Physics GRE Sample Test Solutions

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The Physics GRE Solution Guide

Sample Test

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

November 3, 2009

Author: David S. Latchman

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Preface

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.

David Latchman

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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iv Contents

2.8 AC Circuits . . . 20

2.9 Magnetic and Electric Fields in Matter . . . 20

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

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

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

3 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

4 Thermodynamics & Statistical Mechanics 27 4.1 Laws of Thermodynamics . . . 27

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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

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

4.17 RMS Speed of an Ideal Gas . . . 30

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

4.24 Second Law of Thermodynamics . . . 33

5 Quantum Mechanics 35 5.1 Fundamental Concepts . . . 35

5.2 Schr ¨odinger Equation . . . 35

5.3 Spin . . . 40

5.4 Angular Momentum . . . 41

5.5 Wave Funtion Symmetry . . . 41

5.6 Elementary Perturbation Theory . . . 41

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vi Contents 6.2 Bohr Model . . . 43 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

7 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

7.10 Space-Time Interval . . . 55

8 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

9 Sample Test 61 9.1 Period of Pendulum on Moon . . . 61

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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

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

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

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

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viii Contents

A Constants & Important Equations 87

A.1 Constants . . . 87

A.2 Vector Identities . . . 87

A.3 Commutators . . . 88

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List of Tables

4.22.1Table of Molar Specific Heats . . . 32 9.4.1 Table of Orbits . . . 64 A.1.1Something . . . 87

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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

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 carry-ing wire . . . 83

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Chapter

1 Classical Mechanics

1.1 Kinematics

1.1.1 Linear Motion

Average Velocity v= ∆x ∆t = x2−x1 t2−t1 (1.1.1) Instantaneous Velocity v= lim ∆t→0 ∆x ∆t = dx dt = v(t) (1.1.2)

Kinematic Equations of Motion

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

v= v0+ at (1.1.3) v2 = v2₀+ 2a (x − x0) (1.1.4) x − x0 = v0t+ 1 2at 2 _(1.1.5) x − x0 = 1 2(v+ v0) t (1.1.6)

1.1.2 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

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2 Classical Mechanics Acceleration Centripetal Acceleration a= v 2 r (1.1.7) Angular Velocity ω = v r (1.1.8) We can write eq. (1.1.7) in terms ofω

a= ω2r (1.1.9)

Rotational Equations of Motion

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

ω = ω0+ αt (1.1.10) θ = ω + ω0 2 t (1.1.11) θ = ω0t+ 1 2αt 2 _(1.1.12) ω2 _{= ω}2 0+ 2αθ (1.1.13)

1.2 Newton’s Laws

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

dt = ma (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.

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Work & Energy 3

1.2.2 Momentum

p= mv (1.2.3)

1.2.3 Impulse

∆p = J =w Fdt= Favgdt (1.2.4)

1.3 Work & Energy

1.3.1 Kinetic Energy

K ≡ 1 2mv

2 _(1.3.1)

1.3.2 The Work-Energy Theorem

The net Work done is given by

Wnet= Kf −Ki (1.3.2)

1.3.3 Work done under a constant Force

The work done by a force can be expressed as

W = F∆x (1.3.3) In three dimensions, this becomes

W = F · ∆r = F∆r cos θ (1.3.4) For a non-constant force, we have

W= xf w xi F(x)dx (1.3.5)

1.3.4 Potential Energy

The Potential Energy is

F(x)= −dU(x)

dx (1.3.6) for conservative forces, the potential energy is

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4 Classical Mechanics

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

U(x)= 1 2kx

2 _(1.3.9)

1.4 Oscillatory Motion

1.4.1 Equation for Simple Harmonic Motion

x(t)= A sin (ωt + δ) (1.4.1) 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.

1.4.2 Period of Simple Harmonic Motion

T= 2π

ω (1.4.2)

1.4.3 Total Energy of an Oscillating System

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

KE= 1 2mv 2 = 1 2m dx dt = 1 2mA 2_ω2_cos2_{(ωt + δ)} _(1.4.5)

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Oscillatory Motion 5 The Potential Energy is

U = 1 2kx 2 = 1 2kA 2_sin2_{(ωt + δ)} _(1.4.6)

Adding eq. (1.4.5) and eq. (1.4.6) gives E= 1

2kA

2 _(1.4.7)

1.4.4 Damped Harmonic Motion

Fd = −bv = −b

dx

dt (1.4.8) where b is the damping coefficient. The equation of motion for a damped oscillating system becomes

−_{kx − b}dx dt = m

d2_x

dt2 (1.4.9)

Solving eq. (1.4.9) goves

x= Ae−αtsin (ω0t+ δ) (1.4.10) We find that α = b 2m (1.4.11) ω0 = r k m − b2 4m2 = r ω2 0− b2 4m2 = qω2 0−α2 (1.4.12)

1.4.5 Small Oscillations

The Energy of a system is

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

2_{+ V(x)} _(1.4.13)

We can solve for v(x),

v(x)= r

2

m(E − V(x)) (1.4.14) where E ≥ V(x) Let the particle move in the potential valley, x1 ≤ x ≤ x2, the potential

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6 Classical Mechanics At the points of inflection, the derivative dV/dx is zero and d2_V_/dx2 _{is positive. This}

means that the potential energy for small oscillations becomes V(x) u V(xe)+ 1 2k(x − xe) 2 _(1.4.16) where k ≡" d 2_V(x) dx2 # x=xe ≥ 0 (1.4.17) 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.

1.4.6 Coupled Harmonic Oscillators

Consider the case of a simple pendulum of length, `, and the mass of the bob is m1_.

For small displacements, the equation of motion is ¨

θ + ω0θ = 0 (1.4.18)

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

x= ` cos θ ≈ ` (1.4.19) y= ` sin θ ≈ `θ (1.4.20) eq. (1.4.18) becomes

¨y+ ω0y= 0 (1.4.21)

This is the equivalent to the mass-spring system where the spring constant is k= mω2₀ = mg

` (1.4.22) 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 = 1 2m˙y 2 1+ 1 2m˙y 2 2− ₁ 2ky 2 1+ 1 2κ y2−y1 2+ 1 2ky 2 2 = 1 2m ˙ y12+ ˙y22 − 1 2 ky2₁+ y2₂ + κ y2−y12 (1.4.23) We can find the equations of motion of our system

d dt ∂L ∂ ˙yn ! = _∂y∂L n (1.4.24)

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Oscillatory Motion 7 The equations of motion are

m¨y1 = −ky1+ κ y2−y1 (1.4.25)

m¨y2 = −ky2+ κ y2−y1 (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)

¨y1= −ωy1 ¨y2 = −ωy2 (1.4.27)

Substituting the values for ¨y1and ¨y2into the equations of motion yields

k+ κ − mω2y1−κy2 = 0 (1.4.28)

−κy₁+_k+ κ − mω2_y₂ = 0 _(1.4.29) We can get solutions from solving the determinant of the matrix

k+ κ − mω2 ₋κ −κ _k+ κ − mω2 = 0 (1.4.30) Solving the determinant gives

mω22 − 2mω2_(k_{+ κ) +} k2+ 2kκ = 0 (1.4.31) This yields ω2 ₌            k m = g ` k+ 2κ m = g ` + 2κ m (1.4.32) We can now determine exactly how the masses move with each mode by substituting ω2_{into the equations of motion. Where}

ω2 ₌ k

m We see that

k+ κ − mω2 = κ (1.4.33) 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

m We see that

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.

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1.4.7 Doppler E

ffect

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 vsτ0= vs/ f0. The wavelength is decreased by

λ0 = λ −vs f0 −v − vs f0 (1.4.37) The frequency change is

f0 = v λ0 = f0 _v v − vs (1.4.38)

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 f0 = v 0 λ = f0 1+ vr v (1.4.40)

Moving Source and Moving Observer We can combine the above two equations λ0

= v − vs

f0

(1.4.41) v0 = v − vr (1.4.42)

To give a modified frequency of f0 = v 0 λ0 = _{v − v} r v − vs f0 (1.4.43)

1.5 Rotational Motion about a Fixed Axis

1.5.1 Moment of Inertia

I= Z

R2dm (1.5.1)

1.5.2 Rotational Kinetic Energy

K= 1 2Iω

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Rotational Motion about a Fixed Axis 9

1.5.3 Parallel Axis Theorem

I= Icm+ Md2 (1.5.3)

1.5.4 Torque

τ = r × F (1.5.4) τ = Iα (1.5.5) whereα is the angular acceleration.

1.5.5 Angular Momentum

L= Iω (1.5.6) we can find the Torque

τ = dL

dt (1.5.7)

1.5.6 Kinetic Energy in Rolling

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

K= Krot =

1

2Icontactω

2 _(1.5.8)

Icontactcan be found from the Parallel Axis Theorem.

Icontact = Icm+ MR2 (1.5.9)

Substitute eq. (1.5.8) and we have K = 1 2 Icm+ MR2 ω2 = 1 2Icmω 2₊ 1 2mv 2 _(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.

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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= m1r1+ m2r2+ m3r3+ · · · + mNrN

M (1.6.1)

Velocity Vector of a System of Particles

V= dR

dt

= m1v1+ m2v2+ m3v3+ · · · + mNvN

M (1.6.2)

Acceleration Vector of a System of Particles

A= dV

dt

= m1a1+ m2a2+ m3a3+ · · · + mNaN

M (1.6.3)

1.7 Central Forces and Celestial Mechanics

1.7.1 Newton’s Law of Universal Gravitation

F= −

_GMm r2

ˆr (1.7.1)

1.7.2 Potential Energy of a Gravitational Force

U(r)= −GMm

r (1.7.2)

1.7.3 Escape Speed and Orbits

The energy of an orbiting body is

E= T + U = 1

2mv

2₋ GMm

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Central Forces and Celestial Mechanics 11 The escape speed becomes

E= 1 2mv 2 esc− GMm RE = 0 (1.7.4) Solving for vescwe find

vesc = r 2GM Re (1.7.5)

1.7.4 Kepler’s Laws

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

Second Law A line joining a planet and the sun sweeps out equal areas during 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

R3 = C (1.7.6)

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.

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

1 2v

2₋ GM

r = E < 0 (1.7.7) The Orbital Velocity is

v= r

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.

v= r GM ₂ r − 1 a (1.7.9) where a is the semi-major axis

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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 2v

2₋ GM

r = E = 0 (1.7.10) The Orbital Velocity is

v= vesc=

r 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.

v∞ =

r GM

a (1.7.12)

1.7.6 Derivation of Vis-viva Equation

The total energy of a satellite is

E= 1 2mv

2₋ GMm

r (1.7.13) For an elliptical or circular orbit, the specific energy is

E= −GMm 2a (1.7.14) Equating we get v2= GM ₂ r − 1 a (1.7.15)

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

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Non-inertial Reference Frames 13

1.9.2 Bernoulli’s Equation

P+ 1 2ρv

2_{+ ρgh = a constant} _(1.9.2)

1.10 Non-inertial Reference Frames

1.11 Hamiltonian and Lagrangian Formalism

1.11.1 Lagrange’s Function

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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)

∂L ∂q = d dt ∂L ∂ ˙q ! (1.11.2)

1.11.3 Hamiltonian

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Chapter

2 Electromagnetism

2.1 Electrostatics

2.1.1 Coulomb’s Law

The force between two charged particles, q1and q2is defined by Coulomb’s Law.

F12= 1 4π0 q1q2 r2 12 ! ˆr12 (2.1.1)

where0is the permitivitty of free space, where

0= 8.85 × 10−12C2N.m2 (2.1.2)

2.1.2 Electric Field of a point charge

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

E ≡ F

q0

(2.1.3) The Electric Field of a point charge, q is

E= 1

4π0

q

r2ˆr (2.1.4)

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

E(r)= 1 4π0 n X i=1 qi r2 i ˆri (2.1.5)

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16 Electromagnetism

Electric Fields and Continuous Charge Distributions

If a source is distributed continuously along a region of space, eq. (2.1.5) becomes

E(r)= 1 4π0

Z 1

r2ˆrdq (2.1.6)

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

E(r)= 1 4π0 Z line λ r2ˆrdx (2.1.8)

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

σ = Q A =

dq

dA (2.1.9) The electric field along the surface becomes

E(r)= 1 4π0 Z Surface σ r2ˆrdA (2.1.10)

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

ρ = Q V =

dq

dV (2.1.11) The Electric Field is

E(r)= 1 4π0 Z Volume ρ r2ˆrdV (2.1.12)

2.1.3 Gauss’ Law

The electric field through a surface is Φ = I surface S dΦ = I surface S E · dA (2.1.13) The electric flux through a closed surface encloses a net charge.

I

E · dA= Q

0

(2.1.14) where Q is the charge enclosed by our surface.

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Electrostatics 17

2.1.4 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= q 4π0r2

(2.1.16)

2.1.5 Electric Field due to a line of charge

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

Φ = Z top surface E · dA+ Z bottom surface E · dA+ Z side surface E · dA (2.1.17) 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 (2.1.19)

Thus the flux becomes,

Φ = Z side sirface E · dA = E Z dA (2.1.20) 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 thatΦ = q/0. The electric field becomes

E= λ 2π0r

(2.1.22)

2.1.6 Electric Field in a Solid Non-Conducting Sphere

Within our non-conducting sphere or radius, R, we will assume that the total charge, Q is evenly distributed throughout the sphere’s volume. So the charge density of our sphere is ρ = Q V = Q 4 3πR3 (2.1.23) ©2009 David S. Latchman

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18 Electromagnetism The Electric Field due to a charge Q is

E= Q 4π0r2

(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) where dV = 4πr2_{dr. We can use this to determine the field inside the sphere by}

summing the effect of infinitesimally thin spherical shells E= Z E 0 dE= Z r 0 dq 4πr2 = ρ 0 Z r 0 dr = ₄ Qr 3π0R3 (2.1.26)

2.1.7 Electric Potential Energy

U(r)= 1 4π0

qq0r (2.1.27)

2.1.8 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) And we can see that

V(r)= 1 4π0

q

r (2.1.29) A more proper definition that includes the electric field, E would be

V(r)= − Z

C

E · d` (2.1.30)

where C is any path, starting at a chosen point of zero potential to our desired point. The difference between two potentials can be expressed such

V(b) − V(a)= − Z b E · d` + Z a E · d` = − Z b a E · d` (2.1.31)

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Electrostatics 19 This can be further expressed

V(b) − V(a)= Z b

a

(∇V) · d` (2.1.32) And we can show that

E= −∇V (2.1.33)

2.1.9 Electric Potential due to a line charge along axis

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

V = Z dV = 1 4π0 Z dq r (2.1.34) The charge density is

dq = λdx (2.1.35) 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.

dV = 1 4π0 dq x = _4π1 0 λdx x (2.1.36) This becomes V= λ 4π0 ln _x 2 x1 (2.1.37) where x1and x2are the distances from O, the end of the rod.

Now consider that we are some distance, y, from the axis of the rod of length, `. We again look at eq. (2.1.34), where r is the distance of the point P from the rod’s axis.

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20 Electromagnetism

2.2 Currents and DC Circuits

2

2.3 Magnetic Fields in Free Space

3

2.4 Lorentz Force

4

2.5 Induction

5

2.6 Maxwell’s Equations and their Applications

6

2.7 Electromagnetic Waves

7

2.8 AC Circuits

8

2.9 Magnetic and Electric Fields in Matter

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Capacitance 21

2.10 Capacitance

Q= CV (2.10.1)

2.11 Energy in a Capacitor

U = Q 2 2C = CV2 2 = QV 2 (2.11.1)

2.12 Energy in an Electric Field

u ≡ U volume = 0E2 2 (2.12.1)

2.13 Current

I ≡ dQ dt (2.13.1)

2.14 Current Destiny

I= Z A J · dA (2.14.1)

2.15 Current Density of Moving Charges

J= I

A = neqvd (2.15.1)

2.16 Resistance and Ohm’s Law

R ≡ V

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22 Electromagnetism

2.17 Resistivity and Conductivity

R= ρL A (2.17.1) E= ρJ (2.17.2) J= σE (2.17.3)

2.18 Power

P= VI (2.18.1)

2.19 Kircho

ff’s Loop Rules

Write Here

2.20 Kircho

ff’s Junction Rule

Write Here

2.21 RC Circuits

E − IR − Q C = 0 (2.21.1)

2.22 Maxwell’s Equations

2.22.1 Integral Form

Gauss’ Law for Electric Fields

w

closed surface

E · dA= Q

0

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Speed of Propagation of a Light Wave 23

Gauss’ Law for Magnetic Fields

w closed surface B · dA = 0 (2.22.2) Amp`ere’s Law z B · ds= µ0I+ µ00 d dt w surface E · dA (2.22.3) Faraday’s Law z E · ds = −d dt w surface B · dA (2.22.4)

2.22.2 Di

fferential Form

Gauss’ Law for Electric Fields

∇ · E= ρ 0

(2.22.5)

Gauss’ Law for Magnetism

∇ · B= 0 (2.22.6) Amp`ere’s Law ∇ × B= µ0J+ µ00 ∂E ∂t (2.22.7) Faraday’s Law ∇ · E= −∂B ∂t (2.22.8)

2.23 Speed of Propagation of a Light Wave

c= √ 1 µ00

(2.23.1) In a material with dielectric constant,κ,

c √

κ = c

n (2.23.2) where n is the refractive index.

2.24 Relationship between E and B Fields

E= cB (2.24.1)

E · B = 0 (2.24.2)

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24 Electromagnetism

2.25 Energy Density of an EM wave

u= 1 2 B2 µ0 + 0 E2 ! (2.25.1)

2.26 Poynting’s Vector

S= 1 µ0 E × B (2.26.1)

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Chapter

3 Optics & Wave Phonomena

3.1 Wave Properties

1

3.2 Superposition

2

3.3 Interference

3

3.4 Di

ffraction

4

3.5 Geometrical Optics

5

3.6 Polarization

6

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DRAFT

26 Optics & Wave Phonomena

3.7 Doppler E

ffect

7

3.8 Snell’s Law

3.8.1 Snell’s Law

n1sinθ1= n2sinθ2 (3.8.1)

3.8.2 Critical Angle and Snell’s Law

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 = θcand n2> n1.

n1sin 90= n2sinθc

sinθc =

n1

n2

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Chapter

4 Thermodynamics & Statistical Mechanics

4.1 Laws of Thermodynamics

1

4.2 Thermodynamic Processes

2

4.3 Equations of State

3

4.4 Ideal Gases

4

4.5 Kinetic Theory

5

4.6 Ensembles

6

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28 Thermodynamics & Statistical Mechanics

4.7 Statistical Concepts and Calculation of

Thermody-namic Properties

7

4.8 Thermal Expansion & Heat Transfer

8

4.9 Heat Capacity

Q= CTf −Ti

(4.9.1) where C is the Heat Capacity and Tf and Ti are the final and initial temperatures

respectively.

4.10 Specific Heat Capacity

Q= cmTf −ti

(4.10.1) where c is the specific heat capacity and m is the mass.

4.11 Heat and Work

W= Z V_f

Vi

PdV (4.11.1)

4.12 First Law of Thermodynamics

dEint = dQ − dW (4.12.1)

where dEintis the internal energy of the system, dQ is the Energy added to the system

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Work done by Ideal Gas at Constant Temperature 29

4.12.1 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.

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∆Eintremains the same, thus

∆Q = ∆W (4.12.4) The work done by the system is equal to the heat or energy put into it.

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

∆Eint = 0 (4.12.5)

4.13 Work done by Ideal Gas at Constant Temperature

Starting with eq. (4.11.1), we substitute the Ideal gas Law, eq. (4.15.1), to get W = nRT Z V_f Vi dV V = nRT lnVf Vi (4.13.1)

4.14 Heat Conduction Equation

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

t = kA

TH −TC

L (4.14.1) where k is the thermal conductivity.

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DRAFT

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

R ≈ 8.314 J/mol. K We can rewrite the Ideal gas Law to say

PV = NkT (4.15.2) where k is the Boltzmann’s Constant, such that

k= R NA

≈ 1.381 × 10−23 _J/K

4.16 Stefan-Boltzmann’s FormulaStefan-Boltzmann’s

Equa-tion

P(T) = σT4 (4.16.1)

4.17 RMS Speed of an Ideal Gas

vrms =

r 3RT

M (4.17.1)

4.18 Translational Kinetic Energy

¯ K = 3

2kT (4.18.1)

4.19 Internal Energy of a Monatomic gas

Eint =

3

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Molar Specific Heat at Constant Volume 31

4.20 Molar Specific Heat at Constant Volume

Let us define, CVsuch 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 CV =

1 n

∆Eint

∆T (4.20.3) Substituting eq. (4.19.1), we get

CV=

3

2R= 12.5 J/mol.K (4.20.4)

4.21 Molar Specific Heat at Constant Pressure

Starting with Q= nCp∆T (4.21.1) and ∆Eint = Q − W ⇒_nC_V∆T = nC_p∆T + nR∆T ∴ CV = Cp−R (4.21.2)

4.22 Equipartition of Energy

CV = f 2 ! R= 4.16 f J/mol.K (4.22.1) where f is the number of degrees of freedom.

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DRAFT

Degr ees of Fr eedom Pr edicted Molar Specific Heats Molecule T ranslational Rotational V ibrational T otal (f ) C V C P = C V + R Monatomic 3 0 0 3 3 2 _R 5 2 _R Diatomic 3 2 2 5 5 2 _R 7 2 _R Polyatomic (Linear) 3 3 3n − 5 6 3R 4R Polyatomic (Non-Linear) 3 3 3n − 6 6 3R 4R T able 4.22.1: T able of Molar Specific Heats

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DRAFT

Adiabatic Expansion of an Ideal Gas 33

4.23 Adiabatic Expansion of an Ideal Gas

PVγ= a constant (4.23.1) whereγ = CP

CV.

We can also write

TVγ−1 = a constant (4.23.2)

4.24 Second Law of Thermodynamics

Something.

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Chapter

5 Quantum Mechanics

5.1 Fundamental Concepts

1

5.2 Schr ¨odinger Equation

Let us defineΨ to be Ψ = Ae−iω(_t−x v) (5.2.1)

Simplifying in terms of Energy, E, and momentum, p, we get Ψ = Ae−i(Et−px)

~ (5.2.2)

We obtain Schr ¨odinger’s Equation from the Hamiltonian

H= T + V (5.2.3) To determine E and p, ∂2_Ψ ∂x2 = − p2 ~2Ψ (5.2.4) ∂Ψ ∂t = iE ~Ψ (5.2.5) and H = p 2 2m + V (5.2.6) This becomes EΨ = HΨ (5.2.7)

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DRAFT

36 Quantum Mechanics EΨ = −~ i ∂Ψ ∂t p2Ψ = −~2 ∂2_Ψ ∂x2

The Time Dependent Schr ¨odinger’s Equation is i~∂Ψ ∂t = − ~2 2m ∂2_Ψ ∂x2 + V(x)Ψ (5.2.8)

The Time Independent Schr ¨odinger’s Equation is EΨ = −~2

2m ∂2_Ψ

∂x2 + V(x)Ψ (5.2.9)

5.2.1 Infinite Square Wells

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

for |x|> a,

so that a nonvanishing force acts only at ±a/2. An energy, E, is assigned to the 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 Schr ¨odinger Time Independent Equation, eq. (5.2.9), we substitute V(x) and in the region (−a/2, a/2), we get

− ~

2

2m d2_ψ

dx2 = Eψ (5.2.10)

This differential is of the form

d2_ψ dx2 + k 2_{ψ = 0} _(5.2.11) where k= r 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

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

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DRAFT

Schr¨odinger Equation 37 It shows that

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

kn = nπ

a for n = 1, 2, 3, · · · Substituting this result into eq. (5.2.12) gives

kn= nπ

a = √

2mEn

~ (5.2.16) Solving for Engives

En=

n2_π2

~2

2ma2 (5.2.17)

We cna now solve for B by normalizing the function Z a 0 |_B|2_sin2_kxdx= |A|2 a 2 = 1 So |_A|2 = 2 a (5.2.18) So we can write the wave function as

ψn(x)= r 2 a sin nπx a (5.2.19)

5.2.2 Harmonic Oscillators

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

2kx

2 _(5.2.20)

So the force experienced by this particle is F= −dV

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DRAFT

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

md

2_x

dt2 = −kx (5.2.22)

And the solution of this equation is

x(t)= A cosω0t+ φ

(5.2.23) where the angular frequency,ω0is

ω0=

r k

m (5.2.24) The Quantum Mechanical description on the harmonic oscillator is based on the eigen-function solutions of the time-independent Schr ¨odinger’s equation. By taking V(x) from eq. (5.2.20) we substitute into eq. (5.2.9) to get

d2ψ dx2 = 2m ~2 k 2x 2₋_E ! ψ = mk ~2 x2− 2E k ψ With some manipulation, we get

~ √ mk d2_ψ dx2 =       √ mk ~ x 2₋ 2E ~ r m k      ψ

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

d2_ψ

dξ2 =ξ

2₋λ ψ _(5.2.27)

where the eigenfunctionψ 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 ξ2 max = √ mk ~ A 2 _(5.2.28)

Using the classical connection between A and E, allows us to say ξ2 max= √ mk ~ 2E k = λ (5.2.29)

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DRAFT

Schr¨odinger Equation 39 From eq. (5.2.27), we see that in a quantum mechanical oscillator, there are non-vanishing solutions in the forbidden regions, unlike in our classical case.

A solution to eq. (5.2.27) is ψ(ξ) = e−ξ2_/2 (5.2.30) where dψ dξ = −ξe −ξ2_/2 and d ψ dξ2 = ξ 2_e−_xi2_/2 −_e−ξ2/2=ξ2_{− 1}_e−ξ2/2 This gives is a special solution forλ where

λ0 = 1 (5.2.31)

Thus eq. (5.2.26) gives the energy eigenvalue to be E0= ~ω0

2 λ0 = ~ω0

2 (5.2.32) The eigenfunction e−ξ2_/2

corresponds to a normalized stationary-state wave function Ψ0(x, t) = mk π2_~2 !1₈ e− √ mk x2_/2~ e−iE0t/~ _(5.2.33)

This solution of eq. (5.2.27) produces the smallest possibel result ofλ 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

For the Finite Square Well, we have a potential region where V(x) =( −V0 for −a ≤ x ≤ a

0 for |x|> a We have three regions

Region I: x< −a In this region, The potential, V = 0, so Schr¨odinger’s Equation

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DRAFT

40 Quantum Mechanics 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 Schr ¨odinger’s Equation, eq. (5.2.9), gives − ~ 2 2m d2_ψ dx2 −V0ψ = Eψ or d 2_ψ dx2 = −l 2_ψ where l ≡ p 2m (E+ V0) ~ (5.2.35) 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) This gives us ψ(x) =          Beκx for x < a D cos(lx) for 0 < x < a Fe−κx for x > a (5.2.38)

5.2.4 Hydrogenic Atoms

c

5.3 Spin

3

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DRAFT

Angular Momentum 41

5.4 Angular Momentum

4

5.5 Wave Funtion Symmetry

5

5.6 Elementary Perturbation Theory

6

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DRAFT

Chapter

6 Atomic Physics

6.1 Properties of Electrons

1

6.2 Bohr Model

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 1 4π0 e2 r2 = mev2 r (6.2.1) The Total Energy is the sum of the potential and kinetic energies, so

E= K + U = p

2

2me

− |_{f race}24π₀_r _(6.2.2) We can further reduce this equation by subsituting the value of momentum, which we find to be p2 2me = 1 2mev 2 ₌ e2 8π0r (6.2.3) Substituting this into eq. (6.2.2), we get

E= e 2 8π0r − e 2 4π0r = −_8πe2 0r (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

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DRAFT

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.

L= m3vr (6.2.6)

From eq. (6.2.1) we find v and by substitution, we find L. L= er m3r

4π0

(6.2.7) Solving for r, gives

r= L

2

mee2/4π0

(6.2.8) We apply the condition from eq. (6.2.5)

rn = n2 ~2 mee2/4π0 = n2_a 0 (6.2.9)

where a0is the Bohr radius.

a0 = 0.53 × 10−10m (6.2.10)

Having discreet values for the allowed radii means that we will also have discreet values for energy. Replacing our value of rninto eq. (6.2.4), we get

En = − me 2n2 e2 4π0~ ! = −13.6 n2 eV (6.2.11)

6.3 Energy Quantization

3

6.4 Atomic Structure

4

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Atomic Spectra 45

6.5 Atomic Spectra

6.5.1 Rydberg’s Equation

1 λ = RH ₁ n02 − 1 n2 (6.5.1) where RH is the Rydberg constant.

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.

6.6 Selection Rules

6

6.7 Black Body Radiation

6.7.1 Plank Formula

u( f, T) = 8π~ c3 f3 eh f/kT_{− 1} (6.7.1)

6.7.2 Stefan-Boltzmann Formula

P(T) = σT4 (6.7.2)

6.7.3 Wein’s Displacement Law

λmaxT= 2.9 × 10−3m.K (6.7.3)

6.7.4 Classical and Quantum Aspects of the Plank Equation

Rayleigh’s Equation

u( f, T) = 8π f2

c3 kT (6.7.4)

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DRAFT

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+ h f kT − 1=

h f

kT (6.7.6) Thus eq. (6.7.1) takes the approximate form

u( f, T) ' 8πh c3 f 3kT h f = 8π f2 c3 kT (6.7.7)

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

Quantum

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) ' 8πh c3 f 3_e−_{h f}/kT (6.7.9)

6.8 X-Rays

6.8.1 Bragg Condition

2d sinθ = mλ (6.8.1) for constructive interference off parallel planes of a crystal with lattics spacing, d.

6.8.2 The Compton E

ffect

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

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DRAFT

Atoms in Electric and Magnetic Fields 47 The photon has an associated momentum

E = pc (6.8.3) ⇒ _p = E c = hυ c = h λ (6.8.4) The Relativistic Energy for the electron is

E2 = p2c2+ m2_ec4 (6.8.5) where

p − p0 = P (6.8.6)

Squaring eq. (6.8.6) gives

p2− 2p · p0+ p02 = P2 (6.8.7) Recall thatE = pc and E0 _{= cp}0

, we have

c2p2− 2c2p · p0+ c2p02 = c2P2 E2_{− 2E E}0

cosθ + E02 = E2−_m2

ec4 (6.8.8)

Conservation of Energy leads to

E + mec2 = E 0 + E (6.8.9) Solving E − E0 = E − mec2 E2_{− 2E E}0 + E0 = E2_{− 2Em} ec2+ m2ec4 (6.8.10) 2E E0 − 2E E0 cosθ = 2Emec2− 2m2ec4 (6.8.11) Solving leads to ∆λ = λ0₋ λ = h mec (1 − cosθ) (6.8.12) whereλc = _mh_e_c is the Compton Wavelength.

λc=

h

mec = 2.427 × 10

−12_m _(6.8.13)

6.9 Atoms in Electric and Magnetic Fields

6.9.1 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 (6.9.1)

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DRAFT

48 Atomic Physics or more simply FB = qvB. As this traces a circular path, from Newton’s Second Law,

we see that

FB =

mv2

R = qvB (6.9.2) Solving for R, we get

R= mv

qB (6.9.3) We also see that

f = qB

2πm (6.9.4) The frequency is depends on the charge, q, the magnetic field strength, B and the mass of the charged particle, m.

6.9.2 Zeeman E

ffect

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. 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 andυ0+ 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

υ0 = 1 2π r k me (6.9.5) 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

mev2

r = 4π

2_υ2_rm e

Thus the certipetal force is 4π2_υ2_rm

e= 2πυreB + kr for clockwise motion

4π2_υ2_rm

e= −2πυreB + kr for counterclockwise motion

We use eq. (6.9.5), to emiminate k, to get υ2₋ eB

2πmeυ − υ0 = 0 (Clockwise)

υ2₊ eB

2πme

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DRAFT

Atoms in Electric and Magnetic Fields 49 As we have assumed a small Lorentz force, we can say that the linear terms inυ are small comapred toυ0. Solving the above quadratic equations leads to

υ = υ0+

eB 4πme

for clockwise motion (6.9.6) υ = υ0−

eB 4πme

for counterclockwise motion (6.9.7) We note that the frequency shift is of the form

δυ = _4πmeB

e

(6.9.8) 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 ofυ0−δυ.

6.9.3 Franck-Hertz Experiment

The Franck-Hertz experiment, performed in 1914 by J. Franck and G. L. Hertz, mea-sured the colisional excitation of atoms. Their experiement studied the current 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 mer-cury 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}

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Chapter

7 Special Relativity

7.1 Introductory Concepts

7.1.1 Postulates of Special Relativity

1. The laws of Physics are the same in all inertial frames. 2. The speed of light is the same in all inertial frames. We can define γ = q 1 1 − u2 c2 (7.1.1)

7.2 Time Dilation

∆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= L

0

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.

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52 Special Relativity

7.4 Simultaneity

4

7.5 Energy and Momentum

7.5.1 Relativistic Momentum & Energy

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

Relativistic Momentum

¯p= γm ¯v (7.5.1)

Relativistic Energy

E= γmc2 (7.5.2)

7.5.2 Lorentz Transformations (Momentum & Energy)

p0x = γ px−β E c (7.5.3) p0_y = py (7.5.4) p0_z = pz (7.5.5) E0 c = γ _E c −βpx (7.5.6)

7.5.3 Relativistic Kinetic Energy

K= E − mc2 (7.5.7) = mc2            1 q 1 − v_c22 − 1            (7.5.8) = mc2 _{γ − 1} _(7.5.9)

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Four-Vectors and Lorentz Transformation 53

7.5.4 Relativistic Dynamics (Collisions)

∆P0 x = γ ∆Px−β∆E c (7.5.10) ∆P0 y = ∆Py (7.5.11) ∆P0 z = ∆Pz (7.5.12) ∆E0 c = γ ∆E c −β∆Px (7.5.13)

7.6 Four-Vectors and Lorentz Transformation

We can represent an event in S with the column matrix, s, s=             x y z ict             (7.6.1) A different Lorents frame, S0

, corresponds to another set of space time axes so that s0 =             x0 y0 z0 ict0             (7.6.2) The Lorentz Transformation is related by the matrix

            x0 y0 z0 ict0             =             γ 0 0 iγβ 0 1 0 0 0 0 1 0 −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 matrixL 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 sTs=h x y z ict i             x y z ict             = x2_{+ y}2_{+ z}2₋_c2_t2 _(7.6.5)

Similarly we can show that

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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

b=             bx by 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

p=             px py pz iE/c             (7.6.9) Applying the same transformation rule, we have

p0 = L p (7.6.10) We can also get a Lorentz-invariation relation between momentum and energy such that

p0Tp0 = pTp (7.6.11) The resulting equality gives

p02_x + p02_y + p02_z − E 02 c2 = p 2 x+ p2y+ p2z− E2 c2 (7.6.12)

7.7 Velocity Addition

v0 = v − u 1 − uv_c2 (7.7.1)

7.8 Relativistic Doppler Formula

¯ υ = υ0 r c+ u c − u let r= r c − u c+ u (7.8.1) We have ¯

υreceding = rυ0 red-shift (Source Receding) (7.8.2)

¯

υapproaching = υ0

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Lorentz Transformations 55

7.9 Lorentz Transformations

Given two reference frames S(x, y, z, t) and S0

(x0_{, y}0_{, z}0_{, t}0

), where the S0

-frame is moving in the x-direction, we have,

x0 = γ (x − ut) x= (x0−_ut0₎ _(7.9.1) y0 = y y= y0 (7.9.2) z0 = y y0 = y (7.9.3) t0 = γ t − u c2x t= γ t0+ u c2x 0 (7.9.4)

7.10 Space-Time Interval

(∆S)2 _{= (∆x)}2_{+ ∆y}2_{+ (∆z)}2₋ c2(∆t)2 (7.10.1) Space-Time Intervals may be categorized into three types depending on their separa-tion. They are

Time-like Interval

c2∆t2 > ∆r2 (7.10.2) ∆S2 _{> 0} _(7.10.3)

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

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Chapter

8 Laboratory Methods

8.1 Data and Error Analysis

8.1.1 Addition and Subtraction

x= a + b − c (8.1.1) The Error in x is

(δx)2 _{= (δa)}2_{+ (δb)}2_{+ (δc)}2

(8.1.2)

8.1.2 Multiplication and Division

x= a × b c (8.1.3) The error in x is δx x 2 =δa a 2 + δb b !2 +δc c 2 (8.1.4)

8.1.3 Exponent - (No Error in b)

x= ab (8.1.5) The Error in x is δx x = b δa a (8.1.6)

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58 Laboratory Methods

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

δx = d ln a da ·δa = 1 a ·δa = δa a (8.1.8) Base 10 x= log₁₀a (8.1.9) The Error in x can be derived as such

δx = d(log a) da δa = ln a ln 10 da δa = 1 ln 10 δa a = 0.434δa a (8.1.10)

8.1.5 Antilogs

Base e x= ea (8.1.11) We take the natural log on both sides.

ln x= a ln e = a (8.1.12) Applaying the same general method, we see

d ln x

dx δx = δa ⇒ δx

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Instrumentation 59

Base 10

x= 10a (8.1.14) We follow the same general procedure as above to get

log x= a log 10 log x dx δx = δa 1 ln 10 d ln a dx δx = δa δx x = ln 10δa (8.1.15)

8.2 Instrumentation

2

8.3 Radiation Detection

3

8.4 Counting Statistics

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

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60 Laboratory Methods The fractional uncertainty of our count is δN_N. We can relate this in terms of the count rate. δR R = δN T N T = δN N = √ N N = 1 N (8.4.4) We see that our uncertainty decreases as we take more counts, as to be expected.

8.5 Interaction of Charged Particles with Matter

5

8.6 Lasers and Optical Interferometers

6

8.7 Dimensional Analysis

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, andθ, respectively, each raised to rational powers.

8.8 Fundamental Applications of Probability and

Statis-tics

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Chapter

9 Sample Test

9.1 Period of Pendulum on Moon

The period of the pendulum, T, is

T= 2π s

`

g (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) 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 Te= 2π

s `

ge = 1 s

(9.1.4) and similarly for the moon, the period becomes

Tm = 2π

s ` gm

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62 Sample Test Substituting eq. (9.1.3) into eq. (9.1.5) gives

Tm = 2π s ` gm = √6 Te= √ 6 s Answer: (D)

9.2 Work done by springs in series

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

k1 =

1

3k2 (9.2.2) 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 = −k1x1= F2 = −k2x2 (9.2.3)

where k1 and k2 are the spring constants for the springs S1 and S2 respectively. Thus

we see k1 k2 = x2 x1 = 1 3 (9.2.4) The work done in stretching a spring or its potential energy is

W= 1 2kx 2 _(9.2.5) Thus W1 W2 = 1 2k1x 2 1 1 2k2x 2 2 = k1 k2 · _x 1 x2 2 = 3 (9.2.6) Answer: (D)