = − . E T 3x 3 x

(16.25)

Obviously, dx/x = dV /V because the cross-sectional area of the piston is constant. We can now integrate this equation: V f E f 2 log E = − logV 3 Vi

Ei

Ef = implying Ei

Vi Vf

2/3

=

Tf . Ti

(16.26)

The exponent 23 can be identified with 53 − 1 where 5/3 is γ. For a diatomic molecule, there are five degrees of freedom - three translational and two rotational. In this case, by equipartition, E = 5Ex . Correspondingly, we obtain Tf = Ti

Vi Vf

where 57 − 1 =

2/5

2 5

(16.27) where, one again, we see the connection with γ.

The Kinetic Theory of Gases

16.5

167

Maxwellian Distribution of Velocities of Gas Molecules

Let us consider N molecules of a dilute gas placed initially about a point. The gas molecules spread about this point (let us fix the origin). The components of velocity of a molecule are vx , vy , vz . The number of molecules with x-component between [vx , vx + dvx ] is N f (vx )dvx . Similarly, one can write for the y- and z-components, N f (vy )dvy and N f (vz )dvz respectively. The function f is the same for each direction. Clearly, the components along x, y or z are independent. Hence, the number of particles with velocity in the interval [v, v + dv] is N f (vx ) f (vy ) f (vz )dvx dvy dvz .

(16.28)

Number of molecules per unit volume in velocity space is N f (vx ) f (vy ) f (vz ). As the directions of coordinates are arbitrary, this will depend on the distance from the origin, q v2x + v2y + v2z . That is, f (vx ) f (vy ) f (vz ) = φ (v2x + v2y + v2z )

(16.29)

for some function φ . The solution of this functional equation can be easily verified to be 2

f (vx ) = CeAvx ,

2

φ (v2 ) = C3 eAv .

(16.30)

The constant A must be negative otherwise the number of particles will be infinite with higher velocities. Let us take A to be (−1/α 2 ). Accordingly, the number of particles between vx and vx + dvx is 2 v NC exp − x2 dvx , (16.31) α and similar expressions for the y- and z-components. On normalization of the distribution, α 2 = 23 hv2 i where the mean square velocity is related to temperature by mhv2 i/2 = 3kB T /2. How did Maxwell solve the functional equation? Assume that f and φ are analytic functions. Expand these in Taylor series, giving the distribution in Eq. (16.30).

17

Chapter

Concepts and Laws of Thermodynamics The right understanding of any matter and a misunderstanding of the same matter do not wholly exclude each other. — Franz Kafka

In the earlier chapters, we have discussed mechanics and waves. Mechanics deals with point particles and rigid bodies. However, these idealizations do not allow energy to be stored in them in any form - rotational or vibrational motions. As the constituents of rigid body remain fixed in length w.r.t. each other, energy cannot be exchanged with them. Physics of systems where energy may be stored, converted from one form to another, requires us to understand the laws of thermodynamics. One of the most important concepts is that of entropy. The mystery of this concept has been removed in a series of papers by Leff [84, 85, 87, 88, 103]. We take advantage of these recent essays and follow them rather closely. There are four laws on which thermodynamics is based. Each of the laws, zeroth law to the second law have given a new concept - temperature, energy, and entropy. The third law has established a limitation on thermodynamic functions. We shall explain the ideas and concepts in some detail. The applications of these ideas to real life are everywhere around us. The changes occurring all around from one state to another are consistently described within these laws. To relate a few common instances [100], we have noticed that the valve on a bicycle pump gets hot when we are pumping up a tyre. Why? When we pump and compress the gas, there is no heat transfer from outside. The process is adiabatic. The internal energy of the air increases, leading to an increase in temperature. This hot air heats up the valve. As a second example, again drawn from our experience, is the warm (even hot in certain regions in Northern India) dry wind that blows down from mountains into plains. It is called differently in different parts of the world - Loo (India), Kachchan (Sri Lanka), Santa Ana (Southern California), Chinook (North America), Foehn (Switzerland), Berg Wind (South Africa). As winds come down from the mountains, they move into the regions of greater atmospheric pressure. Hence, the moving air is adiabatically compressed and heated. As the descent is fast, there is no time for any significant exchange with local air. So this wind is warm and dry.

169

Concepts and Laws of Thermodynamics

In this Chapter, we present a formal conceptual discussion of the concepts involved in understanding entropy. We explain the Caratheodory’s formal approach in the next two Sections - the logical origin from the ideas to the laws.

17.1

Adiabatic Transitions and Accessibility of States of a System Empirical Entropy, First and Second Laws

In the myriad of changes that we wish to describe, we see that all states are not equally accessible from a given initial state. Following Buchdahl [94], let us see the statement of the First law: First law: Let us consider a system S where the transition between states J and J 0 is adiabatic. Then the work done is unique, completely given by the terminal states. Accordingly, we can write the work done W0 as a function of the variables x and x0 that specify the terminal states, W0 = f (x, x0 ).

(17.1)

Let J 00 be another state, specified by the variable(s), x00 . Work done in making a transition from J 0 to J 00 is f (x0 , x00 ). Furthermore, the work done in making a transition from J to J 00 is the sum of the work done for the transitions from J to J 0 , and, from J 0 to J 00 : f (x, x0 ) + f (x0 , x00 ) = f (x, x00 ).

(17.2)

Equation (17.2) is valid for any x0 . The RHS of this equation is, in fact, independent of x0 . This implies that the form of f should be of the form f (x, x0 ) = U (x0 ) −U (x).

(17.3)

In other words, there exists a function f such that the work done during an adiabatic transition is just equal to the difference ∆U equal to the RHS of Eq. (17.2). If the change of states occurs without the adiabatic isolation, then the external work done could be measured, denote it by W . The difference between W and W0 is denoted by Q: Q = ∆U −W .

(17.4)

This is called the quantity of heat imparted to the system during the transition. In the above discussion, we haven’t bothered to check if the transition were allowed or not. Empirically, we can see that if an adiabatic transition from J to J 0 is impossible, then the transition from J 0 to J must be possible. That is, there do not exist any mutually inaccessible adiabatically connected states. Instances of inaccessible states can be easily drawn from daily life. Consider a certain volume V of tea contained in a thermos flask (an adiabatic envelope) at a temperature T . On stirring sugar in it (performing external work), the temperature will be increased, rising

170

Mechanics, Waves and Thermodynamics

to T 0 . By stirring in the opposite direction, it is not possible to un-mix or to reach the temperature T . Hence, this original state is inaccessible. Second law: There are limitations upon adiabatic transitions between any two arbitrarily chosen states of a system S . For any adiabatically enclosed S , there are states {Ji0 } that are inaccessible from a state J. Hence, we arrive at a natural logical conclusion that it is possible to put the states in a definite order. It becomes possible to arrange the states as points along a straight line, L analogous to numbers on the number line. To understand this, let us determine whether or not a state J 0 is accessible from J. We associate a point P on the line to the state J. If J 0 is accessible, then we may associate a point P0 on a line to the right of the point P on L. This process of ordering can be shown to be internally consistent [95]. We may now even number the points of L, increasing from left to right. This motivates a defining a function s acting on x, yielding a number. The value of s is called the empirical entropy of the state. The difference in the values of s(x) for two states determines the degree of adiabatic inaccessibility of the states. Now, we may state the following: An adiabatically enclosed system makes a transition from J to an accessible state, J 0 . By definition of s(x), it follows that s0 ≥ s. That is, empirical entropy can’t decrease. It must be immediately noted that this “increase” is due to our arbitrary choice of arranging the accessible states from left to right. The important point is that there is a monotonic variation. Equation (17.4) can be written in a differential form. Since ds(x) is zero whenever dQ is zero. Hence, for any infinitesimal part of a reversible transition, dQ = λ ds

(17.5)

where λ is a function of x. This establishes integrability of ds.

17.2

Sears’ Illustration of Caratheodory’s Treatment

In this Section, we follow very closely the beautiful exposition of Caratheodory’s approach to thermodynamics by Sears [96]. This discussion is a more detailed continuation of the previous Section. To make the discussion instructive and nontrivial, we have to consider a system which has a three-dimensional space to define equilibrium states. The system we consider has two cylinder-piston arrangements attached end-to-end (Fig. 17.1). In equilibrium, the (real) gases in the cylinders are at the same temperature, Θ, however, the pressure may not be the same. The temperature is measured on an arbitrary scale, we call it an empirical temperature. The values are ordered in a way so that. for instance, the temperature of steam is greater than that of ice. An equilibrium state of the system is specified by three coordinates, (Θ,VA ,VB ). Here, there are two deformation coordinates, VA and VB . A non-deformation coordinate could be one of PA , PB , or Θ. In the three-dimensional space, any process may be depicted - be it a reversible or an irreversible one.

Concepts and Laws of Thermodynamics

171

We have seen that between two states, there can be an infinite number of irreversible adiabatic processes. Discussion from Eqs (17.1)–(17.4) leads us to the first axiom of Caratheodory: Work done in all these adiabatic processes is the same. Now let us consider a number of reversible adiabatic processes from an initial state, J to K with final configuration, (VA f ,VB f ). We begin with the state J labelled by (Θ,VA ,VB ) = (Θ0 , 0, 0). Let us consider the changes made along different paths described below:

Fig. 17.1

Two cylinders of volume VA and VB are attached end-to-end with pistons PA and PB on the outer sides. The cylinders may contain different real gases, their walls and pistons providing an adiabatic envelope. The dividing wall is rigid, hence, permeable by radiant heat. We may, in some processes, remove the adiabatic boundary and allow a contact with a heat reservoir.

Path I - we expand A reversibly and adiabatically until the volume of A becomes VA f , keeping VB fixed; followed by reversible expansion of B, keeping VA equal to constant. Each process in this is not adiabatic individually because there is transfer of heat across the wall. The entire system is adiabatic. Path II - we expand B reversibly and adiabatically until the volume of B becomes VB f , keeping VA fixed; followed by reversible expansion of A, keeping VB equal to constant. We could also expand the systems A and B simultaneously to reach the desired final state. The second axiom of Caratheodory requires that all the reversible adiabatic processes terminate at the same state, at the same temperature, Θ f . On the other hand, for a path beginning at the same initial state as above along which the changes are adiabatic and irreversible, the value of Θ will be higher for reaching the final state with (VA f ,VB f ). Since (VA f ,VB f ) is arbitrary, it follows that all states that can be reached from the initial state J by reversible adiabatic processes to K (VA f ,VB f ), lie on a single surface passing through J, which corresponds to Θ = Θ0 . We parametrize this surface by s, and call it empirical entropy of the system. Other surfaces parametrized by different values of s would form for different initial states corresponding to different initial temperatures. We may order s1 , s2 , . . . in such a way that at any configuration they increase monotonically with increasing Θ. Since, the empirical temperature at the end of any irreversible adiabatic process between same initial and final states is greater than for reversible adiabatic process, the same is true of the empirical entropy. This is how the Caratheodory’s second axiom of “adiabatic inaccessibility” of states works.

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A definite value of s is associated with each point of (Θ,VA ,VB )-space. We may thus alternatively consider a state to be defined by (s,VA ,VB ) in terms of which the change in internal energy is ∂U ∂U ∂U ds + dVA + dVB . (17.6) dU = ∂ s VA ,VB ∂VA s,VB ∂VB s,VA In an adiabatic process, dQ = 0, hence dU = −dW . In a reversible process, dW = PA dVA + PB dVB ,

(17.7)

and in a reversible adiabatic process, ds = 0. Using Eq. (17.6), ∂U ∂U , PB = . PA = ∂VA s,VB ∂VB s,VA Hence in any arbitrary reversible process for which dQ = dU + dW , we have ∂U dQ = ds = λ ds. ∂ s VA ,VB

(17.8)

(17.9)

As ds is an exact differential, we may re-write this as dQ = ds; λ

(17.10)

with this, the integrability of dQ is established. Now we prove that the empirical entropy s of a composite system (made of subsystems A and B) is a function of sA and sB , not just a sum of these. The total entropy is s(VA ,VB , Θ, sA , sB ). For a process, any or all of the variables VA ,VB , Θ are varied in some way while sA , sB are constant. The total entropy is s = s(sA , sB ). The heat dQ flowing in the system is simply dQ = dQA + dQB , which can be written as λ ds = λA dsA + λB dsB .

(17.11)

Let us define λ ’s by the equations: λA = λA (Θ, sA ),

λB = λB (Θ, sB ).

(17.12)

The form of λ can be found by considering a process of a system at temperature Θ1 . We may expand A and simultaneously compress B, reversibly and isothermally. Heat flows from B to A, at each stage, dQA = −dQB ,

or at Θ1

λA1 dsA1 = −λB1 dsB1 .

For the whole system, dQ = 0 and ds = 0, and in general, dsA1 6= dsB1 .

(17.13)

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173

Both the systems A and B are taken around Carnot cycles: (i) Isothermal (temperature Θ1 ) and reversible expansion (compression) of A (B), (ii) followed by adiabatic compression of A and B leading to an increase of temperature to Θ2 , (iii) Isothermal (temperature Θ2 ) and reversible expansion (compression) of B (A), (iv) followed by adiabatic expansion of A and B leading to a decrease of temperature to Θ1 . In steps (ii) and (iv), there is no change in entropies. Hence, the change in entropy in (i) and (iii) must be equal and opposite in sign: dsA2 = −dsA1 ,

dsB2 = −dsB1 .

(17.14)

Analogous to step (i), in step (iii), we may write dQA = −dQB ,

or at Θ2

− λA2 dsA2 = −λB2 dsB2 .

(17.15)

In steps (i) and (iii), for the whole system, dQ = 0, ds = 0, adiabatic and isothermal. Combining Eqs (17.13), (17.14), and (17.15), we get λA λA = . (17.16) λB 1 λB 2 That is, λλAB has the same value at temperatures Θ1 , Θ2 . This ratio is independent of temperature. This can happen if the factor related to temperature is common and gets cancelled. The form of λ must be λA (Θ, sA ) = T (Θ) fA (sA ), λB (Θ, sB ) = T (Θ) fB (sB )

(17.17)

where T (Θ) is the same function of Θ for A and B, and hence for all substances. This is called the Absolute temperature. For a reversible process, we saw that λ = dQ/ds; thus we can write dQA = fA (sA )dsA , T (Θ) dQB = fB (sB )dsB . T (Θ)

(17.18)

Clearly, the RHS’s of these equations are exact differentials, let us call denote them by dSA and dSB . SA , SB may be called the Absolute entropies although they are not as “absolute” as the Absolute temperature.

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We have discussed λA,B however, not the one associated with the system as a whole. Dividing Eq. (17.11) by λ , we get ds =

=

λB λA dsA + dsB λ λ T (Θ) fB (sB ) T (Θ) fA (sA ) dsA + dsB . λ λ

(17.19)

We know that s = s(sA , sB ). So, λ must depend on Θ only: λ = T (Θ) f (sA , sB ) = T (Θ) f (s)

(17.20)

dQ = f (s)ds = dS. T (Θ)

(17.21)

and

T (Θ) is the same for A and B, and also for the system as a whole. 17.2.1 Temperature as a property As mentioned earlier, the science of thermodynamics introduces a new concept - absolute temperature. A necessary condition of thermodynamic equilibrium between two systems is attaining an equality of temperatures. Temperature is a property or parameter characterizing a state. If we consider the case where temperature depends on two variables x, y, then we may write dT = Xdx + Y dy,

X=

∂T ∂T , Y= . ∂x ∂y

(17.22)

Evidently, ∂X ∂Y = ∂y ∂x

(17.23)

which is consistent with the exact nature of dT in Eq. (17.22). In other words, I

dT = 0 =

I

∇ × T dxdy.

(17.24)

For three variables, x, y, z, the condition modifies to T ∇ × T = 0.

17.3

Reversible and Irreversible Adiabatic Processes

17.3.1 Reversible process Although ideal, reversible processes play a very important role in thermodynamics. They are infinitely slow, occurring in a large number of small steps which, in principle, could

Concepts and Laws of Thermodynamics

175

be reversed. Hence, during such an evolution, the system passes through equilibrium states. A system that evolves through equilibrium thermodynamic states is quasi-static. All reversible processes are quasi-static. Although, most quasi-static processes are reversible where the surroundings also reverse their path, there are some that are not. For instance, air leaking out of a tyre is irreversible. The slow process is irreversible, converting mechanical energy to internal energy. At constant pressure, if we consider reversible heating [88] by successively bringing it in contact with reservoirs at higher and higher temperatures. In bringing the system from a temperature Ti to Ti+1 , energy Qi+1 is released by the reservoir. The entropy of the reservoir changes by ∆Sres,i+1 = −

Qi+1 , with Ti+1

Qi+1 =

Z Ti+1 Ti

Total entropy change after various steps will be Z Ti+1 1 1 − dT . C p (T ) ∆Stotal = ∑ Ti Ti+1 i Ti

C p (T )dT .

(17.25)

(17.26)

In the limit of infinite steps, the temperature differences will go on reducing; thus, ∆Stotal = 0. This means that the system and the reservoir are in equilibrium states. Extend the above discussion to volume changes. Relate the irreversible compressions and expansions to the increase in entropy (see e.g., [98, 99]). 17.3.2 Carnot cycle One of the well-known reversible processes is the Carnot cycle, Fig. 17.2 shows a temperature-entropy graph of the same. Since the initial and final states are the same, the change in internal energy in going along 12341 is zero. By the First Law, Q = Qin − Qout = W . Along the isotherm 12, dS = δ Q/Th , leading to Qin = Th (Smax − Smin ). Along the isotherm 34, Qout = Tc (Smax − Smin ). Thermal efficiency is η=

Tc W = 1− . Qin Th

(17.27)

Argue that if the cycle is run in reverse order 43214, it is a refrigerator that removes energy Qc from the colder region and delivers energy Qh = Qc + Wext . 17.3.3 Irreversibility An irreversible heat engine operating between two reservoirs has an efficiency less than the corresponding Carnot cycle. The efficiency for maximum power output is [101]. s T− η∗ = 1 − (17.28) T+

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Fig. 17.2

A typical Carnot cycle.

where T− (T+ ) is the temperature of the reservoir at lower (higher) temperature. Problem An ideal gas undergoes a circular cycle on a PV diagram [102]. What is the efficiency of a Carnot cycle operating between the same high and low temperatures for the gas in this circular cycle? Hint Redraw the PV diagram using dimensionless quantities P/P0 and V /V0 where (P0 ,V0 ) are the smallest values from the circular cycle. The circular cycle will now transform to a circumference of a circle of unit radius, centred at (2, 2). To determine the values of maximum and minimum temperatures, draw the hyperbolic isotherms. The points at which the hyperbolic isotherms are tangential on the circle give the P/P0 and V /V0 values - one for the upper tangent and the other for the lower. From these, determine higher and lower temperatures using the ideal gas law. The efficiency should come out to be 77.19 per cent. See [102] for the full solution.

17.4 Order or Disorder A larger volume of a substance has proportionately more entropy; clearly though, just by being larger, it doesn’t become more disordered. Hence, entropy is not a measure of disorder [103]. Another example, given by Styer [104], is the reaction for rust, 4Fe + 3O2 → 2Fe2 O3 . The change in entropy is -549.3 J/K. This decrease is not related to order or disorder in the iron-oxygen system. Because the change in enthalpy is - 1684 kJ/mol, the entropy change is 1684 kJ mol−1 /298.15 K = 5650 J/(K mol). The energy spreads from the system to the environment without any increase in disorder.

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It can be argued that the Boltzmann entropy, S = kB logW is a measure of “missing information”. Out of the possible states that the system could occupy, it is found in a particular state with some probability, pi . Shannon defined the information function as I = − ∑Ni=1 pi log pi . This definition is with an assuring agreement with the case when all the states are equiprobable, and Imax = −N N1 log N1 , i.e., log N. The states are completely specified by energy, and they are equiprobable. To understand this, let us consider two subsystems interacting with each other while they are a part of an isolated system. As the system 1 gains a small amount of energy from 2, the number of energy states increase (decrease) in 1 (2) because entropy increases with energy. Hence, dW1 > 0 and dW2 < 0. For the two subsystems, the total number of states is Wtotal = W1W2 and the total entropy is Stotal = S1 + S2 . Denoting the fractional change of states in system i by fi , the second law of thermodynamics gives [103] dW1 dW2 dStotal = kB − W1 W2

= kB [ f1 − f2 ] ≥ 0.

(17.29)

We see that for this to be valid, f1 ≥ f2 . Energy flows so as to make f1 = f2 . At this point, thermal equilibrium is reached where we can define a quantity which is rate of change of entropy with energy, eventually identified with the inverse of temperature [105].

17.5

How Does Entropy Look Like?

Heat is a process where energy transfer Q occurs due to a change in temperature. For instance, when tea is heated in a microwave oven, there is a temperature difference between the oven and the cup. A special process where Q = 0 (adiabatic) is called pure work. When a gas compresses, work is done by the surroundings on the gas (W < 0). When it expands, work is done by the gas, and we call W > 0. Hence, during an adiabatic change of volume, the work done is Wad , and ∆U = −Wad . In a slow adiabatic expansion where gas does work, the internal energy reduces. Work and heat combine to give ∆U = Q −W .

(17.30)

To understand the definition of Q in terms of measurable work, we consider how two equilibrium states are connected by adiabatic, and, non-adiabatic processes. Work done by an adiabatic process Wad determines the change in internal energy, ∆U. Connecting the same states via a non-adiabatic process requires Q to be ∆U + E = Wad + W . Suppose we heat a gas so that its state changes from A to B. We could either heat it at a constant pressure, or, we could heat it at a constant volume, doing no work, to a state C, and subsequently bring to a state B by an adiabatic expansion. For both the paths, the change in internal energy is the same, ∆U = UB −UA . That is, QAB −WAB = QACB −WACB . Path ACB entails higher pressures than AB, and thus WACB > WAB and QACB < QAB . This illustrates

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that two equilibrium states can be connected by many different paths with different work and heat processes. Point particles and rigid bodies do not possess internal energy. Obviously then, it cannot be redistributed. Hence, the entropy of these systems is zero. Entropy has significance (being nonzero) only when a system can store internal energy. There occurs a spatial redistribution of internal energy. The energy spreads maximally in the sense that an equality of temperature is achieved among the bodies in contact. To calculate entropy, we have the Clausius algorithm (the idea of calling it an algorithm is due to Leff [85]), viz. dS =

δ Qrev T

(17.31)

where the infinitesimal δ Q reminds us of the fact that it is not an exact differential. That is, R2 δ Q 6 = Q2 −Q1 . If the energy transfer is from the system, then δ Q < 0. For an irreversible 1 process, dS > δ Q/T ; thus, an infinitesimal adiabatic free expansion corresponds to δ Q = 0, however, dS > 0. We can actually calculate entropy knowing the data for the specific heat at constant pressure, C p for systems that are heated slowly and reversibly at a constant atmospheric pressure. The energy added to increase the temperature by dT is δ Qrev = C p dT . The entropy change will be ∆S =

Z Tf C p (T ) Ti

T

dT .

(17.32)

It is interesting to make a connection with another thermodynamic function, enthalpy H = U + PV (we discuss thermodynamic functions in the next Chapter). dH = dU + PdV + V dP. First law gives dU = δ Q − δW . For a slow and reversible process, δW = PdV . Then, dH = δ Q + V dP = δ Q = C p (T )dT

(17.33)

for constant pressure. For instance then, to heat a material from zero temperature to room temperature, the energy needed is ∆H. For room temperature solids not undergoing a phase change, as P∆V ∆U, ∆H is an excellent approximation of stored internal energy. A correlation between entropy and enthalpy supplied from zero to room temperature has been illustrated [86], a linear relation is found. This association follows from Eq. (17.31). This points at the significance of entropy being a state function. Energy tends to become equal by spreading from a higher temperature region to a lower temperature region. Hence, a temperature gradient drives the energy flow or the heat process. This spreading of energy in space is accompanied by an increase of entropy in time. Can we argue to a similar conclusion for a pure work process? For instance, stirring of coffee powder in water by a spoon involves the spoon working on various volume elements of water. This leads to an increase in molecular kinetic energy, and, in dissolving the coffee

Concepts and Laws of Thermodynamics

179

powder. This is done until coffee is mixed well, leading at the same time for the energy to distribute uniformly. Thus, we have the mechanical equivalent of heat (Mayer–Joule principle) here as the work led to a change in internal energy which is the same as an isochoric (constant volume) heat process: ∆U = W = Q.

(17.34)

Here increase in entropy occurs in the absence of a heat process. It should be understood that Eq. (17.31) cannot be used for irreversible processes. Using the First Law Eqs (17.30) and (17.31), change in internal energy is dU = δ Q − δW = T dS − PdV . If the process is taking place with constant volume, then dS = dU /T . This implies that

∂S ∂U

V

1 = >0 T

and

∂ 2S ∂U 2

=− V

1 T 2 (∂U /∂ T )V

≤ 0.

(17.35)

The first inequality follows from the assumption that T > 0. The second follows by assuming (∂U /∂ T )V , which is the specific heat at constant volume, must be positive for positive temperature. The equality in the second expression corresponds to a first-order phase transition where S(U ) ∝ U. Expressing entropy as a function of enthalpy and pressure, show that

∂S ∂H

1 = >0 T P

and

∂ 2S ∂ H2

=− V

1 T 2 (∂ H/∂ T )

≤ 0.

(17.36)

P

Note that (∂ H/∂ T )P is the specific heat at constant pressure. Hence, expressed as S(U,V ) or S(H, P), entropy is an increasing, concave downward function for fixed volume and pressure respectively. The slope of each of these curves gives 1/T . Consider two identical systems in states 1 and 2. When they are brought in thermal contact, they become subsystems with energy that is unequally distributed. The subsystem at a lower (higher) temperature absorbs (loses) energy Q, and makes a transition from its state to a final state f . All these states are lying on the S vs H curve. This process is irreversible and the intermediate states are non-equilibrium states, not lying on the curve in Fig. 17.3. The concavity of the curve ensures that the entropy increase in the lower temperature system is greater than the entropy decrease in the higher temperature system. Hence, ∆S1 + ∆S2 > 0

(17.37)

which ensures the Second Law of thermodynamics. The above discussion is modified in the presence of long-range forces as in stars. Discuss this situation with typical data.

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Fig. 17.3

Entropy versus enthalpy is very instructive in understanding the role played by the concavity of this graph.

18

Chapter

Some Applications of Thermodynamics Thermodynamics is a funny subject. The first time you go through it, you don’t understand it at all. The second time you go through it, you think you understand it except for one or two small points. The third time you go through it, you know you don’t understand it, but by that time you are so used to it, it doesn’t bother you any more. — Arnold Sommerfeld

There are a very large number of applications for which we have to content ourselves by referring to classics - articles and books. For example, Rayleigh wrote in the nineteenth century about the thermodynamics of dresses for our body to keep us warm in winter. The energetics from the Sun to all the life forms - plants and humans - allows us to make an elementary however, profound estimate of the total sustainable population on the Earth, as discussed in the famous work, The tragedy of the commons by Garrett Hardin [89]. We briefly visit a few topics, chosen nearly randomly, just so we may appreciate admiringly the paintings mounted on the walls of a gallery while making a quick exit.

18.1

Thermodynamic Potentials

In the description of thermodynamics, we have the possibility of considering variables p, v and T , s for each simple, homogeneous system. The First law states du = T ds − pdv.

(18.1)

The internal energy u = u(s, v) is written in terms of extensive variables is s, v. The quantities written in lower case (e.g., u) denotes that thermodynamic variable (here, internal energy U) per mole. The intensive (extensive) variables are independent (dependent) on the quantity of matter. In Eq. (18.1), T , p are intensive variables which can be written as ∂u ∂u , −p = . (18.2) T= ∂s v ∂v s

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There are four possible combination of mechanical and thermal variables: s, v;

s, p;

T , v;

T , p.

(18.3)

If we would like to replace one of the independent variables, (e.g., s) by its conjugate, it is necessary to subtract from the dependent variable (u, in this example) the product of two conjugate independent variables (T s). This is the Legendre transform. So there can be four possibilities: internal energy, u(s, v); enthalpy, h(s, p) = u + pv; free energy, f (T , v) = u − T s; Gibbs free energy, g(T , p) = u − T s + pv; Table 18.1

(18.4)

Thermodynamic potentials [90]

Potential

Independent variables

U, u

v, s du = T ds − pdv

H, h

p, s

h = u + pv

dh = T ds − vd p

F, f

v, T

f = u−Ts

d f = −sdT − pdv

Thermodynamic relations ∂T = −(∂ p/∂ s)v ∂v

v

s

= ∂ 2 u/∂ v∂ s ∂T = (∂ v/∂ s) p ∂p

p = −(∂ u/∂ v)s T = ∂∂ hs

v = (∂ h/∂ p)s s = − ∂∂ Tf

= ∂ 2 h/∂ p∂ s ∂s = (∂ p/∂ T )v ∂v

p

v

p, T

p = (∂ f /∂ v)T s = − ∂∂Tg

dg = −sdT − vd p

p = (∂ f /∂ v)T

G, g g = u − T s + pv

Conjugate variables T = ∂∂ us

p

s

T

= −∂ 2 f /∂ v∂ T ∂s = −(∂ v/∂ T ) p ∂p T

= −∂ 2 g/∂ p∂ T

these are thermodynamic potentials. We can write down the change of the thermodynamic potentials as follows. dh = du + pdv + vd p = T ds + vd p, d f = du − T ds − sdT = −pdv − sdT , dg = du − T ds − sdT + pdv + vd p,

= −sdT + vd p.

(18.5)

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Table 18.1 summarizes description of an enormous amount of physical phenomena. For example, from the row about F, f , we have ∂s ∂p = . (18.6) ∂T v ∂v T This is the Clausius–Clapeyron equation, which played a very important role in the development ofthesteam engine where p plays the role of vapour pressure at temperature T , and replace ∂∂ qv by ∆q/∆v with ∆q being the heat of vaporization. T We may also find the difference between the specific heats at constant pressure and volume. From du = T ds − pdv, with T held fixed, ∂u ∂s = T −p ∂v T ∂v T

= T

∂p ∂T

− p, using Eq. (18.6).

(18.7)

v

To calculate dq/dT at constant volume or pressure, we need to re-write the First law: dq = du + pdv

=

∂u ∂v

∂u +p + ∂T T

dT .

(18.8)

v

At constant volume, dq ∂u cv = = . dT v=constant ∂T v

(18.9)

At constant p, ∂v ∂u dq ∂u cp = = +p + . dT p=constant ∂v T ∂T p ∂T v

(18.10)

The difference between the specific heats is ∂u ∂v c p − cv = +p ∂v T ∂T p

= T

∂p ∂T

v

∂v ∂T

p

= R for one mole of ideal gas.

(18.11)

184

18.2

Mechanics, Waves and Thermodynamics

Van der Waals Equation for Real Gases

Hitherto, we have discussed properties of an ideal gas, and taken that as a paradigm over which we developed other concepts. However, most of the gases are such that we need to consider size of constituent atoms or molecules, and, their inter-particle interactions. Taking these into account, van der Waals wrote the equation of state as (for one mole) P=

a RT − 2. V −b V

(18.12)

The quantity b is due to the volume of molecules and a is a measure of the interaction. We can re-write Eq. (18.12) by calling a/V 2 as Pa :

(P + Pa )(V − b) = RT .

(18.13)

The coefficient of thermal expansion in this case is 1 α= V

∂V ∂T

= P

V −b V T − 2a R

V −b 2 V

.

(18.14)

The second equality is obtained by differentiating Eq. (18.12) at constant pressure P (i.e., by setting dP to zero). For an ideal gas, we can see that α = 1/T (obtained by setting a = b = 0). The difference for b V , a RTV is 1 1 α− = T VT

2a −b . RT

(18.15)

For gases like O2 , N2 , etc., the right hand side is positive. However, for H2 and noble gases, it becomes negative because cohesive forces are very weak (a < RT b/2). The internal energy of a real gas will consist of kinetic energy and potential energy corresponding to the inter-particle forces. We now show that in contrast to the ideal gas where ∂U /∂V is zero at constant temperature, here it is a/V 2 . Van der Waals applied the ideas of thermodynamics beginning with dS =

dU + PdV T

(18.16)

Considering U (V .T ) for writing dU and Eq. (18.12), we have 1 ∂U ∂U R a dS = dT + dV + − T ∂T ∂V V − b V 2T 1 ∂U 1 = dT + T ∂T T

∂U R a + − 2 ∂V V − b V T

dV .

(18.17)

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185

For this to be an exact differential, 1 ∂ 2U ∂ = T ∂ T ∂V ∂T

1 ∂U R a + − 2 T ∂V V − b V T

.

After simplification, we get ∂U a = 2. ∂V T V

(18.18)

(18.19)

From this, by a partial differentiation w.r.t. T , we see that ∂CV /∂V = 0. That is, CV is just a function of T , which is also the case for an ideal gas. Exercise:

For real gases, show that the difference between the molar specific heats is

−1 2a (V − b)2 . CP −CV = R 1 − RT V3

(18.20)

For this, you may use the expression for the difference of specific heats as

∂U ∂V

+P T

∂V ∂T

.

(18.21)

P

Also, for entropy of real gases, show that Z T ,V T0 ,V0

dS = S − S0 = CV log

V −b T + R log . T0 V0 − b

(18.22)

18.2.1 Liquefaction of gases Enthalpy is a state function for ideal as well as real gases. The change in enthalpy is ∆H = T ∆S + V ∆P. We may express entropy as a function of T , P: ∂S ∂S dS = dT + dP. (18.23) ∂T P ∂P T The first term here can be written in terms of specific heat: 1 ∂Q CP ∂S = = . ∂T P T ∂T P T From Table 18.1, ∂V ∂S =− . ∂P T ∂T P

(18.24)

(18.25)

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Hence, the change in enthalpy is ∂V ∆P. ∆H = CP ∆T + V − T ∂T P

(18.26)

Since enthalpy is constant, we arrive at an important relation connecting the change in temperature to the change in pressure: 1 ∂V VT 1 ∆T = T −V = α− ∆P CP ∂T P CP T 1 = CP

2a − b , for van der Waals gas. RT

(18.27)

When a gas is expanded, ∆P < 0, it will cool (∆T < 0) if 2a/RT > b. The gases can be cooled by repeated expansion until they finally liquefy. The condition Eq. (18.27) can be written also as ∂T VT (α − T −1 ). (18.28) = ∂P H CP In the P − T plane, we can now plot a curve where (∂ P/∂ T )H = 0, called the inversion curve. This is of importance in not only finding parameters to cool gases, but also to avoid accidents. For example, it is well-known that highly compressed hydrogen gas ignites spontaneously when leaking from damaged pipes. This is because (∂ T /∂ P)H is negative at room temperature. Hydrogen can be cooled only after it has been cooled to about 190 K. On the other hand, for air this is positive at room temperature up to 450 atm pressure.

18.3

The Third Law of Thermodynamics

Thermodynamic systems are described by points depicted in phase space, these are connected adiabatically. For an N-dimensional space, these points belong to a volume, enclosed by boundary. The connection of the boundary points and internal points is actually not decided by the laws. An instance of this will be helpful here: if we consider the Carnot cycle with cold reservoir at zero temperature, then the efficiency will be unity; in contradiction with the second law. This shows that the nature of boundary points has to be decided by some new law. The unattainability of absolute zero is implied by the second law. Following [91], Nernst’s theorem states that two configurations x, x0 of a physical system will converge to the same entropy as the absolute temperature is made to approach the absolute zero. The unattainability of absolute zero can be deduced from Nernst’s theorem with some additional assumptions [92]. Hence, third law is a combination of Nernst’s theorem and the unattainability theorem. Third law also expressly states the entropy to tend to a constant value, S0 as the

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187

temperature tends to absolute zero, independent of pressure or phase of the substance. In terms of maximum available work, i.e., change in free energy (F1 − F2 ), Nernst wrote ∂F (18.29) F = U −TS = U +T ∂T V where we have used the expression for entropy from Table 18.1. We can write now: ∂ ∆F ∆F − ∆U = T . (18.30) ∂T V Nernst observed that at absolute zero, ∆F and ∆U are not only equal however, also tangent to each other. Hence, in a limiting sense, we expect d d ∆F = ∆U dT dT

for

T → 0.

(18.31)

Moreover, from ∆F = ∆U − T ∆S, we obtain d d d ∆U − ∆S − T ∆S = ∆U dT dT dT

for

T → 0.

(18.32)

Finally then, we arrive at Planck’s formulation of the third law: ∆S → 0, i.e., S2 → S1

for

T → 0.

(18.33)

Problem Is the entropy of an ideal gas or a van der Waals gas consistent with the Planck’s version of the third law? Why? Show that the coefficient of thermal expansion tends to zero as T → 0.

Exercise:

Exercise: Show that the specific heats at constant volume, and, at constant pressure are zero as T → 0. However, the tendency of specific heat to vanish as temperature tends to absolute zero does not imply the third law [93].

18.4

Gaseous Mixture

18.4.1 Diffusion Let us begin by considering a mixture of gases before going to the case when they react. Air is a mixture of gases, in fact non-reacting gases. Taking them to be ideal, each component occupies the entire volume, V . Hence, we may define partial pressures Pi for each of them, they must add to the total pressure (Dalton’s law): P = ∑ Pi , i

with

PiV = ni RT .

(18.34)

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Immediately, we have PV = nRT with n = ∑i ni . The internal energy is the sum over molar energies ui , U = ∑ ni ui , ui = cvi T .

(18.35)

i

Here cvi denotes the molar specific heat at constant volume for the ith component. Obviously, Cv = ∑ cvi , C p = ∑ c pi , C p −Cv = nR. i

(18.36)

i

The above equations imply that we might imagine each component to be compressed at constant temperature to a volume, Vi = (Pi /P)V such that they sum to give the total volume, V . The mixture of gases is such that each component occupies the same volume, V at a partial pressure Pi , the total entropy of the mixture will be just a sum of the entropies of the individual components. Let ni be the number of moles of the ith component and si be its molar entropy. Then, S(T , P, n1 , n2 , . . .) = ∑ ni si (T , Pi ).

(18.37)

i

We may imagine that the mixture of gases where each component is occupying the entire volume at a partial pressure was brought to this state from an initial state obtained by an isothermal compression of all the components into compartments of volume Vi at a pressure P. The entropy of the components in that state would be S0 (T , P, n1 , n2 , . . .) = ∑ ni si (T , P).

(18.38)

i

Change in entropy due to diffusion is S − S0 . We know this difference per mole to be given by si (T , Pi ) − si (T , P) = R log

P V = R log . Vi Pi

(18.39)

Hence, using equation of state, S − S0 = R ∑ ni log i

P Pi

! n = R ∑ ni log = R n log n − ∑ ni log ni . ni i i This shows that diffusion is an irreversible process.

(18.40)

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189

Problem Consider a vessel with two compartments, each of volume V . One compartment contains one mole of He gas while the other contains one mole of Ar gas. The partition separating the two gases is suddenly removed and the gases diffuse and mix. On reaching an equilibrium state, show that the change in entropy will be 2R log 2. If the gases in the container are identical, then there is no diffusion when the partition is removed, and thus there will be no change in entropy. The paradox - called the Gibbs paradox - is that the change in entropy is non-zero so long as the gases are not identical, and it just becomes zero due to identity. The clarification requires usage of quantum mechanics to systems with large number of particles (remember we have here an Avogadro number of particles on each side. Problem Think about the following situation, presented by Klein [97]. Imagine that the gas A is monoatomic, and gas B is also monoatomic made of the same substance except that the nuclei are in an excited metastable state (called isomeric state). The lifetime of this excited state is longer than the time it takes for the gases to come to equilibrium after diffusion. The entropy just after the partition is removed will change by 2R log 2. Now we wait until the nuclei de-excite to their ground states. the mixing entropy will be zero now. How to now account for the change in entropy? 18.4.2 Law of mass action We now consider the case when chemical reactions are allowed to occur at a constant pressure P and temperature T . For example, steam dissociates into hydrogen and oxygen, all remaining in equilibrium with each other preserving the stoichiometric ratios. Referring to Table 18.1, for the isothermal isobaric process being considered, the appropriate thermodynamic potential is the Gibbs free energy, G, the change of which is given by dG = dU − T dS + PdV as dT = 0 = dP. As the reaction occurs, δ G ≤ 0 until the state where the system comes to an equilibrium where the equality will hold. Suppose we have a chemical equation occurring between n1 moles of A and n2 moles of B to give n3 moles of C. We may write the chemical equation (n1 A + n2 B → n3 C) by taking all the substances on one side (n1 A + n2 B - n3 C), and call the new coefficients as νi . For 2H2 + O2 → 2H2 O, (ν1 , ν2 , ν3 ) = (2, 1, −2). As the system approaches equilibrium along a path in the space of states, it would be such that the Gibbs free energy approaches a minimum, so δ G = 0. The change in number of moles will occur satisfying proportionality relation, δ n1 : δ n2 : δ n3 = ν1 : ν2 : ν3 . The total internal energy and free energy may be written in terms of a sum over molar energies, U = ∑ ni ui (T ), i

H = ∑ ni hi (T ).

(18.41)

i

Total entropy is given by the sum of molar entropies and the entropy of mixing: n S = ∑ ni si (T , P) + R log . ni i

(18.42)

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We can write for the Gibbs free energy in terms of the molar values for each component as n . (18.43) G = ∑ ni gi (T , P) − RT log ni i The system will now proceed towards equilibrium satisfying δ G = 0 with the constraint that n = ∑i ni . Accordingly, with δ log n n , (18.44) δ G = ∑ δ ni gi (T , P) − RT log − RT ∑ ni ni ni i i where δ log n will vanish due to constancy of n, δ G = 0 implies n ∑ νi gi (T , P) − RT log ni = 0. i

(18.45)

From here follows the well-known law of mass action of Guldberg and Waage (1867): n νi 1 i Πi = K where log K = − νi gi (P, T ). (18.46) n RT ∑ i This law constitutes one of the foundations of the subject of Physical Chemistry. It can be re-written in terms of molar concentrations, ci = ni /n. K is called the equilibrium constant. For the case of dissociation of steam, for instance, K = c2H2 cO2 /c2H2 O . To determine ci s, we have ∑i ci = 1, and, the ratio of number of atoms of hydrogen to those of oxygen by measurement: (2c1 + 2c3 )/(2c2 + c3 ).. Exercise: Express the Guldberg-Waage law in terms of partial pressures by eliminating the molar fractions, ci :

∏ Piνi = KP ,

where KP = KP∑i νi .

(18.47)

i

18.5 Chemical Potential If the number of particles or moles ni of constituents is a variable, we need to amend the relation T dS = dU + PdV to T dS = dU + PdV − ∑ µi dni

(18.48)

i

where ni are extensive variables. The quantity −µi are intensive variables conjugate to ni . We can also write the following relations: dU = T dS − PdV + ∑ µi dni , i

dH = dU + d (PV ) = T dS + V dP + ∑ µi dni , i

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191

dF = dU − d (T S) = −SdT − PdV + ∑ µi dni , i

dG = dH − d (T S) = −SdT + V dP + ∑ µi dni .

(18.49)

i

Hence, there appear the following additional formulae to the Table 18.1: ∂H ∂U , µi = , µi = ∂ ni S,V ,n j ∂ ni S,P,n j µi =

∂F ∂ ni

, µi =

T ,V ,n j

∂G ∂ ni

.

(18.50)

T ,P,n j

Exercise: Show that ∂V ∂ µi = . ∂ ni S,P,n j 6=ni ∂ P S,ni

(18.51)

The last relation in Eq. (18.50) is the most important one. We now want to see how the Gibbs free energy can be written as ∑i µi ni . For this we first need to see how it changes when the system is enlarged by a certain factor, λ . All the extensive variables are multiplied by λ , and all the intensive variables are unchanged. Hence, G is a homogeneous function of the first degree in ni ’s: G(P, T , λ ni ) = λ G(P, T , ni ).

(18.52)

Differentiating w.r.t. λ and setting it to unity (Euler’s rule for homogeneous functions), we have ∂G G = ∑ ni = ni µi . (18.53) ∂ ni T ,P,n j 6=ni ∑ i i The µi ’s depend on T , P, ni . The condition of equilibrium can be written as ∂G δG = ∑ δ ni = ∑ µi δ ni = 0, δ P = 0 = δ T . ∂ ni T ,P,n j i i

(18.54)

On comparing Eq. (18.53) with Eq. (18.45), we have µi = gi (T , P) − RT log

n , n = ∑ ni ni i

(18.55)

where we recall that gi is molar Gibbs free energy for a pure component. This shows that chemical potential is different from gi . The term RT log n/ni is the entropy of mixing. All mixtures obeying Eq. (18.55) are called ideal mixtures.

192

18.6

Mechanics, Waves and Thermodynamics

Van’t Hoff Equation of State for Dilute Solutions

A solution is called dilute when the quantity of solvent is much larger than that of solute. Van’t Hoff, the first Chemistry Nobel laureate, discovered a certain similarity between dilute solutions and ideal gases. It is so remarkable that we must discuss this, as an application of thermodynamics. First, some notation has to be carefully understood. Here, pure solvent (e.g., water (W )) and the solution (e.g., water and sugar (W S)) are two subsystems separated by a semi-permeable membrane. The solvent is denoted by subscript, W and solute by S. The solution is denoted by superscript W S and pure solvent by W . Moles of solvent W in solution W S will be denoted S W by nW W . The subsystem W contains no solute, so nS = 0. We assume there is no heat exchange at the wall. So temperature is the same on both the sides, but pressure could be different. Hence, PW S 6= PW , and equilibrium condition is given by δ G = 0. Note the following relations: S W δ nW W + δ nW = 0 conservation of mass of solvent S W δ nW S = δ nS = 0 semi − permeability of wall.

(18.56)

Introducing now chemical potentials, we can re-write WS S W W µW δ nW W + µW δ nW = 0.

(18.57)

Using Eq. (18.56), W WS S WS µW PW , T = µW PW S , T , nW S /nW .

(18.58)

S W W S otherwise the This implies that nW S is non-zero which, in turn, implies that P 6= P equilibrium condition will not be satisfied. This is because the conservation law must hold at equilibrium. The difference, PW S − PW is called osmotic pressure. Equation (18.58) is an exact equation of state for any solution. Using Eq. (18.55), we can write WS n + nW S WS µW = gW (PW S , T ) − RT log W W S S , nW W µW = gW (PW , T ).

(18.59)

S WS Because of diluteness condition, nW S nW , the logarithm can be approximated by W S W S nS /nW . Hence,

gW (PW , T ) = gW (PW S , T ) − RT

S nW S . S nW W

(18.60)

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193

The LHS can now be expanded in a Taylor series about PW S : W

WS

gW (P , T ) = gW (P

W

WS

, T ) + (P − P

∂ gW (PW S , T ) ) + ... ∂P T

(18.61)

Recognizing from Table 18.1, ∂ gW /∂ P = vW , the molar volume of the solvent, we have PvW = RT

S nW S . S nW W

(18.62)

Due to the diluteness condition, the partial volume of the solute is much lesser than the S volume V of the solution, i.e., V = nW W vW . Hence, (18.62) reduces to S PV = nW S RT .

(18.63)

S This can be stated [90] as: “The osmotic pressure of a dilute solution of nW S moles of W solute is equal to the pressure of an ideal gas which would be measured if nS S moles of the gas exerted pressure on the walls of a vessel of total volume V equal to that of solvent and solute.”

Remark

One can understand the equilibrium between different phases of water, the Gibbs phase rule which is so fundamental in understanding various phases, Raoult’s law for dilute solutions, galvanic cells, the phenomenon of magnetism, blackbody radiation, Onsager’s reciprocity relations, and so on.

19

Chapter

Basic Ideas of Statistical Mechanics We avoid the gravest difficulties when, giving up the attempt to frame hypotheses concerning the constitution of matter, we pursue statistical enquiries as a branch of rational mechanics. — Josiah Willard Gibbs

We have already discussed the important concept of entropy in detail. We return once more to this, briefly, to summarize how different phenomenological ideas compare. We also write entropy in terms of the logarithm of the total number of allowed configurations, which is essentially combinatoric in character. The relative importance of a state is embodied in the Boltzmann factor. This very important idea is then applied to understand some aspects of gases, solids, and particularly liquids. Recently, there have been a number of advances [106, 107, 108] making connections between dynamical motion at the microscopic level and statistical mechanics. Eventually, we observe that although laws of classical physics take us far and wide in our understanding of nature, there is a failure of these laws which becomes self-evident.

19.1

Gibbs and Boltzmann Entropies

Boltzmann and Gibbs developed ideas on entropy, we present them here [109]. Let us consider an N-particle system with no internal structure. The N-particle distribution function is denoted by FN (x1 , p1 ; . . . , ; xN , pN ;t ), this specifies the probability density in the whole phase space. The H-function of Gibbs is defined by HG =

Z

FN log FN dτ

(19.1)

and the function defined by Boltzmann is HB = N

Z

f1 log f1 dτ1 .

(19.2)

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195

f1 (x1 , p1 ;t ) is a single-particle probability density defined as a reduced function from F as: f1 (x1 , p1 ;t ) =

Z

FN dτN−1 .

(19.3)

It is understood that R dτ is d 3 x1 d 3 p1 . . . d 3 xN d 3 pN . The distribution functions are R normalized: FN dτ = f1 dτ1 = 1. FN is symmetric with respect to all the arguments, we use Eq. (19.3) to write HB as HB = N

=

Z

Z

FN log f1 (x1 , p1 )dτ

FN log[ f1 (1) . . . f1 (N )]dτ

where (i) stands for (xi , pi ). So, we have the difference, Z f 1 (1) . . . f 1 (N ) dτ. HB − HG = FN log FN (1 . . . N ) Due to the fact that log x ≤ (x − 1), Z f 1 (1) . . . f 1 (N ) HB − HG ≤ FN − 1 dτ = 0. FN (1 . . . N )

(19.4)

(19.5)

(19.6)

Hence, it is proved that HB ≤ HG ,

(19.7)

the equality being when FN (1 . . . N ) = f1 (1) . . . f1 (N )

(19.8)

holds. The difference, (HB − HG ) depends on the distribution function. We assume that the thermal equilibrium distribution function is canonical: FN ∼ e−β H where β = 1/kB T , with N

H=

p2

∑ 2mi + V (x1 , . . . , xN )

(19.9)

i=1

with V as potential energy. With this, we can write " # 3N/2 p2i β −1 Q exp −βV − β ∑ , FN = 2πm i 2m

(19.10)

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where we have the “configuration integral”, Q(β , Ω ) : =

Z Ω

e−βV d 3 x1 . . . d 3 xN = Ω

Z Ω

e−βV d 3 x2 . . . d 3 xN .

The single-particle function 3/2 2 β 1 e−β p /2m . f1 (x, p) = Ω 2πm

(19.11)

(19.12)

From f1 and FN , we have for the ratio in Eq. (19.5), Q f 1 (1) f 1 (2) . . . f 1 (N ) = N eβV . FN (1 . . . N ) Ω

(19.13)

Denoting the average over canonical ensemble by angular brackets, the difference HB − HG = log Q − N log Ω + β hV i.

(19.14)

We can also verify from Eq. (19.11) for average volume and pressure that hV i = −

∂ log Q , ∂β

β hPi =

∂ log Q . ∂Ω

(19.15)

An ideal gas consists of atoms or molecules with no interactions among them. The pressure of an ideal gas is P0 = NkB T /Ω. The difference between HB and HG may be written as d (HB − HG ) = β dhV i + β [hPi − P0 ]dΩ. Exercise:

(19.16)

Using f1 and hP2 i = 3mkB T , show that

3 3 SB = NkB log(2πmkB T ) + NkB log Ω + NkB , 2 2

∂ SB ∂T

∂ SB ∂Ω

dT =

3NkB dT dhKi = , 2T T

dΩ =

NkB dΩ P0 dΩ = Ω T

Ω

T

(19.17)

where hKi is the total kinetic energy, 3NkB T /2. We can now write both the Boltzmann and Gibbs entropies calculated over a reversible path from an initial state, i to a final state, f . We obtain

(SB ) f − (SB )i =

Z f dhKi + P0 dΩ i

T

,

(19.18)

Basic Ideas of Statistical Mechanics

(SG ) f − (SG )i =

Z f dhK + V i + hPidΩ i

T

=

Z f dQ i

T

.

197

(19.19)

These results clearly spell out the difference between the two entropies - Boltzmann entropy does not consider interparticle interactions whereas Gibbs entropy does. 19.1.1 Entropy and “energy-spreading” Suppose, we are given an isolated system with total energy E and volume V . We may divide the entire available region of states with constant energy into discrete energy cells. The entropy of such a system can also be written as S(E ) = kB logW (E,V ).

(19.20)

W (E,V ) is the total number of ways energy can be distributed in discrete energy cells of 20 the system. These are of the order 1010 or so. As there is no way to favour a state, all the states are assumed to be equally likely. Each has the probability, 1/W - this is the principle of equal a priori probabilities. The constant kB is the Boltzmann constant, 1.38×10−23 J/K. W is determined from the allowed energy levels of a system which, in turn, depends on the intermolecular forces and will therefore vary from system to system. A real system will not be isolated from the surroundings. This implies that energy E will be known only within δ E ( E). The interactions with the environment will spread the energy of the system to other accessible states. Over a period of time, the spread will be over different states. S(E ) can be viewed as a temporal spreading function [87]. The timeaveraged energy of the system is the internal energy, U, and S = S(U,V ). The maximum spreading corresponds to equilibrium.

19.2

Boltzmann Factor: Application to “Phases of Matter”

More than twenty years ago, in the ever-instructive section called “Questions and Answers” in American Journal of Physics, there appeared several responses to a question raised by Dwight E. Neuenschwander on presenting simple arguments for the Boltzmann factor [112]. It is interesting to study various interpretations sent to this question (see e.g. [114]). Matter aggregates as solids, liquids, and gases. Liquids are most intriguing, it is the state whose density is that of solids however, with relatively low viscosity. Although the real description is quantum mechanical, we will keep our discussion classical, pushing the arguments with the help of the Boltzmann factor. We will further assume that the constituents of metals and liquids are “atomic”. We will weave the arguments with the help of geometric aspects of atomic thermal motion in a state [110]. We assume atmospheric pressure throughout. Let us denote the binding energy of the solid per atom at T = 0 K by εs . Depending on the nature of binding, the values range from 0.02 eV for Ne to 340 eV for Al, for H2 O it is

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50 eV. The heat of vaporization per atom at atmospheric pressure εb ; the heat of fusion or melting heat per atom, εm . If Tb is the boiling point, then 1 εs ' εb + εm + kB Tb 2

(19.21)

is valid under the classical assumptions consistent with Dulong-Petit law. Usually, kB Tb and εm εs , so we may replace εs by εb . Also, εm εb and εb ∼ εs . Define εm =

εs εb ∼ . a a

(19.22)

The values of εm range from 0.34 × 10−2 eV for Ne to 11.2 eV for Al, and 6.2 eV for water [111]. The ratio of εs to εm is of the order 6 − 40. The values of εm is partly due to thermal excitation, so εm ∼ kB Tm , the ratios of εm /kB Tm is 3 - 5. We need to understand these values and relations. Tb is related to εb by Trouton’s rule: kB Tb =

εb b

(19.23)

where b ∼ 10. 19.2.1 Gases and solids Before understanding liquids better, we briefly discuss gases and solids. For an ideal, dilute gas of N spherical atoms in a volume V (with no internal degrees of freedom), we can write volume per atom as vg =

V kB T = N p

(19.24)

where p is the pressure. On the other hand, solids were first modelled by Einstein by assuming that each atom or molecule with mass m is elastically bound to a rest position of a lattice. The nearest neighbour spacing between lattice points is d. The atom performs harmonic motion about the rest position with frequency ωE . The energy associated with this energy h¯ ωE can be equated to the thermal energy kB TD . Hence, we can write for Einstein frequency kB TD = cωD . (19.25) ωE = c h¯ The frequency ωD is the highest frequency of the lattice in Debye model where the frequency distribution is ω 2 dω. we want to determine square of the average amplitude δ of oscillator. For fixed energy, E in E = 12 mω 2 δ 2 , δ 2 ∝ ω −2 . The average of ω −2 is √ 3ωD−2 . So ωE−2 = 3ωD−2 . This implies that c = 1/ 3 in Eq. (19.25). We will see that the important quantity in the entire discussion is the “available volume” of an atom at a temperature T . The available volume is the volume in which the

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199

motion of an atom actually takes place. This will be proportional to cube of the amplitude of vibration, A given by mωE2 A2 /2 = kB T . So, 3

A ∝

kB T mωE2

3/2 .

(19.26)

Using the Boltzmann factor, we can get more precise expression as it weights the volume element with the potential energy, W (r = mωE2 r2 /2): vs =

Z

d 3 r e−W (r)/kB T =

2πkB T mωE2

3/2 .

(19.27)

With this, the linear length scale can be written as δs =

v1/3 s

=

2πkB T mωE2

1/2 .

(19.28)

To get a feeling for the numbers involved, we can substitute for Na all the parameters: m = 3.8 × 10−23 g, ωE = 1.19 × 1013 s−1 and find that just near the melting point of about 370 K, δs is approximately 0.64 Angstroms. The lattice spacing is 4.3 Angstroms. If we calculate this number for other metals, then δs turns out to be roughly one-tenth of d just below their melting points. In a gas, of course, there is no restriction on the available volume per atom. It might well be a thousand times the cell volume, d 3 . The equilibrium between gas and solid phases suggests a relation between the available volumes in two phases through the corresponding Boltzmann factor: vg = eεs /kB T vs

(19.29)

where εs is the binding energy of an atom in a solid at T = 0. Using Eq. (19.24), we can re-write Eq. (19.29) in terms of vapour pressure pg above a solid: pg =

kB T −εs /kB T e . vs

(19.30)

We can find the vapour pressure near the melting point using Eqs (19.28) and (19.30). For Na, log10 pg turns out to be −7.3, comparing well with experimental value of −7.1. Similar agreement is found for various other metals. For water, the calculated and theoretical values are −1.1 and 0.66 - the disagreement is probably due to the fact that internal degrees of freedom are important there. For Ne, the values also disagree however, for a different reason, the quantum effects are important there as the melting point is about a third of the Debye temperature.

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19.2.2 Liquids In solids, the position of the centre of the supposedly spherical available volume performs harmonic vibrations with an amplitude δs . This is an instance of long-range order. As the temperature approaches the melting point, the structure loosens. The individual atoms may not return to their original centres as they perform anharmonic vibrations of large amplitude. Hence, the analogous length scale in liquids, δ` will be lesser than δs at the melting point, Tm . This is not much as the atoms are still confined between their nearest neighbours. The density also remains roughly the same. As above, we may define an available volume for an atom or a molecule by δ` = (v` )1/3 .

(19.31)

This measures the average back and forth motion, with the mean position shifting. The molecules not necessarily return to their original positions. It is like a hindered random walk; hindering may be quantified by v` /(d 3 ) 1 however, larger than vs /d 3 . Inspired by Eq. (19.29), we introduce another ratio by employing the Boltzmann factor involving the melting heat εm and Tm : 3 δ` v` = = e−εm /kB Tm . (19.32) vs m δs The ratio v` /vs cannot be too large, so we expect εm ∼ kB Tm . For Na, εm is 2.7 × 10−2 eV and kB Tm is 3.2 × 10−2 eV; the respective values for Al are 11.2 × 10−2 eV and kB Tm is 8 × 10−2 eV. However, for water, εm is 6.2 × 10−2 eV whereas kB Tm is 2.35 × 10−2 eV. This disagreement is again due to the unaccounted internal degrees of freedom in water. Recall that for gases, according to Trouton’s rule, boiling heat is about ten times the kB Tb , simply because the available volume is larger: vg v` . We ask: What is the value of d/δs when melting occurs? From Eq. (19.28), we find δs and compare with d; thus we obtain at T = Tm , 1 δs ∼ (19.33) d T =Tm 6 for several metals even when there are large differences in the values of Tm and ωE . For example, for Ne, Ar, and Al, d/δs is respectively 6.8, 7.2, 6.1 - inverse of each one is roughly 1/6. The other ratio is δ` /δs at T = Tm . From Eq. (19.32), we get δ` ∼ 1.5. (19.34) δs T =Tm For Ne, we can calculate this by using the values, εm = 0.34 × 10−2 eV, kB Tm = 0.21 × 10−2 eV. The ratio turns out to be 1.72. For Na, using εm = 2.7 × 10−2 eV, kB Tm = 3.7 × 10−2 eV, the ratio is 1.32.

Basic Ideas of Statistical Mechanics

201

However, for water, the ratio is 2.42. the reason for disagreement is because the melting heat does not include breaking of hydrogen bonds. Equation (19.34) may be summarized as - there is 50 per cent more linear displacement possible in liquids than in solids, still respecting Eq. (19.33). In terms of the lattice spacing, d, at Tm vs ≡ (d/6)3 ∼ d 3 /200, v` ∼ (1.5d/6)3 ∼ d 3 /60.

(19.35)

We notice that v` ∼ 3vs . Had we considered a hypothetical two dimensional case, the values of vs and v` would be d 2 /36 and d 2 /15, so v` ∼ 2vs . We may conjecture that the factors 3 and 2 correspond to the space dimensionality. We can use Eqs (19.28) and (19.33) to get even the temperature dependence of the ratio: r δs 1 T . (19.36) ≡ d 6 Tm With some more arguments (refer to [110]), it can be shown that

δ` δs

T ≡ 1.5 Tm

1/3 .

(19.37)

Hence, we have seen how the Boltzmann factor can be employed to arrive at a whole lot of interesting conclusions. The readers are encouraged to read [110] for more understanding about liquids.

19.3 Failure of Classical Physics We have seen how the internal energy U of N molecules appears in the equation of state: PV = NkB T = (γ −1)U. For monoatomic gas, the kinetic energy is the internal energy. So, γ − 1 = 2/3. For a diatomic gas (like O2 ), the total energy will consist of kinetic energy for each atom, 32 kB T and a potential energy 12 kB T (assuming that there is a harmonic oscillator binding, and, knowing that the mean potential energy is the same as mean kinetic energy). Hence, U = 7kB T /2 implying that kB T = 27 U, hence γ = 9/7 = 1.286. However, if we glance at the values of specific heats for oxygen and hydrogen, it is about 1.4 which doesn’t agree with 1.28. Moreover, γ is seen experimentally to vary with temperature. For O2 and H2 , around zero degree Celsius, it has a value around 1.4 which, with temperature, decreases slowly, seems to approach 1.286. However, in our discussion, which was within the purview of classical physics, γ is a constant, independent of temperature [2]. Maxwell recognized this in 1859 and finally pointed out in 1870 that this presented the “greatest difficulty” encountered by molecular (kinetic) theory. We need quantum mechanics applied to a system of particles to understand this.

202

Mechanics, Waves and Thermodynamics

The other important aspect missing in classical physics is the beautiful phenomenon of magnetism - once again, we need quantum mechanics to understand this. This then seems to be a natural place to close the book!

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207

Index action 29, 32, 36 angular velocity 24, 27, 56–58, 60, 65, 67, 77, 99 Archimedes’ principle 142 beats 114 Bernoulli equation 146, 147 black holes 104 Boltzmann factor 194, 197, 199–201 Caratheodory approach 169, 170 Sears’ illustration 170 Carnot cycle 173, 175, 196 Centre of Mass 51, 60, 160 centripetal acceleration 24, 26, 36 Chandler wobble 68, 69 chemical potential 190, 192 conservation 2, 14 conservative 42, 70, 71 continuity equation 144, 153 Coriolis force 27 dilute solutions 192 analogy with ideal gas 193 osmotic pressure 192, 193 electric circuits 29, 30 energy 1 deformation 82 loss of 70, 71, 124 entropy 168, 170, 177, 178 concavity 179 energy-spreading 197 Gibbs and Boltzmann 194 order or disorder 176 equipartition 160, 162, 166 Euler equation 145, 146, 153 force central 91, 92, 96, 102 friction 70, 72, 74 rolling 74 sliding 70, 72, 74, 76, 86

Galilean invariance 44, 45, 161 gaseous mixture 187 diffusion 187 law of mass action 189, 190 Hamilton’s principle 30, 32 harmonic oscillator 110–112 damped 110–112 driven damped 111 Helmholtz equation 128, 130 impulse 8, 9, 38, 44, 78 insect flight 106 instantaneous centre of rotation 55 interactions 19, 21, 22 jump 13 karate 78, 82, 83 kinetic energy football game 162, 163 loss on impact 80 rotational 58, 109 Lagrangian 30 mass 5 conservation 144 Maxwellian distribution 167 membrane annular 130 confocal parabolic 130 rectangular 62, 129 triangular 64, 130 missing information 177 moment of inertia 57, 60, 61, 64, 66 momentum 7 change in 16, 78, 159 definition 7–9 electromagnetic 9 music 136 notes 138 ragas 136, 139 shruti 138

210

Index

Newton’s laws of motion 13 Newton’s rule collisions 81 Newton’s second law 8, 17, 34, 48, 53, 109, 134, 144 from least action 34 Newton’s third law 48, 125 nodal domains 129, 130 counting 130 nodal lines 129, 130 nonlinear dynamics falling pencil 86, 87 normal modes 119, 121, 122 nuclear explosion 105 orbit 19, 93, 98, 99, 103 oscillators coupled 108, 116, 119, 120, 131 many coupled 120, 131 three coupled 119 pendulum compound 109, 110 simple 108 variable mass 49 phase velocity 122, 126, 127 Poisson’s hypothesis 79 polar coordinates acceleration 26 processes irreversible 175, 178, 179, 189 reversible 172–175, 179 real gases 184, 185 resonance lineshape 113 restitution 78, 80, 83 reversible process 172–175, 178 adiabatic compression 165, 173 Reynolds number 152 rigid body 15, 48, 55, 57, 60–62, 66–68, 168 angular momentum 57 displacement 55

Euler equations 66 rotation 57, 60, 66 rolling 72, 74 rotation curves 101, 102 simple harmonic motion 104, 108, 114, 115 slipping 72, 74, 75, 77 sound 77, 108, 114, 122, 131 brook 150 effect of bubbles 149 Laplace 133 Newton 131 speed 131, 134, 147, 149, 156 speed in a fluid 147 special orthogonal group 28 Stokes’ theorem 44 streamline 146, 147 temperature absolute 173, 174, 187 thermal expansion 184 thermodyamic potentials 181 enthalpy 179 free energy 182 Gibbs free energy 182, 189–192 thermodynamics first and second laws 169 third law 168, 186 Torricelli’s law 147 variable-mass systems 18 water 151 waves capillary 156, 158 deep water 155 gravity 153 membranes 128 reflection and transmission of 127 standing 126, 128, 136, 137, 149 tsunami 155 work-energy theorem 43