PHYSICS 1 Emerson College De Broglie Wavelength Quantum Mechanics Physics Notes

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Physics 1

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Hello I need 8 pages of notes on selected sections of 2 Physics books, these notes are precise, not too good, and for a sophomore in college Level. I will provide the books the sections that you need to write from. I want the notes to be separated into three sections: special relativity, waves, quantum mechanics. The second book is too big to upload here so when you accept I will send it by Google Drive, it’s by the same writer.

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Fundamentals of Physics t h e o p e n y a l e c o u r s e s s e r i e s is designed to bring the depth and breadth of a Yale education to a wide variety of readers. Based on Yale’s Open Yale Courses program (, these books bring outstanding lectures by Yale faculty to the curious reader, whether student or adult. Covering a wide variety of topics across disciplines in the social sciences, physical sciences, and humanities, Open Yale Courses books offer accessible introductions at affordable prices. The production of Open Yale Courses for the Internet was made possible by a grant from the William and Flora Hewlett Foundation. RECENT TITLES Paul H. Fry, Theory of Literature Christine Hayes, Introduction to the Bible Shelly Kagan, Death Dale B. Martin, New Testament History and Literature Giuseppe Mazzotta, Reading Dante R. Shankar, Fundamentals of Physics Ian Shapiro, The Moral Foundations of Politics Steven B. Smith, Political Philosophy Fundamentals of Physics Mechanics, Relativity, and Thermodynamics r. s h a n k a r New Haven and London Published with assistance from the foundation established in memory of Amasa Stone Mather of the Class of 1907, Yale College. c 2014 by Yale University. Copyright  All rights reserved. This book may not be reproduced, in whole or in part, including illustrations, in any form (beyond that copying permitted by Sections 107 and 108 of the U.S. Copyright Law and except by reviewers for the public press), without written permission from the publishers. Yale University Press books may be purchased in quantity for educational, business, or promotional use. For information, please e-mail (U.S. office) or (U.K. office). Set in Minion type by Newgen North America. Printed in the United States of America. ISBN: 978-0-300-19220-9 Library of Congress Control Number: 2013947491 A catalogue record for this book is available from the British Library. This paper meets the requirements of ANSI/NISO Z39.48-1992 (Permanence of Paper). 10 9 8 7 6 5 4 3 2 1 To my students for their friendship and inspiration Deep and original, but also humble and generous, the physicist Josiah Willard Gibbs spent much of his life at Yale University. His father was a professor of sacred languages at Yale, and Gibbs received his bachelor’s and doctorate degrees from the university before teaching there until his death in 1903. The sculptor Lee Lawrie created the memorial bronze tablet pictured above, which was installed in Yale’s Sloane Physics Laboratory in 1912. It now resides in the entrance to the J. W. Gibbs Laboratories, Yale University. Contents Preface xiii 1. The Structure of Mechanics 1.1 Introduction and some useful tips 1.2 Kinematics and dynamics 1.3 Average and instantaneous quantities 1.4 Motion at constant acceleration 1.5 Sample problem 1.6 Deriving v2 = v02 + 2a(x − x0 ) using calculus 1 1 2 4 6 10 13 2. Motion in Higher Dimensions 2.1 Review 2.2 Vectors in d = 2 2.3 Unit vectors 2.4 Choice of axes and basis vectors 2.5 Derivatives of the position vector r 2.6 Application to circular motion 2.7 Projectile motion 15 15 16 19 22 26 29 32 3. Newton’s Laws I 3.1 Introduction to Newton’s laws of motion 3.2 Newton’s second law 3.3 Two halves of the second law 3.4 Newton’s third law 3.5 Weight and weightlessness 36 36 38 41 45 49 4. Newton’s Laws II 4.1 A solved example 4.2 Never the whole story 4.3 Motion in d = 2 51 51 54 55 viii Contents 4.4 4.5 4.6 4.7 Friction: static and kinetic Inclined plane Coupled masses Circular motion, loop-the-loop 56 57 61 64 5. Law of Conservation of Energy 5.1 Introduction to energy 5.2 The work-energy theorem and power 5.3 Conservation of energy: K2 + U2 = K1 + U1 5.4 Friction and the work-energy theorem 70 70 71 75 78 6. Conservation of Energy in d = 2 6.1 Calculus review 6.2 Work done in d = 2 6.3 Work done in d = 2 and the dot product 6.4 Conservative and non-conservative forces 6.5 Conservative forces 6.6 Application to gravitational potential energy 82 82 84 88 92 95 98 7. The Kepler Problem 7.1 Kepler’s laws 7.2 The law of universal gravity 7.3 Details of the orbits 7.4 Law of conservation of energy far from the earth 7.5 Choosing the constant in U 101 101 104 108 112 114 8. Multi-particle Dynamics 8.1 The two-body problem 8.2 The center of mass 8.3 Law of conservation of momentum 8.4 Rocket science 8.5 Elastic and inelastic collisions 8.6 Scattering in higher dimensions 118 118 119 128 134 136 140 9. Rotational Dynamics I 9.1 Introduction to rigid bodies 9.2 Angle of rotation, the radian 143 143 145 Contents 9.3 9.4 9.5 9.6 ix Rotation at constant angular acceleration Rotational inertia, momentum, and energy Torque and the work-energy theorem Calculating the moment of inertia 147 148 154 156 10. Rotational Dynamics II 10.1 The parallel axis theorem 10.2 Kinetic energy for a general N-body system 10.3 Simultaneous translations and rotations 10.4 Conservation of energy 10.5 Rotational dynamics using τ = dL dt 10.6 Advanced rotations 10.7 Conservation of angular momentum 10.8 Angular momentum of the figure skater 159 159 163 165 167 168 169 171 172 11. Rotational Dynamics III 11.1 Static equilibrium 11.2 The seesaw 11.3 A hanging sign 11.4 The leaning ladder 11.5 Rigid-body dynamics in 3d 11.6 The gyroscope 175 175 176 178 180 182 191 12. Special Relativity I: The Lorentz Transformation 12.1 Galilean and Newtonian relativity 12.2 Proof of Galilean relativity 12.3 Enter Einstein 12.4 The postulates 12.5 The Lorentz transformation 194 195 196 200 203 204 13. Special Relativity II: Some Consequences 13.1 Summary of the Lorentz transformation 13.2 The velocity transformation law 13.3 Relativity of simultaneity 13.4 Time dilation 13.4.1 Twin paradox 13.4.2 Length contraction 209 209 212 214 216 219 220 x Contents 13.5 More paradoxes 13.5.1 Too big to fall 13.5.2 Muons in flight 222 222 226 14. Special Relativity III: Past, Present, and Future 14.1 Past, present, and future in relativity 14.2 Geometry of spacetime 14.3 Rapidity 14.4 Four-vectors 14.5 Proper time 227 227 232 235 238 239 15. Four-momentum 15.1 Relativistic scattering 15.1.1 Compton effect 15.1.2 Pair production 15.1.3 Photon absorption 241 249 249 251 252 16. Mathematical Methods 16.1 Taylor series of a function 16.2 Examples and issues with the Taylor series 16.3 Taylor series of some popular functions 16.4 Trigonometric and exponential functions 16.5 Properties of complex numbers 16.6 Polar form of complex numbers 255 255 261 263 265 267 272 17. Simple Harmonic Motion 17.1 More examples of oscillations 17.2 Superposition of solutions 17.3 Conditions on solutions to the harmonic oscillator 17.4 Exponential functions as generic solutions 17.5 Damped oscillations: a classification 17.5.1 Over-damped oscillations 17.5.2 Under-damped oscillations 17.5.3 Critically damped oscillations 17.6 Driven oscillator 275 280 283 288 290 291 291 292 294 294 Contents xi 18. Waves I 18.1 The wave equation 18.2 Solutions of the wave equation 18.3 Frequency and period 303 306 310 313 19. Waves II 19.1 Wave energy and power transmitted 19.2 Doppler effect 19.3 Superposition of waves 19.4 Interference: the double-slit experiment 19.5 Standing waves and musical instruments 316 316 320 323 326 330 20. Fluids 20.1 Introduction to fluid dynamics and statics 20.1.1 Density and pressure 20.1.2 Pressure as a function of depth 20.2 The hydraulic press 20.3 Archimedes’ principle 20.4 Bernoulli’s equation 20.4.1 Continuity equation 20.5 Applications of Bernoulli’s equation 335 335 335 336 341 343 346 346 349 21. Heat 21.1 Equilibrium and the zeroth law: temperature 21.2 Calibrating temperature 21.3 Absolute zero and the Kelvin scale 21.4 Heat and specific heat 21.5 Phase change 21.6 Radiation, convection, and conduction 21.7 Heat as molecular kinetic energy 352 352 354 360 361 365 368 371 22. Thermodynamics I 22.1 Recap 22.2 Boltzmann’s constant and Avogadro’s number 22.3 Microscopic definition of absolute temperature 22.4 Statistical properties of matter and radiation 22.5 Thermodynamic processes 375 375 376 379 382 384 xii Contents 22.6 Quasi-static processes 22.7 The first law of thermodynamics 22.8 Specific heats: cv and cp 386 387 391 23. Thermodynamics II 23.1 Cycles and state variables 23.2 Adiabatic processes 23.3 The second law of thermodynamics 23.4 The Carnot engine 23.4.1 Defining T using Carnot engines 394 394 396 399 403 409 24. Entropy and Irreversibility 24.1 Entropy 24.2 The second law: law of increasing entropy 24.3 Statistical mechanics and entropy 24.4 Entropy of an ideal gas: full microscopic analysis 24.5 Maximum entropy principle illustrated 24.6 The Gibbs formalism 24.7 The third law of thermodynamics 411 411 418 423 430 434 437 441 Index 443 Preface Given that the size of textbooks has nearly tripled during my own career, without a corresponding increase in the cranial dimensions of my students, I have always found it necessary, like my colleagues elsewhere, to cull the essentials into a manageable size. I did that in the course Fundamentals of Physics I taught at Yale, and this book preserves that feature. It covers the fundamental ideas of Newtonian mechanics, relativity, fluids, waves, oscillations, and thermodynamics without compromise. It requires only the basic notions of differentiation and integration, which I often review as part of the lectures. It is aimed at college students in physics, chemistry, and engineering as well as advanced high school students and independent self-taught learners at various stages in life, in various careers. The chapters in the book more or less follow my 24 lectures, with a few minor modifications. The style preserves the classroom atmosphere. Often I introduce the questions asked by the students or the answers they give when I believe they will be of value to the reader. The simple figures serve to communicate the point without driving up the price. The equations have been typeset and are a lot easier to read than in the videos. The problem sets and exams, without which one cannot learn or be sure one has learned the physics, may be found along with their solutions at the Yale website,, free and open to all. The lectures may also be found at venues such as YouTube, iTunes (, and Academic Earth, to name a few. The book, along with the material available at the Yale website, may be used as a stand-alone resource for a course or self-study, though some instructors may prescribe it as a supplement to another one adapted for the class, so as to provide a wider choice of problems or more worked examples. To my online viewers I say, “You have seen the movie; now read the book!” The advantage of having the printed version is that you can read it during take-off and landing. xiii xiv Preface In the lectures I sometimes refer to my Basic Training in Mathematics, published by Springer and intended for anyone who wants to master the undergraduate mathematics needed for the physical sciences. This book owes its existence to many people. It all began when Peter Salovey, now President, then Dean of Yale College, asked me if I minded having cameras in my Physics 200 lectures to make them part of the first batch of Open Yale Courses, funded by the Hewlett Foundation. Since my answer was that I had yet to meet a camera I did not like, the taping began. The key person hereafter was Diana E. E. Kleiner, Dunham Professor, History of Art and Classics, who encouraged and guided me in many ways. She was also the one who persuaded me to write this book. Initially reluctant, I soon found myself thoroughly enjoying proselytizing my favorite subject in this new format. At Yale Universtity Press, Joe Calamia was my friend, philosopher, and guide. Liz Casey did some very skilled editing. Besides correcting errors in style (such as a long sentence that began in first person past tense and ended in third person future tense) and matters of grammar and punctuation (which I sprinkle pretty much randomly), she also made sure my intent was clear in every sentence. Barry Bradlyn and Alexey Shkarin were two graduate students and Qiwei Claire Xue and Dennis Mou were two undergraduates who proofread earlier versions. My family, from my wife, Uma, down to little Stella, have encouraged me in various ways. I take this opportunity to acknowledge my debt to the students at Yale who, over nearly four decades, have been the reason I jump out of bed on two or three days a week. I am grateful for their friendship and curiosity. In recent years, they were often non-majors, willing to be persuaded that physics was a fascinating subject. This I never got tired of doing, thanks to the nature of the subject and the students. chapter 1 The Structure of Mechanics 1.1 Introduction and some useful tips This book is based on the first half of a year-long course that introduces you to all the major ideas in physics, starting from Galileo and Newton, right up to the big revolutions of the twentieth century: relativity and quantum mechanics. The target audience for this course and book is really very broad. In fact, I have always been surprised by the breadth of interests of my students. I don’t know what you are going to do later in life, so I have picked the topics that all of us in physics find fascinating. Some may not be useful, but you just don’t know. Some of you are probably going to be doctors, and you don’t know why I’m going to cover special relativity or quantum mechanics. Well, if you’re a doctor and you have a patient who’s running away from you at the speed of light, you’ll know what to do. Or, if you’re a pediatrician, you will understand why your patient will not sit still: the laws of quantum mechanics don’t allow a very small object to have a definite position and momentum. Whether or not you become a physicist, you should certainly learn about these great strides in the human attempt to understand the physical world. Most textbooks are about 1,200 pages long, but when I learned physics they were around 400 pages long. When I look around, I don’t see any student whose head is three times as big as mine, so I know that you cannot digest everything the books have. I take what I think are the really essential parts and cover them in these lectures. So you need the lectures to find out what’s in the syllabus and what’s not. If you don’t do that, 1 2 The Structure of Mechanics there’s a danger you will learn something you don’t have to, and we don’t want that, right? To learn physics well, you have to do the problems. If you watch me online doing things on the blackboard or working through derivations in the book, it all looks very reasonable. It looks like you can do it yourself and that you understand what is going on, but the only way you’re going to find out is by actually doing problems. A fair number are available, with their solutions, at You don’t have to do them by yourself. That’s not how physics is done. I am now writing a paper with two other people. My experimental colleagues write papers with four hundred or even a thousand other people when engaged in the big collider experiments like the ones in Geneva or Fermilab. It’s perfectly okay to be part of a collaboration, but you have to make sure that you’re pulling your weight, that everybody makes contributions to finding the solution and understands it. This calculus-based course assumes you know the rudiments of differential and integral calculus, such as functions, derivatives, derivatives of elementary functions, elementary integrals, changing variables in integrals, and so on. Sometime later, I will deal with functions of more than one variable, which I will briefly introduce to you, because that is not a prerequisite. You have to know your trigonometry, to know what’s a sine and what’s a cosine and some simple identities. You cannot say, “I will look it up.” Your birthday and social security number are things you look up; trigonometric functions and identities are what you know all the time. 1.2 Kinematics and dynamics We are going to be studying Newtonian mechanics. Standing on the shoulders of his predecessors, notably Galileo, Isaac Newton placed us on the road to understanding all the mechanical phenomena for centuries until the laws of electromagnetism were discovered, culminating in Maxwell’s equations. Our concern here is mechanics, which is the motion of billiard balls and trucks and marbles and whatnot. You will find out that the laws of physics for this entire semester can be written down on the back of an envelope. A central purpose of this course is to show you repeatedly that starting with those few laws, you can deduce everything. I would encourage you to think the way physicists do, even if you don’t plan to be a physicist. The easiest way to master this subject is to follow the reasoning I The Structure of Mechanics 3 give you. That way, you don’t have to store too many things in your head. Early on, when there are four or five formulas, you can memorize all of them and you can try every one of them until something works, but, after a couple of weeks, you will have hundreds of formulas, and you cannot memorize all of them. You cannot resort to trial and error. You have to know the logic. The goal of physics is to predict the future given the present. We will pick some part of the universe that we want to study and call it “the system,” and we will ask, “What information do we need to know about that system at the initial time, like right now, in order to be able to predict its future evolution?” If I throw a piece of candy at you and you catch it, that’s an example of Newtonian mechanics at work. What did I do? I threw a piece of candy from my hand, and the initial conditions are where I released it and with what velocity. That’s what you see with your eyes. You know it’s going to go up, it’s going to follow some kind of parabola, and your hands get to the right place at the right time to receive it. That is an example of Newtonian mechanics at work, and your brain performed the necessary calculations effortlessly. You only have to know the candy’s initial location and the initial velocity. The fact that it was blue or red is not relevant. If I threw a gorilla at you, its color and mood would not matter. These are things that do not affect the physics. If a guy jumps off a tall building, we want to know when, and with what speed, he will land. We don’t ask why this guy is ending it all today; that is a question for the psych department. So we don’t answer everything. We ask very limited questions about inanimate objects, and we brag about how accurately we can predict the future. The Newtonian procedure for predicting the future, given the present, has two parts, kinematics and dynamics. Kinematics is a complete description of the present. It’s a list of what you have to know about a system right now. For example, if you’re talking about a piece of chalk, you will want to know where it is and how fast it’s moving. Dynamics then tells you why the chalk goes up, why it goes down, and so on. It comes down due to the force of gravity. In kinematics, you don’t ask for the reason behind anything. You simply want to describe things the way they are, and then dynamics tells you how and why that description changes with time. I’m going to illustrate the idea of kinematics by following my preferred approach: starting with the simplest possible example and slowly adding bells and whistles to make it more and more complicated. In the 4 The Structure of Mechanics initial stages, some of you might say, “Well, I have seen this before, so maybe there is nothing new here.” That may well be true. I don’t know how much you have seen, but it is likely that the way you learned physics in high school is different from the way professional physicists think about it. Our priorities, and the things that we get excited about, are often different; and the problems will be more difficult. 1.3 Average and instantaneous quantities We are going to study an object that is a mathematical point. It has no size. If you rotate it, it will look the same, unlike a potato, which will look different upon rotation. It is not enough to just say where the potato is; you have to say which way its nose is pointing. The study of such extended bodies comes later. Right now, we want to study an entity that has no spatial extent, a dot. It can move around all over space. We’re going to simplify that too. We’re going to take an entity that moves only along the x-axis. So you can imagine a bead with a straight wire going through it, which allows it to only slide back and forth. This is about the simplest thing. I cannot reduce the number of dimensions. I cannot make the object simpler than a mathematical point. To describe what the point is doing, we pick an origin, call it x = 0, and put some markers along the x-axis to measure distance. Then we will say this guy is sitting at x = 5. Now, of course, we have to have units and the unit for length is going to be the meter. The unit for time will be a second. Sometimes I might not write the units, but I have earned the right to do that and you haven’t. Everything has got to be in the right units. If you don’t have the units, and if you say the answer is 42, then we don’t know if you are right or wrong. Back to the object. At a given instant, it’s got a location. We would like to describe the object’s motion by plotting a graph of space versus time. A typical graph would be something like Figure 1.1. Even though the plot is going up and down, the object is moving horizontally, back and forth along the spatial x-axis. When it is at A, it’s crossing the origin from the left and going to the right. Later, at B, it is crossing back to the left. In the language of calculus, x is a function of time, x = x(t), and the graph corresponds to some generic function that doesn’t have a name. We will also encounter functions that do have a name, like x(t) = t, x(t) = t 2 , x(t) = sin t, cos t, and so on. The Structure of Mechanics 5 Figure 1.1 Trajectory of a particle. The position, x(t), is measured vertically and the time, t, is measured horizontally. Consider v̄, the average velocity of an object, given by v̄ = x(t2 ) − x(t1 ) t2 − t1 (1.1) where t2 > t1 are two times between which we have chosen to average the velocity. In the example in Figure 1.1, v̄ < 0 for the indicated choice of t1 and t2 since the final x(t2 ) is less than the initial x(t1 ). The average velocity may not tell you the whole story. For example, if you started at x(t1 ) and at time t1 ended up at point C with the same coordinate, the average velocity would be zero, which is the average you would get if the particle had never moved! The average acceleration, ā, involves a similar difference of velocities: ā = v(t2 ) − v(t1 ) . t2 − t1 (1.2) Now for an important concept, the velocity at a given time or instantaneous velocity, v(t). Figure 1.1 shows some particle moving a distance x between times t and t + t. The average velocity in that interval is x . What you want is the velocity at time t. We all have an intuitive t notion of velocity right now. When you’re driving your car, if the needle says 60 miles per hour, that’s your velocity at that instant. Though velocity seems to involve two different times in its very definition—the initial 6 The Structure of Mechanics time and the final time—we want to talk about the velocity right now. That is obtained by examining the position now and the position slightly later, and taking the ratio of the change in position to the time elapsed between the two events, while bringing the two points closer and closer in time. We see in the figure that when we do this, both x → 0 and t → 0, but their ratio becomes the tangent of the angle θ , shown in Figure 1.1. Thus the velocity at the instant t is: x dx = . t→0 t dt v(t) = lim (1.3) Once you take one derivative, you can take any number of derivatives. The derivative of the velocity is the acceleration, and we write it as the second derivative of position: a(t) = dv d2 x = 2. dt dt (1.4) You are supposed to know the derivatives of simple functions like x(t) = t n ( dx = nt n−1 ), as well as derivatives of sines, cosines, logarithms, and expodt nentials. If you don’t know them, you should fix that weakness before proceeding. 1.4 Motion at constant acceleration We are now going to focus on problems in which the acceleration a(t) is just a constant denoted by a, with no time argument. This is not the most general motion, but a very relevant one. When things fall near the surface of the earth, they all have the same acceleration, a = −9.8 ms−2 = −g. If I tell you that a particle has a constant acceleration a, can you tell me what the position x(t) is? Your job is to guess a function x(t) whose second derivative is a. This is called integration, which is the opposite of differentiation. Integration is not an algorithmic process like differentiation, though it is governed by many rules that allow us to map a given problem into others with a known solution. If I give you a function, you know how to take the derivative: change the independent variable, find the change in the function, divide by the change in the independent variable, take the ratio as all changes approach zero. The opposite has to be done here. The way we do that is we guess, and such guessing has been going on for three hundred years, and we have become very good at it. The successful guesses The Structure of Mechanics 7 are published as Table of Integrals. I have a copy of such a table at home, at work, and even in my car in case there is a breakdown. So, let me guess aloud. I want to find a function that reduces to the number a when I take two derivatives. I know that each time I take a derivative, I lose a power of t. In the end, I don’t want any powers of t. It’s clear I have to start with a function that looks like t 2 . Well, unfortunately, we know t2 is not the right answer, because the second derivative is 2, while I want to get a. So I multiply the original guess by 12 a and I know x(t) = 12 at 2 will have a second derivative a. This certainly describes a particle with an acceleration a. But is this the most general answer? You all know that it is just one of many: for example, I can add to this answer some number, say 96, and the answer will still have the property that if you take two derivatives, you get the same acceleration. Now 96 is a typical constant, so I’m going to give the name c to that constant. We know from basic calculus that in finding a function with a given derivative, you can always add a constant to any one answer to get another answer. But if you only fix the second derivative, you can also add anything with one power of t in it, because the extra part will get wiped out when you take two derivatives. If you fixed only the third derivative of the function, you can also add something quadratic in t without changing the outcome. So the most general expression for the position of a particle with constant acceleration a is 1 x(t) = at 2 + bt + c 2 (1.5) where b, like c, is a constant that can be anything. Remember that x(t) in the figure describes a particle going from side to side. I can also describe a particle going up and down. If I do that, I would like to call the vertical coordinate y(t). You have to realize that in calculus, the symbols that you call x and y are arbitrary. If you know the second derivative of y to be a, then the answer is 1 y(t) = at 2 + bt + c. 2 (1.6) Let me go back now to Eqn. 1.5. It is true, mathematically, you can add bt + c as we did, but you have to ask yourself, “What am I doing as a physicist when I add these two terms?” What am I supposed to do with b and c? What value should I pick? Simply knowing that the particle has 8 The Structure of Mechanics an acceleration a is not enough to tell you where the particle will be. Take the case of a particle falling under gravity with acceleration −g. Then 1 y(t) = − gt 2 + bt + c. 2 (1.7) The formula describes every object falling under gravity, and each has its own history. What’s different between one object and another object is the initial height, y(0) ≡ y0 , and the initial velocity v(0) ≡ v0 . That’s what these numbers b and c are going to tell us. To find c in Eqn. 1.7 put time t = 0 on the right and the initial height of y0 on the left: y0 = 0 + 0 + c (1.8) which tells us c is just the initial coordinate. Feeding this into Eqn. 1.7 we obtain 1 y(t) = − gt 2 + bt + y0 . 2 (1.9) To use the information on the initial velocity, let us first find the velocity associated with this trajectory: v(t) = dy = −gt + b dt (1.10) and compare both sides at t = 0 v0 = b. (1.11) Thus b is the initial velocity. Trading b and c for v0 and y0 , which makes their physical significance more transparent, we now write 1 y(t) = − gt 2 + v0 t + y0 . 2 (1.12) Likewise for the trajectory x(t) when the acceleration is some constant a, the answer with specific initial position x0 and initial velocity v0 is 1 x(t) = at2 + v0 t + x0 . 2 (1.13) The Structure of Mechanics 9 In every situation where the body has an acceleration a, the location has to have this form. So when I throw a candy and you catch it, you are mentally estimating the initial position and velocity and computing the trajectory and intercepting it with your hands. (The candy moves in three spatial dimensions, but the idea is the same.) Now, there is one other celebrated formula that relates v(t), the final velocity at some time, to the initial velocity v0 and the distance traveled, with no reference to time. The trick is to eliminate time from Eqn. 1.13. Let us rewrite it as 1 x(t) − x0 = at 2 + v0 t. 2 (1.14) Upon taking the time-derivative of both sides we get v(t) = at + v0 (1.15) which may be solved for t: t= v(t) − v0 . a (1.16) Feeding this into Eqn. 1.14 we find 2    1 v(t) − v0 v(t) − v0 x(t) − x0 = a + v0 2 a a = v2 (t) − v02 2a (1.17) (1.18) which is usually written as v2 − v02 = 2a(x − x0 ) (1.19) where v and x are assumed to be the values at some common generic time t. 10 The Structure of Mechanics 1.5 Sample problem We will work through one standard problem to convince ourselves that we know how to apply these formulas and predict the future given the present. Figure 1.2 shows a building of height y0 = 15m. I am going to throw a rock with an initial velocity v0 = 10m/s from the top. Notice I am measuring y from the ground. The rock is going to go up to point T and come down as shown in Figure 1.2. You can ask me any question you want about this rock, and I can give the answer. You can ask me where it will be 9 seconds from now, how fast will it be moving 8 seconds from now, and so on. All I need are the two initial conditions y0 and v0 that are given. To make life simple, I will use a = −g = −10ms−2 . The position y(t) is known for all future times: y = 15 + 10t − 5t 2 . (1.20) Of course, you must be a little careful when you use this result. Say you put t equal to 10, 000 years. What are you going to get? You’re going to find y is some huge negative number. That reasoning is flawed because you cannot use the formula once the rock hits the ground and the fundamental premise that a = −10ms−2 becomes invalid. Now, if you had dug a hole of Figure 1.2 From the top of a building of height y0 = 15m, I throw a rock with an initial upward velocity of v0 = 10m/s. The dotted line represents the trajectory continued back to earlier times. The Structure of Mechanics 11 depth d where the rock was going to land, y could go down to −d. The moral is that when applying a formula, you must bear in mind the terms under which it was derived. If you want to know the velocity at any time t, just take the derivative of Eqn. 1.20: v(t) = 10 − 10t. (1.21) Let me pick a few more trivial questions. What is the height ymax of the turning point T in the figure? Eqn. 1.20 tells you y if you know t, but we don’t know the time t ∗ when it turns around. So you have to put in something else that you know, which is that the highest point occurs when it’s neither going up nor coming down. So at the highest point v(t ∗ ) = 0. From Eqn. 1.21 0 = 10 − 10t∗ which means t ∗ = 1s. (1.22) So we know that it will go up for one second and then turn around and come back. Now we can find ymax : ymax = y(t∗ ) = y(1) = 15 + 10 − 5 = 20m. (1.23) When does it hit the ground? That is the same as asking when y = 0, which is our origin. When y = 0, 0 = 15 + 10t − 5t 2 . (1.24) The solutions to this quadratic equation are t = 3s or t = −1s. (1.25) Why is it giving me a second solution? Can t be negative? First of all, negative times should not bother anybody; t = 0 is when I set the clock to zero, and I measured time forward, but yesterday would be t = −1 day, right? So we don’t have any trouble with negative time; it is like the year 300 BC. The point is that this equation does not know that I went to a building and launched a rock or anything. What does it know? It knows that this particle had a height of y = 15 m and velocity v = 10 m/s at time t = 0, and it is falling under gravity with an acceleration of −10 ms−2 . That’s 12 The Structure of Mechanics all it knows. If that’s all it knows, then in that scenario there is no building or anything else; it continues a trajectory both forward in time and backward in time, and it says that one second before I set my clock to 0, this particle would have been on the ground. What it means is that if you had released a rock at y = 0 one second before I did with a certain speed that we can calculate (v(−1) = 20m/s from Eqn. 1.21), your rock would have ended up at the top of the building when I began my experiment, with the same height y = 15m, and velocity v0 = 10 m/s. So sometimes the extra solution is very interesting, and you should always listen to the mathematics when you get extra solutions. When Paul Dirac was looking for the energy of a particle in relativistic quantum mechanics, he found the energy E was connected to its momentum p, mass m, and velocity of light, c, by E2 = p2 c2 + m2 c4 , (1.26) in accord with a relation we will encounter in relativity. Now, this quadratic equation has two solutions:  E = ± c2 p2 + m2 c4 . (1.27) You may be tempted to keep the plus sign because you know energy is not going to be negative. The particle’s moving, it’s got some energy and that’s it. This is correct in classical mechanics, but in quantum mechanics the mathematicians told Dirac, “You cannot ignore the negative energy solution in quantum theory; the mathematics tells you it is there.” It turns out the second solution, with negative energy, was telling us that if there are particles, then there must be anti-particles, and the negative energy particles, when properly interpreted, describe anti-particles of positive energy. So the equations are very smart. When you find some laws in mathematical form, you have to follow the mathematical consequences; you have no choice. Here was Dirac, who was not looking for anti-particles. He was trying to describe electrons, but the theory said there are two roots to the quadratic equation and the second root is mathematically as significant as the first one. In trying to accommodate and interpret it, Dirac was led to the positron, the electron’s anti-particle. The Structure of Mechanics 13 Returning to our problem, if you were only asking for the maximum height ymax , and not the time t ∗ when it got there, there is a shortcut using v2 = v02 + 2a(y − y0 ). (1.28) Using v = 0, v0 = 10m/s and a = −10ms−2 we find ymax − y0 = 5m (1.29) —that is, the rock reached a maximum height of 20m from the ground. You can find the speed when it hits the ground (y = 0) using v2 = 102 + 2 · (−10)(0 − 15) = 400 which means v = ±20m/s. (1.30) The root we should take for when it hits the ground is of course −20m/s. As mentioned earlier, the other root +20m/s is the speed with which it should have been launched upward, from y = 0 at t = −1, to follow the dotted trajectory in the figure. 1.6 Deriving v2 = v02 + 2a(x − x0 ) using calculus I want to derive Eqn. 1.19, v2 = v02 + 2a(x − x0 ) in another way that illustrates the judicious use of calculus. Start with dv =a dt (1.31) and multiply both sides by v and write v = dx in the right-hand side: dt v dx dv =a . dt dt (1.32) Now I’m going to do something that is viewed with suspicion, which is just to cancel the dt on both sides. Although I agree that you’re not supposed 14 The Structure of Mechanics dy to cancel that d in dx , canceling the dt on both sides gives valid results if interpreted carefully. Doing so here gives us vdv = adx. (1.33)   This equation tells us that in an infinitesimal time interval t, t + dt , the variables v and x change by dv and dx, and these changes are related as above in the limit dx, dv, dt → 0. Now the limit of dx → 0 or dv → 0 (as compared to their ratio) is of course trivial, and Eqn. 1.33 reduces to 0 = 0. However, the way we interpret and use Eqn. 1.33 is as follows. Suppose in the finite time interval [t1 , t2 ], the variable v changes from v1 to v2 , and x changes from x1 to x2 . Let us divide the interval [t1 , t2 ] into a very large number N of equal sub-intervals of  width dt, and let dx and dv be the changes in x and v in the interval t, t + dt . The relation between these changes is given in Eqn. 1.33. If we sum up the N changes on both sides of Eqn. 1.33 as N → ∞, the sums converge to nontrivial limits, namely the corresponding integrals:  v2 v1  vdv = a x2 dx (1.34) x1 1 2 1 2 v − v = a(x2 − x1 ). 2 2 2 1 (1.35) Thus it must be understood that the two sides of a relation like Eqn. 1.33 are to be ultimately integrated between some limits to obtain a useful equality. Eqn. 1.19 follows upon setting v2 = v, v1 = v0 , x2 = x, x1 = x0 . (1.36) chapter 2 Motion in Higher Dimensions 2.1 Review In the last chapter we took the simplest case, of a point particle moving along the x-axis with a constant acceleration a. What is the fate of this particle? The answer is that at any time t, the location of the particle is given by 1 x(t) = x0 + v0 t + at2 , 2 (2.1) where x0 and v0 are its initial position and velocity. If you took the derivative of this, you would get v(t) = v0 + at. (2.2) You can easily check, by taking one more derivative, that this particle does indeed have a constant acceleration a. This equation, which gives the velocity of the object at time t, in terms of its initial velocity and acceleration can be inverted to give t in terms of v: t= v − v0 . a (2.3) 15 16 Motion in Higher Dimensions Feeding this into Eqn. 2.1 we obtain the result that makes no reference to time: v2 = v02 + 2a(x − x0 ). (2.4) It is understood v and x correspond to some common time. I showed you in the end how we can use calculus to derive this result. It is important to brush up on your calculus. When a student says, “I know calculus,” sometimes that means the student knows it, and sometimes that means he or she once met someone who did. One solution for that is to get a copy of a textbook I wrote called Basic Training in Mathematics. This is a little awkward: I don’t want to foist my book on you. On the other hand, I don’t want to withhold relevant information. If you’re going into any science that uses mathematics—chemistry, engineering, or even economics—you should find the contents of that book useful. Don’t wait for the movie: it is not coming. 2.2 Vectors in d = 2 The next difficult thing is to consider motion in higher dimensions. Everything moves around in d = 3. However, I’m going to use only two dimensions for most of the time. Whereas the difference between one dimension and two is very great, that between two and higher dimensions is not. Later we will encounter a few concepts that make sense in d = 3 but not d < 3. String theorists will tell you that actually we need 9 spatial dimensions plus time to describe superstrings, which will be discussed in depth in Chapter 3,498 of this book. Picture some particle that’s traveling in the x − y plane as shown in Figure 2.1. This is not an x versus t plot or a y versus t plot. It’s the actual path the particle traces out on the x − y plane. You might say “Where is time?” One way to mark time is to imagine the particle carries a clock with it, and put markers every second. Four representative markers at t = 1, 2, 33, and 34 are shown. It obviously is going much slower between 33 and 34 than between 1 and 2. The kinematics of this particle requires a pair of numbers x and y. It’s more convenient to lump these into a single entity, called a vector. The simplest context in which one can motivate a vector and the rules for dealing with vectors is to look at movements in the plane. Let’s imagine that when I went camping I walked for 5 km from the base camp on the first Motion in Higher Dimensions 17 Figure 2.1 Path of a particle in d = 2. Equal intervals in time are indicated by markers on the path numbered 1, 2, . . . , 33, and 34. day and another 5 km on the second day. How far am I from the base camp? You cannot answer that, even if I promised to move only along the x-axis. It’s not enough to say I went 5 km. I have to tell you whether I went to the right or to the left. So I could be 10 km, 0 km or −10 km from base. If I say not just that I walked 5 km, but specified whether it was ±5 km, that takes care of all ambiguity in one dimension. But in d = 2 the options are not just left and right, but an infinity of possible directions. For example, on the first day I could leave the base camp at the origin and move along the arrow labeled A to arrive at the point labeled 1 in Figure 2.2(a). The second hike is described by the arrow Figure 2.2 Adding vectors. Part (a) shows how to add vectors and that A + B = B + A. Part (b) illustrates the meaning of multiplying a vector by a number (2 in this example) and the null vector 0. 18 Motion in Higher Dimensions B, which starts where A ended and brings me to 2. These two arrows are examples of vectors and I use them here for describing displacement, or changes in position. Vectors can be used to describe many other physical quantities, as we will see. A vector is an arrow that has got a beginning and an end. This is why one says a vector has a magnitude and a direction. The magnitude is how long it is, and direction is its angle relative to some fixed direction, usually the x-axis. When you refer to a vector A in your notes, you’re supposed to  In textbooks, vectors are in boldface: put a little arrow on top like this: A. A. If you don’t put an arrow on top or do not use boldface, you’re talking about just a number A. When applied to a vector A, A stands for its length. From Figure 2.2(a), we see that there is a very natural quantity that you can call A + B. One day I moved by A and on the next by B. If I want to do it all in one shot, what is the equivalent step I should take from the start? It’s obvious that the bottom line of my two-day trip is this object C. We will call that A + B. It does represent the sum, in the same sense that if I gave you 4 bucks and then I gave you 5 bucks, you have the equivalent of a single payment of 9 bucks. Here, we are not talking about a single number, but a displacement in the plane, and C indeed represents an effective displacement due to A and B. So here is the rule for adding two vectors that comes from a study of displacements: you draw the first one and at the end of that first one, you begin the second one, and their sum starts at the beginning of the first and ends at the end of the second. You can verify, as illustrated in the figure, that A + B is the same as B + A where you first draw B and from where B ends you draw A. You will end up with the same point, 2, as shown by the sum of the dotted arrows. The next thing I want to do is to define the vector that plays the role of the number 0, which has the property that when you add it to any number, it gives the same number. The vector 0 that I want to call the zero or null vector should have the property that when I add it to any vector, I should get the same vector. So you can guess who it is: a vector of no length. I cannot show you the 0. If you can see it, I’m doing something wrong. Look at part (b) of Figure 2.2. What if I draw A, then I add to it another A to get A + A. You have to agree that if there’s any vector that deserves to be called 2A, it is this guy, A, stretched to twice its length. Now we have discovered a notion of multiplying a vector by a number. If you Motion in Higher Dimensions 19 multiply it by 2, you get a vector twice as long and in the same direction. Then we’re able to generalize that and say, if you multiply it by 2.6, you get a vector 2.6 times as long. So multiplying a vector by a positive number means to stretch it (or shrink it) by that factor. Let us keep going. I want to think of a vector that I can call −A. What do I expect of −A? I expect that if I add −A to A, I should get 0, which plays the role of 0 among vectors. What should I add to A so I get the null vector? It’s clear that you want to add a vector that looks like −A in part (b) of Figure 2.2, because, if you go from the start of A to the finish of −A, you end up where you started and you get this invisible 0 vector. So the minus vector is the same vector flipped over, pointing the opposite way. That’s like −1 times a vector. Once you have got that, you can do −7.3 times a vector: just take the vector, rescale it by 7.3 and flip it over. Multiplying a vector by a number is called scalar multiplication, and ordinary numbers are called scalars. You can do more complicated things. You can take one vector, multiply it by one scalar, take another vector, multiply that by another scalar, and add the two of them. We know what all those operations mean now. You don’t have to memorize the rules for all this. The only rule is: “Do what comes naturally.” Do what you normally do with ordinary numbers. 2.3 Unit vectors Let us go back to the same x − y plane. I’m going to introduce two very special vectors. They are the unit vectors: i and j, pointing along the x and y axes and of unit length, as shown in Figure 2.3. If I had a third axis perpendicular to the page, I would draw a k , but we don’t need that yet. I claim I can write any vector you give me as a number Ax times i, plus a number Ay times j. There’s nothing you can throw at me that lies in the plane that I cannot describe as some multiple of i plus some multiple of j. It’s intuitively clear, but I will just prove it beyond any doubt. Here is some vector A. It is clear from the figure that it is the sum of the dotted horizontal vector and the dotted vertical vector, by the rules of vector addition. The horizontal part, parallel to i, has to be a multiple of i. We know that because we can stretch i by whatever factor we like. Call that factor Ax , which happens to be positive in this example. It is called the x-component of A or the projection of A along i or along the x-axis. The vertical part is likewise jAy where Ay is the y-component of A, or the projection 20 Motion in Higher Dimensions Figure 2.3 The unit vectors i and j and an arbitrary vector A = iAx + jAy built out of them. of A along j or along the y-axis. Therefore I have managed to write A as A = iAx + jAy . (2.5) We refer to the pair i and j, in terms of which any vector can be expressed, as basis vectors or as the basis. If you gave me a particular vector A as an arrow of some length A and orientation θ relative to the x-axis, what do I use for Ax and Ay ? You can see from trigonometry that Ax = A cos θ (2.6) Ay = A sin θ . (2.7) Conversely, given the components, the length and angle are  A = A2x + A2y θ = tan−1 Ay . Ax (2.8) (2.9) Eqns. 2.6 to Eqn. 2.9 will be invoked often. So please commit them to memory. Motion in Higher Dimensions 21 If you give me a pair of numbers, (Ax , Ay ), that’s as good as giving me this arrow, because I can find the length of the arrow by PythagoA ras’ theorem and I can find the orientation from tan θ = Axy . You have the option of either working with the two components of A or with the arrow. In practice, most of the time we work with these two numbers, (Ax , Ay ). In particular, if we are describing a particle whose location is the position vector r, then we write it in terms of its components as r = ix + jy. (2.10) The changes in r are the displacement vectors and examples are A and B in Figure 2.2 that described the two hikes. I have not given you any other example of vectors besides the displacement vector, but at the moment, we’ll define a vector to be any object that looks like some multiple of i plus some multiple of j. If I tell you to add two vectors A and B, you have got two options. You can draw the arrow corresponding to A and attach to its end an arrow corresponding to B, and then add them, as in Figure 2.2. But you can also do the bookkeeping without drawing any pictures as follows: A + B = iAx + jAy + iBx + jBy = i(Ax + Bx ) + j(Ay + By ) (2.11) (2.12) so that the sum C is the vector with components (Ax + Bx , Ay + By ). In the above, I have used the fact that vectors can be added in any order. So I grouped the things involving just i and likewise j. Then I argued that since iAx and iBx are vectors along i, their sum is a vector of length Ax + Bx also along i. I did the same for j. In summary if A+B=C (2.13) Cx = Ax + Bx (2.14) Cy = Ay + By (2.15) then which can be summarized as follows: 22 Motion in Higher Dimensions To add two vectors, add their respective components. An important result is that A = B is possible only if Ax = Bx and Ay = By . You cannot have two vectors equal without having exactly the same x component and exactly the same y component. If two arrows are equal, one cannot be longer in the x direction and correspondingly shorter in the y direction. Everything has to match completely. The vector equation A = B is actually a shorthand for two equations: Ax = Bx and Ay = By . 2.4 Choice of axes and basis vectors I have in mind a vector whose components are 3 and 5. Can you draw the vector for me? If you immediately said, “It is 3i + 5j,” you’re making the assumption that I am writing the vector in terms of i and j. I agree i and j point along two natural directions. For most of us, given that the blackboard or notebook is oriented this way, it is very natural to line up our axes with it. But there is no reason why somebody else couldn’t come along and say, “I want to use a different set of axes. The x and y axes or i and j are not nailed in absolute space. They are human constructs and we’re not wedded to any of them.” Quite often, it’s natural to pick the axes in a certain way to suit the problem. If you are studying a cannon ball launched from the earth, it makes sense to pick the horizontal as the x axis and the vertical as the y axis, but, mathematically, you don’t have to. Another set of rotated but mutually perpendicular unit vectors i and j that form another basis can also be rescaled and added to form any given vector A in the plane. For example, when we study objects sliding down an inclined plane, we will choose our axes parallel and perpendicular to the incline. If I draw an arrow A on a blank sheet of paper, it has life of its own without reference to any axes. The same vector A can be written either in terms of i and j, which is the old basis, or in terms of i and j , the new basis. How do the components (Ax , Ay ) in the new basis relate to the components of the old basis? It’s a simple problem, but I just want to do it so you get used to working with vectors. For this we need the very busy Figure 2.4. It shows the old x and y axes and the x and y axes obtained by rotating the x − y axes counterclockwise by an angle φ. The unit vectors i and j are likewise rotated versions of i and j. The components in the two bases are shown by dotted lines and are simply the projections of A along the various axes. We want to relate (Ax , Ay ) to (Ax , Ay ). Motion in Higher Dimensions 23 Figure 2.4 The same vector A is written as iAx + jAy in one frame and as i Ax + j Ay in the other. The dotted lines indicate the components in the two frames. First we express i and j in terms of i and j using the figure: i = i cos φ + j sin φ (2.16) j = j cos φ − i sin φ. (2.17) Here are the details. The vector i has got a horizontal part, which is its length, namely, 1, times cos φ, and a vertical part that is 1 times sin φ. How about j ? It is at an angle φ relative to j. So its y-component is cos φ. Finally, its x or horizontal component is (− sin φ), where the minus sign comes because it is pointing to the left, along the negative x-axis. All that remains now is to eliminate i and j in favor of i and j in A = i Ax + j Ay and equate it to A written in terms of i and j: A = i Ax + j Ay (2.18) = (i cos φ + j sin φ)Ax + (j cos φ − i sin φ)Ay (2.19) = i(Ax cos φ − Ay sin φ) + j(Ax sin φ + Ay cos φ) (2.20) = iAx + jAy . (2.21) 24 Motion in Higher Dimensions When we equate the coefficients of i and j on the right-hand sides of Eqns. 2.20 and 2.21, we obtain the desired expression for Ax and Ay in terms of Ax and Ay : Ax = Ax cos φ − Ay sin φ (2.22) Ay = Ax sin φ + Ay cos φ. (2.23) So, you can pick your basis vectors any way you like and so can I. Your basis is obtained from mine by a counterclockwise rotation by an angle φ. The same entity A, the same arrow which has an existence of its own, independent of axes, can be described by you and by me using different components. Your components with primes on them are related to mine by Eqns. 2.22 and 2.23. This is called the transformation law for the vector components under rotation of basis vectors. Now, you can ask the opposite question. How do I get Ax and Ay in terms of Ax and Ay ? The quickest way is to replace φ by −φ: if we go from the unprimed to the primed system by a rotation φ, then rotation by −φ is the way to go from the primed to the unprimed basis. The result, using cos(−φ) = cos φ and sin(−φ) = − sin φ, is Ax = Ax cos φ + Ay sin φ (2.24) Ay = −Ax sin φ + Ay cos φ. (2.25) That turns out to be the correct answer. But I want you to think about another way to show this, which often seems to bother some students. If I told you 3x + 5y = 21 (2.26) 4x + 6y = 26, (2.27) you certainly know how to solve for x and y, right? You have got to juggle the two equations, multiply the first by 6, the second by 5, and subtract to isolate x and so on. Why is it when some of you see Eqns. 2.22 and 2.23, you don’t realize it’s the same kind of problem, where you can multiply Eqn. 2.22 by cos φ, Eqn. 2.23 by sin φ and add to isolate Ax , for example? For any particular value of φ, sin φ and cos φ are just some numbers. For Motion in Higher Dimensions 25 example, if I pick φ = π3 = 60◦ , cos φ = become (for this angle) 1 2 and sin φ = √ 3 . 2 The equations √ 3  1  A Ax = A x − 2 2 y √ 3  1  A + A. Ay = 2 x 2 y If you multiply the second by Ax + √ (2.28) (2.29) √ 3 and add it to the first you obtain 3Ay = 2Ax which means √ 1 3 Ax = Ax + Ay 2 2 π π = Ax cos + Ay sin 3 3 (2.30) (2.31) (2.32) in accordance with Eqn. 2.24. So go forth and treat sin φ and cos φ as plain numbers and juggle Eqns. 2.22 and 2.23 to derive Eqns. 2.24 and 2.25. Along the way of course you will have to use identities like sin2 φ + cos2 φ = 1. The components of the vector depend on who is looking at the vector. However, there’s one quantity that’s going to come out the same, no matter who is looking at the vector. It is the length of the vector. It is unaffected by the rotation of axes. It is an invariant under rotations. You may verify from Eqns. 2.24 and 2.25 that   (Ax )2 + (Ay )2 = (Ax cos φ + Ay sin φ)2 + (−Ax sin φ + Ay cos φ)2 (2.33) = A2x (cos2 φ + sin2 φ) + A2y (sin2 φ + cos2 φ) (2.34) = A2x + A2y . (2.35) The Ax Ay term is gone since its coefficient is 2(cos φ sin φ − sin φ cos φ). I want to conclude with one important point. We learned that a vector is a quantity that has a magnitude and a direction. A more advanced view of vectors is that they are a pair of numbers (in d = 2) which, under rotation of axes, transform as per Eqns. 2.24 and 2.25. Anything that transforms this way is called a vector. We already know about the position 26 Motion in Higher Dimensions vector r and the changes in it, the displacement vectors (used in describing the hike). How about more vectors? There turns out to be a very nice way to produce vectors, given one vector like the position vector. And that’s the following. 2.5 Derivatives of the position vector r Let’s take a particle in the x − y plane that moves from r at time t to r + r at time t + t as in Figure 2.5. At time t its location is r = i x(t) + j y(t) (2.36) and at t + t it is r + r = i(x(t) + x) + j(y(t) + y) so that, r = ix + jy and by the usual limiting process, v = lim t→0 dx dy r dr = =i +j . t dt dt dt (2.37) (2.38) (2.39) Figure 2.5 The particle moving along some curved path goes from r at time t to r + r at time t + t. The velocity v is the limit of the ratio r as t → 0, and t thus parallel to r, which eventually becomes tangent to the curve. Motion in Higher Dimensions 27 When you move just along the x-axis, you wait a small time t and you move by an amount x, and their ratio gives the velocity in the appropriate limit. When you move in the plane, your position and its change are both vectors. Can you see why the derivative of a vector is also a vector? Because r, the difference in the vector between two times, is itself a vector. Dividing it by t is like multiplying by 1/t, but we know that when we multiply a vector by a number, we simply rescale the vector. So the limit will be some arrow that we call the instantaneous velocity vector. It will be tangential to the curve r(t) and point toward the instantaneous direction of travel. If I gave you the location of a particle as a function of time, you can find its velocity by taking derivatives. For example, if I say a particle’s location is r = t 2 i + 9t 3 j (2.40) then its velocity at time t is v = 2t i + 27t 2 j. (2.41) You can take a derivative of the velocity or the second derivative of the position to get the acceleration vector a(t) = dv d2 r = = 2 i + 54t j (in our example). dt dt 2 (2.42) You can then also multiply a by the mass m, which is a scalar unaffected by rotations, to get a vector ma, which Newton’s law equates to another vector, the force F. Even though we started with one example of a vector r, we’re now finding out that its derivative has to be a vector and the derivative of the derivative is also a vector. When you learn relativity, you will find out there’s again one vector that’s staring at you, the analog of the position vector, but with four components. But more vectors can be manufactured by multiplying vectors by scalars (like mass) or taking derivatives with respect to a parameter that plays the role analogous to time. Here is an illustration of vector addition and differentiation. Imagine an airplane in flight, as depicted in Figure 2.6. Let rpg be the location of a 28 Motion in Higher Dimensions Figure 2.6 The position of the ball relative to (some origin on) the ground rbg is the vector sum of the position of the ball relative to the (tail of the) plane, rbp , and the position of (the tail of) the plane rpg relative to the ground. fixed point in the airplane, say the tail, with respect to a fixed point on the ground. Imagine that in the airplane there is a ball located at rbp as measured from this fixed point in the airplane. By vector addition the location of the ball with respect to the ground is rbg = rbp + rpg . (2.43) Upon taking a time derivative and in the same notation, the law of composition of velocities follows: vbg = vbp + vpg , (2.44) which says the velocity of the ball as seen by a person on the ground is the velocity of the ball relative to the airplane plus the velocity of the airplane relative to the ground. Taking yet another derivative we may relate the accelerations: abg = abp + apg . (2.45) In the special case of a airplane moving at constant velocity, apg = 0. Then we find abg = apg , (2.46) Motion in Higher Dimensions 29 which means, in this case, the acceleration of the ball is the same as measured by an observer on the ground and an observer on the airplane. These results will be recalled in our study of relativity. 2.6 Application to circular motion Now we’ll take a concrete problem where you will see how to take derivatives to obtain very useful results. I’m going to write a particular case of r(t): r(t) = R(i cos ωt + j sin ωt) (2.47) where R and ω are constants. What is going on as a function of time? What’s this particle doing? Look at the length squared of this vector: rx2 + ry2 = R2 (cos2 ωt + sin2 ωt) = R2 . (2.48) That means the particle is going around in a circle of radius R as shown in Figure 2.7. The x component is R cos ωt and the y component is R sin ωt Figure 2.7 The particle moves along a circle of radius R with an angular velocity ω. 30 Motion in Higher Dimensions where ω is a fixed number. As t increases, this angle ωt increases and the particle goes round and round. Let’s get a feeling for ω. As time increases, the angle increases and we can ask how long it will take the particle to come back to the starting point. Suppose the starting point was on the xaxis. As t increases, ωt increases, and the particle will come back at a time T such that ωT = 2π . (2.49) Thus ω is related to the time period T by ω= 2π = 2π f T (2.50) where f = T1 is the frequency or number of cycles per second. It is measured in Hz, which stands for Hertz. Since in every cycle the particle rotates by 2π , and it completes f revolutions per second, ω = 2π f is called the angular velocity and measures the radians swept out per second. Notice that in equating a full cycle to 2π , I am using radians and not degree to measure angles. For those who have not seen a radian, it’s just another way to measure angles, wherein a full circle, which we used to think was worth 360◦ , now equals 2π radians. Since 2π 6.3, a radian is roughly 60◦ . You will see the advantages of using radians later. For now just remember that a half circle, instead of being 180◦ , will now be π radians, and a quarter circle will be π2 , and so forth. How fast is this particle moving? It’s going around a circle, the angle is increasing at a steady rate ω, and so we know it’s going at a steady speed. Let us verify that by computing the velocity dr(t) dt d sin ωt d cos ωt +j =R i dt dt v(t) = = Rω(−i sin ωt + j cos ωt). (2.51) (2.52) (2.53) At t = 0, the velocity is v = Rω cos 0 j = Rωj, so it is moving straight up at speed v = ωR. You may verify that it has the velocity as shown in the figure at later times. The magnitude of the velocity is always ωR although Motion in Higher Dimensions 31 the direction is changing. From the figure we see it remains tangential to the circle. The constancy of the speed v at an arbitrary time may also be established by computing v2 = (ωR)2 (sin2 ωt + cos2 ωt) = (ωR)2 v = ωR. (2.54) (2.55) Remember the tangential velocity is v = ωR. Let’s take the derivative of the derivative to find the acceleration a and its magnitude: a = −ω2 R(i cos ωt + j sin ωt) = −ω2 r (2.56) a = ω2 R. (2.57) That’s a very important result. It tells you that when a particle moves in a circle of radius R at constant speed v, it has an acceleration, called the centripetal acceleration, directed toward the center and of magnitude a = ω2 R = (ωR)2 v2 = . R R (2.58) This acceleration at constant speed reflects the fact that velocity is a vector and you can change the velocity vector by changing its direction. For example, if a car is going on a racetrack and the speedometer says 60 miles per hour, the lay person’s view is that the car is not accelerating. But you 2 will say from now on that it indeed has an acceleration equal to vR even though no one’s stepping on the accelerator or the brake. Suppose the particle is not moving fully around a circle but traversing just a quarter of the circle. When it is traveling the quarter of a circle, it has the same acceleration directed toward the center of that quarter circle. In other words, you don’t have to be moving actually in a circle to have the 2 acceleration vR . At any instant, the curve you are following can be locally 2 approximated as part of some circle, and, in the formula a = vR , the acceleration is directed toward the center of that circle, R is its radius and v the instantaneous tangential velocity. 32 Motion in Higher Dimensions 2.7 Projectile motion I want to consider a particle for which r0 and v0 are the position and velocity at t = 0 and which has a constant vector acceleration a. What is its location at all future times? By analogy with what I did in one dimension 1 r(t) = r0 + v0 t + at 2 . 2 (2.59) Once you know r0 and v0 , you can find the position of the object at all future times. Let’s take one simple example. Somebody in a car has decided to drive off a cliff as shown in Figure 2.8(a). We want to know when and where the car hits the ground. We pick our origin (0, 0) at the foot of the cliff. Let the height of the cliff be h. The car is traveling with some initial speed v0x in the horizontal direction. Equation 2.59 is really a pair of equations, one along x and one along y with a = −jg, v0 = v0x i, and r0 = hj. Separating out the components x(t) = 0 + v0x t + 0 (2.60) 1 y(t) = h + 0 − gt 2 . 2 (2.61) Notice that the evolution of the two coordinates is completely independent. The time t ∗ when the car hits the ground (y = 0) satisfies the Figure 2.8 (a) A car flies off the cliff at (0, h) and lands at (d, 0). (b) A projectile √ is launched with initial velocity v0 = (i + 3j)m/s. The range is R = 0.35 m and the maximum height reached is ymax = 0.15 m. Motion in Higher Dimensions 33 equation 1 0 = h − gt∗2 2 t∗ = (2.62) 2h . g (2.63) This is exactly how long it would take to hit the ground had it simply toppled over the edge from rest. The horizontal velocity does not delay the crash one bit (unless you take into account the curvature of the earth). As to where the car lands, the location is given by (x(t ∗ ), 0) = (d, 0) where d = v0x t ∗ = v0x 2h . g (2.64) Finally the problem of projectile motion is depicted in Figure 2.8(b). You fire a projectile from (0, 0) with some velocity v0 at some angle θ . It will go up and then come down, moving horizontally at the same time. Where is it going to land? What is the maximum height ymax to which it rises? With what speed will it hit the ground? At what angle should you fire your projectile so it will go the furthest? Here are the equations that contain all the answers, namely Eqn. 2.59 written out in component form: x(t) = 0 + v0x t = v0 cos θ · t (2.65) 1 1 y(t) = 0 + v0y t − gt 2 = v0 sin θ · t − gt 2 . 2 2 (2.66) You can solve them but it is good to have an idea of what’s coming. Imagine you have this monster cannon to fire things. It has a fixed muzzle speed, v0 , but allows you to fire at any angle. How do you aim it so the ball goes as far as possible? There are two schools of thought. One says, aim at your enemy and fire horizontally. Then the ball lands on your foot because it has zero time of flight (assuming the cannon is at zero height). The other school says, maximize the time of flight and point the cannon vertically. It goes up, stays in the air for a very long time, and lands on your head. Then it hits you: the correct answer is somewhere between 0 and 90◦ = π2 . The naive guess 45◦ = π4 turns out to be correct. 34 Motion in Higher Dimensions Now I want to show you how to use the equations to prove this. What’s the strategy for finding the range? You see how long the ball is in the air and multiply that by the constant horizontal velocity v0 cos θ . Again, let t∗ be the time when it hits the ground. The y-equation 1 0 = t ∗ v0 sin θ − gt ∗ 2 (2.67) has two solutions: t ∗ = 0, 2v0 sin θ . g (2.68) So the cannon ball is on the ground on two occasions. One is initially. We are not interested in that trivial solution. If the time you are interested in is t ∗ = 0, you’re allowed to divide both sides of Eqn. 2.67 by it and get t∗ = 2v0 sin θ g (2.69) and the range R = v0x t ∗ = v0 cos θ · t ∗ = 2v02 sin θ cos θ v02 sin 2θ = g g (2.70) using sin 2θ = 2 sin θ cos θ . For the greatest range we must make sin 2θ as large as possible, which occurs for 2θ = π2 or θ = π4 = 45◦ . For any smaller range you will find there are two possible angles that work since sin(2θ ) = sin(π − 2θ ). The maximum height is the y-coordinate at the half-way time 12 t ∗ = v0 sin θ : g 1 ∗ 1 1 t ymax = 0 + v0y t∗ − g 2 2 2 2 2  v0 sin θ 1 v0 sin θ v2 sin2 θ = (v0 sin θ ) = 0 − g . g 2 g 2g (2.71) Motion in Higher Dimensions 35 We could equally well find the half-time 12 t∗ by setting the vertical velocity to zero: t∗ 0 = v0 sin θ − g . 2 (2.72) How about the velocity at impact? The horizontal part is of course v0 cos θ since there is no acceleration in that direction. The vertical part starts out at v0 sin θ and decreases at a rate g: vy = v0 sin θ − gt (2.73) so that at t ∗ it has a value v(t∗ ) = v0 sin θ − g 2v0 sin θ = −v0 sin θ , g (2.74) which is just the opposite of the initial vertical speed. The magnitude of the final velocity is the same as the initial one since reversing one of the components of v does not change its length. There are endless variations. You pick some point off the ground at (X, Y) and want the projectile to arrive there. You are given the launch angle θ and have to find the launch speed v0 . How do you do that problem? You assume the projectile arrives at the destination (X, Y) at some time t ∗ . You go to the x equation and demand that x(t∗ ) = X and solve for t ∗ . Plug this time into y and demand y(t∗ ) = Y and solve for v0 . Sometimes the problems are embellished to make everyone feel involved. For example, instead of a ball dropping, nowadays there’s a monkey or horse that is falling down, so people in life sciences can say, “Hey, we should learn physics since it seems to have applications to our subject.” All those gyrating creatures are very interesting and look great in color, but, in the end, you are told, “Treat the horse as a point particle.” If you’re going to treat the horse as a point particle, why include its color picture? But I agree there are times when only a horse will do. When the Godfather wants to get the contract for Johnny Fontaine, he doesn’t tell his consigliere, “Hey, Tom, put half a point particle on Jack’s bed.” That would have been a disastrous approximation. By the way, don’t forget to treat the falling car in Figure 2.8 as a point particle. chapter 3 Newton’s Laws I 3.1 Introduction to Newton’s laws of motion This is a big day in your life: you are going to learn Newton’s laws, in terms of which you can understand and explain a very large number of phenomena. It’s really amazing that so much information can be condensed into three laws. Your reaction may be that you have already seen Newton’s laws, that you have applied them in school. I realized fairly late in life that they are more subtle than I first imagined. It’s one thing to plug in all the numbers and say, “I know Newton’s laws and I know how they work.” But as you get older and you have more spare time, you think about what you are doing. This is something I have had the luxury of doing, and I have realized the laws are more tricky. I want to share some of that understanding. The first law, or the law of inertia, says, “If a body has no forces acting on it, it will maintain its velocity.” In other words, in the absence of external forces, a body at rest will remain at rest and a body in motion will retain its velocity. It is not surprising that a body at rest will remain so if not acted upon by a force. We see that all the time. I place an eraser on the table. It will stay put unless I do something to it. The great discovery that Galileo and Newton made was that you don’t need a force for a body to move at constant velocity. You don’t see that in daily life—everything seems to come to rest unless you keep pushing or pulling it. But we all know that the reason things come to a halt is that there is always some friction or drag bringing them to rest. If you take a hockey puck on an air cushion and give it a push, it seems 36 Newton’s Laws I 37 it can travel for a very long time. Galileo and Newton abstracted from this an ideal situation in which friction was totally eliminated and the bodies kept moving forever with no help. If you go to outer space, you can check for yourself that, if you throw something, it will go on forever without your intervention. It’s in the nature of things to retain a constant velocity. It is not velocity, but a change in velocity, that calls for a force. The law of inertia is not valid for everybody. I’ll give an example from your own life. You go on an airplane and then, after the usual delays, the plane begins to accelerate down the runway. At that time, if you leave anything on the floor, you know it’s no longer yours. It’s going to slide backward and the physicist in the last row is going to collect everything. That is an example of a frame in which the law of inertia does not work: bodies accelerate with no applied force. Once the plane stops accelerating, the law of inertia becomes operative. It fails when it decelerates during landing when everything now slides to the front. If Newton’s law of inertia works for you, you are called an inertial observer and your frame of reference is called an inertial frame. The plane that’s taking off is not an inertial frame, but the one that is cruising is. The earth seems to be a pretty good inertial frame, because if you leave something at some place, it just stays there—unless the thing is your iPod and the place is Grand Central Station. But this is not a violation of Newton’s laws, just the laws of New York City. Although not every frame is inertial, there are plenty of inertial frames to go around. If you find even one inertial observer, namely one person for whom this law of inertia works, I can find for you an infinite number of other people for whom this is true. Who are these people? They are people moving at constant velocity with respect to the first inertial observer. Suppose you are in a train and you’re moving past me with velocity u, and we both look at some object with no forces on it. We will not agree on its velocity or the velocity of anything: things at rest for me will be moving backward for you at velocity −u, and everything at rest in your train will be moving at velocity u according to me. In short, you and I will differ on the velocity of any object by our relative velocity. But we will agree on the acceleration of any object since it is unaffected by adding a constant to velocity. Adding a constant velocity to objects does not change the fact that those which were maintaining constant velocity still maintain a constant (but different) velocity. Thus neither is every observer inertial, nor is an inertial observer unique. 38 Newton’s Laws I You must know the earth is not precisely inertial. The earth has an acceleration. Can you see why? Yes, it’s spinning around itself and going around the sun, both of which constitute accelerated motion. But the acceleration due to motion around the sun at speed v and radius r is 2 a = vr = .006 ms−2 , which is a very small number, say compared to g. The same goes for the acceleration due to the earth’s rotation about its own axis, which is roughly .03ms−2 or roughly g/300 at the equator. The first law might seem tautological because we never see anything that retains its velocity forever, and every time we see velocity change we say that a force is acting. But it’s not a big hoax, because you can set up experiments in free space, far from everything, where objects will, in fact, maintain their velocity forever. It’s a useful concept even on the earth, because the earth is approximately inertial. 3.2 Newton’s second law Newton’s second law says, “If a body has an acceleration a, then you need a force F = ma (3.1) to produce that acceleration.” In this chapter we will focus on one dimension and write F = ma (3.2) where F and a are along the x-axis. A few words about units. Acceleration is measured in ms−2 . Mass is measured in kilograms or kg. So force has units kilogram meters per second squared. But we get tired of saying that long expression, so we call that a Newton, denoted by N. If you had invented mechanics, we’d be calling it by your name, but it is too late for you now. Here is a typical problem that you may have solved in your first pass at Newton’s laws. A force of 36N is acting on a mass 4 kg. What’s the acceleration? You divide 36 by 4 and you find it is a = 9 ms−2 . You say, “Okay, I know Newton’s laws.” It’s actually more complicated than that. Take yourself back to the seventeenth century, when Newton was inventing these laws. You have an intuitive definition of force: when somebody pushes or pulls an object we Newton’s Laws I 39 say a force is acting on it. Suddenly, you are told there is a law F = ma. Are you better off in any way? Can you do anything with this law? What does it help you predict? Can you even tell if it’s true? Here’s a body that’s moving. Is Newton right? How are we going to check? Well, you want to measure the left-hand side and you want to measure the right-hand side. If they’re equal, you will say the law is working. What can you measure in this equation? Let’s start with acceleration. What’s your plan for measuring acceleration? What instruments will you need? If you say a watch and ruler, that would be correct provided by ruler you don’t mean Queen Elizabeth. Here is your ruler and here is a Rolex. Tell me exactly how you plan to measure acceleration. Everyone seems to know the answer. First, let it go a little distance, and take the distance over time. That gives you the velocity now. Let it go a little more, and repeat the velocity measurement. Take the difference of the two velocities and divide by the difference of the two times, and you have got the acceleration. Since the body has moved a finite distance in a finite time, this gives the average acceleration. You want to make these three positional measurements more and more rapidly. In the end, as all the time intervals shrink to 0, you will measure what you 2 can say is the acceleration now, the second derivative ddt2x = a(t) defined in calculus. Back to testing if what Newton told you is right: You see an object in motion, you measure a, and you get a certain numerical value, say 10 ms−2 . But that’s not yet testing the equation, because you still have to find F and m. What’s the mass of this object? One common idea is to take a standard mass and balance the unknown mass on a seesaw by adjusting its position. But suppose you were in outer space. There’s no gravity. Then the seesaw will balance even if you put a potato on one side and an elephant on the other side. What you are doing now is appealing to the notion of mass as something that’s related to the pull of the earth on the object. You have got to go back and wipe out everything you know. If F = ma is all you have, there is no mention of the earth in these equations. You only know how to measure a, but not the other two. So you have a problem. You cannot say that since F = ma, it follows that m = Fa ; that is circular reasoning since you have not told me how to measure F either. Let me give you a hint. How do we decide how long a meter is? You seem to know that it is arbitrary. A meter is not deduced from anything. Napoleon or somebody said, “The size of my ego is one meter.” That’s a 40 Newton’s Laws I new unit of length. Seriously, at the National Bureau of Standards there used to be a rod made of some special alloy in a glass case, and that defined the meter. There are fancier definitions now, but let us stick to this simple one. (See for more details on definitions of units.) Then I ask you, “What is two meters and what is three meters?” We have ways of handling that. You take the meter and attach it to a duplicate, and that’s two meters. You can cut it in half, using dividers and compasses; you can split the meter into any fractions you like. Likewise for mass, we will take a chunk of some material and we will call it a kilogram. That is a matter of convention, just like one second is some convention. I’m going to give you a glass case that contains a block of some metal defined as one kilogram. Then I give you another object, an elephant. What’s the mass of the elephant? Here is a hint: I also give you a spring. We cannot do the seesaw experiment because it requires gravity. A spring, on the other hand, will exert a force even in outer space. Here’s what we do. We hook one end of the spring to a wall and we pull the other end from its equilibrium position by some amount and we attach the one-kilogram mass to it. We don’t know what force it exerts, but it will not matter. We let it go and measure the initial acceleration, a1 . Then we bring the elephant (another point particle) of unknown mass mE , pull the spring by the same amount so it can exert the same force, and find the acceleration aE of the elephant. Assuming only that the same extension produces the same force in the two cases, we have 1 · a1 = mE aE a1 . mE = 1 · aM (3.3) (3.4) Once you have the mass of the elephant you can use it to measure any other mass mo by using mo = mE aE ao (3.5) where aE and ao are produced by the same force. There are subtleties even here. For example, how do we know that when we pull the spring the second time with the elephant, it will exert the same force as the first time when the 1 kg was attached to it? After all, Newton’s Laws I 41 springs wear out. That’s why you change the shock absorbers in your car. So first, we have to make sure the spring exerts a fixed force every time (for a given extension). How are we going to check that? We don’t have the definition of force yet. But we can do the following. We pull the onekilogram mass and let it go, and we note the acceleration. Then, we pull it again, by the same amount, and let it go; we do it ten times. If every time we get the same acceleration, we are convinced this is a reliable spring that is producing the same force under the same conditions. On the eleventh time we pull the spring and attach the elephant. With some degree of confidence, we can say we are applying the same force on the elephant as on the one-kilogram mass. Why is this discussion so important? Because you need to know that everything you or I write down in the notebook or on the blackboard as a symbol is actually a measurable quantity, or, as they say in France, Les Mesurables. You should know at all times how you measure anything that enters your theory or calculation. If not, you are just doing math or playing with symbols. You are not doing physics. This discussion also tells you that the mass of an object has nothing to do with gravitation but with how much it hates to accelerate in response to a force. Newton tells you forces cause acceleration. But the acceleration is not the same on different objects for a given force. Certain objects resist it more than others. They are said to have a bigger mass. We can be precise about how much bigger by saying, “If the acceleration of a body in response to a given force is 101 that of a 1-kilogram mass, then the mass of the body is 10 kilograms.” 3.3 Two halves of the second law We have seen how all objects can be attributed a mass. Now go back to the spring. I want to know how much force F(x) it exerts when I pull it by a certain amount x, which is measured from the point when the spring is neither compressed nor expanded as shown in Figure 3.1. If x is positive, it means the spring is stretched; if it’s negative, it means the spring is compressed. Now I can measure F(x) for any given x because I can measure the acceleration it produces on a known mass m and use F = ma. So I pull it by various amounts, measure F, and draw a graph. It will be a straight line with some slope −k for small values of x: F = −kx (3.6) 42 Newton’s Laws I Figure 3.1 (Left) The mass m is attached to a spring (dotted line) of force constant k. The other end of the spring is attached to a wall. The displacement x is measured from the equilibrium position of the spring. (Right) The force F(x) as a function of x. where k is called the force constant. The minus sign says, if you pull it to the right, so that x is positive, the spring will exert a force in the negative direction. If you compress it, then x is negative and the spring will exert a force in the direction of increasing x. All springs will have a graph like this for small enough distortions from equilibrium. Beyond that the line may bend or the spring may even snap. In any case, we now have a way to measure k for any given spring in the regime when F(x) is linear in x. I want you to think about the two equations F = ma and F = −kx. If the first one is Newton’s law, then what’s the other one? What’s the difference between saying F = −kx and F = ma? Let me paraphrase the answer I usually hear from students: “F = ma is universally true, independent of the nature of the force acting on a body, while F = −kx is only describing the spring.” That is essentially correct, and I will now elaborate. The cycle of Newtonian dynamics has two parts. The first one is to find the acceleration a of a body, the force on which is somehow known, using a = mF . The force F is the cause and the acceleration a is the effect and a = mF is the precise relation between them. The second is to deduce experimentally what force F will be acting on a given body at any given time. For example, if a mass is attached to a spring that has been extended by x, we must do the experiment to find that the spring force is F = −kx. Newton does not tell us that. He never Newton’s Laws I 43 tells us what F is in a given context, with the exception of gravity where he also furnished the left-hand side of F = ma with his law of universal gravitation. If two like charges repel each other, we need Coulomb’s law to tell us that the force varies as 1/r 2 . The nuclear force, say between a proton and neutron, falls exponentially with distance. Surely Newton did not tell us this. But once we have experimentally determined a new force F, we can use his second law to deduce the a it would produce, assuming classical mechanics is applicable. When a new force, like the electric force, is discovered, the F due to it can be measured in one of two ways. One is to compare it to a known force that balances it. For example, to find how the repulsive force between two charged bodies varies with distance, we can glue them to the two ends of a spring of known k and measure by how much it is extended in equilibrium. Another way is to release the charges (of known mass) from rest with some separation, measure the initial acceleration of either, and then compute ma. So physicists are busy either finding a from F (as when computing the orbit of a satellite given the force law for gravity) or F from a (as when stretching springs to find k or dropping apples from trees to measure the force of gravity). Now, a small digression on gravity. Consider a body near the surface of the earth, the force of gravity on it being F = −mg where g = 9.8ms−2 . That’s something you find out by dropping things from a tower. Consider a in the field of gravity. We find it is a= −mg = −g m (3.7) for all bodies. That’s a very remarkable property of the gravitational force— the cancellation of m. If you look at the electrical force, on the proton and electron for example, it’s not proportional to the mass of either object. It’s proportional to the electric charge of the object. Therefore, when you divide by the mass to get the acceleration, the response of different bodies is inversely proportional to the mass. But gravity has a remarkable property that the pull of the earth is itself proportional to the mass of the object. So, when you divide by the same m to find a, it cancels and everything falls at the same rate on the surface of the earth. In fact, that is a property of gravitational fields anywhere, even in outer space. Everything—gold, silver, diamonds, particles, elephants—all accelerate the same way. This 44 Newton’s Laws I remarkable fact was known for a long time, but it literally took an Einstein to figure out why nature behaves in that fashion, why the two masses associated with a body are equal. One is the inertial mass, which is how much a body hates velocity change, how hard it resists acceleration, the mass in F = ma. That quality can be measured far from planets, far from everything. The other is gravitational mass, which is the measure of how much it is attracted to the earth or any other body. There’s no reason why these two attributes had to be equal. Is this just an accident or is it part of a big picture? It turns out that it’s part of a big picture called the general theory of relativity. Here is Albert Einstein’s description of gravity. Imagine a stream and some kids dropping various leaves or paper boats in it. No matter what they drop (within reason) the object’s trajectory will follow the flow lines of the water. The path of all objects is predetermined. This is what gravity does to spacetime—it defines trajectories for objects: anything you release will follow the trajectory etched in spacetime. If you oppose this flow, like a kid holding on to his paper boat, the resistance you feel is the weight of that object. What determines the flow lines at each point? The matter and energy in the universe, as dictated by Einstein’s equations. Here’s a simple example of a complete Newtonian problem. A mass is attached to a spring. It is pulled by a certain amount x and then released. What is it going to do? Newton says F = ma, which in this case becomes m d2 x = −kx. dt 2 (3.8) To proceed, we must know m and k, and I have already discussed how these are measured. Now we have a mathematically complete problem: find the function x(t) whose second derivative is equal to − mk times the function x(t). Then, we go to the math department and say, “What’s the solution to this equation?” This is a problem in mathematics and the answer—that it’s going to be oscillating back and forth—will come from doing the math. Later we will do some of that math ourselves. For now, I am simply pointing out that once we have stated the laws in mathematical form, solving for the consequences is a mathematical problem. Here is another example. Newton discovers a force of gravity acting on everything. Here’s the sun in Figure 3.2, orbited by a planet. At this instant, the planet may be moving at some velocity v. The acceleration of the planet is due to the force of gravity between the planet and the sun, which Newton tells you is directed toward the sun, proportional to the Newton’s Laws I 45 Figure 3.2 The planet is separated from the sun by r. The force F on it was determined to be always opposite to r, i.e., pointing toward the sun and falling as 1 . (The force is slightly offset from r for clarity.) r2 product of the two masses, and which decreases with distance as r12 . This completely specifies the left-hand side of F = ma. That’s the law of universal gravitation. Then again, because the second derivative of the position is connected to the position, you go to the mathematicians and say, “What orbit is the solution to that equation?” and they will tell you it is an ellipse. Of course, Newton did not have mathematicians he could go to. He was the math guy. Not only did he formulate laws of gravitation, he also invented calculus and figured out how to solve the differential equation that came out of his F = ma. There has been no one like him. Here I speak of Newton the scientist; Newton the man had many flaws. 3.4 Newton’s third law The third law says that if there are two bodies, called 1 and 2, then F12 , the force on 1 due to 2, is minus the force on 2 due to 1: F12 = −F21 . (3.9) Action and reaction are equal and opposite. Coulomb’s law and the law of gravity both have this feature. You may assume it for every force in this course. We are now going to put the laws to work. You have to be good at writing down the forces acting on a body. That’s what all these problems are going to boil down to. Do not forget the existing forces and do not make up your own forces. I have seen both happen. Every force, with the exception of gravity, is a force due to direct contact with the body: a rope is pulling it, a rod is pushing it, you are pushing it, you are pulling it, one block is pushing another, and so on. Gravity is one force that acts on a 46 Newton’s Laws I body without the source of the force actually touching it. (Later electromagnetic forces can come in, but not in this book.) That’s it. Once more with feeling: With the exception of gravity all forces we will discuss in this book will be contact forces. We are going to begin with simple problems in mechanics. They will get progressively more difficult. Let’s start with our first triumph. There is some object of mass 5 kilograms and I apply 10 N on it. What’s the acceleration? Everyone knows it is a = 105 = 2ms−2 . You may have done this before, but I hope now you understand how we know the force is 10 N and how we know the mass is 5 kg. The algebra is, of course, very trivial here. Next, I have a 3-kg block placed against a 2-kg block, and I’m pushing the former with 10 N as shown in Figure 3.3. I want to know what happens. One way is just to use your common sense and realize that these two blocks are going to move together. You know intuitively that if they move together, they will behave like an object of mass 5 kg and the acceleration will again be 2ms−2 . What about gravity? What about the force due to the table on which the masses are moving? Imagine that this occurs in outer space where there is no gravity and no need for a table. There’s another way to do this problem, which is to draw free-body diagrams. Here you can pick any one body that you like and apply F = ma to it, provided you identify all the forces acting on that body. We’ll first pick the 3-kg mass. My 10 N is certainly acting on it. What other force is acting? The force of the 2 kg, which has a magnitude f acting to the left. Figure 3.3 Top: A force of 10 N acts on the two blocks, viewed as a single entity. Bottom: The free-body diagram for the two blocks showing all the forces on each block. Notice the third law is being invoked. Newton’s Laws I 47 Do not include the force exerted by the 3 kg on the 2 kg. Next consider the 2-kg mass. There is the same f , but acting to the right by the third law. Here is the mistake some people make: they add to that the 10 Newtons. They feel that the 2 kg will surely feel it since that is what is behind all the acceleration. That will be a mistake. That’s an example of adding a force that you should not be adding. The only force acting on this little guy is this little f . Now we do F = ma for these two guys: 10 − f = 3a (3.10) f = 2a (3.11) 10 = 5a upon adding the previous two equations a = 2. (3.12) (3.13) Notice I’m using the same acceleration for both. I know that if the second mass accelerated faster than the first one, then the picture is completely wrong; it will not feel the force due to the first. If it accelerated less than the first, the first would have plowed into the second. Since that also cannot happen, they’re moving with the same acceleration. There’s only one unknown a. Once you got a = 2, you can go back to Eqn. 3.11 and obtain f = 4 N. Now we know the full story: 4 N acting on 2 kg gives it an acceleration of 2 ms−2 , while (10 − 4) = 6 N acting on the 3 kg gives the same a = 2ms−2 . Here’s another variation shown in Figure 3.4. I have a 3-kg mass attached by a rope to a 2-kg mass, which I pull with a force of 10 N. Again, your common sense tells you that I am pulling something whose effective mass is 5 kg; the answer is 2ms−2 . Let’s confirm that systematically by using the free-body diagrams in the lower half of the figure. Now, there are really three bodies here: the two blocks and the rope connecting them. In all these examples, I assume the rope is massless. We know there is no such thing as a massless rope, so what we mean is a rope whose mass is negligible compared to the two blocks being pulled. We’ll take the idealized limit where the mass of the rope is 0. The 3 kg is being pulled by the rope to the right with a force that I’m going to call T, which stands for tension. The rope is being pulled backward by the 3 kg with a force T by the third law. What is the force on the other end of the rope? What should that be? If you said T, that is right, but not because the rope would 48 Newton’s Laws I Figure 3.4 (Top) A force of 10 N pulls the two blocks connected by a massless rope, all treated as a single entity. (Bottom) The free-body diagram for the two blocks and the rope, which experiences equal and opposite forces of magnitude T called the tension. snap otherwise. Something else will be a problem. If the two forces on its ends don’t cancel, you have a net force. What are you going to divide by to get the acceleration? Zero, right? So, a massless body cannot have a net force on it, because its acceleration would then be infinite. Massless bodies will always have equal and opposite forces on the two ends. In the case of the rope, this is called the tension on the rope. The tension is not 0 just because this T and tha...
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Fundamentals of Physics
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Special Relativity
Galilean and Newtonian relativity
The theory of special relativity, while often associated with Einstein's monumental work of 1905,
can be traced to Galileo and Newton. The Galileo and Newton theory describes the motion of
objects with respect to a reference frame, which is identified as the particular observer. Newton 's
laws hold an inertial frame; it also holds a reference frame moving at a constant velocity. For
example, in a train, an observer who was asleep at the start of the journey and wakes in the middle
of the trip as the train moves cannot observe any changes. As such, if newton’s law is valid, the
inertial observer will not see any changes.
Not all observers and frames are inertial. For example, the equation of 𝑓 = 𝑚𝑎 is not true
in an accelerating train. For instance, in a non-inertial frame, the observer will notice motion or
movement when a train is accelerating or about to stop. The concept of Galilean transformation
considers the movement of objects at a constant velocity. According to this transformation, and
the enacted force causes an object to move. For example, find two trains running at a similar speed.
The force acting on both passengers is in both frames. Since both objects are in an identical
structure, the observer can notice the relative motion but cannot identify which train, in particular,
is moving.
Michelson and Morley and the ether
Michelson and Morley experimented to determine if the speed of light is equated as c—v to assess
the presence of substance referred to as the ether. The experiment obtained a result of a speed of
exactly c. From the research, Michelson and Morley found a negative effect where there was no
significant difference between the speed of light in motion through the presumed ether.

Postulates of Special Relativity
The most significant Postulates Einstein's special relativity include; All inertial observers
are equivalent, and the velocity of light is independent of the state of motion of the source and the
observer. In the first postulate, the concept of "equivalent" takes the total weight of the statement.
In this postulate, each inertial observer has an equal level of privileges in discovering the laws of
nature. For example, person A and person B, both inertial observers, are allowed to make claims.
Person A can claim that he or she is moving, and person B is at rest. Person B, on the other hand,
can equally argue that person A is at rest while he or she is moving. In the second postulate,
Einstein posits that if a light beam is emitted by a moving rocket, it travels and measures at c. As
such, all analysts will get a similar answer.
The Lorentz Transformation
The Lorentz Transformation supersedes the Galilean transformation. The measurement of space
and time actively identifies this concept. This transfor...

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