Physics Gre Sample Test Solutions

The Physics GRE Solution Guide

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Sample Test

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

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November 3, 2009

Author: David S. Latchman

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David S. Latchman

©2009

Preface

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David Latchman

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

Document Changes 05-11-2009

1. Added diagrams to GR0177 test questions 1-25

2. Revised solutions to GR0177 questions 1-25

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04-15-2009 First Version

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David S. Latchman

©2009

Preface Classical Mechanics

i 1

1.1

Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.2

Newton’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

1.3

Work & Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

1.4

Oscillatory Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4

1.5

Rotational Motion about a Fixed Axis . . . . . . . . . . . . . . . . . . . .

8

1.6

Dynamics of Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . 10

1.7

Central Forces and Celestial Mechanics . . . . . . . . . . . . . . . . . . . 10

1.8

Three Dimensional Particle Dynamics . . . . . . . . . . . . . . . . . . . . 12

1.9

Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

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Contents

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1.10 Non-inertial Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . 13 1.11 Hamiltonian and Lagrangian Formalism . . . . . . . . . . . . . . . . . . . 13

2

Electromagnetism

15

2.1

Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2

Currents and DC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3

Magnetic Fields in Free Space . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4

Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.5

Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.6

Maxwell’s Equations and their Applications . . . . . . . . . . . . . . . . . 20

2.7

Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.8

Contents AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.9

Magnetic and Electric Fields in Matter . . . . . . . . . . . . . . . . . . . . 20

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2.10 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.11 Energy in a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.12 Energy in an Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.13 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.14 Current Destiny . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.15 Current Density of Moving Charges . . . . . . . . . . . . . . . . . . . . . 21 2.16 Resistance and Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . 21

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2.17 Resistivity and Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.18 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.19 Kirchoff’s Loop Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.20 Kirchoff’s Junction Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.21 RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

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2.22 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.23 Speed of Propagation of a Light Wave . . . . . . . . . . . . . . . . . . . . 23 2.24 Relationship between E and B Fields . . . . . . . . . . . . . . . . . . . . . 23 2.25 Energy Density of an EM wave . . . . . . . . . . . . . . . . . . . . . . . . 24 2.26 Poynting’s Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Optics & Wave Phonomena

25

3.1

Wave Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.2

Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3

Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.4

Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.5

Geometrical Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.6

Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.7

Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.8

Snell’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

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4

Thermodynamics & Statistical Mechanics

27

4.1

Laws of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.2

Thermodynamic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

David S. Latchman

©2009

Contents v 4.3 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 4.4

Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.5

Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.6

Ensembles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.7

Statistical Concepts and Calculation of Thermodynamic Properties . . . 28

4.8

Thermal Expansion & Heat Transfer . . . . . . . . . . . . . . . . . . . . . 28

4.9

Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.10 Specific Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 4.11 Heat and Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

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4.12 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 28 4.13 Work done by Ideal Gas at Constant Temperature . . . . . . . . . . . . . 29 4.14 Heat Conduction Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.15 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.16 Stefan-Boltzmann’s FormulaStefan-Boltzmann’s Equation . . . . . . . . 30 . . . . . . . . . . . . . . . . . . . . . . . . . . 30

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4.17 RMS Speed of an Ideal Gas

4.18 Translational Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.19 Internal Energy of a Monatomic gas . . . . . . . . . . . . . . . . . . . . . 30 4.20 Molar Specific Heat at Constant Volume . . . . . . . . . . . . . . . . . . . 31 4.21 Molar Specific Heat at Constant Pressure . . . . . . . . . . . . . . . . . . 31 4.22 Equipartition of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.23 Adiabatic Expansion of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . 33

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4.24 Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . 33 5

6

Quantum Mechanics

35

5.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.2

Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 ¨

5.3

Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5.4

Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.5

Wave Funtion Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.6

Elementary Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . 41

Atomic Physics

43

6.1

Properties of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

©2009

David S. Latchman

6.3

Energy Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6.4

Atomic Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6.5

Atomic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.6

Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.7

Black Body Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.8

X-Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6.9

Atoms in Electric and Magnetic Fields . . . . . . . . . . . . . . . . . . . . 47

Special Relativity

51

7.1

Introductory Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7.2

Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7.3

Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7.4

Simultaneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7.5

Energy and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7.6

Four-Vectors and Lorentz Transformation . . . . . . . . . . . . . . . . . . 53

7.7

Velocity Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

7.8

Relativistic Doppler Formula . . . . . . . . . . . . . . . . . . . . . . . . . 54

7.9

Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

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6.2

Contents Bohr Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

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7.10 Space-Time Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Laboratory Methods

57

8.1

Data and Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

8.2

Instrumentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

8.3

Radiation Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

8.4

Counting Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

8.5

Interaction of Charged Particles with Matter . . . . . . . . . . . . . . . . 60

8.6

Lasers and Optical Interferometers . . . . . . . . . . . . . . . . . . . . . . 60

8.7

Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

8.8

Fundamental Applications of Probability and Statistics . . . . . . . . . . 60

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Sample Test 9.1

61

Period of Pendulum on Moon . . . . . . . . . . . . . . . . . . . . . . . . . 61

David S. Latchman

©2009

Contents vii 9.2 Work done by springs in series . . . . . . . . . . . . . . . . . . . . . . . . 62 9.3

Central Forces I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

9.4

Central Forces II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

9.5

Electric Potential I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

9.6

Electric Potential II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

9.7

Faraday’s Law and Electrostatics . . . . . . . . . . . . . . . . . . . . . . . 66

9.8

AC Circuits: RL Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

9.9

AC Circuits: Underdamped RLC Circuits . . . . . . . . . . . . . . . . . . 68

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9.10 Bohr Model of Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . 70 9.11 Nuclear Sizes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 9.12 Ionization of Lithium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 9.13 Electron Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 9.14 Effects of Temperature on Speed of Sound . . . . . . . . . . . . . . . . . . 75

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9.15 Polarized Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 9.16 Electron in symmetric Potential Wells I . . . . . . . . . . . . . . . . . . . . 76 9.17 Electron in symmetric Potential Wells II . . . . . . . . . . . . . . . . . . . 77 9.18 Relativistic Collisions I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 9.19 Relativistic Collisions II . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 9.20 Thermodynamic Cycles I . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 9.21 Thermodynamic Cycles II . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

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9.22 Distribution of Molecular Speeds . . . . . . . . . . . . . . . . . . . . . . . 79 9.23 Temperature Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . 79 9.24 Counting Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 9.25 Thermal & Electrical Conductivity . . . . . . . . . . . . . . . . . . . . . . 80 9.26 Nonconservation of Parity in Weak Interactions . . . . . . . . . . . . . . 81 9.27 Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 9.28 Lorentz Force Law I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 9.29 Lorentz Force Law II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 9.30 Nuclear Angular Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 9.31 Potential Step Barrier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

©2009

David S. Latchman

Contents 87

viii A Constants & Important Equations

A.1 Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 A.2 Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 A.3 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

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A.4 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

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List of Tables

4.22.1Table of Molar Specific Heats . . . . . . . . . . . . . . . . . . . . . . . . . 32 9.4.1 Table of Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

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A.1.1Something . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

List of Tables

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List of Figures

9.5.1 Diagram of Uniformly Charged Circular Loop . . . . . . . . . . . . . . . 65 9.8.1 Schematic of Inductance-Resistance Circuit . . . . . . . . . . . . . . . . . 67 9.8.2 Potential Drop across Resistor in a Inductor-Resistance Circuit . . . . . . 68 9.9.1 LRC Oscillator Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 9.9.2 Forced Damped Harmonic Oscillations . . . . . . . . . . . . . . . . . . . 70

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9.15.1Waves that are not plane-polarized . . . . . . . . . . . . . . . . . . . . . . 76 9.15.2φ = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 9.22.1Maxwell-Boltzmann Speed Distribution of Nobel Gases . . . . . . . . . . 79 9.27.1Hoop and S-shaped wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 9.28.1Charged particle moving parallel to a positively charged current carrying wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

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9.31.1Wavefunction of particle through a potential step barrier . . . . . . . . . 85

List of Figures

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Chapter

1

1.1 1.1.1

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Classical Mechanics Kinematics Linear Motion

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Average Velocity

∆x x2 − x1 = ∆t t2 − t1

(1.1.1)

∆x dx = = v(t) ∆t→0 ∆t dt

(1.1.2)

v=

Instantaneous Velocity

v = lim

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Kinematic Equations of Motion

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

1.1.2

v = v0 + at v2 = v20 + 2a (x − x0 ) 1 x − x0 = v0 t + at2 2 1 x − x0 = (v + v0 ) t 2

(1.1.3) (1.1.4) (1.1.5) (1.1.6)

Circular Motion

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

Classical Mechanics

2 Acceleration Centripetal Acceleration a=

v2 r

(1.1.7)

ω=

v r

(1.1.8)

Angular Velocity

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We can write eq. (1.1.7) in terms of ω

a = ω2 r Rotational Equations of Motion

(1.1.9)

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

(1.1.10) (1.1.11) (1.1.12) (1.1.13)

Newton’s Laws

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1.2

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ω = ω0 + αt ω + ω0 t θ= 2 1 θ = ω0 t + αt2 2 ω2 = ω20 + 2αθ

1.2.1

Newton’s Laws of Motion

First Law A body continues in its state of rest or of uniform motion unless acted upon by an external unbalanced force. Second Law The net force on a body is proportional to its rate of change of momentum. F=

dp = ma dt

(1.2.1)

Third Law When a particle A exerts a force on another particle B, B simultaneously exerts a force on A with the same magnitude in the opposite direction. FAB = −FBA David S. Latchman

(1.2.2) ©2009

Work & Energy

1.2.2

3

Momentum p = mv

1.2.3

Impulse w

∆p = J =

1.3.1

Fdt = Favg dt

Work & Energy Kinetic Energy

1 K ≡ mv2 2

1.3.2

(1.3.1)

The Work-Energy Theorem

The net Work done is given by

Wnet = K f − Ki

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1.3.3

(1.2.4)

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1.3

(1.2.3)

(1.3.2)

Work done under a constant Force

The work done by a force can be expressed as

W = F∆x

(1.3.3)

W = F · ∆r = F∆r cos θ

(1.3.4)

In three dimensions, this becomes

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For a non-constant force, we have

1.3.4

W=

wx f

F(x)dx

(1.3.5)

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Potential Energy

The Potential Energy is dU(x) dx for conservative forces, the potential energy is wx U(x) = U0 − F(x0 )dx0 F(x) = −

(1.3.6)

(1.3.7)

x0

©2009

David S. Latchman

Classical Mechanics

4

1.3.5

Hooke’s Law F = −kx

(1.3.8)

where k is the spring constant.

1.3.6

Potential Energy of a Spring 1 U(x) = kx2 2

1.4.1

Oscillatory Motion

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1.4

(1.3.9)

Equation for Simple Harmonic Motion x(t) = A sin (ωt + δ)

(1.4.1)

1.4.2

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

Period of Simple Harmonic Motion T=

(1.4.2)

Total Energy of an Oscillating System

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1.4.3

2π ω

Given that

x = A sin (ωt + δ)

(1.4.3)

and that the Total Energy of a System is E = KE + PE

(1.4.4)

The Kinetic Energy is 1 KE = mv2 2 1 dx = m 2 dt 1 = mA2 ω2 cos2 (ωt + δ) 2 David S. Latchman

(1.4.5) ©2009

Oscillatory Motion The Potential Energy is

5

1 U = kx2 2 1 = kA2 sin2 (ωt + δ) 2 Adding eq. (1.4.5) and eq. (1.4.6) gives

(1.4.6)

1 E = kA2 2

1.4.4

(1.4.7)

Damped Harmonic Motion

dx (1.4.8) dt where b is the damping coefficient. The equation of motion for a damped oscillating system becomes dx d2 x − kx − b = m 2 (1.4.9) dt dt Solving eq. (1.4.9) goves x = Ae−αt sin (ω0 t + δ) (1.4.10)

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We find that

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Fd = −bv = −b

α=

b 2m r

k b2 − m 4m2

ω0 =

r

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=

1.4.5

(1.4.11)

ω20 −

b2 4m2

q = ω20 − α2

(1.4.12)

1 E = K + V(x) = mv(x)2 + V(x) 2

(1.4.13)

Small Oscillations

The Energy of a system is

We can solve for v(x), r

2 (E − V(x)) (1.4.14) m where E ≥ V(x) Let the particle move in the potential valley, x1 ≤ x ≤ x2 , the potential can be approximated by the Taylor Expansion " # " 2 # dV(x) 1 2 d V(x) V(x) = V(xe ) + (x − xe ) + (x − xe ) + ··· (1.4.15) dx x=xe 2 dx2 x=xe v(x) =

©2009

David S. Latchman

6 Classical Mechanics 2 At the points of inflection, the derivative dV/dx is zero and d V/dx2 is positive. This means that the potential energy for small oscillations becomes 1 V(x) u V(xe ) + k(x − xe )2 2 where

"

d2 V(x) k≡ dx2

(1.4.16)

# ≥0

(1.4.17)

x=xe

As V(xe ) is constant, it has no consequences to physical motion and can be dropped. We see that eq. (1.4.16) is that of simple harmonic motion.

Coupled Harmonic Oscillators

FT

1.4.6

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¨ + ω0 θ = 0

(1.4.18)

RA

We can express this in cartesian coordinates, x and y, where x = ` cos θ ≈ ` y = ` sin θ ≈ `θ

(1.4.19) (1.4.20)

y¨ + ω0 y = 0

(1.4.21)

eq. (1.4.18) becomes

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

(1.4.22)

D

k = mω20 =

This allows us to to create an equivalent three spring system to our coupled pendulum system. The equations of motion can be derived from the Lagrangian, where L=T−V  
2 1 2 1 2 1 1 2 1 2 = m y˙ 1 + m y˙ 2 − ky1 + κ y2 − y1 + ky2 2 2 2 2 2  1  
2  1  2 = m y˙1 + y˙2 2 − k y21 + y22 + κ y2 − y1 2 2

(1.4.23)

We can find the equations of motion of our system ! d ∂L ∂L = dt ∂ y˙ n ∂yn 1

(1.4.24)

Add figure with coupled pendulum-spring system

David S. Latchman

©2009

Oscillatory Motion The equations of motion are

7
m y¨ 1 = −ky1 + κ y2 − y1
m y¨ 2 = −ky2 + κ y2 − y1

(1.4.25) (1.4.26)

We assume solutions for the equations of motion to be of the form y1 = cos(ωt + δ1 ) y2 = B cos(ωt + δ2 ) y¨ 1 = −ωy1 y¨ 2 = −ωy2

(1.4.27)

Substituting the values for y¨ 1 and y¨ 2 into the equations of motion yields   k + κ − mω2 y1 − κy2 = 0   −κy1 + k + κ − mω2 y2 = 0

FT

We can get solutions from solving the determinant of the matrix
−κ k + κ − mω2
= 0 −κ k + κ − mω2 Solving the determinant gives  2   mω2 − 2mω2 (k + κ) + k2 + 2kκ = 0 This yields

(1.4.28) (1.4.29)

(1.4.30)

(1.4.31)

ω2 =

RA

 g  k   =   m ` ω2 =  (1.4.32)  g 2κ  k + 2κ  = +  m ` m We can now determine exactly how the masses move with each mode by substituting ω2 into the equations of motion. Where k We see that m

k + κ − mω2 = κ

(1.4.33)

D

Substituting this into the equation of motion yields y1 = y2

(1.4.34)

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

ω2 =

k+κ We see that m

k + κ − mω2 = −κ

(1.4.35)

Substituting this into the equation of motion yields y1 = −y2

(1.4.36)

Here the masses move out of phase with each other. In this case we see the presence of the spring constant, κ, which is expected as the spring playes a role. It is being stretched and compressed as our masses oscillate. ©2009

David S. Latchman

Classical Mechanics

8

1.4.7

Doppler Effect

The Doppler Effect is the shift in frequency and wavelength of waves that results from a source moving with respect to the medium, a receiver moving with respect to the medium or a moving medium. Moving Source If a source is moving towards an observer, then in one period, τ0 , it moves a distance of vs τ0 = vs / f0 . The wavelength is decreased by vs v − vs − f0 f0

(1.4.37)

  v v = f 0 λ0 v − vs

(1.4.38)

λ0 = λ − The frequency change is

FT

f0 =

Moving Observer As the observer moves, he will measure the same wavelength, λ, as if at rest but will see the wave crests pass by more quickly. The observer measures a modified wave speed. v0 = v + |vr | (1.4.39) The modified frequency becomes

  v0 vr = f0 1 + λ v

RA

f0 =

(1.4.40)

Moving Source and Moving Observer We can combine the above two equations v − vs f0 0 v = v − vr

λ0 =

(1.4.41) (1.4.42)

To give a modified frequency of

  v0 v − vr f = 0 = f0 λ v − vs

D

0

1.5

1.5.1

(1.4.43)

Rotational Motion about a Fixed Axis Moment of Inertia Z I=

1.5.2

R2 dm

(1.5.1)

Rotational Kinetic Energy 1 K = Iω2 2

David S. Latchman

(1.5.2) ©2009

Rotational Motion about a Fixed Axis

1.5.3

1.5.4

9

Parallel Axis Theorem I = Icm + Md2

(1.5.3)

τ=r×F τ = Iα

(1.5.4) (1.5.5)

Torque

1.5.5

FT

where α is the angular acceleration.

Angular Momentum

L = Iω

(1.5.6)

dL dt

(1.5.7)

RA

we can find the Torque

τ=

1.5.6

Kinetic Energy in Rolling

D

With respect to the point of contact, the motion of the wheel is a rotation about the point of contact. Thus 1 (1.5.8) K = Krot = Icontact ω2 2 Icontact can be found from the Parallel Axis Theorem. Icontact = Icm + MR2

(1.5.9)

Substitute eq. (1.5.8) and we have  1 Icm + MR2 ω2 2 1 1 = Icm ω2 + mv2 2 2

K=

(1.5.10)

The kinetic energy of an object rolling without slipping is the sum of hte kinetic energy of rotation about its center of mass and the kinetic energy of the linear motion of the object. ©2009

David S. Latchman

Classical Mechanics

10

1.6

Dynamics of Systems of Particles

1.6.1

Center of Mass of a System of Particles

Position Vector of a System of Particles R=

m1 r1 + m2 r2 + m3 r3 + · · · + mN rN M

(1.6.1)

Velocity Vector of a System of Particles dR dt m1 v1 + m2 v2 + m3 v3 + · · · + mN vN = M

FT

V=

(1.6.2)

Acceleration Vector of a System of Particles

dV dt m1 a1 + m2 a2 + m3 a3 + · · · + mN aN = M

1.7 1.7.1

RA

A=

(1.6.3)

Central Forces and Celestial Mechanics Newton’s Law of Universal Gravitation  GMm rˆ F=− r2

D



1.7.2

1.7.3

(1.7.1)

Potential Energy of a Gravitational Force U(r) = −

GMm r

(1.7.2)

Escape Speed and Orbits

The energy of an orbiting body is E=T+U GMm 1 = mv2 − 2 r David S. Latchman

(1.7.3) ©2009

Central Forces and Celestial Mechanics The escape speed becomes 1 GMm E = mv2esc − =0 2 RE

11 (1.7.4)

Solving for vesc we find r vesc =

1.7.4

2GM Re

(1.7.5)

Kepler’s Laws

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

FT

Second Law A line joining a planet and the sun sweeps out equal areas during equal intervals of time. Third Law The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. T2 =C R3

(1.7.6)

RA

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

1.7.5

Types of Orbits

The Energy of an Orbiting Body is defined in eq. (1.7.3), we can classify orbits by their eccentricities.

D

Circular Orbit A circular orbit occurs when there is an eccentricity of 0 and the orbital energy is less than 0. Thus 1 2 GM v − =E<0 2 r

(1.7.7)

The Orbital Velocity is r v=

GM r

(1.7.8)

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

2 1 GM − r a 

 (1.7.9)

where a is the semi-major axis ©2009

David S. Latchman

12 Classical Mechanics Parabolic 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 1 2 GM v − =E=0 2 r

(1.7.10)

The Orbital Velocity is r v = vesc =

2GM r

(1.7.11)

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

1.7.6

GM a

FT

v∞ =

(1.7.12)

Derivation of Vis-viva Equation

The total energy of a satellite is

RA

1 GMm E = mv2 − 2 r

(1.7.13)

For an elliptical or circular orbit, the specific energy is E=−

GMm 2a

(1.7.14)

Equating we get

2 1 v = GM − r a 



(1.7.15)

D

2

1.8

Three Dimensional Particle Dynamics

1.9

Fluid Dynamics

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

1.9.1

Equation of Continuity ρ1 v1 A1 = ρ2 v2 A2

David S. Latchman

(1.9.1) ©2009

Non-inertial Reference Frames

1.9.2

13

Bernoulli’s Equation 1 P + ρv2 + ρgh = a constant 2

(1.9.2)

1.10

Non-inertial Reference Frames

1.11

Hamiltonian and Lagrangian Formalism

1.11.1

Lagrange’s Function (L)

FT

L=T−V

(1.11.1)

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

1.11.2

Equations of Motion(Euler-Lagrange Equation)

1.11.3

!

RA

d ∂L ∂L = ∂q dt ∂q˙

Hamiltonian

H =T+V ˙ = pq˙ − L(q, q)

D

where

©2009

(1.11.2)

∂H = q˙ ∂p ∂H ∂L =− ∂q ∂x = −p˙

(1.11.3)

(1.11.4)

(1.11.5)

David S. Latchman

Classical Mechanics

D

RA

FT

14

David S. Latchman

©2009

Chapter

2

2.1

Electrostatics

2.1.1

Coulomb’s Law

FT

Electromagnetism

RA

The force between two charged particles, q1 and q2 is defined by Coulomb’s Law. ! q1 q2 1 F12 = rˆ12 4π0 r212

(2.1.1)

where 0 is the permitivitty of free space, where

0 = 8.85 × 10−12 C2 N.m2

Electric Field of a point charge

D

2.1.2

(2.1.2)

The electric field is defined by mesuring the magnitide and direction of an electric force, F, acting on a test charge, q0 . F (2.1.3) E≡ q0 The Electric Field of a point charge, q is E=

1 q rˆ 4π0 r2

(2.1.4)

In the case of multiple point charges, qi , the electric field becomes n 1 X qi E(r) = rˆi 4π0 i=1 r2i

(2.1.5)

16 Electric Fields and Continuous Charge Distributions

Electromagnetism

If a source is distributed continuously along a region of space, eq. (2.1.5) becomes Z 1 1 E(r) = rˆdq (2.1.6) 4π0 r2 If the charge was distributed along a line with linear charge density, λ, λ=

dq dx

(2.1.7)

The Electric Field of a line charge becomes λ rˆdx r2

Z

(2.1.8)

FT

1 E(r) = 4π0

line

RA

In the case where the charge is distributed along a surface, the surface charge density is, σ dq Q σ= = (2.1.9) A dA The electric field along the surface becomes Z 1 σ E(r) = rˆdA (2.1.10) 4π0 r2 Surface

In the case where the charge is distributed throughout a volume, V, the volume charge density is dq Q = (2.1.11) ρ= V dV The Electric Field is Z ρ 1 E(r) = rˆdV (2.1.12) 4π0 r2

D

Volume

2.1.3

Gauss’ Law

The electric field through a surface is I Φ= surface S

I dΦ =

E · dA

(2.1.13)

surface S

The electric flux through a closed surface encloses a net charge. I Q E · dA = 0

(2.1.14)

where Q is the charge enclosed by our surface. David S. Latchman

©2009

Electrostatics

2.1.4

17

Equivalence of Coulomb’s Law and Gauss’ Law

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

q 0

(2.1.15)

From the above, we see that the electric field is E=

2.1.5

q 4π0 r2

(2.1.16)

Electric Field due to a line of charge

FT

Consider an infinite rod of constant charge density, λ. The flux through a Gaussian cylinder enclosing the line of charge is Z Z Z Φ= E · dA + E · dA + E · dA (2.1.17) top surface

bottom surface

side surface

RA

At the top and bottom surfaces, the electric field is perpendicular to the area vector, so for the top and bottom surfaces, E · dA = 0 (2.1.18) At the side, the electric field is parallel to the area vector, thus E · dA = EdA

Thus the flux becomes,

Z

(2.1.19)

Z

Φ=

E · dA = E

dA

(2.1.20)

side sirface

D

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

(2.1.21)

Applying Gauss’ Law, we see that Φ = q/0 . The electric field becomes

2.1.6

E=

λ 2π0 r

(2.1.22)

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 our sphere is Q Q ρ= = 4 (2.1.23) 3 V πR 3 ©2009

David S. Latchman

Electromagnetism

18 The Electric Field due to a charge Q is E=

Q 4π0 r2

(2.1.24)

As the charge is evenly distributed throughout the sphere’s volume we can say that the charge density is dq = ρdV (2.1.25)

2.1.7

Electric Potential Energy

FT

where dV = 4πr2 dr. We can use this to determine the field inside the sphere by summing the effect of infinitesimally thin spherical shells Z E Z r dq E= dE = 2 0 0 4πr Z r ρ = dr 0 0 Qr (2.1.26) = 4 3 π R 0 3

2.1.8

1 qq0 r 4π0

RA

U(r) =

(2.1.27)

Electric Potential of a Point Charge

The electrical potential is the potential energy per unit charge that is associated with a static electrical field. It can be expressed thus U(r) = qV(r)

(2.1.28)

1 q 4π0 r

(2.1.29)

D

And we can see that

V(r) =

A more proper definition that includes the electric field, E would be Z V(r) = − E · d`

(2.1.30)

C

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 Z b Z a V(b) − V(a) = − E · d` + E · d` Z b =− E · d`

(2.1.31)

a

David S. Latchman

©2009

Electrostatics This can be further expressed

19

Z

b

V(b) − V(a) =

(∇V) · d`

(2.1.32)

a

And we can show that

2.1.9

E = −∇V

(2.1.33)

Electric Potential due to a line charge along axis

The charge density is

FT

Let us consider a rod of length, `, with linear charge density, λ. The Electrical Potential due to a continuous distribution is Z Z dq 1 V= dV = (2.1.34) 4π0 r dq = λdx

(2.1.35)

RA

Substituting this into the above equation, we get the electrical potential at some distance x along the rod’s axis, with the origin at the start of the rod. 1 dq 4π0 x 1 λdx = 4π0 x

dV =

This becomes

  λ x2 V= ln 4π0 x1

(2.1.36)

(2.1.37)

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

D

Now consider that we are some distance, y, from the axis of the rod of length, `. We again look at eq. (2.1.34), where r is the distance of the point P from the rod’s axis. Z dq 1 V= 4π0 r Z ` 1 λdx = 4π0 0 x2 + y2
12    12 ` λ = ln x + x2 + y2 4π0 0    12  λ = ln ` + `2 + y2 − ln y 4π0 
1   ` + `2 + y2 2  λ  = ln  (2.1.38)  4π0  d ©2009

David S. Latchman

Electromagnetism

20

2.2

Currents and DC Circuits

2

2.3

Magnetic Fields in Free Space

3

Lorentz Force

4

5

2.6 6

Maxwell’s Equations and their Applications

Electromagnetic Waves

D

2.7

Induction

RA

2.5

FT

2.4

7

2.8

AC Circuits

8

2.9

Magnetic and Electric Fields in Matter

9 David S. Latchman

©2009

Capacitance

2.10

21

Capacitance Q = CV

2.11

(2.10.1)

Energy in a Capacitor Q2 2C CV 2 = 2 QV = 2

Energy in an Electric Field

U 0 E2 = volume 2

(2.12.1)

dQ dt

(2.13.1)

J · dA

(2.14.1)

u≡

2.13

Current

I≡

Current Destiny

D

2.14

2.15

Z I= A

Current Density of Moving Charges J=

2.16

I = ne qvd A

(2.15.1)

Resistance and Ohm’s Law R≡

©2009

(2.11.1)

RA

2.12

FT

U=

V I

(2.16.1) David S. Latchman

Electromagnetism

22

2.17

Resistivity and Conductivity L A

(2.17.1)

E = ρJ

(2.17.2)

J = σE

(2.17.3)

R=ρ

Power

FT

2.18

P = VI

Kirchoff’s Loop Rules

Write Here

2.20

Kirchoff’s Junction Rule

Write Here

RC Circuits

D

2.21

RA

2.19

(2.18.1)

E − IR −

2.22

Maxwell’s Equations

2.22.1

Integral Form

Q =0 C

(2.21.1)

Gauss’ Law for Electric Fields w closed surface

David S. Latchman

E · dA =

Q 0

(2.22.1)

©2009

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

23 w

B · dA = 0

(2.22.2)

closed surface

Amp`ere’s Law

z

d w B · ds = µ0 I + µ0 0 dt

E · dA

(2.22.3)

surface

Faraday’s Law

z

d w E · ds = − dt

B · dA

(2.22.4)

2.22.2

Differential Form

Gauss’ Law for Electric Fields

FT

surface

ρ 0

(2.22.5)

∇·B=0

(2.22.6)

∇·E=

Gauss’ Law for Magnetism

RA

Amp`ere’s Law

∇ × B = µ0 J + µ0 0

Faraday’s Law

∇·E=−

∂B ∂t

(2.22.7)

(2.22.8)

Speed of Propagation of a Light Wave

D

2.23

∂E ∂t

c= √

1 µ0 0

In a material with dielectric constant, κ, √ c c κ = n

(2.23.1)

(2.23.2)

where n is the refractive index.

2.24

Relationship between E and B Fields E = cB E·B=0

©2009

(2.24.1) (2.24.2) David S. Latchman

Electromagnetism

24

2.25

Energy Density of an EM wave 1 B2 u= + 0 E2 2 µ0

2.26

! (2.25.1)

Poynting’s Vector 1 E×B µ0

(2.26.1)

D

RA

FT

S=

David S. Latchman

©2009

Chapter

3

3.1

Wave Properties

1

2

3.3

Interference

D

3

Superposition

RA

3.2

3.4

Diffraction

4

3.5

Geometrical Optics

5

3.6 6

FT

Optics & Wave Phonomena

Polarization

Optics & Wave Phonomena

26

3.7

Doppler Effect

7

3.8

Snell’s Law

3.8.1

Snell’s Law n1 sin θ1 = n2 sin θ2

Critical Angle and Snell’s Law

FT

3.8.2

(3.8.1)

The critical angle, θc , for the boundary seperating two optical media is the smallest angle of incidence, in the medium of greater index, for which light is totally refelected. From eq. (3.8.1), θ1 = 90 and θ2 = θc and n2 > n1 .

(3.8.2)

D

RA

n1 sin 90 = n2 sinθc n1 sin θc = n2

David S. Latchman

©2009

Chapter

4

4.1

FT

Thermodynamics & Statistical Mechanics Laws of Thermodynamics

1

2

4.3

Equations of State

D

3

Thermodynamic Processes

RA

4.2

4.4

Ideal Gases

4

4.5

Kinetic Theory

5

4.6 6

Ensembles

Thermodynamics & Statistical Mechanics

28

4.7

Statistical Concepts and Calculation of Thermodynamic Properties

7

4.8

Thermal Expansion & Heat Transfer

4.9

FT

8

Heat Capacity

  Q = C T f − Ti

(4.9.1)

4.10

RA

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

Specific Heat Capacity

  Q = cm T f − ti

(4.10.1)

D

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

4.11

4.12

Heat and Work Z W=

Vf

PdV

(4.11.1)

Vi

First Law of Thermodynamics dEint = dQ − dW

(4.12.1)

where dEint is the internal energy of the system, dQ is the Energy added to the system and dW is the work done by the system. David S. Latchman

©2009

Work done by Ideal Gas at Constant Temperature

4.12.1

29

Special Cases to the First Law of Thermodynamics

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

(4.12.2)

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

FT

Constant Volume (Isochoric) Process If the volume is held constant, then the system can do no work, δW = 0, thus ∆Eint = Q (4.12.3) If heat is added to the system, the temperature increases. Conversely, if heat is removed from the system the temperature decreases. Closed Cycle In this situation, after certain interchanges of heat and work, the system comes back to its initial state. So ∆Eint remains the same, thus ∆Q = ∆W

(4.12.4)

RA

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 (4.12.5)

4.13

Work done by Ideal Gas at Constant Temperature

D

Starting with eq. (4.11.1), we substitute the Ideal gas Law, eq. (4.15.1), to get

4.14

Vf

Z W = nRT

Vi

= nRT ln

dV V

Vf Vi

(4.13.1)

Heat Conduction Equation

The rate of heat transferred, H, is given by H=

Q TH − TC = kA t L

(4.14.1)

where k is the thermal conductivity. ©2009

David S. Latchman

Thermodynamics & Statistical Mechanics

30

4.15

Ideal Gas Law PV = nRT

(4.15.1)

where n = Number of moles P = Pressure V = Volume T = Temperature and R is the Universal Gas Constant, such that

We can rewrite the Ideal gas Law to say

FT

R ≈ 8.314 J/mol. K

PV = NkT

(4.15.2)

where k is the Boltzmann’s Constant, such that

4.16

R ≈ 1.381 × 10−23 J/K NA

RA

k=

Stefan-Boltzmann’s FormulaStefan-Boltzmann’s Equation P(T) = σT4

RMS Speed of an Ideal Gas

D

4.17

4.18

r vrms =

3RT M

(4.17.1)

Translational Kinetic Energy 3 K¯ = kT 2

4.19

(4.16.1)

(4.18.1)

Internal Energy of a Monatomic gas 3 Eint = nRT 2

David S. Latchman

(4.19.1) ©2009

Molar Specific Heat at Constant Volume

4.20

31

Molar Specific Heat at Constant Volume

Let us define, CV such that Q = nCV ∆T

(4.20.1)

Substituting into the First Law of Thermodynamics, we have ∆Eint + W = nCV ∆T

(4.20.2)

At constant volume, W = 0, and we get

Substituting eq. (4.19.1), we get

1 ∆Eint n ∆T

FT

CV =

4.21

RA

3 CV = R = 12.5 J/mol.K 2

(4.20.3)

(4.20.4)

Molar Specific Heat at Constant Pressure

Starting with

D

and

4.22

Q = nCp ∆T

(4.21.1)

∆Eint = Q − W ⇒ nCV ∆T = nCp ∆T + nR∆T ∴ CV = Cp − R

(4.21.2)

Equipartition of Energy ! f CV = R = 4.16 f J/mol.K 2

(4.22.1)

where f is the number of degrees of freedom. ©2009

David S. Latchman

0 2 3n − 5 3n − 6

3 R 2 5 R 2 3R 3R

CV

5 R 2 7 R 2 4R 4R

CP = CV + R

Predicted Molar Specific Heats 3 5 6 6

FT

RA

Degrees of Freedom 0 2 3 3

Translational Rotational Vibrational Total ( f ) 3 3 3 3

©2009

David S. Latchman

Molecule Monatomic Diatomic Polyatomic (Linear) Polyatomic (Non-Linear)

Table 4.22.1: Table of Molar Specific Heats

D

Thermodynamics & Statistical Mechanics 32

Adiabatic Expansion of an Ideal Gas

4.23

Adiabatic Expansion of an Ideal Gas

where γ = CCVP . We can also write

4.24

33

PV γ = a constant

(4.23.1)

TV γ−1 = a constant

(4.23.2)

Second Law of Thermodynamics

D

RA

FT

Something.

©2009

David S. Latchman

Thermodynamics & Statistical Mechanics

D

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FT

34

David S. Latchman

©2009

Chapter

5

5.1

Fundamental Concepts

1

Schrodinger ¨ Equation

RA

5.2

FT

Quantum Mechanics

Let us define Ψ to be

Ψ = Ae−iω(t− v ) x

(5.2.1)

Simplifying in terms of Energy, E, and momentum, p, we get Ψ = Ae−

i(Et−px) ~

(5.2.2)

D

We obtain Schrodinger’s Equation from the Hamiltonian ¨ H =T+V

(5.2.3)

To determine E and p,

p2 ∂2 Ψ = − Ψ ~2 ∂x2 ∂Ψ iE = Ψ ~ ∂t

and H=

p2 +V 2m

(5.2.4) (5.2.5)

(5.2.6)

This becomes EΨ = HΨ

(5.2.7)

36

~ ∂Ψ ∂Ψ p2 Ψ = −~2 2 i ∂t ∂x The Time Dependent Schrodinger’s ¨ Equation is EΨ = −

i~

Quantum Mechanics

2

∂Ψ ~ 2 ∂2 Ψ + V(x)Ψ =− 2m ∂x2 ∂t

(5.2.8)

The Time Independent Schrodinger’s ¨ Equation is EΨ = −

(5.2.9)

Infinite Square Wells

FT

5.2.1

~ 2 ∂2 Ψ + V(x)Ψ 2m ∂x2

Let us consider a particle trapped in an infinite potential well of size a, such that ( 0 for 0 < x < a V(x) = ∞ for |x| > a,

RA

so that a nonvanishing force acts only at ±a/2. An energy, E, is assigned to the system such that the kinetic energy of the particle is E. Classically, any motion is forbidden outside of the well because the infinite value of V exceeds any possible choice of E. Recalling the Schrodinger Time Independent Equation, eq. (5.2.9), we substitute V(x) ¨ and in the region (−a/2, a/2), we get

This differential is of the form

D

where

~2 d2 ψ = Eψ − 2m dx2

(5.2.10)

d2 ψ + k2 ψ = 0 2 dx

(5.2.11)

r k=

2mE ~2

(5.2.12)

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 ( A cos kx + B sin kx for 0 < x < a ψ(x) = 0 for |x| > a We have known boundary conditions for our square well. ψ(0) = ψ(a) = 0 David S. Latchman

(5.2.13) ©2009

Schr¨odinger Equation It shows that

37 ⇒ A cos 0 + B sin 0 = 0 ∴A=0

(5.2.14)

We are now left with B sin ka = 0 ka = 0; π; 2π; 3π; · · · (5.2.15)

kn =

FT

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

for n = 1, 2, 3, · · ·

Substituting this result into eq. (5.2.12) gives nπ kn = = a

RA

Solving for En gives

√ 2mEn ~

n2 π2 ~2 2ma2 We cna now solve for B by normalizing the function Z a a |B|2 sin2 kxdx = |A|2 = 1 2 0 2 So |A|2 = a En =

(5.2.16)

(5.2.17)

(5.2.18)

D

So we can write the wave function as

5.2.2

r ψn (x) =

  2 nπx sin a a

(5.2.19)

Harmonic Oscillators

Classically, the harmonic oscillator has a potential energy of 1 V(x) = kx2 2

(5.2.20)

So the force experienced by this particle is F=− ©2009

dV = −kx dx

(5.2.21) David S. Latchman

38 Quantum Mechanics where k is the spring constant. The equation of motion can be summed us as d2 x m 2 = −kx dt

(5.2.22)

  x(t) = A cos ω0 t + φ

(5.2.23)

And the solution of this equation is

where the angular frequency, ω0 is r ω0 =

k m

(5.2.24)

With some manipulation, we get

FT

The Quantum Mechanical description on the harmonic oscillator is based on the eigenfunction solutions of the time-independent Schrodinger’s equation. By taking V(x) ¨ from eq. (5.2.20) we substitute into eq. (5.2.9) to get !   d2 ψ 2m k 2 mk 2 2E x − E x − ψ = ψ = dx2 ~2 2 ~2 k

RA

√ r  d2 ψ  mk 2 2E m   ψ =  x − √ 2 ~ ~ k  mk dx ~

This step allows us to to keep some of constants out of the way, thus giving us √ mk 2 x (5.2.25) ξ2 = ~r 2E m 2E and λ = = (5.2.26) ~ k ~ω0

D

This leads to the more compact

 d2 ψ  2 = ξ − λ ψ dξ2

(5.2.27)

where the eigenfunction ψ will be a function of ξ. λ assumes an eigenvalue anaglaous to E. From eq. (5.2.25), we see that the maximum value can be determined to be √ mk 2 2 ξmax = A ~ Using the classical connection between A and E, allows us to say √ mk 2E 2 ξmax = =λ ~ k

David S. Latchman

(5.2.28)

(5.2.29) ©2009

Schr¨odinger Equation 39 From eq. (5.2.27), we see that in a quantum mechanical oscillator, there are nonvanishing solutions in the forbidden regions, unlike in our classical case. A solution to eq. (5.2.27) is ψ(ξ) = e−ξ /2 2

(5.2.30)

where

and

dψ 2 = −ξe−ξ /2 dξ   2 dψ 2 −xi2 /2 −ξ2 /2 2 = ξ e − e = ξ − 1 e−ξ /2 2 dξ

This gives is a special solution for λ where

FT

λ0 = 1

(5.2.31)

Thus eq. (5.2.26) gives the energy eigenvalue to be E0 =

~ω0 ~ω0 λ0 = 2 2

(5.2.32)

The eigenfunction e−ξ /2 corresponds to a normalized stationary-state wave function 2

! 18

√

e−

mk x2 /2~ −iE0 t/~

RA

mk Ψ0 (x, t) = 2 2 π~

e

(5.2.33)

This solution of eq. (5.2.27) 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 the zero-point energy of the system.

5.2.3

Finite Square Well

D

For the Finite Square Well, we have a potential region where ( −V0 for −a ≤ x ≤ a V(x) = 0 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 d2 ψ = Eψ 2m dx2 d2 ψ ⇒ 2 = κ2 ψ √ dx −2mE κ= ~ −

where ©2009

David S. Latchman

Quantum Mechanics

40 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 physically realizable function. So we can drop it to get ψ(x) = Beκx

for x < −a

(5.2.34)

Region II: −a < x < a In this region, our potential is V(x) = V0 . Substitutin this into the Schrodinger’s Equation, eq. (5.2.9), gives ¨

FT

~2 d2 ψ − V0 ψ = Eψ − 2m dx2 d2 ψ or = −l2 ψ 2 dx p 2m (E + V0 ) where l ≡ ~

(5.2.35)

RA

We notice that E > −V0 , making l real and positive. Thus our general solution becomes ψ(x) = C sin(lx) + D cos(lx) for −a < x < a (5.2.36) Region 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

(5.2.37)

for x < a for 0 < x < a for x > a

(5.2.38)

D

This gives us

5.2.4

 κx  Be    D cos(lx) ψ(x) =     Fe−κx

Hydrogenic Atoms

c

5.3

Spin

3 David S. Latchman

©2009

Angular Momentum

5.4

41

Angular Momentum

4

5.5

Wave Funtion Symmetry

5

Elementary Perturbation Theory

FT

5.6

D

RA

6

©2009

David S. Latchman

Quantum Mechanics

D

RA

FT

42

David S. Latchman

©2009

Chapter

6

6.1

FT

Atomic Physics Properties of Electrons

1

Bohr Model

RA

6.2

To understand the Bohr Model of the Hydrogen atom, we will take advantage of our knowlegde of the wavelike properties of matter. As we are building on a classical model 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, orbiting a proton. The cetripetal force is equal to the Coulomb Force. Thus

D

1 e2 me v2 = 4π0 r2 r

(6.2.1)

The Total Energy is the sum of the potential and kinetic energies, so p2 E=K+U = − | f race2 4π0 r 2me

(6.2.2)

We can further reduce this equation by subsituting the value of momentum, which we find to be p2 1 e2 = me v2 = (6.2.3) 2me 2 8π0 r Substituting this into eq. (6.2.2), we get E=

e2 e2 e2 − =− 8π0 r 4π0 r 8π0 r

(6.2.4)

At this point our classical description must end. An accelerated charged particle, like one moving in circular motion, radiates energy. So our atome here will radiate energy

44 Atomic Physics and 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 stationary states take discreet values. 2. The energy of these stationary states are determined by their angular momentum which must take on quantized values of ~. L = n~

(6.2.5)

We can find the angular momentum of a circular orbit.

FT

L = m3 vr

(6.2.6)

From eq. (6.2.1) we find v and by substitution, we find L. r

L=e Solving for r, gives

m3 r 4π0

(6.2.7)

L2 me e2 /4π0

(6.2.8)

n2 ~2 = n2 a0 me e2 /4π0

(6.2.9)

RA

r=

We apply the condition from eq. (6.2.5) rn =

where a0 is the Bohr radius.

a0 = 0.53 × 10−10 m

(6.2.10)

D

Having discreet values for the allowed radii means that we will also have discreet values for energy. Replacing our value of rn into eq. (6.2.4), we get ! me e2 13.6 = − 2 eV (6.2.11) En = − 2 2n 4π0 ~ n

6.3

Energy Quantization

3

6.4

Atomic Structure

4 David S. Latchman

©2009

Atomic Spectra

6.5 6.5.1

45

Atomic Spectra Rydberg’s Equation   1 1 1 = RH 02 − 2 λ n n

(6.5.1)

where RH is the Rydberg constant.

6.6

Selection Rules

6

6.7.1

Black Body Radiation

RA

6.7

FT

For the Balmer Series, n0 = 2, which determines the optical wavelengths. For n0 = 3, we get the infrared or Paschen series. The fundamental n0 = 1 series falls in the ultraviolet region and is known as the Lyman series.

Plank Formula

f3 8π~ u( f, T) = 3 h f /kT c e −1

Stefan-Boltzmann Formula

D

6.7.2

6.7.3

6.7.4

(6.7.1)

P(T) = σT4

(6.7.2)

Wein’s Displacement Law λmax T = 2.9 × 10−3 m.K

(6.7.3)

Classical and Quantum Aspects of the Plank Equation

Rayleigh’s Equation 8π f 2 u( f, T) = 3 kT c ©2009

(6.7.4) David S. Latchman

46 Atomic Physics We can get this equation from Plank’s Equation, eq. (6.7.1). This equation is a classical one and does not contain Plank’s constant in it. For this case we will look at the situation where h f < kT. In this case, we make the approximation ex ' 1 + x

(6.7.5)

Thus the demonimator in eq. (6.7.1) becomes eh f /kT − 1 ' 1 +

hf hf −1= kT kT

(6.7.6)

Thus eq. (6.7.1) takes the approximate form 8πh 3 kT 8π f 2 f = 3 kT c3 hf c

FT

u( f, T) '

(6.7.7)

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

RA

At large frequencies, where h f > kT, quantum effects become apparent. We can estimate that eh f /kT − 1 ' eh f /kT (6.7.8) Thus eq. (6.7.1) becomes

u( f, T) '

6.8.1

(6.7.9)

X-Rays

D

6.8

8πh 3 −h f /kT f e c3

Bragg Condition 2d sin θ = mλ

(6.8.1)

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

6.8.2

The Compton Effect

The Compton Effect deals with the scattering of monochromatic X-Rays by atomic targets and the observation that the wavelength of the scattered X-ray is greater than the incident radiation. The photon energy is given by E = hυ = David S. Latchman

hc λ

(6.8.2) ©2009

Atoms in Electric and Magnetic Fields The photon has an associated momentum

47

E

= pc E hυ h ⇒p = = = c c λ

(6.8.3) (6.8.4)

The Relativistic Energy for the electron is (6.8.5)

p − p0 = P

(6.8.6)

p2 − 2p · p0 + p02 = P2

(6.8.7)

where Squaring eq. (6.8.6) gives

Recall that E = pc and E 0 = cp0 , we have

FT

E2 = p2 c2 + m2e c4

c2 p2 − 2c2 p · p0 + c2 p02 = c2 P2 E 2 − 2E E 0 cos θ + E 02 = E2 − m2e c4 Conservation of Energy leads to

E + me c2 = E 0 + E

(6.8.9)

E − E 0 = E − me c2 E 2 − 2E E 0 + E 0 = E2 − 2Eme c2 + m2e c4 2E E 0 − 2E E 0 cos θ = 2Eme c2 − 2m2e c4

(6.8.10) (6.8.11)

RA

Solving

(6.8.8)

Solving leads to

D

∆λ = λ0 − λ =

where λc =

6.9 6.9.1

h me c

h (1 − cos θ) me c

(6.8.12)

is the Compton Wavelength. λc =

h = 2.427 × 10−12 m me c

(6.8.13)

Atoms in Electric and Magnetic Fields The Cyclotron Frequency

A test charge, q, with velocity v enters a uniform magnetic field, B. The force acting on the charge will be perpendicular to v such that FB = qv × B ©2009

(6.9.1) David S. Latchman

6.9.2

FT

48 Atomic Physics or more simply FB = qvB. As this traces a circular path, from Newton’s Second Law, we see that mv2 = qvB (6.9.2) FB = R Solving for R, we get mv R= (6.9.3) qB We also see that qB (6.9.4) f = 2πm The frequency is depends on the charge, q, the magnetic field strength, B and the mass of the charged particle, m.

Zeeman Effect

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

RA

In the Zeeman experiment, a sodium flame was placed in a magnetic field and its spectrum observed. In the presence of the field, a spectral line of frequency, υ0 was split into three components, υ0 − δυ, υ0 and υ0 + δυ. A classical analysis of this effect allows 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 which the electron motion can be referred. The motion of an electron can be attributed to a simple harmonic motion under a binding force −kr, where the frequency is r k 1 (6.9.5) υ0 = 2π me

D

The magnetic field subjects the electron to an additional Lorentz Force, −ev × B. This produces two different values for the angular velocity. v = 2πrυ

The cetripetal force becomes me v2 = 4π2 υ2 rme r

Thus the certipetal force is 4π2 υ2 rme = 2πυreB + kr

for clockwise motion

4π2 υ2 rme = −2πυreB + kr

for counterclockwise motion

We use eq. (6.9.5), to emiminate k, to get eB υ − υ0 = 0 2πme eB υ2 + υ − υ0 = 0 2πme

υ2 −

David S. Latchman

(Clockwise) (Counterclockwise) ©2009

Atoms in Electric and Magnetic Fields 49 As we have assumed a small Lorentz force, we can say that the linear terms in υ are small comapred to υ0 . Solving the above quadratic equations leads to eB 4πme eB υ = υ0 − 4πme υ = υ0 +

for clockwise motion

(6.9.6)

for counterclockwise motion

(6.9.7)

We note that the frequency shift is of the form δυ =

eB 4πme

(6.9.8)

6.9.3

Franck-Hertz Experiment

FT

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

D

RA

The Franck-Hertz experiment, performed in 1914 by J. Franck and G. L. Hertz, measured the colisional excitation of atoms. Their experiement studied the current of electrons in a tub of mercury vapour which revealed an abrupt change in the current at certain critical values of the applied voltage.1 They interpreted this observation as evidence of a threshold for inelastic scattering in the colissions of electrons in mercury atoms.The bahavior of the current was an indication that electrons could lose a discreet amount of energy and excite mercury atoms in their passage through the mercury vapour. These observations constituted a direct and decisive confirmation of the existence os quantized energy levels in atoms.

1

Put drawing of Franck-Hertz Setup

©2009

David S. Latchman

Atomic Physics

D

RA

FT

50

David S. Latchman

©2009

Chapter

7

7.1 7.1.1

FT

Special Relativity Introductory Concepts

Postulates of Special Relativity

RA

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

(7.1.1)

u2 c2

Time Dilation

D

7.2

1 γ= q 1−

∆t = γ∆t0

(7.2.1)

where ∆t0 is the time measured at rest relative to the observer, ∆t is the time measured in motion relative to the observer.

7.3

Length Contraction L=

L0 γ

(7.3.1)

where L0 is the length of an object observed at rest relative to the observer and L is the length of the object moving at a speed u relative to the observer.

Special Relativity

52

7.4

Simultaneity

4

7.5 7.5.1

Energy and Momentum Relativistic Momentum & Energy

FT

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

7.5.2

(7.5.1)

E = γmc2

(7.5.2)

RA

Relativistic Energy

p¯ = γmv¯

Lorentz Transformations (Momentum & Energy)

E = γ px − β c 0 py = py

D

p0x

7.5.3





= pz   E E =γ − βpx c c p0z 0

(7.5.3) (7.5.4) (7.5.5) (7.5.6)

Relativistic Kinetic Energy K = E − mc2    1 = mc2  q  1−
= mc2 γ − 1

David S. Latchman

(7.5.7)

v2 c2

   − 1 

(7.5.8) (7.5.9) ©2009

Four-Vectors and Lorentz Transformation

7.5.4

53

Relativistic Dynamics (Collisions) ∆E = γ ∆Px − β c 0 ∆P y = ∆P y 

∆P0x

 (7.5.10) (7.5.11)

∆P0z 0

= ∆Pz   ∆E ∆E =γ − β∆Px c c

(7.5.13)

Four-Vectors and Lorentz Transformation

FT

7.6

(7.5.12)

We can represent an event in S with the column matrix, s,    x   y   s =    z  ict

(7.6.1)

RA

A different Lorents frame, S0 , corresponds to another set of space time axes so that  0   x   y0    s0 =  0  (7.6.2)  z   0  ict

D

The Lorentz Transformation is related by the matrix  0   0 0 iγβ  x   γ  y0   0 1 0 0     0  =   z   0 0 1 0  0   ict −iγβ 0 0 γ

    x    y          z    ict

(7.6.3)

We can express the equation in the form s0 = L s

(7.6.4)

The matrix L contains all the information needed to relate position four–vectors for any given event as observed in the two Lorentz frames S and S0 . If we evaluate    x  h i  y  T 2 2 2 2 2   x y z ict s s= (7.6.5)  z  = x + y + z − c t   ict

Similarly we can show that s0T s0 = x02 + y02 + z02 − c2 t02 ©2009

(7.6.6) David S. Latchman

54 Special Relativity We can take any collection of four physical quantities to be four vector provided that they transform to another Lorentz frame. Thus we have    bx   b    b =  y   bz    ibt

(7.6.7)

this can be transformed into a set of quantities of b0 in another frame S0 such that it satisfies the transformation b0 = L b (7.6.8) Looking at the momentum-Energy four vector, we have       

FT

  px  p  p =  y  pz  iE/c

(7.6.9)

Applying the same transformation rule, we have p0 = L p

(7.6.10)

RA

We can also get a Lorentz-invariation relation between momentum and energy such that p0T p0 = pT p (7.6.11) The resulting equality gives

02 02 p02 x + p y + pz −

7.8

(7.6.12)

Velocity Addition

D

7.7

E02 E2 2 2 2 = p + p + p − x y z c2 c2

v0 =

v−u 1 − uv c2

(7.7.1)

Relativistic Doppler Formula r υ¯ = υ0

c+u c−u

r let r =

c−u c+u

(7.8.1)

We have υ¯ receding = rυ0 υ0 υ¯ approaching = r David S. Latchman

red-shift (Source Receding)

(7.8.2)

blue-shift (Source Approaching)

(7.8.3) ©2009

Lorentz Transformations

7.9

55

Lorentz Transformations

Given two reference frames S(x, y, z, t) and S0 (x0 , y0 , z0 , t0 ), where the S0 -frame is moving in the x-direction, we have,

7.10

Space-Time Interval

x = (x0 − ut0 ) y = y0 y0 = y   u 0 0 t = γ t + 2x c

FT

x0 = γ (x − ut) y0 = y z0 = y   u 0 t = γ t − 2x c


(∆S)2 = (∆x)2 + ∆y 2 + (∆z)2 − c2 (∆t)2

(7.9.1) (7.9.2) (7.9.3) (7.9.4)

(7.10.1)

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

c2 ∆t2 > ∆r2

(7.10.2)

∆S2 > 0

(7.10.3)

RA

Time-like Interval

When two events are separated by a time-like interval, there is a cause-effect relationship between the two events. Light-like Interval

c2 ∆t2 = ∆r2

(7.10.4)

S =0

(7.10.5)

c2 ∆t2 < ∆r2 ∆S < 0

(7.10.6) (7.10.7)

D

2

Space-like Intervals

©2009

David S. Latchman

Special Relativity

D

RA

FT

56

David S. Latchman

©2009

Chapter

8

8.1.1

Data and Error Analysis Addition and Subtraction

x=a+b−c

(8.1.1)

(δx)2 = (δa)2 + (δb)2 + (δc)2

(8.1.2)

RA

8.1

FT

Laboratory Methods

The Error in x is

Multiplication and Division

D

8.1.2

x=

a×b c

(8.1.3)

The error in x is

8.1.3



δx x

2

δa = a 

2

δb + b

!2

δc + c 

2 (8.1.4)

Exponent - (No Error in b)

The Error in x is

x = ab

(8.1.5)

  δx δa =b x a

(8.1.6)

Laboratory Methods

58

8.1.4

Logarithms

Base e x = ln a

(8.1.7)

We find the error in x by taking the derivative on both sides, so d ln a · δa da 1 = · δa a δa = a

Base 10

FT

δx =

x = log10 a The Error in x can be derived as such

=

ln a ln 10

δa da 1 δa = ln 10 a δa = 0.434 a

(8.1.10)

Antilogs

D

8.1.5

(8.1.9)

d(log a) δa da

RA

δx =

(8.1.8)

Base e

x = ea

(8.1.11)

ln x = a ln e = a

(8.1.12)

We take the natural log on both sides.

Applaying the same general method, we see d ln x δx = δa dx δx ⇒ = δa x David S. Latchman

(8.1.13) ©2009

Instrumentation Base 10

59

x = 10a

(8.1.14)

We follow the same general procedure as above to get log x = a log 10 log x δx = δa dx 1 d ln a δx = δa ln 10 dx δx = ln 10δa x

FT

8.2

Instrumentation

3

8.4

RA

2

8.3

(8.1.15)

Radiation Detection

Counting Statistics

D

Let’s assume that for a particular experiment, we are making countung measurements for a radioactive source. In this experiment, we recored N counts in time T. The ¯ counting 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. So √ σ= N (8.4.1) Thus we can report our results as Number of counts = N ±

√ N

(8.4.2)

We can find the count rate by dividing by T, so √ N N R= ± T T ©2009

(8.4.3) David S. Latchman

60 The fractional uncertainty of our count is rate.

δN . N

δN T N T

δR = R

Laboratory Methods We can relate this in terms of the count

δN N √ N = N 1 = N =

(8.4.4)

We see that our uncertainty decreases as we take more counts, as to be expected.

Interaction of Charged Particles with Matter

FT

8.5 5

6

8.7

Lasers and Optical Interferometers

RA

8.6

Dimensional Analysis

D

Dimensional Analysis is used to understand physical situations involving a mis of different types of physical quantities. The dimensions of a physical quantity are associated 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.

8.8

Fundamental Applications of Probability and Statistics

8

David S. Latchman

©2009

Chapter

9

9.1

FT

Sample Test Period of Pendulum on Moon

The period of the pendulum, T, is

s

` g

RA

T = 2π

(9.1.1)

where ` is the length of the pendulium string. The relationship between the weight of an object on the Earth, We , and the Moon, Wm , is Wm =

We 6

(9.1.2)

D

From eq. (9.1.2), we can determine the acceleration due to gravity on the Moon and on the Earth; we use the same subscript notation as above. gm =

ge 6

(9.1.3)

On Earth, the period of the pendulum, Te , is one second. From eq. (9.1.1), the equation for the pendulum’s period on Earth is s Te = 2π

` = 1s ge

(9.1.4)

and similarly for the moon, the period becomes s Tm = 2π

` gm

(9.1.5)

Sample Test

62 Substituting eq. (9.1.3) into eq. (9.1.5) gives s Tm = 2π =

√

` gm

6 Te =

√ 6s

Answer: (D)

9.2

Work done by springs in series

FT

Hooke’s Law tells us that the extension on a spring is proportional to the force applied. F = −kx

(9.2.1)

Springs in series follow the same rule for capacitors, see ??. The spring constants are related to each other by 1 (9.2.2) k1 = k2 3

RA

The springs are massless so we can assume that the weight is transmitted evenly along both springs, thus from Hooke’s Law the extension is F1 = −k1 x1 = F2 = −k2 x2

(9.2.3)

where k1 and k2 are the spring constants for the springs S1 and S2 respectively. Thus we see k1 x2 1 (9.2.4) = = k2 x1 3

D

The work done in stretching a spring or its potential energy is 1 W = kx2 2

(9.2.5)

Thus

1 2 k1 x1 W1 2 = 1 2 W2 k2 x 2 2   k1 x1 2 = · k2 x2 =3

(9.2.6)

Answer: (D) David S. Latchman

©2009

Central Forces I

9.3

63

Central Forces I

We are given a central force field where k r

(9.3.1)

L=r×p

(9.3.2)

τ = r × F = r × p˙

(9.3.3)

V(r) = − The Angular Momentum of an object is

From eqs. (9.3.2) and (9.3.3), we see that

FT

and the torque is defined

τ=

dL dt

(9.3.4)

RA

We see that if τ = 0, then L is constant and therefore conserved. This can occur if r˙ = 0, F˙ = 0 or F ∝ r. From 9.3.1, we can determine the force acting on the object since F=−

dV k = 2 dr r

(9.3.5)

As our force is a central force, the force acts in the direction of our radius vector. Thus the torque becomes

D

τ = r × F = rF cos 0 =0

We see that this means that our angular momentum is constant. L = constant

(9.3.6)

A constant angular momentum means that r and v remain unchanged. The total mechanical energy is the sum of the kinetic and potential energies. E = KE + PE 1 k = mv2 + 2 2 r

(9.3.7)

Both the kinetic and potential energies will remain constant and thus the total mechanical energy is also conserved. Answer: (C) ©2009

David S. Latchman

Sample Test

64

9.4

Central Forces II

The motion of particle is governed by its potential energy and for a conservative, central force the potential energy is

V(r) = −

k r

(9.4.1)

FT

we have shown in the above question that the angular momentum, L, is conserved. We can define three types of orbits given k and E.

k

Orbit

Total Energy

Ellipse k>0 Parabola k>0 Hyperbola k > 0 or k < 0

E<0 E=0 E>0

RA

Table 9.4.1: Table of Orbits

From, table 9.4.1, we expect the orbit to be elliptical; this eliminates answers (C), (D) and (E). For an elliptical orbit, the total energy is

D

E=−

k 2a

(9.4.2)

where a is the length of the semimajor axis. In the case of a circular orbit of radius, r, eq. (9.4.2) becomes E=−

k 2r

(9.4.3)

Recalling eq. (9.3.1), we see 1 E = V(r) = −K 2

(9.4.4)

This is the minimum energy the system can have resulting in a circular orbit. Answer: (A) David S. Latchman

©2009

Electric Potential I

9.5

65

Electric Potential I +z

P2 b r2

P1

FT

r1

b

RA

Figure 9.5.1: Diagram of Uniformly Charged Circular Loop The Electric Potential of a charged ring is given by1 V=

Q 1 √ 4π0 R2 + z2

(9.5.1)

where R is the radius of our ring and x is the distance from the central axis of the ring. In our case, the radius of our ring is R = b.

D

The potential at P1 , where z = b is V1 =

Q Q 1 1 = √ √ 4π0 b2 + b2 4π0 b 2

(9.5.2)

The potential at P2 , where z = 2b is V2 =

Q Q 1 1 = q √ 4π0 4π0 b 5 b2 + (2b)2

(9.5.3)

Dividing eq. (9.5.3) by eq. (9.5.2) gives us V2 = V1

r

2 5

(9.5.4)

Answer: (D) 1

Add Derivation

©2009

David S. Latchman

Sample Test

66

9.6

Electric Potential II

The potential energy, U(r), of a charge, q, placed in a potential, V(r), is[1] U(r) = qV(r)

(9.6.1)

The work done in moving our charge through this electrical field is W = U2 − U1 = qV2 − qV1 = q (V2 − V1 )

(9.6.2)

9.7

FT

Answer: (E)

Faraday’s Law and Electrostatics

Gauss’s Law

RA

We notice that our answers are in the form of differential equations and this leads us to think of the differential form of Maxwell’s equations[2]. The electrostatics form of Maxwell’s Equations are[3] ρ 0

(9.7.1)

∇×E=0

(9.7.2)

∇·B=0

(9.7.3)

∇ × B = µ0 J

(9.7.4)

∇·E=

Maxwell-Faraday Equation Gauss’ Law for Magnetism

D

Amp`ere’s Law

Comparing our answers, we notice that eq. (9.7.2) corresponds to Answer: (C) .

Answer: (C)

9.8

AC Circuits: RL Circuits

An inductor’s characteristics is opposite to that of a capacitor. While a capacitor stores energy in the electric field, essentially a potential difference between its plates, an inductor stores energy in the magnetic field, which is produced by a current passing through the coil. Thus inductors oppose changes in currents while a capacitor opposes changes in voltages. A fully discharged inductor will initially act as an open circuit David S. Latchman

©2009

AC Circuits: RL Circuits 67 with the maximum voltage, V, across its terminals. Over time, the current increases and the potential difference across the inductor decreases exponentially to a minimum, essentially behaving as a short circuit. As we do not expect this circuit to oscillate, this leaves us with choices (A) and (B). At t = 0, we expect the voltage across the resistor to be VR = 0 and increase exponentially. We choose (A). L A I V

R

FT

B

Figure 9.8.1: Schematic of Inductance-Resistance Circuit We can see from the above schematic,

V = VL + VR

(9.8.1)

RA

where VL and VR are the voltages across the inductor and resistor respectively. This can be written as a first order differential equation dI R + I dt L

(9.8.2)

V dI R = + I L dt L

(9.8.3)

V=L

Dividing by L leaves

The solution to eq. (9.8.3) leaves

D

Z

  V Rt exp dt + k L L I=   Rt exp L   V Rt = + k exp − R L

Multiplying eq. (9.8.4) by R gives us the voltage across the resistor   Rt VR = V + kR exp − L

(9.8.4)

(9.8.5)

at t = 0, VR = 0 0 = V + kR V ∴k=− R ©2009

(9.8.6) David S. Latchman

Sample Test

68 Substituting k into eq. (9.8.5) gives us    Rt VR (t) = V 1 − exp − L

(9.8.7)

where τ = L/R is the time constant. Where τ = 2 s

7

V(x)

FT

6

Voltage/V

5 4 3

RA

2 1 0

0

5

10 Time/s

15

20

D

Figure 9.8.2: Potential Drop across Resistor in a Inductor-Resistance Circuit

Answer: (A)

9.9

AC Circuits: Underdamped RLC Circuits

When a harmonic oscillator is underdamed, it not only approaches zero much more quickly than a critically damped oscillator but it also oscillates about that zero. A quick examination of our choices means we can eliminate all but choices (C) and (E). The choice we make takes some knowledge and analysis. David S. Latchman

©2009

AC Circuits: Underdamped RLC Circuits

69 L

V

R

A

B C

FT

Figure 9.9.1: LRC Oscillator Circuit The voltages in the above circuit can be written

V(t) = VL + VR + VC dI(t) 1 + RI(t) + q(t) =L dt C

(9.9.1)

which can be written as a second order differential equation

or as

d2 q(t) dq(t) 1 +R + q(t) = V(t) 2 dt dt C

RA L

(9.9.2)

D

dq(t) d2 q(t) + γ + ω20 q(t) = V(t) (9.9.3) 2 dt dt This can be solved by finding the solutions for nonhomogenoeus second order linear differential equations. For any driving force, we solve for the undriven case, d2 z dz + γ + ω20 = 0 2 dt dt

(9.9.4)

where for the underdamped case, the general solution is of the form z(t) = A exp(−αt) sin(βt + δ)

(9.9.5)

where

γ α=− q2 4ω20 − γ2 β= 2

(9.9.6) (9.9.7)

In the case of a step response,    1 V(t) =   0 ©2009

t>0 t<0

(9.9.8) David S. Latchman

Sample Test

70 The solution becomes r   2   R 2  ω −  sin t + δ   0   2L R q(t) = 1 − exp − t 2L sin δ where the phase constant, δ, is cos δ =

(9.9.9)

R 2ω20 L

(9.9.10)

where ω0 ≈ 3.162 kHz and γ = 5 ΩH−1 1.8

V(x)

FT

1.6 1.4 Voltage/V

1.2 1 0.8

RA

0.6 0.4 0.2

0

0

0.5

1 Time/s

1.5

2

D

Figure 9.9.2: Forced Damped Harmonic Oscillations

So in the case of our forced underdamped oscillator, we would expect the voltage to raise, overshoot a little and oscillate while slowly decaying. This resembles choice (C). Answer: (C)

9.10

Bohr Model of Hydrogen Atom

Bohr’s theory of the atom proposed the existence of stationary states by blending new quantum mechanics ideas with old classical mechanics concepts. Bohr’s model of the hydrogen atom starts as a system of two bodies bound together by the Coulomb attraction. The charges and mass of one particle is −e and m and +Ze and M for the other. In the case of our hydrogen atom system, Z = 1, M = mp and m = me . David S. Latchman

©2009

Bohr Model of Hydrogen Atom 71 We will be taking into account the motion of both particles in our analysis. Normally, we expect the mass, M, to be stationary where M/m → ∞ and as the proton-to-electron mass ratio is very large, we approximate to this limiting condition. mp = 1836 me

(9.10.1)

This effect is detectable and should be retained as a small correction. As the effects can be incorporated with little difficulty, we shall do so.2 We take the center of mass to be at th origin,3 . We can reduce our two body system to an equivalent one body description in terms of a single vector given by a relative coordinate. r = r1 − r2 (9.10.2)

FT

As the center of mass is located at the origin

mr1 + Mr2 = 0

(9.10.3)

Solving eqs. (9.10.2) and (9.10.3) gives us

and

M r M+m

RA

r1 =

(9.10.4)

m r (9.10.5) M+m Differentiating eqs. (9.10.4) and (9.10.5), gives us the corresponding velocities r2 = −

v1 =

and

M v M+m

v2 = −

m v M+m

(9.10.6)

(9.10.7)

D

where the relative velocity is

dr dt The total energy can be found from eqs. (9.10.6) and (9.10.7) v=

1 1 K = mv21 + Mv22 2 2 1 mM 2 = v 2M+m

(9.10.8)

(9.10.9)

We can reduce this to the equivalent of a one body system where the reduced mass factor is mM µ= (9.10.10) M+m 2 3

Put figure here as seen in diagram

©2009

David S. Latchman

Sample Test

72 Equation (9.10.9) becomes

1 K = µv2 (9.10.11) 2 As the Coulomb Force is a central force, the total angular momentum of the system will be constant. L = mv1 r1 + Mv2 r2   2  m M 2 vr + M vr =m M+m M+m = µvr

(9.10.12)

The centripetal force of the system is equal to the Coulomb force, thus mv21 r1

=

Mv22 r2

µv2 1 Ze2 = r 4π0 r2

FT

F=

=

(9.10.13)

The potential energy of the system comes from the Coulomb potential energy V=−

1 Ze2 4π0 r

(9.10.14)

RA

The total energy of the system can be found by adding eqs. (9.10.11) and (9.10.14) E=K+V

1 2 1 Ze2 = µv − 2 4π0 r

(9.10.15)

Substituting eq. (9.10.13) into eq. (9.10.15) gives E=−

1 Ze2 2 4π0 r

(9.10.16)

D

We expect the total energy of the system to be negative as it is a bound system. We have, so far, adhered to the principles of classical mechanics up to this point. Beyond this point, we must introduce quantum mechanical concepts. To produce the stationary states he was seeking, Bohr introduced the hypothesis that the angular momentum is quantized. L = n~ (9.10.17) Equating this with eq. (9.10.12) and substitution into eq. (9.10.13) gives discrete values for the orbital radius. 4π0 n2 ~2 rn = (9.10.18) Ze2 µ We can rewrite the above equation rn = David S. Latchman

n2 me a0 µZ

(9.10.19) ©2009

Nuclear Sizes where

73 a0 =

4π0 ~2 e2 me

(9.10.20)

is the Bohr Radius. As the orbital radii is discrete we expect the various orbital energies to also be discrete. Substitution of eq. (9.10.18) into eq. (9.10.16) gives En = −

1 Ze2 2 4π0 rn

Z2 e2 =− 2 n 4π0 En = − where e2 E0 = 4π0

µ 2~2

(9.10.21)

Z2 µ E0 n2 me

(9.10.22)

me = 13.6 eV 2~2

(9.10.23)

FT

or

!2

!2

We see that this analysis eliminates all but one answer.

9.11

RA

Answer: (A)

Nuclear Sizes

We know from electron scattering experiments, the nucleus is roughly spherical and uniform density4 . The Fermi model gives us an expression 1

r = r0 A 3

(9.11.1)

D

where r0 = 1.2 × 10−15 m and A is the mass number. In the case of hydrogen, we recall the Bohr radius to be a0 = 0.0592 nm. So in the case of hydrogen, A = 1, r = r0 = 1.2 × 10−15 m

(9.11.2)

r0 1.2 × 10−15 = a0 0.0592 × 10−9 = 2.02 × 10−5

(9.11.3)

Thus

Answer: (B) 4

Add diagram of nuclear and atomic sizes here

©2009

David S. Latchman

Sample Test

74

9.12

Ionization of Lithium

The ionization energy of an electron is the energy to kick it off from its present state to infinity. It can be expressed as Eionization = E∞ − En Z2 µ = 2 E0 n me

(9.12.1)

where E0 = 13.6 eV and µ is the reduced mass where

FT

µ M = me M + me

(9.12.2)

In the case of atoms, the above ratio is close to one and hence we can ignore it for this case. Lithium has an atomic number, Z = 3 so its electron structure is5 1s2 , 2s1

(9.12.3)

D

Answer: (C)

RA

So the total ionization energy will be the total energy needed to completly remove each electron. This turns out to be # " 2 32 32 3 E = 2 + 2 + 2 13.6 eV 1 1 2 ≈ 20 × 13.6 eV = 272.0 eV (9.12.4)

9.13

Electron Diffraction

We recall that in optics, one of the criteria for diffraction is a monochromatic wave. We expect the electron beam to also have wavelike effects. The de Broglie relations show that the wavelength is inversely proportional to the momentum of a particle and that the frequency is directly proportional to the particle’s kinetic energy. λ=

h p

and E = h f

Thus, for electron we expect the beam to be monoenergetic. Answer: (B) 5

Draw Lithium atom and its electrons

David S. Latchman

©2009

Effects of Temperature on Speed of Sound

9.14

75

Effects of Temperature on Speed of Sound

The speed of sound is determined by its Bulk Modulus and its density s v=

B ρ

(9.14.1)

in the case of gases, the Bulk Modulus can be expressed B = γP

(9.14.2)

For an ideal gas

FT

where γ is the adiabatic ratio and P is the pressure of the gas. PV = nRT

(9.14.3)

Substituting eqs. (9.14.2) and (9.14.3) into eq. (9.14.1) gives us r

nγRT M

RA

v=

(9.14.4)

So we see that

1

v ∝ T2

Answer: (B)

9.15

(9.14.5)

Polarized Waves

D

Given the equations

y = y0 sin (ωt − kx)   z = z0 sin ωt − kx − φ (9.15.1)

For our wave to be plane-polarized, the two waves, y and z must be in phase i.e. when y is a maximum so too is z. So   sin (ωt − kx) = sin ωt − kx − φ ∴φ=0

(9.15.2)

A plane polarized wave will occur when φ = 0. We can also look at the waves below and see that they are not in phase, except for Choice (E). ©2009

David S. Latchman

Sample Test

76 z

z

y(x) z(x)

y(x) z(x) y

y

x

x

(a) φ =

√ 2

z

(b) φ = 3π/2 y(x) z(x)

z

y

y(x) z(x)

x

(c) φ = π/2

FT

y

x

(d) φ = π/4

RA

Figure 9.15.1: Waves that are not plane-polarized

z

y(x) z(x)

D

y

x

Figure 9.15.2: φ = 0

Answer: (E)

9.16

Electron in symmetric Potential Wells I

As our potential is symmetric about the V-axis, then we will expect our wave function to also be symmetric about the V-axis. Answer: (E) David S. Latchman

©2009

Electron in symmetric Potential Wells II

9.17

77

Electron in symmetric Potential Wells II

If the electrons do not interact, we can ignore Pauli’s Exclusion Principle. As a result they will not have spatially antisymmetric states but will have the same spatial wave functions. Answer: (B)

9.18

Relativistic Collisions I

FT

The Relativistic Momentum equation is mv  2 v 1− c

p= r

given that p = mc/2,

(9.18.2)

D

Answer: (D)

RA

mc mv = r  2 2 v 1− c c v ⇒ = r  2 2 v 1− c c ∴v= √ 5

(9.18.1)

9.19

Relativistic Collisions II

Momentum is conserved. So in the horizontal direction, p = 2p f cos 30

(9.19.1)

Solving this shows pf =

mc √ 2 3

(9.19.2)

Answer: (B) ©2009

David S. Latchman

Sample Test

78

9.20

Thermodynamic Cycles I

We have a three stage cyclic process where A(P1 , V1 , T1 )

, B(P2 , V2 , T2 )

Adiabatic Expansion, A −→ B

γ

,

C(P3 , V3 , T3 )

γ

(9.20.1)

P1 V1 = P2 V2

(9.20.2)

P2 = 2−γ P1

(9.20.3)

T2 = 21−γ T1

(9.20.4)

P2 P3 = T2 T3

(9.20.5)

Given that V2 = 2V1 , we have

Isochoric Expansion, B −→ C We have

RA

where T3 = T1 , we have

FT

and

P3 = 2γ−1 P2 1 = P1 2

(9.20.6) (9.20.7)

This becomes

A(P1 , V1 , T1 )

, B(2−γ P1 , 2V1 , 21−γ T1 )

, C(

P1 , 2V1 , T1 ) 2

(9.20.8)

D

On a PV-graph, we see that this makes a clockwise cycle, indicating that positive work is done by the gas on the environment. Answer: (A)

9.21

Thermodynamic Cycles II

We recall Calusius’s Therorem I

dQ =0 T

(9.21.1)

for a reversible cycle, the change in entropy is zero. Answer: (C) David S. Latchman

©2009

Distribution of Molecular Speeds

9.22

79

Distribution of Molecular Speeds

The distribution of speeds of molecules follows the Maxwell-Boltzmann distribution, which has the form # " 3  M 2 2 Mv2 (9.22.1) f (v) = 4π v exp − 2πRT 2RT where R is the gas constant and M is the molar mass of the gas. The speed distribution for noble gases at T = 298.15 K looks like Helium Neon Argon Xenon

0.004

FT

0.0035 0.003 0.0025 0.002 0.0015 0.001

RA

Maxwell Speed Distribution, f(v)

0.0045

0.0005 0

0

500

1000 1500 Molecular Speed, m/s

2000

Figure 9.22.1: Maxwell-Boltzmann Speed Distribution of Nobel Gases

D

Answer: (D)

9.23

Temperature Measurements

All the thermometers won’t be able to survive that high a temperature except for the optical pyrometer. Of course, a little knowledge always helps. Optical Pyrometer Optical pyrometers work by using the human eye to match the brightness of a hot object to a calibrated lamp filament inside the instrument. Carbon Resistor These thermometers are typically used for very low temperatures and not high ones. One of their main advantages is their sensitivity, their resistance increases exponentially to decreasing temperature and are not affected by magnetic fields. ©2009

David S. Latchman

80 Sample Test Gas-Bulb Thermometer May also be known as the constant volume gas thermometer. Doubtful the glass bulb will survive such high temperatures. Mercury Thermometer The boiling point of mercury is about 360 °C. This thermometer will be vaporized before you even had a chance to think about getting a temperature reading. Thermocouple Thermocouples are made by joining two different metals together and produces a voltage that is related to the temperature difference. They are typically used in industry to measure high temperatures, usually in the order ∼ 1800 °C. The metals would most likely start melting above these temperatures.

Answer: (A)

Counting Statistics

RA

9.24

FT

Even if we knew nothing about any of the above thermometers, we could have still take a stab at it. We should probably guess that at that high a temperature we won’t want to make physical contact with what we are measuring. The only one that can do this is the optical pyrometer.

NOT FINISHED Answer: (D)

9.25

Thermal & Electrical Conductivity

D

A metal is a lattice of atoms, each with a shell of electrons. This forms a positive ionic lattice where the outer electrons are free to dissociate from the parent atoms and move freely through the lattic as a ‘sea’ of electrons. When a potential difference is applied across the metal, the electrons drift from one end of the conductor to the other under the influence of the electric field. It is this free moving electron ‘sea’ that makes a metal an electrical conductor. These free moving electrons are also efficient at transferring thermal energy for the same reason. Thermal and electrical conductivity in metals are closely related to each other as outlined in the Wiedemann-Franz Law. κ = LT σ

(9.25.1)

where the Lorenz number, L = 2.44 × 10−8 WΩK−1 and κ and σ are the thermal and electrical conductivities respectively. This corelation does not apply to non-metals due to the increased role of phonon carriers. Answer: (E) David S. Latchman

©2009

Nonconservation of Parity in Weak Interactions

9.26

81

Nonconservation of Parity in Weak Interactions

Of the four interations, electromagnetism, strong, weak and gravity, parity is conserved in all except for the weak interaction. To examine violations of these interactions we must look at the helicity of our particles and see whether they are “left-handed” or “right-handed”. A particle is said to be “right-handed” if the direction of its spin is the same as the direction as its motion. It is “left-handed” if the directions of spin and motion are opposite to each other. Thus the helicity of a particle is the projection of the spin vector onto the momentum vector where left is negative and right is positive.6 (9.26.1)

FT

S·p h ≡ S · p

Particles are not typically characterized as being “left-handed” or “right-handed”. For example, an electron could have both its spin and momentum pointing in the same direction to the right and hence be classified as “right-handed”. But from the reference frame of someone travelling faster than the speed of the electron, would see the electron travelling to the left and hence conclude the electron is “left-handed”.

RA

Neutrinos, on the other hand, travel very close to the speed of light and it would be very difficult to accelerate to a point where one would be able to change the “handedness” of the neutrino. Thus, we say that the neutrino has an intrinsic parity, all of them being left-handed. Anti-neutrinos on the other hand are all right-handed. This causes weak interactions, neutrino emitting ones in particular, to violate the conservation of parity law.7 For the pion decay,

π+ → µ+ + υµ+

(9.26.2)

D

It is very difficult to detect and measure the helicity of the neutrino directly but we can measure it indirectly through the above decay and hence demonstrate nonconservation of parity.

If the pion is at rest and has spin-0, the anti-muon and neutrino will come out in opposite directions.8 In the figure below, the anti-muons are observed with their z-component of angular momentum given by mµ = − 21 . Angular momentum conversation then implies mυ = + 12 for the neutrino. It is very difficult if not impossible to detect neutrinos in a typical laboratory setting but we can detect muons and measure their helicity. Choice (A) The Q-value is the kinetic Energy released in the decay of the particle at rest. Parity deals with mirror symmetry violations and not energy. 6

Draw Helicity Diagrams Add section explaining parity 8 Draw Diagram Here 7

©2009

David S. Latchman

82 Sample Test Choice (B) We measure for violations of parity conservation by measuring the longitudinal polarization of the anti-muon. We choose this answer. Choice (C) The pion has spin-0 and is stationary. So it won’t be polarized. Measuring this gives us no information on our decay products. Choice (D) The angular correlation would be difficult as neutrinos are difficult to detect. Choice (E) Parity deals with spatial assymetry and has nothing to do with time. We can eliminate this choice.

9.27

Moment of Inertia

The moment of inertia is

FT

Answer: (B)

Z

I=

r2 dm

(9.27.1)

RA

In the case of a hoop about its center axis,

I = MR2

(9.27.2)

From eq. (9.27.1), we see that the moment of inertia deals with how the mass is distributed along its axis.9 We see that

A

D

A is equivalent to

Figure 9.27.1: Hoop and S-shaped wire

Thus, see fig. 9.27.1, the moment of inertia of our S-shaped wire can be found from a hoop with its axis or rotation at its radius. This can be calculated by using the Parallel Axis Theorem I = ICM + Md2 (9.27.3) where d2 is the distance from the center of mass. This becomes I = MR2 + MR2 = 2MR2

(9.27.4)

Answer: (E) 9

The moment of inertia of a 1 kg mass at a distance 1 m from the axis of rotation is the same as a hoop with the same mass rotating about its central axis.

David S. Latchman

©2009

Lorentz Force Law I

9.28

83

Lorentz Force Law I + + + + + + + + + + + + + + + + + I FE

u

B

FT

FB

Figure 9.28.1: Charged particle moving parallel to a positively charged current carrying wire The force on the charged particle is determined by the Lorentz Force Equation F = e [E + u × B]

(9.28.1)

RA

where FE = eE and FB = e(u × B). For our charged particle to travel parallel to our wire, FE = FB .10 λ` (9.28.2) E= 2π0 r and the magnetic field can be determined from Amp`ere’s Law I B · ds = µ0 Ienclosed

D

In the case of our wire

B=

µ0 I 2πr

Plugging eqs. (9.28.2) and (9.28.4) into eq. (9.28.1) gives " # µ0 I λ` F=e +u =0 2π0 r 2πr

(9.28.3)

(9.28.4)

(9.28.5)

For the particle to be undeflected, FE + FB = 0 FE + FB = 0 µ0 I λ` −u =0 2π0 r 2πr

(9.28.6)

Now we can go about eliminating choices. 10

Add derivation in a section

©2009

David S. Latchman

84 Sample Test Doubling charge per unit length We see from eq. (9.28.6), halving the current, I, and doubling the linear charge density, λ, will not allow the particle to continue undeflected. µ0 I/2 FB 2λ` −u = 2FE − ,0 (9.28.7) 2π0 r 2πr 2 Doubling the charge on the particle We see from, eq. (9.28.6) that the charge on the particle, e, has no effect on the particle’s trajectory. We would be left with µ0 I/2 λ` FB −u = FE − ,0 2π0 r 2πr 2

(9.28.8)

FT

Doubling both the charge per unit length on the wire and the charge on the particle As shown above, the particle’s charge has no effect on the trajectory. This leaves us with the charge per unit length, λ and as we have seen before, this will change the particle’s trajectory, see eq. (9.28.7). Doubling the speed of the particle If we double the particle’s speed we will get

RA

µ0 I/2 µ0 I λ` λ` − (2u) = −u 2π0 r 2πr 2π0 r 2πr ∴ FE = FB

This is our answer

D

Introducing an additional magnetic field parallel to the wire Recalling eq. (9.28.1), the force due to the magnetic field is a cross product between the velocity and the field. A charged particle moving in the same direction as the field will experience no magnetic force. FB = e [u × B] = uB sin 0 =0

(9.28.9)

Answer: (D)

9.29

Lorentz Force Law II

As we can see from eq. (9.28.5), the forces due to the electric and magnetic fields are equal. " # µ0 I λ` =0 (9.29.1) F=e +u 2π0 r 2πr David S. Latchman

©2009

Nuclear Angular Moment 85 If we move our charged particle a distance 2r from the wire with a speed nu, eq. (9.28.5) becomes " # #  " µ0 I µ0 I λ` 1 λ` e +u =e + nu 2π0 (2r) 2π(2r)r 2 2π0 r 2πr   1 [FE − nFB ] =e 2 =0 (9.29.2) Thus [FE − nFB ] = FE − FB = 0 ⇒n=1

(9.29.3)

FT

The speed of the particle is u. Answer: (C)

9.30

Nuclear Angular Moment

Answer: (B)

9.31

RA

NOT FINISHED

Potential Step Barrier Region I

Ψ(x)

Region II V

D

E

O

x

Figure 9.31.1: Wavefunction of particle through a potential step barrier The point, x = 0, divides the region into two regions, Region I, where classical motion is allowed and, Region II, where classical motion is forbidden. The barrier potential is    0 for x < 0 V(x) =  (9.31.1)  V for x > 0 ©2009

David S. Latchman

Sample Test

86 For, x < 0, the eigenfunction satisfies d2 ψ = −k12 ψ dx2 where k12 =

2m E ~2

(9.31.2)

(9.31.3)

The general form of the eigenfunction is ψI = Aeik1 x + Be−ik1 x

(9.31.4)

For, x > 0, the eigenfunction satisfies

where

FT

d2 ψ = k22 ψ dx2

2m (V − E) ~2 The general form of the eigenfunction becomes k22 =

(9.31.6)

(9.31.7)

RA

ψII = Ce−k2 x + Dek2 x

(9.31.5)

As x → ∞, the ek2 x term blows up. So to allow eq. (9.31.7) to make any physical sense we set D = 0, thus ψII = Ce−k2 x = Ce−αx (9.31.8) We can continue solving for A, B and C but for the purposes of the question we see from eq. (9.31.8) that α is both real and positive.

D

Answer: (C)

David S. Latchman

©2009

Appendix

A

A.1

Constants

Symbol c G me NA R k e 0 µ0 1 atm a0

Value 2.99 × 108 m/s 6.67 × 10−11 m3 /kg.s2 9.11 × 10−31 kg 6.02 × 1023 mol-1 8.31 J/mol.K 1.38 × 10−23 J/K 1.60 × 10−9 C 8.85 × 10−12 C2 /N.m2 4π × 10−7 T.m/A 1.0 × 105 M/m2 0.529 × 10−10 m

D

RA

Constant Speed of light in a vacuum Gravitational Constant Rest Mass of the electron Avogadro’s Number Universal Gas Constant Boltzmann’s Constant Electron charge Permitivitty of Free Space Permeability of Free Space Athmospheric Pressure Bohr Radius

FT

Constants & Important Equations

Table A.1.1: Something

A.2

Vector Identities

A.2.1

Triple Products A · (B × C) = B · (C × A) = C · (A × B) A × (B × C) = B (A · C) − C (A · B)

(A.2.1) (A.2.2)

Constants & Important Equations

88

A.2.2

Product Rules


∇ f g = f ∇g + g ∇ f ∇ (A · B) = A × (∇ × B) + B × (∇ × A) + (A · ∇) B + (B · ∇) A

∇ · f A = f (∇ · A) + A · ∇ f ∇ · (A × B) = B · (∇ × A) − A · (∇ × B)

∇ × f A = f (∇ × A) − A × ∇ f ∇ × (A × B) = (B · ∇) A − (A · ∇) B + A (∇ · B) − B (∇ · A)

Second Derivatives

FT

A.2.3

∇ · (∇ × A) = 0
∇ × ∇f = 0

∇ × (∇ × A) = ∇ (∇ · A) − ∇2 A

Commutators

A.3.1

Lie-algebra Relations

RA

A.3

D

[A, A] = 0 [A, B] = −[B, A] [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0

A.3.2

A.3.3

(A.2.3) (A.2.4) (A.2.5) (A.2.6) (A.2.7) (A.2.8)

(A.2.9) (A.2.10) (A.2.11)

(A.3.1) (A.3.2) (A.3.3)

Canonical Commutator [x, p] = i~

(A.3.4)

Kronecker Delta Function ( δmn =

For a wave function

0 1

if m , n; if m = n;

Z ψm (x)∗ ψn (x)dx = δmn

David S. Latchman

(A.3.5) ©2009

Linear Algebra

89

A.4

Linear Algebra

A.4.1

Vectors

Vector Addition The sum of two vectors is another vector

Associative Zero Vector

|αi + |βi = |βi + |αi

(A.4.2)



|αi + |βi + |γi = |αi + |βi + |γi

(A.4.3)

|αi + |0i = |αi

(A.4.4)

|αi + | − αi = |0i

(A.4.5)

D

RA

Inverse Vector

(A.4.1)

FT

Commutative

|αi + |βi = |γi

©2009

David S. Latchman

Constants & Important Equations

D

RA

FT

90

David S. Latchman

©2009

Bibliography

FT

[1] Stephen Gasiorowicz Paul M. Fishbane and Stephen T. Thornton. Physics for Scientists and Engineers with Modern Physics, chapter 24.2, page 687. Prentice Hall, third edition, 2005. [2] Wikipedia. Maxwell’s equations — wikipedia, the free encyclopedia, 2009. [Online; accessed 21-April-2009].

D

RA

[3] David J. Griffiths. Introduction to Electrodyanmics, chapter 5.3.4, page 232. Prentice Hall, third edition, 1999.

Index

Franck-Hertz Experiment, 49 Gravitation, see Celestial Mechanics

FT

RLC Circuits Sample Test Q09, 68 RL Circuits Sample Test Q08, 66

Kepler’s Laws, see Celestial Mechanics Angular Momentum, see Rotational Mo- Kronecker Delta Function, 88 tion Linear Algebra, 89 Bohr Model Hydrogen Model, 43

Maxwell’s Equations Sample Test Q07, 66 Moment of Inertia, see Rotational Motion

D

RA

Celestial Mechanics, 10 Circular Orbits, 11 Escape Speed, 10 Kepler’s Laws, 11 Newton’s Law of Gravitation, 10 Orbits, 11 Potential Energy, 10 Central Forces Sample Test Q03, 63 Sample Test Q04, 64 Circular Orbits, see Celestial Mechanics Commutators, 88 Canonical Commutators, 88 Kronecker Delta Function, 88 Lie-algebra Relations, 88 Compton Effect, 46 Counting Statistics, 59

Vectors, 89

Doppler Effect, 8 Electric Potential Sample Test Q05, 65 Work Sample Test Q06, 66 Faraday’s Law Sample Test Q07, 66

Newton’s Law of Gravitation, see Celestial Mechanics Oscillations Underdamped Sample Test Q09, 68 Oscillatory Motion, 4 Coupled Harmonic Oscillators, 6 Damped Motion, 5 Kinetic Energy, 4 Potential Energy, 5 Simple Harmonic Motion Equation, 4 Small Oscillations, 5 Total Energy, 4 Parallel Axis Theorem, see Rotational Motion Pendulum Simple Sample Test Q01, 61 Rolling Kinetic Energy, see Rotational Motion Rotational Kinetic Energy, see Rotational Motion

Index Rotational Motion, 8 Angular Momentum, 9 Moment of Inertia, 8 Parallel Axis Theorem, 9 Rolling Kinetic Energy, 9 Rotational Kinetic Energy, 8 Torque, 9

93

Torque, see Rotational Motion

D

RA

Vector Identities, 87 Product Rules, 88 Second Derivatives, 88 Triple Products, 87

FT

Springs Work Sample Test Q02, 62 Subject, 30 System of Particles, 10

©2009

David S. Latchman