GR8677 solutions

DRAFT

The Physics GRE Solution Guide

GR8677 Test

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

November 3, 2009

Author:David S. Latchman

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

DRAFT

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

GR8677 Exam Solutions lxxiii

0.103Motion of Rock under Drag Force . . . . . . . . . . . . . . . . . . . . . . . lxxiii


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Contents vii0.104Satellite Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxiv

0.105Speed of Light in a Dielectric Medium . . . . . . . . . . . . . . . . . . . . lxxiv

0.106Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxiv

0.107Inelastic Collision and Putty Spheres . . . . . . . . . . . . . . . . . . . . . lxxv

0.108Motion of a Particle along a Track . . . . . . . . . . . . . . . . . . . . . . . lxxvi

0.109Resolving Force Components . . . . . . . . . . . . . . . . . . . . . . . . . lxxvi

0.110Nail being driven into a block of wood . . . . . . . . . . . . . . . . . . . . lxxvii

0.111Current Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxvii

0.112Charge inside an Isolated Sphere . . . . . . . . . . . . . . . . . . . . . . . lxxviii

0.113Vector Identities and Maxwell’s Laws . . . . . . . . . . . . . . . . . . . . lxxix

0.114Doppler Equation (Non-Relativistic) . . . . . . . . . . . . . . . . . . . . . lxxix

0.115Vibrating Interference Pattern . . . . . . . . . . . . . . . . . . . . . . . . . lxxix

0.116Specific Heat at Constant Pressure and Volume . . . . . . . . . . . . . . . lxxix

0.117Helium atoms in a box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxx

0.118The Muon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxi

0.119Radioactive Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxi

0.120Schrodinger’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxii

0.121Energy Levels of Bohr’s Hydrogen Atom . . . . . . . . . . . . . . . . . . lxxxii

0.122Relativistic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiii

0.123Space-Time Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiii

0.124Lorentz Transformation of the EM field . . . . . . . . . . . . . . . . . . . lxxxiv

0.125Conductivity of a Metal and Semi-Conductor . . . . . . . . . . . . . . . . lxxxiv

0.126Charging a Battery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxv

0.127Lorentz Force on a Charged Particle . . . . . . . . . . . . . . . . . . . . . lxxxv

0.128K-Series X-Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxv

0.129Electrons and Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxvi

0.130Normalizing a wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . lxxxvii

0.131Right Hand Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxviii

0.132Electron Configuration of a Potassium atom . . . . . . . . . . . . . . . . . lxxxviii

0.133Photoelectric Effect I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxix

0.134Photoelectric Effect II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxix

0.135Photoelectric Effect III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxix


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viii Contents0.136Potential Energy of a Body . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxix

0.137Hamiltonian of a Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xc

0.138Principle of Least Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . xc

0.139Tension in a Conical Pendulum . . . . . . . . . . . . . . . . . . . . . . . . xc

0.140Diode OR-gate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xci

0.141Gain of an Amplifier vs. Angular Frequency . . . . . . . . . . . . . . . . xci

0.142Counting Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xci

0.143Binding Energy per Nucleon . . . . . . . . . . . . . . . . . . . . . . . . . . xcii

0.144Scattering Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcii

0.145Coupled Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcii

0.146Collision with a Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xciv

0.147Compton Wavelength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xciv

0.148Stefan-Boltzmann’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . xciv

0.149Franck-Hertz Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . xcv

0.150Selection Rules for Electronic Transitions . . . . . . . . . . . . . . . . . . xcv

0.151The Hamilton Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcv

0.152Hall Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvi

0.153Debye and Einstein Theories to Specific Heat . . . . . . . . . . . . . . . . xcvii

0.154Potential inside a Hollow Cube . . . . . . . . . . . . . . . . . . . . . . . . xcvii

0.155EM Radiation from Oscillating Charges . . . . . . . . . . . . . . . . . . . xcviii

0.156Polarization Charge Density . . . . . . . . . . . . . . . . . . . . . . . . . . xcviii

0.157Kinetic Energy of Electrons in Metals . . . . . . . . . . . . . . . . . . . . . xcviii

0.158Expectation or Mean Value . . . . . . . . . . . . . . . . . . . . . . . . . . . xcix

0.159Eigenfuction of Wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . xcix

0.160Holograms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . c

0.161Group Velocity of a Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . ci

0.162Potential Energy and Simple Harmonic Motion . . . . . . . . . . . . . . . ci

0.163Rocket Equation I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cii

0.164Rocket Equation II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cii

0.165Surface Charge Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cii

0.166Maximum Power Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . ciii

0.167Magnetic Field far away from a Current carrying Loop . . . . . . . . . . ciii


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Contents ix0.168Maxwell’s Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . civ

0.169Partition Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cv

0.170Particle moving at Light Speed . . . . . . . . . . . . . . . . . . . . . . . . cv

0.171Car and Garage I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cv

0.172Car and Garage II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cvi

0.173Car and Garage III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cvi

0.174Refrective Index of Rock Salt and X-rays . . . . . . . . . . . . . . . . . . . cvi

0.175Thin Flim Non-Reflective Coatings . . . . . . . . . . . . . . . . . . . . . . cvii

0.176Law of Malus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cvii

0.177Geosynchronous Satellite Orbit . . . . . . . . . . . . . . . . . . . . . . . . cviii

0.178Hoop Rolling down and Inclined Plane . . . . . . . . . . . . . . . . . . . cviii

0.179Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . cix

0.180Total Energy between Two Charges . . . . . . . . . . . . . . . . . . . . . . cx

0.181Maxwell’s Equations and Magnetic Monopoles . . . . . . . . . . . . . . . cx

0.182Gauss’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxi

0.183Biot-Savart Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxii

0.184Zeeman Effect and the emission spectrum of atomic gases . . . . . . . . cxii

0.185Spectral Lines in High Density and Low Density Gases . . . . . . . . . . cxiii

0.186Term Symbols & Spectroscopic Notation . . . . . . . . . . . . . . . . . . . cxiii

0.187Photon Interaction Cross Sections for Pb . . . . . . . . . . . . . . . . . . . cxiv

0.188The Ice Pail Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxiv

0.189Equipartition of Energy and Diatomic Molecules . . . . . . . . . . . . . . cxiv

0.190Fermion and Boson Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . cxv

0.191Wavefunction of Two Identical Particles . . . . . . . . . . . . . . . . . . . cxv

0.192Energy Eigenstates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxvi

0.193Bragg’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxvii

0.194Selection Rules for Electronic Transitions . . . . . . . . . . . . . . . . . . cxvii

0.195Moving Belt Sander on a Rough Plane . . . . . . . . . . . . . . . . . . . . cxviii

0.196RL Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxviii

0.197Carnot Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxx

0.198First Order Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . cxxii

0.199Colliding Discs and the Conservation of Angular Momentum . . . . . . cxxii


DRAFT

x Contents0.200Electrical Potential of a Long Thin Rod . . . . . . . . . . . . . . . . . . . . cxxiii

0.201Ground State of a Positronium Atom . . . . . . . . . . . . . . . . . . . . . cxxiv

0.202The Pinhole Camera . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxiv

Constants & Important Equations cxxvii

.1 Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxvii

.2 Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxvii

.3 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxviii

.4 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxix


DRAFTList of Tables

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

0.140.1Truth Table for OR-gate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xci

0.189.1Specific Heat, cv for a diatomic molecule . . . . . . . . . . . . . . . . . . . cxiv

.1.1 Something . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxvii

DRAFT

xii List of Tables


DRAFTList of Figures

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


DRAFTGR8677 Exam Solutions

0.103 Motion of Rock under Drag Force

From the information provided we can come up with an equation of motion for therock.

mx = −mg − kv (0.103.1)

If you have seen this type of equation, and solved it, you’d know that the rock’s speedwill asymtotically increase to some max speed. At that point the drag force and theforce due to gravity will be the same. We can best answer this question through analysisand elimination.

A Dividing eq. (0.103.1) by m gives

x = −g −km

x (0.103.2)

We see that this only occurs when x = 0, which only happens at the top of theflight. So FALSE.

B At the top of the flight, v = 0. From eq. (0.103.2)

⇒ x = g (0.103.3)

we see that this is TRUE.

C Again from eq. (0.103.2) we see that the acceleration is dependent on whether therock is moving up or down. If x > 0 then x < −g and if x < 0 then x > −g. So thisis also FALSE.

D If there was no drag (fictional) force, then energy would be conserved and the rockwill return at the speed it started with but there is a drag force so energy is lost.The speed the rock returns is v < v0. Hence FALSE.

E Again FALSE. Once the drag force and the gravitational force acting on the rock isbalanced the rock won’t accelerate.

Answer: (B)

DRAFT

lxxiv GR8677 Exam Solutions0.104 Satellite Orbits

The question states that the astronaut fires the rocket’s jets towards Earth’s center. Weinfer that no extra energy is given to the system by this process. Section 0.7.5, showsthat the only other orbit where the specific energy is also negative is an elliptical one.

Answer: (A)

0.105 Speed of Light in a Dielectric Medium

Solutions to the Electromagnetic wave equation gives us the speed of light in terms ofthe electromagnetic permeability, µ0 and permitivitty, ε0.

c =1

√µ0ε0

(0.105.1)

where c is the speed of light. The speed through a dielectric medium becomes

v =1√µε0

=1√

2.1µ0ε0

=c√

2.1(0.105.2)

Answer: (D)

0.106 Wave Equation

We are given the equation

y = A sin( tT−

xλ

)(0.106.1)

We can analyze and eliminate from what we know about this equation

A The Amplitude, A in the equation is the displacement from equilibrium. So thischoice is incorrect.

B As the wave moves, we seek to keep the(

tT −

xλ

)term constant. So as t increases, we

expect x to increase as well as there is a negative sign in front of it. This meansthat the wave moves in the positive x-direction. This choice is also incorrect.

C The phase of the wave is given by(

tT −

xλ

), we can do some manipulation to show

2π( tT−

xλ

)= 2π f t − kx = 0

= ωt − kx (0.106.2)


DRAFT

Inelastic Collision and Putty Spheres lxxvOr rather

kx = ωt (0.106.3)

Differentiating eq. (0.106.3) gives us the phase speed, which is

v =λT

(0.106.4)

This is also incorrect

E From eq. (0.106.4) the above we see that is the answer.

Answer: (E)

0.107 Inelastic Collision and Putty Spheres

We are told the two masses coalesce so we know that the collision is inelastic andhence, energy is not conserved. As mass A falls, it looses Potential Energy and gainsKinetic Energy.

Mgh0 =12

Mv20 (0.107.1)

Thusv2

0 = 2gh0 (0.107.2)

Upon collision, momentum is conserved, thus

Mv0 = (3M + M) v1

= 4Mv1

⇒ v1 =v0

4(0.107.3)

The fused putty mass rises, kinetic energy is converted to potential energy and we findour final height.

12

(4M) v21 = 4Mgh1 (0.107.4)

This becomesv2

1 = 2gh1 (0.107.5)

Substituting eq. (0.107.3), we get (v0

4

)2

= 2gh1 (0.107.6)

and substituting, eq. (0.107.2),2gh0

16= 2gh1 (0.107.7)

Solving for h1,

h1 =h0

16(0.107.8)

Answer: (A)


DRAFT

lxxvi GR8677 Exam Solutions0.108 Motion of a Particle along a Track

As the particle moves from the top of the track and runs down the frictionless track,its Gravitational Potential Energy is converted to Kinetic Energy. Let’s assume that theparticle is at a height, y0 when x = 0. Since energy is conserved, we get3

mgy0 = mg(y0 − y) +12

mv2

⇒12

v2 = gy (0.108.1)

So we have a relationship between v and the particle’s position on the track.

The tangential acceleration in this case is

mg cosθ =mv2

r(0.108.2)

where r is the radius of curvature and is equal to√

x2 + y2 .

Substituting this into eq. (0.108.2) gives

g cosθ =v2

r

=gx2

2√

x2 + y2

=gx

√

x2 + 4(0.108.3)

Answer: (D)

0.109 Resolving Force Components

This question is a simple matter of resolving the horizontal and vertical componentsof the tension along the rope. We have

T sinθ = F (0.109.1)T cosθ = mg (0.109.2)

Thus we get

tanθ =F

mg

=10

(2)(9.8)≈

12

(0.109.3)

Answer: (A)3Insert Free Body Diagram of particle along track.


DRAFT

Nail being driven into a block of wood lxxvii0.110 Nail being driven into a block of wood

We recall thatv2 = v2

0 + 2as (0.110.1)

where v, v0, a and s are the final speed, initial speed, acceleration and displacementthat the nail travels. Now it’s just a matter of plugging in what we know

0 = 100 + 2a(0.025)

⇒ a = −100

2(0.025)= −2000 m s−2 (0.110.2)

The Force on the nail comes from Newton’s Second Law

F = ma= 5 · 2000 = 10 000 N (0.110.3)

Answer: (D)

0.111 Current Density

We can find the drift velocity from the current density equation

J = envd (0.111.1)

where e is the charge of an electron, n is the density of electrons per unit volume andvd is the drift speed. Plugging in what we already know

J =IA

I =nAvde

vd =I

nAe

=100

(1 × 1028)π × 2 × 10−4

41.6 × 10−19

(0.111.2)

paying attention to the indices of the equation we get

2 − 28 + 4 + 19 = −4 (0.111.3)

So we expect an answer where vd ≈ 10−4 m s−1.4

Answer: (D)4It also helps if you knew that the electron drift velocity was slow, in the order of mm/s.


DRAFT

lxxviii GR8677 Exam Solutions0.112 Charge inside an Isolated Sphere

You can answer this by thinking about Gauss’ Law. The bigger the Gaussian surfacethe more charge it encloses and the bigger the electric field. Beyond the radius of thesphere, the field decreases exponentially5.

We can calculate these relationships by using Gauss’ Law.∮S

E · dS =Qenclosed

ε0(0.112.1)

where the current density, ρ is

ρ =Q

43πR3

=Qenclosed

43πr3

(0.112.2)

where R is the radius of the sphere.

for r < R The enclosed charge becomes

Qenclosed = ρ(43πr3

)=

Qr3

R3 (0.112.3)

Gauss’ Law becomes

E(4πr2

)=

Qr3

ε0R3 (0.112.4)

The Electric field is

E(r<R) =Qr

4πε0R3 (0.112.5)

This is a linear relationship with respect to r.

for r ≥ R The enclosed charge isQenclosed = Q (0.112.6)

Gauss’ Law becomes

E(4πr2

)=

Qε0

(0.112.7)

The Electric field is

E(r≥R) =Q

4πε0r2 (0.112.8)

The linear increase is exhibited by choice (C).

Answer: (C)

5Draw diagrams.


DRAFT

Vector Identities and Maxwell’s Laws lxxix0.113 Vector Identities and Maxwell’s Laws

We recall the vector identity∇ · (∇ ×A) = 0 (0.113.1)

Thus

∇ · (∇ ×H) = ∇ ·(D + J

)= 0 (0.113.2)

Answer: (A)

0.114 Doppler Equation (Non-Relativistic)

We recall the Doppler Equation6

f = f0

(v − vr

v − vs

)(0.114.1)

where vr and vs are the observer and source speeds respectively. We are told that vr = 0and vs = 0.9v. Thus

f = f0

( vv − 0.9v

)= 10 f0

= 10 kHz (0.114.2)

Answer: (E)

0.115 Vibrating Interference Pattern

Answering this question takes some analysis. The sources are coherent, so they willproduce an interference pattern. We are also told that ∆ f = 500 Hz. This will producea shifting interference pattern that changes too fast for the eye to see.7

Answer: (E)

0.116 Specific Heat at Constant Pressure and Volume

From section 0.65 and section 0.66, we see that

Cp = CV + R (0.116.1)

6Add reference to Dopler Equations.7Add Young’s Double Slit Experiment equations.


DRAFT

lxxx GR8677 Exam SolutionsThe difference is due to the work done in the environment by the gas when it expandsunder constant pressure.

We can prove this by starting with the First Law of Thermodynamics.

dU = −dW + dQ (0.116.2)

Where dU is the change in Internal Energy, dW is the work done by the system and dQis the change in heat of the system.

We also recall the definition for Heat Capacity

dQ = CdT (0.116.3)

At constant volume, there is no work done by the system, dV = 0. So it follows thatdW = 0. The change in internal energy is the change of heat into the system, thus wecan define, the heat capacity at constant volume to be

dUV = CVdT = dQV (0.116.4)

At constant pressure, the change in internal energy is accompanied by a change in heatflow as well as a change in the volume of the gas, thus

dUp = −dWp + dQp

= −pdV + CpdT where pdV = nRdT= −nRdT + CpdT (0.116.5)

If the changes in internal energies are the same in both cases, then eq. (0.116.5) is equalto eq. (0.116.4). Thus

CVdT = −nRdT + CpdT

This becomesCp = CV + nR (0.116.6)

We see the reason why Cp > CV is due to the addition of work on the system; eq. (0.116.4)shows no work done by the gas while eq. (0.116.5) shows that there is work done.

Answer: (A)

0.117 Helium atoms in a box

Let’s say the probability of the atoms being inside the box is 1. So the probability thatan atom will be found outside of a 1.0 × 10−6 cm3 box is

P = 1 − 1.0 × 10−6 (0.117.1)

As there are N atoms and the probability of finding one is mutually exclusive of theother, the probabolity becomes

P =(1 − 1.0 × 10−6

)N(0.117.2)

Answer: (C)


DRAFT

The Muon lxxxi0.118 The Muon

It helps knowing what these particles are

Muon The muon, is a lepton, like the electron. It has the ame charge, −e and spin, 1/2,as the electron execpt it’s about 200 times heavier. It’s known as a heavy electron.

Electron This is the answer.

Graviton This is a hypothetical particle that mediates the force of gravity. It has nocharge, no mass and a spin of 2. Nothing like an electron.

Photon The photon is the quantum of the electromagnetic field. It has no charge ormass and a spin of 1. Again nothing like an electron.

Pion The Pion belongs to the meson family. Again, nothing like leptons.

Proton This ia a sub atomic particle and is found in the nucleus of all atoms. Nothinglike an electron.

Answer: (A)

0.119 Radioactive Decay

From the changes in the Mass and Atomic numbers after the subsequent decays, weexpect an α and β decay.

Alpha DecayAZX→A−4

Z−2 X′ +42 α (0.119.1)

Beta DecayAZX→A

Z+1 X′ +−1 e− + υe (0.119.2)

Combining both gives

AZX→A−4

Z−2 X′ +42 α→

AZ−1 Y +−1 e− + υe (0.119.3)

Answer: (B)


DRAFT

lxxxii GR8677 Exam Solutions0.120 Schrodinger’s Equation

We recall that Schrodinger’s Equation is

Eψ = −~2

2m∂2ψ

∂x2 + V(x)ψ (0.120.1)

Given that

ψ(x) = A exp−

b2x2

2

(0.120.2)

We differentiate and get∂2ψ

∂x2 =(b4x2− b2

)ψ (0.120.3)

Plugging into Schrodinger’s Equation, eq. (0.120.1), gives us

Eψ = −~2

2m

(b4x2− b2

)ψ + V(x)ψ (0.120.4)

Applying the boundary condition at x = 0 gives

Eψ =~2

2mb2ψ (0.120.5)

This gives~2b2

2mψ = −

~2

2m

(b4x2− b2

)ψ + V(x)ψ (0.120.6)

Solving for V(x) gives

V(x) =~2b4x2

2m(0.120.7)

Answer: (B)

0.121 Energy Levels of Bohr’s Hydrogen Atom

We recall that the Energy Levels for the Hydrogen atom is

En = −Z2

n2 13.6eV (0.121.1)

where Z is the atomic number and n is the quantum number. This can easily be reducedto

En = −An2 (0.121.2)

Answer: (E)


DRAFT

Relativistic Energy lxxxiii0.122 Relativistic Energy

The Rest Energy of a particle is given

E = mc2 (0.122.1)

The Relativistic Energy is for a relativistic particle moving at speed v

E = γvmc2 (0.122.2)

We are told that a kaon moving at relativistic speeds has the same energy as the restmass as a proton. Thus

EK+ = Ep (0.122.3)

where

EK+ = γvmK+c2 (0.122.4)

Ep = mpc2 (0.122.5)

Equating both together gives

γv =mp

mK+

=939494

≈940500

(0.122.6)

This becomes

γv =

(1 −

v2

c2

)−1/2

≈ 1.88 (0.122.7)

This is going to take some approximations and estimations but(1 −

v2

c2

)−1

= 3.6 (0.122.8)

which works out tov2

c2 = 0.75 (0.122.9)

We expect this to be close to the 0.85c answer.

Answer: (E)

0.123 Space-Time Interval

We recall the Space-Time Interval from section 0.94.

(∆S)2 = (∆x)2 +(∆y

)2+ (∆z)2

− c2 (∆t)2 (0.123.1)


DRAFT

lxxxiv GR8677 Exam SolutionsWe get

∆S2 = (5 − 3)2 + (3 − 3)2 + (3 − 1)2− c2(5 − 3)2

= 22 + 02 + 22− 22

= 22

∆S = 2 (0.123.2)

Answer: (C)

0.124 Lorentz Transformation of the EM field

Lorentz transformations show that electric and magnetic fields are different aspects ofthe same force; the electromagnetic force. If there was one stationary charge in ourrest frame, we would observe an electric field. If we were to move to a moving frameof reference, Lorentz transformations predicts the presence of an additional magneticfield.

Answer: (B)

0.125 Conductivity of a Metal and Semi-Conductor

More of a test of what you know.

A Copper is a conductor so we expect its conductivity to be much greater than that ofa semiconductor. TRUE.

B As the temperature of the conductor is increased its atoms vibrate more and disruptthe flow of electrons. As a result, resistance increases. TRUE.

C Different process. As temperature increases, electrons gain more energy to jump theenergy barrier into the conducting region. So conductivity increases. TRUE.

D You may have paused to think for this one but this is TRUE. The addition of animpurity causes an increase of electron scattering off the impurity atoms. As aresult, resistance increases.8

E The effect of adding an impurity on a semiconductor’s conductivity depends onhow many extra valence electrons it adds or subtracts; you can either widen ornarrow the energy bandgap. This is of crucial importance to electronics today.So this is FALSE.

Answer: (E)8There are one or two cases where this is not true. The addition of Silver increases the conductivity

of Copper. But the conductivity will still be less than pure silver.


DRAFT

Charging a Battery lxxxv0.126 Charging a Battery

The Potential Difference across the resistor, R is

PD = 120 − 100 = 20 V (0.126.1)

The Total Resistance is

R + r =VI

=2010

R + 1 = 2⇒ R = 1 Ω (0.126.2)

Answer: (C)

0.127 Lorentz Force on a Charged Particle

We are told that the charged particle is released from rest in the electric and magneticfields. The charged particle will experience a force from the magnetic field only whenit moves perpendicular to the direction of the magnetic field lines. The particle willmove along the direction of the electric field.

We can also anylize this by looking at the Lorentz Force equation

Fq = q [E + (v × B)] (0.127.1)

v is in the same direction as B so the cross product between them is zero. We are leftwith

Fq = qE (0.127.2)

The force is directed along the electrical field line and hence it moves in a straight line.

Answer: (E)

0.128 K-Series X-Rays

To calculate we look at the energy levels for the Bohr atom. As the Bohr atom considersthe energy levels for the Hydrogen atom, we need to modify it somewhat

En = Z2eff

1n2

f

−1n2

i

13.6eV (0.128.1)


DRAFT

lxxxvi GR8677 Exam Solutionswhere Zeff is the effective atomic number and n f and ni are the energy levels. For then f = 1 transition

Zeff = Z − 1 (0.128.2)

where Z = 28 for nickle. As the electrons come in from ni = ∞, eq. (0.128.1) becomes

E1 = (28 − 1)2[ 112 −

1∞2

]13.6eV (0.128.3)

This works out to

E1 = (272)13.6eV

≈ (30)2× 13.6eV (0.128.4)

This takes us in the keV range.

Answer: (D)

0.129 Electrons and Spin

It helps if you knew some facts here.

A The periodic table’s arrangement of elements tells us about the chemical propertiesof an element and these properties are dependent on the valent electrons. Howthese valent electrons are arranged is, of course, dependent on spin. So thischoice is TRUE.

B The energy of an electron is quantized and obey the Pauli’s Exclusion Principle. Allthe electrons are accommodated from the lowest state up to the Fermi Level andthe distribution among levels is described by the Fermi distribution function,f (E), which defines the probability that the energy level, E, is occupied by anelectron.

f (E) =

1, E < EF

0, E > EF

where f (E) is the Fermi-Dirac Distribution

f (E) =1

eE−EF/kT + 1(0.129.1)

As a system goes above 0K, thermal energy may excite to higher energy statesbut this energy is not shared equally by all the electrons. The way energy isdistributed comes about from the exclusion principle, the energy an electron myabsorb at room temperature is kT which is much smaller than EF = 5eV. We candefine a Fermi Temperature,

EF = kTF (0.129.2)


DRAFT

Normalizing a wavefunction lxxxviiwhich works out to be, TF = 60000K. Thus only electrons close to this temperaturecan be excited as the levels above EF are empty. This results in a small number ofelectrons being able to be thermally excited and the low electronic specific heat.

C =π2

2Nk

(TT f

)where kT << EF

So this choice is also TRUE.

C The Zeeman Effect describes what happens to Hydrogen spectral lines in a magneticfield; the spectral lines split. In some atoms, there were further splits in spectrallines that couldn’t be explained by magnetic dipole moments. The explanationfor this additional splitting was discovered to be due to electron spin.9

D The deflection of an electron in a uniform magnetic field deflects only in one wayand demonstrates none of the electron’s spin properties. Electrons can be de-flected depending on their spin if placed in a non-uniform magnetic field, as wasdemonstrated in the Stern-Gerlach Experiment.10

E When the Hydrogen spectrum is observed at a very high resolution, closely spaceddoublets are observed. This was one of the first experimental evidence for electronspin.11

Answer: (D)

0.130 Normalizing a wavefunction

We are givenψ(φ) = Aeimφ (0.130.1)

Normalizing a function means ∫∞

−∞

|Ψ(x)|2 dx = 1 (0.130.2)

In this case, we want ∫ 2π

0

∣∣∣ψ(φ)∣∣∣2 dφ = 1 (0.130.3)

and that ∣∣∣ψ(x)∣∣∣2 = ψ∗(x)ψ(x) (0.130.4)

9Write up on Zeeman and anomalous Zeemen effects10Write up on Stern-Gerlach Experiment11Write up on Fine Structure


DRAFT

lxxxviii GR8677 Exam SolutionsSo

⇒

∣∣∣ψ(φ)∣∣∣2 = A2eimφe−imφ

A2∫ 2π

0dφ = 1

A2 [2π − 0] = 1

⇒ A =1√

2π(0.130.5)

0.131 Right Hand Rule

First we use the ‘Grip’ rule to tell what direction the magnetic field lines are going.Assuming the wire and current are coming out of the page, the magnetic field is in aclockwise direction around the wire. Now we can turn to Fleming’s Right Hand Rule,to solve the rest of the question.

As we want the force acting on our charge to be parallel to the current direction, wesee that this will happen when the charge moves towards the wire12.

Answer: (A)

0.132 Electron Configuration of a Potassium atom

We can alalyze and eliminate

A The n = 3 shell has unfilled d-subshells. So this is NOT TRUE.

B The 4s subshell only has one electron. The s subshell can ‘hold’ two electrons so thisis also NOT TRUE.

C Unknown.

D The sum of all the electrons, we add all the superscripts, gives 19. As this is aground state, a lone potassium atom, we can tell that the atomic number is 19.So this is NOT TRUE.

E Potassium has one outer electron, like Hydrogen. So it will also have a sphericallysymmetrical charge distribution.

12Don’t forget to bring your right hand to the exam


DRAFT

Photoelectric Effect I lxxxix0.133 Photoelectric Effect I

We are given|eV| = hυ −W (0.133.1)

We recall that V is the stopping potential, the voltage needed to bring the current tozero. As electrons are negatively charged, we expect this voltage to be negative.

Answer: (A)

0.134 Photoelectric Effect II

Some history needs to be known here. The photoelectric effect was one of the exper-iments that proved that light was absorbed in discreet packets of energy. This is theexperimental evidence that won Einstein the Nobel Prize in 1921.

Answer: (D)

0.135 Photoelectric Effect III

The quantity W is known as the work function of the metal. This is the energy that isneeded to just liberate an electron from its surface.

Answer: (D)

0.136 Potential Energy of a Body

We recall that

F = −dUdx

(0.136.1)

Given thatU = kx4 (0.136.2)

The force on the body becomes

F = −d

dxkx4

= −4kx3 (0.136.3)

Answer: (B)


DRAFT

xc GR8677 Exam Solutions0.137 Hamiltonian of a Body

The Hamiltonian of a body is simply the sum of the potential and kinetic energies.That is

H = T + V (0.137.1)

where T is the kinetic energy and V is the potential energy. Thus

H =12

mv2 + kx4 (0.137.2)

We can also express the kinetic energy in terms of momentum, p. So

H =p2

2m+ kx4 (0.137.3)

Answer: (A)

0.138 Principle of Least Action

Hamilton’s Principle of Least Action13 states

Φ =

∫T

(T(q(t), q(t)

)− V

(q(t)

))dt (0.138.1)

where T is the kinetic energy and V is the potential energy. This becomes

Φ =

∫ t2

t1

(12

mv2− kx4

)dt (0.138.2)

Answer: (A)

0.139 Tension in a Conical Pendulum

This is a simple case of resolving the horizontal and vertical components of forces. Sowe have

T cosθ = mg (0.139.1)T sinθ = mrω2 (0.139.2)

Squaring the above two equations and adding gives

T2 = m2g2 + m2r2ω4 (0.139.3)

Leaving us withT = m

(g2 + r2ω4

)(0.139.4)

Answer: (E)13Write something on this


DRAFT

Diode OR-gate xci0.140 Diode OR-gate

This is an OR gate and can be illustrated by the truth table below.

Input 1 Input 2 Output

0 0 00 1 11 0 11 1 1

Table 0.140.1: Truth Table for OR-gate

Answer: (A)

0.141 Gain of an Amplifier vs. Angular Frequency

We are given that the amplifier has some sort of relationship where

G = Kωa (0.141.1)

falls outside of the amplifier bandwidth region. This is that ‘linear’ part of the graphon the log-log graph. From the graph, we see that, G = 102, for ω = 3 × 105 second-1.Substituting, we get

102 = K(3 × 105

)a

∴ log(102) = a log[K

(3.5 × 105

)]⇒ a ≈ 2 − 5 (0.141.2)

We can roughly estimate by subtracting the indices. So our relationship is of the form

G = Kω−2 (0.141.3)

Answer: (E)

0.142 Counting Statistics

We recall from section 0.98 , that he standard deviation of a counting rate is σ =√

N ,where N is the number of counts. We have a count of N = 9934, so the standarddeviation is

σ =√

N =√

9934

≈

√

10000= 100 (0.142.1)


DRAFT

xcii GR8677 Exam SolutionsAnswer: (A)

0.143 Binding Energy per Nucleon

More of a knowledge based question. Iron is the most stable of all the others.14

Answer: (C)

0.144 Scattering Cross Section

We are told the particle density of our scatterer is ρ = 1020 nuclei per cubic centimeter.Given the thickness of our scatterer is ` = 0.1 cm, the cross sectional area is

ρ =NV

=NA`

⇒ A =Nρ`

(0.144.1)

Now the probability of striking a proton is 1 in a million. So

A =1 × 10−6

1020× 0.1

= 10−25 cm2 (0.144.2)

Answer: (C)

0.145 Coupled Oscillators

There are two ways this system can oscillate, one mass on the end moves a lot and theother two move out of in the opposite directions but not as much or the centermasscan be stationary and the two masses on the end move out of phase with each other. Inthe latter case, as there isn’t any energy transfer between the masses, the period wouldbe that of a single mass-spring system. The frequency of this would simply be

f =1

2π

√km

(0.145.1)

where k is the spring constant and m is the mass.

Answer: (B)14Write up on Binding Energy


DRAFT

Coupled Oscillators xciii0.145.1 Calculating the modes of oscillation

In case you require a more rigorous approach, we can calculate the modes of oscillation.The Lagrangian of the system is

L = T − V

=12

m[x2

1 + 2x22 + x2

3

]−

12

k[(x2 − x2)2 + (x3 − x2)2

](0.145.2)

The equation of motion can be found from

ddt

(∂L∂xn

)=∂L∂xn

(0.145.3)

The equations of motion are

mx1 = k (x2 − x1) (0.145.4)2mx2 = kx1 − 2kx2 + kx3 (0.145.5)mx3 = −k (x3 − x2) (0.145.6)

The solutions of the equations are

x1 = A cos(ωt) x2 = B cos(ωt) x3 = C cos(ωt)x1 = −ω2x1 x2 = −ω2x2 x3 = −ω2x3

(0.145.7)

Solving this, we get (k −mω2

)x1 − kx2 = 0 (0.145.8)

−kx1 +(2k − 2mω2

)x2 − kx3 = 0 (0.145.9)

−kx2 +(k −mω2

)x3 = 0 (0.145.10)

We can solve the modes of oscillation by solving∣∣∣∣∣∣∣∣k −mω2

−k 0−k 2k − 2mω2

−k0 −k k −mω2

∣∣∣∣∣∣∣∣ = 0 (0.145.11)

Finding the determinant results in(k −mω2

) [2(k −mω2

)2− k2

]− k

[k(k −mω2

)](0.145.12)

Solving, we get

ω =km

;km±

√2 k

m(0.145.13)

Substituting ω = k/m into the equations of motion, we get

x1 = −x3 (0.145.14)x2 = 0 (0.145.15)

We see that the two masses on the ends move out of phase with each other and themiddle one is stationary.


DRAFT

xciv GR8677 Exam Solutions0.146 Collision with a Rod

Momentum will be conserved, so we can say

mv = MV

V =mvM

(0.146.1)

Answer: (A)

0.147 Compton Wavelength

We recall from section 0.83.2, the Compton Equation from eq. (0.83.12)

∆λ = λ′ − λ =h

mc(1 − cosθ) (0.147.1)

Let θ = 90, we get the Compton Wavelength

λc =h

mc(0.147.2)

where m is the mass of the proton, mp, thus

λc =h

mpc(0.147.3)

Answer: (C)

0.148 Stefan-Boltzmann’s Equation

We recall the Stefan-Boltzmann’s Equation, eq. (0.61.1)

P(T) = σT4 (0.148.1)

At temperature, T1,P1 = σT1 = 10 mW (0.148.2)

We are given T2 = 2T1, so

P2 = σT42

= σ (2T1)4

= 16T41

= 16P1 = 160 mW (0.148.3)

Answer: (E)


DRAFT

Franck-Hertz Experiment xcv0.149 Franck-Hertz Experiment

The Franck-Hertz Experiment as seen in section 0.84.3 deals with the manner in whichelectrons of certain energies scatter or collide with Mercury atoms. At certain energylevels, the Mercury atoms can ‘absorb’ the electrons energy and be excited and thisoccurs in discreet steps.

Answer: (C)

0.150 Selection Rules for Electronic Transitions

We recall the selection rules for photon emission

∆` = ±1 Orbital angular momentum∆m` = 0,±1 Magnetic quantum number∆ms = 0 Secondary spin quantum number,∆ j = 0,±1 Total angular momentum

NOT FINISHED

Answer: (D)

0.151 The Hamilton Operator

The time-independent Schrodinger equation can be written

Hψ = Eψ (0.151.1)

We can determine the energy of a quantum particle by regarding the classical nonrel-ativistic relationship as an equality of expectation values.

〈H〉 =

⟨p2

2m

⟩+ 〈V〉 (0.151.2)

We can solve this through the substition of a momentum operator

p→~

i∂∂x

(0.151.3)

Substituting this into eq. (0.151.2) gives us

〈H〉 =

∫ +∞

−∞

ψ∗[−~

2m∂2

∂x2ψ + V(x)ψ]

dx

=

∫ +∞

−∞

ψ∗i~∂∂tψdx (0.151.4)


DRAFT

xcvi GR8677 Exam SolutionsWe know from the Schrodinger Time Dependent Equation

−~2

2m∂2Ψ

∂x2 + V(x)Ψ = i~∂Ψ∂t

(0.151.5)

So we can get a Hamiltonian operator

H→ i~∂∂t

(0.151.6)

Answer: (B)

0.152 Hall Effect

The Hall Effect describes the production of a potential difference across a currentcarrying conductor that has been placed in a magnetic field. The magnetic field isdirected perpendicularly to the electrical current.

As a charge carrier, an electron, moves through the conductor, the Lorentz Force willcause a deviation in the carge carrier’s motion so that more charges accumulate in onelocation than another. This asymmetric distribution of charges produces an electricfield that prevents the build up of more electrons. This ‘equilibrium’ voltage across theconductor is known as the Hall Voltage and remains as long as a current flows throughour conductor.

As the deflection and hence, the Hall Voltage, is determined by the sign of the carrier,this can be used to measure the sign of charge carriers.

An equilibrium condition is reached when the electric force, generated by the accumu-lated charge carriers, is equal the the magnetic force, that causes the accumulation ofcharge carriers. Thus

Fm = evdB Fe = eE (0.152.1)

The current through the conductor is

I = nAvde (0.152.2)

For a conductor of width, w and thickness, d, there is a Hall voltage across the widthof the conductor. Thus the electrical force becomes

Fe = eE

=EVH

w(0.152.3)

The magnetic force is

Fm =BI

neA(0.152.4)


DRAFT

Debye and Einstein Theories to Specific Heat xcviieq. (0.152.3) is equal to eq. (0.152.4), thus

eVH

w=

BInewd

∴ VH =BIned

(0.152.5)

So for a measured magnetic field and current, the sign of the Hall voltage gives is thesign of the charge carrier.

Answer: (C)

0.153 Debye and Einstein Theories to Specific Heat

The determination of the specific heat capacity was first deermined by the Law ofDulong and Petite. This Law was based on Maxwell-Boltzmann statistics and wasaccurate in its predictions except in the region of low temperatures. At that point thereis a departure from prediction and measurements and this is where the Einstein andDebye models come into play.

Both the Einstein and Debye models begin with the assumption that a crystal is madeup of a lattice of connected quantum harmonic oscillators; choice B.

The Einstein model makes three assumptions

1. Each atom is a three-dimensional quantum harmonic oscillator.

2. Atoms do not interact with each other.

3. Atoms vibrate with the same frequency.

Einstein assumed a quantum oscillator model, similar to that of the black body radi-ation problem. But despite its success, his theory predicted an exponential decress inheat capacity towards absolute zero whereas experiments followed a T3 relationship.This was solved in the Debye Model.

The Debye Model looks at phonon contribution to specific heat capacity. This theorycorrectly predicted the T3 proportionality at low temperatures but suffered at inteme-diate temperatures.

Answer: (B)

0.154 Potential inside a Hollow Cube

By applying Gauss’ Law and drawing a Gaussian surface inside the cube, we see thatno charge is enclosed and hence no electric field15. We can realte the electric field to

15Draw Cube at potential V with Gaussian Surface enclosing no charge


DRAFT

xcviii GR8677 Exam Solutionsthe potential

E = −∇V (0.154.1)

Where V is the potential.

Gauss’ Law shows that with no enclosed charge we have no electric field inside ourcube. Thus

E = −∇V = 0 (0.154.2)

As eq. (0.154.1) is equal to zero, the potential is the same throughout the cube.16

Answer: (E)

0.155 EM Radiation from Oscillating Charges

As the charge particle oscillates, the electric field oscillates as well. As the field oscillatesand changes, we would expect this changing field to affect a distant charge. If weconsider a charge along the xy-plane, looking directly along the x-axis, we won’t “see”the charge oscillating but we would see it clearly if we look down the y-axis. If wewere to visualize the field, it would look like a doughnut around the x-axis. Based onthat analysis, we choose (C)

Answer: (C)

0.156 Polarization Charge Density

D = ε0E + P (0.156.1)

∇ ·D = ε0∇ · E + ∇ · P

=εD∇ · Eκ

− σp

Answer: (E)17

0.157 Kinetic Energy of Electrons in Metals

Electrons belong to a group known as fermions18 and as a result obey the Pauli Exclu-sion Principle19. So in the case of a metal, there are many fermions present each with

16As we expect there to be no Electric Field, we must expect the potential to be the same throughoutthe space of the cube. If there were differences, a charge place inside the cube would move.

17Check Polarization in Griffiths18Examples of fermions include electrons, protons and neutrons19The Pauli Exclusion Principle states that no two fermions may occupy the same quantum state


DRAFT

Expectation or Mean Value xcixa different set of quantum numbers. The electron with the highest energy state is hasan energy value known as the Fermi Energy.

NOT FINISHED

Answer: (B)

0.158 Expectation or Mean Value

This is a definition question. The question states that for an operator Q,

〈Q〉 =

∫ +∞

−∞

ψ∗Qψdx (0.158.1)

This is the very definition of the expectation or mean value of Q.

Answer: (C)

0.159 Eigenfuction of Wavefunction

We are given the momentum operator as

p = −i~∂∂x

(0.159.1)

With an eigenvalue of ~k. We can do this by trying each solution and seeing if theymatch20

− i~∂ψ

∂x= ~kψ (0.159.2)

A: ψ = cos kx We expect ψ, to have the form of an exponential function. Substitutingthis into the eigenfuntion, eq. (0.159.2), we have

−i~∂∂x

cos kx = −i~ (−k sin kx)

= i~k sin kx , ~kψ

ψ does not surive our differentiation and so we can eliminate it.

B: ψ = sin kx This is a similar case to the one above and we can eliminate for thisreason.

−i~∂∂x

sin kx = −i~ (k cos kx)

= −i~k cos kx , ~kψ

20We can eliminate choices (A) & (B) as we would expect the answer to be an exponential function inthis case. These choices were just done for illustrative purposes and you should know to avoid them inthe exam.


DRAFT

c GR8677 Exam SolutionsAgain we see that ψ does not survive when we apply our operator and so we caneliminate this choice as well.

C: ψ = exp−ikx Substituting this into eq. (0.159.2), gives

−i~∂∂x

e−ikx = −i~(−ike−ikx

)= −~ke−ikx , ~kψ

Close but we are off, so we can eliminate this choice as well.

D: ψ = exp ikx If the above choice didn’t work, this might be more likely to.

−i~∂∂x

eikx = −i~(ikeikx

)= ~ke−ikx = ~kψ

Success, this is our answer.

E: = ψ = exp−kx

−i~∂∂x

e−kx = −i~(−ke−kx

)= −i~ke−kx , ~kψ

Again this choice does not work, so we can eliminate this as well

Answer: (D)

0.160 Holograms

The hologram is an image that produces a 3-dimensional image using both the Am-plitude and Phase of a wave. Coherent, monochromatic light, such as from a laser, issplit into two beams. The object we wish to “photograph” is placed in the path of theillumination beam and the scattered light falls on the recording medium. The secondbeam, the reference beam is reflected unimpeded to the recording medium and thesetwo beams produces an interference pattern.

The intensity of light recorded on our medium is the same as the scattered light fromour object. The interference pattern is a result of phase changes as light is scattered offour object. Thus choices (I) and (II) are true.

Answer: (B)


DRAFT

Group Velocity of a Wave ci0.161 Group Velocity of a Wave

We are given the dispersion relationship of a wave as

ω2 =(c2k2 + m2

) 12 (0.161.1)

The Group Velocity of a Wave is

vg =dωdk

(0.161.2)

By differentiating eq. (0.161.1) with respect to k, we can determine th group velocity

2ωdω = 2c2kdk

⇒dωdk

=c2kω

=c2k

√

c2k2 + m2(0.161.3)

We want to examine the cases as k→ 0 and k→∞.

As k→ 0, we have

dωdk

=c20

√0 + m2

= 0 (0.161.4)

As k→∞, c2k2 >> m2 the denominator becomes√

c2k2 + m ≈ c2k2 (0.161.5)

Replacing the denominator for our group velocity gives

dωdk

=c2kck

= c (0.161.6)

Answer: (E)

0.162 Potential Energy and Simple Harmonic Motion

We are given a potential energy of

V(x) = a + bx2 (0.162.1)

We can determine the mass’s spring constant, k, from V′′(x)

V′′(x) = 2b = k (0.162.2)

The angular frequency, ω, is

ω2 =km

=2bm

(0.162.3)

We see this is dependent on b and m.

Answer: (C)


DRAFT

cii GR8677 Exam Solutions0.163 Rocket Equation I

We recall from the rocket equation that u in this case is the speed of the exaust gasrelative to the rocket.

Answer: (E)

0.164 Rocket Equation II

The rocket equation is

mdvdt

+ udmdt

= 0 (0.164.1)

Solving this equation becomes

mdv = udm∫ v

0dv = u

∫ m

m0

dmm

v = u ln( mm0

)(0.164.2)

This fits none of the answers given.

Answer: (E)

0.165 Surface Charge Density

This question was solved as ‘The Classic Image Problem’. Below is an alternativemethod but the principles are the same. Instead of determining the electrical potential,as was done by Griffiths, we will find the electrical field of a dipole and determine thesurface charge density using

E =σε0

(0.165.1)

Our point charge,−q will induce a +q on the grounded conducting plane. The resultingelectrical field will be due to a combination of the real charge and the ‘virtual’ inducedcharge. Thus

E = −Eyj = (E− + E+) j

= 2E−j (0.165.2)

Remember the two charges are the same, so at any point along the x-axis, or rather ourgrounded conductor, the electrical field contributions from both charges will be the


DRAFT

Maximum Power Theorem ciiisame. Thus

E− =q

4πεr2 cosθ where cosθ =dr

=qd

4πε0r3 (0.165.3)

Our total field becomes

E =2qd

4πε0r3 (0.165.4)

You may recognize that 2qd is the electrical dipole moment. Now, putting eq. (0.165.4)equal to eq. (0.165.1) gives us

σε0

=qd

2πε0r3 (0.165.5)

where r = D, we get

σ =qd

2πD2 (0.165.6)

Answer: (C)

0.166 Maximum Power Theorem

We are given the impedance of our generator

Zg = Rg + jXg (0.166.1)

For the maximum power to be transmitted, the maximum power theorem states that theload impedance must be equal to the complex conjugate of the generator’s impedance.

Zg = Z∗` (0.166.2)

Thus

Z` = Rg + jX`

= Rg − jXg (0.166.3)

Answer: (C)

0.167 Magnetic Field far away from a Current carryingLoop

The Biot-Savart Law is

dB =µ0i4π

d` × rr3 (0.167.1)


DRAFT

civ GR8677 Exam SolutionsLet θ be the angle between the radius, b and the radius vector, r, we get

B =µ0i4π

rd` cosθr3 where cosθ =

br

=mu0i4π

d` cosθr2

=µ0i4π

bd`r3 where r =

√

b2 + h2

=µ0i4π

bd`

(b2 + h2)32

where d` = b · dθ

=µ0i4π·

b2

(b2 + h2)32

2π∫0

dθ

=µ0i2

b2

(b2 + h2)32

(0.167.2)

we see thatB ∝ ib2 (0.167.3)

Answer: (B)

0.168 Maxwell’s Relations

To derive the Maxwell’s Relations we begin with the thermodynamic potentials

First LawdU = TdS − PdV (0.168.1)

Entalpy

H = E + PV∴ dH = TdS + VdP (0.168.2)

Helmholtz Free Energy

F = E − TS∴ dF = −SdT − PdV (0.168.3)

Gibbs Free Energy

G = E − TS + PV∴ dG = −SdT + VdP (0.168.4)


DRAFT

Partition Functions cvAll of these differentials are of the form

dz =

(∂z∂x

)y

dx +

(∂z∂y

)x

dy

= Mdx + Ndy

For the variables listed, we choose eq. (0.168.1) and applying the above condition weget

T =

(∂U∂S

)V

P =

(∂U∂V

)S

(0.168.5)

Thus taking the inverse of T, gives us

1T

=

(∂S∂U

)V

(0.168.6)

Answer: (E)

0.169 Partition Functions

NOT FINISHED

0.170 Particle moving at Light Speed

Answer: (A)

0.171 Car and Garage I

We are given the car’s length in its rest frame to be L′ = 5 meters and its LorentzContracted length to be L = 3 meters. We can determine the speed from eq. (0.87.1)

L = L′√

1 −v2

c2(35

)2

= 1 −v2

c2

⇒ v =45

c (0.171.1)

Answer: (C)


DRAFT

cvi GR8677 Exam Solutions0.172 Car and Garage II

As the car approaches the garage, the driver will notice that things around him, in-cluding the garage, are length contracted. We have calculated that the speed thathe is travelling at to be, v = 0.8c, in the previous section. We again use the LengthContraction formula, eq. (0.87.1), to solve this question.

Lg = L′g

(1 −

v2

c2

)= 4

(1 − 0.82

)= 2.4 meters (0.172.1)

Answer: (A)

0.173 Car and Garage III

This is more of a conceptual question. What happens depends on whose frame ofreference you’re in.

Answer: (E)

0.174 Refrective Index of Rock Salt and X-rays

No special knowledge is needed but a little knowledge always helps. You can start byeliminating choices when in doubt.

Choice A NOT TRUE Relativity says nothing about whether light is in a vacuum ornot. If anything, this choice goes against the postulates of Special Relativity. Thelaws of Physics don’t change in vacuum.

Choice B NOT TRUE. X-rays can “transmit” signals or energy; any waveform canonce it is not distorted too much during propagation.

Choice C NOT TRUE. Photons have zero rest mass. Though the tachyon, a hypothet-ical particle, has imaginary mass. This allows it to travel faster than the speed orlight though they don’t violate the principles of causality.

Choice D NOT TRUE. How or when we discover physical theories has no bearingon observed properties or behavior; though according to some it may seem so attimes.

Choice E The phase and group speeds can be different. The phase velocity is the rate atwhich the crests of the wave propagate or the rate at which the phase of the waveis moving. The group speed is the rate at which the envelope of the waveform


DRAFT

Thin Flim Non-Reflective Coatings cviiis moving or rather it’s the rate at which the amplitude varies in the waveform.We can use this principle of n < 1 materials to create X-ray mirrors using “totalexternal reflection”.

Answer: (E)

0.175 Thin Flim Non-Reflective Coatings

To analyze this system, we consider our lens with refractive index, n3, being coated byour non-reflective coating of refractive index, n2, and thickness, t, in air with refractiveindex, n1, where

n1 < n2 < n3 (0.175.1)

As our ray of light in air strikes the first boundary, the coating, it moves from a lessoptically dense medium to a more optically dense one. At the point where it reflects,there will be a phase change in the reflected wave. The transmitted wave goes throughwithout a phase change.

The refracted ray passes through our coating to strike our glass lens, which is opticallymore dense than our coating. As a result there will be a phase change in our reflectedray. Destructive interference occurs when the optical path difference, 2t, occurs inhalf-wavelengths multiples. So

2t =(m +

12

)λn2

(0.175.2)

where m = 0; 1; 2; 3. The thinnest possible coating occurs at m = 0. Thus

t =14λn2

(0.175.3)

We need a non-reflective coating that has an optical thicknes of a quarter wavelength.

Answer: (A)

0.176 Law of Malus

The Law of Malus states that when a perfect polarizer is placed in a polarized beamof light, the intensity I, is given by

I = I0 cos2 θ (0.176.1)

where θ is the angle between the light’s plane of polarization and the axis of thepolarizer. A beam of light can be considered to be a uniform mix of plane polarization


DRAFT

cviii GR8677 Exam Solutionsangles and the average of this is

I = I0

∫ 2π

0cos2 θ

=12

I0 (0.176.2)

So the maximum fraction of transmitted power through all three polarizers becomes

I3 =(12

)3

=I0

8(0.176.3)

Answer: (B)

0.177 Geosynchronous Satellite Orbit

We can relate the period or the angluar velocity of a satellite and Newton’s Law ofGravitation

mRω2 = mR(2π

T

)2

=GMm

R2 (0.177.1)

where M is the mass of the Earth, m is the satellite mass and RE is the orbital radius.From this we can get a relationship between the radius of orbit and its period, whichyou may recognize as Kepler’s Law.

R3∝ T2 (0.177.2)

We can say

R3E ∝ (80)2 (0.177.3)

R3S ∝ (24 × 60)2 (0.177.4)

(0.177.5)

Dividing eq. (0.177.4) and eq. (0.177.5), gives(RS

RE

)3

=(24 × 60

80

)2

R3S = 182R3

E (0.177.6)

Answer: (B)

0.178 Hoop Rolling down and Inclined Plane

As the hoop rolls down the inclined plane, its gravitational potential energy is con-verted to translational kinetic energy and rotational kinetic energy

Mgh =12

Mv2 +12

Iω2 (0.178.1)


DRAFT

Simple Harmonic Motion cixRecall that v = ωR, eq. (0.178.1) becomes

MgH =12

MR2ω2 +12

(MR2

)ω2 (0.178.2)

Solving for ω leaves

ω =

(ghR2

) 12

(0.178.3)

The angular momentum isL = Iω (0.178.4)

Substituting eq. (0.178.3) gives us

L = MR(

ghR2

) 12

= MR√

gh (0.178.5)

Answer: (A)

0.179 Simple Harmonic Motion

We are told that a particle obeys Hooke’s Law, where

F = −kx (0.179.1)

We can write the equation of motion as

mx − kx where ω2 =km

where

x = A sin(ωt + φ

)(0.179.2)

and x = ωA cos(ωt + φ

)(0.179.3)

We are told that12

= sin(ωt + φ

)(0.179.4)

We can show that

cos(ωt + φ

)=

√3

2(0.179.5)

Substituting this into eq. (0.179.3) gives

x = 2π f A ·√

32

=√

3 π f A (0.179.6)

Amswer: (B)


DRAFT

cx GR8677 Exam Solutions0.180 Total Energy between Two Charges

We are told three things

1. There is a zero potential energy, and

2. one particle has non-zero speed and hence kinetic energy.

3. No radiation is emitted, so no energy is lost.

The total energy of the system is

E = Potential Energy + Kinetic Energy= 0 + (KE > 0)> 0 (0.180.1)

Applying the three condition, we expect the total energy to be positive and constant.

Answer: (C)

0.181 Maxwell’s Equations and Magnetic Monopoles

You may have heard several things about the∇·B = 0 equation in Maxwell’s Laws. Oneof them is there being no magnetic monopoles or charges. There are some implicationsto this. No charge implies that the amount of field lines that enter a Gaussian surfacemust be equal to the amount of field lines that leave. So using this principle we knowfrom the electric form of this law we can get an answer to this question.

Choice A The number of field lines that enter is the same as the number that leaves.So this does not violate the above law.

Choice B Again we see that the number of field lines entering is the same as thenumber leaving.

Choice C The same as above

Choice D In this case, we see that the field lines at the edge of the Gaussian Surfaceare all leaving; no field lines enter the surface. This is also what we’d expect thefield to look like for a region bounded by a magnetic monopole.

Choice E The field loops in on itself, so the total number of field lines is zero. This fitswith the above law.

Answer: (D)


DRAFT

Gauss’ Law cxi0.182 Gauss’ Law

To determine an electric field that could exist in a region of space with no charges weturn to Gauss’ Law.

∇ · E = 0 (0.182.1)

or rather∂∂x

Ex +∂∂y

Ey +∂∂z

Ez = 0 (0.182.2)

So we analyze each choice in turn to get our answer.

Choice A

E = 2xyi − xyk

∇ · E =∂∂x

2xy +∂∂z

(−xz)

= 2y + x , 0 (0.182.3)

Choice B

E = −xy j + xzk

∇ · E =∂∂y

(−xy) +∂∂z

xz

= −x + x = 0 (0.182.4)

Choice C

E = xzi + xz j

∇ · E =∂∂x

xz +∂∂y

xz

= z + 0 , 0 (0.182.5)

Choice D

E = xyz(i + j)

∇ · E =∂∂x

xyz +∂∂y

xyz

= yz + xz , 0 (0.182.6)

Choice E

E = xyzi

∇ · E =∂∂x

xyz

= yz , 0 (0.182.7)

Answer: (B)


DRAFT

cxii GR8677 Exam Solutions0.183 Biot-Savart Law

We can determine the magnetic field produced by our outer wire from the Biot-SavartLaw

dB =µ0

4πd` × r

r3 (0.183.1)

As our radius and differential length vectors are orthogonal, the magnetic field worksout to be

dB =µ0

4πId`rr3

=µ0I4π·

rdθr2

B =µ0I4πr

∫ 2π

0dθ

=µ0I2b

(0.183.2)

We know from Faraday’s Law, a changing magnetic flux induces a EMF,

E =dΦ

dt(0.183.3)

where Φ = BA. The magnetic flux becomes

Φ =µ0I2b· πa2 (0.183.4)

The induced EMF becomes

E =µ0π

2

(a2

b

)dIdt

=µ0π

2

(a2

b

)ωI0 sinωt (0.183.5)

Answer: (B)

0.184 Zeeman Effect and the emission spectrum of atomicgases

Another knowledge based question best answered by the process of elimination.

Stern-Gerlach Experiemnt The Stern-Gerlach Experiment has nothing to do withspectral emissions. This experiment, performed by O. Stern and W. Gerlachin 1922 studies the behavior of a beam atoms being split in two as they passthrough a non-uniform magnetic field.


DRAFT

Spectral Lines in High Density and Low Density Gases cxiiiStark Effect The Stark Effect deals with the shift in spectral lines in the presence of

electrical fields; not in magnetic fields.

Nuclear Magnetic Moments of atoms Close, the splitting seen in the Stern-GerlachExperiment is due to this. Emission spectrum typically deals with electrons andso we would expect it to deal with electrons on some level.

Emission lines are split in two Closer but still not accurate. There is splitting but insome cases it may be more than two.

Emission lines are greater or equal than in the absence of the magnetic field This weknow to be true.

The difference in the emission spectrum of a gas in a magnetic field is due to theZeeman effect.

Answer: (E)

0.185 Spectral Lines in High Density and Low DensityGases

We expect the spectral lines to be broader in a high density gas and narrower in a lowdensity gas ue to the increased colissions between the molecules. Atomic collisionsadd another mechanism to transfer energy.

Answer: (C)

0.186 Term Symbols & Spectroscopic Notation

To determine the term symbol for the sodium ground state, we start with the electronicconfiguration. This is easy as they have given us the number of electrons the elementhas thus allowing us to fill sub-shells using the Pauli Exclusion Principle. We get

1s2, 2s2, 2p6, 3s1 (0.186.1)

We are most interested in the 3s1 sub-shell and can ignore the rest of the filled sub-shells. As we only have one valence electron then ms = +1/2. Now we can calculatethe total spin quantum number, S. As there is only one unpaired electron,

S =12

(0.186.2)

Now we can calculate the total angular momentum quantum number, J = L + S. Asthe 3s sub-shell is half filled then

L = 0 (0.186.3)


DRAFT

cxiv GR8677 Exam SolutionsThis gives us

J =12

(0.186.4)

and as L = 0 then we use the symbol S. Thus our term equation becomes2S 1

2(0.186.5)

Answer: (B)

0.187 Photon Interaction Cross Sections for Pb

Check Brehm p. 789

Answer: (B)

0.188 The Ice Pail Experiment

Gauss’ law is equivalent to Coulomb’s Law because Coulomb’s Law is an inversesquare law; testing one is a valid test of the other. Much of our knowledge of theconsequences of the inverse square law came from the study of gravity. Jason Priestlyknew that there is no gravitational field within a spherically symmetrical mass distri-bution. It was suspected that was the same reason why a charged cork ball inside acharged metallic container isn’t attracted to the walls of a container.

Answer: (E)

0.189 Equipartition of Energy and Diatomic Molecules

To answer this question, we will turn to the equipartition of energy equation

cv =

(f2

)R (0.189.1)

where f is the number of degrees of freedom. In the case of Model I, we see that

Degrees of Freedom Model I Model II

Translational 3 3Rotational 2 2Vibrational 0 2

Total 5 7

Table 0.189.1: Specific Heat, cv for a diatomic molecule


DRAFT

Fermion and Boson Pressure cxvSo the specific heats for Models I & II are

cvI =52

Nk cvII =72

Nk

Now we can go about choosing our answer

Choice A From our above calculations, we see that cvI = 5/2Nk. So this choice isWRONG.

Choice B Again, our calculations show that the specific heat for Model II is larger thanthan of Model I. This is due to the added degrees of freedom (vibrational) that itpossesses. So this choice is WRONG.

C & D They both contradict the other and they both contradict Choice (E).

E This is TRUE. We know that at higher temperatures we have an additional degreeof freedom between our diatomic molecule.

Answer: (E)

0.190 Fermion and Boson Pressure

To answer this question, we must understand the differences between fermions andbosons. Fermions follow Fermi-Dirac statistics and their behavior is obey the PauliExclusion Principle. Basically, this states that no two fermions may have the samequantum state. Bosons on the other hand follow Bose-Einstein statistics and severalbosons can occupy the same quantum state.

As the temperature of a gas drops, the particles are going to fill up the available energystates. In the case of fermions, as no two fermions can occupy the same state, thenthese particles will try to occupy all the energy states it can until the highest is filled.Bosons on the other hand can occupy the same state, so they will all ‘group’ togetherfor the lowest they can. Classically, we don’t pay attention to this grouping, so basedon our analyis, we expect,

PF > PC > PB (0.190.1)

where PB is the boson pressure, PC is the pressure with no quantum effects taking placeand PF to be the fermion pressure.

Answer: (B)

0.191 Wavefunction of Two Identical Particles

We are given the wavefunction of two identical particles,

ψ =1√

2

[ψα(x1)ψβ(x2) + ψβ(x1)ψα(x2)

](0.191.1)


DRAFT

cxvi GR8677 Exam SolutionsThis is a symmetric function and satisfies the relation

ψαβ(x2, x1) = ψαβ(x1, x2) (0.191.2)

Symmetric functions obey Bose-Einstein statistics and are known as bosons. Uponexamination of our choices, we see that21

electrons fermion

positrons fermion

protons fermion

neutron fermion

deutrons Boson

Incidentally, a anti-symmetric function takes the form,

ψ =1√

2

[ψα(x1)ψβ(x2) − ψβ(x1)ψα(x2)

](0.191.3)

and satisfies the relationψαβ(x2, x1) = −ψαβ(x1, x2) (0.191.4)

These obey Fermi-Dirac Statistics and are known as fermions.

Answer: (E)

0.192 Energy Eigenstates

We may recognize this wavefunction from studying the particle in an infinite wellproblem and see this is the n = 2 wavefunction. We know that

En = n2E0 (0.192.1)

We are given that E2 = 2 eV. So

E0 =1n2 E2

=24

eV

=12

eV (0.192.2)

Answer: (C)21You could have easily played the ‘one of thes things is not like the other...’ game


DRAFT

Bragg’s Law cxvii0.193 Bragg’s Law

We recall Bragg’s Law2d sinθ = nλ (0.193.1)

Plugging in what we know, we determine λ to be

λ = 2(3 Å)(sin 30)

= 2(3 Å)(0.5)

= 3 Å (0.193.2)

We employ the de Broglie relationship between wavelength and momentum

p =hλ

(0.193.3)

We get

mv =hλ

⇒ v =h

mλ

=6.63 × 10−34

(9.11 × 10−31)(3 × 10( − 10))(0.193.4)

We can determine the order of our answer by looking at the relevant indices

− 34 − (−31) − (−10) = 7 (0.193.5)

We see that (D) is close to what we are looking for.

Answer: (D)

0.194 Selection Rules for Electronic Transitions

The selection rules for an electric dipole transition are

∆` = ±1 Orbital angular momentum∆m` = 0,±1 Magnetic quantum number∆ms = 0 Secondary spin quantum number,∆ j = 0,±1 Total angular momentum

We have no selection rules for spin, ∆s, so we can eliminate this choice.

Answer: (D)


DRAFT

cxviii GR8677 Exam Solutions0.195 Moving Belt Sander on a Rough Plane

We know the work done on a body by a force is

W = F × x (0.195.1)

We can relate this to the power of the sander; power is the rate at which work is done.So

P =dWdt

= Fdxdt

= Fv (0.195.2)

The power of the sander can be calculated

P = VI (0.195.3)

where V and I are the voltage across and the current through the sander. By equatingthe Mechanical Power, eq. (0.195.2) and the Electrical Power, eq. (0.195.3), we candetermine the force that the motor exerts on the belt.

F =VIv

=120 × 9

10= 108 N (0.195.4)

The sander is motionless, soF − µR = 0 (0.195.5)

where R is the normal force of the sander pushing against the wood. Thus the coefficientof friction is

µ =FR

=108100

= 1.08 (0.195.6)

Answer: (D)

0.196 RL Circuits

When the switch, S, is closed, a magnetic field builds up within the inductor and theinductor stores energy. The charging of the inductor can be derived from Kirchoff’sRules.

E − IR − LdIdt

= 0 (0.196.1)

and the solution to this is

I(t) = I0

[1 − exp

(R1tL

)](0.196.2)


DRAFT

RL Circuits cxixwhere the time constant, τ1 = L/R1.

We can find the voltage across the resistor, R1, by multiplying the above by R1, givingus

V(t) = R1 · I0

[1 − exp

(R1tL

)]= E

[1 − exp

(R1tL

)](0.196.3)

The potential at A can be found by measuring the voltage across the inductor. Giventhat

E − VR1 − VL = 0∴ VL = E − VR1

= E exp(R1t

L

)(0.196.4)

This we know to be an exponential decay and (fortunately) limits our choices to either(A) or (B)22

The story doesn’t end here. If the inductor was not present, the voltage would quicklydrop and level off to zero but with the inductor present, a change in current means achange in magnetic flux; the inductor opposes this change. We would expect to see areversal in the potential at A. Since both (A) and (B) show this flip, we need to thinksome more.

The energy stored by the inductor is

UL =12

LI20 =

12

L(E

R1

)2

(0.196.5)

With S opened, the inductor is going to dump its energy across R2 and assuming thatthe diode has negligible resistance, all of this energy goes to R2. Thus

U =12

L(

VR2

R2

)2

(0.196.6)

The above two equations are equal, thus

E

R1=

VR2

R2

VR2 = 3E (0.196.7)

We expect the potential at A to be larger when S is opened. Graph (B) fits this choice.

Answer: B22If you get stuck beyond this point, you can guess. The odds are now in your favor.


DRAFT

cxx GR8677 Exam Solutions0.197 Carnot Cycles

The Carnot Cycle is made up of two isothermal transformations, KL and MN, and twoadiabatic transformations, LM and NK. For isothermal transformations, we have

PV = nRT = a constant (0.197.1)

For adiabatic transformations, we have

PVγ = a constant (0.197.2)

where γ = CP/CV.

For the KL transformation, dU = 0.

Q2 = WK→L

∴WK→L =

∫ VL

VK

PdV

= nRT2 ln(VK

VL

)(0.197.3)

For the LM transformation,PLVγ

L = PMVγM (0.197.4)

For the MN transformation, dU = 0.

Q1 = WM→N

∴WM→N =

∫ VN

VM

PdV

= nRT1 ln(VN

VM

)(0.197.5)

For the NK transformation,PNVγ

N = PKVγK (0.197.6)

Dividing eq. (0.197.4) and eq. (0.197.6), gives

PLVγL

PKVγK

=PMVγ

M

PNVγN

∴VL

VK=

VM

VN(0.197.7)

The effeciency of an engine is defined

η = 1 −Q1

Q2(0.197.8)


DRAFT

Carnot Cycles cxxiWe get

η = 1 −Q1

Q2= 1 −

−WM→N

WK→L

= 1 −nRT1 ln

(VMVN

)nRT2 ln

(VKVL

)= 1 −

T1

T2(0.197.9)

1. We see that

1 −Q1

Q2= 1 −

T1

T2

∴Q1

Q2=

T1

T2(0.197.10)

Thus choice (A) is true.

2. Heat moves from the hot reservoir and is converted to work and heat. Thus

Q2 = Q1 + W (0.197.11)

The entropy change from the hot reservoir

S =dQ2

T(0.197.12)

As the hot reservoir looses heat, the entropy decreases. Thus choice (B) is true.

3. For a reversible cycle, there is no net heat flow over the cycle. The change inentropy is defined by Calusius’s Theorem.∮

dQT

= 0 (0.197.13)

We see that the entropy of the system remains the same. Thus choice (C) is false.

4. The efficieny is defined

η =WQ2

(0.197.14)

This becomes

η = 1 −Q1

Q2

=Q2 −Q1

Q2(0.197.15)

Thus W = Q2 −Q1. So choice (D) is true,

5. The effeciency is based on an ideal gas and has no relation to the substance used.So choice (E) is also true.

Answer: (C)


DRAFT

cxxii GR8677 Exam Solutions0.198 First Order Perturbation Theory

Perturbation Theory is a procedure for obtaining approximate solutions for a perturbedstate by studying the solutions of the unperturbed state. We can, and shouldn’t,calculate this in the exam.

We can get the first order correction to be ebergy eigenvalue

E1n = 〈ψ0

n|H′

|ψ0n〉 (0.198.1)

From there we can get the first order correction to the wave function

ψ1n =

∑m,n

〈ψ0m|H

′

|ψ0n〉(

E0n − E0

m) (0.198.2)

and can be expressed asψ1

n =∑m,n

c(n)m ψ

0m (0.198.3)

you may recognize this as a Fourier Series and this will help you knowing that theperturbing potential is one period of a saw tooth wave. And you may recall that theFourier Series of a saw tooth wave form is made up of even harmonics.

Answer: (B)23

0.199 Colliding Discs and the Conservation of AngularMomentum

As the disk moves, it possessed both angular and linear momentums. We can notexactly add these two as they, though similar, are quite different beasts. But we candefine a linear angular motion with respect to some origin. As the two discs hit eachother, they fuse. This slows the oncoming disc. We can calculate the linear angularmomentum

L = r × p (0.199.1)

where p is the linear momentum and r is the distance from the point P to the center ofdisc I. This becomes

Lv0 = MR × v0

= −MRv0 (0.199.2)

It’s negative as the cross product of R and v0 is negative.

The Rotational Angular Momentum is

Lω0 = Iω0 (0.199.3)

23Griffiths gives a similar problem in his text


DRAFT

Electrical Potential of a Long Thin Rod cxxiiiAdding eq. (0.199.3) and eq. (0.199.2) gives the total angular momentum.

L = Lω0 + Lv0

= Iω0 −MRv0

=12

MR2ω0 −12

MR2ω0

= 0

Thus the total angular momentum at the point P is zero.

Answer: (A)

0.200 Electrical Potential of a Long Thin Rod

We have charge uniformly distributed along the glass rod. It’s linear charge density is

λ =Q`

=dQdx

(0.200.1)

The Electric Potential is defined

V(x) =q

4πε0x(0.200.2)

We can ‘slice’ our rod into infinitesimal slices and sum them to get the potential of therod.

dV =1

4πε0

λdxx

(0.200.3)

We assume that the potential at the end of the rod, x = ` is V = 0 and at some pointaway from the rod, x, the potential is V. So∫ V

0dV =

λ4πε0

∫ x

`

dxx

=λ

4πε0ln

(x`

)(0.200.4)

Where x = 2`, eq. (0.200.4) becomes

V =Q`

14πε0

ln(2``

)=

Q`

14πε0

ln 2 (0.200.5)

Answer: (D)


DRAFT

cxxiv GR8677 Exam Solutions0.201 Ground State of a Positronium Atom

Positronium consists of an electron and a positron bound together to form an “exotic”atom. As the masses of the electron and positron are the same, we must use a reduced-mass correction factor to determine the enrgy levels of this system.24. The reducedmass of the system is

1µ

=1

me+

1mp

(0.201.1)

Thus /mu is

µ =me ·mp

me + mp

=me

2(0.201.2)

The ground state of the Hydrogen atom, in terms of the reduced mass is

E1 = −µ

meE0

= −12

E0 (0.201.3)

where E0 = 13.6 eV.

Answer: (B)

0.202 The Pinhole Camera

A pinhole camera is simply a camera with no lens and a very small aperature. Lightpasses through this hole to produce an inverted image on a screen. For the photographybuffs among you, you know that by varying the size of a camera’s aperature canaccomplish various things; making the aperature bigger allows more light to enter andproduces a “brighter” picture while making the aperature smaller produces a sharperimage.

In the case of the pinhole camera, making the pinhole, or aperature, smaller producesa sharper image because it reduces “image overlap”. Think of a large hole as a set oftiny pinholes places close to each other. This results in an infinite amount of imagesoverlapping each other and hence a blurry image. So to produce a sharp image, itis best to use the smallest pinhole possible, the tradeoff being an image that’s not as“bright”.

There are limits to the size of our pinhole. We can not say, for example, use an infinitelysmall pinhole the produce the sharpest possible image. Beyond some point diffractioneffects take place and will ruin our image.

24Place cite here


DRAFT

The Pinhole Camera cxxvConsider a pinhole camera of length, D, with a pinhole of diameter, d. We know howmuch a beam of light will be diffracted through this pinhole by25

d sinθ = mλ (0.202.1)

this is the equation for the diffraction of a single slit. As θ is small and we will considerfirst order diffraction effects, eq. (0.202.1) becomes

dθ = λ

⇒ θ =λd

(0.202.2)

The “size” of this spread out image is

y = 2θD

=2λD

d(0.202.3)

So the ‘blur’ of our resulting image is

B = y − d

=2λD

d− d (0.202.4)

We can see that we want to reduce y as much as possible. i.e. make it d. So eq. (0.202.4)becomes

0 =2λD

d− d

∴2λD

d= d

Thus d =√

2λD (0.202.5)

So we’d want a pinhole of that size to produce or sharpest image possible. This resultis close to the result that Lord Rayleigh used, which worked out to be

d = 1.9√

Dλ (0.202.6)

Answer: (A)

25Add image of pinhole camera


DRAFT

cxxvi GR8677 Exam Solutions


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

cxxviii 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 cxxix.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

cxxx Constants & Important Equations


DRAFTBibliography

[1] John J. Brehm and William J. Mullin. Introduction to the Structure of Matter, chapter11-6, pages 567–571. Wiley, first edition, 1989.

[2] David J. Griffiths. Introduction to Electrodyanmics, chapter 3.2.1, pages 121–123.Prentice Hall, third edition, 1999.

[3] Douglas Adams. The restaurant at the end of the universe.

[4] Wikipedia. Spectral line — wikipedia, the free encyclopedia, 2009. [Online;accessed 17-March-2009].

[5] Wikipedia. Term symbol — wikipedia, the free encyclopedia, 2008. [Online;accessed 22-March-2009].



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


[10] David J. Griffiths. Introduction to Quantum Mechanics, chapter 6.1.1, page 249.Prentice Hall, second edition, 2005.


[12] David J. Griffiths. Introduction to Quantum Mechanics, chapter 6.1.2, page 254.Prentice Hall, second edition, 2005.

DRAFTIndex

AmplifiersGR8677 Q39, xci

Angular Momentum, see Rotational Mo-tion

Binding EnergyGR8677 Q41, xcii

Bohr ModelGR8677 Q19, lxxxiiHydrogen Model, lv

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

Cetripetal MotionGR8677 Q06, lxxvi

Circular Orbits, see Celestial MechanicsCommutators, cxxviii

Canonical Commutators, cxxviiiKronecker Delta Function, cxxviiiLie-algebra Relations, cxxviii

Compton Effect, lviiiCompton Wavelength

GR8677 Q45, xcivConductivity

GR8677 Q23, lxxxivCounting Statistics, lxxi

GR8677 Q40, xciCurrent Density

GR8677 Q09, lxxvii

DielectricsGR8677 Q03, lxxiv

Digital CircuitsGR8677 Q38, xci

Doppler Effect, xxiiDrag Force

GR8677 Q01, lxxiii

Elastic ColissionsGR8677 Q05, lxxv

ElectricityGR8677 Q24, lxxxv

Electron SpinGR8677 Q27, lxxxvi

Electronic ConfigurationGR8677 Q30, lxxxviii

Fleming’s Right Hand RuleGR8677 Q29, lxxxviii

Franck-Hertz Experiment, lxiGR8677 Q47, xcv

Gauss’ LawGR8677 Q10, lxxviii

Gravitation, see Celestial Mechanics

Hall EffectGR8677 Q50, xcvi

HamiltonianGR8677 Q35, xc

InterferenceGR8677 Q13, lxxix

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

Laboratory MethodsGR8677 Q40, xci

Linear Algebra, cxxix

DRAFT

Index cxxxiiiVectors, cxxix

Lorentz Force LawGR8677 Q25, lxxxv

Lorentz TransformationGR8677 Q22, lxxxiv

Maximum Power TheoremGR8677 Q64, ciii

Maxwell’s LawsGR8677 Q11, lxxix

MechanicsGR8677 Q07, lxxviGR8677 Q08, lxxviiGR8677 Q37, xc

Moment of Inertia, see Rotational Motion

Newton’s Law of Gravitation, see CelestialMechanics

Nuclear PhysicsRadioactive Decay

GR8677 Q17, lxxxi

Oscillatory Motion, xviiiCoupled Harmonic Oscillators, xx

GR8677 Q43, xciiDamped Motion, xixKinetic Energy, xviiiPotential Energy, xixSimple Harmonic Motion Equation, xviiiSmall Oscillations, xixTotal Energy, xviii

Parallel Axis Theorem, see Rotational Mo-tion

Particle PhysicsMuon

GR8677 Q16, lxxxiPhotoelectric Effect

GR8677 Q31, lxxxixGR8677 Q32, lxxxixGR8677 Q33, lxxxix

Potential EnergyGR8677 Q34, lxxxix

Principle of Least ActionGR8677 Q36, xc

ProbabilityGR8677 Q15, lxxx

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

Satellite OrbitsGR8677 Q02, lxxiv

Schrodinger’s EquationGR8677 Q18, lxxxii

Space-Time IntervalGR8677 Q21, lxxxiii

Special RelativityDoppler Shift

GR8677 Q12, lxxixEnergy

GR8677 Q20, lxxxiiiSpecific Heat

GR8677 Q14, lxxixStefan-Boltzmann’s Equation

GR8677 Q46, xcivSubject, xlivSystem of Particles, xxiv

Thin Film InterferenceGR8677 Q73, cvii

Torque, see Rotational Motion

Vector Identities, cxxviiProduct Rules, cxxviiiSecond Derivatives, cxxviiiTriple Products, cxxvii

Wave EquationGR8677 Q04, lxxiv

Wave functionGR8677 Q28, lxxxvii

X-RaysGR8677 Q26, lxxxv


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