GR9677 Solutions

DRAFT

The Physics GRE Solution Guide

GR9677 Test

http://groups.yahoo.com/group/physicsgre_v2

November 3, 2009

Author:David S. Latchman

http://groups.yahoo.com/group/physicsgre_v2

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2

David S. Latchman ©2009

DRAFTPreface

This solution guide initially started out on the Yahoo Groups web site and was prettysuccessful at the time. Unfortunately, the group was lost and with it, much of the thehard work that was put into it. This is my attempt to recreate the solution guide andmake it more widely avaialble to everyone. If you see any errors, think certain thingscould be expressed more clearly, or would like to make suggestions, please feel free todo so.

David Latchman

Document Changes

05-11-2009 1. Added diagrams to GR0177 test questions 1-25

2. Revised solutions to GR0177 questions 1-25

04-15-2009 First Version

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ii


DRAFTContents

Preface i

Classical Mechanics xv

0.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv

0.2 Newton’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xvi

0.3 Work & Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xvii

0.4 Oscillatory Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xviii

0.5 Rotational Motion about a Fixed Axis . . . . . . . . . . . . . . . . . . . . xxii

0.6 Dynamics of Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . xxiv

0.7 Central Forces and Celestial Mechanics . . . . . . . . . . . . . . . . . . . xxiv

0.8 Three Dimensional Particle Dynamics . . . . . . . . . . . . . . . . . . . . xxvi

0.9 Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxvi

0.10 Non-inertial Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . xxvii

0.11 Hamiltonian and Lagrangian Formalism . . . . . . . . . . . . . . . . . . . xxvii

Electromagnetism xxix

0.12 Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxix

0.13 Currents and DC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv

0.14 Magnetic Fields in Free Space . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv

0.15 Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv

0.16 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv

0.17 Maxwell’s Equations and their Applications . . . . . . . . . . . . . . . . . xxxiv

0.18 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv

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iv Contents0.19 AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv

0.20 Magnetic and Electric Fields in Matter . . . . . . . . . . . . . . . . . . . . xxxiv

0.21 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv

0.22 Energy in a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv

0.23 Energy in an Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv

0.24 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv

0.25 Current Destiny . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv

0.26 Current Density of Moving Charges . . . . . . . . . . . . . . . . . . . . . xxxv

0.27 Resistance and Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv

0.28 Resistivity and Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi

0.29 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi

0.30 Kirchoff’s Loop Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi

0.31 Kirchoff’s Junction Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi

0.32 RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi

0.33 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi

0.34 Speed of Propagation of a Light Wave . . . . . . . . . . . . . . . . . . . . xxxvii

0.35 Relationship between E and B Fields . . . . . . . . . . . . . . . . . . . . . xxxvii

0.36 Energy Density of an EM wave . . . . . . . . . . . . . . . . . . . . . . . . xxxviii

0.37 Poynting’s Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxviii

Optics & Wave Phonomena xxxix

0.38 Wave Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix

0.39 Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix

0.40 Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix

0.41 Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix

0.42 Geometrical Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix

0.43 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix

0.44 Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xl

0.45 Snell’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xl

Thermodynamics & Statistical Mechanics xli

0.46 Laws of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . xli

0.47 Thermodynamic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . xli


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Contents v0.48 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xli

0.49 Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xli

0.50 Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xli

0.51 Ensembles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xli

0.52 Statistical Concepts and Calculation of Thermodynamic Properties . . . xlii

0.53 Thermal Expansion & Heat Transfer . . . . . . . . . . . . . . . . . . . . . xlii

0.54 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlii

0.55 Specific Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlii

0.56 Heat and Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlii

0.57 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . xlii

0.58 Work done by Ideal Gas at Constant Temperature . . . . . . . . . . . . . xliii

0.59 Heat Conduction Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . xliii

0.60 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xliv

0.61 Stefan-Boltzmann’s FormulaStefan-Boltzmann’s Equation . . . . . . . . xliv

0.62 RMS Speed of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . xliv

0.63 Translational Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . xliv

0.64 Internal Energy of a Monatomic gas . . . . . . . . . . . . . . . . . . . . . xliv

0.65 Molar Specific Heat at Constant Volume . . . . . . . . . . . . . . . . . . . xlv

0.66 Molar Specific Heat at Constant Pressure . . . . . . . . . . . . . . . . . . xlv

0.67 Equipartition of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlv

0.68 Adiabatic Expansion of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . xlvii

0.69 Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . xlvii

Quantum Mechanics xlix

0.70 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlix

0.71 Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlix

0.72 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . liv

0.73 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . liv

0.74 Wave Funtion Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . liv

0.75 Elementary Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . liv

Atomic Physics lv

0.76 Properties of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lv

©2009 David S. Latchman

DRAFT

vi Contents0.77 Bohr Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lv

0.78 Energy Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvi

0.79 Atomic Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvi

0.80 Atomic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvi

0.81 Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvii

0.82 Black Body Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvii

0.83 X-Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lviii

0.84 Atoms in Electric and Magnetic Fields . . . . . . . . . . . . . . . . . . . . lix

Special Relativity lxiii

0.85 Introductory Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiii

0.86 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiii

0.87 Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiii

0.88 Simultaneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiii

0.89 Energy and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiv

0.90 Four-Vectors and Lorentz Transformation . . . . . . . . . . . . . . . . . . lxv

0.91 Velocity Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxvi

0.92 Relativistic Doppler Formula . . . . . . . . . . . . . . . . . . . . . . . . . lxvi

0.93 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxvi

0.94 Space-Time Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxvii

Laboratory Methods lxix

0.95 Data and Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxix

0.96 Instrumentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxi

0.97 Radiation Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxi

0.98 Counting Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxi

0.99 Interaction of Charged Particles with Matter . . . . . . . . . . . . . . . . lxxii

0.100Lasers and Optical Interferometers . . . . . . . . . . . . . . . . . . . . . . lxxii

0.101Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxii

0.102Fundamental Applications of Probability and Statistics . . . . . . . . . . lxxii

GR9677 Exam Solutions lxxiii

0.103Discharge of a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxiii


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Contents vii0.104Magnetic Fields & Induced EMFs . . . . . . . . . . . . . . . . . . . . . . . lxxiii

0.105A Charged Ring I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxiv

0.106A Charged Ring II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxiv

0.107Forces on a Car’s Tires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxv

0.108Block sliding down a rough inclined plane . . . . . . . . . . . . . . . . . lxxv

0.109Collision of Suspended Blocks . . . . . . . . . . . . . . . . . . . . . . . . . lxxvi

0.110Damped Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . lxxvii

0.111Spectrum of the Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . lxxvii

0.112Internal Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxviii

0.113The Stern-Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . . . . . lxxviii

0.114Positronium Ground State Energy . . . . . . . . . . . . . . . . . . . . . . lxxviii

0.115Specific Heat Capacity and Heat Lost . . . . . . . . . . . . . . . . . . . . . lxxix

0.116Conservation of Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxix

0.117Thermal Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxix

0.118Mean Free Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxx

0.119Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxi

0.120Barrier Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxii

0.121Distance of Closest Appraoch . . . . . . . . . . . . . . . . . . . . . . . . . lxxxii

0.122Collisions and the He atom . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiii

0.123Oscillating Hoops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiii

0.124Mars Surface Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiv

0.125The Inverse Square Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiv

0.126Charge Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxv

0.127Capacitors in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxvi

0.128Resonant frequency of a RLC Circuit . . . . . . . . . . . . . . . . . . . . . lxxxvi

0.129Graphs and Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxvii

0.130Superposition of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxviii

0.131The Plank Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxix

0.132The Open Ended U-tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . xc

0.133Sphere falling through a viscous liquid . . . . . . . . . . . . . . . . . . . . xc

0.134Moment of Inertia and Angular Velocity . . . . . . . . . . . . . . . . . . . xci

0.135Quantum Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . xcii


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viii Contents0.136Invariance Violations and the Non-conservation of Parity . . . . . . . . . xcii

0.137Wave function of Identical Fermions . . . . . . . . . . . . . . . . . . . . . xciii

0.138Relativistic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xciii

0.139Relativistic Addition of Velocities . . . . . . . . . . . . . . . . . . . . . . . xciii

0.140Relativistic Energy and Momentum . . . . . . . . . . . . . . . . . . . . . xciv

0.141Ionization Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xciv

0.142Photon Emission and a Singly Ionized He atom . . . . . . . . . . . . . . . xcv

0.143Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvi

0.144Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvi

0.145Stoke’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvii

0.1461-D Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvii

0.147High Pass Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvii

0.148Generators and Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . xcviii

0.149Faraday’s Law and a Wire wound about a Rotating Cylinder . . . . . . . xcviii

0.150Speed of π+ mesons in a laboratory . . . . . . . . . . . . . . . . . . . . . . xcix

0.151Transformation of Electric Field . . . . . . . . . . . . . . . . . . . . . . . . xcix

0.152The Space-Time Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcix

0.153Wavefunction of the Particle in an Infinte Well . . . . . . . . . . . . . . . c

0.154Spherical Harmonics of the Wave Function . . . . . . . . . . . . . . . . . c

0.155Decay of the Positronium Atom . . . . . . . . . . . . . . . . . . . . . . . . c

0.156Polarized Electromagnetic Waves I . . . . . . . . . . . . . . . . . . . . . . c

0.157Polarized Electromagnetic Waves II . . . . . . . . . . . . . . . . . . . . . . ci

0.158Total Internal Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . ci

0.159Single Slit Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ci

0.160The Optical Telescope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cii

0.161Pulsed Lasers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cii

0.162Relativistic Doppler Shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . ciii

0.163Gauss’ Law, the Electric Field and Uneven Charge Distribution . . . . . civ

0.164Capacitors in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cv

0.165Standard Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cv

0.166Nuclear Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . cv

0.167Work done by a man jumping off a boat . . . . . . . . . . . . . . . . . . . cvi


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Contents ix0.168Orbits and Gravitational Potential . . . . . . . . . . . . . . . . . . . . . . cvi

0.169Schwartzchild Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cvi

0.170Lagrangian of a Bead on a Rod . . . . . . . . . . . . . . . . . . . . . . . . cvii

0.171Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cvii

0.172Larmor Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cviii

0.173The Oscilloscope and Electron Deflection . . . . . . . . . . . . . . . . . . cix

0.174Negative Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cix

0.175Adiabatic Work of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . cx

0.176Change in Entrophy of Two Bodies . . . . . . . . . . . . . . . . . . . . . . cx

0.177Double Pane Windows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxi

0.178Gaussian Wave Packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxii

0.179Angular Momentum Spin Operators . . . . . . . . . . . . . . . . . . . . . cxii

0.180Semiconductors and Impurity Atoms . . . . . . . . . . . . . . . . . . . . cxii

0.181Specific Heat of an Ideal Diatomic Gas . . . . . . . . . . . . . . . . . . . . cxii

0.182Transmission of a Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxiii

0.183Piano Tuning & Beats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxiii

0.184Thin Films . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxiv

0.185Mass moving on rippled surface . . . . . . . . . . . . . . . . . . . . . . . cxv

0.186Normal Modes and Couples Oscillators . . . . . . . . . . . . . . . . . . . cxv

0.187Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxv

0.188Charged Particles in E&M Fields . . . . . . . . . . . . . . . . . . . . . . . cxvi

0.189Rotation of Charged Pith Balls in a Collapsing Magnetic Field . . . . . . cxvi

0.190Coaxial Cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxvii

0.191Charged Particles in E&M Fields . . . . . . . . . . . . . . . . . . . . . . . cxviii

0.192THIS ITEM WAS NOT SCORED . . . . . . . . . . . . . . . . . . . . . . . cxix

0.193The Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . cxix

0.194Small Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxix

0.195Period of Mass in Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . cxx

0.196Internal Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxi

0.197Specific Heat of a Super Conductor . . . . . . . . . . . . . . . . . . . . . . cxxi

0.198Pair Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxi

0.199Probability Current Density . . . . . . . . . . . . . . . . . . . . . . . . . . cxxii


DRAFT

x Contents0.200Quantum Harmonic Oscillator Energy Levels . . . . . . . . . . . . . . . . cxxii

0.201Three Level LASER and Metastable States . . . . . . . . . . . . . . . . . . cxxiii

0.202Quantum Oscillator – Raising and Lowering Operators . . . . . . . . . . cxxiv

Constants & Important Equations cxxv

.1 Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxv

.2 Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxv

.3 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxvi

.4 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxvii


DRAFTList of Tables

0.67.1Table of Molar Specific Heats . . . . . . . . . . . . . . . . . . . . . . . . . xlvi

0.119.1Table of wavefunction amplitudes . . . . . . . . . . . . . . . . . . . . . . lxxxii

0.181.1Table of degrees of freedom of a Diatomic atom . . . . . . . . . . . . . . . cxiii

.1.1 Something . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxv

DRAFT

xii List of Tables


DRAFTList of Figures

0.201.1Three Level Laser . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxiv

DRAFT

xiv List of Figures


DRAFTClassical Mechanics

0.1 Kinematics

0.1.1 Linear Motion

Average Velocity

v =∆x∆t

=x2 − x1

t2 − t1(0.1.1)

Instantaneous Velocity

v = lim∆t→0

∆x∆t

=dxdt

= v(t) (0.1.2)

Kinematic Equations of Motion

The basic kinematic equations of motion under constant acceleration, a, are

v = v0 + at (0.1.3)

v2 = v20 + 2a (x − x0) (0.1.4)

x − x0 = v0t +12

at2 (0.1.5)

x − x0 =12

(v + v0) t (0.1.6)

0.1.2 Circular Motion

In the case of Uniform Circular Motion, for a particle to move in a circular path, aradial acceleration must be applied. This acceleration is known as the CentripetalAcceleration

DRAFT

xvi Classical MechanicsCentripetal Acceleration

a =v2

r(0.1.7)

Angular Velocity

ω =vr

(0.1.8)

We can write eq. (0.1.7) in terms of ω

a = ω2r (0.1.9)

Rotational Equations of Motion

The equations of motion under a constant angular acceleration, α, are

ω = ω0 + αt (0.1.10)

θ =ω + ω0

2t (0.1.11)

θ = ω0t +12αt2 (0.1.12)

ω2 = ω20 + 2αθ (0.1.13)

0.2 Newton’s Laws

0.2.1 Newton’s Laws of Motion

First Law A body continues in its state of rest or of uniform motion unless acted uponby an external unbalanced force.

Second Law The net force on a body is proportional to its rate of change of momentum.

F =dpdt

= ma (0.2.1)

Third Law When a particle A exerts a force on another particle B, B simultaneouslyexerts a force on A with the same magnitude in the opposite direction.

FAB = −FBA (0.2.2)

0.2.2 Momentum

p = mv (0.2.3)


DRAFT

Work & Energy xvii0.2.3 Impulse

∆p = J =w

Fdt = Favgdt (0.2.4)

0.3 Work & Energy

0.3.1 Kinetic Energy

K ≡12

mv2 (0.3.1)

0.3.2 The Work-Energy Theorem

The net Work done is given byWnet = K f − Ki (0.3.2)

0.3.3 Work done under a constant Force

The work done by a force can be expressed as

W = F∆x (0.3.3)

In three dimensions, this becomes

W = F · ∆r = F∆r cosθ (0.3.4)

For a non-constant force, we have

W =

x fw

xi

F(x)dx (0.3.5)

0.3.4 Potential Energy

The Potential Energy is

F(x) = −dU(x)

dx(0.3.6)

for conservative forces, the potential energy is

U(x) = U0 −

xw

x0

F(x′)dx′ (0.3.7)


DRAFT

xviii Classical Mechanics0.3.5 Hooke’s Law

F = −kx (0.3.8)

where k is the spring constant.

0.3.6 Potential Energy of a Spring

U(x) =12

kx2 (0.3.9)

0.4 Oscillatory Motion

0.4.1 Equation for Simple Harmonic Motion

x(t) = A sin (ωt + δ) (0.4.1)

where the Amplitude, A, measures the displacement from equilibrium, the phase, δ, isthe angle by which the motion is shifted from equilibrium at t = 0.

0.4.2 Period of Simple Harmonic Motion

T =2πω

(0.4.2)

0.4.3 Total Energy of an Oscillating System

Given thatx = A sin (ωt + δ) (0.4.3)

and that the Total Energy of a System is

E = KE + PE (0.4.4)

The Kinetic Energy is

KE =12

mv2

=12

mdxdt

=12

mA2ω2 cos2 (ωt + δ) (0.4.5)


DRAFT

Oscillatory Motion xixThe Potential Energy is

U =12

kx2

=12

kA2 sin2 (ωt + δ) (0.4.6)

Adding eq. (0.4.5) and eq. (0.4.6) gives

E =12

kA2 (0.4.7)

0.4.4 Damped Harmonic Motion

Fd = −bv = −bdxdt

(0.4.8)

where b is the damping coefficient. The equation of motion for a damped oscillatingsystem becomes

− kx − bdxdt

= md2xdt2 (0.4.9)

Solving eq. (0.4.9) govesx = Ae−αt sin (ω′t + δ) (0.4.10)

We find that

α =b

2m(0.4.11)

ω′ =

√km−

b2

4m2

=

√ω2

0 −b2

4m2

=√ω2

0 − α2 (0.4.12)

0.4.5 Small Oscillations

The Energy of a system is

E = K + V(x) =12

mv(x)2 + V(x) (0.4.13)

We can solve for v(x),

v(x) =

√2m

(E − V(x)) (0.4.14)

where E ≥ V(x) Let the particle move in the potential valley, x1 ≤ x ≤ x2, the potentialcan be approximated by the Taylor Expansion

V(x) = V(xe) + (x − xe)[dV(x)

dx

]x=xe

+12

(x − xe)2

[d2V(x)

dx2

]x=xe

+ · · · (0.4.15)


DRAFT

xx Classical MechanicsAt the points of inflection, the derivative dV/dx is zero and d2V/dx2 is positive. Thismeans that the potential energy for small oscillations becomes

V(x) u V(xe) +12

k(x − xe)2 (0.4.16)

where

k ≡[d2V(x)

dx2

]x=xe

≥ 0 (0.4.17)

As V(xe) is constant, it has no consequences to physical motion and can be dropped.We see that eq. (0.4.16) is that of simple harmonic motion.

0.4.6 Coupled Harmonic Oscillators

Consider the case of a simple pendulum of length, `, and the mass of the bob is m1.For small displacements, the equation of motion is

θ + ω0θ = 0 (0.4.18)

We can express this in cartesian coordinates, x and y, where

x = ` cosθ ≈ ` (0.4.19)y = ` sinθ ≈ `θ (0.4.20)

eq. (0.4.18) becomesy + ω0y = 0 (0.4.21)

This is the equivalent to the mass-spring system where the spring constant is

k = mω20 =

mg`

(0.4.22)

This allows us to to create an equivalent three spring system to our coupled pendulumsystem. The equations of motion can be derived from the Lagrangian, where

L = T − V

=12

my21 +

12

my22 −

(12

ky21 +

12κ(y2 − y1

)2+

12

ky22

)=

12

m(y1

2 + y22)−

12

(k(y2

1 + y22

)+ κ

(y2 − y1

)2)

(0.4.23)

We can find the equations of motion of our system

ddt

(∂L∂yn

)=∂L∂yn

(0.4.24)

1Add figure with coupled pendulum-spring system


DRAFT

Oscillatory Motion xxiThe equations of motion are

my1 = −ky1 + κ(y2 − y1

)(0.4.25)

my2 = −ky2 + κ(y2 − y1

)(0.4.26)

We assume solutions for the equations of motion to be of the form

y1 = cos(ωt + δ1) y2 = B cos(ωt + δ2)y1 = −ωy1 y2 = −ωy2

(0.4.27)

Substituting the values for y1 and y2 into the equations of motion yields(k + κ −mω2

)y1 − κy2 = 0 (0.4.28)

−κy1 +(k + κ −mω2

)y2 = 0 (0.4.29)

We can get solutions from solving the determinant of the matrix∣∣∣∣∣(k + κ −mω2)−κ

−κ(k + κ −mω2)∣∣∣∣∣ = 0 (0.4.30)

Solving the determinant gives(mω2

)2− 2mω2 (k + κ) +

(k2 + 2kκ

)= 0 (0.4.31)

This yields

ω2 =

km

=g`

k + 2κm

=g`

+2κm

(0.4.32)

We can now determine exactly how the masses move with each mode by substitutingω2 into the equations of motion. Where

ω2 =km

We see that

k + κ −mω2 = κ (0.4.33)

Substituting this into the equation of motion yields

y1 = y2 (0.4.34)

We see that the masses move in phase with each other. You will also noticethe absense of the spring constant term, κ, for the connecting spring. As themasses are moving in step, the spring isn’t stretching or compressing and henceits absence in our result.

ω2 =k + κ

mWe see that

k + κ −mω2 = −κ (0.4.35)

Substituting this into the equation of motion yields

y1 = −y2 (0.4.36)

Here the masses move out of phase with each other. In this case we see thepresence of the spring constant, κ, which is expected as the spring playes a role.It is being stretched and compressed as our masses oscillate.


DRAFT

xxii Classical Mechanics0.4.7 Doppler Effect

The Doppler Effect is the shift in frequency and wavelength of waves that results froma source moving with respect to the medium, a receiver moving with respect to themedium or a moving medium.

Moving Source If a source is moving towards an observer, then in one period, τ0, itmoves a distance of vsτ0 = vs/ f0. The wavelength is decreased by

λ′ = λ −vs

f0−

v − vs

f0(0.4.37)

The frequency change is

f ′ =vλ′

= f0

( vv − vs

)(0.4.38)

Moving Observer As the observer moves, he will measure the same wavelength, λ, asif at rest but will see the wave crests pass by more quickly. The observer measuresa modified wave speed.

v′ = v + |vr| (0.4.39)

The modified frequency becomes

f ′ =v′

λ= f0

(1 +

vr

v

)(0.4.40)

Moving Source and Moving Observer We can combine the above two equations

λ′ =v − vs

f0(0.4.41)

v′ = v − vr (0.4.42)

To give a modified frequency of

f ′ =v′

λ′=

(v − vr

v − vs

)f0 (0.4.43)

0.5 Rotational Motion about a Fixed Axis

0.5.1 Moment of Inertia

I =

∫R2dm (0.5.1)

0.5.2 Rotational Kinetic Energy

K =12

Iω2 (0.5.2)


DRAFT

Rotational Motion about a Fixed Axis xxiii0.5.3 Parallel Axis Theorem

I = Icm + Md2 (0.5.3)

0.5.4 Torque

τ = r × F (0.5.4)τ = Iα (0.5.5)

where α is the angular acceleration.

0.5.5 Angular Momentum

L = Iω (0.5.6)

we can find the Torque

τ =dLdt

(0.5.7)

0.5.6 Kinetic Energy in Rolling

With respect to the point of contact, the motion of the wheel is a rotation about thepoint of contact. Thus

K = Krot =12

Icontactω2 (0.5.8)

Icontact can be found from the Parallel Axis Theorem.

Icontact = Icm + MR2 (0.5.9)

Substitute eq. (0.5.8) and we have

K =12

(Icm + MR2

)ω2

=12

Icmω2 +

12

mv2 (0.5.10)

The kinetic energy of an object rolling without slipping is the sum of hte kinetic energyof rotation about its center of mass and the kinetic energy of the linear motion of theobject.


DRAFT

xxiv Classical Mechanics0.6 Dynamics of Systems of Particles

0.6.1 Center of Mass of a System of Particles

Position Vector of a System of Particles

R =m1r1 + m2r2 + m3r3 + · · · + mNrN

M(0.6.1)

Velocity Vector of a System of Particles

V =dRdt

=m1v1 + m2v2 + m3v3 + · · · + mNvN

M(0.6.2)

Acceleration Vector of a System of Particles

A =dVdt

=m1a1 + m2a2 + m3a3 + · · · + mNaN

M(0.6.3)

0.7 Central Forces and Celestial Mechanics

0.7.1 Newton’s Law of Universal Gravitation

F = −(GMm

r2

)r (0.7.1)

0.7.2 Potential Energy of a Gravitational Force

U(r) = −GMm

r(0.7.2)

0.7.3 Escape Speed and Orbits

The energy of an orbiting body is

E = T + U

=12

mv2−

GMmr

(0.7.3)


DRAFT

Central Forces and Celestial Mechanics xxvThe escape speed becomes

E =12

mv2esc −

GMmRE

= 0 (0.7.4)

Solving for vesc we find

vesc =

√2GM

Re(0.7.5)

0.7.4 Kepler’s Laws

First Law The orbit of every planet is an ellipse with the sun at a focus.

Second Law A line joining a planet and the sun sweeps out equal areas during equalintervals of time.

Third Law The square of the orbital period of a planet is directly proportional to thecube of the semi-major axis of its orbit.

T2

R3 = C (0.7.6)

where C is a constant whose value is the same for all planets.

0.7.5 Types of Orbits

The Energy of an Orbiting Body is defined in eq. (0.7.3), we can classify orbits by theireccentricities.

Circular Orbit A circular orbit occurs when there is an eccentricity of 0 and the orbitalenergy is less than 0. Thus

12

v2−

GMr

= E < 0 (0.7.7)

The Orbital Velocity is

v =

√GM

r(0.7.8)

Elliptic Orbit An elliptic orbit occurs when the eccentricity is between 0 and 1 but thespecific energy is negative, so the object remains bound.

v =

√GM

(2r−

1a

)(0.7.9)

where a is the semi-major axis


DRAFT

xxvi Classical MechanicsParabolic Orbit A Parabolic Orbit occurs when the eccentricity is equal to 1 and the

orbital velocity is the escape velocity. This orbit is not bounded. Thus

12

v2−

GMr

= E = 0 (0.7.10)

The Orbital Velocity is

v = vesc =

√2GM

r(0.7.11)

Hyperbolic Orbit In the Hyperbolic Orbit, the eccentricity is greater than 1 with anorbital velocity in excess of the escape velocity. This orbit is also not bounded.

v∞ =

√GM

a(0.7.12)

0.7.6 Derivation of Vis-viva Equation

The total energy of a satellite is

E =12

mv2−

GMmr

(0.7.13)

For an elliptical or circular orbit, the specific energy is

E = −GMm

2a(0.7.14)

Equating we get

v2 = GM(2

r−

1a

)(0.7.15)

0.8 Three Dimensional Particle Dynamics

0.9 Fluid Dynamics

When an object is fully or partially immersed, the buoyant force is equal to the weightof fluid displaced.

0.9.1 Equation of Continuity

ρ1v1A1 = ρ2v2A2 (0.9.1)


DRAFT

Non-inertial Reference Frames xxvii0.9.2 Bernoulli’s Equation

P +12ρv2 + ρgh = a constant (0.9.2)

0.10 Non-inertial Reference Frames

0.11 Hamiltonian and Lagrangian Formalism

0.11.1 Lagrange’s Function (L)

L = T − V (0.11.1)

where T is the Kinetic Energy and V is the Potential Energy in terms of GeneralizedCoordinates.

0.11.2 Equations of Motion(Euler-Lagrange Equation)

∂L∂q

=ddt

(∂L∂q

)(0.11.2)

0.11.3 Hamiltonian

H = T + V= pq − L(q, q) (0.11.3)

where

∂H∂p

= q (0.11.4)

∂H∂q

= −∂L∂x

= −p (0.11.5)


DRAFT

xxviii Classical Mechanics


DRAFTElectromagnetism

0.12 Electrostatics

0.12.1 Coulomb’s Law

The force between two charged particles, q1 and q2 is defined by Coulomb’s Law.

F12 =1

4πε0

(q1q2

r212

)r12 (0.12.1)

where ε0 is the permitivitty of free space, where

ε0 = 8.85 × 10−12C2N.m2 (0.12.2)

0.12.2 Electric Field of a point charge

The electric field is defined by mesuring the magnitide and direction of an electricforce, F, acting on a test charge, q0.

E ≡Fq0

(0.12.3)

The Electric Field of a point charge, q is

E =1

4πε0

qr2 r (0.12.4)

In the case of multiple point charges, qi, the electric field becomes

E(r) =1

4πε0

n∑i=1

qi

r2i

ri (0.12.5)

Electric Fields and Continuous Charge Distributions

If a source is distributed continuously along a region of space, eq. (0.12.5) becomes

E(r) =1

4πε0

∫1r2 rdq (0.12.6)

DRAFT

xxx ElectromagnetismIf the charge was distributed along a line with linear charge density, λ,

λ =dqdx

(0.12.7)

The Electric Field of a line charge becomes

E(r) =1

4πε0

∫line

λr2 rdx (0.12.8)

In the case where the charge is distributed along a surface, the surface charge densityis, σ

σ =QA

=dqdA

(0.12.9)

The electric field along the surface becomes

E(r) =1

4πε0

∫Surface

σr2 rdA (0.12.10)

In the case where the charge is distributed throughout a volume, V, the volume chargedensity is

ρ =QV

=dqdV

(0.12.11)

The Electric Field is

E(r) =1

4πε0

∫Volume

ρ

r2 rdV (0.12.12)

0.12.3 Gauss’ Law

The electric field through a surface is

Φ =

∮surface S

dΦ =

∮surface S

E · dA (0.12.13)

The electric flux through a closed surface encloses a net charge.∮E · dA =

Qε0

(0.12.14)

where Q is the charge enclosed by our surface.


DRAFT

Electrostatics xxxi0.12.4 Equivalence of Coulomb’s Law and Gauss’ Law

The total flux through a sphere is∮E · dA = E(4πr2) =

qε0

(0.12.15)

From the above, we see that the electric field is

E =q

4πε0r2 (0.12.16)

0.12.5 Electric Field due to a line of charge

Consider an infinite rod of constant charge density, λ. The flux through a Gaussiancylinder enclosing the line of charge is

Φ =

∫top surface

E · dA +

∫bottom surface

E · dA +

∫side surface

E · dA (0.12.17)

At the top and bottom surfaces, the electric field is perpendicular to the area vector, sofor the top and bottom surfaces,

E · dA = 0 (0.12.18)

At the side, the electric field is parallel to the area vector, thus

E · dA = EdA (0.12.19)

Thus the flux becomes,

Φ =

∫side sirface

E · dA = E∫

dA (0.12.20)

The area in this case is the surface area of the side of the cylinder, 2πrh.

Φ = 2πrhE (0.12.21)

Applying Gauss’ Law, we see that Φ = q/ε0. The electric field becomes

E =λ

2πε0r(0.12.22)

0.12.6 Electric Field in a Solid Non-Conducting Sphere

Within our non-conducting sphere or radius, R, we will assume that the total charge,Q is evenly distributed throughout the sphere’s volume. So the charge density of oursphere is

ρ =QV

=Q

43πR3

(0.12.23)


DRAFT

xxxii ElectromagnetismThe Electric Field due to a charge Q is

E =Q

4πε0r2 (0.12.24)

As the charge is evenly distributed throughout the sphere’s volume we can say thatthe charge density is

dq = ρdV (0.12.25)

where dV = 4πr2dr. We can use this to determine the field inside the sphere bysumming the effect of infinitesimally thin spherical shells

E =

∫ E

0dE =

∫ r

0

dq4πεr2

=ρ

ε0

∫ r

0dr

=Qr

43πε0R3

(0.12.26)

0.12.7 Electric Potential Energy

U(r) =1

4πε0qq0r (0.12.27)

0.12.8 Electric Potential of a Point Charge

The electrical potential is the potential energy per unit charge that is associated with astatic electrical field. It can be expressed thus

U(r) = qV(r) (0.12.28)

And we can see that

V(r) =1

4πε0

qr

(0.12.29)

A more proper definition that includes the electric field, E would be

V(r) = −

∫C

E · d` (0.12.30)

where C is any path, starting at a chosen point of zero potential to our desired point.

The difference between two potentials can be expressed such

V(b) − V(a) = −

∫ b

E · d` +

∫ a

E · d`

= −

∫ b

aE · d` (0.12.31)


DRAFT

Electrostatics xxxiiiThis can be further expressed

V(b) − V(a) =

∫ b

a(∇V) · d` (0.12.32)

And we can show thatE = −∇V (0.12.33)

0.12.9 Electric Potential due to a line charge along axis

Let us consider a rod of length, `, with linear charge density, λ. The Electrical Potentialdue to a continuous distribution is

V =

∫dV =

14πε0

∫dqr

(0.12.34)

The charge density isdq = λdx (0.12.35)

Substituting this into the above equation, we get the electrical potential at some distancex along the rod’s axis, with the origin at the start of the rod.

dV =1

4πε0

dqx

=1

4πε0

λdxx

(0.12.36)

This becomesV =

λ4πε0

ln[x2

x1

](0.12.37)

where x1 and x2 are the distances from O, the end of the rod.

Now consider that we are some distance, y, from the axis of the rod of length, `. Weagain look at eq. (0.12.34), where r is the distance of the point P from the rod’s axis.

V =1

4πε0

∫dqr

=1

4πε0

∫ `

0

λdx(x2 + y2

) 12

=λ

4πε0ln

[x +

(x2 + y2

) 12]`

0

=λ

4πε0ln

[` +

(`2 + y2

) 12]− ln y

=λ

4πε0ln

` +(`2 + y2) 1

2

d

(0.12.38)


DRAFT

xxxiv Electromagnetism0.13 Currents and DC Circuits

2

0.14 Magnetic Fields in Free Space

3

0.15 Lorentz Force

4

0.16 Induction

5

0.17 Maxwell’s Equations and their Applications

6

0.18 Electromagnetic Waves

7

0.19 AC Circuits

8

0.20 Magnetic and Electric Fields in Matter

9


DRAFT

Capacitance xxxv0.21 Capacitance

Q = CV (0.21.1)

0.22 Energy in a Capacitor

U =Q2

2C

=CV2

2

=QV

2(0.22.1)

0.23 Energy in an Electric Field

u ≡U

volume=ε0E2

2(0.23.1)

0.24 Current

I ≡dQdt

(0.24.1)

0.25 Current Destiny

I =

∫A

J · dA (0.25.1)

0.26 Current Density of Moving Charges

J =IA

= neqvd (0.26.1)

0.27 Resistance and Ohm’s Law

R ≡VI

(0.27.1)


DRAFT

xxxvi Electromagnetism0.28 Resistivity and Conductivity

R = ρLA

(0.28.1)

E = ρJ (0.28.2)

J = σE (0.28.3)

0.29 Power

P = VI (0.29.1)

0.30 Kirchoff’s Loop Rules

Write Here

0.31 Kirchoff’s Junction Rule

Write Here

0.32 RC Circuits

E − IR −QC

= 0 (0.32.1)

0.33 Maxwell’s Equations

0.33.1 Integral Form

Gauss’ Law for Electric Fieldsw

closed surface

E · dA =Qε0

(0.33.1)


DRAFT

Speed of Propagation of a Light Wave xxxviiGauss’ Law for Magnetic Fields

w

closed surface

B · dA = 0 (0.33.2)

Ampere’s Lawz

B · ds = µ0I + µ0ε0ddt

w

surface

E · dA (0.33.3)

Faraday’s Lawz

E · ds = −ddt

w

surface

B · dA (0.33.4)

0.33.2 Differential Form

Gauss’ Law for Electric Fields∇ · E =

ρ

ε0(0.33.5)

Gauss’ Law for Magnetism∇ · B = 0 (0.33.6)

Ampere’s Law

∇ × B = µ0J + µ0ε0∂E∂t

(0.33.7)

Faraday’s Law

∇ · E = −∂B∂t

(0.33.8)

0.34 Speed of Propagation of a Light Wave

c =1

√µ0ε0

(0.34.1)

In a material with dielectric constant, κ,

c√κ =

cn

(0.34.2)

where n is the refractive index.

0.35 Relationship between E and B Fields

E = cB (0.35.1)E · B = 0 (0.35.2)


DRAFT

xxxviii Electromagnetism0.36 Energy Density of an EM wave

u =12

(B2

µ0+ ε0E2

)(0.36.1)

0.37 Poynting’s Vector

S =1µ0

E × B (0.37.1)


DRAFTOptics & Wave Phonomena

0.38 Wave Properties

1

0.39 Superposition

2

0.40 Interference

3

0.41 Diffraction

4

0.42 Geometrical Optics

5

0.43 Polarization

6

DRAFT

xl Optics & Wave Phonomena0.44 Doppler Effect

7

0.45 Snell’s Law

0.45.1 Snell’s Law

n1 sinθ1 = n2 sinθ2 (0.45.1)

0.45.2 Critical Angle and Snell’s Law

The critical angle, θc, for the boundary seperating two optical media is the smallestangle of incidence, in the medium of greater index, for which light is totally refelected.

From eq. (0.45.1), θ1 = 90 and θ2 = θc and n2 > n1.

n1 sin 90 = n2sinθc

sinθc =n1

n2(0.45.2)


DRAFTThermodynamics & Statistical Mechanics

0.46 Laws of Thermodynamics

1

0.47 Thermodynamic Processes

2

0.48 Equations of State

3

0.49 Ideal Gases

4

0.50 Kinetic Theory

5

0.51 Ensembles

6

DRAFT

xlii Thermodynamics & Statistical Mechanics0.52 Statistical Concepts and Calculation of Thermody-

namic Properties

7

0.53 Thermal Expansion & Heat Transfer

8

0.54 Heat Capacity

Q = C(T f − Ti

)(0.54.1)

where C is the Heat Capacity and T f and Ti are the final and initial temperaturesrespectively.

0.55 Specific Heat Capacity

Q = cm(T f − ti

)(0.55.1)

where c is the specific heat capacity and m is the mass.

0.56 Heat and Work

W =

∫ V f

Vi

PdV (0.56.1)

0.57 First Law of Thermodynamics

dEint = dQ − dW (0.57.1)

where dEint is the internal energy of the system, dQ is the Energy added to the systemand dW is the work done by the system.


DRAFT

Work done by Ideal Gas at Constant Temperature xliii0.57.1 Special Cases to the First Law of Thermodynamics

Adiabatic Process During an adiabatic process, the system is insulated such that thereis no heat transfer between the system and its environment. Thus dQ = 0, so

∆Eint = −W (0.57.2)

If work is done on the system, negative W, then there is an increase in its internalenergy. Conversely, if work is done by the system, positive W, there is a decreasein the internal energy of the system.

Constant Volume (Isochoric) Process If the volume is held constant, then the systemcan do no work, δW = 0, thus

∆Eint = Q (0.57.3)

If heat is added to the system, the temperature increases. Conversely, if heat isremoved from the system the temperature decreases.

Closed Cycle In this situation, after certain interchanges of heat and work, the systemcomes back to its initial state. So ∆Eint remains the same, thus

∆Q = ∆W (0.57.4)

The work done by the system is equal to the heat or energy put into it.

Free Expansion In this process, no work is done on or by the system. Thus ∆Q =∆W = 0,

∆Eint = 0 (0.57.5)

0.58 Work done by Ideal Gas at Constant Temperature

Starting with eq. (0.56.1), we substitute the Ideal gas Law, eq. (0.60.1), to get

W = nRT∫ V f

Vi

dVV

= nRT lnV f

Vi(0.58.1)

0.59 Heat Conduction Equation

The rate of heat transferred, H, is given by

H =Qt

= kATH − TC

L(0.59.1)

where k is the thermal conductivity.


DRAFT

xliv Thermodynamics & Statistical Mechanics0.60 Ideal Gas Law

PV = nRT (0.60.1)

where

n = Number of molesP = PressureV = VolumeT = Temperature

and R is the Universal Gas Constant, such that

R ≈ 8.314 J/mol. K

We can rewrite the Ideal gas Law to say

PV = NkT (0.60.2)

where k is the Boltzmann’s Constant, such that

k =R

NA≈ 1.381 × 10−23 J/K

0.61 Stefan-Boltzmann’s FormulaStefan-Boltzmann’s Equa-tion

P(T) = σT4 (0.61.1)

0.62 RMS Speed of an Ideal Gas

vrms =

√3RTM

(0.62.1)

0.63 Translational Kinetic Energy

K =32

kT (0.63.1)

0.64 Internal Energy of a Monatomic gas

Eint =32

nRT (0.64.1)


DRAFT

Molar Specific Heat at Constant Volume xlv0.65 Molar Specific Heat at Constant Volume

Let us define, CV such that

Q = nCV∆T (0.65.1)

Substituting into the First Law of Thermodynamics, we have

∆Eint + W = nCV∆T (0.65.2)

At constant volume, W = 0, and we get

CV =1n

∆Eint

∆T(0.65.3)

Substituting eq. (0.64.1), we get

CV =32

R = 12.5 J/mol.K (0.65.4)

0.66 Molar Specific Heat at Constant Pressure

Starting with

Q = nCp∆T (0.66.1)

and

∆Eint = Q −W⇒ nCV∆T = nCp∆T + nR∆T

∴ CV = Cp − R (0.66.2)

0.67 Equipartition of Energy

CV =

(f2

)R = 4.16 f J/mol.K (0.67.1)

where f is the number of degrees of freedom.


DRAFT

xlvi Thermodynamics & Statistical Mechanics

Degrees

ofFreedomPredicted

Molar

SpecificH

eats

Molecule

TranslationalR

otationalV

ibrationalTotal(f)

CV

CP

=C

V+

R

Monatom

ic3

00

332 R

52 RD

iatomic

32

25

52 R72 R

Polyatomic

(Linear)3

33n−

56

3R4R

Polyatomic

(Non-Linear)

33

3n−

66

3R4R

Table0.67.1:Table

ofMolar

SpecificH

eats


DRAFT

Adiabatic Expansion of an Ideal Gas xlvii0.68 Adiabatic Expansion of an Ideal Gas

PVγ = a constant (0.68.1)

where γ = CPCV

.We can also write

TVγ−1 = a constant (0.68.2)

0.69 Second Law of Thermodynamics

Something.


DRAFT

xlviii Thermodynamics & Statistical Mechanics


DRAFTQuantum Mechanics

0.70 Fundamental Concepts

1

0.71 Schrodinger Equation

Let us define Ψ to beΨ = Ae−iω(t− x

v ) (0.71.1)Simplifying in terms of Energy, E, and momentum, p, we get

Ψ = Ae−i(Et−px)~ (0.71.2)

We obtain Schrodinger’s Equation from the Hamiltonian

H = T + V (0.71.3)

To determine E and p,

∂2Ψ

∂x2 = −p2

~2 Ψ (0.71.4)

∂Ψ∂t

=iE~

Ψ (0.71.5)

and

H =p2

2m+ V (0.71.6)

This becomes

EΨ = HΨ (0.71.7)

EΨ = −~

i∂Ψ∂t

p2Ψ = −~2∂2Ψ

∂x2

The Time Dependent Schrodinger’s Equation is

i~∂Ψ∂t

= −~2

2m∂2Ψ

∂x2 + V(x)Ψ (0.71.8)

The Time Independent Schrodinger’s Equation is

EΨ = −~2

2m∂2Ψ

∂x2 + V(x)Ψ (0.71.9)

DRAFT

l Quantum Mechanics0.71.1 Infinite Square Wells

Let us consider a particle trapped in an infinite potential well of size a, such that

V(x) =

{0 for 0 < x < a∞ for |x| > a,

so that a nonvanishing force acts only at ±a/2. An energy, E, is assigned to the systemsuch that the kinetic energy of the particle is E. Classically, any motion is forbiddenoutside of the well because the infinite value of V exceeds any possible choice of E.

Recalling the Schrodinger Time Independent Equation, eq. (0.71.9), we substitute V(x)and in the region (−a/2, a/2), we get

−~2

2md2ψ

dx2 = Eψ (0.71.10)

This differential is of the formd2ψ

dx2 + k2ψ = 0 (0.71.11)

where

k =

√2mE~2 (0.71.12)

We recognize that possible solutions will be of the form

cos kx and sin kx

As the particle is confined in the region 0 < x < a, we say

ψ(x) =

{A cos kx + B sin kx for 0 < x < a0 for |x| > a

We have known boundary conditions for our square well.

ψ(0) = ψ(a) = 0 (0.71.13)

It shows that

⇒ A cos 0 + B sin 0 = 0∴ A = 0 (0.71.14)

We are now left with

B sin ka = 0ka = 0;π; 2π; 3π; · · ·

(0.71.15)

While mathematically, n can be zero, that would mean there would be no wave function,so we ignore this result and say

kn =nπa

for n = 1, 2, 3, · · ·


DRAFT

Schrodinger Equation liSubstituting this result into eq. (0.71.12) gives

kn =nπa

=

√2mEn

~(0.71.16)

Solving for En gives

En =n2π2~2

2ma2 (0.71.17)

We cna now solve for B by normalizing the function∫ a

0|B|2 sin2 kxdx = |A|2

a2

= 1

So |A|2 =2a

(0.71.18)

So we can write the wave function as

ψn(x) =

√2a

sin(nπx

a

)(0.71.19)

0.71.2 Harmonic Oscillators

Classically, the harmonic oscillator has a potential energy of

V(x) =12

kx2 (0.71.20)

So the force experienced by this particle is

F = −dVdx

= −kx (0.71.21)

where k is the spring constant. The equation of motion can be summed us as

md2xdt2 = −kx (0.71.22)

And the solution of this equation is

x(t) = A cos(ω0t + φ

)(0.71.23)

where the angular frequency, ω0 is

ω0 =

√km

(0.71.24)

The Quantum Mechanical description on the harmonic oscillator is based on the eigen-function solutions of the time-independent Schrodinger’s equation. By taking V(x)from eq. (0.71.20) we substitute into eq. (0.71.9) to get

d2ψ

dx2 =2m~2

(k2

x2− E

)ψ =

mk~2

(x2−

2Ek

)ψ


DRAFT

lii Quantum MechanicsWith some manipulation, we get

~√

mk

d2ψ

dx2 =

√mk~

x2−

2E~

√mk

ψThis step allows us to to keep some of constants out of the way, thus giving us

ξ2 =

√mk~

x2 (0.71.25)

and λ =2E~

√mk

=2E~ω0

(0.71.26)

This leads to the more compact

d2ψ

dξ2 =(ξ2− λ

)ψ (0.71.27)

where the eigenfunction ψ will be a function of ξ. λ assumes an eigenvalue anaglaousto E.

From eq. (0.71.25), we see that the maximum value can be determined to be

ξ2max =

√mk~

A2 (0.71.28)

Using the classical connection between A and E, allows us to say

ξ2max =

√mk~

2Ek

= λ (0.71.29)

From eq. (0.71.27), we see that in a quantum mechanical oscillator, there are non-vanishing solutions in the forbidden regions, unlike in our classical case.

A solution to eq. (0.71.27) isψ(ξ) = e−ξ

2/2 (0.71.30)

where

dψdξ

= −ξe−ξ2/2

anddψ

dξ2 = ξ2e−xi2/2− e−ξ

2/2 =(ξ2− 1

)e−ξ

2/2

This gives is a special solution for λ where

λ0 = 1 (0.71.31)

Thus eq. (0.71.26) gives the energy eigenvalue to be

E0 =~ω0

2λ0 =

~ω0

2(0.71.32)


DRAFT

Schrodinger Equation liiiThe eigenfunction e−ξ2/2 corresponds to a normalized stationary-state wave function

Ψ0(x, t) =

(mkπ2~2

) 18

e−√

mk x2/2~e−iE0t/~ (0.71.33)

This solution of eq. (0.71.27) produces the smallest possibel result of λ and E. Hence,Ψ0 and E0 represents the ground state of the oscillator. and the quantity ~ω0/2 is thezero-point energy of the system.

0.71.3 Finite Square Well

For the Finite Square Well, we have a potential region where

V(x) =

{−V0 for −a ≤ x ≤ a0 for |x| > a

We have three regions

Region I: x < −a In this region, The potential, V = 0, so Schrodinger’s Equation be-comes

−~2

2md2ψ

dx2 = Eψ

⇒d2ψ

dx2 = κ2ψ

where κ =

√−2mE~

This gives us solutions that are

ψ(x) = A exp(−κx) + B exp(κx)

As x → ∞, the exp(−κx) term goes to ∞; it blows up and is not a physicallyrealizable function. So we can drop it to get

ψ(x) = Beκx for x < −a (0.71.34)

Region II: −a < x < a In this region, our potential is V(x) = V0. Substitutin this intothe Schrodinger’s Equation, eq. (0.71.9), gives

−~2

2md2ψ

dx2 − V0ψ = Eψ

ord2ψ

dx2 = −l2ψ

where l ≡

√2m (E + V0)~

(0.71.35)

We notice that E > −V0, making l real and positive. Thus our general solutionbecomes

ψ(x) = C sin(lx) + D cos(lx) for −a < x < a (0.71.36)


DRAFT

liv Quantum MechanicsRegion III: x > a Again this Region is similar to Region III, where the potential, V = 0.

This leaves us with the general solution

ψ(x) = F exp(−κx) + G exp(κx)

As x→∞, the second term goes to infinity and we get

ψ(x) = Fe−κx for x > a (0.71.37)

This gives us

ψ(x) =

Beκx for x < aD cos(lx) for 0 < x < aFe−κx for x > a

(0.71.38)

0.71.4 Hydrogenic Atoms

c

0.72 Spin

3

0.73 Angular Momentum

4

0.74 Wave Funtion Symmetry

5

0.75 Elementary Perturbation Theory

6


DRAFTAtomic Physics

0.76 Properties of Electrons

1

0.77 Bohr Model

To understand the Bohr Model of the Hydrogen atom, we will take advantage of ourknowlegde of the wavelike properties of matter. As we are building on a classicalmodel of the atom with a modern concept of matter, our derivation is considered to be‘semi-classical’. In this model we have an electron of mass, me, and charge, −e, orbitinga proton. The cetripetal force is equal to the Coulomb Force. Thus

14πε0

e2

r2 =mev2

r(0.77.1)

The Total Energy is the sum of the potential and kinetic energies, so

E = K + U =p2

2me− | f race24πε0r (0.77.2)

We can further reduce this equation by subsituting the value of momentum, which wefind to be

p2

2me=

12

mev2 =e2

8πε0r(0.77.3)

Substituting this into eq. (0.77.2), we get

E =e2

8πε0r−

e2

4πε0r= −

e2

8πε0r(0.77.4)

At this point our classical description must end. An accelerated charged particle, likeone moving in circular motion, radiates energy. So our atome here will radiate energyand our electron will spiral into the nucleus and disappear. To solve this conundrum,Bohr made two assumptions.

1. The classical circular orbits are replaced by stationary states. These stationarystates take discreet values.

DRAFT

lvi Atomic Physics2. The energy of these stationary states are determined by their angular momentum

which must take on quantized values of ~.

L = n~ (0.77.5)

We can find the angular momentum of a circular orbit.

L = m3vr (0.77.6)

From eq. (0.77.1) we find v and by substitution, we find L.

L = e√

m3r4πε0

(0.77.7)

Solving for r, gives

r =L2

mee2/4πε0(0.77.8)

We apply the condition from eq. (0.77.5)

rn =n2~2

mee2/4πε0= n2a0 (0.77.9)

where a0 is the Bohr radius.a0 = 0.53 × 10−10 m (0.77.10)

Having discreet values for the allowed radii means that we will also have discreetvalues for energy. Replacing our value of rn into eq. (0.77.4), we get

En = −me

2n2

(e2

4πε0~

)= −

13.6n2 eV (0.77.11)

0.78 Energy Quantization

3

0.79 Atomic Structure

4

0.80 Atomic Spectra

0.80.1 Rydberg’s Equation

1λ

= RH

( 1n′2−

1n2

)(0.80.1)


DRAFT

Selection Rules lviiwhere RH is the Rydberg constant.

For the Balmer Series, n′ = 2, which determines the optical wavelengths. For n′ = 3, weget the infrared or Paschen series. The fundamental n′ = 1 series falls in the ultravioletregion and is known as the Lyman series.

0.81 Selection Rules

6

0.82 Black Body Radiation

0.82.1 Plank Formula

u( f ,T) =8π~c3

f 3

eh f/kT − 1(0.82.1)

0.82.2 Stefan-Boltzmann Formula

P(T) = σT4 (0.82.2)

0.82.3 Wein’s Displacement Law

λmaxT = 2.9 × 10−3 m.K (0.82.3)

0.82.4 Classical and Quantum Aspects of the Plank Equation

Rayleigh’s Equation

u( f ,T) =8π f 2

c3 kT (0.82.4)

We can get this equation from Plank’s Equation, eq. (0.82.1). This equation is a classicalone and does not contain Plank’s constant in it. For this case we will look at thesituation where h f < kT. In this case, we make the approximation

ex' 1 + x (0.82.5)

Thus the demonimator in eq. (0.82.1) becomes

eh f/kT− 1 ' 1 +

h fkT− 1 =

h fkT

(0.82.6)


DRAFT

lviii Atomic PhysicsThus eq. (0.82.1) takes the approximate form

u( f ,T) '8πhc3 f 3 kT

h f=

8π f 2

c3 kT (0.82.7)

As we can see this equation is devoid of Plank’s constant and thus independent ofquantum effects.

Quantum

At large frequencies, where h f > kT, quantum effects become apparent. We canestimate that

eh f/kT− 1 ' eh f/kT (0.82.8)

Thus eq. (0.82.1) becomes

u( f ,T) '8πhc3 f 3e−h f/kT (0.82.9)

0.83 X-Rays

0.83.1 Bragg Condition

2d sinθ = mλ (0.83.1)

for constructive interference off parallel planes of a crystal with lattics spacing, d.

0.83.2 The Compton Effect

The Compton Effect deals with the scattering of monochromatic X-Rays by atomictargets and the observation that the wavelength of the scattered X-ray is greater thanthe incident radiation. The photon energy is given by

E = hυ =hcλ

(0.83.2)

The photon has an associated momentum

E = pc (0.83.3)

⇒ p =E

c=

hυc

=hλ

(0.83.4)

The Relativistic Energy for the electron is

E2 = p2c2 + m2e c4 (0.83.5)

wherep − p′ = P (0.83.6)


DRAFT

Atoms in Electric and Magnetic Fields lixSquaring eq. (0.83.6) gives

p2− 2p · p′ + p′2 = P2 (0.83.7)

Recall that E = pc and E ′ = cp′, we have

c2p2− 2c2p · p′ + c2p′2 = c2P2

E 2− 2E E ′ cosθ + E ′2 = E2

−m2e c4 (0.83.8)

Conservation of Energy leads to

E + mec2 = E ′ + E (0.83.9)

Solving

E − E ′ = E −mec2

E 2− 2E E ′ + E ′ = E2

− 2Emec2 + m2e c4 (0.83.10)

2E E ′ − 2E E ′ cosθ = 2Emec2− 2m2

e c4 (0.83.11)

Solving leads to

∆λ = λ′ − λ =h

mec(1 − cosθ) (0.83.12)

where λc = hmec

is the Compton Wavelength.

λc =h

mec= 2.427 × 10−12m (0.83.13)

0.84 Atoms in Electric and Magnetic Fields

0.84.1 The Cyclotron Frequency

A test charge, q, with velocity v enters a uniform magnetic field, B. The force acting onthe charge will be perpendicular to v such that

FB = qv × B (0.84.1)

or more simply FB = qvB. As this traces a circular path, from Newton’s Second Law,we see that

FB =mv2

R= qvB (0.84.2)

Solving for R, we getR =

mvqB

(0.84.3)

We also see that

f =qB

2πm(0.84.4)

The frequency is depends on the charge, q, the magnetic field strength, B and the massof the charged particle, m.


DRAFT

lx Atomic Physics0.84.2 Zeeman Effect

The Zeeman effect was the splitting of spectral lines in a static magnetic field. This issimilar to the Stark Effect which was the splitting in the presence in a magnetic field.

In the Zeeman experiment, a sodium flame was placed in a magnetic field and itsspectrum observed. In the presence of the field, a spectral line of frequency, υ0 wassplit into three components, υ0 − δυ, υ0 and υ0 + δυ. A classical analysis of this effectallows for the identification of the basic parameters of the interacting system.

The application of a constant magnetic field, B, allows for a direction in space in whichthe electron motion can be referred. The motion of an electron can be attributed to asimple harmonic motion under a binding force −kr, where the frequency is

υ0 =1

2π

√k

me(0.84.5)

The magnetic field subjects the electron to an additional Lorentz Force, −ev × B. Thisproduces two different values for the angular velocity.

v = 2πrυ

The cetripetal force becomesmev2

r= 4π2υ2rme

Thus the certipetal force is

4π2υ2rme = 2πυreB + kr for clockwise motion

4π2υ2rme = −2πυreB + kr for counterclockwise motion

We use eq. (0.84.5), to emiminate k, to get

υ2−

eB2πme

υ − υ0 = 0 (Clockwise)

υ2 +eB

2πmeυ − υ0 = 0 (Counterclockwise)

As we have assumed a small Lorentz force, we can say that the linear terms in υ aresmall comapred to υ0. Solving the above quadratic equations leads to

υ = υ0 +eB

4πmefor clockwise motion (0.84.6)

υ = υ0 −eB

4πmefor counterclockwise motion (0.84.7)

We note that the frequency shift is of the form

δυ =eB

4πme(0.84.8)

If we view the source along the direction of B, we will observe the light to have twopolarizations, a closckwise circular polarization of υ0 + δυ and a counterclosckwisecircular polarization of υ0 − δυ.


DRAFT

Atoms in Electric and Magnetic Fields lxi0.84.3 Franck-Hertz Experiment

The Franck-Hertz experiment, performed in 1914 by J. Franck and G. L. Hertz, mea-sured the colisional excitation of atoms. Their experiement studied the current ofelectrons in a tub of mercury vapour which revealed an abrupt change in the currentat certain critical values of the applied voltage.2 They interpreted this observation asevidence of a threshold for inelastic scattering in the colissions of electrons in mer-cury atoms.The bahavior of the current was an indication that electrons could losea discreet amount of energy and excite mercury atoms in their passage through themercury vapour. These observations constituted a direct and decisive confirmation ofthe existence os quantized energy levels in atoms.

2Put drawing of Franck-Hertz Setup


DRAFT

lxii Atomic Physics


DRAFTSpecial Relativity

0.85 Introductory Concepts

0.85.1 Postulates of Special Relativity

1. The laws of Physics are the same in all inertial frames.

2. The speed of light is the same in all inertial frames.

We can define

γ =1√

1 − u2

c2

(0.85.1)

0.86 Time Dilation

∆t = γ∆t′ (0.86.1)

where ∆t′ is the time measured at rest relative to the observer, ∆t is the time measuredin motion relative to the observer.

0.87 Length Contraction

L =L′

γ(0.87.1)

where L′ is the length of an object observed at rest relative to the observer and L is thelength of the object moving at a speed u relative to the observer.

0.88 Simultaneity

4

DRAFT

lxiv Special Relativity0.89 Energy and Momentum

0.89.1 Relativistic Momentum & Energy

In relativistic mechanics, to be conserved, momentum and energy are defined as

Relativistic Momentump = γmv (0.89.1)

Relativistic EnergyE = γmc2 (0.89.2)

0.89.2 Lorentz Transformations (Momentum & Energy)

p′x = γ(px − β

Ec

)(0.89.3)

p′y = py (0.89.4)

p′z = pz (0.89.5)E′

c= γ

(Ec− βpx

)(0.89.6)

0.89.3 Relativistic Kinetic Energy

K = E −mc2 (0.89.7)

= mc2

1√1 − v2

c2

− 1

(0.89.8)

= mc2 (γ − 1)

(0.89.9)

0.89.4 Relativistic Dynamics (Collisions)

∆P′x = γ(∆Px − β

∆Ec

)(0.89.10)

∆P′y = ∆Py (0.89.11)

∆P′z = ∆Pz (0.89.12)∆E′

c= γ

(∆Ec− β∆Px

)(0.89.13)


DRAFT

Four-Vectors and Lorentz Transformation lxv0.90 Four-Vectors and Lorentz Transformation

We can represent an event in S with the column matrix, s,

s =

xyz

ict

(0.90.1)

A different Lorents frame, S′, corresponds to another set of space time axes so that

s′ =

x′

y′

z′

ict′

(0.90.2)

The Lorentz Transformation is related by the matrixx′

y′

z′

ict′

=

γ 0 0 iγβ0 1 0 00 0 1 0−iγβ 0 0 γ

xyz

ict

(0.90.3)

We can express the equation in the form

s′ = L s (0.90.4)

The matrix L contains all the information needed to relate position four–vectors forany given event as observed in the two Lorentz frames S and S′. If we evaluate

sTs =[

x y z ict]

xyz

ict

= x2 + y2 + z2− c2t2 (0.90.5)

Similarly we can show that

s′Ts′ = x′2 + y′2 + z′2 − c2t′2 (0.90.6)

We can take any collection of four physical quantities to be four vector provided thatthey transform to another Lorentz frame. Thus we have

b =

bx

by

bz

ibt

(0.90.7)

this can be transformed into a set of quantities of b′ in another frame S′ such that itsatisfies the transformation

b′ = L b (0.90.8)


DRAFT

lxvi Special RelativityLooking at the momentum-Energy four vector, we have

p =

px

py

pz

iE/c

(0.90.9)

Applying the same transformation rule, we have

p′ = L p (0.90.10)

We can also get a Lorentz-invariation relation between momentum and energy suchthat

p′Tp′ = pTp (0.90.11)

The resulting equality gives

p′2x + p′2y + p′2z −E′2

c2 = p2x + p2

y + p2z −

E2

c2 (0.90.12)

0.91 Velocity Addition

v′ =v − u1 − uv

c2

(0.91.1)

0.92 Relativistic Doppler Formula

υ = υ0

√c + uc − u

let r =

√c − uc + u

(0.92.1)

We have

υreceding = rυ0 red-shift (Source Receding) (0.92.2)

υapproaching =υ0

rblue-shift (Source Approaching) (0.92.3)

0.93 Lorentz Transformations

Given two reference frames S(x, y, z, t) and S′(x′, y′, z′, t′), where the S′-frame is movingin the x-direction, we have,

x′ = γ (x − ut) x = (x′ − ut′) (0.93.1)y′ = y y = y′ (0.93.2)z′ = y y′ = y (0.93.3)

t′ = γ(t −

uc2 x

)t = γ

(t′ +

uc2 x′

)(0.93.4)


DRAFT

Space-Time Interval lxvii0.94 Space-Time Interval

(∆S)2 = (∆x)2 +(∆y

)2+ (∆z)2

− c2 (∆t)2 (0.94.1)

Space-Time Intervals may be categorized into three types depending on their separa-tion. They are

Time-like Interval

c2∆t2 > ∆r2 (0.94.2)

∆S2 > 0 (0.94.3)

When two events are separated by a time-like interval, there is a cause-effectrelationship between the two events.

Light-like Interval

c2∆t2 = ∆r2 (0.94.4)

S2 = 0 (0.94.5)

Space-like Intervals

c2∆t2 < ∆r2 (0.94.6)∆S < 0 (0.94.7)


DRAFT

lxviii Special Relativity


DRAFTLaboratory Methods

0.95 Data and Error Analysis

0.95.1 Addition and Subtraction

x = a + b − c (0.95.1)

The Error in x is(δx)2 = (δa)2 + (δb)2 + (δc)2 (0.95.2)

0.95.2 Multiplication and Division

x =a × b

c(0.95.3)

The error in x is (δxx

)2

=(δaa

)2

+

(δbb

)2

+(δcc

)2

(0.95.4)

0.95.3 Exponent - (No Error in b)

x = ab (0.95.5)

The Error in x isδxx

= b(δaa

)(0.95.6)

0.95.4 Logarithms

Base e

x = ln a (0.95.7)

DRAFT

lxx Laboratory MethodsWe find the error in x by taking the derivative on both sides, so

δx =d ln a

da· δa

=1a· δa

=δaa

(0.95.8)

Base 10

x = log10 a (0.95.9)

The Error in x can be derived as such

δx =d(log a)

daδa

=ln a

ln 10

daδa

=1

ln 10δaa

= 0.434δaa

(0.95.10)

0.95.5 Antilogs

Base e

x = ea (0.95.11)

We take the natural log on both sides.

ln x = a ln e = a (0.95.12)

Applaying the same general method, we see

d ln xdx

δx = δa

⇒δxx

= δa (0.95.13)

Base 10

x = 10a (0.95.14)


DRAFT

Instrumentation lxxiWe follow the same general procedure as above to get

log x = a log 10log x

dxδx = δa

1ln 10

d ln adx

δx = δa

δxx

= ln 10δa (0.95.15)

0.96 Instrumentation

2

0.97 Radiation Detection

3

0.98 Counting Statistics

Let’s assume that for a particular experiment, we are making countung measurementsfor a radioactive source. In this experiment, we recored N counts in time T. Thecounting rate for this trial is R = N/T. This rate should be close to the average rate, R.The standard deviation or the uncertainty of our count is a simply called the

√N rule.

Soσ =√

N (0.98.1)

Thus we can report our results as

Number of counts = N ±√

N (0.98.2)

We can find the count rate by dividing by T, so

R =NT±

√N

T(0.98.3)

The fractional uncertainty of our count is δNN . We can relate this in terms of the count

rate.

δRR

=δNTNT

=δNN

=

√N

N

=1N

(0.98.4)


DRAFT

lxxii Laboratory MethodsWe see that our uncertainty decreases as we take more counts, as to be expected.

0.99 Interaction of Charged Particles with Matter

5

0.100 Lasers and Optical Interferometers

6

0.101 Dimensional Analysis

Dimensional Analysis is used to understand physical situations involving a mis ofdifferent types of physical quantities. The dimensions of a physical quantity areassociated with combinations of mass, length, time, electric charge, and temperature,represented by symbols M, L, T, Q, and θ, respectively, each raised to rational powers.

0.102 Fundamental Applications of Probability and Statis-tics

8


DRAFTGR9677 Exam Solutions

0.103 Discharge of a Capacitor

The voltage of a capacitor follows an exponential decay

V(t) = V0 exp[−

tRC

](0.103.1)

When the switch is toggled in the a position, the capacitor is quickly charged and thepotential across its plates is V. r is small and we assume that the potential differenceacross it is negligible. When the switch is toggled on the b position, the voltage acrossthe capacitor begins to decay. We can find the current through the resistor, R, fromOhm’s Law

I(t) =V(t)

R= V0 exp

[−

tRC

](0.103.2)

At t = 0 V0 = V. Graph B, shows an exponential decay.

Answer: (B)

0.104 Magnetic Fields & Induced EMFs

We have a circuit loop that is placed in a decaying magnetic field where the fielddirection acts into the page. We have two currents in the circuit. The first is due to thebattery and the other is an induced current from the changing magnetic field. We caneasily determine the current of the cell from Ohm’s Law.

Ic =VR

=5.010

A (0.104.1)

The induced EMF from the magnetic field is found from Faraday’s Law of Induction

E = −dΦ

dt(0.104.2)

where Φ is the magnetic flux.dΦ

dt= A

dBdt

(0.104.3)

DRAFT

lxxiv GR9677 Exam SolutionsThe area of our loop, A = 10 × 10cm2. So the induced EMF is

E = −100 × 10−4× 150 = 1.5 V (0.104.4)

The field acts into the page, we consider this a negative direction, it’s decaying, alsonegative. So

negative × negative = positive (0.104.5)

Faraday’s Law of Induction has a negative sign. So we expect our EMF to be negative.Using the Right Hand Grip Rule, and pointing our thumb into the page, our fingerscurl in the clockwise direction. So we see that the current from our cell goes in thecounter-clockwise direction and the induced current in the clockwise direction; theyoppose each other. The total EMF is

V = 5.0 − 1.5 = 3.5 volt (0.104.6)

The current through the resistor is

I =3.510

= 0.35 A (0.104.7)

Answer: (B)

0.105 A Charged Ring I

The Electric Potential is

V =Q

4πε0r(0.105.1)

The distance, r, of P from the charged ring is found from the pythagorean theorem

r2 = R2 + x2 (0.105.2)

Plugging this into the above equation yields

V =Q

4πε0

√

R2 + x2(0.105.3)

Answer: (B)

0.106 A Charged Ring II

The force a small charge, q experiences if placed in the center of the ring can be foundfrom Coulomb’s Law

F =qQ

4πε0R2 (0.106.1)


DRAFT

Forces on a Car’s Tires lxxvIf it undergoes small oscillations, R >> x, then

F = mRω2 (0.106.2)

Equating the two equations, and solving for ω, we get

ω =

√qQ

4πε0mR3 (0.106.3)

Answer: (A)

0.107 Forces on a Car’s Tires

The horizontal force on the car’s tires is the sum of two forces, the cetripetal force andthe frictional force of the road. The cetripetal force acts towards the center, FA, whilethe frictional force acts in the forwards direction, FC. If it’s not immediately clear whyit acts in the forward direction, the tires, as they rotate, exert a backward force on theroad. The road exerts an equal and opposite force on the tires, which is in the forwarddirection3 So the force on the tires is the sum of these forces, FA and FC, which is FB

Answer: (B)

0.108 Block sliding down a rough inclined plane

We are told several things in this question. The first is that the block attains a constantspeed, so it gains no kinetic energy; all its potential energy is lost due to friction.

Answer: (B)

0.108.1 Calculation

If you’d like a more rigorous proof, not something you might do in the exam. Thework done by the frictional force, Fr is

W =

∫Fr dx (0.108.1)

Fr acts along the direction of the incline and is equal to

Fr = mg sinθ (0.108.2)

The distance the force acts isx =

hsinθ

(0.108.3)

3Sometimes the frictional force can act in the direction of motion. This is one such case.


DRAFT

lxxvi GR9677 Exam SolutionsSo the work done is

W = Fr · x

= mg sinθ ×h

sinθ= mgh

Answer: (B)

0.109 Collision of Suspended Blocks

We are told that the ball collides elastically with the block, so both momentum andenergy are conserved. As the ball falls from a height, h, its potential energy is convertedto kinetic energy

mgh =12

mv21

v21 = 2gh (0.109.1)

Momentum is conserved, so

mv1 = mv2 + 2mv3

v1 = v2 + 2v3 (0.109.2)

Energy is also conserved, so

12

mv21 =

12

mv22 +

12

2mv23

v21 = v2

2 + 2v23 (0.109.3)

Squaring eq. (0.109.2) and equating with eq. (0.109.3) gives

v21 = (v2 + 2v3)2

∴ 2v2 = −v3 (0.109.4)

Substituting this into eq. (0.109.2) gives

v1 = v2 + 2v3

= 3v2

⇒ v21 = 9v2

2 (0.109.5)

The 2m block’s kinetic energy is converted to potential energy as it rises to a height ofh2. Thus

mgh2 =12

mv22

∴ v22 = 2gh2 (0.109.6)


DRAFT

Damped Harmonic Motion lxxviiWe see that

2gh9

= 2gh2

⇒h9

= h2 (0.109.7)

Answer: (A)

0.110 Damped Harmonic Motion

From section 0.4.4, we see that the frequency of a damped oscillator is

ω′ =

√ω2

0 −

(b

2m

)2

(0.110.1)

This shows that the damped frequency will be lower than the natural frequency, ω0, orits period, T0, will be longer.

Answer: (A)

0.111 Spectrum of the Hydrogen Atom

The hydrogen spectrum can be found by the emperical Rydberg equation

1λ

= RH

1n2

f

−1n2

i

(0.111.1)

where ni and n f are the intial and final states respectively. The longest wavelength, orthe smallest energy transition, would represent the transition n1 = n f + 1.

For the Lyman series, n f = 1, which lies in the ultra-violet spectrum, we have

1λL

= RH

( 112 −

122

)=

34

RH (0.111.2)

For the Balmer series, n f = 2, which lies in the optical spectrum, we have

1λB

= RH

( 122 −

132

)=

536

RH (0.111.3)

Dividing eq. (0.111.3) by eq. (0.111.2), we get

λL

λB=

536RH

34RH

=527

(0.111.4)

Answer: (A)4

4The other transition, the Paschen series, n f = 3, lies in the infra-red region of the spectrum.


DRAFT

lxxviii GR9677 Exam Solutions0.112 Internal Conversion

We are lucky that they actually tell us what the internal conversion process is. Fromthis we gather that the inner most electron has left its orbit and the most likely outcomewill be for the remaining electrons to ‘fall’ in and take its place. These transitions willresult in the emission of X-ray photons5.

Answer: (B)

0.113 The Stern-Gerlach Experiment

The description of the experiment in the question is the Stern-Gerlach Experiment.In this experiment, we expect the electrons to be deflected vertically into two beamsrepresenting spin-up and spin-down electrons.

Answer: (D)

0.114 Positronium Ground State Energy

The positronium atom consists of an electron and a positron in a bound state. Classi-cally, this atom looks like two planets orbiting a central point or center of mass. Weneed to reduce this system to an equivalent one where an electron cirlces a centralmass. We call this equivalent system the reduced mass of the two body system. This is

µ =meM

M + me(0.114.1)

The energy levels in terms of the reduced mass is defined as

En = −Z2

n2

µ

meE0 (0.114.2)

The reduced mass of positronium is

µ

me=

me

2me=

12

(0.114.3)

and the ground state of this atom, Z = 1 and n = 1. The ground state energy ofHydrogen is 13.6eV.

eq. (0.114.2) becomes

E1 = −12

13.6eV = −6.8 eV (0.114.4)

Answer: (C)5This can also result in the emission of an Auger Electron


DRAFT

Specific Heat Capacity and Heat Lost lxxix0.115 Specific Heat Capacity and Heat Lost

In this question, you are being asked to put several things together. Here, we are told,a heater is placed into the water but the water does not boil or change temperture. Wecan assume that all of the supplied heat by the heater is lost and we infer from thepower of the heater that 100 Joules is lost per second.

The energy to change water by one degree is derived from its specific heat capacity.

E = SHC ×Mass × Temp. Diff. = 4200 × 1 × 1 = 4200 J (0.115.1)

So the time to loose 4200 Joules of heat is

t =EP

=4200100

= 42 s (0.115.2)

Answer: (B)

0.116 Conservation of Heat

Assuming that little to no heat is lost to the environment, the two blocks will exchangeheat until they are both in thermal equilibrium with each other. As they have the samemasses we expect the final temperature to be 50°C. We can, of course, show this morerigorously where both blocks reach a final temperature, T f . The intial temperatures ofblocks I and II are T1 = 100°Cand T2 = 0°C, respectively.

Heat Lost by Block I = Heat gained by Block II

0.1 × 103× 1 ×

(100 − T f

)= 0.1 × 103

× 1 ×(T f − 0

)2(0.1 × 103

)T f = 10 × 103

∴ T f = 50 °C

Thus the heat exchanged is

0.1 × 103× 1 × (100 − 50) = 5 kcal (0.116.1)

Answer: (D)

0.117 Thermal Cycles

We are told the cycle is reversible and moves from ABCA. We can examine each pathand add them to get the total work done.


DRAFT

lxxx GR9677 Exam SolutionsPath A→ B is an isothermal process

WA→B =

∫ V2

V1

P · dV where P =nRT

V

= nRTh

∫ V2

V1

dVV

= nRTh ln[V2

V1

](0.117.1)

Path B→ C is an isobaric process

WB→C =

∫ V1

V2

P2 dV where P2V2 = nRTh

= P2 (V1 − V2) and P2V1 = nRTc

= nR (Tc − Th) (0.117.2)

and Path C→ A

WC→A =

∫P1 dV where dV = 0

= 0 (0.117.3)

Adding the above, we get

W = WA→B + WB→C + WC→A

= nRTh ln[V2

V1

]+ nR (Tc − Th) (0.117.4)

where n = 1 mole.W = RTh ln

[V2

V1

]− R (Th − Tc) (0.117.5)

Answer: (E)

0.118 Mean Free Path

The mean free path of a particle, be it an atom, molecule or photon, is the averagedistance travelled between collisions. We are given the equation as

` =1ησ

(0.118.1)

where η is the number desnity and σ is the collision cross section. The number densityworks out to be

η =NV

(0.118.2)


DRAFT

Probability lxxxiwhere N is the number of molecules and V is the volume. We can determine this fromthe ideal gas law,

PV = NkT

∴ η =NV

=PkT

(0.118.3)

The collision cross section is the area through which a particle can not pass withoutcolliding. This works out to be

σ = πd2 (0.118.4)

Now we can write eq. (0.118.1) in terms of variables we know

` =kTπPd2 (0.118.5)

As air is composed mostly of Nitrogen, we would have used the diameter of Nitrogenin our calculations. This is approximately d = 3.1 Å. Plugging in the constants givenwe have

` =(1.38 × 1023)(300)

π × 1.0 × 105(3.1 × 1010)2

= 1.37 × 10−7 m

As we don’t have a calculator in the exam, we can estimate by adding the indices inour equation,

− 23 + 2 − 5 + 20 = −6 (0.118.6)

So we expect our result to be in the order of 1 × 10−6 m. We choose (B).

Answer: (B)

0.119 Probability

The probability of finding a particle in a finite interval between two points, x1 and x2,is

P(2 ≤ x ≤ 4) =

∫ 4

2|Ψ(x)|2 dx (0.119.1)

with the normalization condition,∫ +∞

−∞

|Ψ(x)|2 dx = 1 (0.119.2)

We can tally the values given to us on the graph


DRAFT

lxxxii GR9677 Exam Solutions

x Ψ Ψ2

1 1 12 1 13 2 44 3 95 1 16 0 0

Total 16

Table 0.119.1: Table of wavefunction amplitudes

The probability of finding the particle between (2 ≤ x ≤ 4) is

P(2 ≤ x ≤ 4) =22 + 32

12 + 12 + 22 + 32 + 12 + 02

=4 + 9

1 + 1 + 4 + 9 + 1 + 0

=1316

(0.119.3)

Answer: (E)

0.120 Barrier Tunneling

Classically, if a particle didn’t have enough kinetic energy, it would just bounce offthe wall but in the realm of Quantum Mechanics, there is a finite probability that theparticle will tunnel through the barrier and emerge on the other side. We expect to seea few things. The wave function’s amplitide will be decreased from x > b and to decayexponentially from a < x < b. We see that choice (C) has these characteristics.

Answer: (C)

0.121 Distance of Closest Appraoch

This question throws a lot of words at you. The α-particle with kietic energy 5 MeV isshot towards an atom. If it goes towards the atom it will slow down, loosing kineticenergy and gaining electrical potential energy. The α-particle will then be repelled bythe Ag atom. Thus

U =1

4πε0

Q1Q2

D= KE (0.121.1)


DRAFT

Collisions and the He atom lxxxiiiWhere D is the distance of closest approach, q1 = ze and q2 = Ze. We are given z = 2for the alpha particle and Z = 50 for the metal atom. Plugging in all of this gives us

D =1

4πε0

(2e)(50e)5 × 106e

=1

4πε0

100e5 × 106

=1

4π × 8.85 × 10−12

100 × 1.6 × 10−19

5 × 106

≈ 0.3 × 10−13 m (0.121.2)

Answer: (B)

0.122 Collisions and the He atom

As the collision is elastic, we know that both momentum and kinetic energy is con-served. So conservation of momentum shows

4uv = (−0.6)(4u)v + MV⇒ 6.4 uv = MV (0.122.1)

Conservation of Energy shows that

12

(4 u)v2 =12

(4 u)(0.6v)2 +12

MV2

4 u[0.64v2

]= MV2 (0.122.2)

Solving for M

M =6.42u

4(0.64)= 16 u (0.122.3)

We see that this corresponds to an Oxygen atom, mass 16 u.

Answer: (D)

0.123 Oscillating Hoops

We are given the period of our physical pendulum, where

T = 2π

√I

mgd(0.123.1)

where I is the moment of inertia and d is the distance of the pivot from the center ofmass. The moment of inertia of our hoop is

Icm = Mr2 (0.123.2)


DRAFT

lxxxiv GR9677 Exam SolutionsThe moment of inertia of the hoop hanging by a nail is found from the Parallex AxisTheorem

I = Icm + Md2 = Mr2 + Mr2 = 2Mr2 (0.123.3)

Plugging this into the first equation gives

T = 2π

√2Mr2

Mgr

= 2π

√2rg

≈ 2π

√2 × 20 × 10−2

10= 4π × 10−1

≈ 1.2 s

Answer: (C)

0.124 Mars Surface Orbit

If a body travels forward quickly enough that it follows the planet’s curvature it is inorbit. We are told that in the case of Mars, there is a 2.0 meter drop for every 3600 meterhorizontal distance. We are also told that the acceleration due to gravity on Mars isgM = 0.4g. So the time to drop a distance of 2.0 meters is

s =12

gMt2

⇒ t = 1 s (0.124.1)

So the horizontal speed is

vx =3600

1m/s (0.124.2)

Answer: (C)

0.125 The Inverse Square Law

Choice A Energy will be conserved. This isn’t dependent on an inverse square law.

Choice B Momentum is conserved. This also isn’t dependent on the inverse squarelaw.

Choice C This follows from Kepler’s Law

mr(2π

T

)2

=Gm1m2

r2+ε

⇒ T ∝ r(3+ε)/2 (0.125.1)


DRAFT

Charge Distribution lxxxvChoice D This is FALSE. This follows from Bertrand’s Theorem6, which states that

only two types of potentials produce stable closed orbits

1. An inverse square central force such as the gravitational or electrostaticpotential.

V(r) =−kr

(0.125.2)

2. The radial Harmonic Oscillator Potential

V(r) =12

kr2 (0.125.3)

Choice E A stationary circular orbit occurs under special conditions when the centralforce is equal to the centripetal force. This is not dependent on an inverse squarelaw but its speed.

Answer: (D)

0.126 Charge Distribution

An inportant thing to keep in mind is that charge will distribute itself evenly throughoutthe conducting spheres and that charge is conserved.

Step I: Uncharged Sphere C touches Sphere A

A B CQ2

QQ2

Step II: Sphere C is touched to Sphere B

A B CQ2

3Q4

3Q4

The initial force between A and B is

F =kQ2

r2 (0.126.1)

The final force between A and B is

F f =k Q

23Q4

r2 =38

F (0.126.2)

Answer: (D)6Add reference here


DRAFT

lxxxvi GR9677 Exam Solutions0.127 Capacitors in Parallel

We have one capacitor, C1 connected to a battery. This capacitor gets charged andstores a charge, Q0 and energy, U0.

Q0 = C1V (0.127.1)

U0 =12

C1V2 (0.127.2)

When the switch is toggled in the on position, the battery charges the second capacitor,C2. As the capacitors are in parallel, the potential across them is the same. As C1 = C2,we see that the charges and the energy stored across each capacitor is the same. Thus

Q1 = C1V1 Q2 = C2V2

Q2 = Q1

U1 =12

C1V21 U2 =

12

C2V22

U2 = U1

We see thatU1 + U2 = C1V2

1 = 2U0 (0.127.3)

We can also analyze this another way. The two capacitors are in parallel, so their netcapacitance is

CT = C1 + C2 = 2C1 (0.127.4)

So the total charge and energy stored by this parallel arrangemt is

Q = CTV1 = 2C1V1

UT =12

2C1V21 = 2U0

Of all the choices, only (E) is incorrect.

Answer: (E)

0.128 Resonant frequency of a RLC Circuit

The circuit will be best ‘tuned’ when it is at its resonant frequence. This occurs whenthe impedances for the capacitor and inductor are equal. Thus

XC =1ωC

and XL = ωL (0.128.1)

When they are equal

XC = XL

1ωC

= XL = ωL

∴ ω =1√

LC(0.128.2)


DRAFT

Graphs and Data Analysis lxxxviiSolving for C,

C =ω2L

=14π2× (103.7 × 106)2

× 2.0 × 10−6

≈ 0.125 × 10−11 F (0.128.3)

Answer: (C)

0.129 Graphs and Data Analysis

It is best to analyse data is they are plotted on straight line graphs of the form, y = mx+c.This way we can best tell how well our data fits, etc.7

A We want a plot of activity, dNdt vs. time, t. If we were to plot this as is, we would

get an exponential curve. To get the straight line graph best suited for furtheranalysis, we take the logs on both sides.

dNdt∝ e−2t

log[dNdt

]= log e−2t

log[dNdt

]= −2t (0.129.1)

We have a Semilog graph with a plot of log[

dNdt

]on the y-axis, t on the x-axis with

a gradient of 2.

B This is already a linear equation we can plot with the data we already have. Noneed to manipulate it in any way.

C We take logs on bot sides of the equation to get

s ∝ t2

log s = 2 log t (0.129.2)

We can plot log s vs. log t. This gives a linear equation with log s on the y-axisand log t on the x-axis and a gradient of 2.

D Again, we take logs on both sides of the equation

VoutVin

∝1ω

log[VoutVin

]= − logω

7This is of course with nothing but a sheet of graph paper and calculator and without the help ofcomputers and data analysis software.


DRAFT

lxxxviii GR9677 Exam Solutions

We have a log-log plot of log[

VoutVin

]on the y-axis and logω on the x-axis with a

gradient of -1. We see that this choice is INCORRECT.

E As with the other choices, we take logs on both sides and get

P ∝ T4

log P = 4 log T

This can be plotted on a log-log graph with log P on the y-axis and log T on thex-axis and a gradient of 4.

Answer: (D)

0.130 Superposition of Waves

As the question states, we can see the superposition of the two waves. For the higherfrequency wave, we see that the period on the oscilloscope is about 1cm. This worksout to be a period of

T =1 cm

0.5 cm ms−1 = 2.0 ms (0.130.1)

The frequency is

f =1

2.0 × 10−3

= 500 Hz (0.130.2)

We can measure the amplitude of this oscillation by measuring the distance from crestto trough. This is approximately (2 − 1)/2, thus8

A = 1 cm × 2.0 V cm−1

≈ 2.0 V (0.130.3)

For the longer period wave, we notice that approximately a half-wavelength is dis-played, is 2(4.5 − 1.5) = 6 cm. The period becomes

T =6.0 cm

0.5 cm/ms= 12.0 ms (0.130.4)

Thus the frequency is

f =1T

=1

12.0 × 10−3 = 83 Hz (0.130.5)

We see that (D) matches our calculations.

Answer: (D)8If you happened to have worked this one first you’ll notice that only choice (D) is valid. You can

stop and go on to the next question.


DRAFT

The Plank Length lxxxix0.131 The Plank Length

This question is best analysed through dimensional analysis; unless of course you’refortunate to know the formula for the Plank Length. We are told that

G = 6.67 × 10−11 m3 kg−1 s−2

~ =6.63 × 10−34

2πJ s−1

c = 3.0 × 108 m s−1

We can substitute the symbols for Length, L, Mass, M and Time, T. So the dimensionsof our constants become

G = L3M−1T−2

~ = ML2T−1

c = LT−1

`p = L

Our Plank Length is in the form`p = Gx~ycz

Dimensional Analysis Gives

L =(L3M−1T−2

)x (ML2T−1

)y (LT−1

)z

We get

L3x + 2y + z = 1

M− x + y = 0

Tz = −3x

Solving, we get

x =12

y =12

z = −32

Thus

`p =

√G~c3

Answer: (E)


DRAFT

xc GR9677 Exam Solutions0.132 The Open Ended U-tube

We recall that the pressure throughout a fluid is equal throughout the fluid. As thesystem is in equlibrium, the pressure on the left arm is equal to the pressure on theright arm. We can set up an equation such that

ρ2g5 + ρ1g (h1 − 5) = ρ1gh2 (0.132.1)

where whater, ρ1 = 1.0 g/cm3, some immiscible liquid, ρ2 = 4.0 g/cm3.

Solving, gives ush2 − h1 = 15 cm (0.132.2)

Let’ call the height of the water column on the left side of the tube, x1. We get

h2 − (x1 + 5) = 15∴ h2 − x1 = 20

We expect the water column to go down on the left side of the tube as it goes up on theright side of the tube; conservation of mass. So we infer the change in height on bothsides is 10 cm. We conclude that since the intial height is 20cm, then h2 = 30 cm andx1 = 10 cm. So

h2

h1=

3015

= 2 (0.132.3)

Answer: (C)

0.133 Sphere falling through a viscous liquid

Our sphere falls through a viscous liquid under gravity and experiences a drag force,bv. The equation of motion can be expressed

ma = mg − bv (0.133.1)

We are also told that the buoyant force is negligible. Armed with this information, wecan analyze out choices and emiminate.

A This statement will be incorrect. We have been told to ignore the buoyant force,which if was present, would act as a constant retarding force and slow our spheredown and reduce its kinetic energy. INCORRECT

B This is also incorrect. In fact if you were to solve the above equation of motion, thespeed, and hence kinetic energy, would monotonically increase and approachsome terminal speed. It won’t go to zero. INCORRECT

C It may do this if it was shot out of a gun, but we were told that it is released fromrest. So it will not go past its terminal speed.


DRAFT

Moment of Inertia and Angular Velocity xciD The terminal speed is the point when the force due to gravity is balanced by the

retarding force of the fluid. Setting ma = 0 in the above equation, we get

0 = mg − bv (0.133.2)

Solving for v yields,

v =mgb

(0.133.3)

We see that our terminal velocity is dependent on both b and m. This choice isINCORRECT

E From the above analysis, we choose this answer. CORRECT

Answer: (E)

0.134 Moment of Inertia and Angular Velocity

The moment of inertia of an object is

I =

N∑i=1

mir2i

where ri is the distance from the point mass to the axis of rotation.

The moment of inertia about point A is found by finding the distances of each of thethree masses from that point. The distance between the mass, m and A is

r =`√

3

Thus the moment of inertia is

IA = 3m(`√

3

)2

= m`2

The Moment of Inertia about B can be found by the Parallel Axis Theorem but it may besimpler to use the formula above. As the axis of rotation is about B, we can ignore thismass and find the distances of the other two masses from this point, which happens tobe `. Thus

IB = 2m`2

The rotational kinetic energy is

K =12

Iω2

So the ratio of the kinetic energies at fixed, ω becomes

KB

KA=

IB

IA=

2m`2

m`2 = 2

Answer: (B)


DRAFT

xcii GR9677 Exam Solutions0.135 Quantum Angular Momentum

The probability is

P =32 + 22

38=

1338

(0.135.1)

NOT FINISHED

Answer: (C)

0.136 Invariance Violations and the Non-conservation ofParity

Electromagnetic and strong interactions are invariant under parity transformations.The only exception to this rule occurs in weak interactions, the β-decay bring one suchexample. It had always been assumed that invariance was a “built-in” property of theUniverse but in the 1950s there seemed to be some puzzling experiments concerningcertain unstable particles called tau and theta mesons. The “tau-theta puzzle” wassolved in 1956 by T.D. Lee9 and C.N. Yang10 when they proposed the nonconservation ofparity by the weak interaction. This hypothesis was confirmed experimentally throughthe beta decay of Cobalt-60 in 1957 by C.S. Wu11.

60Co −−→ 60Ni + e – + υe

The cobalt source was chilled to a temperature of 0.01 K and placed in a magneticfield. This polarized the nuclear spins in the direction of the magnetic field while thelow temperatures inhibited the thermal disordering of the aligned spins. When thedirections of the emitted electrons were measured, it was expected that there would beequal numbers emitted parallel and anti-parallel to the magnetic field, but instead moreelectrons were emitted in the direction opposite to the magnetic field. This observationwas interpreted as a violation of reflection symmetry.

Answer: (D)

9Tsung-Dao Lee is a Chinese-born American physicist, well known for his work on parity violation,the Lee Model, particle physics, relativistic heavy ion (RHIC) physics, nontopological solitons andsoliton stars. He and Chen-Ning Yang received the 1957 Nobel prize in physics for their work on paritynonconservation of weak interactions.

10Chen-Ning Franklin Yang is a Chinese-American physicist who worked on statistical mechanicsand particle physics. He and Tsung-dao Lee received the 1957 Nobel prize in physics for their work onparity nonconservation of weak interactions.

11Chien-Shiung Wu was a Chinese-American physicist. She worked on the Manhattan Project toenrich uranium fuel and performed the experiments that disproved the conservation of parity. Shehas been known as the“First Lady of Physics”, “Chinese Marie Curie” and “Madam Wu”. She died inFebruary 16, 1997


DRAFT

Wave function of Identical Fermions xciii0.137 Wave function of Identical Fermions

The behavior of fermions are described by the Pauli Exclusion Principle, which statesthat no two fermions may have the same quantum state. This is a results in theanti-symmetry in the wave funtion.

Answer: (A)

0.138 Relativistic Collisions

We are told that no energy is radiated away, so it is conserved; all of it goes into thecomposite mass. The relativistic energy is

E = γmc2 (0.138.1)

Given that v = 3/5c

γ =1√

1 − v2

c2

=54

(0.138.2)

So the energy of the lump of clay is

E = γmc2 =54

mc2 (0.138.3)

The composite mass can be found by adding the energies of the two lumps of clay

ET = 2E

Mc2 =104

mc2

∴M = 2.5m = 2.5 × 4 = 10 kg (0.138.4)

Answer: (D)

0.139 Relativistic Addition of Velocities

We recall that the relativistic addition formula

v′ =u + v1 + uv

c2

(0.139.1)

where u = 0.3c and v = 0.6c. This becomes

v′ =0.9c

1 + 0.18=

0.91.18

c ≈912

c = 0.75c (0.139.2)

Answer: (D)


DRAFT

xciv GR9677 Exam Solutions0.140 Relativistic Energy and Momentum

The Relativistic Momentum and Energy equations are

p = γmv E = γmc2 (0.140.1)

We can determine the speed by dividing the relativistic momentum by the relativisticenergy equation to get

pE

=γmvγmc2

=vc2

∴5MeV/c10MeV

=vc2

510c

=vc2

⇒ v =12

c (0.140.2)

Answer: (D)

0.141 Ionization Potential

The Ionization Potential, or Ionization Energy, EI, is the energy required to remove onemole of electrons from one mole of gaaseous atoms or ions. It is an indicator of thereactivity of an element.

24He The Helium atom is a noble gas and has filled outermost electron shells as well as

its electrons being close to the nucleus. It would be very difficult to ionize.

He = 1s2

714N Nitrogen has two outermost electrons.

N = 1s2, 2s2, 2p6, 2s2, 2p2

816O Oxygen has four outermost electrons.

O = 1s2, 2s2, 2p6, 2s2, 2p4

1840Ar Another noble gas, this has filled outermost electrons and is not reactive.

55133Cs We can see that Cs has a high atomic number and hence a lot of electrons. We

expect the outmost electrons to be far from the nucleus and hence the attractionto be low. This will have a low ionization potential.

Answer: (E)


DRAFT

Photon Emission and a Singly Ionized He atom xcv0.142 Photon Emission and a Singly Ionized He atom

The energy levels can be predicted by Bohr’s model of the Hydrogen atom. As aHelium atom is more massive than Hydrogen, some corrections must be made to ourmodel and equation. The changes can be written

En = −Z2

n2

µ

meE0 (0.142.1)

where Z is the atomic number, n is the energy level, E0 is the ground state energy levelof the Hydrogen atom and µ/me is the reduced mas correction factor.

The emitted photon can also be found through a similar correction

∆E =hcλe

= Z2 µ

me

1n2

f

−1n2

i

13.6 (0.142.2)

As Helium’s mass is concentrated in the center, it’s reduced mass is close to unity12.

µ

me=

ZZ + me

≈ 1 (0.142.3)

Plugging in the values we know into eq. (0.142.2), we get

6.63 × 10−34× 3 × 108

470 × 10−9 × 1.60 × 10−19 = 2213.6

1n2

f

−142

(0.142.4)

After some fudging and estimation we get

6.63 × 3470 × 1.6

× 102≈

19470 × 1.6

× 102

≈20

750× 102

= 0.026 × 102 eV (0.142.5)

and2.6

22 · 13.6≈

120

12It is helpful to know that in the case of atoms, the reduced mass will be close to unity and can beignored from calculation. In the case of smaller bodies, e.g. positronium, this correction factor can notbe ignored.


DRAFT

xcvi GR9677 Exam SolutionsSolving for n f , gives13

120

=1n2

f

−142

1n2

f

=120

+1

16=

980

≈19

∴ n f = 3 (0.142.6)

Now we can calculate the energy level at n = 3 from eq. (0.142.1), which gives,

E3 = −22

32 · 13.6

= −6.0 eV (0.142.7)

We get E f = −6.0 eV and n f = 3. This corresponds to (A).

Answer: (A)

0.143 Selection Rules

NOT FINISHED

Answer: A

0.144 Photoelectric Effect

This question deals with the photoelectric effect which is essentially an energy conser-vation equation. Energy of a photon strikes a metal plate and raises the electrons towhere they can leave the surface. Any extra energy is then put into the kinetic energyof the electron. The photoelectric equation is

h f = eVs + K (0.144.1)

As our choices are in electron-volts, our equation becomes

hceλ

= eVs + K (0.144.2)

where K is the kinetic energy of our photoelectrons. Plugging in the values we weregiven and solving for K, we get

K = 0.2 eV (0.144.3)

Answer: (B)13As n f = 3 is only in choice (A), we can forego any further calculation and choose this one.


DRAFT

Stoke’s Theorem xcvii0.145 Stoke’s Theorem

NOT FINISHED

Answer: (C)

0.146 1-D Motion

A particle moves with the velocity

v(x) = βx−n (0.146.1)

To find the acceleration, a(x), we use the chain rule

a(x) =dvdx

=dxdt·

dvdx

= v ·dvdx

(0.146.2)

Differentiating v(x) with respect to x gives

dvdx

= −nβx−n−1

Thus, our acceleration, a(x), becomes

a(x) = βx−n· −nβx−n−1

= −nβ2x−2n−1 (0.146.3)

Answer: (A)

0.147 High Pass Filter

Capacitors and Inductors are active components; their impedances vary with the fre-quency of voltage unlike an ohmic resistor whose resitance is pretty much the same nomatter what. The impedances for capacitors and inductors are

XC =1ωC

XL = ωL

We see that in the case of capacitors, there is an inverse relationship with frequencyand a linear one for inductors. Simply put, at high frequencies capacitors have lowimpedances and inductors have high inductances.

NOT FINSIHED

Answer: (E)


DRAFT

xcviii GR9677 Exam Solutions0.148 Generators and Faraday’s Law

The induced EMF in the loop follows Faraday’s Law

E = −dΦ

dt

In this case, the magnetic field, B, is constant and the Cross Sectional Area, A, throughwhich the magnetic field acts changes. Thus the above equation becomes

E0 sinωt = −BdAdt

(0.148.1)

Let’s say that at t = 0 the loop is face on with the magnetic field,

A = πR2 cosωt (0.148.2)

Substituting this into eq. (0.148.1) gives

E0 sinωt = −B ·dAdt

= −B ·(−ωπR2 sinωt

)= ωBπR2 sinωt (0.148.3)

Solving for ω gives

ω =E0

BπR2 (0.148.4)

Answer: (C)

0.149 Faraday’s Law and a Wire wound about a RotatingCylinder

The induced EMF of our system can be found from Faraday’s Law, where

E = −dΦ

dt(0.149.1)

Here the flux changes because the number of loops enclosing the field increases, so

Φ = NBA (0.149.2)

Substituting this into Faraday’s Equation we get

E = BAdNdt

= BπR2N (0.149.3)

Answer: (C)


DRAFT

Speed of π+ mesons in a laboratory xcix0.150 Speed of π+ mesons in a laboratory

As the π+ meson travels through our laboratory and past the detectors, its half lifeis time dilated in our laboratory’s rest frame. We can also look at things in the π+

meson’s rest frame. In this case, the distance it travels will be length contracted in itrest frame. The speed of our π+ mesons is the length divided by the time dilation in thelaboratory’s rest frame or the length contraction in the π+ meson’s rest frame dividedby its half life. In either case, we get

v =L

T1/2

√1 −

v2

c2

Factorizing we get

v2

1 +L2

c2T21/2

=L2

T21/2

(0.150.1)

We see thatL

T1/2= 6 × 108

Plugging this into eq. (0.150.1), the speed in terms of c

v2

c2

[1 +

369

]=

369

v2

c2 (5) = 4

⇒ v =2√

5c (0.150.2)

Answer: (C)

0.151 Transformation of Electric Field

NOT FINSHED

Answer: (C)

0.152 The Space-Time Interval

We have two events, in the S-frame,

S1(x1, t) and S2(x2, t)

In the S′-frame, the co-ordintes are

S′1(x′1, t′

1) and S′2(x′2, t′

2)


DRAFT

c GR9677 Exam SolutionsThe Space-Time Interval in the S-frame

∆S = ∆x2 = 3c minutes (0.152.1)

In the S′-frame, the Space-Time Interval is

∆S′ = ∆x′2 − c2∆t′2 = 5c minutes (0.152.2)

The Space-Time Interval is invariant across frames, so eq. (0.152.1) is equal to eq. (0.152.2)

(3c)2 = (5c)2− c2∆t2

⇒ ∆t = 4 minutes (0.152.3)

Answer: (C)

0.153 Wavefunction of the Particle in an Infinte Well

The wave function has zero probability density in the middle for even wave functions,n = 2, 4, 6, · · · .

Answer: (B)

0.154 Spherical Harmonics of the Wave Function

NOT FINSIHED

Answer: (C)

0.155 Decay of the Positronium Atom

NOT FINSHED

Answer: (C)

0.156 Polarized Electromagnetic Waves I

We are given an electromagnetic wave that is the superposition of two independentorthogonal plane waves where

E = xE1 exp [i (kz − ωt)] + yE2 exp [i (kz − ωt + π)] (0.156.1)


DRAFT

Polarized Electromagnetic Waves II ciAs we are looking at the real components and E1 = E2, we have

E =<(E1eikz· e−iωt)x +<(E1eikz

· e−iωt· e−iπ)y

=<(E1eikz· e−iωt)x −<(E1eikz

· e−iωt)y

We see that the x and y vectors have the same magnitude but opposite sign; they areboth out of phase with each other. This would describe a trajectory that is 135°to thex-axis.

Answer: (B)

0.157 Polarized Electromagnetic Waves II

NOT FINISHED

Answer: (A)

0.158 Total Internal Reflection

Total internal reflectance will occur when the incident beam, reaches a critical angle,θi, such that the refracted angle just skims along the water’s surface, θr = 90◦. We canfind this critical angle using Snell’s Law

n2 sinθi = n1 sin 90

where n2 = 1.33 and n1 = 1, we have

sinθ2 =1

1.33=

34

(0.158.1)

We know that sin 30° = 1/2 and sin 60° =√

3 /2, so 30° < θ < 60°.

Answer: (C)

0.159 Single Slit Diffraction

For a single slit, diffraction maxima can be found from the formula

a sinθ = mλ (0.159.1)

where a is the slit width,θ is the angle between the minimum and the central maximum,and m is the diffraction order. As θ is small and solving doe d, we can approximate the


DRAFT

cii GR9677 Exam Solutionsabove equation to

d =λθ

=400 × 10−9

4 × 10−3

= 0.1 × 10−3 m (0.159.2)

Answer: (C)

0.160 The Optical Telescope

The magnification of the optical telescope can be found from the focal length of theeyepiece, fe and the objective, fo. Thus

M =fo

fe= 10 (0.160.1)

The focal length of the objective is

fe = 10 × 1.5 = 15 cm (0.160.2)

To achieve this magnification, the lens must be placed in a position where the focallength of the eyepiece meets the focal length of the objective. Thus

D = fe + fo = 15.0 + 1.5 = 16.5 cm (0.160.3)

Answer: (E)

0.161 Pulsed Lasers

Lasers operating in pulsed mode delivers more energy in a short space of time asopposed to delivering the same energy over a longer period of time in a continuousmode. While there are several methods to achieve a pulsed mode, beyond what isneeded to answer this question, we can determine the number of photons deliveredby such a device. The energy of a photon is

E = h f =hcλ

(0.161.1)

The power is the energy delivered in one second. So for a 10kW laser, the total energyin 10−15 seconds is

EL = Pt = 10 × 103× 10−15

= 10 × 10−12 J (0.161.2)


DRAFT

Relativistic Doppler Shift ciiiSo the total number of photons is

n =EL

E

=10 × 10−12

× λhc

=10 × 10−12

× 600 × 10−9

6.63 × 10−34 × 3 × 108

=10 × 6006.63 × 3

× 106≈ 3 × 108 (0.161.3)

Answer: (B)

0.162 Relativistic Doppler Shift

The relativistic doppler shift is

λo

λs=

fs

fo=

√1 + β

1 − β(0.162.1)

The redshift is calculated to be

z =λo − λsλs

=fs − fo

fo(0.162.2)

we can rewrite eq. (0.162.2) as

z =

√1 + β

1 − β− 1 (0.162.3)

In the non-relativistic limit, v << c, we can approximate eq. (0.162.3) to

z ≈ β =vc

(0.162.4)

This, equating eq. (0.162.2) and eq. (0.162.4), we see that

v =∆ ff

c (0.162.5)

Substituting the values given into eq. (0.162.5), we have

v =0.9 × 10−12

122 × 10−9 × 3 × 108

≈ 2.2 m s−1 (0.162.6)

Answer: (B)


DRAFT

civ GR9677 Exam Solutions0.163 Gauss’ Law, the Electric Field and Uneven Charge

Distribution

We can find the electric field in a non-conducting sphere by using Gauss’ Law∮E · dA =

Qenclosed

ε0(0.163.1)

The enclosed charge can be found from the charge density, which is

ρ =Enclosed ChargeEnclosed Volume

=q

43πr3

(0.163.2)

We can find the enclosed charge by integrating within 0 to R/2. The charge density is

ρ =dqdV

=dq

4πr2dr(0.163.3)

∴ dq = ρ4πr2dr

= 4πAr4dr (0.163.4)

Gauss’ Law becomes ∮E · dA =

qenclosed

ε0

E(4πr2

)=

∫ R2

0

dqε0

=4πAε0

∫ R2

0r4dr

=4πAε0

r5

5

∣∣∣∣∣R2

0(0.163.5)

∴ E =Aε0

r3

5

∣∣∣∣∣R2

0(0.163.6)

Solving gives

E =AR3

40ε0(0.163.7)

Answer: (B)


DRAFT

Capacitors in Parallel cv0.164 Capacitors in Parallel

We initially charge both of our capacitors in parallel across a 5.0V battery. The chargestored on each capacitor is

Q1 = C1V Q2 = C2V

where V = 5.0 V, C1 = 1µF and C2 = 2µF.

The battery is then disconnected and the plates of opposite charges are connectedto each other. This results in the excess charges cancelling each other out and thenredistributing themselves until the potential across the new configuration is the same.The charge left after this configuration is

QA = Q2 −Q1 (0.164.1)

The charges on each capacitor becomes

Q1A + Q2A = QA (0.164.2)

where

Q1A = C1V f and Q2A = C2V f (0.164.3)

Solving for V f gives us

V f =(C2 − C1)V

C1 + C2=

53

= 1.67 V (0.164.4)

Answer: (C)

0.165 Standard Model

NOT FINSIHED

Answer: (A)

0.166 Nuclear Binding Energy

Typically a heavy nucleus contains ∼ 200 nucleons. The energy liberated would be thedifference in the binding energies 1 MeV ×200.

Answer: (C)


DRAFT

cvi GR9677 Exam Solutions0.167 Work done by a man jumping off a boat

The work the man does is the sum of the kinetic energies of both the boat and himself.We can find the speeds of the man and the boat because momentum is conserved.

mu = Mv

u =Mvm

(0.167.1)

The total energy of the system, and hence the work the man does in jumping off theboat is

W =12

mu2 +12

Mv2

=12

Mv2[M

m+ 1

](0.167.2)

Answer: (D)

0.168 Orbits and Gravitational Potential

For an attractive potential, such as we would expect for the Gravitational Potential, wehave

Orbit Total Energy

Ellipse E < 0Parabola E = 0Hyperbola E > 0

where E is the total energy of the system; potential and kinetic energy. As the potentialenergy of the system remains unchanged, the only difference is the is the kinetic energy,the orbit will be hyperbolic.

When the spacecraft has the same speed as Jupiter, the orbit will be locked and will beelliptical. If the gravitational potential energy was equal to the kinetic energy, the orbitwill no longer be locked and will be parabolic. We assume that the huge difference willcause the orbit to be hyperbolic.

Answer: (E)

0.169 Schwartzchild Radius

Any mass can become a black hole if it is compressed beyond its Schwarzschild Radius.Beyond this size, light will be unable to escape from it’s surface or if a light beam were


DRAFT

Lagrangian of a Bead on a Rod cviito be trapped within this radius, it will be unable to escape. The Schwarzschild Radiuscan be derive by putting the Gravitational Potential Energy equal to a mass of kineticenergy travelling at light speed.

GMmR

=12

mc2 (0.169.1)

Solving for R yields

R =2GM

c2 (0.169.2)

Plugging in the values given, we get

R =2 × 6.67 × 10−11

× 5.98 × 1024(3 × 108)2

Our indices indicate we will get an answer in the order ≈ 10−3 meters.

Answer: (C)

0.170 Lagrangian of a Bead on a Rod

The Lagrangian of a system is defined

L = T − V (0.170.1)

The rod can move about the length of the rod, s and in circular motion along a radiusof s sinθ. The Lagrangian of this system becomes

L =12

ms2 +12

m(s sinθ)2ω2−mgs cosθ (0.170.2)

Answer: (E)

0.171 Ampere’s Law

We can use Ampere’s Law to tell us the magnetic field at point A.∮B · ds = µ0Ienclosed (0.171.1)

The point A is midway between the center of the two cylinders and as the currents arein opposite directions, thier magnetic fields at A point in the +y-direction. We can usethe right hand grip rule to determine this. This leaves us with choices (A) or (B).


DRAFT

cviii GR9677 Exam SolutionsThe current density, J is

J =I

Area

=Iπr2 (0.171.2)

We draw an Amperian loop of radius, r = d/2, thus the magnetic field becomes,∮B · ds = µ0Ienclosed

B� · (2πr) = µ0J(πr2

)B =

µ0πJr2π

=µ0πJ2π

d2

(0.171.3)

We expect B⊗ to be the same, thus

B = B� + B⊗

=( µ0

2π

)πdJ (0.171.4)

Answer: (A)

0.172 Larmor Formula

The Larmor Formula is used to calculate total power radiated by an accelerating non-relativistic point charge.

P =e2a2

6πε0c3 (0.172.1)

where a is the acceleration. For particles A & B, we are given

A B

Charge qa = q qb = 2qMass ma = m mb = m/2Velocity va = v vb = 3vAcceleration aa = a ab = 4a

From the above, we see that

PA ∝ q2a2 PB = (2q)2(4a2)


DRAFT

The Oscilloscope and Electron Deflection cixThus

PB

PA=

(2q)2(4a2)q2a2

= 64 (0.172.2)

Answer: (D)

0.173 The Oscilloscope and Electron Deflection

As the electron passes through the deflection plates, the electric field, E exerts a forceon the charge, pulling it up. Ignoring gravitational effects, the electric force is

Fe = qE =qVd

= may (0.173.1)

The time it takes to traverse this distance is

t =Lv

(0.173.2)

The deflection angle, θ is determined by

tanθ =vy

vx(0.173.3)

Now

vy = ayt

=qVmed

t

=qVmed

Lv

(0.173.4)

The the horizontal speed is v The angle of deflection becomes

tanθ =

qVmd

Lv

v

=qVLmdv2 (0.173.5)

Answer: (A)

0.174 Negative Feedback

All amplifiers exhibit non-linear behavior of some sort. Negative feedback seeks tocorrect some of these effects by sending some of the output back and subtracting it from


DRAFT

cx GR9677 Exam Solutionsthe input. This results in a decrease in gain. This tradeoff og gain improves linearityand hence the stability of the amplifier. This also allows for increased bandwidthresponse and decreased distortion.

Answer: (A)

0.175 Adiabatic Work of an Ideal Gas

The adiabatic condition states that

PVγ = C (0.175.1)

The work done by an ideal gas is

W =

∫PdV (0.175.2)

Substituting the adiabatic condition into the work equation yields

W = C∫ V f

Vi

dVVγ

= CV−γ+1

1 + γ

∣∣∣∣∣V f

Vi

=C

1 − γ

[V−γ+1

f − V−γ+1i

](0.175.3)

The adiabatic condition is

C = PVγ = PiVγi = P f V

γf (0.175.4)

Substituting this into eq. (0.175.3), we get

W =CV−γ+1

f − CV−γ+1i

1 − γ

=P f V

γf V−γ+1

f − P f Vγf V−γ+1

f

1 − γ

=P f V f − PiVi

1 − γ(0.175.5)

Answer: (C)

0.176 Change in Entrophy of Two Bodies

The change in entrophy of a system is

dS =dQT

(0.176.1)


DRAFT

Double Pane Windows cxiThe change in heat of a system is

dQ = nCdT (0.176.2)

Substituting this into the above equation, we get

dS = mC∫ T f

Ti

dTT

= mC ln[T f

Ti

](0.176.3)

We are told that the two bodies are brought together and they are in thermal isolation.This means that heat is not absorbed from or lost to the environment, only transferredbetween the two bodies. Thus

Heat Lost by Body A = Heat Gained by Body B

mC(500 − T f

)= mc

(T f − 100

)∴ T f = 300 K

The Total Change in Entrophy is the sum of the entrophy changes of bodies A and B.Thus

dS = dSA + dSB

= mC ln(300500

)+ mC ln

(300100

)= mC ln

(95

)(0.176.4)

Answer: (B)

0.177 Double Pane Windows

The rate of heat transer is proportional to the temperature difference across the body,is stated in Newton’s Law of Cooling.

dQdt

= kxdT (0.177.1)

where k is the thermal conductivity, x is the thickness of the material and dT is thetemperature difference across the material.

For Window APA =

dQA

dt= 0.8 × 4 × 10−3dT (0.177.2)

For Window BPB =

dQB

dt= 0.025 × 2 × 10−3dT (0.177.3)

The ratio of heat flow isPa

PB=

0.8 × 4 × 10−3dT0.025 × 2 × 10−3dT

= 16 (0.177.4)

Answer: (D)


DRAFT

cxii GR9677 Exam Solutions0.178 Gaussian Wave Packets

We have a Gaussian wave packet travelling through free space. We can best think ofthis as an infinite sum of a bunch of waves initially travelling together. Based on whatwe know, we can eliminate the choices given.

I The average momentum of the wave packet can not be zero as p = ~k. As wehave a whole bunch of wave numbers present, the average can not be zero.INCORRECT

II Our wave packet contains a bunch of waves travelling together each with a differntwave vector, k. The speed of propagation of these individual wave vectors isdefined by the group velocity, vg = dω/dk. So some waves will travel, someslower than others. As a result of these different travelling rates our wave packetbecomes spread out or ‘dispersed’. This is the basis of our dispersion relation,ω(k), relative to the center of the wave packet. CORRECT

III As we expect the wave packet to spread out, as shown above, the amplitude willdecrease over time. The energy that was concentrated in this packet gets spreadout or dispersed. INCORRECT

IV This is true. This statement is the Uncertainty Principle and comes from FourierAnalysis. CORRECT

We see that choices II and IV are CORRECT.

Answer: (B)

0.179 Angular Momentum Spin Operators

NOT FINISHED

Answer: (D)

0.180 Semiconductors and Impurity Atoms

NOT FINISHED

Answer: (B)

0.181 Specific Heat of an Ideal Diatomic Gas

The formula for finding the molar heat capacity at constant volume can be found by

cv =

(f2

)R (0.181.1)


DRAFT

Transmission of a Wave cxiiiwhere f is the number of degrees of freedom. At very low temperatures, there are onlythree translational degrees of freedom; there are no rotational degrees of freedom inthis case. At very high temperatures, we have three translational degrees of freedom,two rotational and two vibrational, giving a total of seven in all.

For Low Temperatures For High Temperatures

Translational 3 Translational 3Rotational 0 Rotational 2Vibrational 0 Vibrational 2

Total( f ) 3 Total( f ) 7

Table 0.181.1: Table of degrees of freedom of a Diatomic atom

We see that in the case of very low temperatures,

cvl =32

R (0.181.2)

and at very high temperatures,

cvh =72

R (0.181.3)

Thus, the ratio of molar heat capacity at constant volume at very high temperatures tothat at very low temperatures is

cvh

cvl

=72R32R

=73

(0.181.4)

Answer: (D)

0.182 Transmission of a Wave

NOT FINISHED

Answer: (C)

0.183 Piano Tuning & Beats

The D2 note has a frequency of 73.416 Hz and the A4 note has a frequency of 440.000 Hz.Beats are produced when the two frequencies are close to each other; if they were the


DRAFT

cxiv GR9677 Exam Solutionssame, there would be no beat frequency. So we can determine the D2 note where thishappens by setting the beat frequency to zero.

440.000 − n(73.416) = 0 (0.183.1)

This works out to be

n =440.00073.416

≈44072

= 6 (0.183.2)

Thus the closest harmonic will be the 6th one. As we expect this to be very close to theA4 frequency, the number of beats will be small or close to zero. Answer (B) fits this.14

Answer: (B)

0.184 Thin Films

As light moves from the glass to air interface, it is partially reflected and partiallytransmitted. There is no phase change when the light is refected. In the case of thetransmitted wave, when it reaches the air-glass interface, there is a change in phase ofthe reflected beam. Thus the condition for destructive interference is

2L =(n +

12

)λ (0.184.1)

where L is the thickness of the air film and n = 0, 1, 2 is the interference mode. Thus

L =(2n + 1

4

)λ (0.184.2)

We get

L0 =λ4

= 122 nm (0.184.3)

L1 =3λ4

= 366 nm (0.184.4)

L2 =5λ4

= 610 nm (0.184.5)

Answer: (E)14Incidentally, you can multiply the D2 frequency by six to determine the harmonic. This turns out to

be73.416 × 6 = 440.496 Hz (0.183.3)

Subtracting this from the A2 frequency gives

440.496 − 440.000 = 0.496 Hz (0.183.4)


DRAFT

Mass moving on rippled surface cxv0.185 Mass moving on rippled surface

For the mass to stay on the rippled surface, the particle’s horizontal velocity shouldnot be so great that it flies off the track. So as it falls, it must hug the track. The timefor the particle to fall from the top of the track to the bottom is

2d =12

gt2 (0.185.1)

where

t =

√4dg

(0.185.2)

We could also have said,

d − d cos(k ·

2π2k

)=

12

gt2

∴ 2d =12

gt2

but this is the same as eq. (0.185.2).

To stay on the track, the particle must cover a horizontal distance of x = π/k in thesame time. Thus the horizontal speed, v, is

v =xt

=πk

( g4d

)1/2

=π2·

( gk2d

)1/2

(0.185.3)

Any speed greater that 0.185.3 would result in the particle flying off the track. So

v ≤

√g

k2d(0.185.4)

Answer: (D)

0.186 Normal Modes and Couples Oscillators

NOT FINISHED

Answer: (D)

0.187 Waves

NOT FINSIHED

Answer: (B)


DRAFT

cxvi GR9677 Exam Solutions0.188 Charged Particles in E&M Fields

NOT FINISHED

Answer: (B)

0.189 Rotation of Charged Pith Balls in a Collapsing Mag-netic Field

As the magnetic field, B, collapses, it indices a clockwise electric field, E, that causesthe charged pith balls to rotate. The electric field can be found from Faraday’s Law∮

∂SE · d` = −

∂Φ∂t

(0.189.1)

The magnetic flux covers an area, A, of

A = πR2 (0.189.2)

and the electric field makes a loop of circumference

d` = πd (0.189.3)

Substituting eqs. (0.189.2) and (0.189.3) into eq. (0.189.1) gives us

E · πd = πR2 dBdt

(0.189.4)

Solving for E, gives us

E =R2

ddBdt

(0.189.5)

The question answers ask for the angular momentum, we recall that the torque is

τ = r × F (0.189.6)

where F = qE. Substituting eq. (0.189.5) into eq. (0.189.6)

dLdt

= qR2 dBdt

(0.189.7)

So it follows thatL = qR2B (0.189.8)

Answer: (A)


DRAFT

Coaxial Cable cxvii0.190 Coaxial Cable

We expect there to be no magnetic field outside, r > c, the coaxial cable. Choices (D)and (E) make no sense. So choice (B) is our best one left.

We can also use Ampere’s Law to show what the magnetic induction will look like aswe ove away from the center. We recall Ampere’s Law∮

B · d` = µ0Ienclosed (0.190.1)

0 < r < a The current enclosed by our Amperian Loop is

Ienclosed = Iπr2

πa2 (0.190.2)

Ampere’s Law shows us

B (2πr) = µ0Iπr2

πa2

B(0<r<a) =µ0I2π

ra2 (0.190.3)

We see the magnetic induction will increase linearly with respect to r.

a < r < b Within the shaded region, the enclosed current is, Ienclosed = I. Ampere’s Lawbecomes

B (2πr) = µ0I

B(a<r<b) =µ0

2πIr

(0.190.4)

The magnetic induction decreases inversely with respect to r.

b < r < c The area of the outer sheath is

A = π(c2− b2

)(0.190.5)

The enclosed current will be

Ienclosed = I − I[πr2− πb2

πc2 − πb2

]= I

[c2− r2

c2 − b2

](0.190.6)

Ampere’s Law becomes

B (2πr) = µ0I[

c2− r2

c2 − b2

]B(b<r<c) =

µ0I2πr

[c2− r2

c2 − b2

](0.190.7)

This will fall to an inversely with respect to r.


DRAFT

cxviii GR9677 Exam Solutionsr > c We see that Ienclosed = 0, so from Ampere’s Law

B(r>c) = 0 (0.190.8)

The graph shown in (B) fits this15.

Answer: (B)

0.191 Charged Particles in E&M Fields

A seemingly difficult question but it really is not16. All we need to turn to is PythagorasTheorem and relate the centripetal force to the Lorentz Force Law.

From the Lorentz Force Law, we see that

mv2

r= Bqv (0.191.1)

We can simplify this to read,

p = Bqr (0.191.2)

We know B and q. We can determine r in terms of s and ` by using Pythagoras Theorem.

r2 = `2 + (r − s)2 (0.191.3)

Solving for r, gives us

r =12

(`2

s+ s

)(0.191.4)

As s << `, eq. (0.191.4) becomes

r =12`2

s(0.191.5)

Substituting eq. (0.191.5) into eq. (0.191.2), gives us

Bq`2

2s(0.191.6)

Answer: (D)

15This is one of the reasons we use coaxial cables to transmit signals. No external magnetic field fromour signals means that we can, theoretically, eliminate electromagnetic interference.

16Draw Diagrams


DRAFT

THIS ITEM WAS NOT SCORED cxix0.192 THIS ITEM WAS NOT SCORED

0.193 The Second Law of Thermodynamics

This question deals with the Second Lar of Thermodynamics, which states that theentrophy of an isloated system, which is not in equilibrium, will increase over time,reaching a maximum at equilibrium. An alternative but equivalent statement is theClausius statement, “Heat can not flow from cold to hot without work input”. Thusitwould be impossible to attain the 900 K temperature the experimenter needs with asimple lens.

Answer: (E)

0.194 Small Oscillations

We are given a one dimensional potential function

V(x) = −ax2 + bx4 (0.194.1)

We can find the points of stability by differentiating the above equation to get andsetting it to zero.

dVdx

= −2ax + 4bx3 = 0 (0.194.2)

we see that x = 0; a/2b. Taking the second differential gives us the mass’s springconstant, k

k =d2Vdx2 = −2a + 12bx2 (0.194.3)

We can also use this to find the minimum and maximum points of inflection in ourpotential graph. We see that

k =d2V(0)

dx2 = −2a k =d2V(a/2b)

dx2 = 4a

We see that when x = a/2b, we are at a minima and hence at a point of stable equilibrium.We can now find the angular frequency

ω =

√km

=

√4am

= 2

√am

(0.194.4)

Answer: (D)


DRAFT

cxx GR9677 Exam Solutions0.195 Period of Mass in Potential

The total period of our mass will be the time it takes to return to the same point, saythe origin, as it moves through the two potentials.

for x < 0 The potential is

V =12

k2 (0.195.1)

The spring constant, k, is

k =d2Vdx2 = k (0.195.2)

Thus the period of oscillation would be

T = 2π

√mk

(0.195.3)

But this represents the period of the mass could also swing from x > 0. As aresult, our period would just be half this. So

T(x<0) = π

√mk

(0.195.4)

for x > 0 The potential isV = mgx (0.195.5)

We may recognize this as the gravitational potential energy. In this case, the”period” would be the time for the mass to return the origin. This would simplybe

s = v0t −12

gt2 (0.195.6)

where s = 0. Solving for t, we get t = 0 and t = 2v0/g We can solve this in termsof the total energy, E of the mass. The energy is E = 1/2mv2

0. Our period worksout to be

T(x>0) = 2

√2E

mg2 (0.195.7)

Thus our total period is

T = T(x<0) + T(x>0)

= π

√mk

+ 2

√2E

mg2 (0.195.8)

Answer: (D)


DRAFT

Internal Energy cxxi0.196 Internal Energy

NOT FINISHED

Answer: (D)

0.197 Specific Heat of a Super Conductor

Superconductors experience an increase in their specific heat, Cv, at the transitiontemperature, Tc. At high temperatures, the specific heat falls linearly but at lowtemperatures, the specific heat falls below the linear dependence of a degenrate Fermigas and is proportional to exp(−k/T). If the superconductor is placed in a magneticfield that exceeds the critical field strength, B > Bc, the specific heat reverts to thenormal linear behavior.

We see both the linear dependence and exponential decay below a certain value inChoice:(E).

Answer: (E)

0.198 Pair Production

We are given the equation of pair production

γ→ e− + e+ (0.198.1)

Pair production occurs when a γ ray of high energy is absorbed in the vincinity ofan atomic nucleus and particles are created from the absorbed photon’s energy. Thistakes place in the Coulomb field of the nucleus; the nucleus acts as a massive body toensure the conservation of momentum and energy. The nucleus is an essential part ofthis process; if the photon could spontaneously decay into an electron-positron pair inempty space, a Lorentz frame could be found where the electron and positron haveequal and opposite momenta and the photon will be at rest. This is a clear violation ofthe principles of Special Relativity. We choose (A).

Based on what we know, we can determine the maximum wavelength as a matter orinterest. As the nucleus is massive, we will ignore its recoil and consider that all of thephoton’s energy goes into electron-positron creation and the particles’ kinetic energy.

hυ = E− + E+

=(K− + mec2

)+

(K+ + mec2

)= K− + K+ + 2mec2 (0.198.2)

where K− and K+ are the kinetic energies of the electron and positron respectively. Thusthe minimum energy needed to initiate this process is

hυmin = 2mec2 (0.198.3)


DRAFT

cxxii GR9677 Exam SolutionsThe wavelength of this photon is

λmax =h

2mec2 (0.198.4)

which happens to be half the Compton wavelength.

Answer: (A)

0.199 Probability Current Density

The probability current density is defined as17

j(x, t) −~

2im

(Ψ∗∂Ψ∂x−∂Ψ∗

∂xΨ

)(0.199.1)

Thus given the wavefunction

Ψ(x, t) = eiωt [α cos kx + β sin kx]

(0.199.2)

Its complex conjugate is

Ψ∗(x, t) = e−iωt [α∗ cos kx + β∗ sin kx]

(0.199.3)

Differentiating the above two equations yields

dΨ

dx= eiωt [

−kα sin kx + kβ cos kx]

(0.199.4)

dΨ∗

dx= e−iωt [

−kα∗ sin kx + kβ∗ cos kx]

(0.199.5)

Substituting eqs. (0.199.2) to (0.199.5) into eq. (0.199.1) yields

j(x, t) =k~

2im(α∗β − β∗α

)(0.199.6)

Answer: (E)

0.200 Quantum Harmonic Oscillator Energy Levels

Given the potential

V(x) =12

mω2x2 (0.200.1)

17Add derivation in section.


DRAFT

Three Level LASER and Metastable States cxxiiiSolving the Schrodinger’s Equation for this potential

−~2

2md2ψ

dx2 +12

mω2x2ψ = Eψ (0.200.2)

leaves us with

ψn =(mωπ~

) 14 1√

2nn!Hne−ξ

2/2 (0.200.3)

where H(n) are Hermite polynomials and with energy levels of18

En =(n +

12

)~ω (0.200.4)

But we have placed an infinitely large barrier, V = ∞ at x ≤ 0. This serves to constrainψn(0) = 0 at x = 0. This occurs at n = 1, 3, 5, . . . or rather only odd values of n areallowed. Thus

En =32~ω,

52~ω,

72~ω,

112~ω, . . . (0.200.5)

Answer: (D)

0.201 Three Level LASER and Metastable States

Metastability describes a state of delicate equilibrium. Such a system is in a state ofequilibrium but is susceptible to fall into a lower-energy state with a slight interaction.

The laser operation depends on an active medium of atoms whose energy states canbe populated selectively by radiative means.19 In the three-level laser, a dischargeof some sort raises atoms from the ground state, E1, to a higher state, E3. A rapidspontaneous decay then occurs, bringing the excited atoms down to the E2 state. Thisstate is metastabe as it inhibits spontaneous decay back down to the ground state. Itneeds an incident photon of energy hυ = E2−E1 to stimulate the desired laser transitionback down to the ground state, E1.

18Add wavefunctions here19Add laser explanation in one of the sections.


DRAFT

cxxiv GR9677 Exam Solutions

E1

E3

E2 Metastable LevelPu

mpi

ngTr

ansi

tion

Figure 0.201.1: Three Level Laser

The Ruby Laser is an example of a three-level laser. Green light from a flash lamppumps the chromium ions to an excited level and non-radiative de-excitation promptlybrings the ions to a long lived metastable state. Stimulated emission then follows,generating a coherent beam of red light of 694 nm.

Answer: (B)

0.202 Quantum Oscillator – Raising and Lowering Oper-ators

We are given the lowering operator

a =

√mω0

2~

(x + i

pmω0

)(0.202.1)

The raising operator is given by

a† =

√mω0

2~

(x − i

pmω0

)(0.202.2)

So we immediately see thata† , a (0.202.3)

which we see from Choice: III. But what about the other two choices? In reality weonly have to prove/disprove Choice: I but we will go through both.

We know that a is not Hermitian from eq. (0.202.3) and hence not observable. Hencewe can eliminate Choice: II.

This affects Choice: I as a is not observable and hence won’t commute with theHamiltonian, H. We eliminate Choice: I.

Answer: (C)


DRAFTConstants & Important Equations

.1 Constants

Constant Symbol ValueSpeed of light in a vacuum c 2.99 × 108 m/sGravitational Constant G 6.67 × 10−11 m3/kg.s2

Rest Mass of the electron me 9.11 × 10−31 kgAvogadro’s Number NA 6.02 × 1023 mol-1

Universal Gas Constant R 8.31 J/mol.KBoltzmann’s Constant k 1.38 × 10−23 J/KElectron charge e 1.60 × 10−9CPermitivitty of Free Space ε0 8.85 × 10−12 C2/N.m2

Permeability of Free Space µ0 4π × 10−7 T.m/AAthmospheric Pressure 1 atm 1.0 × 105 M/m2

Bohr Radius a0 0.529 × 10−10 m

Table .1.1: Something

.2 Vector Identities

.2.1 Triple Products

A · (B × C) = B · (C ×A) = C · (A × B) (.2.1)A × (B × C) = B (A · C) − C (A · B) (.2.2)

DRAFT

cxxvi Constants & Important Equations.2.2 Product Rules

∇(

f g)

= f(∇g

)+ g

(∇ f

)(.2.3)

∇ (A · B) = A × (∇ × B) + B × (∇ ×A) + (A · ∇) B + (B · ∇) A (.2.4)∇ ·

(f A

)= f (∇ ·A) + A ·

(∇ f

)(.2.5)

∇ · (A × B) = B · (∇ ×A) −A · (∇ × B) (.2.6)∇ ×

(f A

)= f (∇ ×A) −A ×

(∇ f

)(.2.7)

∇ × (A × B) = (B · ∇) A − (A · ∇) B + A (∇ · B) − B (∇ ·A) (.2.8)

.2.3 Second Derivatives

∇ · (∇ ×A) = 0 (.2.9)∇ ×

(∇ f

)= 0 (.2.10)

∇ × (∇ ×A) = ∇ (∇ ·A) − ∇2A (.2.11)

.3 Commutators

.3.1 Lie-algebra Relations

[A,A] = 0 (.3.1)[A,B] = −[B,A] (.3.2)[A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0 (.3.3)

.3.2 Canonical Commutator

[x, p] = i~ (.3.4)

.3.3 Kronecker Delta Function

δmn =

{0 if m , n;1 if m = n;

For a wave function ∫ψm(x)∗ψn(x)dx = δmn (.3.5)


DRAFT

Linear Algebra cxxvii.4 Linear Algebra

.4.1 Vectors

Vector Addition

The sum of two vectors is another vector

|α〉 + |β〉 = |γ〉 (.4.1)

Commutative|α〉 + |β〉 = |β〉 + |α〉 (.4.2)

Associative|α〉 +

(|β〉 + |γ〉

)=

(|α〉 + |β〉

)+ |γ〉 (.4.3)

Zero Vector|α〉 + |0〉 = |α〉 (.4.4)

Inverse Vector|α〉 + | − α〉 = |0〉 (.4.5)


DRAFT

cxxviii Constants & Important Equations


DRAFTBibliography

[1] Wikipedia. Mean free path — wikipedia, the free encyclopedia, 2009. [Online;accessed 23-March-2009].

[2] John J. Brehm and William J. Mullin. Introduction to the Structure of Matter, chapter12-3, page 594. Wiley, first edition, 1989.

[3] John J. Brehm and William J. Mullin. Introduction to the Structure of Matter, chapter15-5, page 772. Wiley, first edition, 1989.

[4] John J. Brehm and William J. Mullin. Introduction to the Structure of Matter, chapter5-10, pages 283–287. Wiley, first edition, 1989.

[5] David J. Griffiths. Introduction to Quantum Mechanics, chapter 5.1.1, pages 203–205.Prentice Hall, second edition, 2005.



DRAFTIndex

Angular Momentum, see Rotational Mo-tion

Bohr ModelHydrogen Model, lv

Celestial Mechanics, xxivCircular Orbits, xxvEscape Speed, xxivKepler’s Laws, xxvNewton’s Law of Gravitation, xxivOrbits, xxvPotential Energy, xxiv

Circular Orbits, see Celestial MechanicsCommutators, cxxvi

Canonical Commutators, cxxviKronecker Delta Function, cxxviLie-algebra Relations, cxxvi

Compton Effect, lviiiCounting Statistics, lxxi

Doppler Effect, xxii

Franck-Hertz Experiment, lxi

Gravitation, see Celestial Mechanics

Kepler’s Laws, see Celestial MechanicsKronecker Delta Function, cxxvi

Linear Algebra, cxxviiVectors, cxxvii

Moment of Inertia, see Rotational Motion

Newton’s Law of Gravitation, see CelestialMechanics

Oscillatory Motion, xviii

Coupled Harmonic Oscillators, xxDamped Motion, xixKinetic Energy, xviiiPotential Energy, xixSimple Harmonic Motion Equation, xviiiSmall Oscillations, xix

GR9677 Q92, cxixTotal Energy, xviii

Parallel Axis Theorem, see Rotational Mo-tion

Rolling Kinetic Energy, see Rotational Mo-tion

Rotational Kinetic Energy, see RotationalMotion

Rotational Motion, xxiiAngular Momentum, xxiiiMoment of Inertia, xxiiParallel Axis Theorem, xxiiiRolling Kinetic Energy, xxiiiRotational Kinetic Energy, xxiiTorque, xxiii

Subject, xlivSystem of Particles, xxiv

Torque, see Rotational Motion

Vector Identities, cxxvProduct Rules, cxxviSecond Derivatives, cxxviTriple Products, cxxv

Documents

GR9677 Solutions