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A textbook developed as part of a writing class for physics majors taught at the University of Massachusetts Amherst
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INTELLIGENT DESIGNSselected topics in physics
x
y
z
Assigned byJuan Pablo FernándezDepartment of Physics
University of MassachusettsAmherst, Massachusetts
INTELLIGENT DESIGNSselected topics in physics
INTELLIGENT DESIGNSselected topics in physics
x
y
z
b
Assigned byJuan Pablo FernándezDepartment of Physics
University of MassachusettsAmherst, Massachusetts
Selected Topics in Physics: Intelligent Designs© University of Massachusetts, October 2008
A U T H O R S
Sam Bingham sbingham @student.umass.edu
Samuel Boone sboone
Morgan-Elise Cervo mcervo
Jose Clemente jaclemen
Adam Cohen afcohen
Robert Deegan rdeegan
Matthew Drake mdrake
Christopher Emma cpemma
Sebastian Fischetti sfischet
Keith Fratus kfratus
Douglas Herbert dherbert
Paul Hughes phughes
Christopher Kerrigan crkerrig
Alexander Kiriakopoulos akiriako
Collin Lally clally
Amanda Lund alund
Christopher MacLellan cmaclell
Matthew Mirigian mmirigia
Tim Mortsolf tmortsol
Andrew O’Donnell anodonne
David Parker dparker
Robert Pierce rpierce
Richard Rines rrines
Daniel Rogers drrogers
Daniel Schmidt dschmidt
Jonah Zimmerman jzimmerm
v
C O N T E N T S
preface xi
i Seeing the Light 1
1 morgan-elise cervo : why is the sky blue? 3
1.1 Introduction 3
1.2 Waves 3
1.3 Electromagnetic Waves 3
1.4 Radiation 5
1.5 Blueness of the Sky 7
1.6 Problems 8
2 matthew mirigian: the physics of rainbows 9
2.1 Introduction 9
2.2 The Primary Bow 9
2.3 The Secondary Bow 10
2.4 Dispersion 11
2.5 Problems 12
3 matthew drake : the camera and how it works 13
3.1 Introduction 13
3.2 Lenses 14
3.3 The Camera Itself 15
3.4 Questions 17
3.5 Solutions 17
4 sebastian fischetti : holography : an introduction 19
4.1 Introduction 19
4.2 The Geometric Model 19
4.3 Types of Holograms 21
4.4 Making Holograms 22
4.5 Applications of Holography 23
4.6 Problems 24
5 colin lally : listening for the shape of a drum 27
5.1 Introduction 27
5.2 Eigenvalues 28
5.3 Problems 29
ii Mind Over Matter 31
6 christopher maclellan: glass 33
6.1 Introduction to Glass 33
6.2 Amorphous Solids 33
6.3 The Glass Transition 34
6.4 Simulating Amorphous Materials with Colloids 35
6.5 Practice Questions 37
7 robert pierce: the physics of splashing 39
7.1 Introduction 39
7.2 Pressure, Surface Tension, and other Concepts 39
7.3 Splashing 42
7.4 Summary 43
7.5 Chapter Problems 44
7.6 Multiple Choice Questions 45
8 sam bingham: freak waves 47
8.1 Introduction 47
vii
viii contents
8.2 Linear Model 48
8.3 Interference 49
8.4 Nonlinear Effects 52
8.5 Conclusion 53
8.6 Problems 54
9 paul hughes: friction 55
9.1 Overview 55
9.2 Amontons/Coulomb Friction 55
9.3 Toward A Conceptual Mesoscopic Model 56
9.4 Summary 58
9.5 Problems 58
10 keith fratus : neutrino oscillations in the standard
model 61
10.1 Introduction 61
10.2 A Review of Quantum Theory 61
10.3 The Standard Model 62
10.4 The Weak and Higgs Mechanisms 64
10.5 The Origin of Neutrino Oscillations 67
10.6 Implications of the Existence of Neutrino Oscillations 71
10.7 Problems 73
10.8 Multiple Choice Test Problems 75
iii Information is Power 77
11 andy o’donnell : fast fourier transform 79
11.1 Introduction 79
11.2 Fourier Transform 79
11.3 Discrete Transform 80
11.4 The Fast Fourier Transform 81
11.5 Multiple Choice 83
11.6 Homework Problems 83
12 tim mortsolf: the physics of data storage 85
12.1 Introduction 85
12.2 Bits and Bytes — The Units of Digital Storage 85
12.3 Storage Capacity is Everything 87
12.4 The Physics of a Hard Disk Drive 88
12.5 Summary 94
12.6 Exercises 94
12.7 Multiple Choice Questions 95
13 tim mortsolf : the physics of information theory 97
13.1 Introduction 97
13.2 Information Theory – The Physical Limits of Data 97
13.3 Shannon’s Formula 99
13.4 The Physical Limits of Data Storage 101
13.5 Summary 103
13.6 Exercises 103
13.7 Multiple Choice Questions 105
14 sam boone : analytical investigation of the optimal
traffic organization of social insects 107
14.1 Introduction 107
14.2 Ant Colony Optimization 109
14.3 Optimization By Hand 109
14.4 Applying ACO Meta-Heuristic to the Traveling SalesmanProblem 110
contents ix
14.5 Solution to the Traveling Salesman Problem Using anACO Algorithm 112
15 christopher kerrigan : the physics of social insects 117
15.1 Introduction 117
15.2 The Electron and the Ant 117
15.3 Insect Current 118
15.4 Insect Diagram 118
15.5 Kirchoff’s Rules (for ants) 119
15.6 Differences 119
15.7 The Real Element 120
15.8 Problems 121
iv What’s Out There 123
16 robert deegan: the discovery of neptune 125
16.1 Introduction 125
16.2 Newton’s Law of Universal Gravitation 125
16.3 Adams and Le Verrier 126
16.4 Perturbation 127
16.5 Methods and Modern Approaches 128
16.6 Practice Questions 129
16.7 Answers to Practice Questions 129
17 alex kiriakopoulos: white dwarfs 131
17.1 Introduction 131
17.2 The Total Energy 132
17.3 Question 133
18 daniel rogers : supernovae and the progenitor the-ory 135
18.1 Introduction 135
18.2 Creation of Heavy Elements 135
18.3 Dispersal of Heavy Elements 136
18.4 Progenitor Theory 138
18.5 A Mathematical Model 138
18.6 Conclusion 139
18.7 Problems 140
19 david parker: the equivalence principle 143
19.1 Introduction 143
19.2 Weak? Strong? Einstein? 143
19.3 Consequences 144
19.4 Example 144
19.5 Problems 145
20 richard rines: the fifth interaction 147
20.1 Introduction: The ‘Four’ forces 147
20.2 The Beginning: Testing Weak Equivalence 147
20.3 A New Force 148
20.4 The Death of the Force 150
20.5 Problems 151
21 douglas herbert : the science of the apocalypse 153
21.1 Introduction 153
21.2 Asteroid Impact 153
21.3 Errant Black Holes 155
21.4 Flood volcanism 155
21.5 Giant Solar Flares 156
21.6 Viral Epidemic 157
22 amanda lund : extraterrestrial intelligence 159
x contents
22.1 Introduction 159
22.2 The Possibility of Life in the Universe 159
22.3 The Search for Intelligent Life 160
22.4 Will We Find It? And When? 161
22.5 Problems 162
22.6 Multiple Choice Questions 163
22.7 Summary 164
a Bibliography 165
Index 173
Physics is to be regarded not so much as the study of somethinga priori given, but rather as the development of methods
for ordering and surveying human experience.— N. Bohr [59]
P R E FA C E
This book has been produced as an assignment for Physics 381,Writing in Physics, taught by the Department of Physics of the
University of Massachusetts Amherst in the Fall 2008 semester.
instructor
Juan Pablo Fernández; 1034 [email protected] or [email protected]
teaching assistant
Benjamin [email protected]
class times and office hours
Most of this class will be taught long distance. I plan to come toAmherst twice a month; on those days we will meet at the usualclass time and hold one-on-one conferences to discuss work inprogress. The best dates and times we will agree upon in class.
A good fraction of our communication will take place via email.Feel free to email me at any time with any questions, comments,requests for help, etc. There is one exception, though: Any ques-tions or comments about end-of-semester grades must be submitted inhard copy.
textbook
Nicholas J. Higham, Handbook of Writing for the Mathematical Sci-ences, Second Edition. Philadelphia, SIAM, 1998.
description
As a professional physicist you will be expected to communicatewith four kinds of audiences, each of which has a direct bearing onyour livelihood: Professionals—including you—that work on yourfield, professionals that work on other fields of physics or science,students of physics or other disciplines, and the general public—i.e., the taxpayer. Most of this communication will be in writing,which in physics includes not just prose but also mathematics anddisplayed material. In this course you will acquaint yourself withthe many different elements that contribute to successful physicswriting and will put them to work in different contexts.
objectives
1. Articulate concepts, methods, and results of theoretical orexperimental physics to other physicists, other scientists,students, and laypeople.
2. Be confident in the use of LATEX and other public-domainproductivity tools for scientists.
3. Appreciate the amount of work that goes into correct, clearwriting.
xi
xii contents
4. Show proper respect to your readers, making sure not towrite above their heads nor “write down at them.”
5. Find the right combination of prose, mathematics, tables, andgraphics that will help you make your points most clearlyand economically.
6. Learn to deal with the limitations of a given medium. By thesame token, learn to appreciate (and not abuse) the marvelsthat technology affords you nowadays.
7. Practice proper attribution when making use of other peo-ple’s work.
8. Collaborate with your classmates in the development ofwritten materials.
9. Have a working knowledge of the peer-review system ofpublication.
evaluation
The grade for this course will be based on five writing projectsassigned in the following order:
1. A journal paper2. A grant proposal3. A textbook4. A science newspaper5. A final project
In due time I will provide more details about each of the projectsand propose a few different topics from which to choose. Ifyou would rather write about something else you must tell mepromptly.
You will hand in two drafts of each project. The first one mustbe submitted in “draft” form and will receive extensive feedback.The second draft will be considered final in terms of content andform and will be assessed as such.
You will have roughly three weeks to complete each project. Thefirst week you can devote to experimenting with the physics andthe technology, the second to producing the first draft, and thethird to producing the final draft.
At least the first two projects will be peer-reviewed: everybodywill (anonymously) evaluate two papers and have theirs evaluatedby two classmates. The evaluations will include a suggested grade,usually an integer from 7 to 10, and will themselves be graded.
For the third and fourth projects you will have to collaborate withyour classmates.
The fifth project will be freestyle and may involve media otherthan paper.
Part I
S E E I N G T H E L I G H T
1M O R G A N - E L I S E C E RV O : W H Y I S T H E S K Y B L U E ?
1.1 introduction
Have you ever looked up at the sky and been amazed by its brilliantshade of blue? Or why the sun changes color at sunset? In this
chapter we will provide an answer to these questions through the studyof electromagnetic waves. We will uncover why sky is blue and notwhite. We will also investigate sunsets and learn why sunsets in somelocations are more beautiful than others.
1.2 waves
To understand what is happening up in the sky we first need to reviewthe general properties of waves. A wave is a disturbance of a continuousmedium that propagates with a fixed shape at a constant velocity [36].A familiar wave equation is that of a sinusoidal wave. This wave can berepresented by the equation,
f (z, t) = A cos[k(z− vt) + δ]. (1.1)
In the above equation the variable A represents the amplitude. Theargument of cosine represents the phase of the wave and δ representsa phase constant. We are familiar with the idea that the wavenumberk = 2π/λ. If we know the velocity v of the wave then we can find theperiod, T (the amount of time it takes for the wave complete a cycle),using the equation,
T =2π
kv. (1.2)
Another useful property of waves to know is the frequency of the wave,which represents the number of oscillations that occur per a unit oftime. The frequency is
f =1T
=kv2π
=vλ
. (1.3)
Frequency can also be solved for in terms of angular frequency. We canfind angular frequency by
ω = 2π f = kv. (1.4)
Now that we understand the different properties of waves it is easy torewrite the equation for a sinusoidal wave as
f (z, t) = A cos(kz + ωt− δ). (1.5)
1.3 electromagnetic waves
Electromagnetic waves are formed when an electric field is combinedwith a magnetic field. The electric and magnetic fields lie orthogonal
3
4 morgan-elise cervo: why is the sky blue?
Figure 1.1: This chart provides wavelengths for the different types of electro-magnetic waves [9].
to motion of the electromagnetic wave. Electromagnetic waves arecategorized by their wavelength. In Figure 1, we have a chart whichgives the wavelengths of familiar electromagnetic waves such as x-raysand microwaves.
In order to understand how the equation for electromagnetic waves isderived we will first review Maxwell’s Equations. Gauss’s law describesthe distribution of electric charge in the formation of an electric field bythe equation,
∇ · E =ρ
ε0(1.6)
where ρ represents the charge density and ε0 is the electric constant.Gauss’s law, in other words, is a mathematical definition of an electricfield. Another equation from Gauss that is useful to know is,
∇ · B = 0. (1.7)
The above equation states that the divergence in a magnetic field is zero;in other words, there are no monopoles in magnetic fields. The thirdMaxwell equation is known as Faraday’s induction law and it forms thebasis of electrical inductors and transformers [62]. The equation reads,
∇× E = −∂B∂t
. (1.8)
In other words, the line integral of the electric field around a closedloop is equal to the negative of the induced magnetic field [8]. The finalMaxwell equation is a correction to Ampere’s law. It states that in astatic electric field, the integral of the magnetic field around a closedloop is proportional to the current flowing through the loop. Rewrittenin mathematical terms we have,
∇× B = µ0 J + µ0ε0∂E∂t
. (1.9)
1.4 radiation 5
Figure 1.2: Notice how from the horizon the color is gradient of color of longestwavelength to shortest.
Now we have the necessary tools to write an equation for an electro-magnetic wave. The equations to describe an electromagnetic wave in avacuum are as follows:
E(r, t) = E0 cos(k · r−ωt + δ)n. (1.10)
B(r, t) =1c
E0 cos(k · r−ωt + δ)(k× n). (1.11)
where k is the propagation vector, n is the polarization and ω is thefrequency.
1.4 radiation
Now that we know electromagnetic waves exist you might question,where do electromagnetic waves come from? The answer is radiation.When a charge is accelerating and therefore changing currents anelectromagnetic wave is produced.
Let’s consider the case of a charge being driven across a wire con-necting to metal spheres. The equation for the charge with respect totime can be written as
q(t) = q0 cos(ωt). (1.12)
The produced electric dipole as a result of the oscillating charge canthen be written as,
p(t) = q0d cos(wt)z, (1.13)
where q0d = p0 is the maximum value of the dipole moment [36]. Theretarded potential of an electromagnetic wave describes the potential foran electromagnetic field of a time-varying current or charge distribution.The retarded potential of the dipole system, for a wave traveling througha vacuum at the speed of light, can be derived using the equation,
V(r, t) =1
4πε0
∫ρ(r′, tr)
γdτ′, (1.14)
where tr represents the retarded time. Using spherical coordinates theequation for the retarding potential can be rewritten as,
V(r, Θ, t) = − p0ω
4πε0c(
cos Θr
) sin[ω(t− r/c)] (1.15)
6 morgan-elise cervo: why is the sky blue?
rdz
q+
q-
Figure 1.3: Problem Geometry
To find the potential vector A, we use the equation for the current inthe wire (the derivative of the z component of charge with respect totime),
I(t) = −q0ω sin(ωt)z. (1.16)
Making reference to Figure 3 we derive,
A(r, t) =µ0
4π
∫d/2−d/2−q0ω sin[ω(t− γ/c)]
γdz. (1.17)
To eliminate the integral’s variable of d, we replace it with the value atthe center leading to,
A(r, Θ, t) = −µ0 p0ω
4πrsin[ω(t− r/c)]z. (1.18)
The equation of the electric field can then be derived by pluggingobtain values into the equation,
E = −∇V − ∂A∂t
(1.19)
Using the values we obtained for the vector potential and potential wefind,
E = −µ0 p0ω2
4π(
sin Θr
) cos[ω(t− r/c)]Θ (1.20)
To find the magnetic field we use B = ∇× A.
B = −µ0 p0ω2
4πc(
sin Θr
cos[ω(t− r/c)]Φ. (1.21)
So far we have described a wave moving radially outward at a frequencyof ω. We have also defined the electric and magnetic fields, which areorthogonal to each other and in the same phase. We now find theintensity of the wave. The energy radiated by an oscillating dipole canbe found by the Poynting vector [36],
〈S〉 =1
µ0(E× B) (1.22)
When we use our values for the magnetic and electric field we find,
〈S〉 =µ0 p2
0ω4
32π2csin2 θ
r2 r. (1.23)
1.5 blueness of the sky 7
Figure 1.4: Plot of three color cones that shows which wavelengths they bestreceive [9]
A visual interpretation of the intensity shows that the function takesthe shape of a donut with the hole along the axis of the dipole. In otherwords there is no radiation along this axis. If we integrate 〈S〉 over asphere of radius r we can find the total power radiated.
〈P〉 =µ0 p2
0ω4
32π2c
∫ sin2 Θr2 r2 sin ΘdΘdφ =
µ0 p20ω4
12πc. (1.24)
Notice that the power of radiation is not dependent on the radius, andtherefore size, of the sphere. However, the power is highly dependenton the frequency, ω.
1.5 blueness of the sky
If we consider the radiation of the sun’s light to be compatible with ourequations for an electromagnetic wave traveling in a vacuum, then wecan say that the large dependence of frequency in the power equationproduces the blueness of the sky. We know already that white light iscomposed of several wavelengths; each wavelength represents a differ-ent color. Shorter wavelengths, or light waves of high frequencies aremore effective in the power equation than light waves of longer lengths.Blue has a shorter wavelength than red for example and consequentlythe sky appears to be blue to us.
example 1 .1: why isn’t the sky violet?We now know that light of greater frequency is radiated most strongly. If welook at Figure 1 we would expect that the sky should be violet instead of bluebecause violet has the shortest wavelength. The reason that the sky looks blueand not purple is because our eyes are able to see some colors more easilythan others. Our eyes have three types of color receptors, or cones. The conesare called blue, green and red; the names of the cones come from the colorthat the receptor most strongly correlates to. The red receptor best sees red,orange and yellow light. The green receptor, best sees green and yellow lightand the blue receptor, blue light. Even though the sky appears blue we knowthat there is strong presence of violet light because we see blue with red tints.If the violet light were absence the sky would appear blue with a green tint.
So why does the sun turn red when the sun is setting? This isespecially puzzling be cause we just explained that red light, with
8 morgan-elise cervo: why is the sky blue?
the lowest frequency, is the least powerful. When the sun is about topass over the horizon, the sun’s light has to travel a greater distancethrough the atmosphere to reach the observer. The particles in theearth’s atmosphere cause more waves of shorter wavelengths to bescattered across this distance. For this reason, the sky is red when thesun is setting. Similarly, sunsets may be more brilliant in coastal areasor very dusty area because there are more particles in the air to scatterlight.
1.6 problems
1. Using the equation for combining waves, A3eiδ3 = A1eiδ1 + A2eiδ2
determine A3 and δ3 in terms of A1, A2, δ1, and δ2 [36].
Solution: If we use the fact that
eiΘ = cos Θ + i sin Θ
we can get that
A3(cos δ3 + i sin δ3) = A2(cos δ2 + i sin δ2) + A1(cos δ1 + i sin δ1).
Now by separating the real and imaginary parts, we have two equations
A3 cos δ3 = A1 cos δ1 + A− 2 cos δ2
A3i sin δ3 = A1 sin δ1 + A− 2 sin δ2
Dividing the second equation by the first and taking the inverse tangent,
δ3 = tan−1( A1 sin δ1+A−2 sin δ2A1 cos δ1+A−2 cos δ2
A3 is found by squaring and adding the two previous equations to get,
A3 =[A2
1 + A22 + 2A1 A2 cos(δ1 + δ2)
]1/2
2. Check that the retarded potentials of an oscillating dipole ((1.15) and(1.18)) satisfy the Lorentz gauge condition.
Solution: The Lorentz Gauge condition is ∇· = −µ0ε0∂V∂t . Where A is the
potential due to the current flowing through a loop of wire,
A = −µ0 pω4πr sin(ω(t− r/c))z.
Then,
∇ · A = 1r2
∂∂r (r2 Ar) + 1
r sin Θ ∂∂Θ(sin ΘAΘ) + 1r sin Θ
∂Aφ∂φ
∇·A = µ0 p0ω4π
[1r2
∂∂r
1r r2 sin(ωt− ωr
c )]
cos Θ−[
ωrc cos(ωt− ωr
c]) cos Θ− 2 sin Θ cos Θ
r2 sin Θ sin(ωt− ωrc )
= µ0 p0ω4π
[sin(ωt− ωr
c )r2 + ωr
c cos(ωt− ωrc )]
cos Θ
= µ0ε0
p0ω4πε0
[1r2 sin ω(t− r/c) + ω
rc cos ω(t− r/c)]
cos Θ
.
Solving for the partial of the scalar potential,
∂V∂t = p0 cos Θ
4πε0r
−ω2
c cos [ω(t− r/c)]− ωr sin(t− r/c)
= p0 cos Θ
4πε0r
omega
r sin(ω(t− r/c) + ω2
c cos [ω(t− r/c)]
= p0ω4πε0
[1r2 sin [ω(t− r/c)] + ω
rc cos ω(r− tc)]
cos Θ
By plugging the solved values into the Lorentz Gauge condition we have,
∇· = −µ0ε0∂V∂t .
2M AT T H E W M I R I G I A N : T H E P H Y S I C S O F R A I N B O W S
2.1 introduction
Rainbows are one of nature’s most beautiful sights. Some mightargue that the rainbow is equally beautiful in its neat and compact
packaging of optical phenomena in the physics of the rainbow. Therainbow is formed by a combination of physical properties of light,namely refraction, reflection and dispersion that occur in a drop of rain.When conditions are right a bright inner bow can be observed as wellas a fainter outer bow. The inner, primary bow, is red on the outsideand violet on the inside, whereas the outer secondary bow is red onthe inside and violet on the outside. In this chapter we will explore thephysics responsible for producing the rainbow.
2.2 the primary bow
To understand how rainbows form we start by analyzing the geometri-cal optics involved. At its most fundamental level, a rainbow is formedby the reflection and diffraction of sunlight in spherical raindrops. Tounderstand how the shape of the rainbow comes about it is usefulto think of the light as monochromatic, made up of one wavelength.Secondly, it should be understood that the sunlight strikes raindrops asrays that are parallel to one another.
Figure 2.1 shows a ray incident on the surface of a drop. The ray isparallel to an axis that, if extended backward, would pass through thesun. The ability of a medium, water in this case, to bend light is calledthe refractive index, designated n. It is the ratio of the speed of light c ina vacuum to the speed of light v in a medium.
n =cv
(2.1)
Snell’s Law summarizes an important experimental observation relat-ing the refractive index of media to the angle that light is refracted. Itsays that if the incident angle i and the refracted angle r are measuredwith respect to the normal, the ratio of the sines of the angles is equalto the inverse ratio of the corresponding indexes of refraction. [60].
sin isin r
=n1
n2. (2.2)
In the drop of rain the light rays that form the primary bow are twicerefracted and once reflected. The angle D1 between the incident lightray and the ray that enters the eye to produce the image of a rainbow isdetermined by the incident and reflected angles, i and r. More preciselyit is given by
D1 = 180 − 2(2r− i). (2.3)
9
10 matthew mirigian: the physics of rainbows
Figure 2.1: This is the path of a ray producing the primary bow. [93]
Figure 2.2: This is the path of a ray producing the secondary bow. [93]
It is clear that D1 is determined from the incident angle, i. If the valuesof D1 are plotted with varying i we see that D1 reaches a minimumof approximately 138. The supplementary angle, 42, corresponds toangle above the horizon that the peak of the primary rainbow is seen,and subsequently accounts for the circular shape of the bow, shown infigure 2.3. This means that as the sun rises in elevation a smaller andsmaller portion of the bow is visible to the observer. This also meansthat an increase in elevation of the observer would allow for a largerpart of the bow to be visible, and it is even possible for a completecircular rainbow to be observed [70]. Figure 2.3 demonstrates wherethe rainbow is visible to the observer. It is also important to note thatthis means that the sun can not be more than 42 above the horizon.
2.3 the secondary bow
The secondary bow, which can be visible above the primary bow, isformed by two internal reflections in the drop of rain shown in figure 2.2.Using the same treatment as for the primary bow we can see the angleof deviation, D2 to be
D2 = 2(180 − 2(3r− i). (2.4)
If we again plot D2 for values of i we see that a minimum occurswhere D2 = 231. This corresponds to the angle of 51 that the observer
2.4 dispersion 11
can view the peak of the secondary rainbow. Notice in figure 2.3 thatthe ray is deviated in the opposite direction as for the primary bow.This explains the reverse in color distribution compared to the primarybow.
2.4 dispersion
What we neglected in the discussion of the bow formation from monochro-matic light we will address in a discussion of dispersion which is cen-tral to the dramatic separation of the visible wavelengths that composewhite light. The previous sections merely explain how sunlight shiningfrom behind the viewer will be diffracted and internally reflected inraindrops so as to produce a bow shape. Dispersion explains why it isthat the viewer sees the light separated into the full visible spectrum.Dispersion is simply the phenomenon in which the change in phasevelocity of waves due to interactions in a medium is related to theirfrequency. A simplified consequence of this effect is that the amountlight is refracted depends on the frequency of the light.
A light ray that passes through a vacuum will arrive at some detector,like our eyes, unchanged with constant velocity. A correct assumptionwould be that each photon that is detected originated from the lightsource. However, this is not true for light that passes through anymedium, such as air, glass, or water. The light is transmitted throughthe medium when the incoming photons are absorbed by the atoms ofthe medium and then immediately reemitted by each of the atoms in theray’s path. The principle that light propagates by successive emissionof wavelets from particles in the beam’s path is known as Huygens’Principle and is seemingly in opposition to Newton’s corpuscular theoryof light. This mechanism of transmission through media results inthe slowing in phase velocity, thus leading to refraction as it passesbetween media of differing refractive properties, like from vacuum intoa medium such as glass. We saw a consequence of this earlier, in ourdiscussion of Snell’s Law.
Newton observed the spectrum of visible light by using a prism. Hedetermined that the index of refraction is related to the the wavelengthof light, what we call dispersion. He observed “that the rays which differin refrangibility will have different limits of their angles of emergence,and by consequence according to their different degrees of refrangibilityemerge most copiously in different angles, and being separated fromone another appear each in their proper colours. [93]" This is themechanism by which white light is split into the colors seen in rainbows.
Trends of dispersion through a wider range beyond visible displaysome consequences of complex interactions on the atomic level likeabsorptions into lattice vibrations of low energy and absorptions dueto electronic excitations of high energy. However, as we consider therainbow, it is enough to know that in the region of visible wavelengthsthe index of refraction of water increases for higher energy.
As sunlight enters a raindrop the violet end of the spectrum isrefract the most and red light the least so the full spectrum of visiblewavelengths is observed. We see that the refractive index should beexpressed as some function of the frequency of light n( f ).
12 matthew mirigian: the physics of rainbows
Figure 2.3: The paths of light forming the primary and secondary bow. [89]
example 2 .1: an example
Light traveling through air (n=1) is incident on the surface of water (n=1.33)at an angle of 30. To determine the direction of the refracted ray we can useSnell’s law, equation 1.2.
11.33
=sin(r)
sin(30)
r = 22.2
2.5 problems
1. A red laser produces light with wavelength 630 nm in air and475 nm in water. Determine the index of refraction water and thespeed of the light in water.
Solution: We know that in any material v = λ f . In vacuum this is c = λ0 f .The frequency is constant in all materials and wavelength changes incorrespondence with the change in velocity. So we can say, f = c/λ0 =v/λw. We can combine this with equation 1.1 and see
λw =λ0n
, n =λ0λw
=630 nm475 nm
= 1.33
Then from n = c/v we can determine the velocity of the light in water.
v =cn
=3.00× 108 m/s
1.33= 2.26× 108 m/s
2. A secondary bow is fainter than the primary bow of a rainbow because
A. It is a partial reflection of the primary bow.
B. Only larger drops produce secondary bows.
C. The sunlight reaching the raindrops is less intense.
D. There is an extra partial reflection in the raindrops.
Answer: D
The secondary bow is produced by two internal reflections inside araindrop shown in figure 2.2.
3M AT T H E W D R A K E : T H E C A M E R A A N D H O W I TW O R K S
3.1 introduction
To examine the camera, we must first look back to earlier times.The first reference to a modern camera was known as a “camera
obscura,” which consisted of a large dark room with a pin-hole on oneside. An image would be formed on the wall opposite the pin-hole asshown in Fig. 3.1.
The camera obscura was first introduced as a model for the eye byAbu Ali Al-Hasen ibn al-Hasan ibn Al-Haytham, or Alhazen for short.Alhazen investigated the camera obscura by placing three candlesoutside of the camera obscura and systematically obstructing eachcandle and observing the effects on the produced image. Alhazenobserved that the image produced was reflected about the pin-holepoint, as seen in Fig. 3.1. To explain why the human eye sees imagesright-side up, Alhazen interpreted these findings to mean that theimage was sensed on the outside of the eye, despite having the nerveand retina structure in his model [43].
The camera obscura was later improved by Girolamo Cardano whenhe suggested putting a convex lens in the pin-hole to help focus the im-age. By inserting a convex lens of proper focal length, the image wouldbecome far more clear. In his original description however, Cardanofailed to explain this mathematically [44].
Take a piece of paper, and place it opposite the lens asmuch removed [from the lens], that you minutely see onthat paper all that is outside the house. This happens mostdistinctly at a certain point, which you find by approachingor withdrawing with respect to the lens, until you find theconvenient place.
Figure 3.1: The inverted image is projected on to the opposite wall from thelight coming in through the pin-hole. Image courtesy of [43]
13
14 matthew drake: the camera and how it works
Principles from the camera obscura are still used in modern cameraswhich we will explore in detail later in this chapter. The camera obscura,along with the lenses will be explained in further sections.
3.2 lenses
To begin our discussion of the camera, we must first understand therole of the lens. Lenses come in two forms, converging and diverging.To use the most simple description, a converging lens is shaped likethe outside of a sphere, while a diverging lens is shaped like the insideof a sphere. To understand a lens’ application to the camera, we onlyneed to understand the thin lens approximation as opposed to moredetailed equations. In general, the thin lens approximation is validwhen the thickness of the lens is small compared to the object andimage distances. The thin lens equation states that
1f
=1s
+1s′
(3.1)
where f is the focal length, s is the distance of the object to the lens,and s′ is the distance of the image that is created by the lens, whereall distances are measured along the optical axis. If the image has apositive distance, then it is a real image. If the image is a negativedistance, then it is a virtual image. Two important properties emergefrom the form of Eq. 3.1.
1. If the object is at the same distance as the focal length, than forEq. 3.1 to hold, 1
s′ must go to zero and thus, the image forms at|∞|.
2. If the object is very far away (approaching |∞|), than for Eq. 3.1to hold, the image forms at the focal length of the lens.
I specify that the distances are at absolute value of ∞ because they canbe towards the positive or negative end of the optical axis. Which endof the optical axis an image is located on is dependent on whether thelens is converging or diverging. By definition, if the lens is a converginglens, than the focal length is positive. If the lens is a diverging lens, thanthe focal length is negative. Figure 3.2 shows a typical image formationfrom a converging lens. The optical axis has the origin located at thecenter of the lens with the left side of the lens typically being thenegative direction.
Along with the placement of the image, we can also examine the sizeof the image that is formed. Because the lens bends the light rays, thesize of the image will be different from the size of the object, except in aspecial case. The magnification of the image comes from the equation
m = − s′
s(3.2)
The magnification tells us the ratio of the size of the image to the size ofthe object. A negative sign in the magnification tells us that the imageis inverted, as is the case in the camera obscura.
3.3 the camera itself 15
Figure 3.2: The rays from point Q are refracted by the thin lens to bend towardthe focal point. Because the lens is thin, the ray is considered to bend at themiddle of the lens. Image courtesy of [99]
example 3 .1: forming an image
Suppose there is an object 1 m away from a converging lens. If the focallength of the lens we use is 20 cm, where will the image be formed? Whatwill the magnification of the image be? What does the placement of the imageand the magnification tell us about the image? To do this, we apply Eq. 3.1and solve for s′. Then we use Eq. 3.2 to find the magnification.
1100
cm−1 +1s′
=120
cm−1 (3.3)
1s′
=5
100cm−1 − 1
100cm−1 =
4100
cm−1 (3.4)
s′ = 25 cm (3.5)
m = − 25 cm100 cm
= −.25 (3.6)
The negative magnification tells us that the image is inverted and one fourthof it’s original size. Since the image distance is positive, we also know that itis a real image.
3.3 the camera itself
Now that we have an understanding of lenses, we may apply thatknowledge to understand how the camera works. The basic methodthat the camera uses is to form a real, inverted image onto a smallscreen using a converging lens. In general, only a real image may beprojected onto a screen, so a converging lens is the natural choice forthis application. The camera is made up of a fully enclosed box, aconverging lens, a shutter to open the lens for a small period of time,and a recording medium. Figure 3.3 shows how an object forms on acamera screen.
The screen of the camera is typically some photosensitive material oran electronic detector, depending on if the camera is a digital cameraor not. When the picture is taken, the shutter opens to allow the lightin for a short period of time and the image is first recorded inverted.
16 matthew drake: the camera and how it works
Figure 3.3: The image of the key is real, inverted, and smaller than the originalobject when it is seen through the camera lens. Image courtesy of [99]
The image will have to be inverted again in order to be viewable in thecorrect orientation. On a digital camera, there is an electronic processwhich will do this. On a film-style camera, this is done when printingthe pictures through a screen process.
Most objects that one would want to photograph would be at adistance considerably larger than the distance of the lens to the screen,and so the focal length of the lens must be accordingly small. To see this,you can work out the Forming the Image example again, but insteadsolve for the focal length with a small image distance. As the objectdistance goes towards infinity, you will see that the image distancebecomes equal to the focal length.
For an image to be properly recorded on to the medium, the correctamount of light intensity needs to reach the screen. If the screen gatherstoo much light, the image will look white from being too bright. If thescreen gathers too little light, the image will be too dim to recognize.The amount of light gathered is controlled by the time the shutter isopen, and also by a property known as the f-number. The f-numberis dependent on the focal length of the lens and on the diameter ofthe aperture as controlled by the diaphragm. The aperture is a circulararea where the lens is placed and the diaphragm is an adjustable piecewhich controls the size of the aperture. The f-number is defined as
f-number =fD
(3.7)
where f is the focal length and D is the diameter of the aperture. Thelight intensity that may reach the film is proportional to the square ofthe inverse f-number, the time that the shutter is open, and also thebrightness of the actual object itself.
Nearly all cameras contain options to zoom in or out on objects. Onemethod of doing this is to arrange two converging lenses as shown inFig. 3.4
The primary lens is a converging lens and brings the light raystogether near it’s focal point and inverted. The image is then seenthrough the secondary lens, another converging lens. Because the imageis inverted a second time, it now has the correct orientation and canbe magnified. To obtain the magnification of the zoom lens, you must
3.4 questions 17
Figure 3.4: The secondary lens projects the image of the primary lens to make amagnified image. Image courtesy of [2]
simply combine the two magnifications of the lenses by multiplyingthem.
mtotal = m1m2 (3.8)
This type of zoom lens is one of the most simple lens and is oftenreferred to as the “astronomer’s lens.”
3.4 questions
1. An image produced by a converging lens of an object at a distancegreater than twice the focal length is
• a) real• b) inverted• c) smaller• d) all of the above
2. If the focal length of the lens on a camera is 1cm, how far shouldthe screen be from the lens to take a picture of an object 10 metersaway? Explain why camera manufacturers would place the screena distance equal to the focal length.
3. Consider two converging lenses in an “astronomer’s lens” setupwith a primary focal length of 30cm and at a separation of 50cmapart. What focal length should the secondary lens have if youdesire a magnification of 2 for an object at a distance of 20m? Youmay use the approximation that for s » f, s′ ≈ f
3.5 solutions
1. Answer: d. If you work out the magnification, it will be negativeand have absolute value less than 1. The image distance will bepositive
2. Answer: The image distance would be s′ = 1000999 cm. This is very
nearly equal to the distance of the focal length and since mostpictures will have an object distance much greater than that of thefocal length, the screen will be placed at a focal length away.
3. Answer: You must use Eq. 3.8 and solve for m1 and m2 and then fitthose to the conditions. Since f2 is only used in the s′2 expression,solve to get s′2 = 2s1s2
f1Continuing from there, you will eventually
end up with
f2 = (f1
2s1s2− 1
s2)−1 (3.9)
4S E B A S T I A N F I S C H E T T I : H O L O G R A P H Y: A NI N T R O D U C T I O N
4.1 introduction
The art of holography, although a relatively recent discovery initself, is based on the same principles of optical interference that
you have undoubtedly studied in your introductory physics courses.A detailed discussion of holography would be too lengthy to includehere; rather, we will only provide an introduction to the mechanisms,production, and applications of holography. The interested student iswelcome to read more detailed books on the subject [47].
Furthermore, because this section assumes a thorough understandingof optical interference, we encourage you to review the topic if you feelthe need; see, for example, any standard introductory physics text [100]or, for a more enjoyable read, the Feynman Lectures on Physics [32].
4.2 the geometric model
Let us begin with a brief review of the mechanism behind classicalphotography. An object is photographed by using an optical setup toproject the three-dimensional object into a two-dimensional image; thisimage, which consists almost always of incoherent light, strikes a platecovered in a chemical emulsion, triggering a reaction that essentiallystores the “brightness” of the light striking it; in this sense, a standardphotograph only records information about the amplitude of the lightwaves striking each section of it, and can only record a two-dimensionalimage (we assume, for simplicity, that the light and photograph aremonochromatic, so we needn’t worry about how the light’s wavelengthis recorded).
How, then, are we able to store a third dimension in a two-dimensionalplate? The answer lies in exploiting the interference effects of coherentlight. If coherent light is emitted from two point sources, it will interferewith itself to yield constructive and destructive interference fringes:you have studied this phenomenon in your lab as Young’s double-slitexperiment. If we were to trace the locations of constructive interferencethroughout space, we would obtain a set of surfaces called interferencefringes. By placing a photographic plate in the setup, we can recordthe locations and directions of these surfaces (see Figure 4.1(a)). Thiswill effectively create a set of “partially reflecting surfaces” within theplate, which, when illuminated with the same type of light used tocreate the image, will reflect light in such a way as to reproduce theoriginal image (see Figure 4.1(b)).
example 4 .1: the shape of interference fringes
Assume coherent light is being emitted from two point sources A and B, asin Figure 4.1(a)). What shape will the resulting interference fringes be?
The interference fringes occur at points of total constructive interfer-ence. Imagine we find a single point P in space where the beams from the
19
20 sebastian fischetti: holography: an introduction
(a) (b)
Figure 4.1: Figure (a) shows the hyperbolic interference fringes produced bytwo point sources of coherent light, and how these fringes are recorded in aphotographic plate; Figure (b) shows how the interference fringes recordedby the holographic plate create a virtual image when illuminated by coherentlight [47]. This is an example of a simple transmission hologram.
two sources interfere constructively; this means that the difference in thepath lengths is an integral number of wavelengths: AP − BP = nλ. Theinterference fringe passing through P must therefore consist of the locus ofpoints such that the difference of the distance to A and B is a constant. This isnone other than the geometric definition of a hyperbola; thus the interferencefringes produced by two point sources of coherent light will be hyperbolicsurfaces. In particular, the total number of interference fringes is limited tothe number of wavelengths that can fit into the distance between A and B.
In order to take holograms of extended objects, we exploit this sameprinciple: we split a beam of coherent light, using one beam (the objectbeam) to illuminate the object to be photographed, and using theother one to illuminate the plate directly (the reference beam). Theinterference between the two beams will be far more complex than inthe case of two point sources mentioned above, but can nonetheless berecorded within a photographic plate. By illuminating the processedplate from the same side as the reference beam, the interference patternswithin the plate will reproduce the image.
Note that as a result, holograms are redundant: every section of theplate contains an image of the entire object. This can be conceptualizedby imagining the plate as a “window” through which we look tosee the holographic image: if all of the plate is covered except for asmall opening, we can still discern the entire image through the theopening, indicating that the uncovered portion of the plate still containsinformation about the entire image (this ceases to be true, however, aswe approach the length scale of the interference fringes themselves).In this sense, holograms store far more information about an imagethan ordinary photographs, where each part of the plate only containsinformation about a small portion of the image.
example 4 .2: diffraction
Based on your knowledge of the propagation of light, what importantphenomenon are we neglecting to take into account in the above geometricalmodel?
We are neglecting to take into account the diffraction of light as itpasses through the various “slits” formed by the interference fringes withinthe plate. For our purposes, the geometrical does give a good enough sense
4.3 types of holograms 21
Figure 4.2: Here is how interference fringes are recorded in a reflection holo-gram; notice that the fringes are approximately parallel to the surface of theplate [47].
of what’s happening, but there are cases where it fails, at which point weneed to take into account diffraction effects with the zone plate model,which will be discussed further in Example 4.3.
4.3 types of holograms
There are two broad types of holograms: transmission and reflectionholograms. A transmission hologram is one in which the referencebeam and the object beam strike the plate from the same side, as inthe simple model shown in Figure 4.1(a). In this case, the plane of theinterference fringes recorded in the plate is more or less perpendicularto the plane of the plate, as shown in Figure 4.1(a), which allows usto explain the image as nothing but the reflection of light off of thevarious “partially reflecting surfaces” produced by the interferencefringes. In fact, two images are produced simultaneously: a virtualimage is produced when the interference patterns within the plate causethe reference beam to diverge, making the image appear behind theplate; this is the image we generally view when looking at holograms,and is the one illustrated in Figure 4.1(b). However, the interferencepatterns can also cause some of the reference beam to converge intothe real image, in which case the image can be projected onto a screen(or, with more difficulty, be viewed directly by placing the eye at itslocation). However, attempting to focus all will be impossible becauseof the image’s depth; this is a result of the hologram’s inherent three-dimensional information content.
In contrast, a reflection hologram is produced by illuminating theplate with the object and reference beams from opposite sides, as shownin Figure 4.2. Now, the plane of the interference fringes is approximatelyparallel to the plate, so that when the plate is illuminated to produce ahologram, the light undergoes Bragg reflection as it penetrates throughthe various interference fringes. As a result, the reflected light, andhence the image, is only visible from a relatively small range of angles.Furthermore, the virtual and object images are not produced simultane-ously; if the plate is illuminated from one side, the reflected light willdiverge, creating a virtual image; if the plate is illuminated from theother side, the reflected light will converge, producing a real image.
example 4 .3: thick vs . thin holograms
Based on the above discussion of how light is reflected from a holographicplate to produce images, explain why thicker holographic plates tend toyield higher-quality images than thinner plates.
22 sebastian fischetti: holography: an introduction
The thicker the emulsion used to store the interference fringes, themore well-defined the shape of the fringes, and the more effectively light canreflect off of them to produce an image. In the case of reflection holograms,the thickness of the emulsion is especially crucial, because it places an upperlimit to how many Bragg-reflecting surfaces can fit within the emulsion. If thethickness of the emulsion is significantly greater than the separation betweensuccessive interference fringes, then the hologram is considered thick. Ifinstead, the emulsion is thinner than the separation between interferencefringes, the hologram is considered thin. A reflection hologram ceases toexist when it becomes too thin because the Bragg interference is lost. A thintransmission hologram can still exist, but in this case is called a surfacehologram, since it essentially consists only of surface striations and yields alower-quality image than its thick counterpart. A surface hologram cannot beexplained using the geometric model we described, and instead requiresuse of the zone plate model described above. In general, thick hologramsare better than thin, as they are capable of containing more informationthroughout their depth.
4.4 making holograms
Because the quality of holograms relies so heavily on the formation ofinterference fringes within the emulsion, the components of the opticalsetup cannot move more relative to each other over a distance greaterthan the wavelength of the light being used. For creating visible-lightholograms, this necessitates the use of an optical workbench dampenedto external vibrations. Furthermore, in order to maintain coherenceof the light, a single source of light must be used, and the scale ofthe image is dictated by the coherence length of the light source. Thetypical light source is a laser, modern versions of which can have verylong coherence length, so there is essentially no limit to the scale ofthe hologram (except, perhaps, for budgetary concerns). Finally, thepossible emulsions to use vary greatly, depending on the wavelengthand intensity of the light used, the type of hologram being produced,budget, and desired exposure time.
Also of importance is the reduction of noise in the image; generally,this is done by adjusting the beam ratio. The beam ratio is defined asthe ratio of the amplitude of the reference beam to that of the objectbeam, and is crucial for filtering out intermodulation noise, whichis caused by the object beam interfering with itself (while we wantit to interfere only with the reference beam). By changing the beamratio, we can change the relative amplitudes of the object and referencebeams, and therefore change the relative amplitudes of the variouspossible interference effects between them. Generally, the ideal beamratio depends on the particular geometry of a setup, and is best foundby trial and error.
To produce a high-quality transmission hologram, we set up an ar-rangement similar to that shown in Figure 4.3. Notice first of all thatonly a single laser is used; all three beams originate from the laser.The reason for this is mentioned above. The particular arrangementillustrated is convenient because it actually uses two object beams to il-luminate the object more uniformly. Generally, transmission hologramsuse a beam ratio greater than 1:1.
4.5 applications of holography 23
Figure 4.3: A typical setup for producing a transmission hologram [47].
Figure 4.4: A typical setup for producing a reflection hologram [47].
To produce reflection holograms, the apparatus used looks morelike Figure 4.4. This particular setup works best with a beam ratio ofapproximately 1:1.
4.5 applications of holography
The applications of holography are varied and complex; here we canonly mention them in passing, but we encourage you to research themon your own if any seem particularly interesting.
The most promising application of holography is data storage. Un-like conventional storage devices (optical disks, magnetic disk drives,etc.), which store information on a two-dimensional surface, holo-grams are capable of storing information throughout their entire three-dimensional volume. As a result, holograms can (theoretically) storeinformation much more densely and efficiently than current means.The current challenges that holographic data storage faces is the lack ofread-write holographic media and the complexity involved in readingholograms via computerized means. As of yet, current progress in thisfield has been limited, but the potential for significant advancementexists.
24 sebastian fischetti: holography: an introduction
Figure 4.5: An example of pattern recognition using Fourier holograms. Tothe left, the transparency of the page in which a certain letter is to be foundis illuminated by a beam of laser light; the leftmost lens creates a Fouriertransform of the transparency, which is projected onto the Fourier hologram ofthe letter to be identified. The lens at right reverses the Fourier transform andprojects an array of dots onto the screen wherever the letter was found.
The “ordinary” holograms discussed so far are lensless – they do notrequire focusing light onto the holographic plate, as conventional pho-tography does. However, it is possible to make a hologram using lenses,made by placing a converging lens between the illuminated object andthe holographic plate such that the object is in the focal plane of thelens. The resulting hologram cannot be viewed via conventional meansbecause the lens destroys the crisp image of the object. Nonetheless,optical information about the object is stored in the hologram. In fact,this configuration produces the Fourier transform of the object at theplate; the resulting hologram is called a Fourier hologram. On its own,a Fourier hologram is not of much use, but can be very useful in patternrecognition. Imagine, for instance, we wish to find all instances of agiven pattern (say, a particular letter) in a page of text. We can do sowith the arrangement like the one illustrated in Figure 4.5. We firstcreate a Fourier hologram of the desired pattern. Then we create anoptical Fourier transform of the page of text and project it onto theFourier hologram of the desired pattern. Finally, we reverse-Fouriertransform the combined beams. The result will be an array of dotsindicating the location of every instance of the desired pattern in theoriginal text. You might have heard of this process of combining twoFourier transforms in your math classes: it is called convolution.
4.6 problems
1. Transmission holograms are visible over a virtually 180 range (aslong as the plate remains in the line of sight between yourself andthe virtual image), while the angular range over which reflectionholograms are visible is very limited. Explain.
Solution: This difference can be understood from the geometric model. Ina transmission hologram, the plane of the interference fringes is essen-tially perpendicular to the plane of the plate; this geometry allows thereference beam illuminating the plate from behind to reflect in virtuallyany direction as it passes through the plate. On the other hand, theplane of the interference fringes in a reflection hologram is parallel tothe plane of the plate, and the image is formed via Bragg reflection. InBragg reflection, a light wave passing through multiple reflective layersis reflected multiple times, interfering constructively for some angles ofincidence and destructively at others. Therefore, a reflection hologramcan only be viewed over a narrow angular range in the vicinity of the
4.6 problems 25
angle of incidence (or reflection) at which the constructive interferencein maximum.
2. Why would a higher beam ratio be preferable to a low one in the makingof a hologram?
Solution: The beam ratio is meant to reduce intermodulation noise, dueto the interference of the object beam with itself. If the beam ratio is high,the reference beam’s amplitude is stronger than the object beam’s, andso the interference of the object beam with itself gives a low amplitudecompared to the high amplitude of the interference of the object beamand the reference beam, drowning out intermodulation noise.
3. Imagine that in the apparatus of Figure 4.5 we replace the screen to theright with a transparency containing an array of dots and we replace thetransparency to the left with a screen; then we project a laser beam fromthe right. What might we expect to happen?
a) Nothing - the apparatus only works in one direction.b) An array of letters will be projected on the screen at left, mirroring
the array of dots.c) It is impossible to tell, since the answer depends on what exactly
the reference hologram is of.
Solution: The correct answer is choice 3b. This is simple symmetry: theFourier transform of the array of dots, when convoluted with the Fourierhologram of the letter, will yield an array of letters.
5C O L I N L A L LY: L I S T E N I N G F O R T H E S H A P E O F AD R U M
5.1 introduction
Aclassic problem in analysis is that posed by Mark Kac in his land-mark paper “Can One Hear the Shape of a Drum?” [45]. We shall
follow his exposition, for a while. Then, we shall become absorbed inthe simpler question “Can one hear the size of a drum?” Along theway, we shall develop the important tool of normal-mode analysis, andhave a quick introduction to asymptotic analysis. For now, consider thefollowing:
A membrane Ω, such as that depicted in Figure 5.1, stationary alongit’s boundary Γ, is set in motion. Its displacement F(x, y; t) ≡ F(r; t) inthe direction perpendicular to its original plane is known to obey thewave equation
∂2F∂t2 = c2∇2F, (5.1)
where c is some constant that we shall normalize to be c =√
1/2.There exist special solutions to the wave equation, of the form
F(r; t) = U(r)eiωt, (5.2)
which are called normal modes. Each of these solutions correspondsto a fundamental frequency at which the membrane can vibrate. Bysubstituting U(r)eiωt into (5.1), we find the corresponding equation forU:
12∇2U + ω2U = 0, U = 0 on Γ (5.3)
An illustration of the solution of (5.3) follows as part of the examplebelow.
example 5 .1: normal modes of a string
This is the one-dimensional limiting case of the general problem being set-upabove
Consider a string of length ` stretched taut between two walls (Figure 5.2).Its end points are fixed; that is, it obeys the boundary conditions
F(0, t) = 0 = F(`, t),
Figure 5.1: A membrane Ω; from [45]
27
28 colin lally: listening for the shape of a drum
where F is the displacement of the string, as given before. We want to findthe normal modes of this string.
Let F = U(x)eiωt be one of the normal modes we seek. Then U(x) is asolution of the equation
d2Udt2 + k2U = 0, (k2 = 2ω2)
which is just (5.3) in one dimension (minus the boundary conditions includedin 5.3). The general solution to this equation is
U = A cos kx + B sin kx.
Note that since the boundary conditions on F involve only x, they are effec-tively boundary conditions on U. Applying them, we have
U(0) = 0 =⇒ A = 0
U(`) = 0 =⇒ B sin k` = 0 =⇒ k =nπ
`,
where n is a positive integer. Thus,
ω =nπ√
2`
and our normal mode (now labeled by the integer n) is
Fn(x, t) = B sin(nπx
`
)einπt/
√2`,
where B is a normalization constant.
A very interesting phenomenon appears in this example: there is adiscrete sequence of normal-mode frequencies ω1 ≤ ω2 ≤ ω3 ≤ . . . .Each of these frequencies corresponds to exactly one Un through therelation between k and ω.
5.2 eigenvalues
It turns out that this result holds generally, regardless of problemdimensionality or geometry. Thus, for any region Ω bounded by asmooth (i.e., differentiable) curve Γ there exists a sequence of eigenvaluesλ1 ≤ λ2 ≤ . . . such that there corresponds to each an eigenfunctionψn(r), which satisfies
12∇2ψn + λ2
nψn = 0.
Naturally, ψn(r) = 0 for any point r that lies on the boundary Γ. Notethat the eigenfunctions are normalized such that∫
Ωψ2
n(r) d2r = 1.
We are now in a position to formulate the problem to which wealluded earlier. We wish to consider two separate regions (Ω1 and
0 l
Figure 5.2: The string considered in 5.1
5.3 problems 29
Ω2) with distinct boundaries (Γ1 and Γ2). Let us then consider thesemembranes’ respective eigenvalue problems:
Ω1 :12∇2U + λU = 0, U = 0 on Γ1 (5.4)
Ω2 :12∇2V + µV = 0, V = 0 on Γ2 (5.5)
Suppose that, for each n, λn = µn. We want to determine whether, if theeigenvalue spectra are identical, the two regions Ω1 and Ω2 necessarily“have the same shape.” Kac more correctly (but less intuitively) couchesthe question in terms of Euclidean congruence [45].
Before we continue, we should note that this particular problem hasrecently been answered in the negative: it is possible to have differently-shaped membranes that possess the same spectrum [35]. This resultwas discovered only in 1992, and is highly mathematical in nature. Weshall therefore do no more than note it, and proceed to examine someinteresting things that can be deduced from eigenvalue spectra. Hence,we shall first see whether one can “hear” the size (really the area) of adrum.
In order to answer this question, we shall use the methods of asymp-totic analysis [5]. Quite briefly, this analysis deals with finding the lim-iting behavior of some expression. If we have two functions f (x) andg(x), then we call them approximately equivalent and write
f ∼ g (x → ∞)
if
limx→∞
f (x)g(x)
= 1.
We can thus find the qualitative behavior of f at large values of x.Proceeding, we would like to know how many eigenvalues (from our
eigenproblem) exist that are less than a given number λ (keep in mindthat we are considering large λ). A result posited by H. A. Lorentz (indifferent form than is used here) and proved by Hermann Weyl is that,for any eigenproblem, the number N(λ) of eigenvalues less than somegiven λ is
N(λ) = ∑λn<λ
1 ∼ |Ω|2π
λ, (5.6)
where |Ω| is just the area of the membrane. It follows that
|Ω| ∼ 2πN(λ)
λ. (5.7)
That is, the area |Ω| of a “drum” (the membrane of Figure 5.1) canbe inferred from a knowledge of the number of small normal-modefrequencies. We can “hear” the area of a drum.
5.3 problems
1. Carry out the analysis done in the example, but use the boundaryconditions
F(0, t) = 0,∂F∂x
∣∣∣x=`
= 0.
These boundary conditions correspond to the physical scenarioof a string with one fixed end at x = 0, and one end free to movein the plane at x = `.
30 colin lally: listening for the shape of a drum
2. Prove the result (5.6)
3. Exam question: Can one hear the shape of a drum?
a) Yesb) No
Part II
M I N D O V E R M AT T E R
6C H R I S T O P H E R M A C L E L L A N : G L A S S
6.1 introduction to glass
Glass is a substance that people encounter frequently in their ev-eryday lives. The average person may assume there is nothing
particularly interesting about glass and would describe it as a brittleclear solid formed from molten sand. Many others believe the commonmisconception that glass is a liquid that flows extremely slowly andthat proof can be found in medieval windowpanes that are thickerat the bottom than at the top. As we will see, this misunderstandingis not entirely rooted in fiction. Glasses are actually a diverse groupof substances with a unique set of properties. Despite the fact thathumans have been making use of glass for thousands of years and itsimportance in our everyday lives, the details that underlie the formationof glasses is still a hotly debated topic in the fields of chemistry andphysics. [16]
In the common sense glasses are often considered to be a group ofsubstances like the ones we encounter in our everyday lives. Theseglasses are usually hard, brittle, and sometimes transparent. In thescientific sense glasses are a much broader group of substances. Forexample, many organic polymers such as polystyrene and polyvinylchloride are technically glasses. Although the exact definition of glassesmay vary slightly from publication to publication, glasses can be de-scribed as amorphous solids that undergo a glass transition. To furtherunderstand the properties of glasses one must understand what itmeans to be an amorphous solid and what a glass transition is. [27]
6.2 amorphous solids
On the macroscopic scale the differences between a solid and liquid areobvious. A commonly unknown fact about solids is that there are twofundamental types: crystalline and amorphous. The distinction betweenthese types can only be seen on the atomic scale. Crystalline solids aremade up of molecules that form a well defined lattice structure thatis repeated throughout the substance. This is called long-range order,or translational periodicity. On the other hand amorphous solids haveno long-range order. Atoms in amorphous solids do show connectivitythat resembles the structural units of crystalline states but it is notregular and does not repeat. This quasi-periodicity is indicative of whatis called short-range order [31]. The difference between the molecularstructures of crystalline solids and amorphous solids is shown in Figure1.
On the molecular level liquids look exactly like amorphous solidsand as one may expect the molecules in an amorphous solid have asignificant degree of freedom. Then what is the difference between andamorphous solid and a liquid, and why don’t they actually flow? Thereare a few different ways to draw the line between amorphous solidsand liquids. One way is to say that a substance is a solid when the timerequired for it to undergo a structural change is much longer than the
33
34 christopher maclellan: glass
Figure 6.1: Atomic Makeup of Crystalline Solids vs. Amorphous Solids [7]
time of the observation [11]. In other words, a substance is a solid whenit doesn’t flow in a reasonable amount of time. In fact, some amorphoussolids would take billions of years or more time than the universe hasexisted before they would show a noticeable structural change due toflow. Another way to describe the difference is to define a liquid asa substance with a viscosity below an arbitrary value. For example,in his book Physics of Amorphous Materials SR Elliot defines a solidas a material with a viscosity above 1014.6 poise. Although these twodefinitions appear different at first glance, they both attempt to quantifythe same obvious difference between liquids and amorphous solids. Ifwe return to the common misconception that glass is a slow flowingliquid we will find that under these definitions what we commonlyconsider to be glass cannot be a liquid; it simply does not flow fastenough. So why are medieval windowpanes thicker at the bottom thanat the top? The answer lies in the method used to create the glass, notin its atomic structure. Now that we have defined amorphous solidsand how they are different from liquids and crystalline solids we mustexamine the glass transition, which is the property that distinguishesglasses from amorphous solids. [27] [16]
example 6 .1: glass in our everyday lives
We have defined glass in a scientific sense, but what about the glass we usein our everyday lives? Glass in the common sense is nothing more than anamorphous solid made up of silica(SiO2). Silica is found in many forms,the most abundant of which is sand. This is not the whole story though.Almost all of what we call glass is not made up of pure silicon dioxide forpractical purposes. Silicon dioxide (crystalline form) has a very high meltingpoint of around 1700oC [27]. Because of this high melting point soda ash(Na2CO3) is often added to the silicon dioxide. This lowers the melting pointto about 1000oC, making the molten glass much easier to handle. To provideextra hardness and chemical stability lime (CaCO3) or dolomite (MgCO3) isadded to the mixture, which is called soda-lime glass. The concentrations ofthese materials can vary, but the makeup is commonly 60-75% silicon dioxide,12-18% soda and 5-12% lime. Other impurities can be added to change theglass properties for aestetic or functional purposes. Soda-lime glass makesup most of the glass we use everyday and can be found in everything fromdrinking glasses to windows.
6.3 the glass transition
When a liquid is cooled it may undergo a first order phase transitionand crystallize at the melting point of the substance. In our everydayexperience the transition from water to ice is a perfect example of this.
6.4 simulating amorphous materials with colloids 35
Figure 6.2: A Thermodynamical View of Liquid-Solid Transitions [7]. The pathAD represents the transition from liquid to glass (for an arbitrary cooling rate),while the path ABC represents the transition from liquid to crystalline solid.Notice the discontinuous change in volume at the melting point during thecrystalline transition . Compare this to the glass transition where there is acontinuous change in volume at the the glass transition temperature.
This is not the only thing that can happen when a liquid is cooled. Ifcooled fast enough it will not crystallize and will remain in a liquidstate below its melting point. This is called a supercooled liquid and isalso indicated by increasing viscosity as temperature is reduced. Even-tually the supercooled liquid cools to the point that it forms a glass.This transition is marked by a glass transition temperature. This tem-perature is defined as the range in which thermodynamical variables(e.g. volume, enthalpy, entropy) begin to change. It is a range becausethe value of the variables changes in a continuous fashion, unlike thechange that occurs at the melting point. It is important to note thata substance does not have a specific glass transition temperature. Asmentioned earlier, supercooled liquids only result from sufficiently fastcooling rates. The cooling rates that will allow a supercooled liquidto form depends on the substance being cooled. This cooling rate alsochanges the region over which a substance can be supercooled. If onereduces the rate of cooling, the range over which the temperature canbe supercooled increases, which effectively lowers the glass transitiontemperature. In fact, the glass transition temperature can vary as muchas 10-20% for very different cooling rates. An odd consequence of theseproperties is that the measurement of the glass transition temperaturevaries by how the glass is prepared. Despite this knowledge of the glasstransition, there are still many questions left to be answered. Physicistsand chemists still debate the details of the glass transition on a regu-lar basis and new methods are constantly being formulated to try toanalyze glass structure and transition. [27]
6.4 simulating amorphous materials with colloids
Some of the most promising ways to study the structure of glassesare being developed using colloids to simulate the molecular structureof amorphous materials. A colloid is a mixture in which a system
36 christopher maclellan: glass
Figure 6.3: A diagram of the experiment. A binary colloid is allowed to settle ina capillary tube and is imaged with a confocal microscope. The larger particlesare found in higher concentrations at the bottom, while the smaller particlesare found in higher concentration at the top [69]
of particles between the sizes of 10−7 and 10−5 cm is dispersed in acontinuous medium [17]. In our case we will consider colloids in whichthe medium is a liquid and the particles are micron sized hard spheres.When colloids are allowed to settle the particles pack together andresemble a continuous solid. In fact, the arrangement of particles isdirectly analogous to the arrangement of molecules in a solid. Becauseof the relatively large size of colloid particles we can measure each one’sposition individually. We exploit this property of colloids to analyzetheir structure and extend our findings to actual solids.
In the laboratory we create a colloid made up of a binary system ofthe aforementioned hard spheres. We use two different sizes because itis well established that monodisperse systems only settle in crystallinearrangements. In contrast, binary systems show areas of amorphousarrangement. In our case we place our colloid is in a capillary tubeso gravity determines the distribution of large and small particles. Asyou would expect the larger particles are more prevalent at the bottomwhile the smaller particles are found in higher frequency at the top.In the middle there is a continuum of different compositions withrespect to sphere size. It is this middle area where a truly binary colloidexists with amorphous structure. A confocal microscope can be used tomeasure the position of each particle in the capillary tube. A diagramof this setup is shown in figure 3. [69]
The particle arrangement is determined by defining a radial distribu-tion function g (r) which is defined as,
g (r) =L2
2π∆rN(N − 1) ∑i 6=k
δ (r− |~rik|) , (6.1)
where g (r) is the ratio of the ensemble average of particles in a regionr ∼= r + ∆r to the average number density N/L2 where L is the squareimage length and δ is the Dirac delta function. This can be thought ofas the relative number of particles present at a distance r from a centralparticle. This single parameter can give a quantitative measurement ofthe degree of long-range order present [69].
The radial distribution function at different heights is shown inFigure 4. You can see that when z = 1.8mm and z = 6.5mm g (r) some
6.5 practice questions 37
Figure 6.4: The radial distribution function g (r) at different heights where σ isthe mean particle diameter. Notice how the oscillatory behavior first decreasesand then increases as r increases indicating a change transition from crystallinestate to glassy state back to crystalline state. Not that the lines graphs are offsetfor clarity. [69]
oscillatory nature exists. This indicates that there is some regular changein density as you get further away from the central particle, which iswhat we would expect from a structure with long-range order. Thisalso agrees with our assertion that crystallinity exists where there isa relatively monodisperse composition of particles. When z = 3.3mmwe would expect more or less equal mixing between the different sizespheres and an amorphous structure. g (r) at this height confirms this,as there is no pattern in how the density of particles changes as rincreases [69].
Using this analysis we have found a way to measure the long rangeorder in a substance. Although the above treatment is mainly quali-tative, more analysis techniques are being developed to describe thepositioning of the spheres in both 2D and 3D. [69] Nevertheless, thismethod has been proven to be a efficient way of creating system thatis analogous to solids that can be experimentally measured and easilymanipulated. Analyzing colloidal system like the one above providesone of the best chances for physicists and chemists to unlock the secretsof how and why glasses form.
6.5 practice questions
1. In a technical sense glasses are identified by their lack of
• a)long range order• b)short range order• c)flexibility• d)color
2. When cooling a liquid to form a gas, raising the cooling rate theglass transition temperature
• a)raises• b)lowers• c)does not change
38 christopher maclellan: glass
3. Colloids are used to simulate the formation of solids becausethey:
• a)are the same thing• b)colloid particles are large enough to have their individual
positions measured• c)never show crystalline arrangements• d)all of the above
7R O B E RT P I E R C E : T H E P H Y S I C S O F S P L A S H I N G
7.1 introduction
Have you ever noticed how the water droplets that come fromyour faucet splash at the bottom of the sink? Have you ever
watched rain droplets splash as they fall and hit your car windshieldwhile driving? The phenomenon known as splashing occurs in manysituations such as these. Splashing is also important in technologicaland industrial situations such as the ignition of fuel in your car, or whenyour printer puts ink onto a piece of paper. This chapter will help youunderstand situations such as these by developing an understanding ofthe physics of splashing.
Scientists have studied splashing for over a hundred years. Becauseof the beauty of the motion involved with splashing, it has been one ofthe most highly praised phenomena studied via the use of high speedphotography. From famous sketches and photographs of scientistssuch as Harold Edgerton and A.M Worthington[96] , the beauty ofsplashing has been available to mainstream society. Yet the physics ofthis tremendous phenomenon is still not fully understood.
The goal of this chapter is to develop an understanding of some ofthe physics involved when a drop of liquid falls and strikes a smoothsurface. We will see that the there are many factors (characteristics ofour system) that determine how the system will evolve in time. We willreview fundamental concepts such as pressure, surface tension, viscos-ity, and others to develop a groundwork for understanding splashingas well.
7.2 pressure, surface tension, and other concepts
Recently, fronted by the research of Dr. Wendy Zhang[97], SidneyNagel[97][98], and others, it has been shown that air pressure has a
Figure 7.1: A photograph taken by Martin Waugh[90] as part of his liquidsculpture images. This is an image of milk splashing on a smooth surface. This isjust another example of the spectacular beauty of splashing, and why splashinghas grabbed the attention of many photographers around the world.
39
40 robert pierce: the physics of splashing
tremendous influence on whether or not a splash will occur when aliquid drop falls onto a smooth surface. The research has shown thatthe lower the gas pressure surrounding the splash zone, the smallerthe splashes become, until they disappear all together. You can seethrough high speed imaging that at a certain critical air pressure, thecorona (a layer of liquid spreading out from the center of impact withthe smooth surface) disappears all together, and there is no splashing.Instead, the liquid just spreads out along the surface in the same waythat one would see spilled milk spread out across the kitchen table. Thisarea of the physics of splashing is a bit too complicated to be addresedhere, and we will focus on something less difficult. We will focus ourattention towards what characteristics of the system produce, not onlya corona, but splashing (when the corona breaks up into thousandsof individual droplets). In order to understand these characteristics ofsplashing to an appropriate degree, we need to review some of thephysics of pressure, viscosity, and surface tension.
7.2.1 Pressure
Pressure is defined as the force per unit area of a surface directedperpendicularly to the surface. In our context of falling liquid drops,we will be focusing on atmospheric pressure due to the force of thegases in the atmosphere above the surface. We have that
P =FNA
, (7.1)
where P is pressure, FN is the normal force on the surface, and A is thearea of the surface. In our system, FN will be equivalent to the weightof the gas particles above the surface, as well as the force these gasparticles exert onto the surface due to their speed. For ideal gases, wemay express the pressure as
PV = NkbT =⇒ P =NV
kbT =⇒ P =ρAMg
kbT, (7.2)
Figure 7.2: Photographs taken by Lei Xu, Wendy W. Zhang, and Sidney R. Nagelin their experiment involving falling alcohol drops onto a smooth surface. Thetop three frames represent alcohonl under an air pressure of 100 kPa, and thebottom three frames represent alcohol under an air pressure of 30 kPa. The leftframes (top and bottom) are the drop just above the surface, at time t = 0 ms,the middle frams are the drop at time t = .276 ms, and the right frames are thedrop at time t = .552 ms. We see that in the middle frames a corona or liquidlayer is spreading out from the center, and by the time expressed in the rightframes, the stress due to air pressure has won, and we see splashing.
7.2 pressure, surface tension, and other concepts 41
where Mg is the molecular mass of the gas g, V is volume, N is thenumber of particles, T is the temperature, kb is the Boltzmann constant,and ρA = Mg/V is the density in the atmosphere. Equation 7.2 is theIdeal Gas Law. From 7.2 we readily find an expression for the gasdensity in the atmosphere to be
ρA =PMg
kbT(7.3)
We will see that after a drop strikes the surface, a liquid layer willspread outwards (the corona), and interact with the atmosphere. Therewill be stress on the liquid layer due to the atmosphere that will beproportional to, among other things, the density of the gas in theatmosphere. We will learn what the other things are later on, butfirst make sure that you are capable of using the above equations toanalyze this system in terms of the atmospherically applied stress.Make sure that you understand pressure in the context of our systemby performing the following example.
example 7 .1: using the ideal gas law
1. When observing a liquid drop spread out after coming into contactwith a smooth surface, we notice that the liquid spreads out veryrapidly. Under these circumstances the stress applied to the spreadingliquid layer due to the atmosphere will be proportional to the densityof the gas in the atmosphere. If our atmosphere is made of air, andif air has a molecular mass of M29 = 29u (u = 1.66x10−23kg), what isthe pressure that the atmosphere may exert on a liquid layer that isspreading outwards? Assume that we observe this at room temperature(T = 295K). The density of air is measured to be ρA = 1.2kg/m3,and Boltzmann’s constant is kb = 1.381x10−23m2kgs−2K−1. Does thispressure look familiar?Solution: Here we may use 7.2 to solve the problem. We are given valuesfor the variables ρA, Mg, and T. Solving for pressure, we find
P =ρ
MgkbT → P =
1.2kg/m3
29 ∗ (1.66x10−23kg)(1.381x10−23m2kgs−2K−1)(295K).
(7.4)
So we find that pressure P is,
P = 101, 500kgm−1s−2→ P = 101.5KPa (7.5)
This pressure is known as atmospheric pressure. It is the pressuremeasured due to the atmosphere at sea level. We get this value becausethe density given in the problem is the density of air molecules at sealevel.
7.2.2 Viscosity and Surface Tension
Viscosity may be defined as the resistance to flow in a liquid. A generalway to view a system involving viscosity is to imagine a flowing liquidsection that is divided into many infinitesimally thick layers fromtop to bottom. Individual layers will move at different velocities fromeachother. This is because the layers at the bottom will interact withthe surface, layers at the top will move with the initial speed of the
42 robert pierce: the physics of splashing
liquid section itself, and the middle layers will move with velocities inbetween the two extremes. Let’s make shear stress more clear: stressis defined as the average force applied to a surface per unit area (seeequation 7.6), and shear means that the stress is being applied parallelto the surface. Shear stress will result as layers move towards or awayfrom other layers. Stress τ is proportional to force F by
τ =FA
. (7.6)
Where A is area. Shear stress is directed parallel to the liquid. Ultimatelywe will see that the stress in the liquid is acting to hold the liquidtogether. We will see that this stress will interact with the stress on theliquid from the atmosphere (due to pressure as described in earliersections), and when the atmospheric stress is stronger, splashing willoccur.
First, let us devote some time to viscosity. In our study of viscositieswe will be interested in what is known as kinematic viscosity. This iswhen we look at the relationship between the resistive force in a liquiddue to viscosity, and the inertial force of the liquid. Kinematic viscosityνL of a liquid is defined as
νL =µ
ρ, (7.7)
where µ is the dynamic viscosity of the liquid, and ρ is the density ofthe liquid. In our system of a liquid drop falling onto a smooth surface,layers of thickness d will advance away from the center (where the dropinitially made contact with the surface). We can estimate the thicknessof the boundary layer, the first layer closest to the surface, to be
d =√
νLt, (7.8)
where t is the time elapsed since the drop struck the surface.Now we are in a position to consider the stress on the expanding
liquid layer due to the liquid layer itself. The stress on the boundaryliquid layer will be due to the surface tension of the liquid striving tokeep the layer intact. With d defined as above, we have that the surfacetensional stress (shear stress) is
ΣL =σ
d→ ΣL =
σ√νLt
(7.9)
And since we have this knowledge of the stress on the expandingliquid layer due to the liquid layer itself (ultimately the force trying tohold the liquid layer together), we can use it in combination with thestress on the expanding liquid layer due to the atmosphere (ultimatelythe force trying to break the liquid layer apart), and observe underwhat combinations we get splashing.
7.3 splashing
Splashing in our system will occur when the stress on the liquid layerdue to the atmosphere is stronger than the stress on the liquid layerdue to the liquid. This is because the stress due to the atmosphere istrying to break the liquid layer apart, and the stress due to the liquid istrying to keep the liquid layer together. We’ll have two equations for the
7.4 summary 43
two different stresses. Earlier we discussed how the stress due to theatmosphere should be dependent, among other things, on the densityof the air. We also know that the stress is proportional to the rate ofexpansion of the liquid layer, and the speed of sound in the atmosphere.If ΣA is the stress due to the atmosphere, then our equation for thestress is
ΣA = (ρA)(CA)(vl) −→ ΣA =PMg
kbT
√γkbTMg
√RV0
2t. (7.10)
Here CA is the speed of sound in the atmosphere, and vl is the velocityof the boundary layer that has undergone shear stress due to thesurface.1 Also in the second form of the equation, R is the initial radiusof the liquid drop, V0 is the velocity of the liquid drop as it lands on thesurface, gamma is the adiabatic constant of the gas in the atmosphere,and t is the amount of time elapsed after the drop impact.
We also saw in equation 7.9 that
ΣL =σ√νLt
. (7.11)
If we take a ratio of ΣA with ΣL, we’ll get
ΣAΣL
=σ√νLt
PMgkbT
√γkbTMg
√RV02t
=√
γMgP
√RV0
2kbT
√νL
σ(7.12)
From this ratio we can see that a more viscous liquid will splashmore easily than a less viscous liquid. This is counter intuitive becauseone would think that a more viscous liquid would stay together moreeasily, however, it is apparent that a more viscous liquid will make thevalue of the ratio in (7.12) larger, which implies that the liquid layeris more likely to break apart and splash. So splashing is clearly not astraightforward phenomenon at all.
7.4 summary
Ultimately we see that splashing is an extremely interesting phe-nomenon that involves some physics that even runs contrary to in-tuition. We see that atmospheric pressure determines whether or not aliquid will splash, and we see that when the pressure is low enoughto allow for splashing, we can quantify when we will and will not seesplashing. We will see splashing when the stress due to the atmosphereis greater than the stress due to the liquid layer. The stress due to theatmosphere is working to pull the liquid layer apart, and the stresswithin the liquid layer is working to hold the layer together, if theatmospheric stress is stronger, the liquid will break apart. Ultimately astrong knowledge of splashing may be used to control when we wantto have splashing with some kind of liquid, and when we don’t wantto have splashing. This, in turn, can be used to improve many forms ofindustry and technology, and may prove to be very important towardsa more efficient future.
1 For a derivation of this equation, see the paper Drop Splashing on a Dry Smooth Surface byLei Xu, Wendy W. Zhang, and Sidney R. Nagel, 2005.
44 robert pierce: the physics of splashing
7.5 chapter problems
1. A liquid drop of alcohol with initial radius R = 1.77 mm falls ontoa smooth surface. After the drop strikes the surface, a corona or aliquid layer moves outward with an initial velocity V0 = 5 m/s,and we observe the moving corona for t = .5 ms. Let’s supposethat our experiment is performed at room temperature (T =295 K, but our atmosphere is made of helium. The pressure dueto the helium atmosphere near our experiment is still equivalentto atmospheric pressure, and the molecular mass of helium isMg = 4u, whereu = 1.66x10−23kg. If the speed of sound in ourhelium atmosphere is 1000 m/s, what will be the stress on thecorona due to the atmosphere ΣA?Solution: Here we may use the given conditions to solve for ΣA using thetwo parts of equation 7.13.
ΣA = (ρA)(CA)(vl) −→ ΣA =PMg
kbT
√γkbTMg
√RV02t
. (7.13)
Looking at the left part of the equation; we are given the speed of soundCA, and we can readily find ρA and vl by looking at their terms in theright part of the equation: Namely:
ρA −→PMg
kbT, (7.14)
vl −→√
RV02t
. (7.15)
Plugging in all of the values given, we find:
ΣA =
((101.3 kPa)(4u)
(1.381× 10−23 m2kg-s−2K−1)(295 K)
)×
(1000 m/s)
(√(1.77 mm)(5 m/s)
2(.5 ms)
), (7.16)
and we arrive at the answer,
ΣA = (1651.1)(1000)(2.97) = 4903.8 kPa. (7.17)
2. Consider the answer found in the previous question. If the shear stress inthe expanding liquid due to the liquid is ΣL = 4500 kPa when observingfor time t = .5 ms, will there be any splashing? Explain, why or whynot?
Solution: From the first question we have that the stress on the expandingliquid layer due to the atmosphere is
ΣA = 4903.8 kPa. (7.18)
We know that if the stress due to the atmosphere is greater than thestress due to the liquid, then the expanding liquid layer will break apart,and we will see splashing. So the ratio from equation 7.12 will be greaterthan 1 when we have splashing, and less than 1 when we don’t havesplashing. In our case,
ΣAΣL
=4903.8 kPa4500 kPa
= 1.09 > 1 (7.19)
We do see splashing. Again, this is because of the fact that the stress dueto the atmosphere (the stress that is trying to break the expanding liquidlayer apart), is stronger than the stress due to the liquid itself (the stressthat is trying to hold the liquid together). Ultimately the atmospherewins, the liquid breaks apart, and we see splashing.
7.6 multiple choice questions 45
7.6 multiple choice questions
1. Which of the following does the atmospheric stress ΣA on theexpanding liquid layer NOT depend on?
(a) The speed of sound in the atmosphere.
(b) The density of the atmosphere.
(c) The density of the expanding liquid.
(d) The speed of the expanding liquiud boundary layer.
(e) None of the above.
Solution: (c) The density of the expanding liquid. See equation 7.10
2. How thick is the boundary layer of an expanding liquid on a smoothsurface of it has been expanding for time t = .4 ms with a kinematicviscosity of νL = 1× 10−6 m2/s?
(a) 2x10−5 m
(b) 2x10−4 m
(c) 2x10−3 m
(d) 2 m
(e) None of the above
Solution: (a) 2x10−5 m
Here we may use equation 7.8 to solve the problem:
d =√
νLt −→ d =√
10−6 m2/s.0004s = 2x10−5m. (7.20)
8S A M B I N G H A M : F R E A K WAV E S
8.1 introduction
In the deep ocean it is common for mariners to see wave heights thatare in the 5-meter to 12-meter range. In the worst storm conditions
waves generally peak at 15 meters (49 ft). Commercial ships have beenbuilt to accommodate waves of such heights. About once a week a shipat sea sinks to the bottom of the ocean [4]. Over the last two decadesmore than two dozen cargo ships measuring over 200 meters in lengthhave been lost at sea [38]. Most of the time these ships will sink withouta mayday signal or any trace left behind of its existence.
For centuries mariners have described waves that appear out ofnowhere that approach 35 meters (115 ft.) in height. However there wasno scientific evidence to back up this anecdotal evidence. It is importantto distinguish that these types of waves are described to appear out ofnowhere in the deep sea. This freakish wave is drastically different thanthat of the tsunami. A tsunami is created by a disturbance in the ocean.The most basic example of this is the example of a stone being droppedin water, which creates waves that are generated radially outward.Likewise, a quick shift of plates making up the sea floor creates a seriesof low waves with long periods. Disturbances known to create tsunamisrange from seismic activity, explosive volcanism, submarine landslides,or even meteorite impacts with the ocean [13]. Tsunamis occur nearland as their name suggests; tsun translates from Japanese in harborand ami into wave (The phenomena known was originally known as atidal wave, however, this name incorrectly implies that they are relatedto the tides).
While tsunamis are much more deadly than the freak waves de-scribed by mariners (The megatsunami of December 26, 2004, killedover 250,000 with waves in the 10-to-30 meter range along the coast ofIndonesia, India, Thailand, Sri Lanka, and elsewhere), their behavior ismuch more understood as about three fourth’s of tsunamis originatefrom submarine seismic disturbances. These disturbances produce agreat amount of energy that translates into the waves as they travel atspeeds approaching 200m/s with wavelengths on the order of hundredsof kilometers [67]. Tsunamis are not felt at the deep sea as they onlyproduce wave heights around a meter or two, which is not detectablewith the typical ocean swell. However, once the waves created by thedisturbance approach areas of shallow water, more energy is packedinto the waves, creating waves in the tens of meters along the shore. Inthe 20th century over 1000 tsunamis have been recorded with 138 beinglabeled as a “damaging” tsunamis [12]. Since the last half of the 20thcentury oceanographers have been able to identify the instigating dis-turbance for the occurrence of a particular tsunami. This is completelydifferent than freak waves out of nowhere described by mariners whichcarried the same type of lore and about as much comprehension asmythical monsters like Loch Ness.
For a long time oceanographers dismissed the possibility of the exis-tence of these types of waves with such a high rate of occurrence. What
47
48 sam bingham: freak waves
Figure 8.1: The crest heights during a 20-minute interval at the Draupner Oilplatform at 15:20 GMT. The peak is 18.6 meters from the surface.
is known as the linear model has been used to predict wave heights outat the deep sea with great accuracy. The commercial shipping industryhas based most of the designs for its ships off of this model because ofthe accuracy of the linear model. The freak waves seen by the marinersare part of the natural wave spectrum but should only occur aboutonce every ten thousand years according to the accepted linear model.Clearly something was wrong.
While oceanographers had started to put more faith and energy intothe belief of such freak waves in the 1970’s the major breakthroughcame with the Draupner wave on January 1st, 1995 at the Draupneroil platform in the North Sea. The Draupner oil platform, which islocated directly west of Norway, had a laser-based wave sensor atan unmanned platform that would record waves in a fashion thatwould be uninterrupted by the platform’s legs [41]. It should be notedthat oceanographers base much of their analysis on what is calledthe significant wave height, or swh, which is the average height fromtrough to crest of one third of the largest waves during a given period.On January 1st at the Draupner Oil platform the swh was in the 11-to-12-meter range. The maximum wave height measured that day was 26
meters and the peak elevation was 18.5 meters, as shown in figure 8.1.The annual probability for a wave of that height in the 12 meter swh is10−2, or once in a hundred years. While the causes for such waves areunderstood with much more confidence since the Draupner wave, it isimportant to determine what waves are freak waves by nature.
8.2 linear model
Now that we understand the Draupner wave and a bit about thesefreak waves let’s take a look at just how odd these waves are whenusing the linear model. In the most simple model of ocean waves,the sea elevation is taken to be a summation of sinusoidal waves ofdifferent frequencies with random phases and amplitudes. In the linearapproximation the wave field is taken as a stationary random Gaussianprocess. The probability density distribution is described as
f(η) =1√2πσ
e−η2/2σ2. (8.1)
8.3 interference 49
In Eq. 8.1, η is the elevation of the sea surface with a zero mean level,〈η〉 = 0, and σ2 is the variance computed from the frequency of thespectrum, S(ω)
σ2 = 〈η2〉 =∫
S(ω)dω. (8.2)
If we take the wind wave spectrum to be narrow, the cumulative prob-ability function of the wave heights will be described by the Rayleighdistribution such that
P(H) = e−H2/8ω2. (8.3)
Equation 8.3 gives the probability that the wave heights will exceed acertain height, H. We can now introduce the significant wave height,Hs, into the equations and begin to see its relevance. Extensive work byStanislaw Massel has shown that Hs ≈ 4σ[58]. From this we can find adirect relation into the probability for a certain wave height to occurunder the conditions that are present by rewriting Eq. 8.3 as
P(H) = e−2H2/Hs2. (8.4)
At this point a mathematical definition for what constitutes a wavebeing a freak wave is necessary. Looking back at the Draupner wave, themaximum wave height would not be completely out of the ordinary forsevere storm conditions that created 15 meter significant wave heights.The fact that the Draupner wave occurred with a significant wave heightof 12 meters is what makes it a freak wave. From this point on we willrefer to these freak waves as rogue waves, which by definition are wavesthat have a wave height that is more than double the significant waveheight.
example 8 .1: shallow water rogue waves
Could a rogue wave occur in shallow water of only 20 meters in depth withthe top third average waves only around 2 meters in height? If one couldwhat would be the most probable rogue wave and with what probabilitywould it occur?Solution: This problem is quite simple but requires careful thought on thedefinition we established for a rogue wave. We stated that a rogue wave is awave with a wave height of twice that of the significant wave height, thus a5-meter wave would be a rogue wave if the swh was 2 but not if it was 4. So ifwe return to the question we can surely have a rogue wave in shallow waterwith a swh of 2 meters if the wave height is greater than 4. This 4-meter wavewould also be the most probable rogue wave with a probability that is foundfrom Eq. 8.4. With our conditions we get P(4) = e−2(4)2/22
= 3.5× 10−4.These types of waves are referred to as shallow water rogue waves and arejust as freakish as the giant ones in the deep sea. The highest recorded ratioH/HS was a wave height of 6m in conditions with a significant wave heightof 2m.
8.3 interference
While the Draupner wave was a tremendous breakthrough for the fieldof oceanography, as it proved that such waves do exist and that there
50 sam bingham: freak waves
were inaccuracies in the standard linear model, it did not show howor why such waves form. In order to examine the possible causes forthese waves researchers looked at the areas in which they occurred. Byplotting areas in which freak wave phenomena were described in thepast a pattern was found. A great number of the waves occur in areaswhere there are fast currents that run counter to the wave direction. Themost extreme example of this is the Agulhas current off of the Capeof Good Hope and the tip of South Africa. The Agulhas current is thefastest current on the planet. Figure 8.3 shows 14 reported freak wavesfound in a study by Captain J. K. Mallory prior to 1975. His studyfound similar results in the Gulfstream current, the Kuroshio currentand any others in the Pacific Ocean.
Figure 8.2: These are the results of a study by Captain Mallory done in 1976.The circles represent cases where abnormal waves were observed, the dottedline is the continental shelf, and the other depths show the continental slope.The abrupt change in depth is what creates the Agulhas current and gives it itsjet like function parallel to the shoreline [49].
Traveling around the tip of South Africa has always been a roughstretch for mariners, but the area of Agulhas current created waves thatmariners said could reach 30 meters in wave height. In this region thewaves are generated off Antarctica and move northward towards SouthAfrica. They run unopposed until they meet the Agulhas current, whichmoves in a southwest direction parallel to the east coast of South Africa.Because of this area of interaction the wind waves are made up of twodifferent systems: the long swell waves coming from the Antarcticaand short/steep waves generated by the sea wind of South Africa. Thesuperposition of these two systems will undoubtedly contribute to anincrease in wave height due to possible constructive interference, butnot enough to account for the rogue waves reported in the area.
The solution to these types of waves relates to geometric optics. Ifwe think about a strong variational current as a lens then it couldfocus opposing waves and create points where enough energy wouldconverge to explain the freak waves. This would be possible becausethe areas of the strong flow of the current would retard the waveadvancement much more than the weaker flow, which would cause thewave crests to bend. This refraction allows for the focusing of the wavesenergy and allows the waves to grow in height as shown in figure 8.3.
8.3 interference 51
The most commonly accepted research on this subject variationalcurrent focusing was done by I.V. Lavrenov in the late 1990’s. His workshowed that the focusing by currents would shorten the wavelengthof storm-forced waves that ran against the current and the wave trainswould press together possibly into a rogue wave. A major concept thatLavrenov derived was that H/H0 would reach a maximum value of2.19 for the conditions of the Agulhas current. In this formula H0 isthe mean height of the swell propagating on still water. From this wecan easily see that the maximum height for a wave caused by currentrefraction would 2.19 times the mean height of the swell.
Since rogue waves are not limited to occurring in areas where thereis a strong counter current, there has to be some explanation for theseother extreme waves. The solution is actually very similar to the onejust discussed. The thought was to look for other possible ways thatwave energy could be focused into one spot. The easiest way to dothis is through the ocean basin. It is not too hard to see that geometricfocusing can occur with the help of the underwater topography. Thebehavior of the rays in the basin, however, is rather complicated as thereal topography creates many caustics.
b
Wavecrests
Variational Current
Waveorthogonals
a
Figure 8.3: a shows the interaction of the two systems that are present off the tipof South Africa [49]. b shows the refraction that waves undergo when flowingagainst a current that has a variable surface speed. This refraction can leadto the build up of rogue waves when the wave orthogonals converge [Figuremade in Inkscape].
52 sam bingham: freak waves
8.4 nonlinear effects
With the understanding of refraction applied to water waves oceanog-raphers could generally predict where the rogue waves would occur.Since the ships in the shipping industry are built to handle 15-meterwaves there was no need to travel through areas in which they couldencounter the rogue waves. The ships could simply travel another routeand there was no need to redesign millions of ships across the world.This apparent success turned out to be rather short lived. In Februaryof 2001 two ships, the Caledonian Star and the Bremen, were nearlysunk when they came into contact with 30-meter waves in a region ofthe South Atlantic. In this region there is no current strong enough tocreate the waves and the basin cannot focus enough energy to accountfor the waves. Once again, something was missing from the deep oceanwave models.
The breakthrough came from the world of quantum mechanics thistime and specifically the non-linear Schrödinger’s equation. There is amodified version of the non-linear Schrödinger’s equation that can beused to describe motion in deep water. The simplified nonlinear modelof 2D deep water wave trains is
i(∂A∂t
+ cgr∂A∂x
) =ω0
8k0
∂2 A∂x2 +
ω0k20
2|A|2 A, (8.5)
and the surface elevation, η(x, t) is given by
η(x, t) =12(A(x, t)ei(k0x−ω0t) + c.c. + ...). (8.6)
Here k0 and ω0 are the wave number and frequency of the carrierwave, c.c. denotes the complex conjugate, (...) determines the weakhighest harmonics of the carrier wave, and A is the complex waveamplitude, which is a function of x and t. At this point what is known asthe Benjamin-Feir instability comes into play. It is very well known that auniform wave train with an amplitude of A0 is unstable to the Benjamin-Feir modulational instability corresponding to long disturbances ofwave number, ∆k, of the wave envelope satisfying the relation
∆kk0
< 2√
2k0 A0. (8.7)
The highest instability will occur at ∆k/k0 = 2k0 A0 which gives amaximum growth a rate of ω0(k0 A0)2/2 [48]. There has been a greatdeal of research on the nonlinear stage of the Benjamin Feir instabilityanalytically, numerically, and experimentally (see Grue and Trulsen(2006) and Kharif (2003)). Through this research it is apparent that wavegroups can appear or disappear on the order of 1/[ω0(k0 A0)2] on thetimescale. This behavior can be explained by breather solutions of thenon-linear Schrödinger equation. Figure 8.4 shows the formation ofwaves that have very high energies and are created by the BenjaminFeir instability. There are many different breather solutions that cancreate various large waves. What is referred to as the algebraic solutionhas a maximal wave height of 3 times the waves around it and somesolutions such as the Ma-breather and the Akhmediev breather areperiodic in time and space.
8.5 conclusion 53
Figure 8.4: The evolution of weakly modulated wave train (numbers - timenormalized by the fundamental wave period). Highly energetic wave groupformed due to the Benjamin Feir instability [48]
In short, these breather solutions are considered as simple analyticalmodels of freak waves. To get more qualitative with the analysis thesemodels say that there are two types of waves in the deep sea. Thereare waves that move around comprised of sines and cosines that canbecome focused to create the special rogue waves, and then there arethese non-linear beasts that can really come out of nowhere. Thesenon-linear waves come from an almost rogue sea where as you watchthe sea state it is random yet tame and controlled. Then one of thesewaves comes up by sucking energy from the waves around it in a verybizarre non-linear fashion. This non-linear fashion also tends to helpdescribe the verticality of the waves described by the mariners. Thewaves become so steep that they actually can tend to break and havebeen likened to the white cliffs of Dover [4]. These breaking waves canlead to as much as 100 tons per square meter, much greater than the15t/m2 that the shipping industry has its ships designed for. It wouldbe practically impossible to design ships to withstand these monstersof the deep, and because of that they will remain untamed monsters.
8.5 conclusion
One area that has received the least amount of research to date comesfrom the focusing potential of wind. Experimental research in thisarea has shown that with out wind waves will focus at a fixed areand create a rather high amplitude wave but with strong wind thehigh amplitude wave will be further downstream and with a higheramplitude. Experimental results has also showed that rogue waves canoccur under the conditions of a strong wind when they typically wouldnot if the wind was not present. This line of research has lead most tobelieve that strong winds can increase the rate of occurrence of roguewaves. The effect the wind has is to weaken the defocussing processwhich leads to a greater chance for the rogue wave. As a result of thisfinding it makes sense that rogue waves are more likely to occur whenstorms with vicious winds are present.
54 sam bingham: freak waves
8.6 problems
1. Under the simplified linear model what is the probability offinding a 30-meter wave in a storm off the coast of Iceland inwhich the average of the highest third of the waves is 11 meters?
Solution: In this problem the height of the wave and the significant waveheight are given so the solution is rather simple and only requires theuse Eq. 8.4. Since H is 30 and HS is 12 we simply plug those into theequation and get
P(H) = e−2H2/Hs2⇒ P(30) = e−2(30)2/122
= 3.73× 10−6.
2. Which of the following is not a possible instigating factor for a tsunami?
a Meteorites landing in the oceanb Underwater earthquakesc Constructive interference of waves caused by wave trains in the deepoceand Underwater volcanoes
3. Which of the following are possible causes for rogue waves?
a Focusing of wave energy by ocean basinsb Focusing of wave energy by variational currents that run counter to theswellc Sudden changes in the depth of the ocean caused by a steep continentalsloped Non-linear effects that lead to a giant wave from a random backgroundof smaller wavesAnswer: a, b, c.
9PA U L H U G H E S : F R I C T I O N
9.1 overview
Friction represents one of the most basic and yet impenetrablephenomena within the purview of classical physics. While its role
in an idealized system can appear deceptively simple, its mechanismsand intrusions into realistic systems are of a complexity well beyondthe scope of any undergraduate course, much less a single textbookchapter.
9.2 amontons/coulomb friction
The quintessential description of friction is that of Amontons at theend of the 17th century, according to which friction is a force linearlyproportional to the normal force and independent of the macroscopicsurface area of contact–nearly a century later, Coulomb (of electrostaticfame) appended this to include that the proportionality of friction tothe normal force is independent of the relative velocity of the surfacesin question. This coefficient of friction µ in the Coulomb/Amontonsformulation does vary in relation to the material compositions of theinterfacing surfaces, however.
In fact, every set of material interfaces has its own coefficient of friction.This is where the matter begins to grow more complicated: applyingthis simple Amontons-Coulomb description of friction by itself wouldrequire a standardized table of friction coefficients for every possiblecombination of materials, as well as sub-tables to account for anylubricants or adhesives applied to the surface of interface. Velocityand temperature also play complicated roles in determining the actualfriction force between two real surfaces.
example 9 .1: simple friction
F
Ff
nw
v
Figure 9.1: The basic Amontons-Coulomb model of friction, ~Ff = µ~F.
55
56 paul hughes: friction
n
m
wv
θ
A block of mass m sits on a conveyor belt inclined at an angle θ from thehorizontal. The conveyor belt surface moves at a velocity ~v such that the blockremains stationary with respect to the surroundings. The coefficient of kineticfriction between the block and the belt surface µk = 0.3. Using only simpleAmontons/Coulomb friction, we can find θ(v).
First, we know that the normal force ~n = −~w · cos(θ): clearly, the magni-tude of ~n should be equal to that of ~w when θ = 0 and 0 when θ = π/2, andwill be in the positive direction rather than the negative like ~w = m~g. Fromthe block’s perspective, it appears to be sliding with velocity ~vbl down alongthe slope of the conveyor belt due to the effective force ~ge f f = m~g sin(θ),where ~vbl = −~v. It is opposite this velocity that we see a friction force arisewhich, for the equilibrium condition stated above,
Ff = ~ge f f = m~g sin(θ) = µ~n (9.1)
m~g sin(θ) = µm~g cos(θ) (9.2)
tan(θ) = µ (9.3)
So we see that θ is in fact independent of velocity ~v, at least in the Coulombrepresentation of kinetic friction. The critical angle θc = arctan(µ).
This direct relationship between incline slope and coefficient of staticfriction is one method of measuring µ values for different materials.Another is the arrangement of a mass m1 on a string connected viapulley to a mass m2 on a flat plane: m2, attached to the upper testsurface, with the lower attached to the level plane, provides the normalforce, and m1 provides the motive force which the friction force willresist [20]. The drawback, of course, is that all µ values must be acquiredexperimentally, and apply only to the specific materials tested.
9.3 toward a conceptual mesoscopic model
In actuality, friction is the result of a number of independent micro-scopic surface phenomena appearing in tangential interactions. Thesephenomena appear on the properties of the surface materials as wellas the substrate materials, the geometries and small-scale topologiesof the interaction surfaces, viscosity and other properties of any inter-stitial substance such as lubrication, relative velocity of the interactionsurfaces, and other influences. Even these individual sources of frictionmay each represent a range of models, and any one model may addressany one or more of the above contributory phenomena [29].
Amontons’s familiar Ff = µFn law evolves on the microscopic scalefrom the assumption that the microscopic contact area A = αFn in-creases as the product of the normal force—i.e., with the force de-
9.3 toward a conceptual mesoscopic model 57
a.
b.
c.
Figure 9.2: a. Interface surfaces near contact showing asperities. b. Interfacesurfaces under normal force Fn, showing some deformation of asperities. c.Interface surface with lubrication, showing fluid space-filling and “coasting”ability.
forming the surfaces to reduce areas of separation—and the plasticdeformability α characteristic of the weakest major constituent of thesurfaces. At the same time, until the characteristic shear strength τ isexceeded, the deformation remains elastic, so that Ff = ταFn. Thus wesee that µ = τα. Taking into account the direction of motion, the resultis the familiar friction law,
Ff = −µ~n · sgn(v), (9.4)
where sgn(v) = |v|/v. This model assumes that what we call “rough-ness” on the macroscopic scale appears, microscopically, as “asperities”or projections from the mean plane of each surface [29].
One way in which this model can break down is in the case ofmicroscopically smooth, geometrically complementary surfaces: themicroscopic contact area reaches its maximum with very little surfacedeformation, meaning that an increased normal force will no longerincrease shearing stress or friction. In such a case, various atomic forces(such as covalent bonds, electrostatic attraction, and “Van der Waals”forces between dipoles) are also particularly likely to play a role: asa noteworthy example, the energy required to separate the interfacesurfaces is equal to the free energy associated with their bonding, minusthe elastic potential energy made available as elastic deformation isrelaxed [28]. Obviously, then, a limit on elastic deformation impliesgreater influence of adhesion forces as surfaces become smoother, andwhen adhesive forces become involved, the friction quickly becomes lessCoulombic–that is, the relationship between friction force and normalforce becomes less linear–as a dependence on the area of interface isintroduced.
In the same way, lubricants serve to fill gaps and minimize the defor-mation necessary for asperities to bypass one another by functioningsomewhat like ball bearings: asperities are permitted to “coast” pastone another, rather than striking, pressing, and producing a deforma-tion on the scale of their geometric overlap—the friction gains, instead,a dependence on the lubricant’s viscosity. Unsurprisingly, then basedon this conceptual model, the highest coefficients of friction are seen
58 paul hughes: friction
in materials which are either pourous and highly elastic (e.g., rubber),or which otherwise exhibit a highly elastic ‘clinging’ action (e.g., vel-cro). Interfaces such as a highly porous, elastic rubber on hard, roughconcrete can yield coefficients of friction well above µ = 1; others in-volving such materials as teflon or ice (especially near the melting point,where the contact surface can melt and self-lubricate) can drop belowµ = 0.04.
9.4 summary
Friction is one of the most familiar, everyday forces, experienced when-ever we rub our hands together to warm them or wrestle with a jammeddoor, but the mechanisms from which it evolves are very complicatedand remain fertile ground for the development of new descriptions.With today’s faster and more powerful computer modeling, highlysophisticated models of friction can be tested more completely thanever before, and the advancement of nanotechnology may play a signif-icant role in the experimental methods employed to verify new models’predictions. The study of friction encompasses not only physics, butalso mechanical engineering, chemical engineering, and even consumerproducts: for example, many cleaning products are specially engineeredso that their residues have frictional properties we associate with a“clean” feeling.
9.5 problems
1. A model elevator of mass m = 1.5kg is allowed to slide freelydown a shaft, starting with v0 = 0 at height h0 = 4m. At h1 = 3ma brake is activated, consisting of a brake pad with coefficient offriction µ = 0.42 on a spring with spring constant k = 1000N/m.The spring, whose natural length `n = 5cm is compressed to` = 1cm. Where will the model elevator stop? If it strikes thebottom of the shaft, what will its velocity be?Fs p = −∆` · k = −(`− `n) · k = (0.04m)(1000N/m) = 40Nv2
1 = v20 + 2g∆h = −2(9.8m/s2)(1m)
v1 = −√
19.6m/s = −4.43m/s|~n| = ~Fsp ergo Ff = µ~n = µ · 40N = 16.8NFtotal = ~Ff + ~w = (16.8 + (1.5kg)(−9.8m/s2))zNatotal = F/m = 16.8
1.5kg − (9.8m/s2))z = 1.4zm/s2
v2 = v1 + atotaltSince we are looking for the time at which v2 = 0,t2 = − v1
atotal= 4.43
1.4 = 3.16sh2 = h1 + v1t2 + 1
2 atotalt22
h2 = 3m + (−4.43m/s)(3.16s) + 12 (1.4m/s2)(3.16s)2
h2 = 0.019m
2. A block of mass m is dropped onto an inclined conveyor belt, withwhich it has a coefficient of kinetic friction µ. The conveyor belt’supper surface is running at a rate v from the lower end towardthe higher end, a total distance of `, and its angle of inclinationfrom the horizontal is θ.a. Assuming that θ is below θc, how long will the box slide on the
9.5 problems 59
conveyor belt?We define the initial velocity vblock of the block relative to theconveyor belt as vblock(t = 0) = −v.Thus Ff = µn = µmg cos(θ), opposed by mg sin(θ)Ftotal = mg[µ cos(θ)− sin(θ)] so a = g[µ cos(θ)− sin(θ)]The box may accelerate to match the conveyor belt’s velocity, inwhich case:v1 = v0 + at, where v1 = v and v0 = 0, so tstop = v
g[µ cos(θ)−sin(θ)]Conversely, the box may still be sliding relative to the conveyorbelt when it reaches the end:s1 = s0 + v0t + 1
2 at2 where s0 = v0 = 0 and s0 = `, so tedge =√2`
g[µ cos(θ)−sin(θ)]Whichever happens first is the time at which the block stops slid-ing and reaches its peak velocity.
b. If m = 1kg, µ = 0.75, v = 20m/s, θ = π12 , and ` = 2m, at
what horizontal distance d from the edge of the conveyor belt willthe the block land?First, tedge =
√4
9.8[0.75 cos(π/12)−sin(π/12)] = 0.936s,
and tstop = 209.8[0.75 cos(π/12)−sin(π/12)] = 4.38s.
Therefore, we see that the block leaves the conveyor belt att = 0.936s with velocityvedge = at = g[µ cos(θ)− sin(θ)]t = 4.27m/s and at heighthedge = ` sin(θ) = 0.518mFrom here, the problem is simply one of projectile motion:hgnd = 0 = hedge + vedge sin(θ)t− 1
2 gt2 so tland = 0.231sd = vedge cos(θ)tland = 0.953m.
3. A very soft, microscopically smooth surface is in contact witha hard, smooth surface. The foremost contributor to the frictionforce is likely to be:a. A very strong normal force.b. Lack of lubrication.c. Elastic deformation.d. Interatomic forces.e. Particulate contamination between the surfaces.
d: Since the surfaces are smooth, increasing normal force willplay relatively little role in increasing the friction force. Thereis little asperity interaction to be mitigated by lubrication or toproduce significant elastic deformations. Because the first surfaceis soft, contamination would tend to deform it and press into thedeformed recess, reducing the contaminating particles’ influence.
10K E I T H F R AT U S : N E U T R I N O O S C I L L AT I O N S I N T H ES TA N D A R D M O D E L
10.1 introduction
Arguably one of the most peculiar aspects of quantum mechanics,that most undergraduates are familiar with, is that of the prin-
ciple of superposition . That is, because a quantum system is treatedin ways very similar to wave-like phenomena, it is possible to taketwo quantum states and add them together in a linear combination,generating another valid quantum state. But as it turns out, this “weird”property of quantum mechanics can be extended to some even morecounter-intuitive realms. It just so happens that a fundamental particlein and of itself can exist as a linear combination of two other particles. Thisamazing property of fundamental particles can lead to some very inter-esting phenomena, such as the one we will discuss in this chapter - thatis, neutrino oscillations. But before we get into just exactly what theseneutrino oscillations are, let’s start with a quick review of quantumtheory, and also touch upon what it is we currently know about thefundamental particles of nature.
10.2 a review of quantum theory
As we all have been taught in our introductory quantum mechanicsclasses, the theory of quantum mechanics deals with the idea of a“state,” which a particle (or system of particles) may happen to finditself in. Every observable property of that particle is represented by an“operator,” or a transformation that acts on the state of the particle. Welearn that the eigenfunctions (or eigenstates, as they are called) of thisoperator represent a set of possible states the particle can attain, andthe corresponding eigenvalues represent the values of the observableproperty associated with that state. If a particle happens to find itselfin a particular eigenstate of an observable’s operator, every time thatobservable is measured, it will return the corresponding eigenvalueassociated with that state [37].
For example, we can have the “spin” operator,
Sz =
(h/2 0
0 −h/2
), (10.1)
which is the operator corresponding to the z-component of intrinsicangular momentum for a particle with a spin of 1/2, in matrix form.Because our operator is represented by a two by two matrix, thisobservable can only acquire two eigenvalues, which correspond tothe spin pointing along or opposite to the z-axis. Actually, differentcomponents of angular momentum never commute in quantum theory,which means their values can never be simultaneously known. Thus, ifthe spin were to be measured to be completely along the z direction, thiswould imply that we know the x and y components to be zero, whichis not possible. So, in reality, these two states represent the projection
61
62 keith fratus: neutrino oscillations in the standard model
of spin along the z axis being either along the positive or negative zaxis [37].
We can represent the two eigenstates as vectors in a complex-valuedHilbert Space:(
1
0
);
(0
1
). (10.2)
The theory makes a restriction on the form of the operators that canrepresent observables, which is that they must be hermitian. Becauseof this fact, all possible eigenvalues must be real, and the eigenstatesof the operator must form a basis of all possible states that the particlecan attain. In our example of spin, this means that the particle’s z-component of intrinsic angular momentum can in general be a linearcombination of the two basis states, or,(
a
b
)= a
(1
0
)+ b
(0
1
), (10.3)
where a and b are some given complex numbers. But the question nowarises, what value will we get when we measure the z-component ofspin, for the particle in this generalized state? The answer is that wecould get either of the two eigenvalues, with some given probability.The probability of returning each eigenvalue is given by the square ofthe norm of the coefficient on the associated eigenstate. This, in essence,is the principle of superposition in quantum mechanics [37].
Because there are often many physical properties that a particle canpossess, there will generally be many different hermitian operatorsthat can act on the state of a particle. Each of these operators will thusdefine a new basis in which we can define the state of our particle,corresponding to the eigenstates of this operator.
10.3 the standard model
The world as we know it today, at least from the perspective of physics,can generally be summed up in a theory dubbed the “Standard Model.”This theory asserts that all of physical phenomena can be traced toa few fundamental particles, interacting through a few fundamentalinteractions. This is often described pictorially, as in Figure 10.1.
There are two types of fundamental particles, fermions, which havehalf-integer spin, and bosons, which have integer spin. The fermionsare generally what we would consider particles that make up matter,and the bosons are the particles that transmit the fundamental “forces”that act on the fermions. The fermions tend to be arranged in threedifferent “families,” or “generations,” with the particles in one familybeing identical to the particles in the next family, except for mass (inthe figure, the families are delineated by the first three columns). Forexample, in the figure, an “up quark” is a fermion, and it interacts withother quarks partly through interaction with what is called the “strongforce.” The strong force is mediated by the gluon, which is a boson,and we can think of it sort of like the two quarks interacting with eachother by passing gluons back and forth. Quarks can combine togetherto make composite particles, such as protons (two up quarks and adown quark) or neutrons (two down quarks and an up quark) [88].
10.3 the standard model 63
Figure 10.1: The Standard Model of elementary particle physics. Image courtesyFermilab.
The fermions come in two types, the quarks, which are in the firsttwo rows, and the leptons, which are in the last two rows, with oneof the biggest differences among them being that the leptons do notinteract with the strong force. In reality, matter particles can formcomposites which have integer spin and are thus bosons, and theparticular bosons represented in the standard model are specificallyreferred to as gauge bosons. But in terms of fundamental particles, wecan safely make the distinction between fermions and bosons withouttoo much consequence. This general analogy of passing bosons backand forth to mediate a force can be extended to the rest of the standardmodel. For each interaction in the standard model, there is a set ofmathematical rules describing how it acts on certain particles, and oneor more bosons that mediate it. In addition to the strong force, there isthe familiar electromagnetic force, mediated by the photon, and alsothe weak force, something we will discuss in more detail later. One ofthe more peculiar aspects of the standard model is that the strong forceactually grows stronger with distance, as if the quarks were attached bysprings. The result of this is that a free quark can never be observed, andthat the existence of quarks must be inferred by indirect means. Thereare some unresolved issues in the standard model, some of which hintat physics beyond the standard model, but aside from this, the standardmodel, for the most part, still represents our current understanding offundamental physics [88].
The origin of these particles is described by something called “Quan-tum Field Theory.” Quantum Field Theory (or QFT) generally statesthat each of the fundamental particles is sort of a “bump” in somethingcalled a quantum field, that can move through space. For example,there is an electron field that extends throughout space, and actualelectrons are like ripples moving across the “surface” of that field (ofcourse, this visual analogy can only be pushed so far). Speaking some-
64 keith fratus: neutrino oscillations in the standard model
what more mathematically, the fields that exist throughout space havecertain normal modes of oscillation, and when energy is input intothese fields, the resulting excitations of the field are what we experienceas fundamental particles [14, 78].
Furthermore, QFT associates with each of these fields somethingcalled a “creation operator” (there are also annihilation operators, butnever mind that for now). Many students have encountered these sortsof operators without ever having been aware of it. A common problemin introductory quantum mechanics is to solve for the possible energyeigenstates of the simple harmonic oscillator. One method for solvingthis problem is to use lowering and raising “ladder” operators, whichcreate lower and higher energy states from an existing one. When wetake an eigenstate of the simple harmonic oscillator, and apply thelowering operator to it, we attain the eigenstate with one less unitof energy. What we are actually modeling when we do this is thecreation of a photon that is emitted from the system, as it carriesaway the difference of the energies of the two states. In QFT, theseoperators “create” all of the possible particles. A given operator actson the vacuum of space, and “creates” an instance of the particleassociated with that operator. For example, the operator associated withthe electron field “creates” instances of electrons. When a photon turnsinto an electron-positron pair, the creation operators for the electronand positron model this process (along with the annihilation operatorfor the photon) [37, 14].
10.4 the weak and higgs mechanisms
While all of this may seem like useless information, seeing as how wedo not intend to pursue all of the associated mathematical details ofQuantum Field Theory, there is an important caveat to all of this. As wementioned before, there are often many different operators associatedwith a particle. This is also true in QFT. Looking at Figure 10.1, wesee a type of boson called a W particle, which is responsible for medi-ating an interaction called the weak force. In essence, the weak forceis responsible for transforming one type of particle into another. TheW boson actually represents two different particles, the W− and W+
particles, which are named according to their electric charge. Becauseof this, a particle must lose or gain a unit of charge when emitting a Wparticle (remember that forces are generally mediated by the exchangeof bosons). Since total electric charge must always be conserved, andbecause every particle has an intrinsic value of electric charge, thismeans that one type of particle must transform into another type ofparticle when emitting a W boson. There is actually another type ofparticle associated with the weak force, the Z0 boson, but it is involvedin a slightly different interaction. It is responsible for the scatteringof one particle off of another, similar to how charged particles canscatter off of one another via the electromagnetic interaction. This re-semblance to electromagnetic phenomena is actually due to a very deepconnection between the electromagnetic and weak forces. Electroweaktheory states that above a certain energy threshold, electromagnetic andweak interactions become indistinguishable, and the two interactionsare “unified.” Of course, this subject could easily form the basis of anentirely separate chapter, but for now we can safely concentrate on just
10.4 the weak and higgs mechanisms 65
the W bosons (though, not to discourage the reader from investigatingthis subject on his or her own!) [76].
As we can see in Figure 10.1, the fermions appear to come in pairs.Each particle in a pair is capable of turning into the other particle in thepair via the weak force. As a matter of fact, the weak interaction actuallypredicted the existence of the charm, top, and bottom quarks, partlydue to this pair behavior. For example, it is possible for a down quarkto turn into an up quark, and for a muon to turn into a muon neutrino.Of course, there is a (somewhat complex) mathematical formalismthat describes how all of this occurs, which involves a set of quantummechanical operators. So in some sense, remembering our previousdiscussion, it is possible to create a “basis” in which we describethe action of the weak force, represented by the eigenstates of thesequantum mechanical operators. These eigenstates are often referred toas the “weak states” of the given fundamental particles. In short, wecan describe the behavior of the fundamental particles of nature bydiscussing them in the context of how the weak force acts on them [76].
As a matter of fact, the weak force represents a sort of tarnishedbeauty in the standard model. It turns out that the weak force iscapable of predicting the existence of the fundamental fermions, but, byitself, it predicts that they should all be massless. Of course, we knowthat particles in the real world have mass, so something else must beresponsible for giving mass to the fundamental particles of nature. Thismechanism is referred to as the Higgs mechanism. It creates “Higgsbosons” out of the vacuum, and these particles interact with the otherparticles in the standard model, creating a “drag” on them which weobserve as inertia. Of course, it also has its own associated mathematicaloperators, which constitute another “basis” in which to describe thedifferent particles of nature. The Higgs boson is not included in thestandard model currently, because, at the time of publication, it has stillnot been definitively detected. However, so much of physical theorydepends on its existence that is essentially assumed to exist [76, 88].
But here’s the crucial point to all of this: it turns out that the manner inwhich the two mechanisms describe the fundamental particles of natureare actually different. In other words, when we approach the theory fromthe standpoint of the weak force, and describe the interactions of theparticles from this perspective, we find that the interactions behavedifferently than if we were talking about them from the standpoint ofthe Higgs mechanism! For example, an up quark defined by the weakforce is not the same particle as an up quark defined by the Higgs mechanism.So we in essence have two different descriptions of the natural world.How are we to reconcile this? [76]
There is actually a simple solution to this dilemma. Quantum mechan-ics postulates that it is indeed possible to describe a physical systemfrom the vantage points of two different mechanisms (or operators, orwhatever you care to call your mathematical devices), and that thesetwo representations can be related via a change of basis, which is verystraightforward from a mathematical perspective, using the languageof linear algebra. To talk about how we would do this, let’s considerthe simplified case where we only consider the first two families offermions (which, as it turns out, is actually a fairly reasonable simplifi-cation). For now, let’s look at the first two pairs of quarks. Because thephenomenon we are dealing with is related to the weak force, whichacts on particles in pairs, we need only consider one particle from each
66 keith fratus: neutrino oscillations in the standard model
pair of quarks, since the weak interaction will take care of how thesecond particle from each pair is affected by the phenomenon. Let’swork with the down and strange quarks, and call them d’ and s’ whenreferring to their representation by the weak force, and d and s whendiscussing them in terms of the Higgs mechanism. Quantum theorysays that these two representations should be related to each other by asort of “change of basis” matrix in some given Hilbert space in whichwe describe the state of our physical system, which will look somethinglike (
d
s
)=
(cos θ − sin θ
sin θ cos θ
) (d′
s′
). (10.4)
The angle θ is a measure of how much of a “rotation” there is betweenthe two bases. If we expand this relation, we get
d = cos θ d′ − sin θ s′ (10.5a)
s = sin θ d′ + cos θ s′. (10.5b)
But look at what we are saying: it is possible for one particle toactually be represented as a linear combination of other particles. Thismay seem incredibly counter-intuitive, and perhaps non-sensical oreven impossible, but the truth of the matter is that in a mathematicalsense, this explains a whole host of phenomena that would appearas inexplicable anomalies otherwise. One example that illustrates thisquite simply has to do with the rate of neutron and lambda beta decay(see the first homework problem) [76].
Actually, there are other scenarios in which we can speak of linearcombinations of particles. It turns out that if we have two particleswhose interactions are identical, it should be possible to interchangethem, and the physics of the situation should be the same overall. Asa matter of fact, we can even take a generalized linear combination ofthese particles, and the physics will be the same. Suppose we have twoparticles in a physical system, called m and n. We can represent this asa vector:(
m
n
). (10.6)
It is then actually possible to apply a “rotation” to this set of particles,just as for the weak and mass states of the down and strange quarks. Ifwe have a unitary matrix M that is two by two and has a determinantequal to one, then applying this matrix to the above vector actuallyrepresents a new set of particles, ones which do not change the overallphysics of the system (matrices of this form are members of a groupof matrices called SU(2), or the special unitary group of degree two,which has tremendous implications for high-energy physics). If youhave objections to the fact that two particles can be added to each otherlike mathematical objects, welcome to the world of quantum mechanics.While it may seem questionable to speak of fundamental particles inthis way, it fits quite naturally into the quantum mechanical descriptionof the universe, in which a given physical system can lend itself to avariety of outcomes upon measurement, each outcome being weightedwith a certain probability. Only now, we are talking about the very typeof particle that we will detect upon measurement [14].
10.5 the origin of neutrino oscillations 67
10.5 the origin of neutrino oscillations
But back to the issue of the weak and Higgs mechanisms. It is oftencustomary, as mentioned previously, to call d and s “mass states,” andd’ and s’ “weak states.” Each of them can be represented as linearcombinations of the others, and are related by the equation givenpreviously. This analogy actually extends to other particles. In particular,we will consider the neutrinos, which are in the bottom row of Figure10.1. Recent evidence suggests that they behave in a manner similarto quarks, in that their mass states and weak states are different (fora while it was believed that neutrinos were massless, and hence werenot subject to this behavior, but this has been shown to not be true).Neutrinos are fermions, and there are three known types of them. Theyeach come paired with another fermion, one that is either the electron(in the case of the electron neutrino), or one that is similar to the electron(either the muon or tau particle, being identical to the electron, exceptfor mass) [14].
So what is the significance of this property of neutrinos? It actuallyjust so happens that this property of neutrinos is responsible for some-thing called neutrino oscillations. That is, it is possible for one typeof neutrino to turn into another type of neutrino as it travels throughspace. This can also occur for quarks, but because of the fact that afree quark can never be observed, it involves the composite particlesmade up of quarks, so the details are somewhat simpler for the case ofneutrinos. The case of neutrinos is also particularly relevant, not onlybecause this phenomenon occurs to a greater extent among neutrinosthan it does for quarks, but because the existence of neutrino oscilla-tions implies that they interact with the Higgs mechanism, and thusindeed have mass, something that was not believed to be true for severaldecades. Neutrino oscillations are also significant in the sense that theyare of great experimental interest. Our current models that describethe nuclear activity in the sun predict that we should be detecting agreater number of electron neutrinos being emitted from it than weactually are. It is believed that the explanation for this discrepancy isthat of neutrino oscillations; that is, electron neutrinos emitted from thesun have the ability to change into other types as they travel betweenthe sun and earth. There are of course many other reasons why thephenomenon is of great interest, but suffice it to say, it is a behaviorthat beautifully demonstrates some of the basic properties of quantummechanics, without too much mathematical complexity [14].
Having discussed the significance of these “oscillations,” let’s see ifwe can get a basic mathematical derivation of them. For the sake ofsimplicity, we’ll consider the case of neutrino oscillations among thefirst two families. That is, oscillation among the electron and muonneutrinos. Once again, this is a somewhat reasonable approximation tothe more complex case of three-family mixing. Because each neutrinocomes paired with an electron-type particle by the weak force, we canconsider the mixing matrix with respect to just one particle in each ofthe pairs. In other words, we can write the mixing equation with respectto just the neutrinos. If we denote the electron and muon neutrinos withrespect to the Higgs mechanism as e and µ, respectively, and denote
68 keith fratus: neutrino oscillations in the standard model
them as e’ and µ’ with respect to the weak mechanism, the generalrelation between the two representations will be given by:(
e′
µ′
)=
(cos θ sin θ
− sin θ cos θ
) (e
µ
), (10.7)
where θ is the angle that describes the “rotation” among the repre-sentations (note that we are now referring to the weak states as theresult of the mass states being rotated through some angle, insteadof the other way around; this is possible because rotations are alwaysinvertible). Let’s consider a muon neutrino that is created through theweak force. Because it is created through the weak force, and the weakforce operates with respect to weak states, the neutrino can definitelybe said to be a muon neutrino with respect to the weak mechanism. Inother words, it is in a pure weak state. This means that it can be givenby a linear combination of mass states, or,
| µ′〉 = − sin θ | e〉+ cos θ | µ〉, (10.8)
in traditional ket notation.We now want to determine how this state evolves with time. A
particle that is moving freely through empty space of course has notime-dependent influences acting on it, so we can treat this as a typicaltime-independent problem in quantum mechanics. If our weak stateis treated as a superposition of mass states, then to describe the time-dependence of our neutrino, we simply add the time-dependent factorto each mass state. So we can write
| ν(t)〉 = − sin θ | e〉e−iEet/h + cos θ | µ〉e−iEµt/h, (10.9)
where Ee and Eµ are the energies of the electron and muon neutrinomass states, respectively. We use the letter ν now, because as our stateevolves over time, it will not necessarily be a muon neutrino, or evenan electron neutrino; in general it will be some linear combination ofthe two.
What we are interested in, of course, is the probability that we willdetect the original neutrino as being an electron neutrino some timeafter it is created. To do this, we follow the standard prescription ofquantum mechanics, which says that the probability of measuring anelectron neutrino is the square of the norm of the projection of ourstate at a given time along the basis vector representing the electronneutrino. Now, when me detect a neutrino, we do so by studying theway it interacts with matter via the weak force. So any neutrino wedetect will be in a definite weak state, since the weak force always actson weak states. This means we need to calculate the projection of ourstate onto the weak state of the electron neutrino. Symbolically, we canwrite this as
P( µ′ → e′ ) = |〈e′ | ν(t)〉|2. (10.10)
If we take this inner product in the basis of the mass states, thenmathematically, our expression for the inner product becomes
〈e′ | ν(t)〉 = ( cos θ , sin θ )
(− sin θ e−iEet/h
cos θ e−iEµt/h
). (10.11)
10.5 the origin of neutrino oscillations 69
The computation of the dot product is straightforward, and leads to
〈e′ | ν(t)〉 = − cos θ sin θ e−iEet/h + sin θ cos θ e−iEµt/h, (10.12)
which allows us to write
P( µ′ → e′ ) = ( cos θ sin θ )2 | − e−iEet/h + e−iEµt/h |2. (10.13)
The square of the norm of a complex number of course is that numbermultiplied by its complex conjugate, so, using this fact, along with atrigonometric identity for the sinusoidal factor, we have
P( µ′ → e′ ) = (12
sin 2θ )2 ( −e−iEet/h + e−iEµt/h ) ( −eiEet/h + eiEµt/h ).
(10.14)
If we expand this expression, we come to
P( µ′ → e′ ) =14
sin2 2θ ( e0 + e0 − (ei(Ee−Eµ)t/h + e−i(Ee−Eµ)t/h)),
(10.15)
which of course we can simplify to
P( µ′ → e′ ) =14
sin2 2θ
(2− 2 cos
(∆Et
h
)), (10.16)
where ∆E is the difference between the two energies, and we have usedthe definition of the cosine function in terms of complex exponentials.If we factor out the two, and use the trigonometric identity
2 sin2(χ
2
)= 1− cos χ, (10.17)
then this ultimately simplifies to
P( µ′ → e′ ) = sin2 2θ sin2(
∆Et2h
)(10.18)
for the probability that we will measure an electron neutrino instead ofthe original muon neutrino. Amazingly, this real world problem with ex-perimental relevance indeed lends itself to such a simple mathematicaltreatment.
One of the most immediately obvious things about this expressionis the factor determined by the rotation angle. Note that this factorcan be any value between zero and one, determined entirely by therotation angle. If the rotation angle happens to be equal to π/4, thenthe factor will be equal to one, and for any angle between zero and π/4,the factor will be less than one. If we consider the second sinusoidalterm, the one that oscillates with time, we can see that the factordetermined by the rotation angle becomes the amplitude of theseoscillations in time. So the probability of measuring an electron willreach a maximum when it is equal to the factor determined by therotation angle, since the maximum value of the oscillation term is ofcourse one, being a sinusoidal function. So immediately we see that theamount of rotation, or “mixing” between the two families of neutrinosis what determines the maximum probability of measuring a switch inthe type of neutrino. Anything less than 45 degrees of rotation impliesthat the muon neutrino will never have a one hundred percent chance
70 keith fratus: neutrino oscillations in the standard model
of being measured as an electron neutrino, implying that the state ofthe neutrino will never be exactly that of an electron neutrino (speakingin terms of weak states, of course).
It is also immediately apparent that the oscillations depend on theenergy difference of the two mass states. This is a feature commonto most quantum mechanical systems, since the Hamiltonian, or totalenergy, is what determines the time evolution of a state. The best wayto explore this phenomenon is with an example, similar to what mightbe encountered when working in an actual neutrino experiment.
example 10 .1 : a relativistic neutrino experiment
Suppose we have a scenario in which muon neutrinos are created with agiven energy at a known location, an energy large enough that they arehighly relativistic. Experimentally, we would like to know the probability ofmeasuring an electron neutrino at a given distance from the source of muonneutrinos. Let’s see how we could go about doing this.
First, we’ll make an approximation when it comes to the energy of theneutrinos. We know from special relativity that the energy of an object canbe given by
E = ( p2c2 + m2c4 )12 , (10.19)
where p is the momentum of the object, c is the speed of light, and m is themass of the object. If we factor out the first term in the square root, we canwrite
E = pc(
1 +m2c2
p2
) 12
. (10.20)
Using a Taylor expansion, the approximation
(1 + x)12 ≈ 1 +
x2
(10.21)
allows us to write the energy as being approximately
E ≈ pc(
1 +m2c2
2p2
), (10.22)
or,
E ≈ pc +m2c3
2p. (10.23)
This approximation is reasonable, because our neutrinos will generally bemoving relativistically, and this implies that the energy attributable to themass of the particle is small compared to the kinetic energy of the particle,and the second term in equation 10.20 is small enough for the approximationto be valid.
If we now consider the energy difference of the two mass states, we canuse the above approximation to write
∆E = Ee − Eµ ≈ pc +m2
e c3
2p− pc−
m2µc3
2p, (10.24)
which simplifies to
∆E ≈ c3
2p(m2
e −m2µ), (10.25)
or,
∆E ≈ c3
2p(∆m2), (10.26)
10.6 implications of the existence of neutrino oscillations 71
0 5 10 15
x 104
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Distance (m)
Pro
babi
lity
to D
etec
t an
Ele
ctro
n N
eutr
ino
Figure 10.2: The probability to measure an electron neutrino as a function ofdistance from the source. See the second homework problem for the parametersused. Image created in MATLAB.
where ∆m2 is the difference of the squares of the two different masses.We have assumed that the momentum of the two mass states is the same,which is a somewhat reasonable approximation. Because the neutrinos moverelativistically, we can assume that the distance x traveled by a neutrino isrelated to the time after creation by
t ≈ xc
. (10.27)
If we now use the approximations given in equations 10.26 and 10.27, ourexpression for the probability of measuring an electron neutrino is
P( µ′ → e′ ) ≈ sin2 2θ sin2(
c2∆m2x4ph
). (10.28)
Once again using a relativistic approximation, we can say that the overallenergy of the neutrino, Eν, is roughly given by
Eν ≈ pc. (10.29)
Also, because the masses of elementary particles are typically measured inunits of eV/c2, it is convenient to introduce a factor of c to the numeratorand denominator of the argument to the time dependent sinusoidal term. Soif we make these last two changes to our expression, our final result is
P( µ′ → e′ ) ≈ sin2 2θ sin2
(c4∆m2x4hcEν
). (10.30)
Figure 10.2 shows the probability to measure an electron neutrino asa function of the distance from the source, using the relation derivedin the example. The parameters used are those given in the secondhomework problem.
10.6 implications of the existence of neutrino oscillations
There are several reasons why our final expression in example 10.1 isof great interest to experimentalists. First, notice that the oscillationsexplicitly depend on the difference of the squares of the two masses. Ifthe difference were to be zero, then our expression would be identically
72 keith fratus: neutrino oscillations in the standard model
zero for all time, implying that there would be a zero probability to mea-sure an electron neutrino. In other words, there would be no oscillation.Because neutrino oscillations have now been experimentally verified,we know that neutrinos must indeed have mass (since the masses can-not all be zero and different from each other at the same time), whichhas tremendous implications for several branches of physics, from highenergy physics to cosmology.
The fact that neutrinos have mass shows a similarity between quarksand leptons. Because quarks and (some) leptons were known to havemass, it was known that quarks and (some) leptons must interact withthe Higgs mechanism in some way. Thus, it was surprising that onlyquarks exhibited oscillation, and not leptons. Now that leptons havebeen shown to exhibit oscillation, it shows a similarity between thetwo groups of fermions. This is crucial for “Grand Unified Theories,”theories which hypothesize that above some threshold energy, all ofthe interactions in the standard model unify into one fundamentalinteraction that acts amongst all of the fermions in the standard modelon an equal footing. The fact that we see slight differences amongst thedifferent fermions, but in general they have the same behavior, suggestsa sort of “broken symmetry” that would presumably be restored abovethis energy threshold. This energy threshold (referred to as the Planckenergy) however is believed to be around 1028 eV, which is much higherthan the energies that can be reached in modern accelerators (the LargeHadron Collider at CERN will only be able to reach energies of 14
TeV) [76, 14].The fact that neutrinos have mass is also of interest to cosmologists,
since it could have an influence on the rate of expansion of the universe.Neutrinos are generally created quite frequently in nuclear processes, soa large number of them flood the universe. Because of this, the fact thatthey have mass has serious implications for the energy-mass density ofthe universe, which, according to the theory of general relativity, affectsthe manner in which the fabric of space-time deforms and evolves withtime [76].
The first real evidence for neutrino oscillations was provided by theSudbury Neutrino Observatory in Sudbury, Ontario, Canada, whichdetected neutrinos from the sun. While previous experiments hadstrongly suggested the existence of neutrino oscillations, the experi-ment in Canada was the first one to definitively identify the numberof neutrinos of each type, and the experimentally determined num-bers agreed with theoretical predictions. Previous experiments hadonly shown a lack of electron neutrinos being detected from the sun,and gave no information regarding the number of neutrinos of othertypes [14].
Amazingly, without resorting to any of the more advanced mathemat-ical tools of Quantum Field Theory, it is possible to get a quantitativeunderstanding of a quantum mechanical effect that has enormousimplications for modern research in physics. Some of the homeworkproblems investigate the subject with specific numerical examples, togive a better idea of some of the numerical values actually involvedin neutrino oscillations. Of course, any reader who is interested isencouraged to research the subject further on his or her own, sinceQuantum Field Theory is a very complex, intellectually rewarding sub-ject that should begin to be accessible to students towards the end oftheir undergraduate study.
10.7 problems 73
10.7 problems
1. A proton, as mentioned in the text, is made up of two up quarksand a down quark, while a neutron is composed of two downquarks and an up quark. Neutron beta decay is the process inwhich a down quark inside of a neutron turns into an up quarkby emitting a W− boson, thus turning the neutron into a proton.There is a certain characteristic rate at which the weak force acts,and we might at first expect the rate of neutron decay to be equalto this characteristic rate. However, the observed rate of neutrondecay is slightly less than this. This can actually be explained bythe fact that the weak states of quarks are not the same as themass states of quarks. The quarks that form composites are massstates, so the quarks inside of the neutron and proton are linearcombinations of weak states. If we simplify the mixing to the firsttwo families, we can write
d = cos θ d′ − sin θ s′ (10.31a)
s = sin θ d′ + cos θ s′, (10.31b)
where, as usual, d and s are the mass states, and d’ and s’ are theweak states. Explain why this might account for the the rate ofneutron decay being less than expected. The mismatch betweenthese two types of quarks also accounts for the existence of an-other type of decay. A lambda particle is one that is comprisedof an up, down, and strange quark. One of the ways it can decayis to a proton, where the strange quark turns into an up quark.This is called lambda beta decay. One might expect this decaytype to be prohibited, since the up and strange quark are in twodifferent families. Explain how this decay might be possible, andhow it is related to the previously mentioned issue of neutronbeta decay [76].
Solution: The key to understanding this problem is realizing that, asmentioned before, the weak force only acts on weak states, and that theweak force is the mechanism that drives the two decay types. Each of themass states in the quark composites is a linear combination of two weakstates, so each has some probability of being measured as one of the twoweak states. In the neutron, the down quark that we expect to decay intothe up quark is actually a linear combination of the two weak states, andthe probability that it will be a down quark weak state is proportional tothe square of the coefficient in the linear combination that is applied tothe down quark weak state. This coefficient is the cosine of the rotationangle, which implies that squaring this term will give us the probabilitythat we will select the down quark weak state. Of course, this impliesthat there is some probability that with respect to the weak force, wewill have a strange weak state, which is not allowed to decay into anup quark. So the probability that the neutron will not be able to decayinto a proton is given by the square of the sine of the rotation angle. Sothe decreased rate is essentially explained by the fact that in some sense,the down quark mass state in the neutron is not always a down quarkwith respect to the weak mechanism, and so the neutron will not alwaysdecay to the proton.
This also explains how a lambda particle can decay into a proton. Thestrange quark in the lambda particle likewise has some probability of
74 keith fratus: neutrino oscillations in the standard model
being a down quark with respect to the weak force, given by the squareof the sine of the rotation angle. So there is a finite probability that thedown quark weak state component of the strange quark mass state willinteract with the weak force, allowing the lambda particle to decay into aproton.
The rate of neutron beta decay will be proportional to the probabilitythat the down quark mass state in the neutron is in the weak down quarkstate, which is given by the square of the cosine of the rotation angle.The rate of lambda beta decay will be proportional to the probabilitythat the strange quark mass state in the lambda particle is in the weakdown quark state, which is given by the square of the sine of the rotationangle. The sum of these two probabilities of course gives unity, so insome sense, the rate of decay of the lambda particle “makes up for” themissing rate of neutron beta decay.
Experimentally, the rotation angle between the first two families is foundto be about 0.22 radians, for a decreased rate of neutron decay of aboutfour percent [76].
2. For the final expression derived in example 10.1, find the spatial wave-length of oscillation.
Solution: The wavelength of the square of the sine function will be thefirst value of the distance that returns the sinusoidal function to zero,since the square of the sine function oscillates back and forth betweenzero and one. The sine of any integer multiple of π will be equal to zero,so if λ is the wavelength of oscillation, and x is the distance from thesource, then the argument that will be equal to π when the distance xfrom the source is equal to the wavelength is simply πx/λ. Equating thisto the expression for the argument found in the example, we have
πxλ
=c4∆m2x4hcEν
. (10.32)
Rearranging this, we have
λ =4πhcEν
c4∆m2 (10.33)
for the spatial wavelength of oscillation. Note that a larger energy im-plies a longer oscillation wavelength, since the neutrino will travel alonger distance in the amount of time it takes to oscillate from one typeof neutrino to another. A larger difference in the masses will cause asmaller wavelength, since the rate of oscillation increases with largermass difference. Note that the wavelength of oscillation is not affected bythe rotation angle among the two families of neutrinos.
3. Using the result derived in the example, find the probability that theemitted neutrino will be measured as an electron neutrino one hundredkilometers from the source if it has an energy of one million electron volts.The experimentally determined value for the mixing among the first twofamilies is 0.59 radians, and the value found for the difference betweenthe squares of the masses is 8 · 10−5eV2 / c4 (see Figure 10.2) [14].
Solution: If we square the sine of twice the rotation angle, then the valuewe attain is
sin2 2θ = sin2(1.18) ≈ 0.8549. (10.34)
If we substitute the value of the constants into the expression for theargument to the second sinusoidal function, we have
c4∆m2x4hcEν
=(c4) · (8 · 10−5 eV2
c4 ) · (105m)(4) · (197.327 · 10−9eV ·m) · (106eV)
, (10.35)
10.8 multiple choice test problems 75
where we look up the value of hc. All of the other constants cancel, andthe value of the argument becomes roughly 10.1355. Thus, the probabilityis approximately
P( µ′ → e′ ) ≈ sin2(1.18) sin2(10.1355) ≈ 0.364. (10.36)
Note that because of the value of the first sinusoidal term, the maximumprobability to measure an electron neutrino at any given time is roughlyeighty-five percent.
10.8 multiple choice test problems
1. Which of the following categories would a neutrino fall under?
a) Gauge bosonsb) Quarksc) Leptonsd) Massless particles
2. Which of the following particles have half-integer spin?
a) Only gauge bosonsb) Only quarksc) All bosonsd) All fermions
3. What is the quark content of a neutron?
a) u, u, db) u, d, dc) u, d, sd) u, u, s
Part III
I N F O R M AT I O N I S P O W E R
11A N D Y O ’ D O N N E L L : FA S T F O U R I E R T R A N S F O R M
11.1 introduction
Fourier Transforms can be found all throughout science and en-gineering. In Physics they are often used in Quantum Mechanics
to convert the wave function in position space into the wavefunctionin momentum space. In Electrical Engineering it is used to converta signal from the time domain into the frequency domain. The toolof the Fourier Transform is essential for analyzing the properties ofsampled signals. However, many times scientists and engineers arenot dealing with known analytic functions but are rather looking atdiscrete data points and therefore they must use the discrete form ofthe Fourier Transform . The Discrete Fourier Transform(DFT) has onemajor drawback, it is computationally slow. It requires N2 computa-tions, where N is the number of samples you have taken and oftenthe frequency of sampling is much higher to lower error. Thereforewe will derive and explore the radix-2 Fast Fourier Transform(FFT) forthe reader. We will also show that the number of operations we haveto perform is N
2 log2(N). This chapter is divided into three sections.The first section will give a quick introduction into the continuousFourier Transform and its inverse. The second section will introducethe Discrete Fourier Transform and talk about its inverse. The thirdsection will talk about the Fast Fourier Transform and a derivation ofthe Radix-2 FFT algorithm will be given.
11.2 fourier transform
The Fourier Transform is given by
X( f ) =∫ ∞
−∞x(t)e−j2π f tdt. (11.1)
Where x(t) is some function of time, j is the imaginary number√−1,
and f is the frequency. The Fourier Transform uses two mathematicalproperties to extract the frequency. In the above, we can think of x(t)as a superposition of sines and cosines. The second term in the aboveequation, e−j2π f t, can be expanded using Euler’s Formula
ejθ = cos(θ) + jsin(θ) (11.2)
Due to the orthogonality of the trig functions, it can be seen that onlyonly at a certain values of 2π f will values of the integral survive.
The equation for the inverse continuous Fourier Transform is givenby
x(t) =∫ ∞
−∞X( f )ej2π f tdt. (11.3)
Given the information on the frequency of a signal, we can then findthe time domain of that signal. [54]
79
80 andy o’donnell: fast fourier transform
11.3 discrete transform
On the computer we do not deal with continous fucntions but ratherwith discrete points. The discrete Fourier Transform and its inverse is
X(m) =N−1
∑n=0
x(n)e−j2πnm/N . (11.4)
x(n) =1N
N−1
∑n=0
X(m)ej2πnm/N . (11.5)
Where m is the frequency domain index, n is the time domain index,and N is the number of input samples. According to Nyquist Criterion,if the sampling rate is greater than twice the highest frequency, thenthe discrete form is exactly the same as the continuous form. So withthat condition in mind, we can easily make the transition from thecontinuous form of Fourier Transform to the discrete Fourier Transform.Here is an example of a discrete Fourier Transform. [54]
example 11 .1 : simple sine function in matlab
Let us look at the Fourier Transform of the function sin(t). If the abovestatement is true about the orthogonality, then we expect that that only onefrequency will survive if we apply for the Fourier Transform to sin(t). Hereis some simple code for evaluating the sin(t) in MATLAB. Here we use theembedded function ’fft’ inside of MATLAB. We can effectively think of it as avery good aproximation to the continuous Fourier Transform.
%MATLAB Code to compute the Foruier Transform of sin(t)
clear all
max=360;
for ii=1:max
x(ii,1)=ii*(pi/180);
y(ii,1)=sin(x(ii,1));
index(ii,1)=ii/360;
end;
Y=fft(y);
figure(1)
hold on;
clf;
title('Fourier Transform of sin(t)')
xlabel('Frequency Hz')
ylabel('Magnitude of X(m)');
plot(index,Y,'.-')
figure(2)
hold on;
plot(index,y)
title('sin(t)')
xlabel('Time in seconds');
ylabel('x(n)')
As seen from the figures of the signal and the Fourier Transform of it, thereis only one position on the frequency space where we find it. This is backedup by out expectations about it.
11.4 the fast fourier transform 81
0 0.2 0.4 0.6 0.8 1−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1sin(t)
Time in seconds
x(n)
Figure 11.1: One period of a sine function.
11.4 the fast fourier transform
As you can see above, for the Discrete Fourier Transform to be useful,there needs to be a faster way to computer it. For example, if you wereto have a value of N = 2, 097, 152, using the Fast Fourier Transformalgorithm it would take your computer about 10 seconds to do it whileusing the Discrete Fourier Transform described above it would takeover three weeks. [3] This next section will get into the details andderivation of the radix-2 FFT algorithm. First, we start off with ouroriginal definition of the discrete Fourier Transform. [46]
X(m) =N−1
∑n=0
x(n)e−j2πnm/N . (11.6)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.5
0
0.5
1
1.5
2
2.5
3
3.5
Fourier Transform of sin(t)
Frequency Hz
Mag
nitu
de o
f X(m
)
Figure 11.2: The FFT of a sine function. Notice how it only has discrete compo-nents .
82 andy o’donnell: fast fourier transform
For the next part of this, we split the sums into evens and odds andget
X(m) =(N/2)−1
∑n=0
x(2n)e−j2π(2n)m/N +(N/2)−1
∑n=0
x(2n + 1)e−j2π(2n+1)m/N .
(11.7)
In the second term, we can easily pull out the phase angle. Doing thatwe find
X(m) =(N/2)−1
∑n=0
x(2n)e−j2π(2n)m/N + e−j2πm/N(N/2)−1
∑n=0
x(2n + 1)e−j2π(2n)m/N .
(11.8)
Next we need to define some notation to simplify the exponentialterms. We shall now use the following notation.
WN = e−j2π/N , WnN = e−j2πn/N , W2
N = e−j2π2/N , WnmN = e−j2πnm/N
(11.9)
Using this notation, we can then replace the above equation with
X(m) =(N/2)−1
∑n=0
x(2n)W2nmN + Wm
N
(N/2)−1
∑n=0
x(2n + 1)W2nmN . (11.10)
Then, through algebraic manipulation we know that W2N = e−j2π2/N =
e−j2π/(N/2) = WN/2. This allows us to change the equations to
X(m) =(N/2)−1
∑n=0
x(2n)WnmN/2 + Wm
N
(N/2)−1
∑n=0
x(2n + 1)WnmN/2. (11.11)
Next let’s consider the X(m + N/2) case and we find that
X(m + N/2) =(N/2)−1
∑n=0
x(2n)Wn(m+N/2)N/2 +Wm+N/2
N
(N/2)−1
∑n=0
x(2n + 1)(N/2)− 1Wn(m+N/2)N/2 .
(11.12)
Next we can use the following expression that
Wn(m+N/2)N/2 = Wnm
N/2WnN/2n/2 = Wnm
N/2(e−j2πn2N/2N) = WnmN/2(1) = Wnm
N/2.
(11.13)
We then call the express in front of the summation the twiddle factorand we can simplify the above as
Wm+N/2N = Wm
N WN/2N = Wm
N (e−j2πN/2N) = WmN (−1) = −Wm
N . (11.14)
We can then plug this in to find that
X(m + N/2) =(N/2)−1
∑n=0
x(2n)W(nm)N/2 −Wm
N
(N/2)−1
∑n=0
x(2n + 1)WnmN/2.
11.5 multiple choice 83
(11.15)
The only different between the X(m + N/2) and X(m) equations isthe minus sign on the second summation. This means we only use thefirst N/2 terms in the DFT and then use those to find the final N/2terms. This is the Fast Fourier Transform because it greatly simplifiesthe computational work that has to be done to solve it. [54]
11.5 multiple choice
Question: The number of multiplication steps in the FFT isA) N2
B) NC) N3
D) N2 log2(N)
Answer: D. See Above text.
11.6 homework problems
1) If N = 4, what are the first four terms of the summation of theDiscrete Fourier Transform? Answer:
Let N=4, compute the 4 terms of the Discrete Fourier Transformoutput.
X(0)=x(0)cos(2 \pi 0*0/4) -jx(0)sin(2\pi 0*0/4)
+ x(1)cos(2 \pi 1*0/4) -jx(1)sin(2\pi 1*0/4)
+ x(2)cos(2 \pi 2*0/4) -jx(2)sin(2\pi 2*0/4)
+ x(3)cos(2 \pi 3*0/4) -jx(3)sin(2\pi 3*0/4)
X(1)=x(0)cos(2 \pi 0*1/4) -jx(0)sin(2\pi 0*1/4)
+ x(1)cos(2 \pi 1*1/4) -jx(1)sin(2\pi 1*1/4)
+ x(2)cos(2 \pi 2*1/4) -jx(2)sin(2\pi 2*1/4)
+ x(3)cos(2 \pi 3*1/4) -jx(3)sin(2\pi 3*1/4)
X(2)=x(0)cos(2 \pi 0*2/4) -jx(0)sin(2\pi 0*2/4)
+ x(1)cos(2 \pi 1*2/4) -jx(1)sin(2\pi 1*2/4)
+ x(2)cos(2 \pi 2*2/4) -jx(2)sin(2\pi 2*2/4)
+ x(3)cos(2 \pi 3*2/4) -jx(3)sin(2\pi 3*2/4)
X(3)=x(0)cos(2 \pi 0*3/4) -jx(0)sin(2\pi 0*3/4)
+ x(1)cos(2 \pi 1*3/4) -jx(1)sin(2\pi 1*3/4)
+ x(2)cos(2 \pi 2*3/4) -jx(2)sin(2\pi 2*3/4)
+ x(3)cos(2 \pi 3*3/4) -jx(3)sin(2\pi 3*3/4)
As you can clearly see, the amount of multiplication that needs to bedone grows very quickly as N increases. The amount of multiplicationthat needs to be done in the discrete Fourier Transform is N2, as can beseen above. [54]2: Prove Linearity in the discrete Fourier Transform:Proof:
Xsum(m) = X1(m) + X2(m) (11.16)
84 andy o’donnell: fast fourier transform
Xsum(m) =N−1
∑n=0
(x1(n) + x2(n))ej2πnm/N . (11.17)
=N−1
∑n=0
(x1(n))ej2πnm/N +N−1
∑n=0
(x2(n))ej2πnm/N . (11.18)
[46]
12T I M M O RT S O L F : T H E P H Y S I C S O F D ATA S T O R A G E
12.1 introduction
In this chapter we are going to apply some physical principles tomodern data storage technology. Digital technology exploded in
the last quarter of the twentieth century with the invention of themicroprocessor. However, without the data storage devices that weredeveloped in parallel, the computer revolution would not have beenpervasive to all areas of society. What good what a computer be to thebanking industry if they could not store your account information?More importantly, what good would your desktop computer be if itwas unable to have a hard disk drive to store your operating system ordesktop applications. The first several sections of this chapter introducesome basic concepts of digital storage, such as binary number systems,that are required to understand the remaining parts of the chapter. Areader experienced with computer science concepts may wish to skipover these sections and start with section 12.4. Hard disk drives arethe most common storage device used by modern computers and thenext section is devoted to explaining the physical principle behindthese devices. In particular, we explain some of the physics behindgiant magnetoresistance, a technology which has enabled disk drivesto exceed storage capacities of up to 1 Terabyte.
12.2 bits and bytes — the units of digital storage
Before we can begin to discuss how data is physically stored, we needto understand the numerical units that are used by computers anddata storage devices. People count using a base-10 number system.This means that each digit of our number system takes on one of tendifferent values, which are the digits 0 through 9. Most scientists believethat we use a base-10 number system because we have 10 fingers andcounting is something that we first learn to do with these 10 fingers.
Digital computers use a base-2 number system. The digits used by acomputer are called binary numbers and these digits can take on onlytwo values — either 0 or 1. The reason that computers use a base-2number system instead of a base-10 number system is because it ismuch easier to design physical devices that have only two states, onand off, as opposed to a device that has ten different states. A singletwo-bit number is called a "bit". Both digital computers and digital datastorage devices operate at the physical level on these binary bits of data.Larger numbers are represented by stringing several bits together. Table12.1 shows how binary numbers with three bits are used to representeight different numeric values.
When N binary bits are combined to form a number, the number ofunique numbers M that they can represent is determined by
M = 2N . (12.1)
85
86 tim mortsolf: the physics of data storage
Table 12.1: Numeric values Represented By 3-Bit Binary Numbers
Binary/Base-2 Number Number
000 0
001 1
010 1
011 3
100 4
101 5
110 6
111 7
If we invert this equation by taking the base-2 logarithm of both sides,then we get a formula to determine the minimum number of binarydigits required to represent a M different values
N = log2 M. (12.2)
A "byte" is a a number formed by collecting eight binary bits toform a single number that can take on 256 different values. The earliestcommon microprocessors operated on 8-bit quantities and had eightseparate wires to signals of these bits between the microprocessor andthe memory. Today’s microprocessors operate on 64-bit numbers butwe still use bytes as the most unit to express the size of a digital storagedevice. Prefixes are used to represent large numbers of bytes like theamount used in data storage systems. Commonly used prefixes are: akilobyte for 2
10 or 1,024 bytes, a megabyte for 220 or 1,048,576 bytes,
and a gigabyte for 230 or 1,073,741,824 bytes.
example 12 .1 : number of bits on a cd-rom
A standard 120 mm CD-ROM contains 703.1 megabytes of data. How manybinary bits of data does a CD hold?
Solution A megabyte is 1,048,576 bytes and each byte consists of 8
bits so we can compute the number of bytes by simply multiplying thesenumbers together,
1 CD = 703.1 megabytes× 1, 048, 576 bytes1 megabyte
× 8 bits1 byte
= 5, 898, 030, 285 bits.
example 12 .2 : using binary digits to store dna
DNA sequences are represented by biochemists with a string of alphabeticletters that represent the primary sequence of a DNA strand. For example,the string AGCTCGAT is a DNA sequence made of eight DNA bases. Foralmost all situations there are only four DNA bases in a DNA sequencethat we represent by 4 letters: A for adenine, G for guanine, C for cytosine,and T for thymine. How many bits of data does it take to store the valueof a DNA base? If we are instead required to use bytes to store thesevalues, what is the most number of DNA bases that we could store into a byte?
Solution Since there are only four different values of the DNA base,
12.3 storage capacity is everything 87
we can simply use formula 12.2 to solve for the number of bits that arerequired
number bits = log2(4) = 2.
There are several different schemes that we could use encode these bits. Hereis one that encodes the bases in alphabetical order of the symbols.
Binary/Base-2 Number DNA base
00 A (adenine)
01 C (cytosine)
10 G (guanine)
11 T (thymine)
A byte has eight bits of information. Since each DNA base requires2 bits for encoding, then a byte can store the value of 8/2 = 4 DNA bases ina sequence. To represent sequences longer than four DNA bases we wouldsimply use additional bytes.
12.3 storage capacity is everything
When customers are purchasing new computer hardware, the mostimportant considerations are performance and price. The manufacturersof computer components are under competitive pressure to regularlycome out with new devices with better performance metrics. When itcomes to data storage devices, and hard disk drives in particular, themetrics that matter are: storage capacity, transfer rate, and access time;of these, storage capacity is by far the most important to the averageconsumer.
Ever since the digital computer has arrived, there have been twoimportant trends with storage capacity. The storage capacity and thetransfer rates have simultaneously increased. This is because the arealdensity of the hard disk drive surfaces have increased ever since theywere invented. Areal density, or bit density, is the amount of bits thatcan be stored in a certain amount of "real estate" of the hard driveplatter’s surface. The units for areal density are BPSI (bits per squareinch). Although we are probably far away from the peak of this trend[25], it is a trend that in the long term appears to oppose the quantummechanical principles of uncertainty. When we consider the future ofdata storage, one limit that that scientists envision is the ability to storea bit of information using a single atom. It might be possible or evenpractical to store more than one bit of information in an atom, but forthe sake of this discussion let’s assume that the one bit per atom limitwill one day be achieved. Quantum computers have the ability to storeone bit of information per degree of freedom and provide a reasonablehope that this is limit can be obtained. As the size of matter used tostore a single bit of information shrinks down to the size of an atom, theenergy of the matter becomes very small. The Heisenberg uncertaintyprinciple relates the spread in energy ∆E of a quantum state to theamount of time ∆t it takes for the spread to evolve to an orthogonalstate with the relation
∆E∆t ≥ h.
88 tim mortsolf: the physics of data storage
Figure 12.1: Transfer rates of magnetic disk drives and microscopic materials.The transfer rates of single bits of information on a magnetic drive has continuedto improve as the density as increased, but we may be approaching a limita trade off is required. The datapoint for the silicon atom shows the transferrate of gold atoms bonded to a silicon surface that were read with an scanningtunneling microscope (STM). The data rate depicted for DNA is presumablyhow fast the DNA is able to be transcribed into an RNA molecule. The rate ofRNA transcription is typically greater than 50 RNA bases per second. (Courtesyof [42])
This has been extended to relate the uncertainty in time to a particlewith average energy E as which we can rearrange to determine theuncertainty in time as [56]
∆t =πh
2∆E.
This physical principle implies that in the future, when the densitiesused to store bits of information approach the quantum mechanicallimits, there will be a trade off between storage density and transferrates. In fact, this is already a limit that we see in microscopic systemssuch as gold atoms on a silicon surface and the decoding of DNA.DNA is very dense, much denser than any man-made storage devicesby several orders of magnitude. DNA uses approximately 32 atomsfor each nucleotide of a DNA sequence; these are the letters A, G, C,and T that one learns about in a introductory biology class. Figure12.1 shows the trend of how storage density and transfer rates haveincreased throughout the history of HDD development but drop offtremendously for atomic systems.
12.4 the physics of a hard disk drive
Hard disk drives (HDDs) are the major data storage devices used bytoday’s desktop computer for permanent storage of data. AlthoughHDDs were introduced into the market place in the 1956 by IBM, the
12.4 the physics of a hard disk drive 89
Figure 12.2: Diagram of the major internal view of a hard disk drive. (Courtesyof [63])
earlier desktop PCs used removable floppy disk drives (FDDs) becauseof the high cost and relatively low data storage of HDDs availableat the time. During the early 1990s, HDDs quickly supplanted FDDsin desktop PCs as technology permitted HDDs to be made at lowercost and with higher storage capacity. In this section we will explainhow hard disk drives work and the physics of giant magnetoresistance(GMR), which is a newer technology that has enabled modern HDDsto have data storage capacities of 1 Terabyte.
12.4.1 Hard Disk Drive Components
Figure 12.2 shows a basic structure of the internal components of aHDD. We are going to focus on the three main components that dealwith how the binary data is magnetically stored inside the hard drive:the platter, the read/write heads, and the magnetic surface. This sectionserves to introduce the reader to how modern disk drives work andcannot begin to fully describe the inter workings and technologies thatthese devices encompass. The references at the end of this chaptercontain suggested books for an interested reader who wishes to learnmore details about hard disk drive technology.
Hard Disk Platters
An HDD is a sealed unit that that uses stores digital bits of data on harddisk platters that are coated with a magnetic surface. These plattersof composed of two types of material — a substrate that forms themechanical structure of the platter, and a magnetic surface that coatsthe substrate and stores the encoded data. The HDDs used in moderndisk drives are composed of ceramic and glass and use both surfaces ofthe platter to store data. Modern HDDs contain more than one platterfor two reasons: first this increases the data capacity of the drive, andsecond this increases data transfer rate since each read/write head
90 tim mortsolf: the physics of data storage
can transfer data simultaneously. The HDD platters are attached to aspindle that rotates very rapidly and at a constant speed. The rotationalvelocity of HDDs is specified in revolutions per minute (rpm) and is animportant parameter for the consumer. As you can guess, the higherthe rpm, the higher the data transfer rate and the lower the data accesstimes. Premium desktop HDDs that are available in November of 2008
rotate at 10,000 rpm and have data transfer rates of about 1 Gbit/s.The information on a hard disk platter is organized at the highest
level in a "track" . A track contains a ring of data on a concentric circleof the platter surface. Each platter contains thousands of tracks andeach track stores thousands of bytes of data. Each track is divided intoa smaller segment called a sector . A sector holds 512 bytes of data plussome additional bytes that that contain error correction codes used bythe drive to verify that the data in the sector is accurate.
Read/write heads
HDD read/write heads read data from and write data onto a hard diskplatter. Each platter surface has its own read/write head. A disk drivewith five platters has ten heads since there are each side of a platter hasa magnetic surface. The heads are attached to a single actuator shaftthat positions the heads over a specific location of the platter. The drivecontroller is the electronic circuitry that controls the rotation of thedrive and the position of the heads. The computer interfaces with thedrive controller to read and write data to the drive. The computer refersto a specific location on a hard drive by using a head number, cylindernumber, and sector number. This drive controller uses the head numberto determine which platter surface the data is on, the cylinder numberto identify what track it is on, and the sector number to identify aspecific 512 byte section of data within the track. To access the data, thedrive controller instructs the actuator to move to the requested cylinder.The actuator moves all of the heads in unison to the cylinder, placingeach head directly over the track identified by the cylinder number.The drive controller then requests the head to read or write data as thesector spins past the read/write head.
When the hard disk drive is turned off, the heads rest on the plattersurface. When the drive is powered up and begins to spin, the airpressure from the spinning platters lift the heads slightly to create avery tiny gap between the heads and the platter surface. A hard drivecrash occurs when the head hits the surface while the platter is spinningand scratches the magnetic media.
Magnetic surface
The magnetic surface, or media layer, of a hard disk platter is onlya few millionths of an inch thick. The earlier hard drive models usediron oxide media on the platter surfaces because of its low cost, easefor manufacturing, and ability to maintain a strong magnetic field.However, iron oxide materials are no longer used in today’s HDDsbecause of its low density storage capacity. Increases in storage densityrequire smaller magnetic fields so that the heads only pick up themagnetic signal of the surface directly beneath them.
Modern disk drives have a thin-film media surface. First a basesurface called the underlayer is placed onto the platter using hardmetal alloys such as NiCr. The magnetic media layer is deposited on
12.4 the physics of a hard disk drive 91
top of the underlayer. This surface is formed by depositing a cobaltalloy magnetic material using a continuous vacuum deposition processcalled sputtering [68]. The magnetic media layer is finally covered witha carbon layer that protects the magnetic from dust and from headscratches.
12.4.2 Hard Disk Magnetic Decoding and Recording
The read/write heads in modern HDDs contain separate heads forthe reading and writing functions. Although the technology for bothheads has improved, most of the increases in data density have comefrom technology improvements of the read head. Over the last decade,the technology for read write heads has evolved through several re-lated by different physical processes. The original heads used in HDDswere ferrite heads that consisted of an iron core wrapped with wind-ings, similar to that used to form a solenoid. These heads were ableto perform both a reading and writing function, but the density anddata rates were terrible by today’s standards. One of the technologiesthat was applied in the 1990s to led to the widespread acceptance ofHDD technology was the use of the magnetoresistance effect. AMR(anisotropic magnetoresistance) read heads utilize the magnetoresis-tance effect to read data from a platter. The next breakthrough was theincorporation of GMR (giant magnetoresistance) read heads that utilizethe giant magnetoresistance effect. We will cover the physical processesof magnetoresistance and GMR technology in the next section; this isan effect that leads to much higher areal densities than AMR and is ledto the explosion of HDD capacities after the turn of the century.
Most HDDs in use today orient the magnetic fields parallel to themagnetic surface along the direction of the track. This magnetizationscheme is referred to as LMR (longitudinal magnetic recording) . Re-cently, the growth rate of HDD capacities have slowed because oflimitations in how dense the thin-film media can reliably store mag-netic bits of information. The fine structure of the magnetic cobalt-alloysin the magnetic media consists of randomly shaped grains that come ina variety of different sizes. Each bit that is written onto the surface mustbe stored in nearly 100 grains in order for the information to be reliablystored [95]. One problem that arises as HDDs encode magnetic infor-mation at higher densities is that thermal energy can excite these grainsand reverse the magnetization of regions on the surface. The amount ofthermal energy required to reverse the magnetization is proportionalto the number of grains used to store the magnetic information. PMR(perpendicular magnetic recording) is a new technology that is beinguse to further push the density envelope on information stored on themagnetic surface. PMR, as its name suggests, encodes the magneticbits in up and down orientations that are perpendicular to the mediasurface. These bits are more resilient to thermal fluctuations, but re-quire a stronger magnetic field to write the information into the grains.Bits encoded with PMR also produce a sharper magnetic response, sonot only can they be more densely encoded into the surface, but theyare also easier for the disk read head to interpret. Figure 12.3 showsa picture of LMR and PMR magnetically recorded bits on a mediasurface. Notice how the waveform produced at the read head is muchsharper for PRM encoding, even though the regions used to store the
92 tim mortsolf: the physics of data storage
Figure 12.3: Depiction of the magnetic bit orientation and read head signalsof media surfaces encoding using LMR (longitudinal magnetic recording) andperpendicular magnetic recording (PMR) technology (Courtesy of [95])
information on the surface are smaller. In the future, all HDDs willprobably use PMR until a newer technology comes along to replace it.
12.4.3 Giant Magnetoresistance
Magnetoresistance is a physical process that causes the resistance of amaterial to change when it is in the presence of a magnetic field. Let’sexplain how a HDD uses the physical process of magnetoresistance todetect the magnetic signals on a HDD. The drive controller applies avoltage to the read heads and detects the current that passes throughthem; in this way the drive controller acts as an ammeter. As themagnetic surface spins beneath a read head, it enters the magnetic fieldinduced by the cobalt alloy magnetic grains on the thin-film mediasurface that is directly beneath the read head. The precise electricalresistance of the read head depends on the magnitude and directionof these magnetic grains. Because the HDD platter is spinning, themagnetic field and hence its electrical resistance changes in directresponse to the magnetic field on the surface. The drive controllerdetects these changes in resistance by measuring the current that passesthrough the read head. Thus, the drive controller is able to interpret themagnetic field information that is encoded on the surface of the disk.The drive controller has very accurate knowledge of which portion ofthe disk is under the read head at any given time and is thus able todetermine the bit values of all the magnetic bits stored in a 512 bytecluster.
12.4 the physics of a hard disk drive 93
As we saw in the last section, the density of HDD information isintrinsically related to how small of an area the HDD is able to reliablystore the magnetic information so that it cannot be modified by thermalenergy. This is one important area where HDD density is increased, butit is also not all that is required. The HDD read heads must also be ableto reliably read back the information that is encoded ever more denselyonto the surface. Just before the turn of the century, one major barrier toimproved HDD storage capacity was that HDD read heads were unableto keep pace at reliably reading magnetic bits at the increased storagedensities that were required. As the density of magnetic informationstored on the disk shrinks, the magnetic fields become smaller and theresistance changes induced by magnetoresistance become weaker. At asmall enough limit, the signal changes become weak enough that thesignal to noise ratio is too large to reliably decode the information. Thebreakthrough that addressed this problem was the incorporation ofGMR (giant magnetoresistance) technology into the HDD read heads.The physics of magnetoresistance has been understood for over a cen-tury so it is not surprising that AMR heads were quickly replaced withtechnologies that work at much high areal densities.
The GMR effect was independently discovered in 1988 by Albert Fertand Peter Grünberg for which they shared a Nobel prize in physicsin 2007 [64]. GMR is a technology that can be used to create largerchanges in electrical resistance with weaker magnetic fields. The term"giant" comes not from the size of the magnets, but from the size of theeffect that is produced. In the explanation of GMR that follows, the thinmagnetic layers in which the GMR effect occurs is are those used in theread/write heads, not the magnetic materials on the thin-media layer.GMR occurs on thin layers of magnetic material, such as cobalt alloys,that are separated by a nonmagnetic spacer that is just a few nanometersthick. This configuration of magnetic materials produces a tremendousreduction of electrical resistance when a magnetic field is applied. Thethin magnetic layers that are used in the read write heads do not havea permanent magnetic dipole associated with them and are quite easilyable to orient themselves in response to an applied magnetic field. Inthe absence of an external magnetic field, the magnetizations of the thinmagnetic layers orient themselves in opposite or antiparallel directions.In this orientation, the electrical resistance is at it highest. When anexternal magnetic field is applied, the thin magnetic become alignedand is accompanied by a tremendous drop in electrical resistance. Agraph of this effect is shown in Figure 12.4.
GMR heads are designed so that one layer is always in a fixedorientation and the other layer is free to reorient itself. A fourth layerthat is a strong antiferromagnet is used to "pin" down its orientation.As the magnetic bits pass under the GMR head, the magnetic fieldfrom the grains on the thin-media layer directly beneath the GMR headreorient the "unpinned" layer which to produce the strong changesin electrical resistance [68]. Since the "pinned" layer is always force inone orientation, grains that are oriented in the same direction producelittle change in resistance but grains that are oriented in the oppositedirection will produce large changes in resistance. Thus, a reliablesignal can be encoded onto the thin-media layer at smaller densitieswith weaker magnetic fields.
94 tim mortsolf: the physics of data storage
Figure 12.4: Graph of the GMR effect of the resistance on thin magnetic alloylayers separated by a thin nonmagnetic spacer. (Courtesy of [85])
12.5 summary
Digital computers and data storage systems store numbers using binarybits because it is practical to build devices that have two states, on andoff. A byte made by combining the values of eight bits into a singlequantity and can represent 28 = 256 different values. Data storage ca-pacity and transfer rates have been increasing since the beginning of thedigital era, but as we scale the devices down to the atomic level, therewill be a trade off between these two metrics. This limit is explainedby the Heisenberg relation which relates the amount of time it takes toforce a physical system into a state to the energy of the state the systemis forced into, ∆E∆t ≥ h. Hard disk drives (HDDs) are the major datastorage devices in use today. They store information by magneticallyencoded physical bits of information onto a hard disk platter that iscoated with a thin-layer media surface composed of magnetic alloys.HDDs read information from the magnetic surface from a change inelectrical resistance that occurs from magnetoresistance. Giant magne-toresistance (GMR) is a recent technology that shows greater responseto changes in magnetic fields by designing the read heads to have thinlayers of magnetic alloys separated by a thin nonmagnetic spacer.
12.6 exercises
1. The human genome contains 3 billion nucleotide base sequences.If we stored each nucleotide in 1 byte, could we store the humangenome on a single 703.1 Megabyte CD?
Solution Using one byte per nucleotide base sequence wouldrequire 3 × 109 bytes of storage. A 703.1 Megabyte CD holds.703× 109 bytes. A CD would not be able to store the humangenome unless a more efficient storage scheme was used. In fact,
12.7 multiple choice questions 95
the human genome has many repeated nucleotide sequences andcan be compressed to fit onto a single CD.
12.7 multiple choice questions
1. Which physical form of magnetic storage has the highest arealdensity?(a) AMR stored with longitudinal encoding(b) AMR stored with perpendicular encoding(c) GMR stored with longitudinal encoding(d) GMR stored with perpendicular encodingSolution (d)
2. The first generation of hard disk drives that were produced, reliedon the magnetoresistance effect to decode the magnetically storedbits of data?(a) True(b) FalseSolution (a)
3. A hard disk drive uses a single read/write head for:(a) All of the platter surfaces in the drive(b) Each platter in the drive(c) Each platter surface in the driveSolution (c)
4. The GMR effect that has been used to increase hard drive densityis primarily used to:(a) Decode bits of information on the magnetic thin-layer media(b) Encode bits of information on the magnetic thin-layer media(c) Decode and encode bits of information on the magnetic thin-layer mediaSolution (a)
13T I M M O RT S O L F : T H E P H Y S I C S O F I N F O R M AT I O NT H E O RY
13.1 introduction
In this chapter we are going to apply information theory to thedata storage concepts presented in the last chapter. The theory of
information is a modern science that is used for the numerical compu-tation of data encoding, compression, and transmission properties. Thechapter is an introduction to information theory and the physical prin-ciples behind "information entropy", a concept which in data storage isanalogous to the role of entropy in thermodynamics.
13.2 information theory – the physical limits of data
We have seen that the technological improvements in data storagedevices over the last 50 years follow a pattern of increased data densitythat roughly doubles every 18 months. At the time this textbook waswritten (2008), researchers announced a new hard drive density recordof 803 Gb/in2 on a hard drive platter surface. This record was achievedusing TMR (Tunneling Magneto-Resistance) read/write heads. Therate law of data storage as described by Moore’s law, is not a law thatcomes from physical principles, but rather a law has been accurate inestimating the engineering advances of computational technology thathave occurred over the past half century.
There are some laws of data storage and computation that can bederived from physical principles. "Information theory" is the branchof science that applies the rules of physics and mathematics to obtainphysical laws that describe how information is quantitized. These lawsare primarily derived from the thermodynamic equations of state for asystem. Entropy is a key player in this arena since entropy is a quantitythat defines the number of available configurations (or states) of aclosed system. In this section, we will introduce some of the lawsof information theory and show how they can be applied to datastorage devices. We will also use these laws to show that there areindeed physical limits to computation that determine the maximuminformation density and computational speed at which theoretical datastorage devices can operate.
13.2.1 Information Theory and the Scientific Definition of Information
Information theory can be informally defined as the mathematical for-mulation of the methods that we use to measure, encode, and transforminformation. "Information" is an abstract quantity that is hard to pre-cisely define. In this chapter we will use a narrow technical term forinformation to mean the encoding of a message into binary bits thatcan be stored on a data storage device. One of the earliest applicationsof information theory occurred when the Samuel Morse designed anefficient Morse code that was used to encode messages for transmissionover a telegraph line. His code used a restricted alphabet of only "dash
97
98 tim mortsolf: the physics of information theory
(long)" and "dot (short)" electrical signals to transmit text messagesover long distances. For example, the letter C is encoded by the Morsesequence "dash, dot, dash, dot". The principles of information theorycan be used to show that the Morse code is not an optimal encodingusing an alphabet of just two characters and the encoding can be furtherimproved by a factor of 15% [3].
The scientific treatment of information began with Hartley’s "Trans-mission of Information" in 1928 [40]. In this paper, he introduced atechnical definition for information .
The answer to a question that can assume the two values ’yes’ or ’no’(without taking into account the meaning of the question) contains one unitof information.
Claude Shannon’s publication "A Mathematical Theory of Communi-cation" in 1948 formed a a solid foundation for the scientific basis forinformation theory [72]. Shannon introduced the concept of "informa-tion entropy", also called the Shannon entropy , that minimum messagelength in bits that must be used to encode the true value of a randomvariable. Today, information theory exists as a branch of mathematicsthat is has its main applications in encryption, digital signal processing,and even biological sciences.
13.2.2 Information Entropy and Randomness
The best way to understand the basics of information entropy is tolook at an example. For our experiment, we want to flip a coin a largenumber of times (let’s say 1,000,000) and encode the results of thisexperiment into a message that we can store on a digital storage device.If the coin used in our experiment is not biased then the two outcomes"heads" and "tails" should occur with equal probability of 1/2. We willencode each coin toss with a bit — let 1 indicate "heads" and 0 indicate"tails". Each coin toss requires exactly one bit of information so for1,000,000 coin tosses, therefore we would need to store 1,000,000 bits ofdata to record the results of our experiment.
Let’s repeat this experiment, but instead let’s assume the coin is nowbiased — let "heads" occur 3/4 of the time and "tails" occur only 1/4
of the time. It might surprise you that the results of this experimentcan be recorded on average with less than 1,000,000 bits of information.Since the distribution of coin flips is biased, we can design an efficientcoding scheme that takes advantage of the biased nature to encode theresults of each coin flip in less than one bit. It turns out that we areable to record each coin flip using only 0.8113 bits of information andcan thus use a total of 813,000 bits of data to record our results. Thissame process explains why we are able to compress text documentsand computer images into a data file that is much smaller than the rawdata that we recover when we decompress the information.
A better example for explaining how we can encode the results ofa coin flip in less than one bit occurs when we let the coin becomeeven more biased — let "heads" occur 99/100 of the time and "tails"occur only 1/100 of the time. If we encoded each coin flip using asingle bit as before, we would still need exactly 1,000,000 bits to recordthe results of our experiment. But for this experiment we can designa variable length encoding algorithm that does much better than this.
13.3 shannon’s formula 99
Since almost all of our coin flips are "heads", the sequence of coin flipswill be a long series of "heads" with an occasional "tail" interspersed tobreak the series. Our encoding scheme takes advantage of the biasednature by recording only those places in the sequence where a "tail" fliphas occurred. Since there are 1,000,000 coin flips, each "tail" requires a20-bit number to record its position in the sequence. Out of 1,000,000
coin flips, the average number of tails will be 1, 000, 000/100 = 10, 000.On average it will take "10, 000× 20 = 200, 000 bits to record the resultsof each experiment. Thus, by using a very simple encoding scheme wehave been able to reduce the data storage of our experiment by a factorof 1/5.
For our initial recording scheme that recorded the values of each coinflip as a single bit, the amount of data required to store the messagewas always 1,000,000 bits, no more and no less. With our new encodingscheme we can only discuss the statistical distribution of what theamount of data required for each experiment will be. If we look at thealmost impossibly unlucky result where each of 1,000,000 coin flipsis "tails" that occurs for (100)1,000,000 of our experiments, the encodingscheme would use "1, 000, 000× 20 = 20, 000, 000 bits of data. This is100 times larger than our average value of 200,000 bits that we expectto use for each trial. These improbable unlucky scenarios are balancedby scenarios where we get lucky and have much fewer coin flips with"tails" than the average result and can encode our experiment in muchless than 200,000 bits of data. The important thing to note is that almostall of experiments will require nearly 200,000 bits of data storage, butthe exact value will be different for each result.
We can design an even better encoding scheme than this one. Insteadof recording the absolute sequence number of each coin flip with a"tails" result, we could just record the difference between the absolutesequence number of each coin flip of "tails". For example, if a coin flipsof "tails" occurred on the coin flip values
99, 203, 298, 302, 390, (13.1)
we could encode these results as the differences in the positions,
99 - 0 = 99, 203 - 99 = 104, 298 - 203 = 95, 302 - 298 = 96, 390 - 302 = 88.
(13.2)
For this encoding scheme, we use 8-bit to encode differences less than256, while our original compression scheme used 20-bit numbers forthose cases where the "tails" result to save its absolute position in thesequence. With our latest encoding scheme, on average it will take"10, 000× 8 = 80, 000 bits to record the results of each experiment. Butcan we do better than this? As you have probably guessed, we canindeed. The limit to how well we can do can be calculated by Shannon’sgeneral formula for information uncertainty that we will develop in thenext section.
13.3 shannon’s formula
We begin our development of Shannon’s formula by introducing theprinciple of information content, also known as information uncertainty.We want to create a definition of information content that representsthe minimum number of bits required to encode a message with M
100 tim mortsolf: the physics of information theory
possible values. Let’s go back to the first experiment of 1,000,000 coinflips of an unbiased coin. Before we start this experiment, we have noidea what the result is going to be. Since each coin flip is can be one oftwo results, namely "heads" or "tails", then there are M = 2n = 21,000,000
possible outcomes for this experiment and each of these outcomes isequally likely since our coin flips are unbiased. It takes n = 1, 000, 000bits of information to record each of these possible outcomes. We definethe uncertainty of the information content I contained in a messagewith M possible different messages as
I = log2(M) (13.3)
We also require that our definition of information content be additive.For example, if we perform an experiment using 500,000 coins followedby a second experiment again of 500,000 coins, then the sum of theinformation content for each experiment should be the same as theinformation content of the experiment for 1,000,000 coins. We cansee that our definition of information content does indeed satisfy ourrequirement. The sum of the uncertainties of two different experimentsperformed n1 and n2 times
I1 = log2 2n1 = n1 I2 = log2 2n2 = n2, (13.4)
equals the uncertainty of an experiment performed n1 + n2 times
I = log2 2n1+n2 = log2 2n12n1 = log2 2n1 + log2 2n2 = I1 + I2. (13.5)
The quantity I is unitless. The unit value of I represents the numberof bits required to encode a message with 2 possible outcomes,
I(1) = log2 2 = 1. (13.6)
We now need to apply the definition for information content tosituations where the values encoded by the message are not equallylikely, such as is the case for our biased coin. Let Pi = 1/Mi be theprobability of getting any message Mi. We define the surprisal ui as the"surprise" that we get when we encounter the ith type of symbol,
ui = − log2 (Pi). (13.7)
If the symbol Mi occurs rarely, then Pi is almost zero and ui becomesvery large, indicating that we would be quite surprised to see Mi.However, if the symbol Mi occurs almost all the time, then Pi is almostone and ui becomes very small, and we would not be surprised to seeMi at all.
Shannon’s definition of uncertainty is the average surprisal of thevalues contained in a message of infinite size. This is a limit that afinite message of length N converges to when the size of the messagebecomes infinite. First let’s determine the average surprise Hn for amessage of length N is encoded by an set of M different symbols, witheach symbol Mi appearing Ni times,
Hn =N1u1 + N2u2 + N3u3 + ... + N4uN
N=
M
∑i
NiN
ui. (13.8)
If we make this calculation in the limit of messages with an infinitenumber of symbols then the frequency Ni/N of symbol Mi convergesto its probability Pi
Pi = limN→∞NiN
, (13.9)
13.4 the physical limits of data storage 101
which we can substitute into (13.8) to get,
H =M
∑i
Piui. (13.10)
We finally arrive at Shannon’s formula for information uncertainty bysubstituting (13.7) into (13.10),
H = −M
∑i
Pi log2 Pi Shannon’s formula (bits per symbol). (13.11)
example 13 .1 : using shannon’s formula
In this section, we indicated that trials from a coin biased with 3/4 "heads"to 1/4 "tails" can be recorded using 0.8113 bits of information. Let’s useShannon’s formula to prove this.
Solution Use Shannon’s formula from 13.11, setting P1 = 3/4 andP2 = 1/4
H = −M
∑i
Pi log2 Pi
= −(3/4× log2 3/4)− (1/4× log2 1/4)
= −(3/4×log10 3/4
log10 2)− (1/4×
log10 1/4log10 2
)
= 0.8113
13.4 the physical limits of data storage
In the closing section to our chapter, we combine Shannon’s law thatwe used to compute the information entropy with the laws of thermo-dynamics. These will be used in a non-rigorous fashion to determinethe physical limits of data storage density and data operations of adevice called the "perfect disk drive". The "perfect disk drive" or PDDis not a real device that exists today, but one we use to determine whatthe ultimate boundaries of data storage capacity and data storage rates.The PDD stores information at the lowest density of matter permittedby physics. Compared to the PDD, today’s computers are extremelyinefficient. Most of the atoms of a highly dense hard drive that weuse today does not store any information, but are instead required tomake the disk drive function. Even if we peel back the layers of a singlesurface of a hard disk drive platter, there are millions (if not more)of atoms required to store the state of just a single bit of information.Let’s consider this problem from a statistical mechanics point of view.A disk drive has an enormous amount of available energy states thatcome from the number of particles that the components of the diskdrive are constructed from. If we try to use quantum mechanics todetermine the number of available energy states for these particles thenwe are attempting to solve something that is impossible with today’sscientific methods. Fortunately, statistical mechanics provides us witha method to approximate the number of available states of a systemfrom the thermodynamic equations of state. These approximations arevery accurate as temperatures increase and the quantum distribution of
102 tim mortsolf: the physics of information theory
energies become nearly continuous and can be treated classically. Theequation
S = kB ln W (13.12)
relates the thermodynamic state variable for entropy S to the quantityW for the number of accessible states of a closed system, where kB isthe value of Boltzmann’s constant. The highest efficiency that our PDDcould achieve for data density is such it is able to precisely encode itsinformation into the number of accessible thermodynamic states. Thatis, we assume the PDD can encode a single bit of information using justone of these accessible states. We are not claiming this is something wewill actually be able to achieve, but we are claiming this is a physicallimit that the device can’t exceed. The number of accessible states W ofour closed system then able to hold exactly W bits of information. In theprevious section we defined the information content as the minimumnumber of bits required to encode a message with M possible values.The physical relation that defines our PDD is
M = W, (13.13)
which means it can encode a message with M possible values usingexactly the number of states W that are accessible to the system. Nowwe can substitute this into (13.12) and use (13.3) to get
S = kB ln M = kB ln 2I = kB I ln 2, (13.14)
which we rearrange to solve for I
I =S
kB ln 2(13.15)
This formula (13.15) relates the information content we can store in thedevice to the thermodynamic entropy of the physical system. Unfor-tunately, there is not a simple analytical formula that we can use tocompute the entropy of an arbitrary device from knowledge of physi-cal properties such as mass, volume, temperature, and pressure. Thecomputation of entropy requires complete knowledge of the quantumenergy levels available to each particle in the device. One author useda method to approximate a lower bound for the maximum amount ofentropy in a 1 kg device that occupies a volume of 1 L and assumesthat most of the energy of the system arises from blackbody photonradiation [52]. Using these simplifications, he arrives at a value ofT = 5.872× 108 K for the temperature at which the maximum entropyoccurs, and S = 2.042× 108 J K−1 for the total entropy. Substitutingthese values into our equation for the information content,
I =2.042× 108 J K−1
1.381× 10−23 J K−1 ln 2= 2.132× 1031 bits (13.16)
How does this compare to today’s hard disk drive technology? Atthe time of this book (November 2008), disk drives with a little over1.5 Terabyte = 1.2× 1013 bits of data can be purchased for desktopcomputer systems.
13.5 summary 103
example 13 .2 : applying moore’s law
Use Moore’s law to see how far away we are from this theoretical limit
Solution Current technology permits us to mass produce desktophard drives with 1.2× 1013 bits of data. We need to find out how long it willtake to achieve 2.132× 1031 bits of data in roughly the same size device.
increased storage ratio =2.132× 1031 bits
1.2× 1013 bits= 1.78× 1018
To find out how many times we have to double the storage capacity toincrease our storage capacity by this much, we take the base 2 logarithm of it.
number times to double = log2 1.78× 1018 =log10 1.78× 1018
log10 2= 60.6 times
Now using Moore’s law of 18 months per doubling of storage capacity, weestimate the amount of time required to be
time required = 60.6 times to double× 18 months1 time to double
× 1 year12 months
time required ≈ 91 years
Thus it will take about 90 years to develop the technology to store data at ourarrived value of the maximum theoretical limit for data storage. Althoughthis calculation does have several simplifying assumptions, it does show thatthere is a lot of room left to increase the data density relative to a device thatprecisely encodes information content into the entropic states of a physicalsystem.
13.5 summary
Information theory is a mathematical formulation of the methods thatwe use to measure, encode, and transform information. Informationentropy is a quantity that measures the randomness of data sequences.Shannon’s formula for information uncertainty is used to calculate thisvalue. The lower the entropy, the less the randomness, and the greaterwe can compress the data. The principles of thermodynamics andinformation entropy can be combined to estimate the maximum amountof information that can be stored at the atomic level per quantum degreeof freedom.
13.6 exercises
1. Your assistant Bob rolls a six-sided dice and you don’t know theresult. What is the minimum number of yes/no questions thatyou could ask Bob to get the answer? Try to use informationtheory to answer this question.
Solution Before posing a set of questions, let’s first use informa-tion theory to try to encode the result. The dice and have sixresults and using a binary coding scheme, we need three bits toencode the results. There are several different schemes that wecould use encode the dice result. Here is one that encodes themin order of lower to higher values.
104 tim mortsolf: the physics of information theory
Encoding Dice Throw
000 1
001 2
010 3
011 4
100 5
101 6
110 Not used
111 Not used
Our encoding scheme is not perfect since it has two unused values.But to record the results of a single dice throw, we cannot designa code more efficient than this. If the results of many dice throwswere recorded we could be able to record the results in less thanthree bits per throw by using a clever compression scheme. Fromthe encoding scheme, we can design a series of questions thatascertain the value of the bits used to encode our results. The firstquestion tests the value of the first bit — "Is the value of the diceroll a five or a six?". The second question tests the value of thesecond bit "Is the value of the dice roll a 3 or 4?". And the thirdquestion tests the value of the last bit — "Is the value of the diceroll an odd number?".
2. Repeat the same problem but this time Bob rolls two six-sideddice and you only want to determine the value of the sum.
Solution The sum of two six-sided dice rolls will have values from2 (when two 1’s are rolled) through 12 (when two 6’s are rolled).Our encoding scheme needs to encode eleven values which takesfour bits. However, since in this case some results are more likelythan others, let’s see if we can design a more clever encodingscheme.
Binary/Base-2 Number Sum of dice Probability
0000 21
36
0001 32
36
0010 43
36
0011 54
36
0100 65
36
0101 6
0110 76
36
0111 7
1000 7
1001 7
1010 85
36
1011 8
1100 94
36
1101 103
36
1110 112
36
1111 121
36
13.7 multiple choice questions 105
This encoding scheme does not have any unused values. Wealready require four bits to encode all of the possible values of thesum. However, instead of having encoding that are unused, theextra codes are assigned to those values of the dice sums that aremore likely to occur, which in this case are the values 6, 7, 8. Thisencoding requires a maximum of four questions to determine thevalue of the dice sums, but for some values it will be less than this,namely the ones for which we have degenerate encoding values.From the encoding scheme, we can design a series of questionsthat ascertain the value of the bits used to encode our results. Butnow we can now test for a value of a six using just two questions.The first question tests the value of the second bit — "Is the valueof the sum a six or a seven?". For an answer of yes, the secondquestion tests the value of the third bit — "Is the value of thesum a six?". Thus, in just two questions we know if the sum has avalue of six. Similarly, with just three questions we will know ifthe value of the sum was a seven or an eight. All the other valueswould require the full four questions. However, on average ourencoding scheme and associated questions require less than fourquestions to determine the result of each set of dice rolls.
Challenge: Use Shannon’s formula to determine the minimumnumber of bits required to encode the results. Then use ourencoding scheme and compare its efficiency to this theoreticallimit.
13.7 multiple choice questions
1. How many bits of information would be required on average torecord the results of an unbiased coin that is flipped 1,000,000
times(a) 813,000
(b) 500,000
(c) 1,000,000
(d) None of the aboveSolution (c)
2. How many bits of information would be required on average torecord the results of 1,000,000 coin flips of a coin that is biased sothat 3/4 of the flips are "heads" and 1/4 are "tails"?(a) 813,000
(b) 500,000
(c) 1,000,000
(d) None of the aboveSolution (a)
14S A M B O O N E : A N A LY T I C A L I N V E S T I G AT I O N O FT H E O P T I M A L T R A F F I C O R G A N I Z AT I O N O FS O C I A L I N S E C T S
Swarm Intelligence is the property of a system where a collec-tion of unsophisticated beings or agents are functionally coherent
through local interactions with their environment allowing for largescale patterns to occur. This organizing principle is named stigmergy.While the term, stigmergy, was dubbed in the 1950’s by Pierre-PaulGrasse to give a name for the relationship between social insects and thestructures they create, such as ant hills, beehives and termite mounds,this anomaly has actually existed for billions of years. The term, stig-mergy, literally means “driven by the mark [86].” The bodies of allmulti-cellular organisms are stigmergy formations. This phenomenonis apparent throughout nature in multiple species of social insects suchas ants, wasps, termites and others. While stigmergy has been provento be a powerful and immensely useful criterion of self-organized opti-mization, it has only been researched in recent years [66]. The researchthat has been conducted has uncovered simple algorithms for trafficcongestion control and resource usage based on local interactions asopposed to centralized systems of control. These powerful algorithmshave already begun to be applied to artificial intelligence and computerprograming.
14.1 introduction
14.1.1 Stigmergy in Nature
Ant trail formation and general large scale organized travel is made pos-sible by the indirect communication between individual ants throughthe environment. The individual ants deposit pheromones while walk-ing. As an ant travels, it will probabilistically follow the path richest inpheromones [22]. This allows ants to adapt to changes or sudden ob-stacles in their path, and find the new shortest path to their destination.This incredibly powerful social system allows entire colonies of ants tomake adaptive decisions based solely on local information permittingants to transport food far distances back to their nest.
The organizing process, stigmergy, is also used in a similar wayby termites. When termites build mounds they start by one termiteretrieves a grain of soil and sets it in place with a sticky glue-like salivathat contains pheromones. In this case the mark referred to in the literalmeaning of stigmergy is the grain of soil. This pheromone filled salivathen attracts other termites who then glue another grain of soil ontop of the last. As this process continues the attractive signal increasesas the amount of pheromones increase [86]. This process is called,stigmertic building. Stigmertic Building allows the termites to constructpillars and roofed galleries. The simplest form of stigmertic building ispillar construction. As the pile of grains of soil accumulate upwards itnaturally forms a pillar. Once the pillar reaches a height approximatelyequal to that of a termite standing on its hind legs the termites begin
107
108 sam boone : analytical investigation of the optimal traffic organization of social insects
to add soil grains laterally. These lateral additions collectively form ashelf extending from the top of the pillar. Eventually, the shelf of onepillar meets with the shelf of another nearby pillar. This connection ofshelves creates a roof.
Figure 14.1: Simulated Stages of Termite Roofed Gallery Construction [86]
Figure 14.2: Common Example of a Termite Roofed Gallery Construction [86]
14.1.2 Applying Stigmergy in Human Life
The stigmergy concept in general has led to simple algorithms that havebeen applied to the field of combinatorial optimization problems, whichincludes routing in communication networks, shortest path problemsand material flow on factory floors [66]. While these applications canaid efforts to make transportation routes become more efficient otherconsiderations must be made, such as, congestion and bottlenecking.This problem applies to user traffic on the internet as well as motorvehicle traffic.
Ant Colony Optimization(ACO) based algorithms have also begun tobe used in multi-objective optimization problems. In recent years videogame designers have started to use ACO based algorithms to manageArtificial Intelligence in their games. In particular, these algorithmsare used to manage functions like multi-agent patrolling systems. Inorder for this function to run properly, all agents must coordinate theiractions so that they can most efficiently visit the areas of most relevanceas frequently as possible. Similar algorithms have been used in robotics,computer network management and vehicle routing. Their has alsobeen recent break-throughs in using ACO to create artificial 3D terrainobjects for video games, as well as engineering, simulations, trainingenvironments, movies and artistic applications.
ACO algorithms can also be very useful for site layout optimization.This proves to be very valuable, whether applied to the constructionof highways and parking garages to manufacturing plants and ports.Any construction site layout that is concerned with the positioning andtiming of temporary objects, machines and facilities that are used tocarry out the actual construction process, as well as the permanentobjects being created with regards to their ultimate purpose that willbe carried out, will greatly benefit from the solutions of these prob-
14.2 ant colony optimization 109
lems. Ultimately these ACO based algorithms will greatly enhance theefficiency while decreasing the cost of these sites.
While ACO research is still very young and fairly scarce, the numberof possible applications seem to be limitless. Almost any field of science,mathematics, engineering, computer design, finance, economics andelectronics can be greatly improved and expedited with the applicationof ACO.
14.2 ant colony optimization
In 1992, Marco Doringo proposed a metaheuristic approach calledAnt Colony Optimization (ACO). Dorigo was inspired by the foragingbehavior of ants. ACO enables ants to find the shortest paths from foodsources to their nest. The decisions of path of travel are probabilisticallythe path of highest pheromone concentration. The ACO algorithms. thatDoringo proposed, are based on a probabilistic model, the pheromonemodel. This model is used to represent the chemical pheromone trails[22]. The pheromone model uses artificial ants that construct solutions inincrements by adding solution components to a partial solution thatis under consideration. The artificial ants perform randomized walkson a graph G = (C, L), called a construction graph, whose vertices arethe solution components C and the set L are the connections [22]. Thismodel is applied to a particular combinatorial optimization problem byhaving the constraints of that particular problem built into the artificialants’ constructive procedure so that in every incremental step of thesolution construction only feasible solution components can be added.Depending on the problem the use of the pheromone trail parameter Tiis beneficial. The set of all pheromone trail parameters is labeled T.This allows the artificial ants to make probabilistic decisions on whichdirection to move on the graph [10].
14.3 optimization by hand
The difficulty of an optimization problem directly correlates with thecomplexity of the problem itself. The more variables that the solutionrelies on the exponentially more difficult the problem becomes. Todrive home this notion we will attempt a simple 2-variable optimizationproblem whose solution can easily be solved by hand.
example 14 .1 :
Problem 1: Find two positive numbers whose sum is 12 and the product ofone number and the square of another is at a maximum.
Solution: Strictly based on the problems design we are left with the factthat:
x + y = 12. (14.1)
By rearranging for y we are left with,
y = 12− x. (14.2)
Our goal is to maximize the product:
P = xy2, (14.3)
which can be rewritten by substituting our value for y to look like,
P = x(12− x)2. (14.4)
110 sam boone : analytical investigation of the optimal traffic organization of social insects
We then take the first order derivative of both sides,
P′ =dx(12− x)2
dx= −2x(12− x)2 + (12− x)2, (14.5)
which simplifies to become,
P′ = (12− x)(−3)(x− 4). (14.6)
Therefore, P′ equals 0 when x = 4or12. We then calculate the values of ourproduct for these values.
P(x = 0, y = 12) = 0, (14.7)
P(x = 4, y = 8) = 256, (14.8)
P(x = 8, y = 4) = 128, (14.9)
P(x = 12, y = 0) = 0. (14.10)
Finally, we are left with our solution:
x = 4, y = 8, (14.11)
P = 256. (14.12)
While this simple optimization problem can be easily solved byhand, a more complex problem with hundreds of variables would beunbelievably difficult to solve and the time to calculate the solutionwould be tremendous. It is for these complex optimization problemsthat ACO based algorithms are enormously helpful.
14.4 applying aco meta-heuristic to the traveling sales-man problem
One of the famous optimization problems that the Ant Colony Opti-mization Meta-Heuristic has been applied to is the Traveling SalesmanProblem. The Traveling Salesman Problem is the problem to find theshortest closed path in which one could visit all the cities in a certain set.While this algorithm or others of the same nature could be tweaked tohandle this problem with different variables we will concern ourselveswith a Traveling Salesman Problem in which there exists a path (edge)between any pair of cities.
14.4.1 Applicable Ant Behaviors Observed in Nature
When a colony of ants forage for food, the ants will initially leave theirnest and head for the foraging area in which the food that they arelooking for lies. Lets assume that their is a natural path from the nestto the foraging area which twice forks and reconnects. Therefore theirare four possible paths to the food in which the ants can travel. Atfirst, there is no information on these paths that could give the ants
14.4 applying aco meta-heuristic to the traveling salesman problem 111
prior information on which path is the shortest. As a result the antswill randomly choose a path to take to the food. Naturally the antsthat arrive at the food first will be the ones that chose the shortest path.These ants will then take the food that they gather and head back totheir nest backtracking on their path that they took to the foraging area.Since these ants will be the first to the food and the first to head back tothe nest their chosen paths will build up higher levels of pheromonesquicker than the paths chosen by their counterparts. Very quickly, theshortest path will by far possess the highest level of pheromone and asa result the vast majority of the colony will begin to use this path. Theamazing aspect of this natural problem solving exposition displayedby these ants is that the natural properties of the paths or obstacles inquestion intrinsically hold the solution to the problem.
Figure 14.3: Double bridge experiment. (a) The ants start to randomly choosepaths to foraging area. (b) The majority of the ants eventually choose theshortest path (Graph) The distribution of the percentage of ants that chose theshortest path in the double bridge experiment [24]
14.4.2 Approach to the Traveling Salesman Problem
While we will use Marco Dorigo’s artificial ants to solve our optimiza-tion problem hands-on we must recall that all of our artificial ants willbe acting based upon natural ant behaviors that have been witnessedin nature. For this problem we will use three ideas that have beenwitnessed in nature by real ants [23]:
1. Preference to choose paths rich in pheromone levels.
2. Increased rate of growth of pheromone levels on shorter paths.
3. Communication between ants mediated by alterations in environ-ment or the trail.
Our artificial ants will move from city to city on the Traveling SalesmanProblem (TSP) Graph based on a probabilistic function that incorporatesboth heuristic values that depend on the length of edges (length oftrails between cities) and trail accumulated on edges [23]. The artificial
112 sam boone : analytical investigation of the optimal traffic organization of social insects
ants will prefer cities rich in pheromone trail and which are connectedby edges (which in our case is all possible cities). Every time step whenan ant travels to another city they modify the pheromone trail on theedge being used. Dorigo calls this local trail updating. Initially we willplace m artificial ants randomly across our (TSP) Graph at randomlyselected cities. Once all of our ants have completed a tour the ant thatmade the shortest journey will modify the edges used in his tour byadding pheromone amounting to a total that is inversely proportionalto the tour length. This step is dubbed global trail updating.
14.4.3 Comparison Between Artificial Ants and Real Ants
As we stated earlier, the artificial ants that we are using act based uponobserved behaviors of real ants in nature, however these artificial antswere given some capabilities that the real ants do not possess. How-ever, all of these differences are necessary adaptations that must takeplace in order for the artificial ants to be able to discovery solutionsto optimization problems in their new environment (the TSP Graph).The fundamental difference between the artificial ants used in the ACOalgorithm and the biological ants their design is based upon is in theway that they modify the trails they travel. Like the ants observed innature who deposit pheromones on the surface of their environment,the artificial ants modify their surroundings by changing some numeri-cal information that is stored locally in the problem’s state [24]. Thisinformation includes the ant’s current history and performance whichcan be read and altered by ants who later visit that region. This informa-tion is called artificial pheromone trail. This artificial pheromone trail is theonly means of communication between each individual member of thecolony. In most ACO algorithms an evaporation mechanism is installed,much like real pheromone evaporation, that weakens pheromone infor-mation over time allowing the ant colony to gradually forget its pasthistory so that it can modify its direction without being hindered toomuch by its past decisions [24].
There are a few traits that the artificial ants possess that the real antsdo not [24].
1. The artificial ants possess an internal state that contains the mem-ory of its past actions.
2. The artificial ants deposit an amount of pheromone that correlatesto the quality of the solution.
3. The artificial ants timing of when they lay their pheromone doesnot reflect any pattern of actual ants. For example, in our caseof the Traveling Salesman Problem the artificial ants update thepheromone trails after they have found the best solution.
4. The artificial ants in the ACO algorithms can be given extra capa-bilities by design like lookahead, backtracking, local optimizationand others, that make the overall system much more efficient.
14.5 solution to the traveling salesman problem using an
aco algorithm
In the Ant Colony System we have designed an artificial ant k is in a cityr and chooses a city s to travel to. This city s will be chosen among those
14.5 solution to the traveling salesman problem using an aco algorithm 113
that are not apart of that ants working memory Mk. The ant choosesthe next city s using the probabilistic formula:
s = argmaxu/∈Mk[τ(r, u)] · [η(r, u)]β, (14.13)
where q ≤ qo. The probabilistic function, “argmax,” stands for the“argument of the maximum.” In other words, this is the value of uwhere [τ(r, u)] · [η(r, u)]β is at a maximum. Otherwise,
s = S, (14.14)
where τ(r, u) is the amount of pheromone on an edge (r, u), η(r, u) isa heuristic function that was designed to be the inverse of the distancebetween cities r and u, β is the parameter that weighs the importanceof the pheromone level and how close the nearest cities are, q is arandom value with a uniform probability [0, 1], qo is a parameter in(0 ≤ qo ≤ 1), and S is a variable that were randomly selected accordingto a probability distribution which favors edges that are shorter andhave higher build-ups of pheromone [23]. The random variable S ischosen based on the probability distribution:
pk(r, s) =[τ(r, s)] · [η(r, s)]β
∑u/∈Mk[τ(r, s)] · [η(r, s)]β
, (14.15)
if s /∈ Mk. Otherwise,
pk(r, s) = 0, (14.16)
where pk(r, s) is the probability of an ant k choosing to move from acity r to a city s [23].
So once all of the artificial ants have completed their respective toursthe ant with the shortest solution goes back and lays pheromone alongthe edges it used in its solution. The amount of pheromone ∆φ(r, s)that the ant deposits is inversely proportional to the length of the tour.This global trail updating formula is:
∆φ(r, s)← (1− α) · φ(r, s) + α · ∆φ(r, s), (14.17)
where ∆φ(r, s) = 1/shortesttour. This global trail updating ensures thatthe better solutions receive a higher reinforcement.
Local trail updating, on the other hand, ensures that not all of theants choose a very strong edge. This is achieved by applying the localtrail updating formula:
τ(r, s)← (1− α) · τ(r, s) + α · τo, (14.18)
where τo is simply a parameter.The beauty of this reinforcement learning system is that the colonies
behavior depends on two constraints, the pairs of formulas (1.1) and(1.2), as well as, (1.3) and (1.4). The first pair allows an ant to exploitthe colony’s accumulated experience, with probability qo, in the meansof pheromone levels, which has been built up on edges that belong toshort tours. The second pair allows the ant to make decisions basedupon exploration, with probability (1 − qo), that is biased towardsshort and high trail edges. This exploration is one of new cities thatare chosen randomly with a probability distribution that is a functionof the heuristic function, the accumulated pheromone levels and theworking memory Mk.
114 sam boone : analytical investigation of the optimal traffic organization of social insects
1. Homework Problem 1: An open rectangular box with a square baseis to be made from 96 f t2 of material. Find the dimensions thatwill allow the box to have the maximum possible volume.
Solution The surface area of the box will be equal to the sum ofthe area of the base and the four sides. This must all be equal to96 f t2,
A = 96 = x2 + 4xy. (14.19)
By rearranging and isolating y we are left with,
y =96− x2
4x. (14.20)
Simplifying this further,
y =24x− 1
4x. (14.21)
The expression for the volume of our box is as follows,
V = x2y. (14.22)
Substituting in our expression for y we are left with,
V = 24x− 14
x3. (14.23)
Now we differentiate both sides,
V′ = 24− 34
x2 (14.24)
V′ =34(32− x2) (14.25)
Therefore, when V′ = 0, x must be equal to +/−√
32. SInce x isa measurement it must be positive. Therefore,
x =√
32 f t. = 5.66 f t., (14.26)
and
y = 2.83 f t. (14.27)
The volume of the box is then,
V = 90.56 f t2. (14.28)
2. Homework Problem 2: In this chapter the only specific ACO algo-rithm that we discussed was the one designed for the TravelingSalesman Problem. We did, however, mention that ACO algo-rithms have been used for many different optimization problemsin many different fields. In general, what aspects of the ACOalgorithm would need to be adjusted for the algorithm to be ablefind a solution for a different problem?
Solution: The major aspects of an ACO algorithm that determinethe problem solving abilities of that algorithm are the probabilisticbiases that the artificial ants’ movements abide to, the way inwhich they lay pheromone and reinforce quality solutions, themovement constraints that exist on the ants due to the problem’sintrinsic design and whether or not the artificial ants possessextra, problem specific abilities that assist them in their search forthe solution.
14.5 solution to the traveling salesman problem using an aco algorithm 115
3. Exam Question 1: The process in which an artificial ant in anACO algorithm for the Traveling Salesman Problem retraces hissteps after every ant has completed its respective tour and addspheromone to the edges he used is called:a = local trail updatingb = the pheromone modelc = stigmertic buildingd = global trail updating
Solution: d = global trail updating
4. Exam Question 2: Which of the following traits is not a differencebetween artificial ant of the ACO algorithm and real ants?a = The amount of pheromone that is deposited by a single antcorrelates to the quality of the solutionb = A preference to choose trails that are rich in pheromonec = The possesion of an internal state that contains the memoryof the owner’s past actionsd = The way in which the ant alters their environment to commu-nicate with the rest of the colony
Solution: b = A preference to choose trails that are rich in pheromone
15C H R I S T O P H E R K E R R I G A N : T H E P H Y S I C S O FS O C I A L I N S E C T S
15.1 introduction
Physics as a whole is an area in which precise measurement andobservation is integral to the expansion of the field. It is perhaps
for this reason that the study of living systems under the framework ofphysics is an oft overlooked application. This chapter therefore concernsitself with attempting to approach one of the groups of living systemsmost closely correlated to preexisting physical concepts and theories:the social insect.
But just what are social insects? This question is best answered bytaking into account the concept of eusociality, but since this involves atleast some digression into the world of sociology, we will leave it to beexplored at will by the reader. For our purposes, the physical aspectsof the social insects are most important. We have a group of hundredsto thousands of insects of the same species who, individually, showalmost no signs of intelligence or survival skills but are a functioninggroup as a whole. The physical dynamics of this group in its entiretyshall be our focus.
It is the author’s experience that nature is a thing of patterns, andthis observation will be the basis for the structure of this chapter.We shall apply to the dynamics of social insects the concepts andtheories derived from a more well-explored avenue of physics, namelyelectromagnetism.
Note: This chapter serves a dual purpose as not only a method ofstudying the dynamics of social insects, but also as an example of howto approach a concept with "‘the physicists mind"’. That is, the chapterexplains how to expand the ideas of physics to cover new areas.
15.2 the electron and the ant
In comparing social insect systems with electromagnetic ones, the firstlogical correlation to draw is of the fundamental constituents of thesystem. In social insect systems, this is the individual insect (whichshall hereafter be referred to by the synecdoche "‘ant"’ to save time). Inelectromagnetism, this is the electron.
The second logical correlation is of the system as a whole. For socialinsects, this is the colony. For electromagnetism, our analogy will bethe circuit.
So what does a circuit do? We know that individual electrons movingthrough a conducting material creates a current. This current is usedto power various components in the circuit which achieve some goal.Likewise, ants move in a constrained group in some sort of cycle toachieve some goal, namely the survival of the group. We can say withconfidence that the movement of the ants is indeed cyclical simplybecause they are members of the colony. If they leave the center of thecolony (to fulfill some goal for the group), they will return at some time(probably having fulfilled said goal). We can define the center of the
117
118 christopher kerrigan: the physics of social insects
colony as the place where more ants are produced. This has an obviousanalogy to the battery (or EMF) in an electric circuit. Also, we can safelysay that the constrained group in which the ants move is probably asingle-file line because of the width of ant tunnels (and from generalobservation.) Knowing that the processes of both systems occur in asimilar fashion, perhaps we can use the equations of electric circuits todescribe the motion of the ants.
15.3 insect current
The current in a circuit is any motion of charge from one region toanother, and can be described as the net charge flowing through somecross-sectional area per unit time. We should be able to define the insectcurrent in the same manner, replacing the charge with the ant. Thiscurrent can be described as the number of ants crossing some width(assuming the ants move along a roughly two-dimensional surface (theground) and do not walk on top of each other) per unit time,
I =Nt
, (15.1)
where N is the number of ants. With this simple and obvious compar-ison, we can draw many conclusions about the motion of a group ofsocial insects.
example 15 .1 : the ant hill
A group of one hundred ants is moving in a single-file line into the ant hill,whose diameter can fit one ant. If ants can travel at a rate of .3 meters persecond and an ant’s size is 10 mm, what is their current through the ant hill?
Answer:
Our line of ants is 10mm × 100 = .1m long. A line of ants ofthis length moving at the given rate will take (.1m)/(.3m/s) = 1/3
seconds. Knowing both the time and the number of ants, we caneasily calculate the current, I = 100/ 1
3 = 300 ants per second.
15.4 insect diagram
The destinations of social insect groups can often be defined in termsof the provisions that the group needs to survive. The most importantfactor for the survival of our ant colony is its food, so we can interpretthe possible destinations of the ants as locations with food to bringback to the center.
We are beginning to make it possible to chart the movement of thecolony. We know that the ants come from a source and move in linestoward a source of sustenance which they collect and return to thecolony center with. This is analagous to the circuit, whose electronscome from a battery and travel through a conductor through variouspathways which, by the definition of a complete electric circuit, mustlead them back to the battery. Since this process can be charted by wayof the circuit diagram, we can chart the progression of the ants in asimilar manner.
Let us first examine the well-known structure of the circuit diagramby referring to Fig. 1. We can see that the diagram depicts a battery and
15.5 kirchoff’s rules (for ants) 119
Figure 15.1: Circuit
two resistors in parallel. If we imagine the battery as being the centerof the ant colony and the resistors as being sources of food, we havewhat we shall refer to as an insect diagram. Ants leave the center fromthe "‘positive side,"’ travel to one of two (for example) food sources,and return to the center on the "‘negative side"’.
15.5 kirchoff’s rules (for ants)
We will now apply two well-known theorems to the notions that wehave already established, namely those of current and diagram. Thetheorems to which we refer are Kirchoff’s junction rule and his looprule. The former states that the algeraic sum of the currents into anyjunction is zero. This would mean, in our insect terms, that the numberof ants going into any point at which they decide which food sourceto travel to is equal to the number of ants at all the subsequent foodsources. This should be not only true, but obvious (assuming no antdies along the way). The loop rule states that the algebraic sum ofvoltages in a closed circuit must be zero. In our terms, this may meanthat there must be both an ant source and a food source in order fora path to exist. This makes logical sense, since ants do not simply goout wandering. We have therefore successfully applied an importanttheorem of electrodynamics to a macroscopic and very real system.
15.6 differences
When applying a theory that works to describe some system to a totallynew system, perhaps the most important relationship between the twois that of their differences. A difference that seems particularly suitedto analysis is the fact that electrons will travel in the direction of lowervoltage, while ants seem to choose an initial path rather randomly andthen perfect this path based on the movement of the returning ants andby sensing the pheromones left behind by ants who have travelled thebetter path. We shall explore this behavior as an exercise in making thedifferences clear.
Assume that two ants start a journey towards food with equal proba-bilities of going on either of two paths (Fig. 2). We will say that one ofthe paths (the "‘lower"’ path in the diagram) is shorter, and thus it takeslonger for the ant not on this path to get to the food (Fig. 3). Following
120 christopher kerrigan: the physics of social insects
Figure 15.2: Ants
Figure 15.3: Ants
this logic, the ant who has taken the shorter path will return to the nestfaster (Fig. 4). This means that just outside the nest, the pheromonedensity is twice as high on the shorter path (because it has been passedtwice, whereas the longer path has only been passed once until the antmakes it back to the nest). Other ants leaving the nest, then, will opt forthe shorter path because of the higher pheremone density. Over manyiterations, the ants further reinforce the shorter path and after a whilethis will be the only path used (Fig. 5).
15.7 the real element
We have managed to describe at least some of the motion of socialinsects in a quantitative manner, which was our initial goal. This methodcan be used to study the social insects in their environment, but it isimportant to remember the constraints and assumptions that we mayhave taken for granted. The behavior of living beings is an extremelycomplicated concept to analyze, and it is therefore vital that we takeour limitations into account.
We can consider, for example, experiments in which ants in an in-duced state of panic and given two doors to escape the "room" theyare in will not leave in a symmetric manner, as a physical analysismay predict. They instead show a tendency to "follow the leader,"which is a complex phenomenon far beyond the reach of this text and
15.8 problems 121
Figure 15.4: Ants
Figure 15.5: Ants
perhaps even biophysics as a whole. This situation is intended to il-lustrate the fact that physical representations of systems are just those:representations, and should be applied as such.
15.8 problems
1. Calculate the probability that one of many ants is at one of n foodsources. The source of the ants and the food sources are all on aline with a distance l between each.Solution:
We know that a given food source is n× l from the ant source. If weassume that the probability of an ant to go either direction at a foodsource is 1/2, we see that for a food source F(n)
P[F(1)] =12
(15.2)
P[F(2)] =12(
12) (15.3)
P[F(n)] = (12)n (15.4)
2. Suppose the distance between two points a and b is dab and Cab is thepheromone density along this line. What is the probability for a single ant
122 christopher kerrigan: the physics of social insects
to go from one point a to another point b?Solution:
We can imagine that the probability of an ant moving in a direction isdirectly proportional to the pheromone density in that direction. Sincedensity along a line is inversely proportional to the length of the line(whereas in three dimensions it is inversely proportional to the volume),we can say that
P =Cabdab
(15.5)
Part IV
W H AT ’ S O U T T H E R E
16R O B E RT D E E G A N : T H E D I S C O V E RY O F N E P T U N E
16.1 introduction
Neptune was first officially discovered on September 23, 1846.While the discovery of any new planet is normally viewed by
society as a rather important event, Neptune’s discovery stands out asnotable in the scientific community as well because of the manner inwhich it was found. Unlike the other seven planets in our solar systemNeptune was not first directly observed, but rather it’s existence wastheorized due to the effects of its gravity, in accordance with Newton’sLaws.
16.2 newton’s law of universal gravitation
Before we can look into exactly how it was that astronomers foundNeptune, we must first understand the laws that govern celestial bodies.The interaction of any two masses was described by Newton with hisLaw of Universal Gravitation which states that every point mass acts onevery other point mass with an attractive force along the line connectingthe points. This force is proportional to the product of the two massesand inversely proportional to the square of the distance between thepoints
F = Gm1m2
r2 (16.1)
where the force is in Newtons and G is the gravitational constant equalto 6.674× 10−11N m2 kg−2[101].
example 16 .1 : celestial gravitational attraction
Using Newton’s Laws we can find the forces acting upon Uranus due to boththe Sun and Neptune, when Neptune and Uranus are at their closest:
Orbital Radius of Neptune (Rn) = 4.50× 1012m
Orbital Radius of Uranus (Ru) = 2.87× 1012m
Rnu = Rn − Ru = 1.63× 1012m
Mass of Sun (Ms) = 1.99× 1030kg
Mass of Neptune (Mn) = 1.02× 1026kg
Mass of Uranus (Mu) = 8.68× 1025kg
Fs−u = GMs Mu
(Ru)2 = 1.399× 1021N (16.2)
Fn−u = GMn Mu
(Rnu)2 = 2.224× 1017N (16.3)
So the force acting on Uranus due to the Sun’s gravity is 4 orders of magnitudelarger than the force due to Neptune’s gravity.
125
126 robert deegan: the discovery of neptune
Figure 16.1: Predicted and Actual Orbits of Uranus and Neptune. Grey =Uranus’ predicted position, light blue = Uranus’ actual position, dark blue =Neptune’s actual position, yellow = Le Verrier’s prediction, green = Adams’prediction [94]
Given that one object is much more massive than the other, thisattractive force will cause the less massive object to accelerate towardsthe more massive object while having little effect on the latter. However,if the less massive object is not at rest but rather has some initial velocityvector and thus an initial angular momentum, then this force will notcause the lesser mass to accelerate directly towards the larger massbut rather cause it orbit about the larger object due to the necessity toconserve the angular momentum of the system. This obviously is thereason the planets in our solar system orbit about the Sun, due to itsrelatively large mass. However, the orbits of the planets in our solarsystem are not just affected by the Sun, in most cases the gravitationalfield of adjacent planets are also strong enough to influence each othersorbit, as we shall see momentarily.
16.3 adams and le verrier
Shortly after Uranus was discovered in 1781 its predicted orbit wascalculated using Newton’s laws of motion and gravitation. However,by looking at data for the position of Uranus from 1754-830(someobservations of the planet were made prior to the discovery that itwas in fact a planet) and specifically over the period from 1818-1826 itwas noticed that there were discrepancies between Uranus’ predictedand observed orbit. Tables predicting Uranus’ orbit had been madeshortly after its discovery and soon comparison to these tables noteddiscrepancies to large to be accounted for by observational error orby the effects of Saturn and Jupiter on Uranus. As these discrepanciesgrew larger and larger in the early 1800’s they became more and moretroubling, and numerous theories to account for this were postulated,one being that there was some unknown body perturbing the orbit. Thistheory appealed to two men, Urbain Le Verrier and John Couch Adams.Both Adams and Le Verrier set out independently to investigate it.
16.4 perturbation 127
Observations had shown that not only was Uranus not at the correctposition in it’s orbit at a given time, but also that its distance fromthe sun was substantially larger than it was predicted to be at somepoints. This evidence to some further supported the hypothesis of theunknown planet, and the increased radius vector clearly showed thatthis mysterious planet must be outside of Uranus’ orbit. This is theassumption Adams made as he began his investigation, but he quicklyran into trouble. The typical way of determining a perturbing bodieseffect on an orbit was to calculate the gravitational effects of a body ofknown mass and position on another planet, and subtract these valuesfrom the observed orbit. After this had been done one could see the"true" orbit of the planet, as it would be without the perturbing body,and then when adding back in the effects of the perturbing planetone gets the actual elliptical orbit followed by the planet and can thuspredict its location with extreme precision.
In this case however, the characteristics of the perturbing body wereunknown which made it impossible to calculate the "true" elliptical or-bit, as this required knowing the perturbative effects. And these effectsin turn could only be found once one already knew the "true" orbit, soAdams’ approach was to solve both problems simultaneously. Usingthis approach Adams soon ran into another problem, that being thatthe perturbation caused by a more massive planet further away fromUranus would be indistinguishable from the effect of a less massiveplanet closer to Uranus. To avoid this issue Adams simply guessed thatthe average distance from the sun of the perturbing planet was twicethat of Uranus, and then to come back and adjust this assumption afterhe had completed his calculations were it necessary to do so. Fromthis point on Adams simply needed to right out a serious of equationsrelating the predicted position at each date to the actual position it wasfound at. Adams wrote 21 such equations, one for every third year forwhich he had observational data, and solved each of these equationsone at a time. Alone these were not enough to give the characteristicsof the perturbing planet, but each narrowed down the possibilitiesfurther and further until after all 21 equations had been solved Adamshad a very accurate model describing the characteristics of this newplanet. Adams then calculated the effect this new planet would have onUranus’ orbit and found that it accounted for all of the discrepanciesbetween Uranus’ actual and predicted orbits. Finally, Adams calculatedthe longitudinal position he expected to find Uranus at and gave thisdata to the Royal Astronomer in order to confirm the existence of thisnew planet there, and indeed the new planet was observed to be withina few arcseconds of Adams’ prediction[77].
At the same time in France, Le Verrier was going through the sameprocesses and calculations and came to an almost identical result bywhich astronomers in France independently discovered this new planetat almost the exact same time that this was being done in England.
16.4 perturbation
It is important to examine how exactly Neptune’s gravitational fieldperturbed the orbit of Uranus. First of all, the gravitational interactionbetween Neptune and Uranus obviously puller Uranus further from theSun and Neptune closer to it. Also, as a result of this increased radiusat certain points in Uranus’ orbit the average distance of Uranus from
128 robert deegan: the discovery of neptune
Figure 16.2: Graviational Perturbation of Uranus by Neptune[18]
the Sun was increased. By Kepler’s third law, P2 = 4π2a3/G(M + m),we know that as a result of this period of revolution of Uranus wasincreased slightly, and obviously the opposite was true for Neptune[19].However, neither of these was the most drastic or noticeable effect;since Uranus is closer to the Sun than Neptune it obviously traversedits orbit faster than Neptune and so there were points where Neptunewas slightly in front of Uranus and points where it was slightly behindit. In the case of the first situation Neptune’s gravitational attractionwould pull Uranus towards it effectively speeding up Uranus’ motionthrough its orbit and pulling it ahead of the predicted location. Theopposite was true when Neptune was behind Uranus, it then pulledit back and slowed its orbit. This effect is what caused the discrepan-cies in Uranus’ longitude that astronomers noticed, as Neptune andUranus got closer together in the early 1800’s the effect became morepronounced since the gravitational force between them was increas-ing, and this is what caused astronomers to finally throw out theircurrent predictions of Uranus’ orbit and try to determine what whatwas causing these discrepancies.
16.5 methods and modern approaches
The method Adams used to predict the characteristics of Neptune wasan incredibly tedious one and at the time was thought to be the onlypossible way to solve the problem. Since then however, another possibleway to solve this problem has been postulated. This process involveslooking at the problem as a three-body system, in which we examine themutual gravitational interactions of the three bodies involved, the Sun,Neptune, and Uranus. This problem is certainly not a trivial one, andthough numerous particular solutions to this problem have been foundthere is still no general solution, and many believe such a solutionis actually impossible as this problem involves chaotic behavior[21].An attempt at a particular solution to this problem for the case ofthe Neptune-Uranus-Sun system is beyond the level of this book andso we shall not attempt it here. It is significant to note though thatdespite numerous advances in the field of celestial mechanics since thetime of Adams and Le Verrier, the method they used is still the only
16.6 practice questions 129
practical one for solving this problem. As mentioned in this sectionthere are other possible ways to solve this problem but they are farmore complicated and no more accurate, so if in the present day therewas a need to find the cause of some unexplained perturbations thesame method that Adams and Le Verrier followed would probably stillbe used(though it would be much faster and more accurate thanks touse of modern computers).
16.6 practice questions
1. Astronomers noted that Uranus was not following its predictedelliptical orbit, but rather traversing a different ellipse. How muchdata is required to determine this, i.e. what is required to charac-terize an ellipse?
2. Assume that the gravitational pull of Neptune increased Uranus’average distance from the sun by 3× 1012 m, how much wouldthis increase Uranus’ period by (in years)?
16.7 answers to practice questions
1. An elliptical orbit is mathematically characterized by six numbers:The average distance between the planet and the sun, the eccen-tricity of the ellipse, three angles to determine the orientation ofthe ellipse in space, and a point in time at which the planet is at aparticular point in the orbit. The average distance of Uranus wasalready known to astronomers, and the eccentricity of the orbitis determined from the length of the semi-major and semi-minoraxises. It requires two reasonably separated data points to deter-mine these axises. In order to determine the orientation of theorbit in space requires three data points as one would expect, andagain some separation is required between these points. Since thepoints used to determine the eccentricity and the orientation of theorbit need not be separate, we can thus conclude that along withthe knowledge of the average distance to the sun(the semi-majoraxis), it requires only three reasonably separated observations todetermine the orbit of a planet.
2. We can apply Kepler’s 3rd Law here to find what effect thisincrease in orbital distance will have on the period:
P2 = 4π2∗a3
G(M+m)a3 = (3× 1012)3 = 2.7× 1037m3
G = 6.67× 10−11s−2m3kg−1
M = 1.99× 1030kgm = 1.02× 1026kg
P2 = 1.066×1039m3
1.327×1020s−2∗m3 = 8.03× 1018s2
P = 2.83× 109s = 89.8 years
17A L E X K I R I A K O P O U L O S : W H I T E D WA R F S
17.1 introduction
White Dwarfs are class D stars based on the Morgan-Keenan spec-tral classification. They fall under extended spectral types and
can further be differentiated by DA (H rich), DB (He I rich), DO (HeII rich), DQ (C rich), DX (indeterminate), DZ (Metal rich), and DC (nostrong spectral lines). The additional letter indicates the presence ofdifferent spectral lines which indicate what elements are present in theatmosphere and do not correspond to other star classes.
They are characterized by low mass about one solar mass [73], smallsizes with characteristic radii of 5000 km [73], and no nuclear fusionreaction in the core. White dwarfs, neutron stars, and black holes arefall under a caterogr of compact objects. These compacts objects do notburn nuclear fuel and have small radii. White dwarfs are thought tobe in late evolutionary stages of less massive stars and from lack ofnuclear reactions in the core are radiating their residual thermal energyslowly over time.
Sirius B a typical white dwarf star has a solar mass of about 0.75 Mto 0.95 M and a radius of about 4700 km [73]. This corresponds to aplanetary sized star with a mass nearing that of the Sun. Because oftheir small radii white dwarfs have higher effective temperatures thanother stars and hence appear whiter. The luminosity varies as R2T4 fora black body. [73].
This make white dwarfs extremely dense stars with mean densitiesof 106 g/cm3, 1, 000, 000 times great than the sun. All four fundamentalforces are actively involved in the dynamics of these stars [73]. Thetremendous density values create immense gravitational forces thatpull the star together. This creates an electron degeneracy pressurethat supports the star from gravitational collapse. According to theuncertainty principal if the electrons positions are all well define thenthey have a correspondingly high uncertainty in their momentum. Evenif the temperature were to be zero there would still be electrons movingabout.
When pressure due to the confinement of the matter exceeds that ofthermally contributed pressure, the electrons are referred to as degen-erate. Black holes however are completely collapsed stars that had nomeans of creating a pressure great enough to support against gravita-tional collapse.
Simplifying this problem into the particle in the box with dimensionsLx, Ly, and Lz and setting the potential to be zero in the box and
infinite outside the box the Schrodinger equation reads − h2
2m∇2ψ = Eψ.ψ factors into 3 functions of X(x), Y(y), and Z(z) and yields threesecond order ordinary differential equations, one for each of the threefunctions. Solving them and setting
kx,y,z =√
2mEx,y,z
h(17.1)
131
132 alex kiriakopoulos: white dwarfs
and taking into consideration the boundary conditions
kx,y,zLx,y,z = nx,y,zπ. (17.2)
The allowed energies of this wave function are
Enx,y,z =h2k2
2m(17.3)
where k is the magnitude of the wave vector k and each state in thissystem occupies a volume π3
V for n = 1. Since the electrons behave asidentical fermions and hence are subject to the Pauli exclusion principleonly two can occupy any given state. Therefore they fill only one octantof a sphere of k-space [37]
18(
43
πk3F) =
Nq2
(π3
V) (17.4)
where k3F = (3ρπ2)
13 is the radius determined by the required volume,
π3
V , the electrons must take, N is the total number of electrons, and q isthe number of free electrons each atom contributes.
Then ρ ≡ NqV is the free electron density. The Fermi surface here is
the boundary separating the occupied and unoccupied states in k-space.The corresponding energy is called the Fermi energy the energy of thehighest occupied quantum state in this system of electrons. The Fermienergy is EF = h2
2m (3ρπ2)23 , and this is the energy for a free electron
gas.
17.2 the total energy
To calculate the total energy in the gas considered a shell of thicknessdk which contains a volume 1
8 (4πk2)dk with the number of electrons
occupying that shell to be Vπ2 k2dk. Each state carries energy h2k2V
2mπ2 k2dk.Thus the total energy must be
Etotalelectron =h2(3π2Nq)
53
10π2mV−
23 (17.5)
This energy is analogous to the internal thermal energy of ordinarygas and exerts a pressure on the boundaries if it expands by a smallamount [37], dV.
dE = − 23 E dV
V resembles that of the work done by this electron pres-sure,
P =2E3V
=(3π2)
23 h2
5mρ
53 (17.6)
Evidently the pressure depends on the free electron density. It iswhen this pressure equals the gravitational pressure from the mutualattraction of the ensemble, do white dwarf conditions occur. This isthe fundamental difference between white dwarfs and "‘normal"’ stars.Stars usually have cores that sustain nuclear fusion of hydrogen intohelium that counteract the gravitational pressure. The white dwarf
17.3 question 133
sustains itself however through the electron degeneracy pressure. Thisis a quantum mechanical result. This is why a cold object would notsimply collapse after being continusouly cooled and allows the whitedwarf to exist.
Expressing the total energy as the sum and the total electron energyplus the gravitational energy of a uniformly dense sphere, U = 3
5GM2
r ,and setting the equation to zero,
Etotal = Etotalelectron + U = 0 (17.7)
reveals the radius function for this minimization is,
R = (9π
4)
23
h2q53
GmM2nuclei N
13
(17.8)
The equation reveals a fundamental nature of these compact objects;as the mass increases the radius decreases in this case we would beincrease N, the number of nucleons. This is how objects such as blackholes or neutrons stars which, are other compact objects like whitedwarfs, can contain so much mass into a smaller and smaller radiusyielding higher densities.
Energy transport within the star occurs via conduction because theparticle’s mean free paths are increased due to density. Particles thenfind it difficult to collide since all the lower states are filled. The coffefi-cent of thermal conductivity becomes very large and the interiors arenot far from being isothermal making the core much hotter. A surfacetemperature of 8000 K will have a core tmeperature of 5, 000, 000 K [80].
The stars itself is composed mainly of carbon and oxygen but highgravitational forces separates these elements from lighter elementswhich are found on the surface. Spectroscopy techinques reveal atmo-sphere to be compose of either helium or hydrogen dominate. Theatmosphere may however contain other elements in some cases such ascarbon and metals.
Magnetic fields in white dwarfs are thought to be due to conservationof total surface magnetic flux. A larger progenitor star generating amagnetic field at one radius will produce a much stronger magneticfield once its radius has decreased according to conservation of mag-netic flux. This explains the magnetic fields on the order of millions ofguass in white dwarf stars. The strength of the field is calculated byobserving the emission of circularly polarized light.
White dwarfs once formed are stable and cool continuously until itcan no longer emit heat or light. Once this happen the white dwarf isreferred to as a black dwarf. No black dwarfs are not thought to existhowever since the time required for a white dwarf to become a blackdwarf is longer then the age of the universe [80].
17.3 question
dE =h2k2V2mπ2 k2dk (17.9)
and
Etotal electron =h2(3π2Nq)
53
10π2mV−
23 (17.10)
134 alex kiriakopoulos: white dwarfs
What is the integral of the above derivative. What is it? What are thelimits of integration? What does the upper limit represent?
Answer:
Etotal electron =h2V
2π2m
∫ kF
0k4dk (17.11)
The limits are 0 to kF the radius of a sphere of k-space. The k-spaceradius, kx,y,z depends on the energy levels, nx,y,z.
18D A N I E L R O G E R S : S U P E R N O VA E A N D T H EP R O G E N I T O R T H E O RY
18.1 introduction
The universe started with a big bang. The temperatures were sohigh in the minutes after this event that fusion reactions occurred.
This resulted in the formation of elements such as hydrogen, deuterium,helium, lithium, and even small amounts of beryllium.[6] This is knownas Big Bang Nucleosynthesis, and nicely explains the presence of theselighter elements in the universe. However, the brevity of this processis believed to have prevented elements heavier than beryllium fromforming.[53] So what is the origin of oxygen, carbon, nitrogen, and themany other heavy elements known to man? And how do we explainthe significant abundances of these elements in our solar system?
18.2 creation of heavy elements
Nuclear fusion is the process by which multiple atomic nuclei jointogether to form a heavier nucleus.[61] As explained before, this waswidespread just after the Big Bang. The result of this process is therelease of considerable amounts of energy; the resultant nucleus issmaller in mass than the sum of the original nuclei, and the differencein mass is converted into energy by Einstein’s equation, E = mc2.[53]
Nuclear fusion also occurs in the cores of stars and is the source oftheir thermal energy. In general, large stars have higher core tempera-tures than small star because they experience higher internal pressuresdue to the effects of gravity.[53] Thus, a star’s mass determines whattype of nucleosynthesis can occur in its core.
In stars less massive then our sun, the dominant fusion process isproton-proton fusion. This converts hydrogen to helium. In stars withmasses between one and eight solar masses (we define our sun as onesolar mass), the carbon cycle fusion process takes place.[53] This con-verts helium into oxygen and carbon once hydrogen is depleted withinthe star. In very massive stars (greater than eight solar masses), carbon
Figure 18.1: Elements are produced at different depths within a star. Thisillustrates the elements that are produced in massive stars (not to scale). Noticethe iron core.[39]
135
136 daniel rogers: supernovae and the progenitor theory
and oxygen can be further fused into neon, sodium, magnesium, sulfurand silicon. Later reactions transform these elements into calcium, iron,nickel, chromium, copper, etc. In a supernova event, neutron capturereactions lead to the formation of elements heavier than iron.[53] Thus,we see that all heavy elements are formed in the cores of stars at variouspoints in their lives as they burn through their thermonuclear fuel. Ingeneral, the mass of a star can be used to determine what elements areformed and the abundances that are produced.
example 18 .1 : conversion in the sun
How much hydrogen is converted to helium each second in the sun? Use thefact that the sun’s luminosity is 3.8× 1026 W, and that 0.7% of the hydrogenmass becomes energy during the fusion process.Solution: We know that the sun produces 3.8× 1026 W, and so 3.8× 1026 Jare emitted each second. Now, simply use Einstein’s equation.
E = mc2 ⇒ m =Ec2 =
3.8× 1026 J(3× 108 m
s )2 = 4.2× 109 kg (18.1)
This is the mass converted to energy in the sun each second. We know thatthis mass is only 0.7% of the mass of the hydrogen that goes into the fusionprocess.
MH =4.2× 109 kg
0.007= 6.0× 1011 kg (18.2)
So we see that the sun fuses about 600 billion kg of hydrogen each second,though about 4 billion kg are converted into energy. The remaining 596 billionkg becomes helium.[6]
18.3 dispersal of heavy elements
A star experiences a constant struggle against collapse due to thegravitational force of its own mass. Throughout the main sequence ofits life, it is able to resist gravity with thermal pressure, as the fusion ofelements in its core heats up the star’s interior gas. The hot gas expands,exerting an outward pressure that balances the inward force of gravity.The life of a star ends when it is completely depleted of thermonuclearfuel, and gravity is able to overcome this outward thermal pressure.[53]
The life span of a star and its final state are determined by themass of the star. Large stars generally live shorter lives than smallstars; although they have more fuel for nuclear reactions, their rate ofconsumption is much greater. When a relatively small star runs out offuel, it collapses because of gravity and becomes a white dwarf. At thispoint, the only outward pressure is due to electron degeneracy. Sincewhite dwarfs succumb entirely to gravity and never explode outward,the elements formed in their cores are never ejected into space.[6]
Stars that are large, however, experience different effects due togreater gravitational forces during collapse. As a massive star beginsto run low on hydrogen fuel, the iron it produces piles up in its core.Iron has the lowest mass per nuclear particle of all nuclei and thereforecannot release energy by fusion. Once all the matter in the core turns toiron, it can no longer generate any energy.[6] This marks the beginningof collapse. As massive stars collapse, reactions take place in whichelectrons and protons are forced together with such great amounts of
18.3 dispersal of heavy elements 137
Figure 18.2: (a) The layered shells of elements in a massive star. (b) Collapsebegins when fuel is depleted. (c) As gravity takes over, the star shrinks signifi-cantly. (d) The red area experiences enormous outward forces as hot gases pileup on the degenerate core. (e) The gases are ejected outward at high speeds. (f)All that remains is the degenerate neutron core.[39]
force that they merge to become neutrons. Quantum mechanics restrictsthe number of neutrons that can have low energy, as each neutronmust occupy its own energy state. When neutrons are tightly packedtogether, as they are in this case, the number of available low energystates is small and many neutrons are forced into high energy states.The resulting neutron degeneracy pressure quickly stops gravitationalcollapse and the matter in the star is subjected to an enormous outwardforce. With gravitational collapse halted suddenly, the outer layers ofgas bounce back upon hitting the degenerate core like a large wavehitting a sea wall.[53] The violent explosion that follows is known as asupernova event.
Most of the energy of a supernova explosion is released in the formof energetic neutrinos. It is this energy that initiates the formation ofelements heavier than iron, as described before. The remaining energyis released as kinetic energy in the ejected matter. The shock wave sendsthe ejected material outward at speeds of over 10,000 km/s.[50] All thatremains is the sphere of tightly packed neutrons, called a neutron star.If the original star was massive enough, the remaining neutron starmay be so large that gravity also overcomes the neutron degeneracypressure and the core continues to collapse into a black hole. Otherwise,it becomes nothing more than the corpse of a star that has depleted itsfuel supply.[53]
The expanding cloud of debris from the supernova explosion isknown as a supernova remnant. The ejected gases slowly cool andfade in brightness, but they continue to move outward at high speeds.Carried with this debris is the variety of heavy elements produced inthe core of the star, as well as those created by the collision of highenergy neutrons during the supernova event.
example 18 .2 : stellar equilibrium
To maintain equilibrium, a star’s outward thermal pressure must balanceinward gravitational forces. This results in enormous pressure at its core.How does the gas pressure in the core of the sun compare to the pressure ofEarth’s atmosphere at sea level? The sun’s core contains about 1026 particlesper cubic centimeter at a temperature of 15 million K. At sea level on Earth, theatmosphere contains 2.4× 1019 particles per cubic centimeter at a temperatureof about 300 K.Solution: All we need to do here is apply the ideal gas law.
Psun
PEarth=
nsunkTsun
nEarthkTEarth=
(1× 1026 partcm3 )(1.5× 107 K)
(2.4× 1019 partcm3 )(300 K)
= 2× 1011 (18.3)
The sun’s core pressure is about 200 billion times greater than the atmosphericpressure on Earth at sea level.[6]
138 daniel rogers: supernovae and the progenitor theory
Figure 18.3: An artist’s depiction of a solar nebula receiving supernovaejecta.[81]
18.4 progenitor theory
Now consider if part of this expelled cloud of isotopes (the ejecta) wereto fall into the gravitational field of a neighboring star that is in theearly stages of its formation. At this point the young star is surroundedby what is known as a protoplanetary disk, which is the disk of gasand dust from which planets and asteroids form. If the ejecta fromthe supernova event collided and blended with this disk, it wouldcontribute to its chemical composition.[83]
This is one of the leading theories that explains the presence ofheavier elements in our own solar system. This is obviously impossibleto prove by experiment, but fortunately there are ways to measure theprobability that this process occurred by studying chemical propertiesof our solar system.
One such property is the concentration of the stable isotope 60Ni.This has been detected and measured in meteorites that have beenuntouched since the formation of our solar system. There is also acorrelation between the amounts of 60Ni and another stable isotope,56Fe, in these meteorites.[65] The laws of chemistry predict that theunstable isotope 60Fe will decay into 60Ni with a half-life of 1.5 millionyears. Since our solar system is 4.6 billion years old, it is theorized thatit was initially 60Fe that was present in the meteorites.[50] It is knownthat 60Fe is one of the isotopes that massive stars form in their cores.[51]Therefore, 60Ni might serve as an indicator as to how much supernovaejecta our solar system collected during its formation.
18.5 a mathematical model
This section will work through a simplified mathematical model of theprogenitor theory. In the end, we will find an expression relating theradius of a young solar system with some of its chemical propertiesand distance from a likely supernova. Remember that there are morefactors that real astronomers take into account, but this is a generalidea of how they hope to prove the validity of the progenitor theory.
We start with defining several variables. Let R be the radius of oursolar system as it was during its formation 4.6 billion years ago. Wewill then call the area of our solar system AR = πR2. Now we define ras the radius of the supernova remnant; here, we want r equal to thedistance between the supernova and our solar system. We will assumethat the material ejected from the supernova is uniformly distributed
18.6 conclusion 139
over a sphere that expands as the ejecta moves outward. Thus we willdefine Ar = 4πr2. Now let Mr60 be the total mass of 60Fe ejected fromthe supernova, and MR60 be the total mass of 60Fe injected into our solarsystem. The amount of ejecta that a solar nebula can receive is inverselyproportional to its square distance from a supernova explosion.[50]
MR60 =Mr60 AR
Ar(18.4)
Now let M be the total mass of our solar system, and MR56 be theamount of 56Fe in our solar system per unit mass. Multiplying themtogether gives the total amount of 56Fe in our solar system. Then wecan define a relationship for the ratio of 60Fe to 56Fe injected into oursolar system.[50]
60Fe56Fe
=MR60
(MR56M)(18.5)
Since it is impossible to know exactly how much 60Fe was initiallypresent in our solar system, we can use an estimated ratio between 60Feand 56Fe based on the 60Ni meteoritic evidence mentioned before. Thisratio has been determined to be on the order of 10−7 based on studiesof the meteorites, though its precise value is still being investigated.[79]
It must be noted that there is a relatively small window of time duringthe formation of a solar system in which supernova ejecta is optimallyreceived. This window is on the scale of a few million years for a starthe size of our sun. Thus, we need to assume that the progenitor starhas a lifetime on this scale to increase the probability of it becominga supernova during the necessary time period. Stars that are around60 solar masses have short lifetimes of about 3.8 million years, makinga star of this size a good candidate.[50] As discussed previously, theamount of iron produced in the interior of a star throughout its lifetimecan be determined from its mass. For a 60 solar mass progenitor star,we can estimate the amount of 60Fe produced to be about 0.0002512
solar masses. This was taken from research by Marco Limongi andAlessandro Chieffi in 2006.[51]
We can use estimates for the other values in the equations aboveas well. For instance, we know that the mass of a protoplanetary diskaround a solar mass sized star is about 0.01 solar masses.[50] Also, theiron in our solar system is thought to comprise roughly 0.014% of themass of the entire system. An estimated 91.57% of this iron consists ofthe isotope 56Fe. This means that the amount of 56Fe in the solar systemcomprises 0.01282% of its mass.[50]
By combining the two equations above, it is possible to find theminimum radius of a solar nebula for it to have received the appropriateamount of the isotopes from a supernova.[50]
R =
√60Fe56Fe 4r2(MR56M)
Mr60
(18.6)
18.6 conclusion
Calculating the necessary radius for the protoplanetary disk of a solarmass star using known properties of our solar system suggests howprobable it is that a supernova event played a role in its formation.
140 daniel rogers: supernovae and the progenitor theory
Protoplanetary disks rarely exceed a few hundred astronomical units(AU) and have never been known to stretch beyond 1,000 AU. Thefarthest we have observed a body orbiting our sun is roughly 47 AU,indicating that it may have been reduced to a few tens of AU beforeplanet formation.[50] Thus, if these calculations yield radii on this scaleacross a wide range of values for r (again, the distance between thesupernova event and our sun), then this will provide support for theprogenitor theory of supernova injected isotopes.
There are, of course, many factors that determined our sun’s distancefrom supernova events. These include the size of the star cluster inwhich it formed and the ratio of the total potential and kinetic energiesin that star cluster (known as the virial ratio). As these propertieschange, the average distances between solar mass stars and likelysupernovae do as well. This is significant since stars that are furtherfrom supernova events will receive fewer ejecta.[50]
There is still one very important assumption that has been made aswell. By calculating the radius we assume that the protoplanetary diskwas face-on relative to the supernova event (to maximize the flux of theejecta through the disk). Very little ejecta would be received by a disk ifit were facing a progenitor edge-on.[50]
These shortcomings are not grounds for abandoning this study, how-ever, as this is still a likely explanation for the presence of heavyelements in our solar system. The results of these calculations are sureto bring us closer to the truth as mathematical techniques and computerprograms evolve over the years to come.
18.7 problems
1. Homework Problem 1
Question: Use the result of Example Problem 1. How many timesdoes the proton-proton fusion reaction occur each second in thesun? The mass of a proton is 1.6726 × 10−27 kg; hydrogen iscomposed of one proton, and helium is composed of four. Themass of a helium nucleus is 6.643× 10−27 kg.
Solution: Here, fusion converts four hydrogen nuclei (protons)into one helium nucleus. Four protons have a mass of 6.690×10−27. When four protons fuse to make one helium nucleus, theamount of mass that disappears and becomes energy is as follows.
6.690× 10−27 kg− 6.643× 10−27 kg = 0.047× 10−27 kg (18.7)
From Example Problem 1 we know that the sun converts a totalof 4.2× 109 kg of mass into energy each second.
mass lost per secondmass lost in each reaction
=4.2× 109 kg
s0.047× 10−27 kg
= 8.9× 1037 reactionss
(18.8)
Thus, nearly 1038 fusion reactions occur in the sun each second.[6]
2. Homework Problem 2
18.7 problems 141
Question: Explain why the ratio of 60Fe to 56Fe is used in theprogenitor theory, rather than the amount of 60Fe.
Solution: Since 60Fe has a half-life of 1.5 million years, nearly allof it that was present during the formation of our solar systemhas decayed into 60Ni. Scientists have been able to determine aratio of 60Ni to 56Fe from samples of meteorites. These meteoritesare indicative of the chemical composition of the early solar sys-tem. Scientists have also been able to roughly determine the totalamount of 56Fe present today, since it has not decayed. Multiply-ing the ratio of 60Ni to 56Fe by the total amount of 56Fe gives anestimate of the total amount of 60Fe injected into our solar systemduring its formation.
3. Test Problem 1
Question: Which of these elements had to be made in a supernovaexplosion? (a) calcium (b) oxygen (c) uranium
Solution: (c) uranium
4. Test Problem 2
Question: Suppose there is a supernova explosion at some dis-tance from a young solar system. If that distance is doubled, bywhat factor would the radius of the solar system need to be mul-tiplied in order to receive the same amount of ejecta? (a) 0.5 (b) 2
(c) 4
Solution: (b) 2
19D AV I D PA R K E R : T H E E Q U I VA L E N C E P R I N C I P L E
19.1 introduction
The basic idea of the equivalence principle is to “assume the com-plete physical equivalence of a gravitational field and a correspond-
ing acceleration of the reference system.”[26] This means that beingat rest on the surface of the Earth is exactly equivalent to being in anaccelerating reference frame free of any gravitational fields. This idea,when originally developed by Einstein in 1907, was the beginning ofEinstein’s search for for a relativistic theory of gravity .
There are three different ways to interpret the equivalence principleeach of which allow or disallow different theories of gravity; Currentlythe only theory of gravity to satisfy all three is general relativity , thisis part of what makes GR peerlessly elegant.
Bear in mind that most discussion of the equivalence principle is inattempts to disprove or provide limitations on it in order to supportan alternate theory of gravity. Because of this much of the discussionrequires you to check you intuitive understanding of the universe evenmore so than GR.
19.2 weak? strong? einstein?
The weak equivalence principle states that “All test particles at thesame spacetime point in a given gravitational field will undergo thesame acceleration, independent of their properties, including theirrest mass.”[92] This form of the equivalence principle is very similiarto Einstein’s original statement on the subject, but is referred to as“weak” because of the extent to which Einstein’s conceptualization ofthe equivalence principle matured. The other two interpretations of theequivalence principle use the weak equivalence principle as a startingpoint, assuming its truth. Very few, if any, theories of gravity contradictthe weak equivalence principle.
The Einstein equivalence principle states that the result of a local non-gravitational experiment in an inertial reference frame is independentof the velocity of location of the experiment. This variation is basicallyan extension of the Copernican idea that masses will behave exactlythe same anywhere in the universe. It grows out of the postulates ofspecial relativity and requires that all dimensionless physical valuesremain constant everywhere in the universe.
The strong equivalence principle states that the results of any localexperiment in an inertial reference frame are independent of where andwhen it is conducted. The important difference between this variationand the first two, weaker, variations is that this is the only variationthat accounts for self-gravitating objects. That is, objects that are somassive as to have internal gravitational interactions. Therefore, thisis an extremely important idea because of the extreme importance ofself-gravitating bodies, e.g. stars, to our understanding of the universe.
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144 david parker: the equivalence principle
Figure 19.1: Bending starlight in a space-time diagram [87]
19.3 consequences
The easiest and most drastic consequence of the equivalence principleis that light will bend in a gravitational field. Imagine an elevator freelyfalling in a gravitational field with a laser on one wall. If the laser shinesthen, according to the laws of relativity, the light will hit the point onthe wall that is directly opposite the laser. an observer in the elevatorwill correctly assert that the light traveled a straight line. However, fromthe perspective of an observer on the surface of the Earth, or whateverbody is causing the gravitational field, the light will still hit the pointon the wall directly opposite the laser, but because of teh elevator’sdownward motion the light will have followed a parabolic path. Thiswas how Einstein originally described the phenomenon of light bendingin a gravitational field, and this is still the best way to describe it.
A sharp student may point out that if you continued this thoughtexperiment by having countless elevators falling side by side withwindows to allow the laser to shine through them all that the lightwould actually bend at a rate twice that of a normal object falling ina gravitational field because the elevators are all accelerating radiallyrather than in the same direction. And with this student I would haveto agree, but Einstein got there first.
19.4 example
1. Imagine that an elevator is falling freely in Earth’s gravitionalfield with a laser mounted in the ceiling pointing directly at thefloor. When the laser shines would an observer at the floor of the
19.5 problems 145
elevator see the light as doppler shifted? Would an observer onthe surface of the Earth?
Answer No the light would not appear redshifted to the observerinside of the elevator, from his perspective he is in a completelymotionless box that is free from any external fields. However forthe observer on the surface of the Earth the light from the laserwould of course be blue-shifted because of its descent into thegravitational field, or conversely, because of the acceleration ofthe observer up to it.
2. Are there any constraints on the equivalence principle, and if so,what are they?
Answer There are constraints, and they actually change the natureof thought about the equivalence principle greatly. The equiva-lence principle is only valid in completely flat space, or a homoge-nous gravitational field. However, since there is no place in theuniverse where we will find either of those things the equivalenceprinciple can only actually be applied in a infinitesimally smallsection of space-time.
19.5 problems
1. An elevator on Earth is accelerating upwards at 6.6 m/sec2. Howlong will it take a rock of 0.2 kilograms to hit the floor of theelevator if it is dropped from a height of 1.5 meters?
Answer 1.5m = f rac12(6.6m/sec2 + 9.8m/sec2)t2 t = .43sec
2. If granite was found to fall at a different rate than water whatconsequences would this have for the principle of equivalence?Answer This is a direct violation of the WEP and therefore wouldinvalidate all of GR and most other theories of gravity in one fellswoop.
3. If the speed of light was measured in an area of flat space, andthen measured again on the surface of the Earth, discountingatmosphere, in which reference frame would the speed of light befaster? Answer They would both be the same. The speed of lightin a vacuum is a constant in all reference frames. The equivalenceprinciple combined with this fact, allows us to predict the bendingof light in a gravitational field.
20R I C H A R D R I N E S : T H E F I F T H I N T E R A C T I O N
20.1 introduction: the ‘four’ forces
Current models of this universe explain all interaction in terms offour fundamental forces. These forces (gravity, electromagnetism,
and the strong and weak nuclear forces), are all defined by unique“sources," or properties of individual particles that determine theirattraction or repulsion in terms of the specific force. For example,observe the known electric potential for a pair of particles:
Ue(r) = − 14πε0
q1q2
r(20.1)
Here it can be seen that the only properties of the matter that determinethe electrical potential are q1 and q2, the respective charges of theparticles. Thus, the “source" of the electric portion of the electromagneticforce is the value of charge. In the case of (non-relativistic) gravity,only the gravitational mass of each particles acts as the source for thepotential:
Ug(r) = −Gm1m2
r, (20.2)
where G is Newton’s gravitational constant1. Notice that in both ofthese cases, the potential falls off as the separation r increases, but hasan infinite range. In the case of the nuclear forces, this is not the case:their ranges are finite.2
example 20 .1 : the resultant force
Forces are often described by their potential. Once such a potential is de-scribed, however, it is simple to determine the force experienced by a systemof two particles:
F12 = − ddr
U(r), (20.3)
where F12 is the force experienced by one particle away from the other. In thecase of gravity:
F12 = − ddr
(−G
m1m2r
)= −G
m1m2
r2 , (20.4)
the negative sign implying the particles move toward one another.
In the early 1980s, some irregularities in experimental data promptedmany to ponder the existence of another, fifth primary force, with apotential determined by particle properties unique from these fourknown forces.
20.2 the beginning: testing weak equivalence
A basis of both Sir Isaac Newton’s Law of Universal Gravitation andAlbert Einstein’s General Theory of Relativity, the weak equivalence prin-
1 Currently accepted to be 6.673 · 10−11 m3
kg·s2
2 Roughly 10−15 m for the strong force and 10−18 m for the weak force
147
148 richard rines: the fifth interaction
ciple is both elegant in its simplicity and profound in its implications.Put simply, the principle states that:
“All test particles at the same spacetime point in a given gravitational fieldwill undergo the same acceleration, independent of their properties, including
their rest mass." [91]
More subtly, this implies that the inertial mass of an object (the propor-tionality between the force acting on the object and its acceleration) isexactly equal to its gravitational mass (the proportionality between theforce an object experiences in a gravitational field and the strength ofthat field). This point is readily seen in observing Newton’s gravita-tional potential (equation (20.4)).
Late in the 19th and early in the 20th century, Physicist Lorand Eötvösset out to verify this equivalence. Between 1885 and 1909, he devised,implemented, and refined an experiment which evidenced the equiva-lence to a much higher degree of accuracy than had been previouslyshown. Put simply, his procedure involved hanging different types ofmasses in a balance along a solid rod. Torque would then be appliedto the rod from two sources: differences in gravitational force on thetwo masses (a measurement of the gravitational mass of the masses),and differences in the centrifugal force each mass experiences (a mea-surement of the inertial mass). Only if these two masses are equivalentwould the rod remain perfectly stationary.
As with any physical procedure, experimental uncertainty and imper-fections plagued Eötvös’s results. These were originally averaged out toprovide fairly precise results favor of weak equivalence. As subsequenttests of the weak equivalence principle, using slightly different exper-imental procedure, did not observe such uncertainty, the variationsin Eötvös’s results were seen as the result of imprecise experimentalprocess.
More recently, however, physicists such as Ephraim Fischbach beganto reexamine these irregularities in terms of a possible fifth interactionbetween elementary particles.
20.3 a new force
Well after the time of Eötvös, physicists were struggling to explaintwo, seemingly unconnected phenomena. The first involved certainviolations of CP-symmetry seen in the decay of K0
L mesons [34]. Suchdecomposition had been observed, but was not expected by any currenttheory or model of elementary particles.
The second was an uncertainty in the gravitational potential: thevalue of the gravitational constant G has much higher experimentaluncertainty than any other physical constant. Recently, experimentshave been conducted providing a range of inconsistent results from−0.1% to +0.6% [1]. A group in based in Russia has measured valueswith a 0.7% fluctuation based on time and position [1]. Furthermore,measurements taken in mineshafts and submarines, though containinga very large range of uncertainty, have consistently provided values thatare greater than the accepted number [34]. This inability to determineprecisely the value of G opened the door to possible modificationsto the gravitational potential, which could in turn, some physicistsargued, explain the aforementioned CP-violations. Such a modification
20.3 a new force 149
on the gravitational potential would then imply a possible fifth particleinteraction.
In this light, Fischbach explained the dissimilarities between Eötvös’sresults and that of his predecessors not as experimental imprecision,but as the result of certain experimental differences between his exper-iments and those of his predecessors. Eötvös’s experiment involvedrelatively small acting distances for gravity (namely that between objectson the surface of the Earth and Earth itself). However, many experi-ments and measurements conducted by later experimentalists in orderto verify the weak equivalence principle relied only on much largerdistances (such as those between celestial bodies). Most relevantly, thetwo subsequent repetitions of the experiment that were found to bemost in conflict with Eötvös relied on the attraction of Earthly objectstoward the sun [33].
To explain the relevance of this discrepancy, Fischbach postulatedthat a new force could be modeled by the Yukawa potential, which hadbeen very effective in modeling the finite-range nuclear forces. In a 1986
paper, he proposed an additive term, in the form of a Yukawa potential,to the gravitational potential to account for the effect of a fifth particleinteraction:
U(r) = −Gm1m2
r
(1 + αe−r/λ
), (20.5)
where α and λ were to be determined experimentally. Clearly, thisterm very quickly approaches unity as radius increases, making it onlyrelevant at closer distances. At these distances, the correction termsacts equivalently to a new, rescaled value of the gravitational constantG. The factor by which the value is scaled is very strongly dependentupon the distance at which the gravity is acting, seemingly explainingthe existence of such a large range of measured values. Furthermore,Fischbach found that with the correct physical parameters3, this modelaccurately predicted the varying published geophysical measurementsof G [33].
Motivated by the potential for modifications on the law of gravityto explain known violations of CP-symmetries [34]4, Fischbach plottedvariations in results between different masses against various elemen-tary properties of the material used (properties which were unknownat the time of Eötvös’s experiments). Such a pattern would imply acompositionally-dependent deviation from the weak equivalence prin-ciple, and therefore the existence of a new, yet unexplained force.
His results were promising: he found evidence of a linear relationshipbetween the error in Eötvös’s results and the difference in a function ofthe baryon number of the particles in balance. The baryon number (B)of a particle is defined as:
B =Nq − Nq
3, (20.6)
where Nq is the number of quarks and Nq the number of antiquarks.Specifically, Fischbach observed the fractional difference (∆κ) betweenthe acceleration of the object and the acceleration of gravity and foundthe following linear relationship:
3 Very roughly, α = −(7.2± 3.6) · 10−3, λ = (200± 50) m4 For a more technical discussion these violations as Fischbach’s (and others’) motivations,
see Franklin [34]
150 richard rines: the fifth interaction
Figure 20.1: The linear relationship observed by Fischbach
∆κ = ∆ag
= α
(Bµ
)+ β (20.7)
Here, a is the acceleration of the object, g is that of gravity, µ is theatomic mass of the object, and α and β are constants5 This relationship,according to Fischbach, proved Eötvös’s results to be compositionallydependent, and this dependence was associated with the fifth funda-mental force. This force would have a short-range Yukawa potential:
U5(r) = −Gm1m2αe−r/λ
r, (20.8)
where α and λ are functions of the baryon numbers of m1 and m2.
20.4 the death of the force
Though Fischbach’s findings were initially quite persuasive, very littleelse came of the fifth force. Of the fourteen subsequent publishedexperiments searching for a force couple to Baryon number, only twohad positive results, and these had not been effectively reproduced[34]. Some of these results, such as those from Washington University’sEöt-Wash group (see figure 20.2), directly contradicted the linear Baryon-coupling results shown by Fischbach. Though some research in the areastill exists, most have accepted that no evidence remains that such a fifthparticle interaction physically exists. In the 1990 Moriond workshop,fewer than ten years after the beginning of the search, it was establishedthat no further experimentation was necessary to disregard the fifthinteraction hypothesis [34].
5 Fischbach found that, very roughly, α = (5.65± 0.71) · 10−6 and β = (4.83± 6.44) · 10−10.
20.5 problems 151
Figure 20.2: New experiments by the Eöt-Wash group provide results contra-dicting Fischbach’s assumption of linearity [34]
20.5 problems
1. Write an expression for the force experienced between two objectsas a result of the fifth interaction.
Answer: F12 = −Gm1m2α(
e−r/λ
r2 + e−r/λ
λr
)2. Assume the additive potential (equation (20.8)) accurately mod-
els the gravitational potential. Using the rough geophysical dataprovided by Fischbach (α ≈ −7.2 · 10−3, λ ≈ 200 m), write anexpression for the fraction by which a measured value of thegravitational constant G on the surface of the Earth (r = 6.38 · 106
m) would differ from that at infinity.
Answer: 11−7.2e−3.19·10−4
3. Two objects of mass 1 kg, with the same composition as Earth, areseparated by a distance of 1 mm. What force do the experience asa result of the fifth interaction? In what direction is this force?
Answer: F = 4.805 · 10−9 N, away from one another
4. Though the existence of a fifth force was eventually shown to lacksufficient evidence, how could its investigation have been usefulto the physicists involved? For further insight, see Franklin [34] fora fascinating discussion of this importance.
21D O U G L A S H E R B E RT: T H E S C I E N C E O F T H EA P O C A LY P S E
21.1 introduction
For 500,000 years Homo sapiens has roamed the Earth, building citiesand creating languages. We’ve gone from stagecoaches to space
travel in the span of one human lifetime, and we’ve sent robotic scoutsto other planets. It’s difficult to imagine it all coming to an end. Yet 99
percent of all species that have ever lived have gone extinct, includingevery one of our hominid ancestors. If humans survive for a longenough time and spread throughout the galaxy, the total number ofpeople who will ever live might number in the trillions. It’s unlikelythat we would be among the first generation to see our descendantscolonize planets, but what are the odds that we would be the lastgeneration of Homo sapiens? By some estimates, the current rate ofextinctions is 10,000 times the average in the fossil record. We may beworried about spotted owls and red squirrels, but the next statistic onthe list could be us.
21.2 asteroid impact
Space is filled with asteroids and comets which can pose a threat tolife on Earth. Fortunately Earth’s atmosphere protects us from thethousands of pebble-sized and smaller asteroids - only weighing a fewgrams - which strike earth every day at speeds of tens of kilometers persecond1. At these high velocities, friction with the upper atmosphereheats the meteoroids white-hot and causes immense deceleration forces.These small meteoroids get destroyed by the heat and deceleration, andare seen from Earth as shooting stars. However, some of the fragments,especially those from iron meteoroids, will reach Earth’s surface. It isestimated that 20 tons of meteorites reach Earth’s surface each day.
So, what if an asteroid hit Earth? Not just the dust that Earth collectsas it sweeps through space, but something serious, like “Armageddon”,or “Deep Impact”? Earth and the moon are heavily cratered fromprevious impacts, the most famous of which happened on Earth at theend of the Cretaceous period, about 65 million years ago. Scientistshypothesize that this impact was the cause of the End-Cretaceous(K-T) extinction, in which eighty-five percent of all species on Earthdisappeared, making it the second largest mass extinction event ingeological history2. Dr. Richard Muller in his book “Nemesis: TheDeath Star” describes the event:
At the end of the Cretaceous period, the golden age ofdinosaurs, an asteroid or comet about 5 miles in diameter(about the size of Mt. Everest) headed directly toward the
1 A meteor is an asteroid which breaches Earth’s atmosphere, a meteorite is one that strikesEarth’s surface.
2 The Permian mass extinction occurred about 248 million years ago and was the greatestmass extinction ever recorded in earth history; even larger than the better known K-Textinction that felled the dinosaurs.
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154 douglas herbert: the science of the apocalypse
Earth with a velocity of about 20 miles per second, morethan 10 times faster than our fastest bullets. Many suchlarge objects may have come close to the Earth, but thiswas the one that finally hit. It hardly noticed the air as itplunged through the atmosphere in a fraction of a second,momentarily leaving a trail of vacuum behind it. It hit theEarth with such force that it and the rock near it weresuddenly heated to a temperature of over a million degreesCelsius, several hundred times hotter than the surface of thesun. Asteroid, rock, and water (if it hit in the ocean) wereinstantly vaporized. The energy released was greater thanthat of 100 million megatons of TNT, 100 teratons, more than10,000 times greater than the total U.S. and Soviet nucleararsenals.
Before a minute had passed, the expanding crater was 60
miles across and 20 miles deep. (It would soon grow evenlarger.) Hot vaporized material from the impact had alreadyblasted its way out through most of the atmosphere to analtitude of 15 miles. Material that a moment earlier had beenglowing plasma was beginning to cool and condense intodust and rock that would be spread worldwide. The entireEarth recoiled from the impact, but only a few hundred feet.The length of the year changed by a few hundredths of asecond3.
In 2028, the asteroid 1997 XF11 will come close to Earth but will missour planet by about two and a half lunar distances, that’s extremelyclose, considering how big outer space is. If something was to changeit’s course, and it did hit Earth, what you would have is a 1.6 km (1 mile)wide meteorite striking the planet’s surface at about 48,000 kph (30,000
mph). The energy released during the impact is related to the kineticenergy of the asteroid before atmospheric entry begins. At typical solarsystem impact speeds of 12 to 20 km/s, energy E is approximatelygiven as one half times the asteroid mass m times the square of theasteroid velocity v , which can be rewritten in terms of the asteroid’sdensity ρ and diameter l , assuming that the asteroid is approximatelyspherical:
E =12
mv2 =π
12ρ l3v2 (21.1)
A kilometer and a half wide asteroid traveling at 20 km/s has anE roughly equal to a 1 million megaton bomb, that’s 10 million timesgreater than the bomb that fell on Hiroshima. A land strike wouldproduce a fireball several miles wide which would briefly be as hot asthe surface of the sun, igniting anything in site, and it would blast tonsof sulfur-rich rock and dust high into the atmosphere, encircling theglobe. As the burning debris rained back down to earth, the soot anddust would blacken the skies for months, if not years, to come. An oceanlanding would be no better, instantly vaporizing 700 cubic kilometers(435 cubic miles) of water, and blasting a tower of steam several milesinto the atmosphere, again benighting the sky. The meteor itself would
3 The impact believed to have caused the extinction of the dinosaurs left a 300 km (186
mile) wide crater on the coast of Yucatán. The impactor had to have been at least 30 km(19 miles) across.
21.3 errant black holes 155
likely crack Earth’s crust at the ocean floor, triggering massive globalearthquakes, sprouting volcanoes at weak spots in Earth’s crust, andcreating a tsunami as high as the ocean is deep, moving at hundreds ofkilometers an hour in every direction. Humans would likely survive,but civilization would not.
The Kuiper belt is an asteroid zone just beyond the orbit Neptune,and it contains roughly 100,000 ice balls that are each more than 80 anda half kilometers (50 miles) in diameter. As Neptune’s orbit perturbsthe Kuiper belt, a steady rain of small comets are sent earthward. If oneof the big ones heads right for us, that would certainly be the end forjust about all higher forms of life on Earth.
21.3 errant black holes
The Milky Way galaxy is full of black holes, collapsed stars about 20 km(12 miles) across. Just how full is hard to say, a black hole’s gravity is sostrong that anything approaching within a certain radius will no longerbe able to escape, including any light which would betray it’s presence,this critical radius is called the “Schwarzchild radius”4. The interior ofa sphere whose radius is the Schwarzchild radius is completely cut offfrom the rest of the Universe, the only way to “see” a black hole is tospot it’s gravitational lensing - the distortion of background light byforeground matter, i.e., the black hole. Beyond the Schwarzchild radius,the gravitational attraction of a black hole is indistinguishable fromthat of an ordinary star of equal mass. Based on such observations,and theoretical arguments, researchers estimate that there are about 10
million black holes in the Milky Way, including one at the core of ourgalaxy5, whose mass is as much as 2 million solar masses.
A black hole has an orbit just like any other star, so it’s not likelythat one is heading toward us. If any other star was approaching us,we would know, with a black hole there would be little warning. A fewdecades before a close approach to Earth, astronomers would probablynotice a strange perturbation in the orbits of the outer planets, as theeffects grew larger, it would be possible to make increasingly preciseestimates of the location and mass of the interloper. A black hole doesn’thave to come very close to Earth to bring ruin, it simply needs to passthrough the solar system. Just as Neptune’s gravity disturbs the Kupierbelt, a rogue black hole could pull Earth’s orbit into an exaggeratedellipse, either too close to the sun, or too far from the sun. It could eveneject Earth from it’s orbit and send the planet off into deep space.
21.4 flood volcanism
In 1783, the Laki volcano in Iceland erupted, pouring out 5 cubickilometers (3 cubic miles) of lava. Floods, ash, and fumes killed 9,000
people and 80 percent of Iceland’s livestock, and the ensuing starvationkilled 25 percent of Iceland’s remaining population. Atmospheric dustlowered winter temperatures by 9 degrees in the newly independentUnited States, and that was minor compared to what Earth is capable
4 In 1916, German astronomer Karl Schwarzchild established the theoretical existence ofblack holes as the solution to Einstein’s gravitational equations. The Schwarzchild radiusis equal to 3 km times the number of solar masses of the black hole, the solar masses aredetermined via Kepler’s laws of planetary motion.
5 Researchers also believe that there is a black hole at the center of every galaxy
156 douglas herbert: the science of the apocalypse
of. Sixty-five million years ago, a plume of hot rock from Earth’s mantleburst through the crust in what is now India, when continental driftmoved that area moved over a “hot spot” in the Indian Ocean. Eruptionsraged century after century, producing lava flows that exceeded 100,000
square kilometers (62 square miles) in area and 150 meters (500 feet) inthickness. - the Laki eruption 3,000 times over. Some scientists believethat this Indian outburst, and not an asteroid, was responsible forthe fall of the dinosaurs (the K-T extinction), since such lava flowswould have produced enormous amounts of ash, altering global climaticconditions and changing ocean chemistry. An earlier, even larger eventin Siberia occurred at the time of the Permian mass extinction 248
million years ago, and is the most thorough extermination known topaleontology. At that time 95 percent of all species were wiped out.
Sulfurous volcanic gases produce acid rains. Chlorine-bearing com-pounds break down the ozone layer6. While they are causing short-termdestruction, volcanoes also release carbon dioxide which yields longterm, greenhouse effect global warming. The last big pulse of floodvolcanism built the Columbia River plateau about 17 million years ago.If the idea of cataclysmic volcanism sounds too unlikely, Tom Bissellnotes in a 2003 Harper’s Magazine article (A Comet’s Tale) that:
...73,500 years ago what is known as a volcanic supere-ruption occurred in modern-day Sumatra. The resultantvolcanic winter blocked photosynthesis for years and verynearly wiped out the human race. DNA studies have sug-gested that the sum total of human characteristics can betraced back to a few thousand survivors of this catastrophe.
21.5 giant solar flares
More properly known as coronal mass ejections, solar flares are theoutbursts caused by enormous magnetic storms on the sun, and theybombard Earth with high speed subatomic particles. A typical solarflare releases the equivalent energy of a billion hydrogen bombs, andejects a hundred billion tons of high energy particles into space. Earth’smagnetic field and atmosphere negate the potentially lethal effects ofordinary flares, knocking the particles back into space, and steeringsome of the particles over the poles, creating Earth’s auroras. But whileexamining old astronomical records, Yale University’s Bradley Schaeferfound evidence that some sunlike stars can brighten briefly by upto a factor of 20. Schaefer believes that these increases are caused bysuperflares millions of times more powerful than those commonlyexperienced by Earth. Scientists don’t know why superflares happen atall, or whether our sun could exhibit milder but still disruptive behavior.A superflare on the sun only a hundred times larger than typical wouldoverwhelm Earth’s magnetosphere and begin disintegrating the ozonelayer (see footnote 6). Such a burst would certainly kill anything baskingin its glow, and according to Bruce Tsurutani of NASA’s Jet PropulsonLabatory, “it would leave no trace in history.”
While too much solar activity could be deadly, too little of it isproblematic as well. Sallie Baliunas of the Harvard-Smithsonian Centerfor Astrophysics says that many sunlike stars pass through extended
6 Without the ozone layer, ultraviolet rays from the sun would reach the surface the earthat nearly full force, causing skin cancer, and killing off the photosynthetic plankton in theocean that provide oxygen to the atmosphere and support the bottom of the food chain.
21.6 viral epidemic 157
periods of lessened activity, during which they become nearly 1 percentdimmer. A similar downturn in our own sun could send us into anotherice age. According to Baliunas, there is evidence that decreased solaractivity contributed to 17 of the 19 major cold episodes on Earth in thelast 10,000 years.
21.6 viral epidemic
Viruses prosper by using the genetic material of a healthy cell to pro-duce more virus. They multiply within the healthy cell, burst out, andattack more healthy cells. They also have the capacity to burst forth insome startling new form, and then to disappear just as quickly. In 1916,people in Europe and America began to come down with a strangesleeping sickness, which became known as encephalitis lethargica . Vic-tims would go to sleep and not wake up, they could be roused withgreat difficulty, but once they were allowed to rest they would fall backinto deep sleep. Some victims continued like this for months beforedying. In ten years encephalitis lethargica killed five million people andthen simply disappeared, the only reason this disease didn’t get muchlasting attention is because the worst epidemic in history was sweepingacross the world at the same time.
The Great Swine Flu, sometimes called the Great Spanish Flu, killedtwenty-one million people in its first four months. Between autumn of1918 and spring of 1919, 548,452 people in the United States died of theflu. In Britain, France, and Germany, the toll was over 220,000 dead ineach country, the global toll was approximately 50 million, with someestimates as high as 100 million. Much about the 1918 flu is understoodpoorly, if at all, one mystery is how it managed to break out seeminglyeverywhere all at once, in countries separated by oceans and mountainranges. A virus can only survive outside a host body for a few hours,so how did this virus appear in Bombay, Madrid, and Philadelphia, allwithin the same week?
Some consider it a miracle that other diseases have not gone rampant.Lassa fever, first detected in 1969 West Africa is extremely virulent. Adoctor at Yale University came down with Lassa fever when he wasstudying it in 1969. He survived, but a technician in a different lab-oratory, with no direct exposure to the virus contracted it and died.Fortunately the outbreak stopped there, but in 1990 a Nigerian livingin New York contracted Lassa fever on a visit home and didn’t developany symptoms until his return to the United States. He died undiag-nosed in a Chicago hospital, without anyone having taken any specialprecautions, or knowing that he had contracted one of the most infec-tious and lethal diseases on the planet. Our lifestyles invite epidemics,air travel makes it possible to spread disease with alarming ease. AnEbola outbreak could begin in Boston, jump to Paris, and then to Tokyobefore anyone ever became aware of it.
We’re accustomed to being Earth’s (and maybe the galaxy’s) domi-nant species, which makes it difficult to consider that we may only behere because of various chance events, or that our continued presencemay be due to the absence of other chance events.
Humans are here today because our particular line never frac-tured - never once at any of the billion points that could haveerased us from history.
158 douglas herbert: the science of the apocalypse
Stephen Jay Gould
22A M A N D A L U N D : E X T R AT E R R E S T R I A LI N T E L L I G E N C E
22.1 introduction
Life, as far as we know, exists only on our own planet Earth. Thereis no evidence of intelligence (or life of any form) in our solar
system, galaxy, or beyond. But are we really alone? The idea that oursmall planet possesses the only biological beings in the universe seemsegotistical and, in light of our recent advances in astronomy and thesciences, unlikely. Many efforts are currently underway in the search forextraterrestrial intelligence, with the hope of finding it still undampedby the lack of success. If intelligent life does indeed exist elsewhere, itremains to be seen whether it will dwell in a region close enough tomake contact with, and even whether or not it will want to be found.
22.2 the possibility of life in the universe
Exploration of our solar system has provided no evidence for theexistence of life anywhere within it except Earth. While ancient Marsmay once have had a life-supporting environment and planetary moonssuch as Europa and Titan might be capable of sustaining life, no life,and certainly no intelligent life, has been observed. Despite this, modernAstronomy suggests that extrasolar planets (planet-star systems like ourEarth and sun) might not be as rare in galaxies as was once believed [55].Many Jovian planets have been detected due to their large mass, butsmaller, rocky planets may even outnumber these gas giants three toone.
The quantity of possible life-supporting planets, along with severalother factors which would determine the amount of intelligent life inthe universe, is expressed concisely in the famous Drake Equation. Thisequation was created by Dr. Frank Drake of the University of California,Santa Cruz in 1960 [55], and defines the total number of civilizations inthe universe with which we might be able to communicate (NC) as
NC = RS fPn fL f I fCL, (22.1)
where
• RS is the formation rate of stars in a galaxy,
• fP is the fraction of stars with planetary systems,
• n is the average number of habitable planets in a planetary system,
• fL is the fraction of habitable planets which develop life,
• f I is the fraction of habitable planets which develop intelligentlife,
• fC is the fraction of planets with life that develop intelligentbeings interested in communication, and
159
160 amanda lund: extraterrestrial intelligence
Figure 22.1: One way of detecting extrasolar planets is by observing a decreasein the brightness of a star as the planet passes in front of it [30].
• L is the average lifetime of a civilization.
As the only term in this equation known with any certainty is the rateof star formation RS (about 2 or 3 per year), the projected values of NCvary greatly [75]. The fraction of stars with solar systems ( fP) has beenestimated between 20% and 50% [55], but the remaining terms dependgreatly on the optimism of the person assigning them. For instance,the average lifetime of an intelligent civilization might be anywherebetween 100 and 1010 years. A belief in the tendency of civilizations toself-destruct would set L at the lower limit, while the opposing viewthat societies can overcome this inclination to eradicate themselveswould let L be as large as the lifetime of their star [75].
As the only known example of fL, f I , and fC is the earth, there isreally no way to obtain an accurate estimate of these probabilities [57].Based on the time it took life to evolve on earth, a rough estimate offL > 0.13 could be assigned. Drake has estimated both f I , and fC to be0.01, though there is not much basis for these assumptions.
22.3 the search for intelligent life
The current quest for intelligent life mainly involve filtering throughmultitudes of electromagnetic emissions in an attempt to "eavesdrop"on possible sentient societies [82]. Since World War II, radio astronomyhas advanced significantly, and most of the searches are being done inthe radio and microwave regions of the spectrum. SETI (the Search forExtraTerrestrial Intelligence), the main organization conducting thesesearches, has been using large radio telescopes in an attempt to pickup alien transmissions. Their receivers are intended to identify narrow-band broadcasts, which are easy to discern even at large distances andcan only be created by transmitters [84].
A particularly important signal for astronomy and SETI is the hy-
22.4 will we find it? and when? 161
drogen 21-centimeter line. This special wavelength of radiation is aquantum effect of the hyperfine structure of hydrogen, which providesa very small correction to the hydrogen energy levels due to the inter-action of the proton and electron spins. A transition from the tripletto the singlet state in the ground state of hydrogen corresponds to thespin of these two particles flipping from parallel to antiparallel. This isa very rare transition, but when it occurs a small amount of energy isreleased at a wavelength of 21 centimeters [15].
example 22 .1 : the 21 centimeter line
Using Planck’s relation E = hν, show that if the energy of the photonemitted in the transition from the triplet state to singlet state of the hydrogenhyperfine structure described above is 5.9 × 10−6 eV, the correspondingwavelength is 21 centimeters.
Solution:First we must convert the energy in eV to joules, using the conversion factor
1 eV = 1.6× 10−19 J. (22.2)
This gives us:
5.9× 10−6 eV = 9.44× 10−25 J. (22.3)
Using both this value of the change in energy in joules and Planck’s constant,
h = 6.626× 10−34 J s, (22.4)
we can determine the frequency of the photon emitted using Planck’s relation.
E = hν (22.5)
ν =∆Eh
=9.44× 10−25 J
6.626× 10−34 J s= 1420 MHz (22.6)
Now, by plugging this frequency into the equation relating frequency andwavelength (where c denotes the speed of light in a vacuum), we can find thecorresponding wavelength.
λ =cν
=3× 108 m/s1420 MHz
= 0.21 m = 21cm (22.7)
Though this type of transition is extremely uncommon, about 90%of the interstellar medium consists of atomic and molecular hydrogen;this greatly increases the probability of observing 21-centimeter radi-ation. This radiation is able to penetrate dust clouds in space, and asfrequencies around it are protected for radio astronomy, there is limitedinterference from Earth [15]. Because it is such an important frequency,some scientists believe it is a likely signal for extraterrestrial intelligenceto send out in a communication attempt. As a result, SETI has radio-telescopes searching for the 21-centimeter line around sun-like starsand is even broadcasting its own 21-centimeter wavelength signals.
22.4 will we find it? and when?
Over the past 50 years, great advances in technology and in our un-derstanding of physics and astronomy have transformed the search forextraterrestrial intelligence from science fiction and Hollywood horrorfilms into an actual science. Despite this, we may still be far from find-ing other life in the universe. If it does indeed exist, it may be nearly
162 amanda lund: extraterrestrial intelligence
Figure 22.2: SETI uses radio telescopes to try to pick up signals from extrater-restrial life [71].
impossible to locate life among the billions of stars and galaxies andlight years of space that surround us, especially if any intelligent life outthere does not want to be found. The search has been likened to tryingto find not a needle but a single atom of a needle in a haystack [74].
Yet there are still optimists who believe we might not be so far offafter all. Notable astronomers including Carl Sagaan and Frank Drake,who are searching through this cosmic haystack themselves, have pre-dicted that extraterrestrial intelligence will be found between 2015 and2026 [74]. SETI even offers a program called SETI@home, where anyonewith a computer can donate disk space to help analyze radio telescopedata and speed up the search. It seems reasonable to believe we are notalone in the universe; the real question, then, must be whether or notwe will see evidence of other intelligence–and if so, when.
22.5 problems
1. Using the Drake Equation (Eq. (22.1)) and given values
a) RS = 2.5,b) fP = 0.35,c) n = 1,d) 100 < L < 1010,e) 0.13 < fL > 1,f) f I = 0.01,g) fC = 0.01,
determine the most optimistic and most pessimistic values of NC.
Solution: Using
NC = RS fPn fL f I fC L, (22.8)
the most optimistic value would use the upper limits in the ranges, andthe result would be
NC = 2.5× 0.35× 1× 1× 0.01× 0.01× 1010 = 8.75× 106. (22.9)
The most pessimistic value, using the lower limits, would be
NC = 2.5× 0.35× 1× 0.13× 0.01× 0.01× 100 = 1.14× 10−3.(22.10)
22.6 multiple choice questions 163
However, this pessimistic estimate is much less than 1, indicating a lackof intelligent civilizations in the universe.
2. Some astronomers have suggested that electromagnetic wavesoutside the radio and microwave regions might be as good away or even a more promising way to succeed in the search forextraterrestrial intelligence [84]. If we were to search for signalswith wavelengths between 750 nm and 10 µm, in what region ofthe electromagnetic spectrum would this be? What would the becorresponding frequencies for these wavelengths?
Solution: These wavelengths are in the infrared region. To find theirfrequencies, we use
ν =cλ
. (22.11)
The upper frequency is
ν =3× 108 m/s7.5× 10−7 m
= 4× 1014 s−1, (22.12)
and the lower frequency is
ν =3× 108 m/s1× 10−5 m
= 3× 1013 s−1, (22.13)
so 3× 1013 s−1 < ν < 4× 1014 s−1.
22.6 multiple choice questions
1. What is an extrasolar planet?
a) A planet in our solar system that is outside the frost lineb) A planet outside our solar system orbiting around another
starc) A massive body that is not a true planet but exists in free
space and does not orbit a stard) A planet on which life exists
Answer: b)
2. How does SETI hope to find extraterrestrial life?
a) By intercepting radio signals from intelligent civilizationsb) By building a spaceship that can travel at close to the speed
of light and sending it toward extrasolar planets in the An-dromeda galaxy
c) By following up on all reports of UFO sightings in hopesthat one of them is valid
d) By using radio telescopes to intercept government intelli-gence transmissions and expose the evidence of extraterres-trials that the government has been concealing
Answer: a)
164 amanda lund: extraterrestrial intelligence
22.7 summary
So far, the only life we know of in the universe exists on Earth. However,given the vastness of the universe and the huge number of galaxies,stars, and planets within it, it is unlikely that we are alone. With ourcurrent technology there is no way to tell how much intelligent lifeexists outside our planet or where it is located; the Drake Equation isone way to estimate the number of intelligent civilizations, but it isbased mainly on guesswork. The SETI institute has set out to find life byattempting to eavesdrop on radio signals from other civilizations–oneimportant wavelength in this search is the 21-centimeter line, a raretransition between two hyperfine states of hydrogen. There is no wayto know when or if we will find anything, though some optimisticastronomers predict as soon as 2015.
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Limongi and Chieffi [51], 138, 139,167
Lloyd [52], 102, 167
Lochner and Gibb [53], 135–137,168
Lyons [54], 79, 80, 83, 168
Marais and Walter [55], 159, 160,168
Margolus and Levitin [56], 88, 168
Mash [57], 160, 168
Massel [58], 49, 168
Mermin [59], xi, 168
Nassau [60], 9, 168
Nave [61], 135, 168
Nave [62], 4, 168
News [63], 89, 168
Palme and Jones [65], 138, 168
Peters et al. [66], 107, 108, 168
Pinet [67], 47, 168
Research [68], 91, 93, 168
Royall et al. [69], 36, 37, 169
SETI [71], 162, 169
Sears and Zemansky [70], 10, 169
Shannon [72], 98, 169
Shapiro and Teukolsky [73], 131,169
Shostak [74], 162, 169
Shu [75], 160, 169
Slansky et al. [76], 65, 66, 72–74,169
Standage [77], 127, 169
Svistunov [78], 64, 169
Tachibana et al. [79], 139, 169
Tayler [80], 133, 169
Than [81], 138, 169
Thomsen [82], 160, 169
Tobin et al. [83], 138, 169
Townes [84], 160, 163, 170
Tsymbal [85], 94, 170
Turner [86], 107, 108, 170
University [87], 144, 170
Veltman [88], 62, 63, 65, 170
Wallace [89], 12, 170
Waugh [90], 39, 170
Wesson [91], 148, 170
Wesson [92], 143, 170
Whitaker [93], 10, 11, 170
William Sheehan and Waff [94], 126,170
173
174 INDEX
Wood et al. [95], 91, 92, 170
Worthington [96], 39, 170
Xu et al. [97], 39, 170
Xu et al. [98], 39, 170
Young and Freedman [100], 19, 171
Young and Freedman [101], 125,171
Young and Freedman [99], 15, 16,171
eot [1], 148, 165
et. al. [28], 57, 166
et. al. [29], 56, 57, 166
of Sciences [64], 93, 168
21-centimeter line, 161
Alhazen, 13
amorphous Solid, 33
Antarctica, 50
Areal density, 87
astronomer’s lens, 17
baryon number, 149
Basis, 62
beam ratio, 22
Benjamin-Feir instability, 52
Binary bit, 85
Binary byte, 86
Bosons, 62
Bragg reflection, 21
breather solutions, 52
Bremen, 52
Caledonian Star, 52
camera obscura, 13
camera screen, 15
Cape of Good Hope, 50
Carl Sagaan, 162
CERN, 72
Change of basis, 66
classical photography, 19
Colloids, 35
converging lens, 14
Creation operator, 64
crystalline Solid, 33
current refraction, 51
diverging lens, 14
DNA, 86
double-slit experiment, 19
Drake Equation, 159
draupner wave, 48, 49
Eötvös, 148
Eigenstates, 61
Einstein equivalence principle, 143
electromagnetic waves, 3
Electromagnetic force, 63
Electroweak theory, 64
Entropy, 102
equivalence principle, 143
extrasolar planets, 159
f-number, 16
Fermions, 62
fifth interaction, 148
Fischbach, 148
focal length, 14
Fourier hologram, 24
Frank Drake, 159, 162
fundamental forces, 147
Gauge Boson, 63
gauss’s law, 4
general relativity, 143
Generations of matter, 62
Giant magnetoresistance (GMR),93
Girolamo Cardano, 13
glass, 33
Glass Transition, 34
Glass Transition Temperature, 35
Gluon, 62
Grand unified theory, 72
gravitation, 147
Gulfstream current, 50
Hamiltonian, 70
Hard disk platter, 89
Hard disk sector, 90
Hard disk track, 90
Hartley, 98
Heisenberg uncertainty principle,87
Hermitian operator, 62
Higgs boson, 65
Higgs mechanism, 65
holographic data storage, 23
holographic convolution, 24
holographygeometric model, 19
Inertia, 65
Information Entropy, 98
Information Theory, 97
Information uncertainty, 100
Information, formal definition of,98
interference fringes, 19
intermodulation noise, 22
John Couch Adams, 127
INDEX 175
Kepler’s Third Law, 128
Kuroshio current, 50
Ladder operator, 64
Lambda particle, 73
Large hadron collider, 72
Lepton, 63
liquid, 33
long-range order, 33
Longitudinal magnetic recording(LMR), 91
Magnetoresistance, 92
magnification, 14
Mass states, 67
Neptune, 125
Neutrino, 67
Neutrino oscillations, 67
Newton’s Law of Universal Gravi-tation, 125
object beam, 20
Pacific Ocean, 50
pattern recognition, 24
Perpendicular magnetic recording(PMR), 91
Photon, 63
Planck energy, 72
Principle of superposition, 61
Quantum field theory, 63
Quantum mechanical operator, 61
Quantum mechanics, 61
Quark, 63
Radial Distribution Function, 36
Rayleigh distribution, 49
real image, 21
redundancy, 20
reference beam, 20
reflection holograms, 21
rogue waves, 49–54
SETI, 160, 162
Shannon, 98
Shannon entropy, 98
Shannon’s formula, 101
short-range order, 33
sine waves, 3
Soda-Lime Glass, 34
South Africa, 50
South Africa, 50
Special relativity, 70
Standard Model, 62
strong equivalence principle, 143
Strong force, 62
Sudbury Neutrino Observatory, 72
Supercooled Liquid, 35
surface hologram, 22
theory of gravity, 143
thick hologram, 22
thin hologram, 22
tidal wave, 47
transmission hologram, 21
tsunami, 47
Uranus, 126
virtual image, 21
weak equivalence principle, 143,147
Weak eigenstate, 65
Weak force, 63
W boson, 64
Yukawa potential, 149
zone plate model, 21, 22
zoom lens, 16
Z boson, 64
