Script Physics Olympiad€¦ · Script Physics Olympiad Second Edition Autumn 2018. PhysicsOlympiadScript 2. Preface ... In physics, vectors are of crucial importance, since many

Script Physics Olympiad

Second Edition

Autumn 2018

Physics Olympiad Script

2

Preface

Every time I gave a lesson at a training camp I opened a book, decided what to tell and took somenotes which often filled a couple of pages. Of course this preparation took me always a lot of timeand sometimes I recognized during the lecture that I had forgotten a topic and had to improvize.I started asking myself if it would not be easier to write a script and every year, look up in thescript what I want to tell at a lesson. Together with some members of the Physics Olympiad Istarted writing a script and what you now hold in your hands is the first edition of it.

As I already mentioned above the script is on the one hand helpful to give lectures. On the otherhand students do not have to write everything down what the teacher tells, they simply can takenotes to complete or stress out important things. Therefore this script does not replace a goodbook. Because in a good book there are often much more details which can not be given in ascript which is written to accompany a two-hours-lecture. Additionally I highly recommend tosolve problems, because then one has to think about what the abstract laws of physics are tellingand how they can be used.

Now the second edition of the script is finished and it somehow reminds me of hiking: Beforethe beginning I was full of good ideas and expectations how it will look like. While I wrote it, Isometimes asked myself why I did all this expenditure because it was a lot of work. When it wasfinished I was proud to have endured all this burden and lucky to have made many interestingexperiences and even learned a lot. In this sense I want to give a special thanks to the otherauthors, namely Cyrill, Levy, David, Quentin, Lionel, Sven and Sebastian. Also a big thanks toall the members who helped, corrected and proofread the script.

If you have any comment on the script, found a mistake or would like to help improving or devel-oping it, please write me a mail at [email protected]

I wish you a entertaining and instructive reading,

Best

Rafael

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Physics Olympiad Script

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Contents

1 Mathematics 111.1 Vector algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.1.1 Scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.1.2 Vector product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.2 Differential calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.1 Derivative of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.2 Differentiation rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.2.3 More derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.2.4 Overview about derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.2.5 Higher derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.2.6 Taylor approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.2.7 The idea of differential equations . . . . . . . . . . . . . . . . . . . . . . . . 28

1.3 Integral calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.3.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.3.2 Integral as an area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.3.3 Fundamental theorem of calculus . . . . . . . . . . . . . . . . . . . . . . . . 331.3.4 More integration rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361.3.5 The idea of multidimensional integrals . . . . . . . . . . . . . . . . . . . . . 42

1.4 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431.4.2 Representation of a complex number, Euler formula . . . . . . . . . . . . . 431.4.3 A first simple application . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451.4.4 Physical examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2 Mechanics 472.1 Getting Ready . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.1.1 Mathematical Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.1.2 Kinematics in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 492.1.3 Describing Uniform Circular Motion . . . . . . . . . . . . . . . . . . . . . . 502.1.4 Choice of the Coordinate Frame . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.2 The Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 532.2.1 Newton’s Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 532.2.2 Dynamics of Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 592.2.3 Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

2.3 Frames of Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.3.1 Choosing a Frame of Reference . . . . . . . . . . . . . . . . . . . . . . . . . 70

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Physics Olympiad Script CONTENTS

2.3.2 Accelerated Frames of Reference . . . . . . . . . . . . . . . . . . . . . . . . 722.4 Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

2.4.1 Newton’s Second Law for Systems of Particles and the Center of Mass . . 752.4.2 Conservation of Linear Momentum . . . . . . . . . . . . . . . . . . . . . . . 792.4.3 Conservation of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . 80

2.5 Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 862.5.1 Kinematics of Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . 872.5.2 How Forces Determine Translational Motion of a Rigid Body . . . . . . . . 892.5.3 How Forces Determine Rotational Motion of a Rigid Body . . . . . . . . . . 912.5.4 Mechanical Energy of Moving Rigid Bodies . . . . . . . . . . . . . . . . . . 962.5.5 Putting Everything Together: Rolling Rigid Bodies . . . . . . . . . . . . . . 992.5.6 Conservation of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . 103

2.6 Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1042.6.1 Newton’s Law of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1042.6.2 Gravitational Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1052.6.3 Energy and Angular Momentum in Gravitational Fields . . . . . . . . . . . 1092.6.4 Two Objects Subject to Mutual Attraction . . . . . . . . . . . . . . . . . . 1112.6.5 Kepler’s Laws of Planetary Motion* . . . . . . . . . . . . . . . . . . . . . 114

3 Thermodynamics 1173.1 Important definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1173.2 The temperature scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1183.3 Zeroth law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1193.4 Thermal energy and heat capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

3.4.1 Molar heat capacities of ideal gases . . . . . . . . . . . . . . . . . . . . . . . 1203.5 Ideal gas law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1213.6 First law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1213.7 Thermodynamic systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1223.8 Equipartition theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1223.9 Thermodynamic processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1233.10 Second law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1253.11 Heat engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1263.12 Kinetic gas theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1283.13 Phase transitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1293.14 Real gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1303.15 Stefan-Boltzmann law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

4 Oscillations 1334.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1334.2 Harmonic Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

4.2.1 Harmonic Oscillations, Spring/Mass Systems and Differential Equations . . 1354.2.2 Further Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1374.2.3 Importance of Harmonic Oscillations . . . . . . . . . . . . . . . . . . . . . . 140

4.3 Beyond Harmonic Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1404.3.1 Example of Forced Oscillating Systems . . . . . . . . . . . . . . . . . . . . . 143

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5 Waves 1455.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1455.2 Harmonic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1465.3 Waves in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

5.3.1 Waves as Functions of 3D Spatial Coordinates . . . . . . . . . . . . . . . . . 1485.3.2 Waves as 3D Functions, Transversal and Longitudinal Waves, Polarization . 148

5.4 Waves Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1495.4.1 Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1495.4.2 Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1505.4.3 Seismic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1505.4.4 Transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1515.4.5 Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

5.5 Waves Propagation at Interfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1535.5.1 Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1535.5.2 Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1535.5.3 Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

5.6 Multi-Waves Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1545.6.1 Same amplitude, frequency and wavenumber . . . . . . . . . . . . . . . . . 1565.6.2 Same amplitude and frequency, opposite wavenumber . . . . . . . . . . . . 1575.6.3 Slightly different frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . 1585.6.4 Fourier Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

6 Fluid dynamics 1616.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

6.1.1 What is fluid dynamics about? . . . . . . . . . . . . . . . . . . . . . . . . . 1616.1.2 How can we model such a fluid? . . . . . . . . . . . . . . . . . . . . . . . . . 161

6.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1626.3 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

6.3.1 Compressible and incompressible fluids . . . . . . . . . . . . . . . . . . . . . 1626.3.2 Hydrostatic pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1626.3.3 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

6.4 Continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1636.5 Bernoulli’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

6.5.1 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1646.6 Surface tension, energy and capillary pressure . . . . . . . . . . . . . . . . . . . . . 1646.7 Friction in fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

7 Electrodynamics 1677.1 Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

7.1.1 Coulomb Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1677.1.2 Electrostatic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1687.1.3 Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1707.1.4 Continuous charge distributions . . . . . . . . . . . . . . . . . . . . . . . . . 1707.1.5 Gauss’ law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1707.1.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

7.2 Potential and Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1757.2.1 Electric potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

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7.2.2 Electric potential of a point charge . . . . . . . . . . . . . . . . . . . . . . . 1767.2.3 Potential of multiple charges . . . . . . . . . . . . . . . . . . . . . . . . . . 1777.2.4 Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1787.2.5 Potential and conducting material . . . . . . . . . . . . . . . . . . . . . . . 1787.2.6 Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1797.2.7 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

7.3 Current and magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1827.3.1 Current and conservation of charge . . . . . . . . . . . . . . . . . . . . . . . 1827.3.2 Magnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1827.3.3 Magnet and electric current . . . . . . . . . . . . . . . . . . . . . . . . . . . 1847.3.4 Lorentz force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

8 Direct current circuits 1878.1 Ohm’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1878.2 Equivalent circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

8.2.1 Wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1888.2.2 Series circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1888.2.3 Parallel circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1888.2.4 Voltage source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

8.3 Electric power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1898.4 Electric components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1908.5 Kirchhoff’s circuit law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

8.5.1 Current law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1908.5.2 Voltage law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1908.5.3 Applying Kirchhoff’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

8.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1928.6.1 Maximal power from a real power source . . . . . . . . . . . . . . . . . . . . 1928.6.2 Charging a capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

9 Electrodynamics 2 1959.1 Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

9.1.1 Magnetic field and Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1959.1.2 Ampere’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1969.1.3 Magnetic field of a moving point charge . . . . . . . . . . . . . . . . . . . . 1969.1.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

9.2 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2009.2.1 Approach and Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2009.2.2 Self induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

9.3 Displacement current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2029.4 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

10 Alternating current (AC) 20510.1 Describing alternating voltage and current . . . . . . . . . . . . . . . . . . . . . . . 205

10.1.1 Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20510.1.2 Usual real notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20610.1.3 Phasor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20610.1.4 Complex notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

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10.2 Impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20810.2.1 Ohmic resistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20810.2.2 Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20910.2.3 Inductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

10.3 Combinations of R,C and L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21210.3.1 Kirchhoff’s laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21210.3.2 Serial and parallel circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21310.3.3 High pass filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21310.3.4 Resonant circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

10.4 Power consideration and effective values . . . . . . . . . . . . . . . . . . . . . . . . 21910.4.1 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22010.4.2 Effective values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22110.4.3 Active, reactive and apparent power . . . . . . . . . . . . . . . . . . . . . . 221

10.5 Three-phase electric power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22210.5.1 Definition and production . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22210.5.2 Star and Delta circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22310.5.3 Advantage of a three-phase system . . . . . . . . . . . . . . . . . . . . . . . 225

11 Special relativity 22711.1 Historical milestones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

11.1.1 Aether and electromagnetic waves . . . . . . . . . . . . . . . . . . . . . . . 22711.1.2 Flying electron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

11.2 Galileo transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22911.2.1 Reference System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22911.2.2 Inertial frame of reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23011.2.3 Galileo Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230

11.3 Lorentz transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23111.3.1 Einstein’s Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23211.3.2 Synchronisation of clocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23211.3.3 Time dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23311.3.4 Lorentz contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23411.3.5 Symmetry of time dilatation and Lorentz contraction . . . . . . . . . . . . . 23611.3.6 Lorentz transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

11.4 Minkowski metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23811.4.1 Definition of Minkowski metric . . . . . . . . . . . . . . . . . . . . . . . . . 23811.4.2 Properties of Minkowski metric . . . . . . . . . . . . . . . . . . . . . . . . . 23911.4.3 Four vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24011.4.4 Note on rigorous derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . 240

11.5 Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24111.5.1 Addition of parallel velocities . . . . . . . . . . . . . . . . . . . . . . . . . . 24111.5.2 Addition of perpendicular velocities . . . . . . . . . . . . . . . . . . . . . . . 24211.5.3 velocity four vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242

11.6 Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24311.6.1 Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24311.6.2 momentum four vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24511.6.3 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24511.6.4 Acceleration and forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

9


11.7 Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24711.7.1 Ladder and barn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24711.7.2 Twin paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24811.7.3 Solution to the flying electron problem . . . . . . . . . . . . . . . . . . . . . 249

12 Quantum Mechanics 25112.1 Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

12.1.1 Black body radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25112.1.2 Photoelectric effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25312.1.3 Double slit experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

12.2 Laws of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25812.2.1 Wavefunction and probability . . . . . . . . . . . . . . . . . . . . . . . . . . 25812.2.2 Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25812.2.3 De Broglie hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25912.2.4 Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25912.2.5 Schrödinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

12.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26112.3.1 Bohr model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26112.3.2 Rigorous example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

13 Introduction to Statistics 26713.1 Location and Spread of a single Set of Data . . . . . . . . . . . . . . . . . . . . . . 267

13.1.1 Bivariate Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26813.2 Uncertainty Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268

13.2.1 Quantification of Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . 26813.2.2 Propagation of Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . 269

13.3 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27013.3.1 The International System of Units (SI) . . . . . . . . . . . . . . . . . . . . . 27013.3.2 Prefixes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27013.3.3 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270

13.4 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27113.4.1 Elements of good graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27113.4.2 Logarithmic Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

Appendices 271

A Further derivations 273A.1 Derivations of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273

A.1.1 Alternative formulations for Variance and Covariance . . . . . . . . . . . . . 273A.1.2 Derivation of the Least Squares Coefficients . . . . . . . . . . . . . . . . . . 274

B Tables 277B.1 List of physical constants (in SI units) . . . . . . . . . . . . . . . . . . . . . . . . . 277B.2 List of named, SI derived units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278B.3 List of material constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

10

Chapter 1

Mathematics

This part of the script presents the most important mathematical tools. Some things will be knownfrom school, other things maybe will be new. The goal is not to present the entire high schoolmathematics, but to give an overview over the concepts, which are needed in the physics part. Theproofs are not really important for the physics, so one can omit them if not interested.

1.1 Vector algebra

In physics, vectors are of crucial importance, since many physical quantities are represented byvectors. The foundation of vector algebra should be known from school. Therefore we will onlyrepeat the two most important concepts for physics. The main reference for this chapter is [1].

1.1.1 Scalar product

There is an intuitive way to add two vectors and to multiply a vector by a number. There areessentially two ways how two vectors can be multiplied. One of them is the scalar product (alsoknown as dot product).

Definition: Let ~a and ~b be vectors. Then the scalar product of ~a and ~b is

~a ·~b =

axayaz

·bxbybz

= axbx + ayby + azbz.

There is a nice connection between the scalar product and the angle ϕ between the vectors ~a and~b (see figure 1.1). We have the following alternative expression for the scalar product:

~a ·~b = axbx + ayby + azbz = |~a| · |~b| · cos(ϕ).

Why are the two expressions for the scalar product the same? The equality axbx + ayby + azbz = |~a| · |~b| · cos(ϕ)

follows from the cosine formula.1 We look at figure 1.1 and calculate the lenght of ~c = ~a −~b in two ways. On one

1The cosine formula tells us, that in an triangle with sidelengths a, b and c we have

c2 = a2 + b2 − 2ab cos(ϕ),

11

Physics Olympiad Script CHAPTER 1. MATHEMATICS

Figure 1.1: ~c = ~a−~b

hand, we have:

|~c|2 = |~a−~b|2 = (ax − bx)2 + (ay − by)2 + (az − bz)2

= a2x + a2y + a2z + b2x + b2y + b2z − 2axbx − 2ayby − 2azbz

= |~a|2 + |~b|2 − 2 (axbx + ayby + azbz) .

On the other hand, it follows by the cosine formula, that:

|~c|2 = |~a|2 + |~b|2 − 2 · |~a| · |~b| · cos(ϕ).

Comparing the two expressions for |~c|2, we find

axbx + ayby + azbz = |~a| · |~b| · cos(ϕ).

Remarks

− The scalar product is often used to calculate the angle ϕ between two vectors ~a and ~b. Usingthe two different expressions for the scalar product, we get

cos(ϕ) =~a ·~b|~a| · |~b|

=axbx + ayby + azbz

|~a| · |~b|.

where ϕ is the angle between a and b.

12

Physics Olympiad Script 1.1. VECTOR ALGEBRA

− The scalar product has an intuitive interpretation: One first projects the vector ~a on thevector ~b to get a vector ~a′ of length

|~a′| = |~a| · cos(ϕ)

(see fig. 1.2). Then the scalar product of ~a and ~b is the product

|~a| · cos(ϕ) · |~b| = |~a′| · |~b|

of the length of ~a′ with the length of ~b. This means that only the part of ~a which is parallelto ~b contributes to the scalar product. The same holds, if one interchanges ~a and ~b.

Figure 1.2: Projection of ~a on ~b

Properties We summarize the most important rules for calculations involving the scalar product.Let ~a, ~b, ~c be three vectors and s a real number. Then we have:

~a · ~a = |~a|2 = a2x + a2

y + a2z

~a ·~b = ~b · ~a

(s~a) ·~b = s ·(~a ·~b

)~a ·(~b+ ~c

)= ~a ·~b+ ~a · ~c.

Furthermore: If ~a,~b 6= ~0, then ~a ·~b = 0 if and only if ~a and ~b are orthogonal.Exercise: Does (~a ·~b) · ~c = ~a · (~b · ~c) hold?

13


1.1.2 Vector product

The scalar product of two vectors is a number (scalar). The vector product (also known as crossproduct) of two vectors is a vector, concretely:

Definition: Let ~a and ~b be vectors. The vector product of ~a and ~b is defined as

~a×~b =

axayaz

×bxbybz

=

aybz − azbyazbx − axbzaxby − aybx

Figure 1.3: Vector product

Properties

1. The vector product ~a ×~b is orthogonal to both ~a and ~b, given that ~a and ~b aren’t parallel(see. fig. 1.3). If ~a and ~b are parallel then ~a×~b = ~0.

2. The vectors ~a, ~b and ~a×~b follow the right hand rule (see. fig. 1.1).

3. If ϕ is the angle between ~a and ~b, we have |~a ×~b| = |~a| · |~b| · sin(ϕ). Therefore the absolutevalue of the vector product is equal to the area of the parallelogram with sides ~a and ~b (seefig. 1.3). This means that only the part of ~a, which is orthogonal to ~b contributes to ~a×~b.

Proof:

1. One calculates ~a ·(~a×~b

)= 0 and ~b ·

(~a×~b

)= 0. So it follows, that ~a and ~b are orthogonal to ~a×~b.

2. This is not really an obvious fact. The idea is to first show the fact for ~a and ~b in the xy-plane and then onereduces the general case to this.

14

Physics Olympiad Script 1.1. VECTOR ALGEBRA

Figure 1.4: Right hand rule [3]

3. One has sin2(ϕ) = 1− cos2(ϕ). Therefore:

(|~a| · |~b| · sin(ϕ)

)2

= |~a|2 · |~b|2 ·(1− cos2(ϕ)

)= |~a|2 · |~b|2 −

(|~a| · |~b| · cos(ϕ)

)2

= |~a|2 · |~b|2 −(~a ·~b

)2

.

On the other hand, one can calculate explicitly, that |~a|2 · |~b|2 −(~a ·~b

)2

= |~a × ~b|2. Therefore one gets

|~a×~b| = |~a| · |~b| · sin(ϕ).

More properties We summarize the most important rules for calculations involving the vectorproduct. Let ~a, ~b, ~c be three vectors and s a real number. Then we have:

~a× ~a = 0

~a×~b = −(~b× ~a

)(s~a)×~b = ~a×

(s~b)

= s ·(~a×~b

)~a×

(~b+ ~c

)= ~a×~b+ ~a× ~c(

~a+~b)× ~c = ~a× ~c+~b× ~c

Exercise: Does (~a×~b)× ~c = ~a× (~b× ~c) hold?

15


1.2 Differential calculus

Physics without calculus is impossible, since most physical laws in their general formulation usederivatives or integrals. In this chapter, we look at differential calculus. The next chapter willtreat integral calculus. The main reference for this chapter is [2].

1.2.1 Derivative of a function

Given a real function f , which maps a real number to another real number, we are often interestedin the slope of the graph of f at some point x0. With the slope of f , we mean the slope of thetangent at the graph of f through f(x0) (see fig. 1.5). How can we calculate this slope? For thiswe first think about how to calculate the slope of the secant through f(x0) and f(x0 +h) for someh > 0 (see. fig. 1.5): This slope is simply the difference in height divided by the difference inlenght, hence

f(x0 + h)− f(x)

(x0 + h)− x0=f(x0 + h)− f(x)

h.

Figure 1.5: Tangent and secant

If we make h smaller, the secant approaches the tangent and the slope of the secant approachesthe slope of the tangent. Therefore the slope of the tangent is the limit of the slope of the secant,when h goes to 0, written as limh→0. We call the slope of the tangent of f at the point x0 thederivative f ′(x0) of f at the point x0.

16

Physics Olympiad Script 1.2. DIFFERENTIAL CALCULUS

Definition: Let f be a real function and x0 a real number. Then we define the derivative of f atthe point x0 as

f ′(x0) = limh→0

f(x0 + h)− f(x0)

h. (1.1)

Remarks

− One often also writes dfdx (x0) for f ′(x0). Then df and dx stand for very small ("infinitesimal")

differences f(x0 + h)− f(x0) and (x0 + h)− x0.

− In physics f is often a function of the time t. Then, we often write

f(t0) =dfdt

(t0)

for the derivative of f at the point t0.

− The limit in the definition of the derivative doesn’t exist for every function. However, wewill only work with functions which are "nice enough" and we will always assume the limitexists everywhere.

− One often calls the derivative f ′(x0) the "instantaneous rate of change" of f at the pointx0, because it tells how fast f is changing at x0. With the help of the derivative, we canapproximate f near x0: For ∆x small, we have

f(x0 + ∆x) ≈ f(x0) + ∆x · f ′(x0), (1.2)

since f(x0) + ∆x · f ′(x0) is the value of the tangent line at the point x0 + h (see fig. 1.6).

Example We calculate the derivative of the function f(x) = x2 at the point x0 = 1. So wecalculate

f ′(1) = f ′(x0) = limh→0

f(x0 + h)− f(x0)

h= lim

h→0

(1 + h)2 − 12

h

= limh→0

1 + 2h+ h2 − 1

h= lim

h→0

2h+ h2

h

= limh→0

(2 + h) = 2.

Mostly, we do not only want to know the derivative of f at some point x0 (as in the example),but generally at an arbitrary point. We define the derivative function of f (mostly, we only sayderivative of f) as the function f ′ = df

dx , which maps every real number x to the derivative f ′(x)of f at the point x. Therefore the derivative f ′ of a function f is again a function. We also say"differentiate" instead of "calculating the derivative".

17


Figure 1.6: Derivative as approximation

Example We calculate the derivative function of f(x) = x2. We proceed just as before, but wewrite x instead of x0 = 1:

f ′(x) = limh→0

f(x+ h)− f(x)

h= lim

h→0

(x+ h)2 − x2

h

= limh→0

x2 + 2hx+ h2 − x2

h= lim

h→0

2hx+ h2

h

= limh→0

(2x+ h) = 2x.

So the derivative of f(x) = x2 is f ′(x) = 2x.

One can show analogously, that for a positive integer n, the derivative of f(x) = xn is the functionf ′(x) = n · xn−1. For example for f(x) = x25, we have f ′(x) = 25 · x24, or for g(x) = x we haveg′(x) = 1 (this can be verified immediately, since the slope of g(x) = x is equal to 1 everywhere).

1.2.2 Differentiation rules

Often it is not necessary to calculate the derivative explicitly using the limit (1.1), since oftenthe function is a combination of simpler functions, whose derivatives are already known. In thischapter we look at the corresponding differentiation rules.

Factor rule: Let s be a real number and g a function. If f(x) = s · g(x), then the derivativef ′(x) = s · g′(x).

18


Sum rule: Let g and k be functions. If f(x) = g(x) + k(x), then f ′(x) = g′(x) + k′(x).

Proof:

− Let f(x) = s · g(x). Then we have

f ′(x) = limh→0

f(x+ h)− f(x)

h= limh→0

s · g(x+ h)− s · g(x)

h

= limh→0

(s · g(x+ h)− g(x)

h

)= s · lim

h→0

g(x+ h)− g(x)

h

= s · g′(x).

− Let f(x) = g(x) + k(x). Then we have

f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

g(x+ h) + k(x+ h)− (g(x) + k(x))

h

= limh→0

g(x+ h)− g(x) + k(x+ h)− k(x)

h

= limh→0

(g(x+ h)− g(x)

h+k(x+ h)− k(x)

h

)= limh→0

g(x+ h)− g(x)

h+ limh→0

k(x+ h)− k(x)

h= g′(x) + k′(x).

Example We calculate the derivative of f(x) = 3x4− 2x2. We set g(x) = 3x4 und k(x) = −2x2,so we have f(x) = g(x) + k(x). Using the factor rule (and the rule for derivatives of powers in thelast section), we get g′(x) = 3 · 4 · x3 = 12x3 and k′(x) = (−2) · 2 · x = −4x. Using the sum rule,we get f ′(x) = g′(x) + k′(x) = 12x3 − 4x.

We are now able to differentiate sums (and also differences using the factor rule). We also wantto differentiate products and quotients. The first guess (g(x) · k(x))′ = g′(x) · k′(x) can be provedwrong, for example by looking at g(x) = k(x) = x. The correct rules are:

Product rule: Let g and k be functions. If f(x) = g(x) · k(x), then we have

f ′(x) = g′(x) · k(x) + g(x) · k′(x).

Quotient rule: Let g and k be functions. If f(x) = g(x)k(x) , then we have

f ′(x) =g′(x) · k(x)− g(x) · k′(x)

(k(x))2 .

19


Proof:

− Let f(x) = g(x) · k(x). Then we have

f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

g(x+ h) · k(x+ h)− g(x) · k(x)

h

= limh→0

g(x+ h) · k(x+ h)− g(x+ h) · k(x) + g(x+ h) · k(x)− g(x) · k(x)

h

= limh→0

k(x) · (g(x+ h)− g(x)) + g(x+ h) · (k(x+ h)− k(x))

h

= limh→0

k(x) · (g(x+ h)− g(x))

h+ limh→0

g(x+ h) · (k(x+ h)− k(x))

h

= k(x) · limh→0

g(x+ h)− g(x)

h+ limh→0

g(x+ h) · limh→0

k(x+ h)− k(x)

h

= g′(x) · k(x) + g(x) · k′(x)

− The quotient rule follows from the product rule: Let f(x) = g(x)k(x)

. Then we have g(x) = f(x) ·k(x). Therefore

by the product rule g′(x) = f ′(x) · k(x) + f(x) · k′(x). If we solve for f ′(x) and plug in f(x) = g(x)k(x)

, we find

f ′(x) =g′(x)− f(x) · k′(x)

k(x)

=g′(x)− g(x)

k(x)· k′(x)

k(x)

=g′(x) · k(x)− g(x) · k′(x)

(k(x))2.

Examples

1. We calculate the derivative of f(x) = x−1x+1 : We set g(x) = x− 1 and k(x) = x+ 1. Then we

have f(x) = g(x)k(x) . Furthermore g′(x) = 1 and k′(x) = 1. Therefore by the quotient rule:

f ′(x) =g′(x) · k(x)− g(x) · k′(x)

(k(x))2 =1 · (x+ 1)− (x− 1) · 1

(x+ 1)2=

2

(x+ 1)2

2. For a positive integer n, let be f(x) = 1xn = x−n. Then we have by the quotient rule

f ′(x) =(1)′ · xn − 1 · (xn)′

(xn)2=

0 · xn − 1 · n · xn−1

x2n

= −n · xn−1−2n = −n · x−n−1 = −n · 1

xn+1

Especially, we have found that the rule (xn)′ = n · xn−1 also holds for negative n.

We are now able to differentiate all polynomial functions and all fractions of polynomial functions.However, functions often appear as compositions of two functions. For example the functionf(x) = (x2−3x+13)17, can be written as f(x) = u(v(x)), where u(y) = y17 and v(x) = x2−3x+13.Theoretically, we could calculate f ′ by expanding (x3 − 3x + 13)17 and applying the above rules,but that would not be a very satisfying solution. We now formulate the chain rule, which dealswhith such compositions.

20


Chain rule: Let u and v be functions and f(x) = u(v(x)). Then

f ′(x) = u′(v(x)) · v′(x).

It is easy to remember the chain rule, using the notation f ′ = dfdx : We formally "expand the

fraction" and get

f ′ =dfdx

=du(v)

dx=

du(v)

dv· dvdx

= u′(v) · v′

Proof: Let f(x) = u(v(x)). Then we have

f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

u(v(x+ h))− u(v(x))

h

= limh→0

(u(v(x+ h))− u(v(x))

v(x+ h)− v(x)· v(x+ h)− v(x)

h

)= limh→0

u(v(x+ h))− u(v(x))

v(x+ h)− v(x)· limh→0

v(x+ h)− v(x)

h

= limh→0

u(v(x+ h))− u(v(x))

v(x+ h)− v(x)· v′(x)

We now define h = v(x + h) − v(x). For h → 0, we also have h → 0 (because we assumed v to be "nice enough").Therefore

limh→0

u(v(x+ h))− u(v(x))

v(x+ h)− v(x)= limh→0

u(v(x) + (v(x+ h)− v(x)))− u(v(x))

v(x+ h)− v(x)

= limh→0

u(v(x) + h)− u(v(x))

h= u′(v(x)).

We get f ′(x) = u′(v(x)) · v′(x).

Example We calculate the derivative of f(x) = (x2 − 3x + 13)17. We have f(x) = u(v(x)),where u(y) = y17 and v(x) = x2 − 3x + 13. We differentiate u and v and get u′(y) = 17y16 andv′(x) = 2x− 3. Therefore f ′(x) = u′(v(x)) · v′(x) = 17 · (x2 − 3x+ 13)16 · (2x− 3).

1.2.3 More derivatives

We want to differentiate more functions. We start with the trigonometric functions sine, cosineand tangent. We will always measure angles in radians and not in degrees.2 The graphs of sineand cosine are presented in fig. 1.7.

We now have a closer look at the graph of the sine function f(x) = sin(x) (see. fig. 1.8). Theslope at the point 0 seems to be approximately f ′(0) ≈ 1. At the point π

2 , the sine function

2Repetition: The magnitude of an angle in radians is the length of the corresponding arc of the unit circle. Forexample 360 correspond to 2π in radians, 180 correspond to π. In general: y corresponds to

y

360· 2π =

y

180· π

in radians.

21


Figure 1.7: The graphs of sine and cosine

obtains its maximum, therefore the derivative is f ′(π2

)= 0. At the point π, the derivative is again

approximately f ′(π) ≈ −1, at the point 3π2 it is again f ′

(3π2

)= 0, etc. We can draw the derivative

function (see fig. 1.8). If we compare the figures 1.7 und 1.8, we conjecture that the derivative ofthe sine function is f ′(x) = cos(x).

Figure 1.8: Derivative of the sine function

One can indeed prove this, but we don’t do this here, since the proof is quite technical. Analogously,one can find the derivative of g(x) = cos(x): It is g′(x) = − sin(x).For the tangent, we use

tan(x) =sin(x)

cos(x)

22


and apply the quotient rule. Then we have (where we use that sin2(x) + cos2(x) = 1):

tan′(x) =sin′(x) · cos(x)− sin(x) · cos′(x)

(cos(x))2

=cos(x) · cos(x)− sin(x) · (− sin(x))

cos2(x)

=cos2(x) + sin2(x)

cos2(x)=

1

cos2(x).

Alternatively we can also write this differently and get:

tan′(x) =cos2(x) + sin2(x)

cos2(x)= 1 +

sin2(x)

cos2(x)= 1 + tan2(x).

Finally, we have

tan′(x) =1

cos2(x)= 1 + tan2(x).

Example We calculate the derivative of f(x) = sin(x2) + (sin(x))2 For g(x) = sin(x2) we getusing the chain rule:

g′(x) = sin′(x2) · 2x = 2x · cos(x2).

for k(x) = (sin(x))2 we get also using the chain rule

k′(x) = 2 · sin(x) · sin′(x) = 2 sin(x) · cos(x).

Using the sum rule, we finally get

f ′(x) = g′(x) + k′(x) = 2x · cos(x2) + 2 sin(x) · cos(x).

Exponential function We now want to find the derivative of the exponential function

f(x) = ax,

where we should have a > 0. We start with the limit (1.1). Using the power rules, we get:

f ′(x) = limh→0

f(x+ h)− f(x)

h= lim

h→0

ax+h − ax

h

= limh→0

ax · ah − ax

h= lim

h→0

(ah − 1

h· ax)

=

(limh→0

ah − 1

h

)· ax (1.3)

We see, that the limit

limh→0

ah − 1

h(1.4)

doesn’t depend on x anymore, so it is simply a number.Exercise: What is the geometric meaning of the limit (1.4)?

23


h = 0.1 h = 0.01 h = 0.0012h−1h 0.717... 0.695... 0.693...

3h−1h 1.161... 1.104... 1.099...

Table 1.1: Calculation of the limit (1.4) for a = 2, 3

We want to understand this limit better and calculate it approximately for a = 2, 3, by pluggingin small numbers for h (see table 1.1). For a = 2, the limit (1.4) is smaller than 1 and for a = 3 itis bigger 1. Therefore there exists a number e, with 2 < e < 3, such that

limh→0

eh − 1

h= 1

For f(x) = ex we therefore have by equation (1.3), that f ′(x) = ex. The derivative of ex is againex. One can calculate that

e = 2.7182...

The number e is called Euler’s number and plays an important role in mathematics.

More about e: We defined e as the number, that satisfies limh→0eh−1h

= 1. We want to find a better representationfor e. By the definition of the limit, we get for very small h

eh − 1

h≈ 1.

Therefore eh ≈ 1 + h, respectivelye ≈ (1 + h)

1h .

If we take the limit for h→ 0, the ≈ gets again a =. If we set h = 1n, we find

e = limn→∞

(1 +

1

n

)n.

This is a widely used representation for e and is often also seen as the definition of e.

We come back to the question about the derivative of ax. For this, we define the natural log-arithm

ln(x) = loge(x).

This means that the number u = ln(x) satisfies the equation

eu = eln(x) = x.

We can use this as follows: Let again be a > 0 and f(x) = ax. With the power rules, we have

f(x) = ax =(eln(a)

)x= eln(a)·x.

With the chain rule:

f ′(x) = eln(a)·x · ln(a) =(eln(a)

)x· ln(a) = ax · ln(a).

The derivative of ax therefore is ln(a) · ax.

An interesting result is the derivative of ln(x). It holds

ddx

ln(x) =1

x.

24


Proof: We use the identity x = eln(x). If we differentiate this on both sides, we get using the chain rule

1 =ddxx =

ddxeln(x) = eln(x) · d

dxln(x) = x · d

dxln(x).

Thusddx

ln(x) =1

x.

Examples

1. We calculate the derivative of f(x) = 23x. Using the chain rule, we get f ′(x) = 3 · ln(2) · 23x.

2. We calculate the derivative of f(x) = xr for any real number r. We notice that

f(x) = xr =(eln(x)

)r= eln(x)·r.

We set u(y) = ey and v(x) = ln(x) · r. Therefore we have f(x) = u(v(x)). By the chain rule,we get

f ′(x) = u′(v(x)) · v′(x) = eln(x)·r · r · 1

x= r · x

r

x= r · xr−1.

1.2.4 Overview about derivatives

We summarize in table 1.2 the most important functions and their derivatives. They appear veryoften and one should know them by heart.

f(x) f ′(x)

xr r · xr−1

ex ex

ax ln(a) · axln(x) 1

xsin(x) cos(x)cos(x) − sin(x)tan(x) 1

cos2(x)= 1 + tan2(x)

Table 1.2: Derivatives of the most important functions

1.2.5 Higher derivatives

We define the second derivative f ′′(x) of a function f(x) as the derivative of f ′(x). Analogously, wedefine the third derivative f ′′′(x) as the derivative of the second derivative f ′′(x), etc. In general, wedefine the n-th derivative f (n)(x) recursively as the derivative of the (n−1)-th derivative f (n−1)(x).

25


Examples

1. Let f(x) = ex. Then

f ′′(x) =ddx

(f ′(x)) =ddx

(ex) = ex.

2. Let g(x) = sin(x). Then

g′′(x) =ddx

(g′(x)) =ddx

(cos(x)) = − sin(x).

3. Let k(x) = 12x

2. Then

k′′(x) =ddx

(k′(x)) =ddx

(x) = 1.

1.2.6 Taylor approximation

We have seen at the very beginning (see equation (1.2)) that the derivative of a function f at apoint x0 can be used to approximate the function f at x around x0: One just calculates the valueof the linear function that is tangent to f at x0, i.e. at x around x0 one approximates

f(x) ≈ f(x0) + (x− x0) · f ′(x0) =: T1,f,x0(x)

This is the best linear approximation of the function f near x0.This is already a special case (and by far the most important!) of a concept called Taylor approx-imation. The idea is the following: Why restrict to linear approximations? How about the bestquadratic approximation? For this we look for a quadratic function T2,f,x0(x) such that at x0 itgoes through f(x0) and has the same first and second derivative as f at x0. It is not hard to showthat T2,f,x0(x) is given by3

T2,f,x0(x) = f(x0) + f ′(x0)(x− x0) +1

2f ′′(x0)(x− x0)2.

Similarly, we can find the polynomial Tn,f,x0(x) of degree n, which approximates f best around x0,i.e. the polynomial of degree n such that for all k ≤ n

T(k)n,f,x0

(x0) = f (k)(x0).

Definition: This polynomial Tn,f,x0(x) is called Taylor polynomial of degree n of f at x0 andgiven by

Tn,f,x0(x) = f(x0) +n∑k=1

1

k!f (k)(x0)(x− x0)k

= f(x0) + f ′(x0)(x− x0) +1

2f ′′(x0)(x− x0)2 + . . .+

1

n!f (n)(x0)(x− x0)n

where n! = n · (n− 1) · . . . · 2 · 1.4

Figure 1.10 shows the Taylor polynomials up to degree 5 for the same function f as above.3This is essentially because a quadratic function has three "degrees of freedom", i.e. three parameters can be

chosen freely.4We additionally define 0! = 1.

26


Figure 1.9: Graphs of f , T1,f,x0and T2,f,x0

Examples

1. Look at f(x) = ex and x0 = 0. Since ddxe

x = ex, we have f (n)(x) = ex for all n. Thusf (n)(x0) = e0 = 1 for all n. Thus

Tn,f,0(x) = 1 +n∑k=1

1

k!xk

is the best polynomial approximation for ex of degree n around 0. Most importantly, we have

ex ≈ 1 + x

for x around 0.

2. Look at f(x) = sin(x) and x0 = 0. Then f ′(x) = cos(x), so f ′(0) = 1 and f ′′(x) = − sin(x),so f ′′(0) = 0. Thus

T2,f,0(x) = f(0) + f ′(0)x+1

2f ′′(0)x2 = x,

in particular for x around 0,sin(x) ≈ x.

3. Look at f(x) = cos(x) and x0 = 0. We have f(0) = cos(0) = 1. Furthermore f ′(x) =− sin(x), so f ′(0) = sin(0) = 0 and f ′′(x) = − cos(x), so f ′′(0) = − cos(0) = −1. Thus

T2,f,0(x) = f(0) + f ′(0)x+1

2f ′′(0)x2 = 1− x2

2,

which means that for x around 0,

cos(x) ≈ 1− x2

2.

27


Figure 1.10: Taylor polynomials of f (solid line) up to degree 5.

One can actually state more precisely how accurate the function f is approximated by Tn,f,x0 : Itholds that (if f satisfies certain conditions5)

f(x) = Tn,f,x0(x) +O(|x− x0|n+1).

O(|x− x0|n+1) means that the approximation error is of order |x− x0|n+1, i.e. there is a constantC > 0 such that when |x− x0| is small enough,

|f(x)− Tn,f,x0(x)| ≤ C · |x− x0|n+1.

1.2.7 The idea of differential equations

Differential equations are functional equations that also contain derivatives of the functions. Theyare crucial for physics since many physical laws can be formulated using differential equations.Let us start with an example:

f ′(x) = f(x).

This equation means that we are looking for a function f such that its derivative f ′ is equal tof . We know already that f(x) = ex is a solution to this differential equation. But actually, forevery c ∈ R, also f(x) = cex is a solution. So the solution to the above differential equation is notunique. To ensure uniqueness of the solution, we also need to fix the value of f at 0. If we thenpresent the differential equation as

f ′(x) = f(x), f(0) = 3,

(a so called initial value problem) the solution is given by f(x) = 3ex. One can show that thissolution is actually the unique solution.

5that we always assume to hold...

28

Physics Olympiad Script 1.3. INTEGRAL CALCULUS

We do not develop the theory of differential equations much further in this script (there are bookson this topic...). We just mention the general result that, if the differential equation does notbehave too bad, then for any given initial value there always exists a unique solution. Specifictypes of differential equations will directly be discussed when needed in the other chapters.

Examples

1. We slightly generalize our first example. Let c ∈ R and f0 > 0, and let

f ′(x) = cf(x), f(0) = f0.

Then the solution is given byf(x) = f0e

cx.

2. Look at the initial value problem

f ′(x) = 1 + f(x)2, f(0) = 0.

We already know thatf(x) = tan(x)

is a solution.

3. We now look at a second order differential equation (i.e. also second derivatives appear). Tomake the solution unique, we need two initial conditions, for example on f(0) and f ′(0). Letα > 0 and f0 ∈ R, and let

f ′′(x) = −α2f(x), f(0) = f0, f′(0) = 0.

This is the differential equation corresponding to a harmonic oscillator. The solution to thisdifferential equation is given by

f(x) = f0 cos(αx).

4. We generalize this equation a bit: Let α > 0 again, and let

f ′′ + 2αf ′ + α′2f = 0.

Then the solution is given byf(x) = (b1 + b2t)e

−αt,

where b1 and b2 are chosen according to the initial conditions. This corresponds to the caseof critical damping of a harmonic oscillator.

1.3 Integral calculus

Roughly speaking, integral calculus deals with the area under the graphs of functions. We will seethat differential and integral calculus are closely related. The main reference for this chapter is [2].

29


1.3.1 Antiderivatives

Definition: An antiderivative of a function f is a function F , which satisfies

f(x) = F ′(x) =dFdx

(x).

Calculating an antiderivative is therefore the reverse of calculating the derivative. For examplean antiderivative of f(x) = x2 is the function F (x) = 1

3x3, since F ′(x) = 1

3 · 3 · x2 = x2 = f(x).

However, the antiderivative F isn’t unique. For example the function G(x) = 13x

3 +42 also satisfiesG′(x) = f(x).In general: If f is a function and F an antiderivative of f , then for any real number c, also F (x)+cis an antiderivative of f . This follows because of

ddx

(F (x) + c) =ddxF (x) + 0 = F ′(x) = f(x).

It even holds that for two arbitrary antiderivatives F and G of f there exists a constant c suchthat

G(x) = F (x) + c.

(If F and G are antiderivatives of f , then the derivative of the difference F (x)−G(x) is

ddx

(F (x)−G(x)) =ddxF (x)− d

dxG(x) = f(x)− f(x) = 0.

Therefore F (x)−G(x) is constant.)

Antiderivatives of many functions can be easily calculated. We can essentially interchange thetwo columns of table 1.2 and get table 1.3. In the table for each function there is only given oneantiderivative. One obtains all the other antiderivatives by adding an arbitrary constant c. We

f(x) F (x)

xr für r 6= −1 1r+1 · x

r+1

1x ln |x|ex ex

ax ax

ln(a)

sin(x) − cos(x)cos(x) sin(x)

Table 1.3: Antiderivatives of the most important functions

have a factor rule and a sum rule for antiderivatives. They are the reversal of the correspondingrules for derivatives.

Factor rule: Let f be a function with antiderivative F and let s be a real number. Then s · F isan antiderivative of s · f .

Sum rule: Let f and g be functions with antiderivatives F and G. Then F+G is an antiderivativef + g.

30


1.3.2 Integral as an area

We define the integral of a function f as "signed area" between the graph of f and the x-axis.

Definition: Let f be a function and let a and b be real numbers with a ≤ b. Then the integral off from a to b ˆ b

af(x)dx

is the area between the graph of f and the x-axis in the region between a and b, where thearea under the x-axis counts negative (see fig. 1.11).

So the integral is a real number.

Figure 1.11: The integral´ baf(x)dx

In the example in figure 1.11, the integral isˆ b

af(x)dx = A−B + C.

Example We calculate ˆ 2

1f(x)dx

for the function f(x) = x. We look at figure 1.12 and see, that the area under the graph between1 and 2 is a trapezoid with height (2 − 1) and side lengths 1 and 2. Therefore the area of thistrapezoid is ˆ 2

1x dx =

1 + 2

2· (2− 1) =

3

2.

For a > b, we define ˆ b

af(x)dx = −

ˆ a

bf(x)dx.

31


Figure 1.12: The integral´ 21x dx

In this case, the area above the x-axis counts negative and the area under the x-axis positive.For arbitrary a, b and c, we then have

ˆ c

af(x)dx =

ˆ b

af(x)dx+

ˆ c

bf(x)dx. (1.5)

This follows because one can simply add the corresponding areas. This property is called "intervaladditivity".

Approximation of integrals using Riemann sums Let us mention a useful way to approx-imate integrals, which can be generalized later. One can calculate the integral

´ ba f(x)dx of a

function f by approximating it using so-called Riemann sums: One divides the interval [a, b] intoa sequence of points a = x0 < x1 < . . . < xn−1 < xn = b and calculates the area of the rectangleswith width ∆xk := xk − xk−1 and height f(xk) for all k = 1, . . . , n (see figure 1.13). The sum ofthe areas of the rectangles is

n∑k=1

f(xk) · (xk − xk−1) =

n∑k=1

f(xk) ·∆xk.

If one makes the partition a = x0 < x1 < . . . < xn−1 < xn = b finer (i.e. increases n) then thesum of the areas of the rectangles will converge to the integral

ˆ b

af(x)dx.

The notation of the integral is motivated by this approximation: In the limit, the∑

becomes anánd the ∆xk becomes a dx.

32


Figure 1.13: Approximation of integral with sums

1.3.3 Fundamental theorem of calculus

In this section, we are going to connect antiderivatives and integrals. The idea is to let vary theupper limit of the integral. In this way, one gets a function of the upper limit of the integral. Letf be a function and a an arbitrary number. Then we define

If,a(x) =

ˆ x

af(t)dt.

Since x already appears in the limits of the integrals, we have to take another integration variablet. The expression If,a(x) is independent of t, which is just a "dummy" variable. Of course, wehave for any real number b that

If,a(b) =

ˆ b

af(t)dt =

ˆ b

af(x)dx.

Example Let f(x) = x. We calculate If,1. We can proceed analogously as in the example above,we only have to replace 2 by x. Then we have

If,1(x) =

ˆ x

1f(t)dt =

ˆ x

1t dt =

1 + x

2· (x− 1) =

x2 − 1

2.

If we differentiate If,1(x) = x2−12 with respect to x, we get

I ′f,1(x) =ddx

(x2 − 1

2

)= x = f(x).

This is not a coincidence, but it’s exactly the statement of the fundamental theorem of calculus:

33


Fundamental theorem of calculus: Let f be a function and a a real number. Then If,a is anantiderivative of f . Concretely, this means:

I ′f,a(x) =ddx

(If,a(x)) =ddx

(ˆ x

af(t)dt

)= f(x).

Proof: We want to calculate

I ′f,a(x) = limh→0

If,a(x+ h)− If,a(x)

h.

For this, we look at If,a(x+ h)− If,a(x) more closely. Using the interval additivity (1.5), we have

If,a(x+ h)− If,a(x) =

ˆ x+h

a

f(t)dt−ˆ x

a

f(t)dt =

ˆ x+h

x

f(t)dt. (1.6)

We look at figure 1.14. Let fh,min be the minimal value of f between x and x+ h, and let fh,max be the maximalvalue of f between x and x+ h. Then we surely have

h · fh,min ≤ˆ x+h

x

f(t)dt ≤ h · fh,max.

If we divide by h, we get

fh,min ≤1

h·ˆ x+h

x

f(t)dt ≤ fh,max.

Using equation (1.6), we have

fh,min ≤If,a(x+ h)− If,a(x)

h≤ fh,max.

If we let go h→ 0, then fh,min → f(x) and fh,max → f(x), hence

f(x) ≤ limh→0


h≤ f(x).

Finally, we get

I ′f,a(x) = limh→0


h= f(x)

and therefore If,a is an antiderivative of f .

The fundamental theorem of calculus gives us a useful tool to calculate integrals. We want to knowthe value of the integral ˆ b

af(x)dx = If,a(b).

Let F be an antiderivative of f . From the fundamental theorem of calculus, we know that If,a(x)also is an antiderivative of f(x). Therefore there exists a real number c such that

If,a(x) = F (x) + c.

But then we have

F (b)− F (a) = If,a(b) + c− (If,a(a) + c) = If,a(b)− If,a(a) = If,a(b),

sinceIf,a(a) =

ˆ a

af(x)dx = 0.

Hence we get ˆ b

af(x)dx = If,a(b) = F (b)− F (a). (1.7)

We emphasize again, that the formula (1.7) does not depend on the choice of the antiderivative.For F (b)− F (a), we also write

[F (x)]ba = F (b)− F (a)

34


Figure 1.14: Proof of the fundamental theorem of calculus

Examples

1. We calculate ˆ 2

1x dx.

An antiderivative of f(x) = x is given by F (x) = 12x

2. So using equation (1.7),

ˆ 2

1x dx =

[1

2x2

]2

1

=1

2· 22 − 1

2· 12 =

3

2.

We indeed get the same result as above

2. We calculate ˆ 1

0x2 dx.

An antiderivative of f(x) = x2 is given by F (x) = 13x

3. So we get

ˆ 1

0x2 dx =

[1

3x3

]1

0

=1

3· 13 − 1

3· 03 =

1

3.

3. We calculate ˆ π

0sin(x) dx.

An antiderivative of sin(x) is given by − cos(x). Therefore,ˆ π

0sin(x) dx = [− cos(x)]π0 = − cos(π)− (− cos(0)) = −(−1) + 1 = 2.

35


The fundamental theorem of calculus motivates the following notation: Let f be a function. Thenwe also write ˆ

f(x)dx

for an arbitrary antiderivative of f . Note that this notation is not really mathematically correct,since

´f(x)dx is not unique, but it should always be clear from the context, what is meant.

1.3.4 More integration rules

Let f be a function. If we are given an antiderivative of f , it is in general easy to check that Fis really an antiderivative of f , since we just have to calculate the derivative F ′. But it can bevery difficult to find an antiderivative.6 In this section we will present more methods to calculateantiderivatives in certain situations. The first one is the method of integration by parts. Inte-gration by parts essentially is the opposite of the product rule. The second one is integration bysubstitution, which is the opposite of the chain rule.

Integration by parts We start with the product rule. Let u and v be two functions andf(x) = u(x) · v(x). Then the product rule says that

f ′(x) =ddx

(u(x) · v(x)) = u′(x) · v(x) + u(x) · v′(x).

If we calculate antiderivatives on both sides, we get

u(x) · v(x) =

ˆ (u′(x) · v(x) + u(x) · v′(x)

)dx

=

û′(x) · v(x)dx+

û(x) · v′(x)dx.

This implies û′(x) · v(x)dx = u(x) · v(x)−

û(x) · v′(x)dx. (1.8)

This equation looks very abstract at first glance and will now be clarified by examples.

Examples

1. We calculate an antiderivative of ex · x. Set u(x) = ex and v(x) = x. Then

ex · x = u′(x) · v(x).

6There even exist functions, where it is impossible: The function f(x) = e−x2

does not have an antiderivativewhich can be expressed by the usual functions.

36


Using equation (1.8) an antiderivative of ex · x is given byêx · x dx =

û′(x) · v(x)dx

= u(x) · v(x)−û(x) · v′(x)dx

= ex · x−êx · 1 dx

= ex · x− ex.

If we differentiate ex · x− ex, we see that it is indeed an antiderivative of ex · x.

2. We calculate an antiderivative of sin2(x). We set u(x) = − cos(x) and v(x) = sin(x). Thenu′(x) = sin(x) and therefore

sin2(x) = u′(x) · v(x).

Using equation (1.8), we calculateˆ

sin2(x) dx =

û′(x) · v(x)dx

= u(x) · v(x)−û(x) · v′(x)dx

= − cos(x) · sin(x)−ˆ

(− cos(x)) · cos(x)dx

= − cos(x) · sin(x) +

ˆcos2(x)dx.

We now usecos2(x) = 1− sin2(x)

and getˆ

sin2(x) dx = − cos(x) · sin(x) +

ˆcos2(x)dx

= − cos(x) · sin(x) +

ˆ (1− sin2(x)

)dx

= − cos(x) · sin(x) +

ˆ1 dx−

ˆsin2(x)dx

= − cos(x) · sin(x) + x−ˆ

sin2(x)dx.

Now we solve for´

sin2(x)dx and getˆ

sin2(x) dx =1

2· (− cos(x) · sin(x) + x)

=x

2− cos(x) · sin(x)

2.

Again, we can check by differentiating that this is indeed an antiderivative of sin2(x).

37


3. We calculate an antiderivative of ln(x). This does not look like a case for integration byparts, but we can write

ln(x) = 1 · ln(x).

Then we define u(x) = x and v(x) = ln(x). Now we have (since v′(x) = 1x)

ln(x) = u′(x) · v(x).

Using equation (1.8) and v′(x) = 1x we getˆ

ln(x)dx =

ˆ1 · ln(x)dx =

û′(x) · v(x)dx

= u(x) · v(x)−û(x) · v′(x)dx

= x · ln(x)−ˆx · 1

xdx

= x · ln(x)−ˆ

1 dx

= x · ln(x)− x.

Integration by substitution The second method we look at is integration by substitution.This is essentially the converse of the chain rule. We start with an example: We want to calculatean antiderivative of

f(x) = 2x · ex2 .We can write f as

f(x) = k′(x) · g′(k(x)),

where k(x) = x2 and g(y) = ey. With the chain rule, we have

f(x) = k′(x) · g′(k(x)) =ddxg(k(x)).

Therefore by definition of an antiderivative,

F (x) = g(k(x)) = ex2

is an antiderivative of f . This is actually the whole idea of integration by substitution.

Integration by substitution (version 1): Let u and v be functions and let U be an antideriva-tive of u. Then U(v(x)) is an antiderivative of u(v(x)) · v′(x). Henceˆ

u(v(x)) · v′(x)dx = U(v(x)).

One checks that U(v(x)) is really an antiderivative of u(v(x)) · v′(x) by differentiating using thechain rule. The simplest application of integration by substitution is the case where v is of theform

v(x) = ax+ b.

If u is a function and U is an antiderivative of u, then an antiderivative of u(ax+ b) is given byû(ax+ b) dx =

1

a

û(ax+ b) · a dx

=1

aU(ax+ b).

38


Remark Integration by substitution can be remembered and applied using a simple trick: Wesimply pretend that we can handle dv, dx and dv

dx just as normal variables. Then we can do thefollowing formal7 calculation. Write v′(x) = dv

dx . Then "v′(x)dx = dv", so

û(v(x)) · v′(x)dx =

û(v)dv = U(v(x)).

Examples

1. We calculate an antiderivative of e3x−2:ê3x−2 dx =

1

3· e3x−2.

2. We calculate an antiderivative of tan(x). For this we write

tan(x) =sin(x)

cos(x)= −− sin(x)

cos(x).

Set v(x) = cos(x). Then dvdx = − sin(x), so we write

− sin(x)dx = dv.

ˆtan(x) dx = −

ˆ1

v(x)· (− sin(x)) dx

= −ˆ

1

vdv = − ln(|v|) = − ln(| cos(x)|).

Definite integrals If one calculates definite integrals using the substitution rule, one gets thefollowing formula:

Integration by substitution (version 2): Let u and v be functions and let a and b be realnumbers. Then ˆ b

au(v(x)) · v′(x)dx =

ˆ v(b)

v(a)u(v)dv.

Example We calculate ˆ π2

0sin(x) · cos(x)dx.

Setv(x) = sin(x).

7Note that here "formal" means that we write things that are not properly defined.

39


Thus dv = cos(x)dx, so

ˆ π2

0sin(x) · cos(x)dx =

ˆ π2

0v(x) · cos(x)dx

=

ˆ v(π2

)

v(0)vdv

=

ˆ sin(π2

)

sin(0)v dv

=

[v2

2

]1

0

=1

2.

Two more elaborate examples At a first glance, functions for which one can apply integrationby substitution seem to have a very special form. But actually, the manipulations seen above canbe applied to a wide range of functions, which we want to illustrate on two examples.

1. We want to calculate an antiderivative of

1

ex + 1.

We set v(x) = ex. Thusdv = exdx.

Using integration by substitution, we haveˆ

1

ex + 1dx =

ˆ1

(ex)2 + ex· exdx

=

ˆ1

v(x)2 + v(x)· exdx

=

ˆ1

v2 + vdv

Now we write

1

v2 + v=

1

v(v + 1)=

(v + 1)− vv(v + 1)

=v + 1

v(v + 1)− v

v(v + 1)

=1

v− 1

v + 1.

We can simply integrate this and get thatˆ

1

v2 + vdv = ln(|v|)− ln(|v + 1|).

40


Therefore ˆ1

ex + 1dx =

ˆ1

v2 + vdv

= ln(|ex|)− ln(|ex + 1|)= ln(ex)− ln(ex + 1)

= x− ln(ex + 1).

If one differentiates x − ln(ex + 1) with respect to x, one can check that it is indeed anantiderivative of 1

ex+1 .

2. We want to find an antiderivative of √1− x2.

This is a case where one can apply integration by substituion "backwards", i.e. one replacesx by some function x(u) depending on some other variable u. With a bit of experience8, onesees that

x(u) = cos(u)

could be a good choice for a substitution. Then we have√1− x2 =

√1− cos2(u) = sin(u).

Note that x = cos(u), sodx = − sin(u)du.

Using integration by substitution, we getˆ √1− x2dx =

ˆ √1− cos2(u) · (− sin(u))du

=

ˆsin(u) · (− sin(u))du

= −ˆ

sin2(u)du

= −(u

2− cos(u) · sin(u)

2

)where we used the antiderivative of sin2(u) which we calculated in the section about integra-tion by parts. Now, we have to write this term as a function of x again. Note that

sin(u) =√

1− cos2(u) =√

1− x2.

Furthermore, we writeu = arccos(cos(u)) = arccos(x).

Thereforeˆ √1− x2dx = −

(u

2− cos(u) · sin(u)

2

)= −

(arccos(x)

2− x ·

√1− x2

2

).

We can differentiate this to see that it is indeed an antiderivative of√

1− x2.9

8One has to guess/see which function is suitable. There are even tables where one can look up the most common"standard substitutions".

9We use thatddx

arccos(x) = − 1√1− x2

.

41


1.3.5 The idea of multidimensional integrals

The idea of integrals is not restricted to functions of one dimension. It can be generalized to higherdimensions. Let’s illustrate this with an example in three dimensions. Assume we have some objectC, at which we look as a subset of the three dimensional space R3. We would like to know themass of C, but we only know the density ρ and the volume V of C. This is easy, since we can justcalculate the mass m of C from this information:

m = ρ · V.

But what if the density ρ is not constant? Let us first assume that it is at least piecewise constant,i.e. we can divide C into disjoint pieces C1, ..., Cn such that C1 ∪ C2 ∪ ... ∪ Cn = C and, oneach of C1, ..., Cn, the density is constant equal to ρ1, ..., ρn. Furthermore, let V1, ..., Vn denote thecorresponding volumes. Then the total mass of C is just the sum of all the masses of C1, ..., Cn,i.e.:

m = ρ1 · V1 + ...+ ρn · Vn =n∑k=1

ρk · Vk.

Now let us assume that the density ρ is actually an arbitrary function of the location ~x in C, i.e.ρ = ρ(~x), where ~x ∈ C ⊂ R3. Now we can still approximate the mass of C by dividing C into alarge number of small disjoint subsets ∆C1, ...,∆Cn, such that ∆C1 ∪ ... ∪ ∆Cn = C. Then foreach of the ∆Ck, we choose a location ~xk ∈ ∆Ck. For k = 1, ..., n, let ∆Vk denote the volume of∆Ck. Then we can approximate the mass m of C by

m ≈n∑k=1

ρ(~xk) ·∆Vk

where the approximation gets better making the partition finer. This value then converges to theactual mass m of C. In analogy to the one dimensional case, we write

m =

˚Cρ(~x)dV (~x).

As in the one dimensional case, the intuition is that in the limit the∑

gets a˝

and the ∆ getsa d. If we replace the density ρ = ρ(~x) by an arbitrary function f = f(~x), we have now alreadydefined ˚

Cf(~x)dV (~x)

for arbitrary10 functions f from R3 to R and subsets C ⊂ R3.

At the moment, we omit the discussion of how one actually calculates those integrals. It is moreimportant to understand the idea. This idea is not restricted to the three dimensional case. If wereplace the volume V by the area A, we can similarly define the integral¨

Cf(~x)dA(~x)

To see this, one differentiates y = cos(arccos(y)) using the chain rule and

cos2(x) + sin2(x) = 1.

10We implicitly assume that f is "nice enough".

42

Physics Olympiad Script 1.4. COMPLEX NUMBERS

for a subset C ⊂ R2 and a function f = f(~x) from R2 to R. One can see that this integral isexactly the (signed) Volume between C and the tow dimensional graph of f .We can generalize this idea even more. For example we can integrate functions on curved twodimensional surfaces in tree dimensional space. In the same spirit, we can also integrate functionsalong one dimensional curves in three dimensional space. The idea stays always the same: Onecuts the set on which one wants to integrate into small parts and then assumes the function f tobe constant on those small parts. Then one just sums over all the small parts to approximate theintegral.

1.4 Complex Numbers

This Chapter is (except for very small modifications) equal to Chapter 5 in [4], which was writtenby Lionel Philippoz.

1.4.1 Introduction

Should you encounter the following equation in a textbook

x2 = 1 (1.9)

and be asked to find its solutions in R, it would not be that difficult to conclude that there aretwo of them, namely 1 and −1. But what happens if one slightly modifies Eq. (1.9) by changinga sign?

x2 = −1 (1.10)

Can you find a real solution? Actually not, since for any real number x, x2 ≥ 0. In the realnumbers,

√−1 is not defined. This equation thus possesses no solutions in R. However, one can

expand the set of real numbers to the so-called set of complex numbers C in which Eq. (1.10)actually has two (complex) solutions.

1.4.2 Representation of a complex number, Euler formula

Complex numbers are not so different from real numbers, and all you actually need to know is thatwe define a new number i ∈ C such that i2 = −1. And that’s it! You can now solve Eq. (1.10) inC and find its two solutions: i and −i.

Any complex number z ∈ C can be written as the sum of two numbers:

z = x+ iy (1.11)

where x, y ∈ R and i as previously defined. x is also called the real part of z (sometimes writtenas <(z) or Re(z)), whereas y is called the imaginary part of z (written as =(z) or Im(z)).

If y = 0, then z is simply a real number, and when x = 0, we say that z is an imaginary number.As you can see, any complex number can now be defined using two “coordinates” x and y, whichmeans it can actually be represented in a plane, the so-called. . . complex plane!

43


R

iR

-1 1 2 3 40

−i

i

2i

3i

z = 3 + 2i

|z|

θ

Figure 1.15: z can be seen as a point in the complex plane, with cartesian coordinates (x, y) or polarones (|z| , θ).

Another possibility to describe the position of a point in a plane consists in using polar coordinates,where one needs to give the distance from the origin as well as the angle between the x-axis (herethe real axis) and the line connecting the origin and the point11. If you know your trigonometrywell, you can then easily relate both coordinate systems by writing:

x = |z| cos(θ)

y = |z| sin(θ)

and z can thus be written as

z = x+ iy

= |z| cos(θ) + i |z| sin(θ)

= |z| (cos(θ) + i sin(θ))

= |z| eiθ (1.12)

The last step is actually performed using the Euler formula:

eiθ = cos(θ) + i sin(θ) (1.13)

which we will take as a definition of a complex exponential function in this script. This is avery important relation which allows one to switch between a trigonometric representation andan exponential function, which is much more easy to handle with. You can actually write bothtrigonometric functions sin(θ) and cos(θ) as a function of complex exponentials! This is done asfollows:

eiθ = cos(θ) + i sin(θ) (1.14)e−iθ = cos(−θ) + i sin(−θ)

= cos(θ)− i sin(θ) (1.15)11If you want to consider z as a vector, then x and y are the components of that vector, |z| the norm of z and θ

the angle between the vector and the x-axis.

44

Physics Olympiad Script 1.4. COMPLEX NUMBERS

If you now add (resp. substract) Eq. (1.14) and Eq. (1.15), you can eliminate the sin part (resp.cos) and solve for cos (resp. sin), which leads to

cos(θ) =eiθ + e−iθ

2(1.16)

sin(θ) =eiθ − e−iθ

2i(1.17)

1.4.3 A first simple application

One useful application of the Euler formula lies in the fact that dealing with trigonometric functionis not always that easy. For instance, do you know by heart how to write sin(2θ) or cos(2θ) withonly functions of the angle θ, not 2θ? If you forgot about those formulas, just switch to the complexworld!

ei(2θ) = cos(2θ)︸︷︷︸<(e2iθ)

+i · sin(2θ)︸︷︷︸=(e2iθ)

ei(2θ) =(eiθ)2

= (cos(θ) + i sin(θ))2

= cos2(θ) + 2i sin(θ) cos(θ) + i2 sin2(θ)

= cos2(θ)− sin2(θ)︸︷︷︸<(e2iθ)

+i · 2 sin(θ) cos(θ)︸︷︷︸=(e2iθ)

where we used the fact that i2 = −1, as you should know by now. It is now easy to conclude byequating the two expressions of the real part (and identically for the imaginary part):

sin(2θ) = 2 sin(θ) cos(θ)

cos(2θ) = cos2(θ)− sin2(θ)

1.4.4 Physical examples

One possible application of complex numbers to physics resides in the representation of any periodicsystem which would require trigonometric functions to be described, such as waves for instance. Ifwe consider a wave function ξ(x, t) 12, it will usually be written as

ξ(x, t) = ξ0 cos(kx− ωt)

or, using the Euler formulaξ(x, t) = ξ0 <

(ei(kx−ωt)

)However, you will never encounter this notation with the real part (or imaginary part if youconsidered a sin-function) in any physics book, where the wave is simply denoted as

ξ(x, t) = ξ0 ei(kx−ωt)

12ξ(x, t) is the displacement of the wave, which depends on the position x and the time t and ξ0 is the amplitude.

45


Physicists simply assume that the real part is taken at the end of their calculations (and that ways,it is shorter to write and much easier to read, don’t you agree?).

What is now the velocity ξ(x, t) ≡ dξ(x,t)dt or the acceleration ξ(x, t) ≡ d2ξ(x,t)

dt2 ? Well, you justneed to differentiate the wave function ξ(x, t), and it is much easier to do it when considering anexponential function!

46

Chapter 2

Mechanics

2.1 Getting Ready

This first section is devoted to the application of Newton’s laws of motion in three dimensionsand to the use of vectorial quantities in classical mechanics. You might already know most ofthe stuff presented in this chapter since the fundamental laws of mechanics are supposed to bediscussed in every high school physics course. However, the extent up to which these laws areapplied in Olympiad problems might be somewhat new. That’s why it could still be useful for youto work at least on the examples presented in this section.

First we will deal with point masses, or point-like particles, that is, with objects of a certain massm occupying a negligible volume in space. They have, however, a definite position, denoted by athree-dimensional vector ~r. This position vector will usually be subject to change in time – that’swhat we call motion. Any change in motion is caused by what we call forces, and Newton’s lawsof motion do nothing but tell you how forces and changes of motion are related to each other.1

Objects consisting of more than just one point-like particle are discussed in section 2.4 and 2.5.You will see there how the laws valid for single point masses can be generalized to extended systemsby introducing just a handful of new concepts.

2.1.1 Mathematical Notation

This paragraph contains a few remarks about mathematical notation used in this chapter. It is notsupposed to provide you any review about mathematical concepts, as they are taught in a differentcourse.

If we want to describe vectorial quantities in three-dimensional space, we have to choose a coordinateframe or coordinate system, that is, a set of three axes perpendicular to each other such that wecan describe any vectorial quantity in terms of its projections along these axes. Usually, we will callthe coordinate frame an xyz-frame you have hopefully been familiar with since your middle schoolgeometry classes. The corresponding unit vectors will be written as x := (1, 0, 0), y := (0, 1, 0) and

1I admit, so far, nothing too new or complicated. Still, I think some useful insight can be gained by reflectingfor a few moments on the most basic concepts once in a while.

47

Physics Olympiad Script CHAPTER 2. MECHANICS

z := (0, 0, 1), respectively. In such a coordinate frame, the vectorial quantity ~b with componentsbx, by, bz can be written as both

~b =

bxbybz

(2.1)

or~b = bx x+ by y + bz z. (2.2)

Personally, I prefer the second variant, as it makes calculations more compact. That’s why youwill find it throughout this chapter.

In general, vectorial quantities will depend on time t, which can be expressed as something like~b(t). If there is no potential misunderstanding, however, dependence on time will not be writtenexplicitly (typing all those brackets and ts becomes quite annoying after a while) and you will findjust ~b instead of ~b(t). If a certain emphasis is put on dependence on time, the notation with (t)will be preferred. Hope this doesn’t bother you too much.

Time derivatives of vectorial quantities are denoted by

d

dt~b =

dbx/dtdby/dtdbz/dt

=

bxbybz

= bx x+ by y + bz z (2.3)

Personally, I find dots over vector arrows quite ugly, that’s why I won’t write anything like ~b butmake use of the differential quotient as on the very left in equation (2.3) instead, that is, somethinglike d~b/dt. The dot might of course be used for scalar quantities, e.g. for the time derivatives ofthe components of a vector as in the second and third expression of (2.3).

An important property of a vector is its magnitude, i.e. its length. The magnitude of the vector~b = bx x+by y+bz z in a given coordinate frame xyz can be determined by Pythagoras’ theoremas ∣∣∣~b∣∣∣ =

√b2x + b2y + b2z. (2.4)

Throughout this chapter, we will use a simple convention: the magnitude of a vector will just bedenoted by the letter symbolizing the vector without the arrow on top. That is, we will just writeb for

∣∣∣~b∣∣∣.In many cases, we will have to discuss the position or motion of objects in space with respect to acertain direction. Of course, introducing a Cartesian frame and expressing everything in terms ofcoordinates would always be a possibility, but it is in general more useful to find descriptions notdependent on any particular coordinate system. Very often the simplest way to refer to a certaindirection in space is to introduce a unit vector, that is, a dimensionless vector of length 1, pointingin that direction2. Unit vectors will be denoted by little hats instead of arrows on top of a letter,which can be either Latin or Greek – e.g. n, ϑ or u.

One more thing, before you get started. Some very important concepts, especially those involvingcircular or rotational motion, are best described in plane polar coordinates. For radial distance

2I know, the length of such a vector can only be calculated if one knows its coordinates with respect to a certainframe, and we just wanted to get rid of frames here. But, if you think of it, the length of a vector is a quantityindependent of choice of the coordinate system, such that it does not depend on any particular coordinate frame.

48

Physics Olympiad Script 2.1. GETTING READY

from the origin we will usually write r, for the polar angle with respect to the x-axis we will useϕ (the Greek letter phi) or ϑ (the Greek letter theta).3 The identities enabling you to convert theCartesian coordinates of a point (x, y) into its polar coordinates (r, ϕ) are then

r =√x2 + y2 + z2

tanϕ =y

x.

(2.5)

With

x = r cosϕ

y = r sinϕ(2.6)

you can convert coordinates the opposite way from polar to Cartesian.

2.1.2 Kinematics in Three Dimensions

We assume that you are familiar with some basic concepts of vector calculus, as they are thesubject of a mathematics course in your Olympiad training. Here you will see how these conceptscan be applied on description of motion of objects in three dimensions, that is, in kinematics.

The position vector of a point-like object in three-dimensional space will be written as

~r = x x+ y y + z z, (2.7)

with the object’s coordinates x, y, z with respect to a given Cartesian frame. The curve describedby ~r with time t in this coordinate frame is called the trajectory of the particle. We will denotevelocity vectors by

~v = x x+ y y + z z (2.8)

and acceleration vectors by~a = x x+ y y + z z. (2.9)

I assume you are familiar with geometrical interpretation of these vectors.

Important properties of kinematic quantities:

− The velocity vector ~v(t) of a particle at time t is always a tangent to its trajectory in ~r(t).

− If the magnitude of the velocity vector of a particle is constant over time, then its accelerationvector is at any time perpendicular to the velocity vector.

3See how this notation is consistent with our notation for the magnitude of vectors: in polar coordinates, thedistance r from the origin is the same as the magnitude of the position vector ~r, which is also denoted by the letterr.

49


− The magnitude of the velocity vector ~v is equal to the rate of change of path length s of theparticle along its trajectory:

|~v(t)| = ds

dt(2.10)

Accordingly, one is able to calculate the path length sA→B between two points A and B alongthe trajectory of a particle by integrating the magnitude of its velocity over time:

sA→B =

ˆ tB

tA

|~v(t)| dt (2.11)

where tA is the time at which the particle is in point A and tB is analogous for point B.

2.1.3 Describing Uniform Circular Motion

To show you how all these mathematical objects and definitions can be applied on mechanics, wewill derive the formulas for uniform circular motion you are most likely already familiar with fromyour high school physics classes.

Consider a point-like particle rotating uniformly along a circular trajectory. Uniformly means thatduring equal times, a radius vector pointing from the center of the circle to the position of theparticle sweeps out sectors of the circle of equal area. In other terms, the angular velocity ω = dϕ

dtof the motion is constant where ϕ denotes the angle of rotation.

First we introduce a Cartesian coordinate frame enabling us to give an analytic description ofmotion of the particle as it rotates along the circle. You might have realized that motion of theparticle is constrained to the plane of the circle, which means it is two-dimensional. It seemsnatural to describe two-dimensional motion in a just two-dimensional coordinate system xy lyingin the plane of the circle. Any other choice would mean we would have to deal with an unnecessarythird coordinate. If the origin of this frame coincides with the center of the circle, we can expectthe expressions and equations involved in mathematical description of the situation to assumequite simple forms. After all, the geometrical symmetries of the situation must be found in theequations describing it.

From your courses in analytic geometry, you might be used to the convention of choosing the axesdescribing circular motion in such a way that anticlockwise motion in the xy-plane correspondsto positive rotation, that is, the polar angle ϕ increases with time as the particle moves along thecircle. Remember that this is just a convention. One could work with a polar angle defined aspositive in clockwise sense, or even with a polar angle with the y-axis as a reference instead of thex-axis. It could be interesting for you to work out the equations presented in this paragraph usingdifferent conventions for the polar angle, just for the sake of practising your geometry skills.

Let R be the radius of the circle along which the particle moves. In the coordinate frame we justchose, the position vector ~r of the particle at time t can be expressed as

~r(t) = (R cos (ωt)) x+ (R sin (ωt)) y, (2.12)

where we just assume that the particle starts on the x-axis at t = 0. We could have made any otherchoice for the starting point, which would have required introduction of a new variable indicating

50

Physics Olympiad Script 2.1. GETTING READY

the initial position of the particle, e.g. an initial polar angle ϕ0. Instead of just ωt, the argumentof the trigonometric functions in the coordinates would be ωt+ϕ0. Again, we want to keep thingsas simple as possible, and we choose ϕ0 = 0 by rotating the coordinate frame in such a way thatthe x-axis passes through the point where the particle lies at t = 0.

Determining the velocity vector is now quite straightforward, as this just involves taking thederivative of the components of ~r(t) with respect to time:

~v(t) =d

dt~r

=d

dt(R cos (ωt)) x+

d

dt(R sin (ωt)) y

= (−Rω sin (ωt)) x+ (Rω cos (ωt)) y

Now let’s look at the vector ~v a bit closer. Its magnitude at time t is given by

v(t) =

√(−Rω sin (ωt))2 + (Rω cos (ωt))2

=√R2ω2

(sin2 (ωt) + cos2 (ωt)

)=√R2ω2 · 1

= Rω,

which is constant over time. Why doesn’t this surprise us? If we look at the particle at twodifferent instants of time, the physical situation is essentially the same except for orientation inspace since we are looking at uniform circular motion. However, orientation in space does notchange the magnitude of any vectorial quantity involved – that’s why the magnitude of ~v mustbe independent of time. In school, you might have seen a much easier derivation for v by a verysimple calculation – according to (2.10) and since v is constant over the whole trajectory, we canexpress it as just the circumference of the circle (i.e. the path length) divided by the time it takesthe particle to cover it:

v =2πR

2π/ω= Rω.

Of course, this derivation is much easier and requires less complex mathematical tools than theone we used before involving vector derivatives. It is my aim, however, to show you how usefulvector calculus can be.

We wish to investigate how the orientation of ~v in the xy-plane changes with time. If you lookcarefully at the expression we derived for ~v, you see that this vector is always perpendicular to theposition vector of the particle:

~v(t) · ~r(t) = Rω (− sin (ωt) x+ cos (ωt) y) · ((R cos (ωt)) x+ (R sin (ωt)) y)

= R2ω− sin (ωt) cos (ωt) x · x− sin2 (ωt) x · y+ cos2 (ωt) y · x+ cos (ωt) sin (ωt) y · y

= 0

at any time t.

51


Acceleration ~a(t) of the particle is also easily determined by taking the time derivative of thevelocity vector ~v(t):

~a(t) =d

dt~v

=d

dt(−Rω sin (ωt)) x+

d

dt(Rω cos (ωt)) y

= −Rω2 sin (ωt)−Rω2 cos (ωt) .

Just take a closer look at this expression. Indeed, we can rewrite

~a(t) = −ω2 ~r(t)

and see that the acceleration of the particle always points radially towards the center of the circle.That’s why it is also called centripetal acceleration. In analogy to the discussion for v, the magnitudeof ~a must also be constant over time:

a = Rω2.

Finally, the laws governing uniform circular motion can be summarized as follows:

Consider a particle moving with uniform angular speed ω along a circular trajectory of radius R.

− The velocity vector ~v of the particle is always tangential to the circle pointing in direction ofmotion of the particle and is of constant magnitude

v = ωR. (2.13)

− The acceleration vector ~a of the particle always points radially towards the center of thecircle and is of constant magnitude

a = ω2R =v2

R, (2.14)

or, in vector notation,~a = −ω2~r (2.15)

with the radius vector ~r pointing from the center of the circle to the instantaneous positionof the particle.

2.1.4 Choice of the Coordinate Frame

In most mechanics problems, one has to find a coordinate frame suitable for the situation. Thereis no recipe for how to choose the orientation of the coordinate axes and the position of the origin.However, one can say that making use of the symmetries of the problem can generally be helpfulin that, as a general rule, these symmetries must be reflected in the equations. For instance, if thesituation is symmetric with respect to a certain line, all equations involved must display symmetrywith respect to this line. If a system is invariant with respect to rotation about a certain axisabout any angle, then all equations involved must also be.

52

Physics Olympiad Script 2.2. THE LAWS OF MOTION

Note that there is no ‘wrong’ choice of coordinate system. The laws of mechanics must not changeunder rotation in space. That is, the solutions of a problem making use of two different coordinateframes must be physically equivalent.

2.2 The Laws of Motion

You are now ready to apply the mathematical concepts of vector calculus you encountered in yourmathematical training together with the notation introduced in the last section on real physics. Fora start, this section will show you how to deal with the most basic laws of mechanics in problemsdealing with motion of single point-like objects.

2.2.1 Newton’s Laws of Motion

Let’s jump straight into matters and work with something you already know from school: New-ton’s laws of motion. By body or object we shall denote a point-like mass. Systems of point-likemasses and extended objects, especially rigid bodies, will be discussed in later sections.

Newton’s Laws of Motion

1. A body does not experience any acceleration if the total force acting on it vanishes.

2. If the total force acting on a body is given by ~F , its linear momentum ~p changes with timet according to the first-order differential equation

d

dt~p = ~F . (2.16)

3. Two bodies always act on each other by means of forces of equal magnitude and oppositedirection, i. e. if we denote by ~FA→B the force by which object A acts on object B and by~FB→A the force by which B acts on A, then

~FB→A = −~FA→B (2.17)

holds.

A few remarks, before you start working on the examples.

− You may have noticed that we did not state the second law in the probably better known,yet less general form

m~a = ~F (2.18)

with the mass m of the body and the acceleration ~a it experiences, but that we used a versioninvolving the linear momentum of the body instead. Recall that the linear momentum of anobject is defined as the product of its mass and its velocity vector ~v:

53


~p := m~v. (2.19)

Equation (2.18) is just a special case of the more general form (2.16) for a body whose massremains invariant over time:

~F =d

dt~p = m

d

dt~v = m~a.

Even if most Olympiad problems deal with objects of constant mass, there have been somenice examples involving bodies whose mass varies during their motion.

− Equation (2.16) is not the definition of a force! In classical mechanics, forces may be defined asinteractions between bodies resulting in acceleration of masses.4 You might have heard thattheir nature is always one of the four fundamental interactions gravity, electromagnetism,weak and strong nuclear interaction. Laws like Coulomb’s law of electrostatic interaction

~FA→B =1

4πε0

qAqB|~rB − ~rA|3

(~rB − ~rA)

and Newton’s universal law of gravity

~FA→B = −G mAmB

|~rB − ~rA|3(~rB − ~rA)

allow you to calculate such forces between particles starting from properties like their massor their electrical charge. In many cases, you may not deal directly with fundamental in-teractions between particles but with macroscopic forces that can be described by means ofempirical formulas like Hooke’s law

~F = −k~x

for ideal springs, or the laws governing friction as described in section 2.2.3.

Equation (2.16) states how a force acts, and does not provide you any information about theorigin of a force. It relates the quantity ~F to the rate of change of the quantity ~p. In otherwords, we know that whenever there is a force acting on a point-like mass, the rate of changeof its linear momentum must be equal to that force.

− Indeed, the Second Law is a vector equation. Consider a body of constant mass m on which atotal force ~F acts. If we define a Cartesian coordinate system and call the body’s coordinatesx, y and z, the statement of equation (2.16) translates as three distinct scalar equations, onefor each coordinate:

md2x

dt2= Fx

md2y

dt2= Fx

md2z

dt2= Fz.

4To be honest, the whole story is way more complicated than this. However, it is sufficient you understand thisstatement for now.

54


These differential equations of second order are known as the equations of motion of the bodysubject to the force ~F . In the case where the components of the force are independent oftime, motion of the body is simple accelerated motion in every dimension.

Note that for a problem in two or three dimensions the Second Law is applied independentlyin each coordinate. Choosing a suitable coordinate system can be of immense importance forthe solution of a problem. The examples you encounter in this chapter will hopefully makethis point clear enough.

− Think carefully about the word total in the statement of the first and second law. A point-like mass can be subject to more than one force and still move at constant velocity, providedthe forces cancel each other in the vector sum.

− It seems pretty obvious that the first law is a consequence of the second – just set ~F = 0 onthe right hand side of (2.16) and you get for an object of constant mass m:

md

dt~v = 0,

meaning that the velocity vector ~v remains constant over time. Note that both the magnitudeand the direction of the velocity vector do not change!

Example 2.2.1.1: (USA 2009 ) A ball of mass m is thrown vertically upward with a speed ofv0. The ball is subject to an air resistance force that is proportional to the velocity F = −kv.The ball rises up to height of h and then returns to the starting point after some total timetf . The acceleration of free fall is g.

Determine the following:

a) An expression for the velocity as a function of time in terms of any or all of the constants.

b) Sketch the graph of velocity vs time. On the sketch indicate the time to rise to thehighest point, tr, and the total time of flight, tf . You do not need to find either tr or tfat this point, but your graph should reflect the relative position of both.

c) Find the time to rise, tr, to the highest point, h, in terms of any or all of the constants.

Solution: For the start, this is a problem in one dimension only. However, the math involvedwill turn out to be quite laborious.

a) We begin by setting up an appropriate coordinate system. Let’s define an x-axis withorigin at the starting point of the motion, positive x-values in upward direction. Withthis choice, the total force acting on the ball is given by −kv −mg, where v := x. TheSecond Law then gives the equation of motion

mdvdt

= −kv −mg,

which is a first-order differential equation in v(t). Separation of variables yields

dvv + mg

k

= − km

dt,

55


a form ready for integration. Before we do so, we need to know between what limits wewant to integrate. Since we are looking for an expression for v(t), where t = 0 correspondswith the start of motion, it makes sense to integrate between t = 0 and an arbitrary timet > 0. The corresponding values for v are then v(t = 0) = v0 and the unknown v(t). Thecalculation becomes

ˆ v(t)

v0

dv′

v′ + mgk

=

ˆ t

0

(− km

)dt′

⇒ ln∣∣∣v(t) +

mg

k

∣∣∣− ln∣∣∣v0 +

mg

k

∣∣∣ = − km

(t− 0)

⇔ v(t) = −mgk

+(v0 +

mg

k

)e−

kmt

where the integration variables v′ and t′ were introduced not to confuse them with thevalues of the integration limits v and t.

b) The graph of v(t) is shown in the picture. tr is the time at which v(tr) = 0, tf the time> 0 at which x(tf ) = 0. The condition for tf is that at that time, the ball reaches x = 0again, or ˆ tf

0v(t) dt = 0

in terms of velocity v(t). This integral can be split up asˆ tr

0v(t) dt+

ˆ tf

tr

v(t) dt = 0,

meaning that tf is the time for which both shaded areas in the figure are equal.

c) From v(tr) = 0, one finds

tr =m

kln

(1 +

kv0

mg

)

Example 2.2.1.2: A parallel-plate capacitor is placed in the horizontal plane such that thenegatively charged plate lies below the positively charged one. Voltage between the plates isU and they lie a distance d apart. A particle of mass m and charge q is injected at a velocityv from a hole in the lower plate under an angle α with the horizontal plane into the spacebetween the plates. The particle is also subject to a uniform gravitational field of acceleration~g pointing downwards.

a) Determine the conditions for U , d, q, m for which, after injection into the capacitor, theparticle eventually returns to the lower capacitor plate on its trajectory.

b) If these conditions are satisfied, determine how much time the particle spends in thespace between the capacitor plates before hitting the lower one.

c) Again, if these conditions are satisfied, determine at what distance from the injection

56


point the particle hits the lower capacitor plate.

Solution: The forces acting on the particle are its weight m~g and an electrostatic force exertedby the uniform electrical field ~E of the capacitor. The equation of motion of the particle lookslike

m~g + q ~E = m~a.

It looks reasonable to introduce a coordinate system with origin at the injection site of theparticle, x-axis on the lower plate and upwards pointing y-axis in such a way that the initialvelocity vector of the particle lies in the xy-plane. Since there are no horizontal force com-ponents acting on the particle, we do not have to worry about orientation of the z-axis. Theequation of motion for the z-component will anyway assume the trivial form z = 0 with initialconditions z(0) = 0, z(0) = 0, from which z(t) = 0 at any time t.

In this coordinate system, the weight of the particle is −mg y. As the positively chargedplate lies on top of the negatively charged one, the electric field points downward in the samedirection as gravity:

~E = −Udy

and the equation of motion of the particle in the capacitor becomes

m~a = −mg y − q Udy.

For the x-component, this means x(t) = 0 for all t, implying that x(t) is a constant over time.The initial velocity vector of the particle is

~v(0) = (v cosα) x+ (v sinα) y

as described in the problem. Therefore,

x(t) = v cosα

and, upon integration and with x(0) = 0 for the initial position of the particle,

x(t) = (v cosα) t.

The y-component is a bit more elaborate: acceleration

y(t) = −g − qU

dm

is a constant, meaning that velocity y(t) is a linear function and position y(t) a quadraticfunction of time. Integration yields

y(t) = −(g +

qU

dm

)t+ v sinα

with the initial velocity y(0) = v sinα. Repeated integration yields position as

y(t) = −1

2

(g +

qU

dm

)t2 + (v sinα) t.

57


The trajectory of the particle in the xy-plane is found by eliminating t from the equation forx(t) and substituting the result into the equation for y(t), leading to

y = −(g +

qU

dm

)1

2v2 cos2 αx2 + (tan)αx.

This equation describes a parabola in the xy-plane passing through the origin. To determinethe shape of the parabola, we rewrite it in the form

y = −Ax2 +Bx = −A(x− B

2A

)2

+B2

4A, (2.20)

with two parameters

A :=

(g +

qU

dm

)1

2v2 cos2 α

B := tanα

a) For A < 0, the parabola opens upwards and the particle never returns to the lower plate.Therefore, the condition A > 0, or

g +qU

dm> 0, (2.21)

must be satisfied. But this is not enough: if the peak of the parabola lies more than dabove 0, the particle will hit the upper capacitor plate before it can return to the lowerone. Therefore, the condition

y|x= B2A< d,

that is, and you are welcome to verify,

v2 tan2 α < 4d

(g +

qU

md

)(2.22)

must apply. As v2 tan2 α > 0 anyway, this condition makes (2.21) superfluous in that(2.21) is anyway satisfied if (2.22) is.

b) Under the conditions of a), the point of return of the particle to the lower capacitor plate issimply the intersection of its parabolic trajectory with the positive x-axis. The conditiony = 0 applies, therefore we can find the time t at which this occurs by computing thepositive zero of y(t):

t =2v sinα

g + qUdm

c) This distance can be obtained in different ways, for example, by setting y = 0 and solving(2.20) for x:

x|y=0 =B

A=v2 sin (2α)

g + qUdm

.

58


2.2.2 Dynamics of Circular Motion

Be aware that this section is extremely important. Problems involving circular motion, besidesbeing a very common subject of Olympiad problems, are some of the best indicators of whethersomeone feels comfortable around Newton’s second law or not.

Uniform Circular Motion

We consider a particle of massm moving uniformly with angular speed ω along a circular trajectoryof radius R. Its acceleration vector is given by (2.15), Newton’s second law then assumes theform

~F = −mω2~r

if the origin of the position vector ~r lies in the center of the circle. From this, we can conclude:

If a particle moves uniformly along a circular trajectory of radius R at angular speed ω, the totalforce acting on it must be directed radially towards the center of the circle, and its magnitudeamounts to

mω2R. (2.23)

The total force acting on a body moving uniformly along a circle is often called a centripetal force.Note that the term ‘centripetal force’ does not tell you anything about the nature of the force,but just something about its effect on motion of the object it acts on. Gravitational attractionof a planet, the tension force by which a rope pulls on a bucket filled with water, a normal forceexerted by the wall of a spinning cylinder – all can keep a particle on a circular path and act ascentripetal forces.

Example 2.2.2.1: (modified from Italy 1999 ) A point-like body of mass m is attached to theend of a massless rigid rod rotating in the vertical plane along a circular path of radius r atangular velocity ω in a uniform gravitational field of acceleration g. Determine the forces bywhich the rod acts on the body at the highest and lowest point of its trajectory and at bothpoints where the trajectory intersects the horizontal axis of symmetry of the circle.

Solution: Introduce a two-dimensional coordinate frame xy containing the circular trajectorywith origin at its center, a horizontal x-axis and a vertical y-axis pointing upward. At anytime, the particle experiences its weight ~G = −mg y and a varying force ~F from the rod. Notthat in this problem, the equations of motion are sort of already solved. We have to workbackwards in order to find the forces.

The body moves uniformly along the circle and according to (2.23), the total force acting onit is given by:

~G+ ~F = −mω2 ~r

− Consider the highest point of the trajectory first. As the total force acting on the bodyhas no tangential component, ~F must point along the y-axis. At first, we cannot tell

59


whether ~F points in positive or negative y-direction. We can just define a component Fyof ~F along the y-axis, its sign will tell us more about its orientation. The y-componentof the equation of motion then takes the form

Fy −mg = −mω2r

meaning~F = Fy y = m

(g − ω2r

)y.

− In the lowest point of the trajectory, the situation is the same, just with an opposite signfor centripetal acceleration:

Fy −mg = mω2r ⇒ ~F = m(g + ω2r

)y

− Let motion be anticlockwise in the xy-plane and consider the point of intersection withthe negative x-axis. Since the tangential acceleration of the body must vanish, Fy mustcounteract the weight of the body, that is, Fy = mg. The x-component of ~F is thenresponsible for centripetal acceleration and we get

~F = mω2r x+mg y.

Similarly, for the point of intersection with the positive x-axis

~F = −mω2r x+mg y.

In both cases, the magnitude of the force is equal to∣∣∣~F ∣∣∣ = m√ω2r2 + g2.

It is tempting to find a general expression for ~F in dependence on the body’s position alongits trajectory – let’s do it. Define ϕ as the angle between the x-axis and the position vector ofthe particle. The equations of motion in the x- and y-direction, respectively, assume the form

Fx = −mω2r cosϕ

Fy −mg = −mω2r sinϕ,

therefore~F = −mω2 cosϕ x+m

(g − ω2r sinϕ

)y.

You might verify that the magnitude of the force vector is∣∣∣~F ∣∣∣ = m

√(ω2r)2 + g2 − 2ω2rg sinϕ.

Its maximum value is attained for sinϕ = −1, that is, when ϕ = −90, at the lowermost pointof the trajectory.

60


Non-Uniform Circular Motion

Olympiad problems often deal with objects moving along circular paths at varying angular velocity.In this paragraph, you will see how the laws governing uniform circular motion can be modified tohandle such situations.

Consider, again, a point-like body of mass m moving along a circular trajectory of radius R. Inthe coordinate frame we already used in paragraph 2.1.3, the position ~r of the particle at time t isgiven by

~r(t) = (R cosϕ(t)) x+ (R sinϕ(t)) y.

To get an expression for the velocity vector ~v, we take the time derivative and introduce theinstantaneous angular velocity you might already know as

ω(t) := ϕ(t), (2.24)

obtaining

~v(t) =d

dt~r

= (−Rϕ sinϕ) x+ (Rϕ cosϕ) y

= Rω (− sinϕ · x+ cosϕ · y) .

So far, this expression does not look too different from the one valid for uniform circular motion.Indeed, the only difference is that ω is now dependent on time and therefore ϕ(t) cannot beexpressed as ωt+ ϕ(0). As for uniform circular motion, the velocity vector ~v is always tangentialto the circle and its now time-dependent magnitude is given by

v(t) = Rω(t). (2.25)

Acceleration of the body is obtained by repeated differentiation:

~a =d

dt~v

=(−Rϕ sinϕ−Rϕ2 cosϕ

)x+

(Rϕ cosϕ−Rϕ2 sinϕ

)y

= −Rω2 (cosϕ · x+ sinϕ · y) +Rα (− sinϕ · x+ cosϕ · y) ,

where we just introduced angular acceleration as the time derivative of angular velocity:

α(t) := ω(t) = ϕ(t). (2.26)

We see that the acceleration vector can be split into two components, one along the time-dependentunit vector

61


ϕ = − sinϕ · x+ cosϕ · y (2.27)

and the other along the time-dependent unit vector

r = cosϕ · x+ sinϕ · y, (2.28)

thus~a = Rα ϕ−Rω2 r.

You are welcome to verify that r is always directed radially along the position vector, that ϕ isalways at a right angle to r pointing in the direction of increasing polar angle ϕ and that bothare actually unit vectors. In other words: The two vectors r and ϕ form a set of orthogonal unitvectors the body carries around as it moves along the circle. The corresponding components of theacceleration vector along these unit vectors are the tangential acceleration

~at = Rα ϕ

and the generalized centripetal acceleration

~ac = −Rω2 r,

such that the total acceleration vector is written as their sum:

~a = ~at + ~ac.

In contrast to the case of uniform circular motion, the acceleration vector now has a tangentialcomponent in addition to the radial one. Recall how, in the uniform case, centripetal accelerationwas only responsible for changing the direction of the velocity vector, thereby keeping the particleon the circle, but had no effect on the magnitude of velocity. In the case of non-uniform motion,however, an additional tangential component is responsible for the change in angular velocity

The expression for centripetal acceleration is just a time-dependent version of the uniform caseand is responsible for changing the direction of the velocity vector, whereas tangential accelerationchanges its magnitude.

Again, a summary will be useful:

Consider a particle moving along a circular trajectory of radius R. At any time, the velocity vectoris given by

~v = Rω ϕ (2.29)

The acceleration vector can be split up into two components:

~a = ~at + ~ac (2.30)

where~at = Rα ϕ (2.31)

62


is the tangential acceleration and~ac = −Rω2 r (2.32)

the centripetal acceleration.

According to the second law of motion, the equation of motion of a particle of mass m movingalong a circular trajectory of radius R is generally given by

~F = mRα ϕ−mRω2 r, (2.33)

where ~F is the total force acting on the particle. As with the acceleration vector, we can decompose~F into a radial component along r and a tangential component along ϕ. Each of these componentswill be responsible for the corresponding acceleration.

Example 2.2.2.2: (USA 2010 ) An object of mass m is sitting at the northermost edge of astationary merry-go-round of radius R. The merry-go-round begins rotating clockwise (as seenfrom above) with constant angular acceleration of α The coefficient of static friction betweenthe object and the merry-go-round is µs.

a) Derive an expression for the magnitude of the object’s velocity at the instant when it slidesoff the merry-go-round in terms of µs, R, α and any necessary fundamental constants.

b) For this problem assume that µs = 0.5, α = 0.2 rad/s2 and R = 4 m. At what angle,as measured clockwise from north, is the direction of the object’s velocity at the instantwhen it slides off the merry-go-round? Report your answer to the nearest degree in therange 0 to 360.

Solution: The total acceleration ~a = ~ac + ~at of the body is provided by static friction alone ofmagnitude µsN , where N is the magnitude of the normal force acting on the body. Since thebody is in equilibrium with respect to the vertical direction, the normal force always equalsthe weight of the body: N = mg. The body starts sliding off the merry-go-round as soon asma exceeds the magnitude of static friction, that is, when

m√a2c + a2

t = µmg.

a) At any instant of time, we have from (2.31) and (2.32)√a2c + a2

t =

√(ωt)2 + (αR)2 =

√α4t4R2 + α2R2

= Rα√

1 + α2t4,

where ω = αt as motion is uniformly accelerated. With this, the condition stated abovegives

t = 4

√1

α2

((µsgRα

)2− 1

)

63


for the time at which the body starts sliding off the merry-go-round, corresponding tothe velocity

v = Rαt = Rα

√1

α

√(µsgRα

)2− 1.

b) As motion is uniformly accelerated, the angle ϕ at time t is just

ϕ =1

2αt2,

In our case,

ϕ =1

2

√(µsgRα

)2− 1,

and, numerically, ϕ = 173 for the angle measured from north.

Example 2.2.2.3: A small point-like particle of mass m is located on the uppermost point ofa sphere of radius R resting on a horizontal plane. The particle is given an infinitesimallysmall poke in such a way it starts to slide down the sphere under the influence of gravitationalacceleration g without friction. Assume m is negligible compared to the mass of the sphere.At what height above the ground does the particle leave the sphere?

Solution: Suppose the particle is located on the sphere at such a position that a segmentpointing from the sphere’s centre to it includes an angle ϑ with the vertical line. The forcesacting on the particle are its own weight in vertical direction, of magnitude mg, and a normalforce N(ϑ) in outward radial direction exerted by the sphere. Since we are dealing with motionalong a circular trajectory, we decompose the resulting force into a tangential component Ftand centripetal Fc component:

Ft = mg sinϑ,

Fc = mg cosϑ−N(ϑ).

The condition for the angle ϑ0 at which the particle leaves the sphere is given by N(ϑ0) = 0.We therefore observe the expression for the centripetal component and apply (2.33), obtaining

mg cosϑ−N(ϑ) = mv(ϑ)2

R,

where the speed v(ϑ) of the particle depends on ϑ. Conservation of energy gives us an expressionfor v(ϑ):

0 = −mgR (1− cosϑ) +1

2mv(ϑ)2,

where the zero level of potential energy is set on on top of the sphere. Solving for v(ϑ)2 andputting this result into the previous expression gives

mg cosϑ−N(ϑ) = 2mg (1− cosϑ) .

64


Our condition N(ϑ0) = 0 leads to

cosϑ0 =2

3,

corresponding to the height

R (1 + cosϑ0) =5

3R

above the ground.

Circular motion also has a very interesting property that can be useful for problem solving:

If, in a given coordinate system, the position and velocity vector of a particle are at any timeorthogonal to each other, then the particle’s trajectory lies in a circle around the origin of thecoordinate frame. Furthermore if the acceleration vector is also perpendicular to the velocityvector at any time, then this circular motion is even uniform.

To see why this holds, consider how the square of the position vector of the particle, which is thesquare of its distance from the center of the coordinate system, changes over time:

ddt|~r|2 =

ddt(x2 + y2 + z2

)= 2 (xx+ yy + zz)

= 2 ~r · ~v = 0

where we made use of the chain rule of differentiation and of the fact that the scalar product ~r · ~vvanishes at any time since the two vectors are assumed to be perpendicular to each other. Thismeans that the distance |~r| of the particle from the origin of the coordinate frame is constantover time as its time derivative vanishes. An analogous calculation shows that with the condition~a · ~v = 0 at any time, the magnitude |~v| of the velocity vector is also a constant, that is, circularmotion is uniform.

Example 2.2.2.4: A charged particle of massm and charge q moves in a homogeneous magneticfield ~B with the initial velocity ~u perpendicular to ~B at time t = 0.

a) The particle will move along a circular trajectory perpendicular to ~B. Knowing this, findthe radius R of the circle and the angular frequency ω of motion.

b) Find equations describing the trajectory of the particle in a suitable coordinate system.

c) What does the trajectory look like if the particle’s initial velocity ~u has a nonvanishingcomponent in direction of ~B?

Solution: This example works in the same way as the ones before, the math is just a bit moreelaborate.

a) In the magnetic field, the charged particle moving at speed ~v experiences a Lorentzforce

FL = q ~v × ~B.

65


At any time t, this force vector is both perpendicular to the velocity ~v and the magneticfield ~B and its direction can be determined with the right-hand rule: for positive chargeq > 0, if the extended thumb of the right hand points along ~v and the extended indexfinger along ~B, then the direction of ~F is given by the middle finger pointing perpendicu-larly to the plane of the thumb and index finger. The second law applied on the particlethen looks like

~a =q

m~v × ~B, (2.34)

such that the acceleration vector ~a is at any time perpendicular to the velocity vector ~v.Under the assumption of circular motion, we know from the discussion we just had priorto this example that ~v is of constant magnitude over time and that ~a must correspondto centripetal acceleration, thus

ω2R = a =∣∣∣ qm~v × ~B

∣∣∣ =quB

m=qRωB

m,

since ~v is assumed to be perpendicular to ~B and |~v| = u at any time. From this,

ω =q

m·B

andR =

mv

qB.

b) We shall choose a coordinate frame reflecting the symmetry of this problem in order tokeep the math as simple as possible. As the magnetic field is homogeneous, we havetranslational symmetry along its direction. That’s why it makes sense to choose one ofthe axes along ~B – let’s say it’s the z-axis. The xy-plane shall contain ~u and the originshall coincide with what we know will be the center of the circular trajectory. From this,the initial position of the particle must be R x and the orientation of the x- and y-axiscan be chosen such that the initial velocity of the particle is u y.

The right-hand side of (2.34) assumes the form

q

m~v × ~B =

q

m(vx x+ vy y + vz z)×B z

=qB

mvx x× z +

qB

mvy y × z +

qB

mvz z × z

= −qBm

vx y +qB

mvy x,

where we used that the scalar product of two different basis vectors is zero. The equationsof motion become quite simple:

vx =qB

mvy

vy = −qBm

vx

vz = 0.

66


From the third equation, we see that vz, the velocity component of the particle parallelto the magnetic field, remains constant in time and therefore equal to 0. The differentialequations for the x- and y-component are coupled, but are easily combined to

vx = −(qB

m

)2

vx

vy = −(qB

m

)2

vy.

Of course, you already know how to solve such differential equations: you encounteredthe same kind when discussing harmonic oscillators. Otherwise, don’t worry – if this wasa real Olympiad problem, their general solution would be stated as a hint:

vx(t) = A1 cos (ωt) +A2 sin (ωt)

vy(t) = B1 cos (ωt) +B2 sin (ωt)

with ω from a) and some constants A1, A2, B1, B2. These can be determined from theinitial conditions of motion: vx = 0, vy = u, ax(0) = vx(0) = −qBu/m (from the secondlaw of motion applied at t = 0), ay(0) = vy(0) = 0. We get:

vx(t) = −u sin (ωt)

vy(t) = u cos (ωt)

vz(t) = 0

Integration over time yields the trajectory of the particle:

x(t) =u

ωcos (ωt) + C

y(t) =u

ωsin (ωt) +D

z(t) = E,

where the constants C, D, E can be determined from the initial condition ~r(0) = R x togive, finally,

x(t) = R cos (ωt)

y(t) = R sin (ωt)

z(t) = 0,

which describe, indeed, uniform circular motion in the xy-plane

c) If uz 6= 0, we see from the calculations of b) that the only equation of motion affected isthe one for vz, such that vz = uz is constant in time, corresponding to linear motion ofthe particle in the z-direction:

z(t) = uzt

As the equations of motion in x- and y-direction are the same as in b), their solutionsremain the same. The trajectory of the particle is uniform circular motion parallel to thexy-plane superposed with uniform linear motion in z-direction, which assumes the formof a helix.

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2.2.3 Friction

Here as well, you already know about friction forces from high school. We won’t get into thedetails of where friction forces come from and into different models describing and quantifyingthem. Instead, we are just concerned with how you can include friction forces into the formalismdescribed in the previous paragraphs to solve Olympiad problems.

A very important concept needed for proper calculation of friction forces is that of normal forces,which can easily be misunderstood. Whenever two surfaces interact, they do so by means of twoforces, which, according to Newton’s third law, must be of equal magnitude and act on eachsurface in an opposite direction. On every point of contact between the two surfaces, each of thetwo forces of interaction can be split up into two components, one tangential and one orthogonalto the surface onto which the force acts. In general, the orthogonal component will be called thenormal force ~N , whereas the tangential one will correspond to the friction force, which in thischapter will just be denoted by a general symbol ~F . The relationship between ~N and ~F generallyassumes very simple forms, where just two cases have to be considered:

− Static friction: The two surfaces do not move relative to each other. The maximum tangentialforce F between the two surfaces is proportional to the normal force N acting between them:

F ≤ µsN, (2.35)

where µs is the coefficient of static friction between the two surfaces, a constant ideallydependent on material properties of the two surfaces only.

− Dynamic/kinetic friction: The two surfaces glide on each other. The friction force betweenthem is given by

F = µdN, (2.36)

where µd is the coefficient of dynamic/kinetic friction between the two surfaces, again aconstant ideally dependent on material properties of the two surfaces only.

Make sure you understand how to deal with friction forces. One very common mistake is that ofintroducing them as additional forces after a tangential force component between two surfaces hasalready been defined or used. It is important that you think of friction as just another name fora tangential force acting between two. . . non-frictionless surfaces. This implies that between twoideal frictionless surfaces, there is no tangential interaction.

Example 2.2.3.1: USA 2008 : A block of mass m slides on a circular track of radius r whosewall and floor both have coefficient of kinetic friction µ with the block. The size of the blockis small compared to the radius of the track. The floor lies in a horizontal plane and the wallis vertical. The block is in constant contact with both the wall and the floor. The block hasinitial speed v0.

a) Let the block have kinetic energy E after travelling through an angle ϑ. Derive anexpression for dE/dt in terms of g, r, µ and E.

b) Suppose the block circles the track exactly once before coming to a halt. Determine v0

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in terms of g, r, and µ.

Solution: Let’s determine the forces acting on the block first. There is of course its weightG = mg. Each of the two surfaces, the horizontal plane and the wall, will exert both a normalforce Nh and Nw, respectively, and dynamic friction Fh and Fw, respectively. We know from(2.36) that Fh = µNh and Fw = µNw.

a) Let v denote the velocity of the block along the track at angle ϑ. The kinetic energyof the block is then equal to E = mv2/2 and it decreases through work done by thefrictional forces Fh and Fw only, as all other forces act in directions perpendicular to thedirection of motion of the block. That is, if the block moves through an infinitesimaldistance ds = rdϑ, the loss of kinetic energy is given by

dE = − (Fh + Fw) · ds.

As the block moves along a circular trajectory, the normal force Nw must account for itscentripetal acceleration at angle ϑ: Nw = mv2/r = 2E/r. Since the block does not movein vertical direction, Nh must equal the block’s weight mg, leading to

dE = −µr(mg + 2

E

r

)dϑ

⇔ dE

dϑ= −µ (2E +mgr)

b) Separation of variables yields

dE

E +mgr/2= −2µdϑ.

which can be integrated between ϑ = 0 with E0 = mv20/2 and general ϑ with correspond-

ing E to give

ln

∣∣∣∣ E +mgr/2

E0 +mgr/2

∣∣∣∣ = −2µϑ,

which provides you with an analytic expression for how v depends on ϑ given the initialspeed v0. Here, we are asked to do the opposite: knowing that v = 0 and therefore E = 0for ϑ = 2π, we obtain

ln

(mgr/2

mv20/2 +mgr/2

)= −4πµ,

leading tov0 =

√gr (e4πµ − 1)

You might have encountered the concept of rolling friction in some of your physics courses, bywhich one denotes friction acting against direction of motion of a body rolling on some surface.Rolling friction basically arises from deformation of the surface by the rolling body itself, leading

69


to dissipation of kinetic energy of the body. In general, rolling friction can be determined by

F = µrN, (2.37)

where the coefficient of rolling friction µr is usually much smaller than µd or µs for the samematerials. One could speak of ‘ideal rolling’ in the case where no such interaction occurs androlling friction vanishes. For a solid body to be able to roll on a surface in the first place, however,static friction must prevent its point of contact with the underlying surface from sliding: in idealrolling, the point of contact is always at rest with respect to the ground and we have the case oftwo surfaces not moving relative to each other. That is, whenever there is a body rolling withoutslipping on a surface, you know that static friction must be acting. In addition, rolling friction canof course be present. Dynamic friction only occurs if slipping is involved.

2.3 Frames of Reference

2.3.1 Choosing a Frame of Reference

You might have realized, or just know from school, that kinematic quantities are not absolute inthat they depend on the point of view from which they are measured. Take, for example, a personwalking the deck of a cruise ship. The velocity of that person with respect to someone leaning onthe railing might be, say, 4 km h−1. With respect to someone walking towards them, to a seagullflying over the ship, or to an island in the sea, the velocity of that person might be very different.In physical terms, we say that velocities and all other kinematic quantities are relative. We saythat they depend on the frame of reference (or reference frame) in which they are measured.

Even though the two terms are very similar, a reference frame is not the same as a coordinateframe. You can think of the reference frame as of a certain point of view, whereas a coordinateframe is more of a mathematical tool associated with a reference frame5.

Take, for example, a situation often encountered in physics competitions: a rigid body, such as asphere, rolling down an inclined plane. As you can read in section 2.5.5, such problems are bestdescribed in a reference frame where the inclined plane is at rest, such that the only object movingin this frame is the rolling body. For the choice of a coordinate system, we have many options. Ona first thought, it seems reasonable to take an immobile coordinate frame with the xy-plane parallelto the ground and the z-axis pointing upwards. On the other hand, it turns out that choosing amoving coordinate frame with xy-plane parallel to the incline such that its origin coincides withthe center of the rolling body and the x-axis points along the direction of rolling is very useful.Remember, the reference frame is still at rest with respect to the inclined plane. It is just the setof coordinates you will use to describe various points inside the rolling body that moves around.You’ll see more about this in section 2.5.

Let’s figure out how kinematic quantities can be converted from one reference frame to another.Consider a point-like particle moving in a reference frame we call A and a coordinate systemassociated with it. We shall write ~rA for the position of the particle with respect to A. Let B be asecond reference frame with associated coordinate system, such that at any time, the origin of thissystem is located at position ~r(A)B with respect to the origin of the coordinate system associated with

5Again, the true story is way more complicated. And again. . . Don’t worry about that.

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Physics Olympiad Script 2.3. FRAMES OF REFERENCE

A. Conversely, the origin of A’s coordinates lies at position ~r(B)A with respect to B’s coordinates.

It is not too difficult to see that~r(B)A = −~r(A)B (2.38)

must hold. The particle will have two different position vectors with respect to the two systems,let’s call them ~r(A) and ~r(B), respectively. From a simple geometrical consideration, it follows that

~r (B) = ~r (A) + ~r(B)A

= ~r (A) − ~r (A)B .(2.39)

Such a relationship is called a coordinate transformation between the two frames of reference A andB. By taking its time derivative, we obtain a transformation rule for the velocity of the particle:

~v (B) = ~v (A) + ~v(B)A

= ~v (A) − ~v (A)B .

(2.40)

Similarly, for accelerations,

~a (B) = ~a (A) + ~a(B)A

= ~a (A) − ~a (A)B .

(2.41)

Make sure you develop an intuitive approach to such transformations by thinking about someexamples. What about transformation of forces? This question is somewhat more difficult. Inour framework, we regard forces as fundamental interactions between bodies, such as gravity orelectrostatic attraction. Therefore it seems natural to assume that a force vector does not dependon the frame of reference. That is, if ~F (A) denotes a force measured in frame A and ~F (B) the sameforce measured in frame B, then

~F (A) = ~F (B) (2.42)

must hold. Let’s say that this force acts on a body of massm, which we also assume to be a quantityindependent of the frame of reference. Newton’s second law would state that ~F (A) = m~a (A) and~F (B) = m~a (B), which just implies ~a (A) = ~a (B). We see from (2.41) that this can only hold if therelative acceleration ~a (B)

A vanishes, that is, if the two systems move with constant velocity withrespect to each other. The case where the relative acceleration does not vanish, however, posesus a bigger problem: either Newton’s second law is not valid or forces must somehow transformbetween frames. Either way, we need to rethink our concepts.

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2.3.2 Accelerated Frames of Reference

Imagine, for example, to be sitting on a bus at night, getting home from an exhausting trainingsession at the Physics Olympiad, as the cutest kitten suddenly appears on the street a coupleof meters ahead of the bus. The driver, who is very alert despite late hours, almost immediatelypushes down the brake pedal. You feel how your body is somehow dragged forward and you almostfall off your seat. It feels as if some magic force acting on your body had appeared at the instantof braking. Anyway, thanks to the fast reflexes of the driver, a fatal accident could be prevented.Some minutes later, as your blood adrenalin levels get back to normal, you start thinking. Bywhat was the force exerted that made you fall forward? Most probably, it was not the bus, as youcould still feel a drag even after your back had lifted off the seat. Was it the air? Guess not, how isthat supposed to work? On the other hand, think about how the mother of the kitten might haveseen it from the roadside. For her, as much as she might have been scared, the whole situation wasnothing but a manifestation of Newton’s first law of motion. The bus would decelerate, and youwould just keep moving at your initial speed along the direction of travel, no magic force involved.Still, from your point of view, the magic force must have existed, as otherwise Newton’s first lawwould have been violated in your reference frame.

Indeed, as we saw at the end of the previous section, it turns out that Newton’s laws of motionare not valid in all frames of reference. Examples like the situation with the bus and the kittenmake us realize that there might be two different kinds of frames of reference: the ones whereNewton’s laws are valid, which we call inertial frames, and those where they aren’t. From theconsiderations at the end of the previous section, we see that non-inertial frames are nothing butjust accelerated frames, or, more precisely, they are frames accelerated with respect to any inertialframe. The frame of reference of the kitten’s mother is an inertial frame as also one associated withthe road or the earth would be, approximately. Your frame of reference on the bus is acceleratedwith respect to these inertial frames, which is why Newton’s first law loses its validity and youfeel an acceleration in forward direction even without any force acting.

Fortunately there is a trick enabling us to rescue Newton’s second law in accelerated frames.Let S be an inertial frame and S′ a frame moving with constant acceleration ~a (S)

S′ with respect toS. Consider a particle moving with acceleration ~a in S and ~a′ in S′. From (2.41), we know that~a′ = ~a − ~a (S)

S . If a force ~F acts on this particle, which is the same force in S′ as in S, we knowfrom Newton’s second law in S that ~F = m~a, where m is the mass of the particle. Therefore,

m~a′ = m~a−m~a (S)S = ~F −m~a (S)

S .

In S′, the second law is satisfied except for an extra term −m~a (S)S having the dimension of a force.

We might as well just cheat and deliberately call this term a force too. This force would only existin S′ and not in S. That’s why we refer to it as a fictious force.

Let’s apply this concept to our example with the bus and the kitten. S shall be the frame of theroad, S′ the frame of the bus, and ~abrake the acceleration of the bus in S as it brakes, pointingopposite to the direction of travel. The fictious force acting on your body is then given by −m~abrake,where m is your mass, and points towards the front of the bus, and your equation of motion inS′ becomes m~a(S′) = −m~abrake for your acceleration ~a(S′) in S′. Finally, the fictious force enablesus to explain from the point of view of S′ the forward drag you could feel in during the brakingmanoeuvre in a fashion more consistent with Newton’s laws.

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Physics Olympiad Script 2.3. FRAMES OF REFERENCE

Let S′ be a frame moving with acceleration ~arel with respect to the inertial frame S. The validityof Newton’s laws of motion in S′ is preserved if you introduce a fictious force given by

~Ff = −m~arel (2.43)

acting on every body in S′.

Example 2.3.2.1: (Germany 1994 ) A train departs from rest with acceleration a on straighthorizontal tracks. At the time the train departs, a girl standing in one of its very long carsthrows a ball at speed v with respect to the tracks under an angle α with the horizontal towardsthe front of the train. Under which conditions does the ball get back to the girl, as long as shedoesn’t move around in the train car? Neglect air friction, take g for gravitational acceleration.

Solution: To apply the concepts developed in this chapter, we solve the problem in the accel-erated frame of reference of the train. As the ball moves in the air, the only force acting onit is its weight m~g, where m is its mass. Obviously, Newton’s second law cannot be applieddirectly, since the train is not an inertial frame. The fictious force due to acceleration of thereference frame of the train is equal to −m~a, where ~a denotes the acceleration of the trainwith respect to the tracks. With the fictious force, we are allowed to pretend that Newton’ssecond law is valid and get the equation of motion

d2

dt2~r = ~g − ~a.

This is interesting: the acceleration of the ball does not depend on its mass. The ball, as anyother object in the frame of the train, seems to behave as it were subject to a tilted accelerationof gravity given by the vector ~geff := ~g−~a. We know what happens with a uniformly acceleratedobject given a nonvanishing initial velocity: in general, it moves along a parabola, which sureis a trajectory that never returns to starting point. We said ‘in general’ because there is justone special case: if the starting velocity is exactly opposite to the direction of acceleration,that is, if ~v is antiparallel to ~geff, the ball will move along a straight line and eventually returnto the girl. You can draw a sketch and convince yourself that this is exactly the case if

tanα =g

a,

a result independent of the magnitude of ~v (which only determines the time it takes the ballto return to the girl).

As you know, the earth is not really an inert object. With respect to the sun, it moves along anelliptical orbit spinning about itself, meaning that any frame associated with the earth is actuallyan accelerated frame. The solar system rotates about the center of the milky way, and the milkyway itself takes part in the expansion of the universe. The reason why we can often treat the earthas an inertial frame is that the fictious forces involved are in general extremely small compared tothe other forces considered in the problem. I’m sure you realize that going through some examplesto quantify this statement would be a nice exercise. Things get somewhat different when space or

73


time scales involved are much larger though. For problems dealing with planetary motion as insection 2.6, the accelerated motion of the earth has to be taken into account.

One more remark about rotational motion. Consider a particle of mass m rotating uniformly withangular velocity ω along a circular trajectory of radius r in some inertial frame. For the particleto move this way, the total force F acting on it must account as a centripetal force for centripetalacceleration, that is, F = mω2/r. Look at what happens if you describe the situation in a framerotating with the particle. This frame is accelerated by ω2/r with respect to any inertial frame,that’s why one has to take into account a fictious force of magnitude mω2/r directed opposite tothis acceleration. This fictious force points in outward radial direction, that’s why we call it thecentrifugal force acting on the particle. In the rotating frame, the total force acting on the particleis just F −mω2/r = 0, which is consistent with the fact that the particle is at rest in this system.

2.4 Systems of Particles

Don’t worry, you won’t spend your whole physics Olympiad career working on problems dealingwith point-like masses and particles only. I bet you are eager to discover how to deal with questionsconcerning more interesting objects such as extrasolar planets, hula hoops, dumbbells, revolvingdoors and spinning tops. In general, such objects can be described as systems consisting of a givennumber n of point masses mi, i = 1, . . . , n, where n may be just equal to 2 or a number so largethat writing∞ instead would save a lot of space. With knowledge from the previous chapter aboutNewton’s laws of motion, one should basically be able to formulate equations of motion for allpoint masses out of which such a system is composed. However, there are two major difficulties:

− Generally, a system is composed of many particles, that is, n is very large. It’s o.k. as longas you are dealing, say, with a system composed of the sun and the earth only, each of whichcan often be treated as a single point mass (we shall see why very soon), i.e. n = 2 and youhave to handle two equations of motion only. But as soon as you start working on somethinglike a wooden sphere rolling down an inclined plane or a figure skater spinning on the rink,you might realize that solving an almost infinite number of equations of motion can give youquite a headache. Therefore, problem number one is the often insanely huge number of pointmasses involved.

− Secondly, let’s take a closer look at the equations of motion we would formulate for such asystem. On each particle of mass mi, a force ~Fi shall act according to

mi~ai = ~Fi, i = 1, . . . , n

Let’s take as an example a billiard sphere rolling freely on a table. The forces ~Fi are ofcourse not the same for all particles inside the sphere. While every single one experiencesgravitational acceleration mi~g from the earth, only the particles in direct contact with thetable will experience friction. That shouldn’t be a great problem, as you know how tocalculate a friction force as soon as you know the normal force and the friction coefficient.There is more to it though. A rolling billiard sphere remains a billiard sphere thanks tointernal forces holding together all of its particles. There is no chance how you would beable to determine these internal forces without exact knowledge of the position and velocityof every single particle in space. That is, problem number two is lack of knowledge aboutinternal forces of the system.

74

Physics Olympiad Script 2.4. SYSTEMS OF PARTICLES

As we will see, we can overcome these difficulties thanks to a set of laws that enable us to treatsystems of point masses as if they were concentrated in one single particle.

2.4.1 Newton’s Second Law for Systems of Particles and the Center of Mass

For a start, we will look at the effect of forces on motion of a system consisting of two point massesm1 and m2 only. We shall later see how our results can be generalized to any arbitrary number nof particles, but let’s keep things simple first.

Let ~F1 and ~F2 denote the total forces acting m1 and m2, respectively. According to Newton’ssecond law, if we assume the masses of the particles to be constant over time,

~F1 = m1d2

dt2~r1

~F2 = m2d2

dt2~r2

.

(2.44)

So far, nothing new. In case we know how to compute the forces ~F1 and ~F2, it is not too difficultto set up the equations of motion. You might have realized that the number of equations involvedis 3 × 2 = 6, one for each of the 3 coordinates of the 2 particles. O.k., this isn’t too much, buthandling six equations simultaneously can take you quite some time – that’s why we might wantto look for some simplification by digging a bit deeper into our formulas.

One idea is to consider the particle system and its surroundings as two different entities undergoingsome kind of interaction. Take the forces ~F1 and ~F2, for instance. In general, the force acting onthe i-th particle can be decomposed into one total internal force ~Fi, int arising from the rest ofthe system itself, that is, exerted by all other particles of the system on the i-th particle, andone total external force ~Fi, ext arising from the surroundings of the system. The internal force canbe gravity, electrostatic attraction, a superposition thereof, a force two particles experience onlyduring a very short time interval as they collide, and many others. The external force can be, forexample, gravitational attraction from the earth, a Lorentz force exerted by a magnetic field, orair friction. For now, we do not care too much about the nature of these forces, but merely aboutthe decomposition

~Fi = ~Fi, ext + ~Fi, int. (2.45)

With this in mind, we now look at what happens if we sum the two equations of motion (2.44)and rearrange stuff using our decomposition:

~F1 + ~F2 =(~F1, ext + ~F1, int

)+(~F2, ext + ~F2, int

)=(~F1, ext + ~F2, ext

)+(~F1, int + ~F2, int

)for the left-hand side and

m1d2

dt2~r1 +m2

d2

dt2~r2 =

d2

dt2(m1~r1 +m2~r2)

= (m1 +m2)d2

dt2m1~r1 +m2~r2

m1 +m2

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as a very fancy-looking version of the right-hand side. Together,(~F1, ext + ~F2, ext

)+(~F1, int + ~F2, int

)= (m1 +m2)

d2

dt2m1~r1 +m2~r2

m1 +m2.

Now stare at the internal forces ~Fi,int in this last equation for some time. In our case of just twoparticles, we might as well write ~F2→1 instead of ~F1,int and something analogous for the secondparticle. From Newton’s third law, we know that ~F1→2+ ~F2→1 = 0, that is, the sum of all internalforces of the system vanishes. The right-hand side also has something to offer: define a point inspace given by

~rCM :=m1 ~r1 +m2 ~r2

m1 +m2(2.46)

and introduce the total mass M = m1 + m2 of the system and the total external force ~Fext, tot =~F1 ext + ~F2 ext acting on it. Our equation can then be rewritten in a rather appealing fashion:

~Fext, tot = Md2

dt2~rCM.

As you can see, by just manipulating the equations of motion of the two particles and by introducinga nice notation, we were able to retrieve a simple equation containing a lot of information aboutmotion of the system. The beauty of the equation lies in the following statement:

Motion of a system of two point-like masses m1 and m2 can be described by Newton’s secondlaw applied on a particle of mass M = m1 + m2 at position of the center of mass ~rCM of the twoparticles:

~Fext, tot = Md2

dt2~rCM, (2.47)

where ~Fext, tot denotes the sum of all external forces acting on the two particles.

In other words, we were able to reduce the system of two particles to motion of one single virtualparticle of mass M at position ~rCM. We know from section 2.2 about Newton’s laws how tohandle motion of single point masses. You will see in the examples how this reduction comes invery handy. But first, let’s take a closer look at our equation.

The vector defined by (2.46) denotes a point in space called the center of mass of the two particles.You can think of the center of mass as a point in space at any time associated with the two particlesin motion. Its position with respect to either particle varies as the particles move.

In order to develop an intuitive approach to the concept of center of mass, you might want to workon the following questions:

− For a one-dimensional system, that is, two particles m1 and m2 moving along a straightline, where does the center of mass lie with respect to the line?

− What is its distance to either particle?

76


− What if m1 > m2? What if m1 m2? What if m1 = m2?

− If you haven’t already done so, introduce a coordinate axis along the line connectingthe two particles with origin at the position of m1. What does the expression for thecoordinate of the center of mass with respect to this axis look like?

It is not too difficult to perform similar calculations for an arbitrary number n of particles of massesmi at positions ~ri. It turns out that one can define the center of mass of the system as

~rCM =

∑imi~ri∑imi

(2.48)

and that the equation of motion of the system assumes the following form:

Newton’s Second Law for a System of Particles

~Fext, tot = Md2

dt2~rCM, (2.49)

where ~Fext, tot is the sum of all external forces acting on the system andM =∑

imi its total mass.Alternatively,

~Fext, tot =d

dt~pCM, (2.50)

where~pCM =

∑i

~pi = M~vCM (2.51)

is at the same time the total momentum of the system and the momentum of its center of mass.

Here again, you might want to spend some more time thinking about the concept of center of massfor a system consisting of more than just two particles.

Take, for instance, n equal point masses m placed on the edges of a regular polygon.

− Where does the center of mass of the polygon lie?

− What happens to its position if the mass of one of the particles increases or decreases,the others kept fixed?

Equation (2.49) has an interesting consequence:

If the sum of all external forces acting on a system of particles vanishes, the speed of their centerof mass remains constant.

Consider, for instance, a radioactive nucleus decaying in flight into three particles going off inseparate directions. If the nucleus is far away from any other object that could interact with it,the speed of the center of mass of the three decay products must be the same as the speed of the

77


nucleus before decay.

Consider a system consisting of various extended objects we call A1, A2, . . ., such as blocks of wood,soccer balls or human beings, of masses m1,m2, . . .. Each of these single objects is itself a systemconsisting of many particles, let’s say A1 consists of point masses mA1,i located at ~rA1,i and so onfor A2 etc. The center of mass of A1 lies at ~rCM, 1 = (

∑imA1,i~rA1,i) /m1, similarly for A2 etc. The

center of mass of the whole system is then obtained as

~rCM =

∑imA1,i~rA1,i +

∑imA2,i~rA2,i + . . .∑

imA1,i +∑

imA2,i + . . .

=m1~rCM,1 +m2~rCM,2 + . . .

m1 +m2 + . . .,

which just means the following:

The center of mass of a system of extended objects A1, A2, . . . of masses m1,m2, . . . whose centersof mass are located at ~rCM,1, ~rCM,2, . . ., respectively, lies at

~rCM =m1~rCM,1 +m2~rCM,2 + . . .

m1 +m2 + . . ., (2.52)

meaning that each of the objects contributes to the position of the center of mass of the wholesystem as if it was a point-like particle concentrated in its own center of mass.

Example 2.4.1.1: Pete (mass mP ) and Dave (mass mD) are in a rowboat (mass mB) on a calmlake. Dave is near the bow of the boat, rowing, and Pete is at the stern, at distance d fromDave. Dave gets tired and stops rowing. Pete offers to row, and they change places. How fardoes the boat move as Pete and Dave change places? Neglect any horizontal force exerted bythe water.

Solution: Consider a system consisting of Dave, Pete and the boat. We look at the situation inthe reference frame of the lake and introduce a coordinate system with x-axis pointing alongthe boat. Take the origin of this axis at the stern of the boat where Pete sits at the beginning.The coordinate of the center of mass of the system is, thanks to (2.52),

xCM =1

M(mP · 0 +mDd+mBxB) ,

whereM = mP +mD+mB is the total mass of the system and xB the coordinate of the centerof mass of the boat at the beginning. Since no horizontal forces act on our system, its centerof mass will not move as Pete and Dave change places. In general, Pete and Dave will not beof the same mass, and therefore the boat must move as well as they change places in order toprevent the center of mass of the whole system from moving. Let ∆x be the distance by whichthe boat moves during the process. In the end, Pete’s coordinate will be that of Dave beforethe changing plus ∆x, that is, d + ∆x. Similarly for Dave, whose center of mass moves fromthe origin to ∆x. The center of mass of the boat will also have moved to xB + ∆x, and theposition of the center of mass of the whole system can be expressed as

xCM =1

M(mD∆x+mB (xB + ∆x) +mP (d+ ∆x)) .

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Combining the two expressions for xCM and solving for ∆x, one finds

∆x =mD −mP

mP +mD +mBd.

Note that the boat moves in direction of the initial position of Pete if he is heavier than Daveand vice versa.

2.4.2 Conservation of Linear Momentum

We already know from the last chapter that the center of mass of a system of particles moves atconstant speed if the sum of all external forces acting on it vanishes. That is,

M~vCM =∑i

mi~vi =∑i

~pi

is constant. What looks very trivial is in fact one of the most fundamental laws in mechanics, the

Law of Conservation of Linear MomentumIf the sum of all external forces acting on a system of particles vanishes, the total momentum ofthe system is constant over time. In particular, the total momentum of an isolated system notundergoing any interaction with its surroundings is constant over time.

I’m pretty sure you already know this law from your high school physics and how you can applyit to problems. In particular, you might know that conservation of linear momentum plays animportant role for problems involving collision, often together with the law of conservation ofenergy.

Just as a reminder: elastic collisions are the ones where kinetic energy, and thus total mechanicalenergy, is conserved. An inelastic collision occurs as soon as kinetic energy is dissipated by theimpact. Whenever particles stick together in a collision, you know that the collision process must beinelastic: Suppose a particle of momentum m1~v1 collides with a particle m2 at rest to build a largerparticle of momentum (m1 +m2)~v′. Conservation of momentum states thatm1~v1 = (m1 +m2)~v′.,from which

~v′ =m1

m1 +m2~v1.

The corresponding kinetic energies are Ekin, start = m1v12/2 before and Ekin, end = (m1 +m2) v′2/2

after the collision, that is,

Ekin, end

Ekin, start=

(m1 +m2) v′2/2

m1v12/2

=(m1 +m2) ·

(m1

m1+m2

)2v1

2

m1v12

=m1

m1 +m2< 1,

meaning that kinetic energy is always dissipated during a collision process after which particlesstick together.

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2.4.3 Conservation of Angular Momentum

Very often, you’ll have to deal with problems taking place in a plane, that is, where all vectorialquantities such as positions, momenta and forces as well as any of their time derivatives can bedescribed as two-dimensional. Moreover, such problems often involve what we call central forces,that is, forces always directed to a fixed point of the plane. For instance, these problems mightinvolve a planet subject to gravitational attraction by a star or an electrically charged particlesubject to Coulomb interaction with a large central charge. Both gravity and the electrostaticCoulomb interaction are central forces. Let ~r be the position of a particle and ~F such a centralforce acting on it in a two-dimensional situation. If we take the origin of a coordinate system atthe center towards which the force ~F is directed, then ~F and ~r must be parallel at any time. Interms of their components in xy-coordinates in this plane,

rx : ry = Fx : Fy,

as long as both ry and Fy are nonvanishing. This condition can be rewritten as

rxFy − ryFx = 0,

and you can verify that this still holds if one of ry or Fy is zero. With Newton’s second law, wecan write

0 = rxFy − ryFx = rxpy − rypx= vxpy + rxpy − vypx − rypx

=d

dt(rxpy − rypx) ,

meaning that the quantity given by rxpy − rypx remains constant over time. You are totally rightif you think that the expression rxpy − rypx looks very familiar: it is nothing but the z-componentof the cross product ~r× ~p. You might verify that the x- and y-component of this cross product areboth 0 at any time, since the position and momentum vector of the particle are always assumedto lie in the xy-plane. Therefore, we can say that the vectorial quantity

~L = ~r × ~p (2.53)

remains constant over time as long as the particle is subject to central forces only. This quantity iscalled the angular momentum of the particle with respect to the origin of the coordinate system.

Example 2.4.3.1: A particle of mass m lies on a table and is attached to a string passingthrough a hole in the table. The particle rotates with velocity v0 on a circle with radius r0

around the hole. How much work is done in shortening the string to r < r0 by pulling itthrough the hole in the table? Neglect friction.

Solution: As the string is shortened, it always pulls on the particle in radial direction. There-fore, angular momentum of the particle is conserved during shortening, and the angular mo-

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menta ~L0 and ~L of the particle at r0 and r, respectively, are the same:

~L0 = ~L

⇔ ~r0 ×m~v0 = ~r ×m~v.

One quickly sees that, since the particle moves on circles both at r0 and r, its correspondingvelocity vectors are always perpendicular to its respective position vectors, making the crossproducts very easy to calculate. That is,

mr0v0 = mrv

⇔ v =r0

v0v

The amount of work done when shortening the string is given by the difference in kinetic energybetween the two radii:

W = Ekin − Ekin, 0 =1

2mv2 − 1

2mv2

0

=1

2mv2

0

((r0

r

)2− 1

).

We might want to generalize this result to particles not necessarily constrained to move in a plane.Consider a particle moving with momentum ~p = m~v and position vector ~r subject to a total force~F . If we take the time derivative of its angular momentum vector ~L = ~r × ~p, we get

d

dt~L =

d

dt(~r × ~p)

=

(d

dt~r

)× ~p+ ~r ×

(d

dt~p

)= ~v × ~p+ ~r × ~F

= ~r × ~F ,

where we used the second law of motion d~p/dt = ~F and the product rule of differentiation forvector products6. Note that the velocity ~v and momentum vector ~p are parallel at any time, theircross product vanishes. The quantity ~r × ~F is called the torque ~τ by which ~F acts on the particlewith respect to the origin. You probably know about the torque and its meaning from school,that’s why we won’t discuss it too extensively here. With this definition, we now have a veryelegant law:

6Given any two vectorial quantities ~A and ~B dependent on time, the time derivative of their vector productsatisfies the rule

d

dt

(~A× ~B

)=

(d

dt~A

)× ~B + ~A×

(d

dt~B

),

which you can verify by writing out the components of ~A× ~B and taking their time derivatives.

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d

dt~L = ~τ . (2.54)

Example 2.4.3.1 with the particle attached to a string shows us how the quantity ~L can in someway be understood as an ‘amount of rotational motion’, similar to linear momentum as a quantitydescribing the ‘amount of linear motion’ – that’s why ~L is called angular momentum. On the otherside, the torque is nothing but a measure for the extent to which a force is able to change angularmotion of a rotating object. It vanishes if the force is radial, that is when ~r and ~F are parallel,and it is maximal for a force at right angles to the position vector.

The relationship of (2.54) somehow reminds us of Newton’s second law, where the force acting ona particle and its linear momentum were linked. Because of this analogy, we call (2.54) Newton’ssecond law for angular motion.

Let’s apply (2.54) to a system of masses mi by adding up the equations for every single particle:

d

dt

∑i

~ri × ~pi =∑i

~ri × ~Fi,

where ~Fi denotes the total force acting on the i-th particle at position ~ri. We now use the sametrick as in section 2.4.1 and divide all forces acting on the particles of our system into an internalcomponent ~Fi,int and an external component ~Fi,ext. The calculation that follows is quite elaborate;you can find it below. Anyway, it turns out that all internal forces cancel out of the picture thesame way they already did for linear motion in section 2.4.1.

The rate of change of the total angular momentum ~L of a system of particles is entirely determinedby the total torque ~τext, tot acting on them due to forces external to the system:

d

dt~L = ~τext, tot (2.55)

Again, this leads us to one of the most central conservation laws of mechanics:

Law of Conservation of Angular MomentumIf the total torque due to external forces acting on a system vanishes, its angular momentumremains constant over time.

In particular:

The total angular momentum of a system of particles subject to radial forces only is constant overtime.

The total angular momentum of an isolated system of particles is constant over time.

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Example 2.4.3.2: Two equal masses m are connected by a rigid rod of negligible mass and oflength a. The center of mass of this dumbbell-like system is stationary in gravity-free space,and the system rotates about the center of mass with angular velocity ω. One of the rotatingmasses strikes a third stationary mass m, which sticks to it.

a) Locate the center of mass of the three-particle system at the instant prior to collision.What is the velocity of the center of mass?

b) What is the angular momentum of the three-mass system about the center of mass atthe instant prior to collision? At the instant following collision?

c) What is the angular velocity of the system about the center of masss after the collision?

d) What are the initial and final kinetic energies?

Solution: Introduce a coordinate system with origin at the center of mass of the dumbbell andxy-plane containing the plane of rotation of the dumbbell prior to collision. choose the x-axisalong the orientation of the dumbbell at the instant prior to collision, such that the free masslies at ~r3 = a x/2 and the two masses of the dumbbell at ~r1 = a x/2 and ~r2 = −a x/2. Ofcourse, ~r1 = ~r3.

a) In this coordinate frame, the center of mass of the three-particle system lies at

~rCM =1

3m(m~r1 +m~r2 +m~r3)

=1

3

(a2x− a

2x+

a

2x)

=1

6a x

right before collision. That is, the center of mass divides the connecting rod into twosegments whose lengths behave as (a/2− a/6) : (a/2 + a/6) = 1 : 2.

Right before collision, the velocity vectors of the two masses of the dumbbell are equaland opposite, that is, ~v1 + ~v2 = 0, and the third particle is still at rest. The velocity ofthe center of mass then is

~vCM =1

3m(m~v1 +m~v2 +m~v3) =

1

3(0 + 0) = 0

b) Let ~Lbefore and ~Lafter be the angular momenta of the three-particle system right beforeand right after collision. Only the particles of the dumbbell contribute to ~Lbefore as thethird mass is still at rest. Their velocities can be written as ~v1 = aω y/2 = −~v2, suchthat

Lbefore = m~r1 × ~v1 +m~r2 × ~v2

= −ma

2· aω

2x× y + m

a

2· aω

2(−x)× (−y)

=1

2mωa2 z

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Note that the vectors ~r1 and ~v1 are perpendicular to each other, as ~r2 and ~v2, whichmakes calculation of the cross products quite straightforward.

The system of three masses is closed, no external force acts on it. Therefore conservationof angular momentum applies, and ~Lafter = ~Lbefore.

c) Right after collision, the dumbbell will keep rotating with the third mass stuck to one ofits masses. The center of mass of the whole system must not move, as no external forceacts on it. That is, the center of rotation of the dumbbell is shifted from the center ofthe rod to the position of the center of mass. If the center of rotation were any otherpoint, then the center of mass of the system would rotate about it together with thethree masses.

Let’s say that rotation about the center of mass after collision occurs at angular velocityω′. For calculation of the velocities of the three masses right after collision one has to takeinto account that the new center of rotation is the center of mass of the three-particlesystem:

~v′1 = ~v′3 =1

3aω′ y

~v′2 = −2

3aω′ y

and the angular momentum becomes

~Lafter = m~r1 × ~v′1 +m~r2 × ~v′2 + ~r3 × ~v′3

= 2 ·ma

2x× 1

3aω′ y + m

a

2x× 2

3aω′ y

=2

3ma2ω′ z.

Conservation of angular momentum with ~Lbefore = ~Lafter yields

ω′ =3

4ω.

d) The collision with the third mass is inelastic, mechanical energy is therefore not expectedto be conserved. Right before the collision, we have

E =1

2mv1

2 +1

2mv2

2

= 2 · 1

2mω2

(a2

)2=

1

4mω2a2

and right after

E′ = 2 · 1

2mv1

′2 +1

2mv2

′2

= m · 1

9ω′2a2 +

1

2m · 4

9ω′2a2

=1

3mω′2a2 =

3

16mω2a2,

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meaning that an amount of energy

E − E′ = 1

16mω2a2

is dissipated during the collision.

This example illustrates an important principle:

Conservation of angular momentum is a useful tool for problems where systems of point massesundergo some process where they rotate in space.

Derivation of Newton’s Second Law for Angular Motion*

Consider a system, not necessarily a rigid body for now, of n particlesmi. We denote their positionsby ~ri, their velocities by ~vi and their linear momenta by ~pi = mi~vi. The total angular momentumof this system is given by

~L =∑i

~ri × ~pi.

We take its time derivative and get

d

dt~L =

∑i

(d

dt~ri × ~pi + ~ri ×

d

dt~pi

)=∑i

~vi × ~pi +∑i

~ri ×d

dt~pi

by the product rule. Now for all i we have that

~vi × ~pi = ~vi × (mi~vi) = mi ~vi × ~vi = 0

and the first sum above vanishes. For the second sum we use Newton’s second law for each singleparticle

d

dt~pi = Fi,

where Fi is the total force acting on the i-th particle. If we divide this force again into an externaland an internal part, ~Fi, ext and ~Fi, int, respectively, we get:

d

dt~L =

n∑i=1

~ri × ~Fi, ext +n∑i=1

~ri × ~Fi, int

and we easily identify the first sum as the external, the second as the internal torque. To get ridof the internal torque, we make the following observation: The forces between the particles aresubject to Newton’s third law. We can split up each ~Fi, int into the components arising from allparticles but the i-th itself:

~Fi, int =

n∑j=1, j 6=i

~Fj→i,

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where ~Fj→i is the force by which particle j acts on particle i. Newton’s third law states that

~Fj→i = −~Fi→j .

Note that every term of the kind~ri × ~Fj→i

inside the double summationn∑i=1

n∑j=1, j 6=i

~Fj→i

has a unique ‘counterpart’~rj × ~Fi→j

within the sum. We can group the sum into such pairs of terms and get

~ri × ~Fj→i + ~rj × ~Fi→j = (~ri − ~rj)× ~Fj→i

for each pair of distinct particles i and j, from the third law. In our purely classical problems, theforces will be of gravitational or electromagnetic nature only. As such, they are known to act alongthe vector ~ri − ~rj joining two particles i and j, meaning that the cross products

(~ri − ~rj)× ~Fj→i

all vanish! The whole internal torque of the system therefore vanishes and we are left with thefinal result

d

dt~L =

n∑i=1

~ri × ~Fi, ext = ~τext,

which was to be shown.

2.5 Rigid Bodies

An example will show you why this section is even necessary.

Example 2.5.0.1: Consider a homogeneous rigid rod of length ` and mass m fixed by meansof a screw to the ground in such a way it can pivot freely in a vertical plane around one of itsends. The rod’s other end is held at an angle ϑ above the ground and let fall freely at a certaintime. Air friction and friction in the pivot are negligible. Calculate the speed of the free endof the rod when it hits the ground, taking g for acceleration of gravity.

At a first glance, this looks like your standard problem involving Newton’s second law in whichyou have to first determine and then solve the equations of motion for an object. You might realizemore or less rapidly that application of Newton’s second law for point masses doesn’t lead youanywhere: the rod is definitely not a point-like particle but an extended system of particles. Maybeone of the tools described in section 2.4 can help. Take (2.54), for instance. In order to apply thisequation, we should calculate the total angular momentum of the rod by summing the angularmomenta of all particles building it. This seems to be quite complicated.

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Physics Olympiad Script 2.5. RIGID BODIES

The rod is a system of point masses of a very special kind in that its particles won’t change theirpositions relative to each other as the rod moves. Such objects are modelled in mechanics as whatwe call rigid bodies. For our purposes, only two properties of rigid bodies are important:

A rigid body is

− a system consisting of many particles

− such that the distance between any two of the particles remains constant over time.

In this section, we will see how the second condition gives rise to laws governing the motion ofrigid bodies following from the laws already stated in the last section about system of particles.

In general, rigid bodies occurring in Olympiad problems will be more or less regular with a more orless symmetric shape, such as spheres, rods, hollow cylinders, prisms, or even various combinationsthereof. Such objects are relatively easy to deal with from a mathematical point of view. Inparticular, competition problems will require you to answer questions concerning rotational motionof rigid bodies – that’s why they usually display some kind of rotational symmetry around someaxis.

2.5.1 Kinematics of Rigid Bodies

First, we need to describe motion of rigid bodies in a useful way by means of kinematic quantities.Unlike point masses, rigid bodies do not only change their position in space by ‘mere displacement’,but they can also rotate. Indeed, general motion of a rigid body satisfies the following rule:

Consider a rigid body moving in space. One can always choose a point P associated with thisbody (and not necessarily a part of it), that is, the relative position of P with respect to any partof the rigid body does not change as the body moves, such that motion of the rigid body can bedescribed as a superposition of

− translation of P in space and

− rotation of the rigid body about an axis passing through P . This axis might change directionin space as the rigid body moves.

This division into translation and rotation is of fundamental importance, as different sets of equa-tions govern these two components of motion.

Choice of the point P is of course not unique. You will see in the examples throughout this sectionthat often P is best chosen as the center of mass of the rigid body, sometimes however as a pointfixed in space during motion of the rigid body such that, in the corresponding set of coordinates,motion will be described as a pure rotation without any translational component. There is nobetter or worse choice of P , since in the end all choices must lead to physically equivalent results.You will find, however, that in certain situations some choices will come with less complicatedalgebra and lead you to the solution of a problem faster than other choices.

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Example 2.5.0.1, continued: We now want to describe motion of the rod as a superpositionof translation and rotation.

If we choose the center of the rod as our point of reference, motion of the rod is described asa superposition of

− translation of the center of the rod along a circular trajectory with center at the pivotand radius `/2 and

− rotation of the rod around its center.

If we choose the pivot as a point of reference instead, translational motion vanishes. Motion ofthe rod is then a pure rotation around the pivot by an angle ϕ measured from the horizontalline.

It is not hard to figure that rotational motion of a rigid body can, at any instant, be described as achange of orientation with respect to a certain axis in space. Imagine a rigid body rotating aroundan axis whose position in space does not change with time. If you look onto the body from eitherup or down along the axis, you can see how every particle of the body moves on a circular trajectoryaround the axis in a plane perpendicular to it. Let ~vi be the speed of the i-th particle and ~r⊥i aposition vector pointing perpendicularly from the axis of rotation to the position of the particle,therefore lying in the plane where the i-th particle moves. The angular speed of rotation in theplane perpendicular to the axis is the same for all particles of the rigid body, as all inter-particledistances do not change over time. If we call this angular speed ω, then the relationship

vi = r⊥iω (2.56)

for the absolute value vi of the speed of the i-th particle holds. As long as the axis of rotationremains invariant in time, description of rotational motion of a rigid body is basically not muchdifferent from circular motion of a single particle in a plane. However, the rigid body is a three-dimensional object and forces acting on it causing changes in its motion will generally not lie inthe same plane as the velocities of its point-like components. Therefore, it sounds reasonable touse a vectorial quantity for description of rotational motion of the rigid body. One could start byintroducing a unit vector n pointing in the direction of the axis of rotation. This enables us togive (2.56) a vectorial character, since

~vi = n× ω ~r⊥iholds. If we introduce the angular velocity vector

~ω := ω n, (2.57)

this expression becomes very compact:

~vi = ~ω × ~r⊥i. (2.58)

Notice how the angular velocity vector ~ω contains a lot of information about rotational motion ofthe body. Not only it shows in direction of the axis of rotation, but it also tells you how fast the

88


body rotates and in which sense: from (2.58), you can deduce that, if you look through the axis ofrotation in direction of ω, the body rotates clockwise. In other words:

If the long fingers of the right hand are curled in the sense of rotation of the body, then ~ω pointsin the direction shown by the stretched right thumb along the axis of rotation.

Something still doesn’t look too elegant in (2.58): that subscript ⊥ for position vectors. Suppose wework in such a coordinate frame that the axis of rotation passes through its origin. Every positionvector ~ri relative to the origin can then be split into ~ri = ~r⊥i + ~r‖i, where ~r‖i is a component of ~rilying on the axis of rotation and ~r⊥i defined as before. With this notation, we see that

~vi = ~ω ×(~ri − ~r‖i

)= ~ω × ~ri − ~ω × ~r‖i

= ~ω × ~ri,

since ~r‖i is parallel to ~ω so that ~r‖i × ω = 0. In other words, if the origin of our coordinate systemlies on the rotation axis,

~vi = ~ω × ~riholds for all particles i.

In physics, one often encounters the concept of degree of freedom, by which one usually means thenumber of parameters of a system that may vary independently. In mechanics, you can think ofthe degrees of freedom of an object as the number of ‘settings’ one has to adjust to give a uniquedescription of its position in space. For instance, a given point particle in space has three degrees offreedom, as each of its three coordinates can be varied independently and all of them are necessaryto locate it. A general system of n particle, accordingly, has 3n degrees of freedom, as long asall particles move independently of each other. A rigid body, no matter how many particles it ismade of, has six degrees of freedom. Three coordinates are needed to describe the position of itscenter of mass, or any other point P defined as previously, and three more variables, like the threecomponents of a unit vector along a certain axis of the body, to determine its orientation in space.

2.5.2 How Forces Determine Translational Motion of a Rigid Body

Now that we have tools for description of motion of rigid bodies, we might as well get to the lawsgoverning it.

The first law we look at is very simple and follows directly from the previous section about systemsof particles:

Given the total sum of external forces ~Ftot, ext acting on a rigid body of mass m, the acceleration~aCM of its center of mass is given by

m~aCM = ~Ftot, ext. (2.59)

For a body to be rigid, a great deal of internal forces must act between its particle-like componentsto hold it together. As we saw from the discussion of systems of particles in section 2.4, theseinternal forces cancel out and we can consider motion of the center of mass only.

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From (2.59), you can easily derive:

If the sum of all external forces ~Ftot, ext acting on a rigid body vanishes, the velocity of its centerof mass ~vCM remains constant.

This is analogous to Newton’s first law for point masses.

Even with (2.59), we have a substantial problem. External forces acting on a rigid body can bemany, especially if a rigid body is subject to a force field like a gravitational field. In such a case,one would have to add all forces acting on every single point-like particle building the rigid body.Take a uniform gravitational field with acceleration ~g, for instance. The i-th particle of mass mi

experiences a weight mi~g, and, according to (2.59):

~aCM due to gravity =1

m

∑i

mi~g =

(1

m·∑i

mi

)~g

= ~g.

This means:

The effect of a homogeneous gravitational field ~g on translational motion of a rigid body of massm can be described as a force m~g acting on the center of mass of the body.In other words: concerning translational acceleration in a homogeneous gravitational field, a rigidbody can be treated as a particle of same mass as the total mass of the rigid body and at positionof its center of mass.

That’s why the center of mass of a rigid body is often called its center of gravity. Note that assoon as one considers an inhomogeneous field, the concept of center of gravity does not make senseany more.

The center of mass of a rigid body can in principle be determined with equation (2.48), that is,

~rCM =

∑imi~ri∑imi

,

provided we know what kind of particles a rigid body is made of, what their masses are and howthey are distributed inside the body. You see, the task of finding the center of mass of a rigidbody does not seem to be very easy. We usually deal with objects in the macroscopic world,meaning that the number n of particles involved is extremely large and the mass mi of one singleparticle extremely small. For once, this immensity is an advantage: it turns out that the exactnumber of particles and the exact mass of every single one of them is not very relevant as long astheir distribution approximates the macroscopic distribution of mass density, which we will denoteby the symbol %, of the rigid body. This quantity can be understood as follows: a very smallvolume dV of the rigid body located at ~r carries the mass % · dV . In general, a rigid body willbe inhomogeneous, meaning that its mass is not evenly distributed across its volume and that %becomes a quantity dependent on position. That’s why we might as well write % (~r). The positionof the center of mass can now be expressed as

~rCM =

∑i % (~ri)~ri dVi

M,

90


with the choice of very small volume elements dVi located at ~ri. In the limit of these volumeelements becoming infinitesimally small, you surely know that a sum such as in the right-handside of the previous equation approaches an integral. That is, we can write

~rCM =1

M

ˆrigid body

% (~r) ~r dV

How this formula can be applied to rigid bodies of different shapes is left to your math courses.For now, it is relevant that you know about some important properties of the center of mass:

− The position of the center of mass always somehow reflects the symmetry of a body. Thecenter of mass of a homogeneous sphere, for instance, is located at its geometrical center.The same holds for a homogeneous plane square or a homogeneous one-dimensional rod.

− In particular, the center of mass of a rigid body can lie outside the body. For example, thecenter of mass of a homogeneous circular hoop lies at its geometrical center.

− For a system consisting of various rigid bodies, equation (2.52) applies.

Example 2.5.0.1, continued: Applied on our example, this means that the weight of the rodcan be described at any time by the vector m~g with starting point at the center of the rod.Choose an xy-coordinate system in the plane of motion of the rod with origin at the pivot,x-axis parallel to the ground and upward pointing y-axis. We have m~g = −mg y. Weight isnot the only external force acting on the rod, there is also an unknown force ~F exerted by thepivot. We cannot apply the second law of motion yet, as there does not seem to be any easyway by which ~F can be determined.

2.5.3 How Forces Determine Rotational Motion of a Rigid Body

In general, (2.59) won’t be sufficient to fully determine motion of a rigid body in space: in general,there are six degrees of freedom and (2.59) involves three coordinates only. In section 2.4, wedeveloped a quantity strongly associated with rotations of systems of particles: it was angularmomentum. Of course, equation (2.55) can also be applied to a rigid body:

d

dt~L = ~τext, tot.

The angular momentum ~L of a rigid body about a certain point turns out to be quite difficult tocompute, as one has add up ~ri × ~vi for all particles inside the body (even turning the sum intoan integral won’t help too much here). However, one can show that for some very symmetricalsituations, the angular momentum of a rigid body rotating about a fixed axis at angular velocity~ω about any point on the axis is just given by

~L = I~ω, (2.60)

where I is the so-called moment of inertia of the body about this axis. O.k., now step by step:

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− When does (2.60) apply? In situations with a high degree of symmetry such as

– a sphere rotating about any axis in space,

– a cylinder rotating about its own axis of rotational symmetry,

– a cylinder rotating about any axis parallel to its rotational axis of symmetry

– a cylinder rotating about any axis perpendicular to its rotational axis of symmetry,

– a cube rotating about one of its edges,

– a single point-like particle rotating about any other point in space,

and many others. You get the principle. Plus, the axis has to be fixed in space. The equationis not valid for axes of rotation moving around.

− What is I? In general, the moment of inertia of a rigid body about a given axis is calculatedas

I =∑i

mir⊥i2, (2.61)

where r⊥i denotes the distance of the i-th particle from the axis of rotation. In manyproblems, rigid bodies won’t be simple rigid systems of point masses, but be extended systemswith continuous mass distributions %, and one can calculate I as

I =

ˆrigid body

r⊥2 dm

=

ˆrigid body

% (~r) r⊥2 dV.

(2.62)

In most cases, you won’t have to calculate moments of inertia by yourself, they will usuallybe given along with the task. The SI unit of the moment of inertia is kg ·m2.

− The moment of inertia is an additive quantity: If a rigid body consists of various partsA1, A2 . . ., the total moment of inertia Itot of the whole body about an axis is just the sumof the moments of inertia Iiof its parts about this axis:

Itot =∑i

Ii. (2.63)

− With respect to what point is ~L calculated? Equation (2.60) is valid for any point along theaxis of rotation as the origin. (You can easily see that ~L must be the same for all pointsalong the axis of rotation from )

92


− From (2.60) follows that in the symmetric cases where this law applies, the angular momen-tum of a rigid body is parallel to the axis of rotation. Note that this won’t be the case ingeneral for situations not displaying a high degree of symmetry.

We won’t give a derivation of (2.60), so feel free to try to figure out how it arises (it is not toodifficult if you make the right assumptions about symmetry of the system). Anyway, textbookslike [5] and [6] provide you full derivations and more advanced discussions of this topic.

For you to get more acquainted with the concept of moment of inertia, here are some examples.

− Let a point-like particle of mass m move along a circle of radius r with angular velocityω. The moment of inertia of the particle about an axis perpendicular to the plane of thecircle passing through its center is equal to mr2. Note how in this case, with origin of thecoordinate frame at the center of the circle, equation (2.60) becomes ~L = mr2~ω, and youmay verify that this is indeed equal to ~r× ~p for a position vector ~r with respect to the centerof the circle and linear momentum ~p of the particle. The second law for rotational motionthen states that

d

dt~L = ~τext, tot

⇒ d

dt

(mr2~ω

)= ~r × ~F

⇒ mr2ω = rFt

⇒ Ftm

= rω

where Ft denotes the tangential component of a force ~F acting on the particle. This is aresult we already derived in section 2.2.2 about non-uniform circular motion.

− Consider a homogeneous ring of mass m and radius r. A small mass element dm of the ringhas a moment of inertia r2dm about an axis perpendicular to the plane of the ring passingthrough its center. The moment of inertia of the whole ring about this axis is obtained bysumming up over all infinitesimal mass elements building it:

´ring r

2 dm = r2´ring dm = mr2.

− Consider a homogeneous flat disc of mass m and radius R and an axis perpendicular to itpassing through its center. To calculate the moment of inertia of the disc about this axis, wecan decompose it into many small rings of radius r, 0 < r < R, and infinitesimal thicknessdr. The area of such a ring is 2πr dr, its mass therefore

2πrdr · MπR2

=2rdr

R2M

and its moment of inertia about the central axis becomes 2Mr dr/R2 · r2 = 2Mr3 dr/R2.The moment of inertia of the disc can be obtained by integrating over the moments of inertiaof all these small rings: ˆ R

r=0

2Mr3

R2dr =

1

2mR2.

− Consider a homogeneous cylinder of mass m, radius R, length L and its axis of rotationalsymmetry. For the corresponding moment of inertia, think of the cylinder as consisting ofmany circular discs of infinitesimal thickness d`, each of mass d` ·m/L, stacked on each other.

93


The moment of inertia of every single disc is mR2d`/ (2L), and the total moment of inertiaof the disc becomes ˆ L

`=0

mR2

2Ld` =

1

2mR2,

a result independent of the length L of the cylinder.

Computation of more complex examples is left to your math courses.

The moment of inertia satisfies a property often coming in very handy:

Parallel Axis Theorem (Steiner’s Theorem)

Let ICM denote the moment of inertia of a rigid body of mass M with respect to an axis passingthrough its center of mass and I ′ be its moment of inertia about a second axis parallel to the firstone at distance d from the center of mass. The following relationship holds:

I ′ = ICM +Md2. (2.64)

Example 2.5.0.1, continued: Let’s use (2.62) to calculate the moment of inertia of the rod Iabout an axis perpendicular to its plane of motion passing through the pivot. Since the rod isa one-dimensional object, the integral to be evaluated will be a one-dimensional one. Choosean x-axis along the rod with origin at the pivot. An infinitesimal rod segment of length dxcarries the mass mdx/`. For I, we get

I =

ˆ `

x=0x2 dm

=

ˆ `

x=0x2m

`dx

=1

3m`2.

We could now use the parallel axis theorem if we wanted to know the moment of inertia Ic ofthe rod with respect to an axis perpendicular to the rod passing through its center:

I = Ic +m ·(

1

2`

)2

⇔ Ic = I − 1

4m`2 =

1

12m`2.

We could have obtained the same result by direct integration as we did for the axis throughthe pivot.

94


With (2.60), equation (2.55) takes the form

Newton’s Second Law for Rotation of Rigid Bodies

Id

dtα = ~τext, tot, (2.65)

where angular acceleration α = ω is introduced. This equation will be your most fundamental toolfor solution of problems involving rotating rigid bodies.

In particular, let’s look at the effect of gravity. Consider a rigid body exposed to a homogeneousgravitational field ~g. The total torque due to the weight mi~g of each particle constituting the bodywith respect to some axis is

~τ =∑i

~ri × ~g =

(∑i

mi~ri

)× ~g =

(∑i

mi

)~rCM × ~g,

that is,

The total torque due to gravity ~g acting on a rigid body of mass M is obtained as

~τ = M~rCM × ~g, (2.66)

where ~rCM is the position of the center of mass of the rigid body.

Together with the result of section 2.5.2, we have a very important result:

For both translational and rotational motion, can treat the weight of a rigid body as if it was onesingle force acting on its center of mass.

Example 2.5.0.1, continued: The external force ~F is still unknown. Yet we can use (2.60) witha special trick. As an axis of reference, we take the line passing through the pivot orthogonalto the plane in which the rod moves. Suppose the rod makes an angle ϕ with the floor attime t > 0. For the right-hand side of (2.60), we need to sum the torques exerted by the twoexternal forces acting on the system. The force and radius vectors involved in computation ofthe torques all lie in the plane of motion of the rod, that is, perpendicular to our axis. Thetorques and angular momenta involved will therefore all be parallel to our axis – that’s whywe might as well calculate with their components along this axis and get rid of vector arrows.That is, we’ll use the symbols τ and L for the components of ~τ and ~L, respectively, along ouraxis of reference. Let’s say coordinates are chosen in such a way that, if we look onto the planeof motion of the rod with the pivot at our left, vectors pointing away from us have a positive

95


component along our axis of reference. The total torque acting on the rod then is

τ = τF + τweight

= 0 · F +1

2` ·mg · sin (90 − ϕ)

=1

2mg` cosϕ.

We see how something very appealing happened: we got rid of F , as it acts on a point throughwhich our axis passes. Equation (2.65) now assumes the form

Iω =1

2mg` cosϕ,

where I is the moment of inertia of the rod about our axis and ω = −ϕ, since, with our choiceof coordinates, ω > 0 corresponds to downward motion of the rod.

With I = m`2/3, the equation of motion of the rod assumes the form

−1

3m`2ϕ =

1

2mg` cosϕ

⇔ ϕ = −3

2

g

`cosϕ.

This differential equation doesn’t look like we know how to solve it. One could of course usesome mathematical trick to retrieve more information (can you think of one?), but we willmove on looking for more useful tools.

2.5.4 Mechanical Energy of Moving Rigid Bodies

We didn’t talk too much about conservation of energy when we looked at systems of particles.That’s because, in general, the kinetic energy of a system of particles is not very easy to calculate,as one would have to sum over all terms mivi

2/2 belonging to each particle:

Ekin =1

2

∑i

mivi2. (2.67)

The situation turns out to look quite simple for rigid bodies though:

The kinetic energy of a rigid body of mass M in motion can be split up into two components, onefor translational motion of its center of mass and one for rotational motion about an axis passingthrough the center of mass:

Ekin =1

2MvCM

2 + Ekin, rot, (2.68)

where vCM denotes the velocity of the center of mass. This result is derived below. In situationsdisplaying a high degree of symmetry, the kinetic energy of rotation about the axis passing throughthe center of mass can be calculated as

Ekin, rot =1

2Iω2, (2.69)

96


where ω is the angular velocity of this rotational motion and I the moment of inertia of the rigidbody about this axis of rotation.

Let’s say motion of a rigid body can be described as pure rotation of its center of mass aboutan axis distant d from it at angular velocity ω. At the same time, one can always describe suchmotion as the superposition of translation of the center of mass at speed vCM = ωd about this axisand rotation of the body about another parallel axis passing through the center of mass at angularvelocity ω. According to (2.68), one can write

Ekin =1

2M (ωd)2 +

1

2Iω2

=1

2

(I +Md2

)ω2

=1

2I ′ω2,

where I is the moment of inertia of the body about the axis passing through its center of mass andI ′ about the other axis with I ′ = I +Md2 according to the parallel axis theorem.

If motion of a rigid body can entirely be described as rotation of its center of mass about an axisat angular velocity ω, its kinetic energy is then given by

Ekin =1

2Iω2, (2.70)

where I denotes the body’s moment of inertia about this axis of rotation.

We already saw how the weight of a rigid body can be treated for both translational and rotationalmotion. We turn now to potential energy of a rigid body in a homogeneous gravitational field. Asusual, we can obtain the potential energy of a rigid body by adding up the potential energies ofall masses mi it is made of:

Epot =∑i

mighi = g ·∑i

mihi = g ·

(∑i

mi

)hCM,

where hi is the height of the i-th particle and hCM the height of the center of mass above the zeroof potential energy. That is,

Potential energy of a rigid body of mass M in a homogeneous gravitational field ~g is given by

Epot = MghCM (2.71)

with the height hCM of the center of mass above the zero level of potential energy.

Example 2.5.0.1, continued: We will try now with conservation of mechanical energy. Whenthe rod is released, its angular speed is 0 and the height of its center of mass above the ground

97


is (` sinϑ) /2, its total mechanical energy therefore amounts to

Estart = Epot, start + Ekin, start =1

2mg` sinϑ+ 0

When the rod hits the ground, we have

Eend = Epot, end + Ekin, end = 0 +1

2Iω2

end,

where I = m`2/3 is the moment of inertia of the rod with respect to the pivot and ωend itsangular velocity as it hits the ground. Conservation of energy reads as Estart = Eend, fromwhich we retrieve

ωend =

√3g sinϑ

`.

The speed of the free end of the rod as it hits the ground is

ωend` =√

3g` sinϑ,

which finally solves our problem.

Derivation of the Kinetic Energy Expression for Moving Rigid Bodies*

The total kinetic energy Ekin of a rigid body moving as in 2.5.4 can be expressed as

Ekin =∑i

1

2miv

2i

where the i-th particle is of mass mi and moving at speed ~vi. If we split up motion into translationof the center of mass of the rigid body and rotation about an axis passing through the center ofmass, we can write

~vi = ~vCM + ~v(CM)i ,

where ~vCM is the velocity of the center of mass of the rigid body and ~v(CM)i the velocity of the i-th

particle with respect to the center of mass. Therefore,

Ekin =1

2

∑i

mi

(~vCM + ~v

(CM)i

)2

=1

2

∑i

mi

(vCM

2 + v(CM)i

2+ 2~vCM · ~v

(CM)i

)

=1

2

(∑i

mi

)vCM

2 +1

2

∑i

miv(CM)i

2+ ~vCM ·

(∑i

mi~v(CM)i

)

=1

2mvCM

2 +1

2

∑i

miv(CM)i

2

98


2.5.5 Putting Everything Together: Rolling Rigid Bodies

One particular kind of motion of rigid bodies is rolling. We consider rigid bodies with a somehowcircular cross section moving on a plane surface. Let’s say that rolling of a rigid body occurs atangular velocity ω about its geometrical axis of rotational symmetry, which is perpendicular to itsdirection of translational motion. You can convince yourself by drawing a sketch that with respectto an axis parallel to the first one but passing through the point of contact with the surface, everypoint of the rigid body also rotates at angular velocity ω. In particular, the geometrical centerof the body moves at speed ωr with respect to the axis through the point of contact if r is theradius of the body. The ever changing, instantaneous point of contact with the underlying surfaceis always at rest with respect to the surface. You will also agree that the condition for rollingwithout slipping to occur is that rω is equal to translational velocity v of the geometrical center ofthe body:

A rigid body of cross-sectional radius r rolls without slipping on a surface if and only if thetranslational velocity v of its geometrical center and its angular velocity ω of rotation about itsgeometrical center satisfy

v = ωr. (2.72)

Rolling motion of a rigid body can be described as the superposition of translational motion of itscenter of mass and rotational motion of the body about the center of mass. The center of massis assumed to lie on the axis of rotational symmetry of the body. We saw in sections 2.5.2 and2.5.3 how to set up the equations of motion for translational and rotational motion, respectively.Translational motion does not pose much of a problem, as application of equation (2.59) is quitestraightforward. Rotational motion is a bit trickier though: equation (2.65) is only valid if theaxis of rotation of the rigid body is fixed in space, but the axis of rotation of a rolling body movesalong with it. Fortunately, it turns out that an equation analogous to (2.65) is always valid in thereference frame where the center of mass is at rest, independently of whether it is accelerated ornot! That is,

Newton’s Second Law for Rolling Bodies

~τ(CM)ext, tot = I(CM)α, (2.73)

where I(CM) is the moment of inertia of the body about its axis of rotational symmetry. Angularacceleration α is the same in a reference frame moving with the center of mass of the body as inany inertial frame. A derivation of this equation is shown below.

Example 2.5.5.1: A rigid homogeneous sphere of mass M and radius R rolls down a plane oflength L inclined by an angle ϑ without slipping, starting from rest.

a) How fast is the center of mass of the sphere at the bottom of the incline?

b) What condition must ϑ satisfy for the sphere to roll without sliding, if the coefficient ofstatic friction between the sphere and the inclined plane is µ?

99


Hint : the moment of inertia of the sphere around an axis passing through its center is 25MR2.

Solution: We wish to determine and solve the equations of motion of the sphere in orderto determine how its speed evolves with time as it rolls down the incline. First, one hasto determine what forces act on the sphere. There is of course its weight ~G = M~g withgravitational acceleration ~g. The force by which the incline acts on the sphere has in generaltwo components, we call them ~N for the normal component at right angle to the incline and~F for the tangential component parallel to the incline. Since the sphere is constrained to movealong the incline, it makes sense to choose a coordinate axis parallel to the incline with thepositive direction pointing downward.

Let’s start with what the second law says for translational motion. If we denote the acceler-ation of the center of mass of the sphere, which coincides with its geometrical center due tohomogeneity, along the coordinate axis by a, where a > 0 shall mean increasing downwardvelocity, the equation of motion for this direction assumes the form

Ma = Mg sinϑ− F,where F > 0 is chosen if ~F points uphill. For the magnitude of F , the relationship |F | ≤ µ|N |holds, where N > 0 means ~N points away from the incline. The sphere does not moveperpendicularly to the incline, and the second law for translational motion in this directionlooks like

N −mg cosϑ = 0.

a) So far, not much can be done to calculate a (except for maybe a lower bound from thecondition |F | ≤ µ|N |). We need more information about the effect of the forces actingon the sphere, and that is what the second law for rolling bodies (2.73) is for. The forcesacting on the sphere will all lie in the plane perpendicular to the axis perpendicular tothe direction of translational motion of the sphere passing through its center. We choosethe orientation of this axis in such a way that the components of angular momentum andtorque in its direction consistent with downhill rotation of the sphere are positive. Withrespect to the center of mass of the sphere, the torque exerted by ~G vanishes, as ~G actsin the center of mass of the sphere itself. The torque exerted by ~N is also zero, since ~N ,which acts on the point of contact between the sphere and the incline, points towards thecenter of the sphere. The only force we are left with is ~F , and the torque exerted by itis, with our convention, equal to FR, since ~F is perpendicular to the line of connectionbetween the center of the sphere and the point of contact with the incline. If we denoteangular acceleration of the sphere by α and take α > 0 for increasing angular velocityrolling down the slope, the second law for rolling of the sphere becomes

Iα = FR,

where I = 25MR2 is the moment of inertia of the sphere with respect to our coordinate

axis through its center. The rolling condition must be satisfied, therefore α = a/R andthe second law for rotational motion becomes Ia = FR2. We can now eliminate F fromthe equation of motion for the center of mass, put it into the equation for rotationalmotion and solve for a to get

a =5

7g sinϑ.

100


From this, the speed of the sphere at the end of the incline is just

√2La =

√10

7gL sinϑ.

Another approach is way more elegant and leads you much more rapidly to a solution:conservation of energy. At the top of the incline, at height L sinϑ above the bottom, thesphere is at rest and its total mechanical energy is

Etop = mgL sinϑ

if the zero of potential energy is taken at the bottom of the incline. There potentialenergy is totally converted into kinetic energy:

Ebottom =1

2mvCM

2 +1

2Iω2

=1

2mvCM

2 +1

2· 2

5mR2 ·

(vCMω

)2

=7

10mvCM

2

The result for vCM can now be determined from conservation of energy Etop = Ebottom.

b) We can solve the equation of motion for translation down the slope to get F :

F = Mg sinϑ−Ma =2

7g sinϑ

At the same time, we have the condition |F | ≤ µ|N | and N = mg cosϑ, so

2

7g sinϑ ≤ µmg cosϑ,

from whichtanϑ ≤ 7

2µ.

Note how we introduced a static friction force, as discussed in section 2.2.3. One can think ofrolling of a rigid body down an incline as follows: in absence of static friction, the body wouldjust slide down the incline. With static friction acting, the point of contact of the body with theincline is prevented from sliding. Therefore the rigid body will start rotating under influence ofits weight exerting a torque due to the inclination of the center of mass with respect to the pointof contact. If the slope of the incline is not too large, static friction can prevent the body fromsliding all along it, resulting in rolling motion.

Derivation of the Equations of Motion for Rolling Rigid Bodies*

Equation (2.73) is a consequence of a much more general principle:

101


Total angular momentum ~Ltot of a system of particles with respect to a certain point of referencecan be expressed as a sum of the total angular momentum of the system with respect of the centerof mass ~L (CM)

tot plus the angular momentum of the center of mass ~LCM with respect to that point:

~Ltot = ~LCM + ~L(CM)tot . (2.74)

To see why this hold, consider the transformations ~ri = ~r(CM)i +~rCM and ~vi = ~v

(CM)i +~vCM for the

position and velocity vector, respectively, of the i-th particle between our point of reference andthe center of mass system. Then

~Ltot =∑i

mi

(~r(CM)i + ~rCM

)×(~v(CM)i + ~vCM

)=

(∑i

mi~r(CM)i

)× ~vCM +

∑i

mi~r(CM)i × ~v (CM)

i

+ ~rCM ×∑i

mi~v(CM)i +

(∑i

mi

)~rCM × ~vCM

= 0 +∑i

mi~r(CM)i × ~v (CM)

i + 0 +

(∑i

mi

)~rCM × ~vCM

= ~L(CM)tot + ~LCM,

where ∑i

mi~r(CM)i = 0

since the center of mass of the system lies at the origin, or 0, of the center of mass frame and∑i

mi~v(CM)i =

∑i

mi (~vi − ~vCM)

=∑i

mi~vi −

(∑i

mi

)~vCM

= 0.

The total external torque in the center of mass frame can be expressed as

~τext, tot =∑

~ri × ~Fi, ext

=∑i

(~rCM + ~r

(CM)i

)× ~Fi, ext

= ~rCM ×

(∑i

~Fi, ext

)+∑i

~r(CM)i × ~Fi, ext

= ~rCM × ~Fext, tot + ~τ(CM)ext .

102


The time derivative of the total momentum with respect to the reference point is, with (2.74),

d

dt~Ltot =

d

dt~LCM +

d

dt~L(CM)tot

=d

dt(~rCM × ~vCM) +

d

dt~L(CM)tot

=

(d

dt~rCM

)× ~pCM + ~rCM ×

d

dt~pCM +

d

dt~L(CM)tot

= 0 + ~rCM × ~Fext, tot +d

dt~L(CM)tot .

With ~τext, tot = ddt~Ltot, it follows that

~τ(CM)ext =

d

dt~L(CM)tot , (2.75)

which is just a more general form of (2.73).

2.5.6 Conservation of Angular Momentum

From (2.59) follows that angular momentum of a rigid body is constant if the total external torqueacting on it vanishes. Of course, this is just the law of conservation of angular momentum for asystem of particles. For rigid bodies, conservation of angular momentum has some very interestingconsequences. Suppose a rigid body rotates at angular speed ω around an axis with a momentof inertia I with respect to this axis. Suppose that this body is not entirely rigid, such that dueto some internal forces its moment of inertia changes from I to I ′. Since only internal forces areconsidered, angular momentum is conserved and the angular speed of the body must change as itsmoment of inertia changes:

Iω = I ′ω′ ⇔ ω′ =I

I ′ω.

Therefore angular velocity increases if the moment of inertia decreases and vice versa. This gives anexplanation for phenomena like figure skaters modulating their spins by holding their extremitiestight to increase spinning velocity or holding them out to slow down.

Example 2.5.6.1: A man sits on a spinning chair, for now at rest, holding a spinning wheelrotating at angular velocity Ω about the vertical in anticlockwise direction as seen from above.The man now turns the spinning wheel upside down. What happens to him? Assume thewheel and the chair to be frictionless, approximate the man with his chair as a rigid body ofmoment of inertia Iman about a vertical axis through the center of the chair.

Solution: Let Iwheel be the moment of inertia of the wheel about its own rotation axis. Considera system consisting of the man, the chair and the wheel. The initial angular momentum of thesysetms before the man turns the wheel is only determined by the angular momentum of thewheel:

~Lstart = Iwheel~Ω.

103


where the vector ~Ω points upwards. The only external forces acting on our system are gravityand forces from the ground on the chair. All of these forces act in vertical direction and aretherefore not able to change the angular momentum of the system. Therefore, total angularmomentum of the system must be conserved as the man turns the wheel. Once the wheel isturned upside down, the angular momentum vector of the wheel reverses its orientation alongwith its angular velocity vector. This means that the man and his chair must start spinningin direction of the initial spinning sense of the wheel in order to compensate and keep totalangular momentum constant. If, after turning the wheel, the man starts spinning at angularvelocity ~ω, total angular momentum of the system can be expressed as

~Lend = Iman~ω − Iwheel~Ω.

From conservation ~Lstart = ~Lend, one gets

ω = 2IwheelIman

~Ω.

2.6 Gravity

2.6.1 Newton’s Law of Gravity

I guess there’s no need to explain you what gravitational attraction is, you’ve surely heard aboutit in school. Let’s get straight to the point by introducing a notation for gravitational forces thattakes their vectorial nature into account:

Newton’s Law of GravityA point-like mass m1 located at ~r1 exerts the gravitational force

~F1→2 = Gm1m2

|~r1 − ~r2|3(~r1 − ~r2) (2.76)

on another point-like mass m2 located at ~r2, where

G = 6.67 · 10−11 m3 s−2 kg−1 (2.77)

is the universal gravitational constant.

In school you might have encountered something more like

F = Gm1m2

r2(2.78)

for the magnitude F of gravitational attraction between the point-like masses m1 and m2 atdistance r apart. This formula looks indeed simpler than (2.76), but it contains no information

104

Physics Olympiad Script 2.6. GRAVITY

about the direction of the force. From (2.76), you see that the force acting from m1 onto m2 is anattractive force, as ~r1 − ~r2 shows from m2 towards m1. We leave it to you to show that formula(2.78) is easily derived from (2.76).

Look at how the third law of motion is contained in (2.76): by swapping the subscripts 1 and 2,you see that

~F2→1 = −~F1→2,

that is, the two bodies 1 and 2 act on each other by means of opposite gravitational forces of equalmagnitude.

2.6.2 Gravitational Fields

Consider a large mass distribution, like a planet or a star, exerting gravitational forces on othersmaller objects in its surroundings, like meteoroids, comets, satellites, or human beings. Theseobjects are assumed to be so small that they do not affect motion of the larger mass distribution.

To calculate the gravitational force by which an extended mass distribution acts on a small mass m,we can think of the large mass distribution as being composed by a large amount of point massesmi, every of which will exert its ‘own’ gravitational attraction ~Fi on m according to

~Fi = Gmim

|~ri − ~r|3(~ri − ~r) ,

where ~ri denotes the position of the i-th point mass mi and ~r the position of m. To calculate thetotal gravitational attraction on m, we make use of the extremely important

Principle of SuperpositionConsider three point masses mA, mB and m. If mA and m were alone, let ~FA be the gravitationalforce mA would exert on m. Analogously, let ~FB be the gravitational force by which mB wouldact on m if mB and m were alone. The gravitational force acting on m if both mA and mB arearound is then given by

~FA + ~FB.

In other words: The gravitational force by which a system of point masses acts on another pointmass is given as the vector sum of all gravitational forces by which each of the point masses of thesystem alone would act on this other point mass.

Applied to our situation, this simply means that we can calculate the total gravitational attractionacting on m as ∑

all masses mi

~Fi =∑i

Gmim

|~ri − ~r|3(~ri − ~r)

= m ·

(G∑i

mi

|~ri − ~r|3(~ri − ~r)

).

The very last term in brackets is a quantity describing how the distribution of the masses mi inspace exerts gravitational forces on a mass located at ~r. We will call this therm the gravitationalfield ~g (~r) produced by this distribution of masses at ~r, such that we can write

105


~F = m~g (~r) . (2.79)

with

~g (~r) = G∑i

mi

|~ri − ~r|3(~ri − ~r) . (2.80)

So far for systems consisting of point masses. Calculation of gravitational fields produced byextended mass distributions turns out to be somewhat more difficult, the same way as determiningthe center of mass of a rigid body by means of finite sums used to be in section 2.5. Exactly aswe did there, we can treat continuous mass distributions as continuous by introducing their massdensity % (~r) and turning the sum into an integral:

The gravitational field produced in point ~r by a continuous mass distribution % confined to theportion of space Ω is given by

~g (~r) = G

ˆΩ

% (~r′)

|~r′ − ~r|3(~r′ − ~r

)dV ′. (2.81)

Again, your math courses will show you such integrals can be computed in various situations. Youmight realize that dealing with the integrand of (2.81) can be a rather tedious affair. Fortunately,a very elegant law exists that makes calculation of gravitational fields in particularly symmetricsituations quite simple:

Gauss’ Law for Gravitational FieldsLet Ω be a volume in space and ∂Ω its outer surface. For gravitational field ~g, the following identityis valid: ˛

∂Ω~g · d ~A = −4πGmΩ, (2.82)

where mΩ denotes the total mass located inside Ω.

You probably already know a very similar law by Gauss for electric fields and charge distributions.Indeed, both the gravitational and the electrostatic versions of Gauss’ law are nothing but a directconsequence of the inverse-square-law nature of Newton’s law of gravity and Coulomb’s law ofelectrostatic interaction, respectively. Again, your math courses will tell you what the integral inthe left-hand side of (2.82) means and how to calculate it.

We won’t discuss Gauss’ law in great detail as you can learn more about it in your electrodynamicscourses. However, we will show you one very important application.

Many Olympiad problems involving gravity deal with planets or stars, which are in most casestreated as spheres of homogeneous mass density. We will use Gauss’ Law to calculate the fieldinduced by such a homogeneous sphere of total mass M and radius R in any point of space. Thissituation shows perfect spherical symmetry in space, that is, one could rotate everything about

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the center of the sphere by any angle and you wouldn’t be able to tell the difference. Therefore,magnitude of the gravitational field must be the same at any point located at the same distancer from the center of the planet. Furthermore, still due to spherical symmetry, the vector ~g mustpoint in radial direction anywhere in space: spherical symmetry would be broken if ~g had at somepoint in space a non-vanishing component not pointing in radial direction. Choose a coordinatesystem with origin at the center of the sphere. The gravitational field at any point in space ~r canbe described as

~g (~r) = g(r) r

with the projection g (~r) of the gravitational field on r. First, we consider a point located insidethe sphere, that is, r < R. Now apply (2.82) on a sphere V (r) of radius r concentric to the planet.The surface integral on the left-hand side is obtained by subdivision of the surface ∂V (r) of thesphere into many infinitesimal area elements dA with normal vectors n and taking the sum overthe scalar products ~g · dA n at any such area element, that is, something like

ˆ∂V

dA ~g (at this area element) · n (at this area element)

=

ˆ∂V

dA (g(r)r) · r

= g(r)

ˆ∂V

dA

= g(r) · surface area of ∂V (r)

= g(r) · 4πr2,

where we used n = r for all surface elements on the sphere. Our result,˛∂V (r)

~g · d ~A = 4πr2 · g(r),

is, according to (2.82), equal to −4πG times the total mass mV (r) = M · (r/R)3 contained withinV (r), that is,

4πr2 · g(r) = −4πG · MR3

r3

⇔ g(r) = −GMR3

r.

Inside the sphere, the magnitude of gravitational field increases linearly with the distance from itscenter.

Now apply the same strategy for any point outside the sphere, that is, r > R. The mass mV (r)

contained inside a spherical surface of radius r around the center is now just the total mass M ofour spherical body. Gauss’ law now yields

4πr2g(r) = −4πGM

⇔ g(r) = −GMr2

,

which is the same as the field produced by a point mass M located at the origin. This is a veryimportant result:

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The gravitational field inside a homogeneous sphere of mass M and radius R increases linearlywith the distance from the center of the sphere:

~g (~r) = −GMR3

r3 r, 0 < r < R (2.83)

Outside the sphere, the gravitational field behaves as if the sphere was a point mass M located atits own center:

~g (~r) = −GMr2

r, r > R. (2.84)

Example 2.6.2.1: (Gravity Train) Consider a planet of homogeneous density % and diameterd in which a straight tunnel of negligible diameter compared to d and length ` is dug connectingtwo points on the surface. Suppose an object of mass m %R3 can move freely along thelongitudinal dimension of the tunnel, without friction, and is dropped at one end into thetunnel with negligible initial speed. After how much time does the object reach the other endof the tunnel?

Solution: In order to find the equation of motion of the object, we need to calculate thegravitational force acting on it at any position in the tunnel. For this, we can take (2.83) tocalculate the gravitational field inside a homogeneous sphere, since neither the tunnel nor mare assumed to affect the mass distribution of the planet. At distance r from the center ofthe planet, you might verify that projection of the gravitational field in radial direction g(r) isgiven by

g(r) = −4

3πG%r.

We introduce an x-axis along the tunnel with origin at its center. The gravitational force actingon the body at position x in the tunnel is of magnitude mg (r(x)), where r is the distance ofthe body to the center of the planet. But, since the body is constrained to move along thelongitudinal direction of the tunnel, only the component of gravitational attraction along thex-axis will be responsible for its acceleration. This component is given by

mg(r) · xr

= −4

3πGm% · x.

This should immediately ring a bell. The force acting on the small body is proportionalto its displacement from the origin and attracts the body towards it. The solution to thecorresponding equation of motion

x = −4

3πG% · x

is a harmonic oscillation of the form

x = A cos (ωt+ ϕ) ,

where

ω =

√4

3πG%

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and the constants A and ϕ depend on the initial conditions of motion. The period of oscillationof the body is T = 2π/ω, the time it takes to reach the other end of the tunnel is T/2 = π/ω,that is, √

3π

4G%.

2.6.3 Energy and Angular Momentum in Gravitational Fields

In school, you might already have encountered the expression

Epot = −G Mm

r(2.85)

for potential energy of a small mass m subject to gravitational field of the larger mass M atdistance r from it. This expression arises from the definition of potential energy of the particleas the amount of work an external force would have to perform to move the mass inside the fieldfrom infinity to a point at distance r from the center of the field.

Consider a point ~r at distance r from the center of the field and another arbitrary point ~r′ atdistance r′ from the center. The amount of work done by an external force when displacing themass from ~r′ to ~r along a path γ connecting ~r′ to ~r is the line integral

Wγ = −ˆγm~g · d~s

Decompose the path γ into very small segments d~s in such a way that their direction is eitherradial or perpendicular to the radial direction. The force vector m~g is always directed radially,therefore only terms with radial d~s will contribute to the integral, since all other terms vanish with~g · d~s = 0. Therefore, the value of Wγ is independent of the exact shape of γ as long as it connects~r′ to ~r. Furthermore, the exact position of both ~r′ and ~r is not relevant for the value of Wγ , theycould be replaced by any points located at distance r′ and r, respectively, from the center of M .For calculation of Wγ , we can therefore replace ~r′ and ~r by the two points (r′, 0, 0) and (r, 0, 0),respectively:

Wγ =

ˆ r

x=r′

(+GMm

x2x

)· (dx x)

= GMm

ˆ r

x=r′

dx

x2

= GMm

(1

r′− 1

r

),

that is,

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Suppose a large mass M generates a gravitational field at the origin of a coordinate system. Thework an external force has to do to bring a small mass m from a point at distance r′ from theorigin to another point distant r from the origin is equal to

GMm

(1

r′− 1

r

). (2.86)

Now back to the concept of potential energy. In a homogeneous gravitational field, one can definepotential energy of an object as the amount of work an external force has to perform to bring theobject from a determined reference height to the height of the object’s current position. For centralgravitational fields, we can apply an analogous definition. It seems reasonable to choose infinityas a distance of reference: if we let r′ → ∞, the corresponding term of (2.86) vanishes. That is,we are left with equation (2.85).

Problems involving gravity can often be solved by considering conservation of mechanical energyof an object moving in a gravitational field:

Total mechanical energy

E =1

2mv2 −G mM

r(2.87)

of an object of mass m moving in a gravitational field produced by a large mass M at the originis constant over time.

Gravitational fields as from point masses or homogeneous spheres are radial. In particular, angularmomentum of particles is preserved as they move in such fields.

Example: A comet of mass m passes the sun of mass M along a hyperbolic orbit. Far awayfrom the sun, the meteoroid moves at speed v. The asymptote of its orbit is at distance d fromthe sun. Determine the closest distance of the meteoroid on its orbit around the sun, that isin the perigee, and the comet’s speed in this point.

Solution: Let a be the distance of the comet from the sun at the perigee and u the correspondingspeed of the comet. The laws of conservation of angular momentum and of energy will give usidentities from which we can retrieve a and u.

Angular momentum: Far away from the sun, the comet moves along an asymptote of thehyperbolic orbit. Its angular momentum about the sun then is of magnitude

L∞ = dmv,

since the asymptote of the hyperbola is distant d from the sun. At the perigee, the velocityvector of the comet must be perpendicular to its position vector with respect to the sun, dueto symmetry of the orbit. The angular momentum of the comet at the perigee amounts to

Lperigee = amu

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and conservation of angular momentum reads

L∞ = Lperigee

dmv = amu.

Energy: Potential energy is 0 far away from the sun and−GMm/a at the perigee. Conservationof total energy gives

Epot, ∞ + Ekin, ∞ = Epot, perigee + Ekin, perigee

0 +1

2mv2 = −G Mm

a+

1

2mu2.

You can do the algebra to verify that these equations yield

u =GM

vd

1 +

√1 +

(v2d

GM

)2

and

a =v2d2

GM

1 +

√1 +

(v2d

GM

)2−1

.

2.6.4 Two Objects Subject to Mutual Attraction

So far, we have only considered small masses subject to an external gravitational field produced bya mass distribution so large not to be affected by the small mass. We turn now to the discussionof systems consisting of two masses m1, m2 of similar orders of magnitudes, such that motion ofeach is influenced by gravitational attraction from the respective other mass. In such a system, ifnot subject to any other external forces, the following conservation laws hold:

− Position of the center of mass (m1~r1 +m2~r2) / (m1 +m2) is constant over time.

− Total energy

E = Ekin, 1 + Ekin, 2 + Epot

=1

2m1v

21 +

1

2m2v

22 −G

m1m2

r12

is a conserved quantity, where r12 denotes the distance between the two particles.

− Total angular momentum~L = m1~r1 × ~v1 +m2~r2 × ~v2

is a conserved quantity

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Note how the expression for total energy consists of three parts: one term for kinetic energy ofeach particle, Ekin, 1 and Ekin, 2, and one for potential energy

Epot = −G m1m2

r12. (2.88)

This expression for potential energy of a system of two point masses looks very similar to theexpression derived in the previous section for a single particle moving in a central gravitationalfield. The two situations are slightly different though. In the last section, we would deal with agravitational field and potential energy of the particle was defined as the amount of work done anexternal force when dragging the particle from infinity to its current position in the field. Here weare dealing with two point masses and using the concept of field as in the last section doesn’t maketoo much sense as either mass influences motion of the other one with its gravitational attraction.We can, however, define the potential energy of the system as the amount of work one would haveto perform to assemble it in its current spatial configuration. This assembling might take placein two steps as follows: take m1 from infinity and put it to ~r1 first, then take m2 from infinityand place it into ~r2. The first step does not require any work, as m1 is all alone in its way to ~r1

and no force acts on it. For the second step, we need to move m2 against gravitational attractionexerted by m1, and the work required is exactly the same as if we thought of m2 being moved ina gravitational field produced by m1 and the amount of work required is therefore just the termon the right-hand side of (2.88). Note that while m2 is put into ~r2, some force must prevent m1

from slipping off its position at ~r1 due to attraction by the approaching m2. This force does notcontribute any mechanical work, since m1 is not displaced under its influence.

Example 2.6.4.1: (USA 2012 ) Two masses m separated by a distance ` are given initial veloc-ities v0 as shown in the diagram. The masses interact only through universal gravitation.

a) Under what conditions will the masses eventually collide?

b) Under what conditions will the masses follow circular orbits of diameter `?

c) Under what conditions will the masses follow closed orbits?

d) What is the minimum distance achieved between the masses along their path?

Solution: For this kind of problem, it never hurts to consider conserved quantities and see howtheir properties influence the dynamics of the system. But before we get to any calculation,let’s just see how much we can retrieve just from the obvious spatial symmetry of the problem.Since the initial conditions of motion, that is both initial position and initial velocity, displaypoint symmetry around the center of their initial connection line and since both bodies areof the same mass, their entire motion will display the same kind of symmetry throughouttime. It makes sense then to define a coordinate system with origin at the center of the initialconnection line of the two masses and x-axis passing through them. if we use two subscripts 1and 2 for the two masses, then

~r1 = −~r2

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and therefore~v1 = −~v2

at any time, same for acceleration.

Now to the conserved quantities. Total angular momentum about the origin is conserved,therefore at any time equal to initial angular momentum

2 ·m`

2v0 = m`v0

The same is valid for total energy

2 · 1

2mv0

2 −G m ·m`

= mv02 −G m2

`.

a) Due to invariance of the position of the center of mass, collision of the particles mustoccur at the origin of the coordinate system. In the instant of collision, total angularmomentum must be equal to zero, since the position vectors of both particles vanish.This is only possible if initial angular momentum is also zero, that is, if v0 = 0 – that is,if the particles are released from rest.

b) In this case, gravitational attraction must act as a centripetal force for a circular orbitof radius `/2 and speed v0:

mv0

2

`2

= Gm ·m(`/2)2

⇔ 2v02` = Gm

c) With a similar argument as in the discussion before, the two particles will remain onclosed orbits if and only if the total energy of the system is negative:

mv02 − 2G

m2

`< 0

⇔ v02 < G

m

`

d) Let d be the minimum separation between the two masses. At their point of minimumseparation, their velocities must both be perpendicular to the position vectors: anyvelocity components parallel to the position vectors would either move the particles nearertowards a point of even closer distance or moving away from such one. This means thattotal angular momentum is then given by 2 ·mv d2 = mvd and conservation of angularmomentum reads mvd = mv0`. This can be used to eliminate v from conservation ofenergy, solving for d:

−Gm2

`+mv0

2 = −Gm2

d+mv2 = −Gm

2

d+mv2

0

(`

d

)2

.

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The resulting equation is quadratic in d and looks quite elaborate to solve. But, if youlook closely, you see that d1 = ` is a solution and you might try to factorize d− ` out ofthe equation to get the second solution

d2 =`

Gmv02`− 1

,

which only makes sense if the denominator is positive. From c), we see that this is alwaysthe case if the orbit is closed. Therefore, in the case of an open orbit the minimum distanceis given by the only solution d1 = `, meaning that such an orbit consists in the two massesmoving away from their starting point. Now, which one is the minimum distance? Thecondition d2 < d1 is equivalent to Gm

v02`> 2. This means:

Gm

v02`> 2 ⇒ minimum distance between the masses is

`Gmv02`− 1

,

Gm

v02`< 2 ⇒ minimum distance between the masses is `.

2.6.5 Kepler’s Laws of Planetary Motion*

Here as well, you most probably already know these laws from your high school physics courses.This section will give you short clarifications about the concepts involved, as they might provide adeeper understanding of the principles governing gravity and Newton’s laws of motion.

Kepler’s Laws of Planetary Motion

1. The orbit of a planet around the sun is a plane ellipse, the sun lies at one of its two foci.

2. Consider the radius vector pointing from the sun to the planet as it moves along its orbit.During equal amounts of time, the radius vector sweeps out segments of the ellipse of equalarea.

3. The third power of the semi-major axis of an elliptical planet orbit is proportional to thesquare of its orbital period around the sun.

The First Law

To understand the first law, we need to clarify some geometry. An ellipse can be defined in differentbut equivalent ways:

A: As the figure you get by stretching a circle by any factor with respect to a straight line.

B: As the locus of all points in the plane such that the sum of their distance to two given, fixedpoints, called the foci of the ellipse, is a constant.

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C: As the locus of all points in the plane whose distance from a given point, called a focus ofthe ellipse, is proportional to their distance from a given straight line, called the directrix,with a constant of proportionality < 1.

D: As the figure you get by intersecting a cone with a plane under such an angle that the figureis closed.

You might find it an interesting exercise in geometry to prove the equivalence of all these definitions.Anyway, for Olympiad problems, the following two parametrizations of an ellipse in the xy-planeare useful: (x

a

)2+(yb

)2= 1 (2.89)

in Cartesian coordinates and

r(ϑ) =p

1 + ε cosϑ(2.90)

in polar coordinates. You might verify that both equations describe the same curve in the planeunder certain conditions for a, b, p, ε.

Equation (2.89) is probably the most useful to understand the concept of major and minor axis.These are defined as the largest and smallest diameter, respectively, of the ellipse. It is easy toshow that their values are given by 2a and 2b, respectively, if the ellipse is described as in (2.89).Furthermore, they coincide with the x- and y-axis. Note how an ellipse is also supposed to havetwo foci, such that the sum of the distances of any point on the ellipse to the two foci is a constant.You might verify that the foci of the ellipse as introduced in definition B are situated at (−s, 0),(s, 0) with s =

√a2 − b2 if a > b or at (0,−s) and (0, s) with s =

√b2 − a2 if b > a. In particular,

this means the foci are always situated on the major axis. Conversely, by starting with definitionB with two foci lying on either the x- or y-axis, you might retrieve equation (2.89) for descriptionof the ellipse in the xy-plane.

In (2.90), the quantities p and ε are called semi-latus rectum and eccentricity of the ellipse, respec-tively. You can verify that the curve described by this equation satisfies the condition of definitionB with one focus at the origin and the other located at

(−2εp/

(1− ε2

), 0)on the x-axis. Defini-

tion C is satisfied for the focus at the origin with a directrix described by the equation x = p/ε,the constant of proportionality between the distance from the points to the origin (which corre-sponds to r) and the distance from the points to the directrix (which corresponds to the points’x-coordinates) is equal to ε. The ellipse intersects the y-axis at two points distant p from theorigin.

I bet you can’t wait to see how an equation like (2.89) or (2.90) can arise from the equationsof motion of a body subject to universal gravitation. The calculations involved are somewhatelaborate and may not be of great help for solving physics competition problems.

At this point, it is just important for you to understand what an ellipse is, what its foci are, andto keep in mind that closed planetary orbits are ellipses with the sun at one focus. In addition, it

115


sure wouldn’t hurt to feel comfortable around the different geometrical definitions of an ellipse, sofeel free to practice your skills in analytic geometry and play around.

The Second Law

Kepler’s second law is a direct consequence of conservation of angular momentum. Put the wholeorbit into a polar coordinate system with the sun at the origin. The position of the particle canbe written as

~r = (r cosϕ) x+ (r sinϕ) y,

at any time. The velocity vector is obtained by taking the time derivative

~v = ϕ ((r cosϕ− r sinϕ) x+ (r sinϕ+ r cosϕ) y) .

The angular momentum vector is then found to be

~L = m~r × ~v = r2ϕ z

if m is the mass of the planet.

During a small amount of time t, let the position vector of the planet sweep out a small area dA ofthe ellipse, thereby covering an angle dϕ = ω dt with ω = ϕ. The small area can be approximatedas a triangle, such that it amounts to 1

2 · r · rdϕ = ωr2dt/2. Due to conservation of angularmomentum, we know that ωr2 is a conserved quantity. This just means that the area dA sweptout during dt is a constant and that it does not depend on the position of the planet along itsorbit. We see that this is also true for any finite time interval of a given length ∆t between anytimes t1 and t2 = t1 + ∆t, as the area swept out by the position vector is just calculated as

ˆtime interval [t1, t2]

dA =

ˆ t2

t1

1

2ωr2(t) dt =

L

2m

ˆ t2

t1

dt =L

2m·∆t,

which does not depend on the exact position of the time interval as long as its length is ∆t – justthe statement of the second law.

The Third Law

It is quite easy to verify the third law for a circular orbit. Suppose the planet of mass m moves ona circular orbit of radius R around the sun of mass M with a time period T . Then gravitationalattraction of the sun on the planet must account for centripetal acceleration, and, by the secondlaw of motion:

m

(2π

T

)2

R = GmM

R2,

which is equivalent toR3

T 2=GM

4π2,

in other words, circular orbits satisfy Kepler’s third law. Calculations for general elliptic orbitsare somewhat more complex.

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Chapter 3

Thermodynamics

Thermodynamics is a phenomenological treatment of classical macroscopic systems and their prop-erties. The systems looked at in thermodynamics involve many degrees of freedom (∼ 1023). There-fore it is impossible to solve the equations of motion exactly.

In this chapter, we will first look at some important definitions, before looking at ideal gases andheat engines. In the end, we will also have a quick look at real gases and the Stefan-Boltzmannlaw.

3.1 Important definitions

Mole: base unit for the amount of substance. One mole of a substance contains exactly NA

particles. (See figure 3.1.)

Avogadro constant NA:the number of molecules in one mole. NA ≈ 6.022× 1023 mol−1.

Boltzmann constant kb:an important proportionality constant in statistical physics to relate energy andtemperature. kb ≈ 1.380× 10−23 JK−1

Gas constant R:proportionality constant to relate energy to a mole of particles at a stated tempera-ture. Molar version of kb: R = kb ·NA

Volume V : space in which the system is confined in.

Temperature T :measured quantity of a thermometer.

Pressure p:force applied per unit area.

Number of molecules N :number of molecules confined in the considered volume.

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Physics Olympiad Script CHAPTER 3. THERMODYNAMICS

Amount of substance n:molar version of N . (See figure 3.1.)

Work W : energy transferred from or to the system by expansion/compression.

Heat Q: transferred energy between systems due to temperature differences.

Internal energy U :also sometimes called just energy, all the energy inside the system.

Ideal gas: a collection of neutral atoms or molecules where the interaction between individualparticles can be neglected.

mass of one particlenumber of particles

amount of substance

Avogadro constant

mass

molar mass

molar volume

density

volume

Figure 3.1: Connection between mass, volume, number of particles and amount of substance. Thevalue for the molar volume is for one mole of an ideal gas at 0 C and 101.325 kPa. Adapted from[10].

3.2 The temperature scale

Since there is no easy way to define temperature, we just state that the temperature is a physicalquantity that tells us in which direction the heat flows. Heat always flows in the direction of thebody with the lower temperature. A way to measure this temperature is by using a thermometer,which uses thermal properties of materials to measure the temperature. Examples are the thermalexpansion of mercury, an electrical resistance that changes with temperature or the volume of agas at constant pressure.

The temperature is a scalar (a number) measure which is constant in an isolated system in thermo-dynamic equilibrium. There are 3 main scales for temperature (Kelvin, Celsius and Fahrenheit),but in physics, only Kelvin and, in rare cases, Celsius are used. In figure 3.2, a comparison betweenthe scales can be seen.

118

Physics Olympiad Script 3.3. ZEROTH LAW OF THERMODYNAMICS

The Kelvin scale starts at absolute zero, which is equal to −273.15 C. At 0 K all thermal motionfreezes. There is no temperature that is < 0 K. The Kelvin scale also uses the same temperaturestep as the Celsius scale and therefore we can easily switch between the two:

T [C] + 273.15 = T [K] (3.1)

Note that most equations in physics only work properly if you calculate everything in the Kelvintemperature scale and with no other.

Figure 3.2: Comparison of different temperature scales related to the energy scale via the Boltzmannconstant kb (e.g. for T=300 K ⇒ kbT = 4.1× 10−21 J ) with the corresponding energies in Zeptojoule(1 zJ = 1× 10−21 J).[11]

3.3 Zeroth law of thermodynamics

The zeroth law of thermodynamics states that if we bring two bodies (A & B) with differenttemperatures (TA & TB) into contact, the two bodies will reach a thermal equilibrium (Teq) aftera certain time. Since the temperature is a measure of thermal movement of atoms/molecules (seechapter 3.12), the equilibrium temperature Teq is between TA and TB:

TA < Teq < TB or TA > Teq > TB, (3.2)

depending on which body had the higher temperature in the beginning.

3.4 Thermal energy and heat capacity

When two bodies with different temperatures touch each other, the heat of one body flows tothe other until they have the same temperature. This thermal energy Q is often (especially inchemistry) measured in calories (cal):

1 cal = 4.1868 J, (3.3)

119


where one calorie is the heating energy one needs to heat up 1 g water at normal pressure (p = 1 bar)from 287.65 K to 288.65 K. Today the SI-unit Joule should be used, but in many textbooks onecan still find calories.

Different materials also take different amounts of energy ∆Q to increase the temperature by acertain extent ∆T . Therefore we define the heat capacity C which is a measure of how muchthermal energy the body needs to heat up.

C =∆Q

∆T⇔ ∆Q = C ·∆T (3.4)

Usually the heat capacity is normalized to the mass of the body (specific heat capacity Cm) or theamount of substance (molar heat capacity CM ), i.e.

Cm =1

m

∆Q

∆T(3.5)

and

CM =1

n

∆Q

∆T. (3.6)

3.4.1 Molar heat capacities of ideal gases

In general there are two different types of heat capacities (for gases). The heat capacity of asubstance depend on how one measures it: If you measure it while leaving the pressure constant(Cp) the specific heat is higher compared to when you leave the volume constant (CV ). This meansthat one needs different amounts of energy to heat the same amount of particles, depending onhow one heats them up. This is because additional work is done when expanding the system whilekeeping the pressure constant. This can be seen from figures 3.3 and 3.4. The difference betweenthe two molar heat capacities is exactly the gas constant

R = Cp − CV . (3.7)

For a derivation look at equation (3.15) and divide both sides by n∆T .

Figure 3.3: Heat capacity with constant volume.[11]

120

Physics Olympiad Script 3.5. IDEAL GAS LAW

Figure 3.4: Heat capacity with constant pressure.[11]

3.5 Ideal gas law

The most important equation in classical thermodynamics is the ideal gas law

pV = nRT, (3.8)

which can also be written as

pV = NkbT. (3.9)

It is a combination of the empirical Boyle’s law (p ∝ 1V ), Charle’s law (V ∝ T ) and Avogadro’s

law (V ∝ n) and is only valid for ideal gases.[12] Such an ideal gas can not be liquefied like a realgas. For real gases there is among others the Van-der-Waals gas equation to describe not only thebehaviour in the gaseous phase, but also the liquid phase as well as the phase transition betweenthem.

3.6 First law of thermodynamics

Temperature is connected to the kinetic energy of the particles in the gas (see chapter 3.12). Thisinternal energy U(p, V, T ) is a function of the parameters of the system: pressure p, volume V andtemperature T . These are state variables like described in 3.7. A change in internal energy onlydepends on the states at the beginning and the end of the process ∆U = UE − UA. This alsomeans that it doesn’t matter how one got from one state to the other.

The first law of thermodynamics states that the energy of a system does change if thermal energyQ is added or when mechanical work W is performed on the system:

dU = δQ + δW (3.10)

This means that for an isolated systems, in which we do not add thermal energy or do work on,the energy is constant. Note that here we introduced the convention that work done onto thesystem, or thermal energy added to the system is positive and the work the system does on it’ssurroundings is negative.

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3.7 Thermodynamic systems

A thermodynamic system is a collection of particles in thermal equilibrium and is characterized byso called state variables like volume V , temperature T , pressure p, entropy S, number of particlesN , amount of substance n and many more. In contrast heat Q and work W are not not statevariables, but process functions, since they do not describe an (equilibrium) state.

The different types of thermodynamic systems are characterized as follows.

Isolated systems

An isolated system has no exchange of matter or energy with its surroundings. It is completelyisolated and will stay in the thermodynamic equilibrium.

Closed systems

Closed systems have, like the isolated system, no exchange of matter with the surroundings. Butthey can exchange energy, for example with a thermal contact or by performing work on eachother.

Open systems

Open systems may exchange energy as well as matter with their surroundings.

3.8 Equipartition theorem

The equipartition theorem states that the mean energy of each molecule with f degrees of freedom(number of independent motions that are allowed, e.g. moving in x,y and z equals to 3 degrees offreedom) is given by

mean energy per molecule =f

2kbT. (3.11)

Thus the energy of an ideal gas with n moles of molecules is given by

U = nNAf

2kbT = n

f

2RT. (3.12)

Using ∆Q = nCV ∆T , we also find that CV = f2R and therefore Cp = f+2

2 R. Due to quantummechanical effects, degrees of freedom can be "frozen" out at low temperatures. This means theycannot be excited and therefore don’t contribute to the inner energy. This can be seen in figure3.5.

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Physics Olympiad Script 3.9. THERMODYNAMIC PROCESSES

Figure 3.5: Specific heat of H2 (schematically). The y-axis corresponds to f2 = CV

R . The vibrationgives two degrees of freedon, since one can store energy in potential or kinetic energy.[11]

3.9 Thermodynamic processes

In this chapter we look at different processes that change the system by doing/extracting workonto it or by heating/cooling it. These processes are mostly characterized by the variables theyleave constant. The different processes in a p-V diagram can be seen in 3.6.

Figure 3.6: Different processes in a p-V diagram. [13]

123


Isobaric processes

In an isobaric process the pressure is held constant. The work done on an expanding gas from theoutside is given by

W =

ˆ Vb

Va

δW = −ˆ Vb

Va

pdV, (3.13)

where Va is the volume at the beginning and Vb at the end of the process. By using that thepressure is constant over the whole expansion, we can take the pressure out of the integral and getthe total work

W = −pˆ Vb

Va

dV = −p(Vb − Va) = −nR(Tb − Ta). (3.14)

By using the equipartition theorem, we can also find the change in internal energy ∆U = nCV ∆Tand therefore, according to the first law of thermodynamics and the ideal gas law, the heat ex-changed is

Q = ∆U −W = nCV ∆T + nR∆T = nCp∆T. (3.15)

Isothermal processes

For isothermal processes the temperature is held constant (T = const.) and therefore, accordingto the equipartition theorem, the internal energy is constant (U = U(T ) = const). Using theideal gas law we also find that the product pV = const as well. Since we know that the internalenergy remains constant, the first law of thermodynamics tells us that the incoming heat is entirelyconverted to work:

Q =

ˆδQ =

ˆ−δW = −W = W (3.16)

Using the ideal gas law we can then calculate the work done by the system as

W =

ˆ Vb

Va

pdV = nRT

ˆ Vb

Va

dV

V= nRT ln

(VbVa

)(3.17)

Isochoric processes

In isochoric processes the volume doesn’t change (V = const.). As a consequence no work is beingdone and therefore the first law of thermodynamics tells us

dU = δQ. (3.18)

Therefore the change in energy is simply given by

∆Q =

ˆ Tb

Ta

nCV dT = nCV ∆T. (3.19)

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Physics Olympiad Script 3.10. SECOND LAW OF THERMODYNAMICS

Adiabatic processes

If during a process the system does not exchange heat with the surroundings, one speaks of adi-abatic processes. This is for example the case when the system is thermally isolated or when theprocess is so fast that the heat exchange with the surroundings is negligible.

We can write down the change in work and energy for the process

δW = −pdV = −nRTV

dV (3.20)

dU = nCV dT. (3.21)

Since we know that there is no heat exchanged ∆Q = 0, we find that for any adiabatic processnCV dT + nRT

V dV = 0. By dividing through nT and integrating we getˆCVTdT = −

ˆR

VdV (3.22)

CV ln(T ) = −R ln(V ) + const. (3.23)

⇒ T = V− RCV + const., (3.24)

which is exactly

TV κ−1 = const. (3.25)

or equivalently

pV κ = const. (3.26)

T κp1−κ = const′. (3.27)

Where κ is the ratio κ =CpCV

which implies RCV

= κ− 1 (using equation (3.7)).

3.10 Second law of thermodynamics

The first law of thermodynamics was about the exchange of energy with the surroundings. Thesecond law of thermodynamics is about the distribution of the molecules within the volume of thesystem. The entropy of a system is a measure for the disorder in the system. There are manyformulations, but the most known one is

S = kb lnW, (3.28)

where S is the entropy and W the probability for the system to be in a given state. (To be moreprecise it is the probability to be in a given macrostate defined by state variables. For a betterexplanation you can have a look at [14].)

The second law of thermodynamics states that the entropy of a system can only increase

dS ≥ δQ

T. (3.29)

There are also other formulations of the second law, like the one from Rudolf Clausius:

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"Heat can never pass from a colder to a warmer body without some other change, connectedtherewith, occurring at the same time." [15]

or Lord Kelvin:

"It is impossible, by means of inanimate material agency, to derive mechanical effect fromany portion of matter by cooling it below the temperature of the coldest of the surroundingobjects." [15]

The second law also splits processes up into those which are reversible and conserve the entropy,and those which are irreversible and do not conserve entropy. For reversible processes, the systemcan be returned to its initial state. For reversible processes we therefore have

dS =δQ

T. (3.30)

There is also the third law of thermodynamics which fixes the entropy at absolute zero S(0 K) = 0.For more information have a look at [14].

3.11 Heat engines

Heat engines do work by transferring heat between two reservoirs at different temperatures. Onecan imagine it a bit like a water mill, where the water from the higher level produces mechanicalwork by running to the lower level.

Heat engines can be characterized by the associated cycle, which is the closed curve on a p-Vdiagram. There are many different heat machines, like conventional car engines, but we will havea look at the theoretically most efficient process, the Carnot cycle. The Carnot process, which isdepicted in figure 3.7, features two isothermal and two adiabatic processes.

To see how efficient the Carnot cycle is we have to write down the equation for every step, shownin table 3.1.

Table 3.1: The Carnot process.

step process T W Q

1: a→b adiabaticcompression

T2 → T1 δW1 = nCV (T1 − T2) 0

2: b→c isothermalexpansion

T1 δW2 = nRT1 ln VcVb

δQ2 = nRT1 ln VcVb

3: c→d adiabatic ex-pansion

T1 → T2 δW3 = nCV (T1 − T2) 0

4: d→a isothermalcompression

T2 δW4 = nRT2 ln VdVa

δQ4 = nRT2 ln VdVa

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Physics Olympiad Script 3.11. HEAT ENGINES

Figure 3.7: The Carnot cycle with its two isotherms and two adiabats. The surface that is enclosedby the Carnot process equals the total work done. [14]

Furthermore we know that TV κ−1 = const. during adiabatic processes

T2Vκ−1a = T1V

κ−1b (3.31)

T2Vκ−1d = T1V

κ−1c (3.32)

and if we divide the two equations, we get the condition

Va · Vc = Vb · Vd. (3.33)

With all of that we can define the Carnot efficiency

ηC =work done

supplied heat=−δW1 + δW2 + δW3 − δW

4

δQ2=T1 − T2

T1. (3.34)

The Carnot cycle defines a thermodynamic cyclic process to produce work with the highest achiev-able efficiency. All reversible heat engines between two heat reservoirs are equally efficient asa Carnot engine operating between the same reservoirs.[15] Therefore all real processes are lessefficient

work donesupplied heat

= ηreal ≤ ηC =Thot − Tcold

Thot. (3.35)

It is also nice to see that the entropy for a fully reversible cycle does not increase:

∆S =

˛dS =

˛δQ

T= 0 (3.36)

This is known as the Clausius equality for reversible processes.[16]

127


3.12 Kinetic gas theory

In this chapter we look at how microscopic motion explains the macroscopic properties of a system.Solving the problem exactly is impossible due to the huge number of particles within a macroscopicvolume, but with some approximations we can model the real system very accurately. The mostimportant approximations are:

− The gas consists of very many, small particles and we approximate them as points, so we canneglect the volume they take.

− All particles are identical.

− The particles move with a constant, random velocity.

A full list of the approximations can be found on Wikipedia[17].

Figure 3.8: Molecules move around in a volume with random direction and speed, indicated byarrows.[18]

We look at a cube of side length L in which the particles are confined, like the one in figure 3.8.When a particle hits the wall, the change in momentum in the direction (e.g. x) is given by

∆P = Pbefore,x − Pafter,x = Pbefore,x − (−Pbefore,x) = 2Pbefore,x = 2mvx, (3.37)

where P is the momentum. Further the particle hits any given wall periodically in time with

∆t =2L

vx(3.38)

between collisions, which gives an average force onto the wall of one particle of

Fparticle =∆P

∆t=mv2

x

L. (3.39)

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Physics Olympiad Script 3.13. PHASE TRANSITIONS

This results in a total force onto the wall of

F = N · Fparticle = Nmv2

x

L, (3.40)

where the bar over the force and the velocity indicates mean values over all particles. Since we donot have a bias in any direction, the average squared speed in any direction should be the same

v2 = v2x + v2

y + v2z = 3v2

x. (3.41)

This leads to a pressure of

p =F

L2=Nmv2

3V, (3.42)

with a volume V = L3.

Comparing with the ideal gas law

pV =Nmv2

3= NkbT, (3.43)

and knowing that kb, m and N are constant, we see that the velocity squared is directly connectedto the temperature

kbT =mv2

3. (3.44)

By using this we can write down the average kinetic energy per particle

1

2mv2 =

3

2kbT, (3.45)

which we recognize as the equipartition theorem with particles having 3 degrees of freedom (theyare able to move in x,y and z).

3.13 Phase transitions

Ideal gases do not show any phase transitions (e.g. condensation), but real gases like described inchapter 3.14 do. For every point in the p-T diagram, there is exactly one phase that minimizes theenergy. The system always tries to minimize this energy and therefore phase transitions happenwhen the system goes from one region to another. During phase transitions the two phases coexist.

When a phase transition happens usually depends on the pressure as well as on the temperature,as can be seen in figure 3.9. In the figure one can see the phase diagram of water, with the freezingpoint at 0 C and the boiling point at 100 C for normal atmospheric pressure. But with changingpressure, the necessary temperature to boil/freeze water gets shifted and one can even get waterthat is under 0 C cold.

129


Figure 3.9: Phase diagram of H2O. On the very left is the solid phase, in the middle the liquid phaseand on the right, extending all the way to the left at low pressure, is the the gaseous phase. [19]

When a system transitions from one phase to the other, heat is either released to the surrounding(e.g. condensation) or taken from the surroundings (e.g. evaporation). This heat is called thelatent heat L and is usually found as specific latent heat Lm which is normalized by the mass.Therefore one can calculate the heat Q needed to evaporate a material only by knowing its massand the specific latent heat

Q = mLm. (3.46)

This is of course only valid if the system is already at the right temperature.

3.14 Real gases

In this script we only looked at ideal gases, but just as an outlook we shall briefly discuss real gases.

As a first approximation we can introduce two empirical parameters to the ideal gas law, bringingus to the van der Waals gas: (

p+an2

V 2

)(V − nb) = nRT. (3.47)

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Physics Olympiad Script 3.15. STEFAN-BOLTZMANN LAW

The two terms are:

1) There is not only the outer pressure, but the interaction of the molecules is also taken intoaccount by the factor a.

2) The volume that the gas can move in is reduced by the volume the molecules occupy. Thisis taken care of by the factor b.

The probably biggest advantage of using the van der Waals equation is that with this model onecan also take phase transitions into account, which is not included in the ideal gas law.

3.15 Stefan-Boltzmann law

Every body radiates power in the form of electro-magnetic waves because of its temperature. Forexample molten metals send out light in the visible range. The power radiated per area by a blackbody (a body that absorbs all incoming light) is given by the Stefan-Boltzmann law

P = σT 4. (3.48)

σ is the Stefan-Boltzmann constant and is given by

σ =2π5k4

b

15c2h3≈ 5.670× 10−8 Wm−2K−4, (3.49)

where c is the speed of light and h is the Planck constant.

131


132

Chapter 4

Oscillations

4.1 Introduction

Important sources for this chapter include [20] and [21].

The term oscillation generally refers to the (periodic) variation of some physical quantity withtime around a point of equilibrium.

There is no strong consensus among physicists for a proper definition of oscillations. Some preferusing broader definitions, to the point of including phenomena that do not "oscillate" in thecommon sense, while some others prefer a narrower definition, with the risk of excluding movementsshowing the characteristic "back-and-forth" variation. Damping is the common example of aphenomenon of the "grey zone", more on this later.

A way of physically defining oscillatory systems (oscillators) is to require that the varying quantityhas an equilibrium value, and that the system naturally tends to go back to that value if taken outof equilibrium by an external perturbation.

From that definition, we already see that oscillators are very common in physics, virtually in anydomains; oscillations, in particular, happen each time the considered quantity (which can be almostany mathematical quantity: a position, the strength of a field, the voltage of an alternating currentcircuit, etc.) is bound to a potential with a local minimum.

In the following, we will take a more precise look at simple oscillations, notably harmonic oscilla-tions and their extensions - damping and resonance.

As a preamble, let us first define periodicity. This term refers to the idea of having some time-dependent quantity that takes the same values over and over, in a regular pattern. For example,considering the time-dependent variable x(t), we call it periodic if we can find some T such that

x(t) = x(t+ T )

for all times t.

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Physics Olympiad Script CHAPTER 4. OSCILLATIONS

As can be seen, if we find such a T , then 2T also satisfies the criterion, as well as any multiple ofT . Thus, we define the period of a periodic system as the (positive) smallest such T .

It is clear from this definition that the period T has the units of an interval of time, thus is givenin s (seconds) in the SI, and measures the time it takes an oscillator to accomplish an oscillationcycle.

We also define the frequency f (also commonly refered to as ν): it is the inverse of the period, i.e.1/T , and it measures how many oscillation cycles an oscillator accomplishes in a time unit. Its SIunit is s−1, also called Hz (Hertz).

4.2 Harmonic Oscillations

Harmonic oscillations are considered the most simple type of oscillation. Despite this (or moreprobably, because of this), it also provides a convenient model for most types of (more complicated)physical oscillations. As is often the case, mathematical and physical simplicity come together.

Mathematically, a quantity x(t) following a harmonic oscillation has the following form:

x(t) = A sin(ωt+ ϕ) , (4.1)

that is, it simply follows a sine function, up to some scaling and shifting factors:

− A is called amplitude, and has the same units as x. It corresponds to the maximal value ofx.

− ω is called angular frequency (sometimes also pulsation) and corresponds to 2πf . Its unitsare that of an angle over a time, i.e. radians per second in the SI. Mathematically, it is ascaling factor that allows transforming the parameter t into an object of the right dimensionfor the sine function to absorb, i.e. ωt should have units of an angle.

− ϕ is called phase and has units of an angle (for the same reason as just explained for ωt),usually given in radians.

Note:

− some textbooks refer to 2A as the amplitude. Be careful!

− another definition can be created using a cosine instead of a sine. Those definitions areequivalent and only differ by a shift of π/2 in the phase.

It is easy to check that harmonic oscillations are periodic, it simply follows from the mathematicalperiodicity of the sine function:

134

Physics Olympiad Script 4.2. HARMONIC OSCILLATIONS

x(t+ T ) = A sin (ω(t+ T ) + ϕ)

= A sin (ωt+ ωT + ϕ)

= A sin (ωt+ 2πfT + ϕ)

= A sin (ωt+ 2πT/T + ϕ)

= A sin (ωt+ ϕ+ 2π)

= A sin (ωt+ ϕ)

= x(t) .

4.2.1 Harmonic Oscillations, Spring/Mass Systems and Differential Equations

At this point, we want to look at a first concrete example of harmonic oscillator and try to relate thephysical equations of such a system with the mathematical form of harmonic oscillations describedabove.

The simplest example of harmonic oscillator is perhaps the spring/mass system, where an extremityof the (massless) spring is fixed and the other is attached to a free mass m. We will only considerhere the 1D case, where the mass can only move along one dimension. We do not take anygravitational force into account.

We will later see that by accounting for further interactions (friction, driving by an external move-ment, etc.) we will be able to generalize this simple example and therefore study various interestingoscillatory cases.

km

Figure 4.1: Spring-mass system. Adapted from [22].

135


Newton’s equation for the mass is as follows:

∑i

Fi = ma .

The only force acting on m is the restoring force of the spring (Hooke’s law):

F = −kx ,

where k is the spring constant (defining its stiffness) in N·m−1 (equivalent to kg·s−2). The originof x is taken at the point of equilibrium of the spring.

Thus we have, by reordering and using that the acceleration is the second time derivative of theposition (a = x):

x+k

mx = 0 . (4.2)

Such an equation is called a homogeneous second-order linear constant coefficient ordinary differ-ential equation:

− differential, because it contains both a function (x(t)) and some of its derivatives (x(t)), andthe goal is to determine x(t);

− ordinary, because the function x depends on only one variable t;

− constant coefficient, because the function and its derivatives only show up with constantcoefficients in the equation;

− linear, because the function and its derivatives only show up with degree 1 in the equation(no square or such);

− second-order, because the highest derivative of x present is the second one;

− homogeneous, because there is no "free" constant or function of t in the equation (all termscontain an x or one of its derivatives at least).

Due to Newton’s law, most of mechanics revolve around finding solutions to second-order differen-tial equations.

Our hypothesis above was that spring/mass systems are harmonic oscillators, i.e. we believe thatsuch systems perform harmonic oscillations (with the variable quantity being the position of themass).

To verify this, we have to check whether 4.1 and 4.2 are compatible, i.e. whether 4.1 is a solutionof 4.2.

136


Let’s insert 4.1 into 4.2:

x(t) = A sin (ωt+ ϕ)

x(t) = ωA cos (ωt+ ϕ)

x(t) = −ω2A sin (ωt+ ϕ)

x+k

mx = −ω2A sin (ωt+ ϕ) +

k

mA sin (ωt+ ϕ)

=

(−ω2 +

k

m

)A sin (ωt+ ϕ)

!= 0 .

For the last equality to hold for all t, we need to have

−ω2 +k

m= 0 ,

i.e.

ω =

√k

m.

Thus, a spring/mass system with the above defined characteristics does behave as a harmonicoscillator with angular frequency

√k/m.

We can also conclude that any system with a single force −Kx for some constant K acting on itis harmonic. Such a system has a potential V (x) = 1

2Kx2, which also suffices to characterize it.

4.2.2 Further Examples

Beyond the spring/mass, other simple systems represent harmonic oscillators:

− Torsion Pendulum

Hooke’s law also holds for the angular position ϕ of an object of moment of inertia J attachedto a torsion spring of angular spring constant κ (with the torque M):

M = −κϕ ,

thus with Newton’s law for rotations:

137


φ

κ

J

Figure 4.2: Torsion pendulum. Adapted from [23].

∑i

Mi = Jα

ϕ+κ

Jϕ = 0 .

And we again find a solution of the form ϕ(t) = Φ sin(√

κ/Jt+ θ).

− Simple Pendulum

A simple pendulum is composed of a (massless) rod of length l fixed at one of its extremitiesto a pivot, and attached at the other end to a moving mass.

In the common case, one restricts oneself to the 2D. Using the angle ϕ of the rod with thevertical, one can then parametrize the system with one single variable - thus the system hasone single degree of freedom.

Two forces act on the mass, the gravity ~FP and the tension ~FT .

Seen from the pivot point, the tension produces no torque (because of its collinearity withthe rod, MT = 0), while the weight of the mass creates

MP = −mgl sin(ϕ)

(the minus sign indicating that the torque opposes itself to the deviation of the rod from thevertical).

138


φ m

lg

Figure 4.3: Simple pendulum.

Thus we get, using Newton’s law in rotationary form (we assume the mass to be point-like,so J = ml2 by Steiner’s theorem):

ϕ+g

lsin(ϕ) = 0 . (4.3)

As can be seen, we again find a homogeneous second-order constant coefficient ordinarydifferential equation, but this time it is not linear, due to the sine!

Therefore the simple pendulum is not harmonic; as can be shown, 4.1 is not a solution of4.3.

However, in the case of small oscillations, the angle ϕ remains itself small and we can linearizethe equation. This is because of the Taylor expansion of sine:

sin(x) = x− x3

6+

x5

120+O(x7) ,

which allows to approximate sin(x) ≈ x for small x.

Thus, for ϕ 1 (with ϕ in radians), we have

ϕ+g

lϕ ≈ 0

and we can say that the pendulum is approximatively harmonic with angular frequency√g/l.

Note that the condition ϕ 1 is necessary for the approximation to hold!

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4.2.3 Importance of Harmonic Oscillations

The fact that non-harmonic oscillations can be approximated in the vicinity of equilibrium pointsby harmonic oscillations is general, and is the reason why harmonic oscillations are so commonand important.

More concretely, if we have a system with potential V (x) we can Taylor-expand (say around 0):

V (x) = V (0) + V ′(0)x+V ′′(0)

2x2 +O(x3) .

If the potential has a local minimum at 0, we have V ′(0) = 0 and V ′′(0) > 0, thus for small x wecan neglect the higher orders and get

V (x) ≈ V (0) +V ′′(0)

2x2 ,

which is the potential of a harmonic oscillator (V (0) is arbitrary and we can redefine V in orderto get rid of it). In terms of force we get

F = −∂V∂x

= −V ′′(0)x

which is obvious linear in x (V ′′(0) is the value of the second derivative at 0 and therefore simplya number).

4.3 Beyond Harmonic Oscillations

Here we will consider our mass-spring system, but accounting for further phenomena. In fact,harmonic oscillations represent an idealized case. In reality, we often have two more ingredients inoscillatory systems:

− Damping

Damping occurs everytime we have friction in the system, opposing the motion.

− Forcing

Forcing (also called driving force) appears when an external element interacts with the os-cillating system. Depending on this interaction, this can typically greatly increase or reducethe oscillation amplitude.

In this script we will consider the following simple situation only: linear damping (with a forceproportional to the velocity, −bv) and sinusoidal forcing (with an external excitation B sin(Ωt)).

140

Physics Olympiad Script 4.3. BEYOND HARMONIC OSCILLATIONS

Newton’s law gives for such a system:

−kx− bv +B sin(Ωt) = ma ,

which leads to the (in general inhomogeneous due to the forcing) differential equation:

x+ 2ζωx+ ω2x =B

msin(Ωt) ,

where ζ = b2√mk

is called the damping ratio.

The full mathematical treatement of this differential equation is beyond the scope of this script (itis partly covered in the analogous case of oscillatory circuits in the AC part), so we only providethe solutions without resolution steps below. We distinguish the following cases:

− B = 0

This is the free case, i.e. there is no forcing. The behavior of the system depends on thedamping ratio:

– ζ = 0

There is no damping here, thus this is the common harmonic oscillator, as discussedabove.

As soon as ζ gets bigger than 0, the oscillator looses energy in the damping (friction)and is not periodic anymore, but rather tends to go back to its equilibrium point andstop there.

– 0 ≤ ζ < 1

This case is called underdamped and corresponds to solutions of the form

x(t) = Ce−ζωt sin(ωt+ ϕ) ,

with C and ϕ arbitrary constants depending on the initial conditions. As can be seen,the movement represents a sinusoidal modulated (squashed) by an exponential decay.It is not periodic, but one can define ω as a so-called pseudoperiod for the oscillation.We have

ω = ω√

1− ζ2 .

– ζ = 1

141


This one is the critically damped case, with solutions

x(t) = (C1t+ C2)e−ζωt ,

where C1 and C2 are again arbitrary constants depending on the initial conditions.

The system here approaches its equilibrium point in the fastest possible way, whichmakes it technically useful in real-world applications, such as springs for automaticallyclosing doors (that should close fast but without oscillating) or shock absorbers onvehicles.

– ζ > 1

Here we have the overdamped case, with the system going back to its point of equilibriumwithout oscillating, but slower that in the critical case described above.

The solutions have the form

x(t) = C1e−(ζ−√ζ2−1

)ωt

+ C2e−(ζ+√ζ2−1

)ωt ,

with C1 and C2 defined in an equivalent way as above.

Figure 4.4: Linear damping with various ζ values. Adapted from [27].

− B 6= 0

In the presence of forcing, we will have to separate to cases:

– We will first have a transitory solution depending on initial conditions.

– Then, we will have a steady-state solution depending only on the forcing.

142

Physics Olympiad Script 4.3. BEYOND HARMONIC OSCILLATIONS

The general solution is a sum of both transitory and steady-state solutions; here we willconsider the steady-state solution only. It has the form

x(t) =B

m√

4ω2Ω2ζ2 + (ω2 − Ω2)2sin(Ωt+ Φ)

with some phase Φ that we won’t discuss in this script. The interesting part is the squareroot Z =

√4ω2Ω2ζ2 + (ω2 − Ω2)2.

For ζ < 1√2, we will have a minimum of Z for a forcing frequency Ωr = ω

√1− 2ζ2, thus

the amplitude of the oscillation for that case of forcing will be strongly increased. In theundamped case (ζ = 0), the amplitude in that case will even diverge to infinity, meaningthat for that forcing frequency Ωr (called resonance frequency) the forcing will continuouslyadd energy to the system.

Ω/ω

x(Ω)x(0)

Figure 4.5: Amplitude resonance with sinusoidal forcing. The maxima line corresponds to Ωr.Adapted from [26].

4.3.1 Example of Forced Oscillating Systems

One of the most common examples of forcing is the swing: one has to furnish energy at the righttime (i.e. with the right frequency) for the swing to increase its oscillation amplitude.

Conversely, any sufficently little damped mechanical system has a resonance frequency that can beused to bring it to strongly oscillating, even with a relatively small amplitude driving interaction.

143


During the engineering of any mechanical product, resonance has to be taken into account in orderto prevent unwanted oscillations that could even make it break. Among others, vehicles partsshould not be brought to resonance by the vibrations induced by the engine.

Buildings are another type of mechanical systems where resonance has to be well controlled. Mod-ern bridges and skyscrapers often contain mechanical parts designed to absorb oscillations.

One of the most well-known examples of so-called resonance disaster is the destruction of theTacoma Narrows Bridge in 1940 due to wind passing through its structure, bringing it to twistingand eventually leading to its collapse.

Figure 4.6: The Tacoma Narrows Bridge collapsing [24]. See also [25].

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Chapter 5

Waves

5.1 Introduction

In the previous chapter, we spoke about oscillations. Now we can ask ourselves the question "whathappens if we couple many oscillators together into a network, so that oscillations on one of themalso influence its neighbors?"

What happens is that the perturbation will little by little, neighbor after neighbor, "travel" acrossthe oscillators network. Such a propagating perturbation is called progressive wave.

Most waves are progressive waves, but there are some counterexamples such as standing waves.

An exact definition of waves is difficult to construct, but one can anyways say that waves areperturbations (variations of some quantity, typically in an oscillating way) in space and time. Thisis the main difference with oscillations, which occur in more or less abstract quantities with time:waves add the idea of space.

Mathematically, waves are often represented as functions of time and space u(~r, t) (u(x, t) in the1D case). Sometimes the oscillations composing the wave also have a determined direction in space,thus the wave in this case is represented by a vector function ~u(~r, t).

We will not go into much details in this script, but the temporal evolution of those functions isfixed by the wave equation, which in 1D has the form

∂2u(t, x)

∂t2= v2∂

2u(t, x)

∂x2. (5.1)

This is a partial differential equation (as opposed to an ordinary DE), meaning that it containspartial derivatives (u is function of multiple variables and gets differentiated - such as in ∂u

∂t - withrespect to one single of those variables at a time, by letting other variables constant). This is adirect consequence of our letting space play a role in the description of waves, in opposition to theordinary differential equations we saw for oscillations, where only time plays a role.

The v in the wave equation can already let us think about the velocity of a wave, which is ingeneral an important property. Sometimes c is used instead of v.

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Physics Olympiad Script CHAPTER 5. WAVES

Waves can be seen as generalizations of oscillations, so in the following we will begin by takinga look at general properties of waves - which may recall us about oscillations. Then we will talkabout their propagation and the media associated. Finally the interference phenomenon (two ormore waves interacting with each other) will be discussed.

5.2 Harmonic Waves

We will provide a more in-depth description of waves using the example of harmonic waves. Thoseare waves that can be described using a sinusoidal function:

u(x, t) = A sin(ωt− kx+ ϕ) . (5.2)

As can be seen, this mathematical form is quite similar to the one we used for harmonic oscillations,with the difference of the argument of the sine gaining a spatial term. Sometimes, the reversedconvention is used, u(x, t) = A sin(kx− ωt+ ϕ′). Both conventions are equivalent, up to a changein the phase from ϕ to ϕ′. Here we use a positive time and a negative space, as it more directlyrelates to the case of oscillations.

As was the case before, u as a function of t (provided that x is held constant) has a sinusoidalform. This means that, at any given point, the "quantity the wave moves in" oscillates with time.But it is also to be noted that the spatial coordinate x has an equivalent place in the formula asthe temporal coordinate t. Thus the other way round also holds: at any given time, the "quantitythe wave moves in" oscillates with space. It is therefore very important, when looking at a plotdescribing the amplitude of a harmonic wave, to check whether it drawn against time or space. Inboth cases the figure will appear sinusoidal.

When talking about oscillations, we defined ω as the angular frequency, a measure of the variation"rapidity" of the quantity with time. In the formula above, x has a similar role as t and it also hassome factor k associated, called wavenumber. And again similarly to the time, where we definedthe period T = 2π/ω as the time interval between (say) two peaks of the oscillation (or wave, at agiven point), we can define the wavelength λ = 2π/k, which is nothing but the space interval (i.e.the distance) between (say) two peaks of the wave (at a given, fixed time).

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Physics Olympiad Script 5.2. HARMONIC WAVES

λ

−A

A

u(x, 0)

u(x,∆t)

x

u

Space

T

−A

A

u(0, t)

u(∆x, t)

t

u

Time

Figure 5.1: A plane wave u(x, t) (with ϕ = 0) represented along each axis, with the other set first to0, then to some small value (dashed plot). Both plots are very similar, up to the choice of the origin.

From this, we can derive a very important equation by inserting 5.2 into 5.1:

∂u

∂t= ωA cos(ωt− kx+ ϕ)

∂2u

∂t2= −ω2A sin(ωt− kx+ ϕ)

∂u

∂x= −kA cos(ωt− kx+ ϕ)

∂2u

∂x2= −k2A sin(ωt− kx+ ϕ)

−ω2A sin(ωt− kx+ ϕ) =∂2u

∂t2

!= v2∂

2u

∂x2

= v2(−k2)A sin(ωt− kx+ ϕ)

⇒ ω2 = v2k2

⇒ v = ω/k = λ/T = λf ,

where we assume v, ω and k to be positive (which is mostly a matter of convention).

This relates the temporal and spatial characteristics of waves through a parameter, the velocity of

147


the wave (which is not very surprising, considering that velocities express in fact the division of aspace interval by a time interval).

5.3 Waves in 3D

5.3.1 Waves as Functions of 3D Spatial Coordinates

When going from waves in one dimension to three, the first element to look at is the fact that nowthe function u(x, t) is modified into u(~r, t).

In order to adapt our formula for harmonic waves, we define ~k, the wavevector and write:

u(~r, t) = A sin(ωt− ~k · ~r + ϕ) . (5.3)

While ~k still indicates the "rapidity" of variation of the wave with space (through its norm k = |~k|),it now also has a direction, which actually is the direction of propagation of the wave itself.

The formula 5.3 creates so-called planar waves: at a given time, all points in planes perpendicularto the wavevector share the same oscillation state.

5.3.2 Waves as 3D Functions, Transversal and Longitudinal Waves, Polariza-tion

Depending on the "quantity the wave moves in", not only the oscillations defining the wave candepend on all the three spatial coordinates, but the oscillations themselves can have a well-defineddirection in space, which is mathematically indicated by a vectorial function ~u(~r, t). In the har-monic case, in general we simply modify 5.3 by using a vectorial amplitude ~A to take this oscillationdirection into account.

As we saw before, waves in 3D have a particular direction determined by ~k, the propagationdirection. For waves whose oscillations themselves have a well-defined direction, we have twopossible cases:

− The oscillation direction can be perpendicular to the propagation direction, in which casethe wave is named transversal.

− Or it can be parallel, and we have a longitudinal wave.

In the first case, even for a fixed propagation direction the oscillation direction is not unique,but can be freely chosen in the plane perpendicular ~k. We thus define the polarization, which isthe oscillation direction. Two transversal waves going in the same direction can nevertheless havedifferent polarization states.

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Physics Olympiad Script 5.4. WAVES PROPAGATION

~k

~k

Figure 5.2: Wave propagation in 3D. Top: transversal, vertically polarized wave, oscillations happenin the vertical direction, thus perpendicularly to the direction of the wave (given by ~k). Bottom:longitudinal wave, oscillations are parallel to ~k.

5.4 Waves Propagation

In the introduction to the present chapter, we said that waves are perturbations travelling througha "network" of coupled oscillators. Later on, we used the expression "quantity the wave moves in"in several sentences. It is now time to try answering the question "what does a wave move in?",i.e. find what is the propagation medium of waves.

This in turn depends on the type of wave, and in particular on the nature of the oscillators.

In the following, we will introduce a couple of examples of wave forms as concrete cases for thedifferent properties that were discussed in the previous sections. This will allow us to also get aninsight into the different propagation media.

5.4.1 Sound

Sound is made of so-called mechanical waves, i.e. waves whose oscillators are positions of matterparticles - be it in a solid, liquid or gas. In the case of sound, those waves are longitudinal, somatter particles get pushed away in the direction of propagation by previous ones, where they hitnext ones, thus propagating the wave further and letting them bounce back (in average) to theiroriginal position.

Like any mechanical waves, sound cannot propagate in the absence of physical medium (like in

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vacuum). Even if an enormous asteroid closely passes by your spacecraft in outer space, you haveno chance of hearing anything! (Unless the ship actually gets hit by some part of the asteroid...)

The frequency of sound is directly related with the perceived height - a higher frequency leadingto a higher pitch. It also leads to a shorter wavelength, as the sound travels at approximately thesame velocity in a given material (round 343 m·s−1 in air) under given conditions, independentlyof its frequency.

5.4.2 Light

While light also composes itself of waves - and while we also can sense it (in some cases), one couldnot imagine anything more different from sound than light!

The first fundamental difference between them is that light is not a material wave, i.e. it is notmade of matter particles brought to oscillating. Moreover, it does not need any matter to propagate(the question on the nature of a hypothetic physical medium - "luminiferous aether" - for lightpropagation actually participated in the development of relativity).

While light does not need any material to propagate, the presence of a material can have an impacton its propagation, such as changing its velocity, direction, etc. We will talk about these generaleffects later on.

We said that waves travel on "networks of oscillators" so, even for light there has to be somethingsupporting it. This "something" is the electromagnetic field, and thus light is actually an electro-magnetic wave. Concretely, this means that light beams are perturbations across space that bringelectric and magnetic fields to locally oscillate.

In most cases (and most importantly in vacuum), light is a transversal wave. Thus, in any pointon the light path, both ~E and ~B are perpendicular to ~k ( ~E and ~B are actually also perpendicularto each other).

As a transversal wave, light is subject to polarization, given by the direction of the electric field. Theexistence of polarization filters, which can block light depending of its polarization orientation, iscrucial for many technologies in both research and engineering; but it is also at the origin of the mostcommon 3D cinema system, where two images are projected on the screen simultaneously, withperpendicular polarizations. Eyeglasses similarly have two lenses with corresponding polarizationfilters, so that each lens exactly lets one of the images pass through while blocking the other one.

Wavelength and frequency of electromagnetic waves cover a huge range of magnitude orders, whichalso leads to very different types of interaction with matter.

5.4.3 Seismic Waves

Like sound, seismic waves - waves created by geological events in the inner of the Earth, such asearthquakes - are material waves. Unlike them, though, they are not necessarily longitudinal. Infact, single events often create a bunch of seismic waves of different types - both longitudinal andtransversal, but also waves travelling through the Earth or across its surface, etc. All those differenttypes come in general together with different wavelengths, frequencies and velocities, which also

150

Physics Olympiad Script 5.4. WAVES PROPAGATION

Buildings Humans Butterflies Needle Point Protozoans Molecules Atoms Atomic Nuclei

104 108 1012 1015 1016 1018 1020

1 K 100 K 10,000 K 10,000,000 K

Penetrates Earth'sAtmosphere?

Radio Microwave Infrared Visible Ultraviolet X-ray Gamma ray

103 10−2 10−5 0.5×10 −6 10−8 10−10 10−12Radiation Type

Wavelength (m)

Approximate Scaleof Wavelength

Frequency (Hz)

Temperature ofobjects at which

this radiation is themost intense

wavelength emitted−272 °C −173 °C 9,727 °C ~10,000,000 °C

Figure 5.3: Electromagnetic spectrum. [29]

depend on the propagation material. In turn, seismology can take advantage of all these propertiesto better understand the Earth’s inner structure as well as to study geological risks.

5.4.4 Transport

While they may transport energy, it is important to note that waves do not transport matter (nottaking into account the small local displacements in material waves).

This may seem counterintuitive, for example in waves on a water surface: the waves peaks moveacross the surface, but the water particles themselves only move up and down. Thus an objectfloating on the water - such as a duck - is itself brought to (almost) only oscillate vertically.

5.4.5 Doppler Effect

Doppler Effect is a phenomenon occurring when the emitter (an object creating a wave) and/orthe receiver (an hypothetic object observing wave oscillations) of a wave move with respect to eachother and, in the case of material wave, with respect to the propagation medium.

In this script we will restrict ourselves to the 1D case for sound waves.

Let’s consider an emitter travelling with velocity vem and creating a sound wave of frequency femand velocity v. Let’s also assume that a receiver travelling with velocity vre is at a distance d fromthe emitter at a given time (which we will set to zero together with the emitter’s initial positionfor the sake of simplicity).

Note that we do not make any assumption on the signs of vem and vre but, as we are consideringthe 1D case, each object is moving either directly towards or away from the other one. We will

151


simply define vem and vre as positive when they are in the same direction as the vector going fromthe emitter to the receiver, as negative otherwise.

Moreover, both vem and vre are relative to the medium.

d

emitter receiver

v v vem re

Figure 5.4: Emitter and receiver. Due to the Doppler effect, the frequency of the sound will bedifferent for the emitter, the receiver and a neutral observer at rest with respect to the medium.

Let t1 be the time taken by the sound emitted at the initial instant to arrive to the receiver. Wehave:

vt1 = d+ vret1 ,

thus t1 = d/(v − vre). Similarly, let t2 be the time taken by the sound emitted after a period ofthe sound (1/fem) to arrive to the receiver, so:

vt2 = d− vem1

fem+ vre

1

fem+ vret2 ,

so we have t2 =(d+ (vre − vem) 1

fem

)/(v − vre).

From t1 and t2 we can calculate the frequency of the sound fre as observed by the receiver, asa function of fem (t2 − t1 representing the difference in periods between the sound received andemitted):

fre =1

t2 − t1 + 1fem

=1

vre−vemv−vre

1fem

+ v−vrev−vre

1fem

=v − vrev − vem

fem .

Note: the signs in the fractions depend on the way we defined the velocities. They vary amongtextbooks and one should be careful when applying the formula.

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Physics Olympiad Script 5.5. WAVES PROPAGATION AT INTERFACES

The Doppler effect typically leads to a higher pitch for approaching source and lower when itrecedes from the observer. An everyday example is the siren of emergency vehicles: one can wellhear the frequency drop when the vehicle passes by.

It can also be used in technical applications, by sending a sound signal and analysing the frequencyof the signal reflected back: radars and blood flow imaging in cardiology are typical uses.

5.5 Waves Propagation at Interfaces

Under this title we understand phenomena where, in contrast to what we saw before with (im-plicitely assumed) homogeneous, isotropic and infinite propagation spaces, there is some elementin the path, preventing the wave from freely propagating further.

An important principle for understanding those phenomena is Fermat’s principle: it states thatlight (and it can be generalized on most waves) always "chooses" the locally fastest way betweentwo points, i.e. the way that it takes the shortest time for the wave to travel. We will see later onwhy this "locally" is important, when studying reflections.

5.5.1 Reflection

Reflection happens when a wave bounces back at an interface between its propagation medium -be it material or not - and some material element.

Supposing that the element’s interface with the medium is sufficiently smooth, we can define theincidence angle ϕi and reflection angle ϕr between the surface normal and the respective pathelements.

Using Fermat’s principle, it can be shown that this is exactly the case when both angles are equal,ϕi = ϕr. We also easily see the local aspect of the principle: of course this path is not the globallyshortest one (a straight line between the extremities of the path would be shorter), but of thecontinuum of paths going from one point to the other while hitting the interface, it is the shortestone.

A straightforward way to see how Fermat’s principle implies equal angles is to imagine that themirror does not exist, but that instead one of the extremities of the path is mirrored, i.e. on theother side of the mirror’s plane. Then it is clear that to minimize the distance, one has to considerthe straight line from that mirrored extremity to the other, unmodified one. And it is also directlyvisible that the equality of angles should be respected in that case.

5.5.2 Refraction

Refraction occurs when a wave arrives at an interface between to propagation media, which typi-cally have different properties, such that the wave has a different propagation velocity in each ofthem.

In order to take these velocities into account, one defines - in case of light - the refractive index nof a medium as the fraction of the vacuum velocity c by the velocity in the medium v: n = c/v.

153


A

B

A

ϕrϕi

ϕ′i

Figure 5.5: Reflection on a plane surface of a wave travelling from A to B. By constructing theimage A of A by symmetry w.r.t. the plane, one sees that ϕi = ϕ′i as well as ϕr = ϕ′i (Fermat’sprinciple requiring AB be a straight line). Hence ϕi = ϕr.

The refractive index is slightly dependent on the light wavelength, which is one of the main reasonsfor chromatic aberration in refractive optical parts. We will nevertheless forget this detail in thisscript.

As can be shown using Fermat’s principle, the angles of incidence ϕ1 and of refraction ϕ2 at theinterface between media of refractive indices n1 and n2 are related by Snell-Descartes law:

n1 sin(ϕ1) = n2 sin(ϕ2) .

5.5.3 Diffraction

Diffraction is what happens when part of the wavefront (the most advanced part of the perturba-tion, with all points in the same oscillation state) hits an object: secondary waves are formed atthe hitting points, which make the wave "turn around" the object, following its curvature a bit.For the diffraction to have an important effect, the wavelength of the wave has to be of roughly thesame scale as the object in the way. This is typically the reason why we can hear a sound withoutactually seeing its source, because the objects in the way are at our scale - which is roughly alsothe sound’s scale - but are much too big for visible light to get diffracted enough to come to us.

5.6 Multi-Waves Phenomena

Until now, we only considered single waves going through space. However, as waves are simplyperturbations, nothing forbids two waves from being at the same place at the same time. Thus

154

Physics Olympiad Script 5.6. MULTI-WAVES PHENOMENA

A

C

ϕ1

ϕ2

n1

n2

A′

ϕ′1

ϕ′2

n1

n2

A′′ϕ′′1

n1

n2

Figure 5.6: Top: Refraction at a medium interface of a wave travelling from A (in medium withrefraction index n1) to C (index n2). Here, n2 < n1 is implied. Middle: limit case with ϕ′2 = π

2 .ϕ′1 is thus called the critical angle. Bottom: For ϕ′′1 > ϕ′1, no refraction is possible anymore, onlyreflection. This phenomenon is therefore called total internal reflection.

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Figure 5.7: Diffraction of a plane wave passing through a slit. Notice how the diffracted wave isapproximately circular and thus also reaches zones (even if attenuated) that are "hidden" sidewaysof the slit, along the barrier. [28]

we have to consider the so-called wave superposition, or wave interference. The main idea here isthe superposition principle, which states that, given two waves u(~r, t) and v(~r, t) of same type (i.e.sharing the same oscillators), the perturbation resulting from their cumulated effect is simply theiralgebraic sum u(~r, t) + v(~r, t). The same goes with vectorial waves as well.

The resulting perturbation can have different forms depending on what the original waves looklike, but the most important property is that the result is again a wave. In the following, we willstudy the most common and interesting interference phenomena for 1D harmonic waves.

In general, we will therefore look at

u(x, t) + v(x, t) = A1 sin(ω1t− k1x+ ϕ1) +A2 sin(ω2t− k2x+ ϕ2)

and try to understand what is going on depending of the relation between A1 and A2, ω1 and ω2,k1 and k2, and ϕ1 and ϕ2.

5.6.1 Same amplitude, frequency and wavenumber

The simplest case is when A1 = A2 = A, ω1 = ω2 = ω and k1 = k2 = k, i.e. the waves are verysimilar but can simply be shifted by some phase. For the sake of simplicity, we will set ϕ1 = 0 andϕ2 = ϕ.

We can use the sine addition formula to rewrite the interference:

156


u(x, t) + v(x, t) = A sin(ωt− kx) +A sin(ωt− kx+ ϕ)

= 2A sin(ωt− kx+

ϕ

2

)cos(ϕ

2

).

As can be seen, the resulting wave is again harmonic, with the same frequency and wavenumber.Depending on ϕ, its amplitude can vary:

− For ϕ = 0, ±2π, ±4π, etc., the amplitude is maximal and equal to 2A. This is the perfectaddition of two identical waves.

− For ϕ = ±π, ±3π, etc., the amplitude is null, i.e. the waves are exactly out of phase andthus completely annihiliate.

−A

A xϕ = 0:

−A

A xϕ = π4 :

−A

A xϕ = 3π

4 :

−A

A xϕ = π:

Figure 5.8: Superposition (thick plot) of two waves u(x, t) (dashed) and v(x, t) (dotted) with sameamplitude, frequency and wavenumber and a phase difference ϕ for v with respect to u. Due to thecoefficient cos(ϕ2 ) in the amplitude, the amplitude of the resulting superposition wave can be biggerthan those of the original waves (positive interference) or smaller or even null (negative interference).

5.6.2 Same amplitude and frequency, opposite wavenumber

This situation corresponds to two identical waves propagating in opposite directions, so A1 = A2 =A, ω1 = ω2 = ω, k1 = −k2 = k. We can set ϕ1 = ϕ2 = 0 without loss of generality.

Again using the addition of sines:

u(x, t) + v(x, t) = A sin(ωt− kx) +A sin(ωt+ kx)

= 2A sin(ωt) cos(kx) .

157


Very interestingly, by summing the waves we get a separation of ωt and kx, which are thereforeuncoupled. This means that the oscillations of the resulting wave are static - some places will con-stantly have zero amplitude due to the cos(kx) ("nodes") while others will have maximal amplitude("anti-nodes"). Similarly, at some times the whole wave will have zero amplitude everywhere dueto the sin(ωt), etc.

Such a wave, that does not seem to travel along its medium, is called a stationary wave or standingwave. It is quite common, as it forms everytime some wave gets reflected back and interferes withitself.

Note that the distance from node to node and from antinode to antinode is λ2 (for space) and T

2(for time).

−2A

2A

−2λ −λ λ 2λ

x

w

Space

w(x, 0)

w(x,∆t)

w(x, 2∆t)

−2A

2A

−2T −T T 2T

t

w

Time

w(0, t)

w(∆x, t)

w(2∆x, t)

Figure 5.9: The resulting stationary wave w(x, t) = u(x, t) + v(x, t) represented in both spaceand time. In both plots, nodes and antinodes remain at the same x, resp. t coordinate, hence"stationary". For our choice of wave equation and ϕ’s, spatial antinodes are at nλ

2 , while temporalnodes are at nT

2 , n ∈ Z. Note that w(x, 0) = 0 for all x’s.

5.6.3 Slightly different frequencies

Another interesting case is A1 = A2 = A, k1 = k2 = k but ω1 6= ω2. We can set both phases tozero without loss of generality.

Using once more the addition of sines:

u(x, t) + v(x, t) = A sin(ω1t− kx) +A sin(ω2t− kx)

= 2A sin

(ω1 + ω2

2t− kx

)cos

(ω1 − ω2

2t

).

What we have here is a conventional harmonic wave of angular frequency ω1+ω22 - i.e. oscillating

more rapidly than the original waves - modulated by an oscillation (not a wave - note the absence

158


of x) of angular frequency ω1−ω22 . In the case where ω1 ≈ ω2, this modulation will be slow with

respect to the oscillation of the wave itself.

The result is the so-called beat phenomenon, with an almost harmonic wave periodically increasingand decreasing in intensity. ω1 ≈ ω2 is needed for the phenomenon to be visible/audible.

−1

1

− πω1−ω2

4πω1+ω2

πω1−ω2

t

−2A

2A

− πω1−ω2

πω1−ω2

t

w

Figure 5.10: The modulated beat wave is the result of the multiplication of the dotted wave by atemporal oscillation (dashed) and an amplitude factor, 2A. Note that both the beat and the dottedlines represent waves, and as such also move in space. In contrast, the dashed line represent a puretemporal oscillation that only "modulates" (constrains the amplitude of) the beat wave. Note alsothat the "enveloppe" of the beat wave is a periodic, non-sinusoidal oscillation with a period half ofthe modulating cosine.

5.6.4 Fourier Analysis

We just saw that the sum of two harmonic waves is again a periodic wave, and in some cases iseven harmonic itself. We can generalize this idea and prove that any superposition of harmonicwaves is periodic.

Conversely, an important result says that any periodic oscillation can be decomposed in a sum of(potentially infinitely many) harmonic oscillations. Even more interestingly, it exists a well-definedmathematical operation, called Fourier-transform, that does exactly that. More precisely, given aperiodical signal, it allows to calculate its spectrum, i.e. the distribution of amplitudes for eachfrequency of fundamental harmonic oscillation.

Fourier Analysis can be thought of as the counterpart of Taylor expansion in terms of periodicalsignals instead of polynoms, and is a crucial instrument in the modern digital world, among otheruses. For example, analogic audio signals can be converted (digitalized) into a sequence of numbersrepresenting a finite approximation of the signal’s spectrum, thus allowing an efficient digitaltreatement and storage.

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160

Chapter 6

Fluid dynamics

This very short chapter is thought more as a guide for both lecturer and students, rather thancomplete lecture notes. For example there are no pictures at all, so feel free to remind the lecturerto draw some pictures if he should forget to do so.

6.1 Introduction

We will first explain the basic assumptions here and then present some important results in thenext sections.

6.1.1 What is fluid dynamics about?

Fluid dynamics describes the dynamic properties of fluids. Dynamic properties (or systems) arethose who can change in time, as opposed to static ones1. The other part of the name is fluid. Afluid is generally something which can flow2 (i.e. it is a gas or a liquid). However the distinctionbetween a fluid and a solid is not that easy in general, as something may look solid if we only lookat it for a short time but it flows on larger timescales (e.g. a glacier or the pitch drop experiment- google it).

6.1.2 How can we model such a fluid?

The first formal concept used is that of a trajectory. Intuitively this is the path of some smallparticle put into the fluid, e.g. a leaf in a stream. For each point in space we can find exactly onetrajectory passing through this point (at any given point in time). The second - more important- concept is that of a velocity field, which is just a function giving us the velocity of a fluid ateach point in space and time. If we have such a velocity field we can define flow lines (analogously

1Normally static systems are seen as an (mostly) easier special case of the corresponding dynamic systems. I.e.we will also look at static properties in this chapter.

2Technically speaking, we could say that in contrary to a solid, in a fluid two initially neighbouring particles canmove arbitrarily far away from each other.

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Physics Olympiad Script CHAPTER 6. FLUID DYNAMICS

to field lines in electrodynamics). This lines are such that they are tangent to the vectors of thevector fields at each point. In the cases we look at here, they are identical to the trajectories3.

6.2 Notation

Before we start a short summary of notations used throughout this chapter:

Velocity ~v: the velocity of the fluid at some point (in space and time).

Speed v = |~v|: the absolute value of the velocity.

Pressure p: force applied per unit area.

Density ρ: the density of a fluid (in mass per volume).

6.3 Pressure

Pressure in fluid dynamics is the same concept as in thermodynamics. It is a force per area whichacts on any surface a fluid touches. The force is always perpendicular to the surface on which itacts.

Note: In contrast to thermodynamics, the pressure here can also depend on the position in thefluid (see below).

6.3.1 Compressible and incompressible fluids

One big difference between water and air is the compressibility. We say air is compressible, thatmeans we can change the density of it e.g. by applying pressure. We cannot do this with water4. Tobe able to describe this we would need some equation relating pressure and density. One examplewould be the ideal gas law (excercise: why?). But we will not calculate compressible flows.

6.3.2 Hydrostatic pressure

Suppose we have a long vertical pipe with radius r, closed at the bottom with a lid. If we nowfill it with water up to some height h, the total volume of water above the lid is V = πr2h. Theforce of all the water pushing on the lid is thus given by F = V ρwaterg, where g = 9.81 m·s−2 is thegravitational acceleration. The pressure on the lid is now given by the force divided by the area,which leads to p = ρgh. The astonishing thing is that the area of the lid cancels. This means thatthe pressure at some depth d (measured from the water surface) is independent of the form of thetube. It also means that if we have connected tubes the pressure at any height h should be thesame in every tube5.

3In general they are not, namely if the velocity field depends explicitly on time.4Of course we can, but the effects are much smaller than in air, and we neglect them here.5Taking tubes with open tops, we can deduce that they have the same water level.

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Physics Olympiad Script 6.4. CONTINUITY EQUATION

6.3.3 Buoyancy

Buoyancy is a phenomenon which is due to the depth-dependence of the hydrostatic pressure. Thisalso means that we need a force like gravity acting on our fluid to have buoyancy. To find a formulafor this effect, consider a small cube of side length a fully submersed in water. Assume the topand bottom of the cube are perpendicular to the gravitational field. Now we look at the forcesdue to the pressure. By symmetry, the two forces for the sides (front,back) and (left,right) cancel.Assume the top is at a depth d then the force on the top is Ftop = pa2 = ρgda2. The bottom isat depth d + a and the force is Fbottom = ρg(d + a)a2. So the total (upward) buoyancy force onthe cube is given by Fup = Ftop − Fbottom = ρgaa2 = ρgVd. Where Vd is the Volume of the waterdisplaced by the cube6. If the cube is only partially submersed, we just set Ftop = 0 and arrive atthe same equation. The same formula holds also for other bodies (i.e. boats), intuitively just thinkof them as being built from small cubes, then also the same formula holds7. A body can now floatin water if Fup > Fg = mg. We can translate this into an equation of of the densities, namely ahomogeneous body floats if ρbody < ρwater.

6.4 Continuity equation

The continuity equation is a consequence from the conservation of mass. In a first step assumewe have water flowing through a tube. We take the tube to have different cross-section areas, sayA at point a and B at point b. The total mass passing point a in a time interval ∆t is givenby8 vaAρ∆t. And analogously at b. Conservation of mass means now, that those two quantitiesneed to be equal, i.e. vaAρ∆t = vbBρ∆t. Time cancels on both sides and we are left with9:vaAρa = vbBρb. Now we look at a different system. Lets say we have some sort of bottle (ofconstant volume) and fill it with air. We can again look at the total mass flowing into the bottlein some ∆t, which is given by ∆M = vAρincoming∆t. As the air cannot escape, the mass insidethe bottle increases and so does the density: ∆M = ∆ρV . We set this equal to the incoming mass(conservation of mass) and get ∆ρV = vAρincoming∆t. We can simplify this by introducing thetime derivative of ρ: dρ

dt = vAV ρincoming. We need to be a bit careful here, as the density of theincoming fluid ρincoming is independent of the change of density inside the bottle dρ

dt . Combiningthe two parts, i.e. looking at a tube with changing density between two points a and b we find:

vaAρa − vbBρb = Vabdρdt, (6.1)

where Vab is the volume between the cross-sections A and B.

6As long as the cube is fully submersed, this is the volume of the water.7You could also do surface integrals over the whole body, which is much more tedious to do and leads to the

same results.8We assume v to be constant over the whole cross-section.9ρ would also cancel here, but as we also want to look at gases, in general ρ is not the same at a and b.

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6.5 Bernoulli’s equation

Let us just start with the full equation and then explain what it says:

v2

2ρ+ ρgh+ p = const. (6.2)

This equation relates the velocity, the gravitational potential and the pressure. Note however thatit is only valid for incompressible fluids and that const. needs to be taken along one flow line (butin most problems here this constant will be the same for all flow lines).

6.5.1 Derivation

Assume we have a tube with a piston at each end, filled with water. Denote the two ends andall quantities there with a and b respectively. Suppose piston i has an area of Ai and is at heighthi above ground. Let the water have pressure pi at the piston. If we now displace piston a bya small distance da, piston b will move db = Aa

Abda = V

Ab. The work needed to displace piston a

is Wa = daAapa = V pa and for piston b: Wb = −dbAbpb = −V pb, where the minus sign comesfrom the fact that the force coming from pressure now points into the other direction (compared topiston a). So the total work we put into the system is given byWtot = Wa+Wb = V (da−db). If welook at the energy of the fluid, we find two effects: We displace a volume V of water from ha to hb,this gives a change in potential energy of ∆Epot = ρV g(hb− ha). The second effect is the increasein kinetic energy ∆Ekin = ρV

2 (v2b − v2

a). If we now combine everything Wtot = ∆Epot + ∆Ekindivide by V and rearrange the terms we get Bernoulli’s equation.

6.6 Surface tension, energy and capillary pressure

Surface tension comes from the fact that molecules at the surface of a liquid have no molecules’above’ them and are attracted by those below. So there is a force acting on them which needs tobe compensated (e.g. by pressure) to have a static surface. Assume that we increase the surface ofa liquid by a small area ∆A (e.g. by bulging it out just a little bit). In general we need to do somework ∆E to achieve this. The surface tension is now given by 10 σ = ∆E

∆A . Capillary pressure isthe pressure forcing water up a thin tube with radius a. This is also a surface effect, as the watermolecules need less energy when they are at the surface to the wall compared to somewhere in theliquid (i.e. they stick to the wall). This pressure is given by:

pc =2γ cos(Θ)

a(6.3)

where γ is the surface tension relative to the wall and Θ is the contact angle. To get the heightthe liquid rises to, equate this pressure with the hydrostatic pressure and solve for h.

10We assume v to be constant over the whole cross-section.

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Physics Olympiad Script 6.7. FRICTION IN FLUIDS

6.7 Friction in fluids

As for now we silently assumed that our flows do not have friction. As friction is in general verycomplicated, we will only state two approximate results which hold for objects moving through afluid (e.g. a submarine, a ball, a car). There are two types of flows, laminar and turbulent, whichhave different formulas for friction. In the laminar case we get a force proportional to the velocity.An exact formula (only valid for a sphere with radius R)11 is:

Fr = 6πηRv (6.4)

where v is the velocity and η is the dynamic viscosity. The turbulent friction is given by:

Fr = cWAρv2

2(6.5)

Where cW is the drag coefficient (a constant depending on the material and form of the object)and A is the area of the object perpendicular to the velocity (i.e. the area you see if you look inthe direction of flow of the fluid).

11A farmer wants to improve the milk production of his cows. He asks a biologist, an economist and a physicistto help him. All three of them come to his farm and observe everything. After a day the economist gives the adviceto fire all cows and outsource the farm to China to improve the production by 2%. The farmer doesn’t like thissuggestion and waits for the other two. After a week the biologist presents the idea of replacing the cows by genemanipulated algae to produce 10% more milk. The farmer decides to wait for the physicist. After several weeks, thephysicist appears with tons of paperwork and claims to have found an idea to improve the production by over 60%.The farmer is really interested and asks the physicist to explain his idea in more detail. The latter starts: "Assumecows to be spherical and in a vacuum."

165


166

Chapter 7

Electrodynamics

Electric attraction and repulsion is a fundamental property of charged particles. It is describedby the electric field. When charged particles move they generate an additional field, the magneticfield1. The theory about electrodynamics describes these fields and how they interact between eachother or with charges and also how they evolve in time. Since the general case of time dependentsystems is pretty complicated, we will focus on static setups. This means that the charges, wiresor whatever we are looking at do not move and have always the same position. The main sourcesfor this chapter are [30] and [31].

7.1 Electrostatics

This chapter examines the interaction between point charges and how this interaction can bedescribed. Knowing the physics of point charges one can easily derive the physics for chargedistributions.

7.1.1 Coulomb Force

The basic experimental observation of electrodynamics is that there exist "things" that can attractor repel each other in a way that is not due to gravity. These "things" we call charges and theattracting or repelling force electric force or Coulomb force. One observes that there exist twotypes of charges, we call them positive and negative charge. A charge which has a very smallspreading compared to the distance to other charges we call a point charge. An ideal point chargeis just a point in space with a charge.

1In relativity, the electric and magnetic field are not two independent fields, they have a strong relation to eachother. That is why it is often called the electromagnetic field. The relativistic treatement is out of scope for thiscourse.

167

Physics Olympiad Script CHAPTER 7. ELECTRODYNAMICS

If we take two point charges q1 and q2 which are separated by a constant distance r we measure aforce ~F acting on q2. This force is related to the charges according to

~F12 = kq1q2

r2~er (7.1)

where ~er = ~r|~r| is the vector that points from q1 to q2 and has length 1. Such a vector with length

1 we call unit vector. k is a constant which depends how we define the basic unit of the electrody-namic theory. There exist different systems of units and as a consequence k is different in each ofthose unit systems. We use the SI system, where the current and the time are defined and thereforealso the charge (see 7.3.1). The unit of charge is Coulomb C (see exact definition in section 9.1.4).One electron has a charge of 1.602 · 10−19C. We then get k = 1

4πε0where ε0 = 8.85 · 10−12 Fm−1.

The reason why k contains a factor 4π is explained in chapter 7.1.5.

The important properties we learn from this formula are:

− The coulomb force always points in the direction of the connecting line of the two charges.The force acting on q1 has the same amount but opposite direction to the force acting on q2

which agrees to Newtons actio equal reactio.

− If both charges are positive or negative the two charges repel each other, if the charges aredifferent they attract each other.

− The force has a 1/r2 dependence as we know it from gravity (see also 7.1.5).

− Opposite the gravitation force, there is also repulsion possible. As a consequence it is possibleto shield a charge from the influence of other charges.

7.1.2 Electrostatic field

In the 18th and 19th century, when the theory about electro-magnetism was developed, there wasa big discussion how the force described by equation (7.1) can act over distances. One pointof view was that the force acts directly and that the two charges interact immediately. Thisview contradicts with some statements from relativity which state that information can maximalpropagate with speed of light. Farady successfully described the Coulomb force by introducingthe electric field: The space has an additional property, the electric field, which is influenced bythe presence of charges and the charges interact with the field. If we apply this field theory tothe example above with the two point charges we get that a charge, for example q1 influences thefield in such a way that the interaction between q2 and the field at the position of q2 results inthe Coulomb force. The simplest way to describe this interaction is defining the electric field ~E asquotient of Force ~F and charge q2

~Eq1 =~F12

q2=

q1

4πε0r2~er (7.2)

168

Physics Olympiad Script 7.1. ELECTROSTATICS

To get the electric field ~E(~r) at the position ~r of a given setup, one calculates the force ~F thatwould act on a very small charge q0 at the position ~r and divides the force through the charge

~E(~r) =~F (~r)

q0

The reason why q0 has to be very small is because otherwise it might influence the other chargesand therefore the electric field. This gets important if one looks at time dependent fields which wedo not treat here.

Fields are often drawn by field lines, see figure 7.1. Putting a very small charge in a field linepicture the force on the small charge points always in the direction of the field line. Thereforein a field line picture the field is always tangential to the field lines and the strength of the fieldis proportional to the density of the field lines. Electric field lines have the following qualitativeproperties: They start and end at the sources, in the electric case they start at the positive chargeor in infinity and end at the negative charges or at infinity. Additionally field lines want to be asshort as possible but they repel each other. This attraction in length and repulsion of each otherdefines a stable state for the field lines which they will take. Of course this is more a qualitativereason for how the field looks like but often one gets a first intuition for a problem.

The advantage of describing the interaction of charges by a field is that it is possible to describethe changing of the field with a finite speed. Therefore two charges interact only with the speedof light and not immediately.

Figure 7.1: Field lines of two charges, at the left both positive, at the right one negative and onepositive. At the left picture some vectors of the electric field are drawn which are tangent to thefieldlines. At the right picture the electric field in the oval with the continuous line is much strongerthan in the one with the dashed line because in the oval with the continuous line the field lines aremuch denser to each other. [32].

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7.1.3 Superposition

The Coulomb force allows us also to examine the field of multiple charges because every pair ofcharges interacts according to equation (7.1). Therefore the total force on a charge q is the sumof all forces between q and the other charges q1...qN . The definition of the electric field is still thesame as defined in equation (7.2) and we get

~E =~F

q=

∑Nj=1

qjq

4πε0r2j~erj

q=

N∑j=1

qj4πε0r2

j

~erj (7.3)

where ~erj is the unit vector pointing from qj to q.

7.1.4 Continuous charge distributions

When there are a lot of point charges involved (1C = 6.2 · 1018 electrons) it is useful to use thecharge density ρ = q

V instead of describing the electric field of every point charge. The chargedensity contains the information how much charge q a volume V contains. Therefore the chargein the volume is given by q = ρV if ρ is constant all over V . It can also happen, that the densitydepends on the position ~y, we then write ρ(~y). To compute the electric field at the position ~x wetreat the charge that is in a very small volume dV (~y) as point charge located at the position ~y.We then integrate the charge over all these dV (~y) which are located inside the total volume V (so~y is inside V ). This leads to the formula

~E(~x) =

˚V

ρ(~y)

4πε0|~x− ~y|2~x− ~y|~x− ~y|

dV (~y) (7.4)

where the ρ(~y) symbolises that one has to take the charge density at the location where the dV (~y)is. It is not important to be able to calculate the electric field for a difficult charge distribution.It is more important to understand the concept (see application 7.1.6). Often the problems havesome nice symmetries which makes it easier to solve.

There might also be two dimensional surface charge distributions or one dimensional charge dis-tributions which can be treated analogously to the three dimensional case discussed above (see7.1.6).

7.1.5 Gauss’ law

The 1/r2 dependence in the Coulomb law (7.1) has an important consequence. When we look ata point charge Q around which we place an imaginary sphere, we observe that the number of fieldlines that pass through the sphere does not depend on the radius of the sphere because field linesstart and end at charges or in infinity. Let’s formulate this more mathematically. The electric fluxΨ for a homogeneous electric field ~E is defined as Ψ = ~E · ~A where ~A is the surface vector for aplane surface. The surface vector points perpendicular to the surface. The length of the vector

170


is equal to the area of the surface. If the sign of Ψ is positive, the electric field goes through thesurface in the same sense as the surface vector is pointing, if it’s negative in the opposite sense.Visually spoken the flux indicates how many field lines pass through the surface. If we now have acurved surface, as it is the case at a sphere, we have to split the surface A of the sphere into smallpieces d~S and we get a small amount of the flux by dΨ = ~E · d~S (see figure 7.2).

Figure 7.2: Curfed surface with a small surface vector d~S and the electric field ~E going through thesurface [33].

Since the surface A of the sphere is always perpendicular to the radius vector, ~E and d~S are paralleland the flux is simply dΨ = ±EdS, where E and dS are the absolute values of ~E and d~S (the± indicates that one has to take into account weather the field lines go into the sphere which isthe case if Q < 0 or if the field lines go out of the sphere for Q > 0). As we stated the wholeflux through the sphere should be independent of the radius and indeed, the calculation also statesthis:

Ψ =

‹AdΨ

=

‹A

~Ed~S

=

‹AEdS

= E

‹AdS =

Q

4πε0r24πr2 =

Q

ε0(7.5)

The two integral symbols indicate that the integral is over a two dimensional surface and the circlein the integral symbolizes that the area is closed (it is the surface of a volume). The step fromthe second last to the last line is right since the electric field is everywhere on the surface parallelto d~S and has the same strength E = Q

4πε0r2(because on the sphere the distance to the charge is

everywhere the same), therefore it does not depend on dS and we can take it out of the integral.The last integral is simply the integral of the small surface areas dS all over A and therefore simplyA = 4πr2 itself.

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Following the idea of field lines it is also clear that in any volume V that has no charge in it thenumber of field lines that enter the volume is the same as the number of field lines that leave thevolume. In mathematical notation this means that the total flux is zero:

Ψ =

‹∂V

~Ed~S = 0 (7.6)

where ∂V is the surface of the volume. If we now combine it with the result from equation (7.5) weget that independent of the volume the flux only depends on the amount of charge that is insidethe volume:

Ψ =

‹∂V

~Ed~S =Qinε0

=1

ε0

˚Vρ(~y)dV (~y) (7.7)

This formula is called Gauss law and its statement is that the flux through a closed surface onlydepends on the electric charge inside that volume and not on the form of the surface. One canconclude that the field lines are produced by the charges. It is pretty useful for example to calculatethe electric field in symmetric problems (see 7.1.6) or to find out how charge is distributed.

7.1.6 Examples

We want now use the laws above to calculate the electric field of some configurations.

infinite plain

Let’s assume there is a thin (height = 0) plate with a surface charge density σ (charge per area) atthe x-y plane (see figure 7.3). We again use Gauss law 7.1.5 by using a small cylinder which hasabove and under the x-y plane a surface parallel to the x-y plane with area S each.

Since the plane is infinite and the charge density is uniform, the electric field has no componentthat points parallel to the x-y plane so the field only points in z-direction. Therefore the wholeflux through the surface of the volume goes through the surface above and under the plane andusing Gauss law one gets

Qinε0

=

¨2S

~Ed~S (7.8)

σA

ε0= E · 2A (7.9)

Therefore the electric field is ~E = ± σ2ε0

~ez. The ± depends weather one calculates the electric fieldabove or under the plate.

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Figure 7.3: Infinite plate whtih the cylinder [35].

infinite long wire

Let’s assume there is an infinite long, straight wire on the x-axis which has a constant chargedensity (per length) λ. The goal is to calculate the electric field at every point.

We’ll use two different approaches:

At the first approach we use Gauss law 7.1.5 and some symmetries to calculate the electric field.To use Gauss law we need some volume and since the problem is rotations symmetric around thez-axis we choose a cylinder with length l and radius r around the wire (see figure 7.4). Since thewire is infinitely long there is no midpoint of the wire and as a consequence the electric field mustpoint radial. As a consequence the flux through the left and right circle S (see figure 7.4) is zero.Furthermore the electric field has the same strength everywhere on the side of the cylinder and ispointing outside, parallel to the surface vector. Therefore the scalar product is the same as themultiplication of the absolute values of the surface vector dS and the electric field E. Now usingGauss law one gets

λl

ε0=Qinε0

=

‹surface of cylinder

~Ed~S

=

ÄEdS

= E

ÄdS

= E2πrl

where A is the area of the side of the cylinder with A = 2πrl.

173


The electric field is therefore

~E =λ

ε02πr~er =

λ

ε02π(x2 + y2)

xy0

Figure 7.4: Cylinder around the wire. The wire is along the z-axis.

The second approach uses the superposition principle for continuous distribution (see 7.1.4) andis more complicated. Since the whole problem is rotation symmetric around the x-axis and alsotranslation invariant along the z-axis the electric field only depends on the distance r from the wireand to simplify the calculation we look at the electric field at the point x = z = 0 and y = r. Thedensity in this problem is per unit length therefore the dV is replaced by a dx in formula (7.4). Sothe integral looks like

~E =

ˆz−axis

λ

4πε0√x2 + y2 + z2

3

xyz

ds

=

∞

−∞

λ

4πε0√r2 + z23

0rz

dz

So we can look at the z and y component of the ~E field separately. For the z Component we get0 because z√

z2+y23 is an odd function (it is point symmetric to the origin) and the integral from

an odd function over a symmetric interval is always zero. Physically this can be interpreted thefollowing way: when we compute the contribution to the electric field from a point on the wireat z0 we find a point −z0 which contributes the same amount to the z-direction of the electricfield but in opposite direction. So the contributions from z0 and −z0 to the z-component cancel

174

Physics Olympiad Script 7.2. POTENTIAL AND VOLTAGE

out. Now let’s calculate the y-component of the electric field which we have to do by solving theintegral. Since λ is constant over the whole wire we can take it out of the integral

Ey =λ

4πε0

ˆ ∞−∞

r√r2 + z23 dz

We now apply the substitution z = r sinh(u) and dz = r cosh(u)du and we get

Ey =λ

4πε0

ˆ ∞−∞

r2 coshu

(r2(sinh(u)2 + 1))3/2du

=λ

4πε0

ˆ ∞−∞

1

r cosh(u)2du

=λ

4πε0

[1

rtanh(u)

]∞−∞

=λ

4πε0r2 =

λ

2πε0r.

This is the same result as we got at the first approach.

7.2 Potential and Voltage

An electric field produces a force on a charged particle. If one displaces the particle work has tobe done. This chapter looks at this work and the energetic properties of the electric field.

7.2.1 Electric potential

The force ~F on a charged particle q in an electric field ~E is given by ~F = q ~E. If one wants tomove the particle from one position P1 to an other position P2 one has to overcome the force ~Ftherefore one has to apply the force ~Fext = −~F = −q ~E. Moving q along a path S one has to effortthe work W

W =

P2ˆ

P1

~Fext · d~s = −P2ˆ

P1

~F · d~s = −qP2ˆ

P1

~E · d~s

The − sign causes the work to be positive if one drags a positive charge q > 0 against the electricfield, therefore one has to apply work (actively). If one pulls a charge q > 0 in direction of theelectric field one gets energy. Therefore if the work from P1 to P2 is positive the charge has at P2

a higher potential energy than at the position P1. If the energy at the reference point P1 is chosen

175


to be E0 then the energy at the point P2 the energy is exactly Ep = W + E0. Since only theenergy difference between the two points can be used, E0 can be set to zero without changing thebehaviour of the system. As a consequence the energy at P2 is proportional to the charge and wedefine a new property, the electrostatic potential ϕ(P2) (often only called electric potential) whichis defined as

ϕ(P2) =Epq

= −P2ˆ

P1

~E · d~s (7.10)

This formula is only true in the electrostatic case, for the dynamic case one can also define an electricpotential but one has to pay attention to more things. The electrostatic potential describes theinfluence of the electric field to the energy of a charged particle q.

7.2.2 Electric potential of a point charge

If we understand the electric potential of a point charge we can easily generalize it to multiplepoint charges or even to a general charge distribution.

Let Q be a point charge at the origin of the 3-dimensional coordinate system and q an other pointcharge which we move. We want to examine the potential energy of q depending on its position.Lets first move q from a point P1 to a point P2 where both points have the same distance r fromthe origin. As path S we chose a path where we have always the same distance r from Q (which isat the origin). Therefore we never move radial and therefore always perpendicular to the electricfield since the electric field points radial (this means in the same direction as the connecting lineof Q and q, see chapter 7.1.1). To calculate the potential at P2 we integrate according to equation(7.10) and since the electric field and the path are always perpendicular to each other the scalarproduct is zero and as a consequence the whole integral. Therefore the energy of q only dependsof the distance to Q, which is reasonable since the whole problem is spheric symmetric.

Let’s now examine the dependence of r on ϕ. The point P1 shall have the distance r1 and P2 thedistance r2 from the origin. Then the potential difference is

ϕ2 − ϕ1 = −r2ˆ

r1

~E · d~s = −r2ˆ

r1

Q

4πε0r2dr = − Q

4πε0

[−1

r

]r2r1

=Q

4πε0

(1

r2− 1

r1

)(7.11)

Since the reference energy can be chosen freely, it is also possible to choose the reference point r1

freely and the simplest way is to choose r1 =∞ and we get.

ϕ(R) = −R

∞

Q

4πε0r2dr =

Q

4πε0

(1

R

)(7.12)

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If we want to calculate the potential difference ∆ϕ between two distances r1 and r2 we can useequation (7.12)

∆ϕ = −r2ˆ

r1

Q

4πε0r2dr

= −∞

r1

Q

4πε0r2dr −

r2ˆ

∞

Q

4πε0r2dr

= ϕ(r2)− ϕ(r1)

which makes it easy to calculate potential differences of point charges.

7.2.3 Potential of multiple charges

We want to examine the potential energy Ep of a point charge q at position ~r in a system of ncharges q1...qn at the positions ~r1... ~rn, where infinity has potential energy zero. Since the force atany point is the sum of the forces between q and one of the other charges the total potential energyis also the sum of the energy between q and each other charge:

Ep(~r) = −~rˆ

∞

n∑j=1

qqj4πε0|~r − ~rj |3

(~r − ~rj)d~s

=

n∑j=1

~rˆ

∞

qqj4πε0|~r − ~rj |3

(~r − ~rj)d~s

=n∑j=1

r′jˆ

∞

qqj4πε0r2

dr =n∑j=1

qqj4πε0r′j

where r′j = |~r− ~rj | is the distance between q and qj . The step from the second last to the last lineis possible because inside the sum we treat the interactions separately and we can therefore applyequation (7.12). Therefore the potential of the n charges is given by

ϕ(~r) =n∑j=1

qj4πε0r′j

177


We also find the potential of a continuous charge distribution ρ in a volume V by applying thesame argument as when we looked at the electric field in chapter 7.1.4 and we find the potentialat the position ~r by

ϕ(~r) =

ˆ

V

ρ(~y)

4πε0|~r − ~y|dV (~y)

where ~y is the position of the infinite small charge dq = ρ(~y)dV (~y) inside the volume.

7.2.4 Voltage

A potential difference is called voltage. The voltage between two points indicates how much energyper charge a charge gains if it moves from one point to the other. The unit of the voltage is voltV.

If one coulomb is moved between two points with one volt then this coulomb charge gains one jouleenergy. Therefore one volt is defined as 1V = 1J/1C. Once the volt is defined one can also definea new energy unit: The energy one electron gains passing one volt is called one electronvolt eV.

The quantity volt plays an important role in electric circuits (see chapter 8).

The fact that the potential is only a function of the place allows us to make an important statementabout the electric field. If we calculate the energy W we have to apply to move a charge q froma starting point ~r0 over a closed path γ with end point equal to the starting point ~r0 we see thatthe applied energy is zero W = 0 since the potential difference between ~r0 and ~r0 is zero. Since Wis proportional to the charge q the integral of the electric field along the path must be zero:

˛γ

~E · d~s = 0 (7.13)

This property makes the electric field an irrotational field (opposite to the magnetic field, seechapter 9.1.2). The statement that the electric field is an irrotational field is only true for thestatic electric field. Therefore the electrostatic field is an irrotational source field which means thatequation (7.13) holds and that the electric field begins and ends at sources which we called charge.

7.2.5 Potential and conducting material

If a body made of conducting material is placed in an external electric field the charges in thatmaterial will move according to the electric field. At the end there will be a stable state with someinteresting properties:

1. The total electric field (sum of external field and field of moved charges in the conductingmaterial) points perpendicular to the surface at every point on the surface. Because if therewould also be a parallel component the charge would accelerate in this direction and it wouldnot be a stable state.

178


2. Inside the body there is no electric field. Otherwise again charge would be accelerated whatcontradicts to the assumption of the stable state.

3. Since the electric field inside the body is zero the potential all over the body is the same.Because if we look at two points of the body and want to calculate the potential differencewe have to apply equation (7.10) and since ~E = 0 we get that the potential difference is zero

4. Inside the body there is no net charge (this means positive and negative charge have thesame density). Otherwise it is not possible to have everywhere in and on the body the samepotential. Therefore all charge is on the surface of the body.

7.2.6 Capacity

Let’s take two bodies made of conducting material which have no net charge on each of them. Wecall them electric neutral (so the negative and positive charge have the same amount). If we takesome charge from the first body and put it on the second body there will be a potential difference∆U between this bodies. Taking a very small charge q (so small that it does not influence thepotential or the electric field) and moving it from the one to the other body we recognize that nomater which way we take, the absolute value of the energy is always |∆Uq|. Therefore the chargeon each body is distributed in a very particular way such that the voltage between two points onthe two bodies is always ∆U . If we put more charge from the first to the second body the voltagewill be bigger but the energy gain displacing q from one body to the other is still independent onthe path. This means that the way the charge is distributed is the same but with a bigger amountof charge. Therefore the direction of the electric field does not depend on how much charge was putfrom the first on the second body, only the strength of the electric field depends on this amountand the sense (if it is pointing in one or the other direction, depends on the sign of the chargewe displaced). Since the electric field is proportional to the charge ∆Q displaced from the firstbody to the second, we have a proportional dependence between the charge on the bodies and thevoltage between them:

C ·∆U = ∆Q

where C is the proportional constant which is called capacity. The capacity only depends on thegeometry of the two bodies.

7.2.7 Example

Let’s look at some examples to get used to the theory.

Plate capacitor

Putting two metallic plates with area A each and both parallel to the y-z plane in a distance d(d A) we get a plate capacitor (see figure 7.5). Let us take the charge Q from the left plateand put it on the right plate. Since d A we model the electric field of this setup as one of

179


infinite spread plates. This means the electric field between the plates has everywhere the samestrength and the same direction2. Such a field is called homogeneous field. Additionally we definethe charge density σ by σ = Q

A .

Figure 7.5: Plate capacitor [36].

We know from chapter 7.1.6 the electric field is perpendicular to the plates and has everywherethe same strength. Therefore we get the voltage U between the two plates by

U =

right plateˆ

left plate

~Etotd~s

=

right plateˆ

left plate

~Eleft plated~s+

right plateˆ

left plate

~Eright plated~s

= Eleft plated− Eright plated

=QA

2ε0d−

−QA

2ε0d =

Qd

Aε0

Therefore the capacity C is given by

C =Q

U=Aε0d.

This problem was easy to solve because of the homogeneous electric field. Because in case of an ho-mogeneous electric field the voltage between two points ~r1 and ~r2 is simply given by U = ~E ·(~r2−~r1).

2This is because the plates are infinite spread out. Therefore the electric field must everywhere look the same(translation and rotation symmetry).

180


If the connection line of the two points lies parallel to the electric field this simplifies even more toU = | ~E|d where d is the distance between the two points.

Additionally we can calculate the energy Epot stored in the capacitor. In principle the energy isgiven as Epot = QU . But we have to pay attention because if we load the capacitor, the voltageand the charge changes and we have to add the potential energy of the different stages of thecharging capacitor by taking the integral

Epot =

Q

0

U(q) dq

=

Q

0

q

Cdq =

[q2

2C

]Q0

=Q2

2C

where we used the relation between the voltage and the charge given by C = QU . Instead of

integrating with respect to the charge we can also do it with respect to the voltage and we get

Epot =

U

0

Q(u) du

=

U

0

Cudu =C

2U2 =

Q2

2C.

There is another approach to calculate the energy stored in the capacitor. Assume we have twocharged plates with charge Q on each. Assume they are separated by a small distance δ. The forcefrom one plate of the other is F = EQ = Q

2ε0AQ where A is the area of the plates. We now can

compute the energy to pull one plate to the distance d to the other plate. This energy is given as

Epot =

dˆ

δ

F ds

=

dˆ

δ

Q2

2ε0Ads =

Q2

2ε0A(d− δ).

If we now set δ = 0 and use C = Aε0d we recover the stored energy from above: Epot

Q2

2C .

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Potential of an infinitely long wire

As we have seen in chapter 7.1.6 the electric field of an infinite long wire with charge density λ is~E = λ

2πε0r~er with ~er the unit vector pointing radial away from the wire. To calculate the voltage

between two distances R1 and R2 we integrate radially from R1 to R2 (as a consequence ~E and d~rare pointing parallel).

ϕ(r) =

R2ˆ

R1

~Ed~r

=

R2ˆ

R1

Edr

=

R2ˆ

R1

λ

2πε0rdr

=λ

2πε0[ln(r)]R2

R1=

λ

2πε0(ln(R2)− ln(R1))

7.3 Current and magnetic field

Current is basically moving charge. A current produces an additional field, called magnetic fieldwitch can be measured. The existence of the magnetic field can be predicted using relativity butthe calculation is far beyond the stuff for this chapter so we make a phenomenological approach.

7.3.1 Current and conservation of charge

If charge is moving one speaks of a current. The unit of the current I is Ampere A. The precisedefinition of the current is I = dQ

dt Where Q is the charge that passes at a certain point.

An important concept of electrodynamics is the conservation of charge. This means that the totalcharge of a closed system can not change. It is for example possible to have a charge neutral atomand take away an electron. But then the atom is positively charged and the total charge is stillzero, i.e. the sum of both charges, the electron and atom. Therefore the charge at a point canonly change if a current flows to that point. On the other hand a current starts and ends at pointswhere the charge changes or the current is a closed circuit.

7.3.2 Magnets

Everybody has already seen a magnet, which looks like small pieces of metal. This magnets arecalled permanent magnets since they are always magnetic. There exist also magnets that work

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Physics Olympiad Script 7.3. CURRENT AND MAGNETIC FIELD

with current and which are called electro magnets (see 7.3.3). A magnet produces a magnetic fieldwhich is somehow similar to the electric field but has some very important differences. A magnethas always two poles, this are the parts of the magnet where the magnetic field leaves or enters themagnet. The north pole is the part where the magnetic field leaves the magnet and the south poleis where the magnetic field enters the magnet (see figure 7.6). Inside the magnet the field linesgo from the south to the north pole, they build therefore closed filed lines (see chapter 9.1.4 andequation (7.14)). The names of the poles come from the fact that the earth has also a magneticfield and the north pole of a magnet is attracted by the geographic north pole and the south poleof the magnet is attracted by the geographic south pole.

Figure 7.6: Magnet with the magnetic field which leaves the magnet on the right, side where thenorth pole is, and enters the magnet on the left side, where the south pole is [38].

As we know from electrostatics field lines want to be as short as possible and they repel each other.If we now put two magnets together (see figure 7.7), we see from the total magnetic field that thesame poles repel each other and two different poles attract each other. Because in the first casethe field lines repel each other and as a consequence also repel the two magnets. In the secondcase the field lines can build nice closed loops which want to get shorter and therefore attract themagnets.

We want to formalize the magnetic field a bit. What we called magnetic field is formally the mag-netic flux density ~B with unit Tesla T. The connection to other SI-units is 1T = 1kg·A−1·s−2 =1W·s·A−1·m−2. One Tesla is a pretty strong field, for example the magnetic field of the earth isabout 5 · 10−5T and a usual magnet produces a field in the order of 0.1T.

One important difference to the electric charge and field is that there exist no magnetic monopoles.This means that it is not possible to separate the north from the south pole, they build always aninseparable pair (see chapter 9.1.4). But this also means that magnetic field lines have no startand no end, therefore the magnetic field is a source free rotational field. If we adapt to Gauss lawin electrostatic (see chapter 7.1.5) we find the law

‹∂V

~Bd~S = 0 (7.14)

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Figure 7.7: Two magnets with the field lines of the total magnetic field. [39].

7.3.3 Magnet and electric current

If one puts a permanent magnet which is freely moveable near a wire where current flows one canobserve that the permanent magnet orient itself in a particular way which is shown in figure 7.8.

Figure 7.8: Permanent magnets oriented along a wire where a current flows. [40].

If we now think the magnetic field being such that the magnets are tangent to the field we get thatthe magnetic field looks like in figure 7.9. The rule is the following: If one takes the right handand places the thumb in the direction of the current then the other four fingers show the directionof the field.

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Physics Olympiad Script 7.3. CURRENT AND MAGNETIC FIELD

Figure 7.9: Magnetfield around a wire. [41].

7.3.4 Lorentz force

Putting a wire in a homogeneous magnetic field and letting flow current trough the wire, thehomogeneous magnetic field and the magnetic field of the wire superpose to a total magnetic fieldwhich is shown in figure 7.10.

Figure 7.10: The homogenous field and the field from the wire are indicated by the dashed line.The total field is indicated by the continuous lines. The cross on the wire shows that to current flowsinto the page. [42].

From the picture it is obvious that on the left side of the wire the magnetic field is pushed moretogether than on the right side. Since magnetic field lines repel each other there is a force pushingon the left side of the wire. On the other hand the field lines on the right side are not straight linesbut a bit curved. Since field lines want to be as short as possible they attract the wire. This forceon the wire in the right direction is called Lorentz force ~F . Formally it is described by the formula

~F = I~l × ~B (7.15)

where ~l is the direction of the current flowing in the wire and ~B is the magnetic field. If we want tocalculate the force on a single charge q we use the mathematically not really precise but intuitivelycorrect equation

185


I~l =dqdt~l =

dq~ldt

= qd~ldt

= q~v

where ~v is the velocity of the charge. Therefore we get that the force on a moving charge is

~F = q~v × ~B (7.16)

It is important that a charge which is not moving has no Lorentz force which is obvious since onlya moving charge has a magnetic field and a magnetic field can only interact with an other magneticfield and not with an electric field. If one looks at electric and magnetic fields which change in timethen a changing electric field produces a magnetic field and a changing magnetic field produces anelectric field. This is described by the Maxwell equations which are too complicated to be treatedhere. Therefore an electric field can only interact with a magnetic field if the electric field changeswith time since then the electric field produces a magnetic field which can interact with an othermagnetic field, or the magnetic field changes and produces an electric field which can interact whichthe electric field. Since we (nearly) always look at static systems (which do not change with time)we will not need the Maxwell equations.

186

Chapter 8

Direct current circuits

Besides the theoretical treatment of electric and magnetic fields there is also a very practicaluse. All the electronic devices base on the laws of electrodynamics. Since the calculation withthe basic equations of electrodynamics is often very complicated one simplifies the calculation ofelectric circuits by introducing electrical components. The laws of electrodynamics then define thebehaviour of the components (overview see chapter 8.4). We will now look at electrical circuitswhich have a constant voltage (and most also a constant current). This sort of circuits are calleddirect circuits. There exist also alternating current where the voltage and the current change withtime but this is not treated here.

8.1 Ohm’s law

Applying a variable voltage U to a body the current I through the body might depend on manyinfluences as temperature or humidity. The quotient U

I is called resistance of a body. The simplest(not trivial) dependence is the proportional dependence: The current is proportional to the voltagewith proportionality constant R with U = RI. R is called ohmic resistance. The linearity is justa model which is valid for many materials and bodies. Mathematically it is not completely wrongto describe the dependence between U and I by a linear function because any (nice) function canbe approximated by a linear function. But there exist also components which have a non lineardependence as the light bulb or the diode. The light bulb is a typical example of the dependenceof the temperature on the resistance of a material. For metallic materials the resistance is higherif the metal is warmer and since the wire of a light bulb gets hotter if a higher voltage is appliedone can recognise that the resistance is higher at the higher voltage.

8.2 Equivalent circuit

Sometimes it is possible and useful to describe a collection of components by a single component.Since there are many possible combinations we want to look at the most important.

187

Physics Olympiad Script CHAPTER 8. DIRECT CURRENT CIRCUITS

8.2.1 Wire

A real wire usually has a little resistance. Since this resistance is spread all over the wire it iscumbersome to describe the wire as a chain of little ohmic resistors. Instead one adds to an idealwire (with no resistance) a single ohmic resistor which describes the total resistance of the wire.

Since two wires also have a capacity one could additionally add a capacitor to a closed circuit. Asone can imagine it is nearly impossible to describe all effects and it is often not necessary to takingall into account but to consider the relevant ones.

8.2.2 Series circuit

If two ohmic resistors are connected one behind the other one talks about a serial circuit (see figure8.1). Since the current I trough R1 and R2 is the same, the voltage drop over both resistors isU = R1I +R2I = I(R1 +R2). Therefore the resistance of both resistors is R = R1 +R2.

By the same argument one can calculate the total resistance Rtot of an arbitrary number of resistorswhich are all connected in series. For n resistors R1, R2, ...Rn the total resistance is given byRtot =

∑nj=1Rj .

Figure 8.1: Left: serial circuit of two resistors. Right: parallel circuit of two resistors.

8.2.3 Parallel circuit

If we put two resistors parallel to each other we get a parallel circuit (see figure 8.1). The voltageU over both resistors is the same and the total current that flows through the two resistors isItot = U

R3+ U

R4. Therefore the total resistance is

R =U

Itot=

11R3

+ 1R4

Similarly the resistance of n parallel connected resistors is given by

Rtot =

n∑j=1

1

Rj

−1

188

Physics Olympiad Script 8.3. ELECTRIC POWER

8.2.4 Voltage source

An ideal voltage source is a device where independent of the current the voltage is always thesame. Since already the wires leaving the ideal voltage source have a resistance one has to addan additional resistance RS in series to the voltage source to describe a real voltage source. RS isusually very small one can often neglect it. Taking RS into account is only important if one wantstake out a lot of energy from the voltage source or if the rest of the electrical circuit has a verylow resistance in the order of RS .

8.3 Electric power

The voltage describes how much energy one Coulomb gets if it passes the voltage. A currentdescribes how much charge passes per unit of time. Therefore the product of a voltage and acurrent describes how much charge gets an energy by the voltage per unit time, therefore theproduct tells us the power that the current performs.

P = UI

where P is the power that the current I emits over the voltage drop U .

189


8.4 Electric components

Table 10.1 gives an overview over the most common symbols.

ideal wire (with no resistance)

switch to open and close the electric circuitohmic resistor

variable ohmic resistor

(ideal) voltage source

ground (reference voltage in infinity), connected to the earth

capacitor (to store charge)inductor (inductive resistor, often a coil)

light bulb

LED (light emitting diode)

diode (lets current only flow in direction of the arrow)

horn

Table 8.1: Overview over different symbols [46]

8.5 Kirchhoff’s circuit law

There are two very important rules which can be used to determine the current and voltage throughany circuit.

8.5.1 Current law

Because there is conservation of charge we have a restriction on the currents: At any point whereno charge is stored the sum of all currents must be zero. It is important to chose the currents thatflow to that point with one sign (so positive or negative) and the currents that flow away with theother sign (negative or positive) (see figure 8.2).

8.5.2 Voltage law

As we have seen in chapter 7.2.4 the sum of all voltages in a closed electric circle is zero. Thereforewe define a direction of summation (in clock wise or anti clock wise) and take the sum over allelectric components of the circuit (see figure 8.3)

190

Physics Olympiad Script 8.5. KIRCHHOFF’S CIRCUIT LAW

Figure 8.2: Knot where many currents flow together. The sum I1 − I2 + I3 − I4 − I5 must be zero.The currents that flow to the knot have positive sign, the currents that flow away negative sign.

Figure 8.3: Here the sence of summation is clock wise. For all the components (which does notnecessarily have to be ohmic resistors) we take the potential difference between the potential at theend of the arrow minus the potential on the start of the arrow. This gives us the voltage in directionof the summation. The sum of all these voltages must be zero U1 + U2 + U3 + U4 + U5 = 0 [44].

8.5.3 Applying Kirchhoff’s law

A knot is a point in the electrical circuit where more than two currents flow. For every knot weapply the current law. This gives us for k knots k independent equations. For every closed circuitone applies the voltage law which gives for every independent closed circuit one more equation.One has to pay attention on the independence of the circuits (see figure 8.4)If one has now n equations (from the current and voltage law) one has to express the voltages bythe currents ore vice versa. Then one should have n equations with n variables which should besolvable.This is a very useful method for complicated circuits. For easier circuit one can apply othermethods, for example simplify complicated systems of resistors by a single resistors (see 8.2) andthen apply the Kirchhoff’s voltage law by saying that the voltage over the voltage source must beequal to the voltage over the single resistor.

191


Figure 8.4: There are three closed circuits: circuit 1: ACDB, circuit 2: AEFB and circuit 3: CEFD.But the three circuits are not independent because circuit 3 is basically circuit 2 minus circuit 1.Therefore the voltage law must only be applied on two of the three circuits (it does not dependon which two). To be sure not doing anything wrong: only take the closed circuits which are thesmallest possible circuits. For example circuit 3 can be contracted to circiut 2 by making a shortcutfrom B to A.[45].

8.6 Examples

8.6.1 Maximal power from a real power source

Let’s look at a real voltage source with voltage U0 and connect it to an ohmic resistor (see figure8.5). Since the voltage source has an ohmic resistance itself it is not possible to get infinite powerout of the source.

Figure 8.5: A real voltage source connected to an ohmic resistor. The real voltage source is the graybox.

We want to calculate the maximal power that one can use at the resistor. The power P on theresistor R is P = UI = I2R. The current is given byI = U0

R+RS. Therefore the power is U2

0R

(R+RS)2

To get the maximal power we consider the power as function of R and set the derivative zero:

0 =dPdR

= U0(R+RS)2 − 2(R+RS)R

(R+RS)4

= R+RS − 2R

R = RS

192

Physics Olympiad Script 8.6. EXAMPLES

Therefore one can take the maximal power out of a real voltage source if the resistor is equal tothe interior resistor of the voltage source. Of course the same amount of power as one can use atthe resistor R is heating up the voltage source because of RS .

8.6.2 Charging a capacitor

A capacitor with capacity C is connected in series with a resistor with resistance R to a voltagesource with voltage U0. At the time t = 0 the switch closes the circuit and the capacity starts tocharge (see figure 8.6).

Figure 8.6: Circuit to charge a capacity

We want to calculate the voltage over the capacity VC as a function of time. We apply Kirchhoff’slaw: U0 = UR +UC with UR = RI the voltage over the resistor and UC = Q

C with Q the charge inthe capacitor. Taking the time derivative we get

0 = RdIdt

+I

C

= RCdIdt

+ I

− 1

RCdt =

dII

− 1

RC

tˆ

0

dt′ =

I(t)ˆ

I0

dII

− 1

RCt = ln(

I(t)

I0)

I(t) = I0e−1RC

t

Where I0 is the current when the switch is closed and it is given by I0 = U0R because at the first

moment no voltage drops over the capacitor (since it is empty) and therefore all the voltage dropsover the resistor. Therefore we get for the voltage over the capacitor

UC(t) = U0 −RI(t)

= U0(1− e−1RC

t)

193


This contains some expected properties: At the beginning when there is no charge in the capacitorthere is no voltage drop over it. For t very big nearly all voltage drops over the capacitor which isalso clear since the capacitor is an interruption of the circuit for a constant voltage.

194

Chapter 9

Electrodynamics 2

In this chapter we continue electrodynamics. First we formalise magnetism, which will lead toAmpère’s Law and Biot Savart’s Law. Then we look at time dependent fields and phenomenarelated to them (for example induction). Finally everything is put together which leads to thefamous Maxwell’s equations.

9.1 Magnetism

In electrodynamics 1 we looked at magnetism more from a phenomenological point of view withoutformal relations. We will catch up on this now. We remember that a wire, where a current is flowingthrough, creates a magnetic field around the wire (see 7.3.3). It is a circular field around the wire.

9.1.1 Magnetic field and Flux

In section 7.3.2 we gave the magnetic field ~B already a unit, namely the Tesla. To be more precisely~B is the magnetic field density1. Visually spoken the field density indicates how much field linespass a certain area. For a given area A, we can therefore define a quantity, which corresponds tothe total number of field line through that area. This quantity is called flux Φ and it is defined as

Φ =

Ä

~B · d ~A

where the differential d ~A is a small area element pointing perpendicular to the surface. If thearea is not curved and the magnetic field is homogeneous through the whole surface the formulasimplifies to

Φ = ~B · ~A

where ~A is the surface vector pointing perpendicular to the surface and has a length equal to thearea of the surface.

1There is also a magnetic field strength, analogous to the electric field strength ~E. This plays a less importantrole in Physics

195

Physics Olympiad Script CHAPTER 9. ELECTRODYNAMICS 2

9.1.2 Ampere’s law

Similar to Gauss’s law (see chapter 7.1.5) where the flux at the surface of a volume only dependson the charge inside we can formulate a law with currents. Instead of the flux though a closedsurface we look at the magnetic field along a closed path (enclosing a surface S). If we have a wirewhere a current I flows through and a surface S (that is not closed) we have that

˛∂S

~Bd~l = µ0Iin =

¨Sµ0~jd~S (9.1)

where ∂S is the boundary line of the surface S and d~l is a infinitesimal short tangent vector onthat boundary line (see figure 9.1). Iin is the current that flows through the surface S and ~j is thecurrent density (current per area) which points in the direction of the current flow. The directionof d~l is given by the right hand rule: if the thumb of the right hand shows in the direction of thecurrent then the other four fingers show the direction of d~l and d~S shows in direction of the current(see also figure 7.9).It is again not important being able to calculate the integrals above for arbitrary cases but it isvery useful to understand the formula.

Figure 9.1: Current flowing through surface S with boundary line along which it is integrated. [43].

The important thing about equation (9.1) is that there exist a quantity connected to the magneticfield (namely the integral on the left hand side) which only depends on the current. In symetriccases this is very useful, see examples 9.1.4

9.1.3 Magnetic field of a moving point charge

An other way to calculate the magnetic field of a current is to look at the explicit dependence ofthe magnetic field on a moving charge. Assume a point charge at the position ~r′ with charge q ismoving with velocity ~v. We want to understand the formula for the magnetic field at a point ~r ofthis point charge. The formula is

~B(~r) = kq

r2~v × ~r − ~r′

|~r − ~r′|(9.2)

196

Physics Olympiad Script 9.1. MAGNETISM

where ~r − ~r′ is the vector pointing from the point charge to ~r and k is a constant. In SI unitsk = µ0

4π with µ0 = 4π · 10−7V·s·A−1·m−1. This formula seems extremely complicated in thefirst moment but looking a bit more precise and comparing it with the electric field of a pointcharge, it gets a lot simpler. The ~B field has a q

r2dependence as the coulomb law. This is

pretty reasonable since as mentioned above the electric field and the magnetic field have astrong connection2. This means that both should have the same dependence to q and to thedistance |~r|. The 4π is part of the constant and is separated with a similar reason as in theelectric case, where 4πr2 is the area of the sphere with radius r around the charge with radius|~r|. Additionally some other laws take a nice form, see equation (9.1). The big differencebetween the formula of the electric and the magnetic field is the vector product. But thevector product fulfils exactly the properties of the magnetic field we stated in chapter 7.3.3:The vector product makes that the magnetic field always points tangent to a circle aroundthe direction of the current (which is here ~v) and since the magnetic field should be zero for ~rin line with ~v the angle between ~v and ~r−~r′ plays also a role and is respected in the vector product.

The formula above is not correct with respect to relativity because in the formula it is assumedthat the moving particle has an immediate influence at the position ~r which is not possible due torelativity. But if we consider |~v| << c, with c the speed of light, the mistake is negligibly small.

To calculate the magnetic field of a current flowing through a wire we use equation (9.2) andredefine some quantities. The charge dq in a short part of the wire with length dl is dq = ρdlwhere ρ is the charge density per unit length. Assume that charge is moving with a (mean) speedv. To pass the length dl the time dt is needed. We therefore have a current I = dq

dt = ρdldt = ρv.

Therefore we have dqv = ρdlv = Idl. Since the charge dq is assumed to be small, and located ata small spot, we can use equation (9.2) to calculate the magnetic field d ~B caused by the currentat ~r′ through the small piece of wire d~l(~r′). We turned the path element dl into a vector in orderto calculate the vector product. The vector has to point in the direction of the current flow. Theformula is then given as

d ~B(~r) =µ0

4πI

d~l(~r′)× (~r − ~r′)|~r − ~r′|3

.

This formula is called Biot-Savart law. The total magnetic field at the point ~r is then the integralof all the d ~B of all the wire, namely

~B =

ˆwire

d ~B =

ˆwire

µ0

4πI

d~l(~r′)× (~r − ~r′)|~r − ~r′|3

.

This integral does basically nothing else than sum up all the contributions of the different partsof the wire. If the current through the wire is constant, the equation above is also relativisticallycorrect, because the magnetic field is constant in time.

2This gets obvious in relativity.

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9.1.4 Example

magnetic field of an infinitely straight long wire

Consider a wire along the x-axis with a current I flowing through in the positive x-direction. Sincethe wire is infinitely long there is no component of the magnetic field pointing in x-direction.Additionally the problem is rotational symmetric around the x-axis. Therefore the strength onlydepends on the distance to the wire. We imagine a circle around the x-axis with radius R. As themagnetic field also makes circles around the wire (see figure 7.9) the line vector of the boundaryline of the circle and the ~B field point in the same direction and therefore ~Bd~s = Bds where Band s are the absolute values of the respective field. Therefore equation (9.1) leads to

µ0I =

˛∂S

~Bd~s

=

˛∂SBds

= B

˛∂S

ds

= B2πR

B =µ0I

2πR.

The B can being taken out of the integral since B is constant at a constant distance r from thewire.

Force on two parallel infinitely long wires

Let’s take a wire along the x-axis and one parallel to the x-axis through the point y = r > 0 andz = 0. Assume that currents I1 and I2 flow through the first and second wire, respectively. Wetake them to be positive if they flow in +x-direction. According to the example above the firstwire produces a magnetic field at the position of the second wire

~B =µ0I1

2πr

001

.

Therefore the force on a length l on the second wire is due to the Lorentz force

~F = I2

l00

× ~B

=µ0I1I2

2πr

l00

× 0

01

=µ0I1I2l

2πr

0−10

.

Therefore the two wires attract each other if both currents flow in the same direction and theyrepel each other if the two current flow in opposite direction.

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Physics Olympiad Script 9.1. MAGNETISM

The formula above is also used to define the SI-unit for electro magnetism:"The ampere is that constant current which, if maintained in two straight parallel conductors ofinfinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, wouldproduce between these conductors a force equal to 2 · 10−7 newton per metre of length." [37]This definition is also the reason why µ0 = 4π · 10−8V·s·A−1·m−1 is a precise constant.

Magnetic field of a coil

If we wind a wire to circles and place them close together we get a coil. We now look at a coilwhose diameter which is much smaller than its length. To calculate the magnetic field of a coil weconsider a rectangle with width L and length l which we place as seen in figure 9.2.

Figure 9.2: Coil with gray rectangle. The surface vector of the red rectangle is pointing into thesheet because the current through the recangle points into the sheet. [43].

Let n be the number of windings per unit length. By amperes law (see equation (9.1)) we get

µ0Itot =

˛∂S

~B · d~s

µ0InL = LB

B = µnI

with Itot = nLI the total current flowing through the rectangle and I the current trough the wire.The step from the first to the second line is because if we look a the sides of the rectangle themagnetic field is nearly perpendicular to the sides of the rectangle and as a consequence ~B ·d~s = 0for the sides. Additionally we make l very large and at the top of the rectangle the magnetic fieldis very weak and therefore the contribution can be neglected.

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The magnetic field of a coil is similar to the one of a permanent magnet. Therefore one couldexplain the magnetic field of a permanent magnet by assuming to have little circular currentsin the magnet. This explanation is not really true since in quantum mechanics there exist alsomagnetic fields which do not occur because of currents. But one can imagine how the magneticfield looks inside a permanent magnet where the field lines go from the south to the north pole. Thefield lines inside and outside the permanent magnet build closed lines as it is stated by equation(7.14).

9.2 Induction

Induction is an important phenomenon in electrodynamics because it gives a (first) connectionbetween the electric and magnetic field. It basically tells us that a magnetic field that changeswith time produces an electric field3.

9.2.1 Approach and Definition

To approach induction we consider a metallic pipe that is falling down. In that pipe a magnet iskept fix (see figure 9.3). In the metal there are charges (positive nucleus and negative electron gas).On these moving charges the Lorentz force is acting4 and accelerates the electrons. They startcircling in the pipe such that the magnetic field they produce opposes the change of the externalmagnetic field. (see also figure 9.4).

Figure 9.3: Left: The pipe is falling and the magnet is kept fix. The white circles with the crossor the point indicate the direction of the current due to the Lorentz force. Right Frame of the pipe:The pipe stands still and the magnet is moving. The curved arrows indicate again the current in thepipe.

3Also the opposite is true, see 9.3.4Since the nuclei are much heavier than the electrons the influence of the magnetic field is much smaller.

200

Physics Olympiad Script 9.2. INDUCTION

In the rest frame of the pipe the pipe itself is not moving. As a consequence there is no movingcharge and therefore no Lorentz force acting. But the current needs to be independent of theframe. The explanation of this is induction: Since the magnet is moving (in the frame of the pipe),the magnetic field at a certain position of the pipe is changing with time. This change causes anelectric field, which causes the movement of the charges.

Figure 9.4: Top of the pipe. The pipe is the gray ring, the current (and the electric field) is indicatedas circle with arrows.

If we integrate the electric field along a closed path (for example the one indicated in figure 9.4)we get a voltage uind. This voltage is connected to the magnetic field by

uind = −dΦ

dt(9.3)

where Φ is the flux that flows through the closed path. The obtained equation (9.3) has some veryimportant properties:− The minus sign corresponds to the fact that the current is such that it opposes a changing

of the magnetic field. It represents therefore energy conservation.− Opposite the electrostatic case, the electric field we obtained above has no starting and no

ending point. The work required to afford to change the position of a charge depends onthe path, therefore one cannot find a potential which is equal to the potential energy of aparticle.

9.2.2 Self induction

In most of the electrical components and in most cases there is a weak interaction between thecomponent and the magnetic field. Nevertheless there is one component that couples very strongto the magnetic field. This component is the coil. Since the different windings of a coil are closeto each other and the same current flows through each winding, there is a lot of charge moving ona small spot. This current produces a strong magnetic field (see section 9.1.4). Since the magneticfield is proportional5 to the current I and since the cross section of a coil does not change withtime, the total flux Φ trough the coil is proportional to the current Φ = LI. The proportionalityconstant L is called inductance of the coil.

5This is not always the case, for example if the magnetic field goes through a ferromagnetic material. TThere socalled saturation can occur.

201


If we apply an alternating current to a coil, the magnetic field through the coil (caused by thecurrent) is also alternating. Therefore induction happens. The induced voltage in the current is

uind = −dΦ

dt= −Ldi

dt.

The coil therefore opposes a change of the current by inducing a voltage. For more details see alsothe AC impedance of an inductor (see section 10.2.3). This phenomenon is called self induction.

9.3 Displacement current

In the previous section we discussed how a time dependent magnetic field causes an electric field. Inthis section we will look at the opposite, namely a time dependent electric field causing a magneticfield.Consider a capacitor that gets loaded with a current I (see figure 9.5). The current causes amagnetic field B around the wire. If we only consider the current, the magnetic field between theplates would be smaller than outside the capacitor. But that’s not the case, because the currentloads the capacitor, therefore the electric field E grows with time. This changing electric field alsocauses a magnetic field. Outside the capacitor, one can not tell from the magnetic field whetherthere is a current or a changing electric field creating that magnetic field.

Figure 9.5: A plate capacitor (grey plains) gets charged by a current I. The current causes amagnetic field B. Between the plates there is no current, but the changing electric field E causesalso a magnetic field.

202

Physics Olympiad Script 9.4. MAXWELL’S EQUATIONS

To get a formal description consider a plate capacitor with capacity C = εAd where each plate hasa cross section area of A and the plates are separated by d. From the basic equation for capacitorswe can relate the current and the changing electric field by

Q = CU,

I =dQ

dt= C

dU

dt

= εA

dd

dE

dt

= εAdE

dt

where we used that the electric field E and the voltage U (for homogeneous electric field) dependas U = Ed. This "current" I between the plates is called displacement current and it has the sameeffect as a usual current.As a consequence we have to take the displacement current also in account in Ampère’s law (seesection 9.1.2). This leads then to

˛∂S

~Bd~l = µ0Itot = µ0

(I + εA

dE

dt

)=

¨Sµ0~j + µ0ε

d ~E

dtd~S.

The last line is the most general description where we assume an arbitrary area S with boundary∂S and a current density ~j. If the current density ~j is zero, the formula above is very similar tothe induction formula: The change of the flux through an area S is proportional to the integral ofthe magnetic field around a closed circle (see also section 10.2.3).

9.4 Maxwell’s equations

In this section we summarize the basic equations we found in electrodynamics. They will form aset of integral equations6 that describe the behaviour of the electric field ~E and the magnetic field~B. This set is called Maxwell’s equations.We use the following notation: With V we denote a volume and ∂V is the surface7 that confinesthe volume V . A small element of the volume is dV . To integrate over the surface of V we need tosplit ∂V into small pieces denoted by d~S where the area of the small piece is equal to the absolutevalue |d~S| and the direction of d~S points perpendicular to the surface outside the volume.With A we denote an area and the border of the area8 is denoted by ∂A which is a closed line. Thearea is again split into small pieces d ~A. We also divide the closed line into small pieces denoted d~l.The length of these pieces is equal to the absolute value |d~l| and it points tangent to the line. Thedirection of the small area pieces d ~A and the small curve pieces d~l need to fulfil the right handrule: If d ~A points in the direction of the thumb of the right hand, the other fingers of that handindicate the direction the d~l have to point.

6One can also write down equivalent differential equations but they are more complicated.7Take surface ∂V as notation for the surface and not as partial derivative or small piece of V .8This are has not to be closed (as surface of a volume), that is why we use another variable than above.

203


With this notation Maxwell’s equations are given as

‹∂V

~E d~S =

˚VρdV Gauss’s law,

‹∂V

~B d~S = 0 no magnetic monopoles exist,˛∂A

~E d~l = − d

dt

Ä

~B d~S Faraday’s law (Induction),˛∂A

~B d~l = µ0

(Ä

~j d ~A+ ε0d

dt

Ä

~B d ~A

)Ampère’s law,

where ρ is the charge density and ~j the current density. The circles at the integrals on the left sidedenote that the surface or the path is closed (therefore the surface of a volume or the boundaryof an area). The number of integral signs denotes the dimensionality of the integral, meaning thedimension of the space the integral as to be taken.Additionally to Maxwell’s equations one has to mention the Lorentz force in order to describe theinteraction between charges and the fields. The Lorentz force is given as

~F = q( ~E + ~v × ~B)

where q is the charge of a particle and ~v its velocity.

With these 5 equations it is basically possible to describe any problem involving charges and theelectromagnetic field.

204

Chapter 10

Alternating current (AC)

When the electric voltage periodically oscillates there are a couple of phenomena which can beobserved but do not appear when a constant voltage is applied. In the following chapter we willdescribe sinusoidal voltages and the behaviour of electric devices when they are connected to it.The main difference to constant voltages is that there exist devices whose resistances depend onthe frequency of the voltage. In particular we will have a look at ohmic resistors, capacitors andinductors and combinations of them.

10.1 Describing alternating voltage and current

Since the voltage (and therefore also the current) oscillates with time the voltage is not simply anumber but a function of time. There are a couple of possibilities to describe this and we will havea look at the most important ones in the next sections.

10.1.1 Fourier series

There is an important theorem in mathematics which states that any periodical function f(t) canbe decomposed in a sum fn(t) of harmonic oscillations1. This theorem is called Fourier’s theoremand the decomposition is called the Fourier series.Let’s formulate this a bit more mathematically: Let f(t) be a periodic function with period T .This means that for any t, f(T + t) = f(t). Then the theorem states that this function f(t) canbe written as

f(t) =

∞∑n=0

An sin(ω0nt) +Bn cos(ω0nt)

where ω0 = 2πT is the angular frequency and An and Bn are constants which depend on the periodic

function f(t). The whole theory about Fourier series is to complicated to be treated here. Theimportant point is that if we know the behaviour of a device for a sinusoidal voltage we canconstruct its behaviour on any other periodic voltage. So from now on we always consider thevoltages to be sinusoidal unless explicitly mentioned.

1Harmonic oscillations are nothing else than sinusoidal oscillations.

205

Physics Olympiad Script CHAPTER 10. ALTERNATING CURRENT (AC)

10.1.2 Usual real notation

A sinusoidal voltage takes the general form

u(t) = U0 cos(ωt+ ϕ)

where U0 is the amplitude of the oscillating voltage, ω = 2πT is the angular frequency and ϕ is the

phase. Of course we could also chose the sine instead of the cosine with a different phase but thisdescription is more consistent with another description (see section 10.1.4). It is convenient in ACnotation to write time dependent quantities with lower-case letters (u(t)) and time independentquantities with capital letters (U0, Ueff ).We assume that the voltage oscillates already a long time. Therefore the whole physics does notchange if we shift time. A shift in time corresponds to an additional phase. It is often easier forcalculation to shift time such that the phase for the voltage or the current is zero. If we take aparticular choice of the phase, it is not guaranteed that other quantities have the same phase.

10.1.3 Phasor

If the voltage oscillates harmonically, one can see the voltage u(t) at a certain time t as cathetusof a rectangular triangle (see picture 10.1) with hypotenuse length U0. Let us now rotate thehypotenuse of the triangle with angular frequency ω in the mathematical positive orentation,which is anti-clockwise. Then the cathetus of the triangle behaves exactly like the alternatingvoltage. This means the whole information of the alternating voltage is encoded in the vector inpicture 10.1: The amplitude, the angular frequency and also the phase are given. Therefore wecan imagine alternating current as a vector which rotates with constant frequency, and by lookingat its projection on the x-axis we get the familiar real notation.

10.1.4 Complex notation

A phasor corresponds to a two dimensional vector which has its starting point at zero and itsendpoint is an x and y coordinate. We can now associate this two dimensional vector to a complexnumber z = x + jy ∈ C where j is the imaginary unit2 and x, y ∈ R are real. Therefore a phasorcan be associated to a complex number with radius r =

√zz =

√x2 + y2 with a turning phase.

This can easily be written as (see section 1.4.2)

z = rej(ωt+ϕ) = r cos(ωt+ ϕ) + jr sin(ωt+ ϕ).

In the case of z representing a voltage, r is the amplitude. Therefore r = U0. The voltage is theprojection of the phasor on the x-axis. Therefore it is the real part of z:

u(t) = Re(z) = U0 cos(ωt+ ϕ).

The complex notation might be a bit strange at the beginning because we use a non-real quantity (acomplex number) to describe something real (for example a voltage). This is not really a problembecause the complex number is just a notation. Nevertheless there are cases where one can really

2Usually i is the imaginary unit. Since i = i(t) is already the time dependent current, j is used.

206

Physics Olympiad Script 10.1. DESCRIBING ALTERNATING VOLTAGE AND CURRENT

Figure 10.1: Phasor u(t) with hypotenuse length U0. The phasor rotates around the origin withthe angular frequency ω. The projection on the real axis corresponds to the measured quantity (forexample the voltage).

think about "turning voltages" and then the phasor and also the complex notation as a descriptionof the phasor get some real properties (see 10.5.1).In this script we use the complex notation, therefore we write a voltage as u(t) = U0e

j(ωt+ϕ) andkeep in mind that the physical property is only the real part. The complex notation allows us alsoto add or subtract complex numbers because the real part of the sum of two complex numbers isequal to the sum of the real parts of the two numbers. Thus for z1 = a1 + jb1 and z2 = a2 + jb2where a1, a2, b1, b2 ∈ R it follows

<(z1 + z2) = <(a1 + a2 + j(b1 + b2))

= a1 + a2 = <(a1 + jb1) + <(a2 + jb2) = <(z1) + <(z2).

This property is important when we formulate Kirchhoff’s laws, see section 10.3.1. It allows us toperform all the (linear) calculations in the complex notation as long as we do not multiply voltage

207


and current. This is the case when we look at the power3.The complex notation is often used and it simplifies the equations a lot. Nevertheless it is alsopossible to do the whole alternating current theory without complex numbers. One then has alwaysto think about the amplitude and the phase and examine the different phenomena according toboth of these quantities. With the complex notation everything can be done with one numbersince the complex number contains the amplitude and the phase.

10.2 Impedance

In this section we examine the relation between the applied alternating voltage and currentfor different electrical components. The components we examine are the resistor, the capacitorand the inductor. Most other electrical devices behave like a combination of these three basiccomponents, we discuss this in section 10.3.

The impedance is a generalisation of the resistance of an electrical device. The resistance R in adirect current circuit is the ratio of the voltage and the current: R = U

I . Since we now examinealternating currents there are more parameters of freedom than only the "amount" of voltageor current which corresponds to the amplitude. We additionally have a phase shift between theapplied voltage and the current. Therefore the impedance not only takes into account the ratio ofthe amplitudes of the voltage and current but also their phase shift. Since the voltage as well asthe current are represented by a complex function, the impedance Z is usually a complex number4

. The absolute value of Z corresponds to the ratio of the amplitude of voltage and current. Theangle between the real axis and Z corresponds to the phase shift.Assume that we have an electrical device and we measure a voltage u(t) = U0e

jωt and a currenti(t) = I0e

j(ωt+ϕ). Then the impedance is given as

Z =u(t)

i(t)

=U0

I0e−jϕ.

From this impedance we conclude that the ratio of the amplitudes is given by |Z| = U0I0

and thatthe cosinus curve of the current follows the cosinus curve of the voltage (see picture 10.2).

10.2.1 Ohmic resistor

An ohmic resistor is a device where the current is always proportional to the voltage. Thereforethere is no phase shift between the voltage and the current and the impedance of an ohmic resistortherefore is a real number R. Of course R is exactly the ohmic resistance known from the directcurrent. This means

R =u(t)

i(t)=U0

I0.

3In this case one has to first take the real part and then multiply the two quantities.4If one describes AC without complex notation one has to consider the ratio of the amplitudes and the phase

shift separately as two real numbers.

208

Physics Olympiad Script 10.2. IMPEDANCE

Figure 10.2: Sinusoidal voltage and current. The amplitude of the voltage is U0 = 9 and the one ofthe current is I0 = 4.5. The absolute value of the impedance is therefore |Z| = 2 and the phase shiftis ϕ = 1rad. The angular frequency is ω = ϕ

10 , the period therefore is T = 2πω = 20.

10.2.2 Capacitor

A capacitor is a device where charge can be stored (see electrodynamics 1 section 7.2.6). The basicequation of a capacitor with capacity C is

C =Q

U(10.1)

where Q is the stored charge and U is the applied voltage. As we have seen, the capacity doesnot depend on the applied voltage. Therefore it is constant for an alternating voltage as well. Ifwe now multiply equation (10.1) with the voltage u(t) = U0e

j(ωt+ϕ) and take the derivative withrespect to time of the whole equation we get

u(t)C = q(t)

Cdu(t)

dt=

dq(t)

dt= i(t)

CdU0e

j(ωt+ϕ)

dt= CU0jωe

j(ωt+ϕ) = jCωu(t) = i(t).

209


As a consequence we get the impedance ZC of a capacitor by

ZC =u(t)

i(t)

=u(t)

Cjωu(t)=

1

jCω= − j

Cω.

In figure 10.3 the phasor and the time evolution of the voltage and the current are shown. The −jin the impedance means that the voltage is rotated 90 clockwise5. This is intuitively clear becausethe capacitor has to be charged in order to have a voltage and it is the current that charges thecapacitor. Therefore the current is first and after that, when the capacitor is already charged a bit,one can measure a voltage. The angular frequency ω in the denominator is intuitively also clearbecause if ω is big, the capacitor gets charged and discharged fast As a consequence the current isbig and therefore the impedance small. With the capacity C in the denominator it is nearly thesame, because a big capacity can store more charge, therefore the current is big.

Figure 10.3: Phasor at time t = 0 and time diagram for a capacity. Be aware that the voltage orcurrent is the projection on the x-axis.

10.2.3 Inductor

An inductor is an electrical device with a high inductance and an ideal inductor has no resistanceand no capacity. The inductance L = Φ

I of a device is the ratio between the magnetic flux Φ andthe current I through a device (causing the magnetic field), see also section 9.2.2. In AC thereis an additional phenomenon, called self inductance. An alternating current causes an alternatingmagnetic field which induces again a voltage in that device. The induced voltage is such that itopposes an additional growth of the current.

5Multiplying a complex number with j rotates that number by an angle of 90 in the positive orientation.Therefore the multiplication with −j rotates it 90 in the negative orientation, therefore clockwise.

210

Physics Olympiad Script 10.2. IMPEDANCE

From electro-magnetism we get the following equation:

u(t) = −uind(t) = Ldi(t)

dt

where L is the inductance of the inductor. The switch of the sign between u(t) and uind(t) comesfrom energy consideration: If we look at an ohmic resistor, the resistance is bigger than zero. Asa consequence u(t) and i(t) are in phase. If we look at a voltage source then the voltage and thecurrent have a phase shift of 180. This is basically a convention but it ensures that Kirchhoff’slaws hold6. The induced voltage is like a source voltage, it therefore has opposite sign to theapplied voltage u(t). Assume that the current is i(t) = I0e

j(ωt+ϕ). Then the voltage is

u(t) = Ldi(t)

dt= L

dI0ej(ωt+ϕ)

dt= LI0jωe

j(ωt+ϕ) = jLωi(t).

The impedance of the inductor therefore is

Z =u(t)

i(t)=jLωi(t)

i(t)= jLω.

The phasor and the time dependence are shown in figure 10.4. The j causes the voltage to be 90

earlier than the current. This is clear: If one applies a voltage at an inductor, the growing currentcauses a growing magnetic field which opposes the current to grow. The current therefore growsslowly whereas the voltage over the inductor drops immediately7. The dependence on the angularfrequency and the inductance is intuitively clear because a large inductance causes a strongermagnetic field and therefore the induced voltage, which opposes the current to grow, is also largerand as a consequence the current smaller. For high frequencies, the magnetic field changes fast.Since the induced voltage is proportional to the change of the magnetic field, the current is smalland therefore the impedance is high.

6If there is no external alternating magnetic field through the circuit, the sum of all voltages over the devices ofa closed path is zero. If we go in the direction of the current then it is obvious that the source and the consumingdevices have different sign. See also 10.3.1.

7This is different to the capacitor where the capacitor has to be charged in order to have a voltage drop

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Figure 10.4: Phasor at time t = 0 and time diagram for an inductor. Be aware that the voltage orcurrent is the projection on the x-axis.

10.3 Combinations of R,C and L

In this chapter we have a look at the most important combinations of basic electrical elements. Forthis the usefulness of the complex notation gets obvious. Nevertheless we calculate one examplewith real impedances as well in order to show how this works.

10.3.1 Kirchhoff’s laws

We can nearly adapt Kirchhoff’s laws with some small modifications from direct current (DC). Thefirst modification is that we use complex currents, voltages and impedances. Since charge cannotbe created or destroyed, we get the continuity equation: For any point in an electrical circuit andfor all time the sum of all currents ik(t) flowing to a particular point has to be equal to the changeof charge q(t) at that point:

∑ik(t) =

dq(t)

dt.

Assuming no charge is stored, as it is the case for (nearly) all electrical elements except thecapacitor, we get that the sum of all currents flowing to a point has to be zero,

∑ik(t) = 0. This

is one of Kirchhoff’s laws.

From electrodynamics we know that for any time, the sum of all voltages uk(t) around a closedpath has to be equal to the change of the external magnetic flux Φext:

∑uk(t) = uind = −dΦext

dt.

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Physics Olympiad Script 10.3. COMBINATIONS OF R,C AND L

Assuming there is no time dependent magnetic field, the equation simplifies to the known Kirch-hoff’s law

∑uk(t) = 0. Be aware that the voltages are complex quantities, which means that the

vectorial sum of all the voltage phasors has to be zero.

10.3.2 Serial and parallel circuit

With the same argument as in the case of DC one gets formulas for the total impedance of electriccomponents connected in series or parallel.If two or more components are connected in series, the current i(t) through the components is thesame everywhere. Multiplying each component by its impedance, one gets that the voltage dropuk(t) = i(t)Zk over each element. The sum of all these voltage drops must be equal to the appliedvoltage u(t) =

∑uk(t)

8. The total impedance is then given by

Z =u(t)

i(t)=i(t)

∑Zk

i(t)=∑

Zk

which is exactly what we expect from DC with the difference, that the impedances Zk are complex.With the analogous argumentation for components which are connected parallel to the voltagesource, one gets that the total impedance is given as

Z =

(∑ 1

Zk

)−1

which is also exactly what we expect from DC.

10.3.3 High pass filter

One important AC circuit is the high pass filter. A high pass filter has a high impedance for lowfrequencies and a low impedance for high frequencies. The high pass filter is often used as anaudiofilter to keep low frequencies away from the tweeters. There are different realisations of a high passfilter with different characteristics. One of the easiest is the RC circuit as shown in figure 10.5.The total impedance is given as

Z = ZR + ZC = R+−jωC

,

1

Z=

1

R− jωC

=ωC

RωC − j=ωC(RωC + j)

(RωC)2 + 1,

|Z| =

√R2 +

(1

ωC

)2

where R is the resistance of the resistor, C the capacity of the capacitor and ω the angular frequencyof the voltage applied. If a voltage u(t) = U0e

jωt is applied to the high pass filter, the voltage dropover the resistor is given by

8Here the voltage over the source is measured in the opposite direction, which causes a change of the sign. It isthen compatible with the notation in the Kirchhoff’s laws.

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Figure 10.5: High pass circuit. An alternating voltage uin(t) is applied to a capacitor and a resistorand the voltage uout(t) over the resistor is measured.

uR(t) = Ri(t) = Ru(t)

Z= Ru(t)

ω2C2R+ jωC

(RωC)2 + 1.

This leads to an amplitude UR0 and phase ϕ of

UR0 =RU0

√ω4C4R2 + ω2C2

(RωC)2 + 1=RU0ωC

√(RωC)2 + 1)

(RωC)2 + 1=

RU0ωC√(RωC)2 + 1

,

tan(ϕ) =ωC

Rω2C2=

1

RωC.

This means that one measures a voltage which has an amplitude UR0 with an angular frequencyω and which leads the voltage of the AC source by the angle ϕ. The voltage over the resistor isleading the voltage of the source: Because the voltage drop over the resistor is proportional to thecurrent; and at the capacitor, the current leads the voltage. In the limit ω →∞ there is no phaseshift and all the voltage drops over the resistor. This is also clear, because at very high frequencies,the capacitor has a very low impedance and therefore almost no influence on the circuit. In figure10.6 the amplitude and the phase is plotted as function of ω.Now let us once do the calculation in real notation in order to see how elegant the complex notationis. For the real notation there are different approaches, the easiest is to do the calculation withphasors by geometrically add impedances. But this is basically the same as we did above in thecomplex notation with vectors instead of complex numbers.For the calculation with real quantities let the current be i(t) = I0 cos(ωt) 9. This does not meanthat the voltage at the voltage source is of the form u(t) = U0 cos(ωt), because there will be aphase shift between current and voltage. We now compute this phase shift as well as the relationbetween U0 and I0.From Ohms law we know that the voltage over a resistor is always proportional to the current,therefore uR(t) = Ri(t) = RI0 cos(ωt). Additionally we know that the voltage over a capacitor is

9We could also assume a different phase because only the phase shift between current and voltage matters. Sowe choose the phase of the current to be zero because the current through both components is the same, so thissimplifies the calculation.

214


Figure 10.6: Amplitude UR0and phase ϕ of the high pass filter. As expected, the output voltage is

higher for high frequencies. For the plot the following values were assumed: U0 = 1V, R = 1000Ωand C = 10−6F. Be aware that the x-axis is a frequency f = ω

2π .

always lagging 90 and therefore uC(t) = I0ωC sin(ωt). The voltage over the voltage source is equal

to the voltage over the RC combination and this is the sum of the two voltages

uin(t) = UR(t) + uC(t) = I0

(R cos(ωt) +

1

Cωsin(ωt)

)= U0 cos(ωt− ϕ).

Using some trigonometric theorems to get

U0 = I0

√R2 +

(1

ωC

)2

,

ϕ = arctan

(1

ωCR

).

The absolute value of the total impedance therefore is U0I0

=

√R2 +

(1ωC

)2, the same as above inthe complex notation. The phase shift is the same as well.

10.3.4 Resonant circuit

There are two different types of resonant circuits, the parallel circuit and the serial circuit. Theybehave differently but the equations to solve the problem are almost the same. So we only look atthe serial circuit.A serial resonant circuit is a circuit where a capacitor, a resistor and an inductor are connected inseries. If there is no resistor (or its resistance is zero) the circuit is called an ideal serial circuit oran LC circuit. Since this is a special case we consider the general case where R 6= 0. Figure 10.7shows the serial resonant circuit. We connect the resonant circuit with a voltage source with anangular frequency Ω.

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Figure 10.7: Serial resonant circuit. An alternating voltage with angular frequency Ω is applied toan inductor, a resistor and a capacitor.

Minimal resistance

The total impedance of the circuit is given as

Z = ZR + ZC + ZL = R+ j

(LΩ− 1

ΩC

).

Obviously the absolute value of the impedance is minimal if the imaginary part vanishes. In thiscase the angular frequency is often called resonance frequency10 Ω = ω0. It then holds

Lω0 =1

ω0C⇒ ω0 =

1√LC

.

In fact this is only the resonance frequency for the ideal LC circuit, because in the case of a notideal circuit it is a bit smaller (see below). Since the impedance of the circuit is minimal, thecurrent from the source through the circuit will be maximal at this frequency.

Natural frequency

Since the Resonant circuit is a very beautiful example of a harmonic oscillator we shortly repeatits properties. As it is an oscillator it can oscillate itself. Consider the following case: We open thecircuit, charge the capacitor with a charge q0 and close the circuit again. As soon as the circuit isclosed, the overall voltage must be zero, and the sum of all voltages is given as

10Resonance is the phenomenon where an oscillating system gets maximally excited by an external excitation.Since maximal excitation is not the same as minimal resistance, the frequency we discussed is often missleadinglycalled resonance frequency although the resonance frequency is something different.

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0 = uL(t) + uR(t) + uC(t) = Ldi(t)

dt+Ri(t) +

1

Cq(t)

= Ld2q

dt2+R

dq

dt+

1

Cq. (10.2)

We therefore search a function q = q(t) which solves the equation above. We try the ansatzq(t) = q0e

λt. Inserting the ansatz in equation 10.2 we get a quadratic equation for λ. Since wesearch for oscillating solutions, the discriminant is constraint by R2 − 4LC < 0 and therefore thesquare root is complex. The general solution is

q(t) = e−δt(Aej√ω20−δ2t +Be−j

√ω20−δ2t

),

δ =R

2L, ω0 =

1√LC

.

The constants A and B depend on the conditions at t = 0. In our case we choose them such thatthe capacitor is maximal charged at t = 0 which leads to

q(t) = q0e−δt cos

(√ω2

0 − δ2t

).

This means that the circuit oscillates with a frequency ω =√ω2

0 − δ2 < ω0 which is called naturalfrequency. The oscillation is damped which is not surprising because energy gets dissipated at theresistor (see also section 10.4.1.

Resonance (maximal current)

If we apply a harmonic oscillating voltage to the circuit we can observe resonance. Assume weapply the voltage u(t) = U0e

jΩt. According to Kirchhoff’s law the applied voltage is equal to thetotal voltage over the consuming devices

U0ejΩt = jLΩi(t) +Ri(t)− j 1

ΩCi(t) =

−jLi(t)Ω

(ω2

0 − Ω2 + 2jδΩ),

δ =R

2L, ω0 =

1√LC

.

i(t) = U0jΩ

L

1

ω20 − Ω2 + 2jδΩ

ejΩt,

1

Z=jΩ

L

1

ω20 − Ω2 + 2jδΩ

,

1

|Z|=

Ω

L

1√(ω2

0 − Ω2)2 + (2δΩ)2

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If we search for the frequency with maximal current, we can take the derivative of 1|Z| with respect

to Ω and set it equal to zero11.

0 =d 1|Z|

dΩ=

1

L

D − Ω 12D

(2(ω2

0 − Ω2)(−2Ω) + 4δ22Ω)

D2,

where D =√

(ω20 − Ω2)2 + (2δΩ)2 is the nasty square root in the denominator of 1

|Z| . Multiplyingthis equation with LD3 one gets

0 = ((ω20 − Ω2)2 + 4δ2Ω2) + 2Ω2(ω2

0 − Ω2)− 4δ2Ω2 = ω40 − Ω4,

Ω = ±ω0.

This result was expected because if the impedance is minimal, the current is maximal.

Resonance (maximal voltage)

Now let us have a look at the voltage drop over the capacitor:

uC = −j 1

ΩCi(t) = U0

1

LC

1

(ω20 − Ω2)− 2jδΩ

ejΩt,

UC0 =U0

LC

1√(ω2

0 − Ω2)2 + (2δΩ)2,

= U0ω2

0√(ω2

0 − Ω2)2 + (2δΩ)2= U0

1√(1− Ω2

ω20)2 + 4 δ

2

ω20

Ω2

ω20

where UC0 is the amplitude of uC(t). Once again let us have a look at the maximal voltage

0 =dUC0

dt= −U0

1

LC

2(ω20 − Ω2)(−2Ω) + 4δ2Ω

2D3,

⇒ Ω2 = ω20 − 2δ2. (10.3)

This means that one can measure the highest voltage at the capacitor at the frequency Ω =√ω2

0 − 2δ2. This frequency is usually called resonance frequency. Picture 10.8 shows the voltageover the capacitor for different frequencies depending on the dumping δ. This frequency dependenceis called resonance curve. Concerning the asymptotic behaviour of the resonance curve we get theexpected result: For very small frequencies, the voltage drop over the capacitor converges towardthe applied voltage and is exactly the applied voltage U0 in case the of DC. This is because whenapplying a constant voltage, the capacity corresponds to an interruption in the circuit and thereforethe whole voltage drops over it. For very high frequencies, the impedance of the capacitor goestowards zero and therefore the voltage drop also goes towards zero.

11Since the first derivative vanishes at the extrema.

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Physics Olympiad Script 10.4. POWER CONSIDERATION AND EFFECTIVE VALUES

Figure 10.8: Voltage drop UC0 over the capacitor as a function of the applied frequency for differentdumpings δ. The y-axis is normed to the excitation and RLC circuit. The x-axis and the dumpingδ are normed with respect to ω0. The dotted line shows the position of the maxima depending onthe resonance frequency (for a given Ω the δ was calculated according to equation 10.3).

Let us make a summery of the different frequencies:

Name Formula Propertynatural frequency

√ω2

0 − δ2 Frequency at which the circuit oscillates withoutexternal excitation.

minimalimpedance ormaximal current

ω0 The two cases are the same since for minimalimpedance the current gets maximal at a givenvoltage

resonance fre-quency

√ω2

0 − 2δ2 Maximal voltage drop over the capacitor

Table 10.1: Overview of different frequencies where ω0 = 1√LC

and δ = R2L .

10.4 Power consideration and effective values

Until now we always considered equations which were linear in the voltage or in the current.Therefore there was no problem when adding complex voltages or currents and then taking the

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real part to get the property one measures at the circuit. Now we look at the power and thereforemultiply current and voltage. So we have to take the real value first and then multiply the twoquantities.

10.4.1 Power

The power consumption at time t of an electrical device is p(t) = u(t)i(t) which is time dependent.In many cases the instantaneous power is not so important and one is more interested in the meanpower consumption12. Since we look at periodic oscillations one only has to consider one periodT to calculate the mean value. Let’s apply a voltage u(t) = U0 cos(ωt) at a device which causesa current i(t) = I0 cos(ωt + ϕ) with phase shift ϕ flowing through that device. The mean powerthen is

P =1

T

T

0

u(t)i(t)dt (10.4)

=U0I0

T

T

0

cos(ωt) cos(ωt+ ϕ)dt

=U0I0

T

T

0

cos2(ωt) cos(ϕ)dt−T

0

cos(ωt) sin(ωt) sin(ϕ)dt

=U0I0

2cos(ϕ) (10.5)

where we use cos(ωt+ ϕ) = cos(ωt) cos(ϕ)− sin(ωt) sin(ϕ) and sin(ωt) cos(ωt) = 12 sin(2ωt). The

first integral13 gives T2 and the second one gives zero. This means that for constant current I0 and

voltage amplitude U0 the mean power depends on their phase shift. To be more precise, the powerwe calculated is the one that flows into that device and gets converted into an other energy, forexample heat at the resistor or rotation in case of an electric motor. The power is maximal forϕ = 0, so for example for a resistor. On the other hand the power is minimal for ϕ ± 90 whichcorresponds to a capacitor or an inductor. In case of the capacitor, the phase shift is ϕ = 90

(see figure 10.3). After the capacitor is completely charged (which is the case for voltage at thecapacitor being maximal, i.e. u(t) = U0) the capacitor gets discharged. This means energy flowsback from the capacitor to the source until there is no charge left in the capacitor. This is the casewhen the voltage is zero, and then it gets charged again. While charging, energy flows from thesource to the capacitor. In the time diagram, the discharging periods are those where the voltageand the current have opposite sign and as a consequence the instantaneous power is negative.Obviously the charging and discharging energy is the same which means that the capacitor has nomean power consumption. A similar argument can be made with a coil where the energy is storedin the magnetic field.

12For example consider a resistor: There the mean heat dispersion is more relevant than its instantaneous powerconsumption because the resistor also has some heat capacity

13One can calculate this using partial integration or using that the cosine and the sine are the same function up toa phase shift and therefore

´ T0

cos2(ωt)dt =´ T0

sin2(ωt)dt = A for a certain number A. Then A =´ T0

cos2(ωt)dt =´ T0

1− sin2(ωt)dt =´ T0

1dt−A = T −A which leads to A = T2

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Physics Olympiad Script 10.4. POWER CONSIDERATION AND EFFECTIVE VALUES

10.4.2 Effective values

In direct current, the power is simply P = UI whereas in AC there are the factors14 2 and cos(ϕ).One defines

Ueff =1√2U0,

Ieff =1√2I0,

as the effective voltage and effective current. Sometimes these quantities are called RMS (rootmean square) of the voltage or current. Using Ueff and Ieff one gets the mean power of a device as

P = UeffIeff cos(ϕ).

In everyday life most of the voltages and currents are indicated with their effective value instead ofthe amplitude. This has the advantage that one can simply multiply the voltage and the currentin order to get the power consumption15. As an example the voltage at the power sockets hasUeff = 230V which means that the amplitude of the voltage is U0 =

√2 · 230V ≈ 325V.

10.4.3 Active, reactive and apparent power

A closer look to the mean power in equation (10.5) allows an interesting view on the phase shift. Ifwe apply a voltage u = U0 cos(ωt) to a device, we can rewrite the current as i(t) = I0 cos(ωt+ϕ) =I0(cos(ωt) cos(ϕ)−sin(ωt) sin(ϕ)) which is basically a superposition of a sine and a cosine oscillationwith amplitude −I0 sin(ϕ) and I0 cos(ϕ) respectively. The cosine oscillation is in phase with thevoltage and therefore it corresponds to the part of the current which dissipates energy at the device.One calls this the active current, which means that is the part of the current that does (useful)work. The sine oscillation does not perform any work, it only causes energy to be transferred fromthe source to the device and back. This current is called reactive current.Analogously one can define active, reactive and apparent power. What we have defined as meanpower is also called active power P = UeffIeff cos(ϕ) since this is the electric power that is used bythe electric device. The apparent power S = UeffIeff is the total power that is transferred betweenthe source and the device. One part of this total power is the past used as active power and theother part is transferred back to the source. The second part is the energy oscillation betweenthe source and the device. It is called reactive power and defined as Q = UeffIeff sin(ϕ). Usingsin2(ϕ) + cos2(ϕ) = 1 one gets the following relation between the three different powers:

S2 = P 2 +Q2.

Generally one wants to have small reactive currents because a reactive current does not transferenergy to the device but it causes a larger current than necessary which leads to more losses in

14The factor 2 is only valid for sinusoidal signals, for other signal forms one has to calculate it from equation(10.4).

15Assuming to have no phase shift ϕ = 0. This is the case in most of the everyday life devices. Nevertheless atbig electric motors cos(ϕ) is indicated.

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the cables. There are different possibilities to avoid reactive current. The simplest is to connecta capacitor or an inductor parallel to the device such that the total impedance suffices ϕ = 0.Then the reactive current only oscillates between the device and the capacitor or inductor and notthrough the cables from the power station to the device.

10.5 Three-phase electric power

The word wide power net, which also provides electricity at home16, is a bit more sophisticatedversion of AC than we have discussed until now. This version is called three-phase electric powerand it has some nice applications we will look at.

10.5.1 Definition and production

Usually, a three-phase power supply consists of three wires and to each wire an AC voltage isapplied. Additionally there is one wire which is called neutral wire, which we consider later. Thethree wires with the voltage are called phases. The amplitude of the voltage in all of the threephases is the same but shifted by 120 = 2π

3 which is shown in figure 10.9.

Figure 10.9: The three voltages of a three-phase system. The wires are labelled by L1, L2 and L3.

There are different ways to create three-phase power, the easiest is the three-phase generator. Itworks as an usual generator but the three coils placed around the rotating magnet are displacedby 120 each, see figure 10.10. The magnet in the center rotates and therefore the magnetic fieldat the coils is changing. This induces a voltage in the coils. Since the coils are displaced by 120,the induced voltage is also shifted by 120. Be aware that the induced voltage at a coil reaches itsmaximum when the magnet changes its polarity because the induced voltage is proportional thevariation of the magnetic field in time. This variation is maximal when the magnet changes itspolarity.The setup described above allows a visualisation of the phaser: Treat the phasor as a vectorperpendicular to the North-South direction of the magnet such that the maximal voltage is induced

16The power is brought to you home with a three-phase system and then the three phases are split up. Thereforeat an usual power socket there is only one phase together with the ground wire and the neutral wire. Only veryfew power sockets are three-phase sockets. They are usually only needed for machines that consume a lot of energy.These sockets (usually) have five poles.

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Physics Olympiad Script 10.5. THREE-PHASE ELECTRIC POWER

in a coil when the phasor is pointing to that coil. Then the induced voltage in each coil is theprojection of the phaser on the axis of the coil.

Figure 10.10: A three-phase generator. The tree coils are rotated by 120. The magnet in the centeris rotating with angular frequency ω and induces an AC voltage in the coil. One pole of each coil isconnected to the neutral wire N , the other poles are the phases L1, L2 and L3.

10.5.2 Star and Delta circuit

A big advantage of the three-phase power circuit is that there are two possibilities of connectinga device. A three-phase device is (usually) connected to all three phases and it (usually) consistsof three independent loads which are drawn as resistors in figure 10.11. However, there are twopossibilities how this can be done, see figure 10.11. Let U0 be the amplitude of all the three phases,the first phase therefore has the voltage u1 = U0 cos(ωt), the second phase hasu2 = U0 cos(ωt+120)and the third one has u3 = U0 cos(ωt+ 240).

Star circuit

The intuitive easier one is the star circuit where each load is connected the same way as the coilsin the three-phase generator. At the first load (connected to phase L1) the voltage u1 drops, asexpected. Analogously the other loads. If all loads have the same impedance then there is nocurrent flowing from the three loads to the generator through the neutral wire. This is because the

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Figure 10.11: Left side: The star circuit where each load is connected to a phase and the neutralwire. Right side: Each load is connected to two phases, the neutral wire is not used at all.

sum of all three currents is zero17. Therefore it is not necessary to connect the common point ofthe three loads with the neutral wire, the loads need only to be connected at one point with eachother. It is often useful to connect the common point with the neutral wire to stabilize the circuit,see also chapter 10.5.3.

Delta circuit

The other possibility is the Delta circuit where each load is connected to two phases. The advantageof this circuit is that the voltage drop over each load is higher. To understand this let’s have alook at the voltage u(t) over one load, for example connected to L1 and L2:

u(t) = u1 − u2 = U0(cos(ωt)− cos(ωt+ 120))

= U0(cos((ωt+ 60)− 60)− cos((ωt+ 60) + 60)

= U0(cos(ωt+ 60) cos(60) + sin(ωt+ 60) sin(60)

− cos(ωt+ 60) cos(60) + sin(ωt+ 60) sin(60))

=√

3U0 sin(ωt+ 60)).

This means that the voltage drop over each load has an amplitude of√

3U0.

To start a big electric motor one can use both, the properties of the star and Delta circuit. Whenstarting the motor it needs a lot of current. If one applies a smaller voltage, the current is alsosmaller18. So one starts the motor in the star circuit and as soon as it rotates with constant speed,one connects it as Delta circuit and the motor has more power because the load is connected to ahigher voltage. In the past this was often used to start large motors and there were special switches

17This can easily be seen if one adds the currents as phasors. Since we assume that all the three impedances arethe same, the three currents are also the same. The three phasors form an equilateral triangle, therefore the startingpoint is equal to the ending point and as a consequence the sum of the three currents is zero.

18This might sound trivial but it is important for the fuse because it should not melt.

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Physics Olympiad Script 10.5. THREE-PHASE ELECTRIC POWER

which allowed to change easily from the star to the Delta circuit. Of course it is a bad idea toconnect a motor which is not build for a Delta circuit to a Delta circuit as it might get damaged.

10.5.3 Advantage of a three-phase system

There are different reasons why our power supply at home is a three-phase system. We nowinvestigate some of them.

Role of neutral wire

As we have seen in the section about the star circuit (see section 10.5.2), there is no current flowingthrough the neutral wire as long as the impedances at the three phases are the same. But if theimpedances of the three consumers are not the same, a current flows through the neutral wire.If the common point in the star circuit is not connected to the neutral wire, this common pointhas a non-zero voltage with respect to the neutral wire. This is because at the common point thetotal current has to be zero19. This is only possible if the three currents are the same20. Sincenot all impedances are the same, the voltage drop over the loads will not be the same and as aconsequence the sum of the voltage will not be zero. This sum of the voltage drops is the same asthe voltage between the common point and the neutral wire. To guarantee that the voltage dropover all the consumers is the same, one connects the common point with the neutral wire and thenthe neutral wire is like a fixing point where current can flow away if there are different impedances.This is exactly the situation at home: The houses are connected to the three-phase net. In thehouse the different sockets are connected to different phases and to the neutral wire. If all thephases are loaded the same (connected with the same impedance), then there is no current flowingout of the house through the neutral wire. But if for example only one power socket is used thenthe current flows through the neutral wire out of the house.

Energy transfer

Concerning the energy transfer the three-phase system offers a very efficient way to transportelectrical energy over a (long) distance. If the impedance at the three phases is the same then(nearly) no current flows through the neutral wire. Therefore we only need the three phases totransport the electrical energy whereas in a simple AC circuit we need two wires for a closedcircuit which corresponds to one single phase. This means in the three-phase system the threewires transport as many energy as three simple AC circuit consisting of six wires.This gets very important if one wants to transport a lot of energy over a long distance because alot of energy means thick cables and long distances mean long cables. Hence, a lot of wire. Thepower net consists of three thick cables where the phases are connected and one thin cable with theneutral wire (to stabilize the net). The different phases are connected then cleverly to the differenthouses (or even different villages) such that the impedance on all three phases is approximatelythe same. In terms of whole villages it is not relevant if someone at home uses a lot of power fromone single phase21. In conclusion the impedances applied to the three phases can be consideredthe same.

19There is no capacitor and Kirchhoff’s law holds.20To distinguish: The three currents are the same if the common point is not connected to the neutral wire. They

are not the same if the common point is connected to the neutral wire.21Additionally each phase at home is connected to a fuse, the power consumption is therefore limited. In terms

of the whole village the consumption of one single house is small/negligible

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Three-phase motor

The three-phase motor is in principle the same as a generator (see figure 10.10) but instead ofturning the magnet and inducing a voltage in the coils, a voltage is applied to the coils whichcreate a magnetic field which causes the magnet to turn around. Since the magnetic field of a coilis proportional to the current flowing through it and the current is proportional to the appliedvoltage22 and the voltage correspond to a rotating phasor, the magnetic field also rotates. But thedirection of the rotation (whether it rotates clockwise or anti-clockwise) depends on how the threecoils are connected to the three phases. Actually, it only depends on how the three phases areconnected to the coils: in clockwise or anti-clockwise direction. This means if the three coils coil1, coil 2 and coil 3 are connected to L1, L2 and L3 respectively or L1, L3 and L2. By swappingthe phases of two coils the motor changes its direction of rotation.This property seems pretty unspectacular but it allows to run a motor in a certain direction whichis not that simple to achieve in a simple AC circuit: In a simple AC circuit there is no rotatingmagnetic field (because there is only one phase). There is only a magnetic field that always pointsin the same direction and changes its strength. As a consequence the motor does not know inwhich direction it should turn so it starts turning in an arbitrary direction. Of course there aresome tricks how one can force the motor of a simple AC circuit to turn in a specific direction butit is more complicated than in case of a three-phase system.

22Maybe with phase shift but this phase shift is the same for the three coils in a three-phase motor.

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Chapter 11

Special relativity

Roughly 100 years ago there were some phenomena in different fields of physics that could not beexplained by the theories of that time. Einstein recognised that (at least some of) these phenomenacan be described using a modification of classical mechanics and taking into account how oneobserves an other reference frame. The complete theory of relativity is very complex and needsa lot of maths. Nevertheless there is a small part of it which gives already a lot of interestinginsights and which gets along with high school maths. This easier part is called special relativityand the more general theory is called general relativity. As long as nothing else is mentioned inthis chapter, we will always do special relativit and sometimes simply call it relativity. The mainsource of this chapter are [65] and [66].In this chapter we will first have a look at classical mechanics. After we understand the mostimportant properties of classical mechanics we will look first at the assumptions of special relativityand then at its conclusions and interpretations. In the end we will resolve some of the most commonparadoxes. Before we start doing physics lets have a look at some historical milestones that leadto relativity.

11.1 Historical milestones

Before Einstein developed relativity there were different experiments and parts of the theory whichwere incompatible. In electrodynamics this inconsistency was especially big. Let’s have a look atthe big questions physicists had roughly 150 years ago.

11.1.1 Aether and electromagnetic waves

In 1864, Maxwell elaborated the famous Maxwell’s equations which describe (nearly) everythingin classical electrodynamics. These equations also predict electromagnetic waves which were thenexperimentally proven by Heinrich Hertz in 1887. If one thinks about waves in every day life onealways imagines a medium where this waves propagate. For example sound waves consist of air (oran other medium) which gets compressed and stretched. But the important thing is that there is amedium in which the wave propagates, eg. if there is no medium, nothing can get compressed andstretched so no sound can propagate. Only in the reference frame where the medium is resting,the wave equations take the simple form, for all other reference systems one has to add additionalterms1.

1To show this mathematically one needs a bit more math, therefore we skip this here

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Physics Olympiad Script CHAPTER 11. SPECIAL RELATIVITY

Until the beginning of the 20th century physicists thought that light would also propagate in amedium called aether. But Michelson and Morley showed in their famous experiment that thiscan not be true. For this they measured the speed of light in the direction of the rotation of theearth and perpendicular to it. They thought that the aether should rotate a bit slower than theearth because it must somehow glue to the rest of the universe which is not rotating. Therefore thespeed parallel and perpendicular to the direction of rotation should differ, which was not the case.Since light propagates as a wave it fulfils the wave equations. In the frame of the medium theytake the simple form given in equation 5.1. If one calculates the transition from one to another(moving) reference frame using classical mechanics, the simple wave equation change. Since lightdoes not travel in a medium, the simple wave equations must hold in all reference frames, whichcontradicts the calculation of classical mechanics. This was a clear hint that something was wrong.At the beginning of the 20th century, famous physicists as Lorentz and Poincaré figured out howone has to calculate the transition from one to the other reference frame. This lead to the Lorentztransformation, which is nowadays the basis of relativity. So the found the equations some yearsbefore Einstein and Einstein surely knew the equations already. But the big success of Einstein wasto give the equations a physical meaning. Because Lorentz and Poincaré (and many others) didn’trealise that these equations contain a deep property of nature, they simply used these equationsto resolve the contradiction in the theory.

11.1.2 Flying electron

We now want to look at an example where classical mechanics leads to a contradiction. Considerthe case where a current flows through a wire and an electron is flying with a constant velocityparallel to the wire, see figure 11.1. The current in the wire creates a magnetic field around thewire. The Lorentz force acting on the moving electron pulls it toward the wire (depending of thedirection of the current and movement of the electron).This is different if we change the reference system. Let’s chose the system moving with the electron.In this system the electron does not move. As a consequence there is no Lorentz force and thereforeit will not be attracted by the wire.

Figure 11.1: In the first reference frame (left) the electron (black dot) is moving parallel to thewire through which a current flows. The current creates a magnetic field which causes a force actingon the electron (Lorentz force). The system left moves with the electron. Since the electron is notmoving, no Lorentz force is acting.

This seems to be a paradox but considering special relativity, one can resolve it. We will have alook at this at the end of this chapter when we understand relativity better.

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Physics Olympiad Script 11.2. GALILEO TRANSFORMATIONS

11.2 Galileo transformations

In order to understand the basic concepts of relativity such as transformation from one referencesystem to another, lets first have a look at these concepts in classical mechanics. There thetransformation from one reference system to another is called Galileo transformation. But beforewe start with the Galileo transformation we have to define what a reference system, also calledframe of reference, is and which kind of frames we will look at2.

11.2.1 Reference System

To describe physics it is often very useful to describe it in a reference system. A reference systemconsists in a choice of an origin and three axis. Usually one takes the axis pairwise orthogonalwhich is then a Cartesian coordinate system3. In this system each object has at every time t acertain position ~x(t). The origin is usually fixed at a certain object such as the sun if one wantsto describe the motion of the planets or one edge of a desk when describing the motion of a ballon the desk. As a consequence the object which is connected to the origin does not move in thatframe of reference because it has for all time the position ~x(t) = ~0. If we fix the origin at a certainobject X (X is here the name of the object) we say that we describe the problem in the referenceframe of X or in the rest frame of X. For example in chapter 11.1.2 we described the problem inthe reference system of the wire (left) and in the system of the electron (right)4.Sometimes we will give our reference frames names, for example we will often denote two framesas Σ and Σ′. One then says that a property, for example time or position, are measured in some(specific) frame. This sounds often pretty abstract and it is sometimes useful to visualize it. Oneconvenient visualization is to think for example Σ as the rest frame of the earth. So measuringa quantity in Σ means that someone (for example you) is standing on the earth and measuresthis quantity. If there are two frames involved it is often easy to think about a space ship as thesecond frame Σ′. For this assume that a colleague of you flies in a space ship and performs his ownmeasurement.There are different names for the most important and most common reference frames.

Rest frame

As already defined above this is the system which is connected to a certain object. This objectwill then rest in that system, therefore it’s the rest frame of that object.

Laboratory frame of reference

This is usually the system where we start describing a problem. If one performs a measurementthis is the rest frame of the experiment.

Centre-of-momentum frame

This is the frame where the total momentum is zero. If the momentum is conserved5, the totalmomentum will stay zero which often simplifies the calculation.

2The description of the transformation between arbitrary reference frames is described by general relativity.3One could also take polar or spherical coordinates. The choice depends on the symmetry of the problem but in

many cases Cartesian coordinates is a good choice (at least to start with).4In that example it was not necessary to chose concrete axes so we left it out.5This means no external forces act on our system.

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Inertial frame of reference

Since this is a very important type of reference frame lets have a closer look at it in the nextchapter:

11.2.2 Inertial frame of reference

Of particular importance is the inertial frame of reference. It is a frame where Newtons laws arevalid6. In particular this means that if no force is acting on an object, this object will not moveor move with a constant velocity.For example the earth is approximately an inertial frame7 because any force acting on an object (forexample gravity) is proportional and parallel to the acceleration. It is not really a inertial framebecause the earth is moving around the sun and rotating around its axis, therefore it’s acceleratedand there are fictitious forces such as the centrifugal force or Coriolis force.If we look at a space station, for example the ISS, orbiting around the earth we might think thisis an inertial system. But in fact it is not because all the masses in the space station seem to beweightless (for example a pen floating), although there is a gravitational force.If we once found an inertial frame of reference then any system moving with a constant velocity tothat system is also an inertial system. In Newtonian mechanics this is obvious. Lets assume theframe Σ is an inertial system and that the system Σ is moving with velocity v(t) measured in Σ.If a force F (measured in Σ) is acting on an object the object gets accelerated with a = F

m , since itis an inertial frame of reference. If in Σ the same force F and the same acceleration a is measured,then Σ is also an inertial system. This gives us restrictions on the velocity v(t) because since

a = a+dv

dt= a

where the first step is simply the transformation of the acceleration between two reference frames(in Newtonian mechanics) and the second step is the assumption that the two accelerations areequal. The consequence is that dv

dt = 0 and therefore the velocity is constant (including v = 0).This means that Σ is an inertial frame if and only if it is moving with constant velocity or it isdisplaced by a constant vector or rotated by a fix angle with respect to Σ (assuming Σ is an inertialsystem).

11.2.3 Galileo Transformation

Now we have enough definitions to look at the Galileo Transformation. This transformation de-scribes how time and space transform between different inertial frames in classical mechanics. LetΣ and Σ′ be two inertial frames of reference8. In classical mechanics there is no reason why thetime in one frame elapses slower or faster than in the other9. This means the time in both framesis the same or mathematically t = t′. This is the transformation of time between the two systems.To examine the transformation of space, assume that the axis in the two systems are parallel to

6This means a force acting on an object is proportional to its acceleration: ~F = m~a.7Neglecting relativistic effects due to gravity.8We denote all quantities measured in Σ′ with a prime ′ and if measured in Σ without.9They might have a different zero point of time and therefore the time differs by a constant. But this constant

has no influence on the the physics so we can without loss of generality set it to zero.

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Physics Olympiad Script 11.3. LORENTZ TRANSFORMATION

each other and that Σ′ is moving with a constant speed along the x axis of Σ (see figure 11.2). InΣ the origin of Σ′ has therefore the coordinates

OΣ′ =

vt00

+

x0

00

where x0 is the position of the origin of Σ′ at t = 0 measured in Σ. This means that a point~r′measured in Σ′ has the coordinates ~r in Σ

~r =

xyz

=

vt00

+

x′

y′

z′

=

vt00

+ ~r′.

This means that the y and z coordinates of the two systems are always the same, eg. y = y′ andz = z′. The x coordinate transforms as x = vt+ x′ or x′ = x− vt.

Figure 11.2: The two frames represented by their axis. One is moving with respect to the otheralong the x axis with the speed v.

If we put all the transformations together we get

t = t′

x = vt+ x′

y = y′

z = z′

where obviously time and space only mix at the x component. Sloppily said: time influencesmeasurements of space but not the opposite. This will be different in relativity.

11.3 Lorentz transformation

In this chapter we will will describe the basic transformation in relativity. Further phenomena willbe discussed in the next chapter.

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11.3.1 Einstein’s Postulates

All the special relativity can be derived by two assumptions about nature. These assumptions arecalled Einstein’s postulates which are:

1. The laws of physics are the same in all inertial frames of reference.2. The speed of light, denoted by c, has a finite value10.

Sometimes the postulates are formulated in a slightly different way but the conclusions remainthe same. In the regime of classical mechanics and considering the definition of inertial framesit is clear that the physics in all inertial frames is the same11. In relativity we have to statethis explicitly because we cannot assume any more that the acceleration in two frames has thesame value. The second postulate needs some more explanation. What is called speed of lightis in fact a much more important speed, namely the speed of information. A precise discussionabout this would go beyond this scope and is not important to understand relativity. So we onlyillustrate this at a small example. Consider two objects that interact with each other, for exampleby electromagnetic or gravitational forces. Then this interaction only happens with the speedof light, which means that one object "sees" or "feels" the other object where it was some timebefore. This is because the information where the objects are spreads out with the speed of light,so the objects have already moved a bit before one sees their position.

Without calculation we can already now state some interesting things.− Since the physics is in all inertial frames the same and the speed of light is a physical property,

the speed of light has the same value in all inertial frames, independent of the movement ofthe light source of the observer.

− Since the speed of light is the maximal possible speed12, the addition of velocities will notbehave the same as in classical mechanics.

In the following we will always look at inertial frames of reference because for all other frames oneneeds one more postulate which leads to general relativity.

11.3.2 Synchronisation of clocks

Since time will not be an absolute quantity13 in relativity we have to find a method how we cansynchronise two clocks. Two clocks are synchronised if they "show" the same time, where "show"means that if we look at two distant clocks we might see two different times because the light fromthe clock further away needs longer to reach our eyes than from the nearer one. If the distance fromthe observer to each clock is the same then the observers sees the same time on the two synchronizedclocks. In fact the "show" is defined more precisely by the process of synchronisation. For thesynchronisation we of course neglect all effects that might happen in a real-life experiment such asimprecise clocks or delayed detectors and so on.To synchronize two clocks we send out one light pulse at the time t0 from one clock, we call itclock 1, to the other, clock 2. There it is (immediately) reflected and it returns to clock 1. Thewhole time from clock 1 to clock 2 and back to clock 1 we denote as ∆t. Clock 2 is now set suchthat at the time the light pulse arrived it would have shown14 the time t0 + ∆t

2 . We therefore have

10The value of c is c = 299′792′545.8 m·s−1

11At least if the speed of light is considered infinite.12Because if something is faster than light it could deliver informations faster than light which is in contradiction

to the fact that informations cannot be delivered faster than light13Absolute time means time does not depend on the position and the reference frame.14This looks like we have to set the time of clock 2 before we actually know which time we have to set. But this

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constructed the synchronisation such that two light pulses which are send out at the two clocks atthe same time meet exactly in the middle of the two clocks. As we will see later, simultaneousnessis not in all frames the same, therefore it does not make sense to synchronize two clocks in differentframes of reference.

11.3.3 Time dilation

Let’s start examine how the different quantities transform from one inertial frame to another. Westart with a time interval. Already now it shall be pointed out that the transformation of time itselfbehaves differently than the transformation of time intervals, which can be seen in section 11.3.6.To investigate the transformation of time intervals we need a clock or even more fundamentalsomething that creates a periodic signal. The simplest clock is the so called light clock, which isshown in figure 11.3. A light clock consists of two mirrors between which a short light pulse istrapped. Assume we observe a light clock moving parallel to the mirrors with a constant velocityv. The frame we observe the moving clock we call Σ and the rest frame of the clock Σ′. One period∆t′ in Σ′ is then the time the light needs to make one complete cycle, which is

∆t′ =2L′

c

where L′ is the distance between the mirrors and c is the speed of light, which is the same in allframes (an therefore there is no ’).

Figure 11.3: Left: the light clock in the rest frame of itself. The light pulse is moving up anddown between the mirrors. Right: The same light clock observed from a frame which is moving withrespect to the clock. The light is not moving straight up and down but makes a zigzag.

In our system we see the same distance between the mirrors15, so L = L′. In Σ, however, the lightdoes not move perpendicularly to the mirrors, as the whole clock is moving. The Period is definedas the time the light needs to go from one mirror to the other and back.

is not necessary because clock 2 can remember the time the light pulse arrived and then comparing with the timet0 + ∆t at clock 1 one can calculate by which amount one has to change the time at clock 2.

15This would follow if one derives relativity more rigorously but this would need more math and would not givemore insight.

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In Sigma this is

∆t = 2L

v⊥

where v⊥ is the velocity perpendicular to the mirrors. Using Pythagoras one can calculate v⊥ andone gets for ∆t:

∆t = 2L1

v⊥,

= 2L1√

c2 − v2,

=2L

c

1√1− β

= γ∆t′,

where β and γ are often used short cuts defined as

β =v

c,

γ =1√

1− β2.

This means that if on a clock in Σ′ one time interval passes, for example one second, in our frameΣ more time passes, for example one and a half seconds. Since both systems are inertial systemsone cannot tell whether one is moving or the other. Therefore it does not depend whether we lookat a light clock or a (Swiss) watch, all will show the same time and therefore show the phenomenondescribed above. This means we see the clocks in a moving frame going slower than ours. Thisphenomenon is called time dilatation.

11.3.4 Lorentz contraction

Similar to time we can have a look at distances. Since we don’t know yet how distances transform,we cannot simply compare an unknown distance with a meter stick in in two different systems16.Since the speed of light is constant in all systems, we can reduce this problem to measuring thetime the light needs to pass a certain distance. First we notice that distances perpendicular tothe direction of motion do not change17. This is because if the light moves perpendicular to thedirection of motion there is no difference in time whether they move in one or the other direction (tounderstand this, keep on reading). A first (wrong) approach to get the transformation for distancesparallel to the direction of monition might be ∆L = c∆T which then would lead to ∆L = γ∆L′

which means that a moving object seems longer than in its rest frame (which is wrong). Themistake in our derivation happened because we did not take into account that in a moving system,the time the light needs for a certain distance depends weather the light propagates in the directionof flight or opposite (see figure 11.4). In fact, moving objects seem shorter than in the in their restframe.

16We don’t know how long a meter seems in a moving system.17We have already implicitly used this at the derivation of the time dilatation.

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Figure 11.4: 1): Sending out a light pulse. 2) Arrival of the light pulse at the mirror after the time∆t1. 3) Reflection of the light at the mirror (this happens immediately). 4) Arrival of the light pulseat the detector, the time the light needs from the mirror to the detector is ∆t2.

To get the right result we have to do the following: We send out a light pulse along the directionwe would like to measure. At the other end of this distance we place a mirror which reflects thelight pulse. The time the light needs to pass the distance source-mirror we denote by ∆t1. Thetime from the mirror to the detector by ∆t2. Measuring the time for twice the distance (∆t1 +∆t2)and dividing it by 2c we get the right length. In the rest frame this is easy because the time tothe mirror and back are the same. We denote this distance ∆L′ (we call this the system Σ′ againsince we want to measure it from our system Σ. Therefore Σ′ is the moving system with respectto our system Σ). The time for the distance in Σ′ is therefore ∆t′ = ∆L′

c . In Σ we get for ∆t1 and∆t2

∆t1 =∆L

c− v,

∆t2 =∆L

c+ v

where the denominators are given this way because the light pulse moves in the same or oppositethe direction of flight. The fact that we simply add or subtract velocities is no contradiction to thestatement above because there is no particle moving with c+ v. The term c+ v appears because

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the light pulse and Σ′ are moving. From time dilatation we know that ∆t1 +∆t2 = γ(∆t′1 +∆t′2) =2γ∆t′ and using some math we get

∆t′ =1

2γ(∆t1 + ∆t2)

∆L′

c=

1

2γ

(∆L

c− v+

∆L

c+ v

)

=2c∆L

2

√1− v2

c2

c2 − v2

=∆L

c

1√1− v2

c2

∆L′ = γ∆L.

Since γ ≥ 1, a moving object seems shorter than it is in its rest frame.

11.3.5 Symmetry of time dilatation and Lorentz contraction

According to the first postulate, the equations (Time dilatation and Lorentz contraction) look thesame in all inertial frames. Assume you are observing a space ship. According to time dilatationyou see the clocks in the space ship going faster than yours with ∆t = γ∆t′. A colleague in thespace ship would observe the same with your clocks, namely that your clocks move slower thanhis. According to the college, the equation ∆t′ = γ∆t holds.One might now argument that from this follow ∆t = γ∆t′ = γ2∆t and therefore γ2 = 1 whichquestions of course all we did (since γ > 1 if v > 0). This argumentation is of course wrong becausewe have to compare same things with same things, what we did not. Because observing a movingclock (which goes slower than yours) one has to measure the time at two different places (whereas the moving clock in its rest frame stays at the same position). This leads to an asymmetry inmeasuring the time one a clock.

11.3.6 Lorentz transformation

Similarly to the Galileo Transformations we can now calculate the transformation of space andtime between moving reference frames. To do this we first have to be aware what we mean bytransforming time and space. From classical mechanics we can keep the definition of a (spatial)reference frame as we have defined it in section 11.2.1 but since space and time will mix, we haveto develop a new intuition for time18. For this we imagine to have at every point in space an(imaginary) clock and each inertial frame of reference has its own clocks. All the clocks belongingto one system show the same time in their rest frame (see also chapter 11.3.2 about synchronisationof clocks). We can then compare the time shown on the clocks of different frames at the sameposition. This is necessary because if we compare the time shown on clocks at different positionswe would have to take into account the time the light needs from the clock to our eyes. In relativity,this union of space and time is called spacetime19.

18In classical mechanics time is absolute, which means that it is a given quantity for all frames (up to a constantshift in time).

19Mathematically it is a 4 dimensional (vector) space but with an important difference to the usual 4 dimension(spatial) space: There is a different measure of distances, see chapter 11.4.

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Let Σ and Σ′ be two inertial frames of reference with their clocks showing the time t and t′. In thetwo frames, a certain point is given as ~r = (x, y, z) in Σ and ~r′ = (x′, y′, z′) in Σ′. For conveniencewe chose the two systems such that the corresponding axis are parallel to each other and thatΣ′ is moving along the x-axis with the speed v. Furthermore we assume that the two origins att = t′ = 0 coincide. Then the transformation is then given as

t′ = γ(t− v

c2x)

x′ = γ (x− vt)y′ = y

z′ = z

where γ = 1√1− v2

c2

is the factor introduced above. Before we look at a (rough) proof, we can rewrite

the transformation above in order make the two equations symmetric. If we associate time t bythe distance ct and rewrite the equations in terms of ct, we get

x′ = γ(x− v

cct)

(11.1)

ct′ = γ(ct− v

cx). (11.2)

To prove this transformations we basically have to repeat the steps we did deducing time dilatationand Lorentz contraction. Hence there will be rather a sketch of proof instead of a rigorous proof.− The transformation of space is in principle the same as in classical mechanics (see Galileo

transformation in chapter 11.2.3). The only difference is that we have to take into accountLorentz contraction which causes the γ factor in equation (11.1).

− For the transformation of time, we first look at the spatial dependence. This transformationdepends only on the coordinate that is in the direction of movement of the two frames. Whenwe look at the synchronisation of clocks in a moving frame (as described in chapter 11.3.2),we observe that the time the light pulse needs in one direction is longer than in the other.We have already observed this in the section about Lorentz contraction (section 11.3.6). Asa consequence we observe for a moving frame that the two clocks (in the moving frame atdifferent positions) are not synchronous which literally means they do not show the sametime20. The time in the moving frame therefore depends on the position.

− When we look at the time dependence in the time transformation we would expect t′ = tγ

according to time dilatation (see section 11.3.3). But there we looked at a slightly differentcase than here: In section 11.3.3 we observed a time interval in the moving frame from our(rest) frame. Here we look at the corresponding clock that passes a certain point ~r at acertain time t. Therefore we have to calculate which clock is at the time t at the position~r. It is the clock that was at time t = 0 at the position ~r0 = ~r − ~vt for which we canperform a clock synchronisation at t = 0 (denoted by t′0 and then calculate how much timein Σ′ is passed until this clock reached the point ~r. The synchronisation at t = 0 leads tot′0 =

x′0c = γ x0c and the time passed is then t′ = t

γ . The time dependence of ~r0 and the timeitself lead then to the given dependence in equation 11.2.

20To avoid any misunderstanding: They show the same time in their rest frame but in our system this frame ismoving and then they do not show the same time any more.

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The transformation from Σ′ to Σ is the same21 with the only difference that in Σ′ the velocitychanges sign and we get

x = γ(x′ +

v

cct′)

(11.3)

ct = γ(ct′ +

v

cx′). (11.4)

Before we finish this chapter about the Lorentz transformation let’s deduce the time dilatationfrom the Lorentz transformation. For this consider two points in time t1 and t2 in Σ and t′1 and t′2in Σ′. For the time dilatation we have to observe one clock in Σ′, which shall be located at x1 attime t1 and x2 = x1 + v(t2 − t1) at t2, whereas x′1 = x′2

22. We then get

∆t′ = t′2 − t′1 = γ(t2 −

v

c2x2

)− γ

(t1 −

v

c2x1

)= γ

(t2 −

v

c2(x1 + v(t2 − t1))

)− γ

(t1 −

v

c2x1

)= γ (t2 − t1)− γ v

2

c2(t2 − t1)

=1− v2

c2√1− v2

c2

(t2 − t1) =1

γ∆t

as we had it in section 11.3.3.

11.4 Minkowski metric

In classical mechanics a distance in the 3 dimensional space is conserved by the Galileo transfor-mations. This means a distance between two points has the same length in all inertial frames.Similarly one can define something like a "distance" in the 4-dimensional time space which will beinvariant under Lorentz transformations. This section will examine this "distance" and define socalled four vectors.

11.4.1 Definition of Minkowski metric

Let Σ and Σ′ be two frames of reference. In Σ we have two points in space and time t1, ~r1 =(x1, y1, z1) and t2, ~r2 = (x2, y2, z2). These points are often called events because an event happensat a certain time at a certain place. In Σ′ we look at the same points which are given accordingto the Lorentz transformation (see section 11.3) and denoted by t′1, ~r′ = (x′1, y

′1, z′1) and t′2, ~r2

′ =(x′2, y

′2, z′2). Then the "distance" in the 4 dimensional time space given by

21This is because the two systems are equivalent according to the first postulate of Einstein and in Σ′ the velocityof Σ is −v instead of v

22Since the y and z-coordinate have no influence on the transformation, we omit them.

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Physics Olympiad Script 11.4. MINKOWSKI METRIC

∆s2 = c2∆t2 −∆~r 2 (11.5)

= c2(t2 − t1)2 − (x2 − x1)2 − (y2 − y1)2 − (z2 − z1)2

= c2(t′2 − t′1)2 − (x′2 − x′1)2 − (y′2 − y′1)2 − (z′2 − z′1)2

= c2∆t′2 −∆~r′ 2 = ∆s′2

is in both frames the same (∆s′2 = ∆s2). This "distance" is called Minkowski metric23 and it hassome properties that are similar to the distances we know from 3 dimensions.To prove that the Minkowski metric is invariant under Lorentz transformations we assume forsimplicity that Σ′ is moving along the x-axis of Σ with velocity v and that the corresponding axisof the two systems are parallel to each other. Therefore we can already state that (y2 − y1)2 +(z2− z1)2 = (y′2− y′1)2 + (z′2− z′1)2. So we only have to look at time and the x-coordinate. We get

c2∆t′2 −∆x′2 = c2(t′2 − t′1)2 − (x′2 − x′1)2

= γ2(

(ct2 − βx2 − ct1 + βx1)2 − (x2 − βct2 − x1 + βct1)2)

= γ2(c2∆t2 + β2∆x2 − 2cβ∆t∆x−∆x2 + c2β2∆t2 + 2cβ∆t∆x

)=

1

1− β2c2∆t2

(1− β2

)+

1

1− β2∆x2

(1− β2

)= c2∆t2 −∆x2

11.4.2 Properties of Minkowski metric

The reason why the Minkowski metric is always denoted as "distance" with quotation marks isthat there is one important difference to distances we know from 3 dimensions. This difference isnot the dimensionality. The reason is the minus sign between the time and space part, if therewould be a plus sign one would also call this a distance. The reason is exactly that minus sign.A distance is something that has always a positive value (or zero). However using the Minkowskimetric, it is also possible to get a negative value if c2∆t2 < ∆~r 2. This change of sign allows us toclassify the "distance" of two events:

light cone

If two events have ∆s2 = 0 we say that they lie on the light cone. This means if at the time t1 at~r1 a light pulse would be send towards ~r2 it would reach ~r2 exactly at t2. This property is obviousbecause ∆s2 = 0 leads to

c∆t = |∆~r|

c =|∆~r|∆t

where the right side is the mean velocity of an object moving from ~r1 to ~r2 in the time ∆t. Thisvelocity is the speed of light c.

23In some books, the Minkowski metric is defined as ∆s2 = ∆~r 2 − c2∆t2. Obviously this definition does notchange the fact that ∆s2 is invariant under Lorentz transformations and the properties are also the same (up to asign)

239


timelike

In case ∆s2 > 0 the "distance" is called timelike. This means the 3-dimensional distance of thetwo events is smaller than the time a light pulse needs to travel from the first to the second event.Therefore there exists a frame of reference where the two events happen at the same position. Thisframe is travelling with the speed v = |∆~r|

∆t < c. Furthermore the condition ∆s2 > 0 means thatthe first event might have caused the second event. This is because if at t1 a light pulse would besend out at ~r1, it would reach ~r2 before t2 and might therefore causes an action at t2.

spacelike

The opposite of timelike is spacelike, where ∆s2 < 0. This means light would have longer betweenthe spatial positions of the two events than ∆t. Since the speed of light is the highest speed forinteraction there is no possibility that the first event could influence the second one. But there isa frame of reference where the two events happen at the same time. Since in this frame ∆t = 0and ∆s2 independent of the frame, ∆~r 2 must be minimal.

11.4.3 Four vectors

A compact notation for an event is the so called four vector. A four vector is a four dimensionalvector which has as first component time multiplied by the speed of light c and the other fourcomponents are the space components. The four vector to the event at t1 and position ~r1 istherefore

r1 =

ct1x1

y1

z1

where we use the convention that a four vector has no arrow above and a spatial (position) vectorhas an arrow. Usually it is clear from context if a symbol represents a four vector or a scalar. Inthis script we will not really use four vectors so you do not have to bother about the distinction offour vectors and scalars.The special feature about four vectors is that one can define a "scalar product" for them. Wehave already seen this "scalar product" and it is nothing else than the Minkowski metric. So the(relativistic) "scalar product" of two four vectors is defined as

ct1x1

y1

z1

·

ct2x2

y2

z2

= ct1ct2 − (x1x2 + y1y2 + z1z2)

11.4.4 Note on rigorous derivation

To derive relativity more rigorous one would start from the Minkowski metric. Since this deriva-tion would need more math but would not give more intuitive insight into relativity we omit it.

240

Physics Olympiad Script 11.5. VELOCITIES

Nevertheless a very rough sketch of the derivation shall be given in order to show the importanceof the Minkowsky metric and the use of four vectors24.From Newtons laws we know that in a inertial frame of reference a free particle25 moves with aconstant velocity. This must be true in all inertial frames because otherwise Einstein’s postulatenumber one would be violated. From this one can conclude that the transformations (with willlead to the Lorentz transformation) between the two frames must be a linear map. Additionallya light pulse should travel with the speed of light in all inertial frames, therefore we search fora transformation that leaves the Minkowski metric invariant26. The task is therefore to find thetransformations for a 4 dimensional (vector) space that leaves the Minkowski metric invariant. Onethen can directly deduce the Lorentz transformation as transformation of four vectors.

11.5 Velocities

In this chapter we will take a closer look at the physical quantity velocity. We will examine howvelocities "add" in relativity which will prepare us for the dynamical part of relativity. Additionallywe will have a look at the four vector of the velocity.Before we start let’s repeat an important property of the velocity between to frames of reference.Assume you are in an inertial frame and you observe a space ship passing with a constant velocityv. From the frame of the space ship you are also moving with the speed −v. This means therelative velocity between two frames is something constant and not relative.

11.5.1 Addition of parallel velocities

In Σ you are observing an other frame Σ′ which is moving with a constant speed v. Assume thatwith respect to Σ′ an object is moving in the same direction as Σ′ with a speed u′ (measured inΣ′)27. The question is, what is the speed of that object in Σ? As already mentioned, it cannot beu = v + u′ because this might be greater than the speed of light c. To calculate the right velocitywe have to go back to the definition of the velocity and apply the Lorentz transformation28. Wedenote by x the position of the object, the velocity is then given as

u =dx

dt=

γ (dx′ + vdt′)

γ(dt′ + v

c2dx′) =

dx′

dt′ + v

1 + vc2

dx′

dt′=

u′ + v

1 + vu′

c2

where we reduced the fraction be dt′ and used the definition of u′ = dx′

dt′ . The obtained result hassome interesting features:− For small speeds (v c and u′ c) the denominator is nearly 1 and as a consequence we

get the classical result.− If v = c or u′ = c we also get u = c.

24If you cannot follow the sketch do not mind, you will see it in full detail if you study physics.25A free particle is a particle on which no force is acting.26Because the Minkowski metric fulfils exactly the condition with the light pulse.27This might be written a bit theoretically. So assume the following example: You observe a space ship (Σ′) which

moves with v. In that space ship some one throws a ball with speed u′ in the same direction as the space ship ismoving.

28Actually one need multidimensional analysis but since the Lorentz transformation is a linear map, the derivative"behaves" like the Lorentz transformation itself.

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The obtained result might lead to the conclusion that the relativistic addition of velocities issymmetric in v and u′. But this is not the case, and would be obvious if one would look atvelocities which are not parallel to each other. The symmetry happens because the γ in thenumerator and denominator drops.

11.5.2 Addition of perpendicular velocities

Assume nearly the same setup as above (section 11.5.1) with the difference that the object inΣ′ moves perpendicular to the relative motion between Σ and Σ′. To make a distinction to thecalculation above we denote the velocities that are measured perpendicular to v with a subscript⊥, therefore u⊥ and u′⊥. Before we calculate the relation between u⊥ and u′⊥ lets state an im-portant fact: Although the object is moving perpendicular to v in the frame Σ′ it is not movingperpendicular in Σ because there it also moves with the speed v in the direction of the relativemotion between Σ and Σ′. Of course it also has a perpendicular component which is exactly theone denoted by u⊥. This means u⊥ is not the speed of the object in Σ, the speed would be givenas√v2 + u2

⊥. Now let’s calculate u⊥ by following the same steps as above but using that there isno Lorentz contraction perpendicular to the direction of motion, so x⊥ = x′⊥:

u⊥ =dx⊥dt

=dx′⊥

γ(

dt′ + vc2

dx′‖

) =u′⊥

γ(

1 + vc2u′‖

) =u′⊥γ

where we used that the parallel component of the velocity if the object in Σ′ is zero29. In caseof u⊥ it is obvious that v and u′⊥ do not enter symmetrically the formula (v enters the formulathrough γ).

11.5.3 velocity four vector

Similarly to the four vector defined in section 11.4.3 one can define four vectors for velocities. Letr = (ct, x, y, z)be the four vector of an object moving with a constant speed. This means at thetime t that object is at the position ~r = (x, y, z). A first (wrong) approach to a four vector for thevelocity u might be

u =d

dt

ctxyz

=

cvxvyvz

where vx is the x component of the velocity and similarly for y and z. But this approach contradictsthe idea of the absolute value of the four vector being invariant under Lorentz transformations.This is because the scalar product of u with itself is given as

u · u = u2 = c2 − v2x − v2

y − v2z .

Depending on the frame the velocities are different so |u|2 is not invariant.29If the object would move in an arbitrary direction (not perpendicular) we would keep u′‖.

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Physics Olympiad Script 11.6. DYNAMICS

To get the right result by an intuitive approach we have to go back at the definition of the derivative.A derivative is basically the fraction of two quantities ∆x

∆t with the limes ∆t going towards zero(and ∆x too). If the numerator is a four vector (as we assume here) and the denominator is aquantity that is constant in all frames, then the whole fraction is again a four vector that has anabsolute value independent of the frame. There is one special time, namely the proper time τwhich is independent of all frames. This is the time in the rest frame of the moving object. Weobtain ∆τ by the time dilatation, which leads to ∆τ = 1

γ∆t. Taking the limit for the derivativewe have do replace the ∆ by the d (roughly speaking) and we get

u =d

1γdt

ctxyz

= γ

cvxvyvz

.

If our construction of the four vector was successful, u · u should be invariant under Lorentztransformations. Lets check this by considering u · u in the rest frame of the object and in anarbitrary frame. In the rest frame the object does not move, therefore vx = vy = vz = 0 andtherefore u · u = c2. In an arbitrary frame it is moving and we get

u · u = u2 = γ2(c2 − v2x − v2

y − v2z)

=1

1− v2

c2

c2(1− v2) = c2

where we used v2x + v2

y + v2z = v2. Obviously this is invariant under Lorentz transformations and

we succeeded in constructing a four vector for the velocity.

11.6 Dynamics

Until now we only did relativistic kinematics, this means we developed a formalism do describe themotion of an object and how this description changes by a Lorentz transformation. Now we willlook at dynamics. A proper description of dynamics in special relativity as we know in classicalmechanics (Newton’s laws) is more difficult. Nevertheless we can define quantities as momentumor energy which will be conserved and which allow to calculate a lot of examples (using theseconservation laws). To derive this we start with the 3-dimensional momentum and then searchfor the four vector of the momentum. From the four vector of momentum one can derive energyconsiderations including Einstein’s famous formula E = mc2. At the end of this section we willbriefly look at acceleration and forces.

11.6.1 Momentum

Looking at the following example one can see that the momentum cannot simply be ~p = m~v wherem is the (rest) mass of the object: We observe a space ship that moves with velocity v along the x-axis. In the space ship two balls with speed v are moving perpendicular to the x-axis towards eachother (see figure 11.5). When they meet the bounce away such that they move in the x direction.From energy and momentum conservation we can deduce that they move also with velocity v alongthe x-axis in the frame of the space ship. From our system (where the space ship is moving), the

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total momentum of the two balls is 2p where p is the momentum in the x-direction of each ball(since the balls are moving towards each other, the total momentum perpendicular to the x-axis iszero). After the collision the total momentum must be conserved. But the ball moving to the leftdoes not move in our frame because in the space ship it moves with −v and the space ship itselfmoves with v, so the overall movement of this ball is zero. The second ball moving to the righthas according to the relativistic addition of velocities a speed of

v2 =v + v

1 + vvc2

=2v

1 + v2

c2

If the momentum would be p = mv this leads to a contradiction since

pbefore = 2mv 6= 2v

1 + v2

c2

m = pafter.

Figure 11.5: Collision and scattering of two balls in the moving space ship. Initially, the two ballsmove with the velocity v perpendicular to the direction of flight of the space ship. After the scatteringthey move in the same direction as the space ship.

Consequently the momentum cannot depend linear from the velocity. The right dependence isgiven as

p = γmv =1√

1− v2

c2

mv.

By a longer calculation one can show that the momentum defined this way fulfils the conservationof momentum. Since there is an other method showing that this is a good candidate for themomentum using four vectors we will look at four vectors.One important note about the mass m: Sometimes a relativistic mass mrel = γm is defined. Thisis a unsuitable definition (see section 11.6.4), when we talk about masses we always mean the (rest)mass, so the mass measured in the rest frame of the mass.

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Physics Olympiad Script 11.6. DYNAMICS

11.6.2 momentum four vector

Since ~p = m~v is wrong, the next approach might be taking the last three components of the fourvector of velocity v and multiply them with the mass. And indeed this leads to the right result.Encouraged from this success we define the four vector for the momentum as

p = mv = γm

cvxvyvz

.

As the absolute value of the velocity four vector is invariant under Lorentz transformations andthe mass is a constant, the absolute value of the momentum four vector is also constant withp2 = m2c2. What we don’t understand yet is the first component of the momentum four vector,namely γmc. This will clarify in the next section:

11.6.3 Energy

The (total) energy of an object can be calculated by

E =

ˆF ds =

ˆdp

dtds; mγc2 (11.6)

where many steps were left out including some longer calculation. Looking at the momentum fourvector, we see that the first component is exactly the energy divided by c. This means we canrewrite the four vector as

p = mv = γm

cvxvyvz

= γ

Ecpxpypz

.

This discovery leads to some interesting conclusions:

Rest Energy

In the rest frame of an object one has ~p = 0 and γ = 1. Since p2 = m2c2 we get for the rest frame

E2

c2= m2c2

E = mc2

which is exactly Einstein’s famous mass-energy equivalence. This means the mass itself is energyand the minimal possible energy of an object is E = mc2 the rest mass (times c2) of that object.

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Energy-momentum relation

For an arbitrary system the absolute value square of p multiplied with c2 leads to

p2c2 = m2c4 = E2 − ~p 2c2

E2 = ~p 2c2 +m2c4. (11.7)

This equation relates the momentum of an object with its energy. Of course this energy is thesame as the one obtained above in equation (11.6) as the following calculation shows:

E =√~p 2c2 +m2c4 = mc2

√γ2β2 + 1 = mc2

√β2

1− β2+ 1 = mc2

√β2 + 1− β2

1− β2.

small speed limit

If we look at small velocities the Energy as written in equation (11.6) can be approximated by aTaylor expansion around v = 0 an one obtains

E = γmc2 ≈(

1 +1

2

dγ

dv2

∣∣∣∣v2=0

v2

)mc2

=

(1 +

1

2

v2

c2

)mc2 = mc2 +

1

2mv2.

The first term is the rest energy, which is always there but the second therm is interesting, sincethis is exactly the kinetic energy we know from classical mechanics. So in the limit for smallvelocities, the kinetic energy of an object is exactly the one from classical mechanics.

11.6.4 Acceleration and forces

According to Newton’s law of motion the force acting on a body is equal to the rate of change ofthe momentum ~F = d~p

dt . Since ~p = mγ~v and m constant, the derivative is only acting on γ~v. Thereare two simple cases one has to consider:

Radial acceleration

If the Force acts perpendicular to the direction of motion, as it is the case at circular motion, theabsolute value of ~v does not change. Therefore γ does not change and we get

~F = γmd~v

dt

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Physics Olympiad Script 11.7. PARADOXES

Linear acceleration

If we accelerate in the direction of motion we also have to take the derivative of γ which isdγdt = γ3 ~v

c2d~vdt . This leads then to

~F = m

(γ3 v

2

c2

d~v

dt+ γ

d~v

dt

)= mγ3 d~v

dt

Since these two cases which scale different with γ an arbitrary force is not parallel to the accelerationany more! That’s why it makes no sense to talk about a relativistic mass because a relativistic massshould be defined as proportionality constant between ~F and ~a. But since these two quantities arenot proportional any more (except in the two cases above), one cannot talk about a relativisticmass. So the only meaningful mass is the rest mass m, as we have defined above (see section11.6.1).

11.7 Paradoxes

In this section we will look at some paradox which might appear when mixing classical mechanicsand relativity. Some can be resolved using special relativity, some need general relativity and wewon’t be able to totally resolve them.

11.7.1 Ladder and barn

A farmer has a ladder and a barn where the ladder is slightly longer than the barn. The farmerwants to put the ladder in the barn but since the barn is shorter this does not work. His idea isto put the ladder on his high speed tractor and them moves with ≈ 0.25c in the barn. Due tolength contraction the ladder should be smaller and therefore fit into the barn (see figure 11.6.The son of the farmer, who read this script and therefore understood relativity very well, redoesthe calculation and points out that in the frame of the moving ladder, the barn seems shorter andtherefore there is absolutely no way how the ladder would fit into the barn. Father and son decideto risk the experiment, who is right?The question "fit into the barn or not" is equivalent to the question weather the head of the ladderpasses the second door before the back of the ladder passes the first door or not. This is thereforea question about simultaneousness, which is of course relative. Or to be more precise: the frameof the tractor and the frame of the barn use different clocks and the question when which part ofthe ladder passes which door depends on the time indicated on the clocks. So the father on thetractor effectively measures 30 that the ladder does not fit into the barn whereas the son standingoutside shortly sees the ladder disappear in the barn (assuming the father drives with constantspeed through the barn).

Since the son was a curious guy he suggests the following modification of the experiment: Heinstalls a photoelectric sensor at each door of the barn and puts a lamp in the middle of the barn.If the front of the ladder passes the photoelectric sensor at the second door before the end of theladder passes the photoelectric sensor at the first door, the lamp gets on. This experiment shouldallow to take a decision whether the ladder fits into the barn or not. What will they observe?

30Measure instead of see because to see something, the light needs some time to pass the corresponding distance

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Figure 11.6: Ladder and barn. There is a door on each side of the barn such that the farmer candrive through the barn. Left: The moving ladder and the barn in the frame of the barn. Right: Theladder and the moving barn in the frame of the ladder.

They indeed observe that the light goes on. In the frame of the son (standing next to the barn)this is obvious since the ladder gets smaller as it moves. In the frame of the father on the tractorthis is not that obvious and we have to consider how the signal from the photoelectric sensor getsto the lamp. Assume there is a cable between the photoelectric sensor and the lamp where thesignal moves with the speed of light c and that there are no effects that delay the signal. Sincethe speed of light is a natural constant, both the father and the son see it move with c. But in theframe of the father the lamp is moving with velocity v. Therefore the overall31 effect of the seconddoor moving with velocity v and the signal moving with c is the same as c− v. This is opposite atthe first door, where the farmer sees an overall velocity of c + v. Although the distance betweenthe lamp and each door is the same, the signal from the second door reaches the lamp first and asa consequence it shines.In this paradox we see very nicely that there are examples in relativity which seem contradictorybut at a closer look there is no mistake in the theory.

11.7.2 Twin paradox

The probably most famous paradox is the twin paradox: One of the two twins makes a space tripin a very fast space ship whereas the other twin stays at home. After several years the first twinreturns and meets the other twin. The one who made the trip is much younger than the one stayedat home.Not knowing relativity this is astonishing because time is usually considered as absolute whichshould not be influenced by movements. Knowing relativity this might be astonishing becauseeach of the twins see the other twin move and therefore the other twin should stay younger. Thepoint in this paradox is that one of the twins has to accelerate in order to return to the other twin.

31This is the addition of two relative (and parallel) velocities measured in the same frame and not the relativisticvelocity addition discussed in section 11.5.1.

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Physics Olympiad Script 11.7. PARADOXES

This acceleration breaks the symmetry of the problem and leads to the fact that one twin staysyounger than the other.The simplest way to calculate the paradox is to assume that the twin which makes his trip moveswith constant velocity v until he changes to another space ship which moves with −v back to earth.The clocks of the two space ships are set such that they have the same time at the moment theymeet and at the position they meet.

11.7.3 Solution to the flying electron problem

We now want to go back to the problem from the beginning, namely the flying electron (see section11.1.2). For simplicity we assume that the electrons in the wire move with the same speed as theelectron above the wire32. Furthermore we assume that in the lab frame (where the electron ismoving) the wire is neutral, this means the distance between two neighbour positive atoms andthe distance between two neighbour negative electrons is the same. Let’s denote this distance withL.First of all we have to state that from the point of the lab we see the distance of two neighbourmoving electrons in the wire with a Lorentz contraction. Therefore their distance (in the directionof flight) in their own frame is L′e = Lγ where the γ = 1√

1−β2and β = v

c the velocity in terms of c.

If we now look at the frame of the moving electron, the electrons in the wire are not moving anymore (by our assumption above), their distance is therefore L′. Additionally the positive atoms aremoving which leads according to the Lorentz contraction to a distance between neighbour atomsof L′p = 1

γL. As a consequence the density of the positive atoms is bigger than the one of thenegative electrons, the wire is therefore electrically positive charged. This positive charge leads toan attraction of the single electron, similarly we observe in the lab frame.This example show how important relativity is in electrodynamics. Without relativity electrody-namics would not be a consistent theory, meaning that different observers would observe differentresult of one experiment. Nevertheless one needs a bit more math to precisely formulate electro-dynamics with relativity.

32One can do the whole calculation without this assumption which leads to the same result but needs a longercalculation respectively is less intuitive.

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250

Chapter 12

Quantum Mechanics

Classical mechanics describes how particles or bodies behave in time. So it characterises in whichway a particle moves. In Newtonian mechanics the main formula that characterises the wayparticles move is ~F = m~a. One then assumes that the particle moves according the the formulaabove on a well defined path. At the end of the 19th century some phenomena did not fit tothe predictions of classical mechanics. A new description was developed which is called quantummechanics.Quantum mechanics describes also how a particle behaves in time but the laws are very differentof those of classical mechanics and intuitively not really understandable. The main idea is that aparticle does not move on a well defined path but the probability to find it at a certain locationis connected to wave properties. This chapter shows the phenomena which did not fit to classicalmechanics and what changed in the description switching to quantum mechanics. Then someconcepts of quantum mechanics are discussed. Since the general description of quantum mechanicsis pretty complicated the concepts are discussed in a more intuitive way. At the end we will lookhow a rigorous calculation in quantum mechanics looks like. This rigorous calculation is surly notnecessarily to know for the Olympiads (not even for the IPhO) but it should show how quantummechanics really looks like. The main sources for this chapter are [47], [48], [49] and [50].

12.1 Experiments

We want to look at some experimental setups and discuss how quantum mechanics changed theunderstanding of physics and what the main statements are.

12.1.1 Black body radiation

A body with a certain temperature T emits electromagnetic waves, called radiation. Dependingon the surface, the body can emit electromagnetic waves for certain frequencies better than forothers. If there are two bodies and one body is hotter than the other, energy can only go fromthe hotter to the colder body. Therefore for any frequency the capability of a body to absorb anelectromagnetic wave is the same as the capability to emit a wave.A black body is a body which absorbs all electromagnetic waves. As a consequence it also emitswaves very well. If a body absorbs an electromagnetic wave it gets heated up by the energy of thewave. The emission of an electromagnetic wave is also due to the thermal energy of a body.A surface which has the property of a black body can by achieved if one takes a closed cavity with

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Physics Olympiad Script CHAPTER 12. QUANTUM MECHANICS

a little hole (see figure 12.1). The area where the hole is behaves like a black body because at thespot where the hole is all the incident waves will go into the cavity and are therefore absorbed bythe body.

Figure 12.1: A cavity with a little hole [51].

Inside the body there are some other interesting properties: Since all the walls of the cavity havethe same temperature there is no net flow of energy from one wall to the other. Therefore theelectromagnetic waves inside the cavity have a intensity such that the emission and absorptionat the walls is in an equilibrium. We assume the hole to be very small, therefore at the holethe radiation is the same as inside the cavity. Since the hole behaves like a black body theelectromagnetic waves inside the cavity look like they were emitted by a black body, although thewalls of the cavity need not to be black. This is because the electromagnetic field and the wallsare in an equilibrium.

At the end of the 19th and beginning of the 20th century there was a big discussion about thequestion how the power that a black body radiates is distributed as a function of the frequency f .This means how much power is radiated between a small interval f and f + df .The classical approach is to model the electric field inside the cavity as standing waves betweenopposite walls. According to thermodynamics any possible standing wave has the same amount ofenergy (equipartition theorem). But for higher frequencies also diagonal standing waves can happentherefore the number of standing waves per frequency is not fix. Since usually the dimension ofsuch a cavity is in the order centimetre and the wavelength of light is in the order of 500 nanometresthe discrete difference between two successive frequencies is very small and we treat the spectrumcontinuous. A rigorous calculation then gives that the number of standing waves between f andf + df is

g(f)df =4πf2V

c3df

where V is the volume of the cavity. We see that the number of standing waves of a certainfrequency is proportional to the square of the frequency (for high frequencies).

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Physics Olympiad Script 12.1. EXPERIMENTS

The equipartition theorem from thermodynamics states that each standing wave has the meanenergy kBT where kB is the Boltzmann constant and T the temperature. Therefore we have anenergy density (Energy per volume) inside the cavity of

ρ(f)df = kBT8πf2

c3df

This law is called Rayleigh-Jeans law (see picture 12.2). The additional factor 2 appears becausethere are two possibilities of polarisation. The big problem is that the energy density for highfrequencies goes to infinity. Therefore in the cavity should be infinite energy which is of coursenot possible.

In 1900, Max Planck introduces an other idea: He said that the energy of light (of a certainfrequency) does not have an arbitrary amount. Instead he claimed that light at a certain frequencyhas a smallest amount of energy which is given by E0 = hf where h is the Planck constant withh = 6.626 · 10−34J·s. He did not believe that his assumption describes a real property of nature,he just tried to do the calculation a bit different. The calculation lead to the formula

ρ(f)df =8πhf3

c3

1

ehfkBT − 1

The additional exponential in the denominator causes the curve to get small for high frequencies.The curve is drawn for different temperatures in figure 12.2. The measurements confirm Planckscalculation.Since for small frequencies: e

hfkBT − 1 ≈ 1 + hf

kBT− 1 = hf

kBTthe Rayleigh-Jeans law is therefore an

approximation for small frequencies.

This was then the beginning of a new theory, the quantum theory. The introduction of the smallestenergy E0 quantises the energy of an electromagnetic field. Therefore electromagnetic waves behavelike waves but have also quantised property which suits more to the model of particles than waves.This particle property gets clearer in the next chapter.

12.1.2 Photoelectric effect

If one shines light on a charged metallic plate one can observe that the plate loses charge with time.To examine this effect let’s look at the experimental setup shown in figure 12.3. We shine light witha frequency f on the left metallic plate. We will see that this plate loses electrons therefore we callit cathode. Parallel to this plate we have an other plate where no light shines on, we call it anode.Between these two plates we connect a voltage source with variable voltage U . We define U > 0 ifthe cathode is connected to the plus pole of the voltage source (so electrons flow from the cathodeto the voltage source) and U < 0 if the cathode is connected to the minus pole. Additionally weput an ammeter on the electric circuit in order to measure the current I that flows.

Dependence of voltage U

We see that for all voltages that are smaller than a certain voltage Ut (also called threshold voltage)there is a current flowing from the anode to the cathode. Since metallic atoms can not really leave

253


Figure 12.2: Radiated intensity for different temperatures. The x-axis is the wavelength and it isobvious that for high frequencies (small wavelength) the intensity according to Plancks calculationgets small. For the callsical approach the intensity goes towards infinity [52].

the metallic plate we conclude that the electrons move. Since the electrons are negative chargedand the current flows from the anode to the cathode, the electrons move from the cathode to theanode. The light seems to hit out electrons at the cathode and they travel to the anode. Theexistence of Ut (for a certain intensity) is reasonable because for higher voltages the electrons needmore energy to fly against the electric field between the anode and cathode.

Dependence of intensity

If the voltage is smaller than the threshold voltage U < Ut we recognise that the current isproportional to the intensity. This seems also reasonable because if we shine with more light onthe cathode there will be more electrons hitting out and reaching the anode (until saturation effectsoccurs).For a certain intensity the existence of a maximal voltage Ut is reasonable because if we raise thevoltage the electrons need more energy to overcome the electric field. The remarkable thing thatthe experiment shows is, that Ut is independent of the intensity. This means that an electron doesnot collect energy from the light until it can leave the cathode and reach the anode. Instead it getsonce a certain amount of energy and if the energy is big enough to leave the cathode and reach theanode the electron will do so. Therefore the energy from the light is concentrated in small packets.We call this energy packets photons.

Dependence of frequency

If we change the frequency of the light we see that the threshold voltage also changes. If we risethe frequency the threshold voltage also rises. The dependence seems to be linear function which

254


Figure 12.3: Light shines on an metal plate (Cathode) which emits electrons. Depending on thevoltage U the electrons reach an other metal plate (Anode). If they reach the anode the electriccircuit is closed and a current flows which is measured at the ammeter (A) [53].

is drawn in figure 12.4. Therefore the energy of a photon is proportional to the frequency. They-intercept can be interpreted as the work that has to be applied to the electron in order to extractit from the metallic plate. The work to leave the plate is therefore V0e where e is the charge of anelectron.

Figure 12.4: Dependence of the cut-off potential Ut on the frequency f [54].

Interpretation

As we already stated, the energy of the light does not flow continuously but in small packets. Wecan read out the amount of Energy per frequency of figure 12.4 by taking the slope of the linearfunction. If we do this, we get that the slope is h, the Planck’s constant. Therefore the energy ofsuch an energy packet is given by E = hf as we already assumed in chapter 12.1.1.The intensity of the light is only a measure of how many Photons arrive at the cathode per second.

Light is therefore on one hand a wave because we have wave phenomena like refraction and diffrac-

255


tion (see chapter 12.1.3). On the other hand it has particle-like properties because its energy (andalso its momentum) travels in small packets. This shows the very unintuitive nature of quantummechanics.

12.1.3 Double slit experiment

The double slit experiment is an experiment which points out the wave property of quantum objects.This experiment (see figure 12.5) consists of a light source which emits light with a wave length λand a wall which absorbs light except at two small slits (small compared to the wavelength) whichare separated by a distance similar to λ. The light can go through these to slits and is measuredon a screen.

Figure 12.5: Double slit experiment: The source shines on the wall which gives a pattern. Thereare two different pattern shown. In the left pattern the intensity I1 from the first slit (upper one) isdrawn if we block the second slit and also the intensity I2 from the second slit if we block the firstone. Additionally the sum I = I1 + I2 is drawn. The right pattern is the wave pattern which oneexpects when a wave hits both slits and gets diffracted [55].

Let’s now discuss what we measure on the screen and what happens if we change the setup slightly:

Normal setup, no changing

The light behaves as a wave and gets diffracted at the two slits. This means that we have to treateach slit as a source of new spherical waves A1 and A2 and then for each point on the screen addthe amplitudes from each slit A = A1 + A2. The measured pattern is then proportional to A2

because we measure the intensity. Therefore we get an interference pattern (the right pattern infigure 12.5).

Closing one slit

If we close one slit the light goes only through the other slit. There a spherical wave is causedwhich hits the screen making a pattern like I1 (or I2 if the first slit is closed).

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Measuring the path

If we have both slits open we get a diffraction pattern as described above. Thinking the lightto be a flow of photons which (should) behave like particles we might be interested in measuringthrough which slit the individual photons pass. Therefore we install a detector which measureswhere the particles come from. As soon as the detector is installed and measures, the light behavesdifferently: The diffraction pattern disappears and we get a pattern as if the particle would flythrough one or the other slit, therefore we get the left pattern in figure 12.5. This (process) iscalled collapse of the wave (function) because the wave property is disappeared. We state that themeasurement influences the outcome of the experiment, which is typically for quantum mechanics.

Very low intensity

If we reduce the intensity until only one photon is emitted at a time we will also get the interferencepattern if we wait long enough (see figure 12.6). Therefore each single photon interferes with itself.

Figure 12.6: The light shines with very low intensity such that single phtons reach the screen. Onthe left side the beginning of the experiment is shown where not many phtons hit the screen. Onthe right side many phtons hit the screen and an interference pattern is visible. [56].

The pattern is then no intensity distribution any more because single photons produce only smalldots. The light dots show that light hits the screen as a small packet (in fact it is not possible tosee a photon hitting an usual screen. But if one replaces the screen by a very sensitive detector itis possible to measure single photons). Therefore the pattern is the sum of all these photons andit is proportional to the probability that a photon interferes to that position.We can also switch on the detector to measure through which slit a photon passes. If we do so wemeasure again a probability distribution like the left pattern in figure 12.5.

Interpretation

Since the behaviour of the experiment does not change if we perform it with single photons wehave to describe the observations by behaviour of single photons (and not by the interaction ofmany photons). We relate the fact that a photon is a light packet to a particle property. Thereforea photon is a particle which travels and if it hits the screen we see a dot like we would expect if aball hits a wall. The interference pattern however is clearly caused by a wave property.

257


If we combine these two properties we get that light is a flow of particles and the probability tofind a particle is connected to a wave like probability distribution. If we describe the light this waywe can also explain intuitively the collapse of the wave when we perform the measurement of thepath. Because in that moment we see the photon going through one slit the probability to findthat photon in the other slit is zero. Therefore the photon starts its wave like behaviour from theslit we observe it and therefore it gets an other probability distribution.

12.2 Laws of Quantum Mechanics

As we have seen in the experiments above quantum systems behave pretty different to what we knowfrom classical mechanics. We now want to examine some basic properties of quantum mechanics.Since the general description is too complicated for this level we will not be able to derive all lawsand understand all connections.

12.2.1 Wavefunction and probability

As we have seen in the double slits experiment the probability P is proportional to the square ofthe superposed light waves P ∝ (E1 +E2)2 where E1 is the electric field coming from the first slitand E2 from the second slit. Therefore the interference is due to the fact that one takes the sumand then squares and not the other way round (E1 + E2)2 6= E2

1 + E22 .

If we describe quantum systems generally we have a function Ψ(x) which is called wavefunctionand which assigns to each place x a value Ψ(x). The probability P to find a particle at a certainplace x is then given by P = |Ψ(x)|2. The absolute value is necessary if the function Ψ is a complexnumber.The probability P is a probability distribution. This means that the probability to find the particleat exact the position x0 is zero vor every x0. It’s like shooting a football on a target: It is impossibleto hit the target exactly in the middle and if it seems one has hit it in the middle you have tolook more precise (perform a more exact measurement) and you will find out that it was not themiddle. Therefore the probability has to be understood a bit different: The probability P12 to finda particle between x1 and x2 is

P12 =

x2ˆ

x1

P (x)dx =

x2ˆ

x1

|Ψ(x)|2dx

Therefore one can understand a probability distribution (of the place) as Pdx is the probability tofind a particle between x and x+ dx.

12.2.2 Measurement

If we perform a measurement the wave behaviour (of the measured quantity) of a quantum mechanicparticle disappears and we get a concrete value (with a certain uncertainty, see chapter 12.2.4). Ifour wavefunction Ψ(x) is related to the probability to find the particle at a certain position x (tobe more precise to find it between x and x + dx) and we measure the position we get a value ofthe position of the particle according to the probability distribution.But if Ψ(x) is the wavefunction to find the particle at a certain position and we measure themomentum of the particle we will not get a momentum distribution according to |Ψ(x)|2. To get

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Physics Olympiad Script 12.2. LAWS OF QUANTUM MECHANICS

the probability distribution |Φ(p)|2 for the momentum p with wavefunction Φ(P ) we have to makesome more calculations which we do not treat here. But we recognise that performing a measure-ment of the momentum knowing the position distribution is similar to transforming the positiondistribution to the momentum distribution. Therefore measurements are associated to operators:The measurement of the momentum is like an operator p which acts as p(Ψ(x)) = Φ(p)Ψp(x)where Ψp(x) is the wavefunction with |Ψp(x)|2 describes the probability distribution to find theparticle with momentum p at the position x (for any momentum p we get a certain distributionΨp(x)). If we measure the momentum we measure a concrete value p0 with wavefunction Ψp0(x).If Ψp0(x) 6= Ψ(x) we get an new wavefunction. This means that the measurement might changesthe behaviour of the quantum system.

12.2.3 De Broglie hypothesis

Until now we only looked at the photon as quantum object. In 1924 De Broglie stated that anyobject is a quantum object which behaves according to quantum laws. Since quantum physicsworks with waves we have to contribute a wavelength to each object. De Broglie stated that thewavelength λ of any object is given by

λ =h

p(12.1)

where p is the momentum of the object and h = 6.626 · 10−34J·s the Planck constant. Thereforebig objects with a high mass have a very small wavelength. This is also the reason why we do notexperience quantum effects in every day life. Because to get the typical properties of a wave thewave has to hit structures which have a similar size as the wavelength (for example diffraction isnearly not observable if slits are very far away from each other). Since the wavelength of objectsaround us (like a football) is very small (≈ 10−34m) there exist no slits or something similar withsuch a small distance (Atoms have a diameter of ≈ 10−10m). Therefore no interference phenomenaoccur with every day objects.But one can take for example electrons or atoms and perform similar experiments as the doubleslits experiment (see chapter 12.1.3) and one get also the wave behaviour of those particles.

A comment to the photon. From relativity there is the equation

E2 = c2p2 + c4m2

with E the energy, p the momentum, c the speed of light and m the mass (to be more preciselythe rest mass). Since the photon has no mass, one gets that E = pc. If we put this in equation(12.1) we get λ = hc

E and using λf = c we have E = hf as we stated at the black body radiationand the photoelectric effect (chapter 12.1.1 and 12.1.2).

12.2.4 Uncertainty Principle

If we want to measure the position of a quantum object we might use light. A limitation of themeasuring precision is the angular resolution: To distinguish two objects which are separated bya distance l we need light with a wavelength λ < l. Otherwise the interference patterns of the twoobjects in our eye (or on the film in a camera) are too close to each other to distinguish them.

259


If we want to measure the position of a quantum object very precisely we need light with verylow wavelength. But light with very low wavelength has a very high momentum and if the lightof our measurement interacts with the quantum object, the quantum object might gain a lot ofmomentum. Therefore we know its position very well but we know nothing about its momentum.Heisenberg stated that the uncertainty of the position σx and the uncertainty of the momentumσp are not independent. There is a natural bound which avoids that we can measure the positionand the momentum very precisely. The Uncertainty Principle is

σx · σp ≥h

4π

where h is again the Planck constant.This inequality does not only hold if we measure with light, it is a fundamental uncertainty andindependent of the measuring method.There is also an uncertainty of the time and energy which is given by

σE · σt ≥h

4π

where σE is the uncertainty about the energy and σt about time.

The uncertainty principle is an interpretation of the diffraction of a quantum object at a very smallslit: The smaller the slit is the more precise we know where the particle is when it crosses the slit.But a small slit produces a wide diffraction pattern, therefore the smaller the slit is the less weknow about the momentum perpendicular to the slit.

12.2.5 Schrödinger Equation

Until now we never discussed how a quantum system changes with time. This has its reasonbecause the time evolution is described by the Schrödinger equation which is a pretty complicateddifferential equation. Therefore this chapter is more to give a complete overview over quantummechanics and it is absolutely not relevant to know or understand the equations.Let Ψ(x, t) be the wavefunction of a quantum particle along the x-axis which also depends on thetime t. The (time dependent) Schrödinger equation is given by

ih

2π

∂Ψ

∂t= − h2

8π2m

∂2Ψ

∂x2+ V (x)Ψ (12.2)

where i is the imaginary unit, m the mass of the particle and V (x) the potential energy. The ∂Ψ∂t

is the derivative with respect to the time. The reason why we used ∂ instead of d is because Ψdepends on the time t and the position x and by using ∂ we indicate that we only differentiateaccording to time. ∂2Ψ

∂x2is the second derivative with respect to x and it is related to the kinetic

energy of the particle.The differential operator i h2π

∂∂t is the operator which is related to the energy distribution. If we

have a constant energy E the energy operator gives us just the energy, therefore i h2π∂Ψ∂t = EΨ.

This leads to the time independent Schrödinger equation

260


EΨ = − h2

8π2m

∂2Ψ

∂x2+ V (x)Ψ (12.3)

In a problem one often searches a function Ψ which solves the equation. As you can imagine thisis pretty complicated.

12.3 Examples

We now want to look at some examples and apply the principles of quantum physics.

12.3.1 Bohr model

One big application of quantum physics is the description of atoms. The simplest atom is thehydrogen atom with one proton as nucleus in the middle and one electron "flying" around theproton. Since the precise calculation of the hydrogen atom is pretty laborious and needs muchmath we will derive the Bohr model. The Bohr model is a semi classical description of the hydrogenatom which is not really correct but gives some nice predictions.The Bohr model assumes the following properties:− The electron orbits around the nucleus. Since an accelerated charge radiates energy, a classical

orbit, as the motion of planets around the sun, is not possible. Instead we assume that theelectron behaves like a wave with frequency f . This wave goes around the nucleus and hasto be a standing wave (this means that the "start" and "end" point of the wave must meeteach other). These standing waves are drawn in figure 12.7.

− If an electron changes the orbit with energy difference ∆E it emits a photon with frequencyν according to ∆E = hν.

Figure 12.7: The nucleus in the middle and two possible electron orbits. The orbits are like closedstanding waves [57].

261


The first assumption leads to the restriction, that circumference of the orbit must be a multiple ofthe wavelength of the electron wave. Therefore

2πr = nλ =nh

mv

where r is the radius of the orbit and n is an integer. We also used the de Broglie wavelengthλ = h

p where the momentum is classically given by p = mv where m is the mass of the electronand v its velocity.

An other equation is given by the classical orbit equation where the centripetal force is due to theelectric attraction. Therefore we get

mv2

r=

Ze2

4πε0r2(12.4)

where Z is the number of protons in the nucleus (which is 1 for hydrogen, but we want to calculateit more generally).

If we combine these two equations we get

rn = n2 ε0h2

πmZe2

where rn labels the different orbits according to n. For the smallest radius of the hydrogen atomwe get r1 = 5.3 · 10−11m

If we now calculate the energy of the electron going from an orbit with radius ra to one with radiusrb we have to take into account two terms: The kinetic energy and the potential energy. Thepotential energy is given by

∆Epot =Ze2

4πε0

(1

ra− 1

rb

)The kinetic energy of the electron on an orbit rn is given by

Ekin rn =1

2mv2

n =1

2

Ze2

4πε0rn

where we used equation (12.4) to express the velocity by the radius.The total energy difference between ra and rb is therefore

∆Etot = ∆Epot + Ekin rb − Ekin ra

=1

2

Ze2

4πε0

(1

ra− 1

rb

)=

1

2

Ze2

4πε0r1

(1

a2− 1

b2

)

262


where we used that ra = a2r1 and rb = b2r1 with a and b integers.If ra > rb the electron loses energy which causes a photon leaving the atom with the frequencyν = E

h respectively with the wavelength λ = chE . For transition from the second a = 2 to the first

b = 1 orbit we get a wave length of 121nm which is clearly in the ultra violet. Since any othertransition to the first orbit has more energy, all transitions to the first orbit are not visible. This isdifferent for transitions to the second orbit b = 2 where we might get visible light. This is shownin figure 12.8. An important property of the calculated energies is that they have discrete values.Therefore a hydrogen atom has some certain frequencies which it can emit light. This frequenciescorrespond to the spectral lines of hydrogen.If we calculate how much energy is needed to take the electron from the first orbit to infinity weget an energy of 13.6eV which agrees to the measurement.

Figure 12.8: Transistions to the second orbit [58].

263


The Bohr model is a first approach to describe the behaviour of atoms with only one electron.The problem with the model is, that it still does not really avoid the problem that acceleratedcharge radiates electromagnetic waves and therefore looses energy. Additionally it violates theHeisenberg’s uncertainty principle. To get the precise description of an atom one has to solveSchrödinger’s equation (see chapter 12.2.5).

12.3.2 Rigorous example

In this chapter we want to look how the calculation with the wave function looks like. This is notrelevant for any selection round, and not even for the IPhO. Since the math is sometimes beyondthis level, not anything can be derived.

The simplest case for a quantum system is when we have a particle in an interval [0, a] where thepotential is zero and infinite anywhere else (see figure 12.9).

Figure 12.9: Infinite potential (V ) for x < 0 and x > a. Additionally the first three standing wavesare shown.

We now search for static wave functions which means that they do not change with time. Thissimplifies the calculation a lot because we can use the time independent equation (12.3) instead ofthe time dependent (12.2). For x < 0 or x > a we get therefore

− h2

8π2m

∂2Ψ

∂x2= (E − V )Ψ

Since E − V is something like minus infinity we see that the particle would need infinite energy tobe in this region which is impossible. Therefore the probability to find the particle there is zeroand as a consequence the wave function too.

264


More interesting is the region 0 < x < a because there we have to solve the equation

− h2

8π2m

∂2Ψ

∂x2= EΨ (12.5)

If the second derivative appears it is always a good idea to think of sin(x) and cos(x). We try tosolve the equation (12.5) by attempting Ψ(x) = A sin(kx)+B cos(kx) where A and k are constantswhich have to be determined. If we formulate equation (12.5) a bit different and use the attemptwe get

−8π2mE

h2Ψ =

∂2Ψ

∂x2

−8π2mE

h2(A sin(kx) +B cos(kx)) = −k2(A sin(kx) +B cos(kx))

k =√

2mE2π

h

To get a restriction on A and B we have to consider the following: at the position x = 0 thewave must have amplitude zero because there the infinite potential begins. Therefore B = 0.Additionally we have also the restriction Ψ(a) = 0 which leads to the fact that the argument inthe sin must be a multiple of π. This means ka = nπ where n is an integer. But this can only bethe case if the Energy E has a certain value, namely

En =n2π2h2

2ma24π2=

n2h2

8ma2

This means that only discrete energy levels are allowed in order to get time independent solutions.The subscript n at En indicates which energy we look at.To get the A we use the condition that the total probability to find the particle between 0 and amust be 1

1 =

aˆ

0

|Ψ(x)|2dx

=

aˆ

0

A2 sin(kx)2dx

= A2

[1

2(−1

ksin(kx) cos(kx) + x)

]a0

= A2a

2

A =

√a

2

265


Therefore the possible wavefunctions are superpositions of the Ψn(x) which are

Ψn(x) =

√2

asin(

nπ

ax)

where the Ψn(x) has the energy

En =n2π2h2

2ma24π2=

n2h2

8ma2

266

Chapter 13

Introduction to Statistics

The main goal of statistical methods is to make inferences based on data. There are three importantsteps in this process: Collecting the data, describing the data and analyzing the data. While wewill focus on descriptive statistics in this introduction, it is important to mention that every stepheavily relies on the previous step. If your experimental setup does not provide good data, youwill not be able to draw any meaningful conclusions (Garbage in - Garbage out).The main sources are [59], [60], [61], [62] and [63].

13.1 Location and Spread of a single Set of Data

Let X = x1, x2, . . . , xn denote a set of n data points. The mean x and the variance σ2 of X aredefined as

x =1

n

n∑i=1

xi (13.1)

σ2 =1

n− 1

n∑i=1

(xi − x)2 (13.2)

By taking the square root of σ2, we find the standard deviation, a common measure for how spreadthe data is, which has the nice property of having the same units as the original data. By dividingthe standard deviation by the mean, we get the coefficient of variation, an indicator for the relativespread of the data.

σ =√σ2 (13.3)

CV =σ

x(13.4)

Another way of better describing how data is located are quantiles. The idea is to divide the datainto equally large groups and indicate where these cuts are. As an example, when calculating thelower and upper quartile, one quarter of the data points are smaller than the lower quartile andone quarter are larger than the upper quartile, with the remaining half being located between thesetwo values. It is important to note, that for this calculation, the data must first be sorted smallestto largest (i.e. xi−1 ≤ xi ≤ xi+1). The general formula for calculating quantiles is then given by(13.5) for a cut at the fraction p (i.e. p of the data points below the cut and (1 − p) above thecut). As an example, for the lower and upper quartiles one would calculate Q(0.25) and Q(0.75).

Q(p) = (1− g) · xj + g · xj+1 (13.5)

267

Physics Olympiad Script CHAPTER 13. INTRODUCTION TO STATISTICS

j = bpn+1

2c g =

(pn+

1

2

)− j (13.6)

(Note that bzc designates the floor of z, i.e. the next lower integer)Common quantiles used are quartiles (p = k

4 ), deciles (p = k10), percentiles(p = k

100) and of coursethe median (p = 1

2), for which we can simplify the calculation:

x =

x[n+1

2] n odd

12 · (x[n

2] + x[n

2+1]) n even

(13.7)

13.1.1 Bivariate Analysis

If we have two sets of data X and Y collected in parallel so that each xi is associated to itscorresponding yi, it might be interesting to measure how the data in these sets might be connected.This can be quantified by calculating the covariance and correlation of these sets:

Cov(X,Y ) =1

n− 1

n∑i=1

(xi − x) (yi − y) (13.8)

Corr(X,Y ) =Cov(X,Y )

σx · σy(13.9)

Finally, if we suspect that there is a linear relationship between X and Y , we can try to calculatea line y = a+ bx that fits our data as much as possible. This is called linear regression and is oftendone using the least squares method. This method minimizes the square of the errors betweenthe calculated line and the observed data points. The coefficients can be calculated as shown in(13.10) and (13.11). The derivation can be found in the appendix.

b =Cov(X,Y )

σ2x

(13.10)

a = y − bx (13.11)

13.2 Uncertainty Propagation

When performing experiments one should always be aware of uncertainties in measurements. Thereare many sources of uncertainties: Noise, minimum resolution, misalignment, calibration, etc. Toasses the impact of uncertain measurements on results of complex calculations, the propagation ofuncertainty has to be analyzed.

13.2.1 Quantification of Uncertainty

There are two different representations of uncertainty for a measurement. Additive uncertaintyis expressed as an absolute range by which the real value might be different from the measuredvalue (e.g. ±0.001m), while relative uncertainty is expressed as a fraction of the measurement (e.g.±0.1%). They are easily convertible into one another:

εrel =εadd|X|

(13.12)

268

Physics Olympiad Script 13.2. UNCERTAINTY PROPAGATION

Since it is often impossible to find definitive upper and lower bounds for the error of a measurement,we usually express uncertainty in terms of the expected standard deviation of the measurement,xreal = xm ± σxm , where σxm can be found either by performing a measurement multiple timesor by carefully assessing the different possible sources of uncertainties (e.g. for a ruler with mmmarkings, it is reasonable to assume an uncertainty of ±0.5mm). For this reason, we will expressuncertainties as additive uncertainties δX in this document.

13.2.2 Propagation of Uncertainty

In this section, we will consider different types of operations R being applied to uncertain variablesX ± δX, Y ± δY and Z ± δZ, and will quantify the uncertainty δR of the result.− Addition of a constant

The addition of a constant will not affect additive uncertainty.

R(X) = X + c (13.13)δR = δX (13.14)

− Addition of uncertain variablesBy adding multiple uncertain variables, the variance of the result is the combined varianceof the individual elements. However, as we express uncertainties with standard deviation, weneed to take the square root of the added variances.

R(X,Y, Z) = X + Y − Z (13.15)

δR =√

(δX)2 + (δY )2 + (δZ)2 (13.16)

− Multiplication with a constantBy multiplying an uncertain quantity with a constant, the uncertainty is simply multipliedwith the absolute value of the constant.

R(X) = a ·X (13.17)δR = |a| · δX (13.18)

− Multiplication of uncertain variables When multiplying uncertain variables, this corre-sponds to adding the relative variances to find the relative variance of the result.

R(X,Y, Z) =X · YZ

(13.19)

δR = |R| ·

√(δX

X

)2

+

(δY

Y

)2

+

(δZ

Z

)2

(13.20)

− General OperationsFor general operations, the formula can be derived by considering the variation of the resultdue to every variable and again adding them just like standard deviations. In fact, all previousrules are just special cases of this rule and can be derived easily.

R(X,Y, . . . ) (13.21)

δR =

√(∂R

∂XδX

)2

+

(∂R

∂YδY

)2

+ . . . (13.22)

269


13.3 Units

A good understanding of the different units and their respective dimensions can be very helpful toavoid careless mistakes.

13.3.1 The International System of Units (SI)

The International System of Units (SI, système international (d’unités)), is the most widely usedsystem of measurement due to its simplicity regarding unit conversions. The system comprisesseven base units from which many other units can be derived (e.g. 1N = 1 kg·m

s2).

Dimension Unit AbbreviationElectric Current Ampere ATemperature Kelvin KTime Second sLength Meter mMass Kilogram kgLuminous Intensity Candela cdAmount of Substance Mole mol

Table 13.1: The seven SI base units

13.3.2 Prefixes

Prefixes can be used to change the order of magnitude of a unit.

Prefix Symbol Factor Prefix Symbol Factorfemto f 10−15 peta P 1015

pico p 10−12 tera T 1012

nano n 10−9 giga G 109

micro µ 10−6 mega M 106

milli m 10−3 kilo k 103

centi c 10−2 hecto h 102

dezi d 10−1 deca da 101

Table 13.2: Metric Prefixes

13.3.3 Dimensional Analysis

We can use the fact that dimensions corresponding to the seven base units cannot be createdfrom any other base dimensions to quickly check equations. If an equality does not have thesame dimension on each side, then surely it cannot be true (however, the opposite cannot besaid, even if the dimensions agree there could be a mistake in form of a dimensionless factor). Toperform dimensional analysis on an equation, we replace all involved variables with their respectivedimension. We then simplify both sides to see if the dimensions cancel each other. As an example,

270

Physics Olympiad Script 13.4. GRAPHS

this technique is applied to Newton’s second Law:

F = am

[M ][L]

[T ]2=

[L]

[T ]2· [M ]

13.4 Graphs

13.4.1 Elements of good graphs

When presenting your results in form of graphs, there are some guidelines that you should respectto make your graph clear and understandable:

1. Complexity: A graph should not be more complex than the data it represents. Avoidirrelevant decoration, 3D effects and distortion.

2. Scaling:(a) The data should not be clumped in one section of the graph(b) The scale should not change along one axis(c) Your axes should include 0 and have no jumps(d) Use simple steps, one square or tick mark could represent 1, 2, 5, 10, etc..

3. Title: Your graphs should have a descriptive title that contains information about the originof the data.

4. Multiple Data Sets: If your plotting multiple sets of data on the same graph, make surethey’re easily distinguishable and include a key/legend (Should not obstruct Data)

5. Labeled Axes: Label your axes with the name of the variable, its unit and the scale (Ticksand Numbers). There are multiple ways of including the units, however we suggest you usethe ISO standard variable_name /unit.

6. Readability: When drawing graphs by hand, use a Ruler.

13.4.2 Logarithmic Plots

Logarithmic and Semi-Logarithmic Plots are useful tool to identify special types of relationshipbetween variables. They are characterized by one or both axis being scaled logarithmically insteadof linearly. For a logarithmic plot, this means that monomials of the form y = axk appear asstraight lines with slope k. This can be seen by applying a log function to both sides of theequation:

log (y) = log (axk) = log (a) + k log (x) (13.23)

In a similar fashion, we can see that relations of the form y = λaγx appear as a line with slope γon a semi-log plot:

loga (y) = loga (λaγx) = γx+ loga (λ) (13.24)

Or using a base 10 log:log y = log (λaγx) = (γ log (a))x+ log (λ) (13.25)

271


Figure 13.1: Different Monomials on a Logarithmic Plot.

Figure 13.2: Different Exponentials on a Semi-Logarithmic Plot.

272

Appendix A

Further derivations

A.1 Derivations of Statistics

A.1.1 Alternative formulations for Variance and Covariance

By applying the sum to individual elements and using∑n

i=1 xi = nx, alternative formulations canbe found that are sometimes more comfortable to apply

σ2 =1

n− 1

n∑i=1

(xi − x)2

=1

n− 1

n∑i=1

(x2i − 2xix+ x2

)=

1

n− 1

(n∑i=1

x2i − 2x

n∑i=1

xi +n∑i=1

x2

)

=1

n− 1

(n∑i=1

x2i − 2nx2 + nx2

)

=1

n− 1

(n∑i=1

x2i − nx2

)

Cov(X,Y ) =1

n− 1

n∑i=1

(xi − x) (yi − y)

=1

n− 1

n∑i=1

(xiyi − xiy − xyi + xy)

=1

n− 1

(n∑i=1

xiyi − yn∑i=1

xi − xn∑i=1

yi +n∑i=1

xy

)

=1

n− 1

(n∑i=1

xiyi − nxy − nyx+ nxy

)

=1

n− 1

(n∑i=1

xiyi − nxy

)

273

Physics Olympiad Script APPENDIX A. FURTHER DERIVATIONS

A.1.2 Derivation of the Least Squares Coefficients

We want to generate a linear regression to predict y for a given x. We will call y = a + bx ourpredictor for y. We define the sum of square errors as a function of the coefficients a and b:

S(a, b) =

n∑i=1

(yi − y)2

=

n∑i=1

(yi − a− bxi)2

Since we want to find a and b that minimize this function, we will set the partial derivatives withrespect to a and b equal to zero and solve for a and b.

∂S

∂a= −2

n∑i=1

(yi − a− bxi) = 0 (A.1)

∂S

∂b= −2

n∑i=1

(xi(yi − a− bxi)) = 0 (A.2)

We will start by simplifying the sum in (26):

n∑i=1

(yi − a− bxi) =

n∑i=1

yi −n∑i=1

a− bn∑i=1

xi (A.3)

= ny − na− bnx (A.4)

In (26):

−2(ny − na− bnx) = 0 (A.5)y − a− bx = 0 (A.6)

a = y − bx (A.7)

We will use this expression for a and simplify the sum in (27):

n∑i=1

(xi(yi − a− bxi)) =n∑i=1

(xi(yi − y + bx− bxi)) (A.8)

=

n∑i=1

xiyi −n∑i=1

xiy +

n∑i=1

xibx−n∑i=1

bx2i (A.9)

=

n∑i=1

xiyi − yn∑i=1

xi + bx

n∑i=1

xi − bn∑i=1

x2i (A.10)

=

n∑i=1

xiyi − nxy + b

(nx2 −

n∑i=1

x2i

)(A.11)

= (n− 1) · Cov(X,Y )− b(n− 1) · σ2x (A.12)

274

Physics Olympiad Script A.1. DERIVATIONS OF STATISTICS

In (27):

−2((n− 1) · Cov(X,Y )− b(n− 1) · σ2

x

)= 0 (A.13)

Cov(X,Y )− bσ2x = 0 (A.14)

b =Cov(X,Y )

σ2x

(A.15)

275

Physics Olympiad Script APPENDIX A. FURTHER DERIVATIONS

276

Appendix B

Tables

B.1 List of physical constants (in SI units)

Name Symbol Value UnitAtomic mass unit u = 1.660 · 10−27 kg

Atomic mass unit uc2 = 931.49 MeV

Avogadro constant NA = 6.022 · 10−23 mol−1

Bohr radius a0 = 5.2917 · 10−11 m

Boltzmann constant kB = 1.3806 · 10−23 J·K−1

Elementary charge e = 1.602 · 10−19 C

Vacuum permittivity / electric constant ε0 = 8.8541 · 10−12 A·s·V−1·m−1

Gravitational acceleration (average) g = 9.807 m·s−2

Universal Gas constant R = 8.3145 J·mol−1·K−1

Gravitational constant G = 6.673 · 10−11 m3·kg−1·s−2

Speed of light c = 2.9979 · 108 m·s−1

Vacuum permeability / magnetic constant µ0 = 4π · 10−7 V·s·A−1·m−1

Normal pressure p0 = 101324 Pa

Planck constant h = 6.626 · 10−34 J·sMass of electron me = 9.109 · 10−31 kg

Mass of neutron mn = 1.675 · 10−27 kg

Mass of proton mp = 1.673 · 10−27 kg

Rydberg constant RH = 1.097 · 107 m−1

Stefan-Boltzmann constant σ = 5.670 · 10−8 W2·m−4·K−1

Wave impedance of vacuum Z0 = 376.7 Ω

Table B.1: List of Phyiscal constants [47]

277

Physics Olympiad Script APPENDIX B. TABLES

B.2 List of named, SI derived units

Unit Symbol Quantity Equivalents SI Equivalenthertz Hz frequency 1/s s-1

radian rad angle m/m 1newton N force kg·m/s2 kg·m·s-2pascal Pa pressure, stress N/m2 kg·m-1·s-2joule J energy, work, heat N·m, W·s kg·m2·s-2watt W power J/s, V·A kg·m2·s-3coulomb C electric charge F·V A·svolt V voltage W/A, J/C kg·m2·s-3·A-1

farad F capacitance C/V kg-1·m-2·s4·A2

ohm Ω resistance, impedance 1/S, V/A kg·m2·s-3·A-2

siemens S conductance 1/Ω, A/V kg-1·m-2·s3·A2

tesla T magnetic field strength V·s/m2 kg·s-2·A-1

henry H inductance V·s/A, Ω·s kg·m2·s-2·A-2

Table B.2: List of named, SI derived units

B.3 List of material constants

278

Physics Olympiad Script B.3. LIST OF MATERIAL CONSTANTSDen

sity

Speed

ofsoun

dLine

arexpa

nsion

coeffi

cient

Specific

heat

capa

c-ity

Melting

tempe

ra-

ture

Heat

con-

ductivity

Specific

electric

resistan

ce(at

20

C)

Magne

tic

perm

eabil-

ity

Nam

eρ/k

g·m−

3c s/m·s−

1α/K−

1C/J·k

g−

1·K−

1Tm/ C

λ/W·m−

1·K−

1ρe/Ω·m−

1µr

Aluminium

2700

524

023.

8·1

0−6

896

660.1

239

2.8

2·1

0−8

1+

2.1·1

0−5

Lead

1134

0125

031.

3·1

0−6

129

327.4

34.8

2.2·1

0−7

diam

agne

tic

Iron

7860

517

012.

0·1

0−6

450

1535

801·1

0−

7≈

5800

Gold

1929

0324

014.

3·1

0−6

129

1063

312

2.2·1

0−8

1−

3.4·1

0−5

Cop

per

8920

390

016.

8·1

0−6

383

1083

390

1.7·1

0−8

1−

6.4·1

0−6

Brass

8470

18·1

0−6

380

905

797.8·1

0−8

Silver

1050

019.

7·1

0−6

235

860.8

428

1.5

9·1

0−8

Tab

leB.3:Prope

rtiesof

diffe

rent

metals[64].

Den

sity

Speed

ofsoun

dVolum

eexpa

nsion

coeffi

cient

Specific

heat

capa

c-ity

Melting

tempe

ra-

ture

Boilin

gtempe

ra-

ture

Entha

lpyof

fusion

Entha

lpyof

vapo

riza-

tion

Nam

eρ/kg·m−

3c s/m·s−

1α/K−

1C/J·k

g−

1·K−

1Tm/

CTb/ C

Lm/J·k

g−

1Lb/J·k

g−

1

Acetone

792

119

01.

49·1

0−3

2160

−94.8

656.2

59.8·1

045.2

5·1

05

Ben

zol

879

132

6.1.2

3·1

0−3

1725

5.5

380.1

1.2

8·1

053.9

4·1

05

Ethan

ol78

9117

01.

1·1

0−3

2430

−11

4.5

78.3

31.0

8·1

058.4·1

05

Oil

≈90

0

Mercu

ry13

546

143

01.

84·1

0−4

139

−38.8

735

6.58

1.1

8·1

042.8

5·1

05

Water

998

148

32.

07·1

0−4

4182

010

03.3

38·1

052.2

56·1

06

Tab

leB.4:Prope

rtiesof

diffe

rent

fluids[64].

279

Physics Olympiad Script APPENDIX B. TABLES

Den

sity

Speed

ofsoun

dMolarehe

atcapa

-city

(pconstant)

Melting

tempe

ra-

ture

Boilin

gtempe

ra-

ture

Van

-der-W

aals

constanta

Van

-der-W

aals

constantb

Nam

eρ/k

g·m−

3c s/m·s−

1Cp/jo

ule/m

ol/K

Tm/

CTb/

Ca/N·m

4·m

ol−

2b/

m3·m

ol−

1

Aargo

n1.

784

-20.9

−77.7

−33.4

0.4

253.7

3·1

0−5

Helium

0.17

8510

0520.9

-−

268.

940.0

034

2.3

6·1

0−5

Carbo

ndioxide

1.97

726

836.8

-−

78.4

50.3

664.2

8·1

0−5

Air

1.29

334

429.1

-−

191.

40.1

353.6

5·1

0−5

Metha

n0.

717

445

35.6

-−

191.

40.2

294.2

8·1

0−5

Neon

0.9

-20.8

−24

8.6

1−

245.

060.0

217

1.7

4·1

0−5

Oxy

gen

1.42

932

629.3

−21

8.7

9−

182.

970.1

383.1

7·1

0−5

Nitrogen

1.25

1310

29.1

−21

0.0

−19

5.82

0.1

373.8

7·1

0−5

Water

vapo

ur-

-33.6

010

00.5

533.0

4·1

0−5

Hyd

rogen

0.08

8913

1028.9

−25

9.2

−25

2.77

0.0

248

2.6

6·1

0−5

Tab

leB.5:Prope

rtiesof

diffe

rent

gases[64].

280

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