Six Ideas That Shaped Physics - All Units [3 ed.] 9780073513942, 0073513946, 2015043352

Table of contents :
Cover
Six IdeasThat Shaped Physics
Unit C
Six IdeasThat Shaped Physics Unit C: Conservation LawsConstrain Interactions
Contents: Unit C Conservation Laws Constrain Interactions
About the Author
Preface
Introduction for Students
Chapter C1 The Art of Model Building
Chapter Overview
C1.1 The Nature of Science
C1.2 The Development and Structure of Physics
C1.3 A Model-Building Example
C1.4 Trick Bag: Unit Awareness
C1.5 Trick Bag: Unit Conversions
C1.6 Trick Bag: Dimensional Analysis
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter C2 Particles and Interactions
Chapter Overview
C2.1 The Principles of Modern Mechanics
C2.2 Describing an ObjectÕs Motion
C2.3 Vector Operations
C2.4 Momentum and Impulse
C2.5 Force and Weight
C2.6 Interaction Categories
C2.7 Momentum Transfer
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter C3 Vectors
Chapter Overview
C3.1 Introduction
C3.2 Reference Frames
C3.3 Displacement Vectors
C3.4 Arbitrary Vectors
C3.5 Seven Rules to Remember
C3.6 Vectors in Two Dimensions
C3.7 Vectors in One Dimension
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter C4 Systems and Frames
Chapter Overview
C4.1 Systems of Particles
C4.2 A SystemÕs Center of Mass
C4.3 How the Center of Mass Moves
C4.4 Inertial Reference Frames
C4.5 Freely Floating Frames
C4.6 Interactions with the Earth
TWO-MINUTE PROBLEMS
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ANSWERS TO EXERCISES
Chapter C5 Conservation of Momentum
Chapter Overview
C5.1 Degrees of Isolation
C5.2 How to Solve Physics Problems
C5.3 Conservation of Momentum Problems
C5.4 Examples
C5.5 Airplanes and Rockets
TWO-MINUTE PROBLEMS
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ANSWERS TO EXERCISES
Chapter C6 Conservation of Angular Momentum
Chapter Overview
C6.1 Introduction
C6.2 Quantifying Orientation
C6.3 Angular Velocity
C6.4 The Angular Momentum of a Rigid Object
C6.5 Twirl and Torque
C6.6 Gyroscopic Precession
C6.7 Conservation of Angular Momentum
TWO-MINUTE PROBLEMS
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Chapter C7 More About Angular Momentum
Chapter Overview
C7.1 First Steps
C7.2 The Cross Product
C7.3 The Angular Momentum of a Moving Particle
C7.4 Rotating Objects
C7.5 Rotating and Moving Objects
C7.6 Torque and Force
C7.7 Why Angular Momentum Is Conserved
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter C8 Conservation of Energy
Chapter Overview
C8.1 Introduction to Energy
C8.2 Kinetic Energy
C8.3 Potential Energy
C8.4 Fundamental Potential Energy Formulas
C8.5 Internal Energy and Power
C8.6 Isolation
C8.7 Solving Conservation-of-Energy Problems
TWO-MINUTE PROBLEMS
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Chapter C9 Potential Energy Graphs
Chapter Overview
C9.1 Interactions Between Macroscopic Objects
C9.2 Interactions Between Two Atoms
C9.3 One-Dimensional Potential Energy Diagrams
C9.4 Relaxing the Mass Limitation
C9.5 The Spring Approximation
C9.6 The Potential Energy Òof an ObjectÓ
C9.7 An Example
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Chapter C10 Work
Chapter Overview
C10.1 The Momentum Requirement
C10.2 The Dot Product
C10.3 The Definition of Work
C10.4 Long-Range Interactions
C10.5 Contact Interactions
TWO-MINUTE PROBLEMS
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Chapter C11 Rotational Energy
Chapter Overview
C11.1 Introduction to Rotational Energy
C11.2 Rotational Energy of an Object at Rest
C11.3 Calculating Moments of Inertia
C11.4 When an Object Both Moves and Rotates
C11.5 Rolling Without Slipping
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Chapter C12 Thermal Energy
Chapter Overview
C12.1 The Case of the Disappearing Energy
C12.2 Caloric Is Energy
C12.3 Thermal Energy as Microscopic Energy
C12.4 Friction and Thermal Energy
C12.5 Heat, Work, and Energy Transfer
C12.6 Specific ÒHeatÓ
C12.7 Problems Involving Thermal Energies
TWO-MINUTE PROBLEMS
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Chapter C13 Other Forms of Internal Energy
Chapter Overview
C13.1 Bonds
C13.2 Latent ÒHeatÓ
C13.3 Chemical Energy
C13.4 Nuclear Energy
C13.5 Modes of Energy Transfer
C13.6 Mechanisms of Heat Transfer
C13.7 A Comprehensive Energy Master Equation
TWO-MINUTE PROBLEMS
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Chapter C14 Collisions
Chapter Overview
C14.1 Types of Collisions
C14.2 One-Dimensional Collisions
C14.3 Two-Dimensional Collisions
C14.4 The Slingshot Effect
C14.5 Using All Three Conservation Laws
C14.6 Asteroid Impacts
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Appendix CA The Standard Model
CA.1 Introduction
CA.2 Matter Particles
CA.3 Fundamental Interactions
CA.4 The Importance of Color-Neutrality
CA.5 Stability and the Weak Interaction
CA.6 Conclusion
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Index
Periodic table
Short Answers to Selected Problems
Unit N
Six Ideas That Shaped Physics Unit N: The Laws of Physics Are Universal
Contents: Unit N The Laws of Physics Are Universal
About the Author
Preface
Introduction for Students
Chapter N1 Newton's Laws
Chapter Overview
N1.1 The Newtonian Synthesis
N1.2 Newton's Laws of Motion
N1.3 Vector Calculus
N1.4 The Formal Definition of Velocity
N1.5 The Formal Definition of Acceleration
N1.6 Uniform Circular Motion
TWO-MINUTE PROBLEMS
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Chapter N2 Forces from Motion
Chapter Overview
N2.1 The Kinematic Chain
N2.2 Classification of Forces
N2.3 Force Laws
N2.4 Free-Body Diagrams
N2.5 Motion Diagrams
N2.6 Graphs of One-Dimensional Motion
N2.7 A Quantitative Example
TWO-MINUTE PROBLEMS
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Chapter N3 Motion from Forces
Chapter Overview
N3.1 The Reverse Kinematic Chain
N3.2 Graphical Antiderivatives
N3.3 Integrals for One-Dimensional Motion
N3.4 Free-fall in One Dimension
N3.5 Integrals in Three Dimensions
N3.6 Constructing Trajectory Diagrams
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Chapter N4 Statics
Chapter Overview
N4.1 Forces from Motion: An Overview
N4.2 Introduction to Statics
N4.3 Statics Problems Involving Torque
N4.4 Solving Force-from-Motion Problems
N4.5 Solving Statics Problems
TWO-MINUTE PROBLEMS
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Chapter N5 Linearly Constrained Motion
Chapter Overview
N5.1 Motion at a Constant Velocity
N5.2 Static and Kinetic Friction Forces
N5.3 Drag Forces
N5.4 Linearly Accelerated Motion
N5.5 A Constrained-Motion Checklist
TWO-MINUTE PROBLEMS
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Chapter N6 Coupled Objects
Chapter Overview
N6.1 Force Notation for Coupled Objects
N6.2 Pushing Blocks
N6.3 Strings, Real and Ideal
N6.4 Pulleys
N6.5 Using the Constrained-Motion Checklist
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Chapter N7 Circularly Constrained Motion
Chapter Overview
N7.1 Uniform Circular Motion
N7.2 Unit Vectors
N7.3 Nonuniform Circular Motion
N7.4 Banking
N7.5 Examples
TWO-MINUTE PROBLEMS
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Chapter N8 Noninertial Frames
Chapter Overview
N8.1 Fictitious Forces
N8.2 The Galilean Transformation
N8.3 Inertial Reference Frames
N8.4 Linearly Accelerating Frames
N8.5 Circularly Accelerating Frames
N8.6 Using Fictitious Forces
N8.7 Freely Falling Frames and Gravity
TWO-MINUTE PROBLEMS
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Chapter N9 Projectile Motion
Chapter Overview
N9.1 Weight and Projectile Motion
N9.2 Simple Projectile Motion
N9.3 Some Basic Implications
N9.4 A Projectile Motion Checklist
N9.5 Drag and Terminal Speed
TWO-MINUTE PROBLEMS
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Chapter N10 Oscillatory Motion
Chapter Overview
N10.1 Mass on a Spring
N10.2 Solving the Equation of Motion
N10.3 The Oscillator as a Model
N10.4 A Mass Hanging from a Spring
N10.5 An Analogy to Circular Motion
N10.6 The Simple Pendulum
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Chapter N11 Kepler's Laws
Chapter Overview
N11.1 Kepler's Laws
N11.2 Orbits Around a Massive Primary
N11.3 Kepler's Second Law
N11.4 Circular Orbits and Kepler's Third Law
N11.5 Circular Orbit Problems
N11.6 Black Holes and Dark Matter
N11.7 Kepler's First Law and Conic Sections
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Chapter N12 Orbits and Conservation Laws
Chapter Overview
N12.1 Overview and Review
N12.2 Conservation Laws and Elliptical Orbits
N12.3 Conservation Laws and Hyperbolic Orbits
N12.4 Solving Orbit Problems
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Appendix NA Differential Calculus
NA.1 Derivatives
NA.2 Some Useful Rules
NA.3 Derivatives and Slopes
NA.4 The Chain Rule
NA.5 Derivatives of Other Functions
HOMEWORK PROBLEMS
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Appendix NB Integral Calculus
NB.1 Antiderivatives
NB.2 Definite Integrals
NB.3 The Fundamental Theorem
NB.4 Indefinite Integrals
NB.5 Substitution of Variables
NB.6 Looking Up Integrals
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Index
Periodic table
Short Answers to Selected Problems
Unit R
Six IdeasThat Shaped Physics Unit R: The Laws of Physics Are Frame-Independent
Contents: Unit R The Laws of Physics Are Frame-Independent
About the Author
Preface
Introduction for Students
Chapter R1 The Principle of Relativity
Chapter Overview
R1.1 Introduction to the Principle
R1.2 Events, Coordinates, and Reference Frames
R1.3 Inertial Reference Frames
R1.4 The Final Principle of Relativity
R1.5 Newtonian Relativity
R1.6 The Problem of Electromagnetic Waves
TWO-MINUTE PROBLEMS
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Chapter R2 Coordinate Time
Chapter Overview
R2.1 Relativistic Clock Synchronization
R2.2 SR Units
R2.3 Spacetime Diagrams
R2.4 Spacetime Diagrams as Movies
R2.5 The Radar Method
R2.6 Coordinate Time Is Frame-Dependent
R2.7 A Geometric Analogy
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Chapter R3 The Spacetime Interval
Chapter Overview
R3.1 The Three Kinds of Time
R3.2 The Metric Equation
R3.3 About Perpendicular Displacements
R3.4 Evidence Supporting the Metric Equation
R3.5 Spacetime Is Not Euclidean
R3.6 More About the Geometric Analogy
R3.7 Some Examples
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Chapter R4 Proper Time
Chapter Overview
R4.1 A Curved Footpath
R4.2 Curved Worldlines in Spacetime
R4.3 The Binomial Approximation
R4.4 Ranking the Three Kinds of Time
R4.5 Experimental Evidence
R4.6 The Twin Paradox
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Chapter R5 Coordinate Transformations
Chapter Overview
R5.1 Overview of Two-Observer Diagrams
R5.2 Conventions
R5.3 Drawing the Diagram tÕ Axis
R5.4 Drawing the Diagram xÕ Axis
R5.5 Reading the Two-Observer Diagram
R5.6 The Lorentz Transformation
TWO-MINUTE PROBLEMS
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Chapter R6 Lorentz Contraction
Chapter Overview
R6.1 The Length of a Moving Object
R6.2 A Two-Observer Diagram of a Stick
R6.3 What Causes the Contraction?
R6.4 The Contraction Is Frame-Symmetric
R6.5 The Barn and Pole Paradox
R6.6 Other Ways to Define Length
TWO-MINUTE PROBLEMS
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Chapter R7 The Cosmic Speed Limit
Chapter Overview
R7.1 Causality and Relativity
R7.2 Timelike, Lightlike, and Spacelike Intervals
R7.3 The Causal Structure of Spacetime
R7.4 The Einstein Velocity Transformation
TWO-MINUTE PROBLEMS
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Chapter R8 Four-Momentum
Chapter Overview
R8.1 A Plan of Action
R8.2 Newtonian Momentum IsnÕt Conserved
R8.3 The Four-Momentum Vector
R8.4 Properties of Four-Momentum
R8.5 Four-Momentum and Relativity
R8.6 Relativistic Energy
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Chapter R9 Conservation of Four-Momentum
Chapter Overview
R9.1 Energy_Momentum Diagrams
R9.2 Solving Conservation Problems
R9.3 The Mass of a System of Particles
R9.4 The Four-Momentum of Light
R9.5 Applications to Particle Physics
R9.6 Parting Comments
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Appendix RA Converting Equations to SI Units
RA.1 Why Use SR Units?
RA.2 Conversion of Basic Quantities
RA.3 Converting SR Unit Equations to SI Unit Equations
RA.4 Energy-Based SR Units
RA.5 Exercises for Practice
Appendix RB The Relativistic Doppler Effect
RB.1 Introduction to the Doppler Effect
RB.2 Deriving the Doppler Shift Formula
RB.3 Astrophysical Applications
RB.4 The Nonrelativistic Limit
RB.5 Doppler Radar
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Index
Periodic table
Short Answers to Selected Problems
Unit E
Six IdeasThat Shaped Physics Unit E: Electric and MagneticFields are Unified
Contents: Unit E Electric and Magnetic Fields Are Unified
About the Author
Preface
Introduction for Students
Chapter E1 Electric Fields
Chapter Overview
E1.1 Introduction to the Unit
E1.2 Charge
E1.3 Electric Force
E1.4 The Field Concept
E1.5 Defining the Electric Field
E1.6 The Field of a Single Particle
E1.7 The Superposition Principle
TWO-MINUTE PROBLEMS
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Chapter E2 Charge Distributions
Chapter Overview
E2.1 Introduction
E2.2 The Dipole
E2.3 Electrical Polarization
E2.4 Other Important Charge Distributions
E2.5 The Field of an Infinite Line
E2.6 The Field of an Infinite Plane
TWO-MINUTE PROBLEMS
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Chapter E3 Electric Potential
Chapter Overview
E3.1 Introduction
E3.2 Review of Energy Concepts
E3.3 Potential Energy and Electrical Potential
E3.4 From Field to Potential
E3.5 From Potential to Field
TWO-MINUTE PROBLEMS
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Chapter E4 Static Equilibrium
Chapter Overview
E4.1 Conductors and Insulators
E4.2 Static Charges on Conductors
E4.3 Capacitance
E4.4 The Parallel-Plate Capacitor
E4.5 The Electric Field as a Form of Energy
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Chapter E5 Current
Chapter Overview
E5.1 Introduction
E5.2 A Model of Current Flow
E5.3 Current Density
E5.4 Flux
E5.5 Electrical Current
TWO-MINUTE PROBLEMS
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Chapter E6 Dynamic Equilibrium
Chapter Overview
E6.1 A Simple Model for a Battery
E6.2 Implications of Equilibrium
E6.3 Surface Charges Direct the Flow
E6.4 Potentials in a Simple Circuit
E6.5 Resistance and Power
E6.6 Discharging a Capacitor
TWO-MINUTE PROBLEMS
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Chapter E7 Analyzing Circuits
Chapter Overview
E7.1 Circuit Diagrams
E7.2 KirchhoffÕs Laws
E7.3 Circuit Elements in Series
E7.4 Circuit Elements in Parallel
E7.5 Analyzing Complex Circuits
E7.6 Realistic Batteries
E7.7 Electrical Safety Issues
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Chapter E8 Magnetic Fields
Chapter Overview
E8.1 The Phenomenon of Magnetism
E8.2 The Definition of the Magnetic Field Direction
E8.3 Magnetic Forces on Moving Charges
E8.4 A Review of the Cross Product
E8.5 Defining the FieldÕs Magnitude
E8.6 A Free Particle in a Magnetic Field
TWO-MINUTE PROBLEMS
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Chapter E9 Currents Respond to Magnetic Fields
Chapter Overview
E9.1 The Magnetic Force on a Wire
E9.2 The Magnetic Torque on a Loop
E9.3 Current Loops Behave Like Bar Magnets
E9.4 The Potential Energy of an Oriented Loop
E9.5 Electric Motors
E9.6 Creating Currents in Moving Loops
TWO-MINUTE PROBLEMS
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Chapter E10 Currents Create Magnetic Fields
Chapter Overview
E10.1 The Magnetic Field of a Moving Charge
E10.2 The Magnetic Field of a Wire Segment
E10.3 The Magnetic Field of an Infinite Wire
E10.4 The Magnetic Field of a Circular Loop
E10.5 All Magnets Involve Circulating Currents
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Chapter E11 The Electromagnetic Field
Chapter Overview
E11.1 A Mystery
E11.2 Relativity and the Electromagnetic Field
E11.3 How the Fields Transform
E11.4 The Mystery Is Solved!
E11.5 The Electromagnetic Field of a Moving Particle
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Chapter E12 GaussÕs Law
Chapter Overview
E12.1 What Is a Field Equation?
E12.2 Divergence
E12.3 The Divergence as a Derivative
E12.4 GaussÕs Law
E12.5 Tools For Applying GaussÕs Law
E12.6 Applications
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Chapter E13 AmpereÕs Law
Chapter Overview
E13.1 Curl
E13.2 The Curl as a Derivative
E13.3 AmpereÕs Law
E13.4 Tools for Applying AmpereÕs Law
E13.5 Applications
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Chapter E14 Integral Forms
Chapter Overview
E14.1 Introduction
E14.2 Integrating GaussÕs Law
E14.3 Using the Integral Form of GaussÕs Law
E14.4 Integrating AmpereÕs Law
E14.5 Using the Integral Form of AmpereÕs Law
E14.6 Which Form Is Better?
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Chapter E15 MaxwellÕs Equations
Chapter Overview
E15.1 Correcting AmpereÕs Law
E15.2 GaussÕs Law Needs No Correction
E15.3 GaussÕs Law for the Magnetic Field
E15.4 FaradayÕs Law
E15.5 MaxwellÕs Equations
E15.6 A Short History of MaxwellÕs Equations
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Chapter E16 FaradayÕs Law
Chapter Overview
E16.1 Applying FaradayÕs Law
E16.2 Magnetic Flux and Induced EMF
E16.3 LenzÕs Law
E16.4 Some Example Applications
E16.5 Superconductivity and Magnetic Flux
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Chapter E17 Induction
Chapter Overview
E17.1 Self-Induction
E17.2 ÒDischargingÓ an Inductor
E17.3 The Energy in a Magnetic Field
E17.4 LC Circuits
E17.5 Transformers
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Chapter E18 Electromagnetic Waves
Chapter Overview
E18.1 An Electromagnetic Disturbance
E18.2 Properties of Electromagnetic Waves
E18.3 Intensity of an Electromagnetic Wave
E18.4 Waves from a Charged Particle
E18.5 MaxwellÕs Rainbow
E18.6 Why the Sky Is Blue
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Appendix EA The Electromagnetic Transformation Law
EA.1 Introduction
EA.2 The Reverse Transformation for the Field
EA.3 The Case of the Transverse Particle
EA.4 Concluding Unrelativistic Postscript
EXERCISE ANSWERS
Appendix EB Radiation from an Accelerating Particle
EB.1 Introduction
EB.2 The Derivation
Index
Periodic table
Short Answers to Selected Problems
Unit Q
Six IdeasThat Shaped Physics Unit Q: Particles Behave Like Waves
Contents: Unit Q Particles Behave Like Waves
About the Author
Preface
Introduction for Students
Chapter Q1 Wave Models
Chapter Overview
Q1.1 What Is a Wave?
Q1.2 A Sinusoidal Wave Model
Q1.3 The Phase Speed of a Sinusoidal Wave
Q1.4 Sound
Q1.5 Energy in Waves
Q1.6 The Doppler Effect
TWO-MINUTE PROBLEMS
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Chapter Q2 Standing Waves and Resonance
Chapter Overview
Q2.1 The Superposition Principle
Q2.2 Reflection
Q2.3 Standing Waves
Q2.4 Resonance
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Chapter Q3 Interference and Diffraction
Chapter Overview
Q3.1 Two-Dimensional Waves
Q3.2 Simple Diffraction
Q3.3 Two-Slit Interference
Q3.4 Two-Slit Interference of Light
Q3.5 Diffraction Revisited
Q3.6 Optical Resolution
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Chapter Q4 The Particle Nature of Light
Chapter Overview
Q4.1 A Short History of Light
Q4.2 The Photoelectric Effect
Q4.3 Idealized Photoelectric Experiments
Q4.4 Predictions of the Wave Model
Q4.5 Confronting the Facts
Q4.6 The Photon Model of Light
Q4.7 Detecting Individual Photons
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Chapter Q5 The Wave Nature of Particles
Chapter Overview
Q5.1 Subatomic Particles as Particles
Q5.2 The de Broglie Hypothesis
Q5.3 Preparing an Electron Beam
Q5.4 The Davisson_Germer Experiment
Q5.5 Modern Interference Experiments
Q5.6 Interference a Quanton at a Time
Q5.7 Implications
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Chapter Q6 Spin
Chapter Overview
Q6.1 Introduction to Spin
Q6.2 Introduction to the Stern_Gerlach Experiment
Q6.3 Gyroscopic Precession
Q6.4 The Stern_Gerlach Experiment
Q6.5 Spin Experiments
Q6.6 Spin Is Genuine Angular Momentum
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Chapter Q7 The Rules of Quantum Mechanics
Chapter Overview
Q7.1 The Game of Quantum Mechanics
Q7.2 The Playing Pieces and the Goal
Q7.3 The Mathematics of Quantum Mechanics
Q7.4 The Rules
Q7.5 Questions and Answers
Q7.6 Examples
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Chapter Q8 Quantum Weirdness
Chapter Overview
Q8.1 Introduction
Q8.2 The EPR Argument
Q8.3 BellÕs Theorem
Q8.4 Superposition and Schr_dingerÕs Cat
Q8.5 The Collapse Problem
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Chapter Q9 The Wavefunction
Chapter Overview
Q9.1 Vectors to Wavefunctions
Q9.2 Wavefunctions and Position Probability
Q9.3 The Collapse of the Wavefunction
Q9.4 The Heisenberg Uncertainty Principle
Q9.5 The Rules Explain Two-Slit Interference
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Chapter Q10 Simple Quantum Models
Chapter Overview
Q10.1 An Introduction to Bound Systems
Q10.2 Energy Eigenfunctions
Q10.3 A Quanton in a Box
Q10.4 The Bohr Model of the Hydrogen Atom
Q10.5 The Simple Harmonic Oscillator
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Chapter Q11 Spectra
Chapter Overview
Q11.1 Energy-Level Diagrams
Q11.2 The Spontaneous Emission of Photons
Q11.3 Spectral Lines
Q11.4 Absorption Lines
Q11.5 The Pauli Exclusion Principle
Q11.6 Conductors and Semiconductors
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Chapter Q12 The Schr_dinger Equation
Chapter Overview
Q12.1 Generalizing the de Broglie Relation
Q12.2 Local Wavelength
Q12.3 Finding the Schr_dinger Equation
Q12.4 Solving the Equation Numerically
Q12.5 Using SchroSolver
Q12.6 Sketching Energy Eigenfunctions
Q12.7 Tunneling
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Chapter Q13 Introduction to Nuclei
Chapter Overview
Q13.1 Introduction to Nuclear Structure
Q13.2 The Size of the Nucleus
Q13.3 The Strong Interaction
Q13.4 Binding Energy and Mass
Q13.5 Questions about Nuclear Stability
Q13.6 A Historical Overview of Radioactivity
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Chapter Q14 Nuclear Stability
Chapter Overview
Q14.1 The Weak Interaction
Q14.2 Why Z Å N
Q14.3 Why N > Z for Large Nuclei
Q14.4 Beta Decay
Q14.5 Alpha Decay
Q14.6 Gamma Decay
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Chapter Q15 Nuclear Technology
Chapter Overview
Q15.1 The Penetrating Ability of Radiation
Q15.2 The Biological Effects of Radiation
Q15.3 Applications of Radioactive Nuclei
Q15.4 Introduction to Nuclear Energy
Q15.5 Fission
Q15.6 Fusion
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Appendix QA Complex Numbers
Appendix Overview
QA.1 Introduction to Complex Numbers
QA.2 The Complex Exponential
QA.3 The Spin Observable Sy
QA.4 The Time Evolution Rule
QA.5 An Application: Spins in a Magnetic Field
QA.6 Momentum Eigenfunctions
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Index
Periodic table
Short Answers to Selected Problems
Unit T
Six IdeasThat Shaped Physics Unit T: Some Processes AreIrreversible
Contents: Unit T Some Processes Are Irreversible
About the Author
Preface
Introduction for Students
Chapter T1 Temperature
Chapter Overview
T1.1 Introduction to the Unit
T1.2 Irreversible Processes
T1.3 Heat, Work, and Internal Energy
T1.4 The Paradigmatic Thermal Process
T1.5 Temperature and Equilibrium
T1.6 Thermometers
T1.7 Temperature and Thermal Energy
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter T2 Macrostates and Microstates
Chapter Overview
T2.1 The Einstein Model of a Solid
T2.2 Distinguishing Macrostates and Microstates
T2.3 Counting Microstates
T2.4 Two Einstein Solids in Thermal Contact
T2.5 The Fundamental Assumption
T2.6 Using StatMech
T2.7 The Emergence of Irreversibility
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter T3 Entropy and Temperature
Chapter Overview
T3.1 The Definition of Entropy
T3.2 The Second Law of Thermodynamics
T3.3 Entropy and Disorder
T3.4 The Definition of Temperature
T3.5 Consistency with Historical Definitions
T3.6 A Financial Analogy
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter T4 The Boltzmann Factor
Chapter Overview
T4.1 The Boltzmann Factor
T4.2 Some Simple Applications
T4.3 The Average Energy of a Quantum System
T4.4 Application to the Einstein Solid
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter T5 The Ideal Gas
Chapter Overview
T5.1 Quantum Particles in a One-Dimensional Box
T5.2 A Three-Dimensional Monatomic Gas
T5.3 Diatomic Gases
T5.4 The Equipartition Theorem and Its Limits
T5.5 The Ideal Gas Law
TWO-MINUTE EXERCISES
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter T6 Distributions
Chapter Overview
T6.1 Counting Quantum States
T6.2 The Maxwell-Boltzmann Distribution
T6.3 The Photon Gas
T6.4 Blackbody Emission
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter T7 Gas Processes
Chapter Overview
T7.1 Work during Expansion or Compression
T7.2 The State of a Gas
T7.3 PV Diagrams and Constrained Processes
T7.4 Computing the Work
T7.5 Adiabatic Processes
T7.6 Ideal Gas Heat Capacities
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter T8 Calculating Entropy Changes
Chapter Overview
T8.1 The Entropy of an Ideal Gas
T8.2 Work and Entropy
T8.3 Constant-Temperature Processes
T8.4 Handling Changing Temperatures
T8.5 Replacement Processes
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter T9 Heat Engines
Chapter Overview
T9.1 Perfect Engines Are Impossible
T9.2 Real Heat Engines
T9.3 The Efficiency of a Heat Engine
T9.4 Consequences
T9.5 Refrigerators
T9.6 The Carnot Cycle
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Chapter T10 The Physics of Climate Change
Chapter Overview
T10.1 Introduction
T10.2 Radiative Equilibrium
T10.3 The Atmospheric Blanket
T10.4 Building a Better Model
T10.5 The Ocean Effect
T10.6 Improving the Model
T10.7 Consequences
T10.8 Concluding Scientific Postscript
TWO-MINUTE PROBLEMS
HOMEWORK PROBLEMS
ANSWERS TO EXERCISES
Index
Periodic table
Short Answers to Selected Problems

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Six Ideas That Shaped Physics Third Edition

Thomas A. Moore

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Some Physical Constants Speed of light Gravitational constant Coulomb’s constant Permittivity constant Permeability constant Planck’s constant Boltzmann’s constant Elementary charge Electron mass Proton mass Neutron mass Avogadro’s number

Standard Metric Prefixes (for powers of 10)

c 3.00  × 108 m/s G 6.67  × 10-11 N·m2/kg2 1/4πε0 8.99  × 109 N·m2/C2 ε0 8.85  × 10-12 C2/(N·m2) µ0 4π   × 10-7 N/A2 h 6.63  × 10-34 J·s kB 1.38  × 10-23 J/K e 1.602 × 10-19 C me 9.11  × 10-31 kg mp 1.673 × 10-27 kg mn 1.675 × 10-27 kg NA 6.02  × 1023

Power Prefix Symbol 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 10-2 centi c 10-3 milli m 10-6 micro µ 10-9 nano n 10-12 pico p 10-15 femto f 10-18 atto a

Commonly Used Physical Data Gravitational field strength g =│g​  ​W│   9.80 N/kg = 9.80 m/s2 (near the earth’s surface) Mass of the earth Me 5.98 × 1024 kg Radius of the earth Re 6380 km (equatorial) Mass of the sun M⊙ 1.99 × 1030 kg Radius of the sun R⊙ 696,000 km Mass of the moon 7.36 × 1022 kg Radius of the moon 1740 km Distance to the moon 3.84 × 108 m Distance to the sun 1.50 × 1011 m † Density of water 1000 kg/m3 = 1 g/cm3 † Density of air 1.2 kg/m3 Absolute zero 0 K = -273.15°C = -459.67°F Freezing point of water‡ 273.15 K = 0°C = 32°F Boiling point of water‡ 373.15 K = 100°C = 212°F Normal atmospheric pressure 101.3 kPa † ‡

At normal atmospheric pressure and 20°C. At normal atmospheric pressure.

Useful Conversion Factors 1 meter = 1 m = 100 cm = 39.4 in = 3.28 ft 1 mile = 1 mi = 1609 m = 1.609 km = 5280 ft 1 inch = 1 in = 2.54 cm 1 light-year = 1 ly = 9.46 Pm = 0.946 × 1016 m 1 minute = 1 min = 60 s 1 hour = 1 h = 60 min = 3600 s 1 day = 1 d = 24 h = 86.4 ks = 86,400 s 1 year = 1 y = 365.25 d = 31.6 Ms = 3.16 × 107 s 1 newton = 1 N = 1 kg·m/s2 = 0.225 lb 1 joule = 1 J = 1 N·m = 1 kg·m2/s2 = 0.239 cal 1 watt = 1 W = 1 J/s 1 pascal = 1 Pa = 1 N/m2 = 1.45 × 10–4 psi 1 kelvin (temperature difference) = 1 K = 1°C = 1.8°F 1 radian = 1 rad = 57.3° = 0.1592 rev 1 revolution = 1 rev = 2π rad = 360° 1 cycle = 2π rad 1 hertz = 1 Hz = 1 cycle/s

1 m/s = 2.24 mi/h = 3.28 ft/s 1 mi/h = 1.61 km/h = 0.447 m/s = 1.47 ft/s 1 liter = 1 l = (10 cm)3 = 10-3 m3 = 0.0353 ft3 1 ft3 = 1728 in3 = 0.0283 m3 1 gallon = 1 gal = 0.00379 m3 = 3.79 l ≈ 3.8 kg H2O Weight of 1-kg object near the earth = 9.8 N = 2.2 lb 1 pound = 1 lb = 4.45 N 1 calorie = energy needed to raise the temperature of 1 g of H2O by 1 K = 4.186 J 1 horsepower = 1 hp = 746 W 1 pound per square inch = 6895 Pa 1 food calorie = 1 Cal = 1 kcal = 1000 cal = 4186 J 1 electron volt = 1 eV = 1.602 × 10-19 J

(  ) T = ​( ___ ​  5K  ​  )​  (T 9°F

T = ​ ____ ​  1K  ​  ​  (T[C] + 273.15°C) 1°C [F]

+ 459.67°F)

(  )

T[C] = ​ ____ ​ 5°C ​  ​(T[F] - 32°F) 9°F T[F] = 32°F + ​ ____ ​  9°F  ​  ​T[C] 5°C

(  )

Six Ideas That Shaped Physics Unit C: Conservation Laws Constrain Interactions Third Edition

Thomas A. Moore

SIX IDEAS THAT SHAPED PHYSICS, UNIT C: CONSERVATION LAWS CONSTRAIN INTERACTIONS, THIRD EDITION Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2017 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2003, and 1998. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 RMN/RMN 1 0 9 8 7 6 ISBN 978-0-07-351394-2 MHID 0-07-351394-6 Senior Vice President, Products & Markets: Kurt L. Strand Vice President, General Manager, Products & Markets: Marty Lange Vice President, Content Design & Delivery: Kimberly Meriwether David Managing Director: Thomas Timp Brand Manager: Thomas M. Scaife, Ph.D. Product Developer: Jolynn Kilburg Marketing Manager: Nick McFadden Director of Development: Rose Koos Digital Product Developer: Dan Wallace Director, Content Design & Delivery: Linda Avenarius Program Manager: Faye M. Herrig Content Project Managers: Melissa M. Leick, Tammy Juran, Sandy Schnee Design: Studio Montage, Inc. Content Licensing Specialists: Deanna Dausener Cover Image: NASA Compositor: SPi Global

Dedication To All My Family, Near and Far whose love conserves what is best All credits appearing on page or at the end of the book are considered to be an extension of the copyright page. The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGrawHill Education does not guarantee the accuracy of the information presented at these sites. Library of Congress Cataloging-in-Publication Data Names: Moore, Thomas A. (Thomas Andrew), author. Title: Six ideas that shaped physics. Unit C, Conservation laws constrain   interactions/Thomas A. Moore. Other titles: Conservation laws constrain interactions Description: Third edition. | New York, NY : McGraw-Hill Education, [2016] |   2017 | Includes index. Identifiers: LCCN 2015043352| ISBN 9780073513942 (alk. paper) | ISBN   0073513946 (alk. paper) Subjects: LCSH: Conservation laws (Physics)—Textbooks. Classification: LCC QC793.3.C58 M66 2016 | DDC 539.7/54—dc23 LC record available at http://lccn.loc.gov/2015043352 www.mhhe.com

Contents: Unit C

Conservation Laws Constrain Interactions About the Author

viii

Preface ix Introduction for Students

Chapter C1 The Art of Model Building C1.1 C1.2 C1.3 C1.4 C1.5 C1.6

Chapter Overview The Nature of Science The Development and Structure of Physics A Model-Building Example Trick Bag: Unit Awareness Trick Bag: Unit Conversions Trick Bag: Dimensional Analysis TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter C2 Particles and Interactions C2.1 C2.2 C2.3 C2.4 C2.5 C2.6 C2.7

Chapter Overview The Principles of Modern Mechanics Describing an Object’s Motion Vector Operations Momentum and Impulse Force and Weight Interaction Categories Momentum Transfer TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

xvi

2 2 2 4 5 7 10 11 12 14 14 17

18 18 18 20 21 23 24 26 28 30 32 33 37

Chapter C3 38 Vectors 38 C3.1 C3.2 C3.3 C3.4 C3.5 C3.6 C3.7

Chapter Overview Introduction Reference Frames Displacement Vectors Arbitrary Vectors Seven Rules to Remember Vectors in Two Dimensions Vectors in One Dimension TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter C4 Systems and Frames C4.1 C4.2 C4.3 C4.4 C4.5 C4.6

Chapter Overview Systems of Particles A System’s Center of Mass How the Center of Mass Moves Inertial Reference Frames Freely Floating Frames Interactions with the Earth TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter C5 Conservation of Momentum Chapter Overview C5.1 Degrees of Isolation C5.2 How to Solve Physics Problems C5.3 Conservation of Momentum Problems C5.4 Examples C5.5 Airplanes and Rockets TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

38 40 40 41 44 45 47 48 49 49 51

52 52 52 54 54 57 59 60 62 64 65 67

68 68 68 70 71 75 76 79 81 82 83

v

Table of Contents

vi

Chapter C6 Conservation of Angular Momentum C6.1 C6.2 C6.3 C6.4 C6.5 C6.6 C6.7

Chapter Overview Introduction Quantifying Orientation Angular Velocity The Angular Momentum of a Rigid Object Twirl and Torque Gyroscopic Precession Conservation of Angular Momentum TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter C7 More About Angular Momentum C7.1 C7.2 C7.3 C7.4 C7.5 C7.6 C7.7

Chapter Overview Introduction to Energy Kinetic Energy Potential Energy Fundamental Potential Energy Formulas Internal Energy and Power Isolation Solving Conservation-of-Energy Problems TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter C9 Potential Energy Graphs C9.1 C9.2

84 86 86 87 88 91 92 93 96 97 99

100 100

Chapter Overview First Steps The Cross Product The Angular Momentum of a Moving Particle Rotating Objects Rotating and Moving Objects Torque and Force Why Angular Momentum Is Conserved TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter C8 Conservation of Energy C8.1 C8.2 C8.3 C8.4 C8.5 C8.6 C8.7

84 84

Chapter Overview Interactions Between Macroscopic Objects Interactions Between Two Atoms

100 102 103 105 106 107 108 110 112 113 117

118 118 118 120 121 122 124 126 126 127 130 131 134

136 136 136 138 138

C9.3 C9.4 C9.5 C9.6 C9.7

One-Dimensional Potential Energy Diagrams Relaxing the Mass Limitation The Spring Approximation The Potential Energy “of an Object” An Example TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

139 144 145 146 146 148 149 152

Chapter C10 154 Work 154 C10.1 C10.2 C10.3 C10.4 C10.5

Chapter Overview The Momentum Requirement The Dot Product The Definition of Work Long-Range Interactions Contact Interactions TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter C11 Rotational Energy C11.1 C11.2 C11.3 C11.4 C11.5

Chapter Overview Introduction to Rotational Energy Rotational Energy of an Object at Rest Calculating Moments of Inertia When an Object Both Moves and Rotates Rolling Without Slipping TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter C12 Thermal Energy C12.1 C12.2 C12.3 C12.4 C12.5 C12.6 C12.7

Chapter Overview The Case of the Disappearing Energy Caloric Is Energy Thermal Energy as Microscopic Energy Friction and Thermal Energy Heat, Work, and Energy Transfer Specific “Heat” Problems Involving Thermal Energies TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

154 156 157 158 161 162 165 166 169

170 170 170 172 172 174 176 177 182 183 185

186 186 186 188 188 190 192 193 194 196 199 200 202

Table of Contents

Chapter C13 Other Forms of Internal Energy Chapter Overview C13.1 Bonds C13.2 Latent “Heat” C13.3 Chemical Energy C13.4 Nuclear Energy C13.5 Modes of Energy Transfer C13.6 Mechanisms of Heat Transfer C13.7 A Comprehensive Energy Master Equation TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

vii

204 Appendix CA 204 The Standard Model 204 206 208 211 212 213 214 216 218 219 221

CA.1 Introduction CA.2 Matter Particles CA.3 Fundamental Interactions CA.4 The Importance of Color-Neutrality CA.5 Stability and the Weak Interaction CA.6 Conclusion TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Periodic table

Chapter Overview Types of Collisions One-Dimensional Collisions Two-Dimensional Collisions The Slingshot Effect Using All Three Conservation Laws Asteroid Impacts TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

222 224 224 227 230 232 234 236 237 239

240 240 240 242 243 245 246 246 246

Index 247

Chapter C14 222 Collisions 222 Short Answers to Selected Problems C14.1 C14.2 C14.3 C14.4 C14.5 C14.6

240 240

255 256

About the Author Thomas A. Moore graduated from Carleton College (magna cum laude with Distinction in Physics) in 1976. He won a Danforth Fellowship that year that supported his graduate education at Yale University, where he earned a Ph.D. in 1981. He taught at Carleton College and Luther College before taking his current position at Pomona College in 1987, where he won a Wig Award for Distinguished Teaching in 1991. He served as an active member of the steering committee for the national Introductory University Physics Project (IUPP) from 1987 through 1995. This textbook grew out of a model curriculum that he developed for that project in 1989, which was one of only four selected for further development and testing by IUPP. He has published a number of articles about astrophysical sources of gravitational waves, detection of gravitational waves, and new approaches to teaching physics, as well as a book on general relativity entitled A General Relativity Workbook (University Science Books, 2013). He has also served as a reviewer and as an associate editor for A ­ merican Journal of Physics. He currently lives in Claremont, California, with his wife Joyce, a retired pastor. When he is not teaching, doing research, or writing, he enjoys reading, hiking, calling contradances, and playing Irish traditional fiddle music.

viii

Preface Introduction This volume is one of six that together comprise the text materials for Six Ideas That Shaped Physics, a unique approach to the two- or three-semester calculus-based introductory physics course. I have designed this curriculum (for which these volumes only serve as the text component) to support an introductory course that combines two elements that rarely appear together: (1) a thoroughly 21st-century perspective on physics (including a great deal of 20th-century physics), and (2) strong support for a student-centered classroom that emphasizes active learning both in and outside of class, even in situations where large-enrollment sections are unavoidable. This course is based on the premises that innovative metaphors for teaching basic concepts, explicitly instructing students in the processes of constructing physical models, and active learning can help students learn the subject much more effectively. In the course of executing this project, I have completely rethought (from scratch) the presentation of every topic, taking advantage of research into physics education wherever possible. I have done nothing in this text just because “that is the way it has always been done.” Moreover, because physics education research has consistently underlined the importance of active learning, I have sought to provide tools for professors (both in the text and online) to make creating a coherent and self-­ consistent course structure based on a student-centered classroom as easy and practical as possible. All of the materials have been tested, evaluated, and rewritten multiple times. The result is the culmination of more than 25 years of continual testing and revision. I have not sought to “dumb down” the course to make it more accessible. Rather, my goal has been to help students become smarter. I have intentionally set higher-than-usual standards for sophistication in physical thinking, but I have also deployed a wide range of tools and structures that help even average students reach this standard. I don’t believe that the mathematical level required by these books is significantly different than that in most university physics texts, but I do ask students to step beyond rote thinking patterns to develop flexible, powerful, conceptual reasoning and modelbuilding skills. My experience and that of other users is that normal students in a wide range of institutional settings can (with appropriate support and practice) meet these standards. Each of six volumes in the text portion of this course is focused on a single core concept that has been crucial in making physics what it is today. The six volumes and their corresponding ideas are as follows:

Unit C: Unit N: Unit R: Unit E: Unit Q: Unit T:

Conservation laws constrain interactions The laws of physics are universal (Newtonian mechanics) The laws of physics are frame-independent (Relativity) Electric and Magnetic Fields are Unified Particles behave like waves (Quantum physics) Some processes are irreversible (Thermal physics)

ix

x

Preface

I have listed the units in the order that I recommend they be taught, but I have also constructed units R, E, Q, and T to be sufficiently independent so they can be taught in any order after units C and N. (This is why the units are lettered as opposed to numbered.) There are six units (as opposed to five or seven) to make it possible to easily divide the course into two semesters, three quarters, or three semesters. This unit organization therefore not only makes it possible to dole out the text in small, easily-handled pieces and provide a great deal of flexibility in fitting the course to a given schedule, but also carries its own important pedagogical message: Physics is organized hierarchically, structured around only a handful of core ideas and metaphors. Another unusual feature of all of the texts is that they have been designed so that each chapter corresponds to what one might handle in a single 50-­minute class session at the maximum possible pace (as guided by years of experience). Therefore, while one might design a syllabus that goes at a slower rate, one should not try to go through more than one chapter per 50-minute session (or three chapters in two 70-minute sessions). A few units provide more chapters than you may have time to cover. The preface to such units will tell you what might be cut. Finally, let me emphasize again that the text materials are just one part of the comprehensive Six Ideas curriculum. On the Six Ideas website, at

𝚠𝚠𝚠𝚙𝚑𝚢𝚜𝚒𝚌𝚜𝚙𝚘𝚖𝚘𝚗𝚊𝚎𝚍𝚞/𝚜𝚒𝚡𝚒𝚍𝚎𝚊𝚜/

you will find a wealth of supporting resources. The most important of these is a detailed instructor’s manual that provides guidance (based on Six Ideas users’ experiences over more than two decades) about how to construct a course at your institution that most effectively teaches students physics. This manual does not provide a one-size-fits-all course plan, but rather exposes the important issues and raises the questions that a professor needs to consider in creating an effective Six Ideas course at their particular institution. The site also provides software that allows professors to post selected problem solutions online where their students alone can see them and for a time period that they choose. A number of other computer applets provide experiences that support student learning in important ways. You will also find there example lesson plans, class videos, information about the course philosophy, evidence for its success, and many other resources. There is a preface for students appearing just before the first chapter of each unit that explains some important features of the text and assumptions behind the course. I recommend that everyone read it.



Comments about the Current Edition

My general goals for the current edition have been to correct errors, enhance the layout, improve the presentation in many areas, make the book more flexible, and improve the quality and range of the homework problems as well as significantly increase their number. Users of previous editions will note that I have split the old “Synthetic” homework problem category into “Modeling” and “Derivations” categories. “Modeling” problems now more specifically focus on the process of building physical models, making appropriate approximations, and binding together disparate formulas. “Derivation” problems focus more on supporting or extending derivations presented in the text. I thought it valuable to more clearly separate these categories. The “Basic Skills” category now includes a number of multipart problems specifically designed for use in the classroom to help students practice basic issues. The instructor’s manual discusses how to use such problems.

Preface

xi

I have also been more careful in providing instructors with more choices about what to cover, making it possible for instructors to omit chapters without a loss of continuity. See the unit-specific part of this preface for more details. Users of previous editions will also note that I have dropped the menulike chapter location diagrams, as well as the glossaries and symbol lists that appeared at the end of each volume. I could find no evidence that these were actually helpful to students. Units C and N still instruct students very carefully on how to construct problem solutions that involve translating, modeling, solving, and checking, but examples and problem solutions for the remaining units have been written in a more flexible format that includes these elements implicitly but not so rigidly and explicitly. Students are rather guided in this unit to start recognizing these elements in more generally formatted solutions, something that I think is an important skill. The only general notation change is that now I use │v​  ​W│    exclusively and W​ .  I still think it is very important universally for the magnitude of a vector v​ to have notation that clearly distinguishes vector magnitudes from other scaW)  notation is too cumbersome to use exclusively, and lars, but the old mag(​v​ mixing it with using just the simple letter has proved confusing. Unit C contains some specific instruction about the notation commonly used in texts by other authors (as well as discussing its problems). Finally, at the request of many students, I now include short answers to selected homework problems at the end of each unit. This will make students happier without (I think) significantly impinging on professors’ freedom.



Specific Comments About Unit C

This unit is the foundation on which a Six Ideas course rests. The current course structure assumes that unit C is taught first, immediately followed by unit N. Unit C contains core material that will be used in all the other units, as well as providing an introduction to the process of model building that is central to the course. Why study conservation laws before Newtonian mechanics? The most important reasons are as follows: (1) Conservation of “stuff” is a concrete idea that is easy to understand. Beginning with such simple ideas helps build student confidence at the beginning of the course. (2) Using conservation laws does not really require calculus, and so helps students polish their algebra skills before getting involved with calculus. (3) Studying conservation of momentum and angular momentum does require vectors, allowing students to use vectors for several weeks in simple contexts before introducing vector calculus. (4) Conservation laws really are more fundamental than even Newtonian mechanics, so it is good to start the course with concepts that are central and will be used throughout the course. I did not intuit these benefits at first: the earliest versions of Six Ideas presented mechanics in the standard order. Rather, this inversion emerged naturally as a consequence of observations of student learning and some reflection about the course’s logical flow. Inverting the order can be a challenge (in both a positive and negative sense) for the student who already has some background in mechanics. Reviewing mechanics from a different perspective can be quite good for such a student because it makes her or him really think about the subject again. The instructor can play a key role in helping such students appreciate this and by emphasizing the power and breadth of the conservation law approach and its importance in contemporary physics, as well as celebrating with them the power one gains by being able to approach situations from multiple angles.

xii

Preface

The momentum-transfer model of interactions (introduced now in chapter C2) is really what makes it possible to talk about conservation laws ­without starting with Newton’s laws. This model will be a new and challenging idea for almost everyone. Instructors should work carefully with students to give them enough practice with the model to ensure they understand it and can talk about it correctly. The payoff is that when students really grasp this model, it by its very nature helps them avoid many of the standard misconceptions that plague students in introductory courses. I have substantially revised this unit from the second edition, focusing my attention on the following specific goals. • I have enlarged the discussion of the model-building process, providing new examples and a more (literally) up-front discussion of tricks and techniques such as unit conversion and dimensional analysis. • I have reorganized the first few chapters to provide a better logical flow. • I have substantially rethought how to present expert problem-solving styles. The second edition’s experiment with cartoon balloons and interaction diagrams was not very successful with my students. I have replaced this approach with checklists that specify tasks to complete and a more flexible solution style that I think will be easier for students to emulate. Colored comments on many of the example solutions help students see the connection between the solutions and checklists, and there is also some opportunity for students to practice writing the comments themselves, and so become more self-reflective about the process. • I have brought the two vector conservation laws (momentum and angular momentum) together, which has several pedagogical advantages (including highlighting how these quantities are similar and how they both contrast with energy). • I have also reorganized the angular momentum chapters so that the basic idea (and what is necessary for later units) appears first and in its own chapter. All the complicated material (involving the cross product) appears in the second chapter, which may be postponed or even omitted. • I have reorganized the material in the conservation of energy chapters to better even out the pace and improve the logical flow. In particular, I have separated the material on potential energy graphs from material on bonds, latent heat, chemical energy, and nuclear energy (this was all just too much for one chapter). • Finally, and most importantly, I have reorganized the energy material to be more consistent with the approach that John Jewett outlined in his series of “Energy and the Confused Student” articles in various issues of The Physics Teacher in 2008. While I don’t completely agree with Jewett on every issue, his insights into student difficulties were consistent with what I have observed in the classroom, and I think his approach is superior pedagogically to what I had been doing. This has meant saying farewell to “k-work,” which now much more correctly appears as the requirement that conservation of momentum imposes on a system. I now also have a complete discussion of “work” that allows a consistent application to deformable systems (including human bodies). Finally, I have sought to provide more flexibility for instructors. Most of the chapters are crucial and should be discussed in order, but, as noted earlier, chapter C7 on the hard parts of angular momentum may be omitted or delayed, because no other chapter depends on it.

Preface

xiii

I have also made chapter C14 optional. I think that is very valuable, particularly as a preparation for the last chapters of unit R, but it is not absolutely necessary.

Appreciation A project of this magnitude cannot be accomplished alone. A list including everyone who has offered important and greatly appreciated help with this project over the past 25 years would be much too long (and such lists appear in the previous editions), so here I will focus for the most part on people who have helped me with this particular edition. First, I would like to thank Tom Bernatowicz and his colleagues at Washington University (particularly Marty Israel and Mairin Hynes) who hosted me for a visit to Washington University where we discussed this edition in detail. Many of my decisions about what was most important in this edition grew out of that visit. Bruce Sherwood and Ruth Chabay always have good ideas to share, and I appreciate their generosity and wisdom. Benjamin Brown and his colleagues at Marquette University have offered some great suggestions as well, and have been working hard on the important task of adapting some Six Ideas problems for computer grading. I’d like to thank Michael Lange at McGraw-Hill for having faith in the Six Ideas project and starting the push for this edition, and Thomas Scaife for continuing that push. Eve Lipton and Jolynn Kilburg has been superb at guiding the project at the detail level. Many others at McGraw-Hill, including Melissa Leick, Ramya Thirumavalavan, Kala Ramachandran, David Tietz, and Deanna Dausener, were instrumental in proofreading and producing the printed text. I’d also like to thank Dwight Whitaker of Pomona ­College and and his Physics 70 students (especially Nathaniel Roy, Eric Cooper, Neel Kumar, Milo Barisof, Sabrina Li, Samuel Yih, Asher Abrahms, Owen Chapman, Mariana Cisneros, Nick Azar, William Lamb, Wuyi Li, Errol Francis, and Cameron Queen) and my students in Physics 71 (especially Alex Hof, ­Gabrielle Mehta, Gail Gallaher and Jonah Grubb) for helping me track down errors in the manuscript. David Haley and Marilee Oldstone-Moore helped me with several crucial photographs and offered useful feedback. Finally a very special thanks to my wife, Joyce, who ­sacrificed and supported me and loved me during this long and ­demanding project. Heartfelt thanks to all! Thomas A. Moore Claremont, California

SmartBook is the first and only adaptive reading experience designed to change the way students read and learn. It creates a personalized reading experience by highlighting the most impactful concepts a student needs to learn at that moment in time. As a student engages with SmartBook, the reading experience continuously adapts by highlighting content based on what the student knows and doesn’t know. This ensures that the focus is on the content he or she needs to learn, while simultaneously promoting long-term retention of material. Use SmartBook’s real-time reports to quickly identify the concepts that require more attention from individual s­ tudents–or the entire class. The end result? Students are more engaged with course content, can better prioritize their time, and come to class ready to ­participate.

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Preface

Learn Without Limits Continually evolving, McGraw-Hill Connect® has been redesigned to provide the only true adaptive learning experience delivered within a simple and easy-to-navigate environment, placing students at the very center. • Performance Analytics – Now available for both instructors and students, easy-to-decipher data illuminates course performance. Students always know how they’re doing in class, while instructors can view student and section performance at-a-glance. • Mobile – Available  on tablets, students can now access assignments, quizzes, and results on-the-go, while instructors can assess student and section performance anytime, anywhere. • Personalized Learning – Squeezing the most out of study time, the adaptive engine in Connect creates a highly personalized learning path for each student by identifying areas of weakness, and surfacing learning resources to assist in the moment of need. This seamless integration of reading, practice, and assessment, ensures that the focus is on the most important content for that individual student at that specific time, while promoting long-term retention of the material.

Introduction for Students Introduction Welcome to Six Ideas That Shaped Physics! This text has a number of features that may be different from science texts you may have encountered previously. This section describes those features and how to use them effectively.

Why active learning is crucial

Why Is This Text Different?

Research into physics education consistently shows that people learn physics most effectively through activities where they practice applying physical reasoning and model-building skills in realistic situations. This is because physics is not a body of facts to absorb, but rather a set of thinking skills acquired through practice. You cannot learn such skills by listening to factual lectures any more than you can learn to play the piano by listening to concerts! This text, therefore, has been designed to support active learning both inside and outside the classroom. It does this by providing (1) resources for various kinds of learning activities, (2) features that encourage active reading, and (3) features that make it as easy as possible to use the text (as opposed to lectures) as the primary source of information, so that you can spend class time doing activities that will actually help you learn.



The Text as Primary Source

Features that help you use the text as the primary source of information

To serve the last goal, I have adopted a conversational style that I hope you will find easy to read, and have tried to be concise without being too terse. Certain text features help you keep track of the big picture. One of the key aspects of physics is that the concepts are organized hierarchically: some are more fundamental than others. This text is organized into six units, each of which explores the implications of a single deep idea that has shaped physics. Each unit’s front cover states this core idea as part of the unit’s title. A two-page chapter overview provides a compact summary of that chapter’s contents to give you the big picture before you get into the details and later when you review. Sidebars in the margins help clarify the purpose of sections of the main text at the subpage level and can help you quickly locate items later. I have highlighted technical terms in bold type (like this) when they first appear: their definitions usually appear nearby. A physics formula consists of both a mathematical equation and a conceptual frame that gives the equation physical meaning. The most important formulas in this book (typically, those that might be relevant outside the current chapter) appear in formula boxes, which state the equation, its purpose (which describes the formula’s meaning), a description of any limitations on the formula’s applicability, and (optionally) some other u ­ seful notes. Treat everything in a box as a unit to be remembered and used together.

What is active reading?



Active Reading

Just as passively listening to a lecture does not help you really learn what you need to know about physics, you will not learn what you need by simply

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scanning your eyes over the page. Active reading is a crucial study skill for all kinds of technical literature. An active reader stops to pose internal questions such as these: Does this make sense? Is this consistent with my experience? Do I see how I might be able to use this idea? This text provides two important tools to make this process easier. Use the wide margins to (1) record questions that arise as you read (so you can be sure to get them answered) and the answers you eventually receive, (2)  flag important passages, (3) fill in missing mathematical steps, and (4)  record insights. Writing in the margins will help keep you actively engaged as you read and supplement the sidebars when you review. Each chapter contains three or four in-text exercises, which prompt you to develop the habit of thinking as you read (and also give you a break!). These exercises sometimes prompt you to fill in a crucial mathematical detail but often test whether you can apply what you are reading to realistic situations. When you encounter such an exercise, stop and try to work it out. When you are done (or after about 5 minutes or so), look at the answers at the end of the chapter for some immediate feedback. Doing these exercises is one of the more important things you can do to become an active reader. SmartBook (TM) further supports active reading by continuously measuring what a student knows and presenting questions to help keep students engaged while acquiring new knowledge and reinforcing prior learning.



Features that support developing the habit of active reading

Class Activities and Homework

This book’s entire purpose is to give you the background you need to do the kinds of practice activities (both in class and as homework) that you need to genuinely learn the material. It is therefore ESSENTIAL that you read every assignment BEFORE you come to class. This is crucial in a course based on this text (and probably more so than in previous science classes you have taken). The homework problems at the end of each chapter provide for different kinds of practice experiences. Two-minute problems are short conceptual problems that provide practice in extracting the implications of what you have read. Basic Skills problems offer practice in straightforward applications of important formulas. Both can serve as the basis for classroom activities: the letters on the book’s back cover help you communicate the answer to a two-minute problem to your professor (simply point to the letter!). ­Modeling problems give you practice in constructing coherent mental models of physical situations, and usually require combining several formulas to get an answer. Derivation problems give you practice in mathematically extracting useful consequences of formulas. Rich-context problems are like modeling problems, but with elements that make them more like realistic questions that you might actually encounter in life or work. They are especially suitable for collaborative work. Advanced problems challenge advanced students with questions that involve more subtle reasoning and/or difficult math. Note that this text contains perhaps fewer examples than you would like. This is because the goal is to teach you to flexibly reason from basic principles, not slavishly copy examples. You may find this hard at first, but real life does not present its puzzles neatly wrapped up as textbook examples. With practice, you will find your power to deal successfully with realistic, practical problems will grow until you yourself are astonished at how what had seemed impossible is now easy. But it does take practice, so work hard and be hopeful!

Read the text BEFORE class!

Types of practice activities ­provided in the text

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C1

The Art of Model Building Chapter Overview Section C1.1:  The Nature of Science One of the main goals of science is the development of imaginative conceptual ­models of physical reality. A model deliberately simplifies a complex reality in such a way that it captures its essence and helps us think more clearly about it. This text’s main purpose is to teach you the art of scientific model building, by helping you not only understand and appreciate the grand models we call theories but also practice building the small-scale models one needs to apply a theory in a given situation. Science is an unusually effective process for generating powerful models of reality that involves four crucial elements coming together: 1. 2. 3. 4.

A sufficiently large community of scholars, who share A commitment to logical consistency as an essential feature of all models, An agreement to use reproducible experiments to test models, and A grand theory rich enough to provide a solid foundation for research.

In the case of physics, the Greek philosophical tradition created a community that valued logical reasoning. Early Renaissance thinkers championed the value of reproducible experiments as being crucial for testing models. But physics was not really launched until 1687, when Newton provided a theory of mechanics grand and compelling enough to unify the community and provide a solid context for research.

Section C1.2:  The Development and Structure of Physics Since the days of Newton, physicists have sought to create models able to embrace originally distinct areas of study and thus cover broader ranges of physical phenomena. The current conceptual structure of physics, illustrated in figure C1.1, rests on two grand theories: general relativity (GR) and the Standard Model (SM) of parti­ cle physics. In practice, though, physicists almost always use five simpler theories (which are approximations valid in various limited contexts): newtonian mechanics, special relativity, electromagnetic field theory, quantum mechanics, and statistical mechanics. This text focuses on these five models. Physicists have recently come to appreciate the role that symmetries play in ­physics. Both GR and the SM acknowledge (as almost any imaginable theory must) certain symmetries (such as the time and position independence of physical laws) that give rise to conservation laws (such as the laws of conservation of energy and ­momentum). Such laws have a validity beyond the specific theories currently in vogue. Indeed, GR and the SM themselves are based on new, nonobvious symmetries.

Section C1.3:  A Model-Building Example To apply a grand theoretical model to any actual physical situation of interest, a scientist must construct an idealized model that simplifies the situation, bringing its essence into focus in such a way that one can easily connect it to the grand model. This course is designed to help you practice this process, which is really the only way to learn how to do it. This section illustrates what is involved in an example situation. In the process, the section describes some useful tricks that can help you simplify situations and think about which simplifications are appropriate:

2

1. Lines or rays from a very distant point are nearly parallel. 2. The length of a gentle curve between two points is almost the same as that of a straight line between those points. 3. The fractional uncertainty of a result calculated by multiplication or division from uncertain quantities is roughly equal to that of the most uncertain quantity involved. The same is true for the sine or tangent of small angles. Part of the art of model building is to develop a bag of such tricks that you can pull out when helpful. The only real way to learn these tricks is by practice, and also by making mistakes that you learn to correct. So be bold and learn from your ­mistakes! Solutions to most physics problems involve three different sections: 1. A model section that describes the simplifications one makes to the situation 2. A math section where one does the mathematics implied by the model 3. A check section where one decides whether the result makes sense Your earlier experience with more trivial problems may lead you to neglect the model and check sections, but I strongly recommend you do not. The model section is particularly important in this course. A good and sufficiently well-labeled diagram is often the core of a sufficient model for problem solutions you prepare.

Section C1.4:  Trick Bag: Unit Awareness One of the most powerful tricks you can put in your bag is being aware of units. Units give meaning to quantities and are essential for correctly communicating that meaning to others. Being constantly aware of units (even when working with symbolic equations) is one of the best and easiest ways to spot mistakes in your work. Here is a list of the things you should know to increase your unit awareness: 1. 2. 3. 4.

Know the basic and derived SI units and SI prefixes (see the inside front cover). Know and/or refer to the SI unit benchmarks in figure C1.2. Know that the units on both sides of an equation must match. Know that you cannot add or subtract quantities with different units, but you can multiply or divide them. 5. Know that you should be aware of units even in symbolic equations. 6. Know that math functions take unitless arguments and yield unitless results.

Section C1.5:  Trick Bag: Unit Conversions In most physics problem solutions, you will need to convert units. My preferred technique for doing this is the unit operator method, where you convert unit equalities such as 1 mile = 1609 meters into a ratio equal to 1 such as 1 = (1 mi/1609 m) or 1 = (1609 m/1 mi). Since anything can be multiplied by 1 without changing it, you can multiply any quantity by such a unit operator and cancel units top and bottom (as if they were algebraic symbols) until only the units you want are left over. For ­example: 23 mi = (23 mi​ ​ ) (1609 m/1 mi​ ​ )  = 37,000 m. This method is foolproof as long as @ @ you pay attention to unit consistency and to canceling units correctly.

Section C1.6:  Trick Bag: Dimensional Analysis Dimensional analysis is a surprisingly powerful trick that often yields good estimates of physics formulas and/or quantities without requiring anything more than the most basic knowledge of a situation. As such, it often represents the simplest model you can construct of a given situation. This trick takes advantage of the facts that (1) units must agree on both sides of any equation, (2) that most formulas in physics are simple power laws, and (3) that most unitless constants appearing in such formulas are within a factor of 10 or so of one. The steps in applying dimensional analysis to a situation are as follows: 1. 2. 3. 4.

Decide what quantities your desired value might depend on. Assume that these quantities appear in a power law formula (e.g., Q = KAmBn). Find the powers by requiring units to be consistent on both sides of the formula. Assume that the unitless constant K in front of the equation is 1.

3

4

Chapter C1

C1.1

Model building occurs at all levels of science

The Nature of Science

By our nature, we humans strive to discern order in the cosmos and love to tell stories that use ideas from our collective experience to “explain” what we see. Science stands firmly in this ancient tradition: stories about how the gods guide the planets around the sky and modern stories about how spacetime curvature does the same have much in common. What distinguishes science from the rest of the human storytelling tradition are (1) the types of stories scientists tell, (2) the process that they use for developing and sifting these stories, and (3) the predictive success enjoyed by the surviving stories. Scientists express their stories in the form of conceptual models, which bear a similar relation to the real world as a model airplane does to a real jet. A good scientific model captures a phenomenon’s essence while being small and simple enough for a human mind to grasp. Models are essential because reality is too complicated to understand fully; models distill complex phenomena into bite-sized chunks that finite minds can digest. Framing a model is less an act of discovery than of imagination: a good model is a compelling story about reality that creatively ignores just the right amount of complexity. Model-making in science happens at all levels. Theories—grand models embracing a huge range of phenomena—are for science what great novels are for literature: soaring works of imagination that we study and celebrate for their insight. But applying such a grand model to a real-life situation requires building a smaller model of the situation itself, simplifying the situation and making appropriate approximations to help us connect it to the grand model. Scientists do this second kind of model-making daily, and one of the main goals of this course is to help you learn that art. Because models are necessarily and consciously simpler than reality, all have limits: the full “truth” about any phenomenon can never be told. Pushing any model far enough eventually exposes its inadequacies. Even so, one can distinguish better from poorer models. Better models are more logical, more predictive in a broader range of cases, more elegantly constructed, and more productive in generating further research than poorer ones are. Science is really a process for building, evaluating, and refining models, one that (since its beginnings in the 1600s) has proved to be an astonishingly prolific producer of powerful and trustworthy models. It owes part of its success to its focus on the natural world, whose orderly behavior at many levels makes finding and testing models easier than in the world of human culture. Scholars of the philosophy and history of science suggest that a discipline becomes a science only when the following four elements come together: 1. 2. 3. 4.

How physics became a science

The Art of Model Building

A sufficiently large community of scholars, who share A commitment to logical consistency as an essential feature of all models, An agreement to use reproducible experiments to test models, and A grand theory rich enough to provide a solid foundation for research.

In the case of physics, the Greek philosophical tradition founded a community of scholars who appreciated the power of logical reasoning: indeed, this community found logic’s power so liberating that it long imagined pure logic to be sufficient for knowing. The idea of using experiments to test one’s logic and assumptions was not even fully expressed until the 13th century, and was not recognized as necessary until the 17th. Eventually, though, the community recognized that the human desire to order experience is so strong that the core challenge facing a thinker is to distinguish real order from merely imagined patterns. Reproducible experiments make what would otherwise be individual experience available to a wider community, anchoring models more firmly to reality. Galileo Galilei (1564–1642) was a great champion of

C1.2 The Development and Structure of Physics

this approach. His use of the newly invented telescope to display features of heavenly bodies unanticipated by models of the time underlined to his peers the inadequacy of pure reason and the importance of observation. A prescientific community lacking a grand theory, however, tends to fragment into schools, each championing its own theory. Rapid progress is thwarted because each school sees any collected data through the lens of its cherished model, making arguments virtually impossible to resolve. This was the situation in physics during most of the 1600s. However, in 1687 Isaac Newton published an ingenious model of physics broad enough to embrace both terrestrial and celestial phenomena. His grand theory captured the imagination of the entire physics community, which turned away from a­ rguing about partial models and toward working together to refine, test, and extend Newton’s basic theory, confident that it would be shown to be universally true and valid. At this moment, physics became a science. The unified community now made rapid progress in constructing powerful subordinate models that greatly extended the reach of Newton’s grand vision, feeding the Industrial Revolution along the way. Ironically, the community that strove energetically to extend Newton’s model universally eventually amassed evidence proving it incomplete! Only a community devoted to a ­theory can collect the kind of detailed and careful evidence necessary to expose its inadequacies, and thus move on to better theories. This irony is the engine that drives science forward.

C1.2

5

The ironic paradox at the heart of science

The Development and Structure of Physics

Unification of apparently distinct models has been an important theme in the development of physics since Newton’s theory unified terrestrial and celestial physics. In the 1800s, work on electricity, magnetism, and light (initially described by distinct partial models) culminated in an “electromagnetic field model” embracing them all, and physicists found how to subsume thermal phenomena into Newton’s model. This process was going so well in the late 1800s that the physicist Lord Kelvin famously claimed that there was probably little left to learn about physics! In the early 1900s, though, physicists began to see that certain experimental results were simply incompatible with Newton’s framework. After what amounted to a period of revolution, the community demoted Newton’s theory and coalesced around two new grand theories—general ­relativity (1915) and quantum mechanics (1926)—which embraced the new results but yielded the same results as Newton’s theory in the appropriate limits. In the 1950s, physicists were able to unify quantum mechanics, electromagnetic field theory, and special relativity (the nongravitational part of general relativity) to create quantum electrodynamics (QED), the first example of a relativistic quantum field theory. In the 1970s, physicists extended this model to create relativistic quantum field theories to describe two new ­(subatomic-scale) interactions discovered in previous decades and integrated them with QED into a coherent theory of subatomic particle physics called the Standard Model. This model has been quite successful, predicting new phenomena and particles that have been subsequently observed. The m ­ odel’s latest triumph was the discovery of the predicted “Higgs boson” in 2012. Currently, general relativity, which covers gravity and other physical phenomena at distance scales larger than molecules, and the Standard ­Model, which works in principle at all distance scales but does not and cannot cover gravity, stand as the squabbling twin grand theories of physics. Though no known experimental result defies explanation by one or the other, physicists

The history of physics since Newton

The current structure of physics

Chapter C1

6

The Art of Model Building

Deep underlying SYMMETRIES in the nature of physical interactions

CONSERVATION LAWS **THE STANDARD MODEL** (interwoven relativistic quantum field theories for everything but gravity)

(Classical) ELECTROMAGNETIC FIELD THEORY (approximately valid for weak fields and large distances)

**GENERAL RELATIVITY** (describes gravity and the motion of objects bigger than molecules)

QUANTUM MECHANICS (approximately valid at low particle energies)

STATISTICAL MECHANICS (describes complex objects constructed of many interacting particles)

SPECIAL RELATIVITY (approximately valid in the presence of weak gravitational fields)

NEWTONIAN MECHANICS (approximately valid for massive objects moving much slower than light in weak gravitational fields)

Figure C1.1 The current grand theories of physics (starred) and the five approximate models more often used in practice.

The importance of symmetries in physics

are dissatisfied with each theory (for different reasons) and especially distressed that we need two deeply incompatible theories instead of one. While many unifying models have been proposed (string theory and loop quantum gravity are examples), these models lack both the level of development and the firm experimental basis to inspire general acceptance. The physics community is thus presently in the curious position of being devoted to two grand theories we already know to be wrong (or at best incomplete). In practice, however, physicists rarely use either to explain any but the most exotic phenomena. Instead, they use one of five simpler theories: newtonian mechanics, special relativity, electromagnetic field theory, quantum mechanics, and statistical mechanics. Each has a more limited range of ­applicability than the two grand theories, but is typically much easier to use within that range. These theories, their limitations, and their relationships to the grand theories and each other are illustrated in figure C1.1. This diagram also emphasizes the importance of symmetry principles in physics. Early in the 1900s, mathematician Emmy Noether showed that, given plausible assumptions about the form that physical laws must have, a symmetry principle stating that “the laws of physics are unaffected if you do such-and-such” automatically implies an associated conservation law. For example, the time-independence of the laws of physics (whatever those laws might be) implies that a quantity that we call energy is conserved (that is, does not change in time) in an isolated set of objects obeying those laws. Conservation laws, therefore, stand independently and behind the ­particular models of physics, as figure C1.1 illustrates. For example, conservation of energy is a feature of newtonian mechanics, electromagnetic field theory, special relativity, quantum mechanics, and statistical mechanics

C1.3 A Model-Building Example

­because all these theories involve physical laws that (1) have forms consistent with Noether’s theorem and (2) are assumed to be time independent. Each theory has a different way of defining energy, but all agree that it is conserved. Symmetry principles are also important because both our current grand models of physics (the Standard Model and general relativity) propose and unravel the consequences of new and nonobvious symmetry principles. The section of this text on special relativity illustrates this by displaying how relativity’s mind-blowing features are in fact simple logical consequences of the symmetry principle that “the laws of physics are unaffected by one’s state of (uniform) motion.” Linking other symmetry principles with their consequences is (unfortunately) not quite so simple (and is beyond the level of this course), but is not qualitatively different. Now, given the structure of physics illustrated in figure C1.1, it might seem logical to begin studying physics by starting with the two fundamental theories (or even the symmetry principles) and then working downward to the five approximate theories. However, this is impractical because the fundamental theories, in spite of their awesome breadth and beauty, are (1) very sophisticated mathematically and conceptually, (2) unnecessarily complicated to use in most contexts, and (3) necessarily expressed using the language and concepts of the five simpler theories. One must therefore start by learning those simpler theories. The other five volumes of this textbook will provide you with a very basic introduction to all five of these simpler theories, as well as exploring many of the supporting models that help broaden their range. This unit begins the process by looking at the conservation laws (in the context of newtonian mechanics) that underlie all these theories. In the remainder of this chapter, though, we will explore the modelbuilding process in more detail and develop some general tools that help us avoid errors and maximize what we can gain from even limited knowledge.

C1.3

7

Why one must learn the five simpler theories first

A Model-Building Example

This book is designed partly to teach you the kind of creative model-building that working scientists do daily. The model-building process cannot be reduced to formulaic procedures that one can follow like a recipe. It is an art that requires knowledge, intelligence, creativity, and most of all, practice. You can no more learn this art simply by reading books or attending lectures than you can learn to play the piano simply by attending concerts. So let’s practice! The exercise below poses a simple question you can answer using some basic trigonometry and geometry, grade-school science, and a bit of creative model building. Spend at least 10 minutes but no more than 15 minutes trying to answer the question before turning the page.

Exercise C1X.1 About 240 B.C.E., Eratosthenes made the first good estimate of the earth’s size as follows. Caravan travelers told him that in the village of Syene, one could see the sun reflected in a deep well at noon on the summer solstice, meaning that it was directly overhead. Eratosthenes noted that at the same time on the same day in Alexandria (5000 Greek stadia to the north, as estimated by camel travel time), a vertical stick cast a shadow about 1/8 of its length. What is the earth’s radius in stadia? (Hint: Draw a picture. In 240 B.C.E., the Greeks knew that the earth was spherical and the sun was very far away.)

One learns the art of model building through practice

An example that illustrates the model-building process

Chapter C1

8

Example C1.1

The Art of Model Building

Model  If the sun is sufficiently far from the earth, light rays traveling from the sun to Alexandria (point A) and Syene (point S) will be almost parallel. Let’s assume they are exactly parallel and that the earth is perfectly spherical. The figure below shows a cross-sectional view of the situation from the east. θ

CLOSE-UP:

earth

shadow

stick parallel rays from the sun

stick

A S

d

θ

θ

C

r

earth’s circumference =c

Because alternate interior angles are equal, the angle θ between the stick and the sun’s rays at Alexandria is the same as the angle θ between lines AC and SC, where C is the earth’s center. The stick is vertical by assumption, so it is perpendicular to its shadow. Thus, the stick, the shadow, and the ray passing the stick’s end form a right triangle (see the “close-up” above). If the shadow is 1/8 the stick’s length, then tan θ = 1/8, so θ = tan-1(1/8) = 7.1°. The distance d between Alexandria and Syene is to the earth’s circumference c as θ is to 360°, so c/d = 360°/θ. Note also that c = 2πr, where r is the earth’s radius. Math  Therefore

(  )

(5000 stadia)(360°) r = ___ ​  c   ​ = ___ ​  d   ​ ​ ____ ​  360°  ​    ​ = _________________    ​      ​ = 40,000 stadia 2π 2π θ 2π(7.1°)

(C1.1)

Check  The exact length of Eratosthene’s stadion is historically ambiguous, but if he meant the “itinerary stadion” (the one used for road trips), then 1 stadion = 0.157 km and r is 6300 km, pretty close to the modern value.

If you got something like this, congratulations! If you had trouble getting a useful result in 15 minutes, that’s normal. Doing a moderately realistic problem like this is hard, not usually because the math or concepts are hard (both are pretty basic here), but because constructing the model is hard. How does one know what approximations to make? How does one create a schematic diagram of a situation (like the one shown) that usefully exposes its essential features? How do you frame things so that the mathematics is simple? You may even be annoyed with my solution: “Well,” you might say, “if I had known it was acceptable to make the false assumption that rays from the sun are parallel, then the solution would have been easy!” That is precisely the point! It is not only acceptable but also usually necessary to make simplifications to solve a problem at all. The trick is simplifying just enough to make the problem tractable without making the result uselessly crude. There are no “correct” answers in such a case, only poorer and better models that yield poorer or better results (and if a poor result is the best one can do, it is still better than nothing!). This is where the creativity and artistry comes in. My goal is to help you learn to simplify (that is, to be productively and creatively lazy) imaginatively, boldly, and exuberantly!

C1.3 A Model-Building Example

With this in mind, let’s examine more closely the simplifications and ­assumptions behind the model in example C1.1. The solution assumes that the sun is sufficiently distant that light rays from it are parallel at the earth: if this is not so, the two angles marked θ in the diagram are not equal. No one knew the distance to the sun in Eratosthenes’s time, so his assumption was quite bold, but we now know that two rays from a single point on the sun that ­arrive at Syene and Alexandria, respectively, are not parallel, but ­actually make an angle of about 0.00035° with each other. Solving this p ­ roblem “more correctly” by taking this into account yields an r that is smaller by about 0.005%. However, this is truly insignificant compared with other simplifications we are making. The sun’s angular diameter when viewed from the earth is about 0.5°, so there is not just one ray that grazes the top of the stick and ­connects it with the shadow on the ground, but rather a bundle of rays that could make angles with each other of as much as 0.53°. This means that the end of the stick’s shadow will be blurred, making its length and thus the angle θ uncertain by about ±0.26°. Also the hills and valleys between Syene and ­Alexandria make the road distance d longer than the distance that would be measured on a perfect sphere. Moreover, Alexandria is not due north of Syene, as the drawing assumes it is. The earth is also not exactly spherical (its polar radius is smaller than its equatorial radius by about 11 km). I could state yet more subtle assumptions, but I think you get the point. Reality is complicated, and the model simply ignores those complications. Now, it turns out if you multiply or divide uncertain or erroneous quantities, the result has (roughly) the same percent uncertainty as the most uncertain of the quantities. This weakest-link rule is also (roughly) true for t­angents or sines of small angles. (Check it out for yourself: see problem C1D.1.) In this case, the uncertainty in θ is roughly ±4% (±0.26°/7.1°) because of the angular width of the sun, and the uncertainty in the distance d is likely to be more than ±10%, since it is determined by camel travel time! The weakestlink rule implies, therefore, that we are not going to know the radius of the earth to better than about ±10% no matter how good our model is. Making a far more complicated model to correct the approximations described above is not ­going to make the slightest bit of practical difference: we are simply not given good enough information to calculate the earth’s radius more precisely. The problem (as stated) therefore does not deserve a better model! Part of the art of model building is knowing when a model is “good enough.” Eratosthenes’s model was not merely “good enough;” it was pure genius at the time, since no other method of determining the earth’s radius was remotely as good. (Sometimes even a crude result is a big step forward!) One learns the art of “good enough” mostly by practice. Indeed, I hope this course will give you (among other things) a bag of useful tricks that are often “good enough.” Treating lines from a distant point as parallel is one such trick. The weakest-link rule about the uncertainty of multiplied or divided quantities is another. Practice with tricks like these puts them into your bag. Another important trick is recognizing the importance of a good diagram. Drawing (and carefully labeling) the drawing in example C1.1 was probably the single most important thing I did to solve the problem. Most of the “Model” in my solution merely restates the diagram verbally. A good diagram is often the most important trick for solving a physics problem. Indeed, solutions to all but the most trivial problems will involve the three sections appearing in the example solution: a model section where one draws a schematic diagram and/or discusses approximations and assumptions, a math section where one does the mathematics implied by the model to solve for the desired result, and a check section where one checks the result to see if it makes sense.

9

The simplifications and ­assumptions involved in ­example C1.1’s model

The “weakest link rule” for uncertain quantities

The art of “good enough”

Good diagrams are essential

The three sections of almost any physics problem solution

10

Chapter C1

Don’t neglect either the “check” or “model” parts

Beginners often neglect the “check” section, but you will learn by ­experience not to (if only to avoid submitting embarrassingly knuckle-headed results). However, in this course (since it is about practicing model-building), the model section is the most crucial part of any solution you submit. A model need not be much more than a well-labeled diagram and/or a few comments about assumptions. I will provide some examples to emulate as we go on. Note that making mistakes (and then correcting them) is an honorable part of the learning process. Werner Heisenberg, a great physicist of the early 1900s, said that “An expert is [someone] who knows… the worst errors that can be made in a subject… and [thus] how to avoid them.” This is true in my experience, and is something one can often only learn by making those ­mistakes.

The value of making mistakes

C1.4 Why unit awareness is important

SI units and prefixes

STANDARD SI PREFIXES Power

Prefix

Symbol

18

exa

E

15

peta

P

12

tera

T

9

giga

G

6

mega

M

10 10 10

10 10

3

10

kilo

k

10–2

centi

c

–3

milli

m

–6

micro

µ

–9

nano

n

–12

pico

p

–15

femto

f

atto

a

10 10 10 10 10

10–18 Table C1.1

Standard SI prefixes for powers of 10. (You can also find this table on the inside front cover.)

The Art of Model Building

Trick Bag: Unit Awareness

An essential item for your bag of tricks is being aware of units. Units attach physical meaning to bare numbers and communicating the magnitude of a measured quantity requires using an agreed-upon unit for that quantity. The importance of this was starkly illustrated in 1999 when NASA’s $125-million Mars Climate Orbiter burned up in Mars’s atmosphere because the spacecraft’s builders had been sending thruster data in English units (pounds) to a NASA navigation team expecting data in metric units (newtons). One commentator to the Los Angeles Times stated, “[This] is going to be a cautionary tale to the end of time.” (He got that right.) To help avoid such catastrophic confusion, in this text I will most o ­ ften use SI units (from the French Système Internationale), the modern version of ­metric units. An international committee has worked since 1960 to provide clear and reproducible definitions for standard physical units. The entire system is based on seven base units, each with a standard ­abbreviation: the ­meter (abbreviation: m) for distance, the second (s) for time, the ­kilogram (kg) for mass, the kelvin (K) for temperature, the mole (mol) for counting molecules, the ampere (A) for electric current, and the candela (cd) for luminous intensity (we will not use the candela in this course). The committee has defined each unit (except the kilogram) in such a way that a scientist can in principle recreate the unit in his or her own laboratory. The SI committee has also defined derived units that are combinations of the base units. The units we will use are the joule (1 J ≡ 1 kg·m2/s2) for energy, the watt (1 W ≡ 1 J/s) for power, the newton (1 N ≡ 1 kg·m/s2) for force, the pascal (1 Pa ≡ 1 N/m2) for pressure, the coulomb (1 C ≡ 1 A·s) for electric charge, the volt (1 V ≡ 1 J/C) for electrical energy per unit charge, the ohm (1 Ω ≡ 1 V/A) for electrical resistance, and the hertz (1 Hz ≡ 1 wave cycle/s). Future chapters will describe what these derived units mean. The SI committee has also defined a set of standard prefixes and prefix abbreviations (see table C1.1) that one attaches to a unit to multiply it by ­selected powers of 10. Thus, 1 millimeter (abbreviation: mm) is equal to 10–3  m, 1 gigawatt (GW, pronounced with the g as in “get”) is 109 W, and 1 nanosecond (ns) = 10–9 s. You may already be familiar with some of the larger prefixes from computer terminology (e.g., TB for terabyte = 1012 bytes, GHz for gigahertz = 109 hertz). My experience is that English-speakers have the most trouble distinguishing milli (= 10–3) from micro (= 10–6) because “milli” sounds like “millionth” even though it means “thousandth.” I recommend memorizing the prefixes at least from pico to tera. Units for angles (the radian and the degree) are standard but (for historical reasons) are not considered formal SI units. I will sometimes also mention English units like the mile (mi), foot (ft), and pound (lb) (a unit of force).

C1.5 Trick Bag: Unit Conversions

11

age of universe = 13.7 Gy = 433 Ps

1054 kg

1012 s

31,700 yr

1042 kg

109 s

31.7 yr

1036 kg

106 s

11.6 days

1030 kg

the sun

103 s

16.7 min

1024 kg

the earth

edge of visible universe ≈1026 m

1018 s

1021 m

galaxy diameter

1018 m

nearer stars

1027 m 1024 m

1015 m 1012 m 109 m 106 m

solar system diameter earth diameter = 12,760 km

1015 s

1s 10–3 s

103 m

10–6 s

1m

10–9 s

1 m = 3.28 ft

10–12 s

10–3 m

period of a typical sound wave light travels one foot

1012 kg

a hill

106 kg

≈ 1000 tons

1 kg

1 kg = 2.2 lb

10–6 kg

a mosquito

10–12 kg

a cell

10–6 m

small cell

10–15 s

10–9 m

atom diameter

10–18 s

light crosses a molecule

10–18 kg

light crosses a nucleus

10–30 kg

10–21 s

10–15 m

nucleus diameter

10–24 s

our galaxy

1018 kg

period of a typical light wave

10–12 m

visible universe

1048 kg

10–24 kg

an atom an electron

Figure C1.2 Some rough benchmarks for distances in meters, times in seconds, and masses in kilograms. (Note that the scales here are logarithmic: equal distances on the scale correspond to multiplication by equal powers of ten.)

Being aware of units is also valuable even when working with symbolic equations. One cannot add or subtract quantities with different units (though multiplying and dividing such quantities are fine). So an expression that (for example) contains 1 + m (where m is a mass) is absurd. If you find yourself writing such an equation, you should know that you have made a mistake. Also, for the record, all mathematical functions [such as sin(x), tan–1(x), ex, ln x] all require a unitless argument and produce a unitless result. For historical reasons, angles in both degrees and radians are considered unitless. One way to spot silly results is to know some SI benchmarks (see ­figure C1.2). For example, if a question asks you how high you can throw a ball and you get a distance larger than a galaxy, maybe something is wrong, hmm? Check figure C1.2 if a calculation’s results seem off. It is also cool to be able to say to your friends at the end of the school year “See you in about 9 megaseconds” and have them understand. (OK, maybe that is just a bit geeky.)

C1.5

Know some benchmarks!

Trick Bag: Unit Conversions

In many realistic physics problems, you are likely to have to change units. One of the most handy tools in your bag of tricks is a foolproof way to convert units. My favorite technique is the unit operator method. For example, suppose that you know that your hair grows at a rate of 6 inches per year and you’d like to know what this is in nanometers per second. We start by writing down the equations that define the relationships between the units (which you can find inside the front cover):

Being aware of units in symbolic expressions is also good

1 nm = 10-9 m, 1 m = 3.28 ft, 1 ft = 12 in, 1 y = 3.16 × 107 s

(C1.2)

The “unit operator” method for unit conversion

12

Chapter C1

The Art of Model Building

Since 1 m is equivalent to 3.28 ft, the ratio of these quantities is 1:

(  )

( 

) (  ) ( 

)

1y 1 ft 1 m 1 nm   ​ ______ ​    ​   ​ = 1  and similarly,  1 = ​ ______ ​  -9    ​   ​ = ​ _____ ​    ​  ​= ​ ___________ ​     ​   ​ (C1.3) 3.17 × 107 s 12 in 10 m 3.28 ft Such a ratio is thus called a unit operator. Since we can multiply anything by 1 without changing it, we can multiply our original rate by 1 in the form of these unit operators. If you then rearrange factors and cancel the units that appear in both the numerator and denominator, we get the desired result:

( 

)

(  ) (  ) (  ) ( 

) ( 

)

1 y​ ​│  1 nm ___________ 6 in​ ​   _____ 6 in 1 ft​ ​⨉  ______ 1 ​ m ​ _______ ____ ​ ​   ​    ​  ·1·1·1·1 = ​ ____ ​    ​    ​ ​ ​    ​  ​ ​ ​    ​   ​ ​ ​  -9   ​   ​ ​ ​     ​   ​ ​│  12  y​ y ​⨉  10 ​m ​ 3.16 × ​10​7​ s ​   3.28 ft​ in​

Why this method is so great

(2) Assume a power-law formula

⧹ 

6 = ______________________ ​         ​ ____ ​ nm  ​  = 4.8 ____ ​ nm  ​    s 12(3.28)(10-9)(3.16 × 107) s

(C1.4)

Note that we can treat all of the units as if they were simply algebraic symbols that we can cancel if they appear in both the numerator and denominator! This unit operator method is great because if I didn’t have the right ­power for one of the unit operators or put it upside down by accident, the units would not cancel and I would have a mess of leftover units that would signal that I had done something wrong. This method is absolutely foolproof as long as you make sure that unwanted units really do cancel out. (Foolproof is good!) How do I know when a unit operator is “right side up?” [After all, 1 = (1 m/3.28 ft) and 1 = (3.28 ft/1 m) both!] “Right side up” for a unit operator is whichever way gets the units to cancel as you need. It’s like the joke where a student asks a sculptor how to make a good statue of a person. The sculptor replies, “Just remove whatever parts of the rock don’t look like the person.” In the unit conversion task, just arrange your unit operators so that they ­remove any units that don’t look like the final units you need.

C1.6

(1) Think about what the desired quantity might depend on

⧹ 

Trick Bag: Dimensional Analysis

Dimensional analysis is a very cool trick because you can often use it to estimate an answer or guess a formula even if you know almost nothing about the physics involved. This will amaze your friends (convincing them that you know more than you do) and is a sign to other physicists that you belong. The trick is based on the fact that the units on both sides of any equation must be consistent. We can often combine this with simple plausibility arguments to find a formula for a desired quantity even if we haven’t a clue about how to actually derive that formula. This often represents the simplest model one can construct (one based on only the most fundamental assumptions). Consider the following problem as an example. The radius R of a black hole is the radius inside which all light is trapped by the black hole’s gravitational field. What is this radius for a black hole with a given mass M? Don’t panic just because we don’t know any general relativity! Think: what could R depend on? It could plausibly depend on the black hole’s mass M, the speed of light c, and the universal gravitational constant G that characterizes the strength of the gravitational field created by a given mass (we will study this constant more later). No other physical quantity appears to be relevant. So let’s assume that only these quantities will appear in the formula for R. This is the first assumption in our model. Secondly, let’s assume that the formula has the form

R = KG j ​M​k​ ​cn​ ​

(C1.5)

C1.6 Trick Bag: Dimensional Analysis

13

where K is some as-yet-unknown unitless constant and j, k, and n are as-yetunknown powers (not necessarily integers). This is a big assumption, but not completely bonkers. For example, as the black hole’s mass M goes to zero, we would expect the trapping radius R goes to zero. Similarly, if G ​ ​  W 0 (meaning that the gravitational field created by a given mass goes to zero), R should also go to zero. Both expectations are consistent with the formula. Let {a} represent the SI units of the quantity a. Thus, {R} = m, {M} = kg, {c} = m/s (since c is a speed), and {G} = m3/(kg·s2) (see the inside front c­ over). So if the units in equation C1.5 are to be consistent, we must have

(3) Require unit consistency

(  ) (  )

3  j n  m = {R} = {K}{G}j {M}k {c}n = ​​ _____ ​  m  2 ​  ​​ ​ (kg)k ​​ __ ​  m ​   ​​ ​= m3j+nkgk-j s-2j-n (C1.6) s kg·s

Since we do not have any units of kilograms on the left side of equation C1.6, we must have k – j = 0, or k = j. Since we also do not have units of seconds on the left, we must have –n – 2j  = 0, or n = –2j. We have one power of meters on the left side, so 3j + n = 1 or (substituting n = –2j from above) 3j – 2j = 1, so j = 1. So our formula must be GM ​   R = KG  jM kc n = KG1M1c-2 = K ​ ____   (C1.7) c2 The constant K we have included (for greater generality) does not have any units, so this method cannot determine its value. If we assume that K = 1, then the radius of a black hole with a mass equal to that of the sun is

( 

)

1.99 × 1030 kg​ ​  m​3​  ​  ​ _______________ -11 _____ R = ____ ​ GM  ​     = ​ 6 .67 × 10 ​  ​         ​ = 1470 m 2 2 8 c ​ · ​s⨉​  (3.00 × 10 m​ kg​  ​@/   s​ ​ )  2 ⨉ @

 

(C1.8)

Therefore, this super-simple model predicts that if our sun were to become a black hole, its radius would be about 1.5 km. Now you might think this result completely untrustworthy because of all the assumptions we have made. Even granting the plausibility of the ­argument that the result must depend only on G, M, and c, what justifies the outrageous assumptions we made in equation C1.5 and in setting K = 1? Such assumptions turn out to be surprisingly trustworthy. The universe seems to prefer simple formulas to complicated ones, and unitless constants appearing in physics formulas usually turn out to be like 1/3, π, or 5/8, not something very large or very small. Therefore taking the unitless constant to be equal to 1 has an uncannily good chance of yielding a reasonable estimate (within a factor of 10 or so). It doesn’t always work, but it often does, and when it does, one can get a good first estimate with spectacularly little effort. In this particular case, the full derivation using general relativity shows that a black hole’s radius is really R = 2GM/c2. Thus, we got everything absolutely right except that K = 2, not 1. Not bad for such a simple model! For historical reasons, this method is called dimensional analysis (long ago, but not today, a quantity’s units were called its “dimensions”). Now you practice. (After making a good effort, check the chapter’s end.)

Exercise C1X.2 When the core of a massive star exceeds about 1.4 solar masses near the end of its life, reactions in its interior suddenly remove the very particles that have been supporting it against its own gravitational field. What is left of the core then falls basically freely inward from rest at a radius of about 10,000 km to a final radius that is negligible in comparison. (This violent collapse ­usually ignites a supernova explosion). How long does this collapse process take?

(4) Assume that any unitless constant is approximately 1

14

Chapter C1

The Art of Model Building

TWO-MINUTE PROBLEMS C1T.1 According to the definition of “science” given in this chapter, astrology is not a science. What does it lack? A. A community of scholars devoted to its study B. Agreement that models must be logically consistent C. Use of reproducible experiments to test models D. A grand theory embracing the discipline C1T.2 According to the definition of “science” given in this chapter, which of the following do you think are sciences? Choose the letter of the first discipline on the list that you think is not a science. (The answer is open to debate!) A. Geology B. Psychology C. Economics D. Anthropology E. Political Science F. Philosophy T. All are sciences C1T.3 Which of the following expressions gives the correct units for the volt in terms of base SI units? A. 1 V = 1 kg·m2C–1s B. 1 V = 1 kg·m2A–1s–3 C. 1 V = 1 kg·m·A–1s–1 D. 1 V = 1 kg·m2s–2C–2 E. 1 V = 1 J/C F. Some other expression (specify). C1T.4 Assume that D and R have units of meters, T has units of seconds, m and M have units of kilograms, v has units of meters per second, and g has units of m/s2. Which of the following equations has self-consistent units? A. D = mR2 B. m = M[1 + R2] C. D = [1 – m/M]gT2 D. g = mv2/R E. D = v2/RT F. None of these can be correct.

C1T.5 One can raise a quantity q to a power a when a has units. T or F? C1T.6 The following formulas are supposed to describe the speed v of a sphere sinking in a thick fluid. C is a unitless constant, ρ is the fluid’s density in kg/m3, A is the sphere’s cross-sectional area, m is its mass, and g is the gravitational field strength in N/kg. Which could be right? A. v = CAρg B. v = Cmg/ρA C. v = (Cmg/ρA)2 D. v = (Cmg/ρA)1/2 E. None of these can be correct. C1T.7 The speed v of sound waves in a gas like air might plausibly depend on the gas’s pressure P (which has units of N/m2), the gas’s density ρ (which has units of kg/m3) and its temperature T (which has units of K), and some unitless constant C. Assuming that no other quantities are relevant, which of the following formulas might possibly correctly give the speed of sound in a gas? A. v = CPρT B. v = CTP/ρ C. v = CP/ρ ____ D. v = C​____ P/ρ ​     E. v = C​ρ/P ​     F. v = C(P/ρ)2 T. None of these can be correct. C1T.8 The two stars in a binary star system revolve around each other with a certain period T. Which of the quantities listed below is not likely to be a part of the formula for this revolution period? A. m1, m2 (the masses of the stars in the system) B. r (the distance between the binary stars) C. ℏ (Planck’s constant, which is generally associated with phenomena involving quantum mechanics) D. G (the universal gravitational constant)

HOMEWORK PROBLEMS Basic Skills C1B.1 If you text to a friend “I’ll be over in 0.50 ks,” how many minutes will your friend wait? (Use unit operators.) C1B.2 What is the speed of light in furlongs per fortnight? (One furlong = _​ 18 ​  mi = length of a medieval farm furrow, and one fortnight = 14 days. Use unit operators.) C1B.3 A light year (1 ly) is the distance that light travels in one year. Find this distance in miles. (Use unit operators.)

C1B.4 The speed of the earth in its orbit around the sun is 18 km/s. Find this speed (a) in miles per hour and (b) in knots, where 1 knot = 1 nautical mile per hour, and 1 nautical mile = 1852 m. (Use unit operators.) C1B.5

What is a month in seconds (approximately)?

C1B.6 Water’s density is 1000 kg/m3. Use unit operators to show that a cube of water 10 cm on a side has a mass of 1 kg, and 1 cm3 of water has a mass of 1 g. (This used to be the definition of the kilogram, but the difficulty of precisely measuring volumes made this standard impractical.)

Homework Problems

C1B.7 A friend says that the range D of a projectile fired at a speed v at an angle θ above the horizontal is D = v sin 2θ/g, where g has units of m/s2. (a) Explain why this can’t be right. (b) Propose a modification so that D still depends on v and g but could be right. C1B.8

Do problem C1T.6 and explain your answer.

C1B.9

Do problem C1T.7 and explain your answer.

Modeling C1M.1 Can “What is justice?” be investigated scientifically? If you think not, does this mean that “justice” does not e­ xist or is not worth thinking about? If you think so, can you ­scientifically verify your belief? Defend your responses. C1M.2 The text describes that historically, physics became a “science” when the physics community accepted Newton’s mechanics as its “grand theory.” What do you think is or was the corresponding grand theory that made biology a science? Chemistry? Geology? Defend your responses. C1M.3 As a fraction of the stick’s length, what is the length of a vertical stick’s shadow in Saint Petersburg, Russia, at noon on the summer solstice? (Saint Petersburg is 3230 km almost due north from Alexandria.) C1M.4 Consider an object of mass m moving in a circle of radius r with a constant speed v. What is a possible formula for the object’s acceleration a (in m/s2)? C1M.5 A planet’s “escape speed” is the speed ve which an object must have at the planet’s surface to be able to escape the planet’s gravitational embrace and coast to an infinite distance. This speed depends on the planet’s mass M, its radius R, and the universal gravitational constant G. Up to an overall constant, what must the formula for ve be? C1M.6 Imagine slicing a thick disk of radius R in half along its diameter. If you stand the half-disk on its curved edge and nudge it, it will rock back and forth. If the rocking is not too extreme, the time T required for a complete back-andforth oscillation turns out to be nearly independent of the angle through which the disk rocks. The only other things that T might plausibly depend on are the disk’s radius R, its mass M, and the local gravitational field strength g (in m/s2), since gravity is what is causing the rocking motion. (If you think about it, the disk’s thickness is only relevant in that a thicker disk has more mass than a thinner one, so we already have this covered if we consider dependence on M.) Use dimensional analysis to find a reasonable formula for this rocking time up to a unitless constant. C1M.7 Consider various pendula, each consisting of a rod hanging in the earth’s gravitational field from a pivot at its end and that is free to swing around that pivot. The rods have identical shapes, but different lengths L (and diameters proportional to those lengths). The rods also might

15

have different densities and thus masses m that don’t scale simply with L. Experimentally, the period T of such a pendulum does not depend on the angle of swing for small oscillations. Find an approximate formula for the period T, and estimate T for a 1-m rod. (Hints: You should not assume, but rather show by dimensional analysis that the rod’s mass is irrelevant. The gravitational field strength g has units of m/s2.) C1M.8 The radius r of a hydrogen atom (which consists of a proton and a comparatively lightweight electron) can only depend on the (equal) magnitudes e of the proton’s and the electron’s charge, the Coulomb constant  1/4πε0  that characterizes the strength of the electrostatic attraction ­between the proton and electron, the mass m of the orbiting electron, and (because quantum mechanics is likely ­involved) Planck’s constant in the form h/2π = ℏ. (Since the proton remains essentially at rest, its mass turns out to be irrelevant.) (a) Assuming that r depends only on the stated quantities, use dimensional analysis and the information given inside the front cover of the text to find a plausible formula for the radius of a hydrogen atom. (b) Calculate a numerical estimate of the radius. (c) Explain why this is only an estimate. C1M.9 Consider an object of mass m hanging from the end of a spring whose other end is attached to a fixed point. The object will oscillate vertically with some period T, which might depend on the object’s mass, the spring’s stiffness constant ks (in N/m), which expresses how much force the spring exerts when it is stretched a certain distance, and the gravitational field strength g (in m/s2) at the earth’s surface (since the object is moving up and down in the earth’s gravitational field). We find e­ xperimentally that the period does not depend on either the object’s ­maximum speed or how far it moves up and down. (a) Up to a unitless constant, what is the formula for the period in terms of ks, m, and g? (b) On the basis of your calculation, if the object oscillates with a period of 1.0 s on the earth’s surface, what is its period on the moon’s surface? C1M.10 The critical density of the universe ρc is the density that the universe must have for the gravitational attraction between its parts to be strong enough to prevent it from expanding forever. This density must depend on the Hubble constant H, which specifies the universe’s current expansion rate (as a fractional expansion per unit time, so its units are 1/s). It might also depend on the speed of light c, which in combination with H tells us the radius of the universe we currently can receive light from, and thus how much of the universe might contribute to the attraction. (a) What is a third quantity that plausibly appears in this formula and why? (b) Up to a unitless constant, what is the formula for the universe’s critical density ρc? (c) Ignoring the unitless constant, what is the critical density for our universe? (H = 2.28 × 10–18 s–1)

Chapter C1

16

The Art of Model Building

Derivations C1D.1 In this problem, you will prove an important result about the calculation of uncertainties. (a) Let a and b be quantities with small errors δa and δb respectively. The error in f = ab is then

δ f = (a + δ a)(b + δ b) − ab (C1.9) Show that if δa 0.25 nm C9T.4 In the hypothetical atomic interaction shown in ­figure C9.13, suppose the atoms initially are separated by a very large distance but are approaching each other with kinetic energy _​ 12  ​   μ│v​  ​W│    2 = 5 × 10-21 J. Roughly how close will the atoms get to each other eventually? A. r = 0 B. r = 0.05 nm C. r = 0.2 nm D. r = 0.4 nm E. Other (specify) C9T.5 In the hypothetical atomic interaction shown in ­figure C9.13, if the system had the appropriate total energy and initial position, could the atoms’ separation remain stable and constant? A. No B. Yes, at a separation of about 0.12 nm C. Yes, at a separation of about 0.25 nm D. Yes, at a separation of either about 0.12 nm or 0.25 nm E. Yes, at some other separation(s) (specify)

The effective potential energy for a hypothetical two-atom system. Veff(r) ​ ​  ​W ∞. W 0 as r  ​ 

C9T.7 In the hypothetical atomic interaction shown in figure C9.13, suppose that initially the system’s kinetic energy is _​ 12 ​   μ│v​  ​W│    2 = 5 × 10-21 J at an initial separation of 0.1  nm. About how much energy would we have to add to the system for the atoms to fly off to infinite separation? A. 1 × 10-21 J B. 6 × 10-21 J C. 11 × 10-21 J D. 16 × 10-21 J E. None. The atoms will do so without adding energy. C9T.8 In the hypothetical atomic interaction shown in figure C9.13, assume that initially the atoms have a total energy of 3 × 10-21 J at r = 0.1 nm. These atoms are A. Unbound (they will separate to infinity). B. Bound (they will not separate to infinity). C. The answer depends on whether the atoms are initially moving toward or away from each other. C9T.9 In the hypothetical atomic interaction shown in ­figure C9.13, assume that the atoms initially have zero radial velocity at a separation of 0.25 nm. If the atoms are jostled slightly so that the system gains a slight amount of energy, what happens to the atoms’ separation afterward? A. It oscillates within a small region about r = 0.25 nm. B. It oscillates between r = 0.25 nm and r = 0.05 nm. C. It directly increases toward infinity. D. It eventually increases toward infinity, maybe directly but maybe after one detour to smaller separations. E. Some other behavior (specify).

Homework Problems

149

HOMEWORK PROBLEMS Energy (10 –21 J) 30 Veff(x)

20 10 0 –10

0.2

0.4

0.6

x (nm)

–20

Figure C9.14 A (simplified) effective potential energy for a hypothetical two-nucleus system.

Basic Skills C9B.1 Figure C9.14 shows a simplified potential energy “curve” for a hypothetical interaction between two atoms. (This “curve” has been approximated by a sequence of straight-line segments to make this problem easier.) Draw a quantitatively accurate graph of the radial component Fx of the force acting on the lighter atom as a function of x. C9B.2 In the hypothetical atomic interaction shown in figure C9.13, assume that the two-atom system has a total energy of -5 × 10-21 J. Where are the turning points and what are the allowed and forbidden regions in this case? C9B.3 In the hypothetical atomic interaction shown in figure C9.13, assume that the two-atom system has a total energy of +5  ×  10-21 J. Where are the turning points and what are the allowed and forbidden regions in this case?

C9B.7 In the hypothetical atomic interaction shown in figure C9.13, assume that the two-atom system has a kinetic energy of about 10 ×  10-21 J when the atoms have a very large separation and that the atoms’ separation is initially decreasing. About how much energy would have to be removed (at the appropriate time) for the atoms to form a “bond” (where the interaction keeps them close together rather than allowing them to separate again to large r)? C9B.8 In the hypothetical atomic interaction shown in ­figure C9.13, assume that the two-atom system has a kinetic energy of about 10 × 10-21 J when the atoms are separated by 0.1 nm. About how much energy would have to be supplied to the system for the atoms to move off to infinity? C9B.9 Estimate the x component of the force that the interatomic potential energy shown in figure C9.13 exerts on the lighter atom when the separation between atoms is 0.3 nm. Describe your reasoning. C9B.10 A spring has a spring constant ks = 1000 J/m2. If its relaxed length is 10 cm, and we stretch it to a length of 12 cm, how much additional energy does the spring store? C9B.11 Suppose a certain spring’s relaxed length is 13 cm. (a) If we must put 50 J of energy into the spring to compress it to a length of 11 cm, what is its spring constant? (b) What additional energy will be required to compress the spring to a length of 9 cm?

C9B.12 Suppose we launch a mug so it slides along a table. We can certainly calculate the mug’s change in kinetic energy in this situation. (a) Explain why we cannot simply use equation C9.1 to compute the potential energy of the interaction between the mug and the table. C9B.4 In the hypothetical atomic interaction shown in­ (b) Where is the mug’s kinetic energy going if not to some figure C9.13, assume that the two-atom system has a form of potential energy? kinetic energy of about 5 ×  10-21 J when the atoms are separated by 0.1 nm and that the atoms are initially moving apart. Describe the subsequent behavior of the atoms’ Modeling separation r as completely as you can. C9M.1 Estimate the approximate spring constant ks of the C9B.5 In the hypothetical atomic interaction shown in potential energy curve shown in figure C9.13 near that ­figure C9.13, assume that the two-atom system has a kinetic curve’s local minimum. Describe your reasoning. energy of about 2 × 10-21 J when the atoms are separated by a very large distance and that the atoms’ separation is C9M.2 A planet’s escape speed │v​  ​W e│ is the minimum speed initially decreasing. Describe the subsequent behavior of an object must have at the planet’s surface to be able to the atoms’ separation r as completely as you can. coast to an essentially infinite distance from the planet. (a) Use a potential energy graph to explain what miniC9B.6 In the hypothetical atomic interaction shown in­ mum total energy E the planet–object system must figure C9.13, assume that the two-atom system has a kinetic have for the object to escape. energy of about 10 × 10-21 J when the atoms are separated (b) Find a symbolic expression for │v​  ​We │ in terms of G, the by a very large distance and that the atoms’ separation is planet’s mass M, and its radius R. initially decreasing. Describe the subsequent behavior of (c) Check that the units of your expression make sense. the atoms’ separation r as completely as you can. (d) What is the earth’s escape speed?

Chapter C9

150

Potential Energy Graphs

C9M.3 Suppose a particle with charge q is confined to the x axis but moves completely freely along that axis. It interacts electrostatically with two charged particles (with the same charge q) anchored to positions y = ±L, respectively, along the y axis. (a) Find an expression for the system’s potential energy V(x) as a function of L, the Coulomb constant 1/4πε0, q, and the sliding object’s position x. (b) Find any equilibrium point or points for this V(x), and specify whether each is stable or unstable. C9M.4 Imagine a binary star system having two stars with equal mass M, and suppose that these stars follow concentric circular orbits of radius R around the system’s center of mass (which is midway between the two stars as shown below), each orbiting once in a time T.

CM

C9M.9 Suppose we hold two air-track gliders (with masses of 0.25 kg and 0.50 kg) together so they compress a spring with a spring constant of 100 J/m2 by a distance of 2.0 cm. The spring is not connected to either glider, so when we release the gliders, the spring falls away. What is the gliders’ final speed relative to each other? (Hint: The fastest solution method does not explicitly require using conservation of momentum, though that also works.)

Derivations

R Star

C9M.8 Suppose a 20,000-kg jet lands on the surface of an aircraft carrier. As it lands, it engages a tailhook wire, a wire which behaves as if it were connected to a spring with a spring constant of ks = 25,000 J/m2. This slows the jet down to rest within 60 m. What was the jet’s landing speed? (Actual tailhooks use hydraulics instead of a spring to absorb the jet’s energy, but let’s pretend.)

Star

(a) Calculate this system’s total energy E assuming that the system’s center of mass is at rest. Express your result in terms of M, R, G, and T. (b) An observer on one star would consider that star to be at rest, and would see the other star orbit once around it in time T in a circle of radius 2R. Show that if we view the situation from this perspective, then equation C9.6 correctly gives the same system energy. C9M.5 Suppose an interaction between two objects has a potential energy of V(r) = -b/r 2, where b is some positive constant. Assuming the objects move in one dimension, sketch a graph of V(r) for this system and qualitatively describe the types of radial motions that might occur in such a system. C9M.6 Consider two quarks with masses m1 and m2. The potential energy of the strong nuclear interaction between two quarks is roughly V(r) = br, where b is a constant. (a) Draw a qualitative sketch of V(r) for this system assuming that the quarks move in one dimension. (b) Describe the possible types of motion for this system. (c) Can the quarks ever have enough energy to be free of each other (that is, to move to infinite separation)? C9M.7 Suppose a bungee jumper of mass m leaps from a bridge over a deep canyon. The jumper has a low-mass bungee cord with relaxed length L0 tied to her feet that when stretched behaves like a spring with spring constant ks. But this flexible cord does not begin to stretch until the jumper has fallen through the distance L0. What is the maximum distance below the bridge that the jumper will reach, in terms of m, ks, L0, and │g​  ​W│   ? (Hint: Set z = 0 to be where the cord begins to stretch. You will have to solve a quadratic equation. One of that equation’s two solutions will be absurd. Which solution is the right one and why?)

C9D.1 Consider the gravitational interaction between two arbitrarily shaped objects that are small compared to their separation r. Can we model these objects as particles in this situation? The total gravitational potential energy for all the inter-particle interactions between the two objects is

mimj V(r) =   ​∑        ​ − G ____ ​​∑ ​   ​     (C9.9) rij i j

where mi is the mass of the ith elementary particle in object 1, mj is the mass of the jth elementary particle in object 2, and rij is the separation of these two particles. But if both object’s sizes are very small compared to their separation r, then rij ≈ r for every pair of particles in this sum. Use this to argue that equation C9.9 reduces to -Gm1m2/r, where m1 and m2 are the total masses of the two objects. C9D.2 (Requires calculus.) A pair of otherwise free objects with masses M and m moving in one dimension participate in a single interaction with potential energy V(x). Assume M >> m. Conservation of energy then implies that

​p 2x​ ​​  E = _​ 12 ​  m​v  2​x​​  + V(x) = ___ ​    ​ + V(x) (C9.10) 2m

where vx, and px  = mvx are the lighter object’s velocity x component and momentum x component, respectively. (a) Take the time derivative of both sides of this equation to show that dpx dV ​  (C9.11) ​ ___ ​ vx = − ​ ___ dt dt (b) According to the chain rule, dV/dt = (dV/dx)(dx/dt). Use this to show that dpx dV ​  (C9.12) ​ ___ ​ = −​ ___ dx dt

Since the interaction characterized by V(x) is the only interaction contributing to the light object’s actual momentum, the x component of the force exerted by that interaction is Fx ≡ [dpx]V/dt = dpx/dt. Therefore, the x-force exerted by the interaction must be Fx = -dV/dx.

Homework Problems

C9D.4 Prove that if m1 ≥ m2, then the smallest value the reduced mass µ can have is _​ 12 ​  m2 (when m1 = m2) and the largest it can have is m2 (in the limit that m1 >> m2). C9D.5 (Requires knowledge of Taylor-series ­expansions.) One can use a Taylor-series expansion to see why the spring approximation is a good match to almost any potential energy function V(x) near a local minimum. Consider an otherwise arbitrary potential energy function V(x) whose derivatives are defined to all orders and that has a true local minimum (meaning it has a zero first derivative but a positive nonzero second derivative) at x = x0. (a) Write out a Taylor expansion for the function V(x) around the point x = x0. (b) Argue that at a local minimum, the second term of the series is automatically zero. (c) Argue that the third term is almost always nonzero. (d) Argue that the fourth and subsequent terms in the expansion will be small compared to the third when x - x0 is sufficiently small (assuming the various higherorder derivatives of V have comparable values.) (e) Under these conditions, then, V(x) is well approximated by the first and third terms alone in the expansion. Argue that this term has the form V(x) = _​ 12  ​   ks (x − x0)2 + C, and explain how ks and C are linked to the first and third terms of the Taylor-series expansion. C9D.6 An object of mass m hangs vertically from the end of a spring with spring constant ks . For simplicity, let z = 0 be the object’s vertical position when the spring is relaxed. (a) Write the total potential energy function V(z) for the combined effect of gravitation and the spring as a function of the hanging mass’s vertical position z. (b) Show that we can write this in the form

V(z) = _​  12 ​   ks(z − z0)2 + C (C9.13)

and determine z0 in terms of ks, m, and │g​  ​W│   . (This means that this system behaves just as it would with no gravity, except that gravity shifts the equilibrium position.) (c) At what vertical position might the object hang at rest?

on the initial rate of expansion of the universe at the Big Bang. This equation is identical in form to a conservation of energy equation: if we think of a as being analogous to a position, then ​ _12  ​   (da/dt)2 would be analogous to the kinetic energy of a unit-mass object, the constant K would be analogous to a total energy, and B ​    − Wa2 (C9.15) V(a) ≡ − ​ __ a



would be analogous to a potential energy. (a) Draw a qualitative graph of V(a). [Hint: Sketch the curves that represent the functions −B/a and −Wa2 and then draw a curve representing their sum.] (b) Assuming that the universe had scale a ≈ 0 (that is, it was infinitesimally small compared to its present size) at the Big Bang and subsequently expanded, read from the graph the possible qualitatively distinct fates of the universe, and describe these behaviors. (Note that the value of K depends on the universe’s “initial conditions,” which we do not know a priori.) C9R.2 In an upper-level mechanics course, you will learn how to define an effective potential energy for two interacting objects (one very massive) moving in three dimensions that treats the kinetic energy of the nonradial aspect of the lighter object’s motion as part of the system’s potential energy. We can thus model the object’s radial motion as if it were a particle moving in one dimension subject to some purely r-dependent potential energy function. Given this, general relativity predicts that the effective potential energy for an object with mass m moving with (conserved) angular momentum L ≡ │L​  ​W│    near a black hole of mass M located at r = 0 is as follows (when M >> m): 2 GMm L2   ​ − ______ Veff (r) = − ​ ______  ​    + ​ _____ ​ GML  ​   2 2 3 2mr mc r r



L2 2mr2

Vmax Veff (r)

r0

0

C9R.1 General relativity implies that the equation of motion for the universe can be written as follows:

(  )

2

B ​   − Wa2 = K (C9.14) 1 ​ ​  ___ ​ __ ​  da ​   ​ −  ​ __ 2 dt a where a represents the “size” of the universe at time t (as a fraction or multiple of its present size), B is a constant that depends on the present density of matter in the universe, W is a constant that depends on the present density of the so-called “vacuum energy” in the universe, and K is a constant (which can be positive or negative) that depends

r

Vmin

-

Rich-Context

(C9.16)

Here is a qualitative graph of this function:

Energy

C9D.3 (Requires calculus.) Suppose an object is moving in one dimension subject to an interaction whose potential energy function is V(x) ≈ _​ 12  ​ ks x2 + C. What is the x component of the force acting on the object at position x?

151

GML2 mc2r3

GMm r

(a) What kinds of qualitative radial motions are possible for this object? Specifically, for what system energies E would the small object’s radial distance from the black hole (1) maintain a fixed value, (2) oscillate between two values, (3) come in from infinity and then go back out to infinity, and (4) be absorbed by the black hole at r = 0? Specify if any of these cases are impossible. (b) The analogous expression for Newton’s theory of gravity is the same except that it lacks the final term. Repeat your analysis for the Newtonian case. (Hint: First sketch the graph of Veff(r) for that case.)

Chapter C9

152

Potential Energy Graphs

C9R.3 Consider an unpowered roller-coaster car of mass m running along a frictionless roller-coaster track that lies in the xz plane (with the x axis horizontal and the z axis vertical). Since the gravitational potential energy of the e­ arth– car system is m│g​  ​W│   z, a graph of V(x) = m│g​  ​W│    z(x), where z(x) is the height of the roller-coaster track at horizontal position x, shows the potential energy of the earth–car system as a function of the car’s horizontal position x. One might be tempted to use such a system as an analogy for an object of mass m moving in one dimension subject to the same potential energy function V(x). In what ways is this a good analogy? In what ways is it misleading? (Hints: Compare turning points, allowed and forbidden regions, stable and unstable equilibrium points, kinetic energy, and forces in the two cases.) C9R.4 Suppose you are on an airless spherical asteroid and you use a large spring gun to launch packets of valuable metals of identical mass toward your company’s central shipping base on another asteroid a considerable distance away (each packet has a beacon so that people on the base can track and retrieve it). Suppose you find that if you compress the spring by a distance D0 when you fire the packet, the packet leaves the end of the gun with a speed │v​  ​W0  │  that is just barely sufficient to get the packet arbitrarily far from the asteroid. How far do you need to compress the

spring to give the packet a final speed of │v​  ​W│    far from the asteroid? Express your answer in terms of │v​  ​W 0│, │v​  ​W│   , and D0. C9R.5 Four identical balls of mass M and radius R are connected by strings in a square arrangement whose side has length L. Between each adjacent ball there is a spring that has a relaxed length 2L and a spring constant ks that is compressed by the string tying the balls together. The diagram below shows the situation. R

Ball, mass M Springs

String

L

(The springs are not attached to the balls, they are s­ imply compressed between them.) The entire system floats in space. Now suppose that we cut the strings simultaneously. The springs will push the balls away from each other. What final speed will each ball have? (Hint: Will the balls have the same final speed? Justify your response.)

ANSWERS TO EXERCISES C9X.1 The light atom will move in toward x3, slowing down as it approaches. At x3, by analogy to the other cases, it will come instantaneously to rest, but because it is still being repelled, it will then begin to move back out toward infinite separation with increasing speed. C9X.2

0 E

x3

allowed V(x)

The calculation goes like this:

( 

)

( 

)

m  m (m + m2) 2 m1 ​m 2​2​​  + m2 ​m 2​1​​  2 = __ ​ 1 ​ ​  ___________        ​​ v ​x​​  + V(r) = __ ​ 1 ​ ____________    ​  1 2 1   ​   ​v ​x​​  + V(r) 2 (m1 + m2)2 2 (m1 + m2)2

forbidden regions x1 x2

C9X.3

m2 vx 2 _1 m1 vx 2 E = _​ 12 ​  m1 ​​ ​ _______   ​   ​​ ​+ ​ 2  ​ m2 ​​ ​ _______   ​   ​​ ​+ V(r) m1 + m2 m1 + m2

Here is the marked-up diagram: Energy

The forbidden regions are 0 ≤ x < x1 and x > x3, and one allowed region between x1 and x3. There is one stable equilibrium point at separation x2.

x

m1 m2 2 = __ ​ 1 ​ ​  _______   ​​   v ​x​​  + V(r) ≡ __ ​ 1 ​   μ  ​v 2​x​​  + V(r) (C9.17) 2 m1 + m2 2 We see that the latter is the same as equation C9.6 if we identify │v ​W​│    = │ vx│.

C10 Work Chapter Overview Introduction In this chapter, we explore what external interactions that exert forces on a system do to that system’s energy, comparing and contrasting what conservation of momentum requires to what conservation of energy requires. We will learn from this discussion how to handle frictionless contact interactions in conservation-of-energy problems. This chapter also lays essential groundwork for future chapters.

Section C10.1:  The Momentum Requirement In this section, we see that conservation of momentum, the definition of force, and the definition of a system’s center of mass require that when external interactions exerting forces act on a system, the change in the system’s center-of-mass kinetic energy is dKCM = │F​  ​We xt││d ​ rW ​C  M│cos θ (C10.4)



where F​ ​W ext is the net external force on the system, d ​ rW ​ CM is a tiny displacement of the system’s center of mass, and θ is the least angle between F​ ​We xt and d ​ rW ​C  M. The displacement W must be small enough so that F​ ​ e xt ≈ constant during the displacement.

Section C10.2:  The Dot Product We can express this relationship more compactly using the dot product: W · w​ W ​ u​ ​  ≡ │u​  ​W│  │w​  ​W│   cos θ (C10.5) • Purpose:  This equation defines the dot product between two vectors u​ ​W and w​ ​W,  W where θ is the least angle between the directions of u​ ​W and w​ ​ .  • Limitations:  This is a definition and so is exact. • Note:  The cross product yields a number with units that are the product of the W units of u​ ​W and w​ ​  (which need not be the same). The dot product has the following convenient properties: W · w​ ​ u​ ​W = w​ ​W · u​ ​W 

(the dot product is commutative)

(C10.6a)

W · (​w​ W + s​ ​ u​ ​W)  = u​ ​W · w​ ​W + u​ ​W · s​ ​W  

(the dot product is distributive) (C10.6b)

W · (b ​w​ W)  = b(​u​ W · w​ ​ u​ ​W) 

(the dot product is linear) (C10.6c)

W · u​ ​ u​ ​W =│u​  ​W│  

(C10.6d)

W · w​ W ⊥ w​ ​ u​ ​W = 0  ⇔ ​ u​ ​W   (assuming│u​  ​W│    ≠ 0 and │w​  ​W│    ≠ 0)

(C10.6e)

2

W The dot product of u​ ​W and w​ ​  in terms of those vectors’ components is as follows: W · w​ ​ u​ ​W = ux wx + uy wy + uz wz (C10.7)

154

Using the dot product, the momentum requirement (equation C10.4) becomes:



W dKCM = F​ ​ e xt · d ​ rW C ​ M (C10.8)

• Purpose:  This equation describes how external interactions that exert a total W external force F​ ​ e xt on a system as its center of mass moves through a tiny displacement d ​ rW ​C  M affect the system’s center-of-mass kinetic energy KCM ≡ ​ _12  ​  M │v​  ​WC  M│2, where M is the system’s total mass and v​ ​WC  M is the velocity of its center of mass. W • Limitations:  The process must have a short enough duration dt so that (1) F​ ​  ext dt is small compared to the system’s total momentum and (2) the forces that we W add to get F​ ​ e xt are each constant during dt. • Notes:  This momentum requirement expresses what conservation of m ­ omentum requires regarding energy transfers.

Section C10.3:  The Definition of Work We define work W to be “energy that crosses a system boundary due to an external interaction that exerts a force on the system during a process.” The section argues that

W i · d ​ rW i ​ W = ∑dWi = ∑ ​F​

(C10.12a)

W W If F​ ​  is constant  ⇒  W = F​ ​  · Δ​ rW  ​ (C10.12b)

• Purpose:  This equation describes the work W done by an external interaction that exerts a force during a process where the point or region on which the force acts moves a certain distance. W • Notes:  If F​ ​  is a constant vector during the process, then equation C10.12b applies, where Δ​ rW  ​is the total displacement of the point or region on which the force acts. If the force changes in magnitude or direction during the process, W then we must divide the process into tiny steps and sum F​ ​ i  · d  rW​  i ​ over all steps, W where F​ ​ i  is the force that acts during the ith step and d  rW​  i ​ is the infinitesimal displacement of the point or region on which the force acts during that step. W • Limitations:  In the latter case, the steps must be small enough that F​ ​ i  is approximately constant during each step. In both cases, these expressions need careful reinterpretation when the affected particles are moving at speeds comparable to that of light.

Positive work means that energy flows into the system; negative work means that energy flows out. Work is to energy what impulse is to momentum and twirl is to angular momentum.

Section C10.4:  Long-Range Interactions This section uses the momentum requirement and the definition of work to explain why long-range interactions rarely affect a system’s internal energy.

Section C10.5:  Contact Interactions The definition of work and the momentum requirement also together explain why any interaction that exerts a force that is always perpendicular to an object’s motion affects neither the object’s internal energy nor its kinetic energy. Such an interaction can be ignored in a conservation-of-energy problem. Such forces include the normal (perpendicular) part (as opposed to the friction part) of a contact interaction between solid surfaces, the lift part (as opposed to the drag part) of the interaction between air and a solid object, and the tension interaction between an inextensible string and an object swinging at its end. We can handle such an internal interaction in a conservation-of-energy problem solution by writing “⊥” next to its name in the interaction list and ignoring it in the master equation.

155

Chapter C10

156

C10.1

Derivation of the momentum requirement dp sinθ

dp

dp cosθ

p p

The Momentum Requirement

The goal of this chapter is to develop some ideas that we need to better apply energy concepts to macroscopic systems (especially deformable systems such as elastic balls or human bodies) and how forces applied to such systems affect the energy they contain. We will begin by examining a useful consequence of conservation of momentum and the definition of force. Suppose we have a system whose total momentum at a certain instant is p​ ​W.  During a very brief subsequent time interval dt, suppose the system particiW​ ′  = p​ W​  + d ​p​ W.  pates in interactions that change the system’s total momentum to p​ Figure C10.1 and the law of cosines then implies that

θ

Work

│p​  ​W′ │2 =│p​  ​W│   2 + 2│p​  ​W│  │ dp​  ​W│   cos θ +│ dp​  ​W│   2 (C10.1)

W.  (If you are not familiar with the law of where θ is the angle between p​ ​W and d ​p​ cosines, you can prove equation C10.1 fairly easily: see problem C10D.1.) If dt is sufficiently small, │ dp​  ​W│    > EB B. EA ≈ EB C. EA m1, b will become very large, so we see that vfx will go to zero and its kinetic energy will be entirely converted to internal energy. This is consistent with a tiny object running smack into an essentially immovable massive object. In the case where m1 = m2, then b = 1 and we see that vfx = _​ 12│ ​   v​  ​W0  │. This certainly satisfies equation C14.1, and it also makes sense for the energy ratio, since the final object has twice the mass but half the speed, and so half the kinetic energy of the original object. So this looks good!

Example C14.1

226

Example C14.2

Chapter C14

Collisions

Problem:  Suppose an object with mass m1 moving with a speed │v​  ​W0  │ in what we can take to be the +x direction elastically collides with another object with mass m2 at rest. After this head-on collision, both objects move along the x axis. What are their x-velocities just after the collision? Solution  Initial and final diagrams appear below. Initial: m1

v0

m2 x (at rest)

Final: v1

v2

x

Known: m1, m2, v​ ​W0  , and   object 2 is initially at rest. Unknown: v1x, v2x Internal interactions: • Contact: ignore because collision is elastic

The system of the two objects is momentarily isolated here, and since we are told that the collision is elastic, the contact interaction has no energy implications. The conservation-of-momentum master equation requires that m m 0  1v1x 2v2x +m  ​W0  │       1│v​      ​   ​       ​ + ​    ​   ​    ​ = ​    ​  ​          ​ + ​    ​      ​ 0 0 0 ​        0 ​     [ ] [  0 ] [    0 0 ] [ 0 ]

(C14.4)

The contact interaction has no potential energy, and since the collision is elastic, the objects’ internal energies do not come into play. Therefore, the conservation-of-energy master equation simply implies that the total system kinetic energy before the collision is equal to that after the collision: ​ _12 ​  m │v​  ​W 0│2 + 0 = _​  12 ​  m1 │v​  ​W 1│2 + _​  12 ​  m2 │v​  ​W 2│2 = _​ 12 ​  m1 ​v ​21 x​​  + _​  12  ​  m2 ​v ​22 x​ ​

(C14.5)

as only the x components of the final velocities are nonzero. Note that the x components of equation C14.4 and equation C14.5 provide two equations in the two unknowns v1x and v1y, so we have enough information to solve the problem. Let’s again try the trick of dividing through by m1 and defining b ≡ m2/m1. If we divide the first equation by m1 and divide the second by _​ 12  ​  m1, we get the simpler equations

│v​  ​W0  │ = v1x + b v2x and │v​  ​W0  │2 = ​v ​21 x​​  + b ​v ​22 x​ ​ 

(C14.6)

We can solve this system of equations by solving the first for v1x and substituting the result into the second:

v1x = │v​  ​W 0│ − b v2x (C14.7)



⇒  │v​  ​W 0│2 = (│v​  ​W 0│ − b v2x)2 + b ​v ​22 x​​  = │v​  ​W 0│2 − 2b │v​  ​W 0│v2x + (b + b2 ) ​v ​22 x​​ 



⇒ 0 = − 2b │v​  ​W0  │v2x + (b + b2 )​v ​22 x​ ​ 

(C14.8)

One of the solutions to the quadratic equation in equation C14.8 is v2x = 0, as if the first object simply passed through the second object without affecting it. While this solution certainly conserves momentum and kinetic energy (which is all that the equations care about), it is not realistic. So if we assume that v2x is not zero, then we can divide both sides by (b + b2) v2x, yielding  

│v​ │v​ v2x = ______ ​  2b 2 ​     ​W0  │ = _____ ​  2   ​     ​W0  │ (C14.9a) 1+b b+b

To find v1x, we simply need to substitute this into equation C14.7:

C14.3 Two-Dimensional Collisions

227

( 

)

2b  2b   ​│ │v​ v1x = │v​  ​W0  │ − b v2x = │v​  ​W0  │ − ​ _____  v​  ​W0  │ = ​ __________ ​  1 + b −  ​    ​│v​  ​W0  │ = _____ ​  1 − b ​    ​W0  │ (C14.9b) 1+b 1+b 1+b Again we can check equations like this by considering units and extreme cases. First note that the units are correct in each equation: since b is unitless, the right sides of both equations C14.9 have the same units as │v​  ​W 0│, which are appropriate units for v1x and v2x . Second, note that if │v​  ​W 0│ = 0, we have no collision, so it makes sense that the formulas predict that v1x and v2x are both zero also. Third, in the limit that the mass of the second object becomes infinite compared to that of the first, then b ≡ m2/m1 ​ W​  ∞ and we see that the equations reduce to v2x = 0 and v1x = −│v​  ​W 0│, meaning that the first object simply bounces off the essentially immovable second object. This is plausible. So these results look credible.

Exercise C14X.1 Another excellent way to check the results of such complicated calculations are to substitute them back into the original equations │v​  ​W0  │ = v1x + b v2x and 2 2 2 │v​ │  ​W0  = ​v 1​  x​​  + b ​v  2​  x​​  (see equation C14.6) to see that they are satisfied. Do this. Note that in drawing the diagram for this problem, I had to choose which direction I thought the first object might have been going after the collision. However, the conservation equations are agnostic about this: v1x and v2x could represent either motions in the +x or −x directions, depending on their signs. We see from the solution that v1x will be positive if m1 > m2 and negative if m1 < m2, and that v2x is always positive. Let me underline two valuable things we can learn from these examples. The first is that in one-dimensional situations, the laws of conservation of momentum and energy provide two equations we can use to solve for two unknowns. In the first case, this allowed us to determine the objects’ final common velocity and the fraction of energy that was converted to internal energy in the collision (note how conservation of momentum constrained the distribution of energy). In the second case, the two equations allowed us to determine the two objects’ final velocities. Second, note how valuable it was in both situations to define the unitless variable b ≡ m2/m1 (think about how you would go crazy in, say, equation C14.9b with all the m1s and m2s flying around). This trick makes solving many collision problems much easier.

C14.3

Comments about directions of final motion in example C14.2

A useful trick

Two-Dimensional Collisions

Now let’s relax the one-dimensional constraint and allow our objects to W​  0 move freely in three dimensions. Consider an object moving with velocity v​ that collides with an object at rest. If the velocities v​ ​W 1 and v​ ​W 2 of the two objects after the collision are not parallel, then they define a plane in space. MoreW​  0 must lie in the same plane: because the system has no component of over, v​ momentum perpendicular to the plane in the final situation by definition, it can have no momentum perpendicular to the plane initially. Therefore, a collision between a moving object and a stationary target is intrinsically a twodimensional problem, even if the objects are free to move arbitrarily. Let’s look at examples of two-dimensional inelastic and elastic collisions.

A collision between a moving object and a stationary one is intrinsically two-dimensional

Chapter C14

228

Example C14.3

Problem:  A croquet player hits a wooden croquet ball so that it strikes an identical ball at rest. Just before the collision, the moving ball had a velocity of v​ ​W0  . After the collision, both balls follow trajectories that make the same angle θ with (but on opposite sides of) the originally moving ball’s initial velocity. What are the speeds │v​  ​W1  │ and │v​  ​W2  │ of the balls after the collision, and how much energy is converted to internal energy in the balls?

y

Initial: v0 Ball, mass m Final:

x Ball, mass m, at rest y

Known: m, v0, θ Unknown: v1 , v2 , ∆U

v1 θ θ

Collisions

x v2

Solution  Initial and final diagrams appear in the margin. Note that I have arbitrarily defined the +y direction so that v1y is positive after the collision. The system in this case is the two balls. That system is poorly isolated by most standards, but because the collision lasts only milliseconds, the system can qualify as isolated if we look at the situation just before and just after the collision. The single internal interaction is the contact interaction between the balls, which we will handle by keeping track of internal energies. Conservation of momentum in this case requires that │v​ │v​ +m  ​W 0│ m    ​W 1│cos θ 0  m │v​  ​W 2│cos θ                  ​   ​       ​      + ​ ​    ​   ​    = ​ ​    ​  ​            + ​ ​    ​         ​ │v​ m │v​  ​W1  │sin θ 0 0 −m   ​W2  │ ​  sin θ        ] [ 0 ] [  0 ] [      [ ] 0 0

(C14.10)

Here the trick of dividing both sides by the moving ball’s mass simply means that the masses cancel out entirely. The second line of equation C14.10 implies immediately that │v​  ​W 1│sin θ = │v​  ​W 2│sin θ  ⇒  │v​  ​W 1│ = │v​  ​W 2 │. Substituting this into the first equation (after dividing through by m) implies that

│v​  ​W  │ │v​  ​W 0│ = 2cos θ │v​  ​W 1│  ⇒  │v​  ​W 1│ = │v​  ​W 2│ = ______ ​  0     ​ (C14.11)

2cos θ

We see that conservation of momentum alone determines the final speeds. Now let’s consider conservation of energy. Since we have no potential energy terms, if Ui and Uf are the combined initial and final internal energies of the balls respectively, the conservation-of-energy master equation implies that ​ _12 ​  m │v​  ​W 0│2 + 0 + Ui = _​ 12 ​  m │v​  ​W 1│2 + _​ 12 ​  m │v​  ​W 2│2 + Uf W 0│2 - ​ _12  ​   m │ ​v​ W 1│2 − ​ _12  ​   m│ ​v​ W 2│2 = ​ _12  ​   m│ ​v​ W 0│2 − 2 · ​ _12  ​   m│ ​v​ W 1│2    ⇒  Uf − Ui ≡ ΔU = ​ _12  ​   m │ ​v​ 2│v​  ​W  │2 2 ​│@ v​  ​W0  │2​  ⇒  ________ ​  1 ΔU 2 ​  = 1 − ______ ​  1 2 ​  = 1 − __________ ​    ​  = 1 − ______ ​  1     ​ (C14.12) _​    ​   m │v​ 2 2 │ │v​    ​W0  │ 1  ​W0 2cos2 θ 4cos  θ​│@ v​  ​W 0│ ​  2 The fraction of the original ball’s kinetic energy is unitless and less than 1, as it must be. Note that as θ goes to zero, the two balls are essentially stuck together, and the fraction goes to _​ 12  ​ , consistent with the result of example _ C14.1. The maximum possible value of θ is 45°, as cos 45° = ​_​ 12    ​ ​, which makes the subtracted term in equation C14.12 equal to 1. In this limit, the collision becomes elastic. Indeed, one can prove quite generally that when an initially moving ball collides elastically with an identical ball at rest, the angle between the balls’ final velocities is always 90° (see problem C14D.1). Angles greater than 90° would require ∆U to be negative, which is not possible in an ordinary collision unless it triggers some kind of explosion.

Note that this example illustrates the continuing pattern for inelastic collisions: conservation of momentum places requirements on conservation of energy that allow us to determine how much energy must be converted to internal energy.

C14.3 Two-Dimensional Collisions

229

Problem:  Suppose that in an atomic physics experiment we fire an atom of mass m with velocity v​ ​W0  at a target atom of mass 2m. After an elastic collision, W​ 2  points at an angle of θ = 45° away from the directhe target atom’s velocity v​ W​ 0  . What is │v​ tion of v​  ​W2  │, and what is the magnitude and direction of the lighter W​ 1 ? atom’s final velocity v​ Solution  Initial and final diagrams appear in the margin. Note that even though the atoms are free to move in three dimensions, we can always W​ 2  . choose our xy plane to be the plane containing the final velocities v​ ​W1  and v​ The system here is the two atoms. Since atoms in a typical atomic experiment are likely to move very quickly (many kilometers a second), the entire experiment, if confined to a laboratory, will only take a fraction of a millisecond, so there is little time indeed for the external force of gravity to affect anything. We are told that the single internal interaction is elastic, so we only need to worry about the atoms’ kinetic energies. Conservation of momentum in this case requires that │v​ 2m│v​  ​W 2│cos θ 0  mv +m  ​W 0│ 1x              │ W 2m v​  ​  2│ ​  sin θ  0  ​   ​     ​     ​+​   0  ​  ​    ​=​   mv ​    ​        + ​ ​       ​      ​ 1y ] [ 0 ] [ 0 ] [ 0 ] [      0

(C14.13)

Though in principle we could write v1x = │v​  ​W 1│cos ϕ and v1y = −│v​  ​W 1│sin ϕ, long experience has taught me that using components for this unknown _ velocity is much easier. Also, as cos θ = sin θ =  ​ _​  12   ​ ​  are simple, easy-to-write unitless numbers here, substituting in their numerical values can save writing in this case. So dividing through by m and substituting in for sin θ and cos θ yields the very simple equations

__

__

│v​  ​W0  │ = v1x +  ​ 2 ​   │   v​  ​W2  │, 0 = v1y +  ​ 2 ​   │   v​  ​W2  │ (C14.14)

These provide two equations for our unknowns, but that is not enough to determine the three unknowns v1x, v1y, and │v​  ​W2  │. The other equation we need comes from conservation of energy, which for this elastic collision requires ​ _12 ​  m │v​  ​W0  │2 = _​  12 ​  m │v​  ​W1  │2 + _​ 12 ​  2m │v​  ​W2  │2  ⇒  │v​  ​W0  │2 = │v​  ​W1  │2 + 2│v​  ​W2  │2 (C14.15) We can solve this system of equations by solving equations C14.14 for v1x and v1y in terms of │v​  ​W0  │ and │v​  ​W2  │ and substituting these results into equa2 │ │ tion C14.15, noting that v​  ​W2  = ​v ​22 x​​  + ​v ​22 y​.​  Doing this yields

__

__

v1x = │v​  ​W 0│ −  ​ 2 ​  │  v​  ​W 2│ and v1y = −​2 ​  │   v​  ​W 2│ (C14.16) __

__

│ v │ │ v │ ⇒​@ ​W​0 │2​  = (│ v​W​0 │ −  ​ 2    ​    v ​W​2 │)2 + 2│ v​W​2 │2 + 2│ v​W​2 │2 = ​@ ​W​0 │2​  − 2 ​ 2    ​    v ​W​0 ││ v​W​2 │+ 2│ v​W​2 │2 + 4│v ​W​2 │2 __

⇒ 0 = −2​2 ​  │  v​  ​W0  ││v​  ​W2  │ + 6 │v​  ​W2  │2 

(C14.17)

One solution to this quadratic equation is │v​  ​W2  │ = 0, which again is the null solution where the atoms don’t actually interact. If we assume that │v​  ​W2  │ ≠ 0,

__

__

_

│v​  ​W2  │ = ___ ​ 2​6 2 ​│ ​    v​  ​W0  │ = __ ​ ​ 32 ​   │ ​   v​  ​W0  │ =  ​ _​  29   ​ ​ │v​  ​W0  │ ≈ 0.47│v​  ​W0  │ (C14.18)   

  

Substituting this back into equations C14.16 yields

v1y v1x = │v​  ​W 0│ − _​ 32 │ ​  v​  ​W 0│ = _​ 31​│    v​  ​W 0│ and v1y = −​ _32​   │v​  ​W 0│ ⇒ ϕ = tan-1​ ___ ​ v  ​  ​ = tan-1 2 = 63°    ⇒ │v​  ​W1  │ =

_____ ________  ​ ​v   ​21 x​​  + ​v ​21 y​ ​ ​  = ​ _​  19   ​ + _​ 94 ​ ​ │  v​  ​W0  │

_ = ​ _​  59   ​ ​ │v​  ​W0  │

| | 1x

≈ 0.75│v​  ​W0  │ (C14.19)

As a check, note that equation C14.15 is satisfied: │v​  ​W0  │2 = _​ 59│ ​   v​  ​W0  │2 + 2 ∙ ​ _29│  ​  v​  ​W0  │2.

Example C14.4

y

Initial: v0 Atom, mass m Final:

Known: m, v0 , θ Unknown: v1 , v2 , ϕ

Atom, mass 2m, at rest y

x

v2 θ = 45° ϕ v1

x

230

Chapter C14

Counting equations and unknowns

In both of these two-dimensional collisions, conservation of momentum and conservation of energy provide three constraint equations on the outcome. However, the two final ball velocities involve four unknown numbers (either two sets of two components, or two magnitudes and two directions). If we know the collision is elastic, we need to specify only one of these unknowns to determine the others: this amounts to specifying how glancing the collision is. If the collision is inelastic, specifying two bits of information about the final velocities allows us to determine the other two bits and how much energy is converted in the collision. Of course, this assumes we know everything about the system initially. If we know everything about the collision outcome, we can use our equations of constraint to infer things about the system’s initial state.

C14.4

Collisions

The Slingshot Effect

A close encounter with a planet can dramatically increase a spacecraft’s speed. NASA space probes often use this slingshot effect to boost their speeds so they can visit more distant planets (this saves money by reducing the fuel they have to carry). How does this work? W​  0 Let’s consider a specific case where the initial velocities of the probe v​ and planet v​ ​W P0 are perpendicular in a reference frame attached to the sun. Imagine we have aimed the probe so the gravitational pull of the planet sweeps the probe’s direction around by about 90° as it passes the planet, so it ends up traveling in the same direction as the planet’s final velocity v​ ​W P, as shown in figure C14.1. What is the probe’s final speed │v​  ​W│  ? Let the probe and planet have masses m and M, respectively. Note that in the solar system frame, we cannot ignore the gravitational effects of the sun; but let’s assume the encounter is brief enough that the energy and momentum that the probe–planet system gets from this external interaction is small. Let’s also assume that both initially and finally the probe is so far from the planet that their gravitational interaction’s potential energy is V = –GMm/r ≈ –GMm/∞ = 0, implying that ∆V = 0. This is the only significant interaction between the probe and planet (assuming the probe does not enter the atmosphere), and gravitational interactions rarely cause internal energies to change, so conservation of energy implies that the system’s total kinetic energy will be conserved: we can model this encounter as an elastic collision. Expressed in the reference frame shown in figure C14.1, the law of conservation of momentum implies that 0 Mv │v​ │v​     M  ​W P│ m  ​W│   Px             ​ + ​   ​ 0 ​  ​      0  │  │v​      ​W ​  Mv ​   m       ​ = ​    ​    ​      ​ + ​    ​  ​    ​ 0 Py        [ 0 ] [     0 ] [ 0 ] [ 0 ]

(C14.20)

Since this is an elastic collision, conservation of energy implies that ​ _12 ​   m │v​  ​W0  │2 + _​  12 ​   M │v​  ​W0  │2 = _​  12 ​   m │v​  ​W│   2 + _​  12 ​   M │v​  ​WP  │2 (C14.21) In this case, dividing all equations by the planet’s mass M and defining b ≡ m/M is advantageous because we will eventually want to take the limit that the planet’s mass is huge compared to the probe’s mass, which means that b (so defined) goes to zero. Doing this to the top two components of equation C14.20 and equation C14.21 yields the simpler equations

│v​  ​W P0│ = b │v​  ​W│    + vPx  ⇒  vPx = │v​  ​W P0│ − b │v​  ​W│   (C14.22a)



b │v​  ​W0  │ = vPy  ⇒  vPy = b │v​  ​W0  │ (C14.22b)

C14.4 The Slingshot Effect

Initial:

231

y

Final:

y v

vP0

x

vP

x

Planet, mass M

Figure C14.1 Initial and final situations for an encounter between a space probe and a planet that will result in the probe’s gaining speed by the slingshot effect.

v0 Probe, mass m



b │v​  ​W0  │2 + │v​  ​WP  0│2 = b │v​  ​W│   2 + │v​  ​WP  │2 = b │v​  ​W│   2 + ​v ​2P x​ ​ + ​v ​2P y​ ​ 

(C14.22c)

We can use the first two equations to eliminate vPx and vPy in the third. After a bit of work, we find that 0 = (1 + b) │v​  ​W│   2 − 2│v​  ​W│  │v​  ​W P0│ − (1 − b) │v​  ​W 0│2 (C14.23)

Exercise C14X.2 Fill in the missing steps between equations C14.22 and C14.23. Applying the quadratic formula to equation C14.23 yields

___________________________



2 │v​  ​WP  0│ ±  ​     4  │v​  ​WP  0│2 + 4 (1 + b) (1 − b)│v​  ​W0  │2 ​ │v​  ​W│   = _____________________________________ ​           ​ 2(1 + b) ___________________



│v​ │  ​WP  0│ ±  ​      v​  ​WP  0│2 + (1 − b2) │v​  ​W0  │2 ​ ____________________________

=     ​      ​ (C14.24) 1+b

Now, the planet’s mass M is typically huge compared to the probe’s mass m, so in the limit that b = m/M goes to zero, this becomes

____________

│v​ │  ​W│    ≈ │v​  ​WP  0│ ±  ​      v​  ​WP  0│2 + │v​  ​W0  │2 ​ 

(if b > m is a good approximation for almost all pairs of objects in the solar system. ­

Section N11.3:  Kepler’s Second Law Conservation of angular momentum implies (1) that the orbit of either object around the system’s center of mass must lie in a plane perpendicular to the fixed angular momentum vector and (2) Kepler’s second law. This section provides a detailed argument supporting both assertions.

Section N11.4:  Circular Orbits and Kepler’s Third Law From now on, we will assume that M >> m unless otherwise specified. In this approximation, the primary basically provides a fixed gravitational field to which the satellite responds. Newton’s law of universal gravitation states that the magnitude of the gravitational force that the primary exerts on the satellite is

170

│F​  W ​ g  │= ______ ​ GMm  ​   



(N11.9)

r2  

W • Purpose:  This equation specifies the magnitude of the gravitational force F​ ​ g  that an object of mass M exerts on an object of mass m (or vice versa) when their centers of mass are separated by a distance r. G = 6.67 × 10-11 N · m2/kg2 is the universal gravitational ­constant. • Limitations:  This equation applies to point masses or spheres, but not to irregularly shaped objects. Also, the equation does not apply to extremely strong gravitational fields (much stronger than any fields in our solar system) or to objects moving at close to the speed of light. Newton’s second law, the law of universal gravitation, and the expression for an object’s acceleration in circular motion imply that for a satellite in a circular orbit ____

2

│v​    ​ ​    W​ │    =  ​ ____ ​  GM      ​  4π   ​   R3 and  T 2 = ____

(N11.11, N11.13)

 

 

R

GM

 W​ │    and period T of a satel• Purpose:  These equations specify the orbital speed │v​ lite in a circular orbit of radius R around a primary with mass M where G is the universal gravitational constant. • Limitations:  The orbit must be circular, and the primary must be much more massive than the satellite. The limitations on equation N11.9 also apply here. • Note:  The second equation is Kepler’s third law for circular orbits.

Section N11.5:  Circular Orbit Problems Circular orbit problems are very much like the constrained-motion problems in ­chapter N7, except that we use equation N11.11 or N11.13 or the magnitude of N ­ ewton’s second law as the master equation. In the conceptual model section, you really only need to (1) describe the interacting objects, (2) check that one is much more massive than the other, and (3) check that the satellite’s orbit is essentially circular.

Section N11.6:  Black Holes and Dark Matter This section discusses how astrophysicists have used Kepler’s third law to show that black holes exist and to discover the existence of dark matter.

Section N11.7:  Kepler’s First Law and Conic Sections In this section, we use the Newton application to show that Newton’s second law and the law of universal gravitation predict that a satellite’s orbit is a conic section (a circle, ellipse, parabola, or hyperbola) described by



b    b     ​ (N11.23) r (θ ) = __________ ​  , with a ≡ ________ ​  │1 − ε 2│ 1 + ε cos θ  

 

 ​    

 

 

• Purpose:  This equation describes a conic section by specifying the distance r the curve is from the origin (the focus) at a given angle θ measured from the line that connects the origin to the curve’s nearest point. The constant ε is the curve’s eccentricity, b (= r at θ = 90°) is a constant, and a is the semimajor axis. • Limitations:  None: this is a definition. • Notes:  The point of the orbit closest to the origin is rc = b/(1 + ε) = a│ 1 − ε│, and for an ellipse, the farthest point is rf = b/(1 − ε) = a(1 + ε). For a parabola and hyperbola, r goes to infinity at angle θ∞ = cos-1 (−1/ε).

171

Chapter N11

172

Kepler’s Laws

N11.1 Kepler’s Laws In the year 1600, Johannes Kepler came to Prague to join the research staff at an observatory operated by Tycho Brahe, an astronomer who was both the official “imperial mathematician” of the Holy Roman Empire and a friend of Galileo. The following year, Brahe died, and Kepler succeeded him as imperial mathematician and as director of the observatory. His new position gave him complete access to Tycho Brahe’s extraordinary collection of careful astronomical observations of the planets, the result of a lifetime of work by perhaps the greatest naked-eye astronomer who ever lived. (The telescope was not invented until about 1608.) In 1609, Kepler published his Astronomia Nova (“New Astronomy”) in which he stated two empirical laws that he induced from Brahe’s planetary observations. In modern language, these laws state that

Kepler’s laws Ellipse

∆t T2 ∝ a3 2a

∆t Sun (at one of the ellipse’s foci)

Figure N11.1 Kepler’s laws illustrated. The ­colored regions show the area swept out by a line connecting the planet and the sun during equal time intervals Δt ­during differ­ent parts of the orbit: their areas are equal by Kepler’s ­second law.

Kepler’s laws are empirical

Newton offered an explanation of these laws

1. The orbits of the planets are ellipses, with the sun at one focus. 2. The line from the sun to a planet sweeps out equal areas in equal times. (I’ll define the focus of an ellipse later.) Kepler offered no theoretical explanation for these laws, he simply presented them as being descriptive of planetary orbits (according to Brahe’s observational data). These laws represented a rather radical departure from the accepted wisdom of the time: up to then, most astronomers assumed that the motions of the planets could be described in terms of combinations of uniform circular motions ­(although it was becoming clear that such schemes had to be extraordinarily complex to fit the best observational data available). Ten years later, in his book Harmonice Mundi (“Harmonics of the World”), Kepler stated a third empirical law: 3. The square of a planet’s period T (the time that it takes to complete one orbit) is proportional to the cube of the orbit’s semimajor axis a. (An ellipse’s semimajor axis is defined to be one-half its widest diameter.) The three numbered laws above are known as K ­ epler’s three laws of planetary motion. Figure N11.1 illustrates these laws. Again let me emphasize that these laws are entirely empirical: they do not so much explain as describe the motion of the planets. However, because Kepler supported them so carefully with observational data of extraordinary quality, these laws became widely accepted in spite of their radical character. For more than six decades after Kepler’s third law was published, the scientific community could not say why these laws were true. Isaac N ­ ewton’s amazing triumph was to show that each of these laws follows directly from his second law of motion and the assumption that the force of gravity between two objects depends on the inverse square of their separation. In other words, Newton offered an explanation of Kepler’s laws in terms of physical principles that applied equally well to terrestrial motion: one simply had to accept that the planets were endlessly falling around the sun. Let me emphasize how radical this suggestion was at the time! Before Newton, scholars had believed that the laws of physics pertaining to the motion of heavenly bodies (where unceasing motion in approximate circles seemed to be the rule) were completely distinct from the laws pertaining to terrestrial motion (where objects generally come to rest rather quickly). It took Newton’s genius not only to see that this very credible division between celestial and terrestrial physics was in fact not necessary at all (which, granted, was beginning to occur to others as well), but also to provide a complete theoretical perspective that unified terrestrial and celestial physics in a manner that demonstrably worked. The simplicity, beauty, and extraordinary

N11.2 Orbits Around a Massive Primary

173

predictive power of Newton’s ideas were so compelling that it brought the physics community to its first real consensus on a grand theoretical structure for physics. This first consensus (as discussed in chapter C1) in some sense marks the birth of physics as a scientific discipline. Our goal in this chapter is to show that Kepler’s laws are a consequence of the laws of mechanics that we have been studying. We will not quite follow the same path Newton did in proving this, since computers and the law of conservation of angular momentum give us more powerful tools than even Newton had at his disposal. (Using these tools will enable us to do in a few pages what it took Newton scores of pages to show in the Principia.)

Our goal is to prove that Kepler’s laws follow from principles we have studied

N11.2 Orbits Around a Massive Primary The first step toward understanding what Newton’s laws say about planetary motion is to take advantage of the simplifications that result when ­(1) our system of interest consists of a very massive object interacting with a much lighter object and (2) we choose the origin of our reference frame to be the system’s center of mass. Consider an isolated system consisting of an object of mass M interacting with a smaller object of mass m (see figure N11.2). If the system is really isolated, then its center of mass will move at a constant velocity and thus can be used as the origin of an inertial reference frame. In practical situations, it actually is more likely that the system is freely falling in some external gravitational field (for example, the earth and moon falling around the sun), but we can still treat the system’s center of mass as the origin of an inertial reference frame if we ignore the external gravitational field (as we saw in chapter N8). If we define the origin to be the system’s center of mass, then the definition of the center of mass means that

W  m  ​   r​W ​ W = −m ​ rW ​   0 = _________ ​ m​  Wr  ​ + M  ​R​  ​     ⇒      M  ​R​   ⇒   R​ ​W = −​ ___   (N11.1) M M+m

This implies that in all circumstances (no matter how the two objects move W and/or interact with each other) the positions R​ ​  and  r​W ​ of the objects relative to the center of mass are opposite and have magnitudes proportional to each other: if r gets bigger or smaller, then so does R (proportionally). Taking the time derivative of both sides of equation N11.1 yields m  ​ ​ v​ W = −​ ___ W   ⇒  │V​ ​ V​  W ​ │   = ___ ​  m ​│    v​  ​W│   (N11.2) M M

m

y r

M R Fg

Fg

x System center of mass

Figure N11.2 An isolated system of two objects interacting gravitationally. Note that if we define the origin of our reference frame to be attached to the system’s center of mass, the positions of the objects are opposite. The gravitational force exerted on each object by the interaction points directly toward the other.

General situation: an isolated pair of interacting objects

Results in a frame based on the system’s CM

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The fact that │V​  ​W│    = (m/M)│v​  ​W│    means that

( ​ Mm ​  v​ ​W  )​​= ​( ​ Mm  ​  )​ ​( ​ 12 ​  m v​ ​W   )​ 

1 ​   M│V​ ​ __  W ​ │   2 = __ ​  1 ​  M ​​ 2 2

 2 ___│   │  ​

___  __  │ │2

m ⇒  KM = ___ ​    ​   Km (N11.3) M

where KM and Km are the kinetic energies of the massive and light objects, respectively. Thus (as we’ve seen before), the more massive object’s kinetic energy in this frame is smaller than that of the lighter object by the factor m/M. W Chapter C7 argued that a particle’s angular momentum L​ ​  about a given W W,  where  rW​  ​  is the particle’s origin is given by the “cross product” L​ ​   =  r​W ​  × m​v​ position relative to the origin, m is its mass, and v​ ​W  is its velocity. If you did not read this chapter, all you need to know for our present purposes is that this cross product produces a vector whose direction is perpendicular to both W  and whose magnitude is │L​ W│  r​W ​  and m​v​  W ​│    = │ r​W │ ​ │m​v​  │sin θ│ (where θ is the angle W​  ). between  rW​  ​ and v​ Given this, we can express the larger object’s angular momentum L​ ​W M W around the system’s center of mass in terms of L​ ​ m  as follows:  

(  )

(  )

m m m mW W M ≡ R​ W = ​ − ​ ___ W)  = ___ ​ L​ ​W × M​V​   ​  ​ ​ rW ​ × M ​ − ​ ___  ​  ​ ​ vW ​= + ​ ___  ​   (​ rW ​ × m ​v​ ​    ​ L​ ​  m (N11.4) M M M M

Implications when M >> m

W m around The system’s total angular momentum L​ ​W  = L​ ​W m + L​ ​W M = (1 + m/M)  ​L​ its center of mass will be conserved if and only if the light object’s angular momentum is conserved around that center of mass. Equations N11.1 through N11.4 apply to any isolated (or freely falling) system of two interacting objects described in a reference frame whose origin is the system’s center of mass. If, in addition, we have M >> m, then 1. 2. 3. 4. 5.

This approximation is very useful in the solar system

W The massive object’s position is R​ ​  ≈ 0 (by equation N11.1). W This object is essentially at rest at the origin: V​ ​  ≈ 0 (by N11.2). Its kinetic energy is negligible: KM ≈ 0 (by N11.3). W Its angular momentum is negligible: L​ ​M   ≈ 0 (by N11.4). The objects’ separation ≈ distance of the lighter object from the origin.

In such a case, the massive object (which we call the system’s primary under these circumstances) is essentially at rest at the origin, and the lighter object (which we call a satellite) orbits it. The primary then provides an essentially fixed origin for the gravitational force exerted on the satellite. This approximation holds very well in the solar system. Even Jupiter’s mass is more than 1000 times smaller than the sun’s mass, and the earth’s mass is more like 330,000 times smaller. Similarly, the moons that orbit the major planets typically have masses much smaller than their primary: even our own moon (the largest in the solar system compared to its primary) has 81 times less mass than the earth. The fact that planetary masses are so small compared to the sun has another important implication. When computing the orbit of one planet, we can ignore the gravitational effects of the others (to an excellent degree of approximation). This is because the sun is so much more massive than anything else that the gravitational force it exerts is by far the greatest influence on each planet’s motion. Therefore, as an excellent approximation, we can pretend that each planet orbits the sun as if it were alone.

N11.3 Kepler’s Second Law W W Equation N11.4 and L​ ​W = L​ ​M   + L​ ​ m  imply that the angular momentum of either object in an isolated interacting pair is conserved along with that of the whole.

N11.3 Kepler’s Second Law

175

This has two important consequences: (1) the object’s orbit lies in a plane, and (2) its position vector sweeps out equal areas in equal times. The first of these consequences follows from the definition of the angular W W W.  momentum, which says that L​ ​   for either object is defined to be L​ ​   ≡  r​W ​  × m​v​ W By definition of the cross product, this means that L​ ​   is always perpendicuW lar to  rW​  ​.  But if L​ ​  has a fixed orientation in space (because it is conserved), then the object’s position vector  rW​  ​ must always lie in the fixed plane perpen­dicular W to L​ ​ .  Therefore, the object’s orbit lies in a certain fixed plane. Conservation of angular momentum also implies K ­ epler’s second law as follows. Imagine that in an infinitesimal time dt, the object moves a certain infinitesimal angle dϕ (shown greatly exaggerated in figure N11.3) as it moves in its orbit around the system’s center of mass. The area swept out by the line between the object and the system’s center of mass is the colored pie slice in figure N11.3. If dϕ is very small, the shape of the slice is very nearly triangular, so its area is dA ≈ _​ 12  ​ (base)(height). Now, as shown in the drawing, the height h of the triangle is very nearly equal to the arc length r dϕ, and this approximation gets better and better as dϕ 0. So if dϕ is small,

The orbit of either object must be in a fixed plane in space

The position vector of either object sweeps out equal areas in equal times

d A ≈ _​  12 ​  r (rdϕ) = ​ _12 ​  r 2 dϕ(N11.5)



 

Dividing both sides by dt and taking the limit as dt and dϕ go to zero yields dϕ d A 1 2 ___ ​ ___ ​ = __ ​   ​  r ​   ​ (N11.6) dt 2 dt  

Now note from the figure that ( v​ ​W  dt ) sin θ = h ≈ r dϕ. The magnitude of the object’s angular momentum is thus  

 

r dϕ dt

dϕ dt

│L​ W│  W ​│   =│ r​W ​│  │m  ​v​  │sin θ│ = r m ____ ​   ​ = mr 2 ___ ​   ​ (N11.7)  



 

This means that

( 

 

)

│L​ dϕ dϕ d A 1 2 ___  W ​│   1 ​ ___ ​ = __ ​    ​  r ​   ​ = ___ ​     ​ ​ m r 2 ___ ​   ​   ​= ____ ​   ​     (N11.8) dt dt 2 dt 2m 2m  

 

 

Since │L​  W ​│    is conserved, dA/dt is constant, meaning that the object’s radius vector sweeps out equal areas in equal times. Note that this applies to either object, independent of the objects’ relative masses or the nature of their interaction. Now, Kepler’s second law actually says that the line between the planet and the sun sweeps out equal areas in equal times, not the line between the planet and the system’s center of mass. But if the sun is much more massive than the planet, it essentially is located at the system’s center of mass, and Kepler’s second law is essentially correct.

Application to Kepler’s second law

Figure N11.3 System center of mass

r

dϕ

Object rˆ

θ vdt

h ≈ rdϕ Orbit

The colored region shows the area swept out by the line between the object and the system’s center of mass as the object moves a tiny angle dϕ around the center of mass (the object’s initial position is the white dot and its final position is the black dot). This area is almost equal to _​ 12  ​r h, and h in turn is approximately equal to the arclength r dϕ.

176

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N11.4 Circular Orbits and Kepler’s Third Law We will assume M >> m from now to the end of the unit

Circular orbits at constant speed are possible

From here through the next chapter, we will (unless otherwise specified) assume the primary-­satellite approximation, in which the mass of the primary is much greater than the mass of the satellite. To simplify notation, when we refer to the satellite’s kinetic energy and/or angular momentum from now W W on, we will drop the subscripts on Km and L​ ​ m  and simply use K and L​ ​ .  If the primary is much more massive than its satellite, then it is possible for the satellite to follow an essentially circular orbit at constant speed around the primary. Let us see whether such an orbit is consistent with what we know about uniform circular motion and the gravitational interaction. We know that an object moving in a circular trajectory at a constant speed is accelerating toward the center of its circular path, and that the m ­ agnitude of its acceleration is │ a​W │ ​   = │v​  ​W│   2/R = a constant, where R is the radius of the object’s orbit and │v​  ​W│    is its constant orbital speed. Newton’s second law tells us that this acceleration must be caused by some force that is directed toward the center of the satellite orbit and which has a constant magnitude. The gravitational force F​ ​Wg  exerted on the satellite by its gravitational interaction with its primary satisfies these criteria. It is directed toward the center of the satellite’s trajectory (to the extent that we can consider the massive object to be at rest). As we saw in chapter N2, its magnitude is given by Newton’s law of universal gravitation:



│F​  W ​ g │= ______ ​  GMm     ​ 

r2

(N11.9)

 

• Purpose:  This equation specifies the magnitude of the gravitational force F​ ​W g that an object of mass M exerts on an object of mass m (or vice  versa) when their centers of mass are separated by a distance r. G = 6.67 × 10-11 N · m2/kg2 is the universal gravitational constant. • Limitations:  This equation applies to point masses or spheres, but not to irregularly shaped objects. Also it does not apply to extremely strong gravitational fields (much stronger than any fields in our solar system) or to objects moving at close to the speed of light.

Implications of Newton’s ­second law for circular orbits

(We derived this in chapter N2 from the gravitational potential energy formula.) Therefore, in the case of a truly circular orbit (where r = R = constant) the magnitude of the force will be GMm/R2, which is a constant, as needed. This means that it is at least plausible that the gravitational force exerted on the satellite due to its interaction with the primary can hold the satellite in a circular orbit. This does not mean that orbits always are circular: indeed, orbits generally are not. But it does mean that a circular orbit is a possibility. In fact, the orbital radii of most major objects in the solar system are constant to within a few percent. Most artificial satellites orbit the earth with approximately circular orbits as well. Therefore, out of all the kinds of possible orbits, this “special case” is a very good approximate model for a broad variety of realistic situations. Let’s see what Newton’s second law can tell us quantitatively about such orbits. If we assume that the gravitational force is the only force acting on the satellite, then │F​  W ​ n  et│ = │F​  W ​ g  │ = GMm/R2. Therefore, the magnitude of Newton’s second law tells us that

N11.4 Circular Orbits and Kepler’s Third Law

177

GMm m│ a​W​  │ = │F​  W ​ n  et│ = ​ ______  ​    R2



(N11.10)

Dividing both sides by m and substituting │v​  ​W 2│/R for │ a​W ​│    yields  



│v ​W​│   GM _____ │v ​W​│ ​      ​ = ____ ​        ​   ⇒     2

R

R2

_____

     ​ ​   = ​ ____ ​  GM R

(N11.11)

• Purpose:  This equation specifies the orbital speed │v​  W​ │    of a satellite in a circular orbit of radius R around a primary with mass M, where G is the universal gravitational constant. • Limitations:  The orbit must be circular, and the primary must be much more massive than the satellite. The limitations on equation N11.9 also apply here. (Note that │v​  ​W│    is appropriately constant.) We can use this information to determine how long it will take the satellite to go once around its circular orbit. The orbit’s period T in this case is the time it takes the satellite to travel a distance 2π R at a constant speed │v​  ​W│  :  

T = ____ ​  2π R ​  (N11.12) │v​  W​ │  



 

If we square equation N11.12 and use equation N11.11, we can show that



2

T 2 = ____ ​  4π   ​   R3(N11.13) GM  

 

• Purpose:  This equation specifies the orbital period T of a satellite in a circular orbit of radius R around a primary with mass M, where G is the universal gravitational constant. • Limitations:  The orbit must be circular, and the primary must be much more massive than the satellite. The limitations on equa­tion N11.9 also apply here.

Exercise N11X.1 Verify equation N11.13. Kepler’s third law states that the square of a planet’s period is proportional to the cube of its semimajor axis. When an orbit is circular, its semimajor axis (one-half of the distance measured across the widest part of the orbit) is its radius R. So we see here that in a few lines we have derived Kepler’s third law (for the special case of circular orbits anyway). Moreover, this derivation even gives us the constant of proportionality between T 2 and R3!  

Exercise N11X.2 The earth orbits the sun in an approximately circular orbit once each year. If it were 4 times as far from the sun, how long would it take to orbit once?

Kepler’s third law for circular orbits

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Chapter N11

Kepler’s Laws

N11.5 Circular Orbit Problems

A checklist for solving circular orbit problems

In this section, we’ll explore some ­examples of how equations N11.11 and N11.13 can answer many questions about orbiting objects and their primaries. Since these examples involve circularly “constrained” motion (although only because most orbits happen to be roughly circular, not because they must be), we can adapt the checklist we used for constrained-motion problems. But we need some modifications. For example, we don’t need to write ­Newton’s laws in component form (equa­tion N11.11 or N11.13 will probably be more helpful), so we don’t need to define coordinate axes. A free-body diagram of an orbiting object will generally show only one force vector, so it is pretty pointless. The only important issues to be addressed in the modeling part are the validity of the circular orbit approximation and the assumption that the primary is indeed much more massive than the satellite. So an adapted problem-solving checklist for orbital motion problems might look as follows: Circular Orbit Problem Checklist ❒ Draw the two interacting objects. ❒ Draw a labeled arrow for the orbiting object’s velocity vector. ❒ Check that one object is very massive. ❒ Check that the satellite’s orbit is circular (or state this as an assumption). ❒ Apply equation N11.11 and/or N11.13 as your master equation(s).

General Checklist ❒ Define symbols. ❒ Draw a picture. ❒ Describe your model and assumptions. ❒ List knowns and unknowns. ❒ Use a master equation. ❒ Do algebra symbolically. ❒ Track units. ❒ Check your result.

The examples that follow illustrate solutions following this outline. You should check for yourself that all of the checklist elements are present in these example solutions.

Example N11.1

Problem:  Consider a spaceship in a circular orbit at an altitude of 250 km above the earth’s surface. What is its orbital speed? Its orbital period? Solution Known: Re = 6380 km M = 6.0 × 1024 kg R = 6380 km + 250 km G = 6.67 × 10–11 N·m2/kg2

v 250 km

R

Earth (mass M)

Unknown: v =?T=? Re

According to the problem statement, the ship is following a circular orbit around the earth. The earth is much more massive than a spaceship, so equations N11.11 and N11.13 should apply.

N11.5 Circular Orbit Problems



179

Equation N11.11 tells us that a circularly orbiting object’s speed is _____________________________________________

( 

)

____ (6.67 × 10-11 N·m2/kg 2)(6.0 × 1024 kg) __________ 1 kg · m/s2 _________________________________  │v​  ​W│    = ​ ____ ​  GM    ​ ​   =  ​      ​              ​​ ​   ​     ​ ​ 6,380,000 m + 250,000 m  R 1N  



= 7770 m/s = 7.8 km/s (N11.14)

If we know the speed, equation N11.12 more quickly yields the period than equation N11.13 does: the orbit’s period is

(  )

2π(6380 km​ ​@ + 250 km​ ​@)  ______ 1 min 2πR _____________________  ​ =    T = ____ ​  ​      ​ ​ ​   ​    ​= 89.4 min (N11.15) │v​ │ 60 s  ​W  7.77 km​ ​@/  s

This is a bit longer than the result we found in example N7.5. But this is plausible: since the ship orbits some distance above the earth’s surface, not only is gravity a bit weaker than at the earth’s surface but the distance around the orbit is a bit longer. Both effects will make an orbit a bit longer than one just barely above the earth’s surface.

Problem: Ganymede, Jupiter’s largest moon, has a nearly circular orbit whose radius is 1.07 Gm. The moon goes around Jupiter once every 7 d, 3 h, and 43 min. What is Jupiter’s mass? Solution Here is a drawing of the situation: Known: T = period of orbit = 7 days, 3 h, 43 min R = 1.07 Gm

Jupiter, mass M

Unknown: M=?

Ganymede v

R

We are told that Ganymede’s orbit is circular, and we can safely assume that Ganymede is much less massive than Jupiter. Since we are given R and T, we can solve equation N11.13 for the mass M of Jupiter: 2

2

3

4π   ​   T 2 = ​ ____ R3  ⇒  M = ______ ​ 4π R2 ​     (N11.16) GM GT



 

   

 

 

The period T, expressed in seconds, is

(  )( 

(  )

)

( 

)

  ______ ⨉ ​    T = 7 d ​ ____ ​  24 h​ ​  ​​ ​  3600 ​ s   ​+ 3 h​​ ​⨉  _____ ​  3600s  ​    ​+ 43 min  ​​ ​ ______ ​  60 s  ​  ​= 6.18 × 105 s ​  1 min​ 1 h​ ​⨉  1d 1h ​⨉​  (N11.17)

Substituting this into equation N11.16 yields

( 

)

1N 4π 2 (1.07 × 109 m)3 _________________________________ M = ​           ​ ​ _________ ​    2 ​   ​= 1.9 × 1027 kg -11 2 2 5 2 1 kg·m/s (6.67 × 10 N·m /kg )(6.18 × 10 s) (N11.18)  

 

 

This seems plausible. Note that we can determine an object’s mass by observing the motion of a smaller object orbiting it (this is indeed the standard method of determining the masses of astronomical objects).

Example N11.2

180

Example N11.3

Chapter N11

Kepler’s Laws

Problem:  Astronomical measurements show that Jupiter orbits the sun in a nearly circular orbit once every 11.86 years. How does the radius of Jupiter’s orbit compare with that of the earth? Solution Known: TJ = period of Jupiter’s orbit = 11.86 y TE = period of Earth’s orbit = 1.0 y Unknown: RJ / RE = ? M = ?

Sun, mass M

vE

Jupiter vJ

RE

RJ

Earth

Jupiter and earth both orbit the sun, which is much more massive than either planet. We are ignoring the comparatively tiny interaction between earth and Jupiter. To do this problem, we have to assume that both the earth’s orbit and Jupiter’s orbit are circular (which is  pretty closely true). If this is so, then equation N11.13 applies to both ­orbits, so 2 4π 2  ​​   ​ T 2​J​  ​= ​ ____ R 3​J​  ​  and   ​T 2​E ​​  = ____ ​  4π   ​​   R 3​E ​​  GM GM  

(N11.19)

 

Since we are really looking for the ratio RJ/RE, the fastest way to find that ratio is to divide the first expression by the second:

(  ) ( 

​T 2​J​  ​ (4π 2/GM)​R ​3J​  ​ RJ ​ ___2   ​ = ____________ ​        ​  ⇒  ___ ​    ​ = ​​ 2 3 RE ​T ​E ​​  (4π /GM)​R ​E ​​   

 

TJ ___ ​    ​  ​​ TE

)

11.86 y 2/3 ​= ​​ _______ ​     ​   ​ ​= 5.20(N11.20) 1.0 y

2/3

This is consistent with the result stated in the inside front cover.

N11.6 Black Holes and Dark Matter

Using Kepler’s third law to find a black hole

In 1994, H. C. Ford and R. J. Harms announced that they had discovered a giant black hole at the center of the galaxy known as M87. This was the first time anyone had presented an apparently iron-clad argument for the existence of a giant black hole at the center of any galaxy. A black hole is an object so dense that light cannot escape it. So how can we hope to “discover” an object that doesn’t emit any light? If you can find something orbiting the black hole and you can measure the orbit’s period and radius, Kepler’s third law allows you to compute the black hole’s mass! Ford and Harms used the Hubble Telescope to take spectra from very small regions of a gas cloud near the center of M87. By measuring the D ­ oppler shifts of the spectra obtained, they were able to determine that gas that was a radius R = 60 light-years (where 1 ly = 0.946 × 1016 m) from the center of the cloud was orbiting the center with a speed of 450 km/s. If we assume that the orbit is roughly circular, T = 2πR/│v​  ​W│   . Plugging the latter equation into Newton’s version of Kepler’s third law and solving for M, we get

(  )

4πR3│v​  ​W│   2 ______ R│v​  ​W│  2 4πR3 2πR 2  ​    ​ ​   ​ _____ ​ = T 2 = ​​ ____ ​  ⇒  M = _________ ​   ​  = ​   ​     (N11.21) 2 │v​  ​W│   GM G(2πR) G  

N11.7 Kepler’s First Law and Conic Sections

181

Substituting in R = 60(0.946 × 1016 m) and │v​  ​W│    = 450 km/s yields 1.7 × 1039 kg, 9 which is about 10 solar masses (1 solar mass = 2.0 × 1030 kg). This is just one of many bits of evidence that Ford and Harms cite. How do we know that this is a black hole? No other explanation works! A cluster containing a billion stars within a radius of 60 ly would emit lots of light that is not seen. No plausible model other than the black hole model explains how a billion solar masses could fit within a sphere 60 ly in radius and yet be consistent with all the available data collected by Ford and Harms. For several decades, astronomers also have been collecting evidence that more than 80% of the mass associated with matter in the universe is dark matter, that is, matter that is not in stars that emit light or in dust clouds that emit radio signals, reflect light, or obstruct light. Consistent evidence for this dark matter comes from a variety of studies of the orbital motion of stars within galaxies, satellite galaxies around large galaxies, and galaxies in galactic clusters. Many of these studies use Newton’s version of Kepler’s third law to determine the mass of the unseen matter by its gravitational effects. In one study, D.N.C. Lin, B. F. Jones, and A. R. Klemona carefully measured the transverse movement of the Large Magellanic Cloud (LMC) relative to background objects. The LMC is a small galaxy that is a companion to and presumably orbits our own much larger galaxy. Lin and his collaborators measured the transverse velocity of the LMC to be about 200 km/s. Since the LMC is about 170,000 ly (1.6 × 1021 m) from our galaxy, if it were in a circular orbit, that would imply that our galaxy’s mass is

( 

Dark matter and Kepler’s third law

)

R│v​  ​W│   2 _________________________ 1N (1.6 × 1021 m)(200,000 m/s)2 _________ M = ______ ​   ​     = ​          ​  ​ ​    2 ​   ​= 1.0 × 1042 kg -11 2 2 G 1 kg·m/s 6.67 × 10 N·m /kg (N11.22) which is about 500 billion solar masses (a more accurate calculation based on the LMC’s real trajectory actually yields 600 billion solar masses). Visible matter in our galaxy amounts to about 100 billion solar masses. What is this dark matter? No one knows! Recent experiments have strongly suggested it cannot be dwarf stars, large planets, or black holes; mounting evidence suggests it cannot be ordinary matter made of protons, neutrons, and electrons at all. The mystery is still unsolved, but part of the point here is that we would not even know this major fraction of the universe existed if it weren’t for Newton’s form of Kepler’s third law!

N11.7 Kepler’s First Law and Conic Sections Kepler’s first law states that planetary orbits are ellipses. An ellipse is one of a family of curves that mathematicians call conic sections (the curves that result from slicing a cone through various angles); the family also includes circles, parabolas, and hyperbolas. The general formula for a conic section is



b     r = __________ ​  ​ (N11.23) 1 + ε cos θ

• Purpose:  This equation describes a conic section by specifying the distance r the curve is from the origin (the focus) at a given angle θ measured from the line that connects the origin to the curve’s nearest point. The constant ε is the curve’s eccentricity and b (= r at 90°) is a constant. • Limitations:  None: this is a definition.

Conic sections

Chapter N11

182

Kepler’s Laws

Figure N11.4 The features of an ellipse. We measure the value of r for any angle θ from the origin, which we call a focus of the ellipse. The constant b is the value of r at θ = π/2, rc = r (0) is the distance from the focus to the closest point on the ellipse, and rf = r(π) is the distance to the farthest point. The major axis is the ellipse’s greatest width: half of that width is a, the semimajor axis. (The ellipse technically has two foci located symmetrically along the ellipse’s major axis.)

r

b 1 + ε cosθ

r(θ) =

b θ

rf

 

2a (major axis)

(second focus)

rc

origin (focus)

ellipse

The eccentricity ε specifies the curve’s shape. If ε = 0, then the expression reduces to r = b, and the curve is a circle of radius R = b. If ε is nonzero but less than 1, then the curve is an ellipse, whose closest point to the origin (focus) is rc = r(0) = b/(1 + ε) and whose farthest point from the origin is rf = r(π) = b/(1 - ε): see figure N11.4. If ε = 1, then the farthest point is at infinity, and the curve becomes a parabola. If ε > 1, then the curve is a hyperbola. Only ellipses have a finite farthest point, but all conic sections have a closest point. So we can calculate a given conic section’s eccentricity ε as follows: b b rc = _____ ​     ​   ⇒  ε = __ ​    ​ − 1 (N11.24) rc 1+ε

An ellipse’s semimajor axis a

I’ll talk more about hyperbolas shortly, but let’s focus for the moment on ellipses. We define an ellipse’s semimajor axis a to be half of its widest diameter. As figure N11.4 illustrates, the widest diameter is rc + rf, so

( 

) ( 

)

b b 1−ε+1+ε b 1 1 b a = __ ​    ​   (rf + rc) = __ ​   ​ ​   ​ _____    ​  + _____ ​     ​  ​= __ ​    ​  ​ ____________ ​     ​   ​= ______ ​     ​   2 2 1+ε 1−ε 2 1 − ε2 1 − ε2  

(N11.25)

 

This means that b = a(1 - ε 2) and therefore that  



b a (1 − ε 2) (​1@ + ε​) (1 − ε) rc = _____ ​     ​  = ________ ​      ​  = a ​  ____________     ​   = a (1 − ε) (N11.26) 1+ε 1+ε 1​@ + ε​   

 

 

 

and similarly rf = a (1 + ε). These results will be useful in what follows. Newton’s second law and the law of universal gravitation implies that if a satellite’s orbital velocity at a given distance r from its primary is in the ______ _______ range  ​ GM/r ​     > m approximation? What other assumptions (if any) do you have to make to solve the problem? (b) Problem N11A.1 discusses a version of Kepler’s third law that applies even when M and m are comparable. Use this result to calculate the distance between the stars’ centers and compare it to the result of part (a).

Answers to Exercises

187

Derivations N11D.1 Suppose two objects with masses M and m are connected by a spring with zero relaxed length, so that the attractive force that each exerts on the other is F = ks r. Assume that M >> m, and that the satellite orbits in a circular orbit of radius R around its primary. Find an expression (analogous to Kepler’s third law) that gives the period of the orbit T as a function of the orbital radius R, the spring constant ks, and whatever else you need.

(a) Suppose that the two black holes have equal masses M and that their orbits are circular and have a radius R0 around the system’s center of mass. Show that the speed of each black hole in its orbit is given by ____

 

N11D.2 Imagine that in a different universe the magnitude of the force of gravity exerted by one object on another were given by │F​  W ​ g  │ = _____ ​  BMm  ​    



(N11.30)

r3  

where r is the separation between the two objects and B is some constant. What would Kepler’s third law for circular orbits be in this universe? N11D.3 The force law for the magnitude of the force between two quarks separated by a distance r turns out to be very roughly │F​  W ​│    = b, where b is a constant. If we pretend that Newton’s laws apply to quarks, and imagine that one quark is much lighter than the other and is in a circular orbit around it, what would be the equation for the lighter quark’s period T as a function of its orbital radius R?

Rich-Context N11R.1 Consider a spherical asteroid made mostly of iron (whose density is 7.9 g/cm3) whose radius is 22 km. Could you run fast enough to put yourself in orbit around this asteroid? N11R.2 Astrophysicists believe that a collision between galaxies might lead to the resulting combined galaxy having a pair of supermassive black holes at its center. Interactions between the black holes and the stars and gas in the galactic center will quickly circularize the black holes’ orbits around their common center of mass.



1 ​  │v​  W​ │    = ​ __ ​ ____ ​ GM    ​ ​    2

R0

(N11.31)

(Hint: Equation N11.11 does not apply in this case because the black holes have comparable masses.) (b) Interactions with the surrounding galactic matter will cause the distance between the black holes to shrink very gradually with time. Imagine that over a period of 60 My (60 million years) the orbital radius shrinks by one-half to R = _​ 12  ​  R0. Compute the change in the blackhole binary system’s total energy (kinetic + potential) during this time (express your answer in terms of G, M, and R0). (c) Has the binary lost energy to or gained energy from its surroundings? Does this make intuitive sense? (d) If M = 2.0 × 1036 kg and the initial separation of the black holes is R0 ≈ 10 ly ≈ 1017 m, roughly what is the average rate at which energy is t­ransferred, in watts? (For comparison, the rate at which the sun radiates energy is about 3.9 × 1026 W.)

N11R.3 Suppose you wanted to put a satellite in such a circular orbit that it appeared on the western horizon every Monday morning at 6 a.m. and was never at any other time on the western horizon. What radius should the orbit have? (Don’t forget to account for the earth’s rotation!)

Advanced N11A.1 Consider an isolated two-object system interacting gravitationally such that M is not necessarily much greater than m. Assume that the smaller object’s orbit around the system’s center of mass is circular and has radius r. Argue that both objects have the same orbital period T around the system’s center of mass and that

2

3

4π D   ​   T 2 = ​ _________ G(M + m)    

 

(N11.32)

where D is the separation of the two objects.

ANSWERS TO EXERCISES N11X.1 Substituting equation N11.11 into equation N11.12 yields ____



____

2πR   ​  T = _________ ​  _______ = 2πR  ​ ____ ​    R   ​ ​  = 2π  ​ ​ ____   R   ​ ​    (N11.33)  GM GM ​ GM/R ​     3

Squaring both sides of this gives us equa­tion N11.13. N11X.2 According to equation N11.13, T ∝ R3/2. If we increase __ 3 3 R by a factor of 4, then R3/2 increases by a factor of (​4 ​   )   =2 = 8. So the orbit would last 8 y.

N11X.3 The formula predicts that the angle where r goes to infinity is θ = cos-1(-1/ε) = cos-1(-1/1.5)  = 132°. To check this, note that as the trajectory’s y value increases from 5.5 AU to 6.5 AU, its x value decreases from -2.3 AU to about -3.2 AU. So the tangent of the angle of this line is opposite over adjacent = 1/(-0.9) = -1.11. For comparison, tan(132°)  = -1.12, so this is pretty close. [Note that if you calculate tan-1(-1.11), most calculators will deliver -48°, but 132° = 180° - 48° is also a valid solution to tan-1(-1.11).]

rbits and N12   OConservation Laws Chapter Overview Introduction In this final chapter of the unit, we will bring many of the tools we have developed in this unit (and in unit C) to bear on the problem that represents the crowning triumph of the Newtonian synthesis. In particular, we will see how we can use conservation of energy and angular ­momentum as tools to help us understand orbital behavior and determine an orbit’s characteristics.

Section N12.1:  Overview and Review This section reviews the conic section results from chapter N11. We see that we generally only need a pair of numbers to completely describe an orbit.

Section N12.2:  Conservation Laws and Elliptical Orbits

Both the energy E and the angular momentum magnitude L ≡ │L​  ​W│    of a primary/satel­ lite system will be conserved throughout the orbit. This section discusses how one can use this fact to link the values of E and L for a given system to the values of a (or b) and ε that determine the size and shape of the orbit.

Section N12.3:  Conservation Laws and Hyperbolic Orbits Similar equations link the values of E and L for a given system to the values of a (or b) and ε for a hyperbolic orbit. One of the most important basic consequences of the equations is that If  E < 0  then ε < 1 (the orbit is elliptical) If  E > 0  then ε > 1 (the orbit is hyperbolic)

Section N12.4:  Solving Orbit Problems

Table N12.1 summarizes the equations that link various features of an orbit with the energy and angular momentum of the system. Given the equations in table N12.1, one can solve an ­astonishing range of orbit problems given only two pieces of information about the orbit (such as the satellite’s distance from the primary and its velocity at a given time, or the points of the orbit that are nearest or farthest from the primary). This section contains a variety of examples that illustrate approaches to solving such problems.

188

Table N12.1 Table of useful equations for solving orbit problems Item

Equation(s)

Formula for a conic section, Formula for a

b   ​  b     ​ r = __________ ​  ,   a = ________ ​  1 + ε cos θ │ 1 − ε 2│

Definitions of E and L Formulas for ε and b

Item Connection between a and E

Location of extremes

Other useful relations

Symbols

r = distance from the primary b = scale constant ε = eccentricity θ = angle from the axis between L ___ __ │ ​W│ │ ​W│ ​  2E   2 − _____ ​ 2GM       and  ​    sin ϕ the primary and the orbit’s m ​ =  v​ r ​ m  ​ = r v​ closest point 2 2 1 L 2E 1 L 2 ______ __ ___ ____ __ E = total system energy ε  = 1 + ​     ​  ​​ ​    ​   ​ ​ ​ ​   ​  ​ and b = ​     ​ ​​ ​    ​   ​ ​ (GM)2 m m GM m L = system’s angular momentum G = gravitational constant m = satellite’s mass M = primary’s mass Elliptical Case Hyperbolic Case │  v​  ​W│    = satellite’s speed at an instant GM GM 2E ​ = + ​ ____ ___ ____ ___ ϕ = angle between  r​W ​ and v​ ​W  ​  2E  ​   = −  ​   ​     ​   ​     m a m a a = semimajor axis (for ellipse, half the widest diameter) rc = a (1 − ε)   a = _​  12 ​  (rf + rc) rc = a (ε − 1) rc = distance from primary to the rf − rc closest point on the orbit rf = a (1 + ε)  ε = ​ ______   ​ θ∞ = cos-1 (−1/ε ) rf + rc rf = distance from primary to the ______ farthest point on the orbit 4π 2  2 T 2 = ​ ____  ​ a3 tan θ∞ = −​ε    − 1 ​  T = orbital period GM ___ 1 L 2E θ = value that θ approaches ____ __ ___ (Kepler’s third law)  ∞         = −  ​     ​ ​ ​    ​   ​ ​ ​     ​ ​   m GM m as r #$ ∞ in a hyperbola

(  ) (  )

(  )

(  )

189

Chapter N12

190

Orbits and Conservation Laws

N12.1 Overview and Review We saw in chapter N11 that orbits around a massive primary are either elliptical or hyperbolic. Armed with this knowledge, we can use conservation of energy and angular momentum to calculate many things about an orbit without having to solve Newton’s second law directly. In this chapter, we will discover just how useful these laws can be in helping us think about, quantify, and understand orbital motion. Before we begin, let’s briefly review what we learned about orbits in section N11.7. We saw that the orbit of an object orbiting a massive primary at the origin (the orbit’s “focus”) is described by the formula for a conic section

b   ​ r = __________ ​    (N12.1) 1 + ε cos θ

where ε and b are two constants that completely specify the orbit’s shape and size. The eccentricity ε describes the orbit’s shape (ε = 0 means the orbit is circular, 0 < ε < 1 that it is elliptical, ε = 1 that it is parabolic, and ε > 1 that it is hyperbolic). Within the elliptical range, the ellipse becomes more elongated as ε increases; within the hyperbolic range, the angle between the hyperbola’s incoming and outgoing legs increases. The constant b specifies something about the orbit’s size for a given shape. In section N11.7, we also defined an orbit’s semimajor axis a to be

b   ​ a ≡ ________ ​    (N12.2) │1 − ε 2│

For an ellipse, this quantity corresponds to half the ellipse’s widest diameter (and reduces to the simple radius of the circular orbit when ε = 0). The semimajor axis is important because Kepler’s third law links the orbit’s period to this quantity. For a hyperbola, it is less clear what the value of a describes physically, but we will see that it is still useful. In both cases, a is linked to rc, the distance from the origin to the orbit’s closest point as follows:

rc = a│1 − ε│ (N12.3a)

and to the distance rf of an elliptical orbit’s farthest point as follows:

rf = a (1 + ε) (N12.3b)

For elliptical orbits, we also have

vc rf

vf

Massive object

Figure N12.1

_​  1  ​ (r − r ) rf − rc f c a = _​ 12 ​  (rf + rc)  and   ε = ________ ​  2  ​     = ______ ​     ​ (N12.4) rf + rc a

so knowing the pair rf and rc is the same as knowing a and ε and vice versa. In what follows, we will see that we can link these orbital shape-and-size constants b, ε, a, rc, and rf to the total energy E and angular momentum L​ ​W  of the orbiting object, and that this will give us powerful insight into the physical meaning and implications of all of these quantities.

rc

An orbiting object’s velocity vectors are perpendicular to the radial direction at the orbit’s extreme points.

N12.2 Conservation Laws and Elliptical Orbits Now consider an object of mass m in an elliptical orbit around a massive primary with mass M. Note that the orbit’s closest and farthest points, the orbiting object’s velocity is entirely perpendicular to the radial direction (see figure  N12.1), since at these extreme points, the object’s velocity is switching from being outward to inward (or vice versa), so its radial component is passing

N12.2 Conservation Laws and Elliptical Orbits

191

through zero. Therefore, the magnitude of the orbiting object’s angular momenW│ tum around the primary is simply L ≡ │L​  W ​│    = │ r​W ​ × m ​v​    = rm│v​  ​W│   sin 90° = r m│v​  ​W│   at these points. Conservation of angular momentum and conservation of energy for these extreme points then imply that L  ​= r │v​    rc│v​  ​Wc │ = __ ​ m f  ​Wf │(N12.5a) │ vc│2 − _____ ​  2GM  ​    = ___ ​ 2E ​ = │v​  ​Wf │2 − _____ ​  2GM  ​      (N12.5b)



rc

m

Conservation of energy and angular momentum for an elliptical orbit

rf

The ratios L/m and 2E/m will appear in many of the equations that follow. I recommend you think about them less as ratios and more as if they were single numbers whose symbols happen to be L/m and 2E/m, respectively. I will place these ratios in parentheses in most of the equations that follow so you can more easily view each as a unit. Multiplying both sides of the left equality in equation N12.5b by ​r 2​c​  ​ = a 2 (1 − ε)2 = a 2 (1 − 2ε + ε 2) (see equation N12.3a) and substituting L/m for rc│v​  ​W c│ (equation N12.5a) yields

(  )

L  ​   2​ ​− 2GMa (1 − ε) = ___ 2 2 ​​ __ ​  m ​  2E m ​ a  (1 − 2ε + ε  ) (N12.6) Similarly, multiplying both sides of the right equality in equation N12.5b by ​r 2​f​  ​ = a 2 (1 + ε )2 = a2(1 − 2ε + ε 2) and using rf│ v​ ​W f│ = L/m (equation N12.5a) yields

(  )

L  ​   2​ ​− 2GMa (1 + ε) = ___ 2 2 ​​ __ ​  m ​ 2E m ​ a  (1 + 2ε + ε  )(N12.7) Subtracting equation N12.7 from equation N12.6 then yields

(  )

(  )

L  ​   2​ ​  − 2GMa (1 − ε) − ​​ __ L  ​   2​ ​+ 2GMa (1 + ε) = ___ 2 ​​ __ ​ m ​ m ​ 2E ​⨉2​ − 1 − 2ε − ε  ​⨉2​   ) m ​ a  (1 − 2ε + ε  GMm 2E ​ a2 4ε  ⇒  E = − ​ ______     ⇒  4GMaε = − ​ ___  ​      (N12.8) m 2a

The link between energy and the semimajor axis

This surprising result tells us that the total conserved energy of an object in an ­elliptical orbit depends on the orbit’s semimajor axis alone (not on the object’s angular momentum or the orbit’s eccentricity), and that the energy gets larger (less negative) as the semimajor axis increases. This very powerful equation is useful in a number of contexts. If we add equations N12.6 and N12.7, we get

(  )

L  ​   2​ ​− 4GMa = ___ 2 2 2 ​​ __ ​ m ​  2E m ​ a  (2 + 2ε  )(N12.9)

If we divide this equation through by 2 and substitute in a = −GM/(2E/m) (from equation N12.8) and rearrange things, we find that

(  ) (  )

2 ε 2 = 1 + ______ ​  1  2   ​​​ __ ​ L  ​ ​ ​​ ___ ​ 2E ​   ​(N12.10) (GM) m m

This equation gives us the relationship between the orbit’s eccentricity and the orbiting object’s conserved energy and angular momentum. Note that E is negative (see equation N12.8), so ε  1), and Massive primary

Figure N12.2 The definition of the asymptotic angle θ∞ for a hyperbolic orbit.

(  )

2 b = ____ ​  1   ​ ​​ __ ​ L  ​ ​ ​ GM m



(N12.14)

also as we found in the elliptical case. Another interesting quantity for ­hyperbolic orbits is the angle θ∞ that the orbiting object’s nearly straight trajectory far from the primary makes with the line connecting the primary with the orbits closest point (see figure N12.2). In section N11.7, we saw that cos θ∞ = −1/ε. Therefore _________



2

______

(  )

___

​ 1  − (1/ε)  ​   2 sin θ∞ ___________ 1 L ___ 2E tan θ∞ = ______ ​     ​  =    ​    ​  = −​ ε    − 1    ​= − ​____      ​ ​ __ ​    ​ ​ ​ ​    ​  ​   (N12.15) cos θ∞

−1/ε

GM m

m

where the last step follows from equation N12.13. Note that equations N12.8, N12.10, N12.12, and N12.13 are consistent in implying that a negative system energy implies an elliptical orbit and a positive system energy implies a hyperbolic orbit. We can summarize this by the following statements: An important result linking an orbit’s energy to its shape



If  E < 0  then  ε < 1  (the orbit is elliptical) If  E > 0  then  ε > 1  (the orbit is hyperbolic)

The case where E = 0 corresponds to the crossover between the two cases. This is of mathematical interest only (no real orbit will have its total energy exactly equal to zero), but the orbit in this case would be a parabola.

N12.4 Solving Orbit Problems

193

Table N12.1 Table of useful equations for solving orbit problems Item

Equation(s)

Formula for a conic section, Formula for a

b   ​  b     ​ r = __________ ​  ,   a = ________ ​  1 + ε cos θ │ 1 − ε 2│

Definitions of E and L Formulas for ε and b

Item Connection between a and E

Location of extremes

Other useful relations

Symbols

r = distance from the primary b = scale constant ε = eccentricity θ = angle from the axis between 2GM L  ​ = r│v​ 2 ___ _____ __ │v​ W│ W│ ​  2E  ​   =   ​   − ​   ​        and  ​   ​   sin ϕ the primary and the orbit’s m r m closest point 2 2 1 L 2E 1 L 2 ______ __ ___ ____ __ E = total system energy ε  = 1 + ​     ​  ​​ ​    ​   ​ ​ ​ ​   ​  ​ and b = ​     ​ ​​ ​    ​   ​ ​ (GM)2 m m GM m L = system’s angular momentum G = gravitational constant m = satellite’s mass M = primary’s mass Elliptical Case Hyperbolic Case │  v​  ​W│    = satellite’s speed at an instant GM GM 2E 2E ___ ____ ___ ____ ϕ = angle between  r​W ​ and v​ ​W  ​  m ​ = −  ​  a ​     ​  m ​ = + ​  a ​     a = semimajor axis (for ellipse, half the widest diameter) rc = a (1 − ε)   a = _​  12 ​  (rf + rc) rc = a (ε − 1) rc = distance from primary to the rf − rc closest point on the orbit rf = a (1 + ε)  ε = ​ ______   ​ θ∞ = cos-1 (−1/ε ) rf + rc rf = distance from primary to the ______ farthest point on the orbit 4π 2  2 T 2 = ​ ____  ​ a3 tan θ∞ = −​ε    − 1 ​  T = orbital period GM ___ 1 L 2E θ = (Kepler’s third law)         = −  ​ ____    ​ ​ __ ​    ​   ​ ​___ ​     ​ ​   ∞ value that θ approaches m GM m as r #$ ∞ in a hyperbola

(  ) (  )

(  )

(  )

The preceding result completely makes sense, if you think about it. The basic distinction between an ellipse and a hyperbola is that a hyperbola goes to infinite r at certain angles. Since the system’s gravitational potential energy goes to zero at infinity, if an orbiting object is to make it to infinity, it must have nonzero kinetic energy there, meaning that E > 0. If E < 0, the object cannot make it to infinity, and the orbit must therefore be elliptical.

N12.4 Solving Orbit Problems The point of the last two sections is that if we know the values of 2E/m and L/m for a system involving a satellite orbiting a massive primary, then we can calculate a and ε (or b and ε), which in turn completely specify the shape of the satellite’s orbit. Table N12.1 summarizes the crucial equations regarding orbits. Note that since E and L are conserved quantities for the orbit, we can calculate these quantities using position and velocity information at any point on the orbit. So what is the minimum information we need to completely determine an orbit’s shape and size? In general, we need to at least know the values of a suitable pair of quantities. The most basic pair are the constants appearing in the conic section equation (b and ε), but Table N12.1 shows that we can calculate these from other value pairs, such as: 2E/m and L/m; rc and rf ; rc and │v​  ​W f│; rf and │v​  ​W f│; a and ε; a and θ∞; and so on. We can also calculate the orbit from the values of r, │v​  ​W│   , and ϕ at any given instant. These equations allow us to solve an astonishing variety of interesting orbit problems. Examples N12.1 through N12.4 are just a sample.

We can use the equations in table N12.1 to solve a variety of orbit problems

194

Example N12.1

Chapter N12

Orbits and Conservation Laws

Problem:  A strange object is discovered 22 AU (3.3 Tm) from the sun. Measurements show its velocity to be 11.2 km/s in a direction 169.7° from the line between the object and the sun (meaning that the velocity vector points mostly toward the sun). Is this a previously unknown member of the solar system, or is it an interloper from outside? How close will it get to the sun? Solution  The drawing below shows the situation. θ

v ϕ

rc

v r ϕ θ

Sun

= 11.2 k m/s, = 22 AU = 10.3° = π – ϕ = 169.7°

?

Since we know the object’s initial distance r, its initial speed │v​  ​W│   , and the angle ϕ between the  r​W ​  and v​ ​W  vectors, we should be able to determine the orbit. Calculating 2E/m will help us determine whether the orbit is elliptical or hyperbolic, which will answer the question about whether the object is a denizen of the solar system or an interloper from outside. The value of GM for the sun is

( 

( 

)

)

N​ ​@ · m2 1 kg​ ​@ · m/s2 30 GM = ​ 6.67 × 10-11 ​ ______  ​    ​  (1.99 × 10 kg​ ​@)  ​ ​ __________  ​     ​ kg ​@2​  1 N​ ​@ 



     = 1.33 × 1020 m3/s2(N12.16) The value of 2E/m = │v​  ​W│   2 − 2GM/r for this object is therefore

( 

)

20

3

2

2E m 2 2(1.33 × 10 m /s ) ___ ​   ​ = ​​ 11,200 __ ​   ​  ​​ ​− __________________ ​         ​= + 4.5 × 107 m2/s2 (N12.17) 12 m

s

3.3 × 10 m

Since this is positive, the orbit is hyperbolic, so the object must be from outside the solar system (assuming it is not powered). Ominous! The system’s angular momentum per unit mass will be

__ ​ L  ​= r│v​  ​W│   sin θ = (3.3 × 1012 m)(11,200 m/s)(sin 169.7°)

m

= 6.6 × 1015 m2/s(N12.18)

Therefore, +GM 1.33 × 1020 m​3​/s​@2​    a = _______ ​     ​  = _______________ ​         ​= 3.0 × 1012 m(N12.19a) 7 2 2 (2E/m) 4.5 × 10 m ​@​/  s​@​  @

 

2 @  ) (4.5 × 107  m ​@2​/   s​@2 ​)(6.6   × 1015 m ​@2​/    ​s​ _______________________________ ε =​​   1 + ​          ​   ​​ ​  = 1.054 (N12.19b) [ ] (1.33 × 1020 m ​@3​/   s ​@2 ​) 2 1/2



and the object’s distance rc from the sun at the orbit’s vertex will be

rc = a(ε − 1) = (3.0 × 1012 m)(0.054) = 1.62 × 1011 m(N12.19c)

Note that rc is nearly the same as the earth’s orbital radius of 1.5 × 1011 m.

N12.4 Solving Orbit Problems

195

Exercise N12X.2 You can see that it is quite handy to know the value of GM. Show that GM for the earth is 3.99 × 1014 m3/s2. (Similarly, GM for the moon is about 4.9 × 1012 m3/s2, and for Jupiter it is about 1.3 × 1017 m3/s2.)

Example N12.2

Problem: A certain asteroid in an elliptical orbit is a distance rc = 3.5 AU from the sun at the closest point in its orbit and a distance rf = 4.5 AU at the farthest point. What is the period of its orbit in years? Solution  The semimajor axis of the orbit is a = _​  12 ​  (rc + rf) = 4.0 AU. The semimajor axis of the earth’s orbit ae ≈ 1 AU (by definition). Since T ∝ a3/2,

(  )

3/2

(  )

3/2

4.0 AU​ ​@  T a __ ​    ​= ​​ __ ​    ​ ​ ​  = ​​ _______ ​   ​    ​​ ​  = 8.0(N12.20)



Te

1.0 AU​ ​@ 

ae

Since the period of the earth’s orbit Te = 1 y by definition, the period of the asteroid’s orbit must be 8.0 y.

Example N12.3

Problem:  At the point of its orbit closest to the earth, a satellite 200 km above the earth’s surface is traveling at a speed of 10.0 km/s. Find the farthest distance this satellite gets from the earth, and determine its orbital p ­ eriod. Solution  See figure N12.3 for a sketch of the orbit. We are given that at the closest point in its orbit, the satellite is 200 km from the earth’s surface, and thus 200 km + 6380 km = 6580 km = rc from the earth’s center, and that its speed │v​  ​Wc │ = 10 km/s at this point. We want to find rf. The earth is extremely massive compared to any human-made satellite, so the massive-primary model we have developed in chapter N11 should work. We can calculate 2E/m and L/m using the values of rc and │v​  ​Wc │ given for the close point, and we calculate anything we want from that. With the given data and a result from exercise N12X.2, we find that

( 

)

14

3

2

2E 2GM m 2 2(3.99 × 10 m /s ) ___  ​Wc │2 − _____ ​      ​ = ​​ 10,000 __ ​   ​  ​​ ​− __________________    ​      ​ ​   ​ = │v​ 3 m

rc

s

6580 × 10 m

7

2



= −2.1277 × 10 m /s2 (N12.21)



GM 3.99 × 1014 m​3​/ s​@2​  __________________ a = − ​ _______   ​  =       ​   ​ = 1.8753 × 107 m(N12.22) (2E/m) −2.1277 × 107 m ​@2​/  s​@2​  @  

which (rounded) is 18,750 km. Since rf + rc = 2a,

rf = 2a − rc = 37,500 km − 6580 km = 30,900 km(N12.23)

and according to the generalized version of Kepler’s third law, _________________



(  )

@ 4π 2(1.88 × 107 m​ ​ ) 3 1h ________________ T = ​​             ​ ​ = 25,600 s ​ ______ ​    ​   ​= 7.1 h (N12.24) 14 3 2 3600 s 3.99 × 10 m ​@​/  s

The units all work out here, and the answers seem plausible.

vc rf

Earth rc

vf

Figure N12.3 The situation described in example N12.3.

Chapter N12

196

Example N12.4 Moon’s orbit

Moon

Probe’s transfer orbit rf

v0

Probe’s initial circular orbit Earth

rc vc

Probe

Figure N12.4 The situation described in example N12.4.

Orbits and Conservation Laws

Problem:  Suppose we have a space probe in a circular orbit 200 km above the earth and we want to put it into an elliptical orbit whose farthest point coincides with the moon. How much do the probe’s rocket engines have to add to its speed to put it into this orbit? Assume the rocket engines fire for so brief a time that the object’s distance from the earth does not change much during the time the engines are firing (that is, during the burn in astronautical language). Solution  Figure N12.4 shows a sketch of the situation. The earth is extremely massive compared to any probe, so the ­massive-primary model should be accurate. If the rocket engines fire only very briefly, the probe will be essentially at the same radius rc = 6580 km both before and just after the engines fire. We also know rf = 384,000 km from the inside front cover. We can use this information to calculate a, use that to calculate 2E/m, and use that to calculate the speed vc the probe must have just after the burn. We can calculate the probe’s original speed │v​  ​W0  │ in its circular orbit by using equation N11.12. Since 2a = rf + rc, we have

2GM GM 2GM  ​  a = __ ​  1 ​  (rf + rc)  ⇒  ___ ​ 2E ​ = │v​  ​W c│2 − ​ _____     ​ = − ​ ____  ​    = − ​ ______  (N12.25) 2 m rc a rf + rc



2GM  ⇒  │v​  ​W c│ = ​   − ​    ______  ​  + _____ ​ 2GM  ​     rf + rc rc

_______________

________________________________________________

 ​

2(3.99 × 1014 m3/s2) 2(3.99 × 1014 m3/s2) __________________________ = ​       − ​            ​ + __________________ ​        ​ ​ 6,580,000 m 6,580,000 m + 384,000,000 m



= 10,920 m/s 

(N12.26)

Since the probe’s original circular orbit speed was

____

________________

rc

6,580,000 m

3.99 × 1014 m3/s2 GM _______________ │v​  ​W 0│ =  ​ ____ ​     ​ ​    = ​ ​             ​ ​ = 7790 m/s

(N12.27)

the engines need to boost the probe’s speed by 3130 m/s. To match the moon’s speed at the far point would require another short burn to boost the probe’s speed. Such a maneuver, which involves two short burns and an elliptical transfer orbit between one circular orbit and another is called a Hohmann transfer: this happens to be the way to move between the two circular orbits that minimizes the probe’s total change in velocity (and thus minimizes the fuel required for the two burns).

Exercise N12X.3 Suppose you are in a spaceship that is orbiting a planet of mass M in a circular orbit of radius R. By what minimum factor would you have to increase your speed to go into a hyperbolic orbit?

Homework Problems

197

TWO-MINUTE PROBLEMS N12T.1 According to the text, an orbiting object having energy E < 0 follows an elliptical orbit. When we specify the numerical value for an object’s energy, we are implicitly comparing it to the energy the object has in some reference situation. So, when we say that E < 0 in this case, we are comparing the energy of the orbiting object to the energy of an object with the same mass that is A. At rest at the primary’s center. B. At rest on the primary’s surface. C. In a circular orbit just above the primary’s surface. D. At rest at r = ∞. E. Some other situation (specify). N12T.2 An object is discovered near the earth with values of 2E/m = 2.12 × 107 m2/s2 and L/m = 7.8 × 1010 m2/s. This object is in what kind of orbit around the earth? A. Elliptical B. Parabolic C. Hyperbolic

A. B. C. D. E. F.

8y 16 y 22.6 y 64 y 226 y Other (specify)

N12T.5 The relationship rc│v​  ​Wc │ = rf│v​  ​Wf │ for an elliptical orbit follows from the fact that the satellite’s velocity is perpendicular to its position vector at the two extreme points and also from A. Conservation of angular momentum. B. Conservation of energy. C. Conservation of momentum. D. Newton’s second law. E. Some other properties of ellipses. F. (The given relationship is false!)

N12T.3 The object described in problem N12T.2 has an orbital eccentricity of: A. 1.08 B. 1.81 C. 0.81 D. 0.90 E. 1.35 F. Other (specify)

N12T.6 Suppose a satellite orbits the earth so closely that it experiences some drag due to the earth’s upper atmosphere. This will drain away some of the orbital energy of this system, converting it to thermal energy. If this happens fairly slowly, the satellite’s orbit will remain nearly circular. What happens to the radius of this satellite’s orbit as time passes in this case? A. It slowly decreases. B. It remains the same: just the satellite’s speed decreases. C. It slowly increases.

N12T.4 A comet is discovered in an elliptical orbit around the sun. Its closest distance from the sun is 1.0 AU, and measurements of its speed at this distance imply that its greatest distance from the sun is 7  AU. About how many years will pass between the comet’s approaches to the sun?

N12T.7 In the situation described in problem N12T.6, what do you think will happen to the satellite’s speed as time passes? A. It slowly decreases. B. It remains roughly the same. C. It slowly increases.

HOMEWORK PROBLEMS Basic Skills N12B.1 An object is discovered near the earth with v ­ alues of E, L, and m such that 2E/m = −8.2 × 106 m2/s2 and L/m = 7.8 × 1010 m2/s. Find the eccentricity of this orbit and the radius of the closest point of the orbit to the earth, and ­ classify the orbit as being e­lliptical, parabolic, or hyperbolic. N12B.2 A space probe near the earth has values of E, L, and m such that L/m = 7.8 × 1010 m2/s and 2E/m = 0. Find the eccentricity of this orbit and the radius of the closest point of the orbit to the earth, and classify the orbit as elliptical, parabolic, or hyperbolic. N12B.3 Imagine you are in an orbit around the earth whose most distant point from the earth is 5 times farther from

the earth’s center than the closest point. If your speed is 10.0 km/s at the closest point, what is your speed at the farthest point? N12B.4 A satellite orbits the earth in such a way that its speed at its point of closest approach is roughly 3 times its speed at the most distant point. How many times more distant from the earth’s center is the farthest point than the closest point?

Modeling N12M.1 A small asteroid is discovered 14,000 km from the earth’s center moving at a speed of 9.2 km/s. Can you tell from the information provided whether this asteroid is in an elliptical or hyperbolic orbit around the earth? Is the direction of its velocity vector important in determining this? Please explain.

198

Chapter N12

Orbits and Conservation Laws

N12M.2 Astronomers spot a mysterious object moving at a speed of 21  km/s at a distance of 3.8 AU from the sun (where 1 AU = earth’s orbital radius = 1.5 × 1011 m). Can you tell from the information provided whether this object is in an elliptical or ­hyperbolic orbit around the sun? Is the direction of its velocity vector important in determining this? Please explain.

N12M.9 A certain comet follows an elliptical orbit around the sun such that rc = re = 1 AU and rf = 17 AU, where re = 1 AU is the radius of the earth’s orbit around the sun. (a) What is the comet’s period in years? (b) Show that the value of 2E/m for this comet is − ​ _19│ ​   v​  W​ 0  │2, ______  where │v​  ​W 0│ = ​ GM/r    ​e   is the earth’s orbital speed. (c) Find the comet’s speed at rc as a multiple of │v​  ​W0  │.

N12M.3 Imagine you are in a circular orbit of radius R = 7500 km around the earth. You’d like to get to a geostationary space station whose circular orbit has a radius of 3R, so you’d like to put yourself in an elliptical orbit whose closest point to the earth has radius R and whose most distant point has radius 3R (that is, a Hohmann transfer orbit). By what percentage would you have to increase your speed while at radius R to put yourself in this transfer orbit? (Assume your rocket engines fire for such a brief time that your distance from the earth remains essentially R throughout the time the engines are on.)

Derivation

N12M.4 Imagine you are an astronaut in a circular orbit of R = 6500 km around the earth. (a) What is your orbital speed? (b) Say that you fire a rocket pack so that in a very short time you increase your speed in the direction of your motion by 20%. What are the characteristics of your new orbit? Calculate a, ε, T, rc, and rf for the new orbit. N12M.5 A certain satellite is in an orbit around the earth whose nearest and farthest points from the earth’s center are 7000 km and 42,000 km, respectively. Find the satellite’s orbital speed at its point of closest approach. N12M.6 An asteroid is in an orbit around the sun whose closest point to the sun has a radius R and whose most distant point has a radius 9R, where R is equal to the radius of the earth’s orbit = 1 AU = 1.5 × 1011 m. (a) What is the asteroid’s speed when it is closest to the sun? (b) How does this compare with the earth’s orbital speed? (c) What is the period of this orbit, in years? N12M.7 A new comet is discovered 6.6 AU from the sun (where 1 AU = earth’s orbital radius = 1.5 × 1011  m) ­moving with a speed of 17 km/s. At that time, its velocW​   makes an angle of 174.3° with respect to its ity vector v​ position vector  r​W ​ (meaning that the comet’s velocity points almost toward the sun). (a) Is this comet in a hyperbolic orbit? (b) What will be its distance from the sun at the point of closest approach? N12M.8 Some recent space probes have made several hyperbolic passes past the earth to help give them the right direction and speed to go to their final destination. Imagine that one such probe has a speed of 12 km/s at a distance of 650,000 km from the earth. If the angle that the probe’s velocity vector v​ ​W  makes with its position vector  r​W​   at that time is 177°, how near will it pass by the earth at its point of closest approach?

N12D.1 Argue that for a given value of angular momentum L, the smallest (most negative) energy that an orbiting object can have is

2

2

3

G M  m E = ​ ________  ​    2L2

(N12.28)

and that the orbit in this case will be circular.

Rich-Context N12R.1 You are the commander of the starship Execrable, which is currently in a standard orbit around a class-M planet whose mass is 4.4 × 1024 kg and whose radius is 6100 km. Your current circular orbit around the planet has a radius of R = 50,000 km. Your exobiologist wants to get in closer (say to an orbital radius of 10,000 km, or R/5) to look for signs of life. Your planetary geologist wants to stay at the c­ urrent radius so that the entire face of the planet can be scanned with the sensors at once. Because you are tired of the bickering, you decide to put the Execrable into an elliptical orbit whose minimum distance from the planetary center is R/5 and whose farthest point is R from the same. Your navigational computers are down again (of course), so you have to compute by hand how to insert yourself in this new orbit. Your impulse engines are capable of causing the ship to accelerate at a rate of 1│g​  ​W│    = 9.8 m/s2. In what direction should you fire your engines, relative to your current direction of motion (­forward or backward)? For how many seconds? N12R.2 You’d like to put your spaceship into a Hohmann transfer orbit between earth and Mars (that is, an elliptical orbit whose closest point to the sun is the same as the earth’s orbital radius and whose farthest point is the same as Mars’s orbital radius). (a) If your spaceship is initially traveling around the sun with the same orbital speed as the earth and at the same distance from the sun as the earth is, by what factor will you need to increase your speed to put you into the transfer orbit? Assume you fire your rocket engines for a short enough time that your distance from the sun is still essentially 1.0 AU when the engines cease firing. (b) How long will it take you to get to Mars following this orbit? (c) Check your work using the Newton app. Set up the probe as having an initial position of 1.0 AU along the x axis, and give it an initial velocity parallel to the y  axis whose magnitude is the value of │v​  ​W c│ you calculated in part (b). I recommend a time step of about

Answers to Exercises

199

0.02 y. Show that running some time steps does indeed produce an elliptical orbit having the desired value of rf = R, and by counting time steps, verify the flight time you predicted in part (a). N12R.3 Imagine you are the orbital engineer for the first NASA space shot to Ceres, the largest known asteroid. Ceres’s nearly circular orbit around the sun has a radius of R = 2.77 AU. After being launched from the earth, the probe will initially be in a circular orbit around the sun with the same radius (rc = 1.0 AU) and the same orbital speed (│v​  ​Wc │ = 6.28 AU/y) as the earth’s orbit. The probe’s rocket engines will then fire briefly to increase the probe’s speed to that speed │v​  ​W c│ needed to put the probe into an elliptical orbit whose initial (and minimum) distance from the sun is rc = 1.0 AU and whose final (and largest) distance from the sun is R = 2.77 AU (this is a Hohmann transfer orbit). (a) How long will it take the probe to get from the earth to Ceres in such an orbit? (b) What is the speed │v​  ​W c│ that the probe has to have just after firing its engines to be inserted into this orbit? (Assume the duration of the boost is short enough that its distance from the sun is still nearly rc just after the engines have been fired.) (c) Check your work using the Newton computer program. Set up the probe as having an initial position of 1.0 AU along the x axis, and give it an initial velocity parallel to the y axis whose magnitude is the value of │v​  ​W c│ that you calculated in part (b). I recommend a time step of about 0.02 y. Show that running some time steps does indeed produce an elliptical orbit having the desired value of rf = R, and by counting time steps, verify the flight time you predicted in part (a). N12R.4 Imagine that an object’s velocity v​ ​W0  far from the earth is such that it would miss the earth’s center by a

distance D if the earth had no gravity. What is the minimum value D can have if the object is to miss hitting the earth’s surface? Use conservation of energy and angular momentum to determine D in terms of the earth’s radius R, │v​  ​W0  │, and GM. (Hint: First explain why the constant value of L/m is equal to D│v​  ​W 0│. Then ensure that rc > R.)

Advanced N12A.1 Consider an object in a hyperbolic orbit around a massive primary. (a) Argue that the object’s speed │v​  ​W│    and angular momentum per unit mass at a point a very long distance from the primary are ___



______

2 __ │v​  ​W│    ≈  ​ ___ ​ 2E    ​ ​      and   ​  L  ​ = │v​  ​W│   a ​ε    − 1 ​  

m

m

(N12.29)

[Hints: The angular momentum formula is the tricky one. First argue that the angle ϕ between the velocity and the radius vector is given by tan ϕ = r dθ/dr, where dθ is a small increment in the angle as the object moves outward and dr is the increment in the distance for that angle increment. Prove then that



r(dθ/dr) _____________ ​ sin ϕ = _______________ ​       2  2 ​ r     (dθ/dr) + 1 ​



(N12.30)

Now evaluate dθ/dr using the basic formula for a conic section (it is easiest to do this by taking differentials of 1 + ε cos θ = R/r). Finally, evaluate r sin ϕ in the limit that cos θ goes to –1/ε and then use the formula linking R to a to eliminate______ the R in favor of a. You should 2 find that r sin ϕ = a ​ε    − 1 ​  .] (b) Use equations N12.29 and the formulas for 2E/m and L/m at r = rc to derive equations N12.12 through N12.14 for a hyperbolic orbit. [Hint: Use the calculation of the corresponding equations for the ellipse as a template, though there are some differences.]

ANSWERS TO EXERCISES N12X.1 Dividing equation N12.9 by two and substituting in a = −GM/(2E/m), we get

( 

(  )

) (  ) ( 

)

2 2 ​  -GM   ​  ​= ​ ___ ​  2E ​  ​ ​​ ______ ​  GM  ​  ​ ​(1 + ε 2) ​​ __ ​  L  ​   ​ ​− 2GM ​ ______ m 2E/m m 2E/m

(  )



(GM)2 L 2 2(GM)2 ______ ⇒  ​​ __ ​    ​   ​ ​+ ​ _______ ​  = ​   ​  (1 + ε 2) 2E/m 2E/m m



2 ​ 2E ​  ​ ​​ __ ​  L  ​   ​ ​+ 2 = 1 + ε 2 (N12.31) ⇒  ______ ​  1  2 ​  ​ ___ (GM) m m

(  ) (  )

Subtracting 1 from both sides yields the desired result. N12X.2 This is just a matter of substituting in numbers for G and M. If you disagree with the answer given, check that you have gotten the right numbers and that you have typed them correctly into your calculator.

N12X.3 To get into a hyperbolic orbit, the spaceship needs to have a speed │v​  ​W c│ at radius R such that 0 < ___ ​ 2E ​ = │v​  ​W c│2 − _____ ​  2GM  ​     ⇒  │v​  ​W c│2 > _____ ​  2GM  ​     (N12.32) m R R



The circular orbit speed is (see equation N11.11) ____



│v​  ​W0  │ =  ​ ____ ​ GM    ​ ​ (N12.33)     

R So the factor by which we must increase the speed is ________

 │v​  ​W c│  ​ 2GM/R ​     __  ​ = __________ ​  _______ ​  = ​ 2 ​     ​ ____ │v​  ​W 0│ ​GM/R ​    

(N12.34)

NA

 Differential Calculus NA.1

The definition of the derivative

Derivatives

The purpose of this appendix is to review (in the relatively informal language used by physicists) some basic principles of differential calculus, principles that we will use repeatedly throughout this unit and in subsequent units. Consider a function f (t), where t is some arbitrary variable (not necessarily time, though functions of time will be our main concern in this particular unit). We define the derivative of f (t) with respect to t as follows: d f   f (t + Δt) − f (t) ​ __ ​ ≡     ​  lim ​  ​ ______________     ​   (NA.1) Δt ​  ​     0 dt Δt W

First example of calculating a derivative from the definition

That is, the derivative is the limit as ∆t goes to zero of the change in f as the variable t changes from t to t + ∆t divided by change ∆t in that variable. The best way to show what we mean by “the limit as ∆t goes to zero” here is to do an example. Suppose that f (t) = t 2. The derivative of f (t) in this particular case is d f d (t + Δt)2 − t 2 ​ __ ​ ≡ __ ​    ​  [t 2] ≡     ​  lim ​​ ____________     ​   (NA.2) Δt ​ ​ dt dt Δt W  0 Multiplying out the binomial in the numerator yields (t + Δt)2 − t 2 ___________________ t 2 − 2tΔt + Δt 2 − t 2 ___________ 2tΔt + Δt 2 ​ ____________     ​   = ​     ​   = ​   ​     = 2t + Δt (NA.3) Δt Δt Δt As ∆t becomes smaller, this clearly gets closer and closer to 2t. So d f if  f (t) = t 2  then  ​ __ ​ ≡     ​  lim ​(2t + Δt) = 2t (NA.4) dt Δt ​ ​ W  0

Second example of calculating a derivative from the definition



Here is a second example. Suppose that f (t) = 1/t. Its derivative is

(  )

( 

)

d f d 1 1 1 1 ​ __ ​ = __ ​    ​ ​ __ ​   ​   ​≡     ​  lim ​ ​  [ ___ ​     ​ ​ ______ ​     ​  − __ ​   ​   ​  ] ​ (NA.5) Δt ​  ​     0 Δt t + Δt t dt dt t W If we put the expression in the brackets over a common denominator, we get

( 

)

( 

)

1 1 1 1 t − {t + Δt} 1 −Δt −1 ​ ___   ​ ​ ______ ​     ​  − __ ​   ​   ​= ___ ​     ​ ​ ​ ___________  ​     ​= ___ ​     ​ ________ ​    ​  = ________ ​  2    ​   (NA.6) Δt t + Δt t Δt {t + Δt}t Δt {t +Δt}t t  + tΔt In the limit that ∆t becomes very small, t 2 + tΔt → t 2, so

1  d f −1 −1 If  f (t) = __ ​   ​     then  ​ __ ​ ≡     ​  lim ​ ​   ________ ​  2    ​    ​= ___ ​  2 ​   [ ] Δt ​  ​     0 dt t t  + tΔt t  W

(NA.7)

This illustrates the general technique: one typically finds that the difference f (t + Δt) − f (t) can be written as something proportional to ∆t, which cancels with the ∆t in the denominator. One then sets Δt = 0 in what is left over.

200

NA.2

Some Useful Rules

In general, one can use the same general approach to show that d f If  f (t) = t n then ​ __ ​ = nt n-1 (NA.8) dt



Rule for simple powers

for any integer n ≠ 0. (Although the proof is more complicated, one can also show that the same applies to non-integer powers n.) Note also that d f 0 c−c If  f (t) = c then ​ __ ​ =     ​  lim ​​  [ _____ ​   ​      ​=     ​  lim ​​  [ ___ ​     ​  ] ​= 0 (NA.9) ] Δt ​  ​     0 Δt ​  ​     0 dt Δt Δt W W



The derivative of a constant

This case makes it clear that the limit of 0/∆t as ∆t approaches zero is not the same as the value of 0/∆t at ∆t = 0. Because 0/∆t is 0 for all nonzero ∆t, the limit is zero (this is what 0/∆t approaches), even though 0/0 is undefined. The following useful theorems follow from the definition of the derivative. Let f (t), g (t), and h (t) be arbitrary functions of t. We then have

d f dg ___ dh If  f (t) = g (t) + h (t) then  __ ​   ​ = ___ ​   ​ + ​   ​  (NA.10) dt dt dt



If  f (t) = g (t) h (t)

d f dg dh ​   ​  h + g ​ ___ ​  (NA.11) then  __ ​   ​ = ___ dt dt dt

The product rule



1 If  f (t) = ____ ​      ​ g (t)

d f −1 ___ dg then  __ ​   ​ = ___ ​  2 ​ ​   ​  (NA.12) dt g  dt

The inverse rule



The sum rule

One can pretty easily prove these rules, as the example below illustrates:

Example NA.1

Problem:  Prove the product rule. Solution  If f (t) = g (t) h (t), then d f g (t + Δt) h (t + Δt) − g (t) h (t) ​         ​  ] ​ (NA.13) ​ __ ​ ≡     ​  lim ​​  [ ___________________________ Δt ​  ​     0 Δt dt W Adding zero to the numerator in the form −g (t + Δt) h (t) + g (t + Δt) h (t) yields d f g (t + Δt) h (t + Δt)− g (t + Δt) h (t) + g (t + Δt) h (t) − g (t) h (t) __ _______________________________________________________ ​        ​  ] ​ ​   ​ ≡     ​  lim ​​  [        dt

Δt

Δt ​ ​ W  0

g (t + Δt) − g (t) h (t + Δt)− h (t) _______________     ​   +    ​   ​   h (t) ] ​   =     ​  lim ​​ [ g (t + Δt) ​ ______________ Δt ​ ​     0 Δt Δt W



(NA.14)

But the limit of a sum is the sum of the limits, and the limit of a product is the product of the limits, so d f g (t + Δt) − g (t) h (t + Δt)− h (t) __ ______________ ​   ​ ≡     ​  lim ​  [ g (t + Δt)] ​     lim ​​      ​   ​    ​+ h (t)     ​  lim ​​   ​  ______________     ​    ​ dt

Δt ​ ​ W  0

[ Δt ​ ​ W  0

Δt

]

dh dg      = g ​ ___ ​ + h ​ ___ ​  dt dt

[ Δt ​ ​ W  0

Δt

]

(NA.15)

as claimed.

201

202

Chapter NA

Differential Calculus

The constant rule

The constant and product rules together directly imply that when c is a constant, d f dc dh dh if  f (t) = ch (t)   then ​ __ ​ = ​ __ ​  h + c ​ ___ ​ = c ​ ___ ​  (NA.16) dt dt dt dt This, in conjunction with the sum and product rules and equation NA.8, make it pretty easy to evaluate the derivative of any polynomial in t.

Example NA.2

Problem:  What is the derivative of f (t) = at 3 + bt + c, where a, b, and c are constants? Solution  By the sum rule, we have d f d d dc d dt ​ __ ​ = __ ​    ​  [at 3 ] + __ ​    ​  [bt] + __ ​   ​ = a ​ __  ​  [t 3] + b ​ __ ​ + 0 (NA.17a) dt dt dt dt dt dt where in the second step, I applied the constant rule. Using equation NA.8, we find that d f ​ __ ​ = a [2t 2] + b [1] = 2 at 2 + b (NA.17b) dt

Exercise NAX.1 Use the rules we have stated so far to calculate the derivative of the function f (t) = (at + b)−1, where a and b are constants.

NA.3 The derivative is the slope of a line tangent to the graph of the function f (t)

Derivatives and Slopes

We can better understand the meaning of the derivative of a function f (t) by considering a graph of the function f (t). This section reviews the graphical interpretation of the derivative. Recall the definition of the derivative (see equation NA.1): d f Δf f (t + Δt) − f (t) ______________ ​ __ ​ ≡     ​  lim ​ ___ ​    ​ ≡     ​  lim ​    ​   ​   (NA.18) Δt ​  ​     0 Δt ​  ​     0 dt Δt W Δt W Figure NA.1 illustrates that on a graph of f (t) plotted versus t, the ratio ∆ f/∆t for a nonzero ∆t is equal to the slope of a straight line drawn between the points on the curve at t and t + ∆t. As ∆t approaches zero, the value of this slope approaches a specific value that equals the slope of a line drawn tangent to the graph at t. We call this the slope of f (t) at that point. Therefore, we can interpret d f / dt evaluated at t as giving the slope of the line tangent to the curve of f (t) at that point t.

NA.4

The Chain Rule

The chain rule is one of the most important and useful theorems of differential calculus. Assume that f (u) is a function of some variable u, which in turn

NA.4 The Chain Rule

f(t)

Tangent to curve at A

203

Curve of f(t)

B C

D ∆fB ∆tB

A

∆fD A ∆tD

t (a)

t + ∆tC

t + ∆tB

t

(b)

Figure NA.1 (a) The ratio ∆fB  /∆tB specifies the slope of the line going through points A and B. The ratio ∆fC  /∆tC specifies the slope of the line going through points A and C. The latter is closer to the slope of the line tangent to the curve of f (t) at point A (light-colored straight line). (b) As ∆t gets very small, the curve and the tangent line become indistinguishable, and ∆f  /∆t specifies the slope of both in the ­neighborhood of A.

is a function u(t) of t. This means that f is also a function of t, and therefore has a meaningful derivative with respect to t. The chain rule claims that d f d f ___ du ​ __ ​ = ___ ​    ​   ​  (NA.19) dt du ​ dt

The statement of the chain rule

This theorem looks relatively pedestrian, but in fact is very useful, as the following example illustrates.

Problem:  Calculate the derivative of f (t) = (at 2 + bt + c)5.

Example NA.3

Solution  We could evaluate this derivative by multiplying out the polynomial and then taking the derivative of each term, using the constant rule and the rule for the derivatives for simple powers. But this would be very tedious. It is much simpler to define u(t) ≡ at 2 + bt + c, which means that f (u) = u5. The chain rule then tells us that

(  ) ( 

d f d f ___ du d ​ __ ​ = ___ ​    ​ ​   ​ = ​ ___ ​    ​ u 5  ​ ​ dt du dt du



)

d __ ​    ​  [at 2 + bt + c]  ​ dt

        = (5u 4 )(2at + b) = 5(at 2 + bt + c)4 (2at + b) (NA.20)

The physicist’s “proof” of this theorem, by the way, is as follows.

d f d f ___ du d f ___ du ​ __ ​ = __ ​   ​ ​   ​ = ___ ​    ​ ​   ​  (NA.21) dt dt du du dt Physicists tend to think of d f, du, and dt as simply being “sufficiently small” numbers, so multiplying and dividing by du seems perfectly natural. Mathematicians worry more about the definitions of the derivatives, the limiting process, and all the things that might go wrong with this cross-­multiplication, but the result is the same. At the very least, you can think of “multiplying top and bottom by du” as a useful mnemonic for the chain rule formula.

A “physicist’s proof”

204

Chapter NA

Differential Calculus

Exercise NAX.2 Use the chain rule d f/dt = (d f/du)(du/dt) to calculate the derivative of the function f (t) = 1/(t 2 − b2).

NA.5 Derivatives of some other useful functions

Derivatives of Other Functions

We will sometimes make use of the following derivatives: d  ​ sin θ = cos θ    and   ___ ​ ___ ​  d  ​ cos θ = −sin θ (NA.22) dθ dθ d  ​ eu = eu  and  ​ ___ d  ​ ln u = __ ​ ___ ​  1  ​ (NA.23) du du u We will often see such functions in contexts where the angle θ or the quantity u depends on t: for example, θ(t) = ωt or u(t) = −bt, where ω and b are constants. In such cases, we can use the chain rule to evaluate the derivatives of such functions with respect to time. For example, if θ(t) = ωt, then d  ​ sin ωt = __ ​ __ ​  d  ​ sin [θ(t)] = ______ ​ d sin θ     ​ ___ ​ dθ ​ = cos θ __ ​  d  ​ ωt = ω cos ωt (NA.24) dt dt dθ dt dt Similarly, if u(t) = −bt, then

(  ) (  )

d  ​ e -bt = ​ ___ ​ __ ​  d  ​  e u  ​​ ___ ​  du ​  ​= e u __ ​  d  ​ (−bt) = e u(−b) = −be -bt (NA.25) dt du dt dt

Exercise NAX.3 Use this technique to find the time derivative of f (t) = A cos (ωt + θ0), where A, ω, and θ0 are constants.

Using WolframAlpha

I strongly recommend you memorize the rules in this appendix: they will be used often enough in the reading that you will go crazy if you have to look everything up. If your calculus skills are rusty, I recommend practicing with some of the homework problems for this appendix (answers are given at the end of the book). However, for derivatives of really nasty functions, you can always use WolframAlpha. For example, one can evaluate the derivative of a function such as f (t) = t 2(eb/t + 1)-3/2 using only the rules described in this section, but if you go to 𝚠𝚠𝚠    𝚠𝚘𝚕𝚏𝚛𝚊𝚖𝚊𝚕𝚙𝚑𝚊    𝚌𝚘𝚖 and type in “𝚝𝚊𝚔𝚎 𝚝𝚑𝚎 𝚍𝚎𝚛𝚒𝚟𝚊𝚝𝚒𝚟𝚎 𝚘𝚏 𝚝^𝟸/(𝚎𝚡𝚙(𝚋/𝚝)+𝟷)^(𝟹/𝟸) 𝚠𝚒𝚝𝚑 𝚛𝚎𝚜𝚙𝚎𝚌𝚝 𝚝𝚘 𝚝” it will say: d f 3beb/t 2t ​ __ ​ = ​ ___________    ​  + ​ __________    ​   b/t 5/2 b/t dt 2(e + 1) (e + 1)3/2

(NA.26)

which would take some considerable time to find by hand (with a low probability of being right). But DO NOT use WolframAlpha for routine calculations of simple derivatives: you will greatly enhance your ability to read physics texts if you practice doing simple integrals so that eventually you can do them in your head. You will only learn do to this with practice.

Answers to Exercises

205

HOMEWORK PROBLEMS Basic Skills NAB.1 Calculate the derivatives of the following functions (where a, b, and c are constants). (a) at 2 + b (b) 1/ct 3 (c) b/(1 − at 2) NAB.2 Use the chain rule to calculate the derivatives of the following functions (where a, b, and c are constants). 3 (a) 3(t _______ + b 3)4 2 2 (b)  ​ t    − b   ​  (c) cos(bt 2) NAB.3 Use the chain rule to calculate the derivatives of the following functions (where a, b, and c are constants). (a) 5b(bt − 1)-4 (b) (bt + 1)3/2 (c) sin(bt + π)

NAD.2 Using the basic approach illustrated in section NA.1, prove that if f (t) = t −2, then d f /dt = −2t −3. (Do not use equation NA.8.) NAD.3 Prove that the sum rule (equation NA.10) follows from the definition of the derivative. (You may assume without proof that the limit as Δt 0 of the sum of two quantities is the same as the sum of the limits of each quantity separately.) NAD.4 Prove that the inverse rule (equation NA.12) follows from the definition of the derivative. (You may assume without proof that the limit as Δt 0 of the product of two quantities is the same as the products of the limits of each quantity separately.) NAD.5 Use the chain rule to prove the inverse rule. NAD.6 Use the chain rule to prove this very useful result: d  ​ (t + c)n = n(t + c)n-1 (NA.27) ​ __ dt

Derivations NAD.1 Using the basic approach illustrated in section NA.1, prove that if f (t) = at (where a is a constant), then d f /dt = a. (Do not use equation NA.8.)

ANSWERS TO EXERCISES NAX.3 If we define θ(t) = ωt + θ0, then the derivative is

NAX.1 Using the inverse rule, we have 1 d −a d f __ ​ __ ​ = − ​ ________    ​  ​    ​ [at + b] = ________ ​     ​   (at + b)2 dt (at + b)2 dt

(NA.28)



NAX.2 Define u(t) ≡ t 2 − b 2. Then 1   ​  __ ​  d  ​ _________ ​  = ___ ​  d  ​ u-3 ___ ​ du ​ = −3u-4 __ ​  d  ​  (t 2 − b 2 ) dt (t 2 − b 2 )3



du

dt

dt

3   ​   6t   ​   = − ​ _________ (2t + 0) = − ​ _________ (t 2 − b 2 )4 (t 2 − b 2 )4

d f dt

(NA.29)

( 

d

d

)

dθ

​ __ ​ = A __ ​    ​cos[θ(t)] = A ​ __ ​    ​ cos θ  ​__ ​    ​ dt

dθ

dt

= A(−sin θ) __ ​  d  ​ (ωt + θ0) = −Aω sin (ωt + θ0) (NA.30) dt

NB

 Integral Calculus NB.1

The definition of the antiderivative

Antiderivatives

The antiderivative F (t) of a function f (t) is defined to be any function whose derivative is f (t). For example, one possible antiderivative for f (t) = at (where a is a constant) is F (t) = _​ 12  ​  at 2 because  

(  )

dF ​ = __ d  ​  (t 2) = __ ​ ___ ​  d  ​  ​  __ ​  1 ​  at 2  ​= __ ​  1 ​   a ​    __ ​ 1 ​   a (2t) = at (NB.1) dt dt 2 2 2 dt  

 

Note that there is not a unique antiderivative for a given function. If F (t) is an antiderivative of f (t), then so is F (t) + C (where C is a constant), since d  ​  [F (t) + C] = ___ ​ __ ​  dF ​ + ___ ​  dC ​ = ___ ​  dF ​ = f (t) (NB.2) dt dt dt dt So the antiderivative f (t) is really a whole family of functions F (t) that differ by an additive constant. We therefore typically write the antiderivative of f (t) as F (t) + C, where C is an unspecified constant we call a constant of integration.

NB.2 The definition of a definite integral

Definite Integrals

Suppose we graph f (t) as a function of t. We define the definite integral of f (t) from a point tA to a point tB on the horizontal axis as follows: tB

​∫ ​  ​  f​  (t) dt ≡ the total area under the curve of f (t) between tA and tB (NB.3a) tA

(considering the area above the t axis to be positive and the area below the axis to be negative). Figure NB.1 shows that we can approximate this area by a set of thin bars of width Δt, where the ith bar’s height is f (ti), where i is an integer index, and ti = tA + (​ i − _​  12  ​  )​  Δt, the value of t halfway across the bar’s width. Perhaps you can convince yourself that as the bar width Δt goes to zero, we can write the area under the function’s curve as Figure NB.1 (a) The definite integral of a function f (t) is the area under a graph of f (t) between t A and t B (see the shaded regions). The area above the axis is positive, and the area below is negative. (b) The area is approximately equal to the total area of the bars shown, where each bar is Δt wide and as tall as the function’s value across the bar.  

206

f (t)

f (t)

Curve of f (t)

 

Curve of f (t)

1

tB tA (a)

t

tA (b)

tB ∆t

2

t

N

tB

​ ∫ ​  ​  f​  (t) dt ≡ ​     lim ​ ​∑    ​ f ​(t i) Δt (NB.3b) Δt ​ W​  0 i = 1

tA

The definite integral as a sum

where N = (tB − tA)/Δt is the number of bars in the set. Indeed, the standard symbol ∫ f dt is meant to be a stylized version of what is on the right, with the S-like integral symbol standing for “sum.” Physicists (more than mathematicians) often think of an integral as being an actual sum of small quantities (as we will see, especially in unit E). The following properties of the integral follow from this definition: tB

tB

tB

​ ∫ ​  ​  ​[  f (t) + g (t)] dt = ​∫ ​  ​  f​  (t) dt + ​∫ ​  ​  g​  (t) dt (NB.4) tA

tA

tB

tA

tB

​∫ ​  ​  c​f (t) dt = c ​∫ ​  ​  f​  (t) dt (NB.5) tA

tA

tC

tB

tC

​∫ ​  ​  f​  (t) dt = ​∫ ​  ​  f​  (t) dt + ​∫ ​  ​  f​  (t) dt (when tC > tB > tA) (NB.6) tA

tB

tA

The sum rule The constant rule The limit rule

The last equation says that the total area under the curve between tA and tC (where tC > tB > tA) is the sum of the areas between tA and tB and between tB and tC (which is not really very surprising). Equations NB.4 and NB.5 are pretty easy to prove from equation NB.3b, if you think about it.

NB.3

The Fundamental Theorem

At first glance, it may not seem that the antiderivative of a function f (t) would have anything to do with the integral of f (t). However, the fundamental ­theorem of calculus asserts that if F (t) is any antiderivative of f (t), then t

F (t) − F (t 0) = ​∫ ​  ​  f​  (t) dt (NB.7)



 

t0

where t 0 is some arbitrary fixed value of the variable t. Note that when we use the variable t as the upper limit of integration, then the value of the integral (the area from t 0 to t) is essentially a function of t. (Some people don’t like to use the same variable symbol for the integral’s upper limit as for the integration variable, but physicists are not usually so finicky.) We can prove this theorem as follows. The definition of the derivative implies that the derivative of the right side of equation NB.7 is

The fundamental theorem of calculus

 

 

f (t)

d  ​​    ​∫ ​  t​  f​  (t) dt   ​= ​     1   ​​    ​∫ ​  t+Δt​  f​  (t) dt − ​∫ ​  t​  f​  (t) dt   ​ (NB.8) ___ ​ __ lim  ​ ​  ] ] t Δt ​ W​  0 Δt [ t dt [ t 0

0

f(t)

0

However, as figure NB.2 shows, the difference between the area under the curve from t 0 to t and that from t 0 to t + Δt (as Δt becomes very small) is essentially the area of the bar, which has width Δt and height f (t). Therefore  

 

d  ​​    ​∫ ​  t​  f​  (t) dt   ​= ​     ​ __ lim ​​   ___ ​  1   ​  f (t)Δt ] ​= f (t) (NB.9) ] Δt ​ W​  0 [ Δt dt [ t 0

t

The integral is thus indeed an antiderivative of f (t): F (t) = ​∫ ​  ​  f ​(t) dt + C. So t0



F (t) − F (t 0) = ​∫ ​  ​  f ​(t) dt + C −​  [ ​∫ ​  ​  f​  (t) dt + C ] ​ = ​∫ ​  ​  f ​(t) dt (NB.10) t

 

t0

t0

t

t0

t0

since the area under the curve from t 0 to t 0 is clearly zero. Q. E. D.  

 

t0

t

t + ∆t

t

Figure NB.2 The difference between the area from t 0 to t and that from t 0 to t + Δt is approximately equal to the area of the colored bar, which has width Δt and height f (t). This approximation becomes exact in the limit that Δt 0.  

 

207

208

Chapter NB

NB.4

Integral Calculus

Indefinite Integrals

The indefinite integral ∫ f (t) dt is any one of the possible antiderivatives of the function f (t) (usually the one with the simplest algebraic form). You will find it useful to memorize the following indefinite integrals, where b, c, and ω are all constants: The integral is the opposite of the derivative



∫ __  ​dt  ​ dt = f (t) for any function f (t) (NB.11)

The constant rule



∫ c dt = ct

Simple powers



∫t

Simple powers of binomials



∫ (bt + c)

Trigonometric functions



1 ∫ sin (ω t + b) dt = − ​ __ ω  ​ cos (ωt + b) (NB.15)



1  ​sin (ωt + b) (NB.16) ∫ cos (ωt + b) dt = ​ __ ω

Exponentials



∫ e 

Logarithms



∫ __ ​ dt ​ = ln t  (assuming t > 0) (NB.18) t

Definite integrals from indefinite integrals

df

n

 

(NB.12)

n+1

t   ​  dt = ​ _____   for n ≠ 1 (NB.13) n+1  

n

bt



(bt + c)n+1 dt = ​ _________ ​   for n ≠ 1 b(n + 1)

dt = __ ​  1 ​  e bt b

(NB.14)

(NB.17)

One can pretty easily prove any of these indefinite integrals by taking the derivative of both sides. Once you know a function’s indefinite integral, you can find its definite integral from t 0 to t by simply evaluating the indefinite integral at these two points and subtracting. For example  

​ ∫ ​  ​  c​dt = [ct​] ​ ​ ​  ≡ ct − ct 0 t

t

 

t0

t0

(NB.19)

This works no matter which member of the family of possible antiderivatives has been selected as the indefinite integral. This is because equation NB.10 in the proof of the fundamental theorem applies to any antiderivative of a given function f (t): whatever arbitrary constant went into the specification of the indefinite integral cancels out in the definite integral.

NB.5 An example of substitution of variables

Substitution of Variables

Beyond the integrals listed above (which every well-educated person should know), calculating integrals can become really hairy. An important method for evaluating more complicated integrals is the technique of substitution of variables. To illustrate the process, consider the following integral (where b is a constant): t dt   ​   ∫ __________ ​  ________ (NB.20)  2 2 ​ b  t + 1 ​  This looks like an appallingly difficult function to integrate. However, imagine that we define a new variable u ≡ b 2 t 2 + 1. Note that this means that  

 

 

 

NB.5 Substitution of Variables

209

du ​ = b 2 (2t) + 0 = 2b 2 t  ⇒  ___ ​ ___ ​  du2  ​ = t dt (NB.21) dt 2b  

 

 

If we substitute these results into equation NB.20, we find that t dt   ​  ∫ __________ ​  ________ = ∫ _______ ​  du   ​  = ___ ​  1   ​ ∫ u 

-1/2



 2   

​ b t 2 + 1 ​ 

2b 2 u 1/2

 

du (NB.22)

2b 2

 

 

where I have used the constant rule in the last step. But the last integral involves a simple power that we can evaluate using equation NB.13: 1/2

+1/2

________

 2 2

+ 1 ​  1   ​  u -1/2 du = ___ ​ ___ ​  1 2 ​ _______ ​  u    ​  = ____ ​  u  2 ​   = __________ ​ ​ b  t 2 ​     2 2b

∫

2b (+ 1/2)

 

 

b

 

(NB.23)

 

b

 

 

Exercise NBX.1 Use substitution of variables to evaluate the integral

______ at + b ​  dt (NB.24)

∫​



  

Check your work by taking the derivative of the result and showing that you recover the integrated function. Please note that if you substitute variables in a definite integral, you must also change the limits of the integration to reflect the new variable. For example, consider a definite integral involving the case just discussed

∫  ​  ​  u​  (NB.25) [ ] u b2 b2 ​ b t + 1 ​  2b 2 u where uA ≡ u (tA) = b 2 ​t 2​A ​ ​+ 1 and uB ≡ u (tB) = b 2 ​t 2​B ​​  + 1. For example, if we are evaluating the integral on the left from t = 0 to t = 2 s, and b = 0.5 s–1, then we would evaluate the integral on the right from u(0) = b 2 0 2 + 1 to u(2 s) = (0.5 s–1)2(2.0 s)2 + 1 = 1 + 1 = 2. Equivalently, one can substitute the definition of u in terms of t back into the final expression for the evaluated integral and then use the original limits: 1 t dt ​​  ​  __________ ​​  ________    ​  = ___ ​     ​ ​ tB

∫ 

tA

 2 2     

uB

1/2

A

 

___

___

​ u  B ​ − ​ u  A ​  1 du = __ ​    ​   u 1/2 ​ ​  ​​  = ___________    ​   ​   uB



A

 

 



 

 

 

[ b  t  + 1 ​  b

When we substitute variables in a definite integral, we must adjust the limits of integration

] 

[

_______

] 

_______

 

________

​ b   ​t ​B ​ ​  +  1 ​   − ​ b   ​t ​A ​ ​ + 1 ​  1 1 t dt    ____________________ ​ ​  _________ ​  ​  ​ _______    ​   = ​ __  ​  u 1/2 ​ ​  ​​   = ​ __  ​   ​ b  2 t 2 + 1 ​​      ​ (NB.26) ​  ​​  = ​    tB

∫ 

tA  2 2     

2

uB

uA

2

 

 

tB

 2 2

 2 2  

 

tA

2

​ b b Both approaches are correct. The point is that you should think carefully about what happens to the limits in definite integrals when you change variables. Another issue that might arise when you change variables is that you might find yourself integrating from a large value of u to a smaller value. In such a case, it is helpful to note that  

 

uB

 

uA

​ ∫  ​  ​  f​  (u) du = −​∫  ​  ​  f​  (u) du (NB.27) uB

uA

Exercise NBX.2 Use substitution of variables to integrate ​ ∫b ​  ∞​  ​ __ ​  b2  ​ e b/t dt (NB.28) t  

Again, check by taking the derivative of the result and showing that you recover the original function. (Note that if t has units of seconds, so does b.)

Swapping the limits changes the integral’s sign

Chapter NB

210

NB.6 No mechanical method exists for determining integrals

Integral Calculus

Looking Up Integrals

While one can calculate the derivative of essentially any function by a diligent but basically mechanical application of the chain rule and other rules, no foolproof mechanical process exists for evaluating integrals. Some very simple-looking integrals cannot be evaluated at all in symbolic form, and others require advanced techniques and clever tricks. A table of integrals provide a list of indefinite integrals for a wide variety of functions. A really good table has a long but well-organized list of functions. I like the table at

Tables of integrals

𝚑𝚝𝚝𝚙://𝚎𝚗   𝚠𝚒𝚔𝚒𝚙𝚎𝚍𝚒𝚊    𝚘𝚛𝚐/𝚠𝚒𝚔𝚒/𝙻𝚒𝚜𝚝𝚜_𝚘𝚏_𝚒𝚗𝚝𝚎𝚐𝚛𝚊𝚕𝚜 because (unlike many other online tables) it has some useful definite integrals that don’t have simple indefinite integrals. But all of the integral tables I can find online are short compared to those one can find in printed books, and I also worry about the accuracy. Because of this, WolframAlpha (𝚠𝚠𝚠   𝚠𝚘𝚕𝚏𝚛𝚊𝚖𝚊𝚕𝚙𝚑𝚊    𝚌𝚘𝚖) is often a better resource. To use WolframAlpha, type in something like “𝚒𝚗𝚝𝚎𝚐𝚛𝚊𝚝𝚎 𝚝/𝚜𝚚𝚛𝚝(𝚋^𝟸*𝚝^𝟸+𝟷) 𝚠𝚒𝚝𝚑 𝚛𝚎𝚜𝚙𝚎𝚌𝚝 𝚝𝚘 𝚝” to evaluate the integral we discussed in the last section. However, using substitution of variables to simplify an integral as far as possible will make it much easier to find an integral in a table or to get something useful out of WolframAlpha (which is easy to confuse, poor thing). Also note that WolframAlpha does not always deliver the result in the nicest form: you may have to do some post-processing on the result. Feel free to use either a table or WolframAlpha for integrals that are more complex than the simple integrals in equations NB.12 through NB.18 (which you really should memorize), but indicate your appreciation of the poor selfless mathematicians who worked out all those integrals by citing your source.

Using WolframAlpha

Feel free to look up integrals! (But cite your source!)

HOMEWORK PROBLEMS Derivations

Basic Skills NBB.1 Integrate the following functions from 0 to t (b is a ______ constant). (a) f (t) = b 2 t 2 (b) f (t) = 1/​bt   + 1 ​ 

NBD.1 Show that equations NB.12 through NB.18 are correct by taking the derivatives of the results.

NBB.2 Integrate the following functions from 0 to t (b and T are constants). (a) f (t) = e -bt + 1 (b) f (t) = b (t − T ) 3/2

NBD.2 Use substitution of variables to evaluate the integral of f (t) = (1 − bt) 3.

NBB.3 Use WolframAlpha to evaluate the integral of

NBD.3 Use substitution of variables to evaluate

 

 

at

∫ ________ ​  te  dt  ​  



(NB.29)

2

(at + 1) 

2s sin ω t dt  ​ ∫0 ​  ​  ​​ ____________     ​ where ω = ___ ​  π  ​   (1 + cos ω t)2 4s

(NB.30)

ANSWERS TO EXERCISES NBX.1 Define the function u ≡ at + b and note that du/dt = a + 0 = a, meaning that du/a = dt. Substituting this into the integral yields

∫​

______ at + b ​  dt

  

3/2

 

 

(  )

2 u(∞) 0 ​ ∫b ​  ∞​  ​​ __b2  ​ e b/t dt = ​∫u(b)   ​  ​  ​​ __b2  ​ e u ​ −__ ​  t  ​   ​  du = −​∫ ​  ​  e​  u du 1 b t t  

=∫

__ ___ du 1 ​u ​     ​     ​ = ​ __ ​ 

a

3/2

a ∫ u 

1/2

du

2u   ​ ____  = __ ​  1 ​​    u   ​ = ​ _____    = ___ ​  2  ​  (at + b) 3/2 (NB.31) a  3/2

NBX.2 Define u ≡ b/t and note that du/dt = -  b/t 2, ­meaning that dt = (-  t 2/b) du. Substituting this into the integral yields

3a

3a

 



 

1

= +​∫ ​  ​  e ​u du = ​​  e u  ​  ​0​​  = e 1 − 1 = 1.718… (NB.32) [ 

1

] 

0

Note how I had to change the limits of integration.

Index NOTE: Entries followed by t and f refer to tables and figures, respectively.

A acceleration average definition of, 12 vs. instantaneous, 12, 28 of center of mass, 14 circularly accelerating frames, 129–130, 129f definition of, 11–12 of freely falling objects, 140 instantaneous vs. average, 12, 28 definition of, 12 kinematic chain and, 22, 40 linearly accelerating frames characteristics of, 127–129, 128f frame alignment in, 78–79 motion diagrams and, 27–29, 27f vs. net force, in Newton’s second law, 7 in nonuniform circular motion, 107–109, 107f in orbits, 176 of particle, definition of, 6–7 in pendulum, 165 from position. See force(s), finding from motion position from. See motion, finding from forces in simple projectile motion, 141–142, 147 as slope of velocity vs. time, 30–31, 30f transformation equations for, 125 in uniform circular motion, 13–14, 15f, 104–105, 107 units of, 12 from velocity. See force(s), finding from motion velocity from. See motion, finding from forces acceleration arrows in constrained motion problems, 79 in motion diagrams, 28–29, 28f in trajectory diagrams, 47, 47f in two-dimensional motion, 29, 29f acceleration of gravity, 45, 140 Almagest (Ptolemy), 4 amplitude, of oscillation calculation of, 162–164 visualization of, 162, 163f angular momentum, in orbits, 174, 175, 191, 193t antiderivatives definition of, 40, 206

finding in free fall, in one dimension, 45 graphical approaches, 40–41 with integrals in one dimension, 42–45 in three dimensions, 45–46 overview of, 40 with trajectory diagrams, 46–48, 47f, 48f fundamental theorem and, 207 anti-lock brakes, 74 Aristotle, 4, 5 Astronomia Nova (Kepler), 172 astronomical units, 183

B banking in airplanes, 109–110, 109f solving problems in, 112 in wheeled vehicles, 111, 111f black holes, 180–181 bob, of pendulum, 164–165 buoyant forces, classification of, 23f, 24

C calculus differential, review of, 200–204 fundamental theorem of, 207 integral, review of, 206–210 invention of, 5 centrifugal force, 129–130, 129f chain rule, for derivatives, 202–203 circularly accelerating frames, 129–130, 129f circular motion analogy to simple harmonic oscillation, 162–164, 163f banking in, 109–111, 109f, 111f nonuniform, 107–109, 107f, 165 reference frames in, 111–112 in orbits. See orbit(s), circular problem-solving framework for, 112 uniform acceleration in, 13–14, 15f, 104–105, 107 definition of, 14 motion diagrams in, 13–14, 15f, 104–105, 104f reference frames in, 111–112 unit vectors, 105–107, 107f

circular orbit, Kepler’s laws, 176–180 coefficient of kinetic friction, 75 for various surfaces, 76t coefficient of static friction, 75 for various surfaces, 76t components of vectors, notation for, 30–31 compression forces, classification of, 23, 23f statics, 57, 57f conic sections, Kepler’s first law and, 181–184, 182f–184f conservation laws elliptical orbits and, 190–192, 190f, 193t hyperbolic orbits and, 192–193, 193t conservation of angular momentum, in orbits, 174–175, 191 conservation of energy, in orbits, 191 constant of integration, 206 evaluation of, 43–44 in finding motion from forces, 44 constant rule for definite integrals, 207 for derivatives, 202 constrained-motion checklist, coupled objects, 95–97 constrained-motion problems definition of, 79 hybrid, 78 problem-solving framework for, 79–81 contact interactions, forces, 22, 23f Copernicus, Nicolaus, 4–5 coupled objects constrained-motion checklist, 95–97 definition of, 88 force notation for, 88–90, 88f pulleys, 94–95 pushed blocks, 90–91, 91f strings, 92–94

D dark matter, 181 definite integrals. See integrals derivatives chain rule for, 202–203 of constant, 202 definition of, 200 of polynomials, 202 rules for, 201–202 and slope, 202, 203f of some useful functions, 204

211

212

derivatives (Continued) time of mathematical function, 8 of position, 9 of vector, 8–9 Descartes, René, 5 differential calculus, review of, 200–204 differential equations, solving, 157–159 directionals. See unit vectors drag, 77 classification of, 23, 23f forces, 23, 23f linearly constrained motion, 77 in simple projectile motion, 140–141, 147–148, 147f, 148f drag coefficient, 77, 147

E eccentricity of orbit, 182, 190, 191, 193t Einstein, Albert, 132–133 electromagnetic forces, classification of, 22, 23f electrostatic forces, classification of, 22, 23f ellipse(s) calculating extremes in, 193t eccentricity of, 193t foci of, 172, 181, 182f calculating nearest and farthest points from, 193t planetary orbits as, 172. See also orbit(s) properties of, 193t elliptical orbit, conservation laws and, 190–192, 190f energy in orbits, 174, 193t conservation of, 191 and orbit shape, 192–193 equation of motion, oscillatory motion, 157–159, 159f equilibrium position, in oscillation, 160 equivalence principle, 133

F fictitious forces, in noninertial reference frames, 122, 130–132 first-law detectors, 126–127, 127f focus (foci) of ellipse, 172, 181, 182 f calculating nearest and farthest points from, 193t force(s). See also frictional forces; gravitational forces vs. acceleration, in Newton’s second law, 7 centrifugal, 129–130, 129f classification of, 22–24, 23f fictitious, 122, 130–132 finding from motion, 22–32 free-body diagram, 25–27 with graphs in one-dimension, 30–31, 30f key to problem-solving in, 72–73 kinematic chain, 22 laws, 24–25

Index

motion diagram, 27–29 with Newton’s second and third laws, 60 problem-solving framework for, 60–61 for constrained-motion problems, 79–81 for coupled-object problems, 95–97 in statics problems, 61–65 quantitative example of, 32 statics, 56 problems, 60–61 finding motion from. See motion, finding from forces normal classification of, 23, 23f notation for, 24 in coupled objects, 88–90, 88f Ford, H. C., 180–181 frame-correction forces, 130–132 frames of reference. See reference frames free-body diagrams banking, 109, 109f, 111f in determining forces, 60 drawing of, 26 purpose of, 25–26 recognizing third-law partners in, 89–90 free fall acceleration in, 140 definition of, 45, 140 in one dimension, 45 freely falling frames, 132–133 frequency, of oscillation, 159 frictional forces. See also drag classification of, 23, 23f kinetic friction, 73–74, 74f classification of, 23, 23f coefficient of, 75 for various surfaces, 76t magnitude vs. normal force, 75 static friction, 73, 74f classification of, 23, 23f coefficient of, 75 for various surfaces, 76t magnitude vs. normal force, 75 fundamental theorem of calculus, 207

H

G

K

Galilean transformation, 122–125, 123f Galilean velocity transformation equation, 120 Galileo Galilei, 5 general relativity, equivalence principle and, 133 gravitational field vector, 45 definition of, 140 variation with distance, 141 gravitational forces classification of, 22, 23f law of universal gravitation, 176 in orbits, 174, 176 in simple harmonic oscillation, 160–162 in simple projectile motion, 142

Kepler, Johannes, 5, 172 Kepler’s laws, 172–173, 172f black holes and dark matter, 180–181 circular orbit, 176–180 first, 172 and conic sections, 181–184, 182f–184f orbits around a massive primary, 173–174, 173f second, 172, 175f proof of, 174–175 third, 172 for circular orbits, 177 and dark matter, 181 proof of, 176–177

Harmonice Mundi (Kepler), 172 Harms, R. J., 180–181 hertz (unit), 159 Hooke, Robert, 5 Hooke’s law, 156 hyperbola, definition of, 184 hyperbolic orbit, 184, 192–193 conservation laws and, 192–193, 192f, 193t

I ideal pulley, 94 indefinite integrals. See integrals inertial reference frames, 126–127, 127f initial conditions, in finding motion from forces, 42, 45 initial phase, of oscillation calculation of, 162–164 visualization of, 163, 163f instantaneous acceleration. See acceleration instantaneous velocity. See velocity integral calculus, review of, 206–210 integrals definite definition of, 206–207 in evaluating constants of integration, 43–44 fundamental theorem and, 207 from indefinite integral, 208 properties of, 207 in finding motion from forces in one dimension, 42–45 in three dimensions, 45–46 indefinite definite integral from, 208 definition of, 208 difficulty of evaluating, 209 substitution of variables in, 208–209 tables of integrals for, 210 interactions contact, types of, 22–23, 23f internal, and momentum, 6 long-range, types of, 22, 23f inverse rule, for derivatives, 201

213

Index

kinematic chain, 22 reverse, 40 kinematics, 22 kinetic energy, in orbits, 174 kinetic friction classification of, 23, 23f coefficient of, 75 for various surfaces, 76t linearly constrained motion, 73–77, 73f, 74f magnitude vs. normal force, 75

L law of universal gravitation, 176 laws of motion. See also specific laws history of development, 4–5 lift forces, classification of, 23, 23f linearly accelerating frames, 127–129, 128f frame alignment in, 78–79 linearly constrained motion accelerated motion, 78–79 constant velocity, 72–73 constrained-motion checklist, 79–81 drag forces, 77 static and kinetic friction forces, 73–77, 73f, 74f LMC (Large Magellanic Cloud), 181

M magnetic forces, 22, 23f magnitude of vectors, notation for, 30 mass, spring oscillatory motion, 156–157, 156f, 160–162, 160f, 161f model step, in force-from-motion problems, 60–61 motion at constant velocity, 72–73 finding forces from. See force(s), finding from motion finding from forces, 30–31, 40–49 in free fall, 45 graphical approaches, 40–41 with integrals in one dimension, 42–45 in three dimensions, 45–46 overview of, 40 reverse kinematic chain, 40 with trajectory diagrams, 46–49, 47f, 48f one-dimensional deriving forces in, 30–31, 30f finding motion from forces in, 42–45 free fall in, 45 of projectile. See projectile motion, simple motion diagrams acceleration in, 27–29, 27f, 28f drawing of, 27–29, 27f in nonuniform circular motion, 107, 107f in two-dimensional motion, 29

in uniform circular motion, 13–14, 15f, 104–105, 104f uses of, 27 velocity in, 27f, 28 multiflash photograph, 27, 27f

N Newton, Isaac, 5 Newton (computer app) applied to orbits, 182–184 basic description, 50 Newtonian synthesis, 4–5 Newton’s first law, 7 in noninertial reference frames, 126–127 Newton’s law of universal gravitation, 25, 176 Newton’s second law, 6–7 for circular orbits, 176 component form of, 72 counterintuitive aspects of, 7 derivation of, 6 in force-from-motion problems, 60–61 free-body diagrams and, 25–26 kinematic chain, 22 law of zero torque, 58–59 in noninertial frames, 126–127, 130 solving for trajectory, 148–149 vector calculus, 8 Newton’s third law, 7 counterintuitive aspects of, 7–8 recognizing in coupled object problems, 88 on free-body diagrams, 89 third-law partners characteristics of, 89 noninertial frames circularly accelerating frames, 129–130, 129f fictitious forces, 122, 130–132 freely falling frames and gravity, 132–133, 133f galilean transformation, 122–125, 123f inertial reference frames, 126–127, 127f linearly accelerating frames, 127–129, 128f nonuniform circular motion circularly constrained motion, 107–109, 107f–108f normal forces classification of, 23, 23f kinetic friction and, 75

O one-dimensional motion deriving forces in, 30–31, 30f finding motion from forces in, 42–45 free fall in, 45 orbit(s) angular momentum in, 174–176, 189t, 190–196 characteristics of, 173–174, 173f circular Kepler’s third law for, 177 Newton’s second law for, 176

possibility of, 176 problem solving in examples of, 178–180 elliptical. See also ellipse(s) energy and momentum in, 191–192 planetary orbits as, 172, 181 energy in, 174, 189t conservation of, 190–193 and orbit shape, 192 hyperbolic. See also hyperbola as possible, 184, 192–193, energy and momentum in, 192–193, 193t and Kepler’s second law, 174–175 parabolic, 181, 190 period of, 177 calculation of, 178–180, 189t semimajor axis and, 172 problem-solving in examples of, 193–196 useful equations, 193t radius of, calculation of, 180 speed of circular, 177 calculation of, 178–180 system energy and shape of, 192 oscillator model, 160 oscillatory motion. See simple harmonic oscillator (SHOs)

P pendulum, simple pendulum equation, 165 period of orbit, 177 calculation of, 178–180, 189t semimajor axis and, 172 of oscillation, 159 phase, initial, of oscillation calculation of, 162–164 visualization of, 162–163, 163f phase rate (angular frequency) of oscillation, 155, 158 Philosophiae Naturalis Principia Mathematica (Newton), 5 position finding, from acceleration. See motion, finding from forces kinematic chain and, 22, 40 in simple projectile motion, 138–139 finding from velocity, 138–139 transformation equations for, 123 pressure forces, 23f primary definition of, 174 massive, orbit around, 173–174, 173f Principia (Newton), 5 product rule, for derivatives, 201 projectile, definition of, 140 projectile motion, simple definition of, 140 drag in, 140–141, 147–148 equations for, 141–142 implications of, 142–144

214

projectile motion (Continued) maximum altitude in, 142–143, 143f parameters in, 140–141 terminal speed in, 147–148, 148f time of impact in, 142–143, 143f trajectory in, 143 Ptolemy, 4, 5 pulley(s), ideal pulley model, 94–95

R radius, of orbit, calculation of, 180 real pulley, 95 reference frames, 122–133 circularly-accelerating, 129–130, 129f in circular motion problems, 111 frame-correction forces in, 130–132 freely falling, 132–133 galilean transformation and, 122–125 inertial, 126–127 linearly-accelerating, 78, 127–129, 128f noninertial, 126–127, 130 fictitious forces in, 122, 130–132 orientation of, in force-from-motion problems, 72 restoring force, in oscillation, 160 reverse kinematic chain, 40

S satellite, definition of, 174 second-law partners (vs. third-law), 89 semimajor axis, 172, 182, 190, 191 SHOs. See simple harmonic oscillators (SHOs) simple harmonic oscillators (SHOs) analogy to circular motion, 162–164, 163f applications of, 160 characteristics of, 160 equation of motion, 157–159, 159f force law for, 156, 156f gravity and, in suspended mass, 160–162 mass, spring, 156–157, 156f, 160–162, 160f, 161f oscillator model, 160 simple pendulum, 164–165, 165f simple pendulum, 164–165, 165f simple projectile motion, 141–142 slope derivatives and, 202, 203f at a point, definition of, 202 of position vs. time, 30–31, 30f of velocity vs. time, 30–31, 31f slope method, of finding motion from forces, 41, 41f

Index

speed instantaneous, definition of, 10 of orbit, 177 terminal, in simple projectile motion, 147–148, 148f spring constant, 156 spring forces. See also simple harmonic oscillators (SHOs) static friction, 73–74, 74f classification of, 23, 23f coefficient of, 75 for various surfaces, 76t magnitude vs. normal force, 75 statics compression, 57, 57f forces from motion, 56, 60–61 problems, 56, 61–65 checklists, 62 torque, 57–59 string(s) in coupled objects, 92–94 ideal, 94 tension on/of, 92, 93, 94 substitution of variables, in indefinite integrals, 208–209 sum rule for definite integrals, 207 for derivatives, 201

T tension forces, classification of, 23, 23f tension on/of string, 92, 94 terminal speed, in simple projectile motion, 147–148, 147f, 148f third-law partners recognizing, 89–90 thrust forces, 23f, 24 time derivative of vector, 8 torque statics, 57–59, 61, 62 trajectory, in simple projectile motion, 46–49, 143 trajectory diagrams construction of, 46–49, 47f transformation equations for accelerations, 125 for positions, 123 for velocities, 123 translation step force-from-motion, 60 in simple projectile motion, 144

U uniform circular motion. See circular motion, uniform unit vectors, 105–107 universal gravitational constant, 176

V vector calculus, 8–9 vector functions, integration of, 45–46 vector(s) components of, notation for, 30 gravitational field, 45 definition of, 140 variation with distance, 141 magnitude of, notation for, 30 time derivative of, 8 velocity average constant, motion in, 72–73 definition of, 10–11 direction of, 10f–11f vs. instantaneous, 10–11, 10f–11f, 12, 14 notation for, 9, 10 definition of, 9 finding acceleration from. See force(s), finding from motion finding from acceleration. See motion, finding from forces instantaneous vs. average, 10–12, 10f–11f, 14 definition of, 9–10 notation for, 9, 10 kinematic chain and, 22, 40 in motion diagrams, 27–29, 27f in simple projectile motion, 138–139, 148 finding from acceleration, 141–142, 147 finding position from, 141–142 as slope of position vs. time, 30–31, 30f slope of vs. time, 30–31, 30f transformation equations for, 123 in uniform circular motion, 13–14, 15f, 105 velocity arrows, in motion diagrams, 27f, 28

W weight definition of, 140 vs. normal forces, 75

X x-axis, in simple projectile motion, 142 x-position, vs. time, in constant x-acceleration, 43 x-velocity definition of, 10 vs. time, in constant x-acceleration, 43

Y y-velocity, definition of, 10

Z z-velocity, definition of, 10

4 Be 9.012

12 Mg 24.31

20 Ca 40.08

38 Sr 87.62

56 Ba 137.3

88 Ra (226)

3 Li 6.941

11 Na 22.99

19 K 39.10

37 Rb 85.47

55 Cs 132.9

87 Fr (223)

89 Ac (227)

57 La 138.9

39 Y 88.91

21 Sc 44.96

3 3B

104 Rf (257)

72 Hf 178.5

40 Zr 91 .22

22 Ti 47.88

4 4B

106 Sg (263)

59 Pr 140.9 91 Pa (231)

58 Ce 140.1 90 Th 232.0

74 W 183.9

42 Mo 95.94

24 Cr 52.00

6 6B

105 Db (260)

73 Ta 180.9

41 Nb 92.91

23 V 50.94

5 5B

92 U 238.0

60 Nd 144.2

107 Bh (262)

75 Re 186.2

43 Tc (98)

25 Mn 54.94

7 7B

1 H 1.008

93 Np (237)

61 Pm (147)

108 Hs (265)

76 Os 190.2

44 Ru 101.1

26 Fe 55.85

8

94 Pu (242)

62 Sm 150.4

109 Mt (266)

77 Ir 192.2

45 Rh 102.9

27 Co 58.93

9 8B

Atomic mass

Symbol

95 Am (243)

96 Cm (247)

64 Gd 157.3

111

110

63 Eu 152.0

79 Au 197.0

47 Ag 107.9

29 Cu 63.55

11 1B

78 Pt 195.1

46 Pd 106.4

28 Ni 58.69

10

97 Bk (247)

65 Tb 158.9

112

80 Hg 200.6

48 Cd 112.4

30 Zn 65.39

12 2B

13 3A

14 4A

15 5A

16 6A

17 7A

98 Cf (249)

66 Dy 162.5

(113)

81 Tl 204.4

49 In 114.8

31 Ga 69.72

13 Al 26.98

5 B 10.81

99 Es (254)

67 Ho 164.9

114

82 Pb 207.2

50 Sn 118.7

32 Ge 72.59

14 Si 28.09

6 C 12.01

100 Fm (253)

68 Er 167.3

(115)

83 Bi 209.0

51 Sb 121.8

33 As 74.92

15 P 30.97

7 N 14.01

101 Md (256)

69 Tm 168.9

116

84 Po (210)

52 Te 127.6

34 Se 78.96

16 S 32.07

8 O 16.00

102 No (254)

70 Yb 173.0

(117)

85 At (210)

53 I 126.9

35 Br 79.90

17 Cl 35.45

9 F 19.00

103 Lr (257)

71 Lu 175.0

86 Rn (222)

54 Xe 131.3

36 Kr 83.80

18 Ar 39.95

10 Ne 20.18

2 He 4.003

Atomic number

1 H 1.008

2 2A

18 8A

1 1A

Periodic Table of the Elements

Short Answers to Selected Problems [Note that most of the derivation (D) problems as well as a number of other problems have answers given in the problem statement. These problems are also useful for practice, but their answers are not reiterated here.]

Chapter N1

B1 (a) The cannon is much more massive than the ball, and so responds less. (b) The earth is even more massive. B3 2600 N backward. B5 (a) a (b) 2 a t − 2 b -2 t -3 (c) 2 a 2 t + 2 a b. B7 [0.5 m/s, 4 m/s, 2 m/s] B9 Zero. B11 5.6 m/s 2 upward. B13 3 m/s 2. M2 54 kg. M3 17,000 m/s2. M5 (a) m and s–1, (b) t = 3π/2ω, (c) m Aω 2 (d) m (Aω 2  + │g​  ​W│   ). M7 (a) 0.067 m│g​  ​W│   , (b) same. R3 Braking is better.  

  

  

Chapter N2

B7 (a) Buoyant force, (b) lift force. B9 To the right. B11 5.2 m/s 2 down and to the right. M1 Banking provides an inward force component. M3b Net force must be inward, not zero. M5 (a) 735 N, (b) 30 hp. M7a M│ aW​  │ ​  /(│g​  W​ │    + │ aW​  │ ​  ). M9c 6000 N. R1 53 mi/h. R3 Greater than 19.6 m/s.

Chapter N7

B1 510 m. B3 10.7 km. B5 [0.49, 0.81, −0.32]. B7 D, E, F. M1 49 m/s. M3 µ s ≥ 0.22. M5 (a) ≥0.74, (b) Not reasonable, (c) 36°. M7 2π [L cos θ/│g​  ​W│   ]1/2. M9 0.21. M11 2.5R. D4c r = │p​  ​W│   /q│B​  ​W│  . R1 Reduce speed to 22 m/s or use a concrete road.

Chapter N8

B1 24 m/s. B3 146 km/h. B5 (a) Zero, (b) 4 m/s 2. B7 9.8 m/s 2. M1 1.94 m/s. M3 (a) 2.2 m/s 2, (b) 140 N, (c) zero, (d) 140 N. M5 152 lbs. M7 0.8 s compared to 1.1 s. M9 2.9 s. D2b 1.4 fm. D3 15 µm/s 2. R1 26th or 27th floor. R3 Yes.

Chapter N9

B1 No, 29° shift in direction. B3 2.24 s. B5 5.1 s. B7 8.5 m/s. M1 7.3 m/s. M3 15.3 m/s. M5 ≥ 9.5 m/s. M7 880 m. M9 (a) 130 m/s, (b) 23 s, (c) drag is significant, (d) 161 m/s, 25.6 s. M11 (a) Range 246 m, height 47 m, time 6.1 s, (b) 134 m, 33 m, 5.1 s, (c) 140.5 m, 38.4 m, 5.8 s, yes. R1 990 m/s. R3 _3​ 2 ​  D.

Chapter N10

Chapter N3

B3 30.6 m. B5 10 cm, 2.45 cm. M3 The first car is faster (25 m/s vs. 18 m/s) but the other is further ahead (90 m vs. 62.5  m). M5 (a) –120 m/min, (b) 2bt, (c) 55 min, –4400 m. M7 (a) s-1 (b) 3.1 m/s, (c) 2.23 m. M9 0.2 s (note that initial ­velocity arrow should be 7 cm long; acceleration arrow 1.1 cm long). D1 (a) _​ 21 ​  bt 2 (b) _​ 21 ​  bt (t − 2T ). R1 I catch the car at 3.6 mi.  

 

Chapter N4

B1 (a) 3.3 s, (b) 0.30 Hz, (c) 1.9 rad/s. B3 20.5 N/m. B5 π. B7 3.6 s. M1 0.45 s. M3 Longer. M5 (a) 6.0 cm, (b) 98 N/m, (c)  0.77  m/s. M7 1.2 Hz. M9 Longer. M11 (a) N, m–1, (b) Hint: sin θ ≈ θ for small θ in radians, (c) 5.0 Hz. R1 56.58 kg. A1 (a) r0 = c 3/b 2, (b) ks = a b 8/c 9, (c) b 4 [a/mc] 1/2/2πc 4.  

 

 

 

 

 

Chapter N11

Roughly 180 lbs, if the person can pull with 100  lbs of force. M3 270 N. M5 No. M7 57 N. M9 _2​ 1 ​  m│g​  ​W│   tan θ. R1b  About 700 N.

B1 44,300 km. B3 42,300 km. B5 29.7 km/s. B7 164 y. B9 83 ms. B11 1.4 y. M1 Yes (CM is 4670 km from the earth’s center). M3 The ratio is 0.0056. M5 (a)Yes, (b) yes. M8 (a) z2.9 × 10 11 m, (b) 1.017 times larger. D1 T = 2π [m/ks] 1/2. 1 D3 T 2 = (4π 2 m/b)R. R1 No. R2 (b) ΔE = − ​ _4  ​ GM 2/R0. 29 (d) 3.5 × 10 W. R3 46,870 ≈ km or 38,660 km.

Chapter N5

Chapter N12

B1 m│g​  W​ │   /cos θ. B3 70 g. B5 (a) 5_​ 1 ​    L│F​  W ​ C  R│ toward viewer, (b) ​ 5_4 ​    L│F​  W ​ C  L│  3 __ │ W away from viewer (c) ​     ​  LM│ g​ ​     t oward the viewer. M1 10

B1 0.68. B3 0.41. B5 45 N. B7 74 m. M1 5.9 m/s 2. M3 8.3 N up the incline. M5 3.6 N. M7 Yes, b = 100 kg/s. M9 1.3 m/s2 upward. M11 ≥ 4.5 s. M13 ≥ 0.21. R1 ≈0.35. R3 Yes, if you have antilock brakes. Drag is not significant.

Chapter N6

B1 (a) yes, (b) no, (c) no. B3 1.5 N. M1 One team can exert a greater static friction force on the ground. M3 86,400 N, 84,000 N. M6 (a) 240 N, (b) 0.43. M7 ≤ 1400 N. M9 0.30 m/s 2 down the incline. R1 The mule can exert a bigger static friction force on the ground than the plow. R3 Maybe Pat can without the backpack, but definitely with the backpack. A1 mn = (0.9)n-1 M, total = 10M[1 - 0.90N]

224

 

 

 

 

B1 0.83, 8300 km, elliptical. B3 2 km/s. M1 Hyperbolic, no. M3 Increase speed _____ by 22.5%. M5 9.90 km/s. M7 (a) Yes, │ (b) 0.072 AU. M9c ​17/9 ​    v​  ​W0 │. R1 Fire engines in the direction of the ship’s motion (to slow it down) for 103 s. R3 (a) 1.29 y, (b) boost speed by 21%.

Appendix NA

B1 (a) 2at, (b) –3c/t 4, (c) 2 a b t(1 – at 2 ) –2. B3 (a) −20b 2 (bt − 1)-5, (b) _​ 32 ​  b(bt + 1)1/2, (c) b cos (bt + π).  

Appendix NB

 

 

 

______

B1 (a) _3​ 1 ​   b 2 t 3, (b) 2b-1( ​bt   + 1 ​   - 1) B3 e at/[a 2(at + 1)]. D3 (2/π)s.      

 

 

Some Physical Constants Speed of light Gravitational constant Coulomb’s constant Permittivity constant Permeability constant Planck’s constant Boltzmann’s constant Elementary charge Electron mass Proton mass Neutron mass Avogadro’s number

Standard Metric Prefixes (for powers of 10)

c 3.00  × 108 m/s G 6.67  × 10-11 N·m2/kg2 1/4πε0 8.99  × 109 N·m2/C2 ε0 8.85  × 10-12 C2/(N·m2) µ0 4π   × 10-7 N/A2 h 6.63  × 10-34 J·s kB 1.38  × 10-23 J/K e 1.602 × 10-19 C me 9.11  × 10-31 kg mp 1.673 × 10-27 kg mn 1.675 × 10-27 kg NA 6.02  × 1023

Power Prefix Symbol 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 10-2 centi c 10-3 milli m 10-6 micro µ 10-9 nano n 10-12 pico p 10-15 femto f 10-18 atto a

Commonly Used Physical Data Gravitational field strength g =│g​  ​W│   9.80 N/kg = 9.80 m/s2 (near the earth’s surface) Mass of the earth Me 5.98 × 1024 kg Radius of the earth Re 6380 km (equatorial) Mass of the sun M⊙ 1.99 × 1030 kg Radius of the sun R⊙ 696,000 km Mass of the moon 7.36 × 1022 kg Radius of the moon 1740 km Distance to the moon 3.84 × 108 m Distance to the sun 1.50 × 1011 m † Density of water 1000 kg/m3 = 1 g/cm3 † Density of air 1.2 kg/m3 Absolute zero 0 K = -273.15°C = -459.67°F Freezing point of water‡ 273.15 K = 0°C = 32°F Boiling point of water‡ 373.15 K = 100°C = 212°F Normal atmospheric pressure 101.3 kPa † ‡

At normal atmospheric pressure and 20°C. At normal atmospheric pressure.

Useful Conversion Factors 1 meter = 1 m = 100 cm = 39.4 in = 3.28 ft 1 mile = 1 mi = 1609 m = 1.609 km = 5280 ft 1 inch = 1 in = 2.54 cm 1 light-year = 1 ly = 9.46 Pm = 0.946 × 1016 m 1 minute = 1 min = 60 s 1 hour = 1 h = 60 min = 3600 s 1 day = 1 d = 24 h = 86.4 ks = 86,400 s 1 year = 1 y = 365.25 d = 31.6 Ms = 3.16 × 107 s 1 newton = 1 N = 1 kg·m/s2 = 0.225 lb 1 joule = 1 J = 1 N·m = 1 kg·m2/s2 = 0.239 cal 1 watt = 1 W = 1 J/s 1 pascal = 1 Pa = 1 N/m2 = 1.45 × 10–4 psi 1 kelvin (temperature difference) = 1 K = 1°C = 1.8°F 1 radian = 1 rad = 57.3° = 0.1592 rev 1 revolution = 1 rev = 2π rad = 360° 1 cycle = 2π rad 1 hertz = 1 Hz = 1 cycle/s

1 m/s = 2.24 mi/h = 3.28 ft/s 1 mi/h = 1.61 km/h = 0.447 m/s = 1.47 ft/s 1 liter = 1 l = (10 cm)3 = 10-3 m3 = 0.0353 ft3 1 ft3 = 1728 in3 = 0.0283 m3 1 gallon = 1 gal = 0.00379 m3 = 3.79 l ≈ 3.8 kg H2O Weight of 1-kg object near the earth = 9.8 N = 2.2 lb 1 pound = 1 lb = 4.45 N 1 calorie = energy needed to raise the temperature of 1 g of H2O by 1 K = 4.186 J 1 horsepower = 1 hp = 746 W 1 pound per square inch = 6895 Pa 1 food calorie = 1 Cal = 1 kcal = 1000 cal = 4186 J 1 electron volt = 1 eV = 1.602 × 10-19 J

(  ) T = ​( ___ ​  5K  ​  )​  (T 9°F

T = ​ ____ ​  1K  ​  ​  (T[C] + 273.15°C) 1°C [F]

+ 459.67°F)

(  )

T[C] = ​ ____ ​ 5°C ​  ​(T[F] - 32°F) 9°F T[F] = 32°F + ​ ____ ​  9°F  ​  ​T[C] 5°C

(  )

Relativistic Units, Conversion Factors, and Benchmarks SR unit system: distance is measured in seconds (so that c = 1) 1 s of distance = 299,792,458 m 1 ns of distance = 0.300 m = 0.984 ft 1 min of distance = 18 × 106 km = 18 Gm 1 h of distance = 1.08 × 109 km = 1.08 Tm 1 day of distance = 2.59 × 1010 km = 25.9 Tm 1 month of distance ≈ 780 Tm 1 year of distance = 1 light-year = 1 ly = 0.946 × 1016 m = 9.46 Pm Average distance between the earth and the moon: 1.28 s. Average distance between the earth and the sun: 8.33 min. Average distance between Mars and the sun: 12.69 min. Average distance between Neptune and the sun: 4.17 h Average distance between Pluto and the sun: 5.53 h. Distance to the nearest star: 4.3 y Distance to the galactic center: ∼30,000 y Diameter of the Milky Way galaxy ∼100,000 y Light travel time from the edge of the visible universe: ∼13.75 Gy  ​W│   = 3.27 × 10-8 s-1 = (0.969 y)-1 │g​

1 kg (energy) = 9.0 × 1016 J Energy released by a 100-kt atom bomb: 4.184 × 1014 J = 4.66 g. 1 J = 1.1 × 10–17 kg (of energy)

Table RA.1

Some important equations and their equivalents in SI units

Equation

SR Version

SI Equivalent

Metric

∆s2 = ∆t2 - ∆x2 - ∆y2 - ∆z2

∆x2 + ∆y2 + ∆z2 ∆s2 = ∆t2 - _______________ ​     ​   c2

Proper time

dτ = dt  ​ 1 -    │v​​  ​W│   ​​ ​  

Lorentz transformations (t and x)

γ = 1/​1  - β2 ​  

γ = 1/​1  - (β/c)2 ​ 



t′ = γ (t - βx)

t′ = γ (t - βx/c2)



x′ = γ (-βt + x)

x′ = γ (-βt + x)

_______

Lorentz contraction

2

______

L=

_______ LR ​1   │v​  ​W │2 ​  

_________

dτ = dt ​1 -    │v​  ​W/   c​│​ ​ ​  2

_________

_________

L = LR ​1   │v​  ​W/   c│2 ​ 

v -β v -β Transformation for x-velocity ​ v ​′ x​​  = _______ ​  x    ​  ​ v ​′ x​​  = __________ ​  x   ​  1 - βvx/c2 1 - βvx Energy in terms of speed

m    E = _________ ​  _______  ​  ​  1 -│v​  ​W│   2   ​

Relativistic momentum magnitude

  ​    ​W │ = _________ ​  _______ │p​

   ​ _________  ​W │ = ___________ ​    │p​

Mass in terms of E and │p​  ​W│  

m2 = E2 - │p​​ ​W│   2​ ​

(mc2)2 = E2 - │p​  ​W c│ ​ 2​ ​

Speed in terms of E and │p​  ​W│  

 ​W│    ​W c│ │p​ │p​  │v​  ​W│     _____ ____ │v​  ​W │ = ____ ​   ​   ​   ​   = ​   ​       c E E

Photon energy in terms of f and λ

E = hf = __ ​  h  ​ λ

m│v​  ​W│  

 ​ 1   │v​  ​W│   2 ​ 

m   ​ _________ E = ___________ ​      ​  1 -│v​  ​W/   c│2   ​ m│v​  ​W│  

 ​ 1   │v​  ​W/   c│2 ​ 

E = hf = __ ​  hc ​  λ

Six Ideas That Shaped Physics Unit R: The Laws of Physics Are Frame-Independent Third Edition

Thomas A. Moore

SIX IDEAS THAT SHAPED PHYSICS, UNIT R: THE LAWS OF PHYSICS ARE FRAME-INDEPENDENT, THIRD EDITION Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2017 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous ­editions © 2003, and 1998. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of ­Mc­Graw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 RMN/RMN 1 0 9 8 7 6 ISBN 978-0-07-760095-2 MHID 0-07-760095-9 Senior Vice President, Products & Markets: Kurt L. Strand Vice President, General Manager, Products & Markets: Marty Lange Vice President, Content Design & Delivery: Kimberly Meriwether David Managing Director: Thomas Timp Brand Manager: Thomas M. Scaife, Ph.D. Product Developer: Jolynn Kilburg Marketing Manager: Nick McFadden Director of Development: Rose Koos Digital Product Developer: Dan Wallace Director, Content Design & Delivery: Linda Avenarius Program Manager: Faye M. Herrig Content Project Managers: Melissa M. Leick, Tammy Juran, Sandy Schnee Design: Studio Montage, Inc. Content Licensing Specialists: Deanna Dausener Cover Image: NASA Compositor: SPi Global

Dedication For My Parents, Stanley and Elizabeth, who taught me the joy of wondering All credits appearing on page or at the end of the book are considered to be an extension of the ­copyright page. The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a ­website does not indicate an endorsement by the authors or McGraw-Hill Education, and M ­ cGraw-Hill Education does not guarantee the accuracy of the information presented at these sites. Library of Congress Cataloging-in-Publication Data Names: Moore, Thomas A. (Thomas Andrew), author. Title: Six ideas that shaped physics. Unit R, The laws of physics are   frame-independent/Thomas A. Moore. Other titles: Laws of physics are frame-independent Description: Third edition. | New York, NY : McGraw-Hill Education, [2016] |   2017 | Includes index. Identifiers: LCCN 2015043354| ISBN 9780077600952 (alk. paper) | ISBN   0077600959 (alk. paper) Subjects: LCSH: Special relativity (Physics)—Textbooks. Classification: LCC QC173.65 .M657 2016 | DDC 530.11—dc23 LC record available at http://lccn.loc.gov/2015043354 www.mhhe.com

Contents: Unit R The Laws of Physics Are Frame-Independent About the Author

vii

Preface viii Introduction for Students

Chapter R1 The Principle of Relativity R1.1 R1.2 R1.3 R1.4 R1.5 R1.6

Chapter Overview Introduction to the Principle Events, Coordinates, and Reference Frames Inertial Reference Frames The Final Principle of Relativity Newtonian Relativity The Problem of Electromagnetic Waves TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter R2 Coordinate Time R2.1 R2.2 R2.3 R2.4 R2.5 R2.6 R2.7

Chapter Overview Relativistic Clock Synchronization SR Units Spacetime Diagrams Spacetime Diagrams as Movies The Radar Method Coordinate Time Is Frame-Dependent A Geometric Analogy TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter R3 The Spacetime Interval

xiv

2 2 2 4 6 8 11 11 15 18 19 21

22 22 22 24 25 27 29 31 32 35 38 39 42

44 44

R3.1 R3.2 R3.3 R3.4 R3.5 R3.6 R3.7

Chapter Overview The Three Kinds of Time The Metric Equation About Perpendicular Displacements Evidence Supporting the Metric Equation Spacetime Is Not Euclidean More About the Geometric Analogy Some Examples TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter R4 Proper Time R4.1 R4.2 R4.3 R4.4 R4.5 R4.6

Chapter Overview A Curved Footpath Curved Worldlines in Spacetime The Binomial Approximation Ranking the Three Kinds of Time Experimental Evidence The Twin Paradox TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

44 46 48 51 53 55 56 58 60 61 64

66 66 66 68 69 73 74 76 78 81 82 85

Chapter R5 Coordinate Transformations

86 86

Chapter Overview R5.1 Overview of Two-Observer Diagrams R5.2 Conventions R5.3 Drawing the Diagram t’ Axis R5.4 Drawing the Diagram x’ Axis R5.5 Reading the Two-Observer Diagram R5.6 The Lorentz Transformation TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

86 88 89 90 92 94 95 99 100 104

v

Table of Contents

vi

Chapter R6 Lorentz Contraction R6.1 R6.2 R6.3 R6.4 R6.5 R6.6

Chapter Overview The Length of a Moving Object A Two-Observer Diagram of a Stick What Causes the Contraction? The Contraction Is Frame-Symmetric The Barn and Pole Paradox Other Ways to Define Length TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter R7 The Cosmic Speed Limit R7.1 R7.2 R7.3 R7.4

Chapter Overview Causality and Relativity Timelike, Lightlike, and Spacelike Intervals The Causal Structure of Spacetime The Einstein Velocity Transformation TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

108 Appendix RA 108 Converting Equations to SI Units 108 110 110 112 114 115 118 120 120 125

126 126 126 128 131 134 136 139 140 143

Chapter R8 144 Four-Momentum 144 R8.1 R8.2 R8.3 R8.4 R8.5 R8.6

Chapter Overview A Plan of Action Newtonian Momentum Isn’t Conserved The Four-Momentum Vector Properties of Four-Momentum Four-Momentum and Relativity Relativistic Energy TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

Chapter R9 Conservation of Four-Momentum R9.1 R9.2 R9.3 R9.4 R9.5 R9.6

Chapter Overview Energy–Momentum Diagrams Solving Conservation Problems The Mass of a System of Particles The Four-Momentum of Light Applications to Particle Physics Parting Comments TWO-MINUTE PROBLEMS HOMEWORK PROBLEMS ANSWERS TO EXERCISES

144 146 147 148 151 153 155 158 158 161

162 162 162 164 166 169 171 174 175 176 177 179

RA.1 Why Use SR Units? RA.2 Conversion of Basic Quantities RA.3 Converting SR Unit Equations to SI Unit Equations RA.4 Energy-Based SR Units RA.5 Exercises for Practice

Appendix RB The Relativistic Doppler Effect RB.1 RB.2 RB.3 RB.4 RB.5

Introduction to the Doppler Effect Deriving the Doppler Shift Formula Astrophysical Applications The Nonrelativistic Limit Doppler Radar HOMEWORK PROBLEMS ANSWERS TO EXERCISES

180 180 180 180 181 182 182

184 184 184 184 186 187 187 188 188

Index 190 Periodic table

193

Short Answers to Selected Problems

194

About the Author Thomas A. Moore graduated from Carleton College (magna cum laude with Distinction in Physics) in 1976. He won a Danforth Fellowship that year that supported his graduate education at Yale University, where he earned a Ph.D. in 1981. He taught at Carleton College and Luther College before taking his current position at Pomona College in 1987, where he won a Wig Award for Distinguished Teaching in 1991. He served as an active member of the steering committee for the national Introductory University Physics Project (IUPP) from 1987 through 1995. This textbook grew out of a model curriculum that he developed for that project in 1989, which was one of only four selected for further development and testing by IUPP. He has published a number of articles about astrophysical sources of gravitational waves, detection of gravitational waves, and new approaches to teaching physics, as well as a book on general relativity entitled A General Relativity Workbook (University Science Books, 2013). He has also served as a reviewer and as an associate editor for A ­ merican Journal of Physics. He currently lives in Claremont, California, with his wife Joyce, a retired pastor. When he is not teaching, doing research, or writing, he enjoys reading, hiking, calling contradances, and playing Irish traditional fiddle music.

vii

Preface Introduction This volume is one of six that together comprise the text materials for Six Ideas That Shaped Physics, a unique approach to the two- or three-semester calculus-based introductory physics course. I have designed this curriculum (for which these volumes only serve as the text component) to support an introductory course that combines two elements that rarely appear together: (1) a thoroughly 21st-century perspective on physics (including a great deal of 20th-century physics), and (2) strong support for a student-centered classroom that emphasizes active learning both in and outside of class, even in situations where large-enrollment sections are unavoidable. This course is based on the premises that innovative metaphors for teaching basic concepts, explicitly instructing students in the processes of constructing physical models, and active learning can help students learn the subject much more effectively. In the course of executing this project, I have completely rethought (from scratch) the presentation of every topic, taking advantage of research into physics education wherever possible. I have done nothing in this text just because “that is the way it has always been done.” Moreover, because physics education research has consistently underlined the importance of active learning, I have sought to provide tools for professors (both in the text and online) to make creating a coherent and self-­ consistent course structure based on a student-centered classroom as easy and practical as possible. All of the materials have been tested, evaluated, and rewritten multiple times. The result is the culmination of more than 25 years of continual testing and revision. I have not sought to “dumb down” the course to make it more accessible. Rather, my goal has been to help students become smarter. I have intentionally set higher-than-usual standards for sophistication in physical thinking, but I have also deployed a wide range of tools and structures that help even average students reach this standard. I don’t believe that the mathematical level required by these books is significantly different than that in most university physics texts, but I do ask students to step beyond rote thinking patterns to develop flexible, powerful, conceptual reasoning and modelbuilding skills. My experience and that of other users is that normal students in a wide range of institutional settings can (with appropriate support and practice) meet these standards. Each of six volumes in the text portion of this course is focused on a single core concept that has been crucial in making physics what it is today. The six volumes and their corresponding ideas are as follows:

viii

Unit C: Unit N: Unit R: Unit E: Unit Q: Unit T:

Conservation laws constrain interactions The laws of physics are universal (Newtonian mechanics) The laws of physics are frame-independent (Relativity) Electric and Magnetic Fields are Unified Particles behave like waves (Quantum physics) Some processes are irreversible (Thermal physics)

I have listed the units in the order that I recommend they be taught, but I have also constructed units R, E, Q, and T to be sufficiently independent so they can be taught in any order after units C and N. (This is why the units are lettered as opposed to numbered.) There are six units (as opposed to five or seven) to make it possible to easily divide the course into two semesters, three quarters, or three semesters. This unit organization therefore not only makes it possible to dole out the text in small, easily-handled pieces and provide a great deal of flexibility in fitting the course to a given schedule, but also carries its own important pedagogical message: Physics is organized hierarchically, structured around only a handful of core ideas and metaphors. Another unusual feature of all of the texts is that they have been designed so that each chapter corresponds to what one might handle in a single 50-­minute class session at the maximum possible pace (as guided by years of experience). Therefore, while one might design a syllabus that goes at a slower rate, one should not try to go through more than one chapter per 50-minute session (or three chapters in two 70-minute sessions). A few units provide more chapters than you may have time to cover. The preface to such units will tell you what might be cut. Finally, let me emphasize again that the text materials are just one part of the comprehensive Six Ideas curriculum. On the Six Ideas website, at

𝚠𝚠𝚠𝚙𝚑𝚢𝚜𝚒𝚌𝚜𝚙𝚘𝚖𝚘𝚗𝚊𝚎𝚍𝚞/𝚜𝚒𝚡𝚒𝚍𝚎𝚊𝚜/

you will find a wealth of supporting resources. The most important of these is a detailed instructor’s manual that provides guidance (based on Six Ideas users’ experiences over more than two decades) about how to construct a course at your institution that most effectively teaches students physics. This manual does not provide a one-size-fits-all course plan, but rather exposes the important issues and raises the questions that a professor needs to consider in creating an effective Six Ideas course at their particular institution. The site also provides software that allows professors to post selected problem solutions online where their students alone can see them and for a time period that they choose. A number of other computer applets provide experiences that support student learning in important ways. You will also find there example lesson plans, class videos, information about the course philosophy, evidence for its success, and many other resources. There is a preface for students appearing just before the first chapter of each unit that explains some important features of the text and assumptions behind the course. I recommend that everyone read it.



Comments about the Current Edition

My general goals for the current edition have been to correct errors, improve the presentation in some key areas, make the book more flexible, and especially to improve the quality and range of the homework problems, as well as significantly increase their number. Users of previous editions will note that I have split the old “Synthetic” homework problem category into “Modeling” and “Derivations” categories. “Modeling” problems now more s­ pecifically focus on the process of building physical models, making appropriate approximations, and binding together disparate concepts. ­ ­ “Derivation” problems focus more on supporting or extending derivations presented in the text. I thought it valuable to more clearly separate these categories. The “Basic Skills” category now includes a number of multipart problems specifically designed for use in the classroom to help students practice basic issues. The instructor’s manual discusses how to use such problems.

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Preface

I have also been more careful to give instructors more choice about what to cover, making it possible for instructors to omit chapters without loss of continuity. See the unit-specific part of this preface for more details. Users of previous editions will also note that I have dropped the menulike chapter location diagrams, as well as the glossaries and symbol lists, that appeared at the end of each volume. There was no evidence that these were actually helpful to students. Units C and N still instruct students very carefully on how to construct problem solutions that involve translating, modeling, solving, and checking, but examples and problem solutions for the remaining units have been written in a more flexible format that includes these elements implicitly but not so rigidly and explicitly. Students are rather guided in Unit N to start recognizing these elements in more generally formatted solutions, something that I think is an important skill. The only general notation change is that now I use │v​  ​W│    exclusively and W​ .  I still think it is very important universally for the magnitude of a vector v​ to have notation that clearly distinguishes vector magnitudes from other scaW)  notation is too cumbersome to use exclusively, and lars, but the old mag(​v​ mixing it with using just the simple letter has proved confusing. Unit C contains some specific instruction about the standard notation that most other texts use (as well as discussing its problems). Finally, at the request of many students, I now include short answers to selected homework problems at the end of each unit. This will make students happier without (I think) significantly impinging on professors’ freedom.

Specific Comments About Unit R Unit R is a relatively short unit that focuses on developing the theory of special relativity as a logical consequence of the principle of relativity. Typically, one spends little time in a traditional introductory physics course exploring relativity, and as a result, few students understand or appreciate the beauties it has to offer. The experience of those of us who have used this text is that if two to three weeks of class time are devoted to the study of relativity using the approach outlined in this unit, students at almost any level can develop a robust and satisfying understanding of the logic and meaning of relativity, and many will become genuinely excited about really being able to understand such a well-known but counterintuitive topic in physics (the intensity of this excitement actually surprised some of our early users). Special relativity is probably the best and most accessible example in all of physics for illustrating how carefully thinking through the consequences of an idea can uncover unexpected truths beyond the realm of daily experience. Therefore, studying relativity can provide students with a glimpse of both the process and the rewards of theoretical physics, as well as help them take an important first step into the world of contemporary physics, where reaching beyond the level of our daily experience requires an increasing reliance on logical reasoning and abstract models. This text therefore emphasizes the logical structure of relativity, clearly showing how well-known and bizarre relativistic effects such as length contraction and time dilation are the inevitable consequences of the principle of relativity. If students come away from this unit feeling that the universe not only is consistent with relativity, but indeed almost has to be, then this unit has been successful. I urge instructors to tailor their efforts toward this goal. This unit should follow a treatment of Newtonian mechanics that ­includes nonrelativistic kinematics, Newton’s second law, conservation of ­momentum and energy, and some study of reference frames. This book can

Preface

be used as a supplement to a traditional introductory text any time after these topics are covered. In a Six Ideas course, this unit should definitely ­follow units C and N. On the other hand, I think it is good to go against history and schedule unit R before unit E for several reasons. First, knowing some relativity can make certain aspects of electricity and magnetism simpler, and unit E takes some advantage of the relativistic perspective in general and Lorentz contraction and the cosmic speed limit in particular (see the preface to unit E for more details). Second, I think it is good for students to get a taste of some exciting contemporary physics between the many weeks of classical physics represented by units C, N, and E. This is especially true if this is the last unit discussed in the first semester: ending the first semester with unit R means that many students will leave the course excited and intrigued about physics and (perhaps) more eager to continue with the second semester. Unit Q uses relativity quite sparingly. The chapters on nuclear physics do draw on the idea that mass can be converted to energy and vice versa, but otherwise, references to relativity are mostly confined to problems. Unit T does not use relativity at all. This unit has been only lightly revised for this edition compared to some of the other units. The most significant change is that I have compressed the material in the second edition’s chapters R1–R3 into the third edition’s chapters R1 and R2. Chapter R3 of the second edition was particularly underweight, and in recent years I have been successfully able to teach the three old chapters in two class sessions. I hope that compressing the text this way will give you more flexibility in allotting time to other topics. In previous editions, I required that the Other Frame must move in the +x direction with respect to the Home Frame (meaning that β was always positive). This was a bit artificial, and the restriction proved tricky for some students. In this edition, I have relaxed this restriction, allowing β to become negative (though I almost always take it to be positive). I have also used hyperbola graph paper more extensively in this edition, and have spent a bit more time showing students how to use such paper. I have included some graph paper at the end of chapter R5 that students can xerox (one can also download graph paper from the Six Ideas website). Otherwise, I have made the notation for vector magnitudes consistent with the other revised units, corrected errors, added some new homework problems, and cleaned up the writing. As a result of the compression, the unit now has nine chapters. The shortest possible treatment of relativity using this book would be to omit chapters R4 and R7 through R9. This would yield a five-session introduction to basic relativistic kinematics (with no dynamics or E = mc2). Adding chapter R4 and/or R7 would provide a richer introduction to pure kinematics. The shortest introduction that includes dynamics would be to omit chapters R4, R6, and R7 and add a single class session devoted to sections R4.1 through R4.4 and section R7.4 (and possibly section R7.1). This would go over everything that is useful for units E and Q within seven class sessions. However, students find the material in chapters R4 and R6 some of the most interesting in the book, and chapter R6 is also where they really test their understanding of relativistic kinematics in the context of tough paradoxes. Therefore, I really recommend doing the whole unit if you have time. I hope that by shortening it by one chapter, I have made this a bit easier. If you want to assign appendix RB on the Doppler shift, you can go over it any time after section R4.2. It can displace some of the latter sections of chapter R4, displace some of the middle sections of chapter R7, or supplement chapter R6 (which involves fewer new ideas than the other chapters).

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Preface

Appreciation A project of this magnitude cannot be accomplished alone. A list including everyone who has offered important and greatly appreciated help with this project over the past 25 years would be much too long (and such lists appear in the previous editions), so here I will focus for the most part on people who have helped me with this particular edition. First, I would like to thank Tom Bernatowicz and his colleagues at Washington University (particularly Marty Israel and Mairin Hynes) who hosted me for a visit to Washingtion University where we discussed this edition in detail. Many of my decisions about what was most important in this edition grew out of that visit. Bruce Sherwood and Ruth Chabay always have good ideas to share, and I appreciate their generosity and wisdom. Benjamin Brown and his colleagues at Marquette University have offered some great suggestions as well, and have been working hard on the important task of adapting some Six Ideas problems for computer grading. I’d like to thank Michael Lange at McGraw-Hill for having faith in the Six Ideas project and starting the push for this edition, and Thomas Scaife for continuing that push. Eve Lipton and Jolynn Kilburg have been superb at guiding the project at the detail level. Many others at McGraw-Hill, including Melissa Leick, Ramya Thirumavalavan, Kala Ramachandran, David  Tietz, and Deanna Dausener were instrumental in proofreading and producing the printed text. I’d also like to thank my students in Physics 70 at Pomona College, Alma Zook, Jim Supplee, David Horner and his students at North Central College, and Frank Van Steenwijk and his students at the University of Groningen for helping me track down errors in the manuscript form and offering useful feedback. Finally a very special thanks to my wife Joyce, who sacrificed and supported me and loved me during this long and demanding project. Heartfelt thanks to all! Thomas A. Moore Claremont, California

SmartBook is the first and only adaptive reading experience designed to change the way students read and learn. It creates a personalized reading experience by highlighting the most impactful concepts a student needs to learn at that moment in time. As a student engages with SmartBook, the reading experience continuously adapts by highlighting content based on what the student knows and doesn’t know. This ensures that the focus is on the content he or she needs to learn, while simultaneously promoting long-term retention of material. Use SmartBook’s real-time reports to quickly identify the concepts that require more attention from individual s­ tudents–or the entire class. The end result? Students are more engaged with course content, can better prioritize their time, and come to class ready to ­participate.

Preface

Learn Without Limits Continually evolving, McGraw-Hill Connect® has been redesigned to provide the only true adaptive learning experience delivered within a simple and easy-to-navigate environment, placing students at the very center. • Performance Analytics – Now available for both instructors and students, easy-to-decipher data illuminates course performance. Students always know how they’re doing in class, while instructors can view student and section performance at-a-glance. • Mobile – Available  on tablets, students can now access assignments, quizzes, and results on-the-go, while instructors can assess student and section performance anytime, anywhere. • Personalized Learning – Squeezing the most out of study time, the adaptive engine in Connect creates a highly personalized learning path for each student by identifying areas of weakness, and surfacing learning resources to assist in the moment of need. This seamless integration of reading, practice, and assessment, ensures that the focus is on the most important content for that individual student at that specific time, while promoting long-term retention of the material.

xiii

Introduction for Students Introduction Welcome to Six Ideas That Shaped Physics! This text has a number of features that may be different from science texts you may have encountered previously. This section describes those features and how to use them effectively.

Why active learning is crucial

Why Is This Text Different?

Research into physics education consistently shows that people learn physics most effectively through activities where they practice applying physical reasoning and model-building skills in realistic situations. This is because physics is not a body of facts to absorb, but rather a set of thinking skills acquired through practice. You cannot learn such skills by listening to factual lectures any more than you can learn to play the piano by listening to concerts! This text, therefore, has been designed to support active learning both inside and outside the classroom. It does this by providing (1) resources for various kinds of learning activities, (2) features that encourage active reading, and (3) features that make it as easy as possible to use the text (as opposed to lectures) as the primary source of information, so that you can spend class time doing activities that will actually help you learn.



The Text as Primary Source

Features that help you use the text as the primary source of information

To serve the last goal, I have adopted a conversational style that I hope you will find easy to read, and have tried to be concise without being too terse. Certain text features help you keep track of the big picture. One of the key aspects of physics is that the concepts are organized hierarchically: some are more fundamental than others. This text is organized into six units, each of which explores the implications of a single deep idea that has shaped physics. Each unit’s front cover states this core idea as part of the unit’s title. A two-page chapter overview provides a compact summary of that chapter’s contents to give you the big picture before you get into the details and later when you review. Sidebars in the margins help clarify the purpose of sections of the main text at the subpage level and can help you quickly locate items later. I have highlighted technical terms in bold type (like this) when they first appear: their definitions usually appear nearby. A physics formula consists of both a mathematical equation and a conceptual frame that gives the equation physical meaning. The most important formulas in this book (typically, those that might be relevant outside the current chapter) appear in formula boxes, which state the equation, its purpose (which describes the formula’s meaning), a description of any limitations on the formula’s applicability, and (optionally) some other u ­ seful notes. Treat everything in a box as a unit to be remembered and used together.

What is active reading?



Active Reading

Just as passively listening to a lecture does not help you really learn what you need to know about physics, you will not learn what you need by simply

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scanning your eyes over the page. Active reading is a crucial study skill for all kinds of technical literature. An active reader stops to pose internal questions such as these: Does this make sense? Is this consistent with my experience? Do I see how I might be able to use this idea? This text provides two important tools to make this process easier. Use the wide margins to (1) record questions that arise as you read (so you can be sure to get them answered) and the answers you eventually receive, (2)  flag important passages, (3) fill in missing mathematical steps, and (4)  record insights. Writing in the margins will help keep you actively engaged as you read and supplement the sidebars when you review. Each chapter contains three or four in-text exercises, which prompt you to develop the habit of thinking as you read (and also give you a break!). These exercises sometimes prompt you to fill in a crucial mathematical detail but often test whether you can apply what you are reading to realistic situations. When you encounter such an exercise, stop and try to work it out. When you are done (or after about 5 minutes or so), look at the answers at the end of the chapter for some immediate feedback. Doing these exercises is one of the more important things you can do to become an active reader. SmartBook (TM) further supports active reading by continuously measuring what a student knows and presenting questions to help keep students engaged while acquiring new knowledge and reinforcing prior learning.



Features that support developing the habit of active reading

Class Activities and Homework

This book’s entire purpose is to give you the background you need to do the kinds of practice activities (both in class and as homework) that you need to genuinely learn the material. It is therefore ESSENTIAL that you read every assignment BEFORE you come to class. This is crucial in a course based on this text (and probably more so than in previous science classes you have taken). The homework problems at the end of each chapter provide for different kinds of practice experiences. Two-minute problems are short conceptual problems that provide practice in extracting the implications of what you have read. Basic Skills problems offer practice in straightforward applications of important formulas. Both can serve as the basis for classroom activities: the letters on the book’s back cover help you communicate the answer to a two-minute problem to your professor (simply point to the letter!). ­Modeling problems give you practice in constructing coherent mental models of physical situations, and usually require combining several formulas to get an answer. Derivation problems give you practice in mathematically extracting useful consequences of formulas. Rich-context problems are like modeling problems, but with elements that make them more like realistic questions that you might actually encounter in life or work. They are especially suitable for collaborative work. Advanced problems challenge advanced students with questions that involve more subtle reasoning and/or difficult math. Note that this text contains perhaps fewer examples than you would like. This is because the goal is to teach you to flexibly reason from basic principles, not slavishly copy examples. You may find this hard at first, but real life does not present its puzzles neatly wrapped up as textbook examples. With practice, you will find your power to deal successfully with realistic, practical problems will grow until you yourself are astonished at how what had seemed impossible is now easy. But it does take practice, so work hard and be hopeful!

Read the text BEFORE class!

Types of practice activities ­provided in the text

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R1

The Principle of Relativity Chapter Overview Introduction In units C and N, we have explored the Newtonian model of mechanics. In this unit, we will explore a different model, called the special theory of relativity, that better ­explains the behavior of objects, especially objects moving at close to the speed of light. This chapter lays the foundations for that exploration by describing the core idea of the theory and linking it to Newtonian mechanics.

Section R1.1:  Introduction to the Principle We can informally state this unit’s great idea, the principle of relativity, as ­follows: The laws of physics are the same inside a laboratory moving at a constant velocity as they are inside a laboratory at rest. The theory of special relativity simply spells out the logical consequences of this idea. This unit is divided into four subsections. The first (this chapter) discusses the principle itself. The second (chapters R2 through R4) explores the relativistic concept of time. The third (chapters R5 through R7) discusses how observers in different reference frames view a sequence of events. Finally, the fourth (chapters R8 and R9) examines the consequences for the laws of mechanics.

Section R1.2:  Events, Coordinates, and Reference Frames What exactly does the principle mean by a “laboratory”? The first step to understanding this better is to describe operationally how we measure a particle’s motion. An event is something that happens at a well-defined place and time. An event’s ­spacetime ­coordinates are a set of four numbers that locate the event in space and time. A ­particle’s motion is a series of events. A reference frame is a tool for assigning spacetime coordinates to events. We can visualize a reference frame as being a cubical lattice with a clock at every intersection. This ensures that there is a clock present at every event, but it also implies that we must synchronize the clocks somehow. An observer is a person who interprets results obtained in a reference frame to reconstruct the motions of particles. A real reference frame does not actually consist of a cubical lattice of clocks, but must be functionally equivalent.

Section R1.3:  Inertial Reference Frames An inertial reference frame is a frame in which an isolated object is always and everywhere observed to move at a constant velocity. We can check whether a frame is ­inertial by distributing first-law detectors around the frame to test for violations of Newton’s first law. A consequence of this definition is that two inertial frames in the same region of space must move at a constant velocity relative to each other. Conversely, if a given frame moves at a constant velocity relative to another inertial frame in the same region of space, the first must be inertial also.

2

Section R1.4:  The Final Principle of Relativity Note that in our original statement of the principle of relativity, the “laboratory moving at a constant velocity” and the “laboratory at rest” are both inertial frames. Moreover, the principle itself implies that there is no physical way to distinguish a frame in motion from one at rest: only the relative velocity between two inertial reference frames is measurable. Our final, polished statement of the principle therefore expresses the core issue without referring to “moving” or being at “rest”: The laws of physics are the same in all inertial reference frames.

Section R1.5:  Newtonian Relativity What does the phrase “the laws of physics are the same” mean? We can examine this issue in the context of Newtonian physics if we temporarily embrace Newton’s hypothesis about time, which is that time is universal and absolute and thus independent of reference frame. Consider two inertial frames that have constant relative velocity β​ ​W  and which are in standard orientation relative to each other; that is, the axes of both point in the same directions in space and the Other (primed) Frame moves along the x axis relative to the Home (unprimed) Frame. The concept of universal time implies that time measured in both frames is the same (t = t ′ ). This, together with some simple vector addition and a little bit of calculus, implies that  

W ​  rW  ​′ (t ′ ) =  rW​   ​ (t) − β​ ​ t  (R1.1)  

 

W W ′ (t ′ ) = v ​ W​  (t) − β​ ​ v​ ​   (R1.3)  

 

​  aW​ ′ (t ′ ) =  aW​ ​  (t) (R1.4)  

 

• Purpose:  These equations describe how to compute an object’s position  rW​  ​′, velocity v​ ​W ′ and acceleration  a​W​ ′ at any given time t ′ in the Other Frame, given W​ ,  and acceleration  a​W​ at the same time t the object’s position, position  r​W  ​, velocity v​ in the Home Frame, where β​ ​W  is the velocity of the Other Frame relative to the Home Frame. • Limitations:  These equations assume that t = t ′, which is not true unless both ​ W W  │Δ ​d​​W│   ​ ​restriction necessary for the derivation of the metric equation. Thus, neither equation really says anything useful about what a clock traveling faster than light would measure. In this section, we will see that there is a deeper problem with traveling faster than light: it violates causality. What do I mean by causality? In physics (and more broadly, in daily life), we know that certain events cause other events to happen (see figure R7.1). For example, even couch potatoes know that if you press the appropriate button on the remote control, the TV channel will change. Causally connected events must happen in a certain order in time: the event being caused must follow the event that causes it. For example, we would be deeply disturbed if the TV channel changed just before we pressed the remote control button! Consider two distinct events (call them P and Q) such that event P causes event Q, or more precisely, Q happens as a direct consequence of the reception of some kind of information that P has occurred. This information can be transmitted from P to Q in any number of ways: via some mechanical effect (such as the movement of an object or the propagation of a sound wave), via a light flash, via an electric signal, via a radio message, etc. Basically, the information can be carried by any object or effect that can move from place to place and is detectable. Let’s consider the TV remote control again as a specific example. Suppose you press a button on your TV remote control handset (event P). The information that the button has been pressed is sent to the TV set in some manner, and in response, the TV set changes channels (event Q). Keep this basic example in mind as we go through the argument that follows. Now let us pretend that the causal influence that connects event P to event Q can flow between them at a constant speed │v​  ​W ci│ faster than the speed of light as measured in your inertial frame, which we will call the Home

Ceiling

String

Firecracker Pivot

Mirror

Fuse Bell

Steel ball Boot

Ring

Bellows

Laser Dominoes

Pull ring to sound bell

Figure R7.1 Some causal connections.

Push button

R7.1 Causality and Relativity

129

t

t

t'

Light-flash worldline

Light-flash worldline

x' Q

Q P

x

P

Parallel to t' axis

Worldline of the causal influence

x

t'Q < 0

Figure R7.2

Figure R7.3

Suppose that events P and Q are connected by a hypothetical causal influence traveling with a speed │v ​  ​Wc  i│ faster than the speed of light (specifically, 5 times the speed of light for the sake of concreteness).

This two-observer spacetime diagram shows the same situation as in figure R7.2, but with added axes for an Other Frame moving with β = _​  25  ​with respect to the Home Frame. Note that in the Other Frame, event Q occurs before event P.

Frame. (Perhaps the TV manufacturer has found some way to convey a signal from the remote to the TV using “Z waves” that travel faster than light.) We will show that this leads to a logical absurdity. Choose event P to be the origin event in that frame, and choose the spatial x axis of the frame so that both events P and Q lie along it. (We can always do this: it is just a matter of choosing the origin and orientation of our reference frame. Choosing the frame to be oriented in this way is purely a matter of convenience.) Figure R7.2 shows a spacetime diagram (drawn by an observer in the Home Frame) of a pair of events P and Q fitting the description above. Note that if the causal influence flows from P to Q faster than the speed of light, its worldline on the diagram will have a slope 1/│v​  ​W ci│  1. │ │ Any Other Frame must travel with   β  Δt) as measured in the Home Frame. Reorient and reposition the axes of the Home Frame so the events in question both occur along the spatial x axis, with the later event ­located in the +x direction relative to the earlier event. (This can be done without loss of generality: we are always free to choose the orientation of our coordinate sysW​ │ tem to be whatever we find convenient.) Once this is done, │Δ d​    =   Δx in the Home Frame. Now, consider an Other Frame in standard orientation with respect to the Home Frame and traveling in the +x direction with x-velocity β with respect to the Home Frame. According to equation R5.11a, the time-­ coordinate ­difference between these events in the Other Frame is then

Δt ′ = γ ( Δt − βΔx )(R7.6)  

Light-flash worldline

B

x' ∆t

A ∆x

x

Figure R7.5 Given any pair of events A and B separated by a spacelike interval (Δx > Δt) in the Home Frame, we can find an Other Frame where the two events are simultaneous. The speed of this Other Frame simply must have the right value so that the diagram x′ axis can connect both points. Since this axis has slope β, this means β must be equal to Δt/Δx.

These events will be simultaneous in the Other Frame (that is, Δt ′ will be equal to zero) if and only if the relative speed of the frames is chosen to W​ │ be │ β│ = │Δt/Δx│ = │Δt│/│Δ d​    . This relative speed │ β│ will be less than W │ │ 1 since  Δ d​ ​    > Δt for our events by hypothesis. In short, given any pair of events that are separated by a spacelike interval in some inertial frame (which we are calling the Home Frame), it is possible to find an inertial Other Frame moving with positive x-velocity β < 1 with respect to the Home Frame in which observers will find the two events to be simultaneous (see figure R7.5). In short, if the spacetime interval between two events is spacelike, then  

1. We can find an inertial frame where these events occur at the same time. 2. The spacetime separation ∆σ is the distance between the events in that special frame. We can measure this with a ruler stretched between the events in that frame. 3. If observers in any other inertial frame use equation R7.4 to calculate ∆σ, they will get the same value as measured directly in the special frame. These statements are directly analogous to statements that can be made about events separated by a timelike spacetime interval. If the spacetime interval between two events is timelike, then 1. We can find an inertial frame in which these events occur at the same place (this is the frame of the inertial clock that is present at both events). 2. The time between the events in this special frame is ∆s. We can measure this with a clock present at both events and at rest in this frame. 3. If observers in any other inertial frame use the ordinary metric equation to calculate ∆s, they will get the same value as measured directly in the special frame. Thus, there is a fundamental symmetry between spacelike and timelike space­time intervals, a symmetry that arises because both reflect the same

How to measure a spacelike spacetime interval

Chapter R7

134

The Cosmic Speed Limit

underlying physical truth: we can describe the separation of any two events in space and time with a frame-independent quantity Δs2 (which we will call  the squared spacetime interval) analogous to the squared distance between two points in plane geometry. It is simply a peculiarity of the geometry of spacetime that the quantity in spacetime that corresponds to ordinary (unsquared) distance on the plane comes in three distinct flavors (the spacetime interval Δs if Δs2 > 0 for the events, the spacetime separation Δσ if Δs2  Δt) cannot influence P or be influenced by it. We say that these events (which inhabit the shaded region of figure R7.6) are causally unconnected to event P. With this in mind, we can relabel the regions in figure R7.6 as shown in figure R7.7. Because every observer agrees on the value of the spacetime interval between event P and any other event, every observer agrees as to which event belongs in which classification. The structure illustrated is thus an intrinsic, frame-independent characteristic of the geometry of spacetime. Now, the boundaries of the regions illustrated in figure R7.7 are lightflash worldlines. If we consider two spatial dimensions instead of one, an omnidirectional light flash is seen as an ever-expanding ring, like the ring of waves formed by the splash of a stone into a still pool of water. If we plot the growth of such a ring on a spacetime diagram, we get a cone. The boundaries between the three regions described are then two tip-to-tip cones, as shown in figure R7.8. Physicists call this boundary surface the light cone for the given event P. To summarize, the point of this section is that the spacetime interval classifications, which are basic, frame-independent features of the geometry of spacetime, have in fact a deeply physical significance: the sign of the squared spacetime interval between two events unambiguously describes whether these events can be causally connected or not. The light cone shown in figure  R7.8 effectively illustrates this geometric feature of spacetime.

Understanding the causal structure of spacetime

The light cone associated with an event

136

Chapter R7

R7.4

The Cosmic Speed Limit

The Einstein Velocity Transformation

In this section, we turn our attention to the relativistic generalization of the Galilean velocity transformation equations R1.3. Imagine a particle that is observed in the Other Frame to move along the spatial x ′ axis with a constant ­x-velocity ​v ′​x ​ ​  . The Other Frame, in turn, is moving with an x-velocity β in the +x direction with respect to the Home Frame. What is the particle’s x-­velocity vx as observed in the Home Frame? Figure R7.9 shows how to construct a two-observer spacetime diagram that we can use to answer this question. After drawing and calibrating both sets of coordinate axes, we simply draw the particle’s worldline so its slope in the Other Frame is 1/​v ′​x ​ ​  . We can then find its x-velocity vx in the Home Frame by taking the inverse slope of that line in the Home Frame, which we can do by picking an arbitrary “rise” (5 ns in figure R7.9), determining the worldline’s “run” for that rise, and then calculating the inverse of the rise/run = run/rise. In the case shown in figure R7.9, where β = ​ _35  ​and ​v ′​x ​​  = ​ _35  ​ , we find that the value of vx is about 0.86, and not _​ 35  ​ + _​ 35  ​ = _​ 65  ​that the Galilean velocity transformation equations would predict. Now let us see if we can derive an exact equation that (like the diagram) allows us to find vx in terms of ​v ′​x ​​  and β. Consider two infinitesimally separated events along the particle’s worldline (which we will assume is moving along the spatial x axis). Let the coordinate differences between these events as measured in the Home Frame be dt and dx. Let the coordinate differences between the same two events as measured in the Other Frame be dt ′ and dx ′. The particle’s x-velocity as it travels between these events is  

Transforming velocities with a two-observer diagram

How to derive the inverse Einstein velocity transformation equations

 

dx γ (β dt ′ + dx ′ ) vx ≡ ___ ​   ​ = ____________ ​    ​ dt γ (dt ′ + β dx ′ )  



 

   

 

(R7.7a)

   

where I have used the infinitesimal limit of the difference version of the inverse Lorentz transformation equations (equations R5.12). Dividing the right side top and bottom by dt ′ and using dx ′/dt ′ ≡ v​  ′​x ​ ​  , we get the following relativistically exact ­equation:  

 

 

β + ​v ′​x ​​  vx = _______ ​     ​   (R7.7b) 1 + β ​v x′​ ​​ 



In a similar fashion, you can derive the y and z component equations as well. The complete set of equations for vx, vy, and vz are _______



_______

2  β + ​v ′​x ​​  ​ 1  − β2   ​ ​v ′​y ​​   ​v ′​z ​​  ​ 1  − β    ​ ___________ vx = ________ ​      vy = ___________ ​      vz =       ​     ​    ​  ​ (R7.8) 1 + β​v ′​x ​​  1 + β​v ′​x ​​  1 + β​v ​′x ​​ 

• Purpose:  These equations describe how to compute an object’s velocity components vx, vy, and vz measured in the Home Frame from its velocity components ​v ​′x ​ ​  , ​v ′y​ ​,​  and ​v ′z​ ​​  measured in the Other Frame, where β is the x-velocity of the Other Frame relative to the Home Frame. • Limitations:  These equations assume that the two frames are inertial and that they are in standard orientation with respect to each other.

Exercise R7X.3 Use the method described above to derive the formula for vz.

R7.4 The Einstein Velocity Transformation

t (ns)

137

t' (ns) Particle worldline 5

5

x' (ns) 3

5

3

5

x (ns)

Figure R7.9 We can use a two-observer diagram to find a particle’s x-velocity in one frame, given its x-velocity in another and the two frames’ relative velocity. For example, if we know that the particle’s Other Frame x-velocity is v​  ​′ x​​  = _​ 35  ​ and that the Other Frame moves at β = _​ 35  ​relative to the Home Frame, then we can construct an appropriate twoobserver diagram for this β, draw the particle’s worldline with a slope of _​ 53  ​relative to the Other Frame axes, and read the slope of this line relative to the Home Frame axes (about 4.3/5 ≈ 0.86 in this case). If we knew the particle’s Home Frame x-velocity, we could just reverse the process.

As an example of the use of equation R7.8a, consider the particular problem illustrated by figure R7.9, where we have v​  ′​x ​​  = β = _​ 35 ​  . The final speed of the particle in the Home Frame (according to equation R7.8a) is

3/5 + 3/5 6/5 30 vx = _____________ ​       ​ = ______ ​    ​  = ___ ​   ​ ≈ 0.88 1 + (3/5)(3/5) 34/25 34

(R7.9)

This is close to the result we read from figure R7.9. We call equations R7.8 the inverse Einstein velocity transformation equations: they express algebraically what a two-observer diagram like ­figure R7.9 expresses graphically. Note that the result is different from what the Galilean velocity transformation predicts: solving equation R1.3a for vx yields

vx = β + ​v ′​x ​   ​  from the Galilean velocity transformation

An example application

These equations reduce to the Galilean equations at low speed . . .

(R7.10)

Note that when the velocities β and v​  ′​x ​​  are very small, the factor β ​v ′​x ​​  that appears in the denominator of equation R7.8 becomes very small compared to 1. In this limit, then, equation R7.8a reduces to the Galilean equation R7.10:

β + ​v ′​x ​​  β + ​v ′​x ​​  vx = _______ ​     ​  ≈ ______ ​   ​   = β + ​v ′​x ​   ​  in low-velocity limit 1 1 + β ​v x′​ ​​ 

(R7.11)

The same kind of argument applies to the other two component equations as well. The Galilean transformation equations are therefore reasonably accurate for everyday velocities, but only represent an approximation to the true velocity transformation law expressed by equations R7.8. Equations R7.8 never yield a Home Frame x-velocity that exceeds the speed of light, even if both β and v​  x​′ ​​  are 1 (their maximum possible value):

vx = ________ ​  1 + 1  ​  = __ ​  2 ​  = 1 1+1·1 2

(R7.12)

Moreover, equation R7.8a (unlike the Galilean equation R1.3a) is consistent with the idea that the speed of light is equal to 1 in all frames: if the x-­velocity of a light flash in the Other Frame is v​  ′​x ​​  = 1, its speed in the Home Frame will be

β+1 β+1 vx = ________ ​    ​  = _____ ​     ​= 1 1+β·1 1+β

independent of the value of β.

(R7.13)

. . . and are consistent with the cosmic speed limit and the invariant speed of light

138

Chapter R7

The Cosmic Speed Limit

In short, the inverse Einstein velocity transformation equations (equations R7.8) provide the answer to the question that we raised in chapter R1 about how the Galilean velocity transformation equations should be modified to be consistent with the principle of relativity. Equations R7.8 reduce to the Galilean transformation equations at low velocities (where the Galilean transformation is known by experiment to be very accurate), but at relativistic velocities they are consistent with both the assertion that nothing can be measured to go faster than light and the assertion that light itself has the same speed in all inertial frames. Equations R7.8 convert Other Frame velocity components v​  ′x​ ​ ​  , ​v ′y​ ​ ​  , ​v ′z​ ​​  to Home Frame components vx , vy , vz . The (direct) Einstein velocity transformation equations transform the velocity components the other way. ______

______





2

2

vz​ 1  − β  ​  vx − β vy ​ 1  − β  ​  ​ v ​′ ​​  = _______ ​     ​  ​   v ′​ ​​  = __________ ​   ​       ​v ′​ ​​  = __________ ​   ​     x

1 − βvx

y

1 − βvx

z

1 − βvx

(R7.14)

• Purpose:  These equations describe how to compute an object’s velocity components ​v ​′x ​  ​  , ​v ′y​ ​  ​  , and ​v ​′z ​​  measured in the Other Frame from its velocity components vx , vy , and vz measured in the Home Frame, where β is the x-velocity of the Other Frame relative to the Home Frame. • Limitations:  These equations assume that the two frames are inertial and they are in standard orientation with respect to each other. These equations can be derived by solving equations R7.8 for the Other Frame components or by using the direct Lorentz transforma­tion equations R5.11 as we used equations R5.12 in the derivation of equations R7.8. But one needn’t do all this work: simply note that the only difference between the Home Frame and the Other Frame is the sign of β. If you compare equations R7.14 with equations R7.8, you will see they are the same, except that I have changed β to -β. Again, these equations themselves are laws of physics that satisfy the principle of relativity.

Two-Minute Problems

139

TWO-MINUTE PROBLEMS R7T.1

Consider the events shown in the figure below: t (s) 5 B E

D C

A

x (s)

For each of the ten event pairs in this spacetime diagram, classify the spacetime interval between them. A. The interval is timelike. B. The interval is lightlike. C. The interval is spacelike. R7T.2 Which pairs of events shown in the spacetime diagram for problem R7T.1 could in principle be causally connected? Which could not? (For each of the ten pairs, answer T if they could be causally connected, and F if not.) R7T.3 Suppose that event A is the origin event in the Home Frame and that event B occurs at t = 1 ns and x = 10 ns in that frame. What would be the minimum x-velocity β that an Other Frame would have to have relative to the Home Frame if B occurred first in that frame? A. 10 B. 1 C. 0.60 D. 0.40 E. 0.10 F. B can’t occur first. R7T.4 Two blinking warning lights are 3000 m apart along a railroad track. Suppose that in the ground frame the west light blinks (event W  ) 5 μs before the east light blinks (event E). Now imagine that both lights are observed by a passenger on the train who passes the west light just as it blinks. It is possible for the train to be moving fast enough that the two blinks are observed (not necessarily seen) by the ­passenger to be simultaneous. T or F? R7T.5 Two blinking warning lights are 1000 ns apart along a railroad track. A train moves from west to east at a speed of 0.5. In the train frame, an observer registers flashes from each light, and after correcting for light travel time, concludes that the east light flashed 200 ns before the west light flashed. Which light flashed first in the ground frame? (Hint: Draw a spacetime diagram.) A. The east light flashes before the west light. B. The west light flashes before the east light. C. The lights flash simultaneously. D. One does not have enough information to say.

R7T.6 Suppose an explosion occurs at x = 0 in the Home Frame (event A). Light from the explosion is detected by a detector at position x = –100 ns (event B) and by a detector at position x = +50 ns (event C). Events B and C are causally connected. T or F? R7T.7 A laser beam is emitted on earth (event A), bounces off a mirror placed on the moon by Apollo astronauts (event B), and then returns to a detector on earth (event C). The detector is 12 ns of distance from the laser. (a) The spacetime interval between events A and B is: A. Timelike B. Lightlike C. Spacelike (b) The spacetime interval between events A and C is: A. Timelike B. Lightlike C. Spacelike (c) Events A and C are definitely causally connected. T or F? R7T.8 If the spacetime interval between two events is timelike, then the temporal order of the two events is the same in every inertial reference frame. T or F? R7T.9 If the spacetime interval between two events A and B is spacelike and event A occurs before event B in some Home Frame, then it is always possible to find an Other Frame where the events occur in the other order. T or F? R7T.10 An object moves with speed vx = +0.9 in the Home Frame. In an Other Frame moving at β = 0.60 relative to the Home Frame, the object’s x-velocity v​  ′ ​x​​  is A. ​ v x′ ​ ​​  = 1.5 B. 1 < ​v x′ ​ ​​  < 1.5 C. 0.9 < ​v x′ ​ ​​  < 1 D. 0.3 < ​v ′ ​x​​  < 0.9 E. ​v x′ ​ ​​  < 0.3 F. ​v x′ ​ ​​  < 0 R7T.11 Harry Potter points his wand at the sun and cries “Spotificus!” Within 8 minutes, observers all over the earth see a large string of sunspots form on the sun. Did Harry create the sunspots? A. Yes. Harry is a very powerful wizard. B. Yes, if magical effects travel at the speed of light. C. No. This is not physically possible, even with magic. R7T.12 An unidentified spaceship cruises past Outpost 11, a space station floating in deep space, at a speed of ​ _35 ​  , without acknowledging requests for identification. Ten seconds after the ship passes (according to Outpost 11 clocks), Outpost 11 sends a message to the departing ship, warning that it will commence firing if the ship does not respond. Fifteen seconds after that, Outpost 11 sensors indicate that the ship has raised its shields. This is clearly a response to the warning. T or F? (Hint: Draw a spacetime diagram.)

140

Chapter R7

The Cosmic Speed Limit

HOMEWORK PROBLEMS Basic Skills R7B.1 At 11:00:00 a.m. a boiler explodes in the basement of the Museum of Modern Art in New York City (call this event A). At 11:00:00.0003 a.m., a similar boiler explodes (call this event B) in the basement of a soup factory in Camden, New Jersey, a distance of 150 km from event A. (a) Why is it impossible for the first event to have caused the second event? (b) An alien spaceship cruising in the direction of Camden from New York measures the Camden event to occur at the same time as the New York event. What is the approximate speed of the spaceship relative to earth? R7B.2 Two balls are simultaneously ejected (event A) from the point x = 0 in some inertial frame. One rolls in the +x direction with speed 0.80 and eventually hits a wall at x  =  8.0 ns (event B). The other rolls in the -x direction with speed 0.40, eventually hitting a wall at x = -8.0 ns (event C). Is the spacetime ­interval between B  and  C spacelike or timelike? Could these events be ­causally connected? R7B.3 A meteor strikes the moon (event A), causing a large and vivid explosion. Exactly 0.47 s later (as measured in an inertial reference frame attached to the earth), a radio telescope receiving signals from the moon goes on the fritz. Could these events be causally related? Explain. R7B.4 Suppose event B happens 3.0 ns after and 5.0 ns east of event A in the Home Frame. In the Other Frame, the events happen at the same time. (a) How fast is the Other Frame moving eastward with respect to the Home Frame if this is true? (b) What is the distance between the events in that frame? R7B.5 Do problem R7T.4. Hand in your spacetime diagram and explain your reasoning. R7B.6 Do problem R7T.12. Hand in your spacetime diagram and explain your reasoning. R7B.7 An object moves with velocity ​v ′ ​x​​  = ​ _25 ​  in an inertial frame attached to a train, which in turn moves with x-velocity β = ​ _45 ​  in the +x direction with respect to the ground. What is the object’s velocity vx with r­ espect to the ground? Evaluate this by reading the velocity from a carefully constructed two-observer spacetime diagram. Check your answer by using the appropriate Einstein velocity transformation equation. R7B.8 An object moves at a speed of 0.80 in the +x direction, as measured in the Home Frame. What is its x-­velocity in an Other Frame that is moving at a speed of 0.60 in the +x direction, also as measured in the Home Frame?

R7B.9 Rocket A travels to the right and rocket B to the 3 _4  , respectively, relative to the left at speeds of _​   ​  and ​   ​  5 5 earth. What is the velocity of A measured by o ­ bservers in rocket B? Answer by reading the velocity from a carefully constructed two-observer diagram. Check your answer, using an appropriate Einstein velocity transformation. R7B.10 Two trains approach each other from opposite directions along a linear stretch of track. Each has a speed of ​ _34  ​relative to the ground. What is the speed of one train relative to the other? Answer this question by using an appropriate Einstein velocity transformation equation. As part of your solution, explain carefully which object you are taking to be the Home Frame, which object you are taking to be the Other Frame, and which is the object whose speed you are measuring in both frames. R7B.11 Two cars travel in the same direction on the freeway. Car A travels at a speed of 0.90, while car B can only muster a speed of 0.60. What is the relative speed of the cars? As part of your solution, explain carefully which object you are taking to be the Home Frame, which object you are taking to be the Other Frame, and which is the object whose speed you are measuring in both frames.

Modeling R7M.1 Suppose we set up a laser that is spinning at a speed of 100 rotations per second around an axis perpendicular to the beam that the laser creates. (a) How far away must you place a screen so the spot of light the laser beam produces on the screen sweeps along it at a speed faster than that of light? (b) Would a spot speed greater than that of light violate the cosmic speed limit? (Hint: If you stand at one edge of the screen and a friend stands at the other, can you send a message to your friend using the sweeping spot? If so, how? If not, why not?)* R7M.2 You are on a jury for a t­ errorism trial. The facts of the case are these. On June 12, 2047, at 2:25:06 p.m. Greenwich Mean Time (GMT), the earth–Mars shuttle Ares exploded as it was being refueled in low earth orbit. (Fortunately, no passengers were aboard, and the refueling was handled by robots.) At 2:27:18 p.m., police record a holoscene (a 3D holographic movie) of a raid on a hotel room on Mars conducted on the basis of an unrelated anonymous tip. In the holo, the defendant is shown with a radio control transmitter in hand. Forensic experts have testi­fied  that the Ares was blown up by remote control, and that the reconstructed receiver was consistent with the transmitter in the defendant’s possession. Just before the explosion, a *Adapted from E. F. Taylor and J. A. Wheeler, Spacetime Physics, San Francisco: Freeman, 1966, p. 62.

Homework Problems

caller predicted the blast and took responsibility on behalf of the Arean Liberation Army; the defendant has links to that organization. Cell-net records show that the ­defendant spoke with someone near the earth less than 10 min before. Hall monitors in the hotel showed that the defendant entered the hotel room at 2:23:12 p.m. and was not carrying the transmitter at that time. The defendant has taken the Fifth Amendment and has offered no defense other than a plea of not guilty. A fragment of the trial transcript f­ ollows. Prosecutor: Is the time shown floating at the top of the holoscene of the raid from a clock internal to the camera? Police witness: No, I am told that the indicated time is computed from signals originating from the master clock on earth that is part of the Solar System Positioning System (SSPS) that defines a solar-system-wide inertial frame fixed on the sun. According to the manual (pulls out the manual), “The signal from the earth master clock is suitably corrected in the camera for the motions of the earth and Mars and the light travel time from the earth, so that the time displayed is exactly as if it were from the clock at the camera’s location, at rest in the solar system frame, and synchronized with the earth-based master clock.” We do this deliberately so as to be able to correlate events on a solar-system-wide basis.

141

10 times the speed of light, relative to earth. After you get to the starbase, you use a similar accelerator to launch you back toward earth at 10 times the speed of light, relative to the starbase. Using a carefully constructed (full-page) two-observer diagram, show that in such a case you will return to the earth before you left. (This is yet another way to illustrate the absurdity of faster-than-light travel.) R7M.4 A flash of laser light is emitted by the earth (event A) and hits a mirror on the moon (event B). The reflected flash returns to earth, where it is absorbed (event C). (a) Is the spacetime interval between events A and B spacelike, lightlike, or timelike? (b) What about the spacetime interval ­between B and C? (c) What about the interval between events A and C? Support your ­answers by describing your reasoning. R7M.5 A solar flare (see the picture of an actual flare below) bursts through the sun’s surface at 12:05 p.m. GMT, as measured by an observer in an inertial frame attached to the sun. At 12:11 p.m., as measured by the same observer, the Macdonald family’s home computer fries a circuit board. Could these events be causally connected? Explain.

Prosecutor: There is no chance that this time (freezes holoscene), which shows the police yanking the device from the defendant here at exactly 2:27:20 p.m. GMT, is in error? Police witness: No, the camera was checked two days previously as part of a normal maintenance program. Prosecutor: We have been told that the Ares blew up at exactly 2:25:06 p.m. GMT. Was that time determined using the SSPS also? Police witness: Yes. Prosecutor: You have testified that the hall monitors show the defendant entering the room at exactly 2:23:12 p.m. This time was also determined using the SSPS? Police witness: Yes. Prosecutor: So the defendant was alone in the room at the time that the Ares exploded? Police witness: Yes, in the SSPS frame. Prosecutor: So this holoscene shows you capturing the defendant red-handed just after the destruction of the Ares, with the incriminating transmitter still in hand . . . Defense: Objection, your Honor! Judge, sighing: Sustained. Guilty or not guilty? Write a paragraph justifying your reasoning very carefully. R7M.3 Starbase Alpha coasts through deep space at a velocity of 0.60 in the +x direction with respect to earth. Let the event of the starbase traveling by the earth define the origin event in both frames. Suppose that at t = 8.0 h a giant accelerator on earth launches you toward the starbase at

R7M.6 The first stage of a multistage rocket boosts the rocket to a speed of 0.1 relative to the ground before being jettisoned. The next stage boosts the rocket to a speed of 0.1 relative to the final speed of the first stage, and so on. How many stages does it take to boost the payload to a speed in excess of 0.95? (Credit: NASA)

Chapter R7

142

The Cosmic Speed Limit

R7M.7 Suppose that in the Home Frame, two particles of equal mass m are observed to move along the x axis with equal and opposite speeds │v​  ​W│    = ​ _35 ​  . The particles collide and stick together, becoming one big particle which remains at rest in the Home Frame. Now imagine observing the same situation from the vantage point of an Other Frame that moves in the +x direction with an x-velocity of β = _​ 35  ​with respect to the Home Frame. (a) Find the velocities of all the particles as observed in the Other Frame, using the appropriate Einstein velocity transformation equations. (b) We have defined the momentum of a particle with W​  to be p​ mass m and velocity v​ ​W = m v​ ​W  . Is the system’s total momentum conserved in the Home Frame? (c) Is the system’s total momentum conserved in the Other Frame? Is this a problem? (We’ll talk more about momentum in chapter R8.) R7M.8 Suppose that in a particle physics experiment in the Home Frame, a particle of mass m0 moving at a velocity v​ ​W0  = _​ 35  ​in the +x direction suddenly decays into a particle of mass m1 moving at a speed of v​ ​W 1 = ​ _45  ​in the +x direction and a particle of m2 at rest. (a) We have defined the momentum of a particle with W​  to be p​ mass m and velocity v​ ​W = m v​ ​W . Assuming that the system’s total momentum (defined this way) is conserved in this decay process in the Home Frame, what must the masses m1 and m2 be? (b) Now let’s look at this decay process in an Other Frame moving along with the initial particle. According to the Einstein velocity transformation, what are the final x-velocities of the decay products in this frame? (c) Is the system’s total momentum conserved in this frame? Is this a problem? (We’ll talk more about momentum in chapter R8.) R7M.9 A train travels in the +x direction with an x-velocity of ​ _45  ​relative to the ground. At a certain time, two balls are ejected, one traveling in the +x direction with an x-velocity of _​ 35  ​relative to the train, and the other traveling in the −x direction with an x-velocity of − ​ _25  ​relative to the train. (a) What are the balls’ x-velocities relative to the ground? (b) What is the x-velocity of the first ball relative to the second? R7M.10 A particle moves with a speed of _​ 45 ​ in a direction 60° away from the spatial +x ′ axis toward the spatial y ′ axis, as measured in a frame (the Other Frame) that is moving in the +x direction with an x-velocity of β = _​ 12  ​relative to the Home Frame. What are the magnitude and direction of the particle’s velocity in the Home Frame?  

Derivation R7D.1 Derive the first of equations R7.14 by solving equation R7.7b for v​  ′ ​x​ ​  . R7D.2 Show that when │ β│  1 in that frame. Show, using a two-­observer spacetime diagram, that it is possible to find an Other Frame moving in the +x direction at an x-velocity β < 1 with respect to the Home Frame in which the object’s worldline lies along the diagram x ′ axis; and │v​ find the value of β (in terms of   ​W│   ) that makes this happen. Why is it absurd for the worldline of any object to coincide with the diagram x ′ axis?  

 

R7D.4 Show, using the Einstein velocity transformation equations R7.8, that a particle traveling in any arbitrary direction at the speed of light will be measured to have the speed of light in all other inertial frames.

Rich-Context R7R.1 You are the captain of a spaceship that is moving through an asteroid belt on impulse power at a speed of _​ 45  ​relative to the asteroids. Suddenly you see an asteroid dead ahead a distance of only 24 s away, according to sensor measurements in your ship’s reference frame. You immediately shoot off a missile, which travels forward at a speed of _​ 45  ​relative to your ship. The missile hits the asteroid and detonates, pulverizing the asteroid into gravel. However, you learned in Starfleet Academy that it is not safe to pass through such a debris field (even with shields on full) sooner than 8 s (measured in the asteroid frame) after the detonation. Are you safe?

 

R7M.11 A train travels in the +x direction with a speed of ​ _45  ​relative to the ground. At a certain time, two balls are ejected so that they travel with a speed of ​ _35  ​(as measured in the train frame) in opposite directions perpendicular to the train’s direction of motion. (a) What are the balls’ speeds relative to the ground? (b) What is the angle that the path of each ball makes with the x axis in the ground frame?

R7R.2 A spaceship travels at a speed of 0.90 along a straight-line path that passes 300 km from a small asteroid. Exactly 1.0 ms (in the asteroid frame) in time before reaching the point of closest approach, the ship fires a photon torpedo which travels at the speed of light. This torpedo is fired perpendicular to the ship’s direction of travel (as measured in the ship’s frame) on the side closest to the asteroid. Will this torpedo hit the asteroid? If not, does it pass the asteroid on the near side or far side (relative to the ship)? Carefully explain your reasoning. R7R.3 Your school gets a letter from one Kent C. M.  Tugadett, a wealthy alumnus who will give a large gift to the

Answers to Exercises

143

Physics Department if someone can successfully explain to him how to resolve the following paradox, which has bothered him since his college physics class. “A train is moving along a straight railroad track at a speed │v​  ​W│    close to the speed of light. Since moving clocks run slow, the train’s clocks all run slow compared to ground-frame clocks. Now, a runner runs inside the train at the same speed │v​  ​W│    relative to the train but in the opposite direction. Since the runner is moving relative to the train, the runner’s watch must run more slowly than the train clocks and so doubly slowly compared to clocks on the ground. But even the Einstein velocity transformation (check this) says that the runner will be at rest relative to the ground. So the runner’s watch must run at the same rate as the ground clocks. How can this be, since the runner’s watch is slower than the train clocks which are slower than the ground clocks?” It falls to you to answer this question. Carefully and diplomatically explain to Mr. Tugadett how to resolve the apparent paradox, and so earn the gift for your school.* (Hint: This is yet another situation where the “moving clocks run slow” idea is misleading.)

Advanced R7A.1 Imagine that in its own reference frame, an object emits light uniformly in all directions. Suppose this object moves in the +x direction with respect to the Home Frame at an x-velocity of β. (a) Show that the portion of the light that is emitted in the forward hemisphere in the object’s own frame is *Adapted from a problem in Taylor and Wheeler, Spacetime Physics, 2nd edition, Freeman, 1992.

observed in the Home Frame to be concentrated in a cone that makes an angle of ϕ = sin-1 __ ​  1  ​(R7.15) γ



with respect to the x axis. (b) Show that if β = 0.99, the angle within which this portion (which amounts to one-half of the object’s light) is concentrated is only 8.1°. (This forward concentration of the radiation emitted by a moving object is called the headlight effect.) R7A.2 Consider a very long pair of scissors. If you close the scissor blades fast enough, you might imagine that you could cause the intersection of the scissor blades (that is, the point where they cut the paper) to travel from the near end of the scissors to the far end at a speed faster than that of light, without causing any material part of the scissors to exceed the speed of light. Argue that (1) this intersection can indeed travel faster than the speed of light in principle, but that (2) if the scissors blades are originally open and at rest and you decide to send a message to a person at the other end of your scissors by suddenly closing them, you will find that the intersection (and thus the message) cannot travel faster than the speed of light. (Hint: The information that the handles have begun to close must travel through the metal from the handles to the blades and then down the blades to cause the intersection to move forward. What carries this information?)†

†See M. A. Rothman, “Things that Go Faster than Light,” Sci. Am., vol. 203, p. 142, July 1960.

ANSWERS TO EXERCISES is spacelike. The interval between B and C is spacelike. R7X.1 Let us take event P to be the origin event in the The interval between D and B is spacelike, but the interval Home Frame. Now, consider figure R7.3. For event  Q to between D and C is timelike. occur below the x ′ axis in the Other Frame, the x ′ axis must have a slope greater than that of the worldline of the causal influence connecting events P and Q. If the causal R7X.3 According to the inverse Lorentz transformation equations, we have influence moves at a speed of 3, then its slope on a two-­ observer diagram like figure R7.3 will be ​ _13  ​ . The slope of dz dz ′/dt ′ dz ′   ​ = ​ ________________ the Other Frame x ′ axis must be therefore greater than ​ _13  ​  . vz = ___ ​   ​ = ____________ ​          ​ dt γ (dt ′ + β dx ′ ) γ [1 + β (dx ′/dt ′ )] Since the slope of this axis is equal to β, the speed at which ______ the Other Frame moves with respect to the Home Frame, 2 ​v ′ ​z​​   ​ 1 − β    ​  ​v ′ ​z​​  1 _ __________ __________ the speed of the Other Frame must be greater than ​ 3  ​  . = ​     ​  = ​   ​      (R7.16) γ (1 + β ​v ′ x​ ​)​  1 + β ​v ′ x​ ​​   

 

 

 

 

 

 

 

R7X.2 The spacetime interval between A and B is timelike, between A and C is lightlike, and between A and D

 

 

   

R8

F  our-Momentum Chapter Overview Introduction Our goal in this final subdivision of the unit is to make the laws of conservation of momentum and energy compatible with the principle of relativity. This chapter introduces a redefinition of the concept of momentum (which turns out to include energy as an integral part), and chapter R9 explores the implications of this idea.

Section R8.1:  A Plan of Action In this chapter, we will see that

W is not consistent 1. Conservation of a system’s total Newtonian momentum  ​p​ W  ≡ m  ​v​ with the principle of relativity. (The slash notation indicates an outmoded definition of momentum.) 2. Four-momentum is a natural relativistic redefinition of momentum. 3. Conservation of four-momentum is consistent with the principle of relativity. 4. The conserved fourth component of four-momentum is energy.

Section R8.2: Newtonian Momentum Isn’t Conserved Consider a collision that conserves Newtonian momentum in the Home Frame. If we use the Einstein velocity transformation equations to calculate the velocities of the colliding objects in another inertial frame, we find that Newtonian momentum is not conserved in the second frame. Conservation of Newtonian momentum is therefore incompatible with the principle of relativity.

Section R8.3:  The Four-Momentum Vector An object’s four-momentum p is a four-­dimensional vector quantity (a four-vector) whose four components in a given inertial reference frame are dt dx dy dz [pt , px , py , pz] =​ [ m ​ ___  ​  ,  m ​ ___  ​ ,  m ​ ___  ​ ,  m ​ ___  ​  ] ​(R8.8) dτ dτ dτ dτ • Purpose:  This equation defines (in a given inertial reference frame) the components [pt , px , py , pz] of a particle’s four-momentum p in terms of its mass m and the rate of change of its spacetime coordinates [t, x, y, z] (in that frame) with respect to the particle’s own proper time τ. • Limitations:  This equation is a definition and so has no limitations. • Notes:  The four-momentum treats time and space coordinates evenhandedly and points tangent to the particle’s worldline in spacetime. Because dτ = (1 − │v​  ​W│   2)1/2 dt, the spatial components of p are indistinguishable from the components of Newtonian momentum p​ ​W when the particle’s speed │v​  ​W│     s is typically an excellent approximation.  

 

Exercise E2X.1 If we were to flip the dipole in figure E2.2 so that its positive charge is closer to point P, would the field vector at P point toward or away from the dipole?

Calculating a dipole’s field at points along its axis

22

Chapter E2

Charge Distributions

The dipole moment

It turns out (see problems E2D.1 and E2A.1) that the magnitude of a dipole’s field at any sufficiently distant point P is proportional to qs/r 3. This means that a dipole with charge magnitude q = q0 and separation s = s0 has the same distant field as that of one with (say) q = _​ 12  ​ q0 but s = 2s0. Since only a dipole’s orientation in space and the value of qs matter for determining its distant field, we can describe a dipole using a dipole moment vector p​ ​W e with magnitude │p​  ​We │ = qs and a direction pointing along the dipole’s axis in the direction of the positive particle relative to the negative particle. In one neat package, p​ ​We  characterizes all that is needed to calculate a dipole’s field at distant points. Using this vector, we can rewrite equation E2.2 as follows:  

We  1 2 ​p​ W E​ ​  (at a point P along a dipole’s axis) = _____ ​     ​ ______ ​     ​  4πε 0 │ rW​  P ​ D│ 3

(E2.3a)

the positive particle’s direction             ˆ​  e =​       ​   ​ where │p​  W​  e│ = qs and p​ ​ [ relative to the negative particle ]

(E2.3b)

 

• Purpose:  This equation specifies the magnitude of the electric field E​ ​W at │ │ a distant point P along a dipole’s axis a distance   rW​  P ​ D  from the dipole’s center, where q is the charge of the dipole’s positive particle and s is the W​ e  is the dipole’s dipole moment. particles’ separation. The vector p​ • Limitation:  This is an approximation valid as long as │ rW​  P ​ D│ >> s.

Check that this gives the right result for both possible dipole orientations along its axis! As we will shortly see, in typical applications we don’t need to know the dipole’s field at points other than along its axis.

E2.3

Figure E2.3 A charged balloon attracting and bending a stream of electrically neutral water(!). (Credit: © McGraw-Hill Education/Mark ­Dierker, photographer)

Electrical Polarization

Understanding dipoles can help us resolve a mystery associated with electrostatic behavior you may have experienced. If you charge a comb, it can attract small bits of paper that are electrically neutral. Similarly, a charged balloon will stick to an uncharged wall or ceiling. A charged balloon can even attract a stream of flowing water (see figure E2.3). How can a charged object attract (but never repel) something that is electrically neutral? An atom’s net charge is normally zero, because the charge of its positive nucleus exactly cancels the negative charge of its electron cloud. Also, that electron cloud’s center usually coincides with the nucleus, as shown in figure E2.4a. Now, we will see shortly that the exterior field of a spherical electron cloud is the same as that of a particle with the same charge located at the cloud’s center. Therefore, the whole atom’s external field is the same as that of a positive and negative particle with zero separation. Since the atom not only has zero charge but zero dipole moment, it has no outside field. Now imagine placing an atom in an external electric field (created by some distant charged particle) whose value is E​ ​W ext at the atom’s location. W  means that the field will pull the positive nucleus in The equation F​ ​W e = q ​E​ W the direction of E​ ​  ext and push the electron cloud in the opposite direction (figure E2.4b), separating the nucleus and the cloud’s center. Of course, the nucleus also attracts the cloud’s center, so before the separation becomes too large, the internal attraction between the cloud and nucleus will balance the external forces trying to push them apart. But in the new equilibrium, the

E2.3 Electrical Polarization

23

Electron cloud center (×)

Electron cloud center (×) Eext

Fe (nucleus) Nucleus (•)

(a)

An atom’s negative electron cloud is normally centered on its positive nucleus.

pe

Eext + Fe (cloud)

Equivalent dipole

Nucleus (•)

(b)

–

(c) This makes the atom behave as an electric dipole oriented with its dipole moment parallel to Eext.

The field of an external charged particle (say, a negative particle located to the left pulls the nucleus in the field direction and pushes the cloud in the opposite direction)

Figure E2.4 (a) An atom far from any charged object. (b) The same atom immersed in an external electric field. (c) Such an atom behaves as if it were an electric dipole.

nucleus and cloud center are separated, and therefore behave as a dipole with a nonzero dipole moment (figure E2.4c). We call this an induced dipole, and the general phenomenon the polarization of matter. Now consider how this induced dipole interacts with the charged particle that creates the external field (see figure E2.5). The external particle is located on the induced dipole’s axis, so equation E2.3 applies. That equation W specifies that the field E​ ​ d  p that the dipole creates at the charged particle’s location has the same direction as the dipole’s dipole moment vector p​ ​We , and thus points to the left in the figure. The electrostatic force on the negative external particle thus points to the right, toward the induced dipole. T ­ herefore, the dipole attracts the external particle. The external charge also attracts the dipole (as required by Newton’s third law) because the external charge’s W field E​ ​ e xt is a bit stronger at the closer positive particle’s location than it is at the farther negative particle’s location, and so the former feels a bit stronger pull to the left than the latter feels pushed to the right.

Polarization explains how a charged object can attract something neutral

Exercise E2X.2 What if the external particle in this scenario were positively charged? Argue that it will still be attracted to the induced dipole. (Hint: What will be the direction of the induced dipole’s moment p​ ​We  in this case?) We conclude that, in the presence of an external charged particle, atoms in normal matter are electrically polarized in such a way so that they attract that particle and vice versa. “Polarized” is not the same as “charged”: the atoms all remain electrically neutral. But the external particle’s field gives them dipole moments that enable them to interact with that particle. This is how a charged balloon can attract an electrically neutral wall. Figure E2.5

r

Edp –

Fe

External charged particle

Dipole axis

pe Eext

+

–

Induced dipole

The atomic dipole induced by an external charged particle will create a field that attracts that particle.

24

Chapter E2

Charge Distributions

How this attractive force varies with separation

Making this quantitative teaches us something interesting. The degree W to which an atom is polarized by an external field E​ ​  is usually proportional to W that field: p​ ​We  = α ​E​,  where α (the Greek letter “alpha”) is a constant specifying the atom’s polarizability. If a charged particle is creating this external field W E​ ​ ,  then │E​  W ​│    = │Q│/(4πε 0r 2), where Q is the external particle’s charge and r is its separation from the atom. Equation E2.3 tells us that the induced dipole W We /(4πε 0 r 3) at the atom’s location. The dipole’s field thus creates a field E​ ​ d  p = 2 ​p​ causes the external charged particle to feel a force of magnitude  

 

 

 

│ Q│  2α│E​  W ​│   2 α Q2 2α ______ W dp│ = │Q│ ​ ______    │F​  ​W e│ = │Q​E​  ​  = │Q│ ______ ​   3 ​  ​     ​  = ________ ​    ​   3 2 4πε 0r 4πε 0r 4πε 0r (4πε 0)2 r 5  

Why water is easily polarized

H O

+

–

H

Figure E2.6 Because oxygen atoms tend to grab electrons more strongly than hydrogen atoms do, a water molecule has a permanent natural dipole.

 

 

 

 

 

 

(E2.4)

 

The force between a charged particle and neutral matter therefore depends on the inverse 5th power of their separation! Forces due to polarization thus increase extraordinarily rapidly with decreasing separation. A typical atom’s polarizability α is on the order of 10-40 C2m/N. It turns out that liquid water has an unusually high value of α compared to most substances, making the effect shown in figure E2.3 much stronger than it would be, say, for a stream of liquid nitrogen. This is because water molecules (like salt molecules) are permanently polarized: the oxygen atom grips the molecule’s electrons more firmly than the two hydrogen atoms do, making the molecule’s oxygen and hydrogen ends negatively and positively charged, respectively (see figure E2.6). The permanent charge separation involved here is much larger than one can typically achieve by polarization. In an external electric field, a water molecule’s atoms polarize as discussed above, but the whole molecule also twists like a compass needle to align its permanent dipole moment with the field (because its positive end gets pulled in the field direction and its negative end pushed away). Thermal jostling keeps disturbing that alignment, but since the molecule’s permanent dipole moment is so large, we get a strong effect even if (on average) the water molecules are only partially aligned with the field.

E2.4

Other Important Charge Distributions

In principle, calculating the electric field at a given point P created by any arbitrary distribution of motionless charged particles is easy: we simply A recipe for calculating the fields of charge distributions

1. Divide the distribution into “bits” small enough to model as particles, 2. Determine each bit’s charge, 3. Calculate the field vector that each bit contributes at P using the formula for the electric field created by a point particle, 4. Sum these contributed field vectors over all bits. This recipe is fairly easy to execute numerically, and computers are great for doing such tedious calculations. Figure E2.7 shows the result of a computer calculation for the electric field vectors above a distribution consisting of 10,000 particle-like “bits” uniformly distributed on a flat square. Numerical calculations can lead to pretty pictures, but we’d really rather have a formula for the distribution’s field: as discussed in chapter E1, only a formula gives us a complete description of the field. For sufficiently simple distributions, one can use calculus to perform the sum in the final step mathematically to arrive at an actual formula. In this section, I will give you formulas calculated this way for three simple charge distributions that also represent useful idealized models for more realistic distributions. We will see how to derive these formulas in the next two sections.

E2.4 Other Important Charge Distributions

25

Figure E2.7 Electric field vectors at selected points above the horizontal center of a planar distribution of 10,000 charged particles. (The distribution’s right and left edges are well beyond the figure’s frame.)

E​ ​W = _____ ​  1   ​ ______ ​  λ   ​   rˆ​   P​ N(E2.5) 2πε 0 │ rW​   P​ N│  • Purpose:  This describes the electric field E​ ​W  created at a point P by an infinite linear charge distribution having a constant charge per unit length λ (in C/m). The vector  rW​   P​ N is the position of P relative to the ˆ​  PN (representing the direction nearest point on the line. The unit vector r​ of  r​W  P​ N) thus points radially away from and perpendicular to that line. • Limitations:  The formula applies exactly only to an infinitely long, infinitesimally thin, and uniformly charged line. It is approximately valid for a uniformly charged straight and finite wire when the distance │   rW​  P​  N│ between the wire and the observation point P is much larger than the wire’s thickness but much smaller than the distance to its nearest end.

The field of an infinite line

 



E​ ​W = ___ ​  σ    ​r​ ​ˆP  N(E2.6) 2ε 0  

W • Purpose:  This describes the electric field E​ ​   created at point P by an infinite plane having a constant charge per unit area σ (in C/m2 ). The ˆ​  PN points perpendicularly away from the source plane at P. unit vector r​ • Limitations:  The formula applies precisely only to a planar charge distribution that is infinitely long, infinitesimally thin, and uniformly charged. It is approximately valid for a uniformly charged finite flat plate as long as the distance │ rW​   P​ N│ between the plate’s nearest point and the point P is much larger than the plate’s thickness but much smaller than the distance to its nearest edge. • Note:  This field is uniform in both magnitude and direction (!). (Note that the field vectors of the finite square plate shown in figure E2.7 point almost perpendicularly away from the plate and do seem to have a fairly constant magnitude.) We will see that the infinite plate has useful applications in spite of its seeming impracticality. The spherical shell is an even more practical charge distribution, and has a surprisingly simple field:

The field of an infinite plane

Chapter E2

26

Charge Distributions

0 inside the shell

The field of a uniformly charged spherical shell



Q 1 ______ W E​ ​  = _____ ​     ​  ​   2 ​  r​ˆ​ PC outside the shell (E2.7) │ │ 4πε 0  r​W P​  C  

• Purpose:  This describes the electric field E​ ​W  created at a point P either outside or inside a thin spherical shell with a uniformly distributed total charge Q. The vector  r​W P ​ C is the position of P relative to the shell’s center, so r​ ​ˆP  C (the direction of  rW​  P ​ C) points radially away from that center. • Limitations: The formula assumes that the shell’s charge is distributed uniformly and has infinitesimal radial thickness. “Inside” the shell means within the hollow space enclosed by the shell: equation E2.7 does not specify the field at points on the shell. • Note: A spherical shell’s external field is the same as that of a simple particle with charge Q located at the shell’s center(!!). The shell also reacts to external fields as if it were such a charged particle.

Finding the field within a spherical distribution’s volume

P

R

We call the last result the shell theorem. Almost any practical method of charging a real spherical object adds (or removes) electrons to (or from) the sphere’s surface atoms, so such an object’s charge distribution is an exceedingly thin shell even if the object itself is solid. This means that we can model a uniformly charged spherical ball of any size as if it were a particle. However, it is not always true that an object’s charge is on its surface. An atom’s electron cloud represents a practical example of a case where charge is distributed smoothly throughout a spherical volume. How might we calculate the field at a point P in the midst of such a volume? If the distribution is spherically symmetric, we can model it as a set of infinitesimally thin uniform shells, and calculate the total field by summing the shells’ fields. Figure E2.8 shows a point P a distance R from the distribution’s center. Shells inside and outside the radius R contribute differently to the field at P, as follows: 1. shells inside R contribute to the field at P as if the shell’s charge were concentrated at the distribution’s center, but 2. shells outside R contribute nothing (since P is inside every such shell).

Spherical charge distribution

Figure E2.8 The electric field at point P within a spherical charge distribution’s volume is as if only the charge inside the radius R were located at the distribution’s center.

P

A

B rA rB

Figure E2.9 Comparing the contributions to the field at P from opposing patches of charge A and B.

So the total field at P is as if only the charge enclosed by the radius R were concentrated at the distribution’s center: any charge outside R doesn’t matter! In the next two sections, I will discuss how we can calculate the fields of an infinite line and plane, but the shell theorem is tougher, and I will defer its proof until chapter E12, when we will have more powerful tools at our disposal. Still, we can understand qualitatively the surprising result that E​ ​W  = 0 at any point inside a shell as follows. Figure E2.9 shows an arbitrary point P inside a positively charged spherical shell. Contributions to the field at point P from particles in the circular patch labeled A will be opposed by those contributed by particles in patch B. For the sake of concreteness, say that P is twice as far from patch B as from patch A: rB = 2rA. The field vector that each charged particle in B contributes at point P is then only one fourth as large as that contributed by a directly opposed particle in patch A. But the area of patch B is four times larger than that of patch A, so (if the shell is ­uniformly charged) there are four times as many charged particles in patch B as in patch A. Therefore, it is plausible that the net contribution to the field at P from the particles in patch B exactly cancels that due to the fewer particles in patch A. This is not a proof, but does make the result more credible. Because the gravitational interaction between particles also goes as 1/r 2, the above arguments also apply to the gravitational field of a spherical object.  

E2.5 The Field of an Infinite Line

y

Ei

27

Ei cos θ

θi P θi

x

rPN

Figure E2.10

rPi (Nearest point to P)

Thin, charged line

xi

E2.5

This diagram defines quantities that will help us calculate the elecW tric field E  ​ ​  i contributed by a given “bit” of the infinite line.

“Bit” ∆x

The Field of an Infinite Line

Here again is the recipe for calculating the electric field vector at any point P created by an arbitrary charge distribution, with a bit more detail: 1. Slice the distribution into tiny particle-like “bits,” labeling each with a different value of some index i. 2. Find the ith bit’s charge qi. W 3. Calculate the electric field vector E​ ​ i  that the ith bit contributes at P using the formula for the electric field created by a point particle. W 4. Sum over all bits to get E​ ​ ,  the total field at P. Let’s implement this recipe in the case of an infinite thin charged line. Let’s take advantage of our freedom to choose coordinates so that the line lies on the x axis and P lies in the xy plane. Figure E2.10 shows this coordinate system and defines some symbols that will help us execute the recipe. We can divide the line up into “bits” having equal length ∆x. Note that xi is the ith bit’s x-position relative to the point P,  rW​  ​ Pi is the observation point P’s position relative to the bit’s position, and that  r​W ​ Pi makes an angle of θi with the vertical. We will assume for the present that the line is positively charged, meaning that its charge per unit length λ is positive. Since each bit’s length is ∆x, each bit’s charge must be qi = λ ∆x. This completes step 2. If ∆x is tiny, then each bit is tiny, and we can model it as a point particle. Using the formula for the field created by a point particle, we see that the electric field contributed by our ith bit has a magnitude of qi 1 1 λΔx │E​  ​W i│ ≈ _____ ​     ​ _____ ​    2 ​  ≈ _____ ​     ​ __________ ​       ​ (E2.8) │ │ │ 4πε 0  r​W P​  i 4πε 0  r​W P​  N│2 + ​x 2i​​  ​ In the coordinate system defined in the figure, this vector has an x component of −│E​  W ​ i │ sin θi, a y component of +│E​  W ​ i │ cos θi, and a z component of zero. Since 2 2 1/2 sin θi = xi/│ r​W P​  i│ = xi/(│ r​W P​  N│ + ​x ​i​  ​  )  and cos θi = │ r​W P​  N│/(│ r​W P​  N│2 + ​x 2​i​  ​  ) 1/2, we have  

 

λ│ r​W ​ PN│ Δx 1 1 −λxi Δx Ex,i ≈ _____ ​     ​ ______________ ​      ​ ,  Ey,i ≈ _____ ​     ​ ______________    ​     ​ ,  Ez,i = 0 (E2.9) 2 2 3/2 4πε 0  (│ r​W ​ PN│  + ​x ​i​  ​ )    4πε 0  (│ r​W ​ PN│ 2 + ​x 2​i​  ​ ) 3/2   

 

This completes step 3. Now we are ready to sum. Note that because the line is infinite, for every bit at position xi to the right of P there is an identical bit at position -xi to the left contributing an x component with the same magnitude but opposite sign. Therefore, in the sum over all bits, the x components of contributions from the bits will cancel in pairs, so Ex = 0. According to equation E2.9,     λ│ r​W ​ PN│    ______________ 1 ______________ λ│ r​W ​ PN│Δ   x Δx Ey = ​∑    ​ ​Ey,i ≈ ​∑    ​ ​_____ ​     ​      ​      ​ ≈ _______ ​   ​     ​∑    ​ ​ ​       ​ 2 2 3/2 2 all i all i 4πε 0  (│ r ​W ​ PN│  + ​x ​i​  ​ )    4πε 0 all i (│ r​W ​ PN│  + ​​x 2​i​  ​ ) ​3/2​   

 

(E2.10)

28

Chapter E2

Charge Distributions

If we take the limit Δx   0, then all the approximations we made earlier become equalities and the final sum becomes an integral. Since the possible values of xi range from -∞ to +∞, we have   λ│ r​W P​  N│  λ│ r​W ​P  N│  +∞ ______________ Δx dx Ey = _______ ​   ​​     lim ​​∑    ​ ______________ ​​       ​ = _______ ​   ​     ​∫  ​  ​  ​​       ​     2 2 2 3/2 Δx​  ​   0 4πε 0 W all i (│ r​W P​  N│ + ​x ​i​  ​  )  4πε 0 −∞ (│ r​W ​P  N│ + x 2 ) 3/2  

(E2.11)

 

Now this is a fairly nasty integral to do from scratch, but that is why God gave us integral tables, which you can easily find online (maybe even on this book’s inside cover). The result in this particular case is +∞ +∞ dx   ​  x   ​   1 2 __ __ ​ ∫  ​  ​  ​ ​ __________ = __ ​  12  ​  ​​    ​ __________  ​ ​ ​ = ​    ​  [1 − (−1)] = ​    ​   2 2 3/2 2 2 1/2 −∞ (a + x   )  a [ (a + x   )  ] −∞ a2 a2

(E2.12)

 

We have already seen that the total field vector must point in the +y ­direction, ˆP  N direction, so E​ which is the same as the + ​r​ ​W = │E​  W ​│   r​  ˆ​ P  N. Combining this with │ │ the earlier results (with a =   r​W  P​ N ) yields equation E2.5: λ│ r​W ​P  N│  ______ 2 1 λ W = _______ ​ E​ ​   ​     ​    2 ​   r​ ​ˆP  N = _____ ​     ​ ______ ​     ​  r​ ​ˆP  N (E2.13) 4πε 0 │ r​W P​  N│  2πε 0 │ r​W ​P  N│   

 

Exercise E2X.3 What if the line is negatively charged? Go through the above argument to see what changes, and show that equation E2.13 is still correct. A crucial part of the recipe: choosing a variable whose value we slice up to get “bits”

Reviewing the section, we see that an important part of using the recipe involves choosing a mathematical variable (in this case, x) whose range we can divide up into tiny equal parts ∆x that slice up our charge distribution into tiny “bits,” the ith of which is uniquely labeled by the variable value xi. Choosing an appropriate variable and expressing the ith bit’s charge qi and W field contribution E​ ​ i  in terms of that variable is key to the recipe’s success, and you should try to do the same thing in applying the recipe in other cases.

E2.6

The Field of an Infinite Plane

Because a planar charge distribution is two-dimensional, uniquely labeling particle-like bits of charge on the plane requires two variables. But the “bits” in our recipe don’t actually have to be like particles, they simply have to be something whose field we know how to calculate. Since we now know the field created at points near an infinite line, we can divide up an infinite plane into “bits” that are infinite line-like stripes. Figure E2.11 shows how we can divide the plane into stripes parallel to the y axis, each with a fixed width ∆x along the x axis. The value xi labels the stripe’s position along the x axis, but​  rW  ​Pi now specifies P’s position relative to the nearest point on the ith stripe and​  rW  ​PN gives P’s position relative to the nearest point on the plane. Here, instead of needing to know the charge qi of the ith particle-like bit, we need to know the charge per unit length λi for our ith stripe if we are to use equation E2.13 to calculate that stripe’s field at P. If σ is the charge per unit area on the plane, then a portion of the ith stripe that has some length L along the y direction will have an area of L ∆x and thus a charge of σL ∆x = λiL. If we divide both sides by L, we see that λi = σ∆x. This completes step 2. As long as ∆x is tiny, the width of the stripe will be tiny, and we can model the stripe as an infinite line. According to equation E2.13, the magnitude of the field contributed at point P by the ith stripe is then

E2.6 The Field of an Infinite Plane

Ei cos θi

θi

Ei

P

29

θi

Figure E2.11

z ∆x

xi

1

λ

y x

(Nearest point to P)



Stripe

1

σ Δx

│E​  ​W i│ = _____ ​     ​ _____ ​  i   ​  = _____ ​     ​ ______________ ​       ​ 2πε 0 │ r​W ​P  i│  2πε 0 (│ r​W ​P  N│ 2 + ​x 2​i​  ​)1/2  

This diagram defines quantities that will help us calculate the elecW tric field E​ ​  i contributed by a given stripe of the infinite plane. All stripes are parallel to the y axis, W​ P  i and r ​ W​ P  N lie in and the vectors r ​ the xz plane.

rPi

rPN

Infinite plane

(E2.14)

 

The field vector at P has an x component of −│ Ei│ sin θi, a zero y component (since each stripe’s field vector is perpendicular to the stripe), and a z component of │E​  W ​ i │ cos θi. Since we still have sin θi = xi/│ r​W P​  i│ = xi/(│ r​W P​  N│2 + ​x 2i​​ ​ )1/2 and cos θi = │ r​W P​  N│/(   │ r​W P​  N│2  + ​x 2i​​ ​ )1/2, we have

σ│ r​W P​  N│ Δx σxi Δx ________________ Ex,i =    ​     ​  ,  Ey,i = 0,  Ez,i = ________________    ​     ​ 2 2 │ │ 2πε 0(  r​W P​  N   + ​x i​​  ​ ) 2πε 0(│ r​W ​P  N│ 2 + ​x 2​i​  ​ )  

(E2.15)

 

This completes step 3. Now we sum over all xi. Note that for every stripe with a positive value of xi on the right, we have an identical stripe at -xi on the left that contributes an opposite x component, so in the sum, the stripes’ x components will canˆ PN cel in pairs. Thus, the total field E​ ​W points totally in the +z direction (the +​r​ direction). Again, converting the sum to an integral that we look up yields

(  )

+∞ σ│ r​W ​P  N│  +∞ __________ σ│ r​W ​P  N│  _____ dx 1 x -1 W = r​  ​   ˆ PN _______  ​ E​ ​ˆ PN _______ ​   ​     ​∫  ​  ​  ​​     ​   = r​ ​ ​   ​     ​    ​  ​  tan ​ ______ ​     ​    ​  ​ ​   ​  │ r​W ​P  N│  ]−∞ 2πε 0 −∞ │ r​W ​P  N│ 2 + ​x 2​i​  ​ 2πε 0 │ r​W ​P  N│ [  

 

(  )

π +∞​ = r​ π __ __     = r​ ​ˆP  N _____ ​  σ   ​​​    ​   ​  − ​ −​   ​   ​   ​ ​ ​ˆP  N ___ ​  σ   ​   2πε 0 [ 2 2 ]−∞ 2ε 0  

(E2.16)

 

This is equation E2.6!

Exercise E2X.4 What if P is below the plane instead of above it? Also what if the plane is negatively charged and P is above or below the plane? Argue that in all three cases, equation E2.16 is still correct.

So a generalized version of our recipe might look like: 1. Divide the range of some variable (call it u) into tiny increments ∆u to slice the distribution into “bits” labeled by ui whose field we can find. 2. Find the ith bit’s charge qi (or whatever charge-related quantity you need to calculate the bit’s field) in terms of the increment ∆u. W 3. Calculate the electric field vector E​ ​ i  that the ith bit contributes at P, W 4. Sum over all bits (that is, all ui) to get E​ ​ ,  the total field at P.

While you have seen examples of its use, to really understand it, you need to practice using it. Working the “Derivations” problems at the end of this ­chapter can give you that practice (many have answers to give you feedback).

Chapter E2

30

Charge Distributions

TWO-MINUTE PROBLEMS DD

A

CC BB

E2T.6 Imagine a spherical shell with a uniformly distributed negative charge, as shown below. What are the directions of the field vectors created by that shell at the two points shown? Choose your answers from figure E2.12.

E = zero F = other

B

(To indicate a doubled letter on the back cover, point to the letter with two fingers.)

C D AA

Shell

Figure E2.12

P1

P2

Direction answer chart. E2T.1 Imagine that we place a positively charged particle a distance r from a neutral atom, as shown below. Choose your answers for parts b through e from figure E2.12. Charged particle

Neutral atom

+ r

(a) The electric field of the charged particle causes the initially neutral atom to become charged. T or F? (b) What is the direction of the neutral atom’s dipole moment after the charged particle is brought near? (c) What is the direction of the neutral atom’s electric field at the location of the charged particle? (d) What is the direction of the force that the interaction between the particle and atom exerts on the particle? (e) What is the direction of the force that the interaction between the particle and atom exerts on the atom? E2T.2 The electric field created by a dipole at an arbitrary point in space around that dipole is always parallel to the dipole’s dipole moment vector. T or F? E2T.3 The dipole moment of an electrically polarized atom always points in the same direction as the external field that has polarized it. T or F? E2T.4 Imagine that we place a dipole a distance r from a neutral atom, as shown below. Neutral atom

Dipole – + r

(a) Will the dipole polarize the neutral atom? Choose E if not or specify the direction of the neutral atom’s induced dipole moment from figure E2.12 if nonzero. (b) Use figure E2.12 to specify the direction of the net force that the neutral atom exerts on the dipole. (Hint: Think of any dipole(s) as independent charged particles.) E2T.5 Imagine a neutral atom near an infinite charged plane. The neutral atom will be polarized by the plane’s field. T or F? Will the atom be attracted to the plate (A), repelled by it (B) or experience no force from it (C)?

(a) What is the direction at point P1 just inside the shell? (b) What is the direction at point P2 just outside the shell? E2T.7 Imagine that we place a spherical shell with a uniformly distributed positive charge Q and a particle with a negative charge -4Q as shown below. What are the directions of the total electric field vectors at the two points shown? Choose your answers from figure E2.12. Shell, charge Q Particle, charge – 4Q P1

– r

r

P2

(a) What is the direction at point P1 at the shell’s center? (b) What is the direction at point P2? E2T.8 Consider two concentric spherical shells, one with radius R and one with radius 2R. Both have the same uniformly distributed charge Q. Let E0 ≡ Q/(4πε 0 R2). Which of the values below is closest to the magnitude of the electric field vector created by these shells at a point P just barely inside the outer shell? A. zero B. ​ _14  ​  E0 C. ​ _12  ​  E0 D. E0 E. 2E0 F. 4E0 T. Other  

E2T.9 Consider a solid sphere with radius R with a charge uniformly spread throughout its volume. The magnitude of the sphere’s electric field is │E​  ​W1  │ at a point P1 on the sphere’s surface and is │E​  ​W2  │ at a point P2 at a distance _​ 21 ​  R from the sphere’s center. In the relationship │E​  ​W1  │ = b│E​  ​W2  │, what is the value of b? A. 0 B. 1 C. 2 D. 4 E. 8 F. Other

Homework Problems

31

E2T.10 The finite wire shown below has a uniformly distributed negative charge. Which of the arrows below is closest to the direction of the field vector that the wire produces at point P? (Point to a letter on the back cover with two fingers to specify a double-letter answer.) BB CC DD

E2T.11 The finite disk shown below has a uniformly distributed positive charge. Which of the arrows in figure E2.12 best indicates the direction of the electric field vector that the plate produces at the points specified? P2

P1

AA Uniformly charged disk

P

A

B

D C

P3

Negatively charged finite wire

(a) At point P1 a small distance above the disk’s center? (b) At point P2? (c) At point P3?

HOMEWORK PROBLEMS Basic Skills E2B.1 A plastic ball and glass marble are given uniformly distributed surface charges of -4.0 nC and +4.0 nC, respectively. If we place them along the x axis so that their centers are 5 cm apart, what is the magnitude and direction of their dipole moment vector? (Hint: The shell theorem allows us to model the ball and marble as point particles.) E2B.2 A CO (carbon monoxide) molecule has an intrinsic dipole moment whose magnitude is 4.0 × 10-31 C·m (which one can measure by examining its electric field). Model this dipole as if the carbon atom has a net positive charge +δ, and the oxygen atom has a net negative charge -δ. If the separation between the atoms is 0.11 nm, what is δ ? E2B.3 A HCl molecule has an intrinsic dipole moment of about 3.5 × 10-30 C·m oriented away from the chlorine molecule and toward the hydrogen molecule. What is the magnitude of the electric field at a point along the dipole’s axis (on the hydrogen side) a distance of 2.0 nm from the dipole’s center? E2B.4 A plastic ball and glass marble are given uniformly distributed surface charges of -2.0 nC and +2.0 nC, respectively, and their centers are separated by 1.0 cm along a line we can call the x axis. What is the magnitude of the force that a rubber ball with a uniformly distributed surface charge of -5.6 nC experiences at a point on the x axis a distance of 10 cm from the dipole’s center? (Hint: The shell theorem allows us to model all these charged spheres as if they were point particles.) E2B.5 Imagine placing a charged particle a variable distance r from a molecule with a permanent dipole moment. Argue that the magnitude of the force that the particle experiences varies in proportion to 1/r 3, and explain why the force is not proportional to 1/r 5, as it would be if the charged particle is interacting with an induced dipole. You don’t have to work out the math in detail: simply explain (physically) why there is a difference.

E2B.6 Consider a sphere with a radius of 4.0 cm having a uniformly distributed surface charge of +33 nC. What is the magnitude of the electric field it creates at a point 5.0 cm from its center? At a point 3.0 cm from its center? E2B.7 Consider a sphere whose radius is 10 cm. If after giving the sphere a uniformly distributed positive charge we find that the electric field at a point just barely outside the sphere is 1000 N/C, what is the sphere’s total charge? E2B.8 Normally, electrons cannot move through air (and thus escape a charged object), but if the magnitude of the electric field exceeds about 3 × 106 N/C, the field accelerates free electrons so violently that they crash into air molecules with sufficient energy to liberate more electrons. This chain reaction allows an object’s charge to be depleted by electrons flowing through air in the form of a spark. Use this information to estimate the maximum charge that one can put on a sphere with a radius of 1 cm. E2B.9 Consider a long thin wire with a constant charge per unit length of +5.6 × 10-7 C/m. What is the m ­ agnitude of the electric field at a point 10.0 cm from the wire (assuming that the point is much closer to the wire’s nearest point than to either of its ends)? E2B.10 Consider a long thin wire with a constant negative charge per unit length. If we measure the electric field to have a magnitude of 300 N/C at a point 12 cm from the wire’s nearest point (and which is much farther from the wire’s ends), what is the wire’s charge per unit length? E2B.11 Consider a disk with a radius of 10 cm and a total charge of +22 nC. Imagine a point P that lies on a line that goes through the disk’s center perpendicular to the disk’s surface. If P is 1.0 cm from the disk’s center, what is the approximate magnitude of the electric field vector at P?

 

 

E2B.12 Consider a uniformly charged horizontal disk with a radius of 20 cm. If the electric field vector at point P located 0.50 cm vertically above the disk’s center has a magnitude of 600 N/C, what is the disk’s approximate charge?

Chapter E2

32

Charge Distributions

Modeling E2M.1 An electron and a neutral carbon atom are floating alone and initially at rest in interstellar space a distance of 100 nm apart. The constant α for a carbon atom in the W is about α = 2.0 × 10-40 in SI units. W​ e  = α ​E​ expression p​ (a) What are the correct SI units for α? (b) Explain physically why the electron will accelerate toward the carbon atom and calculate the magnitude of its initial acceleration. (c) By the time the separation between the two has decreased to 10 nm, by what factor has the electron’s acceleration increased? E2M.2 Imagine that two charged balls placed some distance apart strongly attract each other. Now imagine placing a pane of glass halfway between the two balls. Will insertion of this glass increase or decrease the magnitude of force that each ball feels, or will it have strictly no effect? Use diagrams to explain your reasoning carefully. E2M.3 When dealing with macroscopic matter, physicists are more often interested in a material’s electric ­susceptibility χ (the Greek letter chi) than in the polarization constant α per atom. A material’s susceptibility is defined by W = ε 0 χ ​E​ W ​ P​   (E2.17)

E2M.5 Imagine that a spherical asteroid of radius R has a constant density and a total mass of M. (a) Use the gravitational version of the shell theorem to draw a quantitatively accurate graph of the magnitude of the asteroid’s gravitational field vector as a function of │  ​rW ​ PC│ between 0 and 4R, where │  ​rW ​ PC│ is the distance between the asteroid’s center and the place where the field is evaluated. (b) If you were to drill a tunnel through the asteroid’s diameter, would a person in the tunnel feel more or less gravity than at the asteroid’s surface? E2M.6 Consider a solid sphere of radius R with a hollow region as shown in the cross-sectional diagram below. Hollow region

R

P

Solid sphere

R

The solid portion has a total charge of Q spread uniformly throughout its volume. Find the magnitude of the electric field at point P. (Hint: The superposition principle implies that the field of superposed positive and negative identical objects is the same as if the objects had zero charge.)

 

where P​ ​W  is the polarization density (the total induced W dipole moment per unit volume), E​ ​  is the macroscopic electric field inside the material, and ε0 is the permittivity constant (where 1/4πε0 = 8.99 × 109 N·m2/C2). (a) Argue that χ must be a unitless number. (b) A material’s dielectric constant or relative ­permittivity εr is defined to be εr = 1 + χ . Look online to find the dielectric constant of fused quartz and determine the total dipole moment of a cubic centimeter of quartz in an electric field of 1000 N/C. (c) A material’s permittivity ε ≡ εr ε0. What is the permittivity of quartz? Of a vacuum? (d) Imagine that we hold a cube of Pyrex glass (whose dielectric constant is 5.0) that is 1.0 cm on a side at a point about 20 cm from the center of a sphere with a charge of +56 nC uniformly distributed on its surface. What will be the approximate magnitude of the electrostatic force exerted on the cube? E2M.4 Consider a cube with side L and total charge Q spread uniformly on its surface. Imagine that you find online the following expression for the magnitude of the electric field at a point P outside the cube

1

QL

│E​  ​W│     = ___ ​     ​ ______ ​    ​   2ε 0 │  ​rW ​P  C│2

(E2.18)

 

│  ​rW ​ PC│ is the distance between point P and the cube’s where  center. List at least three independent reasons that this formula can’t possibly be right. (Hints: What should the field be if the cube’s size L s, the angle ϕ ≈ s/r, and the distance between each particle and point P is ≈ r. Use these approximations W We /r 3). to show that in this limit, E​ ​  = (1/4πε 0)(−​p​  

 

E2D.2 In this problem, we will calculate by hand the electric field that a finite line with a positive charge Q uniformly spread over its length L creates at a point P a distance of _5​ 1 ​  L from the line’s center. We’ll do this by dividing the line into 10 equal “bits,” calculating the field vector that each bit contributes to the field at point P by modeling the bit as a point particle, summing the vectors, and comparing the result to equation E2.5. The diagram below illustrates the scheme and (as an example) the field vector contributed by bit 8: P 1 5

1

2 “Bits”

3

4

5

mark each bit’s center. Mark the point P a distance of 4.0 cm from the line’s center, and draw a line from each bit’s center to the point P. Measure the length of each line to the nearest 0.05 cm and write it next to each line. 1 (b) Assume that a particle with charge of ​ __    ​  Q a distance 10 1 _ of ​  5  ​  L (4.0 cm) from point P creates a field vector that we choose to represent by an arrow 5.0 cm long at point P. Explain why a particle with the same charge a distance r from point P creates a field vector that we should represent by an arrow (5.0 cm)(4.0 cm/r)2 long if we use the same scale. (c) Draw correctly scaled arrows (with their tails at P) to represent the field vectors contributed by each of the 10 bits of this line, modeling each bit as a particle located at the bit’s center. Label each vector with its length accurate to the nearest 0.05 cm. (d) Add the vectors graphically to get the total electric field vector E​ ​W  at P. Should this vector point vertically upward? Does it? (e) What is the length of the total electric field vector E​ ​W  in centimeters? Note that 5.0 cm represents the length 1 of the field vector produced by ​ __    ​  Q at a distance of 10 ​ _15  ​  L = │ ​ Wr ​ PN│. What, therefore, is the magnitude of E​ ​W  as a multiple of λ/(4πε 0│ ​ Wr ​ PN│)? (Hint: Remember that λ = Q/L.) (f) Compare this to what equation E2.5 predicts should be the field for an infinite line. Also compare to the result for a finite line given in problem E2D.7.  

E2D.3 There is a theorem (which we might call the “pipe theorem”) that says that the electric field at a point outside an infinite straight pipe with charge uniformly distributed on its surface is as if the pipe’s charge were concentrated at a line along its axis, and that the field at any point inside the pipe is zero. In this problem, we will check this result using a hand calculation where we divide the pipe into 10 “bits,” each of which we will model as an infinite line. We will calculate the field vector contributed by each bit at a certain point P, and add the vectors graphically to determine the total field at P. The diagram below illustrates the scheme (displaying a cross-sectional view of the pipe) and (as an example) the field vector contributed at P1 by bit 8:

rP8

L 6

7

8

8

9

7

10

Line with charge Q

(a) Copy this drawing to a sheet of paper, making the line 20 cm long and putting it at the bottom of the page. Divide the line into 10 “bits,” each 2.0 cm long, and

E+ E–

–q

Derivations

L

ϕ

E

E–

E2M.9 Consider a spherical shell of radius R that has a total charge Q uniformly distributed on its surface. (a) Find the charge per unit area σ on this shell’s surface. (b) At point P just barely outside the shell’s surface, the sphere (like the earth viewed from a height of a few meters) looks like an infinite plane. Use the shell theorem to calculate the magnitude of the shell’s field at such a point in terms of σ and show that it is twice what you’d expect from equation E2.6. (c) Note that a shell’s field just below (that is, inside) its surface is zero (due to distant charges on the other side of the shell), but an infinite plane’s field is not zero below the plane. Imagine that we were to arrange distant charges to contribute a field sufficient to make the total field zero everywhere below an infinite charged plane. What would the magnitude of the field above the plane be in such a case? Is there agreement now? Explain.

E8

r

9

R

10 P2

6 R

1 2

5 4

P1

1 R

R

3

2

E8 “Bit”

(continued)

Chapter E2

34

Charge Distributions

E2D.3 (continued) (a) If your instructor has not provided you with a suitable drawing to work from, draw a circle with a diameter of 10 cm on the left side of a sheet of paper in landscape view. Divide the circle into 10 “bits” as shown, using a protractor to place a dot every 36°. Mark the location of points P1 and P2 as shown in the diagram. (b) Draw lines from each of the 10 dots to point P1 and label each line with its length in centimeters. (c) We are going to model each “bit” as if it were an infinite line going through the numbered dot at each bit’s center. If the pipe’s charge per unit length is λ, then the charge per unit length on each bit is __ ​ 101   ​  λ. Now let’s say that we are going to represent the field vector that an infinite line with charge per unit length λ would create at a point a distance R = 5.0 cm away by a drawn arrow with a length of 20 cm. Argue that the field vector that an infinite line with charge per unit length __ ​ 101   ​  λ would create at a point a distance r away should be (2.0 cm)(5.0 cm/r) = 10 cm2/r long if it is drawn to the same scale. (d) Draw correctly scaled arrows (with their tails at P1) to represent the field vectors contributed by each of the 10 bits. Label each vector with its length. Try to make your measurements accurate to the nearest 0.05 cm. (e) Add the vectors graphically to get the total electric W field vector E​ ​  at P1. (f) What is the length (in centimeters) of the arrow representing the total electric field vector E​ ​W?  What should it be if the pipe theorem is valid? Does it also point in the direction consistent with the pipe theorem? (g) Repeat steps (b) through (e) to show that the field at point P2 inside the pipe is very nearly zero. E2D.4 In this problem, we will use the recipe introduced in this chapter to calculate the electric field at an arbitrary point along the central axis of a thin ring having a total charge Q uniformly distributed around its circumference. The diagram below shows the situation and defines some useful symbols. (Note that I have chosen the z axis here to coincide with the ring’s central axis.) z

y

P θ

rPi x

ϕ R Charged ring

ith “bit”

(a) The first step in our recipe is to choose a variable whose range we can divide into tiny increments to slice the distribution into particle-like bits. Explain why the variable ϕ is a good choice for this role, and explain _______ why the variables x, y, z, θ, and r ≡  ​ x  2 + y 2 ​  are not. (b) Show that the charge of the ith “bit” is  



 

Q qi = ___ ​    ​ Δϕ (E2.19) 2π

(c) Find the magnitude of the electric field │E​  ​W i│ contributed at P by this bit in terms of Q, ε0, ∆ϕ, R, and z (the z coordinate of point P). (d) Explain why the ring’s total electric field must point along the z axis (that is, why the x and y components of the contributed “bit” fields cancel out). (e) Sum up the contributed fields’ z components to show that the total field vector the ring creates at point P is 1 Qz W = _____ ˆ​ ​ E​ ​     ​ __________ ​  2    ​  z​   (E2.20) 4πε 0 (R + z 2)3/2  

 

where z​ ​ˆ is a unit vector pointing in the +z direction. W (f) Argue that this formula yields the correct direction of E​ ​  even if Q and/or z are negative. E2D.5 In this problem, we will use the recipe introduced in this chapter to calculate the electric field at a point a distance r from the end of a finite wire having a total charge Q uniformly distributed along its length. The diagram below shows the situation and defines some useful symbols. r

L Charged wire

rPi

“Bit”

x

P

(a) The first step in our recipe is to choose a variable whose range we can divide into tiny increments to slice the distribution into particle-like bits. There is only one variable that makes sense to use in this case (L and r are fixed values, not variables). What is it and why? (b) Find the charge of the ith bit in this case. (c) Find the magnitude of the electric field │E​  ​Wi │ contributed by the ith bit in terms of Q, L, ε0, r, and the increment in your chosen variable. (d) Convert the sum of E​ ​W i over all i to an integral, and do the integral to show that the total field is given by 1 Q W = _____ ˆ​ ​ E​ ​     ​ _______ ​     ​  x​   (E2.21) 4πε 0 r(r + L)  

where x​ ​ˆ is a unit vector that points in the +x direction, whether Q is positive or negative. (Hint: You may find it helpful to change the variable of integration. You may look up the integral online if that helps.) (e) How would this result change if the point P were to the left of the wire instead of to the right? Explain.

E2D.6 In this problem, we will use the recipe introduced in this chapter to calculate the electric field created by a uniformly charged half ring at the point that would be the ring’s center if it were a full ring. The diagram below shows the situation and defines some useful symbols. y Half-ring with total charge Q

ith “bit” R

rPi θ P

x

Homework Problems

35

(a) The first step in our recipe is to choose a variable whose range we can divide into tiny increments to slice the distribution into particle-like bits. Explain why the variable θ is a good choice for this role, and explain _______ why the variables x, y, and r ≡  ​ x  2 + y 2 ​  are not. (b) Show that the charge of the ith “bit” is  

 

Q qi = __ ​ π ​  Δθ (E2.22)



(c) Show that equation E2.25 is consistent with equation E2.5 in the limit that L >> │  ​rW ​ PN│. Also argue that if │    ​rW ​P  N│  0,

1  __________ Ez = ___ ​  σ   ​​    1 − ____________ ​     ​   ​  (for z > 0)  ] 2ε 0 [ ​ 1  + ( R /z)2 ​ 

(E3.35)

E3R.2 For a certain engineering project, your boss wants you to figure out how to arrange fixed charges to create an electric field that in the xy plane has radial and tangential ˆ​  = _​ 1  ​ar, where a W W components of E r = E​ ​  ·  rˆ​  ​ = a r  and ​E  ​θ​= E​ ​  · θ​ 3 is a constant, r is distance from the origin, the unit vector​  rˆ ​  points radially away from the origin, and the unit vector​ ˆ  is perpendicular to  rˆ​ ​  in the counterclockwise direction. θ​ Sketch the field, and using that sketch, convince your doltish boss that this is physically impossible. E3R.3 (Adapted from Purcell, Electricity and ­Magnetism, 2/e, McGraw-Hill, 1985, problem 2.20.) We can model a gold nucleus as a sphere with a radius of about 7.3  ×  10-15  m  with a charge of 79e uniformly distributed throughout its volume (where e is the charge of a proton). If we define our reference position where ϕ ≡ 0 to be infinity, what is the value of the potential at the center of this nucleus? (Hints: Divide the solid sphere into a series of thin nested shells. You should find that the answer is 23.4 MV.)

 

where σ is the charge per unit area on the disk. (e) Note that if z R, this expression reduces to what we would expect for the field created at a point on the positive z axis by a particle of charge Q at the origin.  

Rich Context E3R.1 For a certain engineering project, your boss wants you to figure out how to arrange fixed charges to create a an electric field that is uniform in a cylindrical region (with​ W  parallel to the cylinder’s axis) but zero outside. Explain E​ to your witless boss why this is impossible.

Advanced E3A.1 Consider a spherical shell of radius R with charge Q uniformly distributed on its surface. Our goal is to prove the shell theorem by calculating the potential ϕ (r) at a point P that is a distance r from the shell’s center. Do this by dividing the shell’s surface into thin rings, where the ith ring is the set of points on the shell’s surface between the “latitudes” θi − ​ _12  ​ Δθ and θi + ​ _12  ​ Δθ, where the shell’s “north pole” is the point on the shell closest to point P and we measure the “latitude” angle θi from that pole (as is conventional in physics) rather than from the equator. First show that the ith ring has charge qi = _​ 12  ​ Q sin θi Δθ. Then find the potential ϕi contributed at P by this ring, sum over all rings, convert the sum to an integral, and evaluate the integral for both r > R and r < R. Finally, determine the shell’s electric field in these regions from the potential field ϕ(r). (Hint: When doing the integral, change variables to _________________ 2 u= ​ R      +  r 2 − 2Rr cos θ ​.)

ANSWERS TO EXERCISES E3X.1 At a given point, the electric field points in the direction of the force a positively charged particle would feel at that point. However, an electron has a negative charge. Therefore, if the field points in the +z direction, the force on an electron is in the -z direction. This means that if the electron is moving in the +z direction, its acceleration is opposite to its motion, meaning that it will slow down. E3X.2

Consider the paths ACB and ADB shown below. A

D

C E

B

Both paths take us from point A to point B, and for any step W along each path’s vertical leg, we have E​ ​  · d ​ rW ​ = 0 (because​ W E​  is perpendicular to any vertical step d ​ rW ​  ). But the sum of E​ ​W  · d ​ rW ​  along the horizontal leg AC is much larger than that along the equal-length horizontal leg DB, because │E​  ​W│    is much larger at points along the upper path. Therefore, W · d ​ rW ​  is not the same for the paths the value of Δϕ = −∫ ​E​ ACB and ADC. But they must be the same for any static electric field. Therefore, the field shown cannot be a valid static electric field. E3X.3 Your sketched arrows should point perpendicular to the equipotentials and have a length inversely proportional to the spacing between equipotentials. The resulting arrows should indeed look like those in figure E2.1 (though the overall scale of the arrows is arbitrary).

E4

Static Equilibrium Chapter Overview Section E4.1:  Conductors and Insulators If charged particles cannot easily move through a substance, we call that substance an insulator, but if at least one kind of charged particle can move easily through a substance, we call the substance a conductor. Metals and salty water are good conductors; glass, rubber, and most plastics are excellent insulators. Excess charge we give a conductor (since it is free to move) usually disperses almost instantly into the environment by flowing through the conductor and the person holding the object. This is why it seems easier to charge insulators.

Section E4.2:  Static Charges on Conductors Any excess charge placed on a conducting object will rapidly redistribute itself W into a  static equilibrium distribution such that E​ ​   = 0 everywhere inside the object’s ­conducting interior. This statement has the following corollaries: in static equilibrium 1. The potential is constant throughout the object’s conducting interior. 2. The electric field just outside the object will be perpendicular to its surface. 3. The object’s conducting interior will be electrically neutral. Charge on a conducting object’s surface will distribute itself in whatever way makes​ W = 0 in its interior (for example, uniformly on the surface of an isolated sphere). In the E​ presence of an external charge, this surface distribution will become polarized in such a way that its field cancels the external field throughout the object’s interior. This is true even when the object is hollow. We call a conducting hollow object that shields something inside it from external electric fields a Faraday cage. The polarization of a conductor’s surface charge distribution is similar to the polarization of insulators discussed in chapter E2 except that it is stronger, because the charged particles are free to move large distances. The section also argues that the electric field will be strongest at the tip of anything that protrudes from a conducting object.

Section E4.3: Capacitance A capacitor consists of two isolated conductors (that we call the capacitor’s plates) having opposite charges Q and -Q. We define its capacitance to be



| |

Q ​ ​(E4.3) C ≡​ ____ ​      Δϕ

• Purpose:  This equation defines the capacitance C of a two-conductor object, where ∆ϕ is the potential difference between them that develops when we move charge Q from one initially neutral object to the other. • Limitations:  This equation assumes that the two conductors are isolated from other charged objects. • Note:  The SI unit of capacitance is the farad, where 1 F ≡ 1 C/V.

56

We can also treat a single isolated object as a capacitor if we assume that the second conductor is an imaginary conducting sphere with infinite radius.

Section E4.4:  The Parallel-Plate Capacitor This section argues that the static equilibrium distribution of charges on closely spaced parallel flat conducting plates is a uniform distribution on the inside surface of each plate. This distribution creates a field with the following characteristics:



│ σ│  │E​  ​W│    ≈ ____ ​   ​   between the plates, │E​  ​W│    ≈ 0 elsewhere

(E4.7)

ε0  

• Purpose:  This equation describes the field between and around identical and oppositely charged parallel conducting plates, where σ is the charge per unit area on either plate and ε 0 is the usual permittivity constant. • Limitations:  These equations apply only if the gap between the plates is very small compared to their size. It also does not apply to points within about a gap’s-width from the plates’ edges. • Notes:  The electric field in the gap is almost exactly uniform and points perpendicular to the plates and away from the positive plate.  

Equation E3.12 implies that the potential difference between the plates in this case is Δϕ = −∫ W ​E​ · d r​W​ = │ W ​E​│   ∫ dr = −│ W ​E​│   s  (between the plates)



(E4.8)

Equation E4.3 then implies that the capacitance of such a parallel-plate capacitor is C = ε 0 __ ​ A s ​  (E4.9)



 

The capacitance of a more general capacitor also increases as its plate area increases and plate separation decreases (though maybe not in such a simple way).

Section E4.5:  The Electric Field as a Form of Energy By considering the energy required to pull oppositely charged parallel plates apart, we find that the density of energy in a region of spaced filled with an electric field is



uE = _​ 12 ​ ε 0│ W ​E│ ​  2 (E4.12)  

• Purpose:  This equation expresses the energy density uE of an electric field in a region where the electric field has the magnitude │E​  W ​│   . The constant ε 0 is the usual permittivity constant. • Limitations:  Though this equation is technically correct even inside materials, its interpretation can be complicated in such circumstances.  

This result implies that the electric field is a dispersed form of energy, as physically real as a particle (which is just a compact form of energy). The electric field in a general charged capacitor stores an energy of



U = _​ 12  ​  C (Δϕ)2(E4.14) • Purpose:  This equation describes the energy U stored in the electric field of a charged capacitor in terms of its capacitance C and the potential difference ∆ϕ between its plates. • Limitations:  None!

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58

Chapter E4

E4.1

Insulators

Conductors

Why it is hard to charge a conducting object

Static Equilibrium

Conductors and Insulators

In the past three chapters, we have developed some powerful tools for describing and calculating electric fields, but many of the concepts have been pretty abstract. In this chapter, we will begin a three-chapter sequence where we will see how these tools help us understand the everyday physics involved in simple electrical circuits. An important first step in moving toward practical applications is to recognize that we can divide most familiar materials pretty unambiguously into two distinct categories that describe how easily charges can move through them. An insulator is a material through which charges cannot easily move: air, glass, and plastic are examples. If we put excess charge on the surface or interior of an ideal insulator, it remains where it is placed. When I have described macroscopic charged objects in the previous chapters, I have been implicitly assuming that these objects were excellent insulators. At the other extreme are conductors, materials through which at least one kind of charged particle can easily move. In most metals, the outermost electron in each atom becomes detached from the atom and becomes free to roam around. This means that most metals (especially silver and copper) are excellent conductors. Salty water is a conductor because it contains many ions that can move around in response to an external electric field. Because your body contains a lot of salty water, it is a pretty decent conductor. These differences can be pretty stark. In a given electric field, the rate that charge flows through a volume of air is about 1022 times smaller than it does through the same volume of metal, and most plastics are even better insulators. There are few common solid materials that are not pretty clearly either conductors or insulators (at least when dry). These differences help explain why some materials are easier to charge than others. Electrons placed on the surface of a good insulator remain fixed even though they strongly repel each other. On the other hand, extra electrons placed on the surface of a metal almost immediately disperse in response to the the repulsive forces that they exert on each other. If you rub a hard rubber comb through your hair, the rubber molecules attract some electrons from your hair molecules, and as you pull the comb away, these electrons remain fixed on the comb’s surface because rubber is an excellent insulator. On the other hand, if you rub a silver spoon through your hair, it will also typically grab some electrons from your hair. But these mutually repelling electrons rapidly move away from each other in the spoon, and indeed ultimately move through your hand and your body back to your head! So a spoon is hard to charge not so much because it is worse at picking up electrons than the comb, but because the extra electrons it picks up rapidly disperse themselves through the spoon into your hand. Because rubbing typically moves only a tiny amount of charge (typically on the order of 10-8 C) from one object to another, a material must be an exceptionally good insulator if this excess charge is not to escape. Natural materials like wood or rock are pretty good insulators compared to most metals, but are still not good enough to hold such tiny charges for a decent amount of time. Almost anything that is the least bit moist also will conduct well enough to quickly dissipate such a tiny charge: this is why it is difficult to charge an object by rubbing it when the air is humid. On the other hand, it is possible to charge a conducting object if that object is isolated from any other conductor by an exceptionally good insulator. A metal sphere on a plastic stand can hold an electrical charge quite well. You can charge a spoon by rubbing it in your hair if you hold the spoon with a plastic glove.

E4.2 Static Charges on Conductors

E4.2

59

Static Charges on Conductors

Because at least some particles (electrons in the case of a metal) are free to move in a region inside a conductor, they in fact must move if there is a nonzero electric field in that region. Charges placed on a conductor will therefore move around until they can settle (typically within a fraction of a microsecond) into a distribution where they can remain at rest (usually trapped against the conductor’s boundaries). We call such a distribution a static ­equilibrium charge distribution. The fundamental characteristic of such a distribution is the following: In static equilibrium, any excess charges placed on a conducting W object will arrange themselves so that E​ ​   = 0 at all points within the object’s conducting interior. W This must be so, because if E​ ​   ≠ 0 in any conducting region, then charges in that region would be free to move in response to the field, which implies that the charges have not yet settled into their final static configuration. This statement has some interesting corollaries: in static equilibrium,

The fundamental condition for static equilibrium

Corollaries of this principle

1. The potential ϕ is the same at all points inside an isolated conducting W · d ​ rW ​ = 0 between region. (Proof: According to equation E3.12, Δϕ = −∫ ​E​ any two points inside the conductor. The conductor’s surface must therefore be an equipotential surface.) 2. The electric field at points on a conductor’s surface must point perpendicular to that surface. (Proof: We saw in section E3.5 that the electric field must point perpendicular to any equipotential surface.) 3. The conducting interior of a conducting object must be electrically neutral in static equilibrium: any excess charge must be located on the object’s surface. (We will see the proof in chapter E12.) Note that these statements only apply to the static equilibrium that results after the charges have finished moving around. In the next chapter, we will discuss dynamic equilibrium in electrical circuits that maintain steady current W flows. In such cases, E​ ​  ≠ 0 in the circuit’s conducting wires, and so corollaries 1 and 2 do not apply (though it turns out that 3 still does). As an example of a static equilibrium distribution, imagine that we give an isolated solid metal sphere an excess negative charge Q. All our conditions for static equilibrium are met if this excess charge ends up uniformly distributed on the sphere’s outer surface (consistent with corollary 3). According to the shell theorem such a surface charge distribution creates an electric field that is zero inside that surface and thus inside the ball (satisfying our fundamental criterion for a static distribution and ensuring compliance with corollary 1). The shell theorem also implies that the electric field at points just outside the surface points radially away from the sphere’s center, and thus perpendicular to its surface, consistent with corollary 2. Indeed, one can prove mathematically that this uniform charge distribution is the only distribution satisfying our criteria for an isolated spherical conductor. Indeed, this makes intuitive sense. Because of their mutual repulsion, the excess electrons that we have placed on the sphere will want to get as far away from each other as possible. They can do this first by migrating to the sphere’s surface. However, they can’t escape the surface into the surrounding air (because air is an insulator), so the only way to further maximize the average distance between the electrons is for them to distribute themselves uniformly on that surface.

The equilibrium distribution on a charged sphere

60

Chapter E4

Polarization of conductors

However, the charge distribution on a sphere will not be uniform if charges outside the sphere create an external electric field. For example, imagine placing a positively charged particle close to an initially neutral metal sphere. We can intuit what will happen: free electrons on the sphere’s surface will crowd toward the side of the sphere nearest to the charge, leaving atoms on the opposite side of the sphere with a deficit of electrons and thus a positive charge (see figure E4.1). (Of course, if the electrons crowd too closely, their mutual repulsion pushes them apart, so their final distribution will reflect a balance between these competing effects.) This polarization of the surface charge distribution is similar to the polarization of insulators discussed in chapter E2, but more extreme: because the electrons are not trapped in atoms, they can move freely, making the charge separation larger. Note that “polarized” is not the same as “charged.” The polarized metal sphere in figure E4.1 still has zero net charge. But its surface charge distribution has become uneven in a way that it creates a nonzero electric field in its interior. The figure shows that this electric field (the black arrow) in fact opposes the external charge’s field in the sphere’s interior. Indeed, the fundamental criterion for static equilibrium implies that the sphere’s final surface charge distribution must be whatever distribution exactly cancels the external charge’s everywhere in the sphere’s interior. This general argument applies to objects of any shape, not just to spheres. It even applies if the conducting object is hollow, as I will now prove. Since the W charge distribution that ensures that E​ ​   = 0 in the object’s interior is located on the object’s surface (by corollary 3), removing the object’s interior does not W in any way change the surface charge distribution, so E​ ​   is still zero in the newly created cavity. Therefore, we can completely shield an object from an external electric field by putting it inside a conducting container, which in this context, we call a Faraday cage. Even time-dependent external fields can be canceled as long as they vary slowly compared to the time it takes the charges to redistribute themselves on the conductor (which is typically a fraction of a microsecond). You may have noticed that when your car crosses a metal-framed bridge, your radio may cut out, particularly if you are tuned to the (relatively low-frequency) AM band. Electrons in the bridge’s metal frame are able to move quickly enough to cancel out the varying electromagnetic field that comprises the radio signal, allowing the bridge to serve as a Faraday cage. The polarization of conductors in an external electric field also explains how bringing a neutral metal sphere near a charged sphere can cause a spark to fly between their closest points. As we bring the spheres closer, they both become increasingly polarized (see figure E4.2). This makes the electric field close to the most crowded regions stronger and stronger. When the electric field exceeds about 3 × 106 N/C at any point in air, it accelerates any free electrons that happen to be in the air (and there are always a few) so vigorously that by the time they reach the next air molecule, a significant fraction have enough energy to knock an additional electron out of that molecule (see problem E4M.8). This creates a “chain reaction” where free electrons beget more free electrons until there are enough free electrons to make air a conductor instead of an insulator. This enables electrons to flow from one sphere to the other along the path where the field is most intense. This flow is a spark. Once the flow gets going, it does not require as strong a field to maintain it, so the flow will continue until the two spheres almost come to the same potential. At that point, there is no net energy to be gained by moving charge from one sphere to the other, so the flow of electrons stops. Your finger is a decent conductor, and so it can play the role of either the charged or uncharged sphere in this case.

External particle

+

Field created at P by the sphere’s surface charge –

– –– – –

– –

–

P

+

ES +

EEP +

+ + + ++ + Field created at P by the external particle

Figure E4.1 Electrons on a neutral metal sphere will crowd toward an external positive particle, leaving the sphere’s opposite side with a positive charge. The final distribution will completely cancel the external particle’s field at any point P in the sphere’s interior.

Why sparks fly

Static Equilibrium

E4.2 Static Charges on Conductors

–

–

61

Figure E4.2

–

Neutral sphere

–

–

–

– –

– –

– +

+

+

+

(a)

(b)

– Charged – sphere –

–

–

– –

–

–

– – –

–

–

–

–

–

– –

–

++ ++ – –– – – –

(c)

–

–

+

+

+

–

–

– –

– –

–

–

Calculating (even numerically) the exact surface charge distribution for a nonspherical charged conductor (or a conductor in an external field) is generally quite difficult. However, there is one other general result that is useful to know. When a charged conductor has a sharp protrusion of some kind, the field is strongest near the protrusion’s tip. Qualitatively, this is because the tip is farther from the bulk of the conductor’s surface than most parts of the surface, so as the excess charged particles on the conductor’s surface strive to get as far away from each other as possible, they will preferentially crowd into the protrusion. The greater charge density in the protrusion creates a stronger electric field in the space nearest the protrusion. We can see that this is so by considering a simplified model for a protrusion. Consider a sphere of radius R connected to a sphere of smaller radius r by a wire, as shown in figure E4.3. Assume that the spheres are sufficiently far apart that each doesn’t much influence the charge distribution on the other, and charges on the thin wire are also insignificant. In static equilibrium, the spheres must have equal potentials. The shell theorem tells us that (assuming that the other sphere is sufficiently distant that we can ignore its potential) the potential of the larger sphere at its surface is as if its charge Q were located at its center: ϕ = Q/4πε 0 R. Similarly, the potential at the surface of the smaller sphere is ϕ = q/4πε 0 r, where q is the small sphere’s charge. Since both spheres must have the same potential relative to infinity, we have

(a) A charged metal sphere creates an electric field that begins to polarize the neutral metal sphere. (b) As the spheres approach each other, the field that each creates increasingly polarizes the other. (c) At a certain separation, the concentration of charges at the spheres’ nearest points becomes large enough to create an electric field that exceeds the breakdown field strength for air, causing a current to flow (a spark) that carries electrons from the charged sphere to the other.

The field is strongest near any protrusions on a charged conductor

r

Small sphere

 

 



q Q r  ​ ______    ​  = ______ ​     ​     ⇒   q = Q ​ __  ​   (E4.1) R 4πε 0 R 4πε 0 r  

 

Wire

So the smaller sphere has a proportionally smaller charge, which makes some intuitive sense. However, the charge per unit area σsm on the smaller sphere is greater than the charge per unit area σ lg on the larger one:

q Q(r/R) _____ Q Q __ R R σsm = _____ ​    2 ​ = _______ ​   ​     = ​     ​  = _____ ​     ​  ​   ​ = σ lg __ ​   ​  (E4.2) 4π rR 4π R 2 r r 4π r 4π r 2  

 

 

The electric field at just outside each sphere is proportional to that sphere’s charge per unit area (for example, │E​  W ​ s m│ = │ q│/4πε 0 r 2 = σsm/ε 0), so the field is  stronger near the smaller sphere as well. Therefore, we see that in this type of protrusion at least, the electric field is stronger at the protrusion’s tip, and the strength increases as the tip becomes “sharper.” So if you are charged, your fingertip is a place where your personal electric field is especially intense! This also explains why lightning tends to strike conducting objects that protrude from the earth.  

 

R

 

Large sphere

Figure E4.3 A simplified model for a ­protrusion on a conducting object.

Chapter E4

62

Example E4.1

Static Equilibrium

Problem:  Suppose you have a metal sphere on an insulating stand and a negatively charged rubber rod. Describe at least one way you could use the rod to give the sphere a net positive charge. (Hint: You can take advantage of the fact that your body is a conductor.) Solution  If you bring the rod close to one side of the sphere while you touch the other side with your finger, the negatively charged rod will repel electrons from the sphere through your body into the environment (see figure E4.4a). Removing your hand then isolates the positive charge on the sphere: it will remain even after you withdraw the rod (figure E4.4b and c).

(a)

(b)

(c)

Figure E4.4 How to use a charged rod to give a conducting sphere the opposite charge.

Exercise E4X.1 Suppose you have a negatively charged rubber rod and two identical metal spheres on insulating stands. Describe a procedure that will give each sphere an exactly equal amount of positive charge.

E4.3 Definition of a capacitor

Definition of capacitance

Capacitance

Consider two isolated and initially neutral conducting objects A and B. Suppose we move a certain amount of charge Q from one to the other, and then allow both objects to come into static equilibrium. Since each conductor has a fixed potential throughout its volume, the potential difference ∆ϕ = ϕB - ϕA between them is clearly defined. We call such a pair of conductors a capacitor and the two conductors the capacitor’s plates (even if they are not shaped like plates). We define the capacitor’s capacitance C to be



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Q ​ ​(E4.3) C ≡ ​ ____ ​      Δϕ

• Purpose:  This equation defines the capacitance C of a two-conductor object, where ∆ϕ is the potential difference between them that d ­ evelops when we move charge Q from one initially neutral object to the other. • Limitations:  This equation assumes that the two conductors are isolated from other charged objects. • Note:  The SI unit of capacitance is the farad, where 1 F ≡ 1 C/V.

E4.4 The Parallel-Plate Capacitor

63

A capacitor with a larger capacitance thus holds more charge on its plates for a given potential difference than one with smaller capacitance. This is the root idea behind the terms capacitor and capacitance. Capacitors are commonly used in electrical circuits to temporarily store electric charge. A capacitor’s capacitance C turns out to depend only on the size, shape, and arrangement of its plates and the nature of the insulating material in which the plates are embedded, not on the magnitude of the potential difference between those plates. Generally, C is hard to calculate for a given arrangement of conductors, though the next section and the homework problems discuss cases where this is possible. One can also talk about the capacitance of a single isolated conductor if we assume that the other “conductor” is essentially a spherical shell with infinite radius. The following example illustrates this idea.

Problem:  What is the capacitance (relative to infinity) of an isolated metal sphere with radius R?

The capacitance of a single conductor (one plate clapping)

Example E4.2

Solution  According to section E4.1, excess charge Q on an isolated spherical conductor will be uniformly distributed on its surface. According to the shell theorem, the potential field of such a distribution outside the sphere will be the same as that of a particle with charge Q at the sphere’s center. So the potential ϕ at the sphere’s surface relative to infinity (and thus the potential difference ∆ϕ between that surface and an imaginary spherical shell at ­infinity) is the same as that evaluated at a distance R from such a particle:

1 Q Δϕ = ϕ (at R) = _____ ​     ​ __ ​   ​  (E4.4) 4πε 0 R  

Substituting this into equation E4.3 yields

C = 4πε 0 R (E4.5)  

Note that the units of this expression work out:

(  ) (  ) (  ) (  )

1 ​J ​/   C ______ C2 1 N·m ______ 1F _____ ​ ​   2 ​  ​m ​ ______ ​   ​    ​ ​ ​   ​    ​ ​ ​     ​   ​= F (E4.6) V​  N·m 1 ​J ​  1 ​ V​  1 C/​

E4.4

The Parallel-Plate Capacitor

A parallel-plate capacitor consists of two identical flat conducting plates with charges Q and -Q separated by a small gap. This section will discuss this important example in some detail. Since the plates are conductors, each will (in static equilibrium) have zero electric field and thus a constant value of ϕ throughout their interior. Indeed (assuming we set ϕ ≡ 0 at infinity), if plates have charges of the same magnitude but opposite signs, their potentials will also have the same magnitudes but opposites signs. Let’s call these potentials ϕ 0 and -ϕ 0. Since the surface of each plate is an equipotential surface, equipotential surfaces just outside the plates should be nearly parallel to the plates’ surfaces. These surfaces also have to go around each plate: they cannot intersect the plate’s surface because it is an equipotential surface with a different value.

An equipotential diagram for a parallel-plate capacitor

Chapter E4

64

+1 V

+2 V

Static Equilibrium

+3 V

+4 V E

Positive plate 0V

Negative plate –4 V –1 V

E –2 V

–3 V

Figure E4.5 This diagram, which is based on a computer calculation, shows a cross-sectional view of the equipotential surfaces around two oppositely charged square conducting plates in static equilibrium. The plane of the drawing is a plane ­perpendicular to the plates that goes through their centers. The plates have potentials of ±4.6 V.

What does this diagram imply about the electric field?

Modeling the plates as infinite planes of charge

If we were to simply guess what the equipotential curves must look like in a plane perpendicular to the parallel plates, we would pretty much be forced to draw something like figure E4.5. Between the plates, the equipotentials going around each plate must be parallel to the plates and close together (as the potential must drop rapidly from ϕ 0 to -ϕ 0 over the short gap from the positive plate to the negative plate), but they will be comparatively quite spread out behind each plate (as the potential in these regions only needs to go from ±ϕ 0 to zero, and has an infinite distance over which to do so). The electric field vectors for this system are everywhere perpendicular to the equipotential lines and have magnitudes that depend inversely on the space between lines. In the gap between the plates, the equipotentials are horizontal, close together, and quite evenly spaced, so the electric field vectors there will be vertical (downward if the upper plate is positive), strong, and astonishingly uniform in strength. These statements remain true until we are within about one gap’s-width of the plates’ edges. Outside the gap, the field is much weaker. The spacing of equipotentials in figure E4.5 (where the plates are still fairly far apart) implies that the field is about 10 times weaker just outside the middle of either plate than it is just inside. If we were to reduce the gap between the plates, the equipotentials in the gap would be compressed even more, so the ratio of the field magnitude behind the plates to that between the plates would become even smaller. So in the limit that the plates’ separation becomes very small compared W to their size, the electric field E​ ​   between the plates becomes very strong and almost exactly uniform, and the electric field outside the plates becomes negligibly small. These are the core features of the parallel-plate capacitor. But what is the magnitude of the nearly uniform electric field between the plates? Ironically, we can answer this very practical question with the help of the infinite-plane model we discussed in chapter E2 (which probably seemed infinitely impractical at the time). To an imaginary microscopic being in the tiny gap between closely spaced plates, both plates would look essentially infinite, so they should have about the same field as that between two infinite, uniformly charged planes separated by the same gap. Supporting this idea

E4.4 The Parallel-Plate Capacitor

65

is the fact that the gap field in our practical case is almost exactly uniform and perpendicular to the plates, just like the field that a pair of infinite planes would create in the space between them. Conversely, the fact that only uniformly charged infinite planes create a uniform field implies that the charge distribution on our conducting plates in this case should likewise be uniform, something that is not obvious (but is supported by computer models). Though free excess charges on an isolated conducting plate would normally crowd toward the plate’s periphery, this tendency is overridden in this case by the charges’ passionate attraction to their opposites on the neighboring plate, to which they want to get as close as possible. Spreading themselves uniformly on the two plates’ facing surfaces allows them to get as close as possible to their opposites while maximizing the distance between their repellent neighbors. Now in the gap, the fields created by each plate add: if the positive plate is above the gap, then the field it contributes points away from that plate (and so downward in the gap), while the negative plate contributes a field that points toward the plate (also downward in the gap). As we saw in chapter E2, the uniform field that each individual infinite plane creates has a magnitude │ E​  W ​│    = │   σ│/2ε 0, where σ is the plane’s charge per unit area, so the two planes together should create a field a magnitude │E​  W ​│    = │σ│/ε 0 in the gap. In the region above or below the two plates, their fields tend to cancel. For example, in the region just above the positive plate, the positive plate’s field points away from the plate and thus upward, while the negative plate’s field points toward that plate and thus downward. In the model where we replace the plates by infinite planes, the field generated by each plane is strictly uniform, and so the fields exactly cancel. In the case of finite plates, the closer plate exerts a slightly stronger field, so the cancellation is not perfect, but this model does explain the weak external field we have already observed. In summary, our idealized model for the parallel-plate capacitor tells us that the electric field created by the plates has the magnitude

Charge is (almost) uniformly distributed on the plates

 

 



│σ│ │E​  W ​│    ≈ ____ ​   ​     between the plates, │E​  W ​│    ≈ 0 elsewhere

ε0  

(E4.7)

The electric field of a ­parallel-plate capacitor

• Purpose:  This equation describes the field between and around identical and oppositely charged parallel conducting plates, where σ is the charge per unit area on either plate and ε 0 is the permittivity constant. • Limitations:  These equations apply only if the gap between the plates is very small compared to their size. It also does not apply to points within about a gap’s-width from the plates’ edges. • Notes:  The electric field in the gap is almost exactly uniform and points perpendicular to the plates and away from the positive plate.  

+

We can now calculate the capacitance of the parallel-plate c­ apacitor. Let the plates’ separation be s. We can calculate the potential ­difference ∆ϕ between W · d ​r​ W for steps along a path c­ onnecting the plates. the plates by calculating ∫ ​E​ Since the uniform electric field in the gap between the plates points directly away from the positive plate, we can make things easy on ourselves by choosing a path that goes from the positive plate to the ­negative plate along a line perpendicular to each (see figure E4.6). Since E​ ​W is parallel to each tiny step d  ​ rW ​   W W │ │ │ │ W W along this path, E​ ​  · d ​ r ​ =  E​  ​    d ​ r ​  . Since the field’s ­magnitude remains constant W · d ​ rW ​  = │E​ as we follow the path, ∫​E​  W ​│   ∫│ d ​ rW ​│   = │E​  W ​│   s. Therefore,

+

+

Path –

Plate +

–

–

dr

E –

+

–

Plate

Figure E4.6 A path for calculating the potential difference between the plates in a parallel-plate capacitor.

66

Chapter E4



Static Equilibrium

W · d ​ rW ​ = │E​ Δϕ = −∫ ​E​  W ​│   ∫ dr = −│E​  W ​│   s  (between the plates)

(E4.8)

(the negative sign indicates that the final destination, the negative plate, is at a lower potential than the positive plate). Now, since the plates have uniformly distributed charge, the charge per unit area σ = Q/A, where A is the area of either identical plate. The definition of capacitance then implies that The capacitance of a parallelplate capacitor



| |

Q Q Q Q A C ≡ ​ ____ ​    ​ ​= _____ ​     ​  = _______ ​     ​  = ε 0 _______ ​     ​  ⇒  C = ε 0 __ ​   ​  (E4.9) │E​ Δϕ (Q/A)s s  ​W│    s (σ/ε 0)s  

 

 

The same results apply qualitatively to more general capacitors, whose capacitance also generally increases as the plate area increases and/or the plate separation decreases (though the dependence is usually not as simple).

Exercise E4X.2 What is the capacitance C of square plates 1 m on a side separated by 1 mm?

E4.5

Calculating the force that one plate exerts on the other

The Electric Field as a Form of Energy

Relativity (see Unit R) tells us that a particle’s mass is a form of energy, meaning that a particle is really just energy in a compact form. In this section, we will see that an electric field is a dispersed form of energy, and we will derive an expression for the density of energy stored in an electric field. This emphasizes that an electric field is a physically real thing, as real as a particle, not just some kind of mathematical abstraction. To see this, let’s focus initially on a single one of the excess electrons on the negative plate. That electron will be trapped on that plate’s surface closest to the positive plate. The plate’s other electrons will exert purely horizontal forces on our electron, so the net electrostatic force on it will be entirely due to the positive plate’s field at its location. This field has a magnitude of │  E​  W ​ +  │ = σ/2ε 0 = Q/(2ε 0 A), where Q is the positive plate’s charge and A is its area. The net force on the one electron is directed toward the other plate and has a magnitude │F​  W ​ 1  │ = e│E​  W ​ +  │ = e(Q/2ε 0 A), where -e is the electron’s charge. Now if the gap between plates is small, the field over virtually all of the plate’s surface will have this same magnitude and direction (only very near the edges will it be much different). Therefore, the magnitude of the total force on the negative plate is the simple sum of all the forces acting on all N excess electrons on the plate. Since the plate has a total charge of -Q = N(-e), we have Ne = Q. Therefore, the total force acting on the plate has magnitude  

 

 



NeQ 2ε 0 A  

Calculating the energy needed to increase the volume that the field occupies

Q2 2ε 0 A

│F​  W ​ -  │ = _____ ​     ​ = _____ ​    ​   (E4.10)  

Now imagine that we slowly pull the negative plate directly away from the positive plate a small displacement Δ ​ rW ​.  To do this, we will have to exert an external force F​ ​We xt that opposes the electrostatic force F​ ​W-  we have just calculated. Since the magnitude of F​ ​W-  will not change significantly during this process (as long as the gap between plates remains small), the work we do on the two-plate system is W = F​ ​We xt · Δ​ rW ​ = │F​  ​We xt││ Δ r​W ​│    = │F​  ​W-  ││ Δ r​W ​│   , because F​ ​We xt is parallel to Δ ​ rW ​ and │F​  ​We xt│ = │F​  ​W-  │. This is energy flowing into the system. But where does this energy go? At the end of the displacement, the plates are again at rest, and nothing has changed except the arrangement of the plates. Even the electric field E​ ​W  in the now slightly larger gap is unchanged.  The

E4.5 The Electric Field as a Form of Energy

67

only thing that has changed is that we have increased the volume of the gap by A│ Δ ​ rW ​│    and filled that volume with an electric field of total magnitude (according to equation E4.7) of │E​  W ​│    = σ/ε 0 = Q/ε 0 A. The energy that entered the system has plausibly gone to creating this region of the electric field! We can calculate the energy density of this electric field by dividing the total amount of energy it contains (which is equal to the work that we put in) by the field volume we have created:  

 

(  ) (  )

│F​ W 1  ​W  ││Δ ​ rW ​│     __     ​  uE = ____ ​    ​ = __________ ​  = ​    ​  ​ A│Δ ​ rW ​│   ΔV A



Q2 ε ​ _____   ​  ​= __ ​  0 ​  ​​ 2ε 0 A 2  

 

Q ε W 2 ____ ​     ​  ​ ​= ​ __0 ​ │E​  ​ │     (E4.11) 2

ε 0 A  

 

2

We have shown this to be true only in the special case where the electric field is uniform, but it turns out that the same result applies in more general cases (see problem E4D.2). Therefore, quite generally,

uE = _​ 12 ​  ε 0│E​  W ​│   2 (E4.12)



 

• Purpose:  This equation expresses the energy density uE of an electric field in a region where the electric field has the magnitude │E​  W ​│   . The constant ε 0 is the usual permittivity constant. • Limitations:  Though this formula is technically correct even inside materials, its interpretation becomes complicated in such circumstances.

The density of energy in an electric field

 

Exercise E4X.3 Check that equation E4.12 has the right units to yield an energy density. This result makes it clear that a charged capacitor stores energy. Indeed, the energy stored in a parallel-plate capacitor whose plates have area A and are separated by s (and thus a volume As where the field is nonzero) is

(  )

│ ​W│ U = uE As = _​  12  ​  ε 0│E​  ​W│   2 As = _​  12 ​  ε 0 ​ __ ​ A   s)2 = _​ 12 ​  C(Δϕ)2 (E4.13) s ​   ​( E​



 

 

where I have used equations E4.8 and E4.12. Again, while we have derived this result for the parallel-plate capacitor, it applies generally:



U = ​ _12 ​  C(Δϕ)2 (E4.14)

• Purpose:  This equation describes the energy U stored in the electric field of a charged capacitor in terms of its capacitance C and the potential difference ∆ϕ between its plates. • Limitations: None!

The fact that capacitors can store energy as well as charge makes them useful in many ways. They are like rechargeable batteries, except that they can be charged and discharged at much higher rates (since no chemical reactions are involved). This makes them useful for applications like electronic camera flashes. Because this would also be ideal for powering electric cars, a number of companies are currently developing “ultracapacitors” that (if they can be made economical) might one day replace batteries in such cars.

Energy stored in a capacitor

Chapter E4

68

Static Equilibrium

TWO-MINUTE PROBLEMS E4T.1 Consider the situation shown below. Object A is a metal sphere with zero net charge, and objects B and C are particles with charges q and -q, respectively. Point P is at the sphere’s center. Answer each question below by choosing one of the arrows to the right of the picture or by choosing answer E. (If you are using the letters on the back cover to display your answer, you can indicate the double-letter choices by pointing to a letter with two fingers.) A DD

P

B

CC

Sphere A

C

BB B +

D

C

AA

–

E: Zero

(a) Which arrow best indicates the direction of the electric field at point P created by particle B? (b) Which arrow best indicates the direction of the electric field at point P created by particle C? (c) Which arrow best indicates the direction of the electric field at point P created by the charge distribution on the sphere’s surface? (d) Which arrow best indicates the position (relative to P) of the greatest concentration of positive charge on the sphere’s surface? E4T.2 Consider the situation below. Object A is a metal sphere with zero net charge, and objects B and C are particles with the same negative charge. Which choice shown to the right best indicates the direction of the electric field at point P? (If you are using the letters on the back cover to indicate your answer, you can indicate the double-letter choices by pointing to a letter with two fingers.) A DD

B

CC

B

–

Sphere A P

C

BB

D

C

AA

–

E: Zero

E4T.3 Are the following statements true (T) or false (F)? (a) The electric field inside a conductor must be zero in all possible circumstances. (b) In static equilibrium, any excess charge on a conductor is uniformly spread on its surface. (c) A closed metal can (in static equilibrium) will perfectly shield a point in its interior from the static field of any external charged particle(s). (d) If the electric field at any point inside a conductor is nonzero, the metal is not in static equilibrium.

E4T.4 Because of electrostatic polarization, a sufficiently charged rubber balloon will attract a neutral horizontal insulating plate strongly enough so that the balloon sticks firmly to the plate’s bottom. If the horizontal plate is a conductor instead of an insulator, the charged balloon will A. stick only if the plate is connected to the earth. B. stick only if the plate is insulated from the earth. C. stick in either case. D. not stick at all. E. repel the plate. E4T.5 Imagine that we have two identical metal spheres on insulating stands that we give the same negative charge. If we slowly bring the spheres closer and closer together, the charge per unit area σ on each sphere will at the point nearest the other sphere A. become increasingly negative. B. become less negative. C. remain constant. E4T.6 Two metal spheres on insulating stands have the same positive charge Q. If we reduce the distance between the spheres’ centers by a factor of two, the magnitude of the repulsive force between the spheres increases by A. a bit less than a factor of two. B. exactly a factor of two. C. a bit more than a factor of two. D. a bit less than a factor of four. E. exactly a factor of four. F. a bit more than a factor of four. E4T.7 Suppose we suspend a negatively charged ball like a pendulum inside a hollow but completely enclosing metal box. Assume that the ball hangs a few inches from the box’s left vertical face. Imagine that we now bring a very strong positively charged object to a point just to the left of the box’s left face, so that it is only a few inches away from the suspended ball (though the box’s left face is between them). The ball will be deflected A. strongly toward the external object. B. strongly away from the external object. C. weakly toward the external object. D. weakly away from the external object. E. not the least little bit. E4T.8 Suppose the two plates of a parallel-plate capacitor have fixed charges Q and -Q. If we double the gap between the plates, the (a) potential difference between the plates, (b) the capacitor’s capacitance, and (c) the energy stored in the capacitor A. quadruples. B. doubles. C. remains the same. D. is reduced by a factor of two. E. is reduced by a factor of four. F. some other result (specify).

Homework Problems

69

HOMEWORK PROBLEMS Basic Skills E4B.1 Suppose we suspend two identical metal balls by insulating threads. One has a negative charge but the other is uncharged. If these balls are brought close together, they will attract each other, touch, and then repel each other. Using diagrams, explain why the balls do this. E4B.2 Two parallel circular conducting plates 30 cm in diameter are separated by 0.5 cm and have charges of +22 nC and -22 nC, respectively. What is the magnitude of the electric field between these plates? E4B.3 Two identical neutral parallel plates, each with an area of 0.20 m2, are separated by 0.5 cm. Imagine that we move 2 µC of charge from one to the other. What is the potential difference between these plates now? E4B.4 The positive and negative plates of a parallel-plate capacitor hold charges of +0.10 µC and -0.10 µC when the potential difference between the plates is 10 V. If the plates are separated by 1.0 mm, what is their area? E4B.5 A capacitor with two square plates 1 m on a side has a capacitance of 1 F. What is the separation between the capacitor’s plates? Will it be easy to build such a capacitor? E4B.6 Argue that the permittivity constant ε0 must have units equivalent to farads per meter. E4B.7 If the atmospheric electric field on a certain cloudless day is 200 N/C (a typical value), how much electric field energy does a cubic meter of atmosphere contain? E4B.8 Imagine that our startup company wants to create a parallel-plate ultracapacitor that stores roughly the same energy as a tank of gasoline (something like 3.0 GJ) when charged to a potential difference of 100 V. It is not plausible that the gap between plates could be less than a few atom’s widths, so let’s assume that the gap is 2 nm. If the plates are in a vacuum, what would their area need to be? (Using an appropriate insulating material to separate the plates might help reduce this area: see problem E4D.3.)

Modeling E4M.1 Imagine that you have two identical neutral metal spheres on insulating stands and a charged rubber rod. Using diagrams, carefully describe a step-by-step procedure that you can execute to give the two spheres exactly opposite charges (all without touching either sphere with the rod). Take care that your procedure ensures that each sphere will end up with charges having opposite signs but exactly the same magnitude, not just approximately the same. (Hint: Start with the spheres touching.)

E4M.2 Imagine a small, initially neutral metal ball hanging as a pendulum bob between two oppositely charged metal spheres on insulating stands. If the spheres have sufficiently large opposite charges, the ball will begin to swing rapidly back and forth, hitting first one sphere and then the other. After a certain time (typically a few seconds) the ball stops. Carefully explain (with the help of appropriate diagrams) what drives the ball to initially start swinging, why it stops, and why it does not make just one swing. E4M.3 Consider a capacitor whose plates are squares 10 cm on a side separated by a 1.0-mm gap filled with air. (a) What is the maximum charge we can put on these plates before a spark will fly? (b) What energy is released by the spark when it does fly? E4M.4 Imagine that we want to construct a 100 µF parallelplate capacitor whose metal plates have an area of 1 m2 and are separated by air. (a) What should the plate separation be? Does this seem realistic? (b) If the potential difference between the two plates is only 1.0 V, what is the magnitude of the field between the plates? Air breaks down and conducts electricity at field strengths above 3 × 106 N/C. Is this a problem? E4M.5 Imagine that a parallel-plate capacitor’s upper and lower plates have a charge density of σ0 and -σ0, respectively. The potential difference between the plates is ∆ϕ0 and the capacitor’s capacitance is C0, and stores an energy U0. Suppose we insert into the gap an uncharged metal slab whose thickness is half of the gap’s thickness, leaving gaps of 1/4 the original gap thickness above and below the inserted metal plate. The inserted slab does not touch either plate during the insertion process. We then allow everything to come into static equilibrium. (a) What (in terms of σ0) are the densities of charge on the upper and lower faces of the inserted slab? (b) What is now the potential difference (in terms of ∆ϕ0) between the original capacitor’s plates? (c) What is the capacitance of the total device after the slab has been inserted? (d) What is the energy stored in the device’s electric fields after the slab has been inserted (as a fraction of U0)? (e) Where has the missing energy gone? Explain your reasoning in each case. E4M.6 What is the total energy in the electric field created by an isolated metal sphere with radius R and charge Q? (Hint: Treat the sphere as a capacitor.) E4M.7 A neutral metal spherical shell with inner radius R2 and outer radius R3 encloses a concentric metal sphere with radius R1 and total charge Q. Calculate the potential (relative to infinity) of the inner sphere’s center in terms of Q, R1, R2, and R3, and explain your reasoning.

Chapter E4

70

Static Equilibrium

E4M.8 In this problem, we will consider some models that might explain the physics of sparks, one of the most interesting everyday electrostatic phenomena. This is a case where careful reasoning and calculation can rule out some naive models about how sparks work. (a) One conceivable model for what happens when a spark flies between two conductors is that electrons in the negative conductor leap all the way to the positive conductor across the macroscopic gap between the conductors. We can rule out this model as follows. A free electron moving through air will eventually collide with any air molecule that is within a distance R of the electron’s path, where R is the effective radius of a typical air molecule (modeled as a sphere). Imagine the electron as a disk of radius R that sweeps out a cylindrical volume as the electron moves (see below). The electron will hit a molecule within distance d if the center of at least one molecule is within the cylinder. d R electron

R molecule

At standard pressure and temperature, 22.4 liters of air contains Avogadro’s number of molecules. A typical molecule might have an effective radius of about 0.15 nm. Use this to show that the average distance an electron might travel before hitting a molecule is d ≈ 0.5 µm. So an electron cannot travel very far in air. (b) If a sufficient fraction of air molecules are ionized into free electrons and positive ions, then individual particles don’t need to move far to transfer a decent amount of charge. Model the free electrons and positive ions as separate cylindrical solid lattices that move a distance of d ≈ 0.5 µm relative to each other in response to the field between conductors. Free electrons emerging from one end of the cylinder would then be collected by the positive conductor, and ions emerging from the cylinder’s other end would be collected by the negative conductor. This would have the same effect as directly transferring charge from one conductor to the other. Show that if all its molecules were ionized, a cylinder of air with a radius of 0.4 mm would be sufficient to transfer 1000 nC of charge, even if no particle moves more than one collision distance d. This seems like a plausible radius for a spark. (Of course, more charge could be transferred if the air molecules get re-ionized after each collision, so that the process can repeat.) (c) So how do a sufficient fraction of air molecules get ionized? We observe that a spark will fly when the field strength in air exceeds about 3 × 106 N/C. Is such a field strong enough to simply rip electrons out of an atom? Let’s model an atom as a single outer electron of charge -e attracted to the rest of the atom (nucleus and other electrons) modeled as a sphere of radius 0.1 nm with charge +e. Show that the external electric field magnitude needed to rip off the outer electron directly would have to exceed about 1.4 × 1011 N/C.

(d) This is about a factor of 50,000 too large, so the atoms must be ionized in some other way. Show that in the atomic model discussed in the previous part, the energy required to pull the outermost electron away from the rest of the atom to infinity is about 2.3 × 10-18 J. (e) Show that in an electric field of magnitude 3 × 106 N/C, an electron would have to travel only about 10 times the typical collision distance d ≈ 0.5 µm to accumulate this much kinetic energy. The accepted model of spark creation is as follows. A body of air always has a few free electrons created by collisions with high-energy particles (cosmic rays) from outer space. In a sufficiently strong electric field, a fraction of these electrons happen by chance to travel far enough to ionize other air molecules. In the absence of other processes, one such free electron hitting a molecule would create two free electrons, which by colliding with other molecules would create four, then eight, and so on. This would grow to an avalanche that ultimately ionizes enough air to transfer a significant amount of charge. Of course, many of the electrons so created collide with molecules and lose energy before they create new electrons. A field of 3 × 106 N/C happens to be strong enough in air so that the number of sufficiently energetic electrons grows by doubling faster than they are rendered ineffective by weakening collisions. Once the avalanche begins, it grows rapidly until a significant fraction of the air is ionized. The energy released by all the collisions heats the air to incandescence, which helps maintain the ionization and thus the flow of charge.

Derivations E4D.1 Consider a capacitor consisting of two concentric spherical metal shells where the inner shell has an outer radius of R1 and the outer has an inner radius of R2. (a) Prove that the capacitance of this capacitor is

R1R2 C = 4πε 0 ​ _______   ​   R2 − R1  

(E4.15)

(b) Compare and contrast this result with equation E4.9 for the parallel-plate capacitor. (c) Show that this result becomes consistent with equation E4.5 in the limit that R2 goes to infinity. E4D.2 Consider the capacitor discussed in problem E4D.1. (a) Use equation E4.15 and equation E4.14 to calculate the energy stored in the capacitor if its inner “plate” has charge Q and the outer charge -Q. (b) Prove that this energy is equal to the energy contained in the capacitor’s electric field according to equation E4.12. (Hint: Since the electric field varies in magnitude in this region, you will need to do an integral.) This shows that equations E4.12 and E4.14 work even for nonuniform fields. E4D.3 Imagine that we place a slab of polarizable insulating material between the plates of a parallel plate capacitor. The uniform electric field between the charged plates will polarize the material, turning its atoms into dipoles.

Answers to Exercises

Plate

+ –

+

+ –

71

+

+ –

Insulator Plate

+

–

–

+ –

+

+ –

+

+ –

Surface charges + –

–

–

+ –

–



C = __ ​  ε  ​  Cvac (E4.17) ε0



where Cvac is the capacitor’s capacitance when the plates are separated by a vacuum. Since ε/ε 0 can be quite a bit larger than 1, this presents a way of dramatically increasing the capacitance of a given arrangement of capacitor plates.

+ –

 

Figure E4.7 Model the insulating slab between the plates as a rigid block of positive charge overlapping a rigid block of negative charge. In the electric field created by the charged plates, the insulator becomes polarized, which is as if the positive block shifts downward and the negative block upward. The bulk of the insulator remains neutral (light color), but leaves the top with a negative surface charge (gray strip) and the bottom with a positive surface charge (colored strip).

(a) Model the positive and negative charges in the slab’s atoms as being two rigid blocks of charge. Before the slab is polarized, these blocks of charge completely overlap, so the slab is neutral. When the slab’s atoms are polarized by an external electric field, it is as if the entire positive block moves a bit in the field direction and the negative block moves a bit opposite the field. In the slab’s interior, these blocks still overlap, so the slab’s interior remains neutral after polarization. But as the blocks move in opposite directions, each slab face adjacent to the capacitor’s plates end up with a surface charge. This is illustrated in figure E4.7. Use the superposition principle to argue that the net effect of these induced surface charges is to reduce the strength of the electric field in the slab. (b) The strength of this effect depends on the degree to which the insulating material is polarized by the original field between the plates. Let’s define an insulating substance’s permittivity ε so that ε W W = __ ​ E​ ​  0 ​  ​E​   (E4.16) ε vac  



where ε0 is the normal (vacuum) permittivity constant,​ W is the electric field inside the insulator, and E​ E​ ​W vac is the electric field that would exist if the capacitor plates were in a vacuum. It turns out that this result applies quite generally, not just when the insulating slab is in a parallel-plate capacitor. We call the ratio ε/ε 0 the insulator’s dielectric constant. Note that since │E​  ​W│     λ is as shown in figure Q3.9 for points far from the slit (compared to a). This single-slit diffraction pattern has bright and dark spots superficially like a two-slit interference pattern. However, note that the central bright spot is much brighter than, and twice as broad as, the flanking bright spots (which physicists call fringes). Note that almost all the wave energy is within the central peak, and thus within the angular range ± sin−1 (λ/a). In the small-angle limit where sin θ ≈ θ, this angular width is simply 2λ/a (in radians). Notice also that when a = λ, the central peak expands to cover the entire angular range ± sin−1(1) = ±90°, consistent with my earlier assertion that if a < λ, the waves are diffracted into circular waves with not much variation in amplitude as a function of angle. Now we can understand the variation in the spot brightness in the interference pattern shown in figure Q3.8c. Light emerging from each slit fans out according to the single-slit diffraction pattern. If the slits are the same width and are close together compared to the distance to the display screen, their single-slit interference patterns almost exactly overlap on the screen. So in the absence of interference between the waves emerging from two slits, the combined intensity pattern would be essentially the single-slit pattern shown in figure Q3.10a. Since a spot’s brightness in the two-slit interference pattern will depend on how much light gets there in the first place to constructively interfere, the brightness of such a spot will be modulated by how bright the light in the ­single-slit pattern is at that point, as illustrated in figure Q3.10b. Figure Q3.9 A graph of intensity versus sin θ for waves going through a single slit of width a > λ. This graph assumes that points where the intensity is measured are very far from the slit compared to a. The inset shows a computer simulation of the approximate diffraction pattern created by waves moving through a slit such that a = 4λ. (Credit: John Wetzel)

I

sinθ –3λ/a

–2λ/a

–λ/a

λ/a

2λ/a

3λ/a

Q3.5 Diffraction Revisited

45

I

Figure Q3.10

(a)

–2λ/a

–λ/a

(b)

–2λ/a

–λ/a

I

λ/a

2λ/a

λ/a

2λ/a

sin θ

sin θ

(c)

(a) A graph of intensity versus sin θ for light emerging from a single slit of width a. (b) A graph of the intensity versus sin θ for light emerging from two slits of width a separated by a distance d = 5a. Since the light required to generate the two-slit interference pattern has to come from the light provided by each single slit. The single-slit pattern provides the maximum possible intensity for the two-slit pattern. (c) A photograph of an actual two-slit interference pattern. (Credit: © David Haley)

Diffraction explains qualitatively why you can hear someone talking around a corner even when you can’t see the person. Figure Q3.9 suggests that longer wavelengths diffract more widely than shorter wavelengths. Sound waves have wavelengths on the order of magnitude of 1 m, so when sound waves go through an open doorway (which is also about a meter wide), they diffract pretty well in all directions, but light (whose wavelength is much shorter) is hardly diffracted at all. This effect also explains why voices around a corner may sound “muffled”: the higher-­frequency components of sound that add sharpness and clarity to a voice may not diffract as well to your position as lower-frequency components do.

Problem:  Light with a wavelength λ = 633 nm goes through a narrow slit of width a = 0.02 mm and is then projected on a screen D = 3.0 m from the slit. How wide (in centimeters) is the central bright spot on the screen? Solution According to figure Q3.9, the total angle spanned by the central maximum is θ = 2 sin−1 (λ/a). Let w be the width of the central bright spot. Using the small-angle approximation, sin θ  ≈  θ, so θ  ≈  2λ/a. Moreover, w ≈ Dθ = the arclength spanned by the angle θ on the screen. Therefore

(  )

2 (633 × 10-9 / ​  ) m​ λ 2λ   ​= 19 cm(Q3.11)   w ≈ Dθ = D2 sin-1 ​ __ ​   ​   ​≈ D ​ ___   ​= (300 cm) ​ _______________       a a 0.020 × 10-3 m​ ​/  The units are right. Note also that λ/a = 0.032, which justifies our use of the small-angle approximation.

Example Q3.3

Chapter Q3

46

Q3.6 Diffraction by a circular aperture

Interference and Diffraction

Optical Resolution

Figure Q3.10 on the previous page displays the diffraction pattern of light going through a rectangular slit that is much longer than its width a. It is more difficult to calculate what happens to light when it goes through a circular aperture, but fig­ure Q3.11a shows the result. We see a bright circular central region flanked by weaker circular fringes. It turns out that the angle of the innermost dark ring where complete cancellation occurs is given by λ ​ (Q3.12) sin θ = 1.22 ​ __ a



(a)

(b) ‒3 ‒2 ‒1 0 1 2 units of λ/a

3

sin θ

Figure Q3.11 A computer simulation of the ­diffraction pattern created by light going through a small circular aperture. (Credit: T. Moore)

Raleigh’s criterion for resolving point sources

where a in this case is the aperture’s diameter. This angle is only a bit larger than the angle that would be predicted for a rectangular slit (sin θ = λ/a). The fact that light is diffracted by a small opening has important implications for the ability of an optical instrument (such as a telescope or an eye) to resolve two distant objects whose angular separation is small. For example, imagine looking at two point sources of light (perhaps two stars) separated by a small angle θ, as shown in figure Q3.12. The light from each source is diffracted somewhat as it goes through your pupil. This means that even if your lens focuses the light as well as possible, the light from each source creates a small but spread-out diffraction pattern on your retina. Figure Q3.13 shows what the resulting diffraction patterns might look like on your retina for different angular separations between two sources. One can just begin to see that the sources are separate objects in the case shown in figure Q3.13b, where the central maximum of one diffraction pattern overlaps the first minimum of the other. Therefore, we can see the sources as separate only when their angular separation θ is such that

( 

)

θ > sin-1 ​ _____ ​  1.22λ  ​(Q3.13) a ​  



• Purpose:  This equation specifies the minimum separation angle θ that sources emitting light at an average wavelength λ can have if they are to be resolved by an optical instrument whose aperture has diameter a. • Limitations:  This equation is an approximation that becomes better if the sensor screen is far away compared to the aperture’s diameter a. • Note:  Equation Q3.13 expresses what is called the Rayleigh ­criterion.

Figure Q3.12 If two stars are separated by a sufficiently large angle, then the diffraction patterns produced on the retina when the light from the stars goes through the pupil do not overlap very much, so the retina will register these images as being separate. (The “bumps” on the retina in this drawing are meant to indicate graphs of the light intensity versus the position on the retina.) If, however, the angle becomes much smaller, the diffraction patterns will begin to overlap, causing the two stars to look like a single blob.

Eye Diffracted image of star B

Iris Light from star A star B

θ θ Pupil

Lens

Diffracted image of star A

Retina

Q3.6 Optical Resolution

47

0.083 0.067 0.050 0.033 0.017 (b)

(a)

(c)

(d) 0.000

Figure Q3.13 A computer simulation showing what the diffraction patterns for two point sources look like as their angular separation decreases. In (b), where the sources are separated according to the Raleigh criterion, the patterns can be barely distinguished as separate. The scale in (d) shows how the intensity values (as a fraction of a single pattern’s peak intensity) relate to the gray levels (fractional intensities above 0.083 are pure white). (Credit: T. Moore)

We call an instrument good enough to resolve sources separated by this angle diffraction-limited. The resolving power of a real instrument may be worse because of other factors, but it cannot be better than this. Since our ancestors depended so much on their eyes for staying alive, evolution has given us eyes that operate pretty close to the diffraction limit in daylight. The Hubble and soon-to-be-launched James Webb space telescopes, by virtue of their position above the atmosphere, also operate at close to the diffraction limit of their apertures. (The resolution of ground-based telescopes is limited more by turbulence in the earth’s atmosphere, though a modern adaptive optics system can correct for almost all of the turbulence.)

Problem:  At night, the pupil of a typical person’s eye opens up to as wide as 8 mm. What would be the smallest possible angular separation between two stars in the sky that the human eye might be able to resolve? Solution  In this case, the aperture that the light goes through is the pupil, so a ≈ 0.008 m. The eye is most sensitive to light whose wavelength is in the center of the visual range, so let’s estimate that λ ≈ 550 nm. If we assume that the eye is diffraction-limited, this gives us enough information to apply the Rayleigh criterion. Since the angle will be pretty small, we can also use the small-angle approximation sin θ ≈ θ. Equation Q3.13 implies that 1.22 (550 × 10-9 m) 1.22λ _____ 1.22λ _________________ θ min ≈ sin-1 _____ ​   ​    ≈ ​   ​    = ​        ​ = 8.4 × 10-5 (rad) 0.008 m a a  

 

( 

)( 

)

as       = (8.4 × 10-5  rad​ ​@) ​ _______ ​  360°  ​  ​​ _______ ​ 3600 ​   ​≈ 20 as (Q3.14) 2π  rad​ ​@  1° where 1 as = 1 second of arc. Note that I have expressed the answer to only one significant digit because the approximations we have made are not more accurate than that. The measured visual acuity of the dark-adapted eye is actually more like 200 as, because the eye averages the response of many retinal cells to respond to the dim light. (In bright light, the eye operates closer to the diffraction limit.) For comparison, earth-based telescopes are capable of resolving stars on the order of 1 as apart. The Hubble telescope can resolve stars on the order of 0.1 as apart, and the Webb telescope about 0.03 as apart at visual wavelengths.

Example Q3.4

48

Chapter Q3

Interference and Diffraction

TWO-MINUTE PROBLEMS Q3T.1 Waves from two slits S and Q will destructively interfere and cancel at a point P if the distance between P and S is larger than the distance between P and Q by A. λ B. nλ (where n is an integer) C.  ​ _12  ​  nλ (where n is an integer) D.   (n + _​ 12  ​)λ (where n is an integer) E. Other (specify) Q3T.2 Consider a two-slit interference experiment like the one shown in figure Q3.7b. The distance between adjacent bright spots in the interference pattern on the screen (A) increases, (B) decreases, or (C) remains the same if (a) The wavelength of the light increases. (b) The spacing between the slits increases. (c) The intensity of the light increases. (d) The width of the slits increases. (e) The distance between the slits and the screen increases. (f) The value of n increases (careful!). Q3T.3 Imagine that sound waves with a certain definite wavelength flow through a partially open sliding door. If the door is closed somewhat further (but not shut entirely), the angle through which the sound waves are diffracted A. Decreases. B. Increases. C. Remains the same. D. Depends on quantities not specified (explain). Q3T.4 Imagine that sound waves with a certain definite wavelength flow through a partially open sliding door. If the wavelength of the waves increases, the angle through which the sound waves are diffracted A. Decreases. B. Increases. C. Remains the same. D. Depends on quantities not specified ­(explain). Q3T.5 Line waves with wavelength λ going through a slit with width a will be diffracted into circular waves with approximately equal amplitude in all forward directions A. Always B. Never C. Only if λ >> a D. Only if a >> λ E. Only if a = λ

Q3T.6 If the two slits in a two-slit interference experiment are so far apart that their diffraction patterns do not overlap, the pattern displayed is consistent with a particle model of light. T or F? Q3T.7 Consider an experiment where we send monochromatic light to a distant screen through a single n ­ arrow slit. The distance between adjacent dark fringes in the diffraction pattern displayed on the screen (A) increases, (B) decreases, or (C) remains the same if (a) The wavelength of the light increases. (b) The intensity of the light increases. (c) The width of the slit increases. (d) The distance between the slit and the screen increases. (e) We look at fringes farther from the central maximum. Q3T.8 In the region where their beams overlap, two car headlights will create a clear interference pattern on a distant screen. T or F? Q3T.9 If we shine white light through two slits onto a distant screen, we will see a clear interference pattern. T or F? Q3T.10 Imagine we have an adjustable slit that we can make wider or narrower. If we shine light of a fixed intensity on the slit, and then crank the slit so its width decreases (while keeping other things the same), will the light at the center of the central peak on the display screen (A) remain the same, (B) get brighter, or (D) get dimmer? Q3T.11 Eagles and other predatory birds have very sharp eyesight. The pupil of an eagle’s eye is roughly twice the size of a human’s under similar (daylight) conditions. A friend claims to have read that an eagle’s eye has 8 times the resolving power of a human eye in broad daylight. This is physically impossible. T or F? Q3T.12 With an optically perfect 200-power telescope with a 1.5-inch-diameter tube, you can resolve objects roughly how many times closer together than you could with your naked eye? (Choose the closest answer.) A. 200 times B. 100 times C. 20 times D. 5 times E. No better at all

HOMEWORK PROBLEMS Basic Skills Q3B.1 Water waves with an amplitude of 0.80 cm and a wavelength of 2.5 cm go through two openings in a barrier. Each opening is 1.2 cm wide, and the openings are separated by 12.0 cm (center to center). Find the angles of the lines along which the waves constructively interfere.

Q3B.2 Sound waves with a frequency of 320 Hz are emitted by two speakers 44 cm wide and 3.5 m apart (­center to center). Find the angles of the lines radiating from the point halfway between the speakers along which the waves coming from the speakers interfere constructively. (Assume that the speakers are synchronized so that they emit wave crests at the same time.)

Homework Problems

Q3B.3 Suppose the distance between two slits in a given experiment is d = 0.050 mm and that the ­distance between the slit mask and the display screen is D = 1.5 m. If the distance between adjacent bright spots (for low n) is 2.0 cm on the screen, what is the wavelength of the light involved? Q3B.4 Suppose the distance between two slits in a given experiment is d = 0.040 mm and that the distance between the slit mask and the display screen is D = 2.5 m. If the ­distance between adjacent bright spots (for low n) is 3.0 cm on the screen, what is the wavelength of the light ­involved? Q3B.5 Suppose the light source for a two-slit interference experiment provides red light with a wavelength of 633 nm. If this light is sent through a pair of slits separated by a distance d = 0.040 mm, find the spacing between adjacent bright spots of the ­interference pattern when it is displayed on a screen 3.2 m from the slits. Q3B.6 Suppose the light source for a two-slit interference experiment provides yellow light with a wavelength of 570 nm. If this light is sent through a pair of slits separated by a distance d = 0.030 mm, show that the angle between the bright spot corresponding to n = 0 and the bright spot corresponding to n = 1 is about 1.1°. If the display screen is a distance D = 2.4 m from the slits, show that the distance between these bright spots on the screen is about 4.6 cm. Q3B.7 Explain in terms of wave concepts why sounds emitted from a person’s mouth can be heard almost equally well in all directions. Q3B.8 As you are just about to round a corner, you hear two people talking some distance around the corner. One has a deep voice, and the other has a high voice. Which voice more easily carries around the corner? Explain. Q3B.9 A stereo speaker 30 cm wide sounds a pure 1250-Hz note. Within what angle from the forward direction would you be able to hear this note (in a perfectly absorbing room)? Q3B.10 Ocean waves with an amplitude of 2.0  m and a wavelength of 15 m go through a 55-m opening in a breakwater that is shaped like a line and protects a body of water shaped like a square 800 m on a side between the breakwater, the beach, and two rocky ridges on either side. Draw a sketch of this situation that accurately and quantitatively illustrates the angle through which the waves are diffracted. (Also show your work in computing that angle.) ­Indicate some points on your sketch where boats within the breakwater will feel little wave motion. Q3B.11 Sound waves from a stereo system inside a house go through a partially open sliding patio door to the yard outside. If the door opening were 12 cm wide, what would be the lowest frequency of sounds that would not be diffracted in essentially all directions through that opening?

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Q3B.12 Light with a wavelength of 633 nm goes through a narrow slit. The angle between the first minimum on one side of the central maximum and the first minimum on the other side is 1.2°. What is the width of the slit? Q3B.13 Light with a wavelength 441 nm goes through a slit. On a screen 2.0 m away, the width of the diffraction pattern’s central peak is 1.5 cm. What is the width of the slit? Q3B.14 The beam of light emitted from a laser has a wavelength of 633 nm and an initial width of 1.0  mm. What is the diameter of the beam when it reaches the moon (384,000 km away)?

Modeling Q3M.1 You are setting up a pair of PA speakers on a field in preparation for an outdoor event. Each speaker is 0.65 m wide, and the speakers are separated by 8.2 m. To test the speakers, your coworker plays a single tone through the speakers whose frequency is 440 Hz. You are standing 52 m directly in front of the speakers and facing them. Roughly how far would you have to walk to your left or right to hear the sound amplitude drop almost to zero? How much farther would you have to go to hear the ­amplitude go back to its original strength? Q3M.2 Two radio antennas 60 m apart broadcast a synchronized signal with a frequency of 100 MHz. Imagine that we have a detector 5.0 km from the a­ ntennas. At this distance, what is the separation between adjacent “bright spots” in the interference pattern along a line parallel to the line between the antennas? Q3M.3 If we hold two flashlights parallel, will they create a clear interference pattern in the region where the two beams overlap? Explain at least two reasons why not. Q3M.4 When you connect stereo speakers to an amplifier, it is important that the speakers be connected in phase, so that if a signal from the amplifier pushes the cone of one speaker out at a given time, it pushes the cone of the other speaker out at the same time. Reversing the two wires connecting one of the speakers to the amplifier will make it so that the same signal pushes one speaker cone out but pulls the other one in. Explain why this could be a problem, or at least undesirable. Would you still be able to hear the music? Q3M.5 Two sets of sinusoidal line waves approach each other from opposite directions. These waves have exactly the same amplitude and wavelength. When they overlap, will they constructively interfere, cancel each other out, or do something else? Describe as carefully as you can what will happen when these waves meet. Q3M.6 How big a speaker would you need to create a directed beam of sound waves with a frequency of 440 Hz whose total width increases by only 5 m for every 100 m the beam goes forward?

50

Chapter Q3

Interference and Diffraction

Q3M.7 Some old-model CRT television sets produce a “whistle” sound at about 15,800 Hz (this is the ­frequency with which the electron beam sweeps across the face of the screen). This sound is audible to most youngsters and some adults who haven’t lost their high-frequency hearing. Imagine that a TV set is on (with the “mute” on) in your sister’s bedroom as you walk by in the hallway. If your sister’s door is ajar, leaving an opening 6.0 cm wide, and your ear is about 1.5 m from the door as you walk by, for about how many centimeters of your walk will you be able to hear the TV “whistle” (if you can at all)? Q3M.8 The headlights on an approaching car are 1.4  m apart. At about what distance could you ­resolve them as being separate? Make appropriate estimates as needed.

 

Q3M.10 Let’s guess that a person can distinguish between letters of the alphabet, if he or she can resolve features of the letter roughly one-fourth the size of the letter. Let’s assume that in bright light, a person’s pupil is about 3 mm in diameter, and that the eye is most sensitive to light with a wavelength of 550 nm. (a) What is the approximate maximum distance that one could read letters that are 3 mm high? (b) The letters in headings (TWO-MINUTE PROBLEMS, HOMEWORK PROBLEMS; exercise numbers, etc.) in this text are about this height: check your calculation by direct observation and ­report the results. Q3M.11 The paintings of Georges Seurat consist of closely spaced small dots (≈2 mm wide) of pure pigment. The illusion of blended colors occurs at least partly because the pupils of the observer’s eyes diffract light entering them. Estimate the appropriate distance from which to view such a painting, considering the fact that art museums are usually very well  lit. (Adapted from Serway, Physics, 3d ed., ­Saunders, Philadelphia, 1990, p. 1096.) Q3M.12 In J.R.R. Tolkien’s The Lord of the Rings (volume 2, p. 32), Legolas the Elf claims to be able to accurately count horsemen and discern their hair color (yellow) 5 leagues away on a bright, sunny day. Make appropriate estimates and argue that Legolas must have very strange-looking eyes, have some means of nonvisual perception, or have made a lucky guess. (1 league ≈ 3.0 mi.)

Derivations Q3D.1 Consider a two-slit interference experiment where you measure the amplitude of the combined wave at points a certain constant distance D >> d from the center of two slits separated by distance d. Using the superposition principle, show that the squared amplitude of the combined wave as a function of θ is

 

)

(Q3.15)

where λ is the waves’ wavelength, and B is some positive constant. (Remember that the wave intensity is proportional to its squared amplitude.) You may assume that the sinusoidal circular waves created by each slit have approximately constant amplitude at all angles. (Hint: An appropriate trigonometric identity might help. Note that if D >> d, the amplitude of the sinusoidal circular waves received from each slit will be roughly equal.) Q3D.2 (Uses differential calculus.) The intensity of waves measured at a point a fixed (large) distance from a single slit of width a as a function of angle θ turns out to be

(  )

sin ψ 2 a sin θ      I (θ) = Imax ​​ _____ ​      ​  ​ ​ where  ψ ≡ π ______   ​        ​ (Q3.16) ψ λ  

Q3M.9 Under ideal conditions and when Mars is closest, estimate the linear separation of two objects on Mars that can barely be resolved (a) by the naked eye and (b) the Hubble telescope (whose main mirror is 1 m in diameter).

( 

d sin θ [A(θ)] 2 = B cos 2 ​ π ​ ______       ​   ​ λ



and λ is the waves’ wavelength. What is the peak intensity of the first fringe to the right or left of the central bright spot compared to that of the central bright spot? (Hint: You will need to solve a simple transcendental equation, either by trial and error or by using www.wolframalpha.com.) Q3D.3 Suppose we have a mask with N identical and equally spaced slits. Argue that the constructive-­interference angles are the same as for the two-slit interference pattern, no matter what N might be. (Increasing the number of slits makes the constructive interference peaks both brighter and narrower, but does not affect their spacing.) Q3D.4 A fairly simple model enables us to predict the multiple angles in the single-slit pattern where we have complete destructive interference. Just as we can model a single very narrow slit as a point-like source that creates circular waves, let’s model a wider slit as an evenly spaced set of N point-like sources that oscillate in unison. ­Figure Q3.14 illustrates a 12-source model for the slit (I chose N  =  12) because it is divisible by many integers). The distance between the sources is a/12, where a is the slit width, and the center of the slit is between source 6 and source 7. Now, consider waves moving from these sources toward some distant point P where the interfering waves are observed. The lines along which these waves move from each source will be almost exactly parallel if the distance to P is much larger than the slit’s width. Let the angle that these lines make with the horizontal direction be θ. Draw a line perpendicular to the lines that go to P that spans the slit’s width as shown in the figure. The angle that this line makes with the vertical axis is then also θ. The distance marked s on the diagram will then be s = a sin θ. Now, suppose the angle θ is such that s = λ. This means that the extra distance a wave has to travel from source 7 to P compared to what a wave has to travel from source 1 to P is ​ _12  ​  λ. Therefore (assuming that the sources are identically strong and oscillating in unison), the waves from source 1 will be exactly canceled by those from source 7. Similarly, the waves from source 2 will be canceled by those from source 8, as well as those from 3 and 9, 4 and 10, 5 and 11, and 6 and 12. So the contribution from each source is exactly canceled by that from another source, meaning there

Answers to Exercises

2 4 a

θ

6 8

To a distant point P

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foghorns are 1.7 km apart, flanking the entrance to a harbor. At a certain time, you notice that you hear the blasts from the foghorns simultaneously. After you have sailed at a steady heading for 22 min at a speed of 8.8 km/h (as measured by your boat’s speedometer), you hear the foghorns exactly out of phase (one honks, then the other, then the first, etc.). Roughly how far are you from the foghorns now? (Hint: Treat the “honks” as if they were wave crests.)

Rich Context

Q3R.2 You are designing a possible low-cost automatic aircraft guidance system for use in remote airstrips. Your idea is to put two radio transmitters 40 m apart on either side of the center of the airstrip (along a line perpendicular to the airstrip). The transmitters emit spherical sinusoidal radio waves that have the same frequency but are exactly out of phase (so that one emits a wave crest at exactly the time that the other emits a trough and vice versa). The transmitters run from batteries that are charged by solar power during sunny days. (a) Argue that the net radio signal detected by a receiver on the airplane will be exactly zero when the airplane is flying somewhere in a vertical plane that includes the airstrip. Therefore, the pilot can determine if the airplane’s direction of motion is aligned with the airstrip by seeing whether the received radio signal is null and remains null as the plane approaches. (b) For this system to be useful, the radio signal must not remain null when the plane is flying in any other direction near the airstrip, and should also increase as rapidly as possible away from zero as distance between the airplane and the vertical plane containing the airstrip increases. Discuss the restrictions that these criteria place on the radio waves’ frequency.

Q3R.1 You are at sea on a foggy night. You are trying to find out how far you are from the shore, but the fog is too thick to see anything and your GPS receiver is not working. After a certain time of aimless sailing, you dimly hear two separate foghorns on your port (left) side, perpendicular to your direction of travel. Each foghorn emits a short blast of sound at a pitch of 120 Hz every 2.00 s exactly, and each sounds about as loud as the other. Looking at your map, you see two foghorn locations marked plausibly near what you guess your location to be, and on the map the

Q3R.3 You are a detective interviewing someone who says that (while sitting on a porch that you know to be 600 ft away) he saw the masked suspect commit the crime in broad daylight. “How do you know it was the suspect?” you ask. “I clearly saw the ‘MOM’ that is tattooed on his upper arm.” “You’re sure it was that particular tattoo and not some other tattoo?” “Yes, I’m sure,” the witness replies. The suspect’s tattoo has letters that are 6 cm tall. Is the witness credible, or might he be framing the suspect? Justify your response and any estimates you might make.

10 12

s

Figure Q3.14 A multiple-source model for waves emerging from a single slit.

is completely destructive interference along the direction where a sin θ = λ  ⇒ sin θ = λ/a. (a) What if θ is such that a sin θ = 2λ? Argue that we again have complete cancellation, and specify the source pairs whose contributions cancel. (b) Do the same if θ is such that a sin θ = 3λ. (c) Argue that if, say, θ is such that a sin θ = _​ 32  ​  λ, though some source pairs will cancel each others’ waves (specify which), waves from other sources will survive, allowing a net wave to be observed at this angle. These results are consistent with figure Q3.9, which shows complete cancellation at angles where sin θ = nλ/a, where n is a nonzero integer (again, this only holds for nλ/a < 1).

ANSWERS TO EXERCISES Q3X.1 The energy a given wave crest contains is spread out over the circumference of a circle, which grows in proportion to the crest’s radius r. If the crest’s conserved energy is E, then the wave’s linear intensity (energy per unit time per unit length) is E/2πr ∝ 1/r. But since the waves’ linear intensity is also proportional to A2, we have A2 ∝ 1/r  ⇒  A = 1/r1/2. Q3X.2 Note that ϕ + α = 90° in figure Q3.6. In the limit that point P is very distant, θ + α will approach 90°. So in this limit, we have ϕ + α ≈ θ + α  ⇒  ϕ ≈ θ.

Q3X.3 Suppose we move point P in figure Q3.6 so that it continues to be the same point of constructive interference in the interference pattern as we vary the distance between the slits. Therefore, we must adjust things so that the distance between Q and R remains some fixed number of wavelengths and thus some fixed distance. So this leg of the triangle ∆QRS must remain fixed as the triangle’s hypotenuse changes. As d gets smaller, ϕ will have to get bigger to maintain a fixed distance between Q and R. Since θ ≈ ϕ if P is far away, the angle θ must also increase.

Q4

T   he Particle Nature  of Light Chapter Overview Section Q4.1:  A Short History of Light This section discusses the history of physical models of light: Newton’s particle model, Maxwell’s model of light as a special case of an electromagnetic wave, and modern quantum electrodynamics. We will see that the study of the nature of light has played a crucial role in every one of the great revolutions in physics.

Section Q4.2:  The Photoelectric Effect In the late 1800s, physicists discovered that ultraviolet light shining on a metal could eject electrons from the metal, a phenomenon physicists call the photoelectric effect. This effect has a number of technological applications in video cameras, image intensifiers, and sensitive photodetectors. Historically, the experimental and theoretical analysis of this effect (carried out by von Lenard and Millikan and Einstein, respectively) proved to be important in establishing a new particle model of light, which in turn led to the development of quantum mechanics.

Section Q4.3:  Idealized Photoelectric Experiments We might quantitatively explore the photoelectric effect by placing two metal plates in close proximity and illuminating one with light. If we connect an ideal ammeter to the plates, we can measure the current that flows between them due to ejected electrons. If we connect an ideal voltmeter across the plates, we can measure the electrostatic potential energy change per unit charge experienced by an electron moving between the plates. As electrons are ejected by the light from one plate (the cathode), some will make it to the other plate (the anode), making both plates charged. This charging process will continue until the potential energy required to travel between the plates is so large that no electrons have enough energy to make it all the way to the other plate. If the voltmeter is truly ideal, the measured potential difference per unit charge (in volts) will be exactly equal in value to the maximum kinetic energy an electron has in electron volts (abbreviated eV) when it leaves the plate, as explained in detail in the section. We can use these two experimental setups to determine how the number and kinetic energy of ejected electrons depend on the intensity and wavelength of the light and/or characteristics of the metal.

Section Q4.4:  Predictions of the Wave Model If light really is a wave, then (as discussed in detail in the section) we can make the following predictions about the outcomes of such experiments: 1. The rate at which electrons are ejected from an illuminated metal will be proportional to the intensity of the illuminating light. 2. At normal intensities, there will be a significant delay between the illumination of the metal and the emission of the first electrons. 3. The rate of electron ejection might depend in a complicated and metal-specific way on the frequency of the illuminating light (due to resonance effects).

52

4. The maximum kinetic energy of ejected electrons will almost certainly increase as the intensity of the illuminating light increases. 5. The maximum kinetic energy of ejected electrons might depend in a complicated way on the frequency of the illuminating light (due to resonance effects), but we would not expect to find any general trend in that dependence.

Section Q4.5:  Confronting the Facts When the experiments are actually performed, the following results emerge: 1. The number of electrons ejected from a metal illuminated with monochromatic light is proportional to the intensity of that light. 2. Electrons are ejected essentially instantly after illumination begins (with delays smaller than nanoseconds), no matter how low the intensity is (!). 3. If the frequency of the illuminating light is below a certain value, no electrons are ejected, no matter how intense the light is! (Physicists call this metal-specific value the metal’s cutoff frequency.) 4. For a fixed frequency of illuminating light, the electrons’ maximum kinetic energy is independent of intensity! 5. Above a metal’s cutoff frequency, the electrons’ maximum kinetic energy is directly proportional to the frequency of the illuminating light, with the same slope for all metals. As discussed in the section, most of these observations contradict the wave model.

Section Q4.6:  The Photon Model of Light In 1905, Einstein proposed a particle model of light to address this problem. In this model, Einstein claimed that light (and indeed all electromagnetic radiation) consists of particles (later called photons) whose energy is given by

E = __ ​ hc ​  (Q4.6) λ



• Purpose:  This equation specifies the energy E carried by a single photon of monochromatic light whose collective wavelength is λ; c is the speed of light = 3.0 × 10 8 m/s and h is Planck’s constant = 6.63 × 10−34 J·s. • Limitations:  This expression only applies to photons. • Note:  The value of hc is 1240 eV·nm (which are convenient units for many applications).  

Since f = c/λ for light, equation Q4.6 also means that E = hf. Einstein proposed that each electron ejected from the metal received its energy by absorbing a single photon, and thus that the kinetic energy K of the ejected photon was simply equal to E − W, where W is the metal’s work function (that is, the energy required to remove the electron from the metal). This model provided a simple and straightforward explanation for all the experimentally observed features of the photoelectric effect. In spite of its simplicity, physicists resisted this model for a long time, because it did not offer an explanation for the wavelike aspects of light discussed in chapter Q3. We will begin to reconcile these models in the next chapter.

Section Q4.7:  Detecting Individual Photons This section describes a modern device called a photomultiplier tube that uses the photoelectric effect to register individual photons of visible light. This device vividly illustrates the particle nature of light.

53

54

Chapter Q4

Q4.1

Development of the wave model of light

Development of the particle model of light

The Particle Nature of Light

A Short History of Light

Few physical phenomena are so commonplace and yet important to our daily experience as light. Yet light is so subtle that understanding its nature has challenged the scientific community for centuries. Just over three hundred years ago, Isaac Newton wrote a book on light (Opticks, 1704) that described a set of optical experiments he performed and the conclusions he drew from those experiments. Though this book made important lasting contributions (particularly with regard to the understanding of color), it is not as famous as his work in mechanics, partly because his particle model of light was (seemingly) contradicted by research done in the early 1800s by Thomas Young, Augustin-Jean Fresnel, and others, which demonstrated fairly conclusively that light was better understood as a wave (chapter Q3 discusses Young’s work on two-slit interference of light). James Clerk Maxwell provided a solid foundation for that wave model in the early 1860s in the course of developing a comprehensive mathematical theory of electric and magnetic fields. Maxwell predicted the existence of electromagnetic waves and was able to show that they had the same characteristics as light. Heinrich Hertz and others provided a firm experimental foundation for Maxwell’s theory by creating and studying electromagnetic waves with wavelengths outside the visible range. At the close of the 1800s, therefore, the physics community had thoroughly accepted Maxwell’s picture of visible light as electromagnetic waves having a certain wavelength range. The compelling beauty of Maxwell’s theory (in combination with experimental work that demonstrated the frame-independence of the speed of light) prompted Einstein to develop the theory of special relativity in 1905. At nearly the same time, however, evidence was beginning to emerge supporting a particle model of light. In 1901, for example, Max Planck showed that modeling a glowing hot object as emitting light energy in discrete bundles (instead of continually as waves) resolved a nasty problem physicists were having matching the observed emissions from such objects with predictions based on the wave model. Planck and most other physicists at first considered this model just a handy trick that approximated some wave processes not yet understood. However, independent support for a particle model of light came from studies of the photoelectric effect (the phenomenon of electron emission from illuminated metal surfaces), studies of the emission and absorption of light by gas atoms, and the discovery of the Compton effect (particle-like collisions between extremely high-frequency electromagnetic waves and electrons). Quantum mechanics (the subject of this unit) arose in 1926 from efforts to make sense of these results in a manner still consistent with the experimental results that seemed to justify Maxwell’s wave model. The successful unification of quantum mechanics, Maxwell’s electromagnetic field theory, and relativity led in the 1940s to the first relativistic quantum field theory, quantum electrodynamics (QED), the physics community’s currently accepted model for light. This theory in turn served in the 1970s as the template for our currently accepted quantum field theories of the weak and strong nuclear interactions. We see that the study of light (and the challenges it has raised) have provided either the impetus or the template for essentially all of the major revolutions in physics since the mid-1800s! It is appropriate, therefore, that we start our investigation of quantum mechanics by studying the nature of light. Chapter Q3 discussed how the phenomenon of two-slit interference firmly establishes the wave nature of light. In this chapter, we will see why the photoelectric effect in particular

Q4.3 Idealized Photoelectric Experiments

55

just as firmly establishes the particle nature of light. Subsequent chapters in this unit will develop the theory of (non-relativistic) quantum mechanics, the first theoretical model big enough to embrace both of these seemingly contradictory pictures.

Q4.2

The Photoelectric Effect

This chapter focuses on the photoelectric effect because it decisively demonstrates the particle nature of light but is easier to understand than Planck’s work on glowing objects and requires less background than the exploration of either the absorption/emission of light by atoms or the Compton effect. Though early photoelectric experiments were quite difficult to perform well enough to justify their radical implications (and were therefore mistrusted by the physics community), modern versions of these experiments clearly and decisively exclude wave models of light. The photoelectric effect was uncovered in bits and pieces in the latter decades of the 1800s, so it is hard to credit any single person with its discovery. In 1887, Hertz (who, ironically, was in the midst of his great work experimentally verifying aspects of Maxwell’s theory of electromagnetic waves) first observed that ultraviolet light shining on metallic electrodes made it easier for a spark to jump between them. Further work by J. J. Thomson and Hertz’s student Philipp von Lenard established by 1902 that many kinds of metal surfaces emit electrons when illuminated by light. Von Lenard’s qualitative observations of the phenomenon provided enough substance for E ­ instein to propose a particle model of light in 1905 to explain the observations. Robert Millikan strongly disbelieved Einstein’s particle model and almost immediately set out to disprove it. His brilliant decade-long effort (which involved, among other things, figuring out how to polish metals in a vacuum) instead quantitatively confirmed every prediction of Einstein’s model. Quite aside from its importance in establishing Einstein’s particle model of light, the photoelectric effect is worth studying because of its many modern technological applications. Electrons ejected by light incident on a surface can be electronically detected and amplified. This is exploited in a wide variety of devices that electronically detect, measure, and record visual images. A standard analog video camera, for example, uses the photoelectric effect to create an electronic “image” of the viewed scene that can be recorded and/or transmitted. Image intensifiers used by astronomers and the military employ the photoelectric effect to convert a very dim image to an electronic equivalent, which is then amplified and converted back into a brighter visual image. The most sensitive photodetectors currently available also exploit the photoelectric effect (as we will see in section Q4.7).

Q4.3

Idealized Photoelectric Experiments

Millikan in particular sought quantitative experimental answers to a set of questions raised by von Lenard’s qualitative observations of the photoelectric effect. How many electrons are ejected in a given time? How does this number depend on the light’s wavelength (that is, the color)? How does it depend on the light’s intensity (that is, the power delivered per unit area on the metal surface)? How energetically are the electrons ejected, and upon what does that ejection energy depend? Are the electrons ejected essentially instantaneously after the surface is illuminated, or is there a measurable time delay?

A basic description of the photoelectric effect

Technological applications

56

Chapter Q4

Figure Q4.1 (a) An idealized apparatus for studying the rate at which electrons are dislodged from a metal plate by light. (b) An idealized apparatus for measuring the kinetic energy of the liberated electrons. (c) A practical photoelectric sensor contains a curved cathode and a central wirelike anode mounted in a vacuum tube. This makes it easy for light to reach the cathode. (Credit: © McGraw-Hill Education/ photo by Chris Hammond)

How to investigate the rate of electron ejection

How to investigate the maximum kinetic energy of ejection

The Particle Nature of Light

Conductor (anode) Light

Electrons

A Ideal Ammeter

(a)

Metallic plate (cathode)

Light

Conductor (anode) – – – – – – – – – – b V a

(b)

+ + + + + + + + + + + Metallic plate (cathode)

Ideal Voltmeter

(c)

Figure Q4.1 illustrates a pair of idealized experimental setups that illustrate the essence of the approach that Millikan used to answer these questions. In both of these setups, monochromatic light (light with a well-defined wavelength) falls on a metallic plate, liberating electrons. (We call this plate a cathode, a word that generally means “something that acts as a source of electrons.”) Some of these electrons are captured by a nearby conducting plate (called an anode, which means “something that collects electrons”). In the first case (figure Q4.1a), these plates are connected by an ideal ammeter. An ammeter measures the electrical current (charge per unit time) flowing through it, which is proportional to the number of electrons flowing through it per unit time. An ideal ammeter does this without resisting the electrons’ flow. Electrons collected by the anode therefore freely flow back to the cathode through the ideal ammeter, and the current the ammeter measures reflects the number of electrons the anode collects every second. While not every electron liberated from the cathode ends up at the anode, the number collected by the anode should at least be proportional to the number liberated. This setup can therefore be used to explore how the rate of electron ejection depends on the intensity or color of the light and how soon after light starts to shine on the cathode a current begins to flow. In the second apparatus (figure Q4.1b), the two plates are connected by an ideal voltmeter. A voltmeter measures the electrostatic potential energy change per unit charge (expressed in joules/coulomb = volts) between the plates. An ideal voltmeter measures this without allowing any charges to flow through it, so that electrons collected by the anode remain on the anode. As light ejects electrons from the cathode, the cathode becomes positively charged, while electrons collected by the anode make the anode negatively charged. This generates an electric field between the plates, thus creating an electrostatic potential energy difference between them that makes it harder for subsequent liberated electrons to reach the anode (just as it would become harder to throw a ball to hit a ceiling whose height is increasing). Whether a given electron can make it to the anode against this electric field depends on how much kinetic energy the electron has to convert to electrostatic potential energy. If, for example, the electrostatic potential energy per unit charge between the plates is 2.0 V at a given time, an electron (which has a charge of 1.6 × 10−19 C) must have an initial kinetic energy of at least (2.0 J/C)(1.6 × 10−19 C) = 3.2 × 10−19 J to possibly make it to the anode with kinetic energy left over. (Even if it does, it might not make it to the anode if its trajectory is too flat. Because of its horizontal motion, the electron marked a in figure Q4.1b has some horizontal velocity at the peak of its trajectory, and

Q4.4 Predictions of the Wave Model

57

so was not able to convert all of its initial kinetic energy to potential energy. The electron marked b has a trajectory better suited for reaching the anode.) We can conveniently express an electron’s kinetic energy in units of electron volts (eV), where we define 1 eV to be the increase in a free electron’s kinetic energy when its electrostatic potential energy per unit charge decreases by 1 V. Since such an electron’s kinetic energy gain comes from a decrease in the electron/plate system’s potential energy, this definition implies that

The electron volt

(  )

1 J/​C​ @  -19 ​)   ​ _____ 1 eV ≡ (1.6 × 10-19 C​ ​@) (1 V​ ​   ​    ​= 1.6 × 10 J (Q4.1) ​ 1 V​ 

In these units, we could say that if the voltmeter registers a potential energy per unit charge of 2.0 V between the plates, then an electron will require at least 2.0 eV of kinetic energy at the cathode to reach the anode. The anode will stop collecting electrons when both plates become so charged that the energy an electron needs to reach the anode becomes barely larger than the maximum kinetic energy that any electron has when it leaves the cathode. The final value (in volts) that the voltmeter displays after we illuminate the cathode thus expresses the maximum kinetic energy (in eV) that electrons have when they are ejected by the cathode. We can therefore use this setup to investigate how the kinetic energy of ejected electrons depends on the intensity and/or color of the light shining on the cathode.

Q4.4

Predictions of the Wave Model

What does the wave model of light predict that the results of such experiments will be? In a wave model, electromagnetic waves falling on a metal plate should wiggle the plate’s electrons back and forth. If waves are intense enough, electrons might be wiggled right out of the plate. This picture would explain the basic phenomenon and is bolstered by the observation that hot metals (which also have violently wiggling electrons) are observed to emit electrons (a process called thermionic emission). Conservation of energy imposes some basic limitations on this model. A typical atom has a diameter d ≈ 0.1 nm. Assume that all the light energy falling on an atom gets channeled to one of its electrons (this the most energy that one electron can possibly receive). Breaking the bonds between the electron and its atom and/or the bulk of the metal (so that it can be ejected) plausibly requires a certain amount of energy E ej. The maximum possible rate for electron ejection from a metal plate illuminated by light will therefore be

A prediction regarding the rate of electron ejection

 

electrons ejected __________________________________ (energy delivered per electron)/(time) ​ _______________     ​   =      ​       ​ time energy needed per electron

( 

)

energy delivered ________________ ​    ​      ​  ​(area per electron) 2 time·area _________________________________ =      ​       ​ ≈ ___ ​ Id  ​   E ej energy needed per electron  

(Q4.2)

 

where I is the light’s intensity. We see that this model predicts that the ejection rate will be proportional to the light’s intensity. The inverse of this calculation yields the minimum time required for a given electron to accumulate enough energy to be ejected: time E ej ​ ________   ​  ≈ ___ ​  2  ​ electron Id  

(Q4.3)

 

In a typical experimental situation, this can take several seconds (see exercise Q4X.1). We would not expect any electrons to be ejected before this time.

Time delay for electron ejection

58

Chapter Q4

Other predictions

To make firm predictions about the average ejected electrons’ kinetic energy or how the ejection rate might depend on frequency, we need a more detailed model of the ejection process. In chapter Q2, we saw that oscillating systems accumulate energy most effectively from a disturbance whose frequency is close to a resonant frequency of the system, so we might expect to see that both the number and energy of ejected electrons depend in a complicated and metal-specific way on the frequency of the illuminating light. As a general trend, though, the ejected electrons’ kinetic energy should plausibly increase as the intensity of the light increases, because more intense light waves would wiggle electrons more violently. As an analogy, we might consider spray thrown around by an ocean wave crashing against a cliff: this spray’s kinetic energy clearly will get larger as the wave amplitude (and thus energy) increases. However, we would not expect the spray’s kinetic energy to change as the waves’ frequency changes: that energy is not plausibly affected by how often the ocean waves arrive, only by the waves’ amplitude. In summary, conservation of energy and the basic principle that energy is spread continuously throughout a wave imply that

Basic predictions of the wave model

The Particle Nature of Light

1. The rate at which electrons are ejected from an illuminated metal should be proportional to the intensity of the illuminating light. 2. At normal intensities, there should be a significant delay between the metal’s illumination and emission of the first electrons. Predictions that may depend on model details

In addition, the following conclusions are probable, but may depend on the microscopic details of how the wave actually ejects the electrons: 3. The electron ejection rate might depend in a complicated and metal-­ specific way on the illuminating light’s frequency (due to resonance effects). 4. The maximum kinetic energy of ejected electrons almost certainly should increase as the intensity of the illuminating light increases. 5. The maximum kinetic energy of ejected electrons might depend in a complicated way on the frequency of the illuminating light (due to resonance), but we would not expect to find any general trend in that dependence.

Exercise Q4X.1 Imagine that a lamp radiates a power of P = 40 W uniformly in all directions. At a distance of r = 1.0 m, the intensity of this light is I = P/4π r 2 = 3.2 W/m2. If a metal plate is placed at this distance from the source, and the metal’s atoms are about d  = 0.1  nm in diameter and it takes several eV to pry an electron out of the metal (as is implied by other observations), estimate the time needed for an atom to accumulate enough energy to eject one electron.  

Q4.5 What we actually observe

 

Confronting the Facts

Actually performing the experiments yields the following results: 1. The number of electrons ejected from a metal illuminated with monochromatic light is proportional to the intensity of that light. 2. Electrons are ejected essentially instantly after illumination begins (with delays smaller than nanoseconds), no matter how low the intensity is (!). 3. If the frequency of the illuminating light is below a certain value, no electrons are ejected, no matter how intense the light is! (Physicists call this metal-specific value the metal’s cutoff frequency.)

Maximum KE (eV)

Q4.6 The Photon Model of Light

3.0

59

Figure Q4.2 Rb

Zn

Cu

Pt

2.0 1.0

1.0 PHz

2.0 PHz

ƒ=

c λ

Plots of maximum electron kinetic energy versus illuminating light frequency for various types of metals (where 1 PHz = 1015 Hz). The slope of each line (h = 4.15 × 10−15 eV·s) is the same for all metals.

4. For a fixed frequency of illuminating light, the electrons’ maximum kinetic energy is independent of intensity! 5. Above a metal’s cutoff frequency, the electrons’ maximum kinetic energy is directly proportional to the frequency of the illuminating light, with the same slope for all metals (see figure Q4.2). Except for the first, these results are troublesome at best and flatly contradict the wave model at worst. Observation 2 (that electrons start coming off within nanoseconds) is particularly troubling in experiments where the delay should be many seconds. If a wave’s energy is truly dispersed continuously across a wavefront (as it must be in any simple wave model), there is no way that an electron can gather the required energy in such a short time. Also troubling are observations 4 and 5. No matter how we model the ­interaction of a light wave with electrons in the metal, we would not expect the electrons’ maximum kinetic energy to be independent of intensity: this is like expecting a 50-ft tsunami crashing on the beach to cause the same damage as a 1-ft ankle-splasher. The strictly linear, substance-­independent frequency dependence of the kinetic energy begs for a simple explanation having to do with the nature of light rather than the character of the metal. Physicists in the early 1900s could not find any way to tweak the wave model to explain these basic observations (and it wasn’t for lack of effort).

Q4.6

E = hf(Q4.4)

where f is the frequency of the light and h is a universal constant (equal to the slope of the graph in figure Q4.2). The value of h measured by Millikan in 1916 was 6.57 × 10−34 J·s. This constant is now called Planck’s constant, and its currently accepted value is 6.63 × 10−34 J·s = 4.15 × 10−15 eV·s. 1

The kinetic energy results are also hard to understand

The Photon Model of Light

Not all of these facts were known in 1905. For example, von Lenard had only described observations 2, 3, and 4, and these only qualitatively. On the basis of this information, Albert Einstein in 1905 proposed a simple alternative to the wave model of light that explained these qualitative observations and predicted the other observations listed earlier. These predictions were experimentally verified by Robert Millikan and described in a paper that Millikan published in 1916. Einstein won the Nobel Prize in 1921, not for special relativity nor general relativity but “for his services to t­heoretical physics and especially for his discovery [sic] of the photoelectric effect.”1 Einstein’s model imagined that light consists of particles (which were later called photons), each of which carries a certain amount of energy. The energy of each photon in a beam of monochromatic light is

Immediate ejection is a particular problem for the wave model

 Quoted in A. Pais, Subtle Is the Lord, Oxford, 1982, p. 527.

The energy carried by a photon in Einstein’s model

60

Chapter Q4

The Particle Nature of Light

We commonly express different colors of visible light in terms of wavelengths λ instead of frequencies f. Since f = __ ​  c  ​ (Q4.5) λ



(where c is the speed of light) we can re-express equation Q4.4 in terms of λ:

Photon energy as a function of wavelength

E = __ ​  hc ​  (Q4.6) λ



• Purpose:  This equation specifies the energy E carried by a single photon of monochromatic light whose collective wavelength is λ; c is the speed of light = 3.0 × 10 8 m/s and h is Planck’s constant = 6.63 × 10−34 J·s. • Limitations:  This expression only applies to photons. • Note:  The value of hc is 1240 eV·nm (in typically convenient units).  

One photon ejects one electron

Einstein’s model further proposed that each electron is ejected as a result of being hit by (and absorbing) a single photon. The maximum kinetic energy K that an electron could have just after leaving the metal must therefore be

K = hf - W = __ ​  hc ​ - W(Q4.7) λ

where hc/λ = hf is the photon’s energy and W is the work function for the metal, defined to be the minimum amount of energy required to liberate the electron from the metal (this is the quantity we called E ej in section Q4.4). This model neatly explains all the experimental evidence listed in section Q4.5, as we shall see. The intensity of light is defined to be the amount of energy it delivers to a unit area in a unit time. If the light has a fixed wavelength, the energy carried by each photon is the same, so the light intensity is directly proportional to the number of photons delivered per unit area per unit time. If each ejected electron was hit by a single photon, then the number of ejected electrons should be directly proportional to the intensity of the light at a given frequency. This was observation 1 on our list of results. Since only a single photon is required to liberate an electron, electrons can start coming out as soon as photons, even if the intensity of the light is so low that only a handful of photons arrive every second. This explains (without violating the conservation of energy) how electrons can be liberated instantly (observation 2) even at very low ­intensities. Equation Q4.7 is also consistent with the graphs shown in figure Q4.2. For light of sufficiently high frequency f, the energy of an ejected electron depends linearly on f, with a metal-independent slope h. On the other hand, if the frequency is so low that hf < W, then absorbing a photon will not provide sufficient energy for the electron to escape the metal, so no ejected electrons will appear. The place at which the graph of K versus f intersects the horizontal axis is simply the point where hf = W, and this point should depend on the type of metal used. This explains observations 3 and 5 on our list.  

The photon model explains all the observed results

Exercise Q4X.2 How does the photon model explain observation 4, that the kinetic energy of ejected electrons is independent of the intensity of the illuminating light?

Q4.6 The Photon Model of Light

61

Einstein’s model thus explains all the results of experiments on the photoelectric effect neatly and easily. Yet even after Millikan verified Einstein’s predictions, most physicists were very reluctant to embrace Einstein’s model. Not until 1923, when Arthur Compton verified that X-rays (electromagnetic waves with very short wavelengths) were scattered by electrons exactly as if they were tiny particles (see problem Q4D.1), did physicists begin to really take the photon model of light seriously. Why was resistance so strong? The problem is that Einstein’s model is completely inconsistent with experimental results that demonstrate the wave nature of light. For example, no simple particle model is able to explain the behavior of light in the Young two-slit interference experiment, which the wave model does easily. When one recognizes that an interference pattern can easily be displayed with simple equipment, while the measurements required to verify the photon model are hard to make (and thus subject to doubt), it is easier at second glance to forgive physicists’ reluctance to accept the photon model of light in the early 1900s. Even the photon model implicitly recognizes the reality of the wave nature of light: the energy of a photon is expressed in terms of the frequency (or alternatively, the wavelength) of the light involved, concepts that would have no meaning in a strict particle model of light. We have to face the fact that neither a wave model nor a particle model is able to explain the behavior of light completely. The problem of reconciling these models will be our focus in the next few chapters.

Problem:  When ultraviolet light of wavelength 250 nm illuminates an aluminum plate, the measured maximum kinetic energy of ejected electrons is 0.68  eV. When light of wavelength 210 nm illuminates the same plate, the corresponding energy is 1.625 eV. Find the value of W for aluminum and the value of hc from this data. Solution  According to equation Q4.7, we have

K1 = __ ​ hc ​ - W  and   K2 = __ ​  hc ​ - W (Q4.8) λ1 λ2

where λ1 and λ2 are the wavelengths of the illuminating light and K1 and K2 are the maximum kinetic energies of the ejected electrons in the two cases, respectively. We thus have two equations for the two unknowns hc and W, so we can solve. Subtracting the first of equations Q4.8 from the second, we get

( 

)

K2 - K1 = __ ​  hc ​ - __ ​  hc ​ = hc ​ __ ​  1  ​ - __ ​  1  ​   ​ λ2 λ1 λ2 λ1

(Q4.9)

Solving this equation for hc yields

(1.625 - 0.680) eV K2 - K1 hc = ​ ___________      ​ = ______________________ ​         ​ = 1240 eV·nm (Q4.10) 1/λ 2 - 1/λ1 (210 nm)-1 - (250 nm)-1

We can now find W from either of equations Q4.8. Choosing the first, we get

W = __ ​ hc ​ - K1 = ___________ ​  1240 eV·nm  ​     - 0.68 eV = 4.28 eV (Q4.11) λ1 250 nm

Note that the result for hc is as expected, and comes out with the correct units. The result for W also has the correct units and a magnitude consistent with the approximate value of “several electron volts” quoted earlier.

The photon model does not explain wave effects

Neither model adequately explains light by itself

Example Q4.1

Chapter Q4

62

The Particle Nature of Light

A photon whose energy is equal to W is barely able to eject an electron. In the case of aluminum, such a photon would have a wavelength such that ​· nm W = __ ​ hc ​   ⇒  λ = ___ ​  hc  ​ = ___________ ​ 1240 eV​  ​     = 290 nm (Q4.12) λ 4.28 eV​ ​  W



Light with λ > 290 nm cannot eject electrons from aluminum at all. This cutoff wavelength for aluminum is in the ultraviolet (shorter than the visible light range of 400 nm to 700 nm). Many metals have cutoff wavelengths in the ultraviolet. Lithium, sodium, and cesium are among the few metals whose electrons can be ejected by visible photons.

Example Q4.2

Problem:  A 20-W LED light bulb uses 20 W of electrical power but radiates only about 5 W of visible light (about the same as what a 100-W incandescent bulb produces). About how many visible-light photons per second hit an 8.5-inch by 11-inch sheet of paper directly facing such a bulb 2 m away? Solution To find the number of photons hitting the page each second, we must determine the light energy hitting the page every second and the energy carried by each photon. We could compute the latter if we knew the wavelength of the light, but the visible light emitted by any light bulb is a mix of wavelengths. Let us estimate that the average wavelength of the bulb’s light is λ ≈ 590 nm, which is in the yellow region of the spectrum. The average energy carried by photons in this light will be about E ph ≈ __ ​  hc ​ (Q4.13) λ



 

To convert this to a number of photons, we need to know the energy per second that falls on the paper. Assuming the bulb radiates uniformly in all directions, the light intensity a distance r from a bulb emitting energy at a rate P is I = _____ ​  P 2 ​  4π r



 

(Q4.14)

 

The energy per unit time Psheet that falls on a sheet of area A will be Psheet = IA. Finally, we can find the number of photons per second falling on this sheet by dividing the energy per second falling on it by the energy per photon: energy/time falling on sheet _____ photons __________________________ Psheet ​ ________  ​     =     ​      ​ = ​   ​  s energy/photon E ph

(Q4.15)

 

Since we know or can calculate all of the quantities we have given symbols above, we can solve. Combining the equations above, we get P IA λA P λ A photons _____ ________ ​   ​     = ​  sheet ​ = _____ ​    ​ = ___ ​   ​ _____ ​     ​ = P ​ __  ​ ​  _____    ​  s



E ph  

(  ) ( 

hc/λ

hc 4π r 2  

) ( 

 

hc 4π r 2  

) ( 

 

) ( 

)

J ​ 2 1 eV​ ​    ​   in × 11 in ​  ​ ​​ _______ ​⨉ = ​ 5 _​    ​  ​  ​ ___________ ​  590 nm  ​   ​  ​ ___________ ​  ​ ​ ____________    ​  8.5    ​  2.54 cm     ​  ​ ​ 2 -19 s 1240 eV​ ​ ·nm 1.6 × 10 J 4π [200 cm​ 1 in ​ ]  ⨉  

≈ 2 × 1016 photons/s.(Q4.16)

I have only quoted this result to one significant digit, because our approximations and estimations (particularly with regard to P and λ) are pretty crude. Still, this is a useful ballpark ­estimate.

Q4.7 Detecting Individual Photons

Q4.7

63

Detecting Individual Photons

The experimental results described in section Q4.5 constitute rather indirect evidence of light’s particle nature. At the time Millikan published his experimental results in 1916, this was the best evidence available. However, we can presently build detectors capable of detecting a single photon. This is no small task, since photons of visible light carry only a tiny amount of energy (roughly 2.0 eV ≈ 3.2 × 10−19 J). Even so, we can do this with a photomultiplier, whose design is illustrated schema­tically in figure Q4.3a. This device works as follows. A single photon strikes a very thin metal plate, called the photocathode. Einstein’s explanation of the photoelectric effect tells us that this photon’s absorption will liberate a single electron from the metal’s surface. A power supply or battery of some kind gives a nearby metal plate (called a dynode) a positive charge relative to the photocathode. The single electron liberated from the photocathode is attracted toward the dynode and strikes it with a substantial amount of kinetic energy (100 eV in the design above, since the electrostatic potential energy difference per unit charge between the plates is 100 V). This electron’s impact dislodges several electrons from that dynode. Another electrostatic potential energy difference between the first dynode and a nearby second dynode attracts these secondary electrons toward the third plate. They strike it, liberating even more electrons, and so on. After several multiplications, the number of electrons becomes large enough that sensitive electronic circuitry can detect the tiny pulse of current that they represent. The circuitry then electronically amplifies this signal sufficiently to send a clear electronic signal to a display. That such a detector can register a single photon does not really provide independent evidence for the photon model of light: my description of the photomultiplier’s operation assumed Einstein’s explanation of the photoelectric effect. However, at very low illumination levels, discrete pulses are observed as individual photons hit the detector, giving vivid (even if not exactly independent) evidence of the particle nature of light.

How a photomultiplier works

Exercise Q4X.3 Find the energy range of visible-light photons. (Hint: The wavelength range of visible light is 700 nm to 400 nm.)

0V

100 V

300 V dynodes

+ + ++ +

Primary electron

Incoming photon

+ + (a)

dynodes

+ + +

+

Very thin photocathode

500 V

Secondary electrons

200 V

+++

+

400 V

(b)

Figure Q4.3 (a) A schematic diagram of a photomultiplier tube. (b) A photograph of an actual photomultiplier tube. (Credit: © McGraw-Hill Education/ photo by Chris Hammond)

64

Chapter Q4

The Particle Nature of Light

TWO-MINUTE PROBLEMS Q4T.1 Sunlight falling on a spaceship in a vacuum will cause the spaceship to become a bit positively charged. T or F? Q4T.2 In the apparatus shown in figure Q4.1a, the value displayed by the ammeter is proportional to A. The rate at which electrons emerge from the cathode. B. The average kinetic energy of the ejected electrons. C. The maximum kinetic energy of the ejected electrons. D. Both A and C. Q4T.3 For experiments involving the setup shown in ­figure Q4.1a, which of the following possible results (if seen) about the value displayed by the ammeter would probably not be consistent with the wave model of light? A. It is zero for some time after light starts shining. B. It increases as the light’s intensity increases. C. It varies as the light’s wavelength changes. D. It is zero if the wavelength is larger than a certain value. Q4T.4 In the apparatus shown in figure Q4.1b, the final value displayed by the voltmeter is proportional to A. The rate at which electrons emerge from the cathode. B. The average kinetic energy of the ejected electrons. C. The maximum kinetic energy of the ejected electrons. D. Both A and C. Q4T.5 For experiments involving the setup shown in figure Q4.1b, which of the following possible results (if seen) regarding the final value displayed by the voltmeter would probably not be consistent with the wave model of light? A. It increases as the light’s intensity increases. B. It is independent of the light’s intensity. C. It varies as the light’s wavelength changes. D. It increases as the rate of electron ejection increases. Q4T.6 In experiments involving the setup shown in figure Q4.1b, the value that the voltmeter displays increases for a short period of time after light starts shining on the cathode, but then quickly settles down to a fixed value. Why? A. The cathode eventually stops emitting electrons. B. The kinetic energy of ejected electrons starts out small but quickly reaches a maximum value. C. The charge difference between the cathode and anode is initially small but with time becomes so large that no electron from the cathode can make it to the anode. D. The cathode eventually runs out of electrons to emit. E. Some other explanation (describe). Q4T.7 In an experiment involving the apparatus shown in figure Q4.1b, suppose that after shining light on the cathode long enough that the voltmeter settles down to a fixed value, we suddenly cut off the light. If the voltmeter is truly ideal, what happens to the value displayed? A. It immediately drops to zero. B. It slowly decays to zero. C. It remains fixed. D. It does something else. (Describe what.)

Q4T.8 A beam of light P has twice the wavelength but the same intensity as beam Q. The number of photons that hit a given area in a given time when it is illuminated by beam P is (A) twice, (B) the same, or (C) one-half of the number that hit when the area is ­illuminated by beam Q? Q4T.9 Though this would be tricky to do in practice, suppose we could increase the frequency of light illuminating the cathode of the apparatus shown in figure Q4.1a while holding the intensity of that light constant. What would Einstein’s photon model predict would happen to the value displayed by the ammeter as the frequency increases? (Assume that the light’s frequency is always above the cathode’s cutoff frequency.) A. It should increase. B. It should remain the same. C. It should decrease. D, It might vary in a complicated way. Q4T.10 The work function W for copper is 4.65 eV. Visible light will dislodge electrons from a copper surface. T or F? Q4T.11 If you want to design a phototube that is able to respond to all visible light wavelengths, what should its cathode’s work function be? A. Below 1.77 eV B. Any value between 1.77 eV and 3.1 eV C. Some specific value between 1.77 eV and 3.1 eV D. Above 3.1 eV E. Some other value or range (specify) Q4T.12 If you want to design a phototube that ignores all visible light, what should its cathode’s work function be? A. Below 1.77 eV B. Any value between 1.77 eV and 3.1 eV C. Some specific value between 1.77 eV and 3.1 eV D. Above 3.1 eV E. Some other value or range (specify) Q4T.13 Violet light with a wavelength of 430 nm falls on a sodium plate (W = 2.75 eV) and a potassium plate (W = 2.30 eV). Which of of the following statements is true? A. Both plates emit electrons, but those emitted by the sodium plate have the larger kinetic energy. B. Both plates emit electrons, but those emitted by the potassium plate have the larger kinetic energy. C. Only the sodium plate emits electrons. D. Only the potassium plate emits electrons. E. Neither plate emits electrons. Q4T.14 A long-wavelength photon, because it is larger, ­carries more total energy than a short-wavelength photon. T or F? Q4T.15 Einstein’s photon model is consistent with the results of two-slit interference experiments involving light. T or F?

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HOMEWORK PROBLEMS Basic Skills Q4B.1 Red light emitted by a standard helium–neon laser has a wavelength of about 633 nm. What is the energy of one photon of such red light? Q4B.2 Yellow light has a wavelength of about 590 nm. What is the energy of one photon of such light? Q4B.3 We find that photons from a certain light source have an energy of 3.5 eV. What is the wavelength associated with this light? Is this light visible? Q4B.4 A typical laboratory helium–neon laser produces about 1 mW (0.001 J/s) of light at a wavelength of 633 nm. How many photons per second does this laser produce? Q4B.5 A certain argon laser produces about 5 mW of light at a wavelength of 514 nm. How many photons per second does this laser produce? Q4B.6 Photons from a certain light source are measured to have an energy of 0.62 eV. What is the wavelength associated with this light? Is this light visible? Q4B.7 The value of W is about 4.24 eV for zinc. What is the maximum wavelength that light falling on a zinc cathode can have if it is to be able to eject electrons? Q4B.8 The value of W is about 2.3 eV for potassium. What is the maximum wavelength that light falling on a potassium cathode can have if it is to be able to eject electrons? Q4B.9 Verify that the intensity of light falling on a surface 1.0 m away from a lamp radiating 40 W of visible light is 3.2 W/m2, as claimed in exercise Q4X.1. Q4B.10 Suppose red and green light-emitting diodes (LEDs) radiate the same amount of power in the form of light at wavelengths of 650 nm and 560 nm, respectively. Which emits more photons per second? By what factor? Q4B.11 Suppose we do an experiment using the setup shown in figure Q4.1b where green light with a wavelength of 540  nm falls on a cesium cathode whose work function is 2.1 eV. What value will the voltmeter display? Q4B.12 Suppose we do an experiment using the setup shown in figure Q4.1b where violet light with a wavelength of 430  nm falls on a cesium cathode whose work function is 2.1 eV. What value will the voltmeter display? Q4B.13 About how many photons per second are broadcast by a FM radio station whose transmit­ter power is 10,000 W and whose frequency is 89.9 MHz? Q4B.14 How does Einstein’s photon model explain the observed result of photoelectric effect experiments indicating

that the rate at which light ejects electrons from a metal surface is directly proportional to the intensity of the light? Q4B.15 How does Einstein’s photon model explain the observed result of photoelectric effect experiments that if the intensity of light falling on a metal plate is held fixed, the rate at which electrons are ejected from the metal decreases as the light’s frequency increases?

Modeling Q4M.1 When we illuminate sodium metal with monochromatic light with a wavelength of 420 nm, suppose we find that the maximum potential difference developed between the plates in the experimental setup shown in figure Q4.1b is 0.65 V; when the wavelength is 310 nm, we find this voltage to be 1.69 V. Check that these experimental results are consistent with a value of 1240 eV·nm for hc (within ±1%), and find the value of W for sodium. Q4M.2 When we illuminate cesium meta with monochromatic light with a wavelength of 500 nm, suppose we find that the maximum potential difference developed between the plates in the experimental setup shown in figure Q4.1b is 0.57 V; when the wavelength is 420 nm, we find this voltage to be 1.04 V. Check that these experimental results are consistent with a value of 1240 eV·nm for hc (within ±1%), and find the value of W for ­cesium. Q4M.3 When we illuminate iron with ultraviolet light with a wavelength of 250 nm, suppose we find the maximum potential difference ­developed between the plates in the experimental setup shown in figure Q4.1b to be 0.46 V. From these data and the accepted value of hc, find the potential difference between the plates if the ultraviolet light wavelength is changed to 220 nm. Also find W for iron. Q4M.4 When we illuminate lead with ultraviolet light with a wavelength of 250 nm, suppose we find the maximum potential difference developed between the plates in the experimental setup shown in figure Q4.1b to be 0.82  V. From these data and the accepted value of hc, find the potential ­difference between the plates if the wavelength is 215 nm. Also find W for lead. Q4M.5 Suppose you are standing in the dark and facing a 20-W LED bulb 100 m away. If the diameter of your pupils is about 8 mm under these conditions, about how many photons of visible light enter your eye every second? Q4M.6 Imagine a model where light consists of photons but each photon carries an energy Eph proportional to the wave’s intensity, not its wavelength. How would this model contradict the expectations of a pure wave model for the photoelectric effect (or would it)? Would it be consistent with the photoelectric effect results actually observed (see section Q4.5)? Explain.

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The Particle Nature of Light

Q4M.7 Imagine a model where electromagnetic radiation consists of photons but each photon carries a fixed amount of energy, say, 4 eV, independent of frequency. This “fixed energy photon” (FEP) model is different than the normal photon model in that an FEP photon’s energy does not depend on wavelength. How would this model contradict the expectations of a pure wave model for the photoelectric effect (or would it)? Would it be consistent with the photoelectric effect results actually observed (see section Q4.5)? Explain by going through each of the five listed predictions and five results and discuss whether the FEP model is consistent with each. Q4M.8 Imagine a model where electromagnetic radiation consists of ball-like photons whose energy is proportional to their volume (like most objects) and that the radius of each ball is proportional to the light’s wavelength. This “energy equals volume” (EEV) model is different than the normal photon model in that an EEV photon’s energy depends on its wavelength in a different way. (Assume, however, that one photon still liberates just one electron, as before.) How would this model contradict the expectations of a pure wave model for the photoelectric effect (or would it)? Would it be consistent with the photoelectric effect results actually observed (see section Q4.5)? Explain by going through each of the five listed predictions and five results and discuss whether the EEV model is consistent with each.

Derivations Q4D.1 (Requires completion of chapter R9.) If we shine X-rays (which are very high-frequency electromagnetic waves) at a target, both the wave and photon models of electromagnetic radiation predict that the X-rays will be scattered in all directions by the target. The wave model predicts that the scattered X-rays will have the same wavelength as that of the incoming X-rays (since the incoming waves will wiggle the electron at the same frequency, so the scattered waves emitted by the electron in all directions must have the same frequency as the bobbing electron that emits them). The photon model, on the other hand, predicts that the wavelength of the scattered X-rays will depend on the angle in a specific way. Our purpose here is to derive this angle dependence. Let’s define our coordinate system so that the initial X-ray photons move in the +x direction. Let the initial energy of the X-ray photons be E 0 and the mass of an electron be m. Imagine that a photon strikes the electron and scatters at an angle θ, while the electron scatters at an angle ϕ, as shown below. BEFORE: photon, energy E0

AFTER: y

x electron at rest

photon, energy E

y θ ϕ

x scattered electron

Since the ratio of any particle’s relativistic momentum │p​  W​ │   to its relativistic energy E is such that   │p​  ​W│   /E =  │v​  ​W│      (in the SR units discussed in unit R), then for a photon, we must have  │p​  ​W│   /E = 1, so  │p​  ​W│    = E in those units. (a) Argue that conservation of 4-momentum implies that E 0       ​ E   ​  0        ​   ​    ​ = ​      0 ​   ​  0

  ] [

[ ] [ ] [

m      ​  0 ​  ​   ​      ​    ​ + ​      ​ 0 ​  0

E              ​ E cos θ    ​   ​              ​     ​ + ​              ​ E sin θ  ​     0

_________

  ]

 2 │ W │2 ​ m       +  p​  ​  e  ​                            ​    ​  │p​ │  ​We  cos ϕ                         ​     ​   ​ (Q4.17) │p​  ​W e│sin ϕ                  ​   ​     0  

where E is the energy of the scattered photon and │p ​W​e │ is the relativistic momentum of the scattered electron. (b) Show that you can square and combine the middle two equations in equation Q4.17 to yield

​ E 2​0​​  - 2 EE 0 cos θ + E 2 = │p​  ​We │2(Q4.18)  

(c) S  how, by isolating the square root and squaring both sides, that the top equation implies that

2mE 0 + ​E 2​0​​  - 2E (m + E 0) + E 2 = │p​  ​W e│2(Q4.19)  

(d) Show that by combining these equations to eliminate │  p​  ​We │, dividing both sides by EE 0 m, and using equation Q4.6, you get (in SI units)  



λ - λ 0 = ___ ​  mhc   ​( 1 - cos θ)(Q4.20)



where λ 0 and λ are the wavelengths of the initial and scattered photons, respectively. (Remember that E and E 0 have units of kilograms in SR units and thus are equivalent to E/c 2 and E 0/c 2, respectively, when the energies are expressed in the SI unit of joules.) In 1923, Arthur Compton verified that the scattered X-rays had exactly this angle dependence, and this is what finally convinced the community that the photon model must be correct.

 

 

 

Rich-Context Q4R.1 Experiments have shown that the nervous system of the human eye effectively takes about 30 “frames” per second (like a movie camera) and that when the eye is fully dark-adapted, it needs to r­eceive only about 500 visible photons per frame from an object to register it. Our sun radiates a power of about 3.9 × 10 26 W at all wavelengths, peaking in the yellow region of the spectrum (but only about one-half of this energy is in the visible range). The pupil of your dark-adapted eye has a diameter of about 8 mm (you might check this experimentally). Estimate how far away a star like the sun could be and still be visible to the naked eye. The nearest visible star is about 4 ly away and most stars we see in the sky are hundreds of light years away. What does this mean for most visible stars’ intrinsic brightness compared to the sun’s intrinsic brightness?  

Q4R.2 Assume that when photographing a scene, the sensor in an electronic camera must collect at least an average of 3 photons per final pixel. Even assuming the camera’s

Answers to Exercises

67

sensor can detect single photons (which is not very likely) and we ignore possible sources of noise in the camera sensor, this will yield a very crude (“grainy”) image where each pixel’s brightness and color will vary in discernible steps depending on the small and discrete count of photons it receives. Therefore, this probably represents the absolute lower limit for the photons per pixel a camera can receive to create a useful image. Assume that your small pointand-shoot camera has a lens about 1 cm in diameter, a roughly square 6-megapixel sensor, and a maximum exposure time of 5 s. Estimate (to about one significant digit) the minimum average intensity of light (at the lens) that your camera must receive to create even such a crude picture as this. How does this compare to 0.5 W/m2 (very roughly) that one might receive from a typical indoor scene? Q4R.3 When a star collapses into a black hole, it appears to a distant observer that the star’s surface “freezes” just barely outside (and asymptotically approaches) a radius of r = 2GM/c 2, which is the radius of the black hole’s “event horizon” within which light cannot escape. Since such a distant observer will always see the star’s glowing surface to be just outside the star’s event horizon, why do we say that a black hole is “black”? Well, it turns out the effects of time dilation and energy conservation imply that the intensity of light received by a distant observer from the surface once it is close to the event horizon decreases exponentially according to  

( 

)

tc 3  ​  ​ (Q4.21) I  ​ ≈ e-​t c​ ​/GM = exp ​ - ​ ____ ​ __ I 0 GM  3

where I 0 is the intensity at time t = 0, c is the speed of light, G is the universal gravitational constant, and M is the star’s mass. Because of a gravitational redshift, the average wavelength of the light received also increases:

( 

3

)

λ R ​ ≈ exp ​ -_____ ​ ___ ​  tc    ​  ​ λ 0 2GM

(Q4.22)

where λ R is the average wavelength received by the distant observer at time t, and λ  0 is the wavelength received at t = 0. Both expressions are approximate, but correct to within an order of magnitude. (a) Show that these expressions imply that the rate R at which the observer’s eye receives photons at time t is

( 

)

tc 3  ​  ​ (Q4.23) R  ​ ≈ exp ​ - ​ _____ ​ ___ R 0 2GM where R 0 is the rate (in photons per second) at which the observer’s eye receives photons at t = 0. (Hint: The intensity is the number of photons received by the observer per unit area per unit time times the typical energy per photon received by the same observer. Assume that the observer’s eye has the same area directed toward the star at both times.) (b) Assume that the collapsing star’s mass is 4 × 10 30 kg (2 solar masses) and that we set t = 0 to be when the received light from the star’s surface has the same average wavelength (about 550 nm) as light from the sun. Physicists estimate that the intensity of light from the surface of such a star will be very roughly 1000 times that of the sun or about 10 6 W/m2 for an observer at the same distance from the forming black hole as we are from the sun. If an observer at this distance looks at the star through pupils about 3  mm in diameter (which will be appallingly dangerous at first), about how much time passes before the rate at which photons enter the eye falls below one per second? (Hint: First calculate R 0 in this case, which you should find to be on the order of magnitude of 1019 photons/s. You should find t to be on the order of a millisecond or so.) (c) Does the answer change much (from a practical point of view) if we set the final rate to one photon per century (= one photon per 3.16 × 10 9 s) or set the star’s initial brightness to be a billion times brighter or a billion times dimmer than the sun? Why not? So summarize: why is a black hole black?



 

 

ANSWERS TO EXERCISES Q4X.1 Let’s take “several eV” to be 3  eV. According to equation Q4.3, the time required is

( 

) ( 

)

E 3.0 eV 1 nm 2 ___________ 1.6 × 10-19 J __________________ T ≈ ​ ___ej2  ​ =    ​     ​ ​​ ______ ​  -9    ​   ​​ ​  ​ ​   ​     ​ 2 2 1 eV Id (3.2 W/​m​ 10  ​m​ ⨉   )(0.1 nm) ⨉   

 



 

(  )

​J​     ​ ​  _____ = 15 ​ ___ ​  1 W  ​  ​= 15 s (Q4.24) W 1 ​J​/   s

Q4X.2 Since each photon’s energy depends only on the illuminating light’s wavelength, not its intensity, and since each ejected electron gets the energy of only one photon,

the ejected electron’s energy depends only on the energy per photon and thus the light’s wavelength. The n ­ umber of photons hitting the plate is therefore irrelevant. Q4X.3 The visible range of photon wavelengths is 700 nm (red) to 400 nm (violet). The energy of a red photon is

E ph = __ ​  hc ​ = ___________ ​ 1240 eV·nm  ​     = 1.77 eV (Q4.25) λ 700 nm

An analogous calculation shows that the energy of a violet photon is 3.10 eV.

Q5

The Wave Nature of Particles Chapter Overview Introduction In chapter Q4, we reviewed some of the evidence supporting a particle model of light. In this chapter, we will discuss some evidence supporting a wave model for entities we normally consider to be particles (for example, electrons and protons). The ultimate point is that neither model is completely sufficient for any form of matter or energy.

Section Q5.1:  Subatomic Particles as Particles Electrons, protons, neutrons, and so on are individually countable and leave definite tracks in particle detectors. A particle model for such objects seems so obvious that it is difficult to imagine that any other possibilities might exist.

Section Q5.2:  The de Broglie Hypothesis Even so, in 1923 Louis de Broglie proposed that a beam of such entities might, like photons, have an associated wavelength given by



│p​  W​ │    = __ ​  h ​  or  λ = ____   (Q5.4) ​  h   ​ │p​ λ  W​  │

• Purpose:  This equation specifies how the magnitude of a particle’s (­relativistic)  W​ │    is related to the de Broglie wavelength λ of a beam of such parmomentum │p​ ticles, where h is Planck’s constant (hc = 1240 eV·nm). • Limitations:  This equation applies to both photons and other kinds of particles. It only applies, though, when the particles have a well-defined value of │p​  W​ │  . • Note:  The relativistic momentum magnitude │p​  W​ │    reduces to the ordinary momentum magnitude m│v​  W​ │    for particle speeds │v​  W​ │